$\mathit{\sigma }$-Approximately module amenable Banach algebras

Abstract

In this paper, we define the notion of sigma-approximate module amenability of Banach algebras and give some properties about this notion. Also for Banach $\mathfrak{A}$-bimodule $\mathcal{A}$, and $\mathcal{J}$, the closed ideal of $\mathcal{A}$ generated by elements of form $(\mathit{\alpha }·a)b-a(b·\mathit{\alpha })$, $(a,b\in \mathcal{A}\text{,}\mathit{\alpha }\in \mathfrak{A})$, we considered some corollaries about $\widehat{\mathit{\sigma }}$-approximate amenability of $\frac{\mathcal{A}}{\mathcal{J}}$ as a Banach $\mathcal{A}$-bimodule, where $\widehat{\mathit{\sigma }}:\frac{\mathcal{A}}{\mathcal{J}}\to \frac{\mathcal{A}}{\mathcal{J}}$ by $\widehat{\mathit{\sigma }}(a+\mathcal{J})=\mathit{\sigma }(a)+\mathcal{J}$ has a dense range.

Introduction

The concept of module amenability for Banach algebras was introduced by Amini [1]. Let $\mathfrak{A}$ and $\mathcal{A}$ be Banach algebras such that $\mathcal{A}$ is a Banach $\mathfrak{A}$-bimodule with the following compatible actions:

$$\begin{array}{c}\hfill \mathit{\alpha }·(ab)=(\mathit{\alpha }·a)b\text{and}(ab)·\mathit{\alpha }=a(b·\mathit{\alpha }),\end{array}$$

for all $a,b\in \mathcal{A}$, $\mathit{\alpha }\in \mathfrak{A}$. Let $\mathcal{X}$ be a Banach $\mathcal{A}$-bimodule and a Banach $\mathfrak{A}$-bimodule with compatibility of actions:

$$\begin{array}{c}\hfill \mathit{\alpha }·(a·x)=(\mathit{\alpha }·a)·x\text{and}a·(x·\mathit{\alpha })=(a·x)·\mathit{\alpha },\end{array}$$

for all $a\in \mathcal{A}$, $\mathit{\alpha }\in \mathfrak{A}$, $x\in \mathcal{X}$, and the same for the other side actions. Then, we say that $\mathcal{X}$ is a Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule. If moreover, $\mathit{\alpha }·x=x·\mathit{\alpha }$, $(\mathit{\alpha }\in \mathfrak{A}$, $x\in \mathcal{X})$, then $\mathcal{X}$ is called a commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule. Note that, when $\mathcal{A}$ acts on itself by algebra multiplication from both sides, it is not in general a Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule because $\mathcal{A}$ does not satisfy $a·(\mathit{\alpha }·b)=(a·\mathit{\alpha })·b$, $(\mathit{\alpha }\in \mathfrak{A},a,b\in \mathcal{A})$ [1].

If $\mathcal{A}$ is a commutative $\mathfrak{A}$-bimodule and acts on itself by algebra multiplication from both sides, then it is also a Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule. Also, if $\mathcal{A}$ is a commutative Banach algebra, then it is a commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule.

Now suppose that $\mathcal{X}$ be an $\mathcal{A}$-$\mathfrak{A}$-bimodule, then a continuous map $T:\mathcal{A}\to \mathcal{X}$ is called an $\mathfrak{A}$-bimodule map, if $T(a\pm b)=T(a)\pm T(b)$ and $T(\mathit{\alpha }·a)=\mathit{\alpha }·T(a)$ and $T(a·\mathit{\alpha })=T(a)·\mathit{\alpha }$, for each $\mathit{\alpha }\in \mathfrak{A},a,b\in \mathcal{A}$. The space of all $\mathfrak{A}$-bimodule maps $T:\mathcal{A}\to \mathcal{X}$ such that $T(ab)=T(a)T(b)$, $(a,b\in \mathcal{A})$, is denoted by ${\text{Hom}}_{\mathfrak{A}}(\mathcal{A},\mathcal{X}).$ Also we denote ${\text{Hom}}_{\mathfrak{A}}(\mathcal{A},\mathcal{A})$, by ${\text{Hom}}_{\mathfrak{A}}(\mathcal{A}).$

Let $\mathcal{A}$ and $\mathfrak{A}$ be as above and $\mathcal{X}$ be a Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule. A bounded $\mathfrak{A}$-bimodule map $D:\mathcal{A}\to \mathcal{X}$ is called a module derivation if

$$\begin{array}{c}\hfill D(ab)=D(a)·b+a·D(b),(a,b\in \mathcal{A}).\end{array}$$

$D$ is not necessary linear, but its boundedness implies its norm continuity, because it preserves subtraction. When $\mathcal{X}$ is commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule, each $x\in \mathcal{X}$ defines a module derivation

$$\begin{array}{c}\hfill {\mathit{\delta }}_x(a)=a·x-x·a,(a\in \mathcal{A}),\end{array}$$

which is called an inner module derivations.

Let $\mathcal{A}$ be a Banach $\mathfrak{A}$-bimodule and $\mathit{\sigma }$$\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{A}).$ A $\mathit{\sigma }$-module derivation from $\mathcal{A}$ into a Banach $\mathcal{A}$-bimodule $\mathcal{X}$ is a bounded $\mathfrak{A}$-bimodule map $D:\mathcal{A}⟶\mathcal{X}$ satisfying

$$\begin{array}{c}\hfill D(ab)=\mathit{\sigma }(a)·D(b)+D(a)·\mathit{\sigma }(b),(a,b\in \mathcal{A}).\end{array}$$

When $\mathcal{X}$ is commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule, each $x\in \mathcal{X}$ defines a $\mathit{\sigma }$-module derivation

$$\begin{array}{c}\hfill {\mathit{\delta }}_x^{\mathit{\sigma }}:\mathcal{A}⟶\mathcal{X},{\mathit{\delta }}_x^{\mathit{\sigma }}(a)=\mathit{\sigma }(a)·x-x·\mathit{\sigma }(a),(a\in \mathcal{A}),\end{array}$$

which is called a $\mathit{\sigma }$-inner module derivation.

$\mathit{\sigma }$-Approximate module amenability

We start this section with definition of sigma-approximate module amenability, then we consider some hereditary properties of this concept.

Definition 1

Let $\mathcal{A}$ be a Banach $\mathfrak{A}$-bimodule and $\mathit{\sigma }\in Ho{m}_{\mathfrak{A}}(\mathcal{A})$. We say that $\mathcal{A}$ is a $\mathit{\sigma }$-approximately module amenable $(\mathit{\sigma }\text{-(AMA)})$, if for each commutative Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule, $\mathcal{X}$, every $\mathit{\sigma }$-module derivation $D:\mathcal{A}⟶{\mathcal{X}}^{\ast }$ is $\mathit{\sigma }$-approximately inner, i.e, there is a net ${(x_i)}_{i\in \mathfrak{I}}$$\in {\mathcal{X}}^{\ast }$ such that $D(a)={lim}_i$${\mathit{\delta }}_{x_i}^{\mathit{\sigma }}(a)={lim}_i\mathit{\sigma }(a)x_i-x_i\mathit{\sigma }(a)$, $(a\in \mathcal{A})$. Also we say that $\mathcal{A}$ is a $\mathit{\sigma }$-approximately module contractible $(\mathit{\sigma }-(AMC))$, if for each commutative Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule, $\mathcal{X}$, every $\mathit{\sigma }$-module derivation $D:\mathcal{A}⟶\mathcal{X}$ is $\mathit{\sigma }$-approximately inner.

The two following results is the $\mathit{\sigma }$-approximate version of [1, Proposition 2.1] and [5], respectively.

Proposition 2

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule and$\mathit{\sigma }\in Ho{m}_{\mathfrak{A}}(\mathcal{A})$.Suppose that$\mathfrak{A}$has a bounded approximate identity and$\mathcal{A}$is$\mathit{\sigma }$-approximately amenable. Then$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}.$

Proof

Let $\mathcal{X}$ be a commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule and $D:\mathcal{A}⟶{\mathcal{X}}^{\ast }$ be a $\mathit{\sigma }$-module derivation. By [1, Proposition 2.1], $D$ is a $\mathit{\sigma }$-derivation, i.e, $D$ is $\mathbb{C}$-linear. Now since $\mathcal{A}$ is $\mathit{\sigma }$-approximately amenable, $\mathcal{A}$ is $\mathit{\sigma }\text{-(AMA)}$. $\square$

Proposition 3

Let$\mathcal{A}$be an essential left Banach$\mathfrak{A}$-bimodule and$\mathit{\sigma }$$\in Ho{m}_{\mathfrak{A}}(\mathcal{A})$.If$\mathcal{A}$is$\mathit{\sigma }$-approximately amenable, then$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}$.

Proof

Let $\mathcal{X}$ be a commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule and $D:\mathcal{A}⟶{\mathcal{X}}^{\ast }$ be a $\mathit{\sigma }$-module derivation. Since $\mathcal{A}$ is an essential left Banach $\mathfrak{A}$-bimodule, $D$ is $\mathbb{C}$-linear [5]. Now since $\mathcal{A}$ is $\mathit{\sigma }$-approximately amenable, $D$ is $\mathit{\sigma }$-approximately inner and thus $\mathcal{A}$ is $\mathit{\sigma }\text{-(AMA)}$. $\square$

Proposition 4

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule and$\mathit{\sigma }\in Ho{m}_{\mathfrak{A}}(\mathcal{A})$.If$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}$,then$\mathcal{A}$is$(\mathit{\lambda }\circ \mathit{\sigma },\mathit{\mu }\circ \mathit{\sigma })\text{-(AMA)}$,for each$\mathit{\lambda },\mathit{\mu }\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{A})$.

Proof

Let $\mathcal{X}$ be a commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule and $D:\mathcal{A}⟶{\mathcal{X}}^{\ast }$ be a $(\mathit{\lambda }\circ \mathit{\sigma },\mathit{\mu }\circ \mathit{\sigma })$-module derivation. Then $\mathcal{X}$ is an $\mathcal{A}$-module derivation with the following module actions:

$$\begin{array}{c}\hfill a\ast x=\mathit{\lambda }(a)·x\text{and}x\ast a=x·\mathit{\mu }(a),(a\in \mathcal{A},x\in \mathcal{X}).\end{array}$$

It is easy to see that $\mathcal{X}$ is a commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule with this product. We have

$$\begin{array}{cc}\hfill D(ab)& =(\mathit{\lambda }\circ \mathit{\sigma })(a)·D(b)+D(a)·(\mathit{\mu }\circ \mathit{\sigma })(b)\hfill \\ \hfill & =\mathit{\sigma }(a)\ast D(b)+D(a)\ast \mathit{\sigma }(b),(a,b\in \mathcal{A}).\hfill \end{array}$$

Thus, $D$ is a $\mathit{\sigma }$-module derivation. So there exists a net $(x_i)\in {\mathcal{X}}^{\ast }$ such that $D(a)={lim}_i{\mathit{\delta }}_{x_i}^{\mathit{\sigma }}(a)$, $(a\in \mathcal{A})$. So we have

$$\begin{array}{cc}\hfill D(a)& =\underset{i}{lim}(\mathit{\sigma }(a)\ast x_i-x_i\ast \mathit{\sigma }(a))\hfill \\ \hfill & =\underset{i}{lim}((\mathit{\lambda }\circ \mathit{\sigma })(a)·x_i-x_i·(\mathit{\mu }\circ \mathit{\sigma })(a)),(a\in \mathcal{A}).\hfill \end{array}$$

Which shows that $D$ is $(\mathit{\lambda }\circ \mathit{\sigma },\mathit{\mu }\circ \mathit{\sigma })$-approximately inner. Thus, $\mathcal{A}$ is $(\mathit{\lambda }\circ \mathit{\sigma },\mathit{\mu }\circ \mathit{\sigma })\text{-(AMA)}$. $\square$

Corollary 5

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule. If$\mathcal{A}$is (AMA), then$\mathcal{A}$is$(\mathit{\lambda },\mathit{\mu })\text{- (AMA)}$, for each$\mathit{\lambda },\mathit{\mu }\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{A}).$

Proposition 6

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule and$\mathit{\sigma }$$\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{A})$.Suppose that$\mathit{\sigma }$is an idempotent epimorphism and$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}$.Then, $\mathcal{A}$is (AMA).

Proof

Let $\mathcal{X}$ be a commutative $\mathcal{A}$-$\mathfrak{A}$-bimodule and $D:\mathcal{A}⟶{\mathcal{X}}^{\ast }$ be a module derivation. So $\tilde{D}=D\circ \mathit{\sigma }$ is a $\mathit{\sigma }$-module derivation, because, for each $a,b\in \mathcal{A}$ and $\mathit{\alpha }\in \mathfrak{A}$ we have

$$\begin{array}{cc}\hfill \tilde{D}(ab)& =D\circ \mathit{\sigma }(ab)=D(\mathit{\sigma }(a)\mathit{\sigma }(b))\hfill \\ \hfill & =\mathit{\sigma }(a)(D\circ \mathit{\sigma })(b)+(D\circ \mathit{\sigma })(a)\mathit{\sigma }(b),\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \tilde{D}(\mathit{\alpha }a)=D(\mathit{\sigma }(\mathit{\alpha }a))=D(\mathit{\alpha }\mathit{\sigma }(a))=\mathit{\alpha }D(\mathit{\sigma }(a)).\end{array}$$

Since $\mathcal{A}$ is $\mathit{\sigma }\text{-(AMA)}$, there exists a net ${(x_i)}_{i\in \mathfrak{I}}\in {\mathcal{X}}^{\ast }$ such that $\tilde{D}(a)={lim}_i(\mathit{\sigma }(a)x_i-x_i\mathit{\sigma }(a))$, $(a\in \mathcal{A})$. Now for each $b\in \mathcal{A},$ there exists $a\in \mathcal{A}$ such that $b=\mathit{\sigma }(a)$. Therefore,

$$\begin{array}{c}\hfill D(b)=D(\mathit{\sigma }(a))=\tilde{D}(a)=\underset{i}{lim}(\mathit{\sigma }(a)x_i-x_i\mathit{\sigma }(a))=\underset{i}{lim}(b{x}_i-x_{i}b),(b\in \mathcal{A}).\end{array}$$

So $D$ is approximately inner and $\mathcal{A}$ is (AMA). $\square$

Proposition 7

Let$\mathcal{A}$and$\mathcal{B}$be Banach$\mathfrak{A}$-bimodules and$\mathit{\sigma }$$\in Ho{m}_{\mathfrak{A}}(\mathcal{A})$and$\mathit{\tau }$$\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{B})$.Suppose that$\mathit{\phi }\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{A},\mathcal{B})$be a surjective map such that$\mathit{\phi }\circ \mathit{\sigma }=\mathit{\tau }\circ \mathit{\phi }.$If$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}$,then$\mathcal{B}$is$\mathit{\tau }\text{-(AMA)}$.

Proof

Let $\mathcal{X}$ be a commutative Banach $\mathcal{B}$-$\mathfrak{A}$-bimodule and $\mathcal{D}:{\mathcal{B}\to \mathcal{X}}^{\ast }$ be a $\mathit{\tau }$-module derivation. Then, $(\mathcal{X},\ast )$ can be considered as a Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule by the following module actions:

$$\begin{array}{c}\hfill a\ast x=\mathit{\phi }(a)·x\text{and}x\ast a=x·\mathit{\phi }(a),(a\in \mathcal{A},x\in X).\end{array}$$

Therefore, $\tilde{D}=D\circ \mathit{\phi }:\mathcal{A}\to ({\mathcal{X}}^{\ast },\ast )$ is a $\mathit{\sigma }$-module derivation, because

$$\begin{array}{cc}\hfill \tilde{D}(ab)& =D(\mathit{\phi }(a)\mathit{\phi }(b))\hfill \\ \hfill & =D(\mathit{\phi }(a))\mathit{\tau }(\mathit{\phi }(b))+\mathit{\tau }(\mathit{\phi }(a))D(\mathit{\phi }(b))\hfill \\ \hfill & =\tilde{D}(a)\mathit{\phi }(\mathit{\sigma }(b))+\mathit{\phi }(\mathit{\sigma }(a))\tilde{D}(b)\hfill \\ \hfill & =\tilde{D}(a)\ast \mathit{\sigma }(b)+\mathit{\sigma }(a)\ast \tilde{D}(b),(a,b\in \mathcal{A}).\hfill \end{array}$$

Since $\mathcal{A}$ is $\mathit{\sigma }\text{-(AMA)}$, there exists a net ${(x_i)}_{i\in \mathfrak{I}}\in {\mathcal{X}}^{\ast }$ such that $\tilde{D}={lim}_i{\mathit{\delta }}_{x_i}^{\mathit{\sigma }}$. So we have

$$\begin{array}{cc}\hfill \tilde{D}(a)& =\underset{i}{lim}\mathit{\sigma }(a)\ast x_i-x_i\ast \mathit{\sigma }(a)\hfill \\ \hfill & =\underset{\mathit{\alpha }}{lim}\mathit{\phi }(\mathit{\sigma }(a))·x_i-x_i·\mathit{\phi }(\mathit{\sigma }(a))\hfill \\ \hfill & =\underset{\mathit{\alpha }}{lim}\mathit{\tau }(\mathit{\phi }(a))·x_i-x_i·\mathit{\tau }(\mathit{\phi }(a)),(a\in \mathcal{A}).\hfill \end{array}$$

Since $\mathit{\phi }$ is a surjective map, so $D(b)={lim}_i\mathit{\tau }(b)·x_i-x_i·\mathit{\tau }(b)$, $(b\in \mathcal{B})$. Hence, $\mathcal{B}$ is $\mathit{\tau }\text{-(AMA)}$. $\square$

Proposition 8

Suppose that$\mathcal{A}$and$\mathcal{B}$are Banach$\mathfrak{A}$-modules and$\mathit{\phi }\in Ho{m}_{\mathfrak{A}}(\mathcal{A},\mathcal{B})$be a surjective map. If$\mathcal{A}$is (AMA), then$\mathcal{B}$is$\mathit{\sigma }\text{-(AMA)}$,for each$\mathit{\sigma }\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{B})$.

Proof

Let $\mathcal{X}$ be a Banach $\mathcal{B}$-$\mathfrak{A}$-bimodule and $\mathit{\sigma }\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{B})$. Then $(\mathcal{X},\ast )$ is a Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule with the following module actions:

$$\begin{array}{c}\hfill a\ast x=\mathit{\sigma }(\mathit{\phi }(a))·x\text{and}x\ast a=x·\mathit{\sigma }(\mathit{\phi }(a)),(a\in \mathcal{A},x\in \mathcal{X}).\end{array}$$

Now, let $D:{\mathcal{B}\to \mathcal{X}}^{\ast }$ be a $\mathit{\sigma }$-module derivation. So $\tilde{D}=D\circ \mathit{\phi }:\mathcal{A}\to ({\mathcal{X}}^{\ast },\ast )$ is a module derivation, because for each $\mathit{\alpha }\in \mathfrak{A}$ and $a,b\in \mathcal{A}$, we have

$$\begin{array}{c}\hfill \tilde{D}(\mathit{\alpha }a)=D(\mathit{\phi }(\mathit{\alpha }a))=D(\mathit{\alpha }\mathit{\phi }(a))=\mathit{\alpha }D(\mathit{\phi }(a)),\end{array}$$

and

$$\begin{array}{cc}\hfill \tilde{D}(ab)& =D(\mathit{\phi }(ab))\hfill \\ \hfill & =D(\mathit{\phi }(a))\mathit{\sigma }(\mathit{\phi }(b))+\mathit{\sigma }(\mathit{\phi }(a))D(\mathit{\phi }(b))\hfill \\ \hfill & =\tilde{D}(a)\ast b+a\ast \tilde{D}(b).\hfill \end{array}$$

So there exists a net ${(x_i)}_{i\in \mathfrak{I}}\in X^{\ast }$ such that $\tilde{D}={lim}_i{\mathit{\delta }}_{x_i}$ and we have

$$\begin{array}{cc}\hfill \tilde{D}(a)& =\underset{i}{lim}{\mathit{\delta }}_{x_i}(a)\hfill \\ \hfill & =\underset{i}{lim}(a\ast x_i-x_i\ast a)\hfill \\ \hfill & =\underset{i}{lim}\mathit{\sigma }(\mathit{\phi }(a))·x_i-x_i·\mathit{\sigma }(\mathit{\phi }(a)),(a\in \mathcal{A}).\hfill \end{array}$$

Since $\mathit{\phi }$ is surjective, for each $b\in \mathcal{B}$, there exists $a\in \mathcal{A}$, such that $b=\mathit{\phi }(a)$. So for each $b\in \mathcal{B}$ we have

$$\begin{array}{c}\hfill D(b)=D(\mathit{\phi }(a))=\tilde{D}(a)=\underset{i}{lim}\mathit{\sigma }(\mathit{\phi }(a))·x_i-x_i·\mathit{\sigma }(\mathit{\phi }(a))=\underset{i}{lim}\mathit{\sigma }(b)·x_i-x_i·\mathit{\sigma }(b).\end{array}$$

Which shows that $D$ is $\mathit{\sigma }$-approximately inner. Thus, $\mathcal{B}$ is $\mathit{\sigma }\text{-(AMA)}$. $\square$

Let $\mathcal{A}$ be a Banach $\mathfrak{A}$-bimodule with compatible actions and $\mathcal{J}$ be the closed ideal of $\mathcal{A}$ generated by elements of form $(\mathit{\alpha }·a)b-a(b·\mathit{\alpha })$, for all $a,b\in \mathcal{A}$ and $\mathit{\alpha }\in \mathfrak{A}.$ Then, the quotient Banach algebra $\frac{\mathcal{A}}{\mathcal{J}}$ is Banach $\mathcal{A}$-bimodule with compatible actions [2]. The following Lemma is proved in [3].

Lemma 9

Let $\mathcal{A}$ be a Banach $\mathfrak{A}$-bimodule and $\mathfrak{A}$ has a bounded approximate identity for $\mathcal{A}$. Suppose that $\mathit{\sigma }\in Ho{m}_{\mathfrak{A}}(\mathcal{A})$ such that $\mathit{\sigma }(\mathcal{J})\subseteq \mathcal{J}$. Then $\widehat{\mathit{\sigma }}:\frac{\mathcal{A}}{\mathcal{J}}\to \frac{\mathcal{A}}{\mathcal{J}}$ by $\widehat{\mathit{\sigma }}(a+\mathcal{J})=\mathit{\sigma }(a)+\mathcal{J}$ is $\mathbb{C}$-linear.

Proposition 10

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule and$\mathfrak{A}$has a bounded approximate identity for$\mathcal{A}$.Let$\mathit{\sigma }$be as in above lemma. If$\frac{\mathcal{A}}{\mathcal{J}}$is$\widehat{\mathit{\sigma }}$-approximately amenable, then$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}$.

Proof

Let $\mathcal{X}$ be a commutative Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule. It is easy to see that $\mathcal{J}·\mathcal{X}=\mathcal{X}·\mathcal{J}=0$. So $\mathcal{X}$ is a Banach $\frac{\mathcal{A}}{\mathcal{J}}$-bimodule with the following module actions;

$$\begin{array}{c}\hfill (a+\mathcal{J})·x=ax\text{and}x·(a+\mathcal{J})=xa,(a\in \mathcal{A},x\in \mathcal{X}).\end{array}$$

Suppose that $D:{\mathcal{A}\to \mathcal{X}}^{\ast }$ be a $\mathit{\sigma }$-module derivation. Define $\widehat{D}:$$\frac{\mathcal{A}}{\mathcal{J}}\to {\mathcal{X}}^{\ast }$ by $\widehat{D}(a+\mathcal{J})=D(a)$, $(a\in \mathcal{A})$. $\widehat{D}$ is well defined [3, Proposition 2.6] and it is $\mathbb{C}$-linear [1, Proposition 2.1]. Also, it is easy to see that $\widehat{D}(ab+\mathcal{J})=\widehat{D}(a+\mathcal{J})$$\widehat{\mathit{\sigma }}(b+\mathcal{J})+\widehat{\mathit{\sigma }}(a+\mathcal{J})\widehat{D}(b+\mathcal{J})$. Moreover according to the above Lemma, $\widehat{\mathit{\sigma }}$ is $\mathbb{C}$-linear. Therefore, $\widehat{D}$ is $\widehat{\mathit{\sigma }}$-derivation. Thus, there exists a net ${(x_i)}_{i\in \mathfrak{I}}\in X^{\ast }$ such that $\widehat{D}={lim}_i{\mathit{\delta }}_{x_i}^{\widehat{\mathit{\sigma }}}$ and we have

$$\begin{array}{cc}\hfill D(a)& =\widehat{D}(a+\mathcal{J})=\underset{i}{lim}(\widehat{\mathit{\sigma }}(a)·x_i-x_i·\widehat{\mathit{\sigma }}(a))\hfill \\ \hfill & =\underset{i}{lim}(\mathit{\sigma }(a)+\mathcal{J})·x_i-x_i·(\mathit{\sigma }(a)+\mathcal{J})\hfill \\ \hfill & =\underset{i}{lim}\mathit{\sigma }(a)x_i-x_i\mathit{\sigma }(a),(a\in \mathcal{A}).\hfill \end{array}$$

Which shows that $D$ is $\mathit{\sigma }$-approximately inner and therefore $\mathcal{A}$ is $\mathit{\sigma }\text{-(AMA)}$. $\square$

In [4], section 3, we stated some properties of $\mathit{\sigma }$-approximate contractibility when $\mathcal{A}$ has an identity and considered some corollaries when $\mathit{\sigma }(\mathcal{A})$ is dense in $\mathcal{A}$. In proof of the following proposition we use those results. Recall that, the Banach algebra $\mathfrak{A}$ acts trivially on $\mathcal{A}$ from left if for each $\mathit{\alpha }\in \mathfrak{A}$ and $a$$\in \mathcal{A}$, $\mathit{\alpha }·a=f(\mathit{\alpha })a$, where $f$ is a continuous linear functional on $\mathfrak{A}$.

Proposition 11

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule with trivial left actions and$\mathit{\sigma }$be as in above lemma. Suppose that$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}$.If$\frac{\mathcal{A}}{\mathcal{J}}$has an identity and$\overline{\widehat{\mathit{\sigma }}(\frac{\mathcal{A}}{\mathcal{J}})}=\frac{\mathcal{A}}{\mathcal{J}}$,then$\frac{\mathcal{A}}{\mathcal{J}}$is$\widehat{\mathit{\sigma }}$-approximately amenable.

Proof

By [4, Corollary 3.3.], we can assume that $\mathcal{X}$ is a $\mathit{\sigma }$-unital Banach $\frac{\mathcal{A}}{\mathcal{J}}$-bimodule. Let $e+\mathcal{J}$ be the identity in $\frac{\mathcal{A}}{\mathcal{J}}.$ So $\widehat{\mathit{\sigma }}(e+\mathcal{J})$ is a unit for $\widehat{\mathit{\sigma }}(\frac{\mathcal{A}}{\mathcal{J}})$. Thus by density of $\widehat{\mathit{\sigma }}(\frac{\mathcal{A}}{\mathcal{J}})$ in $\frac{\mathcal{A}}{\mathcal{J}}$, we see that $\widehat{\mathit{\sigma }}(e+\mathcal{J})=e+\mathcal{J}$. Now let $\widehat{D}:\frac{\mathcal{A}}{\mathcal{J}}\to {\mathcal{X}}^{\ast }$ be a $\widehat{\mathit{\sigma }}$-derivation. By [4, Lemma 3.7], $\widehat{D}(e+\mathcal{J})=0$. Now similar to [3, Proposition 2.7], we can see $\mathcal{X}$ as a commutative Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule and $D=\widehat{D}\circ \mathit{\pi }:{\mathcal{A}\to \mathcal{X}}^{\ast }$ is a $\mathit{\sigma }$-module derivation, where $\mathit{\pi }:\mathcal{A}\to \frac{\mathcal{A}}{\mathcal{J}}$ is the natural $\mathfrak{A}$-module map. Since $\mathcal{A}$ is $\mathit{\sigma }\text{-(AMA)}$, there exists a net ${(x_i)}_{i\in \mathfrak{I}}\in X^{\ast }$ such that $D={lim}_i{\mathit{\delta }}_{x_i}^{\widehat{\mathit{\sigma }}}$ and we have,

$$\begin{array}{cc}\hfill \widehat{D}(a+\mathcal{J})& =\underset{i}{lim}(\mathit{\sigma }(a)+\mathcal{J})·x_i-x_i·(\mathit{\sigma }(a)+\mathcal{J})\hfill \\ \hfill & =\underset{i}{lim}\widehat{\mathit{\sigma }}(a+\mathcal{J})·x_i-x_i·\widehat{\mathit{\sigma }}(a+\mathcal{J}),(a\in \mathcal{A}).\hfill \end{array}$$

Which means that $\widehat{D}$ is $\widehat{\mathit{\sigma }}$-approximately inner and $\widehat{\mathit{\sigma }}$-approximately amenable. $\square$

Let $\mathfrak{A}$ be a non-unital Banach algebra. Then, ${\mathfrak{A}}^{\#}=\mathfrak{A}\oplus \mathbb{C}$, the unitization of $\mathfrak{A}$, is a unital Banach algebra which contains $\mathfrak{A}$ as a closed ideal. Let $\mathcal{A}$ be a Banach algebra and a Banach $\mathfrak{A}$-bimodule with compatible actions. Then, $\mathcal{A}$ is a Banach algebra and a Banach ${\mathfrak{A}}^{\#}$-bimodule with the following actions:

$$\begin{array}{c}\hfill (\mathit{\alpha },\mathit{\lambda })a=\mathit{\alpha }a+\mathit{\lambda }a\text{and}a(\mathit{\alpha },\mathit{\lambda })=a\mathit{\alpha }+a\mathit{\lambda },(\mathit{\alpha }\in \mathfrak{A},\mathit{\lambda }\in \mathbb{C},\mathfrak{a}\in \mathcal{A}).\end{array}$$

Let $\mathcal{A}$ be a Banach algebra and a Banach $\mathfrak{A}$-bimodule with compatible actions and let ${\mathcal{A}}^{\#}=\mathcal{A}\oplus {\mathfrak{A}}^{\#}$. Then $({\mathcal{A}}^{\#},·)$ is a Banach algebra, where the multiplication $·$ is defined by $(a,\mathfrak{u})·(b,\mathfrak{v})=(ab+a\mathfrak{v}+\mathfrak{u}b,\mathfrak{uv})$, $(a,b\in \mathcal{A},\mathfrak{u},\mathfrak{v}\in {\mathfrak{A}}^{\#})$. Also ${\mathcal{A}}^{\#}$ is a Banach ${\mathfrak{A}}^{\#}$-bimodule with the following module actions:

$$\begin{array}{c}\hfill (a,\mathfrak{u})·\mathfrak{v}=(a·\mathfrak{v},\mathfrak{uv})\text{and}\mathfrak{v}·(a,\mathfrak{u})=(\mathfrak{v}·a,\mathfrak{vu})(a\in \mathcal{A},\mathfrak{u},\mathfrak{v}\in {\mathfrak{A}}^{\#}).\end{array}$$

So ${\mathcal{A}}^{\#}$ is a unital Banach ${\mathfrak{A}}^{\#}$-bimodule with compatible actions.

A similar result of [5, Theorem 3.1], for approximate module amenability, is as follows:

Proposition 12

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule,$\mathit{\sigma }$$\in {\text{Hom}}_{\mathfrak{A}}(\mathcal{A})$.Then$\widehat{\mathit{\sigma }}(a,\mathfrak{u})=\mathit{\sigma }(a)\oplus \mathfrak{u}$, $(a\in \mathcal{A},{\mathfrak{u}\in \mathfrak{A}}^{\#})$is in${\text{Hom}}_{\mathfrak{A}}^{\#}({\mathcal{A}}^{\#})$and the following are equivalent;

• (i) $\mathcal{A}$ is $\mathit{\sigma }\text{-(AMA)}$ as an ${\mathfrak{A}}^{\#}$-bimodule.

• (ii) ${\mathcal{A}}^{\#}$ is $\widehat{\mathit{\sigma }}\text{-(AMA)}$ as an ${\mathfrak{A}}^{\#}$-bimodule.

Proof

It is easy to see that $\widehat{\mathit{\sigma }}\in {\text{Hom}}_{\mathfrak{A}}^{\#}({\mathcal{A}}^{\#})$.

$i\Rightarrow 2.$ Let $\mathcal{X}$ be a commutative Banach ${\mathcal{A}}^{\#}-{\mathfrak{A}}^{\#}$-bimodule and $\widehat{D}:{\mathcal{A}}^{\#}{\to \mathcal{X}}^{\ast }$ be a $\widehat{\mathit{\sigma }}$-module derivation. By [4, Lemma 3.1], $\widehat{D}(1)=0$. So $D=\widehat{D}{\mid }_{\mathcal{A}}:{\mathcal{A}\to \mathcal{X}}^{\ast }$ is a $\mathit{\sigma }$-module derivation. Thus by the hypothesis, there exists a net ${(x_i)}_{i\in \mathfrak{I}}\in X^{\ast }$ such that $D={lim}_i{\mathit{\delta }}_{x_i}^{\mathit{\sigma }}$. Note that $\mathcal{X}$ is a commutative Banach $\mathcal{A}$-${\mathfrak{A}}^{\#}$-module and $\widehat{D}(a,0)={lim}_i\widehat{\mathit{\sigma }}(a,0)x_i-x_i\widehat{\mathit{\sigma }}(a,0)$, $(a\in \mathcal{A})$. Also we have

$$\begin{array}{c}\hfill \widehat{D}(a,\mathfrak{u})=\widehat{D}((a,0)+(0,\mathfrak{u}))=\widehat{D}(a,0)+\widehat{D}(0,\mathfrak{u})=\widehat{D}(a,0),((a,\mathfrak{u})\in {\mathcal{A}}^{\#}).\end{array}$$

Thus, $\widehat{D}$ is $\widehat{\mathit{\sigma }}$-approximately inner and therefore ${\mathcal{A}}^{\#}$ is $\widehat{\mathit{\sigma }}\text{-(AMA)}.$

$ii\Rightarrow i.$ Let $\mathcal{X}$ be a commutative Banach $\mathcal{A}$-${\mathfrak{A}}^{\#}$-bimodule and $D:\mathcal{A}\to {\mathcal{X}}^{\ast }$ be a $\mathit{\sigma }$-module derivation. Define $\widehat{D}:{\mathcal{A}}^{\#}\to {\mathcal{X}}^{\ast }$ by $\widehat{D}(a,\mathfrak{u})=D(a)$, $((a,\mathfrak{u})\in {\mathcal{A}}^{\#})$. Thus $\widehat{D}$ is $\widehat{\mathit{\sigma }}$-${\mathfrak{A}}^{\#}$-module derivation, because,

$$\begin{array}{cc}\hfill \widehat{D}((a,\mathfrak{u})(b,\mathfrak{v}))& =\widehat{D}((ab+a\mathfrak{v}+\mathfrak{u}b),\mathfrak{uv})\hfill \\ \hfill & =D(ab+a\mathfrak{v}+\mathfrak{u}b)\hfill \\ \hfill & =D(ab)++D(a)\mathfrak{v}+\mathfrak{u}D(b)\hfill \\ \hfill & =\mathit{\sigma }(a)D(b)+D(a)\mathit{\sigma }(b)+D(a)\mathfrak{v}+\mathfrak{u}D(b)\hfill \\ \hfill & =(\mathit{\sigma }(a)+\mathfrak{u})D(b)+D(a)(\mathit{\sigma }(b)+\mathfrak{v})\hfill \\ \hfill & =\widehat{\mathit{\sigma }}(a,\mathfrak{u})D(b)+D(a)\widehat{\mathit{\sigma }}(b,\mathfrak{v})\hfill \\ \hfill & =\widehat{\mathit{\sigma }}(a,\mathfrak{u})\widehat{D}(b,\mathfrak{v})+\widehat{D}(a,\mathfrak{u})\widehat{\mathit{\sigma }}(b,\mathfrak{v}),(a,b\in \mathcal{A},\mathfrak{u},\mathfrak{v}\in {\mathfrak{A}}^{\#}).\hfill \end{array}$$

and by (ii) is a module $\widehat{D}$-approximately inner. Therefore, $D$ is module approximately inner. So $\mathcal{A}$ is $\mathit{\sigma }$-$(AMA)$. $\square$