Let \[ A=\{0\}\cup\{1/n:n\in\mathbb Z,\ n>1\}, \] and let \(D\) be the open unit disk in \(\mathbb C\). Prove that every bounded holomorphic function on \(D\setminus A\) extends to a holomorphic function on \(D\).
Solution
Let \(f\) be bounded and holomorphic on \(D\setminus A\).
1. For every integer \(n>1\), the singularity of \(f\) at \(1/n\) is removable.
Fix \(n>1\). The point \(1/n\) is isolated in \(A\): there is \(r_n>0\) such that \[ B(1/n,r_n)\cap A=\{1/n\}. \] Thus \(f\) is holomorphic and bounded on the punctured disk \[ 0<\abs{z-1/n}<r_n. \] By the removable singularity theorem, \(f\) extends holomorphically across \(1/n\).
2. The extensions from step 1 define a bounded holomorphic function\(g\) on \(D\setminus\{0\}\) that agrees with \(f\) on \(D\setminus A\).
1. These local definitions agree with\(f\) on punctured neighborhoods, so they define a holomorphic function on\(D\setminus\{0\}\).
If \(\abs{f}\leq M\) on \(D\setminus A\), continuity of the extension at \(1/n\) gives \[ \abs{g(1/n)} = \lim_{z\to1/n}\abs{f(z)} \leq M. \] Hence \(\abs{g}\leq M\) throughout \(D\setminus\{0\}\).
3. The singularity of \(g\) at \(0\) is removable.
By step 2, \(g\) is bounded and holomorphic on the punctured unit disk. The removable singularity theorem therefore gives a holomorphic extension \(G\) of \(g\) across \(0\).
4. The function \(G\) is a holomorphic extension of \(f\) to all of \(D\).
Step 3 gives \(G\in\mathcal O(D)\), and step 2 gives \(G=g=f\) on \(D\setminus A\).
5. Q.E.D.
Step 4 is the required extension.