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.

1 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].

2 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. □

3 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. □

4 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. □

5 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.