Why does there always exist a faithful state in a separable $C^\ast$-algebra?
Existence of faithful state in $C^\ast$-algebras
4
$\begingroup$
functional-analysis
operator-theory
c-star-algebras
-
2user16283 If @t.b.'s comment was sufficient, perhaps you can post (and accept) an answer yourself to check that you understood completely. – 2011-10-28
1 Answers
4
Well, since the unit ball in the dual space is weak-$^*$-separable, there is a separating sequence of states $\omega_n$, then just sum them up $\omega=\sum 2^{-n}\omega_n$ and check that you get a faithful state.
Credits goes to t.b.