Best constants in Markov-type inequalities with mixed Hermite weights

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1 Best constants in Markov-type inequalities with mixed Hermite weights Holger Langenau Technische Universität Chemnitz IWOTA August 15, 2017

2 Introduction We consider the inequalities of the form D ν f C f for all f P n, where P n... algebraic polynomials of degree at most n, D ν f... νth derivative of f. Question What is the best (smallest) constant C such that the inequality holds for all f P n? Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 2

3 Introduction The constant depends on n, ν, and the chosen norms. In this talk, we focus ourselves on the weighted Hermite norm f 2 α = and denote the best constant for f(t) 2 t 2α e t2 dt, D ν f β C f α for all f P n by η n (ν) (α, β). Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 3

4 The story so far the beginning D.I. Mendeleev (1880s): D 4 f for f P 2 on [ 1, 1] A.A. Markov (1889): Df n 2 f for f P n on [ 1, 1] V.A. Markov (1892): D ν f n 2 (n 2 1)(n ) (n 2 (ν 1) 2 ) f (2ν 1)!! for f P n on [ 1, 1] Image source: Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 4

5 The story so far Hilbert space norms E. Schmidt (1944): λ (1) n (0, 0) 2 π n (Laguerre), γ (1) n (0, 0) 1 π n2 (Legendre/Gegenbauer), η (1) n (0, 0) = 2n (Hermite). P. Turán (1960): ( ) λ (1) π 1 n (0, 0) = 2 sin 4n + 2 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 5

6 The story so far higher derivatives L.F. Shampine (1965): λ (2) n (0, 0) n2 µ 2 0, γ n (2) (0, 0) n4, 4µ 2 0 µ (smallest positive solution of 1 + cos µ cosh µ = 0) P. Dörfler (1987/90): bounds for λ (ν) n (0, 0), ν 3 A. Böttcher, P. Dörfler (2009): λ (ν) n (0, 0) K ν n ν, (K ν f)(x) = 1 (ν 1)! Böttcher, Dörfler (2010): x 0 (x y) ν 1 f(y)dy on L 2 (0, 1) λ (ν) n (α, α) L ν,α n ν, γ (ν) n (α, α) Gν,α n 2ν, L ν,α, G ν,α... integral operators on L 2 (0, 1) Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 6

7 The story so far mixed weights Böttcher, Dörfler (2010/2011), motivated by a question of J. Prestin (let ω := β α ν): { λ (ν) 2 ω n (β α)/2 : ω 0, integral, n (α, β) n (ν ω)/2 : ω < 1 2, with γ n (ν) (α, β) L ν,α,β { n ν (L ν,α,β f)(x) = 1 Γ( ω) : ω 0, integral, 2 ω L ν,α,β n ν ω : ω < 1 2, integral, x 0 x α/2 y β/2 (x y) ω 1 f(y)dy. Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 7

8 Determining the best constant Best constant is just the operator norm of the differential operator between the associated spaces. Investigate the matrix representation of this operator with respect to bases of orthonormal polynomials and determine its spectral norm: Ĥ (ν) k k ν (t, α) = j=0 c (ν) jk (α, β)ĥj(t, β), c (ν) jk (α, β): matrix entries, Ĥ j (t, β): jth normalized Hermite polynomial for the parameter β. Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 8

9 Determining the best constant matrix entries I For c (ν) w k (α)h (ν) k k ν (t, α) = j=0 c (ν) jk (α, β)w j(β)h j (t, β): jk (α, β) = w k(α) Γ( k/2 + 1) 2k j+ν w j (β) Γ( j/2 + 1) ( )( ) (j + ν)/2 1/2 β α ν/2 ν k ν /2 + ν k (k j ν)/2 3 F 2 [ ν/2 ν k, (k j ν)/2, β+ j/2 +1/2 β α ν /2 ν k (k j ν)/2+1, j/2 +1/2 ; 1 for k ν j 0 even, zero otherwise, where ν k = kν mod 2 and w k (α) = ( 2 k Γ( k/2 + 1)Γ( k/2 + α + 1/2) ) 1 (normalizing factor for the kth Hermite polynomial). Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 9 ],

10 Determining the best constant matrix entries II The matrix has a chessboard structure. Assuming an even dimension, we can permutate the rows and columns in such a way that a block diagonal matrix is attained. We can work with those two blocks separately. Slightly rewriting of those blocks yields, e.g., c (ν) 2j,2k+ν (α, β) = 2 ν Γ( ν /2 + 1) min{ ν/2,k j} τ=0 Γ(j + β + 1/2) Γ(j + 1) Γ(k + ν /2 + 1) Γ(k + ν /2 + α + 1/2) ( )( j + ν/2 1/2 β + j + τ 1/2 ν /2 τ τ )( β α ν/2 k j τ ) Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 10

11 Determining the best constant In general an exact computation of the constant is not possible. Therefore, we investigate the asymptotic behaviour for n. The methods for determining the spectral norm vary tremendously in dependence of the value of β α, due to the structure of the matrices. Essentially three possible cases: β α N 0, β α (0, ) \ N, β α < 1 2 ( and β α [ 1 2, 0)). Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 11

12 Matrix structure β α N 0 β α (0, )\N β α < 0 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 12

13 Matrix structure β α N 0 β α (0, )\N β α < 0 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 12

14 Matrix structure β α N 0 β α (0, )\N β α < 0 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 12

15 Matrix structure β α N 0 β α (0, )\N β α < 0 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 12

16 Matrix structure β α N 0 β α (0, )\N β α < 0 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 12

17 Integral case, β α ν/2 or β α = 0 Similar to Laguerre/Gegenbauer: banded matrix. Split matrix into diagonals to get upper estimate. Show convergence of (scaled) matrix to Toeplitz matrix to derive the lower bound. Result For integral β α ν/2 or β = α we have for n. η n (ν) (α, β) (2n) (β α+ν)/2 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 13

18 Non-integral case, β α > 0, upper bound To derive an upper bound, we interpolate the operator norm between the parameters shown. We use a lemma which is an application of a theorem of Stein. First, define the operator T : L 2 (R, u(, α)) L 2 (R, u(, γ)), with u(t, α) = t 2α e t2, α, γ > 1/2, like this: T f = D ν f for all f P n (α), T g = 0 for all g P n (α), where P n (α) is the space of all algebraic polynomials, equipped with the norm f 2 α = f(t) 2 u(t, α)dt. Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 14

19 Non-integral case, β α > 0, upper bound Lemma (L., 2016) Under the aforementioned assumptions, we have for any γ > 1 2, T α γ = D ν α γ. If D ν f β C (ν) n f α for all f P n ( ) is true for some β = β satisfying β α Z, and for all β = β + k, k K N, K containing infinitely many numbers, and if C n (ν) (α, β) satisfies C (ν) n (α, β (1 θ)+(β +k)θ) = ( C (ν) n (α, β ) ) 1 θ( C (ν) n (α, β +k) ) θ, θ [0, 1], for all k K, then ( ) holds for all β [β, ). Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 15

20 Non-integral case, β α > 0, lower bound Deriving a lower bound is a bit more complex. The idea is to find a vector which provides a good bound on the norm, is easy to handle. The main contribution to the norm comes from the entries alongside the main diagonal. After flipping the matrix, the largest parts are in the upper left block. Then, we choose the vector in such a way that only the first µ entries are non-zero and let µ grow with the dimension such that the fraction (n µ)/n goes to one. This way we get asymptotically the same result as for the upper bound. Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 16

21 The case β α < 0 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

22 The case β α < 0 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

23 The case β α < 0 n = 5 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

24 The case β α < 0 n = 10 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

25 The case β α < 0 n = 20 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

26 The case β α < 0 n = 40 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

27 The case β α < 0 n = 60 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

28 The case β α < 0 n = 80 Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 17

29 The case β α < 0 Define an integral operator on L 2 (0, 1) by (K N f)(x) = 1 0 k N (x, y)f(y)dy, with k N (x, y) = (A N ) Nx, Ny, with A N being the (modified) matrix. Now we can use a result by Harold Widom which relates the norm of the matrix to the norm of the integral operator: A N = N K N. If we can show that the so defined operator converges in the norm to some integral operator, we get the desired result. Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 18

30 The case β α < 0 For β α < 1 2 the operator ( (0) H ν,α,β f) (x) = 2ν Γ( ν /2 + 1) Γ(α β + ν /2 ) 1 x x β/2 1/4 y α/2+1/4+( ν/2 ν/2 )/2 (y x) α β+ ν/2 +1 ν/2 l=0 ( β l )( ) ( ) β α l x ν/2 l f(y)dy ν /2 l y x is Hilbert-Schmidt and we can show the convergence in the Hilbert-Schmidt norm and thus in the operator norm. Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 19

31 Conclusion We have shown that for α, β > 1/2, the following holds: η n (ν) (α, β) C ν (α, β)n ( β α +ν)/2 with { 2 (β α+ν)/2 : β α 0, C ν (α, β) = 2 (β α ν)/2 max { H (0), H (1) } : β α < 1 2. ν,α,β ν,α,β Holger Langenau Best constants in Markov-type inequalities with mixed Hermite weights 20