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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 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 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) c, and hence 1 is a polynomial.
In general, we have: Theorem 10.2. tion.
Ifliminf,._oo
r.J;) = c
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 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
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 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.
Entire and Meromorphic Functions
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
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
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 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) + ~:(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 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
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 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 -
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)
.~/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)
." 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(~) 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),
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 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)
.(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
(t) =
lot
S
d A. To prove that A" '" A', we must prove that 6(x) = 0(1), where 6(x)
= ).(x) + ).*(x) -
Now, .(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, 1k(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 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 ~ {3L(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, 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 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. (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
en ,
References
14. 15.
16. 17. 18. 19. 20.
21. 22.
23. 24. 25.
26.
27.
28. 29. 30.
31. 32.
183
exponential functions, 282(1) (1984), 1- 32 (and Correction 294 (1) (1986),381). Hille, E., Ordinary Differential Equations in the Complex Domain, John Wiley and Sons, New York, 1976. Hiromi, G., and Ozawa, M., Kiidai Math. Sem. Report, On the existence of analytic mappings between two ultrahyperelliptic surfaces, 17 (1965), 281-306. Kujala, R. 0., Bull. Amer. Math. Soc. Functions of finite A-type in several complex variables, 75 (1969), 104-107. Kujala, R. 0., Thlns. Amer. Math. Soc. Functions of finite A-type in several complex variables, 161 (1971), 327-258. Laine, I., Nevanlinna Theory and Complex DiJJerential Equations, W. de Gruyter, Berlin, 1993. Lang, S., and Cherry, W., Topics in Nevanlinna Theory, Lecture Notes in Math. 1443, Springer-Verlag, New York, 1980. Lindel6f, E., Ann. Scient. Ec. Norm. Sup. Fonctions entieres d'ordre entier, 41 (1905), 369-395. Mahler, K., Lectures on Transcendental Numbers, Lectures Notes in Math. 356, Springer-Verlag, Berlin, 1976. Malliavin, P., and Rubel, L. A., Bull. Soc. Math. France On small entire functions of exponential type with given zeros, 89 (1961), 175206. MarkuSevic, A. I., Entire Functions, American Elsevier Publishing Company, New York, 1966. Miles, J., J. Analyse Math. Quotient Representations of Meromorphic Functions, 25 (1972), 371-388. Miles, J., Bull. Amer. Math. Soc. Representing a meromorphic function as the quotient of two entire functions of small characteristic, 71 (1970), 1308-1309. Nevanlinna, R., Le Thioreme de Picard-Borel lemma et La Theorie des Fonctions Meromorphes, second edition, Chelsea Publishing Company, New York, 1974. Nienhuys, J. W., and Thieman, J. G. F., Proc. Dutch Ar.ademy of Science, Ser. A. On the existence of entire functions mapping countable dense sets on each other, 79 (1976), 331-334. Okada, Y., Science Rep. oj the T6hoku Imperial University Note on Power Series, 11 (1922), 43-50. PainJeve, P .. Lecons sur la Theorie Analytique des Equations DiJJerentielles, Profesees Ii Stockholm, Hermann, Paris, 1897. Palais, R. S., Amer. J. Math. Some analogues of Hartogs' theorem in an algebraic setting, 10 (1978), 387-405. P6lya, G., Math. ZeiL~chrift Untersuchungen tiber Lucken und Singularitatcn von Potenzreihen, 29 (1929), 549-640. Reinhart, G., Schanuel Functions and Algebraic Differential Equations, Ph.D. Thesis, University of lllinois, Urbana, IL, 1993.
184
Entire and Meromorphic Functions
33. Robinson, R., Trans. Amer. Math. Soc. Undecidable rings, 70 (1951), 137-159. 34. Rogers, H., Theory oj Recursive Functions and Effective Computability, McGraw-Hill, New York, 1967. 35. Rubel, L. A., Duke Math. J., A Fourier series method for entire functions, 30 (1963), 437-442. 36. Rubel, L. A., and Taylor, B. A., Bull. Soc. Math. France A Fourier series method for meromorphic and entire functions, 96 (1968), 53-96. 37. Rubel, L. A., and Yang, C. C., Values shared by an entire function and its derivative, Lecture Notes in Mathematics, Springer-Verlag, Berlin, 1977, pp. 101-103; in Complex Analysis Conference in Lexington, Kentucky, 1976. 38. Rudin, W., Real and Complex Analysis, McGraw-Hill, New York, 1974. 39. Schonfield, J. R., Mathematical Logic, Addison-Wesley, Reading, MA, 1967. 40. Sodin, M., Ad. Soviet Math. Value Distribution of Sequences of Rational Functions, 11 (1992). 41. Stoll, W., Proc. Sympos. Pure Math. About entire and meromorphic functions of exponential type, 11 (1968), Amer. Math. Soc., Providence, Rl, 392-430. 42. Stoll, W., Internat. J. Math. Sci. Value distribution and the lemma of the logarithmic derivative in polydisks, (4) (1983),617-669. 43. Taylor, B. A., Duality and entire Junctions, Thesis, University of lllinois, 1965. 44. Taylor, B. A., Proc. Sympos. Pure Math. The fields of quotients of some rings of entire functions, 11 (1968), Amer. Math. Soc., Providence, ru,468-474. 45. Tsuji, M., Japanese J. Math. On the distribution of zero points of sections of a power series, 1 (1924), 109-140. 46. Tsuji, M., Japanese J. Math. On the distribution of zero points of sections of a power series III, 1 (1926), 49-52. 47. van den Dries, L., Pacific J. Math. Exponential rings, exponential polynomials, and exponential functions, 113(1) (1984),51-66. 48. Vitter, A., Duke Math. J. The lemma of the logarithmic derivative in several complex variables, 44 (1977), 89-104. 49. Wilkie, A., On exponentiation-a solution to Tarski's High School Algebra Problem, preprint, ca. 1982, but never published. 50. Wilkie, A., private communication. 51. Zygmund, A., Trigonometrical Ser"ies, 2nd. Ed., Cambridge, 1988.
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-
