In this section, let us examine some pointwise estimates of the rates of convergence of the newly defined Gamma operators. First, the local approximation and the relationship between the local smoothness of g are given. For that, let us describe the following. Let \(s\in (0,1] \) and \(Q\subset [0,\infty ) \). In the circumstances, a function \(g\in C_{b}[0,\infty ) \) can be said \(\operatorname{Lip}_{M_{g}}(s) \) on Q if the following condition holds:
$$ \bigl\vert g(y)-g(x) \bigr\vert \leq M_{g,s} \vert y-x \vert ^{s},\quad y \in [0, \infty ) \text{ and } x \in Q, $$
where \(M_{g,s}\) is fixed just depending on g and s.
Theorem 5
Let \(g \in C_{b}[0,\infty )\cap \operatorname{Lip}_{Mg}(s)\) such that \(s \in (0,1]\) and \(Q \subset [0,\infty )\) given as above. In the circumstances, we have the following inequality:
$$ \bigl\vert \mathcal{T}^{*}_{n}(g,x)- g(x) \bigr\vert \leq M_{g,s} \biggl[ \biggl({ \frac{n+2}{(n-1)(n-2)}}e_{2}(x) \biggr)^{s/2}+2\bigl(d(x,Q)\bigr)^{s} \biggr], \quad x \in (0, \infty ), $$
where \(M_{g,s}\) is defined as above and \(d(x,Q)\) is the distance between x and Q described as
$$ d(x,Q)= \inf \bigl\lbrace \vert y-x \vert , y \in Q \bigr\rbrace . $$
Proof
Let us describe the closure of the set Q as Q̅. Then, one can say that there exists at least one point \(y_{0} \in \overline{Q}\) such that
$$ d(x,Q)= \vert x-y_{0} \vert . $$
Then, utilising the monotonicity properties of \((\mathcal{T}^{*}_{n})_{n \geq 1}\), we deduce that
$$\begin{aligned} \bigl\vert \mathcal{T}^{*}_{n}(g,x)- g(x) \bigr\vert &\leq \mathcal{T}^{*}_{n}\bigl( \bigl\vert g(y)- g(y_{0}) \bigr\vert ,x\bigr)+\mathcal{T}^{*}_{n}\bigl( \bigl\vert g(x)- g(y_{0}) \bigr\vert ,x\bigr) \\ &\leq M_{g,s} \bigl[\mathcal{T}^{*}_{n}\bigl( \vert y-y_{0} \vert ^{s},x\bigr)+ \vert x-y_{0} \vert ^{s}\bigr] \\ &\leq M_{g,s} \bigl[\mathcal{T}^{*}_{n}\bigl( \vert y-x \vert ^{s},x\bigr)+ 2 \vert x-y_{0} \vert ^{s}\bigr]. \end{aligned}$$
In the circumstances, with the help of the Hölder inequality, we obtain the following result:
$$\begin{aligned} \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert &\leq M_{g,s} \bigl[ \bigl( \mathcal{T}^{*}_{n} \bigl( \vert y-x \vert ^{2}, x \bigr) \bigr)^{s/2}+2 \bigl(d(x,Q)\bigr)^{s} \bigr] \\ &=M_{g,s} \biggl[ \biggl({\frac{n+2}{(n-1)(n-2)}}e_{2}(x) \biggr)^{s/2}+2\bigl(d(x,Q)\bigr)^{s} \biggr], \end{aligned}$$
which finalises the proof. □
Let us now calculate the local direct estimate of the new modification of Gamma operators. For this purpose, we need to review the Lipschitz-type maximal function of order s given in [9], that is
$$ \tilde{\omega }_{s}(g,x)= \sup_{0\leq y < \infty , y\neq x} \frac{ \vert g(y) - g(x) \vert }{ \vert y-x \vert ^{s}}, $$
where \(s \in (0,1] \) and \(x \in (0,\infty )\). Now, we can present and prove the theorem.
Theorem 6
Let \(g \in C_{b}[0,\infty )\) and \(s \in (0,1]\), then the following inequality holds:
$$ \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert \leq \tilde{\omega }_{s}(g,x) \mathcal{T}^{*}_{n} \biggl({\frac{n+2}{(n-1)(n-2)}}e_{2}(x) \biggr)^{s/2}, $$
for \(x \in (0,\infty )\).
Proof
Thanks to the definitions of \(\tilde{\omega }_{s}(g,x)\) given above and a well-recognised Hölder inequality, we deduce that,
$$\begin{aligned} \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert &\leq \mathcal{T}^{*}_{n} \bigl( \bigl\vert g(y)-g(x) \bigr\vert ,x\bigr) \\ &\leq \tilde{\omega }_{s}(g,x)\mathcal{T}^{*}_{n} \bigl( \vert y-x \vert ^{s},x\bigr) \\ &\leq \tilde{\omega }_{s}(g,x)\mathcal{T}^{*}_{n} \bigl( \vert y-x \vert ^{2},x\bigr)^{s/2} \\ &\leq \tilde{\omega }_{s}(g,x)\mathcal{T}^{*}_{n} \biggl({ \frac{n+2}{(n-1)(n-2)}}e_{2}(x) \biggr)^{s/2}, \end{aligned}$$
thus, the desired result is obtained. □
Finally, let us consider the following Lipschitz-type space with two parameters, \(\alpha , \beta > 0\), such that
$$ \operatorname{Lip}_{M}^{\alpha ,\beta }(s)= \biggl( g \in C[0,\infty ): \bigl\vert g(y)-g(x) \bigr\vert \leq M \frac{ \vert y-x \vert ^{s}}{(ax^{2}+ bx+y)^{s/2}},x,y \in (0,\infty ) \biggr), $$
introduced in [14], where \(s \in (0,1] \) and M is a positive constant.
Theorem 7
Let us consider \(g \in \operatorname{Lip}_{M}^{\alpha ,\beta }(s)\) and \(x \in (0,\infty )\). Then we have
$$ \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert \leq M \biggl[ \frac{{\frac{n+2}{(n-1)(n-2)}}e_{2}(x)}{ax^{2}+ bx} \biggr]^{s/2}, $$
where \(\alpha , \beta > 0\).
Proof
The proof of this inequality is shown in two steps. First, we take \(s=1\), that is,
$$\begin{aligned} \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert & \leq \mathcal{T}^{*}_{n}\bigl( \bigl\vert g(y)-g(x) \bigr\vert , x\bigr) \\ &\leq M \mathcal{T}^{*}_{n} \biggl( \frac{ \vert y-x \vert }{\sqrt{ax^{2}+bx+y}}, x \biggr) \\ &\leq \frac{M}{\sqrt{ax^{2}+bx}} \mathcal{T}^{*}_{n}\bigl( \vert y-x \vert ,x\bigr), \end{aligned}$$
\(g \in \operatorname{Lip}_{M}^{\alpha ,\beta }(1) \) and \(x \in (0,\infty )\). Here, applying the Cauchy–Schwarz inequality, we deduce that,
$$ \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert \leq \frac{M}{\sqrt{ax^{2}+bx}}\bigl[\mathcal{T}^{*}_{n}\bigl( \vert y-x \vert ^{2},x\bigr)\bigr]^{1/2} \leq M \biggl[ \frac{{\frac{n+2}{(n-1)(n-2)}}e_{2}(x)}{ax^{2}+bx} \biggr]^{1/2}, $$
which confirms the proof of the theorem for \(s=1\). Then, let us consider \(s \in (0,1)\). For \(g \in \operatorname{Lip}_{M}^{\alpha ,\beta }(s)\) and \(x \in (0,\infty )\) we obtain that
$$ \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert \leq \frac{M}{(ax^{2}+bx)^{s/2}}\mathcal{T}^{*}_{n}\bigl( \vert y-x \vert ^{s},x\bigr). $$
With the help of Hölder inequalities, we obtain the following inequality:
$$ \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert \leq \frac{M}{(ax^{2}+bx)^{s/2}}\mathcal{T}^{*}_{n}\bigl( \vert y-x \vert ^{s},x\bigr) \leq \frac{M}{(ax^{2}+bx)^{s/2}} \bigl(\mathcal{T}^{*}_{n}\bigl( \vert y-x \vert ,x \bigr)\bigr)^{s}. $$
Finally, applying the Cauchy–Schwarz inequality, we have,
$$ \bigl\vert \mathcal{T}^{*}_{n}(g,x)-g(x) \bigr\vert \leq \frac{M}{(ax^{2}+bx)^{s/2}} \bigl( \mathcal{T}^{*}_{n}\bigl( \vert y-x \vert ^{2},x\bigr)\bigr)^{s/2} \leq M \biggl[ \frac{{\frac{n+2}{(n-1)(n-2)}}e_{2}(x)}{ax^{2}+ bx} \biggr]^{s/2}, $$
which completes the proof. □