Question on states and nets in a non-unital $C^*$ algebra

Question

Let $A$ be a non-unital $C^*$ algebra, and let $a_\lambda$ be a net such that $a_\lambda b \to 0$ for all $b \in A$. Show that for any state $f, f(a_\lambda) \to 0$.

I tried using approximate units, but to no avail...

Answer 1

Short answer using machinery: By the Cohen-Factorisation theorem, any functional $f\in A^*$ is of the form $f = \omega(-b)$ for some $b \in A$ and $\omega \in A^*$. Then $$f(a_\lambda) = \omega(a_\lambda b) \to \omega(0) = 0.$$

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If the net $\{a_\lambda\}$ is bounded, you can proceed as follows.

Let $(H_f, \pi_f, \xi_f)$ be the GNS-triplet associated with $f$. We claim that $\pi_f(a_\lambda) \to 0$ in the strong topology on $B(H_f)$. Indeed, since the net $\{\pi_f(a_\lambda)\}$ is bounded, it suffices to check the strong convergence on the dense subspace $\pi_f(A)\xi_f$. But $$\pi_f(a_\lambda)\pi_f(b) \xi_f = \pi_f(a_\lambda b)\xi_f \to 0$$ since $a_\lambda b \to 0$ so this is clear. It thus follows that $$f(a_\lambda) = \langle \pi_f(a_\lambda)\xi_f, \xi_f\rangle\to 0$$ since $\pi_f(a_\lambda)\xi_f \to 0$.