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0, ale(lt') = _e"l{) and k«({') = O. Therefore, since k+«({') = 0 for cp E lit,
With at most one exception [namely if 1(0) = ei'P for some ({' E R] we have
n(o'-I - 1e . ) l ",
=0.
Now, by Jensen's Theorem for any constant b, we get (6.2)
(Check the cases where Ibl ~ 1 and Ibl < 1.) Let us integrate (6.1) with respect to ({': (6.3)
-i;
L: {2~ /~ log
I/(rei8 )
-
e""l dO}
dcp
1111' = 21r1111' _,. log lak(cp) I d({' + 21r _ .. N
(r, f _1) dcp - N(r, f). e i ",
Entire and Meromorphic FUnctions
18
Applying Fubini's Theorem to the left-hand side (LHS) and using (6.2) yields
1" 1"-11"
-2 -2 1 1 1r _,.. 1r
log If(re i9 ) - ei
= 211r
1"
_,..
m(r,!) dO = m(r,!).
Hence we obtain the result:
(6.4) mer, f)
r (
1111" 1 1-11" N r, f _1) + N(r, f) = 21r _,.. log lak(
The first integral is a COlll:ltaut auu we are done.
Remark. We could define TJ(r) =
2~D1rN (r, f-~''P) dcp
as the Carbut by a minor abuse of notation we shall write
tan characteristic of f, T(r) = 2~r:,..N (r, f-~''P) dcp since we customarily work modulo bounded functions of r anyway.
Interpretation. Roughly, we have
where
L(r)
= 2~
i:
n (r, f _1
ei
dcp.
What does the function L measure? The function f(re i9 ) may be considered as a mapping of the circumference ODr = {z : Izl = r} into the Riemann sphere.
n(r, f-~''P) counts the number of times the point ei'P
is covered by this map. Hence, J~1rn (r, f-~''P) dcp measures the total arc length of the unit circumference (counting multiplicity) covered by the mapping f. In other words, the more heavily the mapping f covers the unit circumference, the faster T grows. Thus, T measures the covering properties of f. We outline here another characteristic, the Ahlfors-Shimizu characteristic TA(r, f), which behaves much like the Nevanlinna characteristic but which has as an enlightening geometric interpretation. For full details, see, from which our presentation is abstracted. We define where
N(r, a)
r -n(t,ta)- dt,
= 10
6. The Cartan Formulation of the Characteristic
19
as before, but
MA(r,a)
1 = -2111" 1,211" log -[- ] dO, 0 w, a
where
[w, a] =
Iw-al ';1 + lal 2 ';1 + Iwl 2
and
1
[x, a] = -.,r===:=::;:: V
1 + lal 2
Now [w,!l] is the distance on the Riemann sphere between the points on the sphere to which w and a correspond via stereographic projection. It is easy to see that
IT(r,OO)-TA(r,OO)-IOg+lf(a)l\ $
~log2.
By Green's Theorem, one can show that r d 1 2. . - . 211" dr 0 log + If(re ,6 )12 dO+n(r, f)
VI
11r 1211" 1/'(pei6)1 2p dp dO 0 0 [1 + If(pei6)12]2 .
=;:
Denote the right side by A(r), divide by r, and integrate the resulting identity from 0 to r to get
1,. o
A(t)
-
t
1
dt = N(r, f) + -2
1
2 ..-
11" 0
log ./1 +
V
If(rei6 )1 2 dO -log ";1 + 1/(0)1 2 •
H we make a rotation of the Riemann sphere, which corresponds to the transformation _ l+aw w=--
w - a'
where w = I(z), and call the resulting function w = F(z), we can derive the first fundamental theorem for the Ahlfors-Shimizu characteristic.
Theorem. II I is meromorphic in Izl < R, where 0 < R $ 00, then every finite or infinite a and r with 0 < r $ R we have, TA(r,oo)=
10r
lor
A(t) dt=N(r,a)+MA(r,a)-MA(O,a). t
For the geometrical interpretation of T A, note that if S is the Riemann sphere (of diameter 1), do is the element of area in the z-plane near the POint z, and dA is the corresponding element of area on S, then
do
dA
= (1 + Iz12)2 .
Hence 11"A( r) is exactly the area (counting multiplicity) of the image on the .Riemann sphere of {Izl < r} by w = I(z). The rotation described above leaves A(r) invariant and replaces MA(r, 00), N(r,oo) by MA(r, a), N(r, a). We see that TA (r, 00) can be interpreted as an average of the spherical area of the image of disks under the mapping w = I(z).
7 The Poisson-Jensen Formula
The material we present in this chapter is a specialization of some general results of potential theory. Our presentation is in the context of analytic function theory. We consider functions J holomorphic in DR = {z E C : Izl 5 R}. We denote u = ReJ, choose r < R, and write z = re i8 , w = Rei'l'. The Poisson Formula. (7.1)
i8
!
11
___
u(re) -
Pru - 27r
1<
R22 -r
i'l'
u(Re) R2 _,..
-
2r R cos (9 - cp) + r 2 dcp.
The Poisson Kernel.
R2
(7.2)
2
i8 R i
= Iwl 2- Izl2 = Re w + z zl2
Iw -
w-
Z
Proof. Without loss of generality, assume R = 1:
1
1-
Z
- +1-- -zw= w- z
Izl2
(w - z)(1 - zw)
1 1 - Izl2 = -. w Iw - zl2
By the Cauchy integral formula, -
11,
21ri Iwl=l
[1 z] + --_w- z zw
J(w) - -
1-
dw = J(z}.
7. The Poisson-Jensen Formula
21
After parametrizing the integral with respect to the angle ip and taking real parts, we get the Poisson formula.
Remark. For z = 0, Poisson's formula reduces to the Gauss Mean Value Theorem, u(O) = -1 u(el'P) dip. 211' -Ir
11r
The Poisson formula is the "invariant form" of the Gauss Mean Value Theorem in the following sense. Choose zED and define
T%: n
~
][)-
by T.w ~
Let
F
= foT",
= w+z
1 + ZW
U
for
wED.
= uoT%.
H'\ =.J!d!.. then w = l-!w '>'-z and l-!w'
1- zw d.\ dw -= w 1.\ - zl2 .\ • 80
that u(z)
I!
= U(O) = 211'i
dw U(w)-;;
I! 1 1.\ _
= 211'i
u(.\)
Izl2 d.\ Zl2 .\ '
which is the Poisson formula in different notation. This generalization of the Gauss Mean Value Theorem has a natural application which generalizes Jensen's Theorem.
The Poisson-Jensen Formula. Suppose that Then.
I
is meromorphic in the disk DR = {z E C : Izl
(7.3) 1 log If(relll ) 1 = 211'
+
llr -If
R2 r2 log If(Re ill ) 1 R2 - 2rRcos(O - ip)
:E Iz.. I
L
logIBR(z:zlI)l-
+ r2
< R}. r < R.
dip
R
logIBR(Z:wv)l-klog'T"'
Iw.. I
where B is the Blaschke factor defined by
R(z - a) BR(Z : a) = R2 -az and the Zv are the zeros of I. the Wv are the poles of f, and k is the order of the zero or pole at the origin.
22
Entire and Meromorphic Functions
Corollary. II I is holomorphic, then log I/(reill)1 :s;
1
Plog III·
Iwl=R
In other words, if I is holomorphic, then log III is dominated by its Poisson integral. Notice that for z = 0 the Poisson-Jensen formula reduces to Jensen's Theorem. One way to prove the Poisson-Jensen formula is to show that it is the invariant form of Jensen's Theorem. We choose another proof.
Prool of the Poisson-Jensen Formula. IT I is holomorphic and has no zeros, then there is a branch F of log I and log III ReF so that the formula follows as a special case of the Poisson formula. Also, if ..\ denotes the lefthand side and p the right-hand side, notice that ..\(fg) = >.(f) + >.(g) and p(fg) = p(f) + p(g). So it is enough to prove the formula for holomorphic f. Now consider g(z) = l(z)jIIBR(z : z.. ) supposing that 1(0) -# O. Since IBR(w: z.. )1 = 1 for Iwl = R, the formula follows on applying the Poisson formula to g. And if I (0) = 0, consideration of I (z) j zk leads to the general case.
=
8 Applications of T(r)
Theorem. If I is holomorphic and M(r) = sup[{II(z)1 : Izl $ r}, then for any R > r R+r T(r) $ log+ M(r) $ R _ r T(R).
Proof. Since I is holomorphlc, T(r) = mer) and mer)
= -1 /"" log+ If(reis)1 dO $ 271' _,,"
log+ M(r).
Also, by the corollary to the Poisson-Jensen formula, log If(re,s)1 $
2.. f
271' J1z1=R
Plog III.
It is easy to verify that
O
that
R+r R+r log+ M(r) $ - R mer) = - R T(R), -r
-r
and the theorem is proved. We now can prove the following extension of the Liouville Theorem,
which is left 88 an exercise. Exercise. Suppose that f is meromorphic in Izl < bounded. Then I is a constant.
00
and that T(r) is
Entire and Meromorphic Functions
24
Definition. Suppose that A(r) is positive, continuous, and nondecreasing for r > 1 and furthermore is slowly increasing in the sense that ~~2:f is bounded. Let A· be the class of all entire functions f such that log+ M(r) = O(A(r)). It is easy to see that A· is a ring. One of our exercises is to show that f EA· if and only if f E i, where I consists of those entire functions f for which T(r, J) = O(A(r». In the case where A(r) = max(l, r P ), we see that the notions of f being of finite order, of order at most p, and of order at most p exponential type are the same whether defined by the characteristic or the iogarithm of the maximum modulu:s. IL also follows that each A· ring is algebraically closed in the ring of all entire functions. We say that a meromorphic function f in the unit disc is of bounded characteristic to mean that T( r, J) is bounded. Next we characterize the functions of a bounded characteristic in the unit disk D. Theorem. A function f meromorphic in D is of bounded chamcteristic if and only il there exist bounded functions A and B, holomorphic in D, such that f = AlB. Proof. It is easy to see that if I = AlB, then I is of bounded characteristic. In the other direction, suppose that I is of bounded characteristic and, without loss of generality, suppose 1(0) = 1. Since T(r, f) is bounded, it follows that N(r, f) and N(r, are bounded, so that
1)
1
Elog- < rn
00
and
E log..!..Pn <
00.
Here, as before, {rne illn } and {Pneir,pn} are the zeros and poles of f. By the theorem of Chapter 4, there exist bounded functions cp and 1/1, holomorphic in JI), such that the zeros of cp are the zeros of f and the zeros of 1/1 are the poles of I. Thus, 9 = i:s of bounded characteristic and has no zeros or poles. It is enough to show that 9 has a representation 9 = AlB. Note that even though 9 is holomorphic with no zeros and T(r, g) = 0(1), it does not follow that 9 is bounded; witness g(z) = l~Z. Proceeding with the proof, there exists a function h, holomorphic in JI), such that 9 = eh . Writing h = u + iv, we have
f!
j1r lu(reill)1 dO ~ M < 211" -1r
-1
00
for
r < 1,
since m(r,g) and m(r,!) are bounded, and 191 = expu. Now for R < 1, writing
8. Applications of T(r)
25
we have
Now we write u+ R - u-· R'
U R --
Let
{3R(Z)
1111' u~(e''P)-.-. ei'P + z dIP = -2 11" -11' e"P - z
aR(Z)
. ei'P + z = -21111' ui(e''P)-.-- dIP· 11" e"P - Z -11"
It is easy to verify that aR and {3R are holomorphic in D. Let
~:) ,
9R = exp (i'\R where
AR = exp( -aR), BR = exp( -{3R) , and '\R is an appropriate real constant. Now
ReaR 80
~
0 and
Re{3R
~
0
that
IARI ::; 1 and
IBRI::; 1.
Since the fa.milies {AR} and {B R } are uniformly bounded, they are normal families; and since the unit circumference is compact, we may write, for a suitable sequence of R approaching 1,
limAR = A,
limBR
= B, IAI::; 1, IBI:::; 1,
It is easy to see that
limYR = g. We therefore get the required representation
...\A g=e' -
B provided only that B is not identically zero. But
IBR(O)I and
The proof is complete.
= exp( -{3R(O))
lim'\R
=,\.
9
A Lemma of Borel and Some Applications
Definition. A set E of real numbers has length :S f, written lEI :S f, means that there is a countable union of intervals [an, bn], an :S bn, that contains E and such that
lEI = in£{ I. : lEI :S f}. Lemma (trivial). If IEll :S 1.1 and IE21 :S 1.2 • Definition.
then
Borel Lemma. Suppose that JL(r) is defined for all r non decreasing. and that JL(ro) ~ 1. Then for each a > 1 (9.1)
JL
except in a set Ea such that
~
ro. that JL is
(r + p.tr») < ap.(r) lEal :S
a~l.
Remark. The inequality of the Borel Lemma estimates JL at a point greater than r by using the value that p. takes at r. Intuitively, if the inequality fails in "too big" a set, the function p. will become infinite "too soon" and will not be defined for all r.
9. A Lemma of Borel and Some Applications
21
Proof. Let E = Ea = {r > ro : JI. (r +;fry) ;::: 4J1.(r)}. Let f > 0 be given. We proceed to define a sequence {rn} and an allied sequence {r~} by induction. Let rl = inf{r: r E E}. Suppose rb ... ,rn-l have been constructed together with numbers fk ?: 0, k = 1, ... , n-l so that fl +f2 + ... fn-l < f and r" + fk E E. Let ric = rk + fk + ,",h~€/c)" Now define rn = inf{r : r E E and r ?: ~-l}' Choose fn ?: 0 so that fl + ... + fn < f and so that rn + fn E E and proceed. This procedure will terminate after n steps if and only if there does not exist an r E E satisfying r ?: r~. We have (9.2) Now
(r~t r n +!)
n E is empty by construction.
Claim. There exist only finitely many rn or else rn - 00. For otherwise, there would be a finite T such that Tn - T and, by (10.2), ~ - r. However, by the construction we have for all k
which is a contradiction. Claim. E C U:=l[rn,r~l. Pick an arbitrary x E E. Let Tno = max[rn : Tn ~ X]. This makes sense by the previous claim. Now Tno ~ X. Suppose x > T~o' Then Tno+! ~ x, an immediate contradiction. Therefore T no ~ x ~ ~o' which proves the
claim. So we have constructed a countable collection of intervals whose union contains the set E. We estimate }:(r'n - Tn). Notice that
But " 1 < ,,_1_ < ~ _1_ = _a_. L.. ~(rn + En) - L.. JI.(Tn) - ~ a n - 1 a-I
It now follows that
lEI ~
~+
E,
and the lemma is proved.
Entire and Meromorphic Functions
28
Corollary. Under the same hypotheses on p, and a, (9.3)
except for r in a set E~, where E~ has logarithmic length (By this we mean that E~
IE.. I·)
= exp E.. , where IE.. I 5
5 .. ~1 •
i. We write IE~llog
=
Proof. Let J.'l(Y) = J.'(expy). Then p,(exp(y+;;lvy)) 5 4p,(expy) for
Y;' E.. , where IE.. I 5
1l~1 by the Borel Lemma. But
exp 80
1 1 > 1 + --:--....,.. J.'(exp Y) J.'(expy)
that I'
({I
+
J.'(~y)} expy) < 4J.'(expy)
for
y;' E .. ,
and the result follows on writing r = exp 1/.
Application of the Borel Lemma to Nevanlinna Theory We already have proved that if I is entire, then T(r) $;logM(r) 5
! ~ ~T(R)
Choose
R =r ( 1+
if R> r.
T~r) )
to get log M(r) $; (2T(r) + l)T
(r (
1+
T~r») )
.
Unless I is a constant, T(r) -+ 00 80 that 2T(r) + 1 5 ~T(r) for large r. Applying the Borel Lemma we find for entire functions I, (9.4)
logM(r) 53(T(r»2
except for a set of finite logarithmic length. Definition. We say that A(r)
-+ eft"
L means that there is a set E of finite
logarithmic length such that lim A(r) = L as r r~E
-+ 00.
We attach a similar meaning to expressions like A(r) ,..., B(r), A(r)
= B(r) ,etc. eff
We have, in effect, proved the next result.
eff
9. A Lemma of Borel and Some Applications
29
Proposition. If f is a nonconstant entire function, then logloglogM{r) '" loglogT{r). elf
In a certain sense, this says that T{r) and log M{r) have the same size for most of the r. The log log takes a lot of punishment. Propostion. Suppose that a real function f has a continuous, increasing derivative on [1,00], and that lim f{x) = 00. Then %_00
loglog(J{x)) '" loglog(xJ'(x». eff
Proof. Write vet) /(1) = O. Then
= tf'{t), f(x)
=
and suppose, without loss of generality, that :1: /.
J'(t) dt =
Hence f(x) ~ v(x)logx. But for some X large, so that (9.5)
1% v(t)-. dt
l i t
f(x) ~ (v{x))2
f
> 0, we must have vex) 2: EX for
for large x.
In the other direction, if 11 < x, then
1
dt 2: v(y) log f(x) 2: v% v(t)T
(x)y .
Choose x = 11 + 7fu to get
+ t) 2: !t for t near 0 and positive, we have, if y is large, "(11):5 2f(y)f (11 + ~). Applying the corollary of the Borel Lemma, we Since log(1
get (9.6) Equations (9.5) and (9.6) together imply the result.
10
The Maximum Term of an Entire Function
We will give in this chapter a proof that a suitable entire function can grow as fast as we please. Let f (z) = E anz n be an entire function; ao =I 0
For each r, the sequence we can define
.40,
AIr, A 2 r 2 , ••• converges to zero. Therefore
(10.1) B(r) is called the maximum term for r. A term Akrk is a maximum term if Akrk = B(r). Since each Akrk is a nondecreasing function of r (increasing if k =I 0, Ak =I 0), B(r) is nondecreasing. B(r) is also continuous and unbounded. We define the rnnk of the maximum term as (10.2)
J!(r)
= sup(n : Anrn = B(r».
It follows immediately that if n < ~(r), then Anrn ::;; A,.(r)rl'(r), and if n > ~(r), then Anrn < A,.(r)rl'(r). Therefore we also can write the rank of the maximum term as J!(r) = sup(n : Anr" ~ B(r».
10. The Maximum Term of an Entire FUnction
31
Clearly, JL( r) is a non decreasing, integer-Ydlued function of r. Let
{
(10.3) Since
gn = - log An if An ::f. 0 gn = 00 if An = O.
f is entire, we have
(IDA)
lim An-~..
n~oo
= 00
SO
t hat
lim -gn =
n--+oo
n
00.
Let Cn be the point (n,g(n» on the plane. From (lOA) it follows that below any straight line of finite slope there is only a finite number of points c,..
2
3
4
5
6
1bc Newton Polygon
This property of the c,. enables us to construct the Newton polygon 1r(f) of the entire function f. We construct the polygon as follows: Among the segments Coc,., consider those of minimal slope. From these segments of minimal slope choose the one that is the longest; denote this segment by CoCk}' Repeat this selection procedure starting with the point C1c1 to obtain the point Ck 2 , and so on. The vertices of the polygon are 'Yo, 'Y17'" 'Yi,· .. , where 'Yi = Ck. = (~,g(ki» for i = 1,2, ... and '"Yo = cko = Co. The x-coordinates of the vertices '"Yo, '"Y1, . .. are called the principal indices. Let G n be the y-coordin&te of the point on 1r(f) whose x-coordinate is n. Let (10.5)
Entire and Meromorphic Functions
32
A~ is called the logarithmic convexification of An. Since An = exp( -gn), it follows immediately that
{
Gn
= gn
if n is a principal index
Gn
~
for all n.
gn
Given r > 0, we have
logAprP
~
logB(r)
Since {
= log Anrn ,
where
n
= J.£(r).
= -gp logAn = -gn,
log AI'
we have plogr - gp
~
nlogr - gn
or (10.6)
gp ;::: gn
+ (p -
n) log r.
Y = 901
+ (x -
n)logr
But (10.7)
is the equation of the straight line through c,. with slope log r. Equation (10.6) says that all points of 1r(J) lie above the line (10.7); this line is "tangent" to 1r(J). Call this line Dr. Thus, J.£(r) is the rightmost point of contact of Dr with 1r(J) [because J.£(r) is the largest value of n at which we can have equality in (10.6)]. Hence the values of J.£(r) are the principal indices. Since, for x = 0, (10.7) yields y = gn-nlogr = -logAnr" = -log B(r), it follows that Dr cuts the y axis at -logB(r). Two immediate consequences of this are: (a) Given It, 12 entire such that 1r(It) 1r(h), then
=
{
lJ.(r : It) = lJ.(r : h) B(r: It) = B(r : h)
(b) Among all entire functions, h(z) = that has the same IJ. and B as f. Let (10.8)
D
_
G .. -G .. _1
...... - e
_
-
L
A~zn
-A'..-1 A' .
" We call the
R.. the corrected rntios.
is the largest function
10. The Maximum Term of an Entire Function
33
Geometrically, log R,. is the slope of the side of 1r(f) joining the points whose x-coordinates are n - 1 &Ild n. R,. is nondecreasing &Ild R,. - 00
as n- 00.
= =
=
Without loss of generality, assume that Ao ao 1 [note that J.&(O) 0]. From the definition of R,., it follows that eG " = RI . R2 ••• R,.. Since p.(r) runs through the principal indices, g~(r) = Gp(r), it follows that
'Thking logarithms, ~(r) 1 p(r) logB(r) = I'(r)logr+ Elog]f = EIog; .
k
I
But p(r)
T
E log Rr k
1
k
1
r
= 10flog -t
dl'(t),
where pet) = E
1.
Integrating by parts we get:
l
r log -dp(t) = log -r ·I'(t) IT ott 0+ r
IT -I'(t) dt
+
ot
or log B(r)
(10.9)
=
r I'(t)t dt,
Jo
and by the lemma in Chapter 6 it follows that B(r) is a convex function of
logr. Relation between B(r) and M(r) From Cauchy's inequality, 1a..1 :::; r-nM(r), we get Ia..lrn :::; M(r) or B(r):5 M(r).
(10.10)
Remark. M(r):5 F(r) == EA~rn. Note: Choose p > 1'( r). Then R,. > r because log Rp > log r for p > p.(r), as we see on interpreting log R,. as a slope. Then for q ~ p we can write
A'rq = e-G4 r q =
eG,,-lrP - 1
q
r q - p +1
Rp ... Rq
.
Since e-Gp-lrP-l < B(r), we have rq-p+l
A~rq = B(r) ~-P+l :5 B(r)
(
r ) q-p+l
Rp
Entire and Meromorphlc Functions
34
because the slopes of the edges of the Newton polygon are increasing. We have: p-l
F(r) =
00
L e-G"r + L e-G"r n
o
n
p
:5 pB(r) + B(r) ~ (~) q-p+l r
=pB(r) + B(r) R" _ r' Consequently, F(r) :5 B(r)
[p + R"r_ r]
provided
As a heuristic guide, let us try the choice p for the moment that I'(x) > I'(r). Then F(r) :5 B(r) [I'(X)
+R
p> I'(r).
= I'(x) for x > r, supposing
r _ ]:5 B(r) [I'(X)
/£(:.:)
r
+
r/ ].
-''':J:-r
The last inequality is justified as follows: R/£(:.:) is the slope between (p 1, Gp-d and (p, Gp). This slope grows without bound, so there exists an x so that for p > x we have the inequality. Write x = r + y; y > O. Then F(r) :5 B(r) [I'(r + y)
+~] .
Now write y = f so that F(r) :5 B(r) [I'
(r + D+ t] .
We try to make
by choosing t. As a first approximation we choose t = I'(r). For that choice x == r +;f,:). Since we want to guarantee that p > I'(r), our actual choice is p
= I' (r +;fry) + 1. F(r)
We then get:
~ B(r) [I' (r + I'~r») + I'(r) + 1]
10. The Maximum Term of an Entire Function
or
35
JL~1'») + 1].
F(r):::; B(r) [2JL (r+
Using the Borel Corollary (Chapter 9) we find
F(r) :::; 3B(r)JL(r) eff
so that B(r) :::; M(r) :::; 3B(r)JL(r).
(10.11)
eff
For functions of finite order, say order p, we have 00
if p'
> p.
-+ 00
if p'
> p.
logM(r) = o(rP')
88
r -+
log B(r) = o(r P')
88
r
In this case,
We have logB(r)
=
r I-'(t)
10
dt
t
= o(rP')
and we know that
Hence, (10.12)
I'(r) = o(rP ).
Together with (10.11), this implies that if either log M or log B is o(rP), PS p, then both are. Now, from (10.11) we have log M(r) :::; log 3 + log B(r) + log I-'(r) eff
= log 3 + o(r P )
+ plogr.
From B(r) $ M(r) we get
. logB(r) lim sup 1og M() ...... 00 r $ 1.
(10.13)
Also, logM(r) logB(r)
~
log 3 + log B(r) + logp(r) 1 log a + logl-'(r) logB(r) S + logB(r) .
Entire and Meromorphlc Functions
36
Writing F(r) as
> r:
and using the definition of B(r), we get for R F(r) ::; B(R)
L (Iir)" = B(r) R R_ r
so that B(R) ::; M(r) ::; B(R) R ~ r'
(10.14)
Let us choose R = r
+ B(r)
and again apply the Borel Corollary (9.3):
r+
B(r) ::; M(r) ::; B
( r + B~r) )
B(r)::; M(r) ::; B
(r + B~r»)
r
.: (B(r)
+ 1).
Hence, B(r) ::; M(r) ::; 2B(r)2. eff
Taking logarithms twice we obtain
(10.15)
loglogM(r)"" log log B(r). eff
From (10.11) we get B(r) ::; M(r) ::; B(r) ·3· I'
If J1.(r) = o(rP ), p
< 00, then
I'
(r + J1.~r»)
.
(r +;fu) ::; J1.(2r) = o(r so that P)
log B( r) ::; log M (r) ::; log B( r) + p log r + constant. Then
logB(r) <1< logB(r) log M(r) - log M(r)
+
plogr log M(r)'
We shall prove that ~ = 0(1), which implies:
10. The Maximum Term of an Entire Function
Theorem 10.1.
37
If 1 is of finite orner, then
logB(r),...., logM(r). Assertion: lo~l~(~) = 0(1) (unless
1 is a
polynomial).
prool. Otherwise, we can find a sequence {r.. } such that r n
--+ 00
and such
that
logM(rn)
~
clogr.. or M(r.. ) ~
It follows that
ale
= 0 for k
r~.
But Cauchy's inequality says that
> c, and hence 1 is a polynomial.
In general, we have: Theorem 10.2. tion.
Ifliminf,._oo
r.J;) = c <
00,
then
Proof. We again find an increasing sequence {rn}, r ..
1 is
a mtional lunc-
--+ 00
such that
T(r.. ) :5 clogr.. 80
that
N(r.. , f) :5 clog r ... We may suppose without loss of generality that 1(0)
=I- 0,00. Then:
rn net,t f) dt <- clogr..
10 and for
8
> 0 : J:n ..(~,f) dt ~ clog r... Then n(s,f) log rn :5 clogrn or 8
logrn og":
n(s, f) :5 cI -,.-. Let rn --+ 00, and we see that n(s,f) :5 c. Therefore, 1 has at most c poles aI, ... ,ale with k :5 c. Now, multiplying Jby (z - al)'" (z - ak) and applying the previous result to the holomorphic function so obtained, we complete the proof of the theorem. Suppose we are given M'(r) such that log M'(r) is logarithmically con- . We shall assume that M' (r) can be written in the form
M'(r) =
r "Y(t)t dt;
10
"Y(t)
increasing.
38
Entire and Meromorphic Functions
J; np
We already have proved (in Chapter 7) that dt, where 'Y(t) is increasing, is logarithmically convex. Suppose further that log M'(r) = o(rP ) for some p < 00. Then there exists an entire function I such that log M(r, f) '" log M'(r). Thus, for any such M'(r) there is an entire function I whose maximum modulus grows essentially like M'(r). The idea of the proof is the following: Take j.t(t) = 'Y(t). Define B(r) = ~ dt and draw a Newton polygon associated with p. and B. This gives the A~. Let I be the associated function I = A~z". Then
J;
E
10gM(r) '" 10gB(r) = 10gM'(r). Since p.(t) must be integer-valued, for the actual proof we take
pet) = [-y(t»)
[x) =
where
greatest integer not exceeding x.
We have then
'Y(t) - 1 :s; p.(t) Since logM'(r) = o(rP ), it follows that 'Y It remains to be shown that
10gB(r)
But
:s; 'Y(t).
= o(rP) so that p.(r) rvlogB(r).
= logM'(r) + O(logr).
-log ~:5 ro
r
lro
p(t) - 'Y(t) dt:5 0
t
so that
Ilog B(r) -logM'(r)1 = O(logr). Hence log M(r) '" log M'(r), and the proof is complete. Dropping the hypothesis of finite order, we can still get the following result: Given that log M'(r) is logarithmically convex, it is possible to find an entire function I such that M(r) ~ M'(r).
Prool· logB(r)
:s; logM(r)
logB(r) ~ logM'(r)
+ O(logr).
The term O(1og r) is easily disposed of by multiplying by a suitable polynomial. Another result along this line is the following:
10. The Maximum Term of an Entire FUnction
39
Theorem 10.3. Given any continuous function M'(r), there exists an entire function f such that M(r) ~ M'(r).
Proof. It is easy to see that any such function M' (r) has an increasing majorant such that log M'(r) is logarithmically convex, and, indeed,
J;
logM'(r) = ~ dt. Now proceed as in the previous theorem and the proof is complete. Thus, there exist entire functions that grow as rapidly as we please. We conclude this chapter with some more estimates on B(r). We know that:
[r + -.;.-) + 11J
B(r) < F(r) < B(r)
(10.16)
I'~r
and that for R > r F(r)
R
< B(r) R-r'
We can refine our results using the fact that log B(r) Since J.R ~ dt .. t
< rR ~ - Jo t
=
dt ' we get
log B(R) = fR I'(t)) dt
Jo
80
t
r
10
~
I'(t) dt.
lR
t
I'(t) dt
..
t
~ I'(r) log R
r
that
( ) < logB(R)
(10.17)
Choose R
Iog;:R
I' r -
= r + log S(R).
Then we have
logB(r+~) I'(r) :::;
1)
( log 1 + 'fOi1l\rJ
:::; 2 log B
(r + log ~(r») log
From the Borel Corollary it follows that (10.18)
I'(r) :::; 3log 2 B(r). eff
But then (10.11) gives F(r) :::; 3B(r)log2 B(r).
Hence, log F(r) ,.... log B(r). eff
B(r).
11
Relation Between the Growth of an Entire Function and the Size of Its Taylor Coefficients
Let F be an entire function and M(r) be its maximum modulus for Izl = r. Suppose A is a positive continuous increasing function for r ~ 1 such that is bounded.
"it:,}
Definition. If log M(r)
feA
= O(A(r», we say f
is of finite A-type and write
Proposition 11.1. Eanzn is of finite A-type if and only if there exists a constant K such that lanl :5 ~ for each n. K~(r)
Proof. Suppose Eanz" is of finite A-type. Then, since log M(r) = O(A(r», there exists a constant K such that M(r) :5 eK>'(r}. Now the Cauchy inequality gives lanl :5 ~~), which gives the result. eK~(r) "K~(") h ConverseIy, suppose Ia.. I :5 l2rJi' = ""'2n?' so t at lanlrn
:5 TneK>.(r).
Thus
Elanlrn :5 eK>.(r), so that M(r) :5
eK>'(r)
and log M(r) = O(A(r».
Definition. An entire function f is of orrle~ p if for each p' > p there exist constants A = A(p') and K = K(P') such that If(z)1 $
AeKI.zlpl
for all z.
11. The Growth of an Entire Function and Its Taylor Coefficients
41
Definition. An entire function f is of orner p if it is of order $ p but not of order $ Po for any Po < p. We have the following facts immediately from the definitions:
logM(r) $log+ A + Kr P' log+ log+ M(r) $log+ log+ A
+ log+ K + p'log+ r + log+ 2.
\~~: m(r) $ pi + 0(1) as r -+ 00. Now let>. = limsup 106+~: M(r). Then by the above observation we
Thus we have
106+
r .....oo
og
have >. $ p' for all I > p and hence>. $ p. In the other direction, we have !oJ+ Jog-!- M(r) < >. + 0(1) or Jogr
-
log+ log+ M(r) $ (>. + 0(1)) log r. Therefore, Thus, for every>.'
> A and r
> ro) Hence f
large (r
,,' .
we have M(r) $ er
;..'
or, more
generally, for all r, M (r) $ is of order $ A' for all >" and thus f is of order A. We therefore have proved:
Aer
Proposition 11.2.
>A
For any entire function f, log+log+ M(r) p= lim sup . logr
r ..... oo
De8nition. Suppose f is an entire tunction of order p and that If(z)1 $ AeKlzlP for all z. Then we say f is of order p, type at most K. The type T 0/ f is the infimum of those num!lers K such that / is of type at most K.
Definition. We say f is of finite-type if T is finite; we say / is of minimaltype if T = 0; and we say f is of mean-type if it is of finite type but not of minimal-type.
Definition. We say f is of growth (p, T) if either it is of order < p or it is T. A function of growth (I, T) is called a function of
or order p and type $ exponential-type.
Proposition 11.3.
For any entire function T
f of order p,
M(r) . rP
log+ = Iimsup r .....""
The proof is straightforward. In proving the following propositions we shall use the elementary fact
that
RlI
RR/e
max- = 11
yll
(
-!) R/e
=eR/e
Entire and Meromorphic Functions
42
Proposition 11.4.
Given an entire function f, let a
. nlogn = lim sup - - 1 - ' n-oo log l(i;j
Then a =p. Proof. Take u > p. Then lanlrn :5 M(r) :5 er" for large r. Thus lanl :5 .L r-ne r" or log lanl :5 rO' - nlogr. Now choose r = (;;)", which is large for n large. Therefore we have n
n
n
loglan l:5 - - -logu u u
or
1
n
n
n
log- > -log- --. lanl - u u u
Hence nlogn
nlogn
--1-< n n
tT
IT
(1)
as n
--+ 00.
t1
Therefore a :5 u and, since u can be chosen arbitrarily close to p, a :5 p. Now take {1 > a so that ~ < {1 for large n and, without loss of log~
r.!.J
r.!.J
generality, for all n. Thus nlog n :5 (1log or nnlP :5 or lanl :5 n;It,. n Hence lanlr :5 n~ijj and therefore B(r) :5 sup", ",~;jj, where B(r) is the maximum term of the series Ela..lrn (see Chapter 10). If we let y = ~ and R = 1: we have B(r 1IP ) < sup ~ = sup ~ = sup R" = eRIe =
P' '" ",ZIP "lIpll " ,," erlPe . Therefore B(r) :5 exp(rP /e Pe ), logB(r) :5 ~ and loglogB(r) :5 al . log+ log+ /J ogr -log{1e. Hence limsuPr_oo logr B(r) :5 (1. But logM ( r ) logB(r), and so we have p:5 (1 and hence p:5 a. We therefore must have I'V
p=a.
Proposition 11.S.
If f = Eanzn is of order :5 p, then
r =
~ limsupnlanl pln . ep n-oo
Proof. If f is of order p and type r, take r' > r. Using the Cauchy inequality we have lanl :5 ~\;) :5 for large r. We now minimize the expression on the right. Its logarithm is r'r P - n log r, and setting the derivative of this equal
<:P
to zero, r' prp-l - ~ (
~)
11 p.
= 0, we choose r = ( .,.":p )
1/P
• Thus lanl :5 (
"/p e
~
)"'P =
Hence nlan Ipln :5 er' p, so that we have ~.!!P nla..lpln :5
and therefore limsupnlanl P / n :5 erp. n-oo
er p
11. The Growth of an Entire Function and Its Taylor Coefficients
Now take {3 (
> -!p limsup".-oo nla"IP/".
YJ! ) "/P so that
Then for large n we have la,,1 :$
la"lr":$ ( ~ )"/P r". Hence B(r):$ ( ~ )"/P r" or
B(r) :$ maxx (eppr'P Xz P r. If we let y = B(r 1 / p )
43
:$ max (e{3p)YrY pYyY
Y
! and R = e{3r, we have
(e{3r)Y RY = max = max - = eRIe = e Pr • Y Y Y yY
Hence B(r) :$ ePrP and thus limsuPr..... oo log!(r) :$ {3. But logB(r) '" logM(r) so that T :$ {3 and thus T :$ ;P limsup".-oo nla"IP/", which completes the proof. Proposition 11.6. The onlers and types of an entire function f and of its derivative f' are the same. This follows easily from the formulas for order and type in terms of the power series coefficients. Corollary to Proposition 11.S. Write fez) = E'itz". Then f is of esponential-type if and only if Ea"(,, has a finite mdius of convergence.
Proof. We use Stirling's formula n! '" n"e-"../21rn.
nl
Now we have nl'it1 1/" '" ".!"e'!:.n 11/" '" ela"1 1/,,. Furthermore, the radius 0 f convergence 0 f E an../"",18 limBuP _ 1 IQnpln' Hence, smce . Q 11/" '" n I ':it n oo elan I1/ n , f is of exponential-type by Proposition 11.5 if and only if limsuPn.....oo lan l1/ n < 00, as was to be proved. Definition. To the function f given by fez) = E'itzn we associate the function ~(w) = Eanw!+t. Then f is of exponential-type if and only if ~ is holomorphic at 00 and ~ is called the Borel tmnsform of f. Indeed, it is easily seen that we have a onEK>ne linear correspondence between entire functions of exponential-type and functions ~ that are holomorphic near 00 with ~(oo) = O. Proposition 11.7. The Borel tmnsform of f is an analytic continuation ~f the Laplace tmnsform of f. More specifically, ~(w) = fo+ oo f(t)e- tw dt In some right half-plane.
Pr::,f. First, fo+ oo Se- tw dt = w!+t if Rew > O.
Now we estimate tw fo [/(t) - sn(t)]e- dt, where 8,,(t) is the nth partial sum of E~tn. ~ ~ ne have, for ak =~, kOOk :$ M(R) n+1 n+1 n+1 1 (r)n+1 = ( -r)n+1 M(R)--< e R' . , .R --
E laklrk = E lakl Rk Ci) 00
I/(z) -
8n
(z)1 =
R
00
l-i-
R
E Ci)
R-r'
Entire and Meromorphic Functions
44
(!t
where T' > T. Choose R = 2r so that we get I/(z) - sn(z)1 $ e2T 'r, 2r z t so that I/(z) - sn(z)1 $; e 'l l. Now I/(t)1 $; er't and le-twi = e- .. , where u = Rew; thus we have convergence of the integral and we can interchange the summation and integration if we take u > 21". Thus we have
in some right half-planp, n
> 2r', as
was to be provM.
As our final result in this chapter we shall give a direct proof that the order and type of I and f' are the same (Proposition 11.6). Prool. Let M 1 {r) = sUP81f'{rei8 )1. Then we have from the Cauchy integral formula, f'(z) = 2~i ~wl=R (~~):~ if R > r = Izl, that Ml(r) $; M(r)(R~r)2' Now take R = Ar, where A > 1. Then we get M1(r) $
M(Ar)r(,\~1}2 $ M(Ar)('\~l)2 for r> 1. Hence, for r > 1, log+ log+ Ml(r) log+ log+ M(Ar) 10gAr < -log r log Ar log r
+
----'--:--'-----'-~
log+ log+ (>'~1)2 log r
log+ 2
+log r
and thus PI $ p. In the othp.r direction, supposing without loss of generality that 1(0) = 0, we have I (z) = f' (w) dw, whp.re we shall integrate along a ray passing through the origin. It follows that M(r) $ rMl(r). Thus,
J;
log+ log+ M(r) < log+ log+ r log r
+ log+ log+ Mdr) + log+ 2 log r
and hence P $ PI. Therefore P = PI, as desired. Similarly for type, we have
Hence
'1
that log+ M(r} ~
$; AP' for any ,\
< -
10&+ r ~
> 1 and thus
+ log+ MI(r) ~
and hence
Also, M(r) $; rM1(r), so
TI
$
T
-< TI. Therefore T
I.
= TI.
12 Carleman's Theorem
~ 0 and suppose f has no zeros on z = iy. Choose p > 0 so that p < (modulus of the smallest zero of f in Re z ~ 0). Let {zn = rnei6n} be the zeros of f in Re z ~ O. Define the following:
Let f be holomorphic in Re z
(proper multiplicity of the zeros taken into account); /(R)
= I(R : f) = 2~
in (t~
-
~2) log If(it)f( -it)1 dt,
where the integral is taken from ir to iR along the imaginary axis; and
J(R) = J(R: f)
1
= -R 7r
/7
/2
Where the integral is taken along the semicircle of radius R centered at O. Then E(R) = /(R) + J(R) + 0(1).
Proof. Let r be the boundary of the "horseshoe" bounded by the semicircle of radius R, the semicircle of radius p, and the two vertical lines connecting them:
Entire and Meromorphic Functions
46
where we assume that J has no zeros on Izl = R and that log J(z) denotes a branch of 10gJ on r, i.e., 10gJ(z) is some continuous function on r satisfying exp(logJ(z)) = J(z). The proof proceeds by evaluating the contour integral
J=
(12.1)
0(1).
Izl=p Re
z~o
Along the negative imaginary axis z (12.2)
-...!... fRlogJ(_iY) 211'" J p
On z = iy, y > 0, dz (12.3)
2~
J:
[2.._2-] dy=...!...lRIOgJ(-iY) [2-_2..] dy. R2
211'"
y2
y2
p
R2
= i dy, and we have
[(i;)2 + ~2]
10gJ(iy)
= -iy, y > 0, dz = -idy, so
dy =
2~
lR
10gJ(iy)
[:2 - ~2]
dy.
On z = Rei9 (the large semicircle), dz = iRei9 dB, so we have
2. jfr/2 log J(Re i9 )-; (e- 2i9 + 1) eifJ R dB R
211'" -7
(12.4)
=2
2 j7
71"
-fr/2
The sum of the real parts of (12.2) and (12.3) is
2~
lR
(y~ - ~2)
10gIJ(iy)f(-iy)l
dy.
Thus Re S = /(R)
+ J(R) + 0(1).
Now integrate S by parts: u
du
= 10gJ(z) =
dv =
l'(z)
v
J(z) dz
(z1 + ~2) dz 2
z
= R2
1 - -;.
Hence,
[ 1] }l 1lrf ( purely imaginary 1{
finish
z
S = 271"i 10gJ(z) R2 - 2" =
271"i
start Z
-
1lrf ( 1) l'(z) J(z) dz 1) I'(z) fez) dz. 211"i
R2 - -;
Z
R2 - -;
12. Carleman's Theorem
47
On taking real parts and evaluating the remaining integral by the theory of residues we find
L (r~ - ~~) cos On·
=
Re S
Ir"I
Remark 1. There is an obvious extension to meromorphic functions. Remark 2. A formula of Nevanlinna (which stands in the same relation to Carleman's Theorem as the Poisson-Jensen formula to Jensen's formula)
gives the value of I inside a semicircle from its values on the boundary and its zeros and poles. (See [5, p. 2].) Next we present two applications of Carleman's Theorem.
Carlson's Theorem. Suppose I is entire and 01 exponential-type < 1T (i.e., I/(z)1 ::::; Ae B1z1 , B < 1T) and I(n) = 0 lorn = 1, 2, .... Then 1==0.
Rem4rk. Carlson's Theorem also is called the "sampling" theorem. Suppose
= cp(z) =
1 rn=
v 21T
1
1 cp(t)e-'·zt dt = rn=
fB cp(t)e-'
v21T -B
R
·zt
dt.
Then, differentiation under the integral and a simple estimate shows that I is an entire function of exponential-type B < 1T. H I vanishes on the positive integers, then I vanishes everywhere by Carlson's Theorem and therefore cp == 0 as well. By linearity, then, if we know I on the positive integers, we know it everywhere.
Proo/. Suppose I does not vanish id~ntically and satisfies the hypotheses of Carlson's Theorem. Then we may assume that I has no zeros on the imaginary axis. Otherwise, translate the plane z 1--* z - f for an appropriate E> O. Recalling the notation in Carleman's Theorem, we observe that E(r)
~
'" L.J
Ir .. ISR
(.!. - ~) R2 '" log R - 0(1) n
elf
lR (- - -
I(R) < -1 - 1Tp
BR J(R)::::; -R2 1T
1
t2
1 ) Bt dt
R2
lR
1 B B < -log R -< -1Tp -tdt1T
= -2B = 0(1). 1T
Now applying Carleman's Theorem we see B -logR + 0(1) ::::; log R. 1T
But
!- < 1i eventually, this is a contradiction. I
must vanish identically.
In Chapter 22 we will present a theorem of Malliavin-Rubel that gives a Very sharp form of Carlson's Theorem. For the next application, we prove a result that has applications to Polynomial approximation theory.
Entire and Meromorphic Functions
48
Theorem. Let 0' be the difference of two monotone functions on [a, b] and let f(z) = ezt dO'(t). Suppose that f(An) = 0 for each An in a sequence
J:
A of numbers in the right half-plane satisfying ERe (},,) f(z) = 0 foJr all z.
= 00.
Then
Remark. By use of the Hahn-Banach and Stone-Weierstrass Theorems, it can be shown that finite sums of the form Ea n exp Ant, An E A, are dense in the space of all continuous functions on [a, b] in the uniform topology if and only if there is an 0' (actually, we must allow complex a, but there is no significant change) for which exp(Ant) da(t) = 0 for each An E A implies that I exp(zt) da(t) = 0 for all z. Our theorem above then will imply that if Er;l cos On = 00, then the sums EaneXP(Ant) are dense.
J
Proof. It is easily verified that f is an entire function. Unless J vanishes identically, we may suppose that J has no zeros on the imaginary axis, since we could otherwise translate the y axis. Now J is bounded on the imaginary axis, say log IfI ~ M there. Hence 2M I(R) < - -211"
lR (1 1) p
-t 2 - -R2
dt
= 0(1) •
Now, If(z)1 ~ Amax{leztl : a ~ t ~ b}, where A is a constant. If c ;::: max{/al, 161), then If(z)1 ~ Ael:.!: so that J(R) ~ 0(1) also. Hence E(R) = 0(1). We have
- fI! > 1.1. if r n < E. because .1. r.. R - 4r.. - 2 Thus 1 cos On < 00, and on letting rn
L -
rn~f
we get a contradiction.
R
-+ 00;
13 A Fourier Series Method
The idea presented in this chapter is the following: H 1 is a meromorphic function in the complex plane, and if
is the kth Fourier coefficient of log I/(reifJ)l, then the behavior of I(z) is re8ected in the behavior of the sequence {c",(r,/)}, and vice versa. We prove a basic result in Theorem 13.4.5, which characterizes the rate of growth of 1 in terms of the rate of growth of the c"'( r, f) and the density of the poles of I, generalizing Theorem 1 of [35]. We apply this theorem as in [35] to obtain estimates for some integrals involving I/(z)1 and to obtain information about the distribution of the zeros of an entire function from information about its rate of growth. Our presentation follows [36]. By these means, we make a study of certain general classes of meroJnorphic and entire functions that include many of the classically studied classes as special cases. Let A( r) be a positive, continuous, increasing, and lDlbounded function defined for all positive r. We say that the meromorphic function 1 is of finite A-type to mean that there exist positive constants A and B with T(r, I) $ AA(Br) for r > 0, where T is the Nevanlinna characteristic. An entire function 1 will be of finite A-type if and only if there exist positive constants A and B such that
I/(z)1 $ exp(AA(Blzl»
If we choose A(r)
= r P,
for all complex
z.
then the functions of finite A-type are precisely
the functions of growth not exceeding order p, finite exponential-type. We
50
Entire and Meromorphic Functions
obtain here complete answers to certain basic questions about functions of finite A-type. For example, in Theorem 13.5.2 we characterize the zero sets of entire functions of finite A-type. This generalizes the well-known theorem of Lindelof that corresponds to the classical case A( r) = r P • We obtain in Theorem 13.5.3 a corresponding result for merom orphic functions. Then, in Theorem 13.5.4, we give necessary and sufficient conditions on A that each meromorphic function of finite A-type be the quotient of two entire function of finite A-type. In Chapter 14, we give Miles' proof that these conditions always hold. The body of the chapter is divided into five sections, the last two of which contain the main results. The first three sections are concerned with various elementary, although sometllut:!l complicated, results on sequences of complex numbers. The first section discusses the distribution of these sequences. The "Fourier coefficients" associated with a sequence are defined in the second section, and several technical propositions involving these coefficients also are proved there. The third section is concerned with the property of regularity of the function A, which is closely connected with the algebraic structure of the field of meromorphic functions of finite Atype. The fourth section contains the generalizations of the results of [35]. Finally, in the fifth section, the results about the distribution of zeros are proved. We urge that, on a first reading, the reader read §4 first and then §5, referring to §1, §2, §3 for the appropriate definitions and statements of necessary preliminary results. After this, the complex sequence theory of the first three sections will seem much more natural. 13.1. An Analysis of Sequences of Complex Numbers We study here the distribution of sequences Z = {z .. }, n = 1,2,3, ... , with multiplicity taken into account, of nonzero complex numbers z.. such that z.. --+ 00 as n --+ 00. Such sequences Z are studied in relation to so-called growth functions A. We denote by A and B generic positive constants. The actual constants so represented may vary from one occurrence to the next. In many of the results, there is an implicit uniformity in the dependence of the constants in the conclusion on the constants in the hypotheses. For a more detailed explanation of this uniformity, we refer the reader to the remark following Proposition 13.1.11. Let Z = {Zn} be a sequence of nonzero complex numbers such that limz.. = 00 as n -> 00. Definition 13.1.1.
The counting function of Z is the function n(r,Z) =
E IZnlS r
1.
13. A Fourier Series Method
51
Definition 13.1.2. We define N(r, Z) =
proposition 13.1.3.
r n(t,t Z) dt.
10
We have
2:
N(r,Z) =
log
I~I'
IZnl:S;r
proof. Note that
2:
log
1:1 =
Iz.. l:S;r
l
r
log
G) d[n(t, Z)).
The proposition follows from an integration by parts.
Proposition 13.1.4.
We have d n(r, Z) = r dr N(r, Z).
Proof. Trivial.
De&nition 13.1.5. We define, for k S(rik: Z) =
k1
= 1,2,3, ...
L
( 1 Zn
and r
~
0,
)k
Iz.. l:S;r
Deftnition 13.1.6. We define, for k = 1,2,3,. .. and
rb r2
~ 0,
S(rt,ra; k : Z) = S(raik : Z) - S(rli k : Z).
When no confusion will result, we will drop the Z from the above notation and write n(r), S(rj k), etc. l>e&nition 13.1.7. A growth function A(r) is a function defined for < oc that is positive, nondecreasing, continuous, and unbounded.
o< r
Throughout this chapter, A will always denote a growth function. l>etlnition 13.1.8. We say that the sequence Z has finite A-density to Ibean that there exist constants A, B such that, for all r > 0, N(r, Z) :5 AA(Br).
Entire and Meromorphic Functions
52
If Z has finite >'-density, then there are constants
Proposition 13.1.9. A, B such that
nCr, Z) $ A>.(Br). Proof. We have
1
2T
n(r, Z)log 2 $
T
net Z) -;-
dt $ N(2r,Z).
Definition 13.1.10. We say that the sequence Z is >'-balanced to mean that there exist constants A, B such that (13.1.1) for all rl, r2 > 0 and k to mean that
= 1,2,3, ....
We say that Z is strongly >.-balanced
(13.1.2) for all rI, r2
> 0 and k =
1,2,3, ....
Proposition 13.1.11. If Z has finite >.-density and is >'·balanced, then Z is strongly>.-balanced.
Remark. Using this result for illustrative purposes, we make explicit here the uniformity that we leave implicit in the statements of similar results. The assertion is that if Z has finite >.-density with implied constants A, B, and is >'-balanced with implied constants A', B' , then Z is strongly A-balanced with implied constants A", B" that depend only on A, B, A', B' and not on Z or >.. Proof of Proposition 19.1.11. We observe first that, if r > 0, and if we let r' = rk 1/ k , then (13.1.3)
I 3n(r/) IS(r,r jk)1 $ ~.
To prove this we note that
IS(r, r/j k)1 $
liT'
k
T
t1lo dn(t),
from which (13.1.3) follows after an integration by parts. Now, for rI, r2 > 0, let ~ = rIk l /" and r~ = r2kl/". Then
13. A Fourier Series Method
53
On combining this inequality with (13.1.3), Proposition 13.1.9, and the fact that kl/k ::; 2, we have
But, by hypothesis,
for 10 = 1,2, 3, .... Definition 13.1.12. We say that the sequence Z is >'-poised to mean that there exists a sequence a of complex numbers a = {ak}, k = 1, 2, 3, ... such that, for some constants A, B, we have, for k = 1,2,3, ... and r > 0,
(13.1.4)
lak + S(rj k : Z)I :5
A>.(Br) r
k'
If the following stronger inequality:
lak + S(rj k : Z)I:5
(13.1.5)
AA(Br) iorio
holds, we say that Z is strongly A-poised. Proposition 13.1.13. is 'trongly A-poised.
If Z has finite A-density and is >.-poised, then Z
Proof. The proof is quite analogous to the proof of Proposition 13.1.11, based on the substitution r' = rkl/k. We omit the details. Proposition 13.1.14. A sequence Z is >'-balanced if and only if it is ).-poised, and is strongly A-balanced if and only if it is strongly >.-poised. Proof. We prove only the second assertion, since the proof of the first I88ertion is virtually the same. IT it is first supposed that Z is strongly .\..poised, where {a,.} is the relevant sequence, then we have
+ ak - a,. - S(rli k)1 :5 la,. + S(rli k}1 + lak + S(r2i k)l,
IS(rb r2i k)/ = IS(r2; k)
10 that Z is
strongly >.-balanced. Suppose now that Z is strongly >.-balanced, with A, B being the relevant COostants. Let
p(>.) = inf{p = 1,2,3,,,,: liminf A(r) r-oo
rP
= o}.
Entire and Meromorphic Functions
54
Naturally, we let peA) = 00 in the case liminf A(r)rP > 0 as r -+ 00 for each positive integer p. For 1 ~ k < peA), we have infr-kA(Br) > 0 for r > O. Thus, there exist positive numbers rk such that
> 0 and
for r
1:$ k
< p(A).
For k in this range, we define
(13.1.6)
For those k, if there are any, for which k ~ p( l), we choose a sequence with Pj -+ 00 as j -+ 00 such that
o < Pl < P2 < . ..
lim A(BPi) = i-oo pj(>')
o.
For values of k, then, such that k ~ peA), we define (13.1.7)
To show that the limit exists, we prove that the sequence {S(Pi; k)}, j = 1,2, ... , is a Cauchy sequence. Let
We have 1.6..
1 < AA(BPm)
),m -
k~
+
AA(BPi) kJ1j
Since P" 2:: pf'(>') for p 2:: 1, it follows from the choice of the Pi that .6.i ,m as j, m -+ 00. We now claim that
For, if 1 ~ k
~
0
< peA), then
. k)1 Iak+ S( r,. k)1 -IB( rk,r, ~ if k
-+
AA(Br) krk
+
AA(Brk) 3AA(Br). krk ~ krk '
peA), then
lak+S(r; k)1
= ,-00 .lim IS(r,pi; k)1 ~
AA(Br). AA(BPi) k k +~BUP k k r J-OO Pi
=
A>.(Bp) kpk .
13. A Fourier Series Method
55
Definition 13.1.15. We say that the sequence Z is A-admissible to mean that Z has finite A-density and is A-balanced.
In view of Propositions 13.1.11 and 13.1.13, the following result is immediate.
proposition 13.1.16. Suppose that Z has finite A-density. Then the following are equivalent: (i) Z is .\-balanced; (ii) Z is strongly A-balanced; (iii) Z is A-poised; (iv) Z is strongly A-poised; (fJ) Z is A-admissible.
In Proposition 13.3.3, we give a simple characterization of A-admissible sequences in the special case A(r) = r P • 13.2. The Fourier Coefficients Associated with a Sequence We now present the sequence of so-called Fourier coefficients associated with a sequence Z of complex numbers, and study its properties. We will use it in §5 to construct an entire function f whose zero set coincides with Z, and to determine some properties of entire and meromorphic functions whose growth is restricted. The reason for calling them "Fourier coefficients" will become apparent on comparing their definition with Lemma 13.4.2. De8nition 13.2.1.
We define, for k = 1,2,3, ... , 1
S' (rj k : Z)
-
L
=k
(~)
k
.
IZnl$r
Proposition 13.2.2.
We have IS'(rj k : Z)I
~ ~N(er, Z).
Proof. It is clear that IS'(r; k : Z)I ~ n(r)fk, and we also have n(r) ~
l
r
er
n(t)
-t- dt
~ N(er).
Entire and Meromorphic FUnctions
56
Definition 13.2.3. Let a = {ak}, k = 1,2,3, ... , be a sequence of complex numbers. The sequence {ck(rj Z : a)}, k = 0, ±1, ±2, ... , defined by
eo(rjZ: a)
(13.2.1)
= eo(rjZ) = N(r,Z),
rk ck(r; Z: a) ="2{ak (13.2.2)
(13.2.3)
+ S(r; k: Z)}
1 -"2S'(rjk:Z) for k=l,2,3 ... ,
C_k(rjZ:a)=(Ck(r;Zja)) for k=I,2,3 ... ,
is said to be a sequence of Fourier coefficients associated with Z.
on
Definition 13.2.4. A sequence {ck(rj Z : of Fourier coefficients associated with Z is called A-admissible if there exist constants A, B such that AA(Br)
/ck(r : z; a)/ $ /k/ + 1
(13.2.4)
(k
= 0, ±1, ±2, ... ).
Proposition 13.2.5. A sequence Z is A-admissible if and only if there exists a A-admissible sequence of Fourier coefficients associated with Z. Proof. Suppose that Z is A-admissible. Then, by Proposition 13.1.16, Z is strongly A-poised. Let 0 = (ak), k = 1,2,3, ... , be the relevant constants, and form {ck(r; Z : a)} from them by means of (13.2.1)-(13.2.3). Now Definition 13.2.4 holds for k = 0 and some constants A, B since Z has finite A-density. For k = ±1, ±2, ±3, ... , we have ICk(rj Z:
rlkl
1
a)1 $ Tla k + S(rj k)1 + 2IS'(r; k)l.
Then an inequality of the form (13.2.4) holds by Proposition 13.2.2 since Z has finite A-density, and because Z is strongly >.-poised with respect to the constants {Ok}. On the other hand, suppose that (13.2.4) holds. Then N(r)
= eo(r) $
A>.(Br),
so that Z has finite >.-density. Moreover, I r; (ak
+ S(rj k»1
= ICk(rj Z : a)
$
AA( Er) Ikl + 1
+ ~SI(rj k)1
N (er)
+ 2k ::;
2AA( eEr) k '
so that Z is strongly A-poised. By Proposition 13.1.16, it follows that Z is >'-admissible.
13. A Fourier Series Method
57
Proposition 13.2.6. Suppose that Z and a = {aTe} are such that ICk(rj Z : a)1 ~ AA(Br). Then {ck(r; Z : an is A-admissible. In particular, there exist constants A', B ' , depending only on A, B, such that A'A(B'r)
ICk(r; Z : a)l::; Ikl + 11 .
Proof. For k
= 1,2, ... , we have
(13.2.5)
and (13.2.6)
= N(r) ~ AA(Br), Z has finite A-density. Then, by Proposition (13.2.2), IS'(rjk)1 ~ (l/k)O(A(O(r))) uniformly for Ie = 1,2,3, ... , by which we mean that there are constants A", B" for which IS'(r,Ie)1 ::; (ljle)A" A(B"r). From our hypothesis and (13.2.6), it then follows that
Since ~(r)
rklak
+ S(r; k)1 = O(A(O(r)))
uniformly for
k
> O.
Then, by Proposition 13.1.13, we have that 1 rkla,., + S(r; 1e)1 ::; kO(A(O(r)))
uniformly for
Ie = 1,2,3 ....
Then, using (13.2.5), we have 1 ICk(r)1 ~ 'k0(A(O(r)))
uniformly for Ie
= 1,2,3 ....
Since c_k(r) = (ck(r», and since Z has finite upper A-density, the proposition follows immediately. Deftn.ition 13.2.7. The quadratic semi-nonn of a sequence {ck(r; Z: of FOurier coefficients associated with Z is given by
an
Entire and Meromorphic Functions
58
Proposition 13.2.S. The Fourier coefficients {ck(rj Z : a)} are Aadmissible if and only if E 2 (rj Z : a) $ AA(Br) for some constants A, B.
Proof. First, tf
ICk(r; Z : a)1 $ then E 2 (r; Z : a) $ AA(Br), where B
AIA(Blr) Ikl + 1 '
= Bl
and
On the other hand, suppose there are constants A, B for which E2 (rj Z : a) $ AA(Br). Then it is clear that ICk(rj Z : a)1 $ AA(Br), so that by Proposition 13.2.6, {Ck (rj Z : a)} is A-admissible. 13.3. Sequences That Are A-Balanceable In this section, we are concerned with the process of enlarging a sequence Z so that it becomes A-balanced. Growth functions A for which this is always possible are called regular and give rise to associated fields of meromorphic functions with special propertiesj for example, see Theorem 13.5.4. The principal results of this section are Propositions 13.3.5 and 13.3.6, which give the simple condition that A be regular. In addition, we give in Propostion 13.3.3 a simple characterization of A-admissible sequences of the case A(r) = r P • Definition 13.3.1. The sequence Z is A-balanceable if there exists a A-admissible supersequence Z' of Z. Definition 13.3.2. The growth function A is regular if every sequence Z that has finite A-density is A-balanceable. Proposition 13.3.3. Suppose that A(r) = r P , where p > O. Then (i) the sequence Z is of finite A-density if and only if lim sup r-Pn(r, Z) < 00
as r
~
00;
(ii) if p is not an integer, then every sequence of finite A-density is Aadmissible; (iii) if p is an integer, then Z is A-admissible if and only if Z is of finite A-density and S(r;p: Z) is a bounded function ofr; (iv) the function A(r) = r P is regular.
Proof. To prove (i), we have that nCr) = O(rP ) whenever Z has finite A' density. On the other hand, if limsupr-Pn(r) < 00, then nCr) $ ArP for some positive constant A, so that N(r)
= for r1n(t) dt ~ Ap-1rP •
13. A Fourier Series Method
To prove (ii), suppose that N(t)
(13.3.1)
I
T2
Tl
~
59
Atp • Then 50 long as k 1: P, we have
d() (A + _A_) (..\(r IP - kl
~ tk n t ~
1)
r k1
+
..\(r2 r 2k
»)
.
For, on integrating by parts, we have that the integral is equal to
But
and similarly
Moreover,
and the inequality (13.3.1) follows. Hence, so long as p is not an integer, every sequence Z of finite rP-density is rP-balanced. To prove (iii), suppose that Z has finite rP-density and that p is an integer. Then, by (13.3.1), we see that all the conditions that Z be ..\balanced are satisfied except for k = p. For this case, the condition that S(rl,r2; p) be bounded by r;:-P A"\(Brt} + r2"P A"\(Br2) for some A, B is precisely the condition that S( rj p) be bounded, as is quite easy to see. Th prove (iv), we observe first that if p is not an integer, then ..\(r) = r P is trivially regular by (ii). H p is an integer and Z has finite rP-density, let Z' be the sequence obtained by adding to Z all numbers of the form ",-lZ, where wP = 1, but w 1: 1. Then Z' has finite rP-density and SCriP : Z') = 0 for all r > O. Hence, by (iii), Z' is rP-admissible, and it follows that ..\ (r) = r P is regular. The next two results give simple conditions, both satisfied in case ..\(r) = rt', that imply that ..\ is regular. De&nition 13.3.4. We say that the growth function ..\ is slowly increasing to mean that ..\(2r) ~ M..\(r) for some constant M. If ..\ is slowly increasing, it is easy to show that for some positive number P, ACr) = O(r P ) as r -+ 00. Proposition 13.3.5.
If..\ is slowly increasing, then ,.\ is regular.
Proposition 13.3.6.
If log ,.\( e%) is convex, then ,.\ is regular.
The proofs of these results use the next lemma.
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60
Lemma 13.3.7. The growth function A is slowly increasing if and only if there exist an integer Po and constants A, B such that (13.3.2)
1
00
r
A(t)
t-p + 1 dt <
AA(Br)
lor
prp
J'
If A(r) = r P , then we may choose Po = [p]
p -> p0 > O.
+ 1.
Proof. Suppose first that (13.3.2) holds. We may clearly suppose that B ~ 1. Then AA(Br) > prP
-
1
00
r
A(t) dt> fOO A(t) dt> A(2Br) - J2Br tpH - p(2B)PrP
tPH
whenever P ~ Po- Taking P = Po, we have A(2Br) ~ MA(Br), where M = A(2B)Po,
so that A( r) is slowly increasing. Suppose next that A(r) is slowly increasing, say A(2r) ~ MA(r). Then
1
00
r
A(t) 00 -dt=~ tp+l L.-J
k=o
1
211+ 1 r
2"
(M)k
A(t) 00 A(2kHr) A(r) 00 d t < " < M -" k tpH - ~ prp(2 )p prP ~ 2P k=O
r
k=O
Hence, if Po is taken so large that 2pO > M, we have an inequality of the form (13.3.2). In case A(r) = r P , we have M = 2P , and the final assertion follows.
Proof of Proposition 13.3.5. Let A he slowly increasing, and let Z he a sequence of finite A-density. Choose Po as in the last lemma so that, for p~Po,
1
00
r
A(t) d
t ~
tpH
Define
AA(Br) prp '
PO
Z' =
U w-kz, k=O
where no = Po+I, w = exp(271"i/no), and w-kZ = {w-kzn }, n = 1,2,3, ... · Then we have Serb r2; k : Z') = 0 for k = 1,2, ... ,Po, since
1 + wk
+ w2k + ... + wPO k =
0
so long as wk ¥= 1, and this is true for k = 1,2, ... ,Po, Hence, to prove that Z' is A-balanced, we need consider only k > Po. For such k, with r < rl, we have IS(r,r';k:Z}I='k1
I z: ( I$'k 1 1)
Zn
r<JZnJ
k
1
r
r
'
1 dn(t,Z'}. tk
13. A Fourier Series Method
61
On integrating by parts, we have
1~ rl
1
r
d ( Z')I
tk
n t,
< -
n(r', Z')
+
(r,)k
n(r, Z')
rk
+
k
l
rl
r
n(t, Z') d
t.
tk+l
Since Z is of finite A-density and nCr, Z') = (Po + l)n(r, Z), we have n(r,Z') ~ AIA(Blr) for some constants At, Bl by Proposition 13.1.9. Since A is slowly increasing, we have ..\(Blr) ~ A2A(r) for some positive constant A2 > O. To complete the proof of the proposition, we have only to prove that r ' net, Z') A'A(B'r)
1,.
dt ~
tk+1
krk
for some constants A', B'. However,
1 r'
r
net, Z/) dt
t k +1
-
1
00
2
r
A(t) < AA2 A(Br) t k +1 krk
since k > Po· Proof of Proposition 13.3.6. It is no loss of generality to suppose that r-PA(r) --+ 00 as r --+ 00 for each P > 0, since otherwise A is slowly increasing by Lemma 13.3.7, and then Proposition 13.3.5 applies. Now for , = 1,2,3, ... , let R" be the largest number such that
A(Rp) _ inf A(r)
.n: -
r>O
rP
:s
= O.
,
:s
:s ... ,
Then we have that Ro Rl R2 and that Rp --+ 00 Further, by Lemma 13.3.7, r-PA(r) decreases for r R" and increases for r ;::: R". We also have the inequality
and let Ro
88
P
-+ 00.
:s
(13.3.3)
since, by the above remark, A(r) < A(2r) rP-1 - (2r)p-l' Now let Z be of finite A-density. For convenience of notation, we suppose A(r) and nCr) A(r), since we could otherwise replace the function A(r) by the function AA(Br) for suitable constants A, B. We then elabn that
that N(r)
:s
:s
,.'
(13.3.4)
1 r
~ dn(t) < 4..\(r) t'" - rk
Entire and Meromorphic Functions
62
if k ~ 2P and r ~ r' ~ 14. To prove (13.3.4), we first integrate by parts, replacing the integral by
nCr') _ nCr) (r')k rk Now
+
k
J
r'
r
n(t) d tk+l t.
n(r') < ,\(r') < '\(r) (r')k - (r,)k - rio
since r ~ r' and r- k '\(r) is decreasing for r ~
J
r'
r
since rp,\(t)
net) dt < tk+l ~
J
r
'
r P tk+l- p
r
dn(t)
Also,
'\(r) _1_ dt < '\(r) _1_ _1_
r'
~ t ~
r-P'\(r) for r
J t~
14 ~ Rk.
r'
~
-
r P (k-p)r k- p '
Rp. Thus,
~ '\r(~) + '\r(~) + _k_ '\(r) . (k-p)
rk
We have k/k(k - p) ~ 2 since k ~ 2P , and (13.3.4) follows. We now define Z' as follows. For each Zn E Z with Rp-l < we introduce into Z' the numbers 1
IZnl
~
Rp,
1
-zn""'--lzn, wm -
W
where m = m(p) = 2P and w = w(m) following assertions: (13.3.5) nCr, Z')-n(Rp-t, Z')
= exp(21ri/m).
= 2P (n(r)-n(Rp_t})
if Rp-l ~ r :$ 14,
(13.3.6)
nCr, Z')
~ 2P n(r)
if r:$ Rp,
(13.3.7)
N(r, Z') :5 2P '\(r)
if r:5 Rp,
(13.3.8)
l
r
'
r
~ t
dn(t, Z')
~k
l
r
rl
~
t
dn(t)
if k
~2
P
We make the
and r:5 r'
~ 14.
(13.3.9) S(r, r'j k : Z') = 0 if r, r' ~ 14
and k is not a multiple of 21'.
The assertions (13.3.5) and (13.3.6) follow immediately from the definition of Z', while (13.3.7) follows from (13.3.6) and (13.3.8) follows easily
13. A Fourier Series Method
63
from (13.3.5). To prove (13.3.9), it is enough to prove that S(r,r'ik: Z'} = oif Rj-l ~ r ~ r' ~ Rj , j ~ p, and k is not a multiple of 2". But, in this case, we have S(r,r'; k: Z') = -yS(r,r'; k: Z), where
-y = 1 +wk
+w2k
+ ... +w(m-l)k,
where m = m(j) = 2j and w = w(m) = exp(21ri/m). Since k is not a multiple of 2", k is therefore certainly not a multiple of 2j , so that wk i- 1. We then have 'Y =
1_wkm
1-w
k
= 0,
and our assertion is proved. We now prove that Z' is A-admissible. To see that Z' has finite Adensity, let r > 0 and let p be such that Rp-l ~ r ~ Rp. Then, by (13.3.7) and (13.3.3), we have that N(r, Z') ~ 211 A{r) ~ 2A(2r). To see that Z'is .\-balanced, let k be a positive integer and suppose that 0 < r ~ r'. Write k in the form 2P q, where q is odd. Then, by (13.3.9), S(r,r'ik: Z') = 0 if R" ~ r < r'. Suppose that r ~ Rp. Then S(r, r'; k : Z') = S(r, r"; k : Z'), where r" = min(r', Rp), by (13.3.9). However,
IS(r,r"jk: Z')I
~~
1 t~ r"
dn(t,Z').
By (13.3.8), this last term does not exceed
1 ~dn(t), r"
r
t
and this, in turn, does not exceed 4r- k A(r), by (13.3.4). Consequently, we always have IS(r, r'; k : Z)I ~ 4r- k A(r), so that Z' is A-balanced, and the proof is complete. 13.4. The Fourier Coefficients Associated with a Meromorphic Function
In this section, we associate a Fourier series with a meromorphic function and use it to study properties of the function. AF, we mentioned at the besmning, the results of this section are generalized versions of the results of the earlier paper [35), and the proofs are essentially the same. Our .' -ation follows the notation of [35) and the usual notation from the theory Of.lr1eromorphic and entire functions. Our presentation still follows [36]. \Ve first recall the results from the theory of meromorphic functions that 'rill be needed.
Entire and Meromorphic Functions
64
For a nonconstant meromorphic function I, we denote by ZU) [respectively W(f)] the sequence of zeros (respectively poles) of I, each occurring the number of times indicated by its multiplicity. We suppose throughout that 1(0) -:F 0, 00. It requires only minor modifications to treat the case where 1(0) = 0 or 1(0) = 00. By nCr, I) we denote the number of poles of I in the disc {z : Izl ~ r}. By N(r, I) we denote the function
N(r, I) =
r net,t I) dt,
10
and by mer, I) the function
1""
mer, f) = 211' 1 _,," log+ I/(re")1 dB, where log+ x = max(logx, 0). We have, of course, that nCr, I) = nCr, W(I) and N(r, I) = N(r, WU». The Nevanlinna characteristic, which measures the growth of I, is the function
T(r,1) = m(r,1) + N(r,/). Three fundamental facts about T( r, I) are that T(r, I) = T (r,
(13.4.1)
T(r,fg)
(13.4.2)
~
7-) + log 1/(0)1,
T(r,1)
+ T(r,g),
T(r, I + g) ~ T (r, I) + T(r, g) An easy consequence of (13.4.1) is that (13.4.3)
+ log 2.
1""
211' 1 _ .. Ilog I/(re")!! dB ~ 2T(r, I)
(13.4.4)
+ log 1/(0)1·
This follows from (13.4.1) by observing that the first term is equal to mer, /) + m(r, 1//), which is dominated by T(r, /) + T(r, 1/1). For the entire functions I, we use the notation M(r,/) = sup{l/(z)/ : z
= r}.
The following inequality relates these two measures of the growth of I in case I is entire:
R+r
T(r, I) ~ log+ M(r, I) ~ R _ r T(R, I)
(13.4.5)
for 0
~
(13.4.6)
r
~
R. We will use (13.4.5) mostly in the form T(r, f) ~ log+ M(r, I) ~ 3T(2r, I),
which results from setting R = 2r in (13.4.5). The following lemma, which is fundamental in our method, was proved in [9] and [35].t We reproduce the proof of [35] here. tI have a vague memory of seeing this formula ill a paper of Frithiof NevanliDlla published around 1925, but WIllI nnabJe to find it on a recent search.
13. A Fourier Series Method
65
Lemma 13.4.1. III(z) is meromorphic in \z\ $ R, with 1(0) :/: 0, 00, and Z(f) = {zn}, W(f) = {wn}, and iflog(f(z)) = E::o
log If(re i9 ) 1=
(13.4.7)
L
ck(r,!)eikfJ ,
k=-oo
tDhere the ck(r,!) are given by Co(r,!)
=log 1/(0)1 + L
log
I~I -
Illnl:5r
(13.4.8)
= log 1/(0)1 + N
(r, :7) -
L
log
I;kl
l1Dnl::;r
N(r,!).
For k = 1,2,3, ..• ,
(13.4.9)
For k = 1,2,3, ... , (13.4.10) There are appropriate modifications if 1(0) = 0 or 1(0) =
00.
Remork. Observe that in the notation of §1 and §2, formula (13.4.9) becomes
(13.4.11)
1 rk ck(r,!) =2
Proof. We may suppose that I is holomorphic, since the result for melOIDorphic functions then will follow by writing f as the quotient of two boiomorphic functions. We may suppose further that I has no zeros on {z : \zl = r}, since the general case follows from the continuity of both tides of (13.4.8) and (13.4.9) as functions of r. Formula (13.4.8) is, of Jensen's Theorem, and (13.4.10) is trivial since log III is real. To 'iJIrove (13.4.9), write
eourse,
Entire and Meromorpbic Functions
66
for some determination of the logarithm, and k integrating by parts, we have
= 1,2,3,....
Then, by
This may be rewritten as
This last integral may be evaluated as a sum of residues, and on taking real parts we get the kth cosine coefficient of log III. Similarly, considering the integral
Jk(r)
=~
J
pog(f(reis »]sin(k9) diJ,
where the integration is now between 11"12k and 211" + 11" 12k, we get the kth sine coefficient. On combining these, we get (13.4.9). We now define the classes of functions that we shall study. Definition 13.4.2. Let A be a growth function. We say that I(z) is of finite A-type and write I E A to mean that I is meromorphic and that T(r, I) ::::; AA(Br) for some constants A, B and all positive r. Definition 13.4.3. finite A-type.
We denote by AE the class of all entire functions of
Proposition 13.4.4. Let I be entire. Then I is 01 finite A-type il and only il logM(r,1) ::::; AA(Br) lor some constants A, B and all positive r.
Proof This follows immediately from (13.4.6). Note in particuJar that, if A(r) = r P , then I E A if and only if I is of growth at most order p, exponential-type. We also note that by inequalities (13.4.2) and (13.4.4), A is a field and AE is an integral domain under the usual operations. The main theorem of this chapter is the following. Theorem 13.4.5. Let I be a meromorphic function. If f is of finite A-type, then Z(f) and W(f) have finite A-density and there exist constants A, B such that (13.4.12)
Ic",(r,l)l::::;
A'\(Br) Ikl + 1
(k
= 0, ±1, ±2, ... ).
13. A Fourier Series Method
67
In order that / should be 0/ finite )..-type, it is sufficient that Z(f) [or W(f)] have finite )..-density and that the weaker inequality
(13.4.13) hold for some (possibly different) constants A, B. Thus, in order that / should be of finite )..-type, it is necessary and sufficient that Z(f) have finite )..-density and that (13.4.12) should hold. It is also necessary and sufficient that Z(f) have finite )..-density and that (13.4.13) should hold.
proof. The order of the steps in the proof will be as follows. We first show that if / satisfies an inequality of the form (13.4.13), and if either Z(f) or W(f) has finite A-rlfmsity, th('n / must satisfy an inequality of the form (13.4.12). We then show that if / is of finite A-type, then Z(f) and W(f) are of finite A-density, and / satisfies an inequality of the form (13.4.13). Finally, we prove that if Z(f) [or W(f)] has finite A-density and if/satisfies an inequality of the form (13.4.12), then / must be of finite A-type. We shall suppose that /(0) = 1. The case /(0) = 0 or /(0) = 00 causes no difficulty since we may multiply / by an appropriate power of z and the resulting function still will be of finite )..-type. This is because if liminf(A{r)jlogr) = 0 as r -+ 00, then by the exercise in Chapter 8 the class A contains only the constants. Let us suppose that either Z(f) or W(f) is of finite )"-density and that Icre(r,!)1 O(A(O(r))) uniformly for k = 0, ±1, ±2, .... On considering the case k = 0, we see that both Z(f) and W(f) have finite )..-density. It is enough to prove that / satisfies an inequality of the form (13.4.12) for k = 1,2,3, ... , since C-k is the complex conjugate of Ck. We prove this exactly as Proposition 13.2.6 was proved. From (13.4.11), we have
=
(13.4.14)
rk
(13.4.15)
2"lok + S(r; k : Z(!)
- S(r; k : W(f»1 ~ /ck(r,!)1
1
1
+ 2IS'(r;k: Z(fm + 2 IS'(f;k: W(f»I· Then, by Proposition 13.2.2, for Z IS'(r;k: Z)I
= Z(f) or Z = W(f),
= k-10()"(O(r»)
we have
uniformly for k > O.
By (13.4.14), it is therefore enough to prove that 10k + S(r; k :Z(f» - S(r; k : W(f»1 = k-lr-kO()"(O(r») uniformly for k = 1,2,3, ....
Entire and Meromorphic Functions
68
But, we already have from (13.4.15) that ic.t:k
+ S(r; k : Z(f» - S(r; k : W(f»1 = r-kO(,x(O(r)))
uniformly for such k. Replacing r by r' we have that lale
+ S(r'; k : Z(f»
= killer and observing that r' ::; 2r,
- S(r'; k : W(f))1 = k-lr-kO(,x(O(r))).
Thus, the assertion will be proved if we can show that, for Z = Z(f) and Z = W(f), we have IS(r, r'; k : Z)I
= k-1r-leO(,x(O(r))).
This was proved in Proposition 13.1.11 [see (13.1.3)]. Now suppose that f has finite ,x-type. Then N(r, W(fn
= N(r,!)::; T(r,!),
so that W(f) has finite ,x-density. By (13.4.1), the function 1/f also has finite ,x-type. Hence, Z(f) = W(I/ I) also has finite ,x-density. To see that an inequality of the form (13.4.13) holds, note that
Iele (r, 1)1 =12~ ::;
I: I: I
{log If(rei9)1}e-ileB
2~
dol
log If(reiB)11 dO :::; 2T(r, I)
+ log 1/(0)1
by (13.4.4). Finally, suppose that W(f) has finite ,x-density and that (13.4.12) holds. H Z(f) has finite ,x-density, we apply the argument below to the function }. Then N(r,1) = O(,x(O(r))). It remains to prove that mer, f) = O(,x(O(r»). However,
m(r,!):::;
2~
I:
IIOgIJ(reiB)11 dO,
which, by the Schwarz inequality, does not exceed
By Parseval's Theorem, we have, for suitable constants A, B,
Hence, mer, J)
= O(,x(O(r))), which completes the proof of the theorem.
Specializing Theorem 13.4.5 to entire functions, we have the next result.
13. A Fourier Series Method
69
Theorem 13.4.6. Let f be an entire function. If f is of finite A-type, then there exist constants A, B such that ICk(r, f)1 ~
(13.4.16)
A(A(Br» Ikl + 1
(k = 0, ±1, ±2, ... ).
It is sufficient, in order that f be of finite A-type, that there exist (possibly different) A, B such that
ICk(r, f)1 $ A>.(Br)
(13.4.17)
(k
= 0, ±1, ±2, ... ).
Thus, in order that f should be of finite >'-type, it is necessary and sufficient that (13.4.16) should hold. and it is also necessary and sufficient that (13.4.17) should hold.
Proo/.
This result is an immediate corollary of Theorem 4.6 since WU) is empty in case I is entire.
I, we define
DefInition 13.4.7. For a meromorphic function
Eq(r, f) =
{2~
i:
Ilog l/(rei9
)f
.!
dO}
Q
•
Notice that if f is entire with 1(0) = 1, and if a = {ak} is such that CAt(r, f) = cr.:(rj Z(f) : a), k = 0, ±1, ±2, ... , then E2(r, f) = E 2 (rj ZU) : a), where this last quantity is the one defined in Definition 13.2.7.
Theorem 13.4.8. Let and 1 ~ q < 00, then
I
be an entire function. If f is of finite A-type
Eq(r, f)
(13.4.18)
~ AA(Br)
lor 6uitable constants A, B and all r > 0. Conversely, il (19.4.18) holds lor some q ?: 1, then
I
i8
01 finite A-type.
Proof. If I is of finite A-type, then by the Hausdorff-Young Theorem ([51], p. 190), the Lq norm of log I/(re i9 )1, as a function of 9, is bounded by the tit norm of the sequence {Ck}, where (~)+(~) = 1. By Theorem 13.4.6, this tit norm is dominated by an expression of the form AA(Br). Conversely, USing Holder's inequality, ICk(r, 1)1
~ 2~ l"}Og If(rei9 )lldO, ~ ~ {1
1f
Ilog
If(rei8
)fdO}
1
q
~ AA(Br)
for suitable constants A, B, and it follows from Theorem 13.4.6 that I has
finite A-type.
Entire and Meromorphic Functions
70
Theorem 13.4.9. Let f be a meromorphic function of finite A-type, with f(O) :f. 0, 00. Then for each positive number f there exist positive constants a, f3 such that, fOT all r > 0, (13.4.19)
Remark. We have as a consequence that, for all r > 0,
which is somewhat surprising, even in case f is entire, since it is by no means evident that the integral is even finite.
Proof. There is a number f3
> 0 such that
Ic,,(r,f)1 < ~ >'(f3r) - Ikl'" 1 (k Let
F(O) Then
F(O)
= 0, ±1, ±2, ... ).
1
'(J
= F(O, r) = >.(f3r) log If(re' ~
ik(J
= L-t 'Yk e
,
)1·
h ck(r, f) were 'Yk = >.(f3r) .
We may also suppose that the constant M satisfies
2~
I:
IF(O)I dO :5 M
by Theorem 4.9. By a slight modification of [49J (p. 234, Example 4), we know that for any such F there exists a constant a> 0, where a depends only on M and f, such that 1 211'
111' exp(alF(O)1) dO :5 1 + -11'
f,
from which (13.4.19) follows.
13.5. Applications to Entire FUnctions We present in Theorem 13.5.2 a simple necessary sufficient condition on a sequence Z of complex numbers that it be the precise sequence of zeroS of some entire function of finite >'-type. The condition is that Z should be >'-admissible in the sense of Definition 13.1.15. This generalizes a wellknown theorem of Lindelof (see the remarks following the proof of the
13. A Fourier Series Method
71
Theorem 13.5.1) for constructing an entire function with certain properties from an appropriate sequence of Fourier coefficients associated with a sequence of complex numbers). We also prove in Theorem 13.5.4 that A has the property that each meromorphic function of finite A-type is the quotient of two entire functions of finite A-type if and only if A is regular in the sense of Definition 13.3.2. Accordingly) Propositions 13.3.5 and 13.3.6 give a large class of growth functions A for which this is the case, including the classical C88e A( r) = T P • Even this case seems to be unknown. We turn now to our first t88k, the construction of an entire function f from a sequence Z and a sequence {Ck (Tj Z : a)} of Fourier coefficients associated with Z. We recall that we have 88sumed that Z = {zn} is a sequence of nonzero complex numbers such tha.t Zn - 00 as n - 00.
Theorem 13.5.1. Suppose that {ck(r)} = {ck(r; Z : a)}, k = 0, ±1, ±2, ... , is a sequence of Fourier coefficients associated with Z such that for each r > 0, E !ck(r)1 2 < 00. Then there exists a unique entire function f with Z(J) = Z, f(O) = 1, and ck(r, f) = ck(r) for k = 0, ±1, ±2, .... Proof. We define 00
~(pe''P)
L
=
.----' ck(p)e,k'P.
k=-oo
Since E ICk(p) 12 < 00, this defines ~(pei'P) as an element of L2[_1r, 7r] for each p > 0 by the Riesz-Fischer Theorem. For p > 0, we define the following functions:
(
- z) ) -Izn-I p(zn 2 ) Zn P - Zn Z
(13.5.1)
Bp z; Zn =
(13.5.2)
Pp(Z) =
IT
Bp(z; Zn),
'''nl~P
(13.5.3)
(13.5.4)
K(w;z)
w+Z = --, w-z
Qp(Z) =exp{-21, { 7rt
J,wl=p
K(W'Z)~(W)dW}, w
(13.5.5)
We make the following assertions:
(13.5.6) The function /p is holomorphic in the disc {z : Izi < p}, and its zeros there are those Zn in Z that lie in this disc.
Entire and Meromorphic Functions
72
(13.5.7)
If r < p, then cle(r, f)
(13.5.8)
= cle(r).
Now (13.5.6) is clear from the definition of fp. Also,
However,
1. Qp(0)=exp{-2 11'1
~(W)dW}=exP{Co(P)}= IzII -,PI' .. l:5p
f
J1wJ=p
W
Zn
and it follows that fp(O) = 1. To prove (13.5.8), we see by 13.4.1 that it is enough to show that, near Z = 0, 00
logfp(z) = ~alezle, 1e=1
= {a,,}
where the ale are such that a near z = 0,
and
Ck ( r)
= c" (rj Z : a). That is,
(13.5.9)
We now make this computation. First we have that B~(z; zn)
Bp(z; zn)
'Zn/ 2 -
= (zn =
p2
Z)(p2 -
Znz)
=
f (in)" zk-I - f 1e=1
p2
2n 1 p2 - 2nz - Zn -
Z
(z~)1e Zle-l.
1e=1
Thus,
p;,(z) _ ~ u Ie-I P (z) - L...J le,pZ P
where UIe,p =
2: (~)
Iz.. l:S;p
=
near
Z
= 0,
1e=1
For k 0,1,2, ... , we write W definition of CIe(p) we see that
Ie -
2: (Z~)
Ie
Iz.. l:S;p
= peirp and cle(p)ei1cIP = O1cWk. 00
= N(p,Z)
Then by the
13. A Fourier Series Method
and
Then 00
= N(p,Z) + ~:
4>(w)
1e=1
N(p, Z) + so that
-1.
1.
211"1 Iwl=p
t. {
0,•.'+ fi."..
dw = -2.
~(w)K~(w,z)W
where
1.
a
1.
wle dw= k z1e-1 211'i Iwl=p (w - Z)2
and
4>(w)
2
dw,
2w
w-z )2'
uZ
-1
1
211't Iwl=p (w - z)
K~(w,z) = "$lK(w,z) = (
But
(~)
~ f (.!.)1c (w -1 z)2 211'i J1wl=p w
for
k
= 0,1,2, ... ,
dw=O for k=O,1,2, ....
Hence, Q~(z) _ ~ V; Ie-I , Q ( ) - L..J le,pZ p Z
Ie=!
where
Hence, near z = 0 we have /~(Z) _ P,,(z) /p(z) - Pp(z)
Q~(z) _ ~ Ie-! - ~ kale z ,
+ Qp(z)
and (13.5.9) is proved It next follows from (13.5.6)-(13.5.8) that
(13.5.10)
if p' > p, then
/pl is an analytic continuation of /p
Entire and Meromorphic Functions
74
For if we define for
Izl < p J;(z)
F(z)
= Jp(z)'
then Ck(r, F)
= ck(r, Jp') -
ck(r, Jp) = ck(r) - ck(r)
=0
for 0 $ r < p, and therefore IF(z)1 = 1. On the other hand, F(O) = 1, and it follows that F is the constant function 1. We now define the function J of Theorem 13.5.1 by setting J(z) = Jp(z) if p > 14 It is clear that J is entire and. by (13.5.6). that Z(f) = Z. Also, J(O) 1, and c,,(r, J) ck(r, J p ) for p > r, so that ck(r, J) c,,(r). An argument analogous to the one used in proving (13.5.10) proves that J is unique, and the proof of the theorem is complete. We now characterize the zero sets of entire functions of finite A-type.
=
=
=
Theorem 13.5.2. A necessary and sufficient condition that the sequence Z be the precise sequence oj zeros oj an entire Junction J oj finite A-type is that Z be A-admissible in the sense oj Definition 13.1.15, that is, that Z have finite A-density and be A-balanced. ProoJ. H Z = Z(f) for some J E AE, then by Theorem 13.4.6 the sequence {ck(r, is a A-admissible sequence of Fourier coefficients associated with Z and thus Z is A-admissible by Proposition 13.2.5. Conversely, suppose that Z is A-admissible. Then by Proposition 13.2.5 there exists a A-admissible sequence {ck(rn associated with Z. Then by Theorem 13.5.1 there exists an entire function J with Z = Z(f) and {ck(r,J)} = {ck(r)}. Then by Theorem 14.4.7 and the fact that {ck(r)} is A-admissible, it follows that J E As, and the proof is complete.
In
Remark. This theorem generalizes a well-known result of Lindelof [20J, which may be stated as follows.
Theorem 13.5.3. Let Z be a sequence oj compex numbers, and let p > 0 be given. IJ p is not an integer, then in order that there exist an entire function of growth at most order p, finite-type, it is necessary and sufficient that there exist a constant A such that nCr, Z) $ ArP • If p is an integer. it is necessary and sufficient that both this and the following condition be satisfied for some constant B:
L (-l)P IZnl$r
$B.
z"
This result follows immediately from Theorem 13.5.2 and the characterization of rP-admissible sequences given in Proposition 13.3.3. Our result shows that, in general, the angular distribution of the sequence of zeroS
13. A Fourier Series Method
75
of 8. function, and not only its density, is involved in an essential way in determining the rate of growth of the function. We turn now to the second problem of this section, that of determining when A is the field of quotients of the ring A B • We first prove the following
result. Theorem 13.5.4. In order that a sequence Z of complex numbers be the precise sequence of zeros of a meromorphic function of finite >"-type, it is necessary and sufficient that Z have finite A-density.
Proof. The necessity follows immediately from the fact that if J is a meromorphic function, then N(r,J) ::5 T(r,J). For the sufficiency, we remark first that the method used in proving Theorem 13.5.1 can be used to construct suitable meromorphic functions. Indeed, suppose that we are given two disjoint sequences Z, W of nonzero complex numbers with no finite Omit point and constants /'k, k = 1,2,3, ... , such that the coefficients defined by CtJ(r) =N(r, Z) - N(r, W), rll:
ck(r)
="2hk + S(r;k: Z) -
S(rjk: W)}
1 - 2{S'(r;k: Z) - S'(r;k: W)
c-k(r) =(ck(r» satisfy
(k
= 1,2,3, ... )
(k = 1,2,3, ... )
E ICk(r)12 < 00 for every r > O. Dp(z; wn }
=
Then, by defining
IT
Bp(z; wn )
IWnl~P
and
I. ( ) = Pp(z)Qp(z) pZ
Dp(z)
,
one can show, as in Theorem 13.5.1, that the meromorphic function defined by J(z) = Jp(z) for p > Izl has zero sequence Z, pole sequence W, and Fourier coefficients {ck(r)}. It is therefore enough to prove that, given a sequence Z of finite >..-density, there exist a disjoint sequence W of finite A-density and constants 1k, k = 1,2,3, ... , such that the ck(r) satisfy 1~(r)1 :5 AA(Br) for some constants A, B and all r > O. For then, by the first part of the proof of Theorem 13.4.5, the ck{r) must satisfy the stronger inequality
A').{B'r)
ICk{r)l:5 Ikl + 1
(r > 0)
filr SOme constants A', B', so that the function f synthesized from the Ck{ r) lIlust be of finite A-type by Theorem 13.4.5.
Entire and Meromorphic Functions
76
Supposing now that Z = {zn} has finite A-density, we define W = {w n } by Wn = Zn + En, n = 1,2,3, ... , where the En are small complex numbers so chosen that Iwnl = IZnl, n = 1,2,3, ... , all of the numbers Wn and Zk are different, and such that
Then, N(r, W) = N(r, Z) so that W has finite A-density. Hence,
IS'(rj k : Z)I
= k-10(-\(O(r)))
and
I$(rj k: W)I = k-10(-\(O(r»)
k = 1,2,3, ....
We define
It remains to prove that rk
"21'Yk + S(rj k : Z) - S(rjk: W)I uniformly for k rk
= 1,2,3, ....
Now
"21'Yk + S(rj k : Z) - S(rj k : W)I
= O(-\(O(r}))
13. A Fourier Series Method
77
Theorem 13.5.5. The field A of all meromorphic functions of finite >.type is the field of quotients of the rings AE of all entire functions of finite >.-type if and only if >. is regular in the sense of Definition 13.3.2, that is, if and only if every sequence of finite >.-density is >'-balanceable.
Proof. First, suppose that>. is regular and that f E A. Then Z(f) has finite >.-density by Theorem 13.5.3. There then exists a sequence Z' 2 Z(f) such that Z' is >'-admissible. [We may suppose, by the remarks preceding the proof of Theorem 13.4.5, that f(O) f 0, 00]. Then, by Theorem 13.5.2, there exists a function 9 E AE such that Z(g) = ZI. Since we have then tbat Z(g) ~ Z(f), the function h = g/I is entire. However, T(r, h)
~ T(r, g) + T (r,
7) = T(r,g) +T(r,f)
by (13.4.1) and (13.4.2), so that h E AE, and representation.
f
-log 1/(0)1
= g/h
is the desired I
Conversely, suppose that A = AE/ AE. Let Z have finite >.-density. \ Then, by Theorem 13.5.3, there exists a function f E A with Z(f) = Z. We write 1= g/h with g, h E AE • Then Z(g) is >.-admissible, and Z(g) 2 Z(J) = Z, and we have proved that>. is regular.
14 The Miles-Rubel-Taylor Theorem on Quotient Representations of Meromorphic Functions
Let f be a meromorphic function. In this chapter we describe the work of Joseph Miles, which completes the work in the last chapter concerning representations of f as the quotient of entire functions with small Nevanlinna characteristic. Miles showed that every set Z of finite A-density is A-baIanceable. As a consequence of this and the work of Rubel and Taylor in the last chapter, there exist absolute constants A and B such that if I is any meromorphic function in the plane, then f can be expressed as It! 12, where II and 12 are entire functions such that T(r, Ii) ~ AT(Br, f) for i = 1, 2 and r > O. It is implicit in the method of proof that for any B> 1 there is a corresponding A for which the desired representation holds for all I. Miles' proof is ingenious, intricate, and deep. Miles also showed that, in general, B cannot be chosen to be 1 by giving an example of a meromorphic I such that if I = It! 12, where II and 12 are entire, then T(r, h) ¥- O(T(r, J)). We do not give this example here. In the previous chapter, namely in Propositions 13.3.5 and 13.3.6 using Theorem 13.5.2, we have obtained the above theorem for special classes of entire functions. Results in this direction for functions of several complex variables appear in [16], [17], and [10). Quotient representations of functions meromorphic in thf; unit disk are discussed in [2]. The presentation below follows Mile [24]. We state the theorem.
Theorem. There exists absolute constants A and B such that il I is any merom orphic Iunction in the plane, then there exist entire junctions II and
14. The Miles-Rubel-Taylor Theorem on Quotient Representations
h
such that f
= hI h
and such that T(r, fi) ;5 AT(Br, f) for i
79
= 1, 2 and
r> O. Suppose Z = {zn} is a sequence of nonzero complex numbers with IZn I --+ We include the possibility that Zn = Zm for some n 1= m. As in the previous chapter, let 00.
(14.1)
n(r,Z) =
L
1
Iz"ISr
and N(r, Z) =
(14.2)
r n(t, Z) dt.
Jo
t
It was shown in the previous chapter that the following lemma is suffif dent to establish the theorem. .
Lemma. Suppose Z = {zn} is a sequence of nonzero complf!$ numbers with IZnI --+ 00. If A(r) = max(l, N(r, Z», then there exist absolute con.tants A' and B' and a sequence Z = {in} containing Z (with due regard to multiplicities) such that (i) N(r, Z) ;5 5A(4r) r > 0 and, (ii) for j = 1,2,3, . .. and 8 > r > 0,
.J r
.4'A(B'r) ;5
ri
+
A'A(B's) 8i
.
The argument of the last chapter which shows that this lemma is sufficient to prove the theorem may be summarized as follows. Without loss of generality we may assume f has infinitely many poles and that f(O) 1= 00. Let Z be the sequence of poles of f. Recall from the last chapter that condition (i) of the lemma says that Z has finite density with respect to the growth function A(r) and condition (ii) says that Z is balanced with reapect to the growth function A(r). Let AI(r) denote an arbitrary increasing lUlbounded function defined on (0,00). In Theorem 13.5.2 we characterized the zero sets of entire functions 4> such that T(r,4» ;5 aAl(!3r) for some constants a and !3 and all r > 0 18 those sets Z· which both have finite density and are balanced with respect to AI(r). This characterization combined with the above lemma guarantees the existence of constants Al and B and of an entire function 12 having zeros precisely on the set Z (counting multiplicities) such that T(r,f2):5 AIAI(Br) for all r > O. Hence,
(14.3)
Entire and Meromorphic FUnctions
80
for r > ro(f). Letting It = hf, we see that h is entire and that T(r, It) :5 (Al + I)T(Br, f) for r > ro(f). For an appropriate complex constant c, o < Icl < 1, we have for i = 1, 2 that
T(r,cfi)
(14.4)
=0
r
< ro(f)
and (14.5)
=
=
Letting A Al +1, we see that f cltlch is the desired representation. It is implicilin the methods of the last chapter and in the proof of the above lemma that A and B are absolute constants and that to each B > 1 there corresponds an A for which the representation holds for all f. We now prove the lemma. For each integer N we let
If ZN =F ", we relabel the elements of ZN as simply ZII Z2, •.• ,Zk, with each number being listed in this sequence as often as it appears in Z. For 1:5 n:5 k we define Pn E (0,1] and On E (0,211"], SO that Zn = 2N+p"ei6n. We do not indicate in the notation the obvious dependence of k, Z," Pn, and On on N. We let (14.6)
and (14.7)
Clearly,
(14.8)
IhN(O)1 $ 2
t. {t.
2-;('+"")}
< (n(2 N +l, Z) - n(2 N , Z». Letting
we have (14.10)
14. The Miles-Rubel-Taylor Theorem on Quotient Representations
We now define sets ZN for all integers N. If Z N (14.10) that
=1=
81
(14.11)
If [ ] denotes the greatest integer function and LN = [2~ I;1r fN«()) d()], we define for 0 ~ n ~ LN a monotone increasing sequence ~ in [O,2?r] by choosing fJ'n to satisfy
-1 19~ IN«() dO = n.
(14.12)
2?r
0
In addition, we let fJ'LN+l = 2?r. For 0 ~ n :5 LN, we let ~ = 2N-l~ill~ and ZN = {2N-lei6~ : 0 :5 n :5 LN}' AB before, we do not indicate the dependence of e:a and z'n on N. IT ZN = 0, we let ZN = 0 and for that value of N do not define L N or numbers ~ and z~. We let Z' = UNZNand Z = Z U Z'. From (14.11) and the definition of Z it follows that nCr, Z') :5 4n(4r, Z) and hence that nCr, Z) :5 5n(4r, Z) for all r > O. From this fact it is immediate that Z satisfies condition (i) of the lemma. We now consider a positive integer j and a value of N for which Z N =1= 4J. Let ZN = {ZllZ2, ... ,Zk}' From this point until inequality (14.27), we regard j and N as fixed. Although many of the quantities to be defined (S,T,P,Uo,Ue , Va, and Ve ) depend on both j and N, for simplicity we suppress this dependence from the notation. A key step in showing Z satisfies condition (ii) of the lemma is to establish (14.13)
We first observe that (14.14)
L (l)j - +~ f 2?r io k
n=l
Zn
2 ..
e- ij6 2- j (N-l) fN(O) dO
Entire and Meromorphic Functions
82
Thus (14.13) is equivalent to (14.15)
L (: Iz~I=2N-J
n
r-
2~ 121f e- ij6 2- j (N-l) !N«()) d()
< 48j2- j (N-l).
0
That the quantity on the left side of (14.15) is small can be seen intuitively from the observation that the sum is essentially an approximating lliemannn sum for the integral. We first estimate
R{ (14.16)
I
21f L (z~)j - 2~ r e- ij6 2- (N-l) !N(O) dO} ;;' Jo j
I
1-2N-l
n
n -
cosj~ - 2~ 121f !N«())cosjO dO
L
= 2-j (N-l)
Iz:'I=2 N -
1
.
0
[j'lf ,(m;l)"] for m even, 0:::; m :::; 2j - 1, and increases from -1 to Ion [j'lf ,(m;1)'If] for m odd, o :::; m :::; 2j - 1. Let 1m = (j" ,(m;1),..) for 0 :::; m :::; 2j - 1. Define S The function cosjO decreases from 1 to -Ion
to be the set of all n, 0 :::; n :::; L N , such that [~, ~+1] is not contained entirely in some 1m , 0 :::; m :::; 2j - 1. Since each point m1r Ii belongs to [If.., If..+1] for at most two values of n, we see that S has at most 4j elements. Let T be the set of all n, 0:::; n:::; LN, such that either n or n-l belongs to S. Thus T has at most 8j elements. Let P = {O, 1, ... , LN} - T. If n E P, it follows that [If..+1' If..] and [()~, 9'n+1] belong to the same interval 1m for some m, 0 :::; m :::; 2j - 1. Let Uo be the set of all n E P such that [O~+1' 9~] and [O~, If..+1] are contained in 1m for some odd m, and let Ue be the set of all n-l such that n E P and the intervals [()'n-l'~] and [()~, If..+Jl are contained in 1m for some even m. Certainly uonue = 0, for ifn E uonUe , then [If..'~+11 is contained in 1m for both an odd value of m and an even value of m. Letting U = UouUe , we conclude that U and P have the same number of elements. Since 0 ¢ P, we see that -1 ¢ Ue and hence U c {O, 1, ... ,LN}' Thus, {O, 1, ... , LN} - U has at most 8j elements. If n E P is such that u E Uo , then cos j9 is increasing on [If.., ()~+ll and hence, from (14.12), (14.17)
cosj~ :::;
21 16~+J !N(O)cosj() dO. 11"
6~
If n E P is such that n -1 E Ue , then cosjO is decreasing on [O~+l' ~l and thus
(14.18)
cosj~
:::; - 1 i6~ fN(O) cosjO dO. 211"
6:'_1
14. The Miles-Rubel-Taylor Theorem on Quotient Representations
83
The definition of U together with (14.17) and (14.18) implies
(14.19)
~
8j + 8j
= 16;.
We now obtain a lower bound for (14.21)
Let Ve be the set of all n E P such that [9'n-l,o:.] and [0:.,0:.+1] are contained in 1m for some even m, and let Vo be the set of all n - 1 such that n E P and the intervals [O~_l' O~] and [0:.,0:.+1] are contained in 1m for BOrne odd m. Using the same reasoning as before, we see that Ve n Vo = 0 and hence, if V = Ve U Vo, then {O, 1, ... , L N } - V has at most 8j elements. If n E P is such that nEVe, then cosiO is decreasing on [IY,., O~+1] and (14.22)
1
cosifJ'.. ~ -2 11"
18~+1 8~
fNUJ) cosiO dO.
Ifn E P is such that n-l E Vo , then cosiO is increasing on [9'n-l'lY,.] and (14.23)
cosjO~ ~
21 i8~ 11"
fN(O) cosjO dO.
8:'_1
Prom (14.22), (14.23), and the definition of V, (14.24)
Entire and Meromorphic Functions
84
Thus (14.25)
E cosj8~ - 2~ 10 Iz;.I=2 N -
1
2
fN(O)coSjO dO
1<
0
= E cosjO~ + 2: cosjO~ nET
nEP
1 111~+1 1 111~+1 -2:fN(O)cosjOdO fN«(J)cosjOdO- E 211" 211" II'n
nEV
~
E cosj~ - E nET
~
O
I 18~+1
211"
,
O
II'n
!N(O) cosj(J dO
8n
-8j - 8j = -16j.
Combining (14.16), (14.20), and (14.25), we conclude that (14.26)
L (Z~) i - 2~ 102e-
:R {
11:
Iz;.I=2 N -
1
n
ij8 Tj(N-l) !N(O)
d(J}
~ 16j2-
j (N-l).
0
The same discussion applies to the imaginary part of the above quantity. The only minor modification is that we must divide [0,211"] into 2j + 1 subintervals on which sinjO is alternately increasing and decreasing. Since 2j + 1 < 4j, this causes no difficulty. We obtain (14.27)
~ { L (z~)i - 2~ 1 e2
1£
Iz;.I=2 N -
1
n
ij6 2- j (N-l) !N(O)
d(J}
~ 32j2-
j (N-l).
0
Combining (14.26) and (14.27), we obtain (14.15), which in turn establishes (14.13). We are now in position to show that Z satisfies condition (il) of the lemma. Suppose that s ~ Sr. We then have trivially for all positive integers j
-1.
( -1 ~ L... -, J r
(14.28)
<
-
n{8r, Z) jr)
<
-
)j
N{Ser, Z)
.
jr]
<
-
5'x(32er) .. jr]
Suppose that 8r < s. In this case there exist integers q2 - 3, such that (14.29)
ql
and
Q2, Ql ~
14. The Miles-Rubel-Taylor Theorem on Quotient Representations
85
= 1, 2, 3, ... ,
Thus, for j
(14.30)
-j1 r
=
=
We remark that if Zql+" 0 (and hence Z~I+" = 0) for some integer v, then the terms corresponding to that value of 1/ in (14.30) are of course omitted. For any positive integer j we have
~
!
. L.J J r
(14.31)
<
(~)j z..
N(4er,Z)
.'
-
$
-
JrJ
n(4r,Z) jr;
A(4er)
<-.-.-. J Jr
-
From (14.13) we have
92fl ~ (14.32)
(~)j + ~
L
J 2'1+"
.,=2
Zn
'l2-'11- 1
$ 48
L
2- j (ql+.,-I).
Also from (14.13) we have (14.33)
~
L 292 <1"'nl:5a
(~)j + ~ Zn
L
J 1"':'1=2'2- 1
(~)i Z~
L
J 1%!-1=2'1+.. - l
(~)j Zn
Entire and Meromorphic Functions
86
Finally, (14.34)
(~\; =
; L J
I%~ 1=2
q2
Zn)
Combining (14.30) through (14.34), we have
;.J;,. (~y (
14.35
)
$
A()
~e.r
+ (48)
96
< -.-.-+ .( +1) + Jr1 21 Ql
-
2-i (Q,+v-l)
(
+ 2A ~e.8
,,=2
Jr1
A(4er)
L
Q2-q, +1
)
Js1
2A(4es)
.. Js1
A(4er)
96A(r)
< -.-.-+ --.-+ Jr1 r1
-
2A(4es)
.
js1
From (14.28) and (14.35), we see that
(14.36)
< A'A(B'r) A'A(B's) .J r
!
for all positive integers j and all 0 < r completes the proof of the lemma.
< s if A' = 100 and B' = 32e. This
15 Canonical Products
=
We shall suppose that a countable set Z {z..} of complex numbers is given. For convenience, we shall suppose that f/; Z and that Z has no finite limit points. More generally, we consider "sets with multiplicity." This means that some of the Zn may be counted multiple times. It is poesible to make this notion rigorous, but at the price of clumsier notation. For convenience in this chapter, we shall exclude the null function F(z) = 0 from consideration.
°
Definition. H F is an entire function, we write Z(f) for the set of zeros (other than the origin) of F. Definition. H p is a nonnegative integer, we define E(u,p), the Weier.tnu.. primary factor of order p, by E(u,p)
u2 + ... + pUP) = (1- u)exp (u + 2'"
E(u,O) = 1- u.
Lemma 15.1. Given any Z, there exist nonnegative integers An such that the product
f(Z)=gE(~
'An)
converyes uniformly on compact sets to an entire function f such that Z(n Z.
=
We omit the easy proof which follows directly from Estimate A later in An such that An -. 00 will work. The next theorem is a corollary of Lemma 15.1.
this chapter. Any sequence
Entire and Meromorpbic Functions
88
Weierstrass Factorization Theorem. Given an entire function F, there exist nonnegative integers An, a nonnegative integer m, and an entire function g such that
= zm exp(g(z»
F(z)
11
E
(z: 'An).
Definition. Given any Z, the convergence exponent Pl (Z) is defined by
Lemma 15.2.
If E Iz~I"
< 00, then n(t) = o(tO), where n(t) = E
l.
l%,ol$t
Proof. Choose a large number a, and write ""
_1_
> n(t) - n(a) .
1 1° a$lz"l$t Zn L.J
Lemma 15.3.
E Iz:I"
Proof. Write
to
< 00 if and only if fooo ;:.CR dt < 00. 1 r E IZnlo = 10 rO dn(t) Iz.. l
and integrate by parts to prove the lemma. Corollary. Pl(Z) Lemma 15.4.
= inf{a: n(t) = O(tO)}.
Pl(f)
~
p(f).
Proof We must show that if P = p(f), then for each € > 0, n(t) = O(tP+£). But by the Nevanlinna first fundamental theorem, we know that N(t)
=
t n(s)s ds = O(tP+E).
10
a) log 2, by the usual argument, and the result follows. The genus of z, p = p(Z), is defined by p(Z) = inf{q : q is
But N(t) ~ n Definition. an integer,
E~ < 00 }.
Definition. The canonical product of genus p over Z is defined by
15. Canonical Products
89
Definition. The canonical product Pz over Z is defined by
II E (~ ,p),
Pz(z) =
z.,EZ
where p = p(Z).
Theorem 15.5.
The order of Pz is Pl(Z).
The next result is a corollary of Theorem 15.5.
The Hadamard Factorization Theorem. Gi'IJen an entire function f, fez) = Pz(z)zmexpQ(z),
where m is a nonnegati'IJe integer and Q is a polynomial of degree ::; p(f).
Here, Z = Z(f).
Definition. The genus of the entire function f is defined by P(J)
= max(n,p),
where n = degQ.
n
Examples. Suppose Zn = n 2 , n = 1,2,3, .... H I(z) = (1- ~), then I is of genus O. H I(z) = eZ (1- ~), then I is of genus 1.
n
We first shall show how the Hadamard Factorization Theorem follows from Theorem 15.5. Without loss of generality, suppose 1(0) :F O. In fact, 888UIne 1(0) = 1. Let g(z)
fez) = Pz(z)'
Then 9 is an entire function without zeros, so that there exists an entire function Q such that
I
Pz =expQ. We must prove that Q is a polynomial of degree ::; p(f).
Lemma 15.6. IfF and G are meromorphic, thenp (~) ::; max{p(F),p(G)}
Prool. We have 10gT(r, p (/) = lim sup r-oo
f)
logr
T(r,j) =T(r,f) +0(1) T(r, /g) = T(r, f)
+ T(r,g) + 0(1),
Entire and Meromorphic Functions
90
from which the lemma is easily proved.
It then follows that exp Q is an entire function of order at most p. Writing Q = 1£ + iv, we have
for each
tI > p.
Izl =
Since, with
rand R = 2r, we have
1 Q(z) - 1m Q(O) = ~
1,
~1rt
w+zdw u(w)---,
iwj=R
'W -
Z 'W
it follows that
IQ(z) -
1m Q(O)I
= O(rP').
Hence
IQ(z)1 = O(rP'), and it follows that Q is a polynomial of degree :$ p.
Proolol Theorem 15.5. Our proof uses the Fourier series method of Chapter 13. It is shorter and less tedious than the standard proofs. Estimate A. H
Let
1£ =
a:. Now
1z..1 2:: 2Iz/, then
10gE(u,p)
u:. Here 11£1
= - E~l
00
00
p+l
0
:$
~. Hence
Ilog E(u,p)1 :$ L 11£1'" :$ lulP+l L 2-'" = 21u1 P+l. It follows from Estimate A that the product
Pz(z) = lIE
(~
,p)
converges uniformly on compact sets to an entire function 10gl/(z)1
= Elog IE (:.. ,p) I.
We write 00
log I/(z)1 =
E m=-oo
em eimfJ ,
I,
so that
15. Canonical Products
91
where c.,. = c.,.(r) is the mth Fourier coefficient of log If I· We have seen in Lemma 13.4.1 how to calculate c.,.: logE
(~, Zn
80
that logf(z)
=
p) = _!kk f: k=p+l
Hence 1
Zn
f: (- f: ~ (~)k) .
n=O
-an = 2
(~)k
=p+l
{O 1 ~ I - 2I L..J %!"
fork~p
~
lor
Recall the notation from Chapter 13; near
Z
k
~p
+ 1.
=0
Lakzk. 00
logf(z)
=
k=O
, Using the formulas in (13.4.18), we therefore get
(i) CiJ
=
L
log..!:..
1.a"I~r
(ii)c.,.=2~
rn
L (~)m -2~ L (~)m ifm~p ~"I~r
~"I~r
(iii) c.,. = _rm'" (.!.)m _2.. '" (Zn)m 2m L..J 2m L..J r
if m
Zn
1.a.. I>r 1.a"I~r Choose p > p. We must prove that for some M r P'
1c.,.(r)1 ~ M Iml + 1
for
m=
~ p + 1.
= M (p)
0, 1,2, ....
But this estimate is easily derived, as in Chapter 13, from the fact that nCr) = O(r P'}, once we know that the c.,. are given by formulas (i), (li), and (iii). 0 The next theorem follows easily from the Miles-Rubel-Taylor Theorem of Chapter 14 but is included here due to its simple deduction from Theorem 15.5. Theorem. If f is a meromorphic function in the plane with p = p(f) < :5 p and p(h) :5 p, lUch that f = g/h.
00, then there exist entire /unctions 9 and h, with p(g}
Proof. Let W = {Wn } be the poles of I, let PI = PI(W) be the exponent of convergence of W, and let p = peW) be the genus of W. Since N(r, f) :5
Entire and Meromorphic Functions
92
T(r, f) = O(rP+E ) for each ( > 0, we have PI ::; p. By Theorem 15.5, p( Pw) = Pl· We now let 9 = 1 . Pw and observe that 9 is entire. Since p(g) ::; max{p(Pw), pU)} = p, it follows that 1 = g/ Pw is the desired representation. It was proved by similar means by Rubel and Taylor that if A(2r)/A(r) =::: 0(1), then every meromorphic function of finite A-type is the quotient of two entire functions of finite A-type. However, the proof is long and is subsumed in Miles' proof of the last chapter, so we omit it.
Laguerre's Theorem on Separation Zeros. If 1 is a nonconstant entire function with only real zeros, has genus 0 or 1, and is real on the real axis, then the zeros of l' are real and are separated by the zeros of 1, and the zeros of 1 are sepamted by the zeros of f'. Proof. We have either
or
J(z)
(1- ~) ezl"n,
= CzKeazII
where the Zn are real, and c and a are real (possibly a = 0). It follows that either
or
1'(z) l(z) On writing z
k
= -; + a +
L
z Zn(z - Zn)"
= x + iy and recalling that a is real, we get, in both cases,
1m f'(z) _ _ { k J(z) - Y x 2 + y2
+~
1
L.J (x - zn)2 + y2
} '
which does not vanish except for y = 0, so that the zeros of I' are real. Since J is real for real z, the theorems of calculus apply. By Rolle's Theorem. there is a zero of f' between two consecutive zeros of 1. so that the zeros of 1 are certainly separated by the zeros of 1'. To see that the zeros of l' are separated by the zeros of J, note that
(7f')
(x)
= - xk2 -
L
1 (x _ x n )2
< 0,
so that ~gl is decreasing in any interval free of zeros of J, so that l' cannot have two zeros in any such interval. The case of repeated roots is handled by a suitable convention.
16 'Formal Power Series
We consider the formal power series
I(z) = ao + atZ + a2 z2 + ... )
which we usually normalize by ao
= o.
Let In(z : a) = a + alZ + a2z2 + ... + BnZnj In is a polynomial of "degree" n (possibly an = 0). We adopt the convention that In has n zeros %1, Z2, ... , Zn, where, if am =I- 0 but Gm+l = am+2 = ... = an = 0, then Zm+l = Zm+2 = ... = Zn = 00. For later use, we shall make the convention
00/00
= 1.
Let rn(a) = max{lzi : In(z : a) = O}, that is, rn(a) is the modulus of the largest root of In(z : a) = O. Note that if Bn = 0, Tn = 00. In a certain Bense, rn(a) measures the disaffinity of I for the value -a.
Theorem 16.1.
Given any lormal power series I, then
.
(16.1)
rn(a)
lim sUP-(b) :$ 2,
(b # O)j
n-oo rn
ilb = 0, we have r (a)
[
1 ]
limsup~() :$ 1 + lim sup -(-) n-oo rn 0 n-oo Tn a
~,
Where
I(z) =
amZ m
+ am +lZm+ 1 + ... ,Gm # O.
Note that if the roles of a and b are interchanged, then (16.1) yields ! < liminfr,,(G). 2 -
n-oo .... (b)
Entire and Meromorphic Functions
94
Proof. Let 0.,a2,·.·,a n be the zeros of In(z: a) arranged so that la11:s 10'21::; ... ::; lanl· Let f31,{h, ... ,f3n be the zeros of /n(z: b) arranged so that 11311 ::; 11321 ::; ... ::; If3nl· Thus In(z : a) = anll(z - ak) and fn(z : b) = a.ll ll(z - 13k), where we shall take an ':f o. Now,
Suppose there exists a sequence of n for which {an}, {f3n} is such that lanl = -Xnlf3nl with ~n ~ ~ > 1. Otherwise, the conclusion of the theorem is clearly true. Therefore, rn(a) = ~nrn(b). Now,
Letting z =
an in (16.2) gives b -
a = anll(an - 13k). Hence,
Dividing through by b and taking nth roots gives
Now suppose we had
But nlPkl =
~n
- 1 > 1. Then
T£!r, so that supposing ~n -I> 1 implies 11 - il;' ~ (1 + f)j
a contradiction. Hence, limsuP~n ::; 2, so that we have limsup~:~:~ ::; 2. n-oo
Suppose now that b = O. Write I(z)
= amzm+am+lZm+l+ ... ,am ':f O.
n-m
We now have fn(z: b)
-a = an[zm
= zm an IT
n-m
IT
n
(z - Pk) -
k=l
k=1
IT (z -
(z - Pk). Thus, In(z : b) - fn(z : a) == ak)].
k=1
We again have IOn-Pkl
n-m
= (~n-l}IPkl and I-al = lana~ IT
k=1
n-m lanl[rn(a)Jm
IT
(An - 1}IPkl:
k=1
(16.3)
lal~ = [Ian I[rn (a)]m g(~n
_
l)IPkl]
--Ln-m
(Qn-f3k}1 ==
16. Formal Power Series
95
Now suppose for contradiction that
An _
1> lal(t2m)2 (_I_)n-m (_1_) ~. laml rn(a)
Then there exist {En} such that En > 0 and
An
-1 = lal(;;~;;d (_I_)n-m (_1_) ~ (1 + En). laml rn(a) nif I/h:1 = l;al gives
Using this expression in (16.3) and that
11:=1
..
a contradiction. Hence,
Consequently, lim sup
,,-00
r (a) ( 1 )~ n (0) ~ 1 + lim sup - (-) , rn n-+oo r" a
which completes the proof. Theorem 16.1 has the following corollary.
Corollary. If f is holomorphic in a neighborhood of 0 (that is, f has a po.iti"e raditu of con"ergence), then
rn(a) < 2 lim sup-n-+oo rn(b) -
lor all b. Definition. T,. (f)
= -log r n (1) is called the discrete characteristic of I.
Theorem (First Fundamental Theorem) 16.2. Tn(f) = Tn(f -a)+O(I) with at most one exception.
lor all a
The proof is immediate from the definition of T,.(f) and Theorem 16.1. Now let us examine the case of our exception more closely,
(1)
. r,.(1) . limsuP-(O) ~ 1 + lim sup -(1) n-oo rn
n_oo
rn
-ft-m
.
Suppose In(z : 1) = 1 + alZ + ... + a"zn has zeros {3l, ... ,(3n arranged in order of increasing moduli. Thus, anIle -13k) = 1 and hence I1I{3kl = Accordingly, [rn(I»)n ~ and therefore
Jtr.
Ttl,
[rn(l»)~ ~
( 1)*~ or (1)~ $ (lanl"~)-lanl
which gives us the next result.
rn(l)
n-m,
Entire and Meromorphic Functions
96
Corollary. If f is holomofJJhic, then T,,(f)
= Tn(J -
a) + 0(1) for all a.
Theorem (Kakeya) 16.3. Let f be a formal power series not identically zero and R(f) its mdius of convergence. Then R(f) $ lim inf r n $ 2R(f).
"-00
Proof. Choose R' < R(J). The In have only a finite number of zeros in the disk DR' and {fn(z)} --+ {f(z)} uniformly in DR" Hence, past a certain no we have, by Hurwitz's Theorem, that the functions fn have the same number of zeros in DR' as I does. Therefore, the largest zero of In must leave D R'. Thus R(f) $ lim inf r", as desired.
"-00
Alternatively, we have seen that r" 2: lanl-~ and hence liminf rn > R(f). To prove the second inequality in the theorem, we use the following result. Q-fo(XJ
Lemma. Let P(z)
= bo + bIZ + ... + bnzn = O.
-
Then
Proof of Lemma. Clearly, Ibnllzln $1601 + lbol + Ib1llzl + ... + Ibn_1Ilzl n- 1 •
Il!l + Il!-l r + ... + Il!l Suppose now that r > 2 max (lb-'::ll, I 'b: Ii ' ... ,I~I~). Then rn $ 2- nr" + Writing Izi = r, we have
rn
$
r,,-l.
b
2-(n-l)r" + ... + 2- I r n = rn(l and the lemma is proved.
Now let bA: = akRk,P(z) the roots of fn(Rz) = 0 are
rn R $ 2 max
(I
2
+ i + ... + 2~) < r", which is impossible
= f,,(Rz) = 40 + aIRz + ... + a"R!'z".
Then
fi times the roots of fn(z) = O. By the lemma, a,.-2 on-II ' 1 Rna,.Rn
an-In. .
anR"
2
1
ao l i " .. , a,.R" I~) .
Choose R > R(J)j it follows that {lanIRn} is unbounded. Therefore, for 3 suitable subsequence, lanlRn ~ laklRA: for all k $ n. In that subsequence, 1t $ 2. Hence, liminf Tn :5 2R. Therefore, liminf Tn :5 2R(f), as was to ft-+oo be proved. ft~OO
Corollary (Okada [28]). A junction f is entire if and only if lim Tn(f) = n-oo
-00.
16. Formal Power Series
Leta = a
i:beorem (TSUJI.. [45J) 16 .4. ~n
U(f)
= p(f).
= limsup nlogt
Proof. We know that R(f)
97
from Proposition 11.4. Take
n-oo loylanl n~gn 1 ; > Pi then, for large n, pi > l ' Hence, lanl ~ 7r and thus log lanl n pi r: ~ rc6 ~ nn lp', or rn ~ n-:r. Consequently, I~O:::' ~ pi for large nj hence,
(I
~
P RoO; df>Sired.
In the other direction, suppose on the contrary that a < p. Choose tI such that a < tI < p. Then for n large (say n ~ no), rn ~ n-J>r. Choose . 2K ],I > 1 80 that for K = 1,2, ... , no, laKI ~ m--~-. We prove by (K!) I'
induction that for n
~ no, Ian I ~ m
2"..!..
(n!)'"
First,
lanlr: ~ lan_llr:- 1 + ... + lallrn + 1 BJnce In(z : 1) = 1 + aoz + ... + anz n = O. Hence,
Therefore,
IBnI < M {
2,,-1
-
n-J>r[(n-l)!]-J>r
+
2n -
n;-[(n-2)!]:r
2n- 2
2n-l
2'
+ ---",-2'l_T'"}
n7\1I);"
n,.(O!)?T
+ ... +
2
2'
20}
<M { - - + - - + ... + - - + - (n!);' (n!);" (n!)? (n!)?
< _ -M- { 1+2+4+···+2n - I} (n!)?
88
p,
2n (n!);"
~M--,
desired. This now leads to a contradiction of the fact that
~ we have ra:T 1 lor 2: log
1 (nl):r
M
1
2'"
Ian I ~
Co
nsequently,
constant
1
+ pi log n! -
n log 2
f
is of order
Entire and Meromorphic Functions
98
so that nlogn -<
log
nlogn
1i6 - constant +.;, log n! -
,
n log 2
since log n! '" n log n. This implies that p(f) ~ tion.
=p+o
(1)
tI < p = p(f); a contradic-
17 Picard's Theorem and the Second Fundamental Theorem
In this section we shall prove Picard's Theorem, state the second funda.mental theorem of Nevanlinna (leaving the proof for the next section), and derive some of its consequences. We shall use two conventions that will greatly simplify our notation. First, we shall always work "modulo 0(1)". For example, A = B and A :5 B shall mean that A - B is bounded and that A - B is bounded above, respectively. Second, we shall use a notation of Weyl, writing II in front of a statement to mean that the statement holds with the possible exception of a set of finite length.
Picard's Theorem. II f is a nonconstant meromorphic function on the complex plane, then f does not omit three values on the sphere. It may happen that I omits two values. For example, eZ omits 0 and 00. Our proof of Picard's Theorem is patterned after our proof of the second fundamental theorem of Nevanlinna, and illustrates the main features of the method in a simpler context.
Proof. Without loss of generality, we may assume that I omits 0, 1, and
so that f, j, and J~l are entire. The next lemma is referred to as the lemma of the logarithmic derivative.
00,
Lemma. liT (r, f) 00.
::; K'logr+K"log+T(r,f) lor f
omitting 0, 1, and
Entire and Meromorphic Functions
100
From Poisson's formula (7.1), log J(z) Since
J =1= 0 or
1 = -2 1f'
00,
1'"
_".
.
pei'P + z log IJ(pe''P) 1 . dcp;
pe''P -
Z
p> r
= Izi.
we may differentiate with respect to z to get
f'(z) 1 J(z) = 21f'
r
i
2pei'P
L". log IJ(pe 'P)I (pei'P _ z)2 dcp
I~ (p -2pr)2 ~ 1'" If'(z) J(z) 21f' ".
log IJ(pei'P)ldcp
IJ(z) I~ log+ p+ 21og+ P - r
log+ Jf(Z)
_1_
11'"
+ log+ 21f'
". log IJ(pei'P)ldcp (mod 0(1».
Recalling the definition of m(r, f), we find log+
I~(~} I~
log+ p + 2 log + P
~ r + log+ m(r,f) +log+ m (r, -1) .
(r, J)
Since J =1= 0 or 00, we have m(r,f) = T(r,f) and m = T (r, Moreover, by the first fundamental theorem of Nevanlinna, we have
T(r, f)
J).
= T (r, -1) .
Hence,
Since the right-hand side is independent of the argument of z, it follows that the same estimate will hold for the average,
m(r,~) ~log+p+2Iog+ p~r + 2Iog+T(p, f). Now choose p = r+ log+ ~(r,f) and use the Borel lemma on 2 log + T(p, J) to find,
II
T (r,
~)
:5 log+ r
+ 2 log + log+ T(r, f) +
:51og+ r
+ 41og+ T(r,f).
3log+ T(r, J)
17. Picard's Theorem and the Second Fundamental Theorem
This proves the lemma. Now let
1
F(z) = f(z)
101
1
+ f(z) -
1·
IF(z)1 is large whenever f is near 0 or f is near 1. Of course, the set of values of where these occur is disjoint: hence, m( r, F) ~ m( r, f) + m
(r, / . 1).
z
Therefore, (17.1)
Notice that 1 J(z} [f'(z) f'(Z)] F(z) = f(z) J'(z) J(z) + fez) - 1 .
Thus, (17.2)
T(r, F)
~ T (r, j) + T (r, J,) +T (r, ~) +T (r, J~ 1)'
From the first fundamental theorem we have
1
T(r,j) =T(r,f)=T(r,f- )=T(r, J~I) and
T(r,
i,) =T(r, ~).
Therefore, on combining (17.1) and (17.2) we obtain
2T(r,f)
~ T(r,f) + 2T (r, ~) + T (r, f ~ 1) .
Consequently, (17.3) The lemma above shows that
T (r,
~) ~ k'logr + k"logT(r, f).
Applying this lemma to J and f -1 in (17.3) gives T( r, f) ~ k' log r + kill log T( r, f).
or course f
is not a rational function since it omits three values. Consequently, by Theorem 10.2,
TI(r, f) ogr
But we have
-+ 00
as r
-+ 00.
T(r, f), II T(r,logrf) ::; k' + k",log+logr
which is impossible, and the proof by contradiction is complete.
Entire and Meromorphic Functions
102
Theorem. (Second Fundamental Theorem oj Nevanlinna). Suppose J(z) is a nonconstant meromorphic Junction in the plane (or in Izl < R, R < oo). Let at, a2, . .. , a q be q distinct finite complex numbers. Then mer, f)
(17.4)
+
tm
(r, J
~ aJ
5, 2T(r, f) - N(r, f)
+ s(r),
where Nt (r, f)
=N
and
(r, ;,)
+ 2N(r, J) -
N(r,l')
1') ( 71') + m (q r, ~ f _
S(r) = m r,
all
.
Interpretation. It will be shown that " S(r) = O(logT(r, J)
+ O(logr).
Accordingly, we may think of "S" as standing for "small." To interpret the function N1(r,f), suppose fez) = (z - Zo)Kg(z), K E Z, g(z) ¥ 0 in some neighborhood of zoo We analyze the contribution to Nl due to the behavior of f at Zo by examining the counting function nl(r, f) = n (r, + 2n(r, f) - nCr, IT K > 0, then f vanishes at Zo to order K. Consequently, nCr, f) and n(r,l') are unaffected by the behavior of f near ZOo The contribution to nl near z· equals the contribution to n (r, -} ). Of
1').
}I)
f.
course, has a pole of order K - 1 at Zoo IT K < 0, then f has a pole of order - K at Zo and a similar analysis reveals that nl "counts" this pole K - 1 times. The case K = 0 is trivial. In general we find that poles of order K and zeros of order m are "counted" by nl, respectively, K - 1 and m -1 times. The gist of the second fundamental theorem is that for most values of a, the contribution of m to T f~o.) is much smaller than the coD-
(r, /"0.)
(r,
(r, f~o.) to it. Thus, for most values of a, N (r, f~l) make!> the preponderant contribution to T (r, f~o.)' Recall that T (r, /0.) tribution of N
T(r,!)
= 0(1) by the first fundamental theorem .
Remarks. The following weaker form of the theorem also gives us the Picard Theorem: (17.5)
mer,!)
+
Em
v=1
(r,
f
~ av)
$ 2T(r,!) +S(r).
17. Picard's Theorem and the Second Fundamental Theorem
103
For suppose f omits three values, say 0, 1, and 00. Then, since f is entire, m( r, f) = T( r, f) and, since f omits 0 and 1, we have
and
~ 1) = T (r, f ~
m (r, f
J=
T(r,f -1)
= T(r,f).
Thus (17.5) implies that 3T(r,f) :s; 2T(r,f) + S(r) and hence T(r,J) S(r). By a lemma to be proved in the neA-t section we assert that
Hence,
~
II S(r) ~ 6(q+ l)log+T(r,J) +4(q+ 1) log+r
and thus
T(r,f)::; S(r) implies that lim TI(r,f) < 00, r-oo ogr which is only true, by Theorem 10.2, for rational functions. Since f =/: 00, is a polynomial and therefore does not omit three values; a contradiction.
f
Consequences Definition. Let
net, J)
be the number of poles of f in the closed disk Let
fit = {z E C : Izl ~ t} counted once, no matter what the multiplicity. N(r, J)
= J;+ ft(t.[)~fl(o·[)dt.
m(r,_1 ) f - a
N(r
_ 1)
'f - a T(r, f) T(r, f) 6(a) is called the deficiency or defect of the function for the value a. Definition. 6(a) = limr-oo
Definition. (}(a)
= li"' • ....-00
= 1 -limr_oo
[ ( 1) -( 1)] . Nr---Nr-'f - a 'f - a T(r,f)
Definition. The branching index is defined to be (}*(a) where
*
() (a) =
r,-r,--( [( 1)] . f 1) -a. f-a 1 - hIDr_oo T(r, f) lim.. _ oo 1 T(r, f) -.-
N
N
=
Entire and Meromorphic Functions
104
What contributes to O*(a) is a-points that are ta.ken on by I with multiplicity greater than or equal to 2. So a value a that is taken on often with high multiplicity will have a large branching index. Remark. O*(a) O*(a)
O(a) + 6(a). Then,
~
=
lim r-+oo
[1 _fl] = lim [N - fl + 1- N] T T ~ lim [N-N]+ lim [1-N] T T r-oo
r-+oo
= O(a)
r-+oo
+ 6(a).
The next result follows from the second fundamental theorem.
Theorem. Let {a v } be any finite collection 01 diatinct complex numbers po$$iblll including
00.
Then,
(17.6) v
Prool. First suppose I is not a rational function. From the second fundamental theorem we have m(r,f) + vtl m (r,
S(r). Adding N(r, f) (q + I)T(r, f)
+ vtl N
(r,
'':a,,) ~ 2T(r, f) - Nl(r,f) +
,!a,,) to both sides, we get:
~ 2T(r, f) + N(r, f) + ~N (r, f ~ 4V) - Nl(r, f) + S(r).
Hence,
(q-I)T(r,/)
~ ~N (r, f ~ av) -N (r, J,) +N(r,/,)-N(r,f)+S(r).
av
(r, ,':a,,) (r, f. )
Wherever f takes the value with multiplicity k, n -n counts 1. Also, at a pole of I, nCr, /') - nCr, f) counts 1. Therefore, we may write (q -1)T(r,!) Hence, (17.7)
~
t N (r, I ~ v=l
av )
+ N(r,!) + S(r).
17. Picard's Theorem and the Second Fundamental Theorem
105
Now f is not a rational function, and therefore -r-+oo lim T·t}) = 0 since r, IIS(r) = O(logT(r,f) +O(logr) by Lemma 18.3 in the next section. Now, using the fact that lim (A + B) ~ lim A + lim B, we have
(q -1)
<
L q
lim
- v=l r ..... oo
H(
1)
r'r-a: T(r, f)
+
lim N(r,f) r ..... oo T(r, f) ,
from which N
q
_ "lim ~
1) (r,-f - av T(r,f)
-
-lim N(r,f) < 1- q. T(r,f) -
Adding q + 1 to both sides gives
1)] +
q [ H( r'r-a: "I-lim ~ T(r,f)
Hence,
-
I-lim N(r,f) < 2. T(r,p) -
E (r ( av ) ~ 2 holds if 1 is not a rational function. v=l q
Now if 1 is a rational function, the same claims hold up to (17.7). We may suppose I/(z)1 . . . IzIP, where p is a nonzero integer. Then T(r'f) = IPlIog+ r. Also,
Now
m(r,~)
= 2~ J~.,.log+ I~I dO and, for
O. Similarly, for
Izl
large, we have
m
Izl sufficiently large, m(r,~) =
(r, t f:~~:) =
O. Therefore, it
v=l
follows that lim /( (r/ )}) r--oo
r,
= 0, and the rest of the proof follows as before.
Corollary. Let {a v } be any finite collection including 00. Then,
01 complex numbers possibly
Definition. a is called a deficient value of 1 if !S(a) > O. H f never takes on the value B, then !S(a) = 1. The same is true if 1 takes on the value a very infrequently. In general, 6(a) measures the tendency of I to omit the value a and 1 - 6(a) measures the tendency of f to take on the value a.
Entire and Meromorphic Functions
106
Corollary. There are at most countably many deficient values. Corollary. There are at most two values for which 0 > ~. Definition. a is said to be a fully branched value of f if f takes the value a with mllitiplicity either zero or ;::: 2.
Remark 17.1. IT f is entire, f can have at most two fully branched values and this is the best possible, since, for example, ±1 are fully branched values of sinz. Also, if f is entire, then there are at most two values of a for which o(a) > The reader will find it instructive to work out o(a), (I(a), and (I·(a) for some specific functions, such as eZ , tan z, and the Weierstrass p function.
!.
Remark. IT a is fully branched, then
9·(a)
= lim [1 -
N(r' I-a
_ I) ]
T(r,f)
[N (r' I-a
_I)]
> lim 1--
since T;::: N.
I- ) N (, r I-a
1 N(r'f_a) Also, if a is fully branched, then 1 ~! and therefore N(r'f_a)
'
!.
9* (a) ;::: In general, there are at most four fully branched values for a meromorphic function. The next result shows that Nevanlinna theory may be used to study certain types of exponential identities. As an example, consider the functional equation e f +e9 =e\ where j, 9, and h are entire functions. Do there exist j, 9, and h so that the identity holds? What is the relationship between f, 9, and h? We may write e / - h + e9- h = 1. Then the entire function e / - h omits O. But eg - h omits 0 as weD, therefore ef - h must also omit 1 for the identity to hold. By Picard's Theorem, ef - h must be a constant so j = h+ const. Similarly, 9 = h+ const. Lemma (Hiromi-Ozawa [15]). Let ao(z), a, (z), ... , an(z) be memmorphic functions and let 91(Z), ... ,9n(Z) be entire functions. Further, suppose that T(r, aj)
=0
(t mer,
eg
,,»)
j = 0,1, ... ,'"
1/=1
holds outside a set of finite logarithmic length. If an identity n
E al/(z)e
9 ,,(z)
v=1
= ao(z)
17. Picard's Theorem and the Second Fundamental Theorem
107
holds, then we have an identity n
L
c,.,all(z}eg.,(z)
= 0,
v=1
where the constants ell, v
= 1, ... , n
are not all zero.
If
Proof· On writing l' = I, we may use the lemma on the logarithmic derivative to estimate T(r, I'} when I is entire. For notational simplicity, let GII(z) = av (z)e 9 ,,(z). Then we have n
L GII(z) = ao(z).
(17.8)
11=1
By differentiating both sides we obtain n
L G~)(z} = a~I')(z),
(17.9)
v=1
which may be rewritten as g!f')(z) L Gv(z)---:-(z) = ao ..,=1 9
(II)
n
(17.10)
(z), p. = 1, ... , m - 1.
We regard this as a system of simultaneous linear equations in the G v .
Now we have C ..,(II) ( z ) -- P.II ( av, ' a.." ... ,a"(11)' , 9", ... ,9"(II» eg,,(z)
with a suitable polynomial PII in the indicated functions. Thus we have
(17.11)
T (r,
~»)
:;
O(T(r,a v ) + T(r,gv)) =
0
(~m(r,eg,,»)
outside a set of finite logarithmic length. Suppose, for the simultaneous equations (17.8) and (17.10), that the determinant.:l:f. O. By solving (17.10) with respect to Gj,j = 1, ... , n we have by Cramers' rule ~.
Gj = ~, Where 1 C~/G ..
.:l=
Entire and Meromorphic Functions
108
and 1
A;
1
ao a 01
Gj_1 Gj _ 1
G1
a!!:
= G(n-l)
.::..z.=.!....Gj-l
(n-l)
ao
G(n-l)
GCn-l)
.:::i±L.
~
Gn
Gj+l
Since
( dll») (!;m(r,e9")
T r, ~"
(17.12)
=0
n
)
,
we have
T(r,A) for j
= 0 (~m(r,eg,,») ,T(r,A;) = 0 (I: m(r,eg,,»)
= 1, ... ,n outside a set of finite logarithmic length.
Thus we have
m(r,e9") = T(r,e g,,) ::5 T(r,a,,) + T(r,G,,)
~ T(r,a,,) + T(r, A) + T(r, A,,} = (~m(r,eg,,») 0
and hence
outside a set of finite Lebesgue measure. But this is a contradiction. Consequently, the Wronskian A ;: 0 and the result follows. We say that two meromorphic functions I and 9 share the value a (a = 00 is allowed) if I(z) = a whenever g(z) = a and also g(z) = a whenever I(z) = a, counting multiplicities in both cases. A famous theorem of R. Nevanlinna, which will be proven shortly, implies that if two nonconstant meromorphic functions I and 9 on the complex plane share five distinct finite values (ignoring multiplicity), then it follows that I = g, and the number 5 cannot be reduced. We consider here the special case 9 = f', the derivative of I, and prove the following result. Theorem. II I is a non constant entire lunction in the finite complex plane, and il I and I' share two distinct finite values (counting multiplicity), then f' = I· In other words, a derivative is worth two values. We show at the end of the chapter that the number 2 of the theorem cannot be reduced. We
17. Picard's Theorem and the Second Fundamental Theorem
109
do not now know whether there is a result corresponding to our theorem if one ignores multiplicities, or if one considers meromorphic instead of entire functions. Proof of Theorem. To fix the ideas, we suppose that / and f' share the values a and b, where a = 1 and b = 2. Other choices of a and b make no real difference, except if a or b is zero, in which case the analysis becomes easier, and is left to the reader. We may write then
(17.13)
I' -1 k --=e 1
(17.14)
f' -1 k /-2--e ' ,
/-1
where kl and k2 are entire functions. We solve (17.8) and (17.9) for
f to
get (17.15) We now differentiate both sides of (17.10) and substitute (17.8) to get
(17.16) 2e2k \
+ e2k , + (k~ _ k~ _ 3)ek\ +k2 _
e2kt +k,
+ ek\ +2k2 + k~ ekt
- k~ek2 = O.
We shall make repeated use of the le1'l1lI1a of Hiromi and Ozawa. Now, in (17.16), divide by ek2 to get
(17.17) 2e2kt-k2 + i~2
+ (k~ -
k~ - 3)e~ - 22kt
+ ekt +k2 + kjr kt - k2 = k~.
We now apply the lemma to get
(17.18) cle2kt-k2
+ c2e k2 + c3(k~ -
k~ - 3)e kt
+ C4e2kt + c5ektk2 +C(sk~ek\-k, =
0,
Where CI, •.• , Cl) are constants that are not all zero. The hypotheses of the Hiromi-Ozawa Lemma are satisfied because, for example, kl is the logarithmic derivative of e k \ and we may use the lemma of the logarithmic derivative. It follows that T(r, k') = O(T( r, ek outside of a suitably small exceptional set. At any rate, we divide in (17.17) by e k \ to get
»,
and we may use the lemma once more to get
(17.20)
Entire and Meromorphic Functions
110
for suitable dl , • •• , ds . Multiply by ek2 to get (17.21) and apply the lemma yet again to get (17.22) where U1, U2, U3, 1A4 are constants that are not all zero. Now, by successive applica.tions of the lemma, we reach a contradiction, unless possibly one of the five following conditions holds for some constant C:
We now rule out these possibilities unless I' = f. First, it is easy to see that k1 = C (and similarly k2 = C) is consistent with (17.13) and (17.14) unless d = eO = 1, in which case I' = f. For if (/' - 1)/(/ - 1) = d, then f (d - 1) / d + beth: for some constant b, and hence I' = bdedz • This clearly contradicts (17.14) unless d = 1, for we would have
=
(17.23)
If - 2
1-2 =
bde rlz
-
bedz _
2 2
=e
k 2,
which is impossible unless d = 1 (remember that I is not constant, so that b:F 0). Next, we rule out kl = k2 + C. We go back to (17.16) to get (17.24) Apply the lemma again after dividing by e k1 to get (17.25) which implies that kl = 2k2 + C (and similarly, k2 = 2kl + C) is impossible unless 1=1'. From (17.24) we would get (17.26) and apply the lemma for the last time to get (17.27) where flo f2' f3 are constants that are not all zero. In other words, P(e k2 ) === 0, where P is a cubic polynomial, so e k2 is a constant, which we already ha .. e. ruled out unless f = /'. This completes the proof of the theorem.
17. Picard's Theorem and the Second Fundamental Theorem
III
Finally, it is easy to see that there exists a nontrivial entire function that does share one value with its derivative. For example,
fez) = ee
Z
1'" e-e
t
(1 - et)dt
satisfies (I' - 1) / (I - 1) = e.z so that f and f' share the value 1. This shows that the number two of our theorem is the best possible. Now we come, as an application of the second fundamental theorem, to a truly beautiful and surprising theorem of Nevanlinna. Recall that, given two meromorphic functions It(z) and h(z), and a complex number (finite or infinite) w, we say that It and 12 share the value w (ignoring multiplicity) if every z for which It(z) = w also satisfies h(z) = w, and vice versa. We use the same terminology if both functions omit the value
w. Theorem. If two junctions, meromorphic in the whole romplex plane C, share five distinct values, then the two junctions must be the same. Note that e Z and e-.z share 0,
00,
1, -1, so the number five is sharp.
Proof of Theorem. Given a number w, finite or not, let no(r,w) be the number of common roots of It (z) = w and 12 (z) = w contained in the disc {\zl :::; r}, each counted only one time. Then put No(r,w)
r
= 10
no(t w) - no(O w)
'
t
'
dt+no(O,w)logr
and N I2 (r,w)
= N (r, It ~ w) + N (r, 12 ~ w) - 2No(r,w).
Now, taking for WI," . , Wq distinct finite complex numbers and applying the second fundamental theorem to the functions It and 12, we have, off a possible exceptional set of finite length, (q - 2)(T(r,fd
+ T(r, 12)) <
t
(N (r, It ~ W,,) + N (r, 12 ~ W,,))
+ O[logrT(r, It}T(r, h»] q
q
= LNI2 (r,w v ) + 2 LNo(r,w,,) 01=1
01=1
+ O~og rT(r, It)T(r, h)]. Now, if the functions II and 12 are not the same, every common root of the equations It = wv , 12 = W v , is a pole of the function ~. We deduce from this that 12
t
01=1
No(r,w v ) :5 N (r,
II ~
,J
< T(r,1I - h) + 0(1).
Entire and Meromorphic Functions
112
On the other hand,
T(r, it
- h} < T(r, Jd + T(r, h} + 0(1}.
We conclude q
(q - 4)(T(r,Jd
+ T(r, h» < L
N 12(r,wv }
+ O~og(rT(r, it)T(r, h»] off a set of exceptional segments of finite total length. (ll one of the Wv is infinite, just apply a linear fractional transformation to It and h.) Suppose to begin with that one of the two functions, say It, is transcendental. Then among the five given values w = wu , 1/ = 1, 2, ... ,5, there are at least three that are taken by It in an infinite number of points z. By hypothesis, the same is true for h. Hence, 12 is also transcendental. Suppose for a moment that It and 12 are not the same. Apply the last inequality above, which shows that for q = 5, the expression N12 vanishing for the five given values,
T(r, It) + T(r, Ja) < O(log[rT(r, It)T(r, Ja)]). This implies that lim T(r,lt) -+~::.:.. r-oo
logr
<00 and
lim T(r, Ja) <00. logr
r_oo
This is impossible since It and 12 are both not rational functions. The theorem thus is proved by contradiction in the case that one of the given functions is transcendental. But the conclusion is obvious if both of the functions are rational, since a rational function J is uniquely determined by the roots of J(z} = w for any three distinct values of w.
18 A Proof of the Second Fundamental Theorem
We now begin the proof of the second fundamental theorem of Nevanlinna. We continue to use the convention that all equalities are to be read "modulo 0(1)." There are a number of other proofs of the second fundamental theorem, some of them leading to generalizations of it.
Let 1 be a meromorphic junction in the plane, and let be distinct complex numbers. Let .. 1
Lemma 18.1. Gl, ••• ,
an
F(z)
= ~ I(z) - all .
Then m(r,F)?:
~m (r, 1 ~av)'
Prool. We first remark that the lemma is intuitively obvious, for when I(z) is "close to" with j =F II.
all it contributes to m (r, '.!a~)' but not to any m (r, '.!aj)
To prove the lemma in detail, we introduce the following notation. Let 6> 0 be given with 26 $ min{lav - ajl : 1 $ II < j $ n}. We require 6 $1. Let E:, = rl{w: Iw - al'l < t5} = {z: I/(z) - avl < t5} E~(r) = E~ n {z : Izi = r} EII(r) = {O : rei' E ~(r)} ever) = [0,211"1- EII(r).
Entire and Meromorphic Functions
114
alii < 6, Le.,
In other words, EII(r) is the set of all (J for which If(re ill ) those (J's for which
f(re i6 )
contributes significantly to m (r,
clear that EII(r) and Ej(r) are disjoint provided that v Moreover,
11<
where
r
-1 log+ IF(re i9 )ld(J ~ -1 211' _". 211'
U:=l Ev(r).
is over the set
-1 211'
J. LnIl > nIl ~1 =
11=1
-
211'
Ev(r)
211'
E.,(r)
- -
j.
log+ IF(re i9 )ldO,
Now
log+ IF(reill)ldO log+
1
211'
J-
=1=
f!'all).
E,,(r)
If(re,lI). - IdO 1
log+ "
all
1 fJ dO. f(re i ) - a;
n
{;::
#11
Also,
If 211' lE.,(r)
l+~ og
1 {:: f(rei9) _
aj
:::;
n-I () -6- = 0 1.
;i'll
Hence we see that
m(r,F)~~2. f 109+lf(reoA- all Id(J ~ 211' iE (r) v=1
.,
- -1 211'
1
E.,e r )
1 log+ "n d(J ~ f(re i9 ) - aj ;i'''
which proves Lemma 18.1.
It is
18. A Proof of the Second Fundamental Theorem
115
fr)
Definition. N1(r, f) = N (r, + 2N(r, f) - N(r, 1'). N 1 (r, f) is interpreted in the previous chapter after the statement of the second fundamental theorem. Lemma 18.2.
~ m (r, f ~ av) ~ 2T(r, f) -
N 1 (r) - mer, f)
( I')
!')
(n
+m r'l +m r,~ f-a v
.
Proof. Let F(z) be as in Lemma 18.1 and write F z ()
= _1_. fez)
fez) f'(z)
J'{z)
~ fez) n
av '
Then
m(r, F) ~ m (r'l) +m (r, J,) +m (r,t f ~'av) .
(18.1) But m (r,
t) = T(r,f)-N (r, t) andm (r, f,) = m (r, f) +N (r, f)-
(r, f,).
Hence,
m(r,F)
~ T(r, f) -
N (18.2)
+m
(
l) + r,?; f -1') a N (r,
m (r,
n
v
~) + N (r, ~) -
J,)
N (r,
.
Now observe that if we define
cp(r,g)
= N(r,g) -
N (r,
~) ,
then
cp(r,gh)
= cp(r,g) + cp(r, h).
Using this observation in (18.2) we see that (18.3)
m(r,F)
~ T(r,!) -
N (r,j)
+m
(r,
~) + N(r,!,) + N
- N (r, ;,) - N(r,!) + m (r,
~ f !'a
(r,
j)
v )'
Entire and Meromorphic Functions
116
If we now add and subtract 1'( r, I) = m( r, I) + N (r, I) on the right-hand side, we obtain
m(r,F):$ 21'(r,1) - m(r,1) (18.4)
( 1')
+ m r,! + m (r,n ~! 1-,a) v
- {2N(r,1) - N(r,!,)
+ N (r, ;,)}
or
( I')
(n I') -Nt(r).
m(r,F):$2T(r,l)-m(r,l)+m r,! +m r'?;I_av By Lemma 1, m(r, F)
~ E:=t m (r,
,!a.,), and Lemma 18.2 follows.
We are now ready to prove the second fundamental theorem of NevanIinna.
Theorem. Let I be a meromorphic function in the plane and let
f') . ( !') +m (n r'~/-av
S(r)=m r'l Then
mer,!) + ~ m (r, I
~ aJ :$ 2T(r,1) -
Nt(r)
+ S(r).
Proof. This is just Lemma 18.2. We saw previously that in order to deduce the Picard Theorem from the second fundamental theorem, we needed an estimate on the size of S(r). Such an estimate follows from: Lemma (Lemma of the Logarithmic Derivative) 18.3. phic in the plane and 0 < r < p, then m
(r, !') I
:$ 410g+ P + 31og+ _1_ p-r
II I is meromo1'-
+ 41og+ T(p, I).
Proof. Without loss of generality, we suppose that f(O) g(z) = zkf(z),
::f.
0,00 since, if
18. A Proof of the Second Fundamental Theorem
117
Let z = reiD. Then for a suitably defined branch of the logarithm we have by the Poisson-Jensen formula, logJ(z)
(18.5)
1 = -211"
I""
_,,"
+
.
log IJ(pe''P) 1pei'P
L
+ z dcp
pe''P -
Z
L
logBp(z,z.,)-
logBp(z,w.,)+iA,
IW"I
Iz~l
where Bp(z, a) is the Blaschke factor mapping the disk ofradius p onto the unit disk and a onto 0, A is a real constant, and, as usual, z.. and w., are the zeros and poles of J, respectively. [Equation (18.1) holds without any 0(1) terms.] Thus, by differentiating (18.5), we obtain
(18.6) J'(z) J(z)
I""
.
p2 -lz.. 12 = 211" -1\" log IJ(pe''P) 1(peirIJ - z)2 dcp + IJ:~P (z - Z.. )(p2 - z"z) 1
_L IWI' I~p
2pei rIJ
• p2_lw.. 12 (z - w.. )(r - w"w)
Taking simple estimates, we have (18.7) J'(z) 2p 1 itp J(z) (p - r)2 211" -1\" 1log IJ(pe )lIdcp +
I I~
1""
~ -IA,,1 2
I>'~P Iz - A"lIr -
X.. zl'
where A" now runs over both the zeros and the poles of I. Observe that
The last step follows from Ip2 - X"zl ~ p2 - IX"lIzl ~ p2 - fYI" = pep - r). IT we use this estimate in (18.7), take log+ of both sides of (18.7), and apply the additive inequality satisfied by log+, (log+(a, + ... + an) ~ log+ a, + ... + log+ an + log n), we then find that (18.8)
I I
f'(z) ~ log+ p + 2 log + -1log+ -/() z p- r
+ I>.~/Og+
IBp(:,Av)1
1 + log+ -211"
I""
_,,"
II
Ilog+ I/(rei'P) dcp
+ log+ (n(p,f) + n (p,
7)) .
Since Izi = r < p, IBp(~,.\,,)1 ~ 1, we see that log+ IBp(~,.\ .. )1 = log IBp(~,>',,)I'
Entire and Meromorphic Functions
118
Therefore, by Jensen's Theorem, (18.9) 1 1 dO Iog + "fJ 211" _". Bp(ret , >'v)
1".
I
I
{ log...l!... 1>. ... 1 - log ,;, I.>. ... , log
rtr
I>'v I ::; r if I>'vl > r.
if
:1)
To simplify the notation, let nCr) = nCr, f)+n(r, and likewise N(r) = iB N(r, f) + N(r,:1). IT we let z = re and integrate (18.8) over the circle of radius r with center at 0, we obtain (18.10)
1') 1 m ( r, = 211"
f
r log+ I1'(re,B) I L". f(rei6) I dO
1"-
I
::; log+ p + 21og+ -1- + log+ - 1 Ilog+ If(rei'P) I dtp p- r 211" _,.-
+
r
E
~
21 IIOg+ IB ( >. ) II dO + log+ n(p). 11" J-". pre , v I.>." 1< -p
We recognize that O(2T{p, f) and
2~ J':".llog+ If(pe ifJ )IId1p = m(p, f)
+ m
(p':1)
=
Hence, (18.11) m
(r, ~) : ;
log+ p+21og+ p ~ r +log+ T(p,f)+log+ n(p)+N{p)-N(r).
We now will estimate the term log+ n(p). Choose a number p' with pi> r. Then n
r ' dt < _1_
- n(p) (P)-IOg{;')Jp
(pi net) dt
t -log(';i)Jp
N(p') < 2T(p',f)
<--
+C
loge ~ )
- log ( ;.) -
t
,
where C is an appropriate constant. Now 1 _(pi _ p) < p' -
l p
pl
_dt t
= log _p'
1 I - p). < -(p
P - P
18. A Proof of the Second Fundamental Theorem
119
Hence, n( )
< 2T(/,J) +C p;1(, P - P)
p-
so that log+ n(p) :::; log+ pi + log+ _1_ + log+ T(p', J). p' - p Hence, m (r,
ff') :::;log+ p+log+
p' + 2 log + _1_ p-r
+ log+T(p',J) + N(p) Therefore, with r (18.12)
m (r,
ff') : :;
+ log + _pI 1 +log+T(p,f) -p
N(r).
we have
2 log + p' + 2log+ _1_ p-r + N(p) - N(r).
+ log+ _pI 1 + 2Iog+T(P',f) -p
We now want to make the term N(p) - N(r) small. To do this we use the logarithmic convexity of N. [N is logarithmically convex, since net) is increasing a.nd N(r) = ~dt]. Since r < p < p' and N is logarithmically convex, we ha.ve
I;
N(p) - N(r):::; :::;
log(~)
log(~ )
~(p -
(N(/) - N(r»
r) (N(p') _ N(r» :::; p'(p - r) N(p') rep' - r)
~(p' - r)
: :; ~~~ =~j [2T(p', J) + C], .here C is an appropriate constant, C > 1. Now, considering ~ (;-=-~) [2T(p',J) + C] as a function of p, we see that it vanishes for p = r and is greater than 1 for p = I. Since it is a continuous function of p, we may choose p so that p'(p - r) , rep' _ r) [2T(p ,J) + C]
= 1.
For this choice of p, we thus have N(p) - N(r) $ 1 and
(18.13)
rep' - r) ( _ r) _ p - P'[2T(P', f) + C)
Entire and Meromorphic Functions
120
and
Hence, 1 log+ --::; log+ p' +log+T(p',f)
p-r
1
+ log + - - . t-r
Therefore, 1 log + - - ::; log + p' + log + T(p', f)
(18.14)
p-r
Also, we see that +1 ..... 1 + log - - $log' --+log
t -p
t -r
1
1
r
r
1- P'(2T(p'./)+C] 1
$ log+ - - + log+
t -
1 + log + --;;--. p-p
1
1 - 2T(p' ,/)+0
Hence, 1
1
log+ - - < log+ - - .
(18.15)
t-p-
t-r
Using (18.13), (18.14), and (18.15) in (18.12), we have m (r,
(18.16)
~)
::; 4 log + p' + 4 log + T(p', f) + 3log+ t
~ r'
which proves the lemma. We will use the notation IIf(x} ::; g(x) to mean f(x) $ g(x) with Borel exceptions (i.e., a set of finite length). Proposition. IIm(r, Proof· Taking p = r
m
~) ::; 41og+ r + Slog+ T{r,f).
+ log+ ~(r,f)
in the lemma, we have
(r, ff') ::; 4 log+ (r + log+ T{r,f 1 ») + 3Iog+1og+T{r,f) + 41og+ T (r + log+ ;(r,J) , f) ,
Hence, by the Borel Lemma, we get
11m
(r, ~) $ 4Iog+r+Slog+T(r,f).
Corrollary. USer) = O(log+ T(r, f) + o(log+ r). Proof. This follows immediately from the proposition since S(r) is a finite sum of terms of the form mer,
Ii)·
19 "Two Constant" Theorems and the Phragmen-Lindelof Theorems
The Two Constant Theorem. Suppose that f is holomorphic in D = {z: Izl < I} and continuous in D\{l}. Suppose further that If I :5 N in D and If I :5 M in 8D\{1}. Then IfI :5 M in D\{l}. First Proof. Choose a
> 0 and let g(z)
1- z)O = ( -2fez)
~r a suitable branch of (1;zr. Now 9 is analytic in D and continuous in D if we define g(l) = O. By the maximum modulus theorem,
suplg(z)1 = maxlg(z)l· zED
zE8D
But Ig(z)1 :5 M for z E aD. Hence, Ig(z)1 :5 M for ZED, and so
If(z)1 :5 M Now let
Q
1__1° 2
l-z
for
zED.
0 to get the result.
-
Second Proof. This proof uses the Cauchy integral formula. Since there exists a linear fractional transformation U(z) = e,B t::aQz taking an arbitrary POint Zo E D into 0, and mapping fi / {I} conformally onto D/ {1} I we need only prove that 1/(0)1 :5 M. But choosing 00 with 0 :5 00 < 11', 1 1/(0)1 = 1-2 11'
( I(w) dwl :5 21 J1wl=R w 11'
1 (1)
If(reiB)ldO + 21
11'
1 1/ {2}
(re iB )ld8,
Entire and Meromorphic Functions
122
where R < 1 and
But
~ [
<
211" J(2) -
1 {
211" J(t)
Hence, letting R
~
=:; (lv/ + 0(1))
~
(211" - 280 ) 211" •
1- ,
If(O)1 =:; Now letting 80
00 N 11" '
~ N + M (1 - ~) .
0 we again obtain the desired result.
Third Proof. By Jensen's Theorem,
1111" log If(O)1 ~ 211" _11" log If(re ii1 )ldO. The argument of the second proof works here, too.
Fourth Proof. By a conformal mapping, we may replace the unit disk by the upper half-plane. We have If I ~ M on the real axis and If I ~ N above the real axis, and must conclude that If I ~ M above the real axis. Choose A> 0 and let
g(z) = (z
:iA) f(z).
Note that
and that
Ig(x)1 =:; M
for all real x.
Hence, if R is large enough, we see by the maximum modulus theorem tbat Ig(z)l =:; M for z in tbe semicircular region bounded by the real axis and the arc Izl = R, 0 ~ arg z ~ 'IT. Hence, Ig(z)1 :$ M in the upper balf-plane . Letting A -+ 00, we complete the fourth proof.
19. "Two Constant" Theorems and the Phragmen-Lindelof Theorems 123
A Phragmen-Lindelof Theorem. Let fez) be holomorphic inside an angular region of opening 'trIa, where a> 1, and continuous on the closure of the region. Suppose that If(z)1 ~ M on the boundary and that
If(z)1 ~ Ke 1zlIJ in the region, for some constant region.
fJ with
{3
< a. Then If(z)1
~
M in the
Proof. Without loss of generality, we may suppose that the region is given by Iarg zl < ;:. Choose f > 0, and choose "( with {3 < "( < a. Let F(z)
so that For z
= rei8
= e- d ' fez),
IF(re i8 ) I = exp{ -fr'Y cos ,,(9}lfCz)l. in the closed angle, we have exp{ -fr'Y cos"(9}
<
1 so that
IF(z) 1 ~ If(z)1 there. Notice also that by using the estimate on f inside the region we have IF(rei9 )1 ~ exp{-fW cos "(9}Kexp{R.B}, 80
that for R large enough we have
IF(Re i8 ) 1 ~ M. It follows that
If(z)1 ~ M exp{ fR'Y} , and the result follows on letting f -+ O. Taking for simplicity a = 1, the Phragmen-Lindelof Theorem says that if a function is holomorphic and of order less than one in a half-plane and bounded on the boundary, then it is bounded inside by the same bound. The example fez) = e Z in the right half-plane shows that the order 1 is critical. However, the next result shows that the analogous result holds if Vie replace the condition "order less than 1" by "growth at most order 1, zero exponential type." Theorem. If the hypothesis is changed to read that for each 6 > 0 ~
K6 exp{ 6Izl"} in the region, then the conclusion follows as before. Proof. Now let F(z) = exp{-fz"}f(z), If(z)1
and the same proof works.
Remark. Analogous results hold for functions holomorphic in other regions and satisfying appropriate growth restrictions. One useful case is the parallel strip. Calderon has used this case in developing a theory of interpolation of Banach spaces that can be applied in the fields of harmonic analysis and partial differential equations.
20
The P6Iya Representation Theorem
The Polya Representation Theorem plays a central role in the theory of entire functions of exponential-type. We give a somewhat augmented version of this theorem. Before proceeding with the theorem, it will be necessary to discuss convex sets. We say that a set E (in the complex plane) is convex if E contains the line segment joining any two points. That is, if Z1, Z2 E E, then tZl + (1 - t)Z2 E E for all t E [0,1]. The intersection of convex sets is again convex. Definition. Given a set A, the intersection of all half-planes that contain A is called the closed convex hull of A and is denoted by K(A). Definition. A point is an extreme point of a set if it is not the midpoint of any line segment contained in the set. Theorem 20.1. extreme points.
A compact convex set is the convex hull of the set of its
We omit the proof. Definition. For any set E, k(O) support function of E.
= sup{Re(ze-i8)
:
Z
E
E} is called the
We note that k( 0) measures the directed distance from the origin to the most remote point of the projection of E on the ray arg z = 0. Note also that if E is empty, then k = -00. It is easy to show that, for a given set E, K(E) = {z: Re(ze- i8 ) ~ k(O) for all O}. Remark 20.2. If Zo = Xo + iyo TO cos(O - ( 0 ) = Xo cos 0 + Yo sin 9.
= r o ei80
and E
=
{Zo}, then k(O)
===
20. The P6lya Representation Theorem
125
Remark 20.3. Let E be a circle with center at 0 and radius R. Then k(O) = R for all (). Remark 20.4. Let E be the line segment [XO,XI], Xo < Xl. Then k(O) = :tlCOSO if 0:5 and k(O) = xocosO if 0:5 In particular, if Xo = -Xl and Xl = R, then k(O) = RlcosOI. If E is the vertical line segment [-iR,iR}, then k«() = RlsinOI.
-I:5
I
I :5
3;.
&mark 20.5. If El has support function kl and ~ has support function then El + E2 = {ZI + Z2 : ZI E El, %2 E E 2 } has support function + k2 • From Remark 20.3, it follows then that the rect.angle with vertices (±Rl,±iR2 ) has support function k(8) = RllcosOI +r2IsinOI.
"2,
"1
Remark 20.6. max{kt. k2}'
The convex hull of El U E2 has support function k
=
Remark 20.7. H El with support function kl is translated, 80 that the point originally at 0 is moved to Zo = Xo + iyo, then the support function of the translated set is k( 0) + Xo cos 0 + Yo sin O. Definition. Let Ho( 00 )be the class of all functions that are holomorphic near 00 and that vanish at 00. Let f be an entire function of exponential-type and write fez) = The Borel transform ~ of f, defined by ~(w) = E an belongs to Ho. Each function in Ho is the Borel transform of a unique /. We have seen (Corollary to Proposition 11.5) that the type of / is the radius of convergence of the series E an In Proposition 11.7, we saw that ~(w) = oo f(t)e-twdt, in the sense of analytic continuation. We also saw that
til:+! ,
E (t;:t ) ZR.
til:+! . Jo
r is a rectifiable curve that winds once around the singularities of We call this the P61ya integral representation formula.
Where
t.
Definition. Let S(~) be the set of singular points of~, and let k be its SUpporting function. We call S(~) the conjugate indicator diagram of f. Sometimes this name is used for S*(~), the closed convex hull of S(~). We 1rill write D(f) = S·(~). Definition. The indicator function of f is h(9) = hl(9) = limsup r-+oo
1 '(} -log If(re' )1. r
We now state an important part of the P6lya Representation Theorem.
Entire and Meromorphic Functions
126
Theorem 20.S.
h( 8) = k( -8) for all 8.
The proof of Theorem 20.17 is contained in an appendix at the end of this chapter. We now make some remarks to illustrate this theorem. Remark 20.9. If f is of zero-type, then h = 0 so that k is the only possible singularity of cPo
= 0, and hence 0
Remark 20.10. Denoting by r(f) the type of I, we have r(J) = maxh(O).
Remark 20.11. IT I(z) = eBZ , where a is real, then
h( 0)
= lim sup ~ log lea.. i = r-oo
r
lim sup r-+oo
!r ar cos 0 = a cos O.
On the other hand, eP(w) = Ea n w!+i = (w - a)-l and so SeeP) Thus, k( IJ) = a cos IJ and we do have h(IJ) = k( -0).
= {a}.
Remark 20.12. IT I(z) = eiz , we have h(lJ) = - sinlJ and eP(w) = (W-i)-l. Hence, k(lJ) sinO since S(cP) = {i}. Note again that h(O) k(-IJ).
=
=
Remark 20.19. Suppose I(z) = Eane Anz , a finite sum with distinct An and nonzero an. Then eP(w) = Ean(w-An)-l so that SeeP) = {An}. Note that only the extreme points of {An} affect h(O). IT the An lie on a straight line, only the endpoints affect h(IJ). Remark 20.14. It is easy to see that hfg ::; hf+hg, so that D(fg) ~ D(f)+ D(g). An interesting problem that has applications in harmonic analysis is that of finding suitable conditions under which D(fg) = D(f) + D(g). Let Mo be the class of all Borel measures of compact support. Definition. If dp. E Mo, its Laplace transform dp./\ is defined by
It is easy to see that dp./\(z) is an entire function of exponential-type for
(as can be verified on differentiating "by hand") so that dp./\ is entire. And
where R is chosen so that the support of dp. lies in the disk of radius R centered at O.
20. The P61ya Representation Theorem
121
Definition. We write dJ.L "" dll to mean that dJ.L" = dv". It is not hard to show that dJ.L ""' dv if and only if I fdJ.L = J f dv for each entire function f, or for each entire function f of exponential-type. It is clear that '" is an equivalence relation. IT we take dJ.L = dzlr, where r is a circle, then dJ.L"(z) = e-zwdw = 0 by the Cauchy Theorem. Hence, dJ.L '" 0 even though dJ.L =! O. We shall see that for any dJ.L E mo, dJ.L '" dv, where dv = ~(-w)(-dw)lI" where ~ is the Borel transform of dl''' and r is any curve that winds once around
Ir
S(~).
Definition. [dJ.Ll is the class of measures equivalent to dl'. Definition. M~ is the class of all [dl'l for dl' E Mo. Definition. IT f is continuous in the plane and dJ.L in Mo, we define the convolution f * dl' by
(f * dJ.L)(z)
=
J
f(w - z)dl'(w).
Definition. IT dp. and dv are in Mo, we define the convolution dl' * dv by
By means of the Riesz Representation Theorem, it is easy to see that the above definition defines dJ.L * dv as a unique measure in Mo. Indeed, Mo is an algebra over the complex numbers. Proposition. If dJ.LI '" dP.2. then (dJ.LI * dv) "" (d1'2 * dv). It thus makes , .mse to define [dl'l * [dvl = [dp. * dvl. M~ is an algebra over the complex numbers. Definition. Let Eo be the algebra of all entire functions of exponential-
type. Definition. Let E be the space of all entire functions in the topology of uniform convergence on compact sets. Definition. E*, the dual of E, is the space of all continuous linear func-
tionaIs on E. Now E is a locally convex topological linear vector space. It will appear
that each of the spaces M~, Eo, Ho(oo) ''is'' the dual space E·.
l>eftnition. For fEE and [dp.] E M~, define the inner product {FI, [dl'J>
by
Entire and Meromorphic Functions
128
Definition. For FEE and f E Eo, define the inner product (F, f) by (F,!) = (J(D)F)(O),
where D =
E ~F(n)(o).
fz.
This means that if fez) = E~zn, then (F,f) ::::
It is not hard to show that, for each f E Eo and FEE, the series defining (F, f) converges. Indeed, the linear functional A defined by A(F) = (F, f) is a continuous linear functional on E. The same is true for A(F) = (F, dp.).
Definition. For FEE and by
~ E
Ho(oo), define the inner product
(F,~)
1 . f ~(w)F(w)dw, (F,~) = -2 1r&
where
r
lr
is any curve that winds once around
Again, A(F) =
(F,~)
S(~)
defines a continuous linear functional on E.
Theorem 20.15. Let [dp.J E M~, f E Eo, ~ E Ho(oo) be related by f = dlJ." and ~ be the Borel transform of f. Then dlJ., f, and ~ give rise to the same functional in E*. And given any functional in E*, there is a unique (dlJ.J (or f or~) that gives rise to it. To say that f gives rise to A is to say that A(F) = (F, f) for each FEE, with a similar definition for dlJ. or~. We omit the proof of the theorem since much of it is straightforward. Remark. To illustrate the theorem, take f = e" so that dlJ. is the unit point mass at I and ~(w) = (w _I)-I. Then feD) = eD, so that by Taylor's theorem (J(D)F)(O) = F(O)
Note that
+ F'(O) + :,F"(O) + ... =
J F(-z)dlJ.(z) = F(I) also. 1. -2 1U
lrf F(w)~(w)dw =
F(l).
And, by Cauchy's Theorem,
1 . / F(w)ldw = F(l). -2 1rt w-
Summarizing the results of this section so far, we have the following omnibus theorem. We call it the P6lya Representation Theorem, although the name is not entirely appropriate. The P6lya Representation Theorem. Let Eo be the algebra of all entire functions of exponential-type and the convolution algebra of equivalence classes of Borel measures of compact support, where dlJ. ,.." d/l means that J f dlJ. = J f dv for each entire function f· The Laplace transform is defined by dp"(z) = J e-ZWdll(W); dll """ d/l if and only if dlJ." = d/l", so
Mo
20. The P61ya Representation Theorem
129
'it makes sense to talk of [dt£l" = dp.", where [dp.1 is the equivalence class that contains dp.. The Laplace transform is an isomorphism of M~ onto Eo. To invert the Laplace transform, i.e., given f E Eo, to find dp. E Mo such that dt£" = f, take dp.( z) = cl>( - z) ( -dz) Ir, where cl> is the Borel transform of f and r winds once around the set S(cl» of singularities of cl>. The indicator diagmm of f, D{f), is defined as the convex hull of S(cl». If ,,(9) lim sup Jog If~ej8)1 is the indicator function of f, then h(O) k( -0),
=
=
r-oo
where k is the support function of D{f). If E is the space of all entire funcUons in the topology of uniform convergence on compact sets, then both Eo and M~ are the dual space of E, where, if f = dp/', then for FEE {F,I)
= (J(D)F)(O) =
(F,dt£)
=
f
F(-z)dt£(z).
We now tum to some applications of this theorem. (Note. We repeatedly use conventional, but strictly incorrect, phrases 6ke "r winds once around the set S(cl» of singularities of 4>." A more correct wording is "r has winding number -1 around 00, and lies in a connected and simply connected open set containing 00 in which 4> is ana.lytic.") Definition. For an entire function the entire function defined by f>.(z) = fez
I
and a complex number A, let f>. be
+ A}
for all z.
Lemma 20.16. For any entire function I, r(f>.) = r{f). Proof. With M(r: f)
= max{lI(z)1 : Izi = r}, we have M(r : b.) ~ M(r + IAI : f)
80
that
1 1 1
; logM(r : bJ ~ ;M(r + IAI : f)
= (1 + O(I}) r + IAllogM(r + IAI : f).
lienee, r(b.) ::5 r(f).
Similarly, r(f) ~ r(b.)
since f(f>.')-A.
Entire and Meromorphic Functions
130
If f and 9 are entire functions of exponential-type, then if and only if
Lemma 20.17.
D(f)
= D(g)
r(f(z)e az ) = r(g(z)e aZ ) for each complex number a. Proof. Clearly, if DU) = D(g), then r(J(z)e az ) = r(g(z)e az ). To prove the converse, it is enough to compute, say, h(O) from T(J(Z)e aZ ). We show that h(O) = lim TU(z)eo. Z) - a. 0.--+00
Let Do. be the indicator diagram of f( z )e az . Then Do. = a + D(J), that is, Do. is D(J) translated by a. Choose B ~ max (Ihi (I) I, Ihi (-I) I). Now when a is large, Do. lies to the right of the origin, and Do. lies in the strip {z = x + iy : Iyl S B; x S a + hi(O)}. Do. also contains the point (a + h(O),O). Hence, if To. = T(J(Z)e az ), then To. ~
a+h(O)
and so that To. -
(a + h(O»
= 0(1).
o Lemma 20.18. If f is an entire function of exponential-type, then for each complex number A, D(f) = D(I>.).
Proof. We use Lemma 20.17, with 9 = 1>•. Note that
= e-a~(ea(z+~) f(z + A)). So if we let F(z) = eaz J(z), then eo.zg(z) = CF~(z), where C is a nonzero constant. Since T(F~) = T(f), we have T(e aZ J(z» = T(e o.z f~(z», and the eaZg(z) = eaz fez + A)
result follows. The Polya Theorem has the following corollary. Proposition. Ifh(~) <0 andh(-~) <0, thenJ=O.
Proof. h (~) < 0 implies that D(J) lies above the real axis ' - ~) < 0 implies that D(J) lies below the real axis. Hence, D(f) is ~_. d, consequently, IP has no singularities. Since IP( oc) = 0, it follows from Liouville's Theorem that IP = 0, ano hence J = O. The Polya Representation Theorem provides an alternate proof of Carlson's Theorem presented in Chapter 12.
20. The P61ya Representation Theorem
Theorem (F. Carlson, 1914) 20.19. tion of exponential-type with r(f) < n 0, ±1, ±2, .... Then f 0.
=
=
11',
131
Suppose f is an entire funcand suppose that f(n) = 0,
Proof. Consider the function g(z) = f(z)/ sin 1I'Z. It is clear that 9 is entire. Since
T(r,F/G) $ T(r, F)
+ T(r,G),
9 is of exponential-type. Now hg (-~) < 0, and hg (~) < 0, so that by the above proposition, 9 = 0. Hence f = O. Remark 20.20. It is clear from the proof that the hypothesis that r(f) < can be relaxed to hf(±1r/2) < 1r. The condition r(f) < 11' is sharp, since sin 1I"Z is of type 11". The Carlson Theorem says, roughly, that an entire function of exponential-type must grow fast in a direction at right angles to its zeros. Later in this section, we shall get a stronger version of Carlson's Theorem. Chapter 22 is devoted to proving an extremely strong generalization of it.
11"
Definition. Let k denote the set of all entire functions type such that hf(±7r/2) < 11".
f
of exponential-
Definition. A sequence {An}, n = 0,1,2, ... is said to be k-admissible provided that there exists an f E k such that f(n) = An, n = 0,1,2, .... We now give a characterization, due to Buck, of k-admissible sequences.
Theorem (Horseshoe Theorem) 20.21.
The sequence {An} is called kadmissible if and only if F(z) = Anzn is holomorphic at 0 and 00 and in Borne neighborhood of the negative real axis as well.
E
Proof. Suppose first that {An} is k-admissible, SO that An E k. Then, if z is small,
I
r
= f(n) for some
We choose r to be a rectangular path that winds around S(~) such that lies in the strip
S", = {z = x
+ iy: Iyl < 11'}.
It follows that F( z) can be analytically continued in the complement of the image of S(~) under the mapping e- w , and that F(oo) = o. To prove the other half of the theorem, suppose that F satisfies the requirements of the theorem. We may write An
=~ f 2n
J"
F(z) dz, zn z
Entire and Meromorphic FUnctions
132
where A is a circle around O. But A is homotopic, in the complement of S(F), the set of singularities of F, to the path AR illustrated in the figuct> below, where AR is a circle of radius R. y
The Horseshoe Contour
Hence, A_I .. -
Now
r
JAR
-+
0
as
21ri
R
r
-+ 00,
Hence, A_I "--2· 1rl
F(z) dz
JAR --:;;- -;
1 A"
F(oo) =
since
o.
F(z) dz ---, Z"
Z
where A* is any curve that winds around S(F) with multiplicity -1. We thUl3 are led to define
for an appropriate branch of ZW on A*. It is easy to check that f E k, and the proof is complete. We now may prove the following extension of Carlson's Theorem.
20. The P61ya Representation Theorem
Theorem 20.22. 0, 1,2, . . .. Then f
133
Suppose that f E k and that f(n)
= O.
=0
for n
=
Proof. We have
By hypothesis, F
= O.
Hence,
zF(z)
On letting z -+
00,
1 = -2' 11'1
i r
1 z
- ze W
~(w)dw
= O.
we have
2;, Ir
= e.cw~(w)dw, we have 1(-1) = o. Now we use the fact that 1 E k; thus f-l E k, where f-l(Z) = f(z -1). We may apply the same argument to f-l to conclude that f-l(-I) = 0, that is, f( -2) = O. Repeating the argument, we get f(n) = 0 for n = 0,±1,±2, ... , so that f = 0 by Carlson's Theorem. We may prove several theorems about k-admissible sequences by means of the above characterization using classical theorems of function theory that relate the properties of a function to its power series coefficients. Since f(z)
Theorem 20.23. for each n.
If {An,} is k-admissible and IAnll/n
-+
0, then An = 0
The proof is trivial. Theorem (Pringsheim). Let f(z) = E anzn have nonnegative coefficients and radius of convergence R. Then z = R is a singular point of f. Theorem 20.24. 0, I, 2, ... , then An
If {An} is admissible and (-I)nAn ~ 0 for n
= 0 for n = 0, 1, 2, ....
=
Proof. F( -z) = E(-I)n Anzn has nonnegative coefficients, but no positive real number is a singularity of F.
Hadamard Gap Theorem. If f(z) = Lanz n with an = 0 except for = nk, where lim n~;l > 1, then every point of the circle of convergence
n
0/ f
is a singular point of f.
Theorem 20.25. If {An} is k-admissible and An tJJith lim nk+! > 1, then An = 0 for n = 0,1,2, ....
= 0 except for n = nk
nil: This is a simple consequence of the Hadamard Ga.p Theorem.
134
Entire and Meromorphic FunctioDs
Fabry Gap Theorem. If fez) = E anz n with an = 0 except for n = nk and k-1nk - oc, then every point on the circle of converyence of J is a singular point of J. Theorem (Szego). Suppose that J(z) = E anz n , where the an lie in some finite set. Either Izl = 1 is a natural boundary oj J or J is a rational function, and the an are eventually periodic. As a corollary we have Theorem 20.26. If {An} is k-admissible and the An lie in some finite set, then the An are eventually periodic.
Appendix The proof that h(O) = k( -0) is presented in this section. The actual proof of this assertion is fairly simple, but we prefer to give some of the background concerning supporting functions of convex sets. First, we give a simple necessary and sufficient condition that a function h(0) should be the supporting function of a nonempty compact convex set. The condition is that the function should be "subsinusoidal." Next, we prove that if h(O) is the indicator function of an entire function of exponential type, then h( 0) is subsinusoidal. Finally, we show that h( 0) is the supporting function of the conjugate of the conjugate indicator diagram. From now on, when we speak of a "function of 0," we mean a function that is 211"-periodicj and when we speak of a "supporting function," we mean a supporting function of a nonempty compact convex set. Our treatment is a combination ofthe treatments in P61ya [31] and Boas
[5]. Definition. A function H(O) is a sinusoid if it has the form H(O) = acosO+ bsinO. Remark. Given 01 ::/= 82 and real numbers hI and h2' there is a unique sinusoid H such that H(Od = hl and H(8 2 ) = h 2 • We call H the interpolating sinusoid: It is given by (0 < O2 - 8 1 < 11")
(20.1)
H(O) - h sin{02 - 0) 1 sm . ( 8 - 01 ) 2
+
h sinCO - 8d 2 sm • (0 2 - 81 ) .
Definition. Given a function h(8) and 0 1 ::/= 82, we call H the interpolating sinusoid of h if H is given by (20.1) with hI = heed and h2 = h(e2 ). Definition. A function h( e) is subsinusoidal if it is majorized by each of its interpolating sinusoids, that is, (20.2)
20. The P6lya Representation Theorem
135
Remark. The theory of subsinusoidal functions has some similarity to the theory of convex functions. Remark. If h is subsinusoidal, and if H is sinusoidal and H(8d 2:: h(fid, H(8 2) ~ h(02), then H(O) 2:: h(O) if 01 :::; 0:::; 82 with 0 < 82 - 01 < 11". Remark. The sum of two subsinusoidal functions is subsinusoidal. Remark. That h is subsinusoidal is equivalent to the assertion that the point h( 0)ei9 does not lie outside the circle that passes through the points 0, h(Odei91 , and h(82 )ei92 , where 01, O2 , and 8 are in the specified range. This geometric interpretation can be used to supply geometric proofs of some of the subsequent results. Problem. Suppose that hI is subsinusoidal, h2 is supersinusoidal, and ht(O) 2:: h2(9) for all 9. Does there exist a sinusoidal function H such that hI (0) 2:: H( 0) ~ h2 ( 9) for all O? Lemma 20.27. Suppose that h is subsinusoidal, that 81 < 92 < 83 , that 9 < 11", and that H(8) is a sinusoid such that h(91 ) :::; H(8 1 ) and h(82 ) ~ H(8 2 ). Then h(83 ) ~ H(8 3 ).
o < 83 -
Proof. Suppose 6 > 0 and h(83 ) < H(83 )
-
6. Let H6 be the sinusoid such
that
H6(OI) = H(Otl,
H 6(03) = H(8 3) -
o.
Then H 6 (82) < H(02), since
H6(8) = H(O) -Ii
~in(O - ( 1 )
sm(83
-
•
Od
Since h is subsinusoidal and we have
it follows that
which is impossible.
Lemma 20.28. (20.3)
A junction h(8) is subsinusoidal
h(Ol) Sin(03 - 82 )
whenever 81 < 82 < 83 ; O2
+ h(82 ) sin(Ol -
( 3)
if and only if
+ h(83 ) sin(82 -
8t} 2:: 0
81 < 11"; 83 - 82 < 11".
Proof· Clearly, (20.3) is equivalent to (20.2) if 83 - 81 < 11". To prove (20.3) in general, choose 94 so that O2 < 84 < 81 + 11" and let H(O) be the interpolating sinusoidal for h at OJ. ()2. By Lemma 20.27, h«()4) ~ H«()4). Repeating this argument with ()2, ()4, 83 we get h«()3) ~ H«()3). Now h«()d sin«()3 -
()2)
+ h«()2) sin«()}
-
()3)
+ H«(J3) sin(02 -
()l)
but sin«()2 - ( 1 ) > 0 and h«(J3) ~ H(83 ), so the result follows.
= 0,
Entire and Meromorphic Functions
136
Lemma 20.29. If h is subsinusoidal, then it is continuous and even has left and right derivatives at each point. The left derivative is never grenter than the right derivative.
Proof. Choose 0 and suppose without loss of generality that h(O) < O. Otherwise, consider h - H where H is a sinusoid such that H(O) > h(8). Choose f > 0, 8 > 0 with t+8 < 1r. Applying (20.2) in turn to the following triples CPI, CP2, CP3: (i) 0 - f - 8, 0 - f, 8 (ii) 0 - t, D, 0 + t, (iii) 0, 8 + €, 0 + t + 8, we eventually obtain
(20.4)
h(O) - h(8 - f sin(t + 8)
8)
-
< <
h(O) - h(O - t} sint h(O + €) - h(D)
. SlDt
<
h(O + t + 0) - h(O) . ( C) . smt+u
For example, considering the triple (i), we get h(O -
t -
8) sin t
-
h(D - t)sin(f + 8) > -h(O) sin 8
so that
[h(D -
0) - h(O)] sin f - [h(O - f) - h(O)] sin(f + 8) h(O)[sin(f + 8) - Sinf - sin 8]
t -
~
.
€
+8
[
f -
8
= -h(O)2slD -2- cos -2'
-
f + 8] cos -2- > O.
Hence,
[h(O - f - 8) - h(O)] sinf > [h(O - f) - h(8)] Sin(f + 0), which proves the first inequality of (20.4). The others are proved in a similar way. Now (20.4) is precisely the assertion that h(9~~2:h(9) is an increasing function of x when x is small, from which all of the assertions of the lemma follow easily. Lemma 20.30.
If h'(O) denotes the T1ght derivative of h(O). then
h(O) cos(cp - 0)
+ h'(O) sin(cp -
0) - h(cp) SO
< cP < 0 + 1r. Proof. Choose f so that 0 < f < 1r and apply (20.2) to the following triples
for each pair 0, cp with 0 -
1r
CPI, CP2, CP3: (i) cp, e, 0 + f (ii) e, D+ f, cpo This use of (20.2) is admissible if either 8 - 1r S 0 < A or D + f < cP < f + 1r. We get
(20.5)
h(8)sin(cp -
e - f) + h(f} + f)sin(8 -
cp)
+ h(cp)sinf;::: 0,
20. The P61ya Representation Theorem
137
which may be rewritten as (20.6) h(B+f)-h(O). ( 8) h() h(8) sin(
~
o.
Theorem 20.31. The function h(8) is the supporting function of some nonempty compact convex set K if and only if h(8) is subsinusoidal. Proof Suppose first that h supports K, so that for each z = x and any (h. (J3 we have
+ iy E
K
+ ysin8 1 ~ h(8t)
(20.7)
x cos (Jl
(20.8)
xeos 03 + ysin (J3 ~ h(83).
(J2 < (J3, (J2 - (Jl < 11", and (J3 - (J2 < 11", we may multiply (20.7) by the positive quantity sin(03 - 92 ) and (20.8) by the positive quantity sine O2 - ( 1 ) and then add to get
IT we now choose (h, O2 • 03 with {h <
h(/h) sin (83
- (
2) + h(03)Sin(02 - ot}
~
(x cos 82 + ysin(2 )sin(03 - 9t};
but x cos 8 + y sin 8 = h( 8) for some z = x + iy in K, and it follows that h is subsinusoidal. To prove that if h is subsinusoidal, then it is supporting, let Ko
= {z = x + iy: xcosO + ysinO ~ h(O)}
and then let K=nKo.
o Each Ko is a closed half-plane in the direction 8. It follows that K is closed and convex. Further, K is bounded since it is contained in the rectangle Ko n K~ n K'/r n K¥. We now prove that each Ko contains a point of K on its boundary, thus completing the proof of the theorem. Given (J, then, we must prove that there exists a point Zo = Xo + iY6 E K for which (20.9)
Xo
cos 0 + Yo sin 0 = h«(J).
By the theory of envelopes, we also want Xlisin(J - Y6COSO
(20.10)
= -h'«(J),
where we interpret h'(O) as the right derivative. Following this heuristic idea, we let Xli) Yo be the simultaneous solution of (20.9) and (20.10). But on applying Lemma 20.30, we get for all
It follows that
+ Ye sin
h(
Zo E K, and since
h(B) = max{xcosB + ysinB : x + iy E K},
the result is proved.
138
Entire and Meromorphic Functions
Theorem 20.32.
If f is an entire function of exponential type and h(O)
then h(O)
~s
= hf(O) = r_oo lim ! log If(re i6 )1, r
subsinusoidal, and h(O) is consequently a supporting function.
Proof. The proof is a simple application of the Phragmen-Lindelof principle. Suppose that 0 < O2 - 01 < 11', and let hi = h(OI), h2 = h(02), and for 6> 0 let H6 be the interpolating sinusoid for hl + 6, h2 + 6 at Oh O2 , Let A = A6 be the complex number for which H6(O) = ~{Ae-i6}, and let F(z) = f(z)e-A:J so that if z = reiD, then
= If(z)1 exp{ - rH6(fJ)}. is bounded on the rays z = re i61 , z = rei8" IF(z)1
Now F(z) and of order 1 in the angle between. By the Phragmen-LindelOf Theorem, F(z) is bounded in the angle (It :::; 0 :::; ()2, and it follows that h(O) :::; H(O) for each () in this range. Since lim H6(0) = Ho«(), the result is proved. 6-0
Theorem 20.33. If f is an entire function of exponential type and h(0) = hf(O) = lim ~ log If(rei6 ) I, then h(O) = k( -0) where k is the supporting r-oo junction for the conjugate indicator diagram.
=
Proof. Let () ()f be the Borel transform of f and let D = Df, the conjugate indicator diagram of f, be the convex hull of the set of singularities of (). Let C be a rectifiable curve that winds once around D and that stays within an f-neighborhood of D, where € > O. Now 1 . f ()(w)exp(zw) dw, fez) = -2 1I'llc so that if z
= reilJ , then If(re i6 )1:::; Amaxlexp(zw)l, wEC
where A is a constant. Hence,
h(O) :::; maxR(wei8 ). wEC
H we now let € -+ 0, we see that h«() :::; k( -0). In the other direction, it is enough to prove that h(O) ~ k(O) since, if we replace fez) by g(z) = f(zei'P), the general case follows from hg(O) 2: kg(O), since hg(O) = hf(O + cp), ()g(w) = ei'P()/(we-i'P) Dg = ei'PDf and kg (8) = k(O - cpl. Now, as we have seen, ()(w)
=
1
00
f(t)e- tw dt for w > h(O),
so that () has no singularity to the right of the line x inequality h(O) ~ k(O) follows.
= h(O)
and the
21 Integer-Valued Entire Functions
An integer-valued entire function f is one such that fen) is an integer for n = 0, 1,2, .. '. Some examples are (i) sin 1rZ (ii) 2'" (iii) any polynomial with integer coefficients. In this section, we shall mainly follow a paper of Buck [7]. In outline, a certain construction generates a special class RI of integer-valued entire functions. We will be concerned With finding growth conditions on an integer-valued entire function f that imply f E R l • The three examples above belong to R 1 •
Definition. We say that an algebraic number a is an algebraic integer if it satisfies a polynomial equation:
(21.1) where aj E Z, j = 0, 1, ... , n - 1. Notice that the coefficient of zn is 1. Examples. Any n E Z satisfies z - n = O. And ±i satisfies z2 + 1 = O.
It is not hard to prove that the algebraic integers form a ring. If the integer n in (21.1) is minimal, the other roots are called the conjugates of Q. The collection of all of the roots of a minimal polynomial is called a complete set of algebraic conjugates. It is not hard to show that if a is a root of P (where P is not necessarily minimal), then each conjugate of a is also a root. Consider now the polynomial
Entire and Meromorphic Functions
140
where qj E Z for j
= 1,2, ... , n.
We can write n
Q(x)
= II (1 -
/3jX) ,
j=1
where the /3j run over one or more complete sets of algebraic integers. This may be seen from the fact that the {3j are the roots of the polynomial
Now let P be any polynomial with integer coefficients and Q as above. Then P(X) ~ n bnx , where bn E Z. Q(x) =
7
This follows since we can write
Q;X)
= 1+qlX+~ .. +qnxn = 1-~*{X) = I +Q*{x) + [Q*(xW+ .. · = 1 + B 1x + B2X2 + ....
The B j are clearly integers. To express the bj in terms of the /3j, let us first suppose for simplicity that the {3j are distinct, and that P = 1. Using partial fractions and writing
1
00
1- {3jX
n=1
---::-- = E /3~xn , we have
1 m
n{1 - /3jX)
m
J
E.
= E I-P'X ;=1
1
m
00
= EEEjf3jxn, ;=1 n=O
1
where the E; are the coefficients in the partial fraction expansion. Hence, n
bn = EE;/3j, j=1
so that it is natural to take m
fez)
= EE;{3j, ;=1
where
{3i
is suitably defined.
21. Integer-Valued Entire Functions
141
Example Q(z) = 1 + z2 = (1 _1_
Q(z)
+ iz)(1 -
iz)
= ~_1_ + ~_1_ = ~{1 + iz + i2z2 + ... } 21-iz
2
21+iz
1 + 2{1 + (-i)z + (_i)2 Z2 + ... }
bn =
~[in + (-i)nl
2 {bn } = 1,0,-1,0,1,0,-1,0, ...
J(z)
= ~W' + (-Wl = cos
i
z,
In case Q has repeated roots, or if P is not constant, some minor modifications must be made, and in general we have m
bn
= LPj(n)fij, ;=1
so that we ta.ke m
(21.2)
J(z)
= LPj(z).Bj, j=t
where the Pj are suitable polynomials (not necessarily integer-valued).
Definition. Let Rl be the class of functions
J constructed above.
Definition. Let R be the class of integer-valued entire functions exponential-type for which h f (±7r /2) < 7r.
f of
Our problem is to find additional growth conditions on J that imply JERI if fER. The conditions will be phrased in terms of the "mapping radius" of certain sets associated with the indicator diagram of f.
Definition. Let S be a simply connected open set containing 0 such that the complement of S contains at least two points. Let ip be the function (whose existence and uniqueness is guaranteed by the Riemann Mapping Theorem) that maps S conformally one-one onto the unit disk II) = {z : Izl < I} and such that
°
We shall need the following deep theorem of P6lya, which we state without proof. The proof may be found in [8].
142
Entire and Meromorphic Functions
Theorem. If g(z} = E;:"=o bnzn , bn E Z, and if 9 is analytic in a region S containing 0, with peS) > 1, then there exist polynomials P and Q, with integer coefficients, Q(O} = 1, such that P g=
Q'
We also require a simple lemma on polynomials, whose proof we leave to the reader.
Lemma. If P and Q are polynomials with integer coefficients and Q(O) = 1, then there exist polynomials PI and Ql with integer coefficients Ql(O) = 1 such that PI and Q1 hut,f' no c.nmmon factors and PdQ1 = PIQ. Now given
I
E R, let 00
= LJ(n)zfl.
g(z)
n=O
As we have seen earlier,
g(z)
1 = -2' 11"
i
r
1 1 W ~(w)dw - ze
for any curve r that winds once around DU) = S·(~). Now let S be the complement of the image of D(f) UDder the mapping e- w , i.e., S = C\ exp( -D(f». Now 9 is analytic in S, 80 that if peS) > 1, then 9 = where P and Q are polynomials with integer coefficients and Q(O) = 1. By the lemma above, we may suppose that P and Q have no common factors. By the construction that characterizes R I , we can find a fUDction ft E RI such that ft(n) = I(n) for n = 0,1,2, .... By Carlson's Theorem, if we know that hit (± ~) < 11', then we have I = i1, so that I E R. To prove that hit (±~) < 11', we write
5,
m
i1(z)
=L
Pj(z)Pj.
j=l
By construction, the Pi l are the roots of Q, so that the Pi l are the singularities of g, and hence the Pi l are in the complement of S. Hence, we may write Pj = exp(-/Lj), where /Lj E D(f), so that m
ft (z)
=
E Pj(z) exp( -/LiZ), j=l
and it follows that h/J (±~) < 11' since D(f) is interior to the strip We therefore have proved the next theorem.
Iyl <
1('.
21. Integer-Valued Entire Functions
143
Theorem. If fER, and if the complement of the image of D(f) under the map e- W has mapping radius exceeding 1, then f E R l •
For applications, a variant of the foregoing procedure gives a more useful result. We let 6. be the difference operator defined by (6./)(z) = fez
+ 1) -
fez),
defining and Now, using Taylor's Theorem, we may write 6.=e D - l 6.'" = (e D
where
_
1)"',
D=~.
dz We define the functionals T... and T: by T... f
= I(n)
T=I = (6.'" /)(0). To illustrate,
T; = 1(0) Ti I = f(l) -
1(0)
T; 1= f(2) - 2/(1) T; 1= f(3) - 3/(2)
+ 1(0) + 3/(1) -
1(0).
It is easy to show that
T= = (_I)n ~(_I)k (~)Tk'
Tn
=
t (~)Tk' o
For example, the first identity is proved on writing
~n =
(e D _I)'"
=
t (~)ekD(_I)n-k. k=O
Entire and Meromorphic Functions
144
Definition. To say that a sequence {bn}, n is to say that there is an J E R such that
T;:/
n
= bn,
= 0, 1,2, ... , is K* -admissible
= 0, 1,2, ....
For / E R, write 00
g(z)
= ~)T::f)zn. o
It is easily seen that
r
1
g(z)
1
= 21fi if' 1 _ z(w) ~(d)dw,
where (w) = eW - 1 and r is a curve that winds once around D(f). We see that g is analytic outside the image of D(f) under the map (e W _1)-1. The argument may be reversed to prove the next result.
Theorem. A sequence {bOl } is K* -admissible iJ and only if analytic on the segment [-1,0].
E bnz
tl
is
We may now prove the main result of this section.
I E RlJ let E be the complement of the image of D(f) under the mapping (e W - 1)-1 and let E* be the image 0/ D(f) under the mapping eW -1. If p(E) > 1, then I E Rl and J(z) = E Pk(z)(1 + fA)''', where the Pic are polynomials and the Pic run through the complete sets of conjugate algebraic integers lying in E*.
Theorem. If
Prool. Let and let
As we have seen, 1 ( 1 g(z) = 211'i if' ~(w) 1- z(ew _ 1) dw
F(z) = -1.
Ir
21fz r
~(w)
Now, 1 - 1g(z) = 21fi 1 + z
so that g(z)
Irf'
1
1 - ze w
~(w)
dw.
1 dw 1 - l~Zew
= l~zFC;z)'
21. Integer-Valued Entire Functions
145
Similarly, F(w)=_l g(~). l-w l-w
G,
Since p(~] > 1, we see by the P61ya Theorem that g = where P and Q are polynomials with integer coefficients and Q(O) = 1. Thus, 1
F(w)
P( ~)
(1- w)N P( l~w) w)N+l Q( ~)
= 1- w Q( 1~u,) = (1 -
P*(w)
= Q*(w) ,
where we choose N;::: max(degP,degQ). Now, P" andQ* are polynomials with integer coefficients and Q*(O) = 1. Thus, I E Rl . liB berore, we see that
I(z)
= E P.(zhi,
where the 'Y. are the reciprocals of the roots of Q* and the p. are polynomials. If we write 'Y. = 1 +P., we see that Pi 1 is a root of Q, and since the roots of Q are the singularities of g, the theorem is proved. Using this theorem and some facts about algebraic numbers, the next two results can be proved easily. We state them without proof, as illustrative applications. For details, see the paper of Buck [7).
Theorem. II I is an integer-valued function 01 exponential type such that hJ(1f/2) = h J(-1f/2) = 0 (that is, the indicatordiagmm of f is a horizontal line segment), and if L = exphJ(O) - exph/(1f) < 4, then I E R I . Theorem. II, in addition, L po. Ph ... ,Pn , we have fez)
<
v's,
then for some polynomials
= Po{z) + P 1 (z)2 + ... + Pn(z)nz • Z
22 On Small Entire Functions of Exponential-Type with Given Zeros
This chapter is extracted from a paper of the same name by P. Malliavin and L. A. Rubel [22). We obtain here a result that considerably generalizes Carlson's Theorem presented in Chapter 20. For a sequence A of positive real numbers, we denote by F(A) the ideal, in the ring of all entire functions, of those entire functions that vanish at least on A. (We exclude once and for all the null function f = 0 and the ideal containing only the null function.) We introduce an order relation in this system of ideals, F(A) < F(A'), meaning that for each 9 E F(N), there is an f E F(A) such that If(iy)1 ~ Ig(iy) I for every real y. Crudely stated, F(A) < F(A') if it is easier to construct small entire functions that vanish on A than those that vanish on A'. The major problem is to decide, by elementary computations on A and N, whether F(A) < F(A')j we solve this problem here. By specialization, then, we prove as a corollary the following result. Theorem 22.1. There exists a function f If(iy)1 ~ exp7rblYI if and only if
>.(y) - >.(x) ~ blog (;)
+ 0(1),
E
F(A) such that
x ~ y,
where >.(t) is the sum of the reciprocaLs of the elements of A that do not exceed t. Remark. Carlson's Theorem deals with the case b = 1 and A = {I, 2, 3, ... }. The main innovation of our method is to give our entire functions suitable zeros on the imaginary axis, in addition to the required real zeros.
22. On Small Entire Functions of Exponential-Type with Given Zeros 147
We proceed now to the body of the exposition. We study sequences A = {An} of positive real numbers, A :0
and define
L
A(t) = A(t)
< AO ::; Al ::; ... , A;;-l
= L 1= A.. ~t
lt
S
d,x(s).
0
Definition. ,x(t) is called the chamcteristic logarithm of A, and A(t) is called the counting junction of A. For simplicity, we suppose that A is an infinite sequence and that
D(A) = limsup A(t) < 00, t--oo t since the problem is trivial if A is finite or if D(A) = 00. The function W(z) = W(z : A) belonging to F(A) is called the Weierstmss product (over A) and is defined by
= II (1- ~~) .
W(z : A)
We may write log IW(z : A)I where z
= reiB • For 0 #- 0, 11'
=
1
r; e2i81
00
log 11 dA(t}, o t we have, on integrating by parts,
log IW(reiB)1
=r
1
00
o
1
P(t,O)A(rt)- dt,
rt
where P(t 0) = 2 ,
We define, for 0 < b <
00,
1 - t 2 cos 20 . 1 - 2t2 cos 20 + t 4
the arithmetic progression Ar. by
Ar.={~,~,~, ... } and observe that Ab(t) = [btl = bt + 0(1) ).b(t) = blogt + 0(1) (for t;::: 1) W(z : A b ) = sin 1rbz 1I'bz hWb(8) = 1I'blsin81·
We write A C A' to indicate that A is a subsequence of A', and remark that A C A' if and only if ).(y) - >.(x) :5 ).I(y) - ).I(X) for x :5 y.
Entire and Meromorphic functions
148
Definition. A is equivalent to A', written A '" A', shall mean that N(x),\(x) = 0(1). Definition. A' > A shall mean that there exists a sequence A", A" :J A, such that A" '" A'. Definition. A < A' shall mean that there exists a sequence Alii, A'" such that A'" '" A.
c
A',
Although A < A' and A' > A mean two different things, the first corollary of the next lemma resolves this notational difficulty.
Lemma 22.2. A> A' if and only if (22.1)
A(Y) - A(X) $ A'(y) - A/(X) + 0(1);
0 < x $ Y < 00.
Likewise, A < A' holds if and only if (22.1) is satisfied.
> A.
Corollary 22.3.
A < A' if and only if A'
Corollary 22.4.
If A I"V All A' '" A~, and A < A', then Al < A~.
< A2 and A2 < A3 , then Al < A3 • Corollary 22.6. If A < A' and A' < A, then A '" A'. Corollary 22.5.
If Al
Thus, < is a well-defined partial ordering of equivalence classes under
"'. Proof of Lemma 22.2. That A' > A and A < A' each imply (22.1) is trivial. To show that (22.1) implies that A' > A, we define
cp(x)
= inf{A/(S) -
A(S) : S ~ x}.
It follows from (22.1) that cp(x) ~ -K for some constant K. Now cp(x) is constant except for possible jumps at the jumps of >.'(x). Let Xo be a point of discontinuity of cpo Then,
cp(xo - 0)
= A/(XO -
0) - A(Xo - 0)
and
cp(xo
+ 0) $
A/(XO + 0) - A(XO + 0).
We denote by 6,cp(xo) the jump of r.p at Xo. Then (22.2)
6,cp(xo) $ 6,,\I(XO) - 6,'\(xo) $ 6,'\/(XO).
We let A*(t) = [cp(t»),
22. On Small Entire F\mctions of Exponential-Type with Given Zeros 149
where [a] denotes the integral part of a and
cI>(t) =
lot
S
d
and let ).* (t) be the characteristic logarithm of tha.t sequence A* whose counting fundion is A*(t). The function ).*(t) is constant except possibly at the jumps of
FUrthermore, implies
xo~).-(xo)
and
xo~).'(xo)
must be integers, so that (22.3)
(22.4) and this means that A* is a subsequence of A'. We now define )."(x) = ).(x) + ).-(x), so that )'''(x) is the characteristic logarithm of some sequence A" ::> A. To prove that A" '" A', we must prove that 6(x) = 0(1), where 6(x)
= ).(x) + ).*(x) -
Now,
).*(t) =
=
1t
1
o s
).'(x). dcI»(s)
10r!s d[cI>(s)].
An integration by parts shows that ).*(t) -
(~),
so that it is enough to prove that O(x) = 0(1), where O(x) = ).(x) +
= ).(x) -
).'(x) + inf{)"(s) - ).(8) : 8 ~ x}.
But it is clear that 9(x) :::; 0, and (22.1) is simply another way of saying that O(x) ~ 0(1). To prove that (22.1) implies that A < A', we put (22.5)
)."'(x) = ).'(x) - ).*(x).
Since by (22.4) A* is a subsequence of A', there is a subsequence Alii defined by (22.5) and Alii is a subsequence of A'. Since we already have shown that Alii '" A, i.e., 6(x) = 0(1), the proof is complete. We now state the main result.
Entire and Meromorphic Functions
150
Main Theorem. Given A and A', the following three statement are equivalent
(i) F(A) < F(A'). (ii) A < A'. (iii) There exists a single pair, fo, go with fo E F(A), 90 E F(N), Ifo(iy)1 ::; 19o(iy)I for all real y and such that the only zeros of 90 in the open right half-plane belong to A'. Theorem 22.1 is a direct corollary of this result. Given A and b, choose A' = Ab and 90(Z) = sin1l'bz. Since 190(iy)I '" e1rbl ll l, the equivalence of (ii) and (iii) proves Theorem 22.1. Proof of the Main Theorem. We leave the proof that (ii) implies (i) for later. It is clear that (i) implies (iii); a suitable choice for go(z) in (iii) is the Weierstrass product W(z : A'). We now prove that (iii) implies (ii). We write f and g instead of 10 and 90. Now we choose P with 0 < p < Ao so that all the zeros, Zn = rnei6", of 1 in the right half-plane (assuming for convenience that 1 has no zeros on z = iy) satisfy r n > P, and write one form of Carleman's Theorem (Chapter 12), taking y > x > p as (22.6)
E(y) - E(x)
= l(y) -
lex)
+ J(y) -
J(x)
+ 0(1),
where
E(R)
= E(R: f) =
L (r~ - ~~) casOn,
r,,:SR
1JfR (1 1)
J(R)
= I(R: f) =
J(R)
1 = J(R: f) = -R
211'
11'
t 2 - R2
p
I
J
-J
log I/(it)/( -it)1 dt,
10gl/(Re''6 )lcosO dO.
Now (22.7)
L
~ casOn
= 0(1)
since
IL ~~ casOnl ::; L ~~ = ~ L ~ ::; ~n(R) = 0(1), where nCr) counts the number of zeros of I whose modulus does not exceed r. Also, (22.8)
J(R) = 0(1)
22. On Small Entire Functions of Exponential-Type with Given Zeros 151
since
11f~
I:
log If(Re i6 )1 cos OdOI
~ 1f~
I:
Ilog If(Rei6 )11 dO
~ 7r~O(R) = 0(1), since
:7r
I:
Ilog If(Rei6 ) Ii dO = meR, f) + m ( R, 7) ~ 2T(R, f)
and f is of exponential-type. From (22.7), we obtain (22.9)
E(y) - E(x)
~
>.(y) - >.(x) + 0(1),
and using (22.8) and (22.9) in (22.6) we get (22.10)
>.(y) - >.(x) ~ ley) - lex)
+ 0(1).
Now, since If(iY)1 ~ Ig(iy)l, we see that (22.11)
l(y: f) - l(x: f)
~
l(y: g) - l(x : g).
On the other hand, applying Carleman's Theorem now to g, whose only zeros in the right half-plane are the >.~, we see that (22.12)
l(y : g) - l(x : g) = A'(y) - >"(x)
+ 0(1).
Combining (22.12) with (22.11) and (22.10), we get
>.(y) - >.(x) ~ >.'(y) - A'(x) + 0(1), and the proof is complete. To prove now that (li) implies (i), we suppose that >.(y)->.(x) ~ A'(y)>"(x) + 0(1), and we are given 9 E F(A'); we must construct a function IE F(A) with If(iy)1 ~ Ig(iy)1 for all y. By Lemma 23.1, we may suppose that A(t) = >"(t) + 0(1), since A is a subsequence of a sequence A" for which this is true, and F(A"} C F(A). By the Hadamard Factorization Theorem we may write
where 91(Z)=n(1-
92(Z)
~)exp(:~J,
= CzkeQzn (1- ~) exp (~) ,
Entire and Meromorphic Functions
152
where the (n ¥ 0 are the zeros of 9 that are not counted in A'. Writing log 191(iy)1 as a sum of logarithms, and that sum as a Stieltjes integral, we get (22.13)
log 191(iy}1 =
~
1 (1 + ;:) 00
t d>.'(t).
log
The next lemma provides the main tool of our construction; it will enable us to "move the zeros" from the real axis to the imaginary axis.
Lemma 22.7.
Let d.6. be a measure with compact support contained in an interval [E, e l ] for some small f > o. Then there exists a function rp(t) defined on (0,00) such that
(22.14)
and
1
I 0 X1-8 d.6.(s)l·
Irpl < 2 sup x
Proof. By a contour integration, it is easy to see that
21
log 11 + x 2 1 = -
00
log
0
11"
x2 t dt 11- -1--. t2 t2 + 1 t
By Fubini's theorem, (22.15)
J (
X2) d.6.(r)=-;2 log l+r2
J
We therefore are led to define 2
(22.16)
{J
x2 1 logll- w2
'1'( w) = -;
J
t2
w2 + t2
wit } dw w 2jt 2 +1 d.6.(t)-;;.
d.6.(t) t '
and (22.14) asserts (22.13) in another form. The bound on 'I' follows from integrating by parts in (22.15): (22.17)
rp(w) =
~ 11"
roo dLl,(t) _ ~ roo { r d.6.(t)}
10
t
11"
10
10
t
dx (
2x 2). 2
x
Hence, 1
{+ 1
00
1
0
dz;
(
r dA(t) x2 x2)} + w2 s~p IJo -t-I
+w
22. On Small Entire Functions of Exponential-Type with Given Zeros 153 2
since x2~W2 is increasing, and the lemma is proved. We now choose
(22.18)
8(t)
= ~{A/(t) -
A(t)},
d6.(t)
= t d6(t)
but cannot apply Lemma 22.7 to d6. since its support may not be compact. We truncate the support by defining
6 (t) k
6.k(t)
= { 6(t)
8(k)
ift ~ k ift> k
= t tMk(t)
with the same cODvention for A(t) and N(t). We now apply Lemma 22.7 to d~k and conclude that there exist functions r,ok(t) such that
(22.19)
Now, (22.20) where B is a constant that is independent of k, namely, from the bound on Ir,o(t) I in the lemma and the equivalence of A and A',
B = 2 sup IA(t) - A'(t)l.
On putting
(22.21)
Lk(Y)
=
flOg (1 + ~:) ~t dAk(t) + flOg 11 - ~: I dcI>k(t),
where
(22.22) we have (22.23) Rence, by (22.13), (22.24)
klim -00
Lk(Y) = log 191(iy)l·
Entire and Meromorphic Functions
154
At this point, the idea is to find an entire function F for which the hypothetical formula log IF(iY)1
= k-co lim flOg 11 -
Y: I d~k(t) t
holds in some appropriate sense. First, however, the limit need not exist, but a simple argument with normal families will handle this difficulty. Also, the measures d~k(t) Y'k(t)dt are unsuitable since they need not be positive and cannot be discrete. [It is easy to see that all the d~k(t) are positive only in case A ~ A', a trivial case.] But first we show that adding a. constant to 'P/c, in order to make d~k(t) positivp.; does not change L k • Then we show that the resulting measure may be made discrete with little loss of precision. Resuming the construction, we define
=
(22.25) and by (22.20) conclude that (22.26)
A contour integration shows that
J
log 11
(22.27)
-
y2
t2'1 dt = 0,
so that
(22.28)
Lk(y)
=
J ( + t2' log 1
y2) 2 t dAk(t) +
where dWk(t) = ,pk(t) dt. Now let Wk(t) Wk(t), and define (22.29)
L;(y)
Lemma 22.S.
=
J (1 + ~:) ~ log
J
= [Wk(t)],
dAk(t) +
1 t21
log 1 - y2 dWk(t),
J
the integral part of
log 11
-
~: I dW;(t).
There is a constant (3, independent of k, such that for all
y>I
(22.30)
flOg
11 -
~: I dwk(t) :$
J
log 11 -
~: I dw/c(t) + (3log Iyl·
Proof. We apply the next lemma with Wk(t) = vet) and Wk(t) = net). (3 is independent of k because I!tWk(t)1 and IW/c(t) - Wi(t)1 are bounded independently of k.
22. On Small Entire Functions of Exponential-Type with Given Zeros 155
Lemma 22.9. Suppose that v(r) is a continuously differentiable junction jar 0 < r < 00, that 0 :s v'(r) < B < 00, that nCr) is nondecreasing, and that for some constant C
vCr) ~ nCr) > vCr) - C. Then
as y -+ 00. Proof For fixed r, we write L(t) Lebesgue integrable on (0,00):
= 10g!1 -
~I and point out that L is
and that L(t) is decreasing and continuous for t E (0, r) and increasing and continuous for t E (r,oo). We must compare Y = L(t)dn(t) and Z = oo L(t)dv(t). We will prove that Y < Z + O(logr). We assume that vet) ~ p > O. This involves no loss of generality since, if we replace vet) by vet) + t and net) by net) + t, we change Y and Z not at all, because 00 L(t)dt = 0. We may suppose, without loss of generality, that v(O) = 0, since suitably redefining v on the interval [0,1] changes Z only by 0(1), which is small compared to the allowed discrepancy O(log r). With each large r we associate the numbers rl and r2 such that
1:
Io
10
Since vet) 2: p, we will have r - rl the following inequalities hold:
:s r2 -
rl
:s ~.
It is easy to see that
for L(t) dn(t) :s forI L(t) dl/(t),
1
00
It follows that Y = X
L(t) dn(t)
+ Z,
:s 1~ L(t) dv(t).
where
We shall prove that X $ O(logr). Clearly, X $ -
l
r2
r,
t-r log I-I dl/(t). t
Entire and Meromorphic Functions
156
Since r2 - rl ::; ~ and V'(t) ::; B, we have X::; -B
t-r dt::; B(r2 1n~ log-/-I t
so that
x::; BpC log (r+
rd logr2 - B
1~
log-It - rl dt
n
~) +2B.
We now consider polynomials Pk defined by
Lemma 22.10.
There is a constant B ' , independent of k, such that for
all z log IPk(z)1 ::;
(22.31)
Proof. Putting z
B'lzi.
= x + iy, we see that
/ log /1
+ :: / dwk
log /1 + ;: / dWZ(t).
But since Il1 k(t) = [I;{B + IPk(s}}ds] , and since IIPk(S)1 ::; B by (22.20), the proof is immediate on integrating by parts. Since the family {Pk(z)} is therefore uniformly bounded in each disk DR = {z : Izl < R}, it is consequently a normal family, and we may extract a subsequence {Pk'(Z)} that converges uniformly on compact sets to an entire function F: (22.32)
F(z)
= k'--+oo lim Pk'(z).
Because Pk(O) = 1 for each k, it follows that F(O) = 1. By Lemma 22.10, we conclude that F is of exponential-type. Since Pk has only imaginary zeros, so has F. Furthermore, F is an even function since each Pk is. Let ir = {i-rn} be the zeros of F on the positive imaginary axis. Since F is of exponential-type, r has finite upper density. Thus, (22.33)
At last we can define fez). Let (22.34)
fez) = ft(z)F(z)!l2(z),
22. On Small Entire Functions of Exponential-Type with Given Zeros 157
where (22.35) As a consequence of the estimates (22.24) and (22.30) we see that log I/(iy)1 ~ log Ig(iy)1
(22.36)
+ O(log lyD.
Now (22.36) is as good for our purposes as I/(iy)1 ~ Ig(iy)l, since we could otherwise consider r(z) = I(z)a{(iz)-l sin(iz)}b for a suitable choice of a and b. It is not obvious, though. that I is of exponential-type, since It and g2 need not be of exponential-type, although they are certainly of order 1. To prove that I is of exponential-type we appeal to Lindelof's Theorem proven in Chapter 13. Let us denote by an and bn the zeros, other than the origin, of I and g, respectively. Then we see that
0(1)
=
L
b~l =
Ib"I~R
=
L
L
b;;l
+ ,\/(R)
Ib"I~R b,,~A'
b;;l
+ '\(R) + (,\/(R) -
b;;l
+ '\(R) + 0(1)
'\(R»
Ib"I~R
b,,'l-A'
=
L Ib"I~R
bn'l-A'
since ,\/(R) - '\(R) = 0(1) by hypothesis. Now the zeros of I, other than the origin, fall into three categories: (i) those bn not counted in A', (ii) the elements of A, and (iii) the zeros of F. If we consider
S(R) =
L
b;;\
Ib"I~R
the zeros of F contribute nothing to S(R) since F is even. Hence, S(R)
=
L
b~l,
Ibnl~R bn~A'
and it follows that
S(R) = 0(1). This, together with the obvious fact that the zeros of I have finite upper density, implies (via Lindelof's Theorem) that f is of exponential-type. The proof is complete.
23 The First-Order Theory of the Ring of All Entire Functions
The material of this chapter is drawn from the paper [3], "First-Order Conformal Invariants." Let denote the ring of all entire functions as an abstract ring. Much information about the theory of entire functions is present in the theory of e. For example, an entire function f omits the value 7 iff there exists an entire function 9 such that (f - 7)g = 1. We show that even using a restricted logic (first-order logic), a great deal can be expressed in the theory of e. We shall show, indeed, that all of classical function theory can be so expressed. By the ring language we mean the first-order formal language appropriate to the structure e. This language has basic symbols for addition and multiplication of entire functions, as well as the usual logical symbols: 1\ ("and"), V ("or"), -, ("not") and ==> ("implies") as well as quantifier symbols Y ("for all") and 3 ("there exists") together with variables that range over e. For convenience we also include in the ring language a constant symbol which is a name for the constant function i = A. (Otherwise, there would be no way to distinguish between the two solutions of P + 1 = 0.) Formulas and sentences in this language are finite combinations of these basic symbols, arranged according to the obvious formal rules of grammar. A key restriction is that the language is first-order, which means that we can only use quantifiers over elements of C and not over subsets, ideals, relations, etc. Also, the expressions in a first-order language are always finite in length. (See [39] for a general treatment.) The algebra language is appropriate to C as an algebra. This is formed by adding to the ring language a 1-place predicate Const. In e (as an algebra) we interpret Const(f) to mean that f is a constant function. {Thus we are
e
23. The First-Order Theory of the Ring of All Entire Functions
159
identifying C with the field of constant functions in £ and are using the ordinary addition and multiplication of functions in £ to play the role of addition of constants and the scalar multiplication in the algebra.) But in £ it is easy to say that f is a constant (either f = 0, f = 1, or f omits 0 and 1) so that the ring language and the algebra language are equivalent. In dealing with rings and algebras, the expressive power of first-order sentences is reasonably well understood. For example, to say that a ring is commutative is first-order (Vx'Vy(xy yx» but, at least superficially, to say that a ring is simple is not first-order, since it seems to require quantification over subsets (there does not exist a proper two-sided ideal), and in fact does so require. It is first-order to say that a ring has at least two elements (3x3y(x :f. y», but it is not first-order to say that a ring is infinite, since this requires a sentence of infinite length, as one might suspect. Nearly all of the results in this chapter depend on the fundamental definability results for the algebra £. We show that there are formulas in the algebra language which define in £ the set N of nonnegative integers, the set Z of all the integers, the field of rational numbers Q, the field of real numbers lR, the ordering relation ~ on lR, and the absolute value function on C We also show how to interpret in the first-order language of £ the quantifiers that range over countable sets and sequences of constants. (It is striking that second-order concepts can be represented within the restricted first-order language.) An immediate consequence of this is that second-order number theory is interpretable in the first-order theory of £. Other recursive undecidability results are treated later. For example, we show that the first-order theory of the ring of entire functions is recursively isomorphic to second-order number theory. (This improves on a result of Robinson [33], who showed how to interpret first-order number theory in the first-order theory of entire functions. ) Later we extend these definability results even further. We give enough examples to suggest that all of classical analytic function theory on C can be interpreted in the first-order theory of £. This is quite surprising, given the apparent limitations of the first-order algebra language. We also show how to interpret in the first-order language of £ the quantifiers that range Over countable subsets and sequences in £ itself. That is, the first-order language is already as expressive in this context as the restricted secondorder language.
=
The Ring Language
In this section we will begin to explore the expressive power of the first-
order theory of the ring e. To begin, note that the constant functions 0 and 1 are definable using their first-order properties in e. Also, the property that f is a unit in e is first-order expressible: 3g(/g = 1). Thus we can
160
Entire and Meromorphic Functions
express, for any definable constant c, the condition that f omits the value con C by saying that f - c is a unit in e. Since 1 and i are definable in e, so is each Gaussian rational number. This means that for each Gaussian rational q there is a formula Fq(x) in the algebra language such that, for any function fine, Fq (I) holds in if and only if f equals the constant function q. In this section we present a detailed study of certain basic definability questions for the algebra e. These matters are fundamental to the general content of this chapter and are of interest in their own right. We call a function fEe a point function if it has a unique zero on C, of multiplicity one. These will be used to represent the points of C within It is easy to construct a formula P(x) in the algebra language such that P(f) holds in e if and only if f is a point function. That is, P(f) should express
e
e.
f is not a unit, and YgYh[f = gh ~ 9
is a unit or h is a unit].
Using this we can, for example, construct an algebra language formula A(x) such that AU) holds in e if and only if f is It 1-1 conformal mapping on C That is, A(f) should express the condition: For any constant a and any point functions 9 and h, if both 9 and h divide
f - a, then 9 divides h.
[To say that a point function 9 divides f - a is equivalent to saying that f has value a at the unique point of C where 9 is zero. To say that one point function divides another is equivalent to asserting that they are zero at the same point of C. Of course the conformal maps of C are exactly the functions I(z) = az + b, a i= 0.] Next we discuss how to code arbitrary countable or finite sets of constants in C using first-order formulas. It is very striking that we can represent second-order mathematical concepts in a first-order language. Given I, gEe, define V( a; I, g) to mean that a is a constant, 9 i- 0, and there exists Zo E C such that g(zo) = 0 and I(zo) = a. This can be represented by an algebra language formula, since Yea; I,g) is equivalent to the existence of a point function h such that h divides 9 and h divides 1 - a (together with the other conditions, that a is constant and 9 i- 0). We think of the pair (f, g) as coding the set of constants E = {a E C
I ~r(a;f,g) holds
in £}.
If 9 = 0, then this set is empty. If 9 =I 0, then it is a countable or finite subset of C. Moreover, by taking 9 to have an infinite sequence of zeros tending to 00 and by letting 1 vary over £, then we obtain for E all possible countable or finite subsets of C [38].
23. The First-Order Theory of the lung of All Entire Functions
161
We may use this idea, for example, to find a formula IZ(x) in the algebra language such that IZ(g) holds in [ if and only if the zero set of gin C is infinite while g"l O. Namely, let IZ(g) be the formula:
g"l 0/\ 3![V(O; f, g) /\ V'a(V(a; f, g) ===> V(a + 1; f,g})J· This condition asserts the existence of a function f such that the set E ~ N. By the well-known interpolation theorem for entire functions this can happen exactly when g has infinitely many zeros. Theorem 23.1. The following sets and relations are all definable in [ by formulas in the algebra language: The set Z of (positive and ftegativej iniegers,
That set N of nonnegative integers, The set Q of rational numbers, The set Q( i) of Gav.ssian rationals, The ordering relations ~ on Z, N, Q, and The absolute value function on Q.
Proof. This is immediate from the discussion above. To get N we use the Peano axioms: N is the smallest countable subset of C that contains 0 and contains a+ 1 whenever it contains a. Precisely, the formula N(a) defining Nis a is constant /\
V' j\fg[{V(O; f, g) /\ V'/3(V(/3; f, g» ===> V(/3 + 1, f, g)} ===> V(ajJ,g)]. Once having defined N, Z and Q are trivial:
a EZ
{::::::>
a E IIi V -a E N,
a EQ
{::::::>
3/Hr[13 E Z /\ '"Y E N /\ '"Y "I 0/\ Q'"'{ = 13].
To define the Gaussian rationals Q( i) we note that
a E Q( i){::::::> 3/33')'[.8 E Q /\ ')' E Q /\ a = /3 + ')'i]. To define
~,
on Q say, note that
a E /3-¢=:::}3,),36[,), E N /\ 6 E N /\ ')' Finally, for a
"I 0/\ ')'(/3 - a) = 6J.
E Q,
101 =
{J-¢=:::}O ~ /3 /\ ({J = a V /3 = -a).
Note that since N is definable in [, there is an effective procedure for interpreting the first-order theory of (N, +, .) into the first-order theory of
Entire and Meromorphic Functions
162
f. (The operations + and· are just the natural operations on f.) Actually, this interpretation extends to the second-order theory of (fIl, +, .), since we havc a way to discuss countable sets of eonstants in the first-order language of f. It follows that the first-order theory of f is very undecidable in the sensc of recursive function theory. This will be discussed fully below. We caution the reader that at this stage we are only able to define the field of mtional constants in f. It is possible to define the real field lR, as we show below, but there does not seem to be any easy way to do it. Once we have a first-order definition of R in f, no matter what it is, we get immediately as a bonus a first-order definition of ;£ on R and of the absolute value on C:
a ;£ ,8<==>3,),[..,. E R I\. ,8 = a
+ ')'2],
lal = ,8<==>3')'36[..,. E R 1\ 6 E R 1\ a = ')' + 6i 1\ ,8 E R 1\ 0 ;£ ,8 1\ ,82 = ')'2 + 62J. We now turn to the construction of an algebra language formula R(x) that defines the field of real constants in E. The construction of R( x) is complicated, but it is based on an elementary fact: A constant a is a real number if and only if there is a Cauchy sequence S of rationals such that every analytic function that carries S to an equivalent Cauchy sequence of rationals also preserves a. Our problem has two parts: to prove that this equivalence is correct and to construct an algebra language formula that expresses the right side of it. To prove the equivalence we need the following fact: Lemma A. If 13 E R and a E C\R, then there is an entire function f such that /(13) =,8, lea) :j: a, and / maps Q onto Q. Proof. This argument is a simple modification of the proof given in [27]. Fix 13 E R and a E C\R. If 13 E Q, then / is trivial to find, so we may assume 13 rt Q also. We take p(z) = e1zl , so that p(z) ~ elzl for all z E C. Take S = T = QU {f3} and choose /0 E EM so that
/0(13)
=
1
"2 13 , I~(fo(a) -
a)1 ~ 2p(lal),
and let 6 = t. (Here, EM is the set of all entire functions whose restriction to R is a real, monotonically nondecreasing function.) Let us take an enumeration of S (and T) with Xl = f3 (in particular, Xl :j: 0), and let {j > 0 be such that fo(xd
+ 6Xl
E
T
and
18z1 ~
1
"2 p(lzl)
for all
where fo is any function chosen from EM. We define
z
E
C,
23. The First-Order Theory of the rung of All Entire Functions
163
Note that SI = {{J} and h({J) = {J, so that Tl = {{J} also. We now construct the sequences {In}, {Sn}, and {Tn} SO that I~(x) 2: (Tl
+ 2- n )6
and
In(Sn)
= Tn
for all n E N, x E lIt Suppose that In, Sn, and Tn have been constructed and choose a polynomial 9 with real coefficients such that (1) g(z) = O~Z E Sn (z E C) (2) Ig(z)1 ::; 2- n - l p(lzl) (z E C) (3) g'(x) 2: -2- n- 16 (x E JR). [Any polynomial of odd degree with positive leading coefficient for which (1) is valid will also obey (2) and (3) after it has been multiplied by a small enough positive constant. The degree can be chosen odd by adjusting the multiplicity of one of the zeros.] For each M E [0,1] we have
(In
+ Mg)'(x) = I~(x) + Mg'(x) 2: (2- 1 -
2- n- 1 )6
(X
E R),
that In + Mg is strictly increasing on R. Moreover, if we let M vary in [0,1], then for x rt S,,' (In + Mg)(x) varies in an interval of R that contains points of T\T"j and for y rt Tn, (I" + Mg)-I(y) varies in an interval of R that contains points of S\Sn' Now for n odd, let x be the point of S\Sn with smallest index and let ME [0,1] be such that (In + Mg)(x) E T. We define
SO
In+!
= In + Mg,
Sn+!
= Sn U {x}
and
T,,+1
= Tn U {I,,+!(x)}.
For n even, let y be the point of T\Tn with smallest index and let ME [0,1] be such that {In + Mg}-I(y) E S. We define
In+l = In
+ Mg,
Tn+! = Tn U {y}
and
Sn+!
= Sn U {I;;~l (y)}.
The following properties of the constructed sequences are easily verified: (a) Ifn(z) - In-l(z)1 ::; Tnp(lzl) (n E N,z E C) (b)
00
00
n=1
n=l
U Sn = S, U Tn = T, and
From (a) it follows that In converges pointwise to a function f for which I/(z) - 10(z)1 ~ p{lzl)
(z E C).
Entire and Meromorphic Functions
164
Now p(lzl) is a function of z that is bounded on compact subsets of C, so the convergence of {In(z)} is uniform on such sets. From this we conclude that ! is an entire function. For each 11 EN we have I~(x) 2: 46, (x E JR), so the same is true for f. Hence, I is strictly increasing on lR. From (b), it follows that f(Sn) = Tn for each n, and so f(S) = T. Moreover, we have insured that fn(fJ) = {3 for all n E N and therefore l(fJ) = {3. This implies f(Q) = Q also. Finally we have
1(J'(J(a) - a)l2: 1~(Jo(a) - a)I-I~(Jo(a) 2: 2p(lal) - p(la!) which means I(a)
=I a.
l(a»1
> 0,
This completes the proof.
Now it is easy, using Lemma A, to show that our characterization of the real numbers is correct. In one direction, the equivalence is trivial: H a E lR, we need only take S to be a Cauchy sequence of rationals that converges to a. For the converse direction, suppose that a E C\R and S is any Cauchy sequence from Q. Let fJ be the limit of S in lit Use Lemma A to get an entire function I such that l(fJ) = {3, I(a) =I a, and I(Q) ~ Q. Then I maps S to a Cauchy sequence of rationals that is equivalent to S (since both converge to {3) yet I does not preserve a. That is, a E C\lR implies that the right side of our equivalence is false. This completes the proof that our characterization of lR is correct. Now we must express it formally within the algebra language.
Theorem 23.2. There are formulas R(x), L(x,y), and M(x,y) in the algebm language such that for any a, (3 E (i) a E lR<=> R( a) holds in e,(ii) For a, {3 E lR, a ~ {3<=>L(a,{3) holds in C,(iii) For a, {3 E C, lal = {3~ M (a, {3) holds in C.
e:
Proof. We begin by building some machinery for discussing sequences of constants within the first-order language of C. (Earlier we did the same for countable sets of constants.) This is done using a triple of functions (J, g, h)j 9 has infinitely many zeros on C, and on that zero set h takes on the values 0, 1, 2, .... In effect, h lists the 7..eros of g. Then the sequence coded by (f, g, h) is (an), where an is the value !(zn) at the zero of 9 where h take the value n. First let Basis (g, h) be a formula in the algebra language which expresses the fact that h "lists" the zeros of 9 in the manner discussed above: Basis(g,h)~'v'a[V(a;h,g)~a E /I. 'v'a'v'p'v'q[{P(p) /I.
NJ
P(q) /l.pdividesq /I. p divides h - a /I. q divides 9 /I. q divides h - a} ==> p divides qJ.
23. The First-Order Theory of the Ring of All Entire Functions
165
Now we construct an algebra language formula Seq(a, nj /,g, h) which expresses that a is the nth term of the sequence of constants coded by the triple (I,g,h): Seq(o, nj /, g, h){:=::?n E N /\ Basis(l, g, h)
/\ a E C /\ 3p[ P(p) /\ p divides g /\ P divides h - n /\ p divides / - 0].
Using the defining formulas described in Theorem 24.1, we now can say, using a formula in the algebra language, when the sequence coded by (f,g, h) is a Cauchy sequence of rational numbers and when this sequence of rationals converges to O. (Note that since we have the absolute value function only on Q at this point, there is no hope of discussing converging or Cauchy sequences outside Q. This is precisely our difficulty in this entire discussion.) For the first of these: Cauchy Rat
Seq(f,g,h)~Basis(g,h)
/\ "Ia"ln(Seq(a,n,/,g,h)
===> a
E
Q)
/\ "16 E Q+3m E Nrli,j E Nrla,{3 E Q({m ~ i
/\ m
~ j /\
===>
la -
Seq(a, i, /,g, h) /\ Seq({3,j, I,g, h)}
{31 ~ 6].
Here we simply have written down in formal terms the usual version of the concept to be defined. (Q+ is the set {q E Q I 0 < q}, so it is defined by an algebra language formula.) For rational sequences converging to 0, we have an entirely similar formula: Zero Rat Seq (f, g, h)-¢=:;} Basis(g, h) /\ "Ia'v'n(Seq( a, n, I, g, h)
===> a E Q)
/\ 'v'6 E Q+3m E Nrli E N'v'a E Q[{m ~ i /\ Seq(a,i,f,gh)}
===> 101
~
6].
Consider a pair (g, h) for which Basis (g, h) holds and suppose that h, 12 are functions so that (h,g,h) and (12,g,h) code Cauchy sequences of rational numbers, say (an) and ({3n), respectively. Then it is clear that (f1 - 12, g, h) codes the sequence (on - (In). [Here it is essential that the same (g, h) be used.] Therefore, (on) and ({In) are equivalent if Zero Rat Seq(ll - hg,h} holds. Next we face the second major difficulty in formalizing our charactrization of R in the algebra language: We have no direct means of expressing that "the value of / at a equals {J," where f E £ and 0, {J E C. We approach this indirectly by introducing, as a parameter, a 1 - 1 conformal
Entire and Meromorphic Functions
166
mapping a on C. Of course a is an element of C and, as noted above, the property of being a 1 - 1 conformal mapping is expressible by an algebra language formula A(a). Now we define a formula in the algebra language, which we will abbreviate by writing f(a) = P, by CT
f(a)
= p~a is a 1 -
1 conformal map
CT
I\a E range(a) 1\ f(a-l(a» ~A(a) 1\ a -
I\a, pEe 1\ (1
-
=P
a is not a unit a divides f - p.
This formula, which will be quite important in later sections as well, here enables us to express the condition that the composition f 0 (1-1 carries one sequence coded by (ft, g, h) to a second sequence coded by (12, g, h). The formula that expresses this is the following:
"In E Nria,p E c(Seq(a,n,ft,g, h) 1\ Seq(P, n,12,g,h) ==* lea) = P]. CT
Note that for a fixed 1-1 conformal mapping (1, as / ranges over C the composition mapping /0(1-1 ranges over C. We now are ready to give the formula in the algebra language which defines R in C. (It is convenient to define C\R instead.) We see that a E C\R~a E C and there is a 1 - 1 conformal mapping (1 on C, and for any Cauchy sequence (an) of rationals that is f E C and a Cauchy Sequence (Pn) of rationals with the properties (i) f(a n ) = Pn for all n EN); CT
(ii) (an) and (Pn) are equivalent Cauchy sequences of rationals; (iii) /(0.) = a is false. CT It remains only to show that this equivalence is correct. Earlier we proved the equivalence a E C\R{::::}a E C, and for any Cauchy sequence (an) ofrationals there is an entire function g and a Cauchy sequence (Pn) of rationals with the properties (i') g(an ) = Pn for all n; (ii') (an) and (P.. ) are equivalent Cauchy sequences of rationals; (iii') g( a) :j; a. Clearly, if (1 exists for a as above, and if (a .. ), (P.. ) and f satisfy (i), (ii), and (iii), then we need only set 9 equal to /0(1-1. Conversely, the range of every 1 - 1 conformal mapping (1 on C includes a and Q. Given such a a and 9 satisfying (i'), (ii'), and (iii'), just take f to be the composition go
(1.
This finally completes the proof that R is definable in c. As was discussed earlier, from this we get formulas defining ~ on R and the absolute value on C. Thus the proof of Theorem 23.2 is complete. Theorem 23.2 is fundamental to nearly all of our other results.
23. The First-Order Theory of the lling of All Entire Functions
167
More on the Algebra Language In this section we present a variety of results which, when taken together, show that the expressive power of the algebra language is strong enough to include all of the classical mathematical theory of entire functions. It is quite striking that this should be possible, since the first-order algebra language seems to be so limited. These results are not used in any other part of this chapter, and we have not made a great effort to be complete nor to give more than sketchy proofs. Our goal is to give examples that show what is possible. We discussed how to deal with sequences of constants within the firstorder theory of e. Now we will improve this to code sequences of /unctions. The idea is to fix a point Zo in C and a sequence {Zn} in C that converges to Zo. By the uniqueness of analytic continuation, a function lEe is uniquely determined by the sequence of constants {I(Zn)}. Hence, a sequence {1m} of functions can be coded by an infinite matrix of constants {1m (z,,) m, n E N}. Now this is not quite enough, since we cannot refer to the points of C in a direct way. We overcame this earlier by introducing a parameter a that is a 1 - 1 conformal mapping on C. We replace the points Zn by the constants an = a(z,,) and refer to the values Pm.n = Im(z,,) by using the equivalent, definable relationship Im(a n ) = Pm n' Finally, we code the 17 ' sequence (an) and the matrix ({3mn) into a single matrix that has (a - n) as the top row. To code an infinite matrix of constants we proceed as earlier, but use a basis triple (g, hI, h 2 ) instead of a pair (9, h). Here we require that g(=I- 0) have an infinite zero set in C, as before, and that (hI, h 2 ) map this zero set bijectively onto N x N. It is routine to construct a formula M Basis(x, y, z) in the algebra language such that M Basis(g, hI> h2 ) holds in e if and only if this basis condition is satisfied. When M Basis(g, hI, h 2 ) is true, any function I in e determines a matrix of constants (a mn ) by taking am" to be the value of I at the unique zero Zmn of 9 for which hl(zmn) = m and h 2 (zm,,) = n. Earlier we expressed this relation using point functions. Also, by the interpolation theorem for entire functions, every matrix (a mn ) is coded in this way by some I, no matter which basis triple (g, hI, h 2 ) is used. The matrices (a mn ) that arise in coding sequences of functions are not arbitrary, of course. First, the sequence (a mn ) from the top row must be a Cauchy sequence in C. Finally, for each m > 0 there must exist a function 1m in e such that Im(aOn) = a mn
for all
n E N.
0'
Evidently, 1m is uniquely determined by this condition. It is the nth function in the sequence coded by (am,,) and a. As was discussed previously, we take the matrix (a mn ) to be coded by some (J,g, hI, h 2 ), where M Basis(g, hI, h 2 ), holds.
Entire and Meromorphic Functions
168
We can construct an algebra formula Code(f, g, hI, h2' a) that holds in £ if and only if a is a 1- 1 conformal mapping on G and M Basis(g, ht, h2 ) holds and the matrix (amn ) coded by I using the basis (g, ht, h2 ) satisfies the conditions above. [That is, the sequence aOn and its limit lie in the range of a and, for each m > 0, there is an 1m in £ so that Im(ao m ) = a mn for all n EN.] Then we formulate an algebra formula (T
F Seq(k, m, I, g, hi> h2' a) that holds in £ if and only if Code (f, g, hI, h2, a) holds, mEN, and k is the (unique) function that satisfies k(aan) = a mn (T
for all n EN. That is, FSeq(k,m,J,g,h b h2 ,a) holds if and only if k is the mth function in the sequence coded by (f,g, hI, h2' a). Using the absolute value on romp lex constants and t.he "evaluation" of functions I via a (that is, in the form 1(0') = /3), we now may obtain algebra formulas that express various types of convergence of the coded sequence of functions. This can be done for pointwise convergence, uniform convergence on C, or even uniform convergence on compact subsets of C. For this last type of convergence it is not necessary to quantify over arbitrary compact sets, but rather to use a particular exhaustion of C by compact sets. For example, suppose Code (f,g,h 1 ,h2 ,a) holds in £ and we wish to discuss convergence of the coded sequence of functions Urn). Consider the sets G~ ~ range (a) defined by (T
G~ =
{a
EC
110'1 ~ n}.
These sets are definable using an algebra formula from the parameters n and a. Also, (fm) converges uniformly on compact subsets of C if and only if (fm 0 a-I) converges uniformly on each of the compact sets G~. This can be expressed by an algebra formula using FSeq at the end to replace mention of Urn). Tbis method of representing sequences of functions within £ enables us to define many specific sequences-for example, the sequence of powers urn) of a particular function. From this we can find an algebra formula which expresses that 9 is equal to a polynomial in f. Next we discuss a method for interpreting in £ the lattice of open subsets of C (This procedure is also used later, where it is discussed in greater detail.) This is done by associating each open set 0 with the set 1<.(0) of all pairs (q, r), where q is a Gaussian rational, r is a rational> 0, and the disc {a Eel 10' - ql < T} is contained in O. Since 0 is the union of this family of discs, we see that 0 f. D implies 1<.(0) f. 1<.(D) for open sets 0, D. The set R.(O) can be coded by a quadruple (/t, h,g, h) by first writing 1<.(0) as a sequence (qn, Tn) for n E N and then taking (qn, Tn) to be the value of (/t(z), h(z» at the unique z E C for which g(z) = 0 and h(z) = n. Here, (g, h) satisfies the Basis formula described above. We first obtain an algebra formula 0 Basis(/t, h,g, h) that is true in £ if and only if there is an open set 0 such that R.(O) = {an,Pn) I n E N},
23. The First-Order Theory of the lUng of All Entire Functions
169
where (a n ./3n ) is the sequence of pair:; coded by (h, h) using the basis pair (g, h). This must express that every an is in Q( i), every (3n is in Q+; also, it must express that if q E Q( i), r E Q+ and if the disc {"I E IC I h - ql < r} is contained in the union of the disc:; h E IC 11"1 - ani < {3n}, then (q, r) is in the set {(a n ,{3n) n EN}. Of course, 0 is just the union of the disc:; {, Eel h - an I < {3n} for n E N. Thus we can obtain an algebra formula Open(a,h, h,g,h) which expresses that 0 Basis(h, h,g, h) holds and that a is in the open set 0 for which R(O) = {(a n ,{3n) In EN}. Evidently we also can represent sequences of open sets by using matrix pairs (a mn ,{3mn) of constants, each row of which codes an open subset of C. This gives us another way of referring to an exhaustion (G~) of IC by compact sets, referring instead to the sequence C\G~) of open sets. Also, we can develop a way of representing all analytic subsets of C (including all Borel sets) using the Souslin operation applied to sequences of sets. In addition, we can construct an algebra formula that expresses the value of Lebesgue measure applied to these analytic sets using approximation by open and closed sets. These matters are more complicated, and we omit the details. Finally, we discuss how to express integration of functions and the "Nevanlinna characteristic" applied to entire functions using algebra formulas. First we consider integration. Fix fEe and let u be a 1 - 1 conformal mapping on IC. We then can express by using algebra formulas how to integrate f 0 u- 1 along a circular curve "I = {z liz - al = r} contained in the range of 0". That is, we have an algebra formula Int(f,O",a,r,{3) which holds in if and only if 0", a, r are as above and J.., fdz = {3. This can be done by considering a sequence of successively finer subdivisions of "I and by evaluation of f 00"-1 on points of "I using our expressions /(1]) = 6
e
u
as above. We then get upper and lower sums and evaluate the integral by taking limits. (Lebesgue integration can be handled also, but it is more complicated. In any case, our functions are highly continuous.) By a similar procedure we can also treat integration over the inside of the curve "I, with respect to area measure. Now we consider the "Nevanlinna characteristic" on entire functions. For an entire function f this is defined by 1 211"
T(r,f) = -
121r log+ If(re
i8 Id8.
0
We also consider the order a of f defined by _ li log T(r, f) a - msup 1 . r ..... oo ogr Note that when we apply the above approach to integration, the 1 - 1 conformal mapping 0" that appears as a parameter must be affine. Also,
Entire and Meromorphic FUnctions
170
the order of f is equal to the order of f 0 0-- 1 , no matter which affine 0we choose. This shows that there is an algebra formula Ord(f,o} which is true in the algebra of entire functions if and only if the order of f equals o. Obviously there is no completely natural point at which to stop this general discussion of how to express the mathematics of entire functions in the first-order theory of £. As far as we can see, essentially all of the classical theory of entire functions, including topological and measure-theoretic aspects, can be so represented.
Derivatives and Definable Constants 'Ve will call a constant a E C defj:rlfrhlp if there is a formula D(x) in the algebra language such that, for every function f E £,
D(f) holds in £¢=::::?f equals the constant o. Evidently, i is a definable constant, since we have included a name for it in the algebra language. Also, it is clear that 0 is definable if and only if both the real part of 0 and the imaginary part of a are definable. One way to obtain a large number of definable constants is via infinite series and the device for coding sequences of constants that was discussed earlier. Recall that such a sequence is coded by a triple of functions (f,g, h): o is the nth term of the sequence if f(z) = 0, where z is the (unique) zero of 9 for which h(z) = n. It is now a routine matter, given the definability results obtained earlier, to construct an algebra language formula Series(x, y, z, w) such that, for any 0, f, g, hE £, Series(a, f,g, h) holds in £ if and only if (f,g, h) codes a sequence (a .. ) and La.. converges to o. To make this clearer, we take the first step toward the formula: Series(a, f,g,
h)~Basis(g, h) A 0 E
C
A (3k)(Vn E Nrt{J,"I,C E C{Seq({J,n,k,g,h)
A Seq(1', n
+ 1,/, g, h) =:::::> Seq(,8 + "I, n + I, k, g, h»
A'tI{J E C(Seq(,8,O,f,g,h) => Seq({J,O,k,g,h» A 'tic E JR+3n E N'tim E N'tI,8 E C{n ~ m A Seq(,8,m,k,g,h) =:::::>
I{J -
01 ~ 6)].
The middle two clauses of this formula assert that the sequence «(3.. ) coded by the triple (k, g, h) is the sequence of partial sums of the sequence (an) coded by (f,g, h). The third clause asserts that ({In) converges to a. Now we will show, as an example, that e is a definable constant. This follows because we can say in a first-order way that (I, g, h) codes a sequence
23. The First-Order Theory of the Ring of All Entire Functions
171
(an) which satisfies the recurrence relation 0'0 = 1, a n+l = an/en + 1), forcing an = l/n! and so e = E an. Thus the defining formula D(x) for e is obtained from: D(a) <=>3J3g3h[Series(a, f,g, h) A Seq(l,O,f,g, h) A \In E N "113 E
C(Seq(f3, n, f,g, h»
~
Seq(f3/n + 1., n
+ 1, f,g, h)].
[Strictly speaking, we cannot use division, but this is easily eliminated from D(x).] Clearly this same device can be used to obtain most of the familiar transcendental real numbers, as well as many others, for example, Liouville numbers such as E 10"'. All that is required is that the number be the limit of a series whose terms are generated using some recurrence formula.
Theorem 23.3. Let F be the set of all definable constants from C. Then F is a countable algebraically closed field that contains e and 11". Proof. It is clear that F is a subfield of C. For example, if D(x) defines = 0, then the formula
a
3y(D(1I) A xy
= 1)
defines 1/0'. Also, F is countable, since there are only countably many formulas in the algebra language. It is easy to show that F is algebraically closed. For example, suppose aj is definable by Dj(x) for j = 0,1,2,3. We will show how to define a root of the polynomial p(z) = aoz3 + alz 2 + a2Z + 0'3. Consider the linear ordering on C defined by taking
f3<=>!R(a) ~ !R(f3) or (!R(a) = !R(f3) and ~(a):5
a
~
~(f3».
This ordering is definable by an algebra formula B(x, y) in the sense that for any a, 13 E C
B(a, 13) holds in £~a ~ 13. Using this we can define the "smallest" root of p(z) by a formula D(x). Namely, D(a) is equivalent to
3a03a13a23a3[Do(ao) A D 1(al) A D 2(a2) A D3(aa) A a E C A 0'00'3 + 0'10'2 + a2a + aa = 0 A "113(13 E It A 0'0133 + 0'113 2 + 0'213 + 0'3) = 0 ~ B(a, 13)]. Other interesting definability results for the constants come from an indirect treatment of derivatives, which we now discuss. Of course, there is no hope of obtaining a first-order definition of the relation between a
Entire and Meromorphic Functions
172
function and its derivative, since this relation is not conform ally invariant. Our indirect approach comes via the use of a parameter a, which is a 1 - 1 conformal mappin~, as was used earlier to obtain the definition of lit Given such a a and ~iven f, g E £, we will show that the relation (goa-I) = (foa- 1 )' is a first-order property of the triple (f, g, a). Namely, this relation is equivalent to the condition: Tlo: E C 3h3k3{1 E C :Yy E Cq
g = {1 + (a - a)h and
f
[if a - a is a nonunit, then
= 'Y + {1(a -
0:)
+ (a -
a)2k].
Proof. (<==) Let a = a(z) for some z E C and suppose the equations 9 = ,8 + (fT - fY)h and f = 'Y+ ,8(0' - a) + (a - 0:)2k hold over C. Substitute 0'-1 in the second equation and get f 0 O'-I(W) = 'Y + {1(w - a) + (w - 0:)2k(a- l (w» for all wE C. Differentiating with respect to w and setting w = 0: yield (f 00'-1 )'(0:) = (3. Since (g 0 O'-I)(a) = {1, we have the desired equation. (=} ) Suppose that (goa-I) = (f 00'-1), and consider a E C. Expanding goO'-1 and f 0 0'-1 about the point a yields
(g 0 O'-I)(W)
= (3 + (w -
o:)h(w),
(f 0 a-l)(w) = 'Y + (3(w - a) + (w - a)2k(w). Substituting w
= a(z) yields the equations needed.
The exponential function is uniquely determined on any nei~hborhood of 0 by its functional equation = f and f(O) = 1. This leads to a certain definability result for the exponential function:
r
Theorem 23.4. There is a formula E(x, y) in the algebra language such that for any a, {1 E £ E( 0:,,8) holds in £ if and only if 0:, {1 E C and e
Proof. Given a, (3 E C, consider the following condition: E(a,f3)~ there exists a 1 - 1 conformal mapping a and a function f, both in £, such that (i) f(O) = 1, a (ii) (f 0 0'-1), = (f 0 0'-1), (iii) f(o:) = {1.
a
From our previous discussion it is clear that this can be expressed by an al~ebra language formula. Parts (i) and (ii) imply that f 0 a-I is equal to the exponential function, and part (iii) therefore says that e
z
E C.
23. The First-Order Theory of the Ring of All Entire Functions
173
ro
Corollary 23.5. Let be the field of definable constants. Then Fo is closed under the exponential function. Note that Corollary 23.5 gives an alternate proof that Fo contains e. Also, it yields that Fo has infinite transcendence degree. Indeed, by the Hermite-Lindemann Theorem, if a1, .. . ,an are algebraic numbers (hence in Fo) that are linearly independent over Q, then eal , ••• , ean are algebraically independent elements of Fo. (See (21).) We close this section by noting that the other elementary functions, such as sin(z) and cos(z), can be treated in a way that is similar to our discussion of eZ • Recursive U ndecirl;thiHty Earlier there was presented a way of defining N in £ and a method for coding countable sets of constants. This yields an effective interpretation of second-order number theory in the first-order theory of £. This interpretation is an example of a 1 - 1 reduction of one "problem" or set to another. IT 6 1 , 6 2 are sets of sentences in formal languages L17 L 2 , respectively, we say 6 1 is 1-1 reducible to 62, and write 6 1 ~l 62, if there is an effectively computable 1-1 function cp (Le., a recursive function) such that for any sentence S of L17
We say that 6 1 and 62 are recursively isomorphic if there is an effectively computable function cp that maps the sentences of Ll bijectively onto the sentences of L2 and which satisfies cp(6x) = 6 2 • Evidently this means that the problems of deciding membership in 6 1 and in 6 2 are effectively equivalent in a strong way. It is a well-known fact [34] due to Myhill that if 61 ~1 62 and 6 2 ~1 61 , then 6x, 62 are recursively isomorphic. (The converse is obvious.) As an example of what type of undecidability theorem can be proved using the results above, we consider the ring of entire functions £. Robinson [33] showed that the first-order theory of this ring is undecidable by showing that first-order number theory can be interpreted in it. The following result is a substantial improvement of this. Theorem 23.6. The fir.~t-order theory of the ring of entire functions is recursively isomorphic to second-order number theory.
Proof. Let 6 1 denote the first-order theory of £ and let 6 2 denote the second-order theory of (N, +, .). We will show 6 1 ~1 6 2 and 6 2 ~1 6 1 . Let 6~ denote the first-order theory of the algebra of entire functioIL.'i £. Earlier it was shown that the field of constants is definable in £, which yields a direct interpretation of 6~ into 6 1 • In particular, 6~ ~1 6 1 , As was discussed earlier in this section, we also have a direct interpretation of second-order number theory in 6~. This implies 6 2 ~ 1 6~, so that 62 ~1 61 has been proved.
174
Entire and Meromorphic Functions
To show 6 1 ~1 6 2 , we sketch how to give an interpretation of the firstorder theory of £ in second-order number theory. Each entire function f has a power series representation centered at 0,
with an infinite radius of convergence. Each coefficient an = Xn + Yni can be identified with two sequences of integers, say {rn ,; I j E N} and {sn,; I j EN}, where r n,O is the integer part of Xn and r n ,; is the jth decimal digit of X n , and similarly for sn,; and Yn' Therefore, f can ultimately be identified with an infinite matrix of integers M = (mij I i,j EN) obtained by setting mi,2; = ri; and ~,2;+1 = Si; for all z, j E N. Finally, M can be converted to a set of integers M# = {2 i . ai . smii I i,j EN}. Evidently we can recover the function f from the set M*. To show that this provides the desired interpretation, one should prove that the collection of all sets of the form M* is definable in the second-order theory of (N, +, .) and prove the same for the relations on these sets which correspond to addition and multiplication of entire functions. The details are tedious and routine, and we choose to omit them. Corollary 23.7.
Second-order number theory is 1- 1 reducible to To.
24 Identities of Exponential Functions
In this chapter, which is based on [13], we take up some questions prompted by mathematical logic, notably Tarski's "High School Algebra Problem." We study identities between certain functions of many variables that are constructed by using the elementary functions of addition x + y, multiplication x • y, and one-place exponentiation eX, starting out with all the complex constants and the independent variables Zl,"" Zn. We show that every true identity in this class follows from the natural set of 11 axioms of High School Algebra. The major tool in our proofs is the Nevanlinna theory of entire functions of n complex variables, of which we give a brief sketch. It is entirely parallel to the one-variable theory presented in detail earlier in this book. The timid r!:'.ader can take n = 1, at least for a first reading. Tarski's conjecture for a more extended class of terms than those we consider here has been shown to be false by Wilkie (see [50]). The largest class that we are aware of for which Tarski's axioms have been shown to be complete was studied in [11]. We briefly recapitulate the basic Nevanlinna theory. Consider first the case of one variable, n = 1. For meromorphic functions f of one variable defined on the complex plane C, the characteristic function is defined for 0 < r < 00 by
T(r,J) = m(r,J)
+ N(r,J):
this is a sum of two terms: the proximity function
11"
m(r, f) = 21!' _".log+ If(rei/J)ldO,
Entire and Meromorphic Functions
176
which measures how close, on the average, counting function N(r, f) =
I
is to
00,
and the average
fur n(t~ f) dt,
where nCr) is the number of poles of f in the disc Izl :S r. Here, r ranges over the interval 0 < r < 00, and appropriate modifications must be madE' in the definitions of N(r, f) if 1(0) = 0 or if 1(0) = 00. The function log+(t) is defined by setting log+(t) = log(t) for t ~ 1 and log+(t) = 0 for o :S t :S 1. The growth of the characteristic T(r, f) as r -+ 00 gives a very useful measure of the growth of I. The basic properties that we shall use are listed below. (C2.0) T(r, f) is a nondecreasing function of r and a convex function of logr. (C2.1) T(r, I + g) :S T(r, J) + T(r, g) + 0(1). (C2.2) T(r, /g) :S T(r, f) + T(r,g). (C2.3) T(r, 1/(/ - a» = T(r,f) + 0(1) for any complex constant a. (C2.4) T(r, Ilg) :S T(r, f) + T(r, g) + 0(1). (C2.5) T(r, e9 )IT(r,g) -+ 00 as r -+ 00 if T(r, g) is unbounded. The other basic fact we need about the characteristic T(r, f) is the Lemma of the Logarithmic Derivative (LLD): (C2.6) m(/,t If) :S 0 (log(T(r, f) + logr) , except possibly for r lying in a set E of finite length. When it comes to several variables, the theory is substantially the same and the basic properties we need are still expressed in the same form (C2.0)(C2.6). (See [42] for the details of proofs.) Alternatively, one could use a characteristic based on the exhaustion of en by balls, rather than by polydisks on. (See [48] and [10].) For a meromorphic function /(ZI, . .. , zn) defined on en, we define
mer, f) = (2 1)n 11"
I'" I'" ..•
_,..
log+ I/(reirJ>l, ..• reirJ>n )ldcPl ... dcPn,
_'"
where 0 < r < 00. Also, N(r,f) is defined much as before, as an averaged counting function of the poles of f. Note that if / is a holomorphic function on en, then T(r, f) = mer, f). (In [42], a characteristic T(T, f) is developed for a vector variable r = (rl, ... , rn), but we use only the diagonal case rl = ... = rn = r.) The basic properties above, including LLD, are shown to hold in [42] or follow exactly as in the case of one variable (e.g., (C2.5». One thing which needs explanation is the derivative f' that occurs in the LLD. When n 2: 2, we shall take f' to stand for the Euler operator
I
,
= D/ =
8/
ZI8z1
8/ + . + Zn -8z n
.
24. Identities of Exponential Functions
177
This has the useful property that Df = 0 if and only if f is identically constant. (This is because f may be expanded as a nicely converging sum of homogeneous polynomials and because DP = mP for any homogeneous polynomial of degree m.) Our basic tool is a lemma proved in one dimension by Hiromi and Ozawa-see Chapter 17 of this book. The proof in n dimensions is similar to the proof in 1 dimension and depends only on the properties (C2.0)(C2.6) we have listed of the characteristic function T(r, f). (Carlos Berenstein has recently pointed out that the proof in [13] is incomplete. The authors of [13] are preparing a complete proof. More varied Wronskians are needed to establish linear dependence in the N-dimensional version of the Hiromi-Ozawa lemma!. See [4] for a correct version of the proof for the ball-characteristic. )
Lemma 24.1. (Hiromi-Ozawa). Let ao(z}, ... ,an(z) be meromorphic functions and let gl(Z), ... ,9n(Z) be holomorphic /unctions defined on the domain eN. Suppose that these functions satisfy
(a) for each j
= 0, 1, ... ,n; and T( r, ei}
(b)
for at least one i
= 1, 2, ... , n.
::/: O(1og T)
Undt.· these hypotheses, if the identity
71
Laj(z)e9j (z)
= ao(z)
j=1
holds for
z E
eN,
then there exists constants
cb""
en (not all 0) so that
n
L
Cj .
aj(z)e 9j (z) = 0
j=1
for all z E eN. Our use of the Nevanlinna theory tools previously discussed comes entirely through this Hiromi-Ozawa Lemma. Indeed, we use it only in cases where 91, ... ,971 are holomorphic functions, so that T( T, e91 } = m( r, e 91 } (and the prohibition that the function not take the value 0 at the origin is satisfied), and in cases where ao, al,"" an are slowly growing functions. Here we consider expressions that are built up from variables and complex constants using addition, multiplication, and the I-variable exponential function eX (where e is the usual base of the natural logarithm).
Entire and Merornorphic Functions
178
We prove a version of Tarski's High School Algebra Conjecture for these expressions. (This result was proved independently by van den Dries [47] and, for terms containing just one variable, by Wilkie [49]. Their methods are quite different from ours.) We also settle positively a conjecture, due to Schanuel, which asserts that if f is a function on en which is defined by an expression of this type, and if f is nowhere equal to 0, then f = e9 for a function 9 on which is also defined by an expression of the kind considered here.
en
Definition 24.2. E is the smallest class of terms which contains the variables Xl, X2,'" and a constant for each complex number, and which contains the terms s + t, S • t and exp(t) for each .9, tEE. Here we interpret exp(t) to stand for et . We note that if tEE and the variables of t are among X I, ... , X n , then t defines a holomorphic function on all of IT 8 E E also has its variables among Xl, ••. ,Xn., we write t == 8 to mean that t and s define the same function on en. (Various equivalent formulations of this definition are possible in special cases because of the uniqueness of holomorphic functions. For example, if t and 8 contain only real constants, we may be interested only in the functions they define on r. But t == s will hold as long as t and s define the same function on r, or even on where S ~ C is any set with a limit point in C.) One has the additional useful fact that a holomorphic function f on en has the small characteristic T(r, f) = O(log(r» if and only if f is a polynomial. Hence, if f is a polynomial and 9 is any nonconstant holomorphic function on en, then T(r,f) = o(T(r,e 9 » (which is necessary as part of the hypotheses of the Hiromi-Ozawa Lemma as we apply it). See [17, Proposition 4.4ff]. For holomorphic functions this can be proved by estimating the Poisson integral for log Ifl to show that f is of polynomial growth as a function of Xj (when Xi, i =F j, are held fixed), for each j = 1, 2, ... , n. By the Liouville Theorem in one variable, then, f is a polynomial separately in each xi' That f is globally a polynomial now follows from [30]. (There must be many other proofs of our assertion in the literature.)
en.
sn
Theorem 24.3. (Tarski's Conjecture for E). If t, s are any two terms in E and t == s, then the identity t = s is probable from the axioms x + (y + z)
= (x + y) + z,
x+y= y+x, x+O = x, x(y + z) = xy + xz, exp(x + y)
x(yz) = (xy)z, xy = yx, l·x = x, O·x=O,
= exp(x) . exp(y),
together with all axioms giving the facts of addition, multiplication, and exponentiation for constants from C.
24. Identities of Exponential Functions
179
Proof. Because we have included here a constant for -1, the operation of subtraction is available and we need only consider the case where S is O. That is, if tEE, then we must show that t = 0 is formally derivable whenever t == O. Moreover, it is easy to show that for any term tEE there are terms SI, ... , Sk E E and polynomials PI, ... ,Pk in n variables, with coefficients in C (also realized as terms in E) so that the identity t
= PI . exp(st} + ... +
Pk . exp(sk)
is provable from the permitted axiOlDS. We will prove the theorem by induction on the total uumLt:r of :symbois in the sequence 81, ... ,810, showing that Pl!'" ,Pk are polynomials, SI,'" ,Sk E E and PI exP(Sl)+" +PkeXP(S,.) == 0, then P1·exP(St}+·· '+Pk ·exp(sk) = 0 is formally derivable. (Note that we allow Sj to be 0.) First suppose that k = 1: H Pl' exp(SI) == 0, then PI == O. It is well known that PI = 0 is provable from the admitted axioms, since PI is a polynomial. Hence, PI exp(sl) = 0 also is provable. From now on assume k > 1. Assume Pt. ... ,Pk are polynomials, Sl, .. ·, Sk E E and PI ·exp(st} + ... +Pk ·exp(sk) == O. For 1 5. j 5. k, let 11'; be the function on en defined by exp(s;). (Choose n so that all variables in each Pj and 8; are included among Xl!'''' x n .) Note that we may assume each 'If; is nowhere equal to 0 on en. After dividing by 'lfk we have
Suppose first that we can apply the Hiromi-Ozawa Lemma. In this setting, this means T(r, 'If;/1fk) i= O(log(r» for each 1 :5 i 5. k -1. H so, then there exist constants C}, ••• , Ck-l (not all 0) so that
which gives us an identity with k-l exponentials after multiplying through by 'lfk. By the induction hypothesis, the formal identity
is derivable in the allowed system. Now we can use this identity to solve for one of the expressions P; . exp(sj) (1 :5 j 5. k - 1) and eliminate it from the original expression PI exp(st} + ... + Pk exp(sk). The resulting identity (setting this expression = 0) has at most k - 1 exponentials, so it is derivable. From this one deduces the desired identity PI exp(st} + ... + PI; exp(sk)
= o.
Entire and Meromorphic Functions
180
On the other hand, it may happen that for some i(1 5 i 5 k - 1), = O(logr). Since 1I"i, 1I"k are nowhere 0, it follows that 1I"i == C7rk for some constant c. [By (C2.3) the same kind of "big-O" estimate holds for 1I"k/1I"i, and hence both 7rd1l"k and 1I"k/1I"i are polynomials.] That is, exp(si) == c· exp(sk)' so that for some constant dEC, c = ed and Si - Sk == d. Using the induction hypothesis, we therefore get a formal derivation of Si - Sk - d = 0 and, hence, also of exp(si) = c· exp(sk). This allows us to reduce the original identity to one involving only exp(sj) for i :5 j 5 k - 1, which will be derivable by the induction hypothesis. Again this yields a derivation of the identity PI exp(st} + ... + Pk exp(sk) = 0 and completes the proof. T(r,7ri/7rk)
Theorem 24.3 has an interesting corollary for trigonometric functions, which we present next. Consider terms in a language with constants for all the complex numbers, variables Xl, X2, ••• , and function symbols for addition, multiplication, and for sin and cos. Let E* be the set of all these terms.
Corollary 24.4. If t, S are any two terms in E* and t == s, then the identity t = 8 is provable from the axioms X
+ (y + z) = (x + y) + z,
x(yz)
= (xy)z, = yx,
x+y= y+x, x+O=x,
l·x=x,
x(y + z) = xy + xz,
O·x=O,
xy
= sin(x) cos(y) + cos(x) sin(y), sin(-I· x) = -1· sin(x)
sin(x + y)
together with all axioms giving the facts of addition, multiplication, sin, and cos for constants from C. Proof. We use the fact that in the context of the complex plane, ~ is interdefinable with sin and cos. Note that since the allowed axioms include the identities sin(7r/2) = 1 and cos(1I"/2) = 0, we can prove cos (x) = sin(x + 11"/2). This in turn allows us to derive the other addition identity, cos(x + y)
= cos(x) cos(y) -
sin(x) sin(y).
In E*, let EXP(x) be an abbreviation for the term cos ( -i·x)+i ·sin( -i· x). It is easy to verify that from the allowed identities in E* one can prove the exponential identity EXP(x
+ y) =
EXP(x) . EXP(y)
as well as all the numerical facts involving EXP.
24. Identities of Exponential Functions
181
Given any term t in E*, we define a term t# in E by replacing (inductively) each term of the form sin(s) by
-.5i· (exp(i· s) - exp(-i· s)), and eos(s) by .5· (exp(i· s) - exp(-i· s)).
U t is any term in E we define t* in E· by replacing (inductively) each term of the form exp(s) by EXP(s). Note that if t E E*, then the identity t = (t#)* is provable from the axioms allowed in Corollary 24.4. Now suppose t, s E E* and t == s. Then t# = s#, so the identity t# = s# is provable from the axioms allowed in Theorem 24.3. Hence. (t#)* = (s#)" is provable in the system of Corollary 24.4. It follows that t = s is also provable in that system, completing the proof. Remark. Suppose t, S E E* and t, s only contain real constants. We do not know if there is a proof of the identity t = s in the system of Corollary 24.4 in which only real constants appear. Next we settle positively a conjecture of Schanuel.
Theorem 24.5. Let tEE and suppose the function represented by t is nowhere equal to O. Then log(t) is in E, in the sense that t == eS for some sEE.
en
Proof. Let 11" be the function (on say) defined by t. There is some holomorphic function G on so that 11" == eG • We may suppose t is a term of the form PI exp(st} + ... + PkeXP(Sk), and we argue by induction on the number of symbols in Sl,"" Sic as in the proof of Theorem 24.3. Clearly we are done if k = 1. Assume k > 1, and for 1 :::; j :::; k let 1I"j be the function on defined by exp(sj)' Then we have Pt1l"1 + ... + Plc1f1c == e G so that Pl(1I"1e- G ) + ... + plc(1I"/ce- G ) == 1. We may assume the functions Pl11"1e- G , .. . ,PIc1fke-G are linearly independent (otherwise, we could replace t by a simpler term to which the induction hypothesis would apply). Hence, the Hiromi-Ozawa Lemma cannot apply. It follows as argued in the proof of Theorem 24.3 that there must exist 1 :::; i < j :::; k so that 1I"i/1I"j is identically constant. Again this permits us to reduce the complexity of t and to apply the induction hypothesis. This completes the proof.
en
en
We conclude with a related problem.
Problem. Suppose that f is an entire function for which there exists a term tEE such that the function represented by t is equal to f2. Then must there exist a term sEE such that the function represented by s is f? Put more simply (but not as correctly), if f is entire and f2 E E, must fEE? Even if one assumes that j2 and f3 belong to E (and hence fn E E for n = 2, 3, 4, 5, ... ), does it follow that f (assumed to be entire) lies in E?
References
1. Apostol, T., Mathematical Analysis, second edition, Addison-Wesley, Reading, MA, 1974. 2. Beck, W., Efficient quotient representations quotient representations of merom orphic functions in the disk, Ph.D. Thesis, University of lllinois, Urbana, IL, 1970. 3. Becker, J., Henson, C. W., and Rubel, L. A., Annals of Math. Firstorder conformal invariants, 112 (1980), 123-178. 4. Berenstein, C., Chang, D.-C., and Li, B. Q., Complez Variables A note on Wronskians and linear dependence of entire functions in 24 (1993), 131-144. 5. Boas, Entire Functions, Academic Press, New York, 1954. 6. Bucholtz, J. D. and Shaw, J. K., Trans. AMS Zeros of partial sums and remainders of power series, 166 (1972), 269--184. 7. Buck, R. C., Duke Math J. Integral valued entire functions, 15 (1948), 879--891. 8. Dienes, The Taylor Series, Dover, 1957. 9. Edrei, A., and Fuchs, W. H. J, Trans. Amer. Math. Soc. Meromorphic functions with several deficient values, 93 (1959), 292-328. 10. Gauthier, P. M., and Hengartner, W., Annals of Math. The value distribution of most functions of one or several complex variables, 96(2) (1972), 31-52. 11. Gurevic, R. H., Trans. Amer. Math. Soc. Detecting Algebraic (In)Dependence of Explicitly Presented Functions (Some Applications of Nevanlinna Theory to Mathematical Logic), 336 (1) (1993), 1--67. 12. Hayman, W. K., Meromorphic functions, Oxford, at the Clarendon Press, 1964 (Oxford Mathematical Monographs). 13. Henson, C. W., and Rubel, L. A., Trans. Amer. Math. Soc. Some applications of Nevanlinna theory to mathematical logic: identities of
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183
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Index
algebraic integer, 139 Boas, 47 Borel Lemma, 26, 28, 29 Borel transform, 43, 125 B(r),30 branching index, 103 Buck, 131, 139 Calderon, 123 canonical products, i, 87-89, 91 Carleman's Theorem, i, 45,47 Carlson's Theorem, 47, 130 characteristic logarithm, 147 characteristic, Ahlfors-Shimizu, 18, 19 characteristic, Cartan, i, 16-19 characteristic, Nevanlinna, 10, 18 Clunie's Theorem, 14 conjugate indicator diagram, 125 convex, 16 convex hull, 124 convolutioD, 127 corrected ratios, 32 counting function, 7, 50, 147
defect, 103 deficiency, 103 deficient value, 105 c5(a), 103 -+ eff
in words, A( r) approaches L effectively, 28 effectively, 185 = 28 eff'
"', 28
eff
exponential-type, 41 extreme point, 124 Fabry Gap Theorem, 134 finite A-
186
Entire and Meromorphic FUnctions
Gauss Mean Value Theorem, 6, 21
genus, 88 growth function, 51 Hadamard factorization theorem, 89, 151 Hadamard gap theorem, 133 Hahn-Banach Theorem, 48 Hausdorff-Young Theorem, 69 Hayman, 14 12, §1.5-1.6, 18 Hiromi-Ozawa Lemma, 106, 109 Ho(oo), 125 Horseshoe Theorem, 131 h«(}), 125 indicator function, 125 integer valued entire function, 139 interpolating sinusoid, 134 Jensen's Theorem, 6
K* -admissi ble, 144 k-admissible, 131 Kakeya,96 A-admissible, 55 Laguerre's Theorem, 92 Laplace transform, 43, 126 ~-balanced, 52, 58 Lindel6f, 50, 74 Lindelof's Theorem, 157 Liouville's Theorem, 23, 130 logarithmic convexification, 32 Logarithmic Derivative, 116 logarithmic length, 28 logarithmically convex, 17 A-poised, 53
Malliavin, 146 Malliavin-Rubel Theorem, 47 mapping radius, 141 maximum term, 30 Miles, 50 Miles-Rubel-Taylor Theorem, i, 78,91 m(r, f), 10 Nevanlinna, 108, 111 Newton polygon, 31 N(r, f), lU
n(r,J),7 N(r,
j), 7
n(t, J), 103 Okada, 96 order, 40,41 Phragmen-LindelOf Theorems, i, 121, 123 Picard's Theorem, i, 99, 101, 103, 105, 107, 109, 111 7r(f), 31 Poisson Kernel, 20 Poisson-Jensen Formula, i, 20, 21 P6lya Representation Theorem, 124, 128 principal indices, 31 Pringsheim's Theorem, 133 quotient representations, i, 78 rank of maximum term, 30 regular, 58 sampling theorem, 47 Second Fundamental Theorem, i, 99, 101, 103, 105, 107, 109, 111 Second Fundamental Theorem of Nevanlinna, 102 share the value a, 108 sinusoid, 134 S(r), 102 Stone-Weierstrass Theorem, 48
Index
strongly A-balanced, 52 strongly A-poised, 53 subsinusoidal, 134 supporting function, 124 Szego,l34 Tsuji,97
187
Two Constant Theorem, i, 121, 123 type, 41
II, 99 Weierstrass Factorization Theorem, 88
Universitext
(continued)
Sagan: Space-Filling Curves Samelson: Notes on Lie Algebras Schiff: Normal Families Shapiro: Composition Operators and Classical Function Theory Smith: Power Series From a Computational Point of View SmoryIiski: Self-Reference and Modal Logic Stillwen: Geometry of Surfaces Stroock: An Introduction to the Theory of Large Deviations Sunder: An Invitation to von Neumann Algebras Tondeur: Fuuation!) 011 RitllldJUlldJl ?v1anifolds
