On Fatou type convergence of higher derivatives of certain nonlinear singular integral operators

Harun Karsli\(^\ast \) H. Erhan Altin\(^\S \)

June 28, 2012.

\(^\ast \)Abant Izzet Baysal University, Faculty of Science and Arts, Department of Mathematics, 14280, Golkoy Bolu, Turkey, e-mail: karsli_h@ibu.edu.tr.

\(^\S \)Abant Izzet Baysal University, Faculty of Science and Arts, Department of Mathematics, 14280, Golkoy Bolu, Turkey, e-mail: erhanaltin@ibu.edu.tr.

The present paper concerns with the Fatou type convergence properties of the \(r-th\) and \((r+1)-th\) derivatives of the nonlinear singular integral operators defined as

\[ \left( I_{\lambda }f\right) (x)=\int \limits _{a}^{b}K_{\lambda }(t-x,f(t))\, {\rm d}t,\, \, \, \, \, \, \, x\in \left( a,b\right) , \]

acting on functions defined on an arbitrary interval \(\left( a,b\right) ,\) where the kernel \(K_{\lambda }\) satisfies some suitable assumptions. The present study is a continuation and extension of the results established in the paper [ 7 ] .

MSC. 41A35, 47G10, 47H30

Keywords. nonlinear singular integral operator, pointwise convergence, Fatou type convergence.

1 Introduction

Let \(\Lambda \) be a nonempty set of indices with a topology and \(\lambda _{0}\) be an accumulation point of \(\Lambda \) in this topology. By \(\mathcal{U(\theta )}\) we denote the family of all neighborhoods of the neutral element \(\theta \) of \(\mathbb {R}\), and \(x_{0}\) is a fixed accumulation point of \(\mathbb {R}.\) We take a family \(\mathcal{K}\) of functions \(K_{\lambda }:\mathbb {R}\, \)x\(\, \mathbb {R}{\rightarrow }\mathbb {R}{,}\) where \(K_{\lambda }(t,0)=0\) for all \(t\in \mathbb {R}\) and \(\lambda \in \Lambda ,\) such that \(K_{\lambda }(t,u)\) is integrable over \(\mathbb {R}\) with respect to \(t,\) in the sense of Lebesgue measure, for all values of the index \(\lambda \) and second variable \(u\). The family \(\mathcal{K}\) will be called a kernel. In addition, if the kernel function \(K_{\lambda }(t,u)\) is continuous in \(\mathbb {R}\) for every \(t\in \mathbb {R}\), then the kernel function is called Carathéodory kernel function.

In this paper we are concerned with the Fatou type pointwise convergence of \(r-th\) and \((r+1)-th\) derivatives of certain family of nonlinear singular integral operators \(\left( I_{\lambda }f\right) \) of the form

\begin{equation} \left( I_{\lambda }f\right) (x)=\int \limits _{a}^{b}K_{\lambda }(t-x,f(t))\, {\rm d}t,\, \, \, \, \, \, \, x\in \left( a,b\right) ,\label{1}\end{equation}

acting on functions defined on an arbitrary interval \(\left( a,b\right) ,\) where the kernel function \(K_{\lambda }\) satisfies some suitable assumptions. In these theorems the convergence is restricted to some subsets of the plane, i.e. the Fatou type convergence is discussed, whenever the first parameter tends to an accumulation point \(x_{0}\), at which the function \(f\) has finite \(r-th\) and \((r+1)-th\) derivatives, whereas the second one tends to an accumulation point \(\lambda _{0}\) of a given index set \(\Lambda \).

Further results on convergence of the operators (1) and its linear cases can be found in [ 1 ] - [ 12 ] .

In particular, we obtain the rate of the Fatou type pointwise convergence for the nonlinear family of singular integral operators (1) to the point \(x_{0}\), at which the function \(f\) has finite \(r-th\) and \((r+1)-th\) derivatives, as \(\left( x,\lambda \right) \rightarrow \left( x_{0},\lambda _{0}\right) .\) The results presented in this paper are the continuation and extension of those established in [ 7 ] in which the kernel function \(K_{\lambda }\) satisfies

\[ \left[ \tfrac {\partial }{\partial x}K_{\lambda }(t-x,u)-\tfrac {\partial }{\partial x}K_{\lambda }(t-x,v)\right] =\tfrac {\partial }{\partial x}L_{\lambda }(t-x)\left[ u-v\right] \]

for every \(t,u,v\in \) \(\mathbb {R}\) and for any \(\lambda \in \Lambda .\)

Throughout this paper we assume that the function \(K_{\lambda }:\mathbb {R}\) x\(\, \mathbb {R}{\rightarrow }\mathbb {R}\) satisfies the following conditions;
\(a\, )\) Let \(L_{\lambda }(t)\) be an integrable function such that for any fixed \(r\in \mathbb {N}\)

\begin{equation} \left[ \tfrac {\partial ^{r}}{\partial x^{r}}K_{\lambda }(t-x,u)-\tfrac {\partial ^{r}}{\partial x^{r}}K_{\lambda }(t-x,v)\right] =\tfrac {\partial ^{r}}{\partial x^{r}}L_{\lambda }(t-x)\left[ u-v\right] ,\label{c}\end{equation}

holds for every \(t\) \(u,v\in \mathbb {R}\) and for any \(\lambda \in \Lambda .\)
\(b)\) \(\, {\lim \limits _{\lambda \rightarrow \lambda _{0}}\, }\displaystyle {\int \nolimits _{\mathbb {R}\, \backslash \, U}L_{\lambda }(t){\rm d}t}=0, \)for every \(U\in \mathcal{U(}0\mathcal{)}. \)
\(\, c\, )\, {\lim \limits _{\lambda \rightarrow \lambda _{0}}}\left[ \sup \limits _{\left\vert t\right\vert \geq \delta }L_{\lambda }(t)\right] {=0,\, \, }\)for every\({\, \, \delta {\gt}0. }\)
\(d\, )\, {\lim \limits _{\lambda \rightarrow \lambda _{0}}}\displaystyle {\int \nolimits _{\mathbb {R}}L_{\lambda }(t)\, {\rm d}t}=1.\)
According to \(a)\) it is easy to see that \(K_{\lambda }:\mathbb {R}\) x\(\, \mathbb {R}{\rightarrow }\mathbb {R}\) is a kernel function.
We introduce a function \(\overset {\sim }{f}\)\(\in L_{1}(\mathbb {R})\) as

\begin{equation} \overset {\sim }{f}(t):=\left\{ \begin{array}[c]{lll}f(t), & t\in \left( a,b\right), \\ \, 0, & t\notin \left( a,b\right), \end{array} \right. \label{2}\end{equation}

(See [ 6 ] - [ 9 ] ).

Theorem 1

[ 3 ] Let \(1\leq p{\lt}\infty \, \)and assume that \(K_{\lambda }(t,u)\) is a Carathéodory kernel function. If \(f\in L_{p}\left( a,b\right) \), then \(\left( I_{\lambda }f\right) \in L_{p}\left( a,b\right) \) and

\[ \left\Vert I_{\lambda }f\right\Vert _{L_{p}\left( a,b\right) }\leq H(\lambda )\left\Vert f\right\Vert _{L_{p}\left( a,b\right) } \]

for every \(\lambda \in \Lambda .\)

This kind of existence theorem is also valid in general functional spaces (see e.g. [ 3 ] ).

2 Convergence of the derivatives

Let us define, for any constants \(C_{\nu } {\gt} 0 \left( \nu =1,2,...,r\right) ,\) the set

\begin{align} D_{r}:=\left\{ (x,\lambda )\in I\text{~ }x\text{~ }\Lambda :\left\vert x-x_{0}\right\vert ^{\nu }\int \nolimits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert t\right\vert ^{r-\nu }\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert \, {\rm d}t{\lt}C_{\nu }\right\} ,\label{k}\end{align}

where \(I=(a,b)\) is an arbitrary interval in \(\mathbb {R}\).
Now, we are ready to investigate the approximation for finite \(r\)-th derivatives of the operator \(\left( I_{\lambda }f\right) \) in \(L_{1}(I)\).

Theorem 2

Let the function \(L_{\lambda }(t)\) and its derivatives \(\tfrac {{\rm \partial }^{\nu }}{{\rm \partial } t^{\nu }}L_{\lambda }(t)\), \((\nu =1,2,...,r)\) be continuous with respect to \(t\) on \((-\infty ,\infty )\) and \(L_{\lambda }(t)\) be integrable with respect to \(t\) for each fixed \(\lambda \in \Lambda .\) Suppose that the conditions \({\rm c)}\) and \({\rm d)}\) together with

\begin{equation} \sup \limits _{\lambda \in \Lambda }\int \nolimits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert t\right\vert ^{r}\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert \, {\rm d}t<\infty \, \, \, \, {and\, \, \, \, \, \, \, }\lim \limits _{\lambda \rightarrow \lambda _{0}}\sup \limits _{0<\delta \leq \left\vert t\right\vert }\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert =0\label{17}\end{equation}

hold for every \(\delta {\gt}0,\) are satisfied. Suppose that the function \(f\in L_{1}(I)\) has at \(x_{0}\) a finite \(r-th\) derivative.
Then

\begin{equation} \lim \tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }f\right) (x)=f^{(r)}(x_{0})\label{19}\end{equation}

as \((x,\lambda )\rightarrow (x_{0},\lambda _{0})\) and \((x,\lambda )\in D_{r}.\)

Proof â–¼

Supposethat \(a{\lt}x_{0}{\lt}b\) and \(0{\lt}\left\vert x_{0}-x\right\vert {\lt}\delta \) \((\delta {\gt}0).\)
We construct a function

\begin{equation} g(t)=f(x_{0})+(t-x_{0})f^{\prime }(x_{0})+...+\tfrac {(t-x_{0})^{r}}{r!}\, f^{(r)}(x_{0})\label{20}\end{equation}

so that \(g^{(r)}(t)=\, f^{(r)}(x_{0})\).
Firstly we shall prove this theorem for the function \(g(t)\). For this purpose we introduce a function \(\overset {\sim }{g}\)\(\in L_{1}(\mathbb {R})\) as follows;

\begin{equation} \overset {\sim }{g}(t):=\left\{ \begin{array}[c]{lll}g(t), & t\in \left( a,b\right), \\ \, \, 0\, ,& t\notin \left( a,b\right). \end{array} \right. \label{6}\end{equation}

Applying the operator \(I_{\lambda }\) to the function \(g(t)\), we have

\[ \left( I_{\lambda }g\right) (x)=\int \limits _{a}^{b}K_{\lambda }(t-x,g(t)){\rm d}t \]

and according to \((\ref{6})\) we can rewrite the last equality as follows;

\begin{equation} \left( I_{\lambda }g\right) (x)=\int \limits _{\mathbb {R}}K_{\lambda }(t-x,\overset {\sim }{g}(t)){\rm d}t=\left( I_{\lambda }\overset {\sim }{g}\right) (x)\label{7}\end{equation}

and hence

\begin{align*} \tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }g\right) (x) & =\tfrac {\partial ^{r}}{\partial x^{r}}\int \limits _{\mathbb {R}}K_{\lambda }(t-x,\overset {\sim }{g}(t)){\rm d}t =\int \limits _{R}\overset {\sim }{g}(t)\tfrac {\partial ^{r}}{\partial x^{r}}L_{\lambda }(t-x){\rm d}t\\ & =(-1)^{r}\int \limits _{\mathbb {R}}\overset {\sim }{g}(t)\tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x){\rm d}t=\int \limits _{\mathbb {R}}\overset {\sim }{g}^{(r)}(t)L_{\lambda }(t-x){\rm d}t. \end{align*}

In the above, substituting \((\ref{20}),\) we get

\begin{align*} \left\vert \tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }g\right) (x)-f^{(r)}(x_{0})\right\vert & =\left\vert \int \limits _{a}^{b}f^{(r)}(x_{0})L_{\lambda }(t-x){\rm d}t-f^{(r)}(x_{0})\right\vert \\ & =\left\vert f^{(r)}(x_{0})\right\vert \left\vert \int \limits _{a}^{b}L_{\lambda }(t-x){\rm d}t-1\right\vert . \end{align*}

Hence, from condition \(d)\) one has

\begin{equation} \lim \limits _{(x,\lambda )\rightarrow (x_{0},\lambda _{0})}\tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }g\right) (x)=f^{(r)}(x_{0}).\label{21}\end{equation}

We denote \(I_{\lambda }(x):=\tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }g\right) (x)-\tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }f\right) (x).\)
Thanks to (10), for completing the proof of the theorem, it is sufficient to show that

\[ \lim \limits _{(x,\lambda )\rightarrow (x_{0},\lambda _{0})}\left\vert I_{\lambda }(x)\right\vert =0, \]

which gives (6).
It is easy to see that

\begin{align*} \left\vert I_{\lambda }(x)\right\vert & =\left\vert \int \limits _{\mathbb {R}}\overset {\sim }{f}(t)\tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x){\rm d}t-\int \limits _{\mathbb {R}}\overset {\sim }{g}(t)\tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x){\rm d}t\right\vert \\ & =\left\vert \int \limits _{a}^{b}\left[ f(t)-g(t)\right] \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x){\rm }{\rm d}t\right\vert \leq \int \limits _{a}^{x_{0}-\delta }\left\vert f(t)-g(t)\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert {\rm d}t\\ & \quad \! +\! \int \limits _{x_{0}\! -\! \delta }^{x_{0}+\delta }\left\vert f(t)\! -\! g(t)\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert {\rm d}t\! +\! \int \limits _{x_{0}\! +\! \delta }^{b}\left\vert f(t)\! -\! g(t)\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert {\rm d}t\\ & \ =:I_{1}(x,\lambda )+I_{2}(x,\lambda )+I_{3}(x,\lambda ). \end{align*}

To estimate \(I_{1}(x,\lambda )\) and \(I_{3}(x,\lambda ),\) we use the following method:

\begin{align*} I_{1}(x,\lambda ) & =\int \limits _{a}^{x_{0}-\delta }\left\vert f(t)-g(t)\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert {\rm d}t\\ & \leq \sup \limits _{0{\lt}\delta \leq \left\vert t\right\vert }\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert \int \limits _{a}^{b}\left\vert f(t)-g(t)\right\vert {\rm d}t \end{align*}

and

\begin{align*} I_{3}(x,\lambda ) & =\int \limits _{x_{0}+\delta }^{b}\left\vert f(t)-g(t)\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert {\rm d}t\\ & \leq \sup \limits _{0{\lt}\delta \leq \left\vert t\right\vert }\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert \int \limits _{a}^{b}\left\vert f(t)-g(t)\right\vert {\rm d}t. \end{align*}

Now we shall rewrite \(I_{2}(x,\lambda )\) as follows,

\[ I_{2}(x,\lambda )=\int \limits _{x_{0}-\delta }^{x_{0}+\delta }\left\vert \tfrac {f(t)-g(t)}{\left( t-x_{0}\right) ^{r}}\right\vert \left\vert \left( t-x_{0}\right) ^{r}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert {\rm d}t. \]

For every \(\varepsilon {\gt}0\) there exists a \(\delta {\gt}0\) such that;

\[ I_{2}(x,\lambda )\leq \varepsilon \int \limits _{x_{0}-\delta }^{x_{0}+\delta }\left\vert \left( t-x_{0}\right) ^{r}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert {\rm d}t. \]

Let

\[ I_{2,1}(x,\lambda ):\! =\! \int \limits _{x_{0}-\delta }^{x_{0}\! +\! \delta }\left\vert \left( t\! -\! x_{0}\right) ^{r}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t\! -\! x)\right\vert {\rm d}t\! =\! \int \limits _{x_{0}\! -\! x\! -\! \delta }^{x_{0}\! -\! x\! +\! \delta }\left\vert \left( t+x-x_{0}\right) ^{r}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t. \]

It can be rewrite \(I_{2,1}(x,\lambda )\) as follows:

\begin{align*} I_{2,1}(x,\lambda ) & =\int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert \left( t+x-x_{0}\right) ^{r}-t^{r}+t^{r}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t\\ & \leq \int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert \left( t+x-x_{0}\right) ^{r}-t^{r}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t+\int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert t\right\vert ^{r}\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t\\ & =:I_{2,1,1}(x,\lambda )+I_{2,1,2}(x,\lambda ). \end{align*}

From \((\ref{17})\) \(I_{2,1,2}(x,\lambda )\) is finite. So it is sufficient to show that \(I_{2,1,1}(x,\lambda )\) is finite.
Using the obvious identity

\begin{equation} a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})\label{26}\end{equation}

yields

\begin{align} I_{2,1,1}(x,\lambda ) & =\left\vert x-x_{0}\right\vert \int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert \left( t+x-x_{0}\right) ^{r-1}+...+t^{r-1}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t\nonumber \\ & \leq \left\vert x-x_{0}\right\vert \int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert \left( t+x-x_{0}\right) ^{r-1}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t\nonumber \\ & \quad +\left\vert x-x_{0}\right\vert \int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert \left( t+x-x_{0}\right) ^{r-2}t\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t\nonumber \\ & \quad \vdots \label{27}\\ & \quad +\left\vert x-x_{0}\right\vert \int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert \left( t+x-x_{0}\right) t^{r-2}\right\vert \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t\nonumber \\ & \quad +\left\vert x-x_{0}\right\vert \int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert t\right\vert ^{r-1}\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert {\rm d}t.\nonumber \end{align}

Applying the formula (11) successively to the right-hand side of (12), we can see that \(I_{2,1,1}(x,\lambda )\) is less than or equal to the linear combinations of

\[ \left\vert x-x_{0}\right\vert ^{\nu }\int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert t\right\vert ^{r-\nu }\, \left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t)\right\vert \, {\rm d}t,\, \, \, \, \, \, \, \, \, \, (\nu =1,2,...,r). \]

Taking into account \((\ref{17})\) and \((\ref{k}),\) we show that \(I_{2,1}(x,\lambda )\) is finite\(\, \)on any planar set \(D_{r}\).
Hence

\begin{align*} \left\vert I_{\lambda }(x)\right\vert & \leq I_{1}(x,\lambda )+I_{2}(x,\lambda )+I_{3}(x,\lambda )\\ & \leq \varepsilon I_{2}(x,\lambda )+2\sup \limits _{0{\lt}\delta \leq \left\vert t\right\vert }\left\vert \tfrac {\partial ^{r}}{\partial t^{r}}L_{\lambda }(t-x)\right\vert \int \limits _{a}^{b}\left\vert f(t)-g(t)\right\vert {\rm d}t. \end{align*}

Since \(f(t)-g(t)\) belongs to \(L_{1}(a,b)\), in view of \((\ref{17}),(\ref{k})\) and the condition \(c),\) we obtain

\[ \lim \limits _{(x,\lambda )\rightarrow (x_{0},\lambda _{0})}\left\vert I_{\lambda }(x)\right\vert =0, \]

i.e.,

\[ \lim \limits _{(x,\lambda )\rightarrow (x_{0},\lambda _{0})}\tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }f\right) (x)=\lim \limits _{(x,\lambda )\rightarrow (x_{0},\lambda _{0})}\tfrac {\partial ^{r}}{\partial x^{r}}\left( I_{\lambda }g\right) (x). \]

This completes the proof of the Theorem.

Proof â–¼

Secondly, we will investigate the approximation for finite \((r+1)-th\) derivatives of the operator \(\left( I_{\lambda }f\right) \) in \(L_{1}(I)\).

Theorem 3

Let the function \(L_{\lambda }(t)\) and its derivatives \(\tfrac {\partial ^{\nu }}{\partial t^{\nu }}L_{\lambda }(t)\), \((\nu =1,2,...,r,r+1)\) be continuous with respect to \(t\) on \((-\infty ,\infty )\) and \(L_{\lambda }(t)\) be integrable with respect to \(t\) for each fixed \(\lambda \in \Lambda .\) Suppose that conditions \({\rm c)}\) and \({\rm d)}\) are satisfied. Also we assume that the relations

\[ \sup \limits _{\lambda \in \Lambda }\int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert t\right\vert ^{r+1}\left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t)\right\vert \, {\rm d}t{\lt}\infty \, \, \, \, and{\, \, \, \, \, \, }\lim \limits _{\lambda \rightarrow \lambda _{0}}\sup \limits _{0{\lt}\delta \leq \left\vert t\right\vert }\left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t)\right\vert =0 \]

hold for every \(\delta {\gt}0.\) Suppose that the function \(f\in L_{1}(I)\) has at \(x_{0}\) finite derivatives \(f_{+}^{(r+1)}(x_{0})\) and \(f_{-}^{(r+1)}(x_{0})\).
Then

\[ \lim \tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }f\right) (x)=Bf_{+}^{(r+1)}(x_{0})+(1-B)f_{-}^{(r+1)}(x_{0}) \]

as \((x,\lambda )\rightarrow (x_{0},\lambda _{0})\) and \((x,\lambda )\in D_{r+1}\), where

\begin{equation} \underset {(x,\lambda )\rightarrow (x_{0},\lambda _{0})}{\lim }\int \limits _{x_{0}}^{\infty }L_{\lambda }(t-x){\rm d}t=B,\, \, \, \, \, \, \, \, \, \, \, \, 0\leq B\leq 1.\label{a}\end{equation}

Proof â–¼

Let \(a{\lt}x_{0}{\lt}b\) and \(0{\lt}x_{0}-x{\lt}\tfrac {\delta }{2}\) \((\delta {\gt}0).\) Setting

\begin{align} g(t):=\left\{ \begin{array}[c]{c}f(x_{0})+...+\tfrac {(t-x_{0})^{r}}{r!}\, f^{(r)}(x_{0})+\tfrac {(t-x_{0})^{r+1}}{(r+1)!}f_{-}^{(r+1)}(x_{0}),\ \ a{\lt}t{\lt}x_{0},\\ f(x_{0})+...+\tfrac {(t-x_{0})^{r}}{r!}\, f^{(r)}(x_{0})+\tfrac {(t-x_{0})^{r+1}}{(r+1)!}f_{+}^{(r+1)}(x_{0}),\ \ x_{0}\leq t{\lt}b. \end{array} \right. \label{35}\end{align}

Note that

\[ \left( I_{\lambda }g\right) (x)=\int \limits _{a}^{b}K_{\lambda }(t-x,g(t)){\rm d}t=\int \limits _{R}K_{\lambda }(t-x,\overset {\sim }{g}(t)){\rm d}t. \]

Differentiating both sides of the last equality \((r+1)\) times with respect to \(x\), one has

\[ \tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }g\right) (x)=\int \limits _{a}^{b}g^{(r+1)}(t)L_{\lambda }(t-x){\rm d}t. \]

By (14), the last equality can be rewritten in the form

\[ \tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }g\right) (x)=f_{-}^{(r+1)}(x_{0})\int \limits _{a}^{x_{0}}L_{\lambda }(t-x){\rm d}t+f_{+}^{(r+1)}(x_{0})\int \limits _{x_{0}}^{b}L_{\lambda }(t-x){\rm d}t. \]

The hypothesis (13) yields

\[ \underset {(x,\lambda )\rightarrow (x_{0},\lambda _{0})}{\lim }\tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }g\right) (x)=B\, f_{+}^{(r+1)}(x_{0})+(1-B)f_{-}^{(r+1)}(x_{0}). \]

Define

\[ \left\vert I_{\lambda }(x)\right\vert :=\left\vert \tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }g\right) (x)-\tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }f\right) (x)\right\vert . \]

In order to complete the proof of the theorem, it is sufficient to show

\[ \underset {(x,\lambda )\rightarrow (x_{0},\lambda _{0})}{\lim }\left\vert I_{\lambda }(x)\right\vert =0. \]

One has

\begin{align} \left\vert I_{\lambda }(x)\right\vert & =\left\vert \int \limits _{a}^{b}\left[ f(t)-g(t)\right] \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x){\rm d}t\right\vert \nonumber \\ & \leq \int \limits _{a}^{x_{0}-\delta }\left\vert f(t)-g(t)\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & \quad +\int \limits _{x_{0}-\delta }^{x_{0}}\left\vert \tfrac {f(t)-g(t)}{\left( t-x_{0}\right) ^{r+1}}\right\vert \left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & \quad +\int \limits _{x_{0}+\delta }^{b}\left\vert f(t)-g(t)\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & \quad +\int \limits _{x_{0}}^{x_{0}+\delta }\left\vert \tfrac {f(t)-g(t)}{\left( t-x_{0}\right) ^{r+1}}\right\vert \left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & =:I_{1}(x,\lambda )+I_{2}(x,\lambda )+I_{3}(x,\lambda )+I_{4}(x,\lambda ).\label{36}\end{align}

Since \(f\) has at \(x_{0}\) a finite \((r+1)-th\) right and left derivatives, then for every \(\varepsilon {\gt}0\) there exists a \(\delta {\gt}0\) such that

\begin{align} I_{2}(x,\lambda ) & =\int \limits _{x_{0}-\delta }^{x_{0}}\left\vert \tfrac {f(t)-g(t)}{\left( t-x_{0}\right) ^{r+1}}\right\vert \left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & \leq \varepsilon \int \limits _{x_{0}-\delta }^{x_{0}}\left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\label{38}\end{align}

and

\begin{align} I_{4}(x,\lambda ) & =\int \limits _{x_{0}}^{x_{0}+\delta }\left\vert \tfrac {f(t)-g(t)}{\left( t-x_{0}\right) ^{r+1}}\right\vert \left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & \leq \varepsilon \int \limits _{x_{0}}^{x_{0}+\delta }\left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t.\label{39}\end{align}

Thus we get

\[ I_{2}(x,\lambda )+I_{4}(x,\lambda )\leq \varepsilon \int \limits _{x_{0}-\delta }^{x_{0}+\delta }\left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t. \]

Set

\[ I_{2,1}(x,\lambda )+I_{4,1}(x,\lambda ):=\int \limits _{x_{0}-\delta }^{x_{0}+\delta }\left\vert \left( t-x_{0}\right) ^{r+1}\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t. \]

Using the same method as in the proof of Theorem 2 we deduce \(I_{2,1}(x,\lambda )+I_{4,1}(x,\lambda )\) is less than or equal to a linear combination of

\[ \left\vert x-x_{0}\right\vert ^{\nu }\int \limits _{x_{0}-x-\delta }^{x_{0}-x+\delta }\left\vert t\right\vert ^{r+1-\nu }\left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}K(t,\lambda )\right\vert \, {\rm d}t,\, (\nu =1,...,r+1). \]

By virtue of \((\ref{k}),\) the term \(I_{2}(x,\lambda )+I_{4}(x,\lambda )\) is bounded.
Now, we consider the integrals \(I_{1}(x,\lambda )\) and \(I_{3}(x,\lambda ),\) respectively.

\begin{align} I_{1}(x,\lambda ) & =\int \limits _{a}^{x_{0}-\delta }\left\vert f(t)-g(t)\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & \leq M\underset {\tfrac {\delta }{2}{\lt}\left\vert u\right\vert }{\sup }\left\vert \tfrac {\partial ^{r+1}}{\partial u^{r+1}}L_{\lambda }(u)\right\vert \label{40}\end{align}

and

\begin{align} I_{3}(x,\lambda ) & =\int \limits _{x_{0}+\delta }^{b}\left\vert f(t)-g(t)\right\vert \left\vert \tfrac {\partial ^{r+1}}{\partial t^{r+1}}L_{\lambda }(t-x)\right\vert {\rm d}t\nonumber \\ & \leq M\underset {\tfrac {\delta }{2}{\lt}\left\vert u\right\vert }{\sup }\left\vert \tfrac {\partial ^{r+1}}{\partial u^{r+1}}L_{\lambda }(u)\right\vert .\label{41}\end{align}

Using (16), (19) in (15) one has

\begin{align} \left\vert I_{\lambda }(x)\right\vert & \leq I_{1}(x,\lambda )+I_{2}(x,\lambda )+I_{3}(x,\lambda )+I_{4}(x,\lambda )\nonumber \\ & \leq \varepsilon \left[ I_{2,1}(x,\lambda )+I_{4,1}(x,\lambda )\right]\& \quad +2M\underset {\tfrac {\delta }{2}{\lt}\left\vert u\right\vert }{\sup }\left\vert \tfrac {\partial ^{r+1}}{\partial u^{r+1}}L_{\lambda }(u)\right\vert .\label{42}\end{align}

Under the hypotheses of the theorem, (20) yields

\[ \underset {(x,\lambda )\rightarrow (x_{0},\lambda _{0})}{\lim }\left\vert I_{\lambda }(x)\right\vert =\underset {(x,\lambda )\rightarrow (x_{0},\lambda _{0})}{\lim }\left\vert \tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }g\right) (x)-\tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }f\right) (x)\right\vert =0. \]

This completes the proof.

Proof â–¼

Remark 4

If \(B=\tfrac {1}{2}\) , then we have

\[ \underset {(x,\lambda )\rightarrow (x_{0},\lambda _{0})}{\lim }\tfrac {\partial ^{r+1}}{\partial x^{r+1}}\left( I_{\lambda }f\right) (x)=\tfrac {f_{+}^{(r+1)}(x_{0})+f_{-}^{(r+1)}(x_{0})}{2}.\hfil \qed \]

Remark 5

Let \(I=(a,b)\) be an arbitrary bounded interval in \(\mathbb {R}\) and \(f\in L_{1}(a,b)\) be a \((b-a)\)-periodic function. In this case the proofs of the Theorems are similar.â–¡

3 Examples

Example 6

A special case of the function \(K_{\lambda }(t,u)\) satisfying the conditions, is the linear case with respect to the second variable, i.e.,

\[ K_{\lambda }(t,u)=Z_{\lambda }(t)\, u. \]

This case is widely used in Approximation Theory [ 4 ] .â–¡

Example 7

We introduce the function

\[ K_{\lambda }(t,u)=\left\{ \begin{array}[l]{lll}(r+1)\lambda ^{r+1}t^{r}\, u+u, & t\in \lbrack 0,\tfrac {1}{\lambda }],\\ 0, & t\notin \lbrack 0,\tfrac {1}{\lambda }], \end{array} \right. \]

where \(\Lambda =[1,\infty )\) is a set of indices with natural topology and \(\lambda _{0}=\infty \) is an accumulation point of \(\Lambda \) in this topology.
First of all \(K_{\lambda }(t,u)\) is a kernel, i.e., \(K_{\lambda }(t,0)=0\medskip .\)
It is seen that for every \(u\in \mathbb {R},\)

\[ \tfrac {\partial ^{r}}{\partial x^{r}}K_{\lambda }(t-x,u)=\left\{ \begin{array}[l]{lll}(-1)^{r}(r+1)!\lambda ^{r+1}\, u, & t-x\in \lbrack 0,\tfrac {1}{\lambda }],\\ 0, & t-x\notin \lbrack 0,\tfrac {1}{\lambda }]. \end{array} \right. \]

According to (2), we obtain

\[ \tfrac {\partial ^{r}}{\partial x^{r}}L_{\lambda }(t-x)=\left\{ \begin{array}[l]{lll}(-1)^{r}(r+1)!\lambda ^{r+1}, & t-x\in \lbrack 0,\tfrac {1}{\lambda }],\\ 0 , & t-x\notin \lbrack 0,\tfrac {1}{\lambda }]. \end{array} \right. \]

This implies

\[ L_{\lambda }(t-x)=\left\{ \begin{array}[l]{lll}(r+1)\lambda ^{r+1}(t-x)^{r}, & t-x\in \lbrack 0,\tfrac {1}{\lambda }],\\ 0, & t-x\notin \lbrack 0,\tfrac {1}{\lambda }]. \end{array} \right. \]

Moreover

\[ \int \limits _{\mathbb {R}}L_{\lambda }(t){\rm d}t=\int \limits _{[0,\tfrac {1}{\lambda }]}(r+1)\lambda ^{r+1}t^{r}{\rm d}t=1{\lt}\infty . \]

It is easy to see that

\[ {\lim \limits _{\lambda \rightarrow \infty }\, }\int \limits _{\mathbb {R}\, \backslash \, U}L_{\lambda }(t){\rm d}t=0, \]

for every \(U\in \mathcal{U(}0\mathcal{)}\) and

\[ {\lim \limits _{\lambda \rightarrow \infty }}\left[ \sup \limits _{\left\vert t\right\vert \geq \delta }L_{\lambda }(t)\right] {=0,} \]

\({\, }\)for every\({\, \, \delta {\gt}0.}\)â–¡

Acknowledgement

The authors would like to express their sincere
thanks to the referee and to Professor Octavian Agratini for their very careful and intensive study of this manuscript.

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