Let \(f\in C_{B}[0,\infty]\), the space of all bounded and continuous functions on \([0,\infty)\) and \(x\geq\frac{1}{2n}\), \(n\in\mathbb{N}\). Then, for \(\delta>0\), the modulus of continuity of f denoted by \(\omega (f,\delta)\) gives the maximum oscillation of f in any interval of length not exceeding \(\delta>0\), and it is given by
$$ \omega(f,\delta)=\sup_{| t-x|\leq\delta}\bigl\vert f(t)-f(x)\bigr\vert , \quad t\in{}[0,\infty). $$
(4.1)
It is known that \(\lim_{\delta\rightarrow0+}\omega(f,\delta)=0\) for \(f\in C_{B}[0,\infty)\), and for any \(\delta>0\) we have
$$ \bigl\vert f(t)-f(x)\bigr\vert \leq \biggl( \frac{| t-x|}{\delta}+1 \biggr) \omega (f,\delta). $$
(4.2)
Now we calculate the rate of convergence of operators (2.2) by means of modulus of continuity and Lipschitz-type maximal functions.
Theorem 4.1
Let
\(D_{n,q}^{\ast}(\cdot ; \cdot)\)
be the operators defined by (2.2). Then, for
\(f\in C_{B}[0,\infty)\), \(x\geq\frac{1}{2n}\)
and
\(n\in\mathbb{N}\), we have
$$ \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq2\omega ( f;\delta _{n,x} ) , $$
where
$$ \delta_{n,x}=\sqrt{[1+2\mu]_{q}\frac{x}{[n]_{q}}- \frac {1}{4[n]_{q}^{2}} \bigl( 2[1+2\mu]_{q}-1 \bigr) }. $$
(4.3)
Proof
We prove it by using (4.1), (4.2) and the Cauchy-Schwarz inequality. We can easily get
$$ \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq \biggl\{ 1+\frac{1}{\delta} \bigl( D_{n,q}^{\ast}(t-x)^{2};x \bigr) ^{\frac{1}{2}} \biggr\} \omega (f;\delta) $$
if we choose \(\delta=\delta_{n,x}\), and by applying the result (2) of Lemma 2.2, we get the result. □
Remark 4.2
For the operators \(D_{n,q}(\cdot ; \cdot)\) defined by (1.9) we may write that, for every \(f\in C_{B}[0,\infty)\), \(x\geq0\) and \(n \in\mathbb{N}\),
$$ \bigl\vert D_{n,q}(f;x)-f(x)\bigr\vert \leq2\omega (f;\lambda_{n,x} ), $$
(4.4)
where by [19] we have
$$ \lambda_{n,x}=\sqrt{D_{n,q} \bigl((t-x)^{2};x\bigr)}\leq\sqrt{[1+2\mu]_{q} \frac{x}{[n]_{q}}}. $$
(4.5)
Now we claim that the error estimation in Theorem 4.1 is better than that of (4.4) provided \(f\in C_{B}[0,\infty)\) and \(x\geq\frac {1}{2n}\), \(n\in\mathbb{N}\). Indeed, for \(x\geq\frac{1}{2n}\), \(\mu\geq\frac {1}{2n}\) and \(n\in\mathbb{N}\), it is guaranteed that
$$\begin{aligned}& D_{n,q}^{\ast}\bigl((t-x)^{2};x\bigr)\leq D_{n,q}\bigl((t-x)^{2};x\bigr), \end{aligned}$$
(4.6)
$$\begin{aligned}& {}[1+2\mu]_{q}\frac{x}{[n]_{q}}-\frac{1}{4[n]_{q}^{2}} \bigl( 2[1+2\mu ]_{q}-1 \bigr) \leq{}[1+2\mu]_{q}\frac{x}{[n]_{q}}, \end{aligned}$$
(4.7)
which implies that
$$ \sqrt{[1+2\mu]_{q}\frac{x}{[n]_{q}}-\frac{1}{4[n]_{q}^{2}} \bigl( 2[1+2\mu]_{q}-1 \bigr) }\leq\sqrt{[1+2\mu]_{q} \frac{x}{[n]_{q}}}. $$
(4.8)
Now we give the rate of convergence of the operators \({D}_{n,q}^{\ast }(f;x) \) defined in (2.2) in terms of the elements of the usual Lipschitz class \(\operatorname{Lip}_{M}(\nu)\).
Let \(f\in C_{B}[0,\infty)\), \(M>0\) and \(0<\nu\leq1\). The class \(\operatorname{Lip}_{M}(\nu) \) is defined as
$$ \operatorname{Lip}_{M}(\nu)= \bigl\{ f:\bigl\vert f( \zeta_{1})-f(\zeta_{2})\bigr\vert \leq M| \zeta_{1}-\zeta_{2}|^{\nu}\ \bigl( \zeta_{1},\zeta_{2}\in{}[ 0,\infty)\bigr) \bigr\} . $$
(4.9)
Theorem 4.3
Let
\(D_{n,q}^{\ast}(\cdot ; \cdot)\)
be the operators defined in (2.2).Then, for each
\(f\in \operatorname{Lip}_{M}(\nu)\) (\(M>0\), \(0<\nu\leq1\)) satisfying (4.9), we have
$$ \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq M ( \delta_{n,x} ) ^{\frac{\nu}{2}}, $$
where
\(\delta_{n,x}\)
is given in Theorem
4.1.
Proof
We prove it by using (4.9) and Hölder’s inequality. We have
$$\begin{aligned} \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq&\bigl\vert D_{n,q}^{\ast }\bigl(f(t)-f(x);x\bigr)\bigr\vert \\ \leq&D_{n,q}^{\ast} \bigl( \bigl\vert f(t)-f(x)\bigr\vert ;x \bigr) \\ \leq&MD_{n,q}^{\ast} \bigl( \vert t-x\vert ^{\nu};x \bigr) . \end{aligned}$$
Therefore,
$$\begin{aligned}& \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \\& \quad \leq M\frac{[n]_{q}}{e_{\mu,q}([n]_{q}r_{[n]_{q}}(x))}\sum_{k=0}^{\infty}\frac{([n]_{q}r_{[n]_{q}}(x))^{k}}{\gamma_{\mu,q}(k)}\biggl\vert \frac{1-q^{2\mu\theta_{k}+k}}{1-q^{n}}-x\biggr\vert ^{\nu} \\& \quad \leq M\frac{[n]_{q}}{e_{\mu,q}([n]_{q}r_{[n]_{q}}(x))}\sum_{k=0}^{\infty } \biggl( \frac{([n]_{q}r_{[n]_{q}}(x))^{k}}{\gamma_{\mu,q}(k)} \biggr) ^{\frac{2-\nu}{2}} \\& \qquad {} \times \biggl( \frac{([n]_{q}r_{[n]_{q}}(x))^{k}}{\gamma_{\mu,q}(k)} \biggr) ^{\frac{\nu}{2}} \biggl\vert \frac{1-q^{2\mu\theta _{k}+k}}{1-q^{n}}-x\biggr\vert ^{\nu} \\& \quad \leq M \Biggl( \frac{n}{e_{\mu,q}([n]_{q}r_{[n]_{q}}(x))}\sum_{k=0}^{\infty } \frac{([n]_{q}r_{[n]_{q}}(x))^{k}}{\gamma_{\mu,q}(k)} \Biggr) ^{\frac{ 2-\nu}{2}} \\& \qquad {} \times \Biggl( \frac{[n]_{q}}{e_{\mu,q}([n]_{q}r_{[n]_{q}}(x))}\sum _{k=0}^{\infty}\frac{([n]_{q}r_{[n]_{q}}(x))^{k}}{\gamma_{\mu ,q}(k)}\biggl\vert \frac{1-q^{2\mu\theta_{k}+k}}{1-q^{n}}-x\biggr\vert ^{2} \Biggr) ^{ \frac{\nu}{2}} \\& \quad = M \bigl( D_{n,q}^{\ast}(t-x)^{2};x \bigr) ^{\frac{\nu}{2}}. \end{aligned}$$
This completes the proof. □
Let
$$ C_{B}^{2}\bigl(\mathbb{R}^{+}\bigr)=\bigl\{ g\in C_{B}\bigl(\mathbb{R}^{+}\bigr):g^{\prime },g^{\prime \prime} \in C_{B}\bigl(\mathbb{R}^{+}\bigr)\bigr\} , $$
(4.10)
with the norm
$$ \Vert g\Vert _{C_{B}^{2}(\mathbb{R}^{+})}=\Vert g\Vert _{C_{B}(\mathbb{R}^{+})}+\bigl\Vert g^{\prime}\bigr\Vert _{C_{B}(\mathbb {R}^{+})}+\bigl\Vert g^{\prime\prime}\bigr\Vert _{C_{B}(\mathbb{R}^{+})}, $$
(4.11)
also
$$ \| g\|_{C_{B}(\mathbb{R}^{+})}=\sup_{x\in\mathbb {R}^{+}}\bigl\vert g(x)\bigr\vert . $$
(4.12)
Theorem 4.4
Let
\(D_{n,q}^{\ast}(\cdot ; \cdot)\)
be the operators defined in (2.2). Then for any
\(g\in C_{B}^{2}(\mathbb{R}^{+})\)
we have
$$ \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq \biggl\{ \biggl( -\frac {1}{2[n]_{q}} \biggr) +\frac{\delta_{n,x}}{2} \biggr\} \| g \| _{C_{B}^{2}(\mathbb{R}^{+})}, $$
where
\(\delta_{n,x}\)
is given in Theorem
4.1.
Proof
Let \(g\in C_{B}^{2}(\mathbb{R}^{+})\). Then, by using the generalized mean value theorem in the Taylor series expansion, we have
$$ g(t)=g(x)+g^{\prime}(x) (t-x)+g^{\prime\prime}(\psi)\frac {(t-x)^{2}}{2}, \quad \psi\in(x,t). $$
By applying the linearity property on \(D_{n,q}^{\ast}\), we have
$$ D_{n,q}^{\ast}(g,x)-g(x)=g^{\prime}(x)D_{n,q}^{\ast} \bigl( (t-x);x \bigr) +\frac{g^{\prime\prime}(\psi)}{2}D_{n,q}^{\ast} \bigl( (t-x)^{2};x \bigr) , $$
which implies that
$$\begin{aligned}& \bigl\vert D_{n,q}^{\ast}(g;x)-g(x)\bigr\vert \\& \quad \leq \biggl( -\frac {1}{2[n]_{q}} \biggr) \bigl\Vert g^{\prime}\bigr\Vert _{C_{B}(\mathbb{R}^{+})}+ \biggl( [1+2\mu ]_{q}\frac{x}{[n]_{q}}- \frac{1}{4[n]_{q}^{2}} \bigl( 2[1+2\mu]_{q}-1 \bigr) \biggr) \frac{\Vert g^{\prime\prime} \Vert _{C_{B}(\mathbb {R}^{+})}}{2}. \end{aligned}$$
From (4.11) we have \(\| g^{\prime}\| _{C_{B}[0,\infty)}\leq\| g\|_{C_{B}^{2}[0,\infty)}\),
$$\begin{aligned}& \bigl\vert D_{n,q}^{\ast}(g;x)-g(x)\bigr\vert \\& \quad \leq \biggl( -\frac {1}{2[n]_{q}} \biggr) \| g\|_{C_{B}^{2}(\mathbb{R}^{+})}+ \biggl( [1+2\mu ]_{q}\frac{x}{[n]_{q}}-\frac{1}{4[n]_{q}^{2}} \bigl( 2[1+2 \mu]_{q}-1 \bigr) \biggr) \frac{\| g\|_{C_{B}^{2}(\mathbb{R}^{+})}}{2}. \end{aligned}$$
The proof follows from (2) of Lemma 2.2. □
The Peetre’s K-functional is defined by
$$ K_{2}(f,\delta)=\inf_{C_{B}^{2}(\mathbb{R}^{+})} \bigl\{ \bigl( \Vert f-g\Vert _{C_{B}(\mathbb{R}^{+})}+\delta\bigl\Vert g^{\prime\prime }\bigr\Vert _{C_{B}^{2}(\mathbb{R}^{+})} \bigr) :g\in\mathcal {W}^{2} \bigr\} , $$
(4.13)
where
$$ \mathcal{W}^{2}= \bigl\{ g\in C_{B}\bigl( \mathbb{R}^{+}\bigr):g^{\prime},g^{\prime \prime}\in C_{B}\bigl(\mathbb{R}^{+}\bigr) \bigr\} . $$
(4.14)
There exists a positive constant \(C>0\) such that \(K_{2}(f,\delta)\leq C\omega_{2}(f,\delta^{\frac{1}{2}})\), \(\delta>0\), where the second-order modulus of continuity is given by
$$ \omega_{2}\bigl(f,\delta^{\frac{1}{2}}\bigr)=\sup _{0< h< \delta^{\frac{1}{2}}}\sup_{x\in\mathbb{R}^{+}}\bigl\vert f(x+2h)-2f(x+h)+f(x)\bigr\vert . $$
(4.15)
Theorem 4.5
For
\(x\geq\frac{1}{2n}\), \(n\in\mathbb{N}\)
and
\(f\in C_{B}( \mathbb{R}^{+})\), we have
$$\begin{aligned}& \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \\& \quad \leq2M \biggl\{ \omega_{2} \biggl( f;\sqrt{\frac{ ( -\frac{1}{ [n]_{q}} )+ \delta_{n,x}}{4}} \biggr) +\min \biggl( 1,\frac{ ( -\frac{1}{ [n]_{q}} )+ \delta_{n,x}}{4} \biggr) \| f\| _{C_{B}(\mathbb{R}^{+})} \biggr\} , \end{aligned}$$
where
M
is a positive constant, \(\delta_{n,x}\)
is given in Theorem
4.3
and
\(\omega_{2}(f;\delta)\)
is the second-order modulus of continuity of the function
f
defined in (4.15).
Proof
We prove this by using Theorem 4.4
$$\begin{aligned} \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq& \bigl\vert D_{n,q}^{\ast}(f-g;x)\bigr\vert +\bigl\vert D_{n,q}^{\ast}(g;x)-g(x)\bigr\vert +\bigl\vert f(x)-g(x) \bigr\vert \\ \leq& 2 \| f-g \|_{C_{B}(\mathbb{R}^{+})}+ \frac{\delta _{n,x}}{2}\| g \|_{C_{B}^{2}(\mathbb{R}^{+})} + \biggl( -\frac{1}{2 [n]_{q}} \biggr)\| g \|_{C_{B}(\mathbb{R}^{+})}. \end{aligned}$$
From (4.11), clearly, we have \(\| g \|_{C_{B}[0,\infty)}\leq\| g \|_{C_{B}^{2}[0,\infty)}\).
Therefore,
$$ \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq2 \biggl( \| f-g \| _{C_{B}(\mathbb{R}^{+})}+\frac{ ( -\frac{1}{ [n]_{q}} )+ \delta_{n,x}}{4}\| g \|_{C_{B}^{2}(\mathbb{R}^{+})} \biggr), $$
where \(\delta_{n,x}\) is given in Theorem 4.1.
By taking infimum over all \(g\in C_{B}^{2}(\mathbb{R}^{+})\) and by using (4.13), we get
$$ \bigl\vert D_{n,q}^{\ast}(f;x)-f(x)\bigr\vert \leq2K_{2} \biggl( f;\frac{ ( -\frac{1}{ [n]_{q}} )+ \delta_{n,x}}{4} \biggr). $$
Now, for an absolute constant \(Q>0\) in [24], we use the relation
$$ K_{2}(f;\delta)\leq Q\bigl\{ \omega_{2}(f;\sqrt{\delta})+ \min(1,\delta )\| f\|\bigr\} . $$
This completes the proof. □