Let \(f\) be a holomorphic function on the punctured disk \(U = \{z \in \mathbb{C} : 0 < |z| < 1\}\). Suppose that \(|f(z)| \le |z|^{-1/2}\) for all \(z \in U\). Prove that \(f\) has a removable singularity at \(0\).
Solution
1. The function \(g(z)=zf(z)\) extends holomorphically across \(0\) with value \(0\).
For \(0<|z|<1\), \[ |g(z)|=|z|\,|f(z)|\le |z|^{1/2}. \] Thus \(g\) is bounded near \(0\), so the Riemann removable singularity theorem gives a holomorphic extension \(G\) to the unit disk. The displayed estimate also gives \[ G(0)=\lim_{z\to0}zf(z)=0. \]
2. Dividing the zero of \(G\) by \(z\) gives a holomorphic extension of \(f\).
By holomorphic implies analytic, the Taylor series of \(G\) at \(0\) has the form \[ G(z)=\sum_{n\ge1}a_nz^n \] because \(G(0)=0\). Hence \[ F(z)=\sum_{n\ge1}a_nz^{n-1} \] is holomorphic near \(0\) and satisfies \(G(z)=zF(z)\). For \(z\ne0\) in the punctured disk, \[ F(z)=\frac{G(z)}z=f(z). \] Thus \(F\) extends \(f\) holomorphically across \(0\).
3. Q.E.D.
Step 2 constructs the required holomorphic extension, so the singularity at \(0\) is removable.