1 Introduction

Let A be a commutative Banach algebra. Denote by \(\varDelta (A)\) and \({\mathcal {M}}(A)\) the Gelfand spectrum and the multiplier algebra of A, respectively. A bounded continuous function \(\sigma \) on \(\varDelta (A)\) 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 \(\varDelta (A)\) and complex numbers \(c_1,\ldots ,c_n\), the inequality

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

holds. The BSE-norm of \(\sigma \) (\(\Vert \sigma \Vert _{\textit{BSE}}\)) is defined to be the infimum of all such C. The set of all BSE-functions is denoted by \(C_{\textit{BSE}}(\varDelta (A))\). Takahasi and Hatori [16] showed that under the norm \(\Vert . \Vert _{\textit{BSE}}\), \(C_{\textit{BSE}}(\varDelta (A))\) is a commutative semisimple Banach algebra.

A bounded linear operator on A is called a multiplier if it satisfies \(xT(y)=T(xy)\) for all \(x,y\in A\). The set \({\mathcal {M}}(A)\) of all multipliers of A is a unital commutative Banach algebra, called the multiplier algebra of A.

For each \(T\in {\mathcal {M}}(A)\) there exists a unique bounded continuous function \(\widehat{T}\) on \(\varDelta (A)\) such that \(\widehat{T(a)}(\varphi )=\widehat{T}(\varphi )\hat{a}(\varphi )\) for all \(a\in A\) and \(\varphi \in \varDelta (A)\). See [9] for a proof.

Define

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

A bounded net \((e_{\alpha })_{\alpha }\) in a Banach algebra A is called a \(\varDelta \)-weak bounded approximate identity for A if \(\varphi (e_{\alpha })\rightarrow 1\) (equivalently, \(\varphi (e_{\alpha }a)\rightarrow \varphi (a)\) for every \(a\in A\)) for all \(\varphi \in \varDelta (A)\). As is shown in [16], A has a \(\varDelta \)-weak bounded approximate identity if and only if \(\widehat{{\mathcal {M}}(A)}\subseteq C_{\textit{BSE}}(\varDelta (A))\).

A commutative Banach algebra A is called without order if \(aA=\{0\}\) implies \(a=0\) (\(a\in A\)).

A commutative and without order Banach algebra A is called a BSE-algebra (or has BSE-property) if it satisfies the condition

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

The abbreviation BSE stands for Bochner–Schoenberg–Eberlein and refers to a famous theorem, proved by Bochner and Schoenberg [2, 15] for the additive group of real numbers and in general by Eberlein [5] for a locally compact abelian group G, saying that, in the above terminology, the group algebra \(L^1(G)\) is a BSE-algebra (See [12] for a proof).

It is worth to note that the semigroup algebra \(l^1({\mathbb Z}^+)\) (where \({\mathbb Z}^+\) is the additive semigroup of nonnegative integers) is a BSE algebra [17], but for \(k\ge 1\), \(l^1({\mathbb N}_k)\) (\({\mathbb N}_k=\{k,k+1,k+2,\ldots \}\)) is not a BSE algebra.

In [7], we established affirmatively a question raised by Takahasi and Hatori [16] whether \(L^1({\mathbb {R}}^+)\) is a BSE-algebra.

The aim of the present paper is to show that for any totally ordered compact semigroup S with multiplication \(xy=\max \{x,y\}\) and a regular bounded continuous measure \(\mu \) on S, \(L^p(S,\mu )\) (\(1\le p<\infty \)) is not a BSE algebra. However, for any compact abelian group G and \(1\le p<\infty \), the Banach algebra \(L^p(G)\) is BSE [18].

As an application, we will show that the Banach \(L^1(S,\mu )\) is not an ideal in its second dual. However, for a locally compact group G, the Banach algebra \(L^1(G)\) is an ideal in its second dual if and only if G is compact.

Finally, we prove that \(C_{{\textit{BSE}}(n)}((0,1])=C_b((0,1])\) for any natural number n.

2 BSE Property of Totally Ordered Semigroup Algebras

The Banach algebra \(L^p(S,\mu )\;(1\le p<\infty )\), whenever S is a totally ordered compact space with a regular bounded continuous measure \(\mu \) on S, was first introduced by Sapounakis [13, 14] and then extensively studied more by Baker, Pym and Vasudeva [1].

Recall that a totally ordered locally compact space S is a totally ordered space S which is locally compact in its order topology. This has a natural continuous multiplication \(xy=\max \{x,y\}\). For convenience, we adjoin a minimal element 0 to S if it has not already got one. The convolution product \(v_1*v_2\) of two bounded regular measures \(v_1,v_2\) on S is defined (as a linear functional on the space \(C_0(S)\) of continuous functions vanishing at infinity on S) by the usual formula

$$\begin{aligned} \int fdv_1*v_2=\int \int f(xy)dv_1(x)dv_2(y)\;\;\;\;\;(f\in C_0(S)). \end{aligned}$$

By dividing the range of the inner integral on the right into the sets where \(x<y\) and \(x\ge y\) we have

$$\begin{aligned} dv_1*v_2(x)=v_1[0,x[ \;.\; dv_2(x)+v_2[0,x[\;.\;dv_1(x). \end{aligned}$$

In particular, if both \(v_1\) and \(v_2\) are absolutely continuous with respect to some positive measure \(\mu \), say \(dv_1=fd\mu \) and \(dv_2=gd\mu \), then so is \(v_1*v_2\); and if we put \(dv_1*v_2=f*g.d\mu \), then

$$\begin{aligned} f*g(x)=g(x)\int _{[0,x[}fd\mu +f(x)\int _{[0,x]}gd\mu .\;\;\;\;\;\;({\textit{III}}) \end{aligned}$$

By defining the convolution of two measurable functions f and g with respect to \(\mu \) as in (III), one has the following result.

Proposition 1

For \(p=1\), \(L^p(S,\mu )\) is a Banach algebra. For \(1<p\le \infty \), \(L^p(S,\mu )\) is a Banach algebra if and only if \(\mu \) is bounded. Moreover, for \(1\le p\le \infty \) the algebra \(L^p(S,\mu )\) is commutative and semisimple. It has an approximate identity if \(1\le p<\infty \), which is bounded if and only if \(p=1\).

Proof

See [1]. \(\square \)

Remark 1

Here we suppose that \(\mu \) is bounded and it will do no harm to suppose that its total mass is 1. By taking the order completion of S, we may also assume that S is compact, and we lose nothing by taking S to be the support of \(\mu \). In [1] it is shown that all \(L^p-\)algebras associated with continuous measures of mass 1 with support S are algebra isomorphic to the Banach algebra \(L^p([0,1],\lambda )\), where \(\lambda \) is the Lebesgue measure on the interval [0, 1]. Thus, we need only consider \(L^p([0,1],\lambda )\) and this we shall do (See Theorem 1.1 and page 48 of [1]).Then the Gelfand spectrum \(\varDelta (L^p(S,\mu ))\) of \(L^p(S,\mu )\) is equal to the set

$$\begin{aligned} \{\chi _{[0,x]}:x\in (0,1]\} \end{aligned}$$

which may be identified with the half-open interval (0, 1], and the Gelfand transform \(\widehat{f}\) of \(f\in L^p(S,\mu )\) is given by

$$\begin{aligned} \widehat{f}(t)=\int _0^xfd\mu \;\;\;\;\;(t\in [0,1]), \end{aligned}$$

where 0 is the minimal element of S and \(\sigma (x)=t\), where \(\sigma \) is a continuous increasing surjection from S to [0, 1] given in the statement of Theorem 1.1 of [1].

Before turning to the next result we need to recall the following definition from [11].

Definition 1

A real-valued function f defined on [a, b] is said to be absolutely continuous on [a, b] if, given \(\epsilon >0\), there exists \(\delta _{\epsilon }>0\) such that

$$\begin{aligned} \sum \limits _{i=1}^n\vert f(y_i)-f(x_i)\vert <\epsilon \end{aligned}$$

for every finite collection \(\{(x_i,y_i)\}\) of non overlapping intervals with

$$\begin{aligned} \sum \limits _{i=1}^n\vert y_i-x_i\vert <\delta _{\epsilon }. \end{aligned}$$

A complex-valued function \(f=u+iv\) is said to be absolutely continuous if u and v are absolutely continuous.

Theorem 1

(Fundamental Theorem of Lebesgue Integral Calculus). The following conditions on a real-valued function f on a compact interval [a, b] are equivalent:

  1. 1.

    f is absolutely continuous on [a, b].

  2. 2.

    There exists a Lebesgue integrable function g on [a, b] such that

    $$\begin{aligned} f(x)=f(a)+\int _a^xg(t)dt \end{aligned}$$

    for all \(x \in [a,b]\).

Proof

See Page 106 of [11]. \(\square \)

2.1 The Cantor Function

The function \(\varphi :[0,1]\rightarrow [0,1]\) defined by

$$\begin{aligned} \varphi (x)=\sum \limits _{k=1}^{\infty }\frac{t_{xk}/2}{2^k}, \end{aligned}$$

whenever the digit 1 does not appear in the ternary expansion of x, and

$$\begin{aligned} \varphi (x)=\sum \limits _{k=1}^{j_x-1}\frac{t_{xk}/2}{2^k}+\frac{1}{2^{j_x}}, \end{aligned}$$

whenever the digit 1 does appear in the ternary expansion of x, and \(j_x=\min \{k:t_{xk}=1\}\), is called the Cantor function.

Proposition 2

For the Cantor function \(\varphi \) the following statements are valid.

  1. 1.

    \(\varphi \) is continuous and of bounded variation.

  2. 2.

    \(\varphi \) is not absolutely continuous.

  3. 3.

    \(\varphi \) is differentiable almost every where and \(\varphi '=0\) a. e. on [0, 1].

  4. 4.

    \(\int _0^1\varphi (t)dt=\frac{1}{2}.\)

Proof

See [3] and [6]. \(\square \)

Lemma 1

Let \(\varphi \) be the Cantor function. Then the function g defined by \(g(x)=x\varphi (x)\) is not absolutely continuous on [0, 1].

Proof

By Proposition 2, the function g is differentiable almost every where on [0, 1]. Therefore

$$\begin{aligned} g'(x)=x\varphi '(x)+\varphi (x)=\varphi (x)\;\;\;a.e. \end{aligned}$$

holds. Assume towards a contradiction that g is absolutely continuous on [0,1]. So by Theorem 1, we get

$$\begin{aligned} 1=g(1)-g(0)=\int _0^1g'(t)dt=\int _0^1\varphi (t)dt=\frac{1}{2}. \end{aligned}$$

This contradiction completes the proof. \(\square \)

Let \((X,\varSigma ,\mu )\) be a positive measure space. The space \(ba(X,\mu )\) is the Banach space consisting of those bounded and finitely additive signed measures on \(\varSigma \) which vanish on sets of \(\mu \)-measure zero. The norm of an element in \(ba(X,\mu )\) is its total variation. Before stating our next result, we need to recall the following theorem from page 296 of [4].

Theorem 2

Let \((X,\varSigma ,\mu )\) be a \(\sigma \)-finite measure space . There is an isometric isomorphism between \({L^{\infty }}(X,\mu )^*\) and \(ba(X,\mu )\)determined by the identity

$$\begin{aligned} F(f)=\int _Xf(x)d\mu (x) \;\;\;\;\;\;\;\; (f\in L^{\infty }(X,\mu )). \end{aligned}$$

The following theorem is indeed the main result of this paper.

Theorem 3

The Banach algebra \(L^p(S,\mu ) (1\le p<\infty )\) is not a BSE algebra.

Proof

As in Remark 1, we may assume that \(S=([0,1],\max )\) and \(\mu \) is the Lebesgue measure on S.

Case I. \(p=1\). Suppose that \(\sigma \) is a continuous function of bounded variation on [0, 1] and \(\sigma (0)=0\). By Theorem 3.1 of [10], there exists \(\nu \in ba(S,\mu )\) such that \(\sigma (x)=\nu ([0,x])\) for all \(x\in [0,1]\). By Theorem 2, the mapping F given by

$$\begin{aligned} F(f)=\int _Xf(x)d\nu (x) \;\;\;\;\;\;\;\; \left( f\in L^{\infty }(X,\mu )\right) . \end{aligned}$$

is in \(L^{\infty }(S,\mu )^*\). In particular,

$$\begin{aligned} F(\chi _{[0,x]})=\int _S\chi _{[0,x])}(t)d\nu (t)=\nu ([0,x])=\sigma (x)\;\;\;\;\;\;\;\;\;(x\in (0,1]). \end{aligned}$$

Therefore \(F\in L^{\infty }(S,\mu )^* (=L^1(S,\mu )^{**})\) and \(\sigma |_{(0,1]}=F|_{(0,1]}\). Hence by Theorem 4 (ii) of [16], we have

$$\begin{aligned} \sigma |_{(0,1]}\in L^1(S,\mu )^{**}|_{\varDelta (L^1(S,\mu ))(=(0,1])}\bigcap C_b((0,1])=C_{\textit{BSE}}((0,1]). \end{aligned}$$

By Proposition 2 and the above argument, \(\varphi |_{(0,1]}\in C_{\textit{BSE}}((0,1])\), where \(\varphi \) is the Cantor function on [0, 1]. We claim that \(\varphi |_{(0,1]}\notin \widehat{{\mathcal {M}}(L^1(S,\mu ))}\). Suppose towards a contradiction that \(\varphi |_{(0,1]}\in \widehat{{\mathcal {M}}(L^1(S,\mu ))}\). Then for any \(f\in L^1(S,\mu )\), there exists \(h_f\in L^1(S,\mu ))\) such that

$$\begin{aligned} \varphi (x)\widehat{f}(x)=\widehat{h_f}(x)\;\;\;\;\;\;\;\;\;\;\;x\in (0,1]. \end{aligned}$$

Thus

$$\begin{aligned} \varphi (x)\int _0^xf(t)d\mu (t)=\int _0^xh_f(t)d\mu (t)\;\;\;\;\;\;\;\;\;\;\;x\in (0,1], \end{aligned}$$

which is an absolutely continuous function by Theorem 1. In particular, for the constant function \(f(t)=1\; (t\in (0,1])\), we get

$$\begin{aligned} x\varphi (x)=\int _0^xh_f(t)dt\;\;\;\;\;\;\;(x\in (0,1]). \end{aligned}$$

This implies that the function g defined by \(g(x)=x\varphi (x)\;(x\in [0,1])\) is absolutely continuous. This is a contradiction, by Lemma 1. Therefore \(\varphi |_{(0,1]}\notin \widehat{{\mathcal {M}}(L^1(S,\mu ))}\) and consequently, \(L^1(S,\mu )\) is not a BSE algebra.

Case II. \(1<p<\infty \). We prove that \(L^p(S,\mu )\) has no \(\varDelta -\)weak bounded approximate identity. Assume towards a contradiction that \(\{f_{\alpha }\}_{\alpha }\) is a \(\varDelta -\)weak bounded approximate identity for \(L^p(S,\mu )\) with bound C. By the definition of \(\{f_{\alpha }\}\) we have

$$\begin{aligned} \lim _{\alpha }\phi _x(f_{\alpha })=1\;\;\; \;(x\in (0,1]), \end{aligned}$$

for all \(\phi _x\in \varDelta (L^p(S,\mu ))\). By Remark 1,

$$\begin{aligned} \lim _{\alpha }\int _0^x f_{\alpha }(t)d\mu (t)=1\;\;\; \;(x\in (0,1]). \end{aligned}$$

By Holder’s inequality for every \(x\in S\) we have

$$\begin{aligned} \left| \int _0^xf_{\alpha }(t)d\mu (t)\right|\le & {} \int _0^x\vert f_{\alpha }(t)\vert d\mu (t)\\\le & {} \left( \int _0^x\vert f_{\alpha }(t)\vert ^pd\mu (t)\right) ^{\frac{1}{p}}.\left( \int _0^x1^qd\mu (t)\right) ^{\frac{1}{q}}\\ {}\le & {} C. x^{\frac{1}{q}}\;\;\;\;\;\;\;\;\;\;(x\in (0,1]), \end{aligned}$$

This implies that \(1\le C.x^{\frac{1}{q}}\), so \(x\ge \frac{1}{C^q}\) for all \(x\in (0,1]\). This contradiction completes the proof. \(\square \)

It is well known that for a locally compact group G, the Banach algebra \(L^1(G)\) is an ideal in its second dual if and only if G is compact. As a consequence of the above theorem, in the following result, we prove that this is not the case for the compact totally ordered semigroup S.

Corollary 1

The Banach algebra \(L^1(S,\mu )\) is not an ideal in its second dual.

Proof

By Theorem 3, \(L^1(S,\mu )\) is not a BSE algebra, however, it admits a bounded approximate identity. So by Theorem 3.1 of [8], it is not an ideal in its second dual. \(\square \)

Remark 2

Let \(\sigma \) be a function on [0, 1] such that \(\sigma (0)=0\) and \(\sigma |_{(0,1]}\in C_{\textit{BSE}}((0,1])\). By Theorem 4 (ii) of [16], there exists \(F\in L^1(S,\mu )^{**}(= L^{\infty }(S,\mu )^*)\) such that \(\sigma (x)=F(\chi _{[0,x]})\;(x\in (0,1])\). Since \(F\in L^{\infty }(S,\mu )^*\), from Theorem 2, it follows that there exists \(\nu \in ba(S,\mu )\) such that

$$\begin{aligned} F(f)=\int _Sf(t)d\nu (t)\;\;\;\;\;\;\;\;\;\left( f\in L^{\infty }(S,\mu )\right) . \end{aligned}$$

In particular,

$$\begin{aligned} \sigma (x)=F(\chi _{[0,x]})=\int _S\chi _{[0,x])}(t)d\nu (t)=\nu ([0,x])\;\;\;\;\;\;\;\;\;(x\in (0,1]). \end{aligned}$$

From page 190 of [10], we conclude that \(\sigma \) is of bounded variation on [0, 1].

Now if we let \(\sigma (x)=x\sin \left( \frac{1}{x}\right) (x\in (0,1])\) and

$$\begin{aligned} \sigma _1(x)=\left\{ \begin{array}{ll} x\sin \left( \frac{1}{x}\right) &{}\quad x\ne 0\\ 0&{}\quad x=0 \end{array}\right. \end{aligned}$$

Obviously, \(\sigma \in C_b((0,1])\). The function \(\sigma _1\) is not of bounded variation on [0, 1] and so as the above argument, \(\sigma \notin C_{\textit{BSE}}((0,1])\). This implies that \(C_{\textit{BSE}}((0,1])\varsubsetneqq C_b((0,1])\). However, in the sequel we will prove that for any \(n\in {\mathbb N}\), the natural numbers, \(C_{{\textit{BSE}}(n)}((0,1])=C_b((0,1])\), where \(C_{{\textit{BSE}}(n)}(\varDelta (A))\) denotes the set of all complex valued continuous functions \(\sigma \) on \(\varDelta (A)\) which satisfy the following condition: there exists a positive real numbers \(\beta \) such that the inequality

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

holds for any choice of complex numbers \(c_1,\ldots ,c_n\) and \(\varphi _1,\ldots ,\varphi _n \in \varDelta (A)\). For each \(\sigma \in C_{{\textit{BSE}}(n)}(\varDelta (A))\) we denote by \(\Vert \sigma \Vert _{{\textit{BSE}}(n)}\) the infimum of such \(\beta \). Let \(C_{{\textit{BSE}}(\infty )}(\varDelta (A))=\bigcap \limits _{n\in {\mathbb N}}C_{{\textit{BSE}}(n)}(\varDelta (A)\). It is evident that \(\Vert \sigma \Vert _{\textit{BSE}}=\sup _{n\in {\mathbb N}}\Vert \sigma \Vert _{{\textit{BSE}}(n)}\) and

$$\begin{aligned} C_{\textit{BSE}}(\varDelta (A))=\{\sigma \in C_{{\textit{BSE}}(\infty )}(\varDelta (A):\Vert \sigma \Vert _{\textit{BSE}}<\infty \}. \end{aligned}$$

Also we have

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

For more details see [17].

Theorem 4

For \(S=([0,1],\max )\) and \(n\in {\mathbb N}\), we have

$$\begin{aligned} \widehat{L^1}(S,\mu )\varsubsetneqq C_{\textit{BSE}}((0,1])\varsubsetneqq C_{{\textit{BSE}}(n)}((0,1])=C_b((0,1]). \end{aligned}$$

Proof

By Remark 2 and the proof of Theorem 3, we only need to prove the last equality. By Lemma 1 of [17], it is enough to show that for any natural number n, there exists a real number \(\beta _n\) such that for \(c_1,\ldots ,c_n\in {\mathbb C}\) with \(\vert c_i\vert \le 1 \;(1\le i\le n)\) and \(x_1,\ldots ,x_n\in (0,1]\), there exists a function \(f\in L^1(S,\mu )\) such that \(\widehat{f}(x_i)=c_i\;\;(1\le i\le n)\) and \(\Vert f\Vert _1\le \beta _n\). For every \(x\in [0,1]\) we define

$$\begin{aligned} f(x)=\left\{ \begin{array}{cc} \frac{c_1}{x_1} &{} \quad x\in [0,x_1]\\ \\ \frac{c_2-c_1}{x_2-x_1}&{} \quad \;x\in (x_1,x_2]\\ \\ .&{} \quad .\\ .&{} \quad .\\ .&{} \quad .\\ \\ \frac{c_n-c_{n-1}}{x_n-x_{n-1}} &{} \quad x\in (x_{n-1},x_n]\\ \\ 0&{} \quad x\in (x_n,1] \end{array} \right. \end{aligned}$$

Now for every \(1\le i\le n\), we have

$$\begin{aligned} \widehat{f}(x_i)=\int _0^{x_i}fd\mu =\int _0^{x_1}fd\mu +\int _{x_{1}}^{x_2}fd\mu +\cdots +\int _{x_{i-1}}^{x_i}fd\mu =c_i, \end{aligned}$$

and

$$\begin{aligned} \Vert f\Vert _1= & {} \int _0^1\vert f\vert d\mu =\int _0^{x_1}\vert f\vert d\mu +\int _{x_1}^{x_2}\vert f\vert d\mu +\cdots +\int _{x_n}^1\vert f\vert d\mu \\= & {} \vert c_1\vert +\vert c_2-c_1\vert +\cdots +\vert c_n-c_{n-1}\vert \\\le & {} 2n-1. \end{aligned}$$

So if we choose \(\beta _n=2n-1\), then \(f\in L^1(S,\mu )\) is the required function. \(\square \)

Conjecture

It is well known that \(L^{\infty }(S,\mu )\) with the pointwise multiplication is a \(C^*-\)algebra and so it is a BSE algebra. We conjecture that \(L^{\infty }(S,\mu )\) with the convolution multiplication is not a BSE algebra.