Weak-* convergence and trace-class operators

Question

I am reading a proof by Barry Simon, and he makes a statement equivalent to the following:

Let $X$ be the space of compact operators over a Hilbert space $H$. Let $\{y_n\}_n$, $y_n>0$, be a bounded sequence in the dual space $X'$ of trace-class operators over $H$. (In fact we have $1 \leq tr (y_n) \leq g < +\infty$.) By Banach-Alaoglu, there is a weak-* convergent subsequence $\{y_n'\}$, i.e., there exists $y\in X'$ such that for every $A\in X$, $tr(Ay_n')\rightarrow tr(Ay)$.

This is fine. Then he claims, and this I don't understand: "Clearly, $y \geq 0$ and therefore $\liminf_n tr(y_n) \geq tr(y).$"

Why can one conclude $y \geq 0$ easily, and how does the liminf-inequality follow easily? What am I missing? It is unclear from the text whether he considers the weak-* convergent subsequence in the quoted statement.

When I attempt at making my own argument, it becomes much stronger and more involved:

Use the subsequence: Can one argue by setting $A = P_k$, a sequence of projectors, and consider $y_{kn} = P_k y_n$? We have $tr(y_n)=\lim_k tr(P_k y_n)$, so that for any $\epsilon>0$ there exists a $K^n_\epsilon$ such that for all $k>K^n_\epsilon$ and all $n$ sufficiently large,

$$ | tr(y) - tr(P_ky_n)| \leq |tr(y) - tr(y_n)| + |tr(y_n) - tr(P_ky_n)| < \epsilon $$

If this is correct, then $tr(y_n)\rightarrow tr(y) > 0$.

Answer 1

The fact that $y\geq0$: as $y_n'\geq0$ for all $n$ by hypothesis, we get, for any $\xi\in H$, and if we denote by $P_\xi$ the orthogonal projection onto $\mathbb{C}\xi$, $$ \langle y\xi,\xi\rangle=\mbox{tr}(yP_\xi)=\lim\mbox{tr}(y_n'P_\xi)=\lim\mbox{tr}(P_\xi y_n'P_\xi)\geq0, $$ the last equality following from $y_n'\geq0$ and $P_\xi$ selfadjoint (it's actually positive). So $y\geq0$.

Edit:

The assertion $\liminf_n\mbox{tr}(y_n')\geq\mbox{tr}(y)$ is a particular case of the lower semicontinuity of the trace in the weak-$*$ topology.

Indeed, for each finite rank projection $P$ the function $x\mapsto\mbox{tr}(xP)$ is weak-$*$ continuous. And, moreover, if we fix an orthonormal basis $\{\xi_n\}$ and let $P_k=\sum_1^kP_{\xi_n}$, then $$ \mbox{tr}(x)=\sup_k\mbox{tr}(xP_k),\ \ \ \ \ \ x\in X'. $$ This shows that in our situation the trace is a supremum of continuous functions, and thus lower semicontinuous.