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.

  • $\begingroup$ I haven't read it all yet, but just to make sure, those sums are actually integrals, right? $\endgroup$ – Jo Lasker Jun 13 '17 at 7:40
  • $\begingroup$ @JoLasker Sure. I thought that you had asked a question about finite groups. But all that I wrote remains valid if you use an integral. Sorry about my confusion. $\endgroup$ – José Carlos Santos Jun 13 '17 at 7:42
  • $\begingroup$ Thanks. Your answer is very clear. Your proof also shows that a sequence is Cauchy w.r.t to new norm if and only if it is Cauchy w.r.t. the original norm. So, together with the fact that they give the same topology, this means that if the original norm was complete, so is the new one, right? $\endgroup$ – Jo Lasker Jun 13 '17 at 7:47
  • 1
    $\begingroup$ @JoLasker Right. $\endgroup$ – José Carlos Santos Jun 13 '17 at 7:47

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