Compactness of subset of trace-class operators on a Hilbert space

Question

Consider an infinite-dimensional, complex and separable Hilbert space $H$ and let $\mathcal I(H)$ denote the space of trace-class operators.

The set of density operators is defined by $$\mathcal S(H):=\{\rho\in\mathcal I(H)\,|\,\mathrm{Tr}\rho =1\,,\, \rho\geq 0 \}\quad .$$

Question: Is it true that $\mathcal S(H)$ is compact with respect to the weak$^*$-topology?

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Here are my ideas: I think it holds that weak$^*$-closed and norm bounded subsets of $\mathcal I(H)$ are weak$^*$-compact and vice versa, correct? Then, since $\mathcal S(H)$ is (trace-)norm bounded, I tried to prove that it is also weak$^*$-closed to arrive at the desired result.

To this end, I proved that if $(\rho_n)_{n\in\mathbb N}$ with $\rho_n\in\mathcal S(H)$ converges in the weak$^*$ sense ($\mathrm{Tr}\rho_n A\to \mathrm{Tr}\rho A$ for all compact $A$) to some $\rho\in \mathcal I(H)$, then $\rho \geq 0$, which follows from picking one-dimensional projectors $A=|\psi\rangle\langle \psi|$ for $\psi\in H$. I failed however in proving that the trace condition holds.

Perhaps there is a smarter way in proving the assertion (if it is true)?

Answer 1

Nice observations. It is not true that $S(H)$ is compact, though. The element $0$ is in the weak$^*$-closure of $S(H)$, while not being in $S(H)$. Concretely, fix an orthonormal basis $\{\psi_k\}$ and let $e_k$ be the corresponding rank-one projections (that is, $e_k\eta=\langle \eta,\psi_k\rangle\,\psi_k$, or $e_k=|\psi_k\rangle\langle\psi_k|$ if you prefer bra-ket notation). We have that $e_k\in S(H)$ for all $k$, and we also have that $e_k\to0$ in the weak$^*$-topology.

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Edit: I cannot think of an entirely straightforward argument to show that $e_k\to0$ weak$^*$. So here it is. We want to show that $\def\tr{\operatorname{Tr}}\tr(e_kT)\to0$ for all compact $T$.

The first reduction is to see that it is enough to assume that $T\geq0$. This comes from using Hölder's inequality: $$ |\tr(e_kT)|^2=|\tr(Te_k)|^2\leq\tr(e_kT^*Te_k)\tr(e_k)=\tr(e_kT^*T). $$

Next, we may assume that $T$ is positive and finite-rank. Because if $T$ is an arbitrary positive compact operator, by the Spectral Theorem we can find positive finite-rank operators $T_n$ with $\|T_n-T\|\to0$. Then $$ |\tr(e_k(T_n-T))|\leq\|T_n-T\|\,\tr(e_k)=\|T_n-T\|. $$ So given $\def\e{\varepsilon}\e>0$ we can choose $n$ such that $\|T_n-T\|<\e$. And we can choose $k_0$ such that for all $k\geq k_0$ we have that $|\tr(e_kT_n)|<\e$. Then for all $k\geq k_0$ we have $$ |\tr(e_kT)|\leq|\tr(e_k(T-T_n))|+|\tr(e_kT_n)|<2\e. $$

A positive, finite-rank compact operator is a linear combination of rank-one projections. So it is enough to show that $\tr(e_kP)\to0$ for a rank-one projection $P$. The projection $P$ is of the form $P\eta=\langle\eta,\psi\rangle\,\psi$ for a unit vector $\psi$. Then, calculating the trace through an orthonormal basis where $\psi$ is the first element,
$$ \tr(e_kP)=\langle e_k\,\psi,\psi\rangle=|\langle\psi,\psi_k\rangle|^2\to0 $$ by Bessel's Inequality.

Answer 2

$\mathcal S(H)$ need not be weak* compact. Consider the operator sequence $(A_n)$ in $\mathcal S(\ell^2(\mathbb N))$: $$A_n=\begin{pmatrix} \frac{1}{n} & & & & & & \\ & \frac{1}{n} & & & & & \\ & & \ddots & & & & \\ & & & \frac{1}{n} & & &\\ & & & & 0 & &\\ & & & & & 0 & \\ & & & & & & \ddots \end{pmatrix}.$$ For any finite-rank operator $B$, we have $$0\leq |\operatorname{Tr}(A_nB)|\leq \|A_n\|\operatorname{Tr}(|B|)=\frac{1}{n}\operatorname{Tr}(|B|) \to 0 \quad (n\to \infty).$$ For any compact operator $C$ and any $\epsilon>0$, take a finite-rank operator $B$ such that $\|B-C\|<\epsilon$, which follows that $|\operatorname{Tr}(A_nB)-\operatorname{Tr}(A_nC)|\leq \|B-C\|\operatorname{Tr}(|A_n|)<\epsilon$ for ALL $n$. Consequently, $\operatorname{Tr}(A_nC)$ converges to $0$ for any compact operator $C$. Therefore $A_n$ weak* converges to $0\notin \mathcal S(H)$.