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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 $I$.

Thanks!

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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

1 Answer 1

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

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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
1  
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

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