Korovkin type approximation theorems for weighted $\alpha \beta$-statistical convergence

Abstract

The concept of $\alpha \beta$-statistical convergence was introduced and studied by Aktuğlu (Korovkin type approximation theorems proved via $\alpha \beta$-statistical convergence, J Comput Appl Math 259:174–181, 2014). In this work, we generalize the concept of $\alpha \beta$-statistical convergence and introduce the concept of weighted $\alpha \beta$-statistical convergence of order $\gamma$, weighted $\alpha \beta$-summability of order $\gamma$, and strongly weighted $\alpha \beta$-summable sequences of order $\gamma$. We also establish some inclusion relation, and some related results for these new summability methods. Furthermore, we prove Korovkin type approximation theorems through weighted $\alpha \beta$-statistical convergence and apply the classical Bernstein operator to construct an example in support of our result.

1 Introduction, notations and known results

Let $K$ be a subset of $\mathbb {N}$, the set of natural numbers and $K_{n}=\left\{ k\le n: k\in K \right\}$. The natural density of $K$ is defined by $\delta (K)=\lim _{n}\frac{1}{n}|K_{n}|$ provided it exists, where $|K_{n}|$ denotes the cardinality of set $K_{n}$. A sequence $x=(x_{k})$ is called statistically convergent ($st$-convergent) to the number $\ell$, denoted by $st-\lim x=\ell$, for each $\epsilon >0,$ the set $K_{\epsilon }=\left\{ k\in \mathbb {N}:|x_{k}-\ell |\ge \epsilon \right\}$ has natural density zero, that is

$$\begin{aligned} \lim _{n \rightarrow \infty }\frac{1}{n}\left| \{k\le n: |x_{k}-\ell |\ge \epsilon \}\right| =0. \end{aligned}$$

The concept of statistical convergence has been defined by Fast [5] and studied by many other authors. It is well know that every statistically convergent sequence is ordinary convergent, but the converse is not true. For example, $y=(y_{k})$ is defined as follows;

$$\begin{aligned} y=(y _{k})=\left\{ \begin{array}{ll} 1\text {,} \quad k=m^{2}\text {,} \\ 0\text {,} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

It is clear that the sequence $y=(y_{k})$ is statistical convergent to zero but not convergent. The idea $\alpha \beta$-statistical convergence was introduced by Aktuğlu in [7] as follows:

Let $\alpha (n)$ and $\beta (n)$ be two sequences positive number which satisfy the following conditions

• (i) $\alpha$ and $\beta$ are both non-decreasing,

• (ii) $\beta (n)\ge \alpha (n)$,

• (iii) $\beta (n)-\alpha (n)\longrightarrow \infty$ as $n\longrightarrow \infty$

and let $\Lambda$ denote the set of pairs $(\alpha ,\beta )$ satisfying (i)–(iii). For each pair $(\alpha ,\beta )\in \Lambda$, $0<\gamma \le 1$ and $K\subset \mathbb {N}$, we define $\delta ^{\alpha ,\beta }(K,\gamma )$ in the following way

$$\begin{aligned} \delta ^{\alpha ,\beta }(K,\gamma )=\lim _{n \rightarrow \infty }\frac{\left| K\cap P^{\alpha ,\beta }_{n}\right| }{(\beta (n)-\alpha (n)+1)^{\gamma }}, \end{aligned}$$

where $P^{\alpha ,\beta }_{n}$ in the closed interval $[\alpha (n),\beta (n)]$. A sequence $x=(x_{k})$ is said to be $\alpha \beta$-statistically convergent of order $\gamma$ to $\ell$ or $S^{\gamma }_{\alpha \beta }$-convergent, if

$$\begin{aligned} \delta ^{\alpha ,\beta }(\{k:|x_{k}-\ell |\ge \epsilon \},\gamma )=\lim _{n \rightarrow \infty }\frac{\left| \left\{ k\in P^{\alpha ,\beta }_{n}:|x_{k}-\ell |\ge \epsilon \right\} \right| }{(\beta (n)-\alpha (n)+1)^{\gamma }}=0. \end{aligned}$$

In this paper generalizing above idea, we define the weighted $\alpha \beta$-statistical convergence of order $\gamma$, the weighted $\alpha \beta$-summability of order $\gamma$ and the weighted $\alpha \beta$-summability.

2 The weighted $\alpha \beta$-summability

Let $s=(s_{k})$ be a sequence of non-negative real numbers such that $s_{0}>0$ and

$$\begin{aligned} S_{n}=\sum _{k\in P^{\alpha ,\beta }_{n}}s_{k}\longrightarrow \infty , \quad \text {as}\,\,n\longrightarrow \infty \,\,\text {and}\,\,z^{\gamma }_{n}(x)=\frac{1}{S_{n}^{\gamma }}\sum _{k\in P^{\alpha ,\beta }_{n}}s_{k}x_{k}. \end{aligned}$$

Definition 2.1

• (a) A sequence $x=(x_{k})$ is said to be strongly weighted $\alpha \beta$-summable of order $\gamma$ to a number $\ell$ if

  $$\begin{aligned} \lim _{n \rightarrow \infty }\frac{1}{S_{n}^{\gamma }}\sum _{k\in P^{\alpha ,\beta }_{n}}s_{k}|x_{k}-\ell |\longrightarrow 0 \quad \text {as}\,\,n\longrightarrow \infty . \end{aligned}$$

  We denote it by $x_{k}\longrightarrow \ell [\overline{N}^{\gamma }_{\alpha \beta },s]$. Similarly, for $\gamma =1$ the sequences $x=(x_{k})$ is said to be strongly weighted $\alpha \beta$-summable to $\ell$. The set of all strongly weighted $\alpha \beta$-summable of order $\gamma$ and strongly weighted $\alpha \beta$-summable sequences will be denoted $[\overline{N}^{\gamma }_{\alpha \beta },s]$ and $[\overline{N}_{\alpha \beta },s]$, respectively.

• (b) A sequence $x=(x_{k})$ is said to be weighted $\alpha \beta$-summable of order $\gamma$ to $\ell$, if $z^{\gamma }_{n}(x)\longrightarrow \ell$ as $n\longrightarrow \infty$. Similarly, for $\gamma =1$ the sequence $x=(x_{k})$ is said to be weighted $\alpha \beta$-summable to $\ell$, if $z_{n}(x)\longrightarrow \ell$ as $n\longrightarrow \infty$. The set of all weighted $\alpha \beta$-summable of order $\gamma$ and weighted $\alpha \beta$-summable sequences will be denoted $(\overline{N}^{\gamma }_{\alpha \beta },s)$ and $(\overline{N}_{\alpha \beta },s)$, respectively.

This definition includes the following special cases:

• (i) If $\gamma =1$, $\alpha (n)=0$ and $\beta (n)=n$, weighted $\alpha \beta$-summable is reduced to weighted mean summable, and $[\overline{N}_{\alpha \beta },s]$ summable sequences are reduced to $(\overline{N},p_{n})$ summable sequences introduced in [8, 14].

• (ii) Let $\lambda _{n}$ be a none-decreasing sequence of positive numbers tending to $\infty$ such that $\lambda _{n}\le \lambda _{n}+1$, $\lambda _{1}=1$. If we take $\gamma =1$, $\alpha (n)=n-\lambda _{n}+1$ and $\beta (n)=n$ then weighted $\alpha \beta$-summability is reduced to $(\overline{N}_{\lambda }; p)$-summability and $[\overline{N}_{\alpha \beta },s]$ summable sequences are reduced to $[\overline{N}_{\lambda }; p]$-summable sequences introduced in [2].

• (iii) If we take $\gamma =1$, $\alpha (n)=n-\lambda _{n}+1$, $\beta (n)=n$ and $s_{k}=1$ for all $k$ then $\alpha \beta$-summability is reduced to $(V; \lambda )$-summability introduced in [10] and $[\overline{N}_{\alpha \beta },s]$-summable sequences are reduced to $[V,\lambda ]$-summable sequences introduced in [12].

• (iv) Recall that a lacunary sequence $\theta =\{k_{r}\}$ is an increasing integer sequence such that $k_{0}=0$ and $h_{r}:=k_{r}-k_{r-1}$. If we take $\gamma =1$, $\alpha (r)=k_{r-1}+1$ and $\beta (r)=k_{r}$; then $P^{\alpha ,\beta }(r)=[k_{r-1}+1,k_{r}]$. But because of $[k_{r-1}+1,k_{r}]\cap \mathbb {N}=(k_{r-1},k_{r}]\cap \mathbb {N}$, we have

  $$\begin{aligned} \sum _{k\in [\alpha (n), \beta (n)]}s_{k}x_{k}=\sum _{k\in [k_{r-1}+1, k_{r}]}s_{k}x_{k}=\sum _{k\in (k_{r-1}, k_{r}]}s_{k}x_{k}. \end{aligned}$$

  This means that $[\overline{N}_{\alpha \beta },s]$-summable sequences are reduced to weighted lacunary summable, and $[\overline{N}_{\alpha \beta },s]$-summable sequence are reduced to strong weighted lacunary summable sequences.

• (v) If we take $\gamma =1$, $\alpha (r)=k_{r-1}+1$, $\beta (r)=k_{r}$ and $s_{k}=1$ for all $k$ then, weighted $\alpha \beta$-summable is reduced to lacunary summable sequences, and $[\overline{N}_{\alpha \beta },s]$-summable sequences are reduced to $N_{\theta }$ summbale sequences introduced in [6].

Definition 2.2

A sequence $x=(x_{k})$ is said to be weighted $\alpha \beta$-statistically convergent of order $\gamma$ to $\ell$ or $S^{\gamma }_{\alpha \beta }$-convergent, if for every $\epsilon > 0$

$$\begin{aligned} \delta ^{\alpha ,\beta }(\{k:s_{k}|x_{k}-\ell |\ge \epsilon \},\gamma )=\lim _{n \rightarrow \infty }\frac{1}{S_{n}^{\gamma }}\left| \left\{ k\le S_{n}:s_{k}|x_{k}-\ell |\ge \epsilon \right\} \right| =0 \end{aligned}$$

and denote $st^{\gamma }_{\alpha \beta }-\lim x=\ell$ or $x_{k}\longrightarrow \ell [\overline{S}^{\gamma }_{\alpha \beta }]$, where $\overline{S}^{\gamma }_{\alpha \beta }$ denotes the set of all weighted $\alpha \beta$-statistically convergent sequences of order $\gamma$ .

Theorem 2.1

Let $0<\gamma \le \delta \le 1$. Then, we have $[\overline{N}^{\gamma }_{\alpha \beta },s]\subseteq [\overline{N}^{\delta }_{\alpha \beta },s]$ and the inclusion is strict for some $\gamma ,\delta$ such that $\gamma <\delta$.

Proof

Let $x=(x_{k})\in [\overline{N}^{\gamma }_{\alpha \beta },s]$ and $\gamma ,\delta$ be given such that $0<\gamma \le \delta \le 1$. Then, we obtain that

$$\begin{aligned} \frac{1}{S_{n}^{\gamma }}\sum _{k\in P^{\alpha ,\beta }_{n}}s_{k}|x_{k}-\ell |\le \frac{1}{S_{n}^{\delta }}\sum _{k\in P^{\alpha ,\beta }_{n}}s_{k}|x_{k}-\ell | \end{aligned}$$

which gives $[\overline{N}^{\gamma }_{\alpha \beta },s]\subseteq [\overline{N}^{\delta }_{\alpha \beta },s]$. Now, we show that this inclusion is strict. Let us consider the sequence $t = (t_{k})$ defined by,

$$\begin{aligned} t=(t _{k})=\left\{ \begin{array}{ll} 1\text {,} \quad \beta (n)-\sqrt{\beta (n)-\alpha (n)+1}+1\le k\le \beta (n)\text {,} \\ 0\text {,} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

If we choose $s_{k}=1$ for all $k\in \mathbb {N}$, it is clear that

$$\begin{aligned} \frac{1}{(\beta (n)\!-\!\alpha (n)\!+\!1)^{\gamma }}\sum _{k\in P^{\alpha ,\beta }_{n}}|t_{k}-0|\le \frac{\sqrt{\beta (n)-\alpha (n)+1}}{(\beta (n)\!-\!\alpha (n)+1)^{\gamma }}=\frac{1}{(\beta (n)-\alpha (n)+1)^{\gamma -1/2}}. \end{aligned}$$

Since $\frac{1}{(\beta (n)-\alpha (n)+1)^{\gamma -1/2}} \longrightarrow 0$ as $n\longrightarrow \infty$ for $1/2<\beta \le 1$, we have $t=(t_{k})\in [\overline{N}^{\gamma }_{\alpha \beta },s]$. On the other hand, we get

$$\begin{aligned} \frac{\sqrt{\beta (n)-\alpha (n)+1}-1}{(\beta (n)-\alpha (n)+1)^{\delta }}\le \frac{1}{(\beta (n)-\alpha (n)+1)^{\delta }}\sum _{k\in P^{\alpha ,\beta }_{n}}|t_{k}-0| \end{aligned}$$

and $\frac{\sqrt{\beta (n)-\alpha (n)+1}-1}{(\beta (n)-\alpha (n)+1)^{\delta }} \longrightarrow \infty$ as $n\longrightarrow \infty$ for $0<\delta < 1/2$ then, we have $t=(t_{k})\notin [\overline{N}^{\delta }_{\alpha \beta },s]$. This completes the proof. $\square$

Theorem 2.2

Let $(\alpha ,\beta )\in \Lambda$. Then, we have following statements:

• (a) If a sequence $x=(x_{k})$ is strongly weighted $(\alpha \beta )$-summable of order $\gamma$ to limit $\ell$, then it is weighted $\alpha \beta$-statistically convergent of order $\gamma$ to $\ell$, that is $[\overline{N}^{\gamma }_{\alpha \beta },s]\subseteq \overline{S}^{\gamma }_{\alpha \beta }$ and this inclusion is strict.

• (b) If $x=(x_{k})$ bounded and weighted $\alpha \beta$-statistically convergent of order $\gamma$ to $\ell$ then $x_{k}\longrightarrow [\overline{N}^{\gamma }_{\alpha \beta },s]$.

Proof

• (a) Let $\epsilon >0$ and $x_{k}\longrightarrow \ell [\overline{N}^{\gamma }_{\alpha \beta },s]$. Then, we get

  $$\begin{aligned} \frac{1}{S_{n}^{\gamma }}\sum _{k\in P^{\alpha ,\beta }_{n}}|x_{k}-\ell |\!=\! \frac{1}{S_{n}^{\gamma }}\sum _{\begin{array}{c} k\in P^{\alpha ,\beta }_{n} \\ s_{k}|x_{k}\!-\!\ell |\ge \epsilon \end{array}}s_{k}|x_{k}\!-\!\ell |\!+\!\frac{1}{S_{n}^{\gamma }}\sum _{\begin{array}{c} k\in P^{\alpha ,\beta }_{n} \\ s_{k}|x_{k}-\ell |< \epsilon \end{array}}s_{k}|x_{k}-\ell |\ge \frac{\epsilon \left| K^{\alpha \beta }_{s_{n}}(\epsilon )\right| }{S_{n}^{\gamma }}. \end{aligned}$$

  This implies that $\lim _{n \rightarrow \infty }\frac{\left| K^{\alpha \beta }_{s_{n}}(\epsilon )\right| }{S_{n}^{\gamma }}=0$ which means $\delta ^{\alpha ,\beta }(K^{\alpha \beta }_{s_{n}}(\epsilon ),\gamma )=0$, where $K^{\alpha \beta }_{s_{n}}(\epsilon )=\left\{ k\le S_{n}:s_{k}|x_{k}-\ell |\ge \epsilon \right\}$. Therefore, $x=(x_{k})$ is weighted $\alpha \beta$-statistically convergent of order $\gamma$ to $\ell$. To prove $[\overline{N}^{\gamma }_{\alpha \beta },s]\subseteq S^{\gamma }_{\alpha \beta }$ in $(a)$ is strict, let the sequence $x=(x_{k})$ be defined by

  $$\begin{aligned} x_{k}=\left\{ \begin{array}{ll} k\text {,} \quad 1\le k\le [(\beta (n)-\alpha (n)+1)^{\gamma /2}]\text {,} \\ 0\text {,} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

  (2.1)

  Then $x$ is not bounded and for every $\epsilon >0$. Let $s_{k}=1$ for all $k$. Then we have

  $$\begin{aligned}&\frac{1}{(\beta (n)-\alpha (n)+1)^{\gamma }}\left| \left\{ k\le \beta (n)-\alpha (n)+1:|x_{k}-0|\ge \epsilon \right\} \right| \\&\quad =\frac{[(\beta (n)-\alpha (n)+1)^{\gamma /2}]}{(\beta (n)-\alpha (n)+1)^{\gamma }}\longrightarrow 0\,\, \mathrm as \,\,n\longrightarrow \infty . \end{aligned}$$

  That is, $x_{k}\longrightarrow 0[S^{\gamma }_{\alpha \beta }]$. But

  $$\begin{aligned}&\frac{1}{(\beta (n)-\alpha (n)+1)^{\gamma }}\sum _{k\in P^{\alpha ,\beta }_{n}}|x_{k}-0|\\&\quad =\frac{[(\beta (n)-\alpha (n)+1)^{\gamma /2}]\left( [(\beta (n)-\alpha (n)+1)^{\gamma /2}]+1\right) }{2(\beta (n)-\alpha (n)+1)^{\gamma }}\longrightarrow \frac{1}{2}, \end{aligned}$$

  i.e., $x_{k}\nrightarrow 0[\overline{N}^{\gamma }_{\alpha \beta },s]$.

• (b) Assume that $x=(x_{k})$ is bounded and weighted $\alpha \beta$-statistically convergent of order $\gamma$ to $\ell$. Then for $\epsilon >0$, we have $\delta ^{\alpha ,\beta }(K^{\alpha \beta }_{\epsilon },\gamma )=0$. Since $x=(x_{k})$ is bounded, there exists $M>0$ such that $s_{k}|x_{k}-\ell |\le M$ for all $k \in \mathbb {N}$.

  $$\begin{aligned} \frac{1}{S_{n}^{\gamma }}\sum _{k\in P^{\alpha ,\beta }_{n}}s_{k}|x_{k}-\ell |&= \frac{1}{S_{n}^{\gamma }}\sum _{\begin{array}{c} k\in P^{\alpha ,\beta }_{n} \\ s_{k}|x_{k}-\ell |\ge \varepsilon \end{array}}s_{k}|x_{k}-\ell |+\frac{1}{S_{n}^{\gamma }}\sum _{\begin{array}{c} k\in P^{\alpha ,\beta }_{n} \\ s_{k}|x_{k}-\ell |< \varepsilon \end{array}}|x_{k}-\ell |\\&\le \frac{M}{S_{n}^{\gamma }}\left| \left\{ k\in P^{\alpha ,\beta }_{n}:s_{k}|x_{k}-\ell |\ge \epsilon \right\} \right| +\varepsilon , \end{aligned}$$

  this implies that $x_{k}\longrightarrow [\overline{N}^{\gamma }_{\alpha \beta },s]$.

$\square$

3 Application to Korovkin type approximation

In this section, we get an analogue of classical Korovkin Theorem by using the concept of $\alpha \beta$-statistical convergence. Also we estimate, in rates of $\alpha \beta$-statistical convergence. Recently, such types of approximation theorems are proved, [1, 3, 4, 11, 13, 14].

Let $C[a,b]$ be the linear space of all real-valued continuous functions $f$ on $[a,b]$ and let $L$ be a linear operator which maps $C[a,b]$ into itself. We say $L$ is positive operator, if for every non-negative $f\in C[a,b]$, we have $L(f,x)\ge 0$ for $x\in [a,b]$. It is well-known that $C[a,b]$ is a Banach space with the norm given by

$$\begin{aligned} \parallel f\parallel _{C[a,b]}=\sup _{x\in [a,b]}|f(x)|. \end{aligned}$$

The classical Korovkin approximation theorem states as follows (see [7, 9])

$$\begin{aligned} \lim _{n \rightarrow \infty }\parallel L_{n}(f,x)-f(x)\parallel _{C[a,b]}=0 \Leftrightarrow \lim _{n \rightarrow \infty }\parallel L_{n}(f,x)-e_{i})\parallel _{C[a,b]}=0, \end{aligned}$$

where $e_{i}=x^{i}$ and $f\in C[a,b]$.

Theorem 3.1

Let $(L_{k})$ be a sequence of positive linear operator from $C[a,b]$ in to $C[a,b]$. Then for all $f\in C[a,b]$

$$\begin{aligned} \overline{S}^{\gamma }_{\alpha \beta }-\lim _{k \rightarrow \infty }\parallel L_{k}(f,x)-f(x))\parallel _{C[a,b]}=0 \end{aligned}$$

(3.1)

if and only if

$$\begin{aligned}&\overline{S}^{\gamma }_{\alpha \beta }-\lim _{k \rightarrow \infty }\parallel L_{k}(e_{0},x)-e_{0}\parallel _{C[a,b]}=0,\end{aligned}$$

(3.2)

$$\begin{aligned}&\overline{S}^{\gamma }_{\alpha \beta }-\lim _{k \rightarrow \infty }\parallel L_{k}(e_{1},x)-e_{1})\parallel _{C[a,b]}=0,\end{aligned}$$

(3.3)

$$\begin{aligned}&\overline{S}^{\gamma }_{\alpha \beta }-\lim _{k \rightarrow \infty }\parallel L_{k}(e_{2},x)-e_{2})\parallel _{C[a,b]}=0. \end{aligned}$$

(3.4)

Proof

Because of $e_{i}\in C[a,b]$ for $(i=0,1,2)$, conditions (3.2)–(3.4) follow immediately from (3.1). Let the conditions (3.2)–(3.4) hold and $f\in C[a,b]$. By the continuity of $f$ at $x$, it follows that for given $\varepsilon > 0$ there exists $\delta$ such that for all $t$

$$\begin{aligned} |f(x)-f(t)|<\varepsilon ,\quad \mathrm{whenever}\quad \forall |t-x|<\delta . \end{aligned}$$

(3.5)

Since $f$ is bounded, we get

$$\begin{aligned} |f(x)|\le M,\quad -\infty <x,t<\infty . \end{aligned}$$

Hence

$$\begin{aligned} |f(x)-f(t)|\le 2 M,\quad -\infty <x,t<\infty . \end{aligned}$$

(3.6)

By using (3.5) and (3.6), we have

$$\begin{aligned} |f(x)-f(t)|<\varepsilon + \frac{2M}{\delta ^{2}}(t-x)^{2},\,\,\forall |t-x|<\delta . \end{aligned}$$

This implies that

$$\begin{aligned} -\varepsilon -\frac{2M}{\delta ^{2}}(t-x)^{2}<f(x)-f(t)<\varepsilon + \frac{2M}{\delta ^{2}}(t-x)^{2}. \end{aligned}$$

By using the positivity and linearity of $(L_{k})$, we get

$$\begin{aligned} L_{k}(1,x)\left( -\varepsilon \!-\!\frac{2M}{\delta ^{2}}(t\!-\!x)^{2}\right) \!<\!L_{k}(1,x)\left( f(x)\!-\!f(t)\right) \!<\! L_{k}(1,x)\left( \varepsilon \!+\! \frac{2M}{\delta ^{2}}(t\!-\!x)^{2}\right) \end{aligned}$$

where $x$ is fixed and so $f(x)$ is constant number. Therefore,

$$\begin{aligned}&-\varepsilon L_{k}(1,x)-\frac{2M}{\delta ^{2}}L_{k}((t-x)^{2},x)< L_{k}(f,x)-f(x)L_{k}(1,x)\nonumber \\&\quad <\epsilon L_{k}(1,x)+\frac{2M}{\delta ^{2}}L_{k}((t-x)^{2},x). \end{aligned}$$

(3.7)

On the other hand

$$\begin{aligned} L_{k}(f,x)-f(x)&= L_{k}(f,x)-f(x)L_{k}(1,x)+f(x)L_{k}(1,x)-f(x)\nonumber \\&= [L_{k}(f,x)-f(x)L_{k}(1,x)-f(x)L_{k}]+f(x)[L_{k}(1,x)-1].\nonumber \\ \end{aligned}$$

(3.8)

By inequality (3.7) and (3.8), we obtain

$$\begin{aligned} L_{k}(f,x)\!-\!f(x)<\varepsilon L_{k}(1,x)\!+\!\frac{2M}{\delta ^{2}}L_{k}((t-x)^{2},x)+f(x)+f(x)[L_{k}(1,x)-1].\nonumber \\ \end{aligned}$$

(3.9)

Now, we compute second moment

$$\begin{aligned} L_{k}((t-x)^{2},x)&= L_{k}(x^{2}-2xt+t^{2},x)\\&= x^{2}L_{k}(1,x)-2xL_{k}(t,x)+L_{k}(t^{2},x)\\&= [L_{k}(t^{2},x)-x^{2}]-2x[L_{k}(t,x)-x]+x^{2}[L_{k}(1,x)-1]. \end{aligned}$$

By (3.9), we have

$$\begin{aligned} L_{k}(f,x)-f(x)&< \varepsilon L_{k}(1,x)+\frac{2M}{\delta ^{2}}\{[L_{k}(t^{2},x)-x^{2}]-2x[L_{k}(t,x)-x]\\&+\,x^{2}[L_{k}(1,x)-1]\}+f(x)(L_{k}(1,x)-1)\\&= \varepsilon [L_{n}(1,x)-1]+\varepsilon +\frac{2M}{\delta ^{2}}\{[L_{k}(t^{2},x)-x^{2}]-2x[L_{k}(t,x)-x]\\&+\,x^{2}[L_{k}(1,x)-1]\}+f(x)(L_{k}(1,x)-1). \end{aligned}$$

Because of $\varepsilon$ is arbitrary, we obtain

$$\begin{aligned}&\parallel L_{k}(f,x)-f(x)\parallel _{C[a,b]}\le \left( \varepsilon +M+ \frac{2Mb^{2}}{\delta ^{2}}\right) \parallel L_{k}(e_{0},x)-e_{0}\parallel _{C[a,b]}\\&\qquad +\frac{4Mb}{\delta ^{2}}\parallel L_{k}(e_{1},x)-e_{1}\parallel _{C[a,b]} +\frac{2M}{\delta ^{2}}\parallel L_{k}(e_{2},x)-e_{2}\parallel _{C[a,b]}\\&\quad \le R\left( \parallel L_{k}(e_{0},x)-e_{0}\parallel _{C[a,b]}+\parallel L_{k}(e_{1},x)-e_{1}\parallel _{C[a,b]}\right. \\&\qquad \left. +\parallel L_{k}(e_{2},x)-e_{2}\parallel _{C[a,b]}\right) \end{aligned}$$

where $R=\max \left( \varepsilon +M+\frac{2Mb^{2}}{\delta ^{2}},\frac{4Mb}{\delta ^{2}}\right)$.

For $\varepsilon ^{\prime }>0$, we can write

Then, $\mathcal {C}\subset \mathcal {C}_{1}\cup \mathcal {C}_{2}\cup \mathcal {C}_{3}$, so we have $\delta ^{\alpha ,\beta }(\mathcal {C},\gamma )\le \delta ^{\alpha ,\beta }(\mathcal {C}_{1},\gamma )+\delta ^{\alpha ,\beta }(\mathcal {C}_{2},\gamma )+\delta ^{\alpha ,\beta }(\mathcal {C}_{3},\gamma )$. Thus, by conditions (3.2)–(3.4), we obtain

$$\begin{aligned} \overline{S}^{\gamma }_{\alpha \beta }-\lim _{k \rightarrow \infty }\parallel L_{k}(f,x)-f(x)\parallel _{C[a,b]}=0. \end{aligned}$$

which completes the proof. $\square$

We remark that our Theorem 3.1 is stronger than that of classical Korovkin approximation theorem. For this claim, we consider the following example:

Example 1

Considering the sequence of Bernstein operators

$$\begin{aligned} B_{n}(f,x)=\sum _{k=0}^{n}f\left( \frac{n}{k}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) x^{k}(1-x)^{n-k};\,\,x\in [0,1]. \end{aligned}$$

We define the sequence of linear operators as $T_{n}:C[0,1]\longrightarrow C[0,1]$ with $T_{n}(f,x)=(1+x_{n})B_{n}(f,x)$, where $x=(x_{n})$ is defined in (2.1). Now let $s_{k}=1$ for all $k$. Note that the sequence $x=(x_{n})$ is weighted $\alpha \beta$-statistically convergent but not convergent. Then, $B_{n}(1,x)=1$, $B_{n}(t,x)=x$ and $B_{n}(t^{2},x)=x^{2}+\frac{x-x^{2}}{n}$ and sequence $(T_{n})$ satisfies the conditions (3.2)–(3.4). Therefore, we get

$$\begin{aligned} \overline{S}^{\gamma }_{\alpha \beta }-\lim _{n \rightarrow \infty }\parallel T_{n}(f,x)-f(x)\parallel _{C[a,b]}=0. \end{aligned}$$

On the other hand, we have $T_{n}(f,0)=(1+x_{n})f(0)$, since $B_{n}(f,0)=f(0)$, thus we obtain

$$\begin{aligned}\ \parallel T_{n}(f,x)-f(x)\parallel _{\infty }\ge |T_{n}(f,0)-f(0)|\ge x_{n}|f(0)|. \end{aligned}$$

One can see that $(T_{n})$ does not satisfy the classical Korovkin theorem, since $x=(x_{n})$ is not convergent.

4 Rate of weighted $\alpha \beta$-statistically convergent of order $\gamma$

In this section, we estimate the rate of weighted $\alpha \beta$-statistically convergent of order $\gamma$ of a sequence of positive linear operator which is defined from $C[a,b]$ into $C[a,b]$ .

Definition 4.1

Let $(u_{n})$ be a positive non-increasing sequence. We say that the sequence $x=(x_{k})$ is $\alpha \beta$-statistically convergent of order $\gamma$ to $\ell$ with the rate $o(u_{n})$ if for every, $\epsilon >0$

$$\begin{aligned} \lim _{n \rightarrow \infty }\frac{1}{u_{n}S_{n}^{\gamma }}\left| \left\{ k\le S_{n}:s_{k}|x_{k}-\ell |\ge \epsilon \right\} \right| =0. \end{aligned}$$

At this stage, we can write $x_{k}-\ell =\overline{S}^{\gamma }_{\alpha \beta }-o(u_{n})$.

Before proceeding further, let us give basic definition and notation on the concept of the modulus of continuity.

The modulus of continuity of $f$, $\omega (f,\delta )$ is defined by

$$\begin{aligned} \omega (f,\delta )=\sup _{\begin{array}{c} |x-y|\le \delta \\ x,y\in [a,b] \end{array}}|f(x)-f(y)|. \end{aligned}$$

It is well-known that for a function $f \in C[a,b]$,

$$\begin{aligned} \lim _{n\rightarrow 0^{+}}\omega (f,\delta )=0 \end{aligned}$$

for any $\delta >0$

$$\begin{aligned} |f(x)-f(y)|\le \omega (f,\delta )\left( \frac{|x-y|}{\delta }+1\right) . \end{aligned}$$

(4.1)

Theorem 4.1

Let $(L_{k})$ be sequence of positive linear operator from $C[a,b]$ into $C[a,b]$. Assume that

$$\begin{aligned} (i)&\ \parallel L_{k}(1,x)-x\parallel _{C[a,b]}=\overline{S}^{\gamma }_{\alpha \beta }-o(u_{n}),\\ (ii)&\omega (f,\psi _{k})=\overline{S}^{\gamma }_{\alpha \beta }-o(v_{n})\,\,\mathrm where \,\,\psi _{k}=\sqrt{L_{k}[(t-x)^{2},x]}. \end{aligned}$$

Then for all $f\in C[a,b]$, we get

$$\begin{aligned} \ \parallel L_{k}(f,x)-f(x)\parallel _{C[a,b]}=\overline{S}^{\gamma }_{\alpha \beta }-o(z_{n}) \end{aligned}$$

where $z_{n}= \max \{u_{n},v_{n}\}$.

Proof

Let $f\in C[a,b]$ and $x\in [a,b]$. From (3.8) and (4.1), we can write

$$\begin{aligned} |L_{k}(f,x)-f(x)|&\le L_{k}(|f(t)-f(x)|;x)+|f(x)||L_{k}(1,x)-1| \\&\le L_{k}\left( \frac{|x-y|}{\delta }+1;x\right) \omega (f,\delta )+|f(x)||L_{k}(1,x)-1|\\&\le L_{k}\left( \frac{(t-x)^{2}}{\delta ^{2}}+1;x\right) \omega (f,\delta )+|f(x)||L_{k}(1,x)-1|\\&\le \left( L_{k}(1,x)\!+\!\frac{1}{\delta ^{2}}L_{k}\left( (t\!-\!x)^{2};x\right) \right) \omega (f,\delta )\!+\!|f(x)||L_{k}(1,x)\!-\!1|\\&= L_{k}(1,x)\omega (f,\delta )\!+\!\frac{1}{\!\delta ^{2}}L_{k}\!\left( (t\!-\!x)^{2};x\right) \omega (f,\delta )\!+\!|f(x)||L_{k}(1,x)\!-\!1|. \end{aligned}$$

By choosing $\sqrt{\psi _{k}}=\delta$, we get

$$\begin{aligned}&\parallel L_{k}(f,x)-f(x)\parallel _{C[a,b]}\le \parallel f\parallel _{C[a,b]}\parallel L_{k}(1,x)-x \parallel _{C[a,b]}\\&\quad +\,2\omega (f,\psi _{k})+\omega (f,\psi _{k})\parallel L_{k}(1,x)-x \parallel _{C[a,b]}\\&\le H\{\parallel L_{k}(1,x)-x \parallel _{C[a,b]}+\omega (f,\psi _{k})+\omega (f,\psi _{k})\parallel L_{k}(1,x)-x \parallel _{C[a,b]}\}, \end{aligned}$$

where $H=\max \{2,\parallel f\parallel _{C[a,b]}\}$. By Definition 4.1 and conditions (i)–(ii), we get the desired the result. $\square$