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Let $\Omega$ be locally compact topological space. Consider $C_0(\Omega)$ being the space of continuous functions $u$ on $\Omega$ that vanish at infinity, that is, for $r>0$ there is a compacte set $K\subset\Omega $ such that $\sup_{\Omega\setminus \Omega}|u(x)|<r$. The dual of $C_0(\Omega)$ is well-known to be $M_b(\Omega)$ the space of signed bounded Radon measures $\Omega$.

For $u\in C_0(\Omega)$ with $\|u\|_\infty\neq 0$, consider the duality mapping (which is a convex set)

$$J(u) =\{ \mu\in M_b(\Omega)\,\,:\,\, \langle\mu, u\rangle= \|u\|_\infty,\,\,\|\mu\|=1\}$$

I would like to show that if $\mu\in J(u)$ is an extreme point then there exists $a\in \Omega$ such that $$|u(a)|=\|u\|_\infty\qquad \text{and}\qquad \mu= \text{sign}_0(u(a))\delta_a$$ where $\delta_a$ is the Dirac measure at $a\in \Omega$.

I was able to show that $ \mu= \text{sign}_0(u(a))\delta_a$ are extreme points of $J(u)$ using this answer here: \url{https://math.stackexchange.com/q/2303204}

Recall that $\mu$ is an extreme point if and only if $J(u)\setminus\{\mu\}$ is also a convex set.

Any reference or answer is welcome.

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$\newcommand\Om\Omega\newcommand\de\delta$We have to show the following:

Suppose that $\Om$ is a locally compact topological space, $0\ne u\in C_0(\Om)$, and $\mu$ a signed bounded Radon measure over $\Om$ that is an extreme point of $J(u)$. Then $\mu=\pm\de_z$ for some choice of the sign and some $z\in\Om$.

Without loss of generality (wlog), by rescaling, $\max_\Om|u|=\|u\|_\infty=1$. Since $\mu$ is Radon, the topological space $\Om$ must be Hausdorff. Let $(\Om_+,\Om_-)$ be a Hahn decomposition corresponding to the signed measure $\mu$ and let $(\mu_+,\mu_-)$ be the Jordan decomposition of $\mu$. Let $u_\pm:=\max(0,\pm u)$. Then $$1=\int_\Om u\,d\mu=\int_\Om u_+\,d\mu_+ +\int_\Om u_-\,d\mu_- -\int_\Om u_+\,d\mu_- -\int_\Om u_-\,d\mu_+ \\ \le \int_\Om u_+\,d\mu_+ +\int_\Om u_-\,d\mu_-\le1.$$ Therefore and because $1=\|\mu\|=|\mu|(\Om)=\mu_+(\Om) +\mu_-(\Om)$, we have $$u_\pm=\pm1\quad\text{$|\mu|$-almost everywhere on }\Om_\pm.$$

Note: Any nonzero Radon measure $\nu$ over $\Om$ has at least one support point. Indeed, otherwise, for some compact $K\subseteq\Om$ such that $\nu(K)>0$ and for each $z\in K$ there is an open neighborhood $U_z$ of $z$ with $\nu(U_z)=0$. Covering now $K$ by finitely many $U_z$'s, we get $\nu(K)=0$, a contradiction.

So, the support $S_\mu$ of $|\mu|$ is nonempty. Suppose there are two distinct points $x$ and $y$ in $S_\mu$. Let $U$ and $V$ be two disjoint open neighborhoods of $x$ and $y$, respectively. Then $M:=|\mu|(U)>0$, $N:=|\mu|(V)>0$, and $$\int_U u\,d\mu+\int_V u\,d\mu =\int_U |u|\,d|\mu|+\int_V |u|\,d|\mu| =|\mu|(U)+|\mu|(V)=M+N.$$ Take any any $t\in(0,\min(M,N))$ and let $$\mu^\pm(A):=\frac{M\pm t}M\,\mu(A\cap U) +\frac{N\mp t}N\,\mu(A\cap V)+\mu(A\setminus U\setminus V)$$ for Borel $A\subseteq\Om$. Then $\mu^\pm\in J(u)$, $\mu^+\ne\mu^-$, and $\mu=\frac12\,(\mu^+ +\mu^-)$, which contradicts the assumption that $\mu$ is an extreme point of $J(u)$.

So, $S_\mu=\{z\}$ for some $z\in\Om$. If the "restriction" of $|\mu|$ to $\Om\setminus\{z\}$ were nonzero, then, by the Note, $|\mu|$ would have a support point in $\Om\setminus\{z\}$. So, the "restriction" of $|\mu|$ to $\Om\setminus\{z\}$ is zero. Recalling that $\|\mu\|=1$, we get the desired conclusion $\mu=\pm\de_z$. $\quad\Box$.

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