 Solution

Assume the invariant set $C$ is nonempty, as required for any fixed-point assertion. For $x\in C$, put $r(x)={\sup }_{c\in C}\Vert x-c\Vert$ and $r_{\ast }={\inf }_{x\in C}r(x)$. Boundedness makes these finite. The parallelogram identity gives, for $x,y\in C$,The midpoint lies in $C$, so a minimizing sequence $x_n$ satisfiesSince $C$ is closed in a Hilbert space, the sequence converges to $x_{\ast }\in C$. The function $r$ is 1-Lipschitz, hence $r(x_{\ast })=r_{\ast }$. The same midpoint inequality proves uniqueness of this minimizing center.

An affine isometry from the group maps $C$ onto itself, since its inverse also preserves $C$. Therefore it preserves $r$ and sends its unique minimizer to another minimizer. Uniqueness givesThis proof applies to a closed bounded convex subset, the substantive meaning of the printed “convex subspace”; if it is literally a bounded vector subspace, it is just the zero subspace.