Weprove: Ifrl,...,rk are (fixed)positive real numbers with I’lk rj>1,then theonly entiresolutions "C Cof thefunctionalinequality are,(z) czn,wherecisacomplexnumber andn s apositive integer

Internat. J. Math. & Math. Sci.

VOL. 15 NO. 2 (1992) 413-415

413

A

NOTE

ON A FUNCTIONAL

^INEQUALITY

HORSTALZER

Department

^ofPureMathematics Applies Mathematics and

Astronomy

University of SouthAfrica

P.O. Box

392 0001 Pretoria,South Africa

(Received

September25, 1990 andin revisedform March5,

1991)

ABSTRACT.

We

prove: If_{rl,...,rk are}

(fixed)

positive real numbers with

I’l

k rj

>

1,then theonly entiresolutions

"

^{C Cof thefunctionalinequality}

are

,(z)

^cz

n,

^wherecisacomplexnumber andn s apositive integer.

KEY WORDS AND PHRASES.

Functional inequality,entirefunctions.

1991

AMS SUBJECT CLASSIFICATION CODE.

39C05.

1.

INTRODUCTION.

Inspired by^aproblemofH. Haruki, who askedforallentiresolutionsof

I(z + w)

²

+ ^{I(z w)}

²

+ 21 ⁽⁰⁾ 12 ^{> 21 (z)} 12 ^{+ 21 (w)} 12 ^(.x)

J.

Walorski

[1]

provedin1987 thefollowinginteresting proposition:

Let r

>

1 bea

(fixed)

real number. Then theonlyentire solutions qa:C^--,Cof the functional inequality

(z) >_

r

I(z)

are

(z)

^czn

(1.2)

wherecECandnEN.

As

an application ofthis

theorem,

Walorski showed that the only entire functions _0:C C satisfying

(1.1)

^and

o(0)=

^0arethe monomials

(1.2).

^{Theaimofthisnoteis} to proveanextension of Walorski’s result by using a method which is (slightly) different from the two approaches presentedin

[1].

2.

MAIN RESULTS.

Theorem. Let_rl, ^r_k ^be

(fixed)

positive real numberswith I-[ _rj

>

1. Then the onlyentire j=l

414 ^H. ALZER solutions

go:C

Cof

k

]lrj)

I1 I,(,’.z) _> I,(z) : _(2.1)

j= j=

arethefunctionsgo(z)+cz

n,

wherecisacomplex number andnisapositive integer.

PROOF. Simple calculationsreveal that the functions go(z)=^czn

(c

6_C,n^6_

N)

satisfy

(2.1).

Next

weassumethatgo is an entiresolutionofinequality

(2.1).

Becauseof

1"I

k rj

>

1 weconclude from

(2.1)

^{with z 0} that go has at 0a zero. Letn bethe

j=l

order ofthiszero;wedefine

f(z) go(z)/z

ⁿ

(2.2)

then

/

^isanentirefunction with

f(0)

^0.

From (2.1)

^weobtain

H

k

[f(r.z)[ >_( fi rJ.-n) ^{lf(z)l t. (2.3)}

j=l j=l

Wesuppose that

f

^{hasa zeroz}0.

By

inductionitfollows from

(2.3)

^that

zo/rrff

^{is arootof}

f

^{for all}

non-negative integers m. From the identity theorem we conclude

f(z)--0

which contradicts the condition

f(0) #

^{0. Hence}

f

^{hasnozerowhich}implies that thefunction

is entire. From

(2.3)

^weconclude

f(z)

^k

H ^f(,..z)

j=1

(2.4)

rn.

Ig(z)l _<

111

j=

for allz6-

C,

and Liouville’stheorem implies thatg isaconstant. Thereforewehave

f(z) ^K

k

iI f(r.z), ^K

^6_C.

j=l ^/

Since

f(O) #

^{0weget from}

^{(2.5): K}

^1;

hence

/(z) =

j=l

^{I f(,.z).}

^/ Differentiationleads to

Setting

f’(z)

^k

=/E ^f(.z)

kf--

^_- _/

f() _=o

weobtainfrom

(2.7)

^and

(2.8):

E a"zm=E ^{(a.,E rT+X)z,}

rn=O rn=O j=l

(2.8)

(2.9)

FUNCTIONAL INEQUALITY 415

and comparing thecoefficientsofz^myields for allrn

>

0:

k

kam=a

m_j=l

r ^{n+l. (2.10)}

Weassumethat thereexists anintegerm₀

>

0 suchthat

am

0

#

0, ^thenwegetfrom the arithmetic mean-geometricmeaninequality and from

(2.10)"

+ 1]1/k

^{k m0}

jmo

^+I

k

which contradicts the assumption rj

>

1.

Hence,

a_m 0forallm

>

0. Thisimplies that

f

^isa

j=l

constant, saycE

C,

^andthereforeweobtain

qo(z)

cz

n.

It is natural to look for all entire functions

"

^{C--. C which} satisfy the following additive counterpartof inequality

(2.1):

(-

j=l

^{9(rjz)) >} --j=l

^r.I

[o(z)[ ^(2.11)

k

where _rl,...rk are

(fixed)

^{positive real} numbers with

.,rj>k.

^{The monomials}

3=1

(z) czn(c _ ^C,n _ N)

^{aresolutionsof}

(2.11).

^{Indeed,inequality}

(2.11)

^with

(z)

^cznreducesto

k k

E r ^{> E}

^rj,

^(2.12)

j=l

--j=l

k

which follows immediately from

Jensen’s

inequality and the assumption

y _rj>

k.

By

an j=l

argumentation similar to theone we haveused toestablish the theorem it can beshown that the functions

(z)cz" (c.C,n _)

are the only entire solutions of

(2.11).

^This provides another extensionof Walorski’s result.

/

If the expression on the left-hand side of

(2.11)

^{will be} replaced by

I(.)1,

^{then e}

conclude from the triangle inequality that

(z)

^czn

(c _ ^C,

ⁿ

_ ⁾

^{alsosolve 3}

k k

j=l ³

--j=l

k

where _rl,...,rk are

(fixed)

positive real numberswith

y

_rj

>

k. We finish by asking:

more solutionsof

(2.13) (if

^k

> 1)? ^J

(2.13) Are

there

REFERENCE

1.

WALORSKI, J.,

Onafunctionalinequality, Aequationcs

Math.

³²

(1987),

213-215.

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