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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.

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  • $\begingroup$ What do you mean exactly when you sum over uncountably many elements? (See here for a discussion) $\endgroup$ – Silvia Ghinassi Nov 2 '15 at 1:19
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    $\begingroup$ @Silvia: it should mean at most countably many terms in the summation are nonzero. $\endgroup$ – Qiaochu Yuan Nov 2 '15 at 1:21
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    $\begingroup$ 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. $\endgroup$ – Paul Sinclair Nov 2 '15 at 4:38
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Preword

The Hilbert dimension plays no role at all!

Problem

A trace class operator is Hilbert Schmidt. (Decomposition)

A Hilbert Schmidt operator is compact. ([Denseness][2])

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  • $\begingroup$ I will provide a proof of the latter later. In particular it will show how to tackle compactness. $\endgroup$ – C-Star-W-Star Nov 2 '15 at 12:38

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