In MSE post, I asked about the central linear functionals on a Banach algebra. With the help of someone in the chatroom, I was able to answer some lingering questions that hadn't been addressed. In the question I asked about whether the central linear functionals formed a weak$^*$ closed subspace of $B^\ast$. Here's the proof:
The annihilator of the linear span of all commutators in $B$ is precisely the central linear functionals; since the annihilator of any subspace is weak$^*$ closed, it follows that the central linear functionals in $B^\ast$ are weak$^*$ closed.
Now I am wondering about the set of multiplicative linear functionals. Do these form a weak$^*$ closed set? It is easy to show that it is norm closed:
Recall that $\phi_n \to \phi$ in $B^\ast$ with respect to the norm topology if and only if $\phi_n \to \phi$ uniformly on the unit ball in $B$. From this we can deduce that $\phi_n \to \phi$ pointwise over all of $B$, and a simple limit calculation (somewhat similar to the one found in the above link) shows that $\phi$ must also be multiplicative if the $\phi_n$ are multiplicative. Whence it follows the collection of multiplicative linear functionals in $B^\ast$ is norm closed
But what about the weak$^*$ closure of the multiplicative linear functionals?
For any given $a, b \in B$, $\phi \mapsto \phi(a)$, $\phi \mapsto \phi(b)$ and $\phi \mapsto \phi(ab)$ are weak-* continuous functions on $B^*$, and so is $\phi \mapsto \phi(ab) - \phi(a) \phi(b)$. Thus $\{\phi: \phi(ab) - \phi(a) \phi(b) = 0\}$ is weak-* closed. Its intersection over all $a, b \in B$ is the set of multiplicative linear functionals, so this is also weak-* closed.