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$A$ 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.

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    $\begingroup$ By the Cohen--Hewitt factorization $A\times A^*\ni(a,\phi)\mapsto \phi(\,\cdot\,a)\in A^*$ is surjective. $\endgroup$
    – Narutaka OZAWA
    Commented Dec 24, 2018 at 9:45

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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.

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