On the inverse problem of the product of a form by a polynomial: The cubic case
P. Maroni
a,∗, I. Nicolau
baUniversité Pierre et Marie Curie-C.N.R.S., Laboratoire d’Analyse Numérique, 4 Place Jussieu, 75252 Paris cedex 05, France bDepartamento de Matemática, Universidade de Trás-os-Montes e Alto Douro, Apartado 202, 5001-911 Vila Real, Portugal
Abstract
A form (linear functional)u is called regular if there exists a sequence of polynomials{Pn}n0, degPn=n, which is orthogonal with respect to u. On certain regularity conditions, the product of a regular form by a polynomial is still a regular form. In this paper, we consider the inverse problem: given a regular formv, find all the regular formsuwhich satisfy the relationx3u= −λv,λ∈C− {0}. We give the second-order recurrence relation of the orthogonal polynomial sequence with respect tou. Some examples are studied.
2003 IMACS. Published by Elsevier Science B.V. All rights reserved.
Keywords: Inverse problem; Forms; Orthogonal polynomials
Introduction
The product of a linear form by a polynomial, is one of construction process of linear forms. In 1858, Christoffel proved that the product of a positive definite form by a positive polynomial is still a positive definite form [3,4]. This result has been generalized in [7], where it was proved that, under certain regularity conditions, the product of a regular formu by a polynomial R is a regular form. In particular, ifuis a semi-classical form or a second degree form, thenRuis also a semi-classical form or a second degree form, respectively. It is also interesting to consider the inverse problem, which consists of the determination of all regular formsu satisfying Ru= −λv, where v is a given regular form and λ∈C− {0}. In [9], the first author has considered the caseR(x)=x−c, withc∈C, and in [13] the case R(x)=x2. He gave the regularity conditions and the coefficients of the second-order recurrence relation satisfied by the monic orthogonal polynomial sequence (MOPS) with respect tou. The structure relation of the MOPS relatively tou, in the symmetric case, is given in [1] and in the non-symmetric case is given
*Corresponding author.
E-mail addresses: maroni@ann.jussieu.fr (P. Maroni), inicolau@utad.pt (I. Nicolau).
0168-9274/03/$30.002003 IMACS. Published by Elsevier Science B.V. All rights reserved.
doi:10.1016/S0168-9274(02)00250-7
in [2]. Many examples were also treated in [13,2,1]. In this paper, we consider the caseR(x)=x3. In the first section we will give the regularity conditions and the coefficients of the second-order recurrence relation satisfied by the MOPS with respect to u. We will also prove that if v is semi-classical then uis semi-classical and some results concerned to the class of u are given. Section 2 is devoted to the particular case of(u)2=0 and Section 3 to the case (u)2=0: regularity conditions more simple and some examples are given if the formvis symmetric and ifvis symmetric definite positive, respectively.
The regular formsufounded in the examples are semi-classical of classs∈ {1,2,3}and we present there integral representations and the coefficients of the second-order recurrence satisfied by the MOPS with respect tou.
1. The problemx3u= −λv
1.1. The main problem
Let P be the vector space of polynomials with coefficients in C and P its dual. We denote by u, f the action of u∈P on f ∈P. Let us recall that a form u is called regular if there exists a polynomial sequence{Pn}n0, degPn=n, such that u, PnPm =knδn,m, n, m0, kn=0, n0; the left-multiplication hw is defined by hw, p := w, hp. We consider the following problem: given a regular formv, find all regular formsuwhich satisfy the next equation:
x3u= −λv, λ∈C− {0}, (1.1)
with constraints(u)0=1, (v)0=1,where(u)n:= u, xn, n0,are the moments ofu. Equivalently, u=δ−(u)1δ+1
2(u)2δ−λx−3v, (1.2)
where δ, f =f (0), the derivative w=Dwof the formw is defined by w, f := − w, fand the formx−1wby x−1w, f := w, θ0f, with in general
(θcf )(x):= f (x)−f (c)
x−c , c∈C.
So the formudepends on three arbitrary parameters(u)1, (u)2andλ= −(u)3. If we suppose that the formvpossesses the following integral representation:
v, f = +∞
−∞
V (x)f (x)dx, for each polynomialf,
whereV is a locally integrable function with rapid decay, continuous at the origin and with derivative continuous at the origin, then the formuis represented by
u, f = f (0)
1+λPf +∞
−∞
V (x) x3 dx
+f(0)
(u)1+λPf +∞
−∞
V (x) x2 dx
+1 2f(0)
(u)2+λP +∞
−∞
V (x) x dx
−λPf +∞
−∞
V (x)f (x)
x3 dx, (1.3)
where, [5,8]
P +∞
−∞
V (x)
x dx= lim
ε→0+
−ε
−∞
V (x) x dx+
+∞
ε
V (x) x dx
,
Pf +∞
−∞
V (x)
x2 dx= lim
ε→0+
−ε
−∞
V (x) x2 dx+
+∞
ε
V (x)
x2 dx−2 εV (0)
,
and Pf
+∞
−∞
V (x)
x3 dx= lim
ε→0+
−ε
−∞
V (x) x3 dx+
+∞
ε
V (x)
x3 dx−2 εV(0)
.
Let{Sn}n0denote the sequence of orthogonal polynomials with respect tov S0(x)=1, S1(x)=x−ζ0,
(1.4) Sn+2(x)=(x−ζn+1)Sn+1(x)−σn+1Sn(x), n0.
Whenuis regular, let{Zn}n0be the corresponding orthogonal sequence Z0(x)=1, Z1(x)=x−β0,
(1.5) Zn+2(x)=(x−βn+1)Zn+1(x)−γn+1Zn(x), n0.
From (1.1) we know that the sequence{Zn}n0, when it exists, is among the strictly quasi-orthogonal sequences of order three with respect tov[6,10], see also [12, pp. 127, 128], this is
Z0(x)=1, Z1(x)=S1(x)+c0, Z2(x)=S2(x)+c1S1(x)+b0,
(1.6) Zn+3(x)=Sn+3(x)+cn+2Sn+2(x)+bn+1Sn+1(x)+anSn(x), n0,
withan=0, n0.Moreover, the sequence{Zn}n0is orthogonal with respect touif and only if u, Zn =0, n1;
u, xZn(x)
=0, n2,
u, xZ1(x)
=0; u, x2Zn(x)
=0, n3,
u, x2Z2(x)
=0;
(1.7) since the other orthogonality conditions have been used in (1.6). From (1.6) and (1.7) we have
0 = u, Zn+3
= u, Sn+3 +cn+2 u, Sn+2 +bn+1 u, Sn+1 +an u, Sn, n0, 0 =
u, xZn+3(x)
(1.8)
=
u, xSn+3(x) +cn+2
u, xSn+2(x) +bn+1
u, xSn+1(x) +an
u, xSn(x)
, n0, 0 =
u, x2Zn+3(x)
=
u, x2Sn+3(x) +cn+2
u, x2Sn+2(x) +bn+1
u, x2Sn+1(x) +an
u, x2Sn(x)
, n0,
with the initial conditions:
0= u, Z1 = u, S1 +c0,
0= u, Z2 = u, S2 +c1 u, S1 +b0, 0=
u, xZ2(x)
=
u, xS2(x) +c1
u, xS1(x)
+b0(u)1, 0=
u, xZ1(x)
=
u, xS1(x)
+c0(u)1, 0=
u, x2Z2(x)
=
u, x2S2(x) +c1
u, x2S1(x)
+b0(u)2.
(1.9)
If we denote
∆n:=
u, Sn+2 u, Sn+1 u, Sn u, xSn+2(x) u, xSn+1(x) u, xSn(x) u, x2Sn+2(x) u, x2Sn+1(x) u, x2Sn(x)
, n0, (1.10)
then the system (1.8) is equivalent to
∆nan= −∆n+1, n0, (1.11)
∆nbn+1=
u, Sn+2 − u, Sn+3 u, Sn u, xSn+2(x) − u, xSn+3(x) u, xSn(x) u, x2Sn+2(x) − u, x2Sn+3(x) u, x2Sn(x)
, n0, (1.12)
∆ncn+2=
− u, Sn+3 u, Sn+1 u, Sn
− u, xSn+3(x) u, xSn+1(x) u, xSn(x)
− u, x2Sn+3(x) u, x2Sn+1(x) u, x2Sn(x)
, n0. (1.13)
Proposition 1.1. The formuis regular if and only if∆n=0, n0 and(u)2−(u)21=0.In this case, the coefficients of the second-order recurrence relation of{Zn}n0are given by
β0=(u)1, βn+1=ζn+1+cn−cn+1, n0, (1.14)
γ1=(u)2−(u)21, γ2= −∆0 (u)2−(u)21−2
, (1.15)
γ3= −λ∆1[(u)2−(u)21]
∆20 , γn+4=∆n+2∆n
∆2n+1 σn+1, n0.
Proof. Necessity. If {Zn}n0 is orthogonal, it is strictly quasi-orthogonal with respect to v and then an=0, n0. This implies∆n=0, n0. Assuming the contrary, that there exists ann01 such that
∆n0=0, then from (1.11),∆0=0= − u, x2Z2(x) u, xZ1(x) =0, which is a contradiction. Moreover (u)2−(u)21= u, xZ1(x) =0.
Sufficiency. Using (1.2), the conditions u, Z1 =0, u, Z2 =0 and u, xZ2(x) =0 are satisfied for c0=ζ0−(u)1,
c1= {ζ1+ζ0} +λ+(u)1(u)2
(u)2−(u)12, (1.16)
b0=σ1+ζ02−(u)2+
ζ0−(u)1
λ+(u)1(u)2
(u)2−(u)12.
Taking account of (1.16), we also have u, xZ1(x) =(u)2−(u)12=0 and u, x2Z2(x)
=−λ2−2(u)1(u)2λ−ζ0{(u)2−(u)12}λ−(u)23
(u)2−(u)12 = −∆0 (u)2−(u)21−1
=0.
We had just proved that the initial conditions (1.9) are satisfied. Further, the system (1.8) is a Cramer system whose solution is given by (1.11), (1.12) and (1.13). Moreover, from (1.5) forn→n+2 and (1.6) forn→n+1 we can write
Sn+4(x)+cn+3Sn+3(x)+bn+2Sn+2(x)+an+1Sn+1(x)
=(x−βn+3)Zn+3(x)−γn+3Zn+2(x), n0.
Multiplying the above equation byZn+3(x)and applying the formu, we obtain u, Sn+4Zn+3 +cn+3 u, Sn+3Zn+3 =
u, xZn2+3(x)
−βn+3
u, Z2n+3
, n0,
because{Zn}n0is orthogonal with respect tou. Now, we will replaceSn+4(x)for its expression given by (1.4) forn→n+2, and then
u, xSn+3(x)Zn+3(x)
−ζn+3 u, Sn+3Zn+3 +cn+3 u, Sn+3Zn+3
=
u, xZn2+3(x)
−βn+3
u, Zn2+3
, n0.
Finally, replacingSn+3by its expression given by (1.6) u, xZn2+3(x)
−cn+2
u, xSn+2(x)Zn+3(x)
−ζn+3
u, Zn2+3 +cn+3
u, Zn2+3
=
u, xZn2+3(x)
−βn+3
u, Zn2+3
, n0, that is
−cn+2
u, Zn2+3(x)
−ζn+3
u, Z2n+3 +cn+3
u, Z2n+3
= −βn+3
u, Zn2+3
, n0.
As, u, Zn2+3 =0, we have
βn+3=ζn+3+cn+2−cn+3, n0.
Using the same proceeding, we easily prove that βn+1=ζn+1+cn−cn+1, n=0,1,
that is, we have proved (1.14). Let us see now that u, Zn2+3
=
u, xn+3Zn+3(x)
=
x3u, xnZn+3(x)
= −λ
v, xnZn+3(x)
, n0, from (1.1). Using (1.6) this becomes
u, Zn2+3
= −λan
v, xnSn(x)
= −λan
v, Sn2
, n0.
Then
γn+4= u, Z2n+4
u, Z2n+3=an+1 v, Sn2+1
an v, Sn2 =an+1
an
σn+1, n0.
By virtue of (1.11), we successively obtain γn+4=∆n+2∆n
∆2n+1 σn+1, n0, γ1=
u, xZ1(x)
=(u)2−(u)21, γ2= u, x2Z2(x)
u, xZ1(x) = − ∆0
[(u)2−(u)21]2, γ3= u, x3Z3(x)
u, x2Z2(x)= −λa0
−∆0[(u)2−(u)21]−1= −λ∆1
∆20 (u)2−(u)21 .
We have proved (1.15). ✷ 1.2. The computation of∆n
As we had seen in Proposition 1.1, it is very important to have an explicit expression for∆n. In first place, let us note that, from (1.2), we have
u, Sn+1(x)
=Sn+1(0)+(u)1Sn+1(0)+1
2(u)2Sn+1(0)−1 2λ
Sn(1)
(0), n0, (1.17)
with
Sn(1)(x):=
v,Sn+1(x)−Sn+1(ξ ) x−ξ
, n0, u, xSn+1(x)
=(u)1Sn+1(0)+(u)2Sn+1(0)−λ S(1)n
(0), n0, (1.18)
u, x2Sn+1(x)
=(u)2Sn+1(0)−λSn(1)(0), n0, (1.19)
u, x3Sn+1(x)
=0, n0. (1.20)
Using (1.4) and (1.20), we obtain for (1.10):
∆n=
u, xSn+1(x) u, Sn+1 u, Sn u, x2Sn+1(x) u, xSn+1(x) u, xSn(x)
0 u, x2Sn+1(x) u, x2Sn(x)
, n0, (1.21)
that is
∆n =
u, xSn+1(x) u, xSn+1(x) u, xSn(x) u, x2Sn+1(x) u, x2Sn(x)
−
u, x2Sn+1(x) u, Sn+1 u, Sn u, x2Sn+1(x) u, x2Sn(x)
, n0. (1.22)
From (1.17), (1.18) and (1.19), we have u, xSn+1(x) u, xSn(x)
u, x2Sn+1(x) u, x2Sn(x)
=X(1)n−1(0)λ2+
v, Sn2(u)1+ Yn(0)−Xn(0) (u)2
λ+Xn(0)(u)22, n0, (1.23) and
u, Sn+1 u, Sn u, x2Sn+1(x) u, x2Sn(x)
=1 2
Xn(1)−1
(0)λ2+ v, Sn2
−Xn(0)(u)1+1
2 Yn(0)−Xn(0) (u)2
λ +
Xn(0)(u)1(u)2+1
2Xn(0)(u)22
, n0, with
Xn(x)=Sn(x)Sn+1(x)−Sn+1(x)Sn(x), n0, (1.24) Xn(1)(x)=Sn(1)(x)
Sn(1)+1
(x)−Sn(1)+1(x) Sn(1)
(x), n0, (1.25)
Xn(x)=Sn+1(x)Sn(1)−1(x)−Sn(x)Sn(1)(x), n0, (1.26) Yn(x)=
Sn(1)−1
(x)Sn+1(x)−Sn(x) Sn(1)
(x), n0. (1.27)
Therefore,
∆n = Wn(1)−1(0)λ3 +
Sn(1)(0) v, Sn2
+ Wn(0)− Sn(1)
(0) v, Sn2
(u)1+ Tn(0)−Un(0) (u)2
λ2 +
Sn+1(0) v, Sn2
(u)21−(u)2
+ Vn(0)−Rn(0) (u)22 + Sn+1(0)
v, Sn2
+S(1)n (0)Xn(0)+Yn(0)Sn+1(0)
(u)1(u)2
λ
−Wn(0)(u)32, n0, (1.28)
with
Wn(x)=1
2Sn+1(x)Xn(x)−Sn+1(x)Xn(x), n0, (1.29) Wn(1)(x)=1
2Sn(1)+1(x) X(1)n
(x)− Sn(1)+1
(x)Xn(1)(x), n0, (1.30)
Wn(x)=Sn+1(x)Xn(1)−1(x)−Sn(1)(x)Xn(x), n0, (1.31) Un(x)=1
2Sn+1(x) X(1)n−1
(x)−Sn+1(x)X(1)n−1(x), n0, (1.32) Vn(x)=1
2Sn(1)(x)Xn(x)− Sn(1)
(x)Xn(x), n0, (1.33)
Tn(x)=1
2Sn(1)(x) Yn(x)−Xn(x)
− Sn(1)
(x) Yn(x)−Xn(x)
, n0, (1.34)
Rn(x)=1
2Sn+1(x) Yn(x)−Xn(x)
−Sn+1(x) Yn(x)−Xn(x)
, n0. (1.35)
Moreover, if the formuis regular we have for (1.11), (1.12) and (1.13) an= −∆n+1
∆n
, n0, (1.36)
bn+1=σn+2−
u, Sn+2 u, xSn+2(x) u, Sn u, xSn+2(x) u, x2Sn+2(x) u, xSn(x) u, x2Sn+2(x) 0 u, x2Sn(x)
∆−n1, n0, (1.37) cn+2=ζn+2−
u, xSn+2(x) u, Sn+1 u, Sn u, x2Sn+2(x) u, xSn+1(x) u, xSn(x)
0 u, x2Sn+1(x) u, x2Sn(x)
∆−n1, n0, (1.38) by virtue of (1.4).
1.3. Some results on the semi-classical particular case
Let us recall that a formuis called semi-classical if there exists two polynomialsΦandψ such that usatisfies the functional equation
D(Φu)+ψu=0.
The class of the semi-classical form uis s =max(degΦ−2,degψ−1) if and only if the following condition is satisfied
c
ψ(c)+Φ(c)+u, θcψ+θc2Φ>0, (1.39)
wherec∈ {x: Φ(x)=0}[11].
In the sequel the formvwill be supposed semi-classical of classs satisfyingD(Φv)+ψv=0.
From (1.1), it is clear that the formu, when it is regular, is also semi-classical and satisfies D(Φu) + ˜ψu=0,
with
Φ(x) =x3Φ(x) and ψ(x)˜ =x3ψ(x). (1.40)
The class ofuis at mosts˜=s+3.
Proposition 1.2. The class ofudepends only on the zerox=0.
For the proof we use the following lemma:
Lemma 1.3. For all rootcofΦwe have u, θcψ˜ +θc2Φ
=
ψ(c)+Φ(c)
(u)2+c(u)1+c2
−λ v,
θc2Φ+θcψ
(1.41) and
ψ(c)˜ +Φ(c)=c3
ψ(c)+Φ(c)
. (1.42)
Proof. Letcbe a root ofΦ, then we can write
Φ(x) =x3(x−c)Φc(x) withΦc(x)=(θcΦ)(x). (1.43) So from (1.40) and (1.43) we have
u, θcψ˜ +θc2Φ
= u, θc
ξ3ψ +
u, θc
ξ3Φc
. (1.44)
Using the definition of the operatorθc, it is easy to prove that, for two polynomialsf andg, we have
θc(f g)(x)=g(x)(θcf )(x)+f (c)(θcg)(x). (1.45)
Takingg(x)=x3andf (x)=Φc(x), we obtain u, θc
ξ3Φc
=
u, x3(θcΦc)(x)+Φc(c) θcξ3
(x)
= u, x3
θc2Φ (x)
+
(u)2+c(u)1+c2 Φ(c),
because
(θcΦc)(x)=
θc(θcΦ) (x)=
θc2Φ
(x), Φc(c)=(θcΦ)(c)=Φ(c)and θcξ3
(x)=x2+cx+c2. Using (1.2), we obtain
u, θc
ξ3Φc
=
δ−(u)1δ+1
2(u)2δ−λx−3v, x3 θc2Φ
(x)
+
(u)2+c(u)1+c2 Φ(c), hence
u, θc
ξ3Φc
=
(u)2+c(u)1+c2
Φ(c)−λ v, θc2Φ
. (1.46)
Now, takingg(x)=x3andf (x)=ψ(x)in (1.45), we obtain u, θc
ξ3ψ
=
u, x3(θcψ)(x)+ψ(c) θcξ3
(x)
=
u, x3(θcψ)(x) +
(u)2+c(u)1+c2 ψ(c).
Using (1.2), we obtain u, θc
ξ3ψ
=
δ−(u)1δ+1
2(u)2δ−λx−3v, x3(θcψ)(x)
+
(u)2+c(u)1+c2 ψ(c).
Then u, θc
ξ3ψ
=
(u)2+c(u)1+c2
ψ(c)−λ v, θcψ. (1.47)
Replacing (1.46) and (1.47) in (1.44), we obtain (1.41).
From (1.40), we haveΦ(c)=c3Φ(c)andψ(c)˜ =c3ψ(c), hence (1.42). ✷ Proof of Proposition 1.2. Letcbe a root ofΦsuch thatc=0.
If ψ(c)+Φ(c)=0, using (1.41), we have u, θcψ˜ +θc2Φ = 0, since v is semi-classical and so satisfies (1.39).
Ifψ(c)+Φ(c)=0 thenψ (c)˜ +Φ(c)=0,from (1.42).
In any case, we cannot simplify byx−c. ✷
Proposition 1.4. Letvbe a semi-classical form of classs satisfyingD(Φv)+ψv=0, χ1:=(u)1Φ(0)+(u)2
Φ(0)+ψ(0)
−λ
v, θ0ψ+θ02Φ
, (1.48)
and
χ2:=2Φ(0)+(u)1
ψ(0)+2Φ(0) +(u)2
Φ(0)+ψ(0)
−λ
v, θ02ψ+2θ03Φ
. (1.49)
The formugiven by (1.1) is also semi-classical of classs˜satisfyingD(Φu) + ˜ψu=0. Moreover, (1) ifχ1=0 thens˜=s+3,Φ(x) =x3Φ(x)andψ(x)˜ =x3ψ(x);
(2) ifχ1=0 andχ2=0 thens˜=s+2,Φ(x) =x2Φ(x)andψ (x)˜ =x2ψ(x)+xΦ(x);
(3) ifχ1=0, χ2=0 andΦ(0)=0 thens˜=s+1,Φ(x) =xΦ(x)andψ(x)˜ =xψ(x)+2Φ(x).
Proof. (1) Indeed, ψ(0)˜ +Φ(0)=0,
and u,
θ0ψ˜ +θ02Φ (x)
=
u, x2ψ(x)+xΦ(x)
=
δ−(u)1δ+1
2(u)2δ−λx−3v, x2ψ(x)+xΦ(x)
, using (1.2), that is,
u,
θ0ψ˜ +θ02Φ (x)
=(u)1Φ(0)+(u)2
Φ(0)+ψ(0)
−λ
v, θ0ψ+θ02Φ
=χ1.
Therefore, ifχ1=0 it is not possible to simplify, which means that the class ofuiss+3 andusatisfies D(Φu) + ˜ψu=0, withΦ(x) =x3Φ(x)andψ(x)˜ =x3ψ(x).
(2) Ifχ1=0 then letΦ0(x)=x2Φ(x)andψ˜0(x)=x2ψ(x)+xΦ(x). Then ψ˜0(0)+Φ0(0)=0,
and u,
θ0ψ˜0+θ02Φ0
(x)
=
u, xψ(x)+2Φ(x)
=
δ−(u)1δ+1
2(u)2δ−λx−3v, xψ(x)+2Φ(x)
, using (1.2), that is,
u,
θ0ψ˜0+θ02Φ0
(x)
=2Φ(0)+(u)1
ψ(0)+2Φ(0) +(u)2
Φ(0)+ψ(0)
−λ
v, θ02ψ+2θ03Φ
=χ2.
If χ2 =0 it is not possible to simplify, which means that the class of u is s +2 and u satisfies D(Φu) + ˜ψu=0, withΦ(x) =Φ0(x)=x2Φ(x)andψ(x)˜ = ˜ψ0(x)=x2ψ(x)+xΦ(x).
(3) Ifχ1=0 andχ2=0 then letΦ1(x)=xΦ(x)andψ˜1(x)=xψ(x)+2Φ(x). Then ψ˜1(0)+Φ1(0)=3Φ(0).
If Φ(0)=0 it is not possible to simplify, which means that the class of u is s +1 and u satisfies D(Φu) + ˜ψu=0, withΦ(x) =Φ1(x)=xΦ(x)andψ(x)˜ = ˜ψ1(x)=xψ(x)+2Φ(x). ✷
2. The formu=δ−(u)1δ−λx−3v,(u)2=0
Whenvis a symmetric form, we have the following result:
Theorem 2.1. Ifvis a symmetric form,(u)2=0,(u)1=0 and(u)1+λΛn=0, n0, then the formu is regular, where
Λn= n
ν=0
ν µ=0
σ2µ
σ2µ+1
, n0, σ0=1. (2.1)
For the proof we use the following lemmas:
Lemma 2.2. If{yn}n0, {an}n0and{bn}n0are sequences of complex numbers fulfilling yn+1+anyn=bn+1, n0, an=0, n0,
y0=b0, then
yn=(−1)nan−1 n
µ=0
aµ
n
ν=0
(−1)νaν
ν
µ=0
aµ−1
bν, n0.
Lemma 2.3. When{Sn}n0given by(1.4)is symmetric we have S2n+2(0)=(−1)n+1
n µ=0
σ2µ+1, n0, S2n+1(0)=0, n0,
S2n(1)+2(0)=(−1)n+1 n µ=0
σ2µ+2, n0, S2n(1)+1(0)=0, n0, S2n (0)=0, n0,
S2n(1)
(0)=0, n0, S2n(1)+1
(0)=(−1)n n µ=0
σ2µ+1Λn, n0 and S2n(1)+1
(0)=0, n0.
Proof of Lemma 2.3. Asvis symmetric thenζn=0, n0 and so, from (1.4), we have S0(0)=1, S1(0)=0, S0(1)(0)=1, S(1)1 (0)=0, Sn+2(0)= −σn+1Sn(0), n0, Sn(1)+2(0)= −σn+2S(1)n (0), n0, S0(0)=0, S1(0)=1,
S0(1)
(0)=0, S1(1)
(0)=1, Sn+2(0)= −σn+1Sn(0)+Sn+1(0), n0,
Sn(1)+2
(0)= −σn+2
Sn(1)
(0)+Sn(1)+1(0), n0, and
S0(1)
(0)=0, S1(1)
(0)=0, Sn(1)+2
(0)= −σn+2
Sn(1)
(0)+2 Sn(1)+1
(0), n0.
Now, it is enough to use Lemma 2.2 in order to obtain the results. ✷
Proof of Theorem 2.1. Following Lemma 2.3 we have for (1.25), (1.26), (1.30) and (1.31):
Xn(0)=0, X(1)2n(0)=S2n(1)(0) S2n(1)+1
(0), (2.2)
X2n(1)+1(0)= −S2n(1)+2(0) S2n(1)+1
(0), n0, W2n+1(0)=S2n+2(0)S2n(1)(0)
S2n(1)+1
(0), W2n(0)=0, n0, W2n(1)(0)= − S2n(1)+1
(0)2
S2n(1)(0), W2n(1)+1(0)=0, n0.
Therefore we have for (1.28):
∆2n=S2n(1)(0) v, S2n2
λ2, n0,
or
∆2n=(−1)n n µ=0
σ2µ
v, S2n2
λ2, n0, (2.3)
and
∆2n+1 = − S2n(1)+1 (0)2
S(1)2n(0)λ3 + S2n+2(0)S2n(1)(0)
S2n(1)+1 (0)−
S2n(1)+1 (0)
v, S2n2+1 (u)1λ2 +S2n+2(0)
v, S2n2+1
(u)21λ, n0.
As
v, S2n2+1
= −S(1)2n(0)S2n+2(0), n0, (2.4)
we have
∆2n+1= −S2n(1)(0) S2n(1)+1
(0)λ−S2n+2(0)(u)1
2
λ, n0, and using Lemma 2.3, we obtain
∆2n+1=(−1)n+1 n µ=0
σ2µ+1
λΛn+(u)1
2
v, S2n2+1
λ, n0. (2.5)
Under the hypothesis of the theorem it is evident that∆n=0, n0 and thereforeuis regular by virtue of Proposition 1.1. ✷
Let us defineωby
(u)1=ωλ. (2.6)
Therefore we obtain for (2.5)
∆2n+1=(−1)n+1 n µ=0
σ2µ+1{Λn+ω}2
v, S22n+1
λ3, n0. (2.7)
Corollary 2.4. If(u)2=0 andv is a symmetric positive definite form then the form uis regular when ω∈C− ]−∞,0].
Proof. Ifvis positive definite thenσn+1>0, n0.ThereforeΛn>0, n0 and soω+Λn=0, n0, under the hypothesis of the corollary. ✷
Moreover, ifvis symmetric positive definite,(u)2=0 andω∈C− ]−∞,0], we have from (1.16) b0=σ1+ λ
(u)1
=σ1+ 1
ω, (2.8)
c0= −(u)1= −ωλ, (2.9)
c1= − λ
(u)21= − 1
ω2λ, (2.10)