$\sigma$-weak topology versus $\sigma$-strong topology on the bounded operators $B(H)$. 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.
 A: The reason this is looking so complicated is perhaps because of the fancy  definitions of the various topologies that we are using here.  As far as I can tell the "classical" definitions are:

*

*weak topology: seminorms
$$
  x\ni B(H) \mapsto  |\langle x\xi , \eta \rangle |,
  $$
for  $\xi ,\eta \in H$.


*$\sigma $-weak topology: seminorms
$$
  x\ni B(H) \mapsto  \Big|\sum_{n=1}^\infty \langle x\xi _n, \eta _n\rangle \Big|,
  \tag 1
  $$
for  $\{\xi _n\}_n,\{\eta _n\}_n\subseteq H$, with $\sum_{n=1}^\infty \|\xi _n\|\|\eta _n\|<\infty $.


*strong topology: seminorms
$$
  x\ni B(H) \mapsto  \|x\xi \|,
  $$
for  $\xi \in H$.


*$\sigma $-strong topology: seminorms
$$
  x\ni B(H) \mapsto  \big (\sum_{n=1}^\infty \|x\xi _n\|^2\big )^{1/2},
  $$
for  $\{\xi _n\}_n\subseteq H$, with $\sum_{n=1}^\infty \|\xi _n\|^2<\infty $.

Observing that Cauchy-Schwarz implies
$$
  \Big|\sum_{n=1}^\infty \langle x\xi _n, \eta _n\rangle \Big| \leq \big (\sum_{n=1}^\infty \|x\xi _n\|^2\big )^{1/2} \big (\sum_{n=1}^\infty \|\eta _n\|^2\big )^{1/2},
  $$
the answer becomes immediate  (it might also help to observe that, by multiplying $\xi _n$ by $\lambda $, and $\eta _n$ by $1/\lambda $ in
(1), one may assume that $\|\xi _n\| =  \|\eta _n\|$).

Of course one could also ask why are these "classical" definitions equivalent to the ones given by the OP, but
that is another story.
A: 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 $B(H)_*^+$.
A: You have, for trace-class $z$ and using Cauchy-Schwarz, and writing $z=v|z|$ for its polar decomposition,
\begin{align}
|\operatorname{Tr}((x_\alpha-x)z)
&= \operatorname{Tr}(|z|^{1/2}\,(x_\alpha-x)v|z|^{1/2}\,)\\[0.3cm]
&\leq \operatorname{Tr}(|z|^{1/2}|z|^{1/2})^{1/2}\operatorname{Tr}(|z|^{1/2}v^*(x_\alpha-x)^*(x_\alpha-x)v|z|^{1/2})^{1/2}\\[0.3cm]
&=\operatorname{Tr}(|z|)^{1/2}\operatorname{Tr}(zv^*(x_\alpha-x)^*(x_\alpha-x))^{1/2}\\[0.3cm]
&\to0.
\end{align}
