I don't have any idea to prove the following fact. Can anyone help me?
Let $u : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a harmonic function such that
$$ u(x,y) \geq xy, \qquad \text{for every}~~ (x,y) \in \mathbb{R}^2 .$$ Prove that
$$ u(x,y) = xy + c,$$
where $c$ is a non-negative constant.
Thank you.
The function $v(x,y)=xy$ is harmonic, since $\partial_{xx}v+\partial_{yy}v=0$. Now $u-v$ is a positive, harmonic function defined on $\mathbb{R}^2$. We can find a harmonic conjugate of $u-v$, i.e. a harmonic function $h$ such that $f=(u-v)+ih$ is holomorphic in the complex plane. Now since $u-v\geq0$, $f$ maps the complex plane to the upper half plane. If we compose with a Mobius transformation that maps the upper half plane to the unit disk, then we get a bounded, entire function. By Liouville's theorem, this composition is a constant function. But this implies that $f$ is constant. Therefore $u-v$ is constant, hence $u(x,y)=v(x,y)+c=xy+c$.