Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $H$ be a Hilbert space. For $\{e_n\}$ an orthonormal basis of $H$, we call $T\in B(H)$, a Hilbert Schmidt operator if $ \|T\|_2^2:=\sum_n \|Te_n\|^2 <\infty.$

I have seen somewhere before that $ \|ST\|_2 \leq \|S\|_{B(H)} \|T\|_2 $ for all $S\in B(H)$ arbitrary. In other words, the space of Hilbert Schmidt is an ideal in $B(H)$. Where can I find this? Where can I find it for Matrices?

share|cite|improve this question
The concept of a Hilbert–Schmidt operator is not very useful in the finite-dimensional case, because every operator on a finite-dimensional Hilbert space is a Hilbert–Schmidt operator. – Zhen Lin Feb 11 '12 at 19:17
up vote 2 down vote accepted

I don't have a reference, but this is easy to show directly: we have $$ \| ST \|_{HS}^2 = \sum_{n = 1}^{\infty} \| STe_n \|^{2} \leq \| S\|^2 \sum_{n = 1}^{\infty} \| Te_{n} \|^{2} = \| S \|^2 \| T \|_{HS}^2,$$ where $\| \cdot \|_{HS}$ is the Hilbert-Schmidt norm. This, by itself, however is not enough to prove that the set of Hilbert-Schmidt operators form an ideal - assuming you mean two-sided ideal, you also need to show, with $S$ and $T$ as above, that $TS$ is Hilbert-Schmidt, and that the sum of two Hilbert-Schmidt operators is again Hilbert-Schmidt (along with the perfectly obvious statement that the zero-operator is Hilbert-Schmidt). The latter is easy, $$ \| S + S' \|_{HS}^2 = \sum_{n = 1}^{\infty} \| (S + S')e_n \|^2 \leq \sum_{n = 1}^{\infty} (\| Se_n \|^2 + \| S'e_n \|^2) = \|S\|_{HS}^2 + \| S'\|_{HS}^2 < \infty $$ if $S$ and $S'$ are Hilbert-Schmidt, but I expect the former to be more difficult.

Edit: The result follows from the fact that the set of Hilbert-Schmidt operators is $^*$-closed, i.e. if $T$ is H-S, then so is $T^*$. The proof is easy: if $S \in B(H)$ and $T$ is H-S, then $(TS)^* = S^* T^*$ is H-S by the above, and hence $TS$ is H-S also.

Another edit: I'll add that the fact that the class of H-S operators is $^*$-closed follows from the fact that $$ \| S \|_{HS} = \| S^* \|_{HS}. $$ This is a (small) part of exercise 17c) of section XVIII.9 in Real and Functional Analysis by Serge Lang, so it shouldn't be too difficult to show (however, at the moment, I have no idea how to do it; maybe a sign that I need to revisit some of this stuff).

share|cite|improve this answer
Thank you very much! It definitely works! – Mahmood Al Feb 12 '12 at 2:45

$$ \sum_n\|S\,T\,e_n\|^2\le\sum_n\|S\|_{B(H)}^2\,\|T\,e_n\|^2=\|S\|_{B(H)}^2\,\|T\|_2^2 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.