ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I : PRACE MATEMATYCZNE X I I I (1969) ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I : COMMENTATIONES MATHEMATICAE X I I I (1969)
J . K opeć (Szczecin)
On a generalization ol Jensen’s formula
Let f{z) be an analytic function in the disc jz\ < В such that /(0) Ф 0, and let n(r) be the number of zeros of f(z) in the disc И < r < B. The function n(r) is connected with f(z) by means of the Jensen’s formula
2 n r
(1) - i - ( In\f(ret,p)\dtp = In|/(0)| + f - ^ - d x .
2 k J о J x
0
This formula may be treated as a formula defining the coefficient a 0 (r) of the expansion of the function In \f(rel<p)\ in the Fourier series.
We shall prove that calculating the remaining coefficients 2 tc
(2 ) an(r) = ln l/(>^)l<r<w# (n = 1 , 2 , . . . ) 0
we obtain formulae which enable us to simplify the proofs of some clas
sical theorems in the theory of analytic functions, in particular, Hada- mard’s theorem on factorization of an entire function of finite order.
This proof of Hadamard’s theorem does not require any estimations of the canonical product or of the real part of the function f(z).
1. Calculation of the coefficients an(r). If zx,z 2, ... are zeros of f(z) arranged in the order of increasing moduli rt < r 2 < ..., then the function hi\f(rel4,)\ possesses continuous derivatives with respect to r and у in the disc \z\< rx, and in every ring rn_x< r < r n (n = 2, 3, ...), 0 < <p
< 2
k. Applying the identity
we obtain din |/(rÓI
Hence
In I f(re% 4>) I = 1/2 [logf(rel<p) + log/(re^)]
Ее т ±У \ = Г fire19)
fire*) J 4 /(re*) d r—Im Г (re X(p\
f(r e v ) <” 1 d<p.
А [1п|/(те» )П =
( 3 )
78 J . K o p e ć
Writing
Integrating (2) Ъу parts and applying (3) we get
Differentiating (2) under the sign of the integral we have
f u ' w m * r * d * ,
K (0,r)
the last formulae yield the linear differential equation
cC(r)+ y a „ ( r ) = - A J в<* <г<***Р = гП~1<Рп(г),
which gives after integration
Г
(4) an{r) = an(r' 0 )(r' 0 lr)n + r~n j t 271” 1 yn{t)dt
<pn\
{rm < r'o < t < r < rm+1, m = 0 , 1 , .. .; r 0 = 0).
Since /(0) ^ 0 we have in a neighbourhood of zero (5)
and so
/(*)//(«) = c 0 + c 1 z + . .. + cn_ xzn 1+ . . .
fni^) — n{t)
Cn- 1
+£ %v П
\zv\<t
for n — 0 , for n > 0 .
Hence it follows that cpn{t) is a step function having steps at the points tv = rv. Thus the right-hand side of formula (4) is continuous at the points rv. Analogously to [3], p. 149, we prove that the left-hand side of this formula (as defined by (2)) is continuous for 0 < r < B. Hence formula (4) remains valid for 0 < r < E. Setting n = 0 and r '0 = 0 we obtain formula (1). If n > 0 and r '0 0 we have
(6) an{r) = - ^ - 1 rn + r-
2 n ldt (n = 1 , 2 , . . . ) .
Jen sen 's fo r m u la 79
If the function f(z) possesses no zeros, the second term on the right- hand side of this equality is omitted. The integral in this term may be removed integrating by parts. Namely, we have
z7ndt
2 n ___
_ r
2 n 2 n
where z v are zeros of the function/(0) lying on the circle \z\ = rv. Dividing both sides of (6) by rn we thus obtain
(7) r nan{r) = ( l / 2 w ) [ c i _ i + z»n—r П Jtl izvlr)n]-
\z v\<»•
2. Applications.
2.1. From (6) we may immediately obtain formulae expressing the coefficients of the power series by means of the real part of the function.
OO
For this purpose let us write f(z) = exp\h(z)),h(z) = E cnzn. Then
n = 0
/(«)//(») = й'(«) = E n °n Z n S c ’n - i = %си,Ее/г( 0 ) = U { r , <p) = 1п|/(гег?>)|, n= 1
?0, n(t) = 0. From ( 2 тг
— [ U ( r , < p ) e - in 4 < p
TC J
In (/(0)I = B ee 0 ,n (t) = 0 . From (1), (2) and (6) we thus obtain 2 тг
2 Rec0 rncn
for n — 0 , for n > 0
(compare [2], p. 321).
Further applications will be based upon an estimation of the coeffi
cients an(r). Let us write M(r) = max \f(z)\, In \f{rel<p)\ = Ф(г, <p), E x(r)
|Z|<r
= {<p: Ф(г,<р)>0}, E 2 (r) = {(p\ ф(г,<р)< 0}. Then 1 2TT
|an(f)|<— Г |Ф(г,99)|^ 27Г J
= -J— 27 u J Г ф(г, ę>)#+ — J n J — Гф(г, = Ai+J-a.
Ej. E 2
It follows from formula (1) that 0 < A x < In M (г), А г—Л 2 > In |/(0 )|;
hence
(8) A 2 ^ A l - ln|/(0)|, K (r)| < 2 1 n Jf(r) + |ln|/(0)||.
2.2. First let us assume that/(0) is an entire function of finite order q , possessing no zeros. If n > q , then lim r~nIn M (r) = 0, by the definition
r-yoo
of the order. Dividing both sides of (6) by r 11 and taking the limit as r ->■ 00
and n > q we have cn_x — 0, by (8). In view of (6) this means that
80 J . K o p e ć
f'(z)lf(z) is a polynomial of degree к < р—1. Hence f(z) = eQ^\ where Q(z) is a polynomial of degree q < q . In particular, each polynomial possessing no zeros is constant ( q = 0). Thus we obtain the fundamental theorem of algebra.
2.3. Let us assume that f(z) = H(z)P(z), where H(z) is an entire
OO P 1
function without zeros, and P(z) = /7 (1 — 2/2^) exp A;-1 («/»„)*) is the
n = l k = l
canonical product of f(z). Then c* = c*(/) = с*(Я) + с*(Р) and c&(P)
OO
= for k ^ p . Hence we obtain from (7) for n > p
v= l
OO
