# If $S\subseteq A^*$ is separating, does $S$ also separate $M(A)$?

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 $$A=B_0(H)$$.

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 $$l^\infty$$.
• 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 $l^1$? Jul 5 at 21:26
• $l^1$ 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. Jul 5 at 23:09
• 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)? Jul 6 at 7:29
• "You start with functionals on $M(A)$" --- look for the place in my answer where I said this, and realize that I didn't. Jul 6 at 14:12
• 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 $l^\infty$? Jul 6 at 14:16