Let $A$ be a non-unital $C^\*$-algebra. Let $S\subseteq A^\*$ be a set of continuous functionals that separates the points of $A$. Every element $\omega \in A^\*$ extends uniquely to a strictly continuous functional $\omega \in M(A)^\*$, so we can ask: does $S$ also separate the points of $M(A)$? Concretely, if $m \in M(A)$ and $\omega(m)=0$ for all $\omega \in S$, do we have $m=0$?


In some cases, this is automatically true. For instance, if $AS \subseteq S$ or $SA\subseteq S$ this is easily seen to be true. However, it is not clear to me if this is true in general. My intuition tells me that the answer is "no" but I have trouble finding a concrete counterexample. It's not even clear to me what happens if $A= C\_0(X)$ or $A=B\_0(H)$.



If $S\subseteq A^*$ is separating, does $S$ also separate $M(A)$?
You're right, it's not true in general. If $A = c\_0$ then the multiplier algebra is the same as the double dual, $l^\infty$. Then we want a linear subspace of $l^1$ which is weak\* dense (so it separates $c\_0$) but not norm dense (so it doesn't separate $l^\infty$). This can be achieved by taking any $\vec{a} \in l^\infty \setminus c\_0$ and letting $S$ be its kernel in $l^1$. For instance, taking $\vec{a} = 1$, we get $S = \{\vec{b} \in l^1: \sum b\_n = 0\}$. This separates $c\_0$ but not $l^\infty$.


