Consider the vector space of complex Radon measure $M(X)$ on $X$. ($X$ is locally compact hausdroff). There is a result states that $(C_0(X))^*$ is isomorphic to $M(X)$ where $C_0(X)$ is the closure of continuous function with compact support. Now the question is what is $(M(X))^*$.

I have seen a relevant result on Lax's Functional analysis. It states that if $X$ is compact Hausdroff, then the $(M(X)^*)=(C(X))^{**}=L^\infty(X)$. However, the proof is omitted. I guess the similar result also holds when $X$ is LCH. i.e. $M(X)^*= L^\infty(X)$. It is clear that $L^\infty(X)\subseteq M(X)^*$. So my question is how to show that $L^\infty(X)\supset M(X)^*$. Does anyone have any ideas or comments?

Thanks in advance!

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    $\begingroup$ Assuming $M(X)$ carries the usual dual norm induced from $C_0(X)$, I think this is false, even for compact $X$. Take for instance $X=[0,1]$ and let $F \subset M(X)$ be the linear span of all point masses. Then you can check that Lebesgue measure $m$ is at distance 1 from the subspace $F$, so by Hahn-Banach there is a linear functional $\ell \in M(X)^*$ with $\ell(F)=0$ and $\ell(m)=1$. This $\ell$ cannot be represented by any bounded function $f$, because $\ell(F)=0$ means $f(x)=0$ for every $x$. $\endgroup$ Feb 4 at 3:10
  • $\begingroup$ @NateEldredge Thanks for comment. Yeah, it is the total variation norm. The theorem state this is in chapter 8 theorem 14 (ii). It is the following: Let $Q$ be a compact Hausdroff space, $C(Q)$ be the space of continuous real-valued function on $Q$, normed by maximum norm. Then $C^{**}$ is $L^\infty(Q)$. $\endgroup$
    – Zorualyh
    Feb 4 at 3:11
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    $\begingroup$ Hum. Either I am missing something, or else Lax is. Best bet is that it's me, but I don't see what. To see $m$ has distance $1$ from $F$, let $\mu \in F$, which is of the form $m = \sum_{i=1}^n a_i \delta_{x_i}$. For any $\epsilon > 0$, I can easily find a continuous function $g \in C([0,1])$ with $\|g\|_\infty = 1$ which vanishes at $x_1, \dots, x_n$ but for which $\int g\,dm$ is at least $1-\epsilon$. So $\|m-\mu\| \ge 1-\epsilon$. $\endgroup$ Feb 4 at 3:25
  • $\begingroup$ @NateEldredge Great example! The only thing I can think of to explain the theorem is it does not explicitly state the complex measure is radon? But if the underline space is $[0,1]$, I think every finite borel measure is automatically radon. So it still does not explain this counterexample... :( Confused... $\endgroup$
    – Zorualyh
    Feb 4 at 3:33
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    $\begingroup$ There's another mention of this issue at math.stackexchange.com/questions/392249/… $\endgroup$ Feb 4 at 3:47

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