Liouville's theorem for planar harmonic functions

Remarks

Proof idea: $\C$ is simply connected, so by the existence of harmonic conjugates there is an entire function $f$ with $\operatorname{Re} f = u$. If $u \le M$ on $\C$, then $g = e^{f}$ is entire with $\abs{g} = e^{u} \le e^{M}$. By Liouville's theorem, $g$ is constant, so $f' e^{f} = g' = 0$, hence $f' = 0$ and $u$ is constant. If $u$ is bounded below, apply this to $-u$. A one-sided bound is needed: $u(x + iy) = x$ is harmonic, nonconstant, and unbounded in both directions. The theorem is the planar case of Liouville's theorem for harmonic functions on $\R^n$. In the plane even more is true: a subharmonic function on $\C$ that is bounded above is constant. This stronger form fails in $\R^n$ for $n \ge 3$, where $-\min(1, \abs{x}^{2-n})$ is subharmonic, bounded above, and not constant.

Going deeper

For a student

The proof shows a typical planar technique: pass to a holomorphic function with real part $u$ and use complex analysis. The theorem gives quick uniqueness results, for example for harmonic functions on $\C$ with a given one-sided bound. Its subharmonic form says that $\C$ is a parabolic surface in potential theory: it carries no nonconstant negative subharmonic function.