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I am new to this site and i dont really know how to ask questions properly, so i am really sorry if i did something wrong.

My question is if there is a way to prove that a Hilbert-Schmidt operator is compact from the definition of compact operators.

I can prove the result by noticing that: a Hilbert-Schmidt operator is the limit of a sequence of finite-rank operators, the limit of a sequence of compact operators is compact, and a finite-rank operator is compact.

And i was wondering if there is a more direct approach to the problem. Something like taking a bounded sequence $\{x_n\}_{n\in N}$ in a Hilbert space $H$, and getting a convergent subsequence from $\{T(x_n\}_{n\in N}$, where T is a Hilbert-Schmidt operator.

Thanks in advance for your time.

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  • $\begingroup$ I don't think so. Being uniform limit of compact operators is usually the easiest way to prove a an infinite rank operator is compact. $\endgroup$ – Alonso Delfín Aug 1 '15 at 21:08

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