Hilbert-Schmidt operators are compact

Question

I'm trying to show that if we have $T \in B(H,H)$ for some separable Hilbert space $H$ such that for any orthonormal basis $\{e_k \}_ k$ we have $\sum_{k=1}^{\infty }||Te_k||^2 <\infty $, then $T $ compact .

I'm trying to show that $T$ is the operator limit of bounded finite rank (and hence compact) operators.

So I let $T_n x =\sum_{k=1}^{n }\langle x, e_k \rangle T(e_k) .$

But I can't show this converges to $T$.

So $||T_nx-Tx||^2=||\sum_{k=n+1}^{\infty } \langle x, e_k \rangle T(e_k) ||^2 $ — Anonmath101, May 03 '21 at 14:56
In general, $\sum_{k=1}^\infty a_k$ where $a_k\ge 0$ converges iff $\sum_{k=n}^\infty a_k \to 0$ as $n\to\infty$. — Quoka, May 03 '21 at 14:58

score 5 · Answer 1 · answered May 03 '21 at 15:03

5

Note that \begin{align*} \left\|\sum_{k\geq n+1}\left<x,e_{k}\right>Te_{k}\right\|&\leq\sum_{k\geq n+1}\left|\left<x,e_{k}\right>\right|\cdot\|Te_{k}\|\\ &\leq\left(\sum_{k\geq n+1}\left|\left<x,e_{k}\right>\right|^{2}\right)^{1/2}\left(\sum_{k\geq n+1}\|Te_{k}\|^{2}\right)^{1/2}\\ &\leq\|x\|\cdot\left(\sum_{k\geq n+1}\|Te_{k}\|^{2}\right)^{1/2}\\ &\rightarrow 0 \end{align*} as $n\rightarrow\infty$.

answered May 03 '21 at 15:03

user284331

56,315

So: $$T(x)=\sum_{k\ge1}\langle x,e_k\rangle T(e_k)$$For any continuous linear operator $T$ but in general $|T_n-T|$ does not tend to $0$? This is strange – FShrike Mar 27 '22 at 21:19

Hilbert-Schmidt operators are compact

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