Page 114 - Textos de Matemática Vol. 34

104 JOSE´ CARLOS PETRONILHO
(ii) There exist complex numbers rn and sn, with rnsn ̸= 0 and rn ̸= sn for all n = 1,2,···, such that
(8.4) Pn(x) + rnPn−1(x) = Qn(x) + snQn−1(x) for all n = 1,2,···.
Proof. The implication (ii)⇒(i) follows immediately from Theorem 8.1. In fact, if (ii) holds then the hypotheses of Theorem 8.1 are fulfilled since in this case (N = M = 1) we have
2 1 r1 A:=[αij]i,j=1 = 1 s ,
1
hence det A ̸= 0 (by hypothesis) and so (8.3) holds for some a, b, λ ∈ C (with λ ̸= 0). Further, multiplying both sides of (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,··· .
Since (by hypothesis) rnsn ̸= 0 and rn ̸= sn for all n = 1,2,···, thus ⟨u,Pn2⟩ ≠ λ⟨v,Q2n⟩ for all n = 1,2,3,···.

In order to prove that (i)⇒(ii), denote by {βn, γn+1}n≥0 and {βn, γn+1}n≥0
the sets of parameters which appear in the three-term recurrence relations for the MOPS’s (Pn)n and (Qn)n, respectively. Then, by computing the Fourier coefficients of (x − a)Qn and (x − b)Pn with respect to (Pn)n and (Qn)n (resp.), and using the functional equation (8.3) as well as the above mentioned three-term recurrence relations, one get

(8.5) Pn+1 + rn,nPn + rn,n−1Pn−1 = Qn+1 + (βn − a)Qn + γnQn−1 ,
(8.6) Pn+1 + (βn − b)Pn + γnPn−1 = Qn+1 + sn,nQn + sn,n−1Qn−1 ,
where
r
= ⟨u,(x−a)QnPn⟩ =β −a+γ ⟨u,QnPn−1⟩ , ⟨u, Pn2⟩ n n ⟨u, Pn2⟩
n,n
rn,n−1 = ⟨u, (x − a)QnPn−1⟩ = λ ⟨u, Pn2−1⟩
⟨v, Q2n⟩ , ⟨u, Pn2−1⟩
1 ⟨u, Pn2⟩ λ ⟨v,Q2n−1⟩
sn,n=βn−b+γn
⟨v, PnQn−1⟩ ⟨v,Q2n⟩
.
, sn,n−1= (rn,n − βn + b)Pn + (rn,n−1 − γn)Pn−1
Now, subtracting (8.6) from (8.5) we get (8.7)

=
(βn − a − sn,n)Qn + (γn − sn,n−1)Qn−1