On the geometric means of an integral function

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C ^OMPOSITIO M ^ATHEMATICA

K ^ULDIP K ^UMAR

On the geometric means of an integral function

Compositio Mathematica, tome 19, n

^o

4 (1968), p. 271-277

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271

by

Kuldip

^Kumar

Let

f(z)

^be^an

integral

function of order _pand lower order À and

n(r) being

the number of zeros of

f(z)

ⁱⁿ

Izi

^r.^Let

G(r)

^and

g03B4(r)

denote the

geometric

^means

of 1/(z)l,

^defined^as

and

In this paper ^wehave obtained some of the

properties

^of

G(r)

and

g03B4(r).

^Let

and

Using

Jensen’s formula in

(1.2),

^we^have

From

(1.1),

^we

^have,

^forany ^e> 0 and r > ro ⁼

r0(03B5)

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272

Using

^{this in}

(1.6),

^we

have,

for almost all values of r

> ro

Again

^from

(1.5),

^wehave for any ⁸> 0 and r > ro

Substituting

^for

n(r)

^from

(1.6),

^we

^have,

for almost all r > ro

We shall now obtain some of the

properties

^of

gs(r).

^{We may}

write

(1.3)

^as

Now, using (1.4)

^and

(1.6),

^we

get

THEOREM 1. Let

f(z)

^be^an

integral function of

^order

p(O

p

00)

and let

/(0) :A

^0.

Further, (i) if

then

(ii) if

then

PROOF.

(i)

^Since

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therefore,

^forany ^ê> 0 and r > r1 ⁼

r1(03B5),

^we^have

and so from

(2.1)

Taking

^limit^on^both^thesides leads to

(ii) If

we

have,

^forany ^E> 0 and r > r, ^----

r1(03B5), N(r) (ex+E)rP.

Substituting

^{this in}

(2.1), integrating

^and

proceeding

^to

^limits,

the result follows.

THEOREM 2. Let

f(z)

^be^an

integral function of finite non-integral

order _p,and let

then

PROOF. Since

therefore,

Hence,

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274

But,

Hence,

Since n

^is

arbitrary,

^we

^get

Further,

but,

hence,

Therefore ^we

get

Combining (2.2)

^and

(2.3),

^we

get

^the^result.

3.

Combining (1.2)

^and

(1.3),

^we^obtain

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Using (1.6)

ⁱⁿ

^this,

^we

get

Let ^usset

We now prove the

following:

THEOREM 3.

Il f(z)

^is^an

integral function of

^order

p(o

p

~)

and

f(0)

^{=1= 0,}such that

n(r)

^~

cp(r)rPl,

^where

~(r)

^is^a

positive

continuous and

indefinitely increasing function of r and ~(cr) ~ ~(r)

as r - oo

for

every ^constant^c> 0, then

PROOF. Since

n(r)

^,

~(r)r03C11

^wehave for any ^a> 0 and

r ~ ro(s)

or

Now, by

^Lemma

V ([1],

p.

54),

for _every

positive

^a,^and^{so we}

get

Again,

^from

(3.2),

^we^have

giving,

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276

Combining (3.3)

^and

(3.4)

^leads^to

Taking exponentials

^and

proceeding

^to

^limits,

^we

^have,

^since^a

is

arbitrary

^and

n(r)

^~

cp( r )rPl,

THEOREM 4.

1 f f(z)

^has^at^least^{one zero}^and

f(0)

^{=1= 0,}^then

PROOF.

(i) Integrating by parts

^the

integral

ⁱⁿ

(3.1),

^we

get

Since

N(r)

^is

non-decreasing

function of _r,we

get

Since

n(r)

^is

non-decreasing

function of _r,

(3.1) gives

But we know

( [2],

p.

17)

^that

and so

1 wish to express my sincere thanks to Dr. S. K. Bose for his

guidance

^{in the}

preparation

of this paper.

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REFERENCES

G. H. HARDY and W. W. ROGOSINSKI

[1] ^’Note^onFourier Series (III)’, Quarterly Journal of Mathematics, 16, (1945),

pp. ^49-58.

R. P. BOAS, ^JR.

[2] ^’EntireFunctions’, ^NewYork, ^1954.

(Oblatum 9-3-1966 ) Department of Mathematics and Astronomy,

Lucknow University,

Lucknow (India)

On the geometric means of an integral function

C OMPOSITIO M ATHEMATICA

K ULDIP K UMAR