## 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:

$\begin{array}{c}\hfill \mathit{\alpha }·\left(ab\right)=\left(\mathit{\alpha }·a\right)b\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}\left(ab\right)·\mathit{\alpha }=a\left(b·\mathit{\alpha }\right),\end{array}$

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:

$\begin{array}{c}\hfill \mathit{\alpha }·\left(a·x\right)=\left(\mathit{\alpha }·a\right)·x\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}a·\left(x·\mathit{\alpha }\right)=\left(a·x\right)·\mathit{\alpha },\end{array}$

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, $\mathit{\alpha }·x=x·\mathit{\alpha }$, $\left(\mathit{\alpha }\in \mathfrak{A}$, $x\in \mathcal{X}\right)$, 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 $a·\left(\mathit{\alpha }·b\right)=\left(a·\mathit{\alpha }\right)·b$, $\left(\mathit{\alpha }\in \mathfrak{A},\phantom{\rule{1.em}{0ex}}a,b\in \mathcal{A}\right)$ [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 $T\left(a±b\right)=T\left(a\right)±T\left(b\right)$ and $T\left(\mathit{\alpha }·a\right)=\mathit{\alpha }·T\left(a\right)$ and $T\left(a·\mathit{\alpha }\right)=T\left(a\right)·\mathit{\alpha }$, for each $\mathit{\alpha }\in \mathfrak{A},\phantom{\rule{0.166667em}{0ex}}a,b\in \mathcal{A}$. The space of all $\mathfrak{A}$-bimodule maps $T:\mathcal{A}\to \mathcal{X}$ such that $T\left(ab\right)=T\left(a\right)T\left(b\right)$, $\left(a,b\in \mathcal{A}\right)$, is denoted by ${\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A},\mathcal{X}\right).$ Also we denote ${\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A},\mathcal{A}\right)$, by ${\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A}\right).$

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

$\begin{array}{c}\hfill D\left(ab\right)=D\left(a\right)·b+a·D\left(b\right),\phantom{\rule{1.em}{0ex}}\left(a,b\in \mathcal{A}\right).\end{array}$

$D$ 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

$\begin{array}{c}\hfill {\mathit{\delta }}_{x}\left(a\right)=a·x-x·a,\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A}\right),\end{array}$

which is called an inner module derivations.

Let $\mathcal{A}$ be a Banach $\mathfrak{A}$-bimodule and $\mathit{\sigma }$$\in {\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A}\right).$ 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

$\begin{array}{c}\hfill D\left(ab\right)=\mathit{\sigma }\left(a\right)·D\left(b\right)+D\left(a\right)·\mathit{\sigma }\left(b\right),\phantom{\rule{1.em}{0ex}}\left(a,b\in \mathcal{A}\right).\end{array}$

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

$\begin{array}{c}\hfill {\mathit{\delta }}_{x}^{\mathit{\sigma }}:\mathcal{A}⟶\mathcal{X},\phantom{\rule{1.em}{0ex}}{\mathit{\delta }}_{x}^{\mathit{\sigma }}\left(a\right)=\mathit{\sigma }\left(a\right)·x-x·\mathit{\sigma }\left(a\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A}\right),\end{array}$

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 $\mathit{\sigma }\phantom{\rule{0.166667em}{0ex}}\in Ho{m}_{\mathfrak{A}}\left(\mathcal{A}\right)$. We say that $\mathcal{A}$ is a $\mathit{\sigma }$-approximately module amenable $\left(\mathit{\sigma }\text{-(AMA)}\right)$, 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 ${\left({x}_{i}\right)}_{i\in \mathfrak{I}}$$\in {\mathcal{X}}^{\ast }$ such that $D\left(a\right)={lim}_{i}$${\mathit{\delta }}_{{x}_{i}}^{\mathit{\sigma }}\left(a\right)={lim}_{i}\mathit{\sigma }\left(a\right){x}_{i}-{x}_{i}\mathit{\sigma }\left(a\right)$, $\left(a\in \mathcal{A}\right)$. Also we say that $\mathcal{A}$ is a $\mathit{\sigma }$-approximately module contractible $\left(\mathit{\sigma }-\left(AMC\right)\right)$, 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$\mathit{\sigma }\phantom{\rule{0.166667em}{0ex}}\in Ho{m}_{\mathfrak{A}}\left(\mathcal{A}\right)$.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}}\left(\mathcal{A}\right)$.If$\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$\mathit{\sigma }\phantom{\rule{0.166667em}{0ex}}\in Ho{m}_{\mathfrak{A}}\left(\mathcal{A}\right)$.If$\mathcal{A}$is$\mathit{\sigma }\text{-(AMA)}$,then$\mathcal{A}$is$\left(\mathit{\lambda }\circ \mathit{\sigma },\mathit{\mu }\circ \mathit{\sigma }\right)\text{-(AMA)}$,for each$\mathit{\lambda },\mathit{\mu }\in {\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A}\right)$.

### Proof

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

$\begin{array}{c}\hfill a\ast x=\mathit{\lambda }\left(a\right)·x\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}x\ast a=x·\mathit{\mu }\left(a\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A},\phantom{\rule{0.166667em}{0ex}}x\in \mathcal{X}\right).\end{array}$

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

$\begin{array}{cc}\hfill D\left(ab\right)& =\left(\mathit{\lambda }\circ \mathit{\sigma }\right)\left(a\right)·D\left(b\right)+D\left(a\right)·\left(\mathit{\mu }\circ \mathit{\sigma }\right)\left(b\right)\hfill \\ \hfill & =\mathit{\sigma }\left(a\right)\ast D\left(b\right)+D\left(a\right)\ast \mathit{\sigma }\left(b\right),\phantom{\rule{1.em}{0ex}}\left(a,b\in \mathcal{A}\right).\hfill \end{array}$

Thus, $D$ is a $\mathit{\sigma }$-module derivation. So there exists a net $\left({x}_{i}\right)\in {\mathcal{X}}^{\ast }$ such that $D\left(a\right)={lim}_{i}\phantom{\rule{0.166667em}{0ex}}{\mathit{\delta }}_{{x}_{i}}^{\mathit{\sigma }}\left(a\right)$, $\left(a\in \mathcal{A}\right)$. So we have

$\begin{array}{cc}\hfill D\left(a\right)& =\underset{i}{lim}\left(\mathit{\sigma }\left(a\right)\ast {x}_{i}-{x}_{i}\ast \mathit{\sigma }\left(a\right)\right)\hfill \\ \hfill & =\underset{i}{lim}\left(\left(\mathit{\lambda }\circ \mathit{\sigma }\right)\left(a\right)·{x}_{i}-{x}_{i}·\left(\mathit{\mu }\circ \mathit{\sigma }\right)\left(a\right)\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A}\right).\hfill \end{array}$

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

### Corollary 5

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

### Proposition 6

Let$\mathcal{A}$be a Banach$\mathfrak{A}$-bimodule and$\mathit{\sigma }$$\in {\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A}\right)$.Suppose that$\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 $\stackrel{~}{D}=D\circ \mathit{\sigma }$ is a $\mathit{\sigma }$-module derivation, because, for each $a,b\in \mathcal{A}$ and $\mathit{\alpha }\in \mathfrak{A}$ we have

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

and

$\begin{array}{c}\hfill \stackrel{~}{D}\left(\mathit{\alpha }a\right)=D\left(\mathit{\sigma }\left(\mathit{\alpha }a\right)\right)=D\left(\mathit{\alpha }\mathit{\sigma }\left(a\right)\right)=\mathit{\alpha }D\left(\mathit{\sigma }\left(a\right)\right).\end{array}$

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

$\begin{array}{c}\hfill D\left(b\right)=D\left(\mathit{\sigma }\left(a\right)\right)=\stackrel{~}{D}\left(a\right)=\underset{i}{lim}\left(\mathit{\sigma }\left(a\right){x}_{i}-{x}_{i}\mathit{\sigma }\left(a\right)\right)=\underset{i}{lim}\left(b{x}_{i}-{x}_{i}b\right),\phantom{\rule{1.em}{0ex}}\left(b\in \mathcal{A}\right).\end{array}$

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 }$$\in Ho{m}_{\mathfrak{A}}\left(\mathcal{A}\right)$and$\mathit{\tau }$$\in {\text{Hom}}_{\mathfrak{A}}\left(\mathcal{B}\right)$.Suppose that$\mathit{\phi }\in {\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A},\mathcal{B}\right)$be a surjective map such that$\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, $\left(\mathcal{X},\ast \right)$ can be considered as a Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule by the following module actions:

$\begin{array}{c}\hfill a\ast x=\mathit{\phi }\left(a\right)·x\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}x\ast a=x·\mathit{\phi }\left(a\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A},\phantom{\rule{0.166667em}{0ex}}x\phantom{\rule{0.166667em}{0ex}}\in X\right).\end{array}$

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

$\begin{array}{cc}\hfill \stackrel{~}{D}\left(ab\right)& =D\left(\mathit{\phi }\left(a\right)\mathit{\phi }\left(b\right)\right)\hfill \\ \hfill & =D\left(\mathit{\phi }\left(a\right)\right)\mathit{\tau }\left(\mathit{\phi }\left(b\right)\right)+\mathit{\tau }\left(\mathit{\phi }\left(a\right)\right)D\left(\mathit{\phi }\left(b\right)\right)\hfill \\ \hfill & =\stackrel{~}{D}\left(a\right)\mathit{\phi }\left(\mathit{\sigma }\left(b\right)\right)+\mathit{\phi }\left(\mathit{\sigma }\left(a\right)\right)\stackrel{~}{D}\left(b\right)\hfill \\ \hfill & =\stackrel{~}{D}\left(a\right)\ast \mathit{\sigma }\left(b\right)+\mathit{\sigma }\left(a\right)\ast \stackrel{~}{D}\left(b\right),\phantom{\rule{1.em}{0ex}}\left(a,b\in \mathcal{A}\right).\hfill \end{array}$

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

$\begin{array}{cc}\hfill \stackrel{~}{D}\left(a\right)& =\underset{i}{lim}\mathit{\sigma }\left(a\right)\ast {x}_{i}-{x}_{i}\ast \mathit{\sigma }\left(a\right)\hfill \\ \hfill & =\underset{\mathit{\alpha }}{lim}\mathit{\phi }\left(\mathit{\sigma }\left(a\right)\right)·{x}_{i}-{x}_{i}·\mathit{\phi }\left(\mathit{\sigma }\left(a\right)\right)\hfill \\ \hfill & =\underset{\mathit{\alpha }}{lim}\mathit{\tau }\left(\mathit{\phi }\left(a\right)\right)·{x}_{i}-{x}_{i}·\mathit{\tau }\left(\mathit{\phi }\left(a\right)\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A}\right).\hfill \end{array}$

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

### Proposition 8

Suppose that$\mathcal{A}$and$\mathcal{B}$are Banach$\mathfrak{A}$-modules and$\mathit{\phi }\in Ho{m}_{\mathfrak{A}}\left(\mathcal{A},\mathcal{B}\right)$be a surjective map. If$\mathcal{A}$is (AMA), then$\mathcal{B}$is$\mathit{\sigma }\text{-(AMA)}$,for each$\mathit{\sigma }\phantom{\rule{0.166667em}{0ex}}\in {\text{Hom}}_{\mathfrak{A}}\left(\mathcal{B}\right)$.

### Proof

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

$\begin{array}{c}\hfill a\ast x=\mathit{\sigma }\left(\mathit{\phi }\left(a\right)\right)·x\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}x\ast a=x·\mathit{\sigma }\left(\mathit{\phi }\left(a\right)\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A},x\in \mathcal{X}\right).\end{array}$

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

$\begin{array}{c}\hfill \stackrel{~}{D}\left(\mathit{\alpha }a\right)=D\left(\mathit{\phi }\left(\mathit{\alpha }a\right)\right)=D\left(\mathit{\alpha }\mathit{\phi }\left(a\right)\right)=\mathit{\alpha }D\left(\mathit{\phi }\left(a\right)\right),\end{array}$

and

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

So there exists a net ${\left({x}_{i}\right)}_{i\in \mathfrak{I}}\in {X}^{\ast }$ such that $\stackrel{~}{D}={lim}_{i}{\mathit{\delta }}_{{x}_{i}}$ and we have

$\begin{array}{cc}\hfill \stackrel{~}{D}\left(a\right)& =\underset{i}{lim}{\mathit{\delta }}_{{x}_{i}}\left(a\right)\hfill \\ \hfill & =\underset{i}{lim}\left(a\ast {x}_{i}-{x}_{i}\ast a\right)\hfill \\ \hfill & =\underset{i}{lim}\mathit{\sigma }\left(\mathit{\phi }\left(a\right)\right)·{x}_{i}-{x}_{i}·\mathit{\sigma }\left(\mathit{\phi }\left(a\right)\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A}\right).\hfill \end{array}$

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

$\begin{array}{c}\hfill D\left(b\right)=D\left(\mathit{\phi }\left(a\right)\right)=\stackrel{~}{D}\left(a\right)=\underset{i}{lim}\mathit{\sigma }\left(\mathit{\phi }\left(a\right)\right)·{x}_{i}-{x}_{i}·\mathit{\sigma }\left(\mathit{\phi }\left(a\right)\right)=\underset{i}{lim}\mathit{\sigma }\left(b\right)·{x}_{i}-{x}_{i}·\mathit{\sigma }\left(b\right).\end{array}$

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 $\left(\mathit{\alpha }·a\right)b-a\left(b·\mathit{\alpha }\right)$, 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 $\mathit{\sigma }\in Ho{m}_{\mathfrak{A}}\left(\mathcal{A}\right)$ such that $\mathit{\sigma }\left(\mathcal{J}\right)\subseteq \mathcal{J}$. Then $\stackrel{^}{\mathit{\sigma }}:\frac{\mathcal{A}}{\mathcal{J}}\to \frac{\mathcal{A}}{\mathcal{J}}$ by $\stackrel{^}{\mathit{\sigma }}\left(a+\mathcal{J}\right)=\mathit{\sigma }\left(a\right)+\mathcal{J}$ 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$\stackrel{^}{\mathit{\sigma }}$-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;

$\begin{array}{c}\hfill \left(a+\mathcal{J}\right)·x=ax\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}x·\left(a+\mathcal{J}\right)=xa,\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A},x\phantom{\rule{0.166667em}{0ex}}\in \mathcal{X}\right).\end{array}$

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

$\begin{array}{cc}\hfill D\left(a\right)& =\stackrel{^}{D}\left(a+\mathcal{J}\right)=\underset{i}{lim}\left(\stackrel{^}{\mathit{\sigma }}\left(a\right)·{x}_{i}-{x}_{i}·\stackrel{^}{\mathit{\sigma }}\left(a\right)\right)\hfill \\ \hfill & =\underset{i}{lim}\left(\mathit{\sigma }\left(a\right)+\mathcal{J}\right)·{x}_{i}-{x}_{i}·\left(\mathit{\sigma }\left(a\right)+\mathcal{J}\right)\hfill \\ \hfill & =\underset{i}{lim}\mathit{\sigma }\left(a\right){x}_{i}-{x}_{i}\mathit{\sigma }\left(a\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A}\right).\hfill \end{array}$

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 $\mathit{\sigma }\left(\mathcal{A}\right)$ 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}$, $\mathit{\alpha }·a=f\left(\mathit{\alpha }\right)a$, 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$\overline{\stackrel{^}{\mathit{\sigma }}\left(\frac{\mathcal{A}}{\mathcal{J}}\right)}=\frac{\mathcal{A}}{\mathcal{J}}$,then$\frac{\mathcal{A}}{\mathcal{J}}$is$\stackrel{^}{\mathit{\sigma }}$-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 $\stackrel{^}{\mathit{\sigma }}\left(e+\mathcal{J}\right)$ is a unit for $\stackrel{^}{\mathit{\sigma }}\left(\frac{\mathcal{A}}{\mathcal{J}}\right)$. Thus by density of $\stackrel{^}{\mathit{\sigma }}\left(\frac{\mathcal{A}}{\mathcal{J}}\right)$ in $\frac{\mathcal{A}}{\mathcal{J}}$, we see that $\stackrel{^}{\mathit{\sigma }}\left(e+\mathcal{J}\right)=e+\mathcal{J}$. Now let $\stackrel{^}{D}:\frac{\mathcal{A}}{\mathcal{J}}\to {\mathcal{X}}^{\ast }$ be a $\stackrel{^}{\mathit{\sigma }}$-derivation. By [4, Lemma 3.7], $\stackrel{^}{D}\left(e+\mathcal{J}\right)=0$. Now similar to [3, Proposition 2.7], we can see $\mathcal{X}$ as a commutative Banach $\mathcal{A}$-$\mathfrak{A}$-bimodule and $D=\stackrel{^}{D}\circ \mathit{\pi }:{\mathcal{A}\to \mathcal{X}}^{\ast }$ 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 ${\left({x}_{i}\right)}_{i\in \mathfrak{I}}\in {X}^{\ast }$ such that $D={lim}_{i}{\mathit{\delta }}_{{x}_{i}}^{\stackrel{^}{\mathit{\sigma }}}$ and we have,

$\begin{array}{cc}\hfill \stackrel{^}{D}\left(a+\mathcal{J}\right)& =\underset{i}{lim}\left(\mathit{\sigma }\left(a\right)+\mathcal{J}\right)·{x}_{i}-{x}_{i}·\left(\mathit{\sigma }\left(a\right)+\mathcal{J}\right)\hfill \\ \hfill & =\underset{i}{lim}\stackrel{^}{\mathit{\sigma }}\left(a+\mathcal{J}\right)·{x}_{i}-{x}_{i}·\stackrel{^}{\mathit{\sigma }}\left(a+\mathcal{J}\right),\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A}\right).\hfill \end{array}$

Which means that $\stackrel{^}{D}$ is $\stackrel{^}{\mathit{\sigma }}$-approximately inner and $\stackrel{^}{\mathit{\sigma }}$-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:

$\begin{array}{c}\hfill \left(\mathit{\alpha },\mathit{\lambda }\right)a=\mathit{\alpha }a+\mathit{\lambda }a\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}a\left(\mathit{\alpha },\mathit{\lambda }\right)=a\mathit{\alpha }+a\mathit{\lambda },\phantom{\rule{1.em}{0ex}}\left(\mathit{\alpha }\in \mathfrak{A},\mathit{\lambda }\in \mathbb{C},\mathfrak{a}\in \mathcal{A}\right).\end{array}$

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

$\begin{array}{c}\hfill \left(a,\mathfrak{u}\right)·\mathfrak{v}=\left(a·\mathfrak{v},\mathfrak{uv}\right)\phantom{\rule{1.em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathfrak{v}·\left(a,\mathfrak{u}\right)=\left(\mathfrak{v}·a,\mathfrak{vu}\right)\phantom{\rule{1.em}{0ex}}\left(a\in \mathcal{A},\mathfrak{u},\mathfrak{v}\in {\mathfrak{A}}^{#}\right).\end{array}$

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$\phantom{\rule{4pt}{0ex}}\mathfrak{A}$-bimodule,$\mathit{\sigma }$$\in {\text{Hom}}_{\mathfrak{A}}\left(\mathcal{A}\right)$.Then$\stackrel{^}{\mathit{\sigma }}\left(a,\mathfrak{u}\right)=\mathit{\sigma }\left(a\right)\oplus \mathfrak{u}$, $\left(a\in \mathcal{A},{\mathfrak{u}\in \mathfrak{A}}^{#}\right)$is in${\text{Hom}}_{\mathfrak{A}}^{#}\left({\mathcal{A}}^{#}\right)$and the following are equivalent;

1. (i)

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

2. (ii)

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

### Proof

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

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

$\begin{array}{c}\hfill \stackrel{^}{D}\left(a,\mathfrak{u}\right)=\stackrel{^}{D}\left(\left(a,0\right)+\left(0,\mathfrak{u}\right)\right)=\stackrel{^}{D}\left(a,0\right)+\stackrel{^}{D}\left(0,\mathfrak{u}\right)=\stackrel{^}{D}\left(a,0\right),\phantom{\rule{1.em}{0ex}}\left(\left(a,\mathfrak{u}\right)\in {\mathcal{A}}^{#}\right).\end{array}$

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

$ii⇒i.$ Let $\mathcal{X}$ be a commutative Banach $\mathcal{A}$-${\mathfrak{A}}^{#}$-bimodule and $D:\mathcal{A}\phantom{\rule{0.166667em}{0ex}}\to \phantom{\rule{0.166667em}{0ex}}{\mathcal{X}}^{\ast }$ be a $\mathit{\sigma }$-module derivation. Define $\stackrel{^}{D}:{\mathcal{A}}^{#}\to {\mathcal{X}}^{\ast }$ by $\stackrel{^}{D}\left(a,\mathfrak{u}\right)=D\left(a\right)$, $\left(\left(a,\mathfrak{u}\right)\in {\mathcal{A}}^{#}\right)$. Thus $\stackrel{^}{D}$ is $\stackrel{^}{\mathit{\sigma }}$-${\mathfrak{A}}^{#}$-module derivation, because,

$\begin{array}{cc}\hfill \stackrel{^}{D}\left(\left(a,\mathfrak{u}\right)\left(b,\mathfrak{v}\right)\right)& =\stackrel{^}{D}\left(\left(ab+a\mathfrak{v}+\mathfrak{u}b\right),\mathfrak{uv}\right)\hfill \\ \hfill & =D\left(ab+a\mathfrak{v}+\mathfrak{u}b\right)\hfill \\ \hfill & =D\left(ab\right)++D\left(a\right)\mathfrak{v}+\mathfrak{u}D\left(b\right)\hfill \\ \hfill & =\mathit{\sigma }\left(a\right)D\left(b\right)+D\left(a\right)\mathit{\sigma }\left(b\right)+D\left(a\right)\mathfrak{v}+\mathfrak{u}D\left(b\right)\hfill \\ \hfill & =\left(\mathit{\sigma }\left(a\right)+\mathfrak{u}\right)D\left(b\right)+D\left(a\right)\left(\mathit{\sigma }\left(b\right)+\mathfrak{v}\right)\hfill \\ \hfill & =\stackrel{^}{\mathit{\sigma }}\left(a,\mathfrak{u}\right)D\left(b\right)+D\left(a\right)\stackrel{^}{\mathit{\sigma }}\left(b,\mathfrak{v}\right)\hfill \\ \hfill & =\stackrel{^}{\mathit{\sigma }}\left(a,\mathfrak{u}\right)\stackrel{^}{D}\left(b,\mathfrak{v}\right)+\stackrel{^}{D}\left(a,\mathfrak{u}\right)\stackrel{^}{\mathit{\sigma }}\left(b,\mathfrak{v}\right),\phantom{\rule{1.em}{0ex}}\left(a,b\in \mathcal{A},\mathfrak{u},\mathfrak{v}\in {\mathfrak{A}}^{#}\right).\hfill \end{array}$

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