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 :)


3 Answers 3


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 ?

  • $\begingroup$ I got this far, and received that we must have that the inequalities here have to be equalities. Hence by your other answer, there exists some $x_0$ in the support of $\mu$ such that $|f(x_0)|=1$. But I don't know how to proceed further. $\endgroup$ May 27 at 17:07
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    $\begingroup$ First look at the inequality $$\int_K |f(x)| d\mu \leq \int_K 1 d\mu$$ You must have equality here. Try to use the continuity of $f$ to prove that for all $x$ in the support de $\mu$, $|f(x)| = 1$. Then you will look at the equality case of the triangle inequality to understand why it implies that $f(x)=\alpha$ for every $x$ in the support of $\mu$. $\endgroup$ May 27 at 17:11
  • $\begingroup$ @ TheSilverDoe Yes, I see that $|f(x)|=1$ for all $x\in D$, where $D$ is the support of $\mu$ (If there exists $x_0\in D$ such that $|f(x_0)|<1$, then by continuity of $f$, there exists an open neighborhood $V$ of $x_0$ such that $|f|<1$ on $V$. Since $x_0\in D$, $\mu(V)\neq 0$. Then $\int_K |f|d\mu <\mu(V)+ \mu(K\setminus V) =1$, which is a contradiction.) But why would $f$ specifically equal $\alpha$? $\endgroup$ May 27 at 17:26
  • $\begingroup$ I understand that it comes from the fact that convex combinations of points on the circle other than $\alpha$ will not equal $\alpha$. But how to use that in getting the other equality? $\endgroup$ May 27 at 17:34
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    $\begingroup$ I suggest writing the equality proved in the answer as $$\int (1-|f|)\,d\mu=0$$ It s clear now that if $|f|\le 1$ then $|f|=1.$ $\endgroup$ May 27 at 17:39

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

  • $\begingroup$ Thank you for the answer, it does highlight the usage of the fact that the points on the unit circle are extremal in nature. Could you please elaborate on how the last line, i.e, the conclusion follows? $\endgroup$ Jun 2 at 8:17
  • $\begingroup$ Much more helpful answer, +1 $\endgroup$
    – FShrike
    Jun 7 at 12:07

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 $n_x\in \mathbb{Z}$.


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