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I am having trouble proving the following result:

Show that the space $X$ of bounded operators on a separable Hilbert space into itself for which the Hilbert-Schmidt norm is finite, is a Banach space

My attempt: Let $H$ be a separable Hilbert space. Take a Cauchy sequence $\{A_n\}_{n \ge 1}$ of such operators (bounded + Hilbert-Schmidt). Note that $A_n : H \to H$ for each $n$. Being Cauchy means: $$ \forall \epsilon > 0, \exists N \in \mathbb{N} : n,m > N \implies \|A_{n} - A_{m}\| < \epsilon $$ where $\|\cdot\|$ is the operator norm. My idea is to define the limit operator as: $$ A(u) := \lim_{n \to \infty} A_{n}(u) \text{ for each $u \in H$} $$ Then we need to show that $\|A_n - A\| \to 0$ and $A$ is bounded (i.e. $\|A\| < \infty$). This is sufficent due to the inequality $\|A\| \le \|A\|_{H-S}$ .

But I'm not sure how to prove $\|A_n - A\| \to 0$. I also think there's a better way to choose the candidate limit. Can anyone guide me on how to prove the result?

Thank you.

duelspace
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  • Hilbert-Schmidt means that $\sum ||Ae_i||^2 < \infty$ for any orthonormal $e_i$. Now, since $A_n$ is Cauchy it is uniformly bounded in Hilbert-Schmidt norm. Then use point-wise convergence of $A_n(e_i)$ for any $e_i$. – Salcio Apr 28 '22 at 01:18
  • @Salcio I'm not sure why $A_n$ being Cauchy implies it is uniformly bounded. We can say $$|A_n|{HS} = \sum{i} |A_{n}(e_{i})|^2 < \infty$$ for each $n \in \mathbb{N}$ but why do we necessarily have a uniform bound? Regarding the pointwise convergence, I'm not sure how to use this. Are you trying to use it to say $$|A_{n}-A|{HS} = \sum{i} |(A_{n}-A)(e_{i})|_{H} \to 0$$ ? – duelspace Apr 28 '22 at 04:39
  • Since it is Cauchy , there is $n_0$ such that if $n > n_0$ then $||A_n-A_{n_0}||{HS} <= 1$., so, from the triangle inequality $||A_n||{HS} <= 1 +||A_{n_0}||_{HS}$ – Salcio Apr 28 '22 at 20:12

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This is not true for the operator norm. A norm limit (in the operator norm) of Hilbert-Schmidt operators is not necessarily Hilbert-Schmidt; already norm limits of finite-rank operators give you any compact operator. That is, the operator-norm closure of the space of Hilbert-Schmidt operators is $K(H)$, the compact operators.

But that's not what you are being asked. You are asked to show that the Hilbert-Schmidt operators, with the Hilbert-Schmidt norm, form a Banach space.

By choosing a subsequence if necessary, you may assume that $\|A_{n+1}-A_m\|<2^{-n}$. Pointwise, you have $$\tag1 A=A_1+\sum_n(A_{n+1}-A_n). $$ This converges in the Hilbert-Schmidt norm because $\sum_n\|A_{n+1}-A_n\|_{\rm HS}\leq\sum_n 2^{-n}=1$. Which in particular shows that $A$ is Hilbert-Schmidt$^1$. And we have $$ \|A-A_n\|_{\rm HS}=\Big\|\sum_{k\geq n}A_{k+1}-A_k\Big\|_{\rm HS}\leq\sum_{k\geq n}\|A_{k+1}-A_k\|_{\rm HS}\leq\sum_{k\geq n}2^{-k}=2^{-n+1}\to0. $$ So $A$ is the Hilbert-Schmidt norm limit of a sequence of $\{A_n\}$, and as $\{A_n\}$ is Cauchy it has to converge to $A$.

  1. If $T=\sum_kT_k$ exists in norm and $\sum_k\|T_k\|_{\rm HS}<\infty$, then $ T$ is Hilbert-Schmidt. Indeed, fix an orthonormal basis $\{e_j\}$. Then \begin{align} \operatorname{Tr}(T^*T)^{1/2} &=\Big(\sum_j\|Te_j\|^2\Big)^{1/2}=\Big(\sum_j\Big\|\sum_kT_ke_j\Big\|^2\Big)^{1/2}\\[0.2cm] &\leq\sum_k\Big(\sum_j\|T_ke_j\|^2\Big)^{1/2}\\[0.2cm] &=\sum_k\|T_k\|_{\rm HS}<\infty, \end{align} so $T$ is Hilbert-Schmidt. The inequality is Minkowski's Integral Inequality.
Martin Argerami
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  • Oh ok thank you for the clarification. I'm not sure how to argue for the second summation though. So if we fix $k \in \mathbb{N}$ then we can say that $|(A-A_{n})(e_k)|^2 \to 0$ as $n \to \infty$ because of our definition of $A$, so the first sum can be made arbitrarily small (say $< \epsilon /2$). But for the second sum I'm not really sure. I know we have to use the fact $A_n(e_k) \to A(e_k)$ for each $k$ and $\sum_{k} |A_{n}(e_k)|^2 < \infty$ for any $n$ but I can't seem to connect the dots. – duelspace Apr 28 '22 at 16:35
  • I included more details. – Martin Argerami Apr 28 '22 at 21:25
  • "Not sure why you are involving the operator norm." This seems fairly reasonable to me, at least when the problem is phrased as in the question. Paraphrasing: "Here is a subspace of a Banach space. Prove that it's a Banach space." The natural thing to do would be to use the norm on the entire space. – Peter Apr 28 '22 at 22:00
  • Fair enough. I have edited the answer. – Martin Argerami Apr 29 '22 at 02:53
  • @MartinArgerami I have just one doubt. In equation (1) you write $A=A_1+\sum_n(A_{n+1}-A_n)$. I was trying to understand why the $RHS$ is equivalent to $A$. Is it because we can say pointwise: $$A_1+\sum_n(A_{n+1}-A_n) = A_1 + \lim_{K \to \infty} \sum_{n}^{K} (A_{n+1}-A_n) = \lim_{K \to \infty} A_{K+1} = A $$ Have I got that right? – duelspace Apr 29 '22 at 02:54
  • Yes, that's it. $ $ – Martin Argerami Apr 29 '22 at 02:55
  • Perfect, thank you very much! – duelspace Apr 29 '22 at 02:56
  • @MartinArgerami I do not understand why $$\sum \Vert A_{k+1}-A_k \Vert_{HS} \text{ converges } \implies \sum (A_{k+1}-A_k) \text{ converges}$$ is true, since we do not know that the space is complete. Could you explain to me please? – Axel Dec 15 '23 at 21:48
  • @Axel: since $|A_{k+1}-A_k|\leq|A_{k+1}-A_k|_{\rm HS}$, you get first that the series exists in $K(H)$. You then argue that it is Hilbert-Schmidt. – Martin Argerami Dec 16 '23 at 00:10
  • @MartinArgerami Thank you for your quick answer. I get that $A \in K(H)$ since $K(H)$ is complete, but I'm missing the argument that links the convergence of $\sum_k \Vert A_{k+1}-A_k\Vert_{HS}$ to $A$ is Hilbert-Schmidt. – Axel Dec 16 '23 at 11:16
  • @Axel: Sorry for the late reply. I have added some detail. – Martin Argerami Dec 16 '23 at 17:20
  • @MartinArgerami (+1) Thank you for the explanation and for your time. I was not well-versed enough in Hilbert-Schmidt operators to know this, especially Minkowski's integral inequality, so it was great learning about it. I really like this approach as we somehow proved the completeness using the fact that any absolutely convergent series (w.r.t $\Vert{\cdot}\Vert_{HS}$) is convergent. – Axel Dec 16 '23 at 19:46
  • @Axel: glad you find it useful. The proof is modelled on the usual proof that $L^p$ is complete. – Martin Argerami Dec 16 '23 at 19:52
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    By the way, the Hilbert-Schmidt operators actually form a Hilbert space with the inner product $\langle S,T\rangle=\operatorname{Tr}(S^*T)$. So it is literally an $L^2$ space. – Martin Argerami Dec 16 '23 at 20:06