# Are Hilbert-Schmidt operators in non-separable Hilbert spaces compact?

The definition of Hilbert-Schmidt operator should still be valid even when the Hilbert space is not separable: If $e_i$ for $i\in I$ is an orthonormal basis for a Hilbert space, and

$\mbox{Trace}(T)=\sum_{i\in I}\|Te_{i}\|^{2}<\infty$

Then $T$ is Hilbert-Schmidt. Of course if the sum is finite then there can only be countably many non-zero terms in the summation.

However, I am not sure how to show that such operators are compact.

• What do you mean exactly when you sum over uncountably many elements? (See here for a discussion) – Silvia Ghinassi Nov 2 '15 at 1:19
• @Silvia: it should mean at most countably many terms in the summation are nonzero. – Qiaochu Yuan Nov 2 '15 at 1:21
• Doesn't the fact that $T$ is $0$ except on a separable subspace allow you to deduce this from the same result on separable spaces? FYI - Wikipedia gives the definition on arbitrary Hilbert spaces, just as you do here. – Paul Sinclair Nov 2 '15 at 4:38