In this section, we research the approximation properties of \((p,q)\)-gamma operators. The following two theorems show approximation properties about Lipschitz functions.
Theorem 3.1
Let
\(0< q< p\leq 1\)
and
F
be any bounded subset of the interval
\([0,\infty )\). If
\(f\in C_{B}[0,\infty )\cap \mathrm{Lip}_{M}( \gamma , F)\), then, for all
\(x\in (0,\infty )\), we have
$$ \bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert \leq M \bigl( \bigl(B(x)\bigr)^{\frac{\gamma }{2}}+2d^{ \gamma }(x;F) \bigr), $$
where
\(d(x;F)\)
is the distance between
x
and
F
defined by
\(d(x;F)=\inf \{|x-y|:y\in F\}\).
Proof
Let F̅ be the closure of F in \([0,\infty )\). Using the properties of infimum, there is at least a point \(y_{0}\in \overline{F}\) such that \(d(x;F)=|x-y_{0}|\). By the triangle inequality, we can obtain
$$\begin{aligned} \bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert &\leq G_{n}^{p,q}\bigl( \bigl\vert f(x)-f(t) \bigr\vert ;x\bigr) \\ &\leq G_{n}^{p,q}\bigl( \bigl\vert f(x)-f(y_{0}) \bigr\vert ;x\bigr)+G_{n}^{p,q}\bigl( \bigl\vert f(t)-f(y_{0}) \bigr\vert ;x\bigr) \\ &\leq M \bigl(G_{n}^{p,q}\bigl( \vert t-y_{0} \vert ^{\gamma };x\bigr)+G_{n}^{p,q}\bigl( \vert x-y _{0} \vert ^{\gamma };x\bigr) \bigr) \\ &\leq M \bigl(G_{n}^{p,q}\bigl( \vert x-t \vert ^{\gamma };x\bigr)+2d^{\gamma }(x;F) \bigr). \end{aligned}$$
Choosing \(k_{1}=\frac{2}{\gamma }\) and \(k_{2}=\frac{2}{2-\gamma }\) and using the well-known Hölder inequality, we have
$$\begin{aligned} \bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert &\leq M \bigl( \bigl(G_{n}^{p,q}\bigl( \vert x-t \vert ^{k_{1} \gamma };x\bigr) \bigr)^{\frac{1}{k_{1}}} \bigl(G_{n}^{p,q} \bigl(1^{k_{2}};x\bigr) \bigr) ^{\frac{1}{k_{2}}}+2d^{\gamma }(x;F) \bigr) \\ &\leq M \bigl(G_{n}^{p,q}\bigl((x-t)^{2};x \bigr)^{\frac{\gamma }{2}}+2d^{ \gamma }(x;F) \bigr) \\ &=M \bigl(\bigl(B(x)\bigr)^{\frac{\gamma }{2}}+2d^{\gamma }(x;F) \bigr). \end{aligned}$$
This completes the proof. □
Theorem 3.2
Let
\(0< q< p\leq 1\). Then, for all
\(f\in \mathrm{Lip}_{M}(\gamma )\), we have
$$ \bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert \leq MB^{\frac{\gamma }{2}}(x). $$
Proof
Using the monotonicity of the operators \(G_{n}^{p,q}\) and the Hölder inequality, we can obtain
$$\begin{aligned} \bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert &\leq G_{n}^{p,q} \bigl( \bigl\vert f(t)-f(x) \bigr\vert ;x \bigr) \leq MG_{n}^{p,q} \bigl( \vert t-x \vert ^{\gamma };x \bigr) \\ &=MG_{n}^{p,q} \bigl(\bigl( \vert t-x \vert ^{2}\bigr)^{\frac{\gamma }{2}};x \bigr) \leq M \bigl(G_{n}^{p,q} \bigl((t-x)^{2};x\bigr) \bigr)^{\frac{\gamma }{2}}=MB ^{\frac{\gamma }{2}}(x). \end{aligned}$$
□
The third theorem is a direct local approximation theorem for the operators \(G_{n}^{p,q}(f;x)\).
Theorem 3.3
Let
\(0< q< p\leq 1\), \(f\in C_{B}[0,\infty )\). Then, for every
\(x\in (0,\infty )\), there exists a positive constant
\(C_{1}\)
such that
$$ \bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert \leq C_{1}\omega _{2} \bigl(f;\sqrt{B(x)+A ^{2}(x)} \bigr)+\omega \bigl(f; \bigl\vert A(x) \bigr\vert \bigr). $$
Proof
For \(x\in (0,\infty )\), we consider new operators \(H_{n}^{p,q}(f;x)\) defined by
$$ H_{n}^{p,q}(f;x)=G_{n}^{p,q}(f;x)+f(x)-f \bigl(A(x)+x \bigr). $$
Using the operator above and Lemma 2.3, we have
$$ H_{n}^{p,q}(t-x;x)= G_{n}^{p,q}(t-x;x)-A(x)=0 . $$
Let \(x,t\in (0,\infty )\) and \(g\in C_{B}^{2}[0,\infty )\). Using Taylor’s expansion, we can obtain
$$ g(t)=g(x)+g'(x) (t-x)+ \int _{x}^{t}g''(u) (t-u) \,\mathrm{d} u. $$
Hence,
$$\begin{aligned} \bigl\vert H_{n}^{p,q}(g;x)-g(x) \bigr\vert & = \biggl\vert g'(x)H_{n}^{p,q} \bigl((t-x);x \bigr)+H _{n}^{p,q} \biggl( \int _{x}^{t}g''(u) (t-u) \,\mathrm{d} u;x \biggr) \biggr\vert \\ &\leq \biggl\vert H_{n}^{p,q} \biggl( \int _{x}^{t}g''(u) (t-u) \,\mathrm{d} u;x \biggr) \biggr\vert \\ &\leq \biggl\vert G_{n}^{p,q} \biggl( \int _{x}^{t}g''(u) (t-u) \,\mathrm{d} u;x \biggr) - \int _{x}^{A(x)+x}g''(u) \bigl(A(x)+x-u\bigr)\,\mathrm{d} u \biggr\vert \\ &\leq G_{n}^{p,q} \biggl( \int _{x}^{t} \bigl\vert g''(u) \bigr\vert (t-u)\,\mathrm{d} u;x \biggr) + \biggl\vert \int _{x}^{A(x)+x} \bigl\vert g''(u) \bigr\vert \bigl(A(x)+x-u\bigr)\,\mathrm{d} u \biggr\vert \\ &\leq \bigl(B(x)+A^{2}(x)\bigr) \bigl\Vert g'' \bigr\Vert . \end{aligned}$$
Using \(|G_{n}^{p,q}(f;x)|\leq \|f\|\), we have
$$\begin{aligned} &\bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert \\ &\quad = \bigl\vert H_{n}^{p,q}(f;x)+f \bigl(A(x)+x \bigr)-2f(x) \bigr\vert \\ &\quad \leq \bigl\vert H_{n}^{p,q}(f-g;x)-(f-g) (x) \bigr\vert + \bigl\vert H_{n}^{p,q}(g;x)-g(x) \bigr\vert + \bigl\vert f \bigl(A(x)+x \bigr)-f(x) \bigr\vert \\ &\quad \leq 4 \Vert f-g \Vert +\bigl(B(x)+A^{2}(x)\bigr) \bigl\Vert g'' \bigr\Vert +\omega \bigl(f; \bigl\vert A(x) \bigr\vert \bigr). \end{aligned}$$
Taking infimum over all \(g\in C_{B}^{2}[0,\infty )\) and using (3), we can obtain the desired assertion. □
The fourth theorem is a result about the rate of convergence for the operators \(G_{n}^{p,q}(f;x)\):
Theorem 3.4
Let
\(f\in C_{x^{2}}[0,\infty )\), \(0< q< p\leq 1\), and
\(a>0\), we have
$$ \bigl\Vert G_{n}^{p,q}(f;x)-f(x) \bigr\Vert _{C(0,a]}\leq 4C_{f}\bigl(1+a^{2}\bigr)B(a)+2\omega _{a+1}\bigl(f;\sqrt{B(a)}\bigr). $$
Proof
For all \(x\in (0,a]\) and \(t>a+1\), we easily have \((t-x)^{2}\geq (t-a)^{2} \geq 1\), therefore,
$$\begin{aligned} \begin{aligned} \bigl\vert f(t)-f(x) \bigr\vert & \leq \bigl\vert f(t) \bigr\vert + \bigl\vert f(x) \bigr\vert \leq C_{f}\bigl(2+x^{2}+t^{2}\bigr) \\ &=C_{f} \bigl(2+x^{2}+(x-t-x)^{2} \bigr)\leq C_{f} \bigl(2+3x^{2}+2(x-t)^{2} \bigr) \\ &\leq C_{f}\bigl(4+3x^{2}\bigr) (t-x)^{2}\leq 4C_{f}\bigl(1+a^{2}\bigr) (t-x)^{2}, \end{aligned} \end{aligned}$$
(8)
and for all \(x\in (0,a]\), \(t\in (0,a+1]\), and \(\delta >0\), we have
$$ \bigl\vert f(t)-f(x) \bigr\vert \leq \omega _{a+1} \bigl(f, \vert t-x \vert \bigr)\leq \biggl(1+\frac{ \vert t-x \vert }{ \delta } \biggr)\omega _{a+1}(f;\delta ). $$
(9)
From (8) and (9), we get
$$ \bigl\vert f(t)-f(x) \bigr\vert \leq 4C_{f}\bigl(1+a^{2} \bigr) (t-x)^{2}+ \biggl(1+\frac{ \vert t-x \vert }{ \delta } \biggr)\omega _{a+1}(f;\delta ). $$
By Schwarz’s inequality and Lemma 2.3, we have
$$ \begin{aligned} &\bigl\vert G_{n}^{p,q}(f;x)-f(x) \bigr\vert \\ &\quad \leq G_{n}^{p,q}\bigl( \bigl\vert f(t)-f(x) \bigr\vert ;x\bigr) \\ &\quad \leq 4C _{f}\bigl(1+a^{2}\bigr)G_{n}^{p,q} \bigl((t-x)^{2};x\bigr)+G_{n}^{p,q} \biggl( \biggl(1+ \frac{ \vert t-x \vert }{ \delta } \biggr);x \biggr)\omega _{a+1}(f;\delta ) \\ &\quad \leq 4C_{f}\bigl(1+a ^{2}\bigr)G_{n}^{p,q} \bigl((t-x)^{2};x\bigr)+\omega _{a+1}(f;\delta ) \biggl(1+ \frac{1}{ \delta }\sqrt{G_{n}^{p,q} \bigl((t-x)^{2};x\bigr)} \biggr) \\ &\quad \leq 4C_{f}\bigl(1+a ^{2}\bigr)B(x)+\omega _{a+1}(f;\delta ) \biggl(1+\frac{1}{\delta } \sqrt{B(x)} \biggr) \\ &\quad \leq 4C_{f}\bigl(1+a^{2}\bigr)B(a)+\omega _{a+1}(f;\delta ) \biggl(1+\frac{1}{\delta }\sqrt{B(a)} \biggr). \end{aligned} $$
By taking \(\delta =\sqrt{B(a)}\) and supremum over all \(x\in (0,a]\), we accomplish the proof of Theorem 3.4. □
The following three results are theorems about weighted approximation for the operators \(G_{n}^{p,q}(f;x)\).
Theorem 3.5
Let
\(f\in C^{0}_{x^{2}}[0,\infty )\)
and the sequences
\((p_{n})\), \((q_{n})\)
satisfy
\(0< q_{n}< p_{n}\leq 1\)
such that
\(p_{n}^{n}\rightarrow 1\), \(q_{n}^{n}\rightarrow 1\), \([n]_{p_{n},q_{n}}\rightarrow \infty \)
as
\(n\rightarrow \infty \), then there exists a positive integer
\(N\in \mathbb{N_{+}}\)
such that, for all
\(n>N\)
and
\(\nu >0\), the inequality
$$ \sup _{x\in (0,\infty )}\frac{ \vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \vert }{(1+x ^{2})^{\frac{3}{2}+\nu }}\leq 4\sqrt{2}\varOmega \biggl(f;\frac{1}{\sqrt{[n-1]_{p _{n},q_{n}}}} \biggr) $$
(10)
holds.
Proof
For \(t>0\), \(x\in (0,\infty )\) and \(\delta >0\), by the definition and properties of \(\varOmega (f;\delta )\), we get
$$\begin{aligned} \bigl\vert f(t)-f(x) \bigr\vert &\leq \bigl(1+ \bigl(x+ \vert x-t \vert \bigr) \bigr)^{2}\varOmega \bigl(f; \vert t-x \vert \bigr) \\ &\leq 2\bigl(1+x^{2}\bigr) \bigl(1+(t-x)^{2} \bigr) \biggl(1+\frac{ \vert t-x \vert }{ \delta } \biggr)\varOmega (f;\delta ). \end{aligned}$$
Using \(p_{n}^{n}\rightarrow 1\), \(q_{n}^{n}\rightarrow 1\), \([n]_{p_{n},q _{n}}\rightarrow \infty \) as \(n\rightarrow \infty \) and Lemma 2.4, there exists a positive integer \(N\in \mathbb{N_{+}}\) such that, for all \(n>N\),
$$\begin{aligned}& G_{n}^{p_{n},q_{n}}\bigl((t-x)^{2};x\bigr) \leq \frac{2(1+x^{2})}{[n-1]_{p_{n},q _{n}}}, \end{aligned}$$
(11)
$$\begin{aligned}& G_{n}^{p_{n},q_{n}}\bigl((t-x)^{4};x\bigr) \leq 1. \end{aligned}$$
(12)
Since \(G_{n}^{p_{n},q_{n}}\) is linear and positive, we have
$$\begin{aligned} \begin{aligned}[b] \bigl\vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \bigr\vert \leq {}&2\bigl(1+x^{2}\bigr)\varOmega (f; \delta ) \biggl\{ 1+G_{n}^{p_{n},q_{n}} \bigl((t-x)^{2};x \bigr) \\ &{}+G_{n}^{p_{n},q_{n}} \biggl( \bigl(1+(t-x)^{2} \bigr) \frac{ \vert t-x \vert }{ \delta };x \biggr) \biggr\} . \end{aligned} \end{aligned}$$
(13)
To estimate the second term of (13), applying the Cauchy–Schwarz inequality and \((x+y)^{2}\leq 2(x^{2}+y^{2})\), we have
$$ G_{n}^{p_{n},q_{n}} \biggl( \bigl(1+(t-x)^{2} \bigr) \frac{ \vert t-x \vert }{ \delta };x \biggr)\leq \sqrt{2} \bigl(G_{n}^{p_{n},q_{n}} \bigl(1+(t-x)^{4};x \bigr) \bigr) ^{\frac{1}{2}} \biggl(G_{n}^{p_{n},q_{n}} \biggl(\frac{(t-x)^{2}}{ \delta ^{2}};x \biggr) \biggr)^{\frac{1}{2}}. $$
By (11) and (12),
$$ G_{n}^{p_{n},q_{n}} \biggl( \bigl(1+(t-x)^{2} \bigr) \frac{ \vert t-x \vert }{ \delta };x \biggr)\leq \frac{2\sqrt{2}(1+x^{2})^{\frac{1}{2}}}{ \delta [n-1]_{p_{n},q_{n}}}. $$
Taking \(\delta =\frac{1}{\sqrt{[n-1]_{p_{n},q_{n}}}}\), we can obtain
$$ \bigl\vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \bigr\vert \leq 4 \sqrt{2}\bigl(1+x^{2}\bigr)^{ \frac{3}{2}}\varOmega \biggl(f; \frac{1}{\sqrt{[n-1]_{p_{n},q_{n}}}} \biggr). $$
The proof is completed. □
Theorem 3.6
Let the sequences
\((p_{n})\), \((q_{n})\)
satisfy
\(0< q_{n}< p_{n}\leq 1\)
such that
\(p_{n}\rightarrow 1\), \(q_{n}\rightarrow 1\), and
\(p_{n}^{n} \rightarrow \alpha \), \(q_{n}^{n}\rightarrow \beta \), \([n]_{p_{n},q _{n}}\rightarrow \infty \)
as
\(n\rightarrow \infty \). Then, for
\(f\in C^{0}_{x^{2}}[0,\infty )\), we have
$$ \lim _{n\rightarrow \infty } \bigl\Vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \bigr\Vert _{x^{2}}=0. $$
(14)
Proof
By the Korovkin theorem in [14], we see that it is sufficient to verify the following three conditions:
$$ \lim _{n\rightarrow \infty } \bigl\Vert G_{n}^{p_{n},q_{n}} \bigl(t^{k};x\bigr)-x ^{k} \bigr\Vert _{x^{2}}=0,\quad k=0,1,2. $$
(15)
Since \(G_{n}^{p_{n},q_{n}}(1;x)=1\), \(G_{n}^{p_{n},q_{n}}(t^{2};x)=x ^{2}\), then (15) holds true for \(k=0,2\). By Lemma 2.2, we can get
$$\begin{aligned} \bigl\Vert G_{n}^{p_{n},q_{n}}(t;x)-x \bigr\Vert _{x^{2}} &=\sup _{x\in (0,\infty )}\frac{1}{1+x ^{2}} \bigl\vert G_{n}^{p_{n},q_{n}}(t;x)-x \bigr\vert \\ &=\sup _{x\in (0,\infty )}\frac{x}{1+x^{2}} \biggl\vert \frac{\sqrt{p _{n}}-\sqrt{q_{n}}}{\sqrt{q_{n}}}- \sqrt{\frac{p_{n}}{q_{n}}}\frac{p _{n}^{n+1}}{[n+2]_{p_{n},q_{n}}} \biggr\vert \\ &\leq \sup _{x\in (0,\infty )} \biggl\vert \frac{\sqrt{p_{n}}-\sqrt{q _{n}}}{\sqrt{q_{n}}}-\sqrt{ \frac{p_{n}}{q_{n}}}\frac{p_{n}^{n+1}}{[n+2]_{p _{n},q_{n}}} \biggr\vert \rightarrow 0,\quad n \rightarrow \infty . \end{aligned}$$
Thus the proof is completed. □
Theorem 3.7
Let the sequences
\((p_{n})\), \((q_{n})\)
satisfy
\(0< q_{n}< p_{n}\leq 1\)
such that
\(p_{n}\rightarrow 1\), \(q_{n}\rightarrow 1\), \([n]_{p_{n},q _{n}}\rightarrow \infty \)
as
\(n\rightarrow \infty \). For every
\(f\in C_{x^{2}}[0,\infty )\)
and
\(\kappa >0\), we have
$$ \lim _{n\rightarrow \infty }\sup _{x\in (0,\infty )}\frac{ \vert G _{n}^{p_{n},q_{n}}(f;x)-f(x) \vert }{(1+x^{2})^{1+\kappa }}=0. $$
Proof
Let \(x_{0}\in (0,\infty )\) be arbitrary but fixed. Then
$$\begin{aligned} \begin{aligned}[b] \sup _{x\in (0,\infty )}\frac{ \vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \vert }{(1+x ^{2})^{1+\kappa }} \leq{} &\sup _{x\in (0,x_{0}]}\frac{ \vert G_{n}^{p _{n},q_{n}}(f;x)-f(x) \vert }{(1+x^{2})^{1+\kappa }} \\ &{}+ \sup _{x\in (x_{0},\infty )}\frac{ \vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \vert }{(1+x ^{2})^{1+\kappa }} \\ \leq{}& \bigl\Vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \bigr\Vert _{C(0,x_{0}]} \\ &{}+C_{f}\sup _{x\in (x_{0},\infty )}\frac{ \vert G_{n}^{p_{n},q_{n}}((1+t ^{2});x) \vert }{(1+x^{2})^{1+\kappa }} \\ &{}+\sup _{x\in (x_{0},\infty )}\frac{ \vert f(x) \vert }{(1+x^{2})^{1+ \kappa }}. \end{aligned} \end{aligned}$$
(16)
Since \(|f(x)|\leq C_{f}(1+x^{2})\), we have \(\sup_{x\in (x_{0},\infty )}\frac{|f(x)|}{(1+x^{2})^{1+\kappa }} \leq \frac{C_{f}}{(1+x_{0}^{2})^{\kappa }}\). Let \(\epsilon >0\) be arbitrary. We can choose \(x_{0}\) to be so large that
$$ \frac{C_{f}}{(1+x_{0}^{2})^{\kappa }}< \epsilon . $$
(17)
In view of Lemma 2.2, while \(x\in (x_{0},\infty )\), we obtain
$$ C_{f}\lim _{n\rightarrow \infty }\frac{ \vert G_{n}^{p_{n},q_{n}}((1+t ^{2});x) \vert }{(1+x^{2})^{1+\kappa }} =C_{f} \frac{(1+x^{2})}{(1+x^{2})^{1+ \kappa }}=\frac{C_{f}}{(1+x^{2})^{\kappa }}\leq \frac{C_{f}}{(1+x_{0} ^{2})^{\kappa }}< \epsilon . $$
Using Theorem 3.4, we can see that the first term of inequality (16) implies that
$$ \bigl\Vert G_{n}^{p_{n},q_{n}}(f;x)-f(x) \bigr\Vert _{C(0,x_{0}]}< \epsilon ,\quad \mbox{as } n\rightarrow \infty . $$
(18)
Combining (16)–(18), we get the desired result. □
The last result is a Voronovskaja-type asymptotic formula for the operators \(G_{n}^{p,q}(f;x)\).
Theorem 3.8
Let
\(f\in C_{B}^{2}[0,\infty )\)
and the sequences
\((p_{n})\), \((q_{n})\)
satisfy
\(0< q_{n}< p_{n}\leq 1\)
such that
\(p_{n}\rightarrow 1\), \(q_{n}\rightarrow 1\)
and
\(p_{n}^{n}\rightarrow \alpha \), \(q_{n}^{n} \rightarrow \beta \), \([n]_{p_{n},q_{n}}\rightarrow \infty \)
as
\(n\rightarrow \infty \), where
\(0\leq \alpha ,\beta <1\). Then, for all
\(x\in (0,\infty )\),
$$ \lim _{n\rightarrow \infty }[n-1]_{p_{n},q_{n}} \bigl(G_{n}^{p _{n},q_{n}}(f;x)-f(x) \bigr)=\frac{\alpha +\beta }{2} \bigl(-xf'(x)+x ^{2}f''(x) \bigr). $$
(19)
Proof
Let \(x\in (0,\infty )\) be fixed. By Taylor’s expansion formula, we obtain
$$ f(t)=f(x)+f'(x) (t-x)+ \biggl(\frac{1}{2}f''(x)+ \varTheta _{p_{n},q_{n}}(t,x) \biggr) (t-x)^{2}, $$
where \(\varTheta _{p_{n},q_{n}}(x,t)\) is bounded and \(\lim_{t\rightarrow x}\varTheta _{p_{n},q_{n}}(t,x)=0\). By applying the operator \(G_{n}^{p_{n},q_{n}}(f;x)\) to the relation above, we obtain
$$\begin{aligned} G_{n}^{p_{n},q_{n}}(f;x)-f(x)={}&f'(x)G_{n}^{p_{n},q_{n}} \bigl((t-x);x \bigr)+ \frac{1}{2}f''(x)G_{n}^{p_{n},q_{n}} \bigl((t-x)^{2};x \bigr) \\ &{}+G_{n}^{p_{n},q_{n}} \bigl(\varTheta _{p_{n},q_{n}}(t,x) (t-x)^{2};x \bigr). \end{aligned}$$
Since \(\lim_{t\rightarrow x}\varTheta _{p_{n},q_{n}}(t,x)=0\), then for all \(\epsilon >0\), there exists a positive constant \(\delta >0\) which implies \(|\varTheta _{p_{n},q_{n}}(t,x)|<\epsilon \) for all fixed \(x\in (0,\infty )\), where n is large enough, while \(|t-x|\leq \delta \), then \(|\varTheta _{p_{n},q_{n}}(t,x)|<\frac{C_{2}}{\delta ^{2}}(t-x)^{2}\), where \(C_{2}\) is a positive constant. Using Lemma 2.4, we obtain
$$\begin{aligned} &[n-1]_{p_{n},q_{n}} \bigl\vert G_{n}^{p_{n},q_{n}} \bigl(\varTheta (t,x) (t-x)^{2};x \bigr) \bigr\vert \\ &\quad \leq \epsilon [n-1]_{p_{n},q_{n}}G_{n}^{p_{n},q_{n}} \bigl((t-x)^{2};x \bigr) \\ &\qquad {}+\frac{C_{2}}{\delta ^{2}}[n-1]_{p_{n},q_{n}}G_{n}^{p_{n},q_{n}} \bigl((t-x)^{4};x \bigr) \rightarrow 0\quad (n\rightarrow \infty ). \end{aligned}$$
The proof is completed. □