Let $\epsilon>0$ and let $f$ be holomorphic in $\{z \in C : |z| < 1+\epsilon\}$. Suppose that $f(z)$ is real on $|z| = 1$. Prove $f$ is constant.

Question

Let $\epsilon>0$ and let $f$ be holomorphic on the disc $\{z \in C : |z| < 1+\epsilon\}$. Suppose that $f(z)$ is real whenever $|z| = 1$. Prove that $f$ is constant.

My proof:

We consider the maximum modulus principle. We know that the imaginary part achivevs maximum on $|z|=1$ ( max on $|z|\leq1$). But the maximum is then $0$. Thus the imaginary part is $0$ and so by cauchy reimann we know the real part is constant, and so $f$ is a real constant.

IS this correct? Is there a different way to go about this?

Answer 1

It's not really what's usually called "maximum modulus", rather the fact that the imaginary part of an analytic function is a harmonic function.

Answer 2

A different way is to apply MMP to $e^{if(z)}$ and $e^{-if(z)}$. These two functions have modulus $1$ for $|z|=1$ so $|e^{\pm if(z)}| \leq 1$ for $|z| \leq 1$. This gives $|e^{if(z)}| = 1$ for $|z| \leq 1$. Hence $e^{if(z)}$ is a constant. This implies that $f$ is itself a constant.

Answer 3

There is also a geometric proof (by the reflection principle only continuity on the unit circle is needed as that implies that $f$ has a holomorphic extension beyond the disc if it is real on the unit circle).

$f(\mathbb D)=U$ is an open bounded set whose boundary is included (in general not equal) into $f(\partial \mathbb D)=F$; however $F$ is a compact connected subset of the reals, hence a closed finite interval $[a,b]$ and any (plane) open set $U$ with boundary in $[a,b]$ is obviously unbounded. Contradiction!