Page 115 - Textos de Matemática Vol. 34

ORTHOGONAL POLYNOMIALS 105
for all n = 1, 2, · · · . Since Pn and Qn are monic polynomials, this implies
rn,n −βn +b=βn −a−sn,n , n=1,2,··· .
Further, rn,n −βn +b ̸= 0 and βn −a−sn,n ̸= 0 for all n = 1,2,···. In fact,
ifeitherrk,k−βk+b=0orβk−a−sk,k =0forsomek≥1thenfrom(8.7) we get rk,k−1 = γk and so, taking into account the definition of rk,k−1 we get ⟨u,Pk2⟩ = λ⟨v,Q2k⟩ , in contradiction with the hypothesis. As a consequence, (8.7) can be rewritten as (8.4), with
rn := rn,n−1 −γn , sn := γn −sn,n−1 (n=1,2,···). rn,n −βn +b rn,n −βn +b
Itisclearthatrnsn ̸=0foralln≥1,sinceifrk =0orsk =0forsomek≥1 then this imply again ⟨u,Pk2⟩ = λ⟨v,Q2k⟩ . We now show that r1 ̸= s1. If r1 = s1 then, by (8.4), P1 = Q1 and so using (8.3) we get ⟨u, P12⟩ = ⟨(x − a)u, P1⟩ = ⟨(x − a)u, Q1⟩ = λ⟨v, Q21⟩ , which violates again the hypothesis. Finally, we prove that also rn ̸= sn for all n ≥ 2. In fact (as above) multiplying (8.4) by (x − a)Pn−2 and then applying u and taking into account (8.3), we find
rn ⟨u,Pn2−1⟩ = sn λ⟨v,Q2n−1⟩ , n = 2,3,··· .
Therefore, since by hypothesis 0 ̸= ⟨u,Pn2−1⟩ ̸= λ⟨v,Q2n−1⟩ ̸= 0 , and using that
rnsn ̸=0,wegetrn ̸=sn foralln. Remark 8.1. Under the conditions of Corollary 8.2 we can show that the formal
Stieltjes series Su(z) and Sv(z) associated to u and v (resp.) satisfy
(8.8) λ (z − b) Sv(z) + (z − a) Su(z) = λ − 1
(with the the normalization condition v0 = u0 = 1). This gives us a simple relation between the moments of the functionals u and v. Further, if one of the functionals u or v is semiclassical, say, e.g., D(φu) = ρu for some polynomials φ and ρ, then, being a rational modification of u, v is also semiclassical. The class of v can then be easily obtained (by standard methods) from (8.8).
Remark 8.2. The relation (8.3) between the functionals u and v can be written
as
or, according to the results in section 4,
v = v0δb + 1 (x − b)−1(x − a) u , λ
v = −u0 − λv0 δb + 1 u + b − a (x − b)−1u . λλλ
This formula is useful in the study of the positive-definite case, since from it an integral representation for v can be given assuming that u also has an integral