1 Introduction

In 1987, Lupaş [1] introduced the first q-analogue of the classical Bernstein polynomials. After that, with rapid development of q-calculus, new q-analogue of various positive linear operators are introduced by many researchers (for details see [2]). For detail study of q-calculus one can refer to [3, 4].

We begin by recalling certain notations of (p, q)-calculus (for detail see [5–7]). Let \(0 < q < p \le 1 \). The (p, q)-integer \([n]_{p,q}\) and (p, q)-factorial \([n]_{p,q} !\) are defined by

$$\begin{aligned}{}[n]_{p,q} = \frac{p^n-q^n}{p-q} \quad n = 0,1,2\ldots \quad \end{aligned}$$

For integers \(0\le k\le n\), (p, q)-binomial is defined as

$$\begin{aligned} \left[ \begin{array}{c} n \\ k \end{array} \right] _{p,q} = \frac{[n]_{p,q}!}{[k]_{p,q}![n-k]_{p,q}!}. \end{aligned}$$

The (p, q)-polynomials expansion is

$$\begin{aligned} (x+y)_{p,q}^n = (x+y) (px+qy) (p^2x + q^2y)\cdots (p^{n-1} x + q^{n-1}y). \end{aligned}$$

Let \(f : R\rightarrow R\), then the (p, q)-derivative of function f is:

$$\begin{aligned} (D_{p,q}f) = \frac{f(px)-f(qx)}{(p-q)x} \quad x\ne 0, \quad (D_{p,q}f)(0) = f(0) \end{aligned}$$

provided that f is differentiable at 0.

Let \(f : C[0,a]\rightarrow R\), then the (p, q)-integration of a function f is defined by,

$$\begin{aligned} \int _0^a f(t)d_{p,q}t = (q-p)a \sum _{k=0}^\infty \frac{p^k}{q^{k+1}}f\left( \frac{p^k}{q^{k+1}}a\right) ,\quad when \quad \left| \frac{p}{q}\right| < 1, \end{aligned}$$

and

$$\begin{aligned} \int _0^a f(t)d_{p,q}t = (p-q)a \sum _{k=0}^\infty \frac{q^k}{p^{k+1}}f\left( \frac{q^k}{p^{k+1}}a\right) ,\quad when \quad \left| \frac{p}{q}\right| > 1. \end{aligned}$$
(1.1)

Two (p, q)-analogues of the exponential function (see [8]) are as:

$$\begin{aligned} e_{p, q}(x)= & {} \sum _{n=0}^\infty \frac{p^{\frac{n(n-1)}{2}}x^n}{[n]_{p,q}!},\\ E_{p, q}(x)= & {} \sum _{n=0}^\infty \frac{q^{\frac{n(n-1)}{2}}x^n}{[n]_{p,q}!}, \end{aligned}$$

which satisfy the equality \(e_{p, q}(x)E_{p, q}(-x)=1.\) For \(p = 1\), all the notations of (p, q)-calculus are reduced to q-calculus.

Recently, Tuncer Acar [9] introduced a (p, q)-analogue of Szász-Mirakyan operators as: For \( 0<q<p\le 1\), \(n \in \mathbb {N}\) and \(f:[0,\infty )\rightarrow R\):

$$\begin{aligned} S_{n,p,q} (f;x)=\sum _{k=0}^\infty s_n{(p,q;x)}f \left( \frac{[k]_{p,q}}{q^{k-2}[n]_{p,q}}\right) , \end{aligned}$$

where

$$\begin{aligned} s_n{(p,q;x)} =\frac{1}{E_{p, q}([n]_{p,q}x)}q^{\frac{k(k-1)}{2}}\frac{[n]_{p,q}^kx^k}{[k]_{p,q}!}. \end{aligned}$$

Lemma 1

[9, Lemma 2] Let \( 0<q<p\le 1\) and \(n \in \mathbb {N}\). We have

$$\begin{aligned} S_{n,p,q}(1;x)= & {} 1, \\ S_{n,p,q}(t;x)= & {} qx, \\ S_{n,p,q}(t^2;x)= & {} pqx^2 + \frac{q^2x}{[n]_{p,q}}. \end{aligned}$$

2 Construction of (p, q)-Szász-Mirakyan Kantorovich operators

Motivated by Tuncer Acar, we set (p, q)-Szász-Mirakyan Kantorovich operators for \( 0<q<p\le 1\), \(n \in \mathbb {N}\) and \(f:[0,\infty )\rightarrow \mathbb {R}\) as:

$$\begin{aligned} K_n^{(p, q)}(f;x)= [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x){p}^{-k} q^{k-2} \int _\frac{q^{-k+3}[k]_{p,q}}{[n]_{p,q}}^\frac{q^{-k+2}[k+1]_{p,q}}{[n]_{p,q}} f(t) d_{p,q}t. \end{aligned}$$
(2.1)

Lemma 2

Let \( 0<q<p\le 1\) and \(n \in \mathbb {N}\), we have

$$\begin{aligned} q^{k-2}\int _\frac{q^{-k+3}[k]_{p,q}}{[n]_{p,q}}^\frac{q^{-k+2}[k+1]_{p,q}}{[n]_{p,q}} d_{p,q}t= & {} \frac{p^k}{[n]_{p,q}}, \\ q^{k-2}\int _\frac{q^{-k+3}[k]_{p,q}}{[n]_{p,q}}^\frac{q^{-k+2}[k+1]_{p,q}}{[n]_{p,q}} t d_{p,q}t= & {} \frac{p^k q^{-k+2}([k+1]_{p,q} + q[k]_{p,q})}{(p+q)[n]_{p,q}^2}, \\ q^{k-2}\int _\frac{q^{-k+3}[k]_{p,q}}{[n]_{p,q}}^\frac{q^{-k+2}[k+1]_{p,q}}{[n]_{p,q}} t^2 d_{p,q}t= & {} \frac{p^k q^{-2k+4}([k+1]_{p,q}^2 + q[k]_{p,q}[k+1]_{p,q}+ q^2[k]_{p,q}^2)}{(p^2+ pq + q^2)[n]_{p,q}^3}. \end{aligned}$$

Proof

Using the identity \([k+1]_{p,q} = p^k + q[k]_{p,q}\) and Eq. (1.1), lemma can be proved.

\(\square \)

Lemma 3

Let \( 0<q<p\le 1\) and \(n \in N.\) We have

$$\begin{aligned} K_n^{(p, q)}(1;x)= & {} 1, \end{aligned}$$
(2.2)
$$\begin{aligned} K_n^{(p, q)}(t;x)= & {} qx + \frac{q^2}{[n]_{p,q}(p+q)}, \end{aligned}$$
(2.3)
$$\begin{aligned} K_n^{(p, q)}(t^2;x)= & {} pqx^2 + \frac{(2q^4 +3pq^3 +p^2 q^2)x}{(p^2 + pq + q^2)[n]_{p,q}} + \frac{q^4}{(p^2 + pq + q^2)[n]_{p,q}^2}.\qquad \end{aligned}$$
(2.4)

Proof

Using definition of operator, Lemmas 1 and 2, moments can be obtained as follows:

$$\begin{aligned} K_n^{(p, q)}(1;x)= & {} [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x) p^{-k}q^{k-2} \int _\frac{q^{-k+3}[k]_{p,q}}{[n]_{p,q}}^\frac{q^{-k+2}[k+1]_{p,q}}{[n]_{p,q}}d_{p,q}t\\= & {} [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x) p^{-k}\frac{p^k}{[n]_{p,q}} \\= & {} \sum _{k=0}^\infty s_n(p,q;x) = S_{n,p,q}(1;x)= 1. \end{aligned}$$

And using identity \([k+1]_{p,q}=q^k+p[k]_{p,q}\), we get

$$\begin{aligned} K_n^{(p, q)}(t;x)= & {} [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x) p^{-k}q^{k-2} \int _\frac{q^{-k+3}[k]_{p,q}}{[n]_{p,q}}^\frac{q^{-k+2}[k+1]_{p,q}}{[n]_{p,q}} td_{p,q}t\\= & {} [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x) p^{-k}\frac{p^k q^{-k+2}([k+1]_{p,q} + q[k]_{p,q})}{(p+q)[n]_{p,q}^2} \\= & {} \frac{1}{[n]_{p,q}(p+q)} \sum _{k=0}^\infty s_n(p,q;x)q^{-k+2}(q^k+(p+q)[k]_{p,q})\\= & {} \frac{q^2}{[n]_{p,q}(p+q)}\sum _{k=0}^\infty s_n(p,q;x) + \sum _{k=0}^\infty s_n(p,q;x)\frac{[k]_{p,q}}{q^{k-2}[n]_{p,q}}\\= & {} \frac{q^2}{[n]_{p,q}(p+q)}S_{n,p,q}(1;x) + S_{n,p,q}(t;x) \\= & {} \frac{q^2}{[n]_{p,q}(p+q)} + qx. \end{aligned}$$

Finally,

$$\begin{aligned}&K_n^{(p, q)}(t^2;x)\\&\quad = [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x) p^{-k}q^{k-2} \int _\frac{q^{-k+3}[k]_{p,q}}{[n]_{p,q}}^\frac{q^{-k+2}[k+1]_{p,q}}{[n]_{p,q}}t^2d_{p,q}t\\&\quad = [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x) p^{-k} \frac{p^k q^{-2k+4}\left( [k+1]_{p,q}^2 + q[k]_{p,q}[k+1]_{p,q}+ q^2[k]_{p,q}^2\right) }{(p^2+ pq + q^2)[n]_{p,q}^3} \\&\quad = [n]_{p,q} \sum _{k=0}^\infty s_n(p,q;x) \frac{ q^{-2k+4}\left( (p^2 +pq +q^2)[k]_{p,q}^2 + q^k(2p+q)[k]_{p,q} + q^{2k}\right) }{(p^2+ pq + q^2)[n]_{p,q}^3}\\&\quad = \sum _{k=0}^\infty s_n(p,q;x)\frac{[k]_{p,q}^2}{q^{2k-4}[n]_{p,q}^2}+\frac{(2p+q)q^2}{(p^2 + pq + q^2)[n]_{p,q}}\sum _{k=0}^\infty s_n(p,q;x)\frac{[k]_{p,q}}{q^{k-2}[n]_{p,q}}\\&\quad \quad + \frac{q^4}{(p^2 + pq + q^2)[n]_{p,q}^2} \sum _{k=0}^\infty s_n(p,q;x)\\&\quad = S_{n,p,q}(t^2;x) + \frac{(2p+q)q^2}{(p^2+pq+q^2)[n]_{p,q}} S_{n,p,q}(t;x) + \frac{q^4}{(p^2 + pq + q^2)[n]_{p,q}^2} S_{n,p,q}(1;x)\\&\quad = pqx^2 + \frac{q^2x}{[n]_{p,q}} + \frac{(2p+q)q^3x}{(p^2+pq+q^2)[n]_{p,q}} + \frac{q^4}{(p^2 + pq +q^2)[n]_{p,q}^2}\\&\quad = pqx^2 + \frac{(2q^4 +3pq^3 +p^2 q^2)x}{(p^2 + pq + q^2)[n]_{p,q}} + \frac{q^4}{(p^2 + pq + q^2)[n]_{p,q}^2}. \end{aligned}$$

\(\square \)

Corollary 1

Central moments \( \Phi _m^{(p, q)}(x) = K_n^{(p, q)}((t-x)^m;x)\) for \(m=1,2\) are:

$$\begin{aligned} \Phi _1^{(p, q)}(x)=(q-1)x + \frac{q^2}{(p+q)[n]_{p,q}}, \end{aligned}$$
$$\begin{aligned} \Phi _2^{(p, q)}(x)= & {} (pq-2q+1)x^2 +\left( \frac{2q^4 +3pq^3 +p^2 q^2}{(p^2 + pq + q^2)[n]_{p,q}}-\frac{2q^2}{(p+q)[n]_{p,q}}\right) x\\&+ \frac{q^4}{(p^2 + pq + q^2)[n]_{p,q}^2}. \end{aligned}$$

Proof

Using Lemma 3, central moments can be obtained directly. \(\square \)

Remark 1

For \(q\in (0,1)\) and \(p \in (q,1]\), by simple computations \(\lim _{n\rightarrow \infty } [n]_{p,q}=1/(p-q)\). In order to obtain results for order of convergence of the operator, we take \(q_n\in (0,1)\), \(p_n \in (q_n,1]\) such that \(\lim _{n\rightarrow \infty } p_n=1\) and \(\lim _{n\rightarrow \infty } q_n=1\), so that \(\lim _{n\rightarrow \infty } \frac{1}{[n]_{p_n,q_n}}=0\). Such a sequence can always be constructed for example, we can take \(q_n=1-1/n\) and \(p_n=1-1/2n\), clearly \(\lim _{n\rightarrow \infty }p_n^n=e^{-1/2}\), \(\lim _{n\rightarrow \infty }q_n^n=e^{-1}\) and \(\lim _{n\rightarrow \infty } \frac{1}{[n]_{p_n,q_n}}=0\).

Theorem 2

Let \((p_n)_n\) and \((q_n)_n\) be the sequence as defined in Remark 1. Then for each \(f \in C[0,\infty )\), \(K_{n}^{(p_{n}, q_{n})} (f;x)\) converges uniformly to f.

Proof

By Korovkin theorem, it is sufficient to show that \(\lim _{n\rightarrow \infty } \Vert K_{n}^{(p_n,q_n)} (t^m;x) -x^m\Vert _{C[0,\infty )} = 0 \) for \( m=0,1,2\).

Using Eq. 2.2, result for \(m=0\) is trivial. For \(m=1\), result can be obtained using Eq. (2.3), as follows:

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert K_{n}^{(p_n,q_n)} (t;x) -x\Vert _{C[0,\infty )}= & {} \lim _{n\rightarrow \infty } \left| \frac{q_n^2}{(p_n +q_n)[n]_{p,q}} + q_nx -x\right| \\\le & {} \lim _{n\rightarrow \infty } \left| \frac{q_n^2}{(p_n +q_n)[n]_{p,q}}| + \lim _{n\rightarrow \infty } \right| q_n-1|x\\= & {} 0. \end{aligned}$$

Finally, using Eq. (2.4), we get

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert K_{n}^{(p_n,q_n)} (t^2;x) -x^2\Vert _{C[0,\infty )}= & {} \lim _{n\rightarrow \infty } \left| \frac{q_n^4}{(p_n^2 +q_n^2 + p_nq_n)[n]^2_{p_n,q_n}}\right. \\&\left. + \frac{(2q^4_n +3p_n q^3_n +p^2_n q^2_n)x}{(p^2_n + p_n q_n + q^2_n)[n]_{p_n ,q_n }} + p_nq_nx^2 - x^2\right| \\\le & {} \lim _{n\rightarrow \infty } \left| \frac{q_n^4}{(p_n^2 +q_n^2 + p_nq_n)[n]^2_{p_n,q_n}}\right| \\&+ \lim _{n\rightarrow \infty } |p_nq_n-1|x^2\\&+ \lim _{n\rightarrow \infty } \left| \frac{(2q^4_n +3p_n q^3_n +p^2_n q^2_n)x}{(p^2_n + p_n q_n + q^2_n)[n]_{p_n ,q_n }}\right| \\= & {} 0. \end{aligned}$$

\(\square \)

3 Direct results

In this section, we give some local result for the operator. Let \(C_B[0,\infty )\) be the space of all real valued continuous bounded functions defined on \([0,\infty )\). The norm on the space \(C_B[0,\infty )\) is the supremum norm \(\Vert f\Vert =\sup _{x\in [0,\infty )} f(x)\). Further, Peetre’s K-functional is defined by

$$\begin{aligned} K_2(f,\delta ) = \inf _{g \in W^2} \{\Vert f-g\Vert + \delta \Vert g''\Vert \}, \end{aligned}$$

here \(W^2 = \{ g \in C_B[0,\infty ) : g', g'' \in C_B[0,\infty )\}\). By [10, p. 177, Theorem 2.4], there exists a positive constant \(C > 0\) such that \( K_2(f, \delta )\le C\omega _2(f, \delta ^\frac{1}{2}), \delta > 0\), where

$$\begin{aligned} \omega _2(f, \delta ^\frac{1}{2})= \sup _{0<h<\delta ^{\frac{1}{2}}, x\in [0,\infty )} |f(x+2h)-2f(x+h)+f(x)| \end{aligned}$$

is the second order modulus of continuity of function \(f \in C_B[0,\infty )\). Also, for \(f \in C_B[0,\infty )\) the usual modulus of continuity is given by

$$\begin{aligned} \omega (f, \delta ^\frac{1}{2})= \sup _{0<h<\delta ^{\frac{1}{2}}, x\in [0,\infty )} |f(x+h) - f(x)|. \end{aligned}$$

Theorem 3

Let \((p_n)_n\) and \((q_n)_n\) be the sequence as defined in Remark 1. Let \(f\in C_B[0,\infty )\). Then for all \(n \in \mathbb {N}\), there exists an absolute constant \(C>0\) such that

$$\begin{aligned} |K_n^{(p_n,q_n)}(f;x) - f(x)| \le C\omega _2(f,\delta _n(x)) + \omega (f,\alpha _n(x)), \end{aligned}$$

where

$$\begin{aligned} \delta _n(x) = \left\{ \Phi _2^{(p_n,q_n)}(x) + (\Phi _1^{(p_n,q_n)}(x))^2 \right\} ^\frac{1}{2} \end{aligned}$$

and

$$\begin{aligned} \alpha _n(x) = \left| \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} + (q_n-1)x \right| . \end{aligned}$$

Proof

For \(x\in [0,\infty )\), we consider the auxiliary operators \(K_n^*(f;x)\) defined by

$$\begin{aligned} K_n^*(f;x) = K_n^{(p_n,q_n)}(f;x) +f(x) - f\left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx \right) . \end{aligned}$$

Using above operator and Eq. (2.3), we have

$$\begin{aligned} K_n^*(t-x;x)= & {} K_n^{(p_n,q_n)}(t-x;x) - \left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} + q_nx -x\right) \\= & {} K_n^{(p_n,q_n)}(t;x) - xK_n^{(p_n,q_n)}(1;x) - \left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} + q_nx \right) + x \\= & {} 0 \end{aligned}$$

Let \(x\in [0,\infty )\) and \( g\in W^2\). Using the Taylor’s formula

$$\begin{aligned} g(t) = g(x) + g'(x)(t-x) + \int _x^t (t-u) g''(u) du. \end{aligned}$$

Applying \(K_n^*\) to both sides of the above equation, we have

$$\begin{aligned} K_n^*(g;x) - g(x)= & {} K_n^*((t-x)g'(x);x) + K_n^* \left( \int _x^t (t-u) g''(u) du;x \right) \\= & {} g'(x)K_n^*((t-x);x) + K_n^{(p_n,q_n)} \left( \int _x^t (t-u) g''(u) du;x \right) \\&- \int _x^ {\frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx} \left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx - u\right) g''(u)du \\= & {} K_n^{(p_n,q_n)} \left( \int _x^t (t-u) g''(u) du;x \right) \\&- \int _x^ {\frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx} \left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx - u\right) g''(u)du. \\ \end{aligned}$$

On the other hand,

$$\begin{aligned} \left| \int _x^t (t-u)g''(u)du\right| \le \int _x^t |t-u||g''(u)|du \le \Vert g''\Vert \int _x^t |t-u|du \le (t-x)^2 \Vert g''\Vert , \end{aligned}$$

and

$$\begin{aligned}&\left| \int _x^{\frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx} \left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx - u \right) g''(u)du\right| \\&\quad \le \left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx - x\right) ^2\Vert g''\Vert . \end{aligned}$$

Therefore, we can conclude that

$$\begin{aligned} |K_n^*(g;x) - g(x)|= & {} \left| K_n^{(p_n,q_n)} \left( \int _x^t (t-u)g''(u)du;x \right) \right| \\&+ \left| \int _x^{\frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx}\left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} + q_nx -u\right) g''(u)du\right| \\\le & {} \Vert g''\Vert K_n^{(p_n,q_n)}((t-x)^2;x) + \left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} + q_nx -x\right) ^2 \Vert g''\Vert \\= & {} \delta ^2_n(x) \Vert g''\Vert . \end{aligned}$$

Also, we have

$$\begin{aligned} |K_n^*(f;x)| \le | K_n^{(p_n,q_n)}(f;x) | +2\Vert f\Vert \le 3\Vert f\Vert . \end{aligned}$$

Therefore,

$$\begin{aligned}&|K_n^{(p_n,q_n)}(f;x)-f(x)|\\&\quad \le |K_n^*(f-g;x)-(f-g)(x)| + \left| f\left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx\right) -f(x)\right| \\&\quad \quad + |K_n^*(g;x)-g(x)| \\&\quad \le |K_n^*(f-g;x)| + |(f-g)(x)| + \left| f\left( \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +q_nx\right) -f(x)\right| \\&\quad \quad + |K_n^*(g;x)-g(x)| \\&\quad \le 4\Vert f-g\Vert + \omega \left( f;\left| \frac{q^2_n}{[n]_{p_n,q_n}(p_n+q_n)} +(q_n-1)x\right| \right) + \delta ^2_n(x)\Vert g''\Vert . \end{aligned}$$

Hence, taking the infimum on the right-hand side over all \(g\in W^2\), we get

$$\begin{aligned} |K_n^{(p_n,q_n)}(f;x)-f(x)| \le 4K_2(f,\delta ^2_n(x)) + \omega (f,\alpha _n(x)). \end{aligned}$$

By using property of K-functional, we get

$$\begin{aligned} |K_n^{(p_n,q_n)}(f;x)-f(x)| \le C\omega _2(f,\delta _n(x)) + \omega (f,\alpha _n(x)). \end{aligned}$$

Hence the theorem. \(\square \)

We consider following class of functions: \(H_{x^2}[0,\infty )=\{f:[0,\infty )\rightarrow \mathbb {R}:|f(x)|\le M_f(1+x^2) \text{ here } M_f \text{ is } \text{ constant } \text{ depending } \text{ on } \text{ the } \text{ function } f\}\),

\(C_{x^2}[0,\infty )=\left\{ f \in H_{x^2}[0,\infty ):f \text { is continuous}\right\} ,\)

\(C^*_{x^2}[0,\infty )=\left\{ f \in C_{x^2}[0,\infty ):\lim _{|x|\rightarrow \infty }\frac{f(x)}{1+x^2} \text { is finite}\right\} .\) The norm on the space \(C^*_{x^2}[0,\infty )\) is defined as \(\Vert f\Vert _{x^2}=\sup _{x\in [0,\infty )}\frac{f(x)}{1+x^2}\). We denote the modulus of continuity of f on closed interval [0, a], \(a > 0\) as:

$$\begin{aligned} \omega _a(f;\delta ) = \sup _{|t-x| \le \delta ,\text { } x,t \in [0,a]} |f(t)-f(x)|. \end{aligned}$$

Theorem 4

Let \((p_n)_n\) and \((q_n)_n\) be the sequence as defined in Remark 1. Then for \(f \in C_{x^2}[0,\infty )\), \(\omega _{a+1}(f;\delta )\) be its modulus of continuity on the interval \([0, a+1] \subset [0,\infty )\), \(a>0\) and for every \(n> 1\),

$$\begin{aligned} \Vert K_n^{(p_n,q_n)}(f;x)- f\Vert _{C[0,a]} \le 6M_f(1+a^2){\lambda _n}+2\omega _{a+1}(f;\sqrt{\lambda _n}), \end{aligned}$$

here,

$$\begin{aligned} \lambda _n=(1-p_n q_n)a^2+\frac{1}{[n]_{p_n,q_n}(p_n+q_n)(p^2_n+p_nq_n+q^2_n)}\left( \frac{6a}{p_n+q_n}+\frac{1}{[n]_{p_n,q_n}}\right) . \end{aligned}$$

Proof

For \(x\in [0,a]\) and \(t\ge 0\), we have (see [11, Equation 3.3])

$$\begin{aligned} |f(t)-f(x)|\le & {} 6M_f(1+a^2)(t-x)^2\\&+\, \omega _{a+1}(f;\delta _n) \left( \frac{|t-x|}{\delta _n} +1 \right) . \end{aligned}$$

By using above inequality and Cauchy-Schwarz inequality, we have

$$\begin{aligned} \Vert K_n^{(p_n,q_n)}(f(t);x)- f(x)\Vert _{C[0,a]}\le & {} K_n^{(p_n,q_n)}(|f(t)-f(x)|;x)\\\le & {} 6M_f(1+a^2) K_n^{(p_n,q_n)}((t-x)^2;x)\\&+\,\omega _{a+1}(f;\delta _n) \left( 1+\frac{1}{\delta ^2_n} K_n^{(p_n,q_n)}((t-x)^2;x)\right) ^{1/2}. \end{aligned}$$

For \(x \in [0,a]\), using Corollary 1,

$$\begin{aligned}&K_n^{(p_n,q_n)}((t-x)^2;x)\\&\quad =(p_{n}q_{n}-2q_{n}+1)x^2 +\left( \frac{2q^4_{n} +3p_{n}q^3_{n} +p^2_{n} q^2_{n}}{(p^2_{n} + p_{n}q_{n} + q^2)[n]_{p_{n},q_{n}}}-\frac{2q^2_{n}}{(p_{n}+q_{n})[n]_{p_{n},q_{n}}}\right) x\\&\quad \quad + \frac{q_{n}^4}{(p^2_{n} + p_{n}q_{n} + q^2_{n})[n]_{p_{n},q_{n}}^2}\\&\quad \le (q_{n}(p_{n}-1)-q_{n}+1)a^2+\frac{6a}{(p^2_{n} + p_{n}q_{n} + q^2)[n]_{p_{n},q_{n}}}+\frac{1}{(p^2_{n} + p_{n}q_{n} + q^2_{n})[n]_{p_{n},q_{n}}^2}\\&\quad \le (q_{n}(1-p_{n})-q_{n}+1)a^2+\frac{1}{(p^2_{n} + p_{n}q_{n} + q^2)[n]_{p_{n},q_{n}}}\left( \frac{6a}{p_n+q_n}+\frac{1}{[n]_{p_{n},q_{n}}}\right) \\&\quad \le (1-p_n q_n)a^2+\frac{1}{(p^2_{n} + p_{n}q_{n} + q^2)[n]_{p_{n},q_{n}}}\left( \frac{6a}{p_n+q_n}+\frac{1}{[n]_{p_{n},q_{n}}}\right) =\lambda _n. \end{aligned}$$

Taking \(\delta _n=\sqrt{\lambda _n}\), we will get the theorem. \(\square \)

4 Voronovskaya type theorem

Theorem 5

Let \(0<q_n<p_n\le 1\), such that \(p_n\rightarrow 1\), \(p_n\rightarrow 1\), \(p^n_n\rightarrow a\) and \(q^n_n\rightarrow b\) as \(n\rightarrow \infty \). For any \(f \in C^*_{x^2}[0,\infty )\), such that \(f',f'' \in C^*_{x^2}[0,\infty )\), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}|K_n^{(p_n,q_n)}(f;x)-f(x)|=(\alpha x+ 1/2) f'(x)+x(\gamma x+1)f''(x)/2 \end{aligned}$$

uniformly on [0, A] for any \(A>0\). Here \(\alpha =\lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}(q_n-1)\) and \(\gamma =[n]_{p_{n},q_{n}}\lim _{n\rightarrow \infty }(p_nq_n-2q_n+1)\).

Proof

By the Taylor’s formula, we have

$$\begin{aligned} f(t)=f(x)+(t-x)f'(x)+\frac{1}{2}f''(x)(t-x)^2+r(t, x)(t-x)^2, \end{aligned}$$

here r(t, x) is reminder term and \(\lim _{t\rightarrow x} r(t,x)=0\). Therefore,

$$\begin{aligned}{}[n]_{p_{n},q_{n}}(K_n^{(p_n,q_n)}(f;x)-f(x))= & {} [n]_{p_{n},q_{n}}f'(x) K_n^{(p_n,q_n)}((t-x);x)\\&+\,[n]_{p_{n},q_{n}}\frac{f''(x)}{2} K_n^{(p_n,q_n)}((t-x)^2;x)\\&+\,[n]_{p_{n},q_{n}}K_n^{(p_n,q_n)}(r(t, x)(t-x)^2;x). \end{aligned}$$

By the Cauchy-Schwartz inequality, we have

$$\begin{aligned} K_n^{(p_n,q_n)}(r(t, x)(t-x)^2;x)\le \sqrt{K_n^{(p_n,q_n)}(r^2(t, x);x)}\sqrt{K_n^{(p_n,q_n)}((t-x)^4;x)}. \end{aligned}$$

As \(r(t,x) \in C^*_{x^2}[0,\infty )\), therefore by Theorem 2 and fact that \(\lim _{t\rightarrow x} r(t,x)=0\), we get

$$\begin{aligned} \lim _{n\rightarrow \infty } K_n^{(p_n,q_n)}(r^2(t, x);x)=r^2(x, x)=0, \end{aligned}$$

uniformly for any \(x \in [0, A]\). Hence, by using above equality and positivity of linear operator, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}} K_n^{(p_n,q_n)}(r(t, x)(t-x)^2;x)=0, \end{aligned}$$

Therefore,

$$\begin{aligned} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}(K_n^{(p_n,q_n)}(f;x)-f(x))= & {} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}f'(x) K_n^{(p_n,q_n)}((t-x);x) \nonumber \\&+\lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}\frac{f''(x)}{2} K_n^{(p_n,q_n)}((t-x)^2;x).\nonumber \\ \end{aligned}$$
(4.1)

Consider,

$$\begin{aligned} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}K_n^{(p_n,q_n)}((t-x);x)= & {} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}\Phi _1^{(p_n , q_n)}(x)\nonumber \\= & {} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}(q_n-1)x+1/2 \nonumber \\= & {} \alpha x+ 1/2, \end{aligned}$$
(4.2)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}K_n^{(p_n,q_n)}((t-x)^2;x)= & {} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}} \Phi _2^{(p_n, q_n)}(x)\nonumber \\= & {} \lim _{n\rightarrow \infty }[n]_{p_{n},q_{n}}(p_nq_n-2q_n+1)x^2+x\nonumber \\= & {} x(\gamma x +1). \end{aligned}$$
(4.3)

Hence, by using Eqs. (4.1), (4.2) and (4.3), we get the theorem. \(\square \)