# Representations of compact groups on hilbert spaces

Let $G$ be a compact group. I know that every finite dimensional representation of $G$ is unitarizable (start with an artbitrary inner product and average it along $G$).

Now, let $\mathcal{H}$ be a Hilbert space on which $G$ acts continuously and linearly. The averaging process still seems to work, but does the new inner product define the same topology on $\mathcal{H}$ as the old one?

What I'd really like to know is "is every representation of a compact group on a Hilbert space unitarizable", and the above would answer this question.

Let $\langle\cdot,\cdot\rangle$ be the orginal scalar product on $\mathcal H$. So, you have defined a new scalar product by$$(v,w)=\frac1{|G|}\sum_{g\in G}\bigl\langle g.v,g.w\bigr\rangle,$$then this scaler product induces a norm $\|\cdot\|_G$ defined by$$\|v\|_G=\sqrt{\frac1{|G|}\sum_{g\in G}\|g.v\|^2},$$where $\|\cdot\|$ is the norm induced by $\langle\cdot,\cdot\rangle$. So, obviously$$(\forall v\in V):\|v\|_G\geqslant\sqrt{\frac{\|v\|^2}{|G|}}=\frac{\|v\|}{\sqrt{|G|}}.$$On the other hand, if, for each $g\in G$, $\|g\|$ is the norm of the linear map $v\mapsto g.v$, then$$(\forall v\in V):\|v\|_G\leqslant\sqrt{\frac1{|G|}\sum_{g\in G}\|g\|^2.\|v\|^2}=\sqrt{\frac1{|G|}\sum_{g\in G}\|g\|^2}\,\|v\|$$So, the norms $\|\cdot\|_G$ and $\|\cdot\|$ are equivalent and therefore they induce the same topology.