If $T$ is a compact operator in a Hilbert space, then its range is separable since is a set with compact closure. But what can I say about the $\mbox{ker}(T)$ set? Is it also separable? If so, what is the argument to prove it?
I've tried to observe this by projecting $\overline{\mbox{range} (T)}$ over $\mbox{ker}(T)$, but I can't figure out how to build a dense sequence in $\mbox{ker}(T)$, so I don't know if the result is true or false
Best Answer
Of course if the Hilbert space $H$ is separable then the kernel of $T$ is automatically separable. But if $H$ is not separable then $\operatorname{ker} T$ is never separable.
For example, the adjoint $T^*$ is also compact, and the orthogonal complement of its image (which is separable) is the kernel of $T$. If the kernel of $T$ were separable then we would have $H$ as the direct sum of two separable subspaces.