# Does the top left block of an operator converge?

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.

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

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.

• By the way---where in your argument do you use the fact that $V_k$ is an orthogonal projection? It seems the key property you needed is that $(I - V_k) e_j = e_j$ for all $j > k$ and $0$ otherwise. Sep 27 at 17:34
• Yes, absolutely, that's all that's needed. That condition does imply that $V_k$ is a projection, though. Sep 27 at 18:32

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.

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 $$W_k P \to 0$$.

• Doh. I agree. One can just see this by noting that (viewing these as infinite dimensional matrices), trace of $P_k$, $P$ are the diagonal sum of the first $k$ entries, and all entries on diagonal, respectively, and as you note the latter is the limit of the former. Regarding my second question: does this convergence extend to other norms? E.g. do we have $\|P_k - P\| \to 0$? Sep 26 at 18:20
• @Drew Yes: the fact that $\|P_k - P\|_1 \to 0$ allows you to conclude that $\|P_k - P\| \to 0$ simply because $\|A\| \leq \|A\|_1$ for all trace-class operators $A$. Also, convergence relative to the trace norm implies convergence relative to the Hilbert Schmidt norm. Sep 26 at 18:26
• But didn't you only show that $\mathrm{tr}(P_k - P) \to 0$? Sep 26 at 18:27
• Yes, I'm working on the proof that $P_k \to P$ Sep 26 at 18:27
• I see, so you are claiming that additionally we have $\|P_k - P\|_1 \to 0$. (Which is stronger than the trace claim.) Sep 26 at 18:28