# Sections and Chapters

Maryam Aghakoochaki 1 2020 Mathematics Subject Classifcation. Primary: 46J05;
Secondary: 46J10 Ali Rejali 2 Corresponding author

( 1Isfahan University
<EMAIL_ADDRESS>Orcid: 0000-0002-3851-6550
2Department of Pure Mathematics, Faculty of Mathematics and Statistics,
University of Isfahan, Isfahan 81746-73441, Iran
<EMAIL_ADDRESS>Orcid: 0000-0001-7270-665X
)

#  BSE\- property of tensor product Banach algebras

Maryam Aghakoochaki 1 2020 Mathematics Subject Classifcation. Primary: 46J05;
Secondary: 46J10 Ali Rejali 2 Corresponding author

( 1Isfahan University
<EMAIL_ADDRESS>Orcid: 0000-0002-3851-6550
2Department of Pure Mathematics, Faculty of Mathematics and Statistics,
University of Isfahan, Isfahan 81746-73441, Iran
<EMAIL_ADDRESS>Orcid: 0000-0001-7270-665X
)

###### Abstract

Let $A$ and $B$ be commutative semisimple Banach algebras. In this paper, the
BSE and BED\- property of tensor Banach algebra $A{\otimes}_{\gamma}B$ with
respect to the Banach algebras $A$ and $B$ are assesed. In particular, BSE and
BED\- structure of vector-valued group algebra $L_{1}(G,A)$ and $C^{*}$\-
algebra $C_{0}(X,A)$ characterized.

Keywords: Banach algebra, BSE\- algebra, BED\- algebra, tensor product.

##  1 Introduction

In this paper, $X$ is a metric space with at least two elements and
$(A,\|.\|)$ is a commutative semisimple Banach algebra over the scalar field
$\mathbb{C}$. Let $\Delta(A)$ be the character space of $A$ with the Gelfand
topology. $\Delta(A)$ is the set consisting of all non-zero multiplicative
linear functionals on $A$. Assume that $C_{b}(\Delta(A))$ is the space
consisting of all complex-valued continuous and bounded functions on
$\Delta(A)$, with sup-norm $\|.\|_{\infty}$. A continuous linear operator $T$
on $A$ is named a multiplier if for all $x,y\in A$, $T(xy)=xT(y)$; See [13].
The set of all multipliers on $A$ will be expressed as $M(A)$. It is obvious
that $M(A)$ is a Banach algebra, and if $A$ is an unital Banach algebra, then
$M(A)\cong A$. As observed in [20], for each $T\in M(A)$ there exists a unique
bounded continuous function $\widehat{T}$ on $\Delta(A)$ expressed as:

$\varphi(Tx)=\widehat{T}(\varphi)\varphi(x),$

for all $x\in A$ and $\varphi\in\Delta(A)$. By setting $\\{\widehat{T}:T\in
M(A)\\}$, the $\widehat{M(A)}$ is yield.

If the Banach algebra $A$ is semisimple, then the Gelfand map
$\Gamma_{A}:A\to\widehat{A}$, $f\mapsto\hat{f}$, is injective. Define

${\cal{M}}(A):=\\{\sigma\in C_{b}(\Delta(A)):\leavevmode\nobreak\
\sigma.\hat{A}\subseteq\hat{A}\\}.$

Then $\widehat{M(A)}=\cal{M}(A)$, whenever $A$ is a semisimple Banach algebra.
A bounded complex-valued continuous function $\sigma$ on $\Delta(A)$ is named
a BSE function, if there exists a positive real number $\beta$ in a sense that
for every finite complex-number $c_{1},\cdots,c_{n}$, and the same many
$\varphi_{1},\cdots,\varphi_{n}$ in $\Delta(A)$ the following inequality

$\mid\sum_{j=1}^{n}c_{j}\sigma(\varphi_{j})\mid\leq\beta\|\sum_{j=1}^{n}c_{j}\varphi_{j}\|_{A^{*}}$

holds.

The set of all BSE-functions is expressed by $C_{\textup{BSE}}(\Delta(A))$,
where for each $\sigma$, the BSE-norm of $\sigma$, $\|\sigma\|_{\textup{BSE}}$
is the infimum of all $\beta$s applied in the above inequality. That
$(C_{\textup{BSE}}(\Delta(A)),\|.\|_{\textup{BSE}})$ is a semisimple Banach
subalgebra of $C_{b}(\Delta(A))$ is in Lemma 1 proved in [16]. Algebra $A$ is
named a BSE algebra if it meets the following condition:

$\widehat{M(A)}\cong C_{\textup{BSE}}(\Delta(A)).$

If $A$ is unital, then $\widehat{M(A)}\cong\widehat{A}\mid_{\Delta(A)}$,
indicating that $A$ is a BSE algebra if and only if
$C_{\textup{BSE}}(\Delta(A))\cong\widehat{A}\mid_{\Delta(A)}$. If there exists
some $K>0$ in a sense that for each $a\in A$, the following holds:

$\|a\|_{A}\leq K\|\hat{a}\|_{\textup{BSE}}$

then the semisimple Banach algebra $A$ is named a norm-BSE algebra.

The Bochner-Schoenberg-Eberlein (BSE) is derived from the famous theorem
proved in 1980 by Bochner and Schoenberg for the group of real numbers; [7]
and [15]. Later, Eberlein in [10], revealed that if $G$ is any locally compact
abelian group, then the group algebra $L_{1}(G)$ is a BSE algebra. The
researchers in [15],[17],[18] assessed the commutative Banach algebras that
meet the Bochner-Schoenberg-Eberlein- type theorem and explained their
properties.

In 2007, Inoue and Takahasi in [11] introduced and assessed the concept of
BED\- concerning Doss, where Fourier-Stieltjes transforms of absolutely
continuous measures are specified. The acronym BED stands for Bochner-
Eberlein-Doss and refers to a famous theorem proved in [8] and [9]. They
proved that if $G$ is a locally compact Abelian group, then the group algebra
$L_{1}(G)$ is a BED\- algebra. Later [2], the researcher showed that
$l^{p}(X,A)$ is a BED\- algebra if and only if $A$ is so. The authors in [3]
and [4], investigated and assess the correlation between different types of
BSE\- Banach algebras $A$, and the Banach algebras $C_{0}(X,A)$ and
$L_{1}(G,A)$.

The basic terminologies and the related information on BED\- algebras are
extracted from [11], [13], and [20].

The function $\sigma\in C_{BSE}(\Delta(A))$ is a BED\- function, if for all
$\epsilon>0$, there exists some compact set such $K\subseteq\Delta(A)$, where
for all $c_{i}\in\mathbb{C}$ and for all $\varphi_{i}\in\Delta(A)\backslash K$
the following inequality:

$|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})|\leq\epsilon\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$

holds. This definition of BED\- functions is a modification of the definition
in [8]. The set of all BED\- functions is expressed by the
$C_{BSE}^{0}(\Delta(A))$. Clearly, $C_{BSE}^{0}(\Delta(A))$ is a closed ideal
of $C_{BSE}(\Delta(A))$; See[[11], Corollary 3.9].

We will prove that:
If $A$ is a commutative semisimple unital dual Banach algebra and $G$ is a
locally compact group. Then
(i)

$C_{\textup{BSE}}(\Delta(L^{1}(G,A))=\widehat{M(G)}\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}(\Delta(A))$

(ii)

$C_{\textup{BSE}}^{0}(\Delta(L^{1}(G,A))=\widehat{L^{1}(G)}\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}^{0}(\Delta(A))$

(iii)

$C_{\textup{BSE}}(\Delta(C_{0}(X,A)))=\widehat{C_{b}(X)}\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}(\Delta(A))$

(iv)

$C_{\textup{BSE}}^{0}(\Delta(C_{0}(X,A)))=\widehat{C_{0}(X)}\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}(\Delta(A))$

In general, for each commutative semisimple Banach algebras $A$ and $B$, we
show that:
(i)

$C_{\textup{BSE}}(\Delta(A{\otimes}_{\gamma}B))=C_{\textup{BSE}}(\Delta(A))\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}(\Delta(B)).$

(ii)

$C_{\textup{BSE}}^{0}(\Delta(A{\otimes}_{\gamma}B))=C_{\textup{BSE}}^{0}(\Delta(A))\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}^{0}(\Delta(B)).$

##  2  BSE\- properties of Banach algebras

In this section, some of the basic terminologies of BSE\- algebra are reviewed
and proven. Let $A$ be a commutative semisimple Banach algebra and $X$ be a
locally compact Hausdorff space.

######  Lemma 2.1.

If $\sigma\in C_{\textup{BSE}}(\Delta(A))$, $c_{i}\in\mathbb{C}$ and
$\varphi_{i}\in\Delta(A)$ for all $1\leq i\leq n$, then

$\|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})\varphi_{i}\|_{A^{*}}\leq\|\sigma\|_{BSE}.\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}.$

###### Proof.

$\displaystyle\|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})\varphi_{i}\|_{A^{*}}$
$\displaystyle=\textup{sup}\\{\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})\varphi_{i}(a):\leavevmode\nobreak\
\|a\|_{A}\leq 1\\}$
$\displaystyle=\textup{sup}\\{\sum_{i=1}^{n}c_{i}(\sigma.\hat{a})(\varphi_{i}):\leavevmode\nobreak\
\|a\|_{A}\leq 1\\}$

Since $\hat{A}\subseteq C_{\textup{BSE}}(\Delta(A))$, so $\sigma.\hat{a}\in
C_{\textup{BSE}}(\Delta(A))$. Thus the following is the yield:

$\displaystyle\|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})\varphi_{i}\|_{A^{*}}$
$\displaystyle\leq\textup{sup}\\{\|\sigma.\hat{a}\|_{\textup{BSE}}.\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}:\leavevmode\nobreak\
\|a\|_{A}\leq 1\\}$
$\displaystyle\leq\textup{sup}\\{\|\sigma\|_{\textup{BSE}}.\|\hat{a}\|_{\textup{BSE}}.\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}:\leavevmode\nobreak\
\|a\|_{A}\leq 1\\}$
$\displaystyle\leq\|\sigma\|_{\textup{BSE}}.\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$

This implies that

$\|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})\varphi_{i}\|_{A^{*}}\leq\|\sigma\|_{\textup{BSE}}.\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}.$

∎

In the following, we give a simple proof of the result in [11], with another
definition.

######  Proposition 2.2.

Let $A$ be a commutative semisimple Banach algebra. Then
$C_{\textup{BSE}}^{0}(\Delta(A))$ is a closed ideal of
$C_{\textup{BSE}}(\Delta(A))$.

###### Proof.

Assume that $\sigma\in C_{\textup{BSE}}(\Delta(A))$ and $\sigma_{0}\in
C_{\textup{BSE}}^{0}(\Delta(A))$. So for all $\epsilon>0$ there exists some
compact set $K\subseteq\Delta(A)$ where for all $c_{i}\in\mathbb{C}$ and
$\varphi_{i}\in\Delta(A)\backslash K$ the following inequality is holds:

$|\sum_{i=1}^{n}c_{i}\sigma_{0}(\varphi_{i})|\leq\epsilon\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$

Then

$\displaystyle|\sum_{i=1}^{n}c_{i}\sigma.\sigma_{0}(\varphi_{i})|$
$\displaystyle=|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})\sigma_{0}(\varphi_{i})|$
$\displaystyle\leq\epsilon\|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})\varphi_{i}\|_{A^{*}}$

Furthermore, according to Lemma 2.1 we have

$|\sum_{i=1}^{n}c_{i}\sigma.\sigma_{0}(\varphi_{i})|\leq\epsilon\|\sigma\|_{\textup{BSE}}.\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}.$

Therfore $\sigma\sigma_{0}\in C_{\textup{BSE}}^{0}(\Delta(A))$, so
$C_{\textup{BSE}}^{0}(\Delta(A))$ is an ideal of
$C_{\textup{BSE}}(\Delta(A))$. In the sequel, let $(\sigma_{n})$ be a Cauchy
sequence in $C_{\textup{BSE}}^{0}(\Delta(A))$, where $\sigma_{n}\to\sigma$ for
some $\sigma\in C_{\textup{BSE}}(\Delta(A))$. Then for all $\epsilon>0$ there
exists some number $N$ where for all $n\geq N$ such that

$\|\sigma_{n}-\sigma\|_{BSE}\leq\frac{\epsilon}{2}.$

Since

$\sigma_{N}\in C_{\textup{BSE}}^{0}(\Delta(A))$

So for all $\epsilon>0$ there exists some compact set $K\subseteq\Delta(A)$
where for all $c_{i}\in\mathbb{C}$ and $\varphi_{i}\in\Delta(A)\backslash K$,
the following inequality is holds:

$|\sum_{i=1}^{n}c_{i}\sigma_{N}(\varphi_{i})|\leq\frac{\epsilon}{2}\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$

This implies that

$\displaystyle|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})|$
$\displaystyle\leq|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})-\sum_{i=1}^{n}c_{i}\sigma_{N}(\varphi_{i})|+|\sum_{i=1}^{n}c_{i}\sigma_{N}(\varphi_{i})|$
$\displaystyle\leq\|\sigma-\sigma_{N}\|_{BSE}.\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}+\frac{\epsilon}{2}\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$
$\displaystyle\leq\epsilon\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$

Therefore $\sigma\in C_{\textup{BSE}}^{0}(\Delta(A))$ and so
$C_{\textup{BSE}}^{0}(\Delta(A))$ is a closed ideal of
$C_{\textup{BSE}}(\Delta(A))$. ∎

######  Theorem 2.3.

Let $A$ be a commutative Banach algebra. If $A$ has a bounded approximate
identity in $A_{0}$, then $\widehat{A}\subseteq
C_{\textup{BSE}}^{0}(\Delta(A))$.

###### Proof.

Assume that $(e_{r})$ is a bounded approximate identity for $A$ in $A_{0}$ and
$a\in A$. Then for all $\epsilon>0$ there exist $r_{0}$ such that
$\|a-e_{r_{0}}a\|_{A}<\epsilon$. If $P=\sum_{i=1}^{n}c_{i}p_{i}$ where
$p_{i}\notin supp(\widehat{e_{r_{0}}})$, then

$\displaystyle|\hat{a}(P)|$
$\displaystyle=|\hat{a}-{\widehat{e_{r_{0}}}}(P){\hat{a}}(P)|$
$\displaystyle=|(a-e_{r_{0}}a\widehat{)}(P)|$
$\displaystyle\leq\|P\|.\|(a-e_{r_{0}}a\widehat{)}\|_{A}$
$\displaystyle\leq\|P\|.\|a-e_{r_{0}}a\|_{A}<\epsilon\|P\|$

Thus ${\hat{a}}\in C_{\textup{BSE}}^{0}(\Delta(A))$. ∎

######  Corollary 2.4.

Let $G$ be a locally compact group. Then

$C_{0}(G\widehat{)}\subseteq C_{\textup{BSE}}^{0}(\Delta(C_{0}(G)))$

and

$L^{1}(G\widehat{)}\subseteq C_{\textup{BSE}}^{0}(\Delta(L^{1}(G))).$

###### Proof.

The algebras $C_{0}(G)$ and $L^{1}(G)$ have the bounded approximate identity
in $C_{00}(G)$. Thus $C_{0}(G\widehat{)}\subseteq
C_{\textup{BSE}}^{0}(\Delta(C_{0}(G)))$ and $L^{1}(G\widehat{)}\subseteq
C_{\textup{BSE}}^{0}(\Delta(L^{1}(G)))$. ∎

In the sequel, the correlation between the BSE and BED is assessed. The
following theorem is proved in [11] and [17]. We modify their proof as
follows:

######  Proposition 2.5.

Let $A$ be a commutative semisimple BED\- algebra. Then
(i)

$C_{\textup{BSE}}(\Delta(A))\subseteq\widehat{M(A)}.$

(ii) If $A$ has $\Delta$\- weak approximate identity, then $A$ is a BSE\-
algebra.

###### Proof.

(i) Since $A$ is BED\- algebra, so $C_{\textup{BSE}}^{0}(\Delta(A))=\hat{A}$.
Then by applying proposition 2.2, $\hat{A}$ is an ideal of
$C_{\textup{BSE}}(\Delta(A))$. Thus
$C_{\textup{BSE}}(\Delta(A)).\hat{A}\subseteq\hat{A}$. Therefore

$C_{\textup{BSE}}(\Delta(A))\subseteq{\cal M}(A)=\widehat{M(A)}.$

(ii) By using [[16], Corollary 5], and part (i), it is immediate. ∎

In the following Lemma, we consider the corelation between
$C_{\textup{BSE}}^{0}(\Delta(A))$ and $C_{\textup{BSE}}(\Delta(A))$.

######  Lemma 2.6.

Let $A$ be a commutative semisimple Banach algebra. Then
(i) If $a\in A$ and $c\in\mathbb{C}$, then there exists some
$\varphi\in\Delta(A)$ where $\varphi(a)=c$.
(ii) If $\Delta(A)$ is compact, then

$C_{\textup{BSE}}^{0}(\Delta(A))=C_{\textup{BSE}}(\Delta(A))$

(iii) If $\Delta(A)$ is discrete, then $C_{\textup{BSE}}^{0}(\Delta(A))$ is
dense in $C_{\textup{BSE}}(\Delta(A))$.
(iv) If $A$ is unital, then

$C_{\textup{BSE}}^{0}(\Delta(A))=C_{\textup{BSE}}(\Delta(A))$

and $A$ is a BSE\- algebra if and only if $A$ is a BED\- algebra.
(v) $\hat{A}$ is dense in $C_{0}(\Delta(A))$.

###### Proof.

(i) Let $b=a-ce$ where $e=1_{A}$ and $I=\\{bx:x\in A\\}$ be the ideal
generated by $b$. Then there exists a maximal ideal $M$ where $I\subseteq M$.
Put $M=ker(p)$ for some $p\in\Delta(A)$, then $p(a)=c$.
(ii) If $\Delta(A)$ is compact clearly,

$C_{\textup{BSE}}^{0}(\Delta(A))=C_{\textup{BSE}}(\Delta(A))$

(iii) If $\Delta(A)$ is discrete and $p\in\Delta(A)$, then $\chi_{p}\in
C_{\textup{BSE}}^{0}(\Delta(A))$. Thus $C_{00}(\Delta(A))\subseteq
C_{\textup{BSE}}^{0}(\Delta(A))$, but $C_{00}(\Delta(A))$ is dense in
$C_{0}(\Delta(A))$ and $C_{\textup{BSE}}^{0}(\Delta(A))\subseteq
C_{0}(\Delta(A))$. Therefore $C_{\textup{BSE}}^{0}$ is dense in
$C_{0}(\Delta(A))$.
(v) Since $A$ is semisimple , so $A=\hat{A}$. Hence $\hat{A}$ is a subalgebra
of $C_{0}(\Delta(A))$. If $p\neq q$ in $\Delta(A)$, then there exists $a\in A$
wherre $p(a)\neq q(a)$, thus $\hat{A}$ separates the points of $\Delta(A)$.
Therefore by Stone- Weirstrass theorem $\hat{A}$ is dense in
$C_{0}(\Delta(A))$.

∎

The following lemma shows that our definition of BED\- algebra is equivalent
to its definition in [11].

######  Lemma 2.7.

Let $A$ be a commutative Banach algebra where $\Delta(A)\neq\emptyset$,
$K\subseteq\Delta(A)$ be compact set and $\sigma\in
C_{\textup{BSE}}(\Delta(A))$ such that
(i)

$\displaystyle\|\sigma\|_{\textup{BSE},K}:=\leavevmode\nobreak\
\textup{sup}\\{|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})|:\leavevmode\nobreak\
\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}\leq 1,\leavevmode\nobreak\
c_{i}\in\mathbb{C},\leavevmode\nobreak\ \varphi_{i}\in\Delta(A)\backslash
K\\}$

(ii)

$\displaystyle\|\sigma\|_{\textup{BSE},\infty}:=\leavevmode\nobreak\
\textup{inf}\\{\|\sigma\|_{\textup{BSE},K}:\leavevmode\nobreak\
K\subseteq\Delta(A)\leavevmode\nobreak\ is\leavevmode\nobreak\ compact\\}.$

Then

$C_{\textup{BSE}}^{0}(\Delta(A))=\\{\sigma\in
C_{\textup{BSE}}(\Delta(A)):\leavevmode\nobreak\
\|\sigma\|_{\textup{BSE},\infty}=0\\}$

###### Proof.

If $\sigma\in C_{\textup{BSE}}(\Delta(A))$ and
$\|\sigma\|_{\textup{BSE},\infty}=0$, then there exists a sequence of compact
sets $(K_{n})$ in $\Delta(A)$ where $\|\sigma\|_{\textup{BSE},K_{n}}\to 0$. So
for all $\epsilon>0$ there exists some integer number $N$ where for all $n\geq
N$ we have $\|\sigma\|_{\textup{BSE},K_{n}}\leq\epsilon$. Set $K:=K_{N}$. At a
result

$|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})|\leq\|\sigma\|_{\textup{BSE},K}<\epsilon$

where $\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}\leq 1$, $\varphi_{i}\notin
K$ and $c_{i}\in\mathbb{C}$. Define

$F:\overline{<\Delta(A)>}\to\mathbb{C}$

by

$F(\sum_{i=1}^{n}c_{i}\varphi_{i})=\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i}).$

Thus $F$ is continuous and so by Hahn- Banach theorem there exists an
extension $F\in A^{**}$ where $\|F\|=\|\sigma\|_{\textup{BSE}}$. If
$M:=\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$, then
$\|\sum_{i=1}^{n}\frac{c_{i}}{M}\varphi_{i}\|_{A^{*}}\leq 1$ and so

$|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})|<\epsilon\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$

So $\sigma\in C_{\textup{BSE}}^{0}(\Delta(A))$.

Conversely, if $\sigma\in C_{\textup{BSE}}^{0}(\Delta(A))$, then for
$\epsilon=\frac{1}{n}$ there exist some compact set
$K_{n}\subseteq\Delta(A)$ where for all $c_{i}\in\mathbb{C}$ and
$\varphi_{i}\in\Delta(A)\backslash K_{n}$ the following is the yield:

$|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i})|\leq\frac{1}{n}\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}.$

This implies that:

$\|\sigma\|_{\textup{BSE},K_{n}}\leq\frac{1}{n}$

Thus

$\displaystyle 0$
$\displaystyle\leq{\textup{inf}}\\{\|\sigma\|_{\textup{BSE},K}:\leavevmode\nobreak\
K\subseteq\Delta(A)\leavevmode\nobreak\ is\leavevmode\nobreak\ compact\\}$
$\displaystyle\leq\underset{n}{\textup{inf}}\\{\|\sigma\|_{\textup{BSE},K_{n}}:\leavevmode\nobreak\
K_{n}\subseteq\Delta(A)\leavevmode\nobreak\ is\leavevmode\nobreak\
compact\\}=0$

Therefore $\|\sigma\|_{\textup{BSE},\infty}=0$ ∎

######  Definition 2.8.

The Banach algebra $A$ is called BED\- norm algebra, if there exist $M>0$ such
that $\|a\|_{A}\leq M\|\hat{a}\|_{\infty}$ for all $a\in A$.

######  Proposition 2.9.

(i) Let $A$ be a commutative semisimple BED\- algebra. Then it is a BSE\- norm
algebra.
(ii) Let $A$ be a commutative semisimple BED\- norm algebra. Then $A$ is a
BED\- algebra.

###### Proof.

(i) By hypothesis, $\widehat{A}=C_{\textup{BSE}}^{0}(\Delta(A))$ is a Banach
algebra, so $(\widehat{A},\|.\|_{\textup{BSE}})$ is complete. But the map
$a\to\hat{a}$ is a continuous and isomorphism, thus by applying the open
mapping theorem it has a continuous inverse map. As a result, there exists
$M>0$ such that $\|a\|\leq M\|\hat{a}\|_{\textup{BSE}}$.
(ii) If $A$ is a BED\- norm algebra, then

$\|\widehat{a}\|_{\infty}\leq\|\widehat{a}\|_{\textup{BSE}}\leq\|a\|_{A}\leq
M\|\widehat{a}\|_{\infty}$

Thus $A=(\widehat{A},\|.\|_{\infty})$ is a sub- $C^{*}$ algebra with
equivalent norms. Since Property BSE and property BED are stable under
uniformity with equivalent norms, $A$ is a BSE and BED\- algebra. Therefore

$A=C_{\textup{BSE}}^{0}(\Delta(A)).$

∎

######  Proposition 2.10.

Let $A$ be a semisimple commutative Banach algebra. Then
(i) $A$ is a BSE\- norm algebra if and only if
$A\cong(\widehat{A},\|.\|_{\textup{BSE}})$.
(ii) $A$ is a BED\- norm algebra if and only if
$A\cong(\widehat{A},\|.\|_{\infty})$.

######  Definition 2.11.

Assume that $A$ is a commutative Banach algebra with non-empty character
space. If $A_{0}:=\\{a\in A:\leavevmode\nobreak\
Supp(\hat{a})\leavevmode\nobreak\ is\leavevmode\nobreak\ compact\\}$ is norm
dense in $A$, then $A$ is called Tauberian.

Clearly, every unital Banach algebra is Tauberian.

######  Theorem 2.12.

Let $A$ and $B$ be commutative Banach algebras with non-empty character
spaces. Assume that $\epsilon\leq\gamma\leq\pi$ and $A{\otimes}_{\gamma}B$ is
a Banach algebra. Then $A{\otimes}_{\gamma}B$ is Tauberian if and only if $A$
and $B$ are Tauberian.

###### Proof.

Assume that $A{\otimes}_{\gamma}B$ is Tauberian, $\varphi\in\Delta(A)$ and
$a\in A$ where $\varphi(a)=1$. If $b\in B$ and $\epsilon>0$, then there exist
$s\in(A{\otimes}_{\gamma}B)_{0}$ such that

$\|a\otimes b-s\|_{\gamma}<\epsilon.$

Define

$\displaystyle P:\leavevmode\nobreak\ $ $\displaystyle A{\otimes}_{\gamma}B\to
B$ $\displaystyle x\otimes y\mapsto\varphi(x)y$

Put $t:=P(s)$. As a result, the following is the yield:

$\displaystyle\|t-b\|_{B}$ $\displaystyle=\|P(s)-P(a\otimes b)\|_{B}$
$\displaystyle\leq\|P\|.\|s-a\otimes b\|_{\gamma}$
$\displaystyle\leq\epsilon\|P\|$

If

$\displaystyle Q:\leavevmode\nobreak\ $
$\displaystyle\Delta(A)\times\Delta(B)\to\Delta(B)$
$\displaystyle\psi_{1}\otimes\psi_{2}\mapsto\psi_{2}$

then

$\displaystyle\hat{t}(\psi)=\psi(P(s))=\varphi\otimes\psi(s)={\hat{s}}(\varphi\otimes\psi)$

for all $\psi\in\Delta(B)$. As a result

$\displaystyle supp(\hat{t})$ $\displaystyle=Supp(\widehat{P(s)})$
$\displaystyle\subseteq Q(supp(\hat{s}))$

Since $Q$ is continuous, $supp(\hat{s})$ is compact and $supp(\hat{t})$ is
closed, so $t\in B_{0}$ and

$\|t-b\|_{B}<\epsilon\|P\|$

Thus $B_{0}$ is norm dense in $B$ and therefore $B$ is Tauberian. In a similar
way, $A$ is Tauberian.

Conversely, assume that $A$ and $B$ are Tauberian.
(i) If $a\in A_{0}$ and $b\in B_{0}$, then

$supp(a\otimes b\widehat{)}\subseteq supp(\hat{a})\otimes supp(\hat{b})$

Thus

$a\otimes b\in(A{\otimes}_{\gamma}B)_{0}.$

(ii) If $x\in A_{0}\otimes B_{0}$, then

$x=\sum_{k=1}^{n}a_{k}\otimes b_{k}$

for some $a_{k}\in A_{0}$ and $b_{k}\in B_{0}$. Then

$supp(\hat{x})\subseteq\cup_{k=1}^{n}supp(\widehat{a_{k}})\otimes\cup_{k=1}^{n}supp(\widehat{b_{k}})$

Therefore

$A_{0}\otimes B_{0}\subseteq(A{\otimes}_{\gamma}B)_{0}.$

(iii) If $x\in A{\otimes}_{\gamma}B$, then there exist $y\in A\otimes B$ where
$\|y-x\|_{\gamma}<\frac{\epsilon}{2}$. Assume that
$y=\sum_{k=1}^{n}a_{k}\otimes b_{k}$. Thus there exist some sequence
$(a_{k}^{l})\in A_{0}$ and $(b_{k}^{l})\in B_{0}$ where

$a_{k}^{l}\overset{\|.\|_{A}}{\to}a_{k}$

and

$b_{k}^{l}\overset{\|.\|_{A}}{\to}b_{k}.$

Then

$\sum_{k=1}^{n}a_{k}^{l}\otimes b_{k}^{l}\to\sum_{k=1}^{n}a_{k}\otimes b_{k}.$

So there exists $z=\sum_{k=1}^{n}a_{k}^{l_{0}}\otimes b_{k}^{l_{0}}$ where
$\|z-y\|_{\gamma}<\frac{\epsilon}{2}$. Thus

$z\in A_{0}\otimes B_{0}\subseteq(A\otimes_{\gamma}B)_{0}$

and so $z\in(A{\otimes}_{\gamma}B)_{0}$. Therefore

$\overline{(A{\otimes}_{\gamma}B)_{0}}=A{\otimes}_{\gamma}B.$

This implies that $A{\otimes}_{\gamma}B$ is Tauberian. ∎

######  Lemma 2.13.

(i) $C_{0}(X,A)$ is Tauberian if and only if $A$ is so.
(ii) $L^{1}(G,A)$ is Tauberian if and only if $A$ is so.

###### Proof.

By using [19], it is immediate. ∎

######  Lemma 2.14.

Let $A$ be a commutative Tauberian algebra. Then
(i)

$\hat{A}\subseteq C_{\textup{BSE}}^{0}(\Delta(A)).$

(ii) $C_{\textup{BSE}}^{0}(\Delta(A))$ is dense in $C_{0}(\Delta(A).$

###### Proof.

(i) Since $A_{0}$ is dense in $A$, so $\widehat{A_{0}}$ is dense in $\hat{A}$.
Moreover, $C_{\textup{BSE}}^{0}(\Delta(A))$ is closed, thus

$\hat{A}\subseteq C_{\textup{BSE}}^{0}(\Delta(A)).$

(ii) By applying Lemma2.6 and part (i) it is immediate. ∎

######  Corollary 2.15.

Let $A$ be a commutative semisimple Tauberian Banach algebra. Then
(i)

$\displaystyle C_{0}(X,A\widehat{)}\subseteq C_{BSE}^{0}(\Delta C_{0}(X,A)).$

(ii)

$\displaystyle{L^{1}(G,A\widehat{)}}\subseteq C_{BSE}^{0}(\Delta(L^{1}(G,A))$

In the following example, the distinction between BSE and BED\- algebras is
specified; see [17] and [11].

######  Example 2.16.

(i) If $G$ is a non-discrete Abelian group, then $A=(M(G),\star)$ is a
commutative semisimple unital Banach algebra such that $A$ is not a BSE\-
algebra and BED\- algebra.
(ii) If $A=L^{1}(G)\cap L^{p}(G)$ where $1<p<\infty$ where $G$ is an infinite
compact Abelian group, then $A$ with convolution product and the following
norm

$\|f\|=max\\{\|f\|_{1},\|f\|_{p}\\}$

is a commutative semisimple Banach algebra, such that $A$ is a BSE\- algebra,
but is not a BED\- algebra.
(iii) If $S$ is infinite set, then $A=(l_{1}(S),\|.\|_{1})$ with pointwise
multiplication is a Banach algebra, such that $A$ is not a BSE\- algebra but
it is a BED\- algebra.
(iv) If $X$ is a locally compact Hausdorff space and $A=C_{0}(X)$, then $A$ is
a BSE\- algebra and BED\- algebra.

##  3 BSE- properties of tensor product Banach algebra

Let $A$ and $B$ be commutative Banach algebra with nonempty character spaces.
Then for each cross-norm $\|.\|_{\gamma}$ for which
$\epsilon\leq\gamma\leq\pi$, where $\epsilon$ [resp. $\pi$] is injective
[resp. projective] cross-norms on $A\otimes B=<x\otimes y:x\in A,y\in B>$,
where $A\otimes B$ is a subalgebra of bilinear Banach algebra
$BL(A^{*},B^{*},\mathbb{C})$ defined by,

$x\otimes y:A^{*}\times B^{*}\to\mathbb{C}$

for which $x\otimes y(f,g):=f(x)g(y)$ for $f\in A^{*}$ and $g\in B^{*}$. Then
$A\otimes_{\gamma}B$ is the completion of $A\otimes B$ under the cross- norm
$\|.\|_{\gamma}$. Clearly, $A\otimes B$ is an algebra under the product

$(\sum_{i=1}^{n}a_{i}\otimes b_{i}).(\sum_{j=1}^{m}c_{j}\otimes
d_{j}):=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}c_{j}\otimes b_{i}d_{j}$

In general, $A\otimes_{\gamma}B$ need not a Banach algebra; see [14]. But
$A\otimes_{\pi}B:=A\hat{\otimes}B$ is a Banach algebra and each $u\in
A\otimes_{\pi}B$ can represented as $u=\sum_{n=1}^{\infty}a_{n}\otimes b_{n}$
such that $\sum_{n=1}^{\infty}\|a_{n}\|_{A}\|b_{n}\|_{B}<\infty$. Clearly,
$\hat{u}=\sum_{n=1}^{\infty}\hat{a_{n}}\otimes\hat{b_{n}}$ and
$\|\hat{u}\|_{\infty}\leq\|u\|_{\pi}$. In fact,

$(A\otimes_{\pi}B\hat{)}=\hat{A}\otimes_{\pi}\hat{B}$

In particular,

$(a\otimes b\hat{)}=\hat{a}\otimes\hat{b}.$

In general, if $x\in A\otimes_{\gamma}B$, then $x_{n}\to x$ for some sequence
$(x_{n})$ in $A\otimes B$. But

$\|\widehat{x_{n}}-\hat{x}\|_{\infty}<\|x_{n}-x\|_{\gamma}\to 0$

Thus $\hat{x}=lim\widehat{x_{n}}$. Therefore
$(A\otimes_{\gamma}B\widehat{)}=\hat{A}\otimes_{\gamma}\hat{B}$. Let
$A\otimes_{\gamma}B$ be a Banach algebra. Then
$\Delta(A\otimes_{\gamma}B)=\Delta(A)\times\Delta(B)$; See[13]. Let
$\varphi\in\Delta(A)$ and $\psi\in\Delta(B)$. Then

$\varphi\odot\psi:A\times B\to\mathbb{C}$

defined by

$\varphi\odot\psi(a,b):=\varphi(a)\psi(b)$

for $a\in A$ and $b\in B$. Also, $C_{0}(X,A)^{*}=M(X,A^{*})$ as Banach spaces
under the duality,

$U:M(X,A^{*})\to C_{0}(X,A)^{*}$

for which $\mu\mapsto U_{\mu}$ and for $g\in C_{0}(X)$ and $a\in A$

$U_{\mu}(g\otimes a):=\int_{X}\int_{A^{*}}g(x)f(a)d\mu(x,f)$

Also, $L^{1}(G,A)=L^{1}(G)\hat{\otimes}A$ and
$C_{0}(X,A)=C_{0}(X)\check{\otimes}A$ are Banach algebra; see[13].
Furthermore, $L^{1}(G,A)^{*}=L^{\infty}(G,A^{*})$, where $A$ is a separable
Banach algebra.

Under the duality

$\displaystyle T$ $\displaystyle:L_{\infty}(G,A^{*})\to L^{1}(G,A)^{*}$
$\displaystyle g\mapsto T_{g}$

for all $f\in L^{1}(G,A)$ and $g\in L_{\infty}(G,A^{*})$. There exist sequence
$(g_{n})$ in $L^{\infty}(G)\otimes A^{*}$ such that

$\|g_{n}-g\|_{\epsilon}\to 0$

Let $f:=\sum_{m=1}^{\infty}f_{m}\otimes a_{m}$ where
$\sum_{m=1}^{\infty}\|f_{m}\|.\|a_{m}\|<\infty$. Then

$\displaystyle T_{g}(f)$ $\displaystyle:=\underset{n}{lim}T_{g_{n}}(f)$
$\displaystyle=\underset{n}{lim}\sum_{m=1}^{\infty}g_{n}(f_{m}\otimes a_{m})$

where

$\displaystyle g_{n}(f_{m}\otimes a_{m})$
$\displaystyle:=\sum_{k=1}^{l_{n}}h_{k}^{n}\otimes F_{k}^{n}(f_{m}\otimes
a_{m})$ $\displaystyle=\sum_{k=1}^{l_{n}}h_{k}^{n}(f_{m})F_{k}^{n}(a_{m})$

for some $h_{k}^{n}\in L^{\infty}(G)$ and $F_{k}^{n}\in A^{*}$.

######  Lemma 3.1.

Let $A$ and $B$ be commutative Banach algebras. Then
(i) $A\otimes_{\pi}B$ is semisimple if the following two conditions are
satisfied.
(1) $A$ and $B$ are semisimple.
(2) The natural homomorphism $A\otimes_{\pi}B\to A\otimes B$ is injective.
(ii) $A\otimes_{\pi}B$ is unital if and only if both $A$ and $B$ are unital.
(iii) $A\otimes_{\pi}B$ has bounded approximate identity if and only if both
$A$ and $B$ have so.

###### Proof.

(i) In Theorem 2.11.6 in [13] was proved.
(ii) It was shown that in Theorem 2.11. 7 in [13].
(iii) By applying Theorem 2. 8. 10 in [13], it is proved. ∎

######  Definition 3.2.

Let $A$ and $B$ be commutative Banach algebra with nonempty character spaces.
Then for all $F\in A^{**}$ and $G\in B^{**}$
(i)

$F\odot
G(\sum_{i=1}^{n}\sum_{j=1}^{m}c_{i,j}F(\varphi_{i}\otimes\psi_{j})):=\sum_{i=1}^{n}\sum_{j=1}^{m}c_{i,j}F(\varphi_{i})G(\psi_{i})$

where $\varphi_{i}\in\Delta(A)$ and $\psi_{j}\in\Delta(B)$.
(ii)

$\|\sum_{i=1}^{k}F_{i}\odot
G_{i}\|_{\textup{BSE}}:=sup\\{|\sum_{i=1}^{k}F_{i}\odot
G_{i}(p)|:\leavevmode\nobreak\ p\in<\Delta(A)\times\Delta(B)>_{1}\\}.$

(iii) We denote $A^{**}\overset{\textup{BSE}}{\otimes}B^{**}$ as the complete
metric space of $(<A^{**}\odot B^{**}>,\|.\|_{\textup{BSE}})$. In a special
case, since $C_{\textup{BSE}}(\Delta(A))=A^{**}|_{\Delta(A)}\cap
C_{b}(\Delta(A)$ and $C_{\textup{BSE}}(\Delta(B))=B^{**}|_{\Delta(B)}\cap
C_{b}(\Delta(B)$. We define

$C_{\textup{BSE}}(\Delta(A))\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}(\Delta(B)):=\|.\|_{\textup{BSE}}-cl\\{<C_{\textup{BSE}}(\Delta(A)){\odot}C_{\textup{BSE}}(\Delta(B))>\\}$

and

$A^{**}\overset{\textup{BSE}}{\otimes}B^{**}:=\|.\|_{\textup{BSE}}-cl\\{<A^{**}\odot
B^{**}>\\}.$

$\beta X$ is the Ston- Cech compactification of $X$ and

$\beta X:=w^{*}-cl\\{\delta_{x}:x\in X\\}\subseteq{C_{b}(X)}^{*}$

where $\delta_{x}(f)=f(x)$, for $f\in C_{b}(X)$.

######  Lemma 3.3.

Let $X$ be a completely regular Hausdorff space and $Y$ be a compact Hausdorff
space. Then
(i)

$\beta(X\times Y)=\beta X\times Y$

and so

$C_{b}(X\times Y)=C_{b}(X)\check{\otimes}C(Y).$

(ii)

$C_{b}(X\times Y)=C(\beta X\times Y).$

isometrically isomorphism as Banach algebras.

###### Proof.

(i) Since $X\subseteq\beta X$, so $X\times Y\subseteq\beta X\times Y$. Thus
$\beta(X\times Y)\subseteq\beta X\times Y$.

Conversely, if $z\in\beta X$, and $\widehat{x_{r}}\to z$, then
$(x_{r},y\widehat{)}\to(z,y\widehat{)}$. In fact, for each $f\in C_{b}(X\times
Y)$ we have

$(x_{r},y\widehat{)}(f)=f(x_{r},y)=\widehat{x_{r}}(fy)\to
z(fy)=(z,y\widehat{)}(f).$

Then $\beta X\times Y\subseteq\beta(X\times Y)$. Therefore

$\displaystyle C_{b}(X\times Y)$ $\displaystyle=C(\beta X\times Y)$
$\displaystyle=C(\beta X)\check{\otimes}C(Y)=C_{b}(X)\check{\otimes}C(Y).$

(ii) Define

$\displaystyle F:C_{b}(X\times Y)$ $\displaystyle\to C(bX\times Y)$
$\displaystyle f\mapsto f^{{}^{\prime}}$

where $f^{{}^{\prime}}(z,y)=limf(x_{r},y)$ such that
$\widehat{x_{r}}\overset{w^{*}}{\to}z$ in $C_{b}(X)^{*}$. Thus
$f^{{}^{\prime}}(z,y)=z(f_{y})$ where $f_{y}(x)=f(x,y)$. Clearly, $f_{y}$ is
continuous and

$\|f^{{}^{\prime}}\|_{\infty}\leq\|z\|.\|f_{y}\|\leq\|f\|_{\infty}$

Otherwise $\|f\|\geq|f(x_{r},y)|$, so
$\|f^{{}^{\prime}}\|_{\infty}\geq\|f\|_{\infty}$ and $F$ is isometry. Clearly,
$F$ is continuous.

∎

######  Lemma 3.4.

$<\Delta(A\otimes_{\gamma}B)>=<\Delta(A)>\odot<\Delta(B)>$

###### Proof.

If $V:=<\Delta(A)>\odot<\Delta(B)>$, then $V$ is a vector space. Thus
$\Delta(A)\times\Delta(B)\subseteq V$, and so
$<\Delta(A\otimes_{\gamma}B)>\subseteq V$.

Conversely, $p\odot q=p\otimes q$ on $A\times B$, so
$V\subseteq<\Delta(A\otimes_{\gamma}B)>$. This completes the proof. ∎

Furthermore, $\varphi\odot\psi=\varphi\otimes\psi|_{A\otimes B}$, where
$\varphi\otimes\psi:A^{**}\times B^{**}\to\mathbb{C}$ defined by
$\varphi\otimes\psi(F,G)=F(\varphi)\psi(G)$, for all $F\in A^{**}$ and $G\in
B^{**}$. Thus

$\varphi\odot\psi(a\otimes b)=\varphi\otimes\psi(\hat{a}\otimes\hat{b})$

for all $a\in A$ and $b\in B$. Clearly, $\Delta(A)\subseteq
C_{\textup{BSE}}(\Delta(A))^{*}$, by embeding, $\varphi\mapsto\hat{\varphi}$
where $\hat{\varphi}(\sigma)=\sigma(\varphi)$ for all $\sigma\in
C_{\textup{BSE}}(\Delta(A))$. Let $\sigma_{1}\in C_{\textup{BSE}}(\Delta(A))$
and $\sigma_{2}\in C_{\textup{BSE}}(\Delta(B))$. Then

$\sigma_{1}\otimes\sigma_{2}:C_{\textup{BSE}}(\Delta(A))^{*}\times
C_{\textup{BSE}}(\Delta(B))^{*}\to\mathbb{C}$

defined by $\sigma_{1}\otimes\sigma_{2}(F,G)=F(\sigma_{1})G(\sigma_{2})$. We
denote
$\sigma_{1}\odot\sigma_{2}:=\sigma_{1}\otimes\sigma_{2}|_{\Delta(A)\times\Delta(B)}$.
The linear space generated by $\sigma_{1}\odot\sigma_{2}$ is denoted by
$C_{\textup{BSE}}(\Delta(A))\odot C_{\textup{BSE}}(\Delta(B))$.

If there exists a sequence $(\sigma_{n})$ in $C_{\textup{BSE}}(\Delta(A))\odot
C_{\textup{BSE}}(\Delta(B))$ such that $\sigma_{n}\to\sigma$. Then

$\sigma\in
C_{\textup{BSE}}(\Delta(A))\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}(\Delta(B))$

If $p\in<\Delta(A)>$ and $q\in<\Delta(B)>$, then

$\|(p,q)\|=\|p\|_{A^{*}}+\|q\|_{B^{*}}$

Thus, $\sigma\in C_{\textup{BSE}}(\Delta(A\hat{\otimes}B))$ if and only if for
all $\epsilon>0$ there exists $\delta>0$ where if

$\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}+\|\sum_{i=1}^{m}d_{i}\psi_{i}\|_{B^{*}}<\delta$

then

$|\sum_{i=1}^{n}\sum_{j=1}^{m}c_{i}d_{j}\sigma(\varphi_{i}\otimes\psi_{j})|<\epsilon$

######  Theorem 3.5.

Let $A$ and $B$ be commutative Banach algebra with nonempty character spaces.
Then

$C_{\textup{BSE}}(\Delta(A))\overset{{\textup{BSE}}}{\otimes}C_{\textup{BSE}}(\Delta(B))=C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B)).$

###### Proof.

If $\sigma_{1}\in C_{\textup{BSE}}(\Delta(A))$ and $\sigma_{2}\in
C_{\textup{BSE}}(\Delta(B))$, then

$\displaystyle\|\sigma_{1}\otimes\sigma_{2}\|_{\textup{BSE}}\leq\|\sigma_{1}\|_{\textup{BSE}}\|\sigma_{2}\|_{\textup{BSE}}.$

Thus

$C_{\textup{BSE}}(\Delta(A)){\otimes}C_{\textup{BSE}}(\Delta(B))\subseteq
C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B)).$

As a result

$C_{\textup{BSE}}(\Delta(A))\overset{{\textup{BSE}}}{\otimes}C_{\textup{BSE}}(\Delta(B))\subseteq
C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B)).$

Conversely, if $\sigma\in C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B))$, then
there exists some bounded net $f_{s}$ in $A\otimes_{\gamma}B$ such that
${\widehat{f_{s}}}(\varphi\otimes\psi)\to\sigma(\varphi\otimes\psi)$, for each
$\varphi\in\Delta(A)$ and each $\psi\in\Delta(B)$. If

$f=\sum_{i=1}^{n}a_{i}\otimes b_{i}$

then

${\hat{f}}=\sum_{i=1}^{n}{\hat{a}_{i}}\otimes{\hat{b}_{i}}\in{\widehat{A}}\otimes{\widehat{B}}$

where

${\widehat{A}}\otimes{\widehat{B}}\subseteq
C_{\textup{BSE}}(\Delta(A)){\otimes}C_{\textup{BSE}}(\Delta(B)).$

Thus

$\sigma\in T_{p}-cl{\widehat{A}}\otimes{\widehat{B}}\subseteq
C_{\textup{BSE}}(\Delta(A))\check{\otimes}C_{\textup{BSE}}(\Delta(B))$

If $\|\sigma_{n}-\sigma\|_{\epsilon}\to 0$ in
$C_{\textup{BSE}}(\Delta(A)){\otimes}C_{\textup{BSE}}(\Delta(B))$, Set

$S_{n}=\sigma_{n}|_{\Delta(A)\times\Delta(B)}.$

Thus

$\|S_{n}-\sigma\|_{\textup{BSE}}<\|\sigma_{n}-\sigma\|_{\epsilon}\to 0$

Then $\sigma\in
C_{\textup{BSE}}(\Delta(A))\overset{{\textup{BSE}}}{\otimes}C_{\textup{BSE}}(\Delta(B))$.
This completes the proof. ∎

######  Corollary 3.6.

Let $G$ be a locally compact Abelian group and $A$ be commutative unital
Banach algebra. Then:

$C_{\textup{BSE}}(\Delta(L^{1}(G,A)))=\widehat{M(G)}\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}(\Delta(A)).$

######  Corollary 3.7.

Let $X$ be a locally compact Hausdorff topological space and $A$ be a
commutative unital Banach algebra. Then

$C_{\textup{BSE}}(\Delta(C_{0}(X,A))=C_{b}(X\widehat{)}\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}(\Delta(A)).$

######  Lemma 3.8.

Let $A$ and $B$ be commutative Banach algebra and $A\otimes_{\gamma}B$ be a
Banach algebra, where $\epsilon\leq\gamma\leq\pi$. Then
(i)

$\displaystyle C_{\textup{BSE}}(\Delta(A\check{\otimes}B))$
$\displaystyle\subseteq C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B))$
$\displaystyle\subseteq C_{\textup{BSE}}(\Delta(A\hat{\otimes}B)).$

(ii)

$\displaystyle C_{\textup{BSE}}^{0}(\Delta(A\check{\otimes}B))$
$\displaystyle\subseteq
C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))\subseteq
C_{\textup{BSE}}^{0}(\Delta(A\hat{\otimes}B)).$

###### Proof.

(i) If $\sigma\in C_{\textup{BSE}}(\Delta(A\check{\otimes}B))$,since

$\Delta(A\check{\otimes}B)=\Delta(A\otimes_{\gamma}B)=\Delta(A)\times\Delta(B)$

so

$\displaystyle|\sum_{i=1}^{n}c_{i}\sigma(\varphi_{i}\otimes\psi_{i})|\leq
M\|\sum_{i=1}^{n}c_{i}(\varphi_{i}\otimes\psi_{i})\|_{\epsilon}\leq
M\|\sum_{i=1}^{n}c_{i}(\varphi_{i}\otimes\psi_{i})\|_{\gamma}$

Thus $\sigma\in C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B))$. In a similar
way, because $\|.\|_{\gamma}<\|.\|_{\pi}$, so

$C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B))\subseteq
C_{\textup{BSE}}(\Delta(A\hat{\otimes}B)).$

(ii) The proof is similar to part (i). ∎

######  Lemma 3.9.

Let $A$ and $B$ be commutative Banach algebra with nonempty character spaces
such that $A$ or $B$ is unital and $A\otimes_{\gamma}B$ be a Banach algebra,
where $\epsilon\leq\gamma\leq\pi$. Then

$C_{\textup{BSE}}^{0}(\Delta(A))\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}^{0}(\Delta(B))\subseteq
C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))$

###### Proof.

Assume that $B$ is unital, $\sigma_{1}\in C_{\textup{BSE}}^{0}(\Delta(A))$ and
$\sigma_{2}\in C_{\textup{BSE}}^{0}(\Delta(B))$. Then

$\sigma_{1}\odot\sigma_{2}(\varphi\otimes\psi)=\sigma_{1}(\varphi)\sigma_{2}(\psi)$

for all $\varphi\in\Delta(A)$ and $\psi\in\Delta(B)$. Since $\sigma_{1}\in
C_{\textup{BSE}}^{0}(\Delta(A))$, so for all $\epsilon>0$ there exist some
compact set $K_{1}\subseteq\Delta(A)$ such that for all $c_{i}\in\mathbb{C}$
and $\varphi_{i}\in\Delta(A)\backslash K_{1}$ the following is yield:

$|\sum_{i=1}^{n}c_{i}\sigma_{1}(\varphi_{i})|<\epsilon\|\sum_{i=1}^{n}c_{i}\varphi_{i}\|_{A^{*}}$

Since $B$ is unital, then $\Delta(B)$ is compact. Set
$K:=K_{1}\times\Delta(B)$, thus $K\subseteq\Delta(A)\times\Delta(B)$ is
compact such that for all $c_{i}\in\mathbb{C}$ and
$\varphi_{i}\otimes\psi_{i}\notin K$ we have

$\displaystyle|\sum_{i=1}^{n}c_{i}\sigma_{1}\odot\sigma_{2}(\varphi_{i}\otimes\psi_{i})|$
$\displaystyle=|\sum_{i=1}^{n}c_{i}\sigma_{1}(\varphi_{i})\sigma_{2}(\psi_{i})|$
$\displaystyle\leq\epsilon\|\sum_{i=1}^{n}c_{i}\sigma_{2}(\psi_{i})\varphi_{i}\|_{A^{*}}$
$\displaystyle\leq\epsilon\|\sigma_{2}\|_{\textup{BSE}}\|\sum_{i=1}^{n}c_{i}\varphi_{i}\otimes\psi_{i}\|_{(A\otimes_{\gamma}B)^{*}}$

Also

$\|\sigma_{1}\odot\sigma_{2}\|_{\textup{BSE}}\leq\|\sigma_{1}\|_{\textup{BSE}}\|\sigma_{2}\|_{\textup{BSE}}$

So $\sigma_{1}\odot\sigma_{2}\in
C_{\textup{BSE}}(\Delta(A\otimes_{\gamma}B))$. Therefore
$\sigma_{1}\odot\sigma_{2}\in
C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))$. As a result,

$\displaystyle
C_{\textup{BSE}}^{0}(\Delta(A))\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}^{0}(\Delta(B))\subseteq
C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))$

In a similar way, If $A$ is unital, the above inclusion is established. ∎

######  Lemma 3.10.

Let $A$ and $B$ be commutative Banach algebra with nonempty character spaces
and $A\otimes_{\gamma}B$ be a Banach algebra, where
$\epsilon\leq\gamma\leq\pi$. Then

$C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))\subseteq
C_{\textup{BSE}}^{0}(\Delta(A))\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}^{0}(\Delta(B))$

###### Proof.

Assume that $\sigma\in C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))$. Thus

$\displaystyle\sigma\in C_{0}(\Delta(A\otimes_{\gamma}B)$
$\displaystyle=C_{0}(\Delta(A)\times\Delta(B))$
$\displaystyle=C_{0}(\Delta(A))\check{\otimes}C_{0}(\Delta(B))$
$\displaystyle=C_{0}(\Delta(A)\widehat{)}\check{\otimes}C_{0}(\Delta(B)\widehat{)}$
$\displaystyle=C_{\textup{BSE}}^{0}(\Delta(C_{0}(\Delta(A))\check{\otimes}C_{\textup{BSE}}^{0}(\Delta(C_{0}(\Delta(B))$
$\displaystyle=C_{\textup{BSE}}^{0}(\Delta(A))\check{\otimes}C_{\textup{BSE}}^{0}(\Delta(B))$

Then there exists some sequence $(\sigma_{n})$ in
$C_{\textup{BSE}}^{0}(\Delta(A))\otimes C_{\textup{BSE}}^{0}(\Delta(B))$,
where $\sigma_{n}\overset{\|.\|_{\epsilon}}{\to}\sigma$. Assume that
$\sigma_{n}=\sum_{k=1}^{l_{n}}\sigma_{k}^{n}\otimes\eta_{k}^{n}$, where
$\sigma_{k}^{n}\in C_{\textup{BSE}}^{0}(\Delta(A))$ and $\eta_{k}^{n}\in
C_{\textup{BSE}}^{0}(\Delta(B))$. If
$S_{n}=\sum_{k=1}^{p_{n}}\mu_{k}^{n}\odot\nu_{k}^{n}=\sigma_{n}|_{\Delta(A)\times\Delta(B)}$,
then

$\displaystyle\sum_{k=1}^{l_{n}}\sigma_{k}^{n}\otimes\eta_{k}^{n}(\widehat{\varphi},\widehat{\psi})$
$\displaystyle:=\sum_{k=1}^{l_{n}}\widehat{\varphi}(\sigma_{k}^{n})\widehat{\psi}(\eta_{k}^{n})$
$\displaystyle:=\sum_{k=1}^{l_{n}}\sigma_{k}^{n}(\varphi)\eta_{k}^{n}(\psi)$
$\displaystyle:=\sum_{k=1}^{l_{n}}\mu_{k}^{n}(\varphi)\nu_{k}^{n}(\psi)$
$\displaystyle=S_{n}(\varphi,\psi)$

It is clear that ${\widehat{\varphi}}\in{C_{\textup{BSE}}^{0}(\Delta(A))}^{*}$
and ${\widehat{\psi}}\in{C_{\textup{BSE}}^{0}(\Delta(B))}^{*}$. Therefore

$S_{n}=\sum_{k=1}^{l_{n}}\mu_{k}^{n}\odot\nu_{k}^{n}\in<C_{\textup{BSE}}^{0}(\Delta(A))\odot
C_{\textup{BSE}}^{0}(\Delta(B))>$

where
$S_{n}(\varphi\otimes\psi)=\sigma_{n}({\widehat{\varphi}},{\widehat{\psi}})$,
for all $\varphi\in\Delta(A)$, $\psi\in\Delta(B)$ and $n\in\mathbb{N}$. The
following is yield:

$\displaystyle\sigma:=\underset{n}{lim}\sigma_{n}=\underset{n}{lim}(\sum_{k=1}^{l_{n}}\sigma_{k}^{n}\otimes\eta_{k}^{n})$

At a result

$\displaystyle\sigma(\varphi\otimes\psi)=\sigma({\widehat{\varphi}},{\widehat{\psi}})$
$\displaystyle=\underset{n}{lim}\sigma_{n}({\widehat{\varphi}},{\widehat{\psi}})$
$\displaystyle=\underset{n}{lim}\sum_{k=1}^{l_{n}}\mu_{k}^{n}\odot\nu_{k}^{n}(\varphi\otimes\psi)=\underset{n}{lim}S_{n}(\varphi,\psi)$

and

$\|S_{n}-\sigma\|_{\textup{BSE}}\leq\|\sigma_{n}-\sigma\|_{\epsilon}\to 0$

At a result

$\sigma\in
C_{\textup{BSE}}^{0}(\Delta(A))\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}^{0}(\Delta(B))$

Therefore

$C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))\subseteq
C_{\textup{BSE}}^{0}(\Delta(A))\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}^{0}(\Delta(B)).$

∎

######  Theorem 3.11.

Let $A$ and $B$ be commutative Banach algebra with nonempty character spaces
such that $A$ or $B$ is unital and $A\otimes_{\gamma}B$ be a Banach algebra,
where $\epsilon\leq\gamma\leq\pi$. Then

$C_{\textup{BSE}}^{0}(\Delta(A))\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}^{0}(\Delta(B))=C_{\textup{BSE}}^{0}(\Delta(A\otimes_{\gamma}B))$

###### Proof.

By applying Lemmas 3.10 and 3.9, complete the proof. ∎

######  Corollary 3.12.

Let $X$ be a locally compact Hausdorff topological space and $A$ be a
commutative unital Banach algebra. Then

$C_{\textup{BSE}}^{0}(\Delta(C_{0}(X,A))=C_{0}(X\widehat{)}\overset{{\textup{BSE}}}{\odot}C_{\textup{BSE}}^{0}(\Delta(A)).$

######  Corollary 3.13.

Let $G$ be a locally compact Abelian group and $A$ be a commutative Banach
unital algebra. Then

$C_{\textup{BSE}}^{0}(\Delta(L^{1}(G,A)=L^{1}(G\widehat{)}\overset{\textup{BSE}}{\otimes}C_{\textup{BSE}}^{0}(\Delta(A)).$

## References

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