Show that if \(f\) is once complex differentiable at each point of \(\Omega\), then \(f\) is holomorphic.
Solution
It is enough to prove Goursat’s theorem: complex differentiability at every point already implies the Cauchy integral theorem, without assuming continuity of \(f'\).
1. Let \(T\subset\Omega\) be a closed triangle and set\[ I(T)=\int_{\partial T}f(z)\,dz. \]Subdivide \(T\) into four congruent triangles. Since the integrals on internal edges cancel, one subtriangle \(T_1\) satisfies\[ |I(T_1)|\ge\frac14|I(T)|. \]Iterating gives nested triangles \(T_n\) with\[ |I(T_n)|\ge4^{-n}|I(T)|, \]and with diameter and perimeter scaled respectively by \(2^{-n}\). Their intersection is a single point \(z_0\).
2. Differentiability at \(z_0\) gives\[ f(z)=f(z_0)+f'(z_0)(z-z_0)+\eta(z)(z-z_0), \qquad \eta(z)\to0. \]The first two terms have zero integral around \(\partial T_n\). Hence, for every \(\varepsilon>0\) and all sufficiently large \(n\),\[ |I(T_n)| \le \varepsilon\,\operatorname{diam}(T_n)\,\operatorname{length}(\partial T_n) =4^{-n}C\varepsilon \]for a constant \(C\) independent of \(n\). Combining with the lower bound gives \(|I(T)|\le C\varepsilon\). Since \(\varepsilon\) is arbitrary, \(I(T)=0\).
3. Thus the integral of \(f\) around every triangle compactly contained in \(\Omega\) vanishes. On a disk \(D\Subset\Omega\), fix \(z_*\) and define\[ F(z)=\int_{[z_*,z]}f(w)\,dw. \]The triangle integral identity shows that\[ F(z+h)-F(z)=\int_{[z,z+h]}f(w)\,dw, \]so continuity of \(f\) (which follows from complex differentiability) gives \(F'(z)=f(z)\). Therefore \(F\) is holomorphic and hence analytic; consequently \(f=F'\) is analytic on \(D\). Since every point of \(\Omega\) lies in such a disk, \(f\) is holomorphic/analytic throughout \(\Omega\).