Thanks! Would you mind adding a few more details? Why are these two sets disjoint, and why their union is $G$?

Thanks a lot, Ofer! This is very helpful. My question was indeed missing an important qualifier, but it does sound like the idea is correct in spirit.

Thanks again, Ofer! I changed it to $S_n/n$, and I also added the requirement that $K$ is strictly convex. Note that $K$ is always convex as a pointwise limit of convex functions, and thus $K^\star$ is also always convex. When $K$ is strictly convex then so is $K^\star$. I think.

Thanks, Ofer! Note that I did mean my definition of $K_n$. I am thinking of $S_n$ as something that scales with $n$ like a sum of iid. The prime example is indeed $S_n=X_1+\cdots+X_n$ for iid $X_n$. In this case the statement is true as I've written it. I think :). Your comment on convexity is a good point. For this to be true $K$ would have to be strictly convex, at least at an interval.

Thanks again! Is it easy to see that this subgroup is discrete and closed?

Thanks a lot! I'm not sure I understand what you mean by "$F_2$ act on itself $X=F_2\times Q$"

Regarding my first comment in this thread, it should have also said that $H'$ is discrete in $G$.

I see now. But is this free group discrete? If not then there's no contradiction, no?