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Article

# On the Approximation by Bivariate Szász–Jakimovski–Leviatan-Type Operators of Unbounded Sequences of Positive Numbers

by
Abdullah Alotaibi
Operator Theory and Applications Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia
Mathematics 2023, 11(4), 1009; https://doi.org/10.3390/math11041009
Submission received: 31 December 2022 / Revised: 2 February 2023 / Accepted: 3 February 2023 / Published: 16 February 2023

## Abstract

:
In this paper, we construct the bivariate Szász–Jakimovski–Leviatan-type operators in Dunkl form using the unbounded sequences $α n$$β m$ and $ξ m$ of positive numbers. Then, we obtain the rate of convergence in terms of the weighted modulus of continuity of two variables and weighted approximation theorems for our operators. Moreover, we provide the degree of convergence with the help of bivariate Lipschitz-maximal functions and obtain the direct theorem.
MSC:
41A25; 41A36; 33C45

## 1. Introduction and Preliminaries

In 1912, S. N. Bernstein [1] introduced a positive linear operator for the set of all continuous functions on $[ 0 , 1 ]$, which provides the shortest proof of the well-known Weierstrass approximation theorem while, later, another positive linear operator on $[ 0 , ∞ )$, constructed by Szász in 1950, known as the Szász operator (see [2]), is given by
$S r ( f ; u 1 ) = e − r u 1 ∑ k = 0 ∞ ( r u 1 ) k k ! f k r , u 1 ≥ 0 , r ∈ N and f ∈ C [ 0 , ∞ ) .$
Due to development of the Szász operator, another sequence of positive linear operator constructed by the mathematician Jakimovski and Leviatan in 1969 using the well-known Appell polynomial is given by (see [3])
$L r ( f ; u 1 ) = e − r u 1 C ( 1 ) ∑ k = 0 ∞ P k ( r u 1 ) f k r ,$
where the Appell polynomial is given by $C ( v 1 ) e v 1 y = ∑ r = 0 ∞ P r ( y 1 ) v 1 r$, with the following identities $C ( 1 ) ≠ 0$$C ( v 1 ) = ∑ r = 0 ∞ c r v 1 r$$P r ( u 1 ) = ∑ j = 0 r c j u 1 r − j ( r − j ) ! ( r ∈ N )$.
Recently, with the help of the exponential generating function, the Szász operators were introduced by Sucu [4]. These types of ideas are influenced by the generalized Hermite polynomials in terms of the hypergeometric functions (see [5]). Thus, for the classes of all continuous functions f on $[ 0 , ∞ )$ and a parameter $χ ≥ 0$, the Szász operators given in another form, known as Szász Dunkl operators, are given by
$S r ∗ ( f ; u 1 ) : = 1 e χ ( r u 1 ) ∑ k = 0 ∞ ( r u 1 ) k γ χ ( k ) f k + 2 χ θ k r , u 1 ≥ 0 , r ∈ N ,$
where the exponential generating functions are given by
$e χ ( u 1 ) = ∑ k = 0 ∞ u 1 k γ χ ( k ) .$
$γ χ ( 2 ρ ) = 2 2 ρ ρ ! Γ ρ + χ + 1 2 Γ χ + 1 2 ,$
$γ χ ( 2 ρ + 1 ) = 2 2 ρ + 1 ρ ! Γ ρ + χ + 3 2 Γ χ + 1 2 ,$
and for all $ρ = 0 , 1 , 2 , …$ a recursion $γ χ$ is defined as
$γ χ ( ρ + 1 ) ( ρ + 1 + 2 χ θ ρ + 1 ) = γ χ ( ρ ) ,$
where
$θ ρ = 0 if ρ = 2 r , r ∈ { 0 , 2 , 4 , 6 , ⋯ } , 1 if ρ = 2 r + 1 , r ∈ { 1 , 3 , 5 , 7 , ⋯ } .$
Most recently, a new version of the Szász operators studied by Nasiruzzaman and Aljohani [6], and these types of Szász operators, provide the generalized version of some earlier operators, such as Szász operators, Szász–Jakimovski operators and Szász–Dunkl operators, where the operators are given by, for example, all $f ∈ C [ 0 , ∞ ) , u 1 ∈ [ 0 , ∞ ) , r ∈ N , C ( 1 ) ≠ 0$ and $χ ≥ 0$,
$K r , χ ( f ; u 1 ) = 1 C ( 1 ) e χ ( r u 1 ) ∑ k = 0 ∞ P k ( r u 1 ) f k + 2 χ θ k r .$
There are many published articles related to these works, for example, those by Kajla et al. [7], Mursaleen et al. [8,9,10,11], Mohiuddine et al. [12,13,14,15,16], Nasiruzzaman et al. [6,17,18,19,20,21,22], Özger et al. [23]. For studies on Bernstein and Szász types operators involving the idea of Chlodowsky and Charlier polynomials, we refer to [24,25,26,27,28].
The organization of this manuscript is as follows: in Section 2, we present a generalization of the operators defined and studied by Nasiruzzaman and Aljohani [6] in bivariate sense using the unbounded sequences $α n$$β m$ and $ξ m$ of positive numbers, such that $lim n → ∞ α n n = 0 ,$ $lim m → ∞ 1 ξ m = 0$ and $β m ξ m = 1 + O 1 ξ m$ as $m → ∞$, and estimation of moments and central moments. In Section 3, we discuss the rate of convergence by means of the weighted modulus of continuity and weighted approximation properties. In this section, we obtain the degree of convergence using Lipschitz-type maximal functions of two variables, as well as the direct theorem. In Section 4, we close the paper and provide conclusions.

## 2. Construction of New Operators and Estimation of Moments

Here, we construct the operators and prove some auxillary lemmas, which will be used to prove the approximation results.
Let $I α n = { ( x 1 , y 1 ) : 0 ≤ x 1 ≤ α n , y 1 ∈ [ 0 , ∞ ) }$ and $C I α n = { f : I α n → R is continuous }$, such that it satisfies the norm equipped by
$| | f | | C I α n = sup ( x 1 , y 1 ) ∈ I α n | f ( x 1 , y 1 ) | .$
Then, for all $f ∈ C I α n$ and $n , m ∈ N$, and any $ν , τ ≥ 0$, we define
$B n , m ν , τ ( f ; x 1 , y 1 ) : = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n × Q l ( β m y 1 ) f α n ( k + 2 ν θ k ) n , l + 2 τ θ l ξ m ,$
where $α n , β m$ and $ξ m$ are the unbounded sequences of positive numbers, such that $lim n → ∞ α n n = 0 ,$ $lim m → ∞ 1 ξ m = 0$ and $β m ξ m = 1 + O 1 ξ m$ as $m → ∞$. Moreover, the Appell polynomial $D ( u 1 )$ is given by $D ( u 1 ) e u y = ∑ r = 0 ∞ Q l ( y 1 ) u 1 r$ with the following identities: $D ( 1 ) ≠ 0$$D ( u 1 ) = ∑ r = 0 ∞ c r u 1 r$$Q l ( y 1 ) = ∑ i = 0 l c j y 1 l − i ( l − i ) ! ( l ∈ N )$ and $P k ( x 1 )$ and $C ( 1 ) ≠ 0$.
Lemma 1.
For all $y 1 ∈ [ 0 , ∞ ) , P l ( y 1 ) ≥ 0 , τ ≥ 0$ and $D ( 1 ) ≠ 0$, if we define
$D ( α ) e τ ( α y 1 ) = ∑ l = 0 ∞ Q l ( y 1 ) α l ,$
then, for any unbounded sequence $β m$, we have
$∑ l = 0 ∞ Q l ( β m y 1 ) = D ( 1 ) e τ ( β m y 1 ) ,$
$∑ l = 0 ∞ l Q l ( β m y 1 ) = ( D ′ ( 1 ) + β m y 1 D ( 1 ) ) e τ ( β m y 1 ) ,$
$∑ l = 0 ∞ l 2 Q l ( β m y 1 ) = ( D ″ ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) + β m y 1 ( β m y 1 + 1 ) D ( 1 ) ) e τ ( β m y 1 ) ,$
$∑ l = 0 ∞ l 3 Q l ( β m y 1 ) = ( D ‴ ( 1 ) + 3 ( β m y 1 + 1 ) D ″ ( 1 ) + ( 3 β m 2 y 1 2 + 6 β m y 1 + 2 ) D ′ ( 1 ) + β m y 1 ( β m 2 y 1 2 + 3 β m y 1 + 2 ) D ( 1 ) ) e τ ( β m y 1 ) ,$
$∑ l = 0 ∞ l 4 Q l ( β m y 1 ) = ( D ⁗ ( 1 ) + ( 4 β m y 1 + 6 ) D ‴ ( 1 ) + ( 6 β m 2 y 1 2 + 18 β m y 1 + 11 ) D ″ ( 1 ) + ( 4 β m 3 y 1 3 + 18 β m 2 y 1 2 + 22 β m y 1 + 6 ) D ′ ( 1 ) + β m y 1 ( β m 3 y 1 3 + 6 β m 2 y 1 2 + 11 β m y 1 + 6 ) D ( 1 ) ) e τ ( β m y 1 ) .$
Lemma 2.
For all $x 1 ∈ [ 0 , ∞ ) , P k ( x 1 ) ≥ 0 , ν ≥ 0$ and $C ( 1 ) ≠ 0$, if we define
$C ( β ) e ν ( β x 1 ) = ∑ k = 0 ∞ P k ( x 1 ) β k ,$
then, for any unbounded sequence of positive numbers $α n$, we have
$∑ k = 0 ∞ P k n x 1 α n = C ( 1 ) e ν n x 1 α n ,$
$∑ k = 0 ∞ k P k n x 1 α n = ( C ′ ( 1 ) + n x 1 α n C ( 1 ) ) e ν n x 1 α n ,$
$∑ k = 0 ∞ k 2 P k n x 1 α n = ( C ″ ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) C ( 1 ) ) e ν n x 1 α n ,$
$∑ k = 0 ∞ k 3 P k n x 1 α n = ( C ‴ ( 1 ) + 3 ( n x 1 α n + 1 ) C ″ ( 1 ) + ( 3 n x 1 α n 2 + 6 n x 1 α n + 2 ) C ′ ( 1 ) + n x 1 α n ( n x 1 α n 2 + 3 n x 1 α n + 2 ) C ( 1 ) ) e ν n x 1 α n ,$
$∑ k = 0 ∞ k 4 P k n x 1 α n = ( C ⁗ ( 1 ) + ( 4 n x 1 α n + 6 ) C ‴ ( 1 ) + ( 6 n x 1 α n 2 + 18 n x 1 α n + 11 ) C ″ ( 1 ) + ( 4 n x 1 α n 3 + 18 n x 1 α n 2 + 22 n x 1 α n + 6 ) C ′ ( 1 ) + n x 1 α n ( n x 1 α n 3 + 6 n x 1 α n 2 + 11 n x 1 α n + 6 ) C ( 1 ) ) e ν n x 1 α n .$
Lemma 3.
For the operators $B n , m ν , τ$, defined by (9), as demonstrated here:
$R n ∗ ( f ; x 1 , y 1 ) = 1 e ν n x 1 α n C ( 1 ) ∑ k = 0 ∞ P k n x 1 α n f α n ( k + 2 ν θ k ) n , y 1$
$S m ( f ; x 1 , y 1 ) = 1 e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) f x 1 , l + 2 τ θ l ξ m ,$
then
$B n , m ν , τ ( f ; x 1 , y 1 ) = R n ∗ S m ( f ; x 1 , y 1 ) = S m R n ∗ ( f ; x 1 , y 1 ) .$
Proof.
We can easily see that
$R n ∗ S m ( f ; x 1 , y 1 ) = R n ∗ 1 e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) f x 1 , l + 2 τ θ l ξ m = 1 e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ R n ∗ f x 1 , l + 2 τ θ l ξ m Q l ( β m y 1 ) = 1 e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) × ∑ k = 0 ∞ 1 e ν n x 1 α n f α n ( k + 2 ν θ k ) n , l + 2 τ θ l ξ m P k n x 1 α n = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n Q l ( β m y 1 ) × f α n ( k + 2 ν θ k ) n , l + 2 τ θ l ξ m = B n , m ν , τ ( f ; x 1 , y 1 ) .$
Similarly, we prove $S m R n ∗ ( f ; x 1 , y 1 ) = B n , m ν , τ ( f ; x 1 , y 1 )$. □
Let $R + 2 = R + × R +$ with $R + = [ 0 , ∞ )$, and suppose $γ m$ is the sequence, such that $lim m → ∞ γ m = ∞$ and $I α n γ m = { ( x 1 , y 1 ) : 0 ≤ x 1 ≤ α n , 0 ≤ y 1 ≤ γ m }$. We also consider the operators $T n , m ∗ ( f ; x 1 , y 1 )$, such that
$T n , m ∗ ( f ; x 1 , y 1 ) = B n , m ν , τ ( f ; x 1 , y 1 ) , for ( x 1 , y 1 ) ∈ I α n γ m f ( x 1 , y 1 ) , for ( x 1 , y 1 ) ∈ R + 2 \ I α n γ m .$
Lemma 4.
Let $n , m ∈ N$ and $ϕ j , i = u 1 j v 1 i$ for any $j , i = 0 , 1 , 2 , 3 , 4 ,$ then, we have the following identities:
$( 1 ) B n , m ν , τ ( ϕ 0 , 0 ; x 1 , y 1 ) = 1 ; ( 2 ) B n , m ν , τ ( ϕ 1 , 0 ; x 1 , y 1 ) = x 1 + α n n ( C ′ ( 1 ) C ( 1 ) + 2 ν ) ; ( 3 ) B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) = 1 ξ m ( D ′ ( 1 ) D ( 1 ) + β m y 1 ) + 2 τ ξ m ; ( 4 ) B n , m ν , τ ( ϕ 2 , 0 ; x 1 , y 1 ) = x 1 2 + α n n 2 C ′ ( 1 ) C ( 1 ) + 2 ν + 1 x 1 + α n n 2 C ″ ( 1 ) C ( 1 ) + C ′ ( 1 ) C ( 1 ) + 4 ν 2 ; ( 5 ) B n , m ν , τ ( ϕ 0 , 2 ; x 1 , y 1 ) = 1 ξ m 2 [ D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) ] + 2 τ ξ m 2 [ D ′ ( 1 ) D ( 1 ) + β m y 1 ] + 4 τ 2 ξ m 2 ; ( 6 ) B n , m ν , τ ( ϕ 3 , 0 ; x 1 , y 1 ) = α n 3 n 3 [ C ‴ ( 1 ) C ( 1 ) + 3 ( n x 1 α n + 1 ) C ″ ( 1 ) C ( 1 ) + ( 3 n x 1 α n 2 + 6 n x 1 α n + 2 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 2 + 3 n x 1 α n + 2 ) ] + 6 ν α n 3 n 3 [ C ″ ( 1 ) C ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) ] + 12 ν 2 α n 3 n 3 [ C ′ ( 1 ) C ( 1 ) + n x 1 α n ] + 8 ν 3 α n 3 n 3 ; ( 7 ) B n , m ν , τ ( ϕ 0 , 3 ; x 1 , y 1 ) = 1 ξ m 3 [ D ‴ ( 1 ) D ( 1 ) + 3 ( β m y 1 + 1 ) D ″ ( 1 ) D ( 1 ) + ( 3 β m 2 y 1 2 + 6 β m y 1 + 2 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m 2 y 1 2 + 3 β m y 1 + 2 ) ] + 6 τ ξ m 3 [ D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) ] + 6 τ 2 ξ m 3 [ D ′ ( 1 ) D ( 1 ) + β m y 1 ] + 8 τ 3 ξ m 3 ; ( 8 ) B n , m ν , τ ( ϕ 4 , 0 ; x 1 , y 1 ) = α n 4 n 4 [ ( C ⁗ ( 1 ) C ( 1 ) + ( 4 n x 1 α n + 6 ) C ‴ ( 1 ) C ( 1 ) + ( 6 n x 1 α n 2 + 18 n x 1 α n + 11 ) C ″ ( 1 ) C ( 1 ) + ( 4 n x 1 α n 3 + 18 n x 1 α n 2 + 22 n x 1 α n + 6 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 3 + 6 n x 1 α n 2 + 11 n x 1 α n + 6 ) ] + 8 ν α n 4 n 4 [ ( C ‴ ( 1 ) C ( 1 ) + 3 ( n x 1 α n + 1 ) C ″ ( 1 ) C ( 1 ) + ( 3 n x 1 α n 2 + 6 n x 1 α n + 2 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 2 + 3 n x 1 α n + 2 ) ] + 24 ν 2 α n 4 n 4 [ ( C ″ ( 1 ) C ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) ] + 32 ν 3 α n 4 n 4 [ ( C ′ ( 1 ) C ( 1 ) + n x 1 α n ) ] + 16 ν 4 α n 4 n 4 ;$
$( 9 ) B n , m ν , τ ( ϕ 0 , 4 ; x 1 , y 1 ) = 1 ξ m 4 [ D ⁗ ( 1 ) D ( 1 ) + ( 4 β m y 1 + 6 ) D ‴ ( 1 ) D ( 1 ) + ( 6 β m 2 y 1 2 + 18 β m y 1 + 11 ) D ″ ( 1 ) D ( 1 ) + ( 4 β m 3 y 1 3 + 18 β m 2 y 1 2 + 22 β m y 1 + 6 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m 3 y 1 3 + 6 β m 2 y 1 2 + 11 β m y 1 + 6 ) ] + 8 τ ξ m 4 [ D ‴ ( 1 ) D ( 1 ) + 3 ( β m y 1 + 1 ) D ″ ( 1 ) D ( 1 ) + ( 3 β m 2 y 1 2 + 6 β m y 1 + 2 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m 2 y 1 2 + 3 β m y 1 + 2 ) ] + 24 τ 2 ξ m 4 [ D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) ] + 32 τ 3 ξ m 4 [ D ′ ( 1 ) D ( 1 ) + β m y 1 ] + 16 τ 4 ξ m 3 .$
Proof.
Taking $j = i = 0$ and using the operators (9), we have
$B n , m ν , τ ( ϕ 0 , 0 ; x 1 , y 1 ) = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n Q l ( β m y 1 ) = 1 .$
Taking $j = 1$ and $i = 0$, and using the operators (9), we have
$B n , m ν , τ ( ϕ 1 , 0 ; x 1 , y 1 ) = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n Q l ( β m y 1 ) α n ( k + 2 ν θ k ) n = ( α n n e ν n x 1 α n C ( 1 ) ∑ k = 0 ∞ k P k n x 1 α n ) ( 1 e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) ) + ( 1 e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) ) × ( 2 ν α n n e ν n x 1 α n C ( 1 ) ∑ k = 2 s + 1 ∞ θ k P k n x 1 α n ) , s ∈ N ∪ { 0 } = α n n e ν n x 1 α n C ( 1 ) ( C ′ ( 1 ) + n x 1 α n C ( 1 ) ) e ν n x 1 α n + 2 ν α n n = x 1 + α n n ( C ′ ( 1 ) C ( 1 ) + 2 ν ) .$
Taking $j = 2$ and $i = 0$, and using the operators (9), we have
$B n , m ν , τ ( ϕ 2 , 0 ; x 1 , y 1 ) = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n Q l ( β m y 1 ) α n 2 ( k + 2 ν θ k ) 2 n 2 = ( 1 e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) ) [ α n 2 n 2 e ν n x 1 α n C ( 1 ) ∑ k = 0 ∞ k 2 P k n x 1 α n + 2 ν α n 2 n 2 e ν n x 1 α n C ( 1 ) ∑ k = 2 s + 1 ∞ k P k n x 1 α n θ k + 4 ν 2 α n 2 n 2 e ν n x 1 α n C ( 1 ) ∑ k = 2 s + 1 ∞ P k n x 1 α n θ k 2 ] , s ∈ N ∪ { 0 } = α n 2 n 2 e ν n x 1 α n C ( 1 ) [ C ″ ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) C ( 1 ) ] e ν n x 1 α n + 2 ν α n 2 n 2 e ν n x 1 α n C ( 1 ) [ C ′ ( 1 ) + n x 1 α n C ( 1 ) ] e ν n x 1 α n + 4 ν 2 α n 2 n 2 e ν n x 1 α n C ( 1 ) C ( 1 ) e ν n x 1 α n = 1 n 2 [ α n 2 C ′ ′ ( 1 ) C ( 1 ) + ( 2 n α n x 1 + α n 2 ) C ′ ( 1 ) C ( 1 ) + n 2 x 1 2 + n α n x 1 ] + 2 ν α n n 2 [ α n C ′ ( 1 ) C ( 1 ) + n x 1 ] + 4 ν 2 α n 2 n 2 .$
Taking $j = 3$ and $i = 0$, and using the operators (9), we have
$B n , m ν , τ ( ϕ 3 , 0 ; x 1 , y 1 ) = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n Q l ( β m y 1 ) α n 3 ( k + 2 ν θ k ) 3 n 3 = ( 1 e τ ( β m y 1 ) D ( 1 ) α n 3 n 3 e ν n x 1 α n C ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) ) × [ ∑ k = 0 ∞ k 3 P k n x 1 α n + 6 ν ∑ k = 2 s + 1 ∞ k 2 P k n x 1 α n θ k + 12 ν 2 ∑ k = 2 s + 1 ∞ k P k n x 1 α n θ k 2 + 8 ν 3 ∑ k = 2 s + 1 ∞ P k n x 1 α n θ k 3 ] , s ∈ N ∪ { 0 } = α n 3 n 3 [ C ‴ ( 1 ) C ( 1 ) + 3 ( n x 1 α n + 1 ) C ″ ( 1 ) C ( 1 ) + ( 3 n x 1 α n 2 + 6 n x 1 α n + 2 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 2 + 3 n x 1 α n + 2 ) ] + 6 ν α n 3 n 3 [ C ″ ( 1 ) C ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) ] + 12 ν 2 α n 3 n 3 [ C ′ ( 1 ) C ( 1 ) + n x 1 α n ] + 8 ν 3 α n 3 n 3 .$
Taking $j = 4$ and $i = 0$, and using the operators (9), we have
$B n , m ν , τ ( ϕ 4 , 0 ; x 1 , y 1 ) = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n Q l ( β m y 1 ) α n 4 ( k + 2 ν θ k ) 4 n 4 = ( 1 e τ ( β m y 1 ) D ( 1 ) α n 4 n 4 e ν n x 1 α n C ( 1 ) ∑ l = 0 ∞ Q l ( β m y 1 ) ) [ ∑ k = 0 ∞ k 4 P k n x 1 α n + 8 ν ∑ k = 2 s + 1 ∞ k 3 P k n x 1 α n θ k , s ∈ N ∪ { 0 } + 24 ν 2 ∑ k = 2 s + 1 ∞ k 2 P k n x 1 α n θ k 2 , s ∈ N ∪ { 0 } + 32 ν 3 ∑ k = 2 s + 1 ∞ k P k n x 1 α n θ k 2 , s ∈ N ∪ { 0 } + 16 ν 4 ∑ k = 2 s + 1 ∞ P k n x 1 α n θ k 3 , s ∈ N ∪ { 0 } ] = α n 4 n 4 [ C ⁗ ( 1 ) C ( 1 ) + ( 4 n x 1 α n + 6 ) C ‴ ( 1 ) C ( 1 ) + ( 6 n x 1 α n 2 + 18 n x 1 α n + 11 ) C ″ ( 1 ) C ( 1 ) + ( 4 n x 1 α n 3 + 18 n x 1 α n 2 + 22 n x 1 α n + 6 ) C ′ ( 1 ) C ( 1 ) + ( n x 1 α n 4 + 6 n x 1 α n 3 + 11 n x 1 α n 2 + 6 n x 1 α n ) + 8 ν α n 4 n 4 [ C ‴ ( 1 ) C ( 1 ) + 3 ( n x 1 α n + 1 ) C ″ ( 1 ) C ( 1 ) + ( 3 n x 1 α n 2 + 6 n x 1 α n + 2 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 2 + 3 n x 1 α n + 2 ) ] + 24 ν 2 α n 4 n 4 [ C ″ ( 1 ) C ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) ] + 2 ν 3 α n 4 n 4 [ C ′ ( 1 ) C ( 1 ) + n x 1 α n ] + 16 ν 4 α n 4 n 4 .$
Taking $j = 0$ and $i = 1$, and using the operators (9), we have
$B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) = 1 e ν n x 1 α n e τ ( β m y 1 ) C ( 1 ) D ( 1 ) ∑ k , l = 0 ∞ P k n x 1 α n Q l ( β m y 1 ) l + 2 τ θ l ξ m = ( 1 e ν n x 1 α n C ( 1 ) ∑ k = 0 ∞ P k n x 1 α n ) ( 1 ξ m e τ ( β m y 1 ) D ( 1 ) ∑ l = 0 ∞ l Q l ( β m y 1 ) ) + ( 1 e ν n x 1 α n C ( 1 ) ∑ k = 0 ∞ P k n x 1 α n ) × ( 2 τ ξ m e τ ( β m y 1 ) D ( 1 ) ∑ l = 2 s + 1 ∞ Q l ( β m y 1 ) ) , s ∈ N ∪ { 0 } = 1 ξ m ( D ′ ( 1 ) D ( 1 ) + β m y 1 ) + 2 τ ξ m .$
In a similar way, other identities can be easily proved. □
With the help of the above lemma, we can calculate the central moments as follows:
Lemma 5.
Let $C ( 1 ) ≠ 0$ and $D ( 1 ) ≠ 0$. For all $n , m ∈ N$, the operators $B n , m ν , τ ( . ; . )$ have the following central moments:
$( 1 ) B n , m ν , τ ( ( u 1 − x 1 ) ; x 1 , y 1 ) = α n n ( C ′ ( 1 ) C ( 1 ) + 2 ν ) ; ( 2 ) B n , m ν , τ ( ( v 1 − y 1 ) ; x 1 , y 1 ) = ( β m ξ m − 1 ) y 1 + 1 ξ m ( D ′ ( 1 ) D ( 1 ) + 2 τ ) ; ( 3 ) B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) = α n n 1 − 2 ν x 1 + α n n 2 C ″ ( 1 ) C ( 1 ) + C ′ ( 1 ) C ( 1 ) + 4 ν 2 ; ( 4 ) B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) = [ β m 2 ξ m 2 − 2 β m ξ m + 1 ] y 1 2 + [ β m ξ m 2 + 2 β m ξ m 2 − 2 ξ m D ′ ( 1 ) D ( 1 ) + 2 τ β m ξ m 2 − 2 τ ξ m ] y 1 + 1 ξ m 2 [ D ″ ( 1 ) D ( 1 ) + ( 1 + 2 τ ) D ′ ( 1 ) D ( 1 ) + 4 τ 2 ] ; ( 5 ) B n , m ν , τ ( ( u 1 − x 1 ) 4 ; x 1 , y 1 ) = α n 4 n 4 [ ( C ′ v 1 ( 1 ) C ( 1 ) + ( 4 n x 1 α n + 6 ) C ‴ ( 1 ) C ( 1 ) + ( 6 n x 1 α n 2 + 18 n x 1 α n + 11 ) C ″ ( 1 ) C ( 1 ) + ( 4 n x 1 α n 3 + 18 n x 1 α n 2 + 22 n x 1 α n + 6 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 3 + 6 n x 1 α n 2 + 11 n x 1 α n + 6 ) ] + 8 ν α n 4 n 4 [ ( C ‴ ( 1 ) C ( 1 ) + 3 ( n x 1 α n + 1 ) C ″ ( 1 ) C ( 1 ) + ( 3 n x 1 α n 2 + 6 n x 1 α n + 2 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 2 + 3 n x 1 α n + 2 ) ] + 24 ν 2 α n 4 n 4 [ ( C ″ ( 1 ) C ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) ] + 32 ν 3 α n 4 n 4 [ ( C ′ ( 1 ) C ( 1 ) + n x 1 α n ) ] + 16 ν 4 α n 4 n 4 − 4 α n 3 x 1 n 3 [ C ‴ ( 1 ) C ( 1 ) + 3 ( n x 1 α n + 1 ) C ″ ( 1 ) C ( 1 ) + ( 3 n x 1 α n 2 + 6 n x 1 α n + 2 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n 2 + 3 n x 1 α n + 2 ) ] − 24 ν α n 3 x 1 n 3 [ C ″ ( 1 ) C ( 1 ) + ( 2 n x 1 α n + 1 ) C ′ ( 1 ) C ( 1 ) + n x 1 α n ( n x 1 α n + 1 ) ] + 12 ν 2 α n 3 n 3 [ C ′ ( 1 ) C ( 1 ) + n x 1 α n ] − 32 ν 3 α n 3 x 1 n 3 + 6 x 1 4 + 6 α n n 2 C ′ ( 1 ) C ( 1 ) + 2 ν + 1 x 1 3 + 6 α n n 2 C ″ ( 1 ) C ( 1 ) + C ′ ( 1 ) C ( 1 ) + 4 ν 2 x 1 2 − 4 x 1 4 − 4 α n n ( C ′ ( 1 ) C ( 1 ) + 2 ν ) x 1 3 + x 1 4 ; ( 6 ) B n , m ν , τ ( ( v 1 − y 1 ) 4 ; x 1 , y 1 ) = 1 ξ m 4 [ D ′ v 1 ( 1 ) D ( 1 ) + ( 4 β m y 1 + 6 ) D ‴ ( 1 ) D ( 1 ) + ( 6 β m 2 y 1 2 + 18 β m y 1 + 11 ) D ″ ( 1 ) D ( 1 ) + ( 4 β m 3 y 1 3 + 18 β m 2 y 1 2 + 22 β m y 1 + 6 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m 3 y 1 3 + 6 β m 2 y 1 2 + 11 β m y 1 + 6 ) ] + 8 τ ξ m 4 [ D ‴ ( 1 ) D ( 1 ) + 3 ( β m y 1 + 1 ) D ″ ( 1 ) D ( 1 ) + ( 3 β m 2 y 1 2 + 6 β m y 1 + 2 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m 2 y 1 2 + 3 β m y 1 + 2 ) ] + 24 τ 2 ξ m 4 [ D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) ] + 32 τ 3 ξ m 4 [ D ′ ( 1 ) D ( 1 ) + β m y 1 ] + 16 τ 4 ξ m 3 − 4 y 1 ξ m 3 [ D ‴ ( 1 ) D ( 1 ) + 3 ( β m y 1 + 1 ) D ″ ( 1 ) D ( 1 ) + ( 3 β m 2 y 1 2 + 6 β m y 1 + 2 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m 2 y 1 2 + 3 β m y 1 + 2 ) ] − 24 τ y 1 ξ m 3 [ D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) ] − 24 τ 2 y 1 ξ m 3 [ D ′ ( 1 ) D ( 1 ) + β m y 1 ] − 32 τ 3 y 1 ξ m 3 + 6 y 1 2 ξ m 2 [ D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) ] + 12 τ y 1 2 ξ m 2 [ D ′ ( 1 ) D ( 1 ) + β m y 1 ] + 24 τ 2 y 1 2 ξ m 2 − 4 y 1 3 ξ m ( D ′ ( 1 ) D ( 1 ) + β m y 1 ) − 8 τ y 1 3 ξ m + y 1 4 .$
Lemma 6.
Let $x 1 , y 1 ∈ I α n$; then, for sufficiently large $n , m ∈ N$, we can obtain the following inequalities:
$( 1 ) B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) ≤ O 1 n ( x 1 + 1 ) 2 ≤ M 1 ( x 1 + 1 ) 2 a s m , n → ∞ ; ( 2 ) B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) ≤ O 1 ξ m ( y 1 + 1 ) 2 ≤ C 1 ( y 1 + 1 ) 2 a s m , n → ∞ ; ( 3 ) B n , m ν , τ ( ( u 1 − x 1 ) 4 ; x 1 , y 1 ) ≤ O 1 n 2 ( x 1 + 1 ) 4 ≤ M 2 ( x 1 + 1 ) 4 a s m , n → ∞ ; ( 4 ) B n , m ν , τ ( ( v 1 − y 1 ) 4 ; x 1 , y 1 ) ≤ O 1 ξ m 2 ( y 1 + 1 ) 4 ≤ C 2 ( y 1 + 1 ) 4 a s m , n → ∞$
Remark 1.
Let $R n ∗$ and $S m$ be defined by (18) and (19); then, operators $B n , m ν , τ$ satisfy $B n , m ν , τ ( ϕ 0 , 0 ; x 1 , y 1 ) = R n ∗ ( ϕ 0 , 0 ; x 1 , y 1 ) = S m ( ϕ 0 , 0 ; x 1 , y 1 )$ and, for any $i , j = 1 , 2 , 3 , 4$, it follows that
$( 1 ) B n , m ν , τ ( ϕ i , 0 ; x 1 , y 1 ) = R n ∗ ( ϕ i , 0 ; x 1 , y 1 ) ; ( 2 ) B n , m ν , τ ( ϕ 0 , j ; x 1 , y 1 ) = S m ( ϕ 0 , j ; x 1 , y 1 ) .$

## 3. Weighted Approximation and Degree of Convergence

In this section, we discuss the rate of convergence using a weighted modulus of continuity, weighted approximation results and degree of convergence for our operators (9). Let us recall the following:
Let $Θ ( x 1 , y 1 ) = 1 + x 1 2 + y 1 2$ be a weight function defined by $B Θ R + 2 = f : ∣ f ( x 1 , y 1 ) ∣ ≤ M f Θ ( x 1 , y 1 ) for M f > 0$. Suppose the set of r-times continuously differentiable functions on $R + 2 = ( x 1 , y 1 ) ∈ R 2 : x 1 , y 1 in [ 0 , ∞ )$ is denoted by $C ( r ) R + 2$. We can also assume the following classes of functions:
$C Θ R + 2 = f : f ∈ B Θ ∩ C Θ R + 2 ;$
$C Θ k R + 2 = f : f ∈ C Θ R + 2 such that lim x 1 , y 1 → ∞ f ( x 1 , y 1 ) Θ ( x 1 , y 1 ) = k f < ∞ ;$
$C Θ 0 R + 2 = f : f ∈ C Θ k R + 2 such that lim x 1 , y 1 → ∞ f ( x 1 , y 1 ) Θ ( x 1 , y 1 ) = 0 .$
The norm on $B Θ$ is defined as $‖ f ‖ Θ = sup x 1 , y 1 ∈ R + 2 ∣ f ( x 1 , y 1 ) ∣ Θ ( x 1 , y 1 ) .$
For all $f ∈ C Θ 0 R + 2$ and $δ 1 , δ 2 > 0 ,$ the weighted modulus of continuity [29] is given as
$ω Θ ( f ; δ 1 , δ 2 ) = sup x 1 , y 1 ∈ [ 0 , ∞ ) sup 0 ≤ ∣ h 1 ∣ ≤ δ 1 , 0 ≤ ∣ h 2 ∣ ≤ δ 2 ∣ f ( x 1 + h 1 , y 1 + h 2 ) − h ( x 1 , y 1 ) ∣ Θ ( x 1 , y 1 ) Θ ( h 1 , h 2 )$
and, for any $r 1 , r 2 > 0$, the inequality
$ω Θ ( f ; r 1 δ 1 , r 2 δ 2 ) ≤ 4 ( 1 + r 1 ) ( 1 + r 2 ) ( 1 + δ 1 2 ) ( 1 + δ 2 2 ) ω Θ ( f ; δ 1 , δ 2 )$
holds. It also follows that
$∣ f ( u 1 , v 1 ) − f ( x 1 , y 1 ) ∣ ≤ Θ ( x 1 , y 1 ) Θ ∣ u 1 − x 1 ∣ , ∣ v 1 − y 1 ∣ ω Θ f ; ∣ u 1 − x 1 ∣ , ∣ v 1 − y 1 ∣ ≤ ( 1 + x 1 2 + y 1 2 ) ( 1 + ( u 1 − x 1 ) 2 ) ( 1 + ( v 1 − y 1 ) 2 ) × ω Θ f ; ∣ u 1 − x 1 ∣ , ∣ v 1 − y 1 ∣ .$
Theorem 1.
Let $n , m ∈ N$; then, for all $f ∈ C Θ 0 R + 2$, it follows that
$∣ B n , m ν , τ ( f ; x 1 , y 1 ) − f ( x 1 , y 1 ) ∣ ( 1 + x 1 2 + y 1 2 ) 3 ≤ M O 1 n O 1 m ω Θ f ; O 1 n , O 1 m ,$
where $δ n = B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) = O ( n − 1 / 2 )$ and $δ m = B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) = O ( m − 1 / 2 )$.
Proof.
In the view of above inequalities for all $δ n , δ m > 0$, we see that
$∣ f ( u 1 , v 1 ) − f ( x 1 , y 1 ) ∣ ≤ 4 ( 1 + x 1 2 + y 1 2 ) 1 + ( u 1 − x 1 ) 2 1 + ( v 1 − y 1 ) 2 × 1 + ∣ u 1 − x 1 ∣ δ n 1 + ∣ v 1 − y 1 ∣ δ m ( 1 + δ n 2 ) ( 1 + δ m 2 ) × ω Θ f ; δ n , δ m = 4 ( 1 + x 1 2 + y 1 2 ) ( 1 + δ n 2 ) ( 1 + δ m 2 ) × 1 + ∣ u 1 − x 1 ∣ δ n + ( u 1 − x 1 ) 2 + ( u 1 − x 1 ) 2 ∣ u 1 − x 1 ∣ δ n × 1 + ∣ v 1 − y 1 ∣ δ m + ( v 1 − y 1 ) 2 + ( v 1 − y 1 ) 2 ∣ v 1 − y 1 ∣ δ m × ω Θ f ; δ n , δ m .$
Applying the operators $B n , m ν , τ$ in the light of linearity as
$∣ B n , m ν , τ ( f ; x 1 , y 1 ) − f ( x 1 , y 1 ) ∣ ≤ B n , m ν , τ ∣ f ( u 1 , v 1 ) − f ( x 1 , y 1 ) ∣ ; x 1 , y 1 4 ( 1 + x 1 2 + y 1 2 ) × B n , m ν , τ ( 1 + ∣ u 1 − x 1 ∣ δ n + ( u 1 − x 1 ) 2 + ∣ u 1 − x 1 ∣ δ n ( u 1 − x 1 ) 2 ; x 1 , y 1 ) × B n , m ν , τ ( 1 + ∣ v 1 − y 1 ∣ δ m + ( v 1 − y 1 ) 2 + ( v 1 − y 1 ) 2 ∣ v 1 − y 1 ∣ δ m ; x 1 , y 1 ) ( 1 + δ n 2 ) ( 1 + δ m 2 ) ω Θ f ; δ n , δ m = 4 ( 1 + x 1 2 + y 1 2 ) ( 1 + δ n 2 ) ( 1 + δ m 2 ) ω Θ f ; δ n , δ m × ( 1 + 1 δ n B n , m ν , τ ( ∣ u 1 − x 1 ∣ ; x 1 , y 1 ) + B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) + 1 δ n B n , m ν , τ ( ∣ u 1 − x 1 ∣ ( u 1 − x 1 ) 2 ; x 1 , y 1 ) × ( 1 + 1 δ m B n , m ν , τ ( ∣ v 1 − y 1 ∣ ; x 1 , y 1 ) + B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) + 1 δ m B n , m ν , τ ( ∣ v 1 − y 1 ∣ ( v 1 − y 1 ) 2 ; x 1 , y 1 ) .$
Applying the Cauchy–Schwarz inequality, we have
$∣ B n , m ν , τ ( f ; x 1 , y 1 ) − f ( x 1 , y 1 ) ∣ ≤ 4 ( 1 + x 1 2 + y 1 2 ) ( 1 + δ n 2 ) ( 1 + δ m 2 ) ω Θ f ; δ n , δ m × [ 1 + 1 δ n B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) + B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) + 1 δ n B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) B n , m ν , τ ( ( u 1 − x 1 ) 4 ; x 1 , y 1 ) ] × [ 1 + 1 δ m B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) + B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) + 1 δ m B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) B n , m ν , τ ( ( v 1 − y 1 ) 4 ; x 1 , y 1 ) ] .$
In the view of Lemma 6, we can obtain
$∣ B n , m ν , τ ( f ; x 1 , y 1 ) − f ( x 1 , y 1 ) ∣ ≤ 4 ( 1 + x 1 2 + y 1 2 ) ( 1 + δ n 2 ) ( 1 + δ m 2 ) ω Θ f ; δ n , δ m × [ 2 + M 1 ( x 1 + 1 ) 2 + M 2 ( x 1 + 1 ) 2 ] × [ 2 + C 1 ( y 1 + 1 ) 2 + C 2 ( y 1 + 1 ) 2 ] .$
By choosing $δ n = B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 )$ and $δ m = B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 )$, and by using the inequality from Lemma 6, we can obtain the desired result. □
The following result can be obtained from Lemma 7 and Theorem 2, which are given by:
Lemma 7 ([30,31]).
Let $Θ ( x 1 , y 1 ) = 1 + x 1 2 + y 1 2$ be the weight function for all $( x 1 , y 1 ) ∈ R + × R +$, then any positive linear operator ${ J n , m } n , m ≥ 1$ acting from $C Θ → B Θ$ has the following property
$‖ J n , m ( Θ ; x 1 , y 1 ) ‖ Θ ≤ C ,$
where $C > 0$ is a real constant.
Theorem 2 ([30,31]).
For any positive linear operator ${ J n , m } n , m ≥ 1$ acting from $C Θ → B Θ$ and satisfying conditions
$( 1 ) lim n , m → ∞ ‖ J n , m ( 1 ; x 1 , y 1 ) − 1 ‖ Θ = 0 ; ( 2 ) lim n , m → ∞ ‖ J n , m ( u 1 ; x 1 , y 1 ) − x 1 ‖ Θ = 0 ; ( 3 ) lim n , m → ∞ ‖ J n , m ( v 1 ; x 1 , y 1 ) − y 1 ‖ Θ = 0 ; ( 4 ) lim n , m → ∞ ‖ J n , m ( ( u 1 2 + v 1 2 ) ; x 1 , y 1 ) − ( x 1 2 + y 1 2 ) ‖ Θ = 0 ;$
we can obtain, for each $f ∈ C Θ 0$, that
$lim n , m → ∞ ‖ J n , m ( f ) − f ‖ Θ = 0 ,$
and there is another function $g ∈ C Θ \ C Θ 0 ,$ such that
$lim n , m → ∞ ‖ J n , m ( g ) − g ‖ Θ ≥ 1 .$
Theorem 3.
For all $f ∈ C Θ 0 R + 2$, the operators ${ T n , m ∗ } n , m ≥ 1$ defined by (20) satisfy
$‖ T n , m ∗ f − f ‖ Θ = 0 .$
Proof.
Write
$‖ T n , m ∗ ( Θ ; x 1 , y 1 ) ‖ Θ = sup ( x 1 , y 1 ) ∈ R + 2 ∣ T n , m ∗ ( 1 + u 1 2 + v 1 2 ; x 1 , y 1 ) ∣ 1 + x 1 2 + y 1 2 ≤ 1 + sup ( x 1 , y 1 ) ∈ I α n γ m [ 1 1 + x 1 2 + y 1 2 | B n , m ν , τ ( u 1 2 ; x 1 , y 1 ) + B n , m ν , τ ( v 1 2 ; x 1 , y 1 ) | ] = 1 + sup ( x 1 , y 1 ) ∈ I α n γ m [ x 1 2 + α n n 2 C ′ ( 1 ) C ( 1 ) + 2 ν + 1 x 1 + α n n 2 C ″ ( 1 ) C ( 1 ) + C ′ ( 1 ) C ( 1 ) + 4 ν 2 + 1 ξ m 2 { D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) } + 2 τ ξ m 2 { D ′ ( 1 ) D ( 1 ) + β m y 1 } + 4 τ 2 ξ m 2 ] ≤ 1 + max 0 ≤ x 1 ≤ α n ξ n , m ∗ ( x 1 ) + max 0 ≤ y 1 ≤ γ m ζ n , m ∗ ( y 1 ) ,$
where
$ξ n , m ∗ ( x 1 ) = α n 2 + α n n 2 C ′ ( 1 ) C ( 1 ) + 2 ν + 1 α n + α n n 2 C ″ ( 1 ) C ( 1 ) + C ′ ( 1 ) C ( 1 ) + 4 ν 2$
and
$ζ n , m ∗ ( y 1 ) = 1 ξ m 2 { D ″ ( 1 ) D ( 1 ) + ( 2 β m y 1 + 1 ) D ′ ( 1 ) D ( 1 ) + β m y 1 ( β m y 1 + 1 ) } + 2 τ ξ m 2 { D ′ ( 1 ) D ( 1 ) + β m y 1 } + 4 τ 2 ξ m 2 .$
Therefore, if $n , m → ∞ ,$ $0 ≤ x 1 ≤ α n$ and $0 ≤ y 1 ≤ γ m$ then $lim n , m → ∞ ξ n , m ∗ ( x 1 ) = α n 2$ and $lim n , m → ∞ ζ n , m ∗ ( y 1 ) = 0$. Thus, for all $n , m ∈ N$, there is a positive number C such that $ξ n , m ∗ ( α n ) + ζ n , m ∗ ( γ m ) < C$. Finally, we arrived at
$‖ T n , m ∗ ( Θ ; x 1 , y 1 ) ‖ Θ ≤ M .$
From Lemma 7, we have $T n , m ∗ : C Θ R + 2 → B Θ R + 2$. If we can show that the conditions of Theorem 2 are satisfied; then, proof of Theorem 3 is completed. Hence, by the use of Lemma 4 we can obtain: $lim n , m → ∞ ‖ T n , m ∗ ( 1 ; x 1 , y 1 ) − 1 ‖ Θ = 0 ,$ $lim n , m → ∞ ‖ T n , m ∗ ( u 1 ; x 1 , y 1 ) − x 1 ‖ Θ = 0 ,$ and $lim n , m → ∞ ‖ T n , m ∗ ( v 1 ; x 1 , y 1 ) − y 1 ‖ Θ = 0 .$ Finally, using Lemma 4, we can obtain
$lim n , m → ∞ ‖ T n , m ∗ ( ( u 1 2 + v 1 2 ) ; x 1 , y 1 ) − ( x 1 2 + y 1 2 ) ‖ Θ ≤ sup ( x 1 , y 1 ) ∈ I α n γ m [ 1 Θ | T n , m ∗ ( u 1 2 ; x 1 , y 1 ) + T n , m ∗ ( v 1 2 ; x 1 , y 1 ) − ( x 1 2 + y 1 2 ) | ] ≤ max 0 ≤ x 1 ≤ α n B n , m ν , τ ( ϕ 2 , 0 ; x 1 , y 1 ) − x 1 2 + max 0 ≤ y 1 ≤ γ m B n , m ν , τ ( ϕ 0 , 2 ; x 1 , y 1 ) − y 1 2 ,$
which allow us
$lim n , m → ∞ max 0 ≤ x 1 ≤ α n B n , m ν , τ ( ϕ 2 , 0 ; x 1 , y 1 ) − x 1 2 + max 0 ≤ y 1 ≤ γ m B n , m ν , τ ( ϕ 0 , 2 ; x 1 , y 1 ) − y 1 2 = 0 .$
Therefore, $lim n , m → ∞ ‖ T n , m ∗ ( ( u 1 2 + v 1 2 ) ; x 1 , y 1 ) − ( x 1 2 + y 1 2 ) ‖ Θ = 0$, which completes the proof. □
Theorem 4 ([30,31]).
Let the sequence of positive linear operators ${ K n , m } n , m ≥ 1$ acting from $C Θ R + 2 → B Θ R + 2$ be defined as before, and $Θ 1 ( x 1 , y 1 ) ≥ 1$ be the continuous function, such that
$lim ∣ x 1 , y 1 ∣ → ∞ Θ ( x 1 , y 1 ) Θ 1 ( x 1 , y 1 ) = 0 .$
If $K n , m$ satisfy all conditions of Theorem 2; then, for all $f ∈ C Θ R + 2$
$‖ K n , m f − f ‖ Θ 1 = 0 .$
Theorem 5.
Let ${ T n , m ∗ } n , m ≥ 1 : C Θ R + 2 → B Θ R + 2$ and $Θ 1 ( x 1 , y 1 ) ≥ 1$ be the continuous function, such that $lim ∣ x 1 , y 1 ∣ → ∞ Θ ( x 1 , y 1 ) Θ 1 ( x 1 , y 1 ) = 0$. Then, for any $f ∈ C Θ R + 2$ we can obtain the equality
$‖ T n , m ∗ f − f ‖ Θ 1 = 0 .$
Proof.
We prove our results using Theorem 3 and Theorem 4. It is easy to obtain the operators ${ T n , m ∗ } n , m ≥ 1$ acting from $C Θ 1 R + 2 → B Θ 1 R + 2$. Using Lemma 4 we can see
$‖ T n , m ∗ ( Θ ; x 1 , y 1 ) ‖ Θ 1 ≤ 1 + sup ( x 1 , y 1 ) ∈ I α n γ m x 1 2 Θ 1 ( x 1 , y 1 ) + α n n 2 C ′ ( 1 ) C ( 1 ) + 2 ν + 1 sup ( x 1 , y 1 ) ∈ I α n γ m x 1 Θ 1 ( x 1 , y 1 ) + α n n 2 C ″ ( 1 ) C ( 1 ) + C ′ ( 1 ) C ( 1 ) + 4 ν 2 sup ( x 1 , y 1 ) ∈ I α n γ m 1 Θ 1 ( x 1 , y 1 ) + β m ξ m 2 sup ( x 1 , y 1 ) ∈ I α n γ m y 1 2 Θ 1 ( x 1 , y 1 ) + β m ξ m 2 1 + 2 D ′ ( 1 ) D ( 1 ) + 2 τ sup ( x 1 , y 1 ) ∈ I α n γ m y 1 Θ 1 ( x 1 , y 1 ) + 1 ξ m 2 1 + 2 D ″ ( 1 ) D ( 1 ) + ( 2 τ + 1 ) D ′ ( 1 ) D ( 1 ) + 4 τ 2 × sup ( x 1 , y 1 ) ∈ I α n γ m 1 Θ 1 ( x 1 , y 1 ) ≤ 1 + max 0 ≤ x 1 ≤ α n 0 ≤ y 1 ≤ γ m x 1 2 Θ 1 ( x 1 , y 1 ) + max 0 ≤ x 1 ≤ α n 0 ≤ y 1 ≤ γ m y 1 2 Θ 1 ( x 1 , y 1 ) = 1 + μ n , m + ν n , m ,$
where, clearly, $μ n , m = 1 + max 0 ≤ x 1 ≤ α n 0 ≤ y 1 ≤ γ m x 1 2 Θ 1 ( x 1 , y 1 )$ and $ν n , m = max 0 ≤ x 1 ≤ α n 0 ≤ y 1 ≤ γ m y 1 2 Θ 1 ( x 1 , y 1 )$; then, for any $n , m ∈ N$ there is a positive real number C, such that $μ n , m + ν n , m < C$. Therefore, we have
$‖ T n , m ∗ ( Θ ; x 1 , y 1 ) ‖ Θ 1 ≤ 1 + C .$
From Lemma 7, it is obvious that the operators ${ T n , m ∗ } n , m ≥ 1$ acting $C Θ 1 R + 2 → B Θ 1 R + 2$. If $n , m → ∞$, then from Theorem 1, it is easy to obtain $‖ T n , m ∗ ( ϕ 0 , 0 ; x 1 , y 1 ) − 1 ‖ Θ 1 = 0 ,$ $‖ T n , m ∗ ( ϕ 1 , 0 ; x 1 , y 1 ) − x 1 ‖ Θ 1 = 0 ,$ $‖ T n , m ∗ ( ϕ 0 , 1 ; x 1 , y 1 ) − y 1 ‖ Θ 1 = 0 ,$ and
$‖ T n , m ∗ ( ϕ 2 , 0 + ϕ 0 , 2 ; x 1 , y 1 ) − ( x 1 2 + y 1 2 ) ‖ Θ 1 = 0 .$
Hence, operators $T n , m ∗$ satisfy all the conditions of Theorem 1. Therefore, Theorem 4 implies that $‖ T n , m ∗ f − f ‖ Θ 1 = 0$. This completes our proof. □
For our operators $B n , m ν , τ$, we obtain the degree of convergence; therefore, we let $C ( I b c )$ denote the set of all continuous functions on $I b c = [ 0 , b ] × [ 0 , c ] ⊂ I α n$, which endowed the sup-norm $sup ( x 1 , y 1 ) ∈ I b c ∣ f ( x 1 , y 1 ) ∣$.
Theorem 6.
For any $φ ∈ C ( I b c )$ we can obtain the inequality
$∣ B n , m ν , τ ( φ ; x 1 , y 1 ) − φ ( x 1 , y 1 ) ∣ ≤ 2 ω 1 ( φ ; δ n , x 1 ) + ω 2 ( φ ; δ m , y 1 ) .$
Proof.
To prove the result, we use the Cauchy–Schwarz inequality and obtain
$∣ B n , m ν , τ ( φ ; x 1 , y 1 ) − φ ( x 1 , y 1 ) ∣ ≤ B n , m ν , τ ∣ φ ( u 1 , v 1 ) − φ ( x 1 , y 1 ) ∣ ; x 1 , y 1 ≤ B n , m ν , τ ∣ φ ( u 1 , v 1 ) − φ ( x 1 , s ) ∣ ; x 1 , y 1 + B n , m ν , τ ∣ φ ( x 1 , s ) − φ ( x 1 , y 1 ) ∣ ; x 1 , y 1 ≤ B n , m ν , τ ω 1 ( φ ; ∣ u 1 − x 1 ∣ ) ; x 1 , y 1 + B n , m ν , τ ω 2 ( φ ; ∣ v 1 − y 1 ∣ ) ; x 1 , y 1 ≤ ω 1 ( φ ; δ n ) 1 + 1 δ n B n , m ν , τ ( ∣ u 1 − x 1 ∣ ; x 1 , y 1 ) + ω 2 ( φ ; δ m ) 1 + 1 δ m B n , m ν , τ ( ∣ v 1 − y 1 ∣ ; x 1 , y 1 ) ≤ ω 1 ( φ ; δ n ) 1 + 1 δ n B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) + ω 2 ( f ; δ m ) 1 + 1 δ m B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) ,$
by puting $δ n 2 = B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) = δ n , x 1 2$ and $δ m 2 = B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) = δ m , y 1 2$, we get the desired result. □
To obtain our next result, we suppose that, for a positive real number $K$ and any $0 < ρ 1 , ρ 2 ≤ 1 ,$ the Lipschitz maximal function $L$ on space $J × J ⊂ R + 2$ is defined by
$L ρ 1 , ρ 2 ( J × J ) = { φ : sup ( 1 + u 1 ) ρ 1 ( 1 + v 1 ) ρ 2 φ ρ 1 , ρ 2 ( u 1 , v 1 ) − φ ρ 1 , ρ 2 ( x 1 , y 1 ) ≤ K 1 ( 1 + x 1 ) ρ 1 1 ( 1 + y 1 ) ρ 2 } ,$
$φ ρ 1 , ρ 2 ( u 1 , v 1 ) − φ ρ 1 , ρ 2 ( x 1 , y 1 ) = ∣ φ ( u 1 , v 1 ) − φ ( x 1 , y 1 ) ∣ ∣ u 1 − x 1 ∣ ρ 1 ∣ v 1 − y 1 ∣ ρ 2 ; ( u 1 , v 1 ) , ( x 1 , y 1 ) ∈ I b c ,$
where $φ$ is a continuous and bounded function defined on $R + 2$.
Theorem 7.
For any $φ ∈ L ρ 1 , ρ 2 ( J × J )$, there is a real positive number K satisfying
$∣ B n , m ν , τ ( f ; x 1 , y 1 ) − f ( x 1 , y 1 ) ∣ ≤ K { ( d ( x 1 , J ) ρ 1 + δ n , x 1 2 ρ 1 2 ) ( d ( y 1 , J ) ρ 2 + δ m , y 1 2 ρ 2 2 ) + d ( x 1 , J ) ρ 1 d ( y 1 , J ) ρ 2 } ,$
where $0 < ρ 1 , ρ 2 ≤ 1$$( x 1 , y 1 ) ∈ I b c$ and $δ n , x 1$$δ m , y 1$ are defined in Theorem 6.
Proof.
Take $( x 1 , y 1 ) ∈ I b c$; then, for any fixed $( x 0 , y 0 ) ∈ J × J$ suppose $∣ x 1 − x 0 ∣ = d ( x 1 , J )$ and $∣ y 1 − y 0 ∣ = d ( y 1 , J )$, where $d ( x 1 , J ) = inf { ∣ x 1 − y 1 ∣ : y 1 ∈ J }$; thus, we can write
$∣ φ ( u 1 , v 1 ) − φ ( x 1 , y 1 ) ∣ ≤ K ∣ φ ( u 1 , v 1 ) − φ ( x 0 , y 0 ) ∣ + ∣ φ ( x 0 , y 0 ) − φ ( x 1 , y 1 ) ∣ .$
Applying $B n , m ν , τ$ on both sides
$∣ B n , m ν , τ ( φ ; x 1 , y 1 ) − φ ( x 1 , y 1 ) ∣ ≤ B n , m ν , τ ∣ φ ( u 1 , v 1 ) − φ ( x 0 , y 0 ) ∣ + ∣ φ ( x 0 , y 0 ) − φ ( x 1 , y 1 ) ∣ ≤ K B n , m ν , τ ∣ u 1 − x 0 ∣ ρ 1 ∣ v 1 − y 0 ∣ ρ 2 ; x 1 , y 1 + K ∣ x 1 − x 0 ∣ ρ 1 ∣ y 1 − y 0 ∣ ρ 2 .$
For any $p , q ≥ 0$ and $0 ≤ ρ ≤ 1$, we know the inequality $( p + q ) ρ ≤ p ρ + q ρ ;$ thus, we obtain
$∣ u 1 − x 0 ∣ ρ 1 ≤ ∣ u 1 − x 1 ∣ ρ 1 + ∣ x 1 − x 0 ∣ ρ 1$
and
$∣ v 1 − y 0 ∣ ρ 2 ≤ ∣ v 1 − y 1 ∣ ρ 2 + ∣ y 1 − y 0 ∣ τ 2 .$
Therefore,
$∣ B n , m ν , τ ( φ ; x 1 , y 1 ) − φ ( x 1 , y 1 ) ∣ ≤ K B n , m ν , τ ∣ u 1 − x 1 ∣ ρ 1 ∣ v 1 − y 1 ∣ ρ 2 ; x 1 , y 1 + K ∣ x 1 − x 0 ∣ ρ 1 B n , m ν , τ ∣ v 1 − y 1 ∣ ρ 2 ; x 1 , y 1 + K ∣ y 1 − y 0 ∣ ρ 2 B n , m ν , τ ∣ u 1 − x 1 ∣ ρ 1 ; x 1 , y 1 + 2 K ∣ x 1 − x 0 ∣ ρ 1 ∣ y 1 − y 0 ∣ ρ 2 B n , m ν , τ ϕ 0 , 0 ; x 1 , y 1 .$
Using Hölder inequality, we obtain
$B n , m ν , τ ∣ u 1 − x 1 ∣ ρ 1 ∣ v 1 − y 1 ∣ ρ 2 ; x 1 , y 1 = R n ∗ ∣ u 1 − x 1 ∣ ρ 1 ; x 1 , y 1 S m ∣ v 1 − y 1 ∣ ρ 2 ; x 1 , y 1 ≤ B n , m ν , τ ( ∣ u 1 − x 1 ∣ 2 ; x 1 , y 1 ) ρ 1 2 B n , m ν , τ ( ϕ 0 , 0 ; x 1 , y 1 ) 2 − ρ 1 2 × B n , m ν , τ ( ∣ v 1 − y 1 ∣ 2 ; x 1 , y 1 ) ρ 2 2 B n , m ν , τ ( ϕ 0 , 0 ; x 1 , y 1 ) 2 − ρ 2 2 ,$
thus, we have
$∣ B n , m ν , τ ( φ ; x 1 , y 1 ) − φ ( x 1 , y 1 ) ∣ ≤ K δ n , x 1 2 ρ 1 2 δ m , y 1 2 ρ 2 2 + 2 K d ( x 1 , J ) ρ 1 d ( y 1 , J ) ρ 2 + K d ( x 1 , J ) ρ 1 δ m , y 1 2 ρ 2 2 + K d ( y 1 , J ) ρ 2 δ n , x 1 2 ρ 1 2 ,$
which completes the proof. □
Theorem 8.
Suppose $( x 1 , y 1 ) ∈ I b c$. Then, for any function $ψ ( x 1 , y 1 ) ∈ C ′ ( I b c )$, the operators $B n , m ν , τ$ have the inequality
$∣ B n , m ν , τ ( ψ ; x 1 , y 1 ) − ψ ( x 1 , y 1 ) ∣ ≤ ‖ ψ x 1 ′ ‖ C ( I b c ) δ n , x 1 2 1 2 + ‖ ψ y 1 ′ ‖ C ( I b c ) δ m , y 1 2 1 2 ,$
where $δ n , x 1 , δ m , y 1$ are given in Theorem 6 and $ψ x 1 ′ = ∂ ψ ( x 1 , y 1 ) ∂ x 1 , ψ y 1 ′ = ∂ ψ ( x 1 , y 1 ) ∂ y 1$.
Proof.
Take $ψ ∈ C ′ ( I b c )$ and $( x 1 , y 1 ) ∈ I b c$. Then, for any fixed $( u 1 , v 1 ) ∈ I b c$, we see that
$ψ ( u 1 , v 1 ) − ψ ( x 1 , y 1 ) = ∫ x 1 u 1 ψ υ ′ ( υ , v 1 ) d υ + ∫ y 1 v 1 ψ ς ′ ( x 1 , ς ) d ς$
which implies
$B n , m ν , τ ψ ( u 1 , v 1 ) ; x 1 , y 1 − ψ ( x 1 , y 1 ) = B n , m ν , τ ∫ x 1 u 1 ψ υ ′ ( υ , v 1 ) d t ; x 1 , y 1 + B n , m ν , τ ∫ y 1 v 1 ψ ς ′ ( x 1 , ς ) d ς ; x 1 , y 1 .$
Using the equipped sup-norm $I b c$, it is easy to obtain
$∫ x 1 u 1 ψ υ ′ ( υ , v 1 ) d υ ≤ ∫ x 1 u 1 ∣ ψ υ ′ ( υ , v 1 ) d υ ∣ ≤ ‖ ψ x 1 ′ ‖ C ( I b c ) ∣ u 1 − x 1 ∣$
and
$∫ y 1 v 1 C ς ′ ( x 1 , ς ) d ς ≤ ∫ y 1 v 1 ∣ ψ ς ′ ( x 1 , ς ) d ς ∣ ≤ ‖ ψ y 1 ′ ‖ C ( I b c ) ∣ v 1 − y 1 ∣ .$
In light of (24), (25) and (26), we obtain
$∣ B n , m ν , τ ψ ( u 1 , v 1 ) ; x 1 , y 1 − ψ ( x 1 , y 1 ) ∣ ≤ B n , m ν , τ | ∫ x 1 u 1 ψ υ ′ ( υ , v 1 ) d υ | ; x 1 , y 1 + B n , m ν , τ | ∫ y 1 v 1 ψ ς ′ ( x 1 , ς ) d ς | ; x 1 , y 1 ≤ ‖ ψ x 1 ′ ‖ C ( I b c ) B n , m ν , τ ∣ u 1 − x 1 ∣ ; x 1 , y 1 + ‖ ψ y 1 ′ ‖ C ( I b c ) B n , m ν , τ ∣ v 1 − y 1 ∣ ; x 1 , y 1 ≤ ‖ ψ x 1 ′ ‖ C ( I b c ) B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) B n , m ν , τ ( 1 ; x 1 , y 1 ) 1 2 + ‖ ψ y 1 ′ ‖ C ( I b c ) B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) B n , m ν , τ ( 1 ; x 1 , y 1 ) 1 2 = ‖ ψ x 1 ′ ‖ C ( I b c ) δ n , x 1 2 1 2 + ‖ ψ y 1 ′ ‖ C ( I b c ) δ m , y 1 2 1 2 .$
Theorem 9.
Suppose $B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 )$ is as defined by Lemma 4. If, for any $g ∈ C ( I b c )$, we define the auxiliary operators $R n , m ∗$ such that
$R n , m ∗ ( g ; x 1 , y 1 ) = B n , m ν , τ ( g ; x 1 , y 1 ) + g ( x 1 , y 1 ) − g x 1 , B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) ,$
then, for an arbitrary function, $φ ∈ C ′ ( I b c )$, we obtain the following inequality
$R n , m ∗ ( φ ; x 1 , y 1 ) − φ ( x 1 , y 1 ) ≤ [ δ n , x 1 2 + δ m , y 1 2 + { ( β m ξ m − 1 ) y 1 + 1 ξ m ( D ′ ( 1 ) D ( 1 ) + 2 τ ) } 2 ] ‖ φ ‖ C 2 ( I b c ) .$
Proof.
From Lemma 4 and (27), we have $R n , m ∗ ( 1 ; x 1 , y 1 ) = 1 ,$ $R n , m ∗ ( u 1 − x 1 ; x 1 , y 1 ) = 0$ and $R n , m ∗ ( v 1 − y 1 ; x 1 , y 1 ) = 0$. For all $φ ∈ C ′ ( I b c )$, the Taylor series expansion gives us
$φ ( u 1 , v 1 ) − φ ( x 1 , y 1 ) = ∂ φ ( x 1 , y 1 ) ∂ x 1 ( u 1 − x 1 ) + ∫ x 1 t ( u 1 − κ ) ∂ 2 φ ( κ , y 1 ) ∂ κ 2 d κ + ∂ φ ( x 1 , y 1 ) ∂ y 1 ( v 1 − y 1 ) + ∫ y 1 s ( v 1 − ϱ ) ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ ,$
by applying $R n , m ∗$ defined by (27), we can obtain
$R n , m ∗ φ ( u 1 , v 1 ) ; x 1 , y 1 − R n , m ∗ φ ( x 1 , y 1 = R n , m ∗ ∫ x 1 t ( u 1 − κ ) ∂ 2 φ ( κ , y 1 ) ∂ κ 2 d κ ; x 1 , y 1 + R n , m ∗ ∫ y 1 s ( v 1 − ϱ ) ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ ; x 1 , y 1 = B n , m ν , τ ∫ x 1 t ( u 1 − κ ) ∂ 2 φ ( κ , y 1 ) ∂ κ 2 d κ ; x 1 , y 1 − ∫ x 1 x 1 ( x 1 − κ ) ∂ 2 φ ( κ , y 1 ) ∂ κ 2 d κ + B n , m ν , τ ∫ y 1 s ( v 1 − ϱ ) ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ ; x 1 , y 1 − ∫ y 1 β m ξ m y 1 β m ξ m y 1 − ϱ ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ .$
Therefore,
$∣ R n , m ∗ φ ; x 1 , y 1 − R n , m ∗ φ ( x 1 , y 1 ∣ = B n , m ν , τ | ∫ x 1 t ( u 1 − κ ) ∂ 2 φ ( κ , y 1 ) ∂ κ 2 d κ | ; x 1 , y 1 + B n , m ν , τ | ∫ y 1 s ( v 1 − ϱ ) ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ | ; x 1 , y 1 − | ∫ y 1 B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) − ϱ ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ | .$
Therefore, we can easily conclude that
$| ∫ x 1 t ( u 1 − κ ) ∂ 2 φ ( κ , y 1 ) ∂ κ 2 d κ | ≤ ∫ x 1 t | ( u 1 − κ ) ∂ 2 φ ( κ , y 1 ) ∂ κ 2 d κ | ≤ ‖ φ ‖ C 2 ( I b c ) ( u 1 − x 1 ) 2 ,$
$| ∫ y 1 s ( v 1 − ϱ ) ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ | ≤ ∫ y 1 s | ( v 1 − ϱ ) ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ | ≤ ‖ φ ‖ C 2 ( I b c ) ( v 1 − y 1 ) 2 ,$
and
$| ∫ y 1 B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) − ϱ ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ | ≤ ∫ y 1 B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) | B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) − ϱ ∂ 2 φ ( x 1 , ϱ ) ∂ ϱ 2 d ϱ | ≤ B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) − y 1 2 ‖ φ ‖ C 2 ( I b c ) .$
Hence, using equalities (28), (29) and (30), we can obtain
$∣ R n , m ∗ φ ; x 1 , y 1 − φ ( x 1 , y 1 ) ∣ ≤ { B n , m ν , τ ( ( u 1 − x 1 ) 2 ; x 1 , y 1 ) + B n , m ν , τ ( ( v 1 − y 1 ) 2 ; x 1 , y 1 ) + B n , m ν , τ ( ϕ 0 , 1 ; x 1 , y 1 ) − y 1 2 } ‖ φ ‖ C 2 ( I b c ) ,$
this completes the results. □

## 4. Conclusions

The motivation for this research article was to introduce the bivariate Szász–Jakimovski–Leviatan operators using unbounded sequences of positive numbers. We also discussed the rate of convergence and weighted approximation theorems for these operators. With the help of bivariate Lipschitz-maximal functions, we obtained the degree of convergence, as well as the direct theorem for our operators. These results meant that more attention was paid to these researchers, providing them with a new research path in bivariate sense. Moreover, our newly constructed operators generalized some existing operators in the literature (see [2,3,6]). For further research on the above operators, one can study the approximation results using the idea of convergence given in [32,33,34], and also extend our bivariate operators for more than two variables and study their approxmation properties.

## Funding

This research received no external funding.

Not applicable.

Not applicable.

Not applicable.

## Conflicts of Interest

The author declares no conflict of interest.

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