Dense subgroup of an extremely amenble group is extremely amenable?? Recall that a topological group $G$ is called extremely amenable if every continuous action of $G$ on a compact space $K$ admit a fixed point. i.e there is a point $\xi\in K$ such that $g.\xi=\xi$ for all $g\in G$.
Now let $H$ be a dense subgroup of an extremely group $G$. My goal is to show that $H$ is also extremely amenable. 
More precisly, i don't know if this true that in this case every continuous action of $H$ on a compact space can be extend to an action of $G$ on the same compact space or not? Or may be there is another approach to solve this question.
Thank for any help.
 A: I think you should not try to solve this using the fixed-point definition for extremely amenable groups. This because the "inducing action machinery" outside of the world of locally compact groups is not so useful.
So I would like to suggest you to use the multiplicative left-invariant mean definition. Precisely, a topological group $G$ is extremely amenable if it has a multiplicative left-invariant mean on $\mathcal{C}_{ru}^b(G)$, i.e., a positive functional $m$ on $\mathcal{C}_{ru}^b(G)$ such that $m(1_G)=1$, $m(f\cdot h) = m(f)m(h)$ and $m(gf)=m(f)$ for every $f,h\in \mathcal{C}_{ru}^b(G)$ and $g\in G$. (Here $\mathcal{C}_{ru}^b(G)$ is the set of all bounded right-uniformly continuous functions on $G$).
Let now consider the extension map $ext: \mathcal{C}_{ru}^b(H) \longrightarrow \mathcal{C}_{ru}^b(G)$. Note that this map is well-defined as the extension is unique ( see Bourbaki Topologie générale) and it is linear and positive. Moreover we have that


*

*$ext(1_H) = 1_G$

*$ext(f\cdot h)= ext(f)ext(h)$ for every $f,h \in \mathcal{C}_{ru}^b(H)$

*$ext(gf)=g\cdot ext(f)$ for every $g\in H$ and $h\in \mathcal{C}_{ru}^b(H)$
Now it is easy to construct a multiplicative left-invariant mean on $\mathcal{C}_{ru}^b(H)$ using the fact that you know that there is one on $\mathcal{C}_{ru}^b(G)$. Hence, $H$ is extremely amenable. 
I hope it is clear :-)
