On characterizations of classical polynomials

ON CHARACTERIZATIONS OF CLASSICAL POLYNOMIALS R. ´ ALVAREZ-NODARSE Abstract. It is well known that the classical families of Jacobi, Laguerre, Hermite, and Bessel polynomials are characterized as eigenvectors of a second order linear differential operator with polynomial coefficients, Rodrigues formula, etc. In this paper we present an unified study of the classical discrete polynomials and q-polynomials of the qHahn tableau by using the difference calculus on linear-type lattices. We obtain in a straightforward way several characterization theorems for the classical discrete and q-polynomials of the q-Hahn tableau. Finally, a detailed discussion of the Marcelln et. al. characterization is presented. 1. Introduction The classical polynomials (those of Hermite, Laguerre, Jacobi, and Bessel) are the most important instances of orthogonal polynomials. One of the reasons is because they satisfy not only a three-term recurrence relation (TTRR) xPn(x) = αnPn+1(x) + βnPn(x) + γnPn−1(x), γn6= 0, P−1(x) = 0, P0(x) = 1,(1.1) but also other useful properties: they are the eigenvectors of a second order linear differential equation with polynomial coefficients, their derivatives also constitute an orthogonal family, their generation functions can be given explicitly, among others (see for instances [1, 8, 24, 25] or the more recent work [3]). Among all these properties there are very important ones that characterize the classical families. In fact not every property characterizes the classical polynomials. The simplest example is the TTRR (1.1). It is well known (see e.g. [8]) that the TTRR characterizes the orthogonal polynomials if γn6= 0 for all n∈N. This is the so-called Favard Theorem (for a review see [18]). Nevertheless there exist several families that satisfy the TTRR but not a linear differential equation with polynomial coefficients, or a Rodrigues-type formula. In fact only few families of orthogonal polynomials satisfy these properties as we will show. For reviews on the characterization theorems see [1, 3, 8]. 2000 Mathematics Subject Classification. 33C45,33D45. Key words and phrases. classical polynomials, q-Hahn tableau, discrete polynomials, characterization theorems. 1 2 R. ´ ALVAREZ-NODARSE The oldest characterization is the so called Hahn characterization —unless this was firstly observed and proved for the Jacobi, Laguerre, and Hermite polynomials by N. Sonin in 1887—. In [11], Hahn proved the following Theorem 1.1 (Sonin-Hahn [11, 19]).Given a sequence of orthogonal polynomials (Pn)n, it is a classical sequence if an only if the sequence of their derivatives (P0 n)nis an orthogonal sequence. In fact the following theorem holds (see the nice survey paper [1] and also [19, 20]) Theorem 1.2. The following properties are equivalent: (1) (Pn)nis a classical orthogonal polynomial sequence (COPS), (2) The sequence of their derivatives (P0 n)nis an COPS1, (3) (Pn)nsatisfies the second order linear differential equation with polynomial coefficients (Bochner [7]) σ(x)P00 n(x) + τ(x)P0 n(x) + λPn(x) = 0, where deg(σ)≤2,deg(τ) = 1, and are independent of n, and λis a constant independent of x. (4) (Pn)ncan be expressed by the Rodrigues formula (Tricomi [27] and Cryer [9])Pn(x) = Bn ρ(x) dn dxn[σn(x)ρ(x)]. (5) The polynomials are orthogonal with respect to a weight function ρ that satisfies the Pearson differential equation [σ(x)ρ(x)]0=τ(x)ρ(x), where the polynomials σand τare such that deg(σ)≤2,deg(τ) = 1 (Hildebrandt [14]). (6) There exist three sequences (an)n,(bn)n,(cn)n, and a polynomial σ, deg(σ)≤2, such that (Al-Salam & Chihara [2]) σ(x)P0 n(x) = anPn+1(x) + bnPn(x) + cnPn−1(x), n ≥1.(1.2) (7) There exist two sequences (fn)nand (gn)nsuch that the following relation for the monic polynomials holds (Marcell´an et al [19]) Pn(x) = P0 n+1(x) n+ 1 +fnP0 n(x) + gnP0 n−1(x), n ≥1.(1.3) The proof of this theorem can be found in the appendix A. A natural extension of the classical polynomials are the so-called discrete polynomials (those of Charlier, Meixner, Kravchuk, and Hahn, see e.g. [8, 24, 25]) and the q-polynomials (see e.g. [6, 24, 25]). In fact, Hahn in 1949 [13] posed the problem of finding all the orthogonal polynomial sequences that satisfy the conditions 2–5 from theorem 1.2 but instead of using the 1Notice that this is not the Hahn theorem. In the Hahn theorem the orthogonality of both sequences it is impossed whereas here a more restrictive conditions is supposed: (Pn)nor (P0 n)nis a classical family. ON CHARACTERIZATIONS OF CLASSICAL POLYNOMIALS 3 derivatives, he use the linear operator Lq,w Lq,wf(x) = f(qx +w)−f(x) (q−1)x+w, q, w ∈R+. Hahn solved the problem for the case when q∈(0,1) and w= 0, that leads to the q-Hahn tableau (see e.g. [16] and [5]). The case w=q= 1, leads to the classical discrete polynomials of Charlier, Meixner, Kravchuk, and Hahn (see [8, 17, 24]). A complete study of the characterization theorems for these two cases has been performed using a functional approach in the papers [10] (discrete case) and [21] (“q” case). The main aim of the present paper is twice: on one hand to present a very simple and unified approach to the afore said two cases using the theory of difference equations on lattices presented in [24, 25], and on the other hand to complete the study started in [10, 20, 21]. The structure of the paper is as follows: In section 2 we introduce the “linear” lattices x(s) and characterize them. In section 3 the characterization theorem is presented and proved for any linear-type lattice and, as corollaries, the corresponding theorems for the uniform lattice x(s) = sand the q-linear lattice x(s) = c1qs+c2are obtained. Finally, in Section 4, we discuss each case in details as well as the classical case (that can be obtained taking an appropriate limit q→1−). In particular, some problems related with the Marcell´an et al. characterization [19] are discussed. 2. The linear-type lattices x(s) Definition 2.1. We say that x(s)is a linear-type lattice if x(s+ζ) = F(ζ)x(s) + G(ζ),∀s, ζ ∈C, F(ζ)6= 0.(2.1) Obviously for the linear lattice x(s) = swe have F(ζ) = 1 and G(ζ) = ζ. Another important instance of the linear-type lattice is the q-linear lattice, (q6={0,±1}), i.e., the functions of the form x(s) = Aqs+B. In this case x(s+ζ) = F(ζ)x(s) + G(ζ), where F(ζ) = qζand G(ζ) = B(1 −qζ). Proposition 2.2. Let q6={0,±1}. The function x(z)is a q-linear lattice of zif and only if it satisfies x(z+ 1) = qx(z) + C. Proof. A straightforward computation shows that if x(z) is a q-linear function of n, i.e., x(z) = cqz+dthen it satisfies the recurrence formula x(z+ 1) = qx(z) + C, where C=d(1 −q) is a constant. But the general solution of the difference equation x(z+ 1) = qx(z) + Cis x(z) = Aqz+D, where Aand Dare, in general, non-zero constants.  Notice that for the linear-type lattices, if Qm(x(s)) is a polynomial of degree min x(s), Qm(x(s+α)) is also a m−th degree polynomial in x(s), i.e., Qm(x(s+α)) = e Qm(x(s)). Moreover, for the linear-type lattices we have the following 4 R. ´ ALVAREZ-NODARSE Lemma 2.3. Let x(s)be a linear-type lattice and Qm(x(s)) a polynomial of degree min x(s). Then ∆Qm(x(s+α)) ∆x(s+β)=Rm−1(x(s)),∀α, β ∈C, where Rm−1(x(s)) is again a polynomial in x(s)but of degree m−1and ∆f(s) = f(s+ 1) −f(s). Proof. It is sufficient to prove the lema for the powers xn(s). Since x(s) is a linear-type lattice ∆xn(s+α) ∆x(s+β)=∆(F(α)x(s) + G(α))n F(β)∆x(s)= n X k=0 n kF(α)kG(α)n−k F(β) ∆xk(s) ∆x(s). But ∆xk(s)/∆x(s) is a polynomial of degree k−1 in x(s) and therefore ∆xn(s+α)/∆x(s+β) also is.  To conclude this section let point out the following Remark 2.4. From Proposition 2.2 and Definition 2.1 it follows that the only linear-type lattices are those corresponding to F(1) = 1 (the linear lattice x(s) = C1s+C2) and the ones when F(1) = q6={0,±1}(the q-linear lattices x(s) = c1qs+c2). 3. The characterization theorem for classical polynomials In the sequel we will assume that (Pn[x(s)])nis a sequence of orthogonal polynomials on a linear-type lattice x(s). For sake of simplicity we will denote Pn(s) := Pn[x(s)]. Since Pn(s) are orthogonal they satisfy the TRRR x(s)Pn(s) = αnPn+1(s) + βnPn(s) + γnPn−1(s), P−1(s) = 0, P0(s) = 1. (3.1) Let us point out that if γn6= 0, for all n∈N, then the above TTRR defines an orthogonal polynomial sequence. Nevertheless there are several examples for which γn= 0 for some n0∈N(e.g. the Hahn and q-Hahn polynomials). In this case we have a finite family of orthogonal polynomials (see e.g. [8, 25]). In the first case, i.e., when γn6= 0, for all n∈Nwe say that it is a quasi-definite case [8] (also called the regular case) whereas in the second one, we get a weak-quasi-definite case or weak-regular case. Here we will deal with the “classical” polynomials and we will assume that γn6= 0 for all n∈ N where by Nwe denote the set N= 1,2,...,n0for some n0∈Nor N:= N. Here we will use the notation of the theory of difference calculus on nonuniform lattices (for more details see [25, §13] or [24, chapter 3]). Let s=a, a + 1, a + 2,.... We will define the forward and backward differences in x(s) by ∆y[x(s)] ∆x(s),∇y[x(s)] ∇x(s), ON CHARACTERIZATIONS OF CLASSICAL POLYNOMIALS 5 respectively, where ∇f(s) = f(s)−f(s−1), ∆f(s) = f(s+ 1) −f(s). For the operator ∆ we have ∆{f(s)g(s)}=g(s){∆f(s)}+f(s+ 1){∆g(s)}.(3.2) Thus the following formula of summation by parts holds b X s=a f(s)∆g(s) = f(s)g(s) b+1 a − b X s=a∆f(s)g(s+ 1).(3.3) Also we define the k-th forward difference of a function f(s) by ∆(k)f(s) := ∆ ∆xk−1(s) ∆ ∆xk−2(s)... ∆ ∆x(s)f(s), xm(s) = xs+m 2. Remark 3.1. Notice that the differences ∆(k)Pn(s)can be written in the linear-type lattice, up to a constant factor, as (∆/∆x(s))kPn(s). Moreover, the operator ∆/∆x(s)for the q-linear lattice x(s) = c1qsbecomes into the classical Jackson operator Dqdefined by DςP(x) = P(ςx)−P(x) x(ς−1) , ς 6= 0,±1.(3.4) Next we state the Hahn-Lesky theorem: Theorem 3.2. Given a sequence of orthogonal polynomials (Pn)n, it is a classical sequence if an only if •The sequence of their finite differences (∆Pn)nis an orthogonal sequence [17, 10]. •The sequence of their q-differences (DqPn)nis an orthogonal sequence [13, 21]. Notice that since we are deal with linear lattices the statement of the theorem can be replaced by the following equivalent one: Theorem 3.2. A sequence of orthogonal polynomials (Pn)nis classical if and only if the sequence of their finite differences (∆/∆x(s)Pn)nis an orthogonal sequence. The standard proof of this theorem can be found in [17] for the linear lattice x(s) = s, and in [10] using the functional technique developed by Maroni. For the q-linear lattice x(s) = qsit has been done by Hahn in [13] and using a functional approach in [21]. We start with the following Definition 3.3. We say that the sequence (Pn)nis a classical family on the linear-type lattice if they are orthogonal with respect to the discrete measure ρ(s)∇x1(s), i.e., b−1 X s=a Pn(s)Pm(s)ρ(s)∇x1(s) = δnmd2 n,∆s= 1,(3.5) 6 R. ´ ALVAREZ-NODARSE where ρis the solution of the Pearson-type equation ∆ ∆x(s−1/2)[σ(s)ρ(s)] = τ(s)ρ(s),(3.6) and σand τare fixed polynomials on x(s)of degree at most 2 and exactly 1. The function ρis usually called the orthogonalizing weight function of the polynomial family (Pn)n. Now we are ready to enunciate our main result: Theorem 3.4. Let x(s)be a linear-type lattice and let σ(s)and ρ(s)be two functions such that akσ(a)ρ(a) = bkσ(b)ρ(b) = 0, forall k≤0. Then, the following properties are equivalent (1) (Pn)nis a classical orthogonal polynomial sequence (COPS). (2) The sequence of their differences ∆(1)Pnnalso is an COPS. (3) (Pn)nsatisfies the second order linear difference equation with polynomial coefficients σ(s)∆ ∆x(s−1/2) ∇Pn(s) ∇x(s)+τ(s)∆Pn(s) ∆x(s)+λPn(s) = 0,(3.7) where deg(σ)≤2,deg(τ) = 1, are independent of nand λis a constant independent of x. (4) (Pn)ncan be expressed by the Rodrigues-type formula2 Pn(s) = Bn ρ(s) ∇ ∇x1(s) ∇ ∇x2(s)··· ∇ ∇xn(s)[ρn(s)].(3.8) (5) The polynomials are orthogonal with respect to a weight function ρ that satisfies the Pearson-type difference equation (3.6), where deg(σ)≤2,deg(τ) = 1. (6) There exist three sequences (an)n,(bn)n,(cn)n, and a polynomial φ, deg(φ)≤2, such that φ(x)∆Pn(s) ∆x(s)=anPn+1(x) + bnPn(x) + cnPn−1(x), n ≥1. (7) There exist three sequences (en)n,(fn)n,(gn)nsuch that the following relation holds for all n≥1 Pn(x) = en ∆Pn+1(s) ∆x(s)+fn ∆Pn(s) ∆x(s)+gn ∆Pn−1(s) ∆x(s), en6= 0, gn6=γn, where γnis the corresponding coefficient of the TTRR (1.1). As a simple consequence of the above theorem we have the following Corollary 3.5 ([10, 21]).The discrete polynomials on the linear lattice x(s) = sare classical. The q-polynomials in the q-linear lattice (or exponential lattice) x(s) = c1qs+c2are classical. 2The operator ∇ ∇x1(s) ∇ ∇x2(s)· · · ∇ ∇xn(s)in the linear type lattices can be rewritten in the form ∇nfor the linear lattice and q−n(n+1)/2“∇ ∇x(s)”nfor the q-linear ones. ON CHARACTERIZATIONS OF CLASSICAL POLYNOMIALS 7 Proof. It follows from the fact that x(s) = sand x(s) = c1qs+c2are lineartype lattices.  Let us prove the Theorem 3.4. The idea of the proof is summarized in the next figure:  * HH Hj ? 6  1 2 3 4 5 HH Hj  * 6 6 6 7 We start proving that (1)→(2): Proposition 3.6. Let x(s)be a linear-type lattice and let (Pn)nbe a classical family orthogonal with respect to a weight function ρ, solution of the Pearson-type equation (3.6) and such that3 σ(a)ρ(a) = σ(b)ρ(b) = 0.(3.9) Then the sequence ∆(1)Pn(s)n, where ∆(1)Pn(s) = ∆Pn(s) ∆x(s), is also a classical orthogonal family with respect to the function ρ1(s)∆x(s), where the weight function is ρ1(s) = σ(s+ 1)ρ(s+ 1). Proof. Let Qk(s) be an arbitrary k-th degree polynomial on x(s), k < n. The orthogonality conditions for (Pn)nyield, for all k < n, 0 = b−1 X s=a Pn(s)Qk−1(s)τ(s)ρ(s)∇x1(s) (from (3.6)) = b−1 X s=a Pn(s)Qk−1(s)∆(σ(s)ρ(s)) (from (3.3), (3.9)) =− b−1 X s=a ∆(Pn(s)Qk−1(s))σ(s+ 1)ρ(s+ 1) Applying the Leibniz rule (3.2) 0 = − b−1 X s=a (∆Pn(s))Qk−1(s)σ(s+ 1)ρ(s+ 1)+ b−1 X s=a Pn(s+ 1)(∆Qk−1(s))σ(s+ 1)ρ(s+ 1) (s→s−1, and (3.9)) 3This condition leads to the so-called discrete orthogonal polynomials, i.e., polynomials with a discrete orthogonality of the form (3.5). For the q-linear lattices (3.5) becomes into the q-Jackson integral (see e.g. [5, 15, 16]). For the continuous orthogonality see [24, §3.10]. 8 R. ´ ALVAREZ-NODARSE =− b−2 X s=a∆Pn(s) ∆x(s)Qk−1(s)σ(s+ 1)ρ(s+ 1)∇x1(s+ 1/2)+ b X s=a+1 Pn(s) ∆Qk−1(s−1) ∆x(s−1/2) σ(s)ρ(s)∇x1(s) Next, we use Lemma 2.3 as well as the conditions (3.9), then 0 = − b−2 X s=a∆Pn(s) ∆x(s)Qk−1(s)σ(s+ 1)ρ(s+ 1)∇x1(s+1/2)+ b−1 X s=a Pn(s)Rk−2(s)σ(s) |{z } degree ≤n ρ(s)∇x1(s) (from (3.9), (3.5)) =− b−2 X s=a∆Pn(s) ∆x(s)Qk−1(s)σ(s+ 1)ρ(s+ 1)∇x1(s). Thus, ∆Pn(s)/∆x(s) is orthogonal with respect to ρ1(s)∇x1(s+ 1/2) = σ(s+ 1)ρ(s+ 1)∆x(s). We only need now to prove that ∆(1)Pn(s) is a classical family. For doing this notice that the weight function ρ1(s) satisfy the Pearson type equation (see e.g. [24, §3.2.2]) ∆ ∆x1(s−1/2) [σ(s)ρ1(s)] = τ1(s)ρ1(s) where τ1is a first degree polynomial on x(s) given by τ1(s) = σ(s+ 1) −σ(s) + τ(s+ 1)∆x1(s) ∆x(s). Thus ρ1satisfies a difference equation of the form (3.6). This complete the proof.  In the same way, using induction we have Corollary 3.7. Let x(s)be a linear-type lattice and let (Pn)nbe a classical family. Then, the sequence of their k-th finite differences ∆(k)Pn(s), where ∆(k):= ∆ ∆xk−1(s) ∆ ∆xk−2(s)... ∆ ∆x(s), also is a classical family. Now we prove that (1)+(2)→(3): Proposition 3.8. Let x(s)be a linear-type lattice. If the sequences (Pn)n and ∆(1)Pnnare classical, then (Pn)nsatisfies the second order linear difference equation of hypergeometric type (3.7). ON CHARACTERIZATIONS OF CLASSICAL POLYNOMIALS 9 Proof. Let k < n. Then, using the orthogonality of ∆(1)Pn, 0 = b−2 X s=a ∆Pn(s) ∆x(s) ∆Qk(s) ∆x(s)σ(s+ 1)ρ(s+ 1)∇x1(s+1/2) (from (3.9)) = b−1 X s=a ∆Pn(s) ∆x(s)∆Qk(s)σ(s+ 1)ρ(s+ 1) (from (3.3), (3.9)) =− b−1 X s=a Qk(s)∆∆Pn(s−1) ∆x(s−1) σ(s)ρ(s)(∆f(s) = ∇f(s+ 1)) =− b−1 X s=a Qk(s)∆∇Pn(s) ∇x(s)σ(s)ρ(s)(from (3.2)) =− b−1 X s=a Qk(s)σ(s)ρ(s)∆∇Pn(s) ∇x(s)+∇Pn(s+1) ∇x(s+1) ∆[σ(s)ρ(s)] (from (3.6)) =− b−1 X s=a Qk(s) σ(s)∆ ∆x(s−1/2) ∇Pn(s) ∇x(s)+τ(s)∆Pn(s) ∆x(s)!ρ(s)∇x1(s). But, since the lattice x(s) is of the linear type, Q(s) := σ(s)∆ ∆x(s−1/2) ∇Pn(s) ∇x(s)+τ(s)∆Pn(s) ∆x(s) is a polynomial of degree nin x(s). Therefore, it should be, up to a constant factor (in general depending on n) the polynomial Pn(s). Thus Q(s) = −λPn(s).  Remark 3.9. The proof of the last proposition in the linear lattice x(s) = s can be found in the first Russian edition of the book [25]. The last proposition is very important because it gives a very simple method for finding the classical polynomials on the linear-type lattice. In fact, it was the key in the proofs of Hahn and Lesky for proving the Theorem 3.2. The solutions of the difference equation (3.7) have been extensively studied (see e.g. [6, 24, 25]). In particular they can be written by the Rodriguestype formula (3.8) [24, 25], so (3)→(4). Let us mention that from the Rodrigues-type formula (3.8) one can obtain an explicit expression for the classical polynomials in terms of the hypergeometric or basic hypergeometric series as it is shown in several previous works (see e.g. [6, 24]). Another consequence of the Rodrigues formula is the following: Putting n= 1 in (3.8) we obtain P1(s) = B1 ρ(s) ∆ ∆x(s−1/2)[σ(s)ρ(s)] ⇒∆ ∆x(s−1/2)[σ(s)ρ(s)] = ρ(s)τ(s), i.e. the Pearson-type equation (3.6) thus (4)→(5). 16 R. ´ ALVAREZ-NODARSE But now, using the expression (see e.g. [3, page 108]) cn=λnγn/n, we see that for all n≥1, cn6= 0. The condition p+na 6= 0 for all n∈ N is the admissibility condition in this case. Let us now analyze the structure relation (4.12). In this case [3, page 109] gn=−(n−1)aγn p+(n−2)a, therefore in the quasi-definite case gn6= 0. If γn=gnfor all n, then we obtain that p+ (2n−3)a= 0, for all nwhich is in contradiction with the admissibility condition. Remark 4.3. In [10] the condition gn6= 0 for all n∈ N was imposed but not the more restrictive one gn6=γn, from where the first one immediately follows. For the discrete case in [10] the admissibility condition p+na 6= 0 it is assumed and therefore gn6=γnfor all n∈ N . From the above discussion also follows that the classical discrete polynomials are completely characterized by the relation (4.12) with the restriction gn6=γnfor all n∈ N. Moreover, if gn=γnfor all n∈ N, then the corresponding orthogonal polynomial sequence, if such a sequence exists, is not a classical one. 4.3. The classical case. The classical case can be obtained from the q-case taking the limit q→1−. Nevertheless the Theorem 1.2 can be proven using the same scheme section 3. The only difference is that here one uses he standard integral calculus and integration by parts instead of the calculus with the difference operator. Of particular interest is the proof of property 7 so we will provide it here: Taking derivatives of the TTRR (1.1) and using (1.3), we have the expression xP0 n(x) = n n+ 1P0 n+1(x) + (βn−fn)P0 n(x) + (γn−gn)P0 n−1(x),(4.13) from where, if gn6=γn,∀n∈ N, and using the Favard theorem the sequence (P0 n)nis an OPS, and therefore by the Sonin-Hahn Theorem 1.1 Pnis a classical family. Notice again that the condition gn6=γnshould be imposed. Using the formulas in [20] it is easy to see that this condition is equivalent to the condition nσ00/2 + τ0= 0 which is nothing else that the admissibility condition for the classical polynomials [20]. Let us point out that the more restrictive condition γn6=gnfor all n∈Nwas not considered in [19] (they considered only the regular case, i.e., γn6= 0). As in the cases already discussed we conclude that the classical continuous polynomials are completely characterized by the relation (1.3) with the restriction gn6=γnfor all n∈N. Moreover, if gn=γnfor n= 1,2,...,n0, then the corresponding orthogonal polynomial sequence, if such a sequence exists, is not a classical one. 4.4. The Marcell´an et al. characterization. At this point the following question arises: what happens if we do not impose the condition gn6=γn, ∀n= 1,2,...,n0? There is any family of orthogonal polynomials, necessarily non classical, that satisfies the TTRR (1.1) where γn6= 0 for n∈ N, and the relation (1.3) with gn=γnfor all n∈ N? i.e., ON CHARACTERIZATIONS OF CLASSICAL POLYNOMIALS 17 Pn(x) = P0 n+1(x) n+ 1 +fnP0 n(x) + γnP0 n−1(x).(4.14) To answer this question we can use (4.13) but rewritten in the form4 P0 n+1(x) = n+ 1 n(x−βn+fn)P0 n(x), that leads to P0 n(x) = n n−1 Y j=1 (x−βj+fj), n ≥2. Therefore, substituting the last expression in (4.14) we get, denoting ξj= βj−fj, Pn(x) = [(x−ξn)(x−ξn−1) + nfn(x−ξn−1) + (n−1)γn] n−2 Y j=1 (x−ξj). But this implies that for n≥3, two consecutive polynomials have common zeros that is a contradiction. Therefore there is not any family of orthogonal polynomials that satisfy (4.14). For the linear lattices x(s) = sand x(s) = qsthe situation is the same. We present here the computations only for the q-case, the other case is analogous —in fact the final expression for the polynomials Pncoincide with the one in the classical “continuous” case. For the q-case we proceed as before, i.e., we take the q-derivatives of the TTRR (3.1) and use the relation (4.3) where en= 1/[n]q,gn=γn,F(1) = 1, G(1) = 0, we obtain DqPn+1(x) = [n+ 1]q [n]q (x−ξn/q), ξj=βj−fj. Substituting it in (4.3) when gn=γnwe obtain the following expression for the polynomials Pn Pn(x) = [(x−ξn/q)(x−ξn−1/q) + [n]qfn(x−ξn−1/q) + [n−1]qγn] n−2 Y j=1 (x−ξj/q). As before, from this expression follows that for n≥3, two consecutive polynomials has common zeros, that is in contradiction with the fact that they constitutes an orthogonal sequence. From the above discussion follows that the structure relation (3.11) when gn6=γnfor all n∈ N completely characterizes the classical orthogonal polynomials. 4As in section 4.3 we will take the derivative of the TTRR (1.1) but now use (4.14). 18 R. ´ ALVAREZ-NODARSE Acknowledgements. The author thanks L. Cardoso, J.S. Dehesa, A. Dur´an, F. Marcell´an, J. C. Medem, and J.C. Petronilho for stimulating discussions. This work was supported by the Ministerio de Ciencia y Tecnolog´ıa of Spain under the grant BFM-2003-06335-C03, and the Junta de Andaluc´ıa under grant FQM-262. Also the financial support by Acciones Integradas HispanoLusas HP2002-065 & E-6/03 is acknowledged. Appendix A. The classical polynomials In this appendix we will present the proof of the Theorem 1.2. We will follow the same scheme in Section 3 (see figure 1). As starting point we will use the Pearson equation, i.e., we say that the classical polynomials are the polynomials orthogonal with respect to a continuous weight function ρsupported in the interval (a, b), solution of the Pearson equation [σ(x)ρ(x)]0=τ(x)ρ(x),(A.1) where σand τare polynomials of degree at least two and exactly one, respectively, and such that the following boundary conditions hold5σ(a)ρ(a) = σ(b)ρ(b) = 0. (1)→(2): Using the orthogonality of the classical family (Pn)nwith respect to ρwe have that for any polynomial of degree less than or equal to k−1, Qk−1, with k < n, 0 = Zb a Pn(x)Qk−1(x)τ(x) |{z } degree≤k<n ρ(x)dx =Zb a Pn(x)Qk−1(x)[σ(x)ρ(x)]0dx =Pn(x)Qk−1(x)σ(x)ρ(x)b a |{z } =0 −Zb a [Pn(x)Qk−1(x)]0σ(x)ρ(x)dx =−Zb a Pn(x) degree<n z}| { Q0 k−1(x)σ(x)ρ(x)dx | {z } =0 −Zb a P0 n(x)Qk−1(x)[σ(x)ρ(x)]dx. Thus P0 nis orthogonal to any polynomial of degree k−1< n −1, i.e., (P0 n)n is also an orthogonal family. Furthermore, since the weight function for the sequence (P0 n)nis ρ1(x) = σ(x)ρ(x), we have that they satisfy the equation [σ(x)ρ1(x)]0= [τ(x) + σ0(x)]ρ1(x), i.e., a Pearson equation (A.1). 5These conditions follow from the fact that for the classical families the moments µn=Rb axnρ(x)dx,n≥0, of the measure associated with ρ(x) are be finite. ON CHARACTERIZATIONS OF CLASSICAL POLYNOMIALS 19 (2)→(3): We use now that (P0 n)nis an orthogonal family with respect to the weight function ρ1(x) = σ(x)ρ(x). Thus 0 = Zb a P0 n(x)Q0 k(x)τ(x)ρ1(x)dx =P0 n(x)Qk(x)σ(x)ρ(x)b a |{z } =0 −Zb a [σ(x)ρ(x)P0 n(x)]0Qk(x)dx =−Zb a Qk(x){[σ(x)ρ(x)]0 |{z } =τ(x)ρ(x) P0 n(x) + σ(x)ρ(x)P00 n(x)} =Zb a Qk(x)[σ(x)P00 n(x) + τ(x)P0 n(x)]ρ(x)dx. But since the last integral vanishes for every polynomial Qkof degree k < n then σ(x)P00 n(x)+τ(x)P00 n(x) should be proportional to Pn, i.e., σ(x)P00 n(x)+ τ(x)P00 n(x) = −λnPn, where λnis a constant, in general depending on n. (3)→(4): The solution of the above differential equation can be written in the following compact form (see e.g. [25, §2] or [24, §1.2]) usually called the Rodrigues formula Pn(x) = Bn ρ(x) dn dxn[σn(x)ρ(x)], where Bnis a constant. (4)→(5): It follows from the Rodrigues formula just putting n= 1. (4)→(6): From the Rodrigues formula the following expression (see e.g. [25, Eq. (7) page 25]) immediately follows σ(x)P0 n(x) = λn nτ0 nτn(x)Pn(x)−Bn Bn+1 Pn+1(x), τn(x) = τ(x) + nσ0(x), from where, using the three-term recurrence relation for the family (Pn)n the structure relation (1.2) follows. (6)→(2): Suppose that (1.2) holds where deg σ≤2 and (Pn)nis an orthogonal family. Notice that the integral Zb a Qk(x)P0 n(x)σ(x)ρ(x)dx= Zb a Qk(x)ρ(x)[anPn+1(x)+bnPn(x)+cnPn−1(x)]dx vanishes for all k < n −1. Then (P0 n)nis an orthogonal family with respect to the weight function ρ1(x) = σ(x)ρ(x) and therefore by the Sonin-Hahn Theorem 1.1 (Pn)nis a classical family. 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E-18071 Granada, Spain E-mail address:[email protected]