MULTIPLIERS ON THE SECOND DUAL OF ABSTRACT SEGAL ALGEBRAS

Abstract

We characterise the existence of certain (weakly) compact multipliers of the second dual of symmetric abstract Segal algebras in both the group algebra $L^{1}(G)$ and the Fourier algebra $A(G)$ of a locally compact group G.

1 Introduction

Let G be a locally compact group. By a classical result of Sakai [14], G is compact if and only if the group algebra $L^1(G)$ has a nonzero (weakly) compact right multiplier. In [10], Lau showed that an analogous result is true on the dual side, that is, G is discrete if and only if its Fourier algebra $A(G)$ has a nonzero (weakly) compact multiplier. Along this line of research, Ghahramani and Lau proved that G is compact if and only if any symmetric Segal algebra $S^1(G)$ of $L^1(G)$ has a nonzero (weakly) compact right or left multiplier [6].

Moreover, it was shown in [4] that G is amenable if and only if $L^\infty (G)^*=L^1(G)^{**}$ , the second dual of $L^1(G)$ equipped with the first Arens product, has a nonzero (weakly) compact right multiplier. Along the way, Ghahramani and Lau proved that G is compact if and only if $L^1(G)^{**}$ has a (weakly) compact left multiplier $ T $ with $ \langle T(n),1\rangle \neq 0 $ for some $ n\in L^{1}(G)^{**} $ [5]. Dually, G is discrete if and only if $A(G)^{**}$ has a (weakly) compact left multiplier $ T $ with $ \langle T(n),1\rangle \neq 0 $ for some $ n\in A(G)^{**}$ .

It is thus natural to try to determine when the second dual of a symmetric abstract Segal algebra of $L^{1}(G)$ or $A(G)$ has a nonzero (weakly) compact left or right multiplier. We answer this question by proving that if $ \mathcal {B} $ is a symmetric abstract Segal algebra of a Banach algebra $\mathcal {A} $ and $ \varphi $ is a nonzero character on ${\mathcal A}$ , then the existence of a (weakly) compact left or right multiplier on $ \mathcal {B} $ is equivalent to the existence of the same multiplier on $ \mathcal {A}$ .

For a symmetric Segal algebra $S^1(G)$ of the group algebra $L^1(G)$ , we denote by $ K $ the set of all right multipliers $ T $ on $ S^1(G)^{**} $ with rank one such that $ \langle T(n),\varphi _{1}\rangle =1$ whenever $ \langle n,\varphi _{1}\rangle =1 $ , where $ \varphi _{1} $ is the nonzero character on $ L^{1}(G) $ defined by $ \varphi _{1}(f)=\int _{G}f(x)\,dx$ for all $f\in L^{1}(G) $ . We prove that if $ G $ is amenable and noncompact and $ d(G) $ is the smallest possible cardinality of a covering of $ G $ by compact sets, then $ \lvert K\rvert \geq 2^{2^{d(G)}}\kern-1pt$ .

2 Preliminaries

We shall now fix some notation. We denote the closed linear span by $\overline {\langle \cdot \rangle }$ . Let ${\mathcal A}$ be a Banach algebra. Then ${\mathcal A}^*$ is naturally a Banach ${\mathcal A}$ -bimodule with the actions

$$ \begin{align*} \langle f\cdot a, b\rangle=\langle f, ab\rangle,\quad \langle a\cdot f, b\rangle=\langle f, ba\rangle, \end{align*} $$

for all $f\in {\mathcal A}^*$ and $a,b\in {\mathcal A}$ . It is known that there is a multiplication $\Box $ on the second dual ${\mathcal A}^{**}$ of ${\mathcal A}$ , extending the multiplication on ${\mathcal A}$ . The first Arens product in ${\mathcal A}^{**}$ is given as follows. For $m,n\in {\mathcal A}^{**}$ , $f\in {\mathcal A}^*$ and $a\in {\mathcal A}$ ,

$$ \begin{align*} \langle m\Box n, f\rangle=\langle m, n\cdot f\rangle,\quad \langle n\cdot f, a\rangle=\langle n, f\cdot a\rangle. \end{align*} $$

If $ \mathcal {A} $ is a Banach algebra, then a linear mapping $ T:\mathcal {A}\rightarrow \mathcal {A} $ is a right (respectively left) multiplier if $ T(ab)=aT(b)$ (respectively $T(ab)=T(a)b $ ) for all $ a,b\in \mathcal {A} $ . In particular, for each $ a\in \mathcal {A} $ , the multiplication operators $ \lambda _{a}:\mathcal {A}\rightarrow \mathcal {A} $ and $ \rho _{a}:\mathcal {A}\rightarrow \mathcal {A} $ defined by $ \lambda _{a}(b)=ab $ and $ \rho _{a}(b)=ba $ are respectively left and right multipliers on $ \mathcal {A} $ . We also denote by $\Delta ({\mathcal A})$ the set of all nonzero characters on ${\mathcal A}$ .

We recall from Burnham [2] that a Banach algebra $\mathcal B$ is an abstract Segal algebra of $\mathcal A$ if:

1. (i) $\mathcal B$ is a dense left ideal in $\mathcal A$ ;

2. (ii) there exists $M> 0$ such that $\lVert b\rVert _{\mathcal A}\leq M \lVert b\rVert _{\mathcal B}$ for each $b\in \mathcal B$ ;

3. (iii) there exists $C> 0$ such that $\lVert ab\rVert _{\mathcal B}\leq C\lVert a\rVert _{\mathcal A} \lVert b\rVert _{\mathcal B}$ for each $a, b\in \mathcal B$ .

We further say that ${\mathcal B}$ is symmetric if it is also a two-sided dense ideal in ${\mathcal A}$ and for each $a,b \in {\mathcal B}$ ,

$$ \begin{align*} \lVert ba\rVert_{\mathcal B}\leq C\lVert a\rVert_{\mathcal A}\lVert b\rVert_{\mathcal B}. \end{align*} $$

In this case, by [2, Theorem 2.1], $\Delta ({\mathcal A})$ and $\Delta ({\mathcal B})$ are homeomorphic.

Throughout this paper, we assume that $ G $ is a locally compact group with a fixed left Haar measure and let $ L^{1}(G) $ be the group algebra of $ G $ . Then $ L^{1}(G) $ is a Banach algebra with the convolution product defined by

$$ \begin{align*} (f*g)(x)=\int_{G}f(y)g(y^{-1}x)\,dy\quad(f,g\in L^1(G)). \end{align*} $$

A linear subspace $ S^{1}(G) $ of $ L^{1}(G) $ is called a Segal algebra, if:

1. (i) $ S^{1}(G) $ is dense in $ L^{1}(G) $ ;

2. (ii) $ S^{1}(G) $ is a Banach space under some norm $ \lVert \cdot \rVert _{S} $ and $ \lVert f\rVert _{1}\leq \lVert f\rVert _{S} $ for all $ f\in S^{1}(G) $ ;

3. (iii) $ S^{1}(G) $ is left translation invariant and the map $ x\mapsto l_{x}f $ of $ G $ into $ S^{1}(G) $ is continuous;

4. (iv) $ \lVert l_{x}f\rVert _{S}=\lVert f\rVert _{S} $ , for all $ x\in G $ and $ f\in S^1(G) $ .

We note that every Segal algebra is an abstract Segal algebra of $ L^{1}(G)$ by [13, Proposition 1]. A Segal algebra $ S^{1}(G) $ is symmetric if it is right translation invariant, $ \lVert r_{x}f\rVert _{S}=\lVert f\rVert _{S} $ and the map $ x\mapsto r_{x}f $ from $ G $ into $ S^{1}(G) $ is continuous for all $ x\in G $ and $ f\in S^{1}(G) $ . Note that every symmetric Segal algebra is a two-sided ideal of $ L^{1}(G) $ and has an approximate identity in which each term has norm one in $ L^{1}(G) $ (see [13, Section 8, Proposition 1]).

3 Multipliers on the second dual

Let $ \mathcal {B} $ be a symmetric abstract Segal algebra of a Banach algebra $ \mathcal {A} $ . We note that for every $ f\in \mathcal {B}^{*} $ , $ a\in \mathcal {A} $ and $ b\in \mathcal {B} $ , we can define $ f\bullet b\in \mathcal {A}^{*} $ by

$$ \begin{align*} \langle f\bullet b,a \rangle=\langle f,ba\rangle. \end{align*} $$

Hence, for every $ m\in \mathcal {A}^{**} $ and $ f\in \mathcal {B}^{*} $ , we may define the functional $ m\bullet f\in \mathcal {B}^{*} $ by

$$ \begin{align*} \langle m\bullet f,b\rangle=\langle m,f\bullet b\rangle\quad (b\in \mathcal{B}). \end{align*} $$

Thus, for every $ m\in \mathcal {A}^{**} $ and $ n\in \mathcal {B}^{**} $ , we can define the functional $ n\odot m \in \mathcal {B}^{**} $ by

$$ \begin{align*} \langle n\odot m,f\rangle=\langle n,m\bullet f\rangle\quad (f\in \mathcal{B}^{*}). \end{align*} $$

For $ f\in \mathcal {B}^{*}$ and $ a\in \mathcal {A} $ , we also can define $ f\bullet a\in \mathcal {B}^{*} $ by

$$ \begin{align*} \langle f\bullet a,b\rangle=\langle f,ab\rangle. \end{align*} $$

Thus for $ n\in \mathcal {B}^{**} $ and $ f\in \mathcal {B}^{*} $ , we may define the functional $ n\bullet f\in \mathcal {A}^{*} $ by

$$ \begin{align*} \langle n\bullet f,a\rangle=\langle n,f\bullet a\rangle\quad (a\in \mathcal{A}). \end{align*} $$

Therefore, for $ m\in \mathcal {A}^{**} $ and $ n\in \mathcal {B}^{**} $ , we can define the functional $ m\odot n \in \mathcal {B}^{**} $ by

$$ \begin{align*} \langle m\odot n,f\rangle=\langle m,n\bullet f\rangle \quad(f\in \mathcal{B}^{*}). \end{align*} $$

Let $ \iota :\mathcal {B}\rightarrow \mathcal {A} $ be the inclusion map. Then $ \iota $ is an injective Banach $ \mathcal {A} $ -bimodule morphism. Consider the adjoints $ \iota ^{*}:\mathcal {A}^{*}\rightarrow \mathcal {B}^{*} $ and $ \iota ^{**}:\mathcal {B}^{**}\rightarrow \mathcal {A}^{**} $ of $ \iota $ . Since $ \iota $ has a dense range, $ \iota ^{*} $ is injective. It is not hard to see that $ \iota ^{*} $ is in fact the restriction map. The following lemma will prove useful.

Lemma 3.1. Let $ \mathcal {B} $ be a symmetric abstract Segal algebra of $ \mathcal {A} $ . Then for every $ {m\in \mathcal {A}^{**} }$ and $ n,p\in \mathcal {B}^{**}$ , the following statements hold:

1. (i) $\lVert n\odot m\rVert \leq C\lVert n\rVert \,\lVert m\rVert $ ;

2. (ii) $\iota ^{**}(n\odot m)=\iota ^{**}(n)\square m $ ;

3. (iii) $p\odot (m\square \iota ^{**}(n))=(p\odot m)\square n $ ;

4. (iv) $\lVert m\odot n\rVert \leq C\lVert n\rVert \,\lVert m\rVert $ ;

5. (v) $\iota ^{**}(m\odot n)=m\square \iota ^{**}(n) $ ;

6. (vi) $( \iota ^{**}(n)\square m)\odot p= n \square (m\odot p) $ .

Proof. The proofs of (i), (ii), (iv) and (v) are straightforward.

(iii) For $ f\in \mathcal {B}^{*} $ , $ b\in \mathcal {B} $ and $ a\in \mathcal {A} $ ,

$$ \begin{align*} \langle \iota^{**}(n)\cdot(f\bullet b),a\rangle &=\langle \iota^{**}(n),f\bullet ba\rangle=\langle n,\iota^{*}(f\bullet ba)\rangle \\ &=\langle n,f\cdot ba\rangle=\langle n\cdot f,ba\rangle=\langle (n\cdot f)\bullet b,a\rangle. \end{align*} $$

Therefore,

$$ \begin{align*} \langle(m\square \iota^{**}(n))\bullet f),b\rangle&=\langle m\square \iota^{**}(n),f\bullet b\rangle=\langle m,\iota^{**}(n)\cdot(f\bullet b)\rangle\\ &=\langle m,(n\cdot f)\bullet b\rangle=\langle m\bullet(n\cdot f),b\rangle. \end{align*} $$

Thus,

$$ \begin{align*} \langle p\odot (m\square \iota^{**}(n)),f\rangle &=\langle p,(m\square \iota^{**}(n))\bullet f\rangle=\langle p,m\bullet (n\cdot f)\rangle\\ &=\langle p\odot m,n\cdot f\rangle=\langle(p\odot m)\square n,f\rangle. \end{align*} $$

Hence, we obtain $ p\odot (m\square \iota ^{**}(n))=(p\odot m)\square n $ , as required.

(vi) Let $ f\in \mathcal {B}^{*} $ , $ b\in \mathcal {B} $ and $ a\in \mathcal {A} $ . Then

$$ \begin{align*} \langle (p\bullet f)\cdot b,a\rangle &=\langle p\bullet f,ba\rangle=\langle p,f\cdot ba\rangle\\ &=\langle p,(f\cdot b)\bullet a\rangle=\langle p\bullet(f\cdot b),a\rangle. \end{align*} $$

Therefore,

$$ \begin{align*} \langle m\cdot (p\bullet f),b\rangle &=\langle m,(p\bullet f)\cdot b\rangle=\langle m,p\bullet(f\cdot b)\rangle \\ &=\langle m\odot p,f\cdot b\rangle=\langle (m\odot p)\cdot f,b\rangle. \end{align*} $$

Thus,

$$ \begin{align*} \langle ( \iota^{**}(n)\square m)\odot p,f\rangle &= \langle\iota^{**}(n)\square m,p\bullet f\rangle=\langle\iota^{**}(n),m\cdot (p\bullet f)\rangle\\ &=\langle n, \iota^{*} (m\cdot (p\bullet f))\rangle =\langle n, m\cdot (p\bullet f)|_{\mathcal{B}}\rangle\\ &=\langle n,(m\odot p)\cdot f\rangle=\langle n\square(m\odot p),f\rangle. \end{align*} $$

Hence, $ ( \iota ^{**}(n)\square m)\odot p= n \square (m\odot p) $ and the proof is complete.

Theorem 3.2. Let $ \mathcal {B} $ be a symmetric abstract Segal algebra of $ \mathcal {A} $ and let $ \varphi \in \Delta (\mathcal {A}) $ . Then the following statements are equivalent:

1. (i) there is a compact (weakly compact) left (right) multiplier $ T $ of $ \mathcal {B}^{**} $ such that $ \langle T(n),\varphi \rangle \neq 0$ for some $ n\in \mathcal {B}^{**} $ ;

2. (ii) there is a compact (weakly compact) left (right) multiplier $ T $ of $ \mathcal {A}^{**} $ such that $ \langle T(m),\varphi \rangle \neq 0 $ for some $m\in \mathcal {A}^{**} $ .

Proof. Suppose that $ T $ is a compact (weakly compact) left multiplier of $ \mathcal {B}^{**} $ with $ \langle T(n),\varphi \rangle \neq 0 $ for some $ n\in \mathcal {B}^{**} $ . Putting $ p=T(n) $ makes $ \lambda _{p}=T\circ \lambda _{n} $ a compact (weakly compact) left multiplier of $ \mathcal {B}^{**} $ . Now for each $ n\in \mathcal {B}^{**}$ , consider the continuous linear map $l_{n}:\mathcal {A}^{**}\rightarrow \mathcal {B}^{**} $ defined by $ l_{n}(m)=n\odot m $ for all $ m\in \mathcal {A}^{**} $ . Since ${ \iota ^{**}\circ \lambda _{p}=\lambda _{\iota ^{**}(p)}\circ \iota ^{**} }$ , by using Lemma 3.1(ii), $\lambda _{\iota ^{**}(p^{2})}=\lambda _{\iota ^{**}(p)}\circ \iota ^{**}\circ l_{p} =\iota ^{**}\circ \lambda _{p} \circ l_{p} $ is a compact (weakly compact) left multiplier of $ \mathcal {A} ^{**}$ such that

$$ \begin{align*} \langle \lambda_{\iota^{**}(p^{2})}(\iota^{**}(p)),\varphi\rangle=\langle \iota^{**}(p^{3}),\varphi\rangle=\langle p^{3},\varphi\rangle=\langle p,\varphi\rangle^{3}\neq 0. \end{align*} $$

Conversely, suppose that $ T $ is a compact (weakly compact) left multiplier of $ \mathcal {A}^{**} $ such that $ \langle T(m),\varphi \rangle \neq 0$ for some $ m\in \mathcal {A}^{**} $ . Then $ \lambda _{p} $ is a compact (weakly compact) left multiplier on $ \mathcal {A}^{**} $ , where $ p=T(m) $ . Choose $ n_0\in \mathcal {B} $ with $ n_0(\varphi )=1 $ . Using Lemma 3.1(iii), $ n_0\odot (p \square \iota ^{**}(n)) =(n_{0}\odot p)\square n$ for all $n\in \mathcal {B}^{**} $ . Then the map $ \lambda _{n_{0}\odot p}=l_{n_{0}}\circ \lambda _{p}~\circ ~\iota ^{**} $ is a compact (weakly compact) left multiplier of $ \mathcal {B}^{**} $ such that

$$ \begin{align*} \langle \lambda_{n_{0}\odot p}(n_{0}),\varphi\rangle=\langle p,\varphi\rangle\neq 0, \end{align*} $$

as required. The result for a right multiplier T can be proved similarly.

From [4, Theorem 2.1] and the above theorem, we obtain the following corollary.

Corollary 3.3. Let $ S(G) $ be a symmetric abstract Segal algebra of $ L^{1}(G) $ . Then $ G $ is amenable if and only if there is a compact (weakly compact) right multiplier T of $ S(G)^{**} $ such that $ \langle T(n),\varphi _{1}\rangle \neq 0 $ for some $ n\in L^{1}(G)^{**}$ .

From [5, Theorem 4.1] and Theorem 3.2, we also obtain the following result.

Corollary 3.4. Let $ S(G) $ be a symmetric abstract Segal algebra of $ L^{1}(G) $ . Then $ G $ is compact if and only if there is a compact (weakly compact) left multiplier T of $ S(G)^{**} $ such that $ \langle T(n),\varphi _{1}\rangle \neq 0 $ for some $ n\in S(G)^{**}$ .

To state the next corollary, let $A(G)$ be the Fourier algebra of a locally compact group G as defined in [3]. Combining Theorem 3.2 with [5, Theorem 4.3], we obtain the following characterisation of discrete groups.

Corollary 3.5. Let $SA(G) $ be an abstract Segal algebra of the Fourier algebra $A(G)$ . Then $ G $ is discrete if and only if there is a compact (weakly compact) left multiplier T of $ SA(G)^{**} $ such that $ \langle T(n),1\rangle \neq 0 $ for some $ n\in SA(G)^{**}$ .

4 Multipliers with rank one

Let ${\mathcal A}$ be a Banach algebra and let $\varphi \in \Delta ({\mathcal A})$ . Following [8], we call an element $ m\in {\mathcal A}^{**} $ a topologically left invariant $ \varphi $ -mean if $ m(\varphi )=1 $ and $ m(f\cdot a)=\varphi (a)m(f) $ for every $f\in {\mathcal A}^{*}$ and $ a\in {\mathcal A}$ , or equivalently $ a\square m=\varphi (a)m$ . We denote the set of all topologically left invariant $ \varphi $ -means on ${\mathcal A}^{*}$ by $ TLI_{\varphi }({\mathcal A}^{**})$ . We also put $ {I_{\varphi }:=\lbrace a\in {\mathcal A}: \varphi (a)=0\rbrace }$ which is a co-dimension one closed ideal in ${\mathcal A}$ . Recall that a locally compact group G is called amenable if there exists a topologically left invariant mean m on $L^\infty (G)$ , that is, a bounded linear functional with $\lVert m\rVert = m(1)=1$ such that $m(f\cdot a)=a(1)m(f)$ for all $f\in L^\infty (G)$ and $a\in L^1({\Bbb G})$ . Topologically right invariant means and (two-sided) invariant means on $L^\infty (G)$ are defined similarly. It is known that the existence of a topologically right invariant mean and the existence of a topologically invariant mean are both equivalent to G being amenable.

A standard argument, used in the proof of [11, Theorem 4.1] on F-algebras, a class of Banach algebras including group algebras, shows that amenability of G is equivalent to the existence of a topologically left invariant $\varphi _1$ -mean on $L^\infty (G)$ (see also [7, Remark 1.3]).

Theorem 4.1. Let $ S(G) $ be an abstract Segal algebra of $ L^{1}(G)$ . Then G is amenable if and only if there is a nonzero idempotent $m\in S(G)^{**}$ such that $\rho _{m}$ has rank one.

Proof. Suppose that G is amenable. Then by [1, Corollary 3.4], there is a topologically left invariant $\varphi _1$ -mean m on $S(G)^{*}$ . It is clear that m is a nonzero idempotent and the map $\rho _m$ on $S(G)^{**}$ , defined by $ \rho _{m}(n)=n\square m=\langle n,\varphi _{1}\rangle m$ for all $n\in S(G)^{**}$ , has rank one.

Conversely, let $m\in S(G)^{**}$ be a nonzero idempotent such that $\rho _{m}$ on $S(G)^{**}$ has rank one. Then there is a functional $\varphi \in S(G)^{***}$ such that $n\square m=\varphi (n)m$ for all $n\in S(G)^{**}$ . Since m is a nonzero idempotent, we obtain $\varphi (m)=1$ . Moreover,

$$ \begin{align*} \varphi (a*b)m&=(a*b)\square m=a\square (b\square m)\\ &=a\square (\varphi (b)m)=\varphi (b)a\square m\\ &=\varphi (b)\varphi (a)m, \end{align*} $$

for all $a,b\in S(G)$ . This implies that $\varphi (a*b)=\varphi (a)\varphi (b)$ for all $a,b\in S(G)$ . Since the map $n\mapsto n\square m$ on $S(G)^{**}$ is weak $^*$ -weak $^*$ continuous and $\varphi (m)=1$ , it follows that ${\varphi \in \Delta (S(G))=\Delta (L^1(G))}$ . This shows that m is a topologically left invariant $\varphi $ -mean on $S(G)^{*}$ . Hence, ${G}$ is amenable by [1, Corollary 3.4].

Lemma 4.2. Let $ S^1(G) $ be a symmetric Segal algebra of $ L^{1}(G)$ and let $\varphi \in \Delta (L^1(G))$ . Then there is a one–one correspondence between the set of topologically left invariant $\varphi $ -means on $ S^1(G)^{*} $ and on $L^{\infty }(G) $ .

Proof. Let $ \iota :S^1(G)\rightarrow L^{1}(G) $ be the inclusion map. Consider the map $ \iota ^{**}:TLI_{\varphi }(S^1(G)^{**})\rightarrow L^{\infty }(G)^{*}$ . Let $ n\in TLI_{\varphi }(S^1(G)^{**}) $ and $ m=\iota ^{**}(n) $ . It is clear that $ m(\varphi )=1 $ . Moreover, for every $a\in L^{1}(G) $ , there is a sequence $ (a_{i})$ in $S^1(G) $ such that $ \lVert a_{i}-a\rVert _{1}\rightarrow 0 $ . Since $ \Delta (S^1(G))=\Delta (L^{1}(G)) $ , we have

$$ \begin{align*} a\square m &=\lim_{i}(a_{i}\square \iota^{**}(n)) =\lim_{i}\iota^{**}(a_{i}\square n)\\ &=\lim_{i}\varphi(a_{i})\iota^{**}(n) =\varphi(a)\iota^{**}(n)=\varphi(a)m. \end{align*} $$

Therefore, $ \iota ^{**}(TLI_{\varphi }(S^1(G)^{**}))\subseteq TLI_{\varphi }(L^{\infty }(G)^{*})$ . We next show that

$$ \begin{align*} \iota^{**}: TLI_{\varphi}(S^1(G)^{**})\rightarrow TLI_{\varphi}(L^{\infty}(G)^{*}) \end{align*} $$

is injective. In fact, suppose that $m, n\in TLI_{\varphi }(S^1(G)^{**})$ with $m\neq n$ . Then there exists $f\in S^1(G)^*$ such that $m(f)\neq n(f)$ . Let $b_0\in S^1(G)$ be such that $\varphi (b_0)=1$ . Then ${m(f\cdot b_0)=m(f)\neq n(f)=n(f\cdot b_0)}$ . It follows that

$$ \begin{align*} \langle \iota^{**}(m), f\bullet b_0\rangle=\langle m, f\cdot b_0\rangle\neq \langle n, f\cdot b_0\rangle=\langle \iota^{**}(n), f\bullet b_0\rangle. \end{align*} $$

Therefore, $\iota ^{**}(m)\neq \iota ^{**}(n)$ . We now prove that $ \iota ^{**} $ is surjective. Suppose that $ {m\in TLI_{\varphi }(L^{\infty }(G)^*) }$ . Then for each $ f\in S^1(G)^{*} $ and $ a,b\in S^1(G) $ , we have

$$ \begin{align*} \langle m\bullet f,a*b\rangle=\langle m,f\bullet a*b\rangle=\langle m,(f\bullet a)\cdot b\rangle=\varphi(b)\langle m\bullet f,a\rangle. \end{align*} $$

Thus, for $ b\in I_{\varphi } $ , we have

$$ \begin{align*} \langle m\bullet f,a*b\rangle=0. \end{align*} $$

Since $S^1(G)$ has an approximate identity (not necessarily bounded), it follows that $ \overline {\langle S^1(G)*I_{\varphi }\rangle }=I_{\varphi } $ . Thus $ (m\bullet f)|_{I_{\varphi }}=0 $ . As $ a*b-b*a\in I_{\varphi } $ , we obtain

$$ \begin{align*} \langle m, f\bullet(a*b)\rangle=\langle m, f\bullet(b*a)\rangle. \end{align*} $$

Let $ \varphi (b_{0})=1 $ for some $ b_{0}\in S^1(G) $ and consider the functional $ \tilde {m}\in S^1(G)^{**} $ defined by

$$ \begin{align*} \tilde{m}(f)=\langle m,f\bullet b_{0}\rangle, \quad f\in S^1(G)^{*}. \end{align*} $$

Then for each $ b\in S^1(G) $ and $ f\in S^1(G)^{*} $ , we have

$$ \begin{align*} \tilde{m}(f\cdot b)&=\langle m,(f\cdot b)\bullet b_{0}\rangle=\langle m,f\bullet(b* b_0)\rangle\\ &=\langle m,f\bullet(b_0* b)\rangle=\langle m,(f\bullet b_0)\cdot b)\rangle\\ &=\varphi(b)\langle m,f\bullet b_{0}\rangle=\varphi(b) \tilde{m}(f). \end{align*} $$

Furthermore, it is obvious that $ \tilde {m}(\varphi )=1 $ . Hence, $ \tilde {m} \in TLI_{\varphi }(S^1(G)^{**}) $ . We have to show that $ \iota ^{**}(\tilde {m})=m $ . In fact, for every $ g\in L^{\infty }(G) $ , we have

$$ \begin{align*} \langle \iota^{**}(\tilde{m}),g\rangle =\langle \tilde{m},\iota^{*}(g)\rangle=\langle m,\iota^{*}(g)\bullet b_{0}\rangle -\langle m,g\cdot b_{0}\rangle=\langle m,g\rangle, \end{align*} $$

and the proof is complete.

Before giving the next result, recall that the compactness of G is equivalent to the existence of a topologically invariant mean in $L^{1}(G)$ . The following theorem is inspired by [4, Theorem 2.15].

Theorem 4.3. Let $ S^1(G) $ be a symmetric Segal algebra of $ L^{1}(G)$ and $ K $ be the set of all right multipliers $ T $ of $ S^1(G)^{**} $ with rank one such that $ \langle T(n),\varphi _{1}\rangle =1$ whenever $\langle n,\varphi _{1}\rangle =1 $ for $ n\in S^1(G)^{**} $ . Then the following statements hold:

1. (i) $ K\neq \emptyset $ if and only if $ G $ is amenable;

2. (ii) $ \lvert K \rvert =1 $ if and only if $ G $ is compact;

3. (iii) if $ G $ is amenable and noncompact and $ d(G) $ is the smallest possible cardinality of a covering of $ G $ by compact sets, then $ \lvert K\rvert \geq 2^{2^{d(G)}}$ .

Proof. (i) Suppose that $ G $ is amenable. Then by [1, Corollary 3.4], there is a topologically left invariant $ \varphi _{1}$ -mean $ m $ on $ S^1(G)^{*}$ . Since $ \rho _{m}(n)=n\square m =\langle n,\varphi _{1}\rangle m$ for all $n\in S^1(G)^{**}$ , it follows that $\rho _m$ belongs to $ K $ .

Conversely, suppose that $ T\in K $ and $ \langle n,\varphi _{1}\rangle =1 $ for some $ n\in S^1(G)^{**} $ . Putting ${m=T(n) }$ , we have $\langle m,\varphi _{1}\rangle =1 $ . By the same argument as that used in the proof of Theorem 4.1, it is easy to show that m is a topologically left invariant $\varphi _1$ -mean on $S^1(G)^*$ . Thus, G is amenable by [1, Corollary 3.4].

(ii) Let $ T\in K $ and $ n\in TLI_{\varphi _{1}}(S^1(G)^{**}) $ . Putting $ m=T(n)$ , by (i), $ m $ is a topologically left invariant $ \varphi _{1}$ -mean on $ S^1(G)^{*}$ . In particular, for each $ p \in S(G)^{**} $ with $\langle p,\varphi _{1}\rangle =1 $ , we obtain $ p\square m= m$ . Thus,

$$ \begin{align*} \rho_{m}(p)=p\square m=m=T(p). \end{align*} $$

By linearity, we conclude that $ \rho _{m}=T $ and so there is a one–one correspondence between $ K $ and $TLI_{\varphi _{1}}(S^1(G)^{**}) $ . By Lemma 4.2, $ \lvert K\rvert = \lvert TLI_{\varphi _{1}}(L^{\infty }(G)^*)\rvert $ .

Now suppose that G is compact. Then there is a topologically invariant mean m in $L^{1}(G)$ . Thus, for each $n\in TLI_{\varphi _{1}}(L^{\infty }(G)^{*})$ , we have

$$ \begin{align*} m=n(\varphi_{1})m=m\square n=m(\varphi_{1})n=n. \end{align*} $$

This shows that $ \lvert K\rvert =\lvert TLI_{\varphi _{1}}(L^{\infty }(G)^{*})\rvert =1$ .

Conversely, suppose that $ \lvert K\rvert =1 $ . Then $ \lvert TLI_{\varphi _{1}}(L^{\infty }(G)^{*})\rvert =1$ . Therefore, there is a unique topologically left invariant $\varphi _1$ -mean m on $L^\infty (G)$ . It follows that m belongs to $L^1(G)$ (see [9]), whence G is compact.

(iii) Suppose that $ G $ is noncompact. Then by [12, Theorem 1], the cardinality of $TLI_{\varphi _{1}}(L^{\infty }(G)^{*})$ is at least $2^{2^{d(G)}}$ . Therefore, $ \lvert K\rvert = \lvert TLI_{\varphi _{1}}(L^{\infty }(G)^{*})\rvert \geq 2^{2^{d(G)}} $ .