Let $A$ be a Banach algebra with a bounded approximate identity, and let $E$ be a Banach left $A$-module. Suppose neither $A$ nor $E$ has the Schur property.


**Question:** Given a weakly null sequence $(w\_n)$ in $E$, does there exist


* a weakly null sequence $(a\_n)$ in $A$
* a bounded sequence $(x\_n)$ in $E$


such that $w\_n = a\_nx\_n$ for all $n\in\mathbb{N}$?



Do the weakly null sequences in a Banach module factor?
No; here's a counter-example. Let $E$ be some Banach space without the [Schur Property](https://en.wikipedia.org/wiki/Schur%27s_property), give $E$ the zero product, and let $A=E\oplus\mathbb C$ be the unitisation, with natural character $\epsilon:A\rightarrow\mathbb C; a=(x,\lambda)\mapsto\lambda$. Use the character to turn $E$ into a left $A$-module:
$$ a\cdot x = \epsilon(a) x \qquad (a\in A, x\in E). $$
If $(a\_n)$ is weakly-null in $A$ then $\epsilon(a\_n)\rightarrow 0$ and so $(a\_n\cdot x\_n)$ is norm-null. Taking $(w\_n)$ to be any weakly-null sequence in $E$ which is not norm-null (which exists as $E$ fails to have the Schur property) yields the counter-example.


