- Let \(h : \CC \to \RR\) be a harmonic function such that h is bounded below. Prove that h is constant.
Solution
1. Construct an entire function whose real part is \(h\).
Write \(z=x+iy\), regard \(h\) as a function of \((x,y)\), and define on all of \(\RR^2\) \[ v(x,y)=-\int_0^x h_y(t,0)\,dt+ \int_0^y h_x(x,s)\,ds. \] The first and second derivatives used here are continuous, and \(h_{xx}+h_{yy}=0\), since \(h\) is harmonic. Differentiation under these finite integrals gives \[ v_y(x,y)=h_x(x,y), \] and \[ \begin{aligned} v_x(x,y) &=-h_y(x,0)+\int_0^y h_{xx}(x,s)\,ds\\ &=-h_y(x,0)-\int_0^y h_{yy}(x,s)\,ds =-h_y(x,y). \end{aligned} \] Thus \(H=h+iv\) has continuous first partial derivatives and satisfies the Cauchy–Riemann equations on the plane. By Cauchy--Riemann implies holomorphic, it is entire and has real part \(h\).
2. The lower bound makes an entire exponential bounded.
Choose \(m\in\RR\) with \(h(z)\geq m\) everywhere. Then \(E(z)=e^{-H(z)}\) is entire and satisfies \[ \abs{E(z)}=e^{-\operatorname{Re}H(z)}=e^{-h(z)}\leq e^{-m}. \] Liouville's theorem makes \(E\) constant. Since an exponential never vanishes, differentiation of this constant function yields \(0=E'=-H'e^{-H}\) and hence \(H'=0\) everywhere. An entire function with zero derivative is constant on the connected plane, so \(h=\operatorname{Re}H\) is constant.
3. Q.E.D.
Step 2 proves that \(h\) is constant.