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0 and ReIR>0 so that
IARI 1. Then for each a > 1
µ (r +
aA(r)
except in a set Ea such that AEQI < a61
Remark. The inequality of the Borel Lemma estimates p at a point greater
than r by using the value that µ takes at r. Intuitively, if the inequality fails in "too big" a set, the function u will become infinite "too soon" and will not be defined for all r.
9. A Lemma of Borel and Some Applications
27
P r o o f . Let E = E,, = {r > ro : µ (r + ) > ap(r)}. Let c > 0 be given. We proceed to define a sequence {rn} and an allied sequence {r,} by induction.
Let r1 = inf{r : r E E}. Suppose rl,...,rn_1 have been constructed together with numbers Ek > 0, k = 1, ... , n- I so that El +C2 +- .. En- 1 < E and rk + Ek E E. Let r' = rk + Ek + µ(Tk+Ck
Now define rn = inf{r : r E E and r > rn_1}. Choose c,, > 0 so that + En < E and so that rn + En 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 > rn. We have El +
rn < rn + En < rn < ra+l
(9.2)
Now (r,, rn+1) fl E is empty by construction.
Claim. There exist only finitely many rn or else rn - oo. For otherwise, there would be a finite r such that rn -> r and, by (10.2), r, -+ r. However, by the construction we have for all k 1
1
rk - rk = Ek +
0,
A(rk + Ek) - p(r) >
which is a contradiction. Claim. E C Un=1[rn, r ,1.
Pick an arbitrary x E E. Let rno = max[rn : rn < x]. This makes sense by the previous claim. Now rno < x. Suppose x > rno. Then rno+1 < x, an immediate contradiction. Therefore rno < x < rno, which proves the claim. So we have constructed a countable collection of intervals whose union
contains the set E. We estimate E(rn - rn). Notice that IA(rn) = it rn + En +
1
A(rn+En))
-> ap(rn + En) -> al,(rn)
80 that p(rn+1) > ap(rn). Therefore, p(rn+1) > anp(r1) > an. Hence 1
(rn - rn)
-
But 1
µ(rn + En)
En +
U(rn + En) /
0, we have log April < log B(r) = log Anr", Since
where n = µ(r).
log Ap = -gp log An = -9n,
we have
p log r - gp < n log r - 9n or (10.6)
gp?9n+(p-n)logr.
But
y=gn+(x-n)logr
(10.7)
is the equation of the straight line through cn with slope log r. Equation (10.6) says that all points of 7r(f) lie above the line (10.7); this
line is "tangent" to rr(f ). Call this line D,. Thus, µ(r) is the rightmost point of contact of D,. with rr(f) [because µ(r) is the largest value of n at which we can have equality in (10.6)]. Hence the values of p(r) are the principal indices.
Since, for x = 0, (10.7) yields y = gn-n log r = - log Anrn = - log B(r),
it follows that D, cuts the y axis at - log B(r). Two immediate consequences of this are:
(a) Given fl, f2 entire such that rr(fl) = 7r(f2), then
R(r : fl) = p(r : f2) 1 B(r : fl) = B(r : f2) (b) Among all entire functions, h(z) _ that has the same µ and B as f.
Anzn is the largest function
Let (10.8)
Rn
We call the Rn the corrected ratios.
-An_1 An
10. The Maximum Term of an Entire Function
33
Geometrically, log Rn is the slope of the side of 7r(f) joining the points whose x-coordinates are n - 1 and n. Rn is nondecreasing and Rn -0 oc as n --* oo.
Without loss of generality, assume that Ao = ao = 1 [note that p(0) = 0]. From the definition of Rn, it follows that eGn = R1 R2 Rn. Since p(r) runs through the principal indices, gu(r) = G,u(r), it follows that
B(r) =
rn(r)
Ri.R2...Rn
Taking logarithms,
log B(r) = p(r) log rFlog F p(r)
But
Ar) 1og
=
j
log t dp(t),
r
11 (r)
Rk =
log Rk .
p(t) _
where
1.
Rk p(r). Then R,, > r because log R. > log r for p > p(r), as we see on interpreting log Rp as a slope. Then for q > p we can write r9-P+1 A r9 = e-Ggrq = eG,-irP-1
Since e-Go-, rP-i < B(r), we have rq-P+1
`44r4 = B(r)1VrP+1 < B(r) P
9-P+1
r RP
/
Entire and Meromorphic Functions
34
because the slopes of the edges of the Newton polygon are increasing. We have:
P-1
F(r) _
e_c"r"
00
+ : e-c°r" p
0
/
00
p B(r) + B(r) 4 =P
= pB(r) + B(r)
4-Ptl
(k)
Rp r
r.
Consequently,
F(r) < B(r) [p + Rpr rl
provided p > µ(r).
As a heuristic guide, let us try the choice p = µ(x) for x > r, supposing for the moment that µ(x) > µ(r). Then
F(r) < B(r) [µ(x) +
r r] < B(r) I µ(x) + _ Rµ(
1
r1
I
The last inequality is justified as follows: R,,(...) is the slope between (p 1,Gp_1) 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 > 0. Then
F(r) < B(r) Now write y =
[ILr+Y)+].
so that
F(r) < B(r) [µ (r + t) + t] . We try to make
t=µ(r+tr\
r
I
µ
r+ µ
I
rr+f
)
by choosing t.
As a first approximation we choose t = p(r). For that choice z = r + u . Since we want to guarantee that p > µ(r), our actual choice is p = µ (r + n ; ) + 1. We then get:
t
F(r) < B(r) I p f r+
pr)
J
+ $L(r) + 1 J
10. The Maximum Term of an Entire Function
35
or
F(r) < B(r) L2µ 1 \\r + A('.) f + 11 Using the Borel Corollary (Chapter 9) we find /
.
1
F(r) < 3B(r)p(r) eff
so that
B(r) < M(r)
(10.11)
3B(r)p(r). eff
For functions of finite order, say order p, we have
log M(r) = o(r") as
r - oo
if p' > p.
In this case,
log B(r) = o(rP) as r -i oo if p' > p. We have
logB(r)=ptt)
Jr
dt=o(nl )
andwe know that
p(r) log 2
r:
F(r) < B(R)
B(r)RR r
(R)n
so that
B(R) < M(r)
1 RAM dt>µ(r)logR
so that
AN
a limsupn. nlanl1/n. Then for large n we have Ian)
0. Now we estimate f+°°[f(t) - sn(t)]e'tw dt, where sa(t) is the nth partial sum of Proof. First, fo
n!
Eat'.
We have, for ak
=k
00
I/(z) - sn(z)I =
00
Iaklrk = E IaklRk GO k < M(R) E (-r) n+1
n+1
(r
= lRl
n+l
n+1
1
(r
M(R)1- R - \RJ
n+l Rr, R a
R-r'
k
Entire and Meromorphic Rinctions
44
(2e2r'',
where T' > T. Choose R = 2r so that we get I f(z) - sn(z)I Ie-tw I
= e-t", where so that if (z) - s71(z)I < e21 . Now If (t) l < er't and u = Rew; thus we have convergence of the integral and we can interchange the summation and integration if we take u > 2T'. Thus we have IZI
+00
{
(t)e-tw dt =
1
r +oo (E 0
l `
i t') a-tw dt J
J
+00 n
Ea:n
J0
nl a-tw
dt = EWn+l an = fi(w)
in some right half-plane, it > 2r', as was to be proved.
As our final result in this chapter we shall give a direct proof that the order and type of f and f' are the same (Proposition 11.6).
Proof. Let M1(r) = suPe I f'(re'o)I. Then we have from the Cauchy in-
if R > r = IzI, that Mi(r) < tegral formula, f'(z) = tai R M(r) R Now take R = Ar, where A > 1. Then we get Ml (r) < M(ar) r 1} < M(ar) (A a1 for r > 1. Hence, for r > 1, .
a-\
log+ log+ M1 (r) < log+ log+ M(Ar) log AT + log+ log+ la log r log r log .1r log r
-
113
+ log+ 2 log r
and thus p1 < p.
In the other direction, supposing without loss of generality that f (0) = 0, we have f (z) = fo f'(w) dw, where we shall integrate along a ray passing through the origin. It follows that M(r) < rM1(r). Thus, log+ log+ M(r) < log+ log+ r + log+ 1og+ M1 (r) + log+ 2 log r log r
-
and hence p < p1. Therefore p = p1, as desired.
Similarly for type, we have log+ ,'l11 (r) < log+ M(ar) AP + log+ rP 1A -1i2 rP (Ar)P
Hence 71 < APT for any A > 1 and thus T1 < T. Also, M(r) rMl (r), so that to + M(r) < log+ r to + M, r and hence T < T1. Therefore r = T1. rp r, +
-
12
Carleman's Theorem
Let f be holomorphic in Re z > 0 and suppose f has no zeros on z = iy. Choose p > 0 so that p < (modulus of the smallest zero off in Re z > 0). Let {zn = rneie^ } be the zeros of f in Re z > 0. Define the following:
E(R) = E(R : f) = E (r, - R2
cos Bn
(proper multiplicity of the zeros taken into account); n
I (R) = I (R : f) = 1 2rr
r
(t 1
J(R)
J(R : f)
;R
2
f
R12 }
log If (it)f (-it)I dt,
"/2
x/2 log If (ReiB)I cos 0 d9,
where the integral is taken along the semicircle of radius R centered at 0. Then
E(R) = I(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: S
27ri Jr
log f (z) I T2 + R2 L
J
dz,
Entire and Meromorphic Functions
46
where we assume that f has no zeros on IzI = R and that log f (z) denotes
a branch of log f on F, i.e., log f (z) is some continuous function on F satisfying exp(log f (z)) = f (z). The proof proceeds by evaluating the contour integral
r (12.1)
= 0(1).
J
Izj=P
Re z>0
Along the negative imaginary axis z = -iy, y > 0, dz = -idy, so (12.2) R
logf(-iy) [R2
27r
- y2]
dy = 2
[
rR
logf(-iy)
J
- R2]
dy.
On z=iy, y>O, dz=i dy, and we have (12.3) !P
1
tar J R log A W
[)2+] R2
R
dy
2a 0. Recalling the notation in Carleman's Theorem, we observe that
(_j)logR_o(1) n
E(r)
I(R)
_ (R (t2 P
J(R)
BR
aR
\
2=
2B
RZ
Bt dt
0, where T is the Nevanlinna characteristic. An entire function f will be of finite A-type if and only if there exist positive constants A and B such that I f(z)I < exp(AA(BIzI))
for allcomplex z.
If we choose A(r) = rp, then the functions of finite A-type are precisely the functions of growth not exceeding order p, finite exponential-type. We
Entire and Merornorphic Functions
50
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". We obtain in Theorem 13.5.3 a corresponding result for meromorphic 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 sometimes 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 W e study h e r e the distribution o f sequences Z = {zn}, n = 1, 2, 3, ... , with multiplicity taken into account, of nonzero complex numbers z,a such that zn --+ oo as n -f oo. 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
limzn=ocasn -- oo. Definition 13.1.1. The counting function of Z is the function
n(r, Z) = E 1. IZ,IST
13. A Fourier Series Method
51
Definition 13.1.2. We define Lr n(t' Z)
N(r, Z) =
proposition 13.1.3.
t
dt.
We have
N(r,Z) proof. Note that
log ICI
r
E log Ixnl = Jor log (r)t d[n(t, Z)].
Iz+.l 0, S(r; k : Z) _
l 0, S(r1i r2; k : Z) = S(r2; k : Z) - S(r1; k : Z). When no confusion will result, we will drop the Z from the above notation and write n(r), S(r; k), etc.
Definition 13.1.7.
A growth junction A(r) is a function defined for O< r < oc that is positive, nondecreasing, continuous, and unbounded. Throughout this chapter, A will always denote a growth function.
Definition 13.1.8. We say that the sequence Z has finite A-density to mean that there exist constants A, B such that, for all r > 0, N(r, Z) < AA(Br).
Entire and Meromorphic Functions
52
Proposition 13.1.9.
If Z has finite A-density, then there are constants
A, B such that n(r, Z) < AA(Br). Proof. We have Zr
n(r, Z) log 2
0, and if we let
r' = rk1/k, then (13.1.3)
IS(r,r';k)I < 3kr,) k
To prove this we note that I S(r, r'; k) I
k
fr
tk dn(t),
from which (13.1.3) follows after an integration by parts. Now, for r1, r2 > 0, let r'1 = rlkl/k and r2 = r2k1/k. Then IS(ri, r2; k)I _< IS(r', r2; k)I + IS(ri, ri; k)I + I S(r2, rz;
k)I.
13. A Fourier Series Method
53
On combining this inequality with (13.1.3), Proposition 13.1.9, and the fact that k11k < 2, we have I S(rl, r2; k)I
I S(r' , r2i k)I + krk AA(Br1) + krk AA(Br2). 2
1
But, by hypothesis,
IS(ri, r2; k)I < kri AA(Br1) + krz AA(Br2) for k = 1,2,3,....
We say that the sequence Z is A-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,
Definition 13.1.12.
AA(Br)
jai, i S(r; k : Z)
(13.1.4)
rk
If the following stronger inequality:
Iak + S(r; k : Z)I
0 for r > 0. Thus, there exist positive numbers rk such that A(Brk)
0 and 1 < k < p(A). For k in this range, we define
ak = -S(rk; k).
(13.1.6)
For those k, if there are any, for which k > p(A), we choose a sequence with pj ---* oo as j --* oc such that 0 < P1 < p2 < A(Bp')
lim
j moo PjPW
= 0.
For values of k, then, such that k > p(A), we define
ak = jAm -S(pj; k).
(13.1.7)
-00
To show that the limit exists, we prove that the sequence {S(pj; k)}, j = 1, 2,..., is a Cauchy sequence. Let Aj,m = S(Pj; k) - S(pm; k) = S(pm, pj; k) We have
AA(Bpm) kpkm
Since pk
+ AA(Bpj kp;
)
pP(a) for p > 1, it follows from the choice of the pj that
0
as j, m - oo. We now claim that I ak + S(r; k) I
0. Then (i) the sequence Z is of finite A-density if and only if lim sup r-Pn(r, Z)
0.
If A(r) = ro, then we may choose p0 = [p] + 1.
Proof. Suppose first that (13.3.2) holds. We may clearly suppose that B > 1. Then A(2Br) r- Adt > p(2B)PrP
AA(Br) > r0 ,\(t) prP Jr tP+1 dt > 2Br tP+1
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
foo A(t)
00
dt =
rkr+lr +i dt < k-0 J
PrP(2k)P
< M E (M)k.
k=0
k=0
Hence, if po is taken so large that 2PO > M, we have an inequality of the form (13.3.2). In case A(r) = rP, we have M = 2P, and the final assertion follows.
Proof of Proposition 13.9.5. Let A he slowly increasing, and let Z be a sequence of finite A-density. Choose po as in the last lemma so that, for
p>po,
1
co A(+1+i
dt
2P and r < r' < Rp. To prove (13.3.4), we first integrate by parts, replacing the integral by nrk) 1- k
( r')k
Now
r tk+l dt.
/
n(r') < A(r)
A(r)
(1J)k - (ri)k Rp and k is not a multiple of 2p. 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'; k : Z') =
0 if R;_1 < r < r' < R;, j > p, and k is not a multiple of 2P. But, in this case, we have
S(r, r'; k : Z') = 7S(r, r'; k : Z), where
ry=
1+wk+w2k+...+w(m-Uk'
where m = m(j) = 2' and w = w(m) = exp(2iri/m). Since k is not a multiple of 2P, k is therefore certainly not a multiple of 21, so that wk # 1. We then have _ 1 - wkm ti =0,
1-wk
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 4_1 < r < RP. Then, by (13.3.7) and (13.3.3), we have that N(r, Z') < 2PA(r) < 2A(2r). To see that Z' is A-balanced, let k be a positive integer and suppose that 0 < r < r'. Write k in the form 2Pq, where q is odd. Then, by (13.3.9), S(r, r'; k : Z') = 0 if Rp < r < r'. Suppose that r < R.P. Then S(r, r'; k : Z') = S(r, r"; k : Z'), where r" = min(r', RP), by (13.3.9). However, IS(r, r"; k: Z') I< k
J
rI
t dn(t, Z').
By (13.3.8), this last term does not exceed tdn(t), T
and this, in turn, does not exceed 4r-kA(r), by (13.3.4). Consequently, we always have IS(r, r'; k : Z)j < 4r-kA(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. As we mentioned at the beguning, the results of this section are generalized versions of the results of the earlier paper [35], and the proofs are essentially the same. Our notation follows the notation of [35] and the usual notation from the theory 0(meromorphic and entire functions. Our presentation still follows [36]. We first recall the results from the theory of meromorphic functions that gill be needed.
Entire and Meromorphic Functions
64
For a nonconstant meromorphic function f, we denote by Z(f) [respectively W(f)] the sequence of zeros (respectively poles) of f, each occurring the number of times indicated by its multiplicity. We suppose throughout
that f (O) 0 0, oo. It requires only minor modifications to treat the case where f (0) = 0 or f (0) = oc. By n(r, f) we denote the number of poles of f in the disc {z : IzI < r}. By N(r, f) we denote the function
N(r, f) =
J
r n(t, f) dt, t
and by m(r, f) the function
m(r, f) = 2x
I f (re`B) I d6,
J x log+ where log+ x = max(log z, 0). We have, of course, that n(r, f) = n(r, W (f) ) and N(r, f) = N(r, W(f)). The Nevanlinna characteristic, which measures
the growth of f, is the function
T(r, f) = m(r, f) + N(r, f). Three fundamental facts about/T(r, f) are that
T(r,f)=T(r,f/+loglf(0)1,
(13.4.1)
fg) 1, then f is of finite A-type. Proof. If f is of finite A-type, then by the Hausdorff-Young Theorem ([511, p. 190), the L9 norm of log If (reie)I, as a function of 0, is bounded by the P norm of the sequence {ck }, where (p) + (q) = 1. By Theorem 13.4.6, this norm is dominated by an expression of the form AA(Br). Conversely, using Holder's inequality, Ick(r,f)I ykeike,
where yk = ck(r,f) A(Qr)
We may also suppose that the constant M satisfies 21r
IF(0)I dO < M
by Theorem 4.9. By a slight modification of [49] (p. 234, Example 4), we know that for any such F there exists a constant a > 0, where a depends only on M and e, such that x 1
21r
J
exp(aIF(0)I) dO < 1 + e,
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 A-type. The condition is that Z should be A-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 case A(r) = re'. Even this case seems to be unknown.
We turn now to our first task, the construction of an entire function f from a sequence Z and a sequence {ck(r; z : a)} of Fourier coefficients associated with Z. We recall that we have assumed that Z = {zn} is a sequence of nonzero complex numbers such that zn --+ 00 as n - oc.
Suppose that {ck(r)} = {ck(r; Z : o!)}, k = 0, ±1, 12,..., is a sequence of Fourier coefficients associated with Z such that for each r > 0, E Ick(r)12 < oo. Then there exists a unique entire function f with Z(f) = Z, f (0) = 1, and ck(r, f) = ck(r) for k = 0, ±1, ±2,....
Theorem 13.5.1.
Proof. We define ,P(pe"°) = E
ck(p)eikw.
as an element of L2[-w, 7r] Since E Ick(p)12 < oo, this defines for each p > 0 by the Riesz-Fischer Theorem. For p > 0, we define the following functions: (13.5.1)
(13.5.2)
B°(z' Z.)
_
zn p(zn - z) Izn1
p2 - znz
Pp(z) = H B,, (z; zn), I z. I
(13.5.3)
(13.5.4)
(13.5.5)
K(w; z) _
Q(z) = exp p
w+z
w-z dw 1
21ri
J1w1=p
K(w z)4i(w) w
,
fp(z)PP(z)QP(z)
We make the following assertions: (13.5.6)
The function fp is holomorphic in the disc {z : Iz) < p}, and its zeros there are those zn in Z that lie in this disc.
Entire and Meromorphic Functions
72 (13.5.7)
ff(0) = 1.
(13.5.8)
ff r < p, then ck(r,f) = ck(r).
Now (13.5.6) is clear from the definition of fP. Also, Iznl
fP(0) = PP(0)QP(O) = QP(0) R
P
IE^1
(i)C
I .I_P
F o r k = 0,1, 2, ... , we write w = pe"° and ck(p)e'k`° = I kwk. Then by the definition of ck(p) we see that
Po = N(p, Z)
13, A Fourier Series Method and
2nk +
f2k
E
2k
Iz, I{SZkwk + S1kwk}
k=1 00
1ZkP2k N(p, Z) + E {kwk +
(i)k}
k=1
so that 2
27ri KIWI=P
-'(w)KK(w, z) w
27ri
(w
twz)2 dw,
where
But 1
wk
dw = kzk-1
21W-i
I,,.I=p (w - z)2
1
r
and 2arz
rl
IwI=P
1
(w - z)2
w
k = 0,1,2,...,
for
dw = 0 for
k=0,1,2,....
Hence, 00
Q' (z)
QP(x)
L°
= c Vk zk_1, k=1
where Vk,P
= 20k = (xk + k Izn I SP
n) k -
\ p2) k
Hence, near z = 0 we have
f
°(z) fP(z)
00
_ P(z)
()
PP(z) + QP z
kakz k
1
k=1
and (13.5.9) is proved
It next follows from (13.5.6)-(13.5.8) that (13.5.10)
if p' > p, then f,,' is an analytic continuation of fP
Entire and Meromorphic Functions
74
For if we define for Izi < p
F(z) = f,(z) fa(z)' then
Ck(r, F) = ck (r, fP') - Ck(r, fP) = Ck(r) - ck(r) = 0
for 0 < r < p, and therefore IF(z)l = 1. On the other hand, F(0) = 1, and it follows that F is the constant function 1. We now define the function f of Theorem 13.5.1 by setting f (z) = f f(z) if p > Izj. It is clear that f is entire and, by (13.5.6), that Z(f) = Z. Also,
f(0) = 1, and ck(r, f) = ck(r, fp) for p > r, so that ck(r, f) = ck(r). An argument analogous to the one used in proving (13.5.10) proves that f 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 of zeros of an entire function f of finite A-type is that Z be A-admissible in the sense of Definition 13.1.15, that is, that Z have finite A-density and be A-balanced.
Proof. If Z = Z(f) for some f E AE, then by Theorem 13.4.6 the sequence {ck(r, f)} 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(r)} associated with Z. Then by Theorem 13.5.1
there exists an entire function f with Z = Z(f) and {ck(r, f)} = {ck(r)}. Then by Theorem 14.4.7 and the fact that {ck(r)} is A-admissible, it follows
that f E AE, and the proof is complete. Remark. This theorem generalizes a well-known result of Lindelof [201, which may be stated as follows.
Theorem 13.5.3. Let Z be a sequence of compex numbers, and let p > 0 be given. If 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 n(r, Z) _< Are. If p is an integer. it is necessary and sufficient that both this and the following condition be satisfied for some constant B: < B.
Iz,.I 0. Then, by defining
Dp(z; w..) = J B,(z; w..) Iwnl 0)
for some constants A', B', so that the function f synthesized from the ck(r) Must be of finite A-type by Theorem 13.4.5.
Entire and Meromorphic Minctions
76
Supposing now that Z = {zn } has finite A-density, we define W = {wn } by wn = zn + En, n = 1, 2, 3, ... , where the En are small complex numbers
so chosen that Iwn I = Izn I, n = 1,2,3,..., all of the numbers wn and zk are different, and such that IEnI
It remains to prove that
2I'Yk+S(r;k:Z- Sr;k:W uniformly for k = 1, 2,3,.... Now
2
Ilk+S(r;k:Z)-S(r;k:W)I
()k}
rk 2
rk
2
k
n/ k -
nl>r
[
1 [ (wn)k - (zo)k Ir
IznI>r
(wnzn)k
However, I(wn)k - (zn)kl C
2
r
Izn I2k
klEnllznik-1, so that we have
Iyk+S(r;k:Z)-S(r;k:W)I
rk 2
IE
1znl>>r
Iznl
j
r Iznl
< ' A(0) < A(r). 2
13. A Fourier Series Method
77
The field A of all meromorphic functions of finite Atype is the field of quotients of the rings AE of all entire functions of finite A-type if and only if A is regular in the sense of Definition 18.3.2, that is, if and only if every sequence of finite A-density is A-balanceable.
Theorem 13.5.5.
proof. First, suppose that A is regular and that f E A. Then Z(f) has finite A-density by Theorem 13.5.3. There then exists a sequence Z' D Z(f) such that Z' is A-admissible. [We may suppose, by the remarks preceding the proof of Theorem 13.4.5, that f (0) 54 0, oo]. Then, by Theorem 13.5.2,
there exists a function g E AE such that Z(g) = Z. Since we have then that Z(g) C Z(f), the function h = g1 f is entire. However,
T(r,h) 0. 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 f. 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 f such that if f = fl/f2, where fl and f2 are entire, then T(r, f2) i4 O(T(r, f)). 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 the 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 if f is any meromorphic function in the plane, then there exist entire functions fl and
14. The Miles-Rubel-Taylor Theorem on Quotient Representations
79
f2 such that f = f, /f2 and such that T(r, f;) < AT(Br, f) for i = 1, 2 and
>0. Suppose Z = {zn } is a sequence of nonzero complex numbers with jzn I -->
co. We include the possibility that zn = z,n for some n # m. As in the previous chapter, let (14.1)
n(r, Z) IZn Kr
and
(14.2)
N(r, Z) =
f r n(tt Z) dt. 0
It was shown in the previous chapter that the following lemma is suffi f dent to establish the theorem.
Lemma. Suppose Z = {zn} is a sequence of nonzero complex numbers with IzzI -- oo. If A(r) = max(1, N(r, Z)), then there exist absolute constants A' and B' and a sequence Z = {-;n} containing Z (with due regard to multiplicities) such that N(r, Z) < 5A(4r) r > 0 (1) and,
(ii) for j = 1, 2, 3.... ands > r > 0,
(;)'
A'A(B'r)
A'A(B's)
0
8
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 (0) 54 oo.
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 respect to the growth function A(r). Let A1(r) denote an arbitrary increasing unbounded function defined on (0, oo). In Theorem 13.5.2 we characterized the zero sets of entire functions
such that T(r, ¢) < aa1(lr) for some constants a and Q and all r > 0 as those sets Z* which both have finite density and are balanced with respect to Al (r). This characterization combined with the above lemma guarantees the existence of constants Al and B and of an entire function f2 having zeros precisely on the set Z (counting multiplicities) such that T(r, f2):5 A1A1(Br) for all r > 0. Hence, (14.3)
T(r, f2):5 A1N(Br, Z) < A1T(Br, f)
Entire and Meromorphic Functions
80
for r > ro(f). Letting fl = f2 f, we see that fl is entire and that T(r, fl) < (A1 + 1)T(Br, f) for r > ro (f ). For an appropriate complex constant c, 0 < Icl < 1, we have for i = 1, 2 that
T(r,cf=)=0 r 0.
(14.5)
Letting A = Al+1, we see that f = cfl/cf2 is the desired representation. It is implicit in 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
ZN=Zl{z:2N 0. 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 ZN ¢. Let ZN = {z1, z2,. .. , 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, U0, UE7 V0, 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 k
-8j - 8j = -16j. Combining (14.16), (14.20), and (14.25), we conclude that (14.26)
f2-
7 J
27r
1=2Nzn
e- +je2 -j(N-1) fN(8) dB
< - 16j2- j(N-1)
The same discussion applies to the imaginary part of the above quantity.
The only minor modification is that we must divide [0,2w] into 2j + 1 subintervals on which sinjO is alternately increasing and decreasing. Since 2j + 1 < 4j, this causes no difficulty. We obtain
fzx e-'j''2-j(N-1)fN(0) dO
27r
< 32j2-j(N-1).
Combining (14.26) and (14.27), we obtain (14.15), which in turn establishes (14.13).
We are now in position to show that 2 satisfies condition (ii) of the lemma.
Suppose that s < 8r. We then have trivially for all positive
integers j
(i)1 n (2) log 2, by the usual argument, and the result follows.
Definition. The genus of Z, p = p(Z), is defined by p(Z) = inf{q : q is an integer, E zj ; < oo}. Definition. The canonical product of genus p over Z is defined by z
P(z)
H E
z EZ
+P
n
15. Canonical Products
89
Definition. The canonical product PZ over Z is defined by z
PZ(z) = [J EZn- P znE Z
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. Given an entire function f, f (z) = PZ(z)zm expQ(z),
where m is a nonnegative integer and Q is a polynomial of degree < p(f).
Here, Z=Z(f). Deflnition. The genus of the entire function f is defined by
P(f) = max(n,p), where n = deg Q.
Examples. Suppose zn = n2, n = 1, 2,3,... . If f (z) = rj (1 - n ), then f is of genus 0. If f (z) = ez rj (1 n ), then f is of genus 1. We first shall show how the Hadamard Factorization Theorem follows from Theorem 15.5. Without loss of generality, suppose f (O) 96 0. In fact, assume f (O) = 1. Let
f(z)
9(z) = PZ(z) Then g is an entire function without zeros, so that there exists an entire function Q such that
z = exp Q. We must prove that Q is a polynomial of degree < p(f).
Lemma 15.6. If F and G are meromorphic, then p (c) < max{ p(F), p(G)} Proof. We have log T(r, f ) log r r_.oo
P(f) = lim sup
T `r, 1) T(r,f)+O(1) T(r, fg) = T(r, f)+T(r,g)+D(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 = u + iv, we have Iu(Te'B)I dO = 0(r°)
for each p' > p. Since, with IzI = r and R = 2r, we have Q(z) - Im Q(O) =
uw
u('w) w +
2i,
z
it follows that
IQ(z) - Im Q(0)I = 0(r°) Hence
IQ(z)I = 0(r,"), and it follows that Q is a polynomial of degree < p. Proof of 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. If Iznl > 21z1, then log E (.!n , p) 1:5 21
Let u = z- . Now log E(u, p)
IP+1
1. Otherwise, the conclusion of the theorem is clearly true. Therefore, rn(a) = Anrn(b). Now,
Ian - Pkl -Ianl-IPk1=Anl/nl-IPkl - (An-1)IPkl. Letting z = an in (16.2) gives b - a = anH(an -,Ok). Hence, Ib - al >- Ianllrr(an - Pk)l >- IanIfI(An
- 1)IPkI-
Dividing through by b and taking nth roots gives
1- an
[i'n(A_1)IflkJ]
n
>[ Now suppose we had An - 1 > 1. Then L
an Ianl n I1-bil>[IbI(1+E)IIIPkI,". But rrlPkI = aa so that supposing An - 1 > 1 implies I1 - e ° > (1 + E); a contradiction. Hence, lim sup An < 2, so that we have lim sup rn( r. b) < 2. n-oo
Suppose now that b = 0. Write f (z) = amzm +an+lzm+1 +..., am # 0. n-m We now have fn(z : b) = zman jj (z -,6k). Thus, fn(z : b) - fn(z : a) k=1 n-m n
-a = anlzm fT (z - Pk) - f1 (z - ak)) k=1
k=1
n-m We again have Ian-PkI = (An-1)IPkj and I-al = Ianan f1 (an-Pk)I k=1
n-m
lanl[rn(a)]m fl (An - 1)IQkl: k=1
(16.3)
_ Ial
n-m ^-*" Llanlfrn(a)ln` H (An -1)IPA; I
rr
L
k=1
16. Formal Power Series
Now suppose for contradiction that
n-m
\l ml there exist {En} such that En > 0 and An - 1 =
lal(^(laml)
95
(The m
(rn(a)
mss-
(1+En).
n-m n-m
Using this expression in (16.3) and that n 113kl =
S-I gives
k=1
lal ^ m >lal n'tn (1 + En)n-mi a contradiction. Hence,
An-1 R(f); it follows that {IanIRn} is unbounded. Therefore, for a suitable subsequence, IanIRn > IakIRk for all k < n. In that subsequence, < 2. Hence, lim inf rn < 2R. Therefore, lim inf rn < 2R(f), as was to n-oo n-.oo be proved.
Corollary (Okada (28)). A function f is entire if and only ifn-oc lim Tn(f) = -00.
16. Formal Power Series
97
log n =1im log n sup eorem (Tsuji [45]) 16.4. Let a = o(f) = lim sup n-.oo log rn n.oo Tn
I- en a(f) = P(f) proof. We know that R(f) = lim sup
n log n 1
10g Ian
pl > p; then, for large n, p' >
n log1
from Proposition 11.4. Take
I
Hence, IanI
R > n''
, or rn > n o
< p' for large n;
Consequently, o
1r.
hence, a < p as desired.
In the other direction, suppose on the contrary that a < p. Choose p' such that or < p' < p. Then for n large (say n > no ), rn > n or . Choose 2K
M > 1 so that for K = 1,2,..., no, laid < m (K1)
-
We prove by
2n
induction that for n > no, IanI < m
(n!) °
First,
Ianlrn < Ian-l lrn 1 + ... + Iallrn + 1 + anzn = 0. Hence,
since &(z : 1) = 1 + aoz + IanI
m(r, f) + m (r, ff 1 1) . Therefore,
T(r, F) > T I r, 1) +T (r,
(17.1)
f
1
f-1/
Notice that 1
f(z)
I
+
f'(z)
F(z) -
f'(z)
f(z) - 1
f(z)
f(z) f'(z)
since f 54 0, 1.
Thus, (17.2)
T(r,F) 0, then f vanishes at zo to order K. Consequently, n(r, f) and n(r, f') are unaffected by the behavior of f near Of zo. The contribution to n1 near z equals the contribution to n (r, course, j has a pole of order K - 1 at zo. If K < 0, then f has a pole of order -K at zo and a similar analysis reveals that n1 "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 n1, respectively, K-1 and m - 1 times. f).
The gist of the second fundamental theorem is that for most values of a, the contribution of m (r, ffl a) to T (r, f 'a) is much smaller than the con-
tribution of N (r, f 1 ) to it. Thus, for most values of a, N (r, f11) makes the preponderant contribution to T (r, f Q- . Recall that T (r, f 1 a) T(r, f) = 0(1) by the first fundamental theorem. Remarks. The following weaker form of the theorem also gives us the Picard Theorem: (17.5)
m(r,f)+Emlr, -1
\\\
} 0. If f never takes on the value a, then b(a) = 1. The same is true if f takes on the value a very infrequently. In general, b(a) measures the tendency of
f to omit the value a and 1 - b(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 6 > 3 . Definition. a is said to be a fully branched value of f if f takes the value a with muAiplicity either zero or > 2. If 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 sin z. Also, if f is entire, then there are at most two values of a Remark 17.1.
for which 6(a) > 2.
The reader will find it instructive to work out b(a), 9(a), and 9*(a) for some specific functions, such as eZ, tan z, and the Weierstrass p function.
Remark. If a is fully branched, then 1
9* (a) =1im 1 - NTr(r,
> lim 1-
f)
N(r,
Also, if a is fully branched, then
N
1
f -a
since T > N.
N
f ia) 1
N(r,fa)
2. In general, there are
of + e9 = eh,
where f, g, and h are entire functions. Do there exist f, g, and h so that the identity holds? What is the relationship between f, g, and h? We may write
of-h + eg-h = 1.
Then the entire function of -h omits 0. But eg-h omits 0 as well, therefore of -h must also omit 1 for the identity to hold. By Picard's Theorem, ef-h must be a constant so f = h+ const. Similarly, g = h+ const.
Lemma (Hiromi-Ozawa [15]). Let ao(z), a, (z),... , an(z) be meromorphic functions and let g1(z), ... , gn(z) be entire functions. Further, suppose that
`
n
T(r,ai) = o Em(r,e9°)J j =0,1,...,^ v=1
holds outside a set of finite logarithmic length. If an identity n 2av(z)e9I(=)
V=1
= ao(z)
17. Picard's Theorem and the Second Fundamental Theorem
107
holds, then we have an identity n
E c. a.(z)esvW = 0 V=1
where the constants c,,, v = 1, ... , n are not all zero.
Proof. On writing f' = f f, we may use the lemma on the logarithmic derivative to estimate T(r, f) when f is entire. For notational simplicity, Then we have
let
n
E G (z) = ao(z).
(17.8)
V=1
By differentiating both sides we obtain n (17.9)
G(") (z) = a( 0")(z),
which may be rewritten as n
(17.10)
>2 G., (z)
g) (z) = a(µ)
z p --1, o (),
... , m - 1.
(z)
V=1
We regard this as a system of simultaneous linear equations in the G. Now we have G(IJ) (z)
= P,(a,,,a;,,...Ia(" ),
(µ))eg.,(z)
with a suitable polynomial P, in the indicated functions. Thus we have (17.11)
(m(re9))
T
Y-1
outside a set of finite logarithmic length.
Suppose, for the simultaneous equations (17.8) and (17.10), that the determinant A 34 0. By solving (17.10) with respect to G j = 1, ... , n we have by Cramers' rule where 1
GI/G1
... ...
1
G,1,/Gn
0= Gin-1)IGn
.
__
G(n-1)IG.
Entire and Meromorphic Functions
108
and 1
...
1
a0
1
...
1
G1
...
Gj-1 Gj-1
al
G C i+1
...
.1 G.
Gj_-1_
...
0
Aj= GI
Gin l1, GI_'
(n-1)
a0
___ Gj+t
____
...
Go
Since
T (r, C
(17.12)
I=o
m(r, e9) f
,
we have n
T(r,A)=0
=1
m(r,e9°) ,T(r,Aj)=o(
m(r,e9°))
for j = 1, ... , n outside a set of finite logarithmic length. Thus we have
m(r,e9") =T(r,e9°) E m (r Proof We first remark that the lemma is intuitively obvious, for when f (z)
is "close to" a it contributes to m (r, with j # Y.
lam) ,
but not to any m (r,
)
To prove the lemma in detail, we introduce the following notation. Let d > 0 be given with 2b < min{ la - a,1 : 1 < v < j < n}. We require
6 1. Now, considering 2 (PP-r ) [2T(p', f) + C] as a function of p, we see that it vanishes for p = r and is greater than 1 for p = p'. Since it is a Continuous function of p, we may choose p so that P'(P - r) [2T(P', f) + C] = 1. r(p' r) For this choice of p, we thus have
N(p) - N(r) < 1 and (18.13)
r(p' - r)
(p - r) =
P'[2T(P',f)+C]
Entire and Meromorphic Functions
120
and
r
(p'-p)=(P -r) 1- p' ]2T (p', f) + C] Hence, 1og+
< log+ p' + log+ T (p', f) + log+
p-r 1
1
p' - r
Therefore, log+
(18.14)
1
p-r
< log+ p' + log+T(p', f) + log+
1
p' - p.
Also, we see that log+
1
< log+
p' - p -
1
p' - r + log+ + log+
1
1
1
1
r
1
-2Tp',f+C
Hence, log+
(18.15)
< log+
1
1
Using (18.13), (18.14), and (18.15) in (18.12), we have (18.16)
r)
m (r,
< 4 log+ p' + 4 log+ T (p', f) + 3 log+
p'
- r'
which proves thelemma. We will use the notation 11f (x) < g(x) to mean f (x) < g(x) with Borel exceptions (i.e., a set of finite length).
Proposition. I]m(r,
4log+ r + 8log+ T(r, f ).
fl in the lemma, we have
Proof. Taking p\= r +
m (r, f, < 4log+ 1 r +
!
log+
+ 41og+ T r+
7,(r f) +
i
31og+ 1og+ T (r, f)
'f
log+ T (r, f) Hence, by the Borel Lemma, we get 11m
(r,
f i < 4log+ r + 8log+ T(r, f ).
Corrollary. IIS(r) = O(1og+ 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 m(r, 4-).
19
"Two Constant" Theorems and the Phragmen-Lindelof Theorems
The Two Constant Theorem. Suppose that f is holomorphic in D = 1z: IzI < 1} and continuous in D\{1}. Suppose further that if I N in D and if 1< M in oD\{1}. Then If I < M in D\{1}. First Proof. Choose a > 0 and let
(Li) u f (z)
g(z) _
f o r a suitable branch of (1 2 1)". Now g is analytic in D and continuous in D if we define g(1) = 0. By the maximum modulus theorem,
zaD
suupplg(z)I = m xlg(z)I
But Ig(z)I<Mfor zEBID. Hence, Ig(z)I<Mfor zEID,and so 0
If(z)I<MII
zI
for
zED.
Now let a - 0 to get the result. Second Proof. This proof uses the Cauchy integral formula. Since there exists a linear fractional transformation U(z) = e'9 taking an arbitrary Y Point z0 E D into 0, and mapping I)/{1} conformally onto ID/{1}, we need only prove that if (0)I < M. But choosing 00 with 0 < 00 < 7r,
f(w)dwl if(o)i = I27r 1 w=R w
I
l
1 1 If(rei8)Ide+ 27r 21r
I
2)
If(Te`B)Id0,
122
Entire and Meromorphic Functions
where R < 1 and
L)fo 0 If(z)I 5 K6exp{6IzI°} in the region, then the conclusion follows as before. Proof. Now let F(z) = exp{-eza} f (z), 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 Polya 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. A compact convex set is the convex hull of the set of its extreme points.
We omit the proof.
Definition. For any set E, k(8) = sup{Re(ze-'B) : z E E} is called the support function of E. We note that k(O) 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 = -oo. It is easy to show that, for a given set E,
K(E) = {z : Re(ze-'B) < k(0) Remark 20.2.
for all 0}.
If zo = xo + iyo = roe'Bo and E = {zo}, then k(0) =
TO cos(0 - Bo) = zo cos 0 + yo sin 0.
20. The Polya Representation Theorem
125
Remark 20.3. Let E be a circle with center at 0 and radius R. Then k(8) = R for all 0.
Let E be the line segment [xo, x1], xo < x1. Then k(8) _ x1 cos 0 if - a < 0 < 2 and k(0) = xo cos 8 if 2 < 0 < s2 . In particular, if xo = -x1 and x1 = R, then k(0) = RI cos 01. If E is the vertical line Remark 20..¢.
segment [-iR, iR], then k(8) = RI sin 01.
If E1 has support function k1 and E2 has support function k2, then E1 + E2 = {z1 + z2 : z1 E E1, Z2 E E2} has support function Remark 20.5.
k1 + k2. From Remark 20.3, it follows then that the rectangle with vertices (±R1i ±iR2) has support function k(0) = R1I cos e] + r21 sin 01. Remark 20.6.
The convex hull of E1 U E2 has support function k =
max{k1, k2 }.
Remark 20.7. If E1 with support function k1 is translated, so that the point originally at 0 is moved to zp = xo + iyo, then the support function of the translated set is k(O) + xo cos 0 + yo sin 0.
Definition. Let Ho(oo)be the class of all functions that are holomorphic near oo and that vanish at oo.
Let f be an entire function of exponential-type and write f (z) _ () zn. The BoreI transform -6 of f, defined by 4'(w) = > anw mar, belongs to Ho. Each function in Ho is the Borel transform of a unique f. We have seen (Corollary to Proposition 11.5) that the type of f is the radius of convergence of the series >2 an war. In Proposition 11.7, we saw that D(w) = f ow f (t)e- t7°dt, in the sense of analytic continuation. We also saw that
f (z) =
27ri ,
14'(w)ez'°dw,
r
where r is a rectifiable curve that winds once around the singularities of I. We call this the P61ya integral representation formula.
Definition. Let S(4') be the set of singular points of 4', and let k be its anpporting function. We call S(4') the conjugate indicator diagram of f. Sometimes this name is used for S*(4'), the closed convex hull of S(4'). We
will write D(f) =
Definition. The indicator function of f is h(8) = hf(0) = limsup r-.oo
1 log ]f(reie)1.
r
We now state an important part of the Polya Representation Theorem.
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Entire and Meromorphic Functions
Theorem 20.8. h(0) = k(-0) for all 0. 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 = 0, and hence 0 is the only possible singularity of 1. Remark 20.10. Denoting by r(f) the type off, we have r(f) = max h(0).
Remark 20.11. If f (z) = eaz, where a is real, then
h(0) = lim sup 1 log lea` l = lim sup 1 ar cos 0 = a cos 0. r.-.oo
r
r-.oo
r
On the other hand, O(w) a" w- r = (w - a)-1 and so S(O) = {a}. Thus, k(0) = acos0 and we do have h(9) = k(-0). Remark 20.12. If f (z) = eiz, we have h(0) = - sin O and 4)(w) = (w-i)-1. Hence, k(0) = sin 0 since S(4,) _ {i}. Note again that h(O) = k(-O).
Remark 20.13. Suppose f (z) _ > a,,ea^z, a finite sum with distinct A,, so that S((D) = {A,.}. Note and nonzero an. Then O(w) = affect h(0). If the A,, lie on a straight that only the extreme points of line, only the endpoints affect h(0). Remark 20.14. It is easy to see that h f.9 < h f+h9, so that D(fg) C 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 dµ^ is defined by dµ^(z) =
J
e-zwdA(w).
It is easy to see that dµ^(z) is an entire function of exponential-type for
dz
dp^(z) = f(_w)e'd/L(w)
(as can be verified on differentiating "by hand") so that d,i' is entire. And
Idi^(x)I < f Iez `Jjdµ(w)I 5
eRjzj
f Idu(w)I,
where R is chosen so that the support of dµ lies in the disk of radius R centered at 0.
20. The Pblya Representation Theorem
127
Definition. We write dµ - dv to mean that dµ^ = dv^.
It is not hard to show that dp - dv if and only if f f dµ = f fdv for each entire function f, or for each entire function f of exponential-type. It is clear that - is an equivalence relation.
If we take dµ = dzjr, where r is a circle, then dp^(z) = fr e-a'dw = 0 by the Cauchy Theorem. Hence, dµ - 0 even though dµ 0 0. We shall see that for any dµ E mo, dµ - dv, where dv = 4P(-w)(-dw)Ir, where is the Borel transform of dp^ and IF is any curve that winds once around
S(C. Definition. [dp] is the class of measures equivalent to dp. Definition. Mo is the class of all [dp] for dµ E Mo.
Definition. If f is continuous in the plane and dp in Mo, we define the convolution f * dµ by
(f * dµ) (z) = J f (w - z)dp(w). Definition. If dp and dv are in Mo, we define the convolution dµ * dv by
f*(dµ*dv)=(f*dµ)*dv. By means of the Riesz Representation Theorem, it is easy to see that the above definition defines dµ * dv as a unique measure in Mo. Indeed, Mo is an algebra over the complex numbers.
Proposition. If dp1 - dp2, then (dpi * dv) - (dµ2 * dv). It thus makes sense to define [dp] * [dv] = [dµ * dv]. Ma is an algebra over the complex numbers.
Definition. Let Eo be the algebra of all entire functions of exponentialtype.
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 functionals on E. Now E is a locally convex topological linear vector space. It will appear that each of the spaces Mo', Eo, Ho(oo) "is" the dual space E.
Definition. For f E E and [dµ] E Mo', define the inner product (Fl, [dµ]) hY
(F1, [dp]) = (F * dp)(0) = f F(-z)dp(z).
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128
Definition. For F E E and f E Eo, define the inner product (F, f) by
(F,f) = (f(D)F)(0), where D = dZ. This means that if f (z) = E n F(")(0)
z", then (F, f)
It is not hard to show that, for each f E Eo and F E E, 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, dµ).
Definition. For F E E and fi E H0(oo), define the inner product (F, 4b) by
(F,
_1I 0 for n =
0,1,2,..., then An=0 fore=0,1,2,.... Proof. F(-z) = E(-1)"Anzn has nonnegative coefficients, but no positive real number is a singularity of F.
Hadamard Gap Theorem. If AZ) = E anz" with an = 0 except for n = nk, where lim 1, then every point of the circle of convergence off is a singular point of f.
Theorem 20.25. If {An} is k-admissible and An = 0 except for n = nk with lim nk+1 > 1, then An = 0 for n = 0, 1, 2, ... . nk
This is a simple consequence of the Hadamard Gap Theorem.
Entire and Meromorphic Functions
134
Fabry Gap Theorem. If f (z) = > anzn with an = 0 except for n = nk and k-'nk -+ oc, then every point on the circle of convergence of f is a singular point of f.
Theorem (Szego). Suppose that f (z) _ anzn, where the an lie in some finite set. Either IzJ = 1 is a natural boundary of f or f 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(0) = 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(0) 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 27r-periodic; and when we speak of a "supporting function," we mean a supporting function of a nonempty compact convex set. Our treatment is a combination of the treatments in Pd1ya [31] and Boas [5].
Definition. A function H(0) is a sinusoid if it has the form H(O) _ acos0+bsinO.
Remark. Given 01 $ 02 and real numbers h1 and h2, there is a unique sin soid H such that H(01) = hi and H(02) = h2. We call H the interpolating sinusoid: It is given by (0 < 02 - 01 < 7r) sin(02 - 0) (20.1)
H(0) =
sin(0 - 01)
hlsin(02 - 01) + h2sin(02 - 01)
Definition. Given a function h(0) and 01 # 02, we call H the interpolating sinusoid of h if H is given by (20.1) with h1 = h(01) and h2 = h(02).
Definition. A function h(0) is subsinusoidal if it is majorized by each of its interpolating sinusoids, that is, sin(02 - 01)
h(o1)sin(03 - 02) + (20.2)
h(03)
h(02) < sin(03 - 01)
whenever 01 < 02 < 03 with 0 < 03 - 01 < ir.
sin(03 - 01)
20. The P61ya 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(81) > h(81),
H(82)>h(02),then H(8)>h(8) if81 0. Now f (x) = I -b(w) exp(zw) dw,
if
2i c
so that if z = refe, then If (reie)I < A mEax I exp(zw) I,
where A is a constant. Hence, h(0) < max R(weie). WEC
If we now let c - 0, we see that h(0) < k(-0). In the other direction, it is enough to prove that h(0) > k(O) since, if we replace f (z) by g(z) = f (ze'`°), the general case follows from hg(0) >_ kg(0), since hg(0) = h f(0 + gyp), 4Ds(w) = e"P4bf(we-"°) Dg = e"'Df and kg(0) = k(0 - cp). Now, asrwe have seen, -O(w) =
J0 co f (t)e-*w dt
for w > h(0),
so that fi has no singularity to the right of the line x = h(0) and the
inequality h(0) > k(O) follows.
21
Integer-Valued Entire Functions
An integer-valued entire function f is one such that f (n) is an integer for . Some examples are n = 0, 1, 2, (i) sin irz (ii) 2Z
(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 R1. The three examples above belong to R1.
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, I,_ , n - 1. Notice that the coefficient of z" is 1. Examples. Any n E Z satisfies z - n = 0. And ±i satisfies z2 + 1 = 0. 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
Q(x) = 1±q1x+g2z2 +... +qnz",
Entire and Meromorphic Functions
140
where qj E Z f o r j = 1, 2, ... , n. We can write n
Q(x) _ fl (1 -$jz), j=1
where the f3j run over one or more complete sets of algebraic integers. This may be seen from the fact that the f3j are the roots of the polynomial
R(x) =
znQ ( 2) = xn + Q1xn-1 + ... + qn
Now let P be any polynomial with integer coefficients and Q as above. Then
P(x) Q(z)
= E bnxn, where bn E Z.
This follows since we can write 1 4;(,) = 1 +Q`(z) + [Q*(z)]2 + .. .
1
1
Q(x)
1 + qlx +
+ gnxn
1-
= 1+B1x+B2x2+ . The BI are clearly integers. To express the bj in terms of the ,6j, let us first suppose for simplicity that the f3j are distinct, and that P = 1. Using partial fractions and writing 00
= r fjnxn' L1
1 - #jx
n=1
we have
M 00
E.
1
= L.r
11(1 - Qjx)
j=1 1 - f3jx
=j=1Ln=O L
1
where the Ej are the coefficients in the partial fraction expansion. Hence, n
L
bn = [ Ejf31", j=1
so that it is natural to take m
f(z) = where Q. is suitably defined.
j=1
Ej/,,
21. Integer-Valued Entire Functions
141
Example Q(z) = 1 + z2 = (1 + iz)(1 - iz)
Q(z)
21
2{1+iz+i2z2 }
iiz+21+iz
}
+ [in + (-i)n] 2
1,0,-1,0,1,0,-1,0,... f (z) =
2 [iZ
+ (-i)Z] = cos 2 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
b = Ef'.i(n), so that we take (21.2)
Pj(z)Q,,
f(z) _ i=t
where the P3 are suitable polynomials (not necessarily integer-valued).
Definition. Let Rl be the class of functions f constructed above.
Definition. Let R be the class of integer-valued entire functions f of exponential-type for which h f(±ir/2) < 7r.
Our problem is to find additional growth conditions on f that imply f E R1 if f E R. 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 cp be the function (whose existence and uniqueness is guaranteed by the Riemann Mapping Theorem) that maps S conformally one-one onto the unit disk D = {z : IzI < 1} and such that w (O) = 0 and cp'(0) > 0. Then cp is the normalized mapping function of S and p(S) = ) is called the mapping radius of S. We shall need the following deep theorem of P61ya, which we state without proof. The proof may be found in [8].
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Entire and Meromorphic Functions
Theorem. If g(z) = >J 0
b,, E Z, and if g is analytic in a region S containing 0, with p(S) > 1, then there exist polynomials P and Q, with integer coefficients, Q(0) = 1, such that P
9=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(0) = 1, then there exist polynomials P1 and Q1 with integer coefficients Q1(0) = I such that P1 and Q1 have no common factors and PI /Q1 = P/Q. Now given f E R, let f(n)z'.
9(z) _
00
n=0
As we have seen earlier,
f
1
1
9(z) = 2ri 1 - zew -b(w)dw r
for any curve t that winds once around D(f) = S`(4). Now let S be the complement of the image of D(f) under the mapping e''', i.e., S = C\ exp(-D(f)). Now g is analytic in S, so that if p(S) > 1, then g = , where P and Q are polynomials with integer coefficients and Q(0) = 1. By the lemma above, we may suppose that P and Q have no common factors. By the construction that characterizes R1, we can find a function fl E R1 such that f, (n) = f (n) for n = 0, 1, 2,.... By Carlson's Theorem, if we know
that h f, (f 2) < ir, then we have f = fl, so that f E R. To prove that h f, (±M) < r, we write M
fl (z) = E P, (z)f; . =1
By construction, the fji 1 are the roots of Q, so that the ,QJ 1 are the singularities of g, and hence the 131 1 are in the complement of S. Hence, we may write f3j = exp(-,uj), where µj E D(f), so that m
fi (z) _
Pi (z) exp(-FUiz),
j=1
and it follows that hf, (f 2) < a since D(f) is interior to the strip Iyj < 7r We therefore have proved the next theorem.
21. Integer-Valued Entire Functions
143
Theorem. If f E R, and if the complement of the image of D(f) under the map a-' has mapping radius exceeding 1, then f E Rl. For applications, a variant of the foregoing procedure gives a more useful result. We let A be the difference operator defined by (Af)(z) = f(z + 1) - f(z),
defining
A°f = f and
An+lf = A(Anf)
Now, using Taylor's Theorem, we may write
A=eD-1 An = (eD - 1)n where
D dz'
We define the functionals Tn and Tn by
Tnf = f (n) T* f
= (Anf)(0)
To illustrate,
T = f (0)
Tif =f(1)-f(0) T2*f=f(2)-2f(1)+f(0) T3 *f = f(3)-3f(2)+3f(1)- f(0). It is easy to show that
T,a = (-1)n >(-1)k (n) Tk.
0
Tn=E(k)Tk. For example, the first identity is proved on writing An
= (eD - 1)n =
n
k0
()e)(_1y1_1c.
Entire and Meromorphic Functions
144
Definition. To say that a sequence {bn }, n = 0, 1,2,..., is K *-admissible is to say that there is an f E R such that
n=0,1,2,....
T; ,f =b,,,
For f E R, write 00
g(z) = E(Tn*f)zn0
It is easily seen that g(z) = 27ri
Jr 1 - z((w)
-D(d)dw'
where ((w) = e' - 1 and I' 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'° -1)-1. The argument may be reversed to prove the next result.
Theorem. A sequence {bn} is K`-admissible if and only ifEbnzn is analytic on the segment [-1,01. We may now prove the main result of this section.
Theorem. If f E R1, let E be the complement of the image of D(f) under the mapping (e' - 1)-1 and let E" be the image of D(f) under the mapping eL - 1. If p(>) > 1, then f E R1 and f (z) = F, Pk(z)(1 + ak)z, where the Pk are polynomials and the Qk run through the complete sets of
conjugate algebraic integers lying in `. Proof. Let g(z) = E(Tk f)zk,
and let
F(z) = E(Tkf)zk As we have seen, 1
9(z) =
F(z)
21ri
1
Jr(w) 1 - z(e"' - 1)
27ri J r
Now, 1
1
ID(w)1
/
- zew
dw
dw.
1
9(z) = 27ri 1 + z (w)1 - 1+z ew dw r so that
9(z)= 1+zF(1+z}
21. Integer-Valued Entire Functions
145
Similarly,
F(w)=
I
Iwg+ 1
ww )-
Since p(E) > 1, we see by the P61ya Theorem that g = 11, where P and Q are polynomials with integer coefficients and Q(O) = 1. Thus, 1
w)
P(1 w) (1-w)'+1Q(1-WW)
Q*(w)'
where we choose N > max(deg P, deg Q). Now, P* and Q* are polynomials with integer coefficients and Q* (0) = 1. Thus, f E R1. As before, we see
that f (z) = E A(z)'Y; " where the yi are the reciprocals of the roots of Q* and the Pi are polynomials. If we write yi = 1 +- pi, we see that Qs 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. If f is an integer-valued function of exponential type such that hf(7r/2) = hf(-ir/2) = 0 (that is, the indicator diagram off is a horizontal line segment), and if L = exp h f(0) - exp h f(7r) < 4, then f E R1.
Theorem. If, in addition, L < Po, Pi,... , Pn, we have
V5-,
f (z) = Po(z) + Pi (z)2z +
then for some polynomials + P,n (z)nz.
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 g E F(A'), there is an f E F(A) such that If (iy)I < 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 A', whether F(A) < F(A'); we solve this problem here. By specialization, then, we prove as a corollary the following result.
Theorem 22.1.
There exists a function f
E
F(A) such that
If (iy)I < exp Trblyl if and only if
)(y)-)(x) A shall mean that there exists a sequence A", A" D A, such that A" - A. Definition. A < A' shall mean that there exists a sequence A"', A' C A', such that A"' - 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) - J,(x) < A'(y) - A'(x) + D(1);
0 < x < y < oo.
Likewise, A < A' holds if and only if (22.1) is satisfied.
Corollary 22.3. A < A' if and only if A'> A. Corollary 22.4.
If A - A1i A' - Ai, and A < A', then Al < A'1.
Corollary 22.5. If Al < A2 and A2 < A3i then Al < A3. Corollary 22.6. If A < A' and A' < A, then A - A'. 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
W(x) = inf{A'(s) - A(s) : s > x}.
It follows from (22.1) that V(x) > -K for some constant K. Now V(x) is constant except for possible jumps at the jumps of A(x). Let xo be a point of discontinuity of W. Then, W(xo - 0) = A'(xo - 0) - A(xo - 0) and
p(xo + 0) < A'(xo + 0) - A(xo + 0). We denote by i (22.2)
(xo) the jump of
at xo. Then
L 4 (xo) < AA'(xo) - AA(xo) < AA'(xo)
We let A*(t) = [W(t)),
22. On Small Entire Functions of Exponential-Type with Given Zeros 149
where [a] denotes the integral part of a and t -D(t) = f s dcp(s), U
and let A* (t) be the characteristic logarithm of that sequence A* whose counting function is A*(t). The function A*(t) is constant except possibly at the jumps of ap(t), and we have AA* (X0)
x}. But it is clear that 9(x) < 0, and (22.1) is simply another way of saying that 9(x) > 0(1). To prove that (22.1) implies that A < A', we put (22.5) AY"(x) = )'(x) - A*(x). Since by (22.4) A* is a subsequence of A', there is a subsequence A" defined by (22.5) and A" is a subsequence of A'. Since we already have shown that A"' 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), go E F(A'), Ifo(iy)I < Igo(iy) I for all real y and such that the only zeros of go 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 go(z) = Since Igo(iy)I - e'"bl't, 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 fo and go. Now we choose p with 0 < p < ao so that all the zeros, z, = r,,eie^, of f in the right half-plane (assuming for convenience that f has no zeros on z = iy) satisfy rn > p, and write one form of Carleman's Theorem (Chapter 12), taking y > x > p as (22.6)
E(y) - E(x) = I(y) - I(x) + J(y) - J(x) + 0(1),
where
E(R)=E(R: f)= I(R) = I(R f) =
n - Rz )Cos0,,, R
1
1
( t2
:
21r
J(R) = J(R : f) _ 7rR
J
1
R2
) log If (it)f (-it) I dt,
I log If(Rese) I cos a d8.
Now
R2 cos0 = O(1)
(22.7)
since
rn R2
cos0,
0. Then there exists a function V(t) defined on (0, oo) such that (22.14)
flog (1 + t2
d0(t) =
r log I1 - e I w(t) dA(t)
and
z 0 ys1 dL(s) I.
IVI < 2 sup (
1
Proof. By a contour integration, it is easy to see that
log (1+x21_
roo
J0
2
log I1-
tt
By Fubini's theorem, (22.15)
fiog(i+E)
dd(r) =
fiogii_i{Jz1t21 do(t) } dw
We therefore are led to define (22.16)
W(w) _
2
i
f
t2
w2 + t2
d0(t) t
and (22.14) asserts (22.13) in another form. The bound on 'p follows from integrating by parts in (22.15): (22.17)
2
'P(w) _
fJx
J0
0
dx
\x2 +w 2 )
Hence, 1W(W)1
1+ J0 1 I.
1
dx
(
) + w2
I STp I J
x da(t)
22. On Small Entire Functions of Exponential-Type with Given Zeros 153 z
since-xr+w is increasing, and the lemma is proved. We now choose
6(t) = 2 {A'(t) - A(t)},
(22.18)
d0(t) = t d6(t)
but cannot apply Lemma 22.7 to d0 since its support may not be compact. We truncate the support by defining 6k(t)-
6(t) { 6(k)
ift k
Ok(t) = t d6k(t) with the same convention for A(t) and A'(t). We now apply Lemma 22.7 to dOk and conclude that there exist functions Wk(t) such that z1
/
flog 1 1 + tJ dAk(t)
(22.19)
r
log I1 -
jpk(t) dt.
Now,
Iwk(t)f < B,
(22.20)
where B is a constant that is independent of k, namely, from the bound on IV(t)l in the lemma and the equivalence of A and A',
B=2supIA(t)-A'(t)1. On putting (22.21)
/
ya \
Lk(y) = fiog 1 1 + Via)
y2
1
floe1 1
t dAk(t) +
where (22.22)
d4ik(t) = cak(t) dt,
we have (22.23)
/
Lk(y) = 2 flog f 1 +
dA(t).
\\
Hence, by (22.13), (22.24)
kl oo Lk(y) = log I91(iy)I
- t2 I dlbk(t),
Entire and Meromorphic Functions
154
At this point, the idea is to find an entire function F for which the hypothetical formula log I F(iy)l
f log 11 - zR!I dfik(t)
=k
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 d4k(t) = cpk(t)dt are unsuitable since they need not be positive and cannot be discrete. [It is easy to see that all the d4k(t) are positive only in case A C A', a trivial case.] But first we show that adding a constant to Yak, in order to make d 0
A contour integration shows that
fioghi_hhl
E! dt = 0,
(22.27)
so that z
a
(22.28)
Lk(y) = f log
(i + ta) tt dak(t) + flog I1 - t
I
where d'Yk(t) = 1k(t) dt. Now let Wk(t) = [Wk(t)], the integral part of WYk(t), and define (22.29)
Lk(y) =
Lemma 22.8.
y>1
(22.30)
f
f
log (1 + M!) t2)
t 2
z
d)tk(t) + flog I1- tz I dlk(t).
There is a constant fl, independent of k, such that for all
log 11 - to I d`l`k(t) < flog 11
- t I dWk(t) +,Qlog IyI
Proof. We apply the next lemma with %Pk(t) = v(t) and '1 (t) = n(t). ,0
is independent of k because dand ITkt) -'pk()i 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 function for 0 < r < oo, that 0 < V(r) < B < oo, that n(r) is nondecreasing, and that for some constant C
v(r) > n(r) > v(r) - C. Then
J
r log I1- e l dv(t) + 0(logy)
log 11 - t2zI dn(t)
oc.
Proof. For fixed r, we write L(t) = log 1 1 - 2 I and point out that L is Lebesgue integrable on (0, oo): L(0+) = +oo,
L(r-) = L(r+) = -oo,
L(oo) = 0
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 = fo L(t)dn(t) and Z = f °O L(t)dv(t). We will prove that Y < Z + O(log r). We assume that v'(t) > p > 0. This involves no loss of generality since, if we replace v(t) by v(t) + t and n(t) by n(t) + t, we change Y and Z not at all, because f0 L(t)dt = 0. We may suppose, without loss of generality, that v(0) = 0, since suitably redefining v on the interval [0,11 changes Z only by O (1), which is small compared to the allowed discrepancy O(log r). With each large r we associate the numbers r1 and r2 such that
v(rj)=n(r)=v(r2)-C. Since v'(t) > p, we will have r - rl < r2 - r1 < v . It is easy to see that the following inequalities hold:
If r L(t) dn(t) < 0
J
L(t) dn(t)
-2-n-16 (x E R). [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
(fn + Mg)'(x) = fn(x) + Mg'(x) > (2-1 - 2-n-1)b
(x E ]R),
so that In + Mg is strictly increasing on R. Moreover, if we let M vary in [0, 1], then for x V S,,, (fn + Mg)(x) varies in an interval of R that contains points of T\Tn; and for y V Tn, (fn + Mg)-1(y) varies in an interval of JR that contains points of S\Sn. Now for n odd, let x be the point of S\Sn with smallest index and let M E [0,1] be such that (fn + Mg)(x) E T. We define fn+1 = fn + Mg,
Sn+1 = S. U {x}
and
Tn+1 =Tn U {fn+1(x)}.
For n even, let y be the point of T\Tn with smallest index and let M E [0, 1] be such that { fn + Mg}-1(y) E S. We define Tn+1 = Tn U {y}
fn+1 = In + Mg, and
Sn+1 = Sn U {fn+l(y)}.
The following properties of the constructed sequences are easily verified: (a) Ifn(z) - fn-1(z)I n).
From (a) it follows that fn converges pointwise to a function f for which 1 f (z) - fo(z)I 16, (x c ]l2), so the same is true for f. Hence, f is strictly increasing on R. From (b), it follows that f (Sn) = Tn for each n, and so f(S) = T. Moreover, we have insured that fn (p) = /3 for all n E N and therefore f (/3) = /3. This implies f (Q) = Q also. Finally we have
I(f(a) - a)I
I(fo(a) - a)I - I(fo(a) - f(a))I > 2p(I al) - p(IaI) > 0,
which means f (a) 0 a. 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: If a E P., we need only take S to be a Cauchy sequence of rationale that converges to a. For the converse direction, suppose that a E C\R and S is any Cauchy sequence from Q. Let # be the limit of S in R Use Lemma A to get an entire function f such that f (/j) = ,Q, f (a) # a, and f (Q) C Q. Then f maps S to a Cauchy sequence of rationals that is equivalent to S (since both converge to Q) yet f does not preserve a. That is, a E C\R implies that the right side of our equivalence is false. This completes the proof that our characterization of R 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 algebra language such that for any a, ,0 E E:
(i) a E HIS R(a) holds in e; (ii) For a, Q E P, a < fl = L(a, /3) holds in E; (iii) For a, /3 E C, j al = /3-#=:>=:a, 0) holds in E. Proof. We begin by building some machinery for discussing sequences of constants within the first-order language of E. (Earlier we did the same for countable sets of constants.) This is done using a triple of functions (f, g, h); g has infinitely many zeros on C, and on that zero set h takes on the values 0, 1, 2,... . In effect, h lists the zeros of g. Then the sequence
coded by (f, g, h) is (a,,), where a,, is the value f (z,,) at the zero of g 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 g in the manner discussed above:
Basis(g, h)t=da[V(a; h, g)t=a E NJ A daVpVq[{P(p) A P(q) A p divides q A p divides h - a A q divides g A q divides h - a} = pdivides q].
23. The First-Order Theory of the Ring of All Entire Functions
165
Now we construct an algebra language formula Seq(a, n; f, g, h) which expresses that a is the nth term of the sequence of constants coded by the triple (f, g, h):
Seq(a,n;
E NABasis(f,g,h)
A a E C A 3p[P(p) A p divides g A p divides h - n
A p divides f - a].
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 0. (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)4-- Basis(g, h) A VaVn(Seq(a, n, f, g, h) a E Q)
AV6 EQ+3mENdi,j ENVa,Q EQ[{m 1 and log+(t) = 0 for 0 < t < 1. The growth of the characteristic T(r, f) as r --, oo gives a very useful measure of the growth of f. 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 log r. (C2.1) T(r, f + g) < T(r, f) + T(r, g) + 0(1).
(C2.2) T(r, fg)