Infinite-dimensional Gaussian change of variables’ formula

Abstract

In this paper, we study the Banach space $\ell _{\infty }$ of the bounded real sequences, and a measure $N(a,\Gamma )$ over $\left( \textbf{R}^{\infty },\mathcal {B}^{\infty }\right)$ analogous to the finite-dimensional Gaussian law. The main result of our paper is a change of variables’ formula for the integration, with respect to $N(a,\Gamma )$, of the measurable real functions on $\left( E_{\infty },\mathcal {B}^{\infty }\left( E_{\infty }\right) \right)$, where $E_{\infty }$ is the separable Banach space of the convergent real sequences. This change of variables is given by some $\left( m,\sigma \right)$ functions, defined over a subset of $E_{\infty }$, with values on $E_{\infty }$, with properties that generalize the analogous ones of the finite-dimensional diffeomorphisms.

1 Introduction

In this paper, we aim to provide a theoretical basis for calculating the integrals of real measurable functions with respect to an infinite-dimensional Gaussian measure; an example is the following integral, shown with Example 42 in Sect. 4:

$$\begin{aligned} \underset{A}{ {\displaystyle \int } }\exp \left( -\frac{1}{2\gamma _{11}}\left( \left( \underset{i=1}{\overset{\infty }{\sum }}2^{-i+2}x_{i}\right) ^{2}+x_{1}^{2}\right) \right) d\left( \overset{\infty }{\underset{i=1}{\bigotimes }}N(0,\gamma _{ii})\right) (x_{i}:i\in \textbf{N}^{*}), \end{aligned}$$

(1)

where

$$\begin{aligned} A= & {} \left\{ (x_{i}:i\in \textbf{N}^{*})\in E_{\infty }:\underset{i=1}{\overset{\infty }{\sum }}2^{-i+1}x_{i}>0\right\} ,\\ E_{\infty }= & {} \left\{ \left( x_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }:\exists \,\underset{i\rightarrow \infty }{\lim }x_{i} \in \textbf{R}\right\} , \end{aligned}$$

$\gamma _{ii}\in \textbf{R}^{+}$, for any $i\in \textbf{N}^{*}$, and $\sum _{i=1}^{\infty }\gamma _{ii}\in \textbf{R}^{+}$.

With current knowledge of mathematical analysis, this calculation cannot be performed exactly, since the set A is not the product of Borel sets, and so we cannot apply Theorem 44 stated in Sect. 6; moreover, it is evident that, to calculate the integral (1), we need an infinite-dimensional change of variables’ formula for the integration of the measurable real functions, analogous to that used in the finite-dimensional case.

In the mathematical literature, some articles introduced infinite-dimensional measures analogue of the Lebesgue one: see for example the papers of Jessen [12], Léandre [16], Tsilevich et al. [23], Baker [9], and Accardi et al. [1]. Similarly, there is a vast literature about Gaussian measures and their transformations: for example, we can cite the books by Kuo [13] and Bogachev [11]; moreover, in the article of Ustunel et al. [24], the authors prove a change of variables’ formula on abstract Wiener spaces (Theorem 4.1). However, in the theorem proved in the last paper, the set of functions that define the change of variables is quite restricted.

In order to improve the results obtained in [24], in the articles [5,6,7,8], we introduce a family of infinite-dimensional measures that generalize the Lebesgue measure; in particular, in the first two papers, we consider the Banach space $\ell _{\infty }\subset \textbf{R}^{\infty }$ of the bounded real sequences $\left\{ x_{n}\right\} _{n\in \textbf{N}^{*}}$, the $\sigma$-algebra $\mathcal {E}_{\infty }$ given by the restriction to $\ell _{\infty }$ of $\mathcal {B}^{\infty }$ (defined as the infinite product of the same Borel $\sigma$-algebra $\mathcal {B}$ on $\textbf{R}$), and we define a class of functions over an open subset of $\ell _{\infty }$, with values on $\ell _{\infty }$, called $\left( m,\sigma \right)$ standard. Moreover, for any strictly positive integer k, we introduce over the measurable space $\left( \textbf{R}^{\infty },\mathcal {B}^{\infty }\right)$ a family of infinite-dimensional measures $\mathcal {\lambda }_{N,a,v}^{(k,\infty )}$, dependent on appropriate parameters N, a, v, given by the product of Lebesgue measures on $\textbf{R}$ or on compact subsets of $\textbf{R}$. Furthermore, we provide a change of variables’ formula for the integration of the measurable real functions on $\left( \ell _{\infty },\mathcal {E}_{\infty }\right)$; this change of variables is defined by particular $\left( m,\sigma \right)$ standard functions, with properties similar to those of the finite-dimensional diffeomorphisms. In the paper [7], we introduce a class of functions, called $\left( m,\sigma \right)$, that generalizes the set of the $\left( m,\sigma \right)$ standard functions given in [6]. The main result is the definition of the determinant of a linear $\left( m,\sigma \right)$ function, as the limit of a sequence of the determinants of some standard matrices. In the paper [8], we prove that the change of variables’ formula given by the standard finite-dimensional theory and in the papers [5, 6] can be extended by using particular $\left( m,\sigma \right)$ functions. There are several applications of this result; for example, we can generalize the papers [2,3,4], by considering, for any $p\in \textbf{R}^{+}$, a Markov chain $\left\{ X_{n}\right\} _{n\in \textbf{N}}$ on $[0,p)^{\textbf{N}^{*}}$ defined recursively by an affine function. It is possible to prove that, with appropriate assumptions, the sequence $\left\{ X_{n}\right\} _{n\in \textbf{N}}$ converges with geometric rate to a random variable with law ${\bigotimes }_{i\in \textbf{N}^{*}}\left( \left. \frac{1}{p}Leb\right| _{\mathcal {B}([0,p))}\right)$.

In this paper, we generalize the observation that, if U and V are open sets of the euclidean topology on $\textbf{R}$, for any $a,b\in \textbf{R}$, for any $\Gamma ,\Lambda \in \textbf{R}^{+}$, for any diffeomorphism $\varphi :U\longrightarrow V$, for any $B\in \mathcal {B}(V)$ and for any measurable function $f:\left( V,\mathcal {B}(V)\right) \longrightarrow (\textbf{R},\mathcal {B})$ such that $f^{+}$ (or $f^{-}$) is integrable with respect to the Gaussian measure $N(a,\Gamma )$, we have

$$\begin{aligned} \underset{B}{ {\displaystyle \int } }fdN(a,\Gamma )= & {} \underset{B}{ {\displaystyle \int } }f(y)\cdot \frac{1}{\sqrt{2\pi \Gamma }}\exp \left( -\frac{\left( y-a\right) ^{2}}{2\Gamma }\right) dy\nonumber \\= & {} \underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\Lambda }{\Gamma }}\frac{1}{\sqrt{2\pi \Lambda }}\cdot f(\varphi (x))\exp \left( -\frac{\left( \varphi (x)-a\right) ^{2}}{2\Gamma }\right) \left| \varphi ^{\prime }(x)\right| dx\nonumber \\= & {} \underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\Lambda }{\Gamma }}f(\varphi (x))\exp \left( -\frac{\left( \varphi (x)-a\right) ^{2}}{2\Gamma }+\frac{\left( x-b\right) ^{2}}{2\Lambda }\right) \nonumber \\{} & {} \left| \varphi ^{\prime }(x)\right| dN(b,\Lambda )(x). \end{aligned}$$

(2)

An analogous formula is true by considering, for any $m\in \textbf{N}^{*}$, a m-dimensional Gaussian measure with arbitrary parameters. Then, we consider the set $E_{\infty }$ of the convergent real sequences, and the $\sigma$-algebra $\mathcal {B}_{\infty }$ given by the restriction to $E_{\infty }$ of $\mathcal {B}^{\infty }$; moreover, we introduce over the measurable space $\left( \textbf{R}^{\infty },\mathcal {B}^{\infty }\right)$ (and in particular over $\left( E_{\infty },\mathcal {B}_{\infty }\right)$) an infinite-dimensional Gaussian measure $N(a,\Gamma )$, and we prove a change of variables’ formula similar to that given by (2). In this regard, observe that, in this new infinite-dimensional formula, we are forced to use the measure $N(a,\Gamma )$, because the measure $\bigotimes _{i\in \textbf{N} ^{*}}Leb$ does not exist.

In Sect. 2, we recall the definitions of the infinite-dimensional measurable spaces $\left( \ell _{\infty },\mathcal {E}_{\infty }\right)$ and $\left( E_{\infty },\mathcal {B}_{\infty }\right)$; moreover, we expose results on these spaces, and on the infinite-dimensional matrices. In Sect. 3, we recall some properties of the $\left( m,\sigma \right)$ functions defined in [7]. In Sect. 4, we present the main theorem of our paper, that is a change of variables’ formula for the integration, with respect to $N(a,\Gamma )$, of the measurable real functions on $\left( E_{\infty },\mathcal {B}_{\infty }\right)$; this change of variables is defined by particular $\left( m,\sigma \right)$ functions (Theorem 41). Moreover, in Example 42, we solve exactly the integral (1), and we compare Theorem 41 with Theorem 4.1 in [24]. In Sect. 5, we expose some ideas for further study in the probability theory, and Sect. 6 is an Appendix about the differentiation theory over $\ell _{\infty }$, and about the theory of the $\left( m,\sigma \right)$ functions.

2 The measurable spaces $\left( \ell _{\infty },\mathcal {E}_{\infty }\right)$ and $\left( E_{\infty },\mathcal {B}_{\infty }\right)$

Definition 1

Let $n\in \textbf{N}^{*}$ and let $I\ne \emptyset$ be a set; indicate by $\overline{\textbf{R}}$, by $\tau$, by $\tau ^{n}$, by $\mathcal {B}$, by $\mathcal {B}^{n}$, by $\mathcal {B}^{I}$, by $\mathcal {B} (\overline{\textbf{R}})$, and by Leb, respectively, the set $\textbf{R} \cup \left\{ -\infty ,\infty \right\}$, the euclidean topology on $\textbf{R}$, the euclidean topology on $\textbf{R}^{n}$, the Borel $\sigma$-algebra on $\textbf{R}$, the Borel $\sigma$-algebra on $\textbf{R}^{n}$, the $\sigma$-algebra $\bigotimes _{i\in I}\mathcal {B}_{i}$, with $\mathcal {B}_{i}=\mathcal {B}$ for every i, the Borel $\sigma$-algebra on $\overline{\textbf{R}}$, and the Lebesgue measure on $\textbf{R}$. Moreover, for any $A\subset \overline{\textbf{R}}$, for any $B\subset \textbf{R}^{n}$, and for any $C\subset \textbf{R}^{I}$, indicate by $\mathcal {B}(A)$ the $\sigma$-algebra induced on A by $\mathcal {B} (\overline{\textbf{R}})$, indicate by $\mathcal {B}^{n}(B)$ the $\sigma$-algebra induced on B by $\mathcal {B}^{n}$, and indicate by $\mathcal {B} ^{I}(C)$ the $\sigma$-algebra induced on C by $\mathcal {B}^{I}$. Furthermore, for any $A\subset \textbf{R}$, for any countable and infinite set K, and for any $B\subset \textbf{R}^{K}$, indicate the set $A^{K}$ by $A^{\infty }$, and the $\sigma$-algebra $\mathcal {B}^{K}(B)$ by $\mathcal {B} ^{\infty }(B)$. Moreover, for any $\emptyset \ne H\subset I$, define $x_{I}=\left( x_{i}:i\in I\right) \in \textbf{R}^{I}$ and define the projection $\pi _{I,H}:\textbf{R}^{I}\longrightarrow \textbf{R}^{H}$ by $\pi _{I,H}\left( x_{I}\right) =\left( x_{H}\right)$; in particular, in the case $I=\textbf{N}^{*}$ indicate $\pi _{I,H}$ by $\pi _{H}$, and $\pi _{\left\{ i\right\} }$ by $\pi _{i}$, for any $i\in \textbf{N}^{*}$. Furthermore, if $\left\{ a_{i}:i\in I\right\}$ is a countable set whose elements are not ordered, define $\sum _{i\in I}a_{i}=\sum _{i\in I}a_{i}^{+}-\sum _{i\in I}a_{i}^{-}$, if this difference exists in $\overline{\textbf{R}}$ (and so $\sum _{i\in I}a_{i}\in \textbf{R}$ if and only if $\sum _{i\in I}\left| a_{i}\right| \in \textbf{R}$). Moreover, if $\left\{ f_{i}\right\} _{i\in I}:S\longrightarrow \textbf{R}$ is a countable family of functions, for any $U\subset S^{I}$, indicate by $\bigotimes _{i\in I}f_{i}:U\longrightarrow \textbf{R}$ the function defined by $\left( \bigotimes _{i\in I}f_{i}\right) \left( x_{i}:i\in I\right) =\prod _{i\in I}f_{i}(x_{i})$, $\forall \,\left( x_{i}:i\in I\right) \in U$. Finally, if E is a topological space, indicate by $C^{0}(E)$ the set of the continuous functions from E to $\textbf{R}$, and, if E is a topological vector space, indicate by $E^{*}$ the set of the linear and continuous functions from E to $\textbf{R}$.

Definition 2

Let $I\ne \emptyset$ be a set; define the function $\left\| \cdot \right\| _{I}:\textbf{R}^{I}\longrightarrow [0,+\infty ]$ by

$$\begin{aligned} \left\| x\right\| _{I}=\underset{i\in I}{\sup }|x_{i}|, \quad \forall \,x=\left( x_{i}:i\in I\right) \in \textbf{R}^{I}, \end{aligned}$$

and define

$$\begin{aligned} \ell _{I}=\left\{ x\in \textbf{R}^{I}:\left\| x\right\| _{I}<\infty \right\} . \end{aligned}$$

In particular, suppose that $I=\left\{ i_{n}:n\in \textbf{N}^{*}\right\} \subset \textbf{R}$ is a countable set such that $i_{n}<i_{n+1}$, for any $n\in \textbf{N}^{*}$; define the vector space

$$\begin{aligned} E_{I}=\left\{ \left( x_{i}:i\in I\right) \in \textbf{R}^{I}:\exists \,\underset{n\rightarrow \infty }{\lim }x_{i_{n}}\in \textbf{R}\right\} \subset \ell _{I}. \end{aligned}$$

Moreover, indicate by $\left\| x\right\| _{\infty }$ the function $\left\| x\right\| _{\textbf{N}^{*}}$, by $\ell _{\infty }$ the set $\ell _{\textbf{N}^{*}}$, by $E_{\infty }$ the set $E_{\textbf{N}^{*}}$, by $\mathcal {E}_{\infty }$ the $\sigma$-algebra $\mathcal {B}^{\infty }(\ell _{\infty })$, and by $\mathcal {B}_{\infty }$ the $\sigma$-algebra $\mathcal {B} ^{\infty }(E_{\infty })$; furthermore, for any $n\in \textbf{N}^{*} \cup \left\{ \infty \right\}$, define the set

$$\begin{aligned} I_{n}=\left\{ \begin{array}{ll} \left\{ 1,\ldots ,n\right\} &{}\quad \text {if}\quad n\in \textbf{N}^{*}\\ \textbf{N}^{*} &{}\quad \text {if}\quad n=\infty \end{array}\right. . \end{aligned}$$

Finally, for any $p\in \left[ 1,\infty \right)$, define the function $\left\| \cdot \right\| _{p}:\textbf{R}^{\infty }\longrightarrow [0,+\infty ]$ by

$$\begin{aligned} \left\| x\right\| _{p}=\left( \underset{i=1}{\overset{\infty }{\sum }}\left| x_{i}\right| ^{p}\right) ^{\frac{1}{p}}, \quad \forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }, \end{aligned}$$

and define the vector space

$$\begin{aligned} \ell _{p}=\left\{ x\in \textbf{R}^{\infty }:\left\| x\right\| _{p} <\infty \right\} \subset E_{\infty }. \end{aligned}$$

Definition 3

Let $I\ne \emptyset$ be a set, let $x_{0}\in \ell _{I}$ and let $A\subset \ell _{I}$; indicate by $\tau _{\left\| \cdot \right\| _{I}}$ the topology induced on $\ell _{I}$ by the norm $\left\| \cdot \right\| _{I}$, and indicate by $\tau _{\left\| \cdot \right\| _{I}}(A)$ the topology induced on A by $\tau _{\left\| \cdot \right\| _{I}}$; moreover, for any $\varepsilon \in \textbf{R}^{+}$, indicate by $B_{I}(x_{0},\varepsilon )$ the set $\{x\in \ell _{I}:\left\| x-x_{0}\right\| _{I}<\varepsilon \}$, and define the set

$$\begin{aligned} \mathcal {D}_{I}(A)=\left\{ x\in \ell _{I}:\forall \,\varepsilon \in \textbf{R} ^{+}\text {, one has }B_{I}(x,\varepsilon )\cap \left( A\backslash \left\{ x\right\} \right) \ne \emptyset \right\} . \end{aligned}$$

In particular, if $I=\textbf{N}^{*}$ and $A\subset \ell _{\infty }$, indicate by $\tau _{\left\| \cdot \right\| _{\infty }}$ the topology $\tau _{\left\| \cdot \right\| _{\textbf{N}^{*}}}$, by $\tau _{\left\| \cdot \right\| _{\infty }}(A)$ the topology $\tau _{\left\| \cdot \right\| _{\textbf{N}^{*}}}(A)$, and by $\mathcal {D}_{\infty }(A)$ the set $\mathcal {D}_{\textbf{N} ^{*}}(A)$.

Remark 4

Let H and I be two sets such that $\emptyset \ne H\subset I$; then:

• 1. $\pi _{I,H}:\left( \ell _{I},\tau _{\left\| \cdot \right\| _{I}}\right) \longrightarrow \left( \ell _{H},\tau _{\left\| \cdot \right\| _{H}}\right)$ is continuous and open.

• 2. $\pi _{I,H}:\left( \textbf{R}^{I},\mathcal {B}^{I}\right) \longrightarrow \left( \textbf{R}^{H},\mathcal {B}^{H}\right)$ is measurable.

Proof

The point 1 follows from Proposition 6 in Asci [6]; moreover, the point 2 can be easily proved. $\square$

Remark 5

$\ell _{\infty }$ is a Banach space, with the norm $\left\| \cdot \right\| _{\infty }$; moreover, one has $\ell _{\infty }\in \mathcal {B} ^{\infty }$.

Proof

In order to prove that $\ell _{\infty }$ is a Banach space, we can see, for example, the proof of Remark 2 in [5]. Moreover, $\forall \,i\in \textbf{N}^{*}$, the function $\left( x_{i}:i\in \textbf{N}^{*}\right) \longrightarrow |x_{i}|$ is $\left( \mathcal {B}^{\infty },\mathcal {B}(\overline{\textbf{R}})\right)$-measurable, and so the function $\left( x_{i}:i\in \textbf{N}^{*}\right) \longrightarrow \left\| x\right\| _{\infty }$ is $\left( \mathcal {B}^{\infty },\mathcal {B} (\overline{\textbf{R}})\right)$-measurable too, from which $\ell _{\infty } \in \mathcal {B}^{\infty }$. $\square$

Remark 6

$E_{\infty }$ is a separable Banach space, with the topology $\tau _{\left\| \cdot \right\| _{\infty }}\left( E_{\infty }\right)$; moreover, one has $E_{\infty }\in \mathcal {B}^{\infty }$ and $\mathcal {D} _{\infty }(E_{\infty })=E_{\infty }$.

Proof

Let $\left\{ x_{n}\right\} _{n\in \textbf{N}}\subset E_{\infty }$ be a sequence such that $\lim _{n\rightarrow \infty }x_{n}=a$, where $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in \ell _{\infty }$; then, $\forall \,\varepsilon \in \textbf{R}^{+}$, $\exists \,n_{1}\in \textbf{N}$ such that, $\forall \,n\in \textbf{N}$, $n\ge n_{1}$, we have $\sup _{i\in \textbf{N}^{*}}\left| \left( x_{n}\right) _{i}-a_{i}\right| =\left\| x_{n}-a\right\| _{\infty }<\frac{\varepsilon }{3}$; moreover, since $x_{n_{1}}\in E_{\infty }$, $\exists \,n_{2}\in \textbf{N}$ such that, $\forall \,i,j\in \textbf{N}^{*}$ such that $i,j\ge n_{2}$, we have $\left| \left( x_{n_{1}}\right) _{i}-\left( x_{n_{1}}\right) _{j}\right| <\frac{\varepsilon }{3}$, from which

$$\begin{aligned} \left| a_{i}-a_{j}\right| \le \left| a_{i}-\left( x_{n_{1}}\right) _{i}\right| +\left| \left( x_{n_{1}}\right) _{i}-\left( x_{n_{1}}\right) _{j}\right| +\left| \left( x_{n_{1}}\right) _{j}-a_{j}\right| <\varepsilon ; \end{aligned}$$

then, $\left\{ a_{i}\right\} _{i\in \textbf{N}^{*}}$ is a Cauchy sequence, and so, since $\textbf{R}$ is complete, we have $a\in E_{\infty }$; thus, $E_{\infty }$ is closed, and so it is complete, since $E_{\infty }\subset \ell _{\infty }$; thus, $E_{\infty }$ is a Banach space. Moreover, let $I=\left\{ i_{n}:n\in \textbf{N}^{*}\right\} \subset \textbf{R}$ be a countable set such that $i_{n}<i_{n+1}$, for any $n\in \textbf{N}^{*}$, and such that $\lim _{n\rightarrow +\infty }i_{n}\in \textbf{R}$; it is known (see, for example, the Rudin’s book [18]) that the set $C^{0}\left( \overline{I}\right)$ is separable; moreover, since the set $E_{I}$ can be identified with $C^{0}\left( \overline{I}\right)$, $E_{I}$ is separable too, and so this is true for the set $E_{\infty }$, which can be identified with $E_{I}$. Furthermore, $\forall \,i\in \textbf{N}^{*}$, the function $\left( x_{i}:i\in \textbf{N}^{*}\right) \longrightarrow x_{i}$ is $\left( \mathcal {B}^{\infty },\mathcal {B}(\overline{\textbf{R}})\right)$-measurable, and so

$$\begin{aligned} E_{\infty }= & {} \left\{ \underset{i\rightarrow \infty }{\lim \sup }x_{i} <\infty \right\} \cap \left\{ \underset{i\rightarrow \infty }{\lim \inf }\ x_{i}>-\infty \right\} \\{} & {} \cap \left\{ \underset{i\rightarrow \infty }{\lim \sup }x_{i}-\underset{i\rightarrow \infty }{\lim \inf }x_{i}=0\right\} \in \mathcal {B}^{\infty }. \end{aligned}$$

Finally, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$, $\forall \,\varepsilon \in \textbf{R}^{+}$, by setting $y=\left( x_{i}+\frac{\varepsilon }{2}:i\in \textbf{N}^{*}\right)$, we have $y\in B_{\infty }(x,\varepsilon )\cap E_{\infty }\backslash \left\{ x\right\}$, and so $x\in \mathcal {D}_{\infty }(E_{\infty })$, from which $E_{\infty } \subset \mathcal {D}_{\infty }(E_{\infty })$; /0258since $E_{\infty }$ is closed, we have $\mathcal {D}_{\infty }(E_{\infty })\subset E_{\infty }$, and so $\mathcal {D}_{\infty }(E_{\infty })=E_{\infty }$.$\square$

The following two results can be easily proved.

Remark 7

For any $p\in \left[ 1,\infty \right)$, $\ell _{p}$ is a Banach space, with the norm $\left\| \cdot \right\| _{p}$, and one has $\ell _{p} \in \mathcal {B}^{\infty }$. Moreover, $\ell _{2}$ is a separable Hilbert space, with the inner product $\left\langle \cdot ,\cdot \right\rangle :\ell _{2} \times \ell _{2}\longrightarrow \textbf{R}$ defined by $\left\langle x,y\right\rangle =\sum _{i=1}^{\infty } x_{i}y_{i}$, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in \ell _{2}$, $\forall \,y=\left( y_{i}:i\in \textbf{N}^{*}\right) \in \ell _{2}$.

Remark 8

For any $p,q\in \left[ 1,\infty \right)$ such that $p\le q$, one has $\ell _{p}\subset \ell _{q}$.

Definition 9

Let $k\in \textbf{N}^{*}$ and let $B\subset \textbf{R}^{k}$; a function $g:\left( C_{c}(B),\tau _{\left\| \cdot \right\| _{B}}\left( C_{c}(B)\right) \right) \longrightarrow \left( \textbf{R},\tau \right)$ linear and continuous is called Radon measure on B; moreover, if, for any $f\in C_{c}(B)$ such that $f\ge 0$, one has $g(f)\ge 0$, then g is called nonnegative Radon measure on B.

Theorem 10

(Riesz) Let $k\in \textbf{N}^{*}$, let $B\subset \textbf{R}^{k}$ and let $g:\left( C_{c}(B),\tau _{\left\| \cdot \right\| _{B}}\left( C_{c}(B)\right) \right) \longrightarrow \left( \textbf{R},\tau \right)$ be a nonnegative Radon measure on B; then, there exists a unique measure $\mu$ on $\left( B,\mathcal {B}^{k}(B)\right)$ such that

$$\begin{aligned} g(f)=\underset{B}{\int }fd\mu ,\quad \forall \,f\in C_{c}(B). \end{aligned}$$

Moreover, every Radon measure on B can be expressed as a difference between two nonnegative Radon measures on B.

Proof

The proof can be found, for example, in the Rudin’s book [18]. $\square$

Proposition 11

Let $I=\left\{ i_{n}:n\in \textbf{N}^{*}\right\} \subset \textbf{R}$ be a countable set such that $i_{n}<i_{n+1}$, for any $n\in \textbf{N}^{*}$, and such that $i_{\infty }\equiv \lim _{n\rightarrow \infty }i_{n}\in \textbf{R}$; moreover, let $g:\left( C^{0}\left( \overline{I}\right) ,\tau _{\left\| \cdot \right\| _{\overline{I}}}\left( C^{0}\left( \overline{I}\right) \right) \right) \longrightarrow \left( \textbf{R},\tau \right)$ be a linear and continuous function; then, there exists $\left( a_{i}:i\in \overline{I}\right) \in \textbf{R}^{\overline{I}}$ such that $\sum _{i\in \overline{I}} \left| a_{i}\right| \in \textbf{R}$ and such that $g(f)=\sum _{i\in \overline{I}}a_{i}f_{i}$, for any $f=\left( f_{i}:i\in \overline{I}\right) \in C^{0}\left( \overline{I}\right)$.

Proof

Since $\overline{I}=I\cup \left\{ i_{\infty }\right\}$ is compact, we have $C_{c}\left( \overline{I}\right) =C^{0}\left( \overline{I}\right)$; then, g is a Radon measure on $\overline{I}$ and, from Theorem 10, there exist two Radon measures $g_{1}$ and $g_{2}$ nonnegative on $\overline{I}$, and two measures $\mu _{1}$ and $\mu _{2}$ on $\left( \overline{I},\mathcal {B}\left( \overline{I}\right) \right)$ such that, $\forall \,f=\left( f_{i}:i\in \overline{I}\right) \in C^{0}\left( \overline{I}\right)$, we have

$$\begin{aligned} g(f)= & {} g_{1}(f)-g_{2}(f)=\underset{\overline{I}}{\int }fd\mu _{1}-\underset{\overline{I}}{\int }fd\mu _{2}\\= & {} \underset{i\in \overline{I}}{\sum }f_{i}\mu _{1}\left( \left\{ i\right\} \right) -\underset{i\in \overline{I}}{\sum }f_{i}\mu _{2}\left( \left\{ i\right\} \right) =\underset{i\in \overline{I}}{\sum }a_{i}f_{i}, \end{aligned}$$

where $a_{i}=\mu _{1}\left( \left\{ i\right\} \right) -\mu _{2}\left( \left\{ i\right\} \right) \in \textbf{R}$, $\forall \,i\in \overline{I}$; moreover, if we choose $f=1$, we obtain $\sum _{i\in \overline{I}} \mu _{1}\left( \left\{ i\right\} \right) =g_{1}(1)\in \textbf{R}$ and $\sum _{i\in \overline{I}}\mu _{2}\left( \left\{ i\right\} \right) =g_{2}(1)\in \textbf{R}$, from which

$$\begin{aligned} \underset{i\in \overline{I}}{\sum }\left| a_{i}\right| \le \underset{i\in \overline{I}}{\sum }\left( \mu _{1}\left( \left\{ i\right\} \right) +\mu _{2}\left( \left\{ i\right\} \right) \right) <\infty . \end{aligned}$$

$\square$

Corollary 12

Let $I=\left\{ i_{n}:n\in \textbf{N}^{*}\right\} \subset \textbf{R}$ be a countable set such that $i_{n}<i_{n+1}$, for any $n\in \textbf{N}^{*}$, and such that $i_{\infty }\equiv \lim _{n\rightarrow +\infty }i_{n}\in \textbf{R}$; moreover, let $g:\left( E_{I},\tau _{\left\| \cdot \right\| _{I}}\left( E_{I}\right) \right) \longrightarrow \left( \textbf{R},\tau \right)$ be a linear and continuous function; then, there exists $\left( a_{i}:i\in \overline{I}\right) \in \textbf{R}^{\overline{I}}$ such that $\sum _{i\in \overline{I}} \left| a_{i}\right| \in \textbf{R}$ and such that $g(f)=\sum _{i\in \overline{I}}a_{i}f_{i}$, for any $f=\left( f_{i}:i\in I\right) \in E_{I}$, where $f_{i_{\infty }}\equiv \lim _{n\rightarrow \infty } f_{i_{n}}$.

Proof

$\forall \,f=\left( f_{i}:i\in \overline{I}\right) \in C^{0}\left( \overline{I}\right)$, we have $\left( f_{i}:i\in I\right) \in E_{I}$; then, let $h_{g}:C^{0}\left( \overline{I}\right) \longrightarrow \textbf{R}$ be the linear function defined by $h_{g}\left( f_{i}:i\in \overline{I}\right) =g\left( f_{i}:i\in I\right)$; $\forall \,\varepsilon \in \textbf{R}^{+}$, $\forall \,x=\left( x_{i}:i\in \overline{I}\right) \in C^{0}\left( \overline{I}\right)$, $\exists \,\delta \in \textbf{R}^{+}$ such that, $\forall \,y_{I}=\left( y_{i}:i\in I\right) \in E_{I}$ such that $\left\| y_{I}-x_{I}\right\| _{I}<\delta$, we have $\left| g\left( y_{I}\right) -g\left( x_{I}\right) \right| <\varepsilon$; then, $\forall \,y\in C^{0}\left( \overline{I}\right)$ such that $\left\| y-x\right\| _{\overline{I}}<\delta$, we obtain

$$\begin{aligned} \left| h_{g}(y)-h_{g}(x)\right| =\left| g(y_{I})-g(x_{I})\right| <\varepsilon ; \end{aligned}$$

then, $h_{g}$ is $\left( \tau _{\left\| \cdot \right\| _{\overline{I}} }\left( C^{0}\left( \overline{I}\right) \right) ,\tau \right)$-continuous and, from Proposition 11, there exists $\left( a_{i}:i\in \overline{I}\right) \in \textbf{R}^{\overline{I}}$ such that $\sum _{i\in \overline{I}}\left| a_{i}\right| \in \textbf{R}$ and such that $h_{g}(l)=\sum _{i\in \overline{I}}a_{i}l_{i}$, $\forall \,l=\left( l_{i}:i\in \overline{I}\right) \in C^{0}\left( \overline{I}\right)$; thus, let $f=\left( f_{i}:i\in I\right) \in E_{I}$ and let $l\in C^{0}\left( \overline{I}\right)$ such that $l_{I}=f$; we have

$$\begin{aligned} g(f)=h_{g}(l)=\underset{i\in I}{\sum }a_{i}l_{i}+a_{i_{\infty }}l_{i_{\infty } }=\underset{i\in I}{\sum }a_{i}f_{i}+a_{i_{\infty }}f_{i_{\infty }} =\underset{i\in \overline{I}}{\sum }a_{i}f_{i}, \end{aligned}$$

where $f_{i_{\infty }}\equiv \lim _{n\rightarrow \infty }f_{i_{n}}$. $\square$

Corollary 13

Let $I=\left\{ i_{n}:n\in \textbf{N}^{*}\right\} \subset \textbf{R}$ be a countable set such that $i_{n}<i_{n+1}$, for any $n\in \textbf{N}^{*}$, and let $i_{\infty }=\lim _{n\rightarrow +\infty }i_{n}\in \textbf{R}\cup \left\{ \infty \right\}$; then:

• 1. For any function $g:\left( E_{I},\tau _{\left\| \cdot \right\| _{I}}\left( E_{I}\right) \right) \longrightarrow \left( \textbf{R},\tau \right)$ linear and continuous, there exists $\left( a_{i}:i\in I\cup \left\{ i_{\infty }\right\} \right) \in \textbf{R}^{I\cup \left\{ i_{\infty }\right\} }$ such that $\sum _{i\in I\cup \left\{ i_{\infty }\right\} }\left| a_{i}\right| \in \textbf{R}$ and such that

  $$\begin{aligned} g(f)=\underset{i\in I\cup \left\{ i_{\infty }\right\} }{\sum }a_{i}f_{i},\quad \forall \,f=\left( f_{i}:i\in I\right) \in E_{I}, \end{aligned}$$

  (3)

  where $f_{i_{\infty }}\equiv \lim _{n\rightarrow \infty }f_{i_{n}}$.

• 2. For any $\left( a_{i}:i\in I\cup \left\{ i_{\infty }\right\} \right) \in \textbf{R}^{I\cup \left\{ i_{\infty }\right\} }$ such that $\sum _{i\in I\cup \left\{ i_{\infty }\right\} }\left| a_{i}\right| \in \textbf{R}$, the function $g:\left( E_{I},\tau _{\left\| \cdot \right\| _{I}}\left( E_{I}\right) \right) \longrightarrow \left( \textbf{R},\tau \right)$ defined by (3) is linear and continuous.

Proof

• 1. If $i_{\infty }\in \textbf{R}$, the statement follows from Corollary 12. Conversely, suppose that $i_{\infty }=\infty$, and let $J=\left\{ j_{n}:n\in \textbf{N}^{*}\right\} \subset \textbf{R}$, where $j_{n} =1-2^{-n}$, $\forall \,n\in \textbf{N}^{*}$; moreover, let $\rho :E_{J}\longrightarrow E_{I}$ be the function defined by $\left( \rho \left( f\right) \right) _{i_{n}}=f_{j_{n}}$, $\forall \,f=\left( f_{j_{n}}:n\in \textbf{N}^{*}\right) \in E_{J}$, $\forall \,n\in \textbf{N}^{*}$; observe that $\rho$ is linear and $\left( \tau _{\left\| \cdot \right\| _{J} }\left( E_{J}\right) ,\tau _{\left\| \cdot \right\| _{I}}\left( E_{I}\right) \right)$-continuous, and so the function $\left( g\circ \rho \right) :E_{J}\longrightarrow \textbf{R}$ is linear and $\left( \tau _{\left\| \cdot \right\| _{J}}\left( E_{J}\right) ,\tau \right)$-continuous; then, let $f=\left( f_{i}:i\in I\right) \in E_{I}$, and let $h=\left( h_{j}:j\in J\right) =\rho ^{-1}(f)\in E_{J}$; since $j_{\infty }\equiv \lim _{n\rightarrow \infty }j_{n}=1\in \textbf{R}$, from Corollary 12, there exists $\left( b_{j}:j\in \overline{J}\right) \in \textbf{R}^{\overline{J}}$ such that $\sum _{j\in \overline{J}}\left| b_{j}\right| \in \textbf{R}$ and such that

  $$\begin{aligned} g(f)=\left( g\circ \rho \right) (h)=\underset{j\in \overline{J}}{\sum } b_{j}h_{j}=\underset{i\in I\cup \left\{ i_{\infty }\right\} }{\sum }a_{i} f_{i}, \end{aligned}$$

  where $a_{i_{n}}=b_{j_{n}}$, $\forall \,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$,and $f_{i_{\infty }}\equiv \lim _{n\rightarrow \infty } f_{i_{n}}$.

• 2. It is easy to prove that $g:E_{I}\longrightarrow \textbf{R}$ is linear; moreover, let $x_{0}=\left( x_{0,i}:i\in I\right) \in E_{I}$, let $\varepsilon \in \textbf{R}^{+}$ and let $\delta =\varepsilon \left( 2 \sum _{i\in I\cup \left\{ i_{\infty }\right\} }\left| a_{i}\right| \right) ^{-1}\in \textbf{R}^{+}$; $\forall \,x=\left( x_{i}:i\in I\right) \in E_{I}$ such that $\left\| x-x_{0}\right\| _{I}<\delta$, we have

  $$\begin{aligned} \left| g(x)-g\left( x_{0}\right) \right|= & {} \left| \underset{i\in I}{\sum }a_{i}(x_{i}-x_{0,i})+a_{i_{\infty }}\underset{n\rightarrow \infty }{\lim }\left( x_{i_{n}}-(x_{0})_{i_{n}}\right) \right| \\\le & {} \delta \underset{i\in I\cup \left\{ i_{\infty }\right\} }{\sum }\left| a_{i}\right| =\frac{\varepsilon }{2}<\varepsilon , \end{aligned}$$

  and so g is continuous in $x_{0}$; then, since $x_{0}\in E_{I}$ is arbitrary, g is $\left( \tau _{\left\| \cdot \right\| _{I}}\left( E_{I}\right) ,\tau \right)$-continuous.

$\square$

Proposition 14

One has $\mathcal {B}_{\infty }=\sigma \left( g:g\in (E_{\infty })^{*}\right) =\sigma \left( g:g\in C^{0}(E_{\infty })\right) =\sigma \left( \tau _{\left\| \cdot \right\| _{\infty }}\left( E_{\infty }\right) \right)$.

Proof

$\forall \,i\in \textbf{N}^{*}$, let $\pi _{i}:E_{\infty }\longrightarrow \textbf{R}$ be the projection defined by $\pi _{i}(x)=x_{i}$, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$; $\forall \,A\in \mathcal {B}$, $\forall \,i\in \textbf{N}^{*}$, we have $\pi _{i}^{-1}(A)=\left( \textbf{R}^{i-1}\times A\times \textbf{R}^{\infty }\right) \cap E_{\infty }$, from which $\mathcal {B}_{\infty }=\mathcal {B}^{\infty }\left( E_{\infty }\right) =\sigma \left( \pi _{i}:i\in \textbf{N}^{*}\right)$, and so $\mathcal {B}_{\infty }\subset \sigma \left( g:g\in (E_{\infty })^{*}\right)$. Moreover, from Corollary 13, $\forall \,g\in (E_{\infty })^{*}$, there exists $\left( a_{i}:i\in \textbf{N}^{*}\cup \left\{ \infty \right\} \right) \in \textbf{R}^{\textbf{N}^{*}\cup \left\{ \infty \right\} }$ such that $g(x)=\sum _{i\in \textbf{N}^{*}\cup \left\{ \infty \right\} } a_{i}x_{i}$, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$, where $x_{\infty }\equiv \lim _{i\rightarrow \infty }x_{i}$, and so

$$\begin{aligned} g(x)=\underset{n\rightarrow \infty }{\lim }\underset{i=1}{\overset{n}{\sum }} a_{i}\pi _{i}(x)+a_{\infty }\underset{i\rightarrow \infty }{\lim }\pi _{i}(x); \end{aligned}$$

then, g is $\left( \sigma \left( \pi _{i}:i\in \textbf{N}^{*}\right) ,\mathcal {B}\right)$-measurable, from which

$$\begin{aligned} \sigma \left( g:g\in (E_{\infty })^{*}\right) \subset \sigma \left( \pi _{i}:i\in \textbf{N}^{*}\right) =\mathcal {B}_{\infty }, \end{aligned}$$

and so $\mathcal {B}_{\infty }=\sigma \left( g:g\in (E_{\infty })^{*}\right)$; finally, since $E_{\infty }$ is a separable Banach space, from Theorem A.3.7 in [11], we have $\sigma \left( g:g\in (E_{\infty })^{*}\right) =\sigma \left( g:g\in C^{0}(E_{\infty })\right) =\sigma \left( \tau _{\left\| \cdot \right\| _{\infty }}\left( E_{\infty }\right) \right)$. $\square$

The following concept generalizes Definition 6 in [5] (see also the theory in the Lang’s book [15] and that in the Weidmann’s book [25]).

Definition 15

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$ and let $A=\left( a_{ij}\right) _{i\in I_{m},j\in I_{n}}$ be a real matrix $m\times n$ (eventually infinite); define the linear function $A=\left( a_{ij}\right) _{i\in I_{m},j\in I_{n}}:\textbf{R}^{n}\longrightarrow \textbf{R}^{m}$, and write $x\longrightarrow Ax$, in the following manner:

$$\begin{aligned} \left( Ax\right) _{i}=\underset{j\in I_{n}}{\sum }a_{ij}x_{j}\quad , \forall \,x\in \textbf{R}^{n}, \quad \forall \,i\in I_{m}, \end{aligned}$$

(4)

on condition that, for any $i\in I_{m}$, the sum in (4) converges to a real number. Furthermore, if $m=n$ and $a_{ij}=0$ for any $i\ne j$, the real matrix A and the corresponding function are called diagonal; in particular, indicate by $\textbf{I}_{n}=\left( \delta _{ij}\right) _{i,j\in I_{n}}$ the diagonal real matrix defined by

$$\begin{aligned} \delta _{ij}=\left\{ \begin{array}{ll} 1 &{}\quad \text {if}\quad i=j\\ 0 &{}\quad \text {otherwise} \end{array} \right. . \end{aligned}$$

Moreover, for any $\emptyset \ne L\subset I_{m}$, for any $\emptyset \ne N\subset I_{n}$, indicate by $A^{(L,N)}$ the real matrix $\left( a_{ij}\right) _{i\in L,j\in N}$, and, for any$\,l,o\in \textbf{N}^{*} \cup \left\{ \infty \right\}$, indicate $A^{(I_{l},N)}$ by $A^{(l,N)}$, $A^{(L,I_{o})}$ by $A^{(L,o)}$, and $A^{(I_{l},I_{o})}$ by $A^{(l,o)}$. Furthermore indicate by $^{t}A=\left( b_{ji}\right) _{j\in I_{n},i\in I_{m} }:\textbf{R}^{m}\longrightarrow \textbf{R}^{n}$, if it is defined, the linear function given by $b_{ji}=a_{ij}$, for any $j\in I_{n}$ and for any $i\in I_{m}$. Furthermore, if $m=n$ and $A=\left. ^{t}A\right.$, the function A is called symmetric. Finally, if $o\in \textbf{N}^{*}\cup \left\{ \infty \right\}$ and $B=\left( b_{jk}\right) _{j\in I_{n},k\in I_{o}}$ is a real matrix $n\times o$, define the $m\times o$ real matrix $AB=\left( \left( AB\right) _{ik}\right) _{i\in I_{m},k\in I_{o}}$ by

$$\begin{aligned} \left( AB\right) _{ik}=\underset{j\in I_{n}}{\sum }a_{ij}b_{jk}, \end{aligned}$$

(5)

on condition that, for any $i\in I_{m}$ and for any $k\in I_{o}$, the sum in (5) converges to a real number.

Proposition 16

Let $m,n,o\in \textbf{N}^{*}\cup \left\{ \infty \right\}$ and let $A=\left( a_{ij}\right) _{i\in I_{m},j\in I_{n}}$ be a real matrix $m\times n$; then:

• 1. The linear function $A=\left( a_{ij}\right) _{i\in I_{m},j\in I_{n} }:\ell _{I_{n}}\longrightarrow \textbf{R}^{m}$ given by (4) is defined if and only if, for any $i\in I_{m}$, $\sum _{j\in I_{n}}\left| a_{ij}\right| <\infty$.

• 2. One has $\sup _{i\in I_{m}}\sum _{j\in I_{n}}\left| a_{ij}\right| <\infty$ if and only if $A(\ell _{I_{n}})\subset \ell _{I_{m}}$ and A is continuous; moreover, if $A(\ell _{I_{n}})\subset \ell _{I_{m}}$, then $\left\| A\right\| =\sup _{i\in I_{m}} \sum _{j\in I_{n}}\left| a_{ij}\right|$.

• 3. If $A(\ell _{I_{n}})\subset \ell _{I_{m}}$, then, for any $\emptyset \ne L\subset I_{m}$, for any $\emptyset \ne N\subset I_{n}$, one has $A^{(L,N)}(\ell _{N})\subset \ell _{L}$.

• 4. If $B=\left( b_{jk}\right) _{j\in I_{n},k\in I_{o}}:\ell _{I_{o} }\longrightarrow \ell _{I_{n}}$ is a linear function, then the linear function $A\circ B:\ell _{I_{o}}\longrightarrow \textbf{R}^{m}$ is defined by the real matrix AB.

Proof

See the proof of Proposition 2.13 in Asci [8]. $\square$

3 Theory of the $\left( m,\sigma \right)$ functions

The following Definition 17 introduces a class of functions $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$, called $\left( m,\sigma \right)$, that in the case $U\subset E_{\infty }$ will be used to provide a change of variables’ formula for the integration, with respect to an infinite-dimensional Gaussian measures, of the measurable real functions on $\left( E_{\infty },\mathcal {B}_{\infty }\right)$. In fact, the properties of some $\left( m,\sigma \right)$ functions generalize the analogous ones of the standard finite-dimensional diffeomorphisms.

The axioms that define the $\left( m,\sigma \right)$ functions serve to prove Proposition 38, which in turn is fundamental to proving Proposition 40 and Theorem 41. Essentially, the $\left( m,\sigma \right)$ functions and their inverse functions must behave similar to infinite-dimensional diagonal matrices, which, when applied to an infinite-dimensional rectangle, generate a new one (see formula (19), where these rectangles are $\left[ -M,M\right] ^{\infty }$ and $\left[ -O,O\right] ^{\infty }$ respectively).

In order to understand the concept of a $\left( m,\sigma \right)$ function $\varphi$, the reader can consider the particular case where $\varphi$ is linear, and so it is defined by Remark 20; moreover, if $\varphi$ is a linear and $\left( m,\sigma \right)$ function, it is possible define the determinant of $\varphi$ (see Theorem 22 and Definition 23): a concept without sense, if $\varphi$ is not $\left( m,\sigma \right)$. The non linear case of Definition 17 can be seen as a generalization of the linear case. After Definition 23, we provide two examples: Examples 24 and 26; a further example (Example 42) will be exposed at the end of Sect. 4, in order to solve the integral (1).

Definition 17

Let $m\in \textbf{N}$, let $\emptyset \ne U^{(m)} \subset \textbf{R}^{m}$, let $U^{(\infty )}\subset \ell _{\infty }$ such that $\pi _{i}(U^{(\infty )})=\textbf{R}$ for any $i\in \textbf{N}^{*}$, let $U=U^{(m)}\times U^{(\infty )}\subset \mathcal {D}_{\infty }(U)$ (with the convention $U=U^{(\infty )}$ if $m=0$), and let $\sigma :\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\} \longrightarrow \textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\}$ be an increasing function; a function $\varphi =(\varphi _{i}:i\in \textbf{N}^{*}):U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ is called $\left( m,\sigma \right)$ if, for any $i\in \textbf{N}^{*}$ and for any $j>m$, there exist some functions $\varphi _{i}^{(\infty ,m)}:U^{(m)}\longrightarrow \textbf{R}$ and $\varphi _{ij}:\textbf{R}\longrightarrow \textbf{R}$ such that:

• 1. $\forall \,i\in \textbf{N}^{*}$, $\forall \,x=(x_{j}:j\in \textbf{N} ^{*})\in U$, one has

  $$\begin{aligned} \varphi _{i}(x)=\varphi _{i}^{(\infty ,m)}(x_{1},\ldots ,x_{m})+\underset{j=m+1}{\overset{\infty }{ {\displaystyle \sum } }}\varphi _{ij}(x_{j}); \end{aligned}$$

  (6)

  moreover, $\forall \,(x_{1},\ldots ,x_{m})\in U^{(m)}$, one has $\lim _{i\rightarrow \infty }\varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m} \right) =0$.

• 2. $\forall \,i\in \left\{ 1,\ldots ,m\right\}$, $\forall \,x=(x_{j}:j>m)\in U^{(\infty )}$, one has

  $$\begin{aligned} \underset{j=m+1}{\overset{\infty }{ {\displaystyle \sum } }}\left| \varphi _{ij}(x_{j})\right| <\infty . \end{aligned}$$

• 3. $\forall \,i>m$, $\forall \,j\in \left\{ 1,\ldots ,m\right\}$, there exists $\frac{\partial \varphi _{i}^{(\infty ,m)}}{\partial x_{j}}:U^{(m)} \longrightarrow \textbf{R}$; moreover, $\forall \,(x_{1},\ldots ,x_{m})\in U^{(m)}$, one has $\sum _{i=m+1}^{\infty }\left| \frac{\partial \varphi _{i}^{(\infty ,m)}(x_{1},\ldots ,x_{m})}{\partial x_{j}}\right| <\infty$.

• 4. $\forall \,i>m$, $\forall \,j>m$ such that $j\ne \sigma (i)$, $\forall \,t\in \textbf{R}$, one has $\varphi _{ij}(t)=0$; moreover, the function $\varphi _{i,\sigma (i)}$ is constant or derivable such that $\varphi _{i,\sigma (i)}^{\prime }(t)\ne 0$ $\forall \,t\in \textbf{R}$, and, $\forall \,\left( x_{j}:j\in \sigma (\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\} )\right) \in \ell _{\sigma (\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\} )}$, one has $\sup _{i>m}\left| \varphi _{i,\sigma (i)}^{\prime }(x_{\sigma (i)})\right| <\infty$.

• 5. By setting

  $$\begin{aligned} \mathcal {I}(\varphi )=\left\{ i>m:\sigma (h)=h\text {, }\forall \,h\ge i\text {, and }\varphi _{ii}\text { is not constant}\right\} , \end{aligned}$$

  $\forall \,x=(x_{i}:i\in \mathcal {I}(\varphi ))\in \ell _{\mathcal {I}(\varphi )}$, there exists $\prod _{i\in \mathcal {I}(\varphi )}\varphi _{ii}^{\prime }\left( x_{i}\right) \in \textbf{R}^{*}$; moreover, there exists $d(\varphi )\in \textbf{R}^{+}$ such that, $\forall \,i\in \mathcal {I}(\varphi )$, $\forall \,t\in \textbf{R}$, one has $\left| \varphi _{ii}^{\prime }(t)\right| \ge d(\varphi )$.

Definition 18

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function; then:

• 1. $\varphi$ is called $\left( m,\sigma \right)$ standard (or $\left( m,\sigma \right)$ of the first type) if, for any $i>m$ and for any $\left( x_{1},\ldots ,x_{m}\right) \in U^{(m)}$, one has $\varphi _{i}^{(\infty ,m)} (x_{1},\ldots ,x_{m})=0$.

• 2. $\varphi$ is called $\left( m,\sigma \right)$ of the second type if, for any $i\in \left\{ 1,\ldots ,m\right\}$, for any $j>m$ and for any $t\in \textbf{R}$, one has $\varphi _{ij}(t)=0$.

• 3. $\varphi$ is called $\left( m,\sigma \right)$ elementary if it is $\left( m,\sigma \right)$-standard and $\left( m,\sigma \right)$ of the second type.

Definition 19

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function; for any $\emptyset \ne L\subset \textbf{N}^{*}$ and for any $\left\{ 1,\ldots ,m\right\} \subset N\subset \textbf{N}^{*}$, denote by $\varphi ^{(L,N)}=\left( \varphi _{i}^{(L,N)}:i\in L\right) :\pi _{N}(U)\longrightarrow \textbf{R}^{L}$ the function given by

$$\begin{aligned} \varphi _{i}^{(L,N)}(x_{j}:j\in N)= & {} \varphi _{i}^{(\infty ,m)}(x_{1},\ldots ,x_{m})\nonumber \\{} & {} +\underset{j\in N\backslash \left\{ 1,\ldots ,m\right\} }{{\sum }}\varphi _{ij}(x_{j}),\quad \forall \,(x_{j}:j\in N)\in \pi _{N}(U),\nonumber \\{} & {} \quad \forall \,i\in L. \end{aligned}$$

(7)

Furthermore, for any $\emptyset \ne L\subset \textbf{N}^{*}$ and for any $\emptyset \ne N\subset \textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\}$, indicate by $\varphi ^{(L,N)}=\left( \varphi _{i}^{(L,N)}:i\in L\right) :\pi _{N}(U)\longrightarrow \textbf{R}^{L}$ the function given by

$$\begin{aligned} \varphi _{i}^{(L,N)}(x_{j}:j\in N)=\underset{j\in N}{ {\displaystyle \sum } }\varphi _{ij}(x_{j})\text {, }\forall \,(x_{j}:j\in N)\in \pi _{N}(U)\text {, }\forall \,i\in L. \end{aligned}$$

(8)

In particular, suppose that $m=1$, $U^{(1)}=\textbf{R}$; then, indicate $\varphi _{i}^{(\infty ,1)}$ by $\varphi _{i1}$, for any $i\in \textbf{N}^{*}$; moreover, for any $\emptyset \ne L\subset \textbf{N}^{*}$ and for any $\emptyset \ne N\subset \textbf{N}^{*}$, indicate by $\varphi ^{(L,N)}:\pi _{N}(U)\longrightarrow \textbf{R}^{L}$ the function defined by formula (8).

Furthermore, for any $l,o\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, indicate $\varphi ^{(I_{l},N)}$ by $\varphi ^{(l,N)}$, $\varphi ^{(L,I_{o})}$ by $\varphi ^{(L,o)}$, and $\varphi ^{(I_{l},I_{o})}$ by $\varphi ^{(l,o)}$.

Moreover, for any $\left\{ 1,\ldots ,m\right\} \subset L\subset \textbf{N}^{*}$ and for any $\left\{ 1,\ldots ,m\right\} \subset N\subset \textbf{N}^{*}$, define the function $\overline{\varphi }^{(L,N)}:U\longrightarrow \textbf{R}^{\infty }$ in the following manner:

$$\begin{aligned} \overline{\varphi }_{i}^{(L,N)}(x)=\left\{ \begin{array}{ll} \varphi _{i}^{(L,N)}(x_{j}:j\in N) &{}\quad \forall \,i\in \left\{ 1,\ldots ,m\right\} ,\quad \forall \,x\in U\\ \varphi _{i}(x) &{} \quad \forall \,i\in L\backslash \left\{ 1,\ldots ,m\right\} ,\quad \forall \,x\in U\\ \varphi _{i,\sigma (i)}(x_{\sigma (i)}) &{}\quad \forall \,i\in \textbf{N}^{*}\backslash L, \quad \forall \,x\in U \end{array}\right. . \end{aligned}$$

Finally, for any $l,o\in \left( \textbf{N}^{*}\backslash \left\{ 1,\ldots ,m-1\right\} \right) \cup \left\{ \infty \right\}$, indicate $\overline{\varphi }^{(I_{l},N)}$ by $\overline{\varphi }^{(l,N)}$, $\overline{\varphi }^{(L,I_{o})}$ by $\overline{\varphi }^{(L,o)}$, $\overline{\varphi }^{(I_{l},I_{o})}$ by $\overline{\varphi }^{(l,o)}$, and $\overline{\varphi }^{(m,m)}$ by $\overline{\varphi }$.

Remark 20

Let $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:\ell _{\infty }\longrightarrow \ell _{\infty }$ be a linear function, let $m\in \textbf{N}$ and let $\sigma :\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\} \longrightarrow \textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\}$ be an increasing function; then, A is $\left( m,\sigma \right)$ if and only if:

• 1. $\forall \,j\in \left\{ 1,\ldots ,m\right\}$, one has $\sum _{i=1}^{\infty }\left| a_{ij}\right| <\infty$.

• 2. $\forall \,i>m$, $\forall \,j>m$ such that $j\ne \sigma (i)$, one has $a_{ij}=0$.

• 3. There exists $\prod _{i\in \mathcal {I}(A)}a_{ii}\in \textbf{R}^{*}$, where

  $$\begin{aligned} \mathcal {I}(A)=\left\{ i>m:\sigma (h)=h, \, \forall \,h\ge i,\,\text {and}\,a_{ii}\ne 0\right\} . \end{aligned}$$

  (9)

Moreover, if the points 1, 2 and 3 hold, one has $\sup _{i>m}\left| a_{i,\sigma (i)}\right| <\infty$. Finally, A is $\left( m,\sigma \right)$ standard if and only if A is $\left( m,\sigma \right)$ and $a_{ij}=0$, for any $i>m$, for any $j\in \left\{ 1,\ldots ,m\right\}$, and A is $\left( m,\sigma \right)$ of the second type if and only if A is $\left( m,\sigma \right)$ and $a_{ij}=0$, for any $i\in \left\{ 1,\ldots ,m\right\}$, for any $j>m$.

Proof

Since A is linear, formula (6) is alwais satisfied, with $\varphi _{i}^{(\infty ,m)}(x_{1},\ldots ,x_{m})=\sum _{k=1}^{m} a_{ik}x_{k}$, $\forall \,i\in \textbf{N}^{*}$, $\forall \,(x_{1},\ldots ,x_{m})\in \textbf{R}^{m}$; analogously, the point 2 of Definition 17 is always satisfied, by Proposition 16. Moreover, it is easy to prove that the point 3 of Definition 17 is equivalent to the condition $\sum _{i=1}^{\infty }\left| a_{ij}\right| <\infty$, $\forall \,j\in \left\{ 1,\ldots ,m\right\}$, and the point 4 of Definition 17 is equivalent to the condition $a_{ij}=0$, $\forall \,i>m$, $\forall \,j>m$, $j\ne \sigma (i)$, and $\sup _{i>m}\left| a_{i,\sigma (i)}\right| <\infty$; moreover, the point 5 of Definition 17 is equivalent to the condition that there exists $\prod _{i\in \mathcal {I}(A)}a_{ii}\in \textbf{R}^{*}$, where $\mathcal {I}(A)$ is defined by formula (9). Furthermore, if $\sum _{i=1}^{\infty } \left| a_{ij}\right| <\infty$, $\forall \,j\in \left\{ 1,\ldots ,m\right\}$, then $\sup _{i\in \textbf{N}^{*}} \sum _{j=1}^{m}\left| a_{ij}\right| <\infty$, and so $\varphi ^{(\infty ,m)}\left( \textbf{R}^{m}\right) \subset \ell _{\infty }$; then, from Remark 54, we have $\sup _{i>m}\left| a_{i,\sigma (i)}x_{\sigma (i)}\right| <\infty$, $\forall \,\left( x_{j}:j\in \sigma (\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\} )\right) \in \ell _{\sigma (\textbf{N} ^{*}\backslash \left\{ 1,\ldots ,m\right\} )}$, and so $\sup _{i>m} \left| a_{i,\sigma (i)}\right| <\infty$; moreover, we have $\lim _{i\rightarrow \infty }\varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m} \right) =\lim {i\rightarrow \infty } \sum _{k=1}^{m} a_{ik}x_{k}=0$, $\forall \,(x_{1},\ldots ,x_{m})\in \textbf{R}^{m}$. Thus, A is $\left( m,\sigma \right)$, with $U^{(m)}=\textbf{R}^{m}$, if and only if the points 1, 2 and 3 hold.

Finally, since $\varphi _{i}^{(\infty ,m)}(x_{1},\ldots ,x_{m})=\sum _{k=1}^{m} a_{ik}x_{k}$, $\forall \,i>m$, $\forall \,(x_{1},\ldots ,x_{m})\in \textbf{R}^{m}$, and $\varphi _{ij}(t)=a_{ij}t$, $\forall \,i\in \left\{ 1,\ldots ,m\right\}$, $\forall \,j>m$, $\forall \,t\in \textbf{R}$, then A is $\left( m,\sigma \right)$ standard if and only if A is $\left( m,\sigma \right)$ and $a_{ij}=0$, $\forall \,i>m$, $\forall \,j\in \left\{ 1,\ldots ,m\right\}$, and A is $\left( m,\sigma \right)$ of the second type if and only if A is $\left( m,\sigma \right)$ and $a_{ij}=0$, $\forall \,i\in \left\{ 1,\ldots ,m\right\}$, $\forall \,j>m$. $\square$

The following definitions and results (from Proposition 21 to Definition 23) can be found in [7].

Proposition 21

Let $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:\ell _{\infty }\longrightarrow \ell _{\infty }$ be a linear $\left( m,\sigma \right)$ function; then, A is continuous.

Theorem 22

Let $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:\ell _{\infty }\longrightarrow \ell _{\infty }$ be a linear $\left( m,\sigma \right)$ function, and let $A^{(n,n)}=\left( a_{ij}\right) _{i,j\in \left\{ 1,\ldots ,n\right\} }$, for any $n\ge m$; then, the sequence $\left\{ \det A^{(n,n)}\right\} _{n\ge m}$ converges to a real number.

Definition 23

Let $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:\ell _{\infty }\longrightarrow \ell _{\infty }$ be a linear $\left( m,\sigma \right)$ function; define the determinant of A, and call it $\det A$, the real number

$$\begin{aligned} \det A=\underset{n\longrightarrow \infty }{\lim }\det A^{(n,n)}. \end{aligned}$$

Example 24

Consider the linear function $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:\ell _{\infty }\longrightarrow \ell _{\infty }$ given by

$$\begin{aligned} (Ax)_{i}=\left\{ \begin{array}{ll} \underset{j\in \textbf{N}^{*}}{\sum }2^{-j}x_{j} &{}\quad \text {if}\quad i=1\\ x_{1}+\underset{j\in \textbf{N}^{*}}{\sum }2^{-j}x_{j} &{}\quad \text {if}\quad i=2\\ 2^{-i}x_{1}+2^{2^{-i}}x_{i} &{}\quad \text {if}\quad i>2 \end{array}\right. , \quad \forall \,x=\left( x_{j}:j\in \textbf{N}^{*}\right) \in \ell _{\infty }; \end{aligned}$$

observe that A is expressed by the following infinite-dimensional matrix:

$$\begin{aligned} A=\left( \begin{array}{cccccc} \frac{1}{2} &{} \frac{1}{4} &{} \frac{1}{8} &{} \cdots &{} \cdots &{} \cdots \\ \frac{3}{2} &{} \frac{1}{4} &{} \frac{1}{8} &{} \cdots &{} \cdots &{} \cdots \\ \frac{1}{8} &{} 0 &{} 2^{\frac{1}{8}} &{} 0 &{} 0 &{} \cdots \\ \frac{1}{16} &{} 0 &{} 0 &{} 2^{\frac{1}{16}} &{} 0 &{} \cdots \\ \vdots &{} 0 &{} 0 &{} 0 &{} \ddots &{} \ddots \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \ddots &{} \ddots \end{array}\right) ; \end{aligned}$$

then, from Remark 20, A is $\left( 2,\sigma \right)$, where $\sigma (i)=i$, $a_{ii}=2^{2^{-i}}$, $\forall \,i>2$, and so $\mathcal {I} (A)=\textbf{N}^{*}\backslash \left\{ 1,2\right\}$; moreover, $\forall \,n>1$, we have

$$\begin{aligned} A^{(n,n)}= & {} \left( \begin{array}{ccccc} \frac{1}{2} &{} \frac{1}{4} &{} \frac{1}{8} &{} \cdots &{} \frac{1}{2^{n}}\\ \frac{3}{2} &{} \frac{1}{4} &{} \frac{1}{8} &{} \cdots &{} \frac{1}{2^{n}}\\ \frac{1}{8} &{} 0 &{} 2^{\frac{1}{8}} &{} 0 &{} \cdots \\ \vdots &{} \vdots &{} 0 &{} \ddots &{} 0\\ \frac{1}{2^{n}} &{} 0 &{} \vdots &{} 0 &{} 2^{\frac{1}{2^{n}}} \end{array}\right) ,\\ \overline{A}^{(n,n)}= & {} \left( \begin{array}{ccccccc} \frac{1}{2} &{} \frac{1}{4} &{} \frac{1}{8} &{} \cdots &{} \frac{1}{2^{n}} &{} 0 &{}\cdots \\ \frac{3}{2} &{} \frac{1}{4} &{} \frac{1}{8} &{} \cdots &{} \frac{1}{2^{n}} &{} 0 &{}\cdots \\ \frac{1}{8} &{} 0 &{} 2^{\frac{1}{8}} &{} 0 &{} 0 &{} 0 &{} \cdots \\ \vdots &{} \vdots &{} 0 &{} \ddots &{} 0 &{} 0 &{} \cdots \\ \frac{1}{2^{n}} &{} 0 &{} \vdots &{} 0 &{} 2^{\frac{1}{2^{n}}} &{} 0 &{} \cdots \\ 0 &{} \vdots &{} \vdots &{} \vdots &{} 0 &{} 2^{\frac{1}{2^{n+1}}} &{} \ddots \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \ddots &{} \ddots \end{array}\right) . \end{aligned}$$

In order to calculate $\det A$, observe that, $\forall \,n>1$, we have $A^{\left( \{2\},n\right) }=u_{n}+v_{n}$, where $u_{n}=A^{\left( 1,n\right) }=\left( 2^{-j}:j\in \left\{ 1,\ldots ,n\right\} \right) \in \textbf{R}^{n}$, and $v_{n}=\left( 1,0,\ldots ,0\right) \in \textbf{R}^{n}$. Then, we have $\det A^{(n,n)}=\det U^{(n,n)}+\det V^{(n,n)}$, where $U^{(n,n)}=\left( u_{ij}^{(n)}\right) _{i,j\in \textbf{N}^{*}}$ and $V^{(n,n)}=\left( v_{ij}^{(n,n)}\right) _{i,j\in \textbf{N}^{*}}$ are the linear functions obtained by substituting the second row of $A^{(n,n)}$ by $u_{n}$ and $v_{n}$, respectively; moreover, since $\left( U^{(n,n)}\right) ^{\left( 1,n\right) }=\left( U^{(n,n)}\right) ^{\left( \{2\},n\right) }$, we have $\det U^{(n,n)}=0$, from which

$$\begin{aligned} \det A=\underset{n\longrightarrow \infty }{\lim }\det A^{(n,n)}=\underset{n\longrightarrow \infty }{\lim }\det V^{(n,n)}. \end{aligned}$$

(10)

Finally, $\forall \,n>2$, we have

$$\begin{aligned} \det V^{(n,n)}&=(-1)^{n+1}2^{-n}\det V^{\left( n-1,\{2,\ldots ,n\}\right) }+2^{2^{-n}}\det V^{\left( n-1,n-1\right) }\nonumber \\&=2^{2^{-n}}\det V^{\left( n-1,n-1\right) }, \end{aligned}$$

(11)

since the second row of $V^{\left( n-1,\{2,\ldots ,n\}\right) }$ is zero, and so $\det V^{\left( n-1,\{2,\ldots ,n\}\right) }=0$. Then, by recursion, from (11) we obtain

$$\begin{aligned} \det V^{(n,n)}=\det V^{\left( 2,2\right) }\underset{j=3}{\overset{n}{\prod }\ }2^{2^{-j}}, \end{aligned}$$

and so formula (10) implies

$$\begin{aligned} \det A=\det V^{\left( 2,2\right) }\underset{n\longrightarrow \infty }{\lim }\underset{j=3}{\overset{n}{\prod }}2^{2^{-j}}=\det V^{\left( 2,2\right) }2^{\left. \underset{j=3}{\overset{\infty }{\sum }}2^{-j}\right. }=-\frac{1}{4}\root 4 \of {2}. \end{aligned}$$

Lemma 25

Let $k\in \textbf{N}$ and let $a=a_{i}:i>k\in \left( -1,+\infty \right) ^{\infty }$ such that $\sum _{i=k+1}{\infty }\left| a_{i}\right| \in \textbf{R}$; then, one has $\prod _{i=k+1}^{\infty }\left( 1+a_{i}\right) \in \textbf{R}^{+}$.

Proof

Since $\sum _{i=k+1}^{\infty }\left| a_{i}\right| \in \textbf{R}$, we have $\lim _{i\rightarrow \infty } a_{i}=0$; then, since $\lim _{t\rightarrow 0}\frac{\ln (1+t)}{t}=1$, there exists $i_{0}>k$ such that, $\forall \,i\in \textbf{N}$, $i\ge i_{0}$, we have $\frac{\ln (1+\left| a_{i}\right| )}{\left| a_{i}\right| }\le 2$, $\frac{\ln (1-\left| a_{i}\right| )}{-\left| a_{i}\right| }\le 2$, from which

$$\begin{aligned} -2\left| a_{i}\right|\le & {} \ln \left( 1-\left| a_{i}\right| \right) \le \ln \left( 1+a_{i}\right) \le \ln \left( 1+\left| a_{i}\right| \right) \le 2\left| a_{i}\right| \\{} & {} \Longrightarrow \left| \ln \left( 1+a_{i}\right) \right| \le 2\left| a_{i}\right| ; \end{aligned}$$

then, we have $\sum _{i=i_{0}}^{\infty }\left| \ln \left( 1+a_{i}\right) \right| \le 2\sum _{i=i_{0}}^{\infty }\left| a_{i}\right| <\infty$, from which $\sum _{i=i_{0}}^{\infty }\ln \left( 1+a_{i}\right) \in \textbf{R}$, and so

$$\begin{aligned} \underset{i=k+1}{\overset{\infty }{\prod }}\left( 1+a_{i}\right)= & {} \left( \underset{i=k+1}{\overset{i_{0}-1}{\prod }}\left( 1+a_{i}\right) \right) \exp \left( \ln \underset{i=i_{0}}{\overset{\infty }{ {\displaystyle \prod } }}\left( 1+a_{i}\right) \right) \\= & {} \left( \underset{i=k+1}{\overset{i_{0}-1}{\prod }}\left( 1+a_{i}\right) \right) \exp \left( \underset{i=i_{0}}{\overset{\infty }{\sum }}\ln \left( 1+a_{i}\right) \right) \in \textbf{R}^{+}. \end{aligned}$$

$\square$

Example 26

Consider the function $\varphi :\ell _{\infty }\longrightarrow \ell _{\infty }$ given by

$$\begin{aligned} \varphi _{i}(x)= & {} \left\{ \begin{array}{ll} \underset{j\in \textbf{N}^{*}}{\sum }2^{-j}\sin \left( x_{j}\right) &{}\quad \text {if}\quad i=1\\ 2^{-i}\cos \left( x_{1}\right) +\left( 1+\frac{1}{i^{2}}\right) x_{i} +\frac{\sin (x_{i})}{i^{2}} &{} \quad \text {if}\quad i>1 \end{array}\right. ,\\ \forall \,x= & {} \left( x_{j}:j\in \textbf{N}^{*}\right) \in \ell _{\infty }; \end{aligned}$$

moreover, $\forall \,i>1$, let $g_{i}:\mathbf {R\longrightarrow R}$ be the function defined by $g_{i}(t)=\left( 1+\frac{1}{i^{2}}\right) t+\frac{\sin (t)}{i^{2}}$, $\forall \,t\in \textbf{R}$; we have $\left| g_{i}^{\prime }(t)\right| =g_{i}^{\prime }(t)=1+\frac{1}{i^{2}}+\frac{\cos (t)}{i^{2}}\ge 1$, $\forall \,i>1$, $\forall \,t\in \textbf{R}$; moreover, $\forall \,x=(x_{j}:j>1)\in \textbf{R}^{\infty }$, we obtain

$$\begin{aligned} \underset{i=2}{\overset{\infty }{ {\displaystyle \prod } }}g_{i}^{\prime }\left( x_{i}\right) =\underset{i=2}{\overset{\infty }{ {\displaystyle \prod } }}\left( 1+\frac{1}{i^{2}}+\frac{\cos (x_{i})}{i^{2}}\right) ; \end{aligned}$$

(12)

thus, since $\frac{1}{i^{2}}+\frac{\cos (x_{i})}{i^{2}}\ge 0>-1$, $\forall \,i>1$, and $\sum _{i=2}^{\infty }\left( \frac{1}{i^{2}}+\frac{\cos (x_{i})}{i^{2}}\right) \in \textbf{R}$, from formula (12) and Lemma 25, we have $\prod _{i=2}^{\infty }g_{i}^{\prime }\left( x_{i}\right) \in \textbf{R}^{+}\subset \textbf{R} ^{*}$.

Then, $\varphi$ is $\left( 1,\sigma \right)$, where $\varphi _{1} ^{(\infty ,1)}(x_{1})=\frac{\sin (x_{1})}{2}$, $\varphi _{i}^{(\infty ,1)} (x_{1})=2^{-i}\cos (x_{1})$, $\varphi _{1j}(t)=2^{-j}\sin (t)$, $\sigma (i)=i$, $\varphi _{ii}(t)=\left( 1+\frac{1}{i^{2}}\right) t+\frac{\sin (t)}{i^{2}}$, $\forall \,i,j>1$, $\forall \,t\in \textbf{R}$, and so $\mathcal {I} (\varphi )=\textbf{N}^{*}\backslash \left\{ 1\right\}$.

4 Gaussian change of variables’ formula

The following two definitions (Definition 27 and Definition 28) can be found, for example, in [11].

Definition 27

Let E be a vector space on $\textbf{R}$ and let T be a topology on E; a probability measure $\mu$ on $\left( E,\sigma (T)\right)$ is called Gaussian measure if, for any function $g:\left( E,T\right) \longrightarrow \left( \textbf{R},\tau \right)$ linear and continuous, the probability measure $\mu \circ g^{-1}$ on $\left( \textbf{R},\mathcal {B} \right)$ is Gaussian; moreover, a random variable $X:\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( E,\sigma (T)\right)$ is called Gaussian random variable on E if its law is a Gaussian measure.

Definition 28

Let E be a Banach space, with the topology T induced by the norm of E, and let $\mu$ be a Gaussian measure on $\left( E,\sigma (T)\right)$; define

$$\begin{aligned}{} & {} \left| x\right| _{H(\mu )}=\sup \left\{ \left| g(x)\right| :g\in E^{*},Var\left[ g\right] \le 1\right\} \text {, }\forall \,x\in E,\nonumber \\{} & {} \quad H(\mu )=\left\{ x\in E:\left| x\right| _{H(\mu )}\in \left[ 0,\infty \right) \right\} . \end{aligned}$$

(13)

The vector space $H(\mu )$ on $\textbf{R}$ (that can be indicated simply by H) is called Cameron–Martin space associated to $\left( E,\mu \right)$. In particular, if E is separable, the pair $\left( E,\mu \right)$ (indicated by $\left( E,H,\mu \right)$ too) is called abstract Wiener space.

Remark 29

Let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }$, let $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*} }:\textbf{R}^{\infty }\longrightarrow \textbf{R}^{\infty }$ be a linear and continuous function, and let $X=\left( X_{i}:i\in \textbf{N}^{*}\right) :\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( \textbf{R} ^{\infty },\mathcal {B}^{\infty }\right)$ be a Gaussian random variable; then, the random variable $AX+a:\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( \textbf{R}^{\infty },\mathcal {B}^{\infty }\right)$ is Gaussian.

Proof

$\forall \,g\in E^{*}$, we have $g(AX+a)=(g\circ A)(X)+g(a)$; moreover, since $(g\circ A)\in E^{*}$, the random variable $(g\circ A)(X):\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( \textbf{R},\mathcal {B} \right)$ is Gaussian, and so the random variable $g(AX+a):\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( \textbf{R},\mathcal {B} \right)$ is Gaussian too. $\square$

Definition 30

Let $m\in \textbf{N}^{*}$ and let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R}^{\infty }$ be a linear and symmetric function, such that $\gamma _{ij}=0$, for any $(i,j)\notin \left\{ 1,\ldots ,m\right\} ^{2} \bigcup \left( \bigcup _{k=m+1}^{\infty }\left\{ (k,k)\right\} \right)$; we say that:

• 1. $\Gamma$ is m-positive semidefinite if the matrix $\Gamma ^{(m,m)}$ is symmetric positive semidefinite, $\gamma _{ii}\in \left[ 0,\infty \right)$, for any $i>m$.

• 2. $\Gamma$ is m-positive definite if the matrix $\Gamma ^{(m,m)}$ is symmetric positive definite, $\gamma _{ii}\in \textbf{R}^{+}$, for any $i>m$.

Definition 31

Let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R}^{\infty }$ be a linear and diagonal function; we say that:

• 1. $\Gamma$ is 0-positive semidefinite if $\gamma _{ii}\in \left[ 0,\infty \right)$, for any $i\in \textbf{N}^{*}$.

• 2. $\Gamma$ is 0-positive definite if $\gamma _{ii}\in \textbf{R}^{+}$, for any $i\in \textbf{N}^{*}$.

Definition 32

Let $m\in \textbf{N}$, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }$ and let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R} ^{\infty }$ be a linear, symmetric and m-positive definite function; define the probability measure $N(a,\Gamma )$ on $\left( \textbf{R}^{\infty },\mathcal {B}^{\infty }\right)$ by

$$\begin{aligned} N(a,\Gamma )=N\left( (a_{1},\ldots ,a_{m}),\Gamma ^{(m,m)}\right) \otimes \left( \underset{i>m}{ {\displaystyle \bigotimes } }N\left( a_{i},\gamma _{ii}\right) \right) . \end{aligned}$$

Moreover, for any $A\subset \textbf{R}^{\infty }$ such that $\pi _{i} (A)=\textbf{R}$ for any $i\in \textbf{N}^{*}$, indicate by $N(a,\Gamma )$ the measure $\left. N(a,\Gamma )\right| _{\mathcal {B}^{\infty }(A)}$ too.

Remark 33

Let $m\in \textbf{N}$, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }$, let $b=\left( b_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }$, let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R} ^{\infty }$ be a symmetric m-positive definite linear function, and let $X=\left( X_{i}:i\in \textbf{N}^{*}\right) :\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( \textbf{R}^{\infty },\mathcal {B}^{\infty }\right)$ be a random variable with law $N(a,\Gamma )$; then, the law of the random variable $X+b$ is $N(a+b,\Gamma )$.

Proof

Let $\mathcal {A}=\left\{ \prod _{i=1}^{\infty } A_{i}:A_{i}\in \mathcal {B}, \forall \,i\in \textbf{N}^{*}\right\} \subset \mathcal {B}^{\infty }$; $\forall \,A= \prod _{i=1}^{\infty } A_{i}\in \mathcal {A}$, we have

$$\begin{aligned} P(X+b\in A)= & {} P\left( X\in \underset{i=1}{\overset{\infty }{\prod }}A_{i}-b\right) \\= & {} N\left( (a_{1},\ldots ,a_{m}),\Gamma ^{(m,m)}\right) \left( \underset{i=1}{\overset{m}{\prod }}\left( A_{i}-b_{i}\right) \right) \\{} & {} \cdot \underset{j=m+1}{\overset{\infty }{\prod }}N\left( a_{j},\gamma _{jj}\right) \left( A_{j}-b_{j}\right) \\= & {} N\left( (a_{1}+b_{1},\ldots ,a_{m}+b_{m}),\Gamma ^{(m,m)}\right) \left( \underset{i=1}{\overset{m}{\prod }}A_{i}\right) \\{} & {} \cdot \underset{j=m+1}{\overset{\infty }{\prod }}N\left( a_{j}+b_{j},\gamma _{jj}\right) \left( A_{j}\right) \\= & {} N(a+b,\Gamma )(A); \end{aligned}$$

then, the law $\mathcal {L}_{X+b}$ of $X+b$ and $N(a+b,\Gamma )$ coincide on $\mathcal {A}$; moreover, we have $\mathcal {L}_{X+b}(\textbf{R}^{\infty })=N(a+b,\Gamma )\left( \textbf{R}^{\infty }\right) =1<\infty$, $\textbf{R} ^{\infty }\in \mathcal {I}$, and so $\mathcal {L}_{X+b}$ and $N(a+b,\Gamma )$ are $\sigma$- finite on $\mathcal {A}$. Then, since $\mathcal {A}$ is a $\pi$-system on $\textbf{R}^{\infty }$ such that $\sigma (\mathcal {A})=\mathcal {B}^{\infty }$, from Proposition 45, $\mathcal {L}_{X+b}$ and $N(a+b,\Gamma )$ coincide on $\mathcal {B}^{\infty }$. $\square$

Lemma 34

$\forall \,x\in \textbf{R}^{+}$, one has

$$\begin{aligned} \overset{\infty }{\underset{x}{\int }}\exp \left( -\frac{t^{2}}{2}\right) dt\le \frac{1}{x}\exp \left( -\frac{x^{2}}{2}\right) . \end{aligned}$$

Proof

See for example, Lemma 1.1.3 in [11]. $\square$

Theorem 35

Let $\left\{ X_{i}\right\} _{i\in \textbf{N}^{*}}$ be a sequence of independent random variables such that $E\left[ X_{i}\right] =0$, for any $i\in \textbf{N}^{*}$, and $\sum _{i=1}^{\infty } Var\left[ X_{i}\right] \in \left[ 0,\infty \right)$; then, there exists a real random variable Z such that $\sum _{i=1}^{n}X_{i}\overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}Z$.

Proof

See, for example, page 112 in the Williams’ book [26]. $\square$

Proposition 36

Let $m\in \textbf{N}$, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }$ and let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R} ^{\infty }$ be a symmetric m-positive definite linear function, such that $\sum _{i=1}^{\infty }\gamma _{ii}\in \textbf{R}^{+}$; then:

• 1. $\forall \,c\in \textbf{R}^{+}$, one has $N(a,\Gamma )\left( \prod _{i=1}^{\infty }\left[ a_{i}-c,a_{i}+c\right] \right) \in \left( 0,1\right)$.

• 2. If $a\in E_{\infty }$, one has $N(a,\Gamma )(E_{\infty })=N(a,\Gamma )(\ell _{\infty })=1$; in particular, $N(a,\Gamma )$ is a probability measure on $\left( E_{\infty },\mathcal {B}_{\infty }\right)$.

• 3. If $a\in \ell _{2}$, one has $N(a,\Gamma )(\ell _{2})=1$.

• 4. If $a\in \ell _{1}$ and $\sum _{i=1}^{\infty } \sqrt{\gamma _{ii}}\in \textbf{R}^{+}$, one has $N(a,\Gamma )(\ell _{1})=1$.

Proof

We have

$$\begin{aligned}{} & {} N\left( (a_{1},\ldots ,a_{m}),\Gamma ^{(m,m)}\right) \left( \overset{m}{\underset{i=1}{\prod }}\left[ a_{i}-c,a_{i}+c\right] \right) \in \left( 0,1\right) , \quad \forall \,c\in \textbf{R}^{+},\\{} & {} N(a,\Gamma )(\ell _{\infty })=\left( \underset{i>m}{ {\displaystyle \bigotimes } }N\left( a_{i},\gamma _{ii}\right) \right) (\ell _{\infty }),\\{} & {} N(a,\Gamma )(E_{\infty })=\left( \underset{i>m}{ {\displaystyle \bigotimes } }N\left( a_{i},\gamma _{ii}\right) \right) (E_{\infty }),\\{} & {} N(a,\Gamma )(\ell _{2})=\left( \underset{i>m}{ {\displaystyle \bigotimes } }N\left( a_{i},\gamma _{ii}\right) \right) (\ell _{2}),\\{} & {} N(a,\Gamma )(\ell _{1})=\left( \underset{i>m}{ {\displaystyle \bigotimes } }N\left( a_{i},\gamma _{ii}\right) \right) (\ell _{1}); \end{aligned}$$

then, we can prove this proposition by supposing that $\Gamma$ is 0-positive definite.

• 1. $\forall \,c\in \textbf{R}^{+}$, we have

  $$\begin{aligned} N(0,\Gamma )\left( \left[ -c,c\right] ^{\infty }\right)\le & {} N(0,\Gamma )\left( \left[ -c,c\right] \times \textbf{R}^{\infty }\right) \\= & {} N(0,\gamma _{11})\left( \left[ -c,c\right] \right) <1; \end{aligned}$$

  moreover

  $$\begin{aligned} N(a,\Gamma )\left( \left[ -c,c\right] ^{\infty }\right) =\underset{i=1}{\overset{\infty }{\prod }}\left( \overset{c}{\underset{-c}{\int }}\frac{1}{\sqrt{2\pi \gamma _{ii}}}\exp \left( -\frac{x^{2}}{2\gamma _{ii}}\right) dx\right) =\underset{i=1}{\overset{\infty }{\prod }}A_{i}, \end{aligned}$$

  (14)

  where

  $$\begin{aligned} A_{i}=A_{i}(c)\equiv \overset{\frac{c}{\sqrt{\gamma _{ii}}}}{\underset{-\frac{c}{\sqrt{\gamma _{ii}}}}{\int }}\frac{1}{\sqrt{2\pi }}\exp \left( -\frac{t^{2} }{2}\right) dt\in \left( 0,1\right) ,\quad \forall \,i\in \textbf{N}^{*}; \end{aligned}$$

  (15)

  furthermore

  $$\begin{aligned} A_{i}= & {} 2\overset{\frac{c}{\sqrt{\gamma _{ii}}}}{\underset{0}{\int }}\frac{1}{\sqrt{2\pi }}\exp \left( -\frac{t^{2}}{2}\right) dt=2\left( \frac{1}{2}-\overset{\infty }{\underset{\frac{c}{\sqrt{\gamma _{ii}}}}{\int }}\frac{1}{\sqrt{2\pi }}\exp \left( -\frac{t^{2}}{2}\right) dt\right) \nonumber \\= & {} 1-\sqrt{\frac{2}{\pi }}\overset{\infty }{\underset{\frac{c}{\sqrt{\gamma _{ii}} }}{\int }}\exp \left( -\frac{t^{2}}{2}\right) dt\ge 1\nonumber \\{} & {} -\sqrt{\frac{2\gamma _{ii}}{\pi c^{2}}}\exp \left( -\frac{c^{2}}{2\gamma _{ii}}\right) \quad \text {(from Lemma 34)}. \end{aligned}$$

  (16)

  Moreover, $\forall \,x\in \textbf{R}^{+}$, we have $\sqrt{x}\exp (-x)\le \frac{1}{\sqrt{2e}}$; then, since $\lim _{i\rightarrow \infty }\gamma _{ii}=0$, there exists $i_{0}=i_{0}(c)\in \textbf{N}^{*}$ such that, $\forall \,i\ge i_{0}$, we have $\gamma _{ii}\le c^{2}$, $\sqrt{\frac{c^{2} }{2\gamma _{ii}}}\exp \left( -\frac{c^{2}}{2\gamma _{ii}}\right) \le \frac{1}{\sqrt{2e}}$, from which

  $$\begin{aligned} \sqrt{\frac{2\gamma _{ii}}{\pi c^{2}}}\exp \left( -\frac{c^{2}}{2\gamma _{ii} }\right) \le \sqrt{\frac{2\gamma _{ii}}{\pi c^{2}}}\cdot \sqrt{\frac{\gamma _{ii}}{ec^{2}}}=\sqrt{\frac{2}{\pi e}}\cdot \frac{\gamma _{ii}}{c^{2} }\equiv x_{i}, \end{aligned}$$

  (17)

  where $x_{i}=x_{i}(c)\in \left( 0,\sqrt{\frac{2}{\pi e}}\right] \subset \left( 0,1\right)$; then, formulas (14), (16) and (17) imply

  $$\begin{aligned} N(0,\Gamma )\left( \left[ -c,c\right] ^{\infty }\right) \ge \left( \underset{i=1}{\overset{i_{0}-1}{\prod }}A_{i}\right) \underset{i=i_{0} }{\overset{\infty }{\prod }}\left( 1-x_{i}\right) ; \end{aligned}$$

  (18)

  finally, observe that $\sum _{i=i_{0}}^{\infty }\left( -x_{i}\right) =-\frac{1}{c^{2}}\sqrt{\frac{2}{\pi e}} \sum _{i=i_{0}}^{\infty }\gamma _{ii}\in \textbf{R}$; then, from Lemma 25, we have $\prod _{i=i_{0}}^{\infty }\left( 1-x_{i}\right) \in \textbf{R}^{+}$, and so formulas (15) and (18) imply $N(0,\Gamma )\left( \left[ -c,c\right] ^{\infty }\right) >0$, from which

  $$\begin{aligned} N(a,\Gamma )\left( \overset{\infty }{\underset{i=1}{\prod }}\left[ a_{i}-c,a_{i}+c\right] \right) =N(0,\Gamma )\left( \left[ -c,c\right] ^{\infty }\right) \in \left( 0,1\right) . \end{aligned}$$

• 2. Suppose $a\in E_{\infty }$, let $a_{\infty }=\lim _{i\rightarrow \infty }a_{i}\in \textbf{R}$ and let $\left\{ X_{i}\right\} _{i\in \textbf{N}^{*}}$ be a sequence of independent random variables such that $X_{i}\sim N\left( a_{i},\gamma _{ii}\right)$, $\forall \,i\in \textbf{N} ^{*}$; we have $E\left[ X_{i}-a_{i}\right] =0$, $Var\left[ X_{i} -a_{i}\right] =\gamma _{ii}$, $\forall \,i\in \textbf{N}^{*}$, and $\sum _{i=1}^{\infty }\gamma _{ii}\in \textbf{R}^{+}$; then, from Theorem 35, there exists a real random variable $T_{1}$ such that $\sum _{i=1}^{n}\left( X_{i}-a_{i}\right) \overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T_{1}$, from which $\left( X_{i}-a_{i}\right) \overset{a.\,s.}{\underset{i\rightarrow \infty }{\longrightarrow }}0$, and so $P\left( \lim _{i\rightarrow \infty }X_{i}=a_{\infty }\right) =1$; thus, since $\left\{ \lim _{i\rightarrow \infty }X_{i}=a_{\infty }\right\} \subset \left\{ \left( X_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }\right\}$, we obtain

  $$\begin{aligned} P\left( \left( X_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }\right) =1, \end{aligned}$$

  that is $N(a,\Gamma )(E_{\infty })=1$; moreover, since $E_{\infty }\subset \ell _{\infty }$, we have $N(0,\Gamma )(\ell _{\infty })=1$.

• 3. Suppose $a\in \ell _{2}$; since $\frac{X_{i}-a_{i}}{\sqrt{\gamma _{ii}} }\sim N(0,1)$, it is easy to prove that $\left( X_{i}-a_{i}\right) ^{2}=\gamma _{ii}\left( \frac{X_{i}-a_{i}}{\sqrt{\gamma _{ii}}}\right) ^{2}\sim \Gamma \left( \frac{1}{2},\frac{1}{2\gamma _{ii}}\right)$ (the Gamma law with parameters $\frac{1}{2}$, $\frac{1}{2\gamma _{ii}}$), $\forall \,i\in \textbf{N}^{*}$, and so $E\left[ \left( X_{i}-a_{i}\right) ^{2}-\gamma _{ii}\right] =0$, $Var\left[ \left( X_{i}-a_{i}\right) ^{2}-\gamma _{ii}\right] =2\gamma _{ii}^{2}$, $\sum _{i=1}^{\infty } 2\gamma _{ii}^{2}\in \textbf{R}^{+}$; then, from Theorem 35, there exists a real random variable $T_{2}$ such that $\sum _{i=1}^{n}\left( \left( X_{i}-a_{i}\right) ^{2}-\gamma _{ii}\right) \overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T_{2}$, from which $\sum _{i=1}^{n} \left( X_{i}-a_{i}\right) ^{2}\overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T_{2}+ \sum _{i=1}^{\infty }\gamma _{ii}$, and so $P\left( \sum _{i=1}^{\infty } \left( X_{i}-a_{i}\right) ^{2}\in \textbf{R}\right) =1$. Moreover, we have $a_{i}X_{i}\sim N\left( a_{i}^{2},a_{i}^{2}\gamma _{ii}\right)$, $\forall \,i\in \textbf{N}^{*}$, and so $E\left[ a_{i}X_{i}-a_{i} ^{2}\right] =0$, $Var\left[ a_{i}X_{i}-a_{i}^{2}\right] =a_{i}^{2} \gamma _{ii}$, $\sum _{i=1}^{\infty } a_{i}^{2}\gamma _{ii} \in \left[ 0,\infty \right)$; then, from Theorem 35, there exists a real random variable $T_{3}$ such that $\sum _{i=1}^{n}\left( a_{i}X_{i}-a_{i}^{2}\right) \overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T_{3}$, from which $\sum _{i=1}^{n} a_{i}X_{i}\overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T_{3}+\sum _{i=1}^{\infty } a_{i}^{2}$, and so $P\left( \sum _{i=1}^{\infty } a_{i}X_{i}\in \textbf{R}\right) =1$; furthermore, $\forall \,n\in \textbf{N}^{*}$, we have

  $$\begin{aligned} \underset{i=1}{\overset{n}{\sum }}X_{i}^{2}=\underset{i=1}{\overset{n}{\sum } }\left( X_{i}-a_{i}\right) ^{2}+2\underset{i=1}{\overset{n}{\sum }}a_{i} X_{i}-\underset{i=1}{\overset{n}{\sum }}a_{i}^{2}, \end{aligned}$$

  from which $P\left( \sum _{i=1}^{\infty } X_{i}^{2} \in \textbf{R}\right) =1$, that is $N(a,\Gamma )(\ell _{2})=1$.

• 4. Suppose $a\in \ell _{1}$ and $\sum _{i=1}^{\infty } \sqrt{\gamma _{ii}}\in \textbf{R}^{+}$; $\forall \,i\in \textbf{N}^{*}$, we have

  $$\begin{aligned} E\left[ \left| X_{i}-a_{i}\right| \right]= & {} 2\underset{a_{i}}{\overset{\infty }{\int }}\frac{x-a_{i}}{\sqrt{2\pi \gamma _{ii}}}\exp \left( -\frac{\left( x-a_{i}\right) ^{2}}{2\gamma _{ii}}\right) dx\\= & {} 2\underset{0}{\overset{\infty }{\int }}\sqrt{\frac{\gamma _{ii}}{2\pi }} \exp \left( -t\right) dt=\sqrt{\frac{2\gamma _{ii}}{\pi }}\\{} & {} \Longrightarrow E\left[ \left| X_{i}-a_{i}\right| -\sqrt{\frac{2\gamma _{ii}}{\pi }}\right] =0,\\{} & {} \quad Var\left[ \left| X_{i}-a_{i}\right| -\sqrt{\frac{2\gamma _{ii}}{\pi } }\right] =Var\left[ \left| X_{i}-a_{i}\right| \right] \\= & {} E\left[ \left( X_{i}-a_{i}\right) ^{2}\right] -\left( E\left[ \left| X_{i}-a_{i}\right| \right] \right) ^{2}\\= & {} \gamma _{ii}-\frac{2\gamma _{ii}}{\pi }=\left( \frac{\pi -2}{\pi }\right) \gamma _{ii}\\{} & {} \Longrightarrow \underset{i=1}{\overset{\infty }{\sum }}Var\left[ \left| X_{i}-a_{i}\right| -\sqrt{\frac{2\gamma _{ii}}{\pi }}\right] \\= & {} \left( \frac{\pi -2}{\pi }\right) \underset{i=1}{\overset{\infty }{\sum }}\gamma _{ii} \in \textbf{R}^{+}; \end{aligned}$$

  then, from Theorem 35, there exists a real random variable $T_{4}$ such that $\sum _{i=1}^{n}\left( \left| X_{i}-a_{i}\right| -\sqrt{\frac{2\gamma _{ii}}{\pi }}\right) \overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T_{4}$, from which $\sum _{i=1}^{n}\left| X_{i}-a_{i}\right| \overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T_{4}+ \sum _{i=1}^{\infty }\sqrt{\frac{2\gamma _{ii}}{\pi }}$, and so $P\left( \sum _{i=1}^{\infty }\left| X_{i}-a_{i}\right| \in \textbf{R}\right) =1$; furthermore, we have

  $$\begin{aligned} \underset{i=1}{\overset{\infty }{\sum }}\left| X_{i}\right| \le \underset{i=1}{\overset{\infty }{\sum }}\left| X_{i}-a_{i}\right| +\underset{i=1}{\overset{\infty }{\sum }}\left| a_{i}\right| , \end{aligned}$$

  from which $P\left( \sum _{i=1}^{\infty }\left| X_{i}\right| \in \textbf{R}\right) =1$, that is $N(a,\Gamma )(\ell _{1} )=1$.

$\square$

Proposition 37

Let $m\in \textbf{N}$, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$ and let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R} ^{\infty }$ be a symmetric m-positive definite linear function, such that $\sum _{i=m+1}^{\infty }\gamma _{ii}\in \textbf{R}^{+}$; then, the probability measure $N(a,\Gamma )$ on $\left( E_{\infty },\mathcal {B} _{\infty }\right)$ is Gaussian, and one has $H\left( N(a,\Gamma )\right) \supset \ell _{2}(\Gamma )$, where

$$\begin{aligned} \ell _{2}(\Gamma )\equiv \left\{ x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }:\underset{i=m+1}{\overset{\infty }{\sum }}\frac{x_{i}^{2}}{\gamma _{ii}}\in \textbf{R}\right\} . \end{aligned}$$

Proof

Let $X=\left( X_{i}:i\in \textbf{N}^{*}\right) :\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( E_{\infty },\mathcal {B}_{\infty }\right)$ be a random variable with law $N(a,\Gamma )$; we can consider X as a random variable on $\left( \textbf{R}^{\infty },\mathcal {B}^{\infty }\right)$; thus, we have $X_{i}\sim N\left( a_{i},\gamma _{ii}\right)$, $\forall \,i\in \textbf{N}^{*}$; morerover, the random variables $\left( X_{1},\ldots ,X_{m}\right)$ and $\left( X_{i}:i>m\right)$ are independent and such that $\left( X_{1},\ldots ,X_{m}\right) \sim N\left( (a_{1},\ldots ,a_{m} ),\Gamma ^{(m,m)}\right)$; finally, the random variables $\left\{ X_{i}\right\} _{i>m}$ are independent. Moreover, let $g:\left( E_{\infty },\tau _{\left\| \cdot \right\| _{\infty }}\left( E_{\infty }\right) \right) \longrightarrow \left( \textbf{R},\tau \right)$ be a linear and continuous function; from Corollary 13, there exists $\left( c_{i}:i\in \textbf{N}^{*}\cup \left\{ \infty \right\} \right) \in \textbf{R} ^{\textbf{N}^{*}\cup \left\{ \infty \right\} }$ such that $\sum _{i\in \textbf{N}^{*}\cup \left\{ \infty \right\} }\left| c_{i}\right| \in \textbf{R}$ and such that $g(x)=\sum _{i\in \textbf{N} ^{*}\cup \left\{ \infty \right\} }c_{i}x_{i}$, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$, where $x_{\infty } \equiv \lim _{i\rightarrow \infty }x_{i}$, and so $g(X)=\sum _{i=1}^{\infty } c_{i}X_{i}+c_{\infty } \lim _{i\rightarrow \infty }X_{i}$; furthermore, by setting $a_{\infty }=\lim _{i\rightarrow \infty }a_{i}\in \textbf{R}$, from Theorem 35, there exists a real random variable T such that $\sum _{i=m+1}^{n}\left( X_{i}-a_{i}\right) \overset{a.\,s.}{\underset{n\rightarrow \infty }{\longrightarrow }}T$, from which $\left( X_{i} -a_{i}\right) \overset{a.\,s.}{\underset{i\rightarrow \infty }{\longrightarrow }}0$, and so $X_{i}\overset{a.\,s.}{\underset{i\rightarrow \infty }{\longrightarrow }}a_{\infty }$; moreover, observe that $\sum _{i=1}^{\infty } c_{i}X_{i}=\lim _{n\longrightarrow \infty }\sum _{i=1}^{n} c_{i}X_{i}$ and $\sum _{i=1}^{n} c_{i}X_{i}\sim N\left( \sum _{i=1}^{n} c_{i}a_{i},\left\langle \Gamma ^{(m,m)}\left( c_{1},\ldots ,c_{m} \right) ,\left( c_{1},\ldots ,c_{m}\right) \right\rangle + \sum _{i=m+1}^{n} c_{i}^{2}\gamma _{ii}\right)$, $\forall \,n>m$; then, we have

$$\begin{aligned} g(X)\sim N\left( \underset{i=1}{\overset{\infty }{\sum }}c_{i}a_{i}+c_{\infty }a_{\infty },\left\langle \Gamma ^{(m,m)}\left( c_{1},\ldots ,c_{m}\right) ,\left( c_{1},\ldots ,c_{m}\right) \right\rangle +\underset{i=m+1}{\overset{\infty }{\sum }}c_{i}^{2}\gamma _{ii}\right) , \end{aligned}$$

and so $N(a,\Gamma )$ is Gaussian on $\left( E_{\infty },\mathcal {B}_{\infty }\right)$. Finally, let $x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in \ell _{2}(\Gamma )$, let $\lambda \in \textbf{R}^{+}$ be the smaller of the eigenvalues of $\Gamma ^{(m,m)}$, and let

$$\begin{aligned} \alpha _{i}=\left\{ \begin{array}{ll} \lambda &{}\quad \text {if}\quad i\in \left\{ 1,\ldots ,m\right\} \\ \gamma _{ii} &{}\quad \text {if}\quad i\in \textbf{N}^{*}\backslash \{1,\ldots ,m\} \end{array} \right. ; \end{aligned}$$

we have $x_{\infty }=0$, $\sum _{i=1}^{\infty } \frac{x_{i} ^{2}}{\alpha _{i}}\in \textbf{R}$ and $\left| x\right| _{H\left( N(a,\Gamma )\right) }=\sup \left\{ \left| \sum _{i=1}^{\infty } c_{i}x_{i}\right| :c\in \mathcal {A}\right\}$, where

$$\begin{aligned} \mathcal {A}=\left\{ c=\left( c_{i}:i\in \textbf{N}^{*}\right) \in \ell _{1}:\left\langle \Gamma ^{(m,m)}\left( c_{1},\ldots ,c_{m}\right) ,\left( c_{1},\ldots ,c_{m}\right) \right\rangle +\underset{i=m+1}{\overset{\infty }{\sum }}c_{i}^{2}\gamma _{ii}\le 1\right\} ; \end{aligned}$$

moreover, $\forall \,c=\left( c_{i}:i\in \textbf{N}^{*}\right) \in \mathcal {A}$, we have

$$\begin{aligned} \left| \underset{i=1}{\overset{\infty }{\sum }}c_{i}x_{i}\right|= & {} \left| \underset{i=1}{\overset{\infty }{\sum }}c_{i}\sqrt{\alpha _{i}}\cdot \frac{x_{i} }{\sqrt{\alpha _{i}}}\right| \le \sqrt{\underset{i=1}{\overset{\infty }{\sum } }c_{i}^{2}\alpha _{i}}\cdot \sqrt{\underset{i=1}{\overset{\infty }{\sum }} \frac{x_{i}^{2}}{\alpha _{i}}}\\= & {} \sqrt{\left\langle \lambda \left( c_{1},\ldots ,c_{m}\right) ,\left( c_{1},\ldots ,c_{m}\right) \right\rangle +\underset{i=m+1}{\overset{\infty }{\sum }}c_{i}^{2}\gamma _{ii}}\\{} & {} \cdot \sqrt{\underset{i=1}{\overset{\infty }{\sum }} \frac{x_{i}^{2}}{\alpha _{i}}}\\\le & {} \sqrt{\left\langle \Gamma ^{(m,m)}\left( c_{1},\ldots ,c_{m}\right) ,\left( c_{1},\ldots ,c_{m}\right) \right\rangle +\underset{i=m+1}{\overset{\infty }{\sum }}c_{i}^{2}\gamma _{ii}}\\{} & {} \cdot \sqrt{\underset{i=1}{\overset{\infty }{\sum }} \frac{x_{i}^{2}}{\alpha _{i}}}; \end{aligned}$$

then, since $\left\langle \Gamma ^{(m,m)}\left( c_{1},\ldots ,c_{m}\right) ,\left( c_{1},\ldots ,c_{m}\right) \right\rangle + \sum _{i=m+1}^{\infty } c_{i}^{2}\gamma _{ii}\le 1$, we obtain $\left| \sum _{i=1}^{\infty } c_{i}x_{i}\right| \le \sqrt{\sum _{i=1}^{\infty }\frac{x_{i}^{2}}{\alpha _{i}}}<\infty$, from which $\left| x\right| _{H\left( N(a,\Gamma )\right) }<\infty$, and so $x\in H\left( N(a,\Gamma )\right)$. $\square$

Proposition 38

Let $\varphi :U=U^{(m)}\times E_{\infty }\longrightarrow E_{\infty }$ be a bijective and continuous $\left( m,\sigma \right)$ function, such that $\overline{\varphi }$ is bijective and continuous, $\varphi ^{(m,m)}$ is open, $\varphi ^{(n,n)}$ is continuous, for any $n\in \textbf{N}^{*}$, $n\ge m$, and $\varphi _{ij}$ is continuous, for any $i\in \left\{ 1,\ldots ,m\right\}$, for any $j>m$; moreover, suppose that there exists $\varepsilon \in \left[ 0,\infty \right)$ such that $\left| \varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) \right| \le \varepsilon$, for any $i>m$, for any $\left( x_{1},\ldots ,x_{m}\right) \in U^{(m)}$, and the sequence $\left\{ \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\right\} _{n>m}$ converges uniformly to $\varphi ^{-1}$ over the closed and bounded subsets of $E_{\infty }$; then, for any $M\in \textbf{R}^{+}$, there exists $O\in \textbf{R}^{+}$ and there exists a closed subset K of $U\cap [-O,O]^{\infty }$ such that, for any $n\in \textbf{N}^{*}$, $n\ge m$, one has

$$\begin{aligned}{} & {} \varphi ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \subset K,\nonumber \\{} & {} \quad \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \subset K. \end{aligned}$$

(19)

Proof

Since $\overline{\varphi }$ is bijective, from Proposition 65, $\varphi ^{(m,m)}$ and $\sigma$ are bijective; furthermore, $\forall \,i>m$, from Remark 55, we have $\sigma (i)=i$, and, from Proposition 65, $\varphi _{ii}:\textbf{R}\longrightarrow \textbf{R}$ is bijective. Let $M\in \textbf{R}^{+}$, $x_{0}=(x_{0,j}:j\in \textbf{N}^{*})\in U$ and $x=\left( x_{j}:j\in \textbf{N}^{*}\right) \in \varphi ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \subset U$; $\forall \,i\in \left\{ 1,\ldots ,m\right\}$, we have

$$\begin{aligned} \varphi _{i}^{(m,m)}(x_{1},\ldots ,x_{m})+\underset{j=m+1}{\overset{\infty }{ {\displaystyle \sum } }}\varphi _{ij}(x_{j})=\varphi _{i}\left( x\right) \in \left[ -M,M\right] , \end{aligned}$$

and so

$$\begin{aligned} (x_{1},\ldots ,x_{m})=\left( \varphi ^{(m,m)}\right) ^{-1}\left( w_{1} ,\ldots ,w_{m}\right) , \end{aligned}$$

(20)

where

$$\begin{aligned} w_{i}=\varphi _{i}(x)-\underset{j=m+1}{\overset{\infty }{ {\displaystyle \sum } }}\varphi _{ij}\left( x_{j}\right) \text {, }\forall \,i\in \left\{ 1,\ldots ,m\right\} ; \end{aligned}$$

(21)

furthermore, $\forall \,i>m$, we have

$$\begin{aligned} \varphi _{i}\left( x\right)= & {} \varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) +\varphi _{ii}(x_{i})\nonumber \\{} & {} \Rightarrow \varphi _{ii}(x_{i})\in \left[ -M-\left| \varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) \right| ,M\right. \nonumber \\{} & {} \quad \left. +\left| \varphi _{i} ^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) \right| \right] \nonumber \\{} & {} \subset \left[ -M-\varepsilon ,M+\varepsilon \right] \nonumber \\{} & {} \Rightarrow x_{i}\in \left[ \alpha _{i},\beta _{i}\right] , \end{aligned}$$

(22)

where

$$\begin{aligned} \alpha _{i}&=\min \left\{ \varphi _{ii}^{-1}\left( -M-\varepsilon \right) ,\varphi _{ii}^{-1}\left( M+\varepsilon \right) \right\} ,\\ \beta _{i}&=\max \left\{ \varphi _{ii}^{-1}\left( -M-\varepsilon \right) ,\varphi _{ii}^{-1}\left( M+\varepsilon \right) \right\} ; \end{aligned}$$

moreover

$$\begin{aligned} \left| \varphi _{ii}^{-1}\left( M+\varepsilon \right) \right|= & {} \left| \varphi _{ii}^{-1}\left( M+\varepsilon \right) -x_{0,i}+x_{0,i}\right| \nonumber \\\le & {} \left| \varphi _{ii}^{-1}\left( M+\varepsilon \right) -\varphi _{ii} ^{-1}(\varphi _{ii}(x_{0,i}))\right| +\left| x_{0,i}\right| ; \end{aligned}$$

(23)

then, since the function $\varphi _{ii}^{-1}$ is derivable on $\textbf{R}$, the Lagrange theorem implies that, for some $\xi _{i}\in \left( \rho _{i},\tau _{i}\right)$, where

$$\begin{aligned} \rho _{i}= & {} \min \{M+\varepsilon ,\varphi _{ii}(x_{0,i})\},\\ \tau _{i}= & {} \max \{M+\varepsilon ,\varphi _{ii}(x_{0,i})\}, \end{aligned}$$

we have

$$\begin{aligned}{} & {} \left| \varphi _{ii}^{-1}\left( M+\varepsilon \right) -\varphi _{ii} ^{-1}\left( \varphi _{ii}(x_{0,i})\right) \right| \nonumber \\{} & {} \quad =\left| \left( \varphi _{ii}^{-1}\right) ^{\prime }(\xi _{i})\right| \left| M+\varepsilon -\varphi _{ii}(x_{0,i})\right| =\frac{\left| M+\varepsilon -\varphi _{ii}(x_{0,i})\right| }{\left| \varphi _{ii}^{\prime }(\varphi _{ii}^{-1}(\xi _{i}))\right| }; \end{aligned}$$

(24)

thus, from (23), we obtain

$$\begin{aligned} \left| \varphi _{ii}^{-1}\left( M+\varepsilon \right) \right| \le \frac{\left| M+\varepsilon -\varphi _{ii}(x_{0,i})\right| }{\left| \varphi _{ii}^{\prime }(\varphi _{ii}^{-1}(\xi _{i}))\right| }+\left| x_{0,i}\right| . \end{aligned}$$

(25)

Moreover, from Definition 17, we have

$$\begin{aligned} \underset{i>m}{\sup }\frac{1}{\left| \varphi _{ii}^{\prime }(\varphi _{ii} ^{-1}(\xi _{i}))\right| }=\frac{1}{\underset{i>m}{\inf }\left| \varphi _{ii}^{\prime }(\varphi _{ii}^{-1}(\xi _{i}))\right| }\le \frac{1}{d(\varphi )}\in \textbf{R}^{+}; \end{aligned}$$

(26)

moreover, from Remark 54, we have $\sup _{i>m}\left| \varphi _{ii}(x_{0,i})\right| <\infty$, and so formula (25) implies

$$\begin{aligned} \underset{i>m}{\sup }\left| \varphi _{ii}^{-1}\left( M+\varepsilon _{i}\right) \right| \le \frac{M+\varepsilon +\underset{i>m}{\sup }\left| \varphi _{ii}(x_{0,i})\right| }{d(\varphi )}+\left\| x_{0}\right\| _{\infty }\equiv O_{1}\in \textbf{R}^{+}; \end{aligned}$$

analogously, we obtain

$$\begin{aligned} \underset{i>m}{\sup }\left| \varphi _{ii}^{-1}\left( -M-\varepsilon \right) \right| \le O_{1}; \end{aligned}$$

then, from formula (22), $\forall \,i>m$, we have $\left| x_{i}\right| \le O_{1}$, from which $\left\| (x_{j}:j>m)\right\| _{\infty }\le O_{1}$.

Moreover, $\forall \,i\in \left\{ 1,\ldots ,m\right\}$, $\forall \,j>m$, the function $\varphi _{ij}$ is continuous, and so there exists $t_{ij} =t_{ij}\left( \varphi ,M,\varepsilon ,x_{0}\right) \in \left[ -O_{1},O_{1}\right]$ such that $\max _{t\in \left[ -O_{1},O_{1}\right] }\left| \varphi _{ij}(t)\right| =\left| \varphi _{ij}(t_{ij})\right|$; then, we have

$$\begin{aligned} \underset{i\in \left\{ 1,\ldots ,m\right\} }{\max }\underset{j=m+1}{\overset{\infty }{ {\displaystyle \sum } }}\left| \varphi _{ij}\left( x_{j}\right) \right| \le \underset{i\in \left\{ 1,\ldots ,m\right\} }{\max }\underset{j=m+1}{\overset{\infty }{ {\displaystyle \sum } }}\left| \varphi _{ij}\left( t_{ij}\right) \right| \equiv O_{2} \in \textbf{R}^{+}, \end{aligned}$$

and so $\left\| \left( w_{1},\ldots ,w_{m}\right) \right\| _{I_{m}}\le M+O_{2}\equiv O_{3}\in \textbf{R}^{+}$, from (21); then, since the function $\left( \varphi ^{(m,m)}\right) ^{-1}$ is continuous, from (20), we have $\left\| \left( x_{1},\ldots , x_{m}\right) \right\| _{I_{m}}\le O_{4}$, for some $O_{4}=O_{4}\left( \varphi ,M,\varepsilon ,x_{0}\right) \in \textbf{R}^{+}$ such that

$$\begin{aligned} \left( \varphi ^{(m,m)}\right) ^{-1}\left( \left[ -O_{3},O_{3}\right] ^{m}\right) \subset \left[ -O_{4},O_{4}\right] ^{m}, \end{aligned}$$

and so $\left\| x\right\| _{\infty }\le \max \left\{ O_{1},O_{4}\right\} \equiv O$; thus

$$\begin{aligned} \varphi ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \subset \left[ -O,O\right] ^{\infty }. \end{aligned}$$

(27)

Furthermore, let $n\in \textbf{N}^{*}$, $n\ge m$, and let

$$\begin{aligned} z=\left( z_{j}:j\in \textbf{N}^{*}\right) \in \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \subset U; \end{aligned}$$

$\forall \,i\in \left\{ m+1,\ldots ,n\right\}$, since $\varphi _{i}(z)=\overline{\varphi }_{i}^{(n,n)}(z)$, by repeating the previous arguments, we have $z_{i}\in \left[ \alpha _{i},\beta _{i}\right] \subset \left[ -O_{1},O_{1}\right]$; conversely, $\forall \,i>n$, we have

$$\begin{aligned} \varphi _{ii}(z_{i})=\overline{\varphi }_{i}^{(n,n)}(z)\in \left[ -M,M\right] , \end{aligned}$$

and so $z_{i}\in \left[ \gamma _{i},\delta _{i}\right]$, where

$$\begin{aligned} \gamma _{i}&=\min \left\{ \varphi _{ii}^{-1}\left( -M\right) ,\varphi _{ii}^{-1}\left( M\right) \right\} ,\nonumber \\ \delta _{i}&=\max \left\{ \varphi _{ii}^{-1}\left( -M\right) ,\varphi _{ii}^{-1}\left( M\right) \right\} ; \end{aligned}$$

(28)

then, since $\left[ \gamma _{i},\delta _{i}\right] \subset \left[ \alpha _{i},\beta _{i}\right]$, we obtain $z_{i}\in \left[ -O_{1},O_{1}\right]$ again; thus, we have $\left\| (z_{j}:j>m)\right\| _{\infty }\le O_{1}$. Moreover, $\forall \,i\in \left\{ 1,\ldots ,m\right\}$, we have

$$\begin{aligned} \overline{\varphi }_{i}^{(n,n)}(z)=\varphi _{i}^{(m,m)}(z_{1},\ldots ,z_{m} )+\underset{j=m+1}{\overset{n}{ {\displaystyle \sum } }}\varphi _{ij}(z_{j}), \end{aligned}$$

and so

$$\begin{aligned} (z_{1},\ldots ,z_{m})=\left( \varphi ^{(m,m)}\right) ^{-1}\left( w_{1} ^{(n)},\ldots ,w_{m}^{(n)}\right) , \end{aligned}$$

(29)

where

$$\begin{aligned} w_{i}^{(n)}=\overline{\varphi }_{i}^{(n,n)}(z)-\underset{j=m+1}{\overset{n}{ {\displaystyle \sum } }}\varphi _{ij}\left( z_{j}\right) , \forall \,i\in \left\{ 1,\ldots ,m\right\} ; \end{aligned}$$

(30)

furthermore, we have

$$\begin{aligned} \underset{i\in \left\{ 1,\ldots ,m\right\} }{\max }\underset{j=m+1}{\overset{n}{ {\displaystyle \sum } }}\left| \varphi _{ij}\left( z_{j}\right) \right| \le \underset{i\in \left\{ 1,\ldots ,m\right\} }{\max }\underset{j=m+1}{\overset{n}{ {\displaystyle \sum } }}\left| \varphi _{ij}\left( t_{ij}\right) \right| \le O_{2}, \end{aligned}$$

and so $\left\| \left( w_{1}^{(n)},\ldots ,w_{m}^{(n)}\right) \right\| _{I_{m} }\le O_{3}$, from (30); then, since the function $\left( \varphi ^{(m,m)}\right) ^{-1}$ is continuous, from (29), we have $\left\| (z_{1},\ldots ,z_{m})\right\| _{I_{m}}\le O_{4}$, and so $\left\| z\right\| _{\infty }\le O$; thus

$$\begin{aligned} \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \subset \left[ -O,O\right] ^{\infty }. \end{aligned}$$

(31)

Moreover, since $E_{\infty }$ is closed, the set $\left[ -M,M\right] ^{\infty }\cap E_{\infty }$ is closed and bounded, and so the sequence $\left\{ \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\right\} _{n\ge m}$ converges uniformly to $\varphi ^{-1}$ over $[-M,M]^{\infty }\cap E_{\infty }$; furthermore, since $\varphi$ is continuous, the set $\varphi ^{-1}\left( [-M,M]^{\infty }\cap E_{\infty }\right)$ is closed; then, there exist $\delta =\delta (\varphi ,M)\in \textbf{R}^{+}$ and $\overline{n}=\overline{n}(\varphi ,M)\in \textbf{N}$, $\overline{n}\ge m$, such that, $\forall \,h\in \textbf{N}^{*},h>\overline{n}$, $\left( \overline{\varphi } ^{(h,h)}\right) ^{-1}\left( [-M,M]^{\infty }\cap E_{\infty }\right) \subset \varphi ^{-1}\left( [-M,M]^{\infty }\cap E_{\infty }\right) +\overline{B_{\infty }(0,\delta )}\subset U$; thus, $\forall \,n\in \textbf{N}^{*}$, $n\ge m$, we have

$$\begin{aligned}{} & {} \left( \overline{\varphi }^{(n,n)}\right) ^{-1}(\left[ -M,M\right] ^{\infty }\cap E_{\infty })\subset \underset{h\ge m}{\bigcup }\left( \overline{\varphi }^{(h,h)}\right) ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \\{} & {} \quad \subset \left( \underset{h=m}{\overset{\overline{n}}{\bigcup }}\left( \overline{\varphi }^{(h,h)}\right) ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) \right) \\{} & {} \quad \cup \left( \varphi ^{-1}\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right) +\overline{B_{\infty }(0,\delta )}\right) , \end{aligned}$$

and so formula 31 implies

$$\begin{aligned}{} & {} \left( \overline{\varphi }^{(n,n)}\right) ^{-1}(\left[ -M,M\right] ^{\infty }\cap E_{\infty })\nonumber \\{} & {} \quad \subset [-O,O]^{\infty }\cap \left( \left( \underset{h=m}{\overset{\overline{n}}{\bigcup }}\left( \overline{\varphi }^{(h,h)}\right) ^{-1}\left( [-M,M]^{\infty }\cap E_{\infty }\right) \right) \right. \nonumber \\{} & {} \quad \left. \cup \left( \varphi ^{-1}\left( [-M,M]^{\infty }\cap E_{\infty }\right) +\overline{B_{\infty }(0,\delta )}\right) \right) \nonumber \\{} & {} \quad \equiv K=K(\varphi ,M)\subset U\cap [-O,O]^{\infty }. \end{aligned}$$

(32)

Analogously, from formula 27, we have

$$\begin{aligned}{} & {} \varphi ^{-1}(\left[ -M,M\right] ^{\infty }\cap E_{\infty })\\{} & {} \quad \subset [-O,O]^{\infty }\cap \left( \varphi ^{-1}\left( [-M,M]^{\infty }\cap E_{\infty }\right) +\overline{B_{\infty }(0,\delta )}\right) \subset K. \end{aligned}$$

Moreover, by assumption, the functions $\overline{\varphi }$ and $\varphi ^{(h,h)}$, $\forall \,h\in \left\{ m,\ldots ,\overline{n}\right\}$, are continuous; then, from Proposition 59, $\overline{\varphi }^{(h,h)}$ is continuous too, and so formula (32) implies that K is a closed subset of $U\cap [-O,O]^{\infty }$. $\square$

Definition 39

Let $m\in \textbf{N}$, let $n\in \textbf{N}^{*}$, $n\ge m$, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }$, and let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*} }:\textbf{R}^{\infty }\longrightarrow \textbf{R}^{\infty }$ be a linear and symmetric m-positive definite function; define the measurable function $\Psi ^{\left( n,a,\Gamma \right) }:\left( \textbf{R}^{\infty },\mathcal {B} ^{\infty }\right) \longrightarrow (\left[ 0,+\infty \right) ,\mathcal {B} \left( \left[ 0,+\infty \right) \right) )$ by

$$\begin{aligned} \Psi ^{\left( n,a,\Gamma \right) }(x)= & {} \frac{\left\langle \left( \Gamma ^{(n,n)}\right) ^{-1}\left( x_{1}-a_{1},\ldots ,x_{n}-a_{n}\right) ,\left( x_{1}-a_{1},\ldots ,x_{n}-a_{n}\right) \right\rangle }{2},\\ \forall \,x= & {} \left( x_{i}:i\in \textbf{N}^{*}\right) \in \textbf{R}^{\infty }. \end{aligned}$$

Proposition 40

Let $\varphi :U=U^{(m)}\times E_{\infty }\longrightarrow E_{\infty }$ be a bijective and continuous $\left( m,\sigma \right)$ function, such that $\overline{\varphi }$ is continuous and, for any $n\in \textbf{N}^{*}$, $n\ge m$, the function $\overline{\varphi } ^{(n,n)}:U\longrightarrow E_{\infty }$ is a diffeomorphism; moreover, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$, let $b=\left( b_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$ such that $a_{i} =\varphi _{ii}(b_{i})$, for any $i>m$, and let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R}^{\infty }$ and $\Lambda =\left( \lambda _{ij}\right) _{i,j\in \textbf{N}^{*}}:\textbf{R}^{\infty }\longrightarrow \textbf{R}^{\infty }$ some symmetric m-positive definite linear functions, such that $\prod _{i=m+1}^{\infty }\frac{\lambda _{ii}}{\gamma _{ii}}\in \textbf{R}^{+}$, $\sum _{i=m+1}^{\infty }\gamma _{ii}\in \textbf{R}^{+}$, $\sum _{i=m+1}^{\infty }\lambda _{ii}\in \textbf{R}^{+}$, and $\sum _{i=m+1}^{\infty }\left| \frac{\left( \varphi _{ii}^{\prime }\left( x_{i}\right) \right) ^{2}}{\gamma _{ii}}-\frac{1}{\lambda _{ii}}\right| \in [0,\infty )$, for any $\left( x_{i}:i>m\right) \in E_{\infty }$; moreover, suppose that there exists $\varepsilon \in \left[ 0,\infty \right)$ such that $\left| \varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) \right| \le \varepsilon$, for any $i>m$, for any $\left( x_{1},\ldots ,x_{m}\right) \in U^{(m)}$, and the sequence $\left\{ \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\right\} _{n>m}$ converges uniformly to $\varphi ^{-1}$ over the closed and bounded subsets of $E_{\infty }$. Then, for any $M\in \textbf{R}^{+}$ such that $M\ge \left\| a\right\| _{\infty }$, there exists $O\in \textbf{R}^{+}$, there exists a closed subset K of $U\cap [-O,O]^{\infty }$, and, for any $q>m$, there exists a measurable function $g_{q}:\left( [-O,O],\mathcal {B} \left( [-O,O]\right) \right) \longrightarrow (\textbf{R}^{+},\mathcal {B} \left( \textbf{R}^{+}\right) )$ such that $\left( \bigotimes _{q=m+1}^{\infty } g_{q}\right) (x)\in \textbf{R}^{+}$, for any $x\in [-O,O]^{\infty }$, and such that, for any $n\in \textbf{N}^{*}$, $n\ge m$, and for any measurable function $l:\left( E_{\infty },\mathcal {B}_{\infty }\right) \longrightarrow (\left[ 0,\infty \right) ,\mathcal {B}\left( \left[ 0,\infty \right) \right) )$ such that $l(x)=0$ for any $x\in E_{\infty }\backslash \left[ -M,M\right] ^{\infty }$, one has

$$\begin{aligned} \underset{E_{\infty }}{ {\displaystyle \int } }ldN(a,\Gamma )= & {} \underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}l\left( \overline{\varphi } ^{(n,n)}\left( x_{i}:i\in \textbf{N}^{*}\right) \right) \nonumber \\{} & {} \cdot \exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left( x_{i}:i\in \textbf{N}^{*}\right) \nonumber \\{} & {} \cdot |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\nonumber \\{} & {} \underset{q=n+1}{\overset{\infty }{\prod }}g_{q} (x_{q}) dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N} ^{*}\right) . \end{aligned}$$

(33)

Proof

The previous assumptions imply that $\overline{\varphi }$ is bijective; then, from Proposition 65 and Remark 55, we have $\sigma (i)=i$, $\forall \,i>m$; moreover, $\varphi _{ij}$ is continuous, $\forall \,i\in \left\{ 1,\ldots ,m\right\}$, $\forall \,j>m$, and the function $\varphi ^{(m,m)}:U^{(m)}\longrightarrow \textbf{R}^{m}$ is open; thus, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$, let $M\in \textbf{R}^{+}$ such that $M\ge \left\| a\right\| _{\infty }$, and let $O\in \textbf{R}^{+}$ and $K\subset U\cap [-O,O]^{\infty }$ the closed set defined by Proposition 38; moreover, let $n\in \textbf{N}^{*}$, $n\ge m$, and let $B=\prod _{i=1}^{\infty } B_{i}\in \mathcal {B}^{\infty }\left( \left[ -M,M\right] ^{\infty }\right)$; we have

$$\begin{aligned} \underset{B}{ {\displaystyle \int } }dN(a,\Gamma )= & {} \underset{B}{ {\displaystyle \int } }d\left( N\left( a_{I_{n}},\Gamma ^{(n,n)}\right) \otimes \left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( a_{q},\gamma _{qq}\right) \right) \right) \nonumber \\= & {} \underset{\underset{p=1}{\overset{n}{ {\textstyle \prod } }}B_{p}}{ {\displaystyle \int } }dN\left( a_{I_{n}},\Gamma ^{(n,n)}\right) \nonumber \\{} & {} \underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}B_{q}}{ {\displaystyle \int } }d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( a_{q},\gamma _{qq}\right) \right) . \end{aligned}$$

(34)

Furthermore, from Proposition 64 and Proposition 66, $\varphi ^{(n,n)}$ is a diffeomorphism, and so

$$\begin{aligned}{} & {} \underset{\underset{p=1}{\overset{n}{ {\textstyle \prod } }}B_{p}}{ {\displaystyle \int } }dN\left( a_{I_{n}},\Gamma ^{(n,n)}\right) =\underset{\underset{p=1}{\overset{n}{ {\textstyle \prod } }}B_{p}}{ {\displaystyle \int } }\frac{1}{\left( 2\pi \right) ^{\frac{n}{2}}\sqrt{\det \Gamma ^{(n,n)}}}\nonumber \\{} & {} \quad \cdot \exp \left( -\frac{\left\langle \left( \Gamma ^{(n,n)}\right) ^{-1}\left( y_{I_{n}}-a_{I_{n}}\right) ,y_{I_{n}}-a_{I_{n}}\right\rangle }{2}\right) dy_{I_{n}}\nonumber \\{} & {} \quad =\underset{\left( \varphi ^{(n,n)}\right) ^{-1}\left( \underset{p=1}{\overset{n}{ {\textstyle \prod } }}B_{p}\right) }{ {\displaystyle \int } }\frac{1}{\left( 2\pi \right) ^{\frac{n}{2}}\sqrt{\det \Gamma ^{(n,n)}}}\nonumber \\{} & {} \quad \cdot \exp \left( -\frac{\left\langle \left( \Gamma ^{(n,n)}\right) ^{-1}\left( \varphi ^{(n,n)}\left( x_{I_{n}}\right) -a_{I_{n}}\right) ,\varphi ^{(n,n)}\left( x_{I_{n}}\right) -a_{I_{n}}\right\rangle }{2}\right) \nonumber \\{} & {} \quad \cdot |\det J_{\varphi ^{(n,n)}}\left( x_{I_{n}}\right) |dx_{I_{n}}\nonumber \\{} & {} \quad =\underset{\left( \varphi ^{(n,n)}\right) ^{-1}\left( \underset{p=1}{\overset{n}{ {\textstyle \prod } }}B_{p}\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda ^{(n,n)}}{\det \Gamma ^{(n,n)}}}\nonumber \\{} & {} \quad \cdot \exp \left( -\frac{\left\langle \left( \Gamma ^{(n,n)}\right) ^{-1}\left( \varphi ^{(n,n)}\left( x_{I_{n}}\right) -a_{I_{n}}\right) ,\varphi ^{(n,n)}\left( x_{I_{n}}\right) -a_{I_{n}}\right\rangle }{2}\right) \nonumber \\{} & {} \quad \cdot \exp \left( \frac{\left\langle \left( \Lambda ^{(n,n)}\right) ^{-1}\left( x_{I_{n}}-b_{I_{n}}\right) ,x_{I_{n}}-b_{I_{n}}\right\rangle }{2}\right) |\det J_{\varphi ^{(n,n)}}\left( x_{I_{n}}\right) |\nonumber \\{} & {} \quad \cdot dN\left( b_{I_{n}},\Lambda ^{(n,n)}\right) \left( x_{I_{n}}\right) . \end{aligned}$$

(35)

Moreover, we have

$$\begin{aligned}{} & {} \underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}B_{q}}{ {\displaystyle \int } }d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( a_{q},\gamma _{qq}\right) \right) \nonumber \\{} & {} \quad =\underset{p\rightarrow \infty }{\lim }\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left. \underset{B_{q}}{ {\displaystyle \int } }dN\left( a_{q},\gamma _{qq}\right) \right. \text { (by Theorem 43)}\nonumber \\{} & {} \quad =\underset{p\rightarrow \infty }{\lim }\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left. \underset{B_{q}}{ {\displaystyle \int } }\frac{1}{\sqrt{2\pi \gamma _{qq}}}\exp \left( -\frac{\left( y_{q} -a_{q}\right) ^{2}}{2\gamma _{qq}}\right) dy_{q}\right. ; \end{aligned}$$

(36)

furthermore, $\forall \,q>m$, $\varphi _{qq}$ is a diffeomorphism, from Proposition 66; then, formula (36) implies

$$\begin{aligned}{} & {} \underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}B_{q}}{ {\displaystyle \int } }d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( a_{q},\gamma _{qq}\right) \right) \nonumber \\{} & {} \quad =\underset{p\rightarrow \infty }{\lim }\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left. \underset{\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\frac{1}{\sqrt{2\pi \gamma _{qq}}}\exp \left( -\frac{\left( \varphi _{qq}\left( x_{q}\right) -a_{q}\right) ^{2}}{2\gamma _{qq}}\right) \left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| dx_{q}\right. \nonumber \\{} & {} \quad =\underset{p\rightarrow \infty }{\lim }\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left. \underset{\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\left( \sqrt{\frac{\lambda _{qq}}{\gamma _{qq}}}\exp \left( -\frac{\left( \varphi _{qq}\left( x_{q}\right) -\varphi _{qq}\left( b_{q}\right) \right) ^{2}}{2\gamma _{qq}}\right) \right. \right. \nonumber \\{} & {} \quad \left. \cdot \exp \left( \frac{\left( x_{q}-b_{q}\right) ^{2}}{2\lambda _{qq}}\right) \left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \right) dN\left( b_{q},\lambda _{qq}\right) \left( x_{q}\right) , \end{aligned}$$

(37)

since $a_{q}=\varphi _{qq}\left( b_{q}\right)$, $\forall \,q>m$, by assumption, and so $b_{q}=\varphi _{qq}^{-1}\left( a_{q}\right) \in \left[ -O,O\right]$, $\varphi _{qq}^{-1}\left( B_{q}\right) \subset \left[ -O,O\right]$, from Proposition 38. Moreover, let $t\in \left[ -O,O\right]$ and let $q>m$; from Lagrange theorem and since the function $\varphi _{qq}$ is injective, there exists a unique $\xi _{q}(t)\in (\min \left\{ t,b_{q}\right\} ,\max \left\{ t,b_{q}\right\} )\in \left[ -O,O\right]$ such that

$$\begin{aligned} \exp \left( -\frac{\left( \varphi _{qq}\left( t\right) -\varphi _{qq}\left( b_{q}\right) \right) ^{2}}{2\gamma _{qq}}\right) =\exp \left( -\frac{\left( t-b_{q}\right) ^{2}}{2\gamma _{qq}}\cdot \left( \varphi _{qq}^{\prime }\left( \xi _{q}(t)\right) \right) ^{2}\right) ; \end{aligned}$$

(38)

then, consider the measurable function $g_{q}:\left( \left[ -O,O\right] ,\mathcal {B}\left( \left[ -O,O\right] \right) \right) \longrightarrow (\textbf{R}^{+},\mathcal {B}\left( \textbf{R}^{+}\right) )$ defined by

$$\begin{aligned} g_{q}\left( t\right) =\exp \left( -\frac{\left( t-b_{q}\right) ^{2}}{2}\left( \frac{\left( \varphi _{qq}^{\prime }\left( \xi _{q}(t)\right) \right) ^{2}}{\gamma _{qq}}-\frac{1}{\lambda _{qq}}\right) \right) \text {, }\forall \,t\in \left[ -O,O\right] ; \end{aligned}$$

from formula (37), we have

$$\begin{aligned}{} & {} \underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}B_{q}}{ {\displaystyle \int } }d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( a_{q},\gamma _{qq}\right) \right) \nonumber \\{} & {} \quad =\underset{p\rightarrow \infty }{\lim }\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left( \sqrt{\frac{\lambda _{qq}}{\gamma _{qq}}}\underset{\varphi _{qq} ^{-1}(B_{q})}{ {\displaystyle \int } }g_{q}(x_{q})\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| dN\left( b_{q},\lambda _{qq}\right) \left( x_{q}\right) \right) \nonumber \\{} & {} \quad =\left. \underset{i=n+1}{\overset{\infty }{ {\textstyle \prod } }}\sqrt{\frac{\lambda _{ii}}{\gamma _{ii}}}\right. \underset{p\rightarrow \infty }{\lim }\underset{\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\left. \underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left( g_{q}(x_{q})\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \right) \right. \nonumber \\{} & {} \quad \cdot d\left( \underset{q=n+1}{\overset{p}{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) \left( x_{I_{p}\backslash I_{n}}\right) . \end{aligned}$$

(39)

Furthermore, the set $[-O,O]$ is closed and bounded, and from Proposition 66 the function $\varphi _{qq}^{\prime }$ is continuous; then, there exists $z_{1}=\left( (z_{1})_{q}:q>m\right) \in [-O,O]^{\infty }$ such that, $\forall \,p\in \left( \textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} \right) \cup \left\{ \infty \right\}$, $\forall \,(x_{q}:q>m)\in [-O,O]^{\infty }$, we have

$$\begin{aligned} \underset{q=n+1}{\overset{p}{\prod }}g_{q}\left( x_{q}\right)= & {} \exp \left( \underset{q=n+1}{\overset{p}{\sum }}\left( -\frac{\left( x_{q}-b_{q}\right) ^{2}}{2}\left( \frac{\left( \varphi _{qq}^{\prime }\left( \xi _{q} (x_{q})\right) \right) ^{2}}{\gamma _{qq}}-\frac{1}{\lambda _{qq}}\right) \right) \right) \nonumber \\\le & {} \exp \left( 2O^{2}\underset{q=n+1}{\overset{p}{\sum }}\left| \frac{\left( \varphi _{qq}^{\prime }\left( \xi _{q}(x_{q})\right) \right) ^{2}}{\gamma _{qq}}-\frac{1}{\lambda _{qq}}\right| \right) \nonumber \\\le & {} \exp \left( 2O^{2}\underset{q=m+1}{\overset{\infty }{\sum }}\left| \frac{\left( \varphi _{qq}^{\prime }\left( (z_{1})_{q}\right) \right) ^{2} }{\gamma _{qq}}-\frac{1}{\lambda _{qq}}\right| \right) \equiv \alpha <\infty , \nonumber \\ \underset{q=n+1}{\overset{p}{\prod }}g_{q}\left( x_{q}\right)\ge & {} \exp \left( -2O^{2}\underset{q=m+1}{\overset{\infty }{\sum }}\left| \frac{\left( \varphi _{qq}^{\prime }\left( (z_{1})_{q}\right) \right) ^{2}}{\gamma _{qq} }-\frac{1}{\lambda _{qq}}\right| \right) >0, \end{aligned}$$

(40)

and in particular $\prod _{q=m+1}^{\infty } g_{q}\left( x_{q}\right) \in \textbf{R}^{+}$; furthermore, since $\mathcal {I} (\varphi )=\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\}$, from Definition 17, we have $\prod _{q=m+1}^{\infty } \left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \in \textbf{R} ^{+}$, from which

$$\begin{aligned} \underset{q=m+1}{\overset{\infty }{\prod }}g_{q}\left( x_{q}\right) \left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \in \textbf{R}^{+}; \end{aligned}$$

(41)

thus, formula (41) implies $\lim _{p\rightarrow \infty } \prod _{q=p+1}^{\infty } g_{q}\left( x_{q}\right) \left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| =1$. Analogously, there exists $z_{2}=\left( (z_{2})_{q}:q>m\right) \in \left[ -O,O\right] ^{\infty }$ such that, $\forall \,p\in \left( \textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} \right) \cup \left\{ \infty \right\}$, $\forall \,(x_{q}:q>m)\in [-O,O]^{\infty }$, we have

$$\begin{aligned} \underset{q=n+1}{\overset{p}{\prod }}\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \le \underset{q=n+1}{\overset{p}{\prod }}\left| \varphi _{qq}^{\prime }\left( (z_{2})_{q}\right) \right| \equiv \beta _{n,p} \in \textbf{R}^{+}, \end{aligned}$$

and so, $\forall \,(x_{q}:q>m)\in [-O,O]^{\infty }$, we obtain

$$\begin{aligned} \underset{q=n+1}{\overset{p}{\prod }}g_{q}\left( x_{q}\right) \left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \le \alpha \beta _{n,p} \in \textbf{R}^{+}; \end{aligned}$$

(42)

then, formula (42) implies that the measurable function

$$\begin{aligned} \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}g_{q}\left| \varphi _{qq}^{\prime }\right| :\left( \underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q}),\mathcal {B}^{\infty }\left( \underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})\right) \right) \longrightarrow (\textbf{R} ^{+},\mathcal {B}\left( \textbf{R}^{+}\right) ), \end{aligned}$$

is $\left( \bigotimes _{q=n+1}^{\infty } N\left( b_{q},\lambda _{qq}\right) \right)$-integrable. Moreover, the sequence $\left\{ \beta _{n,p}\right\} _{p>n}$ converges to $\beta _{n,\infty }$, and so there exists $\overline{\beta }_{n}\in \textbf{R}^{+}$ such that $\beta _{n,p}\le \overline{\beta }_{n}$, $\forall \,p\in \left( \textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} \right) \cup \left\{ \infty \right\}$, from which $\left( \bigotimes _{q=n+1}^{\infty } g_{q}\left| \varphi _{qq}^{\prime }\right| \right) (x)\le \alpha \overline{\beta }_{n}$, $\forall \,x\in \prod _{q=n+1}^{\infty } \varphi _{qq}^{-1}(B_{q})$.

Then, if $\prod _{q=n+1}^{\infty } \varphi _{qq}^{-1}(B_{q})=\emptyset$, there exists $q>n$ such that $\varphi _{qq}^{-1}(B_{q})=\emptyset$, and so

$$\begin{aligned}{} & {} \underset{p\rightarrow \infty }{\lim }\underset{\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\left. \underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left( g_{q}(x_{q})\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \right) \right. d\left( \underset{q=n+1}{\overset{p}{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) \left( x_{I_{p}\backslash I_{n}}\right) =0\nonumber \\{} & {} \quad =\underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\left. \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}g_{q}\left| \varphi _{qq}^{\prime }\right| \right. d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) ; \end{aligned}$$

(43)

conversely, if $\prod _{q=n+1}^{\infty } \varphi _{qq}^{-1}(B_{q})\ne \emptyset$, from Theorem 44, there exists

$$\begin{aligned} D\in \mathcal {B}^{\infty }\left( \underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})\right) ,\quad D\ne \underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q}), \end{aligned}$$

such that $\left( \bigotimes _{q=n+1}^{\infty } N\left( b_{q},\lambda _{qq}\right) \right) (D)=0$ and such that, $\forall \,y_{\textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} }\in \left( \prod _{q=n+1}^{\infty } \varphi _{qq}^{-1}(B_{q})\right) \backslash D$, we have

$$\begin{aligned}{} & {} \underset{p\rightarrow \infty }{\lim }\underset{\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\left. \underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\left( g_{q}(x_{q})\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \right) \right. d\left( \underset{q=n+1}{\overset{p}{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) \left( x_{I_{p}\backslash I_{n}}\right) \nonumber \\{} & {} \quad =\underset{p\rightarrow \infty }{\lim }\frac{1}{\underset{q=p+1}{\overset{\infty }{\prod }}\left( g_{q}(y_{q})\left| \varphi _{qq}^{\prime }\left( y_{q}\right) \right| \right) }\nonumber \\{} & {} \qquad \underset{\underset{q=n+1}{\overset{p}{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\left. \left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}\left( g_{q}\left| \varphi _{qq}^{\prime }\right| \right) \right) \left( x_{I_{p}\backslash I_{n}},y_{\textbf{N}^{*}\backslash \left\{ 1,\ldots ,p\right\} }\right) \right. \nonumber \\{} & {} \qquad \cdot d\left( \underset{q=n+1}{\overset{p}{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) \left( x_{I_{p}\backslash I_{n}}\right) \nonumber \\{} & {} \quad =\underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } }\left. \underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\left( g_{q}(x_{q})\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \right) \right. d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) \left( x_{\textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} }\right) . \nonumber \\ \end{aligned}$$

(44)

Then, from formulas (39), (43) and (44), we obtain

$$\begin{aligned}{} & {} \underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}B_{q}}{ {\displaystyle \int } }d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( a_{q},\gamma _{qq}\right) \right) \nonumber \\{} & {} \quad =\left. \underset{i=n+1}{\overset{\infty }{ {\textstyle \prod } }}\sqrt{\frac{\lambda _{ii}}{\gamma _{ii}}}\right. \underset{\underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\varphi _{qq}^{-1}(B_{q})}{ {\displaystyle \int } } \underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\left( g_{q}(x_{q})\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \right) \nonumber \\{} & {} \qquad d\left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) \left( x_{\textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} }\right) ; \end{aligned}$$

(45)

thus, since $\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(B)=\left( \varphi ^{(n,n)}\right) ^{-1}\left( \prod _{p=1}^{n} B_{p}\right) \times \left( \prod _{q=n+1}^{\infty } \varphi _{qq}^{-1}(B_{q})\right) \cap E_{\infty }$ and $\left( \bigotimes _{q=n+1}^{\infty } N\left( b_{q},\lambda _{qq}\right) \right) \left( E_{\infty }\right) =1$, from formulas (34), (35) and (45), we have

$$\begin{aligned}{} & {} \underset{E_{\infty }}{ {\displaystyle \int } }1_{B}dN(a,\Gamma )=\underset{B\cap E_{\infty }}{ {\displaystyle \int } }dN(a,\Gamma )=\underset{\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(B)}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda ^{(n,n)}}{\det \Gamma ^{(n,n)}}}\nonumber \\{} & {} \quad \cdot \exp \left( -\frac{\left\langle \left( \Gamma ^{(n,n)}\right) ^{-1}\left( \varphi ^{(n,n)}\left( x_{I_{n}}\right) -a_{I_{n}}\right) ,\varphi ^{(n,n)}\left( x_{I_{n}}\right) -a_{I_{n}}\right\rangle }{2}\right) \nonumber \\{} & {} \quad \cdot \exp \left( \frac{\left\langle \left( \Lambda ^{(n,n)}\right) ^{-1}\left( x_{I_{n}}-b_{I_{n}}\right) ,x_{I_{n}}-b_{I_{n}}\right\rangle }{2}\right) |\det J_{\varphi ^{(n,n)}}\left( x_{I_{n}}\right) |\nonumber \\{} & {} \quad \cdot \left. \underset{i=n+1}{\overset{\infty }{ {\textstyle \prod } }}\sqrt{\frac{\lambda _{ii}}{\gamma _{ii}}}\right. \left. \underset{q=n+1}{\overset{\infty }{ {\textstyle \prod } }}\left( g_{q}(x_{q})\left| \varphi _{qq}^{\prime }\left( x_{q}\right) \right| \right) \right. \nonumber \\{} & {} \quad \cdot d\left( N\left( b_{I_{n}},\Lambda ^{(n,n)}\right) \otimes \left( \underset{q=n+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( b_{q},\lambda _{qq}\right) \right) \right) \left( x_{i}:i\in \textbf{N}^{*}\right) \nonumber \\{} & {} \quad =\underset{\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(B)}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left( x_{i}:i\in \textbf{N}^{*}\right) \nonumber \\{} & {} \quad \cdot |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\left. \underset{q=n+1}{\overset{\infty }{\prod }}g_{q} (x_{q})\right. dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N} ^{*}\right) \nonumber \\{} & {} \quad =\underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}1_{B}\left( \overline{\varphi } ^{(n,n)}\left( x_{i}:i\in \textbf{N}^{*}\right) \right) \nonumber \\{} & {} \quad \cdot \exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left( x_{i}:i\in \textbf{N}^{*}\right) |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\nonumber \\{} & {} \quad \cdot \left. \underset{q=n+1}{\overset{\infty }{\prod }}g_{q}(x_{q})\right. dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N}^{*}\right) \quad \text {(since }\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(B)\subset K\text {).} \end{aligned}$$

(46)

Consider the measures $\mu _{1}$ and $\mu _{2}$ on $\Sigma \equiv \mathcal {B} ^{\infty }\left( \left[ -M,M\right] ^{\infty }\right)$ defined by

$$\begin{aligned} \mu _{1}(B)= & {} \underset{E_{\infty }}{ {\displaystyle \int } }1_{B}dN(a,\Gamma ),\\ \mu _{2}(B)= & {} \underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}1_{B}\left( \overline{\varphi } ^{(n,n)}\left( x_{i}:i\in \textbf{N}^{*}\right) \right) \\{} & {} \cdot \exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left( x_{i}:i\in \textbf{N}^{*}\right) \\{} & {} \cdot |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\left. \underset{q=n+1}{\overset{\infty }{\prod }}g_{q} (x_{q})\right. dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N} ^{*}\right) ; \end{aligned}$$

from formula (46), $\mu _{1}$ and $\mu _{2}$ coincide on the set

$$\begin{aligned} \mathcal {I}=\left\{ B\in \mathcal {B}^{\infty }\left( \left[ -M,M\right] ^{\infty }\right) :B=\underset{i=1}{\overset{\infty }{ {\displaystyle \prod } }}B_{i}\right\} ; \end{aligned}$$

moreover, we have $\mu _{1}\left( [-M,M]^{\infty }\right) =\mu _{2}\left( [-M,M]^{\infty }\right) <\infty$, $[-M,M]^{\infty }\in \mathcal {I}$, and so $\mu _{1}$ and $\mu _{2}$ are $\sigma$- finite on $\mathcal {I}$. Then, since $\mathcal {I}$ is a $\pi$-system on $[-M,M]^{\infty }$ such that $\sigma (\mathcal {I})=\mathcal {B}^{\infty }\left( \left[ -M,M\right] ^{\infty }\right)$, from Proposition 45, $\mu _{1}$ and $\mu _{2}$ coincide on $\mathcal {B}^{\infty }\left( \left[ -M,M\right] ^{\infty }\right)$, and in particular on $\mathcal {B}_{\infty }\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right)$; thus, suppose that $\psi :\left( E_{\infty },\mathcal {B}_{\infty }\right) \longrightarrow ([0,+\infty ),\mathcal {B}\left( [0,+\infty )\right) )$ is a simple function such that $\psi (x)=0$ $\forall \,x\in E_{\infty }\backslash \left[ -M,M\right] ^{\infty }$; $\forall \,x\in E_{\infty }$, we have $\psi (x)=\sum _{i=1}^{k} c_{i}1_{B_{i}}$, for some $k\in \textbf{N}^{*}$, $c_{1},\ldots ,c_{k}\in \textbf{R}$, $B_{1},\ldots ,B_{k}\in \mathcal {B}_{\infty }\left( \left[ -M,M\right] ^{\infty }\cap E_{\infty }\right)$, and so

$$\begin{aligned}{} & {} \underset{E_{\infty }}{ {\displaystyle \int } }\psi dN(a,\Gamma )=\underset{i=1}{\overset{k}{\sum }}c_{i}\underset{E_{\infty } }{ {\displaystyle \int }}1_{B_{i}}dN(a,\Gamma )\\{} & {} \quad =\underset{i=1}{\overset{k}{\sum }}c_{i}\underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}1_{B_{i}}(\overline{\varphi } ^{(n,n)}\left( x_{i}:i\in \textbf{N}^{*}\right) )\\{} & {} \quad \cdot \exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left( x_{i}:i\in \textbf{N}^{*}\right) |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\\{} & {} \quad \cdot \left. \underset{q=n+1}{\overset{\infty }{\prod }}g_{q}(x_{q})\right. dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N}^{*}\right) \\{} & {} \quad =\underset{K}{{\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\psi (\overline{\varphi }^{(n,n)}\left( x_{i}:i\in \textbf{N}^{*}\right) )\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \\{} & {} \qquad \left( x_{i}:i\in \textbf{N}^{*}\right) \\{} & {} \quad \cdot |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\left. \underset{q=n+1}{\overset{\infty }{\prod }}g_{q} (x_{q})\right. dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N} ^{*}\right) . \end{aligned}$$

Then, if $l:\left( E_{\infty },\mathcal {B}_{\infty }\right) \longrightarrow ([0,+\infty ),\mathcal {B}\left( [0,+\infty )\right) )$ is a measurable function such that $l(x)=0$ $\forall \,x\in E_{\infty }\backslash \left[ -M,M\right] ^{\infty }$, and $\left\{ \psi _{i}\right\} _{i\in \textbf{N}}$ is a sequence of increasing positive simple functions over $\left( E_{\infty },\mathcal {B}_{\infty }\right)$ such that $\psi _{i}(x)=0$ $\forall \,x\in E_{\infty }\backslash \left[ -M,M\right] ^{\infty }$, $\forall \,i\in \textbf{N}$, and $\lim _{i\longrightarrow \infty }\psi _{i}=l$, from Beppo Levi theorem we have

$$\begin{aligned}{} & {} \underset{E_{\infty }}{ {\displaystyle \int } }ldN(a,\Gamma )=\underset{i\rightarrow \infty }{\lim }\underset{E_{\infty }}{ {\displaystyle \int } }\psi _{i}dN(a,\Gamma )=\underset{i\rightarrow \infty }{\lim }\underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\psi _{i}(\overline{\varphi } ^{(n,n)}\left( x_{i}:i\in \textbf{N}^{*}\right) )\nonumber \\{} & {} \quad \cdot \exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left( x_{i}:i\in \textbf{N}^{*}\right) \nonumber \\{} & {} \quad \cdot |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\left. \underset{q=n+1}{\overset{\infty }{\prod }}g_{q} (x_{q})\right. dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N} ^{*}\right) \nonumber \\{} & {} \quad =\underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}l(\overline{\varphi }^{(n,n)}\left( x_{i}:i\in \textbf{N}^{*}\right) )\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \nonumber \\{} & {} \quad \left( x_{i}:i\in \textbf{N}^{*}\right) \nonumber \\{} & {} \quad \cdot |\det J_{\overline{\varphi }^{(n,n)}}\left( x_{i}:i\in \textbf{N}^{*}\right) |\left. \underset{q=n+1}{\overset{\infty }{\prod }}g_{q} (x_{q})\right. dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N} ^{*}\right) . \end{aligned}$$

(47)

$\square$

Theorem 41

(Gaussian change of variables’ formula). Let $\varphi :U=U^{(m)}\times E_{\infty }\longrightarrow E_{\infty }$ be a bijective, continuous and $\left( m,\sigma \right)$ function, such that $\overline{\varphi }:U\longrightarrow E_{\infty }$ is continuous and such that, for any $n\in \textbf{N}^{*}$, $n\ge m$, the function $\overline{\varphi } ^{(n,n)}:U\longrightarrow E_{\infty }$ is a diffeomorphism; moreover, let $a=\left( a_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$, let $b=\left( b_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$ such that $a_{i} =\varphi _{ii}(b_{i})$, for any $i>m$, and let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}:E_{\infty }\longrightarrow E_{\infty }$ and $\Lambda =\left( \lambda _{ij}\right) _{i,j\in \textbf{N}^{*}}:E_{\infty }\longrightarrow E_{\infty }$ some symmetric m-positive definite linear functions, such that $\prod _{i=m+1}^{\infty } \frac{\lambda _{ii}}{\gamma _{ii}}\in \textbf{R}^{+}$, $\sum _{i=m+1}^{\infty } \gamma _{ii}\in \textbf{R}^{+}$, $\sum _{i=m+1}^{\infty } \lambda _{ii}\in \textbf{R}^{+}$, and $\sum _{i=m+1}^{\infty } \left| \frac{\left( \varphi _{ii}^{\prime }\left( x_{i}\right) \right) ^{2}}{\gamma _{ii}}-\frac{1}{\lambda _{ii}}\right| \in \left[ 0,+\infty \right)$, for any $\left( x_{i}:i>m\right) \in E_{\infty }$; finally, suppose that there exists $\varepsilon \in \left[ 0,\infty \right)$ such that $\left| \varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) \right| \le \varepsilon$, for any $i>m$, for any $\left( x_{1},\ldots ,x_{m}\right) \in U^{(m)}$, and suppose that the sequence $\left\{ \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\right\} _{n>m}$ converges uniformly to $\varphi ^{-1}$ over the closed and bounded subsets of $E_{\infty }$, and the sequences $\left\{ -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)} +\Psi ^{\left( n,b,\Lambda \right) }\right\} _{n>m}$ and $\left\{ \det J_{\overline{\varphi }^{(n,n)}}\right\} _{n>m}$ converge uniformly over the closed and bounded subsets of U. Then, for any $B\in \mathcal {B}_{\infty }$ and for any measurable function $f:\left( E_{\infty },\mathcal {B}_{\infty }\right) \longrightarrow (\overline{\textbf{R} },\mathcal {B}\left( \overline{\textbf{R}}\right) )$ such that $f^{+}$ (or $f^{-}$) is $N(a,\Gamma )$-integrable, one has

$$\begin{aligned}{} & {} \underset{B}{ {\displaystyle \int } }fdN(a,\Gamma )=\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}f(\varphi )\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left| \det J_{\overline{\varphi }^{(n,n)} }\right| dN(b,\Lambda ). \end{aligned}$$

(48)

In particular, assume that, for any $x\in U$, there exists the function $J_{\varphi }(x):\ell _{\infty }\longrightarrow \ell _{\infty }$; then, one has

$$\begin{aligned}{} & {} \underset{B}{ {\displaystyle \int } }fdN(a,\Gamma )=\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}f(\varphi )\\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left| \det J_{\varphi }\right| dN(b,\Lambda ). \end{aligned}$$

Proof

From Proposition 40, $\forall \,M\in \textbf{R}^{+}$ such that $M\ge \left\| a\right\| _{\infty }$, there exists $O\in \textbf{R}^{+}$, there exists a closed subset K of $U\cap [-O,O]^{\infty }$, and, $\forall \,q>m$, there exists a measurable function $g_{q}:\left( [-O,O],\mathcal {B}\left( [-O,O]\right) \right) \longrightarrow (\textbf{R}^{+},\mathcal {B}\left( \textbf{R}^{+}\right) )$ such that $\left( \bigotimes _{q=m+1}^{\infty } g_{q}\right) (x)\in \textbf{R}^{+}$, $\forall \,x\in [-O,O]^{\infty }$, and such that, $\forall \,n\in \textbf{N}^{*}$, $n\ge m$, and $\forall \,$measurable function $l:\left( E_{\infty },\mathcal {B}_{\infty }\right) \longrightarrow (\left[ 0,\infty \right) ,\mathcal {B}\left( \left[ 0,\infty \right) \right) )$ such that $l(x)=0$ $\forall \,x\in E_{\infty }\backslash \left[ -M,M\right] ^{\infty }$, we have $\int _{E_{\infty }}ldN(a,\Gamma )=\int _{K} h_{n}dN\left( b,\Lambda \right)$, where, $\forall \,n\in \textbf{N}^{*}$, $n\ge m$, the function $h_{n}:\left( K,\mathcal {B}^{\infty }\left( K\right) \right) \longrightarrow ([0,\infty ),\mathcal {B}\left( [0,\infty )\right) )$ is defined by

$$\begin{aligned} h_{n}(x)= & {} \sqrt{\frac{\det \Lambda }{\det \Gamma }}l(\overline{\varphi } ^{(n,n)}\left( x\right) )\\{} & {} \cdot \exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left( x\right) \\{} & {} \cdot |\det J_{\overline{\varphi }^{(n,n)}}\left( x\right) |\underset{q=n+1}{\overset{\infty }{\prod }}g_{q}(x_{q})\text {, }\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in K; \end{aligned}$$

thus, we obtain

$$\begin{aligned} \underset{E_{\infty }}{ {\displaystyle \int } }ldN(a,\Gamma )=\underset{n\rightarrow \infty }{\lim }\underset{K}{ {\displaystyle \int } }h_{n}dN\left( b,\Lambda \right) . \end{aligned}$$

(49)

In particular, suppose that l is $\left( \tau _{\left\| \cdot \right\| _{\infty }}\left( E_{\infty }\right) ,\tau \left( [0,\infty )\right) \right)$-continuous and such that $l\left( E_{\infty }\right) \subset \left[ 0,1\right]$, $l(x)=0$, $\forall \,x\in E_{\infty } \backslash \left[ -M,M\right] ^{\infty }$; by assumption, the sequences $\left\{ -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi } ^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right\} _{n>m}$ and $\left\{ \det J_{\overline{\varphi }^{(n,n)}}\right\} _{n>m}$ converge uniformly over K; then, there exists $\widehat{n}\in \textbf{N}^{*}$, $\widehat{n}>m$, such that, $\forall \,n\ge \widehat{n}$, $\forall \,x\in K$, we have

$$\begin{aligned}{} & {} -\Psi ^{\left( n,a,\Gamma \right) }\left( \overline{\varphi }^{(n,n)} (x)\right) +\Psi ^{\left( n,b,\Lambda \right) }(x)\\{} & {} \quad \le -\Psi ^{\left( \widehat{n},a,\Gamma \right) }\left( \overline{\varphi }^{(\widehat{n},\widehat{n})}(x)\right) +\Psi ^{\left( \widehat{n},b,\Lambda \right) }(x)+1,\\{} & {} \quad |\det J_{\overline{\varphi }^{(n,n)}}(x)|\le |\det J_{\overline{\varphi }^{(\widehat{n},\widehat{n})}}(x)|+1, \end{aligned}$$

and so, $\forall \,n>m$, $\forall \,x\in K$, the following inequalities hold:

$$\begin{aligned}{} & {} \exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\left( \overline{\varphi }^{(n,n)}(x)\right) +\Psi ^{\left( n,b,\Lambda \right) }(x)\right) \nonumber \\{} & {} \quad \le \underset{h=m+1}{\overset{\widehat{n}-1}{\sum }}\exp \left( -\Psi ^{\left( h,a,\Gamma \right) }\left( \overline{\varphi }^{(h,h)}(x)\right) +\Psi ^{\left( h,b,\Lambda \right) }(x)\right) \nonumber \\{} & {} \qquad +e\exp \left( -\Psi ^{\left( \widehat{n},a,\Gamma \right) }\left( \overline{\varphi }^{(\widehat{n},\widehat{n})}(x)\right) +\Psi ^{\left( \widehat{n},b,\Lambda \right) }(x)\right) \nonumber \\{} & {} \quad \le e\underset{h=m+1}{\overset{\widehat{n}}{\sum }}\exp \left( -\Psi ^{\left( h,a,\Gamma \right) }\left( \overline{\varphi }^{(h,h)}(x)\right) +\Psi ^{\left( h,b,\Lambda \right) }(x)\right) ,\nonumber \\{} & {} \quad |\det J_{\overline{\varphi }^{(n,n)}}(x)|\le \underset{h=m+1}{\overset{\widehat{n}-1}{\sum }}|\det J_{\overline{\varphi }^{(h,h)}}(x)|+|\det J_{\overline{\varphi }^{(\widehat{n},\widehat{n})}}(x)|+1\nonumber \\{} & {} \quad =\underset{h=m+1}{\overset{\widehat{n}}{\sum }}|\det J_{\overline{\varphi }^{(h,h}}(x)|+1. \end{aligned}$$

(50)

Moreover, $\forall \,h\in \left\{ m+1,\ldots ,\widehat{n}\right\}$, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in K$, we have

$$\begin{aligned} |\det J_{\overline{\varphi }^{(h,h)}}(x)|=|\det J_{\varphi ^{(h,h)}} (x_{1},\ldots ,x_{h})|\underset{i=h+1}{\overset{\infty }{ {\displaystyle \prod } }}\left| \varphi _{ii}^{\prime }(x_{i})\right| ; \end{aligned}$$

(51)

furthermore, from Proposition 71 and Proposition 61, $\forall \,h\in \left\{ m+1,\ldots ,\widehat{n}\right\}$, $\forall \,i\in \textbf{N}^{*}\backslash \left\{ 1,\ldots ,h\right\}$, the functions $\det J_{\varphi ^{(h,h)}}$ and $\varphi _{ii}^{\prime }$ are continuous; moreover, the sets $\pi _{\left\{ 1,\ldots ,h\right\} }\left( K\right)$ and $\pi _{i}\left( K\right)$ are closed and bounded, and from Definition 17 we have $\prod _{i=h+1}^{\infty } \left| \varphi _{ii}^{\prime }(x_{i})\right| \in \textbf{R}^{+}$, $\forall \,x=(x_{i}:i\in \textbf{N}^{*}\backslash \left\{ 1,\ldots ,h\right\} )\in \textbf{R}^{\infty }$; then, from formulas (50) and (51), there exists $c\in \textbf{R}^{+}$ such that

$$\begin{aligned} |\det J_{\overline{\varphi }^{(n,n)}}(x)|\le c\text {, }\forall \,n>m\text {, }\forall \,x\in K; \end{aligned}$$

finally, from formula (40), there exists $\alpha \in \textbf{R}^{+}$ such that $\prod _{q=n+1}^{\infty } g_{q}(x_{q})\le \alpha$, $\forall \,n>m$, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in K$; thus, since $l\left( E_{\infty }\right) \subset [0,1]$, we obtain $\left| h_{n}\right| \le g$, $\forall \,n>m$, where

$$\begin{aligned} g:\left( K,\mathcal {B}^{\infty }\left( K\right) \right) \longrightarrow ([0,\infty ),\mathcal {B}\left( [0,\infty )\right) ) \end{aligned}$$

is the measurable function defined by

$$\begin{aligned} g(x)= & {} \alpha ce\sqrt{\frac{\det \Lambda }{\det \Gamma }}\nonumber \\{} & {} \cdot \underset{h=m+1}{\overset{\widehat{n}}{\sum }}\exp \left( -\Psi ^{\left( h,a,\Gamma \right) }\left( \overline{\varphi }^{(h,h)}(x)\right) +\Psi ^{\left( h,b,\Lambda \right) }(x)\right) \text {, }\forall \,x\in K, \end{aligned}$$

(52)

and so

$$\begin{aligned}{} & {} \underset{K}{ {\displaystyle \int } }gdN\left( b,\Lambda \right) =\alpha ce\sqrt{\frac{\det \Lambda }{\det \Gamma }}\nonumber \\{} & {} \quad \cdot \underset{h=m+1}{\overset{\widehat{n}}{\sum }}\left. \underset{K}{ {\displaystyle \int } }\exp \left( -\frac{\left\langle \left( \Gamma ^{(h,h)}\right) ^{-1}\left( \varphi ^{(h,h)}\left( x_{I_{h}}\right) -a_{I_{h}}\right) ,\varphi ^{(h,h)}\left( x_{I_{h}}\right) -a_{I_{h}}\right\rangle }{2}\right) \right. \nonumber \\{} & {} \quad \cdot \exp \left( \frac{\left\langle \left( \Lambda ^{(h,h)}\right) ^{-1}\left( x_{I_{h}}-b_{I_{h}}\right) ,x_{I_{h}}-b_{I_{h}}\right\rangle }{2}\right) dN\left( b,\Lambda \right) \left( x_{i}:i\in \textbf{N}^{*}\right) \nonumber \\{} & {} \quad \le \alpha ce\sqrt{\frac{\det \Lambda }{\det \Gamma }}\nonumber \\{} & {} \quad \cdot \underset{h=m+1}{\overset{\widehat{n}}{\sum }}\left. \underset{\pi _{\left\{ 1,\ldots ,h\right\} }(K)}{ {\displaystyle \int } }\right. \exp \left( -\frac{\left\langle \left( \Gamma ^{(h,h)}\right) ^{-1}\left( \varphi ^{(h,h)}\left( x_{I_{h}}\right) -a_{I_{h}}\right) ,\varphi ^{(h,h)}\left( x_{I_{h}}\right) -a_{I_{h}}\right\rangle }{2}\right) \nonumber \\{} & {} \quad \cdot \exp \left( \frac{\left\langle \left( \Lambda ^{(h,h)}\right) ^{-1}\left( x_{I_{h}}-b_{I_{h}}\right) ,x_{I_{h}}-b_{I_{h}}\right\rangle }{2}\right) \nonumber \\{} & {} \quad \cdot dN\left( b_{I_{h}},\Lambda ^{(h,h)}\right) (x_{I_{h}})\cdot \underset{E_{\infty }}{ {\displaystyle \int } }d\left( \underset{i=h+1}{\overset{\infty }{ {\displaystyle \bigotimes } }}N\left( b_{i},\lambda _{ii}\right) \right) ; \end{aligned}$$

(53)

furthermore, $\forall \,x_{I_{h}}\in \textbf{R}^{h}$, we have

$$\begin{aligned}{} & {} dN\left( b_{I_{h}},\Lambda ^{(h,h)}\right) (x_{I_{h}})=\frac{1}{\left( 2\pi \right) ^{\frac{h}{2}}\sqrt{\det \Lambda ^{(h,h)}}}\\{} & {} \quad \cdot \exp \left( -\frac{\left\langle \left( \Lambda ^{(h,h)}\right) ^{-1}\left( x_{I_{h}}-b_{I_{h}}\right) ,x_{I_{h}}-b_{I_{h}}\right\rangle }{2}\right) dx_{1}\ldots dx_{h}, \end{aligned}$$

and $\left( \bigotimes _{i=h+1}^{\infty } N\left( b_{i},\lambda _{ii}\right) \right) \left( E_{\infty }\right) =1$ (from Proposition 36); then, formula (53) implies

$$\begin{aligned} \underset{K}{ {\displaystyle \int } }gdN\left( b,\Lambda \right) \le \alpha ce\sqrt{\frac{\det \Lambda }{\det \Gamma }}\underset{h=m+1}{\overset{\widehat{n}}{\sum }}\left. \underset{\pi _{\left\{ 1,\ldots ,h\right\} }(K)}{ {\displaystyle \int } }\right. t_{h}\left( x_{1},\ldots ,x_{h}\right) dx_{1}\ldots dx_{h}, \end{aligned}$$

(54)

where, $\forall \,h\in \left\{ m+1,\ldots ,\widehat{n}\right\}$, the measurable function

$$\begin{aligned} t_{h}:\left( \pi _{\left\{ 1,\ldots ,h\right\} }(K),\mathcal {B}^{h}\left( \pi _{\left\{ 1,\ldots ,h\right\} }(K)\right) \right) \longrightarrow ([0,\infty ),\mathcal {B}\left( [0,\infty )\right) ) \end{aligned}$$

is defined by

$$\begin{aligned} t_{h}\left( x\right)= & {} \frac{1}{\left( 2\pi \right) ^{\frac{h}{2}}\sqrt{\det \Lambda ^{(h,h)}}}\\{} & {} \cdot \exp \left( -\frac{\left\langle \left( \Gamma ^{(h,h)}\right) ^{-1}\left( \varphi ^{(h,h)}\left( x\right) -a_{I_{h}}\right) ,\varphi ^{(h,h)}\left( x\right) -a_{I_{h}}\right\rangle }{2}\right) ,\\ \forall \,x= & {} \left( x_{1},\ldots ,x_{h}\right) \in \pi _{\left\{ 1,\ldots ,h\right\} }(K); \end{aligned}$$

moreover, $\forall \,h\in \left\{ m+1,\ldots ,\widehat{n}\right\}$, the function $\varphi ^{(h,h)}$ is continuous, and so the function $t_{h}$ is continuous too; then, since the set $\pi _{\left\{ 1,\ldots ,h\right\} }(K)$ is closed and bounded, from formula (54), we obtain $\int _{K} gdN\left( b,\Lambda \right) <\infty$. Moreover, we have $\lim _{n\rightarrow \infty }\overline{\varphi }^{(n,n)}=\varphi$, from the point 1 of Definition 17, and $\lim _{n\rightarrow \infty }\left( \prod _{q=n+1}^{\infty } g_{q}(x_{q})\right) =1$, $\forall \,x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in K$, from which

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }h_{n}= & {} \sqrt{\frac{\det \Lambda }{\det \Gamma } }l(\varphi )\\{} & {} \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \left| \det J_{\overline{\varphi }^{(n,n)} }\right| ; \end{aligned}$$

consequently, from formula (49) and from the dominated convergence theorem, we obtain

$$\begin{aligned} \underset{E_{\infty }}{ {\displaystyle \int } }ldN(a,\Gamma )= & {} \underset{K}{ {\displaystyle \int } }\underset{n\rightarrow \infty }{\lim }h_{n}dN\left( b,\Lambda \right) =\underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}l(\varphi )\nonumber \\{} & {} \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) \nonumber \\{} & {} \quad |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) . \end{aligned}$$

(55)

Moreover, $\forall \,y=(y_{i}:i\in \textbf{N}^{*})\in \ell _{\infty }$, $\forall \,\delta =(\delta _{i}:i\in \textbf{N}^{*})\in \left( \textbf{R} ^{+}\right) ^{\infty }\cap \ell _{\infty }$ such that $\inf _{i\in \textbf{N}^{*}}\delta _{i}>0$, $\forall \,h\in \textbf{N}^{*}$, $\forall \,t\in [0,1]$, let $A_{y,\delta ,h,t}=\prod _{i=1}^{\infty }\left( y_{i}-\delta _{i}\left( 1-\frac{t}{2h}\right) ,y_{i}+\delta _{i}\left( 1-\frac{t}{2h}\right) \right) \subset \ell _{\infty }$, and let

$$\begin{aligned} \mathcal {A}_{M}= & {} \left\{ \overset{\circ }{A_{y,\delta ,1,0}}\cap E_{\infty }:y\in \overset{\circ }{\left( -M,M\right) ^{\infty }},\text { }\delta \in \overset{\circ }{\left( 0,M\right) ^{\infty }},\text { }\overset{\circ }{A_{y,\delta ,1,0}}\subset \overset{\circ }{\left( -M,M\right) ^{\infty }}\right\} \\{} & {} \subset \tau _{\left\| \cdot \right\| _{\infty }}\left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right) \subset \tau _{\left\| \cdot \right\| _{\infty }}\left( E_{\infty }\right) , \end{aligned}$$

where the interior parts of $A_{y,\delta ,1,0}$ and $\left( -M,M\right) ^{\infty }$ are considered with respect to the topology $\tau _{\left\| \cdot \right\| _{\infty }}$, and so $\overset{\circ }{\left( -M,M\right) ^{\infty }}=B_{\infty }(0,M)$; since $\inf _{i\in \textbf{N}^{*}}\delta _{i}>0$, we have $\overset{\circ }{A_{y,\delta ,h,t}}\cap E_{\infty } \in \tau _{\left\| \cdot \right\| _{\infty }}\left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right) \subset \tau _{\left\| \cdot \right\| _{\infty }}\left( E_{\infty }\right)$. Consider the function $l_{y,\delta ,h}:E_{\infty }\longrightarrow [0,\infty )$ defined (with respect to the topology $\tau _{\left\| \cdot \right\| _{\infty }}$) by

$$\begin{aligned} l_{y,\delta ,h}(x)=\left\{ \begin{array}{ll} 1 &{}\quad \text {if}\quad x\in \overset{\circ }{A_{y,\delta ,h,1}}\cap E_{\infty }\\ t &{}\quad \text {if}\quad x\in \partial A_{y,\delta ,h,t}\cap E_{\infty }\\ 0 &{}\quad \text {if}\quad x\in E_{\infty }\backslash \overline{A_{y,\delta h,0}} \end{array}\right. ; \end{aligned}$$

observe that $l_{y,\delta ,h}:E_{\infty }\longrightarrow [0,\infty )$ is a function such that $l_{y,\delta ,h}\left( E_{\infty }\right) \subset [0,1]$, $l_{y,\delta ,h}(x)=0$, $\forall \,x\notin [-M,M]^{\infty }$; moreover, $\forall \,t_{1},t_{2}\in [0,\infty )$ such that $t_{1}<t_{2}$, we have

$$\begin{aligned}{} & {} l_{y,\delta ,h}^{-1}(\left( t_{1},t_{2}\right) )=\left\{ \begin{array}{ll} \emptyset &{}\quad \text {if}\quad t_{1}\ge 1\\ \overset{\circ }{A_{y,\delta h,t_{1}}}\cap E_{\infty } &{}\quad \text {if}\quad t_{1}<1<t_{2}\\ \left( \overset{\circ }{A_{y,\delta h,t_{1}}}\backslash \overline{A_{y,\delta h,t_{2}}}\right) \cap E_{\infty } &{}\quad \text {if}\quad t_{1}<t_{2}\le 1 \end{array}\right. ,\\{} & {} l_{y,\delta ,h}^{-1}(\left[ 0,t_{2}\right) )=\left\{ \begin{array}{ll} E_{\infty } &{} \quad \text {if}\quad t_{2}>1\\ E_{\infty }\backslash \overline{A_{y,\delta h,t_{2}}} &{} \quad \text {if}\quad t_{2}\le 1 \end{array}\right. ; \end{aligned}$$

thus, $l_{y,\delta ,h}$ is $\left( \tau _{\left\| \cdot \right\| _{\infty } }\left( E_{\infty }\right) ,\tau \left( [0,\infty )\right) \right)$-continuous, and so it is $\left( \mathcal {B}_{\infty },\mathcal {B}\left( [0,\infty )\right) \right)$-measurable. Then, $\forall \,B=\overset{\circ }{A_{y,\delta ,1,0}}\cap E_{\infty }\in \mathcal {A}_{M}$, since $\left\{ l_{y,\delta ,h}\right\} _{h\in \textbf{N}^{*}}$ is an increasing positive sequence such that $\lim _{h\longrightarrow \infty }l_{y,\delta ,h}=1_{B}$, from Beppo Levi theorem and formula (55), we have

$$\begin{aligned}{} & {} \underset{B}{ {\displaystyle \int } }dN(a,\Gamma )=\underset{E_{\infty }}{ {\displaystyle \int } }1_{B}dN(a,\Gamma )=\underset{h\rightarrow \infty }{\lim }\underset{E_{\infty }}{ {\displaystyle \int }}l_{h}dN(a,\Gamma )\nonumber \\{} & {} \quad =\underset{h\rightarrow \infty }{\lim }\underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}l_{h}(\varphi )\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) \nonumber \\{} & {} \quad =\underset{K}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}1_{B}(\varphi )\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) \nonumber \\{} & {} \quad =\underset{\varphi ^{-1}(B)}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) \nonumber \\{} & {} \quad \text {(since }B\subset [-M,M]^{\infty }\cap E_{\infty }\text {, and so }\varphi ^{-1}(B)\subset K\text {).} \end{aligned}$$

(56)

Consider the measures $\mu$ and $\upsilon$ on $\left( E_{\infty },\mathcal {B}_{\infty }\right)$ defined by

$$\begin{aligned} \mu (B)= & {} \underset{B}{ {\displaystyle \int } }dN(a,\Gamma ),\\ \upsilon (B)= & {} \underset{\varphi ^{-1}(B)}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\\{} & {} \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) ; \end{aligned}$$

from formula (56), $\mu$ and $\upsilon$ coincide on the set $\mathcal {A}_{M}$; moreover, we have

$$\begin{aligned} \mu \left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right) =\upsilon \left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right) <\infty ,\text { }\overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\in \mathcal {A}_{M}, \end{aligned}$$

and so $\mu$ and $\upsilon$ are $\sigma$- finite on $\mathcal {A}_{M}$; furthermore, since $\mathcal {B}^{\infty }\left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right) \supset \sigma (\mathcal {A} _{M})\supset \tau _{\left\| \cdot \right\| _{\infty }}\left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right)$ and since Proposition 14 implies

$$\begin{aligned} \sigma \left( \tau _{\left\| \cdot \right\| _{\infty }}\left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right) \right) =\mathcal {B}^{\infty }\left( \overset{\circ }{\left( -M,M\right) ^{\infty } }\cap E_{\infty }\right) , \end{aligned}$$

we obtain $\sigma (\mathcal {A}_{M})=\mathcal {B}^{\infty }\left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right)$; then, since $\mathcal {A}_{M}$ is a $\pi$-system on $\overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }$, from Proposition 45, $\mu$ and $\upsilon$ coincide on $\mathcal {B}^{\infty }\left( \overset{\circ }{\left( -M,M\right) ^{\infty }}\cap E_{\infty }\right)$.

Moreover, $\forall \,B\in \mathcal {B}_{\infty }$, $\forall \,k\in \textbf{N}^{*}$, set $B_{k}=B\cap \overset{\circ }{\left( -k,k\right) ^{\infty }} \in \mathcal {B}^{\infty }\left( \overset{\circ }{\left( -k,k\right) ^{\infty } }\cap E_{\infty }\right)$; we have $B_{k}\subset B_{k+1}$, $\varphi ^{-1}\left( B_{k}\right) \subset \varphi ^{-1}\left( B_{k+1}\right)$, $\bigcup _{k\in \textbf{N}^{*}}B_{k}=B$ and $\bigcup _{k\in \textbf{N}^{*}}\varphi ^{-1}(B_{k})=\varphi ^{-1}(B)$; thus, from the continuity property of $\mu$ and $\upsilon$, we have

$$\begin{aligned}{} & {} \underset{B}{ {\displaystyle \int } }dN(a,\Gamma )=\underset{k\rightarrow \infty }{\lim }\underset{B_{k}}{ {\displaystyle \int }}dN(a,\Gamma )\nonumber \\{} & {} \quad =\underset{k\rightarrow \infty }{\lim }\underset{\varphi ^{-1}\left( B_{k}\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) \nonumber \\{} & {} \quad =\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) . \end{aligned}$$

(57)

Then, let $B\in \mathcal {B}_{\infty }$ and let $g:\left( E_{\infty },\mathcal {B}_{\infty }\right)$ $\longrightarrow (\left[ 0,+\infty \right] ,\mathcal {B}\left( \left[ 0,+\infty \right] \right) )$ be a measurable function; by proceeding as in the proof of formula (47), formula (57) implies

$$\begin{aligned}{} & {} \underset{B}{ {\displaystyle \int } }gdN(a,\Gamma )=\underset{E_{\infty }}{ {\displaystyle \int }}1_{B}gdN(a,\Gamma )\\{} & {} \quad =\underset{U}{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}\left( 1_{B}g\right) (\varphi )\\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) \\{} & {} \quad =\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int }}\sqrt{\frac{\det \Lambda }{\det \Gamma }}g(\varphi )\\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) . \end{aligned}$$

Then, for any measurable function $f:\left( E_{\infty },\mathcal {B}_{\infty }\right) \longrightarrow (\overline{\textbf{R}},\mathcal {B}\left( \overline{\textbf{R}}\right) )$ such that $f^{+}$ (or $f^{-}$) is $N(a,\Gamma )$-integrable, we have

$$\begin{aligned}{} & {} \underset{B}{ {\displaystyle \int } }fdN(a,\Gamma )=\underset{B}{ {\displaystyle \int } }f^{+}dN(a,\Gamma )-\underset{B}{ {\displaystyle \int }}f^{-}dN(a,\Gamma )\nonumber \\{} & {} \quad =\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}f^{+}(\varphi )\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) \nonumber \\{} & {} \qquad -\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}f^{-}(\varphi )\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) \nonumber \\{} & {} \quad =\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}f(\varphi )\nonumber \\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\overline{\varphi }^{(n,n)}}|dN\left( b,\Lambda \right) . \end{aligned}$$

(58)

In particular, assume that, $\forall \,x\in U$, there exists the function $J_{\varphi }(x):\ell _{\infty }\longrightarrow \ell _{\infty }$; from Proposition 69, we have

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\left| \det J_{\overline{\varphi }^{(n,n)} }(x)\right| =\left| \underset{n\rightarrow \infty }{\lim }\det J_{\overline{\varphi }^{(n,n)}}(x)\right| =\left| \det J_{\varphi }(x)\right| \text {, }\forall \,x\in U, \end{aligned}$$

and so formula (58) implies

$$\begin{aligned}{} & {} \underset{B}{ {\displaystyle \int } }fdN(a,\Gamma )=\underset{\varphi ^{-1}\left( B\right) }{ {\displaystyle \int } }\sqrt{\frac{\det \Lambda }{\det \Gamma }}f(\varphi )\\{} & {} \quad \cdot \underset{n\rightarrow \infty }{\lim }\exp \left( -\Psi ^{\left( n,a,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,b,\Lambda \right) }\right) |\det J_{\varphi }|dN\left( b,\Lambda \right) . \end{aligned}$$

$\square$

Example 42

Let $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*} }:E_{\infty }\longrightarrow E_{\infty }$ be a linear and diagonal function, such that $\gamma _{ii}\in \textbf{R}^{+}$, for any $i\in \textbf{N}^{*}$, and $\sum _{i=1}^{\infty }\gamma _{ii}\in \textbf{R}^{+}$; then, one has

$$\begin{aligned}{} & {} \underset{\left\{ (x_{i}:i\in \textbf{N}^{*})\in E_{\infty }:\underset{i=1}{\overset{\infty }{\sum }}2^{-i+1}x_{i}>0\right\} }{ {\displaystyle \int } }\exp \left( -\frac{1}{2\gamma _{11}}\left( \left( \underset{i=1}{\overset{\infty }{\sum }}2^{-i+2}x_{i}\right) ^{2}+x_{1}^{2}\right) \right) \nonumber \\{} & {} \quad \cdot dN(0,\Gamma )(x_{i}:i\in \textbf{N}^{*})=\frac{1}{4}. \end{aligned}$$

(59)

Proof

Consider the linear function $\varphi :E_{\infty }\longrightarrow E_{\infty }$ defined by

$$\begin{aligned} \varphi _{1}(x)&=\underset{i=1}{\overset{\infty }{\sum }}2^{-i+2}x_{i} ,\quad \forall \,x=(x_{i}:i\in \textbf{N}^{*})\in E_{\infty },\\ \varphi _{j}(x)&=x_{j},\quad \forall \,j>1,\quad \forall \,x=(x_{i} :i\in \textbf{N}^{*})\in E_{\infty }; \end{aligned}$$

observe that $\varphi$ is $\left( 1,\sigma \right)$ standard, where $\sigma (i)=i$, $\forall \,i>1$; thus, from Proposition 21, $\varphi$ is continuous; moreover, the function $\overline{\varphi }:E_{\infty }\longrightarrow E_{\infty }$ is given by

$$\begin{aligned} \overline{\varphi }_{1}(x)&=2x_{1},\quad \forall \,x=(x_{i}:i\in \textbf{N}^{*})\in E_{\infty },\\ \overline{\varphi }_{j}(x)&=x_{j},\quad \forall \,j>1,\quad \forall \,x=(x_{i}:i\in \textbf{N}^{*})\in E_{\infty }, \end{aligned}$$

and so $\overline{\varphi }$ is $\left( 1,\sigma \right)$ standard, bijective and continuous; then, from Propositions 65 and 67, $\varphi$ is bijective; analogously, $\forall \,n\in \textbf{N}^{*}$, the function $\overline{\varphi }^{(n,n)}:E_{\infty }\longrightarrow E_{\infty }$ is $\left( 1,\sigma \right)$ standard, with $\overline{\overline{\varphi }^{(n,n)}}=\overline{\varphi }$, and so $\overline{\varphi }^{(n,n)}$ is bijective too; moreover, $\varphi$ is $C^{1}$ and Proposition 66 implies that $\overline{\varphi }$ is a diffeomorphism; then, from Proposition 63, $\overline{\varphi }^{(n,n)}$ is a diffeomorphism. Observe that,$\forall \,y=\left( y_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }$, $\forall \,n\in \textbf{N} ^{*}$, $\forall \,i>1$, we have $(\varphi ^{-1}(y))_{i}=(\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(y))_{i}=y_{i}$; then

$$\begin{aligned} y_{1}= & {} (\varphi ^{-1}(y))_{1}+\underset{i=2}{\overset{\infty }{\sum }} 2^{-i+2}y_{i},\\ y_{1}= & {} (\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(y))_{1}+\underset{i=2}{\overset{n}{\sum }}2^{-i+2}y_{i}, \end{aligned}$$

and so

$$\begin{aligned} (\varphi ^{-1}(y))_{1}= & {} y_{1}-\underset{i=2}{\overset{\infty }{\sum }} 2^{-i+2}y_{i},\nonumber \\ \left( \left( \overline{\varphi }^{(n,n)}\right) ^{-1}(y)\right) _{1}= & {} y_{1}-\underset{i=2}{\overset{n}{\sum }}2^{-i+2}y_{i},\nonumber \\ (\varphi ^{-1}(y))_{i}= & {} \left( \left( \overline{\varphi }^{(n,n)}\right) ^{-1}(y)\right) _{i}=y_{i},\quad \forall \,i>1; \end{aligned}$$

(60)

thus, from formula (60), we have

$$\begin{aligned} \left\| \varphi ^{-1}(y)-\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(y)\right\| _{\infty }= & {} \left| (\varphi ^{-1}(y))_{1}-(\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(y))_{1}\right| \nonumber \\= & {} \left| \underset{i=n+1}{\overset{\infty }{\sum }}2^{-i+2}y_{i}\right| \le \underset{i=n+1}{\overset{\infty }{\sum }}2^{-i+2}\left| y_{i}\right| . \end{aligned}$$

(61)

Then, let H a closed and bounded subset of $E_{\infty }$, and let $d\in \textbf{R}^{+}$ such that $H\subset \left[ -d,d\right] ^{\infty }$; $\forall \,y\in H$, formula (61) implies

$$\begin{aligned} \left\| \varphi ^{-1}(y)-\left( \overline{\varphi }^{(n,n)}\right) ^{-1}(y)\right\| _{\infty }\le 2^{-n+2}d, \end{aligned}$$

(62)

and so the sequence $\left\{ \left( \overline{\varphi }^{(n,n)}\right) ^{-1}\right\} _{n>1}$ converges uniformly to $\varphi ^{-1}$ over H; moreover, $\forall \,n\in \textbf{N}^{*}$, $\forall \,x=(x_{i}:i\in \textbf{N}^{*})\in E_{\infty }$, we have $\det J_{\overline{\varphi } ^{(n,n)}}(x)=2$, and

$$\begin{aligned}{} & {} \left( -\Psi ^{\left( n,0,\Gamma \right) }\circ \overline{\varphi } ^{(n,n)}\right) (x)+\Psi ^{\left( n,0,\Gamma \right) }(x)\nonumber \\{} & {} \quad =\frac{-\frac{1}{\gamma _{11}}\left( \underset{i=1}{\overset{n}{\sum }} 2^{-i+2}x_{i}\right) ^{2}-\underset{j=2}{\overset{n}{\sum }}\frac{x_{j}^{2} }{\gamma _{jj}}+\underset{k=1}{\overset{n}{\sum }}\frac{x_{k}^{2}}{\gamma _{kk}}}{2}\nonumber \\{} & {} \quad =\frac{-\left( \underset{i=1}{\overset{n}{\sum }}2^{-i+2}x_{i}\right) ^{2}+x_{1}^{2}}{2\gamma _{11}}; \end{aligned}$$

(63)

as we observed, the sequence of real functions $\left\{ h_{n}\right\} _{n\ge 1}$ on $E_{\infty }$ given by $h_{n}(x_{i}:i\in \textbf{N}^{*})=\sum _{i=1}^{n} 2^{-i+2}x_{i}$, $\forall \,n\in \textbf{N}^{*}$, $\forall \,(x_{i}:i\in \textbf{N}^{*})\in E_{\infty }$, converges uniformly to h over the closed and bounded subsets of $E_{\infty }$, where the real function h on $E_{\infty }$ is defined by

$$\begin{aligned} h(x_{i}:i\in \textbf{N}^{*})=\underset{i=1}{\overset{\infty }{\sum }} 2^{-i+2}x_{i},\quad \forall \,(x_{i}:i\in \textbf{N}^{*})\in E_{\infty }; \end{aligned}$$

then, from formula (63), the sequence $\left\{ -\Psi ^{\left( n,0,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,0,\Gamma \right) }\right\} _{n>1}$ converges uniformly over the closed and bounded subsets of $E_{\infty }$, and

$$\begin{aligned}{} & {} \underset{n\rightarrow \infty }{\lim }\left( -\Psi ^{\left( n,0,\Gamma \right) }\circ \overline{\varphi }^{(n,n)}+\Psi ^{\left( n,0,\Gamma \right) }\right) (x)\\{} & {} \quad =\frac{-\left( \underset{i=1}{\overset{\infty }{\sum }}2^{-i+2}x_{i}\right) ^{2}+x_{1}^{2}}{2\gamma _{11}},\quad \forall \,x=(x_{i}:i\in \textbf{N}^{*})\in E_{\infty }; \end{aligned}$$

moreover, $\forall \,\left( x_{i}:i>1\right) \in E_{\infty }$, $\forall \,i>1$, we have $\frac{\left( \varphi _{ii}^{\prime }\left( x_{i}\right) \right) ^{2}}{\gamma _{ii}}-\frac{1}{\gamma _{ii}}=0$, and so $\sum _{i=2}^{\infty } \left( \frac{\left( \varphi _{ii}^{\prime }\left( x_{i}\right) \right) ^{2}}{\gamma _{ii}}-\frac{1}{\gamma _{ii}}\right) =0$. Then, the assumptions of Theorem 41 are satisfied in the case $a=b=0\in E_{\infty }$, $\Lambda =\Gamma$, $B=\left\{ (y_{i}:i\in \textbf{N}^{*})\in E_{\infty }:y_{1}>0\right\}$ and $f(x)=1$, $\forall \,x\in E_{\infty }$; thus, since $\varphi ^{-1}\left( B\right) =\left\{ (x_{i}:i\in \textbf{N}^{*})\in E_{\infty }: \sum _{i=1}^{\infty } 2^{-i+1}x_{i}>0\right\}$, we obtain

$$\begin{aligned} \frac{1}{2}= & {} \underset{B}{ {\displaystyle \int } }dN(0,\Gamma )\quad \text {(since }N(0,\Gamma )(E_{\infty })=1\text {, from Proposition 36)}\\= & {} \underset{\left\{ (x_{i}:i\in \textbf{N}^{*})\in E_{\infty }:\underset{i=1}{\overset{\infty }{\sum }}2^{-i+1}x_{i}>0\right\} }{ {\displaystyle \int } }2\exp \left( -\frac{1}{2\gamma _{11}}\left( \left( \underset{i=1}{\overset{\infty }{\sum }}2^{-i+2}x_{i}\right) ^{2}+x_{1}^{2}\right) \right) \\{} & {} \cdot dN(0,\Gamma )(x_{i}:i\in \textbf{N}^{*})\quad \text {(from Theorem 41),} \end{aligned}$$

from which (59) follows. $\square$

At this point, we can compare Theorem 41 with Theorem 4.1 in [24]. The last result proves a formula like (48), although written in a formally different way. The function that defines the change of variables is of the type $\varphi :E\longrightarrow E$, where E is a separable Banach space with a Gaussian measure $\mu$, such that $\varphi (x)=x+F(x)$, for any $x\in E$, for some function $F:E\longrightarrow H$, where $H=H(\mu )$ is the Cameron–Martin space of $\left( E,\mu \right)$, with the norm $\left| \cdot \right| _{H}$ defined by formula (13). The function F satisfies some conditions, including the hypothesis

$$\begin{aligned} \left| F(x+h)-F(x)\right| _{H}\le c\left| h\right| _{H}, \end{aligned}$$

(64)

for almost all $x\in E$, for all $h\in H$, where $c\in (0,1)$ (see Theorem 3.3 in [24]).

In Theorem 41, we consider a $\left( m,\sigma \right)$ function $\varphi :U\subset E_{\infty }\longrightarrow E_{\infty }$, with some properties. In general, the two theorems cannot be compared, but, in some particular cases, it is possible to prove that the hypotheses of Theorem 4.1 in [24] imply those of Theorem 41, which can therefore be considered a generalization of the first. For example, consider the function $\varphi :E_{\infty }\longrightarrow E_{\infty }$ and the measure $N(0,\Gamma )$ defined by Example 42; as we have seen, this is a case where Theorem 41 can be applied; conversely, let $h=(h_{j}:j\in \textbf{N}^{*})$ given by $h_{j}=\delta _{j1}$, for any $j\in \textbf{N}^{*}$; since $h\in \left\{ x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }:\sum _{i=1}^{\infty } \frac{x_{i}^{2}}{\gamma _{ii}} \in \textbf{R}\right\} =\ell _{2}(\Gamma )$, and since Proposition 37 implies $\ell _{2}(\Gamma )\subset H\left( N(0,\Gamma )\right)$, we obtain $h\in H\left( N(0,\Gamma )\right)$; moreover

$$\begin{aligned} \left| h\right| _{H\left( N(0,\Gamma )\right) }= & {} \sup \left\{ \left| \underset{i=1}{\overset{\infty }{\sum }}c_{i}h_{i}\right| :c=\left( c_{i}:i\in \textbf{N}^{*}\right) \in \ell _{1},\quad \underset{i=1}{\overset{\infty }{\sum }}c_{i}^{2}\gamma _{ii}\le 1\right\} \\= & {} \sup \left\{ \left| c_{1}\right| :c=\left( c_{i}:i\in \textbf{N}^{*}\right) \in \ell _{1},\text { }\underset{i=1}{\overset{\infty }{\sum }}c_{i} ^{2}\gamma _{ii}\le 1\right\} =\frac{1}{\sqrt{\gamma _{11}}}; \end{aligned}$$

furthermore, we have $\varphi (x)=x+F(x)$, where $F_{1}(x)=x_{1}+\sum _{i=2}^{\infty } 2^{-i+2}x_{i}$, $F_{j}(x)=0$, for any $x=(x_{i}:i\in \textbf{N}^{*})\in E_{\infty }$, for any $j>1$, and so $F(x)\in \ell _{2}(\Gamma )\subset H\left( N(0,\Gamma )\right)$; then, since $F(x+h)-F(x)=F(h)=h$, for any $x\in E_{\infty }$, and since $\left| h\right| _{H\left( N(0,\Gamma )\right) }>0$, the hypothesis given by formula (64) is not satisfied, and so we cannot use Theorem 4.1 in [24].

5 Problems for further study

A natural application of this paper, in the probability theory, is the generalization of Definition 32, by defining a random variable on $\left( E_{\infty },\mathcal {B}_{\infty }\right)$ of the form $X=AZ+c$, where $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:E_{\infty }\longrightarrow E_{\infty }$ is a bijective and linear $\left( m,\sigma \right)$ standard function, $c\in \ell _{1}(\Gamma )\equiv \left\{ x=\left( x_{i}:i\in \textbf{N}^{*}\right) \in E_{\infty }:\sum _{i=m+1}^{\infty } \frac{\left| x_{i}\right| }{\gamma _{ii}}\in \textbf{R}\right\} \subset \ell _{2}(\Gamma )\cap \ell _{1}\subset H\left( N(0,\Gamma )\right)$, $Z:\left( \Omega ,\mathcal {F},P\right) \longrightarrow \left( E_{\infty },\mathcal {B}_{\infty }\right)$ is a random variable with law $N(0,\Gamma )$, where $\Gamma =\left( \gamma _{ij}\right) _{i,j\in \textbf{N}^{*}}$ is a symmetric and m-positive definite linear function, such that $\sum _{i=m+1}^{\infty } \gamma _{ii}\in \textbf{R}^{+}$ and $\sum _{i=m+1}^{\infty } \frac{\left| a_{ii}^{2}-1\right| }{\gamma _{ii}}\in \left[ 0,+\infty \right)$. Since A is continuous and $N(0,\Gamma )$ is a Gaussian measure, then X is a Gaussian random variable; moreover, the assumptions of Theorem 41 are satisfied, with $\varphi (x)=Ax+c$, for any $x\in E_{\infty }$, $a=b=0$ and $\Lambda =\Gamma$; consequently we can determine the density of X with respect to the probability measure $N(0,\Gamma )$. In particular, if $A=\textbf{I}_{\infty }$, this result provides an explicit expression of the Cameron–Martin formula (see Corollary 2.4.3 in [11]). Moreover, it is possible to develop some theoretical results (for example, the convolution between the laws of two independent Gaussian random variables with values in $E_{\infty }$) and some applications in the statistical inference.

Furthermore, in the statistical mechanics, it is possible to describe the systems of smooth hard particles, by using the Boltzmann equation (see, for example, the paper [22]), or the more general Master kinetic equation, described in the papers [19,20,21]. In order to study the evolution of these systems, we can consider the model of countable particles, such that their joint infinite-dimensional density can be determined by composing an infinite-dimensional Gaussian random variable with a $\left( m,\sigma \right)$ function.

Appendix

In this section, we expose some definitions and results about the differentiation theory over infinite-dimensional Banach spaces, and about the theory of the $\left( m,\sigma \right)$ functions, introduced in Sect. 3. We start with the following three results (from Theorem 43 to Proposition 45) of the measure theory.

Theorem 43

Let $I\ne \emptyset$ be a set and, for any $i\in I$, let $\left( S_{i},\Sigma _{i},\mu _{i}\right)$ be a measure space such that $\mu _{i}$ is finite. Moreover, suppose that, for some countable set $J\subset I$ and for any $i\in I\backslash J$, $\mu _{i}$ is a probability measure, and $\prod _{j\in J}\mu _{j}(S_{j})\in \mathbf {[}0\mathbf {,\infty )}$. Then, over the measurable space $\left( \prod _{i\in I} S_{i},\bigotimes _{i\in I}\Sigma _{i}\right)$, there is a unique finite measure $\mu$, indicated by $\bigotimes _{i\in I}\mu _{i}$, such that, for any $H\subset I$ such that $0<|H|<\infty$ and for any $A=\prod _{h\in H} A_{h}\times \prod _{i\in I\backslash H} S_{i}\in \bigotimes _{i\in I}\Sigma _{i}$, where $A_{h}\in \Sigma _{h}$ $\forall \,h\in H$, we have $\mu (A)=\prod _{h\in H}\mu _{h}(A_{h})\prod _{j\in J\backslash H}\mu _{j}(S_{j}).$ In particular, if I is countable, then $\mu (A)=\prod _{i\in I}\mu _{i}(A_{i})$ for any $A=\prod _{i\in I} A_{i}\in \bigotimes _{i\in I}\Sigma _{i}$.

Proof

See the proof of Corollary 4 in Asci [5]. $\square$

Theorem 44

For any $i\in \textbf{N}^{*}$, let $\left( S_{i},\Sigma _{i},\mu _{i}\right)$ be a measure space such that $\mu _{i}$ is finite, and $\prod _{i=1}^{\infty } \mu _{i}(S_{i})\in \mathbf {[} 0\mathbf {,\infty )}$; moreover, let $\left( S,\Sigma ,\mu \right) =\left( \prod _{i=1}^{\infty } S_{i},\bigotimes _{i=1}^{\infty } \Sigma _{i}, \bigotimes _{i=1}^{\infty }\mu _{i}\right)$, let $f\in L^{1}\left( S,\Sigma ,\mu \right)$ and, for any $H\subset \textbf{N}^{*}$ such that $0<|H|<\infty$, let the measurable function $f_{H^{c}}:\left( S,\Sigma \right) \longrightarrow (\textbf{R},\mathcal {B})$ defined by

$$\begin{aligned} f_{H^{c}}(x)=\underset{S_{H}}{\int }f(\cdot ,x_{H^{c}})d\mu _{H}\text {, } \forall \,x=\left( x_{H},x_{H^{c}}\right) \in S, \end{aligned}$$

where $\left( S_{H},\Sigma _{H},\mu _{H}\right)$ is the measure space $\left( \prod _{i\in H}S_{i},\bigotimes _{i\in H}\Sigma _{i},\bigotimes _{i\in H}\mu _{i}\right)$. Then:

• 1. If $\prod _{i=1}^{\infty }\mu _{i}(S_{i})>0$, there exists $D\in \Sigma$ such that $\mu (D)=0$ and such that, for any $x\in D^{c}$, one has $\lim _{n\rightarrow \infty }f_{\left\{ 1,\ldots ,n\right\} ^{c} }(x)=\int _{S} fd\mu$.

• 2. If $\prod _{i=1}^{\infty } \mu _{i}(S_{i})=0$ and f is bounded, for any $x\in S$, one has $\lim _{n\rightarrow \infty } f_{\left\{ 1,\ldots ,n\right\} ^{c}}(x)=\int _S fd\mu$.

Proof

• 1. Suppose that $\prod _{i=1}^{\infty } \mu _{i}(S_{i})>0$; this implies $\mu _{i}(S_{i})\ne 0$, $\forall \,i\in \textbf{N}^{*}$; thus, $\overline{\mu }_{i}=\frac{\mu _{i}}{\mu _{i}(S_{i})}$ is a probability measure, and also $\overline{\mu }=\bigotimes _{i=1}^{\infty }\overline{\mu }_{i}$, $\overline{\mu }_{\left\{ 1,\ldots ,n\right\} } =\bigotimes _{i=1}^{n}\overline{\mu }_{i}$, $\forall \,n\in \textbf{N}^{*}$. We have $\int _{S}\left| f\right| d\overline{\mu }=\frac{1}{\prod _{i=1}^{\infty }\mu _{i}(S_{i})}\int _{S}\left| f\right| d\mu <\infty$; then, define the function $\overline{f}_{\left\{ 1,\ldots ,n\right\} ^{c}}:\left( S,\Sigma \right) \longrightarrow (\textbf{R},\mathcal {B})$ by

  $$\begin{aligned} \overline{f}_{\left\{ 1,\ldots ,n\right\} ^{c}}(x)=\underset{S_{\left\{ 1,\ldots ,n\right\} }}{\int }f(\cdot ,x_{\left\{ 1,\ldots ,n\right\} ^{c}} )d\overline{\mu }_{\left\{ 1,\ldots ,n\right\} }; \end{aligned}$$

  from Theorem 3, page 349, in Rao [17], there exists $D\in \Sigma$ such that $\overline{\mu }(D)=0$ and such that, $\forall \,x\in D^{c}$, we have

  $$\begin{aligned} \underset{n\rightarrow \infty }{\lim }f_{\left\{ 1,\ldots ,n\right\} ^{c}}(x)= & {} \underset{n\rightarrow \infty }{\lim }\left( \underset{i=1}{\overset{n}{\prod }}\mu _{i}(S_{i})\right) \overline{f}_{\left\{ 1,\ldots ,n\right\} ^{c}}(x)\\= & {} \underset{i=1}{\overset{\infty }{\prod }}\mu _{i}(S_{i})\underset{S}{\int }fd\overline{\mu }=\underset{S}{\int }fd\mu ; \end{aligned}$$

  thus, since $\mu (D)= \prod _{i=1}^{\infty } \mu _{i} (S_{i})\overline{\mu }(D)=0$, we obtain the statement.

• 2. If $\prod _{i=1}^{\infty } \mu _{i}(S_{i})=0$ and f is bounded, let $M\in \textbf{R}^{+}$ such that $\left| f\right| \le M$, and let $x\in S$; we have

  $$\begin{aligned} 0\le & {} \underset{n\rightarrow \infty }{\lim \inf }\left| f_{\left\{ 1,\ldots ,n\right\} ^{c}}(x)\right| \le \underset{n\rightarrow \infty }{\lim \sup }\left| f_{\left\{ 1,\ldots ,n\right\} ^{c}}(x)\right| \\\le & {} \underset{n\rightarrow \infty }{\lim \sup }\underset{\underset{i=1}{\overset{n}{\prod }}S_{i}}{\int }Md\left( \underset{i=1}{\overset{n}{ {\displaystyle \bigotimes } }}\mu _{i}\right) =M\underset{i=1}{\overset{\infty }{\prod }}\mu _{i}(S_{i})=0, \end{aligned}$$

  and so

  $$\begin{aligned} \underset{n\rightarrow \infty }{\lim }f_{\left\{ 1,\ldots ,n\right\} ^{c} }(x)=0=\underset{S}{\int }fd\mu . \end{aligned}$$

$\square$

Proposition 45

Let $\left( S,\Sigma \right)$ be a measurable space, let $\mathcal {I}$ be a $\pi$-system on S, and let $\mu _{1}$ and $\mu _{2}$ be two measures on $\left( S,\Sigma \right)$, $\sigma$- finite on $\mathcal {I}$; if $\sigma (\mathcal {I})=\Sigma$ and $\mu _{1}$ and $\mu _{2}$ coincide on $\mathcal {I}$, then $\mu _{1}$ and $\mu _{2}$ coincide on $\Sigma$.

Proof

See, for example, Theorem 10.3 in Billingsley [10]. $\square$

The following definitions and results (from Definition 46 to Definition 53), can be found in [8], and generalize the differentiation theory in the finite-dimensional case (see, e.g., the Lang’s book [14]).

Definition 46

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\subset \mathcal {D}_{I_{n}}(U)$, let $x_{0}\in U$, let $l\in \ell _{I_{m} }$ and let $\varphi :U\longrightarrow \ell _{I_{m}}$ be a function; we say that $\lim _{x\rightarrow x_{0}}\varphi (x)=l$ if, for any $\varepsilon \in \textbf{R}^{+}$, there exists $\delta \in \textbf{R}^{+}$ such that, for any $x\in B_{I_{n}}\left( x_{0},\delta \right) \cap \left( U\backslash \{x_{0}\}\right)$, one has $\left\| \varphi (x)-l\right\| _{I_{m} }<\varepsilon$.

Definition 47

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\subset \mathcal {D}_{I_{n}}(U)$ and let $\varphi :U\longrightarrow \ell _{I_{m}}$ be a function; we say that $\varphi$ is continuous in $x_{0}\in U$ if $\lim _{x\rightarrow x_{0}}\varphi (x)=\varphi (x_{0})$, and we say that $\varphi$ is continuous in U if, for any $x\in U$, $\varphi$ is continuous in x.

Remark 48

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\subset \mathcal {D}_{I_{n}}(U)$, let $V\subset \ell _{I_{m}}$ and let $\varphi :U\longrightarrow V$ be a function; then, $\varphi :\left( U,\tau _{\left\| \cdot \right\| _{I_{n}}}(U)\right) \longrightarrow \left( V,\tau _{\left\| \cdot \right\| _{I_{m}}}(V)\right)$ is continuous if and only if $\varphi$ is continuous in U.

Definition 49

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\subset \ell _{I_{n}}$, let $\varphi :U\longrightarrow \ell _{I_{m}}$ be a function and let $\left\{ \varphi _{k}\right\} _{k\in \textbf{N}}$ be a sequence of functions such that $\varphi _{k}:U\longrightarrow \ell _{I_{m}}$, for any $k\in \textbf{N}$; we say that:

• 1. The sequence $\left\{ \varphi _{k}\right\} _{k\in \textbf{N}}$ converges to $\varphi$ over U if, for any $\varepsilon \in \textbf{R}^{+}$ and for any $x\in U$, there exists $k_{0}\in \textbf{N}$ such that, for any $k\in \textbf{N}$, $k\ge k_{0}$, one has $\left\| \varphi _{k}(x)-\varphi (x)\right\| _{I_{m}}<\varepsilon$.

• 2. The sequence $\left\{ \varphi _{k}\right\} _{k\in \textbf{N}}$ converges uniformly to $\varphi$ over U if, for any $\varepsilon \in \textbf{R}^{+}$, there exists $k_{0}\in \textbf{N}$ such that, for any $k\in \textbf{N}$, $k\ge k_{0}$, and for any $x\in U$, one has $\left\| \varphi _{k}(x)-\varphi (x)\right\| _{I_{m}}<\varepsilon$.

Definition 50

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\subset \mathcal {D}_{I_{n}}(U)$, let $v\in \ell _{I_{n}}$ such that $\left\| v\right\| _{I_{n}}=1$ and let a function $\varphi :U\longrightarrow \textbf{R}^{m}$; for any $i\in I_{m}$, the function $\varphi _{i}$ is called differentiable in $x_{0}\in U$ in the direction v if there exists the limit

$$\begin{aligned} \underset{t\rightarrow 0}{\lim }\frac{\varphi _{i}(x_{0}+tv)-\varphi _{i}(x_{0} )}{t}. \end{aligned}$$

This limit is indicated by $\frac{\partial \varphi _{i}}{\partial v}(x_{0})$, and it is called derivative of $\varphi _{i}$ in $x_{0}$ in the direction v. If, for some $j\in I_{n}$, one has $v=e_{j}$, where $(e_{j})_{k}=\delta _{jk}$, for any $k\in I_{n}$, then the function $\varphi _{i}$ is called differentiable in $x_{0}\in U$ with respect to $x_{j}$; furthermore, indicate $\frac{\partial \varphi _{i}}{\partial v}(x_{0})$ by $\frac{\partial \varphi _{i} }{\partial x_{j}}(x_{0})$, and call it partial derivative of $\varphi _{i}$ in $x_{0}$, with respect to $x_{j}$. Moreover, if there exists the linear function defined by the matrix $J_{\varphi }(x_{0})=\left( \left( J_{\varphi }(x_{0})\right) _{ij}\right) _{i\in I_{m},j\in I_{n}}:\ell _{I_{n} }\longrightarrow \textbf{R}^{m}$, where $\left( J_{\varphi }(x_{0})\right) _{ij}=\frac{\partial \varphi _{i}}{\partial x_{j}}(x_{0})$, for any $i\in I_{m}$, $j\in I_{n}$, then $J_{\varphi }(x_{0})$ is called Jacobian matrix of the function $\varphi$ in $x_{0}$. Finally, if, for any $x\in U$, there exists $J_{\varphi }(x)$, then the function $x\longrightarrow J_{\varphi }(x)$ is indicated by $J_{\varphi }$.

Definition 51

Let $n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\subset \mathcal {D}_{I_{n}}(U)$, let $i,j\in I_{n}$ and let $\varphi :U\longrightarrow \textbf{R}$ be a function differentiable in $x_{0}\in U$ with respect to $x_{i}$, such that the function $\frac{\partial \varphi }{\partial x_{i}}$ is differentiable in $x_{0}$ with respect to $x_{j}$. Indicate $\frac{\partial }{\partial x_{j}}\left( \frac{\partial \varphi }{\partial x_{i} }\right) (x_{0})$ by $\frac{\partial ^{2}\varphi }{\partial x_{j}\partial x_{i}}(x_{0})$ and call it second partial derivative of $\varphi$ in $x_{0}$ with respect to $x_{i}$ and $x_{j}$. If $i=j$, it is indicated by $\frac{\partial ^{2}\varphi }{\partial x_{i}^{2}}(x_{0})$. Analogously, for any $k\in \textbf{N}^{*}$ and for any $j_{1},\ldots ,j_{k}\in I_{n}$, define $\frac{\partial ^{k}\varphi }{\partial x_{j_{k}}\ldots \partial x_{j_{1}}}(x_{0})$ and call it k-th partial derivative of $\varphi$ in $x_{0}$ with respect to $x_{j_{1}},\ldots x_{j_{k}}$.

Definition 52

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\subset \mathcal {D}_{I_{n}}(U)$ and let $k\in \textbf{N}^{*}$; a function $\varphi :U\longrightarrow \textbf{R}^{m}$ is called $C^{k}$ in $x_{0}\in U$ if, in a neighbourhood $V\in \tau _{\left\| \cdot \right\| _{I_{n}}}(U)$ of $x_{0}$, for any $i\in I_{m}$ and for any $j_{1},\ldots ,j_{k}\in I_{n}$, there exists the function defined by $x\longrightarrow \frac{\partial ^{k}\varphi _{i}}{\partial x_{j_{k}}\ldots \partial x_{j_{1}}}(x)$, and this function is continuous in $x_{0}$; $\varphi$ is called $C^{k}$ in U if, for any $x_{0}\in U$, $\varphi$ is $C^{k}$ in $x_{0}$.

Definition 53

Let $m,n\in \textbf{N}^{*}\cup \left\{ \infty \right\}$, let $U\in \tau _{\left\| \cdot \right\| _{I_{n}}}$ and let $V\in \tau _{\left\| \cdot \right\| _{I_{m}}}$; a function $\varphi :U\longrightarrow V$ is called diffeomorphism if $\varphi$ is bijective and $C^{1}$ in U, and the function $\varphi ^{-1}:V\longrightarrow U$ is $C^{1}$ in V.

Henceforth, we will consider the $\left( m,\sigma \right)$ functions.

Remark 54

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function; then, for any $\left( x_{j}:j\in \sigma (\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\} )\right) \in \ell _{\sigma (\textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\} )}$, one has $\sup _{i>m}\left| \varphi _{i,\sigma (i)}\left( x_{\sigma (i)}\right) \right| <\infty$.

Proof

From Definition 17, we have $\varphi (U)\subset \ell _{\infty }$ and $\varphi ^{(\infty ,m)}(U^{(m)})\subset \ell _{\infty }$; then, $\forall \,x=\left( x_{j}:j\in \textbf{N}^{*}\right) \in U$, we have

$$\begin{aligned} \underset{i>m}{\sup }\left| \varphi _{i,\sigma (i)}\left( x_{\sigma (i)}\right) \right|= & {} \underset{i>m}{\sup }\left| \varphi _{i}(x)-\varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) \right| \\\le & {} \underset{i>m}{\sup }\left| \varphi _{i}(x)\right| +\underset{i>m}{\sup }\left| \varphi _{i}^{(\infty ,m)}\left( x_{1},\ldots ,x_{m}\right) \right| <\infty . \end{aligned}$$

$\square$

The following results (from Remark 55 to Proposition 61) can be found in [8].

Remark 55

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function; then, $\sigma$ is bijective if and only if, for any $i>m$, one has $\sigma (i)=i$.

Proposition 56

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function and let $\left\{ 1,\ldots ,m\right\} \subset L\subset \textbf{N}^{*}$, $\left\{ 1,\ldots ,m\right\} \subset N\subset \textbf{N}^{*}$; then, one has $\varphi ^{(L,N)}(\pi _{N}(U))\subset \ell _{L}$, $\overline{\varphi } ^{(L,N)}(U)\subset \ell _{\infty }$, and the function $\overline{\varphi } ^{(L,N)}:U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ is $\left( m,\sigma \right)$. Moreover, if $\sigma$ is bijective, for any $n\in \textbf{N}^{*}$, $n\ge m$, $\overline{\varphi }^{(L,N)}$ is $\left( n,\tau \right)$, where the function $\tau :\textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} \longrightarrow \textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\}$ is defined by $\tau (i)=\sigma (i)=i$, for any $i>n$.

Proposition 57

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function, and let $\emptyset \ne L\subset \textbf{N}^{*}$ and $\emptyset \ne N\subset \textbf{N}^{*}$ such that $\left\{ 1,\ldots ,m\right\} \subset N$ or $N\subset \textbf{N}^{*}\backslash \left\{ 1,\ldots ,m\right\}$, and such that, for any $i\in L$ and for any $j\in N\backslash \left\{ 1,\ldots ,m\right\}$, the functions $\varphi _{i}^{(\infty ,m)}:\left( U^{(m)},\mathcal {B}^{m}\left( U^{(m)}\right) \right) \longrightarrow \left( \textbf{R},\mathcal {B}\right)$ and $\varphi _{ij}:(\textbf{R},\mathcal {B})\longrightarrow \left( \textbf{R},\mathcal {B}\right)$ are measurable; then:

• 1. The function

  $$\begin{aligned} \varphi ^{(L,N)}:(\pi _{N}(U),\mathcal {B}^{N}(\pi _{N}(U)))\longrightarrow \left( \ell _{L},\mathcal {B}^{L}\left( \ell _{L}\right) \right) \end{aligned}$$

  is measurable; in particular, suppose that, for any $i\in \textbf{N}^{*}$ and for any $j>m$, $\varphi _{i}^{(\infty ,m)}$ and $\varphi _{ij}$ are measurable functions; then, $\varphi :(U,\mathcal {B}^{\infty } (U))\longrightarrow \left( \ell _{\infty },\mathcal {B}^{\infty }(\ell _{\infty })\right)$ is measurable.

• 2. If $\left\{ 1,\ldots ,m\right\} \subset L$ and $\left\{ 1,\ldots ,m\right\} \subset N$, then the function $\overline{\varphi }^{(L,N)}:(U,\mathcal {B} ^{\infty }(U))\longrightarrow \left( \ell _{\infty },\mathcal {B}^{\infty } (\ell _{\infty })\right)$ is measurable.

Corollary 58

Let $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:\ell _{\infty }\longrightarrow \ell _{\infty }$ be a linear function, and let $\emptyset \ne L\subset \textbf{N}^{*}$ and $\emptyset \ne N\subset \textbf{N}^{*}$; then:

• 1. The function $A^{(L,N)}:(\ell _{N},\mathcal {B}^{N}(\ell _{N} ))\longrightarrow \left( \ell _{L},\mathcal {B}^{L}\left( \ell _{L}\right) \right)$ is measurable; in particular, $A:(\ell _{\infty },\mathcal {B} ^{\infty }(\ell _{\infty }))\longrightarrow \left( \ell _{\infty },\mathcal {B} ^{\infty }(\ell _{\infty })\right)$ is measurable.

• 2. If A is $\left( m,\sigma \right)$, $\left\{ 1,\ldots ,m\right\} \subset L$ and $\left\{ 1,\ldots ,m\right\} \subset N$, then the function $\overline{A}^{(L,N)}:(\ell _{\infty },\mathcal {B}^{\infty }(\ell _{\infty }))\longrightarrow \left( \ell _{\infty },\mathcal {B}^{\infty }(\ell _{\infty })\right)$ is measurable.

Proposition 59

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function such that $\sigma$ is bijective and $\overline{\varphi }:\left( U,\tau _{\left\| \cdot \right\| _{\infty }}(U))\longrightarrow (\ell _{\infty },\tau _{\left\| \cdot \right\| _{\infty }}\right)$ is continuous; then, for any $n\in \textbf{N}^{*}$, $n\ge m$, $\varphi ^{(n,n)}:(U^{(m)}\times \textbf{R}^{n-m},\tau ^{n} (U^{(m)}\times \textbf{R}^{n-m}))\longrightarrow (\textbf{R}^{n},\tau ^{n})$ is continuous if and only if $\overline{\varphi }^{(n,n)}:(U,\tau _{\left\| \cdot \right\| _{\infty }}(U))\longrightarrow (\ell _{\infty },\tau _{\left\| \cdot \right\| _{\infty }})$ is continuous.

Proposition 60

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function; then, $\overline{\varphi }:U\longrightarrow \ell _{\infty }$ is bijective if and only if the functions $\varphi ^{(m,m)}:$ $U^{(m)}\longrightarrow \textbf{R}^{m}$, $\varphi _{i,\sigma (i)}:\textbf{R}\longrightarrow$ $\textbf{R}$, for any $i>m$, and $\sigma$ are bijective.

Proposition 61

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function. Then, $\overline{\varphi }:U\longrightarrow \ell _{\infty }$ is a diffeomorphism if and only if the functions $\varphi ^{(m,m)}:$ $U^{(m)}\longrightarrow \textbf{R}^{m}$ and $\varphi _{i,\sigma (i)}:\textbf{R}\longrightarrow$ $\textbf{R}$, for any $i>m$, are diffeomorphisms, and $\sigma$ is bijective.

The following two results (Propositions 62 and 63) can be found in [6].

Proposition 62

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ standard function such that the functions $\varphi ^{(m,m)}:$ $U^{(m)}\longrightarrow \textbf{R}^{m}$, $\varphi _{i,\sigma (i)}:\textbf{R}\longrightarrow$ $\textbf{R}$, for any $i>m$, and $\sigma$ are bijective; then, $\varphi$ is bijective.

Proposition 63

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ standard function such that $\varphi _{i}:U\longrightarrow \textbf{R}$ is $C^{1}$, for any $i\in \left\{ 1,\ldots ,m\right\}$, and such that $\overline{\varphi }:U\longrightarrow \overline{\varphi }(U)$ is a diffeomorphism; then, for any $n\in \textbf{N} ^{*}$, $n\ge m$, the functions $\varphi ^{(n,n)}:U^{(m)}\times \textbf{R}^{n-m}\longrightarrow \varphi ^{(n,n)}\left( U^{(m)}\times \textbf{R}^{n-m}\right)$ and $\overline{\varphi }^{(n,n)}:U\longrightarrow \overline{\varphi }^{(n,n)}\left( U\right)$ are diffeomorphisms.

Analogously, it is easy to prove the following results (from Propositions 64–67).

Proposition 64

Let $\varphi :U\subset \ell _{\infty }\longrightarrow E_{\infty }$ be a $\left( m,\sigma \right)$ function and let $\left\{ 1,\ldots ,m\right\} \subset L\subset \textbf{N}^{*}$, $\left\{ 1,\ldots ,m\right\} \subset N\subset \textbf{N}^{*}$; then, one has $\overline{\varphi }^{(L,N)}(U)\subset E_{\infty }$, and the function $\overline{\varphi }^{(L,N)}:U\subset \ell _{\infty }\longrightarrow E_{\infty }$ is $\left( m,\sigma \right)$. Moreover, if $\sigma$ is bijective, for any $n\in \textbf{N}^{*}$, $n\ge m$, $\overline{\varphi }^{(L,N)}$ is $\left( n,\tau \right)$, where the function $\tau :\textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\} \longrightarrow \textbf{N}^{*}\backslash \left\{ 1,\ldots ,n\right\}$ is defined by $\tau (i)=\sigma (i)=i$, for any $i>n$.

Proposition 65

Let $\varphi :U\subset \ell _{\infty }\longrightarrow E_{\infty }$ be a $\left( m,\sigma \right)$ function; then, $\overline{\varphi }:U\longrightarrow E_{\infty }$ is bijective if and only if the functions $\varphi ^{(m,m)}:$ $U^{(m)}\longrightarrow \textbf{R}^{m}$, $\varphi _{i,\sigma (i)}:\textbf{R}\longrightarrow$ $\textbf{R}$, for any $i>m$, and $\sigma$ are bijective.

Proposition 66

Let $\varphi :U\subset \ell _{\infty }\longrightarrow E_{\infty }$ be a $\left( m,\sigma \right)$ function. Then, $\overline{\varphi }:U\longrightarrow E_{\infty }$ is a diffeomorphism if and only if the functions $\varphi ^{(m,m)}:$ $U^{(m)}\longrightarrow \textbf{R}^{m}$ and $\varphi _{i,\sigma (i)}:\textbf{R}\longrightarrow$ $\textbf{R}$, for any $i>m$, are diffeomorphisms, and $\sigma$ is bijective.

Proposition 67

Let $\varphi :U\subset \ell _{\infty }\longrightarrow E_{\infty }$ be a $\left( m,\sigma \right)$ standard function such that the functions $\varphi ^{(m,m)}:$ $U^{(m)}\longrightarrow \textbf{R}^{m}$, $\varphi _{i,\sigma (i)}:\textbf{R}\longrightarrow$ $\textbf{R}$, for any $i>m$, and $\sigma$ are bijective; then, $\varphi$ is bijective.

The following two results (Propositions 68 and 69) can be found in [7].

Proposition 68

Let $A=\left( a_{ij}\right) _{i,j\in \textbf{N}^{*}}:\ell _{\infty }\longrightarrow \ell _{\infty }$ be a linear $\left( m,\sigma \right)$ function, such that A is $\left( m,\sigma \right)$ standard or $\left( m,\sigma \right)$ of the second type. Then, if $\sigma$ is bijective, we have

$$\begin{aligned} \det A=\det A^{(m,m)}\underset{i>m}{ {\displaystyle \prod } }a_{ii}. \end{aligned}$$

Conversely, if $\sigma$ is not bijective, we have $\det A=0$. In particular, if $A=\textbf{I}_{n}$, we have $\det A=1$.

Proposition 69

Let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function and let $x_{0} =(x_{0,j}:j\in \textbf{N}^{*})\in U$ such that there exists the linear function $J_{\varphi }\left( x_{0}\right) :\ell _{\infty }\longrightarrow \ell _{\infty }$; then, $J_{\varphi }\left( x_{0}\right)$ is $\left( m,\sigma \right)$; moreover, for any $n\in \textbf{N}^{*}$, $n\ge m$, there exists the linear $\left( m,\sigma \right)$ function $J_{\overline{\varphi }^{(n,n)}}\left( x_{0}\right) :\ell _{\infty }\longrightarrow \ell _{\infty }$, and one has

$$\begin{aligned} \det J_{\varphi }\left( x_{0}\right) =\underset{n\rightarrow \infty }{\lim }\det J_{\overline{\varphi }^{(n,n)}}\left( x_{0}\right) . \end{aligned}$$

The following two results (Propositions 70 and 71) can be found in [8].

Proposition 70

Let $m\in \textbf{N}$, let $n\in \textbf{N}^{*}$, $n\ge m$, and let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function such that, for any $i\in \left\{ 1,\ldots ,n\right\}$, for any $j_{1}\in \left\{ 1,\ldots ,m\right\}$ and for any $j_{2}\in \left\{ m+1,\ldots ,n\right\}$, there exist the functions $\frac{\partial \varphi _{i}^{(\infty ,m)}}{\partial x_{j_{1}}}:\left( U^{(m)},\mathcal {B}^{m}\left( U^{(m)}\right) \right) \longrightarrow \left( \textbf{R},\mathcal {B}\right)$ and $\frac{\partial \varphi _{ij_{2}}}{\partial x_{j_{2}}}:(\textbf{R},\mathcal {B})\longrightarrow \left( \textbf{R},\mathcal {B}\right)$, and they are measurable; then:

• 1. The function $\det J_{\varphi ^{(n,n)}}:\left( U^{(m)}\times \textbf{R}^{n-m},\mathcal {B}^{n}(U^{(m)}\times \textbf{R}^{n-m})\right) \longrightarrow \left( \textbf{R},\mathcal {B}\right)$ is measurable.

• 2. Suppose that, for any $i>m$, the function $\varphi _{i,\sigma (i)} ^{\prime }:(\textbf{R},\mathcal {B})\longrightarrow \left( \textbf{R},\mathcal {B}\right)$ is measurable; then, for any $x\in U$, there exists the function $J_{\overline{\varphi }^{(n,n)}}(x):\ell _{\infty }\longrightarrow \ell _{\infty }$, and it is $\left( m,\sigma \right)$; moreover, the function $\det J_{\overline{\varphi }^{(n,n)}}:\left( U,\mathcal {B}^{\infty }(U)\right) \longrightarrow \left( \textbf{R},\mathcal {B}\right)$ is measurable.

• 3. Suppose that, for any $x\in U$, there exists the function $J_{\varphi }\left( x\right) :\ell _{\infty }\longrightarrow \ell _{\infty }$; moreover, suppose that, for any $i\in \textbf{N}^{*}$, for any $j_{1}\in \left\{ 1,\ldots ,m\right\}$ and for any $j_{2}>m$, the functions

  $$\begin{aligned} \frac{\partial \varphi _{i}^{(\infty ,m)}}{\partial x_{j_{1}}}:\left( U^{(m)},\mathcal {B}^{m}\left( U^{(m)}\right) \right) \longrightarrow \left( \textbf{R},\mathcal {B}\right) \end{aligned}$$

  and $\frac{\partial \varphi _{ij_{2}}}{\partial x_{j_{2}}}:(\textbf{R},\mathcal {B})\longrightarrow \left( \textbf{R},\mathcal {B}\right)$ are measurable; then the function $\det J_{\varphi }:\left( U,\mathcal {B}^{\infty }(U)\right) \longrightarrow \left( \textbf{R},\mathcal {B}\right)$ is measurable.

Proposition 71

Let $m\in \textbf{N}$, let $n\in \textbf{N}^{*}$, $n\ge m$, and let $\varphi :U\subset \ell _{\infty }\longrightarrow \ell _{\infty }$ be a $\left( m,\sigma \right)$ function such that $\varphi ^{(n,n)}$ is $C^{1}$; then, the function $\det J_{\varphi ^{(n,n)}}:\left( U^{(m)}\times \textbf{R}^{n-m},\tau ^{n}(U^{(m)}\times \textbf{R}^{n-m})\right) \longrightarrow \left( \textbf{R},\tau \right)$ is continuous.