Modified Bernstein Polynomials and Jacobi Polynomials in q-Calculus

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Modified Bernstein Polynomials and Jacobi Polynomials in q-Calculus

Marie-Madeleine Derriennic

To cite this version:

Marie-Madeleine Derriennic. Modified Bernstein Polynomials and Jacobi Polynomials in q-Calculus.

Rendiconti del Circolo Matematico di Palermo, 2005, pp.269-290. �hal-00003016v2�

ccsd-00003016, version 2 - 22 Oct 2004

Modified Bernstein Polynomials and Jacobi Polynomials in q -Calculus

Marie-Madeleine Derriennic

IRMAR, INSA, 20 Avenue des Buttes de Co¨esmes, CS 14315,35043-RENNES, FRANCE E-mail adress: mderrien@insa-rennes.fr

Abstract

We introduce here a generalization of the modified Bernstein polynomials for Jacobi weights using the q-Bernstein basis proposed by G.M. Phillips to generalize classical Bernstein Polynomials. The function is evaluated at points which are in geometric progression in ]0,1[. Numerous properties of the mod- ified Bernstein Polynomials are extended to their q-analogues: simultaneous approximation, pointwise convergence even for unbounded functions, shape- preserving property, Voronovskaya theorem, self-adjointness. Some properties of the eigenvectors, which areq-extensions of Jacobi polynomials, are given.

Keywords: q-Bernstein, q-Jacobi, Bernstein-Durrmeyer, totally positive, simul- taneous approximation.

AMS subject classification: 41A10, 41A25, 41A36

1 Introduction

G.M.Phillips has proposed a generalization of Bernstein polynomials based on the q-integers (cf. [9]). We introduce here a q-analogue of the operators which are often called Bernstein Durrmeyer polynomials and denoted M_n,1^α,β (cf. [3],[2]).

In all the paper, we shall assume that q ∈]0,1[ and, α, β >−1 (part 5 excepted).

For any integer n, and a function f defined on ]0,1[ we set

M_n,q^α,βf(x) = Xn

k=0

f_n,k,q^α,β bn,k,q(x) (1)

where each f_n,k,q^α,β is a mean of f defined by Jackson integrals. The polynomials bn,k,q(x) are q-analogues of the Bernstein basis polynomials and are defined by bn,k,q(x) = hn

k i

qx^k(1−x)^n−k_q , with hn k i

q = [n]_q!

[k]_q! [n−k]_q!, (q-binomial coefficient), k = 0, . . . , n. They verify Pn

k=0bn,k,q(x) = 1 (cf. [9]).

We follow the definitions and notations of ([7]).

For any real a, [a]_q = (1−q^a)/(1−q), Γq(a+ 1) = (1−q)^a_q/(1−q)^a, (if a∈N the q-integer [a]_q is [a]_q = 1 +q+· · ·+q^a−1 and Γq(a+ 1) = ([a]_q)!);

(1 − x)^a_q = Q∞

j=0(1 − q^jx).Q∞

j=0(1−q^j+ax) and consequently (1 − x)^a+b_q = (1−x)^a_q(1−q^ax)^b_q holds for any b, (1−x)^m_q =Qm−1

j=0 (1−q^jx) if m is integer.

The notations will be simplified as much as possible, the superscript α, β and the index q whenq is fixed, will be suppressed in some proofs.

We introduce the positive bilinear form:

hf, gi^α,βq =q^(α+1)(β+1)(1−q) X∞

i=0

qⁱq^iα(1−qⁱ⁺¹)^β_qf(q^i+β+1)g(q^i+β+1), (2)

whenever it is defined. It can be written under the form of two definite q-integrals

hf, gi^α,βq =

Z ^qβ+1

0

t^α(1−q^−βt)^β_qf(t)g(t)dqt and hf, gi^α,βq = q^(α+1)(β+1)

Z 1

0

t^α(1−qt)^β_qf(q^β+1t)g(q^β+1t)dqt.

(the definiteq-integral of a functionf isRa

0 f(x)dqx=a(1−q)P∞

i=0qⁱf(qⁱa) (cf. [7])) Definition 1 We set in formula (1) :

f_n,k,q^α,β = hb_n,k,q, fi^α,βq

hbn,k,q,1i^α,βq = R1

0 t^k+α (1−qt)^n−k+β_q f(q^β+1t) d_qt R1

0 t^k+α (1−qt)^n−k+βq dqt , k = 0, ..., n, (3)

to define M_n,q^α,βf(x) = Xn

k=0

hbn,k,q, fi^α,βq

hb_n,k,q,1i^α,βq bn,k,q(x). (4)

We see that the polynomial M_n,q^α,βf is well defined if there exists γ ≥0 such that x^γf(x) is bounded on ]0, A] for someA∈]0,1] andα > γ−1.Indeed, x^αf(x) is then q-integrable for the weight w^α,β_q (x) = x^α(1−qx)^β_q. We will say, in this case, that f satisfies the condition C(α). Also hf, gi^α,βq is well defined if the product f g satisfies C(α), particularly if f² and g² do it.

In many cases, the limit of M_n,q^α,βf(x) when q tends to 1 is : M_n,1^α,βf(x) =

Xn

k=0

Z 1 0

t^k+α (1−t)^n−k+β f(t) dt Z 1

0

t^k+α (1−t)^n−k+β dt

bn,k(x) with b_n,k(x) = ⁿ_k

x^k(1−x)^n−k.

Numerous properties of the operatorM_n,1^α,β will be extended toM_n,q^α,β in this paper.

2 First properties

For any n∈N, the operator M_n,q^α,β has the following properties.

- It is linear, positive and it preserves the constants so it is a contraction ( sup

x∈[0,1]

M_n,q^α,βf(x)

≤ sup

x∈]0,1[|f(x)|).

- It is self-adjoint: hM_n,q^α,βf, gi^α,βq =hf, M_n,q^α,βgi^α,βq .

- It preserves the degrees of the polynomials of degree ≤n.

The first properties are consequences of the definition. The last one follows after the following proposition since Dqx^p = [p]x^p−1.

Proposition 1 If f verifies the condition C(α), we have :

DqM_n,q^α,βf(x) = [n]_q

[n+α+β+ 2]_qq^α+β+2M_n−1,q^α+1,β+1

Dqf ·

q

(qx), x∈[0,1], (5)

where the q-derivative of a function f isDqf(x) = f(qx)−f(x)

(q−1)x if x6= 0.

(When f^′ is continuous on [0,1], the limit of formula (5) is, when q tends to 1, M_n,1^α,βf′

(x) =n(n+α+β+ 2)⁻¹M_n−1,1^α+1,β+1(f^′) (x)) (cf. [4])).

Proof. We compute Dbn,k(x) = [n] (bn−1,k−1(qx)/q^k−1 − bn−1,k(qx)/q^k) if 1≤k≤n−1 andDbn,0(x) =−[n]bn−1,0(qx), Dbn,n(x) = [n]bn−1,n−1(qx)/qⁿ⁻¹ to get DM_n^α,βf(x) = [n]

n−1P

k=0

bn−1,k(qx)(f_n,k+1^α,β −f_n,k^α,β)/q^k. We denote ψ^α,β_n,k(t) =t^k+α(1−qt)^n−k+β_q ,k = 0, . . . , n.

Recall that theq-derivative ofg1g2isDq(g1g2)(x) =Dqg1(x)g2(qx)+g1(x)Dqg2(x).

The q-Beta functions are Bq(u, v) = R1

0 t^u−1(1−qt)^v−1_q dqt = Γq(u)Γq(v)/Γq(u+v).

The function ψ^α+1,β+1_n−1,k (_q^t)f(q^β+1t), t ∈]0,1[, extended by 0 in 0 is continuous at 0.

Hence we may use a q-integration by parts to write, fork = 0, . . . , n−1 : B_q(k+α+ 2, n−k+β+ 1) [n+α+β+ 2] (f_n,k+1^α,β −f_n,k^α,β) =

−q^k+αR1

0(Dψ^α+1,β+1_n−1,k )(^t_q)f(q^β+1t)dqt=R1

0 q^k+α+β+2ψ^α+1,β+1_n−1,k (t)(Df)(q^β+1t)dqt

−h

ψ^α+1,β+1_n−1,k (_q^t)f(q^β+1t)i1 0.

Hence (f_n,k+1^α,β −f_n,k^α,β) = _[n+α+β+2]^qα+β+2+k

R1

0 t^k+α+1(1−qt)^n−k+β(Df)(q^β+1t)dqt Bq(k+α+2,n−k+β+1) and DM_n^α,βf(x) = [n]q^α+β+2

[n+α+β+ 2]

Xn−1

k=0

hbn−1,k, Df(_q^·)i^α+1,β+1q

Bq(k+α+ 2, n−k+β+ 1)bn−1,k(x).

Theorem 1 The following equality holds for any x∈[0,1] : M_n,q^α,βf(x) =

X∞

j=0

Φ^α,β_j,n,q(x)f(q^j+β+1) (6)

where, Φ^α,β_j,n,q(x) = u_j Xn

k=0

v_kb_n,k,q(q^j+β+1)b_n,k,q(x), uj = (1−q)q^j(α+1)(1−q^j+1)^β_q, j ∈N, v⁻¹_k = q^k(β+1)hn

k i

qBq(k+α+ 1, n−k+β+ 1), k = 0, . . . , n.

Moreover, for any r ∈ N, the sequence Φ^α,β_r,n,q,Φ^α,β_r−1,n,q, . . . ,Φ^α,β_0,n,q is totally positive, that is to say, the collocation matrix

Φ^α,β_r−j,n,q(xi)

i=1,...,m,j=0,...,r is totally positive for any family (xi), 0≤x1 < . . . < xm ≤1.

Proof. We set Φj = Φ^α,β_j,q , bn,k,q = bk, c = q^β+1. The formulae (6) come by writ- ing the definite q-integrals hbk, fi^α,βq as discrete sums in (4) and the Beta integrals hb_k,1i^α,βq =hn

k i

qB_q(k+α+ 1, n−k+β+ 1), k = 0, . . . , n.

For the total positivity of the Φj, we have to prove that, for any m ∈N and any two families (xi)i=1,...,m,(jk)k=1,...,m, with 0≤x1 ≤. . .≤xm ≤ 1, 0≤ jm ≤ . . .≤ j1, the determinant det(Φ_jl(x_i))i=1,...,m,l=1,...,m is non negative. From the multilinearity of

the determinants, there is a basic composition formula for the discrete sums (cf. [8]).

We have det(Φ_jl(x_i))i=1,...,m,l=1,...,m =Qm

l=1u_jlE where E = det(Pn

k=0vkbk(cq^jl)bk(xi))i=1,...,m,l=1,...,m

=Pn

k1=0. . .Pn

km=0vk1. . . vkmdet(bki(cq^jl)bki(xi))i=1,...,m,l=1,...,m

= Pn

k1=0. . .Pn

km=0vk1. . . vkmbk1(x1). . . bkm(xm) det(bki(cq^jl))i=1,...,m,l=1,...,m

=P

0≤k1≤...≤km≤nvk1. . . vkmdet(bkl(xi))i=1,...,m,l=1,...,mdet(bki(cq^jl))i=1,...,m,l=1,...,m. We know that theq-Bernstein basis is totally positive (cf. [6]). Hence we have det(bkl(xi))i=1,...,m,l=1,...,m ≥ 0 and also det(bki(cq^jl))i=1,...,m,l=1,...,m ≥ 0 , since cq^j1 <

cq^j2 < . . . < cq^jm. SoE is non negative and the result follows.

Corollary 1 The number of sign changes of the polynomial M_n,q^α,βf on ]0,1[ is not greater than the number of sign changes of the function f.

Proof. For any r ∈ N, the sequence Φ_r,Φ_r−1, . . . ,Φ₀ is totally positive. We deduce that the number of sign changes of the polynomial Pr

j=0Φj(x)f(q^j+β+1) is not greater than the number of sign changes of the sequence f(q^j+β+1), j = 0, . . . , r, hence not greater than the number of sign changes of the function f in ]0,1[ (cf. [5]).

When r tends to infinity this property is preserved hence is true forMnf.

Corollary 2 Let f be a function satisfying the conditionC(α).

1. Iff is increasing (respectively decreasing), then the functionM_n,q^α,βf is increasing (respectively decreasing).

2. If f is convex, then the function M_n,q^α,βf is convex.

Proof. 1) If f is monotone, for any s ∈ R the function f −s has at most one sign change. Hence M_n^α,β(f−s) = M_n^α,βf−shas at most one sign change andM_n^α,βf is monotone. If f is increasing, Df(_q^·) is positive on ]0, q[.Since the operators M_n^α,β are positive, we obtain for x ∈ [0,1], M_n−1^α+1,β+1

Df

· q

(qx) ≥ 0, and, using (5),

DM_n^α,βf(x)≥0. So the function M_n^α,βf is increasing.

2) Let suppose the function f is convex. Since M_n^α,β preserves the degree of the polynomials, for any real numbers γ₁, γ₂ there exist δ1, δ2 and a function g such that g(x) = f(x)−δ1x−δ2 and M_n^α,βf(x)− γ₁x−γ₂ = M_n^α,βg(x). The number of sign changes of g being at most two, it is the same for M_n^α,βg. Hence M_n^α,βf is convex or concave. Moreover, if a function ϕ is convex (respectively concave), D²ϕ(x) = _(q−1)^q32x²(ϕ(q²x)−[2]ϕ(qx) +qϕ(x)) is ≥0 (respectively ≤0). Hence M_n−2^α+2,β+2

D²ϕ

· q

(q²x)≥0.Using (5) two times we obtainD²M_n^α,βϕ(x)≥0 and

M_n^α,βϕ is not concave.

3 Convergence properties

Theorem 2 If f is continuous on [0,1], M_n,q^α,βf−f

∞≤Cα,β ω



f, 1 q[n]_q



,

where kfk∞ is the uniform norm of f on [0,1] and ω(f, .) is the usual modulus of continuity of f, the constant Cα,β being independent of n, q, f.

Proof. As M_n^α,β is positive, O. Shisha and B. Mond theorem can be applied. It is sufficient to prove that the order of approximation off by M_n^α,βf is O(_[n]¹ ) for the

functions fi(x) = xⁱ, i = 0,1,2. We compute the polynomials M_n^α,βfi , i = 1 and 2, with the help of (5) by q-integrations.

[n+α+β+ 2] (M_n^α,βf1(x)−x) =q^β+1[α+ 1]−x[α+β+ 2] and [n+α+β+ 2] [n+α+β+ 3] (M_n^α,βf2(x)−x²) =

[n] [2]x(q^α+2β+3[α+2] (1−x)−[β+1]x) + [a+β+3] [α+β+2]x²+q^2β+2[α+2] [α+1]. The result follows since 0< q <1 and 0≤[a]≤max(a,1) if a≥0.

Remark 1

In order to have uniform convergence for all continuous functions on [0,1], it is sufficient to have lim

n→∞M_n,q^α,βfi =fi for i= 1,2,hence lim

n→∞1/[n]_q = 0. This is realized if and only ifq=qnand lim

n→∞qn = 1. Indeed, for everyn ∈N, in both casesqⁿ<1/2 and qⁿ ≥1/2, we have 1−q <1/[n]_q≤2 max(1−q,ln 2/n). To maximize the order of approximation by the operator M_n,q^α,β_n, we are interested to have [n]_qn of the same order as n, that is to say to have ρn < [n]_qn ≤ n, for some ρ > 0, property which holds with the following property S for (qn).

Definition 2 The sequence (qn)n∈N, has the property S if and only if there exists N ∈N and c >0 such that, for any n > N, 1−qn< c/n.

Lemma 1

The property S holds if and only if the property P1 (respectively P2) holds where : P1 is ”There exists N1 ∈N and c1 >0 such that, for any n > N1, [n]_qn ≥c1n”, P2 is ”There exists N2 ∈N and c2 >0 such that, for any n > N2, q_nⁿ≥c2”.

Proof. For any n∈N,the functionξ(x) = (1−xⁿ)/(1−x) is increasing on [0,1[.

If S holds, we have, for any n > N1 =N, [n]_qn =ξ(qn)≥ξ(1−c/n)≥n(1−e^−c)/c

andP1 follows.IfP1 holds, we have, for anyn > N =N1,1/(1−qn)≥[n]_qn ≥c1nand S follows. If P₂ holds, we have, for any n > N =N₂, n(1−q_n)≤ −nlnq_n <−lnc₂ and S follows. If S holds, there exists N2 > N such that, if n > N2, 1−qn < 1/2 hence qⁿ_n > e^{−2n(1−qn)} > e^−2c and P2 follows.

Theorem 3 If the function f is continuous at the point x∈]0,1[, then,

n→∞lim M_n,q^α,β_nf(x) =f(x) (7)

in the two following cases :

1. If f is bounded on [0,1]and the sequence (qn) is such that lim

n→∞qn= 1,

2. If there exist real numbers α^′, β^′ ≥ 0 and a real κ^′ > 0 such that, for any x∈]0,1[,

x^α′(1−x)^β′f(x)

≤κ^′, α^′ < α+ 1, β^′ < β+ 1 and the sequence (qn) owns the property S.

Theorem 4 If the function f admits a second derivative at the point x∈]0,1[ then, in the cases 1 and 2 of theorem 3, we have the Voronovskaya-type limit :

n→∞lim [n]_qn(M_n,q^α,β_nf(x)−f(x)) = d

dx x^α+1(1−x)^β+1f^′(x) x^α(1−x)^β . (8) (The limit operator is the Jacobi differential operator for the weight x^α(1−x)^β) For the proofs of theorems 3 and 4 we need some preparation.

Proposition 2 We set, for any n, m∈N− {0} and x∈[0,1], q∈[1/2,1[,

T_n,m,q(x) = Xn

k=0

b_n,k,q(x) Z 1

0

t^k+α (1−qt)^n−k+β_q (x−t)^mdqt Z 1

0

t^k+α (1−qt)^n−k+βq dqt

. (9)

For any m, there exists a constant Km >0, independent of n and q, such that :

sup

x∈[0,1]|Tn,m,q(x)| ≤









Km/[n]^m/2_q if m is even, Km/[n]^(m+1)/2_q if m is odd.

To prove this proposition we consider the lemmas 2 and 3.

Lemma 2 We set, for any n, m∈N and x∈[0,1], T_n,m,q¹ (x) =

Pn k=0

bn,k,q(x) R1

0 t^k+α (1−qt)^n−k+β_q (x−t)^m_q dqt R1

0 t^k+α (1−qt)^n−k+βq dqt .

The following recursion formula holds for any q ∈[1/2,1[ and m≥2 : [n+m+α+β+ 2]_qq^{−α−2m−1}T_n,m+1,q¹ (x) =

(−x(1−x)DqT_n,m,q¹ (x) +T_n,m,q¹ (x)(p1,m(x) +x(1−q) [n+α+β]_q[m+ 1]_qq^1−α−m)

+T_n,m−1,q¹ (x)p2,m(x) +T_n,m−2,q¹ (x)p3,m(x)(1−q), (10) where the polynomials pi,m(x), i= 1,2 and 3 are uniformly bounded with regard to n and q.

Proof. 1) We set ψ_k(t) =t^k+α(1−qt)^n−k+β_q and lk(x) =bn,k(x).R1

0 ψ_k(t)dqt , k = 0, . . . , n and T_n,m,q¹ =T_m¹.

We compute x(1−x)DqT_m¹(x) =x(1−x) [m]Pn

k=0lk(x)R1

0 ψ_k(t)(x−t)^m−1_q dqt +Pn

k=0lk(x)R1

0 ψ_k(t)(qx−t)^m_q ([k]−[n]x)dqt =A+B.

We have A=x(1−x) [m]T_m−1¹ (x) and B =q^−αPn

k=0lk(x)R1

0 (Dqψ_k) (t)t(1−qt)(qx−t)^m_q dqt

−q^1−α−m[n+α+β]Pn

k=0l_k(x)R1

0 ψ_k(t)(qx−t)^m+1_q d_qt +(x([n+α+β]q^2−α−m−[n]) + [−α])Pn

k=0lk(x)R1

0 ψ_k(t)(qx−t)^m_q dqt

=B1+B2+B3.

We q-integrate by parts, setting σ(t) =

t

q(1−t)(qx−q^t)^m_q

. The q-integral in B1 becomes R1

0 Dqψ_k(t)t(1−qt)(qx−t)^m_q dqt= [ψ_k(t)σ(t)]¹₀−R1

0 ψ_k(t)(Dqσ)(t)dqt for eachk = 0, . . . , n,

We expand σ(t) =q^−2m(x− q^t)^m+2_q +q^−2m([3]−q^m+2)x−q^m)(x− ^tq)^m+1_q +q^−2m+1(x(q^m−1+ [m] (1−q)q^m)−x²(1 + [2]q(1−q) [m])(x− _q^t)^m_q

−q^2m+3x²[m] (1−q)(q^m−2−x)(x− ^t_q)^m−1_q .

We obtain B1 =−q^{−α−2m−1}([m+ 2]T_m+1¹ (x)−[m+ 1] ([3]x−q^m+2x−q^m)T_m¹(x)

−q^−α−2m[m]x(q^m−1+ (1−q) [m]q^m−x(1 +q(1−q) [2] [m])T_m−1¹ +q^−α+2−2m[m−1]x²[m] (q^m−2−x)(1−q)T_m−2¹ (x).

Moreover we have B2 =−q^1−α−m[n+α+β] (T_n,m+1¹ (x)−(1−q) [m+ 1]xT_n,m¹ (x)), B3 = (x(qⁿ[β−m+ 2]− [2−α−m]) + [−α])(T_n,m¹ (x)−(1−q) [m]xT_n,m−1¹ (x)).

Lemma 3 For any m ∈ N, q ∈ [1/2,1[, x ∈ [0,1], the expansion of (x−t)^m on the Newton basis at the points x/qⁱ, i= 0, . . . , m−1 is:

(x−t)^m = Xm

k=1

d_m,k(1−q)^m−k(x−t)^k_q, (11)

where the coefficients dm,k, verify |dm,k| ≤dm, k = 1, . . . , m, and dm does not depend on x, t,q.

Proof. For m = 1, it is obvious. If for some m ≥ 1, the relation (11) holds, we write x−t=q^−k((x−q^kt)−(1−q) [k]x) for k= 1, . . . , mand we obtain (x−t)^m+1

=

m+1X

k=1

d_m+1,k(1−q)^m+1−k(x−t)^k_q withd_m+1,k =q^−k(qd_m,k−1−[k]d_m,k) ifk = 1, . . . , m anddm+1,m+1 =q^−mdm,m.Since|dm,k| ≤dmwe have|dm+1,k| ≤dm+1 = 2^−m(m+1)dm, k = 1, . . . , m+ 1.

Proof of the proposition 2

At first we prove that, for any x, |T_m¹(x)| ≤ H_m/[n]^m/2 if m is even (respec- tively ≤Hm/[n]^(m+1)/2 if m is odd), where Hm does not depend on n, q, x. We have T_n,0¹ (x) = 1 and the formulae for M_n^α,βfi, i = 1,2 of the proof of theorem 2 give the result for m = 1 and 2. The product [n+α+β] (1−q) = 1−q^n+α+β is positive and bounded by max(1,

1−2^−(1+α+β)

). If the result is true for some p ≥ 2, p odd (respectively even) and any m ≤ p, the result for p+ 1 follows from the recursion formula (10) of lemma 2.

Then, we write, for anyn, m∈Nandx∈[0,1],using lemma 3, and 1−q <1/[n],

|Tn,m,q(x)| ≤dmPm

k=1(1−q)^m−k|T_k¹(x)| ≤dm(T_m¹(x) +Pm−1

k=1 [n]^−m+kHk[n]^−k/2

≤dmT_m¹(x) +Pm−1

k=1 Hk[n]^−(m+1)/2) and the result follows.

Now the following lemma is the key.

Lemma 4 Let (q_n) be a sequence owning the property S, x ∈]0,1[ and δ ∈]0,1[, δ < min(x,1−x). Let α, β, α^′, β^′ be real numbers such that α^′, β^′ ≥ 0, α > α^′ −1, β > β^′ −1. We set ϕ(t) = t^−α′(1−t)^−β′, t ∈]0,1[ and Ix,δ(t) = 1 if |t−x| > δ, I_x,δ(t) = 0 elsewhere.

The sequence E_n(x, δ) = Pⁿ

k=0

b_n,k,qn(x) Z 1

0

t^k+α(1−qnt)^n−k+β_qn ϕ(q_n^β+1t)Ix,δ(t)dqnt Z 1

0

t^k+α(1−qnt)^n−k+βqn dqnt

. verifies lim

n→∞nEn(x, δ) = 0 for any x and δ such that 0< δ < x < 1−δ.

Proof. Letα (respectively β) be the smallest integer such that α≥α (respectively β≥β) and τ (respectively τ^′) be a real number such that τ > α+β+ 2

α−α^′+ 1 (respectively τ^′ > α+β+ 2 β−β^′+ 1).

For any k = 0, . . . , n, we have R1

0 t^k+α(1−qt)^n−k+β_q dqt ≥ R1

0 t^k+α(1−qt)^n−k+β_q dqt

= ^[k+α]![^n−k+β]^! [^n+α+β+1]^! ≥_n

k

−1

(n+α+β+ 1)^−(α+β+1)). We set I_x,δ⁻ (t) = 1 if 0< t < x−δ and I_x,δ⁺ (t) = 1 if x+δ < t <1, I_x,δ⁻ (t) = I_x,δ⁺ (t) = 0 elsewhere.

We split the interval (0,1) introducingen= 1/n^τ, e^′_n= 1−1/n^τ′, n∈N,and we de- fine, using againψ^α,β_n,k,q(t) =t^k+α(1−qt)^n−k+β_q andln,k,q(x) =bn,k,q(x).R1

0 ψ^α,β_n,k,q(t)dqt of lemma 2, A¹_n =Pn

k=0ln,k,qn(x)Ren

0 t^k+α−α′(1−qnt)^n−k+β_qn dqnt, A²_n =Pn

k=0ln,k,qn(x)R1

ent^k+α−α′(1−qnt)^n−k+β_qn I_x,δ⁻ (t)dqnt, A³_n =Pn

k=0ln,k,qn(x)Re^′_n

0 t^k+α(1−qnt)^n−k+β_qn

(1−q_n^β+1t)^β′ I_x,δ⁺ (t)dqnt, A⁴_n =Pn

k=0ln,k,qn(x)R1

e^′_nt^k+α(1−qnt)^n−k+β_qn

(1−q_n^β+1t)^β′ I_x,δ⁺ (t)dqt.

If t > x, (respectively if t < x) then t^−α′ < x^−α′ (respectively (1 − q_n^β+1t)^β′

>(1−q_n^β+1x)^β′ ≥(1−x)^β′).

Hence, we have En(x, δ) ≤ (1/2)^{−(β+1)α′}((1 −x)^−β′(A¹_n +A²_n) +x^−α′(A³_n +A⁴_n)) if qn≥1/2, and it is sufficient to prove lim

n→∞nAⁱ_n= 0 for i= 1,2,3,4.

If qⁿ_n ≥ c and e_n < 1/2, we have for k = 0, . . . , n, Ren

0 t^k+α−α′(1−q_nt)^n−k+β_qn d_qnt

= q^−k(β+1)n _n

k

−1 qn

Ren

0 bn,k,q(q_n^β+1t)t^α−α′(1−qnt)^β_qndqnt ≤ _n

k

−1

qnγ₁e^α−α_n ^′+1, where γ₁ does not depend onk, n, x,sinceq^(β+1)kn ≥c^β+1,0≤bn,k,qn(q^β+1_n t)≤1,and, (1−qnt)^β_qn ≤1 if β ≥ 0 and t ∈ [0,1], (respectively (1− q_nt)^β_qn ≤ (1−e_n)⁻¹ ≤ 2 if β < 0 and t∈[0, en]). Hence, we have A¹_n≤γ₁(n+α+β+ 1)^α+β+1n^{−τ(α−α′+1)}.The choice of τ and the property S on (qn) give lim

n→∞nA¹_n= 0.

We choose m ∈Nsuch that m > τ α^′+ 1 and we write A²_n ≤n^{τ α′}δ^−2m

Pn k=0

ln,k,qn(x)R1

ent^k+α(1−qnt)^n−k+β_qn (x−t)^2mdqnt

≤n^{τ α′}δ^−2mTn,2m,qn(x)≤K2mδ^−2mn^{τ α′−m}, hence lim

n→∞nA²_n= 0 by the choice of m.

Now we have, if t < e^′_n, (1−q_n^β+1t)^β′ >(1−e^′_n)^β′ ≥n^{−τ′β′}, hence

A³_n ≤n^τ′β′ Pn k=0

ln,k,qn(x)Re^′n

0 t^k+α(1−qnt)^n−k+β_qn I_x,δ⁺ (t)dqnt.We choose m^′ ∈N such that m^′ > τ^′β^′+ 1 to haveA³_n ≤n^τ′β′δ^−2m′ Pⁿ

k=0

ln,k,qn(x)R1

0 t^k+α(1−qnt)^n−k+β_qn (x−t)^2m′dqnt

≤n^τ′β′δ^−2m′Tn,2m^′,qn ≤K2m^′δ^−2m′n^{τ′β′−m′},hence lim

n→∞nA³_n= 0 by the choice of m^′. To finish, we prove that (1−q_n^β−β′+1t)^β_qn^′ ≤(1−q_n^β+1t)^β′ for any t∈[0,1].

If 0 ≤β^′ <1, we use theq-binomial formula (cf. [1]) and the inequalities

[−β^′] ≤ −β^′ and [−β^′+k]/[k] ≥ (−β^′+k)/k for any integer k ≥ 1. In the other cases, if l is the integer such that l ≤ β^′ < l + 1, we use the rules of product of q-binomials to write

(1−q_n^β−β′+1t)^β_qn^′ = (1−q_n^β−β′+1t)^l_qn(1−q_n^{β−β′+1+l}t)^β_qn^′−l and the result follows.

Then, with the same rules, we write, for any k= 0, . . . , n and t∈[0,1], (1−q_nt)^n−k+β_qn = (1−q_nt)^β−β_qn ^′(1−q^β−β_n ^′+1t)^β_qn^′(1−q_n^β+1t)^n−k_qn and

(1−qnt)^n−k+β_qn

(1−q_n^β+1t)^β′ ≤(1−qnt)^β−β_qn ^′(1−q_n^β+1t)^n−k_qn . We deduce if q_nⁿ ≥c and e^′_n >1/2, A⁴_n ≤

Pn k=0

qn^−(β+1)k_n

k

−1

qnln,k,qn(x)R1

e^′nt^α(1−qnt)^β−β_qn ^′bn,k,qn(q_n^β+1t)dqnt

≤ γ₂(n+α+β + 1)^α+β+1(1−e^′_n)^β−β′+1 where γ₂ does not depend on k, n, x. The choice of e^′_n and τ^′ gives lim

n→∞nA⁴_n = 0.

Proof of theorem 3

Supposef is continuous atx∈]0,1[.Letε >0 be an arbitrary real number.There exists δ^′ >0 such that |f(x)−f(t)| < ε for any t ∈ [0,1] such that |x−t|< δ^′. Let δ=δ^′/2 andN^′ ∈Nsuch that (1−q^β+1_n )x < δforn > N^′.Then we have, if|x−t|< δ andn > N^′,the inequalities−δ <−q_n^β+1δ < x−q^β+1_n t =q_n^β+1(x−t)+(1−q^β+1_n )x <2δ and

f(x)−f(q_n^β+1t) < ε.

Hence, we have, if |f| is bounded by κ,

f(x)−f(q^β+1_n t)

< ε+ 2κIx,δ(t) and, in

the case 2,

f(x)−f(q_n^β+1t)

< ε+ (|f(x)|+κ^′(q^β+1_n t)^−α′(1−q_n^β+1t)^−β′Ix,δ(t).

We apply the operator M_n,q^α,β_n at the functionh_x(t) =f(t)−f(x).

We have

M_n,q^α,β_nf(x)−f(x) =

M_n,q^α,β_nhx(x)

≤ M_n,q^α,β_n|hx| (x)

≤









ε+ 2κTn,2,qn(x)/δ² in the case 1,

ε+|f(x)|T_n,2,qn(x)/δ²+κ^′E_n(x, δ, q_n) in the case 2.

The second term (respectively and the third term in the case 2) of the right hand side vanishes when [n]_qn tends to infinity. Since lim

n→∞1/[n]_qn = 0 in both cases (remark 1), the result follows.

Proof of theorem 4

We write Taylor formula at the point x,

f(t) =f(x) + (t−x)f^′(x) + (t−x)²f^′′(x)/2 + (t−x)²ε(t−x) where lim

u→0ε(u) = 0.

We apply the operator M_n,q^α,β_n at the functionf of the variable t to obtain M_n,q^α,β_nf(x)−f(x) =−f^′(x)Tn,1,qn(x) +f^′′(x)

2 Tn,2,qn(x) +Rn(x) where Rn(x) = M_n,q^α,β_nζ_x(x) with ζ_x(t) = (t − x)²ε(t − x). We use lim

q→1[a]_q = a for any a ∈ R and we verify, with the help of the formulae of the proof of theorem 2, that

[n]lim_qn→∞[n]_qnTn,1,qn(x) = (α+β+ 2)x−α−1 and lim

[n]_qn→∞[n]_qnTn,2,qn(x) = 2x(1−x).

So, to obtain the result, we have to prove that lim

[n]_qn→∞[n]_qnRn(x) = 0.We proceed in the same manner as in the proof of theorem 3. For any arbitrary η > 0,we can find δ >0 such that, for n great enough, ε(x−t)< η if

x−q_n^β+1t < δ.

We obtain the inequality |ζ_x(t)| ≤ η(x−t)²+ (ρ_x+|f(t)|)Ix,δ(q^−(β+1)t) for any t∈]0,1[,where ρ_x is independent of t and δ. We deduce

[n]_qn|R_n(x)| ≤









[n]_qn(ηTn,2,qn(x) + (ρ_x+κ)Tn,4,qn(x)/δ⁴) in the case 1,

[n]_qn(ηTn,2,qn(x) +ρ_xTn,4,qn(x)/δ⁴) +κ^′nEn(x, δ))in the case 2.

The right hand side tends to 2ηx(1−x) whenn(hence [n]_qn) tends to infinity and is as small as wanted.

Remark 2

1) We see that the best order of approximation in (8) is in 1/[n]_qn.If 1−qn= 1/n^γ with 0 < γ < 1, then lim

n→∞[n]_qn/n^γ = 1, hence [n]_qn can be replaced by n^γ in (8). If 1−q_n = 1/nlogn or 1/n^γ with γ > 1, then lim

n→∞[n]_qn/n = 1, [n]_qn can be replaced by n and we refound exactly the Voronovskaya-limit property ofM_n,1^α,β(x) (case 1).

2) In the case 2, the theorems 3 and 4 are valid for M_n,1^α,β, if wf is Lebesgue integrable on [0,1], and this result is new. (In the proof the Jackson integrals have to be replaced by Lebesgue integrals)

Theorem 5 If f^′ is continuous on [0,1] and q >1/2, then Dq(M_n,q^α,βf)−f^′

∞ ≤C_α,β^′



ω



f^′, 1 q[n]_q



+ω(f^′,1−q)



+[α+β+ 2]_q

[n]_q kf^′k∞, where C_α,β^′ is a constant independent ofn, q, f.

Hence, if lim

n→∞qn= 1, then lim

n→∞Dqn(M_n,q^α,β_nf) = f^′(x) uniformly on [0,1]. Proof. We write, using (5), for any x∈[0,1],

DM_n^α,βf(x)−f^′(x) = _[n+α+β+2]^[n]

M_n−1^α+1,β+1

Df

· q

(qx)−Df(x) +Df(x)−f^′(x)

+

[n]

[n+α+β+2] −1

f^′(x). Since 0< _[n+α+β+2]^[n] <1, we have

D(M_n^α,βf(x))−f^′(x)

≤

M_n−1^α+1,β+1

Df

· q

(qx)−Df(x)

+|Df(x)−f^′(x)|+^[α+β+2]_[n] |f^′(x)|. The theorem (2) for the function Df

· q

gives

M_n−1^α+1,β+1

Df

· q

(qx)−Df(_q^·)(qx)

≤ Cα+1,β+1ω

Df(_q^·),√¹

[n−1]

. Moreover

|Df(x)−f^′(x)| =|f^′(y)−f^′(x)|for someywithqx < y < xhence|y−x|<1−qand

|Df(x)−f^′(x)| ≤ ω(f^′,1−q). The modulus of continuity of Df

· q

is linked with

the modulus of continuity off^′. Indeed, for anyyi ∈[0,1] andi= 1,2, there existszi, such that, yi < zi < yi/qandDf(_q^·)(yi) =f^′(zi).As|z1−z2| ≤ |y1−y2|/q+(1−q)/q we get ω

Df(_q^·), t

= sup

|y1−y2|<t

Df(_q^·)(y1)−Df(_q^·)(y2)

≤ sup

|y1−y2|<t

(|f^′(z1)−f^′(z2)|

≤2(ω(f^′, t) +ω(f^′,1−q)) and the result follows.

Corollary 3 If f^′ is continuous on [0,1] and 1−qn =o(1/n⁴), then

n→∞lim M_n,q^α,β_nf′

(x) =f^′(x) uniformly on [0,1].

Proof. For anyx∈[0,1],there exists u∈(q_nx, x) such that Dqn(M_n,q^α,β_nf)(x) = (M_n,q^α,β_nf)^′(u) and (x−u)<1−qn.

Hence

Dqn(M_n,q^α,β_nf)(x)−(M_n,q^α,β_nf)^′(x)

≤ (1−qn)

(M_n,q^α,β_nf)^′′(v)

for some v and D_qn(M_n,q^α,β_nf)−(M_n,q^α,β_nf)^′

∞≤n⁴(1−q_n)M_n,q^α,β_nf

∞ ≤n⁴(1−q_n)kfk∞,via Markov inequality.

4 self-adjointness properties

In this part q ∈]0,1[ is independent of n.

On the space of polynomials h., .i^α,βq is an inner product. Let P_r,^α,β_q

r∈N be the sequence of the orthogonal polynomials forh., .i^α,βq such that degree of P_r,^α,β_q =r and P_r,^α,β_q (0) =

r+α r

q

= [α+r]_q. . .[α+ 1]_q

[r]_q! . We set νr =

hP_r,^α,β_q , P_r,^α,β_q i^α,βq

1/2

. We define U_q^α,β which is a q-analogue of the Jacobi differential operator by:

U_q^α,βf(x) = Dq

x^α+1(1−q^−β−1x)^β+1_q Dqf(x q)

x^α(1−q^−βx)^β_q . (12)

Proposition 3 The operator U_q^α,β is self-adjoint for h., .i^α,βq . It preserves the space of polynomials of degree r ∈ N. Consequently, for any r ∈ N, P_r,^α,β_q is eigenvector of U_q^α,β for the eigenvalue µ^α,β_r =−q^−β−r[r]_q[r+α+β+ 1]_q.

Proof.

U^α,β is a q-differential operator of order 2. We compute with q-binomial relations, U^α,βf(x) = −q^α−β[β+ 1]x+ [α+ 1] (1−q^−β−1x

)Df(x) + (1−q^−β−1x)^x_qD²f(^x_q), hence the operatorU preserves the degree of polynomials. If f and g are polynomials hUf, gi is well defined. We write, since the q-integration by parts is valid,hU^α,βf, gi

=h

x^α+1(1−q^−β−1x)^β+1_q Df(^x_q)g(x)iq^β+1

0 −

Z q^β+1

0

(qx)^α+1(1−q^−βx)^β+1_q Df(x)Dg(x)dqx.

and the first term vanishes. We compute U^α,βf(x) for f(x) = x^r to obtain U^α,βf(x) = q^1−r[r] ([α+r]x^r−1(1−x)−q^−β−1[β+ 1]x^r where the coefficient of x^r is the eigenvalue µ^α,β_r .

Proposition 4 The eigenvectors of the operators M_n,q^α,β, n ∈ N, are the polynomials P_r,^α,β_q , r∈N and, if f satisfies c(α), we have

M_n,q^α,βf = Pn r=0

λ^α,β_n,rhf, P_r,^α,β_q i^α,βq P_r,^α,β_q /ν²_r with the eigenvalues λ^α,β_n,r =q^r(r+α+β+1) [n]!

[n−r]!

Γq(n+α+β+ 2)

Γ_q(n+r+α+β+ 2) if r≤n, λ^α,β_n,r = 0 otherwise.

Proof. SinceMnis self adjoint and preserve the degree of polynomials, the orthog- onal polynomials Pr are eigenvectors. The eigenvalue λ^α,β_n,r is obtained by computing Mnf(x) forf(x) =x^r.

We use (5)r times to get D^rM_nf(x) = q^r(α+β+r+1)[r]! [n]. . .[n−r+ 1]

[n+α+β+ 2]. . .[n+r+α+β+ 1]. Corollary 4 1. For any n, m∈N, the operatorsM_n,q^α,β and M_m,q^α,β commute on the

space of functions satisfying C(α).