Let $H$ be a Hilbert space and let $T(H)$ be the traces class operators on $H$. Denote the bounded operators on $H$ by $B(H)$. We define the normal functionals on $B(H)$ by
$$B(H)_*:= \{\omega \in B(H)^*: \exists x \in T(H): \forall y \in B(H): \omega( y) = \text{Tr}(xy)\}$$
Let $B(H)_*^+$ denote the positive normal functionals.
We can introduce the $\sigma$-weak and the $\sigma$-strong topology on the space $B(H)$. The former is the locally convex topology generated by the family of seminorms
$$ B(H)\ni x \mapsto |\omega(x)|, \ \omega\in B(H)_*$$
The latter is the locally convex topology generated by the family of seminorms
$$B(H)\ni x \mapsto \omega(x^*x)^{1/2}, \ \omega \in B(H)_*^+$$
I want to show that the $\sigma$-weak topology is weaker than the $\sigma$-strong topology (as the name suggests!). More precisely, suppose that $x_\alpha \to x$ in the $\sigma$-strong topology. Then I want to show also that $x_\alpha \to x$ in the $\sigma$-weak topology. How can I show this?
We need to show
$$\forall \omega \in B(H)_*: |\omega(x_\alpha)-\omega(x)| \to 0$$
using something about positive normal functionals but I'm unsure how to approach this.
Best Answer
We for $\omega\in B(H)_*^+$ we have, $$|\omega(x-x_\alpha)|=|\omega(1^*(x-x_\alpha)|\leq \omega(1^* 1)^{1/2} \omega((x-x_\alpha)^*(x-x_\alpha))^{1/2}$$.
by applying Cauchy-Schwarz to the sesquilinear form $(x, y) \mapsto \omega(y^*x)$.
For a general $\omega\in B(H)_*$ we pick $x\in B(H)$ such that $\omega=\text{Tr}(x\cdot)$ and write $x$ as a linear combination of positive elements $x=\sum \alpha_i x_i$. Then $\omega=\sum \alpha_i\text{Tr}(x_i \cdot)$ and it easy to see that the functionals $\text{Tr}(x_i\cdot)$ are positive: write $x_i=z^*z$, then we have $\text{Tr}(x_i(y^*y))=\text{Tr}(z^*zy^*y)=\text{Tr}(yz^*zy^*)=\text{Tr}((zy^*)^*zy^*)\geq 0$ for all $y\in B(H)$. We conclude that $\omega$ is a linear combination of elements in $B(H)_*^+$.