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On wikipedia it is mentioned that if we are on some (separable) Hilbert space $H$ and there is an ONB $(e_n)$ such that any compact operator $K$ can be written as $$ K = \sum_{n,m =0}^{\infty} K_{n,m} e_n \otimes e_m.$$

This made me wonder whether the result is still true if $K$ is a trace-class operator and the convergence is assumed to be in trace-norm as well? Does anybody know this?

By the way, the article was this one.

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  • $\begingroup$ general point: when you talk about convergence of a doubly-indexed series, you should specify the actual limiting process encyclopediaofmath.org/index.php/Double_sequence (i.e. do you mean $\lim_{N\to\infty} \sum_{m,n=0}^N$ $\endgroup$
    – Yemon Choi
    Feb 11, 2017 at 23:33
  • $\begingroup$ yes, I mean precisely what you wrote $\endgroup$
    – Kinzlin
    Feb 11, 2017 at 23:37
  • $\begingroup$ so your question can be paraphrase by asking whether every trace-class operator is a trace-norm limit of finite-rank operators? $\endgroup$ Feb 11, 2017 at 23:42
  • $\begingroup$ @paulgarrett no. here the question is the converse, we are first given the ONB and represent the operator in this basis.I am asking whether this is possible. $\endgroup$
    – Kinzlin
    Feb 11, 2017 at 23:43
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    $\begingroup$ Yes, of course. This is just the statement that $e_n\otimes e_m$ is, in its natural order, a shrinking basis for the compact operators on $\ell_2$. Boundedness of the partial sum projections in the trace class norm follows by duality from the boundedness in the operator norm. $\endgroup$ Feb 12, 2017 at 0:15

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