Let $u:\mathbb{C}\rightarrow\mathbb{R}$ be a harmonic function such that $0\leq f(z)$ for all $ z \in \mathbb{C}$. Prove that $u$ is constant.

Question

Let $u:\mathbb{C}\rightarrow\mathbb{R}$ be a harmonic function such that $0\leq u(z)$ for all $ z \in \mathbb{C}$. Prove that $u$ is constant.

I think i should use Liouville's theorem, but how can i do that? Help!

set, $v= u-\inf_{Bbb R^d} u$, $\varepsilon>0,$ and $y\in\Bbb R^d$ there is $x= x(\varepsilon)$ s.t $ v(x)\le \inf v+\varepsilon$. Let $R>\max(|x|,|y|)+1$ therefore, $x ,y\in B_R(0)$ and $ B_R(y)\subset B_{3R}(x) $ By Mean value property $$ v(y) = \frac{1}{|B_R(y)|}\int_{B_R(y)} v(z) dz\ = \frac{3^d}{|B_{3R}(x)|}\int_{B_R(y)} v(z) dz \ \le \frac{3^d}{|B_{3R}(x)|}\int_{B_{3R}(x)} v(z) dz\= 3^d v(x)$$ Hence
$$ v(y)\le 3^d v(x)\le 3^d(\inf v+\varepsilon)~~\forall y\in \Bbb R^d$$

i.e for arbitrary $\varepsilon$ we have, $ \sup_{\mathbb R^d} v < 3^d \varepsilon $ — Guy Fsone, Nov 27 '17 at 13:34

score 10 · Answer 1 · answered Oct 24 '13 at 14:18

10

Since $u$ is entire, and $\mathbb{C}$ is simply connected, there is an entire holomorphic function $f \colon \mathbb{C}\to\mathbb{C}$ with $u = \operatorname{Re} f$. Then

$$g(z) = \frac{f(z)-1}{f(z)+1}$$

is an entire bounded function, hence $g$ is constant. Therefore $f$ is constant, and hence also $u$.

answered Oct 24 '13 at 14:18

Daniel Fischer

206,697

Alternatively, just invoke Picard's theorem :P – Potato Oct 24 '13 at 14:30
$\exp(-f)$ can also be used. – Ayman Hourieh Oct 24 '13 at 14:47
3

@Potato Yes. What's your preferred method to kill mosquitoes? – Daniel Fischer Oct 24 '13 at 15:23
Point taken! By the way, I appreciate having you around the site. You write a lot of good analysis answers. It's inspirational. – Potato Oct 24 '13 at 15:34

Let $u:\mathbb{C}\rightarrow\mathbb{R}$ be a harmonic function such that $0\leq f(z)$ for all $ z \in \mathbb{C}$. Prove that $u$ is constant.

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