A study on pointwise approximation by double singular integral operators

Abstract

In the present work we prove the pointwise convergence and the rate of pointwise convergence for a family of singular integral operators with radial kernel in two-dimensional setting in the following form: $L_{\lambda} ( f;x,y ) =\iint_{D}f ( t,s ) H_{\lambda} ( t-x,s-y ) \,dt\,ds$, $( x,y ) \in D$, where $D= \langle a,b \rangle\times \langle c,d \rangle$ is an arbitrary closed, semi-closed or open region in $\mathbb{R}^{2}$ and $\lambda\in\Lambda$, Λ is a set of non-negative numbers with accumulation point $\lambda_{0}$. Also we provide an example to justify the theoretical results.

1 Introduction

Taberski [1] analyzed both the pointwise convergence of functions in $L_{1} ( -\pi,\pi )$, where $L_{1} ( -\pi,\pi )$ is the collection of all measurable functions f for which $\vert f\vert$ is integrable on $( -\pi,\pi )$ and the approximation properties of their derivatives by a two parameter family of convolution type singular integral operators $U_{\lambda} ( f;x )$ of the form

$$ U_{\lambda} ( f;x ) = \int^{\pi}_{-\pi}f( t ) K_{\lambda} ( t-x )\,dt, \quad x\in ( -\pi ,\pi ) . $$

(1.1)

Here, $K_{\lambda} ( t )$ denotes a kernel fulfilling appropriate conditions with $\lambda\in\Lambda$, where Λ is a given set of non-negative numbers with accumulation point $\lambda_{0}$. Following this work, Gadjiev [2] proved the pointwise convergence of operators of type (1.1) at a generalized Lebesgue point and established the pertinent convergence order. Rydzewska [3] extended these results to approximation at a μ-generalized Lebesgue point. Karsli and Ibikli [4, 5] proceeded to the study of the more general integral operators defined by

$$ T_{\lambda} ( f;x ) =\int^{b}_{a}f ( t ) K_{\lambda} ( t-x )\,dt,\quad x\in \langle a,b \rangle,\lambda\in\Lambda\in \mathbb{R}, $$

(1.2)

with functions in $L_{1} \langle a,b \rangle$ where $\langle a,b \rangle$ is an arbitrary interval in ℝ such as $[ a,b ]$, $( a,b )$, $[ a,b )$ or $( a,b ]$.

The convergence of the other operators have been studied at characteristic points such as a generalized Lebesgue point, m-Lebesgue point, and so on, by other workers: a family of nonlinear singular integral operators [6, 7], a family of nonlinear m-singular integral operators [8], Fejer-Type singular integrals [9], moment type operators [10], a family of nonlinear Mellin type convolution operators [11], nonlinear integral operators with homogeneous kernels [12] and a family of Mellin type nonlinear m-singular integral operators [13].

Taberski [14] stepped up his analysis to two-dimensional singular integrals of the form

$$ T_{\lambda} ( f;x,y ) =\iint_{Q}f ( t,s ) K_{\lambda} ( t-x,s-y ) \,dt\,ds,\quad ( x,y ) \in Q, $$

(1.3)

where Q denotes a given rectangle. His findings were later used by Siudut [15, 16] rendering significant results. Yilmaz et al. [17] replaced $K_{\lambda}$ in (1.3) by a radial function $H_{\lambda}$ as follows:

$$ L_{\lambda} ( f;x,y ) =\int_{-\pi}^{\pi} \int_{-\pi}^{\pi}f ( t,s ) H_{\lambda } ( t-x,s-y ) \,dt\,ds, \quad( x,y ) \in \langle-\pi,\pi \rangle\times \langle-\pi,\pi \rangle. $$

(1.4)

The new operator approaches $f ( x_{0},y_{0} )$ as $( x,y,\lambda )$ tends to $( x_{0},y_{0},\lambda_{0} )$. In [18], the function $f\in L_{1} ( \langle-\pi,\pi \rangle\times \langle-\pi,\pi \rangle )$ became $f\in L_{p} ( D )$ where $D= \langle a,b \rangle \times \langle c,d \rangle$ is an arbitrary closed, semi-closed or open region in $\mathbb{R}^{2}$.

The current manuscript presents a continuation and further generalization of [18]. The main purpose is to investigate the pointwise convergence and the rate of convergence of the operators in the following form:

$$ L_{\lambda} ( f;x,y ) =\iint_{D}f ( t,s ) H_{\lambda} ( t-x,s-y ) \,ds\,dt,\quad ( x,y ) \in D, $$

(1.5)

where $D= \langle a,b \rangle\times \langle c,d \rangle$ is an arbitrary closed, semi-closed or open region in $\mathbb{R}^{2}$, at a μ-generalized Lebesgue point of $f\in L_{1} ( D )$ as $( x,y,\lambda ) \rightarrow ( x_{0},y_{0},\lambda_{0} )$. Here $L_{1} ( D )$ is the collection of all measurable functions f for which $\vert f\vert$ is integrable on D and the kernel function $H_{\lambda} ( s,t )$ is a radial function. As concerns the study of linear singular operators in several settings, the reader may see also e.g. [19–23].

The paper is organized as follows: In Section 2, we introduce the fundamental definitions. In Section 3, we give a theorem concerning the existence of the operator of type (1.5). In Section 4, we prove two theorems about the pointwise convergence of $L_{\lambda } ( f;x,y )$ to $f ( x_{0},y_{0} )$ whenever $( x_{0},y_{0} )$ is a μ-generalized Lebesgue point of f in bounded region and unbounded region. In Section 5, we establish the rate of convergence of operators of type (1.5) to $f ( x_{0},y_{0} )$ as $( x,y,\lambda )$ tends to $( x_{0},y_{0},\lambda_{0} )$ and the paper is ended with an example to support our results.

2 Preliminaries

In this section we introduce the main definitions used in this paper.

Definition 1

A function $H\in L_{1} ( \mathbb{R}^{2} )$ is said to be radial, if there exists a function $K:\mathbb{R} _{0}^{+}\rightarrow \mathbb{R}$ such that $H ( t,s ) =K ( \sqrt{t^{2}+s^{2}} )$ a.e. [24].

Definition 2

A point $( x_{0},y_{0} ) \in D$ is called a μ-generalized Lebesgue point of function $f\in L_{1} ( D )$ if

$$ \lim_{ ( h,k ) \rightarrow ( 0,0 ) }\frac {1}{\mu_{1}(h)\mu_{2}(k)}\int_{0}^{h} \int_{0}^{k}\bigl\vert f ( t+x_{0},s+y_{0} ) -f ( x_{0},y_{0} ) \bigr\vert \,dt\,ds=0, $$

where $\mu_{1}(t):\mathbb{R} \rightarrow \mathbb{R}$, absolutely continuous on $[ -\delta_{0},\delta_{0} ]$, increasing on $[ 0,\delta_{0} ]$ and $\mu_{1}(0)=0$ and also $\mu_{2}(s):\mathbb{R} \rightarrow \mathbb{R}$, absolutely continuous on $[ -\delta_{0},\delta_{0} ]$, increasing on $[ 0,\delta_{0} ]$ and $\mu_{2}(0)=0$. Here $0< h,k<\delta_{0}$ [25].

The following two examples are simple applications to a generalized Lebesgue point and μ-generalized Lebesgue point of some functions that belong to $L_{1}(\mathbb{R} ^{2})$.

Example 1

Let $g:\mathbb{R} ^{2}\rightarrow \mathbb{R}$ be given by

$$ g(t,s)= \left \{ \begin{array}{l@{\quad}l} 1, & \mbox{if } (t,s)=(0,0),\\ \frac{1}{\sqrt{\vert t\vert }(1+\vert t\vert )\sqrt{\vert s\vert }(1+\vert s\vert )}, & \mbox{if } (t,s)\in \mathbb{R} ^{2}\backslash(0,0). \end{array} \right . $$

Now, if $\mu_{1}(t)=t^{\frac{1}{4}}e^{t}$ and $\mu_{2}(s)=s^{\frac {1}{4}}e^{s}$, then the origin is a μ-generalized Lebesgue point of $g\in L_{1}(\mathbb{R} ^{2})$ but not a generalized Lebesgue point.

Example 2

Let $f:\mathbb{R} ^{2}\rightarrow \mathbb{R}$ be given by

$$ f(t,s)= \left \{ \begin{array}{@{}l@{\quad}l} e^{-(t+s)},& \mbox{if } (t,s)\in(0,1]\times(0,1],\\ 0, & \mbox{if }(t,s)\in \mathbb{R} ^{2} \backslash(0,1]\times(0,1].\end{array} \right . $$

If we take $\mu_{1}(t)=t^{\frac{1}{4}+1}$ and $\mu_{2}(s)=s^{\frac {1}{4}+1}$, then the origin is a μ-generalized Lebesgue point of $f\in L_{1}(\mathbb{R} ^{2})$. On the other hand, if we take $\alpha=\frac{1}{4}$ and $p=1$, then the origin is also a generalized Lebesgue point. Clearly, this example shows that generalized Lebesgue points are also μ-generalized Lebesgue points.

Definition 3

(Class A)

Let $H_{\lambda }:\mathbb{R}^{2}\times\Lambda\rightarrow\mathbb{R}$ be a radial function i.e., there exists a function $K_{\lambda}:\mathbb{R} _{0}^{+}\times\Lambda\rightarrow \mathbb{R}$ such that the following equality holds for $(t,s)\in \mathbb{R} ^{2}$ a.e.:

$$ H_{\lambda}(t,s):=K_{\lambda} \bigl( \sqrt{t^{2}+s^{2}} \bigr), $$

where Λ is a given set of non-negative numbers with accumulation point $\lambda_{0}$.

$H_{\lambda} ( t,s )$ belongs to class A, if the following conditions are satisfied:

• (a) $H_{\lambda}(t,s)=K_{\lambda} ( \sqrt {t^{2}+s^{2}} )$ is even, non-negative and integrable as a function of $(s,t)$ on $\mathbb{R} ^{2}$ for each fixed $\lambda\in\Lambda$.

• (b) For fixed $(x_{0},y_{0})\in D$, $K_{\lambda} ( \sqrt{x_{0}^{2}+y_{0}^{2}} )$ tends to infinity as λ tends to $\lambda_{0}$.

• (c) $\lim_{ ( x,y,\lambda ) \rightarrow ( x_{0},y_{0},\lambda_{0} ) }\iint _{\mathbb{R}^{2}}K_{\lambda} ( \sqrt{(t-x)^{2}+(s-y)^{2}} ) \,dt\,ds=1$.

• (d) $\lim_{\lambda\rightarrow\lambda_{0}} [ \sup_{\xi\leq\sqrt{t^{2}+s^{2}}}K_{\lambda} ( \sqrt{t^{2}+s^{2}} ) ] =0$, $\forall\xi>0$.

• (e) $\lim_{\lambda\rightarrow\lambda_{0}}\iint_{\xi\leq\sqrt{t^{2}+s^{2}}}K_{\lambda} ( \sqrt{t^{2}+s^{2}} ) \,dt\,ds=0$, $\forall\xi>0$.

• (f) $K_{\lambda} ( \sqrt{t^{2}+s^{2}} )$ is monotonically increasing with respect to t on $(-\infty ,0]$ and similarly $K_{\lambda} ( \sqrt{t^{2}+s^{2}} )$ is monotonically increasing with respect to s on $(-\infty,0]$ for any $\lambda\in\Lambda$. Analogously, $K_{\lambda} ( \sqrt{t^{2}+s^{2}} )$ is bimonotonically increasing with respect to $(t,s)$ on $[0,\infty )\times [0,\infty)$ and $(-\infty,0]\times(-\infty ,0]$ and bimonotonically decreasing with respect to $(t,s)$ on $[0,\infty)\times(-\infty,0]$ and $(-\infty,0]\times [0,\infty)$ for any $\lambda\in\Lambda$.

Throughout this paper we assume that the kernel $H_{\lambda} ( t,s )$ belongs to class A.

3 Existence of the operator

Lemma 1

If $f\in L_{1}(D)$, then the operator $L_{\lambda} ( f;x,y )$ defines a continuous transformation over $L_{1}(D)$ [26].

4 Pointwise convergence

The following theorem gives a pointwise approximation of the integral operators of type (1.5) to the function f at a μ-generalized Lebesgue point of $f\in L_{1}(D)$ where $D= \langle a,b \rangle \times \langle c,d \rangle$ is a bounded region in $\mathbb{R}^{2}$, which is closed, semi-closed or open.

Theorem 1

If $( x_{0},y_{0} )$ is a μ-generalized Lebesgue point of $f\in L_{1} ( D )$, then

$$ \lim_{ ( x,y,\lambda ) \rightarrow ( x_{0},y_{0},\lambda _{0} ) }L_{\lambda} ( f;x,y ) =f ( x_{0},y_{0} ) $$

on any set Z on which the functions

$$ \int_{x_{0}-\delta}^{x_{0}+\delta} \int _{y_{0}-\delta}^{y_{0}+\delta}K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \bigl\vert \bigl\{ \mu_{1}\bigl(\vert t-x_{0}\vert \bigr) \bigr\} _{t}^{{\prime }}\bigr\vert \bigl\vert \bigl\{ \mu_{2} \bigl(\vert s-y_{0}\vert \bigr) \bigr\} _{s}^{{\prime}} \bigr\vert \,dt\,ds $$

(4.1)

and

$$ K_{\lambda} ( 0 ) \mu_{1}\bigl(\vert x-x_{0}\vert \bigr) \quad\textit{and}\quad K_{\lambda} ( 0 ) \mu_{2}\bigl(\vert y-y_{0}\vert \bigr) $$

(4.2)

are bounded as $( x,y,\lambda )$ tends to $( x_{0},y_{0},\lambda_{0} )$.

Proof

Suppose that $( x_{0},y_{0} ) \in D$ is a μ-generalized Lebesgue point of $f\in L_{1} ( D )$. Therefore, for all given $\varepsilon>0$, there exists $\delta>0$ such that for all h, k satisfying $0< h,k\leq\delta$, the following inequality holds:

$$ \int_{x_{0}}^{x_{0}+\delta} \int _{y_{0}-\delta}^{y_{0}}\bigl\vert f ( t,s ) -f ( x_{0},y_{0} ) \bigr\vert \,dt\,ds< \varepsilon \mu_{1}(h)\mu_{2}(k). $$

(4.3)

If we follow the same strategy as used in the proof of Theorem 4.1 in [18], then we obtain

$$\begin{aligned} \bigl\vert L_{\lambda} ( f;x,y ) -f ( x_{0},y_{0} ) \bigr\vert \leq&\iint_{D}\bigl\vert f ( t,s ) -f ( x_{0},y_{0} ) \bigr\vert K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \,dt\,ds\\ &{}+\bigl\vert f ( x_{0},y_{0} ) \bigr\vert \biggl\vert \iint_{\mathbb{R}^{2}}K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \,dt\,ds-1\biggr\vert \\ &{}+\bigl\vert f ( x_{0},y_{0} ) \bigr\vert \iint_{\mathbb {R}^{2}\backslash D}K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \,dt\,ds\\ =&I_{1}+I_{2}+I_{3}. \end{aligned}$$

In view of conditions (c) and (d) of class A, $I_{2}\rightarrow0$, and $I_{3}\rightarrow0$ as $\lambda\rightarrow\lambda_{0}$, respectively,

$$\begin{aligned} I_{1} =& \biggl\{ \iint_{D\backslash B_{\delta}}+\iint _{B_{\delta}} \biggr\} \bigl\vert f ( t,s ) -f ( x_{0},y_{0} ) \bigr\vert K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \,dt\,ds\\ =&I_{11}+I_{12}, \end{aligned}$$

where $B_{\delta}:= \{ ( s,t ) : ( s-x_{0} ) ^{2}+ ( t-y_{0} ) ^{2}<\delta^{2}, ( x_{0},y_{0} ) \in D \}$.

Since $K_{\lambda} ( \sqrt{t^{2}+s^{2}} )$ is monotonically decreasing on $D\backslash B_{\delta}$, the inequality

$$ I_{11}\leq K_{\lambda} \bigl( (\sqrt{2}-1)\delta/\sqrt{2} \bigr) \bigl( \Vert f\Vert _{L_{1} ( D ) }+\bigl\vert f ( x_{0},y_{0} ) \bigr\vert \vert b-a\vert \vert d-c\vert \bigr) $$

holds. Hence by condition (d) of class A, $I_{11}\rightarrow0$ as $(x,y,\lambda ) \rightarrow (x_{0},y_{0},\lambda_{0} )$.

Now, we prove that $I_{12}$ tends to zero as $( x,y,\lambda )$ tends to $( x_{0},y_{0},\lambda_{0} )$. It is easy to see that the following inequality holds for $I_{12}$, i.e.:

$$\begin{aligned} I_{12} \leq& \biggl\{ \int_{x_{0}}^{x_{0}+\delta}\int_{y_{0}-\delta}^{y_{0}} +\int_{x_{0}-\delta}^{x_{0}} \int_{y_{0}-\delta}^{y_{0}} \biggr\} \bigl\vert f ( t,s ) -f ( x_{0},y_{0} ) \bigr\vert K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \,dt\,ds \\ &{}+ \biggl\{ \int_{x_{0}-\delta}^{x_{0}} \int _{y_{0}}^{y_{0}+\delta}+ \int_{x_{0}}^{x_{0}+\delta} \int_{y_{0}}^{y_{0}+\delta} \biggr\} \bigl\vert f ( t,s ) -f ( x_{0},y_{0} ) \bigr\vert K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \,dt\,ds \\ =&I_{121}+I_{122}+I_{123}+I_{124}. \end{aligned}$$

Let us consider the integral $I_{121}$. In view of (4.3), for every $\varepsilon>0$ there exists $\delta>0$ such that

$$ \int_{x_{0}}^{x_{0}+h} \int_{y_{0}-k}^{y_{0}} \bigl\vert f ( t,s ) -f ( x_{0},y_{0} ) \bigr\vert \,dt\,ds< \varepsilon\mu_{1}(h)\mu_{2}(k) $$

holds for all $0< h,k\leq\delta$.

Let us define a new function by

$$ F ( t,s ) := \int^{t}_{x_{0}} \int _{s}^{y_{0}} \bigl\vert f ( u,v ) -f ( x_{0},y_{0} ) \bigr\vert \,du\,dv. $$

(4.4)

For all t and s satisfying $0< t-x_{0}\leq\delta$ and $0< y_{0}-s\leq \delta$ we have

$$ \bigl\vert F ( t,s ) \bigr\vert \leq\varepsilon\mu _{1} ( t-x_{0} ) \mu_{2} ( y_{0}-s ) . $$

(4.5)

In view of (4.4) and (4.5) and applying the method of bivariate integration by parts to $I_{121}$ (see Theorem 2.2, p.100 in [14]) we have

$$\begin{aligned} I_{121} \leq&\varepsilon \int_{x_{0}}^{x_{0}+\delta}\int_{y_{0}-\delta}^{y_{0}}\mu_{1} ( t-x_{0} ) \mu_{2} ( y_{0}-s ) \bigl\vert dK_{\lambda } \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \bigr\vert \\ &{}+\varepsilon\mu_{2} ( \delta ) \int_{x_{0}}^{x_{0}+\delta} \mu_{1} ( t-x_{0} ) \bigl\vert dK_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( y_{0}-\delta -y ) ^{2}} \bigr) \bigr\vert \\ &{}+\varepsilon\mu_{1} ( \delta ) \int_{y_{0}-\delta}^{y_{0}} \mu_{2} ( y_{0}-s ) \bigl\vert dK_{\lambda } \bigl( \sqrt{ ( x_{0}+\delta-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \bigr\vert \\ &{}+\varepsilon\mu_{1} ( \delta ) \mu_{2} ( \delta ) K_{\lambda} \bigl( \sqrt{ ( x_{0}+\delta-x ) ^{2}+ ( y_{0}-\delta-y ) ^{2}} \bigr) . \end{aligned}$$

Let us define the variations:

$$\begin{aligned}& B_{1} ( u,v ) := \left \{ \begin{array}{@{}l@{\quad}l} \bigvee_{u}^{x_{0}+\delta-x} \bigvee_{y_{0}-\delta-y}^{v} ( K_{\lambda} ( \sqrt{t^{2}+s^{2}} ) ) , & x_{0}-x\leq u< x_{0}+\delta-x, \\ &y_{0}-\delta-y<v\leq y_{0}-y, \\ 0,& \mbox{otherwise}, \end{array} \right .\\& B_{2} ( u ) :=\left \{ \begin{array}{@{}l@{\quad}l} \bigvee_{u}^{x_{0}+\delta-x} ( K_{\lambda} ( \sqrt{t^{2}+ ( y_{0}-\delta-y ) ^{2}} ) ) ,& x_{0}-x\leq u<x_{0}+\delta-x ,\\ 0,& \mbox{otherwise}, \end{array} \right .\\& B_{3} ( v ) := \left \{ \begin{array}{@{}l@{\quad}l} \bigvee_{y_{0}-\delta-y}^{v} ( K_{\lambda } ( \sqrt{ ( x_{0}-x+\delta ) ^{2}+s^{2}} ) ) ,& y_{0}-\delta-y<v\leq y_{0}-y ,\\ 0,&\mbox{otherwise}. \end{array} \right . \end{aligned}$$

Taking the above variations into account and applying the method of bivariate integration by parts to the last inequality, we have

$$\begin{aligned} I_{121} \leq&-\varepsilon \int_{x_{0}-x}^{x_{0}-x+\delta} \int_{y_{0}-y-\delta}^{y_{0}-y} \bigl[ B_{1} ( t,s ) +B_{2} ( t ) +B_{3} ( s ) +K_{\lambda} \bigl( \sqrt{ ( x_{0}-x+\delta ) ^{2}+ ( y_{0}-\delta-y ) ^{2}} \bigr) \bigr] \\ &{}\times \bigl\{ \mu_{1} ( t-x_{0}+x ) \bigr\} _{t}^{\prime } \bigl\{ \mu_{2} ( y_{0}-s-y ) \bigr\} _{s}^{\prime}\,dt\,ds \\ =&\varepsilon ( i_{1}+i_{2}+i_{3}+i_{4} ) . \end{aligned}$$

Remark 1

If the function $g:\mathbb{R} ^{2}\rightarrow \mathbb{R}$ is bimonotonic on $[\alpha_{1},\alpha_{2}]\times{}[\beta _{1},\beta_{2}]\subset \mathbb{R} ^{2}$ then the equality given by

$$\begin{aligned} V\bigl(g;[\alpha_{1},\alpha_{2}]\times{}[ \beta_{1},\beta_{2}]\bigr) =&\bigvee _{\alpha_{1}}^{\alpha_{2}} \bigvee_{\beta_{1}}^{\beta_{2}} \bigl( g(t,s) \bigr) =\bigl\vert g(\alpha_{1},\beta_{1})-g( \alpha_{1},\beta _{2})-g(\alpha _{2}, \beta_{1})+g(\alpha_{2},\beta_{2})\bigr\vert \end{aligned}$$

holds [14, 27].

Splitting $i_{1}$ into two parts yields

$$\begin{aligned} i_{1} =&- \biggl\{ \int_{x_{0}-x}^{x_{0}-x+\delta}\int_{0}^{y_{0}-y}+ \int_{x_{0}-x}^{x_{0}-x+\delta} \int_{y_{0}-y-\delta}^{0} \biggr\} B_{1} ( t,s ) \bigl\{ \mu_{1} ( t-x_{0}+x ) \bigr\} _{t}^{\prime} \bigl\{ \mu_{2} ( y_{0}-s-y ) \bigr\} _{s}^{\prime}\,dt\,ds\\ =&i_{11}+i_{12}. \end{aligned}$$

Using Remark 1 and condition (f) of class A, we can write for $i_{11}$

$$\begin{aligned} i_{11} =&- \int_{x_{0}-x}^{x_{0}-x+\delta} \int _{0}^{y_{0}-y} \Biggl[ \Biggl\{ \bigvee _{t}^{x_{0}+\delta-x} \bigvee_{y_{0}-\delta-y}^{0}+\bigvee_{t}^{x_{0}+\delta-x} \bigvee _{0}^{s} \Biggr\} K_{\lambda} \bigl( \sqrt{u^{2}+ v ^{2}} \bigr) \Biggr] \\ &{}\times \bigl\{ \mu_{1} ( t-x_{0}+x ) \bigr\} _{t}^{\prime } \bigl\{ \mu_{2} ( y_{0}-s-y ) \bigr\} _{s}^{\prime}\,dt\,ds \\ =&\int_{x_{0}-x}^{x_{0}-x+\delta} \int_{0}^{y_{0}-y} \bigl( K_{\lambda} \bigl( \sqrt{t^{2}+ ( y_{0}-\delta-y ) ^{2}} \bigr) -K_{\lambda} \bigl( \sqrt {s^{2}+ ( x_{0}+\delta-x ) ^{2}} \bigr) -2K_{\lambda} \bigl( \vert t\vert \bigr) \\ &{} +K_{\lambda} \bigl( \sqrt{t^{2}+s^{2}} \bigr) +2K_{\lambda } \bigl( \vert x_{0}+\delta-x\vert \bigr) -K_{\lambda} \bigl( \sqrt{ ( y_{0}-\delta-y ) ^{2}+ ( x_{0}+\delta-x ) ^{2}} \bigr) \bigr) \\ &{}\times \bigl\{ \mu_{1} ( t-x_{0}+x ) \bigr\} _{t}^{\prime } \bigl\{ \mu_{2} ( y_{0}-s-y ) \bigr\} _{s}^{\prime}\,dt\,ds. \end{aligned}$$

Using the same method for $i_{12}$, we have

$$\begin{aligned} i_{12} =&\int_{x_{0}-x}^{x_{0}-x+\delta} \int _{y_{0}-y-\delta}^{0} \bigl( K_{\lambda} \bigl( \sqrt{t^{2}+ ( y_{0}-\delta-y ) ^{2}} \bigr) +K_{\lambda} \bigl( \sqrt{s^{2}+ ( x_{0}+\delta-x ) ^{2}} \bigr) \\ &{} -K_{\lambda} \bigl( \sqrt{t^{2}+s^{2}} \bigr) -K_{\lambda } \bigl( \sqrt{ ( y_{0}-\delta-y ) ^{2}+ ( x_{0}+\delta -x ) ^{2}} \bigr) \bigr) \\ &{}\times \bigl\{ \mu_{1} ( t-x_{0}+x ) \bigr\} _{t}^{\prime } \bigl\{ \mu_{2} ( y_{0}-s-y ) \bigr\} _{s}^{\prime}\,dt\,ds. \end{aligned}$$

Making similar calculations for $i_{2}$ and $i_{3}$ and collecting the obtained terms, we may write

$$\begin{aligned} i_{1}+i_{2}+i_{3}+i_{4} =&- \int _{x_{0}-x}^{x_{0}-x+\delta} \int_{y_{0}-y-\delta}^{0}K_{\lambda} \bigl( \sqrt{t^{2}+s^{2}} \bigr) \bigl\{ \mu_{1} ( t-x_{0}+x ) \bigr\} _{t}^{\prime} \bigl\{ \mu_{2} ( y_{0}-s-y ) \bigr\} _{s}^{\prime}\,dt\,ds \\ &{}+\int_{x_{0}-x}^{x_{0}-x+\delta} \int_{0}^{y_{0}-y} \bigl( K_{\lambda} \bigl( \sqrt{t^{2}+s^{2}} \bigr) -2K_{\lambda} \bigl( \vert t\vert \bigr) \bigr)\\ &{}\times \bigl\{ \mu _{1} ( t-x_{0}+x ) \bigr\} _{t}^{\prime} \bigl\{ \mu _{2} ( y_{0}-s-y ) \bigr\} _{s}^{\prime}\,dt\,ds. \end{aligned}$$

Hence the following inequality holds for $I_{121}$:

$$\begin{aligned} I_{121} \leq&\varepsilon \int_{x_{0}}^{x_{0}+\delta}\int_{y_{0}-\delta}^{y_{0}}K_{\lambda} \bigl( \sqrt {(t-x)^{2}+(s-y)^{2}} \bigr) \bigl\vert \bigl\{ \mu_{1} ( t-x_{0} ) \bigr\} _{t}^{\prime} \bigr\vert \bigl\vert \bigl\{ \mu _{2} ( y_{0}-s ) \bigr\} _{s}^{\prime}\bigr\vert \,dt\,ds \\ &{}+2\varepsilon K_{\lambda} ( 0 ) \mu_{1} ( \delta ) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) . \end{aligned}$$

By a similar argument to the evaluation of the integral $I_{121}$, we can easily obtain the following inequalities for $I_{122}$, $I_{123}$, and $I_{124}$:

$$\begin{aligned}& \begin{aligned}[b] I_{122} \leq{}&\varepsilon\int_{x_{0}-\delta}^{x_{0}}\int_{y_{0}-\delta}^{y_{0}}K_{\lambda} \bigl( \sqrt {(t-x)^{2}+(s-y)^{2}} \bigr) \bigl\vert \bigl\{ \mu_{1} ( x_{0}-t ) \bigr\} _{t}^{\prime} \bigr\vert \bigl\vert \bigl\{ \mu _{2} ( y_{0}-s ) \bigr\} _{s}^{\prime}\bigr\vert \,dt\,ds\\ &{}+2\varepsilon K_{\lambda} ( 0 ) \bigl( \mu_{1} ( \delta ) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) +\mu _{2} ( \delta ) \mu_{1} \bigl( \vert x_{0}-x \vert \bigr) \bigr) \\ &{}+4\varepsilon K_{\lambda} ( 0 ) \mu _{1} \bigl( \vert x_{0}-x\vert \bigr) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) , \end{aligned}\\& \begin{aligned}[b] I_{123} \leq{}&\varepsilon\int_{x_{0}-\delta}^{x_{0}}\int_{y_{0}}^{y_{0}+\delta}K_{\lambda} \bigl( \sqrt {(t-x)^{2}+(s-y)^{2}} \bigr) \bigl\vert \bigl\{ \mu_{1} ( x_{0}-t ) \bigr\} _{t}^{\prime} \bigr\vert \bigl\vert \bigl\{ \mu _{2} ( s-y_{0} ) \bigr\} _{s}^{\prime}\bigr\vert \,dt\,ds \\ &{}+2\varepsilon K_{\lambda} ( 0 ) \mu_{2} ( \delta ) \mu_{1} \bigl( \vert x_{0}-x\vert \bigr) , \end{aligned}\\& I_{124} \leq\varepsilon\int_{x_{0}}^{x_{0}+\delta}\int_{y_{0}}^{y_{0}+\delta}K_{\lambda} \bigl( \sqrt {(t-x)^{2}+(s-y)^{2}} \bigr) \bigl\vert \bigl\{ \mu_{1} ( t-x_{0} ) \bigr\} _{t}^{\prime} \bigr\vert \bigl\vert \bigl\{ \mu _{2} ( s-y_{0} ) \bigr\} _{s}^{\prime}\bigr\vert \,dt\,ds. \end{aligned}$$

Hence the following inequality is obtained for $I_{12}$ i.e.:

$$\begin{aligned} I_{12} \leq&\varepsilon \int_{x_{0}-\delta}^{x_{0}+\delta} \int_{y_{0}-\delta}^{y_{0}+\delta}K_{\lambda } \bigl( \sqrt{(t-x)^{2}+(s-y)^{2}} \bigr) \bigl\vert \bigl\{ \mu _{1} \bigl( \vert x_{0}-t\vert \bigr) \bigr\} _{t}^{\prime }\bigr\vert \bigl\vert \bigl\{ \mu_{2} \bigl( \vert y_{0}-s\vert \bigr) \bigr\} _{s}^{\prime} \bigr\vert \,dt\,ds \\ &{}+4\varepsilon K_{\lambda} ( 0 ) \bigl( \mu_{1} ( \delta ) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) +\mu _{2} ( \delta ) \mu_{1} \bigl( \vert x_{0}-x \vert \bigr) \bigr) \\ &{}+4\varepsilon K_{\lambda} ( 0 ) \mu _{1} \bigl( \vert x_{0}-x\vert \bigr) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) . \end{aligned}$$

The remaining part of the proof is obvious by the hypotheses (4.1) and (4.2). Hence $I_{12}\rightarrow0$ as $\lambda\rightarrow \lambda_{0}$. Thus the proof is completed. □

The following theorem gives a pointwise approximation of the integral operators of type (1.5) to the function f at a μ-generalized Lebesgue point of $f\in L_{1}(\mathbb{R} ^{2})$.

Theorem 2

Suppose that the hypothesis of Theorem  1 is satisfied for $D=\mathbb{R} ^{2}$. If $( x_{0},y_{0} )$ is a μ-generalized Lebesgue point of $f\in L_{1}(\mathbb{R} ^{2})$ then

$$ \lim_{ ( x,y,\lambda ) \rightarrow ( x_{0},y_{0},\lambda _{0} ) }L_{\lambda} ( f;x,y ) =f ( x_{0},y_{0} ) . $$

Proof

The proof of this theorem is quite similar to the proof of Theorem 4.2 in [18] and thus is omitted. □

5 Rate of convergence

In this section, we give a theorem concerning the rate of pointwise convergence.

Theorem 3

Suppose that the hypotheses of Theorem  1 and Theorem  2 are satisfied. Let

$$\begin{aligned} \Delta(\lambda,\delta,x,y)={}& \int_{x_{0}-\delta}^{x_{0}+\delta} \int _{y_{0}-\delta}^{y_{0}+\delta}K_{\lambda } \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \\ &{}\times\bigl\vert \bigl\{ \mu_{1}\bigl(\vert t-x_{0}\vert \bigr) \bigr\} _{t}^{{\prime}}\bigr\vert \bigl\vert \bigl\{ \mu_{2} \bigl(\vert s-y_{0}\vert \bigr) \bigr\} _{s}^{{\prime}} \bigr\vert \,dt\,ds \end{aligned}$$

for $\delta>0$ and the following assumptions be satisfied:

• (i) $\Delta(\lambda,\delta,x,y)\rightarrow0$ as $(x,y,\lambda)\rightarrow(x_{0},y_{0},\lambda_{0})$ for some $\delta>0$.

• (ii) For every $\xi>0$

  $$ K_{\lambda}(\xi)=o\bigl(\Delta(\lambda,\delta,x,y)\bigr) $$

  as $(x,y,\lambda)\rightarrow(x_{0},y_{0},\lambda_{0})$.

• (iii) For every $\xi>0$

  $$ \iint_{\xi\leq\sqrt{s^{2}+t^{2}}}K_{\lambda } \bigl( \sqrt{t^{2}+s^{2}} \bigr) \,dt\,ds=o\bigl(\Delta(\lambda, \delta,x,y)\bigr) $$

  as $(x,y,\lambda)\rightarrow(x_{0},y_{0},\lambda_{0})$.

Then at each μ-generalized Lebesgue point of $f\in L_{1}(\mathbb{R} ^{2})$ we have as $(x,y,\lambda)\rightarrow (x_{0},y_{0},\lambda _{0})$

$$ \bigl\vert L_{\lambda} ( f;x,y ) -f ( x_{0},y_{0} ) \bigr\vert =o\bigl(\Delta(\lambda,\delta,x,y)\bigr). $$

Proof

Under the hypotheses of Theorem 1 and Theorem 2 we can write

$$\begin{aligned} &\bigl\vert L_{\lambda} ( f;x,y ) -f ( x_{0},y_{0} ) \bigr\vert \\ &\quad\leq\varepsilon\int_{x_{0}-\delta}^{x_{0}+\delta}\int _{y_{0}-\delta}^{y_{0}+\delta}K_{\lambda } \bigl( \sqrt{(t-x)^{2}+(s-y)^{2}} \bigr) \bigl\vert \bigl\{ \mu_{1} \bigl( \vert x_{0}-t\vert \bigr) \bigr\} _{t}^{\prime}\bigr\vert \bigl\vert \bigl\{ \mu_{2} \bigl( \vert y_{0}-s\vert \bigr) \bigr\} _{s}^{\prime} \bigr\vert \,dt\,ds \\ &\qquad{}+4\varepsilon K_{\lambda} ( 0 ) \bigl( \mu_{1} ( \delta ) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) +\mu _{2} ( \delta ) \mu_{1} \bigl( \vert x_{0}-x \vert \bigr) \bigr)\\ &\qquad{} +4\varepsilon K_{\lambda} ( 0 ) \mu _{1} \bigl( \vert x_{0}-x\vert \bigr) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) \\ &\qquad{}+K_{\lambda} \bigl( (\sqrt{2}-1)\delta/\sqrt{2} \bigr) \Vert f\Vert _{L_{1} ( \mathbb{R} ^{2} ) }+\bigl\vert f ( x_{0},y_{0} ) \bigr\vert \iint_{(\sqrt{2}-1)\delta/\sqrt{2}\leq\sqrt{s^{2}+t^{2}}}K_{\lambda} \bigl( \sqrt{t^{2}+s^{2}} \bigr) \,dt\,ds \\ &\qquad{}+\bigl\vert f ( x_{0},y_{0} ) \bigr\vert \biggl\vert \iint_{\mathbb{R}^{2}}K_{\lambda} \bigl( \sqrt{ ( t-x ) ^{2}+ ( s-y ) ^{2}} \bigr) \,dt\,ds-1\biggr\vert . \end{aligned}$$

From (i)-(iii) and using conditions of class A, we have the desired result i.e.,

$$ \bigl\vert L_{\lambda} ( f;x,y ) -f ( x_{0},y_{0} ) \bigr\vert =o\bigl(\Delta(\lambda,\delta,x,y)\bigr). $$

 □

Example 3

Let $\Lambda=(0,\infty)$, $\lambda_{0}=0$, and

$$ H_{\lambda}(t,s)=\frac{1}{4\pi\lambda}e^{\frac {-(t^{2}+s^{2})}{4\lambda}}. $$

To verify that $H_{\lambda}(t,s)$ satisfies the hypotheses of Theorem 1 and Theorem 2 see [15].

Let $(x_{0},y_{0})=(0,0)$, $\mu_{1}(t)=t$ and $\mu_{2}(s)=s$. Hence we obtain

$$\begin{aligned} \Delta(\lambda,\delta,x,y) =& \int_{-\delta}^{+\delta} \int _{-\delta}^{+\delta}\frac{1}{4\pi\lambda }e^{\frac{-((t-x)^{2}+(s-y)^{2})}{4\lambda}}\,dt\,ds \\ =&\frac{1}{2} \biggl( Erf \biggl( \frac{\delta-x}{2\sqrt{\lambda }} \biggr) +Erf \biggl( \frac{x}{2\sqrt{\lambda}} \biggr) \biggr) \biggl( Erf \biggl( \frac{\delta-y}{2\sqrt{\lambda}} \biggr) +Erf \biggl( \frac{y}{2\sqrt {\lambda }} \biggr) \biggr) . \end{aligned}$$

In order to find for which $\delta>0$ the condition (i) in Theorem 3 is satisfied, let $\Delta(\lambda,\delta ,x,y)\rightarrow0$ as $(x,y,\lambda)\rightarrow (0,0,0)$. Hence

$$ \lim_{(x,y,\lambda)\rightarrow(0,0,0)}\Delta(\lambda ,\delta ,x,y)=0 $$

if and only if $\delta^{2}=o(\lambda)$. Consequently, the following equality holds:

$$ \Delta(\lambda,\delta,x,y)=O(\lambda). $$

Finally, in order to get finite limit values from the expressions

$$\begin{aligned}& \lim_{(x,y,\lambda)\rightarrow(x_{0},y_{0},\lambda_{0})}K_{\lambda} ( 0 ) \mu_{1} \bigl( \vert x_{0}-x\vert \bigr) =\lim_{(x,y,\lambda)\rightarrow(0,0,0)} \frac {1}{4\pi \lambda}e^{\frac{-(x^{2}+y^{2})}{4\lambda}} \vert x\vert , \\& \lim_{(x,y,\lambda)\rightarrow(x_{0},y_{0},\lambda_{0})}K_{\lambda} ( 0 ) \mu_{2} \bigl( \vert y_{0}-y\vert \bigr) =\lim_{(x,y,\lambda)\rightarrow(0,0,0)} \frac {1}{4\pi \lambda}e^{\frac{-(x^{2}+y^{2})}{4\lambda}} \vert y\vert , \end{aligned}$$

the rates of convergence $\frac{1}{4\pi\lambda}e^{\frac{-(x^{2}+y^{2})}{4\lambda}}\rightarrow\infty$ and $\vert x\vert \rightarrow0$ and also $\frac{1}{4\pi\lambda }e^{\frac{-(x^{2}+y^{2})}{4\lambda}}\rightarrow\infty$ and $\vert y\vert \rightarrow0$ must be equivalent. Note that $\vert x\vert =\vert y\vert =O(\lambda)$.

Hence

$$ \bigl\vert L_{\lambda} ( f;x,y ) -f ( x_{0},y_{0} ) \bigr\vert =o \bigl( \Delta(\lambda,\delta,x,y) \bigr) =o ( \lambda ) . $$