Theory and Modern Applications

# Approximation by a power series summability method of Kantorovich type Szász operators including Sheffer polynomials

## Abstract

The main purpose of this paper is to use a power series summability method to study some approximation properties of Kantorovich type Szász–Mirakyan operators including Sheffer polynomials. We also establish Voronovskaya type result.

## 1 Introduction and background

Let $$\mathcal{K}_{m}=\{i\leq m:i\in \mathcal{K}\subseteq \mathbb{N}\}$$. Then the natural density of $$\mathcal{K}$$ is defined by $$\sigma (\mathcal{K})=\lim_{m}\frac{1}{m}{|\mathcal{K}_{m}|}$$ provided the limit exists, where $$|\mathcal{K}_{m}|$$ denotes the cardinality of $$\mathcal{K}_{m}$$. A sequence $$\eta =(\eta _{i})$$ is “statistically convergent” (see ) to $$\mathfrak{s}$$ if for every $$\epsilon >0$$

$$\lim_{m}\frac{1}{m} \bigl\vert \bigl\{ i\leq m: \vert \eta _{i}-\mathfrak{s} \vert \geq \epsilon \bigr\} \bigr\vert =0$$

and we write $$st-\lim_{m}\eta {_{m}}=\mathfrak{s}$$.

Let $$\mathfrak{T}=(\mathfrak{d}_{ij})$$ be an infinite matrix. It is said to be regular if it transforms a convergent sequence into a convergent one with the same limit.

Let $$\mathfrak{T}=(\mathfrak{d}_{ij})$$ be regular matrix. A sequence $$\zeta =(\zeta _{j})$$ is said to be $$\mathfrak{T}$$-statistically convergent (see ) to the number $$\mathfrak{s}$$ if, for any $$\epsilon >0$$, $$\lim_{i}{\sum_{j:|\eta _{j}-\mathfrak{s}|\geq \epsilon }} \mathfrak{d}_{ij}=0$$, and denote $$st_{\mathfrak{T}}-\lim \eta =\mathfrak{s}$$. If

$$\mathfrak{d}_{ij}= \textstyle\begin{cases} \frac{1}{j}, & i\leq j, \\ 0; & i>j.\end{cases}$$
(1.1)

Then it reduces to statistical convergence.

For a sequence of positive real numbers $$(\mathfrak{p}_{j})$$, denote the corresponding power series $$\mathfrak{p}(\mathfrak{y})=\sum_{j=1}^{\infty }\mathfrak{p}{_{j}}\mathfrak{y}^{j-1}$$ which has radius of convergence $$R>0$$. A sequence $$\eta =(\eta _{j})$$ is convergent in the sense of power series method (see [12, 21]) if $$\lim_{\mathfrak{y}\rightarrow R^{-}}{ \frac{1}{\mathfrak{p}(\mathfrak{y})}\sum_{j=1}^{\infty }}\eta _{j}{{ \mathfrak{p}}}_{j}\mathfrak{y}{{^{j-1}}}=\mathcal{L}$$ for all $$\mathfrak{y}\in (0,R)$$. Moreover, the power series method is regular if and only if $$\lim_{\mathfrak{y}\rightarrow R^{-}}{ \frac{{{\mathfrak{p}}}_{j}\mathfrak{y}{{^{j-1}}}}{\mathfrak{p}(\mathfrak{y})}}=0$$ holds for each $$j\in \{1,2,\ldots \}$$ (see ). The power series method is more effective than the ordinary convergence (see [22, 23]). For more summability methods, see [35, 7, 13, 1519].

We study a Korovkin type theorem for the Kantorovich type generalization of Szász operators involving Sheffer polynomials via power series method. We determine the rate of convergence for these operators. Furthermore, we give a Voronovskaya type theorem for $$\mathfrak{T}$$-statistical convergence.

The multiple Sheffer polynomials $$\{S_{k_{1},k_{2}}(x)\}_{k_{1},k_{2}=0}^{\infty }$$ are defined as follows. The generating function is

$$A(t_{1},t_{2})e^{xH(t_{1},t_{2})}=\sum _{k_{1}=0}^{\infty }\sum_{k_{2}=0}^{ \infty }S_{k_{1},k_{2}}(x) \frac{t_{1}^{k_{1}}t_{2}^{k_{2}}}{k_{1}!k_{2}!},$$
(1.2)

where $$A(t_{1},t_{2})$$ and $$H(t_{1},t_{2})$$ have series expansions of the form

$$A(t_{1},t_{2})=\sum_{k_{1}=0}^{\infty } \sum_{k_{2}=0}^{\infty }a_{k_{1},k_{2}} \frac{t_{1}^{k_{1}}t_{2}^{k_{2}}}{k_{1}!k_{2}!}$$
(1.3)

and

$$H(t_{1},t_{2})=\sum_{k_{1}=0}^{\infty } \sum_{k_{2}=0}^{\infty }h_{k_{1},k_{2}} \frac{t_{1}^{k_{1}}t_{2}^{k_{2}}}{k_{1}!k_{2}!},$$
(1.4)

respectively, with the conditions

$$A(0,0)=a_{0,0}\neq 0 \quad\text{and}\quad H(0,0)=h_{0,0}\neq 0.$$

In , one defined the positive linear operators involving multiple Sheffer polynomials for $$x\in [ 0,\infty )$$ as follows:

$$G_{n}(f,x)=\frac{e^{-\frac{nx}{2}H(1,1)}}{A(1,1)}\sum_{k_{1}=0}^{ \infty } \sum_{k_{2}=0}^{\infty } \frac{S_{k_{1},k_{2}} ( \frac{nx}{2} ) }{k_{1}!k_{2}!}f \biggl( \frac{k_{1}+k_{2}}{n} \biggr),$$
(1.5)

provided that the right-hand side of the above series converge, under conditions that:

1. (1)

$$S_{k_{1},k_{2}}(x)\geq 0,k_{1},k_{2}\in \mathbb{N}$$,

2. (2)

$$A(1,1)\neq 0, H_{t_{1}}(1,1)=1,H_{t_{2}}(1,1)=1$$,

3. (3)

Series (1.2), (1.3) and (1.4) are convergent for $$|t_{1}|< R$$, $$|t_{2}|< R$$ and $$(R_{1},R_{2})>1$$.

In  one defined the Kantorovich variant of Szász operators induced by multiple Sheffer polynomials as follows:

$$K_{n}^{(S)}(f,x)=\frac{ne^{-\frac{nx}{2}H(1,1)}}{A(1,1)}\sum _{k_{1}=0}^{\infty }\sum_{k_{2}=0}^{\infty } \frac{S_{k_{1},k_{2}} ( \frac{nx}{2} ) }{k_{1}!k_{2}!} \int _{\frac{k_{1}+k_{2}}{n}}^{ \frac{k_{1}+k_{2}+1}{n}}{f(t)}\,dt, \quad x\in [ 0,\infty ),$$
(1.6)

provided that the right-hand side of the above relation exists.

Example 1.1 of , gives us the following expressions for moments of the Kantorovich variant of Szász operators induced by multiple Sheffer polynomials:

\begin{aligned} &K_{n}^{(S)}(1,x) =1, \\ &K_{n}^{(S)}(t,x) =\frac{1}{2}\cdot \frac{2aa_{0,1}+2aa_{1,0}+aa_{0,0}}{aa_{0,0}n}+x, \\ &K_{n}^{(S)} \bigl(t^{2},x \bigr) = \frac{1}{3}\cdot \frac{3aa_{0,2}+3aa_{0,1}+3aa_{0,2}+6aa_{1,1}+3aa_{1,0}+aa_{0,0}}{aa_{0,0}n^{2}} \\ &\phantom{K_{n}^{(S)} (t^{2},x ) =}{} +\frac{x}{2}\cdot \frac{aa_{0,0}hh_{0,2}+2aa_{0,0}hh_{1,1}+aa_{0,0}hh_{2,0}+2aa_{0,0}+4aa_{0,1}+4aa_{1,0}}{aa_{0,0}n}+x^{2}. \end{aligned}

The central moments of the Kantorovich variant of Szász operators induced by multiple Sheffer polynomials are 

\begin{aligned} &K_{n}^{(S)}(t-x,x)=(2\tilde{a}_{0,1}+2 \tilde{a}_{1,0}+\tilde{a}_{0,0}) \frac{1}{2\tilde{a}_{0,0}n}, \\ &K_{n}^{(S)} \bigl((t-x)^{2},x \bigr)=(3 \tilde{a}_{0,2}+6\tilde{a}_{1,1}+3 \tilde{a}_{1,0}+3\tilde{a}_{2,0}+3\tilde{a}_{0,1}+\tilde{a}_{0,0}) \frac{1}{3\tilde{a}_{0,0}n^{2}} \\ &\phantom{K_{n}^{(S)} ((t-x)^{2},x )=}{}+(\tilde{h}_{0,2}+2\tilde{h}_{1,1}+ \tilde{h}_{2,0}) \frac{x}{2n}, \\ &K_{n}^{(S)} \bigl((t-x)^{3},x \bigr)\\ &\quad=(6 \tilde{a}_{0,2}+4\tilde{a}_{3,0}+6 \tilde{a}_{2,0}+4\tilde{a}_{1,0} +4\tilde{a}_{0,3}+12\tilde{a}_{2,1}+ \tilde{a}_{0,0}+4 \tilde{a}_{0,1}+12 \tilde{a}_{1,2} \\ &\qquad{}+12\tilde{a}_{1,1})\frac{1}{4\tilde{a}_{0,0}n^{3}}+(3 \tilde{a}_{0,0} \tilde{h}_{0,2}+2 \tilde{a}_{0,0}\tilde{h}_{0,3} +6\tilde{a}_{0,0} \tilde{h}_{1,1}+6\tilde{a}_{0,0}\tilde{h}_{1,2}+3 \tilde{a}_{0,0} \tilde{h}_{2,0} \\ &\qquad{}+6\tilde{a}_{0,0}\tilde{h}_{2,1}+2 \tilde{a}_{0,0} \tilde{h}_{3,0} +6 \tilde{a}_{0,1} \tilde{h}_{0,2}+12\tilde{a}_{0,1}\tilde{h}_{1,1} +6 \tilde{a}_{0,1}\tilde{h}_{2,0}+6\tilde{a}_{1,0} \tilde{h}_{0,2} \\ &\qquad{}+12\tilde{a}_{1,0}\tilde{h}_{1,1}+6 \tilde{a}_{1,0} \tilde{h}_{2,0}) \frac{x}{4\tilde{a}_{0,0}n^{2}}. \end{aligned}

Similarly, there exist constants $$C_{di}$$ (dependent only on $$\tilde{a}_{i,j}$$ and $$\tilde{h}_{i,j}$$) such that

\begin{aligned} &K_{n}^{(S)} \bigl((t-x)^{4},x \bigr)= \frac{C_{44}}{4\tilde{a}_{0,0}n^{4}} + \frac{xC_{43}}{2\tilde{a}_{0,0}n^{3}}+\frac{3x^{2}C_{42}}{4n^{2}}, \\ &K_{n}^{(S)} \bigl((t-x)^{5},x \bigr)= \frac{C_{55}}{6\tilde{a}_{0,0}n^{5}} + \frac{xC_{54}}{12\tilde{a}_{0,0}n^{4}}+\frac{5x^{2}C_{53}}{8\tilde{a}_{0,0}n^{3}}, \\ &K_{n}^{(S)} \bigl((t-x)^{6},x \bigr)= \frac{C_{66}}{7\tilde{a}_{0,0}n^{6}} + \frac{xC_{65}}{2\tilde{a}_{0,0}n^{5}}+\frac{5x^{2}C_{64}}{4\tilde{a}_{0,0}n^{4}}. + \frac{15x^{2}C_{63}}{8n^{3}}. \end{aligned}

As a consequence of the above relations, we obtain

\begin{aligned} &\lim_{n\to \infty } nK_{n}^{(S)}(t-x,x)= \frac{2\tilde{a}_{0,1}+2\tilde{a}_{1,0}+\tilde{a}_{0,0}}{2\tilde{a}_{0,0}}, \\ &\lim_{n\to \infty } nK_{n}^{(S)} \bigl((t-x)^{2},x \bigr)= \frac{(\tilde{h}_{0,2}+2\tilde{h}_{1,1}+\tilde{h}_{2,0})x}{2}, \\ &\lim_{n\to \infty } n^{2}K_{n}^{(S)} \bigl((t-x)^{3},x \bigr)=E_{3}x, \qquad\lim _{n \to \infty } n^{2}K_{n}^{(S)} \bigl((t-x)^{4},x \bigr)=E_{4}x^{2}, \\ &\lim_{n\to \infty } n^{3}K_{n}^{(S)} \bigl((t-x)^{5},x \bigr)=E_{5}x^{3}, \qquad\lim _{n \to \infty } n^{3}K_{n}^{(S)} \bigl((t-x)^{6},x \bigr)=E_{6}x^{3}, \end{aligned}

where $$E_{3},E_{4},E_{5},E_{6}$$ are constant dependent on the derivatives of $$A(t_{1},t_{2})$$ and $$H(t_{1},t_{2})$$ up to order three at the point $$(t_{1},t_{2})=(1,1)$$.

## 2 Korovkin type results

The statistical form of Korovkin’s theorem was studied in  and the A-statistical version was considered in  (see also [13, 17] for other summability methods).

Let $$B[0,\infty )$$ ($$C[0,\infty )$$) be “the space of all bounded (continuous) functions” on the interval $$[0,\infty )$$.

### Theorem 2.1

Let $$\mathfrak{T}=(\mathfrak{d}_{ij})$$ be regular matrix and $$K_{n}^{(S)}(f,x)$$ be as in (1.6) on $$[0,M]$$, for any finite M. If

$$st_{\mathfrak{T}}-\lim_{n}{ \bigl\Vert K_{n}^{(S)}(f,x)e_{i}-e_{i} \bigr\Vert }=0 \quad (i=1,2),$$

then

$$st_{\mathfrak{T}}-\lim_{n}{ \bigl\Vert K_{n}^{(S)}(f,x)\mathfrak{h}- \mathfrak{h} \bigr\Vert }=0,$$

$$\mathfrak{h}\in C([0,M])$$, where $$\Vert \mathfrak{h} \Vert =\sup_{t\in [ 0,M]}{ |\mathfrak{h}(t)|}$$.

### Proof

From Example 1.1 of , we have $$st_{\mathfrak{T}}-\lim_{n}{ \Vert K_{n}^{(S)}e_{0}-e_{0} \Vert }=0$$. Now

$${ \bigl\Vert K_{n}^{(S)}e_{1}-e_{1} \bigr\Vert }\leq \biggl\Vert \frac{1}{2}\cdot \frac{2aa_{0,1}+2aa_{1,0}+aa_{0,0}}{aa_{0,0}n} \biggr\Vert .$$

Also $$\lim_{n\rightarrow \infty } \Vert K_{n}^{(S)}e_{1}-e_{1} \Vert =0$$. Moreover,

\begin{aligned} &{ \bigl\Vert K_{n}^{(S)}e_{2}-e_{2} \bigr\Vert }\\ &\quad= \biggl\Vert \frac{1}{3}\cdot \frac{3aa_{0,2}+3aa_{0,1}+3aa_{0,2}+6aa_{1,1}+3aa_{1,0}+aa_{0,0}}{aa_{0,0}n^{2}} \\ &\qquad{}+\frac{x}{2}\cdot \frac{aa_{0,0}hh_{0,2}+2aa_{0,0}hh_{1,1}+aa_{0,0}hh_{2,0}+2aa_{0,0}+4aa_{0,1}+4aa_{1,0}}{aa_{0,0}n} \biggr\Vert \rightarrow 0, \end{aligned}

as $$n\rightarrow \infty$$. Now the proof follows directly from the statistical version of the Korovkin theorem . □

### Example 2.2

()

Under the conditions given in Theorem 2.1, set

$$N_{n}(h,x)=(1+u_{n})K_{n}^{(S)}(h,x),$$

where

$$u_{n}=\textstyle\begin{cases} 1; & m^{2}-m\leq n\leq m^{2}-1, \\ \frac{1}{m^{4}}; & n=m^{2};m\in \mathbb{N}\setminus \{1\}, \\ 0; & \text{otherwise},\end{cases}$$

then

\begin{aligned} &N_{n}(e_{0},x) =(1+u_{n}), \\ &N_{n}(e_{1},x) =(1+u_{n}) \biggl( \frac{1}{2}\cdot \frac{2aa_{0,1}+2aa_{1,0}+aa_{0,0}}{aa_{0,0}n}+x \biggr), \\ &N_{n}(e_{2},x)=(1+u_{n}) \biggl( \frac{1}{3}\cdot \frac{3aa_{0,2}+3aa_{0,1}+3aa_{0,2}+6aa_{1,1}+3aa_{1,0}+aa_{0,0}}{aa_{0,0}n^{2}} \\ &\phantom{N_{n}(e_{2},x)=}{}+ \frac{x}{2}\cdot \frac{aa_{0,0}hh_{0,2}+2aa_{0,0}hh_{1,1}+aa_{0,0}hh_{2,0}+2aa_{0,0}+4aa_{0,1}+4aa_{1,0}}{aa_{0,0}n}+x^{2} \biggr). \end{aligned}

If the matrix $$\mathfrak{T}$$ is as in (1.1), then, by Theorem 2.1 we obtain $$st_{\mathfrak{T}}-\lim_{n}{ \Vert N_{n}h-h \Vert }=0$$, but the operators $$N_{n}(h,x)$$, do not satisfy the conditions of the theorem in .

In the following result we use a power series method as in [20, 24]; the Abel summability method was used.

### Theorem 2.3

Let $$(K_{n}^{(S)})$$ be a sequence of positive linear operators from $$C[0,M]$$ into $$B[0,M]$$ $$(0< M<\infty )$$ such that

$$\lim_{t\rightarrow R^{-}}{\frac{1}{\mathfrak{p}(t)} \Biggl\Vert \sum _{n=0}^{ \infty }{ \bigl(K_{n}^{(S)}e_{i}-e_{i} \bigr){\mathfrak{p}}_{n}t^{n}} \Biggr\Vert }=0, \quad i=0,1,2.$$
(2.1)

Then, for $$\mathfrak{h}\in C[0,M]$$,

$$\lim_{t\rightarrow R^{-}}{\frac{1}{\mathfrak{p}(t)} \Biggl\Vert \sum _{n=0}^{ \infty }{ \bigl(K_{n}^{(S)} \mathfrak{h}-\mathfrak{h} \bigr)\mathfrak{p}_{n}t^{n}} \Biggr\Vert }=0.$$
(2.2)

### Proof

Clearly, from (2.2) follows (2.1). Now we show the converse that (2.1) implies (2.2). Let $$\mathfrak{h}\in C[0,M]$$. Then there exists a constant $$K>0$$ such that $$|\mathfrak{h}(u)|\leq K$$ for all $$u\in [ 0,M]$$. Therefore

$$\bigl\vert \mathfrak{h}(u)-\mathfrak{h}(x) \bigr\vert \leq 2K, \quad u\in [ 0,M].$$
(2.3)

For every given $$\epsilon >0$$, there exists a $$\delta >0$$ such that

$$\bigl\vert \mathfrak{h}(u)-\mathfrak{h}(x) \bigr\vert \leq \epsilon$$
(2.4)

whenever $$|u-x|<\delta$$ for all $$u\in [ 0,M]$$. Define $$\psi \equiv \psi (u,x)=(u-x)^{2}$$. If $$|u-x|\geq \delta$$, then

$$\bigl\vert \mathfrak{h}(u)-\mathfrak{h}(x) \bigr\vert \leq \frac{2K}{\delta ^{2}}\psi (u,x).$$
(2.5)

From (2.3)–(2.5), we have $$|\mathfrak{h}(u)-\mathfrak{h}(x)|<\epsilon +\frac{2K}{\delta ^{2}}\psi (u,x)$$, namely,

$$-\epsilon -\frac{2K}{\delta ^{2}}\psi (u,x)< \mathfrak{h}(t)- \mathfrak{h}(x)< \frac{2K}{\delta ^{2}}\psi (u,x)+\epsilon.$$

By applying the operator $$K_{n}^{(S)}(1,x)$$, $$K_{n}^{(S)}(1,x)$$ is a monotone and linear operator, we obtain

$$K_{n}^{(S)}(1,x) \biggl( -\epsilon -\frac{2K}{\delta ^{2}}\psi \biggr) < K_{n}^{(S)}(1,x) \bigl( \mathfrak{h}(u)- \mathfrak{h}(x) \bigr) < K_{n}^{(S)}(1,x) \biggl( \frac{2K}{\delta ^{2}}\psi +\epsilon \biggr),$$

which implies

\begin{aligned} -\epsilon K_{n}^{(S)}(1,x)-\frac{2K}{\delta ^{2}}K_{n}^{(S)} \bigl(\psi (u),x \bigr)&< K_{n}^{(S)}( \mathfrak{h},x)- \mathfrak{h}(x)K_{n}^{(S)}(1,x) \\ &< \frac{2K}{\delta ^{2}}K_{n}^{(S)} \bigl(\psi (u),x \bigr)+\epsilon K_{n}^{(S)}(1,x). \end{aligned}
(2.6)

On the other hand

$$K_{n}^{(S)}(\mathfrak{h},x)-\mathfrak{h}(x)=K_{n}^{(S)}( \mathfrak{h},x)-\mathfrak{h}(x)K_{n}^{(S)}(1,x)+ \mathfrak{h}(x) \bigl[K_{n}^{(S)}(1,x)-1 \bigr].$$
(2.7)

From (2.6) and (2.7) we get

$$K_{n}^{(S)}(\mathfrak{h},x)-\mathfrak{h}(x)< \frac{2K}{\delta ^{2}}K_{n}^{(S)} \bigl(\psi (u),x \bigr)+\epsilon K_{n}^{(S)}(1,x)+\mathfrak{h}(x) \bigl[K_{n}^{(S)}(1,x)-1 \bigr].$$
(2.8)

Now we estimate the following expression:

\begin{aligned} K_{n}^{(S)} \bigl(\psi (u),x \bigr) &= K_{n}^{(S)} \bigl((x-u)^{2},x \bigr)=K_{n}^{(S)} \bigl( \bigl(x^{2}-2xu+u^{2} \bigr),x \bigr) \\ &= x^{2}K_{n}^{(S)}(1,x)-2xK_{n}^{(S)}(u,x)+K_{n}^{(S)} \bigl(u^{2},x \bigr). \end{aligned}

By (2.8), we obtain

\begin{aligned} K_{n}^{(S)}(\mathfrak{h},x)-\mathfrak{h}(x) < {}& \frac{2K}{\delta ^{2}} \bigl\{ x^{2} \bigl[K_{n}^{(S)}(1,x)-1 \bigr]-2x \bigl[K_{n}^{(S)}(u,x)-x \bigr] \\ &{}+ \bigl[K_{n}^{(S)} \bigl(u^{2},x \bigr)-x^{2} \bigr] \bigr\} +\epsilon K_{n}^{(S)}(1,x)+f(x) \bigl[K_{n}^{(S)}(1,x)-1 \bigr] \\ ={}&\epsilon +\epsilon \bigl[ K_{n}^{(S)}(1,x)-1 \bigr]+ \mathfrak{h}(x) \bigl[K_{n}^{(S)}(1,x)-1 \bigr] \\ &{}+\frac{2K}{\delta ^{2}} \bigl\{ x^{2} \bigl[K_{n}^{(S)}(1,x)-1 \bigr]-2x \bigl[K_{n}^{(S)}(u,x)-x \bigr]+ \bigl[K_{n}^{(S)} \bigl(u^{2},x \bigr)-x^{2} \bigr] \bigr\} . \end{aligned}

Therefore,

\begin{aligned} \bigl\vert K_{n}^{(S)}(\mathfrak{h},x)-\mathfrak{h}(x) \bigr\vert \leq {}&\epsilon + \biggl( \epsilon +K+\frac{2KM^{2}}{\delta ^{2}} \biggr) \bigl\vert K_{n}^{(S)}(1,x)-1 \bigr\vert \\ &{}+\frac{4KM}{\delta ^{2}} \bigl\vert K_{n}^{(S)}(u,x)-x \bigr\vert + \frac{2K}{\delta ^{2}} \bigl\vert K_{n}^{(S)} \bigl(u^{2},x \bigr)-x^{2} \bigr\vert . \end{aligned}

From the above relations and the linearity of $$K_{n}^{(S)}$$, we obtain

\begin{aligned} &\frac{1}{\mathfrak{p}(v)} \Biggl\Vert \sum_{n=0}^{\infty }{ \bigl(U_{n,p}( \mathfrak{h};x)-\mathfrak{h}(x) \bigr) \mathfrak{p}_{n}v^{n}} \Biggr\Vert \\ &\quad\leq \epsilon + \biggl( \epsilon +K+\frac{2KM^{2}}{\delta ^{2}} \biggr) \frac{1}{\mathfrak{p}(t)} \Biggl\Vert \sum_{n=0}^{\infty }{ \bigl(K_{n}^{(S)}(1;x)-1 \bigr))\mathfrak{p}_{n}t^{n}} \Biggr\Vert \\ &\qquad{}+\frac{4KM}{\delta ^{2}}\frac{1}{\mathfrak{p}(v)} \Biggl\Vert \sum _{n=0}^{ \infty }{ \bigl(K_{n}^{(S)}(u;x)-x \bigr))\mathfrak{p}_{n}v^{n}} \Biggr\Vert +\frac{2K}{\delta ^{2}}\frac{1}{\mathfrak{p}(v)} \Biggl\Vert \sum _{n=0}^{ \infty }{ \bigl(K_{n}^{(S)} \bigl(u^{2};x \bigr)-x^{2} \bigr))\mathfrak{p}_{n}v^{n}} \Biggr\Vert . \end{aligned}

Hence, (2.2) follows from the last relation and (2.1). □

## 3 Rate of convergence

The modulus of continuity is defined by

$$\omega (\mathfrak{h},\delta )=\sup_{ \vert h \vert < \delta }{ \bigl\vert \mathfrak{h}(x+h)-\mathfrak{h}(x) \bigr\vert }, \quad \mathfrak{h}(x) \in C[0,M]\cap E.$$

Note that

$$\bigl\vert \mathfrak{h}(x)-\mathfrak{h}(y) \bigr\vert \leq \omega ( \mathfrak{h},\delta ) \biggl( \frac{ \vert x-y \vert }{\delta }+1 \biggr).$$
(3.1)

### Theorem 3.1

Let $$\mathfrak{T}=(\mathfrak{a}_{ij})$$ be regular and $$\mathfrak{h}\in C[0,M]$$. If $$(\alpha _{n})$$ is a sequence of positive real numbers such that $$\omega (\mathfrak{h},\delta _{n})=st_{\mathfrak{T}}-0 ( {\alpha _{n}} )$$, then

$$\bigl\Vert K_{n}^{(S)}\mathfrak{h}-\mathfrak{h} \bigr\Vert =st_{\mathfrak{T}}-0(\alpha _{n}),$$

where

\begin{aligned} \delta _{n} ={}& \biggl\{ \frac{1}{3}\cdot \biggl\Vert \frac{3aa_{0,2}+3aa_{0,1}+3aa_{0,2}+6aa_{1,1}+3aa_{1,0}+aa_{0,0}}{aa_{0,0}n^{2}} \biggr\Vert \\ &{} +M \biggl[ \biggl\Vert \frac{1}{2}\cdot \frac{aa_{0,0}hh_{0,2}+2aa_{0,0}hh_{1,1}+aa_{0,0}hh_{2,0}+2aa_{0,0}+4aa_{0,1}+4aa_{1,0}}{aa_{0,0}n} \biggr\Vert \\ &{} + \biggl\Vert \frac{2aa_{0,1}+2aa_{1,0}+aa_{0,0}}{aa_{0,0}n} \biggr\Vert \biggr] \biggr\} ^{2}, \end{aligned}

for any positive integer n.

### Proof

By (3.1), we see

\begin{aligned} &\bigl\vert K_{n}^{(S)}(\mathfrak{h};x)-\mathfrak{h} \bigr\vert \\ &\quad \leq K_{n}^{(S)} \bigl( \bigl\vert \mathfrak{h}(t)-\mathfrak{h}(x) \bigr\vert ;x \bigr)\\ &\quad\leq \frac{ne^{-\frac{nx}{2}H(1,1)}}{A(1,1)}\sum_{k_{1}=0}^{\infty } \sum_{k_{2}=0}^{\infty } \frac{S_{k_{1},k_{2}} ( \frac{nx}{2} ) }{k_{1}!k_{2}!} \int _{ \frac{k_{1}+k_{2}}{n}}^{\frac{k_{1}+k_{2}+1}{n}}{\omega (\mathfrak{h},\delta ) \biggl( 1+ \frac{ \vert t-x \vert }{\delta } \biggr) }\,dt \\ &\quad \leq \omega (\mathfrak{h},\delta ) \Biggl[ 1+\frac{1}{\delta } \frac{ne^{-\frac{nx}{2}H(1,1)}}{A(1,1)}\sum_{k_{1}=0}^{\infty }\sum _{k_{2}=0}^{ \infty }\frac{S_{k_{1},k_{2}} ( \frac{nx}{2} ) }{k_{1}!k_{2}!} \int _{\frac{k_{1}+k_{2}}{n}}^{\frac{k_{1}+k_{2}+1}{n}} \bigl( \vert t-x \vert \bigr)\,dt \Biggr],\quad \text{see } \\ &\quad =\omega (\mathfrak{h},\delta ) \biggl[ 1+\frac{1}{\delta }K_{n}^{(S)} \bigl( \vert t-x \vert ;x \bigr) \biggr]. \end{aligned}

By applying the Cauchy–Schwartz inequality, we have

$$\bigl\vert K_{n}^{(S)}(\mathfrak{h};x)-\mathfrak{h} \bigr\vert \leq \omega (\mathfrak{h}, \delta ) \biggl[ 1+ \frac{1}{\delta } \bigl( K_{n}^{(S)} \bigl( \vert t-x \vert ^{2};x \bigr) \bigr) ^{ \frac{1}{2}} \biggr].$$

From Example 1.1 of , we obtain

\begin{aligned} &K_{n}^{(S)} \bigl((u-x)^{2};x \bigr)\\ &\quad=K_{n}^{(S)}(e_{2};x)-2xK_{n}^{(S)}(e_{1};x)+x^{2}K_{n}^{(S)}(e_{0};x) \\ &\quad\leq \frac{1}{3}\cdot \biggl\Vert \frac{3aa_{0,2}+3aa_{0,1}+3aa_{0,2}+6aa_{1,1}+3aa_{1,0}+aa_{0,0}}{aa_{0,0}n^{2}} \biggr\Vert \\ &\qquad{}+ M \biggl[ \biggl\Vert \frac{1}{2}\cdot \frac{aa_{0,0}hh_{0,2}+2aa_{0,0}hh_{1,1}+aa_{0,0}hh_{2,0}+2aa_{0,0}+4aa_{0,1}+4aa_{1,0}}{aa_{0,0}n} \biggr\Vert \\ &\qquad{}+ \biggl\Vert \frac{2aa_{0,1}+2aa_{1,0}+aa_{0,0}}{aa_{0,0}n} \biggr\Vert \biggr]. \end{aligned}

By taking

\begin{aligned} \delta _{n} ={}& \biggl\{ \frac{1}{3}\cdot \biggl\Vert \frac{3aa_{0,2}+3aa_{0,1}+3aa_{0,2}+6aa_{1,1}+3aa_{1,0}+aa_{0,0}}{aa_{0,0}n^{2}} \biggr\Vert \\ &{}+ M \biggl[ \biggl\Vert \frac{1}{2}\cdot \frac{aa_{0,0}hh_{0,2}+2aa_{0,0}hh_{1,1}+aa_{0,0}hh_{2,0}+2aa_{0,0}+4aa_{0,1}+4aa_{1,0}}{aa_{0,0}n} \biggr\Vert \\ &{} + \biggl\Vert \frac{2aa_{0,1}+2aa_{1,0}+aa_{0,0}}{aa_{0,0}n} \biggr\Vert \biggr] \biggr\} ^{2}, \end{aligned}

we get $$\Vert K_{n}^{(S)}\mathfrak{h}-\mathfrak{h} \Vert \leq 2\cdot \omega (\mathfrak{h},\delta _{n})$$. Therefore, for every $$\epsilon >0$$, we have

$$\frac{1}{\alpha _{n}}\sum_{ \Vert K_{n}^{(S)}\mathfrak{h}-\mathfrak{h} \Vert \geq \epsilon }\mathfrak{t} {_{nj}}\leq \frac{1}{\alpha _{n}}\sum_{2 \cdot \omega (f,\delta _{n})\geq {\epsilon }} \mathfrak{t} {_{nj}}.$$

From the conditions that are given in the theorem, we have $$\Vert K_{n}^{(S)}\mathfrak{h}-\mathfrak{h} \Vert =st_{\mathfrak{T}}-0(\alpha _{i})$$, as claimed. □

Now, we obtain the rate of convergence for our method.

### Theorem 3.2

Let $$\mathfrak{h}\in C[0,M]$$ and let ϕ be a positive real function defined on $$(0,M)$$. If $$\omega (\mathfrak{h},\psi (u))=O(\phi (u))$$, as $$v\rightarrow R^{-}$$, then we have

$${\frac{1}{\mathfrak{p}(v)} \Biggl\Vert \sum_{n=0}^{\infty }{ \bigl(K_{n}^{(S)}e_{i}-e_{i} \bigr) \mathfrak{p}_{n}v^{n}} \Biggr\Vert }=O \bigl(\phi (v) \bigr),$$

where the function $$\psi:(0,M)\rightarrow \mathbb{R}$$ is defined by the relation

$$\psi (u)= \Bigl\{ \sup_{\substack{ x\in (0,M) \\ n\in \mathbb{N}}} \bigl\{ K_{n}^{(S)} \bigl((u-x)^{2};x \bigr) \bigr\} \Bigr\} ^{\frac{1}{2}}.$$

### Proof

For any $$u\in (0,R)$$, $$x\in (0,M)$$ and $$\delta >0$$, we have

\begin{aligned} & \Biggl\vert \sum_{n=0}^{\infty }{ \bigl[K_{n}^{(S)}(\mathfrak{h};x)- \mathfrak{h}(x) \bigr]\mathfrak{p}_{n}v^{n}} \Biggr\vert \\ &\quad\leq \sum_{n=0}^{\infty }{K_{n}^{(S)} \bigl( \bigl\vert \mathfrak{h}(u)-\mathfrak{h}(x) \bigr\vert ;x \bigr)\mathfrak{p}_{n}v^{n}} \\ & \quad\leq \sum_{n=0}^{\infty }{K_{n}^{(S)} \biggl( \omega \biggl( \mathfrak{h},\frac{ \vert u-x \vert }{\delta }\delta \biggr);x \biggr) \mathfrak{p}_{n}v^{n}} \leq \sum _{n=0}^{\infty }{K_{n}^{(S)} \biggl( \biggl( 1+ \biggl[ \biggl\vert \frac{u-x}{\delta } \biggr\vert \biggr] \biggr) \omega (\mathfrak{h},\delta );x \biggr) \mathfrak{p}_{n}v^{n}} \\ &\quad \leq \omega (\mathfrak{h},\delta )\sum_{n=0}^{\infty }{K_{n}^{(S)} \biggl( 1+\frac{(u-x)^{2}}{\delta ^{2}};x \biggr) \mathfrak{p}_{n}v^{n}} \\ &\quad\leq \omega (\mathfrak{h},\delta )\sum_{n=0}^{\infty }{K_{n}^{(S)} \bigl(e_{0}(u);x \bigr) \mathfrak{p}_{n}v^{n}} +\frac{\omega (\mathfrak{h},\delta )}{\delta ^{2}}\sum_{n=0}^{ \infty }{K_{n}^{(S)} \bigl((u-x)^{2};x \bigr)\mathfrak{p}_{n}v^{n}}\\ &\quad=p(v) \omega ( \mathfrak{h},\delta )+\frac{\omega (\mathfrak{h},\delta )}{\delta ^{2}}\sup _{ \substack{ 0\leq x\leq 1 \\ n\in \mathbb{N}}} \bigl\{ K_{n}^{(S)} \bigl((u-x)^{2};x \bigr) \bigr\} \sum_{n=0}^{\infty } \mathfrak{p} {_{n}v^{n}}, \end{aligned}

$$\Biggl\vert \sum_{n=0}^{\infty }{ \bigl[K_{n}^{(S)}(f;x)-f(x) \bigr]\mathfrak{p}_{n}v^{n}} \Biggr\vert \leq \mathfrak{p}(v)\omega (f,\delta )+ \frac{\omega (f,\delta )}{\delta ^{2}}\sup _{0\leq x\leq 1} \bigl\{ K_{n}^{(S)} \bigl((u-x)^{2};x \bigr) \bigr\} \mathfrak{p}(v).$$

If we set $$\delta =\psi (u)$$, then from the last inequality we have

$$0\leq \frac{1}{\mathfrak{p}(v)} \Biggl\Vert \sum_{n=0}^{\infty }{ \bigl(K_{n}^{(S)}\mathfrak{h}-\mathfrak{h} \bigr) \mathfrak{p}_{n}v^{n}} \Biggr\Vert \leq 2 \omega (\mathfrak{h},\delta ),$$

as required. □

## 4 Voronovskaya type theorems

It is well known that there is a Voronovskaya type theorem for the Kantorovich type generalization of Szász operators involving Sheffer type polynomials and it is stated as follows.

### Theorem 4.1

()

For $$f\in C_{B}[0,\infty )$$,

$$\lim_{n\to \infty }n \bigl[K_{n}^{(S)} \bigl(f(t),x \bigr)-f(x) \bigr]=f^{{\prime }}(x) \biggl[ \frac{2\tilde{a}_{0,1}+2\tilde{a}_{1,0}+\tilde{a}_{0,0}}{2\tilde{a}_{0,0}} \biggr]+ \frac{f^{{\prime \prime }}(x)}{2} \biggl[ (\tilde{h}_{0,2}+2 \tilde{h}_{1,1}+\tilde{h}_{2,0})\frac{x}{2} \biggr],$$

for every $$x\in [0,M]$$ and any finite M.

We extend the Voronovskaya type theorem for the $$\mathfrak{T}$$-statistical method for these operators. Let us consider the following operators.

### Example 4.2

Define the operators

$$NB_{n}(h,x)=(1+u_{n})K_{n}^{(S)}(h,x),$$

where

$$u_{n}=\textstyle\begin{cases} \frac{1}{m^{3}} & m^{2}-m\leq n\leq m^{2}-1, \\ \frac{1}{m^{4}} & n=m^{2};m\in \mathbb{N}\setminus \{1\}, \\ 0 & \text{otherwise}.\end{cases}$$

### Lemma 4.3

Let $$\mathfrak{h}\in C[0,M]$$ such that $$\mathfrak{h}^{{\prime }},\mathfrak{h}^{{\prime \prime }}\in C[0,M]$$, $$x\in [ 0,M]$$. Then we obtain

$$n^{2}NB_{n}^{(S)} \bigl((y-x)^{4};x \bigr)\sim E_{4}x^{2}(st_{\mathfrak{T}}) \quad\textit{on } [ 0,M].$$

### Proof

It follows directly from Remark 2.6 given in . □

### Theorem 4.4

Let $$\mathfrak{h}\in C[0,M]$$ such that $$\mathfrak{h}^{{\prime }},\mathfrak{h}^{{\prime \prime }}\in C[0,M]$$, $$x\in [ 0,M]$$, for any finite M. Then

$$n \bigl[NB_{n}^{(S)}(\mathfrak{h};x)-\mathfrak{h}(x) \bigr] \sim \mathfrak{h}^{{\prime }}(x) \biggl[ \frac{2\tilde{a}_{0,1}+2\tilde{a}_{1,0}+\tilde{a}_{0,0}}{2\tilde{a}_{0,0}} \biggr] + \frac{\mathfrak{h}^{{\prime \prime }}(x)}{2} \biggl(\frac{x(\tilde{h}_{0,2}+2\tilde{h}_{1,1}+\tilde{h}_{2,0})}{2} \biggr) (st_{T}),$$

on $$[ 0,M]$$.

### Proof

Taylor’s formula gives

$$\mathfrak{h}(y)=\mathfrak{h}(x)+(y-x)\mathfrak{h}^{{\prime }}(x)+ \frac{1}{2}(y-x)^{2}\mathfrak{h}^{{\prime \prime }}(x)+(y-x)^{2} \psi (y-x),$$
(4.1)

where $$\psi (y-x)\rightarrow 0$$, as $$y-x\rightarrow 0$$. After applying $$NB_{n}^{(S)}$$ on both sides of Eq. (4.1), we obtain

\begin{aligned} NB_{n}^{(S)}(\mathfrak{h}) ={}&(1+u_{n}) \mathfrak{h}(x)+(1+u_{n}) \mathfrak{h}^{{\prime }}(x) \biggl( (2\tilde{a}_{0,1}+2\tilde{a}_{1,0}+\tilde{a}_{0,0})\frac{1}{2\tilde{a}_{0,0}n} \biggr) \\ &{} +(1+u_{n})\frac{\mathfrak{h}^{{\prime \prime }}(x)}{2} \biggl((3 \tilde{a}_{0,2}+6 \tilde{a}_{1,1}+3\tilde{a}_{1,0}+3\tilde{a}_{2,0}+3 \tilde{a}_{0,1}+\tilde{a}_{0,0})\frac{1}{3\tilde{a}_{0,0}n^{2}} \\ & {}+(\tilde{h}_{0,2}+2\tilde{h}_{1,1}+\tilde{h}_{2,0}) \frac{x}{2n} \biggr)+(1+x_{n})NB_{n}^{(S)} \bigl(\Phi ^{2}\psi (y-x);x \bigr). \end{aligned}

This yields

\begin{aligned} nNB_{n}^{(S)}(\mathfrak{h})={}& n(1+u_{n}) \mathfrak{h}(x)+n(1+u_{n}) \mathfrak{h}^{{\prime }}(x) \biggl( (2\tilde{a}_{0,1}+2\tilde{a}_{1,0}+\tilde{a}_{0,0})\frac{1}{2\tilde{a}_{0,0}n} \biggr) \\ &{} +n(1+u_{n})\frac{\mathfrak{h}^{{\prime \prime }}(x)}{2} \biggl((3 \tilde{a}_{0,2}+6 \tilde{a}_{1,1}+3\tilde{a}_{1,0}+3\tilde{a}_{2,0}+3 \tilde{a}_{0,1}+\tilde{a}_{0,0})\frac{1}{3\tilde{a}_{0,0}n^{2}} \\ &{} +(\tilde{h}_{0,2}+2\tilde{h}_{1,1}+\tilde{h}_{2,0}) \frac{x}{2n} \biggr)+n(1+u_{n})NB_{n}^{(S)} \bigl(\Phi ^{2}\psi (y-x);x \bigr). \end{aligned}

Therefore,

\begin{aligned} & \biggl|n [NB_{n}^{(S)}(\mathfrak{h};x)-\mathfrak{h}(x)- \mathfrak{h}^{{\prime }}(x) \biggl[ (2\tilde{a}_{0,1}+2 \tilde{a}_{1,0}+\tilde{a}_{0,0})\frac{1}{2\tilde{a}_{0,0}} \biggr] \\ &\qquad{} -\frac{\mathfrak{h}^{{\prime \prime }}(x)}{2} \biggl((\tilde{h}_{0,2}+2\tilde{h}_{1,1}+\tilde{h}_{2,0})\frac{x}{2} \biggr)\biggr| \\ &\quad \leq nKu_{n}+nK_{1}u_{n} \biggl\vert \frac{2\tilde{a}_{0,1}+2\tilde{a}_{1,0}+\tilde{a}_{0,0}}{2\tilde{a}_{0,0}n} \biggr\vert \\ &\qquad{} +n\frac{K_{2}}{2} \biggl\vert \frac{3\tilde{a}_{0,2}+6\tilde{a}_{1,1}+3\tilde{a}_{1,0}+3\tilde{a}_{2,0}+3\tilde{a}_{0,1}+\tilde{a}_{0,0}}{3\tilde{a}_{0,0}n^{2}} \biggr\vert \\ &\qquad{} +nu_{n}\frac{K_{2}}{2} \biggl\vert \frac{3\tilde{a}_{0,2}+6\tilde{a}_{1,1}+3\tilde{a}_{1,0}+3\tilde{a}_{2,0}+3\tilde{a}_{0,1}+\tilde{a}_{0,0}}{3\tilde{a}_{0,0}n^{2}}+( \tilde{h}_{0,2}+2\tilde{h}_{1,1}+ \tilde{h}_{2,0}) \frac{x}{2n} \biggr\vert \\ & \qquad{} +n \bigl\vert NB_{n}^{(S)} \bigl((y-x)^{2} \psi (y-x);x \bigr) \bigr\vert +u_{n}n \bigl\vert NB_{n}^{(S)} \bigl((y-x)^{2} \psi (y-x);x \bigr) \bigr\vert , \end{aligned}

where $$K=\sup_{x\in [ 0,M]}{|\mathfrak{h}(x)|}$$, $$K_{1}=\sup_{x\in [ 0,M]}{|\mathfrak{h}^{{\prime }}(x)|}$$ and $$K_{2}=\sup_{x\in [ 0,M]}{|\mathfrak{h}^{{\prime \prime }}(x)|}$$.

Now we have to prove that

$$\lim_{n\rightarrow \infty }{n \bigl\vert NB_{n}^{(S)} \bigl((y-x)^{2}\psi (y-x);x \bigr) \bigr\vert }=0.$$

By applying the Cauchy–Schwartz inequality, we obtain

$$n \bigl\vert NB_{n}^{(S)} \bigl((y-x)^{2}\psi (y-x);x \bigr) \bigr\vert \leq \bigl[ n^{2}NB_{n}^{(S)} \bigl((y-x)^{4};x \bigr) \bigr] ^{\frac{1}{2}}\cdot \bigl[ NB_{n}^{(S)} \bigl(\psi ^{2};x \bigr) \bigr]^{ \frac{1}{2}}.$$
(4.2)

Also, by setting $$\eta _{x}(y)=(\psi (y-x))^{2}$$, we have $$\eta _{x}(x)=0$$ and $$\eta _{x}(\cdot )\in C[0,M]$$. So

$$NB_{n}^{(S)}(\eta _{x})\rightarrow 0(st_{\mathfrak{T}}) \quad\text{on } [ 0,M].$$
(4.3)

Now from the previous relation, (4.2), (4.3), and Lemma 4.3, we obtain

$$n^{2}NB_{n}^{(S)} \bigl((y-x)^{2}\psi (y-x);x \bigr)\rightarrow 0(st_{ \mathfrak{T}}) \quad\text{on } [ 0,M].$$
(4.4)

From the construction of $$(u_{n})$$, it follows that $$nu_{n}\rightarrow 0(st_{\mathfrak{T}})$$ on $$[0,M]$$.

For a given $$\epsilon >0$$, we define the sets

\begin{aligned} A ={}&\biggl| \{n:|n [NB_{n}^{(S)}(\mathfrak{h};x)- \mathfrak{h}(x)-\mathfrak{h}^{{\prime }}(x) \biggl[ (2 \tilde{a}_{0,1}+2\tilde{a}_{1,0}+ \tilde{a}_{0,0}) \frac{1}{2\tilde{a}_{0,0}} \biggr] \\ & {}-\frac{\mathfrak{h}^{{\prime \prime }}(x)}{2} \biggl((\tilde{h}_{0,2}+2\tilde{h}_{1,1}+\tilde{h}_{2,0})\frac{x}{2} \biggr)\biggr|, \\ A_{1} = {}& \biggl\vert \biggl\{ n: \vert nu_{n} \vert \geq \frac{\epsilon }{3K} \biggr\} \biggr\vert , \\ A_{2} ={}& \biggl\vert \biggl\{ n: \bigl\vert nNB_{n}^{(S)} \bigl((y-x)^{2}\psi (y-x);x \bigr) \bigr\vert \geq \frac{\epsilon }{3} \biggr\} \biggr\vert , \\ A_{3} ={}& \biggl\vert \biggl\{ n: \bigl\vert nu_{n}NB_{n}^{(S)} \bigl((y-x)^{2}\psi (y-x);x \bigr) \bigr\vert \geq \frac{\epsilon }{3} \biggr\} \biggr\vert . \end{aligned}

From these relations we obtain $$A\leq A_{1}+A_{2}+A_{3}$$. Hence the result follows. □

### Theorem 4.5

Let $$\mathfrak{h},\mathfrak{h}^{{\prime }},\mathfrak{h}^{{\prime \prime }}\in C[0,\infty )$$. Then

\begin{aligned} & \biggl\vert n \bigl( K_{n}^{(S)}(\mathfrak{h},x)- \mathfrak{h}(x) \bigr) -\mathfrak{h}^{{\prime }}(x) \biggl( (2 \tilde{a}_{0,1}+2\tilde{a}_{1,0}+ \tilde{a}_{0,0}) \frac{1}{2\tilde{a}_{0,0}n} \biggr) \\ &\qquad{} -\frac{\mathfrak{h}^{{\prime \prime }}(x)}{2}\cdot \biggl[(3 \tilde{a}_{0,2}+6 \tilde{a}_{1,1}+3\tilde{a}_{1,0}+3\tilde{a}_{2,0}+3 \tilde{a}_{0,1}+\tilde{a}_{0,0})\frac{1}{3\tilde{a}_{0,0}n^{2}} \\ &\qquad{} +(\tilde{h}_{0,2}+2\tilde{h}_{1,1}+ \tilde{h}_{2,0}) \frac{x}{2n} \biggr] \biggr\vert \\ &\quad=0(1)\omega \bigl( \mathfrak{h}^{{\prime \prime }},n^{-\frac{1}{2}} \bigr), \end{aligned}

as $$n\rightarrow \infty$$, and for every $$x\in [ 0,M]$$, for any finite M.

### Proof

From Taylor’s theorem, we have

$$\mathfrak{h}(u)=\mathfrak{h}(x)+\mathfrak{h}^{{\prime }}(x) (u-x)+ \frac{\mathfrak{h}^{{\prime \prime }}(x)}{2}(u-x)^{2}+R(u,x),$$

where $$R(u,x)= \frac{\mathfrak{h}^{{\prime \prime }}(\theta )-\mathfrak{h}^{{\prime \prime }}(x)}{2}(u-x)^{2}$$, for $$\theta \in (u,x)$$. Now we obtain

$$\biggl\vert K_{n}^{(S)}(\mathfrak{h},x)-\mathfrak{h}(x)- \mathfrak{h}^{{\prime }}(x)K_{n}^{(S)} \bigl((u-x);x \bigr)- \frac{\mathfrak{h}^{{\prime \prime }}(x)}{2}K_{n}^{(S)} \bigl((u-x)^{2};x \bigr) \biggr\vert \leq K_{n}^{(S)} \bigl( \bigl\vert R(u,x) \bigr\vert ,x \bigr).$$

From this we get

\begin{aligned} & \biggl|n \bigl( K_{n}^{(S)}(\mathfrak{h},x)-\mathfrak{h}(x) \bigr) -\mathfrak{h}^{{\prime }}(x) \biggl( \frac{2\tilde{a}_{0,1}+2\tilde{a}_{1,0}+\tilde{a}_{0,0}}{2\tilde{a}_{0,0}} \biggr) \\ &\qquad{}- \frac{\mathfrak{h}^{{\prime \prime }}(x)}{2}\cdot [(3 \tilde{a}_{0,2}+6 \tilde{a}_{1,1}+3\tilde{a}_{1,0}+3\tilde{a}_{2,0}+3 \tilde{a}_{0,1}+\tilde{a}_{0,0})\frac{1}{3\tilde{a}_{0,0}n}\\ &\qquad{}+( \tilde{h}_{0,2}+2 \tilde{h}_{1,1}+ \tilde{h}_{2,0})\frac{x}{2}\biggr| \\ &\quad \leq n\cdot K_{n}^{(S)} \bigl( \bigl\vert R(u,x) \bigr\vert ,x \bigr). \end{aligned}

By the properties of the continuity modulus, we have

$$\biggl\vert \frac{\mathfrak{h}^{{\prime \prime }}(\theta )-\mathfrak{h}^{{\prime \prime }}(x)}{2!} \biggr\vert \leq \frac{1}{2!} \biggl( 1+ \frac{ \vert \theta -x \vert }{\delta } \biggr) \omega \bigl(\mathfrak{h}^{{\prime \prime }}, \delta \bigr).$$

On the other hand

$$\biggl\vert \frac{\mathfrak{h}^{{\prime \prime }}(\theta )-\mathfrak{h}^{{\prime \prime }}(x)}{2!} \biggr\vert \leq\textstyle\begin{cases} \omega (\mathfrak{h}^{{\prime \prime },\delta }); & \vert u-x \vert \leq \delta, \\ \frac{(t-x)^{4}}{\delta ^{4}}\omega (\mathfrak{h}^{{\prime \prime }}, \delta ); & \vert u-x \vert \geq \delta.\end{cases}$$

For $$0<\delta <1$$, we obtain

$$\biggl\vert \frac{\mathfrak{h}^{{\prime \prime }}(\theta )-\mathfrak{h}^{{\prime \prime }}(x)}{2!} \biggr\vert \leq \omega \bigl( \mathfrak{h}^{{ \prime \prime }},\delta \bigr) \biggl( 1+\frac{(u-x)^{4}}{\delta ^{4}} \biggr),$$

which gives

$$\bigl\vert R(u,x) \bigr\vert \leq \omega \bigl(\mathfrak{h}^{{\prime \prime }}, \delta \bigr) \biggl( 1+ \frac{(u-x)^{4}}{\delta ^{4}} \biggr) (u-x)^{2}=\omega \bigl(\mathfrak{h}^{{ \prime \prime }},\delta \bigr) \biggl( (u-x)^{2}+ \frac{(u-x)^{6}}{\delta ^{4}} \biggr).$$

By the linearity of $$K_{n}^{(S)}$$ and the above relation we obtain

$$K_{n}^{(S)} \bigl( \bigl\vert R(u,x) \bigr\vert ,x \bigr) \leq \omega \bigl(\mathfrak{h}^{{\prime \prime }}, \delta \bigr) \biggl( K_{n}^{(S)} \bigl((u-x)^{2},x \bigr)+ \frac{1}{\delta ^{4}}K_{n}^{(S)} \bigl((u-x)^{6},x \bigr) \biggr).$$

Taking into consideration Remark 2.6 in , for every $$x\in [0,M]$$, we have

$$K_{n}^{(S)} \bigl( \bigl\vert R(u,x) \bigr\vert ,x \bigr) \leq \omega \bigl(\mathfrak{h}^{{\prime \prime }}, \delta \bigr) \biggl( O \biggl( \frac{1}{n} \biggr) +\frac{1}{\delta ^{4}}O \biggl( \frac{1}{n^{3}} \biggr) \biggr) =O \biggl( \frac{1}{n} \biggr) \omega \bigl( \mathfrak{h}^{{\prime \prime }}, \delta \bigr).$$

For $$\delta =n^{-\frac{1}{2}}$$, we complete the proof. □

## Availability of data and materials

No data were used to support this study.

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