Let $H$ be a Hilbert space and $u: H \to H$ a Hilbert-Schmidt operator, i.e. for an orthonormal basis $E$
$$\Vert u\Vert_2^2:=\sum_{x \in E} \Vert u(x) \Vert^2 < \infty$$
(this does not depend on the choice of basis).
I want to show that $u$ is compact. I looked this question up but all references I found proved this in the case that $H$ is separable, i.e. has a countable orthonormal basis. Is there an elementary proof that works for general Hilbert spaces?
Best Answer
For $$\sum_{x \in E} \Vert u(x) \Vert^2 < \infty$$ to hold, only countably many terms can be non-zero, i.e. $u$ is zero on all $H$ except a separable subspace $V$. So just look at the restriction $$u|_V: V \to W\subseteq H$$ where $W = \text{im}(u) \cong V$ is also separable. $u|_V$ is compact by any of the proofs you know, so it sends the unit ball of $V$ to a relatively compact set in $W$, which is also relatively compact in $H$. $u$ sends the rest of the unit ball of $H$ to 0 so it also sends the unit ball to a relatively compact set.