text:
\section{Introduction}

Let $\cal G$ be a locally compact group. Then the Banach space
$LUC({\cal G})^*$, the topological dual of the space of all bounded
left uniformly continuous functions on $\cal G$, is a Banach
algebra equipped with the Arens-type product. In fact, the
product of $m$ and $n$ in $LUC({\cal G})^*$ is defined by
\begin{eqnarray}\label{arensluc}
\langle m\odot n,f\rangle =\langle m,n\diamond f\rangle ,
\quad\quad (n\diamond f)(s)=\langle n,{_sf}\rangle ,\quad\quad {_sf}(t)=f(st),
\end{eqnarray}
where $f\in LUC({\cal G})$ and $s,t\in {\cal G}$. In general, this
product is not separately weak$^*$ to weak$^*$ continuous on
$LUC({\cal G})^*$, and in recent years there has been shown
considerable interest by harmonic analysts in the characterization
of the following space
$${\frak{Z}}({\cal G})=\Big{\{}m\in LUC({\cal G})^*:~n\mapsto m\odot n~
{\hbox{is}}~{\hbox{weak}}^*~{\hbox{to}}~{\hbox{weak}}^*
~{\hbox{continuous}}\Big{\}}.$$ As far as we know the subject, the
starting point of the study of the space ${\frak{Z}}({\cal G})$ is
the paper by Zappa \cite{zappa}. In details, Zappa proved that
${\frak{Z}}({\Bbb R})$ is precisely $M({\Bbb R})$, where $\Bbb R$
is the additive group of real number and $M({\Bbb R})$ is the
Banach space of all complex Radon measures on $\Bbb R$. This
result was extended to all abelian locally campact groups by
Grosser and Losert in \cite{grossert}, and to all locally compact
groups by Lau in \cite{lau}.

In this paper, considering $\cal X$ as a locally compact Hausdorff
space on which $\cal G$ acts continuously from the left, following Lau and Chu \cite{chulau} we
introduce the Banach space $LUC({\cal X},{\cal G})$.
Then, we present a left action of $LUC({\cal
G})^*$ on $LUC({\cal X},{\cal G})^*$, as an extension of the
natural action of $M({\cal G})$ on $M({\cal X})$, to make a Banach
left $LUC({\cal G})^*$-module. In
particular, we investigate the faithfulness of the action of $M({\cal
G})$ (resp. $LUC({\cal G})^*$) on $M({\cal X})$ (resp. $LUC({\cal X},{\cal G})^*$)
in relation to the action of $\cal G$ on $\cal X$.
Also, we prove that if $\cal X$ is a transitive $\cal
G$-space, then $M({\cal X})$ is an $LUC({\cal G})^*$-submodule of
$LUC({\cal X},{\cal G})^*$ just when $\cal X$ is compact.
Moreover, the main purpose of this work is to study the
topological centre problem related to this module action. To this end,
we introduce the Banach subspace ${{\frak{Z}}({\cal X},{\cal G})}$
of $LUC({\cal G})^*$, as the topological centre related to this
module action, which contains $M({\cal G})$ as a closed subalgebra.
Furthermore, apart from some characterization of the space
${{\frak{Z}}({\cal X},{\cal G})}$, we extend the main results of
Lau~\cite{lau} from locally compact groups to $\cal G$-spaces.  Finally,
we apply our results  to some
special cases of $\cal G$ and $\cal X$ to give examples with

$\centerdot$ ${\frak Z}({\cal X},{\cal G})=M({\cal G})$;

$\centerdot$ $M({\cal G})\subsetneqq{\frak Z}({\cal X},{\cal G})=LUC({\cal G})^*$;

$\centerdot$ $M({\cal G})\subsetneqq{\frak Z}({\cal X},{\cal G})\subsetneqq LUC({\cal G})^*$.

The motivation of the study presented in this paper comes from
some works of Lau and his
coauthors \cite{chulau,lauucb1,lauucb2,lau,laupym1} and some recent developments of the concept of
regularity of bilinear maps were introduced and first studied by
Arens \cite{arens}. It is known that the Arens regularity of module
actions has been a major tool in the study of Banach algebras. For
example, Arens regularity of module actions as a generalization of
Arens regularity of Banach algebras were introduced and first
studied by Filali and Eshaghi Gordji \cite{filalgordji} and they
used this notion to answer some questions regarding Arens
regularity of Banach algebras raised by Lau and \"{U}lger
\cite{laulger}. Also in \cite{mvishki}, Arens regularity of module
actions were considered by Mohammadzadeh and Vishki to investigate
the conditions under which the second adjoint of a derivation into
a dual Banach module is again a derivation, that extends the
results of Dales, Rodriguez-Palacios and Velasco in
\cite{dalesrpv} for a general derivation.

\section{\normalsize\bf Some prerequisites}

Throughout this paper, $\cal G$ is a locally compact group with
left Haar measure $\lambda$ and identity element $e$,
$\Delta$ refers to the modular function on $\cal G$ and
the notation $C_b({\cal G})$ is used to denote the space of all
bounded complex-valued continuous functions on $\cal G$ with the
supremum norm. Also, the subspaces $C_c({\cal G})$ and $C_0({\cal
G})$ of $C_b({\cal G})$ refer to the space of all functions with
compact support and the space of all functions vanishing at infinity,
respectively. Also, for all $s\in {\cal G}$ and $f\in C_b({\cal
G})$ we define ${_sf}, $ the left translation of $f$ by $s$, as
${_sf}(t)=f(st)$, $t\in {\cal G}$. A function $f$ in $C_b({\cal
G})$ is called left uniformly continuous if the mapping
$s\mapsto{_sf}$ from $\cal G$ into $(C_b({\cal
G}),\|\cdot\|_\infty)$ is continuous. As usual, we mean by
$LUC({\cal G})$ the Banach space of all left uniformly continuous
functions on $\cal G$. Moreover, the notation $M({\cal G})$
is used to denote the measure algebra of ${\cal G}$ consisting of all complex regular
Borel measures on ${\cal G}$ with the total variation norm and the
convolution product ``$\ast$" defined by the formula
 $$
 \langle\mu\ast\nu ,f\rangle=\int_{\cal G}\int_{\cal G} f(st)\;
 d\mu(s)\; d\nu(t)
 $$
for all $\mu, \nu\in M({\cal G})$ and $f\in C_0({\cal G})$.
It is well-known that $M({\cal G})$
is the topological dual of $C_0({\cal G})$ with the pairing
 $$
 \langle \mu,f \rangle:=\int_{\cal G}  f(s)\;d\mu(s)
 $$
for all $\mu\in M({\cal G})$ and $f\in C_0({\cal G})$.
This allows us to view $M({\cal G})$ as a subalgebra of $LUC({\cal
G})^*$.

Throughout the paper the letter $M^+({\cal G})$ means the set of
all positive measures in $M({\cal G})$; $\delta_s$ denotes
the Dirac measure at $s\in{\cal G}$; and as usual, $M_a({\cal G})$ denotes the closed ideal of $M({\cal
G})$ consisting of all absolutely continuous measures with respect
to $\lambda$. Let also $L^1({\cal G})$ denote the group algebra of
${\cal G}$ as defined in \cite{farmonic}.
Then, the Radon-Nikodym theorem can be interpreted as an
identification  of $M_a({\cal G})$ with
${\Big\{}\lambda_\varphi:~\varphi\in L^1({\cal G}){\Big\}}$, where
$\lambda_\varphi$ is the measure in $M({\cal G})$ defined on each
Borel subset $B$ of ${\cal G}$ by
$$\lambda_\varphi(B)=\int_B\varphi\;d\lambda.$$
Recall that, the convolution of two functions $\varphi, \psi\in
L^1({\cal G})$ is the function defined by
\begin{eqnarray*}
\varphi\ast\psi(s)&=&\int_{\cal
G}\varphi(t)\psi(t^{-1}s)\;d\lambda(t)\\
&=&\int_{\cal G}\varphi(st^{-1})\psi(t)\Delta(t^{-1})\;d\lambda(t).
\end{eqnarray*}

\section{\normalsize\bf Definitions and some basic results}

Recall that a locally compact Hausdorff space $\cal X$ is said to be a ({\it
left}) $\cal G$-{\it space} if there is a continuous action map
$${\cal G}\times {\cal X}\rightarrow {\cal X};~(s,x)\mapsto s\cdot x,$$
satisfying $(st)\cdot x=s\cdot(t\cdot x)$
and $e\cdot x=x$, for all $s,t\in {\cal G}$ and $x\in {\cal X}$.
In some published articles the terms ({\it left}) {\it
transformation group}, a {\it dynamical system} or a {\it flow}
are also used for the pair $({\cal X},{\cal G})$. Similarly one
can define a right action and a right $\cal G$-space. Through this
paper, when we refer to $\cal X$ as a $\cal G$-space without any explicit
reference to the left or right action, we mean $\cal X$ is a left
$\cal G$-space.

A $\cal G$-space $\cal X$ is called transitive if for each $x,y\in
{\cal X}$ there exists an element $s\in {\cal G}$ such that
$y=s\cdot x$. Also, in the case where $\cal X$ is a $\cal
G$-space which is topologically isomorphic to ${\cal G}/{\cal H}$,
for some closed subgroup $\cal H$ of $\cal G$, we say $\cal
X$ is a homogeneous $\cal G$-space.

The action of $\cal G$ on $\cal X$ is said to be {\it free}, where
$s\cdot x\neq x$ for every $x\in{\cal X}$ and $s\in{\cal G}$ with
$s\neq e$. It is called {\it effective}, when the following closed normal
subgroup of $\cal G$ is trivial
\[{\frak N}({\cal X},{\cal G})={\Big\{}s\in {\cal G}:\,
 s\cdot x=x{\hbox{~for~all}}\; x\in {\cal X}{\Big\}}.\]

Similar to the previous section, we consider the Banach spaces
$C_b({\cal X})$ and $C_0({\cal X})$. We also denote by $M({\cal X})$ the Banach space of all complex Radon measures on $\cal X$ with total variation norm.
For given measures
$\mu\in M({\cal G})$ and $\sigma\in M({\cal X})$ we may define
 their convolution $\mu\star\sigma$, as an element of $M({\cal X})$, by
\begin{eqnarray*}
\langle\mu\star\sigma,F\rangle=\int_{\cal G}\int_{\cal X} F(s\cdot
x)\;d\sigma(x)\;d\mu(s)\quad\quad\quad(F\in C_0({\cal X})).
\end{eqnarray*}
This allows us to consider $M({\cal X})$ as a Banach left-$M({\cal G})$-module. Now let $\cal A$ be a closed subalgebra of $M({\cal G})$, then as usual, we say that $M({\cal X})$ is a faithful Banach left
$\cal A$-module, if the action of each $0\neq\mu\in{\cal A}$ is
non-trivial; that is, if $\mu\in {\cal A}$ is so that $\mu\star
\sigma=0$ for all $\sigma\in M({\cal X})$, then $\mu=0$.
In the following result, we are concerned with relations between
faithfulness of the action of $M({\cal G})$ on $M({\cal X})$ and
the action of $\cal G$ on $\cal X$. In a special case, it gives some
conditions in relation to Veech's Theorem \cite{veech} in terms of the
faithfulness of the action of $M({\cal G})$ on $M({\cal U}{\cal
G})$ which is a functional property, where ${\cal U}{\cal G}$ is
the largest semigroup compactification of $\cal G$.

Let $(f_\alpha)_\alpha$ be a net in $C_b({\cal G})$. We say $(f_\alpha)_\alpha$ converges to some $f\in C_b({\cal G})$ strictly if $\|f_\alpha g-fg\|_\infty\longrightarrow 0$, for all $g$ in $S_0^+({\cal G})$ the set of all non-negative upper semicontinuous real-valued functions on $\cal G$ which vanish at infinity, see \cite{abel} for more details.

\begin{theorem}\label{210422}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then the following statements hold.

\newcounter{j1212}
\begin{list}%
{\bf(\alph{j1212})}{\usecounter{j1212}}
\item The Banach space $M({\cal X})$ is a faithful Banach left $M^+({\cal
G})$-module if and only if $\cal G$ acts effectively on $\cal X$.
\item If for some $x_0\in {\cal G}$ the stabilizer subgroup
${\frak N}(x_0,{\cal G})=\{s\in {\cal G}:\,s\cdot x_0=x_0\}$ of ${\cal G}$
is trivial, then $M({\cal X})$ is a faithful Banach left $M({\cal
G})$-module. In particular, if  $\cal G$ acts freely on $\cal X$,
then $M({\cal G})$ acts faithfully on $M({\cal X})$.
\item If $M_a({\cal G})$ acts faithfully
on $M({\cal X})$, then $M({\cal X})$ is a faithful Banach left $M({\cal
G})$-module.
\end{list}
\end{theorem}

{\noindent Proof.} {\bf (a)} First, suppose that $M({\cal X})$ is a faithful
Banach left $M^+({\cal G})$-module and $s\in {\frak N}({\cal
X},{\cal G})$. Then for all $\sigma \in M({\cal X})$ we have
$$
\langle \delta_{s}\star \sigma , F\rangle=\int_{\cal X} F(s\cdot
x)\,d\sigma(x) =\int_{\cal X} F( x)\,d\sigma(x)= \langle \sigma ,
F\rangle,\quad\quad\quad {\Big(}F\in C_0({\cal X}){\Big)}
$$
Accordingly, $\delta_{s}\star \sigma=\sigma$ for all $\sigma \in
M({\cal X})$. This together with the fact that $M^+({\cal G})$ acts
faithfully on $M({\cal X})$ implies that $\delta_s=\delta_e$ and therefore $s=e$.

Conversely, assume that ${\frak N}({\cal X},{\cal G})=\{e\}$ and
$\mu$ is a positive measure in $M({\cal G})$ such that $\mu\star\sigma=0$ for all
$\sigma\in M({\cal X})$.
Observe that $\nu=\mu+\delta_e$ is a positive measure in $M({\cal G})$
for which $\nu\star \sigma=\sigma$ for all $\sigma\in
M({\cal X})$. Hence, the proof will be completed by showing that
$\hbox{supp}(\nu)=\{e\}$ or equivalently
$\nu=\delta_e$. To this end, suppose on the contrary that $s_0$ is an element in $\hbox{supp}(\nu)\setminus\{e\}$. This, together with the fact that
${\frak N}({\cal X},{\cal G})=\{e\}$ implies that
there exists some $x_0\in {\cal X}$ such that $s_0\cdot x_0\neq
x_0$. Taking a positive function $F$ in $C_c({\cal X})$ with
$F(x_0)=0$ and $F(s_0\cdot x_0)=1$, we have $$(\nu\star
\delta_{x_0})(F)=\delta_{x_0}(F)=F(x_0)=0.$$
On the
other hand, there is an open neighborhood $U$ of $s_0$ such that
$F(s\cdot x_0)>1/2$ for all $s\in U$. As $s_0\in \hbox{supp}(\nu)$, we
have $\nu(U)>0$ and hence, we can write
\begin{eqnarray*}
0 = (\nu\star \delta_{x_0})(F)= \int_{\cal G} F(s\cdot
x_0)\,d\nu(s)\geq \nu(U)/2,
\end{eqnarray*}
which is a contradiction. Thus, we get $\hbox{supp}(\nu)=\{e\}$. It follows that
$\nu=\delta_e$ and therefore $\mu=0$.

{\bf (b)} Assume that ${\frak N}(x_0,{\cal G})=\langle e\rangle$, for some $x_0\in {\cal X}$.
Obviously, for all $F\in C_b({\cal X})$, the function $r_{x_0}F:{\cal G}\rightarrow \Bbb{C}$ defined by $r_{x_0}F(s)=F(s\cdot x_0)$ belongs to $C_b({\cal G})$. Moreover,
$${\cal A}:={\Big\{}r_{x_0}F:\,F\in C_b({\cal X}){\Big\}},$$ is a self-adjoint subalgebra of $C_b({\cal G})$ such that for each $s\in {\cal G}$, there exists some $F\in C_b({\cal X})$ for which $r_{x_0}F(s)=F(s\cdot x_0)\neq 0$. Furthermore, this subalgebra separates the points of ${\cal G}$, since for all $s_1$ and $s_2$ in ${\cal G}$ with $s_1\neq s_2$ we have $s_1\cdot x_0\neq s_2\cdot x_0$ and therefore there exists $F\in C_b({\cal X})$ such that $$r_{x_0}F(s_1)=F(s_1\cdot x_0)\neq F(s_2\cdot x_0)=r_{x_0}F(s_2).$$  From this, by a generalized version of the Stone-Weierstrass theorem \cite[Theorem~3.2]{abel}, we can conclude that ${\cal A}$ is strictly dense in $C_b({\cal G})$.

Now, let $\mu$ be a measure in $M({\cal G})$ such that
$\mu\star \sigma=0$ for all $\sigma\in M({\cal X})$ and
$g\in C_c({\cal G})$ be given. Then, by taking $\sigma=\delta_{x_0}$, we have
\begin{eqnarray}\label{rm2}
\int_{\cal G} F(s\cdot x_0)\,d\mu(s) =  \langle\mu\star\delta_{x_0} , F\rangle=0,
\end{eqnarray}
for all $F\in C_b({\cal X})$.
Take a neighborhood $V$ of $\hbox{supp}(g)$ with compact closure and some $F'\in C_c^+({\cal X})$ such that
$$F'{\Big|}_{\hbox{supp}(g)\cdot x_0}=1\quad\quad{\hbox{and}}\quad\quad F'{\Big|}_{{\cal X}\setminus\overline{V}\cdot x_0}=0.$$ Therefore, $r_{x_0}F'$ is a non-negative continuous function for which
$$(r_{x_0}F'){\Big|}_{\hbox{supp}(g)}=1\quad\quad{\hbox{and}}\quad\quad(r_{x_0}F'){\Big|}_{{\cal X}\setminus\overline{V}}=0;$$ This is because of, for all $s\in \cal G$, the condition $s\notin \overline{V}$ implies that $s\cdot x_0\notin \overline{V}\cdot x_0$. So, $r_{x_0}F'\in C_c^+({\cal G})$ and hence by taking a sequence  $(F_n)_{n\in \Bbb N}\subseteq  C_b({\cal X})$ for which $(r_{x_0}F_n)_{n\in \Bbb N}$ tends strictly to $g$, we can write $\|(r_{x_0}F_n)(r_{x_0}F')-g(r_{x_0}F')\|_\infty\longrightarrow 0$. Consequently, we have
\begin{eqnarray*}
\int_{\cal G} g(s)\,d\mu(s) &=& \int_{\cal G} g(s)(r_{x_0}F')(s)\,d\mu(s)\\
&=&  \langle\mu ,(r_{x_0}F')g\rangle\\
&=& \lim_{n\rightarrow +\infty} \langle\mu , (r_{x_0}F_n)(r_{x_0}F')\rangle \\
&=& \lim_{n\rightarrow +\infty} \int_{\cal G} F_n(s\cdot x_0)F'(s\cdot x_0)\,d\mu(s)\\
&=& \lim_{n\rightarrow +\infty} \int_{\cal G} (F_nF')(s\cdot x_0)\,d\mu(s).
\end{eqnarray*}
As $F_nF'\in C_b({\cal X})$, by using (\ref{rm2}), we deduce
$\int_{\cal G} g(s)\,d\mu(s)=0$. Now, since $g$ is arbitrary in $C_c({\cal G})$, we can deduce that $\mu=0$ and this completes the proof of this part.

{\bf (b)} Let $\mu$ be a measure in $M({\cal G})$ such that
$\mu\star \sigma=0$ for all $\sigma\in M({\cal X})$.
Recall from \cite[Lemma 2.1. and Lemma 2.2.]{ghweight}
that the algebra $L^1({\cal G})$ possesses a
bounded approximate identity $(\varphi_\alpha)$
such that each $\varphi_\alpha$ is a continuous function
with compact support and $\lambda_{\varphi_\alpha}\longrightarrow\delta_e$
in the weak$^*$ topology of $M({\cal G})$.
As $M_a({\cal G})$ is an
ideal of $M({\cal G})$, we have $\lambda_{\varphi_\alpha}\ast\mu\in
M_a({\cal G})$ and also $$(\lambda_{\varphi_\alpha}\ast\mu)\star
\sigma=\lambda_{\varphi_\alpha}\star (\mu\star \sigma)=0,$$ for all
$\sigma\in M({\cal X})$ and all $\alpha$. Accordingly, from the assumption, we get
$\lambda_{\varphi_\alpha}\ast\mu=0$ for all $\alpha$. It follows that $\mu=0$.$\hfill\blacksquare$\\

When $\cal X$ is a $\cal G$-space, by the use of the action
of $\cal G$ on $\cal X$, we introduce a certain subspace of
$C_b({\cal X})$ which is the main object of study of this work. In
details, following Lau and Chu \cite{chulau} for a given $s$ in
$\cal G$, we define the left translation of a function $F\in
C_b({\cal X})$, by an element $s\in G$, by $l_sF(x)=F(s\cdot x)$ for all $x\in {\cal X}$
and consider
$$LUC({\cal X},{\cal G})=\Big{\{}F\in C_b({\cal X}):~ s\in {\cal G} \mapsto l_sF\in
(C_b({\cal X}),\|\cdot\|_\infty)  {\hbox{~is~
continuous}}\Big{\}}.$$ Then $LUC({\cal X},{\cal G})$ is a closed
subspace of $C_b({\cal X})$ which is invariant under left
translation. We always let $\cal G$ act on itself by left
translation and therefore $LUC({\cal G})=LUC({\cal G},{\cal G})$.
It is well-known that $C_0({\cal G})\subseteq LUC({\cal G})$ (see
\cite[Proposition 2.6]{farmonic}), by the same argument one can
prove the following result, which guarantees that $LUC({\cal
X},{\cal G})$ contains enough elements to separate the points of
$\cal X$.

\begin{lemma}\label{luc0}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then the Banach space $C_0({\cal X})$ is contained in
$LUC({\cal X},{\cal G})$.
\end{lemma}

Moreover, if $LUC({\cal X},{\cal G})^*$ denotes the first dual
space of the Banach space $LUC({\cal X},{\cal G})$, $M$ is an
arbitrary element of $LUC({\cal X},{\cal G})^*$ and $F\in
LUC({\cal X},{\cal G})$, we define the function $MF:{\cal
G}\rightarrow \Bbb{C}$ by
\begin{eqnarray*}\label{maction}
MF(s)=\langle M,l_sF\rangle ,
\end{eqnarray*}
for all $s\in {\cal G}$, which belongs to $LUC({\cal G})$; This is
because of,
\begin{eqnarray}\label{211228}
{_s}(MF)(t)=(MF)(st)
=\langle M,l_{st}F\rangle
=\langle M,l_t(l_sF)\rangle
=(M(l_sF))(t),
\end{eqnarray}
for all $s, t\in {\cal G}$. Also we have the following lemma which
plays a key role in this paper.

\begin{lemma}\label{1}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then the mapping
\begin{eqnarray*}
\left\{\begin{array}{clcr}
LUC({\cal X},{\cal G})^*\times LUC({\cal X},{\cal G})\rightarrow LUC({\cal G})\\
(M,F)\mapsto MF
\end{array}
\right.
\end{eqnarray*}
is a bounded bilinear map with
$\|MF\|_\infty\leq\|M\|\;\|F\|_\infty$.
\end{lemma}

Lemma~\ref{1} paves the way for defining a bounded bilinear
map as follows
\begin{eqnarray*}
\left\{\begin{array}{clcr}
LUC({\cal G})^*\times LUC({\cal X},{\cal G})^*\rightarrow LUC({\cal X},{\cal G})^*\\
(m,M)\mapsto m\cdot M
\end{array}
\right.
\end{eqnarray*}
with $\|m\cdot M\|\leq\|m\|\;\|M\|$, where
\begin{eqnarray}\label{eq1}
\langle m\cdot M,F\rangle=\langle m,MF\rangle,
\end{eqnarray}
for all $F\in LUC({\cal X},{\cal G})$.

\begin{proposition}\label{2}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Under the mapping $(m,M)\mapsto m\cdot M$, defined by
{\rm(\ref{eq1}\rm)}, $LUC({\cal X},{\cal G})^*$ becomes a Banach
left $LUC({\cal G})^*$-module with $\|m\cdot M\|\leq\|m\|\;\|M\|$
and $\delta_e\cdot M=M$.
\end{proposition}
{\noindent Proof.} It is enough to prove that $(m\odot n)\cdot
M=m\cdot(n\cdot M)$, where $m, n\in LUC({\cal G})^*$ and $M\in
LUC({\cal X},{\cal G})^*$. To this end, suppose that $F\in
LUC({\cal X},{\cal G})$. Then, by (\ref{arensluc}) and (\ref{211228}), we can see that for all $s\in {\cal G}$
\begin{eqnarray*}
(n\diamond (MF))(s)=\langle n,{_s(MF)}\rangle =\langle
n,M(l_sF)\rangle =\langle n\cdot M,l_sF\rangle =((n\cdot M)F)(s),
\end{eqnarray*}
which shows that $n\diamond (MF)= (n\cdot M)F$. Therefore,
\begin{eqnarray*}
\langle(m\odot n)\cdot M,F\rangle &=&\langle m\odot n,MF\rangle\\
&=&\langle m,n\diamond (MF)\rangle\\
&=&\langle m,(n\cdot M)F\rangle\\
&=&\langle m\cdot(n\cdot M),F\rangle,
\end{eqnarray*}
and this completes the proof.$\hfill\blacksquare$\\

If now, for a given $\sigma\in M({\cal X})$, we define a linear
functional on $LUC({\cal X},{\cal G})$, denoted again by $\sigma$,
which assigns to each $F\in LUC({\cal X},{\cal G})$ the value
$\int_{\cal X} F(x)\,d\sigma(x)$. Then, $M({\cal X})$ may be
regarded as a subspace of $LUC({\cal X},{\cal G})^*$. Moreover, it
is not hard to check that the inclusion $LUC({\cal G})^*\cdot
M({\cal X})\subseteq M({\cal X})$ can fail even if ${\cal X}={\cal
G}$, where
$$LUC({\cal G})^*\cdot M({\cal X})={\Big\{}m\cdot\sigma:~m\in LUC({\cal G})^*,
~\sigma\in M({\cal X}){\Big\}};$$
 In other word, the Banach space $M({\cal X})$ is not in
general an $LUC({\cal G})^*$-submodule of $LUC({\cal X},{\cal
G})^*$. On the other hand, if $\cal X$ is compact, then $M({\cal
X})$ is an $LUC({\cal G})^*$-submodule of $LUC({\cal X},{\cal
G})^*$; This is because of, in this case $M({\cal X})=LUC({\cal
X},{\cal G})^*$. We do not know if the converse of this fact is
valid in general; here, we prove the converse under an extra assumption. The
notation ${\cal CLS}({\cal X},{\cal G})$ in this proposition
and in the sequel denotes the norm closure of the linear span of
the set
$$LUC({\cal X},{\cal G})^*LUC({\cal X},{\cal G}):=
{\Big\{}MF:~ M\in LUC({\cal X},{\cal G})^*,~ F\in LUC({\cal
X},{\cal G}){\Big\}},$$ with respect to the norm topology of
$LUC({\cal G})$. Also, we say that the action of $LUC({\cal G})^*$
on $LUC({\cal X},{\cal G})^*$ is faithful if $m\in LUC({\cal
G})^*$ is so that $m\cdot M=0$ for all $M\in LUC({\cal X},{\cal
G})^*$, then $m=0$.

\begin{proposition}\label{400}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then the following statements hold.
\newcounter{j12}
\begin{list}%
{\bf(\alph{j12})}{\usecounter{j12}}
\item The action of
$LUC({\cal G})^*$ on $LUC({\cal X},{\cal G})^*$ is faithful if and only if ${\cal CLS}({\cal
X},{\cal G})=LUC({\cal G})$.

\item If the action of $LUC({\cal G})^*$ on $LUC({\cal X},{\cal
G})^*$ is faithful, then $\cal G$ acts effectively on $\cal X$.

\item If $\cal X$ is compact, then $M({\cal X})$ is an $LUC({\cal
G})^*$-submodule of $LUC({\cal X},{\cal G})^*$. The converse is
also true if $\cal X$ is a transitive $\cal G$-space.
\end{list}
\end{proposition}

{\noindent Proof.} {\bf(a)} The proof of this assertion uses only
the Hahn-Banach theorem and so the details are omitted.

{\bf (b)} If the action of $LUC({\cal G})^*$ on $LUC({\cal
X},{\cal G})^*$ is faithful and $s\in {\frak N}({\cal X},{\cal
G})$, then for all $M\in LUC({\cal X},{\cal G})^*$ and $F\in
LUC({\cal X},{\cal G})$ we have
\begin{eqnarray*}
\langle \delta_{s}\cdot M , F\rangle =\langle \delta_{s},
MF\rangle =
 MF(s)=\langle M,l_sF\rangle= \langle M , F\rangle.
\end{eqnarray*}
So, $\delta_{s}\cdot M=M$ for all $M\in LUC({\cal X},{\cal
G})^*$ and hence, $\delta_{s}=\delta_e$; that is, $s=e$.

{\bf (c)} We only need to prove the converse of this assertion, which is the
essential part of it. To this end, suppose that $M({\cal X})$ is a
submodule of $LUC({\cal X},{\cal G})^*$. Then for all $m\in
LUC({\cal G})^*$ and $\sigma\in M({\cal X})$, we have $m\cdot
\sigma\in M({\cal X})$. This implies that the norm of the linear
functional $m\cdot \sigma$ on $LUC({\cal X},{\cal G})$ is obtained by
the following equality
$$\|m\cdot \sigma\|=\sup{\Big\{}|\langle m\cdot \sigma, F_0 \rangle|:\, F_0\in C_0({\cal
X}),\,\|F_0\|_\infty=1{\Big\}}.$$ Therefore, $$|\langle m\cdot
\sigma, F \rangle|\leq \sup {\Big\{}|\langle m\cdot \sigma, F_0
\rangle|:\, F_0\in C_0({\cal X}),\,\|F_0\|_\infty=1{\Big\}},$$ for
all $F\in LUC({\cal X},{\cal G})$ with $\|F\|_\infty=1$. Hence, for
all $m\in LUC({\cal G})^*$, $\sigma\in M({\cal X})$, and $F\in
LUC({\cal X},{\cal G})$ with $\|F\|_\infty=1$ we have
\begin{eqnarray}\label{eq3}
|\langle m, \sigma F  \rangle|\leq \sup {\Big\{}|\langle m, \sigma
F_0 \rangle|:\, F_0\in C_0({\cal X}),\,\|F_0\|_\infty=1{\Big\}}.
\end{eqnarray}
From this, we deduce that, for all $\sigma\in M({\cal X})$ and
$F\in LUC({\cal X},{\cal G})$, $\sigma F$ is in the norm closure
of the linear span of the set
$$\sigma C_0({\cal X}):={\Big\{}\sigma F_0:~ F_0\in C_0({\cal
X}){\Big\}},$$ with respect to the norm topology of $LUC({\cal
G})$; since, otherwise, we can find a $m\in LUC({\cal G})^*$ so that
$\langle m, \sigma F\rangle=1$ and $\langle m, \sigma
F_0\rangle=0$ for all $F_0\in C_0({\cal X})$, which is a
contradiction to (\ref{eq3}). In particular, if $x$ and $F$ are
arbitrary elements of $\cal X$ and $LUC({\cal X},{\cal G})$,
respectively, then, there exists a sequence
$(F_n)_{n\in\Bbb{N}}\subseteq C_0({\cal X})$ so that $\|\delta_x
F_n-\delta_x F\|_\infty\longrightarrow 0$. Now from the
transitivity of the action of $\cal G$ on $\cal X$, we get
\begin{eqnarray*}
\|F_n-F\|_\infty&\leq&\sup_{s\in {\cal G}}
|F_n(s\cdot x)-F(s\cdot x)|\\&\leq&\|\delta_x F_n-\delta_x F\|_\infty\\
&=&\sup_{s\in {\cal G}}
|(\delta_x F_n)(s)-(\delta_x F)(s)|.
\end{eqnarray*}
Thus, $\|F_n-F\|_\infty\longrightarrow 0$, which implies that $F$
is in $C_0({\cal X})$. Therefore, $LUC({\cal X},{\cal G})\subseteq
C_0({\cal X})$. As $LUC({\cal X},{\cal G})$ contains the constant
functions on $\cal X$,
 we deduce that $\cal X$ is compact.$\hfill\blacksquare$\\

\section{ The weak$^*$ continuity of the left $LUC({\cal G})^*$-module action }

As we know, Lau \cite{lau} has shown that a left translation
$n\mapsto m\odot n$ is weak$^*$ continuous on $LUC({\cal G})^*$
for a fixed $m$ in $LUC({\cal G})^*$ if and only if $m$ is in
$M({\cal G})$; that is, ${\frak{Z}}({\cal G})$, the topological
centre of $LUC({\cal G})^*$, is $M({\cal G})$. It follows that, if
$\cal X$ is a $\cal G$-space, then for an arbitrary $m$ in
$LUC({\cal G})^*$ the weak$^*$ continuity of the map $M\mapsto
m\cdot M$ on $LUC({\cal X},{\cal G})^*$ can fail even if ${\cal
X}={\cal G}$. A problem which is of interest is that for which
element $m\in LUC({\cal G})$ the map $M\mapsto m\cdot M$ on
$LUC({\cal X},{\cal G})^*$ is weak$^*$ to weak$^*$ continuous?
Therefore, it seems valuable to define
\begin{eqnarray*}
{{\frak{Z}}({\cal X},{\cal G})}=\Big{\{}m\in LUC({\cal
G})^*:\,M\mapsto m\cdot M~ {\hbox{is weak}}^*~{\hbox{to weak}}^*
~{\hbox{continuous on}}~LUC({\cal X},{\cal G})^*\Big{\}},
\end{eqnarray*}
the topological centre of the module action induced by $LUC({\cal
G})^*$ on $LUC({\cal X},{\cal G})^*$. In the special case that we
let $\cal G$ act on itself by left multiplication, the set ${\frak{Z}}({\cal G},{\cal G})$ coincides with
${\frak{Z}}({\cal G})$, which defined in Section 1.

This section studies the subspace ${{\frak{Z}}({\cal X},{\cal
G})}$ of $LUC({\cal G})^*$ in the case where $\cal X$ is a $\cal
G$-space and, in particular, the question when the subspace
${{\frak{Z}}({\cal X},{\cal G})}$ is $M({\cal G})$ or $LUC({\cal G})^*$. Before
proceeding further in this section, we should note that if $\cal
X$ is a $\cal G$-space, then $LUC({\cal X},{\cal G})^*$ is a left
Banach $G$-module. In fact, it is suffices to define the left action
of $\cal G$ on $LUC({\cal X},{\cal G})^*$ by $(s,M)\mapsto
\delta_s\cdot M$.

\begin{lemma}\label{30}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then the action of $\cal G$ on the unit ball of
$LUC({\cal X},{\cal G})^*$ with the weak$^*$ topology is jointly continuous.
\end{lemma}
{\noindent Proof.} For the proof, it is suffices to note that if
$(s_\alpha)$ is a net converging to $s$ in $\cal G$, then the net
$(\delta_{s_\alpha})$ tends to $\delta_s$ with respect to the
weak$^*$ topology of $LUC({\cal G})^*$.$\hfill\blacksquare$\\

Let $\cal X$ be a $\cal G$-space. Given $F\in LUC({\cal X},{\cal
G})$ and $x\in{\cal X}$, we define $r_xF$ on $\cal G$ by
$$(r_xF)(s)=F(s\cdot x),\quad\quad\quad (s\in {\cal G}),$$ then a
routine computation shows that
$$\|{_{s_\alpha}}(r_xF)-{_{s}}(r_xF)\|_\infty\leq
\|l_{s_\alpha}F-l_{s}F\|_\infty,$$ where $(s_\alpha)$ is a net in
${\cal G}$ which tends to $s$, and therefore $r_xF$ is a function
in $LUC({\cal G})$. Hence, if $m$ is an arbitrary element of
$LUC({\cal G})^*$, then we can define a complex-valued function
$Fm$ on $\cal X$ by
$$Fm(x)=\langle m,r_xF\rangle\quad\quad (x\in {\cal X}).$$
Observe that $Fm$ is a bounded function on $\cal X$ with
$\|Fm\|_{\infty}\leq \|m\|\, \|F\|_{\infty}$. The following
lemma shows that the continuity of the function $Fm$ can fail
even if ${\cal X}={\cal G}$.

\begin{lemma}\label{9000}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then, for all $m\in LUC({\cal G})^*$ and $F\in LUC({\cal X},{\cal G})$, the following assertions hold.
\newcounter{j14}
\begin{list}%
{\bf(\alph{j14})}{\usecounter{j14}}
\item For all $x\in {\cal X}$, $Fm(x)=\langle
m\cdot\delta_x,F\rangle$.
\item If $m\in
{\frak{Z}}({\cal X},{\cal G})$, then $Fm$ is in $C_b({\cal
X})$.
\item If $m\in {\frak{Z}}({\cal X},{\cal G})$, then
$\langle\delta_x,l_s(Fm)\rangle
=\langle(m\odot\delta_s)\cdot\delta_x,F\rangle$ for all $x\in
{\cal X}$ and $s\in {\cal G}$.
\end{list}
\end{lemma}
{\noindent Proof.} Let $m\in LUC({\cal G})^*$, $F\in LUC({\cal
X},{\cal G})$, $x\in {\cal X}$ and $s\in {\cal G}$ be given.

{\bf (a)} First, observe that
\begin{eqnarray}\label{210108}
(\delta_xF)(s)=\langle\delta_x, l_sF\rangle =l_sF(x)=F(s\cdot
x)=r_xF(s),
\end{eqnarray}
for all $s\in {\cal G}$. Hence, we have
\[\langle m\cdot\delta_x,F\rangle =\langle m,\delta_xF\rangle =
 \langle m,r_xF\rangle =Fm(x).\]

{\bf (b)} If $m\in {\frak{Z}}({\cal X},{\cal G})$ and
$(x_\alpha)\subseteq {\cal X}$ converges to $x\in{\cal X}$, then the net
$(\delta_{x_\alpha})$ tends to $\delta_x$ in the weak$^*$
topology of $LUC({\cal X},{\cal G})^*$. Therefore, since
$m\in{\frak{Z}}({\cal X},{\cal G})$, we have
\[(Fm)(x_\alpha)=\langle m\cdot\delta_{x_\alpha},F\rangle \longrightarrow
\langle m\cdot\delta_{x},F\rangle =(Fm)(x).\] This together with
the fact that $\|Fm\|_{\infty}\leq \|m\|\, \|F\|_{\infty}$,
implies that $Fm\in C_b({\cal X})$.

{\bf (c)} In this case, $r_xF$ is a function in $LUC({\cal G})$,
$Fm\in C_b({\cal X})$ and for each $t\in{\cal G}$ we have
\begin{eqnarray*}
\delta_s\diamond(r_xF)(t)=\langle\delta_s,{_t}(r_xF)\rangle =
{_t}(r_xF)(s)=F(t\cdot(s\cdot x))=(r_{s\cdot x}F)(t),
\end{eqnarray*}
From this, by (\ref{210108}), we deduce that
\begin{eqnarray*}
\langle\delta_x,l_s(Fm)\rangle=Fm(s\cdot x)=\langle m,r_{s\cdot x}F\rangle=
\langle m\odot\delta_s,r_xF\rangle
=\langle(m\odot\delta_s)\cdot\delta_x,F\rangle ,
\end{eqnarray*}
as desired.$\hfill\blacksquare$\\

Now, suppose that $m$ is an element of $LUC({\cal G})^*$ such that,
for each $F\in LUC({\cal X},{\cal G})$, the function $Fm$ is in
$LUC({\cal X},{\cal G})$. Then every $M$ in $LUC({\cal X},{\cal
G})^*$ gives a linear functional $M\bullet m$ on $LUC({\cal
X},{\cal G})$ as follows
$$\langle M\bullet m, F\rangle =\langle M, Fm\rangle .$$
In addition, if $M\bullet m=m\cdot M$, for all $M\in LUC({\cal
X},{\cal G})^*$, then $m\in {\frak{Z}}({\cal X},{\cal G})$;
Indeed, if $(M_\alpha)\subseteq LUC({\cal X},{\cal G})^*$ tends to
an element $M$ with respect to the weak$^*$ topology, then we have
$$
\lim_\alpha\langle m\cdot M_\alpha, F\rangle= \lim_\alpha\langle
M_\alpha\bullet m, F\rangle=\langle M, Fm\rangle=\langle
M\bullet m, F\rangle=\langle m\cdot M, F\rangle,
$$
for all $F\in LUC({\cal X}, {\cal G})$. The following theorem
consider the converse of this fact whose proof is inspired by
\cite[Lemma 2.2]{lau}.

\begin{theorem}\label{5}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then ${\frak{Z}}({\cal X},{\cal G})$ is precisely the
set of all $m\in LUC({\cal G})^*$ for which the following
conditions are satisfied
\newcounter{j13}
\begin{list}%
{\bf(\alph{j13})}{\usecounter{j13}} \item $Fm\in LUC({\cal
X},{\cal G})$ for all $F\in LUC({\cal X},{\cal G})$, \item
$M\bullet m=m\cdot M$ for all $M\in LUC({\cal X},{\cal G})^*$.
\end{list}
\end{theorem}

{\noindent Proof.} Suppose that $m\in {\frak{Z}}({\cal X},{\cal
G})$ and $F$ is an arbitrary element of
$LUC({\cal X},{\cal G})$. Then, the map $M\mapsto m\cdot M$ is weak$^*$ continuous on
$LUC({\cal X},{\cal G})^*$ and therefore it is weak$^*$ continuous
on bounded sets. Also, part {\bf(b)} of Lemma \ref{9000} implies that $Fm$ is a function in $C_b({\cal X})$.
Moreover, in light of Lemma \ref{30}, part {\bf (c)} of Lemma \ref{9000}, \cite[Lemma 2.5]{bami} and the weak$^*$ continuity of the map $M\mapsto m\cdot M$ on bounded set, we have
\begin{eqnarray}\label{210157}
\langle\Lambda,l_s(Fm)\rangle
=\langle(m\odot\delta_s)\cdot\Lambda,F\rangle
\end{eqnarray}
for all
$\Lambda\in C_b({\cal X})^*$ and $s\in {\cal G}$. Now, suppose
that $(s_\alpha)$ is a net in $\cal G$ converging to $s$ and for
each $\alpha$ the functional $\Lambda_\alpha\in C_b({\cal X})^*$
is chosen so that $\|\Lambda_\alpha\|=1$ and
$$\|l_{s_\alpha}(Fm)-l_s(Fm)\|_{\infty}={\Big|}
\langle\Lambda_\alpha,l_{s_\alpha}(Fm)-l_s(Fm)\rangle
{\Big|}.$$ Let also, $\Lambda$ be a weak$^*$-cluster point of
$(\Lambda_\alpha)$. By passing to a subnet, if necessary, we may assume
that $\Lambda_\alpha\longrightarrow\Lambda$ in the weak$^*$
topology of $LUC({\cal X},{\cal G})^*$.
Hence
\begin{eqnarray*}
\|l_{s_\alpha}(Fm)-l_s(Fm)\|_{\infty}&\leq&{\Big|}
\langle(m\odot\delta_{s_\alpha})\cdot\Lambda_\alpha-(m\odot\delta_{s})\cdot\Lambda,F\rangle {\Big|}\\&&
+{\Big|}\langle(m\odot\delta_{s})\cdot\Lambda-(m\odot\delta_{s})\cdot\Lambda_\alpha,F\rangle {\Big|}.
\end{eqnarray*}
Now, the weak$^*$ continuity of the map $M\mapsto m\cdot M$ on norm
bounded subsets of $LUC({\cal X},{\cal G})^*$ and Lemma \ref{30}
imply that $Fm$ is in $LUC({\cal X},{\cal G})$.

Finally, to prove the equality $M\bullet m=m\cdot M$ for all $M\in
LUC({\cal X},{\cal G})^*$. First note that $\delta_x\bullet
m=m\cdot \delta_x$, for all $x\in {\cal X}$. Hence, the equality
$M\bullet m=m\cdot M$ holds if $M$ is a convex combination of the
Dirac measures. We now invoke \cite[Lemma 2.5]{bami} to conclude
that $M\bullet m=m\cdot M$ holds for all $M\in LUC({\cal X},{\cal
G})^*$.$\hfill\blacksquare$\\

It is clear that if $m\in {\frak Z}({\cal X},{\cal G})$, then
the map $M\mapsto m\cdot M$ is weak$^*$ continuous on all bounded
parts of $LUC({\cal X},{\cal G})^*$. As we have seen in the proof of Theorem \ref{5}, that
the map $M\mapsto m\cdot M$ is weak$^*$ continuous on all bounded
parts of $LUC({\cal X},{\cal G})^*$ plays a key role in its proof. Hence,
with an argument
similar to the proof of Theorem \ref{5} one
can prove the following result. The details are omitted.

\begin{corollary}\label{13-8-94}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then $m\in {\frak Z}({\cal X},{\cal G})$ if and only if
the map $M\mapsto m\cdot M$ is weak$^*$ continuous on all bounded
parts of $LUC({\cal X},{\cal G})^*$.
\end{corollary}

The following two propositions paves the way for obtaining the wide class of
$\cal G$-space $\cal X$ for which $$M({\cal G})\subsetneqq{\frak Z}({\cal X},{\cal G})=LUC({\cal
G})^*.$$ First let us recall that, if $\cal H$ is a subgroup of $\cal G$, then the {\it
index} of $\cal H$ in $\cal G$, denoted by $[{\cal G}:{\cal H}]$,
is the number of left cosets of $\cal H$ in $\cal G$.

\begin{proposition}\label{0202}
Let $\cal G$ be a locally compact group and $\cal X$ be a
$\cal G$-space. If the index of ${\frak N}({\cal X},{\cal G})$ in
$\cal G$ is finite, then ${\frak Z}({\cal X},{\cal G})=LUC({\cal
G})^*$.
\end{proposition}
{\noindent Proof.} Let ${\frak N}:={\frak N}({\cal X},{\cal G})$
and $m$ be an arbitrary element of $LUC({\cal G})^*$. To prove
that $m\in {\frak Z}({\cal X},{\cal G})$, by Corollary
\ref{13-8-94} above, it will be enough to prove that the map
$M\mapsto m\cdot M$ is weak$^*$ continuous on bounded parts of
$LUC({\cal X},{\cal G})^*$. To this end, suppose that $(M_\alpha)$
is a bounded net in $LUC({\cal X},{\cal G})^*$ which tends to
$M\in LUC({\cal X},{\cal G})^*$ with respect to the weak$^*$
topology. Then, since the index of ${\frak
N}$ in $\cal G$ is finite, we can obtain a partition
${\Big\{}s_1{\frak N},s_2{\frak N},\cdots,s_k{\frak N}{\Big\}}$
for $\cal G$ and therefore for each $s\in{\cal G}$, there exists
$t\in{\frak N}$ and a unique $1\leq i\leq k$ for which $s=s_it$.
It follows that for all $\alpha$, $s\in{\cal G}$ and $F\in LUC({\cal X},{\cal
G})$
\begin{eqnarray*}
M_\alpha F(s)=\sum_{i=1}^kM_\alpha F(s_i)\chi_{s_i{\frak
N}}(s),\quad {\hbox{and}}\quad
MF(s)=\sum_{i=1}^kMF(s_i)\chi_{s_i{\frak N}}(s),
\end{eqnarray*}
 where
$\chi_{s_i{\frak N}}$ denotes the characteristic function of
${s_i{\frak N}}$ on $\cal G$.
On the other
hand, since, for each $F\in LUC({\cal X},{\cal G})$, $M_\alpha
F\longrightarrow MF$ pointwisely, we may choose $\alpha_0$ such
that $|M_\alpha F(s_i)-MF(s_i)|<\varepsilon$ whenever
$i=1,\cdots,k$ and $\alpha\succeq\alpha_0$. Therefore, for each
$\alpha\succeq\alpha_0$, we have
$$\|M_\alpha F-MF\|_\infty={\Big\|}\sum_{i=1}^kM_\alpha F(s_i)\chi_{s_i{\frak
N}}-\sum_{i=1}^kMF(s_i)\chi_{s_i{\frak N}}{\Big\|}_\infty<\varepsilon.$$
This implies that
$$|\big<m\cdot M_\alpha,F\big>-\big<m\cdot M,F\big>|\leq\|m\|\;\|
M_\alpha F-MF\|_\infty\longrightarrow 0.$$ Therefore, $m\in {\frak
Z}({\cal X},{\cal G})$.$\hfill\blacksquare$\\

The following proposition, gives some of the main properties of
the set ${\frak{Z}}({\cal X},{\cal G})$.

\begin{proposition}\label{8}
Let $\cal G$ be a locally compact group and $\cal X$ be a $\cal
G$-space. Then the following assertions hold.
\newcounter{j2}
\begin{list}%
{\bf(\alph{j2})}{\usecounter{j2}} \item ${\frak{Z}}({\cal X},{\cal
G})$ is a subalgebra of $LUC({\cal G})^*$.

\item ${\frak{Z}}({\cal X},{\cal G})$ is closed with respect to
the norm topology of $LUC({\cal G})^*$.

\item $M({\cal G})$ is contained in ${\frak{Z}}({\cal X},{\cal
G})$.
\end{list}
\end{proposition}
{\noindent Proof}. In order to prove {\bf(a)}, assume that $m$ and
$n$ are in ${\frak{Z}}({\cal X},{\cal G})$. Suppose also that $F$
is an arbitrary element of $LUC({\cal X},{\cal G})$. Then, by Theorem \ref{5},
the functions $F':=Fm$ and $F'n$ are in $LUC({\cal X},{\cal G})$.
On the other hand, for each $x\in {\cal X}$, we have
\begin{eqnarray*}
(F'n)(x)&=&\langle n\cdot\delta_x,Fm\rangle\\
&=&\langle(n\cdot\delta_x)\bullet m,F\rangle\quad\quad\quad ({\hbox{since}}~
Fm\in LUC({\cal X},{\cal
G}))\\
&=&\langle m\cdot(n\cdot\delta_x),F\rangle~\quad\quad\quad ({\hbox{since}}~
m\in{\frak{Z}}({\cal X},{\cal
G}))\\
&=&\langle (m\odot n)\cdot\delta_x,F\rangle\\
&=&(F(m\odot n))(x).
\end{eqnarray*}
Hence, $F(m\odot n)$ is in $LUC({\cal X},{\cal G})$. Moreover,
for each $M\in LUC({\cal X},{\cal G})^*$, we have
\begin{eqnarray*}
\langle(m\odot n)\cdot M,F\rangle&=&\langle m\cdot(n\cdot M),F\rangle\\
&=&\langle (n\cdot M)\bullet
m,F\rangle\\
&=&\langle M\bullet n,Fm\rangle\\ &=&\langle M,
F'n\rangle\\&=&\langle M\bullet(m\odot n), F\rangle.
\end{eqnarray*}
Therefore, Theorem \ref{5} implies that
$m\odot n$ is in ${\frak{Z}}({\cal X},{\cal G})$.

Now, for the proof of the assertion {\bf(b)}, let $(m_k)$ be a
sequence in ${\frak{Z}}({\cal X},{\cal G})$ which tends to $m\in
LUC({\cal G})^*$ with respect to the norm topology. Also, suppose that
$F\in LUC({\cal X},{\cal G})$ and $M\in LUC({\cal X},{\cal G})^*$ are given.
From Lemma
\ref{9000}, we observe that
\begin{eqnarray*}
\|Fm_k-Fm\|_\infty&=&\sup_{x\in{\cal X}}|Fm_k(x)-Fm(x)|\\
&=&\sup_{x\in{\cal X}}|\langle(m_k-m)\cdot\delta_x,F\rangle |\\
&\leq&\|m_k-m\|\|F\|_\infty.
\end{eqnarray*}
This together with the fact that
 $m_k\in{\frak{Z}}({\cal X},{\cal G})$ implies that $Fm$ is an element of
$LUC({\cal X},{\cal G})$. On the other hand,
\begin{eqnarray*}
\langle m\cdot M,F\rangle
=\lim_k\langle M\bullet m_k, F\rangle
=\lim_k\langle M, Fm_k\rangle
=\langle M, Fm\rangle
=\langle M\bullet m, F\rangle.
\end{eqnarray*}
We now invoke Theorem \ref{5} to conclude that
$m\in {\frak{Z}}({\cal X},{\cal G})$.

Finally, for the proof of the last assertion, let $\mu$ be a
measure in $M({\cal G})$. Since the
measures in $M({\cal G})$ with compact supports are norm dense in
$M({\cal G})$, without loss of generality, we may
assume that $\mu$ has compact support. Let also $(M_\alpha)$ be a bounded net in
$LUC({\cal X},{\cal G})^*$ such that $M_\alpha\rightarrow M$ with
respect to the weak$^*$ topology, for some $M\in LUC({\cal
X},{\cal G})^*$. Choose $K>0$ such that $\|M_\alpha\|,\,\|M\|\leq
K$ for all $\alpha$. For each $F\in LUC({\cal X},{\cal G})$, $\alpha$ and
$s, t\in {\cal G}$ we have
\[|M_\alpha F(s)-M_\alpha F(t)|=|\langle M_\alpha,l_sF\rangle
-\langle M_\alpha ,l_tF\rangle |\leq K\|l_sF-l_tF\|_\infty.\] Hence, the
family of functions $M_\alpha F$ is equicontinuous. Therefore,
since $M_\alpha
F\longrightarrow MF$ pointwisely, the net
$M_\alpha F$  convergence uniformly to $MF$ on every compact
subset of $\cal G$. Thus,
\[|\langle\mu\cdot M_\alpha,F\rangle -\langle\mu\cdot M,F\rangle|
={\Big|}\int_{\cal G} M_\alpha F\;d\mu-\int_{\cal G}
MF\;d\mu{\Big|}\longrightarrow 0.\] It follows that the map
$M\mapsto\mu\cdot M$ is weak$^*$ continuous on all bounded parts
of $LUC({\cal X},{\cal G})^*$. We now invoke Corollary
\ref{13-8-94} to conclude that
$\mu\in {\frak{Z}}({\cal X},{\cal G})$.$\hfill\blacksquare$\\

The following result gives some necessary
 and sufficient conditions
for the validity of the equality ${\frak{Z}}({\cal X},{\cal
G})=M({\cal G})$.

\begin{theorem}\label{13}
Let $\cal G$ be a locally compact non-compact group and let $\cal
X$ be a $\cal G$-space. Then the following assertions are
equivalent.
\newcounter{j19}
\begin{list}%
{\bf(\alph{j19})}{\usecounter{j19}} \item ${\frak{Z}}({\cal
X},{\cal G})=M({\cal G})$. \item ${\cal CLS}({\cal X},{\cal
G})=LUC({\cal G})$. \item The action of $LUC({\cal G})^*$ on
$LUC({\cal X},{\cal G})^*$ is faithful.
\end{list}
\end{theorem}

{\noindent Proof.} By Proposition \ref{400}, we only need to prove
{\bf (a)}$\Leftrightarrow${\bf(b)}. In order
to prove that {\bf(a)} implies {\bf(b)},
suppose that ${\frak{Z}}({\cal X},{\cal G})=M({\cal G})$. If we
had an element $f$ in $LUC({\cal G})$ not belonging to the
subspace ${\cal CLS}({\cal X},{\cal G})$, then, by the Hahn-Banach
Theorem, we would have a functional $m\in LUC({\cal G})^*$
vanishing on the subspace ${\cal CLS}({\cal X},{\cal G})$ and such
that $\big<m,f\big>\neq 0$. From this, we deduce that $m\cdot M=0$
for all $M\in LUC({\cal X},{\cal G})^*$. It follows that $m\in
{\frak{Z}}({\cal X},{\cal G})=M({\cal G})$. Now, let
$$C_0({\cal G})^\perp={\Big\{}n\in LUC({\cal G})^*:~\big<n,f\big>=0~~{\hbox{for all}}~f\in C_0({\cal G}){\Big\}},$$
and pick any $0\neq m'\in C_0({\cal G})^\perp$
that is right cancellable in $LUC({\cal G})^*$ (such points exists
by \cite[Theorem 4]{filrc}). Note that for $m'$ and each $M\in
LUC({\cal X},{\cal G})^*$, we have
$$\big<(m\odot m')\cdot M,F\big>=\big<m\cdot(m'\cdot M),F\big>=0$$
for all $F\in LUC({\cal X},{\cal G})$. Therefore, $(m\odot
m')\cdot M=0$ for all $M\in LUC({\cal X},{\cal G})^*$ and this
implies that $m\odot m'\in {\frak{Z}}({\cal X},{\cal G})=M({\cal
G})$. From this, by \cite[Lemma 1.1]{glausert}, we deduce that
$$m\odot m'\in M({\cal G})\cap C_0({\cal G})^\perp=\{0\}.$$ This together
with the fact that $0\neq m'$ is right cancellable, implies that $m=0$ which is impossible.

To prove implication {\bf(b)}$\Rightarrow${\bf(a)}, suppose that
${\cal CLS}({\cal X},{\cal G})=LUC({\cal G})$. By part {\bf (c)}
of Proposition \ref{8}, the inclusion $M({\cal
G})\subseteq{\frak{Z}}({\cal X},{\cal G})$ being always true, it
will be enough to prove the reverse inclusion. To this end,
suppose that $m\in {\frak{Z}}({\cal X},{\cal G})$ and $(n_\alpha)$
is a bounded net in $LUC({\cal G})^*$ such that
$n_\alpha\longrightarrow n$ in $LUC({\cal G})^*$ with respect to
the weak$^*$-topology. Hence, for every $M\in LUC({\cal X},{\cal
G})^*$ and $F\in LUC({\cal X},{\cal G})$, we have
\begin{eqnarray*}
\langle m\odot n_\alpha,MF\rangle &=&\langle(m\odot n_\alpha)\cdot
M,F\rangle\\ &=&\langle m\cdot(n_\alpha\cdot M),F\rangle
\\&\longrightarrow &\langle m\cdot(n\cdot M),F\rangle\\ &=&\langle m\odot
n,MF\rangle ;
\end{eqnarray*}
This is because of, the net $(n_\alpha\cdot M)$ is a net in $LUC({\cal X},{\cal G})^*$
which converges to $nM$ in the weak$^*$ topology of $LUC({\cal X},{\cal G})^*$.
We now invoke the equality ${\cal CLS}({\cal X},{\cal
G})=LUC({\cal G})$ to conclude that
$$\langle m\odot n_\alpha,f\rangle \longrightarrow\langle m\odot n,f\rangle ,$$
for all $f\in LUC({\cal G})$. From this, by \cite[Theorem
4.1 and Lemma 2.2]{lau}, we deduce that $m$ belongs to ${\frak{Z}}({\cal
G})=M({\cal G})$. Therefore, ${\frak{Z}}({\cal X},{\cal
G})\subseteq M({\cal G})$. Hence, we have the equality
${\frak{Z}}({\cal X},{\cal G})=M({\cal G})$.$\hfill\blacksquare$\\

We conclude this section with the following result which is a
consequence of Theorem \ref{210422}, part (b) of Proposition \ref{400}
and  Theorem~\ref{13}.

\begin{corollary}\label{500}
Let $\cal G$ be a locally compact non-compact group and $\cal X$
be a $\cal G$-space. If $\cal G$ does not act effectively on $\cal
X$, then $M({\cal G})$ is properly contained in ${\frak{Z}}({\cal
X},{\cal G})$.
\end{corollary}

\section{\normalsize\bf Examples}\label{s3}

This section is devoted to examples with two different purposes.
First, illustrating the results presented
in this paper for certain $\cal G$-spaces $\cal X$. Second, to give some example for which

$\centerdot$ ${\frak Z}({\cal X},{\cal G})=M({\cal G})$;

$\centerdot$ $M({\cal G})\subsetneqq{\frak Z}({\cal X},{\cal G})=LUC({\cal G})^*$;

$\centerdot$ ${\frak Z}({\cal X},{\cal G})$ is neither $M({\cal G})$ nor $LUC({\cal G})^*$.\\ To this end,
we commence this section with the following example which characterize the subalgebra
${\frak Z}({\cal X},{\cal G})$ of $LUC({\cal G})^*$ for certain $\cal G$-spaces $\cal X$.

\begin{example}\label{exmpl1112}
{\rm Let $\cal G$ be a locally compact group and $\cal X$ be a
$\cal G$-space.
\newcounter{q1021}
\begin{list}%
{\bf(\alph{q1021})}{\usecounter{q1021}} \item If ${\cal X}={\cal
G}$, then $M({\cal
G})={\frak{Z}}({\cal X},{\cal G})$.

\item If $\cal G$ is compact, then
\begin{eqnarray*}
M({\cal G})\subseteq {\frak{Z}}({\cal X},{\cal G})\subseteq
LUC(G)^*=M({\cal G})
\end{eqnarray*}
which shows that ${\frak{Z}}({\cal X},{\cal G})=M({\cal G})$.

\item If $\cal X$ is a finite discrete space, then
${\frak{Z}}({\cal X},{\cal G})=LUC({\cal G})^*$; Indeed, the
vector space $LUC({\cal X},{\cal G})$, as a subspace of $C_b({\cal
X})$, is a finite dimensional vector space, and so all locally
convex topologies on finite dimensional space $LUC({\cal X},{\cal G})^*$ are coincide. In
particular, for each $m\in LUC({\cal G})^*$ the linear map
$M\mapsto m\cdot M$ is continuous with respect to the norm
topology of $LUC({\cal X},{\cal G})^*$,
and also it is continuous with respect to the weak$^*$ topology. Therefore,
${\frak{Z}}({\cal X},{\cal G})=LUC({\cal G})^*$.
Specially, if $\cal G$ is non-compact, then $M({\cal G})\subsetneqq
{\frak Z}({\cal X},{\cal G})=LUC({\cal G})^*$.
\end{list}
}
\end{example}

It is worthwhile to mention that when $\cal G$ is a locally
compact group and $\cal H$ is a closed subgroup of $\cal G$, then
the space ${\cal G}/{\cal H}$ consisting of all left cosets of
$\cal H$ in $\cal G$ is a locally compact Hausdorff topological
space on which $\cal G$ acts from the left by
$${\cal G}\times{\cal G}/{\cal H}\rightarrow{\cal G}/{\cal H};~(s,t{\cal
H})\mapsto (st){\cal H}.$$ It has been shown
that if $\cal G$ is $\sigma$-compact, then every transitive $\cal
G$-space is homeomorphic to the quotient space ${\cal G}/{\cal H}$
for some closed subgroup $\cal H$, see
\cite[Subsection~2.6]{farmonic}.

In what follows, the notation ${\frak C}_{\cal G}({\cal H})$ is
used to denote the centralizer of the subgroup $\cal H$ in $\cal
G$; That is, $${\frak C}_{\cal G}({\cal H})={\Big\{}s\in {\cal G}:\,sh=hs,
\,{\hbox{for all}}~h\in {\cal H}{\Big\}}.$$

\begin{example}\label{exmpl111}
{\rm Let $\cal G$ be a locally compact, non-compact group and let
$\cal H$ and $\cal K$ be two non-trivial closed normal subgroups
of $\cal G$.
\newcounter{q1020}
\begin{list}%
{\bf(\alph{q1020})}{\usecounter{q1020}} \item If ${\cal X}={\cal
G}/{\cal H}$, then, by Corollary \ref{500}, $M({\cal
G})\varsubsetneq {\frak{Z}}({\cal X},{\cal G})$; This is because
of, in this case
\begin{eqnarray*}
{\frak N}({\cal X},{\cal G})={\Big \{}s\in {\cal G}:\,st{\cal H}=t{\cal
H},\,t\in {\cal G}{\Big\}}= {\cal H}.
\end{eqnarray*}
In particular, if the index of $\cal H$ in $\cal G$ is finite, then, by
Proposition \ref{0202}, we have ${\frak{Z}}({\cal X},{\cal
G})=LUC({\cal G})^*$.

\item If ${\cal X}={\cal K}$, then, obviously, $\cal G$ acts on
$\cal X$ by conjugation and makes ${\cal X}$ into a $\cal
G$-space; That is, $${\cal G}\times {\cal X}\rightarrow {\cal
X};~(s,x)\mapsto x^s:=sxs^{-1}.$$ In this case, ${\frak N}({\cal
X},{\cal G})={\frak C}_{\cal G}({\cal K})$. Hence, we can say if
either ${\frak C}_{\cal G}({\cal G})$ or ${\frak C}_{\cal G}({\cal
K})$ is non-trivial (for example, if $\cal K$ is abelian), then we
have $M({\cal G})\varsubsetneq {\frak{Z}}({\cal X},{\cal G})$.
\end{list}
}
\end{example}

Now, we consider the following special case of the previous example
which illustrates Proposition \ref{0202} and gives an example of $\cal G$ and $\cal X$
such that $M({\cal G})\subsetneqq
{\frak Z}({\cal X},{\cal G})=LUC({\cal G})^*$.

\begin{example}
{\rm Let $\Bbb Z$ be the discrete group of integer numbers and
let, for each $n\in {\Bbb Z}^+$, $n{\Bbb Z}$ be the subgroup generated
by $n$ consists of all integer multiples of $n$. If ${\cal X}_n={\Bbb Z}/{n{\Bbb Z}}$,
then Corollary \ref{500} together with the fact that
${\frak N}({\cal X}_n,{\Bbb Z})=n{\Bbb Z}$ implies that
$M({\Bbb Z})\varsubsetneq {\frak{Z}}({\cal X}_n,{\Bbb Z})$.
On the other hand, since the index of $n{\Bbb Z}$ in $\Bbb Z$ is $n$, by
Proposition \ref{0202}, we have ${\frak{Z}}({\cal X}_n,{\Bbb Z})=LUC({\Bbb Z})^*$.}
\end{example}

The following example illustrates Theorem \ref{13}.

\begin{example}
{\rm Suppose that $\cal Y$ is an arbitrary locally compact
Hausdorff space and $\cal G$ is a locally compact
group. Then  ${\cal X}={\cal G}\times {\cal Y}$, equipped with the
product topology and the action defined by
$${\cal G}\times{\cal X}\rightarrow{\cal X};~s\cdot(t,y)=(st,y),$$
is a $\cal G$-space for which ${\frak Z}({\cal X},{\cal G})=M({\cal G})$;
Indeed, by part (b) of Example \ref{exmpl1112}, we only need to show that the equality
${\frak Z}({\cal X},{\cal G})=M({\cal G})$
is valid when $\cal G$ is non-compact. To this end, first note that,
if $f\in LUC({\cal G})$ and $F\in C_b({\cal Y})$, then
the function $$(f\otimes F):{\cal X}\rightarrow{\Bbb C};~(t,y)\mapsto f(t)F(y),$$
is a function in $LUC({\cal X},{\cal G})$; This is because of,
$$l_s(f\otimes F)=({}_sf)\otimes F,$$ for all
$s\in {\cal G}$. In particular, if $1_{\cal Y}$ denotes the constant function 1 on $\cal Y$, then, for each
$f\in LUC({\cal G})$, the function $f\otimes 1_{\cal Y}$ is a function in $LUC({\cal X},{\cal G})$ and therefore we can consider $LUC({\cal G})$ as a closed subspace of $LUC({\cal X},{\cal G})$ via the inclusion mapping
$$\iota:LUC({\cal G})\rightarrow LUC({\cal X},{\cal G});~f\mapsto f\otimes 1_{\cal Y}.$$
Then $\iota^*:LUC({\cal X},{\cal G})^*\rightarrow LUC({\cal G})^*$ is the restriction mapping and hence norm decreasing and onto by Hahn-Banach Theorem.
Moreover, if $m \in LUC({\cal G})^*$ and $M\in
LUC({\cal X},{\cal G})^*$ is chosen so that $\iota^*(M)=m$, then
\begin{eqnarray*}
(m\diamond f)(s)&=&\langle m,{}_sf\rangle\\&=&\langle \iota^*(M),{}_sf\rangle\\&=&
\langle
M,({}_sf)\otimes1_{\cal Y}\rangle\\&=&\langle M,l_s(f\otimes1_{\cal Y})\rangle\\&=&(M(f\otimes1_{\cal Y}))(s),
\end{eqnarray*}
for all $f\in LUC({\cal G})$ and $s\in {\cal G}$.
Hence, we have
\begin{eqnarray*}
{\Big\{}m\diamond f:\,m\in LUC({\cal G})^*,\, f\in LUC({\cal
G}){\Big\}} \subseteq {\Big\{}MF:\,M\in LUC({\cal X},{\cal
G})^*,\,F\in LUC({\cal X},{\cal G}){\Big\}},
\end{eqnarray*}
which shows that ${\cal CLS}({\cal X},{\cal G})=LUC({\cal G})$. We now invoke Theorem \ref{13} to conclude that ${\frak{Z}}({\cal X},{\cal G})=M({\cal G})$.}
\end{example}

The proof of Example \ref{0925} below, which gives a $\cal G$-space $\cal X$
such that ${\frak N}({\cal X},{\cal G})$ is neither $M({\cal G})$ nor $LUC({\cal G})^*$, relies on the following example.

\begin{example}\label{rem1}
{\rm Let ${\cal G}_1$ and ${\cal G}_2$ be two locally compact groups. Then we can
consider $LUC({\cal G}_1)$ as a closed subspace of
$LUC({\cal G}_1\times {\cal G}_2)$ via the inclusion mapping
$$\iota_{{\cal G}_1}:LUC({\cal G}_1)\rightarrow LUC({\cal G}_1\times {\cal G}_2);~
f\mapsto f\otimes 1_{{\cal G}_2},$$
where $1_{{\cal G}_2}$ denotes the constant function 1 on ${{\cal G}_2}$.
Suppose also that
${\cal X}_1$ is a ${\cal G}_1$-space and ${\cal X}_2$ is
a ${\cal G}_2$-space. Then
${\cal X}_1\times {\cal X}_2$, with product topology and coordinatewise
left module action, is a ${\cal G}_1\times {\cal G}_2$-space for which
$$
\iota_{{\cal G}_1}^*{\Big(}{\frak{Z}}({\cal X}_1\times {\cal X}_2, {\cal G}_1\times
{\cal G}_2){\Big)}\subseteq{\frak{Z}}({\cal X}_1, {\cal G}_1).
$$
In particular, since the map $\iota_{{\cal G}_1}^*$ is onto, if $${\frak{Z}}({\cal X}_1\times {\cal X}_2, {\cal G}_1\times
{\cal G}_2)=LUC({\cal G}_1\times{\cal G}_2)^*,$$ then ${\frak{Z}}({\cal X}_1, {\cal G}_1)=LUC({\cal G}_1)^*$; Indeed,
if $m\in {\frak{Z}}({\cal X}_1\times {\cal X}_2, {\cal G}_1\times
{\cal G}_2)$, $F_1\in LUC({\cal X}_1,{\cal G}_1)$ and $1_{{\cal X}_2}$ denotes the constant function on ${\cal X}_2$ with value 1, then
$F_1\otimes 1_{{\cal X}_2}\in LUC({\cal X}_1\times {\cal X}_2, {\cal G}_1\times
{\cal G}_2)$ and we have
\begin{eqnarray*}
{\Big(}(F_1\otimes 1_{{\cal X}_2})m{\Big)}(x_1,x_2) &=& \langle m,r_{(x_1,x_2)}(F_1\otimes 1_{{\cal X}_2})\rangle\\
&=& \langle m,(r_{x_1}F_1)\otimes 1_{{\cal X}_2})\rangle\\
&=& \langle \iota_{{\cal G}_1}^*(m),r_{x_1}F_1\rangle\\
&=& {\Big(}F_1\iota_{{\cal G}_1}^*(m){\Big)}(x_1),
\end{eqnarray*}
for all $x_1\in {\cal X}_1$ and $x_2\in {\cal X}_2$ and therefore
$$(F_1\otimes 1_{{\cal X}_2})m=(F_1\iota_{{\cal G}_1}^*(m))\otimes 1_{{\cal X}_2}.$$
From this, by Theorem \ref{5}, we have
$$(F_1\iota_{{\cal G}_1}^*(m))\otimes 1_{{\cal X}_2}\in LUC({\cal X}_1\times {\cal X}_2, {\cal
G}_1\times {\cal G}_2).$$ It follows that $F_1\iota_{{\cal G}_1}^*(m)\in LUC({\cal X}_1,{\cal
G}_1)$ for all $F_1\in LUC({\cal X}_1,{\cal G}_1)$. Moreover, if
$$\iota_{{\cal X}_1}:LUC({\cal X}_1,{\cal G}_1)\rightarrow LUC({\cal X}_1\times {\cal X}_2,{\cal G}_1\times {\cal G}_2);~
F_1\mapsto F_1\otimes 1_{{\cal X}_2},$$
then it is not hard to check that
$$M(F_1\otimes 1_{{\cal X}_2})=(\iota_{{\cal X}_1}^*(M)F_1)\otimes 1_{{\cal G}_2},$$
for all $F_1\in LUC({\cal X}_1,{\cal G}_1)$, where $1_{{\cal G}_2}$ denotes
the constant function one on ${\cal G}_2$. It follows that
$$\iota_{{\cal G}_1}^*(m)\cdot \iota_{{\cal X}_1}^*(M)=\iota_{{\cal X}_1}^*(M)\bullet \iota_{{\cal G}_1}^*(m),$$
for all $M\in LUC({\cal X}_1\times {\cal X}_2,{\cal G}_1\times {\cal G}_2)^*$; In details, if $F_1\in LUC({\cal X}_1,{\cal G}_1)$, then, we can write
\begin{eqnarray*}
\langle \iota_{{\cal G}_1}^*(m)\cdot \iota_{{\cal X}_1}^*(M), F_1\rangle
&=& \langle \iota_{{\cal G}_1}^*(m), \iota_{{\cal X}_1}^*(M)F_1\rangle\\
&=& \langle m,(\iota_{{\cal X}_1}^*(M)F_1)\otimes 1_{{\cal G}_2}\rangle\\
&=& \langle m,M (F_1\otimes 1_{{\cal X}_2})\rangle\\
&=& \langle m\cdot M ,F_1\otimes 1_{{\cal X}_2}\rangle\\
&=& \langle M\bullet m ,F_1\otimes 1_{{\cal X}_2}\rangle~\quad\quad\quad{\Big(}{\hbox{since}}~ m\in {\frak{Z}}({\cal X}_1\times {\cal X}_2, {\cal G}_1\times
{\cal G}_2){\Big)}\\
&=& \langle M, (F_1\otimes 1_{{\cal X}_2})m\rangle\\
&=& \langle M, (F_1\iota_{{\cal G}_1}^*(m))\otimes 1_{{\cal X}_2}\rangle\\
&=& \langle \iota_{{\cal X}_1}^*(M), F_1\iota_{{\cal G}_1}^*(m)\rangle\\
&=& \langle \iota_{{\cal X}_1}^*(M)\bullet \iota_{{\cal G}_1}^*(m), F_1\rangle.\quad
\quad{\Big(}{\hbox{since}}~ F_1\iota_{{\cal G}_1}^*(m)\in LUC({\cal X}_1,{\cal
G}_1){\Big)}
\end{eqnarray*}
Hence, since the map $\iota_{{\cal X}_1}^*$ is onto, we have
$$\iota_{{\cal G}_1}^*(m)\cdot M_1=M_1\bullet \iota_{{\cal G}_1}^*(m),$$
for all $M_1\in LUC({\cal X}_1,{\cal G}_1)^*$. We now invoke Theorem \ref{5} to conclude that $\iota_{{\cal G}_1}^*(m)\in {\frak Z}({\cal X}_1,{\cal G}_1)$.}
\end{example}

\begin{example}\label{0925}
{\rm Let $Q_8$ be the quaternion group, the subgroup of the
general linear group $GL(2, \Bbb C )$ generated by the matrices
\begin{eqnarray*}
\bf{1}=\left(\begin{array}{clcr}
1 & 0\\
0 & 1
\end{array}
\right), \quad \bf{i}=\left(\begin{array}{clcr}
i & 0\\
0 & -i
\end{array}
\right), \quad \bf{j}=\left(\begin{array}{clcr}
0 & 1\\
-1 & 0
\end{array}
\right), \quad \bf{k}=\left(\begin{array}{clcr}
0 & i\\
i & 0
\end{array}
\right),
\end{eqnarray*}
which can be also presented by $$Q_8=\langle a,\,b:\,a^4=1,
\,a^2=b^2,\,b^{-1}ab=a^{-1}\rangle,$$
where $a=\bf{i}$ and $b=\bf{j}$ for
instance. Take ${\cal K}=\langle a\rangle$, which is a normal subgroup of $Q_8$.
If we set ${\cal G}=Q_8\times \Bbb F_2$, the direct product of $Q_8$ and
the free group on a two-element set, then ${\cal
X}={\cal K}\times \Bbb F_2$ is a closed normal subgroup of
discrete topological group $\cal G$ and so, $\cal G$ acts on ${\cal X}$ by $${\cal G}\times
{\cal X}\rightarrow {\cal X}:(s,{x})\mapsto
s{x}s^{-1}.$$ Since the algebraic centre ${\frak C}_{\cal G}({\cal G})$ contains
${\frak C}_{Q_8}(Q_8)=\{{\bf{i}}^2,\bf{1}\}$, by part (b) of Example~\ref{exmpl111}, we have $M({\cal
G})\varsubsetneq {\frak{Z}}({\cal X},{\cal G})$. On the other hand, if ${\frak{Z}}({\cal X},{\cal G})= LUC({\cal G})^*$, then by Example~\ref{rem1}, we would have
${\frak{Z}}({\Bbb F_2},{\Bbb F_2})=LUC(\Bbb F_2)^*$, which is impossible; This is because of, since $\Bbb F_2$ is not compact,
$${\frak{Z}}({\Bbb F_2},{\Bbb F_2})={\frak{Z}}(\Bbb F_2) \varsubsetneq LUC(\Bbb F_2)^*.$$
Hence, for $\cal G$-space $\cal X$, we have
$M({\cal G})\varsubsetneq {\frak{Z}}({\cal X},{\cal G})\varsubsetneq LUC({\cal G})^*$.

}
\end{example}

\footnotesize

redpajama_set_name: RedPajamaArXiv


It is important to realize or remember that coin quality is not a precise science. Opinions about quality will vary between countries, mints and individual collectors. Even expert dealers will grade the quality of a coin differently on different days. As there are no worldwide standards to go by, individual mints can refer to their qualities by whatever name they want. It is therefore the case that some mint's brilliant uncirculated quality is better than other mint's proof quality, for example. However, these are the most commonly used quality standards and their definitions.
An abbreviation for "fleur-de-coin" (flower of the die). Similar to an uncirculated coin, but without marks from the minting process as FDC coins are largely prevented from touching other coins during manufacturing. Very similar to, or sometimes identical with, brilliant uncirculated or specimen coins (see below). Not used much by world mints anymore.
An abbreviation for "brilliant uncirculated". Most mints define these coins as uncirculated coins (see above) that have been washed and specially treated before being placed in sets. Most of the time the minting dies and coin blanks get better polished and treated before the minting process starts. In addition, most mints see to it that these coins actually don't have the marks from the minting process that regular uncirculated coins would have. Some mints, like denmark, norway and portugal, actually mint their brilliant uncirculated coins one-by-one in a proof press using specially prepared uncirculated coin dies, sometimes with more than one strike. Most brilliant uncirculated coins have an appearance closer to uncirculated than proof quality. Sometimes described as "superior normal strike".
Specimen is used to describe any coin produced to a particularly high standard of finish. Used frequently before the 1970s, most of these coins would now be called proof quality. In recent years, some mints have produced especially good versions of its uncirculated coins which they call specimens. These are ordinary uncirculated coins which have been handled individually and with greater than normal care, to avoid most, but not all, of the surface blemishes which occur due to bulk handling. In that regard they would be very similar to brilliant uncirculated or FDC coins.
Typically, these are coins that have been struck on proof quality coin blanks, using brilliant uncirculated minting dies. This leaves the coin with a very reflective surface due to the brush polished blanks, but don't have the characteristics of polished proof coins with mirrorlike surfaces and frosted reliefs because the dies are not proof quality. The process makes a beautiful coin at an affordable price.
Most mints have adopted the british standard where the coins have a mirror surface and frosted reliefs. The blanks are typically polished with brushes before minting to remove residue from soap and polish paste that can interfere with the end result. The dies used in this process have sharper detail and do not last as long as uncirculated dies. In addition, the coins are minted using more tonnage of pressure, so that the metal will flow more and require a one-by-one minting technique. Often, 2-3 strikes are utilized, depending on the hardness of the metal. After each strike the dies are inspected and often re-polished by hand. Platinum can require as many as 10-12 punches before a coin is finished. Each additional strike increases the discard rate for the manufacturer, and increases the price of the finished coin. Proof coins can be expected to be absolutely perfect, without blemishes, when looked at with the naked eye. Some mints, like the norwegian, define its proof quality not according to the amount of mirror surface and frosted relief, but according to the sharpness in the detail of the finished coin. This moves the quality towards describing a process instead of a finished product.


Can I retire at age 67 with 700,000?
How long will my money last in retirement? Person retiring at age 67 starting with $700,000, adding $5,376 every year, while hoping to spend $44,027 every year in retirement. These numbers increase with inflation.


Bowen, James,1979-World according to Bob.
Bowen, James,1979-Street cat named Bob.
Bowen, James,1979-Street cat named Bob., Bowen, James,1979-World according to Bob., Cleopatra Entertainment (Firm),publisher., Froggatt, Joanne,1980-actor., Gedmintas, Ruta,1983-actor., Goodall, Caroline,1959-actor., Head, Anthony Stewart,1954-actor., Hirschfelder, David,music composer., Lowe, Antonia,production designer., Rolston, Adam,1962-film producer., Spottiswoode, Roger,film director., Thompson, Jo(Costume designer),costume designer., Tothill, Paul,film editor., Treadaway, Luke,1984-actor., Wunstore, Peter,director of photography.
ils:.b24963355|DVD|Movies||English|Cleopatra Entertainment,||1 videodisc (103 min.) : sound, color ; 4 3/4 in.


text:
Chiropractic Schools
Become an Events Poster
Chiropractic Research
NUHS Faculty Carlo Guadagno named 2016 Sports Chiropractor of the Year
Chiropractic Economics December 12, 2016
December 9, 2016—National University faculty member Carlo Guadagno, DC, CCSP, ICCSP, FICC, was recently named the American Chiropractic Association Sports Council's (ACASC) 2016 Sports Chiropractor of the Year. He received the recognition in October at the council's Symposium in New Orleans.
"The award came as a surprise," Guadagno said. "It was quite an honor to be selected from among many of my distinguished peers."
About a month prior, Guadagno fulfilled his career-long dream of treating Olympic athletes at the 2016 Olympic Games in Rio de Janiero. Chosen from hundreds of applicants, Guadagno was part of an international team of just 16 DCs who were placed in the Olympic Village Polyclinic.
Carlo Guadagno, DC, stands in front of the Olympic Village Polyclinic where he treated athletes at the 2016 Olympic Games in Rio de Janeiro.
"This honor shows just how dedicated Dr. Guadagno is to the growing field of sports chiropractic," said Joseph Stiefel, NUHS president, MS, EdD, DC. "Through his equally dedicated instruction, our students can learn how to achieve the same level of success."
When presenting the award, the selection committee, which is made up of previous winners, cited Guadagno's international work treating various types of athletes including U.S.A. track and field, Pro Beach volleyball and D1 College Football. In 2011, Guadagno was also named Sports Chiropractor of the Year by the FCA Sports Injuries Council. The award he received this year is the council's highest honor nationwide which is given on an annual basis.
"Dr. Guadagno has represented sports chiropractic for years. His dedication to the teaching of students, attendance at sporting events worldwide and his never ending devotion to the growth of this field of chiropractic has never wavered," said Shane Espinoza, DC, CCSP, ACA Sports Council president. "He continues to represent that which is best in this world we call Sports Chiropractic. We were honored to present this distinguished award to a very deserving individual."
Guadagno currently serves as an ACA delegate for Florida. He previously served as a past president of the Council on Sports Injuries, Rehabilitation and Physical Fitness of the FCA and on the board of directors of the ACA Sports Injury Council.
"NUHS is proud to have a top practitioner like Dr. Guadagno be part of our university," said Daniel Strauss, DC, dean of the College of Professional Studies in Florida, where Guadagno is faculty. "One of the reasons NUHS can offer such a high standard of learning to its students is because of great instructors like Dr. Guadagno."
To Guadagno, athletes are perfect patients because they take to heart everything he recommends and will do anything and everything to improve their health.
"My work is not so much about trauma but helping athletes increase their strength, flexibility and range of motion," he said, adding that the slightest adjustment can shave off crucial seconds in performance time.
At the various international events he travels to, Guadagno said he enjoys serving as an ambassador to chiropractic care all over the world and showcasing the benefits of chiropractic treatment. He is already preparing applications for more international sports games, including the World Games in Poland next year and the next Summer Olympics in Japan in 2020.
Carlo Guadagno, DC, receives the 2016 Sports Chiropractor of the Year award from the American Chiropractic Association Sports Council.
In the following, Guadagno discusses firsthand his experiences in the sports chiropractic field:
Why did you decide to pursue chiropractic as a career?
While studying at the University of Miami as an undergraduate student, I had a local chiropractor who made a profound impression on me. I saw the results firsthand when one of my Father's employee's was helped by him following a traumatic injury. He explained the nature of chiropractic and how we can positively affect our patients safely and conservatively.
What sparked your interest in sports chiropractic?
As an athlete myself, I was naturally inclined to be attracted to sports chiropractic. I joined the ACA Sports Council and networked with amazing colleagues. Also, I had seen what a great experience Jeff Solomon (the doctor responsible for my preceptorship) had when selected to the Winter Olympics in Torino, Italy. While practicing in Miami, I was fortunate to know ACA Sports Council Hall of Fame inductee Dr. Tom Hyde who was the original Miami Dolphins chiropractor and one of our early leaders in this field.
How did you get started working with professional sports teams/athletes?
As a member of the ACA Sports Council you will be offered many chances to network with individuals who are influential in the sports chiropractic world, nationally as well as internationally. You must be circulating at seminars and events to be noticed. Rarely will a DC who spends all his time in his office be asked to manage a team.
Any advice for new and current doctors of chiropractic?
Never stop learning and expanding your toolbox even after you graduate. Get some post-graduate training in your area of interest. For example, if you're interested in sports medicine, National University's Post-Graduate Department offers a chiropractic sports physician certification program. When choosing the right chiropractic school, choose one that's going to prepare you well for the future in our profession and not just look at the history of the profession.
Source: National University of Health Sciences
Colombia appoints Dallas chiropractor to olympic medical staff
WFC announces first prize winner
NUHS to support and sponsor integrative medicine conference for underserved
Texas Chiropractic College attending clinician elected as sports medicine staff for women's Nigerian bobsled team
Life University student athletes playing rugby on a global scale
UWS President Joseph Brimhall appointed to Council for Higher Education Accreditation
Filed Under: News, School News
Life Chiropractic College West uses biomarkers to support vitalistic research
NCMIC launches resources designed for chiropractic students
"Roaring into the Future" at Sherman College Lyceum 2020

redpajama_set_name: RedPajamaCommonCrawl


Prices are in NOK less local taxes that may apply.
Definitions, terms and conditions in full detail can be found in the standard agreement.
As with most vendors, we use the term license so that an user organization commonly need one license for one or more users (as opposed to one license per user).
P1 : Single license, 3 users.
P5 : 2-5 licenses, 15 users.
P10 : 6-10 licenses, 30users.
P20 : 11-20 licenses, 60 users.
P50 : 21-50 licenses, 150 users.
P100 : 51-100 licenses, 300 users.
PX: 100+ licenses, 300+ users.
Contact us for prices of more than 100 licenses.
In most cases one would not purchase AllView system service separately, but have a combined service agreement for the AllView based user system and AllView.
Thus, the service prices in the above table could be seen as the minimum fees for a combined service agreement. They would only apply directly in case of an extremely simple user application.
We are of course happy to make offers to suit your needs.
As with system service, the hosting prices in the above table are minimum prices, and we would have to arrange an offer for an actual AllView based user application.
With hosting you can run your license data space directly from our servers from anywhere with Internet access. The access speed, reliability (percent up-time) and other quality and security measures (backup routines etc) can be leveled to your needs.
Adds 15% plus NOK 8 000 to the prices of base level.
Adds 15% plus NOK 40 000 to the prices of base level.
Adds NOK 15 000 to the prices of base level.
Adds 10% plus NOK 15 000 to the prices of base level.
Adds 15% plus NOK 23 000 to the prices of base level.
Adds 15% plus NOK 60 000 to the prices of base level.
NetStorm reserves the right to charge a small fee for making license changes, but no more than the cost of a half man-hour. The charge is limited to the operations of adding, renaming, removing and disabling users, renaming the licensed company name, and downgrading license levels. On long term, the fee is assumed to be less used, as the process will be automated through the member pages.
NetStorm provides automated routines for extracting and inserting individual datasets of licensees. Examples are moving a local installation up to a hosted installation and vice versa. Although this is not a particularly time consuming task, the customer may still be charged for the time spent on this task according to the prices of NetStorm consultancy services.
There's more info in the product home page.
Please contact us if you need more information or if you feel that the AllView price model does not seem to work smoothly with your business model.
Prices in annual system service for already licensed products may be adjusted annually up to the general price increase, if stated in the license agreement, which we do in most cases. A price increase in hosting services will be published at least 30 days earlier. With these exceptions, all other prices can changes will be published on this page (www.netstorm.no/Products/allview/prices.aspx) without further notice.
The above conditions and prices are effective from 2006-02-11.


Outback Jacks Wollongong will open in a great location within a busy shopping centre. Further details will be provided post enquiry.
Full training provided, including pre-opening training at an existing restaurant, training & support at the restaurant when you open, and ongoing support & training.
- The restaurant will offer lunch and dinner 7 days & will feature a full bar, a self-select meat counter & an area for the kids to play.
Outback Jacks Bar & Grill is an exciting chain of restaurants offering relaxed casual dining with full bar facilities. A key feature of Outback Jacks is that they have one of the largest range of steaks on any menu in the world, which provides great marketing opportunities.
Mentored Business Sales has been engaged to find the right business owner to take on the next exciting opportunity in the growth of this iconic restaurant brand. Outback Jacks Bar & Grill in Wollongong is an exciting business opportunity, and now is the ideal time to register your interest.
For more details complete the contact us form on this page or go to www.mentored.business and complete our online confidentiality agreement. Use the reference ID 1881981 to search for the business.
Due to the confidential nature of the sale, images used may be stock images and not representative of the business for sale. Total earnings, EBIT or EBITDA, includes add backs relevant to the current owners. Add backs may include, owners salaries, directors benefits, personal expenses, non-recurring expenses, one off expenses, capital costs, depreciation and generally any costs relevant to the current owners.


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Discover Latest #Half-Life News, Articles and Videos with Contenting
Half-Life is a classic first-person shooter video game series developed by Valve Corporation. The first game in the franchise was released in 1998, and was followed by two sequels, Half-Life 2 and Half-Life 2: Episode Two, as well as two stand-alone episodes, Half-Life: Opposing Force and Half-Life: Blue Shift. The series has earned critical acclaim for its innovative story-telling, immersive gameplay, and cutting-edge graphics. As of 2021, the series has sold over 25 million copies worldwide. Half-Life news, articles, and videos can be found on a variety of gaming and entertainment websites. Popular gaming sites such as IGN and GameSpot often feature news and reviews related to Half-Life. YouTube is also a great source for gameplay videos and walkthroughs. Fans of the series can also find a variety of fan-made content, including fan art, mods, and maps. Additionally, Valve regularly releases updates and patches for the games, and these can be found on the official Half-Life website.
I installed Windows 98 on Xbox Series X and played a range of classic PC games | Eurogamer.net
Quake! Half-Life! Unreal! Command and Conquer!
Rare Half-Life 2: Episode Three concept art surfaces from 2K producer - Dexerto
Concept art of Half Life 2: Episode Three surfaced online thanks to a Producer at 2K Games and fans are analyzing what it could have meant.
Global Esport News
Half-Life 2, but as the bad guys: Trailer announces release date of Entropy Zero 2 - Global Esport News
In Entropy: Zero 2, a mod for Half-Life 2, you don't slip into Dr. Freeman's suit for once, but into a Combine armour and go on the hunt for rebels and the sc
Long-in-the works Half-Life 2 VR mod gets public beta next month | Eurogamer.net
A fan-made Half-Life 2 VR mod nearly eight years in the making will finally be getting a public release next month in t…
Half-Life breaks Steam record 24 years after release thanks to #RememberFreeman - Global Esport News
Although Valve may never make Half-Life 3, the community surrounding the legendary shooter series is still going strong. Half-Life has a new player record on S
Fans release new Half-Life as a mod - and are showered with praise - Global Esport News
A small team of modders releases the fan expansion Entropy : Zero 2 for Half-Life 2 on Steam. And it goes down incredibly well with a whopping 98 percent positi
Long-in-the-works Half-Life 2 VR mod finally launches on Steam today | Eurogamer.net
After nearly eight years in development, an ambitious VR mod for Valve's seminal shooter Half-Life 2 finally launches o…
Half-Life 2 VR in test: Experience one of the best shooters of all time in a new way - Global Esport News
On its coming of age, the classic shooter takes the next, overdue technological step. Our VR tester raves: nothing since Half-Life Alyx has impressed him so muc
Half-Life 2 VR mod now available - Geeky Gadgets
Half-Life 2 fans equipped with virtual reality hardware may be interested to know that the Half Life 2 VR Mod is now available to play thanks to the hard
DF Retro Time Capsule: revisiting Half-Life 2 and Battlefield 3 on classic PC hardware | Eurogamer.net
DF uses modern performance tools and the benefit of hindsight to revisit a range of PC classics.


Here's the new Lego Marvel Super Heroes trailer along with several LEGO videos covering this LEGO and Marvel franchise.
The Lego Marvel Super Heroes video game will be coming to Xbox 360, PlayStation 3, Wii U, Nintendo DS, Nintendo 3DS, PlayStation Vita, Windows PC, PlayStation 4 and Xbox One in Fall 2013.
This last video shows the LEGO Super Heroes "Marvel Avengers" – Captain America's Avenging Cycle, Hulk's Helicarrier Breakout, and Quinjet Aerial Battle.
What do you think of the Lego Marvel Super Heroes trailer & videos?


text:
Проект 1160 шифр «Орёл» — первый советский проект авианосца с ядерной энергоустановкой, прорабатывавшийся в конце 1960-х — начале 1970-х в Невском ПКБ под руководством А. Б. Морина.
К 1986 году планировалось построить три таких корабля, однако, в 1973 году все работы были отменены сперва из-за решения построить два корабля проекта 1153, а затем и этот план был отменён в пользу дальнейшей постройки тяжёлых авианесущих крейсеров проекта 1143 «Кречет».

Авиагруппа
Авиагруппа крейсера должна была включать 60-70 летательных аппаратов. Изначально планировалось оснастить корабль
истребителями МиГ-23А,
противолодочными самолётами П-42,
бомбардировщиками Су-24К и
вертолётами Ка-252.
Позже состав авиагруппы претерпел изменения, и стал выглядеть следующим образом:
12 истребителей Су-27К или МиГ-29К,
12 штурмовиков Су-28К,
4 разведчика-целеуказателя Су-28КРЦ,
6 самолётов ПЛО П-42,
4 самолёта радиолокационного дозора и наведения (РЛДН) на базе П-42,
8 вертолётов Ка-252.

Технические средства обеспечения полётов должны были включать 4 паровых катапульты, аэрофинишёры и аварийный барьер.

Примечания

Ссылки

1160
Авианосцы по проектам
Неосуществлённые в СССР военно-морские проекты

redpajama_set_name: RedPajamaWikipedia


Like checking the gauges on your car's dashboard, it's a good idea to check on employee safety attitudes and beliefs. A status review helps the organization understand how drivers, supervisors and managers feel about measures and practices already in place. It also helps identify gaps and processes that need repairs.
Whether it's an online survey, focus group, coffee room chat, tailgate meeting or a suggestion box, there are several powerful reasons for talking with employees about occupational road safety.
It's interactive. Consultation involves employees in their safety program. It offers valuable ways for employees to provide input and contribute to developing the policies, procedures and initiatives they will implement.
What you learn yields information that helps identify priorities and inform the focus of the road safety plan. Rather than fixing things that aren't broken, paying attention to road safety issues that employees feel are important fosters buy-in, action and results.
Employee opinions matter. They lend credibility to the resulting plan, and justify investing resources in the initiative.
It's a snapshot of safety attitudes at a given point in time, a benchmark that can be compared to future snapshots to provide evidence of changing attitudes, an evolving safety culture and improving performance.
The Road Safety Snapshot is a great way to "check the gauges" in your company. This 10-question online survey focuses on the key characteristics of effective road safety programs. Once employees have submitted their responses, collate results and compare them to other employers in your sector. Have employees re-take the survey every two or three years, and compare snapshots to identify improvements over time.
Formal discussions are another way to gauge your organization's safety culture. Use these sessions to explore employee perceptions, opinions and beliefs about the road safety climate at your organization. Get tips on conducting a formal discussion group.


text:
'use strict';

angular.module('proj2App')
    .controller('ClientDetailController', function ($scope, $rootScope, $stateParams, entity, Client) {
        $scope.client = entity;
        $scope.load = function (id) {
            Client.get({id: id}, function(result) {
                $scope.client = result;
            });
        };
        var unsubscribe = $rootScope.$on('proj2App:clientUpdate', function(event, result) {
            $scope.client = result;
        });
        $scope.$on('$destroy', unsubscribe);

    });

redpajama_set_name: RedPajamaGithub