A COUNTEREXAMPLE TO A RESULT OF JABERI AND MAHMOODI

Abstract

We show that $\ell ^1(\mathbb {N}_\wedge )$ is $\varphi $-amenable for each multiplicative linear functional $\varphi :\ell ^1(\mathbb {N}_\wedge )\rightarrow \mathbb {C}.$ This is a counterexample to the final corollary of Jaberi and Mahmoodi [‘On $\varphi $-amenability of dual Banach algebras’, Bull. Aust. Math. Soc. 105 (2022), 303–313] and shows that the final theorem in that paper is not valid.

1 Introduction and preliminaries

The cohomological notion of amenability was introduced and studied in the pioneering work of Johnson [5]. A Banach algebra $\mathcal {A}$ is amenable if every continuous derivation from $\mathcal {A}$ into a dual Banach $\mathcal {A}$ -bimodule $E^*$ is inner. A modification of amenability depending on multiplicative linear functionals was introduced and studied by Kaniuth et al. [6]. A Banach algebra $\mathcal {A}$ is called $\varphi $ -amenable if there exists an element m in $\mathcal {A}^{**}$ such that $m(\varphi )=1$ and $m(f\cdot a)=\varphi (a)m(f)$ for every $a\in {\mathcal {A}}$ and $f\in {\mathcal {A}^*}$ , where $\varphi $ is a multiplicative linear functional on $\mathcal {A}$ . For a locally compact group G, the Fourier algebra $\mathcal {A}(G)$ is always $\varphi $ -amenable. Moreover, the Segal algebra $S^{1}(G)$ is $\varphi $ -amenable if and only if G is amenable (see [1, 6]).

Jaberi and Mahmoodi [4] introduced the new concept of $\varphi $ -injectivity for the category of dual Banach algebras, where $\varphi $ is a $wk^*$ -continuous multiplicative linear functional on $\mathcal {A}$ . A dual Banach algebra $\mathcal {A}$ is $\varphi $ -injective if whenever $\pi :\mathcal {A}\rightarrow \mathcal {L}(E)$ is a $wk^*$ -continuous unital representation on a reflexive Banach space E, then there is a projection $\mathcal {Q}:\mathcal {L}(E)\rightarrow \pi (\mathcal {A})^{\varphi }$ such that $\mathcal {Q}(STU)=S\mathcal {Q}(T)U$ for $S,U\in {\pi (\mathcal {A})^{c}}$ and $T\in {\mathcal {L}(E)}$ , where $\pi (\mathcal {A})^{\varphi }=\{T\in {\mathcal {L}(E)}:\;\pi (a)T=\varphi (a)T\quad (a\in {\mathcal {A}})\}$ . They proved that $\varphi $ -injectivity is equivalent to $\varphi $ -amenability [4, Theorem 3.6].

There is an important category of dual Banach algebras, called enveloping dual Banach algebras. Let $\mathcal {A}$ be a Banach algebra and let E be a Banach $\mathcal {A}$ -bimodule. An element $x\in {E}$ is called weakly almost periodic if the module maps $\mathcal {A}\rightarrow {E}$ given by $a\mapsto {a\cdot {x}}$ and $a\mapsto {x\cdot {a}}$ are weakly compact. The set of all weakly almost periodic elements of E is denoted by ${\mathit {WAP}}(E)$ [7, Definition 4.1]. Runde observed that ${\mathit {WAP}}(\mathcal {A}^{\ast })^{\ast }$ is a canonical dual Banach algebra associated to an arbitrary Banach algebra $\mathcal {A}$ [7, Theorem 4.10]. By means of the new notion of $\varphi $ -injectivity, Jaberi and Mahmoodi investigated $\varphi $ -amenability of the enveloping dual Banach algebra ${\mathit {WAP}}(\mathcal {A}^{\ast })^{\ast }$ [4, Theorem 4.8]. In a short final section of the paper, they claimed that ${\mathit {WAP}}(\ell ^1(\mathbb {N}_\wedge )^*)^*$ is not $\tilde {\varphi }$ -amenable, where $\tilde {\varphi }$ is the unique extension of the augmentation character $\varphi $ on the semigroup algebra $\ell ^1(\mathbb {N}_\wedge )$ [4, Theorem 5.4]. From this result, they concluded that $\ell ^1(\mathbb {N}_\wedge )$ is not $\varphi $ -amenable, where $\varphi $ is the augmentation character [4, Corollary 5.5].

On the contrary, we show that $\ell ^1(\mathbb {N}_\wedge )$ is $\varphi $ -amenable for each multiplicative linear functional $\varphi :\ell ^1(\mathbb {N}_\wedge )\rightarrow \mathbb {C}$ and comment on the reason for this counterexample to the result stated in [4].

2 $\varphi $ -amenability of $\ell ^1(\mathbb {N}_\wedge )$

Let $S=\mathbb {N}$ . With the semigroup product $m\wedge n=\min \{m,n\}$ , for $m,n\in S$ , the set S becomes a semigroup. It is known that $\Delta (\ell ^{1}(S))$ consists of all the functions $\varphi _{n}:\ell ^{1}(S)\rightarrow \mathbb {C}$ given by $\varphi _{n}(\sum _{i=1}^{\infty }\alpha _{i}\delta _{i})=\sum _{i=n}^{\infty }\alpha _{i}$ , for $n\in S$ (see [2, page 32]). Suppose that $m=\delta _{1}$ . Then $\varphi _{1}(m)=\varphi _{1}(\delta _{1})=1$ and

$$ \begin{align*}am=a\delta_{1}=\bigg(\sum^{\infty}_{i=1}a_{i}\bigg)\delta_{1}=\varphi_{1}(a)\delta_{1}=\varphi_{1}(a)m,\quad\mbox{where } a=\sum^{\infty}_{i=1}a_{i}\delta_{i}\in \ell^{1}(S).\end{align*} $$

It follows that $\ell ^{1}(S)$ is $\varphi _{1}$ -amenable. For $n>1$ , define $m_{n}=\delta _{n}-\delta _{n-1}$ . Then,

$$ \begin{align*}\varphi_{n}(m_{n})=\varphi_{n}(\delta_{n}-\delta_{n-1})=1-0=1\end{align*} $$

and

$$ \begin{align*} am_{n}=a(\delta_{n}-\delta_{n-1})=\sum^{\infty}_{i=n}a_{i}(\delta_{n}-\delta_{n-1})=\varphi_{n}(a)(\delta_{n}-\delta_{n-1})=\varphi_{n}(a)m_{n},\end{align*} $$

where $ a=\sum ^{\infty }_{i=1}a_{i}\delta _{i}\in \ell ^{1}(S).$ It follows that $\ell ^{1}(S)$ is $\varphi $ -amenable with respect to each multiplicative linear functional $\varphi :\ell ^{1}(S)\rightarrow \mathbb {C}$ . Thus, [4, Corollary 5.5] is not true.

This counterexample to [4, Corollary 5.5] shows that [4, Theorem 5.4] is also not true. The mistake is the assertion in the second sentence of the proof of Theorem 5.4 that ‘there is an isometric isomorphism $\Theta $ from $\rho (\ell ^1(\mathbb {N}_{\wedge })^c$ onto $\rho (\ell ^{1}(\mathbb {N}_{\wedge }))^ {\varphi }$ ’. An example showing that $\Theta $ cannot be isometric can be constructed using [3, Theorem 7.6]. Take $\|\sum _{n=1}^\infty a_n\delta _n\| = \sup _{F } \|\sum _{n\in F} a_n\delta _n\|$ , where F is a finite subset of $\mathbb {N}$ . Take indices $1$ and $2n+1$ so that the corresponding basis elements belong to distinct summands. Set A to be the diagonal matrix having ones at indices $1$ and $2n+1$ and zero otherwise. Set B to have ones at indices $1$ and $2n+1$ in the first row and zeros otherwise. Then $B=\Theta (A)$ and

$$ \begin{align*} \|A\| & = \sup\bigg\{\|a_1\delta_n+a_{2n+1}\delta_{2n+1}\| : \bigg\|\sum a_n\delta_n\bigg\|=1\bigg\} \\ & = \sum\{(a_1^2+a_{2n+1}^2)^{1/2} : (a_1^2+a_{2n+1}^2)^{1/2}=1\} = 1, \end{align*} $$

while

$$ \begin{align*} \|B\| &= \sup\bigg\{|a_1+a_{2n+1}| : \bigg\|\sum a_n\delta_n\bigg\|=1\bigg\} \\ &= \sum\{|a_1+a_{2n+1}| : (a_1^2+a_{2n+1}^2)^{1/2}=1\} = \sqrt{2}. \end{align*} $$

Consequently, $\Theta $ is not isometric. By taking k summands in a similar way, it can be shown that $\Theta $ is unbounded on diagonal elements of finite support.