Let $\mathbb{R}^\infty$ be the product of countably many real lines.


Assume that a finitely generated group $\Gamma$ acts on $\mathbb{R}^\infty$ (linearly and continuously) and there is a nonempty convex compact $\Gamma$-invariant subset $K$.


Is it true that $\Gamma$ has a fixed point in $K$?


(In my case, the group $\Gamma$ is NOT commutative and it is NOT amenable.)


*Please help.*



Fixed point of a group action
I don't think so. Let $\Gamma$ be the free group with 2 generators. It has a natural action on the boundary $\partial\Gamma$ and therefore on the space of probability measures on the boundary as well. The latter action is affine and has no fixed points (no invariant measures on the boundary). Let us now realize this action as an action on $\mathbb R^\infty$ by choosing a sufficiently big $\Gamma$-invariant family of test functions on the boundary (so that they separate probability measures). For instance, for any two distinct group elements $g,h\in\Gamma$ let $f\_{g,h}$ be the indicator function of the shadow of $h$ as seen from $g$. Then any probability measure $\mu$ on the boundary is uniquely determined by the collection of the integrals $\langle f\_{g,h},\mu\rangle$, and the action of the group on $\mu$ amounts to permuting the values of the test integrals. 


