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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?

flag	asked Sep 17 2010 at 1:56 Ricky Demer 6,361●8●31

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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 2010 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 2010 at 9:29

Dick Palais · Accepted Answer · 2010-09-17 04:41:29Z UTC

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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."

answered Sep 17 2010 at 4:41

Dick Palais
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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 2010 at 5:27

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Well, perhaps in general that's so, but here the terms are all positive, and that simplifies things a lot. – Dick Palais Sep 17 2010 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 2010 at 11:45

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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 2010 at 12:49

Bill Johnson · Answer 2 · 2010-09-17 02:14:50Z UTC

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Google hilbert schmidt operator.

answered Sep 17 2010 at 2:14

Bill Johnson
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Well, that certainly answers #3, but the pages that brings up either assume the space is separable or do not conclude that the result is independent of basis, so they don't quite answer #1. (And they certainly don't answer #2.) – Ricky Demer Sep 17 2010 at 2:28

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Look at most any book on operator theory or Banach space theory to find that the answers to (1) and (2) are yes and yes. Your question is pretty elementary for MO and I will now vote to close. – Bill Johnson Sep 17 2010 at 2:35

Are they elementary enough that there is an online reference for (1) and (2)? – Ricky Demer Sep 17 2010 at 2:39

I don't think MO is meant to be the Open University... – Yemon Choi Sep 17 2010 at 5:04

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 2010 at 5:07

“Frobenius-finite” linear operators on a Hilbert Space

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