www.elsevier.nl/locate/cam
On the positivity of some bilinear functionals in Sobolev spaces
Andre Draux∗, Charaf Elhami
LMI - UPRES-A CNRS 6085, INSA de Rouen, Departement de Genie Mathematique, Place Emile Blondel, BP 08, F-76131 Mont-Saint-Aignan Cedex, France
Received 13 May 1998; received in revised form 30 January 1999
Abstract
The positivity of a bilinear functional a a(f; g) =
N
X
m=0
mc(m)(f(m); g(m))
is studied as a function of coecients m. The concerned cases are those of Laguerre, Gegenbauer and Jacobi for c(m)=c(0); m= 1; : : : ; N. The domain{
m}Nm=1 whereais positive denite, is given. As a consequence, whenN= 1, the corresponding Markov–Bernstein inequalities are given. c1999 Elsevier Science B.V. All rights reserved.
MSC:33C45; 42C05; 26D05; 26C10
Keywords:Formal orthogonal polynomial; Laguerre; Gegenbauer; Jacobi polynomial; Laguerre–Sobolev; Gegenbauer– Sobolev; Jacobi–Sobolev polynomial; Denite inner product; Markov–Bernstein inequality; Zeros of polynomial
This paper constitutes the second part of a study devoted to the positivity of bilinear functionals using positive denite classical inner products, that is to say those of Hermite, Laguerre and Jacobi measures. The rst part has solved this problem of positivity for the Hermite measure and for some other connected problems [5]. That study was the simplest one, since the domain D where the bilinear functionals are positive denite, is given by means of explicit equations which dene its boundary. This domain D is polyhedric. It remains to solve the more complicated problems of Laguerre and Jacobi measures. We study them here. In these cases the domain D can not be given by means of explicit equations of its boundary when N¿2, because this one is dened as a limit of a nappe of an algebraic hypersurface. The question was: does it exist a part of the domain D
∗Corresponding author.
E-mail addresses: [email protected] (A. Draux), [email protected] (C. Elhami) 0377-0427/99/$ - see front matter c1999 Elsevier Science B.V. All rights reserved.
where some coecients m are negative; the answer is the same as in the Hermite case: yes, when
N¿2. When N= 1; 1 has to be strictly positive in D. But when N= 1, the successive values 1; n, obtained for the positivity with polynomials of degree n= 1;2; : : : ; give the best coecients of the so-called Markov–Bernstein inequalities (see [1,9]). Except in the Hermite case and Laguerre case for = 0 (see [9]) these values are not known explicitly, but we have given lower and upper bounds in the Laguerre and Gegenbauer cases, which gives the behavior of 1; n as a function of n. The help of Mathematica 3.0 [15] was very useful in that case for some tedious algebraic manipulations.
For replacing this study within the framework of the Sobolev orthogonality, the bibliography, presented by Marcellan and Ronveaux [8], is essential.
1. Introduction
In order to establish the general framework of our paper, we begin to study the general case where a bilinear functional a is obtained from two positive denite inner products. To nd the domain Dn={ ∈ R|a(p; p)¿0∀p ∈Pn− {0}} is equivalent to solve a generalized eigen-value problem. Moreover the smallest eigeneigen-value 1; n of this problem which gives the boundary of the domain Dn, also gives the best coecient in Markov–Bernstein inequality. Our presentation is more or less dierent from that given in [9].
P (resp. Pi) will denote the vector space of polynomials (in one variable) with real coecients (resp. of degree at most i).
Letc andc(1) be two positive denite inner products and{P
i}i¿0 the sequence of monic orthogonal
polynomials with respect to c.
Let a be the bilinear functional dened by
∀p; q∈P; a(p; q) =c(p; q) +c(1)(p′; q′);
where is a real parameter and p′ (resp. q′) is the derivative of p (resp. q).
p∈Pn can be written as p=
Pn
i=0yiPi. Thus p′=
Pn
i=1yiP′i and we get
a(p; p) =yTKn;0y+y˜TKn;(1)−1y;˜ (1)
where y (resp. ˜y) is the vector of Rn+1 (resp. Rn) of components y
i; i = 0; : : : ; n, (resp. i = 1; : : : ; n); Kn;0 is the (n+ 1)×(n+ 1) matrix whose entries Kn;0(i; j) are c(Pi; Pj) for i; j= 0; : : : ; n (therefore this matrix is diagonal) andKn;(1)−1 is then×nmatrix whose entriesKn;(1)−1(i; j) arec(1)(P′
i; P
′
j) for i; j= 1; : : : ; n.
Let Kn;−1 be the n×n matrix deduced from Kn;0 in cancelling the rst row and the rst column.
Then from (1) we have the following obvious result.
Theorem 1.1. For a xed parameter ∈R; then; ∀n¿1 and ∀p ∈ Pn− {0}; a(p; p)¿0 if and
only if Kn;−1+Kn;(1)−1 is positive denite.
Moreover, since Kn;(1)−1 is a positive denite symmetric matrix, it can be decomposed by means of Cholesky algorithm as GnGnT where Gn is a lower triangular matrix. Therefore we have
Kn;−1+Kn;(1)−1=Gn(Gn−1Kn;−1(GTn)
−1+I)GT
Hence Kn;−1+K (1)
n;−1 is positive denite if and only if G−n1Kn;−1(GnT)−1+I is positive denite. Let i; n; i= 1; : : : ; n be the eigenvalues of Gn−1Kn;−1(GTn)
−1 with 0¡
1; n62; n6· · ·6n; n. Then we have another obvious results.
Theorem 1.2. Kn;−1+K
(1)
n;−1 is positive denite if and only if ¿−1; n.
Theorem 1.3. a is positive denite ∀p∈P− {0} if and only if ¿−limn→∞1; n.
Proof. Since G−1
n Kn;−1(GnT)−1 is a symmetric matrix, then from an obvious consequence of the Courant–Fischer theorem (see [14]) we have
0¡ 1; n+161; n; n¿1:
Thus the positive sequence {1; n}n¿1 is decreasing and bounded. Then limn→∞1; n exists and limn→∞1; n¿0. Therefore the result holds.
From Theorem 1.2 we obtain the Markov–Bernstein inequality.
Corollary 1.4.
∀p∈Pn; c(1)(p′; p′)6 1
1; n
c(p; p); n¿1:
1=1; n is the best constant. An extremal polynomial is p=
Pn
i=1yiPi where y˜ is the vector deduced
from an eigenvector (GT
n) ˜y of G
−1
n Kn;−1(GnT)
−1 corresponding to the eigenvalue 1; n.
In conclusion, in order to dene the domain Dn where a is positive denite for any p∈Pn− {0}, we have to give the smallest eigenvalue of G−1
n Kn;−1(GTn)
−1. Moreover the domain of positivity of
a depends on limn→∞1; n. To study the general case can be very dicult. In addition our aim is to
give the domain of positivity of a for any N; N¿1, dened by
∀p; q∈P; a(p; q) = N
X
m=0
mc(m)(p(m); q(m));
where 0 = 1; N 6= 0 and c(m) is a positive denite inner product, ∀m= 0; : : : ; N, and p(m) is the derivative of order m of p.
In the cases where the inner products c(m) are linked and the derivatives p(m) are also linked
like for the classical inner products, it is possible to give some basic results about the domain of positivity as well as the 1; n’s.
In [5] we have considered the case where c(0) is a classical inner product (Hermite, Laguerre or
Jacobi) and c(m) is the classical inner product deduced from c(0) such that {P(m)
n }n¿m is a sequence of polynomials orthogonal with respect to c(m). In the sequel we will study the case where c(m)=
In addition to [5] we will give the Markov–Bernstein inequalities in that dierent cases. We have kept the notation C(1)
n used in [5].1 In the Hermite case Cn(1) was already known (see [9]).
Corollary 1.5 (Markov–Bernstein inequalities). Let c(0) and c(1) be two classical inner products
such that the derivatives of the monic orthogonal polynomials with respect to c(0) are orthogonal
with respect to c(1).
Let k · k(m) be the norm associated to the inner product c(m) respectively for m= 0;1. Then ∀p ∈ Pn − {0}; a(p; p)¿0 if and only if 1¿−1=Cn(1). Thus ∀p ∈ Pn; kp′k(1)6
q
Cn(1)kpk(0)
where
Cn(1)=
2n in the Hermite case; n in the Laguerre case;
n(n+++ 1) in the Jacobi case:
2. Classical inner products
Let L2(;) be the Hilbert space of square integrable real functions on the open set
⊂R for the positive Borel measure supported on .
On this space we have the classical denite inner product:
c(f; g) =
Z
f(x)g(x) d(x) ∀f; g∈L2(;)
and the norm c(f; f) =kfk2=R
f
2(x) d(x).
HN(;) will denote the Sobolev space
HN(;) ={f|f(m) ∈L2(;); m= 0;1; : : : ; N};
where the derivatives f(m) of order m are taken in the distributional sense.
The denite inner product on this Hilbert space is
s(f; g) = N
X
m=0
c(f(m); g(m));
∀f and g∈HN(;):
The corresponding norm, denoted by k · ks, is given by
kfk2
s= N
X
m=0 kf(m)
k2
∀f∈HN(;):
Now we consider the symmetric bilinear functional a a:HN(;)×HN(;)→R
1We want to indicate a mistake in [5]. (m)
n = 2m in the Hermite case instead of 1. Consequently Formula given in the same paper at the bottom of the page 173 has a version in the Hermite case, dierent from the Laguerre case, and that formula has to be multiplied by 2N
(N+1)
dened by
a(f; g) = N
X
m=0
mc(f(m); g(m)); ∀f and g∈HN(;); (2)
where m; m= 0; : : : ; N; are N+ 1 xed real numbers with 0= 1 and N 6= 0.
We look for the formal orthogonal polynomials with respect to a, that is to say, we look for the polynomials Sn; n¿0; such that
degSn=n;
a(Sn; xi) = 0 for i= 0; : : : ; n−1: (3) These polynomials Sn will be called Sobolev formal orthogonal polynomials.
Setting Mfn= (a(xj; xi))ni; j−=01, we have the following obvious result:
Theorem 2.1. The formal orthogonal polynomial Sn exists and is unique up to a normalization for
the leading coecient is xed; if and only if the matrix fMn is regular.
Denition 2.2. The bilinear functional a is called positive denite on HN(;)×HN(;) (resp.
on P×P) if, ∀f∈HN(;)
− {0} (resp. P− {0}); a(f; f)¿0.
Theorem 2.3. The bilinear functional a is positive denite on P×P if and only if all the
for-mal orthogonal polynomials Sn; n∈ N; exist with a positive leading coecient and a(Sn; Sn)¿0;
∀n∈N.
Proof. (i) If a(Sn; Sn)¿0; ∀n∈N, then ∀p∈P (degp=k) we have p=
Pk
i=0iSi and a(p; p) =
Pk
i=02ia(Si; Si)¿0.
(ii) For p= 1; a(1;1)¿0. Thus if S0=(0)0 ¿0; a(S0; S0) = ((0)0 )2a(1;1)¿0. Moreover Mf1 is
regular and detMf1=a(S0; S0)¿0. Therefore S1 exists.
Let us assume that detfMi¿0; i= 1; : : : ; n. Then Sn exists. Let us write Sn as Sn=
Pn ‘=0
(n)
‘ x‘ with (n)
n ¿0. Let us denote by Xbn the vector of Rn of components ‘(n); ‘= 0; : : : ; n−1. fMn+1 is
multiplied on the right by the following (n+ 1)×(n+ 1) regular matrix
In bxn 0Tn (nn)
;
where In is the n×n identity matrix and 0n the zero vector of Rn. Thanks to relations (3) we get
f
Mn 0n
ATn a(Sn; xn)
!
;
where An is the vector of Rn of components a(xj; xn); j= 0; : : : ; n−1. a(Sn; xn) =a(Sn; Sn)=(nn)¿0. Therefore detfMn+1¿0 and the desired conclusion follows.
In the sequel we will study in which domain D ofRN; of components
Table 1
Name Laguerre Jacobi Gegenbauer
]0;+∞[ ]−1;1[ ]−1;1[
xe−x (1−x)(1 +x) (1−x2)−12
kn n! (n++ 1) 2nn! (n) ( n) (n−n) n( (n))2
21−2−2nn! (n+2)
(n++1) (n+) with n=n++ 1
n=n++ 1 n=n+n−1
Restrictions ¿−1 ¿−1; ¿−1 ¿−1
2
to the Laguerre, Jacobi and Gegenbauer orthogonal polynomials (respectively denoted by L n; Pn(; ) and G
n). We recall that the Hermite case has been already studied in [5]. We summarize all the useful informations concerning these three families of monic orthogonal polynomials in Table 1 (see [2,12]), where kn is the square norm in L2(;) of a monic orthogonal polynomial of degree n.
Moreover we have the following relations for the derivatives of these polynomials (see [2,12]):
d dxL
n(x) =nL +1
n−1(x); (4)
d dxP
(; )
n (x) =nP
(+1; +1)
n−1 (x); (5)
d dxG
n(x) =nG +1
n−1(x): (6)
3. Matrix interpretation of some particular relations
Some particular staircase recurrence relations will be of a great interest in order to give a general expression of a(p; q); ∀p; q∈P. These relations are (see [2,12]):
Ln(x) =Ln+1(x) +nLn+1−1(x); (7)
Pn(; )(x) =Pn(+1; )(x) +e(n=+1; )Pn(−+11; )(x)
=P(+1; )
n (x)−
2n(n+) (2n+++ 1)2
P(n−+11; )(x); (8)
Pn(; )(x) =P(n; +1)(x) +e(n; =+1)Pn(; −1+1)(x)
=P(; +1)
n (x) +
2n(n+) (2n+++ 1)2
Pn(; −1+1)(x); (9)
G
n(x) =Gn+1(x) +enGn−+12(x)
=Gn+1(x)− n(n−1) 4(n+)2
and are satised ∀n∈N. (a)j; j∈N, denotes the shifted factorial: (a)j=a(a−1)· · ·(a−j+ 1). These relations are simple consequences of the orthogonality when the weight function is modied respectively by x; 1−x; 1 +x; 1−x2.
By convention we complete the set of the previous relations (7) – (10) by the following ones:
L−k=L−+1k = 0;
P−(; k)=P−(k+1; )=P(−; k+1)= 0; G−k=G−+1k = 0;
∀k ¿0; (11)
Moreover the set of relations (4) – (6) is completed by: d
dxL
−k+1=L
+1
−k; d
dxP
(; )
−k+1=P (+1; +1)
−k ; d
dxG
−k+1=G−+1k ;
∀k ¿0; (12)
For the sake of simplicity let us denote by Pi; ∀i∈N, the monic orthogonal polynomials with respect to any previous inner product c, and by Pb−k; n; ∀k and n ∈ N, the vector of components
Pj; j=−k; : : : ; n, with the convention that Pj= 0, if j ¡0. We will denote by ddxPb−k; n the vector whose components are the derivatives of Pj; j=−k; : : : ; n. When the distinction between the dierent families of orthogonal polynomials will be necessary, then we will use the notation bL
−k; n; Pb
(; )
−k; n and
b
G
−k; n instead of Pb−k; n.
Now, let us give some matrix expressions deduced from relations (7) – (12).
Property 3.1.
b
L
−k; n=E−k; nLb−+1k; n; (13)
b
P−(; k; n)=E−(=k; n+1; )Pb−(k; n+1; )=E−(; =k; n +1)Pb(−; k; n+1) (14)
=E−(; k; n)Pb−(+1k; n; +1); (15)
b
G−k; n=E−k; nGb−+1k; n; (16)
where E−k; n; E−(=k; n+1; ); E
(; =+1)
−k; n ; E
(; )
−k; n and E−k; n are(n+ 1 +k)×(n+ 1 +k) nonsingular matrices
whose the entries (i; j) are in row i and column j; i; j= 1; : : : ; n+k+ 1:
E−k; n(i; j) =
1 ∀i=j;
i−k−1; ∀i=j+ 1; k+ 26i6n+k + 1;
0 everywhere else;
E−(=k; n+1; )(i; j) =
1 ∀i=j;
e(i−=k+1−1; ) ∀i=j+ 1; k+ 26i6n+k+ 1;
E−(; =k; n +1)(i; j) =
1 ∀i=j;
e(i−; =k−1+1) ∀i=j+ 1; k+ 26i6n+k+ 1;
0 everywhere else;
E−(; k; n)(i; j) =
1 ∀i=j;
2‘(−)
(2‘++)(2‘+++ 2) ∀i=j+ 1;
with ‘=i−k −1 and16‘6n;
−(2‘4+‘(‘+−1)()(2‘‘++)(+‘++ 1)) 3 ∀
i=j+ 2;
with ‘=i−k−1 and26‘6n;
0 everywhere else;
(17)
E
−k; n(i; j) =
1 ∀i=j;
e‘=−‘(‘−1) 4(‘+)2 ∀
i=j+ 2; with ‘=i−k−1 and26‘6n;
0 everywhere else;
b
P−k; n=JT−k+1; n+1
d
dxPb−k+1; n+1; (18)
where JT
−k+1; n+1 is a (n+ 1 +k)×(n+ 1 +k) nonsingular matrix obtained by multiplication of
E−k; n, ( resp. E−(; k; n); E−k; n) on the right by the diagonal matrix Db−k+1; n+1 whose the elements are
b
D−k+1; n+1(i; i) =
1; i= 1; : : : ; k;
1
‘; ‘= 1; : : : ; n+ 1with i=k+‘:
Proof. Relations (13), (14) and (16) are the matrix version of relations (7) – (10) for the vectors
b
P−k; n. Eq. (15) is obtained from
b
P−(; k; n)=E−(=k; n+1; )E−(+1k; n; =+1)Pb−(k; n+1; +1)=E−(; =k; n +1)E−(=k; n+1; +1)Pb−(k; n+1; +1):
It is easy to verify that
E−(=k; n+1; )E−(k; n+1; =+1)=E(−; =k; n +1)E−(=k; n+1; +1):
From Eqs. (4) – (6) it is obvious that a vector ˆP−k; n can be written as the product of ˆD−k+1; n+1
and the derivative of a vector ˆP−k+1; n+1 (but with modied parameters ; or and ).
4. The quasi-orthogonal polynomial Ri associated with Pi
Let {Ri}i¿0 be a sequence of monic polynomials which satises the following conditions:
degRi=i; ∀i∈N; di
dxiRi=i!P0=i; NP0; for 06i6N−1; (19)
dN
dxNRi= (i)NPi−N =i; NPi−N; for i¿N: (20) Let ˆRn, be the vector of components Ri; i= 0; : : : ; n, and Dn; N be the (n+ 1)×(n+ 1) diagonal matrix (i; N)ni=0.
The matrix interpretation of relations (19) and (20) is
dN
dxNRˆn=Dn; NPˆ−N; n−N: (21) Remark that Dn; N contains i; N=i! for i ¡ N and that d
N
dxNRi= 0 and P−i= 0 for i ¡ N. Using Eq. (18), Eq. (21) becomes:
dN
dxNRˆn=Dn; NJ
T
−N+1; n−N+1
d
dxPˆ−N+1; n−N+1
=Dn; NJT−N+1; n−N+1· · ·J T
−N+j; n−N+j dj
dxjPˆ−N+j; n−N+j ∀j= 2; : : : ; N: (22) In order to simplify the writing of products with dierent matrices J, we will use the following notation: QNj=−i1J−j; n−j=J−i; n−i· · ·J−N+1; n−N+1 in this order. If i ¿ N−1, then
QN−1
j=i J−j; n−j will be taken equal to 1. Thus Eq. (22) is
dN
dxNRˆn=Dn; N
NY−1
i=N−j
J−i; n−i
T
dj
dxjPˆ−N+j; n−N+j ∀j= 2; : : : ; N: (23)
From Eq. (23) written for j=N, we set
ˆ
Rn=Dn; N
NY−1
j=0
J−j; n−j
T
ˆ
P0; n: (24)
Therefore, from Eqs. (24) and (18), we also have
dN−j
dxN−jRˆn=Dn; N
NY−1
i=N−j
J−i; n−i
T
ˆ
P−N+j; n−N+j: (25)
Denition 4.1. Ann×n matrixA= (aij) is a band matrix ifaij= 0;|i−j|¿ h with h xed (h∈N) and h+ 1¡ n: h is the half width of band.
From the denition of the matrices JT, it is clear that (QjN=0−1J−j; n−j)T is a lower triangular band matrix of width N in the Laguerre case, 2N in the Jacobi and Gegenbauer cases. Moreover all the entries of the last lower diagonal which is the border of the band, are non zero from the second column (the rst column contains a unique non-zero entry in the rst row). Therefore we have the following obvious property about the quasi-orthogonality of the Ri’s (see [2]).
Property 4.2. In the Laguerre case (resp. in the Jacobi and Gegenbauer cases) the polynomials
Ri are quasi-orthogonal of order N (resp. 2N) with respect to the Laguerre (resp. Jacobi or
Gegenbauer) inner product. They are strictly quasi-orthogonal when i ¿ N (resp. i ¿2N) in the Laguerre case (resp. in the Jacobi and Gegenbauer cases).
5. Matrix interpretation of a(p; q)
Let p and q be two polynomials of degree n. Since {Ri}i¿0 is a basis of P, we can write p and
q as
p= ( ˆRn)Ty and q= ( ˆRn)Tz;
where y and z are vectors of Rn+1 which contain the coordinates y
i (resp. zi) of p (resp. q) in the basis {Ri}ni=0. Then
dN−j dxN−jp=
dN−j
dxN−jRˆn
T
y
= ( ˆP−N+j; n−N+j)T
NY−1
i=N−j
J−i; n−iDn; Ny:
Let us dene the (t+ 1)×(t+ 1) matrix Kt; j by Kt; j= (c(Pi; P‘)) t−j
i; ‘=−j. Thus Kt; j is a diagonal matrix (ki)t
−j
i=−j where ki=c(Pi; Pi); i¿0; and ki= 0 if i ¡0. Therefore:
c
dN−j
dxN−jp; dN−j dxN−jq
=zTDn; N
NY−1
i=N−j
J−i; n−i
T
Kn; N−j NY−1
i=N−j
J−i; n−iDn; Ny: (26)
Finally
a(p; q) = N
X
i=0
ic(p(i); q(i))
= N
X
i=0
izTDn; N NY−1
j=i
J−j; n−j
!T
Kn; i NY−1
j=i
J−j; n−j
!
Dn; Ny: (27)
From Eq. (27) a(p; q) can also be written as
a(p; q) = b(n) X
i=0
iFi; n(y; z) =Fb(n); n(y; z) b(n) X
i=0
i
b(Yn)−1
r=i
Fr; n(y; z)
Fr+1; n(y; z)
!
; (28)
Setting Y =Dn; Ny and Z=Dn; Nz, we have Its components are numbered from 0 to n.
Let (Kn; r+1) 1
2 be the diagonal matrix whose entries are the square roots of the elements of Kn; r+1.
The vector (Kn; r+1)
Finally we obtain the following expression of a(p; q)
a(p; q) =Fb(n); n(y; z)
Remark that Jn is independent of N. Indeed, since the n last rows and columns of Kn;1 correspond
to Kn−1;0; we have
which can also be written as
J1T; nKn−1;−1J1; n= (Kn−1;0)
Therefore the matrices Jn corresponding to the dierent cases (Laguerre, Jacobi, Gegenbauer) can be obtained by using Eq. (33).
In the Laguerre case (see also [9]), Jn is an n×n symmetric tridiagonal matrix:
Jn(i; j) =
0 everywhere else:
In the Jacobi case, Jn is an n×n symmetric ve diagonal matrix:
0 everywhere else:
(35)
ki is k
(; )
i in Eq. (35).
In the Gegenbauer case (see also [9]), Jn is an n×n symmetric ve diagonal matrix:
Jn(i; j) =
0 everywhere else:
(36)
From Eq. (32) and the denition of Kn−1;0 and Kn;0, we have the following obvious property
Property 5.1. Jn is a positive denite symmetric matrix.
Property 5.2. If i; n; i = 1; : : : ; n; are the eigenvalues ofJn; corresponding to the three previous
Proof. As an obvious consequence of the Courant–Fischer therorem (see [14]) n; n6n+1; n+1; ∀n. Thus the positive sequence {n; n}n¿1 is increasing.
We use the localization of the eigenvalues given by the Gerschgorin disks. From Eqs. (34) – (36) it is clear that all the centers of these disks are at a nite distance and all the radii are bounded. Thus n; n is bounded ∀n and the result holds.
Remark 5.3. Relation (28) can also be written by means of Horner algorithm
d0= 1;
di=i+
Fi−1; n(y; z)
Fi; n(y; z) di−1; i= 1; : : : ; b(n);
a(p; q) =Fb(n); n(y; z)db(n):
(37)
6. Domain of positivity of a
Let be an element of RN of components
m; m= 1; : : : ; N (N 6= 0). We want to nd the domain D⊂RN such that
D={∈RN|a(p; p)¿0∀p∈P− {0}};
where a is considered as a function depending on .
∀n¿1; Dn will denote the domain of Rb(n) such that
Dn={∈Rb(n)|a(p; p)¿0 ∀p∈Pn− {0}}: (38)
For any xed p ∈Pn− {0}; a(p; p) = 0 is a hyperplane of Rb(n) such that the coecients of the
i’s are positive (see (31)). Thus this hyperplane cuts the axes of Rb(n) in the negative part. Let us begin to prove a property of N.
Theorem 6.1. Let i; n; i= 1; : : : ; n; be the eigenvalues of Jn; (0¡ 1; n6· · ·6n; n). n; n is assumed
to be bounded ∀n.
(i) Iflimn→∞ 1; n= 0 and if = (1; : : : ; N)∈D; then N¿0.
(ii) Ifinfn 1; n¿ ¿0 and if i¿0; i= 1; : : : ; N −1; then there exists ¡ˆ 0 depending on the
i′s; i= 1; : : : ; N −1 such that ∀N¿; ˆ = (1; : : : ; N)∈D.
Proof. (i) In FN−1; n(y; y)=FN; n(y; y) we take w(N; n) as the eigenvector corresponding to the smallest eigenvalue 1; n−N+1 of Jn−N+1. The vector u(N; n) of components u(iN; n); i= 0; : : : ; n, is chosen such that u(NN; n−1)= 0; ui(N; n); i=N; : : : ; n, being dened from the successive components of w(N; n) thanks to
the following relation:
(K(n; N)) 1
2u(N; n)=
0N
w(N; n)
; (39)
where 0N is the zero vector of RN. Then
FN−1; n(y; y)
FN; n(y; y) =(w
(N; n))TJ
n−N+1w(N; n) kw(N; n)k2
2
=1; n−N+1;
Thus limn→∞ FN−1; n(y; y)=FN; n(y; y) = 0. Now we take the vector Y of Rn+1 such that
Yi=
(
u(iN; n) for i=N; : : : ; n;
0 for i= 0; : : : ; N −1:
Since u(r; n)=QN−1
j=r J−j; n−jY, then u
(r; n)
i = 0 for i= 0; : : : ; r−1. Therefore we get
a(p; p) =F(N; n)(y; y)
N
X
i=0
i NY−1
r=i
(w(r+1; n))TJ
n−rw(r+1; n)
kw(r+1; n)k2 2
!
: (40)
Each factor (w(r+1; n))TJ
n−rw(r+1; n)=kw(r+1; n)k22 which is the Rayleigh quotientJn−r(w
(r+1; n)) (see [14]),
is bounded by the largest eignevaluen−r; n−r ofJn−r. Sincen−r; n−r is also bounded, thenFr−1; n(y; y)=
Fr; n(y; y) which is the coecient of dr−1 in (37), is also bounded, ∀r = 1; : : : ; N −1. Since the
coecientFN−1; n(y; y)=FN; n(y; y) ofdN−1 tends to zero when n tends to innity,dN remains positive if and only if N¿0, and the result holds.
(ii) The same technique with the same vectors is used in this second part. Thus we have
FN−1; n(y; y)
FN; n(y; y)
¿inf
n 1; n¿ ¿0: Moreover, Jn−r(w
(r+1; n)) is such that
0¿ 61; n−r6Jn−r(w
(r+1; n))
6n−r; n−r6sup n
n; n¡+∞:
Thus using Remark 5.3
NX−1
i=0
iN−1−i6dN−16
NX−1
i=0
i
sup n
n; n
N−1−i
:
Hence if i¿0; i=1; : : : ; N−1; dN which is such that dN¿N−dN−1 is positive for anyN¿ˆ=
−dN−1. Thus the result holds.
6.1. Case N = 1
In the particular case where N= 1, the domains Dn andD are characterized by means of 1; n and limn→∞ 1; n. We give lower and upper bounds of 1; n in the Laguerre case, a lower bound of 1; n in the Gegenbauer case, and the behavior of an upper bound in the Gegenbauer and Jacobi cases.
Theorem 6.2.
Dn={1∈R; 1¿−1; n};
where 1; n is the smallest eigenvalue of Jn.
Proof. From (31) we have
a(p; p) =kw(1; n) k2
2 1+
k0y20 kw(1; n)k2
2
+(w
(1; n))TJ
nw(1; n)
kw(1; n)k2 2
!
The case where w(1; n)= 0 is without interest, since p=y
0. Therefore the
minimum of the right part of Eq. (41) is obtained for y0 = 0 and w(1; n) being the eigenvector
corresponding to the smallest eigenvalue 1; n of Jn; this minimum is equal to 1; n.
Now we give some properties about the smallest eigenvalue 1; n. A upper bound will be obtained from the value of the Rayleigh quotient which corresponds to a particular vector. A lower bound will be given in the case where the eigenvalues are the zeros of orthogonal polynomials. It is given by the rst step of the Newton method with 0 as starting point.
In the Laguerre case the eigenvalues of Jn are the one’s of a tridiagonal matrix which corre-sponds to a Jacobi matrix of sequence of orthogonal polynomials. These polynomials correspond to co-recursive polynomials of the generalized Pollaczek polynomials (this fact was presented in the doctoral dissertation of Perez [10] and in the paper of Marcellan et al. [7]). See also [9].
Theorem 6.3 (Laguerre case). The eigenvalues i; n; i= 1; : : : ; n; are the zeros of the orthogonal
polynomials An(x) dened from the following three-term recurrence relation:
An(x) =
These zeros are real; positive; distinct and
n=
considered as a polynomial in one variable .
Proof. The rst zeros 1;1 and 1;2 are obtained in an obvious way from (42). The rst part is
obtained from a classical property for tridiagonal matrices:
An(x) = det(xI−Jn)
of the positive denite symmetric matrix Jn which are real and positive, are, besides, distinct. The zeros of An being distinct real, those ofA′n; A′′n, and so on, also are real distinct and are located between1; n andn; n. ThusA′′n has a constant sign for x ¡ 1; n. Therefore ’n=−An(0)=A′n(0)¡ 1; n.
Using Eq. (42) it is easy to verify by recurrence on n that
An(0) = (−1)n(+n)n
From Eq. (34) we get for any vector y ∈Rn
Let us choose the vector y such that
yi=
For the vector y given by (46) we have
n
The corresponding numerator of the Rayleigh quotient Jn can be written as
n−1
In the same way we have
Now we will use the following inequalities. Let a be a xed real. If a+x ¿0 for x ∈[‘−1; k+ 1]; ‘; k ∈N, then
lna+k+ 1
a+‘ =
Z k+1
‘
dx a+x6
k
X
i=‘ 1
a+i6
Z k
‘−1
dx a+x = ln
a+k
a+‘−1: (47) Therefore, in using Eq. (47), we obtain Pj=1n−1i6n and
Pn
j=1j6 n. Finally
1; n= min
y∈Rn−{0}Jn(y)
63(n+ n)
n(22n+ 1)
:
Remark that n= O(n) and n= O(n). Therefore 1; n= O(1=n2).
When = 0, Turan [13] (see also [9]) gave the precise expression of 1; n which is
1; n= 4 sin2
2(2n+ 1):
For any , Dorer [4] gave an upper bound of 1=1; n in using the Frobenius norm of a certain matrix. But the entries of that matrix are wrong. Indeed the coecientcwhich appears in his relations depends on k : c(k) = (1 +=k)−1=2. He wrote all the relations with a coecient c independent of
k. Therefore his upper bound of 1=1; n is false.
As a straightforward consequence of Theorems 6.2 and 6.3 we get the Markov–Bernstein inequality in the Laguerre case.
Corollary 6.4. In the Laguerre case the following Markov–Bernstein inequality is satised.
∀p∈Pn− {0}; kp′k6 1
√ 1; nk
pk¡
s
n(n+ 1) 2(+ 1)kpk: 1=√1; n is the best constant.
An extremal polynomial is
p= n
X
i=1
w(1i ; n)
ipki−1
Ri;
where w(1; n)= (w(1; n)
1 ; : : : ; w(1n; n))T is the eigenvector of Jn corresponding to the eigenvalue 1; n.
Proof. The form of an extremal polynomial is a consequence of relations (41) and (39) with
N = 1.
Corollary 6.5. D=R+.
Remark that this result could be obtained from the region of R which is a region of density of the set of all zeros of all An; ∀n¿1 (see Chihara [2,3]). Since limn→∞(2 +=n) = 2 and limn→∞(1 +
=(n−1)) = 1, then [0,4] is that region. Therefore a is positive denite for 1¿0.
In the Gegenbauer case, the matrix Jn has ve diagonals, but the two diagonals which are here and there of the main diagonal are zero. Therefore Jn can be transformed in a similar matrix which has a structure of 2×2 diagonal block matrix. Moreover the two diagonal blocks are tridiagonal. The matrices used in the transformation are permutation matrices Vij (i6=j) which are such that
Vij(‘; ‘) = 1 for‘= 1; : : : ; n; ‘6=iand ‘6=j;
Vij(i; j) =Vij(j; i) = 1;
0 everywhere else:
Of course VijVij=I. Thus Jn is multiplied on the right and on the left by such matrices having the same indices. They permute rows and columns having the same couple of numbers. Finally we get a [(n+ 1)=2]×[(n+ 1)=2] rst tridiagonal block, denoted by ˆJ[n+1
2 ]; and a [n=2]×[n=2] second
tridiagonal block, denoted by ˜J[n
2]; where [:] denotes the integer part. We have
0 everywhere else:
Using the expression of ki; we get the following sequence of monic orthogonal polynomials for this Jacobi matrix
We get another sequence of monic orthogonal polynomials for this Jacobi matrix ˜
Ai+1(x) = (x−i+1 2
) ˜Ai(x)−i+1 2
˜
Ai−1(x); i¿1; (51)
with ˜A0= 1; A˜1(x) =x−(2+ 1)=(8(+ 2)2); i+1 2
and
i+1 2
are obtained from (49) and (50) in
replacing i by i+ 1=2.
Remark that these polynomials should be those found by Perez in her doctoral dissertation [10], but she gave a wrong expression for i. See also [9].
Theorem 6.6 (Gegenbauer case). The eigenvaluesˆi; n; i= 1; : : : ; n; ofJˆn which are the zeros of the
orthogonal polynomials Aˆn given by (48); are real; positive; distinct. Moreover (1 + 2)
2n(2n2+ 2n−1)(n+)¡ˆ1; n6ˆn= O
1
n4
; n¿1:
Proof. It is still obvious that the ˆi; n’s are real, positive, distinct.
(i) For nding the lower bound, we will use the same technique as in Theorem 6.3. Using Eq. (48) it is easy to verify by recurrence on n that
ˆ
An(0) = (−1)n n!
(2n)!(2n−1 +)n
: (52)
Indeed Eq. (52) is satised for ˆA1(0) =−1=(2(1 +)). If Eq. (52) is assumed to be satised up to
the degree n, we have
ˆ
An+1(0) =−
1 4
2n+ 2
(2n+ 1)(2n++ 1)2
+ 2n
(2n+)2(2n+ 2−1)
(
−1)nn! (2n)!(2n−1 +)n
− 2n(2n+ 2−2)(−1)
n−1(n −1)!
16(2n−1)(2n+ 2−1)(2n+)3(2n+−1)(2n−2)!(2n+−3)n−1
= (−1)n+1 2(n+)2(n+ 1)n!
4(2n+ 1)(2n+ 2)(2n++ 1)2(2n)!(2n+−1)n
− 2n(−1)
nn!
4(2n+)2(2n+ 2−1)(2n)!(2n+−1)n
− 2n(2n+ 2−2)(−1)
n−1(n−1)!
16(2n−1)!(2n+)n+2(2n+ 2−1)(2n+−1)
= (−1)
n+1(n+ 1)!
(2n+ 2)!(2n+ 1 +)n+1
:
On the other hand, we have
ˆ
A′n(0) = (−1)n−1 n!(2n
2+ 2n −1)
(2n−1)!(1 + 2)(2n−1 +)n−1
; n¿1: (53)
ˆ
A′1(0) = 1 satises Eq. (53). If Eq. (53) is assumed to be satised up to the rank n; we have
ˆ
A′n+1(0) =(−1) n
2
n+
(2n+ 1)(2n++ 1)2
+ n
× n!(2n
(ii) In order to obtain the upper bound, we will use the Rayleigh quotient Jˆ of ˆJn. Its numerator is Let us choose the vector y such that
=
The computation of a precise upper bound is so tedious that, rstly this work was made with the help of Mathematica 3.0 [15], secondly the expression of this upper bound using the same technique as in Theorem 6.3 is so long that it is unusable and we have preferred to give an equivalent as a O(:).
where ˆri is a polynomial in the variable i of degree 7 whose coecients also are expressions which contain n. Considered as a polynomial in the two variables i and n; rˆi is of degree 8. Since
The denominator of the Rayleigh quotient is 1
Theorem 6.7 (Gegenbauer case). The eigenvalues˜i; n; i= 1; : : : ; n; ofJ˜n which are the zeros of the
orthogonal polynomials A˜n given by (51); are positive; real; distinct. Moreover
1 + 2
4n(n+ 1)(+n)(+n+ 1)¡˜1; n6˜n= O
1
n4
:
Proof. The nature of the ˜i; n’s is still obvious.
(i) From (51) it is very easy to verify by recurrence on n that
˜
An(0) = (−1)n
(2n+ 2)2n 24nn!(2n+)
2n(n+)n
:
On the other hand, we have
˜
A′n(0) = (−1)n−1 (n+ 1)(2n+ 2)2n−1(+n+ 1)
(n−1)!24n−2(2n+)
2n(+n−1)n−1
; n¿1: (58)
˜
A′1(0) satises Eq. (58) which can be assumed to be satised up to the rank n. Then we have
˜
A′n+1(0) =−1 4
2n+ 1 + 2
2(n+ 1)(2n++ 2)2
+ 2n+ 1
(2n++ 1)2(2n+ 2)
(−1)n−1 (2n+ 2)2n−1(n+ 1)(+n+ 1)
(n−1)!24n−2(2n+)
2n(+n−1)n−1
+ (−1)n (2n+ 2)2n 24nn!(2n+)
2n(n+)n
+ (−1)n−1
(2n+ 1)(2n+ 2−1)(2n+ 2−2)2n−3(n+)
(2n+ 2)(2n++ 1)3(2n+)(n−2)!24n−1(2n+−2)2n−2(+n−2)n−2
= (−1)n (2n+ 2+ 2)2n+1
24n+2(n−1)!(2n++ 2)
2n+2(n+−1)n−1
+ (−1)n (2n+ 1)(n+ 1)(2n+ 2)2n−1(+n+ 1)
(n−1)!24n+1(2n++ 1)
2n+1(+n)n(2n+)
+ (−1)n (2n+ 2)2n 24nn!(2n+)
2n(n+)n
+ (−1)n−1 (2n+ 1)(2n+ 2)2n−1
(n−2)!24n+1(2n++ 1)
2n+1(+n−2)n−2(n+)(2n+)
= (−1)n (2n+ 2+ 2)2n+1
24n+2(n−1)!(2n++ 2)
2n+2(+n−1)n−1
+ (−1)n (2n+ 2+ 2)2n+1
24n+1n!(2n++ 1)
2n+1(+n)n
= (−1)n(2n+ 2+ 2)2n+1(n+ 2)(+n+ 2)
24n+2n!(2n++ 2)
2n+2(+n)n
:
The ratio −A˜n(0)=A˜
′
(ii) The numerator of the Rayleigh quotient J˜ of ˜Jn is
We do the same choice (see (54)) for the vector y as in the previous theorem and the numerator becomes
The results of the formal developments are still obtained thanks to Mathematica 3.0.
where ˜qi is a polynomial in the variable i of degree 6 whose the leading coecient is −29(−
1)(16−24+ 112). Then the sum Pn−1
i=1 ˜i can be represented as 32−32+ 112
128 (n−1) + O(lnn): (59) ˜
i 46
32−32+ 112
128 +
˜
ri
128i(i+)(2i+)(2i++ 1)3(4i++ 2)(4i+ 3−2)
;
where ˜ri is a polynomial in the variable i of degree 7 whose the coecients also are expressions which depend on n. In fact ˜ri is a polynomial of total degree 8 in the two variables iand n. Once again the behavior of ˜ri is that of terms of degree 8 in these two variables:
211(
−(16−24+ 112)ni7+ (16−48+ 332)2ni
6
−4(−8+ 112)3ni5+ 2224ni4):
Thus the behavior of ˜i is 32−32+11
2
128 + O(1) and the one of
P2n−1
i=n ˜i is
32−32+ 112
128 n+ O(n): (60)
The denominator of the Rayleigh quotient is still given by (57).
Therefore, gathering (59), (60) and (57), the result concerning the upper bound holds.
Once again the assumptions of Theorem 6.1 are satised. Thus, in the Gegenbauer case, if ∈D, then N¿0.
On the other hand we get the Markov–Bernstein inequality in the Gegenbauer case.
Corollary 6.8. In the Gegenbauer case the following Markov–Bernstein inequality is satised
∀p∈Pn− {0}; kp′k6max
q1
ˆ
1;n ;q1
˜
1;˜
n
kpk¡ Bnkpk;
where n= [n+12 ]; ˜n= [n2] and
Bn=
(n+ 1)(n+ 2+ 1)(n2+ 2n(+ 1) + 2−1)
4(1 + 2)
!1 2
if n is odd;
n(n+ 2)(n+ 2)(n+ 2+ 2)
4(1 + 2)
1 2
if n is even:
max(1=qˆ1;n;1=q˜1;˜
n) is the best constant.
An extremal polynomial is
p= n
X
i=1
w(1i ; n)
ipki−1
Ri;
where w(1; n) = (w(1; n)
1 ; : : : ; w(1n; n))T is the eigenvector of Jn corresponding to the eigenvalue min( ˆ1;n;˜1;˜
Proof. From the lower bounds given for ˆ1; n and ˜1; n it is easy to see that
Replacing n and ˜n by their respective values, we obtain the dierent expressions of Bn.
When = 1=2 and n¿5, Schmidt [11] (see also [9]) obtained the following expression
max
In the Jacobi case, Jn is a symmetric ve diagonal matrix (see (35)). We are only able to give an equivalent upper bound for 1; n.
Theorem 6.9 (Jacobi case). The smallest eigenvalue 1; n of Jn is such that
0¡ 1; n6= O
1
n4
; n¿1: (61)
Proof. Once again the Rayleigh quotient will be used to give a upper bound for 1; n. The numerator
of this Rayleigh quotient is given by
The same choice (see (54)) of the vector y is still done. When n is even the sequence {yi}ni=1 is
completed by yn+1= 0. The numerator of the Rayleigh quotient becomes
Xn−2
i=1
4
s
(i+)(i+)(i++)
i(2i+++ 1)(2i++−1)
i2
2i+++
(−)(i+ 1)2
(2i++)(2i+++ 2)
− (i+ 2)
2
(i+++ 1)(2i+++ 2)
s
(i+ 1)(i++ 1)(i++ 1)(i+++ 1) (2i+++ 1)(2i+++ 3)
!2
+ 4
s
(n−1 +)(n−1 +)(n−1 ++) (n−1)(2n++−1)(2n++−3)
(n−1)2 (2n−2 ++)
+ (−)
2
n
(2n−2 ++)(2n++)
− (n−1)
2
(n++−1)(2n++)
s
n(n+)(n+)(n++) (2n++−1)(2n+++ 1)
!2
+
2Xn−1
i=n
4
s
(i+)(i+)(i++)
i(2i+++ 1)(2i++−1)
(2n−i)2 2i++
+ (−)(2n−i−1)
2
(2i++)(2i+++ 2)
− (2n−i−2) 2
(i+++ 1)(2i+++ 2)
s
(i+ 1)(i++ 1)(i++ 1)(i+++ 1) (2i+++ 1)(2i+++ 3)
!2
+ 4(n+)(n+)(n++)
n(2n+++ 1)3(2n++)
yn2
= Xn−2
i=1
i+bn−1+ 2Xn−1
i=n
i+
4(n+)(n+)(n++)
n(2n+++ 1)(2n++)2(2n++−1)y 2
n;
where yn= 0 if n is even and yn= 1 if n is odd.
i=
4(i+)(i+)(i++)i3
(2i++)(2i+++ 1)3
+ 4(−)
2(i+ 1)4
(2i++)2(2i+++ 2)2
+4(i+ 1)(i++ 1)(i++ 1)(i+++ 1)(i+ 2)
4
(i+++ 1)2(2i+++ 2)(2i+++ 3) 3
+i+!i+i; where
i=−
8i2(i+ 2)2
(2i+++ 2)3(i+++ 1) s
(i+ 1)(i++ 1)2(i++ 1)2(i+++ 1)2
i(2i++−1)(2i+++ 3) ;
!i=
8(−)i2(i+ 1)2
(2i++)2(2i+++ 2) s
(i+)(i+)(i++)
i=
8(−)(i+ 1)2(i+ 2)2
(2i++)(2i+++ 2)2(i+++ 1) s
(i+ 1)(i++ 1)(i++ 1)(i+++ 1) (2i+++ 1)(2i+++ 3) :
If ¡ , we use the same inequality (concerning √i(i+)) as in the proof of Theorem 6.3; then:
!i6
2(−)i(i+ 1)2
(2i++)(2i+++ 2)4
2i++− (+)
2
2i++
!
× 2i++− (−)
2
2i++
!
2i++− 1
2i++
:
If ¡ , we use the inequality √1−u61−u2. Then
!i6
2(−)i(i+ 1)2
(2i++)(2i+++ 2)4
2i++− (+)
2
2(2i++)
!
× 2i++− (−)
2
2(2i++)
!
2i++− 1
2(2i++)
:
In the same way, if ¡ we get
i6−
2(−)(i+ 1)2(i+ 2)2
(2i+++ 2)(i+++ 1)(2i+++ 3)4
× 2i+++ 2− (+)
2
2(2i+++ 2)
!
2i+++ 2− (−)
2
2(2i+++ 2)
!
×
2i+++ 2− 1
2(2i+++ 2)
:
If ¿ ,
i6−
2(−)(i+ 1)2(i+ 2)2
(2i+++ 2)(i+++ 1)(2i+++ 3)4
2i+++ 2− (+)
2
2i+++ 2
!
× 2i+++ 2− (−)
2
2i+++ 2
!
2i+++ 2− 1 2i+++ 2
:
Finally
i6−
i(i+ 2)2
2(i+++ 1)(2i+++ 3)5
2i++− (−)
2
2i++
!
× 2i++− (+)
2
2i++
!
2i+++ 2− (−)
2
2i+++ 2
!
× 2i+++ 2− (+)
2
2i+++ 2
!
2i+++ 1− 4 2i+++ 1
:
Thanks to Mathematica 3.0, we nd that
i6
10−12+ 52+2
4
+ qi
4(i+++ 1)(2i++)3(2i+++ 2)
3(2i+++ 2)3(2i+++ 3)5
;
where qi is a polynomial of degree 14 in the variable i whose the leading coecient is
−212(−32 + 97−1042+ 423+ 31−64+ 382−82+ 102+ 63) if ¡ ;
−212(−32 + 95−1042+ 383+ 33−64+ 422−82+ 62+ 103) if ¿ :
Therefore the behavior of Pi=1n−2i is that of
1
4(10−12+ 5
2+2)(
n−2) + O(lnn): In the same way we get
i6
10−12+ 52+2
4
+ ri
4i(i+++ 1)(2i++)3(2i+++ 2)
3(2i+++ 2)3(2i+++ 3)5
;
where ri is a polynomial of degree 15 in the variable i, and a polynomial of total degree 16 in the two variables i and n. Thus the behavior of ri is that of
−217(4−8+ 52+2)ni15+ 217(2−12+ 152+ 32)2ni
14
−219(52+2)3ni13+ 218(2 + 4+ 52+2)4ni12 if ¡ or if ¿ :
Finally the term ˆn−16n=2 + (−3−2+ 52−6+2)=4 + O(1n): Gathering all the results we have
= 10−12+ 5
2+2
4 (2n−2) +
n
2 + O(n): Therefore the result (61) holds.
The assumptions of Theorem 6.1 being satised, if ∈D, then N¿0 in the Jacobi case.
6.2. Case N ¿1
The main aim of this subsection is to prove that the boundary of Dn corresponds to a nappe of an algebraic hypersurface of degree n (consequence of Theorem 6.17), and that D contains a set of points = (1; : : : ; N) for which the i; i= 1; : : : ; N −1 are not all nonnegative (Theorem 6.20).
Let us assume that a(p; p)¿0∀p ∈ Pn − {0}: Then the sequence of Sobolev monic formal orthogonal polynomials{Si}ni=0 exists and is unique. Let ˆSnbe the vector of components S0; S1; : : : ; Sn. Then
ˆ
Rn=Tn; NSˆn; (62)
where n; N is a (n+ 1)×(n+ 1) lower triangular matrix whose elements are denoted by i; j with
For any polynomials p and q∈Pn, we have
a(p; q) =zTT
n; NK˜n; Nn; Ny; (63)
where ˜Kn; N is the (n+ 1)×(n+ 1) matrix containing elements equal to a(Si; Sj) for i; j= 0; : : : ; n. Therefore ˜Kn; N is the diagonal matrix whose elements are ˜ki; N =a(Si; Si); i= 0; : : : ; n.
From Eqs. (27) and (63) we get
T
n; NK˜n; Nn; N = b(n) X
i=0
iDn; N
b(Yn)−1
j=i
J−j; n−j
!T
Kn; i
b(Yn)−1
j=i
J−j; n−j
!
Dn; N; (64)
where b(n) = min(n; N). If ( ˜Kn; N)
1
2 denotes the (n+ 1)×(n+ 1) diagonal matrix whose elements are ( ˜ki; N) 1
2, we can
remark that (( ˜Kn; N)
1
2n; N)T( ˜Kn; N) 1
2n; N is the Cholesky decomposition of the matrix in the right
part of (64). Due to the particular properties of this matrix, we immediately get the following result.
Theorem 6.10. T
n; N is a lower triangular band matrix with a width of band equal to N in the
Laguerre case; 2N in the Jacobi and Gegenbauer cases. If the elements of this matrix are denoted by i; j; i; j= 0;1; : : : ; n; then i;0= 0; i= 1; : : : ; n; ‘;‘−N 6= 0 for ‘¿N + 1.
For the sake of simplicity, let us write (64) as
Tn; NK˜n; Nn; N = b(n) X
i=0
iCn; i; N: (65)
Theorem 6.11.
˜
k0; N=k0=Q0; N;
˜
kn; N=
Qn; N(1; : : : ; b(n))
Qn−1; N(1; : : : ; b(n−1))
for n¿1; (66)
where b(n) = min(n; N): Qn; N(1; : : : ; b(n)) is a polynomial in b(n) variables 1; : : : ; b(n) of total
degree equal to n.
Proof. We take the determinant of each part of (65). We get
det b(n) X
i=0
iCn; i; N
!
= (det(n; N))2det ˜Kn; N= det ˜Kn; N = n
Y
i=0
˜
ki; N:
Thus
˜
kn; N =
detPbi=0(n)iCn; i; N
detPbi=0(n−1)iCn−1; i; N
;
and (66) holds in setting
Qn; N(1; : : : ; b(n)) = det
b(n) X
i=0
iCn; i; N
!
Thanks to the Cauchy–Binet formula, it is clear that if n6N; Qn; N contains the monomial
Proof. (i) In this case, from (27), we get
a(p; q) =
minor corresponds to the b(n) rst rows and columns of
(iii) The homogeneous polynomial of degree n is obtained from elements of Pbi=0(n)iCn; i; N in (67).
Theorem 6.13. Except in the Gegenbauer case;the polynomials Qn; N are irreducible for N¿1 and
16n6N.
Proof. The degree in n is only 1 in Qn; N. Moreover Qn; N contains the monomial Qnj=1j. Thus, if
Qn; N can be factorized
Qn; N=qQ∗n; (69)
(i) Let us begin to prove that Qn; N can not be factorized by a polynomial q of total degree equal to 1 and containing n. Expanding the determinant in Eq. (67) with respect to the last row and denoting by Pn‘=0‘Cn; ‘; N(i; j); i; j= 0; : : : ; n the elements of the matrix
Pn
i=0iCn; i; N, we get
Qn; N= n
X
i=0
iCn; i; N(n; n)
!
Qn−1; N +A;
whereA=Pnj=1−1Pin=0−1iCn; i; N(n; j)
∗ corresponding cofactor (n; j).Ais not reduced to zero. Indeed
A contains the monomial Cn; n−2; N(n; n−1)3n−2 Qn−3
j=1 j. It is the only one containing n3−2 Qn−3
j=1 j. MoreoverCn; n−2; N(n; n−1)6= 0. Due to the properties of factorization, used in Property 6:12; Cn; n−2; N (n; n−1)6= 0 is equivalent to Cn; n−2; n(n; n−1)6= 0. Since Cn; n−2; n=Dn; n(J−n+1;1)T(J−n+2;2)TKn; n−2
J−n+2;2J−n+1;1Dn; n, it is easy to get the expression of Cn; n−2; n(n; n−1):
Cn; n−2; n(n; n−1) =
n!(n−1)!k1 in the Laguerre case;
n!(n−1)!2(−)k1
(2 ++)(4 ++) in the Jacobi case:
Hence, if 6=, this coecient is also dierent from zero in the Jacobi case. A and Qn−1; N do not depend on n. Thus Qn−1; N=Q∗n: Qn−1; N contains the monomial Qnj=1−1j and A does not contain this monomial which is only obtained by product of diagonal elements. ThereforeA can not be factorized by Qn−1; N. Hence the factorization (69) is impossible.
(ii) Now let us prove that if Qn; N can be factorized by a polynomial q of degree strictly greater than 1 containing n, we can deduce that there exits k ∈ N; 26k ¡ n, such that Qk; N can be factorized by a polynomial of degree 1 containing k.
q can be written as q=np+r withr independent ofn and degp=degq−1¿1. Indeed degQn∗=
n−degq: Qn; N=np Q∗n+r Q∗n andnpQ∗n must contain
Qn
j=1j. Hence p Qn∗=Cn; n; N(n; n)Qn−1; N and therefore Qn−1; N can be factorized. The degree in n−1 is only 1 in Qn−1; N and Qn−1; N contains the monomial Qnj=1−1j. Thus n−1 belongs to p or to Q∗n. Let us denote by q the polynomial containing
n−1 and by Qn∗−1 the other.
If degree of q= 1, Part (i) gives us the result.
If degree of q ¿1, we restart the study with Qn−1; N and the variable n−1 like in Part (ii). Since
the degree of polynomials decreases in the decomposition, a polynomial q of degree 1 will nally be obtained in a factorization Qk; N =q Q∗k with k¿2, and q containing k. Thus Part (i) will give the result.
Remark that Q1; N(1) is the only irreducible polynomial in the Gegenbauer case. Indeed we have
n; N(i; i+ 2j+ 1) = 0∀i; j in (65). Therefore, applying the same process as the one used for obtain-ing ˆJ[n+1
2 ] and ˜J[
n
2] in Section 6.1, it is clear that any polynomial Qn; N(1; : : : ; b(n)) is reducible,
except Q1; N(1).
Theorem 6.14. In the Laguerre and Jacobi (with 6=) cases; for 26N ¡ n; if Qn; N(1; : : : ; N)
is reducible; then Qn; N−1(1; : : : ; N−1) is reducible.
Proof. A factorization
can also be written as
(Nq˜(1; : : : ; N) +q(1; : : : ; N−1))(NQ˜
∗
n(1; : : : ; N) +Qn∗(1; : : : ; N−1)):
Hence, from Property 6:12(i) we have
q(1; : : : ; N−1)Qn∗(1; : : : ; N−1) =Qn; N−1(1; : : : ; N−1): (71)
The two previous relations imply that
if ˜q6= 0, then deg(Nq˜(1; : : : ; N)) = deg(q(1; : : : ; N−1)),
if ˜Q∗n 6= 0, then deg(NQ˜
∗
n(1; : : : ; N)) = deg(Qn∗(1; : : : ; N−1)).
Because of the symmetric part played by the two polynomials of the factorization (70), it is sucient to study the two cases ˜q= 0 and ˜q6= 0 only.
If ˜q= 0, and since degq¿1, then q(1; : : : ; N−1) is a factor of QN−1; N−1(1; : : : N−1) which is
the coecient of nN−N+1, apart from a constant multiplicative factor (see Property 6:12(ii)). Since
QN−1; N−1(1; : : : ; N−1) is reducible, we have q(1; : : : ; N−1) =QN−1; N−1(1; : : : ; N−1); apart from a
constant multiplicative factor. Hence Qn; N−1(1; : : : ; N−1) =QN−1; N−1(1; : : : ; N−1)Q∗n(1; : : : ; N−1).
If ˜q 6= 0, then, since degq(1; : : : ; N−1)¿1, the factorization of Qn; N−1(1; : : : ; N−1) holds (see
(71)).
Corollary 6.15. In the Laguerre and Jacobi (with 6= ) cases; for N ¿2 and n ¿ N; if
Qn; N(1; : : : ; N) is reducible; then Qn;2(1; 2) is reducible.
Conjecture 6.16. In the Laguerre and Jacobi(with6=)cases; Qn;2(1; 2)is irreducible; ∀n¿1.
Let us write Qn; N as a polynomial in b(n), the coecients of this polynomial depending on
1; : : : ; b(n)−1. This polynomial will be denoted by Qn; N(b(n)) and we will give some properties
about the zeros of Qn; N(b(n)).
Theorem 6.17. ∀(1; : : : ; b(n)−1) ∈ Db(n)−1; the zeros b(n(n)); i; i = 1; : : : ; n−b(n) + 1; of Qn; N(b(n))
are real. Moreover if n ¿ N; these zeros are separated by those of Qn−1; N(N); that is to say; if
N; i(n)¿N; i(n)+1; i= 1; : : : ; n−N; then
N;(n)1¿N;(n1−1)¿(N;n)2¿(N;n−21)¿· · ·¿N; n(n−−1)N¿
(n)
N; n−N+1: (72)
Proof. If n6N, and if (1; : : : ; b(n)−1)∈Db(n)−1, then ˜ki; N¿0 for i6b(n)−1. Thanks to (66), we have Qi; N(1; : : : ; i)¿0 for i6b(n) − 1. The monomial b(n)Cb(n);b(n);N(b(n); b(n))Qb(n)−1; N (1; : : : ; b(n)−1) is the only one containing b(n). Thus the property holds.
If n ¿ N, let us begin to transform the matrix PNi=0iCn; i; N thanks to a diagonal matrix Tn; N dened by
Tn; N(i; i) =
1 for i= 0; : : : ; N −1;
1
p
Cn; N; N(i; i)
Let us denote by ˜Qn the (n+ 1)×(n+ 1) transformed matrix
˜
Qn=Tn; N N
X
i=0
iCn; i; N
!
Tn; N:
We will give the following block structure to ˜Qn
˜
Qn= Q˜N−1 Bn
BTn Gn+NI
!
;
where ˜QN−1 (resp. Gn) is an N×N (resp. (n−N+ 1)×(n−N+ 1)) symmetric matrix independent of N, and Bn is an N×(n−N+ 1) matrix. Using the elements of the block structure of ˜Qn−1, we
can also write ˜Qn as
˜
Qn=
˜
QN−1 Bn−1 u
BT
n−1 Gn−1+NI v
uT vT gn+N
:
Using the Schur complement (see [6]), this new form of ˜Qn has the following decomposition:
˜
Qn=
I 0 0
BTn−1( ˜QN−1)
−1 I 0
uT( ˜QN−1)−1 0 1
˜
QN−1 Bn−1 u
0
Wn 0
with
Wn=
Gn−1−BTn−1( ˜QN−1)
−1B
n−1+NI v−BTn−1( ˜QN−1)
−1u
vT−uT( ˜Q N−1)
−1B
n−1 gn−uT( ˜QN−1)
−1u+
N
!
:
Since
Qn; N=
1
detTn; N
2
det ˜Qn=
1
detTn; N
2
det ˜QN−1detWn;
it is clear that detWn is the characteristic polynomial (in −N) of a symmetric matrix. Therefore the zeros (N; in); i= 1; : : : ; n−N + 1, are real. Moreover Qn−1; N corresponds to the leading principal minor of order n−N of this symmetric matrix. Thus (72) holds (see [14] p. 103).
Qn; N(1; : : : ; b(n))=0 is the equation of an algebraic hypersurface inRN. Remark that the
cancella-tion of the homogeneous polynomial of higher degree in Qn; N(1; : : : ; b(n)) (see Property 6:12) gives
the asymptotic directions of this algebraic hypersurface. Let us denote byFn(1; : : : ; b(n)) the nappe
corresponding to the largest zero of Qn; N(b(n)) when (1; : : : ; b(n)−1) ∈ Db(n)−1. If b(n) is greater
than (resp. greater than or equal to) this largest zero, we will denote the corresponding domain by the notation Fn(1; : : : ; b(n))¿0 (resp.Fn(1; : : : ; b(n))¿0). From (66), (38) and Theorem 6.17, it
is clear that, ∀n¿1
Dn={∈Rb(n)|Fn(1; : : : ; b(n))¿0}:
Corollary 6.18.
D=n∈RN
lim
n→∞Fn(1; : : : ; N)¿0
o
:
Property 6.19. Dn is a convex domain.
Proof. For a given ∈Dn, let us denote by a the corresponding bilinear functional (2).
If and ˆ ∈ Dn, for any ∗=+ (1−) ˆ, with 0661, we have a∗ =a+ (1−)aˆ.
Therefore ∗
∈Dn.
For concluding this subsection we will prove that D can not be reduced to the domain with
i¿0; ∀i.
Let us denote by D+
n (resp. D
+
n) the domain of Rb(n)
D+n ={∈Rb(n); = (1; : : : ; b(n))|i¿0; i= 1; : : : ; b(n)}; (resp:D+n ={∈Rb(n); = (1; : : : ; b(n))|i¿0; i= 1; : : : ; b(n)}): Of course, if ∈D+
n; Qn; N(1; : : : ; b(n))¿0.
Theorem 6.20. D\D+N is non-empty.
Proof. (i) Let us begin to prove that Dn\D
+
n is non-empty for n6N. This property is satised by
D1={1 ∈R; 1¿−1;1}. Qn; N(1; : : : ; n) is a polynomial of degree 1 in the variable n which can be written as
Qn; N(1; : : : ; n) = (n!)2k0Qn−1; n−1(1; : : : ; n−1)n+Qn; N−1(1; : : : ; n−1)
from Property 6:12(i) and (ii). The zero of Qn; N will be denoted by 0n( ˆ), where ˆ= (1; : : : ; n−1).
Qn−1; n−1(1; : : : ; n−1)¿0 if ˆ ∈ Dn−1. Moreover, if the property (Dn−1 \D +
n−1 is non-empty) is
assumed to be true, then Qn−1; n−1(1; : : : ; n−1)¿0 for ˆ∈Dn−1\D +
n−1.
Using the assumption of recurrence at the rank n−1 and thanks to the proof of parts ii) and iii) of this theorem which are satised whenN=n−1, it is clear that ˆDn\D
+
n−1 is non-empty. ˆDn is dened by ˆDn ={ ∈ Rn−1|Fn(1; : : : ; n−1)¿0}. Thus, if ˆ ∈ D
+
n−1, then 0n( ˆ)¡0. If ˆ ∈ Dˆn\D
+
n−1,
then 0
n( ˆ)¡0, and if ˆ∈Dn−1\Dˆn, then 0n( ˆ)¿0, where Dˆn={∈Rn−1|Fn(1; : : : ; n−1)¿0}.
Therefore Dn\D
+
n is non-empty.
(ii) Let us denote by ˆ∈RN−1 of components ˆ
1; : : : ;ˆN−1, any point in DN−1. Thus
QN−1; N−1( ˆ1; : : : ;ˆN−1)¿0.
On the other hand let us consider ∈RN such that= ( ˆ;
N). From Theorem 6.17, the zeros of
Qn; N (n¿N), considered as a polynomial in the variable N, are real. The largest one corresponds to Fn( ˆ1; : : : ;ˆN−1; N).
For showing that D\D+N is non empty, it is sucient to prove that the zeros ofQn; N are bounded,
∀n, when ˆ∈DN−1 with kˆk∞ bounded. Since these zeros are linked with the eigenvalues of a part
N or 2N, it is sucient to prove that all the elements of the rows of ˜Qn are bounded ∀n. Thanks to the Gerschgorin disks, the eigenvalues will be bounded.
(iii) Finally let us prove by recurrence on N that the elements of
˜
Qn; N = NX−1
i=0
ˆ
iTn; NDn; N NY−1
j=i
J−j; n−j
!T
Kn; i NY−1
j=i
J−j; n−j
!
Dn; NTn; N
are bounded.
For N= 1, thanks to Eq. (32),
˜
Qn;1=Tn;1Dn;1JT0; nKn;0J0; nDn;1Tn;1=
k0 0
0 Jn
:
In Property 5:2 we have already used the fact that all the elements of Jn are bounded ∀n. Let us assume that
˜
Qn; N−1= N−2 X
i=0
ˆ
iTn; N−1Dn; N−1
NY−2
j=i
J−j; n−j
!T
Kn; i NY−2
j=i
J−j; n−j
!
Dn; N−1Tn; N−1
has all its bounded elements ∀n. Then ˜
Qn; N =LTn; NQ˜n; N−1Ln; N + ˆN−1Tn; NDn; NJ−TN+1; n−N+1Kn; N−1J−N+1; n−N+1Dn; NTn; N;
where Ln; N =Tn; N−1−1D
−1
n; N−1J−N+1; n−N+1Dn; NTn; N. Since Cn; N; N =Dn; NKn; NDn; N, then
Dn; NTn; N =
DN−1; N 0 0 (Kn−N;0)−
1 2
!
Like in Eq. (32) we obtain
Tn; NDn; NJ−TN+1; n−N+1Kn; N−1J−N+1; n−N+1Dn; NTn; N =
where 0N−1 is the (N−1)×(N−1) zero matrix. Thus this matrix has all his bounded elements ∀n.
Finally let us prove that Ln; N has all its bounded elements ∀n. Indeed we have
Tn; NDn; NJ−TN+1; n−N+1T
−1
n; N−1T
−1
n; N−1J−N+1; n−N+1Dn; NTn; N;
which is a Cholesky decomposition of a band matrix having the following structure
N ˆN ˆ
TN Jn−N+1 !
:
Thus Tn; N−1−1J−N+1; n−N+1Dn; NTn; N is a matrix having all its bounded elements, and a fortiori
D−n; N1−1Tn; N−1−1J−N+1; n−N+1Dn; NTn; N, sinceD−n; N1−1 is a diagonal matrix whose the elements
−1
i; N, dened from Eqs. (19) and (20), tend to 0 when n tends to innity.
