Rend. Lincei Mat. Appl. 23 (2012), 259–265 DOI 10.4171/RLM/627 Complex Variables Functions — Inequality for entire functions involving their maximum modulus and maximum term,byTatyana Shaposhnikova, commu- nicated on 9 March 2012. Dedicated to the memory of Gaetano Fichera Abstract. — An estimate of the Wiman-Valiron type for a maximum modulus on a polydisk of an entire function of several complex variables is obtained. The estimate contains a weight function involved also in the calculation of the radius of the admissible ball. Key words: Entire function, several complex variables, maximal modulus and maximum term. 2010 Mathematics Subject Classification: Primary: 35G15, 35J55; Secondary: 35J67, 35E05. Introduction Let a ¼ða1; ...; anÞ be a multi-index with aj b 0 and let z ¼ðz1; ...; znÞ be a point of the n-dimensional complex space Cn. We consider the entire function X a ð0:1Þ f ðzÞ¼ aaz : a For any r ¼ðr1; ...; rnÞ with rj b 0 we introduce the maximum modulus and the maximum term of f : a ð0:2Þ Mf ðrÞ¼ max j f ðzÞj; mf ðrÞ¼max jaajr : fz:jzj j¼rj g a Clearly, by the Cauchy formula for the coe‰cients aa, mf ðrÞ a Mf ðrÞ: For n ¼ 1 the first result on the comparison between Mf ðrÞ and mf ðrÞ is due to A. Wiman [W] who showed that for any r0 there is an r > r0 such that 1=2þe ð0:3Þ Mf ðrÞ < ðlog mf ðrÞÞ mf ðrÞ 260 t. shaposhnikova with any e > 0. Moreover, if A is the set of r > 0 such that (0.3) holds, then Z dr < l: A r Wiman’s proof was essentially simplified by G. Valiron [V]. The following generalization of inequality (0.3) is due to P. C. Rosenbloom [R]. Theorem 1. Let j be a positive nondecreasing function on ð0; lÞ such that Z Z l t À1=2 ð0:4Þ HðtÞ :¼ jðsÞ ds dt < l t t for all t > 0. Given any point r0 with nonnegative coordinates, the ball pffiffiffi ð0:5Þ fr : jlog r À log r0j < nHðlog Mf ðr0ÞÞg contains at least one point r such that n=2 ð0:6Þ Mf ðrÞ a Cmf ðrÞðjðlog Mf ðrÞÞÞ ; where C is a positive constant and log r ¼ðlog r1; ...; log rnÞ. The article [R] contains a proof of this assertion for functions of one variable and outlines the argument in the general case. Remark. An example of a function j for which the condition (0.4) holds is a function given for large positive s by jðsÞ¼sðlog sÞ2ðlog log sÞ2 ...ðlog ...log sÞ2þe with an e > 0. 1. Main result The aim of this note is the proof of the following generalization of Theorem 1. Theorem 2. Let j and H be the same function as in Theorem 1. Further, let h be a positive continuous increasing function such that ð1:1Þ hð2xÞ a chðxÞ with a constant c > 1. For any point r0 with nonnegative coordinates, far away from the origin, the ball ð1:2Þ fr : jlog r À log r0j < Rg inequality for entire functions involving their maximum modulus 261 with the radius defined by the equation Z R 1 1=2 pffiffiffiffiffi ð1:3Þ h jlog r0jþr dr ¼ c 2nHðlog Mf ðrÞÞ 0 2 contains at least one point r such that n=2 ð1:4Þ Mf ðrÞ a Cmf ðrÞðhðjlog rjÞjðlog Mf ðrÞÞÞ ; where C is a positive constant. Proof. We may replace aa by jaaj in the definition of f since mf ðrÞ won’t change and Mf ðrÞ does not decrease. Following [R], we introduce a random vector x ¼ðx1; ...; xnÞ with probability distribution a eax Pðx ¼ aÞ¼ a ; x a Rn: f ðexÞ Let Exx and Dxx stand for the mathematical expectation and the dispersion of x. We introduce the notation FðxÞ¼log f ðexÞ: Clearly 1 ð1:5Þ ‘FðxÞ¼ ‘ f ðexÞ: f ðexÞ x Furthermore, x 2 1 x ð‘x f ðe ÞÞ ð1:6Þ DFðxÞ¼div ‘FðxÞ¼ Dx f ðe Þ : f ðexÞ ð f ðexÞÞ2 By the definition of the expectation, X ExðxÞ¼ aPðx ¼ aÞ a and X 2 2 Exðx Þ¼ a Pðx ¼ aÞ: a This together with (1.5) gives X 1 ax ExðxÞ¼ x aaae ¼ ‘FðxÞ: f ðe Þ a 262 t. shaposhnikova Further, the definition of the dispersion 2 2 Dxx ¼ Exðx ÞðExðxÞÞ gives X X 2 1 2 ax 1 ax Dxx ¼ a aae À aaae : x x 2 f ðe Þ a ð f ðe ÞÞ a Combining this with X x 2 ax Df ðe Þ¼ a aae a we see that x 2 1 x ð‘x f ðe ÞÞ Dxx ¼ Df ðe Þ : f ðexÞ ð f ðexÞÞ2 Now, by (1.6) Dxx ¼ DFðxÞ: By the Chebyshev inequality, 1=2 À2 Pfjx À Exxj b lðDxxÞ g a l ; l > 1; we have ð1:7Þ Pfjx À ‘FðxÞj b lðDFðxÞÞ1=2g a lÀ2; l > 1: The probability on the left-hand side is equal to X 1 0 ð1:8Þ a eax f ðexÞ a with the sum taken over all multiindices a such that ja À ‘FðxÞj b lðDFðxÞÞ1=2: P Let 00 stand for summation over a for which ja À ‘FðxÞj < lðDFðxÞÞ1=2: By inequalities (1.7) and (1.8) we have X 1 00 1 À a eax a lÀ2: f ðexÞ a inequality for entire functions involving their maximum modulus 263 Hence X x À2 À1 00 ax f ðe Þ a ð1 À l Þ aae : Therefore, n nþ2 x 2 l x n=2 ð1:9Þ f ðe Þ a mf ðe ÞðDFðxÞÞ : l2 À 1 Let R be defined by (1.3) and let jx0j > 4R.IfDFðxÞ a hðjxjÞHðFðxÞÞ in the ball BR ¼fx : jx À x0j < Rg; the result follows. Suppose that the opposite inequality DFðxÞ > hðjxjÞHðFðxÞÞ holds in the ball BR. Since, clearly, jxj b jx0jjx À x0j b jx À x0jþjx0j=2; the following inequality holds in the ball BR as well DFðxÞ > hðjx À x0jþjx0j=2ÞHðFðxÞÞ: Consider the equation ð1:10Þ DuðxÞ¼hðjx À x0jþjx0j=2ÞHðuðxÞÞ in BR. By the maximum principle for the Laplace operator, the inequality FðxÞ a uðxÞ with x a qBR implies the same inequality on the whole BR. Suppose that 1:11 u max F x : ð Þ jqBR ¼ ð Þ x A qBR The solution of the Dirichlet problem (1.10)–(1.11) is unique, therefore, u de- pends only on r ¼jx À x0j. Hence u satisfies the boundary value problem nÀ1 0 0 nÀ1 ð1:12Þ ðr urÞr ¼ r hðr þjx0j=2ÞHðuðrÞÞ; ð1:13Þ uðRÞ¼ max FðxÞ; u0ð0Þ¼0: x A qBR This implies Z r 0 1Àn nÀ1 ð1:14Þ u ðrÞ¼r hðs þjx0j=2Þs HðuðsÞÞ ds: 0 264 t. shaposhnikova Clearly, the function H, given by (0.4) is decreasing. Combining this with the monotonicity of the functions h, we obtain 0 À1 ð1:15Þ u ðrÞ a hðr þjx0j=2ÞHðuðrÞÞn r: By the equation (1.12), n À 1 u00ðrÞ¼hðr þjx j=2ÞHðuðrÞÞ À u0ðrÞ: 0 r This and (1.15) lead to the ordinary di¤erential unequality 00 À1 ð1:16Þ u ðrÞ b n hðr þjx0j=2ÞHðuðrÞÞ for all r a ½0; R. Let us show that the number R satisfies the inequality Z R ffiffiffiffiffi 1=2 p ð1:17Þ ðhðr þjx0j=2ÞÞ dr < c 2nHðuðjx0jÞÞ: 0 Having proved (1.17), the result follows from the estimate uðjx0jÞ b Fðx0Þ and the monotonicity of H. Since h increases, by (1.16) we have R þjx j ð1:18Þ u00ðrÞ b nÀ1h 0 HðuðrÞÞ 2 hi for r a Rþjx0j ; R . We multiply (1.18) by u0 r > 0 and integrate the result over hi2 ð Þ Rþjx0j 2 ; R to obtain Z R þjx j 2 2 R þjx j uðrÞ ðu0ðrÞÞ2 À u0 0 b h 0 HðsÞ ds: 2 n 2 uððRþjx0jÞ=2Þ Using u0ðrÞ > 0 once more, we arrive at Z rffiffiffi uðrÞ À1=2 2 R þjx j 1=2 HðsÞ ds u0ðrÞ b h 0 : uððRþjx0jÞ=2Þ n 2 Integrating this inequality over the interval ½R=2; R, we see that R þjx j 1 R þjx j 1=2 ð1:19Þ H u 0 b pffiffiffiffiffi R h 0 : 2 2n 2 By (1.1), Z Z R jx j 1=2 R r þjx j 1=2 R þjx j 1=2 h r þ 0 dr a c h 0 dr a cR h 0 : 0 2 0 2 2 inequality for entire functions involving their maximum modulus 265 Combining this with (1.19), we find Z R þjx j 1 R jx j 1=2 H u 0 b pffiffiffiffiffi h r þ 0 dr: 2 c 2n 0 2 Since u is nondecreasing (see (1.14)), we have R þjx j uðjx jÞ a u 0 : 0 2 By (0.5) the function H does not increase which implies Z R 1=2 1 jx0j Hðuðjx0jÞÞ b pffiffiffiffiffi h r þ dr: c 2n 0 2 The proof is complete. References [R] P. C. Rosenbloom, Probability and Entire Functions, Studies in Mathematical Analy- sis and Related Topics, pp. 325–332, Stanford Univ. Press, 1962. [V] G. Valiron, Sur la croissance du module maximum des se´ries entie`res, Bull. Soc. Math. France 44 (1916), 45–64. [W] A. Wiman, Uber den Zusammenhang zwischen dem Maximalbetrage einer analytischen Funktion und dem gro¨ssten Gliede der zugeho¨rigem Taylorschen Reibe, Acta Math. 37 (1914), 305–326. Received 16 February 2012.