Yes indeed, thanks @Richard. Do you have a good reference where the author constructs all irreducible (no more unitary) representations of $SL_2(\mathbf{R})$? I know that there is the book from Lang on $SL_2(\mathbf{R})$ but there might be better references...

In general there is no reason why it should be semi-simple but you get the idea... may be there is some kind of Jordan-Hoelder decomposition where the irreducible pieces could be weighted by a certain measure!). So it seems that it might be possible to define some kind of monoid structure on $\widehat{G}$.

Thanks Matthew for the link, I just read it. By the way, it seems that $\widehat{G}$ is more than just a set. For example if I take $V_1$ and $V_2$ two irreducible representations of $G$ then one may look at the $G$ representation $V_1\otimes_{\mathbf{C}} V_2$ and decompose it into a "direct sum" of representations of $G$.

Thanks @Dick. So I agree that the equivalence between $X$ being a proper $G$-space iff the map $G\times X\rightarrow X\times X$ is proper not only answers Q1 but generalizes it. So where can I find a proof of this equivalence?

So in order to get rid of the obvious counter-example for $\mathbf{R}$ we could require "some kind of growth condition". For example, if $\rho$ is an infinite dimensional irreducible unitary representation $\rho:G\rightarrow GL_cont(V)$ then we want something like for all $g\in G$ one has that $\lim_{n\rightarrow\infty} \langle ge_n,e_n\rangle\rightarrow 0$. But this is probably to naive...

Right, that is a good observation, so for $\mathbf{R}$, the trivial representation which is irreducible and unitary in the sense above does not occur in $L^2(\mathbf{R})$. So may be in the non-compact case, $L^2(G)$ is not the right object to look at. So what kind of object should replace $L^2(G)$?

Hi @Charles, thanks a lot for your comment. So by "occur" I simply mean that if $M$ is an irreducible unitary $\mathbf{C}$-representation of $G$ as defined above then there exists a continuous inclusion map of $G$-module $\iota:M\hookrightarrow L^2(G)$ which respects the inner products. And then because of the inner product one may define an orthogonal complement which "seems" to imply indeed that $M$ may be viewed as a direct summand!

Yes @Matthew you were right these 2 sentences were confusing I deleted them. The point is if $X$ il locally compact then the group $Aut(X)$ can be turned into a topological group in a natural way using the compact-open topology.

The appendix of Freedman and Calegari paper (Distortion in transformation groups) is from "Yves de Cornulier". Thanks a lot @Igor for the reference.