Approximation on parametric extension of Baskakov–Durrmeyer operators on weighted spaces

Abstract

In the present manuscript, we define a non-negative parametric variant of Baskakov–Durrmeyer operators to study the convergence of Lebesgue measurable functions and introduce these as α-Baskakov–Durrmeyer operators. We study the uniform convergence of these operators in weighted spaces.

Introduction

In the field of mathematical analysis, Karl Weierstrass established an elegant theorem, the first Weierstrass approximation theorem, in 1885. This theorem has specially a big role in polynomial interpolation corresponding to every continuous function $f(x)$ on interval $[a,b]$. The proof given by Weierstrass was rigorous and difficult to understand. In 1912, Bernstein [1] gave a simple proof of this theorem by introducing the Bernstein polynomials with the aid of the binomial distribution, hence for $f\in C[0,1]$, we have

$$ B_{n}(f;x)=\sum_{k=0}^{n} \mathcal{S}_{n,k}(x)f \biggl(\frac{k}{n} \biggr),\quad n\in \mathbb{N}, 0\leq x\leq 1, $$

(1.1)

where $\mathcal{S}_{n,k}(x)=\binom{n}{k}x^{k}(1-x)^{n-k}$. Many mathematicians researched in this direction and studied various modifications in several functional spaces using different error optimization techniques, i.e., Acar et al. [2,3,4,5,6,7], Acu et al. [8, 9], Barbosu [10], Agrawal et al. [11], Aral [12], Mursaleen et al. [13,14,15,16,17], Srivastava et al. [18,19,20]; for more details see also the references therein and [21,22,23,24,25,26,27,28,29,30].

Construction of the α-Baskakov–Durrmeyer operators and estimation of their moments

Recently, Cai, Lian and Zhou [31] presented a new sequence of α-Bernstein operators with $\alpha \in [-1,1]$. Later, Ali Aral et al. [32] gave a sequence of α-Bernstein operators as follows:

$$ L_{n,\alpha }(f;x)=\sum_{k=0}^{\infty }f \biggl(\frac{k}{n} \biggr) \mathcal{S}_{n,k}^{(\alpha )}(x), \quad n\in \mathbb{N}, x\in [0,\infty ), $$

(2.1)

where $f\in C_{B}[0,\infty )$ which denotes the set of all continuous and bounded functions and

$$\begin{aligned} \mathcal{S}_{n,k}^{(\alpha )}(x) =&\frac{x^{k-1}}{(1+x)^{n+k-1}} \biggl\{ \frac{\alpha x}{1+x}\binom{n+k-1}{k}-(1-\alpha ) (1+x) \binom{n+k-3}{k-2} \\ &{}+(1-\alpha )y\binom{n+k-1}{k} \biggr\} \end{aligned}$$

with

$$ \binom{n-3}{-2}=\binom{n-2}{-1}=0. $$

The operators defined by (2.1) are restricted for continuous functions only. To approximate the functions in Lebesgue measurable space, we design a new sequence of operators:

$$ L_{n,\alpha }^{*}(f;x)=\sum _{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \int _{0}^{\infty }\mathcal{Q}_{n,k}(t)f(t)\,dt, $$

(2.2)

where $\mathcal{Q}_{n,k}(t)=\frac{1}{B(k+1,n)} \frac{t^{k}}{(1+t)^{(n+k+1)}}$. Note that, simply in the case of $\alpha =1$, the operators reduced to Baskakov–Durrmeyer type operators; for details see [33].

For $r\in \{0,1,2,3,4\}$, we consider the test functions and central moments,

$$ e_{r}=t^{r} \quad \mbox{and} \quad \psi _{y}^{r}(t;x)=(t-x)^{r}. $$

(2.3)

Lemma 2.1

([31])

We have

$$\begin{aligned}& L_{n,\alpha }(e_{0};x) = 1, \\& L_{n,\alpha }(e_{1};x) = x+\frac{2}{n}(\alpha -1), \\& L_{n,\alpha }(e_{2};x) = x^{2}+\frac{4\alpha -3}{n}x+ \frac{1}{n^{2}}(n+4 \alpha -4). \end{aligned}$$

Lemma 2.2

Let the test functions $e_{r}$ defined by (2.3), then, for all $L_{n,\alpha }^{*}$, we have

$$\begin{aligned}& L_{n,\alpha }^{*}(e_{0};x) = 1, \\& L_{n,\alpha }^{*}(e_{1};x) = \biggl(\frac{n}{n-1}+ \frac{2(\alpha -1)}{n-1} \biggr)x+\frac{1}{n-1}, \\& L_{n,\alpha }^{*}(e_{2};x) = \biggl(\frac{n^{2}}{(n-2)(n-1)}+ \frac{n(4 \alpha -3)}{(n-2)(n-1)} \biggr)x^{2}+ \frac{(4n+10\alpha -10)}{(n-2)(n-1)}x+ \frac{2}{(n-2)(n-1)}. \end{aligned}$$

Proof

Take $f=e_{0}$, then from Lemma 2.1, we have

$$\begin{aligned} L_{n,\alpha }^{*}(e_{0};x) =&\sum _{k=0}^{\infty }\mathcal{S}_{n,k} ^{(\alpha )}(x) \int _{0}^{\infty }\mathcal{Q}_{n,k}(t)\,dt \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \frac{B(k+1,n)}{B(k+1,n)} \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \\ =&1. \end{aligned}$$

For $r=1$

$$\begin{aligned} L_{n,\alpha }^{*}(e_{1};x) =&\sum _{k=0}^{\infty }\mathcal{S}_{n,k} ^{(\alpha )}(x) \int _{0}^{\infty }t \mathcal{Q}_{n,k}(t)\,dt \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \frac{B(k+2,n-1)}{B(k+1,n)} \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \frac{(k+1)B(k+1,n)}{(n-1)B(k+1,n)} \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \frac{(k+1)}{(n-1)} \\ =& \biggl(\frac{n}{n-1}+\frac{2(\alpha -1)}{n-1} \biggr)x+ \frac{1}{n-1}. \end{aligned}$$

For $r=2$

$$\begin{aligned} L_{n,\alpha }^{*}(e_{2};x) =&\sum _{k=0}^{\infty }\mathcal{S}_{n,k} ^{(\alpha )}(x) \int _{0}^{\infty }t^{2} \mathcal{Q}_{n,k}(t) \,dt \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \frac{B(k+3,n-2)}{B(k+1,n)} \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \frac{(k+2)(k+1)B(k+1,n)}{(n-2)(n-1)B(k+1,n)} \\ =&\sum_{k=0}^{\infty }\mathcal{S}_{n,k}^{(\alpha )}(x) \frac{(k+2)(k+1)}{(n-2)(n-1)} \\ =&\frac{n^{2}+n(4\alpha -3)}{(n-2)(n-1)}x^{2}+ \frac{(4n+10\alpha -10)}{(n-2)(n-1)}x+\frac{2}{(n-2)(n-1)}. \end{aligned}$$

 □

Lemma 2.3

Let the operators given by (2.2). Then we have

$$\begin{aligned}& L_{n,\alpha }^{*}\bigl(\psi _{x}^{0};x\bigr) = 1, \\& L_{n,\alpha }^{*}\bigl(\psi _{x}^{1};x\bigr) = \frac{2\alpha -1}{n-1}x+ \frac{1}{n-1}, \\& L_{n,\alpha }^{*}\bigl(\psi _{x}^{2};x\bigr) = \frac{2n+2(4\alpha -3)}{(n-2)(n-1)}x^{2}+ \frac{2n+2(5\alpha -3)}{(n-2)(n-1)}x+\frac{2}{(n-2)(n-1)}. \end{aligned}$$

Proof

In view of Lemmas 2.1 and 2.2 we can apply the linearity and easily complete the proof. □

Approximation in Korovkin and weighted Korovkin spaces

Take $C_{B}(\mathbb{R^{+}})$ be the space of all bounded and continuous functions defined on the set $\mathbb{R^{+}}$, where $\mathbb{R^{+}}=[0, \infty )$ and a normed defined on $C_{B}$ as

$$ \Vert f \Vert _{C_{B}}=\sup _{x\geq 0} \bigl\vert f(x) \bigr\vert . $$

Let

$$ E:=\biggl\{ f:x\in \mathbb{R^{+}} \text{ and } \lim _{x\rightarrow \infty } \biggl(\frac{f(x)}{1+x^{2}} \biggr)< \infty \biggr\} . $$

Lemma 3.1

For every $f\in C[0,\infty )\cap E$ the operators $L_{n,\alpha } ^{*}$ given in (2.2) are uniformly convergent to f on each compact subset of $[0,A]$, whenever $A\in (0,\infty )$.

Proof

In the view of Korovkin-type property, it is enough to show that

$$ L_{n,\alpha }^{*}(e_{s};x)\rightarrow e_{s}(x), \quad \text{for } s=0,1,2. $$

From Lemma 2.2, obviously $L_{n,\alpha }^{*}(e_{0};y)\rightarrow e_{0}(x)$ as $n\rightarrow \infty$ and for $s=1$

$$ \lim_{n\rightarrow \infty } L_{n,\alpha }^{*}(e_{1};x)= \lim_{n\rightarrow \infty } \biggl(\frac{n+2(\alpha -1)}{n-1}x+ \frac{1}{n-1} \biggr)=e_{1}(x). $$

Similarly, we can prove for $s=2$ that $L_{n,\alpha }^{*}(e_{2};x) \rightarrow e_{2}$, which proves Proposition 3.1. □

Suppose $C[0,\infty )$ is the set of all continuous functions and $f\in C[0,\infty )$ with the weight function $\sigma (x)=1+x^{2}$,

$$\begin{aligned}& \mathfrak{P}_{\sigma }(x) = \bigl\{ f: \bigl\vert f(x) \bigr\vert \leq \mathcal{M}_{f}\sigma (x), x\in [0,\infty ) \bigr\} , \\& \mathfrak{Q}_{\sigma }(x) = \bigl\{ f:f\in C[0,\infty )\cap \mathfrak{P}_{\sigma }(x) , x\in [0,\infty ) \bigr\} , \\& \mathfrak{Q}_{\sigma }^{m}(x) = \biggl\{ f:f\in \mathfrak{Q}_{\sigma }(x), \lim_{x\rightarrow \infty }\frac{f(x)}{\sigma (x)}=m, x\in [0,\infty ) \biggr\} , \end{aligned}$$

where the norm defined on weight function σ such as $\Vert f \Vert _{\sigma }=\sup_{x\in [0,\infty )}\frac{ \vert f(x) \vert }{\sigma (x)}$ and the constant $\mathcal{M}_{f}$ depends only on f.

Theorem 3.2

For all $f\in \mathfrak{Q}_{\sigma }^{m}(x)$ the operators $L_{n,\alpha }^{*}( \cdot\, ; \cdot )$ defined by (2.2) satisfy

$$ \lim_{n\to \infty } \bigl\Vert L_{n,\alpha }^{*}(f;x)-f \bigr\Vert _{ \sigma }=0. $$

Proof

Take $f(t) \in \mathfrak{Q}_{\sigma }^{m}(x)$ with $x\in [0,\infty )$ and $f(t)=e_{\nu }$ for $\nu =0,1,2$. Then from the well-known Korovkin theorem $L_{n,\alpha }^{*}(e_{\nu };x)\rightarrow x^{\nu }$, satisfying the properties of uniformly behaving as $n \to \infty$. Since for $\nu =0$, from Lemma 2.2 $L_{n,\alpha }^{*}(e_{0};x)=1$, thus we have

$$ \bigl\Vert L_{n,\alpha }^{*} (e_{0};x ) -1 \bigr\Vert _{\sigma } =0. $$

(3.1)

For $\nu =1$, we have

$$\begin{aligned} \bigl\Vert L_{n,\alpha }^{*} (e_{1};x ) -x \bigr\Vert _{\sigma } &= \sup_{x \in [0,\infty )}\frac{ \vert L _{n,\alpha }^{*}(e_{1};x)-x \vert }{1+x^{2}} \\ &= \biggl(\frac{n+2(\alpha -1)}{n-1}-1 \biggr)\sup_{x \in [0,\infty )} \frac{x}{1+x ^{2}}+\frac{1}{(n-1)}\sup_{x \in [0,\infty )} \frac{1}{1+x^{2}}. \end{aligned}$$

As $n \to \infty$,

$$ \bigl\Vert L_{n,\alpha }^{*} (e_{1};x ) -x \bigr\Vert _{\sigma } =0. $$

(3.2)

In a similar way for $\nu =2$,

$$\begin{aligned} & \bigl\Vert L_{n,\alpha }^{*} (e_{2};x ) -x^{2} \bigr\Vert _{\sigma } \\ &\quad = \sup_{y \in [0,\infty )}\frac{ \vert L_{n,\alpha }^{*}(e_{2};x)-x ^{2} \vert }{1+x^{2}} \\ &\quad = \biggl(\frac{n^{2}+n(4\alpha -3)}{(n-2)(n-1)}-1 \biggr) \sup_{x \in [0,\infty )} \frac{x^{2}}{1+x^{2}} \\ &\qquad {} + \biggl(\frac{4n+10\alpha -10}{(n-2)(n-1)} \biggr)\sup_{x \in [0, \infty )} \frac{x}{1+x^{2}}+ \frac{2}{(n-2)(n-1)}\sup_{x \in [0, \infty )} \frac{1}{1+x^{2}}, \\ & \bigl\Vert L_{n,\alpha }^{*} (e_{2};x ) -x^{2} \bigr\Vert _{\sigma } =0 \quad \text{when } n \to \infty. \end{aligned}$$

(3.3)

This completes the proof. □

Pointwise approximation properties by $L_{n,\alpha }^{*}$

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. □

Conclusion and observations

The manuscript parametric variant of Baskakov–Durrmeyer operators is a new extension of Baskakov Durrmeyer type operators. In the present investigation in our manuscript in order to get uniform convergence for the operators of the α-type extended version we study the order of approximation, the rate of convergence, the Korovkin-type, the weighted Korovkin-type approximation theorems, Peetres K-functional, Lipschitz functions and a set of direct theorems. It must be noted that we have more modeling flexibility when adding the parameter α to the Baskakov–Durrmeyer operators.