GNS-representation corresponding to a strongly continuous functional is strongly continuous on a bounded set

Question

Given a strongly continuous linear functional $\omega:B(H)\to \mathbb{C}$, I would like to show that the corresponding GNS-representation $\varphi_{\omega}$ of $\omega$ is strongly continuous on the unit ball of $B(H)$.

So I need to show that given a strongly converging sequence $u_k \to u$ in the unit ball of $B(H)$, we have $\varphi_{\omega}(u_k)\to \varphi_{\omega}(u)\in B(H_{\omega})$ strongly where $H_{\omega}:=\overline{B(H)/N_{\omega}}$, $N_{\omega}=\{a\in B(H):\omega(a^*a)=0\}$, the Hilbert space obtained by $\omega$. For this, we need that for all $a\in H_{\omega}$, $||\varphi_{\omega}(u_k)(a)- \varphi_{\omega}(u)(a)||\to 0$ where the norm is obtained by the inner product on $H_{\omega}$. However, I run into issues as if I write this norm expression in terms of inner product, I get that for $a\in B(H)/N_{\omega}$ $$ ||\varphi_{\omega}(u_k)(a)- \varphi_{\omega}(u)(a)||^2=\omega(a^*(u_k-u)^*(u_k-u)a) $$ and we have that the adjoint operator is not necessarily strongly continuous. Now the provided solution has $$ \langle \varphi_{\omega}(u_k)a,b\rangle=\omega(b^*u_ka)\to\omega(b^*ua)=\langle \varphi_{\omega}(u)a,b\rangle $$ and using a $3\epsilon$-argument, concludes that $ \langle \varphi_{\omega}(u-u_k)a,b\rangle \to 0 $ for all $a,b \in H_{\omega}$. I am confused as I thought the condition $\langle \varphi_{\omega}(u_k)a,b\rangle\to\langle \varphi_{\omega}(u)a,b\rangle$ was the definition of (relatively) weak convergence. If I am mistaken, how does this condition imply the first condition $||\varphi_{\omega}(u_k)(a)- \varphi_{\omega}(u)(a)|| \to 0 $ ?

Answer 1

Note that you may assume without loss of generality that $u=0$, since the topologies involved are linear.

Note also that you need to assume that $\omega$ is positive, for otherwise you don't get an inner product in your GNS.

Since $$ \|u_kx\|^2=\langle u_k^*u_kx,x\rangle=\langle |u_k|^2x,x\rangle=\|\,|u_k|x\,\|^2, $$ we have that $u_k\to0$ strongly if and only if $|u_k|\to0$ strongly.

Also, $$ |u_k|^2=|u_k|^{1/2}|u_k|\,|u_k|^{1/2}\leq|u_k|^{1/2}\,\|u_k\|\,|u_k|^{1/2}=|u_k|,$$ so $$ \omega(a^*u_k^*u_ka)=\omega(a^*|u_k|^2a)\leq\omega(a^*|u_k|a)\to0 $$ since $a^*|u_k|a\to0$ strongly.