Need Help in Reasoning with Extremal Points There is a common logic I found in quite a few books and papers, which sounds intuitive, but I am unable to comprehend conceretely. The statement considers the following set-up in general:
Let $f:K\rightarrow \mathbb{C}$ be a continuous function on a locally compact Hausdorff space $K$ such that $|f(x)|\leq 1$ for all $x\in K$, and $\mu$ is a regular Borel probability measure on $K$ (i.e, $\mu\geq 0, \mu(K)=1$), whose support is compact. It is given that $\int_K f(x) d\mu(x) = \alpha$, where $|\alpha|=1$.
Then the author(s) state that "Since any point on the unit circle is an extreme point, and $|\int_K f \ d\mu|=1$, we must have that $f(x)=\alpha$ for each $x$ in the support of $\mu$."
I do not see it immediately, any help is greatly appreciated :)
 A: Hint : By the triangle inequality, and the fact that $|f(x)| \leq 1$ for every $x$, one has
$$1 = \left| \int_K f(x) d\mu\right| \leq \int_K |f(x)| d\mu \leq \int_K 1 d\mu = \mu(K)=1$$
So you must have equality in both inequalities. Can you take it from here ?
A: Given any measurable subset $L\subseteq K$, I claim that
$$
\int_L f(x)d\mu(x) = \alpha \mu(L).
$$
This is obvious in case $\mu(L)=0$ or  $\mu(L)=1$, so let us deal with the case that
$$0<\mu(L)<1.
$$
Consider the  partition $K=K_1\sqcup K_2$, where $K_1=L$ and $K_2=K\setminus L$, and note that
$$
\alpha = \mu(K_1) \frac {\int_{K_1}f(x)d\mu(x)}{\mu(K_1)}+\mu(K_2) \frac {\int_{K_2} f(x)d\mu(x)}{\mu(K_2)}.\qquad (*)
$$
Observing that each
$$
z_i:=\frac {\int_{K_i} f(x)d\mu(x)}{\mu(K_i)}$$
lies in the unit disk, that $\alpha$ is a convex combination of $z_1$ and $z_2$ by $(*)$, and that $\alpha$ is a extremal point of the disk, we deduce that
$$z_1=z_2=\alpha,$$
from where the claim follows, and it immediately implies that
$$
\int_L (f(x)-\alpha )d\mu(x) =0,
$$
for every $L$.  Since $f$ is continuous, this implies that $f(x)=\alpha$ on the support of $\mu$.

EDIT. Here is a Lemma justifying the last step in the above proof.
Lemma.  Let $g$ be a continuous, complex valued function on $K$ such that $\int_L g(x)\,d\mu(x)=0$, for all measurable subsets $L\subseteq K$. Then $g$ vanishes on $\text{supp}(\mu)$.
Proof. Reasoning  by contradiction,  let $x_0\in\text{supp}(\mu)$ be such that $g(x_0)\neq0$. Assuming WLOG that $\Re(g(x_0))>0$, let us replace $g$ with its real part and hence we may assume that $g$ is real valued and $g(x_0)>0$.
Choose some $\varepsilon >0$, and some open neighborhood $U$ of $x_0$, such that $g(x)>\varepsilon$, for all $x$ in $U$, whence
$$
\int_Ug(x)\,d\mu(x) \ge \int_U\varepsilon\,d\mu(x)=$$$$=
\varepsilon\mu(U)>0,
$$
where the last inequality follows from the fact that $x_0$ lies in the support of $\mu$.
This contradicts the hypothesis, so the proof is concluded. QED
A: Okay, thanks to the discussions above, I now have an elementary proof of the fact, as outlined below.
Let $D$ denote the support of $\mu$. Note from the comments above that $|f(x)|=1$ for each $x\in D$. Now, we must have that $$1=\Big|\int_K f d\mu\Big| = \int_K |f| d\mu .$$
But, we have
\begin{eqnarray*}
\Big|\int_K f d\mu\Big| = |\alpha| = 1 &=& \frac{1}{\alpha} \int_K f d \mu\\
&=& \int_K (\alpha^{-1} f)(x) d\mu(x)\\
&=& \int_K \Re(\alpha^{-1} f)(x) d\mu(x)\\
&\leq & \int_K |(\alpha^{-1} f)| d\mu = \mu(K)=1,
\end{eqnarray*}
where the fifth equality follows since $\int_K (\alpha^{-1} f) d\mu =1 \in \mathbb{R}$.
Hence we must have $\int_K \Re(\alpha^{-1} f)(x) d\mu(x) =1 = \int_K 1 d\mu$, i.e, $\Re(\alpha^{-1} f)(x) =1$ $\mu$-almost everywhere on $K$. But $(\alpha^{-1} f)$ is continuous, and hence $\Re(\alpha^{-1} f)(x) =1$ for each $x\in D$.
Now let $\alpha = e^{i\theta_\alpha}$ and for each $x\in D$, let $f(x)= e^{i\theta_x}$. Then $$\Re(\alpha^{-1}f(x))= \cos(\theta_x -\theta_\alpha) =1,$$
for each $x\in D$. Thus $f\equiv \alpha$ on $D$, since $\theta_x = \theta_\alpha + 2n_x\pi $ for each $x\in D$, for some $n_x\in \mathbb{Z}$.
