# “Frobenius-finite” linear operators on a Hilbert Space

Let $H = L_2(S)$ be the complex Hilbert space over $S$ with the counting measure. (There might be another term for this concept, but) I define a continuous linear operator $L$ on $H$ with matrix representation $A$ to be Frobenius-finite if and only if $\displaystyle\sum_{(i,j) \: \in \: S \times S} |a_{i,j}|^2 < \infty$

1.
Is Frobenius-finiteness invariant under unitary similarity?

2.
If yes, is there a categorical (i.e., not using the concept of matrix representation) characterization of when a continuous linear operator is Frobenius-finite?

3.
Is there another term for what I am calling Frobenius-finite?

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I wish people wouldn't use "categorical" as a synonym for "nice & algebraic-looking & co-ordinate invariant"... –  Yemon Choi Sep 17 '10 at 5:05
(In case you were still wondering what happens when $S$ is uncountable -- one of your replies to Bill Johnson seemed to imply you thought this could be an issue -- I left a comment to his answer. Perhaps you might have received a less terse response if you'd given a bit more detail in your question about what courses you have & haven't taken in analysis, since it's rather hard to infer that from your profile page.) –  Yemon Choi Sep 18 '10 at 9:29

Since $\sum_{(i,j)∈S×S}|a_{i ,j}|^2 =$ Trace$(A^*A)$ the answers to 1. and 2. are both affirmative, and as has already been said, the answer to 3. is "Hilbert-Schmidt."

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Although in infinite dimensions, making sure that intuitive calculations with trace actually work can be a little tricky for the first-time user/learner - at least that's my dim recollection –  Yemon Choi Sep 17 '10 at 5:27
Well, perhaps in general that's so, but here the terms are all positive, and that simplifies things a lot. –  Dick Palais Sep 17 '10 at 5:38
@Dick Palais: Down voting because, IMO, such elementary questions are not appropriate for MO and should not be answered. I intentionally gave no argument but only enough info so that the OP could easily find the answers. –  Bill Johnson Sep 17 '10 at 11:45
With all due respect, I would like to differ with you. It was a perfectly natural question for a non-expert and I think it is only polite to give an appropriate hint that will allow an OP to figure out an answer for him or herself rather than telling him to go look it up in some book or Wikipedia. My answer was under two lines--- shorter than your own (and a lot less insulting). –  Dick Palais Sep 17 '10 at 12:49

Any operator satisfying your "Frobenius-finite" condition is supported on $T \times T$ where $T$ is a countable subset of $S$. Hence once you have looked up the separable case, the supposedly inseparable case should follow without undue worry or need for consultation –  Yemon Choi Sep 17 '10 at 5:07