- State whether each of the following two statements is true or false, and give either a proof or counterexample for each.
All holomorphic functions \(f:\mathbb C\setminus\{0\}\to H\) are constant, where \(H=\{z\in\mathbb C:\operatorname{Im}z>0\}\) is the upper half-plane.
All harmonic functions \(h : \mathbb { C } \setminus [ 0 , + \infty ) \to [ 0 , 1 ]\) are constant.
Solution
Assertion (a) is true; assertion (b) is false.
1. A holomorphic map from the punctured plane into \(H\) is constant.
Given such an \(f\), define \[ g(z)=\frac{f(z)-i}{f(z)+i}. \] Its denominator is nonzero because \(f(z)\in H\), and \[ |f(z)+i|^2-|f(z)-i|^2=4\operatorname{Im}f(z)>0. \] Thus \(g\) is holomorphic on the punctured plane with \(|g|<1\). The removable-singularity theorem extends it to an entire function \(G\), with \(|G(0)|\leq1\) by continuity [@SS03]. This entire extension is bounded, so Liouville’s theorem makes it constant. Its value \(c=g(1)\) satisfies \(|c|<1\). Solving the fractional transformation gives \[ f(z)=i\frac{1+c}{1-c}\qquad(z\ne0), \] where \(1-c\ne0\). Hence \(f\) is constant, proving (a).
2. A branch of the argument supplies a counterexample to (b).
On \(\Omega=\mathbb C\setminus[0,\infty)\) choose the logarithm branch \[ L(re^{i\theta})=\log r+i\theta, \qquad r>0,\quad 0<\theta<2\pi. \] Every point of \(\Omega\) has a unique such argument, and this branch is holomorphic there, with derivative \(1/z\) [@SS03]. Therefore its imaginary part is harmonic. The function \[ h(z)=\frac{\operatorname{Im}L(z)}{2\pi} \] is harmonic on \(\Omega\) and takes values in \((0,1)\subset[0,1]\). It is not constant: \(h(i)=1/4\) and \(h(-i)=3/4\). It meets all the hypotheses of (b) and contradicts its conclusion, so (b) is false.