I'm doing Ex 3.15 in Brezis's book of Functional Analysis.
Let $(E, |\cdot|)$ be a reflexive Banach space. In the following, the convex subset $K$ of $E$ is equipped with the weak topology $\sigma (E, E')$ such that $K$ is compact. Let $F = \mathcal C(K)$ with its usual norm $| \cdot |_\infty$. Fix some $\mu \in F'$ with $\|\mu\|_{F'}=1$ and assume that $\mu \geq 0$ in the sense that $$ \langle\mu, u\rangle \geq 0 \quad \forall u \in \mathcal C(K) \text{ such that } u \geq 0 \text { on } K . $$ Prove that there exists a unique element $x_{0} \in K$ such that $$ \langle \mu, f_{\mid K} \rangle = \langle f, x_{0} \rangle \quad \forall f \in E'. $$
In below proof, I do not use the assumption that $\mu \ge 0$. I think I made some mistake but could not recognize. Could you have a check on my attempt?
Consider the map $\varphi:E' \to \mathbb R, f \mapsto \langle \mu, f_{\mid K} \rangle$. It's clear that $f$ is linear. Let $B_E$ be the closed unit ball of $E$. Because $K$ is weakly compact, $K$ is bounded. There is $r>0$ such that $rK \subset B_E$. We have $$ |f_{\mid K}-g_{\mid K}|_{\infty} = \sup_{x\in K} |\langle f-g, x\rangle| = \frac{1}{r} \sup_{x\in rK} |\langle f-g, x\rangle| \le \frac{1}{r} \sup_{x\in B_E} |\langle f-g, x\rangle| = \frac{1}{r} \|f-g\|_{E'}. $$
It follows that $\varphi$ is continuous and thus $\varphi\in E''$. Because $E$ is reflexive, we can identify $\varphi$ with a unique $x_0 \in E$, i.e., $$ \langle f, x_0 \rangle = \langle \varphi, f \rangle = \langle \mu, f_{\mid K} \rangle \quad \forall f \in E'. $$
Assume the contrary that $x_0 \notin K$. Then we can strictly separate $\{x_0\}$ and $K$ by Hahn-Banach theorem. This means there are $a,b \in \mathbb R$ and $f \in E'$ such that $$ \langle f,x \rangle <a <b< \langle f,x_0 \rangle = \langle \mu, f_{\mid K} \rangle \le \|\mu\|_{F'} \cdot |f_{\mid K} |_\infty = |f_{\mid K} |_\infty \quad \forall x\in K. $$
Notice that $\langle f,x \rangle < |f_{\mid K} |_\infty$ for all $x\in K$ is a contradiction because $f \in E'$ and $K$ is weakly compact.
Update: @MaoWao showed that I made a mistake at the end. Hopefully, I have found a fix :v
Let $\lambda := \langle \mu, 1 \rangle$. By positivity condition, we get $$ \|\mu\|_{F'} = \sup_{\substack{u\in \mathcal C(K) \\ |u|_\infty = 1}} \langle \mu, u \rangle \le \sup_{\substack{u\in \mathcal C(K) \\ |u|_\infty = 1}} \langle \mu, 1 \rangle = \lambda. $$
It follows that $\|\mu\|_{F'} = \lambda$ and thus $\lambda = 1$. It follows from $f_{\mid K} <a<b$ that $f(x_0) = \langle \mu, f_{\mid K} \rangle \le \langle \mu, a \rangle < \langle \mu, b \rangle$. This implies $a < b < \langle \mu, a \rangle < \langle \mu, b \rangle$ and thus $a<b<\lambda a<\lambda b$. This in turn implies $\lambda \neq 1$ which is a contradiction. This completes the proof.
We can see that the positivity constraint and the fact that $\|\mu\|_{F'} = 1$ are essential.
MaoWao has pointed out the problem. Here is a completely different proof. We have $$\tag1 \langle \mu,f|_K\rangle=\int_Kf\,d\mu. $$ Let $$ Z=\{\eta\in C(K)':\ \eta\geq0,\ \|\eta\|\leq1\}. $$ This set is weak$^*$-closed (easy to check) so weak$^*$-compact by Banach-Alaoglu. It is also convex. Then by Krein-Milman, $$ Z=\overline{\operatorname{conv}}^{w^*}\operatorname{ext}Z. $$ It is easy to check that the extreme points of $Z$ are the point measures. In summary, there exist nets $\{t_{jk}\}_j\subset[0,1]$, $\{s_{jk}\}_j\subset K$, $k=1,\ldots,r(j)$, with $\sum_kt_{jk}=1$, $t_{jk}\geq0$, and $$ \mu=\lim_j \sum_{k=1}^{r(j)}t_{jk}\,\delta_{s_{jk}} $$ pointwise. Then $$\tag2 \int_Kf\,d\mu=\lim_j\sum_{j=1}^{r(j)}t_j\delta_{s_{jk}}(f) =\lim_j\sum_{j=1}^{r(j)}t_j\,f({s_{jk}}) =\lim_j\,f\Big(\sum_{j=1}^{r(j)}t_j\,{s_{jk}}\Big) $$ Because $K$ is convex, $\sum_{j=1}^{r(j)}t_j\,{s_{jk}}\in K$ for all $j$. As $K$ is weak compact, there exists a weak-convergent subnet that converges to some $x_0\in K$. Thus from $(2)$ we get $$\tag3 \mu(f)=\int_Kf\,d\mu=f(x_0). $$ The uniqueness of $x_0$ follows from the fact that $E'$ separates points in $E$.