TOPICS IN
COMPLEX ANALYSIS
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Mats Andersson
Topics in Complex Analysis
Springer
Mats Andersson Department of Mathematics Chalmers University of Technology Goteborg University S-412 96 Goteborg Sweden
Editorial Board (North America): S. Axler
Department of Mathematics Michigan State University
East Lansing, MI 48824
F.W. Gehring Department of Mathematics University of Michigan Ann Arbor, MI 48109
P.R. Halmos Department of Mathematics
USA
USA
Santa Clara University
Santa Clara, CA 95053
USA Mathematics Subject Classification (1991): 30-01, 30C20, 30C15, 32A35
With 4 figures.
Library of Congress Cataloging-in-Publication Data Andersson, Mats. Topics in complex analysis/Mats Andersson. p. cm. - (Universitext) Includes bibliographical references and index. ISBN 0-387-94754-X (soft: alk. paper) 1. Functions of complex variables. 2. Mathematical analysis. 1. Title. QA331.7.A52
1996
515'.9-dc20
96-11793
Printed on acid-free paper. © 1997 Springer-Verlag New York, Inc.
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987654321 ISBN 0-387.94754-X Springer-Verlag New York Berlin Heidelberg SPIN 10524446
Preface
This book is an outgrowth of lectures given on several occasions at Chalmers University of Technology and Goteborg University during the last ten years.
As opposed to most introductory books on complex analysis, this one assumes that the reader has previous knowledge of basic real analysis. This makes it possible to follow a rather quick route through the most fundamental material on the subject in order to move ahead to reach some classical highlights (such as Fatou theorems and some Nevanlinna theory), as well as some more recent topics (for example, the corona theorem and the HiBMO duality) within the time frame of a one-semester course. Sections 3 and 4 in Chapter 2, Sections 5 and 6 in Chapter 3, Section 3 in Chapter 5, and Section 4 in Chapter 7 were not contained in my original lecture notes and therefore might be considered special topics. In addition, they are completely independent and can be omitted with no loss of continuity.
The order of the topics in the exposition coincides to a large degree with historical developments. The first five chapters essentially deal with theory developed in the nineteenth century, whereas the remaining chapters contain material from the early twentieth century up to the 1980s. Choosing methods of presentation and proofs is a delicate task. My aim has been to point out connections with real analysis and harmonic analysis, while at the same time treating classical complex function theory. I also have tried to present some general tools that can be of use in other areas of analysis. Whereas these various aims sometimes can be incompatible, at times the scope of the book imposes some natural restrictions. For example, Runge's theorem is proved by the "Hahn-Banach method," partly because it is probably the simplest way to do so, but also because it is a technique that every student in analysis should become familiar with. However, a constructive proof is outlined as an exercise. Complex analysis is one of the origins of harmonic analysis, and several results in the latter subject have forerunners in complex analysis. Fatou's theorem in Chapter 6 is proved using standard harmonic analysis, in particular using the weak-type estimate for the Hardy-Littlewood maximal function. How-
Preface
vi
ever, most standard tools from harmonic analysis are beyond the scope of this book, and therefore, the LP-boundedness of the Hilbert transform and the Hr-space theory, for instance, are treated with complex analytic methods. Carleson's inequality is proved by an elementary argument due to B. Berndtsson, rather than using the LP-estimate for the maximal function, and Carleson's interpolation theorem is proved using the beautiful and explicit construction of the interpolating function due to P. Jones from the 1980s. However, a proof based on the 86-equation is indicated in an exercise.
Necessary prerequisites for the reader are basic courses in integration theory and functional analysis. In the text, I sometimes refer to distribution theory, but this is merely for illustration and can be skipped over with no serious loss of understanding. The reader whose memory of an elementary (undergraduate) course in complex analysis is not so strong is advised to consult an appropriate text for supplementary reading.
As usual, the exercises can be divided into two categories: those that merely test the reader's understanding of or shed light on definitions and theorems (these are sometimes interposed in the text) and those that ask the reader to apply the theory or to develop it further. I think that for optimal results a good deal of the time reserved for the study of this subject should be devoted to grappling with the exercises. The exercises follow the approximate order of topics in the corresponding chapters, and thus, the degree of difficulty can vary greatly. For some of the exercises, I have supplied hints and answers. At the end of each chapter, I have included references to the main results, usually to some more encyclopedic treatment of the subject in question, but sometimes to original papers. If references do not always appear, this is solely for the sake of expediency and does not imply any claim of originality on my part. My contribution consists mainly in the disposition and adaptation of some material and proofs, previously found only in papers or encyclopedic texts addressed to experts, into a form that hopefully will be accessible to students. Finally, I would like to take this opportunity to express my appreciation to all of the students and colleagues who have pointed out errors and obscurities in various earlier versions of the manuscript and made valuable suggestions for improvements. For their help with the final version, I would like to thank in particular Lars Alexandersson, Bo Berndtsson, Hasse Carlsson, Niklas Lindholm, and Jeffrey Steif. Goteborg, Sweden
Mats Andersson
Contents
Preface
v
Preliminaries §1. Notation ......................................................................... §2. Some Facts ....................................................................
1 1
2
Some Basic Properties of Analytic Functions §1. Definition and Integral Representation ....................... §2. Power Series Expansions and Residues ....................... §3. Global Cauchy Theorems ..............................................
12 18
2.
Properties of Analytic Mappings §1. Conformal Mappings ..................................................... §2. The Riemann Sphere and Projective Space .................. §3. Univalent Functions ...................................................... §4. Picard's Theorems ..........................................................
28 28 33 35 38
3.
Analytic Approximation and Continuation §1. Approximation with Rationals ..................................... §2. Mittag-Leffler's Theorem and the Inhomogeneous Cauchy-Riemann Equation .......................................... §3. Analytic Continuation ................................................... §4. Simply Connected Domains .......................................... §5. Analytic Functionals and the Fourier-Laplace Transform ...................................................................... §6. Mergelyan's Theorem ....................................................
46 46
1.
4.
Harmonic and Subharmonic Functions §1. Harmonic Functions ...................................................... §2. Subharmonic Functions ................................................
5 5
48 51
53 55 58 67 67 71
Contents
viii 5.
Zeros, Growth, and Value Distribution §1. Weierstrass' Theorem .................................................... §2. Zeros and Growth .......................................................... §3. Value Distribution of Entire Functions .......................
82 82 85 88
6.
Harmonic Functions and Fourier Series §1. Boundary Values of Harmonic Functions .................... §2. Fourier Series .................................................................
97 104
7.
HP Spaces §1. §2. §3. §4.
8.
9.
Factorization in IF Spaces ............................................ Invariant Subspaces of H2 ............................................ Interpolation of H°° ....................................................... Carleson Measures .........................................................
Ideals and the Corona Theorem §1. Ideals in A(12) .................................................................. §2. The Corona Theorem .....................................................
97
112 112 116 118 121
130 130 131
141 H' and BMO §1. Bounded Mean Oscillation ............................................ 141 §2. The Duality of H' and BMO ......................................... 146
Bibliography
151
List of Symbols
153
Index
155
Preliminaries
§ 1. Notation Throughout this book the letters 12 and K always will denote open and compact sets, respectively, in 1R2, and w will denote a bounded open set with (when necessary) piecewise C' boundary 8w, which is always supposed to be positively oriented; i.e., one has w on the left-hand side when passing along Ow. The notation w CC St means that the closure of w is a compact
subset of St, and d(K, E) denotes the distance between the sets K and E. Moreover, D(a, r) is the open disk with center at a and radius r, and U denotes the unit disk, i.e., U = D(0, 1), and T is its boundary aU = {z; Izj = 1}. The closure of a set E E 1R2 is denoted by E and its interior is denoted by int E. The space of k times (real) differentiable (complex valued) functions in St is denoted by Ck(Sl) (however, we write C(1) rather than CO(Q)) and C°°(S2) = nCk(Sl). Moreover, &(D) is the subspace of functions in Ck(S2) whose derivatives up to the kth order have continuous extensions to St, and Co (St) is the subspace of functions in Ck(c) that have compact support in Q. Lebesgue measure in 1R2 is denoted by dA, whereas do, denotes arc
length along curves. We use the standard abbreviation a.e. for "almost every(where)." We also use u.c. for "uniformly on compact sets." If f, q5
are functions, then "f = O(0) when x -+ a" means that f/0 is bounded in a neighborhood of a and "f = o(O) when x -p a" means that f /0 0 a. Sometimes we also use the notation f N g, which means that f is less than or equal to some constant times g. Moreover, f - g stands for f N g and f ? g. We will use standard facts from basic courses in integration theory and when x
functional analysis. Sometimes we also refer to distribution theory (mainly
in remarks), but these comments are meant merely for illustration and always can be passed over with no loss of continuity. In the next section we have assembled some facts that will be used frequently in the text. In the first chapters we refer to them explicitly but later on often only implicitly.
2
Preliminaries
Almost all necessary background material can be found in [F] or [Rul], combined with a basic calculus book. For the facts in item B below see, e.g., [Ho], which also serves as a general reference on distribution theory.
§2. Some Facts A. Some facts from calculus. If f is a map from S2 into R2 that is C1 in a neighborhood of a, then
f(a+x)= f(a)+Dfjax+o(IxI)
x-40, for some linear map x t-- D f J x, i.e., f is differentiable at a. If f is when
considered as a complex valued function, then
of
of
DfIax = xl 8x1 a +x2 8x2 a Let 7(t) = (-y1(t),72(t)), a < t < b be a piecewise C' parametrization of the curve r. If P, Q are continuous functions on r, then l Pdx + Qdy = f (P('71(t),'Y2(t))'Yi(t)+Q('Yi(t),`Y2(t))7i(t))dt, b a
and this expression is independent of the choice of parametrization. Note that
jfdg =
ff
if f, g E C1(1) and r c S2. In particular, for an exact form we have
f dg = g('Y(b)) - g(7(a)) The arc length of the curve r is Ir1 = f
=f
l7'(t)12dt =
r
fb
+ ((t))2dt
a
and
If Pdx +Qdyl
<
f
IPI2 + IQI2da < Irl sup
IPI2 + IQ12.
Green's formula (Stokes' theorem) states that if P, Q E C' (e)), then
f Pdx + Qdy = j(Qz d
L
(uLv - vAu)dA = faW (u 8v
- v M da,
8i1
where 8/871 is the outward normal derivative, i.e., 8u/877 = if 77 = (771i rh) is the outward normal to Ow.
77j (8u/8xj )
§2. Some Facts
3
On some occasions we also refer to the inverse function theorem, see, e.g., [Ho]: If f: S2 -+ R2 is C1 and its derivauive Df 1a at a E f2 is nonsingular, then locally f has a C' inverse g.
B. Existence of test functions. There are "plenty" of functions in Co (0), namely, (i) for any K C 1 there is a 0 E Co (SZ) such that
= 1 in a neighborhood
of Kand 0<0<1.
(ii) (iii)
if f E L' (0) and fn OfdA = 0 for all cb E Co (S2), then f = 0 a.e. if f is continuous on K C 1 and e > 0, there is a 0 E Co (52) such that
SUPK 10 - f J < e.
(iv) if USIa = 0, then there is a smooth partition of unity subordinate to the open cover IL, i.e., there are 0k E Co (SL k) such that 0 < ¢k <_ 1, locally only a finite number of Ok are nonvanishing and >k Ok =_ 1 in R.
C. Integration theory. From integration theory we use the standard convergence theorems, such as Fatou's lemma, Lebesgue's theorem on dominated convergence, and the monotone convergence theorem. Moreover, we frequently use Jensen's and Holder's inequalities and Fubini's theorem, the
duality of LP and LQ for p < oo, and the one-to-one correspondence between the continuous linear functionals on C(K) and the space of finite complex Borel measures on K (usually just referred to as measures on K). Furthermore, we need the Jordan decomposition of a real measure, the Lebesgue-Radon-Nikodym decomposition of a complex measure with respect to a positive measure (the Lebesgue measure in our case) and the weak type 1-1 estimate for the Hardy-Littlewood maximal function. In particular, we frequently will make use of "differentiation under the integral sign": Suppose that X, ,u is a measure space and f (x, t) is a measurable function on X x I, where I is an interval in R, which is continuously differentiable in t. Suppose further that f (x, t) and f'(x, t) are in L1(µ) for each fixed t so that g(t) =
and
h(t)
ff'(x,t)d/2(x)
are well-defined. One may ask whether g'(t) = h(t). Suppose that there is a 0 e L1(µ) such that
If'(x,t)1 <_''(x) Then h(t) is continuous by the dominated convergence theorem. Moreover,
ffxt If'(x, t) ldp(x)dt < oo, so we can use Fubini'ss theorem:
J
6
h(t)dt =
fb
f f'(x, t)dp(x)) dt = f (f b f'(x, t)dt) \I
/
a
4
Preliminaries
and hence b
1
h(t)dt = J (f (x, b) - f (x, a)) dp(x) = g(b) - g(a)
n
for all a, b E I. Since h is continuous, g' = h.
D. Functional analysis. We will use basic results such as orthogonal decomposition and Parseval's equality in Hilbert spaces, the Hahn-Banach theorem, the Banach-Steinhaus theorem, and the open mapping theorem in Banach spaces. Moreover, on some occasions we require Arzela--Ascoli's theorem on locally equicontinuous subsets of C(1l) and Tietze's extension theorem.
We also refer to the Fourier transform, Plancherel's formula, and the inversion formula; see, e.g., Ch. 9 in [Rull.
1
Some Basic Properties of Analytic Functions
§1. Definition and Integral Representation We identify C with R2 by identifying the complex number z = x + iy with the point (x, y) E 1R2. Observe that a (complex-valued) differential form Pdx+Qdy always can be written in the form f dz+gdz, where dz = dx+idy
and d2 = dx-idy (take f = (P-iQ)/2 and g = (P+iQ)/2). This motivates us to introduce the differential operators
az-2(ax-iay)
and
a
1
a =2
a a ax + i ay
so that
df = Lx dx +
y dy
Lz dz +
aj dz.
(1.1)
Note that A = a2/ax2 + a2/ay2 = 4a2/8zaz. 1.1 Proposition. If f is differentiable at the point a, then the limit f(a + z) - f(a) 1i
o
(1.2)
exists if and only if (af/az)Ia = 0; and in that case, the limit equals (af/az)Ia. The limit (if it exists) is denoted by [(a).
Proof. The differentiability off is equivalent to
f(a+z)-f(a)=zaf 1a+za-Ia+o(IzI)
6
1. Some Basic Properties of Analytic Functions
Thus, (1.2) exists if and only if lira
z of
-
Z- .O z az a
exists, and this holds if and only if (of/az)Ia = 0. The last statement
0
follows immediately.
Notice that if f = u + iv, where u and v are real, then the CauchyRiemann equation of/az = 0 is equivalent to
{:: -vx
(1.3)
(identify the real and imaginary parts in the equation (a/ax + i&/(9y)(u + iv) = 0).
Definition. A function f E C'(SZ) is analytic (or holomorphic) in S2 if of /a2 = 0 in I. The set of analytic functions is denoted by A(Q). In view of Proposition 1.1, f E C'(1) is analytic in S if and only if (1.2) exists for all a E Il, but we even have
1.2 Goursat's Theorem. If f is any (complex valued) function in 11 such that (1.2) exists for all a E S2, then f is C1 and hence analytic. The proof appears later on! In most cases, it is advantageous to write the Cauchy-Riemann equation in the complex form (9f /c)2 = 0, rather than as (1.3). For instance, clearly the product rules a a _ of ag
az(f9) az9+fa9 az (f 9) az 9 + f az hold; thus, if f,g E A(S2), one immediately finds that fg c A(S2) and (f g)' = f'g + f g'. Suppose that h(t) is C' on an interval I C R and that f is C1 in a neighborhood of the image of h in C. Then by (1.1), and
d(f o h) dt = d(f o h) = Of dh +
az
dt
8f dh = 8z
off dh
az dt
dt + Of dh dt, az Wt
and therefore we have the chain rule
dfoh_Ofdh dt
afdh
ozdt+azdt
In the same way, if h(C) is C' in some domain in C, then
ofoh Or
afdh
ofah
oz Or + aZ or
and
afoh_afah aT
Ofah
-Oz aT + 54 aT
Thus, if f, g are analytic, then f o g is analytic and (f o g)' = f'(g)g'.
§1. Definition and Integral Representation
7
1.3 Example. If f E A(Q) and f'(z) = 0 for all z E 1, then df = 0 and hence f is locally constant. Suppose now that f E A({p < IzI < R}) and that f (z) = f (reie) only depends on 0. Then by the chain rule __ of 19Z of '92 iB of _ so f ar azar+azar - e-5-z f, and therefore f is constant. The same conclusion holds if f is independent
of
0
of 0.
Here are some other simple consequences of the product rule and the chain rule (and the definition). (a) If f, g E A(1l) and a, /3 E C, then f g E A(Sl) and a f +,3g E A(Sl). (b) z -4 z'^, m being a natural number, is analytic in C.
(c) If f E A(1), then 1/ f E A(Sl \ f f = 0}). (First show that 1/z E A(C \ {0}) !) e' =def e' (cosy + i sin y) is analytic in C.
(d) z
(e) ((9/az)zm = mz'-',
(a/az)e= = ez,
(a/az)(1/f) = -f'/f2.
Exercise 1. Show that (a) of/az = of/49z. (b) if f E A(Sl), then z ,--+ f (z) is analytic in {z; z E 1}. (c) if f E A(1) and f is real, then f is (locally) constant. (d) if f E A(Sl) and If I is constant, then f is (locally) constant. If the curve I is given by r(t) = rl (t) + ir2(t), a < t < b, then, see A in the preliminaries,
J
fdz + gdz =
f (f (r(t))r'(t) + g(r(t))r'(t))dt, b
where of course r'(t) = ri(t) +ir2(t). Moreover,
f' If(r(t))IIr'(t)Idt= f rIfIdo
=jb
ffdz V.
so that (Idzl = do)
f
<-
f IfIIdzI <- Irlsrplf1.
(1.4)
1.4 Proposition. If F E A(S2) and f = F', then
ffdz = F(r(b)) - F(r(a)). In particular, fr, f dz = 0 if I' is closed.
Proof. fdz = (aF/az)dz = (aF/az)dz + (aF/(9z)dz = dF; therefore, f dz is an exact form and thus the proposition follows, cf. item A in the preliminaries.
1. Some Basic Properties of Analytic Functions
8
In complex notation Green's formula` becomes (check!)
(dA(z), /
2iJ w
f,g E Cw).
(1.5)
From th is we immediately get
1.5 Cauchy's Integral Theorem. If f E A(w) fl C' (o), then fdz = 0. w
The next theorem in particular tells us that the values of an analytic function in the interior of a domain are determined by its values on the boundary.
1.6 Cauchy's Formula. If f E C' (w`) and z E w, then f(C)dC
1
f(z)=2-7riJsw ( - z
1 7r
,l
8f
dA(()
In particular, if f E A(w) fl C' (i3), then
1
f(()d( f(z) = 2ir: r8w (-z ,
Proof. Take z c w. For a so small that {C; K - z( < E} C w, (1.5) (with f replaced by f(()/(( - z)) gives that Of da(() = fd _ ( 2i (1.6) w\{IC-z1«} a(( - z w (- z IC-zl=f C- z since ( - 1/(( - z) is analytic in w \ {1( - zl < ell. By (1.5) again, we get
I fd(
fdC
lic
_zl=f ( -
2i
1
f
27ri WE2
IZ E
j
C-zl=c
IC-zI<E
(f() + (-
-zl
fda(()
(f (z) + O(j( - zj)) dA(() = 21rif (z) + O(e).
z)-' is locally integrable, the theorem follows from (1.6) when a tends to zero. O Since
1.7 Some Simple but Important Consequences. (a) By Proposition 1.4 and Cauchy's formula (with f - 1), we get (27ri 41=,e
00
if n = 1
if n#1.
§1. Definition and Integral Representation
9
(b) If the curve r starts at a and ends up at b, then, by Proposition 1.4,
fd__I(]=b-a. (c) If 0 E CG(C), then (by Cauchy's formula)
OW _ -
f (80/,%) ((dA(() = aza
m(()d z(())
(1.7)
where the second equality is obtained by making the linear change of variables C F-+ C+z in the last integral and differentiating under the integral sign. Hence (a/OC)1/7r( = bo (the Dirac measure) in the distribution 47rbo since (a/aC) log I(I2 = 1/C sense; this is equivalent to A log (even in the distribution sense). (d) From Cauchy's formula it follows that analytic functions have the mean value property:
f (z) = 2a
f
2ir
f (z + re't)dt.
0
To see this, one simply makes the substitution ( = z + re", 0 < t < 21r (so that d( = ire`tdt = i(C - z)dt) in the formula f (z) = (e)
1 J 1(-zlsrf C(C)d( 27ri -z
If f is analytic and we differentate under the integral sign in Cauchy's formula, we find that f is C°°, f (m) is analytic, and f(m)(z) =
mt j 27x2
f(C)d(
((- z)m+1
m = 0,1,2,....
(1.8)
Thus in particular we have that A(Q) C C°°(1l) for any Q. 1.8 Proposition. If K C w CC S2, then there are constants Cm = Cm,,,,,K such that for all f E A(1l), sup If(m)I <_ CmIIf II LI(w) K
Proof. Take 0 E C, (w), 0 < 0 _< 1, such that 0 = 1 in a neighborhood of K. Let b = d(K, {z E w; O(z) 34 1}). Since f4' E Co (C), we get by (1.7) 1 00 f(()dA(() z E K. (1.9) f (z) =
a
f e(
(- z
Notice that the integration in this integral is performed only over the strip
{C; 0 < 0 < 1} cc w \ K. Hence for z in a neighborhood of K we can differentiate the integral and obtain
f(-)(Z)
-TILE 7r
f(()d)(() f 04' a( ((- z)m+1'
z E K.
10
1. Some Basic Properties of Analytic Functions
In this integral, IC - zj > 6, and therefore we get the estimate
f If(()IdA(().
sup lf(m)t < ,rb-+i Sup I a
0 The formula (1.9) is a variant of Cauchy's formula where the curve is replaced by a strip. The usual Cauchy formula cannot be used in the preceding proof (since we want L1-estimates), nor in the next one (since it deals with functions only defined a.e.). However, even in some other situations it is more convenient to use (1.9) rather than the usual Cauchy formula, as will be apparent in what follows.
1.9 Proposition (Weyl's Lemma). Suppose that f E L 10l c(Q) and of /oz = 0 weakly, i.e.,
f faqS/az = 0,
E Co (S2).
(1.10)
Then there is a g E A(n) such that f = g a.e. Thus, in particular, if f E C°(S2) and (1.10) holds, then f is analytic. An analogous result is also true (with essentially the same proof) for f E D'(S2) (the space of distributions on 11). Clearly, any f E A(S1) satisfies (1.10).
Proof. Take w cc 1 and 0 E CO '(Q) such that
1 in a neighborhood
of w, and let g(z)
1 f (aolaCf (S) dA((),
z E w.
If f is analytic, then (1.9) says that g = f in w; we are going .to show that (1.10) actually implies that f = g a.e. in w. Since g(z) is analytic in w and w cc SZ is arbitrary, our proof is then complete. To do this, take .b tE CO '(w). By Fubini's theorem f 9(z)V5(z)
_-
aS
7
f
z"(Z)
f
i f z (z))) f (o + JS 4
(_.:'
f
(z)) f (C)
The first of these integrals vanishes by the assumption on f, since z
l - 1 fZ (z) z-
+
is in Co (St) (why?). According to (1.7), the second integral is
= f o(C)+G(C)f(C) =
f(c)f()
§1. Definition and Integral Representation
11
since supp i1' C (0 = 11, and thus we have showed that f gp = f f V) for all z/) E Co(w). This implies that f = g a.e. in w. We say that fj - f u.c. (uniformly on compact sets) if for each K C S2, fj -+ f uniformly on K.
1.10 Proposition. Suppose that fj E A(c). a) If fj -+ f u.c., then f E A(c) and f("`) - f('") u.c. for each in. b) If fj f in L1 r.(c), then there is a g E A(Q) such that f = g a.e. and f"` -+ g("') U. C. for each m.
Proof. The hypothesis in (a) implies the one in (b). To prove (b), observe that f satisfies (1.10) and therefore f = g a.e. for some g c A(S2). Then apply Proposition 1.8 to fj - g. One also can easily verify Proposition 1.10 directly from Proposition 1.8 by locally representing fk by formula (1.9) and taking limits.
1.11 Morera's Theorem. If f E C(ft) and fao fdz = 0 for all triangles A C S2, then f E A(SZ).
Proof. Since analyticity is a local property, we may assume that S2 is convex. Fix a point a E 11 and let F(z) = fia z) f where [a, z] is the straight line from a to z. By assumption, fia
F(z + w) - F(z) = wf (z) + J
[z,z+w)
z)
+ fix z+w] = .f)a,z+wj, giving
(f (() - f (z)) d( = wf (z) + o(IwI)
by 1.7 (b) and (1.4). Thus, F is differentiable in z, 8F/8z = f (z), and 8F/82 = 0; therefore, F is in C1(Q) and hence analytic. Therefore, cf. 1.7 (e), f = F' is analytic.
Proof of Goursat's Theorem. Assume that 0 is convex, take an arbitrary closed triangle Ao in S2, and let I = feoo fdz. Divide the triangle into four new triangles Aj by inscribing a trianile in Ao with corners at the midpoints of the edges of A0. Then I = F_1 feo; fdz, which implies
that I feo; f dz > 1I for some Aj. Denote it by A. By repeating this procedure we get a sequence Ak such that Ak D Ak+l and fdz
> 4-kI.
Choose a point a E nAk. Given e > 0 there is a 6 > 0 such that If(z) - f(a) - (z - a)f'(a)I < elz - al
(1.11)
12
1. Some Basic Properties of Analytic Functions
<6. Taken so large that On C {Iz - al < b}. Then if Iz-allfaa fdzI
(f(z) - f(a) - (z - a)f'(a)) dz1
110A,
-n IaAoI2 E L8O Iz - alldzI < E iaAnI2 = E4
,
which together with (1.11) implies that I < ElaooI2. Thus, I = 0 and the desired conclusion follows from Morera's theorem.
1.12 The Maximum (Modulus) Principle. If Il is connected, f E A(Il), and If (a) I = supra If I for some a E II, then f is constant.
Proof. From the mean value property, cf. 1.7 (d), we get
If (a) l 5? 2,r
f2r
I
f (a + re't) l dt < M= sup I f 1, n
with equality in the last inequality if and only if if I = M on the circle IC - aI = r. Hence, if I f (a) I = M, then I f I = M in a neighborhood of a. Therefore, the set A = {z e 11; If (z)l = M} is open and closed in 0. Since A # 0, A = S2 and therefore If I, and hence f, is constant. One can replace this closed-open argument, and others in what follows, by a more conceptual "chain of circles" argument.
By the uniqueness theorem (Theorem 2.6 below) it follows that f is constant as long as If I has a local maximum.
1.13 Corollary. If f E C(i) n A(1l) and S2 is compact, then supra IfI = supan If I.
Proof. Take a point a E Sl such that (f (a) I = supra If I. If a E 1911, the conclusion is obvious. Otherwise, by the maximum principle, f must be constant in the component of Ii that contains a, but then the supremum also is attained on the boundary. Thus, If I always attains its supremum on the boundary if Sl is bounded. For unbounded domains this is not true in general. Consider, e.g., f (z) = exp(-iz) in the upper half-plane. The reader might hope for a bounded counterexample, but that is not possible, cf. Exercise 41.
§2. Power Series Expansions and Residues For the local study of analytic functions, power series expansions are an important tool, which we are now going to exploit. Suppose that cn E C
§2. Power Series Expansions and Residues
13
1/R, where 0 < R < oo. Then the power series
and
00
f(z) = : Cnzn
(2.1)
n=0
converges uniformly in each domain D(0, r) (the disk with center 0 and radius r), r < R, and diverges if IzI > R (exercise!).
2.1 Proposition. Suppose that f is given by (2.1). Then f is analytic in in D(0, R), and D(0, R), f'(z) = EO° 1 n! f(C)d( r < R. (2.2) n!cn = f (") (0) = 2ni J ICI=* ( n+1 Conversely, if f E A(D(0, R)) and cn are given by (2.2), then 0 cnzn ncnzn-1
,
converges to f u.c. in D(0, R).
Proof. Let FN(z) = >
cnxn. Then FN is analytic in D(0, R), and, . sition 1.10 we get that Fir -- f' u.c. and hence f'(z) = °O 1 ncnzn-1 in since FN converges u.c., lim FN = f is analytic in D(0, R). From Propo-
D(0, R). By induction it follows that f (n)(0) = n!cn. The second equality in (2.2) is just (1.8). For the converse, suppose that IzI < r < R. Then zn
1
x
0
('n+1
converges uniformly for ICI = r and hence z n=0
-
1C1=r
Sn+1
z
z
.
n=0
Since
f(C)d( < r-n sup If(C)I (n+1 1C1=r
ltl=r
for each r < R, we get
2.2 Cauchy's Estimate. If f E A(D(0, R)) and If I < M, then If (-1(0)1 = m!ICrnI 5
m!MR-m_
2.3 Liouville's Theorem. If f is an entire function (i.e., analytic in the entire plane) and I f(x)I 5 C (1 + IzIN) ,
14
1. Some Basic Properties of Analytic Functions
then f is a polynomial of degree at most N. In particular, if f is bounded, then it is constant.
Proof. Suppose that f (z) = Fo czn. By Cauchy's estimate, 1c,j < C(1 + RN)R_ r for all R > 0, and consequently c,,, = 0 if m > N.
0
2.4 The Fundamental Theorem of Algebra. If the polynomial p(z) = + co has no zeros in C, then it is constant, i.e., CNZN + CN_1ZN-1 +
cj=0forj>0.
Proof. If p is nonvanishing in C, then 1/p is an entire function. Moreover,
1/p(z) = O(IzI-N) when z --+ no, and hence is constant by Liouville's theorem.
2.5 Proposition. If f E A(D(a, r)) and f is not identically zero, then for some integer m, f (z) = (z - a)mh(z), where h # 0 in a neighborhood of a. One then says that f (z) has a zero of order or multiplicity m at a.
Proof. Since f can be expanded in a power series, 00
f(z) =
j=m
00
cj(z - a)j = (z - a)m E cj+m(z - a)3 = (z - a)mh(z), j=0
where cm # 0. Then h(a) # 0 and hence h # 0 in a neighborhood of a by continuity.
2.6 The Uniqueness Theorem. If Sl is connected, f E A(1l), and the set Z(f) = {z E S2; f (z) = 0} has a limit point in S2, then f - 0 in Q.
Proof. Let A be the set of limit points of Z(f). Then A is closed in 0 (why?). If a E A, then a is an interior point of A according to Proposition 2.5. Hence A is open. Since 0 is connected and A -76 0, A = 0.
0
2.7 Remarks. The assumption about connectedness is necessary. If ) = {z; Izl < 1 or Jzj > 2}, then the function f that is 0 for IzJ < 1 and 1 for Jzj > 2 is in A(D) and its zero set has limit points in S1 even though f is not identically zero itself. Observe that if f1 is connected, f, g E A(1), and
f = g on a set with a limit point in 1, then f =- g in Q. For instance, if the usual trigonometric functions are extended to complex z in the obvious
way, sinz = (expiz - exp(-iz))/2i and so on, then the usual identities, such as sin 2z = 2 sin z cos z, hold since they hold for real z.
§2. Power Series Expansions and Residues
15
If a E 0 and f is analytic in S2\a, we say that f has an isolated singularity at a. We now shall determine the possible nature of an isolated singularity. Without loss of generality we may assume of course that a = 0.
2.8 Proposition. If f has an isolated singularity at 0 and is bounded in a punctured neighborhood, then the singularity is removable, i.e., f can be defined at 0 so that it is analytic in a neighborhood.
Proof. Define h(z) as z f (z) for z # 0 and h(0) = 0. Then h is continuous and by Morera's theorem (how?) it is indeed analytic in Q. Hence, near
0, h(z) = ciz + c2z2 + ... , and therefore f (z) = cl + c2z + .... Thus, f becomes analytic if we let f (0) = cl.
Notice that the corresponding statement for smooth functions is false; consider, e.g., z i-- sin(1/Izt).
2.9 Proposition. If f is analytic in S2 = {O < Izi < R}, then f has a Laurent series expansion
f(z) _
00 Cn z" _,
that converges u.c. in Q. The coefficients cn are given by f(()d(
1
cn = 21ri fCI =* I
C+i
,
n E Z,
0 < r < R.
(2.3)
Sketch of Proof. If 0 < e < jzj < R - e, then
f(z) =
_ 1 fICI=R-ef(C)d( C-z 27ri
1
21ri
[
f(C)d(
ICI=E C - z
One then proceeds as in the proof of Proposition 2.1 (exercise!).
0
More generally: if f E A({0 < r < Izj < R}), then it can be represen-
tated by a Laurent series f (z) = ° c z' that converges u.c., and the representation is unique since the coefficients must satisfy (2.3).
Definition. Assume that f (z) has an isolated singularity at 0 and Laurent series expansion
.,
c = 0 for n < 0, the singularity is
removable (cf. Proposition 2.8). If c = 0 for n < -N and c_N 0, then f has a pole of order or multiplicity N. Otherwise, f is said to have an essential singularity at the origin.
If f (z) has a pole at 0 of order N, then f (z) - EN ckzk has a removable singularity at 0. The sum is called the principal part of f (z) at 0. Moreover,
16
1. Some Basic Properties of Analytic Functions
Z' f (z) has a removable singularity at 0, so f (z) = z-"'h(z), where h(z) coo when z --. 0, e.g., is analytic and nonvanishing near 0. Hence if (z) J the image of 0 < JzJ < r is contained in the complement of any given disk D(0, R) if r is sufficiently small. However, if f has an essential singularity, then it can omit at most one single value in each punctured neighborhood of 0. This is the so-called big Picard theorem, which is proved in Ch. 2. Right now we restrict ourselves to the following much weaker and simpler result.
2.10 Proposition. Suppose that f is analytic in Il = {0 < IzJ < r} and has an essential singularity at 0. Then its image is dense in C. Proof. If the image is not dense, then there is a w E C and a neighborhood
W of w such that f({0 < JzJ < r})f1W = 0. Then 1/(f(z)-w) is bounded, and hence it has a removable singularity. Therefore, 1/(f (z) -w) = zl"h(z), where h(0) # 0, and thus f (z) = w + 1/zNh(z), i.e., f has a pole of order N at the origin.
2.11 Examples. (a) exp(1/z) = Eo z-1/n!, and therefore exp(l/z) has an essential singularity at the origin and omits the value zero. By Picard's theorem it therefore attains all other values in each domain {0 < JzJ < e}. This also can be verified easily by a direct computation (exercise!).
(b) Suppose that f (z) has an isolated singularity at 0 and that f (z) _ O(lzIm) for some integer m. By Propositions 2.5 and 2.8 there is an integer N > m such that f (z) = zNh(z) and h(z) is analytic and nonvanishing near 0. If N > 0, then f has a zero of order N; and if N < 0, it has a pole of order -N.
Exercise 2. Show that the big Picard theorem implies the little Picard theorem: Each entire nonconstant function attains all values with one possible exception.
Definition. Suppose that f has an isolated singularity at the point a. For small enough r > 0, the number
Res(f, a) 27rz =1f
f (a + ()d(
(2.4)
C I=r
is independent of r and is called the residue of f at a; actually, it is just the coefficient c_1 in the Laurent expansion of f at a. If f has a pole of order m at a, then the residue can be computed by the formula (check!)
Res(f, a) =
(m
1
1)! d
11
a(z - a)'f(z);
§2. Power Series Expansions and Residues
17
but in concrete situations it is often simpler (and, in case of an essential singularity, necessary) to derive c-1 in some more direct way.
2.12 Examples. The residue of cot z = cos z/ sin z at z = mir is 1 since mir) cot z = 1; cf. Remark 2.7.
The function f(z) = z2/sin(1/z) is analytic in {0 < Izi < 1/ir}, has an essential singularity at 0, and the residue is 7/360 since 1
sinw =
1
7
1
3
and hence 2
2
sin Z
11 6z
7
1
1
3
360 z3
6
7
1
360 z
Make this calculation rigorous!
If f EC1(l \jai, ...,a,,,})nA(ii\{a1,...,am}), where a,,..., a, are m points in w, then m
Jaw
fd( = 2iriERes(f,aj). j=1
In fact, if ej is chosen such that D(aj, ej) is contained in 1 and does not contain any other ak, then by Cauchy's theorem (how?)
J f d - j-1
f
ail-E
f d(.
Definition. A function f is meromorphic in S2 if there is a sequence of points ak E Sl that has no limit point in 0, and if f is analytic in SZ \ {ak} and has a pole at each ak.
Thus, f is meromorphic in Sl if and only if locally either f or 1/f is analytic. Also note that if f is meromorphic in a connected set Sl and its zero set has a limit point in Sl, then f - 0.
2.13 Proposition. Suppose that f is meromorphic in 0, w C Sl and f has no poles or zeros on 8w. Let Nf and Pf be the numbers of zeros and poles, respectively (counted with multiplicities). Then 1
2Tri
I , f (()
Nf - Pf.
Observe that f has only a finite number of zeros and poles in w.
Proof. Since f'(()/ f (() is analytic outside the poles and zeros of f, the integral is unchanged if 8w is replaced by a union of circles, one around
18
1. Some Basic Properties of Analytic Functions
each pole or zero a of f. Suppose that f (z) = (z - a)mh(z), where h is analytic and nonvanishing in D(a, 2e). Then f'(z) _ h'(z)
1
h(z) +rnz - a'
f(z) and hence
f,(() d< = m2,ri. JIB-61=f f(() From the above one gets the desired conclusion.
f
O
2.14 R.ouche's Theorem. Assume that ft(z): [0, 1] x S2 , C is continuous, ft E A(ul) for each fixed t E [0, 1), and also that fe(z): [0, 11 x St --i C is continuous. If, moreover, ft is nonvanishing on 8w (w CC 11), then fo and fl have the same number of zeros in w.
Thus, if we change fo continuously, the number of zeros in w cannot change unless some zero arrives or disappears over the boundary 8w. If, for instance, f, g E A(Q) and [gI < If I on 8w, then f and f - g have the same number of zeros in w; we then simply apply the theorem to ft = f - tg. Proof. In view of Proposition 2.13, 1
f fi(()dC
27ri Ja. ft(() is integer valued for each t, but since it depends continuously on t, it then must be constant.
§3. Global Cauchy Theorems We already have seen that if r = 8w, then
f d( =
if z Ew if z Vi
(by Cauchy's formula) (by Cauchy's theorem). If I 1 i ... , I N are (piecewise Cl) closed curves with given orientations, we 1
27ri f r (-- z
r1
t0
can define the formal sum r = N F. Such a sum is called a cycle. If 0 is a differential form (1-form), then we let Jr. 0 = EN fry 0. Note that taw can be interpreted as a sum of curves with given orientations, i.e., a cycle, but the notion of "cycle" also includes other cases, e.g., passing through a given curve more than one lap. Now consider a cycle r and a point z v r. The index of r with respect to z, Indr(z), is defined by /' dC I Ind r( z) = - z. 2rri r
§3. Global Cauchy Theorems
19
For all "reasonable" F it is clear from above that Indr(z) is always an integer. This also follows from the next argument, which reveals the geometrical interpretation of Indr(z) as the winding number. However, we first need a simple lemma.
3.1 Lemma (Local Existence of Logarithm). If V is a convex domain, f E A(V) and f # 0 in V, then there is a (not unique) h E A(V) such that
exp h = f and h' = f'/f. It is natural to set h = log f . Proof. As in the proof of Morera's theorem, we can find a g E A(V) such
that g' = f'/f. Thus, (f exp(-g))' = 0 and hence f exp(-g) = C. Take
0
h = g + a, where the number a is chosen such that exp a = C.
If m is an integer, there is a g E A(V) such that gm = f . To see this, simply let g = exp(h/m).
Exercise 3. Show that log z = log Iz) + i arg z, where the second log
denotes the usual real-valued natural logarithm, and arg stands for some branch of the argument function. Suppose that F is a closed curve (the case when r is a cycle then follows) with a parametrization t - F(t), 0 < t < 1. If 0 # r, there is (exercise!) a sequence of disks D0, Dl,. . . , D,,, not containing 0 and such that the curve tk+,j is contained in Dk, where 0 = to < tj < - - < segment rk = t,,, = 1. We can successively choose branches logk z of the logarithm on Dk so that logk z = 1ogk+1 z on Dk fl Dk+l. Then -
f a _ 1: f m-1
r
k-0
rk
ds
m-1
=
E
(1ogk+1 I
(tk+1) - logk F(tk))
k=O
=i(arg,,,, r(o) - arg0 r(0)), if logk z = log I z I + i argk z, and therefore Indr(0) is the number of laps that IF runs around the origin. The same holds for Indr(z).
3.2 Example. The intuitive meaning of index is illustrated by Figure 1.
Exercise 4. Show that Indr is constant on each component of C \ F and vanishes on the unbounded one. Hint: Notice that it is continuous in C \ I' and O(1/lzl) when jzj --+ oo. Suppose that f is meromorphic and has no poles or zeros on the closed curve 7(t). Then 1:'(t) = f o -y(t) is a closed curve and
Indr(0) =
1 .f'(()dS
j A0
1. Some Basic Properties of Analytic Functions
20
Indr (a) = -2 Indr (b) = 1 Indr (c) = 0
FIGURE 1
Hence, Proposition 2.13 implies that the number of zeros minus the number of poles equals the number of times that f o -y(t) turns around the origin,
if y(t) is a parametrization of 8w. This sometimes is called the principle of argument. We leave it as an exercise to supply a similar geometrical interpretation of Rouche's theorem.
Definition. We say that a cycle r in fl is null-homologous with respect to SZ if Indr(a) = 0 for all a 0 H. Two cycles r'1 and r2 are homologous if r1 - r2 is null-homologous. Notice that if w cc f?, then 8w is null-homologous in Cl.
3.3 Cauchy's (Homology) Theorem and Formula. Suppose that f E A(C) and r is a null-homologous cycle in Cl. Then f(C)d( = 0
(a)
Jr
and
f(z) Indr(z)
z
r
Z E fl\ r.
(b)
'
It follows that Jr.' f (C)d( = frz f (C)d( if r1 and r2 are homologous. Note that (b) follows from (a), since C -+ (f (C) - f (z))/(( - z) is analytic in 12:
Proof of (a). Let H be the union of r and the support of Indr. As Indr is locally constant in C \ r (cf. Exercise 4 above) it must vanish on each component that intersects C\Sl. However, Indr vanishes on the unbounded
§3. Global Cauchy Theorems
21
component, and hence H must be a compact set in Q. Now take 0 E CO '(P)
such that 0 = I in a neighborhood of H. Then a0 f(C)da(()
z E H,
z
and by Fubini's theorem (check!) 27ri
f (z)dz =
7r
f aT'f (T) Indr(()dA(() = 0
since Indr is supported on H and 80/8c = 0 there. 3.4 Remark. The preceding proof can be interpreted in the sense of distributions in the following way. Let the distribution (measure) u be defined by u(q) _ (i/2) fl, q5dS' for 0 E CO. Then u has compact support in Sl and (8/8() Indr = u, cf. 1.7 (c). The assumption on r implies as before that Indr is compactly supported in Q. Thus,
Jr fdC = u(f) = as r
(f) = - Indr
)
= 0-
3.5 Remark. Note that f (z)dz is a closed 1-form, i.e., d(fdz) = 0 in Q. By Stokes' theorem (about homology) Jr. f dz = 0 if I' is null-homologous in Q. However, to obtain Theorem 3.3 one then has to verify the equivalence of the usual (topological) notion of "null-homologous" and the definition used
above, which is natural in this context. In one direction this immediately follows from Stokes' theorem. Conversely, if r is null-homologous (in our sense), then the union of components Vk of C\F where Indr is nonvanishing is contained in S2, and r is the boundary of the 2-chain F-k akVk where ak is the value of Indr on Vk; cf. Example 3.2. Suppose that yo and yi are two continuous closed curves in S2, i.e., continuous mappings [0, 11 IZ such that y;(0) = yj (1). They are homo-
topic if there is a continuous mapping H: [0, 1] x [0, 1) - 0 such that H(s, 0) = yo(s), H(s, 1) = yi (s), and H(0, t) = H(1, t) for t E [0,1]. Intuitively this means that the curve yo can be continuously deformed to y, within Sl, and vice versa.
Definition. If 0 is connected and any closed curve -y in 0 is homotopic to a point (i.e., to a constant mapping 10, 1] St), then S2 is said to be simply connected.
For instance, any convex domain is simply connected; take H(s, t) _ (1 - t)-y(s) + ta, where a is any point in Cl.
1. Some Basic Properties of Analytic Functions
22
3.6 Cauchy's (Homotopy) Theorem. If -to and yl are homotopic closed
(piecewise C1-) curves in l and f E A(12), then f70 fdz = f. fdz. In particular, f7 fdz = 0 if y is null-homotopic (homotopic to a point). This immediately follows from Cauchy's homology theorem and
3.7 Proposition. If yo and -yl are homotopic closed curves, then they are homologous.
However, it is just as easy to prove the homotopy theorem directly. Also note that it immediately implies Proposition 3.6.
Sketch of Proof of Cauchy's Homotopy Theorem. Since H: [0, 112 -+ Sl is continuous, there is a positive distance 2e from 52` to the image of H.
One problem is that s '- yt(s) = H(s,t) is not necessarily piecewise C1. However, for some large integer n, JH(s, t) -H(s', t')J < e if Is-s'J+Jt-t'J < 1/n. Then the rectangle with corners at H ( n n ) H ( n nk) H ( n ' n
and H(n , kn ) is contained in Q. Hence the integral of f over its boundary vanishes. When k = 0, instead of the straight line between H(n , 0) and H( , 0), one takes the curves -' H(s, 0), n < $ < '- , and analogously when k = n - 1. Let ryk be the polygon with corners H(n , ), 0 < i < n.
Then we have that f7k fdz = ffk+, fdz (why?). Since we also have that ri
f70y.
= f and f7ifdz = fin --i fdz, it follows that70f fdz = '7tf fdz. Yi
3.8 Remark. If a closed curve -y is null-homotopic in 52, then it is nullhomologous by Proposition 3.7. However, the converse is not true in general. Let 1 = C \ {a, b} and choose a point 0 E Q. Let a be a closed curve in H that starts and ends at 0 and surrounds a so that Ind, (a) = 1 and Ind,(b) = 0, and let /3 be a corresponding curve for the point b. Let y = a/3-1a-1/3 denote the closed curve obtained by first running through /3, then through a backwards, then through /3 backwards, and finally through a. Clearly Ind7(a) = Ind7(b) = 0, and therefore -y is null-homologous by
definition. However, at least intuitively, it is perhaps clear that it is not null-homotopic in St (a rigorous proof is harder). This reflects the topological fact that the first fundamental group is not commutative. In Figure 2 a curve is drawn that is homotopic to y.
Supplementary Exercises
23
FIGURE 2
Supplementary Exercises Exercise 5. Suppose that f is C' in a neighborhood of the origin. Show that
f(z)=f(0)+ k=1
kIr j=0 j
k-72j_a
aZjlO+
Exercise 6. Show that an open set S2 in C is connected if and only if it is pathwise connected (i.e., each pair of points can be connected by a (piecewise C') curve in SZ).
Exercise 7. Try to show a variant of Morera's theorem where the triangles are replaced by circles. (Hint: First assume that f is C1.)
Exercise 8. Assume that f and g are entire functions and that If (z)I < jg(z)I for all z. What can be said about f and g? Exercise 9. Find all of the zeros of sin z and cos z. Exercise 10. Assume that f is analytic in the annulus 52 = {r < Izi < R}. Show that f has a Laurent series expansion f (z) _ E 00 c,iz" that converges u.c. in 11.
Exercise 11. Assume that f has an isolated singularity at 0. Show that (a) if f(z) = o(1/IzI), then the singularity is removable.
(b) if f = o(1/IzI-0+')), then f has a pole of order at most N. (c) if f(z) = o(IzIN), N _> 0, then the singularity is removable and f has a zero of order at least N + 1.
Exercise 12. Compute f .(1 +x2)`1 exp(ixt)dx for t E R.
1. Some Basic Properties of Analytic Functions
24
Exercise 13. Compute f o (1 + x")dx, n = 2, 3, .... Integrate over an appropriate "piece of cake"!
. zexp(-ixt) exp(-x2/2)dx.
Exercise 14. Compute f °° Exercise 15. Compute f 2ixt - x2 = -(x + it)2. Exercise 16. Compute
sinx eiztdx for t E R.
h(z) = - 1
it
f
Note that t2 -
(( - z)-ld)(C)
CI<1
for various z E C, (a) using polar coordinates and solving the "B-integral" by the calculus of
residues.
(b) by first computing (8/8z)h(z) outside the unit circle and noting that at least h(z) is continuous over the circle. (c) by expanding the integrand in approriate power series.
Exercise 17. Compute F_'(1 +
k2)-1 by integrating g(z)
= n(1 +
z2)-1 cot 7rz over the curves r, which are the rectangles with corners at (n + 1/2)(±1, ±i). Also compute Ei° k-2. Exercise 18. Suppose that f E A(il), I f (z) = lim fn(z)
exists for all z E Q. Show that f - f u.c. Exercise 19. Suppose that f,, E A(Q) and that f -+ f in LP r(Q) for
some p > 1. Show that there is a g E A(il) such that f = g a.e. and f -+ g U.C.
Exercise 20. Suppose that D(a, r) C 0 and that ¢, f E A(il). Suppose that f does not vanish on the circle Iz - al = r. Determine (27ri)-1
f (z) 0(z)dz. Iz-ai=r A-)
Exercise 21. Let f E A(i2), and let yo and yl be two simply closed (piecewise C1-) curves in ft Suppose that there is a homotopy H(s, t): [0, 112 --. Q from yo and y1 such that f has no zero on its image. Show that f has the same number of zeros inside yo and yj . Exercise 22. Give a geometrical interpretation of Rouchr 's theorem. Also
formulate and prove a variant where the functions ft are allowed to be meromorphic.
Exercise 23. Suppose that f E A(S2), f (0) = 0 but f # 0. (a) Show that f (U) contains a neighborhood of 0. Hint: For some e > 0, f 34 0 on the circle jzj = e. Consider then 1
2iri
for w near 0.
f
f'(()d(
C=F f (S) - w
Supplementary Exercises
25
(b) Suppose in addition that f'(0) # 0. Show that f has a local inverse g(w) near 0.
(c) Show that g(w) is in fact analytic. Hint: For w near 0, g(w) =
1j 27ri
f'(()cd(
<1= PO - w
Exercise 24. Suppose that f and g are nonvanishing analytic functions
in U such that (f'/ f)(1/n) = (g'/g)(1/n), n = 1,2,3,.... What is the relation between f and g?
Exercise 25. Suppose that 1 is connected, f E A(SZ) are nonvanishing in Sl, and that f,, -+ f u.c. Show that if f is not identically zero, then f is also nonvanishing in 0.
Exercise 26. Suppose that ) is connected, f,, E A(SZ), and fn(Q) c SZ' for some open n'. Show that if f -+ f u.c., then either f is constant or f(S1) c S2'.
Exercise 27. Suppose that f E A(U) nC(U) and I f (z) I < 1 when Izi = 1. How many solutions does the equation f (z) = z have in U? Exercise 28. Suppose that f E A(U) fl C(U), If (z) I > 1 when IzI = 1 and f (0) = 1. Must f have a zero in U? Exercise 29. How many zeros does f (z) = z5 + 4z2 + 2z + 1 have in the unit disk?
Exercise 30. Suppose that f is analytic and f (z) _ >o akzk + o(I zI -) Determine ak.
Exercise 31. Suppose that f E A(l) and a0,.. , a,n E w CC I. Show that .
1
p(z) = f (z) -
f
)dC
(fl:) C-z
zE
,
is the unique polynomial of degree at most m that interpolates f at the points aj, i.e., such that f (a3) = p(aj), j = 0, ... , M. Exercise 32. Suppose that f c C' (w) fl L' (w). Show that
u(z) __ 1 r f (C)dA(C) ,
zEw
C-z is a solution to the equation 8u/8z = f in w. Exercise 33. Supply the details in the following proof that Ind7(0) is an integer: If -y: [0, 11 -- C is C' and closed, 0 $ 'y, and h(t) = fa y'(t)/y(t)dt, then h'(t) = -y'(t)/-y(t). Hence, y(t) exp(-h(t)) is constant. Therefore, exph(t) = y(t)/y(0), and hence exph(1) = 1. Tr
w
1. Some Basic Properties of Analytic Functions
26
Exercise 34. Suppose that f E A(I) and w CC 11. Show that if log If I < Re g on Ow for some g c A(1), then log if I < Re g in w. Hint: Consider f exp(-g). Exercise 35. Let ft = {x + iy; IxI < it/2}. Show that f (z) = exp(expiz) is unbounded in ft but supan If I = 1. Exercises 36-38 exemplify the Phragmen-Lindelof method, which permits certain extensions of the maximum principle to unbounded domains. As an application, examples of complex interpolation then are given in Exercises 39 and 40.
Exercise 36. Suppose that S2 = {x + iy; Ixi < n/2}, f E A(ft) n C(O), If (z)I < exp(A exp alyl) for some a < 1, A < oo, and that if I < 1 on aft. Show that If I < 1 in Q. Hint: (a) Let h, (z) = exp(-2e cos,i3z), where a < ,3 < 1. Show that Ih, I < 1 on S2. (b) Show that if (z)h, (z)I < exp (Ae°Ivl - e6(e'311 + e3)) on n for b = cos 87r/2. (c) Show by the maximum principle that If hE I < 1 on some large rectangle with corners at :En/2 t iyo. (d) Conclude that If I < 1 in ft.
Exercise 37. Let ft = {z; I arg zI < a}, f E A(ft) n C(St). Find a boundedness assumption on f which implies that If I < 1 in l if supon If I < 1.
Exercise 38. Suppose that ft = {x + iy; a < x < b}, f E A(ft) n C(O), and that if I < C in 1. Let M(x) = supYER if (x + iy)I. Show that Mb-a(x) < M(a)b-xM(b)2-2, a < x < b. Hint:
(a) First suppose that M(a) = M(b) = 1 and verify the statement in this case.
(b) Let g(z) = M(a) z M(b) s - and apply (a) to f (z)/g(z). Exercise 39. Suppose that p and A are (reasonable) positive measures on two measurable spaces, and suppose that T is a linear operator L1(p) L2(A) such that llTfIL <- MIIfII1andIITfII2 S IIfII2 L°°(A),andL2(µ) Then T: LP(p) -+ L'(A) for 1 < p < 2 and IIT f IIq < Mp If Ilp (if q is the dual index). Hint: (a) It is enough to show:
I f(Tf)gdAl < MP V simple f,g with
IIfII, = IIgIIp = 1.
(*)
(A function is simple if it is a finite sum F_cajXE;, where Ej are measurable and have finite measures.)
Notes
27
(b) To show (*), let (f, g simple) O(z) =
fgzP1gT(lfIzP1f)dA.
Show that O(z) is an entire function, apply Exercise 38 with a = 1/2 and b = 1, and draw the conclusion that (*) holds.
Exercise 40. Let f (n) _ (1/2ir) J ' f (t)e-'"`dt for f E L'(T). Show the Hausdorff--Young inequality
_
1/9
EIf(n)IQ(2 fjf(t)jPdt
1/p
1
I1
Exercise 41. Show: If 52 co C is arbitrary, f E A(11) fl C(fi), If I < 1 on 852, and if I < M in 52 for some M, then If I < 1 in H. Hint: a) Show that one may assume that U fl 52 = 0. b) Take a E 52 and apply the maximum principle on f"(z)/z.
Notes Most of the material in this chapter is classical. For historical remarks about the notion of analytic function, see [Hi]. Notice that the homology variant of Cauchy's theorem (Theorem 3.3) follows from Cauchy's formula and hence in turn from Green's formula in a trivial case. Exercises 12-17 exemplify the calculus of residues. For hints and further examples, see an elementary book on complex analysis. The formula in Exercise 31 is called Hermite's formula, and it is fundamental in complex approximation theory. The technique used in Exercise 39 is called complex interpolation. More about this can be found in [S-W].
2
Properties of Analytic Mappings
§1. Conformal Mappings We now will concentrate on the aspect of analytic functions as mappings from domains in 1R2 into 1R2. For an arbitrary C1 mapping the relation
f(a + z) - f(a) =
8z Iaz + 821az + o(IzI)
means that the derivative D f (a of f at a is the real linear mapping z
`-' az IaZ + az Ia2.
(1.1)
In particular, if (Of /e2) (a = 0, then D f (n is just multiplication by the complex number f'(a) = (8 f /8z) (Q and hence nonsingular if (and only if) f'(a) 54 0. The determinant of z H f'(a)z is equal to If'(a)l2 (exercise!).
1.1 Proposition. Suppose that f E A(f2) and f'(a) # 0. Then there are neighborhoods V and W of a and f (a), respectively, such that f maps V bijectively onto W. Moreover, the inverse g is in A(W) and g'(f(z)) _ 1/ f'(z).
Proof. By the inverse function theorem there are open sets V and W such that f maps V bijectively onto W and the inverse g E C1(W). Thus, g(f (z)) = z in V, and differentiation of this equality with respect to 8/8z and 8/82 yields the remaining statements. A proof without any reference to the inverse function theorem is outlined in Exercise 23 in Ch. 1.
Now we are going to show that f (1l) is open if f c A(Sl) and f is nonconstant (on each component of 52). We first consider the function 7rm(z) = z"`, m > 0. If 0 E ir,,,(cI), then 0 is an interior point in 7r,,,(I ) since D(0,rm) = 7r,,,(D(0,r)) C 7r,,,(SZ) for some r > 0, and any other w E 7r1(P) is an interior point by Proposition 1.1. A general analytic f can be represented as a composition, as the following result illustrates.
§1. Conformal Mappings
29
1.2 Theorem. Suppose that St is connected and f E A(i) is nonconstant. Let m be the order of the zero off - f (a) at z = a E 1z. Then there is a neighborhood V of a, 45 E A(V) and r > 0 such that
(a) f(z) = f(a) + (4(z))' in V. (b) 0' # 0 in V and q5: V
D(0, r) is a bijection.
Thus, f - f (a) = T,,, o 0 in V, and hence f is an m-to-one mapping from V \ {a} to {0 < Iw - f (a) I < rm}. In particular, f (a) is an interior point in f (1l).
Proof of Theorem 1.2. We may assume that 1z is convex and that f (z)
f (a) in c \ {a}. Then f (z) - f (a) = (z - a)mg(z) in 1z where g
0.
According to Lemma 3.1 in Ch. 1, there is a function h such that h"` = g. If we take 0 = (z - a)h(z), then (a) holds. Since q5(a) = 0 and 0'(a) 0, Proposition 1.1 provides a neighborhood V of a such that (b) also holds.
0
Suppose that f is a mapping from an open set c in the plane. If there is an e'O E T such that
reie) - f (a) = eiee'm = ei(e+m) lim f (a -r\,o If (a + rese) - f (a) I for all 0, we say that f preserves angles at a. This means that two half-lines from a meeting at an angle 0 are mapped onto two curves meeting at the same angle at f (a), and that the orientation is preserved. Definition. If f preserves angles at each point, it is called a conformal mapping.
If f is differentiable at a and df la # 0 (i.e. Df 1a # 0), then f preserves angles at a if and only if the mapping (1.1) does, and this in turn happens if and only if (8f/82)Ia = 0 (exercise!). Notice, however, that for example zIzI preserves angles at 0 even though the derivative vanishes. Hence z any f E A(1l) with a nonvanishing derivative is a conformal mapping.
Conversely, if f is a conformal C1 mapping, it follows that 8f/82 = 0, where Df 54 0, and therefore 8f/82 - 0 in 0, i.e., f is analytic. However, if f is analytic and has a zero of order m > 1 at a, then it follows from Theorem 1.2 that f blows up angles rn times since z - z' does. Thus we have proved the following proposition.
1.3 Proposition. If f E A(11) and f' 0, then f is a conformal mapping. Conversely, if f is a conformal C1 mapping in c, then f is analytic and
f'#0.
Two domains 0 and 0' in C are conformally equivalent if there is a ¢ E A(c) that maps Sl bijectively onto 1z'. We then find that 0-1 E A(c') and
2. Properties of Analytic Mappings
30
both 0 and 45-1 are conformal mappings. If 11 is conformally equivalent to U, then S2 is homeomorphic to U and hence simply connected. Conversely, we have
1.4 The Riemann Mapping Theorem. If 12 C C is simply connected and C \ Il is nonempty, then c2 is con formally equivalent to U.
It follows from Liouville's theorem that C is not conformally equivalent
to U. The Riemann mapping theorem makes it possible to reduce certain problems in simply connected domains into corresponding problems in U. We postpone the proof for a while and begin by studying conformal mappings of U onto itself. To this end we need
1.5 Schwarz' Lemma. Suppose that f E A(U), If (z)1 < 1, and f (0) = 0. Then If (z)1 < 1z1,
z E U
(1.2)
and If'(0)1 s 1.
(1.3)
If equality holds in (1.3) or for some z in (1.2), then f (z) _ Az for some A E T.
Proof. Set fr(z) = f (rz) for r < 1. Then fr(z)/z is in A(U) fl C(U), and by the maximum principle 1fr(z)/z1 < sup Ifr(()/(I
1,
z E U,
i.e., 1f,_(z)1 < 1z1. Letting r tend to 1 we get (1.2) and thereby (1.3).
Observe that g(z) = f(z)/z (g(0) = f'(0)) is in A(U) and that 1s(z)1 < 1. If 1g(z)1 = 1 for some z E U, then g(z) = A for some A E T, again according to the maximum principle.
Exercise 1. Suppose that f E A(U), If (z) I < 1, and that f (z) has a zero of order m at 0. Show that 1 f (z)1 < 1zIm. What happens if there is equality? A bijective conformal mapping of 1 onto itself is called an automorphism. The set of automorphisms of ft is a group under composition.
1.6 Automorphisms of U. For a E U we let (z) =
z-a
1-az
Then 0« is analytic in the entire plane except for a pole at. 1/6, which is outside U. For 1z1 = 1 we have 1z - a1 = 1z - a11z1 = 11 - az1, and
§1. Conformal Mappings
31
therefore T is mapped into itself, and hence by the maximum modulus principle U is mapped into itself (this also follows from the elementary inequality Iz - aI _< 11 - azI for z c U). However, since &,, o 0_a(z) = z (check!), 45a actually maps T onto itself and U onto itself bijectively. For any A E T and a E U we have that ?P(z) = A&a(z) is an automorphism of U. We claim that every automorphism is of this form. In fact, if f (z) is an automorphism, then f (a) = 0 for some a E U, and thus g = f o O_, is an automorphism such that g(0) = 0. Hence, by Schwarz' lemma, jg(z)I < IzI. However, g-' is also an automorphism, and therefore IzI < Ig(z)I. Hence,
I9(z)I = Izj and thus g(z) = Az for some A E T, i.e., f = A. Exercise 2. Suppose that 0: 0 -* U is a bijective analytic mapping. Show that any automorphism of Sl has the form z/i-1(AOa o 0). Also show that if a, b E Sl, then there is an automorphism such that g(a) = b, i.e., the automorphism group is transitive. Let us now consider the class E of analytic functions that map U into U. For a given a E U, we want to optimize Ig'(a)I. If g E E, then Schwarz' lemma applies to f = 4g(a) o g o 0_a, giving I f'(0) I < 1, and equality holds if and only if f(w) = Aw, i.e., g(z) = (.\O,, (z)). However,
f'(0) = 0'9(a)(9(a))9 (a)*' a(0) _ (1- I9(a)I2)-'9(a)(1- IaI2), giving I9'(0)1 <_
1-Wa)I2 1
- Ia12
Summing up, we get: If g E E, then I9 (a)I <
-1IaI2 1
and the maximum is attained if and only if g(z) = Aoa(z). In particular, any optimizing g is automatically bijective and g(a) = 0.
Exercise 3. Suppose that Sl is conformally equivalent to the unit disk. Let E be the set of all f E A(n) with If (z) I < 1 and fix a E 0. Show that there is an f E E such that If'(a)l = supgEE Ig'(a)I, and that this f is a bijective mapping of Cl onto U with f (a) = 0. Hence, if there is any analytic bijective mapping of Cl onto U, one can find such a mapping by optimizing If'(a)I. This is the idea behind the proof of the Riemann mapping theorem. However, to carry out the proof we need some further preparations.
Definition. A set 4' C A(Sl) is called a normal family if any sequence f from 1 has a subsequence that converges u.c. in Cl (the limit function is not necessarily in '1). Sometimes it is convenient to make the definition less
32
2. Properties of Analytic Mappings
restrictive and accept subsequences that tend to oc u.c. For our purposes one can adopt either definition. 1.7 Proposition. If 4) C A(Q) and 4) is uniformly bounded on compacts, then it is a normal family.
Proof. It follows from Proposition 1.8 of Ch. 1 that the gradients of the functions in 4) are uniformly bounded on compacts and hence 4) is equicontinuous on compacts. The conclusion then follows from the Arzela-Ascoli theorem.
1.8 Lemma. If S2 is connected, f,, E A(S2), f , f u.c., all f" are injective, and f is not constant, then f is also injective.
Proof. Take a E S2 and observe that are nonvanishing in S2 \ {a} and that g --i f - f (a) - g u.c. Since g is nonconstant and S2 \ {a} is connected, g is also nonvanishing in 11 \ {a} (this follows for instance from the maximum theorem or Rouche's theorem; cf. Exercise 25 of Ch. 1) and hence g(z) # g(a) if z 54 a.
1.9 Proposition. If 0 is simply connected, then each f E A(Q) has a primitive, and hence each non vanishing f E A(12) has a logarithm.
Proof. Take a E fl and define
F(z) = f j fd(, r where I'Z is a curve from a to z; since 0 is simply connected, F(z) is independent of the choice of r (why?). As in the proof of Morera's theorem, it follows that F'= f. The last claim then follows; cf. Lemma 3.1 in Ch. 1.
Proof of the Riemann Mapping Theorem. Let E be the set of all f E A(f2) that are injective (or univalent) and map S2 into U. By assumption there is some w E C \ S2, and since S2 is simply connected, there is, by Proposition 1.9, a 0 E A(Q) such that 02(z) = z -- w. If Oz1) = ±O(Z2), then zl = z2, and therefore 0 is injective and O(z1) A -O(z2). Since 0(h) is open, it contains a disk D(a, r) with 0 < r < jai. Thus D(-a,r)flO(h) = 0 and therefore 0 = r/(¢ + a) E E; hence E is nonempty. Choose a E Q. The next step consists in proving that if i E E and O(S2) is not the entire U, then there is a E E such that IL.'(a)l.
(1.4)
To this end take 3 E U \ 7/,(c). Then Op o 'O E E is nonvanishing in S2, and therefore there is a g E A(Q) such that g2 = 0,3 o 0 E E. However,
§2. The Riemann Sphere and Projective Space
33
g is injective (why?) and therefore r = 09(a) o g E E. If ir(z) = z2, then 0 = ¢_A 07r o 4_v(a) obi and therefore *'(a) = (0_0 o 7r o U is not injective, the modulus of its U but since 0_0 o ,r o derivative at 0 is less than one, and hence (1.4) follows. r/)'(a)J. It follows Now only the "soft" part remains. Let rJ = sup from the above that 0 E E is onto U if 1'/"(a)l = n (note that 77 > 0 since
E is nonempty and that i < oo since E is uniformly bounded). Choose 7. By Proposition 1.7, E is a sequence x(i from E such that a normal family, and hence we can extract a subsequence (which we also denote -0n) that converges u.c. to some i . Since lip'(a)J = 77 > 0, 0 is nonconstant and hence injective by Lemma 1.8. Since O,,(11) C U, zrb(SI) C U, but Vi(Q) is open and therefore t/i( 1) C U. Thus, we have found a i/i E E with optimal derivative, and by the preceding step it must be surjective; hence, the theorem is proved.
§2. The Riemann Sphere and Projective Space We compactify the plane C by adjoining a point oo so that the disks D(oo, r) = {IzI > r} U {oo} constitute a basis of neighborhoods at oo and denote it P, the one-dimensional (complex) projective space. By the mapping
1(re'B) = (1 +r2)-'(2rcosO,2rsinO,r2 - 1), b(00)=(0,0,1), we can identify P with the sphere S2 in R3. We leave it as an exercise to verify that 4) is indeed a homeomorphism and to find out its geometrical meaning.
In D(oo, 0), w = 1/z is continuous. By taking w as a complex coordinate we can define such concepts as analyticity, zero, removable singularity, pole,
and essential singularity in this set. This is consistent since in the set 0 < JzJ < oo, the function f(z) is analytic if and only if z --+ f(1/z) is, and so on. Thus, from the point of view of the z coordinate, a function f defined in D(oo, r) is analytic if z h-- f (1/z) is analytic in D(0, 1/r), and it has a zero at oc if and only if w -- f (1/w) has a zero at w = 0 and so on. Since f is meromorphic in SI C P if and only if locally either f or 11f is analytic, a meromorphic function in SI is just an analytic mapping from SI to P.
2.1 Remark. Topologically, P is just S2 via the homeomorphism (D. Anyway, we prefer the notion P since it emphazises the complex structure, even though it is often practical to keep in mind the geometrical picture of the sphere, which in this context is referred to as the Riemann sphere. Moreover, even though P is invariant in a certain sense (see Exercise 32), we mostly think of it as an extension of the usual complex plane, so that for instance the points 0, 1, and oo have a specified meaning.
34
2. Properties of Analytic Mappings
2.2 Remark. If w = O(z) is a bijective analytic function in S2, then concepts like analyticity, pole of order m, essential singularity as well as others
can be defined just as well with respect to the coordinate w. This immediately follows from the chain rule. However, the value of the residue at a point depends on the particular choice of coordinate. To clarify at this point, recall that any 1-form can be written uniquely as f dz + gdt. This decomposition is invariant under any analytic change of coordinate. The first term is a (1,0) -form and the other one a (0, 1)-form,. Since integration along a curve is intrinsically defined for 1-forms, it follows from the definition (cf. formula (2.4) of Ch. 1) that the value of the residue at some point a is invariantly defined for the meromorphic (1, 0)form f (z)dz. (In the same way, the value of the derivative of an analytic f at a certain point depends on the choice of coordinate, whereas the form of = f'(z)dz is invariantly defined.) For instance, dz is a meromorphic form with residue 0 at infinity, whereas the form dz/z has residue 1 at the origin and -1 at oo. It is easily verified (exercise!) that if f is a global meromorphic form on IF, then the sum of all of its residues is 0.
Since P is compact, any global analytic function on P must attain its maximum at some interior point and hence by the maximum principle f is constant, since P is connected. However, there are nontrivial meromorphic functions on P. By the compactness of P a meromorphic function f has
just a finite number of poles. Let 1(z) be the sum of the principal parts at these points. Recall that the principal part at a point a E C is the "negative part" of the Laurent expansion with respect to z at this point. If oo is a pole, we take instead the principal part with respect to w = 1/z. Then f - I is analytic on P and hence constant. Therefore, f = I + c is rational, and we obtain the partial fraction decomposition +n n(k) f(z) = p(z) + F
(z
k=1 j=1 `
ak)j
C,k
,
where n(k) is the order of the pole in ak and p is a polynomial (p - p(O) is the principal part at oo).
2.3 Example. We now can give a new approach to Louiville's theorem. If f(z) is an entire function that is O(Izl-) when JzJ -- oo, then it has a pole of order at most m at oo. Therefore, there is a polynomial p(z) of degree
at most m (the principal part at oo with respect to w = 11x) such that f (z) - p(z) is analytic at oo and hence constant by the maximum principle,
so f(z) = c + p(z).
2.4 Example. Let us determine all automorphisms of the entire plane. Certainly any affine mapping z '- az + b is an automorphism of C, and in fact all automorphisms are of this kind. In fact, if f is an automorphism,
§3. Univalent Functions
35
then it has an isolated singularity at oc; but since f is injective and open, it follows that there is some punctured neighborhood of oo where f avoids the open set f (U). Thus, it follows from Proposition 2.10 in Ch. 1 that the singularity is a pole (or removable). By Louiville's theorem then f is a polynomial and hence linear because of its injectivity. Let M be the set of all linear fractions, az + b O(z) cz + d' where ad 0 bc. Each 0 E M is an analytic bijective mapping from P to P, and one easily checks that M is closed under composition and that each 95 E M has an inverse in M, i.e., M is a group. Exercise 4. Show that if a, ,13,'y are three different points on IP, there is a unique 0 E M such that 0(a) = 0, 0(,3) = 1, and 0(-y) = oo. Exercise 5. Show that 0 E M maps "circles" onto "circles," if "circle" means circle or line in C.
2.5 Automorphisms of P. We claim that M is precisely the automorphism group of P. To begin with, assume that g: P P is an automorphism such that g(0) = 0, g(1) = 1 and g(oo) = oo. We already know that g must be rational, and since it has no pole in C, it is a polynomial. Since it is injective, it then must be a linear polynomial, and thus g(z) = z since g(0) = 0 and g(1) = 1. Now if f: P -+ P is any automorphism, we can compose it with a linear fraction 0 such that g = 0 of has 0, 1, and oo as fixed points. However, then 0 o f is the identity and therefore f = 0-1 E M.
A further discussion about IP and its automorphism group M is outlined in the exercises.
§3. Univalent Functions Suppose that f is analytic in U, f (0) = 0, and f'(0) = 1. Then we know from Proposition 1.1 that f (U) contains a neighborhood of the origin.
Is there a constant r such that rU C f (U) for all such f? With-
out further restrictions on f the answer is no. For instance, the function ff(z) = e(exp(z/e) - 1) satisfies these requirements, but the point -e is not contained in f f (U). However, if we impose the extra condition that f be injective, or univalent, then the situation is different. The principal result is Koebe's theorem.
Definition. Let S be the class of injective analytic functions in U with f (0) = 0 and f'(0) = 1.
2. Properties of Analytic Mappings
36
Notice that each f E S is a conformal equivalence from U to its image
f(U) 3.1 Theorem (Koebe's Theorem). If f E S, then f (U) D D(0,1/4). The main tool in the proof of Koebe's theorem is a result referred to as the area theorem. In the proof we use the following simple observation.
Exercise 6. If w is a bounded domain with C1 boundary, then / I
zdz = 2i ! da(z).
W
aW
3.2 Theorem (The Area Theorem). Suppose that h(z) =
1
1 +c0+c1z+c2z2+c3z3+
is analytic and injective for 0 < IzI < 1. Then 00
1
Notice that if h is as in the theorem and avoids the value a, then 1/(h(z)-
a) E S. Conversely, if f E S, then h = 1/f satisfies the hypothesis of the theorem.
Proof. Take r < 1. Since h is injective and analytic, it is actually an orientation preserving diffeomorphism from U to a certain subset of IP containing oo; hence V,. = IP\ h(D(0, r)) is a an open subset of the plane whose boundary is parametrized by ry: 0 F-+ h(reie) and 0 runs from 2a to 0 (sic!). In virtue of the preceding exercise, the area of V, is
A,. = 2i
r
J
wdw = -2z
f a"
h(re'B)h'(reie)ire'dG.
(3.1)
ry
A simple calculation gives that
h(reie)h'(re:e)re:e
2 + (Icl I2 + 2Ic2I2 + 31C3 12 + ... )r2 +
r
...
,
where the dots at the end denote terms with nonzero integer powers of e'B Since L " ei,RedO = 0 for m 54 0, (3.1) implies that
0 < A,. = .
2 T 1
O°
nIcni2r2
The desired result is obtained by letting r tend to 1.
0
§3. Univalent Functions
37
Notice that equality in the area theorem means that the intersection of
the sets V, is a zero set, i.e., h attains almost all values. The theorem implies in particular that 1c1 J < 1. If we have Icl J = 1, then ck = 0 for k > 1 and hence
h(z) = 1 + co + e0z. z
This function is actually injective (verify!) and maps {0 < JzJ < 1} onto the complement of the line segment between the points -2eie/2 + co and 2ei812 + cp. (Since h is continuous on U and h(U) is open, it is enough to verify that h maps T onto this segment.) In particular, if we choose cl = 1
and co = 2, then h(z) avoids the interval [0, 4] and hence f = 1/h E S avoids the set [1/4, oo), which shows that the constant 1/4 in Koebe's theorem is the best possible.
3.3 Proposition. Suppose that f = z + a2z2 + ... E S. (a) There is a function g c S such that g2(z) = f (z2). (b) Ja2J < 2.
Proof. Since f (z) is injective, f (z)/z is nonvanishing and therefore equal to 02 for a 0 with 0(0) = 1. Thus f (z2) (g(z))2, where
g(z) = zq(z2) = z + 2a2zs + ...
,
so to prove (a) we just have to verify that g(z) is injective. However, if
g(z) = g(w), then f(z2) = f(w2) so that z = ±w; but g(-z) = -g(z) and so z = -w implies that z = w = 0. Hence, in any case z = w. For the second statement, letting g(z) be as above we have that 1
1
g(z)
Z
\= 1z
1
1 - 2 a2z2 + ... /J
and therefore by the area theorem, 1(-1/2)a2J
1
2 a2z + ... 1.
3.4 Remark. If we have equality in (b), then as noted above 1/g(z) _ 1/z + co + eiez. The relation f(z2) = g2(z) then implies that co = 0 and thus f (z) = z/(1+eiez)2. We leave it as an exercise to determine the range of this function. Note that 00
f(z) = e-9 E(-1)n+1nein9Zn i n=1
and hence Ia,, I = n. In fact, for any f = z + a2z2 + E S it is true that JanJ < n. For a long time this was known as Bieberbach's conjecture, and it was proved by de Branges in 1984.
38
2. Properties of Analytic Mappings
We have seen an example of an injective h(z) = 1/z+co+clz+... in U that avoided an interval of length 4. Actually this is optimal, as we have
3.5 Proposition. If h(z) = 1/z+co+clz+... is injective in U and avoids the values w1 and w2i then 1w1 - w21 < 4.
Proof. By assumption, 1/(h(z) - wj) = z + (wj - co)z2 +
E S, so
1w.i - col < 2 by Proposition 3.3 (b). This implies that Iwl - W21:5 4. It is now easy to prove Koebe's theorem.
Proof of Theorem 3.1. If f E S and avoids w, then h = 11f avoids 1/w and 0, and therefore 11/w1 < 4. Thus, w E f (U) if Iw1 < 1/4.
§4 Picard's Theorems Picard's theorem states that if an entire function omits two distinct values, then it is constant; and, more generally, that if an analytic function omits two distinct values in some punctured neighborhood of an isolated singularity, then the singularity is either a pole or removable. The latter statement is usually called the big Picard theorem. The proof presented here proceeds via a classical result called Schottky's theorem. The main tool is Bloch's theorem, which states that there is an absolute constant
f such that if f E A(U) and f'(0) = 1, then f (U) contains a disk with radius e. It should be pointed out that in general this disk will not be centered at f (0); cf. the example at the beginning of the preceding paragraph. However, if one also imposes a boundedness condition, then in fact f (U) contains D(f (0), r) for some fixed r.
4.1 Proposition. Suppose that f E A(U) satisfies that f (O) = 0 and f'(0) = 1. If furthermore If 1 < M, then f (U) j D(0, 1/4M). Notice that if M = 1, then f(U) = U by Schwarz' lemma. The proposition implies that f(D(0, R)) D D(0,1f'(0)12R2/4M) if f is analytic in D(0, R), f (0) = 0, and If I < M.
Proof. If w
f (U), there is a h E A(U) with h(O) = 1 such that
h(z) = (I -f(z)/w)112 = 1 - 2wz+...
§4 Picard's Theorems
39
and Ih(z)12 < 1 + M/Iwl. For any h(z) = ao + a1 z +. - - E A(U) we have 00
lh(re`a)12d0 = E r2ttlan12, 0
r < 1,
0
and therefore 1 + r2/41w12 < 1 + M/IwI, i.e., Iwl > r2/4M. Letting r -+ 1, we get the desired result.
4.2 Theorem (Bloch's Theorem). If f E A(U) and f'(0) = 1, then f (U) contains some disk with radius e, where a is an absolute constant.
If f is analytic in D(a, R), it follows that the image of f contains a disk with radius If'(a)IRe. For an even stronger statement also due to Bloch, see Exercise 31.
Proof. We may assume that f is analytic in some neighborhood of the closure of U since otherwise we can consider f(rz)/r for r near 1. Let w(t) = t sup I f'(z)I
Then w(t) is continuous for 0 < t < 1, w(0) = 0, and w(1) = 1, and
therefore there is a least to > 0 such that w(to) = 1. Choose a such that Ial < 1 - to and I f'(a)I = 1/to. In D(a, to/2), I f'I < 2/to since D(a, t0/2) C D(0, 1 - to/2) and in the latter domain sup l f'I < 1/(to/2) = 2/to. Now g(z) = f(z) - f(a) = 1 f'(T)dr a
is analytic in z E D(a, to/2), Ig'(a)I = 11t0, and Ig(z)I < t0/2.2/to = 1, and therefore g(D(a, t0/2)) contains the disk D(0, 1/16), i.e., f(U) contains the disk D(f (a), 1/16). We now can conclude the little Picard theorem. With no loss of generality
we may assume that f is an entire function that omits the values 0 and 1. Since C is simply connected, there is some function log f in U and furthermore some function
g(x) =log
log f
log f
2iri
27ri
We claim that this g omits the set
n-1)+m27ri,
n>1, m=0,f1,f2,...}.
In fact,
f =exp
(2i rexp9 + 2exp(-9)121 ` /1
(4.1)
2. Properties of Analytic Mappings
40
(let a = log f /27ri; then exp g = f - a --I and exp(-g) = /a + a - 1,
and thus expg+exp(-g) = 2f). Certainly there is a positive number d such that each disk with radius d intersects E, and hence it follows from Bloch's theorem that g(a) = 0 for any a. Therefore, g is constant and hence so is f. To obtain the big Picard theorem, we need Schottky's inequality, the proof of which requires a somewhat refined version of the previous argument.
4.3 Proposition (Schottky's Inequality). There are positive functions m(a, )3, r) and M(a, (3, r) such that if f E A(U) avoids 0 and 1 and a < If (0) 1 < 0, then
m(a,/3,r)
If(z)I < M(a,0,r),
IzI < r < 1.
Proof. Assume that f E A(U) and avoids the values 0 and 1. Since U is simply connected, there is a function g E A(U) such that (4.1) holds. Moreover, we can choose g by first determining the value g(0) so that Ig(0) I <- C'(a, Q)
if 0 < a < If (0) I < Q.
(4.2)
If IzI = r < 1, then the function (S) =
g(z + (1 - r)() (1 - r)g'(z)
is analytic in the unit disk and 0'(0) = 1. By Bloch's theorem, therefore, its image contains a disk with radius e, i.e., some disk with radius (1-r)Ig'(z)It is contained in g(U). However, since g avoids the set E, we must have that
Ig (z)I < f(,d r)" Since g(z) = g(0) + fI g'(r)dr,
we thus obtain the estimate Ig(z)I < Ig(0)I + e(1dlzlzl)
C(a,13)
e(1dlzlxl).
(4.3)
+ From (4.3) and (4.2) we now get that I f (z) I < M(a, /3, r) if IzI < r < 1. Applying it to 1/f we get the lower estimate. 0 <
Notice that one needs both the upper and the lower bound on If (0)I even
to get the estimate If(z)I < M.
4.4 The Big Picard Theorem. If f is analytic near a point a and omits two distinct values in some punctured neighborhood of a, then a is a pole or a removable singularity.
Supplementary Exercises
41
Proof. We have to prove that if f is analytic for (z( > 1/2 and avoids 0 and 1, then either f is bounded or has a pole at infinity. If not, the image of f on each set IzI > R is dense according to Proposition 2.10 in oo such that 1/2 < If (A,,) 1 :5 1. Set Ch. 1, and therefore there are An fn(z) = f(Anz). Then fn is analytic in Iz( > 1/2 and 1/2 < lfn(1)I < 1, and therefore Schottky's inequality yields that c < I fn(z)I < C,
z E D(1, 1/4).
For a E TnD(1, 1/4) we can repeat the argument and hence get constants c' and C' such that
c' < If-WI < C',
z E D(a, 1/4).
After a finite number of steps we have covered the entire unit circle, and hence there are constants k and K such that
k
zET,
i.e., (f(< K on Iz( = A, . By the maximum principle
(f
max (K, sup Ill \
T
for 1 < IzI < (Anl. Hence, f is bounded for 1 < Iz( < oo, which is a contradiction.
0
Supplementary Exercises Exercise 7. Show that the maximum modulus principle follows from the open mapping theorem.
Exercise 8. Show that the determinant of the linear mapping z
az is
(a(2.
Exercise 9. Complete the proof of Proposition 1.3, i.e., show that if df(a # 0 and f preserves angles at a, then (8f/8z)1n = 0.
Exercise 10. Given S2 C C, find an exhausting sequence of compacts ... K C Kn+i .... i.e., such that any compact K C 92 is contained in some Kn. Find an increasing sequence of compacts Kn such that UKn = S2 which is not exhausting.
Exercise 11. Let 4iP = If E A(U); f t,,, If (dA < exp(1 -
Is D a
normal family?
Exercise 12. Show that the function t from P (the extended plane) to S2 is indeed a homeomorphism. Include a drawing.
42
2. Properties of Analytic Mappings
Exercise 13. Suppose that f : U - U is analytic and f (a) = r3. Show that
z-a
f(z) - Q
1 -&z
1-Of (Z)
Exercise 14. Find all automorphisms of the upper half-plane II+. Exercise 15. Suppose that f E A(II+) and If I < 1. How large can I f'(i) be? Which f are optimal?
Exercise 16. Suppose that f E A(U) fl C(U) and If I = 1 on T. Show that f has at least one zero in U unless f is constant. Exercise 17. Find all f E A(U) fl C(U) such that If I = 1 on T. Hint: Note that each finite product of functions ¢a is of this kind.
Exercise 18. Suppose that f, g E A(U), f (0) = g(0), g(U) C f (U), and that f is injective. Show that g(D(0, r)) C f (D(0, r)), 0 < r < 1. Exercise 19. Let (D = If E A(U); Ref > 0 and f (0) = 1}. Show that ' is a normal family. What happens if one removes the assumption f(0) = 1? Exercise 20. Find a homeomorphism from U to U that has no continuous extension to U. Exercise 21. Suppose that Il is connected and symmetric with respect to the real axis, f E A(5l), and f is real on ) fl R. Show that f (z) = f (z). Exercise 22. Suppose that f is analytic in a connected neighborhood of T and that If I = 1 on T. Show that f (z) = 11f (1/z). Exercise 23. Find all meromorphic functions f in C such that If I = 1 on T.
Exercise 24. Let A(1, R) = {z; I < IzI < R}. Show that if A(1, R) is conformally equivalent to A(1, R'), then R = R' by filling out the details in the following arguments. Let f: A(1, R) -+ A(1, R') be analytic and 1 when Izj 1 and I f (zj) I --* R' bijective. First show that I f (z,) I R, or vice versa. In that case, replace f by R'/f. Then show when Izj I
that logIf(z)1 =log
R'
log R
logIzI
and conclude that R = R'. Hint: u(z) = log if (z)I - (log R'/ log R) log IzI is harmonic in A(1, R) and vanishes on the boundary, and hence it vanishes identically; see Ch. 4.
Exercise 25. Find all automorphisms of A(1, R).
Supplementary Exercises
43
Exercise 26. Show that if f (z)dz is a meromorphic form on P, then the sum of all of its residues is 0.
Exercise 27. Show that if f E S, then there is a g E S such that gm(z) _
f (z')Exercise 28. Suppose that f (z) is an injective meromorphic function in U such that f (0) = 0 and f'(0) = 1, and suppose that f avoids w1 and W2. Show that Iw1 - w2I/Iw11Iw21 < 4.
Exercise 29. Determine the image of the function f (z) in Remark 3.4. Exercise 30. Here is a sketch of an alternative proof of Proposition 4.1. If f (z) = z + a2z2 + a3z3 + ... , then lakl < M by Cauchy's estimate. For
Izl=r<1, If(z)I > r - Mr2 - Mr3 - ... = r(1 + M) - 1 M = ON, and 0(r) attains its maximum at p = 1 -
M/(1 + M). However, since
(we may assume that) M > 1,
0(P)=( 1+M-vM) 2 =( 1+M+VM) -2
>
1
1
(1 + / )2M > 6M
Thus, if IzI = r < p, then If(z)1 ? 0(r) >_
r0PP)
as r
0(r)/r is decreasing. Hence, f (z) has just one zero in D(0, p). However, if (zl = p and IwI < 0(p), then If(z)I > 0(p) > Iwl and thus by Rouchr 's theorem, z - f (z) - w has exactly one zero in D(0, p). Thus, f (D(O, P)) D D(0, 0(P)) Exercise 31. Show the following stronger version of Theorem 4.2: There
exists an absolute constant b > 0 such that if f E A(S2) and f'(0) = 1, then some open subset of U is mapped bijectively by f onto some disk with radius b.
The following exercises contain a further discussion about the onedimensional complex projective space P and the Riemann sphere. Exercise 32. Here is the usual formal definition of P: On C2 \ {(0,0)} the points Z = (zo, z1) and W = (wo, wl) are equivalent if and only if w = az for some a E C \ {0}. The set of equivalence classes is denoted by IP and is made into a topological space by requiring that the natural projection 7r: C2 \ (0, 0) -+ P be continuous and open. Since 7rI {Iz1=1 } is surjective, P is compact. In 7r({zo # 0}), one can take z = zl/zo as a local complex coordinate. A nonsingular linear transformation on C2 with matrix
G=
44
2. Properties of Analytic Mappings
gives rise to an automorphism O(z) on P given by the formula az + b
O(z) =
(*)
cz+d
in the coordinate z. It follows that the set M of all automorhisms of P of the type (*) is a group. Exercise 33. From this point of view a natural distance or metric on P is Iwozi - wizil X(Z,W) -+1 Z iI2 -+I WO I2 W 1I2, IzoI2
which in the coordinate z = z1/zo is X(z, w) = VI'l+ IzI2
1I
+ Iw12
X(z, 00)
,
=
1 -+1 Z I2
Actually, X(z, w) _> 0 and X(z, w) = 0 if and only if z = w and the triangle inequality holds, i.e., X(z, w) < X(z, u) + X(u, w). Moreover, X(z, w) < 1 with equality if and only if w = -1/z. The metric X is invariant under the
subgroup U of M, that arises from the group of unitary matrices G, i.e., such that G*G = I. The subgroup U is (also) transitive, i.e., for any pair of points z, w E P there is some ?k E U mapping z on w.
Exercise 34. Each "circle" on IP is of the form {z; X(z, w) = r}. Exercise 35. There is a unique normalized measure dp on IP that is invariant under U, i.e., such that p(E) = p( rl,(E)) for all (measurable) E and 0 E U. Since X(z + h, z) _ 1 lim
1 + Iz12'
IhI
it follows that dp(z) _
dA(z)
(1 + IzI2)2 7r
Exercise 36. Show that if P is identified with the sphere S2 via the mapping 4), then X corresponds to half the cordial distance between points on S2. Moreover, show that the measure du corresponds to the normalized surface measure on S2. Also show that a "circle" on P corresponds to a circle on S2.
Notes The first complete proof of the Riemann mapping theorem is from the end of the nineteenth century. The proof presented here depends heavily on normal families. A more constructive proof was obtained by Koebe.
Notes
45
For the history of the R.iemann mapping theorem and its connection to Dirichlet's problem, see [Hi]. Any doubly connected domain (such that the complement has two components) with reasonable boundary is conformally equivalent to an annulus A(1, R) (for a unique R in view of Exercise 24). See [Al] and [A2] for further results about multiply connected domains. Koebe proved in 1907 that there exists some constant k such that f (U) D
D(0, k) for each f E S. The optimal value k = 1/4 was determined by Bieberbach in 1916. Bieberbach also made the conjecture that n for all n. de Brange's proof of the conjecture is published in Acta Math., vol. 154 (1985).
Picard proved the theorem bearing his name in 1879. The proof presented here is due to Bloch (1924). The result in Exercise 31, as well as Theorem 4.2, is due to Bloch. The best constant in Theorem 4.2 is called Landau's constant L, whereas the best constant in Exercise 31 is called Bloch's constant B. Their exact values still are not determined, but it is
known that 0.4330... < B < 1/2 < L < 0.5433... ; see Liu and Minda, Tans. Amer. Math. Soc., vol. 333 (1992), 325-338.
3
Analytic Approximation and Continuation
§1. Approximation with Rationals In this section we study the possibility of approximating analytic functions
with polynomials and, more generally, by rational functions. The main result is Runge's theorem.
1.1 Runge's Theorem. Suppose that K is compact in 1P and that the set {a? } contains one point from each component of 1P \ K. If f is analytic in a neighborhood of K, then, for given e > 0, there is a rational function
r having poles only in the set {aj } such that if - rl < e on K.
Of course, one of the points may be oo. Note that 1P \ K has at most countably many components.
1.2 Corollary. If K C C and C \ K is connected, i.e., K has no "holes," then any f that is analytic in a neighborhood of K can be uniformly approximated by polynomials on K.
Notice that there is no requirement that K be connected. 1.3 Example. There is a sequence of polynomials pn such that pn(z) -+ 1
for all zEUand pn(z)-+0 forallzVU. In order to see this, let K,, = U U {z; 1+1/n < jzj < n,0 < argz < 27r- 1/n}. Since the complement of each K is connected, Runge's theorem applies to the function f that is I in a neighborhood of U and 0 in a neighborhood of Kn \ U; therefore, there are polynomials pn(z) such that 1pn - 11 < 1/n on U and 1pnI < 1/n on Kn \ U. The sequence pn then has the stated properties.
The requirement of (at least) one point from each component of P \ K is necessary. In fact, if a is a point in the component V of P \ K, then
§1. Approximation with Rationals
47
f (z) = (z - a) - l is analytic in a neighborhood of K (take f (z) = z if oo e V). If f (z) could be approximated arbitrarily well on K by a rational
function r(z) with no poles in V, then (z - a)r(z) - 1 would be near 0 on 8V C K and hence in V by the maximum principle. This leads to a contradiction when z = a.
Proof of Runge's Theorem. We may assume that K C C. The dual space of C(K) is precisely the space of (complex) measures on K. If f IK is not in the closure of the space (of restrictions to K) of rational functions having poles in {aj }, there is (according to the Hahn-Banach theorem) a
measure p on K that vanishes on these rational functions but such that
p(f)
0.
Thus, if At is a measure on K that vanishes on rational functions with poles in each {aj }, we have to prove that µ(f) = 0. To this end, let
Kz-(
h(C) = f dp(z)
CEP \ K.
Let V be a component of P \ K and ak a point in v fl jai I. If ak 0 00, then 1
z -
-
(C - ak)" o
(z - ak)"+l
uniformly for z E K if C E D(ak, r) CC V. Hence, h = 0 in D(ak, r) by the assumption on p, and since h is analytic in V, it must vanish identically there; analogously if ak = oo. Hence, h = 0 in P\ K. Now take 0 E Co (C) such that ¢ = 1 in a neighborhod of K. Then for z E K,
f(z) = f(z)O(z) _ -! f f(S)a'zaSda(s) and, by Fubini's theorem,
µ(f) = fK fdu = ,-I f f(C) a"h(C)da(() = 0, since 84,/85 = 0 on K and h = 0 outside K.
0
We leave it to the reader to give an interpretation of this proof in the sense
of distributions; cf. Remark 3.4 in Ch. 1 and Exercise 18. A constructive proof of Runge's theorem is outlined in Exercise 6. By means of the following simple lemma we obtain some important corollaries of Runge's theorem.
1.4 Lemma. If Sl C C, then there are compact sets ... K. C int(Kn+1) C C Si such that each component of P \ K" intersects P \ Si and any K C Si is contained in some K. Kn+1 C
3. Analytic Approximation and Continuation
48
The first condition on Kn means that it has no "unnecessary holes."
Sketch of Proof. Take a sequence... Hn c int(Hn+l) C H.+1 C . C St such that UHn = Q. Let Kn be the union of Hn and all components of 0 P \ Hn that do not intersect P \ Q. One also can define Kn directly by P \ Kn = D(oo, n) U U.VoD(a, l/n). By the lemma we get the following corollaries:
1.5 Corollary. If 0 C P and f E A(c), there are rational functions rn f u.c. in 0. with poles outside Il such that rn 1.6 Corollary. If St C C, 1P\ 1l is connected and f e A(c), then there are polynomials pn such that pn --+ f u.c. in Q. §2. Mittag-Leffler's Theorem and the Inhomogenous Cauchy-Riemann Equation
Recall that a principal part at a E C is a rational function n
p(z) _
ci
i-1
(z-a)j
2.1 Mittag-Leffler's Theorem. Let aJ be a sequence with no limit point in St C C, and let p,(z) be principal parts at a,. Then there is a meromorphic f in 11 that has principal part pj at aj for each aj and no other poles.
Proof. Take Kn as in Lemma 1.4 and let pi (z) ajEK There is only a finite number of a3 in K,,; therefore, each qn is a rational function and qn+l - qn is analytic in a neighborhood of Kn. Hence by qn(z)
Runge's theorem there are rational rn with poles in IP\S2 such that qn - rnI < 2-n on Kn. Define 00
f = ql + E(qn+l - qn - r'n) 1
For fixed N, N-1 f = qN -
00
rn + E(qn+l - qn - rn), 1
N
§2. Mittag-Leffer's Theorem and the Inhomogenous Cauchy-Riemann ...
49
and the last sum converges uniformly on KN and hence is analytic in int(KN). Since the rn have their poles outside S2, f is meromorphic in int(KN) and has the prescribed principal parts there (and no others). Since N is arbitrary, the proof is complete.
2.2 Remark. The notion of principal part p of f at the point a depends on the coordinate z. However, in any case, f - p is analytic near a, and therefore the principal parts with respect to any two different coordinates differ by an analytic function. Mittag-Lefer's theorem can be stated in the following invariant way: Suppose that we have an open cover S2j of 0 C IF and meromorphic fj in 12j such that f, - fk is analytic in S2j fl f2k. Then there is a global meromorphic function f in S2 such that f = fj in S2j. It is readily verified that Theorem 2.1 follows from this statement, a proof of which is outlined in Exercise 7.
Recall (cf. (1.7) in Ch. 1) that if 0 E Co , then
0(()d z(() f solution to 8u/8z = . By a proof analogous to the preceding
1
u(z)
is a C'
(2.1)
proof of Mittag-Leffler's theorem we get an existence theorem for solutions to the inhomogenous Cauchy-Riemann equation.
2.3 Theorem. If 52 C C and f E C1 (Q), there is a u E Coo (Q) such that
4=
0 'LL
8z
f
(2.2)
in Q.
Proof. Take K. as before and choose 0n E Co (int(K,,+1)) such that 0n = 1 in a neighborhood of K. Now there are un E C°°(S2) such that Bun/8z = inf. Observe that (8/(9z)(u,+i - un) = 0 in a neighborhood
of K. Thus there are rational rn, having their poles outside Sl, such that
Iun+1 - un - r,ll < 2-" on K. Then
u = u1 + F_(un+1 - un - rn) 1
is a C°°-function, since on each fixed compact there is only a finite number of terms that are not analytic and the "tail" converges uniformly. Therefore, we may differentiate termwise, and hence
8u
00 1
/
l
F
50
3. Analytic Approximation and Continuation
2.4 Remark. For a function u let eu = (8u/az)d2. Then 8u is the (0, 1)-part of the form df and hence invariant under analytic changes of coordinate; cf. Remark 2.2 in Ch. 2. If F is a smooth (0, 1) form, then the equation 8u = F has an invariant meaning, even at infinity. We then can formulate the preceding theorem as: If St C IP and f is a smooth (0, 1)form in 1, then there is a smooth function u such that 0% = f in 0. This follows immediately from Theorem 2.3 except in the case 0 = P. However, if F is a global (0, 1)-form on IF, we can solve 8uj = F in Sto = D(0, 2R) and Sl1 = D(oo, R), respectively. Then ul - uo is analytic in the annulus {R < IzI < 2R}, and therefore (using, e.g., the Laurent series expansion)
the function can be written hl - ho, where hj are analytic in Ci. Thus uo - ho = ul - hl in the annulus and, since Co U Stl = IF, we get a global solution.
Theorem 2.3 is a pure existence theorem. It is clear that if uo is a particular solution to (2.2), then any other solution has the form u = uo + h where h E A(D). Often it is important to single out a solution with certain properties. This can be done by L2 methods (see Ch. 8, where some examples of this technique are exploited in the text and in the exercises). Another way to get further information about a solution is to use some explicit formula. If f E L' (C), then
u(z) = _ 1
IT
f (S)dA(()
r S-z
provides a solution to (2.2). [To see this, let X be a cut-off function in Cl that is identically 1 in some neighborhood of K C Cl and write f = X f +(I - X) f . This gives rise to two terms: the first one solves (2.2) on K (cf. (2.1)), and
the second one is analytic near K.) It is also possible to construct such formulas that will work for a larger class of f:
2.5 Example. It follows that
Kf(z)
-
7r
-ICz21 f(e)d z(() lU
\1
C
solves (2.2) if f(1 - I S I2) I f I < oo _ In fact, one can just apply the arguments
above to the function
S_ (1_cl2)
-(f(S)
and let z = w. In this formula, f is thus allowed to grow somewhat at the boundary. However, this solution also has a certain minimality property. Let B(U) be the subset of analytic functions in U that belong to L2(U). From Proposition 1.10 of Ch. 1 it follows that B(U) is a closed subspace
§3. Analytic Continuation
51
of L2(U). Let
u(()d\(2) Pu(z) = 1 lr U (1 - Cz) One can verify (see Exercise 9) that P is a bounded operator from L2(U) to B(U). Furthermore, one can check that
K(Ou/8z) = u - Pu (2.3) for u E C'(U). Since, for instance, the analytic polynomials are dense in B(U) (Exercise.9), (2.3) shows that Pu = u if u E B(U) is analytic, i.e., P is a projection. Moreover, it is indeed the orthogonal projection B(U) since its kernel (1 - (z)-2/7r is hermitian (which in P: L2(U) turn implies that P is self-adjoint). The operator P is called the Bergman projection in U and B(U) the Bergman space. Now, if (2.2) has a solution u in for example L2(U) n C1(U), it follows from (2.3) that Kf is the one with minimal norm in L2(U) since it is orthogonal to B(U). Further examples of integral representation of solutions will be given in the exercises.
§3. Analytic Continuation In Ch. 1 we found that the logarithm function could be defined along any curve in the plane that avoids the origin. This is an instance of analytic continuation, a notion that we now shall make precise. A function element is an ordered pair (f, D), where D is an open disk and f E A(D). Suppose that y: [0, 1) - fZ is a curve and we are given fo E A(D(y(0), ro)). We say that fo (or more precisely (fo, D(y(0), ro))) can be continued along y if there
are numbers 0 = so < s1 < . . < s = 1 and function elements (fj, Dj) .
such that -y(O) is the center of Do, y(1) is the center of Dn, y([sj,sj+lj) C
Dj, and fj = fj+i in Dj n Djt1 (which by the assumption is nonempty). Note that if g is a continuation of f along y, then f is a continuation of g along -y. 3.1 Proposition. Any two continuations of fo along -y coincide, i.e., if we
have 0 = so < sl <
< sn = 1, 0 = ao < a1 <
. < a,, = 1, and
corresponding function elements (f j, Dj) and (9k, Dk), then fn = g n in
DnnDm.
Hence, we can talk about the continuation of fo along -y.
Proof. Assume the contrary. Then there are i and j such that [si, si+1) n [aj, aj+1) 0, fi gj on DinDJ, and i+j is minimal. We also may assume that si > aj. Then i > 1 and s, E [aj, aj+1). Thus, 'y(si) E D;_1 n Di n D1.
52
3. Analytic Approximation and Continuation
Since i + j is minimal and [st_1i si] n [off, o j+11 34 0, necessarily fi- I= gi
on D;_1 nD3. Hence, gj = fi_1 = ft on Di_1 nDinD;, but since DjnD; is connected, we must have gj = fi on the whole Dj n Di, which is a contradiction.
Notice that an analytic function has a continuation from the center of a disk to the entire disk if and only if its power series at the center converges in the whole disk. Hence, the definition of analytic continuation along a curve
also provides, at least in principle, a method to perform the continuation if it exists at all; cf. Exercise 19. In practice, however, one often can use some other representation of the continuation.
3.2 Example (The Gamma Function). The formula F(z) =
tz-Ie-tdt
J0
defines an analytic function for Re z > 0 that is called the Gamma function.
An integration by parts reveals that zF(z) = r(z + 1). Since r(1) = 1, it
follows that r(n) = (n - 1)! for positive integers n and that r(z) has a meromorphic continuation to the entire plane with poles at the points
0,-1,-2,....
3.3 Example. Consider the function f (z) defined in the unit disk by
f (z) _
z". 1
We claim that (f, U) can be continued along any curve from U that avoids the points 1 and 0 sic!. To begin with, as fo exp(-nt2)dt = 7r/n/2, one finds that f (z) coincides with
F(z) =
2 7r
00
o
z
e, -z dt
(3.1)
in the unit disk, and therefore the formula (3.1) provides a continuation of f along any curve that does not intersect the half-axis (I, oe). Let F_ (z) denote F(z) below [1, oo) and F+(z) above. We claim that F_ (z) can be continued across any point in (1, oo). To see this, use Cauchy's theorem and move the path of integration in (3.1) from, say, 1/R to R a little bit into the upper half-plane. If the modified path is sufficiently near the positive real axis, then its image under rr --4 exp(-r2) will run in the upper half-plane, and hence the resulting integral will provide a continuation of F_ across the real axis between exp(1/R2) and exp(R2). In the same way, F+ can he continued a little bit below (1,oo). By Exercise 23 -.o C
F(x + ie) - F(x - ie) =
1
og
(3.2)
§4. Simply Connected Domains
53
and hence, by uniqueness,
F_(z) = F+(z) -
1
v1
og z is the branch that is positive on (1,oo). If it is continued from a point in (1,oo) along a closed curve that surrounds 1 but not 0, then one ends up with its negative. We leave it to the reader to determine the continuation of f (z) along an arbitrary curve in a neighborhood of (1, oo), where
avoiding 0 and 1.
Analytic continuation plays an important role in Section 5 below,
3.4 The Monodromy Theorem. Suppose that SZ C P is simply connected and (f, D) is a function element that can be continued along any curve in c that starts at the center of D. Then there is a g E A(i) such
that f =gin D.
Proof. It is enough to show the following: If r and ro are curves that begin at the center of D and terminate at /3, and g and go are the continuations of f along r and ro, respectively, then g = go. Assume the contrary. Since continuation along a curve is reversible, we thus have a closed curve '7o = -I' + r'o such that go # g is the continuation of g along Yo. Since SZ is simply connected, there is a homotopy ryt, ryt(0) = .yt(1) = /3 (why?) between -yo and -yl: [0, 1] --+ /3. Let gt be the continuation of g along ryt. It is
clear that gl = g. Now fix a t. Then there are 0 = so < sl <
<s=1
and (f),Dj) such that ryt([sj,sj+l]) C D3, fo = g, and f = gt. Let e be the least distance between ryt([sj, s,+1]) and 52 \ Dj, j = 0, ... , n. By compactness there is a 6 > 0 such that c yt(s) -'yu(s)I < E if Iu-t] < 6. Take such a u. Then (fj, Dj) is also a continuation along 7'u and hence gu = gt. Thus It E [0, 1]; gt = g} is open and (trivially) closed and therefore equal to the whole interval [0, 11. Hence go = g, which contradicts our assumption.
3.5 Remark. What we in fact proved was: If (f, D) can be continued along any curve in 9 that starts at the center of D, then the continuation along a closed null-homotopic curve must be f itself. Incidentally, by considering, e.g., og z in 0 = C \ {0, 1}, this proves that the curve in Example 3.8 in Ch. 1 is not null-homotopic.
§4. Simply Connected Domains We now will sum up the various conditions on an open set that we have met thus far, which are equivalent to simple connectedness.
54
3. Analytic Approximation and Continuation
4.1 Theorem. Suppose that 11 Ck C is connected. Then the following statements are equivalent:
(a) l is homeomorphic to the unit disk. (b) f is simply connected. (c) Any closed curve in Sl is null-homologous.
(d) P \ fl is connected. (e) Sl is conformally equivalent to the unit disk. (f) The monodromy theorem holds. (g) Any f E A(Sl) can be approximated u.c. in Sl by polynomials. (h) f fdz = 0 for any closed -y and f E A(Sl). (i) Any f E A(Sl) has a primitive function in 0. (j) Any nonvanishing f E A(Q) has a logarithm in Q. In Ch. 4 we will add an additional equivalent condition, namely, (k) Each real harmonic function u in 12 is the real part of some f E A(Sl).
4.2 Remark. Notice that (a) to (d) are purely topological properties of an open set in 1R2, and they are still equivalent even if we remove the assumption that C \ 1 # 0. It is clear that (a)-+ (b)-* (c); cf. Proposition 3.7 in Ch. 1. However, it is not at all obvious that (c) implies (a). The corresponding statement for domains in higher dimensional R1 is not true. More precisely, there are connected domains with vanishing homology groups that are not contractible, as well as contractible domains that are not homeomorphic to the ball in R^'.
Proof. Clearly, (a) (b) (c). From Cauchy's homology theorem it follows that (c) implies (h), and in Ch. 2 we saw that (h) -+ (i) -+ (j) -+ (e) --b (a); therefore all of them are equivalent. Thus (d), (f), and (g) remain.
The Monodromy theorem states that (f) follows from (b). Conversely, suppose that (f) holds and take w E C \ Q. Any function element (f, D) in S1, where f (z) is some branch of log(z - w) in D, can be continued analytically along any curve in Q. Thus if the monodromy theorem holds, there is a global function g(z) that coincides with f (z) on D. Hence, by uniqueness, expg(z) = z - w in Sl so that g'(z) = (z - w)-1; and hence 1 / dz = 1 / g , (z)dz = 0 Indr(w) = 2 r i
,z-w
21ri
r
for all closed curves r. Thus all closed curves are null-homologous in SZ, and hence (f) is equivalent to the first group of conditions. By Runge's theorem (more precisely, Corollary 1.6), (d) implies (g), which in turn implies (h). Hence, it just remains to see that (d) follows from the others. If (d) is false, then P \ SZ consists of two disjoint compacts H and K, where K is compact in C and H contains oo. Let W = Sl U K.
Since W = P \ H, it is open and K is compact in W. Take 0 E Co (W)
§5. Analytic Functionals and the Fourier-Laplace Transform
55
with 0 = 1 in a neighborhood of K, and choose a E K. Then 1
f 80 d)(()
(4.1)
OC( - a Note that a '/e( has its support in Q. (Intuitively, one should think of 8.o/8C as a curve in S2 with index 1 at o. Then (4.1) immediately contradicts (h).) If (i) holds, there is an analytic f in 0 such that f' (z - a)-i, and hence in
1
f
I.
a"f'(()dA(C)
2" f(()da(C) = f 8 8f-d)(C) = 0,
-49
which is a contradiction.
4.3 Remark. In the proof we used the fact that a compact set K in the plane is connected if and only if it has no nontrivial clopen (relatively closed and open) subset. It is not true that each component of K is clopen; for example, a Cantor set is totally disconnected, i.e., each component consists of just one single point. However, for each a E K the component of K containing a is the intersection of all clopen subsets of K that contain a.
§5. Analytic Functionals and the Fourier-Laplace Transform A continuous linear functional p on the space of entire functions A(C), p E A'(C), is called an analytic functional. It is carried by the compact set K if for each open w ID K there is a constant C, such that f E A(C), Ip(f)I IS CwsupIf1,
and any p E A'(C) is carried by some compact set K. In fact, since the family of sets
VK,,={fEA(C); suplfI<e}, K
e>0, KCC
is a basis for the topology on A(C), p-1(U) must contain some VK,,, and hence p is carried by K. By the maximum principle it follows that if P \ K and P \ K' have the same unbounded component (the one containing oo),
then p is carried by K if and only if it is carried by K'. Hence, it is
natural to restrict to carriers with connected complements. Such compacts are called polynomially convex in C (cf. Exercise 5). For example, the functional
p(f) = f0 f(z)dz is carried by any reasonable curve joining 0 and 1. For another example, let e be analytic in P\ K. Take a cut-off function X that is identically 1 in a
56
3. Analytic Approximation and Continuation
neigborhood of K, or take a set w i K with reasonably smooth boundary, and let
rf0azd)(z)
ffct.dz. ,
J
(5.1)
Since this expression is independent of the particular choice of x or w (why),
p so defined is an analytic functional that is carried by K. Actually, any functional carried by K is of this kind. To see this, first notice that by the Hahn-Banach theorem p can be extended to a functional on C(w) and hence is represented by a measure on 0. Thus we can define dp(() = , q5(z) = lL (z 1
)
f
and by varying w this defines a \' that is analytic in P \ K and 0 at oo. It is easily verified that 0 represents p in the sense (5.1). For p E A'(C) we define the Fourier-Laplace transform
(ez) . This is an entire function and p = 0 if a = 0. In fact, after differentiating and evaluating at 0 (keeping in mind that p can be represented by some measure) it follows that p vanishes on all polynomials, and the polynomials are dense in A(C). If p is carried by K, then (5.2) W D K, J,4(z)l S C.eH,(z) Fe(z)
where H,,,(z) is the 1 homogeous function
H. (z) = sup(Rez(). l,EW
Recall that a compact set K C C is convex if its intersection with each line is connected. If K is convex, then
K = {z; RezS < HK((), (E C).
(5.3)
This follows from the Hahn-Banach theorem since any real linear functional on 1R2 has the form z '--f Re (z; see also Exercise 36. An entire function h(z) that satisfies jh(z) 1 < Cexp(CIzl) is said to be of exponential type. Hence, the Fourier-Laplace transform of each p E A'(C) is of exponential type. However, the converse is also true. More precisely, we have
5.1 Theorem (Polya). If K C C is compact and convex and h(z) is an entire function satisfying (5.2), then h(z) is the Fourier-Laplace transform of a unique p E A'(C) that is carried by K.
Proof. The proof consists in constructing the desired analytic functional by means of the so-called Borel transform. Given an entire function f
§5. Analytic Functionals and the Fourier-Laplace Transform
57
of exponential type, the Borel transform is defined near oo (for z with HK(11Z) < 1) by 1
Bf(z) =z
If
(A/z)e-ada.
This clearly defines an analytic function in a neighborhood of - that vanishes at oo. We claim that if p.t is defined via (5.1) (0 = B f ), then 2 = f.
To this end first notice that B f - 0 implies that f - 0 (differentiate Bf (1/w)/w and evaluate at w = 0). Hence it is enough to show that if 0 is analytic near oo, say for IzI > R/2, and vanishes at oo and p is the w
corresponding functional, then Bµ = 46. However, since
2mµ(() =
JIwI=R q(w)etwdw,
we have for IzI > R that
27riBµ(z) = z f W
(J
0(w)ea",/zdw e--`dA. \ IwI=R By the change of variables w/z F-+ w and Fubini's theorem the expression on the right-hand side equals 0
f
O(zw)
/
O(zw)dw =
(f°°e)(w_1)dA) dw = f
IwI=R/IzI w I=R/Izl 1 - w ) where the last equality holds because 0 is analytic in IzI > R and vanishes at oo. To see that p is carried by K if K is convex, we have to verify that B f has an analytic continuation to the set P \ K. For any real 0 let Fe be the half-line 0 < t -- eiet and let
Bef(z) = f f(t)e-azdA = f
f(etet)e
tzetedt.
Since Re(etez) > HK(eie) if and only if the same equality holds for some w D K instead of K, the condition (5.2) implies that Be f is analytic in the half-plane H9 = {Ree'Bz > HK(e'B)}. For all large z with e'Bz real and positive, a simple change of variable reveals that Be f (z) = B f (z). Hence they coincide on their common set of definition, and hence B f (z) has.a continuation to the union of all the sets H9 which, in view of (5.3),
p
is precisely P \ K.
We conclude with a related result.
5.2 Theorem (Paley-Wiener). Suppose that f (z) is an entire function such that
If W1 < CeAI=I
3. Analytic Approximation and Continuation
58 and
10000 I f(x)I2dx < co.
(5.4)
Then there is a g(t) E L2(--A, A) such that f(z) = f A A
(5.5)
Note that the converse is immediate in view of Plancherel's theorem since
f (-x) is the Fourier transform of g. Proof. As in the preceding proof, we find that the Borel transform B f (z) is analytic in {IzI > A}. However, in view of the assumption (5.4), it follows that in fact Bo f (z) is defined for Re z > 0 and B, f for Re z < 0, and therefore B f (z) is analytic in the complement of the interval [-iA, iA]. Now let fi (x) = f (x)e-`I xl. Then 00 ff (t)e-'`fdt = Bof (e + it) - B,.f (-e + i.), f ( ) = foo
which tends to 0 when a -, 0 if ICI > A. On the other hand, since fE -+ f in L2, Plancherel's theorem implies that fE -+ f in L2, and hence f 0 for ICI > A. If g = j, then (5.5) holds for real z by the inversion formula, and as both sides are entire functions, it holds for all z.
§6. Mergelyan's Theorem Let K be a compact set in the plane and suppose that f is a complex function on K that can be uniformly approximated by analytic polynomials
on K. It then follows that f is continuous on K and analytic in the interior. If any such f can be approximated uniformly by polynomials, then the complement of K must be connected; cf. the discussion after Runge's theorem.
6.1 Mergelyan's Theorem. Let K be a compact-set in the plane such that the complement is connected, and suppose that f is continuous on K and analytic in the interior of K. To each e > 0 there is a polynomial such
that If - pI<e on K. Notice that Runge's theorem applies only if f is analytic in a neighborhood of K, and therefore Mergelyan's theorem is considerably stronger. In
particular, if the interior of K is empty, any continuous function can be approximated uniformly by analytic polynomials. If K is an interval, this
§6. Mergelyan's Theorem
59
is the classical Weierstrass' theorem. This section is devoted to the proof of Mergelyan's theorem.
Proof. To begin with, we can extend f to a continuous function with compact support in C, which we also denote f. This is an application of Tietze's extension theorem. Let w(6) be the modulus of continuity of f, w(6) = sup{ I f (z) - f Wk ; Iz - wI < 6}.
Since f is uniformly continuous, w(6) -' 0 when 6 --+ 0. Hence, it is enough to find, for each 6, a polynomial p such that I f(z) - p(z) I < Cw(6),
z E K,
(6.1)
where C is independent of 6. In what follows, C denotes such a constant, but it can be different in different places. Let ¢ be a smooth positive function with support in D(0, 1/2) such that f OdA = 1, and let 46(z) = 6-24(z/6). The function
4)(z) = f * 06(z) = f 06(z - w)f(w)dA(w) =
f5s(w)f(z - w)dA(w)
is then smooth, and since
f(z) - 4)(z) =
f(f(z) - f(z
w))Ob(w)dA('w),
it follows that If (z) - DI < w(6).
(6.2)
Moreover,
8 =
f
(w)f (z - w)dA(w) = f a 66 (w) (f (z - w) - f (z))dA(w)
(since f (8O6/8zv)(w)d)(w) = 0), and hence we get the estimate 84)l <
C,(6)
8z
(6.3)
6
since f IaO6/ewIdA(w) _< C/6. Thus, we have approximated f so far by the function 4, which at least is analytic at points in K that have distance more than 6 to OK. Let H denote the support of A)/8w. The crucial part of the proof is contained in the following proposition. 6.2 Proposition. There is an open neighborhood f2 of K and a continuous function r((, z) defined for ( E H and z E SZ such that r((, z) is holomorphic for z e fl, C,52
r((, z)
(1
z
,5
IS C,5 - Z13'
60
3. Analytic Approximation and Continuation
and
jr((, z) I
<-
where the constant C is independent of 6. Taking this proposition for granted, it is now easy to conclude the proof of the theorem. By formula (1.7) in Ch. 1 we have that IH (a aC)da(C) (D (z)
Now the function
F(z)
_-
Ix
r((, z)
aS
is analytic in Il D K, and by (6.3) IF(z) - -,P(z)I <
Cw(6) b
f
u
r(( , z) -
1
z
dA(()
In this integral we estimate the integrand by C/6+I(-z1-1 when 1(-zl _< 6
and by C6z/I( - z13 when I( - zI > 6. In any case, we get the estimate Cw(b) where C is independent of 6. In combination with (6.2) we get that
If (z) - F(z)I < Cw(b) on K, and since F is analytic in a neighborhood of K, we can apply Runge's theorem and obtain a polynomial p such that (6.1) holds.
It remains to prove Proposition 6.2, and to this end we need yet another result.
6.3 Proposition. Let D be a disk with radius 6 and E a connected compact subset with diameter at least 6 such that P\ E is also connected. Then
there is a smooth function r((, z) defined for z c P \ E and ( E D that is analytic in z and satisfies
r((, z) -
C62 (1
z
(6.4)
1(G6z13
and
Ir((, z)I < b ,
(6.5)
where the constant C is independent of 6.
Proof. By a change of scale and a translation we may assume that 6 = 1
and that D is the unit disk U. Note that in particular it is necessary to find an analytic function g(z) in P \ E that is bounded, and such that zg(z)
1
when
z -' oo.
(6.6)
§6. Mergelyan's Theorem
61
The latter condition means that g(oo) = 0 and that its derivative at oo is I with respect to the coordinate w = 1/z. It follows from the Riemann mapping theorem and its proof that such functions exist, and that if g(z) is such a function with minimal sup norm, then g(z) is in fact a bijection onto some disk D(0, t). Now Proposition 3.5 in Ch. 2 applies to the inverse of the function z '--b g(z/t)/t, and hence the diameter of the set tE is less than 4; so t < 4 and thus lg(z)l < 4. For fixed ( E U and 1z -- (I > 2 we have 9(z)
Q2(()
x+(z-()2+C 1z(13
If we define r((,z) = g(z) - a2(()g2(z),
(6.7)
then r((,z) -
1
z-
= C
(1z 1Ci3)
when z -+ oo. Now a2(0 =
1 JzJ=R 27ri
(z - ()g(z)dz = b - (,
where b = (1/27ri) f :-R zg(z)dz. We can change the path of integration to the unit circle, and since IgI < 4, we then get that IbI < 4 and hence also the estimate (6.5). Finally, the function
Z- (Z-O' (r((, z) -
z
1
(
is analytic in P \ E (it is bounded when z - oo and hence it has removable singularity at oo), and it is bounded by some constant C in U \ E that is independent of ( E U; therefore, by the maximum principle it is bounded by the same constant in P \ E. Thus (6.4) holds. O
Proof of Proposition 6.2. We can cover H by a finite number of disks D; with radii 26 and centers outside K. Moreover, since the complement of K is connected, in each Dj we can find a set Ej of diameter at least 26 that does not intersect K (there must be a curve from the center to the boundary that does not intersect K). For each Dj let rj ((, z) be the functions given by Proposition 6.3, and let R = f11P \ Ej. Then clearly Il D K, and if 03 is a partition of unity subordinate to the open cover Dj of the compact set H, then r((,z) _
has the required properties.
.0i(()ri((,z)
0
62
3. Analytic Approximation and Continuation
6.4 Remark. Essentially the same proof also gives the following more general result: Suppose that K C P is compact and P \ K has a finite number of components. Choose one point aj from each component. Then any f E C(K)flA(int(K)) can be uniformly approximated by rationals with poles only at the points aj.
Supplementary Exercises Exercise 1. Show that if S2 C C is bounded, then P \ Il is connected if and only if C \ S2 is. Give a counterexample when S2 is unbounded.
Exercise 2. Let S1 = {z; IzI < 1, 12z - lI > 1} and f E A(S2). Show that
there are polynomials p,n such that p - f u.c. in P. What can be said about f if p - f uniformly in S2? Exercise 3. Show that there are polynomials p,, such that limp, = 0 in
C\{0} Exercise 4. Do polynomials p exist such that
is equal to 1 when Imz > 0, 0 when Imz = 0, and -1 when Imz < 0? Exercise 5. For a compact K C C, define the polynomially convex hull K = {z E C; Ip(z)I < SUPK IPI for all polynomials p}. Note that K D K. A compact set K is said to be polynomially convex if k = K. Show that K is polynomially convex if and only if C \ K is connected.
Exercise 6. Supply the details in the following constructive proof of
Runge's theorem. First consider the set A = {a E C \ K; the functions z '-- (z - a)-k, k = 1, 2 .... can be uniformly approximated on K by rationals with poles in {aj} }. A is trivially relatively closed in C \ K. Take
aEA. For /3near a
z-b O (z-a)k+l with uniform convergence on K. Hence (z - 0) -1 can be approximated on K by our rationals and hence also (z for k > 1. Thus A is open. Since A intersects each component of C \ K, it follows that A = C \ K. If f is analytic in a neighborhood of K, as usual one can express f as a Cauchy integral where the integration is performed outside K, and this integral
then can be approximated uniformly on K by a finite Riemann sum in which the terms are functions of the type z - (z - a)-1 for a E C \ K. Exercise 7. Use Theorem 2.3 to prove the statement in Remark 2.2. Hint: Let Oj be a locally finite partition of unity subordinate to the open cover nk (i.e., locally all but a finite number of the qj vanish, E Oj = 1, and to each
j there is an nj such that O j E Co (c,,, )). Let gk = >(fk -
)¢j. Then
Supplementary Exercises
63
gk is smooth in fk and 9k-91 = fk-ft in S2kf1S2t. Now 0 = 9gk/8z defines a global smooth function in ft and if 8u/02 = ?i in St, then f = fk - gk + u is a global meromorphic function with the required properties.
Exercise 8. Show that for each b > 0 and -y > 0 there is a positive constant C7,6 such that
f (1 - I(I)-1+6 U I1-(zl1+6+7
1
-C.y,6(1-IZI)Y
( )
Exercise 9. Here is an outline of a proof of the statements in Example 2.5. (i) Prove that P is bounded on L2(U). By Schwarz' inequality, IPu(z)I2 < CJ (1 - I(12)-1/2 r (1 - ICI2)1/2Iu(()I2 I1-(ZI2 U u I1-(@I2 Then apply (*) to the first integral on the right-hand side and use Fubini's theorem. (ii) Show that the space of analytic polynomials is dense in B(U). Show first that if f E B(U) is orthogonal to all polynomials, then f = 0. (iii) Verify (2.3).
Exercise 10. Are the polynomials dense in Bp = A(U) fl LP(U)? Exercise 11. Suppose that St is bounded, /. ((, z) E C' (f) x 52), z '-- Vi((, z) is analytic for each fixed ( E 0, and El)((, () = 1. Show that
u(z) _ - in '1((, z) f
z(()
(**)
is a solution to 3u/82 = f if f E C1 (S2) and for each K C Si there is a constant CK such that sup f IV,((, z)f (()I dA(() < CK.
zEK
2
Exercise 12. Also show the converse: If 0 E C'(S2 x 0) and (**) is a solution for all f in, for example, CO '(Q), then z IL((, () = 1.
z) is analytic and
Exercise 13. Show that
K«f(z) _
f(()dA(() -1 fICI<1(1-\ 1I(I2)" - (z z
solves au/02 = f in the unit disk if I f (()I < C(1 Exercise 14. Use the operators K,,, to show that there is a solution to (2.2) in U with Iu(z)I < C(1 - IzI)-r if I f(()I < (1 I(I)-r-1, r > 0. I(I2)--°+E-1
64
3. Analytic Approximation and Continuation
Exercise 15. For a > 0, let L' be L2 in U with respect to the weight I(I2)a-1dA((). Let B,, = A(U) r1 LQ and (1 P,u(z) _
u (1 -x(12)° (zu(()dA(() 7r
Try to generalize Example 2.5.
Exercise 16. Let Q = C and take '0 in Exercise 11 as an integer power of (1 + (z)/(1 + 1(12). Are the occurring solutions minimal in some sense? What are the corresponding projection operators? Consider the same questions for Vi = exp((z). Exercise 17. One can obtain Mittag-Leflier's theorem directly from Theorem 2.3: First construct a 0 E C°°(S2\ {aj}) such that Eli -- p3 is analytic in a neighborhood of ai for each j. Show that g = (8/82)( - pj) is a global
C°° function in Q. Take u E C°°(1) such that 8u/8z = g and consider
f =1-u.
Exercise 18. Let f c Co (C) have support K (or, more generally, take a measure µ on K). Let aj be a set of points, one from each component of IP \ K. Show that 8u/82 = f has a solution with support in K if and only if fK r(z) f (z)dA(z) = 0 for all rationals with poles in the set {a3 }. Give an interpretation of the proof of Runge's theorem in the sense of distributions.
Exercise 19. Suppose that a _> 0 and limsup(a,)1/" = 1. Show that 1 is a singular point for f (z) = > a"z", i.e., show that f (z) cannot be continued along the positive real axis to the point 1. Hint: Consider the power series expansion of f around the point 1/2. Exercise 20. Assume that (f, b) is the analytic continuation of (f, D) along some curve -y. Let g be an entire function, and suppose that go f = 0
in D. Show that g o f = 0 in D. Exercise 21. Compute F(m/2) for integers m. What are the residues at the points 0, -1, - 2.... ? Exercise 22. Let ¢ E C°°(U), and define a f(a) =
a
ft l<1 (1
- Kl2Y
Show that f (a) is analytic for Re a > 0 and show that f can be continued to a meromorphic function in the whole plane with (possible) poles only
at a = -1, -2, .... In particular, it has a removable singularity at a = 0. Show that if 0 is holomorphic, then f (a) is constantly equal to 0(0). Exercise 23. Verify the limit (3.2). Hint: Let f (t) be a locally integrable function on R such that f. I f(t)Idt/(l + t2) < oc and define the Poisson integral P f (y, x) = 1 7f
J°°
Then P f (y, x) --* f (x) when y
yf (t)dt y>0 y2 + (x - t)2 0 if f (t) is continuous at t = x.
Supplementary Exercises
65
Exercise 24. Let K be a connected compact set in C. Show that C \ K is connected if and only if K has a neighborhood basis consisting of simply connected (open) sets. A family of open neighborhoods of K is a neighborhood basis if any neighborhood of K contains some set from the family.
Exercise 25. Let D be a connected domain in C. Show that the equation
p(a)u = f has an analytic solution in D for each f E A(D) and each constant coefficient linear differential operator ak
p(a) _
ak k=1
aCk
,
if and only if D is simply connected.
Exercise 26. Suppose that f E A(12), 0 is connected, f 0 0, and that for each positive integer n there is a g E A(Sl) such that g' = f. Show that f has a logarithm in Q. Hint: log f exists if and only if f,, f'dz/ f = 0 for all closed curves -y in Q.
Exercise 27. Find all jc E A'(C) that are carried by the set {0}. Exercise 28. Show that if P \ K is connected, then there is a 1-1 correspondence between the space of functionals carried by K and the set of analytic functions in P \ K that vanish at oo.
Exercise 29. Suppose that n 3 K and that 0 E A(ft \ K). Let {aj } contain one point from each component of P \ K, and let K c w cc f such that w is disjoint from {aj}. Show that 0 is the restriction of a function 1 E A(1) if and only if fd. rq5dz = 0 for all rationals r with poles in {a3}. Hint: Take a function F E C°°(1l) that coincides with 0 in the complement of a neighborhood of K. Modify F to an analytic function by solving 49u./8z = 8F/02 in an appropriate way.
Exercise 30. Let f (z) = Io anzn be of exponential type. Show that the Borel transform is given by
Bf(z) _
00 ann!z-(n+1)
0
for large Izi. Use this to prove directly that E;e = f if (P = Bf and (5.1) holds.
Exercise 31. Suppose that µ E A'(C) is carried by K. Show that in each open w 3 K there is a sequence of points aj E w and numbers cj such that /L = ECjAQ,.
Exercise 32. It is clear tnat {z; HK(1/z) < 1} C P\K. For which convex K is the inclusion strict?
66
3. Analytic Approximation and Continuation
Exercise 33. Use Cauchy's theorem to show that all of the BB(z) in Section 5 coincide on their overlaps.
Exercise 34. Verify the statement in Remark 6.4. Exercise 35. Give an elementary proof of Mergelyan's theorem when K is the closed unit disk (one cannot use the Taylor polynomials directly).
Exercise 36. Assume that K C C is convex. Show that through each point p c C \ K there is some line that does not intersect K. Use this result to conclude (5.3). Hint: Assume that 0 V- K and that each line through 0 intersects K. For each real 0, the line t '-+ eiet intersects K either for some positive or some negative t but not both. Show that the set of 0 for which the intersection occurs for positive t is both open and closed. Deduce a contradiction.
Notes Runge's theorem was published in 1885. The Poisson integral in Exercise 23 solves Dirichlet's problem in the upper half-plane; in Ch. 4, the corresponding integral in the unit disk is studied. Polya's theorem (Theorem 5.1) is from 1929; see [B] for references and further results about functions of exponential type. Mergelyan's theorem is from 1952; see AMS Translations 101 (1954). For a different proof see Carleson, Math. Scand., vol. 15 (1964).
4
Harmonic and Subharmonic Functions
§1. Harmonic Functions A function u E CZ(SZ) is harmonic in Sl if Du = 0. If u, v are harmonic and a, /3 E C, then au + /3v is harmonic. If f E A(SZ), then f, f, Ref , and Im f are harmonic in Q. If, in addition, f is nonvanishing, then log If is harmonic. If Il = C \ {0}, then log IzI is harmonic in 11, but there is no f E A(Sl) such that log I z I = Ref. However, we have
1.1 Proposition. If u is real and harmonic in a simply connected domain Sl, then u = Ref for some f E A(SI).
Proof. Since 28u/8z E A(fl) and Sl is simply connected, there is a g E
A(Q) such that g' = 8g/8z = 28u/8z. Thus 8(g + g)/8z = 28u/8z,
and since u is real, 2u = g + + 2c for some real constant c; therefore,
0
u=Re(g+c).
In particular, we find that harmonic functions are C. Moreover, we have the following uniqueness property: If u is harmonic in a connected set Sl and vanishes in an open subset, then u vanishes identically in Q. In fact, one may assume that u is real. Then the statement follows from the identity theorem for analytic functions applied to 8u/8z. If u is harmonic in a neighborhood of D(z, r), then 2n
u(z) = 2- j u(z + re`e)dO,
(1.1)
0
i.e., u has the mean value property. Again we may assume that u is real, and then (1.1) follows from Proposition 1.1 and the mean value property for analytic functions. From (1.1) and the uniqueness property it follows that a real harmonic u has no local maxima or minima unless it is locally constant. Analogously to Corollary 1.13 in Ch. 1, we also have the following.
68
4. Harmonic and Subharmonic Functions
1.2 The Maximum Principle. If 52 is bounded and u E C(f) is real and harmonic in 52, then u attains its maximum and minimum on 852. In particular, u = 0 if u = 0 on Oh. We now shall derive a generalization of (1.1) that represents the value at an arbitrary interior point in terms of the boundary values on the disk. If u is real and harmonic in a neighborhood of U, then, by Proposition 1.1,
u(z) =
2 E anzn + 1: dn.Zn 0
i Anr1n1eine
(1.2)
= -M
0
(where z = re") with absolute convergence on U. Hence, u(reie) =
u(eie)
1 o
00 00
rInIe.n(e-t)dt'
r < 1
(replace reie with e't in (1.2), plug it into the integral above, and use the fact that f e'mtdt = 0 for nonzero integers m), so that
u(re'B) = 2n
f z, u(e:c)p, (B - t)dt = 0
where Pr(t) _ E kernel is
Pr(B - t)
00
rInle`nt.
27r
fz u(et(e-t))P,(t)dt, 0 < r < 1, 0
If z = re"' and =>rknlein(O-t) = Re
= eat, then the Poisson
(eu +z\ (\ eat - z/}
1 - r2
1 - 1z12
1 - +z12
1 -2rcos(B-t)+r2
11-Cz12
1C-z12
Verify the equalities! The second one reveals that the Poisson kernel is a harmonic function of z = re'0 for fixed eat.
Exercise 1. Show that Pr(t) = Pr(-t), t ti Pr(t) is strictly decreasing on (0,7r), Pr(6) --. 0 when r Z 1 for fixed 6 > 0, and that (1/27r) f Pr(t) = 1 for all r < 1; cf. Figure 1 in Section 1 of Ch. 6.
1.3 Remark. Since harmonicity is preserved under translation and dilation, we get that f2w 1 R2 - r2 u(a + react) = 27r R2 - 2Rr cos(O - t) + r2 u(a + Re't)dt r < R if u is harmonic in a neighborhood of D(a, R).
§1. Harmonic Functions
69
For f E C(T), one defines the Poisson integral Pf of f as P f(reie)
r2w
f (e't)P,(0 - t)dt
=
J0
12, 27r
f (e`(0-t))Pr(t)dt,
r < 1,
which is a harmonic function in U since the Poisson kernel is harmonic.
1.4 Proposition. If f E C(T), then Pf is harmonic in U; and if Pf is defined as f on T, then Pf E C(U). Dirichlet's problem is the following: Given a function on Oil, find a harmonic function F in 11 such that F = f on Oil. To discuss its solvability,
one of course must specify which class of domains and functions f one allows, and the exact meaning of f = F. For bounded domains i2 with reasonable boundary (e.g., piecewise C'), any continuous f on Bit has a continuous harmonic extension F to Il, which in view of the maximum principle is necessarily unique. In this book we essentially will restrict our attention to the unit disk. Proposition 1.4 says that the solution to Dirichlet's problem in U for continuous f is given by the Poisson integral
Pf.
Proof of Proposition 1.4. We shall prove that P f (re") - f (eie) uniformly in 0 when r / 1. Given e > 0, take b > 0 such that If (e'(e-t)) f(eie)I < E if Iti < 6. Now
- f(e'e) =
27r
fw2Pf(Te'e) (f (e'(e-t)) - f(eie))Pr(t)dt.
In the set b _< ItI < -7r, Pr(t) < Pr(b) (cf. Exercise 1), whereas in the set 0 < ItI < 6, I f (e'(o_t)) - f (e'B)I < E. Since Pr(t)/27r is positive and has integral 1, we get the estimate
IPf(reie) - f(eie)I 5 2IIfjI.Pr(b) +E, which is less than 2e if r is sufficiently close to 1.
0
We now consider a converse of the mean value property.
1.5 Theorem. Suppose that u E C(i2) and that, for each z E 0, (1.1)
holds for small r > 0. Then u is harmonic in Q.
Proof. We may assume that u is real. Moreover, it is enough to prove that u is harmonic in each disk D CC D. Given such a D, take the harmonic function h such that h = u on 8D. Then v = u - h also satisfies the hypothesis in D. If v had a positive supremum in D, then it would
70
4. Harmonic and Subharmonic Functions
be attained on a nonempty compact set K C D, and this in turn would violate (1.1) for Z E K with minimal distance to 8D_ Thus, v < 0 in D. For the same reason, v > 0, and therefore u = h in D. The proof above shows that it is enough to assume that for each z E Il there is a sequence rn \ 0 for which (1.1) holds. From Theorem 1.5 it follows that u is harmonic if u -. u u.c. and each un is harmonic.
1.6 Weyl's Lemma. If U E L o_(cl) (or u E D'(0)) and f uE4 = 0 for all q5 E Co , then u is harmonic; more precisely, there is a harmonic function v in fI such that u = v a.e. (or as distributions). We will not rely on this result and leave the proof as an exercise; it is in fact a consequence of Theorem 2.12 below.
1.7 Harnack's Theorem. If u are harmonic in a connected domain 0 and ul < u2 < u3 ... , then either u,, / oo for all z E S2 or u. / u < 00 U.C. in 0.
Proof. Take D(a, R) C ft. We may assume that ul > 0. Since (R - r)2 < R2 - 2rR cos(B - t) + r2 < (R + r)2, we have, by Remark 1.3, 7F-+-r
R-r
r
R.
Thus, if u,(a) J' oo, then u / oo in a neighborhood of a. On the other hand, if u,, (a) J' u(a) < oo, then u = limun < oo in a neighborhood of a. Thus, A = {z; u(z) = oo} is open and closed, i.e., ll or 0. If u J' u < oo, then u satisfies the Poisson formula by monotone convergence, and hence it is harmonic. However, if continuous functions u tend to a continuous limit u monotonically, then the convergence must be u.c. (this is sometimes
called Dini's lemma). In the present situation one also can see that the convergence is u.c. from the representation by the Poisson integral.
1.8 Schwarz' Reflection Principle. Suppose that l is a connected domain, symmetric with respect to the real axis, and that L = Il fl R is an interval. Let S2+ = {z E 11; Im z > 0}. Suppose that f E A(S2+) and that Im f has a continuous extension to S2+ U L that vanishes on L. Then there is a F E A(S2) such that F = f in S2+ and F(z) = f (z) in 11 \ 52+.
Proof. If v = Im f is extended to S2 \!n+ by letting v(z) = -v(z), then v E C(1l) and has the mean value property at each point; therefore, it is harmonic by Proposition 1.5. In a simply connected symmetric neighborhood w of L in f2 there is a g E A(w) such that v = Im g and f = g in
§2. Subharmonic Functions
71
f2+flw (why?). Since g is real on L, its power series at points on L has real
coefficients, and therefore g(i) = g(z). Let F(z) = f (i) in f2 \ (S2+ U L)
and F = f in 52+. Then F is analytic in 1 \ L, but F = g in w \ L, and hence F extends to an analytic function in f2.
Exercise 2. Let f2 be as above. Suppose that f E A(S2+) and that - 1 on L. Formulate the corresponding reflection theorem. Then assume instead that fl is an appropriate neighborhood of some connected subset L of T and 0+ = U fl 0. Formulate the reflection theorem in this case. Hint: Use a conformal mapping from the upper half-plane onto U. Jfj
§2. Subharmonic Functions Subharmonicity is a fundamental concept in complex analysis. Many important properties of analytic functions depend simply on the fact that the logarithms of their modulii are subharmonic. Subharmonic functions are related to harmonic functions in much the same way as convex functions (in one or several real variables) are related to linear functions.
Definition. A function u in fl with values in [-oo, oc) is subharmonic, u E SH(11), if (a) u is upper semicontinuous.
(b) for each K C Sl and h E C(K) that is harmonic in int(K) and > u on OK one has that h > u on K. Sometimes one excludes u - -oo from the definition. It is clear that real harmonic functions are subharmonic. Note that if u is upper semicontinuous on a compact set K, then SUPK u is finite and attained at some point
on K. Hence, if u E SH(fl) and K C fl, then supK u is attained on OK (why?).
2.1 Theorem. Suppose that u is upper semicontinuous in 11 with values in [-oo, oo). Then the following conditions are equivalent: (i) u is subharmonic. (ii) If D = D(z, r) cc 0, h E C(D) is harmonic in D and > u on OD, then
h>uinD.
(iii) If D = D(z, r) cc fl, then 1
u(z) <
2,r
r
u(z + re'8)dO.
(2.1)
0
(iv) For each z E S2, (2.1) holds for small r > 0. As u is bounded from above on compact sets, the integral in (2.1) makes sense and takes values in [-oo, oo). From condition (iv) it follows that sub-
72
4. Harmonic and Subharmonic Functions
harmonicity is a local property. Since a harmonic function u has the mean value property, it follows that Jul satisfies (iii) and hence that Jul is subharmonic. If f is analytic in ft, then log if I is upper semicontinuous, takes values in [-oo, co), and satisfies condition (iv) as it is harmonic outside the zeros of f . We therefore obtain
2.2 Corollary. If f is analytic in fI, then u = log If I is subharmonic in Q.
Proof of Theorem 2.1. Trivially (i) implies (ii). Assume that (ii) holds and take D(z,r) CC ft. Since u is upper semicontinuous, there are continuous functions hi such that h; \ u on OD (prove this!). Let Hj be the harmonic extensions of hj to D. Then z'. u(z) < H3(z) = 2 /
hj (z + re'o) dO,
0
according to (ii). Now (iii) follows by monotone convergence.
Clearly, (iii) implies (iv). Assume that (iv) holds and let K C 0, h E C(K), h > u on OK, and h be harmonic in int(K). Then v = u - h < 0 on 8K and satisfies (2.1) at interior points for small r. As v is upper semicontinuous, the same argument as in the proof of Theorem 1.5 gives
0
that v
2.3 Proposition. If ¢ is increasing and convex on R and u E SH(SZ), then 0 0 u E SH(Q) (where (0(-oo): = limx-_. O(x)). Proof. First notice that ¢ou is upper semicontinuous, since (k is increasing and continuous (any convex function on R is continuous). Moreover, for small r, o u(z) <_ 0
(2 I o
u(z + reie)dO
/
<_
2x I
0 o u(z + rese)d6,
o
where the first inequality holds because 0 is increasing and the second one follows from Jensen's inequality. Now the proposition follows from Theorem 2.1.
0
2.4 Example. If f E A(fl), then not only is log If I subharmonic but, e.g., log' If 1: = max(log If 1, 0) and if 4 a > 0 also are. This follows immediately from Proposition 2.3 by taking O(x) = max(x, 0) and O(x) = exp(ax), respectively.
2.5 Proposition. (a) If u E SH(f2), u = sups u0, < oo, and u is upper semicontinuous (e.g., if {a} is finite), then u E SH(f2). (b) If uj E SH(fl), then ul + U2 E SH(Q) and cut c- SH(Q) if c > 0.
(c) If uj \, u and u, E SH(]), then u E SH(fl).
§2. Subharmonic Functions
73
Proof. (a) and (b) immediately follow (e.g., from Theorem 2.1). For (c), first note that {u
2.6 Theorem. If u E SH(S)) and D(z, R) C Q, then
r -+ m(r) =
-J
r2ir
u(z + re'B)d8,
0
0
is increasing and m(r) \ u(z) when r \ 0. In fact, m(r) is continuous for r < R; see Exercise 32.
Proof. Take rl < r2 and continuous h.; \ u on 8D(z, r2), and let H; be the harmonic extensions. Then u < H1 on D(z,rl) and hence
m(rl) <
1
r
2i
2w
Hj (z + rleie)dO = Hj(z) =
JO
hj(z + r2e4a)d9;
but the right-hand side tends to m(r2) by monotone convergence, and thus
r .-+ m(r) is increasing. If u(z) < a, then u(z + reie) < a for small r > 0 because of the semicontinuity, and thus lim sup,,,o m(r) < u(z). Since u(z) < m(r), the theorem follows.
2.7 Theorem. If Sl is connected, u E SH(1), and u 0- -oo, then u E Lio0(Sl); in particular, u > -oo a.e.
Proof. From (2.1) we get that u(z) <
12
err
f
u(z + ()dA(()
!C<,.
if D(z,r) CC Q. If u(z) > -oo, it thus follows that u is integrable in D since u is bounded from above on D. Let
E _ {z E St; u is integrable in a neighborhood of z}. Then trivially E is open. From the argument above it follows that u - -oo in a neighborhood of each point in Sl \ E, and thus E is closed.
2.8 Corollary. If 0 is connected and u are > -oo for all r > 0.
-oo, then the integrals in (2.I)
Our next objective is to approximate an arbitrary subharmonic function with smooth ones. Let u E SH(Sl), u 05 -oo, and choose a positive 0 E Co (U) such that V)(z) = zP(IzI) and f ipdA = 1. If 7P,(z) = e-2i(i(z/e), then
4. Harmonic and Subharmonic Functions
74
fOdl=land u, = u *?P, = J u(z - e()'+l)(()dA(() is in C°°(II ), where 1 C = {z E 52; d(z, c9Q) > e}.
2.9 Proposition. uE E SH(l) fl C°°(12,) and uE \ u when e \ 0. Proof. Take an arbitrary point in Q. With no loss of generality we may assume that it is 0. Then by Fubini's theorem and (2.1),
f 2r f u(reie - c()dO P(()dA(() 1
u,(0) = f 1
2J
5
2ir
27r
uE(reie)dO,
so (2.1) is satisfied for uf, and hence it is subharmonic. Now 2fr
uE(0) = ZOO f u(-ere'B)dOrz/i(r)dr, 0
which decreases when a decreases and has limit fo 21ru(0)r i(r)dr = u(0) by monotone convergence and Theorem 2.6, i.e., u, \ u when e \ 0.
In particular, if u, v E SH(fl) and u = v a.e., then u - v. 2.10 Lemma. If U E C2(1l), then u is subharmonic if and only if Du > 0 in Sa.
Proof. Suppose that Du > 0. Take a disk D cc Q, an h harmonic in a neighborhood of D, and > u on 8D. For simplicity, suppose that D = D(0, r). If e > 0, then v = u - h + CIz12 - ere is a C2 function such that Ov = Du + 4c > 0 and v < 0 on 8D. If supD v > 0, then v has a local maximum at some p E D, and thus the Hessian at p must be negatively semidefinite, which contradicts the fact that Av= the trace of the Hessian > 0. Thus, v < 0 in D, i.e., u - h < er2 in D, and since a is arbitrary, then u < h in D. This proves that u is subharmonic, according to Theorem 2.1. Conversely, assume that u is subharmonic and take a point, say 0, in 0. By Taylor's formula, u(re'B) = u(0) + 2 Re uZ (0)reie + Re uxz (0)r2e2iB + u-,2 (0)r2 + o(r2).
Integrating this equality and using (2.1) we get 0 < u,j(0)r2 + o(r2)
for small r, which implies that iu(0) = 4u-.t(O) > 0.
§2. Subharmonic Functions
75
2.11 Theorem. If u is subharmonic in Il (and 0 -oo on each component), then u0OdA > 0,
0 < 0 E Co (p).
(2.2)
Thus, 0 -* f u0OdA is a positive linear functional on CO A, and hence it is represented by a positive measure (see Exercise 28) that we denote by
Du or Dud.. Proof. Take u, \ u, which arer C°° and subharmonic. Then fui&cb = lim / U, AO = lim
f
¢Du, > 0.
0
2.12 Theorem. If u E LI°c(l1) (or D'(1)) and (2.2) holds, then there is a unique v E SH(1l) such that u = v a.e. (or in D'(Sl)). In particular, this implies Weyl's lemma for harmonic functions.
Proof. Let u, = u * 0, as before. Since Du, = u * DOE _> 0 by (2.2), u, are subharmonic by Lemma 2.10. However, then u * 0E * 06 < u * 0, * 06' if b < 6', i.e., u * 06 * 0, < u * 06, * (k,. Letting e \ 0, we find that u * O6 is subharmonic and decreases if 6 decreases. By Proposition 2.5 (c), it follows
that v = lim6,,a u * 06 is subharmonic, and since u * 06 - u in Lioc (see Exercise 27), we can conclude that u = v a.e. Thus, roughly speaking, u is subharmonic if and only if Du is a positive measure.
2.13 Example. Let us compute A log If I if f is analytic. Suppose that f E A(Q) has zeros at al , a2.... with multiplicities M1, m2, .... As A log Izl = 21rb (see 1.7 (c) in Ch. 1 or the proof of Jensen's formula below), it follows
that Olog If l = 21r
rnjba,,
where b,,; is the point mass at e 9j.
Let us consider the inhomogenous Laplace equation. If f E C°°(ul), then there is a solution to
Du = f
(2.3)
in C°°(1E). This follows from Theorem 2.3 in Ch. 3 applied twice. If f behaves reasonably well near the boundary, then the Newton potential of
76
4. Harmonic and Subharmonic Functions
f,
u(z) = 27<
flog Iz - (If(()dA((),
(2.4)
is a solution in Q. In general, one is interested in a solution with prescribed
boundary values. If v is a solution to (2.3) in U, which is continuous up to the boundary, then certainly u = v - Pv is the unique solution that vanishes on the boundary.
2.14 The Green Potential. For G((, z) =
z EIU let
27r
log ! 1
- Sz
. 1
For fixed z E U, G((, z) = 0(1 - I(I); thus, if u is a positive measure in U such that
J(1 - I(I)du(() < oo,
(2.5)
we can define the Green potential
gu(z) = f G((, z)du(() We leave it to the reader to verify that if p is any positive measure that satisfies (2.5), thenCp(z) is a negative subharmonic function in U that solves Au = p in the distribution sense and vanishes on the boundary in the sense that
lim 1 r71 27r
f
p
2,r
gp(re`e)dO = 0.
If p = f is a bounded function, then g f has a continuous extension to U, and hence it vanishes on the boundary and is therefore the unique solution to (2.3) in U that vanishes on the boundary. Consequently, for any v in, for example, C2(U), we have the Riesz decomposition
v(z) = G(Av)(z) + Pv(z).
(2.6)
Later on it will turn out that this decomposition holds for all subharmonic functions satisfying (2.8). We conclude this section with Jensen's formula, which expresses the con-
nection between the growth of the mean values m(r) and the size of Du for a subharmonic function u.
Supplementary Exercises
77
2.15 Jensen's Formula. If it E SH(D(a, R)) and it 0 -oo, then u(a)
u(a
2R
+ re'B)dO
= 1 fo 1
27r
Iz_.J
log
(2.7)
r I
z-a
I
Du(z)dA(z),
r < R.
Proof. Let a = 0 and assume first that it is C2. Green's identity applied to the domain {e < IzI < r} yields c
log
r Du(z)dA(z)
zI
2'
J u(re'B)dO + J u(ee`e)dO + O(e loge), 0
0
which gives (2.7) when e \, 0. For a general subharmonic it, let u f = u * 1/i, as before. Since (2.7) holds for each uE, we just have to ensure that we can take limits. If g(z) = log(r/lzJ) is defined as zero outside {IzI < r}, then g is continuous except at the origin. Now observe that
f g(z)ouE(z) = J g(z) E * Du(z) = / g *'pE(z)Au(z). However, g * ip, --+ g uniformly outside a neighborhood of the origin (see Exercise 27) and g * i(iE / g in a neighborhood of the origin (since -g is
subharmonic there); therefore, f giu. - f gOu. Since the convergence of the other terms in (2.7) is obvious, Jensen's formula is proved.
2.18 Corollary. If it E SH(U) and
sup] r<1
27r
u+(re'o)d8 < oc,
(2.8)
0
then y = Du satisfies (2.5). We leave the verification of this as an exercise. Conversely, if p satisfies (2.5), then GA is a solution to (2.3) such that (2.8) holds; cf. 2.14.
Supplementary Exercises Exercise 3. Suppose that it E C2(Sl). Show directly that Du = 0 if (1.1) holds for small r > 0. Exercise 4. Suppose that Sl is connected. Show that if each real harmonic u is the real part of some analytic function in Il, then 0 must be simply
78
4. Harmonic and Subharmonic Functions
connected. (Together with Proposition 1.1, this establishes the equivalence of condition (k) to the other conditions in Theorem 4.1 in Ch. 3.)
Exercise 5. Assume that f and f2 are harmonic (and SZ is connected). Show that f or f is analytic. Exercise 6. Suppose that u and v are real and harmonic. If uv is also harmonic, what is the relation between u and v? Exercise 7. What can be said about {Vu = 0} if u is harmonic? Exercise S. Show that u o f is harmonic if u is harmonic and f is analytic. Exercise 9. Suppose that u E LIoc(SZ) and u = u(a)
1
IDI
fD
for all D = D(a, r) cc Q. Show that u is harmonic. Exercise 10. Suppose that fj E A(ul), Il is connected, f, (a) converges for some a E St, and Re fj converges u.c. Show that fj converges u.c. Exercise 11. Suppose that uj u in L10. (c) and uj are harmonic. Show that uj u U.C. Exercise 12. Prove the maximum principle for harmonic functions by using only the fact that they satisfy Du = 0. Exercise 13. Show that if u is real, harmonic in C, and bounded from above, then it is constant. Exercise 14. Suppose that u is real and harmonic in {0 < IzI < 2}, and u(z) = o(log(1/jzj) when z -+ 0. Show that it has a removable singularity at 0. Hint: Let v be the harmonic function in U that equals u on the unit circle. Note that for any e > 0, u - v - e log IzI < 0 on a{6 < IzI < 1} if 6 is small enough.
Exercise 15. Show that it is enough to verify (ii) in Theorem 2.1 for all h that are harmonic in some neighborhood of D. Exercise 16. Suppose that f is analytic and nonconstant. Use the preceding exercise to show that log Ill is subharmonic. Hint: Suppose that g is analytic in a neighborhood of D cc S2 and that log Ifl < Re g on 8D. Then the maximum principle applied to f exp(-g) implies that the inequality holds in D as well.
Exercise 17. A function u E C2(1l), Sl C R" is harmonic if !1u = 82u/8x = 0. Prove the mean value property and the maximum princi-
ple for such functions.
Exercise 18. A continuous real function u in SZ C R is convex if t F-* u(x + ty) is convex for all x E S2, y E R", and small real t. Show that a
Supplementary Exercises
79
u E C2(1l) is convex if and only if the Hessian (32u/aX,8xk)jk is positively semidefinite for each x E H.
Exercise 19. Show that u is subharmonic if it is convex. Give a counterexample for the converse.
Exercise 20. Show that if u, v > 0 and log u and log v are subharmonic, then log(u + v) is subharmonic. Exercise 21. Supply proofs for the statements about the Green function in 2.14.
(a) Suppose that f E C o (C). Show that 0 f log 1 - zj f (()da(() = 27r f . (b) Show that Cz(z) < 0 and is locally integrable in U. (c) Show that Gu(z) is subharmonic. (d) Show that Lg/2 = Fc. (e) Show that (1/27r) f f' Ggp(re`0)dO / 0 when r (f) Show that 9f (z) is continuous up to the boundary if f is bounded. Hints: For (c), one can consider 9Rp(z) obtained from the kernel GR((, z) _
max(G((, z), -R). For (f), one can observe that 9f has bounded derivatives in the interior.
Exercise 22. Suppose that u E SH(U) is nott identically -oo. Show that
sup J u+(re'B)d9 < oo if sup r<1
1/2
< oo.
Exercise 23. Prove Corollary 2.16! Exercise 24. Formulate Jensen's formula when u = log If I, where f is analytic.
Exercise 25. Suppose that u E SH(U) and that limsupu(z..) < 0 for any sequence zn that converges to a boundary point. Show that u < 0 in U. Exercise 26. Suppose that u is upper semicontinuous on T. Show that there are continuous un such that u \ u. Exercise 27. Take 0 E COI(U) such that f ,bdA = 1. Let Vi, (z) _ f 29)(z/e). Show that x(,E * f -. f in Lio, if f E Ll . and that V), * f - f u.c. if f is continuous. Exercise 28. Show that if A is a positive linear functional on Q02(0), then there is a positive measure µ in 1 such that AO = f ¢dµ for all 0 E C02(0). Exercise 29. Prove Weyl's lemma 1.6; cf. (the proof of) Theorem 2.12. Exercise 30. Show that u(z) = log(1/d(z,1 )) is subharmonic in fl. Exercise 31. Find all subharmonic u in U such that u(z) = u(IzD. Exercise 32. Show that m(r) = f u(re'B)dO is continuous if u is subharmonic. Hint: First suppose that u(z) = u(IzD.
80
4. Harmonic and Subharmonic Functions
Exercise 33. Let S be a doubly connected domain with smooth boundary, and let us assume that the Dirichiet's problem can be solved with continuous boundary data (which is true). Then there is a harmonic u(z) such that u = 0 on the boundary of the bounded component and R > 1 on the other one. Show that if R is appropriately chosen, then u has a harmonic conjugate v in Il that is well-defined modulo an integer multiple of 2-7r. Then h(z) = exp(u(z) + iv(z)) is a well-defined analytic function in f2 whose image is contained in the annulus A(1,expR). Show that h(z) is in fact a conformal equivalence. Hint: First show that h is proper, i.e., h-1(K) is compact for each compact K. Then apply the argument principle to cycles in Q approximating an.
Definition. A set E C C is polar if there is a subharmonic u
that E C (u = -oo}.
- oo such
Exercise 34. Show that each countable set is polar. Also show that any polar set has measure zero, and supply a counterexample for the converse. Exercise 35. Suppose that u, E SH(ft), sup,, u,, < M, where M is locally bounded in 52, and lim sup,,-,,,, u,2 < 0 at each point in 52. Show that for each compact K C S2 and e > 0 there is an N such that u < e on K for n > N. Exercise 36. Show the three-circle theorem: If f c A(A(1, R)), then log If(z)1 S
log
ll
IIQa log b/a logm(a) + tolog z/ logm(b), if 1 < a < Izl < b < R and m(r) = sup,z,_,. l f (z) I.
Notes In 1R2, harmonic functions can be considered either as real parts (locally) of analytic functions, or as homogeneous solutions to a real elliptic equation (the Laplace equation). I have tried to shed light on both aspects, and consequently some facts are proved (at least implicitly) twice. For instance, the maximum principle is derived from the mean value property (1.1) (which in turn follows from the mean value property for analytic functions, as well as from Jensen's formula (2.7)), and also directly from the Laplace equation (see Lemma 2.10). There is a close connection between Dirichlet's problem (DP) and the Riemann mapping theorem. For instance, the latter implies solvability of DP in all simply connected domains (with reasonable boundary). In Exercise 33, it is suggested how one can obtain a conformal equivalence from a solution to DP. Solvability of DP under weak assumptions of the
regularity of the boundary is studied in Potential theory, in which the
Notes
81
Newton and Green potentials, polar sets, and capacity are basic concepts; see [Raj for a nice introduction. The family of functions (?GF),>o in Proposition 2.9 and Exercise 27 is
called an approximate identity (in R2). The Poisson kernel Pr(t) is an approximate identity on the circle T. Previously we have met approximate identities in Section 6 and Exercise 23 in Ch. 3.
5
Zeros, Growth, and Value Distribution
§1. Weierstrass' Theorem Our first aim in this chapter is to prove that any subset of S2 C C that has no limit point in 0 is the zero set of some f E A(fl) (Weierstrass' theorem). To this end we have to consider infinite products.
1.1 Lemma. If z are complex numbers, then N
N
(1 + zn) - 1 <M1+IznI)-1. 1
Proof. Since k+1
k
(1 +
1 = (1 + zk+l) (11(1 + zn) - 11 + zk+1,
it follows by induction over k that k+1
\
J k
(1 +
11
+ IznI) -
< (1 + Izk+1 I)
+ Izk+I I
(11(1 _
k+1 fl(1 +
1.
1
1.2 Proposition. If f E A(12), f 0 -oo on each component of Il, and F 11 - fnI converges u.c., then 00
N
f(z) _ fl fn (z) = limflfn(z) 1
1
§1. Weierstrass' Theorem
83
converges u.c. in Q. Moreover, the zeros off are the union of the zeros of fn, counted with multiplicities.
Proof. Take fixed K C Q. Note that N
N
N fnI
1
< 1 1 (1 + 11 - f n 1 ) < - exp E 11 - f n l < C 1
111JJ
I
on K, independently of N (where we use that 1 + x < expx). If N _> M, then by the lemma (Zn = fn - 1) and (1.1), N
fn -
fn 1
1
N
M
M
11 fn-1
fl fn
M+1
1
N
r N
_
I1-fnI-1
M+1
M+1 1
which tends to 0 uniformly on K when N > M -+ oo; therefore, the product converges uniformly on K. Thus, the first statement is proved. For the second statement, note that each z E S2 has a neighborhood where at most a finite number of fn have zeros, since locally EM+1 11- ffI < 1/2 if M is large enough. Now rj° fn = IIM In fM+I fn and N
N
M+1
M+I
11 fn-1 <exp1: 11-fn1-1<expl/2-1<1, 0
and therefore jlM+i fn has no zeros near z. Let
p
z3 Wo=1-z, Wp(z)=(1-z)exp>-, p> 1. 1
1.3 Lemma. WW(z) = 0 if and onlyifz = 1; moreover, I1-Wp(z)I < jzlp+1 in U.
Proof. If h = 1 - Wp, then h(0) = 0 and h'(z) - zpexpE'z'/j. Hence, O(z) = h(z)/zP+l is analytic and, since 0 has positive Taylor coefficients, IO(z)I < O(IzI) < 4(1) < i for Izj < I. We now are prepared to construct functions with prescribed zeros.
1.4 Weierstrass' Theorem. Suppose that {an} is a sequence in S2 C P with no limit point in i, Il # P, and mn is a sequence of positive integers. Then there is an f E A(fl) that has zeros of orders mn at an and no others.
84
5. Zeros, Growth, and Value Distribution
Proof. We may assume that oo E 0 and that oo # an for each n. Let an be an enumeration such that each an occurs precisely mn times. Take Qn E IP \ S2 such that 10n - an I = d(an, an). We claim that
0
RR
f(z)=HWf(an-Nn)
z - Qn
has the required properties. In fact, since oo is not a limit point of {an}, 0 when n -p oo, and hence there is for a given f3n - an I = d(an, 812)
K C Il an N such that Iz - f3nI > d(K,aSZ) > 2lan - Nnl if n > N and z E K. Thus, fan - f3nl/Iz -#,I < 1/2 if n > N and hence, by Lemma 1.3,
/(-Nn an-0
)I<2_n'
n>N,zEK.
It now follows from Proposition 1.2 that f has the desired properties. Certainly this theorem, as well as the next two corollaries, fail if 12 = P.
1.5 Corollary. If g is meromorphic in Il C C, then g = f /h for some f,h E A(SZ).
Proof. Take an h that has zeros where g has poles, and with the same multiplicities. Then f = gh is analytic.
1.6 Corollary. If aj is a sequence with no limit point in SZ C C, fj is analytic near a and mj are positive integers, then there is an f E A(12) such that f(z) - f, (z) = O(Iz - aj +m,+1) when z - a3. Thus, the Taylor expansion of f can be prescribed up to order mj at each aj. Proof. Take g E A(12), which has zeros of orders mj + 1 at aj. According to Mittag-Leffier's theorem, there is a meromorphic h in SZ having poles
only at aj such that h - fj/g is analytic near aj. Then let f = gh.
1.7 Corollary. There is an f c A(1) that cannot be continued to any larger domain. More precisely: If a E Il, g E A(D(a, r)), and f = g near a, then D(a, r) C S1.
Notice that for a general domain the statement about f is stronger than: If a E SZ, g E A(D(a, r)) and f = g in D(a, r) f1 Il, then D(a, r) C 12. For instance, if f2 = C \ {z; Im z = 0, Re z _> 0}, then log z in SZ satisfies the latter statement but not the one in the corollary.
Proof of Corollary 1.7. Let an be an enumeration of all points in Il with rational coordinates such that each such point occurs an infinite number
§2. Zeros and Growth
85
of times. Let K be an increasing exhausting sequence of compact sets in 11, and choose for each n a w E 12 \ K such that Ia - w,1 < d(a,,,1911).
Since any fixed compact set is contained in some K, the sequence w has no limit point in 1I, and hence there is an analytic function f with zeros w and no others. Now, if a E 0 has rational coordinates, then the largest open disk D(a, r) included in 11 will contain infinitely many points
w,,, and hence no function that equals f near a can be analytic (or even meromorphic) in any larger disk with center a. Prom this the corollary follows.
§2. Zeros and Growth So far we only have considered the existence of analytic functions with prescribed zeros. We now shall study the connection between the "number of zeros" and the growth. As a simple example, note that if an entire function has m zeros (counted with multiplicities), then it must grow at least as fast as lzlm when z oo. This is an instance of a general phenomenon which,
roughly speaking, means that "the more zeros f has, the faster If I must grow." Conversely, if there are not too many prescribed zeros, one can find a function with these zeros that does not grow too fast.
If f is analytic in the disk D(0, R), we can measure its growth by its characteristic function 2n
T(r, f) = 2a f
log
1 + If(re'B)I2dO - log
1 + Jf(o)i2.
1 + If (z) 12 is subharmonic (see Exercise 20 in Ch. 4), T(r, f) is a nonnegative increasing function for r < R and T(0, f) = 0.
Since z +-- log
Note that
T(r, f) =
1
2,T
f log+if(re'B)IdO+0(1) 0
when r / R, since log 1 + x2-log+ xI is less than some constant (log f) for all x > 0. In the same way, one verifies that T (r, f - a) = T (r, f) + 0(1) for each fixed a and that T(r, af) = T(r, f) +0(1) if a 54 0. Clearly, control of T(r, f) puts some growth restriction on f, contrary to f log if (re'o) ado, which is constant in r if f is nonvanishing. If f is analytic in the disk D(0, R) and r < R, we let n(r, f) denote the number of zeros of f on the closed disk { Iz r}; and if f (0) # 0, we let
N(r,f) =
f(f)dR
If f (0) = 0, the expression
N(r,f) - N(ro,f) = f n(s,f) r.
ss
86
5. Zeros, Growth, and Value Distribution
is at least well defined, and its growth when r / R does not depend on the choice of ro. If a1, a2, ... are the zeros of f (counted with multiplicities), we claim that
N(r, f) =
r
log
(2.1)
.
Ia,i I
Ia,i
In fact, if Iai I < la2l < .... we have r f(f)dS =J 1
1
M-1
/ Iak+l I
n(s,f)+ f n(s,.f)s
Iakl
s
s
aMI
r klog lak+1l +Mlog IaMI lakl
M
r
log Iakl
A more suggestive verification can be obtained by noting that dn/ds = Ei bIakl in the distribution sense, since (2.1) then immediately follows from a (formal) integration by parts.
2.1 Jensen's Formula. Let f be analytic in D(0, R), r < R, and let an be the zeros off on D(0, r), ordered such that Ian I < Ian+11. Assume that f (0) # 0. Then 1
N(r,f) _
f2wlogif(re'o)Id9-logif(0)I. (2.2)
log IanI
In fact, this is simply Jensen's formula in Ch. 4 applied to u = log If 1. An independent proof is outlined in Exercise 3. This formula implies a restriction on the amount of zeros of f in terms of its growth. In fact, for any f that is analytic in D(0, R) we get (if f (0) = 0 apply Jensen's formula
to f(z)/zM)
N(r, f) - N(ro, f)
(2.3)
We now shall consider the class of functions in the unit disk with a bounded characteristic function, the Nevanlinna class
M=
((f ll
E A(U); sup
1 j log+ If (re'e)Id9 < oo1 .
r<1 21r
ll
11
The inequality (2.3) implies that N(r, f) is bounded for each f E H. This can be rephrased in the following way.
2.2 Theorem. If a1, a2.... are the zeros, counted with multiplicities, to an f E H, then they satisfy the Blaschke condition
E1-Ia,I
§2. Zeros and Growth
87
Proof. We may assume that f (0) 54 0. The Nevanlinna condition and Jensen's formula (or (2.3)) imply that
C> latil
r logak1 -
(r - lakl),
> lckl
for some constant C and all r < 1, which proves the theorem. We now shall see that, conversely, any sequence satisfying the Blaschke condition occurs as the zero set of some Nevanlinna function. Actually, there is such a function that is even bounded. Let H°° denote the space of bounded analytic functions in U with norm IlfIIH- =SUP If IU
Clearly, H°° C N.
2.3 Theorem. If {an} is a sequence in U such that E 1 - Ian I < oo and each an # 0, then
an - z lanl B(z) = zk rl 1-G,z an CIO
(2.4)
1
is in H°°, IIBIIOO = 1, and B has precisely the zeros an and 0 (counted with multiplicities). Moreover,
B'(e'B) = lim B(re'B)
r,1
exists for a.e. 0, IB'(e'B)I = 1 for a.e. 0, and
Proof. Since z Ianl 1- 1an- -C,z an
z
rj lim 1 r71 2ir
p
log IB(re'a)Id0 = 0.
(an + lanIz)(1 (1 - anz)an
Ianl)
11
<
(2.5)
IzI < r,
it is clear that the product converges u.c., that B has the desired zeros, and that 1131 < 1. For the existence of the radial limits B', see Theorem 2.2 in Ch. 6. The limit (2.5) exists since log IBI is subharmonic and nonpositive. If BN is the product of the N first factors in (2.4), then BN is continuous on U and has modulus 1 on T; therefore, 1
log I(B/BN)(0)I
= r/,'l 1. 27r 1
lim
1J
2x
log I (B/BN)(Te'B)I d9
f2r
logB(reiO)IdO+0 <
r 0
logIB(e'e)IdG
0,
88
5. Zeros, Growth, and Value Distribution
where the next to last inequality follows from Fatou's lemma (keeping in oo, (2.5) mind that log CBI < 0). Since log I (B/BN)(0)I / 0 when N and the equality IB'I = 1 a.e. follow. A product of the form (2.4) is called a Blaschke product. These products will play an important role in subsequent chapters.
2.4 Remark. The problem of finding an analytic function with prescribed zeros and a certain growth can be rephrased as a problem for the inhomogeneous Laplace equation. For the sake of simplicity, let us assume that fl is simply connected. The case when n is finitely connected works in essentially the same way; see Exercise 11. If Au = F_6., (in the distribution
sense), we claim that there is an analytic f with zero set {aj} such that u = log If 1. Thus the problem is to solve the Laplace equation with control of the growth of the solution. In fact, if h(z) is an arbitrary function with the prescribed zeros, then u(z) - log Ih(z)I is harmonic and hence it is Re g(z) for some analytic g(z). Then f (z) = h(z) exp g(z) works. Notice that the Blaschke product is precisely the solution obtained in this way from the solution to Au = >2 aa, given by the Green function; cf. 2.14 in Ch. 4.
One also can construct entire functions with prescribed zeros and with control of the growth. Let a be a sequence (with multiplicities) in C with no limit point. Using the recipe in the proof of Weierstrass' theorem we get the entire function 00
f(z) = H W,,, ('z/an) 1
with zeros a,,, where p = n - 1. However, if one has some control of the sequence a,,, one can choose smaller p,,. For instance, if >2 1/janlp+l < oo, one can take pn = p and form the product 11l° which is O(expIzIp+1) when IzI -4 oo; see Exercises 4 and 5. An entire function that satisifies such a growth condition for some p is said to be of finite order.
§3. Value Distribution of Entire Functions We now turn our attention to value distribution of entire functions. Since T(r, f) = T(r, f - a) + 0(1), Jensen's formula also puts some restrictions on how often the value a can be attained by the function f. If n(r, f, a)
n(r, f - a) and N(r, f, a) = N(r, f - a), then
N(r, f, a) - N(ro, f, a) < T(r, f) + 0(1).
(3.1)
§3. Value Distribution of Entire Functions
89
We will show that an entire function must attain all but a few exceptional values as often as is possible in view of (3.1). In particular, the little Picard theorem will follow.
We first notice that one can get uniform estimates from estimates of the characteristic function. Without loss of generality we may assume
that f (0) = 0; otherwise, we can replace f with f - f (0). Since u(z) _ log
1 + If (z)12 is subharmonic,
f
uda < T (r, f)7rrz
(3.2)
(O,r)
for all r; and since u > 0, then
u(z) <
1 it Z z
f
udA < IT zIz
(z,lz0
J
D(o,zlzl)
ud. < 4T(21zl, f)
(3.3)
for all z if f (0) = 0.
3.1 Lemma. If f is entire and
r' r ) < m, lim inf log
r-oo then f is a polynomial of degree at most m.
Proof. First suppose that rn = 0. Then the condition implies that T(rj, f) < E log rj for some sequence rj -- oo. Hence, by (3.3), sup if(z)I < rye,
IzI
and thus f is constant; cf. the proof of Liouville's theorem. Now one can proceed by induction, noting that T(r, (f (z) - f (0))/z) < T(r, f (z) - f (0))-
0
logr+0(1) = T(r, f) - logy+0(1).
For a nonconstant entire function f, the defect with respect to the value a is defined as N(r, f, a)
6(f, a) = 1 - lim sup r- 00
T(-7-,f )
so that the defect 6(f, a) is a number between 0 and 1 that measures to what extent f fails to attain the value a as often as it possibly can in view of (3.1).
3.2 Example. Consider f(z) = exp(zk) for some positive integer k. Since f (z) has no zeros, N(r, f, 0) = 0 and hence 6(f, 0) = 1. On the other hand, zn
T(r, f) = 2 fo log' exp(rk cos(kO))d9 + 0(1),
90
5. Zeros, Growth, and Value Distribution
and therefore lim, T(r, f)/rk = 1/ir. Since n(r, f, 1) - krk/7r, we have that N(r, f, 1) - rk/ir and hence b(f, 1) = 0. In the same way one can verify that b(f, a) = 0 for all a # 0. This example illustrates a general phenomenon.
3.3 Theorem (The Defect Relation). If f is entire and nonconstant, and al, a2, ... , ak are distinct complex numbers, then k
b(f,ai) < 1. In particular, the defect must be zero for all but a countable set of values. If f totally avoids some value, then the defect is zero for all other values, i.e., the little Picard theorem follows. Moreover, if f attains some value a only a finite number of times, then N(r, f, a) = m log r + 0(1), and therefore the defect with respect to a is 1 unless f is a polynomial of degree m (cf. Lemma 3.1). Hence, any entire f that is not a polynomial attains all values, with one possible exception, infinitely many times. If f is a polynomial, then the defect b(f, a) = 0 for all values a, and thus we have strict inequality in the defect relation.
3.4 Remark. One also can define characteristic function and defect for meromorphic functions; the corresponding result then is that for a nonconstant meromorphic f no sum of defects can exceed 2. Since an analytic function is just a meromorphic one that avoids oo, this statement immediately implies Theorem 3.3.
The rest of this chapter is devoted to the proof of Theorem 3.3. Even though we restrict ourselves to the case when f avoids the value oo, it is natural to adopt an invariant view of P. If it is parametrized by z E C as usual, then the measure 1
dA(z)
dA(z) = 'r (1 + Iz12)2
is invariant with respect to the unitary linear fractional transformations of IP (cf. Exercises 33-36 in Ch. 2), i.e., the measure of a given (measurable) set is not changed if it is moved by such a mapping. Moreover, we have the invariant distance Iz - wi X(z, w) =
VF + Iz12
V1_+
,
Iwi2
X(z, w) = 0 if and only if z = w, and X(z, w) < 1 with equality if and only if z and w are antipodal on P.
§3. Value Distribution of Entire Functions
91
For a meromorphic f in D(0, R) and f (0) 0 a E P we define the proximity function
m(r, f, a) =
1
2a
log
1
dA - log
X(f(0),a) It measures how close to a the values off are on the circle I z I = r. Jensen's formula can be rephrased as (check!) 27r Jo
X(f(re`B), a)
3.5 Proposition. If f is analytic in D(0, R) and a # f (0), then T(r, f) = m(r, f, a) + N(r, f, a),
r < R.
(3.4)
Note that (3.4) is exact, i.e., not modulo 0(1). By the invariance of dil and X it follows (Exercise 17) that fP m(r, f, a)dµ(a) = 0 and hence
f N(r, f, a)dj (a) = T(r, f); therefore, at least N(r, f, a) equals T(r, f) in the mean. Moreover, it follows that the proximity function m(r, f,a) is negative for some values a. Nevertheless, we have
3.6 Lemma. If f is an entire function, then
N(r, f, a) - N(ro, f, a) < T(r, f) + 0(1), where 0(1) is uniform in a E P.
Proof. Without loss of generality we may assume that f (0) = 0. For
a
0,
N(r, f, a)-N(ro, f, a) < N(r, f, a)
= T(r, f) - m(r, f, a) < T(r, f) + log
1
(3.5)
X(f (0), a)
On the other hand, for a near 0 (recall that T > 0) N(r, f,a) - N(ro, f, a) < T (r, f) - (m(r, f, a) 1
- m(ro, f, a))
2a
< T(r,f) + 1 f log X(f(roeie) a) dB
" ST(r,f)+C-f loglf(roeie)-aldO
(3.6)
To see the last inequality, note that the integral is >_ -C2 for a near 0 since f 0 on Izl = ro if ro is small and the integral is an increasing function of ro. From (3.5) and (3.6) the lemma follows.
5. Zeros, Growth, and Value Distribution
92
Proof of Theorem 3.3. Let p(a) be a positive function on P with total mass I with respect to dµ(a), and let
-
f2,
AP(r)
p(f(re'B))(1+f'(reiO0l)I2)2dO.
A principal part of the proof is contained in the following estimate.
3.7 Lemma. For r > 0 outside a set E C (0, oo) of finite measure, log AP(r) < log r + 4 log T(r, f) + 0(1).
Proof. First notice that 1A
1
n(r, f, a)p(a)d1L(a) = - J +a 2)2 = f r AP(t)tdt, fp where the integral in the middle is over(1the multivalued image of D(0, r)
under the mapping f. Hence, in view of Lemma 3.6 we get r ds ° .P(t)tdt s fo rf
=
f(N(r,f,a) - N(ro, f, a)) p(a)d(a)
(3.7)
T (r, f) + C.
To conclude the lemma from (3.7), let
L(s) = f
a
Ap(t)tdt
and
0
K(r) = fL(s). p
Let E' be the set where )P(s) > (L(s))2/s. If rl is such that L(rl) > 0, then E niry oo)
ds <
°° sAP(s)ds L(s)2 r,
r,
dL(s)
L(s)2 <
I
L(rl
< oo.
Similarly, if E" is the set where L(r) > rK(r)2, then dK(r) < dK(r) < 1 < o0 fE,,n(rl,-) dr fE"fl[) r L(r) r, K(r)2 K(rl) For r outside E = E' U E", we have L(r)2 r2K(r)4 < AP(r) < < r(T(r, f) + C)4, and hence the lemma is proved. We now can conclude the proof of Theorem 3.3. By Jensen's inequality,
-f 27r
2w
log tp(f(reie))(1+fIf(Te)I)I2)2)dO
l
Supplementary Exercises
93
and hence 1
zwlog p(f(Teie))de < 4T(r,f) +logA,(r,f) +O(1)
(3.8)
2rr J0
when r -+ oo since log I f'(z)12 is subharmonic. Let ao, a1,. .. , ak be distinct points on IP and set
tog p(a) = 2 E log
- 2 log
1
(1: log
2
I
- C.
X(a, as) Then p(a) is integrable with respect to dp(a), and we can choose C so that x(a, a, j)
the total mass of p is 1. We may assume that f (0) differs from each of a0.
.
.
.
. ak. If we apply (3.8) to this choice of p, we get
2Em(r, f,aj) <4T(r, f) +logA,(r)
+1f
zn
(E
Jensen's inequality and (3.4) (recall that N > 0) yield that 2,r
logX(f(re''), (slog a? 1
o
)
dB
and hence by Lemma 3.7 we get k
log r + Clog T(r, f) + 0(1)
m(r, f, aj) < 2T (r, f) +
(3.9)
2
0
for r outside E. Since we already know that the theorem holds for polynomials, in view of Lemma 3.1 we may assume that lim sup log r r-.oo
T(r, f)
.0.
Letting ao = oo we now obtain Theorem 3.3 from (3.9) since m(r, f, a) N(r, f, a) = lim inf 6(f, a) = 1 - lim sup r-oo T(r, f ) r-.oo T(r,f)
(if f(0) # a) and 6(f, ao) = 1.
Supplementary Exercises Exercise 1. Suppose that f c A(U). One says that a E T is a regular point for f if there are r > 0 and g E A(D(a,r)) such that f = g on and lim sup la.1'1' = 1, then some U n D(a, r). Show that if .f = E point on T is singular (i.e., nonregular). Exercise 2. Let aj be points in the upper half-plane. Find a condition on
{aj } so that fl(z - aj)/(z - a3) defines an analytic function there.
5. Zeros, Growth, and Value Distribution
94
Exercise 3. Prove Jensen's formula (2.2) without any reference to Jensen's formula in Ch. 4. First show that the mean value property for the function
logIf(z)11
r
z
-an
implies the desired equality modulo the sum
E
1 j21r 2x
log I1 - e-'Ban/rldO.
Then show that each of these integrals actually vanishes.
Exercise 4. Suppose that aj is a sequence in C such that
E 1/Iaj)P+' < 00Show that
f (z) _ [J WP(z/aj ) defines an entire function such that I f(z)I < expCIzIP+i
Exercise 5. Show that if f is an entire function of finite order, i.e., log If (z)I = O(Izlm) for some m when IzI -+ oo, then
f (z) = zneg(z) [J WP(z/aj)
for some p and n. What is the connection between m and p?
Exercise 6. Suppose that aj are the zeros of some f E A(U) and that lai I <- Ia2I < . < 1. Show that fo n(r, f, 0)dr = E 1 - laj I. Exercise 7. Suppose that f E HI and f (1 - 1/n) = 0 for all n E Z+. Show that f = 0. Exercise 8. Suppose that 0 < Al < 1\2 < ... < no. Show that there is a bounded analytic function f 0- 0 in the right half-plane that vanishes at each .1j if and only if E 1/aj < no.
Exercise 9. Suppose that 0 < .11 < A2 < ... < no and that E 1/Ak = Show that the space of finite linear combinations of the functions
no.
X
is dense in C([0,1]). Hint: Use the Hahn-Banach theorem
and the preceding exercise.
Exercise 10. Suppose that aj is a sequence in U. Show that the Blaschke product (2.4) converges (if and) only if al satisfies the Blaschke condition.
Exercise 11. Suppose that Sl is finitely connected, h is analytic, and u is a real solution to Au = Olog Ih1. Take one point bj from each bounded
Supplementary Exercises
component of P \ Sl. Show that there are real numbers aj with and a function f (z) with the same zeros as h(z) such that
95
Ian
7r
-
I f(z)I < eu(z)nmIz - b3I
Hint: Determine aj such that u-log Ihi- E aj log Iz-b3 i has a well-defined harmonic conjugate modulo 27r.
Exercise 12. Show that T(r, f) <
27r
2
J
Ilogif(re'B)IIdO+O(1) < 2T(r, f)+O(1)
0
when r / R. Hint: Use the fact that u = log If I is subharmonic, so that 2x u(Tpe:e) < 2 r2x u+(reie)d8 -00 < 2- f 0
0
1f T, r
u(reo)d8
for ro < r < R. Exercise 13. Verify the claims in Example 3.2. Exercise 14. Show that if f is meromorphic in D(0, R) and f (0) # a, b E P, then m(r, f, a) + N(r, f, a) = m(r, f, b) + N(r, f, b), r < R.
Exercise 15. Let f be an entire function that attains each value at most m times. Show that f must be a polynomial of degree at most m. ntial of p as Exercise 16. Define the potflog P(w) =
X(w, a)
p(a)dFu(a),
and show that
mo(r, f) =
2-7r
f p(f (re'0))dO - p(f(0)) 0
Exercise 17. Show that
T(r, f) = fN(rfa)dii(a). Exercise 18. Suppose that f is an entire function that attains two distinct values at most a finite number of times. Show that f must be a polynomial. Exercise 19. Show that
rT'(r, f) = f n(r, f, a)dp(a) =: A(r, f)P
Derive the formula
T(r,f) = f0 rA(t,f)dt
5. Zeros, Growth, and Value Distribution
96
Note that A(r, f) is the area (counted with multiplicities) of the image of the set {IzI < r} under the mapping z H f (z). Exercise 20. Show that f21T
logIf (reie) I dO - logIf (0) I +
,
N(r, f, ese)dO o
if f is analytic D(0, R) and r < R.
Notes A thorough treatment of the relationships between zeros, growth, and Taylor coefficients of entire functions can be found in [B]. The value distribution theory was developed by Nevanlinna in the beginning of the 1920s. Proposition 3.5 is called the first fundamental (or main) theorem, and formula (3.9) is called the second fundamental (or main) the-
orem. The proof given here is due to Ahlfors. A thorough discussion of these matters, including the more general result for meromorphic functions and further references, can be found in [Hi].
Since we assume that f is analytic and not merely meromorphic, we easily get (3.8), which in turn simplifies the proof of (3.9). The converse of the result in Exercise 9 is also true; see, e.g., [Rul] for a proof. This characterization is called the Miintz-Szasz theorem.
6
Harmonic Functions and Fourier Series
§1. Boundary Values of Harmonic Functions If u is a function in U, we let ur(e'o) = u(re`o) for r < 1. Let 1/P 1 2Ib(e;t)lPdt) ,r II0IILa = ( 1 f
be the LP-norm on T (for p < oo); and for 1 < p < oo, let hP denote the space of harmonic functions in U such that IIuIIhP =SUP r<1
IItirIILP < 00-
Note that r IIurIILP is increasing since IuIP is subharmonic, and therefore sup may be replaced by lim. One readily verifies that hP is a Banach space. Let M(T) be the space of complex measures ju on T with norm IIAII =
- f dljl. 2 -7r
T
It is convenient to identify measures (and functions) on T with 2ir periodic measures (and functions) on R, so that, e.g.,
jr 2,f 2
2
J
(t)dp(t),
where the integration is to be performed over, e.g., (0, 21r). For p E M(T) we define the Poisson integral of p as
Pp(reie) =
2a
2, P,.(9 - t)dp(t),
r < 1;
0
if dp = fdt and f E C(T), this definition is consistent with the previous one in Ch. 4. Note that Pp is harmonic in U. To begin with, we shall generalize Proposition 1.4 in Ch. 4.
98
6. Harmonic Functions and Fourier Series
1.1 Theorem.
(a) If f E LP(T), 1 < p < oo, then Pf E hp and
IIPf!IhP = IIfIILP-
If
1 < p < oo, then 1 f1 I I(Pf)r - f(ILP = 0.
f tweak', i.e.,
If f c- LOO(T), then (Pf )r
f(Pf)rc--+JTfq
LI(T).
for all
(b) M14 E M(T), then Pp E h', IIPµllh3 = Ilpll, and (Pp), --* p weak', i.e.,
f(Pz)rc -+ J Ldp
for all ¢ E C(T).
(c) If f E C(T), then Pf E h°O n C(U), and (Pf)r
f uniformly.
Proof. First notice that (c) follows from (the proof of) Proposition 1.4 in Ch. 4. To prove (b) notice that if p E .M (T), then 12,
Pp(re'B)I < I
Pr(B - t)dlpl (t);
21r
and if we integrate this inequality and use Fubini's theorem, we get that IIP,IIh= < IIpi. Then take 0 E C(T). By Fubini's theorem, fzA Pr(9 2ir
f
- t)dp(t)¢(e'9)d9
=2zr f
0
0
zn(P4')r(e")dp(t);
1
= 27r
0
but since (PO), --+ 0 uniformly according to (c), the right-hand side tends to 2" Odp/2ir. Finally (cf. Exercise 4),), 1 f27r 1 2T Odu = sup lim q(Pp)r Illiil = sup Iml_:5I
1
27r
r
I0151
< lim r
27r
0
2n
1
2i
I(Pp),I =
and thus (b) is proved. For (a), notice that (Jensen's inequality) Pr(t) Ii. j27r Pr (e `f(et(e-tl)rdt
t)f(ee)dtf 0
for 1 < p < oo. Integrating this inequality we get II Pf II hP <_ e > 0 and 0 E C(T) such that IIf - tIILP < E. Then
II f II LP
Take
II(Pf)r - fIILP <- II(P(f - O))rlI LP + II(PO)r - .IILP +II0 - f (I LP < If -'IILP + II(P4i)r - 4IIL°°+IIf - OIILP < 3E
§1. Boundary Values of Harmonic Functions
99
if r is near 1 because of (c), and hence (Pf),. -+ f in LP(T); in particular, II Pf II hn = limf II (P f ),. II LN = 11f II LP The case p = oo is proved analogously
to (b).
C3
We now consider the converse of Theorem 1.1.
1.2 Theorem. If u E hp, 1 < p < oo, there is a unique f E LP(T) such that u = Pf. If u E h1, there is a unique p E .M(T) such that u = Pp. Thus, we have a one-to-one correspondence (in fact, an isometric isomorphism) between hp and LP(T) for p > 1 and between h' and M(T). In one direction the correspondence is given by the Poisson integral, p -+ Pp, and in the other direction by u --+ limr u, in the sense described in Theorem 1.1. Note that Theorem 1.2 differs from Theorem 1.1 in the sense that it is an
existence theorem; it states that an object with certain properties exists. The key is the Banach-Alaoglu theorem, a proof of which is outlined in Exercise 5.
1.3 The Banach-Alaoglu Theorem. If B is a separable Banach space, {Aa} C B`, and IIAaII 5 M, then there is a subsequence Aa; and A E B' such that A.,u - Au for all u E B, i.e., Aa, -. A weak'.
1.4 Example. In general, I I A II <_ lim inf II Ai II if Aj -+ A weak' (Exercise 4) and strict inequality can occur. Take, e.g., B = L1(T) and let A = eiie Then, according to the Riemann-Lebesgue lemma (see next section), 0 for all f E LI(T), although IIAfIIL- = 1 Proof of Theorem 1.2. Suppose that p > 1 and u E hp. Then {u,.} is a bounded family of elements in LP(T) and therefore there is a subsequence
u,-, (such that ri / 1) and f E LP(T) such that f u,.,0 -p f f¢ for all 0 E LQ(T). We have to verify that u is the Poisson integral of f. Note that z ' u(rjz) is harmonic in a neighborhood of U and therefore P (B u(rrje'B) = 2n - t)u,., (e")dt. rzn
For fixed r < 1 and 0, t P,(O and therefore we get that
is continuous, in particular, in L9(1'), r2
u(re'i)
2
P,.(9
- t)f(eu)dt,
0
i.e., u = P f . From Theorem 1.1 it now follows that u,. = (P f ),.
f
and so f is unique; therefore, the proof is complete for p > 1. Notice that L1(T) C M(T), which is the dual of C(T); therefore, if u E h', then the Banach-Alaoglu theorem provides a p E M (T) such that f u,.)0 -+ f Odp
100
6. Harmonic Functions and Fourier Series
for all 46 E C(T). Then the proof is concluded analogously to the case
0
p>1.
If we combine Theorem 1.1 with the observations about the Green potential Gp in Remark 2.14 in Ch. 4, we get the following decomposition of a subharmonic function u. The special case when u = log If I for some analytic function f will be fundamental in subsequent chapters.
1.5 Theorem (The Riesz Decomposition). Let u be a subharmonic function such that 1
sup -J r<1 2rr
2n
u+(re'B)dO < oo.
o
Then
f(1 - Kl)Lu < 00,
and there is a unique measure p on T such that u(z) = G(Du)(z) + Pp(z), z E U. Moreover, u,
pt weak' and ur
pe in L'(T) if µ is absolutely continuous.
Proof. According to Corollary 2.16 and Remark 2.14 about the Green potential, G(Au) is subharmonic and its Laplacian is equal to Au. It follows (see Exercise 31) that u - G(Du) is harmonic in U. In view of Exercise 22 in Ch. 4, it is in fact in h1, and by Theorem 1.2 it is therefore the Poisson integral of a measure p on T. 0 We now are going to study pointwise convergence of the Poisson integral. To this end we first recall the Lebesgue-Radon-Nikodym decomposition of a measure into an absolutely continuous and a singular part,
A =Ua+I2o, i.e., where pt, is concentrated on a null set (with respect to dB) and pa = fd9
for some f E L' (T). If
Lt (s)
I 1/2t, Isj < t
s 2t Xf t( ) = 1
0,
Isl > t,
then
f Lt(9 - s)dp(s) =
B+t
1 r 2t J9-t
dp(s) --. f(ese) for a.e. 0 when t \, 0, (1.1)
i.e., the mean values over small intervals around e'8 tend to f (e'') for a.e. 0 when the lengths tend to zero. If dp = f dO and f E C(T), this is trivial
and can be rephrased as Lt (s) --, 6 weak' as measures when t \ 0. We
§1. Boundary Values of Harmonic Functions
101
Pr(s - B)dp(s) of p
now will consider instead the mean values (1/27x) j
when r / 1. Also, P,.(t)/27r - S weak' as measures when r / 1. One can think of Pr(t) as regularized characteristic functions. Figure 1 below shows Pr(t) for r = 1/2, r = 3/4, and r = 7/8.
-1
0
2
1
3
t
FIGURE 1
One may guess that the analogue of (1.1) holds for P,./27r.
1.6 Theorem. If u = p. + f dB, then (Pµ),. -# f for a.e. 0 when
r
(1.2)
In particular,
(Pf)r -ffora.e.0when r/1iffEL"(T), 1
then u' is equal to the absolutely continuous part of /2.
Proof. We first recall that the standard proof of (1.1) depends on only two facts: that (1.1) holds for continuous f and that one has the weak type (1,1)-estimate,
{Mp > all <
1114,
of the Hardy-Littlewood maximal function
M/(0)
sup f Lt(9 - s)dI11I (s)
(1.3)
6. Harmonic Functions and Fourier Series
102
Since we know (1.2) for continuous functions, we just have to show the analogue (1.3)' of (1.3), where My is replaced by the new maximal function
1 f2l Pr(B -
Mll(8) = sup r<1 27r
We shall reduce (1.3)' to (1.3), and we begin with representing Pr(s)/27r as a superposition of the functions Lt(s), 2_Pr(s) = Pr(ir)L.,,(s) + 2
if
-Pr(t)2tLt(s)dt.
(1.4)
0
In fact, fo -P,(t)2tLt(s)dt/27r = f"ICI P,(t)dt/27r; an integration by parts then gives (1.4). By (1.4) we get 1
i(9) =
sup
[Pr(ic)fLit(s - O)dlul(s) + 2n f -Pi (t)2t f Lt(s - e)dirli(s)dt) 0
_< sup Ma(O) [Pr(ix) +
27r
/ -Pr(t)2tdtl = Mp(8), o
J
from which (1.3)' follows, and hence also Theorem 1.6.
O
Recall that for each real harmonic u in U there is a unique analytic function Gu such that u = Re Gu and Gu is real at the origin.
1.7 Theorem. Suppose that 1 < p < oo. If u e hP is real, then its
harmonic conjugate v = Im Gu (which vanishes at the origin) also belongs to hP and IIVIIhP 5 APII'ullhp, where AP depends only on p.
A conformal mapping of U onto the strip {I Re w I < 1} shows that the theorem is not true for p = oo and hence (Exercise 8) not true for p = 1
either.
A finite sum F, a trigonometric polynomial. Any real trigonometric polynomial is the restriction to T of the real part of an analytic polynomial g(z) = a(z) + iJ3(z) with X3(0) = 0. The theorem implies in particular that 1IQ11LP <_ APIIoIILP
If I < p < oc, then
Z"Gu(z)
eit f since the integral is analytic, real at the origin, and its real part equals =
zuM(eie)dt,
lzI < 1,
u = Pu'. If we try to estimate the boundary values of Gu by letting
z = e'8 formally, we get e't - e`9 = O(It - BI) in the denominator, and therefore that integral does not exist in the usual sense; it is a so-called singular integral. There is a theory for such integrals which can be used to deduce Theorem 1.7, but here we supply a more elementary proof.
§1. Boundary Values of Harmonic Functions
103
Proof. First suppose that 1 < p < 2, f = u + iv is analytic, v(0) = 0, and u > 0. Since 2au/az = f', a simple computation gives that AuP = p(p - 1)If'I2up`2 and
olflp = p2If'I2lflp-2; therefore,
AlfI" <
p
(1.5)
AuP 1
(keep in mind that 1 < p < 2). Notice that Green's formula (see A in the preliminaries) gives that
rf 0
-OdO or = J
ISI
From (1.5) we then get
f2,
If(reie)IPd8 <
f2r
p
dr but v(0) = 0, and thus f21T
If(reie)IPd9 < p p
1 dr
l
f
u(reB)IPdO;
zn
Iu(re`9)IPdG
for r = 0 and hence for any r < 1. Thus IIGuflhP < CpIlullhP if 0 < u E hP, since (by the maximum principle) a nonnegative harmonic function is strictly positive unless it is identically 0. If u E hP, then u' is in LP(T) and
hence so are u'+ and u'-. Now u = Pu'+ - Pu", where both terms are nonnegative and in hP. Moreover, Gu = GPu'+ - GPu"-, and therefore we get that IIGullhp < CpIIPu*+IlhP +CpIIPu' IIhl S
2CPIIuIIhp.
This proves the theorem for 1 < p < 2 with AP = 2Cp = 2p/(p - 1). Now suppose that 2 < p < oo and let p-1 + q-1 = 1. Take an arbitrary real trigonometric polynomial a = Re g, where Im g = /3 and /3(0) = 0.
For such a g, Nur + avr = Im gfr is harmonic and vanishes at the origin, and therefore f u(reie)/3(e`B)dO = - f v(reie)a(eie)dO. Thus, Holder's inequality gives 1
z,
1
I2 f 0VradOl = 121
f2, ur/3dO
IIurIILPIIaIILQ < AgllurIILpIIaIILQ,
where we have applied the theorem to a and /3. The (real) trigonometric polynomials a are dense in (real) Lq(T) (see below), and thus it follows that IIVrIILP < AgIIurIILP and so IIVIIhP S Agllullhp
0
104
6. Harmonic Functions and Fourier Series
By virtue of Theorems 1.1, 1.2, and 1.7 we can define a bounded operator, the Hilbert transform,
1 < p < oo, in the following way: For a real 0 in LP(T) we let HO be the boundary values of the unique harmonic conjugate to PO that vanishes at the origin. The operator so defined is certainly real linear, and it then is extended to complex-valued functions by linearity. H: LP(T)
LP(T),
§2. Fourier Series For p E M(T) we define the Fourier coefficients 2x e_intdp(t), _77p(n) = !i(n) = 2rr J
n E Z.
0
Note that I!l(n)I < IIpII, and so we have a bounded operator
I' (Z).
Y: M Moreover,
-F(a)(n) _ Fu(-n).
(2.1)
For an integrable function on T, 2n
.Ff(n) = f(n) =
e-int f(ect)dt;
2 .- J 0
and if f is in C' (T), then
.F (d8) (n) = in-Ff(n).
(2.2)
An easy computation reveals that 00
Pp(re'B) = E rl nl -00
00
00
{z(n)eine = 1:
2'(n)zn
+ #(0) + E zn/1(-n).
1
(2.3)
1
Thus, PM and hence /p are completely determined by the Fourier coefficients
µ(n), and Ppt is analytic in U if and only if j2(n) = 0 for all n < 0. To each measure p e M(T), we associate the trigonometric polynomials N
.FN,u(e'B) = E Fi(n)e n9 -N
In view of (2.3), it is natural to ask whether .Nµ tends to p in any reasonable sense when N --+ oo. It follows from (2.2) that Ff (n) = O(1/InI2) no if f E C2(T), and hence the series in (2.3) converges uniwhen Inl formly even for r = 1, i.e., FN f f uniformly. In fact, it is enough that
§2. Fourier Series
105
f E C1(T); see Exercise 13. Since C2(T) is dense in LP(T) (use, e.g., Theorem 1.1), it follows that the space of trigonometric polynomials is dense in LP(T). In particular, this implies that {e"91 is a complete ON-system in L2(T), and therefore we have Parseval's equality 00
(f, 9) = E f (n)9(n),
(2.4)
where 2ir
1
(f, 9) =
fo and hence the isometric isomorphism 2-7r
f (e")9(e't)dt,
(2.5)
12(Z).
' F: L2(T)
Moreover, 11-FNf - fflL2 -+ 0 when N - oo. It is also true that IIFNf f II LP --+ 0 for f E LP(T) if 1 < p < oo; see Exercise 21. In Exercise 40 in Ch. 1 we saw that .F:LP (T)-3QQ(Z),
1
(the Hausdorff-Young inequality). Furthermore, we have the RiemannLebesgue lemma, which states that 97: L1(T) -+ co(Z),
-
where co(Z) is the space of sequences e. such that c,. -+ 0 when Ink oo. This follows quite immediately from the fact that the trigonometric polynomials are dense in L1(T). It is easily verified that
2-
f
Ei (n)O(n) = 2-N
fdu(e)
(2.6)
for p E M(T), from which it follows that .FNP - p at least in the weak sense that zn
fO 0
NU-
f o
2n
Oda,
0EC2(T).
If 0 E C°°(T), then, by (2.2), F (n) = O(In1-M) for all M when In! -+ 00, and therefore (2.3) converges uniformly even after an arbitrary number of differentiations. Therefore, Pq5 E C°°(U), and in particular FNq 0 in C0(T) (i.e., uniformly with all its derivatives).
2.1 Remark. Let P(Z) be the space of sequences c that are O(InJM) for some M when In! -+ oo. For any distribution f E D'(T), let Tf (n) = f(e-:n,). As T is compact, f has finite order, so .Ff(n) = O(Inj M) for some M. Therefore,
.F: D'(T) --+ P(Z).
(2.7)
106
6. Harmonic Functions and Fourier Series
Since (2.6) extends to distributions f, it follows that FNf --+ f in V(T). Conversely, one easily sees that if c, is any sequence in P(Z), then the corresponding Fourier series converges in IY(T), and therefore (2.7) is actually an isomorphism. Let us express the Hilbert transform in terms of the Fourier coefficients.
Let u be a real function in L2(T), and let v = Hu. As the harmonic extension of v vanishes at the origin, v(0) = 0_ Furthermore, as (the harmonic extension of) u + iv is analytic, u(n) + iv(n) = 0 for n < 0. In view of (2.1), we then must have that n > 0
-iu(n),
0, n=0 tiu(n), n < 0.
Hu(n) = S
l
(2.8)
Notice that (2.8) and Parseval's equality (2.4) imply a very simple proof of Theorem 1.7 for p = 2 and that the best constant A2 is 1. From (2.8) and (2.4) it also follows that
HHf = -f + f (0) and H` = -H,
(2.9)
i.e., (Hf,g) = -(f,Hg) for all f,g E L2(T). We now introduce the spaces H' = hP rl A(U), 1 < p < no, which are closed (why?) subspaces of h', and we first consider the case p > 1.
2.2 Theorem. Suppose that 1 < p< on. If f E HP, then f.(e'e) = lim f(reie) r/1
exists for a.e. 9, f' E LP (T),
r .f'C z(()d( f(z) = (Pf')(z) = 1: 2'r z
(2.10)
UIIHP = Ijf'l LP and for p < no also )If. - f'IILP - 0 when r / 1. A function in LP(T) is f * for some f E HP if and only if its negative Fourier coefficients vanish.
Proof. Everything but (2.10) is already proved. If r < 1, then
f (rz) =
1
f(rC)dC
27ri JT
C-z
and since fr -* f" in L1(T), (2.10) follows.
0
Thus, for p > 1, HP may be identified with the closed subspace {u LP(T); u(n) = 0, n < 0} of LP(T). In particular,
E
H2-12(N)= {aE12(Z); an=0forn<0},
§2. Fourier Series
107
and therefore we have the orthogonal projection H2, S: L2(T)
referred to as the Szego projection. We claim that n > 0
f(n),
SJ(n) =
n<0.
0,
(2.11)
In fact, since Sf is analytic (in the sequel we identify Sf with its Poisson integral) for any f and f - Sf is orthogonal to the analytic functions with respect to the inner product (2.5), (2.11) follows from Parseval's equality. Now
S f(z) _
00
00
2n
1
f(n)zn = 00 2x f f(eit)e-intZndt 0
1 j2w f(e't)dt 27r
1-e-itz
1
f f (()d
2lri T C- z
and therefore the (harmonic extension of) the Szegb projection is given by the Cauchy integral. However, the Cauchy integral makes sense for any
fELP(T),p>1.
2.3 Theorem. Suppose that 1 < p < no. Then (a) the Szego projection extends to a bounded mapping
S:LP(T)-.H". (b) any bounded functional on HP is represented by a unique element h in HQ via the pairing (2.5), and the functional norm is equivalent to the HQ norm of h.
Part (b) can be rephrased as (HP)' = H. Proof. From (2.8) and (2.11) we get the relation
Su = 2 (u + iHu + Pu(0)),
u E L2(T).
(2.12)
Since H is bounded on LP(T), so is S, and hence the first statement is confirmed.
Any bounded functional A on HP extends by the Hahn-Banach theorem to a functional on LP(T) with the same norm, and it therefore is represented by a 0 E L9(T) with IItIILQ = IIAII. In particular, Af = (f,0) for f E HP,
but then also A f = (f, SO) for all f E HP, since S f = f and S' = S.
Thus, g = SO E Hq represents A. Then clearly IIAII < II9IINQ, and by the boundedness of S, 1I91IHQ
only depends on q.
CII04Q = CIIAII for some constant C that
0
6. Harmonic Functions and Fourier Series
108
Let us now take a look at the space H'. From Theorems 1.2 and 1.1 we know that if f E H1 C h', then f*(e'B) exists for a.e. 0 and f = Pµ, where u = M. + f'dO. Since f is analytic, we also know that µ(n) = 0 for n < 0. The crucial fact is that then u. is in fact absolutely continuous and thus equal to f'dO.
2.4 The F. & M. Riesz Theorem. If p E M(T) and u(n) = 0 for n < 0, then p is absolutely continuous, i.e., p = gd8 for some g E L'(T). By an obvious argument we then have the following extension of Theorem 2.2.
2.5 Theorem. If f E H', then f'(e'B) = limy/, f(re") exists for a.e. 0, f' E L1(T), f(z) = (Pf*)(z) = II AH' = II f' Il L' and Il fr - f' II L'
21 7ri
f(()d(
JT 0 when r l 1. A function in L1(T) is
f' for some f E H1 if and only if its negative Fourier coefficients vanish. On the other hand, the F. & M. Riesz theorem follows immediately from
Theorem 2.5. To prove the latter, we must use that f is analytic and not merely harmonic. In the next paragraph we shall see that any f E H1 can be factorized as f = gh, where f, 9 E H2; if f is nonvanishing, this is trivial, we just take g = h = 0, where 02 = f. Now Theorem 2.5 easily follows:
Ilfr - f'ilL' =I1grhr-9*h`IIL'
i.e., fr - f" in LI(T). However, then f = Pf', and everything follows.
Supplementary Exercises Exercise 1. Show that hP is a Banach space, i.e., that II Ilho is a complete norm on hp.
Exercise 2. Show that if u is harmonic and positive in U, then u E hl and u = Pp, where u > 0. Exercise 3. Give an example of a u E hP \ 1i if 1 < p < p' < oo. Exercise 4. Let A3 be a sequence of elements in the dual space B' to a Banach space B. Show that IIAil < liminf IIA,11 if A3 - A weak'.
Exercise 5. Fill in the details in the following sketch of a proof of the Banach-Alaoglu theorem. Let D = {u1, u2, ... } be an enumeration
Supplementary Exercises
109
of a dense subset of B. Take a subsequence A1,J from {AQ} such that limJ_.,, A1,,,ul exists. Extract from {A1,J} a subsequence A2,J such that 1im,.. A2,JU2 exists, and so on. The diagonal sequence AJ,J then converges
at each Uk. One then defines Auk as this limit and extends by continuity to an element in B', which has the desired properties. Exercise 6. Suppose that u = Pp and IIUIIhI < IIu'IIL1. Show that p is absolutely continuous.
Exercise 7. Give an example of a real u E h°° such that its harmonic conjugate is not in h°°. Exercise 8. Show that in general the harmonic conjugate v of a real u E h' is not in h1. Is this true if u = Pu"?
Exercise 9. Give an example of a real u E hl such that its harmonic conjugate is not in h'.
Exercise 10. Let p = p+ - a- be the Jordan decomposition of the real measure p E M(T), and let u = Pp. Show that f u,+dt -+ f dpc+(t) when
r / 1.
Exercise 11. Suppose that u = P(u' +,u.). Show that f u,+ -+ fu*+ if and only if A. < 0.
Exercise 12. Show that the space of trigonometric polynomials is dense in C(T). Exercise 13. Show that the partial Fourier sums .F f tend uniformly to f if f E C1(T) (or if df/d9 is only in L2(T) in the distribution sense). Exercise 14. Verify that Hp is a closed subspace of hp. Exercise 15. Verify (2.12) by means of the integral representations of Gu = P(u + iHu) and Su. Exercise 16. Prove the Riemann-Lebesgue lemma, i.e., that F maps L1(T) into co(Z). Exercise 17. Show that .F: M -+ 1°°(Z) is not surjective. Can you find an explicit example of an a E 1°°(Z) that is not in the image? Exercise 18. Let DN(t) = .FN6(t) = END, e'^t. Show that DN(t) - sin ((N + 1/2)t) sin(t/2) and that IIDNIIL1 -' oo when N -+ oo. Exercise 19. Show that (cf. the preceding exercise) FN f (0) =
-
f (t)DN(-t)dt/21r Zo
110
6. Harmonic Functions and Fourier Series
and that f -+ FNf (0) is a bounded linear functional on C(T) with norm equal to IIDNIILI. Then show that there are functions f E C(T) for which ,FN f (0) does not converge when N ---+ oo. Hint: Use the Banach-Steinhaus theorem and the preceding exercise.
Exercise 20. Show that F: L1(T) ---+ co(Z) is not surjective. Hint: Use Exercise 18 and the open mapping theorem for Banach spaces. Exercise 21. Show that the partial Fourier sums TN f tend to f in LP(T) norm for f E L'(T), 1 < p < oo. Hint: First show that the set of partial sums is bounded in LP(T). Exercise 22. Show that a harmonic function u in U is in hz if and only if
fu (1- I(I2)IVuI2 dA(() < coo. Exercise 23. Show that the Szego projection is unbounded on L°°(T) and
L'(T). Exercise 24. Let a E U. Show that f -+ f (')(a) is a bounded linear functional on Hz and find the g E Hz that represents this functional. How large can If (-)(a)l be if IIfJIH2 :5 1? Exercise 25. Show that if p E M(T) and 11(n) = 0 for all n > N, then µ is absolutely continuous.
Exercise 26. Let p be a real measure on T, and show that
rzi
limsup 2x J
r/1
P,.(0 - s)dp(s) <
Jim SUP C%O
o
r
/ Lt(8 - s)d/.c(s). J
Show that P,u has radial limits at each point where the measure derivative of p exists. Exercise 27.
(a) Show that if p > 0 on T and lim inf
t-0 J
Lt(O - s)dp(s) = 0
for all 0 in an open interval I, then µ(I) = 0. (b) Suppose that u > 0 and is harmonic and that u(re4o) -+ 0 when r / 1 for all e's # 1. Show that u(re'g) = cP,(O) for some constant c > 0. Exercise 28. Show that u(z) = Im[(1 + z)'/(1 - z)2] has radial limits 0 for all 0. Show that u is not Pp for any measure p. Exercise 29. Show that the Hilbert transform is an isomorphism on D'(T).
Exercise 30. Is it true that each f E A(U) is equal to g(m) for some m and some g E Hz?
Notes
111
Exercise 31. Suppose that u and v are subharmonic in f? and that Au = Av. Show that u - v is harmonic in Q.
Notes In this chapter as well as in the later ones we restrict ourselves to the unit disk. Most results have analogues in the upper half-plane lI+. For instance, h.P(II+) is the space of harmonic functions in n+ such that sup,>0 fR iu(x+ iy)IPdx < oo. There is a one-to-one correspondence between LP(R) and hP(II+) for p > 1 and between M(R) and hl(1T+), which is given by the Poisson integral Pf (x, U) = n JR Y2 + (x)_t)2 ; cf. Exercise 23 in Ch. 3. A further discussion about weak* compactness and the Banach-Alaoglu theorem can be found in [Ru2]. Theorem 1.6 was first proved by Fatou in 1906 for p = oo. Therefore, results about pointwise convergence at the boundary often are referred to as Fatou theorems. Any u E h' actually has nontangential boundary values
u': If r(o) = {re" E U; 10 - ti < 1 - r}, then u"(e'B) =
lira u(z) r(ete)3z-.e,e
exists for a.e. 0; see, e.g., [G). See also Exercise 28 in the next chapter. For 1 < p <- oo, there is C, > 0 such that II Mf IILP <_ CPII f IILP, f E LP(T), i.e., M is bounded on LP(T); see, e.g., [G]. Theorem 1.7 was first proved by M. Riesz in 1927. The proof here is due to P. Stein. The Hilbert transform is the simplest example of a singular integral; see [S] for a thorough discussion.
Carleson proved in 1966 that FN f -+ f for a.e. 8 if f E L3(T), Acta Math., vol 116 (1966), 135-157. It was generalized to p > 1 by Hunt, Proc. Conf. Edwardsville, Ill. (1967), 235-255, Southern Illinios Univ. Press, Carbondale, Ill. (1968). For further results about Fourier analysis and periodic distributions, see [Ho].
The theorem of F. and M. Riesz is from 1916. For generalizations to lit", see [S].
7 HP Spaces
§1. Factorization in Hp Spaces We extend the definition of the Hardy spaces Hp to p > 0: r
Hp = ilf E A(U); If II Ho = sup r<1
1f 21r
2
1/p
if (reie) I pd9}
< oo
o
For 0 < p < 1, the triangle inequality does not hold; but Ilf + 9IIHP < IIf
IIHP +
IIgIIH,,,
and therefore Hp is at least a vector space. Moreover,
H°°cHP cHI cN, oo>p>s>0, and therefore the zero set of any f E Hp satisfies the Blaschke condition. In fact, it is possible to divide out all the zeros without affecting the norm.
1.1 Theorem. If f E N is not identically zero and B is the corresponding Blaschke product, then f /B E N: If f E Hp, then f /B E Hp and IIf/BIIHP = IIfIIHP.
Proof. Note that log+ I f/BI < log+ I f I + log+ I1/BI = log+ I f I - log IBI.
Since (Theorem 2.3 in Ch. 5) log IBrI -+ 0 in L' (T) when r / 1, f 1B E
N. Now suppose that f E HP, and let B be the first n factors in B. Then clearly IIf /BnIIHP = IIf II HP since I Bn (reie) I
1 uniformly when
r / 1. Moreover, If /Bn I j If /B I when n -+ oo, and thus by monotone convergence we have that f2lF f1 27r
I (f/B)(re'B)Id9 =
linm 27r
.
I (f/Bn)(re`B)Ipd0
lnmllf/BnIIHP =
IIfIIHP,
and therefore IIf / B I I HP S Il f I I HP Since the converse is trivial, the proof is complete.
Theorem 1.1 is the key to the Hp-theory.
§1. Factorization in H' Spaces
113
1.2 Theorem. If f E HP, 0 < p < oo, then f ` = limrll fr exists for a.e. 8, IIfIIHP = II.f'IILP, and for p < oo it is also the case that Ilf, - f'IILP -. 0
when r/1. Proof. Since the case p > 1 already has been dealt with in Ch. 6, we may
assume that 0 < p < 1. Take f E HP and let B be the corresponding Blaschke product. Then f /B is nonvanishing, and hence it is equal to 01
for some 0 E A(U). From Theorem 1.1 we get that IIOIIH, r = IIf II iiP < oc, and so if pm > 1, then 0' = limn qr exists for a.e. 9 and II¢' 11 LP- = IIOIIHp-
according to Section 1 in Ch. 6. However, f = B(O', and therefore f'
exists a.e. and IIf'IILP = II B*(O*)rII LP = II0'IIi P = II0IIHmP = II fII HP. For 0 < p < 1 , I fr I P + If * I P - I fr - f' I P > 0, and an application of Fatou's O lemma yields the last claim.
The Nevanlinna condition on f means precisely that u = log if I satisfies the hypothesis in Theorem 1.5 in Ch. 6, and hence it admits a Riesz decomposition. Let us repeat the argument anyway. Assume that f E N is not identically zero and B is the corresponding Blaschke product. Then log I f /BI is harmonic and f /B E N; therefore, by the the mean value property,
sup
1 j"jlogI(f/B)(re'O)I1dO
r<1 27r
i.e., log If /BI is in h'. Thus, there is a real measure p on T such that log If /BI = Pp. Now
+
/2xe't z J e" - z dp(t), a_ 1
f (x) = cB(z) exp
c c T,
(1.1)
since both sides are analytic and have the same modulus. Conversely, if p is a real measure and f is defined by (1.1), then it is in the Nevanlinna class and the representation (1.1) is unique. If we make the Jordan decomposition It = p+ - p.-, we find that f = g/h, where II9IIH= 5 1, IIhhhH= < 1,
and h#0. 1.3 Theorem. If f E N and f 0 0, then the radial limits f'(eie) exist for a.e. 8 and log If * I E L' (T). Moreover, f(z) = 1
cB(x)exp
e"+z
If ' (e::)Idt exp
1
27r o J eit - z log where p is a unique real singular measure and c E T. 27r
z,
eit+z eit
-
dp(t)(1.2)
Proof. Decompose p in (1.1) as la = Od8 + p where 0 E LI (T) and p, is singular. Since IB*I = 1 a.e., limlog IfrI = O for a.e. 8. If f E H°°, then f'
114
7. H" Spaces
exists (Theorem 1.2) and consequently log If* I = lim log I fj = ¢ ELI (T),
and so f' # 0 for a.e. 0. However, an arbitrary f E N is f = g/h for some g, h E H°°, and since h` # 0 a.e., f = g*/h' exists a.e. and log if*I = limloglfrI.
1.4 Corollary. If f,g E Al and f = g* on a set with positive measure, then f = g. Proof. If f E N and f' = 0 on a set with positive measure, then log If* I V L' (T) and thus f = 0. From Exercise 2 it follows that f -g E Al if f, g E N, and hence the corollary follows.
1.5 Proposition. If f E HP, 0 < p < oo, then p < 0, if p is defined by (1.2).
The class of all f E N such that p < 0 -is denoted N+. Thus the proposition says that HP C N+. Proof. To begin with, we may assume that f is nonvanishing, i.e., B = 1. Then log if I = Pp, and we have to prove that the singular part of µ is nonpositive. After taking an mth root if necessary, we also may assume that f' in f E H'. Now, I log+ x - log' yI $ Ix - yI (check!), and since f,. L' (T), u+ -+ u'+ in L' (T) if u = log If 1. By the Banach-Alaoglu theorem tends weak' to a positive measure A when r? some subsequence Since u,, - p (Theorem 1.1 in Ch. 6), we get
u'+-A=pp=,a'-,u By the minimality of the Jordan decomposition it now follows that p+ < u'+. Thus, p+ is absolutely continuous and the proof is complete.
Definition. A function M E H' is an inner function if IM' I = 1 for a.e. 0.
1.6 Proposition. A function M E HP (or in N+) is an inner function if and only if it has the form 1
M(z) = cB(z) exp
a_
/ 2n e`t + z
J
est - zdp(t),
(1.3)
0
where p is singular and < 0.
Proof. If M has the form (1.3), then it is clear that IM'I = 1 and II M II xo = 1. Conversely, if M is inner and it is decomposed as in (1.2), then (1.3) follows from Proposition 1.5.
Definition. We say that an inner function S is singular if its Blaschke product is a constant, i.e., if S is nonvanishing.
51. Factorization in H' Spaces
115
Hence, any inner function M has a unique factorization M = BS, where B is its Blaschke factor and S is a singular inner function.
Definition. If >_ 0 and log 0 E L' (T), then 1 z' e" + z Q(z)=exp2rr j0 eu_z1°gO(e't)dt
(1.4)
is called an outer function. Notice that IQ*I = (P for a.e. 9 according to (1.2).
1.7 Proposition. If Q is an outer function, then Q E HP if and only if ¢ E LP(T), 0 < p < oc. In this case, IIQIIHP = IIQ'IILv = II'IILP
Proof. First assume that ' E LP(T). By Jensen's inequality, I Q(re'B)IP = exp
2'
r
zn
Pr(9 - t) log9dt <
0
2 fz P,.(9 - t)
'dt.
0
Integrating with respect to 9 and applying F'ubini's theorem, we get that Q E HP and IIQII Hv <_ II0II L" . Conversely, if Q E HP, then IIQ' II LP = IIQIIHP by Theorem 1.2; but log IQ'I = log 0, and therefore 0 E LP(T) and El
11011D, = IIQ'IILP
We now are prepared for the main theorem.
1.8 Main Theorem. Any f E N+ (in particular, f E HP) has a unique factorization
f = MfQf = B1SfQf, where Mf = Bf Sf is an inner function, Bf and Sf are its Blaschke factor and singular factor, respectively, and Q f is the outer function 1
Q f (z) = exp
If
z" ee" 't + z
+z
log If- Idt.
(1.5 )
Moreover, If' I = IQ' I for a.e. 9 and f E HP if and only if Q f E HP, which holds if and only if f' E LP(T). In that case, IIfIIHP = IIQfIIHP = IIf'IILP.
If f E M, we have f = MfQ f/Sf = B fSfQ f/Sf, where 9f also is a singular inner function.
Proof. The existence and uniqueness of the factorization and the equality 1f1 = (Q ' I follow from Theorem 1.3. In view of Theorem 1.7, Q f E H"
if and only if f' E LP(T). If f E HP, then f' E LP(T) by Theorem 1.2,
116
7. H° Spaces
and thus Qf E H. Since IfI S IQiI if f E N+, it follows that f E Hp if Q f E H9. The norm equalities now follow from Theorem 1.2.
0
If f E N and f' E LP(T), it does not follow in general that f E H. If f = 1/S, where S is a singular inner function, then If *I = 1 a.e.; but nevertheless, f is not in any H9-space, not even in N+, unless it is constant, because of the uniqueness of the factorization.
1.9 Example. Taking u as minus the point mass at e't = 1, we get the singular inner function
S(z)=exp
l+z -1-z
One easily sees that IS*(e't)I = 1 for e't 54 1 but 0 for e't = 1. Thus,
f (z) = 1/S(z) = exp
+ 1
z
is a function in N \ N+ such that I f'(e't)I = 1 for all e't # 1.
§2. Invariant Subspaces of H2 We now are going to use the main theorem in Section 1 to study certain subspaces of H2. If 0 is an inner function, then OH2 = 141f; f E H2} is a closed subspace of H2. In fact, since II41f11HP = IIf1IHA, it is clear that OH2 C H2. If lbfj -+ g in H2, then q5fj is a Cauchy sequence in H2; but II fj - fk II H2 = II 41fj - 41fk II Hz . Hence, f j is a Cauchy sequence in H2, and
therefore fj -- f for some f E H2; therefore, g = Of E 0H2. If (Sf)(z) = z f (z), then clearly S: H2 --. H2
is a bounded operator with norm 1. If 0 is an inner function, then SgH2 C ¢H2, and therefore OH2 is a closed S-invariant subspace of H2. The converse is also true.
2.1 Theorem. If V is a closed S-invariant subspace of H2, then V = OH2 for some inner function 0.
Proof. Take f E V with minimal order of its zero at the origin, i.e., such that g1 f is analytic at the origin for all g E V. Then f V zV = { zg; g E V}, and therefore zV is a proper closed subspace (for the same reasons as above) of V. Thus, there is an element 0 E V such that 0 1 zV and II4IIH2 = 1.
However, since V is S-invariant, zV is also and hence 0 1 z"V for all
§2. Invariant Subspaces of H2
117
n > 0, implying that 2n
1
'
it
10 (e `
)I2a-int dt=0,
n=1,2,3....
The function 10 2 is real and in L'(T), and therefore it follows that it is equal to a constant a.e. Moreover, since II0` lI Lz = 1, 10' 1 = 1 a.e. and hence q5 is an inner function. As Z' O E V for all n > 0, p46 E V for all polynomials p. However, the polynomials are dense in H2 (partial sums of power series converge in the H2-norm; see below), and therefore OH2 C V. It remains to show that this inclusion is an equality. To this end, it is enough to show
that if h E V and h 1 OH 2, then h = 0. If h is such a function, then h l ¢zn for n = 0, 1, 2, ... , i.e., 0
2w
27r
n = 0, 1, 2, ...
.
0
However, since h E V, z"h E zV for n = 1, 2.... and 0 1 W, we therefore have
r 2x
0 27r
n = 1, 2,...
0
Hence, all Fourier coefficients of h'c vanish and therefore 0 a.e. However, since 10' = 1 a.e., it follows that h' = 0 a.e. and hence h = 0.
is an inner function. In We claim that OH2 C ipH2 if and only if particular, OH2 = t'H2 if and only if 46 = ctb for some c E T. In fact, if 46H2 C t/iH2, then ¢ = 1bMhQh for some h E H2, and thus 0 = ipMh because of the uniqueness of the factorization in inner and outer functions. Conversely, if 0/7P is an inner function, then certainly OH2 C (iH2. If OH2 = -OH 2, then both 01V) and t'/0 are inner functions, and hence O/i/i is constant. Notice that c/?/i is an inner function if and only if B4,/By, and S,6/S1,
are analytic, which in turn means that each zero of is also a zero of 0 (counted with multiplicity) and that dµ,6 - dµ,, is a negative singular measure. Thus the family of closed S-invariant subspaces is a partially ordered set with a rather complicated structure.
2.2 Remark. Recall that H2 is isometrically isomorphic to
12(N) _ {a ` (ao,ai,... );
11C,112
_
jai12 < oo}
(cf. Section 2 of Ch. 6) if f = o anzn is identified to (ao, al, ...) E £2(N). The operator S corresponds to the shift operator S(ao,ai,...) = (0,ao,ai....
on 12(N).
)
118
7. H" Spaces
We conclude this section with an approximation theorem. If g c H2, let P(g) be the closure of {pg; p polynomial}, which certainly is a closed S-invariant subspace. We already know that P(1) = H2.
2.3 Theorem. Let f and g be in H2. Then (i) P(g) = M9H2 = P(M9). (ii) f E P(g) if and only if Mj/M9 is an inner function. (iii) P(f) = P(g) if and only if M f = cM9. (iv) P(f) = H2 if and only if f is an outer function. Proof. To begin with, P(g) C M9H2 since g E M9H2 and P(g) is the least S-invariant closed subspace that contains g. By Theorem 2.1, P(g) = ¢H2 for some inner function 0. However, then M9Q9 = g = OMhQh; so M9 = 0Mh and hence M9H2 C ¢H2 = P(g). We therefore obtain the first equality in (i). The second equality follows if g is replaced by Mg. The other statements are immediate consequences.
§3. Interpolation in HO° In Corollary 1.6 in Ch. 5 we saw that for a given sequence of points aj in U with no limit points, and complex numbers f3j, one can find an f E A(U)
such that f (aj) = ftj . One says that aj is an interpolation sequence (for H°°) if for each ft = (i30i ftl, ...) E lOD there is an f E H°° that interpolates, i.e., such that f(aj) = f3j. It is easily verified that any interpolation sequence must satisfy the Blaschke condition (Exercise 20). However, the Blaschke condition is not sufficient. The points also must be sufficiently spread out in U.
3.1 Theorem (Carleson). The sequence aj is an interpolation sequence for H°° if and only if there is a 6 such that for each fixed k
n
,
I
- ak I > 6.
(3.1)
1 - ajak
Proof of the Necessity. Suppose that aj is an interpolation sequence and let T: HO° -+ l°° be the bounded linear operator defined by (T f )j =
f(aj). Now, aj being an interpolation sequence means precisely that T is surjective. The quotient space H°°/ Ker T is a Banach space with the usual quotient norm, and the induced operator T: HO°/ KerT -+ l°° is thus bounded, injective, and surjective; and so by the open mapping theorem there is M > 0 such that kEKf
IIf +kIIn= = If +KerTllxw <_
2
IIf3Ii1-
§3. Interpolation in H°°
119
if T f = /3. Hence, for each /3 E 1°° there is an f E H°° such that T f = /3 and I1fIIH- <- MIIQIIto- In particular, there are fk with IlfkllH- <_ M such that fk(aj) = bjk. Let
'f a'-z
Fk(z) = fk(z)
Iai l
aj z aj
Then IIFkIIH. < M (since we can divide out the zeros without affecting the norm), and if we let z = ak, we get
l aj - ak 111 1 -ajak > 1/M. i#k The hard part of Theorem 3.1 is to prove that the condition (3.1) is sufficient. The proof we give here is elementary and completely constructive. In Exercise 21 in Ch. 8, a proof is outlined where one first takes an appropriate smooth solution to the problem and then modifies it to an analytic solution by solving a 8/8z-equation.
First, we reformulate condition (3.1). We say that the sequence aj is separated if
aj - ak
1-ajak
>
c>0, j#k,
uniformly.
3.2 Lemma. The sequence aj satisfies the condition (3.1) if and only if it is separated and there is C6 > 0 such that for all k j=1
(1 - Ia,12)(1 -Iakl2) < C6. 11 - ajakl
(3.2)
Proof. If 0 < c < x < 1, then 1 - x < - logx < C(1 - x) for some C. If the terms in the sum in (3.2) are denoted Ajk, then
1 - A;k = Ia' . - akl2
I1-ajakl2
Ifc<1-Ajk < 1, then
EAjk <
-
73Ek j#k and hence the lemma follows.
log(I - Ajk)
C j#k
Proof of the Sufficiency. It is enough to find Fk(z) such that Fk(aj) = bjk
0 (3.3)
120
7. H' Spaces
and 00
IFk(z)I < C,
(3.4)
since then f (z) = Ei°,8kFk(z) solves the problem. A first attempt to find such Fk would be z - aj ak-a2 Fk(z) =
jo 1 - ajz/ 1 - ajak'
which converges, since aj satisfies the Blaschke condition; in fact, Fk(z) _ Bk(z)/Bk(ak) if Bk denotes the Blaschke product with zeros in {aj; j # k}. The problem is that these Fk do not satisfy (3.4). Therefore we shall modify the construction with weight factors and let Fk(z)
;
z
aj / ak - aj \ r 1 - IakI2 12
1-ajz 1- ajak/)
1-akz J
W(ak, z),
(3.5)
where W (C, z) is analytic in z and W((, C) = 1. Then Fk still satisfy (3.3), and we are going to show that W can be chosen so that (3.4) also is satisfied.
We assume that ... Iakl 5 Iak+1I < ... and first define Ia;l?IGI
(1±2Z_1±(l_IajI2), 1-djz 1-aj(
Since E 1 - Iaj I < oo, the series converges uniformly for fixed C and IzI < r < 1, and therefore z ,-4 ip((, z) is analytic and ?P(C, C)\= 0. Moreover,
Re,i((, z) _
qC-
(1 -Iaj 12IzII2
I1- ajxl
Ia; 21G1
-
121 12) (1 - Iaj 12). 1I1- Ia' &j(12
Observe that 1 - Iajl2l(I2 < 2(1 - 1(12) if Iajl ? ICI; so if W = exp(-Vi), then by (3.2), IW(ak,z)1 5 exp2C6 exp
- E
(1- IajI2)(1-1ajI2Izl2) 11 - ajzl2
Ia;l?Iakl
and W(ak, ak) = 1. From (3.1), (3.5), and (3.6) we now get that lFk(z)I <
exp C6ckexp
-Ecj j>k
where
ck-
Iakl2
\I1-akzll
is
(3.6)
§4. Carleson Measures
121
If cl, c2.... are any positive numbers, then r-,>k C]
-
ck exp
cj j>k
L
< 1E,>k+lci e
dt;
therefore,
cc
I:IFF(z)I < CEckexp k
k
j>k -Ecl
jCJe-`dt=C, o
<
0
and thus (3.4) holds.
We now are going to study the meaning of the condition (3.1). To begin with, let us assume that all of the points lie on the positive real axis, i.e., ak = rk = 1 - Ek and 0 < r1 < r2 < ... < 1. Notice that 1 - rk 1 - rk and 1 - rkrj max(1 - rj, 1 - rk). If rk is separated, we thus have that Ek - Ek+1 rk - rk+1 1 - rkrk+1 Ek and therefore Ek+1 <- cEk for some c < 1. Conversely, if this inequality holds, then rk actually satisfies (3.1). In fact, then it is certainly separated and since Ek < ck-jej if k > j, we have that c/ <
00 (1r2)(1-rk)\1-rk j=o
(1 - rjrk)2
j
1-rj
j>k 1 - rk
E ck-j+ r c_c<1,
j
jL>ik
independently of k. By Lemma 3.2, (3.1) therefore holds. Notice that, for example, rk = 1-1/k2 satisfies the Blaschke condition but is not separated, and hence it is not an interpolation sequence. In the next section we shall see, however, that if rk satisfies the Blaschke condition, one can always choose Bk such that the sequence ak = rke'Bk satisfies (3.1) and hence is an interpolation sequence.
§4. Carleson Measures The interpolation condition (3.1) does not only involve the modulii of the points ak, but also means that they are spread out in U in a certain sense. In order to understand this further one has to introduce the concept of a Carleson measure. If z, ( E U, then
I1-(zl--max{1-IzI+1 -I(I)+Iargz-arg(I,
(4.1)
122
7. H° Spaces
if the arguments are chosen so that the last term is less than or equal to it. This simple observation will be used repeatedly in this section. Let w(e'B, t) = {C E U; 11 - e-iezl < t},
t>0
be the so-called Carleson tents at 09. A positive measure p in U is a Carleson measure if p(w(e:s, t) < Ct,
uniformly for all t and 0. The infimum of all possible constants is called the Carleson norm of p. 4.1 Proposition. A positive measure p in U is a Carleson measure if and only if
f
1 - I zit 11 - CzI2
dp(C) < C
(4.2)
uniformly in z E U.
The exact shape of the tents is not important. In this context it is often convenient to use dyadic cubes and a dyadic decomposition of U. Let
Qk,t = {re'°; 2-k(t - 1) < 0/7r < 2-ke, 0 < 1 - r < 2-k} , where k = 0, 1, 2, ... and f= -2k + 1, -2k + 2,... - 1, 0, 1, ... , 2k. Then each set w(e`B, t) is contained in the union of at most two dyadic cubes Qk,t
with t < C2 -k for some absolute constant C (cf. (4.1)), and therefore p is a Carleson measure if and only if p(Qk,t) < C2 -k for all k and t. We also introduce the sets
Bk,t = {re'B; 2-k(f - 1) < 0/7r < 2-kt, 2-(k+1) < 1 - r < 2-k which constitute a partition of U; cf. Figure 1.
Proof of Proposition 4.1. First assume that (4.2) holds and take an arbitrary cube Qk,t. If z E Bk,t and C E Qk,t, then 1 - Iz12 - 2-k and 11 - CzI - 2_k (cf. (4.1)), and if we restrict the integration in (4.2) to Qk,t, we thus get that p(Qk.1) < C2-k. Conversely, fix z E U. By invariance we may assume that z E Bko,1. If C E Qka,t, then 11 - (zj - 2-4(l + 111) by (4.1), and hence CZ12dp(C)
(1+11)224p(Qko,t) 5 C,
independently of ko. It remains to estimate the integral over the boxes Bk,t for k < ko. However, on each such box we have the trivial estimate p(Bk,t) < p(Qk,t) <- C2-1 and, moreover, I1-s;zj - 2-k(1+Itj) if C E Bk,t,
§4. Carleson Measures
123
FIGURE 1
and therefore we get 2 -ko
t k>2 2-2k(1 + IPi)2
C2_M1 < C.
D
Condition (3.2) means precisely that p = (1 - i(l) >2 6a, is a Carleson measure. In one direction it is obvious since condition (4.2) means that (3.2) holds for all z; in particular, for z = ak. For the other direction, fix a dyadic cube. Again we may assume that it is Qko,l. If there is some ak E Bko,1, then by (3.2)
(1 -
jajl2)a(1a_2akl2)
<
C
e
aj EQkp,l
which implies that p(Qk.,I) (1 - laj l) 5 C2-ko. If there is no ak in Bko,l, we instead can apply the same argument to the two cubes below Bko,, and so on. In this way we obtain an exhausting sequence of subsets of Qko,I in which the measure of each set is bounded by C2-ko.
4.2 Example. Now suppose that 0 < r2 < r2 < ... < 1 and E(1 -
rj) < oo. We claim that one can choose Bj such that aj = rje:ei is an interpolation sequence. Let Ek = {r;; 2-(k+1) _< 1 - rj < 2-k}, and let ck
be its cardinality. Then Ek 2-kCk < oo. Notice that c points always can be distributed into two boxes so that no box contains more than c/2 + 1/2 points. Now fix a k. We can choose points aj = rjeie., rj E Ek, such that
124
7. HP Spaces
the union of all Bk,t that are contained in the same Qk',l' contains at most 2-` points. In fact, first arrange the points so that at most 2-kick + ck/2 + 1/2 of them lie in each Qo,l,, e' = 0,1; then proceed so that at most (ck/2 + 1/2)/2 + 1/2 points occur in each Ql,e,, and so on. Let T = E 6Q' and u = (1 - lzl)T. If now Qko,,*o is a fixed cube, then 00
T(Bk,t) <
2-k
µ(Qko.co) k=ko
2-k(Ck2-k' + 1) < C2 ko. k=ko
Bk,(CQkp,[o
We leave it as an exercise to verify that one also can arrange things so that the sequence becomes separated and hence an interpolation sequence. A simple way to do this is instead just to distribute aj in Bk,t with positive e when for example k is odd and with negative f when k is even. Furthermore, if the points in each box are appropriately spread out, the sequence will be separated.
We now shall consider another interpretation of condition (4.2). If we write dp(() = (1- I(?2)dv((), where dv is another positive (possibly infinite) measure, then the condition becomes fu K(t;, z)dv < C, where K(C, z) _
(I - Jz12)(1 -
1c12)
11_CZ12
By the proof of Lemma 3.2, K((, z) is of the same size as the Green kernel away from the singularity ( = z. Thus K((, z) can be seen as a regularization of the Green kernel, and condition (4.2) then says that a regularization of dv can be written as the Laplacian of a bounded function. (For measures related to the interpolation theorem, this is the content of condition (3.1).) This point of view leads to a simple proof of the following inequality for Carleson measures.
4.3 Theorem (Carleson's Inequality). Suppose that p is a positive measure in U. Then Fc is a Carleson measure if and only if
f if(()JPdp(() 5 ClI,fJ&,, f E Hp, p > 0.
(4.3)
Clearly, (4.3) holds for all p > 0 if it holds for some fixed p since we can divide out the zeros and then take arbitrary p'th roots. In view of Theorem 1.7 in Ch. 6, it also holds for hp if p > 1 (though only with a constant that depends on p). In Chs. 8 and 9 Carleson measures occur, but only as measures satisfying (4.3).
Proof. First suppose that u satisfies (4.3) for p = 2. Since C r+ (1 - ( z)-1 is analytic for z E U, it follows that (4.2) holds and therefore µ is a Carleson measure according to Proposition 4.1.
Supplementary Exercises
125
Conversely, assume that p is a Carleson measure. It is enough to show
the inequality for say p = 1 and f E A(U) that are smooth up to the boundary. By Proposition 4.1
-f
Mtl(z)
2
11 - LzI2 dA(C)
is a bounded function, -C < Mp < 0, and COO in U. Moreover, a straightforward computation shows that it is subharmonic and, more precisely,
_ 4
ll2
L M`(z) = fu 11- C
a
du(()
By formula (2.6) in Ch. 8 (applied to the disks D(0, r) and letting r increase to 1) we get
I (1- IzI2)If(z)IAM14(z)da(z) <- CIIfIIH,. In view of Fubini's theorem it is therefore enough to show that (1 - IC12)(1 - Iz12)If(z)IdA(z).
If(C)I < C f
11 - CzIa u However, for any h E A(U)/n C1(U),
h(() _ _ J
(1
U
IIz 2)
(I - ()
I
(4.4)
C E U;
see, e.g., Exercise 15 in Ch. 3. Hence (4.4) follows by taking h(z) =
f(z)l(1 - Cz) 4.4 Remark. The inequality (4.4) is true for any positive subharmonic u instead of If 1. In fact, if O(z) = (k((z) z)/(1 - z(), then u o 0-1 is subharmonic and hence
u o g-1(0) < C
JIwI<E
u o q-1(w)dA(w);
and so we get that 2 2
UM 5 C f
IO(z)I (
-IzI)
(1 11
- CzIa
u(z)dA(z).
However, in the set I5(z)I < e, I - ICI - 1 - IzI and hence (4.4) follows.
Supplementary Exercises Exercise 1. Show that f E A(U) is in N if and only if f = g/h for some f, g E HOO where h is nonvanishing.
126
7. Hp Spaces
Exercise 2. Show that N is in fact a linear space.
Exercise 3. Suppose that f e A(U). Show that f E Hp, 0 < p < oo, if and only if there is a harmonic u in U such that If (z)II < u(z). Also show that if there is some harmonic majorant, then there is at least one uf. Finally, show that IIfIIHp = uf(0).
Exercise 4. Suppose that f E A(U). Show that f E N if and only if log+ IfI has a harmonic majorant.
Exercise 5. Show that the polynomials are dense in Hp, p > 0. Exercise 6. Suppose that f E A(U). Show that f is a Blaschke product if and only if limrzl(1/2ir) f IlogIf(re'e)II d9 = 0.
Exercise 7. Suppose that f E A(U) and Ref > 0. Show that f is an outer function. Hint: Notice that arg f is bounded.
Exercise 8. Show that f(z) _ (1-z)-' log(1-z) is in Hp for all 0 < p < 1 and that f has unbounded Taylor coefficients.
Exercise 9. Suppose that f E N and f' E L'(T). Does it follow that .Ff'(n) = 0 for n < 0? (Answer: No!)
Exercise 10. Suppose that f c N. Show that f E N+ if and only if
logI
f logI f*(ce)Ide0
f (Te`B)I d6 = lim 2- f27T Hint: Compare to Exercise 11 in Ch. 6.
.
Exercise 11. Suppose that 0 < r < s < oo. Show that the inclusions N i H'' D H° D H°° are proper. Exercise 12. Suppose that f E A(U) and that f (U) is not dense in C. Show that f' = lim f, exists for a.e. 0. Exercise 13. Suppose that K is a compact proper subset of T. Prove that each f E C(K) can be approximated uniformly by analytic polynomials. Hint: Use the Hahn-Banach theorem.
Exercise 14. Show that if f E N+ and 1/f E N+, then f is an outer function.
Exercise 15. Show that if f,g E H' and Ifl $ IgI, then IMf1 < IM91 and IQiI <_ IQ91.
Exercise 16. Show that if f EN+, then zn
log if (0) 1 < 2?r f log 0
with equality if and only if Mf is constant.
If8 ldt,
(4.5)
Supplementary Exercises
127
Exercise 17. Suppose that f E N and (4.5) holds. Does it follow that f E N+? Exercise 18. Show Theorem 2.3 without referring to Theorem 2.1: Since trivially P(g) C M9H2, we have to prove the opposite inclusion. Since IIM9f - gp1IH2 = Of - Q9PIIH2, it is enough to show that P(g) = H2 if g is outer. If h 1 P(g), then 2w
1
0,
1 f
n < 0,
and therefore g"h' = k' for some k E H1. However, since g is outer, k/g E N+ and (k/g)' = h` E L2(T), and therefore k/g E H2. It now follows that h' is constant and hence h = 0, since otherwise g E H2 and f g' = 0, which would imply that g were divisible by z. Exercise 19. Show Theorem 2.3 for 1 < p < oo instead of just p = 2. Hint: Recall Theorem 2.3 in Ch. 6. Exercise 20. Show directly that any interpolation sequence must be the zero set of some bounded f 0 0. Exercise 21. Show that if a,,3 E U have given modulii, then Ia - 0I/I1 a/3I is minimal if and only if they lie on the same radius. Exercise 22. Let T be the operator in the proof of the necessity part of the interpolation theorem. Show that KerT = {gB; g E H°°}, where B is the Blaschke product with zeros a3.
Exercise 23. Show that the sequence rk = 1-1/k2 is not separated. Does it satisfy (3.2)? Exercise 24. Suppose that aj is a sequence in U with no limit points and
0 E 1'. Show that if there is some f E H°° that interpolates 3 in a then it is unique if and only if E 1 - Ia3I = oo. Exercise 25. Show (4.1). Hint: Notice that 11 -- roe-ielr2e&O2I is the Euclidean distance between rlr2e`(e'-e1) and the point 1. Exercise 26. Let µ be a finite measure in U. Show that one can rearrange the mass on each level {2-(k+1) < IzI < 2- k} = UtBk,t so that the new measure is a Carleson measure. Exercise 27. Fill in the details in the construction in Example 4.2. Exercise 28. Let M f be Hardy-Littlewood's maximal function; see Section 1 in Ch. 6. Use Carleson's inequality to prove that
I IMf'Ipde < CIIfIIHo T
(4.6)
for p > 0. The same holds for hp if p > 1. Hint: First prove it for M instead of M. Choose a piecewise C' curve 0 --+ r(B)e'B such that
128
7. H' Spaces
If (r(O)e'B)J > 1/2Mf (e'B). Define a measure on this curve so that the measure of any segment is precisely the length of its projection onto T. Then this measure is a Carleson measure if it is considered as a measure in U. Now Carleson's inequality implies (4.6). The hyperbolic distance d(z, w) between the points z and w in U is defined in the following way: 1
log 1 + `IZI zl,
d(O,z) = d(z,0) = 2
and
d(z, w) = d(0, q=(w)).
Exercise 29. Show that d(z, w) is well defined for all z, w E U, that d(z, w) = d(w, z), and that d(z, w) = d(t(z), O(w))
for any that is either an automorphism or the conjugate of some automorphism of U.
A hyperbolic line in U is the intersection of U with a circle in C that intersects T orthogonally. Thus, hyperbolic lines through the origin are just ordinary lines. Exercise 30. Show that through each disjoint pair of points there is one and only one hyperbolic line.
Exercise 31. Show that all automorphisms and their conjugates map (hyperbolic) lines onto (hyperbolic) lines.
Exercise 32. Show that d(z, w) < d(z, z') + d(z', z)
with equality if and only if z, z', and w are consecutive points on the same hyperbolic line.
Exercise 33. Show that there are constants c and C such that
d(z, w) < I _ w < Cd(z, w), for all z, w such that d(z, w) < c. The unit disk with this geometry provides a model for non-Euclidean geometry. In fact, all of the Euclidean axioms hold except the axiom of parallels.
Exercise 34. Show that through any point outside some given hyperbolic line there are infinitely many lines that do not intersect the given line.
Notes
129
Notes We recommend (D] and [G] for further results and references. The key to the HP-theory here is Theorem 1.1. There is an analogous
result in the upper half-plane. Instead of Theorem 1.1, one can rely on the fact that if 1P is subharmonic for all p > 0. There is a generalization of the Hr-theory to the upper half-space in R"+' for p > (n - 1)/n, but then there is no analogue to Theorem 1.1. This theory instead is based on subharmonicity; see (S].
Theorem 1.1 is due to F. Riesz, 1923, and the main theorem (Theorem 1.8) is due to Smirnoff, 1929. See [D] for more historical remarks. Theorems 2.1 and 2.3 are due to Beurling, 1949. These theorems have been generalized by Gamelin to p > 0; see TAMS, vol. 124 (1966), 158-167. See also Exercise 19.
The interpolation theorem (Theorem 3.1) was proved by Carleson in 1958. The hard direction is to prove that (3.1) is sufficient. The explicit proof given here was discovered by Jones (Acta. Math., vol. 150 (1983), 137-152).
There is a generalization of the interpolation theorem (Theorem 3.1) to HP, p > 0; see [D]. If the sequence Ck in Example 4.2 satisfies E 42-k < oo, which is more restrictive than the Blaschke condition, then for almost all choices of 9k (interpreted in an appropriate sense) ak = rkeiek is an interpolation sequence; see Rudowicz, Bull. London Math. Soc., vol 26 (1994), 160-164.
The proof given here of Carleson's inequality (Theorem 4.3) is due to Berndtsson. The underlying idea that essentially all Carleson measures are of the form (1 - I(I)o0, where 0 is a bounded subharmonic function, is from Proc. Univ. of Maryland 1985--86, Springer-Verlag (1987). The usual proof of Carleson's inequality proceeds via an LP(T)-estimate for the Hardy-Littlewood maximal function. Incidentally, this estimate follows quite easily from Carleson's inequality; see Exercise 28.
8
Ideals and the Corona Theorem
§1. Ideals in A(S2) We begin with
1.1 Theorem. If gi, ... , gn E A(Q) have no common zeros in S2, then there are u1, ... , u, E A(1l) such that F_i gjuj = 1. One can obtain Theorem 1.1 quite easily from Weierstrass' and MittagLeffier's theorems (see Exercise 1), but we will give a proof ttlat relies on the existence of solutions to the inhomogenous Cauchy-Riemann equation (Theorem 2.3 in Ch. 3). The advantage is that one obtains estimates of the solution (u1, . . . , u,) if one can solve the 8/82-equation with estimates. This will be used in the proof of the corona theorem in the next section.
Proof of Theorem 1.1. Consider g = (gl, ... , g,) as a row matrix and u = (ui, ... , un)t as a column matrix (thus t denotes transpose). We are looking for an analytic u such that gu = 1. If IgI2 =
Igi I2 and
9t
y IgI2
then y E C°O(I2) and gy = 1. We then want to modify y to be an analytic solution, and to this end we set
u = y + wgt, where w is an antisymmetric (n x n)-matrix, i.e., wt = -w. Since gwgt is a scalar (a 1 x 1-matrix), gwgt = (gwgt)t = gwtgt = -gwgt, which gives gwgt = 0, and hence u solves gu = 1. If w is now an antisymmetric solution to (1.1)
§2. The Corona Theorem
131
(alai of course acts componentwise on matrices; observe that the righthand side in (1.1) is antisymmetric), then
aw t _
ay
On
ary
+ az g
t
+ 9 (920
a
= 9` ai (9 Y)`/I9I2 = gt
a1
)
t
g IgI2
a-,
az
1
az lgl = 0'
and hence u is in fact an analytic solution as desired. Let
0
In
[gi,.,gn] _ Ev.igj, vi E A(SZ)} 1
denote the ideal in A(1) generated by gi,... , gn. Theorem 1.1 then says that [gl, , 9n] = [1] = A(S1) if (and only if) gi,... , gn have no common zeros.
1.2 Corollary. For given hl , ... , hn E A(1Z) there is a 0 E A(S2) such that
[hl,...,hn] _ [Y']. Thus, each finitely generated ideal in A(Q) is a principal ideal.
Proof. According to Weierstrass' theorem there is a 0 E A(Q) that has zeros with right multiplicities exactly where all hj have common zeros (this
is a set with no limit points unless all hi vanish on some component). However, then the functions gj = hj /4) have no common zeros, and hence
by Theorem 1.1 there are uj such that >ujhj/4) = 1, i.e., Eujhj = 0,
and therefore 0 E [hi, ... , hn]. Since hj E [¢] for each j, the corollary is
0
proved.
§2. The Corona Theorem Suppose that gl, ... , gh E H' = H°°(U) and there are uj E HOD such that E gjuj = 1. Then by Cauchy-Schwarz' inequality, E Igj I2 > 62, where 1 /6 = f maxi II uj II H= Conversely, we have
2.1 The Corona Theorem. Suppose that gi, ... , g,, E H°° and there is a 6 > 0 such that Igj I2 > P. 1
(2.1)
132
8. Ideals and the Corona Theorem
Then there are u1, ... , u E H°° such that
Egjuj=1. The corona theorem was stated as a conjecture in the 1940s in connection with the study of Banach algebras (this is where the name "corona theorem" originates). It was first proved by Carleson in 1961, and a different proof based on functional analysis was found by Hormander in 1967. Around 1980 this proof was simplified by Wolff. The proof we give here is a slight modification of Wolff's proof. To begin with, one can reduce to
2.2 Proposition. There is C6,,, > 0 such that if g1, .., g E H°O n coo(U)
and b < lgJ < 1, then there are uj E HO° with Jul < C6,,, such that
E9iu.i=1. Proof that Proposition 2.2 Implies Theorem 2.1. In fact, we may assume that b < IgI < 1 in the corona theorem. Then g''(z) = g(rz) satisfies the assumptions in Proposition 2.2, and therefore we get analytic u' with Iu'i < C6,,, and g''ur = E gju7 = 1. Since {ur},<, is a normal family (Proposition 1.7 in Ch. 2), we can extract a-convergent subsequence
u''i - u when ri / 1. Then of course Jul < C6,,, and gu = 1, and thus the corona theorem follows.
In what follows we therefore may assume that g satisfies the hypotheses in Proposition 2.2. We shall reformulate it as a 8/32 problem; but in order to give the precise statement, we need the following formalism. Suppose
that 8V/8z = f, V E C°°(U), and v = VAT. Then, by Stokes' theorem ((1.5) in Ch. 1), 2i
r
JU f hda = fvhdz,
Vh E A(U) n C°°(U).
(2.2)
If f E L1(U), v E L1(T) (or v E M(T)) and (2.2) holds, one says that v solves the 8/02 equation on the boundary and writes 8bv = f.
Note that 8b is linear, and that 8bav = of if a E A(U) n C°°(U) and 8bv = f. To be precise, Zibv defines an element in the quotient space L'(U)/{g E L'(U); ghdA = 0 t1 h c A(U) n C°°(U)}. u We claim that if, for example, f E C°°(U) and 8bv = f, then v is the restriction to the boundary of a unique solution V to 8V/8z = f. In fact, let W be an arbitrary solution in C°°(U) to 8W/8z = f (such a solution always exists; however, the reader may just as well assume that f is smooth in a
§2. The Corona Theorem
133
neighborhood of U; see also Exercise 10). If W = WIT, then ab(v - w) = 0, which means in particular that fT(v - w)z" dz = 0, n = 0, 1, 2, ... , i.e., all negative Fourier coefficients of v - w vanish; and therefore, v - w is the boundary values of an analytic function H E Hi. However, then v is the boundary values of the function V = W + H that solves 8V/8z = f. We shall use the following weak formulation of the a equation.
2.3 Proposition. Suppose that f E CO°(U). The equation abv = f has a solution v with IIvIILO < C if and only if 12
/U fhd.) < C
,h EA(U) n CO'(U).
(2.3)
Proof. Suppose that (2.3) holds; for (the restriction to T of) h E A(U) n C°°(U), let Ah = 2i fu f hdA. U
Then iAhi < C fT. ihidO and by the Hahn-Banach theorem, A can be extended to a bounded linear functional on L'(T) with norm < C. Hence there is an a E L°°(T) such that ilallL= < C and
2i f f hda - Ah = J ha dB, h E A(U) n C°° (U). T
U
If now v(z) _ -i2a(z), then (cf. (2.2)) abv = f. The converse is trivial and
0
we leave it as an exercise.
So far we only have made a preliminary reduction (Proposition 2.2) and expressed it in terms of a system of ab-equations. The next theorem due to Wolff is in fact the heart of the matter.
2.4 Theorem. There is a C > 0 such that if 0, f E C°° (U), 0 real, 101 < 1,
IfI2 < Aq,
(2.4)
az :SAO,
(2.5)
and
then there is a solution v E L°° (T) to Obv = f with IiviiL= < C.
Proof of Proposition 2.2. If y = gt/Igf2, then -y E COD(U), gy = 1, and IyJ < 1/6. Assume that we can find an (antisymmetric) solution to _ abw =
g) - a g
1
g1-2
=F
8. Ideals and the Corona Theorem
134
such that w E L°°(T) and Iwl < Can, where C6,,, depends only on 6 and n. If u = ry + wgt, then (as before) 8bu = 0, gu = 1 on T, and Iu1 < 1/6 + C6,n = C6,n. However, then u is the boundary values of a U E H'; and since IIuIIL- 5 C6,n, U is actually in H°° and IIUIIH- 5 C6,n. Finally, gU - 1 since gU - 1 E H°° and the equality holds on T. Thus, we have to solve 8bw = F with L°° (T)-control of the solution. Note that a + = 8x 191-2g = and therefore Iary/8zi 5 lg'l, which implies that 8ry
-2(81914)9
8z
1912,
IF12 < 19'12 _ 01912.
Here, "<" means "less than or equal to C6,n times... ," where C6,n depends only on n and 6. By a similar computation one finds that
azl0 I91+1 I
(91z2-a,-1
`1912_ 01912.
Therefore, Proposition 2.2 follows from Theorem 2.4 with 1912 = t
.
C1
What remains is to prove Theorem 2.4, and to this end we will apply the following simple but useful lemma.
2.5 Lemma. If g, Eli E C' (U), 0 real, and g is analytic, then
1 f(i - z2)(D+G)19e< f T91edO,
f(i - Iz12)(A 1')IgI <4eIIOIIL-(U) f
T
and
2 1 (1 _ U
19I
f
T
(2.6)
(2.7)
I9IdO,
(2.8)
I91dO.
The inequality (2.7) means that (1 - Izl2)o?p is a Carleson measure; cf. Section 4 in Ch. 7.
Proof. First suppose that g - 1. Then 4
f(i - Iz12)O
ev' < f(i - 1212) fU(1_Iz12)a12
=
e"
-I
((a 1 +
2) e'' 1
f e'`d9<2
I e&d9,
§2. The Corona Theorem
135
where we have used Green's identity in the last equality. To obtain (2.6) in the general case, note that log(IgI2 + e) is subharmonic, then replace z/i by +/, + (1/2) log(IgI2 +e), and finally let e \ 0. One gets (2.7) from (2.6) if 0 is replaced by /2II0IIL (U) From the computation above it also follows
that
f(i-_ I zI2) I az
lz e0
<
2
fT
if 0* > 0, and (2.8) follows if one takes ?p = (1/2) log(IgP2 + f) and lets
e\ 0.
Proof of Theorem 2.4. In view of Proposition 2.3 we have to estimate fu f h, and we first rewrite it:
f fh= f IzI2fh+ f(i u
u
hzaz (IzI2
- 1) =
JU(1 - iz12)ez
(zf h) fu IzI2f h = fu f = fu (1 - Izi2) z zh + fu(1 - Iz12) f a (zh),
where the second equality follows from Stokes' theorem (the boundary integral vanishes). Thus, zh + fu (1 - IzJ2)f (1fu fu ez (zh) +fu - Iz12)fh
fh - (1- IzI2) az =I1+I2+I3i and we have to show that
IliI fTIhI, j=1,2,3.
(2.9)
From (2.7) we immediately get 11, 1 <
f(i - IzI2)A.0 hl < f
T
Ihi.
If g = zh, we have 11212
--
(f(i - IzI2)vlg'I)
< fu (1 - IzI2)o0191
f
2
Ihi\ 2 (1
-
Iz12)
111
I9II2
. (I I91)2 =
CfT
by (2.7) and (2.8), and therefore (2.9) holds for j = 2. The estimation of 13 is left as an exercise.
136
8. Ideals and the Corona Theorem
2.6 Remark. Wolff's original formulation of Theorem 2.4 reads: Suppose that the Carleson norms of (1 - I zI) I f I2 and (1- Izj) ja f/azj are < 1. Then there is a solution v E L0(T) to abv = f with IIvIIL- <- C, where C is an absolute constant. See Exercise 19. The discussion preceding Theorem 4.3 motivates the relation between the two formulations. Notice that, contrary to the case of the interpolation theorem, the Carleson measures that appear in the proof of the corona theorem are already smooth, and therefore we do not need to worry about regularizations. See also Exercises 20 and 22. In this context it is natural to mention that there is a very general result due to Hormander about L2-estimates of solutions to the inhomogeneous Cauchy-Riemann equation(s) in several complex variables. Here is the one-variable case. A proof is outlined in Exercises 23 and 24.
2.7 Theorem. Suppose that n C C is bounded and 1(i is any subharmonic function in S2. For any smooth function f there is a (smooth) solution u to
au/a2 = f such that
f
Iu,I2e-P < C2 r I f I2e-0,
o
n
where C is the diameter of Q.
Supplementary Exercises Exercise 1. Here is an outline of an alternative proof of Theorem 1.1: Suppose that it is true for n - 1 generators. Take a 0 that has exactly the zeros that are common to gl , ... , gn_ 1. Then by the induction hypothesis there are vj such that glv1 + ...9n_1Vn_1 = 0. Since gn is nonvanishing at the zeros of (P, there is by Corollary 1.6 to Weierstrass' theorem a h
such that (1 - gnh)/0 = f is analytic. Hence, 1 = Of - gnh = gl fv1 +
-9nh. Exercise 2. For N = 1, 2, 3, ... , let 00
91v(-)=
2
n(1-n2). n=N
Show that the ideal in the ring of entire functions, generated by 19N), is not a principal ideal.
Exercise 3. Show that the solution -y = gt/Ig12 is the solution to gy = 1 that has the pointwise minimal Euclidean norm.
Exercise 4. Suppose that p is a measure on T. Show that µ = fdO for some f E L2(T) if and only if (1 - ICI2)IVPµ(()I2 is integrable over U.
Supplementary Exercises
137
Exercise 5. Suppose that f is C°° in a neighborhood of U, p E M(T), and abµ = f- Show that p, is absolutely continuous.
Exercise 6. Prove the "only if" part of Proposition 2.3.
Exercise T. Suppose that h, g1, ... , g E H' and IhI < Igl3. Show that there are ul,... , u,a E H°O such that > gjuj = h. Hint: Copy the proof of the corona theorem.
Exercise S. Show that 3 in the preceding exercise may be replaced by 2 + E. First show that l9'i2/92-2e < CeAjg12`
Exercise 9. Suppose that gj E H°O, j = 1,... , n and z) =
- Ki2)a - z /)
a>0.
I9i(9i
E
C11
egg I2 > 62. Let
()V2 JJ
Show that if ¢ E A(U) flC°°(U), then O(z)
- -n
f
z)/((' - z) E C°°(U x U). Give explicit u3 = TjO E and that A(U) fl C°°(U) such that >gjuj = 0. Show that T3q E H' if E Hp,
1
Exercise 10. Show that
f f (()dA(()
u(z)
is a C°°(UU)-solution to 8u/8z = f if f E C°°(U). Hint: First show that if there is some solution in C°°(U) or at least in Cc(U), then u(z) is in C1E-1(U). To this end use Cauchy's formula 1.6 in Ch. 1. (Since f has some Ck-extension (in fact a C°°-extension) to a neighborhood of U, it also has a Ck-solution on U.) Alternatively, one can verify the formula 49-u 8zm
_ _M 7r
f `1-z1 1(12
1
"z
(a"` f/ '")(C)da(C) z
Exercise 11. Show the corona theorem in A = {1 < Iz) < 2}. Hint: Note that the corona theorem is true in any simply connected domain, in particular in D = {0 < Re z < log 2}. There is a one-to-one correspondence between functions in A and functions in D such that g(z + m.2ai) = g(z). If fj e HO°(D) are periodic in this way and IfjI2 > 62, show that there are periodic u3 E H°°(D) such that F_ h u3 = 1.
138
8. Ideals and the Corona Theorem
Exercise 12. Show that if f E C°° (U), then 2f (() (() u(z)
l Ju
z E T,
is a solution to 8bu = f. Exercise 13. Show that if 0 E HP, 1 < p < oe, and f satisfies (2.4) and (2.5), then 1
V (z) (Z)
*7r
r zf(().O(()dA(()
u
1 - (z
z E T,
is a solution to 8bu = f o and IT
P.
T
What happens for p = 1?
Exercise 14. Prove (2.9) for j = 3. Show that
j(i - Iz12)lhI < C J
T
lhldO
for holomorphic h.
Exercise 15. Suppose that f E H°°. Show that (1- IzI)I f'12 is a Carleson measure.
Exercise 16. Assume that g E H2. Show that
2f (1 - Iz12)Ig'I2 < f IgI2dO. U
T
(2.10)
Exercise 17. Show that for the proof of Theorem 2.4 all one needs is (2.10). Note that 04 is actually only (a constant times) Ig'I2 for a (sum of) bounded analytic functions.
Exercise 18. Suppose that a/' is bounded and subharmonic in U. Show that it = (1 - K(I)o,'dA(() is a Carleson measure. Hint: Cf. (2.7). Exercise 19. Prove Wolff's theorem as it is stated in Remark 2.6. Hint: Use Theorem 4.3 in Ch. 7.
Exercise 20. Suppose that µ is a Carleson measure in U. Prove that there is a solution v E L°°(T) to 8bV = µ.
Exercise 21. Here is an outline of a proof of Carleson's interpolation theorem (Theorem 3.1 in Ch. 7) based on the ab equation: The separation hypothesis ensures that we can find Xi E Co (U) with mutually disjoint supports such that Xj = 1 when d(z, aj) < e and vanish when d(z, a3) > 2e (d(z, w) - Iz - wI /II - 2w 1). Then 0 _ E,3IX.i is a smooth solution to the
Supplementary Exercises
139
interpolation problem. Let B be the Blaschke product with zeros at aj. Then
F=j>j az'lB is a Carleson measure (note that f JaXj/a&1 < C(1 - tajl)). Solve abv = F
and take g such that g1T = 0 - v. One can in fact assume that there are only a finite number of points aj since the occurring constants will be uniform. The general case then follows by a normal family argument.
Exercise 22. In Carleson's original proof of the corona theorem one of the main steps was to find a CO° solution -y to g-y = 1 such that day/azJ is a Carleson measure. Given such a -y, complete the proof of the corona theorem.
Exercise 23. If t4' E C2(1l) is real and A > 0, then au az
=f
(2.11)
has a solution (in the distribution sense) such that Iu12e--O < 4 j Jf 12e->v/OVG
Jst
n provided the right-hand side is finite. Sketch of proof: Let (g, X) = fo gXe-0, and let 19
Then
01V
az + Oz X) = (g, '9X) for X E C. 00(Q). Show that (2.11) has a solution
with (u, u) < C2 if and only if J(f, X)J2 < C2('9X, i9 ) for X E Co (St).
Verify the equality
-z9
Use it to prove that theorem.
a
+
, X) J(f.
z9 =
4
Ozli.
12 < 4(29X, t9X)(f /(O4/'), f) and conclude the
Exercise 24. Use the results in the preceding exercise to prove Theorem 2.7.
Exercise 25. Let A = A(U) r1 C(U), the so-called disk algebra. Suppose that g ' , . . . , g,,, E A and E Jgj I > 0. Show that there are uj E A such that 1: gjuj = 1. Hint: First, construct a solution yj to 1: gjryj = 1 such that ryj,aryj/az E C(U). To this end, take preliminarily yj as 1/gj times the characteristic function of the set Ej = {z E U; lgj(z) I > 0} and use a smooth partition of unity. Then proceed as in the proof of Theorem 1.1.
140
8. Ideals and the Corona Theorem
Notes A thorough discussion of the history of the corona problem and various aspects of proofs are given in [G). The best known constant (in Proposilog 6), which is independent of tion 2.2), due to Uchiyama, is C6 = the number of generators n. For the proof and various vector-valued versions of the corona theorem, see [N] Appendix 3, and the references given O(6-2
there. The corona theorem is closely related to the interpolation theorem (Theorem 3.1 in Ch. 7). Both can be reduced to the problem of finding a bounded solution to a certain 8b-problem. In the case of the interpolation theorem, the occurring right-hand side is a Carleson measure (cf. Exercise 21) and is therefore readily solved by Carleson's inequality (Theorem 4.3 in Ch. 7) and Proposition 2.3. However, in the case of the corona problem, the most
obvious right-hand side occurring from the smooth solution y = gt/[g[2 requires the more involved Theorem 2.4 to get a bounded solution. However, in Carleson's original proof a more refined -y occurs such that 8y/82 is actually a Carleson measure. By this choice of -y the corona theorem readily follows; cf. Exercise 22.
In the paper by Jones referred to in the notes to the previous chapter, an explicit bounded solution to 8bu = f is constructed, provided f is a Carleson measure. No such formula is known if f satisfies the hypothesis in Wolff's theorem. The corona theorem holds in, e.g., multiply connected domains (a finite number of holes), but the general case is not known; cf. Exercise 11. The analogue question for A = A(U) fl C(U) is true, i.e., if gj E A has no common zeros, then there are uj E A such that F_ gjuj = 1. This is a simple result in the theory of Banach algebras; see, e.g., Ch. 18 in [Rul]. See also Exercise 25, where another proof is suggested. Hormander's theorem (of which Theorem 2.7 is a special case) was published in Acta Math., vol 113 (1965). This is of particular importance in the theory of several complex variables. If the power 3 in Exercise 7 is replaced by 1, the statement is false. The case with power 2 is unknown; see [G] and [N].
9
H' and BMO
§1. Bounded Mean Oscillation Recall from Ch. 6 that a function u on T is in LP(T) if and only if its Hilbert transform Hu is. By virtue of (2.8) in Ch. 6, we can define the Hilbert transform even for u E L1(T) as a formal Fourier series; in general, it will not belong to L'(T) but merely be a distribution; cf. Remark 2.1 in
Ch. 6. More precisely, Hu is in L1(T) if and only if u = Ref` for some f E H1. Definition. HR is the space of (complex-valued) functions in u E L1(T) such that also Hu E L1(T) with norm IIuIIH; = IIuIIL1 + IIHuIILI
Some observations are immediate: (i) Since H is a real opererator, u is in HR if and only if both Re u and Im u are.
(ii) The Hilbert transform H maps HR into itself. (iii) There are strict inclusions L1(T) D HR' D LP(T)
if
p>1.
(1.1)
(iv) If u is real, then u E Hal if and only if u = Re f' for some f E H1 and there is a constant C > 0 such that 1
IlfllWW <_ CIIuII H,
(1.2)
if f (0) is real.
(v) If functions in H1 are identified by their boundary values, then H1 is a subspace of HR and the Szego projection maps S: HR -+ H'. Thus, HR is a somewhat smaller space than L1(T), but with the advantage that the Hilbert and Szego transforms are bounded. In some respect HR therefore serves as an approriate surrogate for L1(T). Because of the
142
9. H1 and BMO
inclusions (1.1) the dual space of HR has to be something between L°°(T)
and LP(T), and the main objective in this chapter is to determine this dual space. It should be pointed out that the HR condition is not a simple condition on the size of if 1. In fact, if If I is in HR, then f also is, but the converse is not true; see Exercises 6 and 7. 1.1 Proposition. HR is a Banach space and COO (T) is a dense subspace.
Proof. It is enough to consider the subspace of real functions. By (1.2) this subspace is isomorphic to { f E H'; f (0) is real}, which is clearly a Banach space. Moreover, any f in this space can be approximated by, e.g., the C'-functions fr(e't)= f (re'') in H1 norm; cf. Theorem 1.2 in Ch. 7.
If I is an interval on T and u is an integrable function, let u, denote the mean value over I,
u, _ III 1 u(t)dt, where III is the length of I (for simplicity we often write u(t) rather than u(eit))
Definition. BMO is the space of all functions u E L2(T) such that IIuII: = sup IIII
1 J Iu(t) - u,i2dt+27r
f"Iu(t)I2dt
0
(1.3)
is finite, where the supremum runs over all intervals I on the unit circle. It is readily verified that 11 11. is a norm and BMO actually is a Banach space; in fact, any Cauchy sequence in BMO must have a limit at least in L2(T), and it is easy to see that this sequence actually converges to this limit in the BMO norm. (Sometimes the last term in (1.3) is omitted in the definition. Then II II. is just a seminorm since it is independent of additive constants, and BMO then becomes a space of functions modulo
constants.) Moreover,
IIuII < VIItIIL.o, and therefore BMO contains L°°(T). However, the inclusion is proper; one can show directly from the definition that, e.g., log Its (or log Ii - e" 1) is in BMO, whereas on the other hand (t/ItI) log ItI is not; see Example 1.5 below and Exercise 2. However, if f belongs to BMO, then If I also does (Exercise 3).
Exercise 1. For u E L2(T) let ur(e't) = Pu(re't), where Pu is the Poisson integral of u. Prove that IIurli. Ilull when r / 1 if u E BMO.
§1. Bounded Mean Oscillation
143
1.2 Remark. BMO stands for bounded mean oscillation, and in our definition we have measured this mean oscillation in the L2(T)-norm. It is more common to define BMO with L' (T)-norms instead. This definition is as follows: An f E L'(T) is in BMO if
If - flldt +
slip
2
f2, If Idt
III J is finite. It is trivial that f satisfies this definition if it is in BMO in our sense. The converse is also true and is a consequence of the John-Nirenberg theorem, and so these two definitions are equivalent. The reason to use L2 instead of L' is merely practical; it simplifies the proofs of the main results. The main result in this chapter is Fefferman's theorem, which states that BMO is the dual of HR (Theorem 2.1). However, we first shall consider some useful alternative ways to express the BMO norm (Proposition 1.3 below). To this end we need the Riesz decomposition in its simplest form for functions that are smooth up to the boundary; cf. (2.6) in Ch. 4: If e E C°°(U), then
l; (z) = P£(z) + 9(0e)(z), z E U, where P is the Poisson integral and G the Green potential.
(1.4)
1.3 Proposition. Suppose that u E L2(T). Then sup I
1 IFI
I
Iu - u112dt (1.5)
ti sup P(ju - Pu(a)12)(a) = sup -GG(IOPuI2)(a). aEU
aEU
In particular, all of them are finite if one of them is.
1.4 Corollary. The Hilbert transform maps BMO into BMO boundedly.
Proof. Let u E L2(T). Since Pu + iPHu is analytic,
(8/8z)Pu = -i(8/82)PHu; so
IOPHuI2 = IVPuI2
(1.6)
for real u. The complex case then follows. Hence, by Proposition 1.3, u is in BMO if and only Hu is.
1.5 Remark. The function v(t) = log 11 - e'tI is in BMO since it is the Hilbert transform of the bounded function arg(1 - e't).
144
9. H' and BMO
Proof of Proposition 1.3. First suppose that u E C°°(T). Then Pu E C°°(U) and we therefore can apply the Riesz decomposition (1.4) to £ _ IPu - Pu(a)12. If we evaluate at the point z = a, we get that
-Q(IVPul2)(a) = P(lu - Pu(a)12)(a), which shows the second part of (1.5) for smooth u. For the general case we
approximate u with u, as in Exercise 1. Then it is enough to verify that for fixed a E U,
P(Iur - Pur(a)12)(a) -+ P(Iu - Pu(a)12)(a) and
G(IVPurI2)(a) -+ c(IVPuI2)(a)
when r / 1. The first limit is quite trivial, and so we concentrate on the second one. Since the Green kernel is 0(1 - I(I) for fixed z E U, it is enough to verify that Ju (1 - ICI)IVPUr12 -+
f(1 - I(I)IVPu12.
Noting that Pur(() = Pu(ro) and therefore VPur(() = rVPu(ro), this follows by a substitution of variables and monotone convergence. We now prove that the first expression in (1.5) is dominated by the second one. By rotation symmetry it is enough to consider intervals I, centered at the origin. If X. denotes the characteristic function of the interval (-p, p), then there is a constant C, independent of p, such that
ZpXv(t) 5 CPr(t),
where Pr(t) is the Poisson kernel and r = 1 - p. Thus, 1
2p J P
lu(e`1) - Pu(r)I2dt < CP(lu - Pu(r)I2)(r),
and now the < part of - follows by
1.6 Lemma. If u E L2(T) and to each interval I we have a number aj, then
4s1 Ill
Jtlu-uj12 <SUP
III
JflU-aj12.
§1. Bounded Mean Oscillation
145
Proof. Denote the right-hand side by A2. Then
Ail
fu -uI12< Fill J 1u-af12+ ij J Jul-a112
/ lu - aj < 2A. F11
In the next to last inequality we have applied Jensen's inequality.
It remains to show the > part of (1.5). By the variant of Lemma 1.6 where the Poisson kernel and the XP are interchanged, it is enough to find for each a E U a number a' such that SUp0EU P(lu - a'12)(a) is dominated by the left-most expression in (1.5). We may assume that a = r > 0. Then we choose a' = u, = u(_p,p) (again r = 1 - p). Now
P(Iu - _P12)(r) = f
Pr(t)u(t) - ul2dt + f
l
Pr(t)Iu(t) - u12dt_
The first of these integrals is easily estimated by the expression on the left in (1.5) since
Pr(t) 5 C2pXP(t) if Itl < p. For the last one, first note that 1 - r2
Pr(t) =
<
I 1 - re't l2
P
p2 + t2 - t2
if Itl > p_ Thus, the proof is concluded with
1.7 Lemma. If u E L2, then
t
r hu(t) - up12dt < Cslp III J Iu - uIl2 = CB2,
where C is an absolute constant.
Proof of Lemma 1.7. Note that 2
I'I,(,2k--1P - u2kp12 -
1.
2k2k
so that
Jf/ P
1
P
('ll - u2kp)1
itl52k-'P
f
Iu - u2kp12 < 2B2, I<2kp
lu2k-lp - u2kpl < fB. Therefore, lu2kp - Upi < k v/ B
146
9. H' and BMO
and hence
lu - upl2 < 2kp2(1 + 2k2)B2.
(1.7)
J111<24'p
(Here we have the convention that the integration is performed over the whole unit circle if 2kp > 7r.) Hence, 00
Iu( t) - ul2dt
J2 -1p
<
Iu(t) - u2dt
p2(1 + 2k2)2kpB2 (2k_lp)2
B2
and thus the lemma is proved. As noted above, Proposition 1.3 now is also proved.
§2 The Duality of H1 and BMO Now we are prepared for the main theorem. In this chapter the notion of a Carleson measure refers to a positive measure in U that satisfies Carleson's inequality (4.3) in Ch. 7.
2.1 Theorem (Fefferman). Suppose that u E L2(T). Then the following four statements are equivalent: (i) u E BMO. (ii) (1 - I(I2)IVPul2 is a Carleson measure in U. (iii) u operates on HIt in the sense that the functional 2n
(0, u) = 2r
f uq5dt 0
has a continuous extension from C°°(T) to H.
(iv) u=a+Hbfor some a,bEL°°(T).
The BMO-norm of u is equivalent to the functional norm in (iii), and to the square root of the Carleson norm in (ii) plus the L2(T)-norm of u. Before the proof we consider a few consequences and corollaries of this theorem. The equivalence of (i) and (iii) can be rephrased as (HR')* = BMO.
In fact, (i) --+ (iii) means that any u E BMO corresponds to a (unique)
functional on HR and any functional on HR is in particular a functional on L2(T), and so it is given by some u c- L2(T). However, then (iii) -+ (i) means that actually u E BMO.
§2 The Duality of H1 and BMO
147
It also follows that we have a continuous inclusion
BMO C LP(T)
for allp
2.2 Theorem. (a) The Szego projection is a bounded map and S:BMO -* BMOA
S:HR-+H'.
(b) BMOA operates on H1 in the sense that the pairing (f, b) has a continuous extension from H2 to H1 if b c BMOA. Conversely, any bounded functional on H1 occurs in this way, and the functional norm is equivalent to the BMOA norm of b.
Proof of Theorem 2.1. (i) -+ (ii): By Proposition 1.3 C2 = sup -s(IVPul2) IIuII!. Recall that OQX = X if X is smooth up to the boundary. If f is analytic, then (2.7) in Ch. 8 gives (first replace u by u, and then take limits)
1111(1 - I(12)IVPul2 = f If(1- I(12)c(IVPuI2) U
C21TIflde.
(ii) -- (iii): It is enough to show that there is a constant such that I(O, u)I <- C11011HQ Ilull
for real u, 0 E C°°(T). Then the inequality follows for any b E BMO by an approximation as in Exercise 1. By Jensen's formula (2.7) in Ch. 4, (u, 0) = 2
fen uodt
J
= f log Il VPu VPdA(() + Pu(O)P(0).
Since-log(I1-(I for 1/2
1(1 - I(I)IVPuIIVPIdA+C sup lPu11PI. Ic151/2
(2.1)
148
9. H' and BMO
However,
Sup IPuIIP4I
I(I51/2
IIuIIL-110I1L2
and so it remains to estimate the integral in (2.1). Let f be the unique element in H1 such that Im f (0) = 0 and Ref = P4,. Then IIuIIH,, - IIflIHw and IVP4I - I ff'I. Thus, the square of the integral is
< Cfv(I
- IoIf'IIVPuI)2
f (1 - I(I)
I
fII 2
jU(1 - I(I)If IIVPuI2 <
(f Ifld9)
,
where the last inequality follows from (2.8) in Ch. 8 and the assumption (ii).
We now prove that (iii) implies (iv). If u satisfies (iii), then it defines a bounded functional A on Hat, but HR can be isometrically identified with the subspace of L1 (T) x L' (T) of elements of the form (0, HO); and so by the Hahn-Banach theorem, A can be extended to a bounded functional on L1(T) x L'(T). Thus, it is represented by an element (a, -b) E LOO(T) x LOO(T), and for all (4,, H¢) we thus have (0, u) = AO = (0, a) + (HO, -b) = (0, a + Hb) for smooth 0 (cf. formula (2.9) in Ch. 6), and hence u = a + Hb. Also note that IIaIIL. + IIbIIL= is equivalent to the functional norm of u. The implication (iv) (i) follows from Corollary 1.4, and hence Theorem 2.1 is completely proved.
Supplementary Exercises Exercise 2. Show directly from the definition that log ItI is in BMO but that (t/ItI) log Iti is not. Also show that log 11 - e'uI is in BMO.
Exercise 3. Show that 111f 111* 5 211f II.. Exercise 4. Suppose that f E L1(T) (or that f is in M(T)). Show that
if Hf E M(T), then f c H.
Exercise 5. Suppose that f E L'(T) (or that f is any distribution on T). Show that f E HR if and only if I f fbI G CIIbhj. for all b E C' (T). Hint: Use the preceding exercise.
Exercise 6. Show that If I E HR implies that f E H. Hint: Use the preceding exercise and Exercise 3.
Exercise 7. Show that there are f E HR such that If I do not belong to
H.
Notes
149
Exercise 8. Show that there is a continuous inclusion BMO C LP(T) for
allp
Exercise 10. Suppose that h E H. Show that the mapping b -* hb is bounded from BMOA to BMO. Is it still bounded if h is replaced by h? Exercise 11. Let BMOo = If E BMO; f (0) = 0} and define ogously. Show that H: BMO -+ BMOo and H: HR -+ HR o are surjective. Exercise 12. Show that S in fact maps L°°(T) onto BMOA. Exercise 13. Derive Theorem 2.1 from Theorem 2.2. Exercise 14. Let X be a smooth function on T, and let dQ = e-XdG. Show that BMO is the dual of HR with respect to the pairing HR,o-anal-
1
(f, 9) =
21r fgda.
27r
fo
Exercise 15. Let P be the orthogonal projection P: L2(T) - H2 with respect to ( , ). Show that P maps LP(T) onto H9, BMO onto BMOA, and HR onto H'. Moreover, show that H9 is the dual of H9 and BMOA is the dual of H' with respect to this pairing. Also show that P: L°°(T) BMOA is surjective. Hint: Note that SP = P and PS = S, and so
S'P = S' if the star denotes adjoint with respect to ( , ). Hence P = S' + (S" - S)P. First, assume that f is in LP(T), 2 < p < oo or in BMO. Then Pf E L2(T). Exercise 16. Let p be a Carleson measure in U. Show that u(z) _ - 1 / zdjc(() z E T, ir U 1-@z is in BMO and that 8bu = /2.
Notes The space BMO was introduced by John and Nirenberg in Comm. Pure Appl. Math., 14 (1961). The John- Nirenberg theorem states that if f E BMO (defined with L' instead of L2 norms; cf. Remark 1.2), then
I{t E I; If - frt > All < CIII exp (ifCA ) ,
- iI
see [G]. It implies that the two definitions agree; it also immediately implies
that BMO C LP(T). Fefferman's duality theorem is from 1971; see [G].
150
9. H' and BMO
The space HR can be defined by atoms. An atom is either the function 1 or a function a(t) E L°°(T) that has support on some interval I and such that fr adt = 0 and IaI < 1/III. One can prove that HR is the space of all sums
f(t) =>2A,aj(t), where aj are atoms and A3 Eel, and the norm of f is equivalent to the infimum of all > IA3I over all possible representations of f. This definition has considerable advantages; basically, many results are reduced to checking its validity for a single atom; for instance, it is almost immediate that
any b E BMO defines a functional on HR, and also that usual singular integral operators are bounded. This definition with atoms was introduced by Coifman (Studia Math. 269-274 (1974)), and has several important generalizations to Hp spaces for p < 1 in one and several dimensions in lR' and on spaces of homogeneous type.
In general, the operator b --+ hb is not bounded on BMO if h is only a bounded function. There is a precise characterization (due to Stegenga,
Amer. J. Math. 98 (1976), 573-589) of the multiplicators of BMO: A function h is a multiplicator if and only if it is in L°°(T) and III
logIIIf Ih - hJIde
(2.2)
The condition is satisfied if h has some regularity (C1 is more than enough); cf. Exercise 9. An analytic function h is a multiplicator on BMOA if and
only if f E H°° and (2.2) holds. There is no linear operator that provides a bounded solution to i9bu = µ for each Carleson measure µ; however, Exercise 16 shows that the Cauchy integral at least gives a solution in BMO.
Bibliography
[Al]
L.V. Ahlfors, Complex Analysis, 2d ed., McGraw-Hill Book Company, New York, 1966.
, Conformal Invariants, 2d ed., McGraw-Hill Book Company, New
[A2J
York, 1973. [B]
R.P. Boas, Entire Functions, Academic Press Inc., New York, 1954.
(D]
P. Duren, Theory of H9 Spaces, Academic Press Inc., New York, 1970.
(F)
G. Folland, Real Analysis, John Wiley & Sons, New York Chichester Brisbane Toronto Singapore, 1984.
[G]
J. Garnett, Bounded Analytic Functions, Academic Press, 1981.
(Hi]
E. Hille, Analytic Function Theory, vols. I & 11, Ginn and Company, Boston, 1959 & 1962.
(Ho]
L. Hormander, The Analysis of Linear Partial Differential Operators I, 2nd ed., Springer-Verlag, Berlin Heidelberg New York, 1990.
[NJ
N.K. Nikolskii, Treatise on the Shift Operator, Springer-Verlag, Berlin Heidelberg New York Tokyo, 1986.
(Ra]
T. Ransford, Potential Theory in the Complex Plane, Cambridge University Press, Cambridge, 1995.
[Rul] W. Rudin, Real and Complex Analysis, 3rd ed., McGraw-Hill Book Company, New York, 1986. [Ru2]
,
1973.
Functional Analysis, McGraw-Hill Book Company, New York,
152 IS)
Bibliography
E.M. Stein, Singular Integrals and Differentiability Properties of F actions, Princeton University Press, Princeton, NJ, 1970.
[S-W) E.M. Stein & G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Princeton University Press, Princeton, NJ, 1971.
List of Symbols
B f 56
M,u M,u
Bef 57
Mf
A(Q)
Bf
6
BMO 142 BMOA 147
C5
132
ab
0a 30 G((, z) 76 Gu 102 1f 76 H f 104 HP
h'
106, 112 97
HR
141
H.
56
Indr(z)
122
Qf 115 Res(f,a)
Sf Sf
S 35 Sf 116 SH(S2)
71
T(r, f)
85
ur(ei8) u'(e'B)
Z(f)
LP(T), 11 Ilz.a 97 .M (T), 11 II 97 M(r, f, a) 91
16
107 115
WP(z) 19
85
n(r, f, a), N(r, f, a) P f 69, 97 Pr(t) 68 Qk,l
5
104
FLAIL
N+ 114 n(r,f), N(r,f)
P 33
D(oo, r) 33 6(f, a) 89
a/az, a/az .Fp(n) 104
102 115
N 86
115 122
Bk,j
101
97 101
83 14
,4(n)
104
(,)
142 105
88
Index
Ahlfors, 96 analytic continuation, 51 function, 6 functional, 55 area theorem, 36 atom, 150 automorphism, 30
Banach-Alaoglu theorem, 99 Banach algebra, 132 Bergman projection, 51 Bergman space, 51 Berndtsson, 129 Beurling, 129 Bieberbach, 45 Bieberbach's conjecture, 37 Blaschke condition, 86 Blaschke product, 88 Bloch, 45 Bloch's theorem, 39
constant, 45 Borel transform, 56 bounded mean oscillation, 143 de Branges, 37,45 carrier of an analytic functional, 55 capacity, 81 Carleson, 66, 111, 118, 129, 132, 140 measure, 122 tent, 122
Carleson's inequality, 124 Cauchy's estimate, 14 Cauchy's formula, 8 Cauchy's integral theorem, 8, 20, 22 Cauchy-Riemann equation, 6, 49 characteristic function, 85 Coifman, 150 complex interpolation, 26 conformal equivalence, 29 conformal mapping, 29 convex
function, 78 set, 56 corona theorem, 131 defect, 89
defect relation, 90 Dirichlet's problem, 44, 66, 69 disk algebra, 139 dyadic decomposition, 122 essential singularity, 16 exponential type, 56 Fatou, 111 Fefferman's theorem, 146 finite order, 88
first fundamental (main) theorem, 96
form ((1,0)-form and (0,1)-form), 34 Fourier-Laplace transform, 56
156
Index
Main theorem, 115 maximum principle for analytic functions, 12 for harmonic functions, 68 Gamelin, 129 for subharmonic functions, 71 gamma function, 52 mean value property Goursat's theorem, 6 for analytic functions, 9 Green potential, 76 for harmonic functions, 67 Mergelyan's theorem, 58 Hardy--Littlewood maximal function, meromorphic, 17 101 Minda, 45 Hardy space, 112 Mittag-Leffler's theorem, 48 harmonic function, 67 monodromy theorem, 53 Harnack's theorem, 70 Morera's theorem, 11 Hausdorfi-Young inequality, 27, 105 multiplicity of pole, 16 Hermite's formula, 27 multiplicity of zero, 14 Hilbert transform, 104 Miintz-Szasz theorem, 96 holomorphic function, 6 homologous, 20 Nevanlinna class, 86 homotopic, 21 Nevanlinna condition, 87 Hormander, 132, 136, 140 Nevanlinna, 96 Hunt, 111 Newton potential, 76 hyperbolic Nirenberg, 149 distance, 128 nontangential boundary values, I11 line, 128 null-homologous, 20 Fourier coefficients, 104 Fourier series, 104-106 function element, 51
ideal, 131
index of a curve, 19 inner function, 114 interpolation sequence, 118 invariant subspace, 116 isolated singularity, 15 Jensen's formula for analytic functions, 86 for subharmonic functions, 77 John, 149 Jones, 129 Koebe, 44, 45 Koebe's theorem, 36
Landau's constant, 45 Laurent series expansion, 15 Liouville's theorem, 13 Liu, 45 logarithm, 19
outer function, 115
Paley-Wiener's theorem, 57 Picard's theorem, 16, 38-40 big Picard theorem, 16, 38-40 little Picard theorem, 38-39, 90 Poisson kernel, 68 integral, 69, 97 polar set, 80 pole, 16 order of, 16 ,Polya's theorem, 56 polynomial convexity, 55, 62
potential theory, 80 principal part, 16, 34, 48 principle of argument, 20 Phragme n-Lindelof, 26 projective space, 33 proximity function, 91
Index
removable singularity, 15 residue, 16 Riesz decomposition, 76, 100 Riesz, F., 129 Riesz, M., 111 Riesz theorem, F. & M., 108 Riemann-Lebesgue lemma, 105 Riemann mapping theorem, 30 Riemann sphere, 33 Rouche's theorem, 18 Rudowicz, 129 Runge's theorem, 46
Smirnoff, 129 Stegenga, 150 Stein, P., 111 subharmonic function, 71 Szego projection, 107
three-circle theorem, 80 trigonometric polynomial, 102 Uchiyama, 140 uniqueness theorem, 14 univalent, 32, 35
Schottky's inequality, 40 Weierstrass' theorem, 83 Schwarz' lemma, 30 Weyl's lemma for analytic functions, 10 Schwarz' reflection principle, 70 for harmonic functions, 70 second fundamental (main) theorem, 96
separated sequence, 119 shift operator, 117 simply connected, 22 singular inner function, 114
Wolff, 132, 136 zero
order of, 14 set, 14, 82
157
Universitext
Tracts in Mathematics This book provides a concise treatment of topics in complex analy-
sis, suitable for a one-semester course. It is an outgrowth of lectures given by the author over the last ten years at the University of Goteborg and Chalmers University of Technology. While treat-
ing classical complex function theory, the author emphasizes connections to real and harmonic analysis, and presents general tools that might be useful in other areas of analysis. The book introduces all of the basic ideas in beginning complex analysis and still has time to reach many topics near the frontier of the subject. It covers classical highlights in the field such as Fatou theorems
and some Nevanlinna theory, as well as more recent topics, for example, the corona theorem and the H '-BMO duality.
The reader is expected to have an understanding of basic integration theory and functional analysis. Many exercises illustrate and sharpen the theory, and extended exercises give the reader an active part in complementing the material presented in the text.
IseH 0-387-94754-M
ISBN 0-387-94754-X
9
IIII III
780387"947549
