Here, we study the order of approximation of a function f with the aid of positive linear operators \(L_{n,\alpha }^{*}(f;x)\) defined by (2.2) in terms of the classical modulus of continuity, the second-order modulus of continuity, Peetres K-functional and the Lipschitz class. A well-known property is the modulus of continuity of order one and of order two defined as follows. For \(\delta >0\) and \(f\in C[a,b]\) the classical modulus of continuity of order one is given by
$$ \omega (f;\delta )=\sup_{x_{1},x_{2}\in [a,b], |x_{1}-x_{2}|\leq \delta } \bigl\vert f(x_{1})-f(x _{2}) \bigr\vert , $$
and of order two it is given by
$$ \omega _{2}\bigl(f;\delta ^{\frac{1}{2}}\bigr)=\sup _{0< h< \delta ^{\frac{1}{2}}} \sup_{x\in \mathbb{R}^{+}} \bigl\vert f(x)-2f(x+h)+f(x+2h) \bigr\vert . $$
(4.1)
Let \(C_{B}[0,\infty )\) denote the space of all bounded and continuous functions on \([0,\infty )\) and
$$ C_{B}^{2}[0,\infty )=\bigl\{ \psi \in C_{B}[0, \infty ):\psi ^{\prime }, \psi ^{\prime \prime }\in C_{B}[0,\infty ) \bigr\} , $$
(4.2)
with the norm
$$ \Vert \psi \Vert _{C_{B}^{2}[0,\infty )}= \Vert \psi \Vert _{C_{B}[0,\infty )}+ \bigl\Vert \psi ^{\prime } \bigr\Vert _{C_{B}[0,\infty )}+ \bigl\Vert \psi ^{\prime \prime } \bigr\Vert _{C_{B}[0,\infty )}, $$
(4.3)
also
$$ \Vert \psi \Vert _{C_{B}[0,\infty )}=\sup_{x\in [0,\infty )} \bigl\vert \psi (x) \bigr\vert . $$
(4.4)
Lemma 4.1
([31])
Let
\(\{P_{n}\}_{n\geq 1}\)
be the sequence for the positive integer
n
with
\(P_{n}(1;x)=1\). Then for every
\(\psi \in C_{B}^{2}[0,\infty )\)
$$ \bigl\vert P_{n}(\psi ;x)-\psi (x) \bigr\vert \leq \bigl\Vert g' \bigr\Vert \sqrt{P_{n}\bigl((s-x)^{2};x \bigr)}+ \frac{1}{2} \bigl\Vert \psi '' \bigr\Vert P_{n}\bigl((s-x)^{2};x\bigr). $$
Lemma 4.2
([31])
For all
\(f\in C[a,b]\)
and
\(h\in (0,\frac{b-a}{2} ) \), we have the following inequalities:
$$\begin{aligned} (\mathrm{i})&\quad \Vert f_{h}-f \Vert \leq \frac{3}{4} \omega _{2}(f,h), \\ (\mathrm{ii})&\quad \bigl\Vert f_{h}'' \bigr\Vert \leq \frac{3}{2h^{2}}\omega _{2}(f,h), \end{aligned}$$
where
\(f_{h}\)
denotes the second-order Steklov function.
Theorem 4.3
For all
\(f\in C_{B}[0,\infty )\)
and
\(x\in [0,a]\), \(a>0\)
we have
$$ \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq 2\omega \bigl(f; \sqrt{\varTheta _{n}(x)} \bigr), $$
where
\(\varTheta _{n}(x)=L_{n,\alpha }^{*}(\psi _{x}^{2};x)\)
and
\(L_{n,\alpha }^{*}(\psi _{x}^{2};x)\)
is defined by Lemma
2.3.
Proof
In view of the classical modulus of continuity, we have
$$\begin{aligned} \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq & \sum _{k=0}^{\infty }\mathcal{S} _{n,k}^{(\alpha )}(x) \int _{0}^{\infty }\mathcal{Q}_{n,k}(t) \bigl\vert f(t)-f(x) \bigr\vert \,dt \\ \leq & \Biggl\{ 1+\frac{1}{\delta }\sum_{k=0}^{\infty } \mathcal{S}_{n,k} ^{(\alpha )}(x) \int _{0}^{\infty }\mathcal{Q}_{n,k}(t) \vert t-x \vert \,dt \Biggr\} \omega (f;\delta ). \end{aligned}$$
In the light of the Cauchy–Schwartz inequality, we get
$$\begin{aligned} \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq & \Biggl\{ 1+\frac{1}{\delta } \Biggl(\sum_{k=0}^{\infty } \mathcal{S}_{n,k}^{(\alpha )}(x) \int _{0}^{\infty } \mathcal{Q}_{n,k}(t) (t-x)^{2}\,dt \Biggr)^{\frac{1}{2}} \Biggr\} \omega (f;\delta ) \\ =& \biggl\{ 1+\frac{1}{\delta } \sqrt{L_{n,\alpha }^{*} \bigl(\psi _{x}^{2};x\bigr)} \biggr\} \omega (f;\delta ). \end{aligned}$$
Choosing \(\delta = (\varTheta _{n}(x) )^{\frac{1}{2}}=\sqrt{ L_{n,\alpha }^{*}(\psi _{x}^{2};x)}\), we arrive at the desired result. □
Theorem 4.4
For every
\(f\in C[0,a]\), \(a>0\)
the operators
\(L_{n,\alpha }^{*}( \cdot\, ; \cdot )\)
defined by (2.2) satisfy
$$ \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq \frac{2}{a} \Vert f \Vert \delta ^{2}+ \frac{3}{4} \bigl(a+2+h^{2}\bigr)\omega _{2}(f;\delta ), $$
where
\(\delta = (\varTheta _{n}(x) )^{\frac{1}{2}}\)
is defined by Theorem
4.3
and
\(\omega _{2}(f;\delta )\)
is by (4.1) equipped with the norm
\(\Vert f \Vert =\max_{x\in [a,b]}|f(x)|\).
Proof
Consider \(f_{h}\) is the Steklov function define in Lemma 4.2. Using Lemma 2.2, we obtain
$$\begin{aligned} \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq & \bigl\vert L_{n,\alpha }^{*}(f-f_{h};x) \bigr\vert + \bigl\vert f _{h}-f(x) \bigr\vert + \bigl\vert L_{n,\alpha }^{*}(f_{h};x)-f_{h}(x) \bigr\vert \\ \leq & 2 \Vert f_{h}-f \Vert + \bigl\vert L_{n,\alpha }^{*}(f_{h};x)-f_{h}(x) \bigr\vert . \end{aligned}$$
In view of the fact that \(f_{h}\in C^{2}[0,a]\) and using Lemma 4.1, we obtain
$$ \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq \bigl\Vert f_{h}' \bigr\Vert \sqrt {L_{n,\alpha }^{*}\bigl((e _{1}-x)^{2};x \bigr)}+\frac{1}{2} \bigl\Vert f''_{h} \bigr\Vert L_{n,\alpha }^{*}\bigl((e_{1}-x)^{2};x \bigr). $$
(4.5)
From the Landau inequality and Lemma 4.2, we have
$$\begin{aligned} \Vert f_{h} \Vert \leq &\frac{2}{a} \Vert f_{h} \Vert +\frac{a}{2} \bigl\Vert f_{h}'' \bigr\Vert \\ \leq &\frac{2}{a} \Vert f_{h} \Vert +\frac{3a}{4} \frac{1}{h^{2}}\omega _{2}(f;h). \end{aligned}$$
On choosing \(\delta = (\varTheta _{n}(x) )^{\frac{1}{4}}\), one has
$$ \bigl\vert L_{n,\alpha }^{*}(f_{h};x)-f_{h}(x) \bigr\vert \leq \frac{2}{a} \Vert f \Vert h^{2}+ \frac{3a}{4}\omega _{2}(f;h)+\frac{3}{4}h^{2} \omega _{2}(f;h). $$
(4.6)
Combining (4.6), (4.5) and Lemma 4.2, we obtain the required result. □
Theorem 4.5
Let
\(L_{n,\alpha }^{*}( \cdot\, ; \cdot )\)
be the operators defined by (2.2). Then, for every
\(f\in C_{B}^{2}[0,\infty )\),
$$ \lim_{n\rightarrow \infty }(n-1) \bigl(L_{n,\alpha }^{*}(f;x)-f(x) \bigr)=\bigl(1+2 \alpha x-x^{2}\bigr)f'(x)+2 \bigl(x+x^{2}\bigr)f''(x), $$
uniformly for
\(0\leq x\leq a\), \(a>0\).
Proof
Let \(x_{0}\in [0,\infty )\) be a fixed number; all \(x\in [0,\infty )\). Then using Taylor’s series, we have
$$ f(x)-f(x_{0})=(x-x_{0})f'(x_{0})+ \frac{1}{2}(x-x_{0})^{2}f''(x_{0})+ \varphi (x,x_{0}) (x-x_{0})^{2}, $$
(4.7)
where \(\varphi (x,x_{0})\in C_{B}[0,\infty )\) and \(\lim_{x\rightarrow x_{0}}\varphi (x,x_{0})=0\).
By applying the operators \(L_{n,\alpha }^{*}\) on (4.7), we deduce
$$\begin{aligned} L_{n,\alpha }^{*}(f;x_{0})-f(x_{0}) =&f'(x_{0})L_{n,\alpha }^{*}(e _{1}-x_{0};x_{0})+\frac{1}{2}L_{n,\alpha }^{*} \bigl((x-x_{0})^{2};x_{0}\bigr)f''(x _{0}) \\ &{}+L_{n,\alpha }^{*}\bigl(\varphi (x,x_{0}) (x-x_{0})^{2}\bigr). \end{aligned}$$
(4.8)
In view of the Cauchy–Schwartz inequality for the last term of Eq. (4.8), we get
$$ (n-1)L_{n,\alpha }^{*}\bigl(\varphi (x,x_{0}) (t-x_{0})^{2}\bigr)\leq (n-1)^{2}\sqrt{L _{n,\alpha }^{*} \bigl((e_{1}-x_{0})^{2}\bigr)L_{n,\alpha }^{*} \bigl(\varphi ^{2}(x,x _{0})\bigr)}. $$
(4.9)
We have
$$\begin{aligned}& \lim_{n\rightarrow \infty }(n-1) \bigl(L_{n,\alpha }^{*}(e_{0}-x _{0};x) \bigr) = \bigl(1+2\alpha x-x^{2} \bigr)f'(x), \\& \lim_{n\rightarrow \infty }(n-1) \bigl(L_{n,\alpha }^{*}\bigl((e _{0}-x_{0})^{2};x\bigr) \bigr) = 2 \bigl(x+x^{2}\bigr)f''(x), \\& \lim_{n\rightarrow \infty } \bigl(L_{n,\alpha }^{*} \bigl((e_{0}-x _{0})^{4};x\bigr) \bigr) = 0. \end{aligned}$$
This completes the proof. □
Now here we estimate the rate of convergence in terms of the usual Lipschitz class \(\operatorname{Lip}_{M}(\nu )\). Let \(f\in C[0,a )\), \(a>0\) and M be a positive constant, and, for any \(\nu \in (0,1]\), the Lipschitz class \(\operatorname{Lip}_{M}(\nu )\) is as follows:
$$ \operatorname{Lip}_{M}(\nu )= \bigl\{ f: \bigl\vert f(\varsigma _{1})-f(\varsigma _{2}) \bigr\vert \leq M \vert \varsigma _{1}-\varsigma _{2} \vert ^{\nu }\ \bigl( \varsigma _{1}, \varsigma _{2}\in [ 0,\infty)\bigr) \bigr\} . $$
(4.10)
Theorem 4.6
Let
\(f\in \operatorname{Lip}_{M}(\nu )\)
with
\(M>0\)
and
\(0<\nu \leq 1\). Then the operators
\(L_{n,\alpha }^{*}( \cdot\, ; \cdot )\)
satisfy
$$ \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq M \bigl( \varTheta _{n}(x) \bigr) ^{\frac{\nu }{2}}, $$
where
\(n>2\)
and
\(\varTheta _{n}(x)\)
defined by Theorem
4.3.
Proof
From the Hölder inequality and (4.10), we conclude
$$\begin{aligned} \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq & \bigl\vert L_{n,\alpha }^{*}\bigl(f(t)-f(x);x\bigr) \bigr\vert \\ \leq &L_{n,\alpha }^{*} \bigl( \bigl\vert f(t)-f(x) \bigr\vert ;x \bigr) \\ \leq & ML_{n,\alpha }^{*} \bigl( \vert t-x \vert ^{\nu };x \bigr) . \end{aligned}$$
Hence
$$\begin{aligned}& \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \\& \quad \leq M \sum_{k=0}^{\infty } \mathcal{S}_{n,k}^{(\alpha )}(x) \int _{0}^{\infty }\mathcal{Q}_{n,k}(t) \vert t-x \vert ^{\nu }\,dt \\& \quad \leq M \sum_{k=0}^{\infty } \bigl( \mathcal{S}_{n,k}^{( \alpha )}(x) \bigr)^{\frac{2-\nu }{2}} \\& \qquad {} \times \bigl(\mathcal{S}_{n,k}^{(\alpha )}(x) \bigr)^{\frac{ \nu }{2}} \int _{0}^{\infty }\mathcal{Q}_{n,k}(t) \vert t-x \vert ^{\nu }\,dt \\& \quad \leq M \Biggl(\sum_{k=0}^{\infty } \mathcal{S}_{n,k}^{(\alpha )}(x) \int _{0}^{\infty }\mathcal{Q}_{n,k}(t)\,dt \Biggr)^{\frac{2-\nu }{2}} \\& \qquad {} \times \Biggl(\sum_{k=0}^{\infty } \mathcal{S}_{n,k}^{(\alpha )}(x) \int _{0}^{\infty }\mathcal{Q}_{n,k}(t) \vert t-x \vert ^{2} \,dt \Biggr) ^{\frac{\nu }{2}} \\& \quad = M \bigl(L_{n,\alpha }^{*}\bigl(\psi _{x}^{2};x \bigr) \bigr)^{\frac{ \nu }{2}}. \end{aligned}$$
This completes the proof. □
Theorem 4.7
For all
\(\psi \in C_{B}^{2}{}[ 0,\infty )\)
and
\(n>2\),
$$ \bigl\vert L_{n,\alpha }^{*}(\psi ;x)-\psi (x) \bigr\vert \leq \biggl(\Delta _{n}(x)+\frac{ \varTheta _{n}(x)}{2} \biggr) \Vert \psi \Vert _{C_{B}^{2}{}[ 0,\infty )}, $$
where
\(\Delta _{n}(x)= (\frac{2\alpha -1}{n-1}x+\frac{1}{n-1} )\)
and
\(\varTheta _{n}(x)\)
is defined by Theorem
4.3.
Proof
Let \(\psi \in C_{B}^{2}(\mathbb{R}^{+})\); for all \(\varphi \in (x,t)\) a Taylor series expansion is
$$ \psi (t)=\frac{(t-x)^{2}}{2}\psi ^{\prime \prime }(\varphi )+(t-x) \psi ^{\prime }(x)+\psi (x). $$
On applying \(L_{n,\alpha }^{*}\), using linearity,
$$ L_{n,\alpha }^{*}(\psi ;x)-\psi (x)=\psi ^{\prime }(x)L_{n,\alpha } ^{*} \bigl( (t-x);x \bigr) + \frac{\psi ^{\prime \prime }(\varphi )}{2}L_{n,\alpha }^{*} \bigl( (t-x)^{2};x \bigr) , $$
which implies that
$$\begin{aligned}& \bigl\vert L_{n,\alpha }^{*}(\psi ;x)-\psi (x) \bigr\vert \\& \quad \leq \biggl(\frac{2\alpha -1}{n-1}x+\frac{1}{n-1} \biggr) \bigl\Vert \psi ^{\prime } \bigr\Vert _{C_{B}{}[ 0,\infty )} \\& \qquad {}+ \biggl\{ \frac{2n+2(4\alpha -3)}{(n-2)(n-1)}x^{2}+\frac{2n+2(5 \alpha -3)}{(n-2)(n-1)}x+ \frac{2}{(n-2)(n-1)} \biggr\} \frac{ \Vert \psi ^{\prime \prime } \Vert _{C_{B}{}[ 0,\infty )}}{2}. \end{aligned}$$
From (4.3) we have \(\Vert \psi ^{\prime } \Vert _{C_{B}{}[ 0,\infty )}\leq \Vert \psi \Vert _{C_{B}^{2}{}[ 0,\infty )}\), \(\Vert \psi ^{\prime \prime } \Vert _{C_{B}{}[ 0,\infty )}\leq \Vert \psi \Vert _{C_{B}^{2}{}[ 0,\infty )}\).
$$\begin{aligned}& \bigl\vert L_{n,\alpha }^{*}(\psi ;x)-\psi (x) \bigr\vert \\& \quad \leq \biggl(\frac{2\alpha -1}{n-1}x+\frac{1}{n-1} \biggr) \Vert \psi \Vert _{C_{B}^{2}{}[ 0,\infty )} \\& \qquad {}+ {\biggl\{ } \frac{2n+2(4\alpha -3)}{(n-2)(n-1)}x^{2}+\frac{2n+2(5 \alpha -3)}{(n-2)(n-1)}x+ \frac{2}{(n-2)(n-1)} {\biggr\} } \frac{ \Vert \psi \Vert _{{}[ 0,\infty )}}{2}. \end{aligned}$$
This completes the proof. □
In 1968 [34] for investigating the interpolation between two Banach spaces Peetre introduced the K-functional by
$$ K_{2}(f;\delta )=\inf_{C_{B}^{2}{}[ 0,\infty )} \bigl\{ \bigl( \Vert f-\psi \Vert _{C_{B}{}[ 0,\infty )}+\delta \Vert \psi \Vert _{C_{B}^{2}{}[ 0,\infty )} \bigr) :\psi \in C_{B} ^{2}{}[ 0,\infty ) \bigr\} $$
(4.11)
and a positive constant \(\mathfrak{D}\) exists such that \(K_{2}(f; \delta )\leq \mathfrak{D} \omega _{2}(f;\delta ^{\frac{1}{2}})\) with \(\delta >0\) and \(\omega _{2}(f;\delta )\) is the second-order modulus of continuity.
Theorem 4.8
Suppose
\(C_{B}{}[ 0,\infty )\)
is the set of all bounded and continuous functions on
\({}[ 0,\infty )\). Then for every
\(f\in C_{B}{}[ 0,\infty )\)
$$ \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq 2 \mathfrak{D} \bigl\{ \omega _{2} \bigl( f;\sqrt{\mathfrak{K}_{n}(x)} \bigr) + \min \bigl( 1,\mathfrak{K}_{n}(x) \bigr) \Vert f \Vert _{C_{B}{}[ 0,\infty )}\bigr\} , $$
where
\(\mathfrak{K}_{n}(x)=\frac{2\Delta _{n}(x)+\varTheta _{n}(x)}{4}\)
is defined by Theorem
4.7.
Proof
In the light of results obtained by Theorem 4.7, we prove the desired theorem; hence
$$\begin{aligned} \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq& \bigl\vert L_{n,\alpha }^{*}(f- \psi ;x) \bigr\vert + \bigl\vert f(x)-\psi (x) \bigr\vert + \bigl\vert L_{n,\alpha }^{*}(\psi ;x)- \psi (x) \bigr\vert \\ \leq& 2 \Vert f-\psi \Vert _{C_{B}{}[ 0,\infty )}+ \biggl(\frac{ \varTheta _{n}(x)}{2}+ \Delta _{n}(x) \biggr) \Vert \psi \Vert _{C_{B}^{2}{}[ 0,\infty )} \\ =& 2 \biggl( \Vert f-\psi \Vert _{C_{B}{}[ 0,\infty )}+ \biggl(\frac{\varTheta _{n}(x)}{4}+ \frac{\Delta _{n}(x)}{2} \biggr) \Vert \psi \Vert _{C_{B}^{2}{}[ 0,\infty )} \biggr). \end{aligned}$$
If we take the infimum over all \(\psi \in C_{B}^{2}{}[ 0,\infty )\) and we use (4.11), we get
$$ \bigl\vert L_{n,\alpha }^{*}(f;x)-f(x) \bigr\vert \leq 2K_{2} \biggl( f; \biggl(\frac{ \varTheta _{n}(x)}{4}+\frac{\Delta _{n}(x)}{2} \biggr) \biggr). $$
Now from [35] we use the relation for an absolute constant \(\mathfrak{D}>0\)
$$ K_{2}(f;\delta )\leq \mathfrak{D}\bigl\{ \omega _{2}(f; \sqrt{\delta })+ \min (1,\delta ) \Vert f \Vert \bigr\} . $$
This completes the proof. □