Let $u:\mathbb R^2 \to \mathbb R$ be a non-surjective harmonic function.

$(i)$ Show that $u$ is bounded from below or from above.

$(ii)$ Prove that $u$ is constant (and therefore any harmonic function is constant or surjective)

I know how to prove $(i) \implies (ii)$:

Suppose $|u|\leq B$, by hypothesis, there exists an entire function $f$ such that $f(x+iy)=u(x,y)+iv(x,y)$. If I consider $e^{f(z)}$, then $e^{f(z)}$ is an entire function and $$|e^{f(z)}|=|e^{u+iv}|$$ $$=e^u|e^{iv}|$$ $$\leq e^B.1$$ $$=c \in \mathbb R_{\geq 0}$$

This proves $e^{f(z)}$ is bounded so, by Liouville's theorem, $e^{f(z)}$ is constant. In this exercise Application of Liouville's theorem exercise it is proven that $e^{f(z)}$ constant $\implies$ $f(z)$ constant.

I am having problems with the first part ($(i)$) of the exercise. Suppose $u$ is not surjective, then there exists $u_0 \in \mathbb R$ such that $u(x,y) \neq u_0$ for all $(x,y) \in \mathbb R^2$. I have no idea how to deduce $u$ is not bounded from this hypothesis, I would appreciate any hints.



Since $u$ is continuous and $\mathbb{R}^2$ is connected then $u(\mathbb{R}^2)$ is connected. So, if $u$ is not surjective then it must be bounded from below or from above. (Thanks to @Vladimir for clarifying this point).

End edit

If $u$ is bounded from above, say $u(x,y)\le M,$ then $|e^{f(z)}|\le e^M$ and you can conclude.

If $u$ is bounded from below, say $u(x,y)\ge -M,$ then $|e^{-f(z)}|\le e^{M}$ and you can conclude.

  • 1
    $\begingroup$ Are you sure this answers the first part? $\endgroup$ – Cameron Williams Jul 2 '14 at 19:47

$u$ not surjective $\implies$ (because $u$ is continuous) $u$ bounded above OR below $\implies$ (thanks to @mfl 's answer) $u$ is constant $\implies$ $u$ bounded above AND below


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