Prove that $f$ is constant if $f$ is real when $|z|=1$

Question

Let $f$ be a holomorphic function in $\mathbb{C}$. Prove that if $f$ is real when $|z|=1$, then $f$ must be a constant function.

I honestly do not know how to do this problem, consider using Schwarz's lemma, but do not get anywhere.

Same question: http://math.stackexchange.com/questions/226786/let-fz-be-entire-function-show-that-if-fz-is-real-when-z-1-then-f?rq=1 — Spenser, Nov 02 '13 at 21:14

score 11 · Accepted Answer · answered Nov 02 '13 at 21:38

11

Note that $\overline{f(\overline{z})}$ is a holomorphic function that equals $f(z^{-1})$ on the unit circle. Therefore these functions are equal everywhere. In particular $f$ is bounded on $\mathbb{C}$ and therefore constant.

answered Nov 02 '13 at 21:38

WimC

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Nice proof ! :) – Spenser Nov 02 '13 at 21:43
Very good proof! – Ernesto Carro Nov 02 '13 at 22:19

score 5 · Answer 2 · answered Nov 02 '13 at 21:01

5

Hint: If $f=u+iv$ is real, then $v=0$ on the unit circle. Now use the maximum principle for harmonic functions.

answered Nov 02 '13 at 21:01

njguliyev

14,543

Prove that $f$ is constant if $f$ is real when $|z|=1$

2 Answers2