Ultraweak continuity implies norm continuity?

Question

Suppose $\mathcal{H}$ and $\mathcal{K}$ are Hilbert spaces, $\mathcal{U}\subset B(\mathcal{H})$ is a C$^*$-algebra, and $\eta:\mathcal{U}\rightarrow B(\mathcal{K})$ is a linear map that is ultraweakly continuous on the closed unit ball $(\mathcal{U})_{1}$ of $\mathcal{U}$. For $x,y\in \mathcal{K}$, the linear functinal $\omega_{x,y}\circ\eta$ on $\mathcal{U}$ is ultraweakly continuous ($\omega_{x,y}$ is the functional defined by $\omega_{x,y}(A)=\langle Ax,y\rangle$ for $A\in B(\mathcal{K})$). My question is why does this imply that $\omega_{x,y}\circ\eta$ is also norm continuous on $(\mathcal{U})_{1}$? I don't have any reason to think the ultraweak topology is coarser than the norm topology (if $\mathcal{U}$ is a von Neumann algebra then norm continuity would follow from the fact that the unit ball of $\mathcal{U}$ is weak operator compact).

Answer 1

When you made the codomain the scalars, the topology problems on the codomain side disappear.

If $A_n\to A $ in norm, then $A_n\to A$ ultraweakly. Then $\omega_{x,y}(A_n)\to\omega_{x,y}(A)$. Thus $\omega_{x,y}$ is norm-continuous.