# Fourier-Stieltjes transform defined by induced representation on locally
compact groups

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###### Abstract.

In this work we extend the Fourier-Stieltjes transform of a vector measure and
a continuous function defined on compact groups to locally compact groups. To
do so, we consider a representation $L$ of a normal compact subgroup $K$ of a
locally compact group $G$, and we use a representation of $G$ induced by that
of $L$. Then, we define the Fourier-Stieltjes transform of a vector measure
and that of a continuous function with compact support defined on $G$ from the
representation of $G$. Then, we extend the Shur orthogonality relation
established for compact groups to locally compact groups by using the
representations of $G$ induced by the unitary representations of one of its
normal compact subgroups. This extension enables us to develop a Fourier-
Stieltjes transform in series form that is linear, continuous, and invertible.

###### Key words and phrases:

Dimension theory, Poincaré recurrences, multifractal analysis.

###### 1991 Mathematics Subject Classification:

Primary: 58F15, 58F17; Secondary: 53C35.

Yao Ihébami Akakpo

1300 rue Ouimet App7, H4L 2L3 Montréal, Qc, Ca

ICMPA-UNESCO Chair (UAC), 072 BP 50, Cotonou, Benin

Tel +15142294451

V.S. Koffi Assiamoua

University of Lomé, Faculté Des Sience,,

Department of Mathematics (BP 1515 Lomé Togo),

Tel +228 92 77 28 26

Koffi Enakoutsa

California State University, Northridge, Department of Mathematics,

18111 Nordhoff Street, Northridge, CA, USA

Mahouton Norbert Hounkonnou

University of Abomey-Calavi,

International Chair in Mathematical Physics and Applications (ICMPA-UNESCO
Chair),

072 BP 50, Cotonou, Rep. of Benin

Tel +229 95 06 26 89

##  1\. Introduction

The vector measures generalizing scalar measures attracted a great interest in
recent decades due to their numerous applications in functional analysis,
control systems, signal analysis, quantum information, quantum theories, and
many other domains of applications. For more details on vector measure theory,
see for instance, [1, 2] and [3] for some applications on compact groups.
Also, Clarkson [4] used theoretical ideas on vector measure to prove that many
Banach spaces do not admit equivalent uniformly convex norms. In the same
vein, Gel’fand [5] proved that $L_{1}[0,1]$ is not isomorphic to a dual of a
Banach space. Lyapunov [6] showed that the range of a (non-atomic) vector
measure is closed and convex. Lyapunov [6]’ s work ccupies a prominent place
in modern mathematics since it lies at the intersection of the theory of
convex sets and measure theory. The Lyapunov convexity theorem became the
starting point of numerous studies in the framework of mathematical analysis
as well as in the realm of geometric research into the convex sets that are
ranges of non-atomic vector measures [11]. In addition, Bartle [7],
Dinculeanu, Kluvánek [1], Dunford and Schwartz [8], and Lindenstrauss and
Pelczyńki [9] gave many seminal results on vector measure. For instance,
Diestel and Uhl Jr [10] provided a comprehensive survey on vector measures.
Applications of vector measures were discussed in 1980 in the work by Kluvánek
[12]. Fernández and Faranjo [13] studied the Rybakov’s theorem for vector
measures in Fréchet spaces. Curbera and Ricker [14] wrote a survey on vector
measures, integration, and applications. More information on vector measures
can also be found in [10, 14].

To focus on our interest, let $G$ be a locally compact group, $m$ a vector
measure on $G$ into a Banach algebra $\mathcal{A}$, $\lambda$ a left or right
Haar measure on $G,$ and $f\in L_{1}(G,\lambda)$. If the group $G$ is abelian,
the Fourier-Stieltjes transform of $m$ is given by the relation

$\displaystyle\hat{m}(\chi)$ $\displaystyle=$
$\displaystyle\int_{G}\overline{\left<\chi,t\right>}dm(t),$ (1)

while the Fourier transform of $f$ is given by

$\displaystyle\hat{f}(\chi)=\int_{G}\overline{\left<\chi,t\right>}f(t)d\lambda(t),$
(2)

where $\chi$ denotes a character of $G$. If $G$ is compact and
$\mathcal{A}=\mathbb{C}$, then the Fourier-Stieltjes transform of $m$ is a
family of endomorphisms $(\hat{m}(\sigma))_{\sigma\in\Sigma}$ given by

$\displaystyle\left<\hat{m}(\sigma)\xi,\eta\right>=\int_{G}\left<\overline{U}_{t}^{\sigma}\xi,\eta\right>dm(t),$
(3)

and the associated Fourier transform is provided by the relation

$\displaystyle\left<\hat{f}(\sigma)\xi,\eta\right>=\int_{G}\left<\overline{U}_{t}^{\sigma}\xi,\eta\right>f(t)d\lambda(t),$
(4)

where $U^{\sigma}$ denotes a unitary representation of the group $G$. If $G$
is compact and $\mathcal{A}$ is any Banach algebra, Assiamoua [3] defined a
Fourier-Stieltjes transform of a bounded vector measure $m$ on $G$ as a family
$(\hat{m}(\sigma))_{\sigma\in\Sigma}$ of sesquilinear mappings of
$H_{\sigma}\times H_{\sigma}$ with values in $\mathcal{A}$ given by the
relation,

$\displaystyle\hat{m}(\sigma)(\xi,\eta)=\int_{G}\left<\overline{U}_{t}^{\sigma}\xi,\eta\right>dm(t),$
(5)

and the Fourier transform of a function $f\in L_{1}(G)$ as a family of
continuous endomorphisms $(\hat{f}(\sigma))_{\sigma\in\Sigma}$ of sesquilinear
applications of $H_{\sigma}\times H_{\sigma}$ with value in $\mathcal{A}$,
given by the relation

$\displaystyle\hat{f}(\sigma)(\xi,\eta)=\int_{G}\left<\overline{U}_{t}^{\sigma}\xi,\eta\right>f(t)d\lambda(t).$
(6)

In the continuation of previous investigations by [3], the present work
addresses a construction of the Fourier-Stieltjes transform on locally compact
groups from a group representation induced by a representation of a compact
subgroup. For this purpose, we consider a locally compact group $G$, $K$ a
compact normal subgroup of $G$, $\mu$ a $G$-invariant measure on the left
coset space $G/K$, and $L^{\sigma}$ a unitary representation of $K$ into a
separable Hilbert space $H_{\sigma}$. Then, we define the Fourier-Stieltjes
transform of a vector measure on the locally compact group $G$ using the
representation $U^{L^{\sigma}}$ of $G$ induced by $L^{\sigma}$. In this
context usually one uses Gel’fand transform to define Fourier transform. We
consider a locally compact group $G$ with Haar measure $dx$. Let $A$ be a
unitary commutative Banach algebra and $X(A)$ its spectrum. The Gel’fand
transform of $x$, $x\in A$, is the function
$\mathcal{G}_{x}:X(A)\longrightarrow\mathbb{C}$ such that
$\mathcal{G}_{x}(\chi)=\chi(x)$. The mapping
$x\longmapsto\mathcal{G}_{x}:A\longrightarrow\mathbb{C}^{X(A)}$ is called the
transformation of Gel’fand associated with $A$. The spherical Fourier
transform is the Gel’fand transform associated with $L_{1}(G)^{\natural}$, the
space of integrable, bi-invariant functions by a compact subgroup $K$ of $G$
on $G$. In this case, for $f\in L_{1}(G)^{\natural}$, $\mathcal{G}_{x}$ is
denoted $\mathcal{F}f$ or $\hat{f}$ and is defined by

$\displaystyle\hat{f}(\chi)$ $\displaystyle=$
$\displaystyle\int_{G}f(x)\chi(x^{-1})dx.$ (7)

Our method has several advantages over the Gel’fand transform. First, our
Fourier transform is injective while the Gel’fand transform is not necessarly
injective. Second the Gel’fand transform is limited to spherical functions
only. In our case the Fourier transform exists for the whole $L_{1}(G,A)$,
$1\leq p<\infty$. Finally, our method is constructive while in Gel’fand
transformation the Fourier transformation is obtained by induction.

The paper is organized as follows. In Section 1, we recall the definition of a
vector measure and unit representation of a group which are useful in the
sequel. In Section 2, we provide the proof of the Shur’s orthogonality
property in connection with induced representation and we define the Fourier-
Stieltjes transform of a vector measure and the Fourier transform of a
function in $L_{1}(G,\mathcal{A})$.

We also report in Section 2 several properties we discovered for our newly
developed Fourier-Stieltjes transform.

##  2\. Preliminaries

In this section, for the clarity of the development, we briefly recall useful
known main definitions and results, and set our notations. We consider a
locally compact space $G,$ the Banach spaces $\mathcal{A}$ and $\mathcal{F}$
over the field $\mathbb{K},\,(\mathbb{K}=\mathbb{R}$ or $\mathbb{C}),$ and
denote by $\mathcal{K}(G,\mathcal{A})$ the vector space of all continuous
functions $f:G\longrightarrow\mathcal{A}$ with a compact support, and by
$\mathfrak{C}(G,\mathcal{A})$ the space of continuous functions
$f:G\longrightarrow\mathcal{A}$.

For simplification, we write $\mathcal{K}(G)$ instead of
$\mathcal{K}(G,\mathbb{R})$ or $\mathcal{K}(G,\mathbb{C}).$ For each subset
$K$ of $G$, denote by $\mathcal{K}_{K}(G,\mathcal{A})$ the space of functions
with support contained in $K$. $\mathcal{K}_{K}(G,\mathcal{A})$ is a subspace
of $\mathcal{K}(G,\mathcal{A})$.

######  Definition 2.1.

For every function $f\in\mathcal{K}(G,\mathcal{A})$, we define

$\|f\|:=\sup_{t\in G}\|f(t)\|_{\mathcal{A}}.$

The mapping $f\mapsto\|f\|$ is a norm on each space
$\mathcal{K}_{K}(G,\mathcal{A}),$ and defines the topology of uniform
convergence on $G$ over $\mathcal{K}(G)$.

######  Definition 2.2.

On $\mathcal{K}(G,\mathcal{A}),$ the topology of the compact convergence is
the locally convex topology defined by the family of seminorms

$\left\|f\right\|_{K}=\sup_{t\in K}\|f(t)\|_{\mathcal{A}},$

where $K$ takes the elements in the set of compact subsets of $G$.

######  Proposition 1.

The space $\mathcal{K}(G,\mathcal{A})$ is dense in the space
$\mathfrak{C}(G,\mathcal{A})$ for the topology of the compact convergence [1].

######  Definition 2.3.

A vector measure on $G$ with respect to two spaces $\mathcal{A}$ and
$\mathcal{F}$, or an $(\mathcal{A},\mathcal{F})$-measure on $G$, is any linear
mapping $m:\mathcal{K}(G,\mathcal{A})\longrightarrow\mathcal{F}$ having the
property that, for each compact set $K\subset G,$ the restriction $m$ to the
subspace $\mathcal{K}_{K}(G,\mathcal{A})$ is continuous for the topology of
uniform convergence, i.e. for each compact set, there exists a number
$a_{K}>0$ such that

$\left\|m(f)\right\|\leq a_{K}\sup\left\\{\|f(t)\|_{\mathcal{A}},\quad t\in
K\right\\}.$

The value $m(f)$ of $m$ for a function $f\in\mathcal{K}(G,\mathcal{A})$ is
called the integral of $f$ with respect to $m$ also denoted by $\int_{G}fdm$
or $\int_{G}f(t)dm(t)$. A vector measure is said to be dominated if there
exists a positive measure $\mu$ such that

$\left\|\int_{G}f(t)dm(t)\right\|\leq\int_{G}|f(t)|d\mu(t),\quad\quad
f\in\mathcal{K}(G).$

If $m$ is dominated, then there exists a smallest positive measure $|m|,$
called the modulus or the variation of $m$, that dominates it. A positive
measure is said to be bounded if it is continuous in the uniform norm topology
of $\mathcal{K}(G).$ A vector measure is said to be bounded if it is dominated
by a bounded positive measure. If $m$ is bounded, then $|m|$ is also bounded.

Denoting by $M_{1}(G,\mathcal{A})$ the Banach algebra of bounded vector
measures on $G,$ the mapping

$\displaystyle m\mapsto\|m\|=\int_{G}\chi_{G}d|m|$ (8)

is a norm on $M_{1}(G,\mathcal{A})$, where $\chi_{G}$ represents the
characteristic function of $G$.

In the sequel, $K$ will denote a compact subgroup of $G$, $\nu$ and $\lambda$
left Haar measures on $K$ and $G,$ respectively.

######  Definition 2.4.

Let $\mu$ be a Radon measure on $G/K,$ the homogenous space of left $K$-cosets
and $g$ an element of $G$. Define $\mu_{g}$ by $\mu_{g}(E)=\mu(gE)$ for Borel
subsets $E$ of $G/K$. The measure $\mu$ is called $G-$invariant measure if
$\mu_{g}=\mu$, for $g\in G$. See [15, 16] for more details.

Throughout the paper, $\mu$ will denote the $G-$ invariant measure on $G/K$.

######  Theorem 2.5.

For any $f\in\mathcal{K}(G),$ we have [15, 17]:

$\displaystyle\int_{G}f(g)d\lambda(g)$ $\displaystyle=$
$\displaystyle\int_{X}d\mu(\dot{g})\int_{K}f(gk)d\nu(k).$ (9)

The previous formula ( 9) extends also to every $f\in L_{1}(G,\lambda,A)$
[18].

######  Definition 2.6.

A unit representation of $G$ is a homomorphism $L$ from $G$ into the group
$U(H)$ of the invertible unitary linear operators on some nonzero Hilbert
space $H,$ which is continuous with respect to the strong operator topology
satisfying for $g_{1},g_{2}\in G$,

$L_{(g_{1}g_{2})}=L_{g_{1}}L_{g_{2}}\quad\text{and }\quad L_{1}=Id_{H}.$

$H$ is called the representation space of $L$, and its dimension is called the
dimension or degree of $L$.

Suppose $\mathcal{M}$ is a closed subspace of $H$. $\mathcal{M}$ is called an
invariant subspace for $L$ if $L_{g}\mathcal{M}\subset\mathcal{M}$ $\forall
g\in G$. If $\mathcal{M}$ is invariant and $\mathcal{M}\neq\\{0\\},$ then
$L^{\mathcal{M}}$ such that

$L^{\mathcal{M}}_{g}=L_{g}|_{\mathcal{M}}$

defines a representation of $G$ on $\mathcal{M}$, called a subrepresentation
of $L$. If $L$ admits an invariant subspace that is nontrivial (i.e.
$\neq\\{0\\}$ or $H$) then $L$ is called reducible, otherwise $L$ is
irreducible. If $G$ is compact and $L$ irreducible then the dimension of $L$
is finite.

######  Definition 2.7.

Two unit irreductible representations $L$ and $V$ into $H$ and $N,$
respectively, are said to be equivalent if there is an isomorphism
$T:H\longrightarrow N$ such that, $\forall t\in G,$

$T\circ L_{t}=V_{t}\circ T.$

Consider now the subgroup $K$, $\Sigma$ the coset space (called the dual
object of $K$), $\sigma\in\Sigma,$ $L^{\sigma}$ a representative of $\sigma$,
$H_{\sigma}$ a representation space of $L^{\sigma},$ and $d_{\sigma}$ its
dimension.

######  Theorem 2.8.

Let $(L_{ij}^{\sigma})_{1\leq i,j\leq d_{\sigma}}$ be the matrix of
$L^{\sigma}$ in an orthonormal basis $(\xi_{i})_{i=1}^{d\sigma}$ of
$H_{\sigma}.$ Then, (see [15, 17, 20, 19]),

$\displaystyle\int_{K}L_{ij}^{\sigma}(t)\overline{L_{lm}^{\sigma}}(t)d\nu(t)$
$\displaystyle=$ $\displaystyle\dfrac{\delta_{il}\delta_{jm}}{d_{\sigma}}$
(10)

and

$\displaystyle\int_{K}L_{ij}^{\sigma}(t)\overline{L_{lm}^{\tau}}(t)d\nu(t)$
$\displaystyle=$ $\displaystyle 0\text{ if }\sigma\neq\tau.$ (11)

Let $q:G\longrightarrow G/K$ be the canonical quotient map of $G$ into $G/K$
and suppose $H_{\sigma}$ separable. Denote by $H^{L^{\sigma}}_{0}$ the set

$\displaystyle H^{L^{\sigma}}_{0}=\left\\{u\in\mathfrak{C}(G,H_{\sigma})\mbox{
}:\mbox{ }q(\mbox{Supp(u))}\text{ is compact }\text{ and
}u(gk)=L^{\sigma}_{k^{-1}}u(g)\right\\}.$ (12)

######  Proposition 2.

If $\eta:G\longrightarrow H_{\sigma}$ is continuous with compact support, then
the function $u_{\eta}$ such that

$\displaystyle u_{\eta}(g)=\int_{K}L^{\sigma}_{k}\eta(gk)d\nu(k)$ (13)

belongs to $H^{L^{\sigma}}_{0},$ and is uniformly continuous on $G$. Moreover,
every element of $H^{L^{\sigma}}_{0}$ is of the form $u_{\eta}$. See [15] for
more details.

######  Proposition 3.

The mapping:

$\displaystyle(u,v)$ $\displaystyle\longmapsto$
$\displaystyle\left<u,v\right>=\int_{G/K}\left<u(g),v(g)\right>_{H_{\sigma}}d\mu(\dot{g}).$
(14)

on $H^{L^{\sigma}}_{0}\times H^{L^{\sigma}}_{0}$ is an inner product on
$H^{L^{\sigma}}_{0}.$

$G$ acts on $H^{L^{\sigma}}_{0}$ by left translation, $u\longmapsto L_{t}u$,
so we obtain a unitary representation of $G$ with respect to this inner
product on $H^{L^{\sigma}}_{0}$. The inner product is preserved by left
translations, since $\mu$ is invariant. Hence, if we denote by
$H^{L^{\sigma}}$ the Hilbert space completion of $H^{L^{\sigma}}_{0},$ the
translation operators $L_{t}$ extend to unitary operators on $H^{L^{\sigma}}$.
Then the map $t\longmapsto L_{t}u$ is continuous from $G$ to $H^{L^{\sigma}}$
for each $u\in H^{L^{\sigma}}_{0},$ and then, since the operators $L_{t}$ are
uniformly bounded, they are strongly continuous on $H^{L^{\sigma}}.$ Hence,
they define a unitary representation of G, called the representation induced
by $L^{\sigma},$ denoted by $U^{L^{\sigma}}:$

$U_{t}^{L^{\sigma}}u(g)=L_{t}u(g)=u(t^{-1}g).$

The representation space is denoted $H^{L^{\sigma}}$.

######  Remark 1.

  * •

The representations of $G$ induced from $K$ are generally infinite-dimensional
unless $G/K$ is a finite set.

  * •

If induced representation $U^{L^{\sigma}}$ is irreducible then $L^{\sigma}$ is
irreducible[19].
The converse of this statement is false.

  * •

$U^{L^{\sigma}}\in U(H^{L\sigma}),$ where $U(H^{L\sigma})$ designates the
group of invertible unitary linear operators on $H^{L\sigma}$.

Lebesgue Spaces
Let $\lambda$ be a positve measure on the group $G$ into a Banach algebra
$\mathcal{A}$. For $f:G\longrightarrow\mathcal{A}$, put

$\displaystyle N_{p}(f)$ $\displaystyle=$
$\displaystyle\left(\int_{G}^{*}\left\|f(t)\right\|_{\mathcal{A}}^{p}d\lambda(t)\right)^{\dfrac{1}{p}}$
(15)

$1\leq p<\infty$, where $\displaystyle\int_{G}^{*}$ designates the upper
integral [2] and

$\displaystyle N_{\infty}(f)$ $\displaystyle=$
$\displaystyle\inf\left\\{\alpha:\|f(t)\|_{\mathcal{A}}\leq\alpha\quad\lambda-\text{
almost everywhere }\right\\}.$ (16)

$\mathcal{L}_{p}(G,\lambda,\mathcal{A})$ denotes the set of all
$\lambda$-measurable functions $f:G\longrightarrow\mathcal{A}$ such
$N_{p}(f)<\infty$, $1\leq p\leq\infty$. Generally the mapping $f\longmapsto
N_{p}(f)$ is not a norm but a seminorm in
$\mathcal{L}_{p}(G,\lambda,\mathcal{A})$, $1\leq p\leq\infty$.

The relation $\mathcal{R}$ defined by $f\mathcal{R}g$ if and only if $f-g$ is
null $\lambda$ almost everywhere is an equivalent relation in
$\mathcal{L}_{p}(G,\lambda,\mathcal{A})$. Denoting the equivalence class of
$f\in\mathcal{L}_{p}(G,\lambda,\mathcal{A})$ by $[f]$, it follows that
$\|[f]\|=N_{p}(f).$

######  Definition 2.9.

The symbol $L_{p}(G,\lambda,\mathcal{A})$ will denote the set of equivalence
classes $[f]$ of functions $f\in\mathcal{L}_{p}(G,\lambda,\mathcal{A})$.

######  Theorem 2.10.

The space $L_{p}(G,\lambda,\mathcal{A})$ is a normed linear space.

In the sequel the symbol $f$ rather than $[f]$ will be used for an element in
$L_{p}(G,\lambda,\mathcal{A})$.

######  Theorem 2.11.

$N_{p}$ define a norm in $L_{p}(G,\lambda,\mathcal{A})$ noted $\|.\|_{p}$.

######  Theorem 2.12.

(Hölder inequality)) Consider $1\leq p\leq\infty$, $q\geq 1$ such that
$\dfrac{1}{p}+\dfrac{1}{q}=1$ ( if $p=1$, $q=\infty$) $f\in
L_{p}(G,\lambda,\mathcal{A})$ and $g\in L_{q}(G,\lambda,\mathcal{A})$ then the
function $fg$ is $\lambda$-integrable ie $fg\in L_{1}(G,\lambda,\mathcal{A})$,
and

$\displaystyle\|fg\|_{1}$ $\displaystyle\leq$
$\displaystyle\|f\|_{p}\|g\|_{q}.$ (17)

######  Theorem 2.13.

(Minkowski inequality) For $1\leq p\leq\infty$ and $f,g\in
L_{p}(G,\lambda,\mathcal{A})$, we have

$\displaystyle\|f+g\|_{p}$ $\displaystyle\leq$
$\displaystyle\|f\|_{p}+\|g\|_{p}.$ (18)

######  Remark 2.

We deduce by Hölder theorem that for $G$ such that $\lambda(G)<\infty$ and
$1\leq q\leq p\leq\infty$, we have $L_{\infty}(G,\lambda,\mathcal{A})\subset
L_{p}(G,\lambda,\mathcal{A})\subset L_{q}(G,\lambda,\mathcal{A})\subset
L_{1}(G,\lambda,\mathcal{A})$ but if $\lambda(G)=\infty$ there are no
inclusion between $L_{p}(G,\lambda,\mathcal{A})$ and
$L_{q}(G,\lambda,\mathcal{A})$ for $q\neq p$.

######  Proposition 4.

$L_{p}(G,\lambda,\mathcal{A})\cap L_{q}(G,\lambda,\mathcal{A})$ is dense in
$L_{p}(G,\lambda,\mathcal{A})$ and in $L_{q}(G,\lambda,\mathcal{A})$.

######  Remark 3.

If $m$ is a dominated vector measure into a Banach space $F$, then
$\mathcal{L}_{p}(G,m,\mathcal{A})$ is by convention the space
$\mathcal{L}_{p}(G,|m|,\mathcal{A}).$

######  Theorem 2.14.

(Stone-Weierstrass theorem for locally compact spaces) Let $\Omega$ be a
locally compact space, which is non-compact, let
$\mathfrak{C}_{0}^{\mathbb{K}}(\Omega)$ be the algebra of continuous
$\mathbb{K}$-valued functions on $\Omega$ that vanish at infinity equipped
with the supremum norm and let
$\mathcal{P}\subset\mathfrak{C}_{0}^{\mathbb{K}}(\Omega)$ with the following
separation properties:

  * (i)

for any two points $\omega_{1},\omega_{2}\in\Omega$ with
$\omega_{1}\neq\omega_{2}$ there exists $f\in\mathcal{P}$ such that
$f(\omega_{1})\neq f(\omega_{2})$

  * (ii)

for any $\omega\in\Omega$ there exists $f\in\mathcal{P}$ with $f(\omega)\neq
0$.

  * (a)

If $\mathbb{K}=\mathbb{R}$ then $\mathcal{P}$ is dense in
$\mathfrak{C}_{0}^{\mathbb{R}}(\Omega).$

  * (b)

If $\mathbb{K}=\mathbb{C}$ and if $\mathcal{P}$ is a ∗-subalgebra (i.e.
$f\in\mathcal{P}\Longrightarrow\overline{f}\in\mathcal{P}$) then $\mathcal{P}$
is dense in $\mathfrak{C}_{0}(\Omega).$

##  3\. Main results

We have $K$ compact and normal, $H_{\sigma}$ is finite and separable; then
$H^{L^{\sigma}}$ is also separable. Each of these spaces admits a Hilbertian
basis according to Gram-Schmidt process.

In this work, we suppose that $K$ is chosen such that $L^{\sigma}$ and
$U^{L^{\sigma}}$ are both irreducible.

###  3.1. Schur orthogonaltity relations

Let $(\theta_{i})_{i=1}^{\infty}$ be an orthonormal basis of $H^{L^{\sigma}}$
and $(\xi_{i})_{i=1}^{d\sigma}$ be an orthonormal basis of $H_{\sigma}$. We
define

$\displaystyle u_{ij}^{L^{\sigma}}(t)$ $\displaystyle:=$
$\displaystyle\left<U_{t}^{L^{\sigma}}\theta_{j},\theta_{i}\right>_{H^{L^{\sigma}}}$
(19) $\displaystyle=$
$\displaystyle\int_{G/K}\left<\theta_{j}(t^{-1}g),\theta_{i}(g)\right>d\mu(\dot{g})$
(20)

and

$L_{ij}^{\sigma}(k):=\left<L_{k}^{\sigma}\xi_{j},\xi_{i}\right>_{H_{\sigma}}$

As a result, there is a family of mappings $(\alpha_{is})_{s=1}^{d\sigma}$ of
$G$ into $\mathbb{K}with$ ($\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$) such that

$\theta_{i}(g)=\sum_{s=1}^{d\sigma}\alpha_{is}(g)\xi_{s}.$

We have:

$\delta_{ij}=\left<\theta_{j},\theta_{i}\right>_{H^{L_{\sigma}}}=\int_{G/K}\left<\theta_{j}(g),\theta_{i}(g)\right>_{H_{\sigma}}d\mu(\dot{g})=\sum_{s=1}^{d\sigma}\int_{G/K}\alpha_{js}(g)\overline{\alpha}_{is}(g)d\mu(\dot{g}).$

######  Proposition 5.

$\forall\sigma\in\Sigma$ and $\forall i,j\in\\{1,2,...\\},$
$u_{ij}^{L^{\sigma}}\in\mathcal{K}(G).$

###### Proof.

The continuity of $u_{ij}^{L^{\sigma}}$ results from their constrction.
According to 13, for all $u\in H^{L^{\sigma}},$ thereis
$\eta\in\mathcal{K}(G,H_{\sigma})$ such that
$u_{\eta}(g)=\displaystyle\int_{K}L^{\sigma}_{k}\eta(gk)d\nu(k).$ It follows
that $q(supp(u))\subset q(supp(\eta)).$

Let $A$ be the support of $\eta.$ For every $g\in G$ and $k\in K,$ $gk\in
A\Longrightarrow g\in Ak^{-1}.$ $AK=\displaystyle\bigcup_{k\in K}Ak^{-1}$ is
compact because $A$ and $K$ are compact. F Furthermore $\forall g\notin AK,$
$u(g)=0$ then $supp(u)\subset AK.$ Since $supp(u)$ is a closed subset of $AK$
then it is compact.

Now let $B$ and $C$ be $supp(\theta_{j})$ and $supp(\theta_{i}),$
respectively. Assume $D=\left\\{t\in G:g\in C\Longrightarrow t^{-1}g\in
B\right\\}$. According to 19,
$\left<\theta_{j}(t^{-1}g),\theta_{i}(g)\right>=0$ if $t\notin D,$ and then
$u_{ij}^{L^{\sigma}}(t)=0,$ then $supp(u_{ij}^{L^{\sigma}})\subset D.$ In
addition,

$\displaystyle t^{-1}g\in B\text{ with }g\in C$ $\displaystyle\Longrightarrow$
$\displaystyle t\in CB^{-1}.$

Then $supp(u_{ij}^{L^{\sigma}})\subset D\subset CB^{-1}.$ Since $CB^{-1}$ is
compact, $supp(u_{ij}^{L^{\sigma}})$ is also compact. ∎

Therefore, we have

######  Corollary 1.

$\int_{G}\left|u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\sigma}}(t)\right|d\lambda(t)<\infty.$

Since $K$ is normal, we have $K\backslash G=G/K$ then

######  Lemma 3.1.

$\forall u\in H^{L^{\sigma}}$, $u(kt)=L_{k}u(t)$ and
$u(tk)=L_{k^{-1}}^{\sigma}u(t)$.

###### Proof.

There is $\eta\in\mathfrak{C}(G,H_{\sigma})$ such that

$u(t)=\int_{K}L^{\sigma}(\xi)\eta(t\xi)d\nu(\xi).\quad\text{\cite[cite]{[\@@bibref{}{folland}{}{}]}}$
$\displaystyle u(tk)$ $\displaystyle=$
$\displaystyle\int_{K}L^{\sigma}(\xi)\eta(tk\xi)d\nu(\xi),\quad k\in K$
$\displaystyle=$
$\displaystyle\int_{K}L^{\sigma}(k^{-1}\xi)\eta(t\xi)d\nu(\xi)$
$\displaystyle=$ $\displaystyle
L_{k^{-1}}^{\sigma}\int_{K}L^{\sigma}(\xi)\eta(t\xi)d\nu(\xi)$
$\displaystyle=$ $\displaystyle L_{k^{-1}}^{\sigma}u(t).$

There exists $\beta\in\mathfrak{C}(G,H_{\sigma})$ such that

$u(t)=\int_{K}L^{\sigma}(\xi^{-1})\beta(\xi t)d\nu(\xi).\quad\text{
\cite[cite]{[\@@bibref{}{B-R}{}{}]}}$ $\displaystyle u(kt)$ $\displaystyle=$
$\displaystyle\int_{K}L^{\sigma}(\xi^{-1})\beta(\xi kt)d\nu(\xi)$
$\displaystyle=$ $\displaystyle\int_{K}L^{\sigma}(k\xi^{-1})\beta(\xi
t)d\nu(\xi)$ $\displaystyle=$ $\displaystyle
L_{k}^{\sigma}\int_{K}L^{\sigma}(\xi^{-1})\beta(\xi t)d\nu(\xi)$
$\displaystyle=$ $\displaystyle L_{k}^{\sigma}u(t).$

∎

The following theorem shows the Schur orthogonality relation for the the case
of the representation $U^{L^{\sigma}}$ of $G$ induced by the unitary
irreductible representation $L^{\sigma}$ of $K$.

######  Theorem 3.2.

We have

$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\sigma}}(t)d\lambda(t)$
$\displaystyle=$ $\displaystyle\dfrac{c_{ijlm}}{d_{\sigma}}$ (21)

where

$\displaystyle c_{ijlm}$ $\displaystyle:=$
$\displaystyle\int_{G/K}d\mu(\dot{t})\sum_{r,s=1}^{d_{\sigma}}\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha}_{ir}(g)d\mu(\dot{g})\int_{G/K}\alpha_{ms}(t^{-1}h)\overline{\alpha}_{lr}(h)d\mu(\dot{h})$
$\displaystyle=$
$\displaystyle\sum_{r,s=1}^{d_{\sigma}}\int_{(G/K)^{3}}\alpha_{js}(t^{-1}g)\overline{\alpha}_{ir}(g)\alpha_{ms}(t^{-1}h)\overline{\alpha}_{lr}(h)d\mu(\dot{g})d\mu(\dot{h})d\mu(\dot{t})$

and

$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\tau}}(t)d\lambda(t)$
$\displaystyle=$ $\displaystyle 0\text{ if }\sigma\neq\tau.$ (22)

###### Proof.

By straightforward computation, we obtain:

$\displaystyle\int_{G/K}\left<\theta_{j}(gk),\theta_{i}(g)\right>_{H_{\sigma}}d\mu(\dot{g})$
$\displaystyle=$
$\displaystyle\int_{G/K}\left<L^{\sigma}(k^{-1})\theta_{j}(g),\theta_{i}(g)\right>_{H_{\sigma}}d\mu(\dot{g})$
$\displaystyle=$
$\displaystyle\sum_{s,r=1}^{d\sigma}\int_{G/K}\alpha_{js}(g)\overline{\alpha}_{ir}(g)\left<L^{\sigma}(k^{-1})\xi_{s},\xi_{r}\right>_{H_{\sigma}}d\mu(\dot{g})$
$\displaystyle=$
$\displaystyle\sum_{r,s=1}^{d_{\sigma}}L_{rs}^{\sigma}(k^{-1})\int_{G/K}\alpha_{js}(g)\overline{\alpha}_{ir}(g)d\mu(\dot{g}).$

Besides, we get:

$\displaystyle u_{ij}^{L^{\sigma}}(tk)$ $\displaystyle=$
$\displaystyle\int_{G/K}\left<\theta_{j}(k^{-1}t^{-1}g),\theta_{i}(g)\right>d\mu(\dot{g})$
$\displaystyle=$
$\displaystyle\int_{G/K}\left<L^{\sigma}(k^{-1})\theta_{j}(t^{-1}g),\theta_{i}(g)\right>d\mu(\dot{g})$
$\displaystyle=$
$\displaystyle\int_{G/K}\left<L^{\sigma}(k^{-1})\theta_{j}(t^{-1}g),\theta_{i}(g)\right>d\mu(\dot{g})$
$\displaystyle=$
$\displaystyle\sum_{r,s=1}^{d_{\sigma}}L_{rs}^{\sigma}(k^{-1})\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha}_{ir}(g)d\mu(\dot{g})$

and

$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\sigma}}(t)d\lambda(t)$
$\displaystyle=$
$\displaystyle\int_{G/K}\int_{K}u_{ij}^{L^{\sigma}}(tk)\overline{u}_{lm}^{L^{\sigma}}(tk)d\nu(k)d\mu(\dot{t})$
$\displaystyle=$
$\displaystyle\int_{G/K}d\mu(\dot{t})\int_{K}(\sum_{r,s=1}L_{rs}^{\sigma}(k^{-1})\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha_{ir}}(g)d\mu(\dot{g})\times$
$\displaystyle\sum_{p,q=1}L_{pq}^{\sigma}(k^{-1})\int_{G/K}\alpha_{mq}(t^{-1}h)\overline{\alpha_{lp}}(h)d\mu(\dot{h}))d\nu(k)$
$\displaystyle=$
$\displaystyle\int_{G/K}d\mu(\dot{t})\sum_{r,s,p,q=1}\int_{K}(L_{rs}^{\sigma}(k^{-1})\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha_{ir}}(g)d\mu(\dot{g})\times$
$\displaystyle
L_{pq}^{\sigma}(k^{-1})\int_{G/K}\alpha_{mq}(t^{-1}h)\overline{\alpha_{lp}}(h)d\mu(\dot{h}))d\nu(k).$

Hence,

$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\sigma}}(t)d\lambda(t)$
$\displaystyle=$
$\displaystyle\int_{G/K}d\mu(\dot{t})\sum_{r,s,p,q=1}\int_{K}(L_{rs}^{\sigma}(k^{-1})L_{pq}^{\sigma}(k^{-1})d\nu(k)\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha_{ir}}(g)d\mu(\dot{g})\times$
$\displaystyle\int_{G/K}\alpha_{mq}(t^{-1}h)\overline{\alpha_{lp}}(h)d\mu(\dot{h})$
$\displaystyle=$
$\displaystyle\int_{G/K}d\mu(\dot{t})\sum_{r,s,p,q=1}\dfrac{\delta_{rp}\delta_{sq}}{d_{\sigma}}\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha_{ir}}(g)d\mu(\dot{g})\times$
$\displaystyle\int_{G/K}\alpha_{mq}(t^{-1}h)\overline{\alpha_{lp}}(h)d\mu(\dot{h})\text{
according to \ref{sch1} }$ $\displaystyle=$
$\displaystyle\dfrac{1}{d_{\sigma}}\sum_{r,s=1}^{d_{\sigma}}\int_{(G/K)^{3}}\alpha_{js}(t^{-1}g)\overline{\alpha}_{ir}(g)\alpha_{ms}(t^{-1}h)\overline{\alpha}_{lr}(h)d\mu(\dot{g})d\mu(\dot{h})d\mu(\dot{t})$
$\displaystyle=$ $\displaystyle\dfrac{c_{ijlm}}{d_{\sigma}}.$

If $\sigma\neq\tau$ then

$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\tau}}(t)d\lambda(t)$
$\displaystyle=$
$\displaystyle\int_{G/K}d\mu(\dot{t})\sum_{r,s,p,q=1}\int_{K}(L_{rs}^{\sigma}(k^{-1})L_{pq}^{\tau}(k^{-1})d\nu(k)\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha_{ir}}(g)d\mu(\dot{g})\times$
$\displaystyle\int_{G/K}\alpha_{mq}(t^{-1}h)\overline{\alpha_{lp}}(h)d\mu(\dot{h})=0\text{
according to \ref{sch2} }$

∎

In the case of a particular orthonormal basis, the orthogonality relation
reduces to the following:

######  Corollary 2.

Choosing an orthonormal basis $(\xi_{i})_{i=1}^{d\sigma}$ of $H_{\sigma}$ such
that

$\displaystyle\int_{(G/K)^{3}}\alpha_{js}(t^{-1}g)\overline{\alpha}_{ir}(g)\alpha_{ms}(t^{-1}h)\overline{\alpha}_{lr}(h)d\mu(\dot{g})d\mu(\dot{h})d\mu(\dot{t})=\left\\{\begin{array}[]{cl}\dfrac{1}{d_{\sigma}^{2}}&\text{
if }j=m\text{ and }i=l\\\ &\\\ 0&\text{ if not}\end{array}\right.$ (26)

(or, also, $c_{ijlm}=\delta_{il}\delta_{jm}$) leads to

$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\sigma}}(t)d\lambda(t)$
$\displaystyle=$ $\displaystyle\dfrac{\delta_{il}\delta_{jm}}{d_{\sigma}}.$
(27)

###### Proof.

With the conditions 26, we have

$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\sigma}}(t)d\lambda(t)$
$\displaystyle=$
$\displaystyle\small{\dfrac{1}{d_{\sigma}}\sum_{r,s=1}^{d_{\sigma}}\int_{(G/K)^{3}}\alpha_{js}(t^{-1}g)\overline{\alpha}_{ir}(g)\alpha_{ms}(t^{-1}h)\overline{\alpha}_{lr}(h)d\mu(\dot{g})d\mu(\dot{h})d\mu(\dot{t})}$
$\displaystyle=$
$\displaystyle\dfrac{\delta_{il}\delta_{jm}}{d_{\sigma}}\sum_{i,j=1}^{d_{\sigma}}\dfrac{1}{d_{\sigma}^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{\delta_{il}\delta_{jm}}{d_{\sigma}}\dfrac{d_{\sigma}^{2}}{d_{\sigma}^{2}}$
$\displaystyle\int_{G}u_{ij}^{L^{\sigma}}(t)\overline{u}_{lm}^{L^{\sigma}}(t)d\lambda(t)$
$\displaystyle=$ $\displaystyle\dfrac{\delta_{il}\delta_{jm}}{d_{\sigma}}.$

In the sequel, $(\theta_{i})_{i=1}^{\infty}$ will designate an orthonormal
basis of $H^{L^{\sigma}}$ and $(\xi_{i})_{i=1}^{d\sigma}$ that of
$H_{\sigma},$ where $(\xi_{i})_{i=1}^{d\sigma}$ is chosen such that

$\displaystyle\int_{G/K}d\mu(\dot{t})\int_{G/K}\alpha_{js}(t^{-1}g)\overline{\alpha}_{ir}(g)d\mu(\dot{g})\int_{G/K}\alpha_{ms}(t^{-1}h)\overline{\alpha}_{lr}(h)d\mu(\dot{h})=\left\\{\begin{array}[]{cl}\dfrac{1}{d_{\sigma}^{2}}&\text{
if }j=m\text{ and }i=l\\\ &\\\ 0&\text{ if not.}\end{array}\right.$

∎

###  3.2. Fourier-Stieltjes transform on a locally compact group with a
compact subgroup

######  Definition 3.3.

Assume $m\in M_{1}(G,\mathcal{A})$. We shall define the Fourier-Stieljes
transform of an arbitrary measure $m$ as the family
$(\hat{m}(\sigma))_{\sigma\in\Sigma}$ of sesquilinear mappings on
$H^{L^{\sigma}}\times H^{L^{\sigma}}$ into $\mathcal{A}$, given by the
relation

$\displaystyle\hat{m}(\sigma)(u,v)=\int_{G}\left<\overline{U}_{t}^{L^{\sigma}}u,v\right>_{H^{L^{\sigma}}}dm(t)=\int_{G}\int_{G/K}\left<\overline{u}(t^{-1}g),v(g)\right>d\mu(\dot{g})dm(t).$
(29)

The Fourier transform of a function $f\in L_{1}(G,\mathcal{A},\lambda)$, where
$\lambda$ denotes a Haar measure on $G$, is a family
$(\hat{f}(\sigma))_{\sigma\in\Sigma}$ of sesquilinear mappings of
$H^{L^{\sigma}}\times H^{L^{\sigma}}$ into $\mathcal{A},$ given by the
relation

$\displaystyle\hat{f}(\sigma)(u,v)=\int_{G}\left<\overline{U}_{t}^{L^{\sigma}}u,v\right>_{H^{L^{\sigma}}}f(t)d\lambda(t)=\int_{G}\int_{G/K}\left<\overline{u}(t^{-1}g),v(g)\right>f(t)d\mu(\dot{g})d\lambda(t).$
(30)

###  3.3. Properties of the Fourier-Stieljes transform

Let denote by
$\mathcal{S}(\Sigma,\mathcal{A})=\displaystyle{\prod_{\sigma\in\Sigma}}\mathcal{S}(H^{L^{\sigma}}\times
H^{L^{\sigma}},\mathcal{A}),$ where $\mathcal{S}(H^{L^{\sigma}}\times
H^{L^{\sigma}},\mathcal{A})$ is the set of all the sesquilinear mappings of
$H^{L^{\sigma}}\times H^{L^{\sigma}}$ into $\mathcal{A}$. Also, assume
$\displaystyle{\prod_{\sigma\in\Sigma}}\mathcal{S}(H^{L^{\sigma}}\times
H^{L^{\sigma}},\mathcal{A})$ is a vector space for addition and multiplication
by a scalar of mappings.

######  Definition 3.4.

For
$\Phi\in\mathcal{S}(\Sigma,\mathcal{A})=\displaystyle{\prod_{\sigma\in\Sigma}}\mathcal{S}(H^{L^{\sigma}}\times
H^{L^{\sigma}},\mathcal{A})$, note $\|\Phi\|_{\infty}$ the quantity is defined
by

$\displaystyle\|\Phi\|_{\infty}$ $\displaystyle=$
$\displaystyle\sup\\{\|\Phi(\sigma)\|:\sigma\in\Sigma\\}$ (31)

with $\|\Phi(\sigma)\|=\sup\\{\|\Phi(\sigma)(u,v)\|_{\mathcal{A}}\mbox{
}\mbox{ }:\mbox{ }\|u\|_{H^{L^{\sigma}}}\leq 1,\mbox{
}\|v\|_{H^{L^{\sigma}}}\leq 1\\}$. Let define:

  * (i)

$\displaystyle\mathcal{S}_{\infty}(\Sigma,\mathcal{A})$ =
$\left\\{\Phi\in\mathcal{S}(\Sigma,\mathcal{A}):\|\Phi\|_{\infty}<\infty\right\\}$.

  * (ii)

$\displaystyle\mathcal{S}_{00}(\Sigma,\mathcal{A})$ =
$\left\\{\Phi\in\mathcal{S}(\Sigma,\mathcal{A}):\\{\sigma\in\Sigma:\Phi(\sigma)\neq
0\\}\quad\mbox{is finite}\right\\}$.

  * (iii)

$\displaystyle\mathcal{S}_{0}(\Sigma,\mathcal{A})$ =
$\left\\{\Phi\in\mathcal{S}(\Sigma,\mathcal{A}):\forall\varepsilon>0\quad\\{\sigma\in\Sigma\mbox{
}\mbox{ }:\mbox{ }\mbox{ }\|\Phi(\sigma)\|>\varepsilon\\}\quad\mbox{is
finite}\right\\}$.

######  Proposition 6.

Let us consider $\Phi$ in $\mathcal{S}(\Sigma,\mathcal{A}).$ There is the
matrix $(a_{ij}^{\sigma})_{1\leq i,j\leq\infty}$,
$a_{ij}^{\sigma}\in\mathcal{A}$ such that

$\Phi(\sigma)=\sum_{i,j=1}^{\infty}d_{\sigma}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}$

with $\hat{u}_{ij}^{L^{\sigma}}$ the Fourier transform of
$u_{ij}^{L^{\sigma}}$ given by

$\hat{u}_{ij}^{L^{\sigma}}(\sigma)(u,v)=\int_{G}\left<\overline{U}_{t}^{L^{\sigma}}u,v\right>_{H^{L^{\sigma}}}u_{ij}^{L^{\sigma}}(t)d\lambda(t).$

###### Proof.

For all $u,v\in H^{L^{\sigma}}$ such that
$u=\displaystyle\sum_{j=1}^{\infty}\beta_{j}\theta_{j}$ and
$v=\displaystyle\sum_{i=1}^{\infty}\gamma_{i}\theta_{i}$,

$\Phi(\sigma)(u,v)=\sum_{i,j=1}^{\infty}\beta_{j}\overline{\gamma}_{i}\Phi(\sigma)(\theta_{j},\theta_{i}).$

Defining $(\Phi(\sigma)(\theta_{j},\theta_{i}))$ = $(a_{ij}^{\sigma})$, the
matrix of $\Phi(\sigma)$ in the basis $(\theta_{i})_{i=1}^{\infty}$ and we
have

$\displaystyle\hat{u}_{ij}^{L^{\sigma}}(\sigma)(u,v)$ $\displaystyle=$
$\displaystyle\int_{G}\left<\overline{U}_{t}^{L^{\sigma}}u,v\right>_{H^{L^{\sigma}}}u_{ij}^{L^{\sigma}}(t)d\lambda(t)$
$\displaystyle=$
$\displaystyle\sum_{l,m=1}^{\infty}\beta_{m}\overline{\gamma}_{l}\int_{G}\overline{u}_{lm}^{L^{\sigma}}u_{ij}^{L^{\sigma}}d\lambda(t)=\dfrac{\beta_{j}\overline{\gamma}_{i}}{d_{\sigma}}$

.

$\displaystyle\Phi(\sigma)(u,v)$ $\displaystyle=$
$\displaystyle\sum_{i,j=1}^{\infty}a_{ij}^{\sigma}\beta_{j}\overline{\gamma}_{i}=\sum_{i,j=1}^{\infty}d_{\sigma}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}(u,v)$

which yields

$\Phi(\sigma)=\sum_{i,j=1}^{\infty}d_{\sigma}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}.$

∎

######  Corollary 3.

For any $f\in L_{1}(G,\mathcal{A},\lambda)$, there is a matrix
$(a_{ij})_{1\leq i,j\leq\infty}$, $a_{ij}\in\mathcal{A}$ such that

$\hat{f}(\sigma)=\sum_{i,j=1}^{\infty}d_{\sigma}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}.$

######  Lemma 3.5.

Let us denote by $\mathcal{L}_{\sigma}(G)$ the set of finite linear
combinations functions
$t\longmapsto\left<U_{t}^{L^{\sigma}}u,v\right>_{H^{L^{\sigma}}}$ and
$\displaystyle\mathcal{L}(G)=\displaystyle\bigcup_{\sigma\in\Sigma}\mathcal{L}_{\sigma}(G).$
$\mathcal{L}(G)$ is dense in $\mathfrak{C}_{0}(G)$ for the topology of the
uniform convergence, where $\mathfrak{C}_{0}(G)$ denotes the space of
continuous funtions of $G$ into $\mathbb{K}$ vanishing at infinity.

###### Proof.

  * (i)

First we have $\mathcal{L}(G)\subset\mathfrak{C}_{0}(G)$ moreover for
$f\in\mathcal{L}(G)$ we have abviously $\overline{f}\in\mathcal{L}(G).$

  * (ii)

Let $t_{1}$ and $t_{2}$ be two elements of $G$ such that $t_{1}\neq t_{2}$.
Let us suppose by contradiction that for every $f\in\mathcal{L}(G),$
$f(t_{1})=f(t_{2})$. Thus, by chosing $f=f_{k}=u_{ik}^{L^{\sigma}}$ we have
$\left<U_{t_{1}}^{L^{\sigma}}\theta_{i},\theta_{k}\right>=\left<U_{t_{2}}^{L^{\sigma}}\theta_{i},\theta_{k}\right>$
for $i,k\in\\{1,2,...\\}$. Let us fix $i$ and make $k$ running over the set
$\\{1,2,...\\}.$ Then, antilinear mappings
$\phi_{U_{t_{1}}^{L^{\sigma}}\theta_{i}}=\left<U_{t_{1}}^{L^{\sigma}}\theta_{i},.\right>$
and
$\phi_{U_{t_{2}}^{L^{\sigma}}\theta_{i}}=\left<U_{t_{2}}^{L^{\sigma}}\theta_{i},.\right>$
are identically equal in $H^{L^{\sigma}}$ therefore
$U_{t_{1}}^{L^{\sigma}}\theta_{i}=U_{t_{2}}^{L^{\sigma}}\theta_{i}\Longrightarrow\theta_{i}=U_{t_{2}t_{1}^{-1}}^{L^{\sigma}}\theta_{i}\Longrightarrow
U_{t_{2}t_{1}^{-1}}^{L^{\sigma}}\equiv I_{H^{L^{\sigma}}}.$ It means that
$t_{2}=t_{1}$. There is a contradiction; hence, there is $f\in\mathcal{L}(G)$
such that $f(t_{1})\neq f(t_{2})$.

  * (iii)

Let $t$ be any element of $G$; since $U_{t}^{L^{\sigma}}$ is invertible, there
exist $i,k\in\\{1,2,...\\}$ such that
$U_{t}^{L^{\sigma}}\theta_{i}=\theta_{k}$. Therefore, we have
$\left<U_{t}^{L^{\sigma}}\theta_{i},\theta_{k}\right>=\left<\theta_{k},\theta_{k}\right>=1\neq
0$ i.e $f(t)\neq 0$.

According to theorem 2.14 $\mathcal{L}(G)$ is dense in $\mathfrak{C}_{0}(G)$.
∎

######  Theorem 3.6.

The mapping $m\longmapsto\hat{m}$ from $M_{1}(G,\mathcal{A})$ into
$\mathcal{S}_{\infty}(\Sigma,\mathcal{A})$ is linear, injective and
continuous.

###### Proof.

Let $m,n\in M_{1}(G,\mathcal{A})$ such that $\hat{m}=\hat{n}.$ For any $u,v\in
H^{L^{\sigma}}\times H^{L^{\sigma}}$ and any $\sigma\in\Sigma,$ we have:

$\displaystyle\hat{m}=\hat{n}$ $\displaystyle\Longleftrightarrow$
$\displaystyle\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}dn(t)=\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}dm(t)$
(32) $\displaystyle\Longleftrightarrow$
$\displaystyle\displaystyle\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}d(n-m)(t)=0$
(33)

for any $\sigma$ in $\Sigma,$ and $u$,$v$ in $H^{L^{\sigma}}$.

$\mathcal{L}(G)$ is dense in $\mathfrak{C}_{0}(G)$ according to the previous
lemma. Then, $\mathcal{L}(G)$ is dense in $\mathcal{K}(G)$. Thus $n-m$ can be
viewed as a linear map of $\mathcal{K}(G)$ which is identically null. Then
$\displaystyle\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}d(n-m)(t)=0$
$\Longrightarrow n-m\equiv 0$ ie $m=n.$ The mapping $m\longmapsto\hat{m}$ is
therefore injective.

Next, let us now prove the continuity of the mapping $m\longmapsto\hat{m}$.

$\displaystyle\|\hat{m}(\sigma)\|$ $\displaystyle=$
$\displaystyle\sup\left\\{\|\hat{m}(\sigma)(u,v)\|_{\mathcal{A}}:\|u\|_{H^{L^{\sigma}}},\leq
1\text{ and }\|v\|_{H^{L^{\sigma}}},\leq 1\right\\}$ $\displaystyle=$
$\displaystyle\sup\left\\{\left\|\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}dm(t)\right\|_{\mathcal{A}}:\|u\|_{H^{L^{\sigma}}},\leq
1\mbox{ and }\|v\|_{H^{L^{\sigma}}},\leq 1\right\\}$ $\displaystyle\leq$
$\displaystyle\int_{G}\chi_{G}d|m|=\|m\|,$

since $U_{t}^{L^{\sigma}}$ is unitary. Thus, $\|\hat{m}(\sigma)\|\leq\|m\|$,
$\sigma\in\Sigma$ and $\|\hat{m}\|_{\infty}\leq\|m\|$. As a consequence,
$\hat{m}\in\mathcal{S}_{\infty}(\Sigma,\mathcal{A})$ and the mapping is
continuous.

######  Corollary 4.

The mapping $f\longmapsto\hat{f}$ from $L_{1}(G,\lambda,\mathcal{A})$ into
$\mathfrak{S}_{\infty}(\Sigma,\mathcal{A})$ is linear, injective, and
continuous.

Proof Here we consider $f,g\in L_{1}(G,\lambda,\mathcal{A})$ such that
$\hat{f}=\hat{g}$ then

$\displaystyle\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}fd\lambda(t)=\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}fd\lambda(t)$
$\displaystyle\Longrightarrow$
$\displaystyle\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}(f-g)d\lambda(t)=0.$

For the same reasons as the previous proof we have $f-g\equiv 0$ then $f=g$.

$\displaystyle\|\hat{f}(\sigma)\|$ $\displaystyle=$
$\displaystyle\sup\left\\{\|\hat{f}(\sigma)(u,v)\|_{\mathcal{A}}:\|u\|_{H^{L^{\sigma}}},\leq
1\text{ and }\|v\|_{H^{L^{\sigma}}},\leq 1\right\\}$ $\displaystyle=$
$\displaystyle\sup\left\\{\left\|\int_{G}\left<\overline{U_{t}^{L^{\sigma}}}u,v\right>_{H^{L^{\sigma}}}f(t)d\lambda(t)\right\|:\|u\|_{H^{L^{\sigma}}},\leq
1\mbox{ and }\|v\|_{H^{L^{\sigma}}},\leq 1\right\\}$ $\displaystyle\leq$
$\displaystyle\int_{G}\|f(t)\|d\lambda(t)=\|f\|_{1}.$

Thus, $\|\hat{f}(\sigma)\|\leq\|f\|_{1}$, $\sigma\in\Sigma$ and
$\|\hat{f}\|_{\infty}\leq\|f\|_{1}$. Hence,
$\hat{f}\in\mathfrak{S}_{\infty}(\Sigma,\mathcal{A})$ and the mapping is
continuous. ∎

###  3.4. Modified Peter-Weyl theorem

The following theorem gives the inverse formula of Fourier transform in the
context of our work.

######  Theorem 3.7.

For every $f\in L_{2}(G,\mathcal{A})$, there is
$a_{ij}^{\sigma}\in\mathcal{A},$ $1\leq i,j<\infty$, $\sigma\in\Sigma$ such
that

$\displaystyle f$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\displaystyle\sum_{i,j=1}^{\infty}a_{ij}^{\sigma}u_{ij}^{L^{\sigma}}.$
(34)

###### Proof.

With Proposition 6 for $g\in L_{1}(G,\lambda,\mathcal{A}),$ there exists
$(a_{ij}^{\sigma})_{1\leq i,j\leq\infty}$, $a_{ij}^{\sigma}\in\mathcal{A}$
such that

$\hat{g}(\sigma)=\sum_{i,j=1}^{\infty}d_{\sigma}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}$

with
$\hat{u}_{ij}^{L^{\sigma}}(\sigma)(u,v)=\displaystyle\int_{G}\left<\overline{U}_{t}^{L^{\sigma}}u,v\right>_{H^{L^{\sigma}}}u_{ij}^{L^{\sigma}}(t)d\lambda(t)$
and $\hat{g}(\sigma)(\theta_{j},\theta_{i})$ = $a_{ij}^{\sigma}.$
Thus
$\hat{g}=\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\sum_{i,j=1}^{\infty}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}$.

It is well known that $L_{1}(G,\lambda,\mathcal{A})\cap
L_{2}(G,\lambda,\mathcal{A})$ is dense in $L_{2}(G,\lambda,\mathcal{A})$.
Then, for every $f\in L_{2}(G,\lambda,\mathcal{A})$ there exists a sequence
$(f_{n})$ in $L_{1}(G,\lambda,\mathcal{A})\cap L_{2}(G,\lambda,\mathcal{A})$
such
that $f_{n}\xrightarrow[n\to\infty]{}f$ in $L_{2}(G,\lambda,\mathcal{A})$.
$\hat{f_{n}}=\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\displaystyle\sum_{i,j=1}^{\infty}\mbox{
}^{n}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}$ where
${}^{n}a_{ij}^{\sigma}=\hat{f_{n}}(\sigma)(\theta_{j},\theta_{i}).$ We have
$\hat{f_{n}}(\sigma)(\theta_{j},\theta_{i})\xrightarrow[n\to\infty]{}\hat{f}(\sigma)(\theta_{j},\theta_{i})$
ie there exists $a_{ij}^{\sigma}$ in $\mathcal{A}$ such that
${}^{n}a_{ij}^{\sigma}\xrightarrow[n\to\infty]{}a_{ij}^{\sigma}.$
Finally
$\hat{f}=\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\sum_{i,j=1}^{\infty}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}.$

According to the corollary 4 the mapping $f\longmapsto\hat{f}$ is injective;
then
$\hat{f}=\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\sum_{i,j=1}^{\infty}a_{ij}^{\sigma}\hat{u}_{ij}^{L^{\sigma}}$
i.e.
$\hat{f}=\displaystyle\widehat{\left(\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\sum_{i,j=1}^{\infty}a_{ij}^{\sigma}u_{ij}^{L^{\sigma}}\right)}$
$\Longrightarrow
f=\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\sum_{i,j=1}^{\infty}a_{ij}^{\sigma}u_{ij}^{L^{\sigma}}.$
∎

######  Corollary 5.

For every $f\in L_{2}(G,\mathcal{A})$, we have

$\displaystyle f$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{\sigma\in\Sigma}d_{\sigma}\displaystyle\sum_{i,j=1}^{\infty}\hat{f}(\theta_{j},\theta_{i})u_{ij}^{L^{\sigma}}.$
(35)

##  4\. Concluding remarks

In this work, we have developed a formalism for the Fourier-Stieljes transform
on a locally compact group. More specifically, we have:

  * •

constructed the Fourier-Stieljes transform of a bounded vector measure on a
locally compact group into a Banach algebra,

  * •

derived the Fourier transform of an integrable function, and

  * •

discussed corresponding relevant properties including linearilty, injectivity,
and continuity.

Also, using the induced representation and its consequences, we established
the associated Shur’s orthogonality relation on a locally compact group.

## References

  * [1] N. Dinculeanu,  _Integration on locally compact groups_ ,  Noordhoff International Publishing, 1974.
  * [2] N. Bourbaki,  _Integration, Chap. 6_ ,  Hermann, Paris, 1959.
  * [3] V.S.K Assiamoua,  $L_{1}(G,A)$-multipliers,  _Acta. Sci. Math_ , 53 (1989), 309–318.
  * [4] J.A Clarkson,  _Abstrakte funktionen und lineare lperatoren_ ,  Trans. Amer. Math. Soc, 40, 396-414, 1936.
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Received xxxx 20xx; revised xxxx 20xx.