$f : D \rightarrow \mathbb{C} $ analytic. Show $\operatorname{Re}(f(z)) \geq 0$.

Question

Let $f : D \rightarrow \mathbb{C} $, where $D=\{z\in\mathbb{C}: |z|<1\}$, such that $\operatorname{Re}(f(z)) \geq 0$ for all $z \in D$ and suppose $f$ is analytic.

I have to show $\operatorname{Re}(f(z)) > 0 $ for all $z \in D$. Any hints how to go about starting this problem? And also why can't $f$ be constant?

There's a condition missing. $f\colon z \mapsto 0$ satisfies the premises but not the conclusion. — Daniel Fischer, Jan 21 '16 at 23:53
This is from Conway's, Functions of One Complex variable, if you want to take a look in section 6.2 problem 3. I just looked and its written the same as I wrote — user1058860, Jan 21 '16 at 23:56
Okay, Conway forgot the condition. Have you an idea which condition? (Consider the counterexample.) — Daniel Fischer, Jan 21 '16 at 23:58
Yes, that's it. We need $f$ non-constant. [You can easily find other constant counterexamples.] What theorems about non-constant analytic functions would be useful here? — Daniel Fischer, Jan 22 '16 at 00:01
Open-mapping theorem, Maximum modulus principle and maybe try to use Schwarz's lemma since that's the section it's under — user1058860, Jan 22 '16 at 00:02
Open mapping theorem sounds good. (The Schwarz lemma comes into play in the next part of the exercise.) — Daniel Fischer, Jan 22 '16 at 00:04

score 4 · Accepted Answer · answered Jan 22 '16 at 00:11

Conway forgot to state the condition that $f$ be non-constant. For constant $f$ with purely imaginary values, the conclusion clearly doesn't hold.

The condition $\operatorname{Re} f(z) \geqslant 0$ tells us that $f(D)$ is contained in the closed right half-plane. If $f$ is non-constant, the open mapping theorem tells us that if there were a $z_0\in D$ with $\operatorname{Re} f(z_0) = 0$, then $f(D)$ would contain a disk with centre $f(z_0)$, hence not be contained in the closed right half-plane. Therefore we must have $\operatorname{Re} f(z) > 0$ for all $z\in D$.

$f : D \rightarrow \mathbb{C} $ analytic. Show $\operatorname{Re}(f(z)) \geq 0$.

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