Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 separable Hilbert space. Recall that a bounded operator $A : H \to H$ is said to be Hilbert-Schmidt if $$\|A\|_{HS}^2 := \sum_{i=1}^\infty \|A e_i\|^2 < \infty$$ where $\{e_i\}_{i=1}^\infty$ is an orthonormal basis for $H$. (The value of $\|A\|_{HS}$ does not depend on the basis chosen.)

Suppose that $A$ is Hilbert-Schmidt, and let $\{P_n\}$ be a sequence of finite-rank orthogonal projection operators on $H$, such that $P_n \to I$ strongly (i.e. $P_n x \to x$ for every $x \in H$). Does $P_n A P_n \to A$ in the Hilbert-Schmidt norm $\|\cdot\|_{HS}$?

I can prove this under the additional assumption that $\{P_n\}$ is increasing, i.e. $P_n H \subset P_{n+1} H$. In this case, we may choose an orthonormal basis $\{e_i\}$ for $H$ such that for each $n$, $e_1, \dots, e_{d_n}$ is an orthonormal basis for $P_n H$, where $d_n$ is the rank of $P_n$. Then $P_n e_i$ = $e_i$ for $i \le d_n$, and $P_n e_i = 0$ otherwise, so we can write $$\|P_n A P_n - A\|_{HS}^2 = \sum_{i=1}^{d_n} \|(P_n - I) A e_i\|^2 + \sum_{i=d_n+1}^\infty \|A e_i\|^2.$$ As $n \to \infty$, $d_n \to \infty$, and the second term goes to 0 because it is the tail of the convergent series $\sum \|A e_i\|^2 = \|A\|_{HS}^2$. For the first term, we have $$\|(P_n - I) A e_i\| \le \|P_n - I\| \|A e_i\| \le 2 \|A e_i\|$$ which is a square-summable sequence because $A$ is Hilbert-Schmidt. And for each $i$ we have $\|(P_n - I) A e_i\| \to 0$ since $P_n \to I$ strongly. So by the dominated convergence theorem, we conclude the first term also goes to 0.

However, without this assumption, $P_n e_i$ is harder to deal with, and I don't see how to craft a proof (nor a counterexample).

If it helps, I'm most interested in the case where $A$ is skew-adjoint, i.e. $A^* = -A$.

I'd also be interested to know if this statement still holds if we drop the assumption that $P_n$ are orthogonal projections, and only assume that they are finite rank and converge strongly to $I$.


share|cite|improve this question
I guess the same argument works for the more general case, where $Q_n$ are operators of finite rank, and $Q_n\to I$ strongly. Then $Q_nAQ_n^*\to A$ in HS-norm. Let $U_n:=\bigcup_{i=1}^n{\rm Ran}(Q_i)$, and keep on extending an orthonormal base for these. As I see, you didn't really use $P_ne_i=e_i$... – Berci Apr 26 '13 at 0:22
@Berci: I did use it: in the first display, it's used to replace $(P_n A P_n - A) e_i$ by $(P_n - I)A e_i$. Without this, I don't see how to control this term, as we do not know so much about $\sum_i \|A P_n e_i\|$. – Nate Eldredge Apr 26 '13 at 1:02
up vote 3 down vote accepted

Suppose that $\{F_n\}$ is a sequence of finite-rank operators such that $F_n\to I$ strongly. Note that by the uniform boundedness principle the sequence is bounded, i.e. there exists $k>0$ with $\|F_n\|<k$ for all $n$ (thanks julien for reminding me of this). I will assume that all $F_n$ are selfadjoint (I need for my estimates, but didn't think if there is a counterexample or not).

We have $$ \|(F_n-I)A\|_{HS}^2=\sum_j\|(F_n-I)Ae_j\|^2=\sum_{j=1}^n\|(F_n-I)Ae_j\|^2+\sum_{j=n+1}^\infty\|(F_n-I)Ae_j\|^2\\ \leq\sum_{j=1}^n\|(F_n-I)Ae_j\|^2+(2k^2+2)\sum_{j=n+1}^\infty\|Ae_j\|^2 $$ and now we can reason as in Nate's proof to say that this goes to zero (i.e. use that $A$ is HS for the tail, and the pointwise convergence of $F_n-I$ to zero in the first finite sum.

This also implies that $A(F_n-I)\to0$ in HS norm. Indeed, $$ \|A(F_n-I)\|_{HS}^2=\mbox{Tr}((F_n-I)A^*A(F_n-I))=\mbox{Tr}(A(F_n-I)(F_n-I)A^*)\\ =\|(F_n-I)A^*\|_{HS}^2 $$ (as $A$ is HS if and only if $A^*$ is, the above works).

Now $$ \|F_nAF_n-A\|_{HS}^2=\|F_nAF_n-F_nA-(I-F_n)A\|_{HS}^2=\|F_nA(F_n-I)+(F_n-I)A\|_{HS}^2\\ \leq2\|F_nA(F_n-I)\|_{HS}^2+2\|(F_n-I)A\|_{HS}^2\leq2k^2\|A(F_n-I)\|_{HS}^2+2\|(F_n-I)A\|_{HS}^2\to0. $$

share|cite|improve this answer
Nicely done, +1. Just a remark: if the sequence $F_n$ converges strongly ie pointwise, the uniform boundedness principle says that it is bounded in $B(H)$ for the operator norm. – 1015 Apr 26 '13 at 3:19
Good point! I was too lazy to think about that. I'll edit accordingly. – Martin Argerami Apr 26 '13 at 4:33
Thanks very much! In fact, on further inspection, it's unnecessary that $F_n$ be of finite rank; the dominated convergence takes care of the first sum even if it's infinite. Also, the argument can show $F_n A F_n^* \to A$ without needing self-adjointness. Indeed, we can show the more general statement: if $S_n \to S$ and $T_n \to T$ strongly, then $S_n A T_n^* \to S A T^*$ in HS norm. – Nate Eldredge Apr 27 '13 at 13:08
Good points, Nate. If you want me, I'll edit for the answer to say this, but I'll leave the decision to you. – Martin Argerami Apr 27 '13 at 23:45

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.