数理解析研究所講究録
第 1797 巻 2012 年 20-41

20

A Note on the Integral Transforms on a Function Space I,II
By

Bong Jin KIM

Abstract
In this paper we obtain some results of integration by parts formulas involving integral transforms
denotes thefor s-a.e. $y\in C_{0}[0, T]$ , whereof functionals of the form
. Furthermore we obtain the various relationships that exist among
Riemann-Stieltjes integral
the integral transform, the convolution product and the first variation for a class of functionals defined
on $K(Q)$ , the space of complex-valued continuous functions on $Q=[0,S]\cross[0,T]$ which satisfy $x(s,0)=$$x(O,t)=0$ for all $0\leq s\leq S$ and $0\leq t\leq T$ . Also we obtain Parseval‘s and Plancherel‘s relations for the
integral transform of some functionals defined on $K(Q)$ .
$F(y)=f(\langle\theta_{1},y\rangle,$$(\theta_{n},y\})$$\langle\theta,y\rangle$$\ldots,$$\int_{0}^{T}\theta(t)dy(t)$\S 1. Parts formulas involving integral transforms on function spaceof analytic functionals on
In a unifying paper [15], Lee defined an integral transforman abstract Wiener space. For certain values of the parameters and and for certain classesof functionals, the Fourier-Wiener transform [3], the Fourier-Feynman transform [4] and the. In [6], Chang, Kim and YooGauss transform are special cases of his integral transform
established an interesting relationship between the integral transform and the convolution product for functionals on an abstract Wiener space. In this paper we establish several integration
by parts formulas involving integral transforms, convolution products, and the first variations ofdenotes
for s-a.e. $y\in C_{0}[0, T]$ , where
functionals of the form $F(y)=f(\{\theta_{1},y\rangle,$.
the Riemann-Stieltjes integral
Let $C_{0}[0, T]$ denote one-parameter Wiener space; that is the space of all real-valued contindenote the class of all Wiener measurable
uous functions $x(t)$ on $[0,T]$ with $x(O)=0$ . Let
subsets of $C_{0}[0,T]$ and let denote Wiener measure. $(C_{0}[0, T],\mathcal{M},m)$ is a complete measurespace and we denote the Wiener integral of a Wiener integrable functional by$\mathcal{F}_{\alpha\beta}$$\alpha$$\beta$$\mathcal{F}_{\alpha\beta}$$(\theta,y\rangle$$\langle\theta_{n},y\rangle)$$\ldots,$$\int_{0}^{T}\theta(t)dy(t)$$\mathcal{M}$

$m$

$F$

(1.1)

$\int_{C_{0}[0,T]}F(x)m(dx)$.

2010 Mathematics Subject Classification(s): $28C20$ .
Key Words: integral transform, convolution product, first variation, integration by parts formula, Wiener integral,
Yeh-Wiener integral, Parseval’s relation, Plancherel‘s relation

21
be nonzero complex numbers. Next we state the definitions of the integral, the convolution productand the first variation
for functionals
defined on $K=K[0, T]$ , the space of complex-valued continuous functions defined on $[0, T]$
which vanish at $t=0$ .

Let andtransform$\alpha$$\beta$$(F*G)_{\alpha}$$\mathcal{F}_{\alpha\beta}F$

Definition 1.1. Let
is defined by
(1.2)

$F$

be a functional defined on$\delta F$$K$. Then the integral transform$\mathcal{F}_{\alpha\beta}(F)(y)\equiv \mathcal{F}_{\alpha_{:}\beta}F(y)\equiv\int_{C_{0}[0,T]}F(\alpha x+\beta y)m(dx)$,$\mathcal{F}_{\alpha,\beta}F$

of

$F$

$y\in K$

if it exists [6, 12, 13, 15].

Definition 1.2. Let and be functionals defined onof and is defined by$F$$(F*G)_{(\chi}$

$F$

$G$

$K$

. Then the convolution product

$G$

(1.3)

,

$(F*G)_{\alpha}(y) \equiv\int_{C_{0}[0,T]}F(\frac{y+\alpha x}{\sqrt{2}})G(\frac{y-\alpha x}{\sqrt{2}})m(dx)$$y\in K$

if it exists[6, 10, 12, 19, 21].

$\delta F$

Definition 1.3. Let
of is defined by$F$

be a functional defined on$K$and let $w\in K$ . Then the first variation

$F$

(1.4)

$\delta F(y|w)\equiv\frac{\partial}{\partial t}F(y+tw)|_{t=0}$,$y\in K$

if it exists [2, 5, 12, 17].
be a complete orthonormal set of real-valued functions in $L_{2}[0, T]$ and assume that each is of bounded variation on $[0, T]$ . Then for each $y\in K$ and $j\in\{1,2, \ldots\}$ , theRiemann-Stieltjes integral
exists. Furthermore
Let$\{\theta_{1},\theta_{2}, \ldots\}$$\theta_{j}$$\langle\theta_{j},y\}\equiv\int_{0}^{T}\theta_{j}(t)dy(t)$(1.5)$| \langle\theta_{j},y\}|=|\theta_{j}(T)y(T)-\int_{0}^{T}y(t)d\theta_{j}(t)|\leq C_{j}\Vert y\Vert_{\infty}$

with
(1.6)

$C_{j}=|\theta_{j}(T)|+Var(\theta_{j}, [0, T])$,

where $Var(\theta_{j}, [0, T])$ denote the total variation of on $[0, T]$ .
Next we describe the class of functionals which is related to this paper. For $0\leq\sigma<1$ , let
be the space of all functionals :of the form$\theta_{j}$$E_{\sigma}$

(1.7)

$F$

$Karrow \mathbb{C}$$F(y)=f(\{\vec{\theta},y\})=f((\theta_{1},y\rangle, \ldots, \langle\theta_{n},y\})$22for some positive integer , whereof exponential type; that is to say,variables$f(\vec{\lambda})=f(\lambda_{1}, \ldots,\lambda_{n})$$n$$\lambda_{1},$is an entire function of the$n$complex$\ldots,\lambda_{n}$(1.8)$|f(\vec{\lambda})|\leq A_{F}\exp\{B_{F}|\vec{\lambda}|^{1+\sigma}\}$for some po $:;itive$ constants$A_{F}$

and$B_{F}$, where$| \vec{\lambda}|^{1+\sigma}=\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}$.

In addition we use the notation$F_{j}(y)=f_{j}(\{\vec{\theta},y\rangle)$.
for $j=1,$
where
Recently [12], Kim, Kim and Skoug established the results that if and are elements$(F*G)_{a},$and examined varand
are also elements ofofthen.
$(F*G)_{a},$. For related work see
andious relationships holding among[3, 6, 10, 12, 15, 17, 19, 21] and for a detailed survey of previous work see [18].
We introduce the following three existence theorems for the integral transform, the convo.
lution $produ|.t$ and the first variation of functionals in$\ldots,n$$f_{j}(/1)arrow=\tau_{\lambda_{j}}^{\partial_{-f(/l_{1},\ldots,l_{n})}}$’$F$$E_{\sigma}$$\delta F(y|\cdot)$$\delta F(\cdot|w)$$F_{\alpha\beta}F,$$\mathcal{F}_{a\beta}F,$$G$$E_{\sigma}$$\delta F$$\mathcal{F}_{a,\beta}G,$$\delta G$$E_{\sigma}[12]$be given by (1.7). Then the integml transform
Theorenn 1.4. Let
belongs to . and is given by the formula$F\in E_{\sigma}$$\mathcal{F}_{\alpha\beta}F$

exists,$E_{\sigma}$(1.9)$\mathcal{F}_{a\beta}F(y)=h(\{\vec{\theta},y\})$for $y\in K,$$w/^{1}lere$(1.10)where$h( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\alpha\vec{u}+\beta\vec{\lambda})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$$\Vert\vec{u}\Vert^{2}:=\sum_{j=1}^{n}u_{J^{2}}$and$d\vec{u}=du_{1}\cdots du_{n}$.

be given by (1.7) with corresponding entirefunctions $f$ and ,Theoren11.5. Let
exists, belongs toand is given by the fomularespectively. Then the convolution$F,G\in E_{\sigma}$$g$$(F*G)_{\alpha}$(1.11)$E_{\sigma}$$(F*G)_{\alpha}(y)=k(\langle\vec{\theta},y\rangle)$for $y\in K,$ wjlere(1.12)

Theorem 1.6. Let(1.13).

$k( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\frac{\vec{\lambda}+\alpha\tilde{u}}{\sqrt{2}})g(\frac{\vec{\lambda}-\alpha\vec{u}}{\sqrt{2}})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$$F\in E_{\sigma}$be given by (1.7) and let $w\in K$ . Then$\delta F(y|w)=p(\{\vec{\theta},y\})$23
for $y\in K$, where(1.14)$p( \vec{\lambda})=\sum_{j=1}^{n}\{\theta_{j},w\}f_{j}(\vec{\lambda})$.

Furthermore, as afimction of $y\in K,$ $\delta F(y|w)$ is an element of$E_{\sigma}$.

Now we state some observations which we use later in this paper. First of all, equation (1.2)
implies that(1.15)$\mathcal{F}_{\alpha\beta}F(y/\sqrt{2})=\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)$for all $y\in K$ . Next, a direct calculation using (1.4), (1.2), (1.13) and (1.15) shows that$\delta \mathcal{F}_{\alpha,\beta}F(y/\sqrt{2}|w/\sqrt{2})=\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}F(y|w)$(1.16)$= \frac{\beta}{\sqrt{2}}\sum_{j=1}^{n}\{\theta_{j},w\rangle \mathcal{F}_{\alpha\beta/\sqrt{2}}F_{j}(y)$for all and
$y$

$w$

in . Finally, by similar calculations, we obtain that$K$

(1.17)$\mathcal{F}_{\alpha\beta}(\delta F(\cdot|w))(y/\sqrt{2})=\frac{\sqrt{2}}{\beta}\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}F(y|w)$

for all and
$y$

$w$

in

$K$

, and for all $y\in K$ ,(1.18)$(\mathcal{F}_{\alpha\beta}F)_{j}(y)=\beta \mathcal{F}_{\alpha,\beta}F_{j}(y)$

.

Let
$A=$

{

$y\in C_{0}[0,$$T]:y$ is absolutely continuous

on $[0,$$T]$

with $y’\in L^{2}[0,$ $T]$ }.

We note that if we choose $z\in L_{2}[0, T]$ and define $w(t)= \int_{0}^{t}z(s)ds$ for $t\in[0, T]$ , then is anelement of $A,$ $w’=z$ a.e. on $[0, T]$ , and for all $v\in L_{2}[0, T],$ $\{v, w\}=(v,w’)=(v,z)$ , where$(v,z)= \int_{0}^{T}v(s)z(s)ds$ .
The following theorem plays a key role throughout this paper. In this theorem the Wiener
integral of the first variation of functional is expressed in terms of the Wiener integral of
multiplied by a linear factor.
$w$

$F$

Theorem 1.7. Let(1.19)$F\in E_{\sigma}$be given by (1.7) and $w\in A$ , then$\int_{C_{0}[0,T]}\delta F(x|w)m(dx)=\int_{C_{0}[0,T]}F(x)\{z,x\rangle m(dx)$where $w(t)= \int_{0}^{t}z(s)ds$ on $[0, T]$ for some $z\in L_{2}[0, T]$ .

$F$

24
Proof. Let$w(t)= \int_{0}^{t}z(s)ds$

find an orthonormal set

for some $z\in L^{2}[0,T]$ . Using the Gram-Schmit process we can, wherewith$\{\theta_{1}, \ldots,\theta_{n},\theta_{n+1}\}$$\theta_{n+1}=\frac{1}{\Vert z_{n+1}||}z_{n+1}$$z_{n+1}=z- \sum_{j=1}^{n}(\theta_{j},z)\theta_{j}$.

Then by the Wiener integration formula$\int_{C_{0}[0,T]}F(x)\{z,x\}m(dx)$$= \int_{C_{0}[0,7]}f(\langle\vec{\theta},x\})(\sum_{j=1}^{n}(\theta_{j},z)\langle\theta_{j},x\}+\Vert z_{n+1}\Vert\{\theta_{n+1},x\})m(dx)$.

$=(2 \pi)^{-(\mathfrak{l}2+1)/2}\int_{\mathbb{R}^{n+1}}f(u\gamma(\sum_{j=1}^{n}(\theta_{j},z)u_{j}+\Vert z_{n+1}\Vert u_{n+1})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}-\frac{1}{2}u_{n+1}^{2}\}du_{n+1}d\vec{u}$If we evaluate the last integral with respect to$u_{n+1}$, we obtain.

$\int_{C_{0}[0,T]}F(x)\{z,x\}m(dx)=(2\pi)^{-n/2}\sum_{j=1}^{n}(\theta_{j},z)\int_{\mathbb{R}^{n}}f(u\gamma u_{j}\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$On the other hand, since $w\in A\subset K$ , by Theorem 1.6.

$\delta F(x|w)=\sum_{j=1}^{n}\{\theta_{j},w\}f_{j}(\{\vec{\theta},x\})=\sum_{j=1}^{n}(\theta_{j},z)f_{j}(\{\vec{\theta},x\})$Hence by$th|^{\sim}$

Wiener integration formula$(_{C_{0}[0,T]} \delta F(x|w)m(dx)=\sum_{j=1}^{n}(\theta_{j},z)\int_{q)[0,T]}f_{j}(\{\vec{\theta},x\})m(dx)$.

$=(2 \pi)^{-n/2}\sum_{j=1}^{n}(\theta_{j},z)\int_{\mathbb{R}^{n}}f_{j}(\vec{u})\exp\{-\frac{1}{2}\Vert u\Vert^{2}\}d\vec{u}$Note that for each $j=1,$$\ldots,n$, the integration by parts formula yields$\int_{\mathbb{R}}f_{j}(u\gamma\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}$.

$= \lim_{abarrow\infty}\lim_{arrow-\infty}[f(\vec{u})\exp\{-\frac{1}{2}u_{j}^{2}\}]_{a}^{b}+\int_{\mathbb{R}}f(\vec{u})u_{j}\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}$But since$f1\lfloor S$

of exponential type, the double limit in the last equation is equal to$\int_{\mathbb{R}}f_{j}(\hat{u})\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}=\int_{\mathbb{R}}f(u\gamma u_{j}\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}$.

$0$and so25
Hence$\int_{C_{0}[0,T]}\delta F(x|w)m(dx)=(2\pi)^{-n/2}\sum_{j=1}^{n}(\theta_{j},z)\int_{\mathbb{R}^{n}}f(u\gamma u_{j}\exp\{-\frac{1}{2}\Vert u\Vert^{2}\}d\vec{u}$and this completes the proof.

$\square$In our next theorem we obtain an integration by parts formula for the products of functionals

in$E_{\sigma}$.

Theorem 1.8. Let
be given by (1.7) with corresponding entire fiunctions $f$ and$w\inA$, respectively. Then for, we have the following integmtion by parts formula.
$F,$ $G\in E_{\sigma}$$g$(1.20)$\int_{C_{0}[0,T]}[F(y)\delta G(y|w)+\delta F(y|w)G(y)]m(dy)=\int_{C_{0}[0,T]}F(y)G(y)(z,y\rangle m(dy)$,where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ .

Proof. Define $H(y)=F(y)G(y)$ for $y\in K$ and let$h(/t)arrow=f(\vec{A})g(/t)arrow$. Then$H\in E_{\sigma}$and$\delta H(y|w)=\sum_{j=1}^{n}\{\theta_{j},w\rangle f_{j}(\{\vec{\theta},y\})g(\{\vec{\theta},y\rangle)+f(\langle\vec{\theta},y\})\sum_{j=1}^{n}\{\theta_{j},w\rangle g_{j}(\{\vec{\theta},y\})$$=\delta F(y|w)G(y)+F(y)\delta G(y|w)$.

Thus equation (1.20) follows from Theorem 1.7.

$\square$

By choosing $G=F$ in Theorem 1.8, we obtain the following corollary.

Corollary 1.9. Let(1.21)$F\in E_{\sigma}$be given by (1.7). Then for each $w\in A$ ,,$\int_{C_{0}[0,T]}F(y)\delta F(y|w)m(dy)=\frac{1}{2}\int_{C_{0}[0,T]}[F(y)]^{2}\{z,y\}m(dy)$where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ .

As we saw in Theorem 1.6 above if belongs to, thenalso belongs to
as a function of . Thus if we replace $G(y)$ with $\delta F(y|w1)$ in Theorem 1.8, then we have thefollowing corollary.
$F$$E_{\sigma}$$\delta F(y|w_{1})$$y$Corollary 1.10. Let$F\in E_{\sigma}$be given by (1.7). Then for each $w_{1},w2\in A$ ,$\int C_{0}[0,T]^{[F(y)\delta^{2}F(\cdot|w_{1})(y|w_{2})+\delta F(y|w)\delta F(y|w_{1})]m(dy)}2$(1.22)$= \int C_{0}[0,T]^{F(y)\delta F(y|w)\langle Z2,\mathcal{Y}\rangle m(dy)}l$

where$w_{i}(t)= \int_{0}^{t}z_{i}(s)ds$

for some$z\iota\in L_{2}[0, T],$

$i=1,2$ .

$E_{\sigma}$

26
also belongs to
As we saw in Theorem 1.4 above if belongs to , thenin Theorem 1.8, then we have the following corollary.
if we replace with
$G$

$G$

$E_{\sigma}$$\mathcal{F}_{a\beta}G$$E_{\sigma}$. Thus$\mathcal{F}_{a}pG$

Corollary 1.11. Let$F,G\in E_{\sigma}$be given as in Theorem 1.8. Then for each $w\in A$ ,$\int_{C_{0}[0,T]}[F(y)\delta \mathcal{F}_{a\beta}G(y|w)+\delta F(y|w)\mathcal{F}_{\alpha\beta}G(y)]m(dy)$(1.23)$= \int_{C_{0}[0,T]}F(y)\mathcal{F}_{a}pG(y)\langle z,y)m(dy)$,where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0,T]$ .
By replacing andfollowing corollary.
$F$

Corollary 1.12. Let

$G$

by

$\mathcal{F}_{a,\beta}F$$F,G\in E_{\sigma}$and$\mathcal{F}_{a\beta}G$, respectively, in Theorem 1.8, we obtain the

be as in Theorem 2.5. Then for each $w\in A$ ,$\int_{ci)[0,T]}[\mathcal{F}_{\alpha\beta}F(y)\delta \mathcal{F}_{a\beta}G(y|w)+\delta \mathcal{F}_{a\beta}F(y|w)\mathcal{F}_{\alpha\beta}G(y)]m(dy)$(1.24),$= \int_{cb[0,T]}\mathcal{F}_{a\beta}F(y)\mathcal{F}_{a\beta}G(y)(z,y\rangle m(dy)$where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ .

\S 2. Various integration formulas and examples
In this section we establish various integration formulas involving integral transforms, convolution products and first variations. Furthermore we give some examples to illustrate theintegration formulas in this paper.
,
In [12], Kim, Kim and Skoug established various relationships holding among$(F*G)_{a},$and . From these relationships and the results in Section 1 above, we can establish various integration formulas.
From Theorem 1.7 above we know that the Wiener integral of the first variation of functional
is expressed in terms of the Wiener integral of multiplied by a linear factor. On theother hand, some of the formulas, for example, Formulas 3.3, 3.5, 4.1, 4.2, and 5.2 in [12] giveus the expressions of the first variation of various functionals. Hence it is easy to obtain thefollowing formulas (2.1) through (2.7) below. We just state the formulas without proofs.
Let $w\in A$ with $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ throughout this section. The paper[12] was concemed with the class. Bm as commented in Remark 5.6 of that paper, all thein Formula
formulas in [12] still true for functionals in . Hence we will assume that
2.1 through Formula 2.6 and Corollary 2.7 below.
$\mathcal{F}_{\alpha_{I}\beta}F,$$\delta F$$\delta G$$F$$F\in E_{\sigma}$$E_{0}$$E_{\sigma}$$F\in E_{\sigma}$$\mathcal{F}_{a\beta}G$27
Formula 2.1. From Formula 3.3 of [7], we have
(2.1)$\beta\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta}\delta F(\cdot|w)(y)m(dy)=\int_{C_{0}[0,T]}\mathcal{F}_{a\beta}F(y)\{z,y\}m(dy)$.

Formula 2.2. From Formula 3.5 of [12], we have$\sum_{j=1}^{n}\frac{\langle\theta_{j},w\rangle}{\sqrt{2}}l_{C_{0}[0,T]}[(F_{j}*G)_{\alpha}(y)+(F*G_{j})_{\alpha}(y)]m(dy)$(2.2)$= \int_{C_{0}[0,T]}(F*G)_{(\chi}(y)\{z,y\rangle m(dy)$and if $F=G$ ,
(2.3)

.

$\sqrt{2}\sum_{j=1}^{n}\{\theta_{j}, w\}\int_{C_{0}[0,T]}(F*F_{j})_{\alpha}(y)m(dy)=\int_{C_{0}[0,T]}(F*F)_{\alpha}(y)\{z,y\}m(dy)$Formula 2.3. From Formula 4.1 of [12], we have$\int_{C_{0}[0,T]}(\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}G(y|w)+\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}F(y|w)\mathcal{F}_{\alpha\beta/\sqrt{2}}G(y))m(dy)$(2.4)$= \int_{C_{0}[0,T]}(\mathcal{F}_{\alpha,\beta}F(\frac{y}{\sqrt{2}})\delta \mathcal{F}_{\alpha\beta}G(\frac{y}{\sqrt{2}}|\frac{w}{\sqrt{2}})+\delta \mathcal{F}_{r\beta}F(\frac{y}{\sqrt{2}}|\frac{w}{\sqrt{2}})\mathcal{F}_{cx\beta}G(\frac{y}{\sqrt{2}}))m(dy)$$= \int_{C_{0}[0,T]}\mathcal{F}_{\alpha/3}(F*G)_{\alpha}(y)\langle z,y\rangle m(dy)$and if $F=G$ ,$\int_{C_{0}[0,T]}\mathcal{F}_{\alpha,\beta}[F(\frac{y}{\sqrt{2}})][\delta \mathcal{F}_{\alpha\beta}F(\frac{y}{\sqrt{2}}|\frac{w}{\sqrt{2}})]m(dy)$(2.5)$= \frac{1}{2}\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta}(F*F)_{\alpha}(y)\{z,y\}m(dy)$.

Formula 2.4. From Formula 4.2 of [12], we have$\frac{\beta}{\sqrt{2}}\sum_{j=1}^{n}\{\theta_{j},w\}\int_{C_{0}[0,T]}[(\mathcal{F}_{a\beta}F_{j}*\mathcal{F}_{\alpha\beta}G)_{\alpha}(y)+(\mathcal{F}_{\alpha\beta}F*\mathcal{F}_{\alpha\beta}G_{j})_{a}(y)]m(dy)$(2.6)$= \int_{C_{0}[0,T]}(\mathcal{F}_{\alpha\beta}F*\mathcal{F}_{\alpha\beta}G)_{\alpha}(y)\{z,y\}m(dy)$.

Formula 2.5. From Formula 5.2 of [12] we have,$\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta}(\delta F(\cdot|w)G(\cdot)+F(\cdot)\delta G(\cdot|w))(\frac{y}{\sqrt{2}})m(dy)$

(2.7)

.

$= \frac{\sqrt{2}}{\beta}\int_{C_{0}[0,T]}(\mathcal{F}_{(x\beta}F*\mathcal{F}_{\alpha\beta}G)_{i\alpha/\beta}(y)\langle z,y\}m(dy)$28
Using the equations (1.10) through (1.12) we obtain the following integration formula for
the Wiener integral of the integral transform with respect to the first argument of the variation.
Formula 2.6. For(2.8)$F\in E_{\sigma}$

we have.

$\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}(\delta F(\cdot|w))(y)m(dy)=\sum_{j=1}^{n}\langle\theta_{j},w\}\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}F_{j}(y)m(dy)$Proof. By (1.16) and Theorem 1.7 we have$\frac{\beta}{\sqrt{2}}\sum_{j=1}^{n}\langle\theta_{j},w\}\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}F_{j}(y)m(dy)=\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}F(y)\{z,y\rangle m(dy)$.

Similarly by (1.17) and Theorem 1.7 we have$\frac{\beta}{\sqrt{2}}\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta/\sqrt{2}}(\delta F(\cdot|w))(\frac{y}{\sqrt{2}})m(dy)=\int_{C_{0}[0,T]}\mathcal{F}_{l\beta/\sqrt{2}}F(y)\langle z,y\}m(dy)$.

Thus we have the above formula (2.8).

$\square$

We next obtain an integration formula for functionals which is a product of elements ofby some linear factors.

Corollary 2.7. Let
and let$k$$E_{\sigma}$be a natural number and let $z_{j}\in L_{2}[0, T]$ for $j=1,2,$ $\ldots,k+1$ . Let$F\in E_{\sigma}$$F^{[k]}(y)=F^{[k-1]}(y) \langle z_{k},y\rangle=F(y)\prod_{j=1}^{k}\langle Zj,\mathcal{Y}\rangle$.

Then we have the following integml equation.

$\int_{C_{0}[0,T]}F^{[k+1]}(y)m(dy)$(2.9)$= \int_{q)[0,T]^{\delta F^{[k-1]}(y|w_{k+1})\{Zk,\mathcal{Y}\}m(d_{\mathcal{Y}})+\{Zk}},w_{k+1}\rangle\int_{C_{0}[0,T]}F^{[k-1]}(y)m(dy)$where $F^{[0]}=F$ and $w_{k+1}(t)= \int_{0}^{t}z_{k+1}(s)ds$ .

Proof. To prove this theorem we simply take the first variation of the$F^{[k]}(y)=F^{[k-1]}(y)\{z_{k},y\}$Now we have$\delta F^{[k]}(y|w_{k+1})=\frac{\partial}{\partial t}(F^{[k-1]}(y+tw_{k+1})\langle zk,\mathcal{Y}+tw_{k+1}\})|_{t=0}$$=\delta F^{[k-1]}(y)\{z_{k},y\rangle+F^{[k-1]}(y)\{z_{k},w_{k+1}\rangle$

.

.

29
Hence we have$\int_{C_{0}[0,T]}\delta F^{[k]}(y|w_{k+1})m(dy)$,$= \int c_{0[0,T]^{\delta F^{[k-1]}(y)\langle z_{k},y\rangle m(dy)+\{w_{k+1}\}\int_{C_{0}[0,T]}F^{[k-1]}(y)m(dy)}}Zk$

But by Theorem 1.7,$\int_{C_{0}[0,T]}\delta F^{[k]}(y|w_{k+1})m(dy)=\int_{C_{0}[0,T]}\delta F^{[k]}\{z_{k+1},y\}m(dy)=\int_{C_{0}[0,T]}F^{[k+1]}(y)m(dy)$and this completes the proof.

$\square$We finish this section by giving some examples for the illustration of the integration by parts
formulas.
Example 2.8. Let$F(w)$ ,$F(y)= \sum_{j=1}^{n}\langle\theta_{j},y\}$which is an element of$E_{\sigma}$, then we have $\delta F(y|w)=$where $w(t)= \int_{0}^{t}z(s)ds\in A$ . Thus we can obtain.

$\int_{C_{0}[0,T]}\delta F(y|w)m(dy)=\int_{C_{0}[0,T]}F(w)m(dy)=\sum_{j=1}^{n}\{\theta_{j},w\rangle=\sum_{j=1}^{n}\int_{0}^{T}\theta_{j}(s)z(s)ds$

Since the constant functional
1.8 yields the following.

$G\equiv 1$

belongs to$E_{\sigma}$and its first variation equals to zero, Theorem.

(2.10)$\sum_{j=1}^{n}\int_{C_{0}[0,T]}\{\theta_{j},y\rangle\langle z,y\}m(dy)=\sum_{j=1}^{n}\int_{0}^{T}\theta_{j}(s)z(s)ds$

Example 2.9. Let$G(y)= \exp\{\sum_{j=1}^{n}\{\theta_{j},y\}\}$

which is an element of$\delta G(y|w)=\sum_{j=1}^{n}\{\theta_{j},w\}\exp\{\sum_{j=1}^{n}(\theta_{j},y\}\}=\sum_{j=1}^{n}\{\theta_{j},$$E_{\sigma}$, then we have$w\rangle G(y)$,where $w(t)= \int_{0}^{t}z(s)ds\in A$ . Hence by the Wiener integration formula$\int_{C_{0}[0,T]}\delta G(y|w)m(dy)=\sum_{j=1}^{n}\{\theta_{j},w\rangle\int_{C_{0}[0,T]}\sum_{j=1}^{n}\exp\{\langle\theta_{j},y\}\}m(dy)$$=e^{n/2} \sum_{j=1}^{n}(\theta_{j},w\}$.

From Theorem 1.8 we obtain the following Wiener integral.
(2.11).

$\int_{C_{0}[0,T]}\exp\{\sum_{j=1}^{n}\{\theta_{j},y\rangle\}\{z,y\rangle m(dy)=e^{n/2}\sum_{j=1}^{n}\{\theta_{j},w\rangle$

30
Example 2.10. Let$H(y)= \sum_{j=1}^{n}[\{\theta_{j},y\rangle]^{2}$

which is an element of$E_{\sigma}$, then we have,$\delta H(y|w)=2\sum_{j=1}^{n}\langle\theta_{j},w\rangle(\theta_{j},y\}=2\sum_{j=1}^{n}(\theta_{j},z)\langle\theta_{j},y\}$.

where $w(t)= \int_{0}^{t}z(s)ds\in A$ . Thus we can obtain the following by Wiener integration formula.
$\int_{C_{0}[0,T]}\delta H(y|w)m(dy)=2\sum_{/=1}^{n}(\theta_{j},z)\int_{C_{0}[0,T]}\{\theta_{j},y\rangle m(dy)=0$.

By Theorem 1.8 we have the following.
(2.12)$\sum_{j=1}^{n}\int_{C_{0}[0,T]}[\langle\theta_{j},y\rangle]^{2}\{z,y\rangle m(dy)=0$

Example 2.11. Let$L(y)=[ \sum_{j=1}^{n}\langle\theta_{j},y\rangle]^{2}$.

which is an element of$E_{\sigma}$, then we have,.

$\delta \mathcal{F}_{\alpha\beta}L(y|w)=2\beta^{2}\sum_{j=1}^{n}(\theta_{j},w\rangle\sum_{j=1}^{n}\langle\theta_{j},y\rangle$

From the Wiener integration formula, we have the followings.
$\int_{C_{0}[0,T]}\delta \mathcal{F}_{a\beta}L(y|w)m(dy)=\psi^{2}\sum_{j=1}^{n}(\theta_{j},w\rangle\int_{C_{0}[0,T]}\sum_{j=1}^{n}\langle\theta_{j},y\rangle m(dy)=0$

and$\int_{C_{0}[0,T]}\mathcal{F}_{a\beta}L(y)\langle z,y\}m(dy)$$= \int_{C_{0}[0,T]}[n\alpha^{2}+|\beta\sum_{j=1}^{n}\{\theta_{j},y\rangle]^{2}]\{z,y\rangle m(dy)$.

$=0+ \beta^{2}\int_{C_{0}[0,T]}[\sum_{j=1}^{n}(\theta_{j},y\rangle]^{2}\langle z,y\rangle m(dy)$From the Corollary 1.11 we have the following.

(2.13)$\int_{C_{0}[0,T]}[\sum_{j=1}^{n}\langle\theta_{j},y\}]^{2}\{z,y\rangle m(dy)=0$.

31
Examples 2.8 through 2.11 are interesting to note that we can obtain the Wiener integrals on
the left hand side of (2.10) through (2.13) by using Theorem 1.8 or Corollary 1.11 rather thandirect calculation using Wiener integration formula.

\S 3. Integral transforms of functionals on function space of two variables
Recently [12] Kim, Kim and Skoug studied the relationships that exist among the integral
transform, the convolution product and the first variation for functionals defined on $K[0, T]$ , thespace of complex-valued continuous functions on $[0, T]$ which vanish at zero. In this paper we
extend the results in [12] for functionals of two variables.
Let $Q=[0,S]\cross[0, T]$ and let $C(Q)$ denote Yeh-Wiener space; that is, the space of all realvalued continuous functions $x(s,t)$ on with $x(s,O)=x(0,t)=0$ for all $0\leq s\leq S$ and $0\leq t\leq$. Yeh [18] defined a Gaussian measure
on $C(Q)$ (later modified in [20]) such that as astochastic process $\{x(s,t);(s,t)\in Q\}$ has mean $E[x(s,t)]=0$ and covariance $E[x(s,t)x(u,v)]=$$\min\{s,u\}\min\{t,v\}$ .
Letdenote the class of all Yeh-Wiener measurable subsets of $C(Q)$ and we denote the
Yeh-Wiener integral of a Yeh-Wiener integrable functional by
$Q$

$T$

$m_{Y}$$\mathcal{M}$

$F$

(3.1)

$\int_{C(Q)}F(x)m_{Y}(dx)$.

Let $K(Q)$ be the space of complex-valued continuous functions defined on and satisfying$x(s,0)=x(0,t)=0$ for all $0\leq s\leq S$ and $0\leq t\leq T$ . Let and be nonzero complex numbers.
Next we state the definitions of the integral transform, the convolution product$K(Q)$and the first variation
for functionals defined on.
$Q$$\beta$$\alpha$$(F*G)_{\alpha}$$\mathcal{F}_{\alpha\beta}F$$\delta F$

Definition 3.1. Let
of is defined by$F$

be a functional defined on $K(Q)$ . Then the integral transform$\mathcal{F}_{\alpha\beta}F$

$F$

(3.2)

,

$\mathcal{F}_{\alpha fl}F(y)=\int_{C(Q)}F(\alpha x+\beta y)m_{Y}(dx)$$y\in K(Q)$

if it exists [6, 12, 13, 15].

It is obvious that (3.2) implies that$\mathcal{F}_{\alpha\beta}F(\frac{y}{\sqrt{2}})=\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)$for all $y\in K(Q)$ .
Definition 3.2. Let and be functionals defined on $K(Q)$ . Then the convolution productof and is defined by$F$$(F*G)_{\alpha}$

(3.3)

$F$

$G$

$G$

,

$(F*G)_{\alpha}(y)= \int_{C(Q)}F(\frac{y+\alpha x}{\sqrt{2}})G(\frac{y-\alpha x}{\sqrt{2}})m_{Y}(dx)$

if it exists [6, 10, 12, 19,21].

$y\in K(Q)$32Definition 3.3. Let be a functional defined on $K(Q)$ and let $w\in K(Q)$ . Then the firstof is defined byvariation$F$$\delta F$

$F$

(3.4)

$\delta F(y|w)=\frac{\partial}{\partial t}F(y+tw)|_{t=0}$,$y\in K(Q)$

if it exists [2,5, 14, 16].
Now we introduce a concept of the function of bounded variation of two variables, and an
integration by parts formula for a Riemann-Stieltjes integral for functions of two variables. The
concept of bounded variation for a function of two variables is surprisingly complex. There areseveral nonequivalent definitions. The paper [7] by Clarkson and Adams is useful in sortingout many of the relationships between the various definitions. In this paper we will use thedefinition used by Hardy and Krouse [1,11] which we now review.
Let $R=[a,b]\cross[c,d]$ and let be a partition of given by
$R$

$P$

$a=s0<s_{1}<\cdots<s_{n}=b$ ,$c=t_{0}<t_{1}<\cdots<t_{m}=d$

A function $f(s,t)$ is said to be of bounded variation on
provided the following three conditions hold.
(a) There is a constant

(3.5)

$k$

$R$

.

in the sense of Hardy and Krousesuch that$\sum_{i=1}^{n}\sum_{j=1}^{m}|f(s_{i},t_{j})-f(s_{i},t_{j-1})-f(s_{i-1},t_{j})+f(s_{i-1},t_{j-1})|\leq k$

for all partition .
(b) For each $t\in[c,d],$(c) For each $s\in[a,b],$$P$

is a function of bounded variation on $[a,b]$ .
is a function of bounded variation on $[c,d]$ .
The total variation $Var(f,R)$ of $f$ over is defined to be the supremum of the sums in (3.5)
over all partitions of R. $Var(f(\cdot,t), [a,b])$ and $Var(f(s, \cdot), [c,d])$ will denote the total variations
on $[c,d]$ , respectively, as functions of single variable.
on $[a,b]$ andof
The definition of bounded variation used by Hardy and Krouse has the important property
that if is continuous on and $f$ is of bounded variation on then the Riemann-Stieltjes integrals $\int_{R}g(s,t)df(s,t)$ and $\int_{R}f(s,t)dg(s,t)$ both exist and are related by the following integration$f(\cdot,t)$$f(s, \cdot)$

$R$

$P$

$f(s, \cdot)$$f(\cdot,t)$

$R$

$g$

$R$

by parts formula [9].

Theorem 3.4. Let $R=[a,b]\cross[c,d]$ . Let $g(s,t)$ be afunction of bounded variation in thesense of Hardy and Kmuse and let $f(s,t)$ be a continuous function on R. Then the following
integmtion by parts formula holds.
$\int_{R}g(s,t)df(s,t)=[f(s,t)g(s,t)]_{R}-l^{d}[f(s,t)dg(s,t)]_{a}^{b}$(3.6)$- \int_{a}^{b}[f(s,t)dg(s,t)]_{c}^{d}+\int_{R}f(s,t)dg(s,t)$

33
where
$[f(s,t)g(s,t)]_{R}=f(b,d)g(b,d)-f(a,d)g(a,d)-f(b,c)g(b,c)+f(a,c)g(a,c)$ ,$[f(s,t)dg(s,t)]_{a}^{b}=f(b,t)dg(b,t)-f(a,t)dg(a,t)$

for each $t\in[c,d]$ , and$[f(s,t)dg(s,t)]_{c}^{d}=f(s,d)dg(s,d)-f(s,c)dg(s,c)$for each $s\in[a,b]$ .
be an orthonormal set of real-valued functions in $L_{2}(Q)$ . Furthermore
assume that each is of bounded variation in the sense of Hardy and Krouse on . Then foreach $y\in K(Q)$ and $j=1,2,$ , the Riemann-Stieltjes integral
exists.
Furthermore
Let$\{\theta_{1},\theta_{2}, \ldots,\theta_{n}\}$$Q$$\theta_{j}$$\langle\theta_{j},y\}\equiv\int_{Q}\theta_{j}(s,t)dy(s,t)$$\ldots$$| \langle\theta_{j},y\rangle|=|\theta_{j}(S, T)y(S, T)-\int_{0}^{T}y(S,t)d\theta_{j}(S,t)$(3.7)$- \int_{0}^{s}y(s, T)d\theta_{j}(s, T)+\int_{Q}y(s,t)d\theta_{j}(s,t)|\leq C_{j}\Vert y\Vert_{\infty}$

with
(3.8)

$C_{j}=|\theta_{j}(S, T)|+$

Var $(\theta_{j}(S, \cdot), [0, T])+Var(\theta_{j}(\cdot, T), [0,S])+Var(\theta_{j}, Q)<\infty$ .

In Section 4 below, we show that if and are elements of
then
,
and
exist and are also elements of. Also we examine all relationships
involving exactly two of the three concepts of ”integral transform”, ”convolution product” and“first variation” for functionals in. Furthermore we obtain Parseval‘s and Plancherel‘s$E_{0}(Q)$relations for functionals in. For related work, see [3, 6, 10, 12, 14, 15, 16, 19, 21] and for a
detailed survey of previous work, see [17].
We finish this section by introducing a well-known Yeh-Wiener integration formula for functionals:$F$$\delta F(\cdot|w)$$G$$\delta F(y|\cdot)$$E_{\sigma}(Q)$$\mathcal{F}_{a,\beta}F(\cdot),$$(F*G)_{\alpha}(\cdot)$$E_{\sigma}(Q)$$E_{\sigma}(Q)$$f(\{\vec{\theta},x\rangle)$

(3.9)

where

$\int_{C(Q)}f(\{\vec{\theta},x\})m_{Y}(dx)=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\vec{u})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$$\Vert\vec{u}\Vert^{2}=\sum_{j=1}^{n}u_{j}^{2}$

and$d\vec{u}=du_{1}\cdots du_{n}$.

\S 4. Integral transform, convolution product and first variation of functionals in
In our first theorem, we show that ifof exists and is an element of$\mathcal{F}_{\alpha\beta}F$

$F$

$F$

is an element of.

$E_{\sigma}(Q)$$E_{\sigma}(Q)$$E_{\sigma}(Q)$, then the integral transform34
be given by (1.7). Then the integml transform
Theorem 4.1. Let
belongs toand is given by the fomula$F\in E_{\sigma}(Q)$$\mathcal{F}_{a\beta}F$

exists,$E_{\sigma}(Q)$(4.1)$\mathcal{F}_{a\beta}F(y)=h(\{\vec{\theta},y\})$for $y\in K(Q)$ , where(4.2)$h( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\alpha\vec{u}+\beta\vec{\lambda})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$.

Proof. For each $y\in K(Q)$ , using the Yeh-Wiener integration formula (3.9) we obtain$\mathcal{F}_{a\beta}F(y)=\int_{C(Q)}f(\alpha\{\tilde{\theta},x\rangle+\beta\langle\vec{\theta},y\})m_{Y}(dx)$$=(2 \pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\alpha\vec{u}+\beta\langle\vec{\theta},y\})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$$=h((\vec{\theta},y\})$where is given by (4.2). By [8, Theorem 3.15],
inequality (1.8) we have$h$$h(\tilde{\lambda})$is an entire function. Moreover by the.

$|h( \tilde{\lambda})|\leq(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}A_{F}\exp\{B_{F}\sum_{j=1}^{n}|\alpha u_{j}+\beta\lambda_{j}|^{1+\sigma}-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$But since$|\alpha u_{j}+\beta\lambda_{j}|^{1+\sigma}\leq|2\alpha u_{j}|^{1+\sigma}+|2\beta\lambda_{j}|^{1+\sigma}$,

we have$|h( \vec{\lambda})|\leq A_{\mathcal{F}_{a\beta}F}\exp\{B_{\mathcal{F}_{\alpha\beta}F}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$

,

where
$A_{\mathcal{F}_{a\beta}F}=(2 \pi)^{-n/2}A_{F}(\int_{\mathbb{R}}\exp\{B_{F}|2\alpha u|^{1+\sigma}-\frac{u^{2}}{2}\}du)^{n}<\infty$and$B_{\mathcal{F}_{a,\beta}F}=B_{F}(2\beta|)^{1+\sigma}$. Hence$\mathcal{F}_{a\beta}F\in E_{\sigma}(Q)$.

$\square$existsIn our next theorem we show that the convolution product of functionals from. We may assume that and in Theorem 4.2 below can be
and is an element ofexpressed using the same positive integer . For details see Remark 1.4 of [12].
$E_{\sigma}(Q)$$F$$E_{\sigma}(Q)$

$G$

$n$

$g$

Theorem 4.2. Let $F,G\in E_{\sigma}(Q)$ be given by (1.7) with corresponding entirefunctions $f$ andexists, belongs toand is given by the formula. Then the convolution(4.3)$(F*G)_{\alpha}$$E_{\sigma}(Q)$$(F*G)_{\alpha}(y)=k(\{\vec{\theta},y\rangle)$

35
for $y\in K(Q)$ , where(4.4)$k( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\frac{\vec{\lambda}+\alpha\vec{u}}{\sqrt{2}})g(\frac{\vec{\lambda}-\alpha\vec{u}}{\sqrt{2}})\exp\{-\frac{1}{2}\Vert\tilde{u}\Vert^{2}\}d\vec{u}$.

Pmof. For each $y\in K(Q)$ , using the Yeh-Wiener integration formula (3.9) we obtain$(F*G)_{\alpha}(y)= \int_{C(Q)}f(\frac{(\vec{\theta},y\}+\alpha(\vec{\theta},x\}}{\sqrt{2}})g(\frac{\{\vec{\theta},y\}-\alpha\{\vec{\theta},x\rangle}{\sqrt{2}})m_{Y}(dx)$$=(2 \pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\frac{\{\vec{\theta},y\rangle+\alpha\vec{u}}{\sqrt{2}})g(\frac{\langle\vec{\theta},y\}-\alpha\vec{u}}{\sqrt{2}})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$$=k(\{\vec{\theta},y\})$where is given by (4.4). By [8, Theorem 3.15],$k$$k(\vec{\lambda})$is an entire function and$|k( \vec{\lambda})|\leq(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}A_{F}A_{G}\exp\{(B_{F}+B_{G})\sum_{j=1}^{n}(\frac{|\lambda_{j}|+|\alpha u_{j}|}{\sqrt{2}})^{1+\sigma}-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$.

By the same method as in Theorem 4.1, we have$|k( \vec{\lambda})|\leq A_{(F*G)_{\alpha}}\exp\{B_{(F*G)_{\alpha}}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$,where $B_{(F*G)_{a}}=(B_{F}+B_{G})2^{(1+\sigma)/2}$ and.

$A_{(F*G)_{a}}=(2 \pi)^{-n/2}A_{F}A_{G}(\int_{\mathbb{R}}\exp\{(B_{F}+B_{G})(\sqrt{2}|\alpha u|)^{1+\sigma}-\frac{u^{2}}{2}\}du)^{n}<\infty$

Hence$(F*G)_{\alpha}\in E_{\sigma}(Q)$.

$\square$

In Theorem 4.3 below, we fix $w\in K(Q)$ and consider $\delta F(y|w)$ as a function of , while in
Theorem 4.4 below, we fix $y\in K(Q)$ and consider $\delta F(y|w)$ as a function of .
$y$

$w$

Theorem 4.3. Let(4.5)$F\in E_{\sigma}(Q)$be given by (1.7) and let $w\in K(Q)$ . Then$\delta F(y|w)=p(\langle\vec{\theta},y\})$for $y\in K(Q)$, where(4.6)$p( \vec{\lambda})=\sum_{j=1}^{n}\{\theta_{j},w\}f_{j}(\vec{\lambda})$.

Furthermore, as afunction of $y\in K(Q),$ $\delta F(y|w)$ is an element of$E_{\sigma}(Q)$.

36
Proof. For $y\in K(Q)$ ,$\delta F(y|w)=\frac{\partial}{\partial t}f(\{\vec{\theta},y\}+t\{\vec{\theta},w\rangle)|_{t=0}$$= \sum_{j=1}^{n}(\theta_{j},w\}f_{j}(\{\vec{\theta},y\rangle)=p(\langle\vec{\theta},y\})$is an entire function,
where is given by (4.6). Since
By the Cauchy integral formula we have$f(\vec{\lambda})$$p$$f_{j}(\vec{\lambda})$and so$p(\vec{\lambda})$are entire functions.

$f_{j}( \lambda_{1}, \cdots,\lambda_{j}, \cdots,\lambda_{n})=\frac{1}{2\pi i}\int_{|\zeta-\lambda_{j}|=1}\frac{f(\lambda_{1},\cdots,\zeta,\cdots,\lambda_{n})}{(\zeta-\lambda_{j})^{2}}d\zeta$

By the inequality (1.10), for any$\zeta$with$|\zeta-\lambda_{j}|=1$.

, we have$| \frac{f(\lambda_{1},\cdots,\zeta,\cdots,\lambda_{n})}{(\zeta-\lambda_{j})^{2}}|\leq A_{F}\exp\{B_{F}[|\lambda_{1}|^{1+cr}+\cdots+|\zeta|^{1+(r}+\cdots+|\lambda_{n}|^{1+\sigma}]\}$$\leq A_{F}\exp\{2^{1+\sigma}B_{F}[\sum_{j=1}^{n}|,t_{j}|^{1+\sigma}+1]\}$

.

Hence
$|f_{j}( \vec{\lambda})|\leq A_{F}\exp\{2^{1+\sigma}B_{F}\}\exp\{2^{1+\sigma}B_{F}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$,and so$|p( \vec{\lambda})|\leq\sum_{j=1}^{n}|(\theta_{j},w\rangle||f_{j}(\vec{\lambda})|\leq A_{\delta F(\cdot|w)}\exp\{B_{\delta F(\cdot|w)}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$where$A_{\delta F(\cdot|w)}=A_{F} \exp\{2^{1+\sigma}B_{F}\}\Vert w\Vert_{\infty}\sum_{j=1}^{n}C_{j}<\infty$

with$C_{j}$

given by (3.8) and$B_{\delta F(\cdot|w)}=2^{1+\sigma}B_{F}$Theorem 4.4. Let $y\in K(Q)$ and let
(4.7)

.

$F\in E_{\sigma}(Q)$$\square$be given by (1.7). Then$\delta F(y|w)=q(\{\vec{\theta},w\rangle)$for $w\in K(Q)$ , where(4.8)$q( \vec{\lambda})=\sum_{j=1}^{n}\prime t_{j}f_{j}(\{\vec{\theta},y\rangle)$

Furthermore, as afu nction of $w,$ $\delta F(y|w)$ is an element of.

$E_{\sigma}(Q)$.

37
Pmof. Equations (4.7) and (4.8) are immediate from the first part of the proof of Theorem

4.3. Clearly
is an entire function. Next, using the estimation forof Theorem 4.3 above we obtain,$q(\vec{\lambda})$$|f_{j}|$we saw in the proof$|q( \vec{\lambda})|\leq\sum_{j=1}^{n}|\lambda_{j}f_{j}(\langle\vec{\theta},y\rangle)|$$\leq A_{F}\exp\{2^{1+\sigma}B_{F}\}\exp\{2^{1+\sigma}B_{F}\Vert y\Vert_{\infty}^{1+\sigma}(C_{1}^{1+\sigma}+\cdots+C_{n}^{1+\sigma})\}\sum_{j=1}^{n}|\lambda_{j}|$Since$t\leq\exp\{t^{1+\sigma}\}$.

for all $t\geq 0$ ,$\sum_{j=1}^{n}|\lambda_{j}|\leq\exp\{(\sum_{j=1}^{n}|\lambda_{j}|)^{1+\sigma}\}\leq\exp\{n^{1+\sigma}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$and so$|q( \vec{\lambda})|\leq A_{\delta F(y|\cdot)}\exp\{B_{\delta F(y|\cdot)}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$where$B_{\delta F(y|\cdot)}=n^{1+\sigma}$

and$A_{\delta F(y|\cdot)}=A_{F}\exp\{2^{1+\sigma}B_{F}\}\exp\{2^{1+\sigma}B_{F}\Vert y\Vert_{\infty}^{1+\sigma}(C_{1}^{1+\sigma}+\cdots+C_{n}^{1+\sigma})\}<\infty$.

Hence, as a function of $w,$ $\delta F(y|w)\in E_{\sigma}(Q)$ .

$\square$Now, we establish all of the various relationships involving exactly two of the three concepts

of“integral transform”, ”convolution product” and ”first variation” for functionals belonging to. The seven distinct relationships, as well as altemative expressions for some of them, aregiven by equations (4.9) through (4.15) below.
In view of Theorem 4.1 through Theorem 4.4 above, all of the functionals that occur in thissection are elements of. For example, let and be any functionals in. Thenby Theorem 4.2, the functional, and hence by Theorem 4.1, thebelongs tofunctional
also belongs to. By similar arguments, all of the functionals thatarise in equations (4.9) through (4.19) below, exist and belong to.
Once we have shown the existence theorems (Theorems 4.1 through 4.4 above), the proofsof the Formulas 4.5 through 4.11 below are exactly the same as those in [12]. Hence we juststate the formulas without proofs.
$E_{\sigma}(Q)$$F$$E_{\sigma}(Q)$$(F*G)_{\alpha}$$\mathcal{F}_{\alpha\beta}(F*G)_{a}$$G$$E_{C\Gamma}(Q)$$E_{\sigma}(Q)$$E_{\sigma}(Q)$$E_{\sigma}(Q)$Formula 4.5. The integral transform of the convolution product equals the product of theintegral transforms:(4.9)$\mathcal{F}_{\alpha\beta}(F*G)_{\alpha}(y)=\mathcal{F}_{\alpha\beta}F(\frac{y}{\sqrt{2}})\mathcal{F}_{\alpha\beta}G(\frac{y}{\sqrt{2}})=\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)\mathcal{F}_{\alpha,\beta/\sqrt{2}}G(y)$for all in $K(Q)$ .
$y$38
Formula 4.6. A formula for the convolution product of the integral transform of functionals:
from$E_{\sigma}(Q)$$(\mathcal{F}_{a\beta}F*\mathcal{F}_{a\beta}G)_{a}(y)$$=(2 \pi)^{-3n/2}\int_{\mathbb{R}^{3n}}f(\alpha\vec{r}+\frac{\beta}{\sqrt{2}}\{\vec{\theta},y\}+\frac{\beta\alpha}{\sqrt{2}}\vec{u})$(4.10)$g( \alpha\vec{s}+\frac{\beta}{\sqrt{2}}\{\vec{\theta},y\rangle-\frac{\beta\alpha}{\sqrt{2}}\vec{u})\exp\{-\frac{\Vert\vec{u}\Vert^{2}+\Vert\vec{r}\Vert^{2}+\Vert s\neg|^{2}}{2}\}d\vec{u}d\vec{r}d\vec{s}$for all in $K(Q)$ .
$y$

Formula 4.7. The integral transform with respect to the first argument of the variation
times the variation of the integral transform:
equals$1/\beta$(4.11)$w\rangle \mathcal{F}_{tt,\beta}F_{j}(y)$$\mathcal{F}_{\alpha\beta}(\delta F(\cdot|w))(y)=\frac{1}{\beta}\delta \mathcal{F}_{(f}pF(y|w)=\sum_{j=1}^{n}\{\theta_{j},$for all and
$y$

$w$

in $K(Q)$ .

Formula 4.8. The transform with respect to the second argument of the variation equals
times the variation of the functional:(4.12)$\beta$$\mathcal{F}_{\alpha\beta}(\delta F(y|\cdot))(w)=\beta\delta F(y|w)$

for all and
$y$

$w$

in $K(Q)$ .

Formula4.9. A formula for the first variation of the convolution product of functionals:
from$E_{\sigma}(Q)$(4.13)$\delta(F*G)_{\alpha}(y|w)=\sum_{j=1}^{n}\frac{\langle\theta_{j},w\}}{\sqrt{2}}[(F_{j}*G)_{\alpha}(y)+(F*G_{j})_{\alpha}(y)]$for all and
$y$

$w$

in $K(Q)$ .

Formula 4.10. A formula for the convolution product, with respect to the first argument of:
the variation, of the variation of functionals from
$E_{\sigma}(Q)$(4.14)$( \delta F(\cdot|w)*\delta G(\cdot|w))_{a}(y)=\sum_{j=1}^{n}\sum_{l=1}^{n}\{\theta_{j},w\rangle\{\theta_{l},w\rangle(F_{j}*G_{l})_{a}(y)$for all and
$y$

$w$

in $K(Q)$ .

Formula 4.11. A formula for the convolution product, with respect to the second argument:of the variation, of the variation of functionals from
$E_{\sigma}(Q)$(4.15)$( \delta F(y|\cdot)*\delta G(y|\cdot))_{a}(w)=\frac{1}{2}\delta F(y|w)\delta G(y|w)-\frac{\alpha^{2}}{2}\sum_{j=1}^{n}F_{j}(y)G_{j}(y)$for all and
$y$

$w$

in $K(Q)$ .

39Finally, letting $G=F$ in equations (4.9), (4.13), (4.14) and (4.15) above, yields the formulas(4.16)$\mathcal{F}_{\alpha\beta}(F*F)_{\alpha}(y)=[\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)]^{2}$,(4.17)$\delta(F*F)_{a}(y|w)=\sqrt{2}\sum_{j=1}^{n}\langle\theta_{j},w\rangle(F*F_{j})_{\alpha}(y)$(4.18)$( \delta F(\cdot|w)*\delta F(\cdot|w))_{\alpha}(y)=\sum_{j=1}^{n}\sum_{l=1}^{n}\{\theta_{j},w\}\{\theta_{l},w\}(F_{j}*F_{l})_{\alpha}(y)$

,

,

and
(4.19)$( \delta F(y|\cdot)*\delta F(y|\cdot))_{\alpha}(w)=\frac{1}{2}[\delta F(y|w)]^{2}-\frac{\alpha^{2}}{2}\sum_{j=1}^{n}[F_{j}(y)]^{2}$for all and in $K(Q)$ .
Furthermore we can obtain the following Parseval’s and Plancherel’s relation.
Let $H_{0}=H_{0}(Q)$ be the space of real-valued functions $f$ on which are absolutely continuous and whose derivative $Df$ is in $L_{2}(Q)$ . The inner product on is given by
$y$

$w$

$Q$

$H_{0}$$\langle f,g\rangle=\int_{Q}(Df)(s)(Dg)(s)ds$.

is a real separable infinite dimensional Hilbert space. Let $B_{0}=B_{0}(Q)$ be the YehThen
Wiener space $C(Q)$ and equipwith the $\sup$ norm. Then $(H_{0},B_{0},m_{Y})$ is an example of an
abstract Wiener space.
We restrict our attention, in this subsection, to the space $E_{0}(Q)$ rather than
. Now itis well known, see for example [6,15], that for all $F\in E_{0}(Q)$ , all $y\in K(Q)$ and all complex
numbers $a,b$ and ,$H_{0}$$B_{0}$$E_{\sigma}(Q)$$c$(4.20)$\int_{C(Q)}\int_{C(Q)}F(ax+by+cw)m_{Y}(dx)m_{Y}(dy)=\int_{C(Q)^{F(\sqrt{a^{2}+b^{2}}}}z+cw)m_{Y}(dz)$and that(4.21)

provided$\mathcal{F}_{\alpha,\beta}(\mathcal{F}_{(x’\beta’}F)(y)=F(y)=\mathcal{F}_{\alpha’\beta’}(\mathcal{F}_{a,\beta}F)(y)$$\beta\beta’=1$

and$\alpha^{2}+(\beta\alpha’)^{2}=0$.

Theorem 4.12. Let $F,G\in E_{0}(Q)$ and let
Then Parseval’s relation(4.22)$\int_{C(Q)}\mathcal{F}_{a_{:}\beta}F(\frac{\alpha’y}{\sqrt{2}})\mathcal{F}_{\alpha\beta}G(\frac{\alpha’y}{\sqrt{2}})$holds. In particular,(4.23)$\alpha’$be a complex number such that$\alpha^{2}+(\beta\alpha’)^{2}=0$$my$ $(dy)= \int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})G(-\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$, we have$\iota f\beta=i$$my$ $(dy)$$\int_{C(Q)}\mathcal{F}_{\alpha,i}F(\frac{\alpha y}{\sqrt{2}})G(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)=\int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{\alpha,i}G(\frac{\alpha y}{\sqrt{2}})$

.

.

40
Moreover, formula (4.23) induces Plancherel’s relation of the form
(4.24)$\int_{C(Q)}|\mathcal{F}_{a,i}F(\frac{\alpha y}{\sqrt{2}})|^{2}$$my$ $(dy)= \int_{C(Q)}|F(\frac{\alpha y}{\sqrt{2}})|^{2}$ $my$ $(dy)$.

Proof. From Formula 4.5 and Definition 3.1, it follows that the left hand side of (4.22) isequal to$\int_{C(Q)}\mathcal{F}_{a\beta}(F*G)_{a}(\alpha’y)m_{Y}(dy)$$= \int_{C(Q)}\int_{C(Q)}(F*G)_{a}(\alpha x+\beta\alpha’y)m_{Y}(dx)m_{Y}(dy)$But by (4.20) and the fact that
equal to the right hand side of (4.22).
From (4.21) we know that$\alpha^{2}+(\beta\alpha’)^{2}=0$.

, the last integral is equal to $(F*G)_{\alpha}(O)$ , which is$\mathcal{F}_{a,i}(\mathcal{F}_{a,-i}G)(y)=G(y)$and so we have$\int_{C(Q)}\mathcal{F}_{a,i}F(\frac{\alpha y}{\sqrt{2}})G(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$$= \int_{C(Q)}\mathcal{F}_{x,i}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{\alpha,i}(\mathcal{F}_{\alpha,-i}G)(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$$= \int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{a,-i}G(-\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$,where the second equality is obtained by (4.22). But it is easy to see thatand this completes the proof of (4.23).
, by (4.23) we have
Finally, since$\mathcal{F}_{a,-i}G(-\alpha y/\sqrt{2})=$$\mathcal{F}_{a,i}G(\alpha y/\sqrt{2})$$\mathcal{F}_{\alpha,i}F(\alpha y/\sqrt{2})=\mathcal{F}_{\overline{a},-\overline{\iota a}/a}\overline{F}(\alpha y/\sqrt{2})$$\int_{C(Q)}|\mathcal{F}_{\alpha,i}F(\frac{\alpha y}{\sqrt{2}})|^{2}m_{Y}(dy)=\int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{\alpha,i}\mathcal{F}_{\overline{\alpha},-\overline{\iota\alpha}/\alpha}\overline{F}(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$But by (4.20), it is easy to see that
proof of (4.24).

$\mathcal{F}_{a,i}\mathcal{F}_{\overline{a},-\overline{\iota a}/\alpha}\overline{F}(\alpha y/\sqrt{2})=F(\alpha y/\sqrt{2})$.

and this completes the$\square$

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