# Some Properties of Kantorovich-Stancu-Type Generalization of Szász Operators including Brenke-Type Polynomials via Power Series Summability Method

Authors: Naim Latif Braha; Toufik Mansour; Mohammad Mursaleen
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we study the Kantorovich-Stancu-type generalization of Szász-Mirakyan operators including Brenke-type polynomials and prove a Korovkin-type theorem via theT-statistical convergence and power series summability method. Moreover, we determine the rate of the convergence. Furthermore, we establish the Voronovskaya- and Grüss-Voronovskaya-type theorems for T-statistical convergence.


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## Body

## 1. Introduction and Preliminaries

LetK⊆ℕ (set of natural numbers) and Km=i≤m:i∈K. Then, the natural density or we can say asymptotic density of K is defined by σK=limm1/mKm whenever the limit exists, where Km denotes the cardinality of the set Km. A sequence η=ηi is statistically convergent to L if for every ε>0(1)limm1mi≤m:ηi−L≥ε=0,and we write st−limmηm=L.LetT=tnj be a matrix and η=ηj be a sequence. The T− transform of the sequence η=ηn is defined by Tη=Tnη,Tηn=∑jtnjηj if the series converges for every n∈ℕ. We say that η is T− summable to the number L if Tηn converges to L. The summability matrix T is regular whenever limjηj=L=limnTηn.LetT=tnj be a regular matrix. A sequence η=ηj is said to be T-statistically convergent (see [1]) to real number L if for any >0,limn∑j:ηj−L≥εtnj=0, and we write stT−limη=L. If T is Cesàro matrix of order 1, then T-statistical convergence is reduced to the statistical convergence.In this paper, we also use the power series summability method which includes several known summability methods such as Abel and Borel (see [2–9]). Note that the power method is more effective than the ordinary convergence (see [10]).Letpj be a sequence of real numbers such that p0>0, p1,p2,⋯≥0, and the corresponding power series pu=∑j=0∞pjuj has radius of convergence R with 0<R≤∞. If limu→R−1/pu∑j=0∞ηjpjuj=L for all t∈0,R, then we say that η=ηj is convergent in the sense of power series method (see [11, 12]). Define AuBηu=∑k≥0pkηuk, where Au=∑j≥0ajuj and Bu=∑j≥0bjuj are analytical functions such that a0≠0 and bj≠0 for all j≥0 (see [13]). Clearly, pkη=∑j=0kak−jbjηj. Moreover, the power series method is regular if and only if limu→R−pjuj/pu=0 holds for each j∈0,1,⋯ (see [14]).We study a Korovkin-type theorem for the Kantorovich-Stancu-type Szász-Mirakyan operators via power series method. We determine the rate of convergence for these operators. Furthermore, we give a Voronovskaya-type theorem forT− statistical convergence. Such type of operators is widely studied by several authors (see [15–19]).We start by recalling the class of Kantorovich-Stancu-type generalization of Szász-Mirakyan operators, including Brenke-type polynomials. For everyh∈CB0,∞=C0,∞∩E,
(2)Knα,βh,η=n+βA1Bnx∑k=0∞pknη∫k+α/n+βk+1+α/n+βhudu,for n∈ℕ,x∈0,∞ and 0≤α≤β, where E=h:x∈0,∞,limx→∞hx/1+x2<∞. In what follows, we calculate the moments and central moments for Kantorovich-Stancu of Szász-Mirakyan operators. Let us mention some properties of the functions Au and Bu (see [13, 20]).
(1)
A1≠0, ak−rbr/A1≥0, for all 0≤r≤k and k=0,1,2,⋯(2)
B:0,∞⟶0,∞(3)
SeriesAu=∑r=0∞arur, for a0≠0, and Bu=∑r=0∞brur, for br≠0r≥0, are convergent for u<RR>1 and Au,Bu are analytic functionsThe next lemma is followed immediately from the fact thatAuBηu=∑k≥0pkηuk.Lemma 1.
LetD be the operator ud/du. For all m≥0,
(3)∑k≥0kmpkηtk=DmAuBηu.
For example, Lemma1 for m=0,1,2,3,4 gives
(4)∑k≥0pkη=A1Bη,∑k≥0kpkη=uA1uBηu+ηuAtB1ηu,∑k≥0k2pkη=u2A2uBηu+A1uuBηu+2ηu2B1ηu+AuηuB1ηu+η2u2B2ηu,∑k≥0k3pkη=u3A3uBηu+A2u3u2Bηu+3ηu3B1ηu+A1uuBηu+6ηu2B1ηu+3η2u3B2ηu+AuηuB1ηu+3η2u2B2ηu+η3u3B3ηu,∑k≥0k4pkη=u4A4uBηu+A3u6u3Bηu+4ηu4B1ηu+A2u7u2Bηu+18ηu3B1ηt+6η2u4B2ηu+A1uuBηu+14ηu2B1ηu+18η2u3B2ηu+4η3u4B3ηu+AuηuB1ηu+7η2u2B2ηu+6η3u3B3ηu+η4u4B4ηu.Theorem 2.
Letei=eit=ti for all i≥0 and let D be the operator td/dt. Then,
(5)Knα,βei,x=n+β−ii+1A1Bnx∑j=0i+1i+1jα+1i+1−j−αi+1−jDjAtBnxtt=1.Proof.
By the definition of the operators, we have(6)Knα,βei,x=n+βA1Bnx∑k≥0pknx∫k+α/n+βk+1+α/n+βtidt=n+βA1Bnx∑k≥0pknxk+1+αi+1i+1n+βi+1−k+αi+1i+1n+βi+1=n+β−ii+1A1Bnx∑k≥0∑j=0i+1i+1jkjpknxα+1i+1−j−αi+1−j=n+β−ii+1A1Bnx∑j=0i+1i+1jα+1i+1−j−αi+1−j∑k≥0kjpknx.
Thus, by Lemma1, we complete the proof.Lemma 3 (for instance, see [21], equation (1.27)).
LetX,L be two operators on the set of functions defined by Xfu=ufu and Lfu=d/dufu. Then,
(7)XLm=∑j=1mSm,jXjLj.
Moreover,(8)XLmfugu=∑j=1m∑i=0jjiSm,jujdiduifudj−iduj−igu,where Sm,j is the Stirling number of the second kind.
Defineaj=dj/dtjAtt=1 and bj=dj/dtjBtt=nx, for all j≥0. Therefore, Theorem 2 with D=XL and Lemma 3 imply the following theorem.Theorem 4.
Letei=eit=ti for all i≥0. Then,
(9)Knα,βei,x=n+β−ii+1α+1i+1−αi+1+n+β−ii+1∑j=1i+1∑ℓ=1j∑s=0ℓi+1jℓsSj,ℓα+1i+1−j−αi+1−jnℓ−sasbℓ−sa0b0x2ℓ−s,where Sm,ℓ is the Stirling number of the second kind.Example 5.
By applying Theorem4 for i=0,1,2,3,4 with using (2), we obtain
(10)Knα,βe0,x=1,Knα,βe1,x=2α+12n+β+a1a0n+β+nb1b0n+βx,Knα,βe2,x=3α2+3α+13n+β2+2a1α+1+a2a0n+β2+2nb1a0α+1+a1a0b0n+β2x+n2b2b0n+β2x2,Knα,βe3,x=4α3+6α2+4α+14n+β3+6a1α+12+6a2α+1+a1+3a2+2a32a0n+β3+nb16a0α+12+12a1α+1+a0+6a1+6a22a0b0n+β3x+3n2b22a0α+1+a0+2a12a0b0n+β3+n3b3b0n+β3x3,Knα,βe4,x=5α4+10α3+10α2+5α+15n+β4+4a1α+13+6a2α+12+2a1+3a2+2a3α+1+3a2+4a3+a4a0n+β4+2nb12a0α+13+6a1α+12+a0+6a1+6a2α+1+3a1+6a2+2a3a0b0n+β4x+3n2b22a0α+12+2a0+2a1α+1+a0+4a1+2a2a0b0n+β4x2+4n3b3a0α+1+a0+a1a0b0n+β4x3+n4b4b0n+β4x4.Theorem 6.
Lety=α−xn+β, and let D be the operator td/dt. Then,
(11)Knα,βt−xi,x=n+β−ii+1A1Bnx∑j=0i+1i+1j1+yi+1−j−yi+1−jDjAtBnxtt=1.Proof.
By the definitions, we have(12)Knα,βt−xi,x=n+βA1Bnx∑k≥0pknx∫k+α/n+βk+1+α/n+βt−xidt=n+βA1Bnx∑k≥0pknxk+1+yi+1i+1n+βi+1−k+yi+1i+1n+βi+1=n+β−ii+1A1Bnx∑k≥0∑j=0i+1i+1jkjpknx1+yi+1−j−yi+1−j=n+β−ii+1A1Bnx∑j=0i+1i+1j1+yi+1−j−yi+1−j∑k≥0kjpknx.
Thus, Lemma1 completes the proof.
By Theorem6 and Lemma 3, we obtain the following result.Theorem 7.
Lety=α−xn+β. Then,
(13)Knα,βt−xi,x=n+β−ii+1y+1i+1−yi+1+n+β−ii+1∑j=1i+1∑ℓ=1j∑s=0ℓi+1jℓsSj,ℓy+1i+1−j−yi+1−jnℓ−sasbℓ−sa0b0x2ℓ−s,where Sm,ℓ is the Stirling number of the second kind.Remark 8.
By applying Theorem7 for i=0,1,2,3, we obtain
(14)Knα,β1,x=1,Knα,βt−x,x=2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx,Knα,βt−x2,x=3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2,Knα,βt−x3,x=4a0α3+6a0α2+12a1α2+4a0α+24a1α+12a2α+a0+14a1+18a2+4a34a0n+β3−6a0b0nα2+6a0b0α2β−6a0b1nα2+6a0b0nα+6a0b0αβ2a0b0n+β3x−−12a0b1nα+12a1b0nα+12a1b0αβ−12a1b1nα+2a0b0n+2a0b0β2a0b0n+β3x−−7a0b1n+12a1b0n+12a1b0β−18a1b1n+6a2b0n+6a2b0β−6a2b1n2a0b0n+β3x+32a0b0n2α+4a0b0nαβ+2a0b0αβ2−4a0b1n2α−4a0b1nαβ+2a0b2n2α2a0b0n+β3x2+3a0b0n2+2a0b0nβ+a0b0β2−4a0b1n2−4a0b1nβ+3a0b2n2+2a1b0n22a0b0n+β3x2−b0n3+3b0n2β+3b0nβ2+b0β3−3b1n3−6b1n2β−3b1nβ2+3b2n3+3b2n2β−b3n3b0n+β3x3.
Theorem6 for i=0,1,⋯,6 (with the help of mathematical programming), we obtain the following result.Proposition 9.
Let us consider that(15)limn→∞BmnxBnx=1form=0,1,2,⋯,6,where Bmt is the m-th derivative of Bt. Then, we obtain
(16)limn→∞Knα,β1,x=1limn→∞nKnα,βt−x1,x=a1/a0+1/2+α−βxlimn→∞nKnα,βt−x2,x=xlimn→∞n2Knα,βt−x3,x=3α+5/2+3a1/a0−3βx2limn→∞n2Knα,βt−x4,x=3x2limn→∞n3Knα,βt−x5,x=15α+35/2+15a1/a0x2−15βx3limn→∞n3Knα,βt−x6,x=15x3.Example 10.
LetAt=1, Bt=et, α=0, β=1, and ft=t−x2. By the fact that AtBxt=∑k≥0pkxtk, we have that pkx=xk/k!.
Table1 presents the values of the functions nKnα,βf;x and x at x=0,0.1,0.2,⋯,0.6 and n=20, n=100, where we approximated Knα,βf;x as
(17)Knα,βf,x=n+βA1Bnx∑k=03000pknx∫k+α/n+βk+1+α/n+βftdt.
We note that the Korovkin-type theorems are very useful tools in approximation which were studied in several function spaces [3–8, 10, 22–29]. We say that sequence of operators Knα,β converges to L in the sense of power series if
(18)limu→R−1pu∑n=0∞Knα,βf,xpnun=L,for every u∈0,R.Table 1
The values of the functions20K20α,βf;x and 100K100α,βf;x at x=0,0.1,0.2,⋯,1.
x00.10.20.30.40.50.620K20α,βf;x0.015110.101730.189260.277700.367040.457290.54845100K100α,βf;x0.003260.100410.197750.295290.393030.490960.58909


## 2. Main Results

We study hereT− statistical convergence of the operators Knα,β. Note that the Korovkin-type theorem for T− statistical convergence was considered in [24] as follows:Theorem 11.
LetBj be a sequence of positive linear operators on C0,1 and let T=tnj be a nonnegative regular summability matrix such that
(19)stT−limnBjei−ei=0,i=0,1,2.
Then, for anyf∈C0,1(20)stT−limnBjh−h=0,where h=max0≤x≤1hx.
Based on the above theorem, we give the following result.Theorem 12.
LetT=tnj be a regular matrix and Knα,β be as in (2) on C0,M∩E such that limn→∞Binx/Bnx=1, where Binx denotes ith derivative and
(21)stA−limnKnα,βei−ei=0i=1,2.
Then, for anyh∈C0,M∩E(22)stA−limnKnα,βh−h=0,where h=maxt∈0,Mht.Proof.
From Lemma 5, we have thatstT−limnKnα,βe0−e0=0. Now, we will estimate the following expressions:
(23)Knα,βe1−e1≤xnn+β·B′nxBnx−1+A′1n+βA1+2α+12n+β.
Note thatlimn→∞B′nx/Bnx=1. So from the last two relations we have that Knα,βe1−e1=0. Moreover,
(24)Knα,βe2−e2=nn+β2B′′nxBnx·x2+nB′nx2A′1+2α+2A1n+β2A1Bnxx+1n+β2A1A′′1+2α+2A′1α2+α+13A1−x2⟶0.
Now proof follows directly from Theorem11.
This theorem is an extension of some known results for the Kantorovich-Stancu-type Szász-Mirakyan operators.Example 13 (see [6]).
Under conditions given in Theorem12, we define the following operators
(25)Nnh,x=1+xnKnα,βh,x,where the sequence xn is given as follows:
(26)xn=1n3;m2−m≤n≤m2−11n4;n=m2;m∈ℕ\10;otherwise,then
(27)Nne0,x=1+xn,Nne1,x=1+xn2α+12n+β+a1a0n+β+nb1b0n+βx,Nne2,x=1+xn3α2+3α+13n+β2+2a1α+1+a2a0n+β2+2nb1a0α+1+a1a0b0n+β2x+n2b2b0n+β2x2.
By Theorem11 we obtain stT−limnNnh−h=0, but the operators Nnh,x do not satisfy Theorem 12. Hence, the sequence Nn is not statistically convergent but it is T− statistically convergent.Remark 14.
The sequencexn is not statistically convergent and hence not convergent. As an example, consider the Cesáro matrix of order 2.
(28)T=tnk=2n+1−kn+1n+2;0≤k≤n,0;k>n,where
(29)0≤limn∑k:xk−α≥εtnk=limn∑k=m2−m,⋯,m2−1k=m2;m∈ℕ\1tnk=limn2n+1n+21+⋯+n≤limn2n+1n+2·nn+12=1.
This proves thatx=xn is T− statistically convergent. We have
(30)Nne0,x=1+xn,Nne1,x=1+xn2α+12n+β+a1a0n+β+nb1b0n+βx,Nne2,x=1+xn3α2+3α+13n+β2+2a1α+1+a2a0n+β2+2nb1a0α+1+a1a0b0n+β2x+n2b2b0n+β2x2.
By Example13, this shows that Nnh,x does not satisfy Theorem 12.
In [27, 29], Korovkin-type theorems are proved by Abel summability method. Now, we discuss for power series method. Let B0,∞ (C0,∞) be the space of all bounded (continuous) functions on the interval 0,∞.Theorem 15.
LetKnα,β be a sequence of positive linear operators from C0,M∩E into B0,M∩E such that
(31)limt→R−1pt∑n=0∞Knα,βei−eipntn=0,i=0,1,2.
Then, forh∈C0,M∩E,
(32)limt→R−1pt∑n=0∞Knα,βh−hpntn=0.Proof.
Clearly, from (32) follows (31). Now, we show the converse that (31) implies (32). Let h∈C0,M∩E, then there exists a constant K>0 such that hu≤K for all u∈0,M. Therefore,
(33)hu−hx∈0,M.
For every givenε>0, there exists δ>0 such that
(34)hu−hx≤ε,whenever u−x<δ for all u∈0,M. Define ψ≡ψu,x=u−x2. If u−x≥δ, then
(35)hu−hx≤2Kδ2ψu,x.
From (33)–(35), we have that hu−hx<ε+2K/δ2ψu,x, namely,
(36)−ε−2Kδ2ψu,x<ht−hx<2Kδ2ψu,x+ε.
By applying the operatorKnα,β1,x, Knα,β1,x is a monotone and linear operator, we obtain
(37)Knα,β1,x−ε−2Kδ2ψ<Knα,β1,xhu−hx<Knα,β1,x2Kδ2ψ+ε,which implies
(38)−εKnα,β1,x−2Kδ2Knα,βψu,x<Knα,βh,x−hxKnα,β1,x<2Kδ2Knα,βψu,x+εKnα,β1,x.
On the other hand,(39)Knα,βh,x−hx=Knα,βh,x−hxKnα,β1,x+hxKnα,β1,x−1.
From (38) and (39) we get
(40)Knα,βh,x−hx<2Kδ2Knα,βψu,x+εKnα,β1,x+hxKnα,β1,x−1.
Now, we estimate the following expression:(41)Knα,βψu,x=Knα,βx−u2,x=Knα,βx2−2xu+u2,x=x2Knα,β1,x−2xKnα,βu,x+Knα,βu2,x.
By (40), we obtain
(42)Knα,βh,x−hx<2Kδ2x2Knα,β1,x−1−2xKnα,βu,x−x+Knα,βu2,x−x2+εKnα,β1,x+fxKnα,β1,x−1=ε+εKnα,β1,x−1+hxKnα,β1,x−1++2Kδ2x2Knα,β1,x−1−2xKnα,βu,x−x+Knα,βu2,x−x2.
Therefore,(43)Knα,βh,x−hx≤ε+ε+K+2KM2δ2Knα,β1,x−1+4KMδ2Knα,βu,x−x+2Kδ2Knα,βu2,x−x2.
From the above relations and the linearity ofKnα,β, we obtain
(44)1pv∑n=0∞Un,ph;x−hxpnvn≤ε+ε+K+2KM2δ21pt∑n=0∞Knα,β1;x−1pntn+4KMδ21pv∑n=0∞Knα,βu;x−xpnvn+2Kδ21pv∑n=0∞Knα,βu2;x−x2pnvn.
Hence, (32) follows from the last relation and (31).


## 3. Rate of Convergence

Modulus of continuity is defined by(45)ωh,δ=suph<δhx+h−hx,hx∈C0,M∩E.It is not hard to verify(46)hx−hy≤ωh,δx−yδ+1.So, we can state the following.Theorem 16.
LetT=tij be a nonnegative regular summability matrix and h∈C0,M∩E. If αn is a sequence of positive real numbers such that ωh,δn=stT−0αn, then
(47)Knα,βh−h=stA−0αn,where
(48)δn=M2n2B′′nxn+β2Bnx−2nB′nxn+βBnx+1+MnB′nx2′1+2α+2A1n+β2A1Bnx−2A′1n+βA1+22α+12n+β+1n+β2A1A′′1+2α+2A′1α2+α+13A12,for any positive integer n.Proof.
Leth∈C0,M∩E. By positivity and linearity of Knα,β and (46), we see
(49)Knα,βh;x−h≤Knα,βht−hx;x≤n+βA1Bnx∑k=0∞pknx∫k+α/n+βk+1+α/n+βωh,δ1+t−xδdt≤ωh,δ1+1δn+βA1Bnx∑k=0∞pknx∫k+α/n+βk+1+α/n+βt−xdtbyLemma1=ωh,δ1+1δKnα,βt−x;x.
By applying the Cauchy-Schwartz inequality, we have(50)Knα,βh;x−h≤ωh,δ1+1δKnα,βt−x2;x1/2.
Based on Examples5 and Remark 8, we obtain
(51)Knα,βu−x2;x=Knα,βe2;x−2xKnα,βe1;x+x2Knα,βe0;x≤M2n2B′′nxn+β2Bnx−2nB′nxn+βBnx+1+MnB′nx2A′1+2α+2A1n+β2A1Bnx−2A′1n+βA1+22α+12n+β+1n+β2A1A′′1+2α+2A′1α2+α+13A1.
By taking(52)δn=M2n2B′′nxn+β2Bnx−2nB′nxn+βBnx+1+MnB′nx2A′1+2α+2A1n+β2A1Bnx−2A′1n+βA1+22α+12n+β+1n+β2A1A′′1+2α+2A′1α2+α+13A12,we get that Knα,βh−h≤2·ωh,δn. Therefore, for every ε>0, we have
(53)1αn∑Knα,βh−h≥εtnj≤1αn∑2·ωf,δn≥εtnj.
From the conditions that are given in the theorem, we have thatKnα,βh−h=stT−0αi, as claimed.
In the next result, we present the rate of convergence for the power summability method.Theorem 17.
Leth∈C0,M∩E and let ϕ be a positive real function defined on 0,M∩E. If ωh,ψu=Oϕu, as v⟶R−, then we have
(54)1pv∑n=0∞Knα,βei−eipnvn=Oϕv,where the function ψ:0,M∩E⟶ℝ is defined by relation
(55)ψu=supx∈0,Mn∈ℕKnα,βu−x2;x1/2.Proof.
Leth∈C0,M∩E. For any u∈0,R,x∈0,M, and δ>0, we have
(56)∑n=0∞Knα,βh;x−hxpnvn≤∑n=0∞Knα,βhu−hx;xpnvn≤∑n=0∞Knα,βωh,u−xδδ;xpnvn≤∑n=0∞Knα,β1+u−xδωh,δ;xpnvn≤ωh,δ∑n=0∞Knα,β1+u−x2δ2;xpnvn≤ωh,δ∑n=0∞Knα,βe0u;xpnvn+ωh,δδ2∑n=0∞Knα,βu−x2;xpnvn=pvωh,δ+ωh,δδ2sup0≤x≤1n∈ℕKnα,βu−x2;x∑n=0∞pnvn,which leads to
(57)∑n=0∞Knα,βf;x−fxpnvn≤pvωf,δ+ωf,δδ2sup0≤x≤1Knα,βu−x2;xpv.
If we setδ=ψu, then from the last inequality we have
(58)0≤1pv∑n=0∞Knα,βh−hpnvn≤2ωh,δ,as required.


## 4. Voronovskaya-Type Theorems

First, we prove a Voronovskaya-type theorem for the operators under consideration.Theorem 18.
Leth,h′,h′′∈C0,M∩E and limn→∞Binx/Bnx=1, for i∈1,2. Then,
(59)limn→∞n+βKnα,βh;x−hx=A′1A1+12+α−βxh′x+x2h′′x,for every x∈0,M.Proof.
Assume thath′,h′′∈C0,M∩E and x∈0,M. By Taylor’s formula, we have
(60)hy=hx+y−xh′x+12y−x2h′′x+y−x2ψy−x,where ψy−x⟶0 and y−x⟶0. Applying in both sides of the above relation operators Knα,β, we obtain
(61)Knα,βh,x−hx=h′xKnα,βy−x;x+h′′x2Knα,βy−x2;x+Knα,βy−x2ψy−x;x,which implies
(62)n+βKnα,βh,x−hx=n+βh′xKnα,βy−x;x+n+βh′′x2Knα,βy−x2;x+n+βKnα,βy−x2ψy−x;x.
Now, we will estimate this expression:(63)limn→∞n+βKnα,βy−x2ψy−x;x.
Letε>0 and δ>0 such that ψy−x<ε, where y−x<δ. We will split the above relation in two parts:
(64)U1=n+βn+βA1Bnx∑k=0∞pknx∫y−x≤δy−x2ψy−xdyU2=n+βn+βA1Bnx∑k=0∞pknx∫y−x≥δy−x2ψy−xdy.
From the above conditions, we have(65)U1≤εn+β21A1Bnx∑k=0∞pknx∫k+α/n+βk+α+1/n+βy−x2dy=εn+βKnα,βy−x2;x.
On the other hand, from Proposition9, condition (3), we get that U1≤ε1.
Let us denote byN=supxψy−x;y−x≥δ. Then, we obtain
(66)U2≤Nδ2n+βn+βA1Bnx∑k=0∞pknx∫y−x≥δy−x4dy=Nδ2n+βKnα,βy−x4;x.
Condition (5) in Proposition9 tells us that limn→∞U2=0, which completes the proof.Example 19.
LetAt=1, Bt=et, α=1/3, β=1/2, and hx=5xe−x. By AtBxt=∑k≥0pkxtk, we see that pkx=xk/k!. Figure 1 presents the graphs of the functions 10+βK10α,βh;x−hx, 20+βK20α,βh;x−hx, and A′1/A1+1/2+α−βxh′x+x/2h′′x.
We extend the Voronovskaya-type theorem forT− statistical method for these operators. Consider operators Nn from Example 13. We start with the following lemma.Lemma 20.
Leth∈C0,M∩E such that h′,h′′∈C0,M∩E,x∈0,M, and limn→∞Binx/Bnx=1, for i∈1,2,3,4. Then, we obtain
(67)n+β4Kny−x4;x~0stAon0,M.Proof.
The proposition follows directly from Proposition9 (5).Theorem 21.
Leth∈C0,M∩E such that h′,h′′∈C0,M∩E,x∈0,M, for any finite M and let limn→∞Binx/Bnx=1, for i∈1,2,3,4. Then, for x∈0,M,
(68)n+β2Nnh;x−hx−h′x2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx−h′′x2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2~h′′x2·3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2.Proof.
Taylor’s formula gives(69)hy=hx+y−xh′x+12y−x2h′′x+y−x2ψy−x,where ψy−x⟶0, as y−x⟶0. Taking into consideration Remark 8, after applying Nn in both sides of relation (69), we obtain
(70)Nnh=1+xnhx+1+xnh′x2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx+1+xnh′′x23a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2+1+xnKnα,βΦ2ψy−x;x.
This yields(71)n+β2Nnh=1+xnn+β2hx+1+xnh′xn+β22a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx+1+xnh′′x2n+β23a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2+1+xnn+β2Knα,βΦ2ψy−x;x.
Therefore,(72)n+β2Nnh;x−hx−h′x2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx−h′′x2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2−h′′x2·3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2≤Mxnn+β2+M1xn2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx+M2·xn3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2+n+β2Knα,βy−x2ψy−x;x+xnn+β2Knα,βy−x2ψy−x;x,where M=supx∈0,Mhx,M1=supx∈0,Mh′x, and M2=supx∈0,Mh′′x.
Now, we have to prove that(73)limn→∞n+β2Knα,βy−x2ψy−x;x=0.
By applying the Cauchy-Schwartz inequality, we obtain(74)n+β2Knα,βy−x2ψy−x;x≤n+β4Knα,βy−x4;x1/2·Knα,βψ2;x1/2.
Also, by settingηxy=ψy−x2, we have that ηxx=0 and ηx·∈C0,M. So
(75)Knα,βηx⟶0stAon0,M.
Now, from the last relation, (74), (75), and Lemma 20, we obtain that
(76)n+β2Knα,βy−x2ψy−x;x⟶0stAon0,M.
From the construction ofxn, it follows that n+β2xn⟶0stA on 0,M.
For a givenε>0, we define the following sets:
(77)A=n:n+β2Nnh;x−hx−h′x2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx−h′′x2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2−h′′x2·3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2≥ε,A1=n:xnn+β2≥ε5M,A2=n:xn2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx≥ε5M1,A3=n:xn3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2≥ε5M2,A4=n:n+β2Knα,βy−x2ψy−x;x≥ε5,A5=n:xnn+β2Knα,βy−x2ψy−x;x≥ε5.
From last relations, we obtain thatA≤A1+A2+A3+A4+A5. Hence, the result follows.Remark 22.
We see that the operatorsNn (see Example 13) do not satisfy a Voronovskaya-type theorem in the usual sense.Figure 1
The graphs of10+βK10α,βf;x−fx, 20+βK20α,βf;x−fx, and A′1/A1+1/2+α−βxf′x+x/2f′′x when x∈0,3.5.


## 5. Grüss-Voronovskaya-Type Theorems

This kind of result, for the first time, was shown in [30].Theorem 23 (see [31]).
LetE=h:x∈0,∞,hx/1+x2,isconvergentasx⟶∞ and
(78)limn→∞BmnxBnx=1form=1,2.
Ifh∈C0,∞∩E, then
(79)limn→∞Knα,βf,x=fx,and the operators Knα,βh,x converge uniformly in each compact subset of 0,∞.
Now, we are ready to prove the following result.Theorem 24.
Forh,h′,h′′∈C0,∞ and any x∈0,∞, limn→∞Binx/Bnx=1, for i∈1,2,3,4,5,6, where Bi denotes the ith derivative of B. Then,
(80)n+βKnα,βh,x−hx−h′x2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx−n+βh′′x2·3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2=01ωh′′,n−1/2,as n⟶∞.Proof.
From Taylor’s theorem, we have(81)hu=hx+h′xu−x+h′′x2u−x2+Ru,x,where Ru,x=h′′θ−h′′x/2u−x2, for θ∈u,x. Now, we obtain
(82)Knα,βh,x−hx−h′xKnα,βu−x;x−h′′x2Knα,βu−x2;x≤Knα,βRu,x,x.
From which we get that(83)n+βKnα,βh,x−hx−h′x2a0α+a0+2a12a0n+β−b0n+b0β−b1nb0n+βx−n+βh′′x2·3a0α2+3a0α+6a1α+a0+6a1+3a23a0n+β2−2a0b0αn+β−2a0b1nα+1+a0b0n+β+2a1b0n+β−2a1b1na0b0n+β2x+b0n2+2b0nβ+b0β2−2b1n2−2b1nβ+b2n2b0n+β2x2≤n+β·Knα,βRu,x,x.
By the properties of modulus of continuity modulus, we have(84)h′′θ−h′′x2!≤12!1+∣θ−x∣δωh′′,δ.
On the other hand,(85)h′′θ−h′′x2!≤ωh′′,δ,u−x≤δ,t−x4δ4ωh′′,δ,u−x≥δ.
For0<δ<1, we obtain that
(86)h′′θ−h′′x2!≤ωh′′,δ1+u−x4δ4,which gives
(87)Ru,x≤ωh′′,δ1+u−x4δ4u−x2=ωh′′,δu−x2+u−x6δ4.
By the linearity ofKnα,β and the above relation, we obtain
(88)Knα,βRu,x,x≤ωh′′,δKnα,βu−x2,x+1δ4Knα,βu−x6,x.
Taking into consideration Proposition9, we have
(89)Knα,βRu,x,x≤ωh′′,δ01n+1δ401n3=01nωh′′,δ.
Forδ=n−1/2, we complete the proof.Theorem 25.
Leth′x,g′x,h′′x,g′′x,hg′x,hg′′x∈C0,∞∩E, and
(90)limn→∞BmnxBnx=1form=0,1,2,⋯,6,x∈0,M,where Bm is the mth derivative of B. Then,
(91)limn→∞n+βKnα,βhg,x−Knα,βh,x·Knα,βg,x=xh′xg′x.Proof.
We know that(92)n+βKnα,βhg,x−Knα,βh,xKnα,βg,x=n+βKnα,βhg,x−hgx−hg′xKnα,βu−x,x−hg′′x2Knα,βu−x2,x−gxKnα,βh,x−hx−h′xKnα,βu−x,x−h′′x2Knα,βu−x2,x−Knα,βh,xKnα,βg,x−gx−g′xKnα,βu−x,x−g′′x2Knα,βu−x2,x+g′′x2Knα,βu−x2,xhx−Knα,βh,x+h′xg′xKnα,βu−x2,x−g′xKnα,βu−x,xKnα,βh,x−hx.
From Proposition9 and Theorems 23 and 24, we obtain
(93)limn→∞n+βKnα,βhg,x−Knα,βh,xKnα,βg,x=xh′xg′x.



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