A NOTE ON A FUNCTIONAL INEQUALITY

We prove: If r1,…,rk are (fixed) positive real numbers with ∏j=1krj>1, then the only entire solutions φ:ℂ→ℂ of the functional inequality∏j=1k|φ(rjz)|≥(∏j=1krj)|φ(z)|kare φ(z)=czn, where c is a complex number and n is a positive integer.


INTRODUCTION.
Inspired by a problem of H. Haruki, who asked for all entire solutions of I(z + w) 2 + I(z w) 2 + 21 (0) 12 > 21 (z) 12 + 21 (w) 12 (.x)J. Walorski [1] proved in 1987 the following interesting proposition: Let r > 1 be a (fixed) real number.Then the only entire solutions qa: C --, C of the functional inequality (z where c E C and n E N. As an application of this theorem, Walorski showed that the only entire functions 0: C C satisfying (1.1) and o(0)= 0 are the monomials (1.2).The aim of this note is to prove an extension of Walorski's result by using a method which is (slightly) different from the two approaches presented in [1].
Theorem.Let rl, r k be (fixed) positive real numbers with I-[ rj > 1.Then the only entire j=l solutions go:C C of k ]lrj) I1 I,(,'.z)_> I,(z) : (2.1) j= j= are the functions go(z)+ czn, where c is a complex number and n is a positive integer.
PROOF.Simple calculations reveal that the functions go(z)= cz n (c 6_C, n 6_ N) satisfy (2.1).Next we assume that go is an entire solution of inequality (2.1).k Because of 1"I rj > 1 we conclude from (2.1) with z 0 that go has at 0 a zero.Let n be the j=l order of this zero; we define f(z) go(z)/z n (2.2) then / is an entire function with f(0) 0. From (2.1) we obtain k (2.3) j=l j=l We suppose that f has a zero z 0. By induction it follows from (2.3) that zo/rrff is a root of 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 has no zero which implies that the function is entire.From (2.3) we conclude f(z) k H f(,..z) j=1 (2.4) rn.
Ig(z)l _< 111 j= for all z 6-C, and Liouville's theorem implies that g is a constant.Therefore we have k f(z) K iI f(r.z),K 6_ C. j=l / Since f(O) # 0 we get from (2.5): K 1; hence /(z) = I f(,.z).(2.9) and comparing the coefficients of z m yields for all rn > 0: k kam=am r n+l.
(2.10) j=l We assume that there exists an integer m 0 > 0 such that am 0 # 0, then we get from the arithmetic mean-geometric mean inequality and from (2.10)" + 1]1/k k m 0 jmo +I k which contradicts the assumption r j > 1.Hence, a m 0 for all m > 0. This implies that f is a j=l constant, say c E C, and therefore we obtain qo(z) czn.
It is natural to look for all entire functions " C--.C which satisfy the following additive counterpart of inequality (2.1): (-9(rjz)) > r.
The monomials 3=1 (z) czn(c _ C,n _ N) are solutions of (2.11).Indeed, inequality (2.11) with (z) cz n reduces to (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 the one we have used to establish the theorem it can be shown that the functions (z)cz" (c.C,n _) are the only entire solutions of (2.11).This provides another extension of 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) cz n (c _ C, n _ ) also solve 3 k k j=l 3 --j=l k where rl,...,r k are (fixed) positive real numbers with y rj > k.We finish by asking: more solutions of (2.13) (if k > 1)?J (2.13) Are there REFERENCE 1. WALORSKI, J., On a functional inequality, Aequationcs Math. 32(1987), 213-215.