Turán Inequalities for Three Term Recurrences with Monotonic Coefficients

0.29 MB·English

by  Ilia Krasikov

TURA´N INEQUALITIES FOR THREE TERM RECURRENCES
WITH MONOTONIC COEFFICIENTS
1
1 ILIAKRASIKOV
0
2
Abstract. We establish some new Tura´n’s type inequalities for orthogonal

n polynomials defined by a three-term recurrence with monotonic coefficients.
a As a corollarywededuce asymptotic bounds onthe extreme zeros of orthog-
J onal polynomials withpolynomiallygrowingcoefficients of thethree-term re-

7 currence.
1

]
A
1. Introduction
C
Considerafamily p oforthogonalpolynomialsdefinedbytheinitialconditions
. i
h { }
p =0, p =1, and the three term recurrence
1 0
t −
a a
k
m (1) pk =(x bk 1)pk 1 ak 1ck 1pk 2,
ck − − − − − − −
[ where the coefficients a and c are strictly positive for all k besides a which
k k 0
1 will be convenient to set to zero. Thus the choice ck = 1 or ck = ak for all k
v correspondsto the orthonormaland the monic normalizationrespectively. We will
4 use boldface characters to distinguish orthonormal polynomials from these in a
0
different normalization.
2
Althoughalltheinformationaboutanymemberofthefamily p isencodedin
3 { k}
. (1), itturnsouttobenotoriouslydifficulttoextractitfromthe recurrence. Oneof
1
few toolswe possesstodayto dealwiththe problemis so-calledTur´aninequalities.
0
Usually they are written in the form
1
1 (2) p2(x) p (x)p (x)>0
: k − k−1 k+1
v and are valid for many particular families of orthogonal polynomials including all
i
X classical ones as well as some other members of the Askey scheme. They play an
importantrolein manyapplications,e.g. to recoverthe absolutelycontinuouspart
r
a ofthe correspondingorthogonalitymeasureorto boundthe extremezeros(see e.g.
[1], [3], [5], [13], [18]).
In terms of the coefficients of the three-term recurrence very general conditions
forthevalidityof(2)weregivenin[25],providedthatthesupportofthecorrespond-
ing measure is finite or half-infinite, such that the polynomials can be normalized
to one at some point. The symmetric case b 0 was consideredin [2] and [13]. It
i
≡
seems almost nothing is known in the general asymmetric case with the measure
supported on the whole axis.
In a sense, Tur´an inequalities can be viewed as an analogue of the Laguerre
inequality f 2 ff > 0. Since the last one is just the first member of an infinite
′ ′′
−
familyofinequalitiesdiscoveredbyJensen[11](andrediscoveredanumberoftimes
later, see e.g. [21]), it is worth trying to look for higher order generalizations of
1991 Mathematics Subject Classification. 33C45.
1
2 I.KRASIKOV
Tur´an inequalities. Some results in this direction were given in [5] and [13] and
recently this question was raised again by P. Nevai [19].
Another related set of inequalities comes from the Newton inequality stating
that a2 a a >0, where a are the coefficients of a real polynomial
i − i−1 i+1 i
n
n
a xi
i
i
i=0 (cid:18) (cid:19)
X
with only real zeros [5], [20].
It is not clear how far such an analogy can go. For example, an important
differencebetweentheseandTur´aninequalitiesisthatinthelastcasenormalization
ofp plays acrucialrole. Infact, bothLaguerreandNewton type inequalities lead
k
directly to higher order Tur´an inequalities only if the corresponding orthogonal
polynomials have a generating function of a very special type [5], [21], [22]. In
this paper we explore these analogies to establish some new Tur´an inequalities for
threetermrecurrencewithmonotoniccoefficientsbothinthesymmetricandinthe
general case. Thus, unless the coefficients of (1) are bounded, the corresponding
measure has an unbounded support.
As an application we give an upper bound on the largest zero of a polynomial
satisfying (1) with the coefficients a and b of polynomial growth,
k k
a kr, b ks.
k k
∼ ∼
Thefollowingremarkableresultforthesymmetriccasewasobtainedin[16],[17]:
Theorem 1. Let p be a family of monic symmetric polynomials given by
k
p =xp a2 p ,
k k−1− k−1 k−2
where
a =ckr 1+o k 2/3 ,
k −
as k and r > 0. Let x < x (cid:16) < (cid:16),x (cid:17)b(cid:17)e the zeros of p . Then, for a
1,k 2,k k,k k
→ ∞ ···
fixed j,
x =2ckr 1 r2/36 1/3i k 2/3+o k 2/3 ,
k j,k − j − −
− −
where i <i < are positiv(cid:16)e zeros of the Airy function(cid:16). (cid:17)(cid:17)
1 2
···
Thecrucialpointoftheproofofthistheoremisthatthebehaviourofafewlargest
zerosunderanappropriaterescalingmimicsthatoftheHermitepolynomials. Itwas
alsoshown that at leastfor some particularfamilies of symmetric polynomials and
the extreme zeros similar results can be obtained via chain sequences [3]. The last
paper providesalso very sharpbounds for the extreme zerosof certainasymmetric
polynomials. This seems to be the only known result of this type for an infinite
intervaloforthogonality. In[13]we gavesome new Tur´aninequality for symmetric
polynomials and used them to give a non-asymptotic version of Theorem 1, yet
with an unavoidably weaker constant at the second order term, since the obtained
bounds hold for any k.
It turns out that to extend Tur´an inequalities to the general asymmetric case
one has to impose rather severe constrains either on the coefficients of (1) or to
restrict the rang of x for which the inequality holds. In particular, most of the
results of this paper deal with monotonic sequences a and b , where a is strictly
k k k
increasing. Moreover, without loss of generality we will assume that the sequence
POLYNOMIALS 3
b is nondecreasing. To justify this assumption it is enough to notice that the
k
polynomials q (x)=( 1)kp ( x) satisfy the recurrence
k k
− −
a
k
q =(x+b )q a c q .
k k 1 k 1 k 1 k 1 k 2
ck − − − − − −
It is important to stress that to apply the approach adopted in this paper for
bounding the largest zero x (the least zero x ) it is enough to have a Tur´an
kk 1k
inequality which holds for x > x (respectively x < x ). Here we will establish
kk 1k
some inequalities of this type. In particular we use one such (rather technical)
higher order inequality to give a version of Theorem 1 in the general asymmetric
case.
Letusnoticethatforpolynomiallygrowingcoefficientsof(1),a ir, b γis,
i i
≈ ≈
thesoughtasymptoticdependsonthethreeparametersr,s,andγ (seeminglymore
general case a γ ir, b γ is, is just a rescaling, see e.g. formulas (53) and
i 1 i 2
≈ ≈
(54) below). As we don’t know asymptotics of the extreme zeros of any specific
three-parametric family with polynomially growing coefficients one hardly can use
the technics of [3] or [16]. On the other hand we conjecture that our method gives
thecorrectorderforthe secondtermoftheasymptoticyetwithaweakerconstant.
Namely, we prove the following:
Theorem 2. Let
1
m= 2
3 1/r 1 −
2 −
2
ρ= m(cid:0)in(cid:1)1,r s+1 ,
3 { − }
and let
ai =ir+o(ir−ρ), bi =γis+o(ir−ρ),
where r 1 and 0 s<r+1 are fixed numbers, and the constant γ satisfies
≥ ≤
(m+2)r−s+1 , 0<s<1,
2s
γ

≤ (m+3)r−s+1 , 1 s<r+1.
3s ≤
Then for sufficiently large kthe largest zero of pk does not exceed
(3) γks+kr 2 2 4/3δ2/3k ρ +o kr ρ ,
− − −
−
where δ is any fixed number s(cid:16)atisfying (cid:17) (cid:0) (cid:1)
2r, 0 s<r,
≤

δ < (2+γ)r, s=r,

γs, r<s<r+1.

The restrictions on γ and δ in the last theorem are definitely not best possible.
However it seems that some constraints of that type are necessary. Concerning
the least zero of p let us notice that the first term of its asymptotic (or, more
k
precisely a lower bound) is given by γks 2kr, (see e.g. Theorem 11 below). It
−
seems plausible that the secondorder termmay be quite different fromO(kr ρ) in
−
general.
4 I.KRASIKOV
Thepaperisorganizedasfollows. Inthenextsectionweintroducesomepossible
generalizations of Tur´an inequalities. Under appropriate conditions the suggested
inequalities will be proved in section 3. We also provide some examples with the
Stieltjes-Wigert, Al-Salam-Carlitz and the Meixner-Pollaczek polynomials. Sec-
tions 4 and 5 deal with the proof of Theorem 2. It turns out that it is enough
to prove the result for a suitably chosen test sequences a = α , b = β . Such
k k k k
sequences and an appropriate Tur´an type inequality will be given in section 4.
We’ll need some rather tedious calculations to verify that the chosen sequences
α , β satisfy the inequality. Given a Tur´an type inequality, there is a quite
k k
{ } { }
straightforward way to derive the sought bounds on the extreme zeros, provided
some technicalconditions are fulfilled. This will be accomplishedinsection 5, thus
proving Theorem 2.
2. Tura´n inequalities. Preliminaries
Consider the case of monic polynomials defined by
(4) q (x)=(x b )q (x)+a2q (x); q =0, q =1.
k+1 − k k k k−1 −1 0
First we notice that Tur´an’s inequality
(5) q2(x) ξ q (x)q (x)>0,
k − k k−1 k+1
is equivalent to the following one for polynomials in arbitrarily normalization de-
fined by (1),
a c
p2(x) ξ k+1 k p (x)p (x).
k − kakck+1 k−1 k+1
Since
q2(x) q (x)q (x)=
k − k−1 k+1
k 1
−
(b b )x2k 1+ a2 a2 (b b ) b +2 b x2k 2+O(x2k 3),
k− k−1 − k− k−1− k− k−1 k i!! − −
i=0
X
then (5) may hold (and in fact holds) with ξ =1 for all x R only if b are con-
k k
∈
stant and a are nondecreasing. Thus, if we want to deal with generalasymmetric
k
orthogonalpolynomials,wehaveeither tochooseξ <1ofto restrictxto asubset
k
of R or both. On the other hand, for sufficiently small ξ , (5) will be fulfilled for
k
all x. Indeed, if ξ < 1 then q2(x) ξ q (x)q (x) > 0 in the neighbourhood
k k − k k−1 k+1
of , as well as around the zeros of q (x), as there q (x) and q (x) have
k k 1 k+1
±∞ −
opposite signs. This simple argument can be presented in a quantitative form, in
particular it is not difficult to show that one can take
4a2
ξ = k ,
k 4a2+max((x b )2,(x b )2)
k 1k− k kk − k
andthis,inasense,isbestpossible. Since inmanycasesanintervalcontainingthe
extreme zeros is known, this yields an explicit Tur´an inequality.
We will need the following theorem due to Wendroff (see e.g. [9, Th.2.10.1]).
Theorem 3. Given two real sequences a , i N, a > 0, and b , i N,
i i i
{ } ≥ { } ≥
and two sequences of interlacing numbers x < x < ... < x and y < y <
1 2 N 1 2
POLYNOMIALS 5
...<y , such that x <y <x , i=1,...,N 1. There is a family of monic
N 1 i i i+1
− −
orthogonal polynomials q (x) such that
i
{ }
N N1
q (x)= (x x ), q (x)= (x y ),
N i N 1 i
− − −
i=1 i=1
Y Y
and
q (x)=(x b )q (x) a2q (x)
k+1 − k k − k k−1
and a =a , b =b , for k N.
k k k k
≥ e e
Theorem 4. Let q be a family of monic orthogonal polynomials satisfying (4)
e e { n}
and let x <x <...<x be the zeros of q (x). Then for k 2 inequality (5)
1n 2n nn n
≥
holds for
4a2k , x>b 4a2+(x b )2,
4a2k+(xkk−bk)2 k− k kk − k
(6) ξ =
k p
 4a2k+(4xa12kk−bk)2 , x<bk+ 4a2k+(x1k−bk)2.
Moreover, one canchoose ξk = 1, provided eithper x > bk −2ak and xkk ≥ bk, or
x<b +2a and x b .
k k 1k k
≤
The result of (6) is the best possibe in the sense that for any fixed k
≥
2, a > 0, b , x , x ; x < x , b < x , (resp. b > x ), and any
k k 1,k kk 1,k kk k kk k 1k
ξ > 4a2k , (resp. ξ > 4a2k ), there is a family of monic orthog-
k 4a2k+(xkk−bk)2 k 4a2k+(x1k−bk)2
onal polynomials and a point x>x (resp. x<x ) such that
kk 1k
q2(x) ξ q (x)q (x)<0.
k − k k−1 k+1
Proof. We will give a proof for the case where the largest zero x is involved, the
kk
second one is similar.
We set t =t (x)=q (x)/q (x), and h=b x . Using (4) we find
k k k 1 k k kk
− −
q2(x) ξ q (x)q (x)
F(x,tk)= k − kq2k−(x1) k+1 =ξka2kt2k−ξk(x−bk)tk+1.
k
First notice that this quadratic in t is positive for
k
2a 2a
k k
b <x<b + .
k k
− √ξ √ξ
k k
Hence it will be enough to prove the claim for x beyond this interval. By x <
ik
x <x we have
i,k 1 i+1,k
−
k 1
1 − x xi,k 1 1
tk = − − < , x>xkk.
x x x x x x
kk ik kk
− i=1 − −
Y
Hence for x>x we can set t = ǫ , where ǫ=ǫ(x) (0,1).
For ξ = kk4a2k = 4ka2k x−wxek,hkave ∈
k 4a2k+(xkk−bk)2 4a2k+h2
2a
x <b + k =b + 4a2+h2 x,
kk k √ξ k k ≤
k
q
and one finds
4a4ǫ2 4a2(x b )ǫ
F(x,t )= k k − k +1=
k (x x )2(4a2 +h2) − (x x )(4a2 +h2)
− kk k − kk k
6 I.KRASIKOV
2a2ǫ (x+x )h 2+4a2(x x )2(1 ǫ)
k − kk k − kk − >0,
(x x )2(4a2 +h2)
(cid:0) − k(cid:1)k k
and (6) follows.
Next we shall prove that for x > b 2a , that is for h 0, and x b , one
k k kk k
− ≥ ≥
can take ξ =1. Setting x=b +δ, where δ 2a , we obtain
k k k
≥
ǫ(δ2+δh a2ǫ) δ2+δh a2 a2 h
F(x,t )=1 − k >1 − k = k + 0.
k − (δ+h)2 − (δ+h)2 (δ+h)2 δ+h ≥
Let us show that the above choice of ξ in (6) is the best possible, provided
k
x >b .ByWendroff’stheoremforanyfixedk, a >0, b , x <x ,thereisa
kk k k k 1,k kk
correspondingorthogonalpolynomial with these parameterssuch that the product
k 1
− x xi,k 1
− −
x x
ik
i=1 −
Y
is arbitrarily close to one for x > x . Hence we can set t = ǫ with ǫ
arbitrarily close to one. Given a ξ , wekkfind k x−xkk
k
ǫ (x b )(x x ) a2ǫ
F(x,t )=1 − k − kk − k ξ .
k − (x x )2 k
(cid:0) − kk (cid:1)
By the assumption x >b , hence h<0 and we choose x=x 2a2kǫ obtaining
kk k kk− h
4a2ǫ+h2
F(x,t )=1 k ξ .
k − 4a2 k
k
This implies
4a2
ξ < k ,
k 4a2ǫ+h2
k
assuming that F(x,t ) is positive and the result follows. (cid:3)
k
In connection with this theorem it’s worth noticing that x b x for
1k k 1 kk
≤ − ≤
anyorthogonalpolynomial. Indeed,letJ bethetruncatedJacobimatrix(see(51)
k
below). Then for 0 i k 1 we have
≤ ≤ −
x = inf (J v,v) (J e ,e )=b sup (J v,v)=x ,
1k k k i i i k kk
||v||=1 ≤ ≤||v||=1
where(.,.)and . denotethescalarproductandthenorminCk ande isavector
i
in Ck with zero||en||tries except for ith which is equal to 1.
Arguments similar to these above are readily applicable to higher order Tur´an
inequalities of the form
m
(7) p2 + ξ(i)p p >0.
k k k i k+i
−
i=1
X
Indeed, one canstart with the usualTur´an inequality and choose sufficiently small
ξ(i) fori 2.Howeverthisseemsrathermisleadingsincesuchaproofsuggestsvery
k ≥
small values of ξ(i), whereas in fact they can grow exponentially e.g. for Hermite
k
polynomials, as the following result of Jensen shows [11].
POLYNOMIALS 7
A higher order generalization of the Laguerre inequality f 2 ff > 0 which
′ ′′
−
holds, in particular, for the functions of the so-called Polya-Laguerreclass has the
following form (for modern exposition see e.g. [21], [22]).
m
1 2m
(8) 2m(f)= ( 1)j+m f(j)f(2m−j) 0, ,m=0,1,...
L 2 − j ≥
j=0 (cid:18) (cid:19)
X
In particular
(f)=f 2 ff ,
2 ′ ′′
L −
(f)=3f 2 4f f +ff(4).
4 ′′ ′ ′′′
L −
By analogue one can introduce corresponding Tur´an type operators which will be
considered in this paper. The first of of them is just the standard Tur´an one
(9) T (p )=p2 p p ,
2 k k− k−1 k+1
and the second is
(10) T (p )=3p2 4p p +p p .
4 k k− k−1 k+1 k−2 k+2
TojustifytheformoftwofollowingoperatorsS andS whichwillbeconsidered
2 4
in the sequel (the last will be treated for symmetric polynomials only ) we need
some explanations. It was noticed in [2] that in the monic normalization among
the expressions of the form p2 ξp p , the polynomial p2 p p has the
k− k−1 k+1 k− k−1 k+1
minimal possible degree. Allowing an arbitrarily normalization defined by (1) let
us consider the following polynomials:
(11) p2 ξ p p ,
k− k k−1 k+1
(12) p2 µ p p +ν p p .
k− k k−1 k+1 k k−2 k+2
Notice that for b 0,
i
6≡
k 1 k
p (x)= xk xk 1 − b +O(xk 2) ci ,
k − i −
− ! ai
i=0 i=1
X Y
whereas in the symmetric case b 0, we have
i
≡
k 1 k
p (x)= xk xk 2 − a2+O(xk 4) ci .
k − − i − a
! i
i=1 i=1
X Y
Simple calculations readily yield that the degree of (11) is minimal and equal to
2k 1 if at least one of b =0, i=0,1,...,k 1, and 2k 2 if b 0, for
i k
− 6 − − ≡
c a
k k+1
ξ = .
k
c · a
k+1 k
The minimum of the degree of (12) is 2k 4 in the symmetric case and is attained
−
for
µk = cck · ak+1(a2ka+1(a+2a2k−aa22k−1)−a2k−2),
k+1 k k+1− k 2
−
νk = cckckc−1 · aak+a2ak+(a12(a2k−aa22k−1)).
k+2 k+1 k k−1 k+1− k−2
This suggests to define the following two operators:
c a
(13) S (p )=p2 k k+1 p p ,
2 k k− ck+1 · ak k−1 k+1
8 I.KRASIKOV
and
(14) S4(pk)=p2k− ckc+k1 · ak+1(a2ka+k1(a+2k+a2k1−−aa2k2k−12)−a2k−2)pk−1pk+1+
−
ckc+kc2kc−k+11 · aakk+ak2−ak1+(a12k(+a2k1−−aa2k2k−−12))pk−2pk+2.
Notice that S (p )=T (p ) for monic polynomials.
2 k 2 k
SomeadditionalmotivationcomesfromthefactthattheinequalitiesS >0and
2
S >0areinvariantwithrespectto normalization. Namely,forp =d p wehave
4 k k k
S (p )=d2S (p ); S (p )=d2S (p ).
2 k k 2 k 4 k k 4 k
Thus for S and S it will be enough to consider the orthonormalcase only.
2 4
Although we will notuse it here,let us notice that for generalasymmetric poly-
nomials the operator S minimizing the degree of the output is defined by
4
a c (b +b b b )
k+1 k k+1 k k 1 k 2
µk = − − − − ,
a c (b b )
k k+1 k+1 k 2
− −
a a c c (b b )
k+2 k+1 k k 1 k k 1
νk = − − − .
a a c c (b b )
k k 1 k+2 k+1 k+1 k 2
− − −
3. Tura´n’s inequalities. Results
We start with with the usual Tur´an inequality T (p )>0.
2 k
Theorem 5. Suppose that
c c
(15) 4 ai+1 i ai ai ai 1 i−1 >(bi bi 1)2,
(cid:18) ci+1 − (cid:19)(cid:18) − − ci (cid:19) − −
and
a c
i+1 i+1
(16) > , i=1,...,k;
a c
i i
then
T (p )>0.
2 k
Proof. The claim follows from T (p )=1, and the identity
2 0
(17) 4c (a c a c )(a T (p ) a c c T (p ))=
k+1 k+2 k+1 k+1 k+2 k+2 2 k+1 k k k+2 2 k
− −
2(a c a c )p +(b b )c c p 2+
k+2 k+1 k+1 k+2 k+1 k+1 k k+1 k+2 k
{ − − }
c c 4(a c a c )(a c a c ) (b b )2c c p2.
k+1 k+2 k+2 k+1− k+1 k+2 k+1 k+1− k k − k+1− k k+1 k+2 k
(cid:3)
(cid:8) (cid:9)
Finding an appropriate sequence c in Theorem 5 may be far from obvious. On
k
the other hand, rather simple sufficient conditions can be obtained by a suitable
choice of c . For example, choosing c =1 and
k 1
k 1 1
(18) ck =2k−1 − aj+1 + aj − , k =2,3,...,
a a
j=1(cid:18) j j+1(cid:19)
Y
one finds
POLYNOMIALS 9
Corollary 1. Suppose that a is increasing. Let
i
(19) 2(a2 a2)>(b b )2,
2− 1 1− 0
and
(20) a−k2(a2k+1−a2k)(a2k−a2k−1)>(bk−bk−1)2, k =2,3,...,
then T (p )>0.
2 k
Forgrowinga thechoiceofc givenby(18)isclosetothebestpossible. Indeed,
k k
since both factors in the left hand side of (15) must be positive, this yields
a c a c
i i i 1 i
< , i=1,2,...,k; − < , i=2,...,k.
a c a c
i+1 i+1 i i 1
−
Hence,
c c a a a a
i i 1 i+1 i i i 1
4 ai+1 ai ai ai 1 − <4ai+1ai 1 − =
(cid:18) ci+1 − (cid:19)(cid:18) − − ci (cid:19) − (cid:18) ai − ai+1(cid:19)(cid:18)ai−1 − ai (cid:19)
4a−i 2(a2i+1−a2i)(a2i −a2i−1),
for i 2. Similarly, for i=1 the left hand side of (15) is less than 4(a2 a2).
≥ 2− 1
In fact, one can say even more. The following nice observation is not mine, but
I was unable to find out who made it. Basic facts on chainsequences can be found
in [4] and [9].
Theorem 6. The assumptions of Theorem 5 arefulfilled if andonly if thesequence
a is increasing and
i
a2(b b )2
i i− i−1 , i=1,...,k,
4(a2 a2)(a2 a2 )
i+1− i i − i−1
is a chain sequence.
Proof. The sequence a must be increasing as (15) and (16) imply
i
a c a c
i+1 i i+1 i+1
> , > .
a c a c
i i+1 i i
Defining
a ci ai
g = i ci+1 − ai+1 <1,
i ai+1 · 1 a2i
− a2i+1
turns (15) into
a2(b b )2
(1−gi−1)gi ≥ 4(a2 i ia−2)(ai−21 a2 ) >0.
i+1− i i − i−1
Hence
a2(b b )2
i i− i−1
4(a2 a2)(a2 a2 )
i+1− i i − i−1
is a chain sequence.
In the opposite direction, if it is a chain sequence then any smaller one is also a
chain sequence. Therefore one can find g and then the corresponding c . (cid:3)
i i
Sincethesequence1/2, 1/4, 1/4, ...isachainsequenceTheorem6readilyyields
Corollary 1. However there are chain sequences with terms greater than 1/4, (see
e.g. [3],[10],[26],[27]). In particular, a positive constant sequence g n is a chain
{ }1
sequence iff 0<g 1 cos 2 π , [9, Th. 7.2.6].
≤ 4 − n+2
10 I.KRASIKOV
Example 1. Choosing in (1)
ak =q−2k q(1 qk), bk =q−2k−1(1+q qk+1), 0<q <1,
− −
q
one obtains so called Stieltjes-Wigert polynomials S (x;q). The sequence a is in-
k k
creasing and (19) is fulfilled by
2(a2 a2) (b b )2 =q 7(1 q2)(2 q 2q2+2q3)>0.
2− 1 − 1− 0 − − − −
For (20) we obtain
(1 qk)q4k+3
−(1 q)2 a−k2(a2k+1−a2k)(a2k−a2k−1)−(bk−bk−1)2 =
−
(cid:0) (cid:1)
(1 q2)2(1 qk)(1+q+q2 qk+1)+q3k+3 >0.
− − −
Hence in the normalization defined by (18) we have T (S (x;q))>0.
2 k
Example 2. The orthonormal Al-Salam-Carlitz polynomials V(α)(x;q) are defined
k
by
ak =q−k αq(1 qk), bk =(α+1)q−k, 0<q <1, α>0.
−
q
The sequence a is increasing and one finds
k
2(a22−a21)−(b1−b0)2 =q−3(1−q)(2α−(1+α2)(q−q2)).
This expression is positive for
1 1 (q q2)2 1+ 1 (q q2)2
(21) − − − <α< − − .
q q2 q q2
p − p −
For (20) we obtain
(1 qk)q2k+1
−(1 q)2 a−k2(a2k+1−a2k)(a2k−a2k−1)−(bk−bk−1)2 =
−
(cid:0) (cid:1)
(α q)(1 αq)(1 qk)+αq2k+1 >0,
− − −
providedq <α<q 1.Thelastinequalityiseverstrongerthan(21)andweconclude
−
that for the Al-Salam-Carlitz polynomials normalized by (18) and q < α < q 1,
−
T (V(α)(x;q))>0.
2 k
In the symmetric case and the monic normalization the following result was
obtained in [13]:
Theorem 7. Suppose that polynomials p satisfy
i
(22) p =xp a2 p , p =0, p =1.
i i−1− i−1 i−2 1 0
Suppose further that a are strictly increasing and
i
(23) a2 3a2+3a2 a2 0, 1 i k 1.
i−1− i i+1− i+2 ≥ ≤ ≤ −
Then for k 2,
≥
(24) T (p )>0.
4 k