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Suppose $K$ is a linear operator in a separable Hilbert space $H$ such that for any Hilbert basis $\{e_i\}$ of $H$ we have $\lim_{i,j \to \infty} (Ke_i,e_j) = 0$.

Is it true that $K$ is compact?

Thanks in advance for any help.

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Is $K$ a linear operator? –  Tomás May 25 '13 at 20:03
    
To Tomás: yes, it is linear –  user79456 May 25 '13 at 20:12
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To Davide: the exact formulation is the following. For any $\epsilon > 0$ there exists $n$ such that $|(Ke_i, e_j)|< \epsilon$ for every $i,j>n$. –  user79456 May 25 '13 at 20:28
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For a first question, that's a good one, +1. I'll remark that an operator on $H$ is compact iff it is weak-norm continuous, so the property does hold quite strongly for compact operators. Converse? Why I neither saw nor thought about this before puzzles me. How did you come up with the question? Is this an exercise in some book? –  1015 May 25 '13 at 22:56
    
Can we conclude that $\lim_{i,j\to\infty} (K e_i,K e_j)=0$? –  Tomás May 25 '13 at 23:11
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1 Answer 1

up vote 8 down vote accepted

Yes, $K$ is compact.

Note first that we can assume that $K$ is selfadjoint; indeed, if $K$ satisfies the hypothesis, so do $K^*$ and $K+K^*$, and if we prove that the real and the imaginary parts of $K$ are compact, then we have that $K$ is compact.

Using the Weyl-von Neumann-Berg Theorem (II.4.1 or II.4.2 in Davidson's C$^*$-algebras by example) we can write $$ K=A+T, $$ where $A$ is selfadjoint and diagonal, and $T$ is compact. Being compact, $T$ satisfies the hypothesis and so $A=K-T$ does too. Now, since $A$ is diagonal, the hypothesis implies that its diagonal tends to zero, and so it is compact. Then $K$ is compact, being a sum of compacts.

To finish, note that it is essential that the property of small entries of $K$ holds for any basis. If we fix a single basis, it is possible to have operators $K$ satisfying $\langle Ke_i,e_j\rangle\to0$ (in the sense of the question) but not compact. Indeed, let $$ K=\bigoplus_n\begin{bmatrix}1/n&\cdots&1/n\\ \vdots&\ddots&\vdots\\ 1/n&\cdots&1/n\end{bmatrix}. $$ This is an orthogonal sum of pairwise orthogonal projections, so it is an infinite-dimensional projection, thus not compact.

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This is a great answer to a great question, +1. –  1015 May 26 '13 at 14:52
    
Thanks, julien. I've always found the Weyl-von Neumann-Berg Theorem wonderful, and that section in Davidson's book really cool. –  Martin Argerami May 26 '13 at 16:38
    
to Martin: Thanks a lot, especially for the example in the end. –  user79456 May 27 '13 at 11:31
    
You're welcome! –  Martin Argerami May 27 '13 at 14:07
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