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Suppose $P, Q$ are two positive, self-adjoint trace-class operators on $\ell^2$. Let $V_k \colon \ell^2 \to \mathbb{R}^k$ denote the projection onto the first $k$ coordinates. Suppose also that $Q$ is strictly positive in the sense that $\langle Qx, x \rangle > 0$ for all $x \neq 0$.

Consider the sequence of operators: $$ P_k = V_k P V_k^\ast, \quad \mbox{and} \quad Q_k = V_k Q V_k^\ast, \quad \mbox{for}~k \geq 1. $$ I am wondering if it is true that $$ \lim_{k \to \infty} \mathrm{tr}(Q_k (P_k Q_k + I)^{-1}) = \mathrm{tr}(Q (P Q + I)^{-1}). $$ The operators $I$ above are (slightly abusing notation) the identity on $\mathbb{R}^k$ and $\ell^2$, respectively. In the case where $P, Q$ are diagonal I can see that this is true, but otherwise I am unsure.

Drew Brady
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1 Answers1

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I'm not entirely sure how to show $PQ + I$ is invertible in infinite dimensions (Edit: a proof of this fact following @MartinArgerami 's arguments has been added to the end of this answer), but assuming that isn't an issue, the following provides a proof that your proposition is correct:

Since $Q$ is strictly positive, $Q^{-1/2}$ is a densely-defined, closed operator. Then $PQ + I = Q^{-1/2}Q^{1/2}PQ^{1/2}Q^{1/2} + Q^{-1/2}Q^{1/2} = Q^{-1/2}(Q^{1/2}PQ^{1/2} + I)Q^{1/2}$. Observe that $Q^{1/2}PQ^{1/2}$ is positive, hence $Q^{1/2}PQ^{1/2} + I$ is invertible. As such, we may consider the operator $Q^{-1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2}$. We see that it is closed and defined on the range of $Q^{-1/2}(Q^{1/2}PQ^{1/2} + I)Q^{1/2} = PQ + I$, which, as $PQ + I$ is invertible, is the entire $l^2$. Hence, it is bounded and it is then easy to see it is exactly the inverse of $Q^{-1/2}(Q^{1/2}PQ^{1/2} + I)Q^{1/2} = PQ + I$. (In finite dimensions this produces a proof that $PQ + I$ is invertible. Domain issues in infinite dimensions seem to complicate things.) Hence, $Q(PQ + I)^{-1} = QQ^{-1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2} = Q^{1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2}$. The same, of course, holds for $Q_k$ and $P_k$, so we need to show $\lim_{k \rightarrow \infty} \mathrm{Tr}(Q_k^{1/2}(Q_k^{1/2}P_kQ_k^{1/2} + I)^{-1}Q_k^{1/2}) = \mathrm{Tr}(Q^{1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2})$.

From this point onwards the best way I know is through functional calculus. We first note the following: For any fixed polynomial $f$ with $f(0) = 0$, since both $P$ and $Q$ are compact, we see that $Q_k^{1/2}P_kQ_k^{1/2} \rightarrow Q^{1/2}PQ^{1/2}$ in norm, hence $f(Q_k^{1/2}P_kQ_k^{1/2}) \rightarrow f(Q^{1/2}PQ^{1/2})$ in norm. Now, we note that $Q_k^{1/2}P_kQ_k^{1/2}$ are uniformly bounded in norm, so there exists a sequence of polynomials $g_n$ which approximates $(\cdot + I)^{-1}$ in a uniform way, i.e., for each $\epsilon > 0$, there exists an $N$ s.t. for all $n \geq N$, $||g_n(Q_k^{1/2}P_kQ_k^{1/2}) - (Q_k^{1/2}P_kQ_k^{1/2} + I)^{-1}|| \leq \epsilon$ for all $k$ and furthermore $||g_n(Q^{1/2}PQ^{1/2}) - (Q^{1/2}PQ^{1/2} + I)^{-1}|| \leq \epsilon$. Since $(0 + 1)^{-1} = 1$, we may choose $g_n$ s.t. $g_n(0) = 1$. Furthermore, such $g_n$ may be chosen so that they are uniformly bounded on the spectra of all $Q_k^{1/2}P_kQ_k^{1/2}$ as well as $Q^{1/2}PQ^{1/2}$, i.e., there exists $C' > 0$ s.t. $||g_n(Q_k^{1/2}P_kQ_k^{1/2})||, ||g_n(Q^{1/2}PQ^{1/2})|| \leq C'$ for all $n, k$. Let $f_n = g_n - 1$, which are then polynomials satisfying $f_n(0) = 0$. Let $C = C' + 1$, we observe that $||f_n(Q_k^{1/2}P_kQ_k^{1/2})||, ||f_n(Q^{1/2}PQ^{1/2})|| \leq C$ for all $n, k$.

Choose a large enough $N$ satisfying the condition in the preceding paragraph. Now, for any $k$,

$$|\mathrm{Tr}(Q_k^{1/2}(Q_k^{1/2}P_kQ_k^{1/2} + I)^{-1}Q_k^{1/2}) - (\mathrm{Tr}(Q_kf_N(Q_k^{1/2}P_kQ_k^{1/2})) + \mathrm{Tr}(Q_k))|$$

$$= |\mathrm{Tr}(Q_k(Q_k^{1/2}P_kQ_k^{1/2} + I)^{-1}) - \mathrm{Tr}(Q_k(f_N(Q_k^{1/2}P_kQ_k^{1/2}) + I))|$$

$$= |\mathrm{Tr}(Q_k(Q_kP_kQ_k^{1/2} + I)^{-1}) - \mathrm{Tr}(Q_kg_N(Q_k^{1/2}P_kQ_k^{1/2}))|$$

$$= |\mathrm{Tr}(Q_k((Q_kP_kQ_k^{1/2} + I)^{-1} - g_N(Q_k^{1/2}P_kQ_k^{1/2})))|$$

$$\leq \mathrm{Tr}(Q_k)||(Q_kP_kQ_k^{1/2} + I)^{-1} - g_N(Q_k^{1/2}P_kQ_k^{1/2})||$$

$$\leq \mathrm{Tr}(Q)\epsilon$$

Where in the final step we used $\mathrm{Tr}(Q_k) \leq \mathrm{Tr}(Q)$. Similarly,

$$|\mathrm{Tr}(Q^{1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2}) - (\mathrm{Tr}(Qf_N(Q^{1/2}PQ^{1/2})) + \mathrm{Tr}(Q))| \leq \mathrm{Tr}(Q)\epsilon$$

Since $f_N$ is a polynomial satisfying $f_N(0) = 0$, by what we have proved before we have $f_N(Q_k^{1/2}P_kQ_k^{1/2}) \rightarrow f_N(Q^{1/2}PQ^{1/2})$ in norm. We also have $\mathrm{Tr}(Q_k) \rightarrow \mathrm{Tr}(Q)$, so for large $k$ we have $||f_N(Q_k^{1/2}P_kQ_k^{1/2}) - f_N(Q^{1/2}PQ^{1/2})|| \leq \epsilon$ and $|\mathrm{Tr}(Q_k) - \mathrm{Tr}(Q)| \leq \epsilon$. Thus, for large $k$,

$$|\mathrm{Tr}(Q_kf_N(Q_k^{1/2}P_kQ_k^{1/2})) - \mathrm{Tr}(Qf_N(Q^{1/2}PQ^{1/2}))|$$

$$\leq |\mathrm{Tr}(Q_k(f_N(Q_k^{1/2}P_kQ_k^{1/2}) - f_N(Q^{1/2}PQ^{1/2})))| + |\mathrm{Tr}((Q_k - Q)f_N(Q^{1/2}PQ^{1/2}))|$$

$$\leq \mathrm{Tr}(Q_k)||f_N(Q_k^{1/2}P_kQ_k^{1/2}) - f_N(Q^{1/2}PQ^{1/2})|| + |\mathrm{Tr}(Q_k) - \mathrm{Tr}(Q)|||f_N(Q^{1/2}PQ^{1/2})||$$

$$\leq \mathrm{Tr}(Q)\epsilon + C\epsilon$$

Hence, putting everything together, we have, for large $k$,

$$|\mathrm{Tr}(Q_k^{1/2}(Q_k^{1/2}P_kQ_k^{1/2} + I)^{-1}Q_k^{1/2}) - \mathrm{Tr}(Q^{1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2})|$$

$$\leq |\mathrm{Tr}(Q_k^{1/2}(Q_k^{1/2}P_kQ_k^{1/2} + I)^{-1}Q_k^{1/2}) - (\mathrm{Tr}(Q_kf_N(Q_k^{1/2}P_kQ_k^{1/2})) + \mathrm{Tr}(Q_k))| + |\mathrm{Tr}(Q^{1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2}) - (\mathrm{Tr}(Qf_N(Q^{1/2}PQ^{1/2})) + \mathrm{Tr}(Q))| + |\mathrm{Tr}(Q_kf_N(Q_k^{1/2}P_kQ_k^{1/2})) - \mathrm{Tr}(Qf_N(Q^{1/2}PQ^{1/2}))| + |\mathrm{Tr}(Q_k) - \mathrm{Tr}(Q)|$$

$$\leq \mathrm{Tr}(Q)\epsilon + \mathrm{Tr}(Q)\epsilon + \mathrm{Tr}(Q)\epsilon + C\epsilon + \epsilon$$

$$= (3\mathrm{Tr}(Q) + C + 1)\epsilon$$

Letting $\epsilon \rightarrow 0$ concludes the proof.

A proof that $PQ + I$ is invertible in infinite dimensions following @MartinArgerami 's arguments:

We show that $\sigma(PQ) \subseteq [0, \infty)$, which is sufficient since then $-1 \notin \sigma(PQ)$, so by definition of the spectrum we have $PQ + I$ is invertible.

We note that $PQ = P^{1/2}(P^{1/2}Q)$. We note the following general fact: In a unital algebra (for example $\mathbb{B}(l^2)$), for any $a, b$ in the algebra, $\sigma(ab)\setminus\{0\} = \sigma(ba)\setminus\{0\}$. Indeed, by scaling and symmetry, it is sufficient to show that $1 - ab$ is invertible implies $1 - ba$ is invertible. A direct calculation shows that if $c$ is the inverse of $1 - ab$, then $1 + bca$ is the inverse of $1 - ba$. (This proof can be found in Remark 1.2.1 in $C^*$ -algebras and operator theory by Gerard Murphy.) Applying this fact to $a = P^{1/2}, b = P^{1/2}Q$, we see that $\sigma(PQ) \subseteq \{0\} \cup \sigma(P^{1/2}QP^{1/2})$. Since $P^{1/2}QP^{1/2} \geq 0$, $\sigma(P^{1/2}QP^{1/2}) \subseteq [0, \infty)$. This proves the claim.

David Gao
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  • I haven't read your answer yet, but the reason $PQ+I$ is invertible is that $\sigma(PQ)\subset[0,\infty)$ (see proof here) and so $\sigma(PQ+I)\subset[1,\infty)$. – Martin Argerami Oct 05 '23 at 15:22
  • @MartinArgerami I see. You would need a slight alteration noting that in any unital Banach algebra (or any unital algebra, for that matter), $\sigma(ab)\setminus{0} = \sigma(ba)\setminus{0}$ instead of saying $F^(FB)$ and $(FB)F^$ have the same eigenvalues. Other than that, everything should carry over to infinite dimensions. – David Gao Oct 05 '23 at 23:03
  • These are compact operators, so the nonzero part of the spectrum consists of eigenvalues. And in an arbitrary Banach algebra you are not guaranteed to have Spectral Theorem, nor square roots, so I'm not sure what you could do there. – Martin Argerami Oct 06 '23 at 01:31
  • @MartinArgerami You're right. I'm not asserting this holds in an arbitrary Banach algebra or anything like that, just saying applying the general result for unital Banach algebras to our current case seems to be the more natural and easier way for me. But that's just my preference. – David Gao Oct 06 '23 at 01:43
  • Could you elaborate "The same, of course, holds for $Q_k$ and $P_k$"? It looks you started with the assumption "$Q$ is strictly positive", but it is not held for $Q_k.$ – dsh Oct 06 '23 at 09:42
  • @dsh $Q_k$ is strictly positive if $Q$ is. Indeed, for any nonzero $x \in \mathbb{C}^k$ (or $\mathbb{R}^k$), $\langle Q_kx, x\rangle = \langle V_kQV_k^x, x\rangle = \langle QV_k^x, V_k^x\rangle$. Since $V_k^$ is an isometry, $V_k^x \neq 0$, so $\langle Q_kx, x\rangle = \langle QV_k^x, V_k^*x\rangle > 0$. – David Gao Oct 06 '23 at 11:33
  • @DavidGao, $V_k$ is "projection onto the first $k$ coordinates". – dsh Oct 06 '23 at 11:40
  • @dsh In the way it's written in the question, $V_k$ is understood to have range $\mathbb{C}^k$ instead of $l^2$, so I interpret $Q_k$ to be an element of the $k \times k$ matrix algebra. In that sense I'm saying it is strictly positive. – David Gao Oct 06 '23 at 11:45
  • @DavidGao, in that case operators involving $Q_k$ are from $\mathbb{R}^k\to \mathbb{ R}^k$ and those involving $Q$ are from $\ell^2\to \ell^2.$ – dsh Oct 06 '23 at 11:51
  • @dsh Both interpretations would give the same value for $\mathrm{Tr}(Q_k(P_kQ_k + I)^{-1})$, so it doesn't really matter for this question. I also don't really need the assumption that $Q$ or $Q_k$ is strictly positive anyway. You can perturb them to strictly positive trace-class operators. Letting the perturbation go to zero then shows that $Q(PQ + I)^{-1} = Q^{1/2}(Q^{1/2}PQ^{1/2} + I)^{-1}Q^{1/2}$ still work without the strict positivity assumption and the same works for $Q_k$ and $P_k$. After that, the proof never used strict positivity, so everything can proceed as before. – David Gao Oct 06 '23 at 11:52
  • @dsh True, but I don't see why that matters? Traces are defined on both, and the question even mentions that $I$ was supposed to be used for the identities both on $\mathbb{R}^k$ and $l^2$. – David Gao Oct 06 '23 at 11:55
  • @dsh Admittedly I abused notations a lot in the proof. For example I talked about how $f(Q_k^{1/2}P_kQ_k^{1/2})$ converges to $f(Q^{1/2}PQ^{1/2})$ in norm. For these just pretend I meant $V_k^f(Q_k^{1/2}P_kQ_k^{1/2})V_k$ instead of $f(Q_k^{1/2}P_kQ_k^{1/2})$. I was just being lazy and didn't want to write all those $V_k$ and $V_k^$'s. – David Gao Oct 06 '23 at 11:58
  • @DavidGao, If $|\mathrm{tr}(AB)| \le |\mathrm{tr} A| |B|$ for trace-class $A$ and bounded $B,$ then $(A,B)\mapsto AB$ is separately continuous and thus simultaneously continuous due to corresponding theorem for (locally convex) Frechet spaces. As $A\mapsto \mathrm{tr} A$ is continuous linear functional on trace-class then everything should follow from independent convergences $Q_n\to Q$ in trace norm and $(P_k Q_k + I)^{-1}\to (PQ + I)^{-1}$ in the $| \cdot |$ norm. Am I missing something? – dsh Oct 06 '23 at 12:08
  • @dsh That should be mostly right. Technically the second convergence is wrong since the two identities are not the same, but you could argue (as I've implicitly done in the comments) that for the purpose of calculating the traces in the question it doesn't matter so you could just change to the identity on $l^2$. After that it seems right. I'm just unsure whether norm convergence implies norm convergence of the inverses is true in general and I don't know enough about holomorphic functional calculus to prove it... – David Gao Oct 06 '23 at 12:28
  • @dsh So I just tried to manipulate the formulae so that I have a self-adjoint element in the inverse. In that case it's continuous functional calculus and I'm more confident with how everything works. Things after that really are the same as in your argument, I just tried to write down all the analytic details in case they are needed. I suppose it have become too long in my pursuit of that. That was my fault. You should write an answer giving your version of the argument. – David Gao Oct 06 '23 at 12:31
  • @DavidGao Could you share reference for $|\mathrm{tr}(AB)| \le |\mathrm{tr} A| |B|$ or provide simple proof? I have seen similar inequality for Hilbert-Schmidt only. You collected all needed data, I only suggest simplification. If you decide not to simplify, I will write an answer some day later. – dsh Oct 06 '23 at 12:35
  • @dsh Theorem 2.4.16 in $C^$ -algebras and operator theory* by Gerard Murphy gives a proof. (A small correction: it should be $|\mathrm{tr}(AB)| \leq \mathrm{tr}(|A|)||B||$, so $\mathrm{tr}(|A|)$ instead of $|\mathrm{tr}(A)|$.) I'll leave it to you to write an answer then. – David Gao Oct 06 '23 at 12:45
  • @dsh ...And I just now found a reference to the inverse being norm-continuous as well. Theorem 1.2.3 in the same book, in case you also want to cite that. – David Gao Oct 06 '23 at 12:46
  • @DavidGao. Great, thank you! You collected all the data needed;) – dsh Oct 06 '23 at 12:50