On -statistical approximation of wavelets aided Kantorovich -Baskakov
operators

Mohammad Ayman-Mursaleen1,2, Bishnu P. Lamichhane 1, Adem Kiliçman2,∗ and
Norazak Senu2,3

1School of Information and Physical Sciences,

The University of Newcastle, University Drive, Callaghan, New South Wales
2308, Australia
2Department of Mathematics and Statistics, Faculty of Science,

Universiti Putra Malaysia, 43400 UPM Serdang, Selangor, Malaysia
3Institute for Mathematical Research (INSPEM),

Universiti Putra Malaysia, 43400 UPM Serdang, Selangor, Malaysia

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∗Corresponding author

Abstract

The aim of this research is to examine various statistical approximation
properties with respect to Kantorovich -Baskakov operators using wavelets. We
discuss and investigate a weighted statistical approximation employing a
Bohman-Korovkin type theorem as well as a statistical rate of convergence
applying a weighted modulus of smoothness $\omega_{\rho_{\alpha}}$ correlated
with the space $B_{\rho\alpha}(\mathbb{R_{+}})$ and Lipschitz type maximal
functions. Both topics are covered in the article.

_Keywords and phrases_ : -integers, Baskakov-Kantorovich operators; Baskakov
operators, \- -Baskakov-Kantorovich operators; Baskakov operators;  Wavelets;
Modulus of Smoothness, Haar basis, Statistical Convergence, Weighted function

_AMS Subject Classification (2010):_ Primary 41A25, 41A36; Secondary 33C45.

##  1\.  Introduction and preliminaries

In 1995, Agratini [1] introduced a class of Szász-type operators by means of
compactly supported wavelets of Daubechies. Later on in 1997, Gonska and Zhou
[12] used the Daubechies' compactly-supported wavelets to establish a new
class of Baskakov-type operators. This technique of employing wavelets in
modifying the classical operators is very useful which provides a tool to
achieve the local information of approximation by such operators. In [17],
Nasiruzzaman et al further modified the operators of Gonska and Zhou [12] by
defining their $q$-analog to get a better rate of convergence. In this
scholarly article, our focus is to delve deeper into the various approximation
properties exhibited by the operators described in [17]. Our proposed study
aims to further enhance our understanding of these operators and their
potential applications.

Note that that the Bernstein polynomials [6] converge uniformly to the value
$\text{}(x)$ for every continuous function , where $x$ is any real value
between 0 and 1. The following defines the Bernstein polynomials:

$\left(\mathcal{B}_{r,s}^{\ast}\text{}\right)(x)=\sum_{s=0}^{r}\binom{r}{s}x^{s}(1-x)^{r-s}\text{}\left(\frac{s}{r}\right),$
(1.1)

where $\binom{r}{i}$ refers to the binomial coefficients.

The Szász [20] as well as Baskakov [5] operators were formed in approximating
the continuous functions which were defined with respect to the unbounded
interval $[0,\infty).$ Here, the Baskakov operators are written as

$\left(\mathcal{B}_{\text{},s}\text{}\right)(x)=\sum_{s=0}^{\infty}\binom{\text{}+s-1}{s}\frac{x^{s}}{(1+x)^{\text{}+s}}\text{}\left(\frac{s}{\text{}}\right).$

Bernstein operators were modified by Kantorovich [14] and were christened
Bernstein-Kantorovich operators. These operators are utilized in approximating
the functions of broader classes as opposed to continuous functions. Moreover,
the following are the operators that define Bernstein-Kantorovich operators
given:

$\left(\mathcal{K}_{\text{},s}\text{}\right)(x)=(\text{}+1)\sum_{s=0}^{\text{}}\binom{\text{}}{s}x^{s}(1-x)^{\text{}-s}\int_{\frac{s}{\text{}+1}}^{\frac{s+1}{\text{}+1}}\text{}(\text{})\mathrm{d}\text{},$
(1.2)

for functions $\text{}\in L_{p}[0,1]$ ($1\leq p<\infty$).

To determine the $L_{p}$-approximation, Ditzian and Totik [11] expressed
Kantorovich modification with respect to Baskakov operators, which is called
the Baskakov-Kantorovich operators written as

$\left(\mathcal{K}_{m,\text{}}~{}\text{}\right)(x)=m\sum_{\text{}=0}^{\infty}\binom{m+\text{}-1}{\text{}}\frac{x^{\text{}}}{(1+x)^{m+\text{}}}~{}\int_{\frac{\text{}}{m}}^{\frac{\text{}+1}{m}}\text{}\left(\text{}\right)\mathrm{d}\text{}.$
(1.3)

The -calculus application appeared as a relatively new research field in the
approximation theory. Here, the first -analogue of the famous Bernstein
polynomials was established by Lupaş [15] by employing the concept of
-integers. On the other hand, in 1997, Phillips [18] took into consideration a
different -analogue with respect to the classical Bernstein polynomials.
Subsequently, numerous researchers investigated the -generalizations with
regard to a variety of operators by examining their approximation properties.
For instance, the -variant with respect to Baskakov operators [4] may be
written as

$\left(\mathcal{V}_{m,\text{},\text{}}~{}\text{}\right)(x)=\sum_{\text{}=0}^{\infty}B_{m,\text{},\text{}}(x)~{}\text{}\left(\frac{[\text{}]_{\text{}}}{\text{}^{\text{}-1}[m]_{\text{}}}\right),$

where

$B_{m,\text{},\text{}}(x)=\left[\begin{array}[]{c}m+\text{}-1\\\
\text{}\end{array}\right]_{\text{}}\frac{x^{\text{}}}{(1+x)_{\text{}}^{m+\text{}}}\text{}^{\frac{\text{}(\text{}-1)}{2}},$

while the -Baskakov-Kantorovich operator [10] are defined by

$\left(\mathcal{T}_{m,\text{},\text{}}~{}\text{}\right)(x)=[m]_{\text{}}\sum_{\text{}=0}^{\infty}\text{}^{\text{}-1}B_{m,\text{},\text{}}(x)\int_{\frac{\text{}[\text{}]_{\text{}}}{[m]_{\text{}}}}^{\frac{[\text{}+1]_{\text{}}}{[m]_{\text{}}}}\text{}\left(\text{}^{1-\text{}}\text{}\right)\mathrm{d}_{\text{}}\text{}.$
(1.4)

######  Lemma 1.1.

With respect to the test functions given by $e_{j}=\text{}^{j},~{}j=0,1,2$, it
follows that

$\displaystyle(1)\quad\left(\mathcal{V}_{m,\text{},\text{}}~{}e_{0}\right)(x)$
$\displaystyle=$ $\displaystyle 1,$
$\displaystyle(2)\quad\left(\mathcal{V}_{m,\text{},\text{}}~{}e_{1}\right)(x)$
$\displaystyle=$ $\displaystyle x,$
$\displaystyle(3)\quad\left(\mathcal{V}_{m,\text{},\text{}}~{}e_{2}\right)(x)$
$\displaystyle=$ $\displaystyle
x^{2}+\frac{x}{[m]_{\text{}}}\left(1+\frac{x}{\text{}}\right).$

###  1.1. Basics of -Calculus

The -integer $[m]_{\text{}}$, the -factorial $[m]_{\text{}}!$ as well as the
-binomial coefficient are expressed as below (see [13]) :

$\displaystyle[m]_{\text{}}$
$\displaystyle:=\left\\{\begin{array}[]{ll}\frac{1-\text{}^{m}}{1-\text{}},&\hbox{if~{}}\text{}\in\mathbb{R}^{+}\setminus\\{1\\}\\\
m,&\hbox{if~{}}\text{}=1,\end{array}\right.\mbox{for
$m\in\mathbb{N}$~{}and~{}$[0]_{\text{}}=0$},$ $\displaystyle[m]_{\text{}}!$
$\displaystyle:=\left\\{\begin{array}[]{ll}[m]_{\text{}}[m-1]_{\text{}}\cdots[1]_{\text{}},&\hbox{$m\geq
1$,}\\\ 1,&\hbox{$m=0$,}\end{array}\right.$
$\displaystyle\left[\begin{array}[]{c}m\\\
\text{}\end{array}\right]_{\text{}}$
$\displaystyle:=\frac{[m]_{\text{}}!}{[\text{}]_{\text{}}![m-\text{}]_{\text{}}!},$

accordingly. Here, the -analogue with respect to $(1+x)^{m}$ is given by the
polynomial

$(1+x)_{\text{}}^{m}:=\left\\{\begin{array}[]{ll}(1+x)(1+\text{}x)\cdots(1+\text{}^{m-1}x)&\quad
m=1,2,3,\cdots\\\ 1&\quad n=0.\end{array}\right.$

The Gauss binomial formula may be written as

$(x+a)_{\text{}}^{m}=\sum\limits_{\text{}=0}^{m}\left[\begin{array}[]{c}m\\\
\text{}\end{array}\right]_{\text{}}\text{}^{\text{}(\text{}-1)/2}a^{\text{}}x^{m-\text{}}.$

On the other hand, the -derivative $D_{\text{}}\text{}$ of a function  may be
written as

$(D_{\text{}}\text{})(x)=\frac{\text{}(x)-\text{}(\text{}x)}{(1-\text{})x},~{}x\neq
0,$

as well as $(D_{\text{}}\text{})(0)=\text{}^{\prime}(0),$ provided that
$\text{}^{\prime}(0)$ exists. If  is differentiable, then

$\lim_{\text{}\to 1}D_{\text{}}\text{}(x)=\lim_{\text{}\to
1}\frac{\text{}(x)-\text{}(\text{}x)}{(1-\text{})x}=\frac{d\text{}(x)}{dx}.$

For $m\geq 1$,

$D_{\text{}}(1+x)_{\text{}}^{m}=[m]_{\text{}}(1+\text{}x)_{\text{}}^{m-1},~{}D_{\text{}}\left(\frac{1}{(1+x)_{\text{}}^{m}}\right)=-\frac{[m]_{\text{}}}{(1+x)_{\text{}}^{m+1}},$

$D_{\text{}}\bigg{(}\frac{u(x)}{v(x)}\bigg{)}=\frac{v(\text{}x)D_{\text{}}u(x)-u(\text{}x)D_{\text{}}v(x)}{v(x)v(\text{}x)}.$

The -Jackson definite integral is expressed by

$\int_{0}^{\infty/A}f(x)d_{\text{}}x=(1-\text{})\sum_{n=-\infty}^{\infty}f\left(\frac{\text{}^{n}}{A}\right)\frac{\text{}^{n}}{A}\qquad(A\in\mathbb{R}-\\{0\\}).$

###  1.2. -Statistical convergence

The definition of -analog with respect to Cesàro matrix $C_{1}$ is not unique
(see [2], [3]). Here, we may take into consideration the -Cesàro matrix,
$C_{1}(\text{})=(c_{nk}^{1}(\text{}^{k}))_{n,k=0}^{\infty}$ expressed by

$c_{nk}^{1}(\text{}^{k})=\begin{cases}\frac{\text{}^{k}}{[n+1]_{\text{}}}\text{
\ if }k\leq n,\\\ 0\text{ \ \ \ \ \ \ \ otherwise}.\end{cases}$

which is regular with respect to $\text{}\geq 1$.

Suppose $\mathcal{K}\subseteq\mathbb{N}$ (the set of natural numbers).
Therefore, $\delta(\mathcal{K})=\lim_{r}\frac{1}{r}\\#\\{k\leq
r~{}:~{}k\in\mathcal{K}\\}$ is known as the asymptotic density with respect to
$\mathcal{K}$, in which $\\#$ resembles the cardinality of the enclosed set.
Moreover, a sequence $\eta=(\eta_{k})$ is known as statistically convergent to
the number $\mathfrak{s}$ provided that $\delta(\mathcal{K}_{\varepsilon})=0$
for every $\varepsilon>0$, in which $\mathcal{K}_{\varepsilon}=\\{k\leq
r:|\eta_{k}-\mathfrak{s}|>\varepsilon\\}$ (refer to [9]).

In the current years, Aktuğlu and Bekar [3] determined -density as well as
-statistical convergence. The -density may be expressed as

$\delta_{\text{}}(\mathcal{K})=\delta_{C_{1}^{\text{}}}(\mathcal{K})=\lim\inf_{n\rightarrow\infty}(C_{1}^{\text{}}\chi_{\mathcal{K}})_{n}=\lim\inf_{n\rightarrow\infty}\sum_{k\in
K}\frac{\text{}^{k-1}}{[n]},\text{ }\text{}\geq 1.$

A sequence $\eta=(\eta_{k})$ is known to be -statistically convergent with
respect to the number $\mathcal{L}$ provided that
$\delta_{\text{}}(\mathcal{K}_{\varepsilon})=0$, in which
$\mathcal{K}_{\varepsilon}=\\{k\leq
n:|\eta_{k}-\mathcal{L}|\geq\varepsilon\\}$ for every $\varepsilon>0$. In
other words, for each $\varepsilon>0,$.

$\lim_{n}\frac{1}{[n]}\\#\\{k\leq
n:\text{}^{k-1}|\eta_{k}-\mathcal{L}|\geq\varepsilon\\}=0$

We may also write $St_{\text{}}-\lim\eta_{k}=\mathcal{L}$.

Provided that $\delta(\mathcal{K})=0$ for an infinite set $\mathcal{K}$,
$\delta_{\text{}}(\mathcal{K})=0,$. Therefore, statistical convergence [9,
Example 15] Implies that it is -statistical convergence but not conversely
(refer to [Example 15][3]).

##  2\. Wavelets aided -Baskakov-Kantorovich operators

We now recall several basic knowledge with respect to wavelets [7, 16]. Here,
the wavelets notion denotes the set of functions of the form given by

$\Psi_{\mu,\nu}(x)=\mu^{-\frac{1}{2}}\Psi\left(\frac{x-\nu}{\mu}\right)~{}\mu>0,~{}\nu\in\mathbb{R},$

which are formed via translations and dilations with respect to a single
function $\Psi$, which is called the mother wavelet or basic wavelet.
Moreover, following the Franklin-Strom̈berg theory, the constant $\mu$ may be
substituted by $2^{i}$ while $\nu$ may be substituted by $2^{i}\text{}$ having
$i$ and  to be the integers. With respect to an arbitrary function $\text{}\in
L_{2}(\mathbb{R}),$ the wavelets have a crucial part in the orthonormal basis,
in which the $g$ function is given as:

$\text{}(x)=\sum_{-\infty}^{\infty}\sum_{-\infty}^{\infty}\gamma(i,\text{})\Psi_{i,\text{}}(x),$

in which

$\gamma(i,\text{})=2^{\frac{i}{2}}\Psi_{i,\text{}}(x)\int_{\mathbb{R}}f(x)\Psi(2^{i}x-\text{})\mathrm{d}x.$

Daubechies [8] formed an orthonormal basis for $L_{2}(\mathbb{R})$ expressed
by

$2^{\frac{i}{2}}\Psi_{s}(x)(2^{i}x-\text{}),$

where $s$ refers to the non-negative integer, $i,\text{}$ denote the integers
set as well as the support of $\Psi_{s}$ is $[0,2s+1]$. For a positive
constant $\xi$, if $\Psi_{s}$ has $\xi s$ order of continuous derivatives,
then for any $0\leq\text{}\leq s,~{}s\in\mathbb{N},$ we have

$\int_{\mathbb{R}}x^{\text{}}\Psi_{s}(x)\mathrm{d}x=0.$ (2.1)

Evidently, when $s=0$, the system is reduced to the Haar system. Here, with
regard to any $\Psi\in L_{\infty}(\mathbb{R}),$ we now have the conditions
given by: (i) a finite positive $\xi$ having the property
$\sup\Psi\subset[0,\xi],$ while (ii) its first $s$ moment vanishes.
Furthermore, for $1\leq\text{}\leq s,~{}s\in\mathbb{N},$ we have
$\int_{\mathbb{R}}t^{\text{}}\Psi(\text{})\mathrm{d}\text{}=0$ and
$\int_{\mathbb{R}}\Psi(\text{})\mathrm{d}\text{}=1$. Therefore, by employing
the Haar basis, the Baskakov type operators may be expressed by [1]:

$\left(\mathcal{L}_{m,\text{}}~{}\text{}\right)(x)=m\sum_{\text{}=0}^{\infty}\binom{m+\text{}-1}{\text{}}\frac{x^{\text{}}}{(1+x)^{m+\text{}}}\int_{\mathbb{R}}\text{}\left(\text{}\right)\Psi\left(m\text{}-\text{}\right)\mathrm{d}\text{},$
(2.2)

in which the operators $\mathcal{L}_{m,\text{}}$ refer to the extensions with
respect to Baskakov-Kantorovich operators. By considering the
$\sup\Psi\subset[0,\xi],$ the operators $\mathcal{L}_{m,\text{}}$ may be
expressed as written below [1]:

$\left(\mathcal{L}_{m,\text{}}~{}\text{}\right)(x)=\sum_{\text{}=0}^{\infty}\binom{m+\text{}-1}{\text{}}\frac{x^{\text{}}}{(1+x)^{m+\text{}}}\int_{0}^{\xi}\text{}\left(\frac{\text{}+\text{}}{m}\right)\Psi(\text{})\mathrm{d}\text{}.$
(2.3)

The current section mentions the -Baskakov type operators by employing
compactly-supported wavelets of Daubechies constructed in [17]. Let
$\int_{\mathbb{R}}x^{s}\Psi_{k}(x)\mathrm{d}_{\text{}}x=0$ when $0\leq s\leq
k$ for $k\in\mathbb{N}$ as well as $\text{}>0.$

With regard to $\Psi\in L_{\infty}(\mathbb{R}),$ we assume the conditions
given below in terms of wavelets: (i) a finite positive $\xi$ having the
property $\sup\Psi\subset[0,\xi];$ and (ii) its first $k$ moment vanishes.
With respect to $1\leq s\leq k$ and $k\in\mathbb{N}$, we now obtain
$\int_{\mathbb{R}}\text{}^{s}\Psi(\text{})\mathrm{d}_{\text{}}\text{}=0$ as
well as $\int_{\mathbb{R}}\Psi(\text{})\mathrm{d}_{\text{}}\text{}=1$.
Therefore, for all $1\leq s\leq k,~{}k\in\mathbb{N}$ as well as $0<\text{}<1,$
we construct the $\text{}-$analogue of Baskakov-Kantorovich type wavelets
operators given by:

$\left(\mathcal{S}_{\text{},s,\text{}}~{}\text{}\right)(x)=[\text{}]_{\text{}}\sum_{s=0}^{\infty}\text{}^{s-1}B_{\text{},s,\text{}}(x)\int_{\mathbb{R}}\text{}\left(\text{}\right)\Psi\left(\text{}^{s-1}[\text{}]_{\text{}}\text{}-[s]_{\text{}}\right)\mathrm{d}_{\text{}}\text{}.$
(2.4)

Thus, our operators $\mathcal{S}_{\text{},s,\text{}}(\text{};x)$ extends the
-Baskakov-Kantorovich operators expressed in (1.4). With respect to the
choices of $k=0$ as well as $\Psi$ Haar basis, we obtain the exact  -Baskakov-
Kantorovich operators $\mathcal{T}_{\text{},s,\text{}}(\text{};x)$ by (1.4).
Additionally, for the choices $k=0,~{}\text{}=1$ as well as $\Psi$ Haar basis,
we determined the Baskakov-Kantorovich operators
$\mathcal{K}_{\text{},s,\text{}}(\text{};x)$ by (1.3). Considering the
$\sup\Psi\subset[0,\xi],$ the operators
$\mathcal{S}_{\text{},s,\text{}}(\text{};x)$ may be expressed as given below:

$\left(\mathcal{S}_{\text{},s,\text{}}~{}\text{}\right)(x)=\sum_{s=0}^{\infty}B_{\text{},s,\text{}}(x)\int_{0}^{\xi}\text{}\left(\frac{\text{}+[s]_{\text{}}}{\text{}^{s-1}[\text{}]_{\text{}}}\right)\Psi\left(\text{}\right)\mathrm{d}_{\text{}}\text{}.$
(2.5)

It is evident that by choosing $\text{}=1$, we obtain classical Baskakov-
Kantorovich wavelets operators $\mathcal{L}_{\text{},s}$ by (2.2) as well as
(2.3).

We need the following result of [17]:

######  Theorem 2.1.

Suppose $e_{j}=t^{j}$ when $0\leq j\leq k$ and $k\in\mathbb{N}$. Therefore, we
obtain

$\left(\mathcal{S}_{r,s,\text{}}\;e_{j}\right)(x)=\left(\mathcal{V}_{r,s,\text{}}\;e_{j}\right)(x),$

in which $x\in[0,\infty)$ as well as the operators
$\left(\mathcal{V}_{r,s,\text{}}\;g\right)(x)$ defined as above.

##  3\. Weighted -Statistical approximation

This section presents the statistical approximation of Wavelets Kantorovich
-Baskakov operators $\mathcal{S}_{\text{},s,\text{}}$ defined by (2.4)
employing a Bohman Korovkin-type theorem proven in [21].

Suppose $N_{\text{}}$ is the constant depending on the function  and represent
the weighted space of a real valued function  expressed on $\mathbb{R}$ by
$B_{\rho}(\mathbb{R})$ with the property that $|\text{}(x)|\leq
N_{\text{}}\rho(x)$ for all $x\in\mathbb{R}$. Now, we take into consideration
the weighted subspace $C_{\rho}(\mathbb{R})$ of $B_{\rho}(\mathbb{R})$ which
is provided as

$\displaystyle C_{\rho}(x)(\mathbb{R})=\\{\text{}\in
B_{\rho}(\mathbb{R}):\mathrm{\text{}}~{}\mathrm{continuous}~{}\mathrm{in}~{}\mathbb{R}\\}.$

This fits with the norm $\parallel.\parallel$, where
$\parallel\text{}\parallel_{\rho}=\sup\limits_{x\in\mathbb{R}}\dfrac{|\text{}(x)|}{\rho(x)}$
and both $C_{\rho}(\mathbb{R})$ and $B_{\rho}(\mathbb{R})$ are Banach spaces.
By the use of A-statistical convergence, Duman and Orhan [21] proved the
theorem given below, which is useful in proving our main result.

######  Theorem 3.1.

(Duman and Orhan [21]). If $A=(a_{j\text{}})_{j,\text{}}$ is a positive
regular summability matrix, then let $(L_{\text{}})_{\text{}}$ denote a
sequence of positive linear operators from $C_{\rho_{1}}(\mathbb{R})$ to
$B_{\rho_{2}}(\mathbb{R})$, in which $\rho_{1}$ as well as $\rho_{2}$
satisfies $\lim\limits_{|x|\to\infty}\dfrac{\rho_{1}}{\rho_{2}}$ = 0. Then

$\displaystyle st_{A}-\lim\limits_{\text{}}\parallel
L_{\text{}}\text{}-\text{}\parallel_{\rho_{2}}=0,~{}\forall\text{}\in
C_{\rho_{1}}(\mathbb{R})$

if and only if

$\displaystyle st_{A}-\lim\limits_{\text{}}\parallel
L_{\text{}}H_{v}-H_{v}\parallel_{\rho_{1}}=0~{}for~{}v=0,1,2,$

in which $H_{v}=\dfrac{x^{v}\rho_{1}(x)}{1+x^{2}}$.

By examining this result, it is clear that if $\mathbb{R}$ is substituted by
$\mathbb{R_{+}}$, then the theorem holds true. Also, by analyzing Lemma 1.1,
we see that the sequence of operators
$(\mathcal{S}_{\text{},s,\text{}})_{\text{}}$ fail to satisfy the properties
of Bohman-Korovkin theorem. Now, let us take into consideration the weight
functions $\rho_{0}(x)=1+x^{2}$ and $\rho_{\alpha}(x)=1+x^{2+\alpha}$ for
$x\in\mathbb{R_{+}}$ and $\alpha>0$ together with the remark below.

######  Remark 3.2.

It is true that for $\text{}\in(0,1)$, then
$\lim\limits_{\text{}\to\infty}[\text{}]_{\text{}}=0$ or
$\dfrac{1}{1-\text{}}$. Now, we consider the sequence
$(\text{}_{\text{}})_{\text{}}$ for $\text{}_{\text{}}\in(0,1)$ with the
property that $st-\lim\limits_{\text{}\to\infty}\text{}_{\text{}}=1$ and
$st-\lim\limits_{\text{}\to\infty}\text{}_{\text{}}^{\text{}}=1$. Based on
these facts, we have
$\lim\limits_{\text{}\to\infty}[\text{}]_{\text{}}=\infty$. This will help to
check the convergence with respect to the operators expressed by inequality
(2.4). Thus, we now obtain the theorem given below:

######  Theorem 3.3.

Suppose that the sequence $(\text{}_{\text{}})_{\text{}}$ satisfy Remark 3.2
above and $\mathcal{S}_{\text{},s,\text{}}$ is a positive linear operator.
Then, we have:

$\displaystyle
St_{\text{}}-\lim\limits_{\text{}}\parallel(\mathcal{S}_{\text{},s,\text{}}(\text{})-\text{}\parallel_{\rho_{\alpha}}=0,~{}\forall\text{}\in
C_{\rho_{0}}(\mathbb{R_{+}}).$

###### Proof.

Based on Lemma 1.1(i) and Theorem 2.1, we now have:

$\displaystyle\parallel(\mathcal{S}_{\text{},s,\text{}}(\text{})-\text{}\parallel_{\rho_{0}}$
$\displaystyle=\sup\limits_{x\in\mathbb{R}}\dfrac{|(\mathcal{S}_{\text{},s,\text{}_{\text{}}}e_{0})(x)-e_{0}(x)|}{1+x^{2}},$
$\displaystyle=\sup\limits_{x\in\mathbb{R}}\dfrac{|1-1|}{1+x^{2}},$
$\displaystyle=0.$

In other words,

$St_{\text{}}-\lim\limits_{\text{}}\parallel(\mathcal{S}_{\text{},s,\text{}}(\text{})-\text{}\parallel_{\rho_{0}}=0.$

Again, based on Lemma 1.1(ii) and Theorem 2.1, we now have:

$\displaystyle\parallel(\mathcal{S}_{\text{},s,\text{}}(\text{})-\text{}\parallel_{\rho_{0}}$
$\displaystyle=\sup\limits_{x\in\mathbb{R}}\dfrac{|(\mathcal{S}_{\text{},s,\text{}_{\text{}}}e_{1})(x)-e_{1}(x)|}{1+x^{2}},$
$\displaystyle=\sup\limits_{x\in\mathbb{R}}\dfrac{|x-x|}{1+x^{2}},$
$\displaystyle=0.$

Using Lemma 1.1(iii) and Theorem 2.1, we now have:

$\displaystyle\parallel(\mathcal{S}_{\text{},s,\text{}}(\text{})-\text{}\parallel_{\rho_{0}}$
$\displaystyle=\sup\limits_{x\in\mathbb{R}}\dfrac{\left|(\mathcal{S}_{\text{},s,\text{}_{\text{}}}e_{2})(x)-e_{2}(x)\right|}{1+x^{2}},$
$\displaystyle=\sup\limits_{x\in\mathbb{R}}\dfrac{\left|\left(x^{2}+x\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}\left(1+\dfrac{1}{\text{}_{\text{}}}x\right)\right)-x^{2}\right|}{1+x^{2}},$
$\displaystyle=\sup\limits_{x\in\mathbb{R}}\dfrac{\left|\left(1+\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{n}}}-1\right)x^{2}+x\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}\right|}{1+x^{2}},$
$\displaystyle\leq\sup\limits_{x\in\mathbb{R}}\left|\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}x^{2}+x\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}\right|,$
$\displaystyle\leq\sup\limits_{x\in\mathbb{R}}\left(\left|x^{2}\right|\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}+\left|x\right|\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}\right),$
$\displaystyle=\left(\parallel
e_{2}\parallel_{\rho_{0}}\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}+\parallel
e_{1}\parallel_{\rho_{0}}\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}\right),$
$\displaystyle\leq\left(\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}+\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}\right).$

From Remark 3.2, we have
$st-\lim\limits_{\text{}\to\infty}\text{}_{\text{}}=1$. Furthermore, we also
obtained $\lim\limits_{\text{}\to\infty}[\text{}]_{\text{}}=\infty$.
Consequently

$\displaystyle
St_{\text{}}-\lim\limits_{\text{}}\parallel(\mathcal{S}_{\text{},s,\text{}}(\text{})-\text{}\parallel_{\rho_{0}}=0.$

By employing Lemma 1.1 and also selecting $A=C_{1}$, known as the Cesáro
matrix of order one, $\rho_{0}(x)=1+x^{2}$, $\rho_{\alpha}(x)=1+x^{2+\alpha}$
for $x\in\mathbb{R_{+}}$ and $\alpha>0$, the proof may be immediately seen
from Theorem 3.1. ∎

##  4\.
The Convergence Rate

In this sub-section, by means of weighted modulus of smoothness correlated to
the space $B_{\rho\alpha}(\mathbb{R_{+}})$ and Lipschitz type maximal
functions, we present the rates of statistical convergence with respect to the
operators $\mathcal{S}_{\text{},s,\text{}}$ expressed by inequality (2.4). The
weighted modulus with respect to smoothness $\omega_{\rho_{\alpha}}$
correlated to the space $B_{\rho\alpha}(\mathbb{R_{+}})$ of a function  is
defined as:

$\omega_{\rho_{\alpha}}(\text{};\delta)=\sup\limits_{x\geq
0,~{}0<i<\delta}\frac{\left|\text{}(x+i)-\text{}(x)\right|}{1+(x+i)^{2+\alpha}},~{}\delta>0,\alpha\geq
0.$

It satisfies the following three axioms.

  1. (a)

$\omega_{\rho_{\alpha}}(\text{};\beta\delta)\leq(\beta+1)\omega_{\rho_{\alpha}}(\text{};\delta)$
for $\delta>0$ and $\beta>0$.

  2. (b)

$\omega_{\rho_{\alpha}}(\text{};\text{}\delta)\leq\text{}\omega_{\rho_{\alpha}}(\text{};\delta)$
for $\delta>0$ and $\text{}\in\mathbb{N}$.

  3. (c)

$\lim\limits_{\delta\to\infty}\omega_{\rho_{\alpha}}(\text{};\delta)=0$.

The following theorem gives an error estimate of an operator
$\mathcal{S}_{\text{},s,\text{}}$ for the unbounded function $h$ by means of
weighted modulus of smoothness correlated to the space
$B_{\rho\alpha}(\mathbb{R_{+}})$.

######  Theorem 4.1.

Suppose that $\text{}\in(0,1)$ and $\alpha\geq 0$. Then, for any $\text{}\in
B_{\rho\alpha}(\mathbb{R_{+}})$, we have

$\displaystyle\left|(\mathcal{S}_{\text{},s,\text{}}~{}\text{})(x)-\text{}(x)\right|\leq\sqrt{\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)}\left(1+\dfrac{1}{\delta}\sqrt{\mathcal{S}_{\text{},s,\text{}}(\phi_{x}^{2};x)}\right)\omega_{\rho_{\alpha}}(\text{};\delta),$

where $\mu_{x,\alpha}(y)=1+\biggl{(}x+\left|y-x\right|\biggr{)}^{2+\alpha}$ as
well as $\phi_{x}(y)=\left|y-x\right|$ for $y\geq 0$.

###### Proof.

Suppose that $\text{}\in\mathbb{N}$ and $\text{}\in
B_{\rho\alpha}(\mathbb{R_{+}})$. Using inequality (4) and axiom (a) above, we
can write that

$\displaystyle\left|\text{}(y)-\text{}(x)\right|$
$\displaystyle\leq\bigg{(}1+(x+\left|y-x\right|)^{2+\alpha}\bigg{)}\bigg{(}1+\dfrac{1}{\delta}\left|y-x\right|\bigg{)}\omega_{\rho_{\alpha}}(\text{};\delta),$
$\displaystyle=\mu_{x,\alpha}(y)\bigg{(}1+\dfrac{1}{\delta}\phi_{x}(y)\bigg{)}\omega_{\rho_{\alpha}}(\text{};\delta).$

Next, using the Cauchy inequality with respect to the positive linear
operator's yields

$\displaystyle\left|(\mathcal{S}_{\text{},s,\text{}}~{}\text{})(x)-\text{}(x)\right|$
$\displaystyle\leq\left[\text{}\right]_{\text{}}\sum_{s=0}^{\infty}\text{}^{s-1}\upsilon_{s,\text{}}^{\text{}}(x)\int_{\mathbb{R}}\left|\text{}(y)-\text{}(x)\right|\Psi\left(\left[\text{}\right]_{\text{}}\frac{\text{}^{s-1}}{1}y-\left[s\right]_{\text{}}\right)d_{\text{}}y,$
$\displaystyle\leq\biggl{(}\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha};x)+\dfrac{1}{\delta}\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}\phi_{x};x)\biggr{)}\omega_{\rho_{\alpha}}(\text{};\delta),$
$\displaystyle\leq\sqrt{\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)}\left(1+\dfrac{1}{\delta}\sqrt{\mathcal{S}_{\text{},s,\text{}}(\phi_{x}^{2};x)}\right)\omega_{\rho_{\alpha}}(\text{};\delta).$

∎

Now, we introduce the lemma given below, which may facilitate in proving the
primary findings with respect to this research, since it is one of the facts
which ensure that $(\mathcal{S}_{\text{},s,\text{}}\text{})(x)\in
B_{\rho\alpha}(\mathbb{R_{+}})$ .

######  Lemma 4.2.

Suppose that $0<\text{}\leq 1$, then for $i,\text{}\in\mathbb{N}$ and
$x\in\mathbb{R_{+}}$, we obtain

$\displaystyle(\Gamma_{\text{},s,\text{}}e_{i})(x)\leq\dfrac{1}{[\text{}]_{\text{}}^{i-1}(1+x)_{\text{}}^{\text{}}}x+\dfrac{2^{i-1}}{\text{}^{i-1}}x(\Gamma_{\text{}+1,s,\text{}}e_{i-1})(x).$

###### Proof.

For $s\in\mathbb{N}$ as well as $0<\text{}\leq 1$, we have the inequality
given below:

$\displaystyle 1\leq[s+1]_{\text{}}\leq 2[s]_{\text{}}.$ (4.1)

Now, let $i\in\mathbb{N}$. Using Equation (1), we have:

$\displaystyle(\Gamma_{\text{},s,\text{}}e_{i})(x)$
$\displaystyle=\sum_{s=0}^{\infty}\upsilon_{\text{},s}^{\text{}}(x)e_{i}\left(\frac{\left[s\right]_{\text{}}}{\text{}^{s-1}\left[\text{}\right]_{\text{}}}\right),$
$\displaystyle=\sum_{s=0}^{\infty}\upsilon_{\text{},s}^{\text{}}(x)\left(\frac{\left[s\right]_{\text{}}}{\text{}^{s-1}\left[\text{}\right]_{\text{}}}\right)^{i},$
$\displaystyle=\sum_{s=0}^{\infty}\upsilon_{\text{},s}^{\text{}}(x)\frac{\left[s\right]_{\text{}}^{i}}{\text{}^{(s-1)i}\left[\text{}\right]_{\text{}}^{i}},$
$\displaystyle=\sum_{s=1}^{\infty}x\upsilon_{\text{}+1,s-1}^{\text{}}(x)\frac{\left[s\right]_{\text{}}^{i-1}}{\text{}^{(s-1)(i-1)}\left[\text{}\right]_{\text{}}^{i-1}},$
$\displaystyle=\sum_{s=0}^{\infty}x\upsilon_{\text{}+1,b}^{\text{}}(x)\frac{\left[s+1\right]_{\text{}}^{i-1}}{\text{}^{s(i-1)}\left[\text{}\right]_{\text{}}^{i-1}},$
$\displaystyle=\dfrac{1}{[\text{}]_{\text{}}^{i-1}(1+x)_{\text{}}^{\text{}}}x+x\sum_{s=1}^{\infty}\upsilon_{\text{}+1,s}^{\text{}}(x)\frac{\left[s+1\right]_{\text{}}^{i-1}}{\text{}^{s(i-1)}\left[\text{}\right]_{\text{}}^{i-1}}.$

Using Inequality (4.1), we have,

$\displaystyle(\Gamma_{\text{},s,\text{}}e_{i})(x)$
$\displaystyle\leq\dfrac{x}{[\text{}]_{\text{}}^{i-1}(1+x)_{\text{}}^{\text{}}}+x\sum_{s=1}^{\infty}\upsilon_{\text{}+1,s}^{\text{}}(x)\frac{(2\left[s\right]_{\text{}})^{i-1}}{\text{}^{s(i-1)}\left[\text{}\right]_{\text{}}^{i-1}},$
$\displaystyle=\dfrac{x}{[\text{}]_{\text{}}^{i-1}(1+x)_{\text{}}^{\text{}}}+\dfrac{2^{i-1}}{\text{}^{i-1}}x\sum_{s=1}^{\infty}\upsilon_{\text{}+1,s}^{\text{}}(x)\frac{\left[s\right]_{\text{}}^{i-1}}{\text{}^{(s-1)(i-1)}\left[\text{}\right]_{\text{}}^{i-1}}.$

Based on Equation (1), we have that:

$\displaystyle(\Gamma_{\text{}+1,s,\text{}}e_{i-1})(x)=\sum_{s=1}^{\infty}\upsilon_{\text{}+1,s}^{\text{}}(x)\frac{\left[s\right]_{\text{}}^{i-1}}{\text{}^{(s-1)(i-1)}\left[\text{}\right]_{\text{}}^{i-1}}.$

Consequently,

$\displaystyle(\Gamma_{\text{},s,\text{}}e_{i})(x)\leq\dfrac{1}{[\text{}]_{\text{}}^{i-1}(1+x)_{\text{}}^{\text{}}}x+\dfrac{2^{i-1}}{\text{}^{i-1}}x(\Gamma_{\text{}+1,s,\text{}}e_{i-1})(x).$

∎

######  Remark 4.3.

Any positive and linear operator is monotone. Theorem 2.1 and Lemma 4.2 ensure
that $(\mathcal{S}_{\text{},s,\text{}}\text{})(x)\in
B_{\rho\alpha}(\mathbb{R_{+}})$ for any $\text{}\in
B_{\rho\alpha}(\mathbb{R_{+}})$ and $\alpha\in\mathbb{N}_{0}$, where
$\mathbb{N}_{0}=\\{0\\}\cup\mathbb{N}$.

We may state the major outcome of this portion as follows based on the
preparation mentioned earlier:

######  Theorem 4.4.

Let $(\text{}_{\text{}})_{\text{}}$ be the sequence satisfying Remark 3.2
above and $\alpha\in\mathbb{N}_{0}$. Then, for every $\text{}\in
B_{\rho\alpha}(\mathbb{R_{+}})$, we have

$\displaystyle\lim\limits_{\text{}}\parallel(\mathcal{S}_{\text{},s,\text{}_{\text{}}}\text{})(x)-\text{}(x)\parallel_{\rho_{\alpha}}\leq
3C_{\alpha}\omega_{\rho_{\alpha}}(\text{};\delta_{\text{}}),$

where $C_{\alpha}>0$ is a constant and
$\delta_{\text{}}=\sqrt{\dfrac{1}{\text{}_{\text{}}[\text{}]_{\text{}_{\text{}}}}}$
.

###### Proof.

From Lemma 1.1, we have the following:

$\displaystyle\mathcal{S}_{\text{},s,\text{}_{\text{}}}(\phi_{x}^{2};x)$
$\displaystyle=\left(x^{2}+x\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}\left(1+\dfrac{1}{\text{}_{\text{}}}x\right)\right)-x^{2},$
$\displaystyle=\left(1+\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}-1\right)x^{2}+x\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}},$
$\displaystyle=\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}x^{2}+\dfrac{1}{\left[\text{}\right]_{\text{}_{\text{}}}}x.$

Consequently, we have the inequality:

$\displaystyle\mathcal{S}_{\text{},s,\text{}_{\text{}}}(\phi_{x}^{2};x)\leq\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}x^{2}+\dfrac{3}{\left[\text{}\right]_{\text{}_{\text{}}}}x.$
(4.2)

Let $\alpha\geq 0$ a constant and $\text{}\in B_{\rho\alpha}(\mathbb{R_{+}})$.
Using Theorem 4.1 as well as the inequality (4.2) above, we may express the
following:

$\displaystyle\lim\limits_{\text{}}\parallel(\mathcal{S}_{\text{},s,\text{}}\text{})(x)-\text{}(x)\parallel_{\rho_{\alpha}}$
$\displaystyle=\dfrac{\left|(\mathcal{S}_{\text{},s,\text{}}\text{})(x)-\text{}(x)\right|}{1+x^{2+\alpha}},$
$\displaystyle\leq\sqrt{\dfrac{\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)}{1+x^{2+\alpha}}}\left(1+\dfrac{1}{\delta}\sqrt{\dfrac{\mathcal{S}_{\text{},s,\text{}}(\phi_{x}^{2};x)}{1+x^{1+\alpha}}}~{}~{}\right)\omega_{\rho_{\alpha}}(\text{};\delta),$
$\displaystyle\leq\sqrt{\dfrac{\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)}{1+x^{2+\alpha}}}\left(1+\dfrac{1}{\delta}\sqrt{\left|\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}x^{2}+\dfrac{3}{\left[\text{}\right]_{\text{}_{\text{}}}}x\right|}~{}~{}\right),$
$\displaystyle\times\omega_{\rho_{\alpha}}(\text{};\delta),$
$\displaystyle\leq\sqrt{\dfrac{\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)}{1+x^{2+\alpha}}}\left(1+\dfrac{1}{\delta}\sqrt{\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}\parallel
e_{2}\parallel_{\rho_{\alpha}}+\dfrac{3}{\left[\text{}\right]_{\text{}_{\text{}}}}\parallel
e_{2}\parallel_{\rho_{\alpha}}}~{}~{}\right)$
$\displaystyle\times\omega_{\rho_{\alpha}}(\text{};\delta).$

Furthermore,

$\displaystyle\lim\limits_{\text{}}\parallel(\mathcal{S}_{\text{},s,\text{}_{\text{}}}\text{})(x)-\text{}(x)\parallel_{\rho_{\alpha}}$
$\displaystyle\leq\sqrt{\dfrac{\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)}{1+x^{2+\alpha}}}\left(1+\dfrac{2}{\delta}\sqrt{\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}}~{}~{}\right)\omega_{\rho_{\alpha}}(\text{};\delta),$
$\displaystyle\leq\parallel\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)\parallel_{\delta\alpha}\left(1+\dfrac{2}{\delta}\sqrt{\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}}~{}~{}\right)\omega_{\rho_{\alpha}}(\text{};\delta).$

Let
$C_{\alpha}=\parallel\mathcal{S}_{\text{},s,\text{}}(\mu_{x,\alpha}^{2};x)\parallel_{\delta\alpha}$
and choose
$\delta=\sqrt{\dfrac{1}{\text{}_{\text{}}\left[\text{}\right]_{\text{}_{\text{}}}}}$,
we have:

$\displaystyle\lim\limits_{\text{}}\parallel(\mathcal{S}_{\text{},s,\text{}_{\text{}}}\text{})(x)-\text{}(x)\parallel_{\rho_{\alpha}}\leq
3C_{\alpha}\omega_{\rho_{\alpha}}(\text{};\delta_{\text{}}).$

∎

######  Remark 4.5.

Since $(\text{}_{\text{}})_{\text{}}$ satisfy Remark 3.2, then the sequence
$(\delta_{\text{}})_{\text{}}$ is statistically null, that is
$st-\lim\limits_{\text{}}\omega_{\rho_{\alpha}}(\text{};\delta_{\text{}})=0$.
Therefore, Theorem 4.4 above gives the statistical rate of convergence of
$\mathcal{S}_{\text{},s,\text{}_{\text{}}}(x)$ to .

##  5\. Graphical analysis

Using computer software, we will demonstrate some numerical examples with
illustrative graphics.

######  Example 5.1.

Let $\text{}(x)=(x-\frac{1}{5})(x-\frac{4}{9})$, $\text{}=0.95$ and
$n\in\\{10,30,80\\}$. The convergence of the operator towards the function
$\text{}(x)$ is shown in Figure 1.

Figure 1. convergence of the operator towards the function
$\text{}(x)=(x-\frac{1}{5})(x-\frac{4}{9})$

######  Example 5.2.

Let $\text{}(x)=x^{2}-1$, $\text{}=1$ and $n\in\\{10,30,60\\}$. The
convergence of the operator towards the function $\text{}(x)$ is shown in
Figure 2.

Figure 2. convergence of the operator towards the function
$\text{}(x)=x^{2}-1$

######  Example 5.3.

Let $f(x)=x^{2}-4x+3$. For $n=50$ and different values of , the convergence of
the operator towards the function $f(x)$ is shown in Figure 3.

Figure 3. Convergence of the operator for different values of

##  6\. Conclusion

With the facilitation of Bohman Korovkin-type theorem, the investigation on
weighted statistical approximation behavior of Wavelets Kantorovich -Baskakov
operators $\mathcal{S}_{\text{},s,\text{}}$ is discussed under this study.
Moreover, the convergence statistical rate with respect to the operators
$\mathcal{S}_{\text{},s,\text{}}$ is provided in this research with regard to
the weighted modulus of smoothness correlated to the space
$B_{\rho\alpha}(\mathbb{R_{+}})$. The statistical approximation properties
discussed in this study are the same as those of classical -Baskakov operators
defined by equation (1) since they share the same moments.

## Declarations

### Ethical Approval

Not Applicable

### Availability of supporting data

Not Applicable

### Competing interests

Not Applicable

### Funding

Not Applicable

### Acknowledgments

Not Applicable

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