Wiman s Inequality for the Laplace Integrals

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1 Int. Journal of Math. Analysis, Vol. 8, 2014, no. 8, HIKARI Ltd, Wiman s Inequality for the Laplace Integrals A. O. Kuryliak 1, I. Ye. Ovchar 2 and O. B. Skaskiv 3 1,3 Ivan Franko National University of L viv, Ukraine 2 Ivano-Frankivsk National Technical University of Oil and Gas, Ukraine Copyright c 2014 A. O. Kuryliak, I. Ye. Ovchar and O. B. Skaskiv. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract We find unimprovable conditions for the Laplace-Stieltjes type integrals under which the Wiman-type inequality holds. Mathematics Subject Classification: 44A10, 30B50 Keywords: Wiman s inequality, the Laplace-Stieltjes type integral 1 Introduction By Wiman-Valiron s theorem (see [1] [4]) for every nonconstant entire function f(z) = + a n z n and some ε > 0 there exists a set E [1; + ) of finite n=0 dr logarithmic measure (i.e. < + ) such that for all r [1; + ) \ E E r Wiman s inequality M f (r) μ f (r)ln 1/2+ε μ f (r) holds, where M f (r) = max{ f(z) : z = r}, μ f (r) = max{ a n r n : n 0}. Analogues of the Wiman-Valiron s theorem for entire (absolutely convergent in the whole complex plane C) Dirichlet series F (z) = + a ne λnz (z C), 0=λ 0 <λ n + (1 n + ), (1) n=0 can be found in [5]. In particular, it follows from result in [5] (see also [6]) that there exists an entire function f such that lim r + M f (r) μ f (r)ln 1/2 μ f (r) =+,

2 382 A. O. Kuryliak, I. Ye. Ovchar and O. B. Skaskiv i.e. in general, Wiman s inequality in some sense is sharp. An unimprovable Wiman-type inequality for random gap power series was established in [6]. 2 Preliminary Notes In this paper we consider a function F : R R defined by the Laplace-Stieltjes type integral of the form F (x) = f(u)e xu ν(du), R + =[0, + ), (2) R + where f(u) be a positive measurable function on R, ν a countable additive measure on the σ-algebra B(R + ) of Borel sets R + (Borel measure) such that ν({x: 0 x b}) < + for any b>0. Denote by supp ν the support of the measure ν, i.e. a closed set E =: supp ν such that ν(r \ E) = 0 and ν({u R: u u 0 <r}) > 0 for any u 0 E and r>0. Denote by ν(e) = ν(dx) E the ν-measure of a ν-measurable set E R, ν(a, b] :=ν((a, b]). By I(ν) we denote the class of the functions F (x) of the form (2). Let L be the class of positive continuous functions ψ : R + R + such that + 0 For x R and F I(ν) we set dx ψ(x) < +. μ (x) = sup{f(u)e xu : u supp ν}, μ (x) = sup{f(u)e xu : u R}. A Wiman-type inequality for the Laplace-Stieltjes type integrals of the form (2) depending on the small parameter x<0 was considered in [7]. In this paper we obtain sharp Wiman-type inequality for integrals from the class I(ν). 3 Main Results We denote ln k+1 t := ln(ln k t)(k N), ln 1 t ln t. Theorem 1. Let F I(ν). If ( ψ L)( p <+ )( t 0 > 0)( t t 0 ): ν(t ψ(t),t+ ψ(t)] t p, (3) then for every ε>0 there exists a set E R + of finite Lebesgue measure, i.e. meas(e) := dx < +, such that for all x [0, + ) \ E E F (x) Cμ (x)ln p μ (x)ln p+ε 2 μ (x), where C is some constant depending on F and ε. Proof. Without loss of generality assume that supp ν is unbounded, and F (x) + as x +. Then μ (x) + as x +. We fix x>0

3 Wiman s inequality for the Laplace integrals 383 and consider the random variable ξ(u) =u: R + R + on the probability space (R +, B(R + ), P), where P (du) = f(u)exu ν(du) is the probability measure F (x) on B(R + ). Let Eξ,Dξ be the expectation and variance of the random variable ξ. We remark that Eξ = g (x), Dξ = g (x), where g(x) =lnf (x). Using the Bienayme-Chebyshev inequality (see [8, 9]) P{u: ξ(u) Eξ a} Dξ/a 2 with ξ(u) =u, a = cdξ, c > 0 we obtain f(u)e xu ν(du) =F (x) P{u: ξ(u) Eξ a} 1 c F (x). x g (x) cg (x) Thus for any c>1 F (x) c c 1 μ (x)ν(g (x) cg (x),g (x)+ cg (x)]. (4) We denote E(h) ={x >0: h (x) ψ 0 (h(x)), h(x) > 1}. Then for any positive differentiable function h: R + R + and for every function ψ 0 L h (x)dx + meas(e(h)) = dx E(h) E(h) ψ 0 (h(x)) dt 1 ψ 0 (t) < +. Hence meas(e(g )) + meas(e(g )) < +, thus for a function ψ Land for every number ε 1 > 0 ( x / E 1 := E(g ) E(g )): 2g (x) ψ(g (x)), g (x) g(x)(ln g(x)) 1+ε 1, and from (4) by condition (3) we have F (x) 2μ (x)(g (x)) p. Therefore for x [0, + ) \ E 1 we obtain F (x) 2μ (x)(g(x)ln 1+ε 1 g(x)) p, hence g(x) 2lnμ (x) asx + (x / E 1 ), and thus F (x) 2μ (x) ( 2lnμ (x)ln 1+ε 1 (2 ln μ (x)) )p μ (x)ln p μ (x)ln p+ε (ln μ (x)) as x + (x / E 1 ) and ε 1 < ε/p. Sharpness of Theorem 1 follows from such a statement. Theorem 2. Let ψ Lbe such that ψ(t) =O(t ln t ln 2 2 t)(t + ) and a measure ν such that ln ν(0,t]=o(t) (t + ) and ( p >0)( C 1 > 0)( t 0 > 0)( t t 0 ): ν(t ψ(t),t+ ψ(t)] C 1 t p. (5) Then for every ε (0,p) there exists F I(ν) such that F (x) μ (x)ln p μ (x)ln p ε +, (x + ). 2 μ (x) Remark 1. For Lebesgue measure ν(dx) = dx on R conditions (3) with p =1/2+ε, ε > 0 and (5) with p =1/2 are satisfied.

4 384 A. O. Kuryliak, I. Ye. Ovchar and O. B. Skaskiv Proof. We put f(u) = exp{ βuln + 3 u}, h(u, x) =lnf(u)+ux and consider a function F of the form (2). It is easy to see that F I(ν). For enough large x the unique point u(x) of maximum of the function h(u, x), u(x) e e, can be found from the equation h β u(u, x) = β ln 3 u + x =0. Hence ln u ln 2 u u(x) + as x +, thus therefore ln μ (x) h(u(x),x)= βu(x) ln u(x)ln 2 u(x) =lnμ (x) (6) u(x) = (1 + o(1))β ln μ (x)ln 2 μ (x)ln 3 μ (x) (7) as x +. Put t ± = u(x)± ψ(u(x)). Remark that h β uu(u, x) u ln u ln 2 u (u + ), thus by Taylor s formula at the point u = u(x) with Lagrange form of the remainder and the relation (6) we get for u [t,t + ] h(u, x) =h(u(x),x)+ h uu(ũ, x) (u u(x)) 2 ln μ β ψ(u(x)) (x) 2 2u(x)lnu(x)ln 2 u(x) as x +, where ũ = ũ(x) (t,t + ). It follows from condition of Theorem 2 that ψ(t) 2Ktln t ln 2 2 t (t e 2 ),K<+. Then by condition (5) we have F (x) (t,t + ] e h(u,x) ν(dx) μ (x)ν(u(x) ψ(u(x)),u(x)+ ψ(u(x))] exp{ Kβ ln 2 u(x)} C 1 μ (x)(u(x)) p exp{ Kβ ln 2 u(x)} as x +. Let β = ε,ε (0,p). Therefore, from (7) we obtain 3K F (x) C 1 (1 + o(1))β p μ (x)ln p μ (x)(ln 2 μ (x)) p ε/3 ln p 3 μ (x) μ (x)ln p μ (x)(ln 2 μ (x)) p ε/2 as x +. This completes the proof of Theorem 2. Remark 2. If for a Dirichlet series of the form (1) we put f(u) =a n and ν(du) = dn(u), where n(u) = λ n x 1(u 0), then we obtain F (z) = + n=0 a ne λnz = R + f(u)e zu dn(u), and statements for entire Dirichlet series in [5] follow immediately from Theorems 1 and 2.

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