Let $H$ be a Hilbert space and let $B\subset B(H)$ be a C*-subalgebra. Suppose that $T\colon M\to M$ is linear, bounded and operator-weakly continuous, then I want to prove that $\|T\|=\|T|_{B}\|$. Let $M$ be the von Neumann algebra generated by $B$. That is, $M=B''=\overline{B}^{\text{s}}$ (I think), where $B''$ is the double commutant of $B$ and $\overline{B}^{\text{s}}$ is the operator-strong closure of $B$.
I think that I need Kaplansy's density theorem (Theorem 4.3.3 in Murphy's book on C*-algebras). In particular, this theorem tells us that $B_{\leq1}$ is strongly dense in $M_{\leq1}$.
Furthermore, I also think that I have to use Theorem 4.2.7 of Murphy, which states that a convex subset of $B(H)$ is strongly closed if and only if it is weakly closed. I think this allows us to use the weak continuity of $T$.
One clearly has $\|T|_{B}\|\leq\|T\|$ and I think one can use the above results to prove that \begin{align*} \|T|_{B}\|&=\sup\{\|T(b)\|:b\in B_{\leq1}\}\\ &\geq\sup\{\|T(m)\|:m\in M_{\leq1}\}=\|T\|. \end{align*} But I dont know how to connect the dots. Any help will be greatly appreciated!
Let $x$ be in the unit ball of $M$ with $\|Tx\|≥ \|T\|-\epsilon$. Now because the unit ball of $B$ is weakly dense in the unit ball of $M$ you have a net $x_\alpha$ in the unit ball of $B$ with $x_\alpha\to x$ in the weak operator topology, because $T$ is weakly (operator) continuous you have that $T x_\alpha \to Tx$ in that sense. Now let $\xi \in H$ with $\|\xi\|≤1$ so that $\|(Tx)[\xi]\| ≥ \|Tx\|-\epsilon ≥ \|T\|-2\epsilon$. Then:
$$\|(Tx)[\xi]\|\ \|(Tx_\alpha)[\xi]\|≥|\langle (Tx)[\xi], (Tx_\alpha)[\xi]\rangle| \to \langle (Tx)[\xi],(Tx)[\xi]\rangle = \|(Tx)[\xi]\|^2 .$$
This gives: $$\liminf_\alpha\|(Tx_\alpha)[\xi]\|≥ \|(Tx)[\xi]\|≥ \|T\|-2\epsilon,$$
in particular since $\xi$ was in the unit ball of $H$ what you get is $\liminf_\alpha\|Tx_\alpha\|≥\|T\|-2\epsilon$. Since $x_\alpha$ was in the unit ball of $B$ you get $\|T\lvert_B\|≥\|T\|-2\epsilon$ for any $\epsilon$.