# Tag Info

## New answers tagged compact-operators

1

Hint. Your idea is fine (and you are almost done). Just make use of the following fact: If $(x_n)$ is a sequence in some metric space $X$ and there is an $x \in X$ such that for every subsequence $(x_{n_k})$, some sub-sub-sequence $(x_{n_{k_j}})$ converges to $x$, then $x_n \to x$ in $X$. Now apply this to $x_n = Au_n$, $x= 0$ and use your idea to ...

0

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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