0
$\begingroup$

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 $C_c(X)$?

$\endgroup$
1
  • 1
    $\begingroup$ Yes, because $C_0(X)$ is the completion of $C_c(X)$ and continuous linear functionals are uniformly continuous hence extends to completion. $\endgroup$ Feb 28, 2021 at 11:23

1 Answer 1

1
$\begingroup$

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 $X$.

$\endgroup$
2
  • $\begingroup$ So the bijection between the duals is given by associating to the continuous linear functional on M it's extension on X right? $\endgroup$
    – roi_saumon
    Feb 28, 2021 at 12:18
  • $\begingroup$ 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 $\endgroup$ Feb 28, 2021 at 12:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .