Central Linear Functionals form a Closed Subspace Let $B$ be a Banach algebra and let $B^\ast$ denote its corresponding dual space, normed with the supremum norm over the unit ball. It is (I think?) easy to verify that $\phi_n \to \phi$ in $B^\ast$ in norm if and only if $\phi_n \to \phi$ uniformly on the unit ball in $B$. A linear functional $\phi$ is said to be central provided that $\phi (xy) = \phi(yx)$ for every $x,y \in B$.
I want to prove that the subspace of all central linear functionals in $B^\ast$ is a norm closed. I think I was able to do this:

Let $\{\phi_n\}$ be a sequence of central linear functionals in $B^\ast$ converging to $\phi$ in norm. Then, in particular, $\phi_n \to \phi$ on the unit ball. Let $x,y \in B$ be nonzero elements. Because
$$\left\| \frac{xy}{||x|| \cdot ||y||} \right\| = \frac{||xy||}{||x|| \cdot ||y||} \le \frac{||xy||}{||xy||} =1$$
we can write
$$\phi \left(\frac{xy}{||x|| \cdot ||y||} \right) = \lim_{n \to \infty} \phi_n \left(\frac{xy}{||x|| \cdot ||y||} \right) = \lim_{n \to \infty} \phi_n \left(\frac{yx}{||x|| \cdot ||y||} \right) = \phi \left(\frac{yx}{||x|| \cdot ||y||} \right)$$
and therefore $\phi(xy) = \phi(yx)$.

First of all, does this seem right? Also, notice that mere pointwise convergence is used. Is it possible that the subspace of all central linear functionals in $B^*$ is closed with respect to a weaker topology on $B^*$?
 A: First of all your idea is correct but at the end you were wrong to write because at the end you must write that 
$\phi(\frac{xy}{||x||||y||})= \phi(\frac{yx}{||x||||y||})$
You can prove the statement also in this way:
$|\phi(xy)-\phi(yx)|\leq |\phi(xy)-\phi_n(xy)+\phi_n(xy)-\phi(yx)|\leq$
$\leq |\phi(xy)-\phi_n(xy)|+ |\phi_n(xy)-\phi(yx)|$
$= |\phi(xy)-\phi_n(xy)|+ |\phi_n(yx)-\phi(yx)|\leq $
$||\phi-\phi_n|||xy||+||\phi-\phi_n||||yx||\leq $
$\leq 2||\phi-\phi_n||||x||||y||\to 0$ for $n\to \infty$
so
$\phi(xy)=\phi(yx)$
The space is also closed with respect the weak* Topology on $B^*$ because that Topology is first countable and so each set is closed if and only if each convergent sequence in the set converge to an element of the set. 
So if you consider an element $\phi$ of the set of central linear operator and a sequence $\{\phi_n\}$ of central linear operator that converge to $\phi$ then you have that 
$\phi_n(z)\to \phi(z)$ for each fixed $z\in B$.
Then 
$|\phi(xy)-\phi(yx)|\leq | \phi(xy)-\phi_n(xy)|+ | \phi(yx)-\phi_n(yx)|\to 0$ for $n\to \infty$ 
