Let \(u\) be a positive harmonic function on \(\mathbb R^2\): \[ u_{xx}+u_{yy}=0. \] Prove that \(u\) is constant.
Solution
1. There is a \(C^1\) function \(v:\RR^2\to\RR\) such that\[ v_x=-u_y, \qquad v_y=u_x. \]
Consider the one-form \[ \omega=-u_y\,dx+u_x\,dy. \] Its exterior derivative is \[ \begin{aligned} d\omega &= (u_{xx}+u_{yy})\,dx\wedge dy\\ &= 0 \end{aligned} \] because \(u\) is harmonic.
The plane \(\RR^2\) is simply connected, so every closed \(C^1\) one-form on it is exact. Hence there is a function \(v\) with \[ dv=\omega, \] which is precisely the pair of displayed derivative identities.
2. The function\[ F(z)=u(x,y)+iv(x,y), \qquad z=x+iy, \]is entire.
The identities in step 1 are the Cauchy–Riemann equations \[ u_x=v_y, \qquad u_y=-v_x. \] Thus \(F\) is holomorphic on all of \(\CC\).
3. The entire function\[ H(z)=e^{-F(z)} \]is bounded.
Since \(u\) is positive, \[ \abs{H(z)} = \abs{e^{-u(x,y)-iv(x,y)}} = e^{-u(x,y)} \leq 1 \] for every \(z\in\CC\).
4. The function \(F\) is constant.
By step 3 and Liouville’s theorem, \(H\) is constant. Since an exponential never vanishes, \[ H(z)\neq0 \] for every \(z\). Differentiating \[ H=e^{-F} \] gives \[ H'=-F'e^{-F}. \] Because \(H'=0\) and \(e^{-F}\neq0\), one has \[ F'=0. \] Therefore \(F\) is constant on the connected plane.
5. The harmonic function \(u\) is constant.
4. Hence \(u\) is constant.
6. Q.E.D.
Step 5 is the required conclusion.