How do i show that the set of Hilbert-Schmidt operators $HS(H) = \{T \in B(H) \; : \; \sum^{\infty}_{n=1}\|Te_n\|^2 < \infty \}$ for some countable ONB $\{e_n\}$, on a separable Hilbert Space $H$, form a Banach space when equipped with the HS norm?

$$\|T\|_{HS} = \sqrt{\sum^{\infty}_{n=1}\|Te_n\|^2}$$


Please let me know if this (doesn't) makes sense. I could stand to think about it a little more carefully.

Say you have a Cauchy sequence, $T_k$, of Hilbert-Schmidt operators. Then the the sequence of sequences $$(T_1(e_1), T_1(e_2), T_1(e_3),...) , (T_2 (e_1), T_2(e_2), T_2(e_3),...), ...$$ is a Cauchy sequence in the Banach space $$\big(H^\mathbb{N}, \sqrt{\sum_{n=1}^\infty ||\cdot||^2} \hspace{3mm} \big) $$

So it converges in this norm to some $(f_1, f_2, f_3,...)$. It is not hard to show that the sequence $T_k$ converges pointwise to some linear operator, $T$. Since also $$\lim_{k\rightarrow \infty}T_k(e_n)=f_n$$

$T$ is a Hilbert-Schmidt operator and $T_k$ converges to $T$ in the Hilbert-Schmidt norm.

  • $\begingroup$ Why does this mean $T_k$ converges to $T$ in the Hilbert-Schmidt norm? Certainly it converges in the $B(H)$ norm. $\endgroup$ – Jake Nov 18 '17 at 15:40

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