A question on Banach space involving an invariant mean on $\mathbb Z$ Let $E$ be a Banach space and suppose $T\in Aut(E)$ satisfies the condition $\|T^n\|\leq C$ for all $n\in \mathbb Z$ for some fixed $C\in \mathbb R$. I have to show that there is an equivalent norm on $E$, say $|\ |$, such that $T$ is isometric with respect to $|\ |$.
As a hint, it is given that, for $x\in E $, define $f(n)=\|T^n(x)\|$. Then $f\in l^{\infty}(\mathbb Z)$ and then it is advised to use invariant mean on $\mathbb Z$.
I am not sure about how to proceed from there. Any help would be appreciated. Thanks in advance.
If you want to recall the definition of invariant mean, you can see here

It is known that $\mathbb Z$(and all abelian semigroups) are amenable. Let $\mu$ be an invariant mean on $\ell^\infty(\mathbb Z)$. $|x|=\mu(f)$ defines a norm on $E$ (why?). This norm is equivalent to $\|.\|$. In fact
$$|x|=\mu(f)\leq \|\mu\|\|f\|\leq C\|x\|.$$
Note that $\|\mu\|=1$ by definition of a mean. At the other hand, for each $n\in \mathbb Z$,
$$\|x\|=\|T^{-n}T^nx\|\leq C\|T^nx\|.$$
By positivity of means, we have
$$\|x\|\leq C\mu(\|T^nx\|)=C\mu(f)=C|x|.$$
Also $T$ is isometric with respect to $|.|$, since
$$|Tx|=\mu(\|T^{n+1}x\|)=\mu(\|T^nx\|)=|x|.$$
Note that for each invariant mean $\mu$, $\mu_n(f(n))=\mu_n(f(n+1))$.
