I am reading Stephane Attals Book Open Quantum Systems 1, chapter Elements of Operator Algebras and Modular Theory. And have issues understanding the proof given here.
- First of all in the "introduction" why is $\Psi(M_1)$ compact in $\mathbb{C}$?
- Second: Is there an easy way to see why the $\sigma$-weak and weak topology coincide ?
- Third: in the proof (i) we should show $f\in M_*$ by showing that $f$ is actually weakly continuous. Since he already argued (my first point) that $M_* \subseteq M^*$ and the defined sequence $f_n \in M_*$ converges in $M^*$, $f\in M^*$ as well and because the topological dual of a space is the set of the continuous linear functionals, $f$ is continuous on $M$. Since $M_1 \subset M$, $f$ is contiuous on $M_1$. Continuity is stronger than weak continuity. So why does weak continuity has to be shown and why not (if anything at all) continuity, which is required by the definition of $M_*$
Thanks a lot in advance.
Because continuous functions map compact sets to compact sets.
Yes. A net $\{A_j\}$ converges weakly to $0$ if $A_jx\to0$ weakly for all $x\in H$. Meanwhile, $\sigma$-weak convergence is the fact that $\def\tr{\operatorname{Tr}}\tr(A_jR)\to0$ for all trace-class $R$. When $R$ is rank-one, say $Rx=\langle x,y\rangle z$, then $$ \tr(A_jR)=\langle A_jx,y\rangle\,\|z\|\to0. $$ When $R$ is finite-rank, it is sum of rank-one operators, and the above gives $\tr(A_jR)\to0$ by linearity. In general, $R=\lim_nR_n$ with $R_n$ finite rank. Then, if $\|A_j\|\leq c$ for all $j$, $$ |\tr(A_jR)|\leq|\tr(A_jR_n)|+|\tr(A_j(R-R_n)|\leq|\tr(A_jR_n)|+c\,\|R-R_n\|_1. $$ This gives $$ \limsup_j|\tr(A_jR)|\leq c\|R-R_n\|_1. $$ We are still free to choose $n$, so $\limsup_j|\tr(A_jR)|=0$, which shows that the limit exists and is zero.
Continuity is stronger than weak continuity. No. It's harder to be continuous in the weaker topology, not easier. You already know that $M_*\subsetneq M^*$, which in particular says that there are norm-continuous functionals which are not normal.