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

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1 Answer 1

up vote 4 down vote accepted

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.

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Thanks! $y\geq 0$ was easy, I see that now ... But there is something I don't get about your discussion of the second fact: Here, you actually prove that $tr(y) \geq \liminf tr(y_n'')$: your assumption that leads to a contradiction is $tr(y)<\liminf tr(y_n'')$, or am I horribly mistaken? Also, I don't quite get the finite-rank-projection-step. When I try to do that, I only get $\geq 0$ for the last equation due to a $-\epsilon$ for the finite-rank convergence, $tr((y''_n-y)p) \in ]tr(y''_n-y)-\epsilon,tr(y''_n-y)+\epsilon[$. –  Simen K. Sep 13 '12 at 7:45
    
You are right, that argument is completely wrong. I'm running out to teach now, so I'll erase it for now and I'll check it in a few hours. –  Martin Argerami Sep 13 '12 at 14:48
    
Thanks! Your effort is appreciated. I will answer my own question if I find it out. –  Simen K. Sep 14 '12 at 7:26
    
I made a new attempt. I hope it works! –  Martin Argerami Sep 14 '12 at 18:55
    
Thanks, I understand! The pointwise sup of l.s.c. functions is l.s.c. My google fu also led me to Advanced Real Analysis by Knapp, Proposition 4.13. scribd.com/doc/64737846/Advanced-Real-Analysis#page=140 –  Simen K. Sep 15 '12 at 20:33

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