We find the exact upper estimate for the upper density of zeros of entire
functions of exponential type whose indicator diagram is contained in a given
interval.Comment: 7 pages, 1 figur

Eremenko, Alexandre

Yuditskii, Peter

English

arXiv

Alexandre Eremenko

Peter Yuditskii

Comptes Rendus Mathématique

C. R. Acad. Sci. Paris, Ser. I 346 (2008) 825–828http://france.elsevier.com/direct/CRASS1/Complex AnalysisAn extremal problem for a class of entire functionsAlexandre Eremenko a,1, Peter Yuditskii b,2a Purdue University, West Lafayette, IN 47907, USAb J. Kepler University, Linz A-4040, AustriaReceived 29 May 2008; accepted 18 June 2008Available online 25 July 2008Presented by Paul MalliavinAbstractLet f be an entire function of the exponential type, such that the indicator diagram is in [−iσ, iσ ], σ > 0. Then the upper densityof f is bounded by cσ , where c ≈ 1.508879 is the unique solution of the equationlog(√c2 + 1 + c) = √1 + c−2.This bound is optimal. To cite this article: A. Eremenko, P. Yuditskii, C. R. Acad. Sci. Paris, Ser. I 346 (2008).© 2008 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.RésuméUn problème extrêmal pour une classe de fonctions entières. Soit f une fonction entière d’indicatrice contenue dans l’inter-valle [−iσ, iσ ], σ > 0. Alors la borne supérieure des zéros de f ne dépasse pas cσ , où c ≈ 1,508879 est la solution d’équation,log(√c2 + 1 + c) = √1 + c−2.Cette borne est exacte. Pour citer cet article : A. Eremenko, P. Yuditskii, C. R. Acad. Sci. Paris, Ser. I 346 (2008).© 2008 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.We consider the class Eσ , σ > 0 of entire functions of exponential type whose indicator diagram is contained in asegment [−iσ, iσ ], which means thath(θ) := lim supr→+∞log |f (reiθ )|r σ | sin θ |, |θ |  π. (1)An alternative characterization of such functions follows from a theorem of Pólya [6]:f (z) = 12π∫γF (ζ )e−iζz dζ,E-mail addresses: eremenko@math.purdue.edu (A. Eremenko), Petro.Yudytskiy@jku.at (P. Yuditskii).1 Supported by NSF grant DMS-0555279.2 Supported by Austrian Fund FWF P20413-N18.1631-073X/$ – see front matter © 2008 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.doi:10.1016/j.crma.2008.06.009826 A. Eremenko, P. Yuditskii / C. R. Acad. Sci. Paris, Ser. I 346 (2008) 825–828where F is an analytic function in C̄ \ [−σ,σ ], F (∞) = 0, and γ is a closed contour going once around the segment[−σ,σ ]. In other words, the class of entire functions satisfying (1) are Fourier transforms of hyperfunctions supportedby [−σ,σ ], see, for example, [2] and [3].Let n(r) be the number of zeros of f in the disc {z: |z|  r}, counting multiplicity. We are interested in the upperdensity:D = lim supr→∞n(r)r. (2)If f satisfies the additional condition:∞∫−∞log+ |f (x)|1 + x2 dx < ∞, (3)then the limit (density) in (2) exists and equals (2π)−1∫ π−π h(θ)dθ . For example, if f (z) = sinσz, then f ∈ Eσ andD = 2σ/π ≈ 0.6366σ . The existence of the limit follows from a theorem of Levinson [5,6]. Much more preciseinformation about n(r) under the condition (3) is contained in the theorem of Beurling and Malliavin [1].In the general case, the density might not exist as was shown by examples in [4,10]. Moreover, it is possible thatD > 2σ/π , see [2]. An easy estimate using Jensen’s formula gives D  2eσ/π ≈ 1.7305σ . This estimate is exact inthe larger class of entire functions satisfying the condition h(θ)  σ , but it is not exact in Eσ .In this Note we find the best possible upper estimate for the upper density of zeros of functions in Eσ .Theorem. The upper density of zeros of a function f ∈ Eσ does not exceed cσ , where c ≈ 1.508879 is the uniquesolution of the equation:log(√c2 + 1 + c) = √1 + c−2, on (0,+∞). (4)For every σ > 0 there exist entire functions f ∈ Eσ such that D = cσ .Proof. Without loss of generality we assume that σ = 1. Moreover, it is enough to consider only even functions. Tomake a function f even we replace it by f (z)f (−z), which results in multiplication of both the indicator h and theupper density D by the same factor of 2.Let tn → +∞ be such sequence that limn(tn)/tn = D. Consider the sequence of subharmonic functions vn(z) =t−1n log |f (tnz)|. Compactness Principle for subharmonic functions [3, Theorem 4.1.9] implies that one can choose asubsequence that converges in D ′ (Schwartz’s distributions). The limit function v is subharmonic in the plane, andsatisfies:v(z)  |Im z|, z ∈ C, and v(0) = 0. (5)Let μ be the Riesz measure of this function. We have to show thatμ({z: |z|  1})  c. (6)First we reduce the problem to the case that the Riesz measure μ is supported by the real line. We havev(z) = 12∫log∣∣∣∣1 − z2ζ 2∣∣∣∣dμζ .Let us compare this withv∗(z) = 12∞∫0log∣∣∣∣1 − z2t2∣∣∣∣dμ∗t ,where μ∗ is the radial projection of the measure μ: it is supported on [0,+∞) and μ∗(a, b) = μ({z: a <|z| < b}), 0  a < b. It is easy to see thatv∗(z)  σ ′|Im z|, z ∈ C, and v∗(0) = 0, (7)A. Eremenko, P. Yuditskii / C. R. Acad. Sci. Paris, Ser. I 346 (2008) 825–828 827with some σ ′ > 0. We claim that one can choose σ ′  1 in (7). Let σ ′ be the smallest number for which (7) holds.Then, by the subharmonic version of the theorem of Levinson mentioned above (see, for example, [9]), the limitlimr→∞ r−1v∗(rz) = σ ′|Im z|exists in D ′ and thuslimr→∞1rr∫0nv∗(t)tdt = limr→∞12πrπ∫−πv∗(reiθ)dθ = 2σ′π,wherenv∗(r) = μ∗[0, r] = μ{z: |z|  r}. (8)Similar limits exist for v, and we have nv = nv∗ , from which we conclude that σ ′  1.From now on we assume that v is harmonic in the upper and lower half-planes, and thatv(iy) ∼ y, y → +∞. (9)Let u be the harmonic function in the upper half-plane such that φ = u + iv is analytic, and φ(0) = 0. Then φ is aconformal map of the upper half-plane onto some region G of the form:G = {x + iy: y > g(x)}, (10)where g is an even upper semi-continuous function, g(0) = 0. Moreover,φ(iy) ∼ iy, as y → +∞, (11)which follows from (9), andφ(−z̄) = −φ(z), (12)because both the region G and the normalization of φ are symmetric with respect to the imaginary axis. Finally, wehave:μ([0, x]) = 2πu(x). (13)For all these facts we refer to [7,8].Remark. The function Reφ(x) = u(x) might be discontinuous for x ∈ R. We agree to understand u(x) as the limitfrom the right u(x + 0) which always exists since u is increasing.Inequality (5) implies that v(x)  0, thus g(x)  0, in other words, G contains the upper half-plane.Thus we obtain the following extremal problem: Among all univalent analytic functions φ satisfying (12) andmapping the upper half-plane onto regions of the form (10) with g  0, g(0) = 0 and satisfying φ(0) = 0 and (11),maximize Reφ(1).We claim that the extremal function g for this problem is:g0(x) ={−∞, 0 < |x| < πc/2,0, otherwise,where c > 1 is the solution of Eq. (4). The corresponding region is shown in Fig. 1. For the extremal function we haveφ0(1) = πc/2 − i∞.To prove the claim, we first notice that for a given G the mapping function is uniquely defined. Let a = φ(1), andb = Rea. Next we show that making g smaller on the interval (0, b) results in increasing Reφ(1) and for g largeron the interval (b,+∞) also results in increasing Reφ(1). The proofs of both statements are similar. Suppose thatg1  g, g1 	= g, and g1(x) = g(x) outside of the two intervals p < |x| < q , where 0 < p < q < b. Let G1 be the828 A. Eremenko, P. Yuditskii / C. R. Acad. Sci. Paris, Ser. I 346 (2008) 825–828Fig. 1. Extremal region.region above the graph of g1, and φ1 the corresponding mapping function normalized in the same way as g. ThenG ⊂ G1, and the conformal map φ−11 ◦ φ is defined in the upper half-plane and maps it into itself. We have:φ−11 ◦ φ(x) = x + 2x∞∫0w(t)t2 − x2 dt,where w 	= 0 is a non-negative function supported on some interval inside (0,1). Putting x = 1 we obtain:φ−11 (a) = 1 + 2∞∫0w(t)t2 − 1 dt,so φ−11 (a) < 1, that is Reφ1(1) > b. This proves our claim.It remains to compute the constant b in the extremal domain. We recall that φ0(1) = b − i∞ and assume thatb = φ0(k) for some k > 1. Here φ0 is the extremal mapping function. Then by the Schwarz–Christoffel formula wehave:φ0(z) = 12z2∫0√ζ − k2ζ − 1 dζ. (14)To find k, we use the condition thatImp.v.k2∫0√ζ − k2ζ − 1 dζ = 0.Denoting c = √k2 − 1 and evaluating the integral, we obtainlog(√c2 + 1 + c) = √1 + c−2.Finally the jump of the real part of the integral in (14) occurs at the point 1 and has magnitude π√k2 − 1 = πc. Thiscompletes the proof of the upper estimate in Theorem 1.To construct an example showing that this estimate can be attained, we follow the construction in [2, Sections 9–10].The role of the subharmonic function u1 there is played now by our extremal function v0 = Imφ0. References[1] A. Beurling, P. Malliavin, On Fourier transforms of measures with compact support, Acta Math. 118 (1967) 291–309.[2] A. Eremenko, D. Novikov, Oscillation of Fourier integrals with a spectral gap, J. Math. Pures Appl. 83 (3) (2004) 313–365.[3] L. Hörmander, Analysis of Linear Partial Differential Operators, vols. I, II, Springer, Berlin, 1983.[4] P. Kahane, L. Rubel, On Weierstrass products of zero type on the real axis, Illinois Math. J. 4 (1960) 584–592.[5] P. Koosis, Leçons sur le théorème de Beurling et Malliavin, Publ. CRM, Montréal, 1996.[6] B. Levin, Distribution of Zeros of Entire Functions, Amer. Math. Soc., Providence, RI, 1980.[7] B. Levin, Subharmonic majorants and some applications, in: Complex Analysis, Birkhäuser, Basel, 1988, pp. 181–190.[8] B. Levin, The connection of a majorant with a conformal mapping, II, Teor. Funktsii Funktional Anal. i Prilozhen. 52 (1989) 3–21 (in Russian).English translation in: J. Soviet Math. 52 (5) (1990) 3351–3364.[9] V. Matsaev, M. Sodin, Distribution of Hilbert transforms of measures, Geom. Funct. Anal. 10 (2000) 1, 160–184.[10] C. Roumieu, Sur quelques extensions de la notion de distribution, Ann. Sci. École Norm. Sup. 77 (1960) 41–121.

An extremal problem for a class of entire functions

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An extremal problem for a class of entire functions of exponential type

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