Hankel Operators plus Orthogonal Polynomials. Yield Combinatorial Identities

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1 Hanel Operators plus Orthogonal Polynomials Yield Combinatorial Identities E. A. Herman, Grinnell College Abstract: A Hanel operator H [h i+j ] can be factored as H MM, where M maps a space of L functions to the corresponding moment sequences. urthermore, 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 h i+j and the coefficients of the orthogonal polynomials. Mathematics Subject Classification: 47B35, 33D45, 5A19. Keywords: Hanel operators, Hausdorff moments, orthogonal polynomials, combinatorial identities, Hilbert matrices, Legendre polynomials, ibonacci numbers. 1. Introduction The context throughout this paper will be the Hilbert spaces l of square-summable sequences and L α (, 1 of square-integrable functions on [, 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 Hanel operators in Section are exactly as in [9]. That is, we study the infinite matrices H [h i+j ] i,j, where h j The following result in [9] will be essential: x j dα(x, j, 1,,... Widom s Theorem: The following are equivalent: (a H represents a bounded operator on l. (b h j O(1/j as j. (c α(1 α(x O(1 x as x 1, α(x α( O(1 + x as x +. 1

2 In Theorem 1 we show that H MM, where M : L α (, 1 l is the bounded operator that maps each function f L α(, 1 to its sequence of moments: M(f x f(x dα(x,, 1,,... We also show that, for u u l, we have the formula M u(x u x In Section 3, we explore the space M α of Hausdorff moment sequences given by the range of the operator M. Extending an idea in [], we derive in Theorem the following necessary and sufficient condition for a sequence to belong to M α. If u u j l and p is an orthogonal sequence of polynomials in L α(, 1 with p (x a j x j, a,, 1,,... (1 define the sequence c c by c a j u j,, 1,,... ( Then u u j M α if and only if c / p <. inally, in Theorem 3 we use Theorems 1 and to produce combinatorial identities as follows. Suppose v v l and u Hv. If p and c are the sequences described above in (1 and (, we obtain one type of identity by equating two ways of expressing the c s. On the one hand, c a j (Hv j a j h j+i v i i On the other hand, the c s are ourier coefficients with respect to the orthogonal polynomials p for some function f L α (, 1. Since Hv MM v Mf, where f(x M v(x v x, we obtain a second formula for c in terms of the components v of v. Other identities come from Parseval s equation. In Section 4, we explore several examples.

3 . A factorization of Hanel Operators The Hanel operators we investigate are given by the infinite matrices H [h i+j ] i,j, where h j x j dα(x, j, 1,,... and α is nondecreasing on [, 1]. We assume H represents a bounded operator on l, which means we have the equivalent conditions on h j and α in Widom s theorem. Theorem 1: or all f L α(, 1, let M(f u f u f, where uf x f(x dα(x,, 1,,... (3 or all v v l, let N(v f v, where f v (x v x, 1 < x < 1 (4 (a N is a bounded linear operator from l into L α(, 1. (b M is a bounded linear operator from L α (, 1 into l. (c M N. (d H MM. (e M is injective. Proof : (a We use the Cauchy criterion to show that f v L α (, 1: n ( 1 n ( n n n v x x v j x m mv j dα(x v v j jm m jm n n n n p v v j h j+ v v j + j m jm m jm x +j dα(x for some p >, where the inequality uses the fact that h j O(1/j. Hence, by Hilbert s inequality, v x converges in L α(, 1. 3

4 To show that N is bounded, we use the change of variables g(x dα(x Thus f v g( y dβ(y, where g is integrable and β(y α( y (5 f v (x dα(x + f v ( y dβ(y and the matrices A and B defined below are bounded Hanel operators on l by Widom s Theorem. A [a j+ ], B [b j+ ], where a j Therefore f v (x dα(x x j dα(x, b j y j dβ(y v v j x +j dα(x v v j a +j Av v A v and f v ( y dβ(y v v j ( y +j dβ(y ( v ( j v j b +j B v The fact that v x converges in L α (, 1 justifies the above interchanges of limits. Therefore N(v ( A + B v. (b Again using the change of variables (5, we have, for any f L α(, 1, u f x f(x dα(x x f(x dα(x + ( y f( ydβ(y Let a denote the first of the two integrals on the right and b denote the second. The following argument is an adaptation of the proof in [5, Theorem 34]. Let v v l. The interchange of limits below is again justified by the fact that v x converges in L α (, 1: v a v x f(x dα(x Nv(xf(x dα(x 1 4

5 Hence, by the Cauchy-Schwarz inequality and part (a, v a N(v f N v f Therefore, by the so-called converse of Hölder s inequality [5, Theorem 15], a N f. Similarly, b N f, and so M(f N f. (c and (d are straightforward verifications. (e Suppose M(g. Then, for all v l, M(g v g, M v g, f v Since the functions f v include all polynomials, and since the polynomials are dense in L α (, 1, it follows that g. Although the factorization H MM appears to be new, it is closely related to the wellnown fact [7, Chapter ] that H is unitarily equivalent to the integral transform T on L α (, 1 defined by f(y Tf(x 1 xy dα(y Specifically, one can show that T M M and hence that H and T are unitarily equivalent whenever M is injective. or M to be injective, it suffices to assume that α has infinitely many points of increase with a cluster point in the open interval (, Orthogonal polynomials and moment sequences In this section we use a complete orthogonal sequence p of polynomials in L α (, 1 to characterize the elements of M α, where M α denotes the space of moment sequences given by the range of M (see [8]. Although 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 [], where it was applied to the shifted Legendre polynomials. 5

6 Theorem : Let p be an orthogonal sequence of polynomials in L α (, 1 with p (x a j x j, a,, 1,,... (6 Let u u j l and define c c by c a j u j,, 1,,... (7 Then u M α if and only if urthermore, when (8 holds, then u M(f, where c p < (8 f(x c p p (x (9 Proof : If u u M α, then there exists f L α(, 1 such that u M(f; that is, u x f(x dα(x,, 1,,... Hence, since p is a complete orthogonal sequence in L α (, 1 (see [8], f(x b p (x where b 1 f(xp p (x dα(x Therefore, by equations (7 and (6, c a j x j f(x dα(x f(xp (x dα(x p b So, by Parseval s equation, c p p b f < urthermore, (9 holds since b c / p. 6

7 On the other hand, if u u l and c / p <, let f(x c p p (x Since c / p <, the above series converges in L α (, 1 and c f(xp (x dα(x,, 1,,... This equation can also be written, using (6 and (7, as a j u j which implies, since a, u j f(x a j x j dα(x,, 1,,... f(xx j dα(x, j, 1,,... Therefore u u j M(f M α. inally, we use Theorems 1 and to produce combinatorial identities corresponding to each choice of α. Theorem 3: Suppose α is a nondecreasing function on [, 1] satisfying and let α(1 α(x O(1 x as x 1, α(x α( O(1 + x as x + h j x j dα(x, j, 1,,... Suppose also that p is an orthogonal sequence of polynomials in L α (, 1 with p (x a j x j, a,, 1,,... and x b j p j (x,, 1,,... 7

8 Then, for all non-negative integers and m, { p b m if m a j h j+m if m < (1 and m i urthermore, for all v v l, 1 p i 1 p i i a ij h j+m a i δ m (11 a j h j+i v i v v j h +j (1 and p v j b j j v v j h +j (13 Proof : Let H [h i+j ] i,j, v v l, and u Hv MM v. Then u M α, and we use Theorem to find two expressions for the coefficients c in equation (7. On the one hand, c a j u j a j (Hv j a j h j+i v i (14 On the other hand, i c f(xp (x dα(x where f(x c p p (x Using Theorem 1, we may also write the last expression for c in terms of the components of v. Note that Hv M(f and Hv MM v M(f v. Since M is injective, we have f f v a. e. and so c v j x j p (x dα(x v j i j b ji p i (xp (x dα(x p v j b j (15 j 8

9 Equating formulas (14 and (15 for c and letting v δ m, we obtain the identity (1. Then, solving for b m, (m in (1 and substituting into m i b ma i δ mi (m i, we obtain the identity (11. urthermore, by Parseval s equation and MM H, we have c p f v M v M v, M v Hv v v v j h +j Substituting formulas (14 and (15 for c in this result yields the identities (1 and (13. Another perspective on Theorem 3 The analysis below shows that, with a different set of assumptions, the identities in Theorem 3 are quite natural. Thus, some instances of them are liely to be nown. Suppose, instead of a distribution dα as in Theorem 3, we are given the following assumptions about a matrix H and a sequence of polynomials p. Let H [h i+j ] i,j be any infinite Hanel matrix with real entries; that is, do not assume h j O(1/j as j. Also let p (x a j x j, a,, 1,,... be a sequence of polynomials with real coefficients that are orthogonal in the sense that { m if m a i h i+j a mj (16 if m i Note that, under the assumptions of Theorem 3, property (16 is the statement that p, p m if m and p, p. So we will denote the left side of (16 by p, p m even though the matrix H does not necessarily generate a true inner product. Then (16 can be written as AHA D, where A denotes the lower-triangular matrix whose (, j entry, for j, is a j and D denotes the diagonal matrix whose th diagonal entry is p, p. Since the diagonal entries of A are nonzero, A has a formal inverse B. Hence AHA D implies AH DB, which is the identity (1. Of course, p must be interpreted as p, p, which is nonzero but possibly negative. Identity (11 follows from (1, as in the proof of Theorem 3. The left side of identity (1 can be expressed as D / AHv, where D / is diagonal and the norm is the usual l norm. However, now we must assume that the sequence v has only finitely many nonzero terms. Then (1 may be derived as follows: D / AHv (D / AHv (D / AHv (D DB v (AHv v Hv 9

10 Identity (13 may be derived similarly. 4. Examples In the examples below, some of the instances of identity (1 have been independently confirmed using the WZ methods in [6]. On the other hand, instances of identities (1 and (13 do not yet seem to be susceptible of confirmation by such algorithmic methods. Example 1 (Hilbert matrices and binomial coefficients: or any r >, let α(x x r if x 1 and α(x if x. Then h j r/(j +r and so H [r/(i+j+r] i,j. In [3], Berg shows that the corresponding orthogonal polynomials may be written as p (x ( ( + j + r 1 ( j x j j j It follows that p r/( + r and x 1 ( + r ( +r ( j ( + r j (j + rp j (x Hence equations (1, (11, (1 and (13 become, respectively, ( ( ( + j + r 1 ( j m+r j j + m + r ( m ( m+r (m + r m i i ( ( ( ( i i + j + r 1 i i + + r 1 ( j+ (i + 1 δ m j i i j + m + r where v j l. ( ( + j + r 1 ( j v i v v j ( + r j j + i + r + j + r i ( j+r j vj ( + r (j + r ( v v j j+r + j + r j j m Example (Legendre polynomials: We begin by considering all Jacobi polynomials: p (α,β (x (α (, α + β + + 1; α + 1; (1 x/! 1

11 where 1 denotes the Gauss hypergeometric series, (α + 1 denotes a shifted factorial, and α, β >. The Jacobi polynomials are orthogonal on [, 1] relative to the weight function w(x (1 x α (1 + x β, and p (α,β α+β+1 Γ(α + + 1Γ(β + + 1! Γ(α + β + + 1(α + β Using Euler s integral representation of Gauss hypergeometric series [1, Theorem..1], we find the moments to be h j x j (1 x α (1 + x β dx ( j! ( j 1 ( α, 1 + j; + β + j; + 1 ( β, 1 + j; + α + j; (1 + β 1+j (1 + α 1+j Rather than writing out identities (1, (11, (1, and (13 for this general case, we write the more compact versions for the special case α β. Then h j /(j +1 if j is even, h j if j is odd, and p p (, is the th Legendre polynomial; hence p /(+1 and ( ( ( ( j/ + j p (x j x j x j+ even j+ even!(j + 1 ( j/ ( 1 ( j!(j + + 1!!p j(x Note: The double factorial n!! is defined to be 1 when n and, for positive integers n, n!! n(n (n 4 (1 or where the terminal factor in this product is 1 when n is odd and when n is even. Hence equations (1, (11, (1, and (13 become, respectively, ( ( { + j ( ( j/ m! j (j + m + 1 if m + even, m (m / ( 1 (m!(+m+1!! otherwise j+ even j+m even m i i+ even i i+j even j+m even ( ( i (j+/ i i i j (j + m ( ( ( i + j i i + i δ m i i

12 ( ( ( j ( j/ v i j i (j + i + 1 j+ even j+i even ( + 1 j!v j (j / ( 1 (j!( + j + 1!! j where v j l. j+ even j+ even j+ even v i v j + j + 1 v i v j + j + 1 Example 3 (ibonacci numbers and ibonomial coefficients: We begin by considering all little q-jacobi polynomials [4, Section 7.3]: p (x; a, b; q n (q ; q j (abq +1 ; q j (xq j (q; q j (aq; q j where (a; q j denotes the q-shifted factorial. We assume that q < 1, aq < 1, and b 1, which suffices to show that these polynomials satisfy the orthogonality property where i p (q i ; a, b; qp m (q i ; a, b; q (bq; q i (q; q i (aq i δ m h m (a, b; q h m (a, b; q (abq; q m(1 abq m+1 (aq; q m (aq; q (aq m (q; q m (1 abq(bq; q m (abq ; q We can express the orthogonality property (17 in terms of an integral on [, 1] as follows. Let p (x p (φx; a, b; q, where φ > 1, and define the discrete signed measure µ by (17 µ i (bq; q i (q; q i (aq i δ q i /φ (18 Then (17 becomes The corresponding moments are h j p (xp m (x dµ(x x j dµ(x i 1 δ m h m (a, b; q (bq; q i (q; q i (aq i ( q i /φ j

13 When b 1, this sum is a geometric series: h j 1 φ j i ( aq j+1 i 1 φ j (1 aq j+1 The following interesting special case was found by Berg [3]. or any α N, let b 1, φ 1 + 5, q , a qα, and µ α (1 q α q iα δ q i /φ, α 1,, 3,... i Note that µ α is the signed measure (18 multiplied by 1 q α so it becomes a measure with total mass 1. Berg shows that h j α / α+j, j, 1,,..., that p (α (x p (φx; q α, 1; q ( ( ( j (j α + + j 1 j x j and p (α (xp m (α (x dµ α (x δ m ( α α α+ where j denotes the jth ibonacci number (with, 1 1 and ( ibonomial coefficient j i+1 i1 i. j denotes the When α is even, µ α is a probability measure that, together with the polynomials p (α (x, satisfies the hypotheses of Theorem 3. Note that p (α α / α+. Hence, equations (11 and (1 become m i i ( ( i(j+ (j ( i j α+ i ( ( i α + i + j 1 i ( ( ( j (j α + + j 1 j ( α + i + 1 vi α+j+i i α+i α+j+m δ m v v j α++j where α, 4, 6,... and v j l. urthermore, the analysis following Theorem 3 shows that both identities also hold for α 1, 3, 5,..., provided v j for all but finitely many values of j. or such α, note that p (α, p (α < when is odd. 13

14 Example 4 (finite dimensional spaces: Let α be a step function with positive jumps J at the distinct points z (, 1,, 1,..., r. Then L α(, 1 has dimension r + 1, and r h j z j J, j, 1,,... (19 In what follows, we will frequently use the following submatrices of the infinite Hanel matrix H [h i+j ] i,j : Let H denote the upper-left ( + 1 ( + 1 submatrix of H. rom equation (19, we see that H V D r V T,, 1,..., r ( where V is rows through of the Vandermonde matrix whose row 1 is [z, z 1,...,z r ], and where D r is the diagonal matrix with diagonal entries J, J 1,...,J r. In seeing an orthogonal basis of polynomials for L α(, 1, we choose their leading coefficients to be 1: p (x x + a j x j,, 1,..., r The remaining coefficients are then uniquely determined; we compute them as follows. Let h(p, q denote the vector with components h p,...,h q, and let a( denote the vector with components a,...,a,. We claim that which may also be written H a( h(, 1, 1,...,r (1 h m+j a j h m+, 1,...,r, m,..., 1 Since H is nonsingular when, 1,...,r, equation (1 uniquely determines the coefficients a j. One may confirm (1 by a straightforward calculation to show that the resulting polynomials are mutually orthogonal. By Cramer s rule, we then have the explicit formula p (x x 1 det(h (j,h(, 1x j,, 1,..., r ( deth where M(j,w denotes the matrix M with its jth column replaced by the vector w. A similar calculation shows that p deth deth,, 1,..., r 14

15 where we assign the value 1 to deth. Next we express each monomial x m as a linear combination of the basis polynomials p, p 1,...,p r : r x m b m p (x, m, 1,,... By equations (1 and (, b m 1 p a j h j+m ( deth 1 h +m det(h (j,h(, 1h j+m deth deth ( 1 (deth h +m det(h (j,h(, 1h j+m deth det(h (,h(m, m + deth To confirm the last step above, evaluate det(h (,h(m, m+ by the Laplace expansion down the last column of the matrix. or the jth entry of that column, j, 1,..., 1, the expansion formula assigns the sign ( +j. We must also move the vector h(, from column j to column 1, and the corresponding transposes change the sign of the determinant by the factor ( j. Thus, the overall sign factor is ( +j+ j. Having confirmed the above computation of b m, we conclude x m r det(h (,h(m, m + deth p (x, m, 1,,... Hence equation (13 becomes r 1 (deth (deth v j det(h (,h(j, j + j v v j h +j where v j l. Acnowledgment: I than my colleague C. rench for suggesting the analysis following Theorem 3. 15

16 References 1. G. E. Andrews, R. Asey, R. Roy, Special unctions, Cambridge University Press, R. Asey, I. J. Schoenberg, A. Sharma, Hausdorff s moment problem and expansions in Legendre polynomials, J. Math. Anal. Appl., v. 86 (198, C. Berg, ibonacci numbers and orthogonal polynomials, J. Comput. Appl. Math. (to appear. 4. G. Gasper and M. Rahman, Basic Hypergeometric Series, nd ed., Cambridge University Press, G. H. Hardy, J. E. Littlewood, G. Pólya, Inequalities, Cambridge University Press, M. Petovše, H. Wilf, D. Zeilberger, A B, A. K. Peters, Ltd., S. C. Power, Hanel Operators on Hilbert Space, Pitman Publishing, Inc., G. Szegö, Orthogonal Polynomials, 4th ed., Amer. Math. Soc., H. Widom, Hanel Matrices, Trans. Amer. Math. Soc., v. 11 (1966,

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