Showing that the space of Hilbert-Schmidt operators form a Banach space.

Question

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?

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

Answer 1

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.

Answer 2

I will only show that $HS(H)$ is a complete space (is easy to see that is a vector space). Let $(T_{n}) \subset HS(H)$ be a Cauchy sequence.

Lemma. $\|T\| \leq \|T\|_{HS}$.

Proof. Recall that the Hilbert-Schmidt norm is well defined. Take $x \in H$ be a unitary vector, and construct an orthonormal base of the form $\{x\} \cup \{e_{n}\}_{n}$. Thus, $$\|Tx\|^{2} \leq \|Tx\|^{2} +\sum_{n}\|Te_{n}\|^{2}=\|T\|_{HS}^{2}.$$ This proves the lemma.

As a corollary of the lemma, we conclude that $(T_{n})$ is Cauchy in $B(H)$. Since $B(H)$ is complete, there exists $T \in B(H)$ such that $T_{n} \to T$ in $\| \cdot\|$. We will show that $T \in HS(S)$ and that $T_{n} \to T$ in $\| \cdot\|_{HS}$.

1. $T_{n} \to T$ in $\| \cdot\|_{HS}$: Let $\varepsilon>0$. Since $(T_{n})$ is Cauchy w.r.t. $\|\cdot\|_{HS}$, we have that $\|T_{m}-T_{k}\|_{HS}^{2}<\varepsilon$ for $m,k>n_{0}=n_{0}(\varepsilon)$. Let $\{e_{n}\}$ a orthonormal base of $H$. For every $A \subset \mathbb{N}$ finite, we have $$\sum_{n \in A}\|T_{m}e_{n}-T_{k}e_{n}\|^{2} \leq \|T_{m}-T_{k}\|^{2}_{HS}<\varepsilon.$$ Letting $k \to \infty$ we find $$\sum_{n \in A}\|T_{m}e_{n}-Te_{n}\|^{2} \leq \varepsilon.$$ Since this is true for any finite set, we can take $A=A_{\ell}:=\{1,\ldots,\ell\}$ and let $\ell \to \infty$ to get $$\|T_{m}-T\|_{HS}^{2}=\sum_{n}\|T_{m}e_{n}-Te_{n}\|^{2}\leq \varepsilon.$$ This shows that $T_{m}-T \in HS(H)$ and that $T_{m} \to T$ w.r.t. $\|\cdot\|_{HS}$.
2. $T \in HS(S)$: The previous step shows that $T_{m}-T \in HS(H)$. Since by hypothesis $T_{m} \in HS(S)$, and using the fact that $HS(H)$ is a vector space, we conclude that $T=T_{m}-(T_{m}-T) \in HS(H)$.

This finishes the proof.