Geometric theory of functions of a complex variable (Translations of Mathematical Monographs, Vol. 26)

S

F' (p (C)) p' (C)11 r dr dcf.

If we set re is co

F' (` (C))

(C) _

anCn,

n=0

ao = p' (0) = f' (0) '

we have m

(F)

I an ISP8n-h9

='t aims

n}

I

>'t ao

9

0

with equality holding only if al = a2 = ... = 0, that is, only for the function F(z) = f(z)/f'(0). This completes the proof of assertion b). The minimal property b) can then be used for an explicit construction of the mapthat is, 21 (F) > nr I

a012,

ping function.

§3. The correspondence of boundaries under conformal mapping Since the boundary points of a domain do not belong to the domain, we were concerned in the preceding section with the establishment of a correspondence between interior points of a domain and its image. To study the correspondence of boundary points, we need to take up the more difficult question of investigating the behavior of a mapping function as we approach the boundary of the domain in question. This investigation leads to a number of remarkable results. We first establish a few auxiliary propositions. Lemma 1 (Koebe). Let lzn I and {zn I denote two sequences of points belonging to the disk IzI < 1 that converge to distinct points a and b respectively on the circle Izi = 1. For every n, let k,, denote a Jordan arc lying in the disk I:I < 1 but outside some fixed neighborhood of the point z = 0. Let f(z) denote a function that is regular and bounded in the disk Izl < 1 and that converges uniformly to 0 on the sequence {hn I. Then f(z) = 0.

32

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

The uniform convergence of the function f (z) to 0 on the arcs X. is to be understood in the sense that, for every c > 0, there exists an N > 0 such that the inequality n > N on An implies the inequality I f (z I < E. Proof. Let us suppose that f(z) 10. We may as well assume that f(0) 0 because, in the opposite case, we could consider the function f(z)/zP, where p denotes the order of the zero at z = 0 of the function f (z), and one can easily verify that this function also satisfies. all the conditions of the lemma and is nonzero at z = 0. On An, we have

If(z)I cmax If(z)I=En, an

where c n

-+ 0 as n

. Let us partition the disk Izl < 1 into n equal sectors

such that the points a and b do not lie at the ends of the boundary radii. For m sufficiently large, the points a and b will lie on the boundaries of distinct sectors. If we now discard these sectors, the remaining portion of the disk will be partitioned into two groups of adjacent sectors, in each of which there will be at least two sectors. In each of these groups, let us choose one radius that serves as the boundary of two adjacent sectors (see Figure 1). One of these radii, which

Figure 1

we.denote by or, intersects infinitely many of the curves An. Let us index these . Beginning with some point of intersection intersected curves Ank, for k = 1, 2,

§3. CORRESPONDENCE OF BOUNDARIES

33

of the curve Ank with a, let us proceed along the curve \nk in either direction till we get to the first point at which this curve meets one of the other boundary radii, which obviously belongs to the boundary of the sector adjacent to a. We denote by Ank the arc of the curve knk from this point to the first point at which hnk touches or as we proceed along knk in the opposite direction. Except for , lie entirely in the two sectors for k = 1, 2, their endpoints, the arcs gdjacent to or. Let S denote whichever of the two sectors contains an infinite IG,et of arcs Ank. For these arcs we keep the same notation . for it = 1, 2, On Ank, we have If(z)I <Enk. We may assume that the sector S is symmetric the real axis since we can always arrange this by considering not only the function f(z) but the function f(eL°'z) with suitable choice of real a, which also satisfies conditions analogous to the conditions of the lemma. Let a;;k denote that portion of the arc Ank extending from the endpoint of lying above the real axis to the first point at which A;,k meets the real axis. Let Ank denote the arc symmetric to hnk about the real axis. On X nk we have I f (z )I < Enk. If we assume that I f (z)I < 11 in B, we can now conclude that the function tA(z) ,f(z) (z--), which is regular in IzI < 1 satisfies the inequality IV'(z)I <MEnk on the arc \;;kU A, k since one of the factors whose product is the function O(z) is !always no greater than 4 and the other is less than Enk. On the other hand, the function 1`' (z) _ (z) ii (,,z) ...

(77m 1Z)+

is regular in the disk IzI < 1. Then, on the closed Jordan curve consisting of the arc A, k U A;;k and the arc obtained from it by means of a rotation about z = 0 through angles 2a/m, 4a/m, , 2(m - 1) rr/m, we have, on the basis of similar considerations, I F (z) I

EnkMIM-1

The same inequality holds inside this curve and, in particular, at the point z = 0. But F(0) = I f(0)12m. Therefore, If(0)I2m <EnkM2in-1 Since Enk -+0 as k -+80, it follows that f(0) = 0, which contradicts the assumption made at the beginning of the proof. This completes the proof of the lemma. Lemma 2 (Lindelof). Suppose that the following holds: 1) The function f(z) is regular in a domain B and I f(z)1 <M in B. 2) zo is a finite point of the domain B and r is a positive number such that the circle 1z - zo I = r contains an arc

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

34

of length (277/m)r, where m is an integer, lying outside B. 3) For arbitrary e > 0, as we let z approach boundary points of B lying in the disk I z - zo I < r through points belonging to B, all the limiting values of I f(z)I are less than or equal to E.

Then If(zo)I`m

-am-'

.

Proof. Obviously, we may take zo = 0. The common portion of the domain B and the domains obtained from it by a rotation about the origin through angles , 2 (m - 1)77/m is an open set including the points z = 0 but not 2rr/m, 4rr/m, including points of the circle IzI = r. It consists of domains among which is a domain B1 including the points z = 0. This last domain lies entirely in the disk IzI < r. Consider the function 2,ct

F(z)=f(z)f('qz)...f(,'4-'Z), (,l=e'"), which is regular in B 1. If we let z approach the boundary of B 1 through points of B1, then, in accordance with condition 3) of the lemma, all the limiting values

of IF(z)I will be no greater than N". But then, we shall have IF(z)I < eM"`-1 at all points of the domain B1. To see this, let M1 denote the least upper bound of IF(z)I in B1. If M1 were greater than (M' -1 there would exist a sequence of points zk C B1 such that IF(zk)I --, M1 as k -- oe, Let z0' denote a cluster point of this sequence. It cannot be a boundary point of B1 because then M1 e M'_ 1. Therefore, IF(zo )I = M1; that is, IF(z)I attains a maximum at the point z = z0', which is impossible. Thus, IF(z)I <efm-1 in B1. In particular, since F(0) =[f(0)]'", we have If(0)Im <EMni-1, which completes the proof of the lemma.

As a corollary of this lemma, we have the following Lemma 3. If a function f(z) is regular and bounded in a simply connected domain B that has more than one boundary point and if there exists a neighborhood of a finite boundary point a such that the function f(z) approaches a constant c as z approaches boundary points lying in this neighborhood through

points belonging to B, then f(z) = c in B. Proof. Let us first suppose that a is also a cluster point of the set of exterior points of the domain B and let the disk Iz - al < p denote the neighborhood referred to in the lemma. Let zo denote an arbitrary point of the disk

§3. CORRESPONDENCE OF BOUNDARIES

35

l z - al < p/2. Then, there exists a disk l z - zo i < r contained in the disk l z - al < p such that there are exterior points of B on the circle lz - zoI = r and hence there exists an arc, lying outside B. Lemma 2 with t = 0 can be applied to the function f(z) - c. Consequently f (zo) = c. Since zo is an arbitrary point in the disk l z - al < p/2, we have

f(z) = c in B. On the other hand, if the point ao is not a cluster point of the set of exterior points of the domain B, let us denote by b a finite boundary point distinct from a. Then, just as in the proof of Riemann's theorem, let us use the function z* = (z - a)/ z - b) to map the domain B univalently onto the domain B*, which, in an arbitrary neighborhood of the point z* = 0, will have exterior points. The function f(z) - c will now be regular as a function of z* in B. and it will have the property mentioned in Lemma 3 at the point z = 0. Therefore, using what we have already proved, we again obtain the result that f(z) = c in B. This completes the proof of the lemma. Accessible boundary points. Let us now examine the behavior of the function J= f(z), which maps a given simply connected domain B univalently onto the disk Ill < 1 close to the boundary of the domain B. Since such a domain can be mapped univalently by means of elementary functions onto a bounded domain, it will be sufficient in what follows to consider the case of a bounded domain B and it is for such a domain that we shall prove all the theorems.. First of all we note that, as points z E B approach the boundary of the domain B, the corresponding points C in the disk ICJ < 1 approach the circle ICI = 1 uniformly in the sense that, for every t > 0, there exists a S > 0 such that, if a point z C B lies at a distance less than S from the boundary of B, the corresponding point C lies in the annulus 1 - t < ICI < 1. Specifically, we may take for S the distance between the boundary of the domain B and the closed set constituting the preimage of the disk ICI < 1 - t. Analogously, we can show that 1, then the corresponding if points of the disk ICI < 1 approach the circle points z C B approach the boundary of the domain B uniformly. For the next step in our investigation, we introduce the concept of accessible boundary points of a domain. A boundary point c of a domain B is said to be an accessible boundary point if it can be joined with an arbitrary interior point of the domain B by a continuous curve 1: z = z (t) for a < t < b that lies entirely in B except for the endpoint c = z (b). Such a curve l can be replaced by a Jordan curve or even by a broken line with no multiple points that consists of a finite or infinite number of segments, in the latter case with endpoints clustering only about c. This is true because the arc z = z (t) for b - tk _1 < t < b - tk (where rk > 0 for

36

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

k = 1 , 2, .. , where Ek > Ek +11 and where Ek -. 0 as k --+ oe is a closed set and

hence lies at a positive distance 8k from the boundary of the domain B. Therefore, it can be partitioned into a finite number of arcs sufficiently small that the distance between any two points of any one of these arcs is less than 8k. Then, the broken line consisting of segments joining successive points of this partition will lie entirely in B. If we apply this for k = 1, 2, , we see that the curve I can be replaced with a broken line consisting of an infinite number of segments. If we discard any loops, we obtain a broken line without multiple points. This broken line is a Jordan curve.

In connection with the concept of accessible boundary points, we have an important supplement to Jordan's theorem (cf. introduction), presented by Schoenflies:' All points of a closed Jordan curve are accessible boundary points of the domains defined by it. For continua of a more complicated type, there may not be an analogue of Jordan's theorem and its supplement. For example, the curve shown in Figure 2 and representing a spiral-shaped curve without multiple points that approaches the circle C asymptotically partitions the plane into three simply connected domains, and, for the two bounded domains, the

circle C consists entirely of inaccessible boundary points. Analogously for the domains shown in Figures 3 and 4 and representing squares with rectilinear cuts that Figure 2 cluster about the segment cd, all points of the segment cd (with the exception of the point d in Figure 4) are inaccessible boundary points. In these last two examples, the point e (see Figures 3 and 4) can be reached along two continuous curves lying in the domain, one on either side of the cut. To distinguish two boundary points at a point e, we identify the accessible boundary point of the domain B not only by its position in the plane but also by the continuous curve in B that joins this point with an interior point. Let 11 and 12 define respectively accessible boundary points at z1 and z2 . If z1 z2, these accessible boundary points are distinct. On the other hand, if zl = z2, we take the view that two curves 11 and 12 lying in B define at zl two distinct accessiblepoints if and only if there exists a neighborhood of the point z1 in which 11

1) See, for example, Caratheodory [19121.

§3. CORRESPONDENCE OF BOUNDARIES

37

and 12 cannot be connected by a continuous curve lying both in B and in that neighborhood. If two curves li and 12 that terminate at a single boundary point d

Figure 3

Figure 4

have common points in an arbitrary neighborhood of z 1, these curves necessarily define the same accessible boundary points. Therefore, accessible boundary points defined by a continuous curve and a broken line inscribed in that curve coincide, as shown above. This enables us to define accessible boundary points by means of Jordan curves only. We have the following theorem 1) on the correspondence of accessible boundary points: Theorem 1. Given a univalent conformal mapping of a domain B in the zplane onto the unit disk ICI < 1, to each accessible boundary point z1 of the dozi

main B we can assign a specific point Ci on the circle ICI = 1 such that, as z approaches zi along an arbitrary curve 11 defining z1, the corresponding point C approaches Ci, and also such that two distinct accessible boundary points of the domain B correspond to two distinct points on ICI = 1 and the set F of points on ICI = 1 corresponding to all accessible boundary points of the domain -B is everywhere dense on fI = 1. Proof. We shall carry out the proof in several stages. 10, Let us show that, as a point z approaches an accessible boundary point z 1 of the domain B by moving along a curve l is z = z (t), for a < t < b, that defines the accessible boundary point, the corresponding point C moves along a continuous curve A1: J= f(z(t)) in the disk 1 and approaches a definite point on the circle KI = I.

1) Koebe [19151.

38

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

Let us suppose the opposite. Then, there must exist on the circle IcI = 1 two distinct points S1 and S2 such that there is on Al a sequence of arcs with endpoints that converge to S1 and S2 and the inverse mapping function z =' converges uniformly on these arcs to z1. Since /(C) is bounded in the disk ICI < 1, it follows from Lemma 1, as applied to the function 40(C) - z1, that

(W) - zl - 0; that is, -0 (0 a const in ICI < 1, and this is impossible. Consequently, as C moves along A1, it approaches a definite point on IcI = I. 2°. Let us show that, as z describes two curves 11 and l2 defining the same accessible boundary point z1 of the domain B as it approaches z1, the corresponding point describes curves Al and A2 that terminate in a single point on the circle ICI = I. Let us suppose that the curves Al and A2 terminate at distinct points and S2 on the circle ICI = 1. In an arbitrary neighborhood of z1, the curves l1 and 12 can be joined by a continuous curve and hence by a Jordan curve contained both in the domain B and in this neighborhood. Therefore, as the neighborhood is constructed to a point, the endpoints of the corresponding Jordan curves in the

disk ICI < 1 approach `'1 and CZ. On these curves, -O(C) approaches z1 uniformly. In accordance with Lemma 1, we again obtain (W) __ const in ICI < 1, which is impossible. Consequently, S1 = CZ. 3°. Let us now show that two distinct points `'1 and S2 on the circle ill = 1 correspond to two distinct accessible points z1 and zZ of the domain B. Suppose that z1 and z2 are defined by the Jordan curves 11 and 12. We may assume that they issue from a single point of the domain B and that they have no other points in common in B. Let Al and A2 denote the corresponding curves in the disk ICI < 1. Let us suppose that they terminate at a single point S1 on ICI = 1. The closed curve Al U A2 partitions the disk into two domains, of one of which it constitutes the entire boundary. Suppose that corresponding to this domain in B is a domain B1. The curve II U 12 is mapped into its boundary. If the complex numbers z1 and z2 are distinct, there exists on the boundary of the domain B1 a boundary point a of the domain B that is distinct from z1 and z2 because, otherwise, the boundary of the domain B1 would be the open Jordan curve 11 U 12 and thus would not be bounded, in contradiction with the assumption of boundedness of B (see p. 35). In a sufficiently small neighborhood of the point a, all boundary points of the domain B 1 are also boundary points of the domain B. Consequently, as z C B approaches these points, the corresponding point C must approach the circle ICI = 1, that is, the point 1. In accordance with Lemma 3, 1

39

§3. CORRESPONDENCE OF BOUNDARIES

this yields f(z) - S1 in B, which is impossible. On the other hand, if the complex numbers z1 and z2 are equal, 11 U 12 is a closed Jordan curve which is mapped into the boundary of the domain B1. In this case, if the domain B1 has a boundary point distinct from points of the curve 11 U 12, this point must lie inside the curve 11 U 12. Consequently, the hypothesis of Lemma 3 is satisfied in a sufficiently small neighborhood of this point and we may again conclude that coast in B, which again is impossible. This proves that the entire boundary of B1 is the curve 11 U lZ. But then, in an arbitrary neighborhood of the point z1, the curves 11 and 12 can be joined by a continuous curve lying in the domain B1 and hence in the domain B, and the curves 11 and 12 do not define distinct accessible boundary points of the domain B. This proves that S1 CZ.

4°. Finally, let us show that the set F referred to in the theorem is everywhere dense on the circle ICI = 1. Let us suppose that the circle JC1 = 1 contains an arc y no points of which correspond to accessible boundary points of the domain B. Let Co denote an interior point of this arc and let IC,, {, for n = 1, 2, ... , denote a sequence of points of the disk I I < 1 that converges to CO. The corresponding points zn E B have at least one cluster point z0, which necessarily lies on the boundary of the domain B. Let {znkI denote a subsequence that converges to z0. Let us denote by Ink the straight line segment connecting znk with the closest point of the boundary of the domain B. This segment defines an accessible boundary point. From 1°, corresponding to this segment in the disk JCj < 1 is the curve An extending from the point Cn to some point on the circle !CI = 1 not on the arc y. Also, -O(C) approaches z0 uniformly on knk as k 00. On each of the curves Ank, we can choose an arc such that the sequences of endpoints of these arcs converge to the point Co in the one case and to a point on the circle ICI = 1 not on the arc y or to one of the endpoints of that arc in the other. Then, we can apply Lemma 1 to the function O(C) - z0, and we conclude that CI) - z6 in the disk IC` < 1, which is impossible. Consequently, the arc y described above does not exist and the set F is everywhere dense on the circle This completes the proof of Theorem 1.

Prime ends. To make a complete investigation of the correspondence of boundaries under univalent mapping of a domain B onto the disk I < 1 , we introduce the concept of a prime end. Suppose that we make a sequence of trans1

verse cuts Ck, for k = 1 , 2, ... , in the domain B. The cut C1 connects two distinct accessible boundary points of the domain B and partitions it into two

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

40

domains, one of which we denote by B I. The cut C2 lies in the domain B 1, connects two distinct accessible boundary points of the domain B that are not endpoints of C1, and partitions the domain B1 into two domains. Let B2 denote whichever one of these domains has no points of the curve Cl on its boundary. In general, the cut Cry+1 lies in Bn, connects two distinct accessible boundary points of the domain B that are not endpoints of Cn, and partitions the domain Bn into two domains. We denote by Bn+l whichever of these domains does not have points of the curve Cn on its boundary. The closed domains Bn, for n = 1, 2, . , which are nested in each other, have a nonempty intersection, which we denote by E. Let us assume that the cuts Cn are always such that E contains only boundary points of the domain B. We denote by Gn the domains in the disk ICI < 1 corresponding to the domains Bn. The closed domains Gn` also have a nonempty intersection. If this intersection consists of a single point Co on the circle C I = 1, the set E is called the prime end of the domain B corresponding to the point Co

1)

It is easy to show that a prime end corresponding to a point CO on ICI = 1 is independent of the sequence of cuts defining it. Let IC, ; and IC, I denote two sequences of cuts defining two prime ends E' and E" corresponding to the same point Co. Then the domains C' and Gn corresponding to the domains Bn and B" defined by these cuts have a unique common point Co. Consequently, each of the domains LR contains all C" beginning with some m; conversely, each of the domains Gn contains all G' beginning with some m. This property carries over

to the domains B' and B", but then, E" C B' and E' C k' for all n; that is E" C E' and E' C Ell. This means that the prime ends E' and E" determined by the cuts C' and Cn consist of the same points, and this in turn means that a prime end is completely determined only by the point Co corresponding to it. A prime end can also be characterized independently of the sequence of cuts C defining it. Specifically, let us show that the prime end E of the domain B corn responding to the point C on ICI = 1 is the geometric locus of all cluster points of sequences of points of the domain B corresponding to all possible sequences

1) (A prime end can be defined without bringing in the concept of conformal mapping, as was done in the definition of accessible boundary points (cf., for example, Carathcodory [1912]).

41

§3. CORRESPONDENCE OF BOUNDARIES

of points of the disk I CI < 1 that converge to Co. Suppose that corresponding to the points C. --' S0, where I' < 1 and 141 = 1, is a sequence of points zn E B. Since any one of the domains Gn considered abodt)ontains all t beginning with some m, any one of the domains Bn has the same property with regard to the Consequently, every cluster point of the sequence (znI belongs to all points $n, that is, belongs to E. Now suppose, conversely, that zo E E. Then z0 lies on the boundary of all the domains Bn. In the domain B, let us choose a sequence z0 such that zn E Bn, where n = 1, 2, .... Corresponding to of points zn .them in the disk ICI < 1 are the points t;n E Gn where n= 1, 2,.... It follows t ; that is, z0 is a cluster point of the set of points zn E Bn correthat n Co Isponding to the points n The question of the correspondence of boundaries under conformal mapping is completely answered in terms of a prime end (see Caratheodory [1913a]): Theorem 2. Under a univalent conformal mapping of a domain B onto the unit disk ICJ < 1, there exists a one-to-one correspondence between points of the circle ICI = 1 and the prime ends of the domain B. Proof. It follows from the definition of a prime end that to every prime end of a domain B there corresponds a unique point on the circle ICI = I. Let us prove the converse, namely, that for every point Co on ICI = 1 there corresponds a prime end of the domain B. Let us take two sequences (t n } and k 1, where n = 1, 2, ... , of points on I C I = 1 distinct from C. which converge from opposite sides to Co and to which correspond sequences of accessible boundary points zn and C.,'J zn of the domain B. We may assume that, for each n = 1 , 2, ... , the points and lie inside the arc (n (n on Cl I= 1 which contains Co (see Figure 5). Suppose zn and zn are determined by the curves In and In . We denote the corresponding curves in the disk ICJ < 1 by An and For every n, we choose on the curves A n' and A"n arcs with endpoints Cn' and Cn such that all their points are closer to the straight line passing through Cn and Cn than they are to the line passing through Sn - 1 and C _ 1 or to the straight line passing through Cn+1 and Cn+1 Let us connect by a straight line segment the ends of these arcs which lie in the disk Cl < 1. As we move from C, along An and from Cn along An to the first points of contact with this segment, we replace our original arcs with those just mentioned and we replace the straight line segment with the line segment joining the endpoints of these arcs, which lie in the disk ICI < 1. The cuts thus obtained in ICI < 1 are constricted, as n --, - , to the I

Cn+1

An.

42

11. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

point Co and have as preimages in the domain B the cuts C,, determining the prime end corresponding to the point Co. This completes the proof of the theorem.

Figure 5

As a supplement to this, we have Theorem 3. Let E denote the set of boundary points of a domain B defined by the cuts CR as above. For E to be a prime end of the domain B, it is necessary and sufficient that E contain no more than one accessible boundary point of the domain B. Proof. It follows from Theorems 1 and 2 that, if E is a prime end, it cannot contain more than a single accessible boundary point. Suppose now that E contains no more than one accessible boundary point. If E were not a prime end, the intersection of the domains GR used to define E would contain an arc of the circle (1 = 1. On this arc, let us choose two points S1 and S2 corresponding to the admissible boundary points z1 and z2 of the domain B. Corresponding to the curves 11 and 12 defining these points are curves Al and A2 in the disk W < 1 that terminate at (1 and Portions of the curves Al and A2 in sufficiently small neighborhoods of the points S1 and S2 lie in C,,. This means that portions of the curves II and 12 in neighborhoods of the points z1 and z2 lie in B,,. This in turn means that the points z1 and z2 belong to all the BR and hence to the set E, which contradicts the hypothesis of the theorem. This completes the proof of the theorem. In Figures 3 and 4, the segment cd is a prime end of the domain in question C2.

§3. CORRESPONDENCE OF BOUNDARIES

43

since it is easy to construct a sequence of curves Cn such that the intersection of the corresponding domains Bn is just this segment. In Figure 4 the segment cd contains only the single accessible boundary point d; in Figure 3 it does not contain any accessible boundary point at all. Analogously the spiral-shaped domain in Figure 2 has the entire boundary circle as its prime end, which does not contain a single accessible boundary point. Special types of domain. Up to now, we have considered the question of the correspondence of boundaries of arbitrary bounded simply connected domains under univalent mappings. For domains of special types, we can draw a number of supplementary conclusions. Thus, it follows from the preceding theorems that, if all the boundary points of the domain B are accessible, then a univalent mapping of this domain onto the disk ICI < 1 defines a one-to-one correspondence between boundary points of the domain B and points of the circle ICI = 1 because, in such a case, every prime end of the domain B consists of a single boundary

point. Let us show that, in this case, the inverse function z = 0(C) is continuous in the closed disk ICI < 1 if we define this function at a point Co on the circle ICI = 1 as the limit of its values as ( approaches Co from the open disk JCI < 1. We need only show that 4(t) is continuous on ICI = 1. If a sequence {Sn1 of points Cn of the closed disk ICI < 1 that converges to Co, where ICol = 1, contains only finitely many points of the circle ICI = 1, then the relation limT(Cn)=p(Co) n-+oo

follows from the definition of ¢(40). On the other hand, if infinitely many of the %o lie on the circle ICI = 1, then these points themselves constitute a sequence of points Tnk -+ CO and we need to show that for them limp (Cnk) _ p*(Co)

Let c denote any positive number. Consider, in the disk IcI < 1, a sequence of points Cnk satisfying the conditions I tnk - Cnk I C T

,

I'f l`ank) - 'P

IC

Since nk --. Co, there exists a K > 0 such that, for k> K, we have -q5(CO)I < c/2. It follows that, for k > K, (Snk) - (CO) I <E, and this means that limk_-'0(Cnk) the point Co.

This completes the proof of the continuity of -0(T) at

11. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

44

If the boundary of the domain B also has the property that, for any two distinct boundary points zI and z2, the complex numbers zI and z2, are not equal, then the function f(z) is continuous in the closed domain B. To show this, let zo denote a point on the boundary of the domain B. If zn E B, where n= 1, 2, ... , and zn _` z0 , then the corresponding sequence {cn } converges to a specific point Co on the circle ICI = 1 because, otherwise, the sequence ICn E would-have two subsequences that converge to distinct points on 141 = 1 and then the corresponding sequences of the points zn would converge to distinct points on the boundary of the domain B. It follows that Co is independent of the choice of sequence zn-a zU and this proves the continuity of f(z) on B.

If we apply what has been said to domains bounded by Jordan curves, we ob-

tain by virtue of the properties of such curves Theorem 4.1) A univalent con formal mapping of a domain B bounded by a

closed Jordan curve onto the disk ICI < 1 maps B bicontinuously onto the closed disk ICI < 1. Since the above considerations are of a local nature, we can conclude immediately that the following more general theorem is valid: Theorem 4 '. If a simply connected domain B has on its boundary a Jordan arc y, no interior point of which is a cluster point for the set of boundary points not belonging -to y, then the function C = f(z) mapping the domain B onto the disk ICI < 1 is continuous in B including the interior points of the arc y and it maps the arc y bicontinuously onto an arc y' of the circle ICI = 1. In the case of domains containing analytic arcs on the boundary, we have Theorem 5. If a simply connected domain B has on its boundary an analytic Jordan arc y, no interior point of which is a cluster point of the set of boundary points not belonging to y, then f(z) is regular at interior points of the arc y and at such points f 0(z) l 0; that is, the mapping is conformal. Proof. Suppose that the equation of the arc y is z = z (t) , where a < t < b.

Let to denote a point in the open interval (a, b). Let z0 denote z(to). The function z(t) can be expanded in a power series in a neighborhood of the point t = to. If we admit complex values for 1, this series defines a regular function in

1) Caratheodory [1913].

§3. CORRESPONDENCE OF BOUNDARIES

45

a sufficiently small neighborhood of the point to. Since z'(to) L 0 (by the definition of an analytic curve), we can choose a neighborhood sufficiently small that it will be mapped univalently by the function z = z(t). The part gt of this neighborhood lying in the upper or lower half of the t-plane is then mapped into the part g s of the neighborhood of the point zo lying in B and having a common boundary arc with y. The function C= f(z(t)) is regular in gt and it maps gt univalently into the part of the neighborhood of the point Co = f(zo) lying in the disk JCJ < 1 and adjacent to the circle JC1 = 1. By Theorem 4', f(z(t)) is then continuous in gt up to the real axis. By the symmetry principle, it can be extended in a neighborhood of to across the real axis and, in particular, it is regular at the point to itself and has a nonzero derivative at to. If t = t(z) is the inverse of the function z = z (t) , it is regular at zo and t'(zo) 0. Consequently, the function f(z(t(z))) = f(z) is also regular at zo and f'(zo) 31 0. This completes the proof of the theorem. In particular, if a domain B is bounded by a closed analytic Jordan curve, a 1 is regular in B and its inverse function f(z) that maps it onto the disk is regular in the closed disk 1cj < I. We conclude our investigation of special types of domain by noting that in

Chapter X, §1, we shall use the theory of functions of a real variable to investigate in addition the correspondence of boundaries under conformal mapping of domains bounded by rectifiable and smooth boundaries. l)

We also point out that the preceding theorems lead immediately to theorems on the correspondence of boundaries under univalent mapping of various domains onto each other. For example, a univalent mapping of one simply connected domain bounded by a closed Jordan curve onto another is single-valued and continuous, including their boundaries; in addition, if the boundaries of these domains are closed analytic Jordan curves, the functions representing the direct and inverse mappings are regular in the closed regions. In conclusion, we present another uniqueness theorem dealing with conformal mappings. We saw above that if all the boundary points of a bounded domain B Qre accessible, then a function z = '(C) that maps the open disk ICI < 1 1) Profound investigations of the correspondence of boundaries under conformal mapping have been made by the Moscow school of the theory of functions, which includes N.N. Luzin, I. I. Privalov, M. A. Lavrent'ev and others. A detailed survey of the corresponding results can be found in the book Thirty years of Mathematics in the USSR, pp. 324-332

(1948).

46

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

< 1. In this univalently onto the domain B is continuous on the closed disk case, the boundary Cof the domain B has the representation z = (eit) = z(t), for 0 < t < 2n, where z(t) is a continuous function, so that C is a continuous curve. Furthermore, since log[0(() - O(0)]/C is obviously a regular function in

Jt;J < 1 and is continuous on ItI'< 1, it follows that arg[O(i) - x(0)1/4 is a harmonic function in JCJ < 1 and is continuous on ICI < 1. This means that the increment in this argument as C moves around the circle JCJ = 1 is zero. But this means that, as C moves around the circle JCJ = 1 in the positive direction, the increment in arg[95(0 - 0(0)] is equal to the increment in argC, that is, equal to 227. We take as the positive direction around the curve C the direction in which arg(z - z0), where z0 = 0(0), acquires an increment of 27r when z completes one encirclement. By virtue of the continuity of the increment in dz (arg (z - zo))c =

z

)

zoJ

depending on z0 C B, the positive direction around C is determined independently of the choice of z0 and hence independently of the choice of mapping function 95 (0 -

Theorem 6. Suppose that all the boundary points of a simply connected bounded domain B are accessible. Then there is a unique univalent mapping of B onto the disk ICI < 1 such that three given points z1, z2, and z3 on the boundary C of the domain B indexed in order of their occurrence as one proceeds in the positive direction are mapped into three given points C1, CZ, and C3 on the circle JCJ = 1 similarly indexed. Proof. Suppose that the function (J' = O(z) maps the domain B univalently onto the disk KK'I < 1. Corresponding to the points z1, z2, and z3 on IC1 = 1 and t;3 arranged in order of occurrence as one proceeds are three points around the circle in the positive direction. If we now map the disk JC' I < 1 univalently onto the disk ICI < 1 in such a way that the points X1,.42, and C3 are mapped into 1, CZ, and C3 respectively (which we know is possible and is achieved by means of a fractional-linear function), then the composite function will yield the required mapping. The uniqueness of such a mapping is proved, as uniqueness is usually proved, by contradiction. This completes the proof of the theorem.

§4. DISTORTION THEOREMS

47

§4. Distortion theorems A considerable portion of the theory of conformal mappings deal with the investigation as to what restrictions the requirement that a mapping be univalent imposes on certain quantities connected with such a mapping (for example, the

modulus of a function or the modulus and argument of its derivative), that is, on the degree of distortion produced by that function at various points in a domain, etc. These restrictions are expressed in various inequalities, the simplest of -which will be given in the present section. Our starting point for the derivation of a number of inequalities dealing with univalent functions is Theorem 1 (the area theorem)'). Suppose that the function F(C) = C+ co, Q'1/C+ - = C+ Y-k=1 Rk/Ck is regular in the finite plane, has a pole at C= and is univalent in the domain ICI > 1. Then, ro

E

(1)

n=1

The name "area theorem" is due to the fact that this theorem is the analytic expression of the following geometric fact: under a mapping of a domain ICI > P, where p > 1, by a function w = F(C) , a part of the plane w = u + iv of positive area remains uncovered (the area principle). This geometrical interpretation indicates the course of the proof. Specifically, the circle ICI = p, where p > 1, is mapped by the function w = F(t;) = u + iv into the closed analytic curve w = F(pei°) = w(t) , for 0 < t < 277, which serves as the boundary of a domain of area S calculated from the formula 2x

S=

2n

w(t)+w(t) 2

uv'dt

tv

2is

2

rPeit + Pe-it

XL

2

+

2i

00

ane Int+anetnt

n=1

2p

rpest+Pe tt lj1 L

(t)TP1'(t)dt

00

n1 =

1) Gronwall [1914-19151.

Wane-int + naneint 29A

Co

9

'CP -

I

n=1

nlan12 pen

48

11. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

and this area is positive for arbitrary p > I. If we take the limit as p -i 1, we obtain (1). A similar theorem holds for the corresponding p-valent functions, that is, functions that assume no value in the domain ICI > 1 at more than p points.`)

An important consequence of inequality (1) is the fact that I all < 1, with equality holding that is, with al = et°' if and only if a2 = a3 = - = 0, that is, if F(C) = C+ e'1/C: This last function maps the domain ICI > 1 univalently onto the plane with a rectilinear cut from the point w = 2e"i "2 to the point w = -2e0,1'2. Suppose now that the function on

f (Z) = z + c2z2 + ...

,

Cnzn

(2)

n=1

is regular and univalent in the disk I zl < 1. Then, the function

fP(Z)=P f(ZP)=z+ Z'9" +...

(3)

(for p = 2, 3, ) is also regular and univalent in IzI < 1. The univalence follows from the fact that, if we had ff(z1) = ff(z2), where z1 and z2 are distinct num= f(zP) and hence zP = zP; that is, z1 = bers in I z I < 1, we would have f(zP) 1 I 2 2 iz2, where i]P = 1. But then, since the series (2) contains only terms with powers znp+1, where n = 0, 1, 2, , we would have ff(z1) _ iiff(z2); that is i] = 1, and hence, z1 = z2, which contradicts the assumption. It follows that the function

F(C)=

111 =C- 2 C + f° c l

is univalent in the domain ICI > 1. Then, as a consequence of the area theorem, we obtain the inequality Ice/21 _< 1, so that Ic21 _< 2 with equality holding, that is, with c2 = 2eia, if and only if F(C) = C+ e`°' /C, that is, if and only if

w =f (z) = (I + ealz)a

= Z - 2earz9 -I- ...

(4)

The function (4) maps the disk Izi < 1 onto the plane with a cut along the ray which issues from the point a-"`/4, and which would include the point w = 0 if produced

in the opposite direction. Thus we have obtained the result 2) 1) See Goluzin [1940] and Alenicyn [1947, 19501. 2) Bieberbach [1916].

49

N. DISTORTION THEOREMS

If the function (2) is regular and univalent in the disk izi < 1, then Ic2I < 2. This is a sharp inequality. With this bound for the coefficient c2, we can now easily obtain a covering theorem for the same functions. Suppose that the function (2) is regular and univalent in the disk I zl < 1 and suppose that c does not belong to the image of that disk under the function (2). Then c # 0 and the function

f1(Z)=Ceff (ZL-=z+C9+ 1 Z9+... is also regular and univalent in Izi < I. Hence

Thus, Icl > % with equality holding only when Ice + 1/cl = I1/c I - Ic2I = 2 and 1C21 = 2, that is, C2 = -2e`o'. However, this will be the case only for the function (4). Thus, we have obtained Koebe's covering theorem: Theorem 2. If the function (2) is regular and univalent in the disk zI < 1, then the disk Iwl < % (but not always a larger disk) is entirely covered by the image of the disk IzI < 1 under that function. The circle Iwi = % contains points not belonging to the image if and only if f(z) is the function (4).1) Koebe's theorem can be put in a different form: If a function f(z) such that f(0) = 0 is regular and univalent in I zI < 1 and does not assume the value c in Izi < 1, then I

(0) l

4c,

This follows from the application of the preceding theorem to the function

f(z)/f'(0),

which does not assume the value c/f'(0) in Izi < 1. If the, function (5)

maps the domain ICJ > 1 univalently, if c does not belong to its image, then the function

1) In this formulation, the theorem was proved by Bieberbach [1916]. Koebe[1907] proved only the existence of such a disk as that asserted in the theorem. This covering also holds for suitable p-valent functions (see Goluzin [19481).

50

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

Ft (z)=

=z+(c-ao)zs+...

F(i)-c 11

is regular and univalent in IzI < 1 and, consequently, Ic - a0I < 2. From this we obtain Theorem 3. If the function (5) maps the domain ICI > 1 univalently, then the

entire boundary of its image is contained in the disk Iw - aoI < 2. Another important application of a bound on the coefficient c2 is the deriva-

tion of bounds on the modulus and argument of the derivative of a univalent function.

Suppose that the function (2) is regular and univalent in IzI < 1. Then, for arbitrary z in IzI < 1, the function I(+ b (C) = f f(f'I(z)

(1C)

_f

Z)

I I)

= C -}- r2' -}- .. .

is also regular and univalent in ICI < 1. Calculation shows that

s=2 ((1-z2)f,(())-22). z Consequently, since I02I < 2; we have zf' (z) f, (z)

2rs

1-rs

c 14r-rs'

r= I z

(6)

If we replace the modulus with the real and imaginary parts in this inequality and note that

at (f' ((z)) = r d_ log I.f' ( z )

f (z))) = r Fr argf' (z),

we obtain the two inequalities

4±rs,

14-r'rdr 4 1

a

rs Car

(7)

4

arg f' (z)

rs .

If we integrate with respect to r from 0 to r, we obtain the inequalities (11+

), <

(z) I < (I

-

r) a,

arg f' (z) I <- 2 log 1 + r

(8) (9)

§4. DISTORTION THEOREMS

51

These conditional inequalities hold for arbitrary z in I zI < 1. Let us see precisely when equality can hold in them. Since they were obtained by integrating inequalities (7) with respect to r from 0 to r, for equality to hold in (8) and (9) for some function f(z) and some z in the disk IzI < 1, it is necessary that the corresponding equality hold in (7) along the entire straight line segment from 0 to z, and this means that equality must also hold in (6) on the same segment. In particular, it follows that i C2 = % I f"(0)/f' (0)I = 2. But this can be true only for the function (4). By direct verification, we see that equality does indeed hold in the case of this function for suitable real a. Consequently, inequality (8) gives sharp bounds for W WI as f(z) ranges over the class of functions that are univalent and regular in I zi < 1. This inequality constitutes a distortion theorems' for this class of functions. With regard to inequality (9), which constitutes a rotation theorem, 2) we have, for the function (4) with z # 0,

argf (z)=arg

_IZI

-e¢zac4aresinizi +etz) (1

4dr

<

IZI

4dr=2log1+r

that is, for this function, equality cannot hold in (9) at any z # 0. Later, we shall also give a sharp bound for Iarg f'(z)I (§1, Chapter IV). By starting with inequality (8), we can easily obtain upper and lower bounds for If(z)j. To obtain an upper bound, we integrate the second of inequalities (8) over the line segment connecting the point 0 with the point z. We obtain z

If (z)IcSI.f'(z)IIdzI 0

lf(z)Idlxllzl b

1

r dr = (1 --r),

(l

z

I

.

I z I)s

To obtain a lower bound, we take the point on the image of the circle I zI = r

that is closest to w = 0 and we draw a line segment between it and the point w = 0. If we then integrate the first of inequalities (8) along the curve L constituting the preimage of that line segment, we obtain l)Bieberbach [19161. 2) Bieberbach [1919].

52

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS Izl

If(z)I=S If'(z)IIdzI: S a

0

1-r dr= (l+IzlIz

(1+r)e

Thus, in I zI < 1, we have

r

(1+:r)a cI1(z)I_ (1 rr)a,

r=I zI.

(10)

Here, equality can hold for z # 0 only if f(z) is the function (4), as is the case for suitable choice of a. So far, we have considered functions that are univalent in a disk. Let us now prove the general distortion theorem. Theorem 4.1) Let B denote a domain and let E denote a closed subset of B. Then, there exists a finite positive constant M such that, for every function f(z)

that is regular and univalent in B and for arbitrary z1 and z2 in E,

1-
f (z.)

Proof. Without loss of generality, we may assume that E is a closed domain contained in B. Specifically, every closed set can easily be included in some closed domain contained in B and, if we prove inequality (11) for this domain, it must also hold (with the same M) for any set E contained in it. Let us first suppose that E is a closed bounded domain. Let d denote the distance between E and the boundary of B. This distance is necessarily positive. Let us cover the z-plane with a grid of squares of side d/4. The union of the closed squares of the grid that contain points in E is a closed domain B'. We have E C B' C B. Let z1 and z2 denote any two points in E. In the domain B', we can find an n-tupl a of points 61= z 1, 621 * * " I n = z 2 such that any two consecutive points lie in adjacent squares of the grid and the number n of them does not exceed the number N of all squares constituting B'. The distance between the points Sk and 6k+l (for k = 1, 2, . - , n - 1) does not exceed d/\12 < 3d/4. The disks , n, all lie in the domain B. Therefore, the function I z - ek I < d, for k = 1, 2, f (Ek + A) - f (Ek)

is, for each k, regular and univalent in ICI < 1. Consequently, in accordance with (8) we have in I z I < 1 1)(Koebe [19101.)

§4. DISTORTION THEOREMS

1+ICI

f (Ek+dC)

53 (12)

f (Ek)

I

In particular for t; = (6k 11 - Sk)/d, with

we have 3

f (Ek+1)

f (Ek)

Now, if we set k = 1, 2,

c

1 + 43113 = M.

11 4/ , n - 1 and multiply the inequalities obtained, we find

.

f (z9)

f (z')

c MI-A c m N-1= M.

The number M depends only on the domain B and the set E; it is independent of the form of the fraction f (z) and the choice of points z1 and z2. If we reverse the roles

of z1 and z2, we obtain the other inequality of the theorem. If the domain B contains the point oc, we can, by using the fact that inequality (11) holds for arbitrary points z1 and z2 on the circle I zI = R, where R is sufficiently large, show that this inequality also holds for arbitrary z1 and z2 in the domain IzI > R. This, in conjuction with what we have already proved, completes the proof of the theorem. Analogously, we have

Theorem 51) Let B denote a domain and let E denote a closed set contained in B. Then, there exists a finite positive constant L such that, for every function f(z) that is regular and univalent in B and for arbitrary points z1 and z2 in E,

I f (z9) I- I f (zl) I+ L I f (zl) I

(13)

Proof. We use the constructions made in the proof of the preceding theorem, again assuming E to be a closed bounded domain. Integrating inequality (12) and using (11), we have 3

If (4,1) -

{ J (4)I

(

\1

1) CKoebe [1910].)

4

41

54

II. CONFORMAL MAPPING ON SIMPLY CONNECTED DOMAINS

If we successively set k = 1, 2, ..., n - 1 and add the resulting inequalities, we obtain

I f(z9) -f(zi)I

12Md(n- 1)I f'(zi)I12MNdif'(zt)1,

from which (13) follows with L = 12MNd. If B contains oo, we again use the maximum modulus principle.

From this last theorem we have a Corollary. Let M1 and M2 denote two positive numbers. Let zo denote a point in a domain B. Suppose that every function in a family of functions that are regular and univalent in B is such that J f(zo)I < M1 and If'(zo)1 < M2. Then, this family uniformly bounded in B and the condensation principle can be applied to it. §5. Convergence theorems on the conformal mapping of a sequence of domains

An important question in the theory of conformal mappings is the question of convergence of a sequence of univalent functions. We present the following definitions. Suppose that we have a sequence of

univalent domains B1, B2,... in the z-plane, each including z = 0. If there exists a disk Jzj < p, where p> 0, that belongs to all the domains in Bn, we define the kernel of this sequence of domains as the largest domain containing z = 0 such that an arbitrary closed subset of it belongs to all the domains Bn from some n on. By "largest domain" is meant the domain containing any other domain possessing this property. If such a disk does not exist, the kernel of the sequence of domains B1, B2,... is defined to be the pointz = 0. We shall say that the sequence of domains B1, B2 , ... converges to the kernel B, and we shall denote this by writing Bn --.. B, if every subsequence of these domains has B as its kernel. In particular, if a sequence of simply connected domains B1, B29 ..., Bn, ... that include z = 0 converges to the limiting domain B (also including z = 0) in the sense that all boundary points of the domains Bn from some n on are arbitrarily close to the boundary of the domain B and all points of the boundary of the domain B are arbitrarily close to the boundaries of the domains B,,, then this sequence has the domain B as its kernel and it converges to that kernel. This applies also for a sequence of domains Bn that include z = 0 and satisfy the condition B1 C B2 C B3 C ... or for a sequence of domains Bn that

§5. CONVERGENCE THEOREMS

55

contain a neighborhood of the point z = 0 and satisfy the condition B1 3 B2

B33

.

Theorem 1 (Carathi6odory).1) Suppose that we have a sequence of functions fn(C), where n = 1, 2, . , that are regular in the disk ICI < 1. Suppose that z fn(0) = 0 and fn '(0) > 0 for n = 1, 2, .. . Suppose that, for each n, the function fn(C) maps the disk CI < 1 onto a domain Bn.

For the sequence ;fn(C)} to converge in ICI < 1 to a finite function, it is necessary and sufficient that the sequence }Bn} converge to the kernel B, which is either the point z = 0 or a domain having more than one boundary point. When convergence exists, it is uniform inside the disk ICI < 1. If the limit function f(C) i const, it maps the disk ICI < 1 onto the kernel B, and the sequence ton(z)} of inverse functions On(z) converges uniformly inside B to the function d7(z) inverse to f(C). in ICI < 1. Let us consider Proof. 1°. Necessity. Suppose that fn(S) -+ first the case f(C) - 0. We shall show that then B is the point z = 0 and that B

n

.-. B. Let us suppose that this is not the case. Then, there exists a disk

'I zi < p, p > 0, that belongs to all the domains Bn. Consequently, by the Schwarz lemma as applied to the disk IzI < p, we obtain I0, (0)l < 1/p, where 0n(z) is the inverse of the function fn(C). This means that fn (0) > p. But then, according to inequality (10) of §4 as applied to the function fn(S)/fn(0) in the disk ICI < 1, we have p ICI/(1 + IC!)z < I fn(S)I. Consequently, the sequence of the functions fn(C) does not converge to f(C) - 0, which contradicts the assumption. Since the same reasoning can be applied to every subsequence of domains among the B1, B2, ... , we have B n _, B. Now, let us suppose that f (C) J 0. Then, it follows from the inequality

fn(0)

IIICD'CIfn(C)ICfn(0) (1

(1

III

I)y

that the numbers fn(0) are bounded and that the functions fn(C) are uniformly

1) Caratheodory [1912]; Bieberbach [1913]. Markusevic [1936] showed that the conditions of 'Theorem 1 are also necessary and sufficient for convergence in mean of }fn(z)} to

f (z), that is, necessary and sufficient for 11m

n - ,:.

S

I fn (z) -f'(z) I' da = 0,

where fn(z) is taken equal to 0 outside the domain 8

.

56

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

bounded inside the disk Kl < 1. Therefore, in accordance with Vitali's theorem, the sequence of the functions fn(4) converges uniformly inside the disk ICI < I to the function f(C), which is univalent in'KI < 1. Consequently, the function z = f(C) maps the disk ICI < 1 univalently onto a domain B containing z = 0. Let us show that B possesses the property that an arbitrary closed domain B" is contained in all the domains Bn from some n on. Let S > 0 denote the distance from B" to the boundary of B. Let us cover the z-plane with a grid of squares of side 8/2 and let us consider the domain B, constituting the union of all (closed) squares containing points of B". Then, B" C B, and B, C B. Let A" and A" denote respectively the preimages of the domains B" and B, under the mapping z = in ICI < 1. Then, A" C A,, and A, is contained in the disk ICI < 1. Let S1 > 0 denote the distance from B" to the boundary of the domain B". Then, on the boundary of the domain A,, we have I f(C) - zol > S1 for any point zo C B". On the other hand, there exists an N > 0 such that, for n > N, we have l fr(y) - f(C)l < S1 on the boundary of A, since the sequence of the functions fn(C) converges uniformly inside the disk lCl < 1. Consequently, for every n > N, Rouche's theorem can be applied to the function

f, (Q - z0 = V n

(W + VP - Zn).

In accordance with this theorem, fn(C) - z0 has a zero in the domain A,. This means that the image of the domain A, under the mapping z = f(C), where n > N, contains an arbitrary point z0 C B". Thus, every closed domain contained in B and hence every closed subset of B is contained in all the domains Bn from some n on. Let us show that B is the largest domain possessing this property. Let B' denote any domain other than B that contains z = 0 and possesses this property. Then, from some n on, each function 0n(z) is bounded on an arbitrary closed subset of the domain B' and the set of such functions is uniformly bounded. It follows that the condensation principle can be applied to the sequence lon(z) 1. Let us choose from the sequence 10n(z)l an arbitrary subsequence lonk(z)1 that converges uniformly inside the domain B' to a regular function O(z)

such that 0(0) = 0 and 0'(0) > 0. Since 1

cp' (0) =urn p, (0) =1im n-m n-.mfn (0)

_

1

f (0)

it follows that 0'(0) > 0, that is, that OW is nonconstant and, consequently,

§5. CONVERGENCE THEOREMS

57

univalent in B', Let us show that c(z) is the inverse of the function f(C). Let Co denote an arbitrary point in the disk IQ < 1. Let Co (C B) denote the point f(C0). Let t denote a positive number such that the circle IC- CoI = e is con1. Let m denote a positive number such that 1g) - z0I > m tained in the disk on that circle. On the other hand, for k > K, we have I fnk(S) - f(01 < in on the circle K - coI = c. We again conclude from Rouche's theorem that the functions fn zo for k > K each have a zero, which we denote by Ck, in the disk '?- CoI <e. We have fnk(Ck) = zo, and C. = 0n (z0) for k > K. As k -- oc, the Ck -+ O(z0), so that O(z0) belongs to the disk JC- CoI <e. Since a is an arbitrary positive number, we see that O(z0) = Co. Thus, we have shown that z0 =

f(r;o) implies Co = /(z0); that is, O(z) is the inverse of the function f(c). Since this line of reasoning can be applied to any convergent subsequence of the sequence {on(z)} and the limit function is always the function O(z) inverse to

f(r;), the sequence {0n(z)I itself converges in the domain B' to the function 0 W. The function J = 0 (z) maps the domain B ' onto a domain A. Since ICbn(z)I < 1 in Bn, it follows that JO(z)1 < 1 in B '; that is, A is contained in the disk ICI < 1. This last fact enables us to show that B' C B. If z0 C B' and SD = o(z0), then ICoI < 1 and, consequently, z0 = f(Co) C B; that is, every point in the domain B' belongs to the domain B also. Thus, B is the largest domain containing z = 0 that possesses the property that any closed subset of it belongs to all domains Bn from some n on; that is, B is the kernel of the sequence of domains B

n

.

It follows that B has more than one boundary point.

Let us now take an arbitrary subsequence Bni, BnZ, .. , of the domains Bn. If we apply the conclusion just obtained to the corresponding subsequence fn1(0, fn2(J), of the functions fn(S), which converges to the function f(4) , we see that the function z = f(C) also maps the disk 1 onto the kernel of the subsequence JB }, and therefore this kernel coincides with B. This shows that --* B, so that we have proved the necessity of the condition of the theorem and have also proved the supplementary conclusions mentioned in the theorem. B

n

2°. Sufficiency. Now, suppose that Bn _., B and suppose that the kernel B either consists of the single point z = 0 or is a domain having more than one boundary point.

sy

First, let us suppose B consists only of the point z = 0. If f'n (0) did not converge to 0, there would be a subsequence ifnk(S)l such that, for each k,

58

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

f ' (0)I > E, e > 0, and

>(1+IC ),

throughout. the disk I I < 1. Consequently, the kernel B would not consist of a

single point. But if f'(0) -. 0, then, for all ( in the disk ICI < 1,

Ifn(t)ICIfn(0)I(l.l Consequently, fn(C) -+ 0 in ICI < 1. This proves the convergence of the sequence }fn(()}.

Now, let us look at the case when the kernel B is a domain having more than one boundary point. In this case, the numbers Q0) are bounded, because otherwise we would have a subsequence {fnk(0)} that approaches oo and we would conclude from the inequality

'

Iflk()I:Ifnk(0)I(1+Icl

I{I)I.

that the images of the disk 141 < 1 under the mappings z = fnk(c), from some k on, contain an arbitrary given disk I zI < R, which would contradict the convergence of 1B.I to B. It follows from what we have proved that the functions fn(0 are uniformly bounded inside the disk ICI < 1. Let us suppose now that the sequence if,,(C) I does not converge at the point Co such that ItOI < 1. Then, there

exist two subsequences }fn,(01 and If ,,(0} that converge in ICI < 1 to two distinct functions f+(C) and f«+(C). If neither f+(0 nor f+«(0 is identically equal to zero, then, from the proof of necessity as applied to the sequences and }f ,,(()}, we obtain the result that the functions z = f+(0 and z = map the disk ICI < 1 onto the kernel of the sequences }B.,I and {BR,, }, that is, onto the domain B. Note that we have f+(0) = 0, f; (0) > 0; f++(0) = 0, f;+(0) > 0. Therefore, by virtue of the uniqueness asserted in Riemann's theorem, we have f+(C) and this contradicts our assumption. On the other hand, if one of the functions f. (C), is identically equal to zero, the other cannot be identically equal to zero or they would again be identically equal to each other. By applying the necessity of the condition of the theorem, we conclude that the kernel of one of the sequences {B,, }, }B,,,,}

§5. CONVERGENCE THEOREMS

59

consists of the point z = 0 and that the kernel of the other sequence is some domain, so that the sequence {Br) does not converge to a kernel. Again, we have reached a contradiction. Thus, the sequence {fY)I must converge in the disk ICI < 1 to a finite function. Furthermore, in accordance with Vitali's theorem, the convergence is uniform inside the disk CI < I. This completes the proof of the theorem. Theorem 1 gives the conditions for convergence of univalent functions only in.the open disk ICI < 1. For the convergence of univalent functions in the closed disk I CI < 1, we give the following theorem, confining ourselves to domains of the Jordan type. Theorem 2.1) Let {Bn } n= 1, 2, ... , denote a sequence of simply connected domains each including the point z = 0 and each bounded by a Jordan curve. Denote the boundary of Bn by Cn. Suppose that the sequence 1B n' converges to a domain B (its kernel) bounded by a Jordan curve C. Let Ifn W I denote a sequence of functions fn(C) such that, for each n, fn(0) = 0, f'(0) > 0, and fn(O maps the open disk C1 < 1 onto the domain Bn. For the sequence 1fn(C)} to con1 to a function z = f(0 that vanishes at verge uniformly on the closed disk 0, has positive first derivative at 0, and maps the open disk ICI < 1 onto the do-

main B, it is necessary and sufficient that for every t > 0 there exists a number N > 0 such that, for n > N, there exists a continuous one-to-one correspondence between the points of the curves Cn and C such that the distance between any point of Cn and-the corresponding point of C will be less than E. f(C) uniformly on ICI < 1, then, by conProof of the necessity. If fn(C) sidering the points zn = fn(C) and z = f(C) as corresponding points on the curves C and C for each value of C on the circle ICI = 1, this establishes a one-toone continuous correspondence between Cn and C with the property stated in the theorem.

Proof of the sufficiency. We first prove that under the condition of the theorem, the set of functions fn(C) is equicontinuous on the circle ICI = 1. Let us suppose the opposite. Then, there exists a S > 0, a sequence of positive integers nk, k = 1 , 2, ... , increasing without bound as k increases, and two sequences of points on the circle 1 such that when k and

1) Rado [1922-1923]. In the case of domains with arbitrary boundaries, the question has been thoroughly investigated by Marlcusevic [1936].

H. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

60

C'

k

r 0, and I fnk (Ck) _-fnk (Ck) I > 8.

We may assume that the sequences Irk I and I converge to a common limit a such thaattI al = 1 because, if this is not the case, we can shift to a subsequence Ifnk(( )I of functions satisfying this condition and keep the same notation for it. Analogously, we may assume that the sequences Iak I and 16k }, where

ak = fnk(Sk) and bk = fnk(converge, let us say, to points a and b. The

points ak and bk lie on Cnk and the points a and b lie on C. On each curve Cnk, we denote by ak the arc akbk corresponding to the arc Ckk , (which is constricted to the point a as k -+ o) and we denote by rk the remainder of the curve C. Since we are assuming that, for sufficiently large k, the distance between ak and bk exceeds S, there exists for such k a point ck on ak at a distance no less than 8/2 from ak and bk. We may assume that ck --+ C, where c E C because otherwise, we could shift to a subsequence Ifnk(C)M We have I c - al > 6/2 and Ic - bI > 8/2. Consequently, c is an interior point of one of the arcs into which the points a and b partition the curve C. We denote this arc by a and the remainder of the curve C by r. Let a' denote an arc on C that includes c and that is contained, together with its endpoints, inside the arc a. Let 77 denote a positive number that is less than the

distance from c to C - a' and less than the distance from a' to r. Let K denote a positive number such that, for k > K, there exists a continuous one-toone correspondence between Cnk and C such that the distance between corresponding points of Cnk and C is less than 77/2 and

Ia-ahI<2, Ib-bkI<2, Ic-ckI<2.

(1)

Then, for k > K, there are no points rk in the open disk I z - cI <77/2. To show this, let us suppose that, for some k > K, there exists a point dk E rk that lies in that disk. Let ck and dk denote the points on C corresponding to the points ck and dk. We have

Ck - C I CI Ck- CkI+I Ck - C \11

dk- cl

dk - dk I + dk-C <71,

that is, c' and dk lie on the arc a'. Since Ck a ak and dk E rk, it follows that ck and dk separate the points ak and bk on C. Therefore, corresponding to

61

§5. CONVERGENCE THEOREMS

the part of the arc a ' between ck and dk on Cnk is an arc containing one of the points ak or bk, let us say, ak. Consequently, the point on C corresponding to ak lies on the arc a' and its distance from a exceeds q, which contradicts the first of inequalities (1). Thus, for k > K, every point of the curve Cnk in the disk z - cl < rl/2 is a point of the arc ak. For k > K, we define the function Ok(z) = Onk(z) - a, where Onk(z) is the inverse of the function fnk(0. This function is regular in Bnk. Setting Ek = we see that Ek -+ 0 as k -+ o. Now, let zo denote an arbitrary point of B that lies inside the disk lz- cl < rl/4. Around this point, let us draw the circle lz - z0l = 71/2. On this circle there are exterior points of the domain B and hence an arc y lying outside B and having a central angle of 2n/m, where m ,js a positive integer. For k greater than some number K1 > K, this arc lies outside Bnk and z0 lies in the interior of Bnk. The limiting values of tk(z)l as z approaches boundary points of the domain Bnk that lie in the disk i z - z0l < rl/4 do not exceed Ek since there are no points of the arc rk in the disk l z - z0l < q/4. Lemma 2 of §3 is now applicable. In the present case, M = 2 and r = Ek. Therefore, for k > K1, we have l

l

7k z0) l C m Ek`Zm-1.

But as k -+ o, the sequence fik(zO) converges to O(z0) - a, where O(z) is a function mapping the domain B onto the disk l zl < 1. Consequently, q(z0) - a = 0; that is, ¢(z) = a for all points z C B in the disk l z - cl < 77/2 and hence everywhere in B, which is impossible. This contradiction proves the equicontinuity of 1. The functions fn(C) are also uniformly the functions fn(c) on the circle bounded on ICI = 1. Reasoning as in subsections 2° and 3° of the proof of the :condensation principle, let us show that the sequence (fn(0 l contains a subseltence that converges on some set that is everywhere dense on the circle 1. Such a And then let us prove that this sequence converges uniformly on oice can be made from an arbitrary subsequence of the sequence (fn(C){. This Obables us to show that the sequence Ifn(O} itself converges on Kl = 1. Specifically, if the sequence (fn(C)I did not converge at some point on !, = 1, we could choose two subsequences of it that converge uniformly on the !8 rcle 1 and consequently in the disk Kl < 1 to two distinct functions that Ste regular in the open disk ICI < 1 and continuous on the closed disk Out this is impossible because, by Theorem 1, the sequence ffn(o I converges in `!

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

62

ICI < 1 to a unique function. Having proved the convergence of the sequence (fn(01 on the circle ICI = 1, we again keep in mind the equicontinuiry and uniform boundedness of the functions fn(s) to show, just as in subsection 3 of §2 of Chap. ter I, that the sequence (fn(0j converges uniformly on the circle ICI = 1 and hence on the closed disk ICI < 1. Obviously, the limit function is f(C). This completes the proof of the theorem. §6. Modular and automorphic functions

Among the different functions that map simply connected domains onto each other, of great theoretical importance are the functions that map certain circular polygons, that is, domains whose boundaries are arcs of circles, either onto a disk or onto a halfplane. In the present section, we shall study the nature of such mappings.

Let us consider first the case in which the domain B is the interior of a regular circular triangle with zero angles inscribed in the circle Izl = 1 (see Figure 6). The sides of this triangle are obviously orthogonal to the circle I zl = 1. Suppose that a function z = halfplane NO > 0 univalently onto the domain B in such a way that the points 0, 1, and oa are mapped into the

vertices of the triangle. Let C= µ(z) denote the inverse of that function. Without knowingly the explicit expression for the functions A(C) and EL(z), we can, by using the principle of symmetry, ascertain the essential properties of these functions in a purely geometrical manner. We denote the sides of the triangle by L1, L2, and L3

Figure 6

and the segments of the real axis corresponding to them by L, L2, and L' . The function ft (z) , which is regular in B and continuous on B except at one of the vertices, assumes real values on the boundary of the domain B. Consequently, in accordance with the symmetry principle, it can be extended past the sides L1, L2, and L3 into the domains (which we name B1, B2, and B3 respectively) obtained from B by an inversion about the circles of which L1, L2, an d L3 are arcs. The domains B1, B2, and B3 are also circular triangles with zero angles inscribed in Izl = 1. Each of them is adjacent to B along one of the sides

1) An explicit expression for the function k( t) will be given in § 1 of Chapter III.

§6. MODULAR AND AUTOMORPHIC FUNCTIONS

63

L1, L2, L3. This is true because under these inversions, an arbitrary circle is mapped into a circle and the circle I zl = 1, being orthogonal to all the circles across which the extension is made, is mapped into itself. In the domains B 1, B2, and B3, the function 4=µ(z) is regular and it maps each of these domains onto the halfplane (4) <0. Consequently, Et (z) is regular in the domain B' ,constituting the union of the domains B, B17 B2 and B3. However, it is no longer univalent in this domain since it assumes at three points of the domain B' every value of the halfplane (C) < 0. If we think of three halfplanes ``(C) < 0 obtained by mapping the three domains B1, B2 and B3 as lying one above the other and connected with the halfplane Z(C) > 0 along segments L11, LZ and L' respectively, we obtain a Riemann surface onto which the function C+ Et (z) maps the domain B ' bijectively. On the boundary of B ', the function C= t (z) is real. The domain B' is a circular hexagon with zero angles inscribed in the circle JzJ = 1. The sides of this hexagon are orthogonal to that circle. Let us repeat this procedure with the domain B'; that is, let us extend the function Et(z) past the six sides of B' into the domains obtained from the domains B1, B2 and B3 by inversions about the corresponding sides. The function µ(z) is regular in each of the six newly obtained domains and it maps them onto the halfplanes ` (c) > 0.Let us think of these halfplanes as being distinct and joined along suitable segments L 1, L 2' and L 3 of the real axis with the halfplanes NJ (C) < 0 that we already have. Consequently, the function C= Et (z) is regular in the newly obtained domain B ", which is a circular duodecagon. It maps B " objectively onto the Riemann surface consisting of seven upper and three lower planes connected in a suitable manner along L1, L2' and L. If we continue this extension process indefinitely, we get, on the one hand, a "modular" grid in the z-plane composed of an infinite set of circular triangles with zero angles all contained in the disk I zl < 1 and having sides orthogonal to the circle Izl = 1 and, on the other hand, a Riemann surface lying over the c-plane and consisting of an infinite set of halfplanes connected with each other along

suitable segments Li, LZ and L3 of the real axis. Let us show that the domain B' consisting of the union of all triangles constituting the modular grid coincides with the entire disk I zl < 1. To do this, it will be sufficient to show that the set of vertices of the triangle of the grid is everywhere

dense on the circle Izl = 1. Let us suppose the opposite, namely, that this circle contains an arc y that does not include any vertex of any of the triangles constituting the grid. Then there exists a sequence i l n j , n = 1, 2, ... , of arcs of the

64

H. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

grid, each arc resting on a y-containing arc of the circle I zl = 1, that converges to an arc L of that circle. Obviously, the arc L is orthogonal to the circle I zl = 1 and it rests on an arc y' containing y. Then, there are no points of the domain B* between y' and L. But this is impossible because, for sufficiently large n, the inversion about the arc In of the entire polygonal domain Bck> whose boundary contains this arc maps the point z = 0 into a point between y 'and L, and hence there are points of B' between y' and L. This shows that the set of vertices of the modular grid is everywhere dense on the circle Izl = 1 and hence that the domain B* coincides with the entire disk I zl < 1. Therefore, the function µ(z) is regular in the disk Izl < 1 and it maps that disk bijectively and conformally onto the infinite-sheeted Riemann surface mentioned above. The function µ(z) assumes every value other than 0, 1, and oe infinitely many times in the disk Izl < 1. On the other hand, it does not assume these three values anywhere in that disk. With regard to the points of the circle I zl = 1, these are all singular points of the function µ(z) because an arbitrary neighborhood of each such point contains infinitely many triangles of the modular grid and hence µ(z) assumes in the subset of this neighborhood contained in I zl < 1 all values other than 0, 1, and °°. Consequently, µ (z) cannot be extended outside the disk I zl < 1 and this disk constitutes its entire domain of definition. In this case, we say that the circle Izl = I is the natural boundary for µ(z). The inverse function A((), defined by any element, can be extended to the (-plane along an arbitrary path that does not pass through 0, 1, or oo. Here, IA(t) < 1 everywhere, though A(C) is obviously a multiple-valued function. The function µ(z) is called a modular function and the Riemann surface onto which it maps the disk I zl < 1 bijectively is called a modular Riemann surface. This name is due to the connection between µ(z) and certain quantities encountered in the theory of elliptic functions and called moduli. The modular function µ(z) possesses the remarkable property of invariance under a certain group of fractional-linear transformations of the argument. Speciffically, fractional-linear transformations composed of an even number of inversions of the type encountered above constitute a group of transformations of the argument under which µ(z) does not change its values. Single-valued analytic functions that remain unchanged under a certain group of fractional-linear transformations of the argument are called automorphic functions, Examples of automorphic functions with infinite groups are trigonometric and elliptic functions. Next to elliptic functions, a modular function is the simplest nonelementary example of an

§6. MODULAR AND AUTOMORPHIC FUNCTIONS

65

automorphic function with an infinite group. We obtain other very simple automorphic functions if we consider a mapping onto the halfplan e ` (C) > 0 of a domain B contained in the disk Izl < 1, where B is an arbitrary circular triangle all sides of which are orthogonal to the circle I zI = 1 and all interior angles of which are nonzero. If the interior angles of this triangle have values it/ml, rr/m2, an d

rr/m3, where ml, m2, and m3 are positive integers, then, just as in the case of modular functions, the function C _ 0 (z) , which maps the domain B univalently onto the halfplane ` (C) > 0, can be extended to the entire disk I zl < 1. This extension is regular in I zl < 1 except at points that are mapped into - and are poles of 0(z), and it maps the disk IzJ < 1 onto a many-sheeted Riemann surface lying over the entire Cplane. To prove the extendability of 0(z) to the entire disk I z I < 1, let us prove first that the vertices of the triangular grid, all of which now lie in the disk I zI < 1, have

no cluster points inside the disk Iz) < 1 (see Figure 7). Let us suppose, to the contrary, that zo is a cluster point of the set of vertices of the grid and let {zk i,

Figure 7

for k = 1, 2, , denote a sequence of distinct vertices of the grid that converges to z0. Since these points zk are vertices of triangles Ak, for k = 1, 2, .. , obtained from B by an even number of inversions, that is, by certain fractional-linear transformations z ' = Tk(z) , where k = 1 , 2, , of the disk IzI < 1 into itself included in the group of transformations that leave the function S6(z) invariant, it will be sufficient to show that the functions Tk(z) map the vertices of the domain B into a set of points that has no cluster points in I zI < 1. We may assume that the zk are such that the triangular grids with vertices at the points zk for different values of k have no points in common, since otherwise we could shift to a subsequence of the sequence {zk { for which this would be the

66

If. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

case. Let a denote one of the vertices of the domain B and suppose that the disk I z - al < p, p > 0, lies entirely in a domain composed of triangles of the grid with vertices at a. The functions z ' = Tn(z), n = 1, 2, . . , map this disk onto disjoint disks Kn contained in I zI < 1. Since the functions 7ID- T. (a+pz)-Ta (a) =.Z-+- .. . pTn (a)

are univalent in Izl < 1, the disk Iwl < % lies entirely in domains onto which these functions map the disk I zl < 1. This means that, for each n, the disks of radius not less than I Tn(a)/4I p lie in Kn. Since these disks are disjoint and are contained in I z'I < 1, the sum of their areas is finite and, consequently, the series 10=1I Tn (a)I 2 converges. Then Tn(a) -. 0 as n -. 00. If a = T (a), the n n function z' = T. (z) is obviously obtained from the equation

Z' -an `eten Z-a 1 - anz'

1 - az '

where a n is a real constant. Therefore Tn (a)

eian

1

so that 1 and --. 1 as n --+ oo. This proves that the sequence lan I has no cluster points in I zl < 1. The situation is analogous with the remaining vertices of the domain B. Since each of the points zk is the image of a vertex of the triangle B under one of the functions z' = Tn(z), it follows that the sequence lzk l also has no cluster points in

I z I < 1.

Let us show now that any point z in the disk I z( < 1 belongs to one of the triangles of the grid or lies on its boundary. Let us suppose the opposite, that is, that there exists a point z in the disk Iz1I < 1 that does not belong to any triangle of the grid. Let us draw a line segment from zi to some point of the domain B. We conclude that this segment contains points of an infinite set of triangles of the grid since it would otherwise be possible to get from B to zI by means of a finite number of inversions and the point zi would belong to some triangle of the grid. From this infinite set of triangles, we choose a sequence of triangles such that the sequences of their vertices converge to limits a, /3 and y, which must necessarily lie on the circle I zl = 1. If all three points a, 8, and y are distinct, the sequence of the triangles themselves must converge to a circular triangle with vertices a, 0, and y and with sides orthogonal to I zl = 1, that is, to a triangle 1

§7. NORMAL FAMILIES OF ANALYTIC FUNCTIONS

67

with zero angles, which is impossible. On the other hand, if only two of the points a, 0, and y are distinct, the sequence of these triangles converges to some arc of a circle orthogonal to the circle I z+ = 1, which again is impossible. Finally, if a, 0, and y coincide, the triangles must be constricted to a point. But this too is impossible since all these triangles contain points of the segment joining z 1 with some point of the domain B. This contradiction shows that the triangular grid entirely covers the disk I zl < 1. §7. Normal families of analytic functions. Applications The results obtained above on conformal mapping owe their generality in many ways to the principle of condensation of analytic functions. But these results in turn lead to considerable further development of questions on the convergence of analytic functions. To show this, we introduce the concept, proposed by Montel, of a normal family of analytic functions (cf. Montel [1936]). An infinite family Yk = If (z)f of functions that are regular in a domain B is said to be normal in B if every sequence of functions belonging to that family has a subsequence that converges uniformly inside B to a regular function or to m. A family is said to be normal at a point z0 E B if it is normal in some neighborhood of z0. If a family 911 = jf(z)} is normal in a domain B, it is obviously normal at each point z0 in B. Let us prove the opposite, namely that if a family 9R = If(z)1 of functions that are regular in B is normal at each point z0 E B, it is normal in the domain B. We may assume that B does not include DC. Let us take the set of all points of the domain B with rational coordinates and arrange it in a sequence Irk j, k= 1, 2, . , and let us denote by Pk the radius of the largest disk with center at rk in which the family 911 is normal. One can easily see that the subsequence {p,, k } can converge to 0 only when the corresponding points cluster r"k only on the boundary of B. Now, let (ff(z)I, n = 1, 2, - , denote an arbitrary sequence of functions of the family R. Since 911 is normal in the disk Iz - r11 < p1, the sequence jfn(z)j has a subsequence {fR 1(z)j that converges uniformly in Iz - r1I < p1/2 to a regular function or to oo. Furthermore, the sequence (f 1(z)} in turn has a subsequence jf2(z) j that converges uniformly in I z - r21 < p2/2 to a regular function or to w. If we continue the process indefinitely, the diagonal sequence (fR R(z)1, n = 1, 2, , converges in the entire domain B to a regular function or to - and the convergence is uniform in each of the disks Iz - rkl <

Pk/2, k = 1, 2,. .. Let us show that it also converges uniformly in B. If E is a

68

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

closed set contained in B, it is entirely covered by the disks 1z - rk I < pk/2. By the Heine-Borel theorem, we can choose a finite set of these disks that provide the same covering. But the sequence E f,, n(z)1 converges uniformly in each of these disks. Consequently, it also converges uniformly on the closed set consisting of points of all these disks and, in particular, on E. This proves the assertion. If all functions of a family 9 that is normal in a domain B are bounded at a single point z0 E B, they are uniformly bounded inside B. This is true because all convergent sequences of functions belonging to 9 can converge only to regular functions, and then the assertion follows according to the condensation principle stated at the end of §2 of Chapter I. Let us now give some tests for the normality of a family. Theorem 1 (fundamental test for normality (Montel)). 1) If a family l of functions that are regular in a domain B is such that there exist two distinct values a and b which no function of the family assumes anywhere in B, then 9 is normal in B. Proof. We may assume that a = 0 and b = 1 since, otherwise, we could consider not the family 1 _ {f(z)I but the family 1(f(z) - a)/(b - a) I the normality of which would imply the normality of R. Furthermore, it will be sufficient to prove the theorem for the case in which B is a disk because, by virtue of what was said above, we need only show the normality of l at each point of the domain B. Thus, let us suppose that the domain B is a disk and that the functions f(z) E R do not assume the values zero and 1 in B. Let w = X(C) denote the function defined in §6 that maps the halfplane ) (C) > 0 onto a regular circular triangle with zero angles inscribed in the circle I zl = 1. The composite functions )f(z) = X (f(z)) , where f(z) E 9, can be extended to B along an arbitrary path since the only singular points of the function X(0 are 0, 1, and ac and f(z) does not assume any of these values in B. Consequently, the functions of are regular in B and they satisfy in B the inequality I0/z)1 < 1. They do depend, however, on the choice of branch of the multiple-valued function X(C). Therefore, let us normalize Of (z) in such a way that the point Of (ZO) will, for given z0 E B and an arbitrary function f(z) E fit, lie in the basic (regular) circular

1) Mantel [19271.

§7. NORMAL FAMILIES OF ANALYTIC FUNCTIONS

69

triangle or in one of the adjacent triangles. We apply to the functions 0 f(z) the principle of condensation of analytic functions. Now consider an arbitrary sequence [fn(z)j of functions fn(z) E R. Let us look first at the case when the set of numbers fn(zo) has a cluster point a other

than 0, -1, or e. Let jfn,(z)I denote a subsequence of the functions fn(z) such that fn, (zo) --, a as n' -. -. The sequence (difn (z)), where 0fn, (z) = X(fn,)) has a subsequence {(kfnk (z)) that converges uniformly inside B to a regular function O(z).

We have I(k(z)I < 1 in B. Here, equality cannot hold at any point of the domain B because we would then have O(z) = const = e" Y and, in particular, O(z0) = e`y, that is, I.0 fn (z0)l ---+ 1, which in turn can be the case only if [fnk(z0)j converges k

to 0 or I or 00. Thus, I¢i(z)I < 1 in B. Now, let E denote an arbitrary closed subset of B. On this set, JO(z)1 K, we have ,cf (z) - O(z)1 K and z E E, we have nk IOfn WI
Iwi < 1. Consequently, µ(w) is bounded for Iwi K, we have IFL(X(f,, (z)) I <M on E; that is, I fnk(z)I <M. This proves the uniform boundedness of the functions fnk(z) in the interior of B. But now we can choose from the sequence If W I a new subsequence that connk verges uniformly in the interior of B to a regular function. Thus, in the present case, we have succeeded in choosing from the sequence i fn(z) f a subsequence that converges uniformly in the interior of B to a regular function. Now, let us look at the case in which the sequence ifn(z0)I has no cluster points other than 0, 1 and 00. Then, if 1 is a cluster point of the sequence, let us consider a sequence of functions Fn(z) = fn (z) defined (by virtue of the double value of the radical) in such a way that Fn(zo) *-1 as n --* oo. The functions Fn(z) are regular in B and they do not assume in B the values 0 and 1. Consequently, they come under the case just considered with a = -1. Therefore, the sequence IFn(z)) contains a subsequence (Fnk (z)I that converges uniformly in the interior of B to a regular function F(z). But then the sequence of the functions [Fnk(z)]2 = fnk(z) also converges uniformly in the interior of B to a regular function. Furthermore, if 0 is a cluster point of the set of numbers fn(zo), then the functions Fn(z) = 1 - fn(z), which are regular in B, do not assume in B the values 0 and 1, and the limit of the sequence of their values

70

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

at the point z0 is the number 1. Consequently, these functions come under the case considered above, and we can choose from their sequence a subsequence }Fn (z)} that converges uniformly in the interior of B to a regular function. The k same is true of the sequence of the functions fnk(z). Finally, if oo is the only cluster point of the set of numbers fn(z), then the functions Fn(z) = 1/fn(z) are regular in B, they do not assume in B the values 0 and 1, and the sequence of their values at the point zo approaches zero. Consequently, the sequence of these functions contains a subsequence }Fnk (z)} that converges uniformly in the interior of B to a regular function F(z). Furthermore, F(zo) = 0. It follows that the function F(z) is identically equal to zero because otherwise, just as in the proof of Theorem 2 of §1 of Chapter I, we could show on the basis of Rouche's theorem that all the functions Fnk(z) beginning with some n have zeros in an arbitrary neighborhood of the point zo, which is impossible. Consequently, fnk(z) -+ oo uniformly in the interior of B; that is, }fnk(z)} is the desired subsequence of the functions fn(z). This completes the proof of the fundamental normality test. Theorem 2 (generalized normality test).') Let -9 denote a family of funcP tions f(z) that are regular in a domain B. Suppose that there exist two distinct finite numbers a and b such that not a single function in the family 111 assumes the value a anywhere in B and no single function in t assumes the value b at more than p points in B. Then 2 is normal in B. Proof. Again, we may assume that the numbers a and b are 0 and 1. Let I nf (z)} denote an arbitrary sequence of functions belonging to l P . Consider the functions gn(z) = P+1 fn (z) with arbitrary choice of value of the radicals. These functions are regular in B and do not vanish anywhere in B. Let ql, 'q2, ... , . . , ?P+1 denote the different values of the (p + 1) st root of unity. Then none of the functions gn(z) assume any of these values in B because otherwise some of the functions fn(z) would assume the value 1 at more than p points of the domain B. Suppose that gn(z) 'qnz in B. The functions h(z) = gn(z)/><,Rn, which n are regular in B, do not assume in B the values 0 and 1. Consequently, in accordance with the fundamental normality test, the sequence }hn(z) } contains a subsequence {hnk(z)} that converges uniformly in the interior of B to a regular function or to o. The corresponding sequence } fnk(z) } also converges uniformly 1) Montel [1927].

§7. NORMAL FAMILIES OF ANALYTIC FUNCTIONS

71

in the interior of B to a regular function or to oo. This completes the proof of the generalized normality test. These two normality tests have a number of remarkable applications in questions regarding the behavior of analytic functions in a neighborhood of essential singularities and in covering questions. We present some of these applications. Theorem 3.1) Let z = 0 denote an essential singularity of the function f(z). Then the sequence lfn(z)I, where fn(z) = f(z/2"), for n = 1, 2, . , is not a normal family in any neighborhood of z = 0 or even at the point z = 0 itself. Proof. Let us take an arbitrary domain 0 < IzI < c in which f (z) is regular, and let us suppose that the functions fn(z) = f(z/2") constitute a normal family in that domain. If {fn(z) } contains a subsequence (fnk(z) I , for k = 1 , 2, . , that converges uniformly inside the domain 0 < I zI < c to a regular function, then the functions fnk(z) are uniformly bounded inside that domain and, in particular, on the circle Izj = c/2. Suppose, for example, that I fnk(z)I < M on IzI = c/2. But the functions fnk(z) assume on IzI =c/2 the same values as f(z) on IzI =E/2"k+1 n+k1+1 ; Consequently, the inequality I f(z)I <M holds on all the circles IzI = /2 , and hence, it holds between them, that is, in the entire domain 0 < I zI < c/2 this is impossible for the essential singularity z = 0. On the other hand, if all convergent subsequences of ifn(z)l converge only to D, let us show that the point z = 0 is not a cluster point of the zeros of the function f(z). Let us suppose that, to the contrary, for any c > 0, there exists a sequence of distinct numbers zk . We may assume that such that 0 < I zk I < c/2 and f(zk) = 0 for all k = 1, 2, nk, where c/4 < I zk I < c/2. The sequence {fnk(z) I cannot have a subzk = zk %2 sequence that converges in 0 < IzI < c to oc since there are no zeros in the annulus c/4 < I zI < c/2. Consequently, a convergent subsequence of this function, which is also a subsequence of the sequence {fn(z)1, converges to a regular function and this in turn contradicts our assumption. Therefore, the functions (b n(z) _ l/fn(z) are, beginning with some n, regular in the domain 0 < I zl <,E , and the subsequences {gnk(z)I converge to the identically-zero function. This implies, 1

as above, that the function O(z) = l/f(z) , 0 as z

0, so that f(z)

which is impossible. This completes the proof of the theorem.

1) Montel [1927].

72

R. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

Two other theorems are consequences of this one: Theorem 4 (Picard). If zo is an essential singularity of a function f(z), then f(z) assumes in an arbitrary neighborhood of zo all finite complex values with one possible exception. 1)

Proof. Let us assume that zo = 0. Suppose that there exist two finite numbers a and b and a neighborhood of zero in which f(z) does not assume either the value a or the value b. Then the functions fn(z) = f(z/2n), n = 1, 2,. , also do not assume the values a and b in that neighborhood. Consequently, they constitute a normal family, which, by Theorem 3, is impossible. Theorem 5.2) Let zo denote an essential singularity of the function f(z). Then, there exists a ray 1 issuing from zo such that, for every neighborhood of the point zo and every sector, with vertex at zo that is bisected by 1, the function f(z) assumes all finite values with one possible exception in the portion of that neighborhood lying in that sector. Proof. Again we can assume that zo = 0. Let I zj
U.-

This completes the proof of the theorem.

Let us now turn to the covering theorems. In §4, we proved Koebe's covering theorem, which asserts that, if a function f(z) = z + c2z2 + is regular and univalent in I zI < 1, then the image of the disk I zI < 1 under this function completely covers the disk JwI < 1; that is, the disk is one and the same for all 1) That there can exist one finite value never assumed by f(z) is illustrated by the function f(Y) = e1/ z, which has a singularity at z = 0 and yet f(z) /0 for z 0. 2) Julia [19241.

§7. NORMAL FAMILIES OF ANALYTIC FUNCTIONS

73

functions of this kind. The corresponding assertion for the entire class of functhat are regular in I zl < 1 is not valid; that is, we tions f(z) = z + c2z2 + cannot exhibit a disk of positive radius with center at the origin that is entirely covered by the image of the disk I zj < 1 under the mapping of each such function because, for example, the function f(z)=l-(l-z)R n

=z+c$ z$-}-...

is regular in I zj < 1 for arbitrary given n 0 but it does not assume in I zl < 1 the value 1/n, which can be made arbitrarily close to zero for n sufficiently great. On the other hand, we do have the following covering theorems: that Theorem 6.1) For the family of all functions w = f(z) = z + c2z2 + Izj < 1 and assume the value 0 at no more than p points of the are regular in disk I zj < 1, there exists a p > 0 such that the image of the disk I zj < 1 under an arbitrary function of the family completely covers the disk JwI < p. Proof. Suppose that no p > 0 with the property described in the theorem exists. Then there must exist a sequence of numbers an that approaches 0 as , of the family in ques-. and a sequence of functions fn(z), for n = 1, 2, n .lion such that, for every n, fn(z) 3L an. The functions on(z) = fn(z)/an, for n = 1 , 2, ... , constitute in i zj < 1 a family of functions that do not assume the value 1 at all and do not assume the value zero at more than p points of the disk Izl < I. Consequently, in accordance with the generalized normality test, this family is normal in I zI < 1. Since n(0) = 0, it is also uniformly bounded in the interior of the disk Izj < 1. Let {onk(z)} denote a subsequence that converges uniformly in ' (0) _ the interior of the disk jzi < I to a regular function (k(z). Then, 9 nk 1/ank which contradicts the fact that ak -. 0. This completes the proof of the theorem. Theorem 7.2) For arbitrary c in the open interval (0, 21r), there exists a pF > 0 such that the image of the disk I zj < 1 under any function w = f(z) = z + that is regular in j zl < 1 covers some circular sector of radius pE with c2z2 + center at z = 0 and with central angle 2rr - e. .

1) Fekete [1925]. For p = 1, the theorem was proved by Caratheodory with an indication of the best value for P. For more on this, cf. end of §1, Chapter III. 2) Valiron [1927]. Valiron proved only the existence of p . Bermant and Lavrent' ev [1935] found by a different procedure the best values for pE depending on E. This value is given in terms of an extremum of an expression consisting of a function inverse to the modular function.

74

II. CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

Proof. Obviously, we may assume that c < n. Suppose that no pE > 0 with the property described in the theorem exists. Then there exist number sequences {an I and ((3n ( that approach zero as n -+ oo such that the difference On = argon - arg an lies, for each n, between c and it and a sequence of functions fn(z) = z + . , for n = 1, 2, ... , that are regular in I zI < 1 and such that, for every n = 1 , 2, .. and every z in I z+ < 1, we have fn(z) L an and fn(z) An . if On < n/2, then, by dropping a perpendicular from the point an to the line segment connecting the points 0 and fan, we see that lien - and is greater than the length of that perpendicular, which is equal to I an' sin (S and therefore not less than Ianl sin a. Consequently, we have INn - ani > Ianl sin c; that is, Ian/( fan - an)I < 1/sin E. On the other hand, if On > n/2, it follows from this same construction that Ion - an' > Ianl. Consequently, we again have an/((3n - an)I < 1 < 1/slot. Let us now consider the functions on(z) _ [f(z) - an] /(on - an), where n = 1, 2,.... These functions do not assume the values 0 and 1 in the disk IzI < 1. Consequently, they constitute a normal family. Furthermore, the numbers On(o) _ -an/((n - an) are bounded. This means that the family is uniformly bounded in the interior of the disk IzI < 1. Let 10nk(z) denote a subsequence that converges uniformly in the interior of the disk I zI < 1 to a regular function O(z). Since On k(0) --. 0'(0), it follows that 1,(nk ark) q'(0), which is impossible since ark and (ink approach zero. This completes the proof of the theorem. Theorem 8.1) Let K denote an annulus R1 < IwI < R2 and let p denote a positive number. Then, there exists a positive number p such that the image of the disk IzI < 1 under each of the functions w = f (z) = z + c2z2 + that is regular in IzI < 1 either covers the disk IwI < p or assumes every value w in the annulus K at no fewer than p points of the disk IzI < 1. Proof. Let us suppose that the theorem is not true. Then, there exist sequences IanI and tonI such that an --, 0 as n - and INnI belongs, for each n, to the interval (R1, R2) and a sequence ifn(z)l of functions fn( z) = z + - . that are regular in I zI < 1 such that, for every z in IzI < 1, we have fn(z) # an for every n and fn(z) = An for no more than p - 1 values of z. But the functions

On(z) = [fn(z) - a]/((3n - an) do not assume the value 0 in IzI < 1 and they assume the value 1 in that disk at no more than p - 1 points. Consequently, 1) Montel [1933].

§7. NORMAL FAMILIES OF ANALYTIC FUNCTIONS

75

these functions constitute a normal family in I zj < 1. Since 0n(0) t

- an/(,8n - an) , 0 as n -+ or they are uniformly bounded in the disk

zl < 1. n(z){ contains a subsequence tonk(z)I that converges uniformly

I

It follows that in the interior the disk Izl < 1 to a regular function O(z) such that '(0) = 0. The function o(z) must be identically equal to zero because otherwise, we could show, by using Rouche's theorem, that the functions on (z) have, from some k k on, zeros in I zj < 1, which, however, is not the case. Consequently, 0' k(0) = an) --. 0, which is impossible. Thus, the positive number p mentioned in the theorem must exist, and the theorem is proved. As a consequence of this theorem, we obtain Theorem 9. 1) There exists a positive number p such that every function that is regular in I zj < 1 covers entirely some neighborhood f(z) = z + cZz2 + of radius no less than p with center at the origin. I:urthermore, we have Theorem 10. For the family of functions f (z) =.z + c2z2 + . . . that are

regular in Izl < 1 and that do not assume in that disk agiven finite nonzero value a, there exists a positive number p such that the disk I zI < p is completely covered by the image of the disk Izl < 1 under an arbitrary function in the family. This theorem also follows from Theorem 8 if we take for K an annulus containing the point a. 2)

1) Landau [1922]. 2) A more profound study of the modular function led Bermant [1944] to a number of

generalizations and sharpenings of Theorems B-10.

CHAPTER III REALIZATION OF CONFORMAL MAPPING OF SIMPLY CONNECTED DOMAINS

§1. Conformal mapping of domains bounded by rectilinear and circular polygons

Although in Riemann's theorem we had a geometrical principle for forming analytic functions (specifically, to every simply connected domain was assigned a univalent function mapping that domain onto some canonical domain), neither in this principle nor in the rest of Chapter I, did we touch on the question of more or less explicit construction of these functions. In the present chapter, we shall look at some questions of this nature. Let us now look at domains of certain particular forms, for example, domains bounded by rectilinear or circular polygons for which we can examine the analytic nature of the mapping functions rather extensively and can obtain appropriate explicit analytic expressions for them. The basic tool for studying these functions in the special cases mentioned is the principle of symmetry. Domains bounded by rectilinear polygons. Assuming that this case is already familiar to the reader, we confine ourselves to formulating the generalized conditions under which the final results hold and to deriving the results themselves. Specifically, suppose that a simply connected domain B in the z-plane does not contain o and is bounded by a rectilinear n-sided polygon with vertices at the points A1, A2, , A. Suppose that the angles at these vertices are respectively aa.l, rra2, . , Iran. Let us regard these vertices as accessible boundary points of a domain B. Therefore, if several angular boundary points fall at a certain point in the z-plane, there will be just as many vertices. We shall also consider the case when one or several vertices lie at cc . Here, if the sides of the corre76

§1. DOMAINS BOUNDED BY POLYGONS

77

sponding angle are parallel, we assume the value of the interior angle to be 0. On the other hand, if the sides diverge at a vertex Ak = oc but their extensions meet at a finite point A k forming an angle of rrak directed toward the vertex A k, we take nak = -nak . With these conventions, ak can also assume values from - 2 to + 2.

Of course, the cases ak = 2, - 1, and - 2 should be understood as limiting cases. Under these conditions, if the domain B is mapped univalently onto the halfplane -`- (C) > 0 and if points ak, k = 1, 2, - - , n, on the real axis of the C_ Plane are mapped onto the vertices Ak , then, in the case in which these ak are all finite, we can show by the reasoning usually employed herel) that the inverse function z = f(c) is determined by the Schwarz-Christoffel formula n

.f lt) = C J rl

(C - ak)ak

dC + C;

(1)

k=1

where C and C' are constants. In the case in which one of the vertices, let us the same formula holds for say Ako, is mapped by this inverse mapping into f(O except that now, the product constituting the integrand is over all k = 1, 2, - , n excluding k = ko. Of course, formula (1) remains valid when the domain B is mapped onto the disk I zl < 1. In this case, the points ak, the images of the vertices Ak , are points on the circle C1 = 1. We note that, in accordance with Theorem 6 of §3, Chapter II, we can choose only three of the points ak arbitrarily. The remaining points ak and also the constants C and C' are uniquely determined by the domain. 2) The absence of a general method for determining the remaining unknown constants decreases the practical significance of the Schwarz-Christoffel formula. Furthermore, it is only n exceptional cases that we can evaluate the integral (1) in closed form in terms k,of elementary functions. We cite as verysimple examples of the application of ;formula (1):

(a) Suppose that B is the interior of a regular n-sided polygon with center at

z = 0 and with a vertex at z = a > 0. Suppose that z = f (C) is a function mapping the disk IC1 < 1 onto the domain B. Suppose that f(0) = 0 and f'(0) > 0. Then,

1) Of course, we use here the results established in Chapter II on the existence of such an inverse function and its behavior on the boundary. 2) (We have Kufarev's method [1947b] for determining the points ak, which is convenient in certain cases.)

78

III. REALIZATION OF CONFORMAL MAPPING

the constants ak are easily determined from symmetry conditions. Specifically, we have ak = e2nkc/n for k = 0, 1, , n - 1. From (1), we obtain C

-

dC

f G) = C

v)2/41

(2)

where C is found from the condition f (1) = a. (b) Suppose that the domain B is the z-plane with cuts along the rays issuing from the points ae2rrkt/n for k = 0, 1, , n - 1, where a is a positive number, in the direction opposite to the point z = 0. Suppose that z = f(C) is a function mapping the disk I CI < 1 onto the domain B and that f(0) = 0 and f'(0) > 0. Then, we can again determine the constants ak from symmetry conditions. From (1) we have

fly)=C

-

0(l-{-

)i

dC=

(3)

2

n

and C=a Above, we have assumed that the domain B does not include oo. If the domain B bounded by an n-sided polygon does include 00, then, for the function z= that maps the disk ICI < 1 onto B in such a way that f(0) we obtain the formula

with analogous values of the constants ak . Here, the rrak are the values of the interior angles (with respect to the domain B). Domains bounded by circular polygons. Now, let B denote a simply connected domain in the z-plane with boundary consisting of a finite number n of circular arcs or straight line segments. In the latter case, the segments can be regarded as degenerate circular arcs. We denote the endpoints of the arcs, that is, the , An and we denote the values of vertices of the polygonal domain, by A1, A2, the interior angles (with respect to B) at these vertices by tra 1, ,rat, ... , nan. We may assume that the domain B is bounded. We can always arrange for this to be the case by means of a suitable fractional-linear transformation. The domain B can be mapped conformally onto the disk ICI < 1, Let us examine in detail the function z = f(0) inverse to one of the mapping functions.

§1. DOMAINS BOUNDED BY POLYGONS

79

, n, points on the circle I CI = 1 corresponding to the We denote by ak, k = 1, , n. The function f (C) is regular in the open disk C l < I vertices A k, k = 1, and continuous on the closed disk I CI _< 1. On an arbitrary arc y of the circle I CI = 1, the values that this function assumes between adjacent points ak all lie on some circular arc. Consequently, in accordance with the generalized symmetry principle, the function f(O can be extended past y into the domain ICI>1 in such a way that f(C) is regular (with nonzero derivative) on y. Now, let us look at the expression

- f' (C)

(4)

21f

which is known as the Schwarz invariant or the Schwarzian. Obviously, it is a regular function in I CI < 1 and also can be extended past an arbitrary arc y of the type described above into ICI > 1. Furthermore, one can easily show that it possesses the important property of invariance under an arbitrary fractional-linear that is, if transformation of the function f f (C) + d

then

If,, 6=1f, C}.

Let us examine the behavior of If, Ci at the points ak, k = 1, 2, - , n. Let us suppose first that the numbers ak do not include the numbers 0, 1, 2 and that no circular arc of the boundary degenerates into a straight line segment. The circles on which the boundary arcs lie, forming angles with vertex Ak, have a second point of intersection Ak. The function t = ((z - Ak)/(z maps .the part of a neighborhood of the point Ak that lies in B onto the part of a neighborhood of the point t = 0 in the t-plane that lies to one side of a line passing through t = 0. According to the symmetry principle, the function ((f(O - Ak)/ ak Ak))1 is therefore extendable from the part of the neighborhood of the point ak lying in ICI < 1 to the part of the neighborhood lying in S I > I. In particular, it is regular at the point ak itself and has a nonzero derivative at that point. Consequently,in a neighborhood of J= ak, we have rf (C) - Ak ak

f(C)-Ak)

- b1 (C - ak) + ...

,

bl # 0,

from which we get f (C) -AA

f (C) -Ak -

(C - ak)ak (Cl

C1 # 0.

80

III. REALIZATION OF CONFORMAL MAPPING

If we set up the expression (4) for this last function and keep in mind that it is equal to (f, Ji we get, after some simple calculations, the expansion in aneighborhood of eJ = ak :

(1 -a"t)

if, C}=

1

{fC

(C-ak)9

k

(5)

+C-ak-}-...,

where Ck is an unknown constant. This expansion shows that If, C1 has secondorder poles at the points ak.1) Therefore, the function If, JJ, originally defined in the disk ICI < 1 and then extended to the entire plane, represents a meromorphic function with poles only at the points ak . The principal parts are determined at these poles from formula (5). The difference

G) =V, C}-

n1

Ck

(C-ak)' +C-ak)

(6)

k-1 has no finite singular points at all. Let us examine its behavior at oo. Since a suitably chosen function

A R)= f(C)+dEA const is real on each of the arcs y considered above and since f 1(0)

it follows from the symmetry principle that this function can be extended from the disk ICI < 1 to the domain ICI > 1 by taking f 1 (S) = f1 In particular, it is regular at the point oe; that is, in a neighborhood of S = oo, it has an expansion oo ,

f, (C) _ co + C + ... , 0 (by virtue of the conformality of f 1(C) at the point the corresponding expansion for If, J) has the form where c 1

if, C}= i; -F-...

= 0). But then,

(7)

This means that { f, I has a zero of at least fourth order at oo . But then the difference (6) has a zero at the point oo; that is, this difference is regular on the entire plane including oo . Therefore, it is identically equal to zero, so that we have the identity

1) We note that, if a finite point to, I to, > 1, is such that f(C o) = oo, then in a neighborhood of t o we have f (C) = b_ 1/(C- C o) + bo + , b-1 ji 0 so that If, t l is regular at t = t o.

1. DOMAINS BOUNDED BY POLYGONS n 2

( _ a)

81

C k

!.

(8)

k=I

Our assumption up to now has been that none of the ak is 0, 1, or 2 and that there are no straight line segments on the boundary of the domain B. However, if these assumptions are not satisfied, we can still show that (8) is valid by replacing the domain B with suitable circular domains and then using the theorem on the convergence of univalent functions (Theorem 1, §5, Chapter II). Thus, equation (8) is proved for an arbitrary domain bounded by a circular polygon. It represents a third-order differential equation in f(C). Its right-hand imember includes the unknown constants ak and Ck , for k = 1 , 2, , n. There are certain relationships between these constants. Specifically, since If, C1 has an expansion of the form (7) in a neighborhood of w, the coefficients of 1/c, 1/ and 1/C3 must be equal to 0 in the corresponding expansion of the right-hand member of (8). This gives us the three conditions

21 Ck=0,

k=1 n

(1 - ak) + yy

2

kt n

Ckak = 0,

(9)

k=1 n

2] (1 - ak)+ 2] Ckak=0,

k-1

k=1

that the constants ak and Ck must satisfy. Furthermore, it should be noted that three of the constants ak can be chosen arbitrarily on C1 = 1 by virtue of Theorem 6, §3, Chapter II. In the case of a circular triangle, this is sufficient to deter.mine all the constants ak and Ck . Similarly we can also show that, when the domain B is mapped onto the upper halfplane (c) > 0, the inverse function z = f(C) also satisfies the differential equation (8), where, this time, the ak are points on the real axis corresponding to the vertices Ak. The only difference is that if one of the vertices A k , let us say A k o, is mapped into oc, then the summation in (8) is over all k = 1, 2, , n excluding k = ko. Now, let us look at the nature of the differential equation (8). Since it is a third-order equation, its general solution has three arbitrary constants. On the other hand, if f (C) is a particular solution of equation (8), then, obviously, the I

Ill. REALIZATION OF CONFORMAL MAPPING

82

fraction (af, (C) + b)/(cf, (C) + d) which already contains three arbitrary constants

(a/c, b/c, and d/c), is also a solution. Consequently, this is the general solution of equation (8). Although we need to know only one particular solution of equation (8) to find the desired mapping, the finding of that one solution is difficult because of the nonlinearity of the equation. However, the problem of solving equation (8) can be reduced to solving a second-order linear equation, owing to a single remarkable property of the function If, C1.

Let w 1 and w2 denote two linearly independent solutions of the linear differential equation

w"-}-q(C)w0

(10)

with given q(t). It is easy to show that w1w2 - w2w1 = const = c (the left-hand member is the Wronskian of the two solutions). Therefore, if we set ri w2/w 1, we have C

(C) =Vii,

C} _

-

!-L

d,

2V; dC

Wt

2q

()

It follows that the ratio of any two linearly independent solutions of equation (10) is a solution of the equation {,q, C} = 2q (C).

(10')

This shows that, if we denote the right-hand member of equation (8) by R(C), set q (C) = R(C)/2, and solve the differential equation (10) with this q (C), then the ratio of the two linearly independent solutions yields a solution of equation (8). 1) Equation (10) with this R(C) belongs to the most intensively studied class of linear differential equations with variable coefficients, known as equations of the Fuchsian type. However, the use of this advantage is made difficult by the fact that we do not have a general method for determining the constants ak and Ck.

1) Also, equation (10' ) can be reduced to equation (10) in another way, specifically, by remembering that

-2

12 (I )

and setting 1/\(-i' = w. Then, if w1 is a particular solution of equation (10), the corresponding particular solution 711 of equation (10') is obtained from -1 = 1/w 1.

§ 1. DOMAINS BOUNDED BY POLYGONS

83

Example. In conclusion, we present a simple but important example when we can make a complete determination of the mapping function. Suppose that we are required to map the disk I C < 1 onto the domain B consisting of the interior of an n-sided circular polygon with center at z = 0, with one vertex at the point z = 1, and with all the interior angles equal to nq, where 0 < q < 2. Suppose that the mapping function z = f(C) is normalized by the conditions f(0) = 0 and f '(0) > 0. To do this, we consider a mapping of the sector I C1 < 1, 0 < arg C < 277/n onto the circular triangle bounded by the two straight line segments extending from the origin z = 0 to the points z = e2n`/n and the circular arc contained in the boundary of the domain B. According to the symmetry principle, the function executing this mapping will provide the required mapping of the disk <1 onto the domain B. It maps the points a 2nkd", for k = 1, 2, , n, onto themselves so that ak = e for k = 1, 2, ' 's A. Furthermore, by virtue of the uniqueness of the mapping, we have the functional equation f (e2ni/n y_ e 2nilnf(C) in JCJ < 1. We see from this equation that the expansion of f(C) in a neighborhood of C= 0 is of the form >

W

f(C)

= k=0 I CkCnk+1,

Ct > 0,

and that the expansion of If, iI is of the form dkCnk

If, C} - i

k-1 On the other hand, since, in the present case, n

k=1

2 (1 -ak)

( - ak)'

_d o q)2 dC =C - ak 1

2

- - 2 (1 - q2)

k1

d C Cn n-11

n 2

an-s02

(1 - q2)

(n Cn 1) Ci '1

it follows from (8) that the summation A

Ck

k=1

C-ak

in that equation must include only terms with powers ok-2; that is, this sum reduces to CCn-2/(0'" - 1). But If, J I has a zero of at least fourth order at 0 . Therefore, C = -(n/2) (1 -JJ q2). Therefore, equation (8) takes the form ifs

C}= 2 ( l -q2) Pnn pl)s

(11)

III. REALIZATION OF CONFORMAL MAPPING

84

The problem of finding f tion

thus reduces to solving the linear differential equa-

w" +

4

(1

- q9) (ca- I )a w - 0.

(12)

In this equation, we make the substitution

=V, '2=(En- 1)p v, where p is a constant. We obtain 11 (E-1)9Ev"+2(E-1)(E(2p+1- n)-(1- n))v'

+(P (P-

nE+P(n-1)-}-I 4q )v-0.

Let us choose p in such a way that the coefficient of v in (13) has a root This yields p = (1 - q)/2. With this choice, equation (13) takes the form

(2 -q- n)-(1-

n))v'

+

or, if we set

2

2

(

= 1-q , T- -n, I-q = 2 -n, I

GE

1--q 1-q- 1n )v=U

p

2

1

1

(14)

it takes the form

W_ - 1) v"+(((Z+ (i+ 1) E - T) v'+aPv= 0.

(15)

This is the familiar differential equation of Gauss. Solution of it with the aid of a power series v = In' _o cn6n shows that it is satisfied by the so-called hypergeometric series 1)

F(a, , T, E)=1+ 1a7 E+a(1 21T(T

} 1)I)E9+

(16)

which converges in the disk IKj < 1 for all a and P and all y> 0. To obtain a second solution of equation (15), we make the substitution v = eI-'Yvi. The resulting equation can also be put in the form of the Gauss equation if we set a- y + 1 = a', 13 _ y + 1 = 13' , and 2 - y = y'. Therefore, this equation is satisfied by v1 = F(a , 13 ', y, ). Then we obtain for equation (15) the second solution

v-Et-1F(a-T+1, P-T+1, 2-7, E),

1) For more on the hypergeometric series, see, for example, Smimov [1953, p. 381 of English translation.

§ 1. DOMAINS BOUNDED BY POLYGONS

85

and this solution is, as one can easily see, linearly independent of (16) for 0 < y < 1. Thus, we have for equation (15) the two linearly independent solutions

-q__ l 1-q n' 2 ' 2 1-q 1- q

and

+n'

1- 1n'

1

(

2

2

1

+n1

E

E

)

and w, we conclude that the function

Returning to the variables F

2

-t

q+ n'

2 q

1q

f(C)=C F(1 -q--1

n

2

1+ n' sl

2

- -1n

(17)

Cn )

is a solution of equation (11). The geometric series in (17) can also be expressed in terms of definite integrals. Consider the integral -Et)_dt,

1(1

1=S0-1 (1 --t) 0

< 1. Let us.expand the factor which converges for 13 > 0, y - 13 > 0, and et)-" in a binomial series and then integrate term-by-term. We obtain (1 m

t

J=

tP-1(1--t)-P-1

1

1) 2

k=0

0

00

X1)ktktkdt= k=0

k

...(-a-k+1)

-a(-

a(a+1)...(a+k-1) 1.2...k 00

a(a+l)...(a+k-1)

X S tP+k 1(1 -t)1 P Idt.tk=1 J

1

k=0

0

X r(P+k)r(1r-p),k=r@)r (1-P) F(a r(1) r(1+k) And, consequently,

F

(a' R'

T' )

r r (T) )

S tP-1(1 - t)1

P-1 (1

-

U)" dt.

(18)

0

Let us use this formula to transform the function (17). Then, for 0 < q < 1 and t;'I < 1, we obtain

86

111. REALIZATION OF CONFORMAL MAPPING

1

q+1 2

= (1 -t) q

1

1

n (1

2

Sl

-Cnt)

q -1 2

I

n

dt

f(C)=cC0

1

q+i

0-1

I

(1 -t)

2

St

(I - tnt)

it

.-1 2

(19) 1

a dt

Since the desired mapping function satisfies the conditions `(0) = 0 and f(1) = 1, it is obviously contained in formula (19) with c = 1. This is the final result. Furthermore, by making the substitution r= (1 - t)/(1 - Jnt) in the integrals in (19), we obtain another form of the function f (C) for real values C and hence for all complex values of C in the disk I C1 < 1: q 2

2

1 + n dc q+1

S

f(C)=C-

2

q-1

1

(20)

ds

2

q--1

s

((1_.C)(1-C"T)) 2

in which the dependence on n is simpler. We note that none of the integrals in formulas (19) or (20) can be evaluated in closed form in terms of elementary functions. The most interesting case is that of q = 0. In this case, equation (20) becomes 1

tl/ndt (1 -Cnt)

fn(C)=Co

- nds !

1

(21)

J with f(i) now being denoted by fn Let us point out some very important properties of the function (21) and of its inverse /n 1 (z). On the basis of the same considerations as in §6 of Chapter II, the function C= [n1(z), defined inside the n-sided polygon with zero angles, can be extended to the entire disk z < 1. It represents in that disk a meromorphic function that assumes only values in the domain Bn, which is the z-plane

with the points ak = eZnk'/n, k = 1, 2, , n, deleted. More precisely, the

§1. DOMAINS BOUNDED BY POLYGONS

87

function J= `,',1(z) maps the disk Izl < 1 bijectively onto the many-sheeted Riemann surface constructed from the interiors and exteriors of the circle I z I = 1 in a manner similar to the construction of the modular surface in §6 of Chapter II. On the other hand, the function z = f (C) can be extended in B along an arbitrary path and it assumes, when this is done, only values in the disk I z I < 1. Let us see what we get if we take the limit as n -.-, oe. The function can, if we begin at the point C = 0 with an initial element of the form c C+ - , be extended along an arbitrary path in the cplane that does not pass through the points 0, 1, or no. From formula (21), we have .

n

I

T In dT 1

oS

LfR G1/n)]" = C

n

T(1--T)(1-CT) i l/n dc T(1-T)(1-CT)

=C (1 + n) ,

(22)

where Jim

Jim An= n

ao

n (t l1n - ti 11n) dT

2 log z dT

yT(I-T)(1-CT) n- ao

T(1-T)(1 -CT)

0

'In dT

dt o YL (1 - T) (1 - CT) (((\\\1'''\

T (1 - T) (1 - CT)

This limiting operation is uniform on an arbitrary finite chain of closed disks not containing 0, 1, or oo . In view of this, formula ( 22) yields the results that the function 1

S

2Iogzdi z (1-T) (1 -Cz)

0 1

d2

f(C)= lim

(23)

A- CO

being regular in the disk ICI < 1, can be extended from CI < 1 along an arbitrary path in the C -plane that does not pass through 0, 1, or no and that the extended function satisfies the inequality If(C)I < 1. Let us denote by B' the image of the disk ICI < 1 under the mapping z = )'(C). This image is a domain in the disk IzI < 1. Let us show that the inverse function C_ `-1 (z), which is necessarily regular in B' , is regular in fact in the entire disk I z I < 1. To do this, let us clarify the boundary of the domain B'. The function z = 1`n (C) maps the disk

88

111. REALIZATION OF CONFORMAL MAPPING

CI < 1 onto a domain one arc of the boundary of which is given by the equation z - 1 - i tan (a/n)I = tan (a/n) with I z I < 1. Consequently, the function z = fn (C I /n) maps the disk C I < I univalently onto a domain bounded by the curve IT- 1 - i tan (a/n)I = tan (a/n) with I z I < 1. As we let n approach no, this curve becomes the curve I log z - is I = a with I z I < 1, which is also the boundary of the domain B. If we shift to the t-plane by setting t = log z, then B' is mapped single-valuedly into a domain B" with boundary determined by the system of equations I t - ink a, k = t 1, t 3, . - . with R (t) < 0, that is, into the left halfplane with cuts along semicircles. The function t = log f (C), which is regular in 0 < I C1 < 1 maps the circle I CI = 1 onto the boundary of the domain B" in such a way that an arbitrary continuous image of the circle with the point C = 1 excluded is the boundary semicircle of the domain B". The inverse function C= f-I(er), which is regular in B", is, by virtue of the symmetry principle, extendable, just as in §6 of Chapter II, to the entire halfplane 9t (t) < 0. This shows that the function C= f -I (z) is regular in IzI < 1. Also, it follows from the extension process that C = f (z); does not assume the value C = 1 in I z I < 1 and that it assumes the value C = 0 only at z = 0. The function f (C), on the other hand, can, by beginning with an element defined in a neighborhood of C= 0, be extended along an arbitrary path that does not pass through the points 0, 1, and 00. But along a path passing through C= 0, its extension through C= 0 is possible only by means of an original element of the function. We easily obtain from (23) the expansion 932

f(t)=16+

+...

(24)

in a neighborhood of C = 0.1)

The following is a simple application of this last function f(C) to one of the covering questions considered in §7 of Chapter II. Suppose that the function F(z) = z9 + c1Z9+I + , for q> 1, is regular in IzI <1 and that it does not assume the value 0 in 0 < I z I < 1. Let c denote a number such that F(z) c in IzI < 1. Then, the composite function 4(z) = f(F(z)/c) = z9/16c + can be 1) In (23), we need to make the substitution r = sin20 for 0
a logsinydcp-- z log 2. 0

§2. PARAMETRIC REPRESENTATION OF UNIVALENT FUNCTIONS

89

extended in I zi < 1 along an arbitrary path. Therefore, it is regular in zJ < 1 and ,O(z)1 < 1 in Iz) < 1. But then, just as in the Schwarz lemma, we easily obtain the result that 11/16cl < 1. From this we obtain the inequality I cl > 1/16 with

.equality holding only in F(z) = cf(el z1). This result, which is the analogue of Koebe's covering theorem for nonunivalent functions, can be formulated as

Theorem 1.1) Suppose that the function F(z) = z9 + c lzq+l + , where q > 1, is regular in the disk I zi < 1 and has no zeros in the domain 0 < I zI < 1. Then the image of the disk Izi < 1 under the mapping t;'= F(z) completely covers the disk IC1 < 1/16 but not always a greater disk with center at C = 0. A more profound study of the function (21) leads to the following theorem on the covering of segments: Theorem 2.2) Suppose that the function F (z) = z q + . , for q > 1, is regular in the disk Izi < 1. Then the image of that disk under the mapping = F(z) completely covers some segment of arbitrary prenamed slope that contains the point = 0 and is of length no less than A = 8n2/F(%)4 = 0.45 - - . The number A cannot be increased without additional restrictions on F W.

§2. Parametric representation of univalent functions

In the present section, we shall give a parametric representation exhibited by Loewner for an important class of univalent functions.3) Thanks to this representation, we can find the complete solution of a number of extremal problems in the theory of univalent functions. This we shall do in the following chapter. Let B denote a simply connected domain in the w-plane including the point w = 0 but not the point w = -a. Consider the set of domains obtained from B by making a cut along any Jordan curve that is contained in B, that has one end point on the boundary of B, and that does not pass through the point w = 0. Let us call these domains the domains (s). If the domain B is unbounded (that is, if -0 is one of its boundary points), this cut can extend to 00 . Then, in contrast with the preceding situation, we understand by "cut" a curve defined parametrically as follows: w = w(t) for a < t < b, where w (t) is continuous in a < t < b, w (t) L w (t") for a < t' < t < b, and w (t) oo as t b. One can easily show that it 1) Caratheodory [1907]; see also Goluzin [.1948]. 2) Bermant [1944] and Goluzin [1948]. 3) Loewner [1922]; see also Goluzin [1939d, 1949c].

90

III. REALIZATION OF CONFORMAL MAPPING

is possible to approximate an arbitrary simply connected domain B' contained in B and including the point w = 0 with domains (s) in the sense that a sequence of functions w = f(z), f (0) = 0, f' (0) > 0 that map the disk Izi < 1 univalently onto the approximating domains converges to a function mapping the domain I zI < 1 onto the domain B'. To see this, note that the domain B' can be approximated first by domains bounded by Jordan curves contained in B by mapping the disk Izi < 1 onto B' and considering the circular images in B'. Assuming B' bounded by a Jordan curve C contained in B, we connect the boundaries of the domains B and B' by a straight line segment contained in B but lying entirely outside B'. We continue along it from the boundary of B to a point c E C and then proceed from c along the curve C to a variable point d. If we let d proceed around C until it returns to the point c, then, in accordance with Theorem 1 of §5 of Chapter II, the domains (s) so obtained approximate the domain B'. By applying a diagonal process, we can obtain the existence of the required sequence of domains (s) approximating an arbitrary domain B' C B including the point w = 0. Now, suppose that we are given a domain (s) obtained from the domain B by making a cut L. By shortening the cut L, beginning with the endpoint lying in B, we obtain a family of simply connected domains BL depending on a real parameter t that varies continuously in the interval 0 < t < to. The domain Bo is the domain B with the complete cut L; Bto is the domain B without the cut;

and, for t' < t", the domain B,, is properly contained in Be,,. Let za,=== g(z,t),

g(0,t)-0, g' s(0,t)>0, 0
denote a function that maps the domain I zI < 1 univalently onto the domain B,. For every z in the disk I z I < 1, this function is a continuous function of t in the interval 0 < t < to by virtue of the theorem on the convergence of univalent functions. Therefore, if we set

g(z, t) = P (t) (z + a, (t)z'+ ...),

(1)

in I z I < 1, the function 8(t) is a positive, strictly increasing, continuous func-

tion of t in the interval 0 < t < to. The strict monotonicity follows from the Schwarz lemma as applied to the function c(z) = g-1(g(z, t'),t") for t' < t", according to which 0' (0) < 1. The form of the parameter t is at our disposal. Let us choose it in such a way that

P(t)=Pet

in 0`t
§2. PARAMETRIC REPRESENTATION OF UNIVALENT FUNCTIONS

91

which, on the basis of what has been said, is possible. Along with the function g(z, t), we define the function .f (z, t) _ g'-t (g (z, 0), t) = et (z + .. - ).

(2)

Like the function g(z, t), the function f(z, t) is a continuous function of t in the interval 0 < t < to. Furthermore, one can easily see that, for every t in the interval 0 < t < to, the function w = f(z, t) maps the disk z I < 1 onto the disk w I < 1 with a cut along a Jordan arc, that is, onto a domain that, with respect to the disk I w I < 1, is a domain (s). Let us show that, for every z in the domain z I < 1, the function f(z, t) for 0 < t < to, satisfies the following differential I

equation, known as Loewner's equation: Of

_

dt -

{1 +kf -'

(3)

1 - kf'

where k = k(t) is a continuous complex function of t such that I k(t)I = 1. Then, remembering that the equation g (f (z, t), t) = g (z, 0) implies (ag/af) (a flat) + ag/at = 0 we see that the function g(z, t) satisfies, for all z in the disk IzI < 1 and all t in the interval 0 < t < to, a second equation of Loewner: ag

_ ag

I + kz

(4)

dt - az z 1 - kz'

To prove (3), consider the function g'(g(z,et'-t"(z-- ...).

For arbitrary t' and t" such that t' < t", this function maps the disk

(5) I zI < 1

univalently onto the disk IwI < 1 with a cut along a Jordan arc. Consequently, for such t' and I", it is continuous in the disk I zI < 1. Let us adopt the following notation: let k(t) denote the point on the circle zI = 1 mapped by the function w = g(z, t) into that endpoint of the cut lying in B. Let S,,,,,, denote the cut in the disk IwI < 1 obtained under the mapping of

the disk IzI < 1 by the function w = h(z, t', t') and let Bt, t denote the arc on the circle IzI = 1 corresponding to it. Then, the mapping w = h(z, t', t") obviously maps the point h(t) into the endpoint of the cut S,,,t in IwI < 1, and the point Mt") is the endpoint of the cut St, . on the circle IwI = 1. It follows immediately from what has been said that

1) for fixed t' = t, the arc Bt , ,t is constricted to the point h(t) as t" --- t; 2) for fixed t" = t, the arc S, I, t is constricted to the point h (t) as t --a t.

III. REALIZATION OF CONFORMAL MAPPING

92

Let us show that under both these limiting operations, the arcs Btu i and St, it,, are constricted simultaneously to the point A(t). If we extend the function h(z, tt") past the arc B:, N, complementary to Bt,,t. with respect to the circle z I = 1 onto the domain Izl > 1, we see that the function w = h(z, t, t ") maps the entire z-plane with a cut along the arc Bt, t onto the entire w-plane with a cut along the arc St, t and along the arc S, ,tt symmetric to St , t about the circle I w I = 1. Since one of the arcs Bt , , t and St #.t if U S, is constricted to the point A(t) in both limiting operations, the other must also be constricted to the same

point. This is proved on the basis of the principle of condensation of analytic functions and the fact that, if the z-plane with a single excluded point is mapped univalently by means of a function w = 0(z), where ,(0) = 0 and 0`(0) = 1, then O(z) = z. On the basis of what has been said, under both limiting operations, the function h(z, t `, t") converges to z uniformly on each closed bounded subset of the z-plane that does not include the point A(t). In particular, it follows that A(t) is continuous throughout the interval 0 < t < to. What we are concerned with at the moment is only that, under both limiting operations, the arc Bt, t,, is constricted to the point A(t). If we denote the endpoints of this arc by eiaand a°Q, where a< < a+ 2rr, then the function (z) = log h (z, f" 0 (0) < 0, (6) (D is regular in the open disk I z I < 1 and continuous in the closed disk Izl < 1 and its real part is zero everywhere on the circle I z I = 1 except on the arc (a, 0), on which it is negative. On the basis of what has been said, if we apply the Schwarz formula to the function (6), we have in I z I < 1 log

h (z, t',

( (e`o)) eie ± d9.

2n a

If we now replace z with the value [(z, t P

log f (z, t") f (z, t')

to

we obtain

eie ±f (z, t'

(q (e )) eie - f (z, t') d0.

2n

(7)

a

For z = 0,

this yields

Y - t'

((1? (eie)) dB.

2n S

If we now apply the integral theorem of the mean to the real and imaginary

(8)

§2. PARAMETRIC REPRESENTATION OF UNIVALENT FUNCTIONS

93

parts of the integral in formula (7), we obtain log f (z, t"

(2, t') r re+f l+ -L1eIe'f(,t')

reiB" "+f (2,

)

f(, e )l

eie

1

X 2n S it ((D (e`)) d9,

(9)

a

where 0' and

8

are points in the interval (a, /3 ). If we divide (9) by ( 8), we

obtain log f

(z, t") -logf (z, t')

t" - t'

_

r

ete'-f-f(z, t')

\e10'-f(z, t')1+

t')']. l (e,e"---f(z, 1el t'))J

(10)

Since a and /3 approach the same limlt, arg A(t), under both limiting operations that we are considering, the same will be true of 0' and 8". Consequently, by taking the limit in (10), we obtain the differential equation

dlogf_ of

that is,

X(t)+f

a (t) - f,

tf

kf, dt - f '+ l - kf' k - k (t) _ (t) 1

Thus, equations (3) and (4) hold throughout the interval 0 < t < to. This result shows that corresponding to every curve L contained in B is a function k (t) that is continuous and equal to unity in absolute value in a suitable interval 0 < t < to such that the function g (z, 0), where g(0, 0) = 0 and g= (0, 0) > 0, which maps the disk jzI < 1 univalently onto the domain B with cut L is defined as the limit

g(z, 0)=limg(z, t), 1-+0

where g(z, t) is that solution of the differential equation dg dt

k (t) z _ dgdz z1-}1-k(t)z

that satisfies the conditions g (0, t) = 0 and gZ (0, t) > 0. If the domain B is not the entire finite w-plane, the function

W=g(z, to)= lim g(z, t) 1-+10 maps the disk I z I < 1 onto the entire domain B. On the other hand, if the domain B has as unique boundary point w = oc, then, for any M > 0, the closed disk

I wI < M lies entirely in all the domains Br fort sufficiently close to to and, consequently, for such values oft on the circle Izi = 1. Therefore, we also have

94

III. REALIZATION OF CONFORMAL MAPPING

I z/g (z, t)I < 1/M in I zI < 1 since, in this case,

g(z, t)-+oo, g(0, t) ->oo to when 0 < I zl < 1. This means that to has to be oo. Now, let us look at the particular case in which the domains B, subjected to the similarity transformation w' = e-tw, converge as t oo to a domain B' as kernel (in the sense of §5, Chapter II). Then, the functions w = e-tg(z, t) converge as t - oo to a function ((z) mapping the disk Izl < 1 onto the domain B`. For example, if the cut L is such that, as we move the point w along it toward oo, this point apas t

proaches asymptotically a ray issuing from the point w = 0, then the function g(z, t)

that satisfies equation (4) also satisfies the condition

lime tg(z, t)_(1

IiI=1.

1- 00

(11)

With regard to the inverse function z = g'1 (w, t), let us show that, for any curve L that extends to oo, we have, for arbitrary finite w, lim petg t (w, t)=w. (12) I -. 00

In accordance with inequality (10) of §4 of Chapter II as applied to the function

/3-le`g(z, t), we have in the disk Izi < 1 (1P+IZI)2Clg(x,

t)I(l_Pet

Z09.

Then, by setting z = g-1 (w, t), we obtain for arbitrary w E Bt

Set(w, t)I

(1-I-(w,

-w

petIg- ` (w, t)1

(1-I g' (w,

t)1)$.

(13)

The first of these inequalities yields Pet lg-1(w, t)l < 4lwl and consequently,

limIg'(w, t)

0.

t -.co

But then it follows from the two inequalities (13) that, for arbitrary finite w, lim pet g 1 (w'

t)

= 1.

(14)

t -. 00

Now, if (12) fails to hold for some finite point w0, there exists a sequence

{ tv } such that t1, - oa and Se'vg-1 (w, t,)/w0 - a

If we choose from the a subsequence that converges everysequence of the functions where in the w-plane to a regular function OW, we conclude on the basis of (14) that I 0 (w)I = 1 and, since 0(0) = 1, O (w) = 1. But O(w0) = a 1. This contradiction proves (12) for arbitrary finite w. Net"9-1 (w, tv)/w

1.

§2. PARAMETRIC REPRESENTATION OF UNIVALENT FUNCTIONS

95

Let us turn now to the function f(z, t). Since f(z, 0) = z, we conclude from what was said above that f (Z' t) is a solution in the interval 0 < t < to of the differential equation (3) that satisfies the initial condition f(z, 0) = z. If B has more than a single boundary point, then, in addition, we have in the disk I zI < 1 li m f I-*10

(x, t) = f (z, to) = g ' (g (z, 0),

to).

(15)

In particular, if B is the disk Iwl < 1, then g(z, t0) = z and it follows from what was said above that the function g(z, 0) can be represented as the limit g (Z' 0) = li m f (z, t), t - io

where f(z, t) is a solution of the differential equation (3) satisfying the initial condition. This was Loewner's result. On the other hind, if the domain B is the entire w-plane excluding the point oc, then, in addition to the differential equation (3) and the initial condition f (z, 0) = z, we have in accordance with (2) and (12) g(z, 0) = 1im pe`f (z, t). This result can be immediately generalized somewhat. Specifically, suppose that the simply connected domain Bp has no exterior points and is, in fact, the wplane with cuts along a finite number of Jordan arcs that do not pass through the point w = 0 and have no points in common with each other except possibly endpoints. Suppose also that at least one of these arcs extends to oo. Then the equation for the boundary L of the domain Bp can be represented as a single equation w = w (t), for 0 < t < 00, where w (t) is a piecewise continuous function of a real

parameter t such that, as t varies from 0 to o, the point w(t) successively describes all the boundary curves in such a way that, for any value of t, the curve corresponding to higher values of t bounds a domain B, of the same form as Bo. Under these conditions, all our reasoning above obviously remains valid and equations (3) and (4) with all the supplementary conclusions hold for the corresponding functions f(z, t) and g(z, t). The function k(t), however, admits a finite number of points of discontinuity of the first kind corresponding to values of t at which the point w(t) jumps from one boundary arc to another. These results are vivid examples of the reduction of geometrical problems to purely analytical ones. For example, the conclusion that we have just drawn can be formulated in the form of the following theorem, which will play an important role in what follows:

96

III. REALIZATION OF CONFORMAL MAPPING

Theorem 1. Let B denote a simply connected domain without exterior points that includes the point w = 0 but not the point w = -. Suppose that the boundary of B consists of a finite number of Jordan arcs. Then there exists a complex function k(t) of real argument t for 0 < t <+ -o that is continuous except at a finite number of points of discontinuity of the first kind. Suppose that k (t)I = 1. Suppose that the function w = f (z), where f(0) = 0 and f '(0) > 0, which maps the disk I z' < 1 onto the domain B, can be represented in the form

f (z)= lim pet f (z, t),

(16)

where f (z, t) denotes that solution of the differential equation

=-f'

d f 1 + kf (0
(17)

Conversely, we have

Theorem 2. Let k(t) denote a function that either is continuous or has a finite number of points of discontinuity of the first kind in the interval 0 < t < Suppose that jk (t)I = 1. Then that solution of the differential equation 1- k (t) w

dt

(18)

that satisfies the initial condition w 1t=0 = z is a function w = f (z, t) such that f (0, t) = 0 and fZ (0, t) = e -t which, for each t in the interval 0 < t < 00 is a regular univalent function of z in the disk IzI < 1.1) Furthermore, the limit

f(z)= lim etf(z, t), f(0)=0, f' (0) = 1,

(19)

t- 00

exists and is also a function with these properties. Proof. Equation (18) with initial condition w W o = z is obviously equivalent to the integral equation t -SI_kw

w = ze

1-kw

0

dt

(20)

1) However, as Kufarev [1947c] has shown with an example, the function w = f(z, t) does not always map the disk Izl < 1 onto the disk Iwl < 1 with a cut.

§2. PARAMETRIC REPRESENTATION OF UNIVALENT FUNCTIONS

97

obtained from (18) by dividing by w and integrating with respect to t from 0 to t. Let us solve equation (20) by making successive approximations: I + hwn-1 dt

wo=z, wn=wn(z, t)=ze

0

,

n=1, 2, ....

(21)

Since l ((1 + kw)/(1 - kw)) > 0 for IwI < 1, it follows from (21) that Iwnl < This means that all the w = wn(z, t) are regular func. IzI for all n = 0, 1, w (0, t) = e-t. Let us tions of z in the disk IzI < 1 and that show that the sequence of the functions w converges uniformly inside the disk IzI < 1 and hence in the interval 0 < t < to, where t0 < oo. From (21), we have awn

own

_ -- w

at u

Therefore,

I + kwn-1 - kttln_.,

n1

'

0w'_1

at - at

(wn - W.-I) I ±

kwn-1 - 27?1n-1 (1

- kwn 11)

kwn-.)

CIW.-'wn_tI I-IZI TIWn-1-Wn-SI(1 )IIZI = A I W. - Wn -.i + B I W.-1 that is,

al wn-wn-1

-

wn-a I,

W. - P1R-1

B I wn.-1 - wn-S Ir where A and B depend only on z. If we multiply this by a-At and integrate, we

at

`4

obtain, for 0 < t < t0 (where t0 < oo), t

wn - wn_1 I C eAto B

7FJn-1 - TQJn_9 dt. 0

Integration leads to the following inequality (with n = 2, 3,

wn-wn-1C

):

(eAtoBt)n - I

(t1-1)1

+

C-C(z),

from which the asserted uniform convergence of the sequence (wn} follows. The limit function w = f (z, t) is regular in I z I < 1 and continuous in 0 < t < oc. Obviously, it satisfies equation (18) and the conditions wlt=0 = z. Furthermore, I f (z, t) I < I z 1. Let us show that the function f (z, t) is univalent in I zI < 1. Let us suppose the opposite. Suppose that, for some II in the interval 0 < t 1 < 00 and for two points z I and z2 in the disk I z I < 1, we have f (Z 1, 11) = AZ 2, t I).

Then the functions f(z1, t) and f(z2i t) have the same value at t = t1. From an

III. REALIZATION OF CONFORMAL MAPPING

98

easily proved theorem 1) on the existence of solutions of equation (18), it follows that f (z 1, t) = f (z 2, t) everywhere on the interval 0 < t < t 1. For t = 0, this yields z1 = z2. It remains for us to prove formula (19). To do this, we substitute into equation (18) the function w = f(z, t) and write it in the form

a(logf+t) at

-

1 - kj'

-

I-kfdt.

2kf

Integration yields

1ogL(Z)+t-

(22)

But the function etf(z, t) C S. Consequently, for I z I < 1, we have I f(z, t) I < MZe-t, where Mz depends only on z for z I < 1. This shows that the integral J 'o (k f/(1 - k/)) dt converges uniformly inside the disk I z I < 1. From this we conclude on the basis of (22) that, as t --, -, e° f (z, t) converges in the disk I zI < 1 to a univalent regular function f(z) such that f(0) = 0 and f' (0) > 0. This completes the proof of the theorem.2) In conclusion, let us show how to obtain the coefficients in the expansion of the function f(z) referred to in Theorem 2. To do this, we consider first the function f(z, t) for 0 < t < to, where to < uo. We define I

gio (z, t) =f (1'-1(z, t), t0)= et -'o (z -f- c2 (t, to) zA + ...).

This function is regular at the point z = 0. From gt0 (f (z, t),

t) =f (z, to)

we have agt0 of

of dt

agto

at

=0.

Consequently, in accordance with (17), we obtain the equation ag5o

at

_ agio z 1l + k (t) z dz

- k (t) z

1) One gets the result from a suitable inequality of the form at

where fl = f(z1, t), f2 = f(z2, t), and Mz depends on z for I z I < 1. 2) Equations (17) and (18) were studied by Kufarev [1946, 1947, 1947a, 1947c, 1947d, 19481. In the article [19431, he studied the question of parametric representation from a more general point of view.

§3. VARIATION OF UNIVALENT FUNCTIONS

99

in a neighborhood of z = 0. If we substitute into this equation the expansion for gto (z, t) and equate coefficients of like powers, we obtain the system of equations n-1 c., (t, to) _ (n - 1) cn (t, to) __f- 2 E Pt cµ (t, to) kn "(t),

=1

n = 2, 3, From this system, we obtain, in view of the fact that g,o(z, to) = z and hence

cn (to, to)=O, for n = 2, 3, ... , to

2e(n-1) e f e (n t)t

C. (t, to)

nl 1cµ (t, to) kn-µ it

eJ

This formula enables us to calculate successively the coefficients Furthermore, since gt0 (z, 0) = f (z, to), by setting

to).

f(z, to)=e-0o(z+c9(to)z9+...),

.f(z)=z+c9z9+...,

we obtain

Cn (t0) = Cn (0, t0), cn = Jim c n (to) = Jim cn (0, t0). io-oo to-o0

In particular, we obtain co

cg=-2S a'k(r)dt,

(23)

0 00

c.,

2 S e-9' (k9 ('t) + 2k ('t) c9 (t, oo )) d r 0

=

00

e'k (T) dt)a.

2 S e-"W (2) d2 + 4 0

(24)

0

These are Loewner's formulas, which we shall need in what follows.

§3. Variation of univalent functions In solving extremal problems of conformal mapping, we often use different variations of univalent functions. A fairly general type of such variations is given by the

Theorem. 1) Suppose that a function w = f(z) such that f(0) = 0 is regular and univalent in the disk I z I < 1. Suppose that the function w* = F (z, A) is a regular function of both z and A in the region r < z I < 1, 1 A I < Ao and that, for 1)

Goluzin [1946f].

III. REALIZATION OF CONFORMAL MAPPING

100

each A in the interval 0 < A < Ao, it is univalent in the annulus r < I z I < 1. Suppose that, for every z in the domain r < z I < 1 and small A, F (z, A) =f (z) -}- Xq (z) -F 0 (XI).

(1)

If we combine the image of the annulus r < I z I < 1 under the function w* _ F(z, A) with the domain interior to the image of the circle z I = r, we obtain, for small A, a simply connected domain BA including the point w` = 0. For the function w* = r (z), where r (0) = 0, which maps the disk I z I < 1 univalently onto BA, we have in i z I < 1 I

f* (z) =f (z) + X q (z) - Azf ' (z) S (z) + Azf' (z) S ( z)

+

0(X$),

(2)

where S(z) is the sum of terms with negative powers of z in the Laurent expansion of the function q(z)/zf' (z) in the annulus r < I z I < 1. The symbol 0(A2) in equations (1) and (2) denotes a quantity whose ratio to A2

is uniformly bounded for all z belonging to an arbitrary closed subset of the disk

IzI <1. Proof. Our problem will be to find the function w* = f * (z) referred to in the theorem, such that f * (0) = 0 (depending on the value of A). We can arrange for this by making a suitable substitution for the variable z in the function w* _ F(z, A). Specifically, we set 00

zz'e"=I

a"w, w)

(3)

where we assume the functions 0,,,( z ' ), v = 1, 2, - -, have been chosen so that 1) they are regular in the annulus r < z ' I < 1/r; 2) 2 (Ov (z' )) = 0 on the circle 1z' = 1 for v= 1, 2, . - ; Av0v(z') converges uniformly in the annulus r < z 3) the series 1/r for sufficiently small A; 4) when we substitute (3) into F (z, A) for sufficiently small A, we obtain a function that is regular in the disk I z' < I and that vanishes at z' = 0. When these conditions are observed, for small positive values of A the circles I z' I = p, where r < p < 1/r are mapped by the function (3) into curves close to the circle I z I = p and the derivative dz/dz' differs on I z' I = p from 1 by a quantity of the order of A. Therefore, for small positive values of A, the function (3) maps the annulus r < I z' I < 1/r in such a way that the subannuli corresponding

§3. VARIATION OF UNIVALENT FUNCTIONS

to z

101

Iz'

I < 1 and I z' I > 1 are mapped respectively into the disks I z I < 1 and > 1 since I z = 1 on I z' 1. But then, for small positive A, the function 00

21 a"'P" lz')

'cv*=F(zre"=1 , A) (4) maps a rather narrow annulus r' < I z' I < 1 univalently onto a doubly connected domain whose outer boundary is the boundary of the domain BA. Since the function (4) is regular in z' I < 1, we conclude on the basis of a familiar theorem that it maps the disk I z' I < 1 univalently onto BA. Thus, the problem of determining the function w"` = f *(z) which maps the disk I z I < 1 univalently onto the domain I

BA will be solved if we can construct functions 0,, W) satisfying conditions 1)4).

To construct the functions (kn (z' ), we substitute the function (3) into F(z, A) and expand the result formally in a series of powers of A:

-

00

E )p " z,

n=0

where ID 0 (z') = f W ), and, for n > 0, Do

D

1

do

(z.) = n! W

_

1

"EIXp

(s')

F (z 'e

A) I

n-1 a'v

da

k-0

(z')

-nl d)lnF(z'e(1 +XV'P. (z)), 1`)I a-0 n--1

_nl

a"

d

(s), 1`)`a-a zf W)'P.(z) n-1

_ xf (z) [n! I

E ""

- d+ Tn(z)J

do F(z'e" dtn z7' (z') 1

(6)

In particular, for n = 1, this gives tilt

/

(Z) = z f (z)

[Fi(zO) Z, f, (Z,)

P, (Z')] .

-J

(7)

But, for r < Iz' I < 1, we have the convergent expansion F)

(Z" 0)

1, C(l)z k.

v k

Z'f' (z')

k--oo

-t -

-1

If we define T, (,Z'')

' C( )z'k l_ Y

k=-co

(8)

I

kco

(9)

III. REALIZATION OF CONFORMAL MAPPING

102

we satisfy conditions 1) and 2) with n = 1. We also satisfy in part condition Iz, I < 1 since the 4): specifically, the function '1(z') is regular in the disk bracketed expression in (7) has no negative powers of z' and $1 (0) = 0. (We note that the function 01 (z') is precisely determined up to a purely imaginary constant term by the conditions listed.) Let us suppose now that the functions for v = 1, 2, , n - 1, are defined in such a way that'conditions 1) and 2) are satisfied and the (Dv(z'), for v = 1, 2, - -, n - 1, are regular in I z' I < I with 0v(0) = 0. Then, in r < I z' I < 1, we have the expansion n-1 0v(zI),

(z')

Fi

do n! dXn !

P. (z'e- 1

,X)

z-o^

z 'f' (z'1

If we take

-1

n (Z)

Ckn1 z k.

-1

C(k.)Z'k + Y h__-oo

(10)

k=-oo C(n)

i,

k--co

then the bracketed expression in (6) will contain no negative powers of z' and, hence, On W) will be regular in z ' I < 1 and On (0) = 0. Furthermore, conditions 1) and 2) are satisfied for On (z' ). (Here, and in general, (bn (z') is determined up to a purely imaginary constant term.) If we choose On (z ') successively, for it = 1, 2, , according to the rule given above, conditions 1) and 2) will be completely satisfied and condition 4) will be formally satisfied. We still need to investigate the convergence of the series appearing in condition 3), on which the satisfaction of condition 4) depends. To do this, we need to prove the I

Lemma. Suppose that the function O(A) has the expansion O(A) _ Yn=1an,1n the coefficients in which satisfy the inequalities Ian I < 1/(n + 1)2 for n = Suppose that the kth power of this function has the expansion [0(A)]k = F,°°_Iank)An where k is a positive integer. Then the coefficients in this last expansion , where C is an absolute satisfy the inequalities la(k)I < Ck/(n + 1)z, n = 1, 2, constant.

Proof. If the coefficients in the expansions of the two functions m

cp (X) = 2] a,A", nc1

satisfy the inequalities

00

4 (X) = f bnX' n_1

§3. VARIATION OF UNIVALENT FUNCTIONS B A 1an1 (n+l)s, bnIC(n+1)s, n= 1 ,

103

2,

and if co

u (A)

(4) _ CE Cn4n n=1

then cn1

alb,,-1+aibn-s+ ... +an_lb1 3,

AB

AB

AB

2AB (n_1)s'I- ... + ns2s(n+1)z [22 1

n_],e + ... + n+1 s 1

1

s

n+1J

1

8AB

).

n

(fl[n+1j+2y]

(n+ 1) 2 where C = 8 (2-2 + 3-2 +.

1

(

/

1

22

CAB 3! + ... = (n+ 1

1)s

If we apply this inequality successively to the

functions 95 (AA)k, we get the conclusion of the lemma.

Let us make a convention on notation. For any Laurent series X(z) _ cnzn we define

IIX(z)II=

n=

Icnl Izln. 00

When such majorizing series converge, they possess the obvious properties IIXI(z)+X9(z)11c11X1(z)II+11X2(2)11, (12)

Let us investigate the convergence of the series that the inequalities DM-1

v=1, 2, ...,

Let us show

(13)

hold in the annulus r < I z' I < 1/r for suitably chosen constants D and M. To prove the validity of (13) for v = 1, we need only take for D the maximum of the quantity 4 11 01 (z' )1I in the annulus r < I z ' I < 1/r.. Let us seek, by choosing M suitably, to satisfy inequality (13) for all v. Let us suppose that inequality (13) holds for v = 1, 2, - , n - 1 and let us find a bound for 11 On (z') 11. To

III. REALIZATION OF CONFORMAL MAPPING

104

do this, we consider the function (10), which we denote by r(z'). We define

F (:, ).) = 21

AklzkXl.

(14)

k=-00, 00 I -0, 00

This series converges for r < I z I < 1 and 0 < A < A0. From (10) and (12), we have

n-l n

II Tn (z) II C I

Aklz

nl dkn

zf (Z)

kale

k

YWy (z/) l

y

I

k, I I

1

zf'(z')

II

d' jIAk!I

x=0

II

n-I kXlelklYEIx°II'Py(Z')Ilr

x

Iz'I

k, I

, n - 1, we obtain on the circle I z'

Then, by (13) for v = 1, 2,

0

r,

zI A'DM°-'

n

do

IITn(x')II
l

Ikl

r-l

(v+I)Y

x-o

k, I

IkID

B

B

Mn-l

I Akl I rk k, I

,

I

I dln

ale

A kl I rkMn l 1 1 k J/DJ l

k,I

J-o

n-l

M

c

y

Y

a=0

do l r jl

jIM1 n. dlR

(if we replace M f with A), where B = maxi

A'

t (V+I)

l

l L.i

A'

-1)'

\1 f

x=o

(15)

=rlI 1/z ' f' (z' )II.

YI

Now, the terms with j = 0 and j = I for l = 0 and the terms with j = 0 for l > 1, 1 n are equal to 0. Therefore, when summing with respect to k, 1, and j, we may assume that 1 + j > 2 and, obviously, that 1 < n. Furthermore, since I

-if

n-l

do

( \1 1v (v-)-1)`/ XV

l

_

11,

1

do

A=o

n-l .-1

1-0

is the coefficient of An-l in the expansion

rn-i cuix" V=I we obtain, by applying the lemma,

for z 0 on the circle I z'

r,

cJ/(n - 1 + 1)2. Therefore (15) yields,

§3. VARIATION OF UNIVALENT FUNCTIONS

IIT; (z) II C B

k,t,J

I Ak1 I rkMn

=B

1

I k J1DhC1

II

Aki (f+ 1)9

(n_1+1)

1)2

I k -JDICI

rkMn-l-I

J!((t-f 1)(n-1+01i

k. l.!

NM" I +

105

I Akr I (1- l)9

rkM-1-11 k jIDIC!

(16)

k, [.J

where N is independent of z . If in the last sum we drop the requirement that 1 + j > 2, this sum will be equal to /

IkIDC

AN I (1-I- 1)Q rkM-h! k. l

and this series converges for sufficiently large values of M since the series (14) is a regular function for r < I z I < 1 and I AI < A0. Therefore, this last sum in (16) with large values of M is of the order of 1/M2. Consequently, the right-hand member in (16) is, for sufficiently large M (though independent of n), less than DMn-1/2(n + 1)Z. Thus, for such M on the circle I z' I = r, we have II T.

In particular,

)II2( +1)s

C(klllz kc2D(17) n-1

k:.-oo Furthermore, since the majorant series of the second sum in (11) does not exceed the majorant series of the first sum, for Iz ' I < 1, it follows that (17) yields the inequality DMn-t

(z) II ` ( + 1)

(18)

on the circle Iz'

r and hence in the annulus r < I z I < 1/r. For M chosen independently of n, we conclude by induction that inequality (13) holds for all v in the annulus r < I z ' I < 1/r. This shows that the series converges uniformly in the annulus r< Iz ' i < 1/r if IAI < 1/M, and this proves the validity of condition 3) and hence of condition 4). Thus the functions 0v(z ') satisfy all conditions 1)- 4) and function (4) constructed from them does indeed map the disk W I < 1 univalently onto the domain BA. This function f *(z') is, for every z' in the disk I z ' < 1, an analytic I

III. REALIZATION OF CONFORMAL MAPPING

106

function of A in a neighborhood of the point A = 0. We have the expansion DO

Ia (z) = 1

(19)

.=o

where (Do (z ') = f (z '). Furthermore, keeping (1) in mind, we obtain from formulas (7) and (9)

411(z')=q(z')-zf'(z')(S(z)-SCZI

)).

,

From (19), we obtain (2). This completes the proof of the theorem. Let us give some examples of the application of this theorem. Let us consider first the function

I in

w*=w+Aw(A+

L ryAkYplk

\1

(20)

k=1

wm are finite points in the

where A and A k are complex constants and w I, w-plane.

For arbitrary given p > 0 and for sufficiently small A > 0, the function (20) is univalent in the infinite domain consisting of the exteriors of the m circles I W - wk = p, k = 1, , in, because, for distinct points w' and w " in this domain,

I w* (w') - w* (w') I

m

= I (w' - w') (1 + A), + k I Ak m, u,' -f - task (tai' + t°'')) k=1

('

M

Wk) W - wk)

'W'I i wklklIIw'+W'I

Iw -w'I(1-A%-Xmax a' wk-1

W'-W

where the maximum, taken over all pairs w' and w", is obviously finite. Therefore, for sufficiently small A, the last expression in parentheses is positive, and hence w" (w") / w" (w "). Now, suppose that the function w = f (z), where f(0) = 0 and f'(0) = 1, is regular in the disk 1 z I < 1 and that it maps that disk onto the domain B. If the , wm are all exterior points to the domain B then, for any positive p points w l, less than the distance between any of the points wk and the boundary of B, we conclude that, for small A > 0, the function (A = 0) M

=f* (z) =J (z) +

k-

f

( z)

(z)

(Z),s

(21)

§3. VARIATION OF UNIVALENT FUNCTIONS

107

is regular and univalent in the disk I z < 1 and that it is normalized by\,the condition f* (0) = 1. On the other hand, if the points w1, all lie in the interior of the domain B, then, for sufficiently small A> 0, the function (21) is regular and univalent in some annulus r < I Z I < 1. If Z1, , zm are points of the disk jz I < 1 such that wk = f (z k), k= 1, 2, - , m, then (21) can be rewritten -

in

Akf (Z)' w* =f (z) + A k _ f (z) -f (zk)

The theorem that we have proved is applicable to this function. Since the function m

_IA

f (z)' zf (z) `k=1 k T77 (z) f (z) -f (Zk) 4 (z)

1

has only simple poles in the disk z I < 1 at the points zk, k = 1, )2 , k = 1, residues Ak f(zk) 2 /zkf , m, it is obvious that

, m, with

m

Ak r f (Zk) `9

S+ (z) _

l Consequently, the formula yields

k=1

in

J * (Z)

=J{

Zk

(zkf' (zb)) Z - Zb in

AjLf

(Z) + x I

k=1 T

() (Zk) - AZf' (Z) 11 Ak (Zkf (Zk) ll9 Z ZkZk k=1

m XZp

(z )

aAk

Zkf

(Zk) Q

(X9 )

1

(22)

Here, we have

f*

m

(0)

I +X

L Ak (zkf((zk) 19 + °

(23)

k=1 1

If we again wish to obtain a variational function f= (z) satisfying the condi-

tion f' (0) = 1, we need only divide (22) by (23). We then have in

f* (z) =.f (z) + A I f (z) k=1

XP Z) I Ak (1 zk f (zk) /l $ f (zk) - AJ (z) k-1 m

- aZ f' (z))

k1

f (Zk) Ak (.Zkf

I

Zk

119

Z - Zk

(Zk)

in

+ Xz! J (Z)

k-1

Ak k

(Zk)

`I9

( Zkf' (Zk))

-*kZ

1 - 2kz + 0

(k2).

(24)

III. REALIZATION OF CONFORMAL MAPPING

108

Formulas (21) and (24) are the desired variational formulas. 1) From these formulas we can easily obtain the corresponding variational formulas for the function

> 1 except for the pole which is univalent and regular in the domain C _ 00 and which does not vanish in that domain. To do this, it is sufficient to apply (21) and (24) to the function

f(z)= and set

(C)=

1

(We note that F, (')

0 in

+ ...,

=C+ao

I I

> 1.)

Simple transformations lead to the following formulas: in

F* (C) = F (C)

I Ak k=1

m F* (C) = F (C) + A

A k

)F (C) (C(C-Wk

Fwk

+ O (X9),

(25)

(C)k:) F((tk)

k-1 M

+ AF (C)

M

Ak (C,yF((Ck)l Je - CF (C) k=1

+ ),CF C

m

A(

I

F ltk)

Ak )s

(

l C

i

lr (Xi),

(26)

CkF' GO

where the wk in formula (25) are arbitrary fixed points all exterior to the image of the domain ICI > 1 under the mapping by the function w = F(C). As our next example, illustrating the theorem on the variation of a univalent function, let us consider a function w = f(C) that is regular and univalent in z < 1 and that does not assume in I z I < 1 certain given finite values al. am, where m > 1. Following the lead of the preceding variational formulas, we can construct univalent varied functions with the same property, that is, not assuming the values a 1i , am in I z I < 1. Specifically, if B is the image of the disk I z I < 1 under the mapping 1) Formula (24) was first stated by Schiffer [1943].

§3. VARIATION OF UNIVALENT FUNCTIONS'

109

w = f (z), then, for arbitrary wk either lying inside B or external to B, for arbitrary A and for sufficiently small A > 0, the function m

II (w - ak)

w*=w+AAwkM1

(27)

11 (w - wk)

k=1

will, as was shown above, map the portion of some neighborhood of the boundary of B that lies in B onto a doubly connected domain and it will map the boundary itself into the boundary of some simply connected domain Bx that includes w* = 0. Furthermore, the doubly connected domain referred to above will lie in BX. Since , am do not lie in B and are mapped by (27) into themselves, the points al, it follows that, for sufficiently small A, they will lie outside BX. Therefore, the function w* = f* (z), which maps the domain z I < 1 onto the domain Bk does not assume in I z I < 1 the values al, - , am . This function is constructed in I

the same way as before. In the case in which the wk all lie inside B, that is, when

Wk=f(zk), IxkI (1, k=1, ..,, nr, it is convenient to put (27) in the form m

I

zm*-zee-{-Xw(A0+

W A

,k),

k=1 M

77

(wk -a,)

11

Ao=A, Ak=A" m

(28)

I

11 (wk-w,) Then, we find the formula for f * (z): M

II (f (Z) - ak)

P (z) =f (z) + AAf (z)

Mk-1

IT (f (Z) --f (--k))

k=l

m

- AAzf' (z)

Akf (zk)

k_1Zkf (Zk)' (Z - Zk) m

+ AAz2f (z) J

Akf (Zk)

k=1 i kf' (Zk)' (1 -ZkZ)

(29)

CHAPTER IV EXTREMAL QUESTIONS AND INEQUALITIES HOLDING IN CLASSES OF UNIVALENT FUNCTIONS

1. Rotation theorems As we noted in Chapter II and as will again follow from Chapter V, extremal questions for various classes of univalent functions are of prime importance in the theory of univalent conformal mappings. When these questions are answered, it turns out that the requirement of univalence of the mapping influences the increase in various quantities characterizing to some degree or other the amount of distortion of the domain being mapped. In §4 of Chapter II, a number of extremal problems dealing with inequalities involving the initial coefficients of the mapping functions, and covering and distortion questions were solved by an elementary procedure. In the present chapter, we shall concern ourselves specifically with these questions, this time using more powerful methods, for example, the method of parametric representation of univalent functions and the method of variations. 1) For convenience, let us adopt some notational conventions regarding certain classes of univalent functions with which we shall be dealing in what follows. Specifically, let us denote by S the class of functions ((z) = z + czz2 + ... that are regular and univalent in the disk I z I < 1. Let us denote by I the class of

functions FW = C+ ao + a1(-1 + .. that are meromorphic (with pole at Do) and univalent in the domain ICI > 1. As a supplement to the sharp bounds on the modulus of a function belonging

to the class S and the modulus of its derivative that were established in §4 of 1) The reader wishing to become more familiar with extremal questions regarding univalent functions is referred to the author's articles [1939d, 1949c] .

110

§1. ROTATION THEOREMS

111

Chapter II, we shall give in the present section the corresponding sharp bounds on their arguments. In particular, we shall establish the final formula for the rotation theorem for the class S consisting of a sharp bound on the argument of the derivative. The parametric representation of univalent functions given in §3 of Chapter III will provide us with these bounds.') Let us denote by S' the subclass of the class S that consists of functions ((z) having a representation of the form

f (z)=

lim etf (z, t),

(1)

t-m where [(z, t) is a solution of the differential equation df dt

_ _f 1-F kf 1 -kf'

(2)

satisfying the initial condition

f(z, 0)=z

(3)

that, for fixed t, is a regular function of z in I z I < 1 such that If (z, t) I < 1 in that interval, f(0, t) = 0, and [' (0, t) > 0. Here, k = k(t) is an arbitrary piecewisecontinuous function that is equal to unity in absolute value in the interval 0 < t < no. In order to find a bound for any of the quantities that we shall consider below for the entire class S, it will be sufficient for us to do this for the subclass S' because, as was shown in §3,of Chapter III, an arbitrary function ((z) of the class S can be approximated by functions [, (z) such that [, (0) = 0 and [' (0) > 0 each of which maps the disk I z I < 1 univalently onto the w-plane with cut along a Jordan arc that goes out to no and does not pass through w = 0, and consequently by functions [, (z)/[n (0) E S'_. When this approximating procedure is carried out, the sequence of bounded quantities for the approximating functions converges to the same quantity for the function 1(z). A bound for arg [I (z)/z] and arg [zI' (z)/I(z)]. We divide the identity df (z t) at

- f (z' t) 1

k (t) f (z, t)

'

by I(z, t) and separate the real and imaginary±parts. This gives us the two identities t)I1-If(z, t)I4 dIP2, t)1=-If(z, at

1 1-- k (t) f (z, t) 1s

1) See Goluzin [1936] and a number of subsequent articles by the author in Mat. Sbornik, for example, [1939c, 1946c].

(4)

IV. EXTREMAL QUESTIONS

112

It follows from (4) that

dargf(z, t)

23(k(t)f(z, t))

at

1- k (t) f (z, t)

I f (z, t) I

(5)

decreases monotonically from

IzI

to 0 as

t

increases from 0 to o c. This enables us to take as our independent variable the quantity I f (z, t) I instead of t. Now, from (5), we have 12' k f12 dt

I di argf 1

(where we write for brevity f and k instead of f(z, t) and k(t)) or, eliminating dt with the aid of (4) and keeping in mind that arg f = arg erf,

1I 2d

I dt arg e f I

l

l 1,

If we integrate this with respect to t from 0 to - or, consequently, with respect to I f I from I z I to 0, we obtain, by virtue of (1) and (3) argfZZ) I

(6)

Here, the argument is defined by the formula 00

arg

f i Z ) _ dr argf Z, t)

Consequently, it represents that branch of this multivalued function that approaches

0 as z---- 0. Inequality (6), which we have proved for the class S `, holds also for the entire class S by virtue of what was said above. Let us show that the bounds for arg Lf(z)/z] given by inequality (6) are sharp. To do this, we show that there exists a function k (t) of absolute value I which is continuous in the interval 0 < t < 00 and for which f (z, t) defined by equation (2) and condition (3) will also satisfy the condition (k (t) f (zo, t)) _ - I f (zo, t) I for arbitrary t such that 0 < t < oo and given zo, where Izol < 1. If (7) is satisfied, then from equations (4) and (5), which are equivalent to equation (2), we have for z = zo

dlfl

if 11 at - Ifl1+IfP,

dargf

-

21f1 1+lfj'.

tt

If we integrate the first of these equations, taking If(zo, 0)I = Izol, we obtain

(7)'

(8)

113

1. ROTATION THEOREMS

IffI'=eel-I go Zo18

I

(9)

From this we determine I f I as a positive function o ft that decreases monotonically from I zo I to 0 as t increases from 0 too . Furthermore, by eliminating dt from equations (8), we have

2difI l dargf=- 1-if from which we get ar

g

f(zo, t)-lo I+Iz°I -log

zo -

g

1-IzoI

g 1-If1'

(10)

which, on the basis of (9), can also be expressed in terms of t. Then, k(t) is determined from

argk(t)-}-argf(zo, t)_- 2 Since all these operations are reversible, we have equation (10) for the function f(z, t) defined from the function k(t), which is a solution of equation (2) evaluated at the point z = zn Z. Here, as t -4 a the rieht-hand member approaches log (1 + Izo j)/(1 - Izo I) . Consequently, the upper bound for arg [f (z)/z] given by inequality (6) is the least upper bound. In the same way we can show that the lower bound for arg [f(z)/z] given by inequality (6) is the greatest lower bound.

Since the function

zC) -f (z) f (C+z l l -}-

g(,)= f'(z)(I-Iz12) =C } ...,

(11)

where z is an arbitrary number in the disk I z I < 1, belongs to the class S when f(z) does, by applying inequality (6) to this function with J_ -z, we obtain the following inequality: 1) (z) arg zf f(z)


(12)

.which holds for an arbitrary function belonging to the class S and for an arbitrary point z in I z I < 1. By arg [zf '(z)/f (z)] we mean the branch defined by the equation 1) Inequalities (6) and (12) were first given by Grunsky [1932]. For information on

their generalization to univalent functions in multiply connected domains, see Goluzin E19371.

1V. EXTREMAL QUESTIONS

114

=-z

f iz) argzfO

Cf

d arg

`1±zi --f z) f O( I)

=

I

o

d arg

f(C)-z(z) '

S

that is, the branch that approaches 0 as z -- 0. The bounds given by inequality (12) are also sharp because, when we apply (12) to the function (11) at the point z, we again obtain (6). A bound for arg f '(z). In an equally simple manner, we can find a bound for arg f'(z). Specifically, from (2) we have df'___

f-

=df(z,dzt) f 1 (1+2kf-k'f _k f)a .

,

from which we get d arg f

(1+2kf- k'f91

at -

_ 2,1 ((1-- kf)')

(l-kf)' J- I1-kfI'

l

and, consequently,

-dlfl

dtargf,= 2 ((1-kf)2)

Il -kfI' IfI(1-Ifl$)'

But

1-kf12

=I sin2arg(1-kf)I sin 2 arc sin f1=2

for for

1

Therefore,

4dlfI

_

l-IfP

for

ifI
Ifl(1-IfI2)

for

If1>_y2

2difl

Integrating this with respect to t from 0 to Izl

argf'(z)I<

4difI

1-Ifr

we obtain for I z I < 1/V

-4 arc sin IzI,

and, for I z 11 1/\/-2, I

Ir-2

If I IS+

-If

I fl

2d If Y

)r2

Ifl=)'

§1. ROTATION THEOREMS

115

that is, 9

log

Iargf'(z)J-iiv

1

Izlzl'

Consequently, for functions f (z) in the class S, we have 1)

for JzJ
4aresin IzI arg f" (z)

s

I

-x -- log

zllz

l

for

JzI

Y2 .

(13)

I9

That this inequality gives a sharp bound on I arg f'(z)I for I z I < 1// is obvious from the fact that, for a function z

f(z) in the class S, we have

1-e'az dz=z+csz2--I-...> a=arccosp,

0

Iargf'(p),I=4aresin p for every p < The sharpness of the bound (13) (for arbitrary z in I z I < 1 in fact) can be seen by showing that there exists a function k(t) of absolute value equal to 1 that is continuous in (0 <.t < oo) for which f (z, t) defined by equation (2) and the condition f (z, 0) = z satisfies the condition

-2If1Y l-IfI' for IfI
-sin2arg(1-kf)^

1

for If 1 2

-1

or, what amounts to the same thing, the condition 3 (kf)

- sin ar (1 - bf) -

- Ill 1

Y2

for

for

(14)

for arbitrary t in the interval 0 < t < oo and given z = zp in the disk I z I < 1. From (14), we obtain by simple calculations

IfI(IfI- l

1 -Ifls)

for

IfIv-2,

kf= I:t 21f1'-1 _i1+y21fI"-1 for IfI> 1 2 2

1) Inequality (13) was proved by the author (see Goluzin [19361 ). That it represents a sharp bound for I z I > 1/\/Z was shown by Bazilevic [1936a1.

IV. EXTREMAL QUESTIONS

116

Here, we can take either of the signs ±. Now that we have found k f expressed in terms of I f I, let us substitute the result into (4) and (5). Then, from (4) we obtain t as a continuous monotonic function of if i and, consequently, I f I as a continuous and positive function of t. From (5) we then have a representation of arg [f(zo, t)/zo] as a function of t. Knowing arg kf, we find arg k(t) and thus k (t) itself. For the function (1) constructed from this k(t), we have, as we can see from the proof of inequalities (13), I

for I z

4arc sin z0

argf'(zo) = it

+ log

01

I

for IzoI,V-2

z0I' t-IZoI'

This shows that the upper bound for arg f ' (z) given by inequalities (13) is the least upper bound for arbitrary z in I z I < I. In the same way we can show that the lower bound given by inequalities (13) is the greatest lower bound. We note that, reasoning as above, we can prove the inequality arg f(2)') I

-

log (1 - I z I9),

(15)

which gives sharp bounds for arg Lz 2f ' (z)/f (z)Z] at an arbitrary point z in

z I < 1. Therefore, by shifting from f(z) to the function F(t) = 1/f(1/0 + for arg F '(c): const, we immediately obtain sharp bounds in the class

IargF'P

I<`log(l-

(16)

We note that, by modifying slightly the preceding derivations, we can sometimes obtain stronger results. For example, if we divide equation (2) by f, difIf, we obtain ferentiate with respect to z, and then divide by f '

dlogf

2kf

kf)"

from which we get

Eliminating dt here and in equation (4), we arrive at the equation

Idlogf f

---l-IfI" 2dlfl

§1. ROTATION THEOREMS

117

On the basis of this equation, we have for arbitrary real 0, which we assume independent of t, e-" d log

-

f)

1-jf'

If we integrate both sides of this last inequality with respect to t from 0 to oo, we obtain e 1B to

g

zf'(z) Elzl 2dif_.-lo I+IzI gI-IzI 1-IfI f(Z))

Consequently, for all f (z) E S we have in I z I< 1

(e!° log

f (z))) - log I +

z

.

(17)

Since 0 is arbitrary, this is equivalent to the inequality log

Zf (z) C log I + I z

f(z)

If we apply (17) to the function (11) with

l

l-IzI'

(18)

= -z, we obtain another inequality'

logLZZl--log(1-IzI')IC1og1+1zI

(19)

which also holds for all f(z) E S and all z in I z I < 1. We note that we can show by a procedure like the one used above that inequality (17) is sharp for arbitrary 0. In the same way we can show that for the functions F (C) E E in the domain > 1 we have the sharp inequality I log F* G) I <--log (1 - IIIIt) .

(20)

Furthermore, this inequality, from which one can get the inequality 1

I

1

1-IOI2

9

(21)

constituting the distortion theorem for the class 1, will be derived later by a different procedure.

(We give the following definition: suppose that a functional 1g is defined

on a certain class of functions g(z). The set D of values w that the functional w = Ig assumes on the given class of functions is called the range of values of the functional Ig in that class of functions. 1) Inequalities (18) and (19) were obtained by Grunsky [1932] by a different procedure.

IV. EXTREMAL QUESTIONS

118

As will be shown in §3, p. 133, inequality (20) determines the range of values of the quantity log F '(C) in the class 1. By the same reasoning, we can show that inequalities (18) and (19) determine the ranges of values of log [z f '(z)/ f (z)] and log [ f (z)/z] respectively in the class S. For more on this point, see also Lebedev [1955a].) §2. Sharpening of the distortion theorems As another application of the method of parametric representation, we now present sharpened forms of certain inequalities constituting distortion theorems for the classes S and 1 .1) Let us first show a number of relationships that hold in the parametric representation of univalent functions and that play an important role in questions concerning bounds. Consider an arbitrary function f(z) E C ' defined in accordance with formula

(1) of §1, where f (z, t) satisfies in the interval 0 < t < o the differential equation (2) and initial condition (3) of §1. For the function f (z, t), we have a number of interesting relationships. Specifically, let us take in I z I < 1 arbitrary points zvfor v = 1, , n and let us write for brevity fv = f(zv, t). By using the equations

()f" we obtain by direct calculation the relations d

dt log

e tt f"-f", [f"f z"

2

kf" kf 1 - kf, - 17- kf (

kf

(1) N

Nlog(1-ffv,)=21 kfkf" 11-kf".) (where k = k(t)). If zv = zv,, the expression (fv- f,.,,)/(z, - z,,) in (1) should be understood to mean f (zv). If we integrate equations (1) with respect to t from 0 to - and keep (1) and. (3) of §1 in mind, we obtain the following integral formulaslog

F(C")-F(C C"

C",

W

2

kf"

kf

f 1- kf, ' 7- kf", dt

,

0

log g

I `

(2)

kf 2 f fkf, (1-kf k

C"C"'

1) Goluzin [1948b, 1951a].

co 0

1

yr

dt,

§2. SHARPENING OF DISTORTION THEOREMS

119

where we set Z, }.

(-1-)

We introduce the expansions log

F(C)

-

00

F (C')

ICI>1,

ak,:C_

IC'I>1,

,

k, 1=1 (3)

00

I bk(t)zk, z<1.

k(t)f(z, t)

l-k(t)f(z,t)

k=1

When we substitute these expansions into (2) and identify the resulting equations with respect to C. and we arrive at the formulas 00

(k, 1=1, 2, ...),

ak.1= - 2 S bk (t) b, (t) dt

(4)

0

0

Co

bk (t) bt (t) dt0

2k

for

1,

for k -

(k, 1=

1,

2, ...).

(5)

Formulas (5) express the important property of orthogonality of the functions

bk(t) for k= 1, 2,

.

We shall now prove two theorems dealing with the applications of formulas (2) to inequalities. for Theorem 1. Let F denote a function belonging to 1. Let av. In v, v' = 1, 2, .. , n, where n > 1, denote real numbers such that ,, mIa'v v'x v' is a positive quadratic form. Then, for arbitrary C., v = 1, 2, , n, the inequalities n

I1 Vl=1

f,

V

V

V,

it

F(Q-F(tv) CV - Cy,

(6) 1

n

ay 1

1

vl

CCy, Y, V' = 1

hold throughout the domain I I > 1, where we should understand the factor IF' (Cv)I in the second product when Cv = Cv'

Proof. Let us first prove inequalities (6) for the functions F(t;) = 1/f(110, where f(z) E S'. In this case, if we set

IV. EXTREMAL QUESTIONS

120

kfY

1- kfY

=XY+IYY, , =1, ..., n,

and take out the real parts in (2), we obtain log

F(CY)-F(CY,) CY - CY

I

60

= - 2 f (X.,X,, - Y., Y.,,) dt, 0

log 1-CC,I

2 f (XYXY

rY 0

Then, by adding and subtracting, we arrive at the following formulas log

F(C,)-F(C,,) CY - C,

CO

=- log

-4f XYXv dt, 0

logI

F(C,)-F(C, ) c - CY,

+ 4f YYr' dt.

= log

0

We multiply these formulas by the numbers a v, and sum each of them with with respect to v, v ' = 1, , n. Then, remembering that for numbers av v, that satisfy the conditions of the theorem, we have n

r.y VIXYXYI ' O, Y V'=t

"`Y, V' YYYY J O Y

We then immediately obtain inequalities (6). But these inequalities, which were proved for the functions F(C) E represented in the form F(c) = 1/f(1/0, where f (z) E S', obviously remain valid for arbitrary functions F(C) E that can be represented in the form F(C) 1/f(11C) + roost, where f(z) E S, and consequently they remain valid for the entire class . This completes the proof of the theorem.

Theorem 2. Let n denote a positive integer. For v = 1, , n, let yv denote arbitrary complex numbers and let Cv denote complex numbers in the domain I C) > 1. Then T'TY' log

F(Cr)-F(C,) C -C n

1YT: log (1

} - Y1 C CY,/

(7)

§2. SHARPENING OF DISTORTION THEOREMS

121

for F(C) E E.1) Proof. Just as above, it will be sufficient to prove (7) for the functions F (C) = l/ f (1/C), where f (z) E S'. But, in the present case, we have from the first of formulas (2) F (CY) - F (C,.)

n y

IYTY' log Co

n

IYTY' 0

=1

n

00

kf"

1

2

kfY'

1 - kfY 1-kfY,

dt

kfY

2

TY

\

19

1 - kfY l

dt

V

and, consequently, n

F (C") - F (CY,) TYT:Iog CY-CY,

n

9

kfY

_< 2

TV I

V=1

M = 2

- kf

dt

Y

n

Y

'1 ' 1--M,kfY

Rj"'

1 - kfY,

dt !

From the second of formulas (2) applied to the last integral, this yields (7) and completes the proof of the theorem. As will be shown in §3, inequality (7) is sharp for arbitrary yr, and and determines the range of values for the summation on the left . In an analogous manner, we can prove the more general Theorem 3. Let n and n' denote two positive integers. Let yr,, v = 1, , n, , v = 1, , n, , n ' , denote n + n' complex numbers . Let and v'= 1, > 1. and complex numbers in the domain , denote n + n ' v r = I, n) . Let F denote a function belonging to E. Then, A6.d

J,TYT,'IogF(C")-F(C") V,

Y-1 Y'=1

CY' - C

1) The expression log F (C ) -F (C) with t and C in the domain I C I > 1 always refers to that branch (single-valued in this domain) of the multivalued function log

F (C)

-

F (C')

that approaches zero when at least one of the points

approaches

co. For I = C' , this quantity becomes equal to log FP(C) and then it can be understood as the value of that branch of the multivalued function that vanishes at

C = W.

IV. EXTREMAL QUESTIONS

122 n

17,T,' log 11 -

7;,1og

1

/

(8)

(If we write inequality (8) for F2(t;) E 12, replace n and n' with 2n and 2n', and set yn +y = - yy and Sn+v = - Sv for v= 1, , n and y.,+,, = Yv and ,,,}v = , n', 'we obtain the - Cv for v = 1, Corollary. Let n and n' denote positive integers. Let yy for v = 1, ,n for v' = 1, , n' denote complex numbers. Let Cv, for v = 1, . , n and tv, for v'= 1, , n', denote complex numbers in the domain C] > I. Let F2 denote a function belonging to 1 2. Then, and y,;

tJ`n1'

v

F,(Cv)-F2(Cv')Cv+Cr' 7riv log F. (Cv) + F. (Cr,) Cr l r'=1 `J1

c

"

7r7r, log

n1

CrCr, +

IX- -

CSC,, +

7,7" log C'XY, 1

-1

(9)

Theorem 3 and its corollary were first obtained by Lebedev [19511, who started with the area principle.) We mention some special cases of Theorem 3. In the first place, if n = n'= 1 and S I = I S I = P > 1, then (8) gives the sharp inequality (the subscripts are omitted for t;)

Ilog F(CC-C (C')

<-log (1-(10)

with equality holding for the function F(C) = t;+ e`L/t;.E E, where a is a real constant (Goluzin [1947]). Second, if n = n' = 2, YI=Y2 =1, Y2 YI 1, C2 C1, and t`Z =-SI, we obtain the inequality log (F (C) - F (Cl)) (F (-C}-F-(- C')) (C + C')'

2 log 11-+ Pe 1

(F(C)-F(-C'))(F(-C)-F(C'))(C-C')s

which, if F(t;) is an odd function, yields the sharp inequality of Bazilevic [19511 and Milin (see Lebedev and Milin [19511): log (F (C) - F (C')) (C + C') g (F (C) + F (C')) (C - C')

log I

l+ 1

Pp . 1

P'

(11)

§2. SHARPENING OF DISTORTION THEOREMIS

123

with equality holding for the same function F()). (Inequality (11) also follows

= n' =1 and from(9)for n=n'

I b1 I = 161

I =P> 1.)

The applications of formulas (4) and (5) are effectively used.

E S. If we set

Theorem 4. Suppose that f(z) = In ((Ill

F(C)=

ICI>landlog F(C

;(C)'

«') C

00

E ak, t (:

ICI / l

k, 1=1

I,I

l

where the ak,t are rational functions of the coefficients ck, then, for arbitrary , n with n > 1), values of the complex variables xk (where k = 1, 2, n

n

,

ak,I xkxr

( 12)

k l9

k,1=1

k=1

In the case of real values of xk, equality holds in (12) for the functions f(z) _ z/(1 ±z)2. Proof. We can confine ourselves to functions f (z) E S'. In this case, the ak,1 have an integral representation (4) with the functions bk(t) subject to conditions (5). Therefore, n

oo

n

oo

n

E ak, txk,xt = - 2 E bk (t) b, (t) xkxldt = -2S (E bk (t) xk\2 dt k, 1=1

0 k, 1=1

0

h=1

/J

and, consequently, n

Y ak,1XkXI k, 1=I

00

2S 0

n

2

ao n

E bk (t) Xk dt = 2 S E b, (t) bt k=1

(t)xk_xidt.

(13)

0 k, 1=1

On the basis of (5), this yields inequalities (12). In the case of functions f(z) _ z/(1 ± Z)2, we have k (t) = ± 1 for 0 < t < so that the functions b k (t) for k = 1, 2, - -, corresponding to them are real. Therefore, equality holds in the conditional inequality (13) for these functions when the values of xk, k = 1,2, , n, are real. This completes the proof of the theorem. Schiffer [19481 proved the sharpness of inequalities (12) for arbitrary complex

values of xk. As a generalization of inequality (12), let us prove, under the same conditions,

IV. EXTREMAL QUESTIONS

124

M

n

J Ixkl'

I Yak, exkxe AMW

k=I1=1

k

k=1

x18 1

l=1

where in > 1, n > 1, and xk and x1' are arbitrary complex numbers (for k = 1,

2,---,m and l= 1,

n.)

Let us prove two more theorems for functions in the class

Theorem 5. For functions F (,'') E I and arbitrary n > 2, on the circle I C1 = p, p > 1, we have

, n, where

v = 1,

2

}(1-:)2(n_I)II

F (C,) - F (C,,)

H Y$Y

.

CY - CY

n

(14)

CYCyr

Y#Yr

In the derivatives, v and v

range over all pairs of values v, v' = 1, except coincidental ones (v = v' ).

,n

Proof. In (6), let us take for a v, two systems of numbers av v, and av v, satisfying the conditions of Theorem 1 and let us divide the resulting inequalities one by the other. We then obtain as one of the inequalities the following:

n

1 I,% -Y /

F(CY)

R

aY' Yr -

Yr

II 1

CY __ CVr

Yr Vr=I

Y, Yro1

aY' Y

V,

(15) 1

CYCY.

In particular, since the forms xi--+...+xng

+xjx8-,-...+xn_1xn.= (xl+...-}, xn)

X; + ... + x'R-n-1 (xtx9 + ... + xR_lxn) = 2 (n

I

1

)

I (xV

xY,)

Y$Y'

are both positive, we can set

= 1,

v,

v' = 1, ..., n,

for v=v',

1

n.

aY, Y _

-

for

1

n-1

v -A v;

We then obtain the inequality

11 V # V.

F (CY) - F (Cv,) CY - CYr

-

Ycl

I

\l

l

a

)2-

2

2

nil YY

n

(Y(, `

1

I

(16)

125

§2. SHARPENING OF DISTORTION THEOREMS

, n, this yields inequality (14). This completes the

with I v I = p, v = 1,

proof of the theorem.

Theorem 6. Let n denote an integer > 2. Let Ev denote distinct numbers , n. Let F denote a function belonging to I. Then, such that Ie, = 1, v = 1, on the circle k 1 = p, p > 1, I

max

(17)

n

p

where A is positive and depends only on n and cv, v = 1, 2,

, n.

, n, and Proof. We apply inequality (14) for Sv = ev C, v = 11 Denoting by M the maximum on the left-hand side of (17), we have

F ( t) I

f' 5'r

I

I = P.

2n(n t)M4(n 1).

Y#Y

Furthermore, on the circle I I = p, we have I=Arpn(n-t),

rjIE,C-e.CI=p'(n-1)TTIEv-`v V:;1-V'

Y#r

I1. 1r$r

_1 evev P

f

1 min

lI 1--

I
1 VEV'P

Here, A 1 and A 2 are positive and they depend only on n, a 1, Keeping all this in mind, we obtain from (14) 2n(n- t)Mn(n-t)

AjAQpn(n-1) (I

)S(n-t)

19 P

, Ev.

/

which yields inequality (17). This completes the proof of the theorem. We now have the following sharpened bounds on the modulus of a function belonging to the class S:

Theorem 7. Let n denote an integer > 2. Let cv, v = 1, . . . , n, denote distinct numbers such that IEvI = 1. Then, for functions f(z) E S on the circle z I = r, where r < 1, min If (EYz) I C (] V=I,.. n

Ar r)2/n'

-

(1S)

where A is finite and depends only on n and Ev. Proof. Inequality (18) is obtained from (17) if we apply the latter to the funcEv = l/zv and then replace C. with F We note that the exponent 2/n in the right-hand member of inequality (18) cannot be replaced with a smaller number without additional restrictions on the

tion F(C) = l/f(z), where J= 1/z, and to the points V.

IV. EXTREMAL QUESTIONS

126

function f (z). To see this, consider the function z

f (z) _

n

11 (1-a-0) 2/a k=1

In I z I < 1, we have

dargf(z) dR

-+ 2 k-

n

n

tkz I/

I

2 (1_z)= k=1

a

n

1+

2

z

1-Iz12

>0

Ekz l

I

r, where z re `s. Consequently, as z moves around an arbitrary circle I z where 0 < r < 1, in the positive direction, the function arg f (z) increases monotonically by an amount 2n; that is the point w = f (z) describes (one time around) a closed Jordan curve. But then the function w = f (z) maps the disk I z I < r univalently for any r in the interval 0 < r < 1. This means that f (z) E S. Now, for 0 < r < 1, we have

r

Ar

Ii I I - -8.,r I21n

(1 - r)2/a

If(Evr) I- n km1

where A is a positive number depending only on n and e, v = 1, - - -, n. This proves what was said regarding the sharpness of inequality (18). We note that, for n = 2, we obtain from inequality (14) the following inequality for F (C) E 1: F(C1)-F(ra) (19) C1-C,

where CI and C2 are arbitrary points on the circle ICI p, where p > 1. Inequality (19) is called the theorem on the distortion of chords under mapping of the domain CI > 1 by functions belonging to the class 1. This theorem, which" was proved by the author, l) has found numerous applications. Here, we mention the direct elementary derivation (given by Milin) 2) of inequality (19) and also of inequality (10).

Specifically, the function'() = log [F(O - F(io)], where ICO I > 1, is regular and univalent in I< I CI < I CO I, and its expansion in that annulus has 1) Goluzin [1946e]. 2) Lebedev and Milin[1951], inequality (11.1).

§2. SHARPENING OF DISTORTION THEOREMS

127

the form

log

F(C)-F(Co) C

+log(-Co)+ log (1-

_ - ` Cn

an (Co)

)

Co

Co

}- log (- Co)

n=l

-

Co

1

n=l

C

n

a (Co

Therefore, the area S of the bounded domain bounded by the image CP of the circle I z I = p, where 1 < p < 1 Co I, is 4\qual to 00

S- a\ Il

n I an p

2nl

fn

l

n (

1

1

10

)

0

CIO

log(1-Iol9}

11

n=1

n=1

ICoI/2n

If we now let p approach 1, this yields Co

nan(C0)I9c-log(1- IC )

n=1

Now we have, for 1 < I

1 < Co 1, 00

00

F(C)-'F(C0)

l I

log

t5:_

lan(Co)I

C - Co

I C In

n an

nIa_(Co)14' n-

nJC1 n

n-

(1

which, as we let

I Co I

log F (CC

_ _I C In

n-l

-

C Is) log (1

- C7

approach I I = p, yields

_F

(Co)

c- log (1- P9),

ICI=ICoI=P>1.

If we now replace the absolute value with the real part taken with a minus sign, we obtain inequality (19). We note that inequality (11) can be derived in the same way.

IV. EXTREMAL QUESTIONS

128

§3. Extrema and majorizations of the type of the distortion theorems

The method of variations also enables us to establish a number of results of the distortion-theorem type and to calculate completely the extremal functions. 1) Furthermore, in the investigation of extrema of certain quantities, we are able to ascertain their ranges of values. We shall stop here to go into only one theorem in detail.

Theorem 1. Let n denote a positive integer. Let y,,, v = 1, , n, denote n complex numbers, not all zero. Let S1, Cn denote distinct complex numbers in the domain ICI > 1. Let F denote a function belonging to E. Then, we have the sharp inequality n

n /

I\

j -1

7,7y, log (1

7Y7 109 .,--1

> 1 by the

Equality holds in (1) only for functions defined in the domain equation n

(1)

jjn

logF(C -C (C.)_-L7,

1

.=1

(2)

1og11-

y

> 1 onto the entire w-plane with a cut along

These functions map the domain the curve defined by the equation

(2]n 7, log (zap - F (C )))=const.

(2')

Furthermore, for I CI > 1, we have the inequality n

.,=1 .'

log F (C') - F (C,.) Cy - C,

n

<

-t

7y7V-log

1

_

l ,), C,t,

(3)

which defines the range of values for the summation on the left. Thus, for arbi, n, this range of values is a disk. trary given yv and t;',,, v = 1, 2, Proof. Let us find the maximum of the quantity n

IF

=

for given y and [;v, v = 1, 2,

class 1. 1) Goluzin [19471.

(L 7'7y' log F (CIJ _ 1 (C 1)) ., v'=I

C'

C,

, n, and for F(t) ranging over the entire

(4)

§3 EXTREMA AND MAJORIZATIONS

129

Since the quantity IF has the same value for functions that differ from each other by a constant amount, this maximum coincides with the maximum IF in the subclass E0 of functions F(J) of the class I such that F(O 0 in ICI > 1. That this maximum is attained follows from the normality of the class 7.0. Suppose

that w = F(t) E !0 is one of the a'tremal-functions. Let us suppose that the image B of the domain I CI > 1 under the mapping w = F (C) has an exterior point w0 . Then, let us form the varied function F. (0 E 10 by using formula (25) of §3 of Chapter III with m = 1 and with w1 replaced with w0. For this function, we have

(F,-Fy,)(1-AA1

n

IF*= at( E 7r7r' log

C - C' I

V, V-1

IF-X

7r7r'

( Al

(Fv -- W0)

(P"'

-

tD0)

. )+ (4)

[Here and in what follows we shall write for brevity, Fv instead of F(Cv).] Since w0 can always be chosen in such a way that the last sum is nonzero, we obtain IF. > IF for small ,A > 0 and suitable arg A 1. But this contradicts the extremality of the function F(J). Consequently, the domain B cannot have exterior points. Let us now turn to the varied function (26) of §3 of Chapter III with m = 1 and with t 1 replaced with CO , where ICI > 1. For this function we have a

Al

IFS =1F

F'2

17r7r'

v,r'-1 It

Al

(Fv - FO) (Fr'

7r7,' (C0 F 0

_

4

7,7" J - Al (C F v' Lr r' - 1 0

+

0

0

n

r,r'-1

7r7r'

r

v

C"F",

COC, -1 9

Y

F-F.

C,F,

A1(Ca

C''F'y'

C'FF

n

F

l9

FO)

1

toC,, -1 I+ 0 (A2).

Fy - Fr.

l

1

(Here, all the ratios representing indeterminate forms must be replaced with the ratios of derivatives.) If we replace the last term in the large parenthetical expression following $2 with its complex conjugate, we conclude on the basis of the arbitrariness of arg A 1 and the extremality of F (C), that the quantity in the parentheses following R after we divide by the common factor A 1 must be equal

IV. EXTREMAL QUESTIONS

130

to zero because otherwise, for suitable choice of A 1, we would have IF. > 1F, which would contradict the extremality of the function F(C). This leads to the condition n

it

(Fv - Fo)

00

-1 n

-}- N

TT,

v

vFr C2 1F. C, - Cr

F,

v

V r'_1

Cv"F"r"

Co-Cr,

- Fv, Cr,F;,

C. FY

-

r _1

TrTr'

Cc-I

Coc,, - 1

Fr-F,"

(5)

Since 4 in this equation is an arbitrary point in the domain I CI > 1, if we replace Co with C, we obtain the differential equation that the extremal function F (C) must satisfy. We denote the right-hand member of this equation by A (c). Let us show that A(C) > 0 everywhere on ICI = 1. To show this, let us consider a function t = z/i(Z) where 1/1(00) = 00 and 1/i' (oo) > 0, that maps the domain I CI > 1 univalently onto the domain t I > 1 with a small radial cut issuing from I

the point t = e'`B. This function is obviously obtained from the equation teie-}-

to=(1 +h) (Cete-+ -'Ceie)

+2h

with h > 0. From this we obtain, for small h,

1-Ceie+0("' COO)=1+h. By means of the function l/i(C), let us define the following function in the class Fl (C) _

(

F (C) - h (F (C) + CF (C) I,

0 w).

For this function, n

IF.=1F-h(Y

TvTv n

+

CYF'r+Y -Cr"F'r" C+Cy" C-C 7r7y

Fr - Fr.

,) + 0(h2),

r, r1 = 1

where r=

do.

Then, by virtue of the extremality of FW, we obtain, for all t on the circle

1

§3. EXTREMA ANP MAJORIZATIONS

r

131

C F' C+CY -Cy,F',,C

n

n

V, Y1= 1

V, V1

+C,,

IVIV

If we denote the left-hand member of this inequality by B (0, we have B on the circle I CI = I. On the other hand, on this circle we have n

A (C)-B (C)=12

TvTy'(1Y, 1

C"F',

-C"F'y,))

F, - F, ,

JJ

0

(6)

Let us show that the right-hand member of this equation is equal to zero. To do this, let us again consider the function F0 = e-`BF(e`B/') E E0 for different real values of 0. The quantity IF, attains its maximum on the interval 0 < 0 < 2rr at 0 = 0. Therefore, (dIF6/d0)I B=o=0, which, in expanded form is equivalent 1, we have to vanishing of the right-hand member of ( 6). Thus, on the circle 0.

It follows from this result that, if the rational function A (C) has zeros on the circle CI = 1, all these zeros are of even multiplicity because the increase in arg A (4) as C moves around a small circle with center at any of these zeros will, by virtue of the principle of symmetry, be equal to twice the increase in arg A (C) as C moves around an arc lying in ICI > 1 and this last increase is equal to an integral multiple of 27r. Therefore, the function VrjT?) is regular and real on ICI = 1. Turning now to the differential equation (5) (with CO replaced with C), we note that twice the sum in its left-hand member can be represented in the form n TY

( 1

Consequently, this equation can be written in the form n

C. I F, T' F = A (C) . V=1

If we solve it we get the following dependence of F(O on

j_

n

v.1

7, log

which holds in I CI > 1.

(Fy -F)

--

S

A (C) CC = C (C),

(7)

IV. EXTREMAL QUESTIONS

132

If we take an arbitrary boundary wo of the domain B and denote by Co the corresponding point on ICI = 1 under the mapping C = F -1(w), we conclude by passing to the limit that equation (7) holds (with wo instead of F) holds between 1. Also, since o and wo. In this sense, equation (7) also holds on

51 (C(C)) _ ( ae)) _

(1CC (-)) _ f (I i/A (C)) = 0,

1, where C= ee, it follows that the right-hand member of on the circle equation (7) and, hence, the right-hand side of the transformed equation

rsl lT,log Cy-C +7ylog(1 finI

(

C (C) -

V=1

(7, log G.,

- ) - 7ylog \ 1 -

11

(8)

will have a constant real part on ICI 1. But this means that the sum on the left side of (8) has a real part that is constant everywhere in the domain I C1 > 1 because it is a harmonic function in ICI > 1 and hence it cannot attain its maximum or minimum in I C1 > 1. From this we conclude that the left-hand side of (8) is a constant everywhere in I C1 > 1. Then, by considering its behavior at we conclude that this constant is equal to zero.

> 1 and Thus, for an extremal function, we have proved equation (2) in we have proved equation (2') for the boundary of the domain B. and sum over Let us now set C= Sv, in (2), multiply both sides by v' = 1, , n. We then obtain for the extremal function n

/

1p=-

log (I

1

(The sum on the right is, of course, real.) This proves inequality (1) with the supplementary conclusion regarding the equality sign. Furthermore, if we replace yv for v = 1, . , n in (1) with yve`B, where 0 is an arbitrary real number, we obtain the following sharp inequality, n

T-a" log F (Cy) - F (C,,)

e

r, .1=1

C'-Cy%

/

-

n

L+ 7y7y' V. V-1

log

CyC y

which, by the virtue of the arbitrariness of 0, implies inequality (3). But this also shows that the set of values of the quantity

§3. EXTREMA AND MAJORIZATIONS

WF

log F (C°) -

Gr V.

v = 1, 2, for given yv and 1, completely covers the circle

.

133

F (CY ")

Cy

r'=l

, n, and for F(C) ranging over the entire class

I wl=-

7Y7,'

log(1

-

V,

(9)

CyC,,)

n

WF,p= I ii,'1ogF(pC)

pCy(PC,,)

(10)

r'ml

with the same yv and Cv and with given p > 1. From what we have proved, as FM ranges over the entire class F,, the values W = WF.P cover the entire circle

7J,'log( 1

IW V, %'ml

-

1

P'CYCY,

)

But from the construction of the quantity (10), we can easily see that WF(PC)'

WFtC),p

where F(pC)/p E F, if F(O E I . This shows that the set of values of WF as

F(C) ranges over the entire class F, also covers the circle (11). As p ranges from 1 to oo, the corresponding set of circles (11) covers the entire disk

WI-

I n

7y7Y'log(1-C

l

V. V'= l

which, therefore, is the range of values for WF. This completes the proof of the theorem.

For n = 1, inequality (3) takes the form

IlogF'(C)I-1og(1-IClla),

(12)

and inequality (1) takes the form, with yl = e'a

ffI(e9=alogF(C))c-log (1-I

- 2 cac 2 .

I2

(13)

1

Here, equality holds in (13) with C= S1 only when Lcf. (2')] the function F(O maps the domain I CI > 1 onto the w-plane with cut along an arc of the curve defined by the equation ffi (e' log (w - F (Ci)) = const.

(14)

134

IV. EXTREMAL QUESTIONS

But equation (14) defines a logarithmic spiral with asymptotic point at w = F(CI) such that rays issuing from that point intersect the spiral at an angle 77/2 - a. In

the special cases a= 0 and a= rr/2, this spiral degenerates respectively into a circle with center at the point F(C1) and a ray issuing from F(C1). For a= 0 and a= rr/2, we obtain from inequality (13) the distortion theoreml)

for the class E:

I-

F'(C)I<

(15)

1

I C12

From what has been said above, the least upper bound is attained for = S1 only by a function w = F (C) that maps the domain 141 > 1 onto the w-plane with cut along an arc of the circle with center at the point w = F(CI) and the greatest lower bound is attained only by a function F(C) that maps the domain I CI > 1 onto the plane with cut along a segment of the ray issuing from the point F(J1). Here, F(C1) obviously can be chosen arbitrarily. We give yet another application of Theorem 1.

Let us apply inequality (1) in the case n = 2, first for y j = e and Y2 - ei a and then for yj = ie`a and y2 = -ie", where a is a real number. Then let us add the results. We obtain 1 (e2" log

F (C3 r

F(tl)log

8

=e(C.))c - 2 log [(1

(CQ) I<- 2 1og[(1

IC

r i, )

l(( - I I$1 ) !

Cl i,)

Cl

9

1

I C, 19)]

(16)

.

Equality can hold in (16) only if the functions F(C) E E for which equations of the type (2) hold simultaneously in the domain I CI > 1. If we add and subtract these equations, multiplied respectively by 1 and i, we obtain the following two equations:

F(C)-F(C,)_(1_ C-C.

1

-e-2ia 1,2,

CC'

or 1

F(C)=F(CY)+(C-CY)(1-

Cc,)

-e - 2ia ,

v=1, 2.

In particular, from (16) with a= rr/2 and IC1I = IC2I = P, where p> 1, we again obtain the sharp inequality 1) Loewner [19191J.

§3 EXTREMA AND MAJORIZATIONS

135

F(C2)

(17)

C, - Cs

Here from what was said above, equality holds only for the function QiO F(C) +const,

=C-

argC1-}-argC2.

i and C1 4 C2, inequality (16) will not be sharp because, Furthermore, for a if it were, the extremal function F(C) would, as follows from the corresponding equations of the type (2'), map the circle I CI = 1 onto an arc of a logarithmic spiral with asymptotic point simultaneously at F(CI) and at F(C2) or ( for a= 0) onto an arc of the circle with center simultaneously at F(C1) and F(C2), which is impossible. With regard to the complete answer of the question that we have touched on here, we present the following theorem:

Theorem 2. Let a denote a real number and let C1 and t 2 denote distinct complex numbers outside the disk ICI > 1. Then, among the functions F(O E Y, the function maximizing the quantity legta iog

F

(18)

(L8) 1

(C )- F

maps the domain ICI > 1 onto the entire w-plane with cut along the curve defined by the equation

w _ F1±Fs+ F1-F2 (e(c+ioe to 2

2

+

C

(c+il)e-`a

(19)

where t is a real parameter defining points of the curve. For a= 0, this will be a cut along the ellipse with foci at F1 and F2. For a= n/2, it will be a cut along the hyperbola with foci at F 1 and F 2, Proof. Following the same reasoning as in the proof of Theorem 1, let us show that extremal functions exist, that each of them maps the domain ICI > 1 onto a domain B without exterior points, and that each of them in I I > 1 satisfies a differential equation of the form efia

Cg F e

s) (F- F=

Q C(F-F1)

where Q(C) is a rational function that is real and nonnegative on I4I=1. When we solve this equation, we get the following dependence of w = F(C) on C:

IV. EXTREMAL QUESTIONS

136

which also holds on the circle IC I = 1. From this we conclude, just as above, that all boundary points of the domain B satisfy the equation

eialog( w-Fl- w-F9)=c+2lt, where c = const and t is real. When we solve this equation for w, we obtain a parametric representation of the boundary of the form (19). This completes the proof of the theorem. We mention without proof 1) that the quantity W- =log

F (Cl) -F (Cs) Cl - CA

has, for arbitrary C1 and C2, a disk for its range of values. But neither the radius nor the center of that disk can be expressed in terms of S1 and S2 by the use of elementary functions.

Theorem 3.2) For functions F(C) = C + a1/C+... E E, we have the sharp inequality F(C) -}- 1

-

I

-2

C A-

-<:: 2-2

(20)

in I Cl > 1. Here we use the notation

K (k) _ c

dx

(1- x$) (1- A'x')'

-

(20')

E (k) = SJ

1 - k9x9 1 - x9

dz.

0

Inequality (20) defines the range of values for the quantity F(0/C for arbitrary given C in the domain ICI > 1. Consequently, this range is a disk. Proof Let us look at the problem of maximizing the quantity IF = Rt (C)) (e91a F

1) Goluzin [1947 I. 2) Theorems 3 and 4 were proved by the author (see Goluzin [1943a1 ).

§3. EXTREMA AND MAJORIZATIONS

for given

F(C) = C+

137

>.h.nd for F(C) ranging over all the functions

where E aI/C+

Obviously, this maximum coincides with the maximum

of the quantity

rF = R (ells (F (Ci)

- an)) among all the functions F(0 E E. ( Here a0 is the constant term in the La urent expansion of F(C).) But I`" has equal values for nonconstant functions F(0 EE. Therefore, among the extremal functions of this last problem there are functions without zeros in Icl > I. Let F(C) denote one of them. By applying the varied function (25) of §3 of Chapter III, let us show that the image B of the domain ICI > 1 under the mapping w = F (C) does not have exterior points. If we then apply the varied function (26) of the same section, we get the following differential equation for F (C) in ICI > 1: C9F'4 Fe9i F1= Q (C),

where Q (0 is a rational function such that Q (C) > 0 on ICI = 1. If we now solve this equation, we conclude that the extremal function F(C) satisfies on J C I = 1 the equation Rt(e'a

F(C)-F(C1))=c, c=const.

(21)

Resting on this circumstance, we can construct an explicit expression for the function F(C). We need only consider the case 6 > I. Then, we have, on 16 = I. I4=(C-C1) (I-C,C)

IC-C,

C

If we divide (21) by I4- S1 I, we get

t(e ,a

F

C (F (C)

(C-C,)(1-C1C))-c

(22)

(C-C1)(1-C,C)

Since the function V (C-C1)(I-C1C)

is regular in CI > 1 and, obviously, continuous on I CI > 1 and since its real part on ICI = 1 is given by formula (22), this function itself in ICI > 1 is determined by the Schwarz formula (the integral is taken in the counterclockwise direction): e

C(F(C)-F(C1))

to

V

(C - C1) (1-C1C)

c

ttIt'I S

C'

C'+ Cdt'

(C'-C,)(1-C,C')C'-C C'

-{ ict

(23)

IV. EXTREMAL QUESTIONS

138

From this we obtain F(C). To determine the real constants c and c1 and also F(C) - ao, we use the normalization of F(O. Specifically, if we understand the radical appearing in (23) as referring to that branch that approaches 1 as V o and equate the constant terms and coefficients of 1/C in the expansions of the two sides of (23) about = we have e1'

t

dC'

c 2xl`,

C,

=1

(24)

k1

C7(C1-C')(1-C1C')

is (

2i

+'ao(Ct))=-

1C1-{-C1

IC'I=t

C'(C,

(25)

CC)(1-C1C')

Let us make cuts along the real axis of the C'-plane from 0 to 1/C1 and from C1 to 0o and if we take for YC' (C1_00 - C1C') that branch that is positive on the right-hand side of the cut (0, 1/r;'1). Since the integrands in (24) and (25) are regular in the cut planes, we can deform the paths of integration and, in particular, we can constrict them to the path extending from 1/C1 to 0 along the upper side of the cut and then from 0 to along the

lower side of the cut (0, 1/C1). As a result, we obtain ti

ele I

dC'

c

n

C1

JC'

(1 -C1C')

+ 1C

,

(26)

I

era

2i

2c '

II

C'dC'

C'+ + -ao- F (C1))=n

(27)

C'(C1-C')

where the radicals are understood to mean their positive values. By equating the. real (and imaginary), parts of the two sides of (26), we determine c and c I. We then obtain from (27) an expression for F (C) - ap : C1

`

,F (CI)-Qo=CI-

1

C1+2(e

9ia - 1)

C'dC'

O

VC'(C1-C')(1-C1C')

1

dC'

3/ C' (C1- C') (1- C1C')

§3. EXTREMA AND MAJORIZATIONS

139

If we make the substitution -xs SjS' in these two integrals and use the notation (20 ') for complete elliptic integrals, we obtain r1 +2C,(e-sta-1)r 1-E1C/

F(C,)-ao=Ci+C

T,

K(

For the maximum that we are seeking, we have IF=OR

(e2'%

E (1

(Ci

.+2Ct

CI)+2CI

1

K

_E

1

K \C,

From From this it follows that for all the functions F(C) _

+ aI/C+ E E and arbi-

trary given a, we have es;a

(F)+C-

-E(C1l

1

1

2C

(28)

1

K1C/

for t> 1. By virtue of the arbitrariness of a in (28), we obtain inequality (20). That this inequality also holds for complex C in the domain ICI > 1 follows from examination of the function The remaining conclusion regarding the range of values of F(C)/C is proved just as in Theorem 1. This completes the proof of the theorem.

Theorem 4. For functions F(0 E I in CF"(C)

P(C) +

41C 1--2 IC19-1

41CIO

E(ICI

> 1, we have the sharp inequality

c

4101'

/1-E\ICI/ Ir

K`ICII

I

(29)

ICI'-1 K11Cl/ Inequality (29) determines the range of values of the function V"(0/P(C) for arbitrary given C in ICI > 1. Consequently, this range is a disk .

Proof. When the function F(C) belongs to the class F ) 0 r FI ( C ) F 1±.ICLC I') + 2 (1 - I Ct Is) \

so does the function

=c -I

+...

for every CI in ICI > 1 and conversely. If we apply inequality (20) to FI(C) at the point C= ZI , we obtain inequality (29) for C_ CI . Conversely, from (29) we can obtain (20). To do this, note first that, if f(z) E S, then, by applying (29) to the function F(O ^ 1/f(11C), we obtain in I z I < 1 the inequality

IV. EXTREMAL QUESTIONS

140

z f " (z)

f' (Z)

2

' 4 - 2 1 z 1' + z f ' (z) + 4 E (I z 1) 2 f(Z) l -IzI' 1-IZ13K(IZU

4Iz12(1-K(Izl) Second, if f(z) = z + c2z2 +

(30)

E S, then, by applying (30) to the function f1fc+z

g(c)=

...Es

f-(Z)(1-Iz12)

(where I z I < 1) at the point t; = - z, we obtain in I z I < 1 the inequality

If(z)+c2z+1-Izj2-2

Finally, if F(C) = C+ a1/C+

(LO

(31)

E I and F(0 1 c in C1 > 1, then, by apply-

ing (31) to the function f (z) = 1/[F (1/z) - c] E S, we obtain inequality (20) in CI > 1. From this we get the conclusions on the range of values. This completes the proof of the theorem. §4. Application of the method of variations to other extremal problems

In all the cases considered in §3, the method, of variations led to complete solution of the extremal problems. This method can also be applied to various other extremal problems. However, final solution of the problem frequently depends on the determination of a certain number of unknown constants. We shall now stop to look at certain problems of this type. l) Let us now apply the method of variations to the investigation of the extremum of the average diameter of the boundary of the image of the domain > 1 under functions of the class . .2)

Theorem 1. Let F(0 denote a function belonging to I. Let n denote an integer > 2. Let PF denote the least upper bound of the product

Ia, - avI V$YI

(1)

over all systems of n points av, v = 1, -, n, located on the boundary of the image of the domain ICI > 1 under the function w = F(O. (The product (1) is 1) See Goluzin [1946f, 1947, 1951d]. 2) Goluzin [1947, 1949c].

§4. METHOD OF VARIATIONS

141

, n.) Let Pdenote taken overall pairs of distinct-numbers v, v' taken from 1, the least upper bound of the quantity PF over all functions F E Y'. Then P is attained by some function w = F (C) that maps the domain I ' > I onto the entire w-plane with analytic cuts and that satisfies the, differential equation [F1p12

C2

in

1

1

1

(F(C) - a") (F (C) - a",)

-n1 (n -

)

(2)

1.

Proof. In addition to the quantities Pp and P mentioned in the statement of the theorem, we define the following: for a given function F(C) E L_ we denote by PP ' F the maximum of the product [1 I F (C P) - F (C",. P)

(3)

V:f V,

over all possible systems of points C"P1v = 1,

, n, on the circle

( = P,

where p > 1, and we denote by PP the maximum of PPP. F over all F (C) E 1.

By virtue of the normality of the class 1, the maximum PP obviously is attained. Let FP(C) denote one of the extremal functions. We denote by C"'P, for v = 1, , n, the points on the circle p for which the quantity (3), set up for the function FP(C), attains its maximum PP. For the system of these points Cv, P, let us consider the extremal problem of Theorem 1 of §3. One of the extremal functions in that problem is obviously FP(C). It follows from the proof of Theorem 1 (equation (5) with y,,,, v = 1 for v v' and y v, = 0 for v = v') that the function FP(S) must satisfy in the domain > 1 the differential equation [CF (C)]9

I

V

(F(C) - F", P) (F(C) - F"',P) :Pl v

=n(n- I)+2

F' C'V.FF"P 1

11

(F",P - Fv'. P) (C - Cv,P) CV, PF".

P

v,P-Fv, p) (ft,, , - I)

(4)

where F,"P= FP(CvI P).

The right-hand member of this equation can be simplified further. Note that, if we replace one of the values Cv, P with Cv, Peta in the expression (3) written for FP(C) and for the values Cv, P, where v = 1, , n, we obtain a function of

IV. EXTREMAL QUESTIONS

142

a defined in the interval - rr < a< rr that attains its maximum at a= 0. Therefore, the derivative of the logarithm of this function vanishes at a= 0 and this leads to the condition Cv,P.F,

P

Fv,P-Fv'.P V1=1

_ o, v,

where the summation is over all v' = 1, , n excluding = v. Since equation holds for all v = 1, , n, we conclude that the numbers

this

n

v Cv, PFv, F, P V'=1 ,

v=1,

n,

-P

are all real. Therefore, the last term on the right side of equation (4) is equal to Cv, PFV. P v

v'

(Fv,p-Fv, P)(CCr,p-1)

Then, it can easily be combined with the preceding term. As a result, equation (4) takes the form 1

[CF' (C)]'

VI (F (C) - F, p) (F (C) - Fv',P)

=n(n-1)-2C 1 }, (FY vt-v'

Fv,

1)

) (C - Cv, p)'pFr'P (C-C,

p -1)

(5)

for w = FP(C) in ICI > 1.

Now, by virtue of the normality of the class 1, the condensation principle can be applied to the functions FP(C), treated as functions of the parameter p > 1. Thus, we can choose a sequence of functions that has the following properties as p --> 1:

1) FP(C) converge in > 1 to a function F(C) E 1, , n; 2) v = 1, , n, converge to some Cv, v= 1, Sv, P, , n, converge to av (points on the boundary of the image of 3) Fv, P, v = 1, the domain > 1, under the mapping w = F(J)); 4) 1) IF, (Cv,dI converge to certain finite numbers. This last is possible, because, for example, from the second of inequalities (15) of §3 for F (C) e I in I C1 > 1 we have (1 - 1/I CI ) I F'( C) I < 1. We note that the points av, for v = 1, , n, are all distinct because, otherwise, for the chosen p ---, 1

143

§4. METHOD OF VARIATIONS

we would have PP --+ 0, which is impossible since it is easy to find a positive lower bound for PP that does not depend on p (calculating, for example, PP, F, where F (C) _ e . If, for the sample made, we take the limit in equation (5), written for F =

FP(C), as p -- 1, we obtain the following differential equation for F(C) in

ICI>1: [CF'V'2

\

1 Y#Y

(F(C)-a,,)

n

=n(n- 1)+C

I

C, 9,

cY = const.

(6)

Y=I

It is now easy to show that cv = 0 for v = 1,

- -

, n: We write equation (6) in inte-

gral form: 1

S

(F- aY) (FY#Y

a,,) dF

nn-1

A

I

cY

]

Y=1

the left-hand member of If some cv 10, then, in a neighborhood of the point which (7) is bounded whereas the right-hand member behaves like log is impossible. Thus equation (7) reduces to the form (1). We need also to show that the function FW that we have obtained is the extremal function in the problem mentioned in the formulation of the theorem. To do this, we note that, for given c > 0 and fixed pI sufficiently close to 1 but greater than 1, we have P > PP1 F - c. Furthermore, for a sequence p' -- 1, p' > 1, such that FP, (J) --+ F(C) we have, from some point on PP1,F - c > PPI,FPS 2c > PP, F , - 2c = PP, - 2c. Consequently, P > PP, - 2E. On the other hand, for these same values of p ' , we have PP, _> P. This shows that PP, --+ P; that is, the function F(J) exhibited above is extremal. > 1 onto the entire It remains to show that w = F(C) maps the domain w-plane with analytic cuts. The analyticity of the arcs constituting the boundary of the image follows from equation (2). We denote this image by B. That B does not have any exterior points follows from the fact, that, if it did, we could, without changing PF, add on to B a two-dimensional segment and then for the function w = F. (c), with IF'(-) I < 1, mapping I CI > 1 onto the modified domain B' we would have I F. (..) i < 1, by Lindelof's principle. But then the function w =

144

IV. EXTREMAL QUESTIONS

F' (01F*(-) E I would give PF a higher value than the function F (c), in contradiction to the extremality of F(C). This completes the proof of the theorem. The problem considered in Theorem l with n = 2 constitutes the problem of the extremum of the ordinary diameter of the boundary of the image of the domain 16 1 > 1 under the mapping w = F (C) E Y.. Its solution is well known: P = 4, and the extremal function is the function F (C) = £+ ??IC + const, where l1l = 1. This same result can easily be obtained by solving equation (2) with n = 2. The complete solution of this problem for arbitrary n is laborious. We note that the function F(0 = C(1 + rl/C")21n E E, satisfies equation (2) for arbitrary n > 2 also, for this function av = 22/net n"il" for v = 1, - , n. However, for the case n = 3, we can carry the solution through to the end.

Theorem 2. Let F(C) denote a function belonging to E. Then, for arbitrary points a1, a2, a3 on the boundary of the image of ICl > 1 under the mapping

w=F(C),

l (at-as)(at-aa)(as-aa)

12,

(8)

with equality holding for the function 2

F()=C(I+to

IT11=1.

(9)

Proof. Let us consider the extremum of the left-hand member of inequality (8) over the points a1, a2, and a3 mentioned in the theorem and all functions F(C) E E. We arrive at the problem considered in Theorem 1 with n = 3. One of the functions providing this extremum must, in accordance with Theorem 1, satisfy the differential equation (2), which for n = 3 can be written in the form F

(C)-a1+a,+a8 3

`2 [F (Q]9 (F(C)-a1)(F(C)-a9)

(F(C)-a.)=1.

(10)

Here, a 1, a2 , and a3 are boundary points providing the corresponding maximum for the left-hand member of (8). The fraction on the left side of (10) cannot be reduced further. To see this,

suppose, for example, that (a1 + a2 + a3)/3 = a1, that is, that (a2 + a3)/2 = a1. Then, since lag - a3 l < 4, we would have j al - a2 l < 2, la, - a3 1 < 2, and, consequently, P < 16, whereas for the function (9), which maps the disk lCl > 1 onto the w-plane with three rectilinear cuts extending from the point w = 0 to the points 22/3i71/3e2rrv`/3 for v = 1, 2, 3, the maximum of the right-hand mem-

145

§4. METHOD OF VARIATIONS

her of (8) is equal to 12/3, which exceeds 16. But if the fraction cannot be reduced, the function F(C) must have expansions of the form

F(C) = av + c9( - C + ...

,

c2/:O,

for v = 1, 2, 3, on ICI = 1 corresponding to the points a 1, a2, a3. These expansions show that av , for v = 1, 2, 3, are the endpoints on the boundary of the image of the domain I z I > 1 under the mapping w = F(C) and the point (aI + a2 + a3)/3 is a multiple boundary point out of which issue the boundary arcs extending to av for v = 1, 2, 3. Since the extremality of the function F (C) is not lost when we add a constant to it, let us normalize the function in such a way that a1 + a2 + a3 = 0. If this extremal function has an expansion in neighborhoods of the points

,

then, by substituting it into equation (10) (with a1 + a2 + a3 = 0), we find that

ay= a0=0' a t= - - alas+alas+a9aa 4

a,a,aa. 6

(11)

Suppose now that x3 + px + a is a polynomial with zeros x= a 1, a2, a3. Then,

p = alai + alas + alas = - 4alr

q= - ala2as = - 6;. With regard to the left-hand member of (8), its square is the absolute value of the discriminant of the equation x3 + px + q = 0, which is equal to 4p3 + 27g2, so that we have (at - as)(at - as)(aa - aa) I9 44,al + 30.4al I < 4(431 a, Is + 311 w 19)

(12)

Now, let us find a bound for

Q=41 IaiI1+311as11

(13)

over all the functions + in the class Y.0 to which our exC+ aI tremal function belongs. Let us denote by C' the image of the circle ICI = p, for p > 1, under the mapping w = F(C). From inequality (4) of §7, we obtain, by constricting the curve C to the point w = 0,

SR14): 0, x > o.

C'

146

1V. EXTREMAL QUESTIONS

From this we get, just as in §7, for A> 0 and p > 1, 2x

I F(pei) Iado -- 0. dP

Let us apply this result with A = 3. We set F(C)3

C3/3 (I

c

c

3 3 cg=2a1, c8=2aQ..

For p > 1, 00

9 -2 (n- 2

3

r

\\

Zp.

n-2

If we take only the first two terms of the series, we get, by letting 3,

Iel11+31c911

that is, Ia112+31a2 12

Therefore, 31 a2 12 < 4/3 - I a 1 12. Consequently, (681

QC481a',18+33.4-31 1al11=38.4-4a11 4-Ia11 Since I al 1 < 1, so that the parenthetical expression on the right is positive, we

have Q < 33 4. Therefore, in accordance with inequality (12), we have (8) with equality holding for the function (9) with suitably chosen a1, a2, and a3 in (8). This completes the proof of the theorem. This theorem leads to a number of corollaries.- We present one of them that

deals with the class S. Corollary. Let f (z) denote a function belonging to S. Then, for an arbitrary triple of points c1, c2, c3 that lie on three rays issuing from w=0 forming equal angles with each other and that belong to the boundary of the image of the disk I z I < 1 under the function w = f (z), we have

cl c9 c3 >

(14)

4

with equality holding for the function

f(z)=

z

(1 +, za) a/a

III=1.

(15)

§4. METHOD OF VARIATIONS

147

Proof. We may assume that all the cv are finite. Consider the function

F(0 = 1/f(1/t) E E0. Then, the points av = 1/cv, v = 1, 2, 3, lie on the boundary of the image of the domain I CI > 1 under the function w = F (C) and, in fact, on three rays issuing from w = 0 at equal angles with each other. From Theorem 2, we have inequality (8). On the other hand, for v v' (v, v' = 1, 2, 3, we have,

a,-a,

-I a,, I9-2cos23 Ia,a,, a,12+ Ia,,11+Ia,a,'I =31a,a,,I

and, consequeently, (a1 - a9) (a1 - a3) (a3 - a3) I i 3 V3 I a1aSa3

This last inequality, together with (8), yields I a 1 a2 a3 l < 4, which is equivalent to inequality (14). We verify directly that equality holds for the function (15) and the points cv = 2-2/3q-1/3e217vi/3, v = 1, 2, 3. Without going into detail, we mention the following. By using Schiffer's studies [19381, one can show first that equation (10) holds in the domain ICI > 1 for an arbitrary extremal function of the type mentioned in Theorem 2 and then that equality holds in (8) and (14) only for the functions (9) and (15) respectively.') Let us look at the following problem on the extrema of the coefficients in

the class S. Theorem 3. Among all the functions w = f(z) = X,°=1cvz' E S, the maximum of the quantity (16)

Tyco

for arbitrary given n > 2 and arbitrary given complex numbers y,, for v = 1, , n, where y,, X 0, is attained only by functions that map the disk I z I < I onto the entire w-plane with analytic cuts that extend to w = oe without finite multiple points and that form an angle less than n/4 with the corresponding radial direction at each point.

Proof. The existence of extremal functions is obvious. Suppose that f (z) is one of these. The function f (z) also maximizes the quantity

JI = t (e`,

T,cy)

r=1

(17)

1)(Theorems 1 apd 2 and the more general corollary of Theorem 2 have recently been proved by Reich and Schiffer [19641 with the additional assertion that the extremal functions are unique.)

IV. EXTREMAL QUESTIONS

148

where 0 is a real number, out of all functions of the class S. Let us suppose that the image B of the disk I z I < 1 under the mapping w = f (z) has an exterior point w0 . Let us form the varied function f (z) in accordance with formula (21) of §3 of Chapter III with m = 1. Let us denote by (00}n the coefficient of z' in the expansion of an arbitrary function 96(z) that is regular at z = 0 in a neighborhood of the point z = 0, taking AA I = h. We have "-1

f (E)

{f» (E)}" = c" + h

f ()

Tao

J + O (hs) = c" + h I

if (Ek

k=l

and, consequently, n

"-I

i,» = Jj + FR llh"=11 k=1

tyk

(hs).

J

°

+ 0 (hs),

°

`+

7"1 {f(E)k+,}

+1 }

Since the expression in parentheses is a polynomial in 1/w0 and the coefficient of 1/w0"'1 in that expression is equal to y / 0, this polynomial is, for the value of w0 that we are considering and wo arbitrarily close to w0, .also nonzero for the corresponding function f. (z) with small I h I and suitable arg h, J1- > Jf, which is impossible. Thus B has no exterior points. Let us now consider the varied function (24) of §3 of Chapter III with m= 1 and arbitrary z0 in the disk I z < 1. We/ have

tf# ()}" - "

+hf

fI(E)'

l f (E) - f (Zo)

-h

(

" -hf

'C

fo ` Zo

)

"

fo 1' (zof' (E)1 _f _ h-( fo ) ' Z 'A 1 zafo J 1 E- Z. f"

1,

zOE'f (E) 1

1- 2°E J" + Q (I h Is)

and consequently, n

Jf.=Jl+51 (eteh I17"(E)

-

n

eiBh Zofo (fo 1' f

n

((ef(zo)1"eieh Goofo )"Y1T"e" n

" Jzo`f' (E)1 + e'oh. (foo 1'"=f" J oE'f' (E)l) 1 Ezo J" 3 EJ + 0(1 hIs) Z' ( 1

r=1

1

The condition of extremality of the function f(z) requires that the inequality Jir. < Jf hold, that is, that

§4. METHOD OF VARIATIONS

(ereh

7V

re fo 1' e h (zto

f (E) IRE) -f(zoJ.,

l

I1TVc,

149

fo 1' - e reh (z too /

a

z Ef

71{E - zo J,

+eie)i,(

T,l For small

IhI

1oE'

(E)

and arbitrary arg h, this can be true only if

ere

zT f (E)'

If

f(E)-f(zo)

}

v

n

l9

(r

f

(ere,(Yc. + et0T

/ 11

IzE

I zo)1Y-ereTV { i Eef'(E) }Y) =

0.

(18)

Since (18) holds for arbitrary zo in I z I < 1, we can by replacing z0 with z in equation (18), obtain a differential equation for f(z): n

ere (zf' (z)1'

J

f(z)

f (E)'

f(E)-f(z)

1

n (ere,(YcY

.=t

11

+

et0

JzEf (E)1 1

E-Z V

f zE'f (E) 11

ere

7°

l 1-3E lYl

(19)

It follows from this differential equation that the boundary of the domain B consists of a finite number of analytic arcs and that equation (19) holds for w = f(z) on the circle I z I = 1. Let us show that the right-hand side of formula (19) is negative for z I. We note that a function w = O(z) that satisfies the conditions t/i(0) = 0 and i/i' (0) > 0 and that maps the disk z I < 1 univalently onto the disk w < 1 with radial cut of length r issuing from the point w = e-`q6, can be determined from the equation I

werI-{- were - (1

zero -}-

h)

)+2h, sere

1

9

h - 4 T__8

For h, we obtain

yr(z)=z-hz i+zer ; +0(h9), 4+' (0)=1-h, h>0. w

Let us use this function to define a function in the class S: fk (z) =f ,+(((z)) =f (z) + hf (z) - hzf' (z)1 +

zzeil

erlp + 0 (h9).

IV. EXTREMAL QUESTIONS

150

For this function we have n

n

Jf.=Jf+hUt(e;e T+c+-e

1- Eet

0(h9). y

Consequently, by virtue of -the extremality of f (z), we have on I z I = 1 n

t (eie Y

n

+ efe

,

+-1

17+

v=1

{Ef' (E)

i

l

±E

< 0.

(20)

The difference between the right-hand side of (19) and the left-hand side of (20) is equal to n n ye+

ete

l.

+=1

But this quantity is equal to zero, as is shown by comparing its value for the function f (z) with its values for the functions e-"-"f (e `az) E S with real a. This proves that the right-hand side of (19) is negative for z I = 1. Let us now show that, if our extremal function w = f (z) does not map the disk I z I < 1 onto the w-plane with rectilinear cuts that would include the point w = 0 if extended in the opposite direction, then the real part of the quantity n

Q zm=

e;eI

J

f(E)°

+-I which appears in equation (19), is positive at all finite boundary points of the domain B. Let w0 denote an arbitrary finite point on the boundary of B. The function

-f( )(Z

0

f*(z)=wo1Of (z)=f (z)

'w0

(21)

belongs to the class S and for this function we have Jr. = Jf - ;7t (Q (two))

.

This shows that 92 (Q (w0 )) > 0. If 92 (Q (w0)) were equal to zero, the function (21) would also be extremal both for the quantity (17) and for the quantity (16).

This requires that Q(w0) be equal to zero. Let us write for the function f* (z) the equation corresponding to (19): n f*(E)*(E)

*(z)

R* (Z),

(22)

4. METHOD OF VARIATIONS

151

This R* (z) must again be negative on I z I = 1. But

_

f* (E)2

and, consequently,

f

P02

f (e), f (E) - wo

(z)

n

e"I

fs (E)'

} f* (E)-f* (z) = Q (f (z)) - Q (wo) = Q (f(z)) v=1 Therefore, if we replace f* (z) in (22) with its expression in terms of f (z), we J

7y

1

obtain

1'(Zf

\ww

1

If we divide (23) by (19), then, for wo

P

(23)

)QQf)=R*(z).

z

= 1, we have

=,t:- 0,

1-Y=

wo -f (z)1

117

-

where T is a real number. This last equation shows that all boundary points lie on the straight line passing through w = 0. Since this was excluded, the contradiction that we have obtained shows that 5 (Q (w )) > 0 at all finite points on the boundary of B. But then we also have l2 (1/Q (w)) > 0 and we see from equation (19) that Zf (Z)1°1< o

(24)

f (_) J J

at all points on I z 1 at which f (z) is finite. The additional conclusions on the boundary of the domain B follow from equation (19) and inequality (24). We can see this as follows. If the boundary B consists only of radial rays, everything is proved. On the other hand, if this is not the case, then Q (w) 0 at all finite boundary points w and it follows from equation (19) that none of these points can be a multiple or an angular point of the boundary. Consequently, the boundary of the domain B consists of analytic arcs that extend out to ° . Furthermore, it follows from inequality (24) that, for a suitable value of the argument, we have arg

zf (z) + < 71 f (z)

2

4

on the entire boundary of the domain. But the quantity on the left-hand side of this inequality measures the angle between the tangent to the boundary of B at the point w = f(z) and the ray passing through the points w = 0 and w = f(z). Consequently, this angle must be less than it/4. This completes the proof of the

IV. EXTREMAL QUESTIONS

152

theorem.

Let us now apply the method of variations to the investigation of a particular extremal problem for the class S. of functions

f

(z)=clz+c%Z%-{-...,

that are regular and univalent in the disk I z I < 1 and that do not assume m given , a,,, , where m > 1. Let us look at the problem of maximizing finite values a 1, for given n > I with respect to functions of the class Sa (see Goluzin [1946f]). Icn I -

Theorem 4. A function w = f(z) = c1 z + c2 z2 + C Sa for which I c I, where n is a given integer > 1, attains its maximum maps the disk I z I < 1 onto the entire w-plane with cuts along a finite number of analytic arcs and satisfies in the disk I z I < 1 the differential equation _

n.-1

R((1(z)) z2f''(z)9=n+ om l/ \Cv Z -v L..I (n - V) Cn V( )) l V - I

CCnv zv),

(25)

where m

m-3

k=1

k=-n-1

P(w)= II(w-ak), R(w)_ E bkwk, the bk being constants. In equation (25), the right-hand member is real and nonnegative on the entire circle I z I = 1, and the boundary cuts referred to are with a suitable parametric representation w = w(t) (t real), the integral curves of the differential equation R(n')

P (w)

(26)

dt

Proof. The existence of extremal functions is obvious. Specifically, the quantity I c I has a finite upper bound because, if f(z) = I' k=1 ck zk E Sa, we k have the well-known inequality I c I _< en for the function f (z)/c 1 =1 c k z and consequently, I c I < en I c 1 I . If we apply the familiar inequality If (z)/c 11 > I Z I/(1 + I Z I ) 2 to a point z on the circle I z I = r at which If(z)I < I a11, we obtain c 1 I< 4 I a 1 a s r -+ 1. ConseI C 1 I< ((1 + r) 2/r) I a I. It follows from this that quently, I

Icnl<4enIa1 By virtue of the normality of the class Sa , this upper bound is attained. If f (z) is an extremal function, the function f (e C1L z) for arbitrary real a, which has the same image B, will also be an extremal function. Therefore, there

§4. METHOD OF VARIATIONS

153

exist extremal functions such that c > 0 and they exhaust all extremal functions in the sense of the characteristic of the images B. Suppose that f(Z) = 17_1 Ck Zk is one of the extremal functions such that cn > 0. Let B denote the corresponding image. By using the varied function obtained from equation (27) of §3 of Chapter III, when we replace w with f(z) and take w I = w2 = .. - = wn = 0, we can show, following the same reasoning as in the proof of Theorem 3, that the domain B does not have exterior points. Nov\We can easily obtain a differential equation for the function f(z). Since the functioa\(29) of §3 of Chapter III is a function in the class S. for arbitrary distinct z 1, j , z,, in the disk I z I < 1 and sufficiently small A, we have, setting AA = /Y in that formula m

t{g E

-CC.+h {(E) k-l

1J

! k=1

(f (E) - ak)

m

hI Akf(Zk) k=

(f (E}-f(n

'(E lZkf' (Zk)9 E - "Z n

m

+

+ 0 Q h 9) 1 -2kE n where (0())In denotes the coefficient of z in the expansion of the function O(z) about z =0. But since $2((f"
k= Zkf'(Zkr

m

(f (E) - ak) f (E) km I

h

II (f(C)-f(Zk)) n

k-1

Akf (zk) IEf' (E) I

E - Zk n k-1 kf (k)9 m

+W J Akf(Zk) k=l zkf'(Zk)9 1- ZkCJJn If we replace the next to the last term with its conjugate (which leaves the real part unchanged), we obtain m

R (f (E) - ak)

lh

f (E)

k=1

IT (A0 -f(Zk)) n k=1

m

Akf(zk) zkf' (Zk)s

k=1

Ef' (E)

(it - Zk In

-{1'f 2kEIn/J/+0(1 h19)=0. Because of the arbitrariness of arg h, this leads to the condition

IV. EXTREMAL QUESTIONS

154 m

J1 (f (E) k=1

f

- ak)

m

II (f (E) f(zk)) ln

m

-

IE'f' (E)

Akf (zk) (fEf' (E)

ll-

kf' (zk)' 1

l = 0.

(27)

kEIn)

Since the first term on the left does not vanish identically for all values of z1, , zm, we can choose_ z 1, , zm such that this term is nonzero. Now, fixing considering it a variable z. Then, the A k take let us change z 1 , ,zm, z 2, the form Al __ P (f (z)) Q (f (z))

Ak

- f (z)

(k = 2 , ... , m) ,

Bkf (zk)

where m

in

P (w) = lI (w - ak), Q (w) _ fl (w -f (xk)), k-2

k=1

and the Bk are constants. Therefore, condition (27) can now be written in the form

m

k=1

f(E)

(f (E) - ak)

_ P (f (z))

f (z) Q (f (z)) zf' (z)9

'"

R MO -f (zk))

k=1

Ef' (E)

l (a - 2

n in

Il1f, (01")

- 3E

+1 f(z) -f (zk) k=2 Ck

0, ,

or, in view of the fact that m

(f (E) - ak) J (E)

in

°1

=

-1

H (f (E) -f (z))) Jn k=_R

k=1

we have JEE ' ( In

-

in

Bkf (z)k,

Ch

h-2f (z) -f (zk) m-2

Q (f (z)) k Dkf

(Z) k'

JiYfE In

z [zrtl + i so that we finally obtain

+ ncn + (n - 1) cn_1Z -}- ... + ctx"-t] ,

155

§4. METHOD OF VARIATIONS

R-

P(f(z))

-z$f'(z)'= Rl, z-

(28)

where

m-3

R (w) _ }J

bkwk.

k=-n-1

Thus, the function f(z) satisfies the differential equation (28) in the disk z I < 1. Admittedly, this differential equation contains the unknown constants ck aril bk, We conclude, therefore, from the analytic theory of differential equations,that f(z) is regular not only in the disk I z I < 1 but also on the circle I Z I = 1 except at a finite number of points. Consequently, the boundary of the domain B consists of a finite number of analytic arcs. The differential equation for the boundary of the domain B can be put in a simpler form. To do this, let us consider a function w = 1r(z) where 4i(O) = 0 and 0'(0) > 0, that maps the disk I z I < 1 univalently onto the disk I w I < 1 with a small radial cut issuing from the point w = e-'8. For this function, we have the expression in I z I < 1 zeie W=t (z) = z - hz 1+ 1-zero `} 0 (w).

The function

1-I zeie f,

h ()z + O(g)

obviously belongs to the class S,,. By setting f 1 (z) = k=1 c (1) zk , we have n-1

C(,) = cn - h (ncn +2 E k-1

kcke(n-l)re) + 0 (h9). /

From the extremality of the function f (z) we have Y (c;11) < c,, , which,

because of the smallness of h, leads to the condition n-1

ncn + 251(E kcker(n -k) e 1 ! k=1

0.

If we now set e`8 = z, we obtain an inequality expressing the nonnegativeness of the right-hand member of equation (28). Thus, on the circle z 1, R (f (z)) z1 f, (z)9 ':- 0.

P (f (z))

It follows that, with a suitable parametric representation w = w(t), where t is

IV. EXTREMAL QUESTIONS

156

real, of the boundary of the domain B, we have for this boundary the differential equation R (tai

where

P(w)

I

dw\ ,+ 1 _0'

L

dt m-3

m

P (w) = ft (w - ak), R (w) _ 2] k=1

bk wk.

n-1

Consider the function f (z) with c = I c I e-`0. Let us apply the above reasoning to f (e `B/6 Z). If we replace z with a-`0" z In the formulas obtained, we obtain the assertion of the theorem (I c I is included in the coefficients bk ). Let us stop for the important special case of n = 1; that is, let us find a bound for I c 11 = I f ' (0) 1. This quantity represents the radius of the disk onto which the domain B containing the point z = 0 is mapped univalently by the function z = 95(w) normalized by the conditions 95(0) = 0 and 0'(0) = 1. It is called the con formal radius of the domain B (cf. Chapter II, §2). The results obtained above concerning the bounds for the coefficients with n = 1 thus apply to the extremum of the conformal radius for domains B not containing oe and a given finite system of points a1, , am. The corresponding extremal function belonging to the class S. satisfies the differential equation obtained from (25) with n = 1: Rt ( f ' (z))

P (f' (z))

_

(Z )

f (z) 1

1,

where R 1(w) = lk =0 bk wk. If Z = O(w) is the inverse function, we obtain the differential equation p' (w) -_ ,f (w)

R, (w) w'P (w)

in B. Consequently,

incp(w)= S

bo+b,w+...+bm_rwm-r I/-W1 - (w - ar) (w - a9) ... (w - am) dw --f- b,n.

(29)

This yields an explicit analytical expression for the function O(w). The constants B0, B 1, , Bm remain undetermined. A formula of the form (29) was proved earlier by Lavrent ev (1934] by starting

with the principles of Lindelof and Montel.l) These principles lead to the

1) Grotzsch L1930] studied the extremum of the conformal radius for the class S. on the basis of other considerations. He also pointed out the analytic form of the extremal function in a certain particular case.

§4. METHOD OF VARIATIONS

157

uniqueness of the extremal function f (z) normalized by the condition f ' (0) > 0.

We note that, in the case of the extremum of I c I for n > 1, the function f (z) normalized by the condition c > 0 does not always have a unique extremum. Let us show this with an example. Suppose that ak = ae21rki/m (k = 1, 2, , n), where n > 1 and n i 1 (mod m). If f (z) is one of the normalized extremal functions, then the function f1(z) = f(ze2ni/m) e-err`/m belongs to the class S. and is -also a normalized extremal function for I cn I. This function is distinct from f(z)

bec"µse f'(0) =

e-2n(n-1)i/m f'(0)

f'(0).

Ijt is also easy to show that Lindelof's principle does not apply to Icn I for n > 1, that is, that I c i does not necessarily increase when the domain B is I

broadened.

In conclusion, let us examine the following problem on disjoint domains: Let a, a2 , , an, where n > 2, denote n distinct finite points in the w-plane. Functions w = fk (z), k = 1, 2, , n, that are regular in the disk I z I < 1 map that disk univalently onto disjoint domains Bk including the points ak , for k = 1, 2, , n, respectively in such a way that fk (0) = ak for k = 1, 2, - , n, The question arises: can we speak of the maximum of the product rt

J_HIfk(0) I k=l

n? over all possible functions fk (z), for k = In the case n = 2, the answer to this question was given by Lavrent rev L19341 (on the basis of his variational method). It consisted in the sharp inequality

If (p)f;(0)Icla,-al I9,

in which equality holds only for the functions f 1(z) and f 2,(z) that map the disk I z I < 1 onto the halfplanes about the common boundary of which the points a1 and a2 are symmetric. The existence of extremal functions is easily proved. Specifically, since the function gk (z) = (fk (z) - ak )/ fk (0), belongs to S for k = 1, 2, , a, and does not assume in the disk z I < 1 the values (ak 1 - ak )/fk (0), where k 1 L k, it follows from Koebe's theorem that (ak 1 - a0 f; (0) > %; that is, I fk (0)I 4I ak 1 - akI . This proves that the quantity I has a bound independent of the form of the functions f k (z) for k = 1, 2, - , n. From this and the normality of the family of functions fk(z) that we are considering, we conclude that the least upper bound J0 of the quantity J is attained. I

IV. EXTREMAL QUESTIONS

158

, n, denote a system of extremal funcNow, let w = fk (z), for k = 1, 2, tions and let Bk denote the corresponding extremal domains. On the basis of Lindelof's principle ( p. 30 ), the domains Bk are such that the union of their closures Unk=1Bk coincides with the entire w-plane. Let us prove certain variational formulas that we shall need. Let n denote an integer greater than 1. For fixed points ak, where k = 1, 2, .. , n, consider the function n77

11 (w-a,)

w*=w+h v=1

kon,

1

n

II (w-w") V=1

v#ko

where A is a complex number and the wv, for v = 1, 2, points. This function can be represented in the form

.

, n, v / k0 are distinct

n

w*=w+h w+ xo+ I

au-1

v#ko

From this we easily see that, for sufficiently small the function w` is univalent on the entire plane from which sufficiently small neighborhoods of all points wv, for v = 1, 2, , n, v / k0 are removed. Therefore, if wv = fv(z,,), for v = 1, 2, , n, where z v I .< 1, then the function I

n

f k (z) =fk (z) + h

1I (fk (Z) - av) -I

1

n"

kon,

(30)

11 Vk (Z) -A (Z,))

V=1

v#ko

is, for k = k0, regular and univalent in the disk I z I < 1. For k / k0 , the function (30) has a simple pole z = zk in the disk z I < 1. Consequently, the theorem in §3 of Chapter III can be applied to it. As a result, we obtain the function I

H (fk (z) - av) f *k (z) =fk (z) 4-h

n

II

V=1 v= ko

`

(fk (Z)

-

-fv (zv))

§4. METHOD OF VARIATIONS

159

n

17

1 (A (4) - av)

zfk (z)

V=1

1

n

Zk V; (zk))9

z

v=1 v#k, ko

(fk (zk) -fv (Z9))

n

17

+h

1 (fk (zk) - av)

i°fk (z) xk

1

n

(fk (zk))2

1

v=1 -,#k, ko

Zkz

-I T 0 (I h I9)

(31)

(fk (zk) -fv (zv))

which is regular and univalent in the disk I z I < 1 for small values of I A I: also , n, thus defined now map the f k (0) = ak . The functions w = f k(z), k = 1, 2, disk I z I < 1 for small I A I onto disjoint domains and f k(0) = ak ,. k = 1, 2, ; n. Equations (30) and (31) are the desired variational formulas. From the functions f k (z) for k = 1, 2, , n; k ko , let us now construct the varied functions (31) with z 0 for v = 1, 2, . , n, v / k0 , which, together with the function fk0 (z) constructed in accordance with formula (3), are also admissible functions of the extremal problem that we are considering. Now, we easily obtain the equation n v=1

(ak - av)

ko

J*_ ft Ifk (0)I=Jo 1+9Z I h k=1

y111 (ak -fv (zv)) of k n 7 11 (fk (zk) - av)

n

+21 Zk (f(zk))9

+ O (I h Is)

n

k-1

.

(fk (zk) - fv (zv))

k*ko

v=l

v$k, ko

By virtue of the arbitrariness of arg A and the extremality of the functions fk (z) for k =.I, 2, , n, this last relation can hold only if the expression in the large square brackets is equal to zero, that is only if n

k=1

11 (ak vl v$k

lI _ '

9 -/'ko

av)

(ak ---A (Z9))

n

11 (fk (zk) - av)

1

I

T

zk (fk (zk))9

k_ 1

v=1 11 v=1

k - k °

v#k. ko

= 0.

(fk (Zk) -A (zv))

(32)

160

IV. EXTREMAL QUESTIONS

Consequently, this condition must be satisfied for an arbitrary choice of the points z k , for k= 1, 2, - . , n in the disk I z I < 1 and for arbitrary k0 = 1, 2, - - - , n. It follows from (32) that the functions fk (z) are piecewise-analytic on the circle I z I = 1. To see this, let us fix all the points zk, k = 1, 2, - -, n; k k0, with the exception of one of them zk 1 = z (where 1 < k 1 < n). We see that condition (32) becomes a differential equation for fk 1(z): z9 (f"k, (z))9 = R (.fk, (z)),

where R (w) is a rational function. Our assertion then follows from the analytic theory of differential equations. Let us look at some particular cases. The case n = 2 can be carried through to its completion without difficulty and we shall not stop to do this. Now let us look at the case n = 3. In this case, equation (32) with 'k0 = 1 yields (a,- a9) (a, - aa) _L (a3 - a,) (a, - a,) (a, - f, (z8)) (a, f a (za)) (a9 -f2 (23)) (a8 -f8 (23))

1=1( a

(aa-a,) (as -f9 (za)) (aa -fa (za))

+

v2 z 9)- a,)

f, (z9) -fa (za)

3 s (f2 (z9))'

l

II (fa (za) - a,)

fa (za) - f9 (z9)

zi v a (za))a

- ()

If we multiply through by the difference f2 (z2) - f3 (z3) and decompose the first three fractions into partial fractions, we arrive at the form A (fs(za)) = A(fa (z3)),

(33)

where A is an operator defined by

A (f (z)) - (a, - a9) (a, - as) _L (a, - a,) (aa - as) a, -f (z) a9 -f (z)

a

(a8-al) (aa-a9) _l as -f (z)

T

II (f (z) -a,,)

y-1

za V, (z))9

Analogously, for k0 = 2, we obtain from equation (32) A (f, (z1)) = A (fs (za))

(34)

Since equations (33) and (34) hold for arbitrary Z1 , Z2, and z3 in the disk Iz I < 1, it follows that in the disk I z I < 1 we have A (fk (z)) = C = const for

161

§4. METHOD OF VARIATIONS

for k = 1, 2, 3; that is, the functions w = fk (z) for k = 1, 2, 3 are each a solution of the differential equation a

II (ak-w)

k=1

z°w'9

_ (a,-a,) (a,-a,) + (a,-a) (a,-as) a, _W a, -w + (a, - a,) (a, - a+ C, a, - w

(35)

C=const. 1

Of cdurse, the constant C may depend on al, a2 and a3 However, we are not able 'to determine its value. Let us consider separately the two cases C / 0 and C = 0. In the first case, equation (35) reduces to the form

YP(w)w'

Z,

3

7J=±'1,

(36)

11 (ak - w) k=1

where p (w) is a polynomial of exactly third degree, and its zeros are distinct from a1, a2, and a3. For extremal domains Bk, the union UJ=1 Bk coincides with the entire w-plane. Therefore, the. point w = oe lies on the boundary of certain of these domains Bk , let us say, for example, on the boundary of B 1 . Let us 1 corresponding to it under the mapping w = denote the point on the circle I z f 1(z) by z = z 1. Then, in a neighborhood of z = z 1, we have

f; (A+ B-f-...) _'z fi

A t- 0.

3

f,2 Consequently, 1(A'-}-+...)=_71ogz_

fa

f,

z-z,+

z,

z,

c,(z-z1)", v-2

This means that, in a neighborhood of z = z 00

ray =fi (z) =

L b, (z -- z1) v--2

b_9 t- 0.

(37)

If the point w = oe were also a boundary point of some other domain Bk (k 1), the same would be true about the corresponding function fk (z). But this is impos-

sible since the domains Bk, k = 1, 2, 3, are disjoint. Thus, in the present case, the point w = oo lies on the boundary of only one of the domains Bk. The other

IV. EXTREMAL QUESTIONS

162

two domains must be bounded.

Let us turn now to the case C = 0. Let us suppose first that the points ak are vertices of an equilateral triangle, that is, that ak = aek, k = 1, 2, 3, where f = eZnl13. Then, equation (35) (with C = 0) reduces to the form a'ww'

3

z'

w8-a°

(36)

When we solve it, we obtain explicit expressions for the fk (z):

fk(Z)=askl -+ckz)3, ckz \\\

1

ck=const,

k=1, 2, 3.

For the functions fk(z) to be regular in the disk I z I < I and for the union U jc=1 Bk to coincide with the entire w-plane, we must have ck = e 1k for k = 1, 2, 3, so that we finally obtain 2

fk (Z) =

I +e i¢kz

l

\ I - etakz

k=1,2,3.

3

(39)

From this we conclude that each of the domains Bk is the domain between two rays with an angle of 2n/3 between them with vertex at w = 0, and with bisector passing through the point ak . Calculation then shows that J = (64/27)1 a 13 or, what amounts to the same thing, Jo = 816

3

I (at - a.) (al - as) (a2 - as) I

(40)

Thus, for an arbitrary system of functions fk (z), for k = 1, 2, 3, that satisfy the condition fk (0) = ak , that are regular in the disk I z I < 1, and that map this disk univalently onto disjoint domains, we obtain the sharp inequality 3

fk (0) c

gl

Y 8 I (a1- a3) (at -- as) (as - as) I

(41)

Now, let ak, k = 1, 2, 3, be arbitrary. Equation (35) with C = 0 has the property that if we subject the unknown function w and the constants ak to the same fractional-linear transformation, that is, if we set W

cw

+d,

ak

cak + d ,

k =1, 2, 3,

(42)

ti then w as a function of z satisfies the same,differential equation a

k=l

(a1-a9) (al-as) I_ (a,- a1) (a3-a8) zew'4

a1-w

,

aq-w

(a]-al) (a8-a9) ay - tat

163

§4. METHOD OF VARIATIONS

This is easy to verify directly by substituting the expressions (42) for w and the ak into equation (35). Using this property, we choose the transformation (42) in such a way that the points ak, k = 1, 2, 3, are mapped into the points ak = atk, where e = e 2n`13. Then, the equation for w reduces to the form (38). Consequently, ti ti its solutions are functions w = fk(z), with f k (0) = a k of the form (39) that map the disk I z I < 1 onto the three angles mentioned above. The functions w = fk (z), with Ak (0) = ak , that satisfy equation (35) with C = 0 are then of the form %

fk (z) =

k = 1, 2, 3,

afk (z) -F b

cfk(z)+d

(43)

and the domains Bk corresponding to them will possess the property (obvious but important for what follows) that the point w = o is a boundary point of at least two of the three domains Bk. Since we can verify directly that 3

3

fk (0) k=1

(a, - a.) (a, - al) (a. - a3).

-

H fk(0)

_

k=1

(a. - a2) (a, - a2) (a2 - a.)

Ip can be calculated in the present case from formula (40). Consequently, again in this case we have inequality (41). The case C = 0 does not always hold. For this case to hold, it is necessary and sufficient that the point w = -d/c corresponding to the point w = oo underti the transformation (42) lie on the boundary of one of the corresponding domains Bk , for k = 1, 2, 3, because it is only then that regularity in the disk I z I < I is ensured for all the functions fk (z) defined by formulas (43). However, inequality (41) always holds, that is, even when the ak are such that C 10. To show this, let us examine the question of the maximum of the quantity 3

I_

11 fk

(0)

k=1

(fl (0) -f2 (0)) (fi (0) -fa (0)) (f2 (0) -fa

(0))

over all possible systems of functions fk (z), for k = 1, 2, 3, that are regular and univalent in the disk I z I < 1 and that map this disk onto disjoint domains. Since I does not change when we replace the functions fk(z) with cj:k(z), where c is a constant, we can confine ourselves to functions fk(z) that satisfy the condition Ifk (0) I < 1. Now, just as above, let us show that I ff (0) I

4 Ifr (0) -fa (0) I,

Ifa (0) J C 4 Ifs (0) -f3 (0)

If9(0)1-4If3(0)-fr(0)I

IV. EXTREMAL QUESTIONS

164

From this we conclude that the family of functions fk (z) that we are considering is normal and that I < 83. Consequently, the least upper bound of the quantity I is finite and is attained by some system of functions f k (z), for k = 1, 2, 3. Let us set f k(0) = ak , for k = 1, 2, 3, and let us look at the question of the maximum of the quantity

J=

6 fk (0)

k=1

over all systems of functions fk (z), for k = 1, 2, 3, that satisfy the condition fk (0), that are regular in the disk I z < 1, and that map that disk univalently onto disjoint domains. According to what was said above, this maximum is attained by the functions f k(z). Therefore, on the basis of the exposition above, we conclude that the images Bk of the disk z I < 1 under the mappings w = f k(z) are such that the union Uk=lBk coincides with the entire w-plane. Therefore, there

exists a point a that lies on the boundary of at least two of the domains Bk. Let us define new functions k = 1, 2, 3.

These functions also map the disk I z < 1 onto disjoint domains, and the quantity I has the same value for them as for the functions f `(z) for k = 1, 2, 3 since I is invariant under a single fractional linear transformation performed on all the functions fk (z) for k = 1, 2, 3. This means that the functions f (z), for k = 1, 2, 3, are extremal under search for the maximum of the quantity 1. If we set rr f k (0) = 1/(ak - a) = ak , we conclude that the functions f k`(z) are extremal also for the maximum of the quantity

J'= Ik=1

I

with respect to all of the functions fk (z), k = 1, 2, 3, that satisfy the condition fk(0) = ak , that are regular in the disk Izj < 1, and that map this disk univalently onto disjoint domains. Since the point w = o lies on the boundary of at least two of the three domains onto which the disk I z I < 1 is mapped by w = f k* (z), it follows in accordance with the analysis made above that the functions w = f%`(z) are solutions of a differential equation of the type (35) with C = 0. But then, the maximum Jp of the quantity J is calculated, in accordance with (40), from the formula

4

64

81 Y g

1 (a; -- Q (al -

as) (a; - ay) I

§5. LIMITS OF CONVEXITY AND STARLIKENESS

165

This fact, in conjunction with the fact that the maximum of the quantity l is attained by the .functions f ` (z), leads us to the conclusion that this maximum is equal to 64/81 /. But then, for an arbitrary system of functions fk (z), k = 1, 2, 3, that are regular in the disk I z I < 1 and that map this disk univalently onto disjoint domains, we have I < 64/81 vl_, which is equivalent to inequality (41). We note, finally, that, since l is invariant under an arbitrary fractional-linear transformation of the functions fk (z) (but the same transformation for all three values of I), inequality (41) holds also when one of the functions fk (z) is meromorphic

in the disk IzI <1. We formulate the results of this section in two theorems:

Theorem 5. If points a1 , a2, and a3 are vertices of an equilateral triangle, then the maximum of the quantity j = ink-lfk (0)1 out of all possible systems of functions fk (z) for k = 1, 2, 3 that satisfy the condition fk (0) = ak, that are regular in the disk I z < 1, and that map that disk univalently onto disjoint domains is attained by the functions w = fk (z), k = 1, 2, 3, each of which maps the disk z I < 1 onto the domain between two rays at an angle 21r/3 with vertex at the center of the triangle and with bisector passing through the point ak . On the

other hand, if the points al, a2 and a3 are not all situated in such a position, then the same maximum is attained by the functions fk(z), for k = 1, 2, 3, one of which is bounded in the disk I z I < 1.

Theorem 6. For functions fk (z), k = 1, 2, 3, that map the disk I z I < 1 univalently onto disjoint domains, we have the sharp inequality

4

k=1

6 fk (0)

81

3

I (fr (0) -.ff (0)) (ft (0) -fa (0)) (f9 (0) -f9 (0)) I

(44)

Equality holds in (44) only for functions (39) and for the functions obtained from them by any fractional-linear transformation (the same transformation for k = 1, 2, 3).

§5. Limits of convexity and starlikeness The degree of distortion of a conformal mapping can be indicated in another manner, one that is easier to grasp, by means of what are called the limits of convexity, the limits of starlikeness, etc. These tell us which of the disks z I < r < 1 are mapped by an arbitrary function of the class S into domains of a

IV. EXTREMAL QUESTIONS

166

specified type, for, example, convex domains, starlike domains, etc.

Limits of convexity. We define the limit of convexity for the class S as the least upper bound of the radii r of the disks I z I < r that are mapped by every function of the class S onto a convex domain, that is, onto a domain such that the tangent at the boundary points of the domain always turns in the same direction as we move around the boundary. The limit of convexity proves to be nonzero. We denote it by R,. To determine its value, we note that if we take a point z = re`' on the circle z = r, then the tangent to I z I = r issuing from that point forms an angle 77/2 + q with the real axis. Consequently, the corresponding tangent to the image of the circle z I = r under a function w = [(z) E S forms an angle 77/2 + 0 + arg ['(z) with the real axis. It follows that I z I < r is mapped into a convex domain if and only if I

'(2 +,p+argf'(z))

0

on the circle I z I = r or, what amounts to the same thing, if and only if

zf " (z)1 + 1 : 0 f' (z) / on that circle. But inequality (6) of §4, Chapter II yields, when we replace the absolute value on the left with the real part taken with minus sign (

1-4IzI+Izl2 It follows- that

I

Z 11, f (Z) z I < r is indeed mapped onto a convex domain if r2 - 4r + 1 > 0,

that is, if r < 2 - v. Since this is true for an arbitrary function w = f(z) E S, we see that R, > 2 - g. On the other hand, for the function 1(z) = z/(1 - z)2 E S, we have

zf'(z)+l= 1-4z+z2

1 - z2 f (z) This means that the disks z I _< r, where r > 2 - J3 are not mapped by every function w = [(z) E S onto convex domains. Thereforel) I

R,-2-Y3=0.26....

(1)

Analogously, if we denote by R' the greatest lower bound of radii of circles = p, where p > 1, that are mapped by an arbitrary function w = F(O E I onto convex curves, we can show by the same reasoning, but using only Theorem 4 1) R. Nevanlinna [1919-1920].

§5. LIMITS OF CONVEXITY AND STARLIKENESS

167

of §3, that R, is a root of the equation

E( ) KI 1)

7

8P,=0. 1

T

lP

Calculations show that R' = 1.78

.

The limit of starlikeness. We define the limit of starlikeness for the class S as the least uupperr bound R., of radii r of disks I z I < r that are mapped by an arbitrary function td,= f(z) of the class S onto domains that are starlike about w = 0, that is, onto 4omains possessing the property that arg w always increases as a point w moves around the boundary of such a domain in the positive direction. To determine R5 , we note that the disk I z I < r is mapped onto a starlike domain if and only if 0argf(z) =s1 (zf'z(z)10

fO

where z = re `(P, everywhere on the circle thing,

Iz

= r, or, what amounts to the same

zf (z) C a

arg f (z)

By virtue of inequality (12) of §1, which gives a sharp inequality for arg (zf ' (z)/f(z))I for arbitrary z in I z I < 1, it now follows that R5 is a solution of the equation log [(1 + r)/(1 - r)] = rr/2; that is, 1) x

R5 = 1 - e

2

1+e

2

_ tanh 4 = 0.65... .

(2)

Limits of generalized starlikeness. It is also possible to show what analytic properties are possessed by mappings of disks of the form I z I < r by means of functions w = f(z) that belong to the class S when r exceeds R5 .2) To do this, we introduce the concept of generalized starlikeness. We shall say that a closed domain is of the type D (a) if every point of it can be connected with a point a by a broken line that lies entirely inside it and that consists of no more than n rectangular segments. Thus, a domain of the type D1(a) is a domain that is starlike about a point a. 1) Grunskyj1934]. 2) Goluzin 19351.

IV. EXTREMAL QUESTIONS

168

Now, let us denote by D. (where n = 1, 2, ) the least upper bound of all p such that an arbitrary function of the class S maps the disk I z I < p into a region of the type D (0). Obviously, from the definition itself, we have D1-Ds
Furthermore, Dn < 1 for arbitrary n because there exist functions of the class S that map the open disk z I < 1 onto spiral-shaped domains. Furthermore, DI = RS and, consequently, we have from (2) I

D1= tank 4 .

(3)

The exact values of D,, for n > 1 are unknown. We shall give the greatest possible lower bounds for L), . We use the following

/ d1 d.2jTll9)' IEIC1, do
Lemma. The closed disk

n

(4)

is mapped by an arbitrary function w = f(z) C S into a domain of the type Dn(f(.))

Proof. If f(z) belongs to the class S, so does the function

(C+El fl1+EC-f(E) g(C)=

=c+b9C9-}-...

f (E) (IC0

for arbitrary 6 in tI < 1.. Consequently, this function maps the closed disk ICI : do , where d < D,, , into a region of the type D. (0). Therefore, the function w = f(z) maps the closed disk

z=1+EC,

Itl=d,,,

that is, the disk I(z - l;)/(1 - Tz)I < d,, onto a domain of the type From (5), we get in succession

Iz-EI9
r $- 2 Z1-dJIE (1-da 1

+

E1'(1-d,i)

1-dnjEl-

(5)

§5. LIMITS OF CONVEXITY AND STARLIKENESS

d'-IEI' 1-d,EI'

169

dn(1-IEfs)s

IE1'(1-d:)'

(1-dnIEI')s,

(1-dnIEI2)'

from which (4) follows immediately.

We shall now use this lemma to prove the inequalities

\\ 1- Din < (1- Dn \ 1+Dsn_ \1+DnI ' 1

1- Dsn+m 1+Dsn+m

< f l - Dn

1I +DnI

I - Dm 1+Dm

(6)

for arbitraril m > 1 and n > 1.

1 and To derive the first inequality, we set 6= adn in (4), where I a n > 0. We then see that every function w = f (z) C S maps the closed disk adn z-1+ dnI C

do +d.-,

dn< Dn

(7)

1

onto a domain of the form Dn(f (adn)). The disk (7) includes the point z = 0 on its boundary. If we let a move around the circle I a I = 1, then the disks (7) cover completely the disk I z < 2dn/(1 + An arbitrary point in the image of this disk can be connected with the point f (0) = 0 by a broken line that lies entirely in it and that consists of no more than 2n rectilinear segments because there exists an a of modulus unity such that the image of the disk (7) includes both these points, with the result that it is possible to connect them with the point f(ade) by broken lines lying entirely in it and consisting of not more than n rectilinear segments. We conclude from this that 2dn/(1 + d 2 ) < D2n . Since

this is true for arbitrary d < Dn, by taking the limit as do -+ Dn, we see that 2Dn/(1 + Dn) < D2n and hence that 2Dn

1-Din 1+Dt 2Dn 1+DsnC1 + 1 +D'n

(1-D. l9 1+Dn1 '

which proves the first of inequalities (6). To prove the second of inequalities (6), we set do < r < 1, and n > 0. Since

= ar in (4), where

I al

1,

I E I (1-dx) }dn(1-IEII) _1 _ IEI do 1-djIEls - 1-dnjEj' i-dnIEI '

we conclude on the basis of geometrical considerations that the disk (4) is contained in the annulus IEI - dn C I z C IEI+dn 1 - dnIEI

I

- 1+d aIEI'

(8)

170

IV. EXTREMAL QUESTIONS

and is tangent to both its boundary circles. If we now let a move around the circle I a I = 1, the disks ( 4) completely cover this annulus. Let us now choose 6 in such a way that (161 - dn)/(1 - d,, 16I) = D. , where m > 0. This yields 161 = (Dm + dm)/(1 + d,Dm) < 1. Then, by the same reasoning used to derive the first of inequalities (6), we see that the image of the disk I z I < (161 + do )/(1 + do 161) is of the type Den+m (0) and consequently that (I 61 + dn)/(1 + do 19) < Den,.This yields

1-Dm_1+dnl1+Dn-I-dn1+

1-D2n+m

I+Dnn+mC1+dn1+IEI .

If we now take the limit as do -+ Dn and divide the inequality by the equality, we obtain the second of inequalities (6). From these inequalities with n = 1, we have

I-D

((1-D

1+D,

< 1+D,)

'

r1-D l+Dm-s al-D , m >2 . I+DC `1+D,) 1 -Din

From these we find by iteration that, for arbitrary in,

(I -D m 1 -D 1+DmC 11+D, )

(9)

Noting that D1 = tanh (n/4) we obtain from (9) for arbitrary m > 0

I- Dm<e T2 I+Dm that is, D. :-:- tanh MIC - >

am

1 - 2e

2

This is the desired lower bound for Dm. In particular, D2 > 0.91 and D3 > 0.98.

§6. Covering of segments and areas In §4 of Chapter II, we showed that the image of the disk I z I < I under an arbitrary function of the class S completely covers the disk I w I < 1/4. We now pose the question of determining the length of the longest of the segments with ends at w = 0 that lie on n arbitrary rays issuing from w = 0 at equal angles from each other and that belong entirely to the same image. The answer to this

§6. COVERING OF SEGMENTS AND AREAS

171

question is given by the theorem on p. 174. The proof of that theorem is based on two lemmas, which we shall now prove.

Lemma 1 (Grotzsch's principle)." Suppose that we have a set of n disjoint , n, that are contained in the simply connected domains Sk, for k = 1, 2, annulus r
sides of lengths ak, then

n

T ak 1ikC

kt

2n

log

(1)

k r

with equality holding only if the Sk are domains of the form r < I arg z < ck+1 completely filling the annulus.

z

I < R, ok <

Proof. Let us cut the annulus r < I z I < R along an arc belonging to the boundary of one of the strips Sk and let us map the simply connected domain obtained onto the plane

t = log z = u + iv (see Figure 8). This mapping maps the strips Sk into strips S1," that extend from the line u = log r to the line u = log R (see Figure 9). Let z = gk (t) denote the functions inverse to those mentioned in the lemma. These functions map the strips Sk onto the rectangles Rk : 0 < 6 < bk , 0 < 71 < ak (see Figure 10). Then the functions t =

log gk(t) = fk(') map Rk onto the strips Sk . The

Figure 8

images of the sides of length ak under this mapping lie on the straight lines u = log r and u = log R. We denote by Ik the area of the strip Sk . Then, we obviously have r

I Ikc2rtlogR.

k=t

1) Grorzsch[19281.

r

IV. EXTREMAL QUESTIONS

172

On the other hand,

jk- S

Ifk(C)I9dEd

i=

ak bk S ( S Ifk(C)I'dE) d-

ak

bk

ek(

IfkG)I

dE)9A,

the inequality being obtained on the basis of the Schwarz inequality. But the integrals bk S

I A (c) I dt

0

are the lengths of the images of the sides parallel to the sides of length bk ; that is , they are the lengths of certain curves extending from the line u = log r to the line u = logg R. Therefore f 0k f k (C)d> log (R/r ), and, consequently, Ik > (ak /bk) (log (R/r))2. If we substitute this into ILI Ik < 2 n log (R/r) and divide by (log (R/r))2, we obtain (1). For equality to hold in (1), it is obviously necessary that the strips Sk completely fill the annulus r < I z I < R and that the images (referred to above) of all the lines parallel to the sides of length bk be lines parallel to the real axis, that is, that the Sk be of the form r < I z I < R, ck < arg z < ck+1. A simple check shows that in this case equality does indeed hold in (1).

R,

lOgr

Figure 9

a

logR Figure 10

Lemma 2. Suppose that we have a set of n disjoint simply connected domains

Sk, for k = 1, . , n, contained in a rectangle with sides of lengths A and B (see Figure 11). Suppose that the boundary of each of these domains is a closed Jordan curve that contains a nondegenerate segment (that is, a segment not consisting of a single point ) of each side of length A. (The domains Sk can be regarded as strips extending from one side of length A to the other.) If we map

173

§6. COVERING OF SEGMENTS AND AREAS

these domains onto rectangles in the ic-plane with sides equal respectively to ak and bk in such a way that the line segments referred to are mapped onto sides of

length ak, then n

Y bk_

(2)

kal E.gv,ality holds only if the Sk are rectangles entirely filling the rectangle of sides

A and A, Proof `(analogous to the preceding). Let z = fk (C) denote the functions inverse to those mentioned in the lemma. These functions fk(C) map the Sk onto rectangles Rk : 0 < e < bk , 0 < q < ak . Let 1k denote the area of Sk . Then, on the one hand, we have

kl

Ik -,c-- AB

and, on the other, ak

d'i

Ik =SSA (C)19 A

S bk

Rk

0

bk

ak

d l = ak B9 1)0 fia (C) dE)9 d > B9 bk bk

Consequently, Y,k=1akB2/bk < AB, from which we obtain (2). That equality holds in (2) only when the Sk themselves are rectangles completely filling the rectangle

of sides A and B is proved just as in Lemma 1. The possibility of conformal mapping of the strips asserted in Lemmas 1 and 2 follows from the fact that every simply connected region having more than a single boundary point can be mapped onto some rectangle in such a way that four given accessible points on the boundary are mapped into 01

z

8

the vertices of the rectangle. We prove this last Figure 11 assertion as follows: Let us first map our domain onto the upper half plane and let us denote by a 1i a2 , a 3 , and a4 the points on the real axis into which the four given boundary points are mapped. (We can arrange for all the ak to be finite.) We then map the upper halfplane onto the rectangle by using the function

-

C

dC

(C - al) (C - a1) (C - a1) (C -- a4)

Then, the points a I , a2, a 3, and a4 are mapped into the vertices of the rectangle.

174

IV. EXTREMAL QUESTIONS

As a consequence of Lemma 2, we obtain the result that the ratio of the sides of the rectangle is determined uniquely by the mapped domain and the points that are mapped into the vertices.

Theorem. 1) Let f(z) denote a member of S. Consider n rays issuing from the point w = 0 forming equal angles each with the next. On each of these n rays consider that point of intersection of the ray with the boundary C of the image of the disk z I < 1 under the mapping w = f(z) which is closest to the point w = 0. I

Then, at least one of these n points is at a distance no less than l/4 5/7-

from

w=0. Proof. Let us suppose first that the boundary C is an analytic curve and that the theorem is not valid. Then there exists a system of rays Lk, k = 1, , n, that issue from w = 0 at equal angles such that the boundary points on them that are closest to w = 0 are all at a distance not exceeding d < n 1/4 from w = 0. The function

w=fo(z)=dY4 -Z2

=az+...,

a=di/4<1,

(1 +etazn)n

for suitable a maps the disk I z I < 1 onto the plane with cuts in the form of n halflines lying on the rays L k issuing from the points Ak = e-(a+2kn)i/n for k = 1, . , n that include the point w = 0 on their extensions. Since a < 1, it follows that for r sufficiently small, the image s of the circle I z I = r under the function w = f (z) contains in its interior the image so of a circle I Z I = Qr, where Q > 1, under the function w = fo W. Let us now partition the annulus Qr < I z I < 1 into a system of strips by drawing an arbitrary set of rays issuing from z = 0 in such a way that the following three conditions are satisfied: 1) The set includes all rays lk containing segments that the function w = fo (z) maps into segments OAk and all rays that are bisectors of the angles between adjacent lk (see Figure 12); 2) it includes all rays that the function w = fo(z) maps into curves tangent to C; 3) if this set includes a certain ray, it also includes the ray symmetric to it about the closest of the rays lk. The system of strips Sk constructed in this way can be partitioned into pairs of symmetric

v

1) Lavrent'ev and Sepelev[1930, 1937](Rengel [1933 ]).The proof presented here was given by the author [1939d] . The theorem can be generalized to the case of p-sheeted functions (see Goluzin [1948]).

§6. COVERING OF SEGMENTS AND AREAS

175

strips Sk, and Sk"" whose images Ik' and Ik " under the function w = fo (z) will have a common boundary segment on the rays Lk and will together constitute

Figure 12

a strip Ik ' U lk

which extends from so and returns to so . The entire w-plane exclusive of so and, consequently, the image of the annulus r < I z I < 1 under the function w = f (z) is partitioned into subsets contained in the strips Ik'U 1k" Also, by hypothesis, the number of such subsets of the image in question that lie in each strip 1k' U Ek" will not be less than two. Consider the pairs ak , and ak

of these subsets contained in the strips Ik'U Ik" and adjacent to s. Let

us denote by ak ' , bk ' and ak, bk " the sides of the rectangles into which they are mapped in accordance with Lemma 1 and let us denote by sk ' and sk their preimages under the mapping w = f(z) lying in r < z I < 1. Then, in accordance with Lemma 1, we have a'kr

bk,

ak

+ bkl

2a 1

(3)

1ogr Now, let us denote by Ak' and Bk' (Ak "" and Bk ") the lengths of the sides of the rectangles into which the strip Sk' (Sk ), and hence the strip 1A,

(ak 11) is mapped (as described in Lemma 1). Obviously, we may assume that AA, , = AV and B k ' = B k . Then, the strip I k ' U E k " is mapped into the rectangle Rk ' k with sides Ak ' and 2Bk ' such that corresponding to sides of length Ak ' are the parts of the boundary Ik ' U Yk lying on so. These last

IV. EXTREMAL QUESTIONS

176

mappings ak , and ak are mapped into strips contained in Rk' k and extending from one side of length 2Bk to the other. If we map these strips into rectangles as indicated in Lemma 2 and if we take for the continua on

the sides of length 2Bk, segments on £k' U IV' corresponding to the segments that, in turn, correspond under the mapping w = f (z) to the parts of the boundaries s k and s k N contained in Qr < I z I < 1, then, from Lemma 2 we have bk,

bk

2Bk,

ak,

ak

A h'

On the basis of this, when we twice apply the inequality (a + b)/2 > a_b, where a and b are both positive, we obtain

ak, + a,:,--,, 2

ak, a ,,

_

2

bk,

bk

V ak, akr, 4

2Ak,

Ak,

Ak

bk dkr + ak

Bk,

Bkr

B4r,

bk

But, in accordance with Lemma 1, we have Ak.)

(Ak,

Bk,+Bk"

_

2a

log Qr

Consequently, (g_L,

bk,

+

ak"

2S

log

1

Qr

When we compare this with (3), we obtain 2z'

log

1

2n 1

log r Qr But this contradicts the fact that Q > I. This proves the theorem for the case in which C is an analytic curve. In the general case, we need only apply this result to the function f(pz)/p C S, where p < 1, and then take the limit as p --' 1. By the same method we can

also prove the following theorem on the covering of the segments.

§6. COVERING OF SEGMENTS AND AREAS

177

Theorem 2.1) Let f(z) denote a member of S. Then, in the image of the disk z I < 1 under this function there are n rectilinear segments issuing from w = 0 at equal angles (2rr/n) to each other the sum of the lengths of which can be made arbitrarily close to n.

Proof. Let us look at the image B of the disk z I < R, where R < 1, under the function w = f(z). We define the star of the domain B as the largest domain cdgttained in B that is starlike about w = 0, that is, that has the property that a straight line segment containing an arbitrary point in it with w = 0 is contained in it.', Corresponding to the annulus r < (z I < R in the w-plane is a doubly connected domain the inner boundary s of which will, for small r, lie outside the disk J w < r (1 - Sr) = p, where Sr -- 0 as r --p 0. Let us partition the part of the star lying outside s by drawing a finite system of rays issuing from w = 0 1) the system contains all tangents to the that have the following properties: boundary C of the domain B: 2) it is mapped into itself by a rotation about w = 0 through an angle 21rk/n for k = 1, -, n; and 3) the fluctuations in I w I on the portion of C between two adjacent rays (though not on the rays themselves) is less than a given t > 0. Since C is an analytic curve, such a system does exist. There are nm such rays, where m is an integer. Let us denote them in order by ll , 12, , l nm . Here, lm k+ q is obtained from lq by a rotation through an angle 2rrk/n (see Figure 14). Let us denote by pk (k = 1 , 2, , nm) the greatest distance between w = 0 and points of the part gk of the star lying between lk and lk+1 and outside s (for k = 1, 2, , nm, gnm+i = gl) By means of the function J= log (w/p), we transform this system of domains gk into a system of domains hk lying respectively in rectangles with sides of length log (pk/p) and ak . I

-

lK,F

Figure 14 1) Goluzin [1936a1. A generalization of this theorem was obtained by Bermant [1938, 19441.

IV. EXTREMAL QUESTIONS

178

Here, Gk"` , ak = 2n where amk+ 1 = a'1. On the horizontal sides of each of these rectangles is a boundary segment (see Figure 15).

I

dK

Figure 15

When the domain hk is mapped onto a rectangle with sides ak and bk such that the boundary segments mentioned above are mapped into the sides of length bk, we have, in accordance with Lemma 2,

ak : ak

bk J log Pk P

Therefore,

nm

Yk

Aftj bk k-l

m n-1

n-I

m

amk+1

at

J l-lk=olog Pmk+1 1=1 k=o log P

Pmkl PT

But the system of domains gk is the image of a system of strips contained in the annulus r < I z I < R under the mapping w = ((z) and the rectangles mentioned above with sides ak and bk are the rectangles mentioned in Lemma 1. Consequently, from what was said above, we obtain on the basis of Lemma 1 M

2n

log R

al 1=1

n-1 i1 k=o log

Pmk+1

Therefore, since 1'1=1 a1 = 27r/n, we have n-1 2

log

n-1

in

min L R ' k-o log-P 1=I r Pmk+1

If we now use twice the familiar inequality

2n U1

mI n

k=olog

Pmk+l P

(4)

§6. COVERING OF SEGMENTS AND AREAS

n`

n n

I Ck k_1

n Ck, Ck

Y

179

0,

(5)

k

we obtain n-I nz

1

I k=0 log Pmk+i

n-1n

P

n-1

I1. log

k=o

log Pmk+l

k=0

P

Consequently, (4) yields n log

I

Pmk+i

P

n-I

R

log Pmk+i

max 1

P

k-0

or

(f n/ < max l

n-1 Pmk+l

k=0

Therefore, by taking the limit as r ---p 0, we obtain for some l1 in the closed interval 1 < l1 < m n-1

Rn c II Pmk+ii

(6)

k=0

or, again using (5),

n-1 nR c I Pmk+1ik=0

Let us now take an arbitrary system of n rays issuing from w = 0 at equal angles (2a/n) with each other and intersecting respectively gi 1 , 9m+11 , ' , 9(n-1)m 1 1 . The parts of these rays lying inside the star are of lengths X1

, A2 , .

. , an and hence are respectively greater than P11 - e, Pit - e,

' P(n-1)m+11

E.

,

Consequently, n

n-1

E 4 -- E Pmk+l, - nE ! n (R - E)

k=1

k=0

For R sufficiently close to 1 and a sufficiently small, this is arbitrarily close to n. This completes the proof of Theorem 2. As a special case (with n = 2) of this theorem, we obtain the result that there exists an interval of length arbitrarily close to 2 that passes through the point w = 0 and that lies entirely in the image of the disk I z I < 1 under the

IV. EXTREMAL QUESTIONS

180

mapping w = f (z) E S. Furthermore, when we rewrite inequality (6) in the n-1

log Pmk+1,

2n

form

log R

k=0

and take the limit as n

--1, 00,

we obtain the following inequality: 1) 1 2a

log R,

S

where w = pe`g' and the integral is over the boundary of the star of the image of the disk I z < R under the mapping w = f(z). On the other hand, if we raise (6) to the power 2/n and if we apply inequality (5) to the right side of the resulting inequality, then, when we take the limit as n = oe, we obtain the inequality 1

2

S p'd

where the integral is over the boundary of the same star. This integral is equal to the area of the star. By letting R approach 1, we obtain Theorem 3.2) if f (z) is a member of S, then the area of the star of the image of the disk I z I < 1 under the mapping w = f(z) is no less than rr. Starting with other considerations, we can obtain the stronger result:3)

Theorem 3 '. If f (z) is a member of S, then the product of the area of the star of the image of the disk I z I < 1 under the mapping w = f (z) and the area of the preimage of this star is not less than n2. Proof. Consider the star of the image of the disk I z I < R, where R < 1, under the mapping w = f(z) and let h denote its boundary and l the curve corresponding to it in I z < 1. This last curve is a closed Jordan curve. The function

F (x) =log f Z) = U-}- l V, F(0)=0i is regular in I z I < 1. Consequently, from Green's formula, we have, taking z = x + iy and using the Cauchy-Riemann conditions

UdV= S Uadx+Uydy a

1

_

SR

I

S,

1) Goluzin_[1917a . 2) Goluzin 3) Bermant[1938a].

/du av l az ay

OU dV`

Ty dz dz dy =

5 IF (x) I9 dx dy,

I

S,

s

§6. COVERING OF SEGMENTS AND AREAS

181

d0

that is, by setting z = re`d` and f (z) = pe`g', log

r

or

rd +logpdy.

log pd + log x

A

1

(7)

1

On',the other hand, if we integrate by parts, we obtain

0= 3 (1S

= S log P- d'f +

ZZ) dz)

log I dp -

S log

`P) d log r

rd(

log p dy -

log r

i.e.

logpdp=S logrd+. But this last integral is equal to 2a log R because v' = const on the radial segments lying on A and log R on the remaining arcs. On the basis of this, inequality (7) takes the form

log pdq+log rdp:-:- 4nlogR.

(8)

Let us now consider the doubly connected domain B. bounded by the curve

A and a small circle y: w = e. From Green' s formula, we have, just as above log p

dt

- J=2

du dv log p

5

WI

(9)

P

I

with a double integral taken over the domain BE . Now, if we denote by SR the area of the domain B bounded by the curve A (that is, the area of the star) and note that the integrand in. the last integral in (9) is no less in the part of the disk w < SR/7r not belonging to the domain B than in the part (of equal area) of the domain B lying outside that disk, we conclude that du dv

SS B`

du dv

I'-

SS <

i'

i

Pdp1

SS h
R

= 2n Remembering that

( log

- log e) .

IV. EXTREMAL QUESTIONS

182

we obtain from (9)

logpd

it

logs.

(10)

Analogous reasoning is applicable to a doubly connected domain bounded by a curve l and a small circle I z I = E . Thus, if we denote by s R the area of the

domain bounded by the curve 1, we have

logrdylog -R S

7C.

From (9)-(11), we obtain is log

that is,

-}- r log s l 4a log R, SR . SR > OR'.

As we let R approach 1, the area SR approaches the area S referred to in the star theorem and sR approaches the areas of the image of the star, and we obtain S. s 7t9, which completes the proof of the theorem. As a consequence of this result and of the property of minimization of the area under a mapping (see §2 of Chapter II), we obtain a supplement to Theorem 3: the area of the star mentioned in that theorem is equal to it only if f(z) = z. We note that all these theorems except Theorem 1 remain valid for an arbithat is regular in I z I < 1. 1) The only difference is trary function f(z) = z + that, in this case, instead of the univalent image of the disk I Z I < 1, we need to consider the Riemann surface onto which the disk I z I < 1 is mapped bijectively by the function w = f(z) and, consequently, the star located on that Riemann surface. All the above reasoning is valid word for word in this case.

§7. Lemmas on the mean modulus. Bounds for the coefficients

Let us now investigate the influence of the univalence of a mapping on the values of the coefficients of the mapping functions. To do this, we need to 1) Goluzin [1937a] and Bermant [1947].

§7. LEMMAS ON THE MEAN MODULUS

183

know bounds for certain moduli. These bounds are provided in the following lemmas.

Lemma 1.1) Let f(z) denote a function belonging to S. For A > 0 and 0 <

r<1, 2a

r

M PP)' dp,

If (reie) Ix A A

27C

(1)

where

M (p)= max I f (z) I

IzI=P

Proof. We have 2%

d

2a

2x

d1f

S If (re") Ix d9

d

ie) I'

d9 = l

If (reie) 1a

d log I f(rets) I

A

But, from the Cauchy-Riemann conditions,

-

d log I f (reie) t_ d arg f (re`B)

dlogr

dB

Therefore, when we set f (z) = Re`' , we obtain 2w

If(reie)I" d9

dr

r

RAdI,

(2)

6r

where Cr is the image of the circle I z I = r under the mapping w = f(z). Now, let C and C' denote two closed analytic Jordan curves in the w-plane, where

w = u + iv, such that C encircles the point w = 0 but is itself encircled by C. Denote the polar coordinates of points in the w-plane by R and 0. If we apply Green's formula to the doubly connected domain B bounded by these curves and then use the Cauchy-Riemann conditions, we obtain successively Rad(D 1

C

Rad(D = S RX du du+Ra" dv' c--c C S XRa rd log R d J LOU dv B

__

CBS

_ d log R O'I du

du l

duR )s + ('0V

Ra [ (d-'O

du dv

dil dv= l cBc Ra-s du dv;

aJ

`l JJ

1

that is, we obtain the formula

1) Prawitz [1927-19281. See also Goluzin [1949d].

IV. EXTREMAL QUESTIONS

184

SRAdI-S R'd(D =ASSR'-9dudv. C'

(3)

B

C

From this formula with A > 0 (it obviously holds also with A < 0) we get the inequality (4)

SR'd(Dc5R'd(D.

C'

C

In particular, if C' is the circle w I = R0 , then (4) yields R'd(D _2itRo.

(5)

C

If we apply inequality (5) to the case in which C is C, and R0= M(r) + e, where c > 0, and then let a approach 0, we obtain from (2) the inequality 29

2aa

I f (re") I' dO

dr

M rr)l

,

integration of which with respect to r from zero to r leads to inequality (1). This completes the proof of the lemma. Lemma 2.1) Let f(z) E S. For 0
I' 2a

"If (T era)

..

-

I f (re'°) I de'

1

2p(1±a)-2

r

e

21+

1-r

P

1//TM(1)(M(P))2

a

dP,

(6)

8

where

M(P)=maxIf(x)I,

p

IZI-a is a positive integer, and A + 1 - p-1 > 0. Proof. Let us consider the function

fv (Z) = P

w

Z5 _

1,

n-0

which, as we showed in §4 of Chapter II, belongs to the class S. From ( 7), we have

1) Goluzin[1949c].

(7)

185

§7. LEMMAS ON THE MEAN MODULUS

G (z) = z" f (zP)P - 1 f (z"), so that

f(z) = zP -1 f (z)1

P fa (zP).

Let us denote the left-hand side of ( 6) by I, replacing f' (z) in it with the above expression. We obtain 1 -1 2x rp 2a

I+

If (rel8)

1

-1

P I f; (z P) I d9,

from which we get 21+2- 2

2x

I< r

b)

2x

2

(8)

P d9.2 us rP If (zP) Is d9.

I f (Tete) I

Let us find a bound for each of the integrals under the radical. Lemma 1 provides a bound for the first of these: 2x

da+2-

If (re')

2a

12,+2-

2

P dO < (2a + 2 -

l

2A+2- 2

cr

p2) .)

M (P)

p

(9)

dp

0

With regard to the second integral, we have from equation (7) 2x

2a

00

1

Ifp(ZP)I'd9=

Y(nA+1)$IcR'I9r2a.

(10)

a-0

But, for 0<x<1 and m>0, xm(1

mm

x)----0 max (xm(1 -x))= 5M-+1)m+i 1

-m

1

1

m+1Ct'1

I

(1+ ml

from which we get

O<x0.

m< ex-(I- x)

If we now set m = (np + 1)/p and x = r, we obtain P

np+1 Equation (10) yields

1

nP+ l

er

P

(1-r)

(11)

M. EXTREMAL QUESTIONS

186

c2i

2

1

m

nP+I

o1cPc(P)nisr P -}- ) I a (1- r) P

A \ Iff(zP)IsrPdO j5

(12)

But the last summation is equal to (1/7r) times the area of the image of the disk z I < r1 /'2p under the mapping w = fp (z). Consequently, it does not exceed 2/p . I f(z)I2lp (max,z!=r1/2pIfP(z)I)2; that is;.it does not exceed max Thus, it follows from (12) that 2n

M(J '

P

Ifp(zI1P)Isr2/PdA< T U -

2n

)2/1,

(13)

r)

Using inequalities (9) and (13), we obtain inequality (6) from (8). This completes the proof of the lemma.

Lemma 3. Let f(z) denote a member of S such that, in the disk I z I < 1, MIZI

If(z)I where M and a are independent of r and where a> 1/2. Then, for 0 < r < 1,

(14)

C(1-Iz1Y,

2n

2a

I f'(re'e) I drr< (1M' -r)a

(15)

where M 1 depends only on M and a. Proof. Since, by (14),

M(P)=ImIxPIf(z)I<

O
(1MPP)"'

by applying Lemma 2 with A = 0, we obtain, for 0 < r < 1, 2m+1

1

2+c

2a ` I f Crete) I A<

r Mere

1

,

1- 2

2p

(1 -

r) p

(1

a P P -p \ 2--

dP

(16)

P)

Let us now choose p > 1 in such a way that (2 - 2/p) a > 1 (which is possible since a > %). We then have r

2 P

(1-P)(2

2

P

-

d p -< r

t-P

P)a

where M' depends only on a and p.

2-2

r

f 0

M'r

dp

0-P)(2

P)

a

P

(1-r)(2-P) -1

187

§7. LEMMAS ON THE MEAN MODULUS

By using this last inequality, we obtain from (16)

29

_

2a+1

2+c 1

f'(re )Id9<

r

MIM12 v

p

:e

1-r

r)fa

(1

1

p

1

2p

-(1-r)a

Here, M" depends only on a and p. Inequality (15) then follows for % < r < 1. But (15) obviously holds also for 0 < r < % when M1 is suitably chosen. This completes the proof of the lemma. Now, we have some theorems on bounds for the coefficients of functions

belonging to the class S. Theorem 1.1) For functions f(z)

=

z7=1 c z" E S, we have

cnI < en,

n = 2, 3, .. .

(17)

Proof. Beginning with the integral representation of the coefficients

Cn-2ai

n = 2, 3, ..., 0 < r < 1,

y(z) dz,

(18)

Izl=r

we have

2c

cC.

I c 2nrn

If (reie) I A

If we now apply Lemma 1 with A = 1, we obtain 1

CnIC rn

M (P) P

dP

But, from inequality (10) of §4 of Chapter II with 0 < p < 1, we have M(p) < p/(1 - p) 2. Consequently, r

IC.I<

dp

1 1

b

Here, r is an arbitrary number in the interval 0 < r < 1 . If we set r = 1 - 1/n, we have

CnIC(nnn

n-1

1)n-1=n (i+n

1

1)

<en.

Thus, we have obtained inequality (17). This completes the proof of the theorem.

1) Littlewood[1925].

IV. EXTREMAL QUESTIONS

188

Theorem 2.1) For an odd function f(z) = 1

1

I

Y-n=lc2n_1z2n-1 F S

Ic,,\<2432e2=3.39....

(19)

Proof. We have

ncn = Tai1

'

f (z) Zn

n=2, 3, ..., O
dz,

(=1=r

From this we get 2n

nI cnl

S I1'(fefe)I dA.

2Rrn

0

Consequently, from Lemma 2 with A = 0 and p = 4, we have n I C. I C

n I6e

kf (/' r )1/l

r

(20)

M (P) '/A dP'

r

p

But if the odd function f (z) is a member of S, then so is the function g(z) f (z 54)2. To see this, suppose that g (z 1) = g (z 2), where z 1 and z 2 are points in the disk I z I < 1. We can easily show then that z 1 = z2. If we now apply inequality (10) of §4 of Chapter II to g (z), we obtain If(z) I < I z I/(1 - I z I2) for I z I < 1. Consequently, for 0 < p < 1, we have

M(P)c

-p P

1

When we substitute this into (20), we obtain

fICnI 1r t

r

1

Crn ,

p9dP

(1-r)'/,

e

(1 -pS)e/. __

66

(1-r)°1t

e

6 e

dp

r' /s.

(1-p,)s/,

r

r`/4

rn--2(l -r) 6

l1/B

l 1

T/ rn- 2(1 -r) e

(1 l1/a 41

-

2" 31' a ins l

rn

2(1-r)

1) The existence of an absolute constant bounding `cn I was proved by Littlewood and Paley [19321. For the numerical value of the constant, see Levin [19351.

§7. LEMMAS ON THE MEAN MODULUS

189

Let us take r = (2n - 1)/(2n + 1). Then, remembering that 1

1

r

(

_

2(1-r)

l1 + 2n/

(2n + 1)'"+'

1

-T

l1- 1

T2n

l

nen(1+2n)log(1+2n) = ne" (1

1

1

2n)( Tn

1

1

l

c net + 3 (2n)9 + 6 (2n) 1

1

n_ a

2n)

n(1-2n)log(1_in)

I \ 2 (2n)y+ 3 (2n)s + ".)+(1 1

2

1

1

l l 2n)( 2n + 2 (2n)y + 3 1

+...)

1

1

1

l

1

(2n)s+"').

1

- (2 (2n)9 + 4 (2n)' + ...)

en,

we obtain

n{c,,K2`/, 3'/ 2 e"

n,

that is, we have obtained inequality (19). This completes the proof of the theorem. The functions 00

z

(1-z)'

nzn,

(21)

_ Ltl zsn-1,

(22)

n=1 00

1- z8

n=1 which belong to the class S, show that the bounds on the coefficients given by Theorems 1 and 2 with regard to order are sharp. However, in many extremal problems dealing with the functions that we have been considering, the functions (21) and (22) are extremal. In connection with this, mathematicians came to feel that these functions are also extremal in the problem of sharp bounds on the coefficients, that is, that the inequality Icn I < n provided the answer to the question in the case of the entire class S and that the inequality Icn I < 1 provided the answer in the case of the subclass of odd functions. The first inequality with arbitrary n for the entire class S has not at the present time been proved either true or false. This question constitutes a widely known problem today. Except for the inequality Ice I < 2, we have only succeeded in proving so far that C 3 I <- 3.1) In the general case, all that we have are various improvements in the 0

1) Loewner [1923] . (At the present time, it has been shown that I cn l< e (cf. §2,

3 of the supplement))

IV. EXTREMAL QUESTIONS

190

inequality, for example, the inequalities 1c,, I < % en and I c,,, I < (e/2)n + 2. With regard to odd functions in the class S, it has now been proved that there exist functions for which Ic2n_1I exceeds unity for given n> 3 (for Icy I see §9). Furthermore, there even exist functions of this kind with real coefficients. 2)

§8. The relative growth of coefficients of univalent functions Now, let us look at the question of the relative growth of the coefficients of univalent functions. As a preliminary, let us prove the Lemma. All the minors of the µxµ Vandermonde determinant, where it > 2, µ are nonzero. composed of the numbers q = e2 , where v Proof. We set 7jj1

... ,

We denote by Av rV, the cofactor of the element

V 'q v'

of this determinant. In

accordance with the properties of determinants,

(0 for k=v',

L 71k

l 0 for k

=t

v'.

k= 1,

.,µ

if we multiply these equations respectively by rlkv, for k = 1, - , µ, and add the resulting equations, we get, the properties of the roots of unity, p.Av, = O,1.r

Since A L 0, it follows that A v v the proof of the lemma.

0 for v, v ' = 1,

, µ . This completes

Theorem 1.33 Let f(z) denote a function belonging to S. Let it and v denote a pair of natural numbers with 1 < v < µ. Suppose that the inequality 1) Goluzin [1948a]. Bazilevic [1951] and, independently, Milin (see Lebedev and Milin [19511) have given a sharp bound for Icnl in the class S. The bound given in the text was found by Milin (see Lebedev and Milin [1949] ). (In 1964, Milin showed that Icn I < 1.243 n(see §2 of the supplement).) For an improvement in the bounds for the coefficients of odd functions, see Alenicyn [19511. (In 1951, Gong Sheng (Kung Sheng = Kung Sun) showed that, for the coefficients of odd functions, Icn I < 2.55 (see §2 of the supplement).) 2) Schaeffer and Spencer [ 1943] and also Goluzin [1949c].

3) Goluzin [1948b].

§8. THE RELATIVE GROWTH OF COEFFICIENTS

191

I < cna, where a> 0, holds for all n = µk + v for k = 0, 1, . Then, the inequality Icn I < c ' na, where c ' depends only on c, a and ft, holds for all . For a< 0, this assertion does not hold for all p and v. n = 2, 3, Ic n

Proof. We define co

1al i- "

n=0

Then, J (x)

V, (z).

(1)

v!=l

Now, from the conditions of the theorem, we have in the disk I z I < 1 Co

00

I Cµn +. I rµn + v C cr ' (µn + V)a rµn

I fw. I (z) I C

n=0

n=0 `°O

ooo

7, (n+ 1)arµn
where r =

zI.

fµ, y (z) I

cr

1-rµ

_ clr

1-r

On the other hand, if a> 0, when we apply inequality (11) of §7 with in

=

µn/2a and x = r, we obtain (2a)a

a -1

µn

r2(1-r)a Consequently, (z)

cr (µa

jr W

2saQa .

- r)a

2`

n=1

)

C

Mlr (1 - r)'+a

Thus, in the disk- I z I < 1 we have M,r

(z)IC

Ifµ

-r)'+Q

(2)

where M 1 depends only on a, c, and It. Let us show now that satisfaction of inequality (2) in the disk satisfaction of the inequality (z)

with the same M2. We set 77,

e

2 277V

,

v' = 1,

z I < 1 implies

(3)

Mgr `'1

I

p.

For fixed z in the

192

IV. EXTREMAL QUESTIONS

disk z I< 1, let us look at the values of I f(r7,, z)I, v' = 1, , 1L. Suppose that the greatest of these absolute values is obtained for v ' = vI . Then, from Theorem 7 of §2 ( with n = 2) applied to I f(rit,I z) I and I f(riv, z) 1, where v,i i I

we obtain the .t - 1 inequalities Mr If (,q." Z) I <

(4)

Vt,

V

1

where M depends only on- µ It follows from (1) that

f (i . z)=Tlv'fµ.I (z)- Ylyfµ.a(z)+ ... +mmfP. µ Therefore, inequalities (4) can be written in the form

17Wfµ. t (z) +a (z) + ...

+

(z)

Mrr

1

(5)

Now, let v, denote any of the numbers 1, , µ distinct from v. Consider the system of it - 1 linear equations with the P. - 1 unknowns y,,, , where v = 1, , iC and v' j v I :

.'#.1

7y ply _

I

for

0 for

S

V*,

S= 1, ...,

p.,

S

V.

(6)

According to the lemma, this system has solutions. Let us denote these solutions by y,, . For each v' , we multiply inequality (5) by I y , and add, replacing the sum of absolute values on the left with the absolute value of the sum. We then obtain the inequality I

fµ. V. (z) + afµ. (z) I C 1Ml

r

(7)

where Mil and a depend only on P. . Using (2), we obtain from (7) the inequality Ifµ. V. (z) I `

M (1

)l+a

(8)

where M 1 depends only on µ, a, and c. This is true for all v, = 1, , P. but v, # v. But in accordance with (2), such an inequality holds for v. = v. In view of this, we obtain from (1) inequality (3) with M2 depending only on µ, a, and c. But now that we have proved inequality (3), it follows immediately on the basis of Lemma 3 of §6 that Icn I < c' na, where c' is the c' of the theorem. Thus, the assertion of the theorem is proved when a> 0. That this assertion is not valid for all P. and v when a < 0 is shown by the example of the function . Consequently, when a< 0, the (22) of §7 for which c2n = 0 for n = 1, 2,

193

§8. THE RELATIVE GROWTH OF COEFFICIENTS

inequalities Ic,, I < cna are satisfied for arbitrary even .t and v for all n = µk + , whereas the coefficients c do not even approach zero for v with k = 0, 1, odd n. This completes the proof of the theorem. We note that, in the simplest case of µ = 2 and a = 0, this theorem is a direct sharpening of Theorem 2 of §7 in the sense that it establishes the relative boundedness of the even and odd coefficients of functions in the class S. We now give bounds on the relative growth of successive coefficients of functions in the class S. Theorem 2.1) For functions f(z) = E7=1 c z" belonging to the class S, we have 1

n = 2, 3, ...,

Ilcn+1

(9)

where c is an absolute constant. Proof. Here, we shall need a more definite form of sharpening of the distortion theorem than that provided by Theorem 7 of §2 (with n = 2). Specifically, let f (z) denote a member of S and suppose that the maximum value of I f (z)I on the circle I z I = r, where 0 < r < 1, is attained at the point z 1 . Then, by applying inequality (16) of § 2 to the function F (C) = l/ f (1/0 E E and to the points

C,=zl, Cs=

I

, IzI=Iz1I=r,

we have

Ix, -zI 1-rs r2

1

1

l

1

2

f (Z,)

T (Z)

T f (2,)1

If (2) I

I f (Z) I

that is, we have

I z -- Z,IIf(z)I2rr,

(10)

This is the sharpened form of the distortion theorem that we shall need. Let us now prove inequality (9). From

(z - zt)f' (z) = - z, +

[- (n + 1) cn+izt + nc5l zn n=1

we have

- (n + 1) cn+1z1 + ncn -tat

S

(z - zi)f' (z)

*

and, consequently, 1) Goluzin [1946e]. (This result has been improved by Hayman (see §2 of the supple-

ment)-)

IV. EXTREMAL QUESTIONS

194

I(n+ I)Icn+tlr-nIcnll --- 27c n

I(z`z,)f' (z) I d0.

S

Izl=r

Let A denote any number in the interval -1 < A < 0 and set µ = - A, so that A + µ = 0. From (10), we have 2a

S

Izl=r

I (z - zt)f (z) I d0

1S Iz-

IIf(z)I"'If(z)I'If (z) I dA

Izl=r 2t-µ

(12rsr2)

If (z) I ' If- (z) I A.

. 2n

J/

obtain``

I z 1= r

If we now apply Lemma 2 of §7, we

r"I(n+1)Icn+iIr_-n lcnI I

(1-r)µ

M(Yr)P

r

M(P)

1-r

I

2

2x+2- 2

_ 2r

vi

dp,

p

(11)

where A and p are connected by the condition A + 1 - 1/p > 0. But M (P) C (l

P

p)e,

Vr)9 <(1

-2

R1 (Y r) c 0

4 r)p, ,

so that the right-hand member of inequality (11) does not exceed 2

2

VP

P

r(1-r) ++P

p

r0

P

4(1+t_P\

(1-p)

\\

dp

(12)

)

Now, let us choose A so that 4(A + 1 - 1/p) = 1; that is, let us set A = -% + 1/p. Then, the quantity (12) takes the form 21+ P

6+1

r(1-r)4

log

1 1-+Yrr

(13)

P

Furthermore, if we set p = [log (1/(1 - r))] and assume r to be such that p > 1, (13) reduces to the form Cr

r(1-r)4

log 1

1

r,

195

§8. THE RELATIVE GROWTH OF COEFFICIENTS

with I cT I < c, where c is an absolute constant. As a result, inequality (11) yields (n+1)Icn+IIr-nIcnIIrn+I (1c r)6/,log

l

1 r.

(14)

Finally, if we set r = n/(n + 1), we obtain inequality (9) from (14) for n sufficiently large and hence for all n = 2, 3, - - . This completes the proof of the theorem.

Theorem 2 '. For odd functions f (z) = En =I c zn_I z 2n-I belonging to the

class S, we have

Ilcn+tl-Icn_iIICcn 4 logn, n=2, 4, ...,

(15)

where c is an absolute constant.

Proof. Let us represent f(z) in the form f(z) = g(z2), where g(z) _ [f(z%)]2 E S. Suppose that the maximum value of Ig(z)I on I z I = r, 0
00

- zt)f'(z) = -zt + Ei I- (n+ 1) cn+tzt + (n n=2

-

1) cn--tl zn+t,

from which we get

- (n + 1) cn+Izt + (n - 1) cn_t (Z9-zI)f,

-2x1

(z)

S

ZsZ9

Isl=Yr 2n!

(z9 - zi) g (_)

Isll, S

!G (z'j i so

Let X denote an arbitrary number belonging to -1 < A < % and set µ = -Jt that A+ µ = -%. Then, I(n+l)lcn+Ilr-(n-1)Icn III

1

S

z'

ztIIg(z9)IV--Ig(z$)I'Ij(z2)Id1

2ar2 Isl=l'r n 2

g(z)(z)IdO.

S

2ar2(1--r)w IzI=r

N. EXTREMAL QUESTIONS

196

We get a bound for the last integral just as in the proof of Theorem 2. As a result, for r sufficiently close to 1, we have

r

(n+l)1cn+lI r

1

= (n - 1)/(n + 1), this yields inequality

(15), proving the theorem.

The inequalities given by Theorems 2 and 2 ' are undoubtedly not sharp, but they are the best that have been obtained.

§9. Sharp bounds on the coefficients As we stated above, in addition to a sharp bound for the second coefficient, we have a sharp bound for the third coefficient for functions in the class S. To illustrate the nontrivial nature of solving certain problems on the extrema of the coefficients, we now present a more general result. Its derivation is based on Lemma 1. Let A(r) denote: an arbitrary real function that is continuous for 0 < r < o. except possibly at a finite number of points at which there are discontinuities of the first kind. Suppose that JA(r)J <e-r for r> 0. Then, by setting 00

>0,

1,2 (,t)dT=(v

(1)

we have 00

SX(t)dv

(v+l)a

(2)

0

Inequality (2) is sharp for each given v with equality holding only for the func-

tion a(r) = ±z(r), where p(r) = e-" for 0 < r< v and p(r) = e-r for r> v. [We note that a v satisfying condition (1) always exists because the lefthand member does not exceed f p e- Zrdr = %, and the right-hand member decreases

from % to 0 as v increases from 0 to oe .] Proof. That the function µ(r) belongs to the type of functions A(r) with given v is easily verified. Specifically, 00

CO

l19(T)dr- eAVdT+ a 9`dt=(v+ 2)e-1"_ JD

b)

00

X (T)dr.

(3)

r

Now, for an arbitrary function A(r) of the type in question, we have for r > 0

§9. SHARP BOUNDS ON THE COEFFICIENTS

197

(EL(T)-IA(r)I)(2e`"-p(t)-IA(r)1):0, because the left-hand memher is, for 0 < r( v, equal to e_v r> v, we have > µ(r) > I A(r)I. From this we obtain

(e_v

- IA(r)I)2 and, for

E.'('r)-A'(t)<2e-' (p.('c)--IA(c)I).

(4)

Integrating equation (4) from 0 to c-, we obtain in accordance with (3) m

co

00

SI (r)IdT<S (t)dt=Se-"dr+ 0

0

0

Se`dt=(v+1)e-',

r

Consequently we obtain inequality (2). From what was said above, we easily see that equality holds in (2) only for the function A(r) = ±p(r). This completes the proof of the lemma. Theorem 1. 1) For functions

f (z)

00

_ 21 C,,z" nel

belonging to the class S and an arbitrary real number a in the interval 0 < a < 1, we have the sharp inequality 26

Cs -act'

2e

I--- + 1.

(5)

In particular I C31 < 3.

Proof. It will be sufficient for us to consider a function f (z) C S' (see § 1). For such functions, formulas (23) and (24) of §2 of Chapter III are valid. On the basis of these formulas, we have co

CS - acy = 4 (1 - a) r10S e 'k (c) dr - 2 S e-' k9 (r) dc. 0

(6)

0

Since the function e`13 f(e-` 3z) for arbitrary real 13 belongs to S' whenever f(z) does, the maximum value of Ic3 - acZ I in the class S coincides with the maximum of the quantity R (c 3 - acZ). Now, we have from (6), by setting k(r) = e`0(r), co

DI(ca-aqj =4(1-a)K S e 1cos0(t)d-rJ9 0 00

co

-(S a sin 6(t)dt)', -4 S e-9' cos'6(t)ds+1 0

0

1) Fekete and Szego [1933]. See also Bazilevic [1936].

IV. EXTREMAL QUESTIONS

198

00

"0 e cos 6 (t) dr )t - 4 4 (1 - a) (S e-2" cos' 0 (t) dr + 1' 0

0 00

= 4 (1 - a) \

co

k (T) dT)t 0

- 4 10 kt (T) dr +

(7)

1,

where X(r) = e-T Cos 0(r). Since IA(r)I < e-T, by representing the last integral in the form (1), we have from (2)

(c3-aes)<4(1 -a)(v-f-1)2e 2'-4 (v+ 2)e +1

=4 [(1 -a)vt-(1 -2a)v+ '-a] e-t"+1.

(8)

It remains for us to find the maximum of the right-hand member of inequality

(8) for v in the interval 0 < v < o . This maximum is attained at v = a/(1 - a). This maximum is attained at v = a/(1 - a). Consequently, we obtain the inequality 2a

31(c8-ac;)<2e I-a+l,

(9)

and, hence, inequality (5). Now, for equality to hold in (9), it is obviously necessary and sufficient that k (r) be such that 00

Se' sin6(r)dT=0'.

(10)

0

and that A(r) = e-rcos 0(r) coincide, in accordance with the lemma, with the

function ft(r) (with v = a/(1 - a)), that is, that k (t)

I

e-('--0 ±j

S`

1-e-2('-z)

for rev, for

1

t ] v.

(11)

It is easy to show that we can satisfy condition (10) by choosing the sign ± properly in (11). Specifically, if we take the plus sign in (11) for 0 < r < S < v and the minus sign for 13 < r < v, then, by virtue of ( 11), condition (10) can be written in the form e-`

1-e

"

dT -- e` Yl--e

dt=0.

(12)

0

For f3 = 0, the left-hand member of (12) is negative and, for f3 = v, it is positive

Consequently, there exists a 8 in the interval 0 < 0 < v such that (12) is satisfied for the corresponding function k(r) in (9) and equality holds in (9). This completes the proof of the theorem.

Corollary I. For functions

§9. SHARP BOUNDS ON THE COEFFICIENTS

199

00

fP (z)

21 c k=0

`

+ lzk+1 E S,

where p > 1, we have the sharp inequality

ap=

IC e - P+1 +p 2

In particular, Ic ;2)1 < e-2 /3 + 2'1 = 1.013

(13)

.

Proof. The function f, (z) can be represented in the form f, (z) = n f(zp), where f(z) _ l,°, =1 c z" E S. We have

c2P+1=

P

(c3-PZPI c9

(14)

But, in accordance with Theorem 1, we have the sharp inequality

c3-P2P c9 1

2e

-2p-1 P+1+

This together with (14) yields (13). For certain special subclasses of functions of the class S, sharp bounds for the coefficients c" have been obtained. For this, we need Lemma 2. Suppose that a function f(z) = 1 + a1z + a2z2 + ... is regular in the

disk I z I < 1 and satisfies in that disk the condition l(f (z)) > 0. Then, Ian I < 2

for arbitrary n = 1, 2,

.

Proof. The condition 32 (f (z)) > 0 is equivalent to the inequality I 1 + f (z)I 11 - f(z)j. It follows that the function

z)f(z)-}-12 z+... B(f (Z)- 1

al

is regular in the disk Iz I < 1 and satisfies the condition I g(z)I < 1 in that disk. It then follows on the basis of the Schwarz lemma that Ia, I < 2. If we now denote by rlk, for k = 1, . . , n the nth roots of unity, we have 1

n

Consequently, the function n

n

f `Ikz0

k=t I

jf(1kzn)=1 k=1

t

az

200

IV. EXTREMAL QUESTIONS

also satisfies the conditions of the lemma. Therefore, Ian 2, . This completes the proof of the lemma.

2 for all n = 1,

Theorem 2.1) For functions f (z) £'n =1 cn Zn C S with real coefficients, we have the sharp inequalities

IC.I_n, n=1, 2, ....

(15)

Proof. By virtue of the univalence of f (z) in I z I < 1, we have, for arbitrary

z1 and z2 in IzI <1

f(zi)-f(z.) zl - z,

00

=1+

nil

Cnz"-4 zl-z,

0.

In particular, for z 1 = re `o and Z2 = re-'O , where 0 < r < 1, this shows that the expression 00

I (r, p) =1 + I ckrk-1 sin sin kcp tb

k-2is real and nonzero, that is, that it retains sign for 0 < r < 1 and arbitrary values

of 0. Consequently, the function

2 sin' TO (r, p) = I + c5r cos y + (car' - 1) cos 2p

+ (ckr' - c9) r cos 3p +.. -k- (ckr' - ck_2) rk-a cos (k- 1) cp -j- .. also retains sign. Since this expression is positive at r = 0, the sign is plus. -

On the basis of this, the function F (z) =1 + csrz + (car' - 1) z' + (ckr9 - c9) rza + (Cnr' - cn_9) rn-8zn-1 +. , is regular in the disk I z I < 1 and satisfies the condition l (F(z)) > 0. In accordance with the lemma, we have

Ic9I<-2, Icar' -II<2, Ic4r'-c511 2, ..., or, taking the limit as r

Icnr'-cn_5I2,

...

1,

Ic9I-2, Ica-II<2, Ic4-c5lc2, ..., Icn-crt-9I<2.

(16)

Itfollows by induction that I cn I < n for arbitrary n. This completes the proof of the theorem.

1) Dieudonne [19311; Rogosinski [19321.

§9. SHARP BOUNDS ON THE COEFFICIENTS

201

Theorem 3.1) For odd functions co

f(z) _

C9n_1z9n

n-1

-1

ES

with real coefficients,

_1

4logn, n =2, 4, ...,

(17)

where c is an absolute constant.

Proof. Since inequalities (16) are applicable both to the function f(z) and to the function

0 i f (Lz).=

I (- 1)n-'

9n-1z9n

1ES

n=1

we have

Consequently,

Icn±cn-s1-2, n=3, 5, ...

.

cn.1+ICn-%1<2, n=3, 5, ....

(18)

On the other hand, Theorem 2' of §8 can be applied to f (z). According to this theorem,

_

1

Cnl-ICn-21I
(19)

Adding (18) and (19), we obtain (17).

In connection with Theorem 3, we state without proof 2) that there exists a function f(z) satisfying the conditions of this theorem for which Ic2n+1 1 with arbitrary given n > 2 is greater than unity.

In §5, we gave definitions of convex and starlike domains for the case of domains with analytic boundaries since we were then speaking of the images of disks I z I < r with r < 1 udder functions of the class S. We shall now give definitions of convex and starlike domains that do not involve properties of the boundary of a domain.

A one-sheeted domain is said to be convex if the segment connecting any two points in it is entirely contained in that domain.

1) Goluzin [1949c]. 2) This was mentioned on page 190 with bibliographic references.

202

IV. EXTREMAL QUESTIONS

A one-sheeted domain is said to be starlike about a point if the segment connecting an arbitrary point in the domain with the point in question is contained in that domain. Let us now look at functions f(z) = z + c2 z2 + E S that map the disk z I < 1 onto domains that are starlike about w = 0. Let f (z) denote such a function and let B denote the image of the disk I z < 1 under the mapping w = f (z). Let us show that, in this case, the domain B, of the disk I z I < r, where 0 < r < 1, is also starlike about w = 0. For any point w that belongs to B, the point tw, where 0 < t < 1, also belongs to B. Consequently, the function f -1(t f (z)) = i/i(z) with 0 (0) = 0 is defined and regular in the disk I z I < 1 and I(A (z)I < 1 in that disk. Therefore, ItA (z) I < I z I in I z I < 1. Now, let w1 denote a member of of B,. Then, w 1 = f (z 1), where z 1 is a point in the disk I z 1 I < r. Consequently,

V(z1)I
that is, tw1 E B,. This shows that the domain B, is starlike about w = 0. A consequence of this is that, if a function w = f (z) maps the disk I zI < 1 onto a starlike domain, then, as z moves around the circle I z I r, 0 < r < 1, in a specified direction, arg f (z) changes in the same direction, and, in accordance with §5, this is equivalent to satisfying the inequality Zf (Z)

(20)

_ f(Z))>0,

in the disk I z I < 1. In accordance with the maximum principle of harmonic functions, equality cannot hold here. We now have

Theorem 4. For functions f(z) =I

n=i

ES

onto starlike domains, we have the sharp inequalities

that map the disk

IzI<1

1)

IcnICn, n=2, 3,...,

(21)

and under the additional assumption that f (z) is an odd function, we have the sharp inequalities 2)

n=3, 5,...

(22)

1).Loewner [1917]. 2) Privalov [1924]. The requirement that the function f (z) be odd can be replaced with the condition c2 = 0 (see Goluzin [1938]).

203

§9. SHARP BOUNDS ON THE COEFFICIENTS

Proof. If we set zf (z)

atz

7 (2)

(23)

we have, in accordance with the lemma, I a.,, I < 2 for n = 1, 2, hand, comparison of the coefficients in (23) yields

. On the other

(n - 1) C. = an_t + an-acs -I" ... + atcn_t,

from which we get

IcnI°n-sl

an-,I

1C21+...+[I

n-1

I'Icn-il:

_21+II+...

n-1

For n = 2, it follows that then conclude that

I

C2

< 2. Let us assume that I ck I < k for k < n. We

ICnIc21+

2+...+(n-1)_n, n-1

so that inequality (21) holds for all n. Inequality (22) is proved in an analogous manner.

E S that map the disk I Z 1 Consider the functions f(z) = z + C2 z2 + onto convex domains. Let ((z) denote such a function and let B denote the image of the disk. I z I < 1 under the mapping w = f W. From the definition given above of a convex domain, for two points w 1 and w 2 belonging to B, all points t w 1 + (1 - t)w2, for 0 < t < 1, that is, all points of the line segment connecting w 1 and W2, also belong to B. Let us show that, in this case, the image Br of the disk z I < r, for 0 < r < 1, is also a convex domain. If w 1 = f (z 1) and w2 = f (z2), where

IZ1I

_<

I z 2 I < r, we easily see that the points O (Z) = tf (z 1 z/z 2) +

(1 - t) [(z) for I z I < 1 and 0 < t < 1 all belong to B. Therefore, the function O(z) = f -1 ((k (z)) is regular in I z I < 1, t&(0) = 0, and ItA (z)I < 1 in the disk I z I < 1. From the Schwarz lemma, we have kt'(z)I < I z I. For z = z2, this yields Il-1 (tw1 + (1 - t)w2)I < r. But twl + (1 - t)w2 = f(zo), where Izo I < 1. Therefore, I z o I < r, so that tw 1 + (1 - t )W2 E Br . This proves that the domain Br is convex.

A consequence of this is the fact that, if a function w = f (z) E S maps the disk I z I < 1 onto a convex domain, then, as z moves in the positive direction around an arbitrary circle I z I = r, where 0 < r < 1, the corresponding point w describes a curve the tangent to which always turns in a counterclockwise direction, and, in accordance with §5, this is equivalent to satisfying the condition

204

IV. EXTREMAL QUESTIONS

(24) (z)

in I z I < 1. By virtue of the maximum principle of harmonic functions, equality

cannot hold here. It is easy to see that the converse is also true, that is, if a function f (z) = z + c2 z2 + - - is regular in I z I < 1 and satisfies condition (24) in that disk, it maps the disk I z I < 1 univalently onto a convex domain. Analogously, if a function f (z) = z + c 2 z2 + is regular in the disk I z I < 1 and satisfies condition (20) in that disk, it maps the disk I z I < 1 univalently onto a domain that is starlike about wo. These conclusions are in accordance with the definitions of convexity and starlikeness that we gave here and in §5. If two functions f 1(z) = z + I z I < 1 and are connected by

and f2 (z) = z +

are regular in the disk

f9 (z) = zfi (z),

(25)

then it follows from the equation (( zfi (z)l __

(zfi (z)l +

1A(z)1 f.(z)I 1 that the function w = f 1 (z) maps the disk I z I < 1 univalently onto a convex domain if and only if the function w - f2 (z) maps the disk I z I < 1 univalently onto a domain that is starlike about w = 0. We immediately see from this that,. if a function w = f (z) = z + c 2 z 2 + maps the disk I z I < 1 univalently onto a convex domain, then, for all n = 2, 1; this is a sharp inequality since equality is attained by , we have I c,, 3, the function Go

Y, zn' z

n-1.

which maps the disk I z I < 1 onto the halfplane $2 (w) > -4

(26)

CHAPTER V UNIVALENT CONFORMAL MAPPING OF MULTIPLY CONNECTED DOMAINS

1. Univalent conformal mapping of a doubly connected domain onto an annulus

In the present chapter, we shall consider the problem of univalent conformal mapping of multiply connected domains onto various domains of special forms known as canonical domains. We make one important general remark regarding such

domains. Specifically, it is easy to show that, if a function C= f (z) maps a domain B univalently, then at each isolated boundary point of the domain B either f (z) is regular or has a pole and is univalent in the domain obtained from B by adjunction of all isolated boundary points of B. To see this, let z 0 denote a finite isolated boundary point of the domain B. Then, z o can be only a singlevalued isolated singularity of the function f W. Let p denote a positive number sufficiently small that the disk Iz - z0I < p includes no boundary points of B other than z 0. We denote by B the portion of the domain B contained in the disk Iz - z0I < p. B* is a doubly connected domain that the function J= f(z) maps univalently onto a domain in the plane containing exterior points. Let a denote one of the exterior points. Then the function F(z) = Of (z) - a) is bounded in B that is, z o is a removable singularity of F (z). If we define the value of the function F(z) at z = z0 as the limit of its values as z -- z0, we render F (z) regular at z 0 also. If this limit is nonzero, then z 0 is a regular point of the function f (z) analogously defined at z 0. On the other hand, if the limit is 0, then f (z) has a pole at z = z 0. In this case, we take f (z 0) = oc. In both cases, the function C = f (z) maps a neighborhood of the point z 0 onto a domain including the point Co = f (z0). Now, if f (z) is not univalent in the domain obtained from B by adjoining to it the point z 0, there exists a point z 1 C B such that f (z 1) = Co. A neighborhood of the point z t is mapped by the function 205

206

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

C= f (z) onto a domain including the point 0. Consequently, to every point of a sufficiently small neighborhood of the point Co there correspond no fewer than two points of the domain B. However, this contradicts the univalence of f (z) in B. Thus, our remark is proved. It shows that, when we are investigating questions of univalent conformal mapping, isolated boundary points can be adjoined to the domain being mapped without causing any complication in the problem. Let us now begin with the simplest case of the problem posed, namely, the case of doubly connected domains. Let us show that every doubly connected domain can be mapped univalently onto some circular annulus, whose boundary circles may degenerate to points. Let B denote a doubly connected domain in the z-plane. If it has an isolated boundary point z0, then, by adjoining it to the domain B, we obtain a simply connected domain that we can then map univalently either onto the disk 11 < I or onto the cplane with the point C= oo excluded. We can do this in such a way

that the point z = z0 is mapped into C= 0. The domain B is thus univalently mapped either onto the annulus 0 < IzI < 1 or onto the annulus 0 < IzI <.. Suppose now that the boundary of the domain B consists of two continua K1 and K2. One of these, let us say K1 is necessarily bounded. The complement of K1 in the z-plane is an open set consisting of two disjoint domains. One of these domains, let us say B 1 contains the domain B. The domain B 1 is simply connected. Therefore, it can be mapped conformally onto the disk 1z 'I < 1. Under this mapping, the continuum K2 is mapped into a continuum K2 contained in 1z 'I < 1, and the domain B is mapped into a domain B'. Let us now map whichever of the simply connected domains complementary to K2 contains B' onto the domain Iz "I > 1 in such a way that z '= oQ is mapped into z "= oa. Under this mapping, the circle 1z 'I = 1 is mapped into an analytic Jordan curve contained in the domain Iz "I > 1 and the domain B' is mapped into a doubly connected domain B" that does not include -o and that is bounded by this curve and by the circle Iz "I = 1. The composite of these two mappings constitutes a univalent mapping of the domain B onto the domain B ". Furthermore, on the basis of §3 of Chapter II, we conclude that this mapping sets up a one-to-one correspondence between the prime ends of the domain B and the boundary points of the domain B ". Here, just as in § 3 of Chapter II, by the prime end of the domain B that corresponds to the point z o on the boundary of the domain B ", we mean the set of all cluster points of all sequences of points of the

§ 1. MAPPING A DOUBLY CONNECTED DOMAIN ONTO AN ANNULUS

2 07

domain B corresponding in the domain B " to sequences that converge to z o . Let us now show that the domain B " can be mapped univalently onto a circular annulus 1 < I < h. We first map the domain Iz "I > 1 by means of the function t = log z onto the halfplane 3 ( t ) > 0. This mapping is not one-to-one because corresponding to each point z " in the domain Iz "I > 1 there correspond infinitely many points of the form t + 2kiri, for k = 0, + 1, + 2, . - . belonging to the halfplane $2(t) > 0. The domain B " is thus mapped into a vertical curvilinear strip in the t-plane bounded by the imaginary axis and an infinite analytic curve contained in the halfplane R (t) > 0. This strip is a simply connected domain B 1 with the property that, if a point t C B 1, then all points t + 2k ni, for k = 0, ± 1, + 2, - , also belong to B 1. Let us now map the domain B 1 univalently onto a vertical strip B 2 of the form 0 < R (w) < h in such a way that the boundary points t = - oo i, 0, and + oo i are mapped into the boundary points w = - oo i, 0, and +ooi. 1) Under this mapping, the point t = 21ri is mapped into a point wo on the positive half of the imaginary axis. By means of a similarity transformation in the w-plane if necessary, we can arrange to have w0 = 2ni. Then, if we denote by w = f (t) the mapping function and if we denote by t = q (w) its inverse, we conclude that these two functions satisfy the relations f (t+2ar1) = f (t) -}- 2it1, p (w+21r1) = p (w) -}- 2n1.

(1)

To see that this is the case, note that the functions f (t + 277i) and f (t) + 27ri both map the domain B 1 univalently onto the strip B 2 in such a way that the boundary points - oo i, 0, and + oa i are mapped in both cases into the points - oo i, 2iri, and + oa i respectively. Consequently, by virtue of the uniqueness, these functions coincide, so that we have the first of equations (1). The second of equations (1) is obtained in an analogous manner. Finally, let us map the strip B2 by means of the function C= ew onto the annulus 1 < et`. Again, the mapping is not one-to-one. If we now express : in terms of z, we obtain the function = ef(1°g z"), which maps the domain B " onto the annulus 1 < Ii < eh. This mapping is univalent because, by virtue of properties (1), the functions [;'= of (log Z") and z "= a 000g 0 are single-valued respectively in the domain B " and in the annulus 1 < I 1 < et`. Furthermore, it follows from what we said above that the function (;'= of ('Og is analytic in 1) Every strip has two infinitely distant boundary points, which we denote by ±ai.

208

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

the closed region B" and that its inverse is analytic in the annulus 1 < Ii < eh. Summarizing these results, we have Theorem 1. Every doubly connected domain B can be mapped univalently onto a circular annulus. This mapping defines a one-to-one correspondence between the prime ends of the domain B and the points of the boundary of the annulus. If the domain B is bounded by Jordan curves, the mapping is one-to-one and continuous exclusive of the boundary. On the other hand, if the domain B is bounded by analytic curves, then, in addition, the mapping function is analytic in B and the inverse function is analytic in the closed annulus. As a supplement to this existence theorem, let us prove a uniqueness theorem, one which reveals a significant difference between univalent mappings of simply and multiply connected domains. Theorem 2. The circular annulus onto which a given doubly connected domain B is mapped univalendy is unique up to a linear transformation: that is, if the domain B is mapped univalently onto two distinct circular annuli, then each of these annuli can be obtained from the other by means of a linear transformation. Consequently, the ratio of the radius of the larger boundary circle to that of the smaller is the same for the two annuli. Furthermore, the function f (z) that maps the domain B onto an annulus in such a way that a given point on the boundary of the annulus corresponds to a given point on the boundary of B is uniquely determined.

Proof. If a domain B is mapped univalently onto two distinct circular annuli, these annuli can be mapped univalently onto each other. Consequently, it will be sufficient for us to show that, if two circular annuli are mapped univalently onto each other, they can be obtained from each other by a linear transformation. And to do this, it is obvious that we need only show that, if two annuli r < Iz I < R and r < 14 < R' can be mapped univalently onto each other in such a way that IzI = r is mapped into W = r, 1) this last mapping is of the form C= e"z, where a is a real constant. This can be shown as follows: A function f (z) can be extended, by the symmetry principle, beyond the circle IzI = R into the annulus R < I z I < R 2/r and maps this annulus univalently into the annulus R '< I ci < R 121r. Then f(z) can be extended further into the annulus R2/r < Izi < R4/r3, etc. 1) We can always arrange for this by an inversion the radii of the boundary circles does not change.

under which the ratio of

§ 1. MAPPING A DOUBLY CONNECTED DOMAIN ONTO AN ANNULUS

209

Analogously, we can extend f (z) past the circle IzI = r into the annulus r2/R < IzI
Z=p(`)_ E C4C" n=-ao

in the annulus. If G and g are the areas of the finite domains bounded by the preimages of the circles I1 = r i and 11 = R , where r ' < r l' < R i < R ', we have

G=

g=

SCI S

-Rl

S ICI:= r,

uv'd9=a n=-oo

ttv'de^,r

'I Ic

co

n1

I9R'2n,

I

n--oo

(where C = re`B), so that G

nR'2

g

nr,2

ari2Ri E

n=-OD

n cn I' co

(Ri2n-2

` r'1 n-2 ) 0,

2] n I Cn 1' r12n n--00 since n (R i 2"-2 - r 12n-2) > 0 for all integers n except 0 and 1. If we now let ri approach r' and Ri approach R', we obtain ar12

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

210

nRA

gxr9

with equality holding only when sequently, from (2) we have

(2)

c0 + c 1S. But G < rrR2 and g > rrr2. Con-

RJR' r--

(3)

with equality holding, obviously, only for 0 (0 = c 1 C, where Ic 1 I = 1.

The assertion of the theorem follows because, if we map the extended domain B ' onto the annulus r < Iz I < R, the domain B is mapped onto a subdomain of that annulus, to which (3) is applicable with inequality holding. Then, the ratio on the right is the modulus of the domain B. This completes the proof of the theorem.

In conclusion, we note that, for the first part of the existence theorem given above, we can give a proof based (like the proof of Riemann's theorem) on the solution of an extremal problem: Let B denote a doubly connected domain that is contained in the domain Iz I > 1 and that has as one of its boundary continua the circle Izi = 1. Out of the family l of all functions J= f(z) that map the domain B univalently onto domains of the same type in such a way that the circle Iz I = 1 is mapped onto itself, find the one for which the quantity

M (f) = sup If (z) I zEB

has the greatest value. The function representing the solution of this extremal problem maps the domain B univalently onto a circular annulus. §2. Univalent mapping of a multiply connected domain onto a plane with parallel rectilinear cuts Let us now investigate univalent conformal mapping of arbitrary multiply connected domains onto canonical domains of various kinds. The simplest such domain is a plane with parallel rectilinear cuts. In this case, our investigation will be based on the solution of certain extremal problems. Lemma 1. Out of all functions F(z) = z + al/z + that are univalent in R(e_2iea1), where 0 is a given real number, is maximized IzI > R, the quantity by a function mapping the domain IzI>R onto the plane with rectilinear cut that

makes an angle 0 with the real axis, and only by that function. For it,

§2. MAPPING ONTO A PLANE

211

R(e-2i6a1) = R2. Proof. It follows from the area theorem of § 4 of Chapter II as applied to the

functions

RF(RC)=C+' +-..., that lalI < R2 with equality holding only when F(z) = z + R2eiu/z. Therefore,

R(e 9iea,)cR$, with equality holding, obviously, only for the function F (z) = z + R 2e 2 /z, which maps the domain IzI > R onto the plane with cut at an angle 0 to the real axis. is univalent in Lemma 2. If the function C= F(z) = z + ao + al/z + IzI > R, then IF(z) - aol _< 21zl in the domain IzI > R and the entire boundary of the image of the domain IzI > R under the mapping t = F(z) is contained in the

disk IC- aol < 2R. Proof. If I z 0 > R, then the function

C=F,(z)= I F(zuz)- zo is univalent in IzI > 1. Consequently, in accordance with §4 of Chapter II, the entire boundary of the image of the domain IzI > 1 under this function is contained in the disk 11 < 2. In particular, IF1(1)I < 2; that is, If(z) - a0l < 21zl, where IzI > R. Then, if we shift from IzI > R to Izl = R, we obtain the second part of the lemma. Theorem 1.1) Every domain B in the z-plane can be mapped univalently onto a domain B' in the extended cplane such that an arbitrary continuum in the complement of the domain B' with respect to the plane is a straight line segment of given inclination 0 to the real axis. Also, this mapping is such that a given point a of the domain B is mapped into C = oa and the expansion of the mapping function about z = a is of the form +'a,(z-a)+... or z+ Z ..., z a

according as a is finite or infinite. Proof. It will be sufficient to prove the theorem for the case in which a because, in case of finite a, we can map the domain B by means of the preliminary transformation z" = 1/(z - a) onto the domain B' containing z * = oo, 1) Beginning with the extremal property of the mapping function, which was indicated in the proof given below, this theorem was proved by de Possel [19311 and Gtotzsch [19321.

212

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

which reduces the problem to the original case. In the case of a simply connected domain B, the theorem is easily proved. Specifically, if the boundary of the domain B has more than a single boundary point, then it follows from Riemann's theorem that it can be mapped univalently onto an arbitrary domain constituting the entire c'-plane with cut at inclination B to the real axis in such a way that z = o is mapped into z '= oo and the expansion of the function C'= C'(z) about z = oc is of the form a_ 1z + a0 + a1/z + . , where a_I > 0. Then, the function C= (C'(z) - a0)/a_1 provides the required mapping of the domain B. Let us turn to the general case. Consider the family 11 of all functions f (z) that map the domain B univalently and have the expansion f (z) = z + a1/z + in a neighborhood of z = oo. An example of such a function is f (z) = z. Let us pose the extremal problem: Out of all the functions of the family R, find the one that

maximizes the quantity R(e-2`Ba1) Let us first prove that this problem has a solution. Suppose that the entire boundary of the domain B is contained in the disk Izj < R. Then all functions of the family R are univalent in the domain jzj > R. Consequently, in accordance with Lemma 1, for these functions we have R (e-2` Ba1) _< R 2; that is, the numbers R(e-206a1) are uniformly bounded. Let A denote the least upper bound of these numbers. If this upper bound is not attained, there exists a sequence of functions fn(z) E Dl for which the sequence {R(e-2`Ba1)) converges to A. But, from Lemma 2, for functions of the family Dl we have if (z)/z1 < 2 in jzl > R. This shows that the condensation principle can be applied to the functions

ff(z)/z, that is, that the sequence { f,(z)/z} contains a subsequence (f,,k(z)/zI that converges uniformly in the interior of the domain Izj > R to a regular funcsuch that R(e-2`Bai0)) = A. But in accordance tion f0(z)/z = I + aJO)/z2 + with Lemma 2, we have l fnk(z)l < 2R' for functions fnk(z) on Iz = R', where R '> R. By virtue of the univalence of flk(z) in B, the same inequality holds also at points of the domain B that belong to the disk 1zI < R'. Consequently, the functions fnk(z) are uniformly bounded in the interior of the domain B0 obtained from B by removing the point z = oc. In accordance with Vitali's theorem, the sequence of these functions converges inside B0 to a regular function f 0(z), which, in accordance with Theorem 2 of § 1 of Chapter 1, is univalent in B0 and hence in B. Therefore, f0(z) E 2, which contradicts our assumption that the least upper bound A is not attained by the numbers R(e-2cea1). Thus,

§2. MAPPING ONTO A PLANE

213

the existence of a solution of the extremal problem posed is proved. Let us now show that the extremal function f0(z) provides the mapping referred to in the theorem. Let us suppose that the complement of the domain B with respect to its image B ' under the function C= fo(z) contains a continuum other than a straight line segment of inclination 9 with the real axis. Let B 1 denote whichever of the simply connected domains complementary to this continuum in the plane contains the point C= cc. Suppose that the function w = w(C) maps B 1 univalently onto the w-plane with rectilinear cut of inclination 0 with the real axis and that it has the expansion t;+ 01/C+ in a neighborhood of 4_ cc. Let us show that $2(e-2`61) > p. RiFemann's In accordance with theorem, the domain B 1 can be mapped univalently onto the domain It > R' in such a way that the function C_ C(t) inverse to the mapping function has the form C(t) = t + yo + y1/t + . The function w = w(C(t)) - yo = t + (131 + yl)/t + maps the domain It` > R ' univalently. onto the w-plane with rectilinear cut of inclination 0 with the real axis, and the function r= C(t) - yo = t + yl/t + . maps ItI > R ' onto a different domain. Therefore, on the basis of Lemma 1 we have at (e-0 71) < (e 9`B (N1 + Ti))

that is, R (e-z` 681) > 0. Now, let us look at the function w = w (f 0(z)). This function maps the domain B univalently and has the expansion z + (ai0> + O1)/z + in a neighborhood

of z =cc. Consequently, w(f 0(z )) E 9 and we have at (e-20 (4) + r1)) = R (e-20 a()) + at (e9te 1'1) > A,

which, however, is impossible. This shows that the function f0(z) meets all of the requirements of the theorem. At the same time, we have established an important extremal property of the mapping function, namely, the fact that it maximizes the quantity R(e-2`9a1). In the case of finitely connected domains, this theorem can be formulated as follows :

Theorem 1 '(Hilbert). Every n-connected domain B in the z-plane can be

mapped univalently onto the aplane with n parallel finite cuts of inclination a with the real axis in such a way that a given point z = a is then mapped into

214

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

C_ oc, and the expansion of the mapping function about z = a has the form

z-a -}-at(z-a)4-... 1

or

z4-i'-4-...,

according as a is finite or not. Some of the cuts referred to may consist of single points.

Let us now look at the question of the correspondence of boundaries under univalent conformal mapping of multiply connected domains. We shall confine ourselves here to finitely connected domains that have only accessible boundary points. We may assume that the point oc is in the interior of the domain. Let B denote such a domain and let K 1, K2, - - , Kn denote its boundary continua. Let us map the domain B univalently onto a domain bounded by closed analytic Jordan curves. This can be done as follows: we map the simply connected domain including z = ea and bounded by the continuum K 1 onto the interior of a circle. Under such a mapping, the domain B is mapped into a domain B' bounded by a circle Ki and the continua K21 , K,, . Then, we map the simply connected domain containing the domain B' and bounded by the continuum K2 onto the interior of a circle. Under this mapping, the domain B' is mapped into a domain B ° bounded by a circle KZ , an analytic curve K1 and the continua K3, , Continuing in this way, after n steps we arrive at an n-connected domain B(n) whose boundary consists of n closed analytic Jordan curves. The univalent mapping of the domain B onto the domain B(n) is a composite of successive mappings of univalent domains. Since these mappings define a one-to-one correspondence between the boundaries, the correspondence between the points of these boundaries defined by the mapping of B onto B( n) is also one-to-one. Furthermore, as in § 3 of Chapter II, we can show that the function inverse to the mapping function is continuous in B(") except at the point -a. If the domain B is bounded only by closed Jordan curves, we can show in addition that the mapping of B

onto B is also bicontinuous. If we now map the domain B(n) onto the plane with rectilinear cuts, then, just as in the proof of Theorem 5 of §3 of Chapter II, we can show that the mapping function is regular on the entire boundary of the domain B("). Combining all that we have said, we conclude that, in the present case, the mapping mentioned in Theorem 1' is one-to-one except for boundary points of the domain B. An important supplement to the existence theorem proved above is the following uniqueness theorem:

§ 2 MAPPING ONTO A PLANE

215

Theorem 2. There exists only one function performing the mapping mentioned

in Theorem I'. Proof. Let us suppose first that B is a bounded domain with boundary con, K,,. Let us suppose that there sisting, of closed analytic Jordan curves KI, C'= fl(z) and t;'"= f2(z) that map the domain B univalently are two functions onto the plane with parallel rectilinear cuts of inclination 0 and normalized as indicated in Theorem 1 '. Then both these functions are regular in B except at , n, they assume values that lie respectively the point z = a. On K, v = 1, c and Ne-` 9C°) = c v, for v= 1, , n; on certain straight lines -J(e-` c,, and -J(e-`Bf2(z)) = cv, where cv and is, on K we have that , n, are constants. It follows from this that the difference t; = c v = 1, F (z) = f 1(z) - f 2(z) is regular in B and assumes on K (for v = 1, , n) values lying respectively on certain straight lines d Consequently, if we take an arbitrary point Co that does not lie on any of the straight lines dv, we conclude that arg [F (z) - Col does not change as we move around any of the curves K. Therefore, on the basis of Cauchy's familiar theorem on the zeros of analytic functions, it follows that the function F (z) - Co has no zeros in B, that is, that F(z) does not assume the value Co in B. Thus, all values that F(z) assumes in B lie on the straight lines dv. However, this is possible only when F(z) = const in B. Since F(a) = p, it follows that F(z) a p, that is, f 1(z) = f2(z), so that the theorem is proved in this case. Now, let B denote an arbitrary finitely connected domain. Here, we may assume that it has no isolated boundary points. If there were two distinct functions performing the mapping described in Theorem 1 ', we could, by mapping the domain B univalently onto the bounded domain B0 with boundary consisting of closed analytic Jordan curves, find two distinct functions mapping the domain B0 univalently as indicated in Theorem 1 ', which, on the basis of what was said above, is impossible. This completes the proof of the theorem. In conclusion, we present an important relationship for the functions mentioned in Theorem 1' for various values of 0. Let B denote an n-connected domain bounded by closed analytic Jordan curves KI, , K and let C= jg(z, a) denote a function that maps the domain B univalently as indicated in Theorem 1 '. Let us show that the following relationship holds for arbitrary 0:

216

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

2

jo (z, a) = ei0 (cos 6 jo (z, a) -1 sin 6 j (z, a)).

(1)

The difference d(z) between the two sides of this asserted equation is a function that is regular in the domain B and that vanishes at z = a. Furthermore, all the values that d(z) assumes on any of the curves Ki, lie on a straight line 9 (e-' 64) = const.

Reasoning as in the proof of Theorem 2, let us show that d(z) = 0; that is, let us prove (1). This relation, which has been proved for a domain bounded by closed analytic Jordan curves is also valid for an arbitrary finitely connected domain B. For this, it is sufficient to map the domain B onto a domain B* bounded by closed analytic Jordan curves, apply(1) to B*, and then return to the domain B. This yields the relation (1) for B. Equation (1) yields jg(z, a) for arbitrary 0 as soon as we know j0(z, a) and l n/ 2 (z, a).

§3. Univalent mapping of a multiply connected domain onto a helical domain

In an analogous manner, we shall find the answer to the question of univalent mapping of multiply connected domains onto a plane with cuts along arcs of logarithmic spirals and, as limiting cases, onto the plane with radial cuts and with cuts along circular arcs of concentric circles. For constant 9 and c, the equation -9 (e ° Blog ') = c defines a logarithmic spiral in the plane with asymptotic point at the origin. This spiral has the property that it is intersected by an arbitrary ray issuing from the origin at an angle 0. This last follows, for example, from the fact that, if we shift to the plane t = log C, this logarithmic spiral is mapped into the straight line 9 (e-` 6t) = c with inclination 9 to the real axis, and the ray referred to is mapped into a straight line parallel to the real axis. For 9 = 0, the logarithmic spiral degenerates into a ray issuing from the origin. For 0= n/2, it degenerates to a circle with center at the origin. If we hold 0 constant and vary c, we obtain various curves constituting the family of logarithmic spirals of inclination 0. In all that follows, when we speak of logarithmic spirals of inclination 9, these are what we mean. Let us show that, for any simply connected domain B that has boundary points, it is possible to map B onto the plane with cut along an arc of a logarithmic spiral of inclination 0 in such a way that given points a and b of the domain B are mapped into 0 and oo and the expansion of the mapping function

§3. MAPPING ONTO A HELICAL DOMAIN

217

about z = b has the form z

b

+aa+at(z-b)-} ...

or

x+ao+ z

+-

(1)

according as b is finite or infinite. In the case in which the domain B has a single boundary point, this is obvious, and then the arc of the logarithmic spiral referred to degenerates to a point. On the other hand, if the boundary of the domain B is a continuum, let us first map B conformally onto the domain 1z 'I > 1 in such a way.that the point z = b is mapped into z '= 00. Then, the mapping function z '= q(z) has in a neighborhood of z = b, the expansion

za=tb +as+at(z-b)+...

or

a_tz+ao-}

y -{-...,

according as b is finite or infinite. This is possible by virtue of Riemann's theorem. Suppose that the point z = a is mapped into a point z'=a'. It now remains to establish the possibility of mapping the domain 1z 'I > 1 onto the plane with cut along a logarithmic spiral of inclination 0. But this possibility follows from the solution of the problem on the minimum of the quantity

R (e-2i 'log F'(a')) in the class 1. This problem was studied in § 3 of Chapter IV (the first application of Theorem 1 with a= 7r/2 - 0). It was shown there that this problem has a solution and that the extremal function C= F(z') normalized by the condition F (a') = 0 provides the required mapping. As a result, we see that the function C= cF(O(z)) = f (z) with f (0) = 0 and f'(') = oo with suitably chosen c provides a mapping of the domain B onto the plane with cut along a logarithmic spiral and has an expansion of the form (1). This result for simply connected domains can be generalized to multiply connected domains. For this generalization, we need the analogue of the minimal property proved in §3 of Chapter N, which we formulate as a Lemma. Of all functions J= F(z) = z + ao + al/z + that map the domain IzI > R univalently in such a way that a given finite point a and the point oo are mapped into the points 0 and oo respectively, the quantity 82 (e-Z` 6 log F'(a)) is minimized by the function F0(z) that maps the disk IzI > R onto the cplane with cut along an arc of a logarithmic spiral of inclination 0. Here log F'(z) means the branch that approaches 0 as z --. oo. We now have

Theorem 1.1) Every domain B in the z-plane can be mapped univalently onto 1) Grotzsch [1931].

218

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

a domain B' in the cplane that includes the points 0 and oo such that an arbitrary continuum of the complement of the domain B' with respect to the cplane is an are of a logarithmic spiral of given inclination 0. This mapping maps given points a and b of the domain B into 0 and oo and the expansion of the mapping or z + ao + function about z = b has the form (z - b) 1 + ao + a1(z - b) + according as b is finite or infinite. a1z-1 + Proof. We may assume that b = 00 because, if this is not the case, we can use the transformation z* = 1/(z - b) to switch from B to a new domain B. and we can then prove the theorem for this last domain, replacing a and b with 1/(a, b) and oo.

In the case of simply connected domains, the theorem was proved at the beginning of this section. To prove it in the case of a multiply connected domain B, let us consider the family 3R of all functions f (z) that map B univalently in such a way that a and oo are mapped respectively into 0 and oo and that have an in a neighborhood of z = 00. An example of expansion f (z) = z + ao + al/z + such a function is f (z) = z - a. We pose the extremal problem: Out of all functions of the family 9, find the one that minimizes the quantity R(e-2`Blog f'(a)), where log f'(z) denotes the single-valued branch in the domain B that approaches

0 as z Let us show that this problem has a solution. To do this let us choose a simply connected domain B' contained in B that includes a and oo. In this domain, all functions f (z) E 9 are univalent. Consequently, for these functions, the quantity R (e- 2` S log f '(a)) is bounded below by the same number as was calculated for a function that maps the domain B' univalently onto the plane with cut along an arc of a logarithmic spiral of inclination 0 in such a way that a and 00 are mapped into 0 and oc. Let us denote the greatest lower bound of the R(e-2c0log quantity f'(a)) for functions in 3R by A.

Let us show that this greatest lower bound is attained. We assume the opposite. Then, there exists a sequence of functions f"(z) E 9, for n = 1, 2, such that, as n -- 00, R (e-9`B log fn (a)) - A.

Suppose that the entire boundary of the domain B lies in the disk Izl < R. According to Lemma 2 of § 2, it follows that the entire boundary of the image of the domain Iz I > R under the function w = f"(z) = z + ao") + al" )/z + (where n = 1, 2, ) lies in the disk JC- ao")l < 2R and that 1f"(z) - ao")l < 2Izl in the

§3. MAPPING ONTO A HELICAL DOMAIN

219

domain IzI > R. If Ial > R, this last inequality with z = a implies that Iao")I _< 21al. On the other hand, if lal < R, then the point Co = &(a) = 0 does not belong to the image of the domain IzI > R under the mapping C= f"(z} hence, it belongs to the disk IC- a("'I < 2R. Therefore, lao"Il < 2R. Thus, in all cases, lao"))l < 2max (lal, R) = M. But then, If" (x)/zl < 2 + ;11/R in IzI > R. Consequently, the condensation principle can be applied the sequence 1f"(z)/z{, so that it contains a subsequence {f"k(z)/z1 that converges uniformly in IzI > R to a regular function. The corresponding sequence tf"k(z)} converges uniformly in an arbitrary closed bounded subset of the domain IzI > R to a function f0(z) that is univalent in IzI > R. But since the values of C= f"k(z) corresponding to points of the domain B that lie in IzI < 2R belong to the disk a((0"'I < 4R and hence to the disk l l < 4R + M, the set of functions f"k(z) is ICuniformly bounded inside the domain B with the point z = oo excluded. Therefore the uniform convergence, mentioned above, to f0(z) holds also in the interior of the domain B, and f0(z) C M. The function f0(z) thus proves to be a solution of the extremal problem posed, so that this problem does have a solution. Let us show that the extremal function f0(z) provides the mapping indicated in the theorem. Let us suppose that the complement of the image of the domain B under the mapping C = f0(z) contains a continuum other than an arc of a logarithmic spiral of inclination 0. Let B1 denote whichever of the simply connected domains in the plane that are complementary to this continuum contains 0 and o-. Suppose that the function w = q (0 maps B 1 univalently onto the w-plane with cut along an arc of a logarithmic spiral of inclination 0 in such a way that 0 (0) = 0 and QS (oo) = 00 and suppose that 4 (0 has an expansion C + go + f31/C + in a neighborhood of C = -o. Let us show that

T(e l" log y'(0)) < 0. In accordance with Riemann's theorem, the domain B 1 can be mapped univalently onto a domain Itl > R in such a way that the inverse function has an expansion of the form J= C(t) = t + y0 + yl/t + . Suppose that the point t = a' corresponds to the point C = f 0(a). The function w = .0 (C(t)) maps the domain ItI > R univalently onto the w-plane with cut along an arc of a loga-

rithmic spiral of inclination 0 in such a way that the point t = a' is mapped into. w = 0. From the preceding lemma, we have sf (e-2:e (log gyp' (0) + log C (d))) < 51 (e-211 log C' (d)).

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

220

Therefore, R (e-2` Slog 0'(0)) < 0. Let us now look at the function w = q5(fp(z)) = O(z). This function maps the domain B univalently in such a way that a and oc are mapped into 0 and DC in a neighborhood of respectively and it has the expansion z + c 0 + c I/z +

_ . Consequently, it belongs to the family 9k and M (e2 ° log ' (a)) _ ,T (e-111 log ?'(0)) + R (e2d log fY (a)) < A,

which, however, contradicts the definition of A. This completes the proof of the theorem.

At the same time, we have established the following extremal property of the mapping function, which includes the distortion and rotation theorems for the family.of univalent functions that we are considering. 1) Of all univalent functions i F (z) defined on a domain B including w that have an expansion of the form

F(z)=z+ao+ z +..., in a neighborhood of z = o, the quantity R(e-208log F'(z)) with given z e B and given real 0 is minimized by the function defined in Theorem 1. In the case of multiply connected domains, the existence theorem that we have proved can be formulated in a different way:

Theorem 1'.2) Every n-connected domain B in the z-plane can be mapped univalently onto the t; plane with n cuts along arcs of logarithmic spirals of inclination 0 in such a way that given points a and b of the domain B are mapped into 0 and - respectively, and the expansion of the mapping function about z = b has the form z

- b+ao+a,(z-b)+... or z+ ao-}-i' +...,

according as b is finite or infinite. Some of these arcs of logarithmic spirals may degenerate into points. From Theorem 1', we obtain as special cases with 0 = 0 and 0 = 77/2 theorems on the existence of a univalent mapping of an n-connected domain onto the plane with radial cuts and onto the plane with cuts along arcs of concentric circles. 1) Gr&zsch [1931a], Goluzin [1937, 19471.

2) (Koebe [1918].

§3. MAPPING ONTO A HELICAL DOMAIN

221

The correspondence of the boundaries under the mappings referred to in Theorem

is studied in the same way.as in §2. In particular, let us show that, if the domain B has only accessible boundary points, then the mapping is one-to-one in this case also except for points on the boundary of the domain B. On the other band, if the domain B is bounded by closed Jordan curves, the mapping function is continuous in B except at the point z = b. Finally, if the domain B is 1'

bounded by analytic Jordan curves, the mapping function is analytic on the boundary of the domain B. We also have a uniqueness theorem: Theorem 2. There exists only one function carrying out the mapping mentioned in Theorem 1'.

Proof. Let us assume first that the domain B is finite and bounded only by closed analytic Jordan curves K 1, , K,,. Let us suppose that there exist two functions C= f 1(z) and C= f 2(z) that carry out the mapping of the domain B mentioned in Theorem 1 '. These functions are regular on Ki, for v = 1, ,n and, on each of these curves, we have 3 (e

i0 logf1 (z)) = c,

and

(ete log ff (z));= C"'

where c v and c are constants. Therefore, it we set = F (z) = log (f I(z)/f 2(z)), then the function F(z) is regular in B and the values that it assumes on the curves K. for v = 1, , n, lie on a straight line segment. We then see, just as

in the proof of Theorem 2 of §2, that F(z) = const, that is, fl(z)/f2(z) const. But f1(z)/f2(z) - 1 as z b. Consequently, f 1(z) = f 2(z). Now, if B is a finitely connected domain, we can prove the theorem by reasoning analogous to that used in the proof of Theorem 2 of §2. We mention one relationship regarding the mapping functions which is analogous to (1) of § 2. Suppose that B is an n-connected domain bounded by closed analytic Jordan curves K1, ... , K,, and that C= jg(z, a, b) is a function that maps the domain B univalently as indicated in Theorem 1 '. Let us show that the following relationship holds in B for arbitrary 0: log j0 (z, a, b) = ei6 (cos 0 log j0 (z, a, b) - I sin 0 logj (z, a, b)) (2)

222

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

where suitable branches of the logarithms are chosen. The difference d(z) between the two sides of (2) is a regular function in B and the values that it assumes on , n, all lie on some straight line -J(e-Dgt) = const. the curves K,,, for v = 1, Reasoning as in the proof of Theorem 2 of §2, let us show that d(z) const = c, where c is equal to 0 for a suitably chosen branch of d(z). This yields equation (2), which is now proved for a particular form of the domain B. To see this, let us map the domain B onto a domain B. bounded by closed analytic Jordan curves and isolated points. We apply formula (2) to B. and then return to the domain B. As a result, we obtain (2) for the domain B. Some theorems on the existence of univalent mappings of multiply connected domains onto various other canonical domains will be given in §6. §4. Some relationships involving the mapping functions

Equation (2) of §3 enables us to find, for a given n-connected domain B a function je (z, a, b) with arbitrary 6 if we know the functions jo (z, a, b) and jn/2(z, a, b). But we can also establish relationships between these last functions and the functions jo(z, a) and jn/2(z, a) of §2 which correspond to the same domain B. To put these relations in the most symmetric form, we set

f' (z, a) = 2 (J x (z, a) - j o (z, a)), (1)

Q (z, a) = 2 (J x (z, a) + Jo (z, a)) and, analogously,

P (z, a, b) = 2 (log J,, (z, a, b) - log Jo (z, a, b)), 2 Q (z, a, b) =

(2)

(log j x (z, a, b) + log Jo (z, a, b)). 2

These relations then take the forms dz

d

P (z, a, b) = P (b, z) - P (a, z),

-dz-Q(z,

(3)

a, b)=Q(b, z)-Q(a, z).

On the basis of the same principles as in the preceding sections, we can

§4. RELATIONSHIPS INVOLVING THE MAPPING FUNCTIONS

assume that the domain B is finite and bounded by analytic curves KI, K21 K,,. Since (lo (z, a)) = const, 2 (11 w (z, a)) = const, T

on K. (v= 1,

223 ,

, n) for fixed a e B, that is, since jo (z, a) = jo (z, a) + const

j(z, a)_- j, (z, a)+const, 2

we also have on the basis of (1) P (z, a)

Q (z, a) + k, (a),

(4)

on Kv, where kv(a) is independent of z C K. Analogously, since 3 (log jo (z, a, b)) = const,

-3 (11og J. (z, a, b)) = const,

-

5-on Kv for fixed a, b C B, it follows on the basis of (2) that

P (z, a, b) = - Q (z, a, b) + k,, (a, b),

(5)

on K v, where k v (a, b) is independent of z C K v. Now, let us look at the integral 1' - tat } P (C, z) PP (C, a, b) dC,

taken over the entire boundary K = U'= I K, of the domain B in the positive direction. In accordance with (1) and (2), the functions F(j, z) and P (C, a, b) are, for arbitrary fixed z, a, b C B, regular in B. Therefore, in accordance with Cauchy's theorem, 11 = 0. On the other hand, applying (4) and (5), we have n r

1' - r=1 tat J

=

P (C, z) dP (C, a, b)

n

(-Q(C z)+k, (z))d(-Q(C, a, b) + k, (a, b))

I n

v=1

tat

Q (C, z) d Q (C, a, b)

Q (C, z) Qc (C, a, b) dC

224

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

(Here, we used the single-valuedness of the function Q (C, a, b) on each of the curves K u.) Since Q (C, z) and Q'(C, a, b) have simple poles in B at the point z and the points a and b respectively, by applying the residue theorem to the last integral we obtain

Q: (z, a, b) - Q (a, z)+Q(b, z). Remembering that I1 - 0, we then arrive at the second of formulas (3). Let us examine the integral 19 = 2nt - C Q (C, Z) Pi (C, a, b) dC

in an analogous fashion. On the it is obviously equal to PZ (z, a, b). On the other hand, we can show, just as above, that it is equal to P (b, z) P (a, z). Thus we obtain the first of formulas (3). One consequence of formulas (3) is that the derivative Qz (z, a, b) is an analytic function in B of the arguments a and b, whereas Pz (z) a, b) is an analytic function of a and W. Conversely, we conclude from these formulas that

the difference Q (z, a) - Q (z', a), where z and z ' belong to B, is an analytic function of a in B and that P (z, a) - P (z ', a) is an analytic function of a. If we let z' approach z, we see that analogous conclusions hold for the derivatives QZ (z, a) and PZ (z, a).

The last result can be derived in a different way, specifically, from the relations

Pp (z, a) = Pa (a, z), Q- (z, a) = Q. (a, z), which we can prove by examining the integrals 19

r = tat 1 P (C, Z) Qc (C, a) d`, la =tai } P (C, Z) P (C, a) dC

(6)

the same way as above. That the functions P (z, a) and Q (z, a) themselves are not necessarily analytic functions of a is shown by the example when B is the domain Jzl > 1. In this case, we easily see that

_

P(z, a)-1-1 1

z-a

a111-az,

Q(z, a)=1

1 -az

aJ'z-a

§4. RELATIONSHIPS INVOLVING THE MAPPING FUNCTIONS

225

This method of deriving various relationships between functions that we do not know explicitly is called the method of contour integration.') Relations (4) and (5) on the boundary of the domain play the significant role in it. We now point out a curious analytic relationship with the preceding values of the function mapping the domain B onto a circular annulus with concentric circular cuts. Specifically, let us denote by J= f (z) a function that maps a domain B not containing - and bounded by analytic Jordan curves K,,, v = , n, onto the annulus q < ICI < 1 with cuts along arcs of concentric v = 1, circles (see Chapter V, §6, the case 0n/2 of Theorem 7). Suppose that the curves K1 and K2 are mapped onto the circles q and Just as before, we calculate each of the integrals I6

P (C, z, z') (logf (C))i a,

=tai

Ifi = 2x1

S P (C, z) (log f (C)) dC,

(7) (8)

h

where z, z 'E B, by two different methods. Keeping in mind this fact, together with (6) and (7), and the fact that log f (C) _ ]og f (C) +const on K, and

tats (1ogf (C))i dC= 1, tat J (logf (Q) dC=- 1, Kt

we can prove the following formulas:2) f (z) =f (z) eks (z, z') - ki (z,

f (z) = ej kt (z) dz - k2 (z) dz 1) Garabedian and Schiffer [19491 2) In the process, we need to evaluate the integral 2x1

Q G, z, z)(logf (C))i dC,

and, to do this, we need, by virtue of the multiple-valuedness of the function Q(C, z, z to make a cut in B connecting the points z and z ' and then replace the integral over K with an integral over a closed curve y encircling this cut. We then constrict the curve y around the cut. Integration by parts then leads to an integral that can be calculated in accordance with the residue theorem.

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

226

We note that z ' appears in (9) as an apparent parameter since f (z) is, in accordance with Theorem 7 of §6 of Chapter V, determined up to a constant factor. In conclusion, we derive an integral formula involving the function P(z, a, b). For any two functions f (z) and g (z) that are regular in B, we define their scalar product as

(f, g)= S

(11)

Sf'(z)g'(z)dxdy.

B

In accordance with Green's formula

Sf(z)g'(z)dz=sf(z)g' (z)dx -lf(z)g (z)dy K

K

SB S - t

d f (=a g' (z)

-

df(Za

g

Z)

(y

) dx dy = - 21 S S f' (z) g' (z) dx dy, B

J

where z = x + iy. For (f, g) we also have the contour representation g)

=

2i

S f (x)

(z) dz.

(12)

If we apply (12) to the case in which g(z) = P(z, a, b) and remember that f (z) P', (z, a, b) dz

= f (z) d 2 (log j, (z, a, b) -- log j0 (z, a, b)) f (z) d 2 (- log j x (z, a, b) - log jo (z, a, b)) V

_-Sf(z)dQ(z, a, b)=21cl(f(b)-f(a)), K

we obtain the desired formula (f (z), P (z, a, b)) = a (f (a)

- f (b))

We point out two applications of this formula. The first of these is the

(13)

§4. RELATIONSHIPS INVOLVING THE MAPPING FUNCTIONS

227

Theorem (minimization of area). 1) Out of all functions f (z) that are regular in B and satisfy the conditions f (a) = q f (b) = 1, where a, b E B, the minimum value of the integral

(f, f)=SSIf'(z)I9do, B

which is equal to n/P(b, b, a), is attained only by the function (z, b, a)

fo (z) = P

P (b, b, a)

Proof. If we represent an arbitrary function f (z) of the type described in the theorem in the form f (z) = f0 (z) + (z), we conclude that the function 0(z) is regular in B and that 0 (a) _ o (b) = 0. For this function, we have in accordance

with formula (13) (gyp (z), P (z, b, a)) = 0.

Therefore,

(f, f) _ (fo, fo) + (p, fo) + (P, fo) + (y, F) _ (fo, fo) + (,P,

P)

It follows that

(f, f) with equality holding only when completes the proof of the theorem.

(fo, fo),

0, that is, only when f (z) - f o(z). This

As our second application of formula (13), consider a sequence of functions

k(C) such that kt,(b) = 0 for v = 1, 2,

,

that are regular in $ and satisfy

the orthogonality conditions 1

for v=v' v

l 0 for v

v' = 1

2 .. .

v'

and have the property that an arbitrary function f (z) with f (b) = 0 that is regular in B can be expanded in a convergent Fourier series in B: 1) (The problem of the minimum area of the image of a finitely connected domain has also been solved for the class of all functions that are regular in the domain that have at given points of it given initial segments of their Taylor expansions. See Alenicyn [19641).

228

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS Co

f (z) = 2] cyky (z), c,

V-j

_ (f,

k0)

(It can be shown that such sequences of functions

exist.) Let us apply this to the function f (z) = P(z, a, b), where a and b belong to

B. Since, in accordance with (13), we have

cy = (P (z, a, b), k0 (z)) _ ik., (a), we obtain for P(z, a, b) the constructive formula 00

P (z, a, b) ='t }J k, (a) k, (z)

§5. Convergence theorems for univalent mapping of a sequence of domains Let us now consider the question of the convergence of a sequence of functions that are univalent in multiply connected domains in connection with suitable convergence of these domains and their images.

As a preliminary, we introduce the following concepts. Let {BJ, for n = 1, 2, , denote a sequence of domains in the z-plane that include the point z = on. We define the kernel of the sequence { Bn { as the largest domain B including z = on every closed subset of which is contained in each Bn from some n on. Obviously, for a kernel to exist, it is necessary and sufficient that all the domains Bn contain a fixed neighborhood of the point z = -a. We shall say that a sequence {Bn} of domains Bn converges to its kernel B (and we shall write Bn --+ B) if an arbitrary subsequence of these domains has the same kernel B. , then the sequence of the domains Bn has a For example, if B1 C B2 C B3 C kernel and converges to it. 'Theorem 1. Let {An) denote a sequence of domains An, n = 1, 2, . . , in the z-plane that include the point z = on. Suppose that this sequence converges to a kernel A. Let { fn(z)} denote a sequence of functions C= fn(z) such that, for each n = 1, 2, , the function fn(z) maps the domain An onto a domain Bn including the point C_ on in such a way that fn(o) = on and fn W = 1. Then, for the sequence .{fn(z)1 to converge uniformly in the interior of the domain1) A to a 1) The existence of a pole at z = co is not significant in the present case because uniform convergence of the sequence of the functions fn(z) is equivalent here to uniform convergence of the sequence of the functions fn(z) - z.

§5 CONVERGENCE THEOREMS

229

univalent function f(z), it is necessary and sufficient that the sequence IB"I have a kernel and converge to it, in which case the function C= f(z) maps A univalently onto B. We note that although the functions are defined at certain points z E A only from some n on, this does not destroy the sense of convergence at such points. Proof of the necessity. Suppose that the sequence i fn(z)y converges uniformly inside the domain A to a univalent function f W. Then f (z) is regular in A except at the point z = 00, where f (z) This maps the domain A onto a domain B' containing 00. has a kernel and that B' is contained Let us show that the sequence in that kernel. It will be sufficient to show that an arbitrary closed bounded domain B * contained in B' belongs to every Bn from some n on. Let 8 > 0 denote the distance from R* to the boundary of B'. Let us construct a grid of squares with sides of length 8/2. Consider the domain B. constituted by the union of all squares containing points in B. Then B` C B. and $. C B'. Let 81 > 0 denote the distance from $* to the boundary of B.. Corresponding to the domains B' and B. under the function = f (z) in A are domains we have A* and A.. Here, A* C A. and A. C A. But, for any point CO E If U) - Col > 81 on the boundary of the domain A.. On the other hand, it follows from the uniform convergence of the functions fn(z) in J. that there exists an n > 0 such that Ifn (z) - f (z)I < 81 on the boundary of the domain A. for n > N. Consequently, from the equation fn (z) - CO = V. (z) - f (z)) + V (z) - co)

we conclude on the basis of Rouche's theorem that the function fn(z) - Co has one zero in A., that is, that for n > N the image of the domain An under the mapping C= fn(z) contains an arbitrary point Co E B. This in turn shows that the sequence I Bn I has a kernel, which we denote by B, containing B' Now, let us consider the functions z = (An(C) inverse to the functions C- fn(z). For any closed subset of the domain B, these functions are defined on it from some n on. For sufficiently large positive r, the functions On(0 are all defined and univalent in the domain ICI > r and in that domain they have the expansions

230

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS Vin)

/

J

Consequently, from Lemma 2 of §2, we have in ICI > I'Pn(C)-Q( n)I{2ICI, 0

so that Or. C2-F Ia)I

But the convergence of the sequence (&(z)} to a function f (z) to a function f (z) implies that the sequence {a0")} converges to a finite limit. Therefore, the functions On(O/C are regular and bounded in the domain ICI > r and the condensation principle can be applied to them, so that the sequence has a subsequence (0"k (0/0 that converges uniformly in the interior of the domain ICI > r to a limit function O(C)IC. On the other hand, the sequence 10"k (:)} converges uniformly to O(C) in an arbitrary bounded subset of the domain 11 > r. On the other hand, for all C E B" belonging to the disk ICI < 2r, we have, in accordance with .Lemma 2 of § 2, 1 yn (C) - m(on) I< 4r.

Consequently, the functions q"(t) are uniformly bounded in every closed subdomain

of the domain B that is contained in the disk ICI < 2r. Therefore, in accordance with Vitali's theorem, the sequence { 011k (0} converges uniformly to O (C) in the interior of the domain B exclusive of the point C = oo, and 0 (c) is univalent in B. Reasoning word for word as on p. 57, we can show that the function z = c¢(C) is the inverse of the function C= f W. It follows that 4(c) is independent of the sequence chosen, that is, that the sequence h itself converges in B. Thd function z = O(C) maps the domain B onto a domain A'. Reasoning as at the beginning of the proof but reversing the roles of the domains A" and we can show that A 'C A.

Now, it is easy to show that B '= B. If Co E B, then z o = O(to) C A and, consequently, Co = f (z o) E B '. This shows that B C B '. But we showed earlier that B 'C B. Consequently, B '= B. Thus, the function C= f (z) maps A univalently onto B. If we now take an arbitrary subsequence of the sequence 1Bn} and the corresponding subsequence of the sequence I f4 (z)}, which also converges to f (z), then by applying to these subsequences the conclusions derived above, we see

§ 5. CONVERGENCE THEOREMS

231

that the function C = f (z) maps A onto the kernel of this subsequence, which again is the domain B. Thus, Bn - B. This proves the necessity of the condition in the theorem. B. Let p denote a positive Proof of the sufficiency. Now suppose that Bn number such that the domain IzI > p is contained in each of the domains An and z + (n) + > r is contained in each Bn. If suppose that the domain ain)/z + in a neighborhood of z = 00 and if c does not belong to the domain Bn, then Ic - a(On'I _< 2p by virtue of Lemma 2 of § 2. On the other hand, I c I < p. Consequently, Iaon I _< 3p; that is, the coefficients a0(n) are uniformly bounded

with respect to n. Again in accordance with Lemma 2 of §2, we have 1fn(z) - ao")I < 2IzI in IzI > p, so that Ifn(z)I < 2IzI + 3p. In particular, Ifn(z)I < 5p on IzI = p. By virtue of the univalence of the fn(z), this last inequality remains valid at all points of An belonging to the disk IzI < p. Comparing the bounds on Ifn(z)I in the disk IzI < p and in the domain IzI > p, we conclude that Ifn(z)I < 2IzI + 5p in every domain An. This shows that for every closed bounded subset of A, all the functions fn(z) from some n on are defined and uniformly bounded on it. Consequently, the condensation principle can be applied to an arbitrary sequence of these functions. Let us show now that the sequence { fn(z)} converges in A. Suppose that there exists a point z0 E A such that 1fn(z0)} diverges. Then, there exist two subsequences of functions that converge at z = z0 to distinct limits. Furthermore, these two subsequences themselves contain subsequences { fn,(z)} and { fnu(z)1 that converge uniformly in an arbitrary closed subset of A to functions f.(z) and f..(z), that are regular and univalent in A except at z = 00, and that have different values at z0. It follows from our proof of the necessity of the condition of the theorem as applied to the sequences {Bn,} and {Bnu} thai the functions f .(z) and f ..(z) map the domain A onto the kernels of these last sequences, that is, onto the same domain B. The function

z' =f* ' (f,

,

(z)) = + (x)

maps the domain A into itself and ifi(z)i z. Suppose that >fi(z) has, in a neighborhood of z = 00, the expansion

where a,,, for v > 0, is the first nonzero coefficient (which necessarily exists).

232

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

Consider the functions 02(z), 03(z), the expansions

defined in a neighborhood of z = oo by

W_3(z)_ (2(z))=z+3y+ .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

These functions also map the domain A univalently into itself. Since the domain IzI > p is contained in A, the functions t/i(z), for n = 2, 3, , are, in particular, univalent in IzI > p and the functions !(pz)/p = z + na,/p"+lzv are valent in IzI > 1. It then follows as a consequence of the area theorem that V+I I < 1, n = 2, 3, , which can be the case only if a = 0. This conIna ,,/p

tradiction shows that ,(z) = z, that is, that f.(z) = f..(z). This last fact in turn shows that the sequence 1 f(z)] converges in z0. Thus, the sequence 1f (n)1 converges everywhere in the domain A. The uniform convergence of this sequence inside the domain A now follows from Vitali's theorem. This completes the proof of the sufficiency of the conditions of the theorem and hence the theorem itself. One can also prove other theorems on the convergence of sequences of univalent functions on the basis of the manner in which they are normalized. We shall give one of these theorems. Just as above, we shall first introduce the following concepts: Let 1B,I , where n = 1, 2, , denote a sequence of domains contained in the z-plane: Suppose that a given point z0 belongs to each of the domains B. We define the kernel of this sequence with respect to zo as the largest domain B containing the point z = z0 and having the property that an arbitrary closed domain contained in it belongs to every B from some n on. Convergence to a kernel is defined just as before. Obviously, the kernel of the sequence of images of the domains B under a fractional-linear mapping of the z-plane is the image of the kernel B (of course, with respect to the image of the point z0). This does not violate convergence to the kernel. Furthermore, if we perform similarity transformations and then let p approach 1, the z '= pnZ on the domains B for n = 1, 2, kernel of the sequence 1B} remains unchanged and convergence to it is not violated. We shall now prove a convergence theorem. Theorem 2. Suppose that a point z = z0 belongs to each of a sequence of

§5. CONVERGENCE THEOREMS

233

domains An, for n = 1, 2, , in the z-plane. Suppose that the sequence }An} converges, with respect to z0, to the kernel A. Suppose that the functions C= fn(z) map the domains An univalently onto domains Bn in such a way that fn(zo) = Co and fR (zo) > 0 for n = 1, 2, . For the sequence {fn(z)} to converge uniformly inside the domain A to a univalent function, it is necessary and sufficient that the sequence (Bn} have a kernel with respect to 4 and converge to it. Then the limiting function f (z) maps A univalently onto B. We may assume that z 0 = Co = 0 because we can always arrange for this by making

translations of the z- and

planes. The proof of the first part of the

theorem is now based on Theorem 1.

Proof of the necessity. If fn(z) -' f (z) uniformly in the interior of A and f (z) is univalent in A, then f (0) f'(0) 0. Consequently, the sequence of the functions In (0)

Fn(z) =

1

In (z)l which are univalent in the domains An obtained from An by means of the mapping z '= 1/z, converges uniformly inside the corresponding kernel A'. Therefore, on the basis of Theorem 1, the sequence of the images Bn of the domains An under the mappings C= F,(z) converges to the kernel B'. Consequently, the sequence }BnI converges to a nondegenerate kernel B connected with B' by the relationship J'= The conclusion on the nature of the mapping C= f (z) follows. Proof of the sufficiency. Suppose that Bn --. B. Then, in the same notation, the sequence {Bn} has the kernel B' and it converges to that kernel. Consequently the sequence {Fn(z)} defined by

F.

(z)

=

fn (0)

fn(Z)

converges uniformly in the interior of the domain A ' to a univalent function F (z) such that F(oQ) _ o and F'(00) = 1. But then the sequence { fn(z)} defined by

fn z) =

fn (0)

Fn(z)

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

234

converges uniformly in the interior of the domain A to the univalent function

f(z)=

Ci

F(Z)

,

C/Q,

which maps A onto the kernel B of the sequence of the domains B. This completes the proof of the Theorem. §6. Univalent mapping of multiply connected domains onto circular domains. The continuity method

In §§2 and 3, when we were solving certain extremal problems, we proved theorems on the existence of a univalent mapping of multiply connected domains onto canonical domains. In the case of finitely connected domains, an incomparably stronger method of proof is the continuity method. The proofs that use this method are extremely similar to each other. They are based on the uniqueness of a univalent mapping of finitely connected domains onto the canonical domains in question and on a single topological theorem of Brouwer. We shall illustrate this method by proving an important theorem on univalent mapping onto circular domains, that is, domains bounded by a finite number of complete circles with no points in common (some of the circles may degenerate to points). We first prove a uniqueness theorem for circular domains. Theorem 1. There exists no more than one function that maps a given finitely connected domain B in the z-plane including - univalently onto a circular

domain in the -plane, also including -, and that has the expansion C= z + in a neighborhood of the point z = --. One can easily see that this theorem is equivalent to 't'heorem 1 '. Every function that maps a circular domain B (finitely connected) univalently onto another circular domain B' is a fractional-linear funcaiz-1

+

tion.

Since the two theorems are equivalent, we need only prove Theorem 1

,.

Proof. Suppose that the domain B is contained in the z-plane and is bounded by the circles K1, K21 , K,,, that the domain B' is contained in the t; plane and is bounded by circles K I', K2' , , K, and that the mapping in question maps Kv onto K v (v = 1, 2, - , n). Without loss of generality, we may assume that the domains B and B' contain the point w and that this point is mapped into itself.

§6. MAPPING ONTO CIRCULAR DOMAINS

235

Denoting the mapping function by 4 _ f (z), we then have f (oo) = -. The function f (z) can be extended across the circle K,, (v = 1, 2, , n) into the domain B,, obtained from B by inversion about K,,. It maps B univalently onto the domain B, obtained from B' by inversion about K. Consequently, the function , B unif (z) maps the circular domain consisting of the domains B, B1, valently onto the circular domain consisting of the domains B', B j , , B,. We reason analogously with these extended domains and continue the process indefinitely. As a result of such extensions in the z-plane, we obtain a limit domain G, which the function C= 1(z) maps univalently onto a limit G' in the 4 -plane. Let z '= S (z) denote a fractional linear function composed of an even number of inversions about the circles K1, K29 , K,, taken in any order. Then, the function f (S(z)) also maps the domain G onto the domain G' univalently since the function

S(z) maps the domain G into itself. Consequently, [(z) is invariant under the group of fractional-linear transformations z '= Sk(z), for k = 1, 2, . . , composed , K,,. The functions of a finite number of inversions about the circles K1, K21 z '= Sk(z), for k - 1, 2, .. , exclusive of the function z ' - z map the domain B , that have no common onto disjoint circular domains B(k), for k = 1, 2, , K,,. boundary points and lie each inside one of the circles K1, K21 Consider a disk Iz - al < r contained in B. Then, in accordance with Koebe's covering theorem, the domain B(k) (for k = 1, 2, - ) contains a disk of radius (r/4)ISk (a)I Consequently, the area of the domain B(k) is not less that Since the sum of the areas of all the domains B(k ), for k = n (rI Sk 1, 2, , is finite, we see that the series (a)1/4)2.

I Isk(a)I9

k=1

(1)

converges. If we now denote by Kk, for k = n + 1, n + 2, , all the circles that serve as boundaries for the domains B(k), for k = 1, 2, , and if we denote their radii by Rk for k = n + 1, n + 2, , we can show that the series

I Rk

(2)

k-n+ t converges. To see this, let us delete from the domain G the images of the point z = 00 under the mappings z '= Sk(z), k = 1, 2, . Then all the functions Sk(z) are regular and univalent in the remaining portion G* of the domain G. Therefore, from the general distortion theorem (see §4 of Chapter II), we have on K

236

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

(v= 1, 2,---,n) for all k= 1,

ISk(z)IM,ISk(a) where .tl depends on v but not on k or z. Therefore, we get the following expression for the circumference 2rrRk v of the circle Kk v onto which the function z '= Sk(z) maps Kv: 21rRk., =S Ky

K,

from which it follows that

Rky-MA IS,(a)I. If we square both sides of this inequality and then sum over all k and all v, recalling that each circle Kk is the image of a boundary circle of the domain B under exactly one of the functions z '= Sk(z), we obtain co

n

m

E R42 E M.Rv- Y Sk(a)I®

k=n+1

k=I

v=1

This proves the convergence of the series (2). Analogous conclusions hold for G'. Consequently, if we denote by Rk the radii of the circles into which the circles Kk, for k = n + 1, n + 2, , are mapped by the function C= f(z) defined on G, we conclude that the series 00

Y

Rk

k=n+I converges. It then follows, in particular, that the series co

E Al

k=n+l

(3)

converges, where Ak is the oscillation of the function f(z) on Kk, that is, Ak

max

If(z')-f(z,)I.

Let us now denote by G. a circular domain contained in G that contains B and that is bounded only by the circles Kk,, Kk,,, ... , Kk(,,,)

with indices k', k ", .. , k(-) sufficiently large that

§6. MAPPING ONTO CIRCULAR DOMAINS

237

v=I

where a is a given positive number. This is always possible by virtue of the convergence of the series (2) and (3). Now, for a sufficiently large circle K: lzl = R, we have, in accordance with Cauchy's formula, f (z')

1

1

z'-dz z - v=1 tai

f(z)=2ni

f (z')

z'-z dz' Kk(r)

(4)

for any point z belonging both to G. and the disk lzl < R. But f (z) dz' =

1

J z' - z

2ni

1

2tci

k(,)

C f (z') -f (z) dz' J

z' - z

k(v)

Now let d denote the distance from z to the boundary of the domain G.. It follows from what was just shown that m

m f (z

1

z' -z

I.d 2ni

v=1

Kk(v)

dz'

C

I "Al" d v=1

R,,(v)

m

M

Rk(r) `

okh) v-1

d'

v=1

But the left-hand member and the first term of the right-hand member of (4) are independent of the domain G. chosen and hence of e. Therefore,

f (z) =tai1 ` z'-z f (z) dz'. This shows that the function f(z) is regular in the disk lzl < R for arbitrarily large R, that is, that f(z) is an entire function. Since this function has a simple pole at z = -, we conclude that f(z) is a linear function. This completes the proof of Theorem I'. This existence theorem which we now wish to prove consists in the following:

Theorem 2.1) Every n-connected domain B in the z-plane can be mapped 1) This theorem was proved by Koebe. The continuity method was developed in his article [1918a] It has been simplified by the present author (see Goluzin [1938a]} Another proof has been given by Keldys [1939].

238

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

univalently onto a circular domain in the cplane. Among these mappings there is only one normalized mapping that maps a given point z = z E B into C= - and has an expansion in a neighborhood of the point z = a of the form z

a+al(z-a)+... or z+ z +...,

according as a is finite or infinite. Before proceeding with the proof of this theorem, let us recall a few concepts from n-dimensional geometry. Consider n-dimensional Euclidean space, in which a point is an n-tuple of finite real numbers (x1, x2, , xn). We denote this space by Rn and we denote a point in it by a single letter, for example, x or by an expression of the form x(xl, ... , xn) indicating in parentheses the specific real numbers defining it, these numbers being called the coordinates of the point x. Two points are thought of as coinciding if and only if their corresponding coordinate are equal. The , x(')) and x(Z)(x(Z), distance between two points x(1)(x(1), , x(2)) in the n n 1

1

-x(z))z]1/z An space Rn is defined as the number p(x(l)x«)) ' k1 (xCI) k k open ball of radius p with center at a is defined as the set of all points x E Rn whose distance from a is less than p. A neighborhood of a point a E Rn is defined as any open ball with center at a. A neighborhood is sufficiently small if the radius of the ball is sufficiently small. A set E C Rn is said to be bounded if it is contained in some ball. A point a E R is said to be a cluster point of a subset E of Rn if every neighborhood of the point a includes at least one point of E other than a. A sequence of points 5(m) E Rn, for m = 1, 2, , is said to converge to a point x(0) c R if p( (m), x(0)) --p 0 as in --. oc. A point a belonging to a set E C Rn is called an interior point of E if some sufficiently small neighborhood of a is contained in E. A set E C Rn is said to be closed if it contains all its cluster points. A set E C R is said to be open if it consists only of interior points. It is said to be connected if any two points in it can be connected by a continuous curve contained in E. By a continuous curve in Rn, we mean a set of points in the space R the coordinates of which can be repre, n, where each 01, (t) sented in the form xk = q,k (t) for a < t < b and k = 1, is a continuous function of t in the interval a < t < 6. All the properties analogous to the properties of point sets in a plane that were expounded in the geometrical introduction can be established for point sets in the space Rn. Now, consider a mapping of a subset E of Rn onto a subset E' of another

§6. MAPPING ONTO CIRCULAR DOMAINS

239

n-dimensional Euclidean space R,,. This means that to every point in the set E is assigned a point in the set E'. The mapping is said to be one-to-one or bijective if to every point of E there corresponds exactly one point of E' and vice versa. The mapping is said to be continuous if, for every sequence lx(')[ of points in E that converges to a point x(') in E, the sequence of images of these points converges in E' to the image x '(0) of the limit of the original sequence. It is easy to show that the inverse of a bijective continuous mapping of a closed set E C Rn onto a set E 'C Rn is also continuous. A one-to-one continuous mapping is also called a topological mapping. Obviously, the image of a closed set under a topological mapping is a closed set. The following theorem of Brouwer on topological mappings of arbitrary sets is known as the theorem on preservation of a domain. Theorem 3 (Brouwer). The image of an interior point of a set E C Rn under a

topological mapping is always an interior point of the image set E'; equivalently, the image E' of an open set E C Rn under a topological mapping is an open set. We postpone the proof of this important theorem to the following section. The proof of Theorem 2 by the continuity method is based on it. Proof of Theorem 2. By virtue of the results of § 2, any finitely connected domain can be mapped univalently onto the plane with parallel straight-line cuts at an arbitrary given inclination 0 and with suitable normalization. Therefore, it will be sufficient to prove Theorem 2 for domains constituted by the z-plane with rectilinear cuts parallel to the real axis and for a = .. Let us denote by 9 the family of all n-connected domains B consisting of the z-plane with n finite rectilinear cuts parallel to the real axis (numbered from 1 to n) and let us denote by 9R' the set of all n-connected circular domains B' in the cplane with boundary circles numbered from 1 to n. Two identical domains B and B' wi th boundaries numbered in a different manner are considered as distinct. Let us consider normalized (as indicated in the theorem) univalent mappings of domains B E 9R onto domains B 'C 91' such that identically indexed boundary curves of the domains B and B' are mapped onto each other. We consider domains in the families 911 and 9' as corresponding to each other if they can be mapped one onto the other as indicated. Then, from Theorem 1 of the present section and Theorem 2 of § 2, we conclude that

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

240

(a) to every domain B E 9 there corresponds no more than one domain B' E 9R',

(0) to every domain B' E 91' there corresponds a unique domain B E 9. Our problem is to show that to every domain B E 9R there corresponds exactly one domain WE W. The families 9 and 21 consist of domains defined by 3n parameters. For the parameters of the domain B we may take, for example, the coordinates of the origins of the cuts and their lengths. For the parameters of B' we may take the coordinates of the centers and the radii of the boundary circles. If we represent the domains B and B' by points in 3n-dimensional Euclidean spaces Ran and R3n and denote these points also by B and B', we conclude immediately from geometrical considerations that the point sets B and B' corresponding both to the domains B E 9 and to the domains B'E 9R' (and we shall denote these also by .9 and 9R') are connected open sets in Ran and Ran respectively. If we set up between the points of the sets 11 and 9' the same correspondence as exists between the domains of the families 3 and 9R', we conclude on the basis of (a) and 09) that to every point B 'E R' there corresponds exactly one point B E 2 and that to every point B E 2 there corresponds no more than a single point B' E W. Let us show that this correspondence is continuous, specifically that B1 --. B as Bi B' (that is, as the boundary circles of the domain B1' ap,proach coincidence with the boundary circles of the domain B '). Here, B1 corresponds to B11 and B corresponds to B'. If a sequence of the domains B' did not converge to a limit domain, we could choose from it. two subsequences }Bn 1}

and IBn2} that converge to the domains B and B0, where B I B0. On the basis of the convergence theorem, we conclude that the domain B' is mapped normwisc onto B and also onto B0, which, in accordance with (3) implies that B = B0. Here, we assumed that the boundaries of all the domains B1 are bounded, but this follows from Lemma 2 of §2. We mention also the following property of the limit domains: if B. E 9 and B. and the B,, if there exists a sequence of domains Bn E 91 such that Bn can be mapped normwise onto Bn E 9', then the domain B. can be mapped normwise onto B: E 99'. This is true because the sequence of the domains B converges to a domain B.' E 9R' since otherwise would there would be two subsequences }B,,'1} and }B'2} that converge respectively to domains B . and Bo,

§6. MAPPING ONTO CIRCULAR DOMAINS

241

where B: Bo. It then follows on the basis of the same convergence theorem that B; = Bo. But if B --> B. and B,,' B:, then we conclude, again on the basis of the convergence theorem, that the domain B. can be mapped normwise onto the domain B., which proves the assertion made. From this point, the proof of the existence theorem is fairly simple. We are

now dealing with a one-to-one continuous mapping of the open set Dl' of points in 3n-dimensional space R3- onto a subset (proper or otherwise) of the set Dl in another 3n-dimensional space R 3n- Consequently, Brouwer's theorem can be applied. According to this theorem, the image Do of the set 9R' is an open set in 3n-dimensional space. Let us show that Dto coincides with 9. Let B 1 denote a domain in the family Di that can be mapped onto the domain B' in the family V'. Such a domain can be obtained by a mapping of an arbitrary domain in 9' onto the domain in Dl corresponding to it. On the other hand, let B 2 denote an arbitrary domain in V for which we have not yet determined whether it can be mapped. Let us connect the point in the space Ran representing the domain B2 with the point representing the domain B1 by a continuous curve L contained in Dl . Let us proceed along this curve from the point B 1 to the point B2. If the domain B2 could not be mapped onto a domain in Dt', then, as we proceed along L from B1 to B21 we would encounter a point B. such that all the domains in DI corresponding to points of the curve L between B1 and B. could be mapped onto domains in D2' and the domain B. either could not be mapped onto domains in DI' or would be a cluster point of the set of points of L corresponding to domains that cannot be mapped onto domains in Df'. But this is impossible because, by virtue of the property of the limit domains mentioned above, the domain B. can be mapped onto a domain DI' and hence the point B. belongs to Dio, but then it cannot be a cluster point of domains that cannot be mapped onto domains in Dl' because go is an open set. Thus, all points of the curve L from B1 to B2 belong to the set D10. But then, again on the basis of the property of limit domains, B2 E DI0.

Thus, we have shown that the set go coincides with 9R and, consequently, for every domain in R there exists a domain in Dl' onto which the first can be mapped univalently and normwise. This completes the proof of Theorem 2. Since the proof that we have given used only the uniqueness theorem and a few trivial properties of circular domains, the same procedure can be used with

242

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

equal success in proving theorems on the existence of a univalent mapping of finitely connected domains onto various other canonical domains as soon as we have proved the corresponding uniqueness theorem for them. For example, we can prove in this way the following theorems of Koebe, 1) which generalize Theorems 1 and 1 ' of § 2 and Theorems 1 and 1 ' of §3:

Theorem 4. For a given point a E B, there exists a unique function mapping an arbitrary given n-connected domain B in the z-plane with boundary continua K1, , K univalently onto the cplane cut by n rectilinear cuts with given inclinations B1, , On to the real axis in such a way that the continua K,,, for , n, are mapped respectively into the cuts with inclinations 8,,, the v = 1, point is mapped into C= oo, and the expansion of the mapping function about z = a has the form z

1 a+a1(z-a)+...

or

z+ z +

according as a is finite or infinite. Theorem 5. Every n-connected domain B in the z-plane with boundary con, K can be mapped univalently onto the -plane with n cuts along tinua K1, arcs of logarithmic spirals of inclinations 01, , On respectively to the radial , n, are mapped directions in such a way that the continua K,,, for v = 1, respectively into arcs of inclination B,, given points a and b in B are mapped into 0 and - respectively, and the expansion of the mapping function about z = b has the form

Z-b

ao+at(z-b)+...

or

z+aro+ Z -...,

according as b is finite or infinite. The uniqueness of the mapping functions referred to in these theorems is proved in just the same way as Theorem 2 of §2 and Theorem 2 of §3 were proved.

By the continuity method, Koebe [1918a] proved, in particular, the following theorems: Theorem 6. Every n-connected domain B contained in the disk JzJ < 1 that 1) See Koebe [1918a] and Goluzin [1938a] Krylov [1938] has given a different proof of these theorems.

§6. MAPPING ONTO CIRCULAR DOMAINS

243

includes the point z = 0 and has the circle IzI = 1 as one of its boundary continua

can be mapped univalently onto the disk I < 1 with cuts along arcs of logarithmic spirals of given inclination B in such a way that zl = 1 is mapped into I = 1 and the mapping function C= f (z) is normalized by the conditions f (0) =0 and Oo) > 0. This mapping is unique. Theorem 7. Every n-connected domain B contained in the annulus q < IzI <1 that has as two of its boundary continua the circles Izl = 1 and zl = q can be mapped univalently onto an annulus q '< ICI < 1 with cuts along arcs of logarithmic spirals of given inclination B in such a way that zI = 1 and IzI = q are mapped into ICI = 1 and ICI = q '. This mapping function is unique up to a constant factor. We note that Grotzsch succeeded in proving analogous uniqueness theorems for extremely broad classes of canonical domains and at the same time established by the continuity method theorems on the existence of a univalent mapping of arbitrary finitely connected domains onto such domains (cf. Grotzsch [19350. We conclude our investigation of questions concerning univalent mappings of multiply connected domains with a consideration of the possibility of univalent mapping of any two given n-connected domains onto each other. In the case of simply connected domains, Riemann's theorem tells us that, with very few exceptions, this is always possible. The situation is different in the case of n-connected domains, where n > 1. For it to be possible to map two such domains univalently onto each other, it is obviously necessary and sufficient that there exist two circular domains onto which they can each be mapped univalently and which can be univalently mapped one onto the other. We can take these circular domains

of a "special" form, specifically, annuli of the form q < z( < Q with n - 2 circular cuts in the interior since an arbitrary circular n-connected domain can always be mapped onto a domain of such a form by means of a suitably chosen fractional-linear function. But if two circular domains of the special form can be mapped univalently onto each other, then by Theorem 1 ', the mapping function is fractional-linear and, according to the symmetry principle, it maps the points 0 and = into each other. Consequently, this function is of the form z '= az or z '= a/z. In the first case, the following quantities must coincide for the two circular domains that we are considering: the ratios of the radii of all corresponding boundary circles, the angles between the rays extending from the origin to the centers of the circular cuts, and the ratios of the distances of these centers

244

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

from the origin; that is, the values of one quantity in the case n = 2 and of all 3n - 6 independent quantities determining these domains in the case n > 2 must be the same for the two domains. In the second case, an analogous situation holds for one of the circular domains in question and for the domain obtained from the other by the transformation z '= l/z. Conversely, if these requirements are satisfied, the two domains can obviously be mapped univalently onto each other. Thus, for it to be possible to map two n-connected circular domains of a special form onto each other, it is necessary and sufficient that, in the case n = 2, one, and in the case n > 2, 3n - 6 suitable real quantities determining one of these domains coincide with real quantities determining the other. These quantities are called the moduli of the circular domains in question and of all the domains that are mapped univalently onto them.

If we partition all n-connected domains into classes, putting into a single class all domains that can be mapped univalently onto each other, then each such class will contain circular domains of a special form and, in accordance with what was said above, all classes of n-connected domains constitute a set depending in the case n = 2 on one and in the case n > 2 on 3n - 6 real parameters. §7. Proof of Brouwer's theorem I)

To prove Brouwer's theorem as stated in the preceding section, we need some additional information regarding the space Rn. , x satisfying a given linear A set of points x E R. with coordinates x 1, equation a,x, + a,x, +... + anxn = b,

where al, a2, .. , an are not all zero, is called an (n - 1)-dimensional plane or simply the plane Rn_1. The set of all points common to two (n - 1)-dimensional planes the left-hand members of the equations of which are linearly independent constitutes an (n - 2)-dimensional plane in Rn. In general, the set of all points common to k planes under the same condition constitutes an (n - k)-dimensional plane in Rn. An (n - k)-dimensional plane in Rn can be regarded as an (n - k)dimensional Euclidean space. Let x(k) (x ik ), 1) (Sperner [1928].)

,

xnk )), k = p, 1,

, n, denote n + 1 points in Rn that

§7. PROOF OF BROUWER'S THEOREM

do not lie in a single plane. The set of points X (X 1,

. ,

245

x,) whose coordinates

are determined from the formulas n

t =1,2,...,n,x1 o,

n

1j=1,

(1)

is called an n-dimensional simplex and the points x(k), 4=0, 1, , n, are called its vertices. 1) Let us denote this simplex by 2. The numbers Ai in (1) are single-valued continuous functions of x1, , xn. To see this, remember that, since the points x(k), k = 0, 1, , n, do not lie in a single plane, the determinant

cannot be equal to 0. This means that the system of n + 1 equations in (1) is a system of n + 1 linear equations in the Al that has a unique solution. Consequently, the Al are obtained as linear functions of the coordinates xi, for , n; hence, they are continuous functions of the xi. The set of points i = 1, of the simplex E that are obtained from (1) when one of the Aj is equal to 0 is called an (n - 1)-dimensional face or simply a face of the simplex E. In general, the set of points of the simplex E that are obtained from (1) if k of the numbers Ai are equal to 0 is called an (n - k)-dimensional face. An (n - k)-dimensional face of a simplex E is itself a simplex in (n - k)dimensional Euclidean space, specifically the (n - k)-dimensional plane passing through those vertices of the simplex E that lie on that face. The point x E obtained from (1) for Ao = Al = -' . = An = 1/(n + 1) is called the center of the , xn simplex E. By virtue of the continuity of the Ai as functions of x 1, (which we showed above), it follows that an arbitrary point x E E. obtained from (1) for all nonzero A is an interior point of the simplex E. Let us show that a simplex I can be partitioned into a finite number of simplexes the diameters of which (the diameter of a figure being the greatest distance between any two points of it) is less than an arbitrary number e > 0 in 1) In the case n = 1, the simplex is a line segment; for n = 2, it is a triangle; for n = 3, it is a tetrahedron.

246

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

such a way that, if two subsimplexes have points in common, then the set of all common points is a common face of some dimension or else a vertex of these simplexes. Such a partition can be made as follows, for example: let us look at all possible permutations jo, j 1, , jn of the numbers 0, 1, , n. Let us the set of points x E I that have the representation (1) denote by with ai0 < ai 1 < . < ain. _Let us show that Yio,i 1,...,in is a simplex. Let 6(k) (6 ik ), , n) denote the center of an (n - k)-dimensional , Cnk )) (k = 0, 1, face of the simplex E with vertices at x0k), xUk+1), En , x(in). Then,

Elk]

_ vE= kXV')

n-k+1

1= 1, ...

,

(2)

n.

From (2), we have XiUv1=(n

(n-v)E(v+11,

n-1'

Xi'n1 Consequently,

Xi

X

=(n-{-1)X10,

X, = (n -o -{- 1) (A/v - A1, 1), n

21 XV = v-0

' = 1, ...

,

n,

n

v=0

X,

v-0

It follows that a point x belongs to i o.i I i.. tin if and only if it belongs to the simplex with vertices 6(0), 6(1), . . . , 6(n). Therefore, the set 2io-i 1,...,i n coincides with this simplex. The simplex I is therefore partitioned into (n + 1)!

simplexes lio.il,...'in.

Two simplexes and lio.i{,...,in have in common only those points x for which there are equal numbers among the numbers A0, A1, ' ' ' , An, that is, for which certain of the All are equal to 0. These points constitute the common faces of different dimensions of the simplexes

Jn and

Tn particular, all these simplexes contain the

§7. PROOF OF BROUWER'S THEOREM

247

point 60). Let us now look at the question of the dimensions of the simplex keeping the notation given above for it. Let d denote the greatest ')I for i = 1, , n. We then have , n and of the numbers jxi') 0, n

(xi1 -xi1'))I_n+l'

n,

1=0

, n, must be equal to 0). (since one of the numbers x(j) - zip ' ), for j = 0, 1, , n, then , xn) E I and xi = In=oaixi' ), i = 1, Therefore, if x (x 1, EjO)

- xiI=

n

n

a/(Et)-xt/))

I

n+1

/=0

In particular, for points 6(k), k = 1, 2, I

Eio)

-

J=0

n+

, n, we have

Elk) I c n+ l.

Reasoning analogously with the (n - 1)-dimensional face of the simplex , 6("), we can show that with vertices at 6"),

IEII)-E,k)Icnn

n -}-1

nd do n+1

for the points 6(k), k = 2, 3, , n. We proceed in this way with faces of fewer dimensions. As a result, we see for do j, j'=0, that all the numbers not exceed nd/(n + 1). Consequently, the quantities corresponding to the quantity d for the simplexes Yio,i 1,...,i do not exceed nd/(n + D. If we now repeat this partitioning process for the subsimplexes with each of the simplexes Y'io,i 1'"''in , then with the simplexes obtained from these, etc., the simplex is partitioned after p stages into subsimplexes having in common only faces of various dimensions and having the property that the difference between corresponding coordinates of the vertices of each of them does not exceed (n/(n + i))"d. in absolute value. Let V denote any of these subsimplexes with vertices y(k)(y(k), , n. Then, if , ynk)) for k= 0, 1, n

y, y' E E', yi =

ajyii),

/=o

y;

_

'LV i

i=o

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

24 8

we have n

n

vc-y;l=J=0Oi-a;))'ill= J=0

Xi)(YV'-yi° n

(n+1)PdI (xj /

(n+1)Pd

i=0

/

and n

P (.Y, y')=

i=I

(yi-y;)9_

n28(,i+1)

eP

d'

2y

(n+1)P

n

d.

(3)

It follows that the diameters of all the subsimplexes after a sufficient number p of stages in this partitioning process will not exceed an arbitrary given number e > 0. This is the desired partition of the simplex 1. To prove Brouwer's theorem, we need first to prove two lemmas.

Lemma 1. Let I denote an n-dimensional simplex with vertices x(j) for j = 0, 1, , n. Suppose that the points of I are distributed among n + 1 closed sets Ei, for .j = 0, 1, , n, in such a way that the following three conditions are satisfied: (1) each point x E I belongs to at least one of the Ei; (2) x(') E Ei; (3) Ei does not contain a single point on the face si with vertices in x(k) for k # j. Then, the sets Ei, for j = 0, 1, , n, have at least one common point. Proof. Let us partition the simplex I into a finite number of subsimplexes Ql, a2P ' + UP in such a way that any two of these simplexes have in common only faces of some dimension or other. Let us show that among the ak there exists a simplex containing points of each set Ei for j = 0, 1, , n. Each vertex of the simplex ak belongs to one or more of the sets E1. If it belongs to the sets where v1 < vZ < < vy , let us assign to it the number vl as coordinate. To the simplex ak let us assign as coordinates the n + 1 numbers which are the coordinates of its vertices and to each face let us assign the n numbers that are the coordinates of the vertices of ak in it. Let us show that the number of simplexes ak of which all the coordinates are distinct is an odd number. This assertion is true in the case n = 1 because then I is a line segment which is divided into subsegments ak by a finite number of points.

§7. PROOF OF BROUWER'S THEOREM

249

The coordinates of the vertices of these subsegments are the numbers 0 and 1. If we proceed from the vertex of E with coordinate 0, the first of the ak that we encounter with different coordinates has coordinates (0, 1), the next has coordinates (1, 0), then (0, 1), etc. (Here, we are indicating the coordinates of points met successively as we move along I.) The last of these ak must necessarily have coordinates (0, 1) because the second end of I must have coordinate 1. This proves that the number of segments ak with distinct coordinates is odd. Let us now suppose that the assertion has been proved for an (n - 1)-dimensional simplex and let us prove it for an n-dimensional simplex. Let tk denote the , n - 1). If tk > 0, number of faces of the simplex ak with coordinates (0, 1, then the simplex ak has coordinates (0, 1, , n - 1, a), where 0 < a < n. Here, if a = n, we obviously have tk = 1. On the other hand, if a < n, then tk = 2 because the coordinates of all n + 1 faces of the simplex ak are obtained from , n - 1, a) by discarding a single number and we obtain the numbers (0, 1, , n - 1) only by deleting the number a when encountered twice. Let e (0, 1, , n) and let f denote the number of all simplexes ak with coordinates (0, 1, denote the number of all ak for which tk = 2. Then, from what was said above, P

E tk=e+2f.

k=1

(4)

On the other hand, if a face of a subsimplex ak lying inside I has coordinates (0, 1, . . , n - 1), it appears twice in the sum Y.k=1tk because each such face belongs to two of the simplexes ak Let g denote the number of such faces belonging to distinct ak. A face of a simplex ak lying on one of the faces of the , n - 1) only if simplex 1, on the other hand, will have coordinates (0, 1, this face lies on sn. But sn is an (n - 1)-dimensional simplex the points of which belong to the n intersections sn n E;, j = 0, 1, , n - 1, that satisfy conditions (1), (2), and (3) of the lemma. Our assertion is considered proved for the case of (n - 1)-dimensional simplexes. Consequently, Sn will have an odd number h of faces sk with coordinates (0, 1, , n - 1). Thus, we have P

2] tk = 2g+ h.

k=1

(5)

Comparing (4) and (5), we obtain e = 2 (g - f) + h; that is, e, the number of simplexes ak with coordinates (0, 1, , n), is an odd number. Consequently,

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

250

there exist simplexes with coordinates (0, 1,

-

points belonging to all the sets E1, for j = 0, 1,

,

n). Each such simplex contains , n, because its vertices

belong to distinct E1.

Now, suppose that we are given a sequence of positive numbers Ek, for k = 1, 2, , that approaches 0 as k --+ oo. Let us partition Y. into a finite number of simplexes having in common only faces of different dimensions with diameters less than E1. Suppose that among the subsimplexes al is a simplex containing points of all the sets E;, for j = 0, 1, , n. From what we have shown, such a simplex exists. Let us now partition into a finite number of simplexes of the

same kind but with diameters less than E2- And suppose that aZ is a simplex containing points of all the sets E. We continue the process. As a result, we obtain a sequence of simplexes all a'Z, . each of which is contained in I and the sequence of diameters approaches 0. In accordance with Weierstrass's theoren generalized to the space Rn, there exists in Y. a point a an arbitrary neighborhood of which contains infinitely many of the simplexes ak. Since the sets E;

are closed, the point a belongs to all the sets E1, for j = 0, 1,

, n. This com-

pletes the proof of the lemma.

Let us point out one way of partitioning the simplex I into sets E; as indicated in Lemma 1 such that the sets E; have a unique common point a, which is an interior point of 1. Suppose that Y. has the representation (1) and that corresponding to the point a are the values of the parameters A; equal to A;0), where 0, for j = 0, 1, , n. Let us form closed sets E1, for j = 0, 1, , n, by A1°) assigning to E; all points of the set E for which A; > in the representation lies in at least one of the sets E1 because, other(1). Each point of the set wise, we would have Ai < 0) for all j = 0, 1, - , n, which would contradict the equations

For the same reason, the sets E;, for j = 0, 1,

, n have the point a as their

unique point in common. Furthermore E; obviously contains the point x(j) and We shall use this fact in what follows. does not contain points of the face

Lemma 2. Let E denote a closed bounded subset of R. Suppose that E is covered by a finite number of closed sets Ek such that each point of the set E belongs to at least one of them. If E includes a unique point a belonging to no

§7. PROOF OF BROUWER'S THEOREM

251

fewer than n + 1 sets Ek and this point is a boundary point of E (that is, not an interior point of E), then, by changing the sets Ek in an arbitrarily small neighborhood of the point a, we can arrange to have every point of the set E belong to no more than n of the sets Ek. Proof. Suppose that E is a simplex of arbitrarily small diameter and that a is an interior point of L. We can get the simplex E by taking an arbitrary simplex, displacing it in R in such a way that some interior point of it occupies the position of a, and then performing a similarity transformation on it about a. Let S

denote the boundary of I (that is, the set of points on all of the faces of 1). Define D = S n E. The set D does not include any points common to more than n of the sets Ek. Consequently, every point x E D has a neighborhood whose intersection with E can be covered by no more than n of the Ek. In accordance with the Heine-Borel theorem, generalized to Rn, it then follows that there exists an c > 0 such that every ball of radius c with center at an arbitrary point x C D consists of points belonging to no more than n of the sets Ek. Let us now cover S with a finite number of closed sets Fi, for j = 1, 2, - , with diameter less than c such that each point of S will belong tp no more than n of the sets F. Let us show that such a covering exists. Let us partition S into (n - 1)-dimensional simplexes with diameters less than i2n and then let us represent each such simplex as the union of n closed sets with a unique point in common, each of which contains one of the vertices of the simplex. Let us take one of the vertices of the subsimplexes, for example b. Suppose that it belongs to the simplexes li, for j = 1, 2, - , h. The point b belongs to only one of the closed sets into which each of the simplexes Ej is divided. Consequently, b is contained in h such sets, let us say, Fill for j = 1, 2, , h, that lie respectively in Yi. Let us set -

h

Fk=U F,. We reason in the same way for all the vertices of the subsimplexes. Then, the set S is completely covered by the sets Fk, and each point x E S belongs to no more than n sets Fk because an interior point of a subsimplex belongs to no more than n of the sets Fk into which it is partitioned and, consequently, to no more than n of the sets Fk. On the other hand, a point on a face of a subsimplex can belong to only those Fk that contain the vertices of that face and hence to no more than n - 1 of the sets Fk. Furthermore, one can easily show that the

diameters of all the sets Fk is less than e. Thus, the sets Fk provide the

252

V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS

required covering for S. Let us now combine the sets Fk with the sets Ek as follows: 1) If Fk does not contain points of D, we combine it with any one of the sets Ek. 2) On the other hand, if Fk does contain points of D, we combine Fk with one of the sets Ek to which such points belong. Let us show that each point x E S still belongs to no more than n of the sets Ek. This is true because a point x E S that does not belong to D belongs to no more than a of the sets FA, and hence now belongs to no more than n of the sets El, On the other hand, if x E D, it belongs to no more than n of the original sets, that is, to E1, EZ, , E f, where f < n. On the other hand, x belongs to certain of the sets Fk, let us say to F1, F2, , Fg, where g < h. The sets F1, . . . , Fg are contained in the ball of radius E with center at x. But, from our construction, this ball contains points belonging to no more than n of the original sets Ek, let us say to the sets E1, E2, , Eh, where 1 < h < n. From 2), the sets F1, . . , Fg are combined with certain of the sets E1, , Eh, Consequently, a point x E S can belong to no more than n of the modified sets Ek, namely to certain of these sets that contain the original sets E1, , Eh. Thus, after we adjoin the sets FA, to the sets Ek, every point x E S belongs to no more than n of the sets Ek. Since a is a boundary point of the set E, the simplex I contains an interior point c that does not belong to E. Let us distribute the points of the simplex E and hence let us redistribute the points of the set K = E U I among the sets Ek as follows: If a point x E S belongs to the sets EvI, E 2 , - . . , Evp, where p < n, then the entire segment cx (one-dimensional simplex) belongs to all these sets and only to them. With this new covering of the set E U E, the point c is the only point that can belong to more than n of the sets Ek. But c does not belong to E. Consequently, E does not contain points belonging to more than n of the sets Ek. This completes the proof of the lemma. Brouwer's theorem. A topological mapping of a set E C R onto a set VC R maps interior points of E into interior points of E . Proof. Let a denote an interior point of the set E and let I denote an arbitrarily small simplex contained in E and including a as an interior point. Corresponding to the point a is a point a' in E ', and corresponding to the simplex is a set ' C V. Let us show that a ' is an interior point of the set V. Let us subdivide I into n + 1 closed subsets E;, j = 0, 1, - , n, in such a way that each E; contains only one vertex and no point of the opposite face and in such a way that a is the only point belonging to all the sets E;. The mapping of E

§7. PROOF OF BROUWER'S THEOREM

253

onto E' assigns to the sets E; closed sets E;' in V that have the unique comcommon point a'. If a' were a boundary point of E' and hence of 2', then, in accordance with Lemma 2, by modifying the sets Ej in a neighborhood of a', we , n, to have no point in common. With could arrange for the sets E;, j = 0, 1, such a covering, the corresponding modified closed sets E; have no common points and at the same time satisfy the conditions of the lemma, which, however, is impossible. The contradiction proves Brouwer's theorem.

CHAPTER VI MAPPING OF MULTIPLY CONNECTED DOMAINS ONTO A DISK

§1. Conformal mapping of a multiply connected domain onto a disk

In §1 of Chapter II, we showed that no regular function can map a multiply connected domain bijectively onto the disk JCI < 1. Let us now consider, in addition to regular functions defined on a multiply connected domain B, functions that can be extended into B along an arbitrary path. Here, by such a function we mean an initial element of a function and all possible analytic continuations of that element into B. As we shall now show, among these functions there is one with the property that all values that it assumes in B lie in the disk I C1 < 1 and that each value in the disk I C1 < 1 is assumed at exactly one point of the domain B. We shall call the correspondence that this function sets up between points of the domain B and points of the disk I C1 < 1 a conformal mapping of the domain B onto the disk ICI < 1. The conformality follows from the fact that all elements of the mapping function are univalent in the corresponding domains and, consequently, their first derivatives do not vanish anywhere (that is, at any finite point). The mapping function, although it is regular at every point of the domain B, is not univalent in B because, if it were, it would set up a one-to-one correspondence between the domain B and the disk I i1 < 1, and we have seen that this cannot be. In the case of a finite domain B, the inverse function can be extended into the disk I I < 1 along an arbitrary path and hence is a regular function in that disk. Before proceeding to a proof of the possibility of such a mapping of a multiply connected domain onto a disk, let us first mention the exceptional case when such a mapping does not exist. This is the case when the domain B B. constitutes the

entire z-plane with two exceptional points a and b. To see this, let us assume 254

§1. MAPPING OF A MULTI PLYCONNECTED DOMAIN

255

without loss of generality that a and b are finite and let us observe that, in this case, the function t = log ((z - a)/(z - b)) can be extended into B along an arbitrary path and it maps the domain B onto the t-plane with the point -0 excluded. The inverse function z = 0(t) is regular in that plane. Therefore, if there existed a function C= f(z) that maps the domain B conformally onto the disk ICI < 1, the function C = f (0 (t)) would be regular in t I < Do and it would not exceed unity in absolute value, and this cannot be the case. I

Let us turn now to the proof of the basic theorem. Theorem.1) Every domain B in the z-plane with more than two boundary points can be mapped con formally onto the disk I I < 1 in such a way that corresponding to a given point z0 E B and a given direction at that point there are (for a basic element of the mapping function) a point C= 0 and the positive direction of the real axis. This mapping is unique. Proof. We note first of all that the domain B can be mapped onto a domain B* contained in I CI < 1 in such a way that to each point of B* corresponds only

One point of B. To see this, let a, b, and c denote three boundary points of the domain B. By means of a fractional-linear mapping, let us map the z-plane onto the t-plane in such a way that these three points are mapped respectively into 0, 1, and o-. The domain B is then mapped into a domain B* that does not include the three points 0, 1, oo. Therefore the function C= A(t), inverse to the modular function t = g(z) (see §6 of Chapter II) can be extended into B* along an arbitrary path and the values that it assumes all lie in the disk I CI < 1 and occupy a domain B*. Under this mapping, to every point of the domain B* there corresponds as preimage only one point of the domain B* and hence only one point of B. By means of a fractional-linear transformation of the disk I CI < 1 into itself, we can always arrange for the point C= 0 and the positive direction of the real axis to correspond to the point z0 and a given direction at z0. ( Other points in the disk 1CI < 1 may also correspond to z0.) Proof of the theorem is now reduced to proving it for the case of a domain B contained in the disk z I < 1. The given point z o E B is the point z = 0 and the given direction at that point is the direction of the positive half of the real axis. I

1) This theorem was discovered by Poincare. Rado noted [1922-1923a] that the idea of the proof of Riemann's theorem that was given in §2 of Chapter II can be used to prove this theorem also.

256

VI. MAPPING ONTO A DISK

In this case, let us consider the set 9 = (f(z)} of all functions f(z) with the following properties: 1) they are defined by basic elements about z = 0 for which f (0) = 0 and f ' (0) > 0; 2) they can be extended into B along an arbitrary path; 3) the values that these functions assume in B as a result of analytic continuation all belong to the disk I CI < 1; 4) every value of 4 in the disk < 1 is assumed I

I

by every function ar no more than one point of the domain B.

An example of such -a function is the function f (z) = az, where 0 < a < 1. We pose the following extremal problem: Out of all functions f (z) E R, find the one maximizing f'(0), calculated for the basic element of the function. Since there exists a disk I z I < p, where p > 0, that lies entirely in B, the basic element of each of the functions f (z) C 9 is regular and less than unity in absolute value in this disk. Therefore, by the Schwarz lemma, we have f' (0) < 11p. This shows that the set of numbers f'(0) is bounded above. We denote the least upper

bound by a. Obviously, a> 1. There exists a sequence of functions fn (z) E R, n = 1 , 2, , such that if, (0)} converges to a. According to the condensation principle, the sequence of basic elements of the functions fn (z) has a subsequence that converges uniformly in the interior of the disk z I < p to a regular function fo (z), which is not a constant, since f 0' (0) = a> 0. If we now extend these elements into B along different paths, we see step by step in accordance with Vitali's theorem that the sequences of analytic continuations obtained converge to the corresponding analytic continuations of the function fo (z). Consequently, the function fo (z) can be extended into B along an arbitrary path and IA (z)I < 1 everywhere. Let us show that C= fo (z) assumes in B every value in the disk I CI < 1 at no more than one point. Let us suppose, to the contrary, that there exist in B two distinct points z I and z2 such that fo (z 1) = fo (z2) = c. Here, fo (z 1) and f0 (z2) are the values at the points z I and z2 of certain elements of the function f0 (z) that are obtained from the basic element by analytic continuation along certain paths 11 and 12 contained in B. Suppose that the disks I z - z I I < e and I z - z 21 < E have no common points, that they are contained in B, and that these elements assume the value c neither in the interiors of these disks, except at the center, nor on their boundaries. Suppose that Ifo(z) - c I > m > 0 on the circles I z - z I I = r and I z - z 21 = e. Let k denote an integer such that I fnk (z) - fo (z)I < m on these two circles. Then from I

fnk(Z)-C^ (Jnk (Z) - A

(z))+VO(z)-c)

§ 1. MAPPING OF A MULTIPLY CONNECTED DOMAIN

257

it follows, on the basis of Rouche's theorem, that the function 1 k (z) assumes the value c at two points of the domain B, which is impossible. This proves our assertion. Let us show now that the function f0 (z) provides the mapping mentioned in the theorem; specifically, that it assumes in B every value C in I CI < 1. Suppose, to the contrary, that there exists a number c such that I c I < 1 and f (z) c everywhere in B. Let us denote by t = q(C) the function used in §2 of Chapter II. We note that this function can be extended into I CI < 1 along an arbitrary path that does not pass through c and that 0 (0) = 0 and q ' (0) > 1. We see that the function F (z) _ 0 (fo (z)) C 3 and F' (0) = f6' (0) 0 (0) > a., which is impossible. It remains to prove the uniqueness of the mapping mentioned in the theorem. Suppose that C = fI (z) and C= f2 (z) are two functions providing this mapping. Then the function C' = f 1 (fZ I (C)) = OW and its inverse C= f2 (I1 (C )) are regular in the disks ICI < 1 and IC' I < 1, respectively. Therefore, the function 4' = q(c) maps the disk ICI < 1 bijectively onto itself in such a way that 9(0) = 0 and 0' (0) > 0. But then, 0(C) _ C; f 1 (z) = f2 (z). This completes the proof of 0

the theorem. Now let C= f(z) denote the function providing the mapping indicated in the preceding theorem. Let C= F W denote any other function mapping the domain B onto a disk but not satisfying the same normalization conditions as C = f (z). Then,

the function C' = F(f-1(C)) maps the disk 141 < 1 bijectively onto itself. Consequently, it must be a fractional-linear function. Thus, F(z) = (af(z) + 13)/(y f(z) + 8); that is, any other mapping function can be expressed as a fractional-linear function of f (z). In particular, if we take two distinct elements of the same multiplevalued mapping function f (z) at the point z0 and extend them along all possible paths contained in B, we obtain two functions, that is, two branches of f (Z), both of which map the domain B conformally onto the disk I C.I < 1. It follows from what has been said that these branches can be expressed each as a fractionallinear function of the other. Thus, all branches of a multiple-valued mapping func1tion can be represented as a fractional-linear function of just one of them and the corresponding fractional-linear functions map the disk ICI < 1 into itself. This conclusion enables us to study more deeply the nature of the multiplevaluedness of the mapping function f(z). Let f0(z) denote one of the branches of the function f (z). We denote by C' = S (C) = (a.C + 0)/(yC + 8) or more briefly, by S the fractional-linear function mapping the branch f0 (z) into the branch

258

VI. MAPPING ONTO A DISK

f. (z); that is, f. (z) = S fo (z). Let us show that the set G of all "transformations" S corresponding to all possible branches of the function f (z) constitutes a group. Let us extend the branch S 1 f0 (z), S 1 E G, along a closed path such that extending the branch fo (z) along this path yields the branch S2 fo (z), S2 E C. We thus obtain a new branch S 1S2 f0 (z) of the function f (z). Therefore S 1S2 E C. Analogously, if we extend the branch fo (z) along a closed path the, extension of fo (z) along which yields a branch S f0, S E G, though in the opposite direction, we obtain a branch f. (z) = S f0 (z), where S. E G and we again obtain from Sf0(z) the branch fo (z) and at the same time the branch SS, f0 (z); that is, we have

f0 (z) = SS, fo (z), so that S. = S -1. Consequently, S -1 E G. Thus, G is a group. We note that the neutral element of this group is the identity transformation With regard to the function z = 0(0 inverse to C= f (z), when B does not < 1 along an arbitrary path and hence is contain -, it can be extended into regular in I I < 1. From this we conclude that, in the general case, its only singularities in the disk CI < 1 are poles; that is, that it is a meromorphic function in I CI < 1.

It follows from the relation z = 0(f(z)) that 0(Sfo(z)) = 0(f0(z)) or 0(SC) _ 95 (c), where S E G. This shows that the function 0(e) remains invariant under transformations belonging to the group G. Therefore, it is an automorphic function with group C. We noted above that all the transformations belonging to the group G leave the circle ICI = 1 unchanged. Let us show now that the fixed points of all these transformations lie on the circle ICI = 1. By fixed points of the transformation S, we mean points C satisfying the equation SC= C. A fractional-linear transformation other than the identity transformation has no more than two fixed points. Let us suppose that a given transformation S E G has a fixed point a such that

< 1. Then, 1/a will also be a fixed point of S. But the equation Sa= a implies that SC a in a sufficiently small domain 0 < ICI al < S. Suppose that I SC - al > m > 0 on the circle I C - al = S. Let 0 denote an arbitrary number in the domain 0 < IC - °'I < m. Since S(C) - f3 = (S (C) - a) + (a- 0), it follows from Rouche's theorem that the function S (C) - j3 has a zero y in the disk C - al <.5. Here, y is distinct from a and 13. It follows from S (y) = 0 that 95 (y) _ 95 (9 ). But, for sufficiently small S and m, this last cannot be since 95 '(a) 0. Consequently, the function z = '(C) maps a sufficiently small neighborhood of the IaI

point J= a univalently. This contradiction leads to the conclusion that there can be no fixed points of the transformations S E G in the open disk ICI < 1 and hence that there can be none in the domain I CI > 1. Therefore, all the fixed points lie on the circle I I = 1.

§1. MAPPING OF A MULTIPLY CONNECTED DOMAIN

259

If a transformation S E G other than the identity transformation has two fixed points on the circle ICI = 1, it is called a hyperbolic transformation. If it has only one, it is called a parabolic transformation. Let us show that, if S is either a parabolic or a hyperbolic transformation, the sequence of transformations S, S2, S3, . . does not include the identity transformation. Let us suppose first that S is a hyperbolic transformation with fixed points SI and 6. If we define the transformation t = T(C) = (C- CI)/(C- Cz), we see that J' = TST-1(C) is a fractional-linear transformation with fixed points 0 and oo. Therefore, C' - aC and, consequently, TS(e) = aT (C). This last shows that the transformation J' = S(C) is obtained from the equation

-s-a

Cs, a=const.

(1)

in (1) lies on the circle ICI = 1, then the point J must Now, if the point also lie on the circle ICS I = 1. Therefore, the arguments arg((t; ' - 0/'- 0) and arg ((C- CI) (C- Cz)) indicating the angles subtended by the segment blb2 at the points e' and C are equal. From this we conclude that a > 0. Furthermore, a 1 because we would then have C' = 4 which cannot be. Thus, the transformac f. Here, a tion S can be obtained from equation (1) by solving that equation for is positive and different from unity. The transformation J' = S"t; (where n = ± 1, ± 2, . ) can obviously be obtained from the equation

_

C'-C, C'-Cf-a

. C-C2

(2)

Therefore, it is always different from the identity transformation. Furthermore, it follows from (2) that the sequence of the functions PC and S-"J converge as n - + = to the constants SI and C2. This convergence is uniform on an arbitrary closed set not including the points z1 and z2. This last conclusion will be necessary for §2. We proceed analogously in the case in which S is a parabolic transformation with fixed point S1. By making the transformation t = T (C) = 1/(C- SI) we see that t' = TST-1(t) is a transformation with unique fixed point oo. Therefore, 7 , = C + a and, consequently, TS (C) = T (C) + a, where a = const. This shows that the transformation C' = SC is obtained from the equation C'-C,

C-C1a' +

(3)

260

VI. MAPPING ONTO A DISK

where a X 0 because, in the opposite case, we would have ' = C, which is impossible. On the other hand, the transformation C' = S' C (for n = ±1, ±2,

)

is defined by the equation

_ +na --C1 1

C'

CI

(4)

Cr

Therefore, once again, it eannot be the identity transformation for any n. We note that the sequences of the functions S"C and S-' converge uniformly to 6 on an arbitrary closed set including the point CI as n +o. This property shows, in particular, that G is an infinite group, that is, that it contains an infinite number of distinct transformations and that f (z) is an infinitely-many-valued function in B. On the other hand, the function 0(0 assumes in ICI < 1 every value in B at an infinite number of points. All points in the disk ICI < 1 at which the function O(C) assumes the same value are called equivalent points. For every point a in the disk I a I < 1 there is an infinite set of points equivalent to a. These are all obtained from each other by transformations of the group G. Since these points are zeros of the function O(C) - 0(a), they do not have points of accumulation in I I < 1. It follows, in particular, that the group G is countable. Since the function f(z) is not single-valued in B, the question arises as to what conditions must be satisfied by the path 1 contained in B and issuing from and returning to a point z0 in such a way that as we proceed along it with some element of the function f (z) we return to the point z0 with the same element. Consider the case in which B does not include oo. If, as we proceed along 1 from zo with some element of f(z), we return to z0 with the same element, the corresponding point C= f(z) issuing from some point CO returns to Co after describing a closed path l' in the disk ICI < 1. The path l' can, by a continuous (for example. similarity) deformation be reduced to a. point that does not leave the disk ICI < 1. Obviously, under this deformation, the preimage 1 as it is continuously deformed also reduces to a point without getting outside B. Let us show that, conversely, if the path l can, by a continuous deformation, be reduced to a point without ever getting outside B, then, by proceeding along it with an arbitrary element of f(z), we always return to the original point with the same element of f (z). This is true because, otherwise, under such a deformation, the corresponding path 1', while being continuously deformed in I CI < 1, will go from the point Co to some other

§1. MAPPING OF A MULTIPLY CONNECTED DOMAIN

261

point equivalent to it and cannot be reduced to a point. Thus, for us to proceed in B along a closed path with some element of the function f (z) and return to the original point with the same element of f (z), it is necessary and sufficient that it be possible to reduce this path by a continuous deformation to a point without getting outside B.1) In conclusion, let us pause to construct a Riemann surface connected with the mapping that we have been considering. We confine ourselves to the case of a finitely connected domain B. If the domain B is n-connected, we can render it simply connected by means of n - 1 suitable transverse cuts (broken lines). We then take an infinite set of copies of the domain B so cut and consider them as lying one above another. We choose any one of them as the basic domain and denote it by Bo. We adjoin to Bo along each edge of each cut one of the remaining copies. We make this con-

nection-this gluing-along opposite edges of the cuts. As a result, we obtain a (2n - 1)-sheeted surface B I lying over the domain B and having (2n - 3) (2n - 2) free edges of cuts, that is, edges along which no gluing has been made. Then, we again adjoin along each of these edges one of the as yet unused copies of the cut domain B in such a way that we again perform the gluing along opposite edges of the cuts. We then obtain a new surface B2 with a finite number of sheets and some number of free edges of cuts. If we continue this way indefinitely, we finally obtain an infinitely-sheeted Riemann surface B.. If we consider the initial element fo (z) of the mapping function f(z) defined in the domain B0, the analytic continuation of this element into B is identical to its analytic continuation onto the surface B00 . As the result of all possible analytic continuations of f (z) onto the surface B., we obviously obtain a singlevalued function on that surface. Consequently, the function C = f (z) that we studied above also provides a one-to-one mapping of the surface B. onto the disk ICI < 1. The surface B. is called the universal covering surface of the domain B. The modular surface constructed earlier (see §6 of Chapter II) is, in this terminology, the universal covering surface of the domain consisting of the plane with the points 0, 1 and os excluded.

1) If B includes os, an analogous condition must be satisfied on the Riemann sphere.

262

VI. MAPPING ONTO A DISK

§2. Correspondence of boundaries under a mapping of a multiply connected domain onto a disk

In § 1, we examined a function f (z) that maps a given multiply connected domain B conformally onto the disk CI < 1 only inside B. If we now let the point z approach the boundary of the domain B, we can easily show that the sequence of the corresponding points C approaches the circle ICI ="1 uniformly. In other words, for arbitrary e > 0, there exists a S > 0 such that, if the distance

from the point z E B to the boundary of B is less than S, the distance of the corresponding point C from the circle ICI = 1 is less than c. To see this, let E denote the set of all points z E B corresponding to points of the disk I Cl < 1 Then we need merely take for S the distance from the set E to the boundary of the domain B. When we make a more thorough study of the behavior of the function f (z) as z approaches the boundary of the domain B, we shall confine ourselves to the case of a finitely connected domain B. Such a domain can easily be mapped univalently onto a domain bounded by closed analytic Jordan curves and isolated points. The correspondence of the boundaries under such a mapping has already been studied. Therefore, for our purposes, it will be sufficient in the future to consider a domain B with a boundary consisting of closed analytic Jordan curves and isolated points.

Let a denote an isolated boundary point of the domain B, which we may assume to be finite. Suppose that the closed disk I z - a I < r contains no boundary points of the domain B other than a. As we go around the circle 1: I z - a I = r one time in the positive direction, the branch fo (z) of the function f (z) becomes the branch So fo (z), where So / 1 is one of the transformations belonging to the group G considered in §1. Furthermore, if we go n times around l not necessarily in the same direction each time), we end up at the original point with the element So" fo(z). Let us show that the transformation So is a parabolic transformation. Let us suppose, to the contrary, that So is a hyperbolic transformation with two distinct fixed points C1 and S2 on ICI = 1. If we choose a point on l and move around l an infinite number of times with branch fo (z), the corresponding point C describes a curve A in the disk ICI < 1 that approaches the circle I CI = 1 and, more specifically, approaches the points S1 and S2 If we continue fo (z) in the domain 0 < I z - al < r along an arbitrary path, we see that all values that the function assumes in the process lie in one of the domains, let us say B 1, into which the curve A partitions the disk CI < 1. If we let r

§2. CORRESPONDENCE OF BOUNDARIES

263

approach 0 continuously, the corresponding curves A, all passing through S1 and CZ, cover the domain B This shows that the inverse function 0(C), which is regular and bounded in B approaches a constant uniformly, on a sequence of arcs connecting the points 6 and S2 If we map the domain B 1 onto the disk It I < 1 and use Lemma 1 of §3 of Chapter II, we see that the O(C) = const, which, however, is impossible. This shows that S must be a parabolic permutation. If Co is its fixed point on ICI = 1, the curve A that we considered above must be a closed curve in Cl I< 1 whose only point in common with the circle ICI = 1 is Co. For values of z inside 1, the corresponding points C lie inside the curve X. But as r 0, the curve A is constricted to the point Co. Consequently, as we let z approach a, the branch fo (z) approaches Co uniformly. If we approach a with another branch S fo (z) of the function f (z), the corresponding point C = S fo (z) approaches the point SCo. Thus, as we let z approach a, the only limiting values of the function f(z) can be the points SCo, where ICo = 1 and S E G. When the entire boundary of the domain B consists of isolated points, it follows from what was said above that, as we let z approach the boundary of the domain B, the function f(z) can have as its limiting values only values belonging to a countable set of numbers, which are points on ICI = 1. It follows that, as C describes a path in the disk ICI < 1 that terminates on the circle ICI = 1 at some point different from these points, the corresponding point z = O(C) describes a path that cannot terminate at a definite point. Thus, in the present case, the function (k(C) does not have definite limits on ICI = 1 as r approaches any but a countable set of points on that circle from ICI < 1. Now, let K1 denote a closed curve constituting part of the boundary of B1. Let us encircle K1 with a Jordan curve l contained in B and defining, together with K1, a doubly connected domain B ' consisting only of points belonging to B. If we go around the curve l with branch fo (z) once in the positive direction, we arrive at the original point with a branch Silo (z). On the other hand, if we go around l several times (possibly some times in one direction and some times in the other), we arrive at the original point with branch S i ° fo (z). Here, S is a transformation of the group C other than the identity transformation. If we proceed around the curve infinitely many times with the branch fo (z), the corresponding point C describes in the disk ICI < 1 a Jordan curve A that terminates at the fixed points S1 and C2 of the transformation S (which, for all we know as yet, may coincide). If we extend f 0 (z) inside the domain B ', we obtain a multipleI

1

1

264

VI. MAPPING ONTO A DISK

valued function in that domain. If, by means of a cut, we map the domain B ' into a simply connected domain, all the branches of the multiple-valued function will be single-valued in that domain and will map it bijectively onto domains adjacent to each other and filling up completely one of the domains, let us say B 1, into which the curve A partitions the disk ICI < 1. There is a continuous one-to-one correspondence of the boundaries under these mappings. Therefore, the arcs corresponding to the curve K 1 lie on the circle I CI = 1 and together they cover the arc s1C2 In addition to the arc S1S2 there are other arcs on ICI = 1 and these arise in the same way from other branches of the function f (z) than S i f o (z) and from other curves constituting parts of the boundary of the domain B (with one exception to be mentioned below). These arcs cannot overlap because, if they did, domains analogous to the domain B 1 would have common points whereas either domains analogous to the domain B' have no common points or the branches of the function f(z) considered in them do not coincide at a single point in them. It which does not reduce to a point, cannot cover comfollows that the arc pletely the circle I CI = 1; that is, S1 Cz. The only exception is the case when B is a doubly connected domain one boundary of which is an isolated point. In this case, the group G consists of the powers of one of its transformations which is a parabolic transformation. Thus, when we exclude this last one, we may assert that the transformation S 1 considered above is hyperbolic. If the point z C B approaches a point on K1 while describing a continuous curve, the corresponding point ' defined by an arbitrary branch of f(z) describes in ICI < 1 a curve that terminates at an interior point of one of the arcs obtained according to the type of the arc C1C2. On the other hand, as the point z approaches K along a curve that encircles K infinitely many times, the corresponding point C describes in < 1 a curve that terminates at one of the endpoints of these arcs. It follows from what has been said that, if the boundary of a multiply connected domain B does not consist entirely of isolated boundary points, then points on the circle I C1 = 1 fall into four classes: 1) interior points of all arcs defined according to the type of the arcs b1 C2 Corresponding to each such point is a unique nonisolated boundary point of the

domain B, and at each of these 0(4) is regular. 2) points of the type S1 and S2 are fixed points of hyperbolic transformations of the group G . There are countably many of these points. 3) points of the type of Cp, that is, fixed points of parabolic transformations

§2. CORRESPONDENCE OF BOUNDARIES

265

of the group G. There are also countably many of these.

4) all remaining points on the circle I4I = I. This classification was made from the point of view of the behavior of the function (k (C) in a neighborhood of the circle ICI = 1. If the function 0(C) approaches a definite limit as the point C on the circle IC1 = 1 approaches a point a on that circle, the point a belongs to class 1) or 3) because then the point z = O(C) approaches some point on the boundary of the domain B as Ca and, consequently, the point C corresponding to it approaches either an interior point of one of the arcs of the type 66 or one of the points of the type Co. Let us show that almost all points on the circle IC1 = 1 are points of class 1), that is, that the sum of the lengths of all arcs of the type Cl 2 is equal to 2rr. . Let . Let uk (c), for us arrange these arcs in a sequence {.(lk) y(k){, for k= 1, 2, k = 1, 2, , denote a harmonic function that is bounded in I CI < 1, that is equal = 1. and that is equal to 0 at all other points on to 1 on the arc S lk) This function can be constructed in accordance with Poisson's formula' I

1- r'

1

ttk(C)-2a

C==re'.

(1)

((k%(ak)

Consider the series

i

00

Z Uk (C).

(2)

y k=l For C= 0, this series obviously converges. Furthermore, if we denote by sn(S) the sum of its first n terms, we see that sn W < s n+1(J) in I C1 < I. It follows, in accordance with Harnack's theorem, that the series (2) converges uniformly < 1. inside the disk I tI < 1 and that its sum s (C) is a harmonic function in Furthermore, it follows from (1) and (2) that 2a

r 1-r' ed0=1 S(C)C2n of 1-2rcos(0-T)+r 1

< 1. If we now let C approach an interior point of the arc S 1k1S 2k 1, the 1 as 1 and, since sk(C) < s (C) < 1, it follows that s sequence sk(C) C approaches an interior point of that arc. Furthermore, if we map the disk I Cl < 1 (k) y ( sk) are into itself by means of transformations of the group G, the arcs ybvalues only on then mapped into each other (because O(C) has definite limiting

in

1) This formula will also be proved in Chapter IX.

VI. MAPPING ONTO A DISK

266

these arcs). Consequently, the function s (C) is invariant under transformations of the group G. This last shows that, if we use the function z = O (C) to map the disk C1 < 1 onto the domain B, we obtain in B a single-valued bounded harmonic function s (f (z)) that is equal to unity at all nonisolated boundary points. Such a function is identically equal to unity in B because, otherwise, it either would attain a maximum or a minimum inside B or would approach an extreme value as z approaches isolated boundary points, which, in accordance with the maximum principle and the beginning of the present section, is impossible. Consequently, 1 in the disk 1. But

S (0)=

LEI 2a

S

k=1 t(k)t(k)

so that I' (length

de = 2n

length (Clk)C9kl),

k=1

27r. This completes the proof. §3. Dirichlet's problem and Green's function

A finite real function that, together with its first two derivatives,') is singlevalued and continuous in a domain B of the z-plane and satisfies Laplace's equation in B is said to be harmonic in B. Such a function can be regarded as the real part of some analytic function that can be extended in B along an arbitrary path. A function is said to be harmonic at a point if it is harmonic in some domain including that point. We note that, if u (z) is a harmonic function in a sufficiently small neighborhood of a point z = a with the point a itself deleted and if it is bounded below in that neighborhood, that is, if u (z) > m (or bounded above, that is, if u (z) < M), it will also be harmonic at z = a if its domain of definition

is extended to include the point z = a in a suitable manner. To see this, let v(z) denote the harmonic conjugate of U(z) and let t,) denote the increment in the function v (z) as z moves once around a sufficiently small circle I z - al = t (or a R in case a = ce). The function sufficiently large circle I z 2w

F(z)=e

(U + iv)

1) The derivatives here refer only to derivatives at finite points. In what follows, the harmonic functions mentioned may at times be multiple-valued, but in such cases an arbitrary branch of such a multiple-valued function will be single-valued in an arbitrary sufficiently small neighborhood of each point in the domain B.

§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION

267

with either sign chosen is then regular and bounded in the neighborhood in question of the point z = a, and, with suitable choice of the value for F (a), regular at the point a itself. Obviously, F (a) is nonzero. Therefore, the function u + iv is also regular at z = a and the function u (z) is harmonic at z = a. It follows that a function u(z) that is harmonic and bounded in a domain B is harmonic at all isolated boundary points when it is extended in a suitable manner to these points. The maximum principle of harmonic functions has an important generalization: -Jf a function u(z) is harmonic and bounded above in a domain B (that is, its values in B do not exceed some finite constant) and if all its limiting values, as its argument approaches any except finitely many boundary points of B through values in B, are no greater than M, u (z) < M everywhere in B. An analogous generalization holds for functions bounded below.

Proof. We may assume that the domain B includes the point o. Let al, an denote the exceptional points referred to and let H denote the diameter of the boundary of the domain B. For arbitrary c > 0, consider the function a

v(z)=M-u(z)+e Iloglz Haktl' , an, which is harmonic in B. As z approaches boundary points other than a 1, this function approaches nonnegative limits since each term in the sum above has a nonnegative value. As z approaches one of the points ak, this sum approaches + oo, but u(z) is bounded above. Consequently, the limit of v(z) as z approaches any of these points is +Do. Since v(z) is obviously not a constant, it cannot have minima inside the domain B. Therefore, v (z) > 0 in B; that is, n

it(z)M+e

log

tl IZ -ak

k=l This holds for arbitrary c > 0. By letting c approach 0, we obtain u (z) < M, which completes the proof. Let us now pass on to Dirichlet's problem. Suppose that the domain B has a boundary K consisting of a finite number of closed Jordan curves without common points. The Dirichlet problem for such a domain consists in finding a func-

tion u(z) that is harmonic and bounded in B, that is continuous in B except possibly at countably many points on the boundary K; and that assumes prestated values at points of continuity on the boundary K. The boundary values are subject

268

VI. MAPPING ONTO A DISK

only to the condition that they constitute a function on K that has a finite or countable set of points of discontinuity. Let us show that this problem always has a unique solution. Let us map the domain B onto the disk < 1, just as in §1. Consider the

set of points V on the circle I(' I = 1 belonging to the class (1) of §2. Let U W) denote a function whose values coincide with the values u (z ') at points z ' of the boundary K that correspond under this mapping to the points V on

I = 1. This real function U is defined, bounded, and continuous almost everywhere I) on the circle I C' 1 and satisfies almost everywhere on that circle the condition U (S (C )) = U('), where S is an arbitrary transformation belonging to the group G of §2. Thus, we arrive at the problem of finding a function U ((') that is harmonic and bounded in the disk ICI < 1 and that coincides with U(C') at all points of continuity of the function on ICI = 1. Consider the function I C'

UP=

2x

1- r'

ie

1

U ( e ) 1- 2r cos (0 - , f ) -}- r'

2a

tP dO C = re,

where the integral is in the sense of Riemann. By virtue of the known properties of a Poisson integral, this function is harmonic and bounded in I I < 1 and, under the shift to C' I = 1, it coincides with U (a') at all points of continuity of the function U(CH). On the other hand, any function U(C) possessing these properties is unique. To see this, suppose that there exists another such function U1(O. Then, for arbitrary C in 41 < 1, we have, in accordance with Poisson's formula I

2x

U1 (C)=2

e

U1(Ret)

RH-2RrR9

H

ret', cos(er 'p) -}-r' de' C=

where r < R < 1. If we now take the limit as R -' 1, remembering that we can take the limit under the integral sign (Lebesgue's theorem), we see that U1(C) U(C) in ICI < 1. It follows from the uniqueness property that U (C) remains invariant under transformations of the group G. In fact, the functions U(S(C)) and U(C) are both solutions of the problem that we have just examined and hence they are 1) In the present case, with the possible exception of a countable set of points on ICI= 1.

§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION

269

identical in CI < 1. If we now return to the z-plane, the function U(5) becomes the function u(z), which is single-valued and bounded in B and which satisfies the Dirichlet problem posed above. It also follows from what was said above that this problem has a unique solution. We note that if certain of the curves constituting the boundary K degenerated into points, the Dirichlet problem for the domain obtained would not always have a solution. To see this, note that the function u(z) which provides a solution to the Dirichlet problem for such a domain is, by virtue of what was said above, a harmonic function at all isolated boundary points. *Therefore, it is uniquely determined by its values on the remaining boundary curves. This shows that the values of the function u(z) at isolated boundary points cannot be chosen arbitrarily. Thus, we have proved the existence and uniqueness of the solution to the Dirichlet problem for an arbitrary finitely connected domain B bounded by Jordan curves and for arbitrary given values on the boundary K that constitute either a continuous function or a function with a finite or countable set of points of disContinuity. To find the solution of an arbitrary Dirichlet problem for a given domain B, it is sufficient to know the solution of any one particular Dirichlet problem. Here, it is a question of finding the Green's function. The Green's function for a domain B is defined as a real function g (z, C) satisfying the following three conditions: 1) for every C E B, it is a harmonic function of z in the domain B except at the point z = J; 2) g(z, C) approaches +oo as z - C in such a way that the difference g (z, C) - log 111z remains bounded if is finite and in such a way that the difference g (z, 00) - log I z is bounded if C = o.; 3) as (z, C) approaches the boundary K, the function g(z, C) approaches 0. In accordance with the maximum principle, g(z, C) must be positive everywhere in B. If a Green' s function g(z, C) exists, then, for finite S the difference g(z, C) log (14 z -el) when assigned the appropriate value at z = J, at the point z = is harmonic in B and continuous in B and assumes on the boundary K the values -log(1/jz - CI); that is, this difference is a solution of a particular Dirichlet problem for the domain B. Conversely, if u(z) is a solution of this Dirichlet problem, then the function log (1/I z - CI),+ u(z) is obviously a Green's function for the domain B. An analogous situation holds for C= 00. Under a singlesheeted mapping of the domain B by means of a function z* = f(z) onto another ,domain B* bounded by Jordan curves, a Green's function g(z, C) is transformed I

into the function g(f-1(z*), fwhich, obviously, is a Green's function for the domain B*. This shows that, if we know a Green's function for one domain,

270

VI. MAPPING ONTO A DISK

we can, by means of univalent mappings, obtain Green's functions for various other domains of the same order of connectedness. In the case of a simply connected domain B, there is no Green's function, as the function -log I f(z)I indicates, where t = f (z) is a function that maps a domain B univalently onto the disk It I < 1 in such a way that the point z = C is mapped into t = 0. For multiply connected domains, there is no such simple relationship with conformal mapping.

In the case of a simply connected domain B bounded by a closed Jordan curve, an analytic function P (z, C) the real part of which is a Green's function g(z, C) for a domain B is obviously continuous on B except at the point z = . If, in addition, the boundary of the domain B is an analytic curve, then p (z, is regular on that curve. It turns out that these properties also hold in the case of finitely connected domains. Let us consider first a finitely connected domain B bounded by arbitrary closed Jordan curves. Let g(z, C) denote a Green's function of B and let h(z, C) denote its harmonic conjugate. As we shall see, the last function is non-single-valued in B. Consider the function F(z) = g - ih. Let us take an arbitrary points z0 on one of the boundary curves of the domain B and let us describe a circle y about it sufficiently small that it intersects` the curve and there are no other boundary curves either inside the circle y or on it. The part of the domain B lying inside y consists of simply connected domains. We denote by B0 whichever one of these domains has the point z0 on its boundary. Let us map B0 onto the halfplane `fi(t) > 0 in such a way that z = z0 is mapped into t = 0. If we denote the mapping function by z = 0(t), the function F(OW) is regular in the domain (t) > 0 and its real part u (t) is also continuous in (t) > 0 and is nonnegative. Noting that u (t) = 0 at points on the real axis sufficiently close to t = 0, we describe a circle t I = p sufficiently small that u(t) = 0 on the diameter lying on the real axis. Let us now define a function NO on t = p as follows: U (t) coincides with the function u (t) on the upper semicircle and it assumes at points in the lower semicircle the same values as the function u (t) assumes at the conjugate points in the upper halfplane with with opposite sign. The function U(t) is obviously continuous on the circle I = P. Consider the integral I

U(t)2i

2+c U(Peie)p'-2prps-rs

It defines a function U (t) that is harmonic in

cos(B-cp)+r'dA,

I

t I < p and continuous on I t I < p .

§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION

271

It is obviously equal to 0 at points on the real axis and it coincides with u (t) on the upper boundary semicircle. Consequently, U (t) coincides with it (t) on the 'entire boundary of the upper half-disk and therefore throughout this entire halfdisk. If t(t) is an analytic function in It 1 : p and its real part is the function U(t), then the difference $(t) - F(O(t)) has real part equal to zero in the upper half-disk. Therefore, this difference must be a constant in t I < p. This last fact shows that the function F(0(t)) can be extended in a neighborhood of t = 0 beyond the real axis and, in particular, it is a continuous function at the point ,it = 0. Returning to the z-plane, we conclude that the function F(z), considered in 9, is continuous at the point z0. Since z0 is an arbitrary point on the boundary of the domain B, this means that F(z) is continuous on B except, of course, at the point C. Furthermore, since t ' (0) is obviously nonzero, it follows that f' (z) = $ (t)/ (t) 0 in a part of a sufficiently small neighborhood of the point z0 contained in B. This means that the zeros of the function F' (z) cannot cluster on the boundary of the domain B. Consequently, F' (z) and hence pa (z, have only finitely many zeros. Finally, since the function w = fi(t) maps the part of the neighborhood of the point t = 0 lying above the real axis into the part of the neighborhood of the point w = CO) that lies to the right of a vertical line passing through the point w = (N0), it follows that, as we move along the boundary of B in a neighborhood of z0, the function h(z, C) always changes in the same direction. More precisely, as z describes the boundary of the domain B in ,the positive direction, h(z, C) increases. Now, if z0 is an interior point of an analytic arc on the boundary of B, then the function 0 (t) is regular at the point t = 0. But then, the function F (z) and hence the function p (z, C) is regular at I

the point z 0.

By analogous reasoning, we can obtain a generalized theorem on harmonic functions: Let u(z) denote a function that is harmonic in a domain B bounded by closed

analytic Jordan curves. Suppose that u(z) is continuous in B and that its values .on the boundary of B constitute an analytic function of a parameter determining points on the boundary (that is, a function that can be represented in a neighborhood of each boundary point in the form of a series of powers of the parameter in question). Then an analytic function F(z) the real part of which is u(z) is regular in B.

Proof. Just as above, we shift to the t-plane. Then, the values of u (z) on

272

V1. MAPPING ONTO A DISK

the boundary of the domain B in a neighborhood of a point z0 can be represented by a Maclaurin series of powers of the real. parameter t. In the complex t-plane, this series defines a function that is regular at t = 0 and real on the'real axis (in a neighborhood of t = 0). When we return to the z-plane, this fast function becomes a function T W that is regular at the point z o and equal to u (z) on the boundary of the domain B. in a neighborhood of the point z o . The difference F(z) - 'Y (z) is regular in the portion of the neighborhood of the point z0 lying in B, and its real part is continuous, including boundary points of the domain B, at which points its value is 0. Resting on this last remark, we can again show that this difference is regular at the point z0. But then, the function F(z) itself is regular at the point z o. Let us now show how, once we know a Green's function for the domain B, we can find the solution of an arbitrary Dirichlet problem for that domain. Let us first consider the case of a domain B with boundary K consisting of closed analytic curves. Let u(z) denote a solution of a Dirichlet problem for the domain B with given boundary values which define on K an analytic function of a real parameter determining points on K. Green's formula 1) S S (utul-u1Ltu)dxdy=- S lud- uldn)dsBi

is applicable to the functions u(z) and u1(z) = g(z, 0 and to the domain B. obtained from B by subtracting the disk I Z - CI < E.2) Here, the integral on the left is over the domain B. i that on the right is over the boundary KE of the domain Be in the positive direction; ds denotes an element of'arc on KE; and d/dn denotes the derivative with respect to the inner normal. Since Au = Au 1 = 0 in BE and since u 1 = 0 on K, the preceding formula reduces to u gdRds=O. dgds(1) I zHere, the first integral is independent of c. Remembering that the function g (z, C) is of the order of magnitude of log c on the circle I z - CI = E, we easily u

dgds-+ -

tf

1) This formula is also valid for a domain B including z = o., as we can see if we apply it first to the domain Be obtained from B by subtracting the domain IzI > 1/E, let C approach 0, and then proceed as described in the text. 2) For finite C. In the case t = a., we take the domain Izt > 1/E.

§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION

273

conclude that the last integral approaches 0 as a --b 0 . With regard to the middle integral, we have d log

lim

.-.o

u dg ds = lim

It

1

ds

z

do

ds`

u

lim C

Iz

27cu (C). i

Thus, if we take the limit in ( 1) as a -+ 0, we obtain u (C)

2a

u (z) do ds.

(2)

This formula is known as "Green's formula". It expresses the values of the function u(z) inside the domain B in terms of its values on the boundary of B. Since the function p (z, C) = g - ih is regular on K and hence dg/dn = dh/ds (one of the Cauchy-Riemann conditions), we can put the preceding formula in a different form:

u (C) = 21-

it (z) dh (z, Q.

(3)

Let us show that formula (3) holds for a finitely connected domain B bounded by arbitrary closed Jordan curves and for an arbitrary given function u (z) on the boundary K, where the integral on the right should be understood as a Stieltjes integral. Let g(z,. C) denote a Green's function for the domain B and let 8 > 0 denote a number less than the distance from any of the zeros of the function p'(z, t;) (where p = g - ih) or the point C to the boundary K and also less than 'half the distance between different boundary curves constituting K. On the set consisting of all points of the domain B whose distance from the boundary K is 8 or greater, the function g (z, C) has a positive minimum A0. Consider the set KA of points of the domain B defined by the equation g(z, C) = A, where 0 < A < Ao . This set consists of analytic curves. To see this, let zo denote a member of KA. The points of KA in a neighborhood of zo have a parametric representation z = z (t) that can be obtained by solving the equation p (z, t;) = A + it. Since pi (z, 0 for z = z0, this parametric representation defines an analytic arc in a neighborhood of zo. Furthermore, the subset K,, of the set KA consisting of points at a distance no greater than 8 from a given boundary curve L of the domain B consists of a single closed analytic Jordan curve surrounding L because,

274

VI. MAPPING ONTO A DISK

otherwise, B would contain a domain in which g (z, t;) would be a harmonic function and on the boundary of which g (z, would be equal to A, which is impos-

sible since we would then have g(z, ) = A in B. It follows that the inequality g(z, t;) > A defines in B a domain Bk including the point z = J with boundary KA consisting of closed analytic Jordan curves approximating the boundary of the domain B. Now, suppose that u (z) is a function that is harmonic in B and continuous on B. Let us apply formula (3) to the domain BA. Since g(z, t;) - A is obviously a Green's function for the domain BA and since the function a = h(z, t;") increases as z moves around KA in the positive direction, so that we may take a as parameter determining points on KA, it follows that it (C) = 2 5 it (z) dh (z, () = 2

it (za (a)) da.

(4)

ltx

But u(z) is continuous on B. Therefore, if we also express the points on K in terms of a = h (z, t;), we conclude that, for every e > 0, the values of the function u(z) at points on K and KA corresponding to the same value of a will, for sufficiently small positive A, differ from each other by an amount less than e. Consequently, the integral in the right-hand member of (4) approaches the integral

over K as A -. 0. By taking the limit, we obtain it (C) =

(z,, (a)) da, 2a 1c u

that is,

u ) - 2>< 5 it (z) dh (z, t)

(5)

Finally, formula (5) remains valid when the function u (z) has a finite number of points of discontinuity on the boundary K but remains bounded (Iu(z)I < M) in B. To see this, let us include the points of discontinuity on K in the interiors of small arcs (z (ak ), z (ak )), such that I I ak - ak I < t/2M and let us denote by K. the remainder of the boundary K when these arcs are removed. Then, l

(u (za (a)) - it (zo (a))) do I

2

+

I it (z). (a)) - u (zo (a)) I do.

But off these arcs, u (zA(a)) -- u (zo (a)) uniformly. Consequently, for sufficiently small A, the last integral is also less than a/2. This proves formula (5) in the present case.

§4. APPLICATION TO A UNIVALENT MAPPING

275

We note that univalent mappings of the domain B transform formulas (3) and (5) into the corresponding formulas for the image. In conclusion, let us prove the symmetry of a Green' s function; that is, let us prove that g (z, C) = g (C, A. Let us apply formula (5) to the function u (z) _ g(z, C) + log 1z - Cj, which is harmonic in B and continuous on B. Since g (z, C) = 0 on K, we obtain

g(z,C)+ logI z-CI=Z 1ogI z'-CIdh(z, z). It Follows from this formula that g (z, C) is a harmonic function of C in B except at the point C= z, in a neighborhood of which it also has a singularity of the same type as the function g (C, z). Therefore, the difference d(z, C) = g (z, C) g(J, z) is a harmonic function of each argument throughout the entire domain B. g(C, z) < 0; that is, But, as we let z approach K, we have lim d(z, C) d(z, C) < 0 for every in B. On the other hand, as we let C approach K, we have lim d (z, J) = lim g (z, ) > 0; that is d (z, C) > 0 in B. Consequently,

d (z, C) = 0, that is g (z, C) = g (t z) for every z and C in B. §4. Application to a univalent mapping of multiply connected domains We can use the solution of the Dirichlet problem to give new proofs of theorems on the existence of a univalent mapping of finitely connected domains onto 'Certain canonical domains, for example, onto the plane with rectilinear parallel cuts or onto the plane with cuts along arcs of logarithmic spirals of equal inclination. Let us stop here to prove Theorem 1 ' of §2 of Chapter V on the mapping of an n-connected domain B in the z-plane onto the c-plane with finite parallel straight-line cuts of inclination 0 to the real axis. As we know, it will be sufficient to consider the case when the domain B does not include z = oo and is bounded by closed analytic Jordan curves K 1, K2, , K,,, where K is assumed to be the external boundary. Consider n - 1 functions Uk (z), for k = 1, , n - 1, that are harmonic in B. Suppose that each uk (z) is equal to 1 on Kk and equal to zero on all Kk, for k' k. For each it, let vk (z) denote the harmonic conjugate of uk (z). The functions vk(z) are in general multiple-valued in B but, in accordance with §3, they are continuous on B. Let us denote by r'k( the increment in the function vk(z)

276

VI. MAPPING ONTO A DISK

as z moves around KI in the positive direction (with respect to the domain B). One can easily show that the determinant jck,1I of the numbers a k.t is nonzero. Let us suppose, to the contrary, that this determinant is zero. Then there exist real constants kk , for k = 1, . , n - 1, not all equal to zero, such that Ik=1Akcok,1 = 0, 1= 1, ... , n- 1. The function

n-1

f(z)= YjXk(uk+1vk), k_l which can be extended into B along an arbitrary path, returns to its original value as z moves around each of the curves K, encircling K. But then, this function is single-valued in B and hence is regular in B. The real part of the function f (z) is constant on each of the curves Kk, for k = 1, , n. Then, reasoning as in the proof of Theorem 2 of §2 of Chapter V, we can show that f (z) = const in B. But this is impossible since the real part of the function f (z) vanishes on the curve K and is nonzero on at least one of the curves Kk for k = 1, , n - 1. This contradiction proves that Icuk.,I 0. Now, suppose that u0 is a harmonic function in B that is equal to -3`(e'`B/(z - a)) on the boundary K = Uk=1Kk and that v0 is its harmonic conjugate. Let us choose real constants Ak, k = 1', , n - 1, such that the function n-1

ve +

Xkvk

will be single-valued in B. This is possible since 1cok.,i / 0. Let us show that the function

f(z)=z

1

a +el01

-L

n-I

(110+1VO+ Z `k(uk+lvk)) k=1

is univalent in B. This function is regular in B except at the point z = a and assumes values on Kt (for l = 1, - , n) that lie on some straight line d, at an inclination 0 with the real axis. This shows that, if Co does not lie on the straight lines dl, l = 1, - - , n, the increase in the function arg (f(z) - CO) as z moves around the boundary of the domain B in the positive direction is equal to zero. On the other hand, this increase must be equal to 2n times the difference between the number of zeros and the number of poles of the function f(z) - Co in the domain B. Since f (z) - Co has a single simple pole in B, namely, at z = a,

it follows that f (z) assumes the value 4 at exactly one point in B. If the function f (z) assumed any value C1 at more than one point in B, this would also be true of points sufficiently close to S1, which, by virtue of what was shown above,

§5. MAPPING OF AN n-CONNECTED DOMAIN

277

cannot be the case. Thus, the univalence of f (z) in B is proved and so is the fact that the image of the domain B under the function C = f (z) is a plane with parallel rectilinear cuts of inclination 0 with the real axis. The function f (z) + c yields the same mapping, and for some c we have, in a neighborhood of z = a, the expansion I

z-a -{--al (z----a)+..., if a is finite and the expansion z + a1z-t +- - ' if a = oe. By analogous reasoning, we can prove Theorem of §3 of Chapter V regarding the mapping of a finitely connected domain B onto a plane with cuts along arcs of logarithmic spirals of equal inclination. §5. Mapping of an n-connected domain onto an n-sheeted disk

In conclusion, let us look at the possibility of mapping an n-connected onesheeted domain B onto an n-sheeted disk C1 < 1. By an n-sheeted disk, we mean here a Riemann surface lying entirely over the disk CI < 1 and having over each point of that disk exactly n points (a branch point of order p is considered as p points). This mapping is assumed to be effected by a bijective regular function. If we do not wish to resort to the concept of a Riemann surface, the mapping should be understood in the sense that, if 4 = f (z) is the function performing the mapping, then the function f (z) - Co has exactly n zeros in B for arbitrary Co such that I C0I < 1 (with a zero of order m counted m times) and has no zero at all in B for any Co such that I Co I > 1. Theorem. Every n-connected domain B in the z-plane that has no isolated boundary points can be put in one-to-one correspondence with the n-sheeted disk ICI < 1 by means of a regular function. If B is bounded by closed Jordan curves K1, , K,,, then the function defining this correspondence is continuous in B and it can be normalized in such a way that, at n given points al, , any where , n, it assumes a single given value a such that I a.I = 1 ak a Kk, for k = 1, and, at two given points b and c of the curve K that are distinct from the an and that are distributed along with the an in a definite order on K., assumes given values f3 and y distinct from a and distributed, along with a, in the same order on the circle I CI 1. Under this normalization the mapping is unique. (Here, "order" refers to the direction of motion around the boundary of the domain B and the boundary of the n-sheeted disk.)

278

VI. MAPPING ONTO A DISK

Proof. As always, it will be sufficient to consider a domain B that does not include - and that is bounded by closed analytic Jordan curves Kk, k = 1, - , n, where K,, is the outer boundary. Suppose that the curve Kk (where k = 1, -- , n) is defined by the equation z = zk (t), for 0 < t < 1. Suppose also that zk (0) = ak and that the point zk (t) moves along Kk in the positive direction as t increases. Instead of the disk I < 1 and the n-sheeted disk mentioned in the theorem, we can obviously choose an arbitrary simply connected domain G that can be mapped onto the disk and the n-sheeted surface over it respectively. For such a domain lying over the (w = a + iv)-plane, we take the strip G : 0 < R W < 1 and seek a function w = f(z) = u + iv that is regular in the domain B and that maps B onto the n-sheeted strip lying over G and having the property that w --, + - i as z approaches points ak through points in B. To construct the function w = f(z), let us choose on each of the curves Kk, for k = 1, - , n, a point bk = zk (tk), where 0 < tk < I. These points, together with the points ak partition the curves Kk into arcs Kk : z = zk (t) for 0 < t < tk and Kk : z = zk (t) for tk < t < 1. Suppose that u (z) is a bounded harmonic func-

tion in B that is equal to zero on the arcs Kk and equal to 1 on the arcs Kk . Let v(z) denote the harmonic conjugate of u(z) (defined up to a constant term). In general, the function v(z) is not single-valued in B. Let wk, for k = 1, -, n, denote the increase in v(z) as z moves around Kk in the positive direction. Here, in a neighborhood of the points ak and bk, the motion is along small arcs contained in B. The numbers wk depend on the choice of the points bk. If we succeed in choosing the points bk in such a way that all the numbers wk are equal to zero, then the function f (z) = u + iv will be single-valued and hence regular in B. Let us show that, in such a case, the function f W serves as the required mapping. To do this, let us investigate first the behavior of f (z) in a neighborhood of the points ak and bk. Let us map the simply connected domain bounded only by the curve Kk and containing the domain B onto the halfplane J) > 0 in such a way that the points ak and bk are mapped respectively into The mapping function z = qS(t;) is regular at ; = 0 and has an expan0 and 0, in a neighborhood of that point. , where cl sion z = 0 (C) = ak + c 1t; + This transformation maps a(z) into a function that is harmonic and bounded in a domain of the c-plane that contains the real axis on its boundary and that has the values 0 and 1 on the positive and negative halves of the real axis respectively. But the function (1/rr) arg C has this same property. Therefore, the difference f(0(C)) - (1/77i) log C can, in accordance with the symmetry principle, be continued

§5. MAPPING OF AN n-CONNECTED DOMAIN

.

279

past the real axis and it is a function that is regular on that axis except at the point 1= 0. It follows that the function f(z) = (1/7ri) log (z -ak) is regular in a neighborhood of the point z = ak. In the same way we can prove that the function f(z) + (1/7ri) log (z - bk) is regular in a neighborhood of the point z = bk. It follows from what was said above that the function v(z) approaches +00 uniformly, at the same speed as the function -n-1log I z - akl, as z approaches ak through values in B, and that v(z) approaches -- uniformly at the same speed as the function -n-'log (111z - bkl), as z approaches bk through values in B. This .property enables us to show that the function w = f (z) maps the domain B onto the' n-sheeted strip 0 < $2 (w) < 1. Let w0 denote any point in the strip 0 < Y (w) < 1. Since the function f(z) does not assume the value w0 in sufficiently small neighborhoods of the points ak and bk for k = 1, , n, the number of .zeros of the function f(z) - w0 in B is equal to (21r)-1 times the increment in arg [f(z) - w0] as z moves around the boundary K = Uk=IKk of the domain B in the positive direction, where, in neighborhoods of the points ak and bk, the path is along small arcs contained in B. But as z moves around the curve Kk (for k = 1, , n) in the positive direction, beginning at some point zo a Kk, the corresponding point w = f(z) first moves downward along the imaginary axis and then moves along a sufficiently distant curve to the line S% (w) = 1 and then moves upward along that line. It then moves along a sufficiently distant curve back to the imaginary axis and returns along it to the initial point z0. This shows that the

increment in arg [f(z) - w0] as z moves around the curve Kk in the positive direction is equal to 2n. Then, the increment in arg [f(z) - w0] as z moves around the entire boundary of K in the positive direction is equal to 27rn. Consequently, the number of zeros of the function f (z) - w0 in B is equal to n no matter what the value of w0; that is, the function w = f (z) does indeed map the domain B bijectively onto the n-sheeted strip 0 < R(w) < 1. We assume that the function f (z) is single-valued in B. , G) n Let us show that values t 1, , t,, for which all the numbers rv 1. are equal to zero do exist. Choosing tR arbitrarily in the interval 0 < t < 1, let us consider the different tk such that 0 < tk < 1 for k = 1, 2, , n - 1. , n, are functions of t 1, Then u (z), v (z), and &)k, k = 1, , t,-I- We set

tt(z)=tt(z, t1, ..., t,-t), v(z)=v(z, ti, ..., to-), Dk=(lk/4, .. ., to - r), k = 1, ..., n.

280

VI. MAPPING ONTO A DISK

Let us show that there exist numbers t1,

,

t,,-, such that

k=1, 2, ..., n.

Wk(t1, .. .,

(1)

To do this, we mention first a few properties of the functions W k (t 1, '

to-1): are continuous functions of each of the

1) The functions CJk(t1, ,

variables tv for 0 < tY < 1, where v = 1, accordance with §3,

, n - 1. This is true because, in

a

tt (z, t1, ... , to -1) = 2a

A

S dh (C, z) = 2a

k=1Kk

(h (bk, z) - h (ak, z)), k=1

,

and hence u(z, t1i au/ax, and au/dy are continuous functions of z, , as z varies in B and each tv varies in the interval 0 < tv < 1 for ti,

v=1, ,n-1. But wk (t 1,

,

1) is determined from the formula

dv- -aydx+azdy

wk

k

where the integral is over an arbitrary closed analytic curve Ck contained in B and encircling only the curve Kk. Consequently, each wk (t1, , t,,_1) is a continuous function of tv in.the interval 0 < tv < 1, for v = n - 1. 2) Each wk (t 1, , is an increasing function of the parameter tv for v k. To see this, note that, for tv > lv, the difference A", k = wk 01, ... , t", ... , to-1) - wk 01, ... , t", ..., to-1) is the increment in the function vv(z) as z moves around the contour :Kk. This function is the conjugate of the function uv(z), which is harmonic in B and vanishes on K except on the arc of the curve K,, between the points zv(ty) and zv(tv), on which it is equal to 1. Since uv > 0 in B, we have auv/an > 0 on Kk. This is a strict inequality because vanishing of the above derivative, together with vanishing of the derivative auv/as, would mean that the derivative of the function fv(z) = uv + iv,,, vanishes at some point on Kk and hence the image of the domain B under the function w = fv(z) would get outside the strip 0< 92(w)
".k=

dv"=a ds=- a k

k

k

ds

§5. MAPPING OF AN n-CONNECTED DOMAIN

,

that ev,k <0, and this means that wk (tl,

t

for v

281

k.

3) We have

n

wk (t1,

..., to 1) = 0.

(2)

k -- 1

To see this, note that if the domain B can be converted into a simply connected domain by means of suitable cuts, the function v(z) will be singlevalued in it and consequently its increment as z moves around the entire boundpry of the domain is equal to 0. On the other hand, this increment is equal to them left-hand member of equation (2).

, t,_1) (for k = 1,

4) It follows f r o m properties 2) and 3) that &)k (t 1, -

-

, n - 1) is a decreasing function of the parameter tk.

5) The function wk (t 1, is positive for tk = 0 and negative for , , In- 1 are different .tk = 1 (under the assumption that all the other numbers t 1, -

from both 0 and 1). To see this, note that for tk = 0, the function u is equal to unity on Kk and less than unity in B. Consequently, au/an < 0 on Kk. Also, we can show just as in 2) that this has to be a strict inequality. Therefore it follows on the basis of the formula

wk= dv=- 1 ands 5.

(3)

k

that wk > 0. Analogously, for tk = 1, the function uk is equal to 0 on Kk and ,positive in B. Consequently, au/an > 0 on Kk and, on the basis of equations (3), we have a) k < 0.

Now, by using properties 1)-5), let us show that the system (1) has exactly one solution. It follows from properties 1) and 5) that, for every system of values , t_), 0 < t l < 1, such , 1, there exists a unique value t 1 = r1(t2, 92; is continuous with respect to 0 with rl (t2i , , that w 1 (rl, t2, -

-

-

t2i

. , In- 1 Since, in accordance with property 2),

w 1 (rl, t2,

,

In-,) increases

with increase in t2, , In- 1 and, in accordance with property 4), decreases with increasing rl, it follows from the identity w1 [T1 (t9, ... , to-1), {y, ..., tn_r] = 0

that rl (t2i -, tn_0 increases with increasing ti, for v / 1. Let us define the functions wk(t9, ..., to-1)=wk(r1(t9, ...,

tn_1), t9,

..., tn_1), k=2, ...,

lt.

282

VI. MAPPING ONTO A DISK

These functions are continuous with respect to t2, , tn_ 1. For each k, the obviously increases with increasing tv for v k. function wk (t2, - , Furthermore, it follows from (2) that, for all t2, , tr_1,

wk (t2, ..., to-1)=0, k-2

,

n - 1) increases with It then follows that wk (t2, or k = 2, 1) is posiincreasing tk. Finally, it follows from property 5) that (Jk (t2, , tive for tk = 0 and negative for tk = 1. Consequently, the functions wk(t2, , satisfy conditions analogous to conditions 1)-5), with the number of functions and the number of variables decreased by one. We reason in the same way with these functions: there exists a unique value t2 = r2 1) n the interval , 2 (t3, 0 < t2 < 1 that satisfies the equation w2'(t2, , 1) = 0. Here, r2 (t;, , tn_1)

is continuous and increases with increasing t for v>2. Now let us consider the functions Wk (t3, ... , to-1) - ok ('t8 (t3, ... , trt_1), t3i ... , tn_1), k = 3, ... , h, and continue in the same way. After n - 1 steps, we arrive at the uniquely defined system of values tn_1 = 'tn_I = const, tn-9 = 'cn-2 (tn_1),

, n - 1. But then, in accordance with equation (2), we also have wn(r1, ..., rii_1) = 0. This proves that there exists a system of values t1i satisfying equations (1) and hence that there exists a function w = f(z) mapping the domain B bijectively onto the n-sheeted strip such that wk (r1,

, rn_ 1) = 0 for k = 1,

,

0 < Pw < 1 in such a way that the given points ak E Kk, k = 1, 2, . , n, are mapped into w = + i-. Let us now prove the uniqueness of the mapping. In the treatment above, the only arbitrariness was in the choice of the value of t in the interval 0 < t < 1 and the constant term up to which the single-valued function f (z) was defined. Suppose now that, in addition to the point an, two arbitrary points b and c are defined on the contour Kn and suppose, for definiteness, that the points an , b, and c occur in that order as one moves in the positive direction around Kn. If c = z (tn), for 0 < to < 1, we fix that value t = to as indicated above and from it we can find the system of values t1, , tn_1 which determines up to a constant term the single-valued function v (z). We choose this term in such a way that the point b C Kn is mapped into a given point 8 on Rw = 0. Then, the points an, b,

§6. SOME IDENTITIES

283

and c on K are mapped respectively into + ioo, S, and - ioo and the function w = f (z) is uniquely determined. This completes the proof of the theorem. The proof that we have given is due to Grunsky.1) The question arises whether the idea of the proofs of numerous existence theorems can be used in the present case to establish the existence of functions mapping an n-connected domain B onto an n-sheeted disk by using some extremal property of these functions. It turns out that this is possible and that in this connection the problem centers around the maximum value of If' (a) 1, where a is a finite member of B, out of all the functions f (z) that satisfy the condition f (a) = 0, that are regular *in B, and that do not exceed unity in absolute value. The proof is given in §3 of Chapter XI. It uses a number of boundary properties of analytic functions which will be explained in Chapters IX and X. §6. Some identities connecting a univalent conformal mapping and the Dirichlet problem

In §4 of Chapter V, by using the method of contour integration, we established certain relationships connecting various functions that map a given finitely connected domain univalently onto canonical domains. In the present section, we shall continue the application of this method2) by introducing into our study appropriate harmonic functions defined as the solutions of certain Dirichlet problems. Specifically, suppose that B is an n-connected domain that does not include oo and that is bounded by n closed analytic Jordan curves K. for v = 1,--. , n. Suppose that a)v(z) (for v = 1, . , n) is a function that is harmonic in B, equal to 1 on Kv, and equal to zero on all Kµ, for µ v. We note that the functions ojv(z), for v = 1, - , n, satisfy the relation n

a Ya

(z) =1,

(1)

because the sum on the left is a harmonic function in B and it is equal to 1 everywhere on the boundary K = U v=I K. The harmonic function iv(z) conjugate to wv(z) (v = 1, , n) is not in general single-valued in B. Let (Z). (z))K, as usual denote the increment in the function tvv(z) as z moves around the curve Kµ in the positive direction (that is, positive with respect to the domain). Sup-

1) Grunsky [19371.

2) See Garabedian and Schiffer [19491.

VI. MAPPING ONTO A DISK

284

pose that

( (z))K = 2aPl,

µ, v=1, ..., n.

If we now set wv(z) = cov(z) + iwv(z) (for v = 1, . , n), then the analytic funcsatisfies the conditions

tion

1) W.(z)_-W.(z)+c,,,, on Kµ, cµ

const,

(2)

2) (W, (z))K = 2it1Pt,.

(3)

3) WY(Z) is regular in B. Turning to the contour-integration method, we shall use the functions P(z, u), Q (z, u), P (z, u, v), and Q (z, uy v) introduced in §4 of Chapter V. With respect to these functions, we recall that they satisfy on K. (v = 1, . , n) the conditions

P (z, u) = - Q (z, it) + k. (u),

(4)

P(z, u, v)_- Q(z, u, v)+k,(u, v),

(5)

where kv(u) and kt,(u, v) are independent of z in K. We note that the functions kt,(u) and kv(u, v) will appear explicitly in the relationships that we derive below.

Now consider the integrals

t = tai 5 P (C, u) WY (C) dC,

Is =tat

P (` u, v) w, (C) dC.

(6)

5

In accordance with Cauchy's theorem, each of these integrals is equal to 0. If we use formulas (2) and (4) to transform the first of these integrals, we obtain n

II =

J P (C,

u) dw, (Q (C, u) -+ kµ (u)) dw., (C) c

c KK

IL

KKµ

n

l

(- Q (C, u) + k, (u)) dW, (C)

W=

_-

I,

a

l

271

Q (G u) Wy (C) dC + YI kl, (u) Pµ. µ=1 n

W (u) + Y, kµ (u) PI,. µ-1 As a result, we arrive at the formulas (for u E B)

W;(u)_ I k,,(u)PIL. M

µe1

(7)

§6. SOME IDENTITIES

285

Transforming the second of the integrals (6) in the same way, we obtain

+ 1'

n

Q (C. U, v) %'(Q A -

= - tat

kµ (u, v) P1, W=1

KKK

To evaluate the remaining integral, we make a cut y in B from u to v. Since the integrand in the domain so obtained is single-valued, the integration over K can be replaced with integration over an arbitrary closed analytic Jordan curve C contained in B and containing the cut y but no point of the boundary of B. Integration by parts then leads to the following: n 19

k (a, v) Pµ,

(C, u, v) w. (C) dC + µ=1

n

w (u) - w, (v) +

kN, (u v) Pµ. V.

µ=1

As a result, we obtain the formulas n

w (v) - w, (u) _ E kµ (u, v) Pµ,

v = 1, ... ,

n.

(6)

µ=1 Formulas (7) and (B) are interesting in that they provide expressions for the functions wv(u) and wv(v) - wv(u) (defined as solutions of Dirichlet problems)

in terms of the functions kAL (u) and kIL (u, v) (defined in terms of univalent mappings). With an eye to finding inverses of these formulas, let us show that the rank of the matrix I1PIL,vII is equal to n - 1 and let us at the same time clarify

the nature of the indeterminateness of the inverse problem. To do this, let us consider the system of linear equations n

E Pµ .kµ=0, µ-l If this system has nonzero solutions AL, for R = 1,

(9)

, n, then the function

n

I) wµ (z) j

µ=1

is single-valued and consequently regular in B. Also, it assumes on each curve K. (for v = 1, , n) values lying on some rectilinear segment. By the same reasoning as was used in the proof of Theorem 2 of §2 of Chapter V, we can show that this function must be constant in B. But then the function µ=1 " A (D (z) µk will also be a constant. Since this constant assumes the value av on K. (for

286

VI. MAPPING ONTO A DISK

v= 1,

, n), it follows that Al= A2 = .. = An. This shows that the only possible system of solutions of equations (9) is the system Al = A2 = ... = An .

Now, on the basis of (1), we have n

µ=1

wµ (z) = const in B,

(10)

so that the increment in the sum on the left as z moves around any of the curves in question is equal to 0; that is n

Ei Pµ,0, µ=1, ..., n.

(100)

µ=I

From this we conclude that the numbers Al = A2 = .. = An do indeed satisfy the system (9). This shows that the rank of the matrix lI Pµ, vll is equal torn - 1. Furthermore, it follows from what we have shown that the functions kµ(u), for µ = 1, . , n, can be determined from formulas (7) and that the functions kµ(u, v), for p = 1, , n can be determined from formulas (8), in both cases up to a com-

mon term. On the otter hand, the differences kµ(u) - kµ, (u) and kµ(u, v) kµ, (u, v), for p, p' = 1, , n, are uniquely determined by these formulas. This last fact is significant, for example, for formulas (9) and (10) of §4 of Chapter V. Let us now turn to the Green's function g(z, C) for the domain B. We denote the corresponding analytic function of z by p(z, t), so that g(z, 0 = 2 (p (z, Obviously, the function p (z, C) has a logarithmic singularity at the point z = t and is not single-valued in B. The multiple-valuedness is easily shown: (p (z,

C))K _ ` dp (z, C) _ t 12`

V

V

a' (P (z, Q) as

ds

-

ag (z, C) ds = - 21tlw (C).

t AS

Thus, the function p (z, t) possesses the properties C10) 1) p (z, C) = - p (z, C) + const on KY; (10,11) 2) (p (z, C))KV = - 21tw (C); 3) pE (z, C) is regular in B except at the simple pole z = t with residue -1. Consider the integrals 18

=tat

P (C, u) pz (C, z0) dC and 1 = tat S P (C, u, 'U) pz (C, z(,) 4

§6. SOME IDENTITIES

287

with u, v E B. If we evaluate these integrals first by using the residue theorem directly and then by using (4), (5), and (10 as above, we arrive at the following formulas: A

pz (u, zo) = Q (zo, u) + P (z0, u) -

kµ (u) wµ (zo), µ=1

P (v, zo) - P (u, zo) = Q (zo, u, v)

(11)

+ P (zo, u, v) -

kµ (u, v) wµ (zo). IL-1

we separate the real parts in (11), we obtain the curious formula n

g (v, zo) - g (rt, zo) =1 og j , (zo, u, v)

og pµ (u, v) . wµ (zo)

(12)

Here pµ(u, v), for p = 1, . , n, are the radii of circles on which lie respectively , n, under the mapping w = jn/Z(z, u, v). the images of the curves K., for p = 1, By using the symmetry of the Green's function and replacing zo with z E B, we can also rewrite (12) in the form g (z, v) - g (z, u) _ R (log j n (z, u, v)) -

a

log Pµ (u, v) . wµ (z)),. µ=1

Then, together with (11) we have the formula

p(z, v)-p(z, n)=logj (z, it, v)- Elog pµ(u, v) w,(Z)j-

µ=1

The purely imaginary constant term on the right is not written because we have not normalized the imaginary parts of the functions p (z, C) and w. (C) in any way. The actual result is obtained as a relationship of the analytic representation of functions that are meromorphic in the domain B and real on the boundary K. These functions are known as Schottky functions. Let f(z) denote such a func-

tion that has in B only simple poles ak, for k = 1, , N, with residues rk. Furthermore, suppose that f(z) is regular on K. Consider the integral Ia

=2

P (C, u, v)f (C) A.

From the residue theorem we have N

15 = }J rkPZ (ak, u, v). k=1

On the other hand, by using (5) and the relationship

288

VI. MAPPING ONTO A DISK

-

f(z)=f(z) we obtain n

16 = I r=1

2aI

t

on K, (v=1, ..., n),

(- Q (C, it, v) + kr (u, v)) df (C) r

N rk Q: (ak, tt,

Q (C, u, v) f' (C) dC' =

= tat

v) +f (u) -f (v)

k=1

(The last integral is calculated by use of the cut y connecting u and v, just as before.) As a result, we obtain the relationship N

f (it) -f (u) = E (rkQ; (ak, it, v) - rkPZ (ak, u, v)).

(13)

A-1

By using formulas (3) of §4 of Chapter V, we transform (13) to the form N

f (u) - Z (rk Q (u, ak) - rkP (it, ak) k=1

N

=f (v) - E (rkQ (v, ak) - rkP (v, ak)) k=1

Since the left-hand member depends only on u and the right-hand member only on v, we conclude that the function f (u) has a representation in the domain B of the form

f (u) _

(rkQ (u, ak) - rkP (u, ak)) + A,

(14)

k=1

where A is a constant. , N, cannot be arbiIt turns out that the numbers ak and rk, for k = 1, trary. It is easy to establish certain relationships between them. For z E Kv, we obtain from (14) on the basis of (4) f (u)

_

N

}J (rkQ (u, ak) + rkQ (u, ak) - rkkr (ak)) + A k=l N

N

=2Z'T(rkQ(it,ak))- Z rkkr(ak)+A k=1

k-1

(15)

Since f (z) is real on K., we have N

Erkkr (ak) -A)=O, v=1, ..., n. k=1

(16)

289

§6. SOME IDENTITIES

If we multiply (16) by Psum with respect to µ = 1,

, n, and use relation-

ships (7) and (10), we obtain the conditions k=1

rkw"(ak))=0, =1, ..., n.

(17)

It is a remarkable fact that conditions (17) on the numbers ak and rk are not only necessary but also sufficient for the function (14) constructed from them to be, for suitable A, a Schottky function. To prove that conditions (17) are sufficient, suppose that they are satisfied. Then, by using (7), we can transform them to the form 0

y

nn

E PP. .3

P=1

k=1

rkkµ (ak)) = 0, y ^ 1, ... ,

n.

(18)

If we consider (18) as a system of equations of the type (9), we conclude on the basis of what was proved above that this system has only equal solutions, and this leads to n relations (16) with suitable A. But then, it follows from the identity (15) as applied to z C K. that the function f(u) defined in accordance with formula (14) is real on K,, for all v = 1, - - , n, so that f (z) is a Schottky function. , n, Let us stop for an application of this result. Let ak, for k = 1, denote arbitrary points in a domain B. From the identity n

w,(z)0, zEB,

v=1

we conclude that the system of n linear homogeneous equations n

(etewY

(ak)) = 0, k = 1,

. . .

has, for arbitrary 0, the nonzero system of solutions A 1 = .. = An = 1. Consequently, the rank of the matrix IIg (e`Bwv(ak))II is less than n. But then the system of equations n

rk (e;eW

(ak)) = 0, Y = 1, ... ,

n,

also has nonzero real solutions in the rk, for k = 1 , - , n. Consequently, conditions (17) are satisfied for the numbers rke `9 and ak , for k = 1 , , n. Therefore, there exists a constant A such that the corresponding function f(u) defined in accordance with formula (14), which, in the present case, is the function n

t

! (z) = E rk k=1

n

(e l1 Q

(z, ak) -

e-'OP

(z, ak)) + A =

k=1

rke1B1-e (z, ak) + A

290

VI. MAPPING ONTO A DISK

[this last on the basis of formula (1) of §2 of Chapter V], is a Schottky function. The only poles that this function can have in B are included among the ak. If the number of these poles is m, where m < n, then, by evaluating the increment (arg [f(z) - c]) K, we easily conclude that f(z) assumes every value not lying on suitable intervals of the real axis at exactly m points of the domain B. Consequently, the Riemann surface consisting of m sheets of the w-plane with suitable cuts lying over the real-axis is the single-valued image of the domain B under the bijective function w = f(z). In conclusion, let us look at functions that are regular in the region B with modulus equal to unity everywhere on the boundary K. It follows easily from Cauchy's theorem on the number of zeros that if such a function F(z) has exactly N zeros in B, it maps the domain B onto the N-sheeted disk I C1 < 1. Suppose now that the function F(z) has in B only simple zeros ak, for k = N, and that lim s-.alt z

Then, the functions

F(z) ak

-

= , k=l,..., N. rk 1

F (z) + F (z) and I (F (z)

(19)

-F

(z))

are obviously Schottky functions in the domain B. If we apply formula (14) to them, keeping (19) in mind, we obtain the following formulas: N

F(z)+F(z) = Y, (rkQ(z, ak)-rkP(z, ak))+A, k=1

I

(20)

N F (Z) - (z)Z) "? (rkQ (z, ak) + rkP (z, ak)) + B,

where A and B are constants. Furthermore, from (20) we obtain N

F(z)=- Zrkp(z, ak)+C, k=I

N F (z) _

L rkQ (z, ak) _r 'jh-1

In the present case, conditions (17) take the forms N

N

2(Erkw,(ak))=O, 9(Zrkwy(ak))=0, v=1, ..., n, k-I

k-1

(21)

§6. SOME IDENTITIES

291

and are equivalent to the complex conditions N

I rkw (ak)=0, v=1, ..., n.

(22)

k=1

The question regarding the sufficiency of conditions (22) for the possibility of construction of the corresponding function F(z) by means of formulas (21) is not answered here. Remark. The functions that we nave been considering include, for example, the functions of §5 that map the domain B onto an n-sheeted disk. However it is possible to exhibit examples of simpler functions of this kind. Specifically, &ch of the two functions Pi (z, u) Ps (z, u, o) { J 1 (z) = Qs (z, u) , A (z) = Qz (Z' 1, v) ,

(23)

where u, v E B, is such a function and each of them maps the domain B onto the 2n-sheeted disk CI < 1. We confine ourselves to proving this assertion only for the first of the functions (23), assuming, as above, that the domain B is bounded by closed analytic , n. Jordan curves K,,, v = 1, In the first place, it follows from formula (1) of §2 of Chapter V that, for z, u C B.

je (z,_ u) = ei0 (cos 6j; (z, u) - I sin 9j' (z, u)) # 0. Therefore,

T Jo (z, u)

J'. (z, u) # I tan'9,

(24)

0

for all real 0. This means that, for all z, u C B, the ratio on the left side of (24) cannot assume purely imaginary values. And since this ratio approaches 1 as z --* u, we conclude that it assumes only values in the right halfplane. But then, we have in B

Qz (z, u) = 2 (jo (z, u) +j'a, (z, u)) # 02

Thus, the function Qz (z, u) has in B a unique double pole z = u and has no zeros at all. In the second place, if z = zv(s) is the equation of the curve K. and s is the length of arc on K. measured in the positive direction with respect to the domain B, then, when we differentiate (4) with respect to s , we obtain

VI. MAPPING ONTO A DISK

292

P (Zr (S), U) zv (z) _ - Qz (z,, (S), U) z (S).

(25)

It follows that on Kv, Pi (z, u)

If, (z)

(26)

Qi (z, U)

On this basis, we conclude that the functions PZ (z, u) and QZ (z, U) have no zeros on K. To see this, suppose, for example, that QZ (a, u) = 0, where a E K. Then, in accordance with (26), we have Pz (a, u) = 0 and hence jo (a, u) _ .0 (a, u) = 0. Therefore, in accordance with formula (1) of §2 of Chapter V, we have 1B (a, u) = 0 for all 0. This means that, for all 0, the mapping £= 1e(z, u) maps the point a into the endpoint of the boundary cut of the image of the domain B. On the other hand, if in a neighborhood of z = a we set

J (z, u)=cl(z-a)+..., f,a(z, u)=dl(z-a)+... we conclude from the geometrical interpretation of the argument of the derivative of an analytic function that the ratio c 1/d1 is purely imaginary. Therefore, since je (z, u) = e18 (c1 cos 9 - Id1 sin 9) (z - a) + ... ,

it follows that the derivative 1e (z, u) has at the point z = a a zero of no less than second order for some real 0, and this contradicts the univalence of the function je (z, u) in the domain B. This proves the assertion. Thirdly, from (25) we have(( arg

Q'(z,

)/K-(argz'(s))K-2(argz'(S))K=4a (n-2).

Since (arg Q. ,(z, u))K = - 4n, it follows from what we have proved that (arg Pz (z, u))K = 4rr (n - 1).

This last equation shows that the function PZ (z, u), which is regular in the region B, has exactly 2n - 2 zeros in B. But then the function f1 (z) has exactly 2n zeros in B. This information, Cauchy's theorem on the number of zeros, and property (25) enable us to conclude that the function 4 = f 1(z) assumes in the domain B every value in the disk I CI < 1 at exactly 2n points and it does not assume any other values at all; that is, it maps the domain B onto a 2nsheeted disk. This completes the proof of the assertion.?) 1) In a similar way, Garabedian and Schiffer [19501 have constructed functions that map the domain B onto an n-sheeted disk.

CHAPTER VII THE MEASURE PROPERTIES OF CLOSED SETS IN THE PLANE

§1. The transfinite diameter and tebygev's constant In contemporary questions in the theory of functions of a complex variable, a significant role is played by certain specific ways of measuring closed sets in the complex plane. We begin with one of these ways, which was proposed by Fekete. 1) Let E denote a closed bounded infinite set of points in the z-plane. For n points z 1, z 2, .. , z. C E, we define n

V (z1' zy, ... ,

za) = rl (zk - zl),

n ,:- 2-

(1)

k, 141

k
As we know, the product V(z 1, . , z) is equal to the Vandermonde determinant of the numbers z 1, , zn Let V. = Vn(E) denote the maximum value of ,V(z1, . , zn)I as z1, , zn range over all systems of n distinct points belonging to the set E. Such a maximum exists by virtue of the fact that V(z 1 , .. , zn) is a continuous function of its arguments and the fact that the set E is closed. Finally, let us define 2

d5= Vn("-).

(2)

Let us show that do does not increase with increasing n. Let el' e2' , n+1 denote a system of points belonging to the set E such that C IV(S1, S2' ,' Sn+I)I = V, +1* Then, since V 01, ... , to+1) = (1

W (F1 - `a) ... (1 - -, t'i+1) V (Eg, ... , to+1)

we obtain 1) Fekete [ 19231. 293

(3)

VII. MEASURE PROPERTIES OF CLOSED SETS

294

IEt-Es1 ... IEt-En+tl Vn.

(4)

Analogously, Vn+1`IEl-ElI'IE2-EaI ... IEl - En+tl Vn, .

.

.

.

.

.

.

.

Vn+t<_IEn+t-Eil

.

.

.

.

.

.

.

.

.

.

.

E.4 -En

IEn+t-E91 ...

(5)

.

V.

Multiplying (4) and (5), we obtain Vnn f11

<

n.+1 r

so that 2

2

Vn+

VR+1

for

Vn+llln C

Vnln-1)

(6)

which was what we wished to show. Furthermore, since do obviously does not exceed the diameter of the set E no. This limit is for any n, it follows that do approaches a finite limit as n called, in accordance with Fekete, the trans finite diameter of the set E and is

denoted by d = d(E), d = limn, dn. In the case of a set E consisting of a finite number of points, it is natural to set d(E) = 0. There is a connection between the transfinite diameter d of a set E and another constant, known as Z`ebysev's constant. For fixed n, consider all nth+ cn and consider the maximum of degree polynomials pn (z) = zn + cIzn-1 + the modulus of each of them on E. Let us show that there exists a polynomial whose maximum modulus is minimum. Let mn = mn(E) denote the greatest lower bound of these maxima. Then, there exists a sequence of polynomials pn.k(z), k = 1, 2, , -, of the type described such that the sequence of their moduli on E converges to mn. Let us represent these polynomials in accordance with Lagrange's formula: n+1 (z

Pn, k (x) =

1z, v

1) (z - Z2) ... (Z -

- ()z (4,

(Z- Z1+1) ... (z

ZY-J ( Y 9) ... (Z, - Y-1)

Y+1) ... (zY

ZA+0

n+l)

Pn. k (zY),

where z 1, , zn t1 are points in E (the same points for all the polynomials). It follows from this representation that the polynomials pn. k (z) are uniformly bounded in absolute value on an arbitrary bounded subset of the z-plane. Consequently, in accordance with the condensation principle, the sequence of them contains a subsequence of polynomials that converges uniformly on every bounded

§1. TRANSFINITE DIAMETER AND CEBYSEV'S CONSTANT

295

subset of the z-plane. Since the sequences of their coefficients converge when the sequence of these polynomials themselves converges, it follows that the limit + cn, whose maximum modulus on E is function is a polynomial t. (Z) = zn + obviously mn. This is the polynomial with least maximum modulus on E. It is called ?ebyrev's polynomial for E. Together with to (z), let us also consider the polynomial tn (z) = Zn + + cn, which has the smallest maximum modulus on E out of all the polynomials p(z) = zn + + cn, which have zeros only on E. In this case, we denote the smallest maximum modulus by mn. We shall also call the polynomial 1 n (z) ? ebysev's polynomial for E. Let us define r n = and let us show that the sequence of the numbers n converges. We note that this sequence is bounded since, for rn, for n = 1, 2, z 0 E E,

T,-
maxIz-zoD, zEE

where D is the diameter, in the usual sense, of the set E. Define

llrr 2n-a, n-co

urn

Tn=P.

n-o0

We have a < P. We shall show that the opposite inequality, namely 8 < a, also holds. Let a denote any positive number and choose n such that rn < a + c. Then, on E, we have I tn(z)I < (a } e)n. If k and l are positive integers and z0 is a fixed point belonging to E, then, for all z in E (z - zo)l [tn (z)1k

DI (a +

_

T

It follows that mnk+t _< Dl (a + e)nk and I

Tnk+l C Dnk+t (a +

E)nk

nk

e)nk+l.

(7)

Now, if rnv - /3 as n,---+., by setting n v =nkv + l), where 0 <1 v < n, we conclude from (7) that /3 < a + e. Since a is an arbitrary positive number it follows that 0 < a. Thus, S = a; that is, the sequence Jr. I converges. The number r to which it converges is called the '2eby3eev constant of the set E. Let us show that r = d. Here, we may assume that the set E is infinite. As a preliminary, let us prove the inequalities

VII. MEASURE PROPERTIES OF CLOSED SETS

296

l' < (n + 1) M. < VV n

(8)

In the first place, we have

(x-Ei)(x=E9) ...

(X-En)_

V(X, E ..., En) V (El, ..:

En)

Therefore, if the points x , 61, .. , e n belong to E and if 61, I V(61,

,

6. are such that

, Sn)I = Vn, we conclude that the modulus of the right-hand member does not ex-

ceed V +1/ V . For x E E, the modulus of the left-hand member is no less than mn. Consequently, M. < Vn+1 IV . Furthermore, 1

,V (X1, ... , xn+1) =

1

1

x1

X2

xln--1

x2

' tn (X3)

to (X2)

...

VI

Xn+1

n-l

... Xn+l ... to (Xn+1)

If we expand the determinant in terms of elements of the last row, we obtain V (X1, ..., Xn+1) I < I to (XI) I I. I V (X9,

....,

Xn+1) I

+1 to (X9) I I V (X 1, x3, ... , Xn+1) I + ... + I to (X.+]) I I V (XI, ... , xn) I

Now, if x1, .. , xn+1 are points on E such that I V (x1, we obtain

,

xn+1)I = V +1'

Vn+i<(n+1)m.Vn, i.e. V nl<(n-}-1)mn. Thus, inequalities (8) are proved. They can be written in the form n

Now for n let us set the successive numbers 2, 3, sulting inequalities. We obtain 2

, n and multiply the re-

2

(T2... zn)n(n+1).(Vy)n (n+l)<(tn}1 2 1

I n(n+l) T2

2

Tn n( +l)

But (V2) 2/n(n+1) -, 1 and [(n + 1)!]2/n(n+1) _. 1 as n---,, co. Cons

ently, to

show that d = r it remains only to show that 2

(C1r 4

...

C:)n(n+l)_..

(9)

§1. TRANSFINITE DIAMETER AND CEBYSEV'S CONSTANT

297

, in as n -' oo. Since rn -+ r as n -+ w, the sequence r1, r2, r2, r3, r3, r3, which rn is repeated n times, also converges to r and the sequence of their logarithms converges 1) to log r. The arithmetic mean of the first n(n + 1)/2 of

these logarithms is the logarithm of the left-hand member of (9). That it converges to log r follows from converges to a finite Lemma I. If a sequence of real numbers al, a2, limit a, then the sequence of their arithmetic means (al + a2 + ' , + an)/n also converges to a as n - m.

Proof. Let c denote a positive number. Choose m > 0 such that Ian - al < 42 for n > m. Then, for n > m,

` (a, -a) L' n

ai+a2+ ... +an n

_a

n

r-1

(a, - a) n

The second term in the expression on the right is less.than a/2. But, for sufficiently large n, the first term is also less than a/2. This proves the convergence asserted in the lemma. The above reasoning remains valid when we are considering polynomials with zeros only on E. This means that if we set r = m, we also obtain the result

that rn

-b r = d as n - - - + oo.

Summarizing what has been said, we have

Theorem 1. For an arbitrary closed bounded set E of points in the complex plane, the trans finite diameter d and the &bysev constants r and r coincide : ti

d=r=r.

From this theorem follows the simple fact that, if the entire boundary of some bounded domain belongs to the set E, then the transfinite diameter does not change if we adjoin to E the interior points of that domain. Resting on the identity of the transfinite diameter and Cebysev's constant, We shall give two simple examples of the calculation of the transfinite diameter. 1. Suppose that E is the disk l zl < R. Since 2x

2n

1) KFor r

0. It is easy to show that (9) is also valid when r = 0.)

298

VII. MEASURE PROPERTIES OF CLOSED SETS

where z = Re`B, for an arbitrary polynomial pn (z) = z' + - + cn, it follows that mn > Rn. But for pn(z) = zn, we have maxZEE Ipn(z)I = Rn. Consequently, mn = Rn and d = r = R. Since the transfinite diameter does not vary under parallel translation of the set E, it follows that the trans finite diameter of an arbitrary disk is equal to its radius.

2. Suppose that E is the interval -a< x < a. For an arbitrary polynomial + cn, we have

pn(z) = zn + .

mExIPn(z)I=0:x_-It max IPn(acosx)I max 0_-X_-?C

Pn 2 (etx -}-

a tx) l

I

-

max I qn (x) I,

O'* x:* t

where qn(x) is the trigonometric polynomial an q4(x)=a0 +al cos x---...+an cos nx, aK=n=i

Since 2t 1

n

Ign(x)I9dx=2Ia0I9+I Iak

I9aan

21(A-1)

k=d

it follows that

2 max O-

2a

q4(x)I>. 29

a9n (n-i)

so that max

Izl=l

IP4(z)Ian

2

n-2I

This shows that m n > an/2n- and r> a/2. On the other hand, since maxzEE Ipn(z)I = an/2n-I for the polynomial n

Pn (z) = 2n-i cos n arc COs a)=:z-+...+c,,

it follows that d = r = a/2. From this we conclude that the trans finite diameter of an arbitrary interval is equal to one-fourth its length.

299

§ I. TRANSFINITE DIAMETER AND CEBYSEV'S CONSTANT

Theorem 2. 1) Let E denote a closed bounded set and let p(z) denote a + cn. Let E* denote the set of all points z such that w = polynomial zn +

p(z)6E. Then d(E*)=" "-d(-E).

(10)

Proof. Let t. (z) and tn(z) denote the &bysev polynomials for E and E Itn(z)I, * = maxzEEs Itn(z)I. Then, we have the obvious Define in n = inequalities mnr = max

t,*,r (z)

max I zEP*

zEE*

tr (p(z)) I < max zEE'I t, (z)

mr

n

nr

we have the inequality mnr < jlmr, . If v d(E*) < . On the other hand, if, for arbitrary fixed w0 E E, we denote by z 1, , zn the roots of the equation p (z) - w0 = 0 and if we denote by zi, .

and consequently,

zv the roots of the equation tv(z) = 0, then it follows from the equation

,

n

r

r

n

H H (z!-z%)I=k=11=1 H jj(zt-z!)I 1=1k=1 that

1=1

(P(Zk)-Wo)I.

tv (Z,)I=I JJ

k-1

Since

4r (w) _ fl (w - P (ZZ)) k=1

is of with degree in w, we have from (11) my-,< max

(z) zEE*

WEE

and consequently, n

1) Fekete [1930].

V nor

Vm

11

300

VII. MEASURE PROPERTIES OF CLOSED SETS

As v -, 00, we have l d(E) < d(E"). Therefore, equation (10) must hold. This completes the proof of the theorem. If we apply Theorem 2 to the case in which E is a disk, we obtain the

Corollary. The set of points z defined by the inequality Ip(z)I < R, where + cn has transfinite diameter d = V R p (z) = zn + §2. -Bounds for the transfinite diameter

The question arises as to the relationship between the transfinite diameter and the usual measures that are employed in analysis. Pdlya proved two theorems of this nature, dealing with outer Jordan'content. Let us review the definition of this content. Let E denote a bounded set in the z-plane. Let us cover the plane with a grid of squares of side 1/n and let us denote by sn the sum of the areas of those closed squares that contain at least one point belonging to E. The limit of sn as s -p 00 is called the outer Jordan content (two-dimensional) of the set E. Analogously, if E is a bounded subset of a straight line, we partition this line into intervals of length 1/n and we denote by In the sum of the lengths of the closed intervals that contain at least one point in E. The limit of In as n ---f no is called the one-dimensional outer Jordan content of the set E. Let us turn now to the theorems mentioned above. Theorem 1. 1) The two-dimensional outer Jordan content of a closed bounded set E does not exceed 7rd2, where d is the transfinite diameter of E. Proof. Let us first prove the theorem for the special case when the set E is defined by the inequality Ip(z)I < R, where p(z) = zn + + c In this case, in accordance with §l, d(E) = 7-R-. Therefore, if we denote by SR the sum of .,We areas of the domains covered by the set E, we need to prove the inequality 2

SR=ICBM

(1)

For this, along with C = p (z), let us consider the inverse algebraic function z = f([;). This function has no more than n - 1 branch points (that is, points C= p(z) such that p'(z) = 0). Let us suppose that the circle ICI = p does not pass through any of these points. Then, the set of points z defined by the equation Ip(z)` = p consists of disjoint closed analytic Jordan curves K1, ... , K. These must lie outside each other because inside each of them we have I p (z)I < p by virtue of the maximum principle. There are no more than n of these curves 1)Polya [1928].

§2. BOUNDS FOR THE TRANSFINITE DIAMETER

301

(that is, m < n) because inside each of them there is at least one root of the polynomial p W. This last fact follows from the fact that otherwise we could conclude, on the basis of the maximum principle as applied to I1/p(z)I and to the interiors of the Kk, that p(z) = const. Then, as the point z moves once around the curve Kk, the point t; = p (z) moves around the circle ICI = p exactly mk times and the ) moves once around the circle p 1 im k. Consequently, one point t; = of the branches of the function z = f (Cm k) is regular on t; = pl/mk and has on that circle,the Laurent expansion

f

ao

(Cmk) = Fj a k)rv.

r=-ao

Therefore, on IC I = p, we have for one of the branches of the function f (C) co

v

Z =f (C) _ E a(k)C"'k.

(2)

Equation (2), with C = pe`46, is a parametric equation of the curve Kk. Here, c varies from 0 to 2rrmk. Therefore, the area of the interior of the curve K. is equal to 2zmk

xdy-ydx= 2 5(zdz)= 2 1k

dyl

dC

kk I

-z

/

2smk ao

l

(

a\7

v

ivy

ao

l avklpmke Wk l

L

!rq

V

ayk)pmkemkdp\

L M-A V-00

V=-00

J

00

I

- it

2v

vIa(rk)I9pmk.

Therefore, if we denote by SP the sum of the areas of the interiors of the curves K 1,

)

Km, we obtain m

2r

2

_LP

02/6' k-l v-_ao v I ask) I9 pmk

n

(3)

Here, the right-hand member is a nondecreasing function of p since each term in the summation is a nondecreasing function of p. This last is obvious for v > 0. The term with v = 0 is equal to 0. A term with v < 0 is negative and the exponent

VII. MEASURE PROPERTIES OF CLOSED SETS

302

2v/mk - 2/n in such a term is also negative. Therefore, SP/rrp2/" does not decrease with increasing p and is obviously a continuous function of p, including

values of p equal to the moduli of the branch points of the function f (c). But for p sufficiently large, we have m = 1 and m 1 = n because the function f(C) on ICI = p is then defined by the expansion

f which is the inverse of the polynomial for large p,

ap%n

p (z) about

This means that,

= 1 - I as 19F- 2/n - ... ` 1.

(4)

Consequently we have SP/rrp 2 "" < 1, that is, SP < np 2"" for all p. In particular, SR < rrR 2/n, which is what we had to prove. (We note that equality holds here only when al = a2 = - - = 0, that is, only for the polynomial p(z) = (z - a)".) Let us turn now to the general case of the theorem. For this, we use the coincidence of the transfinite diameter with &byslev's constant. Let tn (z) denote the &bysev polynomial and let r denote &bysev's constant for E. Then, for any e > 0, if n is sufficiently great we have I t" (z)I < (r + e)" on E. Consequently, the set E lies entirely in the interior of finite domains bounded by the curves defined by the equation I t" (Z) l = (r + e) " Therefore, the two-dimensional outer Jordan content of the set E does not exceed the sum of the areas of these domains and by virtue of what was shown above, this sum in turn does not exceed n(r + E) . Letting c -p 0, we obtain the assertion of the theorem. This completes the proof of the theorem. Theorem 2. The one-dimensional outer Jordan content of the orthogonal projection of a closed bounded set E onto an arbitrary straight line does not exceed 4d, where d is the transfinite diameter of E. Proof. Obviously, it will be sufficient to prove the theorem for the projection onto the real axis. Again, we shall consider first the case in which E is the set defined by the inequality 1p (z)1 < R, where p (z) = z" + + c". Suppose that n

P(z)_ k-1 [J (z-Tk), z=x+ly, Tk=ak+lbk.

§2. BOUNDS FOR THE TRANSFINITE DIAMETER

303

If P(z) = IIk1(z - ak), we obviously have I P(x)I < Ip(x)I. Consequently, if z E E, then IP (x)I < R. Let us denote by P the set of all real numbers x at which I P (x)I < R. Then, the projection of the set E onto the real axis is contained in P. This set P consists of a finite number of segments at the ends of which P (x) = ± R. Let us show that each of these intervals contains at least one root of the polynomial P (x). Consider the set of points z in the complex plane that satisfy the inequality IP(z)I < R. This set consists of a finite number of domains on the boundaries of which IP(z)I = R. Each of these domains contains no more than one of the segments in question. To see this, suppose that one of the. domains contains at least two segments. Then, by virtue of its symmftry about the real axis, it must be a multiply connected domain. Since IP(z)I = R on the exterior boundary curve of this domain, it follows on the basis of the maximum principle that I P (z)I is less than rather equal to R on the interior boundary curves. Furthermore, each of these domains contains one of the segments because the interior of any one of them that did not would not contain roots of the polyno-

mial P(z) (since all the roots of P(z) are real), so that, again by virtue of the maximum principle, we conclude that P (z) = const. Now, if there were no roots of P(z) on one of these segments, there would also be none anywhere in the entire domain containing- it and, consequently, we would again conclude that P (z) coast. This proves the assertion made. Let 1 denote the farthest right of all the segments containing P and suppose that 1 contains m roots of the polynomial P(z) (without consideration of their multiplicity). We have 1 < m < n with m = n only in the case in which P consists of a single segment. In addition to P (x), let us construct polynomials P 1 (x), P 2(x), of degree n with real coefficients and in particular with the coefficient of x' equal to 1 in all cases. By Pk, IV and mk, we mean the same thing for Pk(x) as we mean by P, 1, and m for P (x). Let us show that, if m < n, there exists a polynomial P (x) such that m 1 > m and the sum of the lengths of the segments constituting P 1 is greater than the corresponding sum for P. To see this, suppose that P(x) = Q(x)R(x), where Q(x) is a polynomial with zeros on 1 and R W is a polynomial with zeros outside 1. Furthermore, let d denote the distance from I to P - 1. Let us define P 1(x) = Q(x + d)R (x). If x E P-1 then I Q(x + d)I < I Q(x)I because each linear factor in the expression for Q(x + d) is less in absolute value than the corresponding factor in Q(x) by an amount d. Consequently, if x E P - 1, then I P 1(x) I < I P (x)I < R. This shows that 1

304

VII. MEASURE PROPERTIES OF CLOSED SETS

P_ I E P1. On the other hand, if x E 1, we have analogously I P 1(x - 41 = IQ (x)R (x - d)I 0 and sufficiently great n, we have I tn(z) I < (r + C) n on E. But, by virtue of what was said above, the projection of the set of points defined by this inequality onto an arbitrary straight line does not exceed 4 (r + e). Consequently, the projection of the set E onto the same straight line has one-dimensional outer Jordan content not exceeding 4 (r + E), that is, by virtue of the arbitrariness of c > 0, not exceeding 4r. This completes the proof of the theorem. If the set E is a continuum, we easily obtain from Theorem 2 the result that the ordinary diameter of the set E does not exceed 4d. As we shall see in §3, this result is equivalent to a bound on the diameter of the boundary of the image of the domain ICI > R under a univalent mapping by the function z = t; + ap + a 1/t,' + . and it follows directly from §4 of Chapter II. Thus, Theorem 2 is a generalization of this property of simply connected domains to multiply connected domains if by the sets E we mean the boundaries of domains. The question arises, is it possible in the case of an arbitrary closed bounded set to conclude on the basis of its transfinite diameter whether there exists an upper bound for its ordinary diameter, as is the case for a continuum? The following example shows that the answer to this question is negative. Let E denote the closed bounded set of points defined by the inequality

n

§2. BOUNDS FOR THE TRANSFINITE DIAMETER

(z-(X)(z-

305

Q)I_l, a>1.

Obviously, the transfinite diameter of this set is equal to 1. The boundary of the domain B complementary to E and containing oo is defined by the equation

j(z - a) (z - 1/a)l = 1, that is, by the equation

z-

2a

22

-1

and, for a2 - 1 > 2a, consists of two ovals without points in common. On the boundary of the domain B lie the roots of the equation (z - a) (z - 1/a) = 1, that is, the points 0 and a + 1/a. The difference between these two roots, which is equal to a +'1/a, can be made arbitrarily great by a suitable choice of a. In this case, the ordinary diameter of the set E can also be made arbitrarily great. We mention some consequences of Theorems 1 and 2 for a set E with zero transfinite diameter. In this case, the two-dimensional Jordan content for E is, by virtue of Theorem 1, equal to zero. It remains equal to 0 in the case in which all complementary domains not containing oo are adjoined to E because this does not change the transfinite diameter. It follows that E has no complementary domains not containing oo. Thus, if d(E) = 0, the complement to the set E is the unique domain containing oo and every point of the plane is either an interior or a boundary point of that domain. Theorem 2 proves more than this, namely, that E does not contain any continuum because if it did there would be a straight line such that the projection of the set E onto it would have a positive one-dimensional outer Jordan content. These are necessary conditions for d(E) to be equal to 0. They are not, however, sufficient. As Nevanlinna 1) has shown, there exist sets that do not contain a continuum but that have a positive transfinite diameter. We shall not stop here to construct such sets. In' conclusion, we present another interesting theorem. Theorem 3. 1) Let E denote a closed bounded set in the z-plane and let B denote the domain complementary to E that contains oo. If f(z) is regular in B and if it has the expansion f(z) = aIZ-1 + a2z-2 + in a neighborhood of then, by setting

1) KNevanlinna [1936].

2)Polya [1929].

VII. MEASURE PROPERTIES OF CLOSED SETS

306

An=

al

as

... a,

a9

a3

... an+t (5)

an

an+1

we have litn

ns

... a1,-1

d,

x-00

(6)

where d is the transfinite diameter of E. Proof. It follows from §1 that, for arbitrary c > 0, there exists a closed bounded set Et containing E and having in its complement the unique domain with boundary consisting of a finite number of closed analytic Jordan curves. We denote this boundary by K. Suppose that E* has transfinite diameter d' < d + Furthermore, it follows from §1 that, for the same c > 0, there exists an N > 0 such that, for n > N and for arbitrary points zk, where k = 1, . , n, in the set E, we have n(n-I)

n

k,t=1

(zk - zt)
2

(7)

k<1

If we now keep in mind the fact that

-

n=1, 2, ... ,

an = tai S f (z') z' n-1 dz , J 1

K

and substitute this expression into (5), we obtain A,,

= (2,c1)n J

...

f (z,).f (z9) ... f (z,,) ziz9ze ... zn-'

R

I

x

1

z1

I....

1

z9 ... zn

dzldz9

zj -I z ' -1

...

dzn,

(8)

n-1

. . . zn

where zk denotes the variable of integration in the integrals in the kth column. If in the integral in (8) we make all the possible n! permutations of the z 1, z 2'. , z,,, which does not change the value of the integral, and if we then sum over all these integrals, we obtain

§2. BOUNDS FOR THE TRANSFINITE DIAMETER

307

0 1

nlAn --

1

(tai) n KSK

... f(zi) ... f(Z.)

...

2

1

Zi ... Zn

dzi ... dzn.

(9)

n-i

n-l

... zn

zi

But we have inequality (7) giving a bound for the Vandermonde determinant in (9). we have Consequently, if we set M = maxZEK 1f(z)1,

A. I < (2a)n -'

-n(d nl

-2E) n (n-1)

n (n-1)

where L is the length of K. Therefore, lim nV An l C d + 2P.

n-oo

Since c is an arbitrary positive number, this yields (6). This completes the proof of the theorem.

Theorem 3, like Theorems 1 and 2, gives a lower bound for the transfinite diameter of a given set E. As an example of its application, let us look at a bounded continuum E containing the points 0 and 1. Since the projection of this continuum E onto the real axis is an interval of length no less than 1, Theorem

2 leads to the inequality d> 1/4. Let us show that this inequality can also be obtained from Theorem 3. The function

f (z)

log

z

-1 z

1

Z

I

1

...

1

T 2 ze

- n Zn - .. . 1

1

is regular in the domain B complementary to E that contains oc. Consequently, Theorem 3 can be applied to E and f(z). The determinant (5) in this case is 1

1

An= I

2

1

1

2

3

I

I

...

n

"'

n --- 1

1

1

I

n n+1 ' 2n-1 I

To evaluate this determinant, we subtract the last column from the first, second, , (n - 1)st columns. Then, after we remove common factors from the individual rows and columns, we subtract the last row in the resulting determinant from the first, second, , (n - 1)st rows. When we again remove common factors from the

308

VII. MEASURE PROPERTIES OF CLOSED SETS

rows and columns, we finally obtain the recursion formula A

n=[ 1-2...(n-1)

n (n+ 1) ... (2n -1) ]'A '_1

From this formula we get 2/n 1 2 ... (n- 1) = lim Ln(n+1) ... (2n-1)1 a1 n-.oo

An

lim n-+oo

(10)

16

This last limit can be calculated with the aid of Stirling's formula. We conclude from (10) that, for arbitrary positive c, there exists a positive number N such that, for n > N, (

`16+F If we now give to n the values N + 1, N + 2, . ities obtained, we get r

1

(T6

-t)

P(2N+P+1) 2

AN+p

,

N + p and multiply the inequal-

16 +E)

< AN

p(2N+P+1) 2

or 1

1 16 -

P(2N+p+1)

1

4

s) 2(N+P)' ANN+P)
P(2N+P +11

1

1 N+PP)

9

16

+

E)

1

a

? IN+P) AN +Pl

By virtue of the arbitrariness of e > 0, when we let p approach oo, we get n' limVAn=4 1

Consequently, in accordance with Theorem 3, we have d > 1/4. This is .a sharp bound for d because, if E is the segment connecting the points 0 and 1, we have d = 1/4 in accordance with §1. As will be pointed out in §5 of Chapter XI, inequality (6) remains a sharp inequality for an arbitrary set E consisting of a finite number of disjoint closed analytic Jordan curves. 1) This means that, if we replace d in (6) with an arbitrary number d' < d, this inequality will fail to hold for some function f(z) that is regular in the domain B. In that section, we shall present a theorem on the distribution of values of polynomials in the complex plane expressed in terms of the transfinite diameter. Furthermore, we mention briefly an actual application of Theorem 3 to power series. Specifically, we have

1)Goluzin [1946b].

§3. CAPACITY OF A CLOSED BOUNDED SEA'

309

alz-1 the theorem of P61ya: If two coefficients in the expansion f(z) = + are integers and if the function f(z) is regular in the complement of + a2Z_ 2

a closed bounded set E such that d(E) < 1, then it is a rational function. To see this, note that since, in the present case, all the determinants (5) are integers, it follows that (6) with d < 1 can hold only under the condition that these determinants, beginning with some number n, are all equal to zero. This last is a necessary and sufficient condition for f(z) to be rational (Kronecker's theorem). We shall not stop to prove this assertion.

§3. The capacity of a closed bounded set The importance of the constants d and r defined in the preceding section follow from their association with the Green's function. Let us again look at a closed bounded set E of points in the z-plane. The complement to this set in the z-plane consists of a finite or countable set of disjoint domains. We denote by B that one of these domains that includes oo. Let us show that one can exhaust this domain by means of domains B("), where n=1, ; each of which is bounded by a finite number of closed Jordan curves. Here, 2, exhaustion is understood in the sense that B(")CBI"+>, B(") C B, and every point z e B beginning with some n lies in B("). Let us cover the domain B with a grid of squares of side e/2 and let us consider the set consisting of points of the closed squares in the grid that contain at least one point belonging to E. The complement of this set consists of domains one of which is the domain Be containing oo. This domain, together with its boundary, lies in B. The distance from its boundary points to the set E does not exceed e. By extending the domain Be, we can arrange for its boundary to consist of disjoint closed Jordan curves. To do this, we need only adjoin to Be sufficiently small neighborhoods of the multiple points of the boundary of B. By giving c suitable values e" such that E" - 0 as n -- oo, we obtain a sequence of domains B(n). where n = 1, 2, , , that exhaust B. One can easily show that if the domains B(") exhaust B, then, for an arbitrary given e > 0, when n is sufficiently great the distance between all points of the boundary of the domain B(n) to the set E is less than C. Suppose now that the domain B defined above is exhausted by domains B(") with boundaries K("). The Green's function g"(z, oo) for the domain B(") is, as we know, a harmonic function in Bw except at the point z = oo, it is continuous in B("), it assumes the value 0 on K("), and it behaves in a neighborhood of z = - in such a way that the limit

310

VII. MEASURE PROPERTIES OF CLOSED SETS

lim g, (Z, 00) - log I Z l) = T.

exists and is finite. The quantity y" is called Robin's constant for the domain Bl"I. Consequently, in a neighborhood of z = -, we have ga (Z, 00)

log I Z + T. + U. (z),

where u" (z) is a harmonic function in Bi"' including the point z = 00, and approaches 0 as z -, 00. Since B(n) C B("+l), it follows that the function g" +l (z, -) - g" (z, 00) is harmonic in B (n ), including the point oc, and it is nonnegative on the boundary K'"). In accordance with the maximum principle, this last statement also holds everywhere in the domain B("); that is, for all z e B("), gn+1 (z, co):,:- gn (z, oo), Tn+1 :-:,.Tn

It follows in accordance with Harnack's theorem that the sequence of the functions y" + u"(z) either converges everywhere to + - in B or converges to a harmonic

function y + u(z) such that u(oo) = 0. In the latter case, the quantity y is called Robin's constant for the domain B, the quantity C = e-7 is called the capacity of the set E, and the function g(z, -) = log Iz I + y + u(z) is called the Green's function of the domain B. Obviously, C and G(z, o) are independent of the choice of exhausting domains B("). Green's function g(z, -) does not assume negative values anywhere in B. Consequently, as we let z approach the boundary B, all its limiting values are nonnegative. However, we cannot assert that these limiting values are always equal to zero. For example, at isolated boundary points of the domain B, the value of g(z, -) will indeed be positive because, in a neighborhood of any point, the function g(z, -) is a harmonic function that is' bounded below, so that it is also harmonic at that point itself. We note that, in the case in which Y. + u"(z) -i 'a for z in B, it is also true that y" -+ oc because u" (oo) = 0. In this case, the capacity of the set E is taken equal to zero. It should be noted that the concept of the Green's. function g(z, -) that we have given here coincides, when the boundary of the domain B consists of a finite number of closed Jordan curves, with the Green's function defined in Chapter VI. Specifically, if g(z, oc) is a Green's function in the previous sense, for the domains B(n) that exhaust the domain B let us take domains defined respectively by the inequalities g(z, oc) > e", where e" -, 0 as n - oc. The Green's functions for the domains B(n) are respectively the functions g(z, 00) - en , and

they converge in B to g(z, -). The result that we obtained above can be formulated somewhat differently as

§3. CAPACITY OF A CLOSED BOUNDED SET

311

Theorem I. For a domain B containing oo to have a Green's function g(z, it is necessary and sufficient that the capacity of its boundary be positive: With regard to capacity, we shall now prove Theorem 2. 1) The capacity of an arbitrary closed bounded set E of points in the plane is equal to its transfinite diameter; that is C(E) = d(E). Proof. Consider first a domain B containing oc with boundary K consisting of a finite number of disjoint closed Jordan curves. Suppose that g(z, C) is its Green's function and that h (z, -) is the conjugate function to g (z, 0. For arbitrary finite z C B, the function

it(C, z)=g(C, z)-g(C,

0o)-filogIC-z1,

as a function of ', is harmonic in B. If we apply formula (5) of §3 of Chapter VI to it, keeping in mind the fact that u (C, z) = log z I on K, we obtain, for t;,

z C B, U

(C, z) = 2 log 1 C' - z 1dh (C', Q. K

If we now set _ oc and note that u(oo, z) = g(z, ) - y, where y is Robin's constant for the domain B, we have

g (z, 00) - 7 =

2a

log I C' - z 1 dh (C', oo).

(1)

Making use of this formula, let us show that r = e-7, where r is tebysev's constant for K. Let h denote a positive number sufficiently small that the inequality g(z, 00) > X defines in B a domain Bh with boundary Kh consisting of closed Jordan curves that approximate the corresponding curves constituting parts of K. Since

1a dh(C',co)=1, let us partition the boundary K by points ck, for k = 1, . , n (numbered in order of occurrence as one moves around K, into subsets lk such that

1) Szego [1924].

VII. MEASURE PROPERTIES OF CLOSED SETS

312

2a

S

A (C', oo) = n ,

k = 1, ..., n.

!k

Let us compare the integral (1) with the sum

)logZ n

n

k=1

For z E KA, the oscillation of log IC- zI on each of the subsets lk consisting of only one arc will, for sufficiently large n, be less than prenamed e > 0 indepently of z. On the other hand, the number of subsets lk consisting of several arcs does not exceed the connectedness of the domain B. Therefore, the difference

- n },loglCk- zI n

$n(z)=

2a

C logl C'-zldh(C', oo)

k=1 n

S(loglC'-zl-loglCk-zI)dh(C', oo) k=1 1k

approaches zero uniformly with respect to z E KA as n - oo. Consequently, in accordance with (1), if z E Kx and n is sufficiently great, we have n

a-T- n loglCk-z+a.(z) k=1

where IS.(z)I
attains its maximum in the complement to BA with respect to the z-plane somewhere on KA, it follows that (2) also holds at all points of this complement and, in particular, on K. This means that < e2A-7 for n sufficiently great, where mn (see §1) is the quantity corresponding to the set K, and consequently, r < e2A_7. By virtue of the arbitrariness of A >'O, it follows that r < e-7. (3) I

I

On the other hand, if ' (z) is the Cebysev polynomial for K (with zeros on and Mn = max=EK Itn W1 then the function

vn (z) = 9(z, oo) - n log

I tm(Z) n

K)

(4)

§3. CAPACITY OF A CLOSED BOUNDED SET O

313

is harmonic in B and assumes nonnegative limiting values as we let z approach K Consequently, vn (z) > 0 everywhere in B. In particular vn (o) =y + (1/n) log mn > 0; that is.

> V m nNow as n -+ oo, we obtain r> e

e- '.

(5)

and this, together with (3) yields the re-

sult r = e - Y. Proceeding now to proof of the theorem for the set E, consider the domain B complementary to E and containing Do. Suppose that B is exhausted by domains

B(" bounded by closed Jordan curves. Let yn denote the Robin's constant for the domain B(n). We denote by En the complement of B(n) with respect to the z-plane. We note that the Cebysev polynomials tv,N (z) of degree v for the set Enn are also Cebysev polynomials for the boundary of the domain B(">. Let us dey fine rn = limvyM mv.n . In accordance with what we proved above, t = e-7".

But since E C En, it follows that r < rn, where r is Cebysev's constant for E. Consequently, r < e-yn. By letting n approach oc, we see that r < a-y. On the other v hand, it. follows from (5) that V > e y". This shows that

e-7n - max

zEEn

(z) I,

v = 1, 2, .. .

where E v(z) is the Cebysev polynomial for E. This means that, for every n, there exists a point zn on the boundary of the domain B(n> such that an < Itv (z,,)I. All the cluster points of the sequence (zn , for n = 1, 2, -, lie on E. Let z0 denote one of these points. Then, this last inequality implies a-y < v v Itv (z0)I and consequently, a-y
v,

theorem.

An interesting result is obtained when E is a continuum. Then, the domain B is simply connected and the Green's function g(z, oc) for B coincides with the logarithm of the modulus of the function C = `(z) which maps the domain B univalendy onto the domain ICI > 1 in such a way that the point z = on is mapped into C = no. Therefore, Robin's constant for the domain B is equal to log If'(-)I, and the capacity C(E) = 1/If'(oo)I = R. Here, R is the conformal radius of the domain B (with respect to no), that is, that number R such that the domain B is mapped univalently onto the domain ICI > R by means of a normalized

314

VII. MEASURE PROPERTIES OF CLOSED SETS

function J= F(z) that satisfies the conditions F(oo) = oo and F'(-) = 1. Thus, we have Theorem 3. 1) The capacity, and hence the trans finite diameter, of a bounded continuum E are equal to the conformal radius of the domain complementary to E and containing oo. In conclusion, we present one more theorem connecting Green's function with iebyit'ev polynomials.

Theorem 4. Suppose that a domain B with boundary K contains oo and has Green's function g(z, oo). Let tn(z) denote the l,ebysev polynomials for K with zeros on K. Then n

y

m i in 00

I tnmz) I

_ eg (z

mE

I

K

t n (x) I ,

this limit being uniform in the interior of B. Proof. Consider again the functions v,, (z), for n = 1, 2, . , defined by formula (4). Since these functions are harmonic and nonnegative in B, the condensation principle can be applied to them. According to this principle, the sequence {vn(z)I itself and any subsequence of it contain subsequences that converge uniformly in the interior of B to harmonic functions. Here, the limit functions are always nonnegative in B and, in accordance with Theorem 1 they vanish at so. Consequently, these functions vanish identically in B. This means that the sequence Ivn(z)j itself converges uniformly in the interior of B to 0. This is equivalent to the assertion of the theorem. §4. Harmonic measure of closed bounded sets Together with the methods considered above for measuring closed bounded subsets of a plane, R. Nevalinna 2) introduced another measure, which has proved very important in the theory of functions.

Let a denote a closed bounded set of points in the z-plane and let B denote any of the domains complementary to a. Let f3 denote a closed Jordan curve in the domain B that has no points in common with a. Let us remove 0 and its interior from B and denote the remaining portion of B by B18. Let B(n), for 1) Fekete [19231.

2)R. Nevanlinna [1936a]. All the material of the present section is taken from that article.

§4. HARMONIC MEASURE OF CLOSED BOUNDED SETS

315

n = 1, 2, , denote a sequence of domains contained in B and containing (3. Suppose that their boundaries an consist of a finite number of closed Jordan curves and that the sequence exhausts the domain B. Finally, let B('), for n = 1, 2, , denote the domain obtained from B (n) by removing the curve S and the domain bounded by it. We denote by to (z, an, B( )) a function that is harmonic in B I , equal to unity on a., and equal to zero on P. Obviously, this function is nonnegative and less than unity in B(n). Furthermore, the difference to(z, a,n B()`, 18 to (z, an+1' Bn+1) since it is nonnegative on the boundary of the domain BP(n) 18 is positive in the interior of BQ I . This means that for fixed z E B'6, the quantity to(z, a BQ I) is defined, beginning with some n and decreases with increasing n. We can now apply Harnack's theorem, concluding from it that (n)) = !o (z, a, BQ) exists everywhere in B18 and is a harmonic limn oo o. (z, an, BP function in BR. Obviously, 0 < to (z, a, BR) < 1 in B'8. We note that to (z, a, BR) does not depend on the choice of sequence B1°I because, if there are two sequences B(n' ) and B(n" of this kind, then, for arbitrary n', there exists an n" such that B(R ) C B I and conversely. Consequently, in the first case we have the inequality to(z, a ,, B') < (o (z, a ,,, and, in the second case, we have the opposite inequality. The quantity &j (z, a, BP) is called the harmonic measure of the set a with respect to the domain B, the curve (3, and the point z E BR. We note that, as we let z approach points of the set a through values in BR, the harmonic measure to (z, a, BR) does not always approach unity. For example, it will not approach unity as z approaches isolated points of the set a because at each of these points to(z, a, BR) is a harmonic function. However, we can show that, if to (z, a, BR) j 0, then, as we let z approach points of the set a through values in BQ, the limiting values of the harmonic measure oo (z, a, BQ) are bounded below by a positive number. To see this, let go denote a closed Jordan curve in B that surrounds the curve A and does not contain any point of the set a. On P0, the function &)(z, a, BR) has a positive minimum 9 such that 0 < B.< 1. Let zQ denote an arbitrary point of the domain BR that lies outside

go and let n be sufficiently great that 00 and zo lie in B(nI. Since in B 1, it follows that a) (z, a., BQ I) > 9 on 00 Furthermore, to(z, an, B I) = I on an. Consequently, in accordance with the

(0 (Z, a, BQ I) > a) (z, a, B

maximum principle, we obtain ao (z0, an, B18 (n)) > B for all n sufficiently great. As we let n approach oc, we see that &o(z0) a, BQ) > B > 0. It follows that all

the limiting values of &j (z, a, BR) as z approaches a exceed 0.

316

VII. MEASURE PROPERTIES OF CLOSED SETS

The question arises as to ways of telling when the harmonic measure of a given set a is zero or positive. The vanishing or not of a harmonic measure co(z, a, BA) holds simultaneously for all points of a domain BR because, if it vanishes at a single point in Bp, it must vanish everywhere in BA by virtue of the maximum principle. However, the same phenomenon holds with regard to the curves S chosen in B. To see this, suppose that for some curve S, we have denote another curve of the same kind. Let y and y' co (z, a, BR) = 0 and let such that y' denote closed Jordan curves lying in B and enclosing S and

lies inside y and there are no points of the set a between y and The curves y and S' together form the boundary of a domain B0 contained in B. Set 0 = max Z E y, w (z, y, B o), 0 < 0 < 1. Now, let a denote any positive number. For

n sufficiently large and for z E y, we have w(z, an, BA (4) < e because the convergence of the sequence (w(z, an, B ))E to zero in BR implies its uniform conA vergence to 0 inside BA. Let us denote by A and A' the maxima of the function Co (z, an , Bp;) on y and y'. Since the difference w (z, a n, B $ ) ) - W (z, an, B 4 >) is a harmonic function in the portion Bin> of the domain B(n) that is bounded by the curves an and

y' and since it is equal to 0 on an and does not exceed A' on y', it follows that A < A' + e. Furthermore, since the difference w (z, an, B(n) - Aw (z, y, B 0) is a.

harmonic function in Bo that vanishes on S` and is less.than 0 on y, it is less than zero in Bo and, in particular, on y', where the maximum of the function w(z, an, B 4' ) is equal to V. Consequently, on y' we have A'
Therefore, on y' we have w(z, an, BQ,) <(1

Since a is an arbitrary posi-

tive number, it follows that w(z, a, Be.) = 0 on y' and hence everywhere in BR.. Thus, for a given domain B complementary to a, vanishing of the harmonic measure

w (z, a, B,8) is independent of the choice of curve S in that domain. Let us show now that if a contains a continuum, then w (z, a, BA) > 0 for any domain B complementary to a and for any curve S C B. Just as in the proof of Riemann's theorem in §2 of Chapter II, we can, by a suitable transformation, arrange for the domain B to have exterior points. Let B0 denote the domain bounded by the curve S and some other closed Jordan curve lying outside B. Then, B( n) C B0. Consequently, just as above, we can show that, for z E BA(n) w (z, a,

By letting n approach oo, we see that

B 0) > 0.

§4. HARMONIC MEASURE OF CLOSED BOUIpED SETS

317

co(z, a, BQ)>a)(z, a0, B0)>0. It follows from this last assertion that, if the harmonic measure of the set a with respect to some domain complementary to a is equal to zero, then a does not contain a continuum and the complement to a is the unique domain including oo. We now point out the following simple properties of sets of zero harmonic measure.

Every closed subset of a closed bounded set of harmonic measure zero is also a set of harmonic measure zero. This follows immediately from examination of the differences between corresponding functions of the type W (z, an, B( )). 18

The sum of two closed bounded sets of zero harmonic measure is a set of harmonic measure zero.

To show this, denote the two sets by a1 and a2 and let S denote a curve in the complement to a = a1 + a2. Let {B(1)} and {BZ"I} denote two sequences that exhaust the complements to a1 and a2 respectively and that contain S. Then, for a given point z in the domain B complementary to a and containing S, we have (0 (z, at n, Bln)) < 7t,

w (z, as, .,

B(2n))

\2'

for sufficiently large n. Here a 1,n and a 2, n are the boundaries of the domains B(1 and B(z) . Let B(n) denote that one of the domains in the intersection of B in and B (z) that contains S . For suitable choice of the domains B (n) and Ben) (they can, for example, be constructed from squares as at the beginning of §2), the boundary of the domain B(n) consists, for every n = 1, 2, , , of a finite number of closed Jordan curves. Therefore, the domains B(n), for n = 1, 2, exhaust the domain B. The function

,

0 - co( z a

w (z,r an, y B(n))

r

1-

' 1, n B`'n)

w

(z, as, n a9B(n') 9 r

which is harmonic in B(n) is equal to zero on S and does not exceed zero on an . 18 Consequently, in Bp we have w (z, a, Bp) < w (z, at, B1) + w (z, as, B9),

that is, w (z, a, BR) = 0. We shall now prove the theorem of R. Nevanlinna, which is fundamental in the theory of harmonic measure.

318

VII. MEASURE PROPERTIES OF CLOSED SETS

Theorem I. A closed bounded set has zero harmonic measure if and only if its capacity is equal to zero. Proof. As a preliminary, we shall make a necessary observation concerning Green's function. Let B denote a domain containing oo and bounded by a finite number of closed Jordan curves. Let_g(z, 00) denote its Green's function and let h(z, 00) denote the conjugate function. Let y denote Robin's constant. Formula (1) of §3 is then applicable and from it we obtain the inequalities

logd,(z)cg(z, oo)-7clog dg (z),

(1)

where d1(z) and d2(z) are the minimum and maximum distances from the point z to the set of boundary points of the domain B. We shall very shortly have occasion to use these inequalities. Turning to the proof of the theorem, we may assume that the closed bounded set referred to in the statement of the theorem is contained in the disk I z I < 1 because, otherwise, we could arrange for this by means of a similarity transformation without affecting the property to be proved. Let us denote this set by a and let us denote by B the domain complementary to a that includes *a. Let B' denote a domain including 00 that is bounded by a finite number of closed Jordan curves contained in I z I < I and satisfying the condition B' C B. Let us denote by g(z, 00) and g'(z, 00) the Green's functions for B and B'. Let us denote by co P (z, a', B'P) the harmonic measure of the boundary a' of the domain B' with respect to the subset BP of the domain B contained in I z I < p, where p > 1. (1) If a has positive capacity, then the constant y = limz.,00(g(z, -)-log Iz1) is finite. Let p denote a fixed number > 2 and let MP and mP denote the maximum and minimum values of the function g' (z, oo) on I z I = p. The difference g, - (1 - w p)mP is a harmonic function in BP and it is nonnegative on the boundB'P. ary of Consequently, this difference is also nonnegative in B that is, g, > (1 - a) P)mP

(2)

in B P . If z 1 is a point on the circle I z I at which g' (z, oo) has a minimum, which we denote by m1, then inequality (2) yields

mt

(1 - (OP (z1, a', BP)) m

But m1=min jz1=1(g'-logI z1)
(3)

N. HARMONIC MEASURE OF CLOSED BOUNDED, SETS

319

MP > log (p - 1) + y'. Consequently, we obtain from (3) T > (1 - WP (z1, a', B'P)) (log (P - 1) + 7'),

so that

log (P -1)

7

wP (z, a', BP):1 - T +log(9-1)

7+1og(9-1)

> 0,

since y' < y. This shows that if the capacity of the set a is positive, the harmonic measure of a is also positive. (2) Now suppose that the capacity of the set a is equal to zero. Since the difference (1 - a)P)MP- g' is a harmonic function in BP and since it is nonnegative on the boundary of B P, it follows that this difference is also nonnegative in B o and this yields

g' < (1 - 0P)MP.

(4)

For a point z 2

on I zj = 1 at which g' attains a maximum, which we denote by M 1, inequality (4) yields

Mi C (1 - WP (z9, a', BP)) MP. But

MI -max(g'-log IzI):7, Izl=1

and, in accordance with inequality (1),

MP` log (P+1)+T. Consequently,

so that

T c(I

wP(z9i

min w c w (Zs, IzI=I1, P

a, BP))(log (P+1)+T'),

a', B') c 1P

T"

7'+1og(P+1)

If we fix p > 1 and let B' (a sequence of domains exhausting B) approach B, then y' approaches oc and consequently min,,,=161 (z, a', BP) approaches 10 zero. Since the function o (z, a', BP) on I zj = I then converges uniformly to (a (Z' a, BP) where BP is the portion of the domain B lying in the disk I zj < p, it follows that t) (z, a, B ") - 0 because the function o U (z, a', B'P) would otherwise have a positive bound on I z = 1. Therefore, the harmonic measure of

320

VII. MEASURE PROPERTIES OF CLOSED SETS

a is zero. This completes the proof of the theorem. As simple applications of the theory of harmonic measure, we shall now prove two theorems, one of which deals with removable singularities of harmonic functions and the other with the maximum principle for harmonic functions.

Theorem 2. Let a denote a closed bounded set of points in the z-plane that is contained in a domain B. If the harmonic measure of a is zero, then every function u(z) that is harmonic and bounded in B except at points of the set a is harmonic in the entire domain B if we define it in a suitable way on a. On the other hand, if the harmonic measure of a is positive, this property does not always hold.

Proof. Suppose that the harmonic measure of a is zero. Then we can assume without loss of generality that B is a bounded domain with boundary consisting of a single closed Jordan curve 0 and that u(z) is a harmonic function on 8 because otherwise we could consider not B but a suitable domain of the type described contained in B and containing a. Let u0(z) denote a function that is bounded and harmonic in the domain B and coinciding with u(z) on the boundary

of B. Let us show that u0(z) a u(z) in B. For a point z0 C B and given c> 0, there exists'a domain Bp I, contained in B including z0 but not containing a, the boundary f3 + an of which consists of a finite number of Jordan curves and which has the property that w(z0, an, B (n)) <E. If (u(z)I M in B - a, then Iu0(z) I < M in B. Consequently, for the difference v(z) = u(z) - u0(z) in B - a, , we have Iv(z) I < 2M. Now, let us consider the function v(z) + 2M6)(z, an, B18(n) ).. This function is harmonic in B( ), equal to zero on S, and nonnegative on a n .

Consequently, v(z) + Ma (z, an, B( )) > 0 in B( )

Therefore, v(z0) >

- 2Mc, > - 2Me. Since c is an arbitrary positive number, v(z0) > 0. Treating the function v(z) - 2Mco (z, an, BP (n)) , we can show that v(z0) < 0. Consequently, v(z0) = 0; that is, u0(z) = u (z) in B - a. It follows that if we define u (z) on a equal to u0(z), then u (z) . becomes harmonic in B. On the other hand, if a has positive harmonic measure, then, for example, the domain of definition of w(z, a, B), which is harmonic and bounded in B - a cannot be extended to a in such a way as to make w (z, a, B) a harmonic function in B because this function, since it is equal to zero on S, would be identically equal to zero in B - a and then the set a would be of zero harmonic measure. Theorem 3. Let u(z) denote a function that is harmonic and bounded above in a domain B. If lim:. Z u(z) < m for all points Z* on the boundary except for a set of harmonic measure zero, then u(z) < m everywhere in B.

§5. APPLICATION TO MEROMORPHIC FUNCTIONS

321

Proof. Suppose that u (z) < M in B. Let us suppose that there exists a point z0 in B at which u(z) > in. It follows from the hypotheses of the theorem that there exists a point zl E B such that u(z1) < u(zo). Let us describe a small disk S around z 1 in which u (z) < m 1, where m < m 1 < u (z 0). From the hypothesis of the theorem, the set a can be contained in the interior of a finite number of Jordan curves lying in the complementary domain containing B. Let us denote the union of these Jordan curves by an, so that, for the domain B( n) with boundary an U 8, we have w(z 0 , a n , B (1))
v(z)=u(z)-m-(M-m)a)(z, an, BAS),

which is harmonic in the intersection of the domains B and B.T. As we let z approach boundary points of the domain B that are distinct from points of the set a, its limiting values do not exceed m - in - (M - m) < 0 and as we let z approach points of an, they do not exceed M - in - (M - m) = 0. Finally, on 8 we have v(z) < rn - m. It follows that v(z) < m1 - m in B() n B. In particular, for z = z0, this yields

it (zo)Cm1+(M-m)w(zo, an, Bpn))cml+(MSince a is an arbitrary positive number, we obtain u(z0) < m1, which contradicts the assumption that u (z 0) > m 1. This completes the proof of the theorem. §5. An application to meromorphic functions of bounded form A function f(z) is called a meromorphic function of bounded form in Izl < 1 if it can be represented in that disk as the ratio of two bounded regular functions . f (z) _ T(z) (z)

(1)

We may assume that I,(z)I < 1 and IVi(z)I < 1 in Iz I < 1. A test of whether f(z) is a meromorphic function of bounded form is given by Theorem 1. 1) Let f(z) denote a function that is meromorphic in the disk Iz I < 1 such that f(0) no. For f(z) to be a function of bounded form, it is necessary and sufficient that the quantity T (r) = 2n \ log If (re") I dO + 1 log kkr

if

1)R. Nevanlinna [1922].

I

e

bk

322

VII. MEASURE PROPERTIES OF CLOSED SETS

be bounded above for 0 < r < 1. Here, the bk are the poles of the function f(z) other than the point z = 0. These zeros are indexed in such a way that a pole of multiplicity n is counted n times. The summation is over all poles bk in the disk I z I < r. The symbol log a is defined by

(logs if a>1

+

log a=j

if0
L0

Proof. If the zeros of the function f(z) are denoted by ak and the poles by bk, then, by setting Jr (z)

rr r (z - ak) :

fl r' - akz Iakl
r (z - bk)

Ibkl
O
r' , bkz

we conclude that the function f,(z) has the same zeros and poles in I z I < r as does the function f(z). Furthermore, fr(z) I = 1 on the circle IzI = r because, for arbitrary a such that I aI < r, we have I r(z - a)/(r2 - az)I = 1 on I z I = r. Therefore, if r is such that there are no zeros or poles of f (z) on I z I = r, then the function F(z) = f (z)/fr(z) is regular and has no zeros on I z I < r and log F(z) is a regular function on Iz I < r. If we apply the Schwarz formula to the function log F (z) and to the disk IzI < r, we have a:

tog F() z =- 12n 1ctog I F re'° ) I reie+z rest-Z d6

IC

in I z I < r, where C is a real constant. Keeping in mind the value of F(z)

and

the fact that IF(z)I =If(z)I on the circle IzI = r, we obtain from the above equation the formula to

g.fz )-

log r (z

ak)

r' - akz

lakI
-

to r (z - bk)

g

Ibkl
r'-bkz

2s

-}- 2a

log If (re`) I r

reteie-+

z d9 + W.

(2)

This formula remains valid when there are zeros and poles of the function f (z) on the circle I z I = r. That this is true can be proved by a passage to the limit with respect to r. In the particular case in which f(0) / 0, if we set z = 0 in (2) and equate the real parts of the two sides of the equation, we obtain the familiar formula of Jensen

§5 APPLICATION TO MEROMORPHIC FUNCTIONS

logIf(0)1=

I

log lbkl

k

logl

323

IakI
IbkI
-}

2n

log j f (re') I A

(3)

Formula (2), which is a generalization of Jensen's formula, is called the Jensen-Schwarz formula.

Keeping in mind the fact that

l

log I f (re1B) I = o g If (re") I- log i

f (e11)

,

we obtain from formula (2) the following representation of the function f(z) in the disk I zI < r:

f(z)`Ir(z) +r (z)'

(4)

where 2+c

I

'Pr (z)

H r (Z - aA'

I a, < r

r9 - 8kz

2s

Sto 0

red±y A+IC

1

} (reie)

rep

e

2+c

r (z - bk) e-

4,. (z) Ibkl
2z

z

iB

log 11(reie) 11e dB 0

re

-z

rs_ bk z

Here the functions Or(z) and Or(z) are regular in the disk I z I < r, also, Itr(z)I < 1 and I fr(z)I < 1 in that disk and 10r(O)l = e-T(r). Suppose now that T (r) is less than some constant M for 0 < r < 1. Let (rn), where rn < 1, for n = 1, 2, , denote a sequence that converges to 1 as n approaches oo. Then, the sequences I/rn(z)) and {1n(z)( contain subsequences (i/in (z)} and ion (z)} that converge uniformly in the disk I z I < 1 to regular functions VI (z) and k 0 (z) such that Ir (z)I < 1 and 10 (z)I < 1 in I zI < 1. As a result, we obtain the representation (1). Conversely, suppose that the representation (1) holds also for i/i(0) / 0 (to which we can easily reduce the representation (1)). Then since log l f (z) 1- log

in the disk I zI < 1, we have

l + (z) Il

clog l 4' (z)1

- log l 4' (z) I,

324

VII. MEASURE PROPERTIES OF CLOSED SETS

O
I

and, consequently, 2%

T (r) ` - 2n

log I

(re') I d0 + )

log r l

IbkI
(5)

where the b'k, for k = 1, 2, , constitute all the zeros of the function tA(z) in I z I < 1. On the other hand, in accordance with Jensen's formula (3) as applied to ,i (z), we have 2a

log I

(0) I =

71,

log I bk I + 221 S log

(re.") I d9.

(6)

0

bk l
On the basis of (6), the right-hand member of (5) is equal to -log (0)I. Consequently, for 0 < r < 1, we have T (r) < - log I tfi (0)I; that is, T (r) is bounded. This completes the proof of the theorem. If we replace the requirement that T (r) be bounded for 0 < r < I with the requirement that its limiting values be bounded as r - 1, Theorem 1 will remain valid without the assumption that f (0) oc. One can show this by modifying the proof given above. Now, by using the material of §3, we shall prove a sufficient condition for a meromorphic function f(z) to be bounded form. Theorem 2. 1) Suppose that a function f(z) is meromorphic in the disk IzI < 1 and that it does not assume in that disk a set of values w that is a closed bounded set E of positive capacity. Then, f(z) is of bounded form.

Proof. It will be sufficient to prove the theorem for a function f(z) that is oo because in the opposite case, we could consider not f (z) but the function f(z + a) /1 + az) for suitable a such that I al < 1. For arbitrary c # f (0) and 0 < r < 1, if we apply Jensen's formula (3) to the function f(z) - c, we have not bounded in I z I < 1. We may assume that f (0)

2a

logl akl-

logIf(0)-c I akI
loglbkl+

1

2n

log I f (rei) - c I de.,

IbkI
1) With regard to Theorems 2 and 3, see R. Nevanlinna [ 1936a].

(7)

§5. APPLICATION TO MEROMORPHIC FUNMONS

325

Here the ak and bk are the zeros and poles respectively of the function f(z) - c
r

2n

c

= 2a

S

L 2a

\ r log f (re`B) - c I dhr (c, oo) dt+

log

I bkl
r bt

,

since fg dhr(c, ) = 2n. If we now apply formula (1) of §3, we obtain r 2n

gr (f (0), 00) = 2n

gr (f (re'),

co) d8 + I Ibkl
log I bk

(8)

Now let R > 1 denote a number such that, for all r in the interval 0 < r < 1, the set Kr is contained in the disk Iw I < R. Then, for i C Kr and w such that I w I > 2R, we have

log W-C ,::- log(1Consequently, from the formula (again see (1) of §3)

gr (w, °o) - Tr = 2a 5 log C - w I dhr (C, oo) r

= log l w + 2 5 Iog f w W C as applied to w such that I w I > 2R, we obtain gr(w,

dhr (C, oo),

r

109 1 WI`I-Tr-log 2R.

VII. MEASURE PROPERTIES OF CLOSED SETS

326

Furthermore, since Kr lies entirely in the disk I w I < R, we have d(Kr) = e-yr - log R. Therefore, the above inequality yields log 3R. (9) A. (w, 00) > log I W Inequality (9) also holds for all points of the domain Br lying in the disk I W I < 2R because the left-hand member is positive at these points whereas the right-

hand member is negative. Using inequality (9), which is valid in the entire domain Br, we obtain from formula (8)

gr(f(0), 00):-2n

gIf(rete)Id9-log3R+ I log

21

IbkI
Ibkl

Or

7'(r)-

zR

loglf(re°B)Id6

2

+ I log I bk l
bk c g,

(f(0), o0) -1-- log 3R.

(10)

Furthermore, since the domain Br is contained in the domain B complementary

to E that includes a, we have gr(f(0), -) < g(f(0), 00), where g(w, 00) is the Green's function for B. Therefore, we obtain from (10)

T(r)
Theorem 3. Let w = f(z) denote a function that maps the disk Izl < 1 conformally onto a multiply connected domain B containing 00 (see §1 of Chapter VI). For this function to be a meromorphic function of bounded form, it. is necessary and sufficient that the capacity of the boundary of the domain B be positive. Proof. The function w = f (z) does not assume in I z I < 1 values constituting the boundary K of the domain B. Therefore, if this boundary has positive capacity, then f(z) is, in accordance with Theorem 2, a function of bounded form. Now, suppose, conversely, that f(z) is a function of bounded form. Let c be a member of 8 other than 0 or oc. The function f(z) - c has an infinite

327

§5 APPLICATION TO MEROMORPHIC FUNCTIONS

, number of zeros in I z j < 1. Let us denote these zeros by ak for k = 1, 2, indexing them in order of increasing absolute value. These zeros are the poles of

the function 1/(f(z) - c), which is also a function of bounded form. If we apply Theorem 1 to this last function, we conclude that the quantity 2 + 2n , log I

f

(z)- c

d9+

log

a.I
r lakl.

and, in particular, the second term in it is bounded above. Suppose that log (r/I akI) < A for 0 < r < 1, where A is independent of r. Since the 1I akl < r terms in this summation are positive, it follows that, for arbitrary n and for r sufficiently close to 1, n

k=1

If we now take the limit first as r equality

1 and then as n

we arrive at the in-

00

k=1

loglakl'A'

which means that the series on the left converges. Therefore, keeping in mind the fact that log 1

x-n1 1-x we conclude that the series

akI) also converges. Now, for I zI < 1,

we have

1 -a,pz log I

11og (1 ---- akz) - log (1 - ak/) - log I ak I

z - ak 00

CI v

1

v

dvz

v

X00

v=l

I

00

I-loglakl= V

zy lalaklv 1

loglak

00

2(1-Iakl)

v al

lY-,loglakl- 2(1-Iakl) - loglakl 1-IQk

328

VII. MEASURE PROPERTIES OF CLOSED SETS

Therefore it follows from what was said above that the series 00

42 S (z) = I log Z-dk k-1 1

I

converges absolutely in the disk I z I < 1 and defines in that disk a function that is harmonic except at the points ak, where it has a logarithmic singular term - log I z - akl. The function s (z) is invariant under fractional-linear transformations in the group G (see §1 of Chapter VI) corresponding to the domain B because these transformations map the set of zeros ak into itself. Therefore, if we by means of the function z = f(-"(w), then s(z) is transshift to the formed into the function s I (w) = s (f -1 (w)), which is single-valued, nonnegative, and harmonic in B except at the point w = c, where it has a singular term of the form -log1w- cI. Suppose now that the disk Iw - cI < e is contained in B. Define

me= min

w-e I='

s1(w).

If the capacity of the boundary of the domain B were equal to zero, then Theorem 3 of §4 would be applicable to the portion B I of the domain B in which I w - cl > e and to the function -s I W. According to that theorem, - s I (w) < - mE in B I; that is, s 1(w) > mE in B I. But mE + oc as e - 0. Consequently, s 1(w) _ +

in B, which contradicts the finiteness, proved above, of the function s W. ibis contradiction proves the theorem.

In connection with other results dealing with the application of the theorem on harmonic measure to the study of analytic functions, we refer the reader to the monographs of R. Nevanlinna [1936a] and Privalov [1950].

CHAPTER VIII MAJORIZATION PRINCIPLES AND THEIR APPLICATIONS

§1. An invariant form of the Schwarz lemma

In §2 of Chapter II, we established a simple but extremely important property of bounded functions, known as the Schwarz lemma. A very simple and widely used form of this lemma is as follows:

Theorem ('the Schwarz lemma). Let f W denote a function that is regular in the disk I z I < 1. Suppose that f (0) = 0 and that if (z)I < 1 in I z I < 1. Then z) < I z I throughout the disk I z < 1 and f' (0)j < 1. Also, in the inequality I f (z)j -}z 1, the equality holds, for points z 0, only if f (z) = rz with I e If we now discard the requirement that f (0) = 0, that is, if we consider an arbitrary function f(z) that is regular in the disk z < 1 and satisfies the ine-

quality IfWj < 1 in that disk, then, if we set

f I

C+

)

z

(C1+2C) =c1C+ I+zC+Z) ...,

-f(z)f

C1T I-If(Z)I, for fixed z in the disk z < 1, we see that the function g(C) satisfies the conditions of the Schwarz lemma. Therefore, the inequalities mentioned in the conclusion of that lemma hold. From this we obtain a further generalization of the I-IZ18f(z),

I

Schwarz lemma.

Theorem 2. If a function f (z) is regular in the disk I z I < 1 and if l1 (z) I < 1 in that disk, then, for arbitrary C and z in the disk I z I < 1,

f (C) -f (Z) Ic C- Z 1 -f (Z)1(C) 329

1 - zC

f

(1)

330

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

and, for arbitrary z in the disk I z I < 1,

If,(x)1=11If(Z)I

(2)

with equality only for f(z) = E (z + a)/(1 + a z), where I E (= 1 and a I < 1. We note that, if we set z = 0 in 1, we obtain the result that, in ICI < 1,

f(C)-f(0)

(3)

I -f (0)f (C)

After some simple manipulations, as in §4 of Chapter IV, we can reduce this inequality to the form

f(C)-f(o)1-Ifo)CIIaICI' 'I`II-If(o)II'ItI''

(4)

From this in turn we get the inequality I + if If(t)I I+C, IfOIICI 1

)

(5)

I

Here, for C 0, equality holds only when f() = E (C+ a)/(1 + a t;')., where

I E I = 1,

Ial <1, and arg C=arga if a geometrical interpretation. To do this, we define a

special metric in the disk I z I < 1. Let z I and z2 denote two points of the disk I z I < 1 and let z3 and z4 denote the points of intersection of the circle I z 1 with the circle orthogonal to it passing through the points z1 and z2. Let us assume that we have numbered the points in such a way that the point z3 is closer to z I than to z2. Let us form the anharmonic ratio of these four points: zl - za . za - za

(z3, z1, z2, z4) = z1-Z4 Z2-24'

(6)

As one can easily verify, this ratio is invariant under an arbitrary fractional-linear mapping of the z-plane; that is, if such a mapping maps the points z 1, z 2, z 3 have and Z4 into points C1, S2, C3 Land '4 i (C3. C1,

(7)

CA) = (z3, zi, z9, z4). CA)

Furthermore, the anharmonic ratio (6) is positive and less than unity for arbi-

trary zI and z2 in the disk I z I < 1. To see this, note that the function C = (z - z I)/(1 - a Iz) maps the disk I z I < 1 into itself in such a way that the points z I and z2 are mapped into the points SI = 0 and S'2 = (z - zI)/(1 - a Iz2) = re`s, where 0 < r < 1, and it maps the circle orthogonal to the circle

Iz

= 1 and passing

1. INVARIANT FORM OF THE SCHWARZ LEMMA

331

through z1 and z2 into a straight line passing through C= 0. It follows that the points z3 and z4 are mapped into the points t3 = -e`6 and tq = e`o. Then, by using (7), we obtain eiP

(z8, zi, z9, zt) =

e-VIP

'

reiv+eiPll +,-rr'

1(z8, z1, za, z4) = -

(8)

1+

Formula (8) gives a convenient expression for the anharmonic ratio in question. In particular, the assertion made above follows from it. We now define the hyperbolic distance between two points z 1 and z2 of the disk I z I < 1 as the quantity 1

D (z1, z2) = -

2

log (za, z1, z2, zi) =

i

I

2

log

Zl - Zy 1 -Z1Zy

zl -z'

(9)

O"uusly, D(z1, z2)=D(z2, z1) and D(z1, z2)>0. Also, D(z1, z2)=0 only if z 1 =`sZ Furthermore, for arbitrary points z 1, z 2, and z 3 in the disk D (z1, z9) + D (z2, za) > D (zt, z3)

I z I < 1,

(9)

(the triangle inequality),. with equality holding only when the points z 1, z2i and Z3 lie on a single circle orthogonal to the circle I z I = 1. This property follows from (9) if we note that, by virtue of the invariance of the hyperbolic distance under fractional-linear transformations, we can always have z2 = 0 and that, by (C- z)/(1 - TIC) at virtue of inequality (5) as applied to the function

the point J= z3, we have Za

Zl

za+ zll

zlza

+IzliIzal

with equality holding only if arg z 1 = arg z 2.

We shall not stop to construct a plane geometry based on this metric, a geometry that gives a Euclidean interpretation to the non-Euclidean geometry of Lobacevski% but point out that the hyperbolic distance is invariant under fractional-linear mappings of the disk I z I < 1 into itself and that the role of straight lines in this geometry is played by circles orthogonal to I z I = 1. Furthermore, for any fixed z 1, the distance D of z 1 and z2 approaches + oo as the point Z2

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

332

approaches the circle I z I = 1. This follows from (9). The set of points z in the disk I z I < 1 whose distance from a given point z0 in that disk does not exceed a given quantity p, where p > 1, that is, points for which D (z, z0) < p, will be referred to as the hyperbolic disk with hyperbolic center z0 and hyperbolic radius p. It follows from (9) that this hyperbolic disk is defined by the inequality Z-Z0 C 1 -e'P 4

+ eiP = tan g

1 - zoz

and, consequently is a Euclidean disk but with center distinct from z0. It also follows from (9) that A

D (z, z -}- Az)

i z I,

-{- e),

where c --, 0 as Az --a 0. This shows that the linear differential element doe in the hyperbolic metric is given by the formula

das= 1

I diz

(10)

la ,

is the Euclidean differential element. The hyperbolic length L of the curve z = z (t) for a < t < b in the disk z I < 1 is defined as the least upper bound of the sums Ik_a D(zk, zk+1) for all possible systems of points zk = z (tk), where a = t0 < t 1 < . . . < t, = b. In the case of a piecewise-smooth curve, one can easily show that this length is finite and given by the formula where

I dz I

6

L= 3 1-IIz'(t)I z(t)12 a

dt.

In view of these definitions, we can reword Theorem 2 in the following form: Theorem 3(the invariant form of the Schwarz lemma). 1) if a function f(z) is regular in the disk I z I < 1 and if If(z)I _< 1 in that disk, then, for arbitrary points z 1 and z 2 in it, D (f (x1), .f (zs)) <- D (z1, zs)

Furthermore, in that disk, do f

dal

(12)

(the differential form of the Schwarz lemma). Equality holds in (11) and (12) only if f(z) is a fractional-linear function mapping the disk I z I < 1 into itself. 1) Pick [19161.

§1. INVARIANT FORM OF THE SCHWARZ LEMMA

333

Inequality (11) shows that, when the disk I z I < 1 is mapped by a function w = f(z) that is regular in the disk and satisfies there the inequality If (z)I < 1, the hyperbolic distance between the images of two points zI and z2 in the disk does not exceed the hyperbolic distance between the points zI and z2 themselves and is equal to that distance only in the case of a mapping of the disk I Z I < 1 into itself. From inequality (12) it follows that such a mapping maps an arbitrary piecewise-smooth curve l into a curve of no greater hyperbolic length. We now present some extensions of the Schwarz lemma.

Theorem 4. Let f(z) denote a function that is regular and satisfies the inequality if W I < 1 in the disk I z I < 1. Suppose that this function has the representation f (Z)

Ckzk, where n > I. k=n Then, if (z)I < I Z In in I z I < 1 with equality holding only when f (z) = ez°, where

I(I=I. The proof of this theorem is the same as the proof of the Schwarz lemma if we consider the function 0(z) = f(z)/z n, q(0) = en. Theorem 5.1) Suppose that, in the disk I z I < 1, a function f (z) is regular and satisfies there the inequality if (z) < 1 and has the representation f (z) _ CO + Fkan CkZk, where n > 1. Then, in the disk I z I < 1, we have

If- (z)IC

l

11Z I', ZIsn

(1-If(z)I9)

(13)

or

(14)

do, C daan,

with equality holding only when

Proof. By virtue of the invariance of. do= under a fractional-linear transformation of the disk I z I < 1 into itself, we have If (Z) I

where 1) Goluzin [19451.

1-If(Z)1'

=

I fi (Z) I

I,

(15)

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

334

00

1f(zc,

fl (Z) -

Y TkZk

)of (

k-n

Since the function 0 (z) = [1(z)/z" does not exceed unity in absolute value in the disk I z I < 1, inequality (2) can be applied to it. This yields r

r'n - I ft (Z) 1'

f, (z) I

& (Z)

zn - n zn+i

r2n (I

`

- r°)

(16)

Therefore,

9n- 9

If1(Z)I,n r

rn(1-ra)

and, consequently,

'

P=jf1(z)

,

r'5-P3 " r + rn (1- ra) P

I fl (z)

1-IA(Z)1' I

(17)

1-P'

Here p= I(I (z)I
function of p, increases in the interval 0 < p < r". Its derivative with respect to p has the same sign as the expression n -ran rt rp -2r4(1-r2)P-}-r. 9

(18)

Since the product of the roots of the polynomial (18) (in p) is equal to 1, this polynomial has no more than one root in the interval 0 < p < r". But (18) is positive at p = 0. Let us show that it is positive also at p = r'. We have n

r

r9n-2 -ran n 1-ra + r I = rn-1 [n (rn + Vin) -2(r'-'+ T ) - 2 (r_3 + r

1

- ...J.l

Since x + x'I decreases in the interval 0 < x < 1, we have in the interval 0 < r<1

rn+rn frk-}- r

k=0, 1,

..., n- i.

Consequently (18) is positive. This shows that the right-hand member of (17) is an increasing function of p in the interval 0 < p < r". If we replace p in (17) with the quantity r", which is at least as great, we obtain (13). Furthermore, it follows from what was said above that equality holds in (13).only when ((z) = cz ", where I c I = 1, that is, only when [(Z) = f (z" + a)/(1 + i z "), with I c I = 1 and Ia I < 1. By direct verification, we can show that equality does indeed hold in (13) for this last function. This completes the proof of the theorem.

§ 1. INVARIANT FORM OF THE SCHWARZ LEMMA

335

Theorem 5 can be strengthened further if we take into account the values of the initial coefficients of the function f (z). Thus, in the proof that we have given,

p=I.fa(z)I
IzI <1

days 1 -I (()I'

Idzl

dop(,),

(19)

where r= IzI and 0(r)=r n (r +Iy"I)/(1 + rjyi), where in turn y" = c,,41 - Ico12). Equality holds only when co+zn

.f (z) =

z+Tn 1 + Tnz

1 + cozn z + Tn

= co + cnz" +....

1 + Tnz We Tote in conclusion that, by virtue of what was said above, we immediately get a bo nd for the derivative of a bounded function f(z). Specifically, from Theorems 2 an.,5, we have the following theorem.

Theorem 7. If f(z) is regular in the disk z I < 1 and if I f(z)I < 1 there then

with equality holding at a point z = z0,I z0I < 1, only if f(z) _ e(z Ik="ekzk, where n > 1, then, in IzI -- zo)/(1 - To z). If, in addition f(z) = co + I

Iz

< 1, (21)

with equality holding at a point z = zo, where

lel

= 1.

1) Goluzin [ 1945].

I zo I

< 1, only if f(z) = e(z" - z"p)/(1- z 0"z"),

336

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

§2. The hyperbolic metric principle Let us turn now to the generalization of the invariant form of the Schwarz lemma to the case of multiply connected domains. Let B denote a multiply connected domain with more than two boundary points in the z-plane. This domain can be mapped conformally (as indicated in § 1 of Chapter VI) onto the disk I CI < 1. Here, the mapping function is not unique and it is not single-valued in B. However, we note that every branch of a single mapping function is connected with any branch of the same function or any other mapping function by means of a fractional-linear transformation. This circumstance will be extremely important in the present case.

The hyperbolic metric in the domain B is now defined as the hyperbolic metric of the disk I C1 < 1 carried over into B by this mapping. In other words, its differential element is given by the formula do,

-

-

IdCI -dat - 1-ICI'.-

C'(z)IIdzI

I

1

-IC(z)I9

(1)

Since the last fraction is invariant under fractional-linear transformations of the function C(z ),. the hyperbolic metric does not depend either on the choice of mapping function or on the choice of branch of it, and it is completely defined by the domain B and the position of the points in B. For any piecewise-smooth curve 1: z = z (t), where a < t < b, contained in B, we define the length L of 1 by the formula

-s

L-C do -

b

C'(z)IIdzl

s

1-IC(z)I9 -

S

a

IC'(z(t))IIz'(t)Idt 1-IC (z(t))I'

(2)

Under these conditions, we now have the Hyperbolic metric principle. Let B denote a domain in the z-plane and let G denote a domain in the w-plane. Suppose that each of these two domains has at least three boundary points. Let w = f(z) denote an analytic function defined by any of its elements and continued into B along an arbitrary path. Suppose that this function has the property that all the values that it assumes in B lie in the domain G. Then, an arbitrary piecewise-smooth curve l in the domain B is mapped by w = f(z) into a curve X the hyperbolic length of which relative to the domain G does not exceed the hyperbolic length of the curve l relative to the domain B. These two lengths will be equal only when f (z) = w(C(z)). Here, the function C= C(z) maps the domain B conformally onto the disk ICI < 1 and the

§2. THE HYPERBOLIC METRIC PRINCIPLL'

337

function w = w(0 maps the disk ICI < 1 conformally onto the domain G (both mappings in the sense of § 1 of Chapter VI). Furthermore, if do,. is the differential element at the point z E B and dow is the corresponding differential element at the image w E G, then daw < da= with the same condition for equality.

Proof. Suppose that functions C = jz) and t = t (w) map the domains B and 'G respectively onto the disks I CI < 1 and I t I < 1 in the sense of § 1 of Chapter VI. Let us denote by z = z (C) the function inverse to C = C(z) and let us define 'xhe composite function t = t (f(z (c))) = X(C). Obviously, this function is regular tin

I C I < 1; also, I X (C) < 1 in I C I < 1. From Theorem 3 of § 1, .we have IX'(C)IIdCI

1-ILL()Is or

< I -IdtI IC Is

1 -(Z()))Iz'(C)I t2 _141< ItV(z(C))I1f

JAI 1

-It is

If we now shift from the disk I I < 1 to the domain B, we obtain in B the inequality It'(f(z))IIf(z)IIdzI IC'(z)IIdzI (3)

l-It(f(z))Is

< l - IC(Z)Is

that is, tj daw < das , where w = f (z). Furthermore, if we integrate (3) with respect to 1, we' see that the length of the curve A does not exceed the length of the curve 1. Equality holds in (3) only when X(C) maps the disk ICI < 1 univalently into itself, so that the function f(z) = t-I(X(C(z))) has the form indicated in the wording of the principle. This proves the hyperbolic metric principle.

Corollary. Let G denote a domain in the z-plane and let B denote a proper sub domain of G. Then, throughout the entire domain B, we have dGai < dBai; that is, broadening of the domain decreases the hyperbolic distance between two points. This result is obtained from the hyperbolic metric principle by applying the latter to the function f (z) = z. We can use it to obtain a bound for dBaz by replacing the domain B with simpler domains that we know how to map onto a disk. As an example of the application of the hyperbolic metric principle, let us now prove a theorem.

Theorem.') Suppose that a function w = f(z) = co + clz + 1) For the case q = 2, see Landau [19041 and Schottky [19041.

is regular in

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

338

I z I < R and that it fails to assume in that disk q given finite values a1, , aq, where q > 2. Then, if c 1 0, the number R is bounded by a finite quantity that depends only on aI, - , aq, co, and cI (Landau's theorem). Furthermore, the -

quantity I f WI is bounded in the disk I z I < OR, where 0 < 0 < 1, by a finite quantity depending only on a1, , aq, CO (Schottky's theorem).

We shall prove this theorem and at the same time find an exact bound for R.

Proof. The hyperbolic metric in the disk I z I < R is given by the differential element do, = R I dz I/(R2 - I Z 12). In particular, at the point z = 0, we have duz = I dz I /R. On the other hand, let us denote by t = 0 (w, a 1 , , aq) a function that maps the domain G constituting the entire w-plane exclusive of the points a1, , aq onto the disk t I <1 in such a way that the point w = c o is mapped into t = 0. Then, for w = co , we have I

doas, = I p' (co, at, ..., aq) I

I dam, I.

From the hyperbolic metric principle, do,,, < due, which, for z = 0, yields cp' (co, a1,

...,

R<

aq) I If (0) I C

I I Ci I

I V (Co, at, ... , aq) I

Here, equality holds only in the case in which f (z) = 0 -1(z/R, a I , , aq). Furthermore, for any z0 in the disk I z I < 0 R, where 0 < 0 < 1. the hyperbolic distance between the points f (0) and f (z0) does not exceed the hyperbolic distance between the points 0 and z0 in the disk Iz I < R; that is, it does not exceed the quantity OR

L=

Rdr

5

R,-r, = 2 log 1-}-0 1-e

0

If E is the set of all points of the domain G such that the hyperbolic distance between them and the point co does not exceed L, then this set is closed and bounded and is completely determined by the values of c0 , a 1, , aq. Consequently, if M denotes the maximum of the ordinary distance from the set E to the point w = 0, then I f (z0) < M = M (co , aI , - - , aq). Here, equality holds for the same function as before. This completes the proof of the theorem. Also, the first part of the theorem can be obtained directly from the Schwarz lemma if we apply that lemma to the function 0 (f (z), a 1, , aq). -

§3. LINDELOF'S PRINCIPLE

339

§3. Lindelof's principle With the aid of Green's function, we can give another generalization of the Schwarz lemma.

Lindelof's principle.') Suppose that B and G are two finitely-connected domains bounded by Jordan curves lying respectively in the z- and w-planes. Suppose that w = f(z) is a function that is regular in the domain B and that the values that it assumes in that domain lie in the domain G. Then, for any two points

z and zo in B, we have (1)

g0(f(z), f(zo)):gB(z, zo),

where gB (z, zo) and 9G (w, wo) are Green's functions for the domains B and G.

If equality holds in (1) for one pair z, zo E B, it holds for all such pairs. Proof. We may assume that zo - because, in the opposite case, we can arrange for this by means of a fractional-linear transformation of the z-plane. Consider the two functions

g0 (f (z), f (zo)) + log If (z) - f (zo) I and gB (z, zo) + log I z -

zo

Both these functions are harmonic in the domain B punctured by the point z = zo. Theref re, their difference, that is, the function g0 (,f (z), f (zo)) - gB (z, zo) + log I

f (z) --f (zo)

z-zo

is also harmonic in B. This means that the difference gG (((z), f(zo)) - gB (z, z0) is a harmonic function in the domain B after the zeros of the function ((z) - f(zo) are deleted. (This difference may approach m as z approaches any of these zeros.) On the other hand, as z approaches the boundary of the domain B, all the limiting values of this difference are nonnegative. Therefore, in accordance with a property of harmonic functions, we have everywhere in B g0 (f (z), f (zo)) with equality holding only when

gB (z, zo)

0,

90(f(z), f(zo))-gB(z, zo)=0. This proves Lindelof's principle.

We recall that a Green's function g(z, z0) of a simply connected domain is equal to the negative of the logarithm of the modulus of a function that 1) Lindelof [1908].

340

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

maps this domain univalently onto the disk I CI < 1 in such a way that z = z0 is mapped into C= 0. Therefore, in the case of simply connected domains B and G, we can give Lindelof's principle a different formulation. Specifically, if B, and G, are the images of the disk I C I < r under mappings of the disk I C I < 1 onto the domains B and G respectively such that C= 0 is mapped respectively into z0 and w0 = f(z0), then Lindelof's principle asserts that the values assumed in the domain B, by the function w = f(z) all lie in the domain Cr. Furthermore, if one of these values lies on the boundary of GT, then the function w = f (z) maps B univalently onto G.

Let us give some examples of the application of Lindelof's principle. 1. In the case in which B and G are the disks I z I < 1 and I w I < 1, inequality (1) coincides with inequality (1) of §1. 2. Suppose that B is the disk I z I < 1 and that G is the halfplane l(w)> G. Suppose also that z0 = 0 and w0 > 0. In this case, a function mapping the disk CI < 1 univalently onto the domain B so that the origin is preserved is z = C, and a function mapping the disk ICI < 1 univalently onto the domain G so that C= 0 is mapped into w0 = f(0) has the form w = w0 (1 + C)/(1 - C). Consequently, all the values assumed in the disk I z I < r, where r < 1, by a function f (z) that is regular in B, has positive real part in I z I < 1, and is equal to a positive number w0 at the origin lie in the disk I (w - w0)/(w + w0)I < r. This disk is symmetric about the real axis and its boundary circle intersects the real axis at the points w0 (1 + r)/(1 - r) and w0 (1 - r)/(1 + r). This means that the radius of the disk I(w - w0)/(w + wo)l < r is equal to 2wor/(l - r2). In particular, in the disk I z I < r, r < 1, we obtain the inequalities f(0)11 -r +rC8j I

(f (z)) I

(f(z))
l 2r

r"

f(0)1+ cIf(z)ICf(0)I+r. 3. Suppose that B is the disk I z I < 1 and that G is the strip - 1 < l (w) < 1. Suppose that z0 = 0 and w0 = 0. A function that maps the disk Cl < 1 into G univalently in such a way that C = 0 is mapped into w = 0 is of the form +C

w= 2 log

1

Therefore, for functions f(z) regular in I z I < 1 and having the property that

§4. HARMONIC MEASURE. SIMPLEST APPLICATIONS

341

-1 < R (f (z)) < 1 in I z I < 1 and f (0) = 0, we have the following inequalities in

IzI
arc tan r, 79

(f(z))I C log 1 -r,

If (z)I<2log

i+r

We can also apply the idea of the proof of Lindelbf's principle to generalize it in a particular case. Suppose that the function f(z) is regular in a finitely, p., each connected domain B bounded by Jordan curves except for poles p 1, of which is numbered as many times as its multiplicity. Suppose also that the. , nI-L in B (these may or may not be all the zeros function f (z) has zeros n 1, of f(z) in B). Finally, suppose that all the limiting values of the quantity I f(z)I as z approaches the boundary of the domain B are bounded by a number M. Then, the function log

I M+

µ

gB (z, nk) k=1

gB (z, Pk) k-I

is harmonic in the domain B except possibly at the zeros of the function f(z). As z approaches any of these zeros, this function may approach -m. Furthermore,

this function has nonpositive limiting values as z approaches the boundary of the domain B. Consequently, in accordance with the maximum principle, µ g(z,%)+k18(Z,PA,)

If(z)I <Me k-1

(2)

everywhere in B. In the particular case in which B is the disk IzI < 1, this inequality takes the form

µ

V

z - nk

If(z)I < 111

k_I 1-nkz

:1

H

z - pk

k_I 1-PkZ

(3)

§4. Harmonic measure. The simplest applications Let B denote a domain in the z-plane bounded by Jordan curves.1) Again 1) The definitions given below and the results remain valid for domains with nonJordan boundaries, specifically, for domains obtained from domains of the type indicated by means of univalent mappings that are continuous up to the boundaries of these domains.

342

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

let us assume for simplicity that B is finitely connected. Consider a finite number of arcs (closed or open) of the boundary K of the domain B. Let us denote by a the set of all points belonging to these arcs and let us consider a function that is harmonic and bounded in B, that assumes the value 1 on a, and that assumes the value 0 at all other points of the boundary K. Here, we exclude the endpoints of the arcs constituting a since the function is not continuous at such points. By §3 of Chapter VI, this function exists and is unique. Let us denote it by rv (z, a, B) and let us call it the harmonic measure of the part a of the boundary K of the domain B about the point z. Obviously, 0 < c (z, a, B) < 1 everywhere in B. If we partition the boundary K into parts a 1, ... , an of the type indicated, we have

E w (z,

(1)

k=1

everywhere in B because the function on the left is a harmonic bounded function in B that assumes the value 1 on the boundary K except at a finite number of points. Under a univalent mapping of the domain B onto a domain B 1, the harmonic measure of an arbitrary part of the boundary K is mapped into the harmonic measure of its image in B1. An important property of harmonic measure is included in the following extension principle: The extension principle.l) If the boundary K of the domain B is partitioned into two parts a and S of the type indicated above, then, when the domain B is extended by changing only the part fi, the harmonic measure w (z, a, B )increases and when B is extended by notifying only a, the measure ta(z, a, B) decreases. Here when we speak of extending the domain B by modifying f3, for example, we mean replacing the domain B with a domain B' containing B and with bound-

ary containing the entire set a . Proof. Suppose that B' was obtained from B by extending the latter set in such a way that only 0 is modified. Then, the function u (z) = U) (z, a, B') - w (z, a, B), which is harmonic and bounded in B, is equal to 0 on a and is nonnegative on 0. Consequently, this function is nonnegative everywhere in B, which proves

1)Carleman [1921]

§4. HARMONIC MEASURE. SIMPLEST APPLICA'IRONS

343

the first part of the assertion. The second part of the assertion now follows from the formula o (z, a, B) = 1 - ca (z, S, B). The significance of the extension principle consists in the fact that it enables us to strengthen various inequalities by replacing complicated parts of the boundary K with simple ones for which the harmonic measure is easily calculated. Another important property of harmonic measure is given by the following theorem.

Theorem 1.1) Suppose that a function f (z) is regular and bounded in a domain

B with boundary K and suppose that K is partitioned into two parts a1 and a2. Suppose that all the limiting values of f (z)I as z approaches interior points of arcs constituting the ak (for k = 1, 2) through values in B do not exceed Mk. Then, in A log I f (z) I < w (z, at, B) log M1 + w (z, %, B) log M9.

(2)

Furthermore, if equality holds in (2) at a point z E B, it holds for all points in B. Proof. The function

u(z)=1oglf(z)I-w(z, %, B)1ogM1-w(z, as, B)logM2

(3)

is a harm nic function at all points of the domain B except at the zeros of f (z). As z ap roaches all points of the boundary K except the endpoints of the arcs constituting the ak from inside B, all the limiting values of the function u (z) are nonpositive. On the other hand, u(z) approaches -oe as z approaches the zeros of f (z). Consequently u (z) < 0 everywhere in B; that is, inequality (2) holds and the remark made in the theorem about equality is obvious. This completes the proof of the theorem. From the theorem just proved, we get

Theorem 2. Suppose that a function f(z) is regular and bounded in a domain B and the boundary K of B is partitioned into two parts a1 and a2. Suppose that the limiting values of I f(z)I do not exceed Mk as z approaches interior points of the arcs constituting the ak (for k = 1, 2) from inside B. Then, for every closed set E C B, there exist positive constants Al and A2 independent of the function f(z) such that )LI + A2 = 1 and I f (z) I

on E. 1) F. and R. Nevanlinna [1922].

Mi1M9'

(4)

344

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

Proof. Suppose that M 1 and M2 are numbered in such a way that M I < M2. For the function ((z) in the domain B, we have inequality (2) which, by virtue of (1), we can write in the form If(x)I<(MM,)m(Z.at,B)

,

M.

(5)

If we set Al = min 2EE(O(z, a1, B) and )L2 = 1 - A1, then AI and A2 are obviously positive and we obtain from (5) ( If(Z)I_I")

1\x

on E, which completes the proof of the theorem. Inequality (2) will find more than one application in what follows. Let us give an example. Suppose that the domain B is the annulus rI < z I < r2. Suppose that a1 is the circle I z rl and a2 is the circle I z = r2.

Then, as is easily seen,

_ log r/r, r _-IzI'

_ log r/r,

w (z, at, B) - log rjr, ' w (z, al, B) - log r,/r, '

(6)

By setting M (r) = maxi z is r I f (z) r I < r < r2, we obtain from (2) log r/r,

log r/rl

M (r) C MIlogrl/rr . M2logr2/ri

(7)

This inequality is known as Hadamard's "three-circle theorem". It can also be obtained directly. Specifically, consider the function z °ff(z), where a is an arbitrary real constant. The maximum value of Iz°f(z)I in the annulus rl
1. Suppose that B is the halfplane Z (z) > 0 and that a is a segment of the real axis. Then, as one can easily verify, we have (in the case of a halfplane, the harmonic measure on the boundary is defined as above)

§4. HARMONIC MEASURE. SIMPLEST APPLICAWONS

345

w(z,a,B)=1, it

(8)

where (k is the angle subtended by the segment a at the point z. 2. Suppose that B is the half-disk I z I < 1, a (z) > 0 and that a is the boundary semicircle. Then, one can again verify easily that w (z,

(9)

B)

where
Theorem 3. Suppose that a function f (z) is regular and bounded in the half(z) > 0, and continuous except on the positive half without the point z = 0

plane

of the real axis. Suppose that f (z) , a as z approaches the origin along this 0 in the angle half axis. Then, f (z) --> a uniformly for arbitrary 8 > 0 as z

0<argz 0. Let us choose arbitrary c in (0, 1) and let us choose r > 0 such that If (z) < e on the interval a, of the real axis from 0 to r. If we apply inequality (2) to the halfdisk B,.:

z I < r, 3 (z) > 0, we obtain log I f(z)I ---- w(z,a,.,B,)loge

(10)

in B,. Let\us find a lower bound for w (r, a,, B,). To do this, let us find an explicit expression for this quantity. The function C= rz/(r - z)2 maps the halfdisk B, into the halfplane (C) > 0 in such a way that the point z = 0 is mapped into the point C= 0 and the interval a, is mapped into the positive half of the real axis. We denote this last by a. If z and C are corresponding points, then Also, by virtue of (8),

w (z, a B,) = w (C, a, B)-

_71-11R

, 0<argC
Thus,

w(z,a,.,B,)=1-

I 79

arg(rrzz)'-1- a argr LL

lZ)I 1- rl J

It follows that all limiting values of the quantity co (rz, a,, B,) are at least equal to 1 - (tr - 8 )/tr = 8/tr as z - 0 in the angle 0 < arg z < tr - 8. This means that there exists an r' in the interval (0, r) such that co(z, a., B) > 8/2tr= 8

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

346

in the domain I z I < r' , 0 _< arg z < n - 8. Consequently [cf. (10)], I f (z) _< E S. Since S does not depend here on r or c, this means that f(z) --' 0 uniformly as z -+ 0 in the angle 0 < arg z < n - 6. This completes the proof of the theorem.

Theorem 4. Suppose that a function f(z) is regular and bounded in the half plane (z) > 0 and 1 is a Jordan curve lying in that halfplane that terminates at z = 0. Suppose also that f(z) --, a as z --+ 0 along 1. Then, f(z) --, a uniformly as z --, 0 in the angle 0< arg z < n - 6 for arbitrary 8> 0. Proof. We can again assume that a = 0 and I f (z)I < 1 in (z) > 0. Let c denote a number in the interval 0 < e < 1. Let us choose r sufficiently small that If (01 .5 e on the part of the curve 1 lying in the half-disk B,: I z I < r, (z) > 0. Let us denote by a, the part of 1 extending from z = 0 to the first point at which 1 intersects the circle I z I = r. The arc a, partitions the half-disk B, into'two domains B, and BT. Suppose that B, is adjacent to the negative half of the real axis. By applying inequality (2) to B,. , we obtain log E.

Here again, it remains only to find a lower bound for -(z, ar, Br ). Let us denote by 0, the portion of the boundary of the domain Br complementary to a,, and let us denote by ar the portion of the boundary of the domain B r complementary to f3, . Finally, let us denote by a" the interval (0, r) of the real axis. Then, from the extension principle, we have w (z, ar, B,.) . w (z, a Br) : w (z, ar, Br) But, as follows from the proof of Theorem 3, the quantity a) (z, 0-,", Br) has a positive lower bound S1 in the sector I z I < r', 0 < arg z < n -6 with sufficiently

small r' in (0, r). ,

e

Thus I f (z) I < Esl in the part of B , lying in this sector. Treating B,s in

the same way, we can show that there exists a S2 > 0 such that I f (z)I < E 2 in that part of B,. lying in the sector I z I < r" , 0< arg z < n with sufficiently small r" in (0, r). Suppose now that 0 < $ < n/2. If a is the greater of the numbers S1, 82 and if r1 is the smaller of the two numbers r' , r" , then If(z)I <eS for I z I < r1 and h < arg z < n -$ . This proves the uniform convergence of f (z) to 0 as z - 0 in the angled < arg z < n - $. This completes the proof of the theorem.

Theorem 5. Suppose that a function f (z) is regular and bounded in the half (z) > 0 and is continuous except at finite points of the real axis. Suppose

plane

§4. HARMONIC MEASURE. SIMPLEST APPLICATIONS

347

that f(z) approaches limiting values a and b as z approaches - and +along the real axis. Then a = b and f(z) -+ a uniformly as z

cc in the half

plane Z(z) > 0.

Proof. If we apply Theorem 3 to the function f (-1/z) - b, we conclude that this function approaches zero uniformly as z -+ 0 in the angle 0 < arg z < n - $, where $ > 0. Furthermore, we can show, in a manner analogous to the proof of Theorem 3, that the function f (- I/z) - a approaches zero as z --+ 0 in the angle $< arg z < n. For $< n/2, we see from this that the function f (z) approaches both a and b as z. -+ cc in the angle $ < arg z < n - $. Consequently, a = b and f(z) , a uniformly as z -+ oo through values in the halfplane (z) > 0. This completes the proof of the theorem. By means of a conformal mapping, we obtain from Theorem 3 the following two analogues of this theorem.

Suppose that a function f(z) is regular and bounded in the angle $1 < arg z < $Z and that it is continuous except at finite boundary points of that angle. Suppose that it has definite limits a and b as z approaches o along both rays of that angle. Then, a = b and f(z) -+ a uniformly as z -+ oc in that angle.. Theorem 5". Suppose that a function f(z) is regular and bounded in a simply connec ed domain B bounded by a Jordan curve, that it is continuous on B except at a point z = on the boundary K of the domain B, and that it has finite limits b as' z approaches c along K from the two sides. Then, a = b and f(z) is continuous on B if we define f(c) = a. Theorems 3-5 are due to Lindelof1) and they apply to bounded functions. We shall now give another theorem on bounded functions, this one due to Milloux.2) It is of a somewhat different nature from the preceding ones. Theorem

.

Theorem 6. Suppose that a function f(z) is regular in a simply connected domain B that is contained in the disk I z I < 1 and has boundary K consisting of an arc a of the circle I z I = 1 and of a continuum f3 belonging to the disk z I < 1 and containing z = 0. Suppose that I f(z) I _< 1 in B and that none of the limiting values of If(z)I exceed e, where 0 < e < 1, as z approaches points on JS through values in B. Then, in B, If(z)I<e11 1) Lindelof [19151. 2) Milloux [19241.

areslnl+1t1/EIC (1-IZII. (11)

348

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

For any zo in the disk Iz I < 1, there exists a domain B containing zo and in it the function f(z) is such that equality holds in the first of inequalities (11) at

z=zo. Proof." When we apply inequality (2) to the function f(z) for z E B, we obtain log If (x) I < w (x, P, B) log e.

(12)

Here, the question consists in finding a bound for the harmonic measure cw(z, 8, B).

We shall first need to prove an inequality for univalent functions. Suppose that a function g(C) is regular and univalent in the disk ICI < 1 and that it does not assume the value zero in that disk. Then, the function

g` h(Q)

C)-g(C.) 0

g'(CO)(1-ICoj)

-c+...

will, for arbitrary Co such that ICo I < 1, be regular and univalent in IKI < 1 and will not assume in that disk the value -g(to)/g'(t;') (1 - I Co12). By Koebe's theorem it then follows that I g(Co)/g '(co) (1 - K0!2)! > 4; that is, 4 g(Ca)

1-ICoI

(13)

Equality holds in (13) in the case of the function g(') = [(1 + 77.0/(1 - 770] 2 where 1771 = 1. If we set Co = rei9 in (13) and integrate with respect to r from 0 to r, we obtain, for arbitrary C in the disk I CI < 1 the inequality llog

with equality holding under the same condition. Finally, if g(C) is such that g(0) = 1, then for z = g(t;) we have in I z I < 1

-ylzl

1±ylzl'

(14)

This is the desired inequality. Let us now see about finding a bound for w (z, 0, B). We map the domain B onto the half-disk B `: I CI < 1, (C) > 0 in such a way that the arc a is mapped into the boundary diameter of the half-disk. Suppose that the point zo E B is 1) The proof given here with the sharp inequality was given by E. Schmidt [1932].

*4. HARMONIC MEASURE. SIMPLEST APPLICATIONS

349

mapped into a point Co, Idol < 1. By means of a supplementary mapping of the half-disk B ' into itself, we can arrange to have the point Co lie on the imaginary axis. Specifically, let us draw through Co a circle orthogonal to the circle ICI = I and to the real axis. Denote by a the point of intersection of this circle with the real axis. When we map the upper half of the C-plane into itself in such a way

that the points - 1, a, and + 1 are mapped into the points -1, 0, and + 1 respectively, the half-disk B' is mapped into itself and the orthogonal circle mentioned above is mapped into the imaginary axis. In particular, the point Co is mapped into a point on the imaginary axis. Thus we may assume that Co = ito for to > 0. We denote by z = g(C) the inverse function providing the mapping mentioned. We have zo = g(ito). The function g(C) can, according to the symmetry principle, be extended beyond the real axis, and it maps the complete disk ICI < 1 univalently onto a domain that does not contain the points 0 and oo. Inequality (14) is therefore applicable to it. If we denote by j9 ' the boundary semicircle in B , we now have by using (9) 4 4 .w (zo, B) = to (Co, P', B') = arc tan to = arc tan Co R which, by (14), yields (zo, P, B}

arc tan

tan = i1-

+

IzoI

IzoI'

4

arc tan

I --

zo I

TT

zo I

(15)

,+,-,l z° l Zo-,

Sin

2tan 4 _i-IzoI =f -{-lane

i --1-1 zo I

and, consequently, I zo I -2I arc sin r-1+1zol*

Therefore (15) can be written in the form (o (z

> a arc sin i

zoI

(16)

where zo is an arbitrary point of the domain B. Inequalities (12) and (16) also lead to the first of inequalities (11). Equality holds in it only when B is a disk with cut along a suitable radius and the function f(z) is such that, as z assumes values on the half-disk B', the function log I f (z) I is a harmonic function that is equal to 2 (1 - 96/n) loge on the boundary, where 96 is the same as in formula (9). From the inequality

350

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

arc sin

1-Izl' 1-IzI> 1-IzI I+IzI

1+Iz1

2

we get the second of inequalities (11). This completes the proof of the theorem.

Now, we shall finally give an application of the theory of harmonic measure to a single question associated with the principle of the maximum modulus. We recall that, if a function f (z) is regular in a domain B and if none of the limiting values of its modulus exceed M as z approaches any of the boundary points of the domain B, then if (z) I < M throughout the domain B. The situation is quite different when the requirement of boundedness of the limiting values of I f(z)I is violated even at a single point of the boundary. Then, the modulus principle no longer holds. For example, the function f (z) = e-" is regular in the upper halfplane and its modulus is equal to 1 at finite points of the real axis. Furthermore,

I f(z)I > 1 in the halfplane (z) > 0. The reason for this is the fact that the limiting values of the function are not all finite as z approaches the point z = o. in (z) > 0. Now, we have

Theorem 7.1) Suppose that a function f(z) is regular in the halfplane (z) > 0 and that all the limiting values of its modulus are bounded above by 1 as z approaches any finite point on the real axis. Then, either (1) If(z)I < 1 in (z) > 0 or (2) there exist positive numbers A and R such that

M(r)>eAr

(17)

where M(r) denotes the maximum value of I f(z)I on the semicircle Izi = r,

(z)>0 for r > R. Proof. Let us denote by B the half-disk I z I < r, (z) > 0 and let us denote by a. its boundary semicircle. From inequality (2), we have 1

log I f (z) I C w (z, a, B) log M (r).

(18)

But by (9), if z = x + iy,

w(z,a,B)= 22

-azc tan r y Xt. ac tan r+X 11- 2

1)Phragmen and Lindelof [19081.

(arc tan

x

arc tan

§5. ON THE NUMBER OF ASYMPTOTIC VALC"ES

351

Consequently, for fixed z and sufficiently large r, cu

(z,a,

B ti 2 (l r-yx + ry--}- x ) = 2r. re2yr - x° 7c

4y

.

nr '

(19)

Suppose now that the inequality I f(z)I < 1 does not hold everywhere in B. Let zo = x0 + iyo denote a point in B such that I f(zo)I > 1. Then, from (18) and (19), we have log If (zo) I < 11M log M (r)

r which shows,that, for suitable A > 0 and sufficiently large r, inequality (17) holds. This completes the proof of the theorem. By performing a conformal mapping, we obtain from Theorem 7 a generalizing theorem that will be used in the following section. 4yo

r _, co

Theorem 7'. Suppose that a function f (z) is regular in an angle of magnitude no with vertex at z = 0. Suppose that the limiting values of I f(z)I do not exceed 1 at finite boundary points. Then either (1) If WI < 1 in the entire angle or (2) there exist positive numbers A and R such that

M (r) > e''110

(20)

where M (r) is the maximum value of I f(z)I on an arc of the circle I z I = r lying in th angle in question, where r > R.

§5. On the number of asymptotic values of entire functions of finite order Let us give an application of the theory of harmonic measure to a more complex question, one dealing with entire functions. Suppose that f (z) is an entire function and that l is a continuous curve in the z-plane that extends from some finite point to oo . If, as we let z move along

1 towards 00, the function f(z) approaches a definite limit a, then a is called the asymptotic value of the entire function f(z) defined by the curve 1. Suppose that two curves ll and 12 of the type described define finite asymptotic values a1 and a2. If a1 a2, these values are considered distinct; if a1 = a2, then let us agree to consider the curves ll and 12 as defining distinct asymptotic values only in the case when there is some R > 0 such that the curves 11 and 12 cannot be connected by a continuous curve lying in the domain I z I > R and having

352

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

the property that the oscillation of the function f (z) on it is arbitrarily small. In the theory of entire functions, this convention identifies the asymptotic values of the entire function f(z) with nonalgebraic (transcendental) singularities of the inverse function. If an entire function f (z) is such that, for some finite k > 0 and sufficiently large r, we have M (r) < e rk , where M (r) = maxe Zl =r I f (z)j then f (z) is called an entire function of finite order. The greatest lower bound p of the numbers k that possess this property is called the order of the entire function f (z). It was assumed that the number of asymptotic values of an entire function of finite order is finite and is bounded by a quantity depending only on the order p. The complete answer to the question was given by Ahlfors.1) We shall give his solution, using the method of strips, which we have already used in §6 of Chapter IV.2) Lemma. Suppose that there are a finite number n of nonoverlapping radial strips Sk, for k = 1, 2, , n, in the annulus ro < I z I < r that extend from the circle I z I = ro to the circle I z I = r. Let ak (rk) denote the arc of the circle z = ro (I z I = r) that serves as part of the boundary of the strip Sk. Suppose that none of these arcs degenerate to a point. If we map these strips conformally onto nonoverlapping domains F,k: ro <

< rF, 'k <arg S < 'k+1

that completely fill the annulus ro < I CI < r' and if this mapping maps the arcs o, k and rk into the arcs tk < arg C < Ok+1 on the circle I I = ro and on the

circle ICI = r', then r' > r. Proof. The assertion follows from Lemma 1 of §6 of Chapter IV. Specifically the function t = log C maps the domains Ek onto the rectangles Tk: log ro < , n, with sides of lengths ak = R(t) _< log r', 'k < (t) < 95k+l, k = 1, Ok+1 -'k' bk = log(r'/ro). But the strips Sk are also mapped onto these rectangles and in the way required by the lemma of Chapter IV referred to. Therefore, in accordance with that lemma, n

r, < log r log !o

Pk+i - Pk

k-1

2+c

. ,

ro

that is, r' > r. This completes the proof of the lemma. Theorem. An entire function f(z) J const of order p has no more than 2p

1) Ahlfors [19301 2) Goluzin [1946d1

§5. ON THE NUMBER OF ASYMPTOTIC VALUES

353

finite asymptotic values.

Proof. We note that if two curves 11 and 12 that extend to infinity define two distinct asymptotic values of the function f (z), then, outside some disk I z I < R they have no points in common. Therefore, we can in general assume them to have no points in common. Without changing the asymptotic values, we can replace these curves with broken lines that extend to infinity and that consist of a finite or countable number of segments whose endpoints cluster only at infinity. To see this, let us partition, for example, the curve l1: z = z 1(t) for t0 < t < t1, where lim4,1 z 1(t) = -, with the points t" tot, . . , (n), and let us inscribe in the part of I lying between the points z(t(n)) and z(t("+I)) a broken line An such that the values of the function f (z) on each segment of this broken line will differ from the values on the arc beside it by an amount less than En, where each e. > 0 and c. --+ 0 as n --' a-. Then, on the broken line 11, consisting of A1, A2, . , the function f (z) has the same asymptotic value as on 11. We can even assume that the broken line 11 has no multiple points because, as we move along it, we can bypass any closed polygons constituting part of it. , In defining a distinct finite asymSuppose that we have n curves 11, ptotic values of the function f(z). On the basis of what has been said, we can assui a that these issue from z = 0, that they have no other points in common and that ach is a broken line without multiple points. These curves partition the z, n, indexed in such a way that each Bk is plan into domains Bk, for k = 1, boun ed by 1k and 1k+1 (here, 1n+1 = 11). In each of the domains Bk, the func" tion f z) must be unbounded; otherwise, in accordance with Theorem 5 of §4 (applied to the function f(z_1 + c)) its value would converge uniformly to a limit as z --* - in one of the domains Bk and then the curves bounding this domain would not define distinct asymptotic values. Therefore, the set of points in each of the domains Bk, where

If(z)I>N= IVsup In If(z)I is nonempty. This sets consists of domains that extend to infinity. This is true because, if one of them were bounded, inside it we would have If (z)I > N, which the maximum principle asserts is impossible. For every k, we shall consider not Bk but one of the domains lying in Bk just mentioned and we shall now denote it by Bk. Let r0 denote a positive number such that the circle I z I = r0 contains points belonging to all the domains B k, for k = 1, , n. The maximum value of

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

354

I f(z)I in the part of Bk contained in I z I < ro is attained at some point zk on the circle I z I = ro which does not lie on the boundary of Bk. Let us now look at the part of the domain Bk lying in the domain I z I > ro. It consists of domains. We denote by Bk that domain containing the point zk on its boundary. This Bk must extend to - because otherwise the maximum principle would again lead us to a contradiction. If o k- is the longest arc of the circle I z = ro that contains a point z k and constitutes part of the boundary of the domain B k , then, by replacing the domain Bk with the domain Bk consisting of Bk, vk, and that domain that is included in the intersection of the domain Bk with the inversion of Bk about the circle z I = ro which is adjacent to ak. Since zk is an interior point of the arc ak, then, if we denote by Nk the maximum value of I f(z)I on the boundary of the domain Bk, we have Nk < I f(zk)I. These operations are carried out for all Let us now consider the annulus ro < I z I < r for arbitrary r > ro. We denote by Sk the connected subset of the domain Bk that is contained in that annulus and that is adjacent to the arc vk. We denote one of its boundary arcs on the , n, as strips of circle I z I = r by rk . We treat the domains Sk, for k = 1, 2, the type mentioned in the lemma. As in the lemma, we map these strips onto strips Ek filling the annulus ro < I C1 < r'. We denote by C = CA: (z) the functions carrying out this mapping and we denote their inverses by z = z k (C). Then, we have r' > r. But, in accordance with the symmetry principle, the function C = Ck (z) is, for each k = 1, - , n, extendable to the entire domain Bk and maps that domain univalently onto some domain Gk contained in the sector ICI < r', Ok < arg C < Ok+1. Let ak denote the image of the arc rk under this mapping and let 13k denote the complement to ak on the boundary of Gk. On ak , I

and on PA,,

If(zk(c))I -M(r)=maxIf(z)I, zI=r

If(zk(C))IGNk. If we apply inequality (2) of §4 to the domain Gk, to the function f (zk (c)), and to the point C k = Ck (zk), we obtain

loglf(Zk)I=loglf(zk(Ck))I W(Ck, ak, Gk)log M(r)+w(Ck, Pk, Gk)logNk M N(k) +logNk. U) Gk, ak, Gk)log

Let us now assume that r is sufficiently great that M(r)/Nk > 1 for all k.

(1)

§5. ON THE NUMBER OF ASYMPTOTIC VAIIUES

355

Let us find an upper bound for the harmonic measure co(Ck, ak, Gk). By the extension principle, it is no greater than the harmonic measure of the arc ak with respect to the sector obtained from lk by adjoining suitable points of the disk C1 < ro. It follows from the mapping of the sector by the function cr _ C 1/8k , where '&k = (Ok+1 - ck)/rr, that this last measure is equal in the half-disk (C' i r* = r' rlok,3 (C') > 0 to the harmonic measure of the boundary semicircle C' J = r*, 0 < arg C' < n relative to some point CA, = ek + irlk on r1/8k; that is, in accordance with (9) of §4, it is equal to

-

(arc

n

tan

77k

r*-Ek

+ arc tan

Thus we obtain W

/

lCk, ak, GO

- (arc 2 n \

tan

2

'k

'1k k r*-Ek + arc tan r*+ek

4

'Ik

7 \r*-Ek +r*+Ek)

'R r'2/8k-Ek 4

and (1) yields

r'1/8krc/8k

x r'2/ek

c a r2 8

1

- r2/4k o

4

r1/$rol$k

a r2/ek

- r2/ek o

1/8k l/8k

log I f (k) I k

ro

k- r2/eo

k

log M (r)

4

a

Ark

r;/kkr- i/ek log M (r)

Nk

, n and let Iret us denote by A the smallest of the numbers Nk for k = 1, , n. Then, us de ote by eB the smallest of the ratios I f (zk)I/Nk for k = 1, sufficiently large r, for M (r)

Aecr, kBr118k, lim Cr, k r co

i/ek, A> 0, B> 0.

4r0

But Y.%=1 $k = 2n, $k > 0. Consequently, there exists at least one k such that $k < 2/n, that is, such that 1/ak > n/2. If we choose such a k in the last inequality, we obtain, for sufficiently large r, M (r) > Aecr Brn/2, lim cr> 0,

and this means that f (z) is of order no less than '-con/2. Consequently, n/2 < p; that is, n < 2p. This proves that f(z) cannot have more than 2p finite asymptotic values. This completes the proof of the theorem. The bound for the number of finite asymptotic values of entire functions given

in the theorem is a sharp bound when p is an integer. This is shown by the examples of the entire functions

356

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS z

sin zP zP

and

s'n zP dz zP

of order p. The first of these has 2p distinct asymptotic values 0 defined by the rays arg z = krr/p, for k = 1, 2, , 2p; the second has 2p asymptotic values e 217ki/ P, for k = 1, 2,

,

2p , which are defined by the same rays.

§6. The hyperconvergence of power series

The theory of harmonic measure also finds significant applications in the theory of power series. Let us now look at the question of what is known as hyperconvergence of a power series. Suppose that we have a power series W

f(z)==

E cnzn n=0

(1)

with radius of convergence R > 0. If, for a suitable grouping of its terms without changing their order, this series converges on some set of points z outside the closed disk I z < R, this property is called the hyperconvergence of the power series. For example, consider the series W

f (z) = > [z (I + z)]An,

(2)

n=0

where the X. are integers such that io < Al < h2 < ... and hn+1 > 2,1,, for n = 1, 2, . .. If, in each term of this series, we expand the parentheses and write the term in the form of a polynomial arranged in order of increasing powers, we obtain a power series of the form (1) with radius of convergence given by the equation

R (1 + R) = 1, so that R = (''S- 1)/2. For this power series, the series (2) is a grouped series and it converges on the set E of points satisfying the inequality I z (1 + z) I < 1. Since E contains points in the domain I z > R, this means that the power series in the present case possesses the property of hyperconvergence. We have a number of theorems due to Ostrowski 1) on hyperconvergence. A,

Theorem 1. Suppose that a power series f(z) = En=pcz ", where 0 < Ao < h1 <,12, . , has radius of convergence 1 and that it is a lacunary series, that is, there exists a sequence of positive integers nk, for k = 1, 2, - , such that ik+1>gJ`np,

1) Ostrowski [1927].

§6. THE HYPERCONVERGENCE OF POWER SERIES

for k = 1, 2,

,

357

where q > 1 and is independent of k, then the grouped series W

f (z)=panr (z)+ E pan+l, k=1

Panr (z) = Coza° +

k

Xn

k

(3)

(z),

... + CnrzXnr,

Zn + 1

pank+I,knk+1(z)=Ck 1 z

b+I an

--... +Cnk+Iz k+1, k=1, 2,

.. ,,

converges in a sufficiently small neighborhood of an arbitrary point on the circle z = 1 at which f (z) is regular. 1) Proof. Suppose that the function f (z) is regular at the point z = 1 (if it is regular at a point z = e` °', we could consider the function f (e-taz)). Let us set

z = a(t;) = Q' (1 + C)/2, where p is a positive integer. Since u(1) = 1 and la (C) I < 1 for I C1 < 1 but C 1, the function F ([;) = f (a (O) is regular on CI < 1. From (3) we now obtain the transformed series OD

F(C)=f(a(C))=panr (a(C))+

(4)

k_1

k

Here, the two quantities constituting the subscript to Q are the lowest and highof C that appear in the corresponding polynomial. If we take p > est 1/(q -J 1), so that (p + 1)/p < q, we have (p + 1)Ank < plink+I In this case, the series (4) is a grouped series with respect to the power series representation of < p, where F(C) about 0. Since this last series converges in some disk p > 1, 't also converges on the set of points C into which a sufficiently small neighborhood of the point z = 1 is mapped. But then, if we shift back to the variable z in (4), we conclude that the series (3) converges throughout a sufficiently small neighborhood of the point z = 1. This completes the proof of the theorem. In a sense, the following theorem is a converse to Theorem 1. Theorem 2. Suppose that a power series f (z) =

F+n=ocnzn

has radius of con-

1) In particular, if the inequality hn+1 > qhn, where q > 1, is satisfied for all n = 1, , then the power series itself can be regarded as grouped. In this case, pi,n, Xn+1(z) = enzxn+ 1. In accordance with Theorem 1, it will therefore converge in a sufficiently small neighborhood of every regular point on I z I = 1. But this last is impossible because a power series cannot converge outside its circle of convergence. This means that, in the present case, the function f(z) cannot be continued beyond the circle of convergence. This is a familiar theorem of Hadamard.

2,

358

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

vergence 1 and that corresponding to it is a grouped series that converges uniformly in a sufficiently small neighborhood of some point z0 such that I z0I > 1. Then, f(z) can be represented as the sum of a lacunary series and a series with radius of convergence exceeding 1. Proof, It follows from the conditions of the theorem that there exists a neighborhood y contained in I z I > 1 on which the grouped series converges uniformly. This means that there exists a sequence of partial sums snk (z) = L.nk0 CnZn, k= 1, 2, , of the original power series that converges uniformly on A. We may assume that nk+1 > 2nk for k = 1, 2, . Consider the function -

uk(z)=

klogls"k(z)I-log Izl

(k = 1, 2, ...),.

which is harmonic in the z-plane except at the zeros of the polynomial snk (z). Obviously, there exists a positive number Ka such that, for k > K1, the function uk (z) < - q on y, where q is a positive number independent of z and It. Now, let r denote a positive number. Define S = e'"Z , so that S < 1. Since the maximum value of Isnk (z)/z"kI in z I > S is attained for Iz I = S, if we set mS.k = max I-S ISnk (z)I, then, in j z > S and, in particular, in I Z I > 1, we have

Uk(Z)criklogma,k-logs=riklogma,k+ 2 Since

lim ma k = 0

k -.co

it follows that there exists a positive number K2 such that uk (z) < E in I zI > for k > K2. Inequality (2) of §4 can be applied to the domain bounded by the circles I z I = 1 and y and to the closed set E representing the circle Iz I = R, where R > 1, lying inside this domain. (By means of a conformal mapping, inequality (2) of §4 can obviously be carried over to the present case.) As a result, we see that, for k > K = max (K1, K2), we have uk (z) < - A1q + A2E on z I = R, where Al and A2 are positive numbers depending only on R (note that K depends on e). If we choose a sufficiently small that -A1q + A2E = -rl <0, we conclude R the inequality uk (z) < -77, or, that, for k > K, we have on the circle z what amounts to the same thing, I

I Snk (z) I

(Re --'1)4k.

By using this inequality, we obtain from the formulas

§6. THE HYPERCONVERGENCE OF POWER SERIES

cm

Snk (z)

('

1

ai

359

dz, m ` nk,

zm

IzI - R

bounds for the coefficients:

IC.I<

(Re-1)"k

m < nk.

Rm

In particular, for q ' nk < m < nk, where 1/2 < q' < 1, this yields CmIC Rq, )l ((Re'1

nk (5)

If we choose q' sufficiently close to 1 that Re '1/R'' = p < 1, we obtain from (5)

IcmI Cpnk
(6)

Inequality (6) shows that the series 1k=1 Pq, nk+l, nk(z), treated as a power series converges in the disk I z I < 1/p, where 1/p > 1. The remaining part of the power series is obviously a lacunary series. This completes the proof of the theorem.

Theorem 3. Suppose that a power series f (z) = EE CnzZn n-0

has a radius of convergence equal to unity and that there exist positive numbers nk, for k = 1, 2, such that Ank+l/Ank m as k --, o. Then the grouped series co

f (z)=P (z)+ EPxn+1, 1

k=1

k

xn

(z)

k+1

(7)

converges at all finite points of the z-plane at which f(z) is regular. The function f(z), treated in the sense of Weierstrass, is single-valued in the z-plane. Proof. Suppose that the function f (z) can be continued along some curve C to the point z0, beginning with an initial element defined by the given powers series. Then, there exists a simply connected though not necessarily one-sheetedl) domain B lying over the z-plane and containing the points 0 and z0 such that f (z) is regular on the closure of that domain.

1) With regard to Riemann surfaces and mappings of them onto a disk, see §2 of Chapter XI of the present book.

360

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

The domain B can be put in one-to-one correspondence with the disk ICI < 1 by means of an analytic function. The inverse of this function z = (k(C ) is regular in ICI < 1. Suppose that z o = q (Co ), where I Co I = p < 1. The function 0 (0 can be uniformly approximated on an arbitrary closed subset of the disk ICI < 1 by means of polynomials. Therefore, there exists a polynomial a(C) with a(0) = 0 of some degree p such that the values assumed by it in the closed disk I CI < \IP will be continuously distributed on the domain B. The transformed series (7) co

F

f (a )) = Qpa, (C) + I Qxn+ 1, pan

(C)

(8)

k-1 k k+1 is obviously, after we discard a finite number of initial terms, a grouped series for some power series. Since F (z) is regular on I CI < 1, the series (8) converges uniformly on the disk I C I < J and in particular it converges at the point Co . Returning now to the variable z, we shift from the series (8) back to the series (7), which consequently converges at the point zo. Since the value of the function f(c) is defined at the point zo as the sum of the same power series independently of the method of analytic continuation, it follows that f(z) has a one-sheeted domain of definition. This completes the proof of the theorem. §7. A nonanalytic generalization of the Schwarz lemma. A theorem on covering of disks We pause now for a nonanalytic generalization of the Schwarz lemma with its application to covering questions.1)

Suppose that a generalized metric is defined in a domain B in the z-plane, where z = x + iy. Suppose that the differential element of this metric is defined by ds = A(z) I dz 1, where A(z) is a nonnegative function. Here, the function A(z) can vanish only at isolated points of the domain B. The metric is said to be continuous in the domain B (or at a point zo E B, respectively) if the function A(z) is continuous in B (or at zo). It is said to be regular in B (at the point zo E B) if A(z) is continuous and has continuous first and second partial derivatives with respect to x and y in the domain B (in some neighborhood of zo). In what follows, of especial importance are those regular metrics for which the function A(z) satisfies at all points of B at which it is nonzero the inequality Alog X

1) Ahlfor [1938].

4l'

(1)

§7. NONANALYTIC GENERALIZATION OF SCHW1tZ LEMMA

361

(where A denotes the Laplacian operator) or, if we set u(z) = log A(z), the inequality

Au ,- 4e'u.

We note that in the case of the disk

zI

(2)

< I and the hyperbolic metric dQ =

Idzl/(1-Iz12), if A(z)=1/(1-1z12), we have in

IzI<1

A log l = U2.

(3)

Sometimes, a metric will be defined not in the entire domain B but only in some neighborhood. of a given point zo E B. Then, we shall speak of a metric defined in that neighborhood. It is a remarkable fact that an analytic mapping maps a metric satisfying condition (1) (or (3), respectively) into a metric satisfying con-

dition (1) (or (3)). To see this, note that if a function z = z(C) is regular in a neighborhood of a point C_ CO and a metric ds = A(z) ldzl satisfying condition (1) (or (3)) is defined in a neighborhood of the point z0 = z (Co) E B, then the transdtI also satisfies condition (1) formed metric ds = A(z (C))I z' (C)I IdCI

(or (3)) because, as one can easily verify, A log ki = I z' (C) i9 A, log X.

If two metrics ds = k(z)ldzl and ds* A* (z)I dzI are defined in some neighborhood'of the point zo, if A5(zo) = A(zo), and if a' (z) < A(z) everywhere in that neighborhood, then we shall call the second metric the support metric with respect

to tl a first and we shall say that the metric ds = A(z)I dzI admits at the point zo t e support metric ds* = A (z)I dz 1. With these definitions at our disposal, we have the following generalization

\

of the Schwarz lemma, given by Ahlfors:

Theorem 1. Suppose that a continuous metric ds = A(z)1 dzI is defined in the disk I z I < 1 and that it admits a regular support metric satisfying condition (1) at every point at which A(z) 0. Then ds < dg, where dcr = Idz I/(1 - I z I is the hyperbolic metric, throughout the disk I z I < 1. Proof. We set

u(z)=logl(z), v(z)=log R,

11

,

O(R<1.

Let us show that u(z) < v(z) in the disk I z I < R. Let us suppose the opposite. Then, the difference u(z) - v(z) assumes positive values in the disk +a and hence u (z) - v (z) -. -to as z approaches the I z I < R. Since v (z) circle I z I = R, the difference u(z) - v(z) attains a positive maximum at some

362

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

point z0 in the disk I z I < R. The metric ds = A(z)I dz I has at the point z0 a regular support metric ds* = A* (z)I ds I that satisfies condition (1) or, if we set u*(z) = log A*(z), condition (3). Since u*(zo) - v(zo) = u(zo) - v(z0) > 0, it follows that, in a sufficiently small neighborhood of the point z0i we have u*(z) v(z) > 0. But u*(zo) - v(z0) = u(z0) - v(z0) and, in a sufficiently small neighborhood of z0, we have u*(z) - v(z) < u(z) - v(z). This means that the difference U(z) = u*(z) - v(z) has at the point z0 a positive maximum with respect to some neighborhood Iz - z0I < r0 of the point z0 and in that neighborhood it is positive. Let us show that this last is impossible. The relations Au* > 4e2a* , Av = 4e2°, and u* > v imply that, in the disk Iz - z0l < r0, A(u*-v)r4(e2U*`e2,,)>0,

that is, AU > 0. On the other hand, in accordance with Green's formula we have for r < r0 , I z-zo I=r

Tr-

rde-

AUdxdy>O, Iz-zo
where we set z = z0 + ret; that is, dr

U(zo+reiB)d6>0.

If we now integrate with respect to r from 0 to r, where 0 < r < r0, we obtain 2+c

U (zo)

2

U (zo + re'°) d9.

Since the function U(z) attains its maximum at the point z0, this last inequality is possible only when U(z) - U(z0) on Iz - zol = r for all r in the interval 0 < r < r0; this would mean U(z) = U(z0) = const in z - z0l < ro. This is incompatible with the inequality AU > 0. Thus, we have shown that u (z) < v (z) = log [R/(R 2- I z l 2)] in I z I < R. As

R , 1, we see that A(z) < 1/(1 - I Z I2), that is, that ds < da. This completes the proof of the theorem. Now, following Ahlfors, let us give an application of this theorem to covering questions. In §7 of Chapter II, it was pointed out that there does not exist a disk IzI 0, that is completely covered by the image of the disk I z I < 1 under

§7. NONANALYTIC GENERALIZATION OF SCHWARA, LEMMA

363

that is regular in I z i < 1. However, it follows from Theorem 7 of §7 of Chapter II that there exists a positive number p such that the image of the disk I z I < 1 under an arbitrary function of the type indicated completely covers some disk of radius p because it completely covers, for example, some half-disk with center at the origin and radius independent of the form of f (z). Bloch has proved more than this: there exists a positive that is regular in number p such that, for an arbitrary function w = f(z) = z + z I < 1, the w-plane contains a disk of radius p that is a one -sheeted image of some portion of the disk I z I < 1 under the mapping w = f W. Let us now introduce some new notation. We denote by L f the radius of the largest disk in the w-plane that is completely covered by the image of the disk z I < 1 under a function w = f(z) = z + that is regular in I z I < 1. We denote by B r the radius of the largest disk in the w-plane that is the one-sheeted image of a subdomain of the disk I z I < 1 under the same mapping. (In other words, Bf is the radius of the largest one-sheeted disk lying on the Riemann surface onto which the function w = f (z) maps the disk I z I < 1 bijectively.) Furthermore, we an arbitrary function w = f (z) = z + c 2z 2 +

define

B

lnf Bf, 1(z)

L = inf

Lf,

f (z)

that are regular in the disk Here, hese infima are over all functions f(z) = z + z I < ; here B is called the Bloch constantly and L is the Landau constant. The n merical values of these two constants are as yet unknown: we have only upper and lower bounds for them. 2) The theorem proved above enables us to obtain the best results as regards lower bounds for the constants B and L. We begin with a lower bound for B. is regular in I z I < 1 and has Suppose that the function w = f (z) = z +

finite Bf. Let z0 denote an arbitrary point in the disk I z I < 1. If f' (z0) 0, then the inverse function z = f-1(w) is regular in a neighborhood of wo = f(z0). We denote by p(w0) the radius of the largest circle with center at w0 in which the function z = f -I(w) is regular and the image of which under the mapping z = f - I(w) is contained in I z I < 1. Obviously, B f is an upper bound for the numbers p(w0) for all possible z0 such that IzOI < 1.

1) Bloch [1925] showed that B > 0. 2) On the basis of elementary considerations, Landau [1929] gave some numerical bounds for B and L.

364

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

If we denote by G 0 the domain in I z I < 1 that is mapped by w = f (z) onto the disk 1w - wol < p(wo), then Gao must contain on its boundary either points of the circle I z i = 1 or points at which f '(z) = 0. Furthermore, we can easily see that p (w), where w = f (z), is a continuous function of z in I z I < 1 and we

have, close to the point zo, where Izol < 1 and f'(zo) = 0, p(w) = I w - wol, where wo = f(zo). Let us now define a metric ds = A(z)I dz I in I z I < 1, taking a (z)

(4)

2((A-p(w))

where A is a constant exceeding B1 The function A(z) is obviously continuous at all points of the disk I z I < 1 at which f '(z) 0. On the other hand, if a point zo such that Izol < 1 has the property that

f (z) =f (r0) + LJ 00

Ck (z - z0)k,

0, n> 1,

C,n

k=n in a neighborhood of it, then, in a neighborhood of zp, we have p(w) _ I f(z) - f(zo)I and, consequently, (z)

Y -A I ncn (Z

^ 2 I cn (Z - zor

n-1

=Iz-zpIa

-

zo)n-'

+ ...

I

... I'12 (A - I cn (z - zo)n+ ... )'"A-

I)

I

2lcn+... I'll (A--Icn(z-z(,)"-I-... U

This shows that, for n = 2, the metric in question is continuous and, in fact, regular at zo and that, for n > 2, we have A(zo) = 0. (Here, A(zo) is defined as limn-=oA(z).)

Now, let us consider the point z0 such that Izol < 1 and f '(z0) 0. On the boundary of Gao is a point a which either lies on the circle I z I = 1 or has the properties that f '(a) = 0 and I a I < 1. In Gao, we define the metric ds* = A*(z)ldzl, where

z) --

(

2

>r If (z) I p`(w)(A-p (w))

P*(w)=lw-f(a)I, w=f(z)

(5)

This metric is regular in Gao and condition (3) holds for it in Gao because it is obtained from the hyperbolic metric in the disk I 'l < 1 by means of the analytical mappings C = w - f a A, w = f W. Furthermore, throughout the entire

§7. NONANALYTIC GENERALIZATION OF SCHWARZ LEMMA

365

domain Gzo, we obviously have p*(w) > p(w) and p' (wo) = p(wp). Therefore, in Gro we have A' (w) <X(w) and A* (wo) = A(wo) though only on condition that the function It (A - t) increases in the interval 0 < t < B f. Consequently, when this last condition is satisfied, the metric (5) is a support metric for the metric (4) and it satisfies condition (1). With regard to points at which f '(z) = 0, as was shown above, at such points A(z) = 0 or else the metric ds = A(z)I dz I is regular and hence can, itself, serve as support metric. What has been said leads to the conclusion that if A is such that the function f (A - t) increases in the -interval 0 < t < B f, then, for the metric (4), the condi-

tions of the theorem established above are satisfied. But f (A - t) increases for 0 < t < A/3. Therefore, if we take A > 3Bf, we have in I z I < 1 the inequality ds < da, that is,

V If'(z)I

1

Iz

2

(6)

Since O (A - t) increases in 0 < t < Bf, if we replace p (w) with Bf in (6), we obtain

V_A If'(z)I<2}(zl,B,.) Therefore, as A --. 3Bf, we have

I f' (z) I < 4 B1 Thi last inequality holds everywhere in I z I < 1. In particular, with z = 0, it yrellt s Bf > f /4. Since the function f (z) is any function that is regular in ,I z I < 1, we have the bound

B>4=0.43.... In an analogous manner, we can find a lower bound for L. For every z o such that Izo I < 1, let us denote by p(wp) the radius of the largest circle with center at wo = f(zo) that is completely covered by the image B of the disk I z <1 under the mapping w = f(z) and let us define on I z I < 1 the metric ds = A(z)I dzl, where

(z) -

if (2) I

=f (z),

2p (w) log p w

C being a constant greater than Lf.1) This metric is continuous in I z I < 1. 1) Again, it will be sufficient to consider functions f (z) with finite Lf.

(7)

366

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

Now, let us denote by Gao the largest domain contained in z I < 1 that includes z0, where Iz0 I < 1, and that has the property that its image is contained in the disk Iw - w0I < p(w0), where w0 = f(zo). Finally, let a denote a point that is both on the circle Iw - w0 I = p(w0) and on the boundary of B. Let us define on Gao a metric ds* = A' (z)I dz I, where I

k"(z)=

-

If'(z)I

2p' (w) logp*(W)

p`O-I

I, w=f(z).

(8)

This metric is regular in GZo and satisfies condition (3) in Go because it is obtained from the hyperbolic metric in the disk ICI < 1 by the successive analytic mappings: C_ (1 - t;')/(1 + c'), ' = log (C/(w a)), w = f W. Since obviously p` (w) > p(w) everywhere in GZo and since p* (w0) = p(w0), it follows that A* (w) < A(w) everywhere in GZ0 provided the function t log (C/t) increases in the interval 0 < t < p(w0). Under this condition, the metric (8) will be a support metric for the metric (7) at the point z 0 . If t log (C/t) increases in the interval 0 < t < Lf, then the conditions of the theorem that we have proved are not satisfied for the metric ( 7). But t log (C/t) increases in the interval 0 < t < C/e. Therefore, if C > eLf, then in the disk I z I < 1 we have ds < da, that is,

If'(z)Ic

l- IZi

2p (W) l og P (w)

If we now replace p(w) with Lf and take the limit as C -+ eLf, we obtain in I z I < 1 the inequality If, (z) I

2L,

1-Izl'

For z = 0, this yields L f > 1/2. Thus

L2. Let us now find some upper bounds for B (Ahlfors and Grunsky [1937]) and L.

In § 1 of Chapter III we constructed a function that maps the disk I ]I < 1 onto a regular circular triangle Aa inscribed in the disk I w I < 1 with interior angles of size rra. If this function maps the points C = 0 and C = 1 into the is points w = 0 and w = 1 respectively, then this function w = fn(0 = aIC+ obtained in accordance with formula (19) of § 1 of Chapter III (with c = 1). It follows that al can be expressed in terms of the beta function and hence in terms of the gamma function as follows:

§7. NONANALYTIC GENERALIZATION OF SCHWARZ LEMMA

B(1-a ai =f. (0) = B/1-a 2 11

2

'

1+a

+

2 + l+a

2

+3\r`3}

2

3/l

2

-3)

1

367

r2

Keeping this in mind, let us construct the function w = F(z) = f113(11/6(z)), which maps the triangle Al/6 univalently onto the triangle Al/3 in such a way that the point z = 0 is mapped into w = 0 and the vertex z = 1 is mapped into the vertex w = 1. By the symmetry principle, the function F(z) can be extended from Al/6 to the entire disk I z I < R, which is orthogonal to the circles of which the sides of the triangle Al/6 are arcs, and is single-valued and regular in that disk. It is easy to find the radius R of that disk. Specifically, the chords connecting the endpoints of the sides of the triangle A1/6 are of length VT and they make angles of n/12 with the corresponding arcs of the triangle Al/6 On the other hand, a circle y containing one of the boundary arcs of the triangle Al/6 that terminates at the point z = 1 intersects the real axis at an angle of 702. Consequently, the segment of the real axis lying inside y, also is of length VF3_. Therefore points of intersection of the circle y with the real axis are 1 and f + 1 and their product must be equal to R2. Therefore R = j + 1. It follows from the prcess of the analytic continuation of the function F(z) that the Riemann surfacF F, which is the single-valued image of the disk z I < R under the func= F(z), is composed of an infinite set of triangles congruent to O1/3 tion and each vertex of each triangle is a second-order branch point of that surface. This last fact shows that a one-sheeted disk of radius greater than 1 cannot enlist on the surface F. I

i

Now, consider the function W

(z) = F (Rz) = z -}-

... .

This function is regular in the disk I z I < 1 and it maps that disk onto a Riemann surface that does not contain a one-sheeted disk of radius greater than 1/RF ' (0). Therefore, we have B < 1/RF ' (0), and, consequently,

Bfs(o) 1- r13)r(12J fg(0)

V3 +lr(4)

Thus, we finally have for B the inequalities

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS.

368

r(1)r fti

4 =0.43 ...
3

12 1 )

+1r(4)

=0A7 ...

(9)

By considering the function w = F (z) = fI 13 (fI (z)), we can show that the

-- F.

function

F (Z) (0)

=Z+...

is regular in the disk I z I < 1 and maps that disk onto a Riemann surface F consisting of a finite set of triangles obtained from A1/3 by means of a similarity transformation with similarity coefficient 1/F' (0) but with the vertex of none of these triangles belonging to the surface F. Consequently, this surface F cannot covet any disk of radius greater than 11F'(0). It follows that

- ' - r( r( )r( g {2

L

1

`1'I (0)

e,

_

3)

-f. (0)

5

(0)

3

)

3

217(`

Thus, we have the inequalities for L:

1
3

2r (3)

=0.56 ..

(10)

§8. Majorization of subordinate analytic functions Suppose that a function f (z) is meromorphic in the disk I z I < 1 and that a f unction F (z) is meromorphic and univalent in I z I < 1. Suppose that f (0) = F (0). If the image of the disk I z I < 1 under the mapping w = f (z) is contained in the image of that disk under the mapping w = F(z), we say that the function f(z) is subordinate to F(z) in the disk I z I < 1 and that the function F(z) is a univalent majorant of f (z). We denote this by writing f (z) -< F (z). This obviously is equivalent to regularity of the function F-1(f (z)) _ 95(z) in the disk I z I < 1, where ,(0) = 0 and I,(z)I < 1 in I z I < 1. It follows that the set of all functions f (z) that are subordinate in the disk I z I < R to a given univalent majorant F(z) is defined by the formula f (z) = F ((p (z)),

where 96(z) is an arbitrary function satisfying the conditions of the Schwarz

(1)

§8. MAJORIZATION OF SUBORDINATE ANALYTIC FUNCTIONS

369

lemma; that is, it is regular in z.1 < 1, 0 (0) = 0, and 195 (z) I < 1 in I z I < 1. In such a form, the concept of subordination is generalized to the case of nonunivalent majorants: a function f (z) is said to be subordinate in I z I < 1 to a majorant F(z) if it can be represented in the form (1) in I z I < 1. I

We shall prove a number of theorems 1) regarding subordinate functions.

Theorem 1. If two functions f(z) and F(z) such that f(0) = F(0) = 0 are regular in I z I < 1 and if f(z)- 0 and 0 < r < 1, 2a

2a

S If(rei) IpdOS IF(re1B)IpdA.

(2)

0

0

Proof. Let r0 denote an arbitrary number in the interval 0 < r0 < 1 such that there are no zeros of F(z) on the circle I z I = r0. Let us write F(z) in the form where cm

F (Z) = 0. Then, let us define

b (z) = ro/ Z

m

if F W has no zeros in 0 < I z I < r0,

n

fror I/

km1

r(Z

9

Zki)

if F(z) has zeros in 0 < I z I
z 1, , z denote the zeros (all of them) of F (z) with each zero counted in acc dance with its multiplicity. Then, the function F(z)/b(z) has no zeros in Z17 r0. Consequently, the function (F WA (z))p is regular in I z I < r0 for arbitrary p > 0. By Poisson's formula, we have

Here, I

rrF(z) p_ 1 `b (z))

2%

2,c

F(z) 1p(z'+z de, z b (z') J \z' - z

rpete

(3)

in the interval I z < r0 . Therefore, if we represent f (z) in accordance with formula (1), we obtain in z I < r0 tic

f (z) IPC 1 ('I F (z') Ip, (z' + T (z) d9 . 0 b (z) l b ( (z))

(4)

Furthermore, since Ib(z')I = 1 on Iz'I = r0 and Ib(z)I < 1 in IzI
1) Rogosinski [1943].

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

370

2x

if (z)Ipc2- c IF(z'I°t (z'+f1z) Z'

)dA.

(5)

0

If we integrate (5) over the circle I z I = r, 0 < r < ro, and keep in mind the fact that, by virtue of the theorem of the mean, 2x

('t (z,z + (z) ) dt (z)

=

(z, + T (0)) = 27; z (0)

(6)

we obtain

where z = re

2x

2%

I F (roete) )P dO,

` I f (rest) I P dt

(7)

1

0

and, as r -, r0, this yields (2). The case in which r0 is such that the function F(z) has zeros on I z I = r0, is obtained from this one by means of a passage to the limit. This completes the proof of the theorem.

Theorem 2. If the functions f(z) = Lk_lakzk and F(z) = Ek=1/lkzk are regular in I z I < 1 and if f (z) -
n

Z Iakl'` Ek=1 IAkI9.

(8)

k=1

Proof. We set n

3n (z)

n

00

2] akzk, S. (z) = LJ Akzk, R. (z):--- E Akzk' k-a+1 k-1 k-1

If f (z) = F (96 (z)), then, obviously (since 96 (0) = 0), C0

f(z) = Sn ((P (z)) + Rn ((P (z)) = Sn (p (z)) + E akzk

k-n+1

and, consequently, cc

S. (p (z)) = 3a (z) + But

akzk.

in I z I < 1. Therefore, by Theorem 1, we may write 2% S

that is, by (9),

k=s+I

2x

IS.(p(re`B))I'dO

IS.(ree) I9d9,

(9)

§8. MAJORIZATION OF SUBORDINATE ANALYTIC FUNCTIONS

k=1

ak

I9r9k+

I Ak

E [a l I9r9k

k=n-}-1

371

1'rik

k=1

From this we obtain M

a

1 l ak 12r218 - E

As r

I AA, I9r'k

1, this yields (B), and completes the proof of the theorem. Theorem 3. Under the conditions of Theorem 2, we have lat 1 < I Al I and

1a2 l < max (I A 1 I, I A2 I)

Proof. Suppose that f(z) = F(/(z)) and 96(z) = Ek_lakzk. Then, al = a1A1,

a9 = a1A9 + a9A1.

I a 1 1 _< 1 and, from inequality (2) of § 1 as applied to the function 96(z)/z at z = 0, we find I a2 l < 1 - I a 1 12. Therefore, from what was said above, we obtain

But

I«.1i
Ia9I
This completes the proof of the theorem.

Theorem 4.1) If two functions f(z) and F(z) such that f(O) = F(0) = 0 and F ' (0) = 1 are defined and regular in the disk I z I < 1, if F(z) is one-sheeted, and if f(z)..< F(z), then,

If'(z)I<(1+1z1

If(z)I<(1 IZI IZD..

(10)

These inequalities are sharp with equality holding only for a function of the form

f (z) =

(1

gzeZ),

,

1711=161=1.

(11)

Proof. From the representation of f(z) in the form (1), we have if '(z)I = IF '(96 (z))I - 10' (z)I. But inequality (8) of §4 of Chapter II holds for If' (z)1, and inequality (2) of §1 of the present chapter holds for 10' (z)l, so that

IF'(,p(z))I< 1+Iy(z)I (z)I)

,

1-IT(z)I' IT (z)I< 1-IzI

Consequently, '1+I,(z)I'

If'' (z) I < 1-IR(z)J) 1) Schiffer [1936), Lohin [19491.

I 1-Iz=I

(12)

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

372

Since JO(z)1 < I z I, we have the first of inequalities (10). We obtain the second of inequalities (10) by integration of the first. Since equality holds in the relationship 1 0W1 < I z I only for 0 (z) = 71z, where 177I = 1 and since equality holds in the first of inequalities (12) only for the function F(O) = O/(1 - 7110)2, where 17711 = 1, equality holds in the relations (10) only for a function of the form (11). This completes the proof of the theorem. Theorem 5. Let f(z) = Y-' k= 1akzk denote a function that is defined and regular in I z I < 1. Let F(z) denote a function that is regular and univalent in the

same disk. Suppose that F(0) = 0 and F `(0) = 1 and suppose that f(z) is subordinate to F(z). Suppose that the image of the disk I z I < 1 under F(z) is a starlike domain. Then, we have the sharp inequalities I an I < n, for n = 1, 2, <1, If the starlike domain referred to is convex, we have the sharp inequalities for n = 1, 2, . Proof. Let us consider first the second case, in which the image B of the disk z I < 1 under the mapping w = F(z) is a convex domain. Since wk = f(11kz) it follows that belongs to B for z < 1, where 71 = e2n`/" for n = 1, 2 (w I + . + w,n)/n. E B. But this means that the function " n

Y

(71kz1/")

= a"z -}- .. .

k-1

is subordinate in I z I < 1 to the majorant F (z). But then, it follows on the basis of Theorem 2 (with n = 1) that Ia" I < 1 for all n= 1, 2, . If the image of the disk I z I < 1 under the mapping w = F(z) is a starlike domain, then the function F1(z) defined by F(z) = zF'(z), FI(0) = 0, F1, (0) = 1 maps the disk I z I < 1 onto a convex domain (see §9 of Chapter IV). Let us represent f(z) in accordance with formula (1). In the disk I z I < 1, we have 10 (z) < I z 1. Therefore, for arbitrary C in ICI < 1 we have F I (CO (z)) -< F I (Jz) in I z < 1. Therefore, if we set m

FI (CAP (z)) _ E A. (C) zh,

At (C) =1,

(13)

k='1

we have on the-basis of what we have already shown IA,,(C)I < 1 for I I < 1 and n = 1, 2, .. But A,, (C) is an nth-degree polynomial in C such that A,, (0) = 0. Therefore, in accordance with Cauchy's formula, we have for I I < 1 An (C)

-

TA n (CC))'

2.I

1

S

IC'I=1

dC =

§s. MAJORIZATION OF SUBORDINATE ANALYTIC FUNCTIONS

373

co

1' dCr

n (C')L

2al

l

=0

IC'1=1

n-1

_ 2ni and, consequently,

I A. (C) I c

k

[1

le

, dC

r

k=0

1C'I=I

-I A.(Cll I I I k=0

2a

IC'I=1

$IC

n-I

C 2, IC'1=1

Hence as

A. (C')

n-I

Ik

I

k=0

14' I=

I C 10.

k=0

1, we see that IA,'(1)I < n. Now from A. (C) = gal

we obtain An (1)

-gal 1

C

,J

(4

Fi 9+,z)) dz, S

O

T

1,

Izlar

(z) Fi ('P (z)) zn+t

1

(

dx - gal J IzI-r

F (p (z)) zn+t

dz = an.

I z I=r Consequently, Ian = IAn(1)I
To\prove the next theorem, we shall need a lemma, due to Rogosinski, which is a sharpening of the Schwarz lemma.

Lemma. Let q(z) denote a function that is defined, regular, and bounded above in modulus by 1 in the disk I z I < 1 such that 0(0) - 0 and 0'(0) > 0. Let zo denote a point in IzI < 1. Let Azo denote the domain contained in the disk I z I < I zo I, containing the disk I Z I < I zo 12, and bounded by an arc of the circle I Z I = IzoI2 and the arcs of two curves passing through zo and tangent to the circle I Z I = I z012. Then wo = /(zo) belongs to Azo. Proof. In accordance with inequality (3) of §1 as applied to the function f(z) = z-IO(z), the point wo = ,(zo) lies in the disk with boundary defined by 1) This inequality can also be proved if we replace the assumption that the image of the disk j Z I < 1 under the mapping w = F(z) is starlike with the assumption that it is symmetric about the real axis. For more information on this, see Littlewood [19441 .

374

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

the equation I(w - a1zo)/(zo - alw)I = I zo1, where a1 = 0'(0) and 0 < a1 < 1. If we now take a= al in the interval 0 < a< 1, let us find the envelope of all the disks obtained. These disks are obviously contained in the disk Iw I < I zo I and their centers lie on the segment (0, zp). The equation of the envelope is determined from the equations

F (w, a)=., (log z - aw )= log I zo 1, 0 z°

Fa(w,

C

w - azo

-

w

zo - atm

' =0.

(14)

If we replace the first fraction in the second of these equations with its complex conjugate, the equation takes the form OR

r

1

\ (i-V-ago) (za-aw))

=0 or

tm - azp zo-arw0.

This equation together with the first of equations (14) leads to the conclusion that (w - azo)/(zo - aw) = ± i Izo 1, i.e. w = [(a ±ilzo 1)/(1 ±ialzo I)]zo on the envelope. As a varies in the interval 0 < a < 1, this determines the arcs of the circles passing in one case through the points zo, i Izo Izo, and -zo (corresponding to the values a= 1, 0, and -1) and the other passing through the points

zo, -iIzo Izo, and - zo. It follows that the circles I(w - alzo)/(zo - alw)I = 0 < a 1 < 1, cover the domain ; tzo mentioned in the lemma. This completes the proof of the lemma.

Izo 1,

Theorem 6.1) Let f (z) and F (z) denote two functions defined and regular in the disk I z I < 1 and suppose that F(z) is univalent. Suppose that f(0) =

F(0) = 0, that arg f'(0) = arg F '(0) [this condition is dropped if f'(0) = 0], and that f(z) -< F(z). Finally, suppose that f(z) / F(z). Then,

If(z)I
(15)

in the disk I z 1 < %. (The number % is not the best constant.2' Proof. To prove the validity of inequality (15) at a given point zo such that Izo I = %, it will be sufficient to show that

IF(z)I
§8. MAJORIZATION OF SUBORDINATE ANALYTIC PUNCTIONS

375

at all points z zo in the domain Azo mentioned in the lemma. This is true because the function F '(f(z)) = a1z + satisfies the conditions of the lemma and hence, by applying (16) to the point z = F-'(f(zo)) Azo, we obtain (15). To prove inequality (16) with Izol = %, we define the number p C (0, 1) as a root of the equation

P=0.123 ... >Izol'= 16

z,

P

1

(I±IzoI)a; Since the function F(z) is univalent in (1

P)'

I

z I

I I

(17)

< 1, we have

IF(zo)I>IF(o)I

(I-41Z.1

-

-lZ )l

(18)

for z E Azo. If I z I < p, inequality (16) follows immediately from (18) and (17). Let us now suppose that inequality (16) is not satisfied everywhere in the domain Azo . Theta, on the boundary of the domain Azo there exists a . 11Z point z at which I F (z) I "attains the value m = max=E esa I F(z)I . Obviously, the point z I lies on one of the circular boundary arcs passing through zo (see Figure 16). Let us denote this arc by 1. The tangent to the image of the arc l under the map1

ping w = F (z) at the point w 1 - F (z 1) makes an angle co = ..

` \\

n/2 w th the radial direction at that point. Let us calculate ` At this le in a different way. We first find the angle y formed Figure 16 by the ngent to l at the point z I (directed toward the origin) and the ra ial direction at the point z 1. Let us denote by R the radius of the circle containing the arc 1. Then, from the right triangle with vertices at the points 0, zo, and the center c of that circle, we obtain RZ = Izol z + (R-Izol Y. Therefore, R = (1 + Izo I Z)/2. Furthermore, from the triangle with vertices at the points 0, z 1, and c, we have

(R-IzoI9)'=R"+Izi1'-2RIz11 sin i. Therefore,

sin I-

I z, I' + I z0 I'

Izil(I-f-Izo1') I= arc sin 161z,12+1

16 I z, I' -}- I 1712,1

17112,1

Now, the angle of inclination of the tangent to the arc l at the point z 1 (with respect to the real axis) is equal to -y + arg z 1. Furthermore, the angle of inclination of the tangent to the image of the arc l at the point w 1 = F (z 1) is equal

376

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

to -y + arg z 1 + arg F' (z 1). Finally, the angle between this last tangent and the radial direction at the point w 1 = F (z 1) is equal to y - arg z 1 - arg F'(z 1) + arg F (z 1). As a consequence of all this, we have W

7

-

arg z'F F((z'z) 1) = arc sin

16 1

z,

1

-F 1

17 z, I

- arg z, F F('

z1)

(zi)

1

But inequality (12) of §1 of Chapter IV holds for this last argument. Therefore, we obtain

+log'+I21I IwI<aresin 161_'12+1 17Iz,1 1-Iz,1 < arc sin

16p=+1

17p

+ log 1+Izol 1-Iz o

1

x

2'

However, this contradicts our previous conclusion that co = n/2. This contradiction proves the validity of inequality (16) and, consequently, inequality (15) on the circle i z I = %. But since the maximum value of If(z)1 F(z)I in the disk z < % is attained on the circle I z I = %, it follows that (15) holds also in I z < %. This completes the proof of the theorem. Remarks. We note that, using more complicated inequalities, we can show that the radius p of the largest disk in which inequality (15) holds, under the and conditions of Theorem 6, lies in the interval 0.35 < p < (3 - f 5)/2 = 0.382 that under the additional assumption that F(z) is starlike in I z I < 1, we have 1) p = (3 - J)/2. If we also assume that f (z) is univalent the number p is defined as a root of the transcendental equation2) log 1 + p

+2 arc tan p==%,,

P

that is, p = 0.39 . In conclusion, we shall prove one more theorem: Theorem 7.3) Suppose that f(z) and F(z) are defined and regular in I z I < 1

and that F(z) is univalent in that disk. Suppose that f(0) = F(0) = 0, that arg f'(0) = arg F'(0), and that f (z) -< F (z). Then, in the disk I z < 0.12. . . , we have I f ' (z)I < IF '(z)I with equality holding only for f(z) = F(z).4)

1) Goluzin [1951b]. <See editor's footnote on p. 374).) 2) Biernacki [193t and Goluzin [1939c] 3) Goluzin [1951b].

4) (Xia Dao-xing has shownthat the best constant here is the number 3 -y (see §3, subsection 30 of the supplement))

§8. MAJORIZATION OF SUBORDINATE ANALYTIC FUNCTIONS

Proof. Since the function F(z) = clz +

377

is regular and univalent in I z I< 1,

the same is true of the function C+Z

it1-z)

F( =C+..., F G(C)=F F' _(z) (1- I Z J')

(19)

for arbitrary z such that z I < 1. Therefore, in I Cl < 1, we have (1I +It

1JCI' If we substitute equation (19) into this inequality and set c_ (x - z)/(1 - F x), where I x I < 1, we obtain

F'(x) F (z)

1x-z

G-I

1+I1-zxl 1-lzls xx=z 1JxJ9'

(20)

Now, let f(z) and F(z) denote the pair of functions mentioned in the theorem and suppose that f(z) = F(O(z)). If we represent the function cb(z) in the form

p(z) Z' 1+a ((z)), a= (0)-0, we easily conclude that co(z) satisfies the conditions of the Schwarz lemma. Lei's\us set x = O(z) in (20). Keeping in mind the fact that

r(1-a)I1-wI

(!I-,Z

1 -z z)I-I 1-ar°-ico (a-r')I

r( 1-a)

1

IwI'-2

(1-ar')'--1_I w I' (a-r')°+2 (1-art) (a - ra) X

= r and R(w(z)) = A, and also the fact that the numerator of the derivative of the radicand with respect to A can be reduced to the form where

z

(1-r')(1+a) (l__IcuI9 a-r9\

1-ar

and hence is negative for

`

1-ar

a < 1 and r < 1, we obtain

r(1-a) (1+I)I)

(p(z)-z

1-up(z) C1-ar'-Ito (a-r') r(1-a)(1--fir) _ r(1-a) C 1-ar'-ar+r° - 1+r,-r(l+a) for

x I< I co I< r. Consequently, (20) yields

'

r(1-a) F'

(z))

F'(z) C

1+ 1 1

r(1-a) - 1+r'-r(1+a)

/

1-r'

VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS

378

that is F' (gyp (z)) I

F (z)

1-r'

(1--r'-2r)

1-Ip19'

On the other hand, we have the familiar inequalities

IzI

< 1, where 01(z) = z-10(z) = a.... On the basis of these inequalities,

we have, in i z I < 1,

+1-r r

1-r

9

a

rI,,U a + r

r+1+ar 1-ITI' 1+r a+r 1-r'

1-ICI 1+r1'P,I

1 +ar

that is,

(z) < a 1+ r' + 2arr 1

1

I r'

(22)

If we multiply (21) and (22) and set a = (1 + r2)/2r, a > 1, we obtain f' (z) F'(z)

I

(a-¢ ' aa+1 C`a-1) a+a

(23)

Here, the sign- of the derivative with respect to a of the right-hand member coincides with the sign of the polynomial

p(a)=a8-3a-(3a8+1)a-2aas. For 0
so that, if a Z - 4a - 1 > 0, that is, if a > 2 + V5, we have p (a) > 0 in 0 < a < 1 and, consequently, the right-hand member of (23) is an increasing function of a. But it is equal to unity for a= 1. This shows that I j' (z)j < I F ' (z)I if a > 2 + J5, that is, if r < 0.12. Equality holds only when a= 1, that is, only when O(z) = z, so that f(z) = F(z). This completes the proof of the theorem. Remark. Consider the two functions

F(z)=(lz) f(z)=F(zi+ z), 0
§8. MAJORIZATION OF SUBORDINATE ANALYTIC FUNCTIONS

r

_1-r2 as+1 4r8

(a + a)e,

379

l+r2

a = 2r

The function f' (r) as a function of a has a derivative equal to (a - 3)/(a + 1)4 at a= 1. If a < 3, that is, if r > 3 - V1*81 this derivative is negative for a= 1. Consequently, for a less than but sufficiently close to unity, we have I f (r) I > I F '(z) I . From this we conclude that the inequality If' (z) I < IF '(z) I is not always satisfied under the conditions of Theorem 7 if z is such that I z I > 3 V8 = 0.17 . If we make the additional requirement that the function f (z) in Theorem 7 be univalent in I z I < 1, then the inequality I j'(z) I < I F '(z) I holds in the disk I-z I < 3 - fe though not always in a larger disk. l)

1) Goluzin [19511].

CHAPTER IX BOUNDARY VALUE PROBLEMS FOR ANALYTIC FUNCTIONS DEFINED ON A DISK 1)

§1. Limiting values of Poisson's integral As we know, an integral of the form 2z

u(r, 9)=2_ f(t)P(r, t - e)dt,

(1)

where co

P (r,

B)=1-2rcos9+r'=I+2 I

r"cosn6

(2)

is called Poisson's integral. In the case in which a real function f(t) is continuous for all real values of t and has period 2n, the integral (1) yields the solution of the classical Dirichlet problem for the disk ICI < 1, where c = re' This solution is a harmonic function in the open disk ICI < 1, it is continuous in the closed disk ICI < 1, and it assumes the values f(O) on the circle ICI = I. In this section, we shall study Poisson's integral under less stringent conditions imposed on the function f(t). Let f(t) denote an arbitrary real function that is summable in (0, 2n) and that has period 2n. Obviously, the function u(r, 0) defined 2) by the integral (1) is again a harmonic function in the disk ICI < 1. Then we have 1) Strictly speaking, this chapter is not a part of the geometric theory of functions. However, the information contained in it will be used to a considerable extent in the following chapter. 2)In this and the following chapters, a familiarity with the fundamentals of the theory of functions of a real variable is assumed. Thus the integral (1) is understood in the sense of Lebesgue. Also, because of the periodicity of f(t), we can displace the interval (0, 277) and hence we can interpret the integral as being over the circle I CI = 1.

380

§1. LIMITING VALUES OF POISSON'S INTEGRAL

381

Theorem I. If the function f (O) is continuous at 0 = 00, then u (r, 0) f(00) as C= rei0 approaches the point e`00 by any mode of approach through < 1. points in the open disk Proof. For given e > 0, we take 8 in the interval 0 < S < n such that f(t) - f(00)I < E/2 whenever I t - 001 < S. Suppose that 10 - 00I < 8/2. Then, since 2n

c P(r, t-O)dt=1,

(3)

2x a'

we have 2n

u (r, 6) --f (00) = - (f (t) -f &)) P (r, t - e) dt. I

Let us partition the integral on the right into two integrals I1 and I2' the first over he arc (00 - S, 00 + 8) and the second over the remaining arc A of the circleIC I = 1. Noting that P(r, 0) > 0 in the disk ICI < 1, we obtain Ap+b

I1tI=4n S

2n

P(r, t-6)dt_

P(r, t-0)dt= 2.

00-a

Since I t - 01 > 8/2 on A and hence, P(r, t - 0) < P(r, 8/2), we have

1121-

1-ra 1-- 2r cos

2

+ r'

2n S

If(t)I dt.

This shows that I2 0 as r -s 1. Consequently, there exists an rt > 0 such that 1121 < e/2 whenever r > 1 - ri. Consequently, for r > 1 - n and 10 - 001 < 8/2, we have I u (r, 0) - f (00)1 < c, which completes the proof of the theorem. As a consequence of Theorem 1, we obtain the familiar theorem that, if the function f(t) is continuous for all t, then the function u(r, 0) defined on the circle ICI = 1 as the limit of u(r, 0) when the approach is through points inside the disk I CI < 1 is continuous on I C I < 1 and u (l, 0) = f(0). On the other hand, if f (0) is continuous on some arc of the circle I C I = 1, then the function u (r, 0) defined on that arc as the limit of u (r, 0) when the approach is through points inside the disk I C I < 1 will be continuous in the union of the open disk I < 1 and the I

set of interior points of that arc. Theorem 2. If a function f(0) has a finite derivative at 0 = 00, then

au(r, 0)/a0 - f'(00) as t;'= re`' approaches the point e`00 through points in

382

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

the open disk JCI < 1 along any nontangential path (that is, a path that lies in ie0 some angle with vertex at e that, sufficiently close to eie0, lies entirely inside the disk ICI < I). Proof. Let us first prove the theorem for the case of a function f(0) that is linear in the interval (a, a + 27r): f(0) = A0 + B and that is extended as a periodic function outside that interval. Then, from (1) we obtain

:du(, 0)= 1

2f(t)dp(r, t--0)dt=-

a a

0)

Sexf(t)dP(r,

t

2 a

dt

a

(f (a -}- 21r - 0) -f (a -{- 0)) P (r, a - 0) a+2x

f'(t)P(r, t-0)dO=-AP(r, a.-O)+A.

} 2n a

It follows that, for a < 00 < a + 2n, the derivative au (r, 0)1x0 --. A as rei0 e`00 along any path leading to the point e`60 from inside Icl < 1. Let us turn now to the general case. It follows from (1) that 2x

du (r, 0) d9

1

= 2a

sin (t - 0) f (t) P (r, t - 0) I - 2r2rcos (t - 0)

r'

dt.

(4)

Let us denote by uI(r, 0) the function obtained from (1) if we replace f(t) with the linear function f(00) + (t - 00)f'(00 ) in the interval (a, a + 277), where eai e` e 0 and extended as a periodic function outside that interval. From what we have shown, au, (r, 0)100 - f'(00) as re` a -__, Now, from (4) and the same relationship applied to u 1(r, 0) we obtain, ree`0o.

membering the periodicity of f (o), du (r, 0)

de -

du, (r, 0) de

-

a+ 2x 1

T7C

S

r f (t) -f (00) l t - 0o

- f' (00))

a

(t - 00) sin (t -- 0) X P (r, t - 6) 1-2r2r (5) cos (t - 0) -f- r' d t Let us show that the last factor in the integrand in this equation is bounded for t E (a, a + 277) if rece is sufficiently close to e`0b and lies in an angle with vertex at a `90 and sides making angles `0.0, where 0 < a 0 < n/2, with the tangent to the circle Cl I= 1 at the point a We write this factor as follows:

§1. LIMITING VALUES OF POISSON'S INTEGRAL sin

Q= 2r (t

383

(t

et 00). ret

(6)

0)

°

But j," - re`BI = Ie'('-'9) - rl > Isin(t - 0)I (this last is obtained by replacing the absolute value of the expression as a whole with the absolute value of the imaginary part). Consequently,

2rIt-00I[

F9_'1__ Well

Furthermore, if we assume for the moment that I t - 001 < a.0 and note that x

reie=etB we have

2

se

0, a0
e

x

g°+.+2)lei

{e'r-re`BI-Iett-eIBO-seI sin t

x

sin

t-00

2

00

-E

I

2 1 sin (a -

2sin (a02)

2

t

2 00 1

2

It2e°l=

sin

It -Ool

(again the same sort of substitution). Consequently,

QL sin. 2

that is, Q is bounded for It- 001 < a0. If re`B is sufficiently close to e`BO, that is, if e < E0, then Q is bounded on the remaining part of the interval a < t < a + 21r because the denominator in (6) is then bounded below. Thus, for C < co and all t, we have IQ I < Q0. Therefore, we obtain from (5) du (r, 0) d0

-

a+ 2,g

I

du, (r, 0) d0

C

f (t) -f (00) -f (go) I

Q0 21c

S

a

t-00

(r, t - 8) dt. I

But Theorem 1 is applicable to this integral because f' (00) exists and hence the first factor in the integrand approaches 0 as t - 00. Thus, this integral itself

approaches 0 as a -' 0, that is, as rei0--. e'B0 in the angle in question. Therefore, by passing to the limit, we have lira

du

(r' 0) =

de

(r' 0) =f' (00), lim du, do

384

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

which completes the proof of the theorem.

Theorem 3. If the value of the function f(0) in (1) at 0 = 00 is finite and equal to 0 = 00:(d/de) fof Wdt = f(0) 100 then u(r, 0) --+ f(0o) as t = re`0 approaches the point eie0 along an arbitrary nontangential path from inside Iii < 1. Proof. Suppose that .0 0 - 2n < a < 00. We define the function F(0) in the interval a < 0 < a + 21r by the formula a

F(8)=Sf(0)A a and then extend it as a periodic function. Then, F' (00) = f(00). If we now integrate (1) by parts, we obtain

tt(r, 0)=2-F(t)P(r,

-

t-0)l a+ 27c

a+2n 2rz

C F (t) aP (r, t-0) dt = 1 P (r, a-0) a

a+2:

2,c

X S f (0) d0 + ae an

F (t) P (r, t - 0) dt,

and Theorem 3 is obtained on the basis of Theorem 2.

Since f(0) is equal almost everywhere in (0, 20 to the derivative of F(0), Theorem 3 leads to the important Corollary 1. Poisson's integral (1) has almost everywhere on the circle JCI = 1 limiting values equal to f(0) along all nontangential paths. An analogous result holds for the more general integral, known as the PoissonStieltjes integral 2x

P (r, t - 0) da (t),

it (r, 0) = 2'

(7)

d'

where a(t) is a function of bounded variation in the interval (0, 27T), which can always be extended as a periodic function outside that interval. "Specifically, Specifically, 1) The integral (1) can be regarded as the integral (7) with a(t) = fo f (t) dt. This follows from the properties of a Stieltjes integral.

§2. REPRESENTATION OF HARMONIC FUNCTIONS

385

repeating the same reasoning as above, we can prove Theorem 4. If the function a(t) in (7) has a finite derivative at 0 = 00 where 0 < 00 "< 2n, then u(r, 0) --s a'(00) as C= rei9 approaches e`B0 along any nontangential path in JCI < 1. Since a(t) has a derivative almost everywhere in (0, 27r), we obtain, as before,

Corollary 2. The Poisson-Stieltjes integral (7) has almost everywhere on JCJ = 1 limiting values equal to a'(0) along all nontangential paths. All the theorems of this section are due to Fatou, 11 who initiated the investigation of the boundary properties of analytic functions. §2. The representation of harmonic functions by means of Poisson's integral and the Poisson-Stieltjes integral

In this section, we shall take up questions of the possibility of representing harm nic functions by means of Poisson's integral and the Poisson-Stieltjes integra. The answers to these questions involve special classes of harmonic functions tr which we shall introduce special notation. Specifically, we denote by h, where p > 0, the class of functions u (r, 0) that are harmonic in the disk < 1 and have the property that the integral 2x

S I it (r, 0) Ip dO

(1)

0

is bounded for 0 < r < 1. It follows from the obvious inequality x" < 1 + x', which holds for 0 < p' < p and all x > 0, that, if u (r, 0) E hp, then a (r, 0) E hp so that the class hp is broadened with decreasing p. We begin with the representation of harmonic functions by means of a Poisson-Stieltjes integral. To do this, we need as a lemma the following Theorem 1 (Helly). Every uniformly bounded infinite family { f (x){ of real functions that is of uniformly bounded variation 21 in the interval (a, b) contains a sequence that converges at all except possibly countably many 3) points in (a, b) to a function of bounded variation. Proof. Obviously, it will be sufficient to consider the case in which all the 1) Fatou [1906].

2) This means that the total variations of all the functions in the interval (a, b) are bounded by the same finite number. 3)Helly's theorem still holds even without this exception.

386

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

functions f(x) are nondecreasing in the interval (a, b). 1) Consider the set of all rational points in (a, b) and arrange them in a sequence r1, r2, . By using a diagonalization process, we can select from the family (f(x)1 a . sequence fk (x), for k = 1, 2, , that converges at all the points r1, r2, Let us show that this sequence converges not only at these points but everywhere in (a, b) except possibly at countably many points. Let a denote a positive number. We assert that the set of points x in (a, b) for which

lim fk(x) - limfk(x):e

k-.oo

(2)

k_oo

is a countable set. If this were not the case, the set of these points would have a cluster point and hence it would contain a monotonic, let us say increasing, sequence of points xn, where n = 1, 2, .. Between each pair (xv, x,+1)' v = 1, 2, , there would be a rational number rnv at which the sequence ifk(x)1 converges. By setting limk,. fk (rnv ) = f(rnv ), we obtain

f(r) -f (rnv_)= Jim fk (rny) - limk-oo fk (rni_1) k-.oo

lira fk (x') - lira fk (x,): E,

k_oo

k-oo

oo as v = 1, 2, . Consequently, f(r) > f(r) + (v + 1) e, that is, f(r) v --+ oo. But this is impossible because of the uniform boundedness of the func-

tions f(z). By giving e a sequence of values ev that approach zero and choosing all points for which (2) holds with different ev , we see that fk (x) converges everywhere in (a, b) except possibly at a countable set E of points. The limiting function f (x) is defined and monotonic on the set (a, b) - E. Consequently, at points of the set E it has left- and right-hand limits. If we define the value of the function f(x) at points x in E as the arithmetic mean of these limits, we obtain a function f (x) that is defined and monotonic throughout the interval (a, b). Also, fn (x) -+ f(x) in (a, b) except possibly at countably many points. This completes the proof of the theorem.

Regarding the representability of harmonic functions by means of a PoissonStieltjes integral, we have Theorem 2. ') For a function u(r, 0) that is harmonic in the disk JCJ < 1 to 1)This is true because every function of bounded variation can be represented as the difference of two nondecreasing functions and the total variation of each of these functions does not exceed the total variation of the given function. 2)Plessner [1923].

§2. REPRESENTATION OF HARMONIC FUNCTIONS

387

be representable in that disk by means of the Poisson-Stieltjes integral (7) of §1, where a(t) is a function of bounded variation, it is necessary and sufficient that u(r, 0) belong,to h1. For u(r, 0) to have such a representation in the disk <1 when the function a(t) is nondecreasing, it is necessary and sufficient that u(r, 0) > 0 in I[; I < 1. Proof. 1°. If equation (7) holds for u(r, 0) in ICI < 1, by representing the function a(t) as a difference of two nondecreasing functions, we also represent the function u(r, 0) in that disk as a difference of two nonnegative harmonic functions. -Let us denote these by u1(r, 0) and u2(r, 0). Then, for arbitrary r in the interval 0 < r < 1, we have tic

2'

2%

I u (r, 0) I A 0

S u1(r, 0) de -F- S u2 (r, 0) de = 2-n (ut (0) + U2 (0)),.

(3)

0

or

Hers and in what follows, we shall write u(r, 0)I=U = u(0). Then it follows that

u(r,Eh1.

2b. Conversely, let us suppose that u(r, 0) E h1. Consider the function a, (0)401 u(r, t) dt which depends on the parameter r, where 0 < r < 1. Since

Ia,(e) S u(r, t)IdtCM 0

and

k=1

Ok+1

n

n

ar (0k) - Qtr (0k+1) I c

1

h=1

S

tt (r, t) dt 1 c 2ivM,

ek

0-01<0$<...<0,,,+1=2v, where M is finite and independent of r and 0, the set of functions a,(0), for 0 < r < 1, is uniformly bounded and of uniformly bounded variation with respect to r for 0 < r < 21r. Consequently, if we take a sequence of functions a (0), Pk k = 1 , 2, , pk -+ 1, we can, in accordance with Helly's theorem, take a subse-

quence aP,k (0) that converges everywhere in (0, 2rr) except possibly at countably many points to a function a(0) of bounded variation. Let us now fix the point C = re`6 and let us suppose that r < pk < I for k > K. Then, in accordance with Poisson's ordinary formula we have, for k > K, 2x

u (r, 0)

Zn

Pk''

u (pk, t) pk9 - 2p kr cos (t- 0) -}- r9 A

388

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

If we integrate by parts, we obtain 27ru (r,

Pk' - r'

e) = aPk (27t) Pks - 2pkr cos 9 -}- r° 2x

q

Pk - l

aPk (t)

!

at pill -26 r cos (t - 9) dt.

0

If we now take the limit as k -+ o and remember that aP,k (2n) = 277u(0) = a(2n), we have 2x

2nu (r, 0) = a (2a) P (r, 0) - a (t) aP (rdt - e) dt. Inverse integration by parts now yields formula (7) of §1. This completes the proof of the first part of the theorem. The second part is proved analogously.

Corollary 1. For a function u(r, 0) to belong to the class hi, it is necessary and sufficient that it have a representation in ICI < 1 in the form of the difference of two nonnegative harmonic functions.

Proof. Suppose that u (r, h) E h I Then, u (r, h) can be represented in ICI < 1 in accordance with formula (7) of §1. Consequently, as was shown above, it can be represented in the form of the difference required. The converse follows from (3). Corollary 2.I If a function u(r, 0) belongs to the class h I, it has definite' limiting values constituting a boundary function u(9) almost everywhere on the circle ICI = 1. This follows from the theorem just proved and Theorem 4 of §1. To establish conditions for representability of harmonic functions by means of Poisson's integral and for many other questions that will concern us later.on, we need to have some auxiliary inequalities, which we shall now prove. We shall use the generally accepted notation. Specifically, we shall write Ax) E L" on E (where p > 0 and E is a linear space) to indicate that I f(x)I" is summable on E. For p = 1, we shall write simply L instead of L I.

Holder's inequality. If g (x) E L" and h (x) E L 9 on E, where p > 1, q > 1,

and 1/p,+ 1/q = 1, then g(x)h(x) CL on E and

Sg(x)h(x)dxI _(SIg(x)IPd.c)h/P(SIh(x)Igdx) '1Q.

(4)

389

§2. REPRESENTATION OF HARMONIC FUNCTIONS

Proof. We note that, if a, b, a and ig are all positive and a + 8 = 1, then aa'b,8 < as + 13 b.

(5)

To see this, note that the function 0 (x) = b,8xa - ax - 3b increases in the interval 0 < x < b since

(x)-o;((X)

- 1,>o.

Since S6(b) = 0, it follows that O(x) < 0 for 0 < x < b; this proves (5) for a > b.

By reversing the roles of a and b, we can prove (5) in an analogous manner for

a < b. Let us replace a, 8, a and b in (5) with l/p, l/q, ap and bg. Then, for a, b, p and q all positive with 1/p + l/q = 1, we have the inequality

p ap + bq

.ab

(6)

Now, since the functions I g (x)I 1 and I h (x)I q are summable on E, it follows from 'o) that g(x) h(x) is summable. Furthermore, from (6) we have 11

I g (x) I

g(x)IPdx)'IP

I h (x) I

((Ih(x)Igdx)'lq

Y

dx

.+ 1

i g (x) IP

I h (x) I g

dx,

SIg(x)IPdx

E

that is, we obtain inequality (4). It is easy to show that equality holds in (4) only when I g (x) I P = A I h (x)I ' everywhere on E for some constant A.

Minkowski's inequality. If g(x) E LP and h (x) E LP on E, where p > 1, we have E

I g(x)+h(x)I"dx)

(7)

"n

E

E

Proof. By using (4), we have 51 9(x) +h (x) IPdx- S I g(x) I E

I g(x)+h (x)

+Sih(x)I

IP-t dx

g(x)+h(x)IP-'dx

(SIg(x)IPdx)'

(SIg(x)+h(x)Iq(p-1)dx) '1q

+(S I h (x) Ipdx)''P (S i g(x)+h (x) Iq(p-t) dx)11q. E

E

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

390

Since q(p - 1) = p, this leads after some simplification to inequality (7). In accordance with the remark concerning inequality (4), it follows that equality holds in (7) only when g (x) = kh (x) on E for some constant X. Using Holder's and Minkowski's inequalities, let us establish a test for the possibility of taking the limit under the Lebesgue integral sign. Specifically, If a sequence of functions fn (X)' for n = 1, 2, , defined on E approaches a limit function f(x) almost everywhere on E and if

SIf, (x)1 dxcM, F,

where p > 1 and M is finite and independent of n, then Jim S fn (x) dx = S f (x) dx. n-oo E E

(8)

Proof. In accordance with a familiar lemma of Fatou, we have the inequality

SIf(x)V'dxc lim SIfn.(x)IPdx, n-+00 E

E

from which it follows that f(x) C LP on E. Furthermore, by applying the inequalities that we have proved, we see that, for an arbitrary set e C E, S (f (x) - f (x)) dx l C S I fn (x) - f (x) I dx e

(S I fn

(x) -f (x) IP dx)"P (mes e)li4

(mes e)'/a [(S I fn (x) I P dx)' IP 0

+ (S I f (x) I P dx)'/p11

(mes e) 11q . 2M 1/P,

By making mes e sufficiently small, we can make the right-hand member of this inequality arbitrarily small independently of n. And this, in accordance with a familiar test for the possibility of taking the limit under the Lebesgue integral sign, yields (8). Let us now turn to the representation of functions that are harmonic in 4 < 1 by means of Poisson's integral (1) of §1. We note first of all that not every function u(r, 0) that can be represented in the form (7) of §1 can also be represented in the form (1) of §1. For example, the function

§2. REPRESENTATION OF HARMONIC FUNCTIONS

391

u(r, 0)= (1+1) which is harmonic and nonnegative in ICI < 1, by Theorem 2 can be represented in

the form (7) of §1. But its limiting values on the circle It; l = 1 are equal to zero everywhere except at the point C= 1. Therefore, if it could be represented in the form (1) of §1, we would have f(t) = 0 in (0, '2n) in accordance with Theorem 1 of §1 and hence we would have u(r, 0) = 0 in ICI < 1, which is impossible. We do have 1)

Theorem 3. For a function u(r, 0) that is harmonic in the disk ICI < 1 to have a representation in the disk ICI < 1 by means of a Poisson integral (1) of §1, where f(t) E L°, where p > 1, in (0, 2n), it is necessary and sufficient that

u(r,0)Elhp. Proof of the sufficiency. If u (r, 0) E hp, where p > 1, then u(r, 0) E h and, in accord ce with Corollary 2 to Theorem 1, we conclude that u (r, 0) has almost everywhere on the circle ICI = 1 limiting values along all nontangential paths, and in particular along the radii. Furthermore, these limiting values define a function u (0) C L° cause, by Fatou's lemma, 2x

2x S

I u (6) Ip dA - lim S I u (r, 0) Ip d0.

r-1 0

0

Consider Poisson's formula

'-r'p'-2prcos(t-e)--r'dt, r
In

u(r,

0)zi

u(p, t)

(9)

Since the fraction in the integrand is bounded as p -, 1 and since 2x

S Iu(p, 0)IpdOcM 0

for 0 < p < 1, we may take the limit in (9) under the integral sign as a result, we obtain

p-.1.

As

2x

tt (r, 0) = 2

u (t) P (r, t - 0) dt,

I) The question of the representability of u(r, B) by means of Poisson's integral (1) of §1 with f(t) E L has also been solved, but the complete result is complicated to formulate.

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

392

in ICI < 1; that is, we have the representation (1) of §1 with f(t) = u(t) 6 L. Proof of the necessity. Suppose that equation (1) of §1 holds in the disk ICI < 1 with f(t) C GP, where p > 1. Then, by applying Holder's inequality (again, 1/p + 1/q = 1), we have 2x

2x

I u (r, 6) IP de 2z

_

2x

If (t) I P (r, t - B) dt]9 d6

{2a 2x

(21)P

f (t) I P11P (r, t - 0). PI /Q (r, t 2x

8) dt] P d6

2x

2x

If(t)I"P(r, t-0)dtde;--=

2a

f(t) j" dt,

in ICI < 1. Consequently, u(r, 0) E hp. This completes the proof of the theorem. Finally, let us pose the question: What conditions must a function u(r, 0) that is harmonic in the disk ICI < 1 satisfy for its harmonic conjugate v(r, 0) to have a representation in accordance with Poisson's formula in ICI < 1? A sufficient condition can be obtained from Theorem 2 and the following theorem.

Theorem 4. I) If u(r, 0) E hP, where p > 1, then v(r, 0) E hp. The proof of this theorem is rather complicated if p / 2. Therefore, we shall not present the proof in the general case. 2) In the case p = 2, it follows easily from Parseval's formula applied to the expansions

u (r, 8) = ao + Z (anrn cos n8 - barn sin n8), n=1 00

v (r, 8) = bo + Z (burn cos n8 + anrn sin n8), n=1

which constitute the real and imaginary parts of the Taylor expansion 00

a + Iv = E (Qn + tbn) (n. n=0

Specifically, in this case, we have

1)M. Riesz [1927]. 2) A proof can be found, for example, in the book by Zygmund [ 1959].

§3. LIMITING VALUES OF ANALYTIC FUNC'T'IONS

393

2a

21c

Iv(r, 8)I1 d8= S Iu(r, 6)11 d8-2,t(a09 -bo). 0

0

In conclusion, we note that the representation (1) of §1 is simply Poisson's formula 2a

it (r, 0) = 2n

u (t) P (r, t - 0) dt,

(10)

which gives the value of a harmonic function u(r, 0) in ICI < 1 in terms of the limiting values of u (t) along nontangential paths. This follows from the fact that, in accordance with Theorem 3 of §1, we have f (O) = u(0) almost everywhere on the circle ICI = 1. Therefore, Theorem 2 can also be regarded as giving a sufficient condition for applicability of Poisson's formula (10) to u(r, 0). This condition is expressed in terms of the membership of u(r, 0) in the class hP, where

p>1. §3. The limiting values of analytic functions Using the preceding results regarding harmonic functions, we can easily derive a very simple result regarding the existence of limiting values of analytic functions. Specifically, suppose that a function f(C) is regular and bounded in I C I < 1. Then, its real and imaginary parts represent bounded harmonic functions in I C I < 1 which belong, in particular, to the class h1. In accordance with §2, these functions have definite limiting values almost everywhere on ICI = 1 as C approaches this circle along nontangential paths. Therefore, f(C) also has definite limiting values along nontangential paths almost everywhere on that circle. These limiting values define a function on ICJ = 1, which we shall denote by f(C) or by

f(e`0) where J= e`0. This very simple result can now be generalized to the class N (introduced by R. Nevanlinna) of functions f(C) that are regular in the disk ICI < 1 and that can be represented in that disk in the form of the ratio of two bounded functions:

f(C)=`f ) We can always assume that unity in the disk < I. I

(1)

and fi(C) are bounded in absolute value by

I

Another way of characterizing the class N is on the basis of Nevanlinna's theorem:

394

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

Theorem 1. For a function f(c) / 0 to belong to the class N, it is necessary and sufficient that the integral 2x

+

log If (rei) I dO

(2)

0

be bounded by some constant M for 0 < r < I.

Proof. 1) If the function f(C) / 0 belongs to the class N, that is, if it can be represented in the form of the ratio (1) with Io(t;)I < 1 and I(C)I < 1 in the disk ICI < 1, then, since If(C)I < 1/Ii/,(C)I in that disk, we have 2x

+

2x

log If (rei°) I A

log 14' (re'°) I d6.

(3)

0

0

Cm+1 + ...,where m> 0, then in accordance with Jensen's formula as applied to the function ,/i(C)/C'", we have, for 0 < r < 1, Now, if ,/i (C) = c.C' + cm

+1

2

2

r logICkl+mlogr.

S log(retB)Id6-log Ic,,, +

0
0

(4)

r

Here, the C. denote the zeros of the function O(C) in 0 < It; I < 1 and the sum is over all the zeros of 4,(C) that lie in 0 < I C I < r. Since the right-hand member of (4) is a nondecreasing function of r in 0 < r < 1, the right-hand member., of (3) is a nonincreasing function of r and, consequently, is bounded above in 0 < r < 1. This shows that the integral (2) is bounded in the interval 0 < r < 1. Conversely, suppose now that the function f (C) J 0 is such that the integral (2) is bounded for 0 < r < 1. Then, in accordance with the Jensen-Schwarz formula, we have in ICI < 1 2w

log

logf (C) _ ICkI
r 9C - Ck)

r -CkC

+ 2a

S log If (re") I reiT C d6 -}- IC, 0

where C is a real constant. This can be represented in the form f (C) _ _r (r)

(5)

9r (C) '

where we set

1) The proof given here differs only slightly from the proof of the more general Theorem 1 of §5 of Chapter VII.

§3. LIMITING VALUES OF ANALYTIC FUNCTIONS zx

-k r log+ r (C-Ck) e CC ICkI
-

0

1

{

reie +C

J(re10)

I

rata-C

395

2x

S log If(ree) I rate dC

The functions Or(C) and '/!r (C) are regular in ICI < r, also 1 and (C) < 1 in that disk. If we take a sequence of numbers rk in the interval 0 < rk < 1 that approaches 1 as k - Do, we can, on the basis of the condensation principle, choose a subsequence {q,rk# (C)} from the sequence that converges uniformly inside the disk ICI < 1 to a regular function 0 i( ). Also, I S(C)I < 1 in I C I < 1. Since the values of , are bounded r (0)I, for k = 1, 2, below by a positive quantity independent of k, it follows that i# (C) A 0. It follows from (5) that the sequence JO k,(C)I converges in ICI < 1 to some function (C is regular in ICI < 1; also (C)I < 1 in ICI < 1. This completes the proof of the theorem. I

I

From the representation (1) for the functions f(c) r < 1, we have S I log I f (reie) I 0

2.

2x

21c

0 in the class N for 0 <

I log l y (re') I I dB -}- 5 I log I' (re") I

I dO 0

I d9

0

2x

2x

s log I y (rei°) I d9 - s log I (re'B)I d6, 0

0

and, as in the first part of the proof of Theorem 1, the last two integrals do not decrease in the interval 0 < r < 1. Therefore, not only the integral (2) but also the integral 2x

S I log If (reie) I I de

(6)

0

for 0 < r < 1 is bounded by a finite quantity independent of r. This property of functions belonging to the class N will now be of use to us. Theorem 2. If a function f(c) J 0 belongs to the class N, it has definite limiting values f(e`9) almost everywhere on the circle ICI = 1 along all nontangential paths. Furthermore, Ilogl f(e`9)II is summable on ICI = 1. Proof. If a function f(c) E N, not identically zero, is bounded in 1,

396

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

then, as was noted above, it has almost everywhere on ICI = 1 definite limiting values f(ei0) along all nontangential paths, and, in particular, along radial paths. In accordance with Fatou's lemma, we have 2a

2a

S I log If (e') I dO Cr urn S I log If (re's) I I A. I

(7)

In accordance with what was said regarding the integral (6), the right-hand member of this inequality is bounded. Consequently, Ilogl f(e`0)II is summable on ICI = 1. But then, the values of 1; that is, the values of f(e`B) log I f(e`g) I are finite almost everywhere on are almost everywhere nonzero on I cI = 1. Now, if f(C) J 0 is an arbitrary function belonging to the class N, then in its representation (1) the functions q(C) and O(C) have almost everywhere on ICI = I definite limiting values along nontangential paths and these limiting values are nonzero almost everywhere. But then the function f (C) has almost everywhere on ICI = I definite limiting values f(e`0). If we again apply Fatou's lemma to the integral (6), we conclude that Iloglf(e`B)II is summable on Ill = 1. This completes the proof of the theorem. The finite limiting values of the function f(C) E N along nontangential 1, will now be called its boundary paths, which exist almost everywhere on values. As a consequence of Theorem 2, we have the following uniqueness theorem.

Theorem 3. I) If two functions fI(C) and f2(C) E N have the same boundary values on a set E of positive measure on the circle ICI = 1, they are identically equal in the disk ICI < 1. Proof. Obviously f(C) = fl(C) - fZ(C) E N. If f(C) 0 in ICI < 1, then, by Theorem 2, Iloglf(e`B)II is finite almost everywhere on CI = 1; that is, f(e`B) is finite and nonzero almost everywhere on It;I = 1. This contradicts the condition according to which f(ei9) = 0 almost everywhere on E (except at points at which f, (C) = f 2 (0 = °O)' In connection with the boundary values of functions in the class N, we also have the following convergence theorem:

1) This theorem is a special case of a general theorem of Privalov, which will be treated in §2 of Chapter IX.

§3. LIMITING VALUES OF ANALYTIC FUNCTIONS

397

Theorem 4. 1) Let If,, (C)j denote a sequence of functions that are regular in the interval ICI < 1. Suppose that there exists a single finite constant M such that for every n and for 0 < r < 1, 2a +

S log I fn (re`°) I dO

M.

(8)

0

Suppose that the sequence Ifn(C)I converges at all points belonging to a subset E of positive measure of the circle ICI = 1. Then this sequence converges uniformly in the disk ICI < 1 to a function belonging to the class N. Proof. If we take the real parts in formula (5) and note that the first two terms o% oethe the right are nonpositive for ICI < r, we obtain in any disk ICI < r, where

r < 1, \

logIf(C)I<2s

log If(re`) I51 1ret±

d9.

(9)

If we apply this inequality to the functions f, (C) and use inequality (8), we obtain the result that, in such a disk

r+ItI r ItlM

loglfa(C)

If we fix C in the disk I I < I and let r approach 1, we obtain the inequality 1+1CIM

If.(C)I<el-ICI This inequality shows that the functions ff (C) are uniformly bounded inside the disk ICI < 1. Let us suppose now that they do not converge throughout ICI < 1. Then, just as in the proof of Vitali's theorem (§2 of Chapter I), they contain two subsequences) &,k(C)1 and for k = 1, 2, .. that converge in ICI < 1 to distinct functions that are regular in that disk. By hypothesis, the sequence of the differences fk (0 = f° 'k (0 - fn"k (0 converges on the set E to zero and it converges in ICI < 1 to a function q (c) A 0. Furthermore, since 2) 1) This theorem was proved by Hincin [1924] for uniformly bounded functions and later generalized by Ostrowski. 2)This is true because, for arbitrary positive numbers a1 and a2,

log (a, + a2) C log (2 max ak) = k=1, 2

max log (2ak)
*=- 1. 2

k-t, 2

E log a, } log a9 ± log 2.

398

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

log I

(c) I < log Ifnk(C) I + log l l n t (C) I + log 2,

and 0 < r < 1,

we obtain from condition (8), for k = 1, 2, 2s

2x S

log lpn (reiB) I dA ` 2M -}-log 2.

(10)

0

On the basis of Fatou's lemma, we conclude from this, in particular, that 0(c) C Let us now apply inequality (9) to the functions 0n(C). Since for arbitrary real a we have a= 2a - l a 1, it follows that, for l l < r, log I'Pn (C) l <2. 2u

2s

log I pn (re") I

fresh) A

2n

log Ipn (re') I I U

( reie

dA 2%

r+iCi2(2M-+ -log2)r+ICI 2u ) Then, for any fixed C in using Fatou's lemma,

IlogIpn(fe`B)IIdA.

1, we obtain, letting r approach 1 and again 2n

logIT.(C)I_i±ICI2(2M-+ -log2)-1+ICl

loglpn(e,e)IIdA.

2n

S

0

If we let n approach -o in this inequality, the integral on the right approaches by virtue of Fatou's lemma because (b(e`0) = 0 on E and mes E > 0. This shows that 0,(C) 0 in the entire disk ICI < 1. But this contradicts what was said earlier about / (C). Thus the sequence { fn (t) S converges uniformly in the disk 1C1 < 1. This completes the proof of the theorem. We now consider another very important question associated with the zeros of analytic functions. Let f(C) denote a function that is regular in ICI < 1. The question arises whether there exists a function b(C) that is regular in the disk ICI < 1, that satisfies the inequality Ib(C)l < 1 in that disk, that satisfies the equation Ib(C)I = 1 almost everywhere on the circle ICI = 1, and that has in the disk ICI < 1 the same zeros and of the same multiplicity as f(C). If such a function b(C) exists, it is called the Blaschke function for f(C).

§3. LIMITING VALUES OF ANALYTIC FUNC'T'IONS

399

The complete answer to this question is given by the following theorem of Ostrowski. Theorem 5. For a function f(C) that is regular in the disk ICI < 1 to have a Blaschke function, it is necessary and sufficient that the integral 2c

S log If (rete) I d8 0

be bounded for 0 < r < 1. [We point out that, in accordance with Jensen's formula, the integral (11) is a nondecreasing function of r for 0 < r < 1.1 Proof. First of all, let us show that the condition in the theorem is a necessary and sufficient condition for the function f(C) to have a representation in ICI < 1 of the form (12)

is gular in I [; I < 1 and I b ([;) < 1 in I [; I < 1 ana where h ([;) is where b regular and has no zeros in ICI < 1. In accordance with formula (5), we have in the disk ICI < 1

f(C)_ r"Pr (C) hr (C), where m = 0 if 'f (0) 0 and m = multiplicity of the zero z = 0 of the function f (C) if f (0) = 0 and where

Pr (Q =

r (C - Ck)

L ryo
it C

2c

1 S log l 1tr ) I - +t de

hr(C)=e.

°

is regular and < 1, and also Moreover, in the disk II < r we have: r denote a sequence of hr(C) is regular and hr(C) 0. Let (rk1, for k = 1, functions in the interval 0 < rk < 1 that approaches 1 as k - 00. Then, the se-

quence (crk (C)( contains a subsequence that converges uniformly in the disk C1 < 1 to a regular function O (C). Here, either 0(0) = 0, in which case O (C) 0 in I I < 1, or 0 (0) 0. In the first case, the corresponding subsequence of the functions hr k (C) approaches 00 in I I < 1 and, in particular, at C= 0, which means that the integral (11) is not bounded for 0 < r < 1. In the second case, on the other hand, this same sequence converges to a regular function with no zeros in ICI < 1, and then the integral (11) is bounded for 0 < r < 1.

400

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

This shows that, if the integral (11) is bounded, we have in ICI < 1 the required representation (12) (where b (C) = C m.0 (0). Conversely, if (12) holds, then, for 0 < r < 1, 2:

2:

29

Slog lf(re:e)IdO=S logIb(re')Ido 0

0

0

logIh(re')Ide

2,tlogIh(0)1;

that is, the integral (11) is bounded for 0 < r < 1. It follows, in particular, that the condition in the theorem itself is necessary because, if b (C) is a Blaschke function.for f (C), then we have the representation f(C) = b

in ICI < 1, where the function h(C) is regular in ICI < 1 and has no zeros in that disk. To prove the sufficiency of the condition in the theorem, it remains, as before, to prove that the bounded function O (C) has a Blaschke function. If the function q(C) j 0 is regular in ICI < 1 and Io(C)( < 1 in ICI < 1, then when O(C) has only finitely many zeros in that disk, the representation (12) is immediately obtained with

b(C)=LmJT

C_-Ck 1 .- CkC '

where the product is over all the zeros of the function O(C) in 1Cl < 1. On the , indexed in such a way other hand, if there are infinitely many zeros, Cl' C2, . , we form the infinite product that ICII < ICZI < 0o

1 - C Ck

b(C)=Cm IIICk k=l

(13)

1

and we denote by b. (C) the finite product obtained by replacing Do with n. We note that the function Ibn(C)I approaches 1 uniformly as J approaches the circle ICJ = 1. Considering the function O(C)/b,,(0, we easily arrive at the inequalities 1.0 (C) I < I bn (4)l < 1

in ICI < 1. Now, the sequence Ibn (C)( contains a subsequence that converges in ICI < 1 to a regular function b (C) / 0 that has the same zeros as and satisfies in ICI < 1 the inequalities

§3. LIMITING VALUES OF ANALYTIC FUNCTIONS

401

I.O(C)I < ib(C)I < 1.

Here, in accordance with (13), we have b(m)(0)/m! = IIk_1 IQ' If the sequence 1b,,(C)I itself did not converge in ICI < 1, then, in addition to b(C), there would be another function b* (C) with these properties. Then, in accordance with what was said above, it follows that in ICI < 1 we have the inequalities I b* (C)I < b (C) I and I b (C)I < I b` (C)I; that is, we have b` (C) = b (J). This shows that the infinite product (13) must converge in ICI < 1. Having the function (13), we immediately conclude that there exists the representation (12) with h(C) 0 in 16 < 1. It remains to show that I b (C)I = 1 almost everywhere on I C I = 1. To do this, we note first that, if the function g(C) is regular and bounded in the disf 1, then, by §2

\

2s

2a 1

g(e") p(r, t-O)dt, (=reie,

in IcI < 1, so that 2a

g (re") I

Ig

2a

(eit) I

P (r, t- 0) dt.

0

Integrating with respect to 0 from 0 to 2rr, we obtain 2n

2rz

l g (reie) I d& _ S I g (eu) I dt.

(14)

0

0

By applying inequality (14) to the function mind the fact that I bn(e`r)I = 1, we obtain

b(e)/bn(C) and keeping in 2E

b (reie)

2' .1 o

I

bn Crete)

A

I b (e'`) I dt.

(15)

0

But the sequence Ibn(J)I converges uniformly on ICI = r, where 0 < r < 1, to b(J). Therefore, if we take the limit under the integral sign in (15) as r we obtain 2x

0

that is, 2,c

S(1-Ib(e`')I)dt0.

(16)

0

But it follows from the inequality I b (t) < 1 in ICI < 1 that I b (ei0 )I < 1 almost

402

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

everywhere on the circle ICI 1. This, together with (16) shows that Ib(eit)I = 1 almost everywhere on ICI = 1. This completes the proof of the theorem. In particular, since the boundedness of the integral (2) in 0 < r < 1 implies boundedness of the integral (11), we obtain Corollary I. Every function f(t;) 10 belonging to the class N has a Blaschke function.

§4. Boundary properties of functions in the class HP

Now, let us denote by HP, for p > 0, the class of functions f(C) that are regular in ICI < 1 and have the property that the integral 2

S If(re'e)Ipd9

(1)

0

is bounded for 0 < r < 1. Obviously, all functions that are regular and bounded in ICI < 1 belong to the class H P for arbitrary p > 0. Furthermore, the inequalities xP ' < xP + 1, 0 < p' < p and x > 0 and the inclusion relation f (C) C HP imply that f (t) C HP1; that is,

HP C HP, for all p and p', where 0 < p' < p. Finally, from the inequality logx = (1/p) log XP < xP/p, which holds for arbitrary p > 0 and x > 0, we conclude that f (C) C N if f (t;) C HP; that is, HP C N for all p > 0. It follows from this last fact and §3 that a function f (C) E HP, for p > 0, has, almost everywhere on the circle ICI = 1, definite limiting values along nontangential paths and these form a limit function f(e`B). In accordance with Fatou's lemma applied to the integral (1), by letting r approach 1 we conclude that the function Ado) C LP in (0, 20. With regard to functions in the class HP, we have a number of theorems.

Theorem 1. 1) If f(C) E HP, for p > 0 and if f(C)

0, then

f (C) = b (C) h (C),

(2)

in ICI < 1, where b(e) is a Blaschke function and h(C) is a function in H P that is nonzero in IC I < 1. Proof. Let us show first that the integral (1) is a nondecreasing function of

r in 0 < r < 1. For arbitrary fixed p in the interval 0 < p < 1, the function fp(C) = f(pC) is regular in It;I < 1 and consequently has in ICI < 1 the representation f p(C) = bP(C)

1) F. Riesz [1923].

§4. BOUNDARY PROPERTIES OF FUNCTIONS IN CLASS Hp

403

where bP(C) is its Blaschke function and hP(C) 0 in ICI < I. The function [hP(t;)]P is regular in ICI < 1. Consequently, from Poisson's formula applied to this function,-we have in ICI < 1 2x

hp (refe)IPC

1

` Ihp(elf)IPP(r, t - A)dt.

By integrating this inequality with respect to 0 from 0 to 2rr, we obtain 2x .

2x

I hp (re'°) IP dA 0

I hp (e+e) I P dA.

S

(3)

0

1, it follows that Bu since I bP(C)I < 1 in ICI < 1 and I bP(C)I = 1 on fp(0I < hP(t )I in ICI < 1 and I fP(t)I = I hp(C)I on ICI = 1. Consequently, in I

accordance with (3), we have a fortiori 2%

2x

S If (rpe") I P dA - S

I f (pets)

IP dA,

0

0

which is now proved, for arbitrary r and p in 0 < r < 1. If we now replace r with p'/p, where p' < p, we obtain an inequality proving that the integral (1) is nondecreasing in 0 < r < 1. Turning now to the proof of the theorem, we note that, in accordance with §3, the representation (2) holds in the disk ICI < 1 with the function h(C), which is regular and has no zeros in ICI < 1. Let us show that h(C) E HP. Suppose that the least upper bound of the integrals (1) in the interval 0 < r < 1 is equal to M. Let us denote by bn([;) the product of the first n factors in the representation (13) of §3 of the Blaschke function b(C) and, for fixed a in the interval 0 < e < 1 and fixed n, let us choose rr > 0 such that I bn (C)I > 1 - e for It; I > 1 - rl, which it is possible for us to do. For 1 - 77 < r < 1, we have 2x

re b(((ee

fP )j

M

dOc(1-r)P

(4)

But since this integral is a nondecreasing function of r in the interval 0 < r < 1, inequality (4) also holds for 0 < r < 1 - 77, that is, throughout the entire interval 0 < r < 1. If we fix r and let n approach oc, we obtain, remembering the arbitrariness of e > 0, the result that, in 0 < r < 1,

404

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS 2x

S Ih(rera)IPdO<M.

0

This proves that h (C) C Hp and, in fact, that the least upper bounds of the integrals (1) for f(C) and h(C) in the interval 0 < r < 1 are equal. This completes the proof of the theorem.

Theorem 2. 2) If f(C) E Hp, where p > 0, we have 2a

2x

im S If(red) Ipd9=S If(e10)19d9

(5)

and 2+c

f(e'e) IP d0 = 0. lim S I f (rere) r_1 0

(6)

Proof. It will be sufficient to assume that f (C) j 0. If we represent f (C) in accordance with formula (2), we conclude from Theorem 1 that the function [h(C)]P"2 C H2. Consequently, by virtue of Theorem (2) of §2, the real and imaginary parts of this function can be represented in ICI < 1 -in terms of their limiting values by means of Poisson's integral. But then, the function [h (J)]P / 2 itself will'have such a representation in ICI < I. 2a

hp/2 (Tet°) = 2n

hP12 (eu) P (r, t -

9) dt.

(7)

In accordance with the Bunjakovskii-Schwarz inequality, this yields 2cc

I h (rele) IP C [S I V12 (e") I P'/2 (r, t - 0).pl/2 (r, t - 0) dt]2 0 2

CSI h(err)I'P(r, t-0)dt, 0

so that 21; S

0

21t

I h(re") I"dO<S I h(e")IPdt. 0

But since If(C)I < Ih(C)I in ICI < 1 and I f(O = I h(C)I almost everywhere on ICI = 1, we obtain from (8), for 0 < r < 1, 1) F. Riesz [1923].

(8)

§4. BOUNDARY PROPERTIES OF FUNCTIONS IN CLASS HP 2x

405

.2%

S If(1eia)I"dO

S

0

0

I f(e')IPdo.

(9)

Therefore, 2w

2w

limy s if (re") I P dA C S If (e") IP dA. 0

(10)

0

[The existence of the limit on the left follows from the monotonicity of the integral (1).]

On the other hand, by applying Fatou's lemma to the integral on the left in (8), we obtain the opposite inequality. Therefore, equation (5) holds. To pr\ ove (6), let c denote any positive number. Then, let us choose 8 > 0, such that, or any set e with mes e < S we have fe I f(eee)I P dO < e in the interval (0, 2a). Sine I f(reee)I -- I f(eie)I almost everywhere in (0, 2a) as r--+ 1, by taking an arbi y sequence of numbers r n < 1 that approaches 1 as n c, we conclude on the basis of Egorov's theorem that there exists in (0, 20 a set el such that mes eI < 8 and I f(rn eoe)I -' If(ei9)I uniformly outside e1 as n Therefore, if we denote the complement to e 1 with respect to (0, 20 by E, we have

3Jim SIf(rneie)IPdA=SIf(eie)IPdA.

(11)

It follows from (5) and (11) that lim S I f n-oo e,

(rnece) Ia

dA= S I f (ete) I P dA. el

(12)

But the right-hand member of (12) is less than E. Therefore, there exists a positive number N such that fe 1 I f (rnefe)I P dO < e for n > N. Now, let us choose N sufficiently great that we also have the inequality I f(rne`0) - f(eie)I P <e on E whenever n > N. Then, since

Ia-bl"C(Ia(+IbI)P=(max(2Iai, 2Ib1))P =2amax(1aIP, IbIP)C2P(IaIP+IbIP), for arbitrary complex numbers a and b, we have 2%

S If (r.ere) -f (e") I P dO c S I f (rnete)

-

f(eie)1P dO

E

+S2°(If(rnere)IP+I.f(ei) I")dO--- 217E+2p2a=(27-, +2P+I) e. el

(12)

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

406

Since c is an arbitrary positive number, this means that 2%

lim S

n-mp

I f (rnete) -f (e") la d9 = 0.

Also, because of the arbitrariness of the sequence (rn1, equation (6) follows. This completes the proof of the theorem. Theorem 3. 1) For a function f(C) that is regular in ICI < 1 to have a representation in It; I < 1 in terms of Poisson's integral, it is necessary and sufficient that f(t;) E H1. If this condition holds in ICI < 1, we have Poisson's formula 2+i

f(C)=2 f(e")P(r, t-0)dt.

(13)-

Proof. If f (C) can be represented in the disk ICJ < 1 by the formula 2x

f(C)=2 p (t)P(r, t-0)dt, where 9(t) E L for t E (0, 27r), then it easily follows that, for 0 < r < 1, 2x

2x

S If(reta)Ide<S I p(t)I dt, 0

0

that is f(t) E H1 Conversely, if f (C) E H I, then, for. 0 < r < p < 1, we have 2x

f (ref)= 2a f(pe")p2 - 2pr cos (t Therefore, f (Tets)

-

211

1

r' e) + r d(.

f (vtt) P (r, t - 0) dt 2%

<2 \ If

(Pete) -f (ete) I

pi -2pr cos (tr' 0) +r' dt

2x

-}-2n 1) Fihtengol'c [1929].

If(ett)I p'-2prcos(t-

0)+r,_

P(r,t-0)I dt.

(14)

§4. BOUNDARY PROPERTIES OF FUNCTIONS IN CLASS HP

407

But in the first of the integrals on the right, the Poisson kernel is, for fixed r in 0 < r < 1, bounded as p -+ 1. Therefore, in accordance with (6) with p = 1, we conclude that this integral approaches zero as p -, I. Furthermore, in the second of the integrals on the right, the integrand approaches zero almost everywhere in (0, 27r) and at the same time it does not exceed the function M if (e`B) I for some finite M independent of p. Therefore, we can take the limit under the in-

tegral sign in this integral as p -p 1. This means that it approaches zero as p -p 1. We conclude from inequality (14) that its left-hand member, being independent of p, is egdal to zero and this yields formula (13). This completes the proof of the theorem. Theorem 4. Ii If f (t;') C HP, where p > 0 and if I f(e`9)I < M almost every-

where on ICI = 1, then I f(t;')I < M in ICI < 1. Furthermore, if f(C) C H , where P p > 0, and if f(e`0) C LQ, where q > p, then f(C) E H Proof. If f(C) C HP, where p > 0, then, in the representation (2), we have h(C) C HP. Consequently, Poisson's formula (13) 4.

2s

h9

h"(e")P(r, t-0)dt

T1-

S 0

is applicable in I I < 1 to the function [h()]" C H I. Since I f (0I < I h (C)I in ICI < 1 and I f(e`0)I = I h(e`0)I almost everywhere on ICI = 1, this formula leads to the inequality 2n

If(C)IP2- If(e")IPP(r, t-0)dt. Therefore, if If(e`0)I < M almost everywhere on ICI = 1, it follows that < 1. On the other hand, if f (e `B) C L Q, where q > p, then, by I f (C) I < M in using Holder's inequality, we have 2x

9-P

If(C)IP
Qv

r 2x

2x

if (e it) 19 p (r, t - 0)

71

(r,t -0)dt

dt)P/9

,

a

4-P

P (r, t - 0) dt)

9

S

2x dt)919,

(Zs)P/9

1) V. I. Smirnov [1929].

I f (eu) I Q P (r, t - 0)

408

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

that is, 2n

I f () I°

If (err) I q P (r, t- 9) dt.

2

If we integrate this over the circle ICI = r, we obtain, for 0 < r < 1, 2x

2x

f If(Te1) I'dO

If(e")Igdt.

0

0

This proves that f (C) E H.. This completes the proof of the theorem. Theorem 5. 1)1f the function f(C) is regular in the disk ICI < 1 and if f(re`B) --y f(ei0) almost everywhere on the circle ICI = 1 as r- 1 and if

f(e`9) E L in (0, 2n), then, for Cauchy's formula 2s 1

f(C)=2ai

f

(elf)

it

to hold in the disk ICI < 1, it is necessary and sufficient that f(C) EH1. Furthermore, for an arbitrary function f(C) E H1, Cauchy's theorem 2

f(elf)dett=0

(16)

S

0

holds.

Proof. If f(C) E H1, then the proof of formulas (15) and (16) is the same as the proof of formula (13).

Suppose, conversely, that formula (15) holds for the function f(t;') in ICI < 1. The integral on the right can be regarded as a linear combination of four Cauchy integrals of the form 2n

y (t) dt

I

tai

.

elf

(17)

G'

0

where 9(t) is a real nonnegative function in (0, 2a). Now, if we denote by F(C) the integral (17), we have OR

I

2n

(F (C)) = 2a

1) F. and M. Riesz [1916].

1- r cos (t - e) p (t) I- 2r cos (t - 9) + r3 dt:

0,

C - retO.

§5. FUNCTIONS CONTINUOUS ON A CLOSED DISK

409

in I C I < I. If we rule out the case F(C) = 0 (when what we are trying to prove is

obvious), we have 92(F([;)) > 0 in ICI < I. Therefore, if F(C) = Rethen 101 < tr/2 in ICI < 1. Now, suppose that 0 < p < 1. If we set u,(r, 0) =R([f(,'')1 )= RP cos pb, we have u. (r, 0) > 0 in ICI < 1. Therefore RP = uD (r, e)

cospO

C

u. (r, e)

cosp

2

If we integrate this over the circle ICI = r, for 0 < r < 1, we have 2z

4,r

'

F (re") I P dO s cos

up (r, 0)0- 2a(!),

2

cos

7C

Consequently, F(C)' E H I) On the basis of this, we conclude that the Cauchy P integral appearing in formula (15) belongs to the class HP for arbitrary' p in the interval 0 < p < 1; that is, f ([;) E HP. But f (e`9) E L. It follows from Theorem 4 that f(C) E H. Consequently, we have proved the necessity of the condition in the theorem. This completes the proof of the theorem. §5. Functions that are continuous on a closed disk In conclusion, we point out some properties of some very important types of functions that are regular in the open disk < 1 and continuous on the closed

disk 16 < 1. Theorem 1. 2) For a function f(C) that is regular in the open disk 141 < 1

to be continuous on the closed disk IC I < 1 and absolutely continuous on the circle ICI = 1, it is necessary and sufficient that f'(z) E HI. If f'(z) E then df(e`9)/d0= ie`Bf`(e`B) almost everywhere on the circle ICI = 1. In the equation above, f' (e`0) means the limiting values of the derivative f' (z) as z approaches ei9 along nontangential paths, and df(e`B)/de means the derivative HI,

of f(e`9) with respect to 0. Proof. Suppose that the function f (4) is continuous on I C I < 1 and absolutely continuous on I4I = 1. In ICI < 1, we have 2z

f(re1A)=2a J f (eu)P'(r, t - 0) dt

1)This property of Cauchy integrals was proved by Smirnov [1929]. 2)F. Riesz [1923].

(1)

410

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

and, consequently, 2%

lretef (reee) - 2a C f (e1) dP (rat - 0) dt.

(2)

If we integrate (2) by parts, we see that the function i J'(c) can be represented

in ICI < 1 by a Poisson integral. Consequently, iJf'(J) CHI; that is, f'(OEH,. Conversely, let f' (C-) denote a member of H 1 Then, the function icf ' (0 can be represented in ICI < 1 in terms of its limiting values ie`Bf'(e''B) bymeans of Poisson's integral: 2n

le1e f

The function

(eat)P(r,t-0)dt.

(3)

t (4) 0

is absolutely continuous in (0, 217) and, in accordance with Caucchy's theorem, g(2n) = 0. Integrating (3) by parts, we have 2n

1Cf'(C)_

g(t)dP(r, t -0) dt,

that is, elf (reie) de

=d

2:

ae6J

g (t) P (r, t - 0) dt.

Therefore, in ICI < 1, 2n

f(C)=S g(t)P(r, t-0)dt+c(r), 0

where c (r) depends only on r. But c (r), being the difference between two functions that are harmonic in ICI < 1, is itself a harmonic function. Since

Ac() r

d'c (r)

dr'

1

+r

do (r)

dr

'

we easily obtain the result that c (r) = alog r + b, where a and b are constants. Since c(r) is continuous at r = 0, it follows that a = 0 and hence c(r) = coast=c. Thus, in I C I < 1,

§5. FUNCTIONS CONTINUOUS ON A CLOSED DISK

411

2x

2x

f(Q=5 g(t)P(r, t-0)dt+c=S (g(t)+c)P(r, t-0)dt,

(5)

that is, the function f(C) can be represented in the disk ICI < 1 by means of a Poisson integral of an absolutely continuous function. Consequently, it is continuous in the disk ICI < 1 and it is absolutely continuous on the circle ICI = 1. Obviously, f(e`t) = g(t) + c and equation (5) becomes Poisson's formula (1). Furthermore, it follows from (1) that lim,. ,1(df(re`B)/dB) = ie`Bf'(e`B) almost everywhere on ICI = 1. This completes the proof of the theorem. Theorem 2. If a function f(C) is regular in the open disk ICI < 1 and continuous on the closed disk ICI < 1 and if it is a function of bounded variation on the circle ICI = 1 (that is, if the real and imaginary parts of the function f(C) are functions of bounded variation on that circle), then f(C) is absolutely continuous on the circle ICI = 1. Proof. Since f(C) can be represented in ICI < 1 in accordance with formula (1), we find, just as at the beginning of the proof of Theorem 1, that the real and imaginary parts of the function if'() can be represented in C1 < 1 by means of a Poisson-Stieltjes integral and, consequently, f'() E H 1. It follows on the

basis of Theorem 1 that f(C) is absolutely continuous on the circle ICI = 1. Theorem 3 (Hardy-Littliewood). For a function f(C) that is regular in the open disk It;'I < 1 to be continuous on the closed disk It;'I < 1 and to satisfy a Lipschitz condition

f(ete) -f

I c K I 0 - 8' I°`,

a

1,

(6)

on the circle ICI = 1, it is necessary and sufficient that the inequality

where M is a finite constant, be satisfied in ICI < 1. Proof. For a = 1, the theorem is obvious. Let us consider the case when 0 < a < 1. Suppose that f(C) is continuous on ICI < 1 and that inequality (6) is satisfied on ICI = 1. Since f(C) can be represented in ICI < 1 in accordance with Cauchy's formula 2"

f(L)--2a fe it follows that, for

I C I < 1,

tr

it dt'

412

-

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS 29 1

f (0=2n

2a

f (eet) eit

S

e(" - C)a dt = 2n

f(elt) -f(ell)

-(ell- 2 e tt dt,

where C = re". Consequently, 2+c

f(C)I

it

.0

.1

2a 7t

But since sinx> (2/7r)x for 0 < x < a/2, we have

1 -2rcosp+r2=(1-r)2+4rsin' 2 J(1

I

+ n;P$.

Therefore, on the basis of what was said above and also inequality (6),

m k i p la d4

If (C) I -- 2n

kyad n J (1- r)'+: 4r 1P'

(1-r)2+4y,2

0

it

l+a

kna r

00

ylld_

2 r)I-a

.

1 +'P'

0

This shows that inequality (7) holds in 1/2 < ICI < I for suitable finite M independent of r. If we take M greater than the least upper bound of the quantity (1 - r)1-aI f'(C)I in ICI < 1/2, then (7) will hold in ICI < 1. Conversely, suppose that inequality (7) is satisfied in the disk ICI < 1. Then, the integral fQ f'(re`B)dr converges for every 0 and hence, the limit f(e`o) = limr,1 f(re°B) exists fo? every 0. Furthermore, if we integrate (7) with respect to r from 0 to r, where 0 < r < 1, we see that f(C) is bounded in ICI < 1 and, consequently, can be represented by means of a Poisson integral in terms of its limiting values f(e`0). Let us show that (6) holds. It will then follow by virtue of §1 that f(C) is continuous on the closed disk ICI < 1. Obviously, it will be sufficient to prove (6) for 10 - 0' I < I. We have f(ete)-

f(ell')=Sf'(C)dC,

where we take for 1 a curve consisting of the radial segments (e`0, hei0) and In (e`e0, he`e) and the arcs (he`0, he`0) on ICI = h, where h = 1

this case, we have

413

§5. FUNCTIONS CONTINUOUS ON A CLOSED DISK 8'

1

I f(ete)-f(eie')J

SIf'(1ete)I dr+I ShIf'(he")Idt

I

+

e,

1

I f (re"') I dr < 2 ll

-M

2

(1-h) a

1

Mdr

(1- r)'- +

Mh

(1-

by-a

hI8-8'I

-M ( + (1-h)1^a)-

dt

2

h )I 8-9`

I'

that is (6) with k = M(2/a + 1). This completes the proof of the theorem. Theorem 4 (Hardy-Littlewood). If a function f(C) that is regular in the open disk It,'I < 1, is continuous on the closed disk ICI < 1, and satisfies a Lipschitz condition

f(e,e)_

(8)

on the circle ICI = 1, then it satisfies the complex Lipschitz condition

if(t)-f(c')I
(9)

on the closed disk ICI < 1. Proof. It is easy to see that (9) will be proved if we can prove the two particular inequalities

If(rei°)-f(re"')IKI reie-rele'I°, If(re`-e)-f(r'ei) I _--KIre's

--r'e'el-

(10) (11)

for arbitrary points re`e' re`B , r'e`B in the disk ICI < 1. To prove (10), we may assume that 10 - 0'I < n. Now, the function 95e' t3 (t; )= (f (t; e`0) - f (t e `8')) / C is regular in I C I < 1 and continuous on the closed disk I C I < 1. Therefore, the maximum of its absolute value on the closed disk ICI < 1 does not exceed the maximum of the quantity I f(e°(t+0) - f(e`(t+e' )I in [0, 2n). In accordance with (8), this maximum is no greater than k 10 - 0'I a Consequently,

I f(reie)-f(reie') I

kr I0-0'I

But a

e

2

2

f sin

9

26

=-4I

ere

- eie'I'

(12)

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

414

so that (10) with K = (7ra/2°)k follows from (12). Let us now prove (11). Suppose that 0 < r < r' < 1. In accordance with Theorem 3, inequality (7) holds in ICI <. 1. From this inequality we get r

r'

I f (re°e) -f (re')

dt

If'(te;) I dt

I

r If

r

r' < (1 + r)/2 and hence r' - r < 1 - r', we have

-f(re') If (rete)

r

M

M

dt

r') ) I-&

(1

I

r

(r- r)t_a=M.r - r)a.

On the other hand, if r' > (1 + r)/2, so that r' - r > 1 - r', then

(I - t)a rlr M t CQ (reiB)-f(re°)I<_M <M (1-r')a a a

li-rla.

In both cases, we have obtained (11) for suitable K. This completes the proof of the theorem.

Theorem 5. 1) if a function f(C) = u(r, 0) + iv(r, 0) is disk ICI < 1 and if u(r, 0) is continuous on the closed disk fies the Lipschitz condition

the open

1 and satis-

0
(13)

on the circle ICI = 1, then f(C) satisfies a complex Lipschitz condition on the closed disk ICI < I. Proof. Since f(C) can be represented in the open disk ICI < 1 in accordance with the Schwarz formula 2x

f(C)=2n

tt(t)e + Cdt+IC,

it follows that 2x

f (C) = 2

1

2x

(e"- C), dt ==u (t)7- u(p(8)a rr dt. 2u (t) eir

1

Consequently, just as in the proof of Theorem 3, we obtain the inequality

IF'(r\I< 1) Privalov [1919].

M (1-r)1--a

3

(14)

§5. FUNCTIONS CONTINUOUS ON A CLOSED DISK

415

in I CI < 1; that is, f (c) satisfies the condition of Theorem 3. But then, in accord-

ance with Theorems 3 and 4, it follows that equation (14) holds. This completes the proof of the theorem. Theorem 6. If a function f(C) = u(r, 0) + iv(r, 0) is regular in the open disk 1 and satisfies on the ICI < 1 and u(r, 0) is continuous on the closed disk circle ICI = 1 the condition

Iu(9)-u(9')I
(15)

where the function A(t) is nondecreasing and is such that the integral f (A(t)/t)dt converges, then f(C) is bounded in the open disk ICI < I. Proof. Formula (14) holds in ICI < 1. We transform this formula to the form 2x

f(C)=2 (u(t)-u(9)) ett±t dt+u(8)+IC where (t = rei9). Consequently, if f (C) I <_

-x

I u (t + 9) - u (e) I 1 re« +

M,

where M is a constant. Therefore, in accordance with (15), we have x

2

a (t)

If(C)Ic a -reteIdt+M

2

a (t)

1-rett1dt+Mt,

where M1 is also a constant. If we replace the absolute value in the denominator

of the integral on the right with its imaginary part and use the inequality sin t > (2/7r)t, we obtain x

It

2

2

If (C) I < 2 S

r sin)t dt + M

r

' lt) dt + Mt,

which proves the boundedness of f(C) in ICI < 1. Theorem 7. ')If a function f(C) = u(r, 0) + iv(r, 0) is regular in the open disk ICI < 1, and if u(r, 0) is-continuous on ICI < 1, then E H P for all

p>0.

Proof. Let p and c denote positive numbers such that p( < n/2. Since u(r, 0) is continuous on the circle ICI = 1, it follows that, in accordance with the familiar theorem on the approximation of periodic functions, there exists a 1)Smirnov [19321.

416

IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS

trigonometric polynomial m

s (9) = as + Z (an cos nO + bn sin nO

n=l

,

such that Iu(6) - s(O)I < E in the interval (0, 27r). If we set in

p (C) = as + Ej (an -1bn) Cn n-1

(so that 92(p(e`0)) = s(0)) and 0(e) = f(C) - p(C), we have, for 0 < r < 1,

'S I t l=r

ta

e1Pv

(tC

= e1Pp (0).

If we now separate the real and imaginary parts, we obtain 2x

e'PP (t) I cos p (u (9) - s (9)) dO - 9R (e'P? (o)). 2a S Therefore, since Cos P (U (9)

- S (9)) ':- Cos pr > 0,

we have 2x C"(etP9v (o

1

C I e'PP (t) I d9

2n J

cospe

(16)

Inequality (16) has been proved for all r in the interval 0 < r < 1. This means that e`O(C) E H If we now keep in mind the fact that the function a-`P (C) is bounded in IcI < 1, we obtain the result that e`f(C) E HP also. This completes the proof of the theorem.

CHAPTER X BOUNDARY QUESTIONS FOR FUNCTIONS THAT ARE

ANALYTIC INSIDE A RECTIFIABLE CONTOUR

§ 1. The correspondence of boundaries under conformal mapping

In the present chapter, we shall study various questions associated with the boundary values of functions that are regular in a domain B that is bounded by a rectifiable closed Jordan curve. In these investigations, an important role is played by certain boundary properties of functions that map the domain B univalently onto a disk. We shall begin with an exposition of these properties, which were established by N. N. Luzin and L L Privalov. 1) A continuous curve C: z = z (t) = x (t) + iy (t) for a < t < b is said to be rectifiable if, for an arbitrary number n + 1 and an arbitrary sequence of values

t1, t2,...,tn+1 such that

the sum 'I

21 1 zz (tk+l)

k=1

- x (tk)

remains bounded. The least upper bound of these sums out of all possible com-

binations of values t2, ... , t is called the length of the curve C. It is obvious from the definition that a necessary and sufficient condition for a Jordan curve C to be rectifiable is that the functions x (t) and y (t) be of bounded variation in (a, b). Furthermore, it is obvious that, if. the curve C is rectifiable, any arc of it is also rectifiable. We denote the length of an arc z = z (t) for a < t < t ', which is a function of t ', by s (t '). Let us prove the following lemma regarding s(t). Lemma. If x(t) and y(t) are absolutely continuous in (a, b), then

s(t)=S z'(r)Idwr, act-< b. a 1) See Privalov [1919].

417

(1)

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

418

Proof. Suppose that a < t < b and that a = t 1 < t 2 < . . . < 1, + 1 = t. Then, the function

Pk('0=1z('t)-z(tk)I=V(x(T)-x(tk))9+(y(T)-y(tk))' (for k = 1, , n) is absolutely continuous in (a, b) because the absolute value of the difference between its values at any two points of the interval (a,1b) does not exceed the sum of the absolute values of the corresponding differeces between functions x(t) and y(t). Consequently, Ok(r) has a derivative almost everywhere in (a, b) and (x (T) - X (tk)) X' (T) + (y

-y (tk)) y' (1) (X (T) - .r (tk))' + (y (ti) -y (tk))

(X (T) - X (tk))2 + (y (I) -y (tk))' (,)'+y (T)9 = I Z' (T) I Y (X ('t) - X (tk))' + (Y (T) -Y (tk))'

almost everywhere. Therefore, for tk < r < b, we have

IIfk(T)IC S Iz,(T)Idt. tk

In particular, for r = tk + 1, tk+1

z (tk+l) - z (tk) I<

I z' (T) L dT.

S

tk

Therefore

s(t)GSIz'(T)Idt.

(2)

On the other hand, since s(t) is obviously a nondecreasing function, it is continuous in (a, b) except possibly at countably many points. Excluding these points, we have in (a, b) by Fatou's lemma t

t

s (ti + hh - 3 (T) dT

lim

h > 0,

s' (T) dT,

and since h

S (T + h) - 3 (T) dc =--

1

t

s (T) dT -

s (T) dT

a

=h [

t

A

a h

t+ h

a

t

s(T)dTs(r)dT,
s(T)dc<S(t+h),

§1. CORRESPONDENCE OF BOUNDARIES

419

it follows that s (t)

S s' (T) dT. a

But at points at which all the derivatives s'(r), x'(r), and y '(r) exist, we have s' (T) = lira

As

c'C-.o AT

Jim c-C-.o

AX, + Ay,

= x' (T)9 +y' (T)a = z' (T) I,

A%,

Consequently, at these points and hence at all points t E (a, b), (T)IdT.

(3)

a

Inequalities (2) and (3) lead to equation (1). This completes the proof of the lemma. Suppose now that a finite simply connected domain B in the z-plane is bounded by a rectifiable closed Jordan curve C. Suppose that a function 6 = X(z) maps the domain B conformally onto the disk ICI < 1. Let z = &)(0 denote the inverse function. In accordance with Theorem 4 of §3 of Chapter II, the function X(z) is continuous on B and the function co(O is continuous on ICI < 1. Also, the curve C can be represented parametrically in the form z = cu(ett) for 0
E H 1;

3) dtj(e`t)/dt = iett&)'(e`t) almost everywhere on the circle

4) the length s(t', t ") of the arc z = a )(e") for t' < t < t " is given by the formula ti,

s (r, r = S

IW

(ett) dt.

(3 ')

N

Let us now look at various sets of points on the boundary C. A set E C C is said to be closed on C if it is closed as a set of points in the plane. A set

420

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

E C C is said to be open on C if its complement is closed on C. One can easily show that an open set on C is the union of a finite or countable set of arcs. We define the measure of an open set on C as the sum of the lengths of the arcs of which it is composed. Now, let E denote an arbitrary set of points on C. The outer measure of the set E is defined as the greatest lower bound of the measures of all open sets on C containing E. We denote this outer measure by m E. We define the inner measure, which we denote by miE, of the set E by t formula miE = s - me CE, where CE is the complement of E with respect to C and s is the length of C. We say that the set E is measurable on C if meE = miE. When this is the case, we denote the common value of meE and miE by mes E and call it the measure of the set E. If E is measurable on C, then it follows from the relations

miE= s - meCE, miCE= s - meE, meE=miE that CE is also measurable on C. In particular, measurability of open sets implies measurability of closed sets. In general, one can show that all the theorems regarding Lebesgue measure can be carried over to the study of curves and, for this reason, we shall treat them as proved. Theorem 2. The mapping mentioned in Theorem 1 maps a set of points of

measure zero (of positive measure) on the circle ICI = 1 onto a set of points of measure zero (of positive measure) on C, and conversely. Proof. 1. Let E denote a set on ICI = 1 such that mes EC = 0 and let EZ denote the corresponding set on C. For any e > 0, there exists on the circle I t I= 1 an open set 0 C such that E C C 0 C and mes 0< E. Corresponding to it on C is an open set OZ that contains E. We have

meEz<mes Oz= S

I w'(ei') I dt.

oc

But meEz does not depend on a and the last integral can be arbitrarily small for sufficiently small c. Consequently, me E. = mes Ez = 0. 2. Before proving the converse, we make the following remark: Since &)'(0 0 almost everywhere on ICI = 1, the measure of the set E(Ij S(C)I < of points on CI = 1 at which < e approaches zero as c - 0, it follows that when we set mes E (Icj'(C)I < E) = S(c), we have the limit relation S(e) - 0 as c--b 0. Therefore, if E is a measurable set on the circle ICI = 1 such that

1. CORRESPONDENCE OF BOUNDARIES

421

S w (e``)ldt<e, then

e5

; (err) dt >,-

I a (e'`) I dt

(mes E

so that

mesE
)=ry(e).

This shows that mes E can be made arbitrarily small by making a sufficiently small and >f(E) depends only on e. Now, let EZ denote a set of points on C such that mes EZ = 0. Let E. denote the set of points on JC = 1 corresponding to it. For given e> 0, there exists on C an open set OZ containing EZ such that mes OZ < e. In accordance with the remark just made, we have mes 0 < ri(e), and consequently, meEz < rj(E)

does not depend on c and q (c) - 0 as t -- 0. Therefore, me E _ mes E = 0. But me E

3. To prove that corresponding to a set of positive measure on JCJ = 1 there is always a measurable set of positive measure on C, we note that, if E is a 1 and mes E t > 0, then on I I = 1 there are closed sets measurable set on G (k), for k = 1, 2, , such that G(") C E. and mes G (k) -s mes E t. Let us Then, G C C E and mes G t> mes G (k), for k 1, 2, define G t= U k 1 GC so that mes G C = mes E,. Consequently, the set N = E - G t is a set of measure zero on ICI = 1, and we have therefore represented E as the union of the set N of measure 0 and the closed sets G (k), for k = 1, 2, But corresponding to the set N t is the set NZ of measure zero on C and, obviously, corresponding to the sets Gk) are the closed sets G=k) on C. Consequently, the set E. corresponding on C to the set E, is the set NZ U Uk 1Gik), which is a measurable set. Its measure is positive because otherwise, E t would, from what we have C

proved, be of measure zero. In an analogous way, we can show that corresponding

to a set of positive measure on C is a measurable set of positive measure on ICI = 1. This completes the proof of the theorem. Remark. It follows from part 2 of the proof of Theorem 2 that a function _ ).(z) that maps the domain B onto the disk JC1 < 1 is absolutely continuous on

422

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

C if we regard it as a function of arclength on C because the inequality Elsk - s kI < c implies I I X (z (Si)) - X (z (S;)) I C i7 (e). 11

Theorem 3. The mapping z = 0)(C) in Theorem 1 is conformal almo t everywhere on ICI = 1 or, respectively, conformal almost everywhere on C.

Proof. Obviously, we can consider the conformality of the mapping only at points on ICS = 1 in the one case and on C in the other. Since the derivative dw(e`t)/dt exists and is equal to zero almost everywhere on the circle CI = 1, there exists almost everywhere on C a unique tangent such that in a neighborhood of a point of tangency, the curve C lies on both sides of the normal to C at that point. Let us show that angles are preserved at all points on ICI = 1 that are mapped into points of the curve C with this property. Let 4U denote one of these points on ICI = 1 and let zo denote the point on C corresponding to it. Then, every branch of the, function %

u (C,

CO) = arg

M - (to) C-Co

is continuous in the closed disk ICI < 1 except at the point Co. But since the curve C has a tangent with the property described at the point zo, the function u (C, CO) approaches a definite limit a as C approaches Co along the circle ICI = 1 (this value being the same for approach from the two sides). We take this limit as the value of is (C, C.) at the point Co. Obviously, the oscillation of the function is (C, CO) in the portion of the disk ICI < 1 contained in a sufficiently small neighborhood of the point Co is less than 3n. This shows that the function u(C, CO) is a bounded harmonic function in ICI < 1 and it can be represented in that disk in terms of its values on the circle ICI = 1 in accordance with Poisson's formula. These values define a continuous function on ICI = 1. Consequently, the function u (C, CO) is continuous on the closed disk ICI < 1 including the point C., where it is equal to a. Now, consider an arbitrary curve A lying in the disk ICI < 1 with endpoint Co, and let l denote the corresponding curve in the domain B. Since u (C, CO) -- a as C, t through values belonging to the disk ICI < 1, if the curve A has a definite tangent at the point Co, the curve l will have a definite tangent at the

§1. CORRESPONDENCE OF BOUNDARIES

423

point z0 and vice versa. Also, if these curves have tangents at the points indicated, a will be the angle of rotation of the tangent in the transition from the curve A to the curve 1. Since this angle of rotation is the same for all pairs of curves A and 1, this is equivalent to conservation of angles at the point Co in the shift from the disk ICI < 1 to the closed domain 8 or the other way around. This completes the proof of the theorem. Let us now look at the question of the correspondence of boundaries under conformal mapping of domains bounded by smooth Jordan curves. By a smooth Jordan curve C: z = z(t), where a < t < b, we mean a curve possessing at every

point a tangent that rotates continuously as z moves along C, that is, a curve for which the angle 0(z) of inclination of its tangent to the real axis is a continuous function of the point of tangency. In the case of a closed curve, we must also have 0(z(b)) = 0(z(a)) + 21r. Obviously, a smooth curve is a rectifiable curve. Therefore, 0(z) can also be regarded as a function of the arclength s on C: 0 = 0(s). We now have

Theorem 4.1) Let z = a) (4) denote a function that is regular in the disk < 1 and that maps that disk univalently onto a domain B bounded by a smooth closed Jordan curve C. Let L'= a. (z) denote its inverse. Then arg &)%) is a continuous function in the closed disk ICI < 1 and arg a.'(z) is a continuous function in the closed domain B. Furthermore, on the circle ICI = 1,

argni (()=9(Q-argC- 2.

(4)

Proof. As a preliminary, we point out that, by virtue of the smoothness of the curve C, for any point a E C and any number t > 0, there exists an arc on C containing the point a as an interior point such that the angle of inclination of the chord connecting any two points of that arc to the real axis differs from the angle of inclination of the tangent at the point a by an amount less than t. This follows from the fact that the angle of inclination of the chord is equal to the angle of inclination of the tangent at some interior point of the arc intercepted by that chord.

Now let us assume that the condition of the theorem is satisfied and let us consider the function 1) Lindelof [1916].

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

424

u (G i) = arg

Celt

-C

(C) ,

which is, for arbitrary fixed r> 0, a harmonic function in the closed disk Icl < 1. We assert that there exists a finite constant K such that lu (C, r)I < K for all C in the closed disk ICI < 1 and all r in the interval 0 < r < n. If this were not the case, there would be a sequence for n = 1, 2, , such that Mn = as n -, oc. Obviously, r, -4 0 as n --, oc. Furthermore, max 1 CIS1 lu (C, we can, by shifting to a subsequence of the arcs r, if necessary, arrange to have m converge to a the sequence 1 ,1 of points on i C 1= 1 at which lu (t,,, point Co on the circle ICI = 1. But because of the smoothness of C, the sequence converges to the angle formed by the tangent to C at the of the values is ((, point zo = w(Co) and the tangent to the circle I = 1 at the point Co. This contradiction proves the assertion made.

Now let a= e4 denote an arbitrary point on the circle I = 1 and let 0 denote the angle between the tangent to C at the point a = w(a) and the tangent to the circle I C I = 1 at the point a. Then, for given e > 0, there exist q > 0 and S > 0 such that, for Irl < 8, we have

lu(C T)-?I< a on an arc of length q containing points on each side of a. Let us show that this inequality also holds in that portion A of some neighborhood of the point a that lies in the disk C1 < I. If we set U r 0 and I k I< 1 in accordance with Poisson's r) formula 2n

U (G

r) = 2L V

= 2n

U (e "t, r) P (r,

t - B) dt

a

U(e", 2) P (r, t - 9) dt + 2n

U(e", r) P (r, t - 6) dt,

where C= reie and I is the portion of the circle ICI = 1 complementary to the arc (>/i - q , t/i + t?). Therefore, U (C,

r) l

P (r, t - O) dt +

47C

4-17

§1. CORRESPONDENCE OF BOUNDARIES

+X+

TI

S P(r, t-6)dt-<2+K2I7I

C P(r,

425

t-6)dt.

27

But this last integral converges uniformly to zero as a through values in the disk ICI < 1. Consequently, there exists an nl > 0 such that the inequality IC- aI < rll, where ICI < 1, implies I NC, r)l < t, that is Iu (c., r) - 01 < t for all r such that Irl < S. Then, for CC A, where A is the portion of a neighborhood of the point a mentioned above, we have the limit relation u (C, r) ..._, arg to '(c) as

r -- 0. Therefore in A,

Iargw (C)-pI<e. This shows that arg a)' (C) is continuous in the closed disk ICI < 1 if we define this function at the points a such that IaI = 1 as being equal to the angle that we have been considering.

The continuity of arg X'(z) in 9 now follows from the fact that X'(z) _ 1/w'(c). Since 0 = B(s) - Vi - n/2, formula (4) also follows from what we have shown. This completes the proof of the theorem. Theorem 5. Under the conditions of Theorem 4, we have log w%) C H1 and

w'(0 C Hp for all p > 0. Proof. From the inequality between the arithmetic and geometric means, we

have, for 0 < r < 1, 2%

I n

log I w' (re1t) I dt - lim t 2n 1 b n.oo n

zni

log l w' (re n k) I 1

n

lim log n-ao

n

)I

k- 1 1

imo log

d k

I w (re n

n

n

2at 1

k

I w' (Ten

2cn

log

2a } I w (re`t) I dt.

(4 ')

Consequently, in accordance with Theorem 1, the first interval is bounded in the interval 0 < r < 1. But since the integral 2n S

0

I arg w' (re`°) I d6

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

426

is, by Theorem 4, bounded in the interval 0 < r < 1, it follows that log co' (i) E H I. The second assertion of the theorem follows from Theorem 7 of § 5, Chapter IX asapplied to the function u(p, 0) = arg co'(re`e), which, in accordance with Theorem This completes the proof of the 4, is continuous in the closed disk theorem.

Theorem 6 (Kellogg). Let z = to(G) denote a function that is regular in the disk JAI < 1 and maps that disk univalently onto a domain B bounded by a smooth closed Jordan curve C. Suppose that the angle of inclination 0(s) of the tangent to C as a function of the arclength s on C satisfies the Lipschitz condition Ja,

O
(5)

Then, to'(C) is continuous on the disk JCJ < 1 and the following Lipschitz conditions (6)

log w' (ete) - log w

(ei0') I

k2 l 0

- 0' Ja

(7)

where kI and k2 are constants, hold on the circle JCI = 1. Proof. By Theorem 4, we have

argw'(Q-0(()-argC- 2,

(8)

and, by Theorem 5, we have a) 1(0 E HP for all p > 0. On the basis of this fact and formula (3'), we have, for 0'< 0,

Is-s'IIai(etl)IA

w(eie)I9A A-,--k,10-0'Ji/2, e,

so that

(S)-0(S) -I-- le-e'I k9 Is-s'I2 f I0-0'1
largw

It follows on the basis of Theorem 5 of § 5, Chapter IX, that log c,'() and hence to'(c) are continuous functions in the closed disk ICI < 1. But then, e

I S- s' J- S I w (et°) I d6 - k3.10-6'j, k3 = max I w (() I, e,

and, consequently,

ICI-1

§ 1. CORRESPONDENCE OF BOUNDARIES

427

Iargw'(ei) -argw' (ete')I-k4I0-0'1", so that, again by Theorem 5 of § 5, Chapter IX, inequality (7) follows. Furthermore, for arbitrary w and w' in the disk Jwl < M, we have 00

Iem_ew'1=

ca

wn

ft

k-1 If we apply this to w = log w'(e'e) and w'= log

L n _1=kMIW-w'I.

k®1

nl

(Ae`Bi)

and use (7), we obtain

This completes the proof of the theorem. Theorem 7.1) Let z = to(C) denote a function that is regular in ICI < 1 that maps that disk univalently onto a domain B bounded by a smooth Jordan curve C

such that the angle of inclination O() of its tangent to the real axis, considered as a function of arclength on C, is an absolutely continuous function and 9'(s) E LP, where p > 1. Then cj"(C) E Hp. Proof. From the condition of the theorem, we find, by using H61der's inequality s

le(s)-0(s)I=S0'(s)ds<(S0'(s)pds) 9(Sds)iiQMIs-s 111q, s'

where 1/p + 1/q = 1. From this we conclude, by Theorem 6 (with a= 1/q), that co'(0 is a continuous function on the closed disk ICI < 1. Since dO (s) dA

-

d9 (s) I w (e;8) I, 1 ds

we conclude that d6(s)/d9 E L"; that is, d

(argw'(e'e)+e+2)ELP.

But, by Theorem 3 of §4, Chapter IX, and Theorem 2 of § 1, Chapter IX, we have almost everywhere on the circle ICI = 1 d log co' (e1e) dA

-

(ell)

lee `", m' (et )

so that ! (ec0to "(e`B)/w'(e`°)) E L". It now follows from the representation of the function 1) Smirnov [1932].

by means of Poisson's integral that

428

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR Cam, (C)

E hp

Then, by Theorem 4 of § 2, Chapter IX, Cw "(C)/w'(c) E Hp. Consequently, w "(c) E Hp. This completes the proof of the theorem.

§2. Privalov's uniqueness theorem Let us agree on some terminology. Suppose that B is a domain bounded by a Jordan curve C and that z0 is a point on C at which there is a unique tangent to C and in a neighborhood of which the curve C lies on both sides of the normal. We shall refer to a continuous curve l contained in B and ending at point z0 as a nontangential path if that part of 1 in a neighborhood of z 0 lies inside some angle of magnitude less than rr with vertex at the point z0 and with bisector coinciding with the inner normal to C. When we speak of a set of points on the boundary of the domain B, we shall usually mean only points at which there are tangents to C possessing the property just described. Furthermore, if a function f (z) that is regular in B approaches a value a as z approaches a point z0 on C along every nontangential path contained in C, we shall say simply that f (z) assumes (or has) the value a along nontangential paths. We turn now to the investigation of functions that are regular in a domain bounded by a rectifiable Jordan curve. We begin with the exceedingly important uniqueness theorem of Privalov. 1) 'Theorem 1. Let f (z) denote a function that is regular in a domain B bounded by a rectifiable closed Jordan curve C. Suppose that f (z) assumes the value 0 along tangential paths on some set E of positive measure on C. Then f (z) = 0 in B.

Proof. Let us consider first the case in which B is the disk Iz I < 1. About every point z0 E E, let us construct a right angle with vertex at z0, with rays directed into IzI < 1, and with bisector directed along the radius to z = 0. Let us cut from this angle sectors S('), for n = 1, 2, , of radii 1/n. Consider a 0 sequence of functions fn(z), for n = 1, 2, , that are measurable 2) on E. 1) Privalov [1919]. 2) The measurability of fn(z) follows from the fact that we can regard fn(z) as the limit of a sequence of functions fn,r(z) constructed in the same way but issuing not from sectors but from the parts of them lying in Iz I < r as r ---' 1. The functions fn r(z) are, for r sufficiently close to unity, defined and continuous on E.

§2 PRIVALOV'S UNIQUENESS THEOREM

429

at the point zo 6 E as the maximum value of in the sector SO). Obviously, this sequence converges to zero on E. In accordance with 0 Egorov's theorem, the set E contains a closed subset P of positive measure on converges uniformly to zero. The complement of P which the sequence with respect to the disk IzI = 1 is an open set and hence consists of finitely or countably many disjoint arcs. We replace those arcs that are less than a semicircle with the sides of the right angles intercepting them from inside the disk IzI < 1 in such a way.that the sides of the angles intersect the circle IzI = 1 at an angle zr/4. On the other hand, we replace an arc greater than a semicircle with the two radii that pass through its endpoints. As a result, we obtain in place of the circle IzI = 1 a rectifiable closed Jordan curve C1 bounding the domain B1 and contained in the disk IzI < 1. Here, the set P lies entirely on the boundary C1. The function f (z) is continuous in B if we assign to it the value 0 at points of the set P. Let us now map the domain B1 onto the disk ICI < 1. Let z = &)(C) denote the inverse of the function executing this mapping. Corresponding to the set P is a measurable set P1 of positive measure on the circle 141 <1, The function f 1(C) = f (w([;')) is regular in the open disk ICI < 1, continuous on the closed disk I4I < 1, and equal to zero at points of the set P1. Then, in accordance with 3 of Chapter IX, f 1(4) = 0; that is, f (z) = 0 in IzI < 1. Thus, the theorem is proved for the case of a circle. Let B denote an arbitrary domain defined by a rectifiable closed Jordan curve C. And let us map it onto the disk ICI < 1. We denote by E ' the measurable set of positive measure of ICI = 1 into which E is mapped. The function f (4) is now transformed into a function fl(4) that is regular in the disk ICI < 1 and that assumes the value zero along nontangential paths on the set E ' after deletion of a set of measure zero at points in which this mapping does not conserve angles. In accordance with what has been proved, f 1(4) = 0 in 141 < 1. Consequently, f (z) - 0 in B. This completes the proof of the theorem. Define

The following theorem 1) is of great significance in applications of this unique uniqueness theorem. Theorem 2. There does not exist a function that is regular in a domain B bounded by a rectifiable Jordan curve and that assumes the value oc along nontangential paths on a set of points E of positive measure on the boundary C. 1) Privalov [1919].

430

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

Proof. As above, it will be sufficient to prove the theorem for the case of the disk IzI < 1. Let us suppose that a function, f (z) having the properties mentioned in the theorem does exist. We follow the same reasoning as above except that now' we define the functions frt(z) on E according to the formula fn(z) = mins(,)If (z)I, and we apply.Egorov's theorem to the functions On(z) = 1/(1 + &W). As a result, we can show that the function f (z) is regular in the domain B 1 (cf. preceding proof) and approaches - on a set P of positive measure contained in the boundary B 1. But the function f (z) has only a finite number of zeros in B 1. This is true because otherwise, these zeros would have a cluster point which must lie on IzI = 1 and the only boundary points of the domain BI on IzI =4 are points of the set P such that f (z) -' 00 as z approaches them from inside B 1. We denote by p (z) a polynomial having the same zeros in B 1 as f (z) has. Then, the function p(z)/f (z) is regular and bounded in B 1 and it has limiting values equal to zero on the set P. But (cf. again proof of Theorem 1) this function would then have to be the function identically equal to the zero in B 1 but in the case that we are considering it does not have any zeros at all in B 1. This contradiction proves the theorem. As a direct consequence of Theorems 1 and 2, we have Theorem 3. If two functions that are regular in a domain B bounded by a rectifiable closed Jordan curve on some set of positive measure on the boundary have equal values along nontangential arcs, they are identically equal in B. §3. On the limiting values of Cauchy's integral Let C denote a Jordan curve, closed or otherwise, in the z-plane. Since every point z' on it is completely determined by the arclength s measured along C from an initial point z o to z ', we can take s as parameter. Then, the parametric representation of C is z ' = z '(s) for 0 < s < S. Obviously, the function z '(s) satisfies the inequality I'z

(S1)-x

(S9)I_S1-s9, 0<s9<s1

S,

and hence is an absolutely continuous function of s. Therefore, the derivative dz'(s)/ds exists almost everywhere on C. Furthermore, in accordance with the Lemma of § 1, we have Idz '(s)/dsl = 1 almost everywhere on C. But the curve C has a unique tangent at every point zo E C at which dz'(s)/ds exists and is nonzero and C lies on both sides of the normal in every neighborhood of zo. At

3. THE LIMITING VALUES OF CAUCHY'S INTIrGRAL

431

such points, we may speak of nontangential paths to C. We shall say that a function f (z') defined on C is summable on C if the integral fc f (z')dz', meaning the Lebesgue integral S

f V(8)) a ds S) ds'

exists. Now consider the Cauchy integral (1)

where f (z') is a summable function on C. For values of z not on C, this integral obviously exists and is a regular function. On the other hand, if z lies on C, the integral (1) can be understood only in a conventional sense. Specifically, let z o = z '(s o) denote an interior point of the curve C and let CE denote the part of C obtained after deletion from C of the smaller arc from the point z '(s o - e) to the point z '(s o + e). Then, the limit of the integral f (z) Z' - zo dz

(2)

as c ---p 0, if this limit exists, is called Cauchy's singular integral, and this is

what we shall mean by the integral (1) when z = zo. The following theorem 1) establishes a relationship between the limiting values of the integral (1) as z approaches points on C and the existence of the singular integral (2). 'Theorem 1. If Cauchy's integral (1) has almost everywhere on C definite limiting values along all nontangential paths lying on one side of C, then Cauchy's singular integral (2) exists almost everywhere on C and limiting values of the integral (1) along nontangential paths lying on the other side of C also exist almost everywhere on C. Conversely, if Cauchy's singular integral (2) exists almost everywhere on C, then definite limiting values of Cauchy's integral (1) along all nontangential paths on each side of C exist almost everywhere on C. Here, in both cases the formula urn u

1) Privalov [1919].

z z) dz = z

(Z zo

dz' ± if (zo)

(3)

432

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

holds almost everywhere on C. In this formula, the limit in the left-hand member is understood to mean a limit along nontangential paths. In the right-hand\,,member, the positive sign is taken when the nontangential path lies to the left of tie tangent to C directed in the direction of the integration; the negative sign is taken when the path lies to the right of that tangent. Proof. It will be sufficient to prove that the difference

C

f

dz'-

(z

zf (z i, dz',

a -Iz - zoo

(0)

z

approaches ± ni f (z o) for all z o E C as z approaches z o along all nontangential paths. Here, the sign is plus when the nontangential path lies to the left of the tangent to C at the point zo, and it is minus when this path lies to the right of that tangent. If we can prove this assertion about the difference (4), then, almost everywhere on C and in particular at points at which f (z') is finite, the existence of one of the integrals (4) will imply the existence of the corresponding limit for the other integral. This assertion is easily proved if f (z')

on C. Let z o = z'(s o) denote an arbitrary

point on C at which Idz'(s)/dsI = 1. The first integral in (4) gives the increment in log (z '-z) for one circuit around the curve C. The imaginary part of this increment is the increment in the angle of inclination of the vector extending from the point z to the point z '. But (log (z '- z))c does not, change if we replace the arc (z '(s o - e), z '(s o + e)) on C with a semicircle y with the same endpoints located on the other side of C from z. On the other hand, the increment in log (z'- z) for one circuit around the new path will differ from the increment in log (z ' - z o) for one circuit around this new path by an amount that approaches zero as a -- 0. Obviously, the value of the subtrahend in (4) is equal to this last increment decreased by the increment in (log (z '- z o))y, which, for sufficiently small c, can be made arbitrarily close to + vi (the sign depending on the position of the point z with respect to the direction of integration on C). This proves that in the present case, the difference (4) approaches ± vi with the required choice of sign as a 0. To prove this same assertion in the general case, it will now be sufficient to show that the quantity

F(z,zo)=

z zadz'- Czf(z)dz'-f(zoI) L7

C

[ ,dz z-z-S

dz,

§3. THE LIMITING VALUES OF CAUCHY'S INTEGRAL

f (zz - z(zo) dz - S f--(z-1 1

) 0

433

e = z - z'

dz',

a

(5)

approaches zero for all points z o E C as z approaches z o. Now, suppose that the point z o = z '(s o) on C is such that the function f (z ') is finite, and the derivative dz'(s)/ds exists and is equal to unity in absolute value. Let yo denote the angle of inclination of the tangent to C at the point z 0 to the real axis. By letting z approach z o along a nontangential path 1 and setting z = zo+ icei(Xo++), we conclude that, for sufficiently small e, we have III < qo or I-- qI < qio, where qio < p/2. We may assume that this last condition is satisfied for all points z on the curve 1. Now, suppose that we are given Tj > 0. Let h = h (q) > 0 denote a number such that, for Is - so 1 < h,

z'(s)-zo S - so

=etxo+s, IaI`r)

Let a denote an arbitrary number in the interval 0 < e < h. We have

f (z) -f (zn ) dz 2

'

- 2

-

(f (z' (s)) -f (zo))

-

dz' (s) ds ds I(Xo+'h)

iXo

If we now set s - s 0 = a, Se-`X0 = m = m (a), (f (z'(s) - f (z,,))(dz'(s)/ds) x = q (a), where m c Tj for I o I < h and 0 (0) = 0, we have

e- 'X0

S

f(z')-f(zo)dz'=

z'-2

9P (a) da

a-eie'

Therefore, we can write (5) as follows:

F

(z,

x°)_-

q (a) do

etc' aicp (a) do

((1+m)a-aie'+) (1+m)a +

S

(1+m)a-iee'V'

(6)

We require also that the point z0 be such that urn. 0c(a)/a= 0, where a

' (a) = S I IF (a) I do = 5 I f(z' (So + a))

(so)) I do.

Let us show that almost all points on C satisfy this requirement. Consider the set (r 1, r2, 1 of all points in the plane with rational coordinates. Note that,

434

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

for each rk, we have by Lebesgue's theorem, s+a

s

If(z'(s))-rkIds=oim

ds

S

a

S

for almost all s in 0 < s < S. Therefore, if Ek is the set of points in the interval 0 < s < S at which the above assertion does not hold or at which f(z '(s)) = oo, then mes Ek = 0. Let us now set

1

00

E= U Ek, k=1

Then mes E = 0. If s 0 does not belong to E, then, for arbitrary o > 0, we can choose rk such that

If (z'(so))-rkI< a . For a sufficiently small, e

a

If (z' (so + o)) -f (z' (so)) I da o

I f (z' (so + o)) - rk I do.

o

Pf

+c

If(z'(so))-rkIds< 2 + 2 =s.

But this means that for so outside E,

0 as a-.0.

(a) G

Under the assumption that we have made, let us find a bound for each of the integrals in (6). We have (a) do

Rr=l J

a

(1+m)a-eiet

I

'4 (a)

do

C SeI (I+ m) a-ciel0'('

or, replacing the modulus in the denominator with the imaginary part,

Rte

cos+o -1

a

Scos4+o

_e

2 mx

()I.

I
I

Let us now break the first integral in (6) into three integrals over the arc Ch and over the other two arcs, which together we denote by Ch. We have g1Ve1cp (a) da

R9 = ICs ((1 -}- m) a - e ie .A

(l -{- m) a

c [

eh

a

cp (a)

da

ae'0-clI (I -,)a

§4. CAUCHY'S FORMULA

435

or, replacing the modulus in the denominator with the real part,

R9<

r 1 , f (a)

5

_

I do

a

1_ (a)

ar(cos`0-X1)(1-1j)-(cos`0-X1)(1-,)

as

A

1

dd.

an

Integration by parts yields t(;) L do= 11 a) I -{- 2. bah

ah

max

Consequently,

ah

(a) 13

I

lal
do

L

h++h+

(cos o 1.1)

(1- I) k I
12

J

max I

() as

I

'

As a consequence of these inequalities, we have etet4'q (a) do

IF(z,zo)I
+R

Ch

where

R (0 _ (2 -}- 1

12

-

cos

-+I

I

Amax " (a) ,

it aR (q)

0.

Finally, we, note that the last integral approaches zero as c - 0 because we can take the limit under the integral sign. Therefore, we conclude that none of the limiting values of IF (z, z0')I as c --+ 0 exceed the number R(,1). But since these limiting values are independent of 71 and since R(rl) ---+ 0 as 71 - 0, this means that F(z, zp) -, 0 as c -+ 0, that is, as z --+ zp along the curve 1. This proves the assertion and with it the theorem. §4. Cauchy's formula The results expounded in the preceding section enabled Privalov 1) to give a complete answer to the question of the representability of analytic functions in accordance with Cauchy's formula in the case of a simply connected domain bounded by an arbitrary rectifiable closed Jordan curve. Theorem 1. Let f (z) denote a function that is regular in a bounded domain B, the boundary of which is a closed rectifiable Jordan curve C. Suppose that f (z) has definite limiting values almost everywhere on C along nontangential paths. Suppose that these limiting values define a symmetric function f (z ') on C. Then for Cauchy's formula 1) Privalov [1919].

436

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR,

f(Z) = tai z

(z

d z'

( 1)

z

to hold in B, it is necessary and sufficient that S

C

f(zr)z,"di =0,

n=0, 1,...,

(2)

or, what amounts to the same thing, that outside B

z

(z'z

i

dz' z 0 .

(3)

Proof. Consider the function

(z) =f (z)'-

tai

zf

(z z dz' .

(4)

If the function f (z) can be represented in the domain B in accordance with Cauchy's formula (1), that is, if ,(z) = 0 in B, then Cauchy's integral 1

2ni

f -) dz' z'-z

(5)

has limiting values equal to f (z') along all nontangential paths in B almost everywhere on C. Conversely, if Cauchy's integral (5) has almost everywhere-on C limiting values equal to f (z') along all nontangential paths in B, then 0 (z) - 0 almost everywhere on C along all nontangential paths in B. Consequently, by the uniqueness theorem of § 2, V,(z) e 0 in B; that is, the function f (z) can be represented in B in accordance with Cauchy's formula (1). Thus, the function f (z) has a representation in B by.Cauchy's formula (1) if and

only if Cauchy's integral has almost everywhere on C limiting values equal to f (z') along all nontangential paths contained in B. But, by formula (3) of § 3 with the positive sign taken, this last will hold if and only if Cauchy's singular integral 1

2ni

f(z),dz' z'-zo

exists almost everywhere on C and is equal to f (zo)/2. This, in turn, is by formula (3) of §3 with the negative sign taken, equivalent to the requirement that Cauchy's integral (5), which represents a function that is regular outside C, has almost everywhere on C limiting values equal to zero along all nontangential

§4. CAUCHY'S FORMULA

-

437

paths lying outside C or, by virtue of the uniqueness theorem, equivalent to the requirement that equation (3) hold outside C. The expansion of the left-hand side of (3) about z = - in powers of liz shows that condition (3) is equivalent to the system of conditions (2). This completes the proof of the theorem. Theorem 2. Let f (z') denot e a summable function defined on a closed rectifiable Jordan curve C. Then, for there to exist a function f (z) that is regular in the finite domain B bounded by the curve C, that has limiting values almost everywhere on C equal to f (z') along nontangential paths, and that can be represented in B in accordance with Cauchy's formula (1), it is necessary and sufficient that the system of conditions (2) be satisfied. Proof. If the function f (z) mentioned in the theorem exists, then, by Theorem 1, the system of conditions (2) must be satisfied. Conversely, if the conditions (2) are satisfied, then outside B we have (3). Consequently, Cauchy's integral dZ

gal) z (z z

has almost everywhere on C limiting values equal to zero along all paths lying outside B. Therefore, from the theorem of §3, we conclude that Cauchy's singular integral exists almost everywhere on C and that

tai

f (z') dz' z' - zo

- f (zo) 2 '

(6)

If we now set

f (z) = tat

z (z

dz' z

in B, we conclude from the theorem of §3 that the function f (z) has almost everywhere on C limiting values among all nontangential paths lying outside B equal to 1

27ti

f (z')

z' - zo dz'

+f (zo) 2

or, in accordance with (6), equal to f (z a). Consequently, the function f (z) answers the requirements of the theorem. This completes the proof of the theorem.

438

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

§ 5. Classes of functions. Cauchy's formula Suppose, just as above, that B is the finite domain in the z-plane bounded by a rectifiable closed Jordan curve C. Let = X (z) denote a function that maps the domain B conformally onto the disk ICI < 1. Let z = a) (C) denote its inverse function. Let Cr denote circular images in the domain B, that is, curves in B corresponding to circle 1 I = r under the mapping z Let us denote by EP, where p > 0, the class of all functions f (z) J 0 that are regular in B and have the property that the integral

S If (z) I°ds cr

E

(1)

is bounded for 0 < r < 1 (V. I. Smirnov). By changing to the variable C in this

integral, ;e asil. see that the function f (z) belongs to the class EP if and only '(t;) E Hp. Thus, by § 4 of Chapter IX and § 1 of the present chapter, if the function f (z) belongs to EP, then f (z) has definite limiting values f (z ') almost everywhere on C, over all nontangential paths; If (z ')I is summable on C; and if f (co

limy S I}(z)Ipds` If(z')IPds. Cr

(2)

C

The class EP for p > p, can also be defined in this way, but it can also be defined independently of conformal mapping, namely, as the class of functions f (z) that are regular in B and have the property that, for each of them, there exists in B a sequence of rectifiable closed Jordan curves Cn, for n = 1, 2, , which converge to C and for which the integrals

S If(z)Ids C.

are bounded by a finite quantity independent of n. 1) Obviously, every function f (z) E p satisfies this condition. Let us prove the converse. Suppose that, for

a function f (z) that is regular in B, there exists a sequence of curves Cn with the property described. Let z = con(C )denote a function that maps the disk ICI < 1 conformally onto a finite domain bounded by the curve Cn in such a way arg w' (0). Then, by the theorem on the conthat con (p) = w (0) and arg vergence of univalent functions, the sequence of the functions con(C), n = 1, 29 , 1) Keldys and Lavrent rev [1937].

§5. CLASSES OF FUNCTIONS. CAUCHY'S FORMULA

439

converges uniformly in the interior of the disk ICI < 1 to the function w (C). Let

us set ,P (C) =f (w (C))p w' (C), T. (0 = f (w (C)) ° W. (C).

Since, for every n, the function f (wn(C)) is bounded in ICI < 1, it follows that On(:) E HP. Therefore, for 0
2x

If(z)I°dsA

-S I cpn(e)IPA 0

0

CA

where M is independent of r. But the sequence ( On (j)1 converges uniformly in the interior of the disk ICI < 1 to the function Consequently, for 0 < r < 1, we have 2x

M or S If(z)Ipds<_M,

S tcP(reie)Ipd0 0

Cr

which proves that the function f(z) belongs to the class Ep.

We use the notation adopted above in the following two theorems :

Theorem 1. If f(z) E E1, then Cauchy's theorem holds:

Sf(z) dz'=0.

(3)

C

This theorem follows from Theorem 5 of §4, Chapter IX if we apply the latter theorem to the function f(&) (C ))w' (j). Theorem 2. 1) Let f (z) denote a function that is regular in B. For this function to have almost everywhere on C definite limiting values along nontangential paths and for Cauchy's formula

f(z)-2l

z'(Z

dz'

(4)

to hold everywhere in B, it is necessary and sufficient that f (z) be a member of E1.

Proof. Let f (z) denote a member of E 1. Then, in accordance with what was said above, the function f (z) has almost everywhere on C definite limit values f (z `) along nontangential paths and these values are summable on C. Consider the integral 1) Smirnov [1932a].

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

44 0

2n

I-

f (w (ere)) w' (e19)

1

de's

(5)

We set

When we substitute this into (5), we obtain, remembering that f (rv(C))t,j' (C) 6 HIP 2%

1=f (w (C)) -{- 2L

f (w (ere)) w' (ere) R (ere, C) deie.

(6)

But, for fixed C in the disk ICI < 1, the function R(C', C) as a function of is regular in IC'I < 1 and continuous on 10 < I. Therefore, the function f (w(r;'))W'(C')R (C', C) is a member of H1 and consequently, by Cauchy's theorem, the integral on the right side of (6) is equal to zero. Thus, we have 2x 1

f (w (ell)) w' (ere) Me

2x1

w (ete) - w (b

-f (w (C))

If we now shift to the z-plane, we obtain Cauchy's formula (4). 1)

Now suppose, conversely, that the function f (z) has almost everywhere on C definite limiting values f(z') along nontangential paths, that these values are summable on C, and that Cauchy's formula (4) holds in B. Consider the integral

F_

1

2ni

We set 1

c

f (z') dz' X (z') - X (z) 1

X (z') - X (z) - X' (z) -W- z) + Q (z ' Z)'

Then, for Z E B,

1,_`X'()+2ni )f(x)Q(z, z)dz'.

(7)

But for fixed z 6 B, the function Q (z ', z) as a function of z' is regular in B and continuous on B. Consequently, in accordance with Walsh's theorem,2) it can be approximated uniformly in 1 by means of a polynomial to an arbitrary degree of accuracy. Since by Theorem 1 of §4, 1) The proof given has the advantage that it does not depend on §§ 3 and 4 and can be used to prove Cauchy's formula for domains with rectifiable boundaries other than Jordan curves. On the other hand, if we do use Theorem I of §4, and keep in mind the fact

that, by Theorem 1, fC(f (z')/ (z'_ z))dz'= 0 for all z outside B, we immediately obtain the validity of formula (4). 2) Walsh [1960], p. 47.

§6. ON THE EXTREMA OF MEAN MODULI

441

S f(x)z'ndz'=0, n=O 1, C

we conclude that the integral in the right-hand member of (7) is.equal to sequently, in B 1

2ai

f (z) dz'

zero. Con-

_ f (z)-

X (z') - X (z)

X' (z) '

If we now shift to the disk ICI < 1, we obtain 2

I

N-1

f (6) (eie)) u' (ei0) de"

a -C

=f (w (C)) (0' (C),

that is, Cauchy's formula holds for the function f (w (C))w '(c), which is regular in ICI < 1. Consequently, f(wC AI; that is, f (z) E EI. This completes the proof of the theorem.

§6. On the extrema of mean moduli We keep the notation used in § 5 regarding the domain B except that we now make the additional assumptions that the point z = 0 belongs to B, that 6j(0) = 0,

and that ca '(0) > 0. Let us show that, in this case, the function yin, Fp(z)= rX'(z) P>O, 1X' (0) )

(1)

and only this function minimizes the integral

I=SIf(z)J'ds

(2)

c

out of all functions F(z) belonging to the class EP that satisfy the condition F (0) = 1. We shall also show that this minimum is equal to 27rw '(0).

Shifting to the disk ICI < 1, we see that the problem of minimizing the integral (2) reduces to minimizing the integral 2n

2n

0

0

f IF(w(e'))w'(efe)I1PIPds=S I f(e")I'd9

(3)

with respect to the functions f ( j) = F (w (t;) I P belonging to the class Hp that satisfy the condition [(0) = w '(0)1" P. Now, the only possible extremal function for the integral (3) is a function f (C) without zeros in ICI < 1 because, otherwise, by dividing f (t;) by a Blaschke function b (C) we would get a function

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

442

f1(C) = b (0)f (C)/b (C) E HP for which f 1(0) = w'(0)11 P and for which the,, integral

(3) is equal to the same integral for the function f (C) except that it is mdltipliedby Ib (o)I P, where, obviously, Ib (0)I < 1. If we now consider the function' f (C) E HP such that f (0) = a)'(0)1/ P without zeros in ICI < 1 and if we set f (QP/z

n=0

C0 = Ui (0),

we have 2s z><

S

2s

I f (e!9) IPd6

0

cc

n m 2a l I f (re16)2/pj9d6 -'ym 1 icn 9rF > I c0 y 1

6

n=0

Here, equality can hold only when f() = c o P = cv '(0)1" P; that is, F(z) is given by formula (1). This was what we wished to show. Along with all the functions F W E EP, let us now consider the set E(') of polynomials qn(z) of degree not exceeding n such that qn(0) = 1 and let us pose the problem of minimizing the integral (2).

Let us show that extremal polynomials exist. Suppose that the integral (2) is equal to M for one of the polynomials qn(z) C Er"). Consider the set of all polynomials q(n)(z) C E(") for which the integral (2) does not exceed M. If qn(z) is one of these polynomials and b (C) is Blaschke's function for qn(w (J)), then f (z) _ (qn(z)/b(x(z)))P C E1. Therefore, in accordance with Cauchy's formula, we have

f (z) = 1 z'f (Z) -z dz'. Now, let us denote by 6 the distance between the point z and the curve C. For Iz - z'I > 8/2, we have

If(z)I c e If(z')Ids and, consequently,

9n (z) Ip C xb 5 I q,, (z') 1-0 ds


,

But the coefficients in the polynomial qn(z) do not exceed certain finite

§6. ON THE EXTREMA OF MEAN MODULT

443

quantities depending only on S and M. Therefore, if we apply to the coefficients that in these polynomials the condensation principle, we find a polynomial yields the smallest value for the integral (2). This proves the existence of extremal polynomials in E("). We denote these extremal polynomials by We now define the function 2x 2a

d (C) = e

1s

logy U1 lets)

°

eie+- do

(4)

which will play a significant role in all that follows. We have 2x 1

[

2x

Id(C)I=e

log l w' (eie) I P(r. a -'P) de (5)

°

where C= re`O. But, from Holder's. inequality, we have, for 0 < A < 1, Rx

w (eie)IaP(r, A - cp)A 2.

2%

c(S Iw (e1B)I P(r, A-'P)dO)X.(SP(r, 0

e-p)dA)1-1.

0

that is, 2x

21;

2a 1 I w (eie)

Ix

P (r, A - y) A

I cup (ete) I p

2

de a

If we now expand both sides in powers of A, subtract 1 from both sides, divide

by A, and then let A approach 0, we obtain the inequality 2x

2%

S log I

(eie) P (r, A - cp) dA <- log [2- ` I w (eie) I P (r, e - p) A 7C

0

On the basis of this inequality, we obtain from

(5)

2x

Id(C)Ic2 Iw'(e!e)IP(r, A-`p)dA, so that, for 0 < r < 1, 2x

2x

SId(rei°)Ido;SIw'(ei°)Ido. 0

0

This shows that d(C) C- H 1. Furthermore, it follows from (5) that

.

444

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR,

Id (e '19)1 = I(o'(e`B)1 almost everywhere on ICI = 1

It is obvious from (4) that the identity d(C) _ a)'(c) in ICI < 1 holds only when the function log Ia)'(0I, which is harmonic in ICI < 1, can be represented in that disk by Poisson's integral: 2x

logIw:(()I=2n

log I w (ei°)IP(r, 0-p)d0.

(6)

0

Condition (6) is satisfied, for example, when C is a smooth curve because then, by Theorem 5 of § 2, we have log tz)'(c) E H I and consequently, the function log a,'(%) itself can be represented in C1 < 1 by Poisson's integral. It follows from the proof of Theorem 5 of §2 that condition (6) is also satisfied in the more general case in which arg a)'(c) is bounded in 1. This will be the case, in particular, for starlike domains and also for domains bounded by piecewisesmooth curves that make nonzero interior angles with each other. 1) However, there exists an example of a rectifiable closed Jordan curve C for which condition (6) is not satisfied.2) We shall refer to Condition (6) in connection with the curve C as condition (S), following Smirnov.3) We note that condition (S) is equivalent to the condition 2x

logIw'(0)I=2a

log Iw (ete)Id6,

(6 ')

that is, to the condition d(o) _ ('(0). This is true because (6) is obtained from (6) with J= 0. To prove that (6') implies (6), we note that (6) can be written in the form of the limiting relationship 2x

2x

log I cu' (rei°) I dO = 2x C log I w (e'e) I d9. lim12n

(6 N )

On the other hand, the inequality log x < x'/p for p > 0 implies that the integral

1) Keldys and Lavrent'ev [1937] exhibited a number of other cases in which (6) is satisfied. 2) Keldys and Lavrent'ev [1937]. 3) Smirnov [ 1928].

.

§6. ON THE EXTREMA OF MEAN MODULI

445

2x + S (log I w (ret) 1)2 dO 0

is bounded for 0 < r < 1. Therefore, in accordance with the familiar test for taking the limit under the Lebesgue integral sign, we also have 2%

'im

2n

2x

log I w (reie) I d6 =

2a

C 1+0g I w (0) I d9.

(6

If we double this equation and subtract it from (6 "), we get 2x

Jim

2x

2n .) I log I w (Ce`e) I

IA=

I log w (el

I d2

Keeping this in mind and reasoning as in the proof of equation (6) of § 4 of Chapter IX, we obtain the limit relation 2x

urn S I log I w (reie) I - log I w' (eie) I I de = 0.

r-l0

Finally, from this we obtain equation (6) just as in the proof of Theorem 3 of §4, Chapter IX. We now have the

Theorem.') In the case of an arbitrary rectifiable closed Jordan curve C that encircles the point z = 0, we have imo S I P., p (z') j

R-

ds = 2-m d (0),

(7)

where d (o) is given by formula (4). Proof. Since the family of polynomials E(n ) becomes broader with increasing n, the integrals in (7) do not increase. Consequently the limit in (7) exists. Let bn,p(z) denote the Blaschke function for pn,p([v((;)) and suppose that

P.,p(w()) =b..p(C)h.,p(Q. Then, the equation bn,p(0)hn,p(0) = 1 implies that Ihn.p(0)I > 1. We have 2x

S IPn,p(z)Ipds=S

1) Keldys and Lavrent 'ev [1937].

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR'

446

2a

d(ei0)I dO

S Ihn, 0

2s

I hn/2 (e') d1/2 (ei°) S

I9 dt

2,m I hn.

0

p (0) I P d (0)

2,,d (0)

and, consequently,

l im n-00

It remains now to construct a polynomial q(z) E E(") for which the \alue of the integral (2) will be arbitrarily close to 217d(0). Since, by (5), log Id(i)I can be represented in ICI < 1 in accordance with Poisson's formula, we have, in particular, 2a

log d (0) = 2 S log I d (eie) I A, 0

Consequently, by setting g(g)

d((0))

we conclude that the function g(0) is summable in (0, 2n) and that 2n

S logg(0)de=0. 0

On the basis of this, the interval (0, 2n) contains a measurable set P on which the functions g (0) and 1/g(6) are bounded, whereas on its complement CP, 2n S I log g (8) I dO < log (1 + e), CP

2-

g (B) d9 < e , P

where c is a given positive number.

Let us denote by 0(0) the function that is equal to 1/g(9) on P and 1 on CP. Define 2a

2xp

S log y (e) e+ C e

-C

e

de

=e

de

- 2 cp S t°g S (e) eie

P

e

C

We have

1

9T) 1

almost everywhere on P, almost everywhere on CP.

§6. ON THE EXTREMA OF MEAN MODUCI

447

Now, 2n

(eye)IPId(ete)IdO= S

I

g(e)Id(e+B)Ide+

=

Id(e`B)Ide CP

d(0)Ide+Id(0)I 5g(O) A P

c 21rd (0) + 2ire d (0) = 2ad (0) (1 + e).

(8)

Furthermore, 2a

'I(0)=e

I 2xp

logy 0

(0) dB

=e

- Zip S 109 B (B)

dO

P

2a

=e

log 8 (b A pCP

Consequently, 1 (0) > 0 and (9)

But it follows from the theory of trigonometric series that corresponding to a bounded measurable function 1,(e'9) is a sequence of polynomials an (e`B) (Fejer's sums 1)) in e`B, uniformly bounded in (0, 2nd that converges almost

everywhere in (0, 270 to t,(e`9) with an(0) = i/i(0) Obviously, for these polynomials, 2a

2a

lim S Io, (em)IPId(e18)Id9=S

n-_oo 0

0

For sufficiently large n, we therefore have 2a

S Ia.(e")IPId(et0)IdO. 21rd(0)(1 +2e) 0

or

(10)

S Io, (X(z'))IPds2ad(0)(1 +2e) C

and furthermore, 1

Qn(0)>

I + 2f

Since the function an (Y (z)) is continuous on B it can, in accordance with 1) Specifically, if gr(e`e

2n00=0ene one then

o-,

(e"9) =

Yn-1 ((n k= 0

- k)/n)e `k 0

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

448

-B by a polynomial p(z) that also

Walsh's 1) theorem, be uniformly approximated

j

satisfies inequalities (10) and (11). 1ben,, P(0) = 1, we have

the polynomial P(z) = p(z)/p(0),

S I P(z)Ipds <2ad(0)(1 +2s)9=tad(0)(1 +a), C

where c' and a are arbitrarily small. This completes the formula of (7),

and hence

the theorem.

§7. Approximation in mean and the theory of orthogonal polynomials

As an application of the results of §6, let us now look at questions of approximation in the class E2- 2) Theorem 1. The sequence of the extremal polynomials pn 2(z) defined in §6 converges in mean on C to the function F2

that is,

(z)

d (x (z))

F2 (0)

(1)

1,

limoSIF1(z)-Pn.2(z)I9ds=0.

(2)

Proof. Shifting to the disk Ill < 1, we have a 21t

F1 (z) - P..9 (z) I2 ds

wet

Pn.9(

2A

= S IPn,s(w(ei°)) 1

( )- d) -d (O)

d(et)-

ie

j

I d (e)I d6'

d(0)Isd6

0 2n

=

S

I Pn, 9 (w

(e'e)) I21 d (e'e) I d6 + 2ad (0)

0

2n

- 25?

(

d (0) S Pn. 9 (w (e;e))

d (e;5 d6)

0

and, since

2((0 (C))

d (C) E H21 this is equal to 5

I Pn, s (z) I9 ds - 2ad (0).

1) Cf. Walsh [1960] p. 47 of 1965 ed. 2) Smirnov [19281 We note that the theorems that follow have been generalized by

Keldys and Lavrent'ev to the case of the class Ep.

§7. APPROXIMATION IN MEAN

449

Equation (2) then follows on the basis of the theorem in §6. This completes the proof of the theorem. Since F a (z)

d (x(0)(z))

=

= d (0) X (z) ,

almost everywhere on C, condition (S) is a necessary and sufficient condition for F'2(z) to be identically equal to F2(z), where F2(z) is defined by equation (1) of §6. Therefore, the preceding theorem leads to Theorem 2. For the limit relation nl m S I F2 (z) - Pn. 2 (z) l ds - 0

c.

(3)

to hold, it is necessary and sufficient that the curve C satisfy condition (S) (cf. p. 444). We turn now to approximation in mean. We shall say that the family of all

polynomials in z is complete in a domain B_if, for any function F(z) E E2 and any e > 0, there exists a polynomial P(z) such that 5

I F (z) - P (z) Is ds < e.

C

Theorem 3. For the family of all polynomials in z to be complete in a domain

B, it is necessary and sufficient that the curve C satisfy condition (S). Proof of the sufficiency. Suppose that condition (S) is satisfied. Then, in accordance with Theorem 2, we have equation (3). Now,' if F(z) E E2, then E H2 and, consequently, in accordance with Theorem 2 of f (C) = F (W (C)) W § 4 of Chapter IX, 2x

iii or

5

I f (ete) - f (rete) J2 d6 = 0

r lim S r-1 C IF (z) -f (rX (z))

I s ds -- 0.

This means that, for given e > 0, there exists an r in 0 < r < 1 such that

5IF(z)-f(rX(z)) C

X'(,&)I9ds<e.

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

450

But the function f(r)((z)) is continuous on B. Consequently, in accordance with Walsh's theorem, l) there exists a polynomial Q(z) such that, in B, If (rx (z)) X' (0) - Q (z) I < e. Let us form the polynomials Pn(z)= Q(Z)P". 2(Z)' We have

I F(z)-Pn(z)I$dsCs(IF(z)-f(rX(z))Y X ()I _C

C

p ..

+ If (rX (z)) (Y X (z)

, , 9 (z) "

iC (0)) I

+ I Pn. 2 (z) (f (rX (z)) Yx' (0) - Q (z)) I )gds. Of

Consequently, by applying Minkowski's inequality, we obtain

(IF(z)_Pn(z)I2ds )1/9 _(SIF(z)-f(rX(z)) VX7 (z) I'ds)

1/2

C

+ max If (rX (z)) I zFB

(S IF2(z)-Pn.g(z)I'ds)1/9 C

+ e \ CS I Pn, g (z) 11 ds)19 Here, the right-hand member is, for sufficiently. large n, less that 2f. Since e is an arbitrary positive number, this proves the sufficiency of the condition of the theorem.

Proof of the necessity. Suppose that the system of polynomials is complete for the domain B. Then, for the function f7(z) E E2, there exists a sequence of polynomials Pn(z) such that lim

(z) - Pn (z) I2 ds = O.

JI

n-00 C

(4)

If we now apply Minkowski's inequality twice, we have 5 I Pn (z) I' ds)1/' = (S I P. (z) C

(5 I x '(z) I ds)1I9

Ig d$)1'9 + (S I x '(Z) I ds)1'9 ,

C

C

X' (z) - Pn (z)12 ds

(5 I

(S I P,, (z) I2

which, in accordance with (4), yields lim 1) Cf. Walsh [1960], p. 47.

IPn(z)Igds=2Tr

C

(5)

§7. APPROXIMATION IN MEAN

451

On the other hand, we obtain from (4), by use of Cauchy's formula, the result that

litn P.(z)=

n-.oo

in B. 1) In particular, limn-+cPn(0)=

Therefore, if we set pn(z) = Pn(z)/Pn(0), pn(0) = 1, we have, in

accordance

with (5), n

i oo

S I P,, (z) I9 ds = 21rw (0).

In accordance with §6, this yields d(o) = &)'(0); that is, condition (S) is satisfied. This completes the proof of the theorem. The results that we have obtained are closely connected with the theory of orthogonal polynomials for a domain B. By a system of orthogonal polynomials, we mean here a sequence of polynomials for n = 0, 1, - , such that, for each n, K,,(z) is of degree n and

SKm(z)Kn(z)ds=

1,

if

01

if

m=n,

(6)

min.

Obviously, an arbitrary polynomial in z can be represented as a linear combination of the polynomials Now, if F(z) E E1, then the numbers

cn=SF(z)K,,(z)ds, n=0, 1, ...,

(7)

are called the Fourier coefficients of the functions F(z). Theorem 4. For the family of all polynomials in z to be complete in a domain B, it is necessary and sufficient that an arbitrary function F(z) in E2 satisfy the closedness condition

F(z)11ds= C

k=1

Ck

where the ck are the Fourier coefficients for F(z). Proof. Let p(z) denote an arbitrary nth-degree polynomial and let 1) See proof of Theorem 6.

(8)

use

452

X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR

represent it in the form P (z) _

k=0

r)

c Kk (z).

(8

For f(z)EE2,

SIF(x)-P(x)I9ds= SIF()12ds+SIP(z)I'ds C

- 2,9R

F (x) P (x) ds) = S I F (z) l i ds + C

- 2(Ek-0

n

CkCk)=SlF(z)I9ds- E ICkI9+

k=0

C

I ck I9

k=0

C

E

ICk-C;12.

k-0

It follows that the minimum for the integral (9)

I F (x) -P (x) I e dt C

yields a polynomial p(z) determined in accordance with formula (8) with ck = e k. Here, this minimum is equal to

JF(z)12ds-

IckI'.

k=0

(10)

For the system of all polynomials in z to be complete, it is obviously necessary and sufficient that the difference (10) approach 0 as n oo for an arbitrary function F(z) E E2, that is, it. is necessary and sufficient that equation (8) hold. This completes the proof of the theorem. By combining Theorem 4 with Theorem 3, we immediately obtain Theorem 5. For an arbitrary (unction f (z) E E2 to satisfy the closedness formula, it is necessary and sufficient that C satisfy condition (S). We now have

Theorem 6. When condition (S) is satisfied, an arbitrary function F(z) E E2 can be expanded in B in a Fourier series 00

F (x)

ckKk (z),

k=0

ck = S F (z) Kk (z) ds, C

§7 APPROXIMATION IN MEAN

453

that converges uniformly inside B.

Proof. Since the minimum of the integral (9) out of all polynomials in z yields a polynomial defined in accordance with formula (8) with ck = ck, for k = 0, 1, , n, it follows that, for these polynomials pn(z), lim

IF(z)-Pn(z)I9ds=0,

(12)

if. condition (S) is satisfied. Now, by Cauchy's formula, we have in B I

F (x) -Pn (z) = 2z1

F (z') -Pn (z') dz .

z' - z

If E is a closed set in the domain B and if S is the distance from E to C, then, for

I F (z) -Pn (z) I

I F (z') -Pn (z) I ds 2sg

F(z)-'Pn(z)I

C

From this.we conclude in accordance with (12) that pn(z) - F(z) uniformly on E. Since pn(z) is a segment of the Fourier series (11), this proves the theorem. In particular, if we apply Theorem 6 to the function f (z) = X '(z) and keep in mind the fact that then 2x

c,=S YT X (z)Kn(z) ds=SK.(w(ete))

dO

2rrKn (0) ylw (0),

we obtain

'Theorem 7. For curves C satisfying condition (S), we have in B the formula LJ Wk (0) Kk (z))

X'(z)=2a

k

(13)

2' 1K, (0) 1.2

k=0

where the convergence of the series in the numerator is uniform inside B.

CHAPTER XI SOME SUPPLEMENTARY INFORMATION

§ 1. Gluing theorems Gluing theorems establish the existence of analytic functions that obey certain relationships on the boundaries of certain domains. The nature of these

relationships is clear from the statements of the theorems we now present. Theorem 1. Suppose that a function x'= g(x) sets up a continuous one-to-one mapping of the interval 1: - 1 < x < 1 into itself, mapping each endpoint onto itself. Suppose also that, as a function of a complex argument, g (z) is regular and univalent in a sufficiently narrow circular tune with vertices at the points ± 1 and containing the interval 1. Under these conditions, there exist two functions w = f 1(z) and w = f2(z) that map the semidisks B 1: IzI < 1, (z) > 0, and B2: Izl < 1, Nz) < 0, respectively onto two disjoint domains G1 and G2 obtained from the disk JwJ < 1 by making it a smooth cut A that is analytic at all interior points. 1) Furthermore, the mappings are such that, on 1,

A (x) =fs (g (x)), that is, such that any two points x, g(x) E I that lie on the boundaries of the domains 81 and B2 are mapped into the same point X. Proof. It follows from the conditions of the theorem that the function z g(z) is regular and univalent in a sufficiently narrow segment 8(1) bounded by the interval l and the circular arc contained in ` (z) > 0 and that, on this segment, `(G(z)) > 0. We can take the segment B(1) in such a way that the values of the interior angles at the vertices are equal to n/2n, where n is an integer. Let B(2) denote the segment symmetric to B(1) about the x-axis (see Figure 17). In B11) we consider the function w = g1(z) = g(z), and in B12> the function w = 92W = z. For arbitrary x E 1, these functions map the domains B") and 1) That is, such that any interior arc of it is an analytic curve. 454

§ 1. GLUING THEOREMS

455

B(2) in such a way that the points x and g(x) in the interval l are mapped into the same point w of that interval (see Figure 18). Let us now map the univalent domain made up of the images G(1) and C(2) of the domains B(1) and B(2) under these mappings and of the interval l univalently onto the lune Bc' U B(2) U 1 in the w '-plane in such a way that the points ± 1 are mapped into themselves. Under this mapping, the domains G(1) and G(2) are mapped into domains adjacent to some smooth curve Al that is analytic at all interior points (see Figure 19). If a funcFigure 17 tion w'= f (w) provides this mapping, then the function w '= f (gk(z)) = gk +2(z), k = 1, 2, will provide mappings of the domains B(1) and 8(2) onto these same adjacent domains. In accordance with the symmetry principle, the functions w'= gk+2(z), for k = 1, 2, also map the

Figure 18

Figure 19

"double segments" B(3) and BW bounded by the interval I and the circular arcs inclined to l at the angle rr/2"-1 in an analogous manner. Let G'3) and G(4) denote the images under these mappings and then let us map the domain C(3) U G4 )U kl univalently onto the lune B(3) u B4 )U l in the w "-plane in such a way that the points ± 1 are mapped into themselves. Under this mapping, the domains G(3) and G(4) are mapped into domains that are tangent along some smooth curve A2 that is analytic at all interior points. By applying the symmetry principle n - 1 times as indicated, we obtain the mappings mentioned in the

XI. SOME SUPPLEMENTARY INFORMATION

456

theorem. This completes the proof of the theorem. Theorem 1 is a modification of a gluing theorem of Lavrent 'ev. 1) By using

Lavrent'ev's theory of quasiconformal mappings, Volkovyski%Z) succeeded in obtaining several other gluing theorems that play an important role in the theory of Riemann surfaces. In these theorems, the requirements imposed on functions of the type g(x) are weaker. Another procedure for obtaining gluing theorems is based on the use of the boundary properties of analytic functions that were obtained in Chapters IX and X. Specifically, let us prove Theorem 2.3) Let a and b denote two distinct points on the circle jzj and let yl denote one of the arcs connecting them. Let g(z) denote a function defined on yl that possesses the following properties: 1) g(z) is regular and has nonzero derivative at all interior points of the arc yl; 2) the function z '= g(z) maps the arc yl bijectively onto the complementary arc Y2 on the circle Izl = 1 and maps the endpoints into themselves. Under these conditions, there exists a function

W=F(z)= z +alz+ ...,

(1)

that is regular in the closed disk jzj < 1 except at the points 0, a, and b and that satisfies the equation4)

F(z)=F (g (z))

(2)

at all interior points of the arc yl. Proof. Consider the family IR of functions F(z) of the form

i

F(z)= +z+f(z) where f (z) E H2 and f (0) = 0 (with regard to the class H21 cf. § 4 of Chapter IX) that satisfy almost everywhere on yl the relationship f (F(g (z))) = at (F (z)).

(3)

1) Lavrent'ev [1935] 2) VolkovyskiT [1946] 3) Schaeffer and Spencer [1947] 4) In a monograph by the same authors [1950] the existence of a function F(z) that is univalent in Iz) < 1 is proved.

§ 1. GLUING THEOREMS

457

An example of such a function is the function

F(z)= _1_Z. We pose the extremal problem: Out of all functions F(z) E 9, find the one minimizing the integral (4)

IF =S iZ S If (z) I9 ds. 1

Let m denote the greatest lower bound of the integrals (4) in the family P. Let us show that, for some function F(z) E 9 , we have IF = m. Let us suppose the opposite. Then, there exists a sequence of functions F,,(z) = z-1 + z + for n = 1, 2, , such that IF. m as n - - . Let us define CD

fn (z) _ E Ckn1,Zk. k=1

Then, 00 (nl 9

(") 9

l-.1k=1

r 9k

n, v=1, 2, ...

Therefore, the functions fn(z) are uniformly bounded inside the disk Izl < 1 and

the condensation principle is applicable to them. Let } f,,k(z)} denote a subsequence of functions that converges uniformly inside the disk Izl < 1 to the function fo(z). Obviously, fo(z) E Hz. Furthermore, in accordance with Minkowski's inequality, we have, for

0
S

I

f0

(e'2 -fn (e ,B) 12 de)i/9

0

Ifo (rei°) -f,, (re'°

`

(

+1/2

j

9 dB

5 1A (efl) _ f0 (re') j9 A)

1/2

n

2s

) /' --' (S If (re1B) -fn (e,a) I9 d6)1/°. n

(5)

For given c > 0, we choose r in the interval 0 < r < 1 in such a way that the first and third integrals of the right side of inequality (5) are less than e. Then, we can choose N > 0 in such a way that the second integral is, for n > N, less than e. Consequently, for n > N, the integral on the left side of inequality (5) is less than 9c. This means that the sequence of values of this integral converges

458

XI. SOME SUPPLEMENTARY INFORMATION

to zero as n --,

Then, as we know, the sequence of values of converges almost everywhere on I zI = 1 to f0(e`0). But equation (3) holds almost everywhere on yl for the functions Consequently, this will also be true for F0(z) = z-1 + z + fo(z). This means that the function F0(z) E 11. Furthermore, since IsSS

r

Ifo(z)'ds= nJim

li.

f

z

for 0 < r < 1, it follows that IF _< m, which means that IF = m. This proves 0 0 that our extremal problem does indeed have a solution. Let F0(z) denote any of

the extremal functions. Let X(eiO) denote an arbitrary real function of 0 that is twice differentiable on the arc yl and is constant in a sufficiently small neighborhood of each of the points a and b. Let us define X(e`B) on the arc y2 by the equation

(5)

x (g (z)) = X (z).

Let us now denote by t(z) a function that vanishes at z = 0, is regular in the disk jzl < 1, and has on Izi = 1 a real part equal to X(z). This function is constructed in accordance with the Schwarz formula, on the basis of which, when we set 00

(z) = E akzk,

(6)

ka0

we have

2a

ak = 2a

X (e')

a-k,e d9,

k=

1, 2,

... .

If we integrate the right-hand member by parts twice, we see that jak 1 < M/k 2, for k = 1, 2, . Here M is finite and independent of k. This shows that the function (6) is continuous on the closed disk 1z I < 1, and by (5) satisfies on yl the relation -

a' ('P (g (z))) _ OR ('P (z))

Let us set IF(z) _ (z) - a 0 and let us define F* (z) = Fo (z) + e(D (z).

(7)

1. GLUING THEOREMS

459

Obviously, this function belongs to the family 2 for arbitrary real C. We have IFS = IFe -}- 2 e

T }o (z) ' (z)) ds + s9

IZI<1

11

IzI<1

10' (z) )s ds.

By virtue of the extremality of F0 and the arbitrariness of e, this leads to the condition

J1 MVo(z)V(z))ds=0.

(8)

Izl<1

But, if we set fo(z) = u o + iv o and IF (z) = u + iv, we have, by Green's formula and the Cauchy-Riemann conditions, for 0 < r < 1,

v0d1=Iz °r

Liu voaXdx+zoddy ay7

S

IzI mr du duo')

_S S ` Izl
ax

ax ax 1

ds =

JJ I'I
Co (z) ' (z)) ds.

Therefore, condition (8) can be written in the form lim

vo (z) du (z) = 0

and also in the form lim

vo (rele) de u (rei°) dO= 0.

r-I

From Poisson's formula, we see (integrating by.parts) that the function c9u(re`B)/aO is bounded in Izl < 1 and the integral a,c

vo (rel) 19 dB 0

is bounded in 0 < r < 1. Therefore, in equation (9), we can take the limit under the integral sign as r - 1. When we do this, we obtain Za

vo(e'e)WeX(e')

A=0,

that is, vo (ele) dX (ere) = 0. 1

0

By using equations (2) and (7), we see that, if a = e`a and b = e`,8, then

460

XI. SOME SUPPLEMENTARY INFORMATION S (vo (et°) - vo (B (e °B)) dX

(e') = 0.

(10)

If we now set

V(ere)=S(vo(e°)-vo(B(ei))dA and integrate by parts, we obtain

V (e

(11)

d9, X (e-4) dA = 0.

This equation is established for an arbitrary function X(0), that satisfies the conditions mentioned above.

Now let Xl(e`B) denote an arbitrary function defined on yl that has a continuous.first derivative with respect to 0 and vanishes in sufficiently small neighborhoods of the points a and b. Then the function X (e') = S X, (eis) do

satisfies the conditions mentioned above. Therefore, equation (11) yields V(eA)doXt(eb0)d9=0.

We now have the conditions of a well-known lemma of du Bois-Reymond in the

calculus of variations. Therefore, we conclude that V (e`B) = const on YI It then follows from (10) that vo (g (z)) = vo (z), and hence

2 (F0 (g, (z))) = 2 (F0 (z))

almost everywhere on yl. This, together with the relationship T (Fo (g (z))) = OR (Fe (z)),

implies the validity of the relation Fo (9(z)) = Fo (z)

(12)

almost everywhere on yI.

Now, let zo denote an interior point of the arc yl. The function w = g(z) is

§2. MAPPING OF SIMPLY CONNECTED RIEMANN SdRFACES

461

univalent in a sufficiently small neighborhood U. o of the point z 0 and it maps U20 into some neighborhood Uw0 of the point wo = g(zo) E Y2. Here, the subset UZ of the neighborhood UZ0 that is contained in the domain jzj > 1 is mapped 0 into the subset U,,,o of the neighborhood U,,,0 contained in the disk IwI < 1. Thus the function Fo(g (z )) is regular in UZ 0 Consequently, the neighborhood UZp is partitioned by the circle in I = 1 into two subsets, in one of which the function F0(z) is regular, and in the other the function Fo(g (z)) is regular. These two functions are equal almost everywhere on the common arc y of the circle. Furthermore, the values of these functions on the boundaries of the corresponding domains or on the boundaries of suitable subdomains are such that their absolute values are square-integrable. But then, Cauchy's formula and theorem can be applied. When we use these, we can show by the usual line of reasoning that each of the functions F0(z) and Fo(g (z)) is an analytic continuation of the other across the arc y. In particular, these functions are regular and equal to each other on the arc y itself. Since zo is arbitrary, we conclude that the function Fo(z) is regular at all interior points of the arc yI and that equation (2) holds for it everywhere on yI. This completes the proof of the theorem. §2. Conformal mapping of simply connected Riemann surfaces Let us now consider the question of the conformal mapping of simply connected Riemanp surfaces onto a one-sheeted disk. In the constructive definition

of Riemann surfaces, we take for the simplest plane figures not circles, as is sometimes done, but triangles, including curvilinear triangles. Specifically, suppose that there are over the w-plane and parallel to it a . By "triangle" we mean here finite or countable set of triangles A I, A2, either a finite domain bounded by a closed Jordan curve consisting of three analytic arcs that meet each other at nonzero angles or a domain obtained from such a domain by inversion about one of the "vertices". 1) We shall refer to the three boundary arcs as sides and we shall refer to their endpoints as the vertices of the triangle. These triangles are connected, i.e. glued, along the sides in such a way that, for each k the triangle dk is glued along some or all of its sides to another triangle in the set which has no other points in common with Ak. The triangles together may overlap but their points lying over the same point in the w-plane 1) The difference between the two cases mentioned in the definition of triangles disappears if we shift to the Riemann sphere.

XI. SOME SUPPLEMENTARY INFORMATION

462

are treated as different points if they are not identified under the gluing described above. We shall denote the points of the triangles by the same letters as the points in the w-plane over which they lie. Consider a Riemann surface F consisting of (1) the interior points of all the triangles Ak, (2) the interior points of all those sides of each triangle Ak that are also sides of other triangles adjacent to Ak, (3) the vertices of the triangles Ak in which a finite number of triangles Ak adjacent to each other are connected in circular order. We note that the sum of the angles of the triangles at each vertex is a multiple of 2n. If this sum exceeds 27r, the vertex is called a branch point of the Riemann surface. A characteristic property of connectedness for surfaces is assumed satisfied for Riemann surfaces, namely, if any two points lying on F lie in the triangles Ak and Ak', F contains a sequence of triangles Ak1' , Akn, each adjacent to the next, such that Ak 1 = Ak, and Akn = Ak . Ak 2, Let us suppose that we can assign to the surface F a one-sheeted simply connected domain B consisting of triangles A, of the same kind. Suppose also that there exists between the triangles Ak and Ak a one-to-one correspondence that always assigns to two adjacent triangles on F two adjacent triangles in B. In this case, the Riemann surface is said to be simply connected. With regard to simply connected Riemann surfaces, we have Poincare's theorem, which constitutes the subject of the present section: Theorem 1. An arbitrary simply connected Riemann surface F is in one-toone correspondence with a disk Iz I < r, where 0 < r < w, or the entire z-plane, where the correspondence is defined on the disk or plane by some meromorphic function.

1)

Proof. Suppose that the triangles Ak in the composition of the surface F are in the relationship described above with the triangles ok constituting a onesheeted simply connected domain B. Let us construct in B a finite or countable set of simply connected domains B,,, for n = 1, 2, , such that B C Bn + 1, for n > 1, is obtained from Bn either by adjoining to Bn one of the triangles A, 1) If r < m, F is called a surface of the hyperbolic type: if r = m, it is called a surface of the parabolic type; in the case of the entire plane, it is called a surface of the elliptic type. (In the last case, the meromorphic function in question is obviously a a rational fraction.) A simply connected Riemann surface F is always of a definite type. This is a simple consequence of Liouville's theorem.

§2. MAPPING OF SIMPLY CONNECTED RIEMANN SURFACES

463

that do not appear in Bn and one side of Ok or by adjoining to B. the common side not belonging to Bn of the two triangles appearing in B. Furthermore, let us assume that, in the case of a finite number of domains Bn, this last coincides with B. In the case of an infinite number of domains B, these constitute a sequence exhausting the domain B. Obviously, such a construction can always be done beginning with B 1 = A, . The triangles Ak corresponding to the triangles Ok appearing in B (for n = 1, 2, . ) constitute a simply connected Riemann surface F contained in Fn + 1, and an arbitrary point of the surface F lies in Fn beginning with some n. Obviously, the surface F 1 = A 1 is the image of some disk Iz I < r 1, where r 1 < DO, under a suitable univalent function w = f 1(z) such that f 1(0) is a point (fixed) in the interior of the triangle Al and f'(0) = 1. Let us now suppose that, for some n, the surface Fn is in one-to-one correspondence with a disk Iz I < rn, where rn < oo, the correspondence being defined by a meromorphic function w = fn(z) such that fn(o) = 0 and fn'(0) = 1. Let us show that this will also be true for Fn + 1 (with the disk in question possibly replaced with the domain Iz I < or the entire z-plane). Here, we need to consider two cases: 1. Suppose that the surface Fn + 1 is obtained by adjoining to Fn the triangle Ak along one of its sides 1. In such a case, Fn and Ak can be regarded as the single-valued images of the semidisks B 1: Iz I < 1, N z) > 0 and B2: Iz I < 1, 'Az) < 0 under suitable meromorphic functions w = g1(z) (see Figures 20 and 21) G

Figure 20 Figure 21 Figure 22 and w = c2(z) in such a way that both mappings map the arc 1 common boundary diameter. Obviously, the function g(x) = 024-1 1W), satisfies the conditions of Theorem 1 of § 1. On the basis of this theorem, there exist functions z '= Vi 1(z) and z '= 02(z) that map B 1 and B2 univalently onto two complemen-

tary domains G1 and G2 of the disk Iz 'I < 1 (see Figure 22) in such a way that the points x and g(x) (for - 1 < x < 1) on the boundaries of B 1 and B 2 are mapped into the same point of the disk Iz 'I < 1. But then the functions

XI. SOME SUPPLEMENTARY INFORMATION

464

w = 01(0 i 1(z)) and w = map the domains G1 and G2 onto Fn and Ak respectively in such a way that the common portions of the boundaries of these functions coincide. In accordance with Riemann's theorem on analytic continua-

tion, we conclude that both these functions define a meromorphic function w = F(z) on the disk IzI < 1 that maps that disk bijectively onto the surface F + 1. By a suitable normalization, we easily obtain the function w + 1(z), fn + 1(0) wo, f + 1(0) = 1, which yields such a mapping of the disk IzI < r + 1, where rn+1 < oo on Fn+1.

2. Let us now consider the case when the surface to Fn, of two triangles Ak and

F

Ak' contained in Fn. If a and b are the endpoints of the arc 1 and a lies in Fn + 1' we remove from the surface Fn a sufficiently small one-sheeted or multiple-sheeted disk K: Iw - al < E constituting a neighborhood of the point a on Fn + 1 but cut along 1. The surface Fn with points belonging to K deleted from it and the disk K, with a sufficiently small arc VC 1 with endpoint at a adjoined can obviously be regarded as the single-valued images of the semidisks B1: IzI < 1, %z) > 0 and B2: IzI < 1, %z) < 0 (see Figure 23) under meromorphic

Figure 23 functions. Here, both functions map the common boundary diameter onto the portions of the boundaries lying over the circle Iw - al = E. Therefore, reasoning

as in case 1, we can show that the surface Fn U 1' is the image of a finite disk under a suitable meromorphic function. When we note that the surface Fn U 1' is obtained from Fn by shortening of the cut 1, we continue this shortening process indefinitely so that as a result we exhaust the entire cut 1. If we now use the

§2. MAPPING OF SIMPLY CONNECTED RIEMANN SURFACES

convergence of the corresponding normalized functions w =

465

where

which was proved above for the surfaces F , we conclude that the surface F,, + 1 is the image themselves, where n = 1, 2, of the disk Izi
wo and 0v'(0)=1 for v = 1, 2,

,

z-plane, the correspondence being defined by a suitable meromorphic function. On the other hand, if the number of surfaces is infinite, we shall have a sequence of functions f,,(z) and we need to investigate what happens when we take the limit.as n --. oo.

To do this, consider the functions

C = n (z) =fn+ 1 (fn (z)), in (0) = 0,

4n (0) =1, n =1, 2,, .. .

The function C= is regular in the disk IzI < r,, and the values that it assumes in that disk all lie in the disk ICI < r + 1. By the Schwarz lemma, we have ,b'(0) = 1 < r + 1/r,a. This means that r does not decrease with increasing

n. Therefore, the sequence Irj approaches a finite or infinite limit r as n We now set

(Pn. , (z) =fn' (f, (z)) = z + ... ,

v < n, n ^ 1, 2,

We see that, for every fixed v (v = 1, 2, - ), the functions c,,, (z), n = 1, 2, - - , are regular and univalent in IzI
function

Here, we may assume that the sequence of the numbers n k > is contained in the sequence of the numbers n(k1) (with v = 1). Since. 1(z)) ow v) (0y. 1(z)) = 0n (V) 1 (z) in the disk IzI < r l, so that k

it. follows that

k

.01(fl 1(w)) in A1.

This shows that the function w = f (0v 1(C)), inverse to C= 0 (f v 1(w)) is independent of v. Here, if C is the image of the disk zI < r,, under the mapthen the function ping L= I

XI. SOME SUPPLEMENTARY INFORMATION

466

00

w=

onto the surface F1, It follows that G. C Gv+19 maps the one-sheeted domain v = 1, 2, . It remains to show that the limiting domain G of the sequence of , coincides with the entire open disk ICI < r. the domains G, v= 1, 2, Suppose that r = oc. Since the image Gv of the disk Iz I < rv under the map-

ping '_ (hv(z) is contained in the disk ICI < r,/4 for all v = 1, 2,

(by

Koebe's theorem) the domain G contains in this case an arbitrary finite disk and consequently must coincide with the entire finite plane. Now, suppose that r is finite. Then, we have Icv(z)I < r in the disk Iz I < rv. Consequently, by the Schwarz lemma, we have in Iz I < rv

I(P,(z)I`TIzI Let us now take arbitrary p, P11 P21 0 < P < P1 < 02 < r. Let v denote a number such that P2 < rv < r. If we set co ( akv)Zk,

21 k-2

we have

0 ZR

-)

4

I py (P9e`e) 19 dO = P8 + I I a" 19 P2k k.=2

lbj

T9 Pg. Y

Consequently,

j w

L.i

(Y) I

sk

I9 P2 c

r- - ry - P9

ra

k=2

Therefore,

Iak(Y)

_rs

-

rY

1

Pik=2, 3, ... .

rY

Using this inequality, we have on the circle IzI = p1,

, IakY)IPi=Pt- 'I

00

Ip,(zW I.Pi-

k.7

00

k=2

r9

r,-

rY

k

P` P9-I

-Pt

rt-rY9

r,

Pi

P9-P1'

3. EXTREMUM FOR BOUNDED FUNCTION

467

that is,

v(z)Pi-

T' -Tv rv

Pi

Pa - PL

Since for sufficiently large v the right-hand member of this inequality is greater than p, it follows that, for such values of v, the domain G,,, being the image of the disk Iz(
Thus, in the case of an infinite number of triangles Ak, the surface F is the single-valued image of the disk ICI < r, where r < oc, under the meromorphic function w = qi (C). This completes the proof of the theorem. Poincare"s theorem assigns to each simply connected Riemann surface F a function that is meromorphic in some disk (or on the entire plane) that maps that disk bijectively onto F.

It is easy to prove the converse, namely, that to every function f (z) that is meromorphic in the disk Iz I < r, where r < oc, can be assigned a simply connected Riemann surface onto which the function f (z) maps the disk Iz I < r bijectively. Specifically, we partition the disk Iz I < r into small triangles, in each of which

the function f (z) is univalent, and consider the surface F composed of the images of these triangles. Thus, simply connected Riemann surfaces can be regarded as the geometric equivalents for functions that are meromorphic in a disk (or on the entire-plane). This conclusion is completely analogous to the conclusion that follows from

univalent functions when we use the existence theorems established in Chapters II and V. §3. An extremum for bounded functions in multiply connected domains

In § 2 of Chapter II and ' 1 of Chapter VI, we considered questions dealing with an extremum of the quantity f'(0) in certain classes of analytic functions defined in a given domain B. This led to important existence theorems dealing with mappings of the domain B onto a disk. We now answer one of those questions.

468

XI. SOME SUPPLEMENTARY INFORMATION

Theorem 1.1) Let IR denote the family of all functions f (z) that are regular in a given n-connected domain B with nondegenerate boundary continua, that are bounded by unity in modulus in B and that satisfy the condition f (a) = 0 at a given finite point a E B. Then the quantity I f '(a)I is maximized by a function mapping the domain B onto the complete n-sheeted disk (see §5 of Chapter VI). Proof. Without loss of generality, we may assume that B is a finite domain bounded by n closed analytic Jordan curves K1, , K,,: K = U k=1Kk. That the extremal problem posed in the theorem has a solution is obvious. Let f o(z) denote one of the extremal functions. Since I f U (a)1 > 0, the point z = a is a simple zero of the function f o(z). Suppose that the function fo(z) has, in addition to a, zeros at the points al, , z,,, in the domain B. (Of course, it is not assumed that these are all the zeros of the function fo(z) in B.) Consider the Green's function g(z, C) for the domain B and the harmonic function h(z, C) associated with it. In addition to fo(z), consider the varied func-

tion f1(z) E 9, defined by 109 1fi (Z) 1=109 I.fo(z) I +u(z),

(1)

where m

U

(z)

k=1

(s (g (z, zk) - g (z, zk)),

where, in turn, zk E B for k= 1, . , m. Obviously, If 1(z)I < 1 in B. Also, for f 1(z) to be single-valued in B, it is necessary that the conjugate function m

v (z) = E (h (z, zp) - h (z, zk)) A-1 be single-valued.

Since, in accordance with (10") of §6 of Chapter VI, (h (z, zk))Kl = - 2Ttiw1(zk), (h (z, zk))Kl = - 2nlwl (zk), l=1, ... , n,

where wl(z) is the harmonic measure of the curve K1 with respect to the domain

B, it follows that single-valuedness of v(z) in B leads to the n conditions on 1) Ahlfors [1947].

§3. EXTREMUM FOR BOUNDED FUNCTIONS

469

the zk: m

F1 (z;, ... , Zm) _ E1 (w1 (zk) -- W 1(Zk)) = 0, 1- 1, ... , n.

Since 17j 1W1(z) = 1 so that the function f (z i ,

(2)

.. , z;n) can be expressed as a

linear combination of the functions Fk(z , , z,F,), k = 1, , n - 1, the nth of conditions (2) is a consequence of the remaining conditions and we can drop it. When conditions (2) are satisfied, we have by virtue of the extremality of the 10

function f 0(z) log If, (a) I

-

m

log I.fo (a) I

= E (g(a, zk) - g(a, zk) k-1

k-1

(g (zk, a) - g (zk, a)) - 0.

This leads to the conclusion that the function m

Fo (zi,

where the zk belong to B, k = 1,

..., zm)= E g(zi,, a),

ks1 , m, and satisfy conditions (2) for

, n - 1, is minimized when zk = zk for k = 1, the matrix (zk = xk + ix,, +k, zk = xk + ixm +k) 1,

OF,

dx,,ll,

(3) 1=

, m. But then, the rank of

k=1, ..., 2m, 1=0, 1, ..., n-1,

(4)

with xk = xk, k = 1, , 2m, does not exceed n - 1. This is true because if the rank of this matrix were equal to n (which can be the case only when 2m > n - 1), then the functions 61 = F1(z i , , for 1= 0, , n - 1, would map a neighborhood of the point (x1, , xvm) in 2m-dimensional Euclidean space onto an open set in n-dimensional space containing the point (0, 0, , 0). The intersection of this set with the S0-axis (61 = 62 = . '= 6n-1 = 0) is an open segment containing the point (0, 0, , 0). Consequently, the function F0(z 1 , ... , z,) would not be minimized when z k = zk, k = 1, , m. From what we have shown, we conclude that there exist real numbers A0, X1, , an-1, not all equal to 0, such that the 2m equations hold:

as =0, k=1, ..., 2m. kt

(5)

470

XI. SOME SUPPLEMENTARY INFORMATION

Keeping (2) and (3) in mind, we conclude from (5) that, for the values Ak mentioned above, the partial derivatives with respect to x and y (where z = x + iy) of the function

- Agg(z, a)

vanish at z = zk, for k = r,

ill (z) T ... -1-

(z)

, m. This function is obviously not a constant

since the Ak are not all zero. Let us define analytic functions p(z, C) and wk(z), k = 1,

, n, by

8 (z, C), = at (P (z, C)), wk (z) _ M (Wk (z))

(see §6, Chapter VI) and let us define n-1 w (.z) _ - Aop (z, a) + 1=1

A1w1(z).

From the latter equation, we conclude that the derivative w '(z) is not identically zero but that it has zeros at all the points z 1, , Z M. Since the real part of the function w(z) is constant on each boundary curve K1, for l = 1, , n, in the case in which w'(z) has no zeros on K, we obtain the result n

(arg

(z))K

t=1

(arg

(z))K,

n

[(arg

)K1- (arg

W

--

_

2a (n - 2),

where ds is the element of arc length on the boundary K. We note that the.function w' (z) is regular at a when A0 = 0 and we note that w '(z) has a simple pole z = a in B when A0 0. Therefore, we conclude from Cauchy's theorem on the number of zeros and poles that the number of zeros of the function w'(z) in B distinct from z = a does not exceed n - 2 if A0 = 0 and is exactly equal to n - 1 if A0 0. This result will also hold when w ' (z) has zeros on K, as follows from the same theorem of Cauchy but this time with consideration of the zeros on

§3. EXTREMUM FOR BOUNDED FUNCTIONS

471

the boundary. Here, the zeros on K are counted with half multiplicity. I) This result leads to the first important conclusion regarding the extremal function fp(z). The zeros of the function fo(z) in the domain B are also zeros of the function w ' W. The number of these zeros does not exceed n - 1. If there are exactly n - I of them, then A0 i 0.2) , zm, m < n - 1, constitute all the zeros of the Now suppose that z 1, function fo(z) inside B. Then, the function

U(z)=log Ifo (z) I +g(z, a) +

k-l

g(z, Zk)

(6)

is harmonic in B and U(z) < 0 in B. Since this last inequality is satisfied in the part of the domain B contained in a small neighborhood of an arbitrary point of the boundary K, if we shift to the disk by means.of a univalent mapping and use the boundary properties of nonnegative harmonic functions in a disk (see §2 of Chapter IX), we conclude that U(z) has definite limiting values almost everywhere on K along nontangential paths and these values define a summable function on K. Now, let f2(z) denote the varied function obtained from the relationship log I fs (z) I =log I fo (z) I + u (z),

(7)

where u(z) is a harmonic function in B. To make f2(z) single-valued in B, we require that the conjugate v(z) to u(z) be single-valued in B. For If2(z)I not to exceed 1 in B, it will be sufficient to require that u(z) be bounded in B and that U(z) + u(z) be nonpositive almost everywhere on K. This is true because, in the first place, by virtue of (6), and (7), m

If2(z)I=e U(z)+u (z) e

-B (z, a) - E 8 (z. Zk) k=1

I.fo (z) I

(8)

in B. Furthermore, under the assumptions made, u(z) also has definite limiting 1) For a direct derivation of this theorem for the function w '(z), consider neighborhoods not only of its zeros lying in B but also of its zeros lying on K. Then Cauchy's theorem can be applied to the logarithmic derivative of w'(z). If we then let the neighborhoods of the zeros shrink to a point, the difference between the number of zeros and the number of poles of the function w '(z) in B will be equal to (argw'(z))K. This last expression means the sum of the increments in argw '(z) over the set of arcs constituting K and not encircling the zeros of w '(z). 2) (A more complete study of the zeros of the function fo(z) has been made by Garabedian [19491)

472

XI. SOME SUPPLEMENTARY INFORMATION

values almost everywhere on K. From Green's formula

0(z)+u(z)-2a J

z)ds,

where K. is the set of curves g(z ', z) = e (where a is a sufficiently small positive number), we obtain Green's formula for the domain B 2a J

U(z)+u(z)-

(U(z) 'u(z))

dg (z', z) ds

(9)

by taking the limit as c --p 0. (As always, the normal n is directed inside the domain B.) It follows that U(z) = u(z) < 0 in B. Therefore, in accordance with (8), Ifz(z)I < I in B. Thus, f2(z) E IR. If we rewrite (7) in the form log ff (z)

z-a

I

= log

fo (z)

I z-a

I + u (z),

and let z approach a, we obtain log'I f9 (a) I = log I fo (a) I + u (a).

Then, by virtue of the extremaliry of the function f o(z), we have u(a) < 0. By using Green's formula (as derived above), we see that this leads to the inequality

u(a)= 2n c u(z)

dg (z', a

(10)

Furthermore, if we denote by KE.1,1 a curve in B that is close to Kt and defined by the equation i1(z) = 1 - e, where a is a sufficiently small positive number, and if we denote by KE,I,k, where k 1, a curve in B close to Kk and defined by the equation &)I(z) = e with the same e, then the single-valuedness of v(z) in B can be expressed in the form of n - 1 equations (for sufficiently small e): (v (Z)) K,.

dv (z) _ - S

1, 1= S K., 1,1

k

K 1.l (m1(z)-e)

s

I

Kr11, k

do

do

ds = -

ds+2a

Td u

S

(w1 (z) + e)

K+ 1,1

S

duds=

n

ds

§3. EXTREMUM FOR BOUNDED FUNCTIONS-

u

S

n

(z) !L do z) ds + 2e

do

s

473

ds = 0,

K,0 1, 1

UK,

1, k

kal

1=1, ..., n-1. Here, Fo is a fixed small positive number. This is equivalent to satisfaction of the n - 1 limiting (as E -- 0) equations

u(z)d!-

ds=0, I=i, ..., n-'1.

Turning now to the function (6), let us suppose that, for some q > 0, there exists a subset E of K of positive measure such that U(z) < - q for z E E. Then, for u(z) we may -take the function defined by the boundary values

U (z)

-

(0 on K outside E n

(

I

s ll- °

dg (z, a)

d'k`

do + k=1 k^ do J

on E,

where the Ak, for k = 1, ... , n - 1, are constants and F is a constant of either sign but sufficiently small absolute value. (To construct u(z), we need only shift from the domain B to the disk ICI < 1, as was done in § 3 of Chapter VI, then use Poisson's formula to construct a harmonic function on that disk from the new boundary values, and finally return to the domain B with it.) For this function u(z), conditions (11) take the forms n-1 dg dm1

do do ds +

kk

k=1

c duk 1

do . do ds = 0'

(12)

CCC

1=1,...,n-1, and, by virtue of the choice of the Ak, for k = 0, 1, , n - 1, they must be satisfied. But conditions (12) constitute a system of n - 1 homogeneous linear equations with n unknowns Ak, k = 0, 1, , n - 1. Therefore, this system has nonzero solutions. Let us now denote by Ak, k 0,...,n - 1, any nonzero system of solutions and let us construct from them the function u(z) in (10). We then obtain an inequality in which the left-hand member includes the factor c, which maybe of either sign. This leads to the condition

474

XI. SOME SUPPLEMENTARY INFORMATION

'Xo)

n-1

(dgdn

a))2ds+

I

dwk do

a

k=l

dg (z a) do

ds = 0.

(13)

If we multiply equation (13) by -A0 and equations (12) by Al (for each 1 = , n - 1) and then add all the resulting equations, we obtain 1,

(_X6 11, a)

+X1

do

d&)

+...+kn_t

do

dwn_y

do

) ds=0.

It follows that, almost everywhere on E, nom)-11

- o do + Lj Aff

Xk

k-1

do = 0.

Since, in addition, everywhere on K we have

-0+

n-I

d

dw

k_1

0,

kk dSk

we conclude that the function n

w'(z)

AoP1 (z,

a) + 2] Akwk (z), k-1

which is regular on K, vanishes on E. But then w '(z) __ 0 in B and hence the function

- kog (z, a) +

Akwk (z)

k-l

is constant in B. However, this cannot be the case because of the choice of the , n. This contradiction shows that U(z) = 0 almost everyAk for k = 0, 1, where on K. But then, if we apply Privalov's uniqueness theorem (cf. §2 of Chapter X) to a suitable subset of the domain B, we conclude that U(z) - 0 in B; that is, in B we have m

IogI/o(z)I =-g (z, a)- Z g(z, zk) k-1

and

m

I p (z, z ) ffe (z) = e

k

I

§3. EXTREMUM FOR BOUNDED FUNCTIONS

475

Now, the conditions under which f0(z) is single-valued in B take the form 2a

(argfo (z))K, = wt (a) + I wl (zk) = V1,

l =1, ... ,

(14)

n,

k=1

where the vi are integers. Obviously, vl > 1, for 1 = 1, over l = 1, . . . , n, we obtain

m+1=

n

r= l

, n. If we sum (14)

vl:n.

But m
- , n.

Summarizing what we have said, we conclude that the function f0(z) has

exactly n zeros in the domain B and that it is equal to unity in modulus on K. Then, by evaluating (arg (f 0(z) - w0))K, we prove that the function f 0(z) - w 0

has, for each w 0 in the disk Iw0I < 1 exactly n zeros (counting each one according to its multiplicity) in B, and this means that the function w = f0(z) maps the domain B onto the n-sheeted disk Iwl < 1. This completes the proof of the theorem. By means of a third, more complicated variation of the function f0(z), one can

show 1) that the extremal function is unique up to a constant factor ei°. However, we shall not stop here to do this. Let us denote by F(z, a) that one of the extremal functions of Theorem 1 for which the derivative is positive at the point z = a, thus emphasizing its dependence on a. This functions also appears in the construction of extremal functions of certain other extremal problems associated with the same class of bounded functions. 2) We present one theorem of this kind: Theorem 2. Suppose that a function f (z) is regular in a finitely connected domain B with nondegenerate boundaries and that If (z)I < 1 in B. Then, in B,

If'(z)Ic(I-If (z)I9)F(z, z),

(f'(z, z)=OF(dt

z)

c=zl'

(15)

with equality holding at the point z = a E B only for functions of the form

f(z)=el+aF(z, a), IEI=1, IalZ 1. 1) Cf. Ahlfors [1947]. 2) Alenicyn [195oa]. (See also Alenicyn [19611.>

(16)

, XI. SOME SUPPLEMENTARY INFORMATION

476

Furthermore, in B,

I f' (z) I c F' (z, z),

(17)

with equality holding at z = a E B only for ((z) = EF(z, a), where

IEI = 1.

Proof. If a is any point in the domain B, then the function g(z) _ (f (z) - f (a))/(1 - f (a) f (z)) satisfies the conditions of Theorem 1, and conversely, if g(z) satisfies the conditions of Theorem 1, then the function 1(z) = (g(z) + a)/(1 + ag(z)), where a is a constant such that la I < 1, is of the type in question. By Theorem 1, we have Ig'(a)I < F'(a, a), which is inequality (15) with z = a. Obviously, equality holds only for the function (16). The second part of the theorem follows immediately from the first. This completes the proof of the theorem.

§4. The three-disk theorem If a function f(z) is regular in the closed annulus r1 < IzI
log r8

If (Z) I C M1

tog rtr'

log

M9

r rl ry

]ogrl

(1) ;

for r1 < IzI < r2, with equality holding for some zo in the open annulus rl < Izol < r2 only in the case of a function f (z) = czwhere c and A are constants depending on Izol, r1, r2, M1, and M2, and where A is real. Since this last function is regular in the annulus only for the exceptional values of Izol at which A is an integer, it is only in these cases that inequality (1) is sharp. The problem arises of finding a sharp inequality in the remaining cases. We shall now investigate this question. 1) We note first of all that, by means of simple transformations, we can reduce the problem posed to the case when r1 = q, r2 = 1, M1 = p, M2 = 1. Hence, we can consider functions f (z) that are regular in the closed annulus q < Iz I < 1 and satisfy the inequality I1(z)I < 1 on the circle IzI = 1 and the inequality If(z)I < p on the circle IzI = q. The question then deals with the maximum value of I1(z0)I 1) Robinson [1943].

§4. THE THREE-DISK THEOREM

-

477

at some point zo in the open annulus q < Izol < 1. We can also assume that zo is positive and that the maximum sought may be found only among the functions f (z) for which f (z0) > 0. Let us denote the extremal functions by f (z, p), thus pointing out their dependence on p. Then, we can easily see that between suitable extremal functions we have the relationship f (z, pq) = zf (z, p). This last means that the case of arbitrary p can be reduced to the case in which q < p < 1. We shall now consider this case. We note that the case p = 1 is trivial because we then have f (z, p) _- 1.

Lemma. If the function f (z) is regular in the annulus q < Izl < 1 except possibly at a single simple pole on the segment - 1 < z <- q and if l1(z)I < 1 on the circles Iz I = 1 and Iz I = q, then If U) I < 1 on the open segment q < z < 1 with equality holding only in the case f (z) = const. Proof. We may confine ourselves to the case when f (z) actually does have a pole on the interval - 1 < z < - q. Suppose that q < z 0 < 1. We may assume that f (z 0) > 0. Let us define

F (z) =

2

(f (z)+f (z))>

This function also satisfies the conditions of the lemma and it assumes real values for real values of z. If F(z) is regular in the open annulus q < Iz I < 1 and F(z) coast, then obviously F(z0) < 1. Let us now consider the case in which F(z) has a simple pole on - 1 < z < - q. Since IF(- 1)I < 1 and IF(-q)I < 1, it follows that F(z) assumes on the interval -1 < z < - q every real value of absolute value greater than 1. On the other hand, it easily follows from Cauchy's theorem on the number of zeros and poles that the function w = F(z) assumes every value w such that Iw I > 1 at exactly one point of the annulus q < Iz I < 1. Consequently, IF(z)I must not exceed 1 anywhere on the interval q < z < 1 and, in particular, F(z o) < 1. If equality holds here, then F'(z 0) = 0 and a neighborhood of the point z 0 would be mapped by w = F(z) into a multiple-sheeted neighborhood of the point w = 1 or of the point w = - 1. But then, F(z) would assume certain values w such that Iwi > 1 at certain points of the annulus q < IzI < 1. Thus, we have shown that F(z0) < 1 with equality holding only when F(z) = const. But F(z 0) = f (z 0) and, consequently, f (z 0) < 1. Now, if f (z 0) = 1, we have F(z 0) = 1 and then F(z) - 1. On q < z < 1 and - 1 < z < - q, this yields R (f (z)) ° 1, Consequently, the image of the interval - 1 < z < - q will be the

XI. SOME SUPPLEMENTARY INFORMATION

478

entire line R (f (z)) = 1, and the image of the interval q < z < I is a segment of that line. Also, the line R(w) = I lies entirely inside lwl > 1. On the other hand, it follows just as above that the function f (z) assumes every value w exceeding I

in modulus at exactly one point of the open annulus q < zl < 1. This con-

tradiction proves that f (zo) must be less than 1. This completes the proof of the lemma.

Theorem. Out of all functions f (z) that are regular in the closed annulus q< lzl < 1 such that l f (z)l < 1 on the circle lz l = 1 and i f (z)l