The Bochner–Schoenberg–Eberlein Property for Vector-Valued $\ell ^p$-Spaces

Abstract

Let X be a non-empty set, ${\mathcal {A}}$ be a commutative Banach algebra, and $1\le p<\infty$. In this paper, we establish some basic properties of $\ell ^p(X,\mathcal {A})$, inherited from ${\mathcal {A}}$. In particular, we characterize the Gelfand space of $\ell ^p(X,\mathcal {A})$, denoted by $\Delta (\ell ^p(X,{\mathcal {A})})$. Mainly, we investigate the BSE property of the Banach algebra $\ell ^p(X,\mathcal {A})$. In fact, we prove that $\ell ^p(X,\mathcal {A})$ is a BSE algebra if and only if X is finite and $\mathcal {A}$ is a BSE algebra. Furthermore, in the case that $\mathcal {A}$ is unital, we show that for any natural number n, all continuous bounded functions on $\Delta (\ell ^p(X,{\mathcal {A}}))$ are n-BSE functions. However, through an example, we indicate that there is some continuous bounded function on $\Delta (\ell ^p(X,{\mathcal {A}}))$ which is not BSE. Finally, we prove that if $\ell ^1(X,{\mathcal {A}})$ is a BSE-norm algebra, then $\mathcal {A}$ is so. We also prove the converse of this statement, whenever $\mathcal {A}$ is a supremum norm algebra.

1 Introduction

The notion of BSE algebras and BSE functions was first introduced and studied by Takahashi and Hatori in 1990 [18] and subsequently by several authors for various kinds of Banach algebras, such as Fourier and Fourier–Stieltjes algebras, semigroup algebras, abstract Segal algebras, etc. The interested reader is referred to [5, 8, 11,12,13,14, 19, 20]. Moreover, in a recent work, Dabhi and Upadhyay proved that $\ell ^1({\mathbb {Z}}^2,\max )$ is a BSE algebra [4]. Furthermore, in [1], we investigated the BSE property for vector-valued Lipschitz algebra $\mathrm{Lip}_\alpha (X,\mathcal {A})$, and proved that for unital commutative semisimple Banach algebra $\mathcal {A}$, $\mathrm{Lip}_\alpha (X,\mathcal {A})$ is a BSE algebra if and only if $\mathcal {A}$ is so.

The acronym BSE stands for Bochner–Shoenberg–Eberlein famous theorem which characterizes the Fourier–Stieltjes transforms of the bounded Borel measures on locally compact abelian groups; that is, in fact, the BSE property of the group algebra $L^1(G)$ for a locally compact abelian group G; see [2, 7, 17]. This has led the Japanese mathematicians to introduce the BSE property for an arbitrary commutative Banach algebra as follows:

Let ${\mathcal {A}}$ be a commutative Banach algebra. Denote by $\Delta ({\mathcal {A}})$ to be the Gelfand space of $\mathcal {A}$; i.e., the space consisting of all nonzero multiplicative linear functionals on $\mathcal {A}$.

A bounded continuous function $\sigma$ on $\Delta ({\mathcal {A}})$ is called a BSE function if there exists a constant $C>0$, such that for every finite number of $\varphi _1,\ldots ,\varphi _n$ in $\Delta ({\mathcal {A}})$ and complex numbers $c_1,\ldots ,c_n$, the inequality:

$$\begin{aligned} \left| \displaystyle \sum _{j=1}^nc_j\sigma (\varphi _j)\right| \le C \;\left\| \displaystyle \sum _{j=1}^nc_j\varphi _j\right\| _{{\mathcal {A}}^*} \end{aligned}$$

holds. The BSE norm of $\sigma$ ($\Vert \sigma \Vert _{\mathrm{BSE}}$) is defined to be the infimum of all such C. The set of all BSE functions is denoted by $C_{\mathrm{BSE}}(\Delta ({\mathcal {A}}))$. Takahasi and Hatori [18] showed that under the norm $\Vert . \Vert _{\mathrm{BSE}}$, $C_{\mathrm{BSE}}(\Delta ({\mathcal {A}}))$ is a commutative semisimple Banach algebra, embedded in $C_b(\Delta ({\mathcal {A}}))$ as a subalgebra.

Here, we provide some preliminaries, which will be required throughout the paper. See [15] for more information. A bounded linear operator on a commutative Banach algebra ${\mathcal {A}}$ is called a multiplier if it satisfies $xT(y)=T(xy),$ for all $x,y\in$. The set $M({\mathcal {A}})$ of all multipliers of ${\mathcal {A}}$ is a unital commutative Banach algebra, called the multiplier algebra of ${\mathcal {A}}$. Set:

$$\begin{aligned} \widehat{{ M}({\mathcal {A}})}=\{\widehat{T}:T\in {M}({\mathcal {A}})\}. \end{aligned}$$

Remark 1.1

Let $\mathcal {A}$ be a commutative semisimple Banach algebra. Suppose that $\Phi :\Delta ({\mathcal {A}})\rightarrow {\mathbb {C}}$ be a continuous function, such that $\Phi .\widehat{\mathcal {A}}\subseteq \widehat{\mathcal {A}}$. We call $\Phi$ a multiplier of $\mathcal {A}$. This is another definition of a multiplier of a Banach algebra. In the presence of supersimplicity, this definition is equivalent to the above definition, by considering $\Phi =\widehat{T}$; see [16] for more details. Define:

$$\begin{aligned} {\mathcal {M}}({\mathcal {A}})=\{\Phi :\Delta ({\mathcal {A}})\rightarrow {\mathcal {C}}:\Phi \;\text {is continuous and}\; \Phi .\widehat{\mathcal {A}}\subseteq \widehat{\mathcal {A}}\}. \end{aligned}$$

When $\mathcal {A}$ is semisimple, $\widehat{M({\mathcal {A}})}={\mathcal {M}}({\mathcal {A}})$.

A commutative Banach algebra ${\mathcal {A}}$ is called without order if $a{\mathcal {A}}=\{0\}$ implies $a=0$ ($a\in {\mathcal {A}}$). A commutative and without order Banach algebra ${\mathcal {A}}$ is called a BSE algebra (or has the BSE property) if it satisfies the condition:

$$\begin{aligned} C_{\mathrm{BSE}}(\Delta ({\mathcal {A}}))=\widehat{{ M}({\mathcal {A}})}. \end{aligned}$$

Furthermore, ${\mathcal {A}}$ is called a BSE algebra of type I if:

$$\begin{aligned} C_{\mathrm{BSE}}(\Delta ({\mathcal {A}}))=\widehat{{ M}({\mathcal {A}})}=C_{b}(\Delta ({\mathcal {A}})). \end{aligned}$$

By Remark 1.1, in the case that $\mathcal {A}$ is a semisimple commutative Banach algebra, the BSE property of $\mathcal {A}$ is equivalent to the following equality:

$$\begin{aligned} C_{\mathrm{BSE}}(\Delta ({\mathcal {A}}))=\mathcal {M}({\mathcal {A}}). \end{aligned}$$

It is worthy to note that all semisimple Banach algebras are without order.

Let X be an arbitrary non-empty set and consider the Banach algebra $\ell ^1(X)$ with pointwise multiplication. In [19], the authors proved that the Banach algebra $\ell ^1(X)$ is BSE if and only if X is finite.

Throughout the paper, let X be a non-empty set and $\mathcal {A}$ be a commutative Banach algebra. In this paper, at first, we investigate several properties of the vector-valued Banach algebra $\ell ^p(X,\mathcal {A})$ ($1\le p<\infty$), inherited from $\mathcal {A}$. Moreover, we characterize the Gelfand space of the Banach algebra $\ell ^p(X,\mathcal {A})$ by the set X and the Gelfand space of $\mathcal {A}$. Then, we present necessary and sufficient conditions for $\ell ^p(X,\mathcal {A})$ to be a BSE algebra. In fact, we prove that $\ell ^p(X,\mathcal {A})$ is a BSE algebra if and only if X is finite and ${\mathcal {A}}$ is a BSE algebra. Furthermore, we show that for any $n\in {\mathbb {N}}$ and a unital Banach algebra ${\mathcal {A}}$, the Banach algebra $C_{\mathrm{BSE}(n)}(\Delta (l^p(X,{\mathcal {A}})))=C_{\mathrm{BSE}(n)}(X\times \Delta ({\mathcal {A}}))$ is equal to the Banach algebra $C_b(X\times \Delta ({\mathcal {A}}))$. However, with an example, we show that this result is not true for $C_{\mathrm{BSE}}(X\times \Delta ({\mathcal {A}}))$, even for a unital Banach $\mathcal {A}$. Moreover, we investigate BSE-norm property for $\ell ^1(X,{\mathcal {A}})$ and prove that if $\ell ^1(X,{\mathcal {A}})$ is a BSE-norm algebra, then $\mathcal {A}$ is so. We also prove that the converse of this results is valid, whenever $\mathcal {A}$ is a supremum norm algebra.

Finally, we present a different proof, from abstract Segal algebras point of view, to show that $\ell ^p(X)$ is a BSE algebra if and only if X is finite.

2 Some Basic Properties $\ell ^p(X,\mathcal {A})$ Inherited from $\mathcal {A}$

Let X be a non-empty set, $\mathcal {A}$ be a commutative Banach algebra, and $1\le p<\infty$. Let:

$$\begin{aligned} \ell ^p(X,\mathcal {A})=\left\{ f:X\rightarrow \mathcal {A}: \displaystyle \sum _{x\in X}\Vert f(x)\Vert ^p<+\infty \right\} . \end{aligned}$$

It is easily verified that $\ell ^p(X,\mathcal {A})$ is a commutative Banach algebra, endowed with the norm:

$$\begin{aligned} \Vert f\Vert _p=\displaystyle \left( \sum _{x\in X}\Vert f(x)\Vert ^p\right) ^{1/p}\quad (f\in \ell ^p(X,\mathcal {A})) \end{aligned}$$

and pointwise product. In this section, we investigate some elementary and basic properties about $\ell ^p(X,\mathcal {A})$, which will be useful for further results. Let us first introduce some noteworthy vector-valued functions on X, which play an important role in our results. For any finite subset F of X and nonzero element $a\in \mathcal {A}$, we define the function $\delta _a^F$ as follows:

$$\begin{aligned} \delta _a^F(t)=\left\{ \begin{array}{lr} a &{}\;\;\; t\in F\\ 0 &{}\;\;\; t\not \in F. \end{array}\right. \end{aligned}$$

These functions belong clearly to $\ell ^p(X,\mathcal {A})$. In the case that F is a singleton, namely $F=\{x\}$, then we simply rewrite $\delta _a^F$ as $\delta _a^x$.

Proposition 2.1

Let X be a set, $\mathcal {A}$ be a commutative Banach algebra, and $1\le p<\infty$. Then, $\ell ^p(X,\mathcal {A})$ is unital if and only if X is finite and $\mathcal {A}$ is unital.

Proof

At first, suppose that X is finite and $\mathcal {A}$ has an identity e. It is not hard to see that the constant function:

$$\begin{aligned} I:X\rightarrow A,\;\;\;x\mapsto e \end{aligned}$$

is the identity element of $\ell ^p(X,\mathcal {A})$. Conversely, suppose that $I\in \ell ^p(X,\mathcal {A})$ is the identity element of $\ell ^p(X,\mathcal {A})$. Then, for each $f\in \ell ^p(X,\mathcal {A})$, we have:

$$\begin{aligned} f(x)I(x)=f(x)\;\;\;\;(x\in X). \end{aligned}$$

Specially:

$$\begin{aligned} \Vert I(x)\Vert =\Vert I(x)I(x)\Vert \le \Vert I(x)\Vert \;\Vert I(x)\Vert \quad (x\in X). \end{aligned}$$

Thus, for each $x\in X$, $I(x)=0$ or $\Vert I(x)\Vert \ge 1$. Note that since $I\in \ell ^p(X,\mathcal {A})$, $I(x)=0$, except for finitely many $x_1,\ldots , x_n\in X$. Now, we show that:

$$\begin{aligned} X=\{x_1,\ldots ,x_n\}. \end{aligned}$$

Suppose on the contrary that there exists $x\in X$, such that $x\notin \{x_1,\ldots ,x_n\}$. Take $a\in \mathcal {A}$ to be nonzero and consider the function $\delta _a^x(t)$. Thus:

$$\begin{aligned} 0=I(x)\delta _a^x(x)=\delta _a^x(x)=a, \end{aligned}$$

which is impossible. It follows that $X=\{x_1,\ldots ,x_n\}$. In the sequel, we show that $\mathcal {A}$ is unital. For all $x\in X$ and $a\in \mathcal {A}$, we have:

$$\begin{aligned} I(t)\delta _x^a(t)=\delta _a^x(t)\quad (t\in X). \end{aligned}$$

Consequently:

$$\begin{aligned} I(x)a=a\quad (x\in X, a\in \mathcal {A}). \end{aligned}$$

It follows that I is a constant function. Indeed, for all $x,y\in X$ with $x\ne y$:

$$\begin{aligned} I(x)I(y)=I(x)=I(y). \end{aligned}$$

Therefore, I(x) is the identity element of $\mathcal {A}$. $\square$

Proposition 2.2

Let X be a set, $\mathcal {A}$ be a commutative Banach algebra, and $1\le p<\infty$. Then, $\ell ^p(X,\mathcal {A})$ is without order if and only if $\mathcal {A}$ is without order.

Proof

First, suppose that $\mathcal {A}$ is without order and $0\ne f\in \ell ^p(X,\mathcal {A})$. Then, there exists $x_0\in X$, such that $f(x_0)=a\ne 0$. By the hypothesis, there exists $b\in \mathcal {A}$ such that $ab\ne 0$. It follows that:

$$\begin{aligned} f(x_0)\delta _b^{x_0}(x_0)=ab\ne 0, \end{aligned}$$

and so, $f\;\delta _b^{x_0}\ne 0$. Consequently, $\ell ^p(X,\mathcal {A})$ is without order. Conversely, suppose that $\ell ^p(X,\mathcal {A})$ is without order and $0\ne a\in \mathcal {A}$. For any $x_0\in X$, we have:

$$\begin{aligned} \delta _{x_0}^a(x_0)=a\ne 0. \end{aligned}$$

By the hypothesis, there exists $f\in \ell ^p(X,\mathcal {A})$, such that $f\;\delta _{x_0}^a\ne 0$. Thus:

$$\begin{aligned} f(x_0)a=f(x_0)\delta _{x_0}^a(x_0)\ne 0. \end{aligned}$$

Take $b:=f(x_0)$. It follows that $ba\ne 0$. Therefore, $\mathcal {A}$ is without order. $\square$

Theorem 2.3

Let X be a set, $\mathcal {A}$ be a commutative Banach algebra, and $1\le p<\infty$. Then, the Gelfand space of $\ell ^p(X,\mathcal {A})$ is homeomorphic to $X\times \Delta (\mathcal {A})$.

Proof

Define the function $\Theta$ as:

$$\begin{aligned} \Theta : X\times \Delta (\mathcal {A})\rightarrow & {} \Delta (\ell ^p(X,\mathcal {A}))\\ (x,\varphi )\mapsto & {} \Theta _{(x,\varphi )}, \end{aligned}$$

where:

$$\begin{aligned} \Theta _{(x,\varphi )}(f)=\varphi (f(x))\quad (f\in \ell ^p(X,\mathcal {A})). \end{aligned}$$

It is obvious that $\Theta _{(x,\varphi )}\in \Delta (\ell ^p(X,{\mathcal {A}}))$ and so $\Theta$ is well defined. Now, we show that $\Theta$ is injective. Suppose that $\Theta _{(x,\varphi )}=\Theta _{(y,\psi )}$, for some $x,y\in X$ and $\varphi ,\psi \in \Delta (\mathcal {A})$. Then, for any $f\in \ell ^p(X,\mathcal {A})$, we have $\varphi (f(x))=\psi (f(y))$. For each $a\in \mathcal {A}$, consider the function $\delta _a^{\{x,y\}}$. Thus:

$$\begin{aligned} \varphi (\delta _a^{\{x,y\}}(x))=\psi \left( \delta _a^{\{x,y\}}(y)\right) . \end{aligned}$$

It follows that:

$$\begin{aligned} \varphi (a)=\psi (a)\quad (a\in A), \end{aligned}$$

and we obtain $\varphi =\psi$. Moreover, the equality $\varphi (f(x))=\varphi (f(y))$ ($f\in \ell ^p(X,\mathcal {A})$), which implies that $\varphi (f(x)-f(y))=0$ for each $f\in \ell ^p(X,\mathcal {A})$. If $x\ne y$, then:

$$\begin{aligned} \varphi (\delta _a^{x}(x)-\delta _a^{x}(y))=0\quad (a\in \mathcal {A}). \end{aligned}$$

This implies that $\varphi (a)=0$ for all $a\in A$ and so $\varphi =0$. This contradiction implies that $x=y$. Consequently, $\Theta$ is injective. To prove the surjectivity, let $\Phi \in \Delta (\ell ^p(X,\mathcal {A}))$. Since $\Phi$ is nonzero, there exists $f=\sum _{t\in X}\delta _{f(t)}^t$, such that $\Phi (f)\ne 0$. It follows that $\Phi (\delta _a^{x_0})\ne 0$, for some $x_0\in X$. Such $x_0\in X$ is unique. Indeed, let there exists $x\ne x_0$, such that $\Phi (\delta _a^x)\ne 0$. Since $\delta _a^{x_0}.\delta _a^x=0$, we have:

$$\begin{aligned} 0=\Phi (\delta _a^{x_0}.\delta _a^x)=\Phi (\delta _a^{x_0})\Phi ( \delta _a^x)\ne 0, \end{aligned}$$

which is a contradiction. Now, define:

$$\begin{aligned} \varphi _0: \mathcal {A}\rightarrow & {} {\mathbb {C}}\\ \varphi _0(a)= & {} \Phi (\delta _a^{x_0}). \end{aligned}$$

We show that $\Phi =\Theta _{(x_0,\varphi _0)}$. For $f\in \ell ^p(X,\mathcal {A})$, we may rewrite f as:

$$\begin{aligned} f(x)=\displaystyle \sum _{t\in X}\delta _{f(t)}^t(x). \end{aligned}$$

Thus, we obtain:

$$\begin{aligned} \Theta _{(x_0,\varphi _0)}(f)=\varphi _0(f(x_0))=\Phi \left( \delta _{f(x_0)}^{x_0}\right) =\Phi (f). \end{aligned}$$

This implies that $\Theta$ is surjective. To prove the continuity of $\Theta$, consider the net $\{(x_{\alpha },\varphi _{\alpha })\}_{\alpha \in \Lambda }$ converges to $(x,\varphi )$, in the topology of $X\times \Delta (A)$. So that there exists $\alpha _0\in \Lambda$, such that for all $\alpha \ge \alpha _0$, $x_{\alpha }=x$. Moreover, for each $f\in \ell ^p(X,A)$, we have:

$$\begin{aligned} \lim _{\alpha }\Theta _{(x,\varphi _{\alpha })}(f)=\lim _{\alpha }\varphi _{\alpha }(f(x))=\varphi (f(x))=\Theta _{(x,\varphi )}(f). \end{aligned}$$

Thus, $\Theta$ is continuous. For openness, let $\Theta _{(x_{\alpha },\varphi _{\alpha })}$ tends to $\Theta (x,\varphi )$, in the Gelfand topology of $\Delta (\ell ^p(X,\mathcal {A}))$. It follows that for any $f\in \ell ^p(X,\mathcal {A})$:

$$\begin{aligned} \lim _\alpha \Theta _{(x_{\alpha },\varphi _{\alpha )}}(f)=\Theta _{(x,\varphi )}(f), \end{aligned}$$

and so:

$$\begin{aligned} \lim _\alpha \varphi _\alpha (f(x_\alpha ))=\varphi (f(x)). \end{aligned}$$

In particular, for $a\in A$ with $\varphi (a)\ne 0$, we have:

$$\begin{aligned} \lim _\alpha \varphi _\alpha (\delta _a^x(x_\alpha ))=\varphi (\delta _a^x(x))=\varphi (a). \end{aligned}$$

Consequently, for $\varepsilon =\dfrac{\vert \varphi (a)\vert }{2}$, there exists $\alpha _0\in \Lambda$, such that for all $\alpha \ge \alpha _0$:

$$\begin{aligned} \left| \varphi _\alpha (\delta _a^x(x_\alpha ))-\varphi (a)\right| <\frac{\vert \varphi (a)\vert }{2}. \end{aligned}$$

We show that there exists $\alpha _1\in \Lambda$, such that for any $\alpha \ge \alpha _1$, $x_{\alpha }=x$. Suppose on the contrary that for any $\alpha \in \Lambda$, there exists $\beta _{\alpha }\ge \alpha$ such that $x_{\beta _{\alpha }}\ne x$. It follows that there exists $\beta _{\alpha _0}\ge \alpha _0$, such that $x_{\beta _{\alpha _0}}\ne x$. Thus:

$$\begin{aligned} \left| \varphi _\alpha (\delta _a^x(x_{\beta _{\alpha _0}}))-\varphi (a)\right| <\frac{\vert \varphi (a)\vert }{2}. \end{aligned}$$

Since $\delta _a^x(x_{\beta _{\alpha _0}})=0$, this implies that $\varphi (a)=0$, which is a contradiction. So that there exists $\alpha _1\in \Lambda$, such that $x_{\alpha }=x$, for any $\alpha \ge \alpha _1$. This means that the net $\{x_\alpha \}_{\alpha \in \Lambda }$ tends to x, in the discrete topology of X. Furthermore, for $\alpha \ge \alpha _1$, $\delta _a^x(x_{\alpha })=a$, which implies that $\lim _{\alpha }\varphi _{\alpha }(a)=\varphi (a)$. Consequently, $\{\varphi _{\alpha }\}_{\alpha \in {\Lambda }}$ tends to $\varphi$, in the Gelfand topology of $\Delta (\mathcal {A})$. This completes the proof. $\square$

Proposition 2.4

Let X be a set, $\mathcal {A}$ be a commutative Banach algebra, and $1\le p<\infty$. Then, $\ell ^p(X,\mathcal {A})$ is semisimple if and only if $\mathcal {A}$ is semisimple.

Proof

Let $\ell ^p(X,\mathcal {A})$ be semisimple and $0\ne a\in \mathcal {A}$. Then, for any $x\in X$, $\delta _x^a\ne 0$. Define:

$$\begin{aligned} \Theta :X\times \Delta (A)\rightarrow & {} \Delta (\ell ^p(X,\mathcal {A}))\\ (x,\varphi )\mapsto & {} \Theta _{(x,\varphi )}, \end{aligned}$$

where

$$\begin{aligned} \Theta _{(x,\varphi )}(f)=\varphi (f(x))\quad (x\in X, \varphi \in \Delta (\mathcal {A})). \end{aligned}$$

Then, there exists $\varphi \in \Delta (A)$, such that $\Theta _{(x,\varphi )}(\delta _x^a)=\varphi (\delta _x^a(x))\ne 0$. This follows that $\varphi (a)\ne 0$ and $\mathcal {A}$ is semisimple. Conversely, suppose that $\mathcal {A}$ is semisimple and $0\ne f\in \ell ^p(X,A)$. There exists $x_0\in X$, such that $f(x_0)\ne 0$. Since $\mathcal {A}$ is semisimple, there exists $\varphi \in {\mathcal {A}}$, such that $\varphi (f(x_0))\ne 0$. This means that $\Theta _{(x_0,\varphi )}(f)\ne 0$. $\square$

A bounded net $(e_{\alpha })_{\alpha \in I}$ in a Banach algebra $\mathcal {A}$, satisfying the condition:

$$\begin{aligned} \lim _\alpha \varphi (xe_\alpha )=\varphi (x), \end{aligned}$$

for every $x\in \mathcal {A}$ and $\varphi \in \Delta (\mathcal {A})$, is called $\Delta$-weak bounded approximate identity for $\mathcal {A}$, in the sense of Jones-Lahr; see [6, 9].

Remark 2.5

Let X be a set. In [19, Theorem 5], it has been proved that $\ell ^1(X)$ has a $\Delta$-weak bounded approximate identity if and only if X is finite. One can follow the exact arguments to prove this result for $\ell ^p(X)$, where $1\le p<\infty$.

In the following result, we generalize [19, Theorem 5], for the vector-valued case.

Theorem 2.6

Let X be a set, $\mathcal {A}$ be a commutative Banach algebra and $1\le p<\infty$. Then, $\ell ^p(X,\mathcal {A})$ has a $\Delta$-weak bounded approximate identity if and only if X is finite and $\mathcal {A}$ has a $\Delta$-weak bounded approximate identity.

Proof

First suppose that $X=\{x_1,\ldots ,x_n\}$ is finite and $(e_{\alpha })_{\alpha \in I}$ is a $\Delta$-weak bounded approximate identity for $\mathcal {A}$ with $\sup _{\alpha \in I}\Vert e_{\alpha }\Vert \le \beta$. For any $\alpha \in I$, define the constant function $f_\alpha$ on X as:

$$\begin{aligned} f_{\alpha }(x)=e_{\alpha }\quad (x\in X). \end{aligned}$$

It is easily verified that $f_{\alpha }\in \ell ^p(X,\mathcal {A})$, for all $\alpha \in I$. Moreover, for all $i=1,\ldots ,n$ and $\varphi \in \Delta (\mathcal {A})$:

$$\begin{aligned} \displaystyle \lim _{\alpha }(x_i,\varphi )(f_{\alpha })=\lim _{\alpha }\varphi (f_{\alpha }(x_i))=\lim _{\alpha }\varphi (e_{\alpha })=1. \end{aligned}$$

It follows that $(f_{\alpha })_{\alpha \in I}$ is a bounded $\Delta$-weak approximate identity for $\ell ^p(X,\mathcal {A})$. Conversely, suppose that $(f_{\alpha })_{\alpha \in I}$ is a bounded $\Delta$-weak approximate identity for $\ell ^p(X,\mathcal {A})$ with $\sup _{\alpha \in I}\Vert f_{\alpha }\Vert _p\le \beta$. We first show that $\ell ^p(X)$ has a bounded $\Delta$-weak bounded approximate identity. For a fixed element $\psi \in \Delta (\mathcal {A})$, define:

$$\begin{aligned} g_{\alpha }:=\psi \circ f_{\alpha }:X\rightarrow {\mathbb {C}}\quad (\alpha \in I). \end{aligned}$$

Then, we have:

$$\begin{aligned} \displaystyle \sum _{x\in X}\vert g_{\alpha }(x)\vert ^p= & {} \displaystyle \sum _{x\in X}\vert \psi \circ f_{\alpha }(x)\vert ^p\\= & {} \displaystyle \sum _{x\in X}\vert \psi ( f_{\alpha }(x))\vert ^p\\\le & {} \Vert \psi \Vert ^p\displaystyle \sum _{x\in X}\Vert f_{\alpha }(x)\Vert ^p\\\le & {} \beta ^p\Vert \psi \Vert ^p. \end{aligned}$$

It means that $g_{\alpha }\in \ell ^p(X)$, for all $\alpha \in I$ and:

$$\begin{aligned} \sup _{\alpha }\Vert g_{\alpha }\Vert _p\le \beta \Vert \psi \Vert . \end{aligned}$$

Furthermore, for any $x\in X$, we have:

$$\begin{aligned} \lim _{\alpha }\varphi _x(g_{\alpha })=\lim _{\alpha }g_{\alpha }(x)=\lim _{\alpha }\psi (f_{\alpha }(x))=\Theta _{(x,\psi )}(f_{\alpha })=1. \end{aligned}$$

It follows that $(g_{\alpha })_{\alpha \in I}$ is a $\Delta$-weak bounded approximate identity for $\ell ^p(X)$. Therefore, X is finite by Remark 2.5. Now, take $x_0\in X$ to be fixed and for any $\alpha \in \Lambda$, and define $e_{\alpha }:=f_{\alpha }(x_0)\in \mathcal {A}$. Thus, we have:

$$\begin{aligned} \Vert e_{\alpha }\Vert =\Vert f_{\alpha }(x_0)\Vert \le \Vert f_{\alpha }\Vert _p\le \beta . \end{aligned}$$

Moreover, for any $\varphi \in \Delta (\mathcal {A})$:

$$\begin{aligned} \lim _{\alpha }\varphi (e_{\alpha })=\lim _{\alpha }\varphi (f_{\alpha }(x_0))=\lim _{\alpha }\Theta _{(x_0,\varphi )}(f_{\alpha })=1. \end{aligned}$$

Therefore, $(e_{\alpha })_{\alpha \in I}$ is a $\Delta$-weak bounded approximate identity for $\mathcal {A}$. $\square$

3 The BSE Property for $\ell ^p(X,\mathcal {A})$

In this section, we state the main result of the present paper. In fact, we provide a necessary and sufficient condition for $\ell ^p(X,\mathcal {A})$ to be a BSE algebra.

Theorem 3.1

Let X be a set and $\mathcal {A}$ be a commutative semisimple Banach algebra. Then, $\ell ^p(X,\mathcal {A})$ is a BSE algebra if and only if X is finite and $\mathcal {A}$ is a BSE algebra.

Proof

First, suppose that $\ell ^p(X,\mathcal {A})$ is a BSE algebra. Then, by [18, Corollary 5], $\ell ^p(X,\mathcal {A})$ admits a $\Delta$-weak bounded approximate identity. Proposition 2.6 implies that X is finite and $\mathcal {A}$ has a $\Delta$-weak bounded identity. Again, by [18, Corollary 5], we have:

$$\begin{aligned} {\mathcal {M}}(A)\subseteq C_{\mathrm{BSE}}(\Delta (\mathcal {A})). \end{aligned}$$

(3.1)

Now, we prove the reverse of inclusion (3.1). Suppose that $\sigma \in C_{\mathrm{BSE}}(\Delta (\mathcal {A}))$. Then, there exists a bounded net $(a_{\lambda })_{\lambda \in \Lambda }\subseteq \mathcal {A}$, such that for each $\varphi \in \Delta (\mathcal {A})$, $\lim _{\lambda }\widehat{a_{\lambda }}(\varphi )=\sigma (\varphi )$. For any $\lambda \in \Lambda$, define the constant function $f_{\lambda }: X\rightarrow \mathcal {A}$ by $f_{\lambda }(x)=a_{\lambda }$, $(x\in X)$. Since X is finite, then $f_{\lambda }\in \ell ^p(X,\mathcal {A})$, for all $\lambda \in \Lambda$. Moreover:

$$\begin{aligned} \lim _{\lambda }\Theta _{(x,\varphi )}(f_{\lambda })=\lim _{\lambda }\varphi (f_{\lambda }(x))=\lim _{\lambda }\varphi (a_{\lambda })=\sigma (\varphi )\quad (x\in X,\varphi \in \Delta (\mathcal {A})). \end{aligned}$$

Define the function $\sigma '$ as follows:

$$\begin{aligned} \sigma ':X\times \Delta (A)\longrightarrow & {} {\mathbb {C}}\\ \sigma '(x,\varphi )= & {} \sigma (\varphi ), \end{aligned}$$

for all $x\in X$ and $\varphi \in \Delta (\mathcal {A})$. Thus, we have:

$$\begin{aligned} \sigma '(x,\varphi )=\sigma (\varphi )=\lim _{\lambda }\widehat{a_{\lambda }} (\varphi )=\lim _{\lambda }\widehat{f_{\lambda }}\Theta _{(x,\varphi )}\quad (x\in X,\varphi \in \Delta (\mathcal {A})). \end{aligned}$$

This implies that $\sigma '\in C_{\mathrm{BSE}}(\Delta (\ell ^p(X,\mathcal {A})))$. Since $\ell ^p(X,\mathcal {A})$ is a BSE algebra, it follows that $\sigma '\in {\mathcal {M}}(\ell ^p(X,\mathcal {A}))$. Now, take $a\in {\mathcal {A}}$ and consider the constant function $f:X\rightarrow \mathcal {A}$ defined by $f(x)=a$$(x\in X)$. Then, $f\in \ell ^p(X,\mathcal {A})$, and so there exists $g\in \ell ^p(X,\mathcal {A})$, such that $\sigma '\widehat{f}=\widehat{g}$. Consequently for all $x\in X$ and $\varphi \in \Delta (\mathcal {A})$:

$$\begin{aligned} \sigma '(x,\varphi )\widehat{f}(x,\varphi )=\widehat{g}(x,\varphi ). \end{aligned}$$

It follows that:

$$\begin{aligned} \sigma (\varphi )\varphi (f(x))=\sigma (\varphi )\varphi (a)=\varphi (g(x)), \end{aligned}$$

(3.2)

and so:

$$\begin{aligned} \varphi (g(x))=\varphi (g(y))\quad (x,y\in X,\varphi \in \Delta (\mathcal {A})). \end{aligned}$$

Semisimplicity of $\mathcal {A}$ implies that $g(x)=g(y)$, for all $x,y\in X$. So that g is a constant function and thus $g(x)=b$$(x\in X)$, for some $b\in \mathcal {A}$. Now, the equality (3.2) implies that:

$$\begin{aligned} \sigma (\varphi )\widehat{a}(\varphi )=\sigma (\varphi )\varphi (a)=\varphi (g(x)) =\varphi (b)=\widehat{b}(\varphi ). \end{aligned}$$

Therefore, $\sigma \;\widehat{a}=\widehat{b}$, and so, $\sigma \in {\mathcal {M}}(\mathcal {A})$, as claimed.

Conversely, suppose that X is finite and $\mathcal {A}$ is a BSE algebra. By [18, Corollary 5], $\mathcal {A}$ has a $\Delta$-weak bounded approximate identity. By proposition 2.6, $\ell ^p(X,\mathcal {A})$ also has a $\Delta$-weak bounded approximate identity. Again, by [18, Corollary 5], we have:

$$\begin{aligned} {\mathcal {M}}(\ell ^p(X,\mathcal {A}))\subseteq C_{\mathrm{BSE}}(\Delta (\ell ^p(X,\mathcal {A}))). \end{aligned}$$

For the reverse of the above inclusion, suppose that $\sigma \in C_{\mathrm{BSE}}(\Delta (\ell ^p(X,\mathcal {A})))$. We show that $\sigma \in {\mathcal {M}}(\ell ^p(X,\mathcal {A}))$. To that end, take $h\in \ell ^p(X,\mathcal {A})$. We find $g\in \ell ^p(X,\mathcal {A})$, such that $\sigma \;\widehat{h}=\widehat{g}$. By [18, Theorem 4], there exists a bounded net $(f_{\lambda })_{\lambda \in \Lambda }$ in $\ell ^p(X,\mathcal {A})$, such that $\sup _{\lambda }\Vert f_{\lambda }\Vert _p \le \beta$, for some $\beta >0$, and:

$$\begin{aligned} \lim _{\lambda }\widehat{f_{\lambda }}(x,\varphi )=\sigma (x,\varphi )\quad (x\in X,\varphi \in \Delta (\mathcal {A})). \end{aligned}$$

Thus:

$$\begin{aligned} \lim _{\lambda }\varphi (f_{\lambda }(x))=\widehat{f_{\lambda }(x)}(\varphi )=\sigma (x,\varphi ). \end{aligned}$$

For each $x\in X$, we define the function $\sigma _x$ as follows:

$$\begin{aligned} \sigma _x:\Delta (\mathcal {A})\longrightarrow & {} {\mathbb {C}}\\ \sigma _x(\varphi )= & {} \sigma (x,\varphi )=\lim _{\lambda }\widehat{f_{\lambda }(x)}(\varphi ). \end{aligned}$$

This follows from [18, Theorem 4] that $\sigma _x\in C_{\mathrm{BSE}}(\Delta (\mathcal {A}))$. Since $\mathcal {A}$ is a BSE algebra, $\sigma _x\in {\mathcal {M}}(\mathcal {A})$. Thus, for each $x\in X$, there exists $a_x\in \mathcal {A}$, such that $\sigma _x\widehat{h(x)}=\widehat{a_x}$. Now, define the function g on X as follows:

$$\begin{aligned} g:X\longrightarrow & {} A\\ x\mapsto & {} a_x. \end{aligned}$$

For all $x\in X$ and $\varphi \in \Delta (\mathcal {A})$, we have:

$$\begin{aligned} \sigma (x,\varphi )\widehat{h}(x,\varphi )= & {} \sigma _x(\varphi )\varphi (h(x))\\= & {} \sigma _x(\varphi )\widehat{h(x)}(\varphi )\\= & {} \widehat{a_x}(\varphi )\\= & {} \widehat{g(x)}(\varphi )\\= & {} \widehat{g}(x,\varphi ). \end{aligned}$$

Thus, $\sigma \;\widehat{h}=\widehat{g}$. So that $\ell ^p(X,\mathcal {A})$ is a BSE algebra. $\square$

Remark 3.2

In [19, Theorem 5], it is shown that for an arbitrary non-empty set X:

$$\begin{aligned} \widehat{\ell ^1(X)}=C_{\mathrm{BSE}}(X)\subset \mathcal {M}(\ell ^1(X))=C_b(X). \end{aligned}$$

Moreover, $C_{\mathrm{BSE}}(X)$ and $C_b(X)$ coincide if and only if X is finite. In fact, $\ell ^1(X)$ is a BSE algebra of type I if and only if X is finite. Note that, by some similar arguments as in the proof of [19, Theorem 5], we can deduce the same results for $\ell ^p(X)$$(1<p<\infty )$, as well. Moreover, it is easily verified that if X is finite and $\mathcal {A}$ is a unital BSE algebra, then:

$$\begin{aligned} \widehat{\ell ^p(X,\mathcal {A})} =\mathcal {M}(\ell ^p(X,\mathcal {A}))=C_{\mathrm{BSE}}(X\times \Delta (\mathcal {A}))\subseteq C_{b}(X\times \Delta (\mathcal {A})). \end{aligned}$$

However, these equalities are not valid in general. For instance, take X to be a finite set and $\mathcal {A}$ to be a non-unital BSE algebra. Then:

$$\begin{aligned} \widehat{\ell ^p(X,\mathcal {A})}\subsetneqq \mathcal {M}(\ell ^p(X,\mathcal {A}))=C_{\mathrm{BSE}}(X\times \Delta (\mathcal {A}))\subseteq C_{b}(X\times \Delta (\mathcal {A})). \end{aligned}$$

It is worth to note that even in the case that X is finite, $C_{\mathrm{BSE}}(X\times \Delta (\mathcal {A}))$ may not be equal to $C_{b}(X\times \Delta (\mathcal {A}))$, as the following example shows.

Example 3.3

Let X be a finite set with $card(X)=n>1$ and $\mathcal {A}=\ell ^\infty (X)$. Then, $\Delta (\ell ^\infty (X))=X$ and since $\ell ^\infty (X)$ is a unital BSE algebra, it follows that $\ell ^p(X,\ell ^\infty (X))$ is also a unital BSE algebra. Consequently:

$$\begin{aligned} \widehat{\ell ^p(X,\ell ^\infty (X))}=C_{\mathrm{BSE}}(X\times X). \end{aligned}$$

Suppose on the contrary that:

$$\begin{aligned} C_{\mathrm{BSE}}(X\times X)=C_b(X\times X). \end{aligned}$$

(3.3)

It follows that $\ell ^p(X,\ell ^\infty (X))$ is a BSE algebra of type I, and so, by [18, Theorem 3], $\ell ^p(X,\ell ^\infty (X))$ is a $C^*$-algebra. Consider the function $f\in \ell ^p(X,\ell ^\infty (X))$, defined by $f(x)=\mathbf {1}$$(x\in X)$, where $\mathbf {1}\in \ell ^\infty (X)$ is the constant function $\mathbf {1}(x)=1$$(x\in X)$. Then:

$$\begin{aligned} \Vert f\;\bar{f}\Vert _p=\Vert f^2\Vert _p=n^{1/p}\ne \Vert f\Vert _p^2=n^{2/p}. \end{aligned}$$

This contradiction indicates that the equality (3.3) is not satisfied and:

$$\begin{aligned} C_{\mathrm{BSE}}(X\times X)\subsetneqq C_b(X\times X). \end{aligned}$$

In other words, there are continuous bounded functions on $\Delta (\ell ^p(X,\ell ^\infty (X)))$ which are not BSE.

For a natural number n, a function $\sigma \in C_b(\Delta (\mathcal {A}))$ is called a n-BSE function, if there exists positive real numbers $\beta$ (depending only on n), such that for any choice of $\varphi _1,\ldots ,\varphi _n$ in $\Delta ({\mathcal {A}})$ and complex numbers $c_1,\ldots ,c_n$, the inequality:

$$\begin{aligned} \left| \displaystyle \sum _{j=1}^nc_j\sigma (\varphi _j)\right| \le C \;\left\| \displaystyle \sum _{j=1}^nc_j\varphi _j\right\| _{{\mathcal {A}}^*} \end{aligned}$$

holds. The set of all n-BSE functions on $\Delta (\mathcal {A})$ will be denoted by $C_{\mathrm{BSE}(n)}(\Delta (\mathcal {A}))$. We denote by $\Vert \sigma \Vert _{\mathrm{BSE}(n)}$, the infimum of such $\beta$. By [19, Lemma 1]:

$$\begin{aligned} C_{\mathrm{BSE}(n)}(\Delta (\mathcal {A}))=C_b(\Delta (\mathcal {A})) \end{aligned}$$

if and only if there exists a positive real numbers $\beta _n$ (depending only on n), such that for any choice of $\varphi _1,\ldots ,\varphi _n$ in $\Delta ({\mathcal {A}})$ and complex numbers $c_1,\ldots ,c_n$ in the closed unit disk ${\mathbb {C}}_1$, there exists $x\in \mathcal {A}$, such that $\Vert x\Vert \le \beta _n$ and $\hat{x}(\varphi _i)=c_i$.

Let:

$$\begin{aligned} C_{\mathrm{BSE}(\infty )}=\bigcap _{n\in \mathbb {N}}C_{\mathrm{BSE}(n)}(\Delta (\mathcal {A})). \end{aligned}$$

Evidently, $\Vert \sigma \Vert _{\mathrm{BSE}}=\sup _{n\in \mathbb {N}}\Vert \sigma \Vert _{\mathrm{BSE}(n)}$ and:

$$\begin{aligned} C_{\mathrm{BSE}}(\Delta (\mathcal {A}))=\{\sigma \in C_{\mathrm{BSE}(\infty )}:\;\Vert \sigma \Vert _{\mathrm{BSE}}<\infty \}. \end{aligned}$$

Moreover, we have the following inclusions:

$$\begin{aligned} \widehat{\mathcal {A}}\subseteq & {} C_{\mathrm{BSE}}(\Delta (\mathcal {A}))\subseteq C_{\mathrm{BSE}(\infty )}(\Delta (\mathcal {A}))\subseteq \cdots \\\subseteq & {} C_{\mathrm{BSE}(2)}(\Delta (\mathcal {A}))\subseteq C_{\mathrm{BSE}(1)}(\Delta (\mathcal {A}))\\= & {} C_b(\Delta (\mathcal {A})). \end{aligned}$$

See [19], for more information.

In Example 3.3, we observe that a continuous bounded function on the Gelfand space $\ell ^p(X,{\mathcal {A}})$ needs not be a BSE function. However, in the sequel, we prove that for any unital commutative Banach algebra $\mathcal {A}$ and natural number n:

$$\begin{aligned} C_{\mathrm{BSE}(n)}(\Delta (\ell ^p(X,\mathcal {A})))=C_b(\Delta (\ell ^p(X,\mathcal {A})))=C_b(X\times \Delta (\mathcal {A})). \end{aligned}$$

In fact, all continuous bounded functions on $\Delta (\ell ^p(X,\mathcal {A}))$ are n-BSE functions.

Proposition 3.4

Let X be a set and $\mathcal {A}$ be a commutative semisimple and unital Banach algebra with unit e. Then, $C_{\mathrm{BSE}(n)}(\ell ^p(X,\mathcal {A}))=C_b(X\times \Delta (\mathcal {A}))$, for each $n\in \mathbb {N}$.

Proof

To prove, we use [19, Lemma 1]. Take $c_1\ldots ,c_n\in \Delta$, $x_1,\ldots x_n\in X$, and $\varphi _1,\ldots ,\varphi _n\in \Delta (\mathcal {A})$. Define the function f on X as:

$$\begin{aligned} f(x)=\left\{ \begin{array}{ll} c_ie &{}\quad x\in \{x_1,\ldots ,x_n\}\\ 0 &{}\quad otherwise. \end{array}\right. \end{aligned}$$

Then, for each $i=1,\ldots ,n$ we have:

$$\begin{aligned} \hat{f}(x_i,\varphi _i)=\varphi _i(f(x_i))=\varphi _i(c_ie)=c_i. \end{aligned}$$

Moreover:

$$\begin{aligned} \Vert f\Vert _p=\left( \sum _{i=1}^n\Vert f(x_i)\Vert ^p\right) ^{1/p}=\left( \sum _{i=1}^n|c_i|^p\right) ^{1/p}=n^{1/p}. \end{aligned}$$

Thus, it is sufficient to take $\beta _n=n^{1/p}$, and so, the proof is completed. $\square$

It is known that in any commutative Banach algebra $\mathcal {A}$, $\Vert \hat{x}\Vert _{\mathrm{BSE}}\le \Vert x\Vert$, for all $x\in \mathcal {A}$. In [20], the authors were interested in a class of commutative Banach algebras which satisfy the condition $\Vert \hat{x}\Vert _{\mathrm{BSE}}=\Vert x\Vert$, for each $x\in \mathcal {A}$. These algebras are called BSE norm algebras. All function algebras on a locally compact Hausdorff space, endowed with the supremum norm, and also the algebra $\ell ^1(X)$ belong to such a class. In the sequel, we show that under some circumstances, $\ell ^1(X,\mathcal {A})$ also belongs to this class. To that end, we require the following elementary lemma.

Lemma 3.5

Let X be a set and $\mathcal {A}$ be a commutative semisimple Banach algebra. Suppose that $c_1,\ldots c_n$ and $(x_1,\varphi _1),\ldots ,(x_n,\varphi _n)$ are disjoint elements of $\mathbb {C}$ and $X\times \Delta (\mathcal {A})$, respectively, such that $x_{k_1}=\cdots =x_{k_m}$, where $1\le k_1,\ldots ,k_m\le n$. Then:

$$\begin{aligned} \left\| \sum _{i=1}^{m}c_{k_i}\varphi _{k_i}\right\| _{{\mathcal {A}}^*} \le \left\| \sum _{i=1}^{n}c_i(x_i,\varphi _i)\right\| _{{\ell ^1(X,\mathcal {A})}^*}. \end{aligned}$$

Proof

Let $x_{k_1}=\cdots =x_{k_m}=x$. Then, we have:

$$\begin{aligned} \left\| \sum _{i=1}^{m}c_{k_i}\varphi _{k_i}\right\| _{{\mathcal {A}}^*}= & {} \sup \left\{ \left| \sum _{i=1}^{m}c_{k_i}\varphi _{k_i}(a)\right| :\;\Vert a\Vert \le 1\right\} \\= & {} \sup \left\{ \left| \sum _{i=1}^{m}c_{k_i}\varphi _{k_i}(\delta _{a}^{x}(x_{k_i}))\right| :\;\Vert a\Vert \le 1\right\} \\= & {} \sup \left\{ \left| \sum _{i=1}^{n}c_{i}\varphi _{i}(\delta _{a}^{x}(x_{i}))\right| :\;\Vert \delta _{a}^{x}\Vert _1\le 1\right\} \\= & {} \sup \left\{ \left| \sum _{i=1}^{n}c_{i}(x_i,\varphi _{i})(\delta _{a}^{x})\right| :\;\Vert \delta _{a}^{x}\Vert _1\le 1\right\} \\\le & {} \sup \left\{ \left| \sum _{i=1}^{n}c_{i}(x_i,\varphi _{i})(f)\right| :\;\Vert f\Vert _1\le 1\right\} \\= & {} \left\| \sum _{i=1}^{n}c_i(x_i,\varphi _i)\right\| _{{\ell ^1(X,\mathcal {A})}^*}. \end{aligned}$$

Thus, the proof is completed. $\square$

Recall from [16] that a Banach algebra $\mathcal {A}$ is called a supremum norm algebra if $\Vert \hat{a}\Vert _\infty =\Vert a\Vert$, for each $a\in \mathcal {A}$. For example, all $C^*$-algebras are supremum norm algebra.

Theorem 3.6

Let X be a set and $\mathcal {A}$ be a commutative semisimple Banach algebra. If $\ell ^1(X,\mathcal {A})$ is a BSE norm algebra, then $\mathcal {A}$ is so. The converse is true if $\mathcal {A}$ is a supremum norm algebra.

Proof

Suppose that $\ell ^1(X,\mathcal {A})$ is a BSE norm algebra. Thus, for each $f\in \ell ^1(X,\mathcal {A})$, we have:

$$\begin{aligned} \Vert f\Vert _1=\Vert \hat{f}\Vert _{\mathrm{BSE}}. \end{aligned}$$

It follows that:

$$\begin{aligned} \Vert a\Vert =\Vert \delta _a^x\Vert _1=\Vert \widehat{\delta _a^x}\Vert _{\mathrm{BSE}}\quad (x\in X,a\in \mathcal {A}). \end{aligned}$$

(3.4)

Let $a\in \mathcal {A}$ and take $x\in X$ to be fixed. Then, for any finitely many complex numbers $c_1,\ldots ,c_n$ and the same number of elements $(x_1,\varphi _1),\ldots ,(x_n,\varphi _n)$ of $X\times \Delta (\mathcal {A})$ with $x_{k_1}=\cdots =x_{k_m}=x$, we have:

$$\begin{aligned} \left| \sum _{i=1}^{n}c_i\widehat{\delta _a^x}(x_i,\varphi _i)\right|= & {} \left| \sum _{i=1}^{n}c_i\varphi _i(\delta _{a}^{x}(x_i))\right| \\= & {} \left| \sum _{i=1}^{m}c_{k_i}\varphi _{k_i}(\delta _{a}^{x}(x_{k_i}))\right| \\= & {} \left| \sum _{i=1}^{m}c_{k_i}\varphi _{k_i}(a)\right| \\= & {} \left| \sum _{i=1}^{m}c_{k_i}\hat{a}(\varphi _{k_i})\right| \\\le & {} \Vert \hat{a}\Vert _{\mathrm{BSE}}\left\| \sum _{i=1}^{m}c_{k_i}\varphi _{k_i}\right\| _{{\mathcal {A}}^*}\\\le & {} \Vert \hat{a}\Vert _{\mathrm{BSE}}\left\| \sum _{i=1}^{n}c_i(x_i,\varphi _i)\right\| _{{\ell ^1(X,\mathcal {A})}^*}, \end{aligned}$$

where the last inequality is obtained from Lemma 3.5. Consequently:

$$\begin{aligned} \left| \sum _{i=1}^{n}c_i\widehat{\delta _a^x}(x_i,\varphi _i)\right| \le \Vert \hat{a}\Vert _{\mathrm{BSE}}\left\| \sum _{i=1}^{n}c_i(x_i,\varphi _i)\right\| _{{\ell ^1(X,\mathcal {A})}^*}. \end{aligned}$$

(3.5)

Note that if all $x_1,\ldots ,x_n$ are different from x, then the inequality 3.5 is obviously satisfied. Thus, we have:

$$\begin{aligned} \Vert \widehat{\delta _a^x}\Vert _{\mathrm{BSE}}\le \Vert \hat{a}\Vert _{\mathrm{BSE}}. \end{aligned}$$

(3.6)

Now, the equality (3.4) and inequality (3.6) imply that:

$$\begin{aligned} \Vert a\Vert \le \Vert \hat{a}\Vert _{\mathrm{BSE}}\quad (a\in \mathcal {A}). \end{aligned}$$

Therefore, $\mathcal {A}$ is a BSE norm algebra.

Conversely, suppose that $\mathcal {A}$ is a supremum norm algebra. We show that $\ell ^1(X,\mathcal {A})$ is a BSE norm algebra. Take $f\in \ell ^1(X,\mathcal {A})$ to be nonzero. It is enough to show that $\Vert f\Vert _1\le \Vert \hat{f}\Vert _{\mathrm{BSE}}$. For $\varepsilon >0$, there exists $N\in \mathbb {N}$, such that:

$$\begin{aligned} \Vert f\Vert _1-\varepsilon <\sum _{k=1}^N\Vert f(x_k)\Vert . \end{aligned}$$

By the hypothesis:

$$\begin{aligned} \Vert f(x_k)\Vert =\Vert \widehat{f(x_k)}\Vert _\infty =\sup _{\varphi \in \Delta (\mathcal {A})}|\varphi (f(x_k))|, \end{aligned}$$

for each $k=1,\ldots ,N$. Since $\mathcal {A}$ is unital, $\Delta (\mathcal {A})$ is compact and so all $\widehat{f(x_k)}$$(k=1,\ldots ,N)$ take their supremum on $\Delta (\mathcal {A})$. It follows that there exists $\varphi _k\in \Delta (\mathcal {A})$, such that:

$$\begin{aligned} \Vert \widehat{f(x_k)}\Vert _\infty =|\varphi _k(f(x_k))|\quad (k=1,\ldots , N). \end{aligned}$$

Now let:

$$\begin{aligned} C_k=\dfrac{\Vert f(x_k)\Vert }{\varphi _k(f(x_k))}\quad (k=1,\ldots , N). \end{aligned}$$

Then, $|C_k|=1$ and:

$$\begin{aligned} \left| \sum _{k=1}^NC_k\hat{f}(x_k,\varphi _k)\right|= & {} \left| \sum _{k=1}^NC_k\varphi _k(f(x_k))\right| \\= & {} \Vert f(x_1)\Vert +\cdots +\Vert f(x_N)\Vert \\= & {} \sum _{k=1}^N\Vert f(x_k)\Vert \\> & {} \Vert f\Vert _1-\varepsilon . \end{aligned}$$

Thus:

$$\begin{aligned} \Vert f\Vert _1-\varepsilon <\left| \sum _{k=1}^NC_k\hat{f}(x_k,\varphi _k)\right| . \end{aligned}$$

(3.7)

Moreover:

$$\begin{aligned} \left\| \sum _{k=1}^NC_k(x_k,\varphi _k)\right\|= & {} \sup _{\Vert h\Vert _1\le 1}\left| \sum _{k=1}^NC_k(x_k,\varphi _k)(h)\right| \\= & {} \sup _{\Vert h\Vert _1\le 1}\left| \sum _{k=1}^NC_k\varphi _k(h(x_k))\right| \\\le & {} \sup _{\Vert h\Vert _1\le 1}\sum _{k=1}^N|C_k|\Vert \varphi _k\Vert \Vert h(x_k)\Vert \\= & {} \sup _{\Vert h\Vert _1\le 1}\sum _{k=1}^N\Vert h(x_k)\Vert \\\le & {} 1. \end{aligned}$$

The last inequality together with (3.7) yields that:

$$\begin{aligned} \Vert f\Vert _1-\varepsilon< & {} \left| \sum _{k=1}^NC_k\hat{f}(x_k,\varphi _k)\right| \\\le & {} \Vert \hat{f}\Vert _{\mathrm{BSE}}\left\| \sum _{k=1}^NC_k(x_k,\varphi _k)\right\| \\\le & {} \Vert \hat{f}\Vert _{\mathrm{BSE}}. \end{aligned}$$

Since $\varepsilon$ is arbitrary, it follows that $\Vert f\Vert _1\le \Vert \hat{f}\Vert _{\mathrm{BSE}}$, as claimed. $\square$

4 The BSE Property of $\ell ^p(X)$

Let X be a nonempty set. By [19, Theorem 5], $\ell ^1(X)$ is a BSE algebra if and only if X is finite. Note that this result remains valid for $\ell ^p(X)$, where $1\le p<\infty$. In this section, we provide another proof for this result, which is interesting in its own right. We first recall the definition of abstract Segal algebras; see [3] for more information.

Let $(\mathcal {A},\Vert .\Vert _{\mathcal {A}})$ be a commutative Banach algebra. A commutative Banach algebra $(\mathcal {B},\Vert .\Vert _{\mathcal {B}})$ is an abstract Segal algebra with respect to $\mathcal {A}$ if:

• (i) $\mathcal {B}$ is a dense ideal in $\mathcal {A}$.

• (ii) There exists $M>0$, such that $\Vert b\Vert _{\mathcal {A}}\le M\Vert b\Vert _{\mathcal {B}}$, for all $b\in B$.

• (iii) There exists $C>0$, such that $\Vert ab\Vert _{\mathcal {B}}\le C\Vert a\Vert _{\mathcal {A}}\Vert b\Vert _{\mathcal {B}}$, for all $a, b\in B$.

Moreover, $\mathcal {B}$ is called essential if:

$$\begin{aligned} \mathcal {B}=\{ab:\;a\in \mathcal {A},b\in \mathcal {B}\}. \end{aligned}$$

Our new proof for [19, Theorem 5] is based on [10, Theorem 3.1], which is described below:

“If $(\mathcal {B},\Vert .\Vert _{\mathcal {B}})$ is an essential abstract Segal algebra with respect to the BSE algebra $\mathcal {A}$, then $\mathcal {B}$ is a BSE algebra if and only if it has a $\Delta$-weak bounded approximate identity.”

For this purpose, we remind the reader of some known spaces. Recall that $c_0(X)$ is the space, consisting of all functions vanishing at infinity. Moreover, $c_0(X)$ is a Banach algebra under pointwise product and supremum norm, defined as:

$$\begin{aligned} \Vert f\Vert _\infty =\{|f(x)|:\;\;\;x\in X\}\quad (f\in c_0(X)). \end{aligned}$$

The subspace $c_{00}(X)$ of $c_0(X)$, consisting of all finite support functions on X, is dense in $c_0(X)$. Moreover:

$$\begin{aligned} c_{00}(X)\subseteq \ell ^p(X)\subseteq c_0(X) \end{aligned}$$

and $\Vert f\Vert _\infty \le \Vert f\Vert _p$, for all $f\in \ell ^p(X)$.

Lemma 4.1

Let X be a set and $1\le p<\infty$. Then, $\ell ^p(X)$ is an essential abstract Segal algebra with respect to $c_0(X)$.

Proof

Since $\ell ^p(X)$ contains $c_{00}(X)$ and $c_{00}(X)$ is dense in $c_0(X)$, it follows that $\ell ^p(X)$ is also dense in $c_0(X)$. Moreover, $\ell ^p(X)$ is an ideal in $c_0(X)$ and for each $f\in \ell ^p(X)$ and $g\in c_0(X)$, we have:

$$\begin{aligned} \Vert f\;g\Vert _p=\left( \sum _{x\in X}|f(x)g(x)|^p\right) ^{1/p}\le \Vert f\Vert _p\Vert g\Vert _\infty <\infty . \end{aligned}$$

Consequently, $\ell ^p(X)$ is an abstract Segal algebra in $c_0(X)$. In the sequel, we show that $\ell ^p(X)$ is essential. To that end, note that the collection $\mathcal {F}$, consisting of all finite subsets of X, is a directed set by the upward inclusion; that is:

$$\begin{aligned} F_1\le F_2\;\;\text {if and only if}\;\; F_1\subseteq F_2. \end{aligned}$$

It is easily verified that the net $(\chi _F)_{F\in \mathcal {F}}$ is a bounded approximate identity for $c_0(X)$, where $\chi _F$ is the characteristic function on X at F. To establish the essentiality of $\ell ^p(X)$, by applying Cohen factorization theorem, it is sufficient to show that $(\chi _F)_{F\in \mathcal {F}}$ is an approximate identity for $\ell ^p(X)$; that is:

$$\begin{aligned} \Vert f\chi _F-f\Vert _p\rightarrow _F 0\quad (f\in \ell ^p(X)). \end{aligned}$$

(4.1)

Suppose that $f\in \ell ^p(X)$ and take $\varepsilon >0$ to be arbitrary. There exists $N\in \mathbb {N}$, such that:

$$\begin{aligned} \sum _{i=N+1}^\infty |f(x_i)|^p<\varepsilon ^{p}. \end{aligned}$$

Let $F_0=\{x_1,\ldots ,x_n\}$. Since $f\in c_0(X)$, there exists finite subset $F_1$ of X, such that $|f(x)|<\varepsilon$, for all $x\not \in F_1$. Set $F_2:=F_0\cup F_1$. Thus, for each $F_2\le F$, we have:

$$\begin{aligned} \Vert f\chi _F-f\Vert _p=\left( \sum _{x\not \in F}|f(x)|^p\right) ^{1/p}\le \left( \sum _{i=N+1}^\infty |f(x)|^p\right) ^{1/p}<\varepsilon , \end{aligned}$$

and so, (4.1) is satisfied. This completes the proof. $\square$

Note that $c_0(X)$ is a $C^*$-algebra, and so, it is a BSE algebra by [18, Theorem 3]. Now, Theorem 2.6 and Lemma 4.1 together with [10, Theorem 3.1] yield the following result.

Theorem 4.2

Let X be a set and $1\le p<\infty$. Then, $\ell ^p(X)$ is a BSE algebra if and only if X is finite.

Proof

By Lemma 4.1, $\ell ^p(X)$ is an essential abstract Segal algebra in $c_0(X)$. Since $c_0(X)$ is a BSE algebra, [10, Theorem 3.1] implies that $\ell ^p(X)$ is a BSE algebra. It follows that $\ell ^p(X)$ has a $\Delta$-weak bounded approximate identity, and so, X is finite by Theorem 2.6. The converse is obvious. $\square$