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. [duplicate]
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!
Solution 1:
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 $u$.