REMARK ON HE FORMULA BY RAKHMANOV AND SEKLOV S CONJECURE S. DENISOV Abstract. he conjecture by Steklov was solved negatively by Rakhmanov in 1979. His original proof was based on the formula for orthogonal polynomial obtained by adding point masses to the measure of orthogonality. In this note, we show how this polynomial can be obtained by applying the method developed recently for proving the sharp lower bounds for the problem by Steklov. 1. Introduction: Steklov s conjecture and recent development Consider the weight ρx on the interval [ 1, 1] and the sequence of polynomials {P n x} n=, which are orthonormal 1 P n x P m x ρx dx = δ n,m, n, m =, 1, 2... 1 1 with respect to ρ. Assuming that the leading coefficient of P n x is positive, these polynomials are defined uniquely. he Steklov conjecture dates back to 1921 [13] and it asks whether a sequence {P n x} is bounded at any point x 1, 1, provided that ρx is positive on [ 1, 1], i.e., ρx δ, δ >. 2 his conjecture attracted a lot of attention check, e.g., [3, 4, 5, 7] and a survey [14]. It was solved negatively by Rakhmanov in [1]. All existing proofs use the following connection between the polynomials orthogonal on the segment of the real line and on the unit circle. Let ψ, x [ 1, 1], ψ 1 = be a non-decreasing bounded function with an infinite number of growth points. Consider the system of polynomials {P k }, k =, 1,... orthonormal with respect to the measure dψ supported on the segment [ 1, 1]. Introduce the function σθ = { ψcos θ, θ π, ψcos θ, π θ 2π, which is bounded and non-decreasing on [, 2π]. Consider the polynomials φ k z, σ = λ k z k +..., λ k > orthonormal with respect to measure dσ, i.e., φ n e iθ φ m e iθ dσ = δ n,m, n, m =, 1, 2... 4 hese polynomials can be though of as polynomials orthonormal on the unit circle with respect to a measure σ given on as well. Later, we will use the following notation: for every polynomial Q n z = q n z n +... + q of degree at most n, we introduce the operation: Q n z Q nz = q z n +... + q n his depends on n. hen, we have the Lemma. 1 3
Lemma 1.1. [6, 15] he polynomial φ n is related to P k by the formula where x=z+z 1 /2. P k x, ψ = φ 2kz, σ + φ 2k z, σ 2π [ 1 + λ 1 2k φ 2k, σ ] z k, k =, 1,..., 5 his reduction also works in the opposite direction: given a measure σ, defined on and symmetric with respect to R, we can map it to the measure on the real line and the corresponding polynomials will be related by 5. For every δ, 1, let S δ denote the class of probability measures σ which satisfy σ θ δ/2π, a.e. θ [, 2π. 6 We will call S δ the Steklov class. he version of Steklov s conjecture for the unit circle then reads as follows: Given σ S δ, is it true that the sequence {φ n z, σ} is bounded for every z? he normalization dσ = 1 is not restrictive because of the scaling: φ n z, σ = α 1/2 φ n z, ασ, α >. he negative answer to this question see [1] implied the solution to Steklov s conjecture on the real line due to Lemma 1.1. Besides the orthonormal polynomials, we can define the monic orthogonal ones {Φ n z, σ} by requiring coeffφ n, n = 1, Φ n e iθ, σφ m e iθ, σ dσ =, m < n, where coeffq, j denotes the coefficient in front of z j in the polynomial Q. he original argument by Rakhmanov was based on the following formula for the orthogonal polynomial that one gets after adding several point masses to a background measure at particular locations on the circle. Lemma 1.2. Rakhmanov s formula, [1] Let µ be a positive measure on with infinitely many growth points and n K n ξ, z, µ = φ j ξ, µφ j z, µ be the Christoffel-Darboux kernel, i.e., j= P ξ = P z, K n ξ, z, µ L 2,µ, P : deg P n. hen, if {ξ j }, j = 1,..., m, m < n are chosen such that then where Φ n z, η = Φ n z, µ η = µ + K n 1 ξ j, ξ l, µ =, j l 7 m k=1 m k Φ n ξ k, µ 1 + m k K n 1 ξ k, ξ k, µ K n 1ξ k, z, µ 8 m m k δ θk, ξ k = e iθ k, m k. k=1 2
It is known [1] that for every ẑ, the function K n 1 ξ, ẑ, µ has exactly n 1 different roots {ξ j ẑ}, j = 1,..., n 1 and they all lie on. Moreover, K n 1 ξ i, ξ j, µ = for i j see [1]. he limitation that {ξ j } must be the roots is quite restrictive and the direct application of this formula with background dµ = δ2π 1 dθ yields only logarithmic lower bound in the following variational problem: M n,δ = sup σ S δ φ n z, σ L Cδ log n, n > n δ 9 he straightforward iteration of this fixed-n, varying σ construction gave the negative solution to the original conjecture of Steklov [1]. Remark. It is known [12] that for probability measures σ in the Szegő class, i.e., those σ for which log σ dθ >, we have 1 exp log2πσ θdθ Φ n z, σ 4π φ n z, σ 1, z C hus, for measures in Steklov class, i.e., those satisfying 6, the following estimate holds Φ n z, σ δ φ n z, σ 1, z C so, is Φ n or φ n grow in n, they grow simultaneously. he upper bound for M n,δ is easy to obtain M n,δ Cδ n 1 and the corresponding result for fixed σ S δ and n is contained in the following Lemma. Lemma 1.3. If σ S δ, then φ n z, σ L = o n, n. 11 his result follows from, e.g., [8], heorem 4 see also [9],[16] for the real-line case. he gap between log n and n was nearly closed in the second paper by Rakhmanov [11] where the following bound was obtained: under the assumption that δ is small. M n,δ Cδ n log 3 n In the recent paper [2], the following two heorems were proved. heorem 1.1. [2] If δ, 1 is fixed, then and M n,δ > Cδ n. 12 heorem 1.2. [2] Let δ, 1 be fixed. hen, for every positive sequence {β n } : lim n β n =, there is a probability measure σ : dσ = σ dθ, σ S δ such that for some sequence {k n } N. φ kn z, σ L β kn kn 13 3
hese two results completely settle the problem by Steklov on the sharpness of estimates 1 and 11. he method used in the proof was very different from those of Rakhmanov. In the current paper, we will show that it can be adjusted to the cover construction by Rakhmanov. his new modification is interesting in its own as it contains certain cancellation different from the one used in [2]. he structure of the paper is as follows. he second section contains the explanation of the main idea used in [2] to prove heorem 1.1. In the third one, we show how it can be used to cover the Rakhmanov s construction. We will use the following notation. he Cauchy kernel Cz, ξ is defined as Cz, ξ = ξ + z ξ z, ξ. he function analytic in D = {z : z < 1} is called Caratheodory function if its real part is nonnegative in D. Given a set Ω, χ Ω denotes the characteristic function of Ω. If two positive functions f 12 are given, we write f 1 f 2 if there is an absolute constant C such that f 1 < Cf 2 for all values of the argument. We define f 1 f 2 similarly. Writing f 1 f 2 means f 1 f 2 f 1. 2. Method used to prove heorem 1.1 In this section we explain an idea used in the proof of heorem 1.1. We start with recalling some basic facts about the polynomial orthogonal on the unit circle. With any probability measure µ, which is defined on the unit circle and have infinitely many growth points, one can associate the orthonormal polynomials of the first and second kind, {φ n } and {ψ n }, respectively. {φ n } satisfy the following recursions [12], p. 57 with Schur parameters {γ n }: { φn+1 = ρ 1 n zφ n γ n φ n, φ = 1 φ n+1 = ρ 1 n φ n γ n zφ n, φ = 1 14 and {ψ n } satisfy the same recursion but with Schur parameters { γ n }, i.e., { ψn+1 = ρ 1 n zψ n + γ n ψn, ψ = 1 ψn+1 = ρ 1 n ψn + γ n zψ n, ψ = 1 15 he coefficient ρ n is defined as ρ n = 1 γ n 2 he following Bernstein-Szegő approximation is valid: Lemma 2.1. [6],[12] Suppose dµ is a probability measure and {φ j } and {ψ j } are the corresponding orthonormal polynomials of the first/second kind, respectively. hen, for any N, the Caratheodory function F N z = ψ N z φ N z = Cz, e iθ dµ N θ, where dµ N θ = dθ 2π φ N e iθ 2 = has the first N aylor coefficients identical to the aylor coefficients of the function F z = Cz, e iθ dµθ. dθ 2π φ N eiθ 2 In particular, the polynomials {φ j } and {ψ j }, j N are the orthonormal polynomials of the first/second kind for the measure dµ N. We also need the following Lemma which can be verified directly: 4
Lemma 2.2. he polynomial P n z of degree n is the orthonormal polynomial for a probability measure with infinitely many growth points if and only if 1. P n z has all n zeroes inside D counting the multiplicities. 2. he normalization conditions dθ 2π P n e iθ 2 = 1, coeffp n, n > are satisfied. Proof. ake 2π P n e iθ 2 dθ itself as a probability measure. he orthogonality is then immediate. We continue with a Lemma which paves the way for constructing the measure giving, in particular, the optimal bound 12. It is a special case of a solution to the truncated moment s problem. Lemma 2.3. Suppose we are given a polynomial φ n and Caratheodory function F which satisfy the following properties 1. φ nz has no roots in D. 2. Normalization on the size and rotation φ nz 2 dθ = 2π, φ n >. 16 3. F C, Re F > on, and 1 2π Re F e iθ dθ = 1. 17 Denote the Schur parameters given by the probability measures µ n and σ dθ dµ n = 2π φ ne iθ 2, d σ = σ dθ = Re F e iθ dθ, 2π as {γ j } and { γ j }, respectively. hen, the probability measure σ, corresponding to Schur coefficients γ,..., γ n 1, γ, γ 1,... is purely absolutely continuous with the weight given by σ = 4 σ φ n + φ n + F φ n φ n 2 = 2 Re F π φ n + φ n + F φ n φ n 2. 18 he polynomial φ n is the orthonormal polynomial for σ. he proof of this Lemma is contained in [2]. We, however, prefer to give its sketch here. Proof. First, notice that { γ j } l 1 by Baxter s heorem see, e.g., [12], Vol.1, Chapter 5. herefore, σ is purely absolutely continuous by the same Baxter s criterion. Define the orthonormal polynomials of the first/second kind corresponding to measure σ by { φ j }, { ψ j }. Similarly, let {φ j }, {ψ j } be orthonormal polynomials for σ. Since, by construction, µ n and σ have identical first n Schur parameters, φ n is n-th orthonormal polynomial for σ. Let us compute the polynomials φ j and ψ j, orthonormal with respect to σ, for the indexes j > n. By 15, the recursion can be rewritten in the following matrix form φn+m ψ n+m Am B φ n+m ψn+m = m φn ψ n C m D m φ n ψn 19 5
where A m, B m, C m, D m satisfy A B 1 = C D 1 Am B m 1 = C m D m ρ... ρ m 1, z z γ m 1 1 γ m 1 z γ... z γ 1 and thus depend only on γ,..., γ m 1. Moreover, we have φm ψm Am B φ m ψ m = m 1 1. C m D m 1 1 hus, A m = φ m + ψ m /2, B m = φ m ψ m /2, C m = φ m ψ m/2, D m = φ m + ψ m/2 and their substitution into 19 yields 2φ n+m = φ n φ m ψ m + φ n φ m + ψ m = φ m φ n + φ n + F m φ n φ n 2 where F m z = ψ mz φ mz. Since { γ n } l 1 and {γ n } l 1, we have [12], p. 225 F m F as m and φ j Π, φ j Π as j. uniformly on D. he functions Π and Π are the Szegő functions of σ and σ, respectively, i.e., they are the outer functions in D that satisfy on. In 2, send m to get Π 2 = 2πσ, Π 2 = 2π σ 21 2Π = Π φ n + φ n + F φ n φ n 22 and we have 18 after taking the square of absolute values and using 21. In [2], to prove 12 with small δ, the polynomial φ n and F were chosen to satisfy extra conditions see Decoupling Lemma in [2]: φ n 1 > C n 23 and φ nz + F zφ nz φ n z C Re F z, z 24 23 yields the n growth claimed in heorem 1.1. he last inequality guarantees that σ belongs to Steklov class due to 18 and 21. However, as will be made clear in the next section, 24 is not necessary for polynomials to have large uniform norm. 3. Rakhmanov s construction via new approach Our goal in this section is twofold. Firstly, we use the method explained in section 2 to reproduce Rakhmanov s polynomial and polynomials with the similar structure that have large uniform norm and which are orthogonal with respect to a measure in Steklov class. Secondly, we show that the last condition in the Decoupling Lemma [2], formula 3.6, or, what is the same, the bound 24 above is not really necessary for the orthogonal polynomial to have large uniform norm. Instead, that can be achieved by a different sort of cancellation which might be of its own interest. 6
We start with recalling the construction by Rakhmanov [1]. In Lemma 1.2, take the Lebesgue measure µ : dµ = dθ/2π. We have the following expression for the kernel K n 1 ξ, z, µ = n 1 j= ξ j z j = z ξ n 1 z ξ 1 Given two parameters ɛ, < ɛ < 1 and m, m < n 1, we add the mass m k = ɛm 1 to each of the points ξ k = e i2πk/n, k =,..., m 1. hen Lemma 1.2 gives and therefore Φ n z, µ = z n Φ nz, µ = 1 d l = ɛm 1 1 + ɛnm 1 m 1 j= hus, if n is even and m = n/2, we have m 1 ξ j n 1 z n 1 +... + ξ j z + 1 j= ɛm 1 1 + ɛnm 1 d 1z + d 2 z 2 +... + d n z n 25 m 1 ξj n l = j= ξ l j, l = 1,..., n d n = m, d l = 1l 1 e i2πl/n, l = 1,..., n 1 26 1 and d n l = d l, l = 1,..., n 1. hen, 1 + 3ɛ Φ n Φ n = 1 z n, Φ n Φ n 1 + 2ɛ L < C 27 Since e i2πl/n 1 = i 2πl l 2 n + O n 2, l <.1n, 28 it is clear that Φ n L 1 + ɛ log n and this growth occurs around the points z = 1 and z = 1. he choice of {m j } can be rather arbitrary and does not have to be given by equal mass distribution to provide the logarithmic growth. Since η = 1 + ɛ and η = 2π 1, the normalized measure η/ η S δ, δ = 1 + ɛ 1 φ n z, η/ η L 1 + ɛ ln n his argument proves 9. he next theorem is the main result of the paper. It is not new see, e.g., [1]. However, we give a different proof. In particular, it explains how the polynomial of the structure similar to 25 can be obtained by the method described in the previous section. heorem 3.1. For every ɛ, 1, there is σ = σ dθ: and dσ = 1, σ θ 2π 1 ɛ φ n z, σ L ɛ log n Proof. We will consider an analytic polynomial M n of degree n 1 satisfying two conditions Re M n e iθ dθ =, Re M n e iθ L < C, Im M n e iθ L log n 29 7
his M n is easy to find. Consider lθ = χ <θ<π χ π<θ<2π and take Lz = Cl = 1 2π Cz, e iθ le iθ dθ = 1 2π e iθ + z e iθ z leiθ dθ Re C is the Poisson kernel so Re Le iθ = lθ, θ, π. hen, we take M n = F n L, where F n is the Fejer kernel. Since F n is real, nonnegative trigonometric polynomial of degree n 1 and F n L 1 [,2π] = 1, we have Re M n = F n l 1, z ; Re M n e iθ dθ =, M n = he logarithmic growth of M n around the points θ = and θ = π is a standard exercise, e.g., Im M n e iθ log n, θ < Cn 1 3 with arbitrary large fixed C. Now, take a small positive ɛ and define F = 1 2ɛM n, D n = M n + b, φ n = a1 + ɛd n + D n 31 where a and b are positive parameters to be chosen later so that all conditions of the Lemma 2.3 are satisfied. We have φ n = az n + ɛd n + Dn Notice that φ n φ n = a1 z n and compare it with 27. Since D n = b and deg D n = n 1, we have Dn = and then φ n = a1 + ɛb >. Let us check other normalization conditions for these functions. Re F e iθ = 1 + Oɛ >, Re F e iθ dθ = 2π 32 Choose b such that Re D n [C 1, C 2 ] with C 1 >. For example, if b = 2, then Re D n [1, 3]. We can write 1 + ɛd n + Dn = D n ɛ1 + e inθ 2Θn + D 1, z = e iθ where Θ n = arg D n. Notice that D n is zero free in D since it has positive real part on. Since Re 1 + e inθ 2Θn, Re Dn 1 = Re D n D n 2 > we have that φ n is zero free in D. hen, for z, and 1 + ɛd n + D n Re D n D n 1 + ɛd n + D n 2 dθ n D n 1 since D n L 2 b + L L 2 1. On the other hand, 1 + ɛd n + Dn 1 L 2 2ɛ D n L 2 ɛ D n 2 dθ 1 33 and so 1 + ɛd n + D n L 2 = 2π + Oɛ. From Cauchy-Schwarz inequality, we get and so 2π 1 + ɛd n + D n L 2 1 + ɛd n + D n 1 L 2 2π 2π + Oɛ 1 + ɛd n + D n 1 L 2 1 8
Let us choose a so that φ n 2 dθ = 2π which implies a 1. We satisfied all conditions of the Lemma 2.3. Consider the formula 18. We can write φ n + φ n + F φ n φ n = 2φ n 2ɛM n φ n φ n = 34 2φ n 2aɛM n M n z n = 2a 1 + ɛd n + Dn ɛm n M n z n = 2a 1 + ɛd n + Dn ɛm n M n + M n M n z n = 2a 1 + ɛb1 + z n + 2ɛz n Re M n Let us control the deviation of σ from the constant. We get 2πσ = 42a 2 Re F 1 + Oɛ 2 = a 2 1 + Oɛ 1 + Oɛ 2 where we used Re M n 1 and 32. Since a 1, we have that the deviation of 2πσ from a 2 is at most Cɛ. Since σ is a probability measure, this implies a = 1 + Oɛ. We are left to show that φ n L log n. By construction, it is sufficient to prove Indeed, as follows from 3. M n + z n M n L log n M n z n + z n n M n z n log n, z n = e iπ/n Remark. Our analysis covers the polynomial 25 constructed by Rakhmanov too. If d = m, we can rewrite 25 as with Φ n = 1 + ɛ ɛm 1 1 + ɛnm 1 1 + ɛnm 1 d + d 1 z + d 2 z 2 +... + d n z n 35 = 1 + ɛ 1 + 2ɛ ɛb + bzn + M n + Mn b = 1 1 + 2ɛ, M n = m 1 21 + 2ɛ d 1z + d 2 z 2 +... + d n 1 z n 1 he straightforward analysis shows that 26 implies 29. he formula 35 differs from 31, in essence, only by the negative sign and the normalization factor. Different sign makes checking conditions 1 and 2 in the Lemma 2.3 harder when compared to the argument in the proof of heorem 3.1. However, in this particular case, this can be done directly by analyzing the polynomial d + d 1 z +... + d n z n around points z = 1 and z = 1. Indeed, we have N sinjθ j < C j=1 uniformly over θ and N. hen 28,26, and 25 imply Re Φ n = 1 + Oɛ, z. herefore, and the opposite estimate Φ ne iθ 2 dθ < C 1 Φ ne iθ 2 dθ > C 2 > 9
follows from the analysis of Φ n away from z = ±1, i.e., on the arcs z = e iθ, ɛ < θ < π ɛ. Now, we can normalize Φ n and define φ n = aφ n so that φ ne iθ 2 dθ = 2π For the constant a, we then have a 1. Next, to check that φ n corresponds to a Steklov measure, one only needs to modify the choice of F by changing the sign in front of ɛ: and repeating 34 with properly chosen C. F = 1 + CɛM n Remark. As one can see from the proof of heorem 3.1, the different sort of cancellation has been used to show the Steklov condition of the measure. In particular, the estimate 24 is violated as φ n φ n = a1 z n does not provide the strong cancellation around z = 1. Acknowledgement. he work on section 3 was supported by RSF-14-21-25. he research done in the rest of the paper was supported by the grant NSF-DMS-1464479. References [1] M. U. Ambroladze, On the possible rate of growth of polynomials that are orthogonal with a continuous positive weight Russian, Mat. Sb. 182 1991, no. 3, 332 353; English translation in: Math. USSR-Sb. 72 1992, no. 2, 311 331. [2] A. Aptekarev, S. Denisov, D. ulyakov, On a problem by Steklov, to appear in Journal of the AMS. [3] Ya. L. Geronimus, Some estimates of orthogonal polynomials and the problem of Steklov, Dokl. Akad. Nauk SSSR, 236 1977, no. 1, 14 17. [4] Ya. L. Geronimus, he relation between the order of growth of orthonormal polynomials and their weight function. Mat. Sb. N.S. 61 13, 1963, 65 79. [5] Ya. L. Geronimus, On a conjecture of V.A. Steklov. Dokl. Akad. Nauk SSSR, 142, 1962, 57 59. [6] Ya. L. Geronimus, Polynomials orthogonal on the circle and on the interval, GIFML, Moscow, 1958 in Russian; English translation: International Series of Monographs on Pure and Applied Mathematics, Vol. 18 Pergamon Press, New York- Oxford-London-Paris, 196. [7] B. L. Golinskii, he problem of V.A. Steklov in the theory of orthogonal polynomials. Mat. Zametki, 15 1974, 21 32. [8] A. Máté, P. Nevai, V. otik, Extensions of Szegő s theory of orthogonal polynomials. II, III. Constr. Approx. 3 1987, 1, 51 72, 73 96. [9] P. Nevai, J. Zhang, V. otik, Orthogonal polynomials: their growth relative to their sums. J. Approx. heory 67 1991, 2, 215 234. [1] E. A. Rahmanov, On Steklov s conjecture in the theory of orthogonal polynomials, Matem. Sb., 1979, 1815, 581 68; English translation in: Math. USSR, Sb., 198, 36, 549 575. [11] E. A. Rahmanov, Estimates of the growth of orthogonal polynomials whose weight is bounded away from zero, Matem. Sb., 1981, 114156:2, 269 298; English translation in: Math. USSR, Sb., 1982, 42, 237 263. [12] B. Simon, Orthogonal polynomials on the unit circle, volumes 1 and 2, AMS 25. [13] V. A. Steklov, Une methode de la solution du probleme de development des fonctions en series de polynomes de chebysheff independante de la theorie de fermeture, Izv. Rus. Ac. Sci., 1921, 281 32, 33 326. [14] P. K. Suetin, V.A. Steklov s problem in the theory of orthogonal polynomials, Itogi Nauki i ech. Mat. Anal., VINII, 1977, 15, 5 82; English translation in: Journal of Soviet Mathematics, 1979, 126, 631 682. [15] G. Szegő, Orthogonal Polynomials, Amer. Math. Soc. Colloq. Publ., 23, Providence RI, 1975 fourth edition. [16] R. Szwarc, Uniform subexponential growth of orthogonal polynomials. J. Approx. heory 81 1995, 2, 296 32. University of Wisconsin Madison Mathematics Department 48 Lincoln Dr., Madison, WI, 5376, USA denissov@math.wisc.edu and 1
Keldysh Institute for Applied Mathematics, Russian Academy of Sciences Miusskaya pl. 4, 12547 Moscow, RUSSIA 11