Jtdylktuy

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!

Suppose that ${F_n}$ is a sequence of finite-rank operators such that $F_nto I$ 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
$$
|(F_n-I)A|_{HS}^2=sum_j|(F_n-I)Ae_j|^2=sum_{j=1}^n|(F_n-I)Ae_j|^2+sum_{j=n+1}^infty|(F_n-I)Ae_j|^2\ leqsum_{j=1}^n|(F_n-I)Ae_j|^2+(2k^2+2)sum_{j=n+1}^infty|Ae_j|^2
$$
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
$$
|F_nAF_n-A|_{HS}^2=|F_nAF_n-F_nA-(I-F_n)A|_{HS}^2=|F_nA(F_n-I)+(F_n-I)A|_{HS}^2\ leq2|F_nA(F_n-I)|_{HS}^2+2|(F_n-I)A|_{HS}^2leq2k^2|A(F_n-I)|_{HS}^2+2|(F_n-I)A|_{HS}^2to0.
$$