Suppose that \(f: \CC\to\CC\) is continuous everywhere and analytic on \(\CC\setminus \RR\) and prove that \(f\) is entire.
Solution
Just reproducing the proof of holomorphicity in the Schwarz reflection theorem.
Note \(f\) is continuous on \(\CC\) since analytic implies continuous (\(f\) equals its power series, where the partials sums uniformly converge to it, and uniform limit of continuous is continuous).
Strategy: take \(D\) a disc centered at a point \(x\in \RR\), show \(f\) is holomorphic in \(D\) by Morera’s theorem.
Let \(\Delta \subset D\) be a triangle in \(D\).
Case 1: If \(\Delta \intersect \RR = 0\), then \(f\) is holomorphic on \(\Delta\) and \(\int_\Delta f = 0\).
Case 2: one side or vertex of \(\Delta\) intersects \(\RR\), and wlog the rest of \(\Delta\) is in \(\HH^+\).
Then let \(\Delta_\eps\) be the perturbation \(\Delta + i\eps = \theset{z+ i\eps \suchthat z\in \Delta}\); then \(\Delta_\eps \intersect \RR = 0\) and \(\int_{\Delta_\eps} f = 0\).
Now let \(\eps\to 0\) and conclude by continuity of \(f\).
Parametrize \(\Delta\) by a piecewise-smooth closed curve \(\gamma: [a,b] \to \CC\), and write \(\gamma_\eps(t) \da \gamma(t) + i\eps\) for the corresponding parametrization of \(\Delta_\eps\). Then \(\gamma_\eps'(t) = \gamma'(t)\), so \[ \int_{\Delta_\eps} f = \int_a^b f(\gamma_\eps(t)) \gamma_\eps'(t)\,dt = \int_a^b f(\gamma(t) + i\eps) \gamma'(t)\,dt. \]
We claim this converges to \(\int_\Delta f = \int_a^b f(\gamma(t))\gamma'(t)\,dt\) as \(\eps \to 0\). The image \(\gamma([a,b])\) is compact, so \(f\) is uniformly continuous on a compact neighborhood of it: for every \(\eta > 0\) there is \(\eps_0 > 0\) such that \(|f(\gamma(t) + i\eps) - f(\gamma(t))| < \eta\) for all \(t \in [a,b]\) and all \(0 < \eps < \eps_0\). Since \(\gamma'\) is bounded on \([a,b]\) (it is continuous on a compact interval), say \(|\gamma'(t)| \le M\), we get \[ \left| \int_a^b f(\gamma(t)+i\eps)\gamma'(t)\,dt - \int_a^b f(\gamma(t))\gamma'(t)\,dt \right| \le \int_a^b |f(\gamma(t)+i\eps) - f(\gamma(t))|\,|\gamma'(t)|\,dt \le \eta M (b-a), \] which tends to \(0\) as \(\eta \to 0\). Hence \(\int_{\Delta_\eps} f \to \int_\Delta f\), and since each \(\int_{\Delta_\eps} f = 0\), we get \(\int_\Delta f = 0\).
Case 3: \(\Delta\) intersects both \(\HH^+\) and \(\HH^-\).
- Break into smaller triangles, each of which falls into one of the previous two cases.