# Why is $M(X)=C_c(X)^*=C_0(X)^*$?

Let $$X$$ be a a locally compact, separable and complete metric space.and $$M(X)=\{$$finite signed measures on $$X\}$$. Now I learned that $$M(X)=C_c(X)^*$$ i.e. it is the dual of continuous functions compactly supported in $$X$$. Now it seems $$M(X)=C_0(X)^*$$ as well (dual of continuous functions vanishing at $$\infty$$ as well. But why is that? Does it have something to do with the fact that $$C_0(X)$$ is the set of uniform limits of elements of $$C_c(X)$$?

• Yes, because $C_0(X)$ is the completion of $C_c(X)$ and continuous linear functionals are uniformly continuous hence extends to completion. Feb 28, 2021 at 11:23

If $$X$$ is a normed linear space and $$M$$ is dense subspace of $$X$$ then $$M$$ and $$X$$ have the same dual. This is because any continuous linear functional on $$M$$ has a unique extension to a continuous linear functional on $$X$$.
• Yes, the inverse of the unique extension is the restriction map: Any continuous linear functional on $X$ restricted to $M$ is a continuous linear functional on $M$. @roi_saumon Feb 28, 2021 at 12:20