Representation in Banach space and norms 'induced' by representation By $G$ we denote some compact group, $X$ stands for some Banach space. Suppose $\pi\colon G\longrightarrow \mathrm{GL}(X)$ to be some representation in $X$. I'm trying to prove that there is an equivalent norm on $X$ such that each operator $\pi(g)$ becomes an isometry. 
I hope there is a nice way without Haar measure (I don't know how to prove its existence). Could you give me a hint?

I think the averaging technique using the Haar measure is crucial to such an argument. For the proof using Haar measure, proceed as follows:
Let $\Vert . \Vert$ be the norm on $X$. Define the new norm
$$\Vert v \Vert_{G}=\frac{1}{\vert G \vert} \int_{G}\Vert \pi(g)v \Vert\ d\mu$$
One can easily verify that this is a norm and is equivalent to $\Vert . \Vert$. 
Now since the Haar measure is left invariant, we have
$$\Vert \pi(h)v \Vert_{G}=\frac{1}{\vert G \vert}\int_{G} \Vert \pi(h)\pi(g)v \Vert \ d\mu=\frac{1}{\vert G \vert}\int_{G} \Vert \pi(hg)v \Vert \ d\mu=\frac{1}{\vert G \vert}\int_{G} \Vert \pi(u)v \Vert \ d\mu=\Vert v \Vert_{G}$$
Where in the last step $u=hg$ and we have used left invariance of Haar.
It follows that $\pi(h)$ is unitary for each $h$.
