ON THE SIZE OF THE POLYNOMIALS ORTHONORMAL ON THE UNIT CIRCLE WITH RESPECT TO A MEASURE WHICH IS A SUM OF THE LEBESGUE MEASURE AND P POINT MASSES.

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1 ON HE SIZE OF HE POLYNOMIALS ORHONORMAL ON HE UNI CIRCLE WIH RESPEC O A MEASURE WHICH IS A SUM OF HE LEBESGUE MEASURE AND P POIN MASSES S DENISOV Abstract For the measures on the unit circle that are equal to the sum of Lebesgue measure p point masses, we give an estimate on the size of the corresponding orthonormal polynomials As a simple corollary of the method, we obtain a bound for some exponential polynomials 1 Introduction Let δ 0, 1 Define S δ to be the class of the probability measures σ on the unit circle that satisfy the following condition: σ θ δ/π for ae θ [ π, π We denote the n-th orthonormal polynomial by φ n z, σ n-th monic orthogonal polynomial by Φ n z, σ he problem of Steklov consists in estimating the size of φ n for σ S δ he sharp bounds for φ n L were obtained in [1] he following variational problem played an important role in the proof Consider In [1], the following estimate was established M n,δ = sup σ S δ φ n z, σ L 1 M n,δ Cδ n for fixed δ Moreover, it was proved that a maximizer exists that every σ arg max φ n z, σ L σ S δ can be written as N σ = δµ/π + m j δθ θ j, where N n, dµ = dθ is the Lebesgue measure, m j > 0, 0 < θ 1 < < θ N π No further information on σ was obtained that motivated us to study the following variational problem Given large n integer p : p < n/, we consider the following set of probability measures P δ,p = {σ = δµ/π + m j δθ θ j } define heorem 11 For every κ > 15, we have M n,p = max σ P δ,p φ n z, σ L M n,p Cδ, κ min{α n,p p, p}, αn,p = log κ n/p 1

2 Remark Notice that p < α n,p p for p < pn, κ κ pn, κ log n Corollary 11 If N is the number of point masses in, then N n he proof of the heorem is given in the next section It contains several auxiliary variational problems In the third section, we write the dual formulation for one of them obtain a slight improvement of the estimate for exponential polynomials discovered recently by Kós [9] Erdélyi [6] he last section contains a discussion about the sharpness of the obtained estimated Proof of the main heorem Proof heorem 11 Notice that σ P δ,p implies see [15], formula 31 1 γ j = δ j=0 by the Szegő sum rule In this formula, {γ j } denote the parameters of recursion Since 1/ n 1 Φ n = φ n 1 γ j we have M n,p Cδ max Φ n z, σ L σ P, δ,p we can consider instead the variational problem for Φ n Recall the following characterization of the monic orthogonal polynomials Φ n z, σ = arg min z n + a n 1 z n a 1 z + a 0 dσ, {a j } j=0 which immediately follows from the orthogonality condition If σ P δ,p, then min z n + a n 1 z n a 1 z + a 0 dσ = min 1 a n 1 z + + a 1 z n 1 + a 0 z n dσ {a j } {a j } n 1 = δ + min δ a j + m l 1 z l Q n 1 z l where {a j } j=0 n 1 Q n 1 z = a j z j, z l = e iθ l herefore, if Φ n z, σ = z n + Q n 1 z, σ σ P δ,p, then Q n 1 z, σ Q n 1 z + δ 1 m l 1 z l Q n 1 z l for every Q n 1 so, by Cauchy-Schwarz, j=0 Φ n z, σ L 1 + Q n 1 z, σ L 1 + n Q n 1 z, σ 1 + n Q n 1 z + nδ 1 m l 1 z l Q n 1 z l

3 for every Q n 1 Since δ 0, 1 is fixed, it is left to show that D n,p = max min Q n + m l 1 z l Q n z l ɛ n,p, ɛ n,p = min{ɛ n,p, ɛ n,p} {z l },{m l },z l,m l 0, m l 1 =1 Q n with ɛ n,p = p αn,p/n, ɛ n,p = p /n hat would follow immediately from max min m l 1 z l Q n z l ɛ n,p with some C If {z l },{m l },z l,m l 0, m l 1 =1 Q n C ɛ n,p I n = I 1 n = min deg Q n n, Q n C ɛ n,p min deg Q n+1 n+1, Q n+1 C ɛ n,p m l 1 z l Q n z l, m l 1 Q n+1 z l, then I n I 1 n Moreover, if I n is reached on a a n+1 z n+1, then a 0 C ɛ n,p I n = m l 1 a 0 z l a a n+1 zl n C ɛ n,p m l a z l a a n+1 zl n m l 1 z l a a n+1 z n l C ɛ n,p + I 1n Since ɛ n,p ɛ n+1,p, we only need to show that there is some constant C so that 1 If p p, then min m l 1 Q n z l ɛ n,p 4 Q n C ɛ n,p for any choice of {z l }, {m l }; If p > p, then min Q n C ɛ n,p for any choice of {z l }, {m l } For the first case, we will show that min Q n C ɛ n,p m l 1 Q n z l ɛ n,p 5 m l 1 Q n z l = 0 6 his will follow from the analysis of a different minimization problem Consider We need to prove that E n,p = min Q n 7 Q n:q nz l =1,,,p E n,p C ɛ n,p 3

4 In [10], the following generalization of the Halász result [8] was obtained It was shown that min π π m+1 m L π m:deg π = cos m m,π m0=1,π m1=0 m + 1 herefore, taking the product of p polynomials of degree m = n/p we can assume without loss of generality that p divides n with zeroes at {z l }, l = 1,, p, we construct P n, deg P n = n, such that P n z l = 0, l = 1,, p, P n 0 = 1, [ ] πp p+n P n L cos n + p Letting Q n = 1 P n, we get Q n = π Q n z l = 1, l = 1,, p Re P n dθ + P n dθ = P n dθ π by the Mean Value heorem for harmonic functions hus, [ ] πp p+n Q n π cos 1 n + p E n,p π [ cos ] πp p+n 1/ 1 = ζ n,p 8 n + p Notice that our construction does not guarantee that this estimate is sharp If n p = o n, then ζ n,p Ĉ p n, Ĉ > π3 / 9 for n > n 0 Ĉ 6 is proved We are left to consider p > p prove 5 Recall the following simple Lemma attributed to Havin Lemma 1 Let measurable E 0 < τ 1 hen, there is a function f H D such that fz 1 τ, z E; f H D ; 10 f H D E 1 + log τ Proof ake φ = χ E θ define fe iθ = 1 exp φθ + i φθ log τ, where φ is the harmonic conjugate Since Φ = φ + i φ defines the function analytic in D, f is the boundary value of the function analytic in D he estimates 10 are satisfied by construction hen, f 1 + τ, z E f = 1 e i φ log τ min{1, log τ φ }, e iθ E c herefore f E + log τ φ = E 1 + log τ 4

5 since φ = φ Had we tried to find the analytic function, not a polynomial, that would give an estimate 5, we would have taken E = p [θ j 1/n, θ j + 1/n] τ = β n,p p/n hen, taking f from the Lemma above, we have f p 1 + log β n,p p/n p log n = β p n,p n n p n = τ, if we let β n,p = logn/p his would have allowed us to take κ in the formulation of the heorem 11 equal to 1 However, we need to produce a polynomial of order n for this purpose we will need to modify the construction a bit ake E = p where = [θ j µ, θ j + µ], µ > 1/n will be adjusted later aking τ = α n,p p/n, we use the Lemma to construct f which satisfies f pµ1 + log τ p n α n,p; 1 fz τ, z E provided that We make the following choice µ n 1 log n κ 1 p µ = n 1 log n p κ 1 he function f H D we take Q n = f F n, where F n is defined as follows Given arbitrary γ 0, 1, consider an even function gx such that ĝω, its Fourier transform, is supported on 1, 1 see, eg, [14] ĝ0 = 1, gx e x γ, γ < 1 Let g n x = ngnx define F n x = j Z g n x jπ For this π-periodic function, we have F n l = ĝl/n by the Poisson summation formula so Q n is a polynomial of degree at most n Let us show now that it satisfies the bounds similar to those that hold for f We trivially have Q n f it is left to check that Q n z j 1 τ, j = 1,, p Notice that F n x n e nγ x πj γ ne nγ + ne nγ x γ, x 1 j From the Lemma, we get fz, z fz 1 τ, z E 11 First, notice that f Fn Fn 1 1, z Second, since F nθdθ = 1, one can write F n θ j θfθdθ 1 = 5 fθ j θ 1F n θdθ

6 nµ θ >µ F n θ dθ + θ <µ F n θ fθ j θ 1 dθ 1 he last integral is bounded by Cτ due to 11 the choice of E he first term in 1 is bounded by p n ne nγ + e ξγ dξ e nµγ nµ 1 γ = e logn/pκ 1γ logn/p κ 11 γ < Cτ = C n logκ p provided that κ 1γ = 1 C is large Since γ can be chosen as an arbitrary number smaller than 1, we have κ > 15 Remark One can repeat the argument used for the comparison of I 1 I to show that D n,p D n+1,p, D n,p = max min Q n + m l 1 Q n z l 13 {z l },{m l },z l,m l 0, m l 1 =1 Q n he following formula attributed to Geronimus [7] needs to be mentioned Lemma Geronimus, [7] Consider σ t = 1 tσ + tδθ where t 0, 1 hen, Φ n z, σ t = Φ n z, σ t Φ n1, σk n 1 1, z, σ 1 t + tk n 1 1, 1, σ 14 When p = 1, this Lemma gives an explicit formula whose analysis confirms the estimate we obtained above However, if p is large, its recursive application is complicated Instead, the application of the heorem 11 to this formula gives the following corollary Lemma 3 If σ P δ,p p < n/, then sup φ n z 1, σk n 1 z 1, z, σ 1 + K n 1 z 1, z 1, σ min{α n,p p, p} z 1,z In the case when p n {z j } {1 1/n }, Rakhmanov [13] wrote the exact formula for Φ n z, σ which implies φ n z, σ L log n for an arbitrary distribution of masses this bound is saturated for some {m j } {z j } 3 he Dual Problem some inequalities for Exponential Polynomials Let us start with the following stard Lemma from Linear Algebra Lemma 31 Suppose p < n a linear A : C n C p is surjective Given fixed v C p, consider the following variational problem: min At=v t hen, the minimizer t pseudo-inverse is unique t = sup x C p,a x 0 x, v A x Proof We can assume that v 0 Since C n = kera rana, any solution to At = v can be written as t = t + ξ where ξ is arbitrary in ker A t rana he Pythagorean theorem then implies that t is the pseudo-inverse If t = A η, then gives AA η = v t = AA η, η = v, η 6

7 Dividing by A η, we have On the other h, by Cauchy-Schwarz t = η, v A η sup x C p,a x 0 x, v A x x, v A = x, At x A = A x, t x A t x For given points {θ j }, j = 1,, p, take A as follows: { n+1 } A : t = {t l } C n+1 t l e iθ jl 1 For E n,p, defined in 7, we can write E n,p = π t,,p where v = 1,, 1 he inequality 8 the previous Lemma give n x j ζ n,p π x j e iθ jl Now, given λ 1 < < λ p, define the following exponential polynomial we get x = 0 ζ n,p π l=0 x j e iλ jx n l/n, l=0 C p where θ j = λ j /n n is large enough for {θ j } If, for fixed p {λ j }, we take n, then thanks to 9 hus, we have 0 π 4 p L [0,1] Lemma 3 If λ 1 < < λ p are arbitrary p real numbers x = p x je iλ jx, then 0 πp L [0,1] Many interesting estimates for the exponential sums its derivatives were recently obtained in [9, 6] see also [, 3, 4, 11, 1] Our estimate has a better constant π/ if compared to the bounds in [9, 6] We believe that the duality argument can be used to replace π/ by the optimal constant 1 after the analysis of orthogonal polynomials for measures with Fisher-Hartwig singularities [5] 7

8 4 Sharpness of some results We do not know whether the estimates obtained in heorem 11 are sharp However, we can discuss other results mentioned in the proofs above First, notice that the lemma by Havin is essentially sharp Indeed, suppose we can find f H D such that f ɛ, 1 f ɛ on θ I, I ɛ ake P = 1 f We have P 0 = π 1 P dθ = 1 + Oɛ By the subharmonicity of log P z, we get Oɛ = log P 0 π 1 log P θ dθ = J 1 + J J 1 = π 1 log + P θ dθ, J = π 1 log P θ dθ Since log + x x 1, x > 1, we have by triangle s inequality J 1 1 f 1dθ For J, P >1 J this gives contradiction as ɛ 0 I P >1 f dθ ɛ log 1 f dθ I log ɛ ɛ log ɛ Lemma 41 Consider p < n/ spread p point {θ j }, j = 1,, p evenly on the whole interval [ π, π] Assume that is a trigonometric polynomial, such that hen, deg = n, θ j 1, j = 1,, p Proof he proof is by contradiction Assume that p n ɛp/n, where ɛ is small Let be an interval centered at θ j of length π/p so [ π, π] = p { } are disjoint We have dθ ɛp/n herefore, there is a subset S {1,, p} so that S > p/, j S : dθ ɛ/n Consider j S assume without loss of generality that θ j = 1 We have dθ ɛ/n herefore, by a simple contradiction argument, there is a point θj such that θj < 1/ θ j θ j 8ɛ/n 8

9 hen, the Cauchy-Schwarz inequality yields θ 1/ < θ j θj j = θdθ θ j θj 1/ dθ dθ > n 3ɛ Summing over S gives π dθ dθ > n S π j S 3 ɛ np 64 ɛ On the other h, for every trigonometric polynomial of degree n, we have which gives hat leads to contradiction for ɛ < 1/8 he last Lemma implies that θ j n, np 64ɛ ɛpn D n,p > Cδp/n 1/ Open problem Can one improve α n,p in the heorem 11? In the regime when p is fixed n, we can show that the multiple p in the estimate is sharp D n,p p /n Lemma 4 Given any fixed p, we have nd n,p p, n > np Proof We only need to prove one direction We assume the opposite By 13, this can be written as lim inf nd n n,p Λp, Λp = op, p ake m j = 1/p he points {θ j } will be chosen later By our assumption, there is a subsequence {n} such that there exists a polynomial Q n satisfying the properties: 1 Q n z j < C pλp n, Q n < C Λp n for n > np Denote δ j = 1 Q n z j, j = 1,, p fix them hen, we have min r n < C Λp deg r n n,r nz j =1+δ j n hen, applying the dual formulation, we get inequality x j 1 δ j C Λp n n x j e iθ jl 9 l=0

10 Now, let θ j = λ j /n where λ 1 < < λ p are arbitrary Fix {x j } {λ j } send n Since lim n δ j = 0 for each j = 1,, p, we get 0 < CΛp L [0,1], for every x = x j e iλjx However, it was proved in [6] see also the remark after heorem 7711 in [16] that sup deg p this gives a contradiction for large p 0 L [0,1] Cp 5 Acknowledgement he author is grateful to F Nazarov who gave him references to the results by Havin Halasz explained some of them he author thanks S Kupin for interesting discussion Erdélyi for sharing the preprint [6] with him for many useful remarks he research of SD was supported by NSF grant DMS , RScF , by the IdEx Bordeaux Visiting Professor scholarship 014 References [1] A Aptekarev, S Denisov, D ulyakov, On a problem by Steklov, submitted [] P Borwein, Erdélyi, Nikolskii-type inequalities for shift invariant function spaces Proc Amer Math Soc , no 11, [3] P Borwein, Erdélyi, Polynomials polynomial inequalities, Springer, New-York, 1995 [4] P Borwein, Erdélyi, A sharp Bernstein-type inequality for the exponential sums, J Reine Angew Math 476, 1996, [5] P Deift, A Its, I Krasovsky, Asymptotics of oeplitz, Hankel, oeplitz+hankel determinants with Fisher-Hartwig singularities Ann of Math , no, [6] Erdélyi, Inequalities for exponential sums, preprint [7] Ya L Geronimus, Polynomials orthogonal on the circle on the interval, GIFML, Moscow, 1958 in Russian; English translation: International Series of Monographs on Pure Applied Mathematics, Vol 18 Pergamon Press, New York- Oxford-London-Paris, 1960 [8] G Halász, On the first second main theorem in urán s theory of power sums, Studies in Pure Mathematics, 59 69, Birkhauser, Basel, 1983 [9] G Kós, wo urán type inequalities Acta Math Hungar , no 3, 19 6 [10] M Lachance, E Saff, R Varga, Inequalities for polynomials with a prescribed zero Math Z , no, [11] D Lubinsky, L p Markov-Bernstein inequalities on arcs of the circle J Approx heory , no 1, 1 17 [1] B Nagy, V otik, Bernstein s inequality for algebraic polynomials on circular arcs Constr Approx , no, 3 3 [13] E A Rahmanov, On Steklov s conjecture in the theory of orthogonal polynomials, Matem Sb, 1979, , ; English translation in: Math USSR, Sb, 1980, 36, [14] W Rudin, Real Complex Analysis, McGraw-Hill, 1987 [15] B Simon, Orthogonal polynomials on the unit circle, volumes 1, AMS 005 [16] G Szegő, Orthogonal polynomials, Vol 3, Amer Math Soc Colloq Publ, Providence, Rhode Isl, 1975 University of Wisconsin Madison Mathematics Department 480 Lincoln Dr, Madison, WI, 53706, USA denissov@mathwiscedu 10