On the functional equation for classical orthogonal polynomials on lattices
aa r X i v : . [ m a t h . C A ] J a n ON THE FUNCTIONAL EQUATION FOR CLASSICALORTHOGONAL POLYNOMIALS ON LATTICES
K. CASTILLO, D. MBOUNA, AND J. PETRONILHO
Abstract.
Necessary and sufficient conditions for the regularity of solutionsof the functional equation appearing in the theory of classical orthogonal poly-nomials on lattices are stated. Moreover, the functional Rodrigues formula anda closed formula for the recurrence coefficients are presented. Introduction
Let P be the vector space of all polynomials with complex coefficients and let P ∗ be its algebraic dual. P n denotes the space of all polynomials with degree lessthan or equal to n . Define P − := { } . A simple set in P is a sequence ( P n ) n ≥ such that P n ∈ P n \ P n − for each n . A simple set ( P n ) n ≥ is called an orthogonalpolynomial sequence (OPS) with respect to u ∈ P ∗ if h u , P n P k i = h n δ n,k ( n, k = 0 , , . . . ; h n ∈ C \ { } ) , where h u , f i is the action of u on f ∈ P . u is called regular, or quasi-definite, ifthere exists an OPS with respect to it. It is well known that u is regular if andonly if det (cid:2) u i + j (cid:3) ni,j =0 = 0 for each n = 0 , , . . . , where u n := h u , z n i . In 1940 Ya.L. Geronimus proved the following result [7, Theorem II]: A sequence of polynomials and the sequence of its derivatives are or-thogonal with respect to sequences of moments ( u n ) n ≥ and ( v n ) n ≥ ,respectively, if and only if (1.1) ( na + d ) u n +1 + ( nb + e ) u n + ncu n − = 0 for each n , where a, b, c, d, e are complex numbers such that na + d = 0 and det (cid:2) u i + j (cid:3) ni,j =0 = 0 , and (1.2) v n = au n +2 + bu n +1 + cu n . Date : February 2, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Functional equation, regular functional, classical orthogonal polyno-mials, lattices, Racah polynomials, Askey-Wilson polynomials. Recall that given a sequence of complex numbers ( u n ) n ≥ , ( P n ) n ≥ is said to be orthogonalwith respect to ( u n ) n ≥ if it is an OPS with respect to the functional u ∈ P ∗ defined by h u , f i := n X k =0 a k u k , f ( z ) = n X k =0 a k z k ∈ P . Given u ∈ P ∗ and g ∈ P , the derivative of u and the left multiplication of u by g are the functionals Du ∈ P ∗ and g u ∈ P ∗ defined by h Du , f i := −h u , f ′ i , h g u , f i := h u , f g i ( f ∈ P ) . This allows us to rewrite the difference equation (1.1) as a functional equation (1.3) D ( φ u ) = ψ u , where φ ( z ) = az + bz + c and ψ ( z ) = dz + e , and (1.2) as v = φ u , being u and v the linear functionals on P given by h u , z n i = u n and h v , z n i = v n . It is worthmentioning that Geronimus’ result was motivated by Hahn’s characterization ofthe classical OPS of Hermite, Laguerre, Jacobi, and Bessel —the only OPS suchthat their sequences of derivatives are also OPS [8]. We call a functional u ∈ P ∗ classical in Hahn’s sense , or, simply, D − classical , if the corresponding OPS is suchthat the associated sequence of derivatives is also an OPS. Accordingly, Geronimus’theorem can be rewritten as follows : Theorem A. u ∈ P ∗ is D − classical if and only if there exist φ ∈ P and ψ ∈ P such that u satisfies (1.3) and the conditions (1.4) na + d = 0 , H n := det (cid:2) u i + j (cid:3) ni,j =0 = 0 ( n = 0 , , , . . . ) hold, where a := φ ′′ / , d := ψ ′ , and u n := h u , z n i . The condition na + d = 0 for every n means that ( φ, ψ ) is an admissible pair [16].Note that, concerning the applicability of Theorem A, although it is trivial to checkthe admissibility condition for a given pair ( φ, ψ ) ∈ P × P , the same does notholds (in general) for the regularity condition H n = 0, because the order of H n grows with n . Theorem B in bellow improves Theorem A, showing that conditions(1.4) may be replaced by rather simple ones —see (1.6) in bellow—, as well as howto compute the monic OPS ( P n ) n ≥ with respect to u using only the pair ( φ, ψ ).This is achieved from the explicit formulas for the coefficients B n and C n appearingin the three-term recurrence relation (TTRR) for ( P n ) n ≥ , namely(1.5) zP n ( z ) = P n +1 ( z ) + B n P n ( z ) + C n P n − ( z ) ( n = 0 , , . . . ) , P − ( z ) = 0 . Theorem B. u ∈ P ∗ \{ } is D − classical if and only if there exist polynomials φ ( z ) := az + bz + c and ψ ( z ) := dz + e such that (1.3) holds and (1.6) na + d = 0 , φ (cid:18) − nb + e na + d (cid:19) = 0 ( n = 0 , , . . . ) . Under such conditions, the monic OPS ( P n ) n ≥ with respect to u fulfils (1.5) with (1.7) B n = ne n − d n − − ( n + 1) e n d n , C n +1 = − ( n + 1) d n − d n − d n +1 φ (cid:16) − e n d n (cid:17) ( n = 0 , , . . . ) Usually (1.3) is called Pearson (or Pearson-type) functional (or distributional) equation.Without wanting to get into nitpicking about the nature of the name, it seems appropriate tocall (1.3) Geronimus-Pearson (functional) equation. As pointed out in [3], this characterization was indeed stated before by Sonine [21]. As an application of this result, Geronimus also proved that a monic OPS ( P n ) n ≥ isD − classical if P n = n +1 P ′ n +1 + b n P ′ n + c n P ′ n − ( n = 0 , , . . . ) , P − = 0 , where b n and c n are complex numbers (see [7, (42)]). Of course, the converse of this sentence wasknown long before. N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 3 where d n := na + d and e n := nb + e . Moreover, the following functional Rodriguesformula holds: P n u = k n D n (cid:0) φ n u (cid:1) , k n := n − Y j =0 d − n + j − ( n = 0 , , . . . ) . Theorem B was stated in [13]. Observe that conditions (1.6) mean that ( φ, ψ )is an admissible pair and ψ + nφ ′ ∤ φ for each n = 0 , , . . . . It is well knownthat if there exist u ∈ P ∗ and nonzero polynomials φ ∈ P and ψ ∈ P suchthat (1.3) holds, then the regularity of u implies that ( φ, ψ ) is an admissible pair(see e.g. [16, 13]). Therefore, the above definition of D − classical functional isequivalent to the following statement: u ∈ P ∗ \{ } is D − classical if and only ifit is regular and there exist nonzero polynomials φ ∈ P and ψ ∈ P such that u satisfies (1.3). This statement is indeed the definition of D − classical functionaladopted nowadays within the algebraic theory of orthogonal polynomials, developedby Maroni [14, 15, 16] (see also [20]).Recently, an analogue of Theorem B in the framework of OPS with respect to thediscrete Hahn operator has been proved in [2]. In order to give here the statementof this analogue, we need to recall several definitions. Given complex numbers q and ω subject to the conditions(1.8) | q − | + | ω | 6 = 0 , q (cid:8) , e ijπ/n | ≤ j ≤ n − n = 2 , , . . . (cid:9) , the (ordinary) Hahn operator D q,ω : P → P , studied by W. Hahn in [9], is D q,ω f ( x ) := f ( qx + ω ) − f ( x )( q − x + ω ( f ∈ P ) .D q,ω induces an operator on the dual space, D q,ω : P ∗ → P ∗ , given by (see [5]) h D q,ω u , f i := − q − h u , D ∗ q,ω f i ( u ∈ P ∗ , f ∈ P ) , where D ∗ q,ω := D /q, − ω/q . These definitions are on the basis of the following notionof classical OPS: u ∈ P ∗ is ( q, ω ) − classical if it is regular and there exist nonzeropolynomials φ ∈ P and ψ ∈ P , such that (1.9) D q,ω ( φ u ) = ψ u . Define also operators L q,ω : P → P and L q,ω : P ∗ → P ∗ by (see [5]) L q,ω f ( x ) := f ( qx + ω ) , h L q,ω u , f i := q − h u , L ∗ q,ω f i (cid:0) f ∈ P , u ∈ P ∗ (cid:1) , where L ∗ q,ω := L /q, − ω/q . Finally, recall the definition of the q − bracket:[ α ] q := q α − q − , if q = 1 α , if q = 1 ( α, q ∈ C ) . The ( q, ω ) − analogue of Theorem B, stated in [2, Theorem 1.2], reads as follows: The explicit formulas (1.7) appeared earlier in Suslov’s article [22], where they have beenderived by a different method, although the regularity conditions for u had not been discussedtherein. K. CASTILLO, D. MBOUNA, AND J. PETRONILHO
Theorem C.
Fix q, ω ∈ C fulfilling (1 . . u ∈ P ∗ \{ } is ( q, ω ) − classical if andonly if there exist polynomials φ ( z ) := az + bz + c and ψ ( z ) := dz + e such that u satisfies the functional equation (1.9) and the conditions d n = 0 , φ (cid:16) − e n d n (cid:17) = 0 ( n = 0 , , , . . . ) hold, where d n ≡ d n ( q ) := dq n + a [ n ] q and e n ≡ e n ( q, ω ) := eq n + ( ωd n + b )[ n ] q .Under these conditions, the monic OPS ( P n ) n ≥ ≡ ( P n ( · ; q, ω )) n ≥ with respect to u satisfies the TTRR (1.5) , where B n = ω [ n ] q + [ n ] q e n − d n − − [ n + 1] q e n d n , C n +1 := − q n [ n + 1] q d n − d n − d n +1 φ (cid:16) − e n d n (cid:17) ( n = 0 , , . . . ) . In addition, the Rodrigues formula P n u = k n D n /q, − ω/q (cid:16) Φ( · ; n ) L nq,ω u (cid:17) ( n = 0 , , . . . ) holds in P ∗ , where k n := q n ( n − / n − Y j =0 d − n + j − , Φ( x ; n ) := n Y j =1 φ (cid:0) q j x + ω [ j ] q (cid:1) . At this stage, a natural question arises, asking for analogues of theorems B andC for OPS on lattices in the sense of Nikiforov, Suslov, and Uvarov [18] (see also[3, 10, 11, 1]). The aim of this work is answering to this question. The structureof the paper is the following. In Section 2 we review some basic definitions andresults focusing on the theory of OPS on lattices. Section 3 contains a functionalRodrigues formula for classical OPS on lattices. In Section 4 we state our mainresults, giving the analogues of theorems B and C for OPS on lattices (for linear, q − linear, quadratic, and q − quadratic lattices). Finally, as a straightforward appli-cation of the main results, in Section 5 we revisit the Racah polynomials and theAskey-Wilson polynomials, computing their recurrence coefficients directly fromthe functional equation fulfilled by the associated regular functional.2. Preliminary results on lattices
In this section we review the definition of lattice. In addition, we derive a Leibnizformula (on lattices) for the left multiplication of a functional by a polynomial, aswell as some preliminary results needed.2.1.
Definitions and basic properties.
A lattice is a mapping x ( s ) given by(2.1) x ( s ) := ( c q − s + c q s + c if q = 1 , c s + c s + c if q = 1( s ∈ C ), where q > c j (1 ≤ j ≤
6) are (complex) constants, that maydepend on q , such that ( c , c ) = (0 ,
0) if q = 1, and ( c , c , c ) = (0 , ,
0) if q = 1.In the case q = 1, the lattice is called quadratic if c = 0, and linear if c = 0; andin the case q = 1, it is called q − quadratic if c c = 0, and q − linear if c c = 0 (cf.[3]). Notice that x (cid:0) s + (cid:1) + x (cid:0) s − (cid:1) αx ( s ) + β , N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 5 where α and β are given by(2.2) α := q / + q − / , β := ( (1 − α ) c if q = 1 , c / q = 1 . The lattice x ( s ) fulfills (cf. [3]): x ( s + n ) + x ( s )2 = α n x n ( s ) + β n ,x ( s + n ) − x ( s ) = γ n ∇ x n +1 ( s )( n = 0 , , . . . ), where x µ ( s ) := x (cid:0) s + µ (cid:1) , ∇ f ( s ) := f ( s ) − f ( s − α n ) n ≥ ,( β n ) n ≥ , and ( γ n ) n ≥ are sequences of numbers generated by the following systemof difference equations α = 1 , α = α , α n +1 − αα n + α n − = 0(2.3) β = 0 , β = β , β n +1 − β n + β n − = 2 βα n (2.4) γ = 0 , γ = 1 , γ n +1 − γ n − = 2 α n (2.5)( n = 1 , , . . . ). The explicit solutions of these difference equations are α n = q n/ + q − n/ , (2.6) β n = β (cid:18) q n/ − q − n/ q / − q − / (cid:19) if q = 1 β n if q = 1 , (2.7) γ n = q n/ − q − n/ q / − q − / if q = 1 n if q = 1 . (2.8)These formulas may be easily checked (alternatively, see [3]). We point out thefollowing relations: γ n +1 − αγ n + γ n − = 0 , (2.9) α n + γ n − = αγ n , (2.10) (2 α − α n + ( α − γ n − = αα n +1 , (2.11) γ n = 2 α n γ n , (2.12) α n + ( α − γ n = α n = 2 α n − , (2.13) α n − − αα n = (1 − α ) γ n , (2.14) α + α n γ n = α n − γ n +1 , (2.15) 1 + α n +1 γ n = α n γ n +1 (2.16)( n = 0 , , . . . ), with the conventions α − := α and γ − := −
1, consistently with(2.6) and (2.8).
Definition 2.1.
Let x ( s ) be a lattice given by (2.1) . The x − derivative operator on P , D x , and the x − average operator on P , S x , are the operators on P defined for K. CASTILLO, D. MBOUNA, AND J. PETRONILHO each f ∈ P so that deg(D x f ) = deg f − , deg(S x f ) = deg f , and D x f ( x ( s )) = f (cid:0) x ( s + ) (cid:1) − f (cid:0) x ( s − ) (cid:1) x ( s + ) − x ( s − ) , (2.17) S x f ( x ( s )) = f (cid:0) x (cid:0) s + (cid:1)(cid:1) + f (cid:0) x (cid:0) s − (cid:1)(cid:1) . (2.18)The relations (2.17) and (2.18) appear in [18, (3.2.4)-(3.2.5)] up to a shift in thevariable s . The operators D x and S x on P induce two operators on the dual space P ∗ , namely D x : P ∗ → P ∗ and S x : P ∗ → P ∗ , via the following definition (cf. [6]): Definition 2.2.
Let x ( s ) be a lattice given by (2.1) . For each u ∈ P ∗ , the func-tionals D x u ∈ P ∗ and S x u ∈ P ∗ are defined by (2.19) h D x u , f i := −h u , D x f i , h S x u , f i := h u , S x f i ( f ∈ P ) . We call D x u the x − derivative of u and S x u the x − average of u . Hereafter, z := x ( s ) being a lattice given by (2.1), we consider two fundamentalpolynomials, U and U , introduced in [6], defined by U ( z ) := ( α − z + β ( α + 1) , (2.20) U ( z ) := ( α − z + 2 β ( α + 1) z + δ , (2.21)where δ ≡ δ x is a constant with respect to the lattice, given by (2.22) δ := (cid:18) x (0) + x (1) − β ( α + 1)2 α (cid:19) − x (0) x (1) . A straightforward computation shows that(2.23) δ = ( ( α − (cid:0) c − c c (cid:1) if q = 1 , c − c c if q = 1and(2.24) U ( z ) = ( ( α − (cid:0) z − c (cid:1) if q = 1 , c if q = 1 . Hence(2.25) U ( z ) = ( ( α − (cid:0) ( z − c ) − c c (cid:1) if q = 1 , c ( z − c ) + c if q = 1 . Finally, we recall the following useful relations which can be easily proved (see [6]):D x U = α − , S x U = α U , (2.26) D x U = 2 α U , S x U = α U + U . (2.27) The constant δ x appears in [6] without an explicit expression; a full expression, different fromthe ones provided here, is given in [12]. N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 7
Properties of the x − derivative and x − average operators. We start bypointing out some useful properties.
Lemma 2.1.
Let f, g ∈ P and u ∈ P ∗ . Then the following properties hold: D x (cid:0) f g (cid:1) = (cid:0) D x f (cid:1)(cid:0) S x g (cid:1) + (cid:0) S x f (cid:1)(cid:0) D x g (cid:1) , (2.28) S x (cid:0) f g (cid:1) = (cid:0) D x f (cid:1)(cid:0) D x g (cid:1) U + (cid:0) S x f (cid:1)(cid:0) S x g (cid:1) , (2.29) S x D x f = α D x S x f − D x (cid:0) U D x f (cid:1) , (2.30) S x f = α − S x (cid:0) U D x f (cid:1) + α − U D x f + f , (2.31) f S x g = S x (cid:16)(cid:0) S x f − α − U D x f (cid:1) g (cid:17) − α − U D x (cid:0) g D x f (cid:1) , (2.32) f D x g = D x (cid:16)(cid:0) S x f − α − U D x f (cid:1) g (cid:17) − α − S x (cid:0) g D x f (cid:1) , (2.33) D x ( f u ) = (cid:0) S x f − α − U D x f (cid:1) D x u + α − (cid:0) D x f (cid:1) S x u , (2.34) S x ( f u ) = (cid:0) S x f − α − U D x f (cid:1) S x u + α − (cid:0) D x f (cid:1) D x ( U u ) , (2.35) S x ( f u ) = (cid:0) α U − α − U (cid:1) (D x f ) D x u + (cid:0) S x f + α − U D x f (cid:1) S x u , (2.36) D x (cid:0) U u (cid:1) = α S x u + D x (cid:0) U S x u (cid:1) − α u , (2.37) D x (cid:0) U u (cid:1) = (2 α − α − ) S x u + α − U D x S x u − α u , (2.38) D x S x u = α S x D x u + D x (cid:0) U D x u (cid:1) . (2.39) Proof.
The reader may encounter properties (2.28)–(2.35) in [12, Propositions 5–7].To prove (2.36), set f = U in (2.34) and then use (2.27) to obtain D x ( U u ) = (cid:0) α U − U (cid:1) D x u + 2 U S x u . Replacing this expression in the right-hand side of (2.35) we obtain (2.36). Next,taking arbitrarily f ∈ P , we have h D x (cid:0) U u (cid:1) , f i = h u , U D x f i = h u , α S x f − S x ( U D x f ) − αf i = h α S x u + D x (cid:0) U S x u (cid:1) − α u , f i , where the second equality holds by (2.31). This proves (2.37). Setting f = U in(2.34) and replacing therein u by S x u , and taking into account (2.26), we deduce D x (cid:0) U S x u (cid:1) = α − U D x S x u + ( α − α − ) S x u . Substituting this into the right-hand side of (2.37) we obtain (2.38). Finally, (2.39)follows easily from (2.19) and (2.30). (cid:3)
Proposition 2.1.
For the lattice x ( s ) = c q − s + c q s + c , the following holds: D x z n = γ n z n − + u n z n − + v n z n − + · · · , (2.40) S x z n = α n z n + b u n z n − + b v n z n − + · · · (2.41)( n = 0 , , . . . ) , where α n and γ n are given by (2.6) and (2.8) , and u n := (cid:0) nγ n − − ( n − γ n (cid:1) c , (2.42) v n := (cid:0) nγ n − − ( n − γ n (cid:1) c c (2.43) + (cid:0) n ( n − γ n − − n ( n − γ n − + ( n − n − γ n (cid:1) c , b u n := n ( α n − − α n ) c , (2.44) b v n := n ( α n − − α n ) c c + n ( n − α − α n − c . (2.45) K. CASTILLO, D. MBOUNA, AND J. PETRONILHO
Proof.
The proof is given by mathematical induction on n . For n = 0, we haveD x z = D x x z = 1. Since γ = 0 and α = 1, we see that (2.40)–(2.41)hold for n = 0. Next suppose that relations (2.40)–(2.41) are true for all integernumbers less than or equal to a fixed nonnegative integer number n (inductionhypothesis). Using this hypothesis together with (2.28)–(2.29), we obtainD x z n +1 = D x z n S x z + S x z n D x z = ( αz + β )D x z n + S x z n = ( α n + αγ n ) z n + ( αu n + b u n + βγ n ) z n − + ( αv n + b v n + βu n ) z n − + · · · . Similarly,S x z n +1 = U ( z )D x z D x z n + S x z n S x z = U ( z )D x z n + ( αz + β )S x z n = (cid:0) αα n + ( α − γ n (cid:1) z n +1 + (cid:0) ( α −
1) ( u n − γ n c ) + α b u n + βα n (cid:1) z n + (cid:0) ( α − (cid:0) v n − u n c + ( c − c c ) γ n (cid:1) + α b v n + β b u n (cid:1) z n − + · · · . Therefore, using relations (2.9)–(2.16), we obtain (2.40)–(2.41) with n replaced by n + 1. Consequently, (2.40)–(2.41) holds for each nonnegative integer n . (cid:3) A Leibniz formula.
Here we state a useful version of the Leibniz formula,involving the x − derivative operator, for the left multiplication of a functional by apolynomial. We need to state some preliminary results. Lemma 2.2.
Let f ∈ P . Then D nx S x f = α n S x D nx f + γ n U D n +1 x f ( n = 0 , , . . . ) . (2.46) Proof.
Once again, we use mathematical induction on n . Clearly, (2.46) holds for n = 0. Suppose that (2.46) is true for all positive integers less than or equalto a fixed integer n . Then by using successively (2.46) firstly for n (inductionhypothesis) and secondly for n = 1 (already proved) with f replaced by D nx f , andapplying (2.28) to D x ( U D n +1 x f ), and also taking into account (2.26), we deduceD n +1 x S x f = D x (D nx S x f ) = D x (cid:0) α n S x D nx f + γ n U D n +1 x f (cid:1) = α n D x S x (D nx f ) + γ n D x (cid:0) U D n +1 x f (cid:1) = α n (cid:0) α S x D n +1 x f + U D n +2 x f (cid:1) + γ n (cid:0) ( α − x D n +1 x f + α U D n +2 x f (cid:1) = (cid:0) αα n + ( α − γ n (cid:1) S x D n +1 x f + ( α n + αγ n ) U D n +2 x f. Finally, using properties (2.14), (2.3), (2.10), and (2.9), we see that (2.46) is truewhenever n is replaced by n + 1. Hence (2.46) is true for all n . (cid:3) The next result is a functional version of (2.46).
Lemma 2.3.
Let u ∈ P ∗ . Then (2.47) α D nx S x u = α n +1 S x D nx u + γ n U D n +1 x u ( n = 0 , , , . . . ) . Proof.
We prove (2.47) by mathematical induction on n . Since α = α and γ = 0,then (2.47) is trivial for n = 0. For n = 1, (2.47) is obtained multiplying bothsides of (2.39) by α and taking into account that, by (2.34) and (2.26), the equality α D x (cid:0) U D x u (cid:1) = U D x u +( α − S x D x u holds, and recalling also that α = 2 α − γ = 1. Suppose now that property (2.47) holds for a fixed integer n ∈ N (induction hypothesis). Then, we have(2.48) α D n +1 x S x u = D x (cid:0) α D nx S x u (cid:1) = α n +1 D x S x D nx u + γ n D x (cid:0) U D n +1 x u (cid:1) . N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 9
Considering (2.47) for n = 1 and replacing therein u by D nx u , we obtain(2.49) D x S x D nx u = α − α S x D n +1 x u + α − γ n U D n +2 x u . Moreover, using again (2.34) and (2.26), we deduce(2.50) D x (cid:0) U D n +1 x u (cid:1) = α − U D n +2 x u + α − ( α − S x D n +1 x u . Putting (2.49) and (2.50) into the right-hand side of (2.48) and taking into account(2.10) and (2.11), we obtain (2.47) with n replaced by n + 1. This proves (2.47). (cid:3) Next, we introduce an operator T n,k : P → P ( n = 0 , , . . . ; k = 0 , , . . . , n ),defined for each f ∈ P as follows: if n = k = 0, setT , f := f ;(2.51)and if n ≥ ≤ k ≤ n , define recurrentlyT n,k f := S x T n − ,k f − γ n − k α n − k U D x T n − ,k f + 1 α n +1 − k D x T n − ,k − f , (2.52)with the conventions T n,k f := 0 whenever k > n or k <
0. Note thatdeg T n,k f ≤ deg f − k . We are ready to state the following
Proposition 2.2 (Leibniz’s formula) . Let u ∈ P ∗ and f ∈ P . Then D nx (cid:0) f u (cid:1) = n X k =0 T n,k f D n − kx S kx u ( n = 0 , , . . . ) , (2.53) where T n,k f is a polynomial defined by (2.51) – (2.52) .Proof. The proof is done by mathematical induction on n . Clearly, (2.53) is true if n = 0. Suppose now that (2.53) holds for a fixed nonnegative integer n . Then(2.54) D n +1 x (cid:0) f u (cid:1) = D x (cid:0) D nx ( f u ) (cid:1) = n X k =0 D x (cid:0) T n,k f D n − kx S kx u (cid:1) . Notice that, by (2.47),(2.55) S x D n − kx S kx u = 1 α n +1 − k (cid:16) α D n − kx S k +1 x u − γ n − k U D n +1 − kx S kx u (cid:17) . Therefore, using successively (2.34), (2.55), (2.10), and (2.52), we may write D x (cid:0) T n,k f D n − kx S kx u (cid:1) = (cid:0) S x T n,k f − α − U D x T n,k f (cid:1) D n +1 − kx S kx u + α − D x T n,k f S x D n − kx S kx u = (cid:16) S x T n,k f − γ n +1 − k α n +1 − k U D x T n,k f (cid:17) D n +1 − kx S kx u + D x T n,k fα n +1 − k D n − kx S k +1 x u = (cid:16) T n +1 ,k f − D x T n,k − fα n +2 − k (cid:17) D n +1 − kx S kx u + D x T n,k fα n +1 − k D n − kx S k +1 x u . Substituting this expression in the right-hand side of (2.54) and then applying themethod of telescoping sums, we get D n +1 x (cid:0) f u (cid:1) = n X k =0 T n +1 ,k f D n +1 − kx S kx u + D x T n,n fα S n +1 x u − D x T n, − fα n +2 D n +1 x u . Finally, since T n, − f = 0 and α D x T n,n f = T n +1 ,n +1 f (this last equality followsfrom (2.52) taking therein k = n and then in the resulting expression shifting n into n + 1), we obtain (2.53) with n replaced by n + 1. Thus (2.53) is proved. (cid:3) Corollary 2.1.
Consider the lattice x ( s ) := c q − s + c q s + c . Let u ∈ P ∗ and f ∈ P . Write f ( z ) = az + bz + c , with a, b, c ∈ C . Then D nx ( f u ) = (cid:18) aαα n α n − ( z − c ) + f ′ ( c ) α n ( z − c ) + f ( c ) + 4 a (1 − α ) γ n c c α n − (cid:19) D nx u (2.56) + γ n α n (cid:18) a ( α n + αα n − ) α n − ( z − c ) + f ′ ( c ) (cid:19) D n − x S x u + aγ n γ n − α n − D n − x S x u ( n = 0 , , . . . ) . In particular, D nx (cid:0) ( bz + c ) u (cid:1) = (cid:18) b ( z − c ) α n + b c + c (cid:19) D nx u + bγ n α n D n − x S x u . (2.57) Proof.
Since (2.57) is the particular case of (2.56) for a = 0, we only need to prove(2.56). This can be proved combining the Leibniz formula (2.53) and identities(2.34) and (2.47). Alternatively, we may apply induction on n , as follows. Define g ( z ) := f ( z − c ) = a ( z − c ) + b ( z − c ) + c . We need to show that(T n, g )( z ) = g (cid:18) z − c α n + c (cid:19) + aγ n α n − U (cid:18) z − c α n + c (cid:19) , (2.58) (T n, g )( z ) = γ n α n (cid:18) a ( α n + αα n − ) α n − ( z − c ) + b (cid:19) , (2.59) (T n, g )( z ) = aγ n γ n − α n − (2.60)for each n = 0 , , , . . . , where T n,k f is defined by (2.51)–(2.52). Note that(T n, g )( z ) = αaα n α n − ( z − c ) + bα n ( z − c ) + c + 4 a (1 − α ) γ n α n − c c . (2.61)We proceed by induction on n . Setting n = 0 in (2.58)–(2.60), we obtain T , g = g and T , g = 0 = T , g . This agrees with (2.51)–(2.52). Next suppose that (2.58)–(2.60) hold for all positive integers up to a fixed n . Then, by (2.52), we haveT n +1 , g = S x (T n, g ) − γ n +1 α n +1 U D x (T n, g ) , (2.62) T n +1 , g = S x (T n, g ) − γ n α n U D x (T n, g ) + 1 α n +1 D x (T n, g ) , (2.63) T n +1 , g = S x (T n, g ) + 1 α n D x (T n, g ) . (2.64)Using the identitiesS x (cid:0) ( z − c ) (cid:1) = (2 α − z − c ) + 4(1 − α ) c c , D x (cid:0) ( z − c ) (cid:1) = 2 α ( z − c ) , S x (( z − c )) = α ( z − c ) , N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 11 we find S x (T n, g )( z ) = α (2 α − aα n α n − ( z − c ) + αbα n ( z − c ) + c + 4 a (1 − α )( α + α n γ n ) α n α n − c c , S x (T n, g )( z ) = γ n α n (cid:18) αa ( α n + αα n − ) α n ( z − c ) + b (cid:19) , D x (T n, g )( z ) = 1 α n (cid:18) α aα n − ( z − c ) + b (cid:19) , D x (T n, g )( z ) = aγ n ( α n + αα n − ) α n α n − . Therefore, from (2.62) and using (2.9)–(2.16), we obtain(T n +1 , g )( z ) = αaα n α n − (cid:18) α − α (1 − α ) γ n +1 α n +1 (cid:19) ( z − c ) + b ( αα n +1 + (1 − α ) γ n +1 ) α n α n +1 ( z − c ) + c + 4 a (1 − α )( α + α n γ n ) α n α n − c c = αaα n α n +1 ( z − c ) + bα n +1 ( z − c ) + c + 4 a (1 − α ) γ n +1 α n c c . Hence (2.58) holds for all n . Similarly, from (2.63), and using again (2.9)–(2.16)and the identity α n +1 γ n ( α n + αα n − ) + 2 α α n = α n − γ n +1 ( α n +1 + αα n ), we get(T n +1 , g )( z ) = aα n α n − (cid:18) γ n ( α n + αα n − ) α n + 2 α α n +1 (cid:19) ( z − c ) + bα n (cid:18) γ n + 1 α n +1 (cid:19) = γ n +1 α n +1 (cid:18) a ( α n +1 + αα n ) α n ( z − c ) + b (cid:19) . Hence (2.59) holds for n = 0 , , . . . . Finally, from (2.64) it is obvious that (2.60)holds with n replaced by n + 1 and, consequently, it holds for all n . Therefore(2.58)–(2.60) hold and so (2.56) is proved. (cid:3) Classical OPS and Rodrigues formula
This section concerns classical OPS on lattices and their associated regular func-tionals. We start by reviewing some basic definitions and then we derive a functionalversion of the Rodrigues formula.
Definition 3.1.
Let x ( s ) be a lattice given by (2.1) . u ∈ P ∗ is called x − classicalif it is regular and there exist nonzero polynomials φ ∈ P and ψ ∈ P such that (3.1) D x ( φ u ) = S x ( ψ u ) . An OPS with respect to a x − classical functional will be called a x − classical OPS(or a classical OPS on the lattice x ). Definition 3.1 appears in [6], and extends the definition of D − classical functional(cf. Section 1). We will refer to (3.1) as x − Geronimus–Pearson functional equation on the lattice x , or, simply, x − GP functional equation . As mentioned in the intro-ductory section, our principal goal in this work is to state necessary and sufficientconditions, involving only φ and ψ (or, equivalently, their coefficients), such thata given functional u ∈ P ∗ satisfying the x − GP functional equation (3.1) becomesregular. In order to move on we need to introduce some notation and to prove somepreliminary properties.We denote by P [ k ] n the monic polynomial of degree n defined by P [ k ] n ( z ) := D kx P n + k ( z ) Q kj =1 γ n + j = γ n ! γ n + k ! D kx P n + k ( z ) ( k, n = 0 , , . . . ) . (3.2)Here, as usual, it is understood that D x f = f , empty product equals one, and γ ! := 1 , γ n +1 ! := γ ...γ n γ n +1 ( n = 0 , , . . . ) . Definition 3.2.
Let φ ∈ P and ψ ∈ P . ( φ, ψ ) is called an x − admissible pair if d n ≡ d n ( φ, ψ, x ) := γ n φ ′′ + α n ψ ′ = 0 ( n = 0 , , . . . ) . This is an analogous for lattices of the corresponding definitions for the D − classicaland ( q, ω ) − classical cases (cf. [16, 13, 2]).3.1. Preliminary properties.
Following [6], given u ∈ P ∗ , φ ∈ P , and ψ ∈ P ,we define recursively polynomials φ [ k ] ∈ P and ψ [ k ] ∈ P by φ [0] := φ , ψ [0] := ψ , (3.3) φ [ k +1] := S x φ [ k ] + U S x ψ [ k ] + α U D x ψ [ k ] , (3.4) ψ [ k +1] := D x φ [ k ] + α S x ψ [ k ] + U D x ψ [ k ] , (3.5)and functionals u [ k ] ∈ P ∗ by(3.6) u [0] := u , u [ k +1] := D x (cid:0) U ψ [ k ] u [ k ] (cid:1) − S x (cid:0) φ [ k ] u [ k ] (cid:1) ( k = 0 , , . . . ). u [ k ] may be seen as the higher order x − derivative of u . Next, weprovide explicit representations for the polynomials φ [ k ] and ψ [ k ] . Proposition 3.1.
Consider the lattice x ( s ) := c q − s + c q s + c . Let φ ∈ P and ψ ∈ P , so there are a, b, c, d, e ∈ C such that φ ( z ) = az + bz + c , ψ ( z ) = dz + e . Then the polynomials φ [ k ] and ψ [ k ] defined by (3.3) – (3.5) are given by ψ [ k ] ( z ) = (cid:0) aγ k + dα k (cid:1) ( z − c ) + φ ′ ( c ) γ k + ψ ( c ) α k , (3.7) φ [ k ] ( z ) = (cid:0) d ( α − γ k + aα k (cid:1)(cid:0) ( z − c ) − c c (cid:1) (3.8) + (cid:0) φ ′ ( c ) α k + ψ ( c )( α − γ k (cid:1) ( z − c ) + φ ( c ) + 2 a c c , for each k = 0 , , . . . .Proof. Set(3.9) φ [ k ] ( z ) = a [ k ] z + b [ k ] z + c [ k ] , ψ [ k ] ( z ) = d [ k ] z + e [ k ] , where a [ k ] , b [ k ] , c [ k ] , d [ k ] , e [ k ] ∈ C . Clearly, by (3.3), a [0] = a , b [0] = b , c [0] = c , d [0] = d , e [0] = e . In order to determine the coefficients a [ k ] , b [ k ] , c [ k ] , d [ k ] , and e [ k ] for each k = 1 , , . . . ,we proceed as follows. Firstly we replace in (3.4) and in (3.5) the expressions of N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 13 φ [ k ] , φ [ k +1] , ψ [ k ] , and ψ [ k +1] given by (3.9); and then, in the two resulting identities,using (2.40) together with (2.20) and (2.21), after identification of the coefficientsof the polynomials appearing in both sides of each of those identities, we obtain asystem with five difference equations, namely a [ k +1] = (2 α − a [ k ] + 2 α ( α − d [ k ] , (3.10) b [ k +1] = αb [ k ] + ( α − e [ k ] + 2 β (2 α + 1) a [ k ] + β ( α + 1)(4 α − d [ k ] , (3.11) c [ k +1] = c [ k ] + b v a [ k ] + βb [ k ] + β ( α + 1) e [ k ] + (cid:0) β ( α + 1) + αδ (cid:1) d [ k ] , (3.12) d [ k +1] = 2 αa [ k ] + (2 α − d [ k ] , (3.13) e [ k +1] = b [ k ] + αe [ k ] + 2 βa [ k ] + β (2 α + 1) d [ k ] (3.14)( k = 0 , , . . . ). The explicit solution of this system is a [ k ] = d ( α − γ k + aα k , (3.15) b [ k ] = ψ ( c )( α − γ k + φ ′ ( c ) α k − c (cid:0) d ( α − γ k + aα k (cid:1) , (3.16) c [ k ] = φ ( c ) + 2 a c c − c (cid:0) ψ ( c )( α − γ k + φ ′ ( c ) α k (cid:1) (3.17) + ( c − c c ) (cid:0) d ( α − γ k + aα k (cid:1) ,d [ k ] = aγ k + dα k , (3.18) e [ k ] = φ ′ ( c ) γ k + ψ ( c ) α k − c (cid:0) aγ k + dα k (cid:1) . (3.19)This can be easily proved by induction on k . The representations (3.7)–(3.8) areobtained by inserting (3.15)–(3.19) into (3.9). (cid:3) Lemma 3.1 in bellow is proved in [6]. We point out that the statement of theresult given in [6] assumes a priori that u is a regular functional. However, inspec-tion of the proof given therein shows that the result remains unchanged withoutsuch assumption. Lemma 3.1.
Let u ∈ P ∗ . Suppose that there exist φ ∈ P and ψ ∈ P such that (3.1) holds. Then u [ k ] satisfies the functional equation (3.20) D x (cid:0) φ [ k ] u [ k ] (cid:1) = S x (cid:0) ψ [ k ] u [ k ] (cid:1) ( k = 0 , , . . . ) . The next result gives some additional functional equations fulfilled by u [ k ] . Lemma 3.2.
Let u ∈ P ∗ and suppose that there exist φ ∈ P and ψ ∈ P suchthat u satisfies the x − GP functional equation (3.1) . Then the relations D x (cid:0) u [ k +1] (cid:1) = − αψ [ k ] u [ k ] , (3.21) S x (cid:0) u [ k +1] (cid:1) = − α (cid:0) αφ [ k ] + U ψ [ k ] (cid:1) u [ k ] , (3.22) 2 U u [ k +1] = S x (cid:0) U ψ [ k ] u [ k ] (cid:1) − D x (cid:0) U φ [ k ] u [ k ] (cid:1) (3.23) hold for each k = 0 , , . . . .Proof. Using (2.38) and (3.20), we deduce D x (cid:0) U ψ [ k ] u [ k ] (cid:1) = (2 α − α − ) S x (cid:0) ψ [ k ] u [ k ] (cid:1) + α − U D x S x (cid:0) ψ [ k ] u [ k ] (cid:1) − αψ [ k ] u [ k ] = (2 α − α − ) S x D x (cid:0) φ [ k ] u [ k ] (cid:1) + α − U D x (cid:0) φ [ k ] u [ k ] (cid:1) − αψ [ k ] u [ k ] = D x S x (cid:0) φ [ k ] u [ k ] (cid:1) − αψ [ k ] u [ k ] , where the last equality follows from (2.47) for n = 1 and taking into account that α = 2 α − γ = 1. Therefore, by the definition of u [ k +1] , we obtain D x u [ k +1] = D x (cid:0) U ψ [ k ] u [ k ] (cid:1) − D x S x (cid:0) φ [ k ] u [ k ] (cid:1) = − αψ [ k ] u [ k ] . This proves (3.21). Next, by (2.34) and (2.35), we may write D x (cid:0) U φ [ k ] u [ k ] (cid:1) = (cid:0) S x U − α − U D x U (cid:1) D x (cid:0) φ [ k ] u [ k ] (cid:1) + α − (cid:0) D x U (cid:1) S x (cid:0) φ [ k ] u [ k ] (cid:1) , S x (cid:0) U ψ [ k ] u [ k ] (cid:1) = (cid:0) S x U − α − U D x U (cid:1) S x (cid:0) ψ [ k ] u [ k ] (cid:1) + α − (cid:0) D x U (cid:1) D x (cid:0) U ψ [ k ] u [ k ] (cid:1) . After subtracting these two equalities and taking into account (3.20), as well as therelation α − D x U = 2 U (cf. (2.27)), we obtain (3.23). To prove (3.22), note firstthat, by the definition of u [ k +1] ,(3.24) α S x u [ k +1] = α S x D x (cid:0) U ψ [ k ] u [ k ] (cid:1) − α S x (cid:0) φ [ k ] u [ k ] (cid:1) . Using again (2.47) for n = 1, we have α S x D x (cid:0) U ψ [ k ] u [ k ] (cid:1) = α D x S x (cid:0) U ψ [ k ] u [ k ] (cid:1) − U D x (cid:0) U ψ [ k ] u [ k ] (cid:1) and by (2.38), we also have α S x (cid:0) φ [ k ] u [ k ] (cid:1) = − U D x S x (cid:0) φ [ k ] u [ k ] (cid:1) + α φ [ k ] u [ k ] + α D x (cid:0) U φ [ k ] u [ k ] (cid:1) . Substituting these two expressions into the right-hand side of (3.24), we get(2 α − S x u [ k +1] (3.25) = α D x S x (cid:0) U ψ [ k ] u [ k ] (cid:1) − α D x (cid:0) U φ [ k ] u [ k ] (cid:1) − U D x u [ k +1] − α φ [ k ] u [ k ] . Next, by taking f = U and replacing u by u [ k +1] in (2.34), and then using (2.26)and (3.21), we derive D x (cid:0) U u [ k +1] (cid:1) = U ψ [ k ] u [ k ] + ( α − α − ) S x u [ k +1] . Multiplying both sides of this equality by 2 α and combining the resulting equalitywith the one obtained by applying D x to both sides of (3.23), we deduce(2 α − S x u [ k +1] = α D x S x (cid:0) U ψ [ k ] u [ k ] (cid:1) − α D x (cid:0) U φ [ k ] u [ k ] (cid:1) + 2 α U ψ [ k ] u [ k ] . (3.26)Finally, subtracting (3.26) to (3.25), and taking into account (3.21), (3.22) follows. (cid:3) A functional Rodrigues formula.
Here we prove a functional version ofthe Rodrigues formula on lattices, extending results stated in [13, 2].
Theorem 3.1 (Rodrigues’ formula) . Consider the lattice x ( s ) := c q − s + c q s + c . Let u ∈ P ∗ and suppose that there exists a x − admissible pair ( φ, ψ ) such that u fulfills the x − GP functional equation (3.1) . Set (3.27) d n := φ ′′ γ n + ψ ′ α n , e n := φ ′ ( c ) γ n + ψ ( c ) α n ( n = 0 , , . . . ) . Then R n u = D nx u [ n ] ( n = 0 , , . . . ) , (3.28) N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 15 where u [ n ] is the functional on P defined by (3.6) and ( R n ) n ≥ is a simple set ofpolynomials given by the TTRR R n +1 ( z ) = ( a n z − s n ) R n ( z ) − t n R n − ( z ) ( n = 0 , , . . . ) , (3.29) with initial conditions R − = 0 and R = 1 , and ( a n ) n ≥ , ( s n ) n ≥ , and ( t n ) n ≥ are sequences of complex numbers defined by a n := − α d n d n − d n − , (3.30) s n := a n (cid:18) c + γ n e n − d n − − γ n +1 e n d n (cid:19) , (3.31) t n := a n α γ n d n − d n − φ [ n − (cid:18) c − e n − d n − (cid:19) , (3.32) φ [ n − being given by (3.8) . (It is understood that a := − αd and s := αe .)Proof. We apply mathematical induction on n . If n = 0, (3.28) is trivial. If n = 1,(3.28) follows from (3.21), since R = − αψ . Assume now (induction hypothesis)that (3.28) holds for two consecutive nonnegative integer numbers, i.e., the relations R n − u = D n − x u [ n − , R n u = D nx u [ n ] (3.33)hold for some fixed n ∈ N . We need to prove that R n +1 u = D n +1 x u [ n +1] . Noticefirst that, by (3.7) and (3.27), we have(3.34) ψ [ k ] ( z ) = d k ( z − c ) + e k ( k = 0 , , . . . ) . By (3.21) and the Leibniz formula in Proposition 2.2, we may write D n +1 x u [ n +1] = D nx D x u [ n +1] = − α D nx ( ψ [ n ] u [ n ] )= − α T n, ψ [ n ] D nx u [ n ] − α T n, ψ [ n ] D n − x S x u [ n ] . From (2.57) we have T n, ψ [ n ] = d n γ n /α n , and so, using also (3.33), D n − x S x u [ n ] = − α n αd n γ n (cid:16) D n +1 x u [ n +1] + α (cid:0) T n, ψ [ n ] (cid:1) R n u (cid:17) . (3.35)Shifting n into n −
1, and using again the induction hypothesis (3.33), we obtain D n − x S x u [ n − = − α n − αd n − γ n − (cid:16) R n + α (cid:0) T n − , ψ [ n − (cid:1) R n − (cid:17) u . (3.36)Next, using (3.21), (2.34), and (3.22), we deduce D n +1 x u [ n +1] = − α D nx (cid:0) ψ [ n ] u [ n ] (cid:1) = − α D n − x (cid:0) D x ( ψ [ n ] u [ n ] ) (cid:1) (3.37) = − D n − x (cid:16)(cid:0) α S x ψ [ n ] − U D x ψ [ n ] (cid:1) D x u [ n ] + D x ψ [ n ] S x u [ n ] (cid:17) = D n − x (cid:0) ξ ( · ; n ) u [ n − (cid:1) , where ξ ( · ; n ) is a polynomial of degree 2, given by ξ ( z ; n ) = α (cid:0) ψ [ n − S x ψ [ n ] + φ [ n − D x ψ [ n ] (cid:1) ( z ) . (3.38)The following identities may be proved by a straightforward computation: d n − − αd n − = a [ n − ,d n − (cid:0) e n − α c d n (cid:1) + d n (cid:0) b [ n − + αe n − (cid:1) = 2 d n − ( αe n − c d n ) for each n = 1 , , . . . . (The second one is achieved by using equation (3.14).) Usingthese relations, together with (3.9), (3.34), (2.40), and (2.41), we deduce ξ ( z ; n ) = α d n d n − z + 2 α d n − ( αe n − c d n ) z (3.39) + α (cid:0) d n c [ n − + ( e n − − c d n − )( e n − α c d n ) (cid:1) . Since deg ξ ( · ; n ) = 2, using again Proposition 2.2, we may write D n − x (cid:0) ξ ( · ; n ) u [ n − (cid:1) = T n − , ξ ( · ; n ) D n − x u [ n − + T n − , ξ ( · ; n ) D n − x S x u [ n − (3.40) + T n − , ξ ( · ; n ) D n − x S x u [ n − . Since, by (2.57), T n − , ξ ( · ; n ) = α γ n − γ n − d n d n − /α n − , combining equations(3.40), (3.37), (3.36), and (3.33), we obtain D n − x S x u [ n − = α n − α γ n − γ n − d n d n − (cid:16) D n +1 x u [ n +1] − (cid:0) T n − , ξ ( · ; n ) (cid:1) R n − u (3.41) + α n − T n − , ξ ( · ; n ) αγ n − d n − (cid:16) R n + α (cid:0) T n − , ψ [ n − (cid:1) R n − (cid:17) u (cid:19) . On the other hand, by (3.22), S x u [ n ] = η ( · ; n ) u [ n − , η ( z ; n ) := − α (cid:0) αφ [ n − + U ψ [ n − (cid:1) ( z ) . (3.42)Taking into account that η ( · ; n ) is a polynomial of degree at most two, by theLeibniz formula and (3.33), we may write D n − x S x u [ n ] = D n − x (cid:0) η ( · ; n ) u [ n − (cid:1) (3.43) = (cid:0) T n − , η ( · ; n ) (cid:1) R n − u + T n − , η ( · ; n ) D n − x S x u [ n − + T n − , η ( · ; n ) D n − x S x u [ n − . Note that η ( · ; n ) is given explicitly by η ( z ; n ) = α ( αd n − − d n ) z − α (cid:16) αb [ n − + ( α − e n − − c d n − ) (cid:17) z (3.44) − α (cid:0) αc [ n − + β ( α + 1)( e n − − c d n − ) (cid:1) . Hence, using (2.57), T n − , η ( · ; n ) = αγ n − γ n − ( αd n − − d n ) /α n − . Therefore,substituting (3.35), (3.36), and (3.41) in (3.43), we obtain(3.45) D n +1 x u [ n +1] = (cid:0) A ( · ; n ) R n + B ( · ; n ) R n − (cid:1) u , where A ( · ; n ) and B ( · ; n ) are polynomials depending on n , given by ǫ n A ( z ; n ) = α n (cid:0) T n, ψ [ n ] (cid:1) ( z ) γ n d n − α n − (cid:0) T n − , η (cid:1) ( z ; n ) αγ n − d n − (3.46) + α n − ( αd n − − d n ) (cid:0) T n − , ξ (cid:1) ( z ; n ) α γ n − d n d n − d n −
2N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 17 and ǫ n B ( z ; n ) = (cid:0) T n − , η (cid:1) ( z ; n ) − α n − (cid:0) T n − , ψ [ n − (cid:1) ( z ) (cid:0) T n − , η (cid:1) ( z ; n ) γ n − d n − + ( d n − αd n − ) (cid:0) T n − , ξ (cid:1) ( z ; n ) αd n d n − (3.47) + α n − ( αd n − − d n ) (cid:0) T n − , ξ (cid:1) ( z ; n ) (cid:0) T n − , ψ [ n − (cid:1) ( z ) αγ n − d n d n − d n − , where ǫ n := d n − αd n − αd n d n − − α n αγ n d n = − d n − αγ n d n d n − . Note that γ n d n − ( α + α n ) d n − = − d n − ( n = 0 , , . . . ). We claim that A ( z ; n ) = − α d n d n − d n − ( z − c ) + αγ n d n d n − e n − d n − d n − − αγ n +1 d n − e n d n − (3.48) = a n z − s n ,B ( z ; n ) = α γ n d n d n − d n − φ [ n − (cid:18) c − e n − d n − (cid:19) = − t n (3.49)( n = 0 , , . . . ), where a n , s n , and t n are given by (3.30)–(3.32). Indeed, by (3.15)and (3.16), a [ n − = d n − − αd n − , (3.50) b [ n − = e n − αe n − − c a [ n − , (3.51) d n − αd n − + d n − = 0(3.52)( n = 1 , , . . . ). By (3.39), (3.44), and (3.34), and applying (2.56)–(2.57) togetherwith (3.51), we obtain(T n, ψ [ n ] )( z ) = d n α n ( z − c ) + e n , (3.53) (T n − , ξ )( z ; n ) = γ n − α n − (cid:16) α d n d n − ( α n − + αα n − ) α n − ( z − c )(3.54) + 2 α e n d n − (cid:17) , (T n − , η )( z ; n ) = γ n − α n − (cid:16) α ( αd n − − d n )( α n − + αα n − ) α n − ( z − c )(3.55) + α ( e n − − αe n ) (cid:17) . Similarly,(T n − , η )( z ; n ) = α ( αd n − − d n ) α n − α n − ( z − c ) + α ( e n − − αe n ) α n − ( z − c )(3.56) + η ( c ; n ) + 4 α (1 − α ) γ n − ( αd n − − d n ) α n − c c , (T n − , ξ )( z ; n ) = α d n − d n α n − α n − ( z − c ) + 2 α e n d n − α n − ( z − c ) + ξ ( c ; n )(3.57) + 4 α (1 − α ) γ n − d n − d n α n − c c . Using (3.53), (3.54), and (3.55) together with the identity α n + αγ n = γ n +1 (thislast one follows from (2.9) and (2.10)), we deduce from (3.46) that ǫ n A ( z ; n ) = 1 γ n ( z − c ) − e n − d n − + α n e n γ n d n + αe n αd n − − d n d n d n − = 1 γ n ( z − c ) − e n − d n − + γ n +1 e n γ n d n ( n = 0 , , . . . ). This gives (3.48). From (3.53)–(3.57) it is straightforward to verifythat (3.47) reduces to ǫ n B ( z ; n ) = η ( c ; n ) + ( d n − αd n − ) ξ ( c ; n ) αd n d n − + α ( αe n − e n − ) e n − d n − + 2 α ( αd n − − d n ) e n e n − d n d n − . Moreover, from the definitions of ξ ( . ; n ) and η ( . ; n ) given in (3.38) and (3.42), wededuce ξ ( c ; n ) = α ( e n e n − + φ [ n − ( c ) d n ) , η ( c ; n ) = − α φ [ n − ( c )( n = 0 , , . . . ). Consequently, by (3.50), we show that ǫ n B ( z ; n ) = − α d n − d n − φ [ n − ( c ) + αe n − (cid:18) e n d n − − e n − d n − (cid:19) . Therefore, using successively (3.50) and (3.51), we obtain B ( z ; n ) = α γ n d n d n − d n − (cid:18) d n − e n − d n − − e n e n − d n − + φ [ n − ( c ) (cid:19) = α γ n d n d n − d n − (cid:18) ( a [ n − + αd n − ) e n − d n − − ( b [ n − + αe n − + 2 c a [ n − ) e n − d n − + φ [ n − ( c ) (cid:19) = α γ n d n d n − d n − φ [ n − (cid:18) c − e n − d n − (cid:19) , and (3.49) is proved. It follows from (3.48) and (3.49) that (3.45) reduces to D n +1 x u [ n +1] = R n +1 u , which completes the proof. (cid:3) N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 19
Regularity of u [ k ] . The next result is virtually proved in [13, Theorem 2].
Lemma 3.3.
Let u ∈ P ∗ be regular. Suppose that there is ( φ, ψ ) ∈ P ×P \{ (0 , } so that (3.1) holds. Then neither φ nor ψ is the zero polynomial, and deg ψ = 1 . Lemma 3.4 in bellow gives the regularity of u [ k ] . The result appears in [6,Proposition 4]. However the proof of the x − admissibility condition given therein isincorrect. We present a proof following ideas presented in [13, 2]. Lemma 3.4.
Let u ∈ P ∗ . Suppose that u is regular and satisfies (3 . , where ( φ, ψ ) ∈ P × P \ { (0 , } . Then ( φ, ψ ) is a x − admissible pair and u [ k ] is regularfor each k = 1 , , . . . . Moreover, if ( P n ) n ≥ is the monic OPS with respect to u ,then (cid:0) P [ k ] n (cid:1) n ≥ is the monic OPS with respect to u [ k ] .Proof. Set φ ( z ) = az + bz + c and ψ ( z ) = dz + e . By Lemma 3.3, both φ and ψ are nonzero polynomials, and d = 0. If deg φ ∈ { , } then d n = dα n = 0 for each n = 0 , , . . . , and so the pair ( φ, ψ ) is x − admissible. Assume now that deg φ = 2.Then d n = aγ n + dα n , with a = 0. To prove that d n = 0, we start by showing that(3.58) (cid:10) u , (cid:0) U ψ D x P [1] n + φ S x P [1] n (cid:1) P n +2 (cid:11) = − (cid:10) u [1] , (cid:0) S x P n +2 + α − U D x P n +2 (cid:1) P [1] n (cid:11) for each n = 0 , , . . . . Indeed, we have (cid:10) u , (cid:0) U ψ D x P [1] n + φ S x P [1] n (cid:1) P n +2 (cid:11) = (cid:10) U ψ u , P n +2 D x P [1] n (cid:11) + (cid:10) φ u , P n +2 S x P [1] n (cid:11) = (cid:10) U ψ u , D x (cid:0) (S x P n +2 − α − U D x P n +2 ) P [1] n (cid:1) − α − S x (cid:0) P [1] n D x P n +2 (cid:1)(cid:11) + (cid:10) φ u , S x (cid:0) (S x P n +2 − α − U D x P n +2 ) P [1] n (cid:1) − α − U D x (cid:0) P [1] n D x P n +2 (cid:1)(cid:11) = − (cid:10) u [1] , (cid:0) S x P n +2 − α − U D x P n +2 (cid:1) P [1] n (cid:11) − α − (cid:10) S x ( U ψ u ) − D x ( U φ u ) , P [1] n D x P n +2 (cid:11) , where the second equality holds by (2.33) and (2.32). Therefore, using (3.23) for n = 0, we obtain (3.58). Now, on the one hand, U ψ D x P [1] n + φ S x P [1] n is a polynomialof degree at most n + 2, being the coefficient of z n +2 equal to ( α − dγ n + aα n .Hence, since the relations( α − dγ n + aα n = d n +1 − αd n = αd n − d n − ( n = 1 , , . . . )hold, we get U ψ D x P [1] n + φ S x P [1] n = (cid:0) αd n − d n − (cid:1) z n +2 + (lower degree terms)for each n = 1 , , . . . . Consequently,(3.59) (cid:10) u , (cid:0) U ψ D x P [1] n + φ S x P [1] n (cid:1) P n +2 (cid:11) = ( αd n − d n − ) h u , P n +2 i ( n = 1 , , . . . ) . On the other hand, since S x P n +2 + α − U D x P n +2 = P n +2 j =0 c n,j P [1] j for some coeffi-cients c n, , . . . , c n,n +2 ∈ C , and using the next equation for k = 1 (cid:10) u [ k ] , P [ k ] n P [ k ] m (cid:11) = α d [ k − n γ n +1 h u , ( P [ k − n +1 ) i δ n,m (0 ≤ m ≤ n ; n = 0 , , . . . )(3.60)(see [6, Proof of Theorem 5 – step 1.1]), we obtain(3.61) (cid:10) u [1] , (cid:0) S x P n +2 + α − U D x P n +2 (cid:1) P [1] n (cid:11) = αc n,n d n γ n +1 h u , P n +1 i ( n = 1 , , . . . ) . Substituting (3.59) and (3.61) into (3.58), and since C n +2 = h u , P n +2 i / h u , P n +1 i ,we deduce α (cid:18) c n,n γ n +1 C n +2 (cid:19) d n = d n − ( n = 1 , , . . . ) . Therefore, since d = d = 0, we conclude that d n = 0 for each n = 0 , , . . . andso ( φ, ψ ) is a x − admissible pair. Consequently, by (3.60) for k = 1, ( P [1] n ) n ≥ isthe monic OPS with respect to u [1] . This proves the last statement in the lemmafor k = 1. Since d [ k ] n = a [ k ] γ n + α n d [ k ] ( n, k = 0 , , . . . ), it is easy to see, using(3.15)–(3.18), that d [ k ] n = d n +2 k for all n, k = 0 , , . . . . Thus the last sentence in thelemma follows from (3.60). (cid:3) Main results: regularity conditions
In this section we state our main results, giving the analogues of Theorems B andC for OPS on lattices. Thus, once a lattice as in (2.1) is fixed, we state necessaryand sufficient conditions so that a functional u ∈ P ∗ satisfying (3.1) is regular.Furthermore, the monic OPS with respect to u is described.4.1. The lattice x ( s ) = c q − s + c q s + c . We start by considering a lattice x ( s )with q = 1, so that x ( s ) is a q − quadratic or a q − linear lattice. Theorem 4.1.
Consider the lattice x ( s ) = c q − s + c q s + c . Let u ∈ P ∗ \ { } and suppose that there exist ( φ, ψ ) ∈ P × P \ { (0 , } such that (4.1) D x ( φ u ) = S x ( ψ u ) . Set φ ( z ) = az + bz + c and ψ ( z ) = dz + e ( a, b, c, d, e ∈ C ), and let d n and e n bedefined by (3.27) , and φ [ n ] and ψ [ n ] be given by (3.7) – (3.8) . Then, u is regular ifand only if ( φ, ψ ) is a x − admissible pair and ψ [ n ] ∤ φ [ n ] for each n = 0 , , . . . ; thatis to say, u is regular if and only if the following conditions hold: (4.2) d n = 0 , φ [ n ] (cid:18) c − e n d n (cid:19) = 0 ( n = 0 , , . . . ) . Under such conditions, the monic OPS ( P n ) n ≥ with respect to u satisfies (4.3) P n +1 ( z ) = ( z − B n ) P n ( z ) − C n P n − ( z ) ( n = 0 , , . . . ) , with P − ( z ) = 0 , where the recurrence coefficients are given by B n = c + γ n e n − d n − − γ n +1 e n d n , (4.4) C n +1 = − γ n +1 d n − d n − d n +1 φ [ n ] (cid:18) c − e n d n (cid:19) (4.5)( n = 0 , , . . . ) . Moreover, the following functional Rodrigues formula holds: P n u = k n D nx u [ n ] , k n := ( − α ) − n n Y j =1 d − n + j − ( n = 0 , , . . . ) . (4.6) N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 21
Proof.
Suppose that u is regular. By Lemma 3.4, ( φ, ψ ) is a x − admissible pair,meaning that the first condition in (4.2) holds, and (cid:0) P [ n ] j (cid:1) j ≥ is the monic OPSwith respect to u [ n ] , for each fixed n . Write the TTRR for (cid:0) P [ n ] j (cid:1) j ≥ as(4.7) P [ n ] j +1 ( z ) = ( z − B [ n ] j ) P [ n ] j ( z ) − C [ n ] j P [ n ] j − ( z ) ( j = 0 , , . . . )( P [ n ] − ( z ) := 0), being B [ n ] j ∈ C and C [ n ] j +1 ∈ C \ { } for each j = 0 , , . . . . Let uscompute C [ n ]1 . We first show that (for n = 0) the coefficient C ≡ C [0]1 , appearingin the TTRR for ( P j ) j ≥ , is given by(4.8) C = − dα + a φ (cid:16) − ed (cid:17) = − d φ (cid:18) c − e d (cid:19) . Indeed, taking k = 0 and k = 1 in the relation h D x ( φ u ) , z k i = h S x ( ψ u ) , z k i , weobtain 0 = du + eu and au + bu + cu = − dαu − ( eα + dβ ) u − eβu , where u k := h u , z k i ( k = 0 , , . . . ). Therefore,(4.9) u = − ed u , u = − dα + a h − ( b + eα ) ed + c i u . On the other hand, since P ( z ) = z − B [0]0 = z − u /u , we also have(4.10) C = h u , P i u = u u − u u = u u − (cid:18) u u (cid:19) . Substituting u and u given by (4.9) into (4.10) yields (4.8). Since equation (3.20)fulfilled by u [ n ] is of the same type as (4.1) fulfilled by u , with the polynomials φ [ n ] and ψ [ n ] in (3.20) playing the roles of φ and ψ in (4.1), C [ n ]1 may be obtainedreplacing in (4.8) φ and ψ ( z ) = dz + e by φ [ n ] and ψ [ n ] ( z ) = d n ( z − c ) + e n ,respectively. Hence,(4.11) C [ n ]1 = − d n α + a [ n ] φ [ n ] (cid:18) c − e n d n (cid:19) = − d n +1 φ [ n ] (cid:18) c − e n d n (cid:19) . Since u [ n ] is regular, then C [ n ]1 = 0, and so the second condition in (4.2) holds.Conversely, suppose that conditions (4.2) hold. Define a sequence of polynomials,( P n ) n ≥ , by the TTRR (4.3)–(4.5), with P − ( z ) := 0. According to the hypothesis(4.2), C n +1 = 0 for each n = 0 , , . . . . Therefore, Favard’s theorem ensures that( P n ) n ≥ is a monic OPS. To prove that u is regular we will show that ( P n ) n ≥ isthe monic OPS with respect to u . For that we only need to prove that u = 0 , h u , P n i = 0 ( n = 1 , , . . . ) . We start by showing that u = 0. Indeed, suppose that u = 0. Since (4.1) holds,then h D x ( φ u ) − S x ( ψ u ) , z n i = 0 ( n = 0 , , . . . ). This implies d n u n +1 + n X j =0 a n,j u j = 0 ( n = 0 , , . . . )(4.12)for some complex numbers a n,j ( j = 0 , , . . . , n ). Since u = 0 and d n = 0 for all n = 0 , , . . . , from (4.12) we deduce u n = 0 for each n = 0 , , . . . . This implies u = ,contrary to the hypothesis. Therefore u = 0. Next, notice that P n ( z ) = k n R n ( z ) ( n = 0 , , . . . ), where R n is defined by (3.29) and k − n := ( − α ) n Q nj =1 d n + j − .Therefore, by the Rodrigues formula (3.28) given in Theorem 3.1, we obtain h u , P n i = k n h u , R n i = k n h R n u , i = k n (cid:10) D nx u [ n ] , (cid:11) = ( − n k n (cid:10) u [ n ] , D nx (cid:11) = 0for each n = 1 , , . . . . Hence u is regular. The remaining statements in the theoremfollow from Theorem 3.1. (cid:3) Remark 4.1.
It is important to highlight that the Rodrigues formula holds fora functional u satisfying a x − GP functional equation, even if u is not regular,provided that the pair ( φ, ψ ) appearing in that x − GP functional equation forms a x − admissible pair. This fact is a consequence of Theorem 3.1. The lattice x ( s ) = c s + c s + c . We consider now a lattice x ( s ) for q = 1,so that x ( s ) is a quadratic or a linear lattice. Recall that, here, c = 4 β . We justgive the results for this situation, since the techniques and computations are verysimilar to the ones presented for the case q = 1. For the lattice x ( s ) = c s + c s + c the system of difference equations (3.10)–(3.14) becomes a [ n +1] = a [ n ] ,d [ n +1] = 2 a [ n ] + d [ n ] ,b [ n +1] = b [ n ] + 6 β ( a [ n ] + d [ n ] ) ,e [ n +1] = e [ n ] + b [ n ] + β (2 a [ n ] + 3 d [ n ] ) ,c [ n +1] = c [ n ] + β ( b [ n ] + 2 e [ n ] ) + β d [ n ] + (cid:16) β − β c + c (cid:17)(cid:0) a [ n ] + d [ n ] (cid:1) . with the initial conditions a [0] = a , b [0] = b , c [0] = c , d [0] = d and e [0] = e . Thesolution of this system is a [ n ] = a , b [ n ] = b + 6 βn ( an + d ) , d [ n ] = 2 an + d ,e [ n ] = bn + e + 2 dβn + βn (2 an + d ) ,c [ n ] = φ ( βn ) + 2 βnψ ( βn ) − n (cid:18) β c − c (cid:19) ( an + d )( n = 0 , , . . . ). Thus, applying a limiting process (as q →
1) on Theorem 4.1, wemay infer and then to prove rigorously the following
Theorem 4.2.
Consider the lattice x ( s ) = c s + c s + c . Let u ∈ P ∗ \ { } and suppose that there exist ( φ, ψ ) ∈ P × P \ { (0 , } such that (4.13) D x ( φ u ) = S x ( ψ u ) . Set φ ( z ) := az + bz + c and ψ ( z ) := dz + e . Then u is regular if and only if (4.14) d n = 0 , φ [ n ] (cid:18) − βn − e n d n (cid:19) = 0 ( n = 0 , , . . . ) , where d n := an + d , e n := bn + e + 2 βdn , and φ [ n ] ( z ) = az + ( b + 6 βnd n ) z + φ ( βn ) + 2 βnψ ( βn ) − n (cid:0) β c − c (cid:1) d n . N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 23
Under these conditions, the monic OPS ( P n ) n ≥ with respect to u satisfies (4.3) with B n = ne n − d n − − ( n + 1) e n d n − βn ( n − , (4.15) C n +1 = − ( n + 1) d n − d n − d n +1 φ [ n ] (cid:18) − βn − e n d n (cid:19) (4.16)( n = 0 , , . . . ) . Moreover, the following functional Rodrigues formula holds: P n u = k n D nx u [ n ] , k n := ( − n n Y j =1 d − n + j − ( n = 0 , , . . . ) . (4.17) Remark 4.2.
For x ( s ) = c we immediately recover Theorem B. Two examples: Racah and Askey-Wilson polynomials
In this section we derive the recurrence coefficients for the Racah and the Askey-Wilson polynomials, using as departure point their associated x − GP functionalequations. It is worth mentioning that Theorem 4.1 allow us also to answer posi-tively to a conjecture posed by Ismail [10, Conjecture 24.7.8]. Since the proof of thisconjecture is rather technique and involves long computations, it will be presentedin a different work [4].5.1.
The Racah polynomials.
Consider the quadratic lattice x ( s ) = s ( s + a + b + 1) , where a, b, c, d ∈ C . Let u ∈ P ∗ be a functional satisfying (4.13), where φ ( z ) := 2 z + [( a + b + 2 c + 3) d + c ( a − b + 3) + 2( a + b + ab + 2)] z + (1 + a )(1 + d )( a + b + 1)( b + c + 1) ,ψ ( z ) := 2( d + c + 2) z + 2(1 + a )(1 + d )( b + c + 1) . According to (4.14) in Theorem 4.2, u is regular if and only if { a, c, d, d + c − , b + c, c + d − a, d − b } ∩ Z − = ∅ , where Z − := {− , − , . . . } . Under these conditions, by (4.15)–(4.16), the recurrencecoefficients for the monic OPS ( P n ) n ≥ with respect to u are given by B n = − ( n + a + 1)( n + d + 1)( n + b + c + 1)( n + d + c + 1)(2 n + d + c + 1)(2 n + d + c + 2) − n ( n + c )( n + d + c − a )( n + d − b )(2 n + d + c )(2 n + d + c + 1) ,C n +1 = ( n + 1)( n + a + 1)( n + c + 1)( n + d + 1) × ( n + d + c + 1)( n + b + c + 1)( n + c + d − a + 1)( n + d − b + 1)(2 n + d + c + 1)(2 n + d + c + 2) (2 n + d + c + 3)( n = 0 , , . . . ). Therefore, P n ( z ) = R n ( z ; d, c, a, b ) ( n = 0 , , . . . ) , ( R n ( . ; d, c, a, b )) n ≥ being the sequence of the monic Racah polynomials [11, p.190]. The Askey-Wilson polynomials.
Consider the q − quadratic lattice x ( s ) = c q − s + c q s + c . Let u be a linear functional on P satisfying (4.1), where φ and ψ are given by φ ( z ) = 2(1 + abcd )( z − c ) − √ c c ( a + b + c + d + abc + abd + acd + bcd )( z − c )+ 4 c c ( ab + ac + ad + bc + bd + cd − abcd − ,ψ ( z ) = 4 q / q − (cid:16) ( abcd − z − c )+ √ c c ( a + b + c + d − abc − abd − acd − bcd ) (cid:17) , with a, b, c, d ∈ C . The parameter d n given by (3.27) reads as d n = − (cid:16) q / − q − / (cid:17) − q − n/ (cid:0) − abcdq n (cid:1) . By Theorem 4.1, u is regular if and only if c c (1 − abcdq n )(1 − abq n )(1 − acq n ) × (1 − adq n )(1 − bcq n )(1 − bdq n )(1 − cdq n ) = 0 ( n = 0 , , . . . ) . Under these conditions, using formulas (4.4)–(4.5) in Theorem 4.1, we obtain therecurrence coefficients for the monic OPS ( P n ) n ≥ with respect to u : B n = c + 2 √ c c (cid:20) a + 1 a − (1 − abq n )(1 − acq n )(1 − adq n )(1 − abcdq n − ) a (1 − abcdq n − )(1 − abcdq n ) − a (1 − q n )(1 − bcq n − )(1 − bdq n − )(1 − cdq n − )(1 − abcdq n − )(1 − abcdq n − ) (cid:21) (if a = 0, define B n by continuity, taking a → C n +1 = c c (1 − q n +1 )(1 − abcdq n − ) × (1 − abq n )(1 − acq n )(1 − adq n )(1 − bcq n )(1 − bdq n )(1 − cdq n )(1 − abcdq n − )(1 − abcdq n ) (1 − abcdq n +1 )( n = 0 , , , . . . ). Hence P n ( z ) = 2 n ( c c ) n/ Q n (cid:18) z − c √ c c ; a, b, c, d | q (cid:19) ( n = 0 , , . . . ) , where ( Q n ( · ; a, b, c, d | q )) n ≥ is the monic OPS of the Askey-Wilson polynomials (see[11, (14.1.5)]). Acknowledgements
This work is supported by the Centre for Mathematics of the University ofCoimbra – UID/MAT/00324/2019, funded by the Portuguese Government throughFCT/MEC and co-funded by the European Regional Development Fund throughthe Partnership Agreement PT2020. DM is also supported by the FCT grantPD/BD/135295/2017.
N THE FUNCTIONAL EQUATION FOR CLASSICAL OPS 25
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