Assume $f$ is harmonic on $\mathbb{R}^2$. I want to prove that if there exists a constant $M$ such that $f(x,y) \geq M$ for all $(x,y)\in \mathbb{R}^2$, then $f$ must be a constant fuction.
I'm having trouble because usually when we have to prove existence of a constant inequality with a function, usually we have to prove that a function is bounded with $\leq$, etc. But here I want to show that the function is bigger than equal to a constant, and I kinda have no idea where to start. Can someone help me out? Thanks!
Xiao
Here's one way to do it. Let $g$ be a harmonic conjugate of $f$, and let $F(z) = \exp(-(f+ig))$. Then $F$ is entire and $|F| = \exp(-f) \le \exp(-M)$, so $F$ is bounded.
By Liouville's theorem $F$ is constant, and consequently so is $f$.