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 $C_c(X)$?
Why is $M(X)=C_c(X)^*=C_0(X)^*$
functional-analysismeasure-theory
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Your argument is correct, thanks to the following result.
Theorem. For a locally compact Hausdorff space $X$ the following are equivalent:
- $X$ is Lindelöf.
- $X$ is $\sigma$-compact.
- There is a sequence $\langle K_n:n\in\Bbb N\rangle$ of compact subspaces of $X$ such that $K_n\subseteq\operatorname{int}K_{n+1}$ for each $n\in\Bbb N$ and $X=\bigcup_{n\in\Bbb N}K_n$.
- The point at infinity in the one-point compactification of $X$ has a countable local base.
Proof. Suppose that $X$ is Lindelöf. For each $x\in X$ let $U_x$ be an open nbhd of $x$ with compact closure, and let $\mathscr{U}=\{U_x:x\in X\}$. $\mathscr{U}$ is an open cover of $X$, so it has a countable subcover $\{V_n:n\in\Bbb N\}$. Clearly $X=\bigcup_{n\in\Bbb N}\operatorname{cl}V_n$ is a representation of $X$ as a countable union of compact sets, so $X$ is $\sigma$-compact.
Now assume that $X$ is $\sigma$-compact, and let $\{C_n:n\in\Bbb N\}$ be a countable cover of $X$ by compact sets. Let $K_0=C_0$. If the compact set $K_n$ has already been defined, for each $x\in K_n$ let $U_x$ be an open nbhd of $x$ with compact closure, and let $\mathscr{U}=\{U_x:x\in K_n\}$. $K_n$ is compact, so $\mathscr{U}$ has a finite subcover $\{V_1,\dots,V_m\}$. Let $K_{n+1}=C_n\cup\bigcup_{k=1}^m\operatorname{cl}V_k$; then $K_{n+1}$ is compact, and $K_n\subseteq\operatorname{int}K_{n+1}$. Finally, $C_n\subseteq K_n$ for $n\in\Bbb N$, so $\bigcup_{n\in\Bbb N}K_n=X$.
Now assume that there is a sequence $\langle K_n:n\in\Bbb N\rangle$ as in (3). Let $X^*$ be the one-point compactification of $X$, and let $p$ be the point at infinity. For $n\in\Bbb N$ let $B_n=X^*\setminus K_n$, and let $\mathscr{B}=\{B_n:n\in\Bbb N\}$; then $\mathscr{B}$ is a countable local base at $p$. To see this, let $U$ be any open nbhd of $p$ in $X^*$. Then $C=X^*\setminus U$ is a compact subset of $X$. $\bigcup_{n\in\Bbb N}\operatorname{int}K_n=X$, so $\{\operatorname{int}K_n:n\in\Bbb N\}$ is an increasing open cover of $C$, and therefore there is an $n\in\Bbb N$ such that $C\subseteq\operatorname{int}K_n\subseteq K_n$, whence $B_n\subseteq U$.
Finally, suppose that $X^*$ is first countable at $p$, the point at infinity, and let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a nested local base at $p$. For $n\in\Bbb N$ let $K_n=X^*\setminus B_n$; then each $K_n$ is compact, and $X=\bigcup_{n\in\Bbb N}K_n$. Let $\mathscr{U}$ be an open cover of $X$. For each $n\in\Bbb N$ let $\mathscr{U}_n$ be a finite subset of $\mathscr{U}$ covering the compact set $K_n$, and let $\mathscr{V}=\bigcup_{n\in\Bbb N}\mathscr{U}_n$; then $\mathscr{V}$ is a countable subcover of $\mathscr{U}$. $\dashv$
Of course a second countable space is Lindelöf, so you get all of (1)-(4).
$C_0(\Omega)$ is the closure of $C_c(\Omega)$ only with respect to the sup topology. The inductive limit topology on $C_c(\Omega)$ gives you a complete topological vector space.
The dual of $(C_0(\Omega),\tau_{sup})$ is the space of bounded (also called finite) Radon measures. The dual of $(C_c(\Omega),\tau_{ind})$ is the space of locally bounded Radon measures. At least, if $\Omega$ is $\sigma$-locally compact. The two spaces coincide if $\Omega$ is compact.
The space $(C_0(\Omega),\tau_{ind})$ does not make sense, because the inductive limit topology depends on the sequence of spaces you use to cover the entire functional space and, in general, you cannot cover $C_0(\Omega)$ through $C_c(U)$ spaces with $U$ included in $\Omega$.
Finally, the space $(C_c(\Omega),\tau_{sup})$ makes sense but bounded linear functionals on this space are naturally identified to bounded linear functionals on $(C_0(\Omega),\tau_{sup})$.
If you want to go deep into these questions you have to take a look to these references:
- Irene Fonseca, Giovanni Leoni. Modern Methods in the Calculus of Variations: $L^p$ Spaces. Springer Science & Business Media, 2007.
- Charalambos D. Aliprantis, Kim C. Border. Infinite Dimensional Analysis: A Hitchhiker's Guide. Springer-Verlag Berlin and Heidelberg GmbH & Company KG, 2013.
- Nelson Dunford, Jacob T. Schwartz. Linear operators, part 1: general theory. Vol. 10. John Wiley & Sons, 1988.
Best Answer
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 $X$.