Let \(S\definedas \theset{z\in \DD\suchthat \Im(z) \geq 0}\). Suppose \(f:S\to \CC\) is continuous on \(S\), real on \(S\intersect \RR\), and holomorphic on \(S^\circ\).
Prove that \(f\) is the restriction of a holomorphic function on \(\DD\).
Solution
Define a function \[ F(z) \da \begin{cases} f(z) & \Im(z)\geq 0 \\ \bar{f(\bar z)} & \Im(z) < 0 \end{cases} .\] Then \(F\) is holomorphic on \(\tilde S\da \ts{z\in \DD\st \Im(z) < 0}\) – write \(w_0\in \tilde S\) as \(w_0 = \bar{z_0}\) for some \(z_0\in S\), then \[ f(z) &= \sum_{k\geq 0}c_k (z-z_0)^k \\ \implies \bar{f(\bar z)} &= \sum_{k\geq 0}\bar{c_k} \bar{\qty{\bar z - z_0}^k} \\ &= \sum_{k\geq 0}\bar{c_k} \qty{z - \bar{z_0}}^k \\ &= \sum_{k\geq 0}\bar{c_k} \qty{z - w_0}^k ,\] which yields a power series expansion of \(F\) about \(w_0\). So \(f\) is analytic at every point in \(\tilde S\) and thus holomorphic. Since \(\bar{f(\bar z)} = f(z)\) for \(z, f(z)\in \RR\), \(F\) is a continuous extension of \(f\) to \(\DD\). By the symmetry principle, \(F\) is holomorphic, and \(\ro{F}{S} = f\).