If $\$ $yx_n\to 0$ for all $y$ in a C$^*$-algebra, Is it true that $x_n$ is weakly convergent to $0$?

$$A$$ is a C$$^*\!$$-algebra and $$(x_n)_{n\in \mathbb{N}} \subseteq A$$. If $$\$$ $$yx_n\to 0$$ for all $$y\in A$$, Is it true that $$x_n$$ is weakly convergent to $$0$$ ?

For unitals this is trivial. For characters like $$w\in \Omega (A)$$ we have $$w(x_n)\to 0$$ but if for all functionals, I don't know.

• By the Cohen--Hewitt factorization $A\times A^*\ni(a,\phi)\mapsto \phi(\,\cdot\,a)\in A^*$ is surjective. – Narutaka OZAWA Dec 24 '18 at 9:45

Yes, it's true. By the GNS construction, every bounded linear functional on $$A$$ is of the form $$A\ni a\mapsto \langle \pi(a)\xi,\eta \rangle$$ for some non-degenerate *-representation $$\pi$$ on $$H$$ and $$\xi,\eta\in H$$. By the Cohen--Hewitt factorization theorem, $$H=\pi(A)H$$ (no need to take the closure). Consequently, $$A\times A^*\ni (a,\phi)\mapsto \phi(a,\,\cdot\,)\in A^*$$ is surjective.