# Statistical Approximation ofq-Bernstein-Schurer-Stancu-Kantorovich Operators

Authors: Qiu Lin
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We introduce two kinds of Kantorovich-typeq-Bernstein-Schurer-Stancu operators. We first estimate moments of q-Bernstein-Schurer-Stancu-Kantorovich operators. We also establish the statistical approximation properties of these operators. Furthermore, we study the rates of statistical convergence of these operators by means of modulus of continuity and the functions of Lipschitz class.


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## Body

## 1. Introduction

In 1987, Lupaş [1] introduced a q-type of the Bernstein operators and in 1997 another generalization of these operators based on q-integers was introduced by Phillips [2]. Thereafter, an intensive research has been done on the q-parametric operators. Recently the statistical approximation properties have also been investigated for q-analogue polynomials. For instance, in [3] q-Bleimann, Butzer, and Hahn operators; in [4] Kantorovich-type q-Bernstein operators; in [5] q-analogue of MKZ operators; in [6] Kantorovich-type q-Szász-Mirakjan operators; in [7] Kantorovich-type discrete q-Beta operators; in [8] Kantorovich-type q-Bernstein-Stancu operators were introduced and their statistical approximation properties were studied.The main aim of this paper is to introduce two kinds of Kantorovich-typeq-Bernstein-Stancu operators and study the statistical approximation properties of these operators with the help of the Korovkin-type approximation theorem. We also estimate the rate of statistical convergence by means of modulus of continuity and with the help of the elements of the Lipschits classes.Before proceeding further, let us give some basic definitions and notations. Throughout the present paper, we consider0<q<1. For any n=0,1,2,…, the q-integer [n]q is defined as (see [2])
(1)nq=1+q+⋯+qn-1,n=0,1,2,…,0q=0
and the q-factorial [n]q! as
(2)nq!=1q2q⋯nq,n=1,2,…,0q!=1.For the integersn, k, n≥k≥0, the q-binomial or the Gaussian coefficient is defined as
(3)nkq=nq!kq!n-kq!.For an arbitrary functionf(x), the q-differential is given by
(4)dqfx=fqx-fx.Theq-Jackson integral in the interval [0,b] is defined as (see [9])
(5)∫0b‍ftdqt=1-q∑j=0∞‍fqjbqj,0<q<1,
provided that sums converge absolutely.Suppose0<a<b. The q-Jackson integral in a general generic interval [a,b] is defined as
(6)∫ab‍ftdqt=∫0b‍ftdqt-∫0a‍ftdqt,0<q<1.


## 2. Construction of the Operators

For anyn∈N, p a fixed nonnegative number and α, β a real parameters satisfying the conditions 0≤α≤β, we introduce the Kantorovich-type q-Bernstein-Schurer-Stancu operators Kn,q(α,β)(f;x):C[0,p+1]→C[0,1] as follows:
(7)Kn,qα,βf;x=∑k=0n+p‍p¯n,kq;x×∫01‍ftn+1+βq+qk+αqn+1+βqdqt,kkkkkkkkkkkkkkkkkkkkkkkkkkkkx∈0,1,
where p¯n,kq;x=n+pkqxk∏s=0n+p-k-1(1-qsx).The moments of these operatorsKn,q(α,β)(f;x) are obtained as follows.Lemma 1.
ForKn,q(α,β)(ti;x), i=0,1,2, one has
(8)Kn,qα,β1;x=1,(9)Kn,qα,βt;x=n+pqn+1+βqqα+1x+1n+1+βq12q+qαq,(10)Kn,qα,βt2;x=n+pqn+p-1qn+1+βq2q2α+3x2+n+pqn+1+βq2×22qqα+1+q2+α2αq+qαx+1n+1+βq213q+2qαq2q+q2αq2.Proof.
It is obvious that(11)∫01‍1dqt=1,∫01‍tdqt=12q,∫01‍t2dqt=13q.
Fori=0, since ∑k=0n+p‍p¯n,k(q;x)=1, (8) holds.
Fori=1,
(12)Kn,qα,βt;x=∑k=0n+p‍p¯n,kq;x×∫01‍tn+1+βq+qk+αqn+1+βqdqt=∑k=0n+p‍p¯n,kq;x∫01‍tn+1+βqdqt+∑k=0n+p‍p¯n,kq;x∫01‍qk+αqn+1+βqdqt=12qn+1+βq+∑k=0n+p‍p¯n,kq;xqk+αqn+1+βq.
Using the properties of the generalized q-Schurer-Stancu operators ([10, Lemma 2])
(13)∑k=0n+p‍p¯n,kq;xk+αqn+βq=qαn+pqn+βqx+αqn+βq,
we have
(14)Kn,qα,βt;x=n+pqn+1+βqqα+1x+1n+1+βq12q+qαq.
Fori=2,
(15)∫01‍tn+1+βq+qk+αqn+1+βq2dqt=1n+1+βq2×∫01‍t2dqt+2qk+αq∫01‍tdqt+q2k+αq2∫01‍1dqt=1n+1+βq213q+2qk+αq2q+q2k+αq2,
we obtain
(16)Kn,qα,βt2;x=1n+1+βq2×∑k=0n+p‍p¯n,kq;x13q+2qk+αq2q+q2k+αq2=13qn+1+βq2+n+βqn+1+βq22q2q×∑k=0n+p‍p¯n,kq;xk+αqn+βq+q2n+βq2n+1+βq2∑k=0n+p‍p¯n,kq;xk+αq2n+βq2.
In view of (13) and [10, Lemma 2]
(17)∑k=0n+p‍p¯n,kq;xk+αq2n+βq2=n+pqn+p-1qn+βq2q2α+1x2+n+pqqαn+βq22αq+qαx+αq2n+βq2,
by simple calculation we can get the stated result (10).Lemma 2.
Forδn(q;x):=Kn,q(α,β)((t-x)2;x), one has
(18)δnq;x≤n+pqn+1+βqqα+1-12+21+αn+pqn+1+βq2+1+α2n+1+βq2.Proof.
In view of Lemma1, we have
(19)δnq;x=Kn,qα,βt2;x-2xKn,qα,βt;x+x2=n+pqn+p-1qq2α+3n+1+βq2-2qα+1n+pqn+1+βq+1x2+n+pqn+1+βq222qqα+1+q2+α2αq+qαkkkkkkkkn+pqn+1+βq222qqα+1+q2+α2αq+qα-2n+1+βq12q+qαqx+1n+1+βq213q+2qαq2q+q2αq2≤n+pqn+1+βqqα+1-12+21+αn+pqn+1+βq2+1+α2n+1+βq2.Now, we consider a sequenceq=qn satisfying the following two expressions:
(20)lim⁡n→∞qn=1,lim⁡n→∞1nqn=0.
By Korovkin’s theorem, we can state the following theorem.Theorem 3.
LetKn,qn(α,β)(f;x) be a sequence satisfying (20) for 0<qn<1. Then for any function f∈C[0,p+1], the following equality
(21)lim⁡n→∞Kn,qnα,βf;·-fC[0,1]=0
is satisfied.Proof.
We know thatKn,q(α,β)(f;x) is linear positive. By Lemma 1, if we choose the sequence q=qn satisfying (20) and using the equality
(22)n+pqn=nqn+qnnpqn,n+1+βqn=nqn+qnnβ+1qn
we have
(23)lim⁡n→∞Kn,qnα,βti;x-xiC[0,1]=0,i=0,1,2.
Because of the linearity and positivity of Kn,qn(α,β)(f;x), the proof is complete by the classical Korovkin theorem.We now redefineKn,q(α,β)(f;x) as
(24)K~n,qα,βf;x=∑k=0n+p‍p¯n,kq;x×∫01‍ftn+1q+β+qkq+αn+1q+βdqt.
Let us give some lemmas as follows.Lemma 4.
ForK~n,q(α,β)(ti;x), i=0,1,2, one has
(25)K~n,qα,β1;x=1,(26)K~n,qα,βt;x=n+pqn+1q+βqx+1n+1q+β12q+qα,(27)K~n,qα,βt2;x=n+pqn+p-1qn+1q+β2q3x2+n+pqn+1q+β22q2q+q22α+1x+1n+1q+β213q+2qα2q+q2α2.Proof.
It is obvious that (25) holds.
Fori=1,
(28)K~n,qα,βt;x=∑k=0n+p‍p¯n,kq;x×∫01‍tn+1q+β+qkq+αn+1q+βdqt=∑k=0n+p‍p¯n,kq;x∫01‍tn+1q+βdqt+∑k=0n+p‍p¯n,kq;x∫01‍qkq+αn+1+βqdqt=12qn+1q+β+∑k=0n+p‍p¯n,kq;xqkq+αn+1q+β.
Taking into account [11, Lemma 1]
(29)∑k=0n+p‍p¯n,kq;xkq+αnq+β=n+pqnq+βx+αnq+β,
we have
(30)K~n,qα,βt;x=n+pqn+1q+βqx+1n+1q+β12q+qα.
Fori=2,
(31)∫01‍tn+1q+β+qkq+αn+1q+β2dqt=1n+1q+β2×∫01‍t2dqt+2qkq+α×∫01‍tdqt+q2kq+α2∫01‍1dqt=1n+1q+β2×13q+2qkq+α2q+q2kq+α2,
we obtain
(32)K~n,qα,βt2;x=1n+1q+β2×∑k=0n+p‍p¯n,kq;x13q+2qkq+α2q+q2kq+α2=13qn+1q+β2+nq+βn+1q+β22q2q×∑k=0n+p‍p¯n,kq;xkq+αnq+β+q2nq+β2n+1q+β2∑k=0n+p‍p¯n,kq;xkq+αnq+β2.
From (29) and [11, Lemma 1]
(33)∑k=0n+p‍p¯n,kq;xkq+αnq+β2=n+pqn+p-1qnq+β2qx2+n+pqnq+β22α+1x+α2nq+β2,
by simple calculation we arrive at the desired result (27).Lemma 5.
Forδ~n(q;x):=Kn,q(α,β)((t-x)2;x), one has
(34)δ~nq;x≤n+pqn+1q+βq-12+21+αn+pqn+1q+β2+1+α2n+1q+β2.Proof.
From Lemma4, it is immediately seen that
(35)δ~nq;x=K~n,qα,βt2;x-2xK~n,qα,βt;x+x2=n+pqn+p-1qq3n+1q+β2-2qn+pqn+1q+β+1x2+n+pqn+1q+β222qq+q22α+1kkkkkkkkn+pqn+1q+β222qq+q22α+1-2n+1q+β12q+qαx+1n+1q+β213q+2qα2q+q2α2≤n+pqn+1q+βq-12+21+αn+pqn+1q+β2+1+α2n+1q+β2.We can give the following result, a theorem of Korovkin type.Theorem 6.
Letq=qn be a sequence satisfying (20) for 0<qn<1. Then for any function f∈C[0,p+1], the sequence K~n,qn(α,β)(f;x) converges to f(x) uniformly on [0,1].The proof of the above theorem follows along Theorem3; thus we omit the details.


## 3. Statistical Approximation of Korovkin Type

Further on, let us recall the concept of statistical convergence which was introduced by Fast [12].Let us setK∈N and Kn={k≤n:k∈K}; the natural density of K is defined by δK∶=lim⁡n→∞1/nKn if the limit exists (see [13]), where Kn denotes the cardinality of the set Kn.A sequencex=xk is called statistically convergent to a number L if, for every ɛ>0, δ{k∈N:xk-L≥ɛ}=0. This convergence is denoted as st-lim⁡kxk=L. It is known that any convergent sequence is statistically convergent, but its converse is not true. Details can be found in [14].In approximation theory by linear positive operators, the concept of statistical convergence was used by Gadjiev and Orhan [15]. They proved the following Bohman-Korovkin-type approximation theorem for statistical convergence.Theorem 7 (see [15]).
If the sequence of linear positive operatorsAn:C[a,b]→C[a,b] satisfies the conditions
(36)st-lim⁡nAnei;·-eiC[a,b]=0
for ei(t)=ti, i=0,1,2, then, for any f∈C[a,b],
(37)st-lim⁡nAnf;·-fC[a,b]=0.In this section, we establish the following Korovkin-type statistical approximation theorems.Theorem 8.
Letq=qn, 0<qn<1, be a sequence satisfying the following conditions:
(38)st-lim⁡nqn=1,st-lim⁡nqnn=aa<1,st-lim⁡n1nqn=0;
then for f∈C[0,p+1], one has
(39)st-lim⁡nKn,qnα,βf;·-fC[0,1]=0.Proof.
From Theorem7, it is enough to prove that st-lim⁡nKn,qn(α,β)(ei;·)-eiC[0,1]=0 for ei=ti, i=0,1,2.
By (8), we can easily get
(40)st-lim⁡nKn,qnα,βe0;·-e0C[0,1]=0.
From equality (9) we have
(41)Kn,qnα,βe1;·-e1C[0,1]≤qnα+1n+pqnn+1+βqn-1+1n+1+βqn12qn+qnαqn≤qnα+1n+pqnn+1+βqn-1+1+αnqn.
Now for a givenɛ>0, let us define the following sets:
(42)U=k:Kn,qkα,βe1;·-e1C0,1≥ɛ,U1=k:qkα+1k+pqkk+1+βqk-1≥ɛ2,U2=k:1+αkqk≥ɛ2.
From (41), one can see that U⊆U1∪U2, so we have
(43)δk≤n:Kn,qkα,βe1;·-e1C[0,1]≥ɛ≤δk≤n:qkα+1k+pqkk+1+βqk-1≥ɛ2+δk≤n:1+αkqk≥ɛ2.
By (22) and (38) it is clear that
(44)st-lim⁡nqnα+1n+pqnn+1+βqn-1=0,st-lim⁡n1+αnqn=0.
So we have
(45)st-lim⁡nKn,qnα,βe1;·-e1C[0,1]=0.
Finally, in view of (10), one can write
(46)Kn,qnα,βe2;·-e2C[0,1]≤n+pqnn+p-1qnn+1+βqn2qn2α+3-1+n+pqnn+1+βqn222qnqnα+1+qn2+α2αqn+qnα+1n+1+βqn213qn+2qnαqn2qn+qn2αqn2.
Using (22),
(47)22qnqnα+1+qn2+α2αqn+qnα≤2+2α,13qn+2qnαq2q+q2αq2≤1+α2,qnn+p-1qn=n+pqn-1,
we can write
(48)Kn,qnα,βe2;·-e2C[0,1]≤n+pqnqn2α+2n+1+βqn2+qn2α+2n+pqn2n+1+βqn2-1+2+2αn+pqnn+1+βqn2+1+α2n+1+βqn2≤qn2α+2n+pqn2n+1+βqn2-1+3+2αnqn1+pqnnqn+1+α2nqn2=∶θn+γn+ηn.
Then, from (22) and (38), we have
(49)st-lim⁡nθn=st-lim⁡nγn=st-lim⁡nηn=0.
Here for a given ɛ>0, let us define the following sets:
(50)T=k:Kn,qkα,βe2;·-e2C0,1≥ɛ,T1=k:θk≥ɛ3,T2=k:γk≥ɛ3,T3=k:ηk≥ɛ3.
It is clear that T⊆T1∪T2∪T3. So we get
(51)δk≤n:Kn,qkα,βe2;·-e2C0,1≥ɛ≤δk≤n:θk≥ɛ3+δk≤n:γk≥ɛ3+δk≤n:ηk≥ɛ3.
By (49), we have
(52)δk≤n:Kn,qkα,βe2;·-e2C0,1≥ɛ=0,
which implies that
(53)st-lim⁡nKn,qnα,βe2;·-e2C[0,1]=0.
In view of (40), (45), and (53), the proof is complete.Theorem 9.
Letq=qn, 0<qn<1, be a sequence satisfying (38); then for f∈C[0,p+1], one has st-lim⁡nK~n,qn(α,β)(f;·)-fC[0,1]=0.Proof.
From Theorem7, it is enough to prove that st-lim⁡nK~n,qn(α,β)(ei;·)-eiC[0,1]=0 for ei=ti, i=0,1,2.
Using (25), we can easily get
(54)st-lim⁡nK~n,qnα,βe0;·-e0C[0,1]=0.
From equality (26) we have
(55)K~n,qnα,βe1;·-e1C[0,1]≤qnn+pqnn+1qn+β-1+1+αnqn.
Now for a givenɛ>0, let us define the following sets:
(56)U~=k:K~n,qkα,βe1;·-e1C0,1≥ɛ,U~1=k:qkk+pqkk+1qk+β-1≥ɛ2,U~2=k:1+αkqk≥ɛ2.
From (55), one can see that U~⊆U~1∪U~2, so we have
(57)δk≤n:K~n,qkα,βe1;·-e1C[0,1]≥ɛ≤δk≤n:qkk+pqkk+1qk+β-1≥ɛ2+δk≤n:1+αkqk≥ɛ2.
By (38) it is clear that
(58)st-lim⁡nqnn+pqnn+1qn+β-1=0,st-lim⁡n1+αnqn=0.
So we have
(59)st-lim⁡nK~n,qnα,βe1;·-e1C[0,1]=0.
Finally, in view of (27), one can write
(60)K~n,qnα,βe2;·-e2C[0,1]≤n+pqnn+p-1qnn+1qn+β2qn3-1+n+pqnn+1qn+β222qn+qn2α+1+1n+1qn+β213qn+2qnα2qn+qn2α2.
Using qn[n+p-1]qn=[n+p]qn-1, then we can write
(61)n+pqnn+p-1qnn+1qn+β2qn3-1=n+pqnn+1qn+βqn2-1+n+pqnqn2n+1qn+β2.
So,
(62)K~n,qnα,βe2;·-e2C[0,1]≤n+pqnn+1qn+βqn2-1+3+2αn+pqnn+1qn+β2+1+α2nqn2=∶θ~n+γ~n+η~n.
Then, from (38), we have
(63)st-lim⁡nθ~n=st-lim⁡nγ~n=st-lim⁡nη~n=0.
Here for a given ɛ>0, let us define the following sets:
(64)T~=k:K~n,qkα,βe2;·-e2C0,1≥ɛ,T~1=k:θ~k≥ɛ3,T~2=k:γ~k≥ɛ3,T3=k:η~k≥ɛ3.
It is clear that T~⊆T~1∪T~2∪T~3. So we get
(65)δk≤n:K~n,qkα,βe2;·-e2C0,1≥ɛ≤δk≤n:θ~k≥ɛ3+δk≤n:γ~k≥ɛ3+δk≤n:η~k≥ɛ3.
By (63), we have
(66)δk≤n:K~n,qkα,βe2;·-e2C0,1≥ɛ=0,
which implies that
(67)st-lim⁡nK~n,qnα,βe2;·-e2C[0,1]=0.
In view of (54), (59), and (67), the proof is complete.


## 4. Rates of Statistical Convergence

Letf∈C[0,p+1] for any t∈[0,p+1] and x∈[0,1]. Then we have ft-fx≤ω(f,t-x), so for any δ>0, we get
(68)ωf,t-x≤ωf,δ,t-x<δ,ωf,t-x2δ,t-x≥δ.Owing toω(f,λδ)≤(1+λ)ω(f,δ) for λ>0, it is obvious that we have
(69)ft-fx≤1+δ-2t-x2ωf,δ
for any t∈[0,p+1], x∈[0,1] and δ>0.Next we will give the rates of convergence of bothKn,q(α,β)(f;x) and K~n,q(α,β)(f;x) in terms of the modulus of continuity.Theorem 10.
Letq=qn, 0<qn<1 be a sequence satisfying (38); then for any function f∈C[0,p+1], x∈[0,1], one has
(70)Kn,qnα,βf;x-fx≤2ωf,δn,
where
(71)δn=n+pqnn+1+βqnqnα+1-12+21+αn+pqnn+1+βqn2+1+α2n+1+βqn2.Proof.
Using the linearity and positivity of the operatorKn,q(α,β)(f;x) and inequality (69), for any f∈C[0,p+1] and x∈[0,1], we get
(72)Kn,qα,βf,x-fx≤Kn,qα,βft-fx;x≤1+δ-2Kn,qα,βt-x2;xωf,δ.
In view of Lemma 2, take q=qn, 0<qn<1 as a sequence satisfying (38) and choose δ=δn in (72); the desired result follows immediately.Theorem 11.
Letq=qn, 0<qn<1 be a sequence satisfying (38); then for any function f∈C[0,p+1], x∈[0,1], one has
(73)K~n,qnα,βf;x-fx≤2ωf,δn*,
where
(74)δn*=n+pqnn+1qn+βqn-12+2+2αn+pqnn+1qn+β2+1+α2n+1qn+β2.Proof.
Using the linearity and positivity of the operatorK~n,q(α,β)(f;x) and inequality (69), for any f∈C[0,p+1] and x∈[0,1], we get
(75)K~n,qα,βf,x-fx≤K~n,qα,βft-fx;x≤1+δ-2K~n,qα,βt-x2;xωf,δ.
In view of Lemma 5, take q=qn, 0<qn<1 as a sequence satisfying (38) and choose δ=δn* in (75); the desired result follows immediately.Finally, we give the rates of statistical convergence of bothKn,q(α,β)(f;x) and K~n,q(α,β)(f;x) with the help of functions of the Lipschitz class. We recall a function f∈Lip⁡M⁡(λ) on [0,p+1], if the inequality
(76)ft-fx≤Mt-xλ,t,x∈0,p+1
holds.Theorem 12.
Letf∈Lip⁡M⁡(λ) on [0,p+1], 0<λ≤1. Let q=qn,0<qn<1 be a sequence satisfying the condition given in (38). If we take δn as in (71), then one has
(77)Kn,qnα,βf;x-fx≤Mδnλ/2,x∈0,1.Proof.
Letf∈Lip⁡M⁡(λ) on [0,p+1], 0<λ≤1. Since Kn,qn(α,β)(f;x) is linear and positive, by using (76), we have
(78)Kn,qnα,βf;x-fx≤Kn,qnα,βft-fx;x≤Kn,qnα,βt-xλ;x.
If we take p′=2/λ, q′=2/2-λ and apply the Hölder inequality and Lemma 2, then we obtain
(79)Kn,qnα,βf;x-fx≤MKn,qnα,βt-x2;xλ/2≤Mδnλ/2.Theorem 13.
Letf∈Lip⁡M⁡(λ) on [0,p+1], 0<λ≤1. Let q=qn, 0<qn<1 be a sequence satisfying the condition given in (38). If we take δn* as in (74), then one has
(80)K~n,qnα,βf;x-fx≤Mδn*λ/2,x∈0,1.Proof.
Letf∈Lip⁡M⁡(λ) on [0,p+1], 0<λ≤1. Since K~n,qn(α,β)(f;x) is linear and positive, by using (76), we have
(81)K~n,qnα,βf;x-fx≤K~n,qnα,βft-fx;x≤K~n,qnα,βt-xλ;x.
If we take p′=2/λ, q′=2/2-λ and apply the Hölder inequality and Lemma 5, then we obtain
(82)K~n,qnα,βf;x-fx≤MK~n,qnα,βt-x2;xλ/2≤Mδn*λ/2.



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