Show that a harmonic function is a polynomial.

Question

I am trying to solve:

Let $u$ be a harmonic function in $\mathbb{R}^2$, with $u(0,0)=0$ and $u(x,y)\leq{x^2}-y^2$ for all $(x,y)$ in $\mathbb{R}^2$, show that $u$ is a polynomial.

My approach:

If $u$ is a polynomial it would be of the form $u(x,y)=a_1x+a_2y+a_3xy$ and its second derivative should vanish. I also think the mean value property, since $u(x)$ and $x^2-y^2$ are both harmonic will be needed. But I don't know how to properly proceed here.

I would really appreciate any help.

Answer 1

Hint: Let $v(x,y) = x^2-y^2-u(x,y)$. What properties does $v$ satisfy? Consider Liouville's theorem/the Harnack inequality.

The function $v$ is harmonic in $\mathbb R^2$, non-negative in $\mathbb R^2$, and $v(0,0)=0$. There is a version of Lioville's theorem that states that if $v$ is harmonic in $\mathbb R^2$ (or $\mathbb R^n$) and is bounded from below then it is a constant. Since $v(0,0)=0$, this implies that $v$ is in fact equal to $0$, so $u(x,y)=x^2-y^2$.

If you weren't familiar with this version of Lioville's theorem then you could instead have used the Harnack inequality: Since $v$ is nonnegative and harmonic, for all $\rho>0$, $$0\leqslant \sup_{B_\rho} v \leqslant C \inf_{B_\rho} v =0,$$ so $v=0$.