Assume $\Gamma$ acts by isometries on a separable Hilbert space $H$, and $$\operatorname{diam} H/\Gamma\le 1.$$ Is it true that $H/\Gamma$ is compact?
Comment.
At the moment I do not have an answer even if $\Gamma$ acts by translations. Here is a related question:
Let $L$ be a lattice in $\mathbb R^q$ ($q$ is any positive integer). Assume $$\operatorname{diam} \mathbb R^q/L>1000.$$ Is it true that there is a midpoint $m$ of two points in $L$ such that $|m-x|>1$ for any $x\in L$?
If the answer to the this question is "YES" then the answer to my original question is "NO".
Stupid example. Assume the action of $\Gamma$ on $H=\ell_2$ is generated by coordinate translations $x_n\mapsto x_n+\epsilon_n$. Then $$\operatorname{diam} H/\Gamma=\tfrac12\cdot\sqrt{\sum_{n=1}^\infty\epsilon_n^2}.$$ Thus, if $\operatorname{diam} H/\Gamma\le 1$ then $H/\Gamma$ is a quotient of Hilbert cube, and has to be compact.
