Question 4. Let \(u ( z ) > 0\) be a positive harmonic function in the punctured plane \(0 < | z |\) Show that u must be constant.
Solution
1. Lifting by the exponential map gives a positive harmonic function on the whole plane.
Define \[ v(w)=u(e^w),\qquad w\in\mathbb C. \] Since the exponential map is holomorphic and never vanishes, \(v\) is defined on all of \(\mathbb C\). Composition of a harmonic function with a holomorphic map is harmonic; explicitly, writing \(w=s+it\) and using the chain rule gives \[ \Delta v(w)=|e^w|^2(\Delta u)(e^w)=0. \] Moreover \(v(w)>0\) everywhere because \(u>0\) on the punctured plane.
2. Every positive harmonic function on the plane is constant.
Because \(\mathbb C\) is simply connected, the harmonic function \(v\) has a global harmonic conjugate, so there is an entire function \(H\) with \(\operatorname{Re}H=v\). Then \[ E(w)=e^{-H(w)} \] is entire and \[ |E(w)|=e^{-v(w)}<1. \] Liouville’s theorem makes \(E\) constant. Since an exponential never vanishes, \(0=E'=-H'e^{-H}\) implies \(H'=0\), hence \(H\) and therefore \(v\) are constant.
3. Constancy descends to the punctured plane.
The exponential map is onto \(\mathbb C\setminus\{0\}\). For any \(z\ne0\), choose \(w\) with \(e^w=z\). If \(v\equiv c\), then \[ u(z)=u(e^w)=v(w)=c. \] Thus \(u\) is constant on the punctured plane.