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 $tr(y_n)\rightarrow tr(y) > 0$.