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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 $A=B_0(H)$.

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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 $l^\infty$.

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  • $\begingroup$ Thanks for your answer. Maybe some short follow-up questions: What do you mean with the weak$^*$-topology on $l^1$? Does it come from the natural embedding $l^1 \hookrightarrow (l^\infty)^*$? Why is $S$ weak$^*$-dense in $l^1$? $\endgroup$
    – Andromeda
    Commented Jul 5, 2022 at 21:26
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    $\begingroup$ $l^1$ as the dual of $c_0$. A codimension 1 subspace of $l^1$ is weak* closed iff it is the kernel of the map evaluating on some $\vec{a} \in c_0$; since our $\vec{a}\not\in c_0$ its kernel cannot be weak* closed. $\endgroup$
    – Nik Weaver
    Commented Jul 5, 2022 at 23:09
  • $\begingroup$ Thanks for the reply. One more thing: How are we sure that the elements $\ell^1 \subseteq (\ell^\infty)^*$ are continuous w.r.t. the strict topology on $\ell^\infty = M(c_0)$? Maybe the strict topology on $\ell^\infty$ admits a nice description relating it with some weak$^*$-topology (on bounded subsets)? $\endgroup$
    – Andromeda
    Commented Jul 6, 2022 at 7:29
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    $\begingroup$ "You start with functionals on $M(A)$" --- look for the place in my answer where I said this, and realize that I didn't. $\endgroup$
    – Nik Weaver
    Commented Jul 6, 2022 at 14:12
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    $\begingroup$ It seems that you're having difficulty with abstraction. Can you look at the specific example $S = \{\vec{b} \in l^1: \sum b_n = 0\}$ and verify directly that it separates $c_0$ but not $l^\infty$? $\endgroup$
    – Nik Weaver
    Commented Jul 6, 2022 at 14:16

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