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Hankel Operators Plus Orthogonal Polynomials Yield Combinatorial Identities

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Hankel Operators Plus Orthogonal Polynomials Yield Combinatorial Identities Hankel Operators plus Orthogonal Polynomials Yield Combinatorial Identities E. A. Herman, Grinnell College ∗ Abstract: A Hankel operator H = [hi+j] can be factored as H = MM , where M maps a space of L2 functions to the corresponding moment sequences. Furthermore, a necessary and sufficient condition for a sequence to be in the range of M can be expressed in terms of an expansion in orthogonal polynomials. Combining these two results yields a wealth of combinatorial identities that incorporate both the matrix entries hi+j and the coefficients of the orthogonal polynomials. 2000 Mathematics Subject Classification: 47B35, 33D45, 05A19. Keywords: Hankel operators, Hausdorff moments, orthogonal polynomials, combinatorial identities, Hilbert matrices, Legendre polynomials, Fibonacci numbers. 1. Introduction The context throughout this paper will be the Hilbert spaces l2 of square-summable se- 2 quences and Lα( 1, 1) of square-integrable functions on [ 1, 1] with the distribution dα, α nondecreasing.− Their inner products are the usual dot− product for the former and the usual integral inner product for the latter. The Hankel operators in Section 2 are exactly as in [9]. That is, we study the infinite matrices H = [hi+j]i,j≥0, where 1 j hj = x dα(x), j =0, 1, 2,... Z−1 The following result in [9] will be essential: Widom’s Theorem: The following are equivalent: (a) H represents a bounded operator on l2. (b) h = O(1/j) as j . j →∞ (c) α(1) α(x)= O(1 x) as x 1−, α(x) α( 1) = O(1 + x) as x 1+. − − → − − → − 1 In Theorem 1 we show that H = MM ∗, where M : L2 ( 1, 1) l2 is the bounded operator α − → that maps each function f L2 ( 1, 1) to its sequence of moments: ∈ α − 1 k M(f)k = x f(x) dα(x), k =0, 1, 2,... Z−1 We also show that, for u = u l2, we have the formula h ki ∈ ∞ ∗ k M u(x)= ukx Xk=0 In Section 3, we explore the space α of Hausdorff moment sequences given by the range of the operator M. Extending an ideaM in [2], we derive in Theorem 2 the following necessary and sufficient condition for a sequence to belong to . If u = u l2 and p is an Mα h ji ∈ h ki orthogonal sequence of polynomials in L2 ( 1, 1) with α − k p (x)= a xj, a =0, k =0, 1, 2,... (1) k kj kk 6 j=0 X define the sequence c = c by h ki k ck = akjuj, k =0, 1, 2,... (2) j=0 X Then u = u if and only if ∞ c 2/ p 2 < . h ji∈Mα k=0 | k| k kk ∞ Finally, in Theorem 3 we use TheoremsP 1 and 2 to produce combinatorial identities as 2 follows. Suppose v = vk l and u = Hv. If pk and ck are the sequences described above in (1) and (2),h wei obtain ∈ one type of identityh i by equatinh i g two ways of expressing the ck’s. On the one hand, k k ∞ ck = akj(Hv)j = akj hj+ivi j=0 j=0 i=0 X X X On the other hand, the ck’s are Fourier coefficients with respect to the orthogonal poly- 2 ∗ nomials pk for some function f Lα( 1, 1). Since Hv = MM v = Mf, where h ∗ i k ∈ − f(x) = M v(x) = k vkx , we obtain a second formula for ck in terms of the compo- nents vk of v. Other identities come from Parseval’s equation. P In Section 4, we explore several examples. 2 2. A factorization of Hankel Operators The Hankel operators we investigate are given by the infinite matrices H = [hi+j]i,j≥0, where 1 j hj = x dα(x), j =0, 1, 2,... Z−1 and α is nondecreasing on [ 1, 1]. We assume H represents a bounded operator on l2, which means we have the equivalent− conditions on h and α in Widom’s theorem. h ji Theorem 1 : For all f L2 ( 1, 1), let ∈ α − 1 M(f)= uf = uf , where uf = xkf(x) dα(x), k =0, 1, 2,... (3) h ki k Z−1 For all v = v l2, let h ki ∈ ∞ k N(v)= fv, where fv(x)= v x , 1 <x< 1 (4) k − Xk=0 (a) N is a bounded linear operator from l2 into L2 ( 1, 1). α − (b) M is a bounded linear operator from L2 ( 1, 1) into l2. α − (c) M ∗ = N. (d) H = MM ∗. (e) M is injective. 2 Proof : (a) We use the Cauchy criterion to show that fv L ( 1, 1): ∈ α − 2 n 1 n n n n 1 k k j k+j vkx = vkx vjx dα(x)= vkvj x dα(x) −1 ! j=m ! j=m −1 kX=m Z kX=m X kX=m X Z n n n n p = v v h v v k j j+k ≤ | k|| j|k + j j=m j=m kX=m X kX=m X for some p> 0, where the inequality uses the fact that hj = O(1/j). Hence, by Hilbert’s inequality, v xk converges in L2 ( 1, 1). k k α − P 3 To show that N is bounded, we use the change of variables 0 1 g(x) dα(x)= g( y) dβ(y), where g is integrable and β(y)= α( y) (5) − − − Z−1 Z0 Thus 1 1 2 2 2 fv = fv(x) dα(x)+ fv( y) dβ(y) k k | | | − | Z0 Z0 and the matrices A and B defined below are bounded Hankel operators on l2 by Widom’s Theorem. 1 1 j j A = [aj+k], B = [bj+k], where aj = x dα(x), bj = y dβ(y) Z0 Z0 Therefore 1 1 ∞ ∞ ∞ ∞ 2 k+j 2 fv(x) dα(x)= v v x dα(x)= v v a = Av · v A v | | k j k j k+j ≤k kk k 0 0 j=0 j=0 Z Z Xk=0 X Xk=0 X and 1 ∞ ∞ 1 2 k+j fv( y) dβ(y)= v v ( y) dβ(y) | − | k j − Z0 k=0 j=0 Z0 X∞ X∞ = ( 1)kv ( 1)jv b B v 2 − k − j k+j ≤k kk k j=0 Xk=0 X k 2 The fact that k vkx converges in Lα( 1, 1) justifies the above interchanges of limits. Therefore N(v) 2 ( A + B ) v 2. − k Pk ≤ k k k k k k (b) Again using the change of variables (5), we have, for any f L2 ( 1, 1), ∈ α − 1 1 1 uf = xkf(x) dα(x)= xkf(x) dα(x)+ ( 1)kykf( y)dβ(y) k − − Z−1 Z0 Z0 Let ak denote the first of the two integrals on the right and bk denote the second. The 2 following argument is an adaptation of the proof in [5, Theorem 324]. Let v = vk l . k h i ∈ The interchange of limits below is again justified by the fact that k vkx converges in 2 Lα( 1, 1): − ∞ 1 ∞ 1 P k vkak = vkx f(x) dα(x)= Nv(x)f(x) dα(x) 0 0 Xk=0 Z Xk=1 Z 4 Hence, by the Cauchy-Schwarz inequality and part (a), ∞ 2 2 2 2 2 2 vkak N(v) f N v f ≤k k k k ≤k k k k k k k=0 X Therefore, by the so-called converse of H¨older’s inequality [5, Theorem 15], a kh kik ≤ N f . Similarly, b N f , and so M(f) 2 N f . k kk k kh kik≤k kk k k k ≤ k kk k (c) and (d) are straightforward verifications. (e) Suppose M(g) = 0. Then, for all v l2, ∈ ∗ 0= M(g) · v = g, M v = g, fv h i h i Since the functions fv include all polynomials, and since the polynomials are dense in L2 ( 1, 1), it follows that g = 0. α − 2 Although the factorization H = MM ∗ appears to be new, it is closely related to the well- known fact [7, Chapter 2] that H is unitarily equivalent to the integral transform T on 2 Lα( 1, 1) defined by − 1 f(y) T f(x)= dα(y) 1 xy Z−1 − Specifically, one can show that T = M ∗M and hence that H and T are unitarily equivalent whenever M ∗ is injective. For M ∗ to be injective, it suffices to assume that α has infinitely many points of increase with a cluster point in the open interval ( 1, 1). − 3. Orthogonal polynomials and moment sequences In this section we use a complete orthogonal sequence p of polynomials in L2 ( 1, 1) h ki α − to characterize the elements of α, where α denotes the space of moment sequences given by the range of M (see [8]).M AlthoughM these sequences may be finite or infinite, to simplify notation, the details below will be expressed as if the sequences are infinite. The conclusions are valid in all cases, and the finite case will be treated as one of the examples in Section 4. The following theorem extends an idea in [2], where it was applied to the shifted Legendre polynomials. 5 Theorem 2 : Let p be an orthogonal sequence of polynomials in L2 ( 1, 1) with h ki α − k p (x)= a xj, a =0, k =0, 1, 2,... (6) k kj kk 6 j=0 X Let u = u l2 and define c = c by h ji ∈ h ki k ck = akj uj, k =0, 1, 2,..
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