An Approximation Formula of the Integral Transform on Wiener Space

Abstract

In this paper, we introduce an integral transform with a kernel defined on the Wiener space. We first establish the existence of the integral transform and present several illustrative examples. As the main result, we derive an approximation theorem for the integral transform. Our approach demonstrates that the integral transform can be effectively computed even in cases where direct calculation is difficult or infeasible.

1. Introduction

An integral transform maps a function from its original function space into another, often more analytically tractable, function space through an integration process. In many cases, certain properties of the original function become more apparent or manageable in the transformed domain. Typically, the original function can be recovered via an appropriate inverse transform. The classical integral transform on Euclidean space is defined by the formula

$$T\left(f\right)\left(v\right)={\int }_{\mathbb{R}}f\left(u\right)g(u,v)du$$

for some appropriate functions f and g [1]. The function g is referred to as the kernel, integral kernel, or nucleus of the transform. Research on integral transforms in Euclidean spaces has a long and rich history, dating back to ancient times, and continues to be an active area of mathematical study. Mathematical notation aside, the motivation behind integral transforms is easy to understand. There are many classes of problems that are difficult to solve or at least quite unwieldy algebraically in their original representations. Depending on the structure of the kernel function, the integral transform is categorized under various classical types. Notable examples include the Fourier transform, the Laplace transform, the Hartley transform, and the two-sided Laplace transform [1,2,3,4,5].

For $T>0$, let $C_0[0,T]$ denote the classical Wiener space with the associated Wiener measure $m_w$. On the Wiener space, many research results on integral transform have been obtained, and they are being generalized in various ways. However, research on the form in which the integral transform has a kernel in its definition itself is lacking [5,6,7,8,9,10,11,12,13,14,15]. Recently, several authors have published papers on integral transforms whose kernels involve the Fourier transform and the two-sided Laplace transform [3,16,17].

In this paper, we define an integral transform ${\mathbb{T}}^G\left(F\right)$ of a functional F defined on $C_0[0,T]$ given a functional G defined on $C_0^2[0,T]\equiv C_0[0,T]\times C_0[0,T]$, and establish the existence of the integral transform ${\mathbb{T}}^G\left(F\right)$ with some examples. Our integral transform ${\mathbb{T}}^G\left(F\right)$ shares certain similarities with the classical integral transform $T\left(f\right)$. However, there are many differences in results and properties between ${\mathbb{T}}^G\left(F\right)$ and $T\left(f\right)$ on Euclidean space because, primarily because the Wiener measure $m_w$ is a probability measure. As a main result, we give an approximation formula of the integral transform ${\mathbb{T}}^G\left(F\right)$. The main theorem of this paper demonstrates that the integral transform ${\mathbb{T}}^G\left(F\right)$ can be effectively approximated or computed even in cases where direct evaluation is analytically intractable or computationally prohibitive.

This paper is structured as follows. In Section 2, we introduce the fundamental definitions and notations necessary for understanding the subsequent developments. Section 3 presents an approximation of the integral transform, along with illustrative examples. Finally, Section 4 concludes the paper.

2. Preliminaries and Definitions

In this section, we introduce the fundamental definitions and notations necessary for understanding the subsequent development of this paper. We then define an integral transform on Wiener space $C_0[0,T]$. Furthermore, we establish the existence of the proposed integral transform and present several illustrative applications.

We state a well-known integration formula which is used later in this paper.

Theorem 1.

Let $(t_1,\dots ,t_n)$ be an n-tuple of $[0,T]$ with $0=t_0<t_1<\dots <t_n=T$. Let $f:{\mathbb{R}}^n\to \mathbb{C}$ be Lebesgue measurable and let F a functional of the form

$$F\left(x\right)=f(x\left(t_1\right),\dots ,x\left(t_n\right)).$$

Then,

$$\begin{gathered} {\int }_{C_0[0,T]}F\left(x\right)d{m}_w\left(x\right)\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} \\ ={\int }_{C_0[0,T]}f(x\left(t_1\right),\dots ,x\left(t_n\right))d{m}_w\left(x\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\left(\prod _{j=1}^n{\displaystyle \frac{1}{2\pi (t_j-t_{j-1})}}\right)}^{{\displaystyle \frac{1}{2}}}{\int }_{{\mathbb{R}}^n}f\left(\overrightarrow{u}\right)exp\left\{-\sum _{j=1}^n{\displaystyle \frac{{(u_j-u_{j-1})}^2}{2(t_j-t_{j-1})}}\right\}d\overrightarrow{u}, \end{gathered}$$

(1)

where $u_0=0$, in the sense that if either side of (1) exists, then both sides exist and the equality holds [17,18,19,20,21].

We are now ready to state the definition of the integral transform of functionals on $C_0[0,T]$.

Definition 1.

Let F be a functional on $C_0[0,T]$ and let G be a functional on $C_0^2[0,T]$. The integral transform ${\mathbb{T}}^G\left(F\right)$ of F given G is defined by the formula

$${\mathbb{T}}^G\left(F\right)\left(y\right)={\int }_{C_0[0,T]}F\left(x\right)G(x,y)d{m}_w\left(x\right),y\in C_0[0,T]$$

(2)

if it exists. In this case, the functional G is called the kernel of integral transform ${\mathbb{T}}^G$.

Let $E^{\left(n\right)}$ be the class of all functionals F on $C_0^n[0,T]$ of the form

$$F(x_1,\dots ,x_n)=f(x_1\left(T\right),\dots ,x_n\left(T\right)),$$

(3)

where f is an infinitely differentiable function on ${\mathbb{R}}^n$ with

$$|f\left(\overrightarrow{u}\right)|\le A_{f}exp\left\{B_f\sum _{j=1}^n\left|u_j\right|\right\}$$

(4)

for some real numbers $A_f>0$ and $B_f\ge 0$.

Remark 1.

Note that $E^{\left(n\right)}$ is a very rich class because $E^{\left(n\right)}$ has many unbounded functionals. In fact, if F is given by (3), then the function f is bounded if and only if it is a constant function.

In Theorem 2 below, we shall establish the existence of integral transform ${\mathbb{T}}^G\left(F\right)$ of F given the kernel G.

Theorem 2.

Let $F\in E^{\left(1\right)}$ and $G\in E^{\left(2\right)}$ with $F\left(x\right)=f\left(x\right(T\left)\right)$ and $G(x,y)=g\left(x\right(T),y(T\left)\right)$. Then, the integral transform ${\mathbb{T}}^G\left(F\right)$ of F given the kernel G exists, belongs to $E^{\left(1\right)}$ and is given by the formula

$${\mathbb{T}}^G\left(F\right)\left(y\right)=\Gamma \left(y\left(T\right)\right),$$

(5)

where

$$\Gamma \left(v\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}f\left(u\right)g(u,v)exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du.$$

Proof. 

Using Equations (1) and (2), we have

$$\begin{gathered} {\mathbb{T}}^G\left(F\right)\left(y\right)={\int }_{C_0[0,T]}F\left(x\right)G(x,y)d{m}_w\left(x\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\int }_{C_0[0,T]}f\left(x\left(T\right)\right)g(x\left(T\right),y\left(T\right))d{m}_w\left(x\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}f\left(u\right)g(u,y\left(T\right))exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du \\ =\Gamma \left(y\left(T\right)\right). \end{gathered}$$

Furthermore, the function $\Gamma$ is infinitely differentiable since g is an infinitely differentiable function, and using Equation (4), we see that

$$\begin{gathered} |\Gamma \left(v\right)|\le {\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}\left|f\left(u\right)\right|\left|g(u,v)\right|exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\le {\displaystyle \frac{A_f{A}_g}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}exp\{B_f\left|u\right|+B_g\left(\right|u|+\left|v\right|\left)\right\}exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\le {\displaystyle \frac{A_f{A}_g}{\sqrt{2\pi T}}}exp\{B_g\left|v\right|\}{\int }_{\mathbb{R}}exp\{(B_f+B_g)\left|u\right|\}exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du \\ =A_{\Gamma }exp\{B_{\Gamma }\left|v\right|\} \end{gathered}$$

where

$$A_{\Gamma }={\displaystyle \frac{A_f{A}_g}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}exp\{(B_f+B_g)\left|u\right|\}exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du<\infty$$

and $B_{\Gamma }=B_g$. These tell us that ${\mathbb{T}}^G\left(F\right)$ is in $E^{\left(1\right)}$, and hence, we complete the proof of Theorem 2 as desired. □

We conclude this section by presenting several examples that illustrate the utility of Theorem 2, using meaningful kernel functions analogous to those appearing in classical integral transforms on Euclidean space.

Example 1

(The kernel of the Fourier-transform). Let F and G be as in Theorem 2. Let $g(u,v)=exp\{-iuv\}$. Then $\left|g(u,v)\right|\le 1=A_{g}exp\{B_g\left(\right|u|+\left|v\right|\left)\right\}$, where $A_g=1$ and $B_g=0$. Hence $G(x,y)=g(x\left(T\right),y\left(T\right))\in E^{\left(2\right)}$. Thus, by Theorem 2, the integral transform ${\mathbb{T}}^G\left(F\right)$ exists, and using Equation (5), we have

$$\begin{gathered} {\mathbb{T}}^G\left(F\right)\left(y\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}f\left(u\right)exp\{-iuy\left(T\right)\}exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du \\ =\widehat{h}\left(y\left(T\right)\right) \end{gathered}$$

where $h\left(u\right)=f\left(u\right)exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}$ and $\widehat{f}$ denotes the Fourier-transform of f.

Example 2

(The kernel of the Hartley transform). Let F and G be as in Theorem 2. Let $g(u,v)={\displaystyle \frac{cos\left(uv\right)+sin\left(uv\right)}{\sqrt{2\pi }}}$. Then $\left|g(u,v)\right|\le \sqrt{{\displaystyle \frac{2}{\pi }}}=A_{g}exp\{B_g\left(\right|u|+\left|v\right|\left)\right\}$, where $A_g=\sqrt{{\displaystyle \frac{2}{\pi }}}$ and $B_g=0$. Hence $G(x,y)=g(x\left(T\right),y\left(T\right))\in E^{\left(2\right)}$. Thus, by Theorem 2, the integral transform ${\mathbb{T}}^G\left(F\right)$ exists, and using Equation (5), we have

$${\mathbb{T}}^G\left(F\right)\left(y\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}f\left(u\right){\displaystyle \frac{cos\left(uy\right(T\left)\right)+sin\left(uy\right(T\left)\right)}{\sqrt{2\pi }}}exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du.$$

In fact,

$$|{\mathbb{T}}^G\left(F\right)\left(y\right)|\le {\displaystyle \frac{{\parallel f\parallel }_1}{\pi \sqrt{T}}}<\infty$$

provided $f\in L_1\left(\mathbb{R}\right)$.

Example 3

(The kernel of the Weierstrass transform). Let F and G be as in Theorem 2. Let $g(u,v)={\displaystyle \frac{1}{\sqrt{4\pi }}}exp\left\{-{\displaystyle \frac{{(u-v)}^2}{4}}\right\}$. Then $\left|g(u,v)\right|\le 1=A_{g}exp\{B_g\left(\right|u|+\left|v\right|\left)\right\}$, where $A_g=1$ and $B_g=0$. Hence $G(x,y)=g(x\left(T\right),y\left(T\right))\in E^{\left(2\right)}$. Thus, by Theorem 2, the integral transform ${\mathbb{T}}^G\left(F\right)$ exists and using Equation (5), we have

$${\mathbb{T}}^G\left(F\right)\left(y\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}f\left(u\right)exp\left\{-{\displaystyle \frac{u^2}{2T}}-{\displaystyle \frac{{(u-y\left(T\right))}^2}{4}}\right\}du.$$

In fact,

$$|{\mathbb{T}}^G\left(F\right)\left(y\right)|\le {\displaystyle \frac{{\parallel f\parallel }_1}{\sqrt{2\pi T}}}<\infty$$

provided $f\in L_1\left(\mathbb{R}\right)$.

Example 4

(The kernel of the exponential weighted transform). Let F and G be as in Theorem 2. Let $g(u,v)=exp\{u+v\}$. Then $\left|g(u,v)\right|\le A_{g}exp\{B_g\left(\right|u|+\left|v\right|\left)\right\}$, where $A_g=1$ and $B_g=1$. Hence $G(x,y)=g(x\left(T\right),y\left(T\right))\in E^{\left(2\right)}$. Thus, by Theorem 2, the integral transform ${\mathbb{T}}^G\left(F\right)$ exists and using Equation (5), we have

$${\mathbb{T}}^G\left(F\right)\left(y\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}f\left(u\right)exp\left\{u-{\displaystyle \frac{u^2}{2T}}\right\}duexp\left\{v\right\}.$$

In fact,

$$|{\mathbb{T}}^G\left(F\right)\left(y\right)|\le {\displaystyle \frac{{\parallel f\parallel }_1{\parallel k\parallel }_{\infty }}{\sqrt{2\pi T}}}exp\left\{\right|v\left|\right\}$$

provided $f\in L_1\left(\mathbb{R}\right)$ where $k\left(u\right)=exp\left\{u-{\displaystyle \frac{u^2}{2T}}\right\}$ because $k\in L_p\left(\mathbb{R}\right)$ for all $1\le p\le \infty$.

3. Approximation Formula of the Integral Transform

In this section, we give an approximation of the integral transform with examples.

As shown in the previous section, the existence of an integral transform can be established once a kernel function is given. However, depending on the structure of the kernel, proving the existence or evaluating the transform may become challenging or even infeasible through direct methods. In such cases, it becomes necessary to develop alternative approaches for the computation of the integral transform.

In order to express simply, we need a notation as below. Let f be an infinitely differentiable function on ${\mathbb{R}}^n$ with

$$|f\left(\overrightarrow{u}\right)|\le A_{f}exp\{B_f\sum _{j=1}^n|u_j|\}$$

for some real numbers $A_f>0$ and $B_f\ge 0$. For a $s\in \mathbb{N}\cup \left\{0\right\}$, let

$$\mathcal{L}(f,u^s)={\int }_{\mathbb{R}}{u}^{s}f\left(u\right)exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du.$$

(6)

Then $\mathcal{L}(f,u^s)$ always exists for all $s\in \mathbb{N}\cup \left\{0\right\}$ because

$$|\mathcal{L}(f,u^s)|\le A_f{\int }_{\mathbb{R}}{\left|u\right|}^{s}exp\left\{-{\displaystyle \frac{u^2}{2T}}+B_f\left|u\right|\right\}du<\infty .$$

(7)

In Lemma 1 below, we establish an integration formula.

Lemma 1.

Let F be an element of $E^{\left(1\right)}$ with $F\left(x\right)=f\left(x\right(T\left)\right)$. Then for a non-negative integer s, we have

$${\int }_{C_0[0,T]}{x}^s\left(T\right)F\left(x\right)d{m}_w\left(x\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}\mathcal{L}(f,u^s).$$

(8)

Proof. 

Using Equations (1) and (6), we have

$$\begin{gathered} {\int }_{C_0[0,T]}{x}^s\left(T\right)F\left(x\right)d{m}_w\left(x\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}{\int }_{\mathbb{R}}{u}^{s}f\left(u\right)exp\left\{-{\displaystyle \frac{u^2}{2T}}\right\}du \\ ={\displaystyle \frac{1}{\sqrt{2\pi T}}}\mathcal{L}(f,u^s). \end{gathered}$$

From Equation (7), we see that (8) always exists. □

Remark 2.

We shall explain the concept of the Taylor series at $(0,0)$ of two variable function. Let g be an infinitely differentiable function on ${\mathbb{R}}^2$ with the convergence radius set to infinity. Then we have the following expression:

$$g(u,v)=\underset{n\to \infty }{lim}{g}_n(u,v),$$

(9)

where

$$g_n(u,v)=\sum _{k=0}^n\sum _{m=0}^k{a}_{k-m,m}{u}^{k-m}{v}^m$$

(10)

and

$$a_{k-m,m}={\displaystyle \frac{1}{(k-m)!m!}}{\displaystyle \frac{{\partial }^{k}g}{\partial u^{k-m}\partial v^m}}{|}_{(u,v)=(0,0)},0!=1,$$

for more detailed see [1,2,4].

To obtain the main result, we need two lemmas. In the first lemma, we give a formula that the integral transform ${\mathbb{T}}^{G_n}\left(F\right)$ of $F\in E^{\left(1\right)}$ given $G\in E^{\left(2\right)}$ exists.

Lemma 2.

Let $F\in E^{\left(1\right)}$. Let $g_n$ be as in Equation (10) and let $G_n(x,y)=g_n(x\left(t\right),y\left(T\right))$. Then G is in $E^{\left(2\right)}$ and

$${\mathbb{T}}^{G_n}\left(F\right)\left(y\right)=\sum _{k=0}^n\sum _{m=0}^k{b}_{k-m,m}\mathcal{L}(f,u^{k-m})y^m\left(T\right)$$

(11)

where $b_{k-m,m}={\displaystyle \frac{1}{\sqrt{2\pi T}}}{a}_{k-m,m}$.

Proof. 

Since $F\in E^{\left(1\right)}$ and $G\in E^{\left(2\right)}$, from Theorem 2, the integral transform ${\mathbb{T}}^{G_n}\left(F\right)$ of F exists and it is an element of $E^{\left(1\right)}$. We shall now prove equality in Equation (11) holds. Using Equations (1), (2) and (10), we have

$$\begin{gathered} {\mathbb{T}}^{G_n}\left(F\right)\left(y\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\int }_{C_0[0,T]}f\left(x\left(T\right)\right)g_n(x\left(T\right),y\left(T\right))d{m}_w\left(x\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}=\sum _{k=0}^n\sum _{m=0}^k{a}_{k-m,m}{y}^m\left(T\right){\int }_{C_0[0,T]}{x}^{k-m}\left(T\right)f\left(x\left(T\right)\right)d{m}_w\left(x\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}=\sum _{k=0}^n\sum _{m=0}^k{b}_{k-m,m}\mathcal{L}(f,u^{k-m})y^m\left(T\right), \end{gathered}$$

which completes the proof of the Lemma 2. □

We next give a convergence formula for the kernel functionals.

Lemma 3.

Let $G(x,y)=g\left(x\right(T),y(T\left)\right)$ and let $G_n(x,y)=g_n(x\left(T\right),y\left(T\right))$, where g is as in Remark 2 and $g_n$ is as in Lemma 2. Then

$$G(x,y)=\underset{n\to \infty }{lim}{G}_n(x,y)$$

(12)

in the sense of $L_1\left(C_0^2[0,T]\right)$.

Proof. 

First, we note that

$$\begin{gathered} |g(u,v)-g_n(u,v)|\le |g(u,v)|+|g_n(u,v)| \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\le A_{g}exp\{B_g\left(\right|u|+\left|v\right|\left)\right\}+\sum _{k=0}^n\sum _{m=0}^k|a_{k-m,m}{\left|\right|u|}^{k-m}{\left|v\right|}^m. \end{gathered}$$

Now, let

$$h(u,v)=A_{g}exp\{B_g\left(\right|u|+\left|v\right|\left)\right\}+\sum _{k=0}^n\sum _{m=0}^k|a_{k-m,m}{\left|\right|u|}^{k-m}{\left|v\right|}^m.$$

Then we see that

$${\int }_{\mathbb{R}}{\int }_{\mathbb{R}}h(u,v)exp\left\{-{\displaystyle \frac{u^2+v^2}{2T}}\right\}dudv<\infty .$$

We next use Equations (9), (10) and (1) to establish Equation (12). Then one can see that

$$\begin{gathered} {\int }_{C_0[0,T]}{\int }_{C_0[0,T]}|G(x,y)-G_n(x,y)|d{m}_w\left(x\right)d{m}_w\left(y\right) \\ \hspace{1em}\hspace{1em}={\displaystyle \frac{1}{2\pi T}}{\int }_{\mathbb{R}}{\int }_{\mathbb{R}}|g(u,v)-g_n(u,v)|exp\left\{-{\displaystyle \frac{u^2+v^2}{2T}}\right\}dudv. \end{gathered}$$

Thus, using the Dominated convergence theorem, we can conclude that

$$G(x,y)=\underset{n\to \infty }{lim}{G}_n(x,y)$$

in the sense of $L_1\left(C_0^2[0,T]\right)$. Hence we have the desired result. □

The following Theorem 3 below is the main result in this paper. Equation (13) is called the approximation formula of the integral transform ${\mathbb{T}}^G\left(F\right)$ of F given G.

Theorem 3

(Approximation formula of the integral transform). Let F be as in Lemma 2 and let G and $G_n$ be as in Lemma 3. Then the integral transform ${\mathbb{T}}^G\left(F\right)$ of F given G exists and is given by the formula

$${\mathbb{T}}^G\left(F\right)\left(y\right)=\underset{n\to \infty }{lim}{\mathbb{T}}^{G_n}\left(F\right)\left(y\right)$$

(13)

in the sense of $L_1\left(C_0[0,T]\right)$.

Proof. 

Using Equations (2) and (12), we have

$$\begin{gathered} {\int }_{C_0[0,T]}|{\mathbb{T}}^G\left(F\right)\left(y\right)-{\mathbb{T}}^{G_n}\left(F\right)\left(y\right)|d{m}_w\left(y\right) \\ \le {\int }_{C_0[0,T]}{\int }_{C_0[0,T]}\left|F\left(x\right)\right||G(x,y)-G_n(x,y)|d{m}_w\left(x\right)d{m}_w\left(y\right) \\ \to 0 \end{gathered}$$

as $n\to \infty$. Hence we have the desired result. □

We give a few examples to explain the usefulness of Theorem 3 above.

Example 5.

Let $g(u,v)=sin\left(u+v+{\displaystyle \frac{\pi }{2}}\right)$ and let $G(x,y)=g\left(x\right(T),y(T\left)\right)$. Then $G\in E^{\left(2\right)}$ and we see that

$${\displaystyle \frac{{\partial }^{k}g}{\partial u^{k-m}\partial v^m}}{|}_{(u,v)=(0,0)}=\left\{\begin{array}{cc}0,\hfill & k=2q+1\hfill \\ -1,\hfill & k=4q+2\hfill \\ 1,\hfill & k=4q\hfill \end{array}\right.$$

for $q=0,1,2,\dots$. Hence using Equation (13), we have

$${\mathbb{T}}^G\left(F\right)\left(y\right)=\underset{n\to \infty }{lim}{\mathbb{T}}^{G_n}\left(F\right)\left(y\right)$$

in the sense of $L_1\left(C_0[0,T]\right)$, where

$${\mathbb{T}}^{G_n}\left(F\right)\left(y\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}\sum _{k=0}^n\left({\displaystyle \frac{{(-1)}^k+1^k}{2}}\right)\sum _{m=0}^k{\displaystyle \frac{{(-1)}^{s\left(k\right)}\mathcal{L}(f,u^{k-m})}{(k-m)!m!}}{y}^m\left(T\right),$$

where

$${(-1)}^{s\left(k\right)}=\left\{\begin{array}{cc}-1,\hfill & k=4r-2\hfill \\ 1,\hfill & otherwise\hfill \end{array}\right.$$

for $r=1,2,\dots$.

Example 6.

Let $g(u,v)={\displaystyle \frac{1}{1-u-v}}$ and let $G(x,y)=g\left(x\right(T),y(T\left)\right)$. Then $G\in E^{\left(2\right)}$ and we see that

$${\displaystyle \frac{{\partial }^{k}g}{\partial u^{k-m}\partial v^m}}{|}_{(u,v)=(0,0)}={\displaystyle \frac{(k+m)!}{k!m!}}.$$

Hence using Equation (13), we have

$${\mathbb{T}}^G\left(F\right)\left(y\right)=\underset{n\to \infty }{lim}{\mathbb{T}}^{G_n}\left(F\right)\left(y\right)$$

in the sense of $L_1\left(C_0[0,T]\right)$, where

$${\mathbb{T}}^{G_n}\left(F\right)\left(y\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}\sum _{k=0}^n\sum _{m=0}^k{\displaystyle \frac{\mathcal{L}(f,u^{k-m})(k+m)!}{(k-m)!k!{(m!)}^2}}{y}^m\left(T\right).$$

Example 7.

Let $g(u,v)=sin\left(2u\right)+cos\left(v\right)$ and let $G(x,y)=g\left(x\right(T),y(T\left)\right)$. Then $G\in E^{\left(2\right)}$ and we see that using Equation (13), we have

$${\mathbb{T}}^G\left(F\right)\left(y\right)=\underset{n\to \infty }{lim}{\mathbb{T}}^{G_n}\left(F\right)\left(y\right)$$

in the sense of $L_1\left(C_0[0,T]\right)$, where

$${\mathbb{T}}^{G_n}\left(F\right)\left(y\right)={\displaystyle \frac{1}{\sqrt{2\pi T}}}[1+\sum _{k=1}^n\sum _{m=0}^k({(-1)}^{l\left(k\right)}{2}^k+{(-1)}^{l\left(k\right)+1}){\displaystyle \frac{\mathcal{L}(f,u^{k-m})}{(k-m)!m!}}{y}^m\left(T\right)]$$

where

$$l\left(k\right)=\left\{\begin{array}{cc}1,\hfill & k=4r-2\hfill \\ -1,\hfill & k=4r\hfill \\ 0,\hfill & otherwise\hfill \end{array}\right.$$

for $r=1,2,\dots$.

4. Conclusions

As illustrated in the examples of Section 2 and Section 3, there are numerous instances in which the structure of the kernel renders direct computation of the integral transform difficult or impractical. Nevertheless, by employing the theorem presented in Section 3, we have demonstrated that a wide class of integral transforms can be expressed in the form of convergent series representations.