Open problems. Christian Berg a a Department of Mathematical Sciences, University of. Copenhagen, Copenhagen, Denmark Published online: 07 Nov 2014.

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1 This article was downloaded by: [Copenhagen University Library] On: 4 November 24, At: :7 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: Registered office: Mortimer House, 37-4 Mortimer Street, London WT 3JH, UK Integral Transforms and Special Functions Publication details, including instructions for authors and subscription information: Open problems Christian Berg a a Department of Mathematical Sciences, University of Copenhagen, Copenhagen, Denmark Published online: 7 Nov 24. To cite this article: Christian Berg (25) Open problems, Integral Transforms and Special Functions, 26:2, 9-95 To link to this article: PLEASE SCROLL DOWN FOR ARTICLE Taylor & Francis makes every effort to ensure the accuracy of all the information (the Content ) contained in the publications on our platform. However, Taylor & Francis, our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content. Any opinions and views expressed in this publication are the opinions and views of the authors, and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information. Taylor and Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content. This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden. Terms & Conditions of access and use can be found at

2 Integral Transforms and Special Functions, 25 Vol. 26, No. 2, 9 95, Open problems Christian Berg Downloaded by [Copenhagen University Library] at :7 4 November 24 Department of Mathematical Sciences, University of Copenhagen, Copenhagen, Denmark The following open problems were presented at OPSFA-2, Sousse, Tunisia, March 29, 23. Christian Berg, University of Copenhagen, has collected these open problems. Mohamed J. Atia: Conjecture about a symmetric system of orthogonal polynomials We consider the weight w(x) = x + 2 / + x + x 2 / x on [, ], and claim that there exits a sequence of polynomials {P n } n orthogonal with respect to w(x) on [, ] and which fulfils P (x) =, P (x) = x () P n+2 (x) = xp n+ (x) γ n+ P n (x), n such that γ = 2, γ 2 = 4, γ 3 = 7 3, γ 4 = 4 5, that is to say γ 3 + γ 4 = 2, γ 5 = 4, γ 6 = 2 49, γ 7 = 25 98, γ 6 + γ 7 = 2, γ 8 = 4, γ 9 = , γ = , γ 9 + γ = 2, γ = 2, γ 3n+2 = 4, γ 3n+3 + γ 3n+4 = 2. This conjecture was presented, first, in Leuven, Belgium 29, second, in Madrid, Spain 2 and, third, in Sousse, Tunisia 23. The weight w(x) is the solution of a second-order differential equation that is why we call the sequence {P n } n a sequence of second category. berg@math.ku.dk c 24 Taylor & Francis

3 Integral Transforms and Special Functions 9 References [] Atia MJ, Leffet S. On orthogonal polynomials: semiclassical and of second category. Integral Transforms Spec Funct. doi:.8/ University of Gabès, Tunisia jalel.atia@gmail.com Downloaded by [Copenhagen University Library] at :7 4 November 24 Howard Cohl: Hypergeometric series Given a hypergeometric series ( ) a,..., a p p+f p ; x, b,..., b p where the parameters in the hypergeometric series satisfy the relations a + = a + b = =a p + b p, then the series is called well-poised. If one has a well-poised hypergeometric series, where there exists some i p, so that a i = b i + = a 2 +, with a C \{, 2, 4,...}, then the hypergeometric series is called very-well-poised. I have a terminating very-well-poised 9 F 8 () hypergeometric series, a, a, a, n, n, 9F 8 2,2a + n,2a + n, a 2 +,, + n + a,+ n + a, a + 2, n a, n a, a ;. 2 Does anybody have any ideas on how I can sum this in closed form? National Institute of Standards and Technology, USA hcohl@nist.gov Anbu Swaminathan: Pick functions The function which is analytic in the slit domain C\[, ) and whose Taylor coefficients are the Hausdorff moment sequences also belong to the set of Pick functions [] which are related to moment problems. I am interested in the following problem. Problem To characterize the members of this class of Pick functions. The generalized polylogarithm [3] is defined by the normalized function p,q (a, b; z) = k= (( + a)p ( + b) q /(k + a) p (k + b) q )z k, z <, where, k + a, k + b, p, q are complex numbers such that Re p >, Re q >. Note that for a = b =, p + q = r C, Re r >, p,q (a, b; z) = L r (z), is the polylogarithm or Jonquière s function that further reduces to Riemann zeta function for z =. Numerical evidences suggest the following conjecture.

4 92 C. Berg Conjecture For p + q p 2 + q 2, the ratio p,q (a, b ; z) p2,q 2 (a, b ; z) belongs to the class of Pick functions. Downloaded by [Copenhagen University Library] at :7 4 November 24 Note that for p + q = a and p 2 + q 2 = b, a b, with a = b = a 2 = b 2 =, this result that L a (z)/l b (z) is in the class of Pick functions, is obtained in [4] using duality techniques of function theory. It would be interesting to see if the ratio given in the conjecture has the g-fraction as there exist a relation between the completely monotone sequences and the g-fractions, see [6]. Hence, it is important to identify the corresponding g-sequence for this ratio. Particular cases of this conjecture are proved in [5]. Further generalized polylogarithms are particular cases of the well-known generalized hypergeometric function [3] and since certain ratios of Gaussian hypergeometric functions [2] and their q-analgoues [5] can be written in terms of g-fractions, I propose Problem 2 Characterize the ratios of generalized (basic) hypergeometric functions so that the corresponding g-fractions lead to the geometric properties of Pick functions. References [] Berg C, Pedersen HL. Pick functions related to the gamma function. Rocky Mountain J Math. 22;32(2): [2] Küstner R. On the order of starlikeness of the shifted Gauss hypergeometric function. J Math Anal Appl. 27;334(2): [3] Mondal SR, Swaminathan A. Geometric properties of generalized polylogarithms. Integral Transforms Spec Funct. 2;2(9):69 7. [4] Roth O, Ruscheweyh S, Salinas L. A note on generating functions for Hausdorff moment sequences. Proc Am Math Soc. 28;36(9): [5] Swaminathan A, Tsygvintsev A. Pick functions and chain sequences for hypergeometric type functions. Preprint. [6] Wall HS. Analytic theory of continued fractions. New York: D. Van Nostrand Company; 948. Indian Institute of Technology, Roorkee, India mathswami@gmail.com Walter Van Assche: Minkowski s question mark function In 94, Hermann Minkowski introduced an interesting function, which he called the question mark function and he denoted its values by?(x). This notation with a question mark is somewhat confusing, so instead I will denote the function by q and I will only consider it on the interval [, ]. If < x <, then we can write x as a regular continued fraction x =, a i N {}. a + a 2 + a

5 Integral Transforms and Special Functions 93 Downloaded by [Copenhagen University Library] at :7 4 November 24 The Minkowski question mark function at x is then defined as ( ) k+ q(x) = 2. 2 a +a 2 + +a k k= If x is a rational number, then the continued fraction is terminating and q(x) is given by a finite sum (see Figure ). By setting q() = and q() = one can show that q : [, ] [, ] is a continuous and increasing function, so that q is a probability distribution function on [, ] which defines a probability measure on [, ]. This distribution function has the property that q (x) = almost everywhere on [, ] so that the corresponding measure is singular and continuous. See [7] for some information about this function and references. The definition via regular continued fractions is not convenient for integration. A more convenient way is to define it as a fixed point of an iterated function system with two rational functions. One has 2 q q(x) = ( x 2 q x ( ) x, x x 2, ), 2 x and one can show that the sequence of probability distribution functions (q n ) n N, with ( ) x 2 q n, x x 2, q n (x) = ( ) x 2 q n, x 2 x, and q any probability distribution on [, ], converges uniformly to Minkowski s question mark function. This allows us to compute integrals by a limit procedure Salem s problem f (x) dq(x) = lim n f (x) dq n (x). A first open problem is about the Fourier coefficients of Minkowski s question mark function: α n = e 2inπx dq(x). The Riemann Lebesgue lemma tells us that Fourier coefficients of an absolutely continuous measure on [, ] tend to zero. The Minkowski question mark function is singularly continuous, so one cannot use the Riemann Lebesgue lemma. Nevertheless, the support of q is the full interval [, ] and it was proved by Salem [5] that q is Hölder continuous of order α = log 2/(2log(( 5 + )/2)) =.722. Furthermore, Salem showed that n n α k = O(n α/2 ), k= so that α n converges to zero on the average and there is the possibility that α n. This is the problem posed by Raphaël Salem [5,]: Do the Fourier coefficients of the Minkowski question mark function converge to? If not, what are lim inf n α n and lim sup n α n?

6 94 C. Berg.8.6 Downloaded by [Copenhagen University Library] at :7 4 November Figure The Minkowski question mark function. Orthogonal polynomials Consider the orthonormal polynomials for the Minkowski question mark function: with recurrence relation x p n (x)p m (x) dq(x) = δ m,n xp n (x) = a n+ p n+ (x) + b n p n (x) + a n p n (x), n, with p = and p =.Thesymmetryofqaround the point 2 implies that b n = for all 2 n N. Rakhmanov s theorem [4] tells us that for an absolutely continuous measure μ on [, ] for which μ > almost everywhere, one has a n 4 as n. In our case q = almost everywhere, so one cannot use Rakhmanov s theorem to deduce the asymptotic behaviour of the recurrence coefficients. However, it is known (see, e.g., [2,3,6]) that there exist discrete measures and singular continuous measures (on [, ]) for which the recurrence coefficients have the behaviour b n 2 and a n as n, so that they are in the Nevai class (for the interval 4 [, ]). The open problem is whether the Minkowski question mark function is such a singular continuous function for which the recurrence coefficients are in the Nevai class (for the interval [, ]), i.e. Is lim n a n = 4? If not, then what are lim inf n a n and lim sup n a n? Acknowledgements I like to thank S. Yakubovich for bringing the Minkowski question mark function and Salem s problem to my attention.

7 Integral Transforms and Special Functions 95 References Downloaded by [Copenhagen University Library] at :7 4 November 24 [] Alkauskas G. The Minkowski?(x) function and Salem s problem. C R Math Acad Sci Paris. 22;35(3 4):37 4. [2] Lubinsky DS. Singularly continuous measures in Nevai s class M. Proc Am Math Soc. 99;: [3] Van Assche W, Magnus AP. Sieved orthogonal polynomials and discrete measures with jumps dense in an interval. Proc Am Math Soc. 989;6: [4] Rakhmanov EA. On the asymptotics of the ratio of orthogonal polynomials. Mat Sb. 977;3(45): (in Russian); translated in Math. USSR Sb. 977;32: Rakhmanov EA. On the asymptotics of the ratio of orthogonal polynomials, II. Mat. Sb. 982;8(6):4 7 (in Russian); translated in Math. USSR Sb. 983;47:5 7. [5] Salem R. On some singular monotonic functions which are strictly increasing. Trans Am Math Soc. 943;53: [6] Totik V. Orthogonal polynomials with ratio asymptotics. Proc Am Math Soc. 992;4: [7] s_question_mark_function (March 29, 23). University of Leuven, Belgium walter@wis.kuleuven.be