Every norm-decreasing algebra morphism $L_1(G)\to\mathcal{B}(E)$ comes from a group representation

Question

In section 8 of this paper http://arxiv.org/abs/math/0611833v3 the author proves the following: If $E$ is a reflexive Banach space, $G$ a locally compact group and $\pi:L_1(G)\to\mathcal{B}(E)$ a norm-decreasing algebra morphism, then there exists a one-complemented subspace $F$ of $E$ and a group representation $\sigma:G\to\mathcal{B}(F)$ such that $\pi(f)$ restricted to $F$ is given by $\int f(s)\sigma(s)\,ds$.

I have two questions concerning the proof:

1) He takes an approximate identity $(e_{\alpha})$ in $L_1(G)$ of bound 1 and defines $\sigma:G\to\mathcal{B}(E)$ by \begin{equation} \langle\mu,\sigma(s)x\rangle=\lim_{\alpha}\langle\mu,\pi(s\cdot e_{\alpha})x\rangle\qquad (x\in E, \mu\in E^*) \end{equation} where $(s\cdot f)(t)=f(s^{-1}t)$.

Why does that limit exist for every $x\in E$?

2) Then there is a subspace $F$ such that, by restriction, $\sigma:G\to\mathcal{B}(F)$ becomes a group representation. Moreover, there exists a contractive projection $P:E\to F$ such that $P\pi(f)P=\pi(f)$ for all $f\in L_1(G)$.

Why does this contractive projection exist?

(He later concludes that $F=\{\pi(f)x\ :\ f\in L_1(G), x\in E\}$ thanks to the Cohen factorisation theorem)

Answer 1

1) As Yemon said, just pass to a subnet (via compactness) where it converges.

2) Follow the suggestion in the paper and use Cohen Factorisation. In the form I like, if $A$ is a Banach algebra with a bai and $E$ is a left Banach $A$-module, set $F$ to be the closed linear span of $\{a\cdot x:a\in A,x\in E\}$, then $F = \{ a\cdot x : a\in A,x\in E \}$ (no linear span or closure).

So, if $F = \{\pi(f)(x) : f\in L^1(G), x\in E\}$ then $F$ is equal to its own closed linear span, and is (clearly) equal to the collection of $x\in E$ such that $\pi(e_\alpha)x \rightarrow x$. Taka said as much as this in his comment.

Now let $P$ be a weak$^*$-limit point of the net $\pi(e_\alpha)$ in $B(E)$. Then $P(x)=x$ for $x\in F$. Also $P\pi(f) = \pi(f) = \pi(f)P$ for all $g\in L^1(G)$, and so for $y=\pi(f)(x)\in F$ we have that $P(y)=y$. Thus $P$ is a (contractive) projection onto $F$.

Now prove an identity like $(s\cdot f)g = s\cdot(fg)$ for $s\in G, f,g\in L^1(G)$. It follows that $\sigma(s)\pi(f) = \pi(s\cdot f)$. Hence also $\sigma(s)P=\sigma(s)$. Similarly, $\sigma(s)\sigma(t) = \sigma(st)$. So $\sigma$ restricts to a group representation on $F$.