Suppose \(f\) is holomorphic on \(\Delta^*\) and \(\Re(f) \geq 0\). Show that \(f\) has a removable singularity at \(z=0\).
Solution
We have \(f: \Delta^* \to \ts{\Re(z) > 0}\), so let \(T: \ts{\Re(z) > 0}\to \Delta\) be the rotated Cayley map \(T(z) = {z-1\over z+1}\). Then \(G\da \inverseof{T} \circ f: \Delta^* \to \Delta\), and since \(\abs{T} < 1\), \(z=0\) is a removable singularity for \(F\) and \(F\) extends holomorphically to \(G:\Delta\to \CC\), since a priori \(G(0)\) may not be bounded by 1. Supposing \(G\) is not constant, \(G(\Delta) \subseteq \overline{\Delta}\) by continuity and \(G(\Delta)\) is open by the open mapping theorem, and \(\Delta = f(\Delta) \subseteq G(\Delta)\), so \(G:\Delta\to \Delta\) is a map of the disc. Then \(F \da T\circ G: \Delta\to \ts{\Re(z) > 0}\) is an extension of \(F\) which is bounded in neighborhoods of \(z=0\), making zero a removable singularity for \(f\).