Does weak*-convergence imply convergence of the operator norms?

Question

Let $\mathcal A$ be a unital C*-algebra with topological dual $\mathcal A^*$ and denote the unit ball as $B_1^*:=\{\phi \in \mathcal A^* : \vert\vert \phi \vert\vert_{sup}\leq 1\}$.

If $\phi_n \rightarrow \phi$ is a weak*-convergent sequence with $\vert\vert \phi_n \vert\vert_{sup}= 1, \forall n \in \mathbb N$, does this imply that $\vert\vert \phi \vert\vert_{sup} = 1$?

Answer 1

No. Consider the Fourier coefficient functionals on $C[0,2\pi]$.

Or if $H$ is a separable Hilbert space with orthonormal basis $(e_n)$, consider the functionals $T\mapsto\langle Te_1,e_n\rangle$ on $B(H)$.

In general you have $\|\phi\|\leq \liminf_n\|\phi_n\|$, but the inequality can be strict as these examples show.