Does the top left block of an operator converge?

Question

Let $P$ be a bounded linear operator on the sequence space $\ell^2$. Assume it is trace class, positive, and self adjoint.

Let $P_k$ denote the map $V_k P V_k^*$ where $V_k \colon \ell^2 \to \ell^2$ denotes the restriction to the first $k$ coordinates. (One can think of $P_k$ as zero-ing out all but the top left corner of $P$)

Questions:

(a) Is it true that $\mathrm{tr}(P_k) \to \mathrm{tr}(P)$ as $k \to \infty$?

(b) Is it true that $\| P_k - P\| \to 0$ as $k \to 0$?

I can see that this true with the additional assumption that $P$ is diagonal, but am wondering if it holds more generally.

Answer 1

$\def\tr{\operatorname{Tr}}$

It is true for any trace-class $P$ that $\|P-P_k\|_1\to0$. This also answers the second question; for the operator norm of a normal compact operator is its largest eigenvalue in absolute value. Hence for any trace-class operator we have $$ \|T\|=\|T^*T\|^{1/2}=\|\,|T|^2\,\|^{1/2}=\|\,|T|\,\|\leq\tr(|T|)=\|T\|_1. $$ In particular, $\|P-P_k\|\leq\|P-P_k\|_1$.

So we have to prove that $\|P-P_k\|_1\to0$. I don't think there is an entirely elementary argument (I'll be very happy to be shown one!).

Note that $V_k$ is an orthogonal projection, that is $V_k^*V_k^2=V_k$. By the Polar Decomposition, $|P-P_k|=U^*(P-P_k)$ for a partial isometry $U$. We can write $U^*(P-P_k)=U^*P(I-V_k)$. Since $$\|P-P_k\|_1=\tr(U^*P(I-V_k)),$$ it is enough to show that $\tr(S(I-V_k))\to0$ for a trace-class operator $S$. Expanding on the canonical basis, $$ \tr(S(I-V_k))=\sum_j\langle S(I-V_k)e_j,e_j\rangle=\sum_{j>k}\langle Se_j,e_j\rangle\to0, $$ this latter convergence because the series for $\tr(S)$ converges.

Answer 2

The answer is yes. It suffices to show that $$ \operatorname{tr}(P_k) = \sum_{j=1}^k \langle e_j,P e_j\rangle, $$ where $\{e_1,e_2,\dots\}$ denotes the standard basis. Because $P$ is trace class, the sum $\operatorname{tr}(P) = \sum_{j=1}^\infty \langle e_j,P e_j\rangle$ is (absolutely) convergent, so its partial sums $\operatorname{tr}(P_k) = \sum_{j=1}^k \langle e_j,P e_j\rangle$ converge to the overall sum, which was what we wanted.

And yes, the operators $P_k$ will necessarily converge to $P$, as is noted here.

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Attempted proof that $P_k \to P$:

Denote $W_k = \text{id} - V_k$. We have $$ \|P-P_k\|_1 = \|W_kP + PW_k - W_kPW_k\| \leq \|W_kP\| + \|PW_k\| + \|W_kPW_k\|. $$ Thus, it suffices to show that $W_kP \to 0$ and $PW_k \to 0$. Because $(PW_k)^* = W_kP^*$, it suffices to show that $W_k P \to 0$.