3750

# On Fourier algebra of a hypergroup constructed from a conditional
expectation on a locally compact group

A. A. Kalyuzhnyi  Institute of Mathematics, National Academy of Sciences of
Ukraine, 3 Tereshchenkivs’ka, Kyiv, 01601, Ukraine<EMAIL_ADDRESS>G. B.
Podkolzin  National Technical University of Ukraine ”Kyiv Polytechnic
Institute”, 37 Prospect Peremogy, Kyiv, 03056, Ukraine
<EMAIL_ADDRESS>Yu. A. Chapovsky  National Technical University of
Ukraine ”Kyiv Polytechnic Institute”, 37 Prospect Peremogy, Kyiv, 03056,
Ukraine<EMAIL_ADDRESS>
(Date: 13/11/2016; Revised 28/12/2016)

###### Abstract.

We prove that the Fourier space of a hypergroup constructed from a conditional
expectation on a locally compact group has a Banach algebra structure.

###### Key words and phrases:

Hypergroup, Fourier algebra, conditional expextation, comultiplication.

###### 2000 Mathematics Subject Classification:

43A62

##  1\. Introduction

It is well known [4] that, for a locally compact group $G$, the Fourier-
Stieltjes space $\mathscr{B}(G)$ and the Fourier space $\mathscr{A}(G)$, make
commutative Banach algebras.

For a hypercomplex system [2] or a DJS-hypergroup [1], one can also define the
Fourier-Stieltjes and the Fourier spaces. It is well known that, being Banach
spaces, they do not, in general, make Banach algebras. Conditions for this to
be the case were considered in [11, 12] for commutative hypergroups and
special type hypergroups called ultraspherical. In [9], a general sufficient
condition for the Fourier and the Fourier-Stieltjes spaces of a locally
compact hypergroup to be Banach algebras was obtained. It was also proved
there that there exists the Fourier Banach algebra on the double coset
hypergroup constructed from a locally compact group.

In this paper we prove that the Fourier space of a DJS-hypergroup that is
constructed from a locally compact group and an expectation, see [6, 8], also
has a Banach algebra structure. This result is a generalization of the one
obtained in [12], where the author imposes additional conditions on the
hypergroup.

The paper is organized as follows. In Section 2, we recall a traditional
definition of a DJS-hypergroup $Q$ in terms of the convolution algebra
structure on bounded measures on $Q$. Then, in Propositions 1 and 2, we give
an equivalent form for a definition of a DJS-hypergroup in a dual setting in
terms of a comultiplication on the $C^{*}$-algebra of continuous functions on
$Q$ vanishing at infinity and its multiplier algebra of continuous bounded
functions on $Q$. This turns out to be more convenient for purposes of this
paper.

Section 3 gives a main result of the paper. In Theorem 1, we describe
construction of a DJS-hypergroup from a conditional expectation on a locally
compact group. It follows from proofs that the construction of a conditional
expectation is similar to the construction of orbital morphisms [5], except
that the corresponding orbital mapping need not be open. Using results from
[9] we prove in Theorem 2 the main result that the Fourier space of the
hypergroup constructed from a conditional expectation on a locally compact
hypergroup is a Banach algebra.

##  2\. Preliminary

###  2.1. Hypergroups

Let $Q$ be a Hausdorff locally compact topological space.

The linear space of complex-valued continuous functions (resp., bounded
continuous functions, continuous functions approaching zero at infinity) is
denoted by $\mathscr{C}(Q)$ (resp., $\mathscr{C}_{b}(Q)$,
$\mathscr{C}_{0}(Q)$). The space $\mathscr{C}_{b}(Q)$ is endowed with the norm
$\|f\|_{\infty}=\sup_{t\in Q}|f(t)|$. Support of a function
$f\in\mathscr{C}(Q)$ is denoted by $\mathrm{supp}\,(f)$. For an open subset
$U$ of $Q$, $\mathscr{K}(U)$ denotes the subspace of functions
$f\in\mathscr{C}(Q)$ such that $\mathrm{supp}\,(f)\subset U$, and
$\mathscr{K}_{+}(U)$ (resp., $\mathscr{K}_{+}^{*}(U)$) denotes the subset of
functions $f\in\mathscr{K}(U)$ such that $f\geq 0$ (resp., $f>0$). For $p\in
U$, we denote by $\mathscr{K}_{+}^{p}(U)$ the subset of $\mathscr{K}_{+}(U)$
such that $f(p)>0$. By $1_{Q}$, we denote the constant function, $1_{Q}(s)=1$
for all $s\in Q$.

A measure is understood as a complex Radon measure [3] on $Q$. The linear
space of complex Radon measures, over the field $\mathbb{C}$ of complex
numbers, is denoted by $\mathscr{M}(Q)$. The subspace of $\mathscr{M}(Q)$ of
bounded (resp., compactly supported) measures is denoted by
$\mathscr{M}_{b}(Q)$ (resp., $\mathscr{M}_{c}(Q)$) and, for
$\mu\in\mathscr{M}_{b}(Q)$, its norm is $\|\mu\|=\sup_{f\in\mathscr{K}(Q),\
\|f\|_{\infty}\leq 1}|\mu(f)|$. The subset of $\mathscr{M}(Q)$ of nonnegative
(resp., probability) measures is denoted by $\mathscr{M}_{+}(Q)$ (resp.,
$\mathscr{M}_{p}(Q)$). For a measure $\mu\in\mathscr{M}_{+}(Q)$, its support
is denoted by $\mathrm{supp}\,(\mu)$. If
$\mu\in\mathscr{M}_{+}(Q)\cap\mathscr{M}_{b}(Q)$, then $\|\mu\|=\mu(1_{Q})$.
The Dirac measure at a point $s\in Q$ is denoted by $\varepsilon_{s}$. The
integral of $f\in\mathscr{K}(F)$ with respect to a measure $\mu\in\mathscr{M}$
is denoted by
$\mu(f)=\langle\,f,\mu\,\rangle=\int_{F}\langle\,f,\varepsilon_{t}\,\rangle\,d\mu(t)=\int_{F}f(t)\,d\mu(t)$.

The set of all nonempty compact subsets of $Q$ is denoted by $\mathcal{K}$. We
consider $\mathcal{K}$ endowed with the Michael topology [10]. Recall that
this topology is generated by the subbasis of all
$\mathcal{U}_{V}(W)=\\{F\in\mathcal{K}:F\subset W\ \hbox{and}\ F\cap
V\neq\emptyset\\}$ for open subsets $V$ and $W$ of $Q$.

A (locally compact) hypergroup is a locally compact Hausdorff topological
space $Q$ such that $\mathscr{M}_{b}(Q)$ is endowed with a multiplication,
called convolution and denoted by $*$, satisfying the following conditions
[1]:

  * $(H_{1})$

$(\mathscr{M}_{b}(Q),*)$ is an associative algebra over $\mathbb{C}$.

  * $(H_{2})$

For all $s,t\in Q$, $\varepsilon_{s}*\varepsilon_{t}\in\mathscr{M}_{p}(Q)$ and
$\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})$ is compact.

  * $(H_{3})$

The mapping $(s,t)\mapsto\varepsilon_{s}*\varepsilon_{t}$ of $Q\times Q$ into
$\mathscr{M}_{p}(Q)$ is continuous with respect to the weak topology
$\sigma(\mathscr{M}_{p}(Q),\mathscr{C}_{0}(Q))$ on $\mathscr{M}_{p}(Q)$.

  * $(H_{4})$

The mapping $(s,t)\mapsto\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})$ of
$Q\times Q$ into $\mathcal{K}$ is continuous with respect to the Michael
topology on $\mathcal{K}$.

  * $(H_{5})$

There exists a (necessarily unique) element $e\in Q$ such that
$\varepsilon_{e}*\varepsilon_{s}=\varepsilon_{s}*\varepsilon_{e}=\varepsilon_{s}$
for all $s\in Q$.

  * $(H_{6})$

There exists a (necessarily unique) homeomorphism $s\mapsto\check{s}$ of $Q$
into $Q$ such that $\check{\check{s}}=s$ and
$(\varepsilon_{s}*\varepsilon_{t})\,\check{}=\varepsilon_{\check{t}}*\varepsilon_{\check{s}}$,
where $\check{\mu}$ denotes the image of the measure $\mu$ with respect to the
homeomorphism $s\mapsto\check{s}$, i.e.,
$\langle\,f,\check{\mu}\,\rangle=\langle\,\check{f},\mu\,\rangle$, where
$\check{f}(s)=f(\check{s})$.

  * $(H_{7})$

For $s,t\in Q$, $e\in\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})$ if and
only if $s=\check{t}$.

For a measure $\mu\in\mathscr{M}(Q)$ and $h\in\mathscr{C}(Q)$, the measure
$h\mu$ is defined by $\langle\,f,h\mu\,\rangle=\langle\,fh,\mu\,\rangle$ for
$f\in\mathscr{K}(Q)$; it is clear that $h\mu\in\mathscr{M}_{c}(Q)$ for
$h\in\mathscr{K}(Q)$.

Everywhere in the sequel, we assume that the hypergroup possesses a left
invariant measure, denoted by $m$, which means that

$\varepsilon_{s}*m=m$

for all $s\in Q$. The Banach space $L_{p}(Q,m)$, $1\leq p\leq\infty$, is
denoted by $L_{p}(Q)$.

For $\mu\in\mathscr{M}_{b}(Q)$, denote by $\mu^{\star}$ the bounded measure
defined by $\mu^{\star}(f)=\overline{\check{\mu}(\bar{f})}$ for
$f\in\mathscr{K}(Q)$. It follows from the axiom $(H_{6})$ of a hypergroup that
⋆ is an involution on the algebra $(\mathscr{M}_{b}(Q),*)$. It is well known
[1] that $(\mathscr{M}_{b}(Q),*,{}^{\star})$ is an involutive Banach algebra.

We denote $L_{1}(m)=\\{fm:f\in L_{1}(Q)\\}$. It is well known that
$(L_{1}(m),*,{}^{\star})$ is a closed two-sided ideal of
$(\mathscr{M}_{b}(Q),*,{}^{\star})$.

Identifying each $f\in L_{1}(Q)$ with the measure $fm\in L_{1}(m)$ yields an
involutive Banach algebra structure on $L_{1}(Q)$, denoted by
$(L_{1}(Q),*,{}^{\star})$, where

$\displaystyle(f*g)(s)$ $\displaystyle=$
$\displaystyle\int_{Q}f(t)\langle\,g,\check{\varepsilon}_{t}*\varepsilon_{t}\,\rangle\,dm(t),$
$\displaystyle f^{\star}(s)$ $\displaystyle=$
$\displaystyle\kappa^{-1}(s)\overline{f}(\check{s}),$

where $\kappa:Q\rightarrow\mathbf{R}$ is the modular function,
$m*\varepsilon_{s}=\Delta(s)m$, see [1]. Also denote by

$f^{\dagger}(s)=\overline{f(\check{s})}.$

It is wellknown that a function $f\ast f^{\dagger}$ is a positive definite on
hypergroup $Q$, see [1].

Let $(Q,*,e,\check{}\,)$ be a locally compact hypergroup. Define a linear map
$\tilde{\Delta}$, homomorphisms $\epsilon$ and $\,\check{}\,$ as follows:

$\begin{array}[]{c}\tilde{\Delta}\colon\mathscr{C}_{b}(Q)\rightarrow\mathscr{C}_{b}(Q\times
Q),\qquad(\tilde{\Delta}f)(p,q)=\langle\,f,\varepsilon_{p}\ast\varepsilon_{q}\,\rangle,\\\\[4.2679pt]
\epsilon\colon\mathscr{C}_{b}(Q)\rightarrow\mathbb{C},\qquad\epsilon(f)=f(e),\\\\[4.2679pt]
\check{}\,\colon\mathscr{C}_{b}(Q)\rightarrow\mathscr{C}_{b}(Q),\qquad\check{f}(p)=f(\check{p}),\end{array}$
(1)

where $\mathscr{C}_{b}(Q)$ is considered as a commutative $C^{*}$-algebra.

The following lemma is clear.

###### Lemma 1

Let $Q$ be a locally compact space,
$\mu\in\mathscr{M}_{b}(Q)\cap\mathscr{M}_{+}(Q)$, and $F\subset Q$ is a closed
set. Then $F\cap\mathrm{supp}\,(\mu)=\emptyset$ if and only if there is
$f\in\mathscr{C}_{+}(Q)$, $\|f\|_{\infty}=1$, such that $f(r)=0$ for all $r\in
F$ and $\langle\,f,\mu\,\rangle=\|\mu\|$.

###### Proposition 1

Let $(Q,*,e,\check{}\,)$ be a locally compact hypergroup. Then the maps
defined by (1) have the following properties:

  * $(\widetilde{H}_{1})$

the map $\tilde{\Delta}$ is linear and coassociative, that is,

$(\tilde{\Delta}\times\mathrm{id})\circ\tilde{\Delta}=(\mathrm{id}\times\tilde{\Delta})\circ\tilde{\Delta};$

  * $(\widetilde{H}_{2})$

    * (a)

the map $\tilde{\Delta}$ is positive, that is,
$\tilde{\Delta}(\mathscr{C}_{b,+})\subset\mathscr{C}_{b,+}(Q\times Q)$, where
$\mathscr{C}_{b,+}(Q)$ (resp., $\mathscr{C}_{b,+}(Q\times Q)$) is the cone of
bounded nonnegative continuous functions on $Q$ (resp., $Q\times Q$),

    * (b)

$\tilde{\Delta}\,1_{Q}=1_{Q\times Q}$,

    * (c)

for any $s,t\in Q$ there is $f\in\mathscr{K}_{+}(Q)$ such that
$\|f\|_{\infty}\leq 1$ and $(\tilde{\Delta}\,f)(s,t)=1$;

  * $(\widetilde{H}_{4})$

if $F\subset Q$ is closed, $(s_{0},t_{0})\in Q\times Q$ and
$f_{0}\in\mathscr{K}_{+}(Q)$ is such that $\|f_{0}\|_{\infty}=1$, $f_{0}(r)=0$
for all $r\in F$ and $(\tilde{\Delta}\,f_{0})(s_{0},t_{0})=1$, then there
exist an open neighborhood $U\subset Q\times Q$ of $(s_{0},t_{0})$ and
$f\in\mathscr{K}_{+}(Q)$ such that $\|f\|_{\infty}=1$, $f(r)=0$ for all $r\in
F$ and $(\tilde{\Delta}f)(s,t)=1$ for all $(s,t)\in U$;

  * $(\widetilde{H}_{5})$

the homomorphism $\epsilon$ is a right and left counit, that is,

$(\epsilon\times\mathrm{id})\circ\tilde{\Delta}=(\mathrm{id}\times\epsilon)\circ\tilde{\Delta}=\mathrm{id};$
(2)

  * $(\widetilde{H}_{6})$

the homomorphism $\,\check{\,}$ satisfies
$\mathscr{K}(Q)\,\check{}\,\subset\mathscr{K}(Q)$ and the following
identities:

$\begin{array}[]{c}\check{}\,\circ\,\check{}\,=\mathrm{id},\\\\[4.2679pt]
\tilde{\Delta}\circ\,\check{}\,=\Sigma\circ(\,\check{}\,\times\,\check{}\,)\circ\tilde{\Delta},\end{array}$
(3)

where $\Sigma:\mathscr{C}_{b}(Q\times Q)\rightarrow\mathscr{C}_{b}(Q\times Q)$
is defined by $(\Sigma f)(p,q)=f(q,p)$;

  * $(\widetilde{H}_{7})$

if $\Gamma\,\check{}\,\subset Q\times Q$ is a graph of the homeomorphism
$\,\check{}\,$, that is,

$\Gamma\,\check{}\,=\\{(p,\check{p})\in Q\times Q:p\in Q\\},$

then

$\Gamma\,\check{}\,=\bigcap_{f\in\mathscr{K}_{+}^{e}(G)}\mathrm{supp}\,\bigl{(}\tilde{\Delta}(f)\bigr{)},$
(4)

where $\mathscr{K}_{+}^{e}(G)=\\{f\in\mathscr{K}_{+}(G):f(e)>0\\}$.

###### Proof.

First of all note that $\tilde{\Delta}f\in\mathscr{C}(Q\times Q)$ if
$f\in\mathscr{C}(Q)$, as directly follows from ($H_{3}$). It also immediately
follows from ($H_{2}$) that $\tilde{\Delta}$ is positive and
$\tilde{\Delta}\,1_{Q}=1_{Q\times Q}$. To see that $\tilde{\Delta}\,f$ is
bounded for bounded $f\in\mathscr{C}(Q)$, we have that
$-\|f\|_{\infty}\,1_{Q}\leq f\leq\|f\|_{\infty}\,1_{Q}$ and, hence,

$-\|f\|_{\infty}\,1_{Q\times
Q}\leq\tilde{\Delta}\,f\leq\|f\|_{\infty}\,1_{Q\times Q},$

which means that $\tilde{\Delta}\,f$ is bounded. It is also clear that
$\check{f}$ is continuous if $f$ is such, and that
$\|\check{f}\|_{\infty}=\|f\|_{\infty}$ . This means that the maps in (1) are
well defined.

To prove ($\widetilde{H}_{1}$), let $f\in\mathscr{C}_{b}(Q)$ and consider

$\bigl{(}(\tilde{\Delta}\times\mathrm{id})\circ\tilde{\Delta}\,f\bigr{)}(s_{1},s_{2},s_{3})=\langle\,\tilde{\Delta}\,f,(\varepsilon_{s_{1}}*\varepsilon_{s_{2}})\otimes\varepsilon_{s_{3}}\,\rangle=\langle\,f,\varepsilon_{s_{1}}*\varepsilon_{s_{2}}*\varepsilon_{s_{3}}\,\rangle.$

Similarly,

$\bigl{(}(\mathrm{id}\times\tilde{\Delta})\circ\tilde{\Delta}\,f\bigr{)}(s_{1},s_{2},s_{3})=\langle\,\tilde{\Delta}\,f,\varepsilon_{s_{1}}\otimes(\varepsilon_{s_{2}}*\varepsilon_{s_{3}})\,\rangle=\langle\,f,\varepsilon_{s_{1}}*\varepsilon_{s_{2}}*\varepsilon_{s_{3}}\,\rangle.$

It has already been mentioned, that all statements in ($\widetilde{H}_{2}$)
except for (c) hold true. To prove (c) we use compactness of
$F=\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})$ in ($H_{2}$) a function
$f\in\mathscr{K}_{+}(Q)$ such that $f(p)=1$ for all $p\in F$. Then

$\displaystyle(\tilde{\Delta}\,f)(s,t)$
$\displaystyle=\langle\,f,\varepsilon_{s}*\varepsilon_{t}\,\rangle=\int_{Q}f(p)\,d(\varepsilon_{s}*\varepsilon_{t})(p)=\int_{F}f(p)\,d(\varepsilon_{s}*\varepsilon_{t})(p)$
$\displaystyle=\int_{F}d(\varepsilon_{s}*\varepsilon_{t})(p)=\langle\,1_{Q},\varepsilon_{s}*\varepsilon_{t}\,\rangle$
$\displaystyle=\langle\,\tilde{\Delta}\,1_{Q},\varepsilon_{s}\otimes\varepsilon_{t}\,\rangle=1_{Q\times
Q}(s,t)=1.$

Let us prove ($\widetilde{H}_{4}$). Denote
$E_{0}=\mathrm{supp}\,(\varepsilon_{s_{0}}*\varepsilon_{t_{0}})$. If $F$ is
closed and $f_{0}$ is such that $\|f_{0}\|_{\infty}=1$, $f_{0}(r)=0$ for all
$r\in F$ and
$(\tilde{\Delta}\,f_{0})(s_{0},t_{0})=\langle\,f_{0},\varepsilon_{s_{0}}*\varepsilon_{t_{0}}\,\rangle=1$,
then it follows from Lemma 1 that $F\cap E_{0}=\emptyset$, hence $E_{0}$ is a
subset of the open set $Q\setminus F$. Let $V\subset Q$ be an open
neighborhood of $E_{0}$ such that $\overline{V}\subset Q\setminus F$. By
($H_{4}$) there is an open neighborhood $U\subset Q\times Q$ of
$(s_{0},t_{0})$ such that
$\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})\subset V$ for all $(s,t)\in
U$. Let $f\in\mathscr{C}_{+}(Q)$, $0\leq f(r)\leq 1$, be such that $f(r)=0$
for all $r\in F$ and $f(r)=1$ for all $r\in\overline{V}$. Then, for any
$(s,t)\in U$, we have

$(\tilde{\Delta}\,f)(s,t)=\langle\,f,\varepsilon_{s}*\varepsilon_{t}\,\rangle=\langle\,1_{Q},\varepsilon_{s}*\varepsilon_{t}\,\rangle=1.$

It is easy to see that ($\widetilde{H}_{5}$) holds. Indeed. let
$f\in\mathscr{C}_{b}(Q)$ and $s\in Q$. Then

$\bigl{(}(\epsilon\times\mathrm{id})\circ\tilde{\Delta}\,f\bigr{)}(s)=\langle\,f,\varepsilon_{e}*\varepsilon_{s}\,\rangle=\langle\,f,\varepsilon_{s}\,\rangle=f(s).$

The second part of (2) is proved similarly.

Property ($\widetilde{H}_{6}$) is a direct consequence of ($H_{6}$).

Let us prove ($\widetilde{H}_{7}$). Let $f\in\mathscr{K}_{+}^{e}(G)$. Then

$(\tilde{\Delta}\,f)(s,\check{s})=\langle\,f,\varepsilon_{s}*\varepsilon_{\check{s}}\,\rangle>0,$

since $e\in\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{s}})$ by
($H_{7}$). Thus $(s,\check{s})\in\mathrm{supp}\,(\tilde{\Delta}\,f)$ and

$\Gamma\,\check{}\,\subset\bigcap_{f\in\mathscr{K}_{+}^{e}(G)}S(\tilde{\Delta}\,f).$

Conversely, let $s_{0}\neq t_{0}$, hence
$(s_{0},\check{t}_{0})\notin\Gamma\,\check{}\,$. By ($H_{7}$),
$e\notin\mathrm{supp}\,(\varepsilon_{s_{0}}*\varepsilon_{\check{t}_{0}})$.
Since $\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{t}})$ is compact
there are open neighborhoods $U$ of $e$ and $V$ of
$\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{t}})$ such that $U\cap
V=\emptyset$. Let $f\in\mathscr{K}_{+}(U)$ be such that $f(e)=1$. Hence
$f\in\mathscr{K}_{+}^{e}(G)$ and, as it follows from ($H_{4}$), there is an
open neighborhood $W\subset Q\times Q$ of $(s_{0},\check{t}_{0})$ such that
$\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{t}})\subset V$ for all
$(s,\check{t})\in W$, hence
$\mathrm{supp}\,(f)\cap\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{t}})=\emptyset$
for such $(s,\check{t})$. Thus

$(\tilde{\Delta}\,f)(s,\check{t})=\langle\,f,\varepsilon_{s}*\varepsilon_{\check{t}}\,\rangle=0,$

and $(s_{0},\check{t}_{0})\notin\mathrm{supp}\,(\tilde{\Delta}\,f)$. This
means that

$(s,\check{t})\notin\bigcap_{f\in\mathscr{K}_{+}^{e}(G)}\mathrm{supp}\,(\tilde{\Delta}\,f),\quad(s,\check{t})\in
W,$

which finishes the proof. ∎

######  Remark 1.

It immediately follows from ($\widetilde{H}_{2}$) (a) and
($\widetilde{H}_{2}$) (b) that ($\widetilde{H}_{2}$) (c) is equivalent to the
condition that there is $f\in\mathscr{K}_{+}(Q)$ such that
$\tilde{\Delta}\,f=\|\tilde{\Delta}\|$.

###### Proposition 2

Let $Q$ be a Hausdorff locally compact space, and let a linear map
$\tilde{\Delta}\colon\mathscr{C}_{b}(Q)\rightarrow\mathscr{C}_{b}(Q\times Q)$,
$*$-homomorphisms
$\,\check{}\,\colon\mathscr{C}_{b}(Q)\rightarrow\mathscr{C}_{b}(Q)$ and
$\epsilon\colon\mathscr{C}_{b}(Q)\rightarrow\mathbb{C}$ be given and satisfy
the properties ($\widetilde{H}_{1}$) – ($\widetilde{H}_{7}$) in Proposition 1.
For $\mu_{1},\mu_{2},\mu\in\mathscr{M}_{b}(Q)$, define
$\mu_{1}*\mu_{2}\in\mathscr{M}_{b}(Q)$ and $\check{\mu}\in\mathscr{M}_{b}(Q)$
by

$\begin{array}[]{rcl}\langle\,f,\mu_{1}*\mu_{2}\,\rangle&=&\displaystyle\int_{Q\times
Q}(\tilde{\Delta}f)(s_{1},s_{2})\,d\mu_{1}(s_{1})d\mu_{2}(s_{2}),\\\\[11.38109pt]
\langle\,f,\check{\mu}\,\rangle&=&\langle\,\check{f},\mu\,\rangle,\par\end{array}$
(5)

where $f\in\mathscr{K}(Q)$, and let $e=\mathrm{supp}\,(\epsilon)$. Then
$(Q,*,e,\,\check{}\,)$ is a locally compact hypergroup.

###### Proof.

First consider ($H_{1}$) and show that $(\mathscr{M}_{b}(Q),*)$ is indeed an
associative algebra, if the convolution of measures is given by the first
formula in (5).

If $\mu_{1},\mu_{2}\in\mathscr{M}_{b}(Q)$, then the measures $|\mu_{1}|$ and
$|\mu_{2}|$ are also bounded. Moreover, by (5) and using
($\widetilde{H}_{2}$), we have

$\displaystyle|\langle\,f,\mu_{1}*\mu_{2}\,\rangle|$
$\displaystyle=\bigl{|}\int_{Q\times
Q}(\tilde{\Delta}\,f)(s_{1},s_{2})\,d\mu_{1}(s_{1})d\mu_{2}(s_{2})\bigr{|}$
$\displaystyle\leq\int_{Q\times
Q}(\tilde{\Delta}\bigl{|}f|)(s_{1},s_{2})\,d|\mu_{1}|(s_{1})d|\mu_{2}|(s_{2})$
$\displaystyle\leq\int_{Q\times
Q}\tilde{\Delta}\bigl{(}\|f\|_{\infty}\,1_{Q}\bigr{)}(s_{1},s_{2})\,d|\mu_{1}|(s_{1})d|m_{2}|(s_{2})$
$\displaystyle=\|f\|_{\infty}\int_{Q\times
Q}1_{Q}(s_{1})1_{Q}(s_{2})\,d|\mu_{1}|(s_{1})d|\mu_{2}|(s_{2})$
$\displaystyle=\|f\|_{\infty}\,\|\mu_{1}\|\,\|\mu_{2}\|,$

which shows that $\|\mu_{1}*\mu_{2}\|\leq\|\mu_{1}\|\,\|\mu_{2}\|$, hence
$\mu_{1}*\mu_{2}\in\mathscr{M}_{b}(Q)$.

Associativity of the algebra $(\mathscr{M}_{b}(Q),*)$ is immediately implied
by ($\widetilde{H}_{1}$).

Consider ($H_{2}$). Since $\tilde{\Delta}$ is positive by
($\widetilde{H}_{2}$), the measure $\varepsilon_{s}*\varepsilon_{t}$ is
nonnegative for $s,t\in Q$. Moreover,

$\langle\,1_{Q},\varepsilon_{s}*\varepsilon_{t}\,\rangle=\langle\,1_{Q}\otimes
1_{Q},\varepsilon_{s}\otimes\varepsilon_{t}\,\rangle=1,$

hence $\varepsilon_{s}*\varepsilon_{t}\in\mathscr{M}_{p}(Q)$.

Let us show that $\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})$ is compact
for any $s,t\in Q$. Indeed, using ($\widetilde{H}_{2}$) (c) there is
$f\in\mathscr{K}_{+}(Q)$ such that $\|f\|_{\infty}\leq 1$ and
$(\tilde{\Delta}\,f)(s,t)=1$. This means that
$\langle\,f,\varepsilon_{s}*\varepsilon_{t}\,\rangle=1$. Denote
$F=\mathrm{supp}\,(f)$ and let $g\in\mathscr{K}_{+}(Q)$ be arbitrary such that
$\|g\|_{\infty}\leq 1$ and $\mathrm{supp}\,(g)\subset Q\setminus F$. Then
$f+g\leq 1_{Q}$, that is, $g\leq 1_{Q}-f$, and

$0\leq\langle\,g,\varepsilon_{s}*\varepsilon_{t}\,\rangle\leq\langle\,1_{Q}-f,\varepsilon_{s}*\varepsilon_{t}\,\rangle=\langle\,1_{Q},\varepsilon_{s}*\varepsilon_{t}\,\rangle-\langle\,f,\varepsilon_{s}*\varepsilon_{t}\,\rangle=0.$

This shows that $\langle\,g,\varepsilon_{s}*\varepsilon_{t}\,\rangle=0$ for
any such $g$, which means that
$\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})\subset F$. Since $F$ is
compact and support of a measure is closed,
$\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})$ is compact.

Property ($H_{3}$) is immediate, since
$\tilde{\Delta}\,f\in\mathscr{C}_{b}(Q\times Q)$ by definition for any
$f\in\mathscr{C}_{0}(Q)$.

Consider ($H_{4}$). Let $(s_{0},t_{0})\in Q\times Q$, and denote
$F_{0}=\mathrm{supp}\,(\varepsilon_{s_{0}}*\varepsilon_{t_{0}})$, which is a
compact set by ($H_{2}$). Let $V,W\subset Q$ be open such that
$\mathcal{U}_{V}(W)$ is an open neighborhood of $F_{0}$ in the Michael
topology, that is, $F_{0}\subset W$ and $F_{0}\cap V\neq\emptyset$. Choose
$p\in F_{0}\cap V$ and let $f\in\mathscr{K}_{+}^{p}(V)$. Then, since $p\in
F_{0}$,

$(\tilde{\Delta}\,f)(s_{0},t_{0})=\langle\,f,\varepsilon_{s_{0}}*\varepsilon_{t_{0}}\,\rangle>0,$

and, by continuity of $\tilde{\Delta}\,f$, we can find a neighborhood
$U_{1}\subset Q\times Q$ of $(s_{0},t_{0})$ such that
$(\tilde{\Delta}\,f)(s,t)>0$ for all $(s,t)\in U_{1}$. This would imply that
$\langle\,f,\varepsilon_{s}*\varepsilon_{t}\,\rangle>0$, hence
$\mathrm{supp}\,(f)\cap\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})\neq\emptyset$,
in particular,
$V\cap\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})\neq\emptyset$ for all
$(s,t)\in U_{1}$.

To proceed, let $W\subset Q$ be open and
$\mathrm{supp}\,(\varepsilon_{s_{0}}*\varepsilon_{t_{0}})\subset W$. Let
$F=Q\setminus\mathrm{supp}\,(\varepsilon_{s_{0}}*\varepsilon_{t_{0}})$. Then
there is a function $f_{0}\in\mathscr{C}_{+}(Q)$, $0\leq f(r)\leq 1$, $r\in
Q$, such that $f(r)=0$ for $r\in F$ and $f(r)=1$ for
$r\in\mathrm{supp}\,(\varepsilon_{s_{0}}*\varepsilon_{t_{0}})$. We have
$\langle\,f_{0},\varepsilon_{s_{0}}*\varepsilon_{t_{0}}\,\rangle=\langle\,1_{Q},\varepsilon_{s_{0}}*\varepsilon_{t_{0}}\,\rangle=1=\|\varepsilon_{s_{0}}*\varepsilon_{t_{0}}\|$.
Hence, there is a neighborhood $U\subset Q\times Q$ of the point
$(s_{0},t_{0})$ and a function $f\in\mathscr{C}_{+}$, $\|f\|_{\infty}=1$, such
that $f(r)=0$ for all $r\in F$ and
$\langle\,f,\varepsilon_{s}*\varepsilon_{t}\,\rangle=1$ for all $(s,t)\in U$.
By Lemma 1, $F\cap\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})=\emptyset$,
hence $\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{t})\subset W$ for all
$(s,t)\in U$.

($H_{5}$) (resp., ($H_{6}$)) is immediately implied by ($\widetilde{H}_{5}$)
(resp., ($\widetilde{H}_{6}$).

Let us prove ($H_{7}$). Let $s\neq t$ and show that
$e\notin\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{t}})$. Indeed,
$(s,\check{t})\notin\Gamma\,\check{}\,$, hence there exists
$f\in\mathscr{K}_{+}^{e}(G)$ such that
$(s,\check{t})\notin\mathrm{supp}\,(\tilde{\Delta}\,f)$. This means that

$(\tilde{\Delta}\,f)(s,\check{t})=\langle\,f,\varepsilon_{s}*\varepsilon_{\check{t}}\,\rangle=0.$

This means that
$f\restriction_{\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{t}})}=0$
but $f(e)>0$. Hence
$e\notin\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{t}})$.

Conversely, let us show that
$e\in\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{s}})$ for any $s\in
Q$. Indeed, if $U$ is an arbitrary open neighborhood of $e$, choose a function
$f\in\mathscr{K}_{+}^{e}(U)$. Since $(s,\check{s})\in\Gamma\,\check{}\,$, it
follows from (4) that $(s,\check{s})\in\mathrm{supp}\,(f)$, which means that

$(\tilde{\Delta}\,f)(s,\check{s})=\langle\,f,\varepsilon_{s}*\varepsilon_{\check{s}}\,\rangle>0,$

and $e\in\mathrm{supp}\,(\varepsilon_{s}*\varepsilon_{\check{s}})$. ∎

###  2.2. Fourier and Fourier-Stieltjes algebras

######  Definition 1.

Let $Q$ be a locally compact hypergroup and $H$ a Hilbert space. A
representation $\pi$ of the hypergroup $Q$ on the Hilbert space $H$ is an
involution homomorphism $\pi:\mathscr{M}_{b}(Q)\to B(H)$ of the involution
algebra $(\mathscr{M}_{b}(Q),*,{}^{*})$ into the $C^{\ast}$-algebra $B(H)$ of
all linear bounded operators on $H$. The set of all representations of $Q$
will be denoted by $\Sigma$.

Let us remark that definition 1 can be rewritten in terms of the
comultiplication in (1) as follows [7]: a weakly continuous map $\pi:Q\to
B(H)$ is ca;;ed a representation of the hypergroup $Q$ if the following
conditions are satisfied:

  * (i)

$\pi(e)=I$, where $I$ is the identity operator on $H$;

  * (ii)

$\pi(p)^{\ast}=\pi(p^{\ast})$ for all $p\in Q$;

  * (iii)

for any $\xi,\eta\in H$, we have
$\tilde{\Delta}(\pi(\cdot)\xi,\eta)_{H}(p,q)=(\pi(p)\pi(q)\xi,\eta)_{H}$.

######  Definition 2.

Let $\tilde{m}$ be a left Haar measure on the hypergroup $Q$. A left (resp.,
right) regular representation is a mapping $\lambda:Q\ni p\mapsto L_{p}\in
B(L_{2}(Q))$ (resp., $\mu:Q\ni p\mapsto R_{p}\in B(L^{2}(Q))$) defined, for
$f\in\mathscr{K}(Q)$, by $(L_{p}f)(q)=(\tilde{\Delta}f)(p^{\ast},q)$ (resp.,
$(R_{p}f)(q)=(\tilde{\Delta}f)(p,q)\kappa^{\frac{1}{2}}(p)$) and then extended
to $L^{2}(Q)$ by continuity in virtue of the estimates
$\|L_{p}f\|_{2}\leq\|f\|_{2}$ and $\|R_{p}f\|_{2}\leq\|f\|_{2}$ [7].

Let us recall that each representation of a hypergroup can be uniquely
continued to a nondegenerate representation of the Banach $*$-algebra
$L^{1}(Q)$ and, conversely, each nondegenerate representation of the
$\ast$-algebra $L^{1}(Q)$ gives rise to a representation of the hypergroup $Q$
[7]. By $C^{\ast}(Q)$, we denote the full $C^{\ast}$-algebra of the hypergroup
$Q$, that is the closure of $L^{1}(Q)$ with respect to the $C^{*}$-norm
$\|f\|=\mbox{sup}_{\pi\in\Sigma}\pi(f)$.

The reduced $C^{*}$-algebra of the hypergroup $Q$ is denoted by
$C^{\ast}_{r}(Q)$ that is the $C^{*}$-closure of the $*$-algebra generated by
the family $\\{L_{p}:p\in Q\\}$ of operators on $L^{2}(Q)$ with respect to the
norm of $B(L^{2}(Q))$.

######  Definition 3.

The Banach space dual to the full $C^{*}$-algebra $C^{*}(Q)$ will be called
the Fourier-Stieltjes space and denoted by $\mathscr{B}(Q)$. The Banach space
dual to the reduced $C^{*}$-algebra of the hypergroup $Q$, $C^{*}_{r}(Q)$,
will be denoted by $\mathscr{B}_{\lambda}(Q)$.

######  Definition 4.

Let $(H_{1},\pi_{1})$ and $(H_{2},\pi_{2})$ be two representations of a
hypergroup $Q$. The representation $\pi_{1}$ is said to be weakly contained in
the representation $\pi_{2}$ if the kernel of the representation $\pi_{1}$
contains the kernel of the representation $\pi_{2}$.

It is well known, see [11] that for any $\alpha\in\mathscr{B}(Q)$
(respectively, $\alpha\in\mathscr{B}_{\lambda}(Q)$) there is a representation
$(H,\pi)$ of the hypergroup $Q$ (respectively, a representation $(H,\pi)$
weakly contained in the left regular representation) and two vectors
$\xi,\eta\in H$ such that the function $a\in C_{b}(Q)$ given by

$a(p)=(\pi(p)\xi,\eta)_{H},\quad p\in Q,$

defines a linear functional $\alpha$ by

$\alpha(f)=\int_{Q}a(p)f(p)dm(p),\quad f\in L_{1}(Q),$

such that its norm is given by

$\|\alpha\|=\sup_{f\in
L_{1}(Q),\|f\|_{\Sigma}^{\prime}=1}\left|\int_{Q}a(p)f(p)dm(p)\right|=\|\xi\|_{H}\|\eta\|_{H},$

where $\Sigma^{\prime}=\Sigma$ (respectively, $\Sigma^{\prime}=\lambda$).

######  Definition 5.

The Fourier space of a hypergroup $Q$ will be called the closure of the space
generated by the elements $f\ast f^{\dagger}$, $f\in\mathscr{K}(Q)$ with
respect to the norm of the space $\mathscr{B}_{\lambda}(Q)$, and denoted by
$\mathscr{A}(Q)$.

###  2.3. Tensor product and conditional expectation

Let $A$ and $B$ be $C^{*}$-algebras. The tensor product of $A$ and $B$,
denoted by $A\otimes B$, is the completion of the algebraic tensor product
$A\odot B$ with respect to the min-$C^{*}$-norm on $A\odot B$,

$\|\sum_{i=1}^{n}a_{i}\otimes
b_{i}\|_{\textup{min}}=\sup_{\pi_{A}\in\Sigma_{A},\pi_{B}\in\Sigma_{B}}\Bigl{\|}\sum_{i=1}^{n}\pi_{A}(a_{i})\otimes\pi_{B}(b_{i})\Bigr{\|},$

where $\Sigma_{A}$ (resp., $\Sigma_{B}$) is the set of all representations of
the $C^{*}$-algebra $A$ (resp., $B$) [14].

For a $C^{*}$-algebra $A$, the $C^{*}$-algebra of multipliers of $A$ is
denoted by $M(A)$, see [14] for details.

Let $A$ be a $C^{*}$-algebra and $B\subset A$ a $C^{*}$-subalgebra of $A$. A
bounded linear map $P\colon A\rightarrow B$ is called a conditional
expectation if it satisfies the following properties [16]:

  * (i)

$P$ is a projection onto and has norm $1$, that is, $P^{2}=P$ and $\|P\|=1$;

  * (ii)

$P$ is positive, that is $P(a^{*}a)\geq 0$ for any $a\in A$;

  * (iii)

$P(b_{1}ab_{2})=b_{1}P(a)b_{2}$ for any $a\in A$ and $b_{1},b_{2}\in B$;

  * (iv)

$P(a^{*})P(a)\leq P(a^{*}a)$ for all $a\in A$.

It follows from (ii) and the polarization identity that

  * (v)

$P(a^{*})=P(a)^{*}$, $a\in A$.

It also follows from [15] that (i) implies (ii), (iii), and (iv).

##  3\. Main results

Let $G$ be a locally compact group with a Haar measure $m$, $A$ denote the
$C^{*}$-algebra $\mathscr{C}_{b}(G)$, $A_{0}$ its $C^{*}$-subalgebra
$\mathscr{C}_{0}(G)$, and let $\mathscr{K}(G)$ be the ideal of $A$ consisting
of functions with compact supports. Let $P\colon A\rightarrow A$ be a
conditional expectation. The restriction of $P$ to $\mathscr{C}_{0}(G)$ is
still denoted by $P$.

###### Proposition 3

Let $B$ be a $C^{*}$-subalgebra of $\mathscr{C}_{0}(G)$,
$P\colon\mathscr{C}_{0}(G)\rightarrow B$ be a conditional expectation. Let $Q$
denote the spectrum of the commutative $C^{*}$-algebra $B$, and identify $B$
with $\mathscr{C}_{0}(Q)$ via the Gelfand transform. Let
$\iota\colon\mathscr{C}_{0}(Q)\cong B\rightarrow A$ denote the embedding map.
Then we have the following.

  1. (i)

There is a unique continuous surjection $\varphi\colon G\rightarrow Q$ such
that $\iota(\tilde{f})(p)=\tilde{f}(\varphi(p))$ for any
$\tilde{f}\in\mathscr{C}_{0}(Q)$ and any $p\in G$. If $p,q\in G$ are such that
$\varphi(p)=\varphi(q)$, then $\iota(\tilde{f})(p)=\iota(\tilde{f})(q)$.

  2. (ii)

Denote by $P^{*}\colon\mathscr{M}_{b}(Q)\rightarrow\mathscr{M}_{b}(G)$ the
linear map adjoint to $P$,

$\langle\,f,P^{*}(\tilde{\mu})\,\rangle=\langle\,P(f),\tilde{\mu}\,\rangle,\qquad\tilde{\mu}\in\mathscr{M}_{b}(Q),\quad
f\in\mathscr{C}_{b}(G),$

and let $\varphi_{*}\colon\mathscr{M}_{b}(G)\rightarrow\mathscr{M}_{b}(Q)$ be
the linear map induced by $\varphi$,

$\langle\,\tilde{f},\varphi_{*}(\mu)\,\rangle=\langle\,\tilde{f}\circ\varphi,\mu\,\rangle,\qquad\tilde{f}\in\mathscr{C}_{b}(Q),\quad\mu\in\mathscr{M}_{b}(G).$

Then

$\varphi_{*}\circ P^{*}=\mathrm{id}_{\mathscr{M}_{b}(Q)}.$

  3. (iii)

Let $s\in Q$, and denote

$O_{s}=\\{p\in G:\varphi(p)=s\\}.$

If $P(\mathscr{K}_{+}(G))\subset\mathscr{K}_{+}(G)$ and, for $p\in G$ and
$f\in\mathscr{K}_{+}(G)$, $P(f)(\varphi(p))>0$ whenever $f(p)>0$, then $O_{s}$
is compact.

  4. (iv)

For any $s\in Q$,

$\mathrm{supp}\,(P^{*}(\varepsilon_{s}))\subset O_{s}.$

If $P(f)(\varphi(p))>0$ whenever $f(p)>0$ for $p\in G$ and
$f\in\mathscr{K}_{+}(G)$, then

$\mathrm{supp}\,(P^{*}(\varepsilon_{s}))=O_{s}.$

###### Proof.

(_i_) Regarding $G$ as $\mathrm{Spec}\,(\mathscr{C}_{0}(G))$, the spectrum of
the commutative $C^{*}$-algebra $\mathscr{C}_{0}(G)$, we define

$\varphi(p)(\tilde{f})=p\\!\\!\upharpoonright_{B}(\tilde{f})$

for $\tilde{f}\in B$ and $p\in\mathrm{Spec}\,(\mathscr{C}_{0}(G))$. Hence, by
the definition, for $\tilde{f}\in\mathscr{C}_{0}(Q)$ and $p\in G$, we have

$\iota(\tilde{f})(p)=\tilde{f}(\varphi(p))=p\\!\\!\upharpoonright_{B}(\tilde{f}).$

If $s\in Q=\mathrm{Spec}\,(B)$, then it follows from [13] that $s$ can be
extended to an element $p\in\mathrm{Spec}\,(\mathscr{C}_{0}(G))$ such that
$p\\!\\!\upharpoonright_{B}=s$, hence $\varphi$ is onto.

The last statement in the item is a direct implication of the definition of
$\varphi$.

(_ii_) If $\tilde{f}\in\mathscr{C}_{0}(Q)$, then
$(P\circ\iota)(\tilde{f})=\tilde{f}$, since $P$ is a projection. Hence, for
$\tilde{f}\in\mathscr{C}_{0}(Q)$ and $\tilde{\mu}\in\mathscr{M}_{b}(Q)$, we
have

$\langle\,\tilde{f},(\varphi_{*}\circ
P^{*})(\tilde{\mu})\,\rangle=\langle\,\tilde{f}\circ\varphi,P^{*}(\tilde{\mu})\,\rangle=\langle\,\iota(\tilde{f}),P^{*}(\tilde{\mu})\,\rangle=\langle\,(P\circ\iota)(\tilde{f}),\tilde{\mu}\,\rangle=\langle\,\tilde{f},\tilde{\mu}\,\rangle.$

(_iii_) Indeed, for $p\in O_{s}$, let $f\in\mathscr{K}_{+}(G)$ such that
$f(p)=1$. Then, by the assumption, $P(f)(s)=P(f)(\varphi(p))>0$. Hence, for
any $q\in O_{s}$, $P(f)(q)=P(f)(s)>0$, and
$O_{s}\subset\mathrm{supp}\,(P(f))$. Since $O_{s}=\varphi^{-1}(s)$ is closed
and $\mathrm{supp}\,P(f)$ is compact, $O_{s}$ is compact.

(_iv_) We first prove that $\mathrm{supp}\,(P^{*}(\varepsilon_{s}))\subset
O_{s}$. Let $p\in\mathrm{supp}\,(P^{*}(\varepsilon_{s}))$ and assume that
$p\notin O_{s}$, hence $\varphi(p)\neq s$. Choose an open neighborhood $V$ of
$\varphi(p)$ such $s\notin V$, and let $U=\varphi^{-1}(V)$. Then $p\in U$ and,
for any $f\in\mathscr{K}_{+}(U)$, $\mathrm{supp}\,(P(f))\subset V$, hence
$P(f)(s)=0$. Thus

$\langle\,f,P^{*}(\varepsilon_{s})\,\rangle=\langle\,P(f),\varepsilon_{s}\,\rangle=0,$

and $p\notin\mathrm{supp}\,(P^{*}(\varepsilon_{s}))$.

It remains to prove, with the assumption made, that
$O_{s}\subset\mathrm{supp}\,(P^{*}(\varepsilon_{s}))$. Let $p\in O_{s}$ and
$U$ be an open neighborhood of $p$. Choose a function $f\in\mathscr{K}_{+}(U)$
such that $f(p)=1$. By the assumption, $P(f)(s)>0$. Hence

$\langle\,f,P^{*}(\varepsilon_{s})\,\rangle=\langle\,P(f),\varepsilon_{s}\,\rangle=P(f)(s)>0,$

and $p\in\mathrm{supp}\,P^{*}(\varepsilon_{s})$. ∎

######  Remark 2.

In the sequel, we will identify elements of $\mathscr{C}(Q)$ with those in
$B$, according to Proposition 3 (i).

###### Theorem 1

Let $G$ be a locally compact group,
$P\colon\mathscr{C}_{b}(G)\rightarrow\mathscr{C}_{b}(G)$ be a conditional
expectation. Assume that $P$ satisfies the assumptions in Proposition 3 (iii),
and let the following hold:

$\begin{array}[]{c}\displaystyle\bigl{(}(P\times\mathrm{id})\circ\Delta\circ
P\bigr{)}(f)=\bigl{(}(\mathrm{id}\times P)\circ\Delta\circ
P\bigr{)}(f)=\bigl{(}(P\times P)\circ\Delta\bigr{)}(f),\\\\[5.69054pt]
P(\check{f})=\bigl{(}P(f)\bigr{)}\check{},\end{array}$ (6)

for all $f\in\mathscr{C}_{b}(G)$, where
$\Delta\colon\mathscr{C}_{b}(G)\rightarrow\mathscr{C}_{b}(G\times G)$ is the
group comultiplication on $\mathscr{C}_{b}(G)$ defined by $(\Delta
f)(p_{1},p_{2})=f(p_{1}p_{2})$, and $\check{f}(p)=f(p^{-1})$,
$p_{1},p_{2},p\in G$.

Let $B=P(\mathscr{C}_{0}(G))$, denote by $Q$ the spectrum of the commutative
$C^{*}$-algebra $B$, and identify $B$ with $\mathscr{C}_{0}(Q)$. Define
$\tilde{\Delta}\colon\mathscr{C}_{b}(Q)\rightarrow\mathscr{C}_{b}(Q\times Q)$
by

$\tilde{\Delta}(\tilde{f})=\bigl{(}(P\times
P)\circ\Delta\bigr{)}(\tilde{f}),\quad\tilde{f}\in\mathscr{C}_{b}(Q).$

For $s\in Q$, define $\check{s}\in Q$ by

$\check{s}(\tilde{f})=\check{\tilde{f}}(s),\quad\tilde{f}\in B.$

Let $\tilde{e}\in Q$ be defined by

$\tilde{e}(\tilde{f})=\tilde{f}(e),\quad\tilde{f}\in B,$

where $e$ is the identity of the group $G$.

Then $(Q,\tilde{e},\,\check{}\,)$ is a locally compact hypergroup with
comultiplication $\tilde{\Delta}$.

If $m\circ P=m$, where $m$ is a left Haar measure on the group $G$, then
$\tilde{m}\in\mathscr{M}(Q)$ defined by
$\tilde{m}=\varphi_{*}(m)=m\\!\\!\restriction_{B}$ is a left Haar measure on
$Q$.

###### Proof.

To prove the theorem, using Proposition 2, we will check the conditions
($\tilde{H}_{1}$)—($\tilde{H}_{7}$) in Proposition 1.

($\tilde{H}_{1}$) We have

$\displaystyle(\tilde{\Delta}\times\mathrm{id})\circ\tilde{\Delta}$
$\displaystyle=(P\times
P\times\mathrm{id})\circ(\Delta\times\mathrm{id})\circ(P\times P)\circ\Delta$
$\displaystyle=(P\times P\times P)\circ(\Delta\times\mathrm{id})\circ\Delta.$

On the other hand,

$\displaystyle(\mathrm{id}\times\tilde{\Delta})\circ\tilde{\Delta}$
$\displaystyle=(\mathrm{id}\times P\times
P)\circ(\mathrm{id}\times\Delta)\circ(P\times P)\circ\Delta$
$\displaystyle=(P\times P\times P)\circ(\mathrm{id}\times\Delta)\circ\Delta.$

Now, the result follows from coassociativity of $\Delta$.

($\tilde{H}_{2}$) (b) Since $P(1_{G})=1_{Q}$ and $\Delta 1_{G}=1_{G\times G}$,
the claim is immediate.

($\tilde{H}_{2}$) (a) It is clear that $\tilde{\Delta}$ is positive, being a
composition of the positive maps $\Delta$ and $P\times P$. It is also
immediate from positivity and ($\tilde{H}_{2}$) (b) that
$\tilde{\Delta}(\tilde{f})\in\mathscr{C}_{b,+}(Q\times Q)$ for
$\tilde{f}\in\mathscr{C}_{b,+}(Q)$.

($\tilde{H}_{2}$) (c) For any $\tilde{f}\in\mathscr{K}_{+}(Q)$ and $s,t\in Q$,
we have

$\displaystyle(\tilde{\Delta}\,f)(s,t)$
$\displaystyle=\langle\,\tilde{\Delta}\,\tilde{f},\varepsilon_{s}\otimes\varepsilon_{t}\,\rangle=\langle\,\bigl{(}(P\times
P)\circ\Delta\bigr{)}(\tilde{f}),\varepsilon_{s}\otimes\varepsilon_{t}\,\rangle$
$\displaystyle=\langle\,\Delta\,\tilde{f},P^{*}(\varepsilon_{s})\otimes
P^{*}(\varepsilon_{t})\,\rangle=\langle\,\tilde{f},P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t})\,\rangle,$

where $*_{G}$ is the convolution of measures on $G$ defined by

$\langle\,f,\mu*_{G}\nu\,\rangle=\int_{G^{2}}f(pq)\,d\mu(p)d\nu(q),\qquad
f\in\mathscr{C}_{b}(G),\quad\mu,\nu\in\mathscr{M}_{c}(G).$

It follows from Proposition 3 that $P^{*}(\varepsilon_{s})$ and
$P^{*}(\varepsilon_{t})$ are compact, hence the set
$F=\mathrm{supp}\,(P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t}))$ is also
compact. It also follows from the first identity in (6) that, if
$\tilde{f}=P(f)$, $f\in\mathscr{C}_{b}(G)$, then

$\langle\,\tilde{f},P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t})\,\rangle=\langle\,f,P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t})\,\rangle.$

Hence, let $f\in\mathscr{K}_{+}(G)$ be such that $f(p)=1$ for all
$p\in\mathrm{supp}\,(P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t}))$.
Then, by the assumption, $\tilde{f}=P(f)\in\mathscr{K}_{+}(Q)$, and

$\displaystyle\langle\,\tilde{f},P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t})\,\rangle$
$\displaystyle=\langle\,f,P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t})\,\rangle=\langle\,1_{G},P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t})\,\rangle$
$\displaystyle=\langle\,1_{Q\times
Q},\varepsilon_{s}\otimes\varepsilon_{t}\,\rangle=1.$

($\tilde{H}_{4}$) Let $F\subset Q$ be closed, $(s_{0},t_{0})\in Q\times Q$,
and $\tilde{f}_{0}\in\mathscr{K}_{+}(Q)$ be such that $\|\tilde{f}_{0}\|\leq
1$, $f_{0}(r)=0$ for $r\in F$ and
$(\tilde{\Delta}\,\tilde{f}_{0})(s_{0},t_{0})=1$. Since
$\tilde{\Delta}\,\tilde{f}_{0}\in\mathscr{C}_{+}(Q\times Q)$ there is a
neighborhood $U\subset Q\times Q$ of the point $s_{0},t_{0}$ such that
$(\tilde{\Delta}\,\tilde{f}_{0})(s,t)>\frac{1}{2}$ for all $(s,t)\in U$. This
means that

$\langle\,\tilde{f}_{0},P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t})\,\rangle>\frac{1}{2},\quad(s,t)\in
U.$

This means that the set $E$,

$E=\bigcup_{(s,t)\in
U}\mathrm{supp}\,(P^{*}(\varepsilon_{s})*_{G}P^{*}(\varepsilon_{t}))\subset\mathrm{supp}\,\tilde{f}_{0},$

is compact. Choose $\tilde{f}\in\mathscr{K}_{+}(Q)$ to be such that
$\tilde{f}(r)=1$ if $r\in E$, and $\tilde{f}(r)=0$ if $r\in F$. This function
satisfies the condition in ($\tilde{H}_{4}$).

($\tilde{H}_{5}$) For $\tilde{f}\in\mathscr{C}_{0}(Q)$, we have

$\displaystyle(\epsilon\times\mathrm{id})\circ\tilde{\Delta}(\tilde{f})$
$\displaystyle=(\epsilon\times\mathrm{id})\circ(P\times
P)\circ\Delta(P(\tilde{f}))$ $\displaystyle=(\epsilon\times
P)\circ\Delta(P(\tilde{f}))=P(P(\tilde{f}))=\tilde{f}.$

($\tilde{H}_{6}$) First of all note that $\check{\tilde{f}}\in\mathscr{K}(Q)$
for $\tilde{f}\in\mathscr{K}(Q)$, since
$P(\check{\tilde{f}})=P(\tilde{f})\,\check{=}\check{\tilde{f}}$ by (6).

It is clear that the first identity in (3) holds. To prove the second identity
in (3), consider

$(\tilde{\Delta}\check{\tilde{f}})(s,t)=\bigl{(}(P\times
P)\circ\Delta\check{\tilde{f}}\bigr{)}(s,t)=\bigl{(}(P\times
P)\circ\Sigma\circ(\ \check{}\ \times\ \check{}\
)\circ\Delta\bigr{)}(\tilde{f})(s,t)\\\ =\bigl{(}(P\times
P)\circ\Delta\bigr{)}(\tilde{f})(\check{t},\check{s})=(\tilde{\Delta}\,\tilde{f})(\check{t},\check{s})$

for $s,t\in Q$, $\tilde{f}\in B$.

($\tilde{H}_{7}$) We will prove that ($H_{7}$) holds true. To this end, it is
sufficient to prove that $e\in O_{s}\cdot O_{\check{t}}$, $s,t\in Q$, if and
only if $s=t$, since the set $O_{s}\cdot O_{\check{t}}$ is compact because
such are $O_{s}$ and $O_{\check{t}}$. Since $O_{\check{t}}=O_{t}^{-1}$,
$p^{-1}\in O_{\check{s}}$ for $p\in O_{s}$, hence $e\in O_{s}\cdot
O_{\check{s}}$. On the other hand, if $e\in O_{s}\cdot O_{\check{t}}$, then
there is $p\in O_{s}\cap O_{t}$, that is, $s=t$.

Let us prove that $\tilde{m}=\varphi_{*}(m)$ is a left invariant measure on
$Q$, that is, for any $\tilde{f},\tilde{g}\in\mathscr{K}(Q)$ the following
relation holds:

$(\mathrm{id}\times\tilde{m})\bigl{(}\tilde{\Delta}(\tilde{f})\cdot(1\otimes\tilde{g})\bigr{)}=(\mathrm{id}\times\tilde{m})\bigl{(}(1\otimes\tilde{f})\cdot(\,\check{}\,\times\mathrm{id})\circ\tilde{\Delta}(\tilde{g})\bigr{)}.$
(7)

Consider the left-hand side of (7), use the definitions of $\tilde{m}$ and
$\tilde{\Delta}$, and let $\tilde{f}=Pf$, $\tilde{g}=Pg$,
$f,g\in\mathscr{K}(G)$,

$(\mathrm{id}\times m)\bigl{(}((P\times P)\circ\Delta(Pf))\cdot(1\otimes
Pg)\bigr{)}\\\ =(\mathrm{id}\times
m)\bigl{(}(P\times\mathrm{id})\circ\Delta(Pf)\cdot(1\otimes Pg)\bigr{)}\\\
=P\bigl{(}(\mathrm{id}\times m)(\Delta(Pf)\cdot(1\otimes Pg))\bigl{)}.$

Now using a relation similar to (7) that holds for $m$ on $G$ and that
$P\circ\,\check{}\,=\check{}\,\circ P$ we get

$P\bigl{(}(\mathrm{id}\times m)(\Delta(Pf)\cdot(1\otimes Pg))\bigl{)}\\\
=P\bigl{(}(\mathrm{id}\times m)((1\otimes
Pf)\cdot(\,\check{}\,\times\mathrm{id})\circ\Delta(Pg))\bigr{)}\\\
=(\mathrm{id}\times m)\bigl{(}(1\otimes Pf)\cdot(\,\check{}\,\circ
P\times\mathrm{id})\circ\Delta(Pg)\bigr{)}\\\ =(\mathrm{id}\times
m)\bigl{(}(1\otimes Pf)\cdot(\,\check{}\,\circ P\times
P)\circ\Delta(Pg)\bigr{)}\\\ =(\mathrm{id}\times m\circ P)\bigl{(}(1\otimes
Pf)\cdot(\,\check{}\,\times P)\circ\Delta(Pg)\bigr{)}\\\
=(\mathrm{id}\times\tilde{m})\bigl{(}(1\otimes
Pf)\cdot(\,\check{}\,\times\mathrm{id})\circ\tilde{\Delta}(Pg)\bigr{)}.$

∎

###### Lemma 2

Let $\langle\,Pf,m\,\rangle=\langle\,f,m\,\rangle$ for all
$f\in\mathscr{K}(G)$. Then the conditional expectation
$P\colon\mathscr{K}(G)\rightarrow B$ can be extended by continuity to an
idempotent on $L_{1}(G,m)$ and an orthogonal projection on $L_{2}(G,m)$, still
denoted by $P$.

###### Proof.

Let $f\in\mathscr{K}(G)$. Since $-|f|\leq f\leq|f|$ and $P$ preserves the cone
of nonnegative functions, we have $|P(f)|\leq P(|f|)$. Hence,

$\displaystyle\|P(f)\|_{1}$
$\displaystyle=\int_{G}|P(f)|(p)\,dm(p)\leq\int_{G}P(|f|)(p)\,dm(p)$
$\displaystyle=\int_{G}|f|(p)\,dm(p)=\|f\|_{1},$

which shows that $P$ is continuous with respect to the $L_{1}$-norm.

Now, since $|P(f)|^{2}\leq P(|f|^{2})$, we have

$\displaystyle\|P(f)\|^{2}_{2}$
$\displaystyle=\int_{G}|P(f)(p)|^{2}\,dm(p)\leq\int_{G}P(|f|^{2})(p)\,dm(p)$
$\displaystyle=\int_{G}|f(p)|^{2}\,dm(p)=\|f\|_{2}^{2}.$

Hence, $P$ is continuous with respect to the $L_{2}$-norm on $\mathscr{K}(G)$.

It is clear that the continuous extension of $P$ to $L_{2}(G,m)$ is a
projection. To see that it is orthogonal, let $f_{1},f_{2}\in\mathscr{K}(G)$
and consider

$(Pf_{1},f_{2})_{L_{2}(G,m)}=\int_{G}(Pf_{1})(p)\overline{f_{2}(p)}\,dm(p)=\int_{G}P\bigl{(}(Pf_{1})(p)\overline{f_{2}(p)}\bigr{)}\,dm(p)\\\
=\int_{G}(Pf_{1})(p)\overline{(Pf_{2})(p)}\,dm(p)=\int_{G}P(f_{1}(p)\overline{(Pf_{2})(p)})\,dm(p)\\\
=\int_{G}f_{1}(p)\overline{(Pf_{2})(p)}\,dm(p)=(f_{1},Pf_{2})_{L_{2}(G,m)}.$

∎

######  Remark 3.

In what follows, we will identify the space $L_{1}(Q,\tilde{m})$ (resp.,
$L_{2}(Q,\tilde{m})$) with a closed subspace of $L_{1}(G,m)$ (resp.,
$L_{2}(G,m)$).

For a hypergroup $Q$, consider the product hypergroup $Q\times Q$ [1], and let
$\delta\colon\mathscr{M}_{b}(Q)\rightarrow\mathscr{M}_{b}(Q\times Q)$ denote a
linear extension of the map defined by

$\delta(\varepsilon_{s})=\varepsilon_{s}\otimes\varepsilon_{s},\quad s\in Q,$
(8)

that is, for $\mu\in\mathscr{M}_{b}(Q)$ and $F\in\mathscr{K}(Q\times Q)$,

$\langle\,\delta(\mu),F\,\rangle=\int_{Q}F(s,s)\,d\mu(s).$

######  Definition 6.

Let $B$ be a $*$-algebra and $A$ a $C^{*}$-algebra. A linear map $\phi\colon
B\rightarrow A$ will be called _positive_ , if $\phi(b^{*}b)\geq 0$ for all
$b\in B$, and _completely positive_ , if the map $\mathrm{id}\otimes\phi\colon
M_{n}(\mathbb{C})\otimes B\rightarrow M_{n}(\mathbb{C})\otimes A$ is positive
for all $n\in\mathbf{N}$, where $M_{n}(\mathbb{C})$ is the $C^{*}$-algebra of
$(n\times n)$-matrices over $\mathbb{C}$.

It follows from [14] that a map $\phi:B\to A$ from a $*$-algebra $B$ into a
$C^{\ast}$-algebra $A$ is completely positive if and only if, for any
$n\in\mathbf{N}$, $b_{i}\in B$ and $a_{i}\in A$, $i=1,\dots,n$, we have

$\sum_{i,j=1}^{n}a_{j}^{*}\phi(b_{j}^{\star}b_{i})a_{i}\geq 0.$ (9)

###### Theorem 2

Let $P$ be a conditional expectation on $A_{0}={\mathcal{C}}_{0}(Q)$ that
satisfy all the conditions in Theorem 1. Let $Q$ be the corresponding
hypergroup. Denote by $\lambda_{Q}$ the left regular representation of
$\mathscr{M}_{b}(Q)$ on $L_{2}(Q)$,

$(\lambda_{Q}(\tilde{\mu}).\,\tilde{f})(t)=\int_{Q}\langle\,\tilde{f},\check{\varepsilon}_{s}*_{Q}\varepsilon_{t}\,\rangle\,d\tilde{\mu}(s),\qquad\tilde{\mu}\in\mathscr{M}_{b}(Q),\quad\tilde{f}\in
L_{2}(Q),\quad t\in Q,$

and let $\delta$ be defined by (8). Then the linear map

$(\lambda_{Q}\otimes\lambda_{Q})\circ\delta\colon L_{1}(\tilde{m})\rightarrow
B(L_{2}(Q)\otimes L_{2}(Q))$

is completely positive.

###### Proof.

We will identify $L_{2}(Q)$ with the closed subspace $P(L_{2}(G))$ of
$L_{2}(G)$, and the Banach $*$-algebra $L_{1}(Q,\tilde{m})$ with the Banach
$*$-subalgebra $P(L_{1}(G,m))$ of $L_{1}(G,m)$. With such an identification,
we have

$\begin{array}[]{c}(\tilde{f},\tilde{g})_{L_{2}(Q)}=(\tilde{f},\tilde{g})_{L_{2}(G)},\quad\tilde{f},\tilde{g}\in
L_{2}(Q),\\\\[5.69054pt]
\lambda_{Q}(\tilde{\mu}).\,\tilde{f}=\lambda_{G}(P^{*}(\tilde{\mu})).\,\tilde{f},\quad\tilde{\mu}\in\mathscr{M}_{b}(Q),\quad\tilde{f}\in
L_{2}(Q),\\\\[5.69054pt]
\delta(P^{*}(\varepsilon_{s}))=P^{*}(\varepsilon_{s})\otimes
P^{*}(\varepsilon_{s}),\quad s\in Q.\end{array}$ (10)

To prove the theorem, using (9), it is sufficient to show that

$\displaystyle\sum_{i,j=1}^{n}$
$\displaystyle\bigl{(}\,A^{*}_{j}\cdot(\lambda_{Q}\otimes\lambda_{Q})\circ\delta(\tilde{\mu}_{j}^{\star}*\tilde{\mu}_{i})\cdot
A_{i}.\,F\,\bigm{|}\,F\,\bigr{)}_{L_{2}(Q)\otimes L_{2}(Q)}$
$\displaystyle=\sum_{i,j=1}^{n}\bigl{(}\,(\lambda_{Q}\otimes\lambda_{Q})\circ\delta(\tilde{\mu}_{j}^{\star}*\tilde{\mu}_{i})\cdot
A_{i}.\,F\,\bigm{|}\,A_{j}.\,F\,\bigr{)}_{L_{2}(Q)\otimes L_{2}(Q)}\geq 0$

for any $A_{i}\in B(L_{2}(Q)\otimes L_{2}(Q))$, $\tilde{\mu}_{i}\in
L_{1}(\tilde{m})$, $F\in L_{2}(Q)\otimes L_{2}(Q)$, $i=1,\dots,n$.

Hence letting $A_{i}F=F_{i}\in L_{2}(Q)\otimes L_{2}(Q)$, and
$\mu_{i}=P^{*}(\tilde{\mu}_{i})\in L_{1}(m_{G})$, and using (10), we have

$\displaystyle\sum_{i,j=1}^{n}$
$\displaystyle\bigl{(}\,(\lambda_{Q}\otimes\lambda_{Q})\circ\delta(\mu_{j}^{\star}*\mu_{i}).\,F_{i}\,\bigm{|}\,F_{j}\,\bigr{)}_{L_{2}(Q)\otimes
L_{2}(Q)}$
$\displaystyle=\sum_{i,j=1}^{n}\int_{G}\bigl{(}\,(\lambda_{G}(P^{*}(\varepsilon_{\varphi(p)}))\otimes\lambda_{G}(P^{*}(\varepsilon_{\varphi(p)})).\,F_{i}\,\bigm{|}\,F_{j}\,\bigr{)}_{L_{2}(G)\otimes
L_{2}(G)}$ $\displaystyle\qquad\qquad d({\mu_{j}}^{\star}\ast\mu_{i})(p).$

Let $\mu_{i}=f_{i}m$, $f_{i}\in P(L_{1}(G))\subset L_{1}(G)$, $i=1,\dots,n$.
Then $\mu_{j}^{\ast}=f^{\ast}_{j}m$ and

${\mu_{j}}^{\star}\ast\mu_{i}=(f_{j}^{\ast}\ast f_{i})m.$

Using left invariance of the measure $m$ we thus have

$\displaystyle\sum_{i,j=1}^{n}$
$\displaystyle\bigl{(}\,(\lambda_{Q}\otimes\lambda_{Q})\circ\delta(\mu_{j}^{\star}*\mu_{i}).\,F_{i}\,\bigm{|}\,F_{j}\,\bigr{)}_{L_{2}(Q)\otimes
L_{2}(Q)}$
$\displaystyle=\sum_{i,j=1}^{n}\int_{G}\bigl{(}\,(\lambda_{G}(P^{*}(\varepsilon_{\varphi(p)}))\otimes\lambda_{G}(P^{*}(\varepsilon_{\varphi(p)}))).\,F_{i}\,\bigm{|}\,F_{j}\,\bigr{)}_{L_{2}(G)\otimes
L_{2}(G)}$ $\displaystyle\qquad\qquad\cdot(f_{j}^{\star}*f_{i})(p)\,dm(p)$
$\displaystyle=\sum_{i,j=1}^{n}\int_{G^{2}}\bigl{(}\,(\lambda_{G}(P^{*}(\varepsilon_{\varphi(p)}))\otimes\lambda_{G}(P^{*}(\varepsilon_{\varphi(p)}))).\,F_{i}\,\bigm{|}\,F_{j}\,\bigr{)}_{L_{2}(G)\otimes
L_{2}(G)}$ $\displaystyle\qquad\qquad\cdot
f^{*}_{j}(q)f_{i}(q^{-1}p)\,dm(q)dm(p)$
$\displaystyle=\sum_{i,j=1}^{n}\int_{G^{2}}\bigl{(}\,(\lambda_{G}(p)\otimes\lambda_{G}(p)).\,F_{i}\,\bigm{|}\,F_{j}\,\bigr{)}_{L_{2}(G)\otimes
L_{2}(G)}$ $\displaystyle\qquad\qquad\cdot
f^{*}_{j}(q)f_{i}(q^{-1}p)\,dm(q)dm(p)$
$\displaystyle=\sum_{i,j=1}^{n}\int_{G^{2}}\bigl{(}\,(\lambda_{G}(qp)\otimes\lambda_{G}(qp)).\,F_{i}\,\bigm{|}\,F_{j}\,\bigr{)}_{L_{2}(G)\otimes
L_{2}(G)}$ $\displaystyle\qquad\qquad\cdot
f^{*}_{j}(q)f_{i}(p)\,dm_{G}(q)dm_{G}(p)$
$\displaystyle=\sum_{i,j=1}^{n}\int_{G^{2}}\bigl{(}\,(\lambda_{G}(p)\otimes\lambda_{G}(p)).\,F_{i}\,\bigm{|}\,(\lambda_{G}(q^{-1})\otimes\lambda_{G}(q^{-1}).\,F_{j}\,\bigr{)}_{L_{2}(G)\otimes
L_{2}(G)}$ $\displaystyle\qquad\qquad\cdot
f^{*}_{j}(q)f_{i}(p)\,dm_{G}(q)dm_{G}(p)$
$\displaystyle=\sum_{i,j=1}^{n}\int_{G^{2}}\bigl{(}\,(\lambda_{G}(p)\otimes\lambda_{G}(p)).\,F_{i}\,\bigm{|}\,(\lambda_{G}(q^{-1})\otimes\lambda_{G}(q^{-1}).\,F_{j}\,\bigr{)}_{L_{2}(G)\otimes
L_{2}(G)}$
$\displaystyle\qquad\qquad\cdot\overline{f}_{j}(q^{-1})f_{i}(p)\,dm_{G}(q^{-1})dm_{G}(p)$
$\displaystyle=\Bigl{\|}\sum_{i=1}^{n}\int_{G}f_{i}(p)(\lambda_{G}(p)\otimes\lambda_{G}(p))\,dm_{G}(p).\,F_{i}\Bigr{\|}^{2}_{L_{2}(G)\otimes
L_{2}(G)}\geq 0.$

∎

###### Corollary 1

The Fourier space $\mathscr{A}(Q)$ is a Banach algebra.

###### Proof.

The statement of this corollary follows from the results of [9]. ∎

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