- Show that if \(f\) is once complex differentiable at each point of \(\Omega\), then \(f\) is holomorphic.
Solution
Goal: Show that if \(f\) is once complex-differentiable at each point of a domain \(\Omega\), then \(f\) is holomorphic (i.e. analytic) on \(\Omega\).
1. \(f\) is continuous on \(\Omega\).
Differentiability at each point implies continuity at each point.
2. (Goursat) \(\int_T f = 0\) for every closed triangle \(T\) contained (with interior) in \(\Omega\).
2.1. Assume, toward a contradiction, that \(\abs{\int_T f} = M > 0\) for some triangle \(T\).
Otherwise done.
2.2. Subdivide \(T\) into four congruent subtriangles \(T_1, \ldots, T_4\); then \(\abs{\int_{T_j} f} \geq M/4\) for some \(j\).
The sum of the four integrals equals \(\int_T f\) (interior edges cancel), so by the triangle inequality at least one has modulus \(\geq M/4\).
2.3. Iterating, get a nested sequence of triangles \(T_n\) with \(\abs{\int_{T_n} f} \geq M/4^n\) and diameters \(\to 0\).
Repeat 2 indefinitely; the diameters shrink by a factor \(1/2\) each step.
2.4. The nested triangles have a common point \(z_0 \in T\) (indeed \(z_0 \in \Omega\)).
Nested compact sets with diameters tending to \(0\) intersect in a single point.
2.5. Write \(f(z) = f(z_0) + f'(z_0)(z - z_0) + (z - z_0)\varepsilon(z)\) with \(\varepsilon(z) \to 0\) as \(z \to z_0\).
This is the definition of differentiability at \(z_0\).
2.6. \(\int_{T_n} \qty[f(z_0) + f'(z_0)(z - z_0)]\, dz = 0\).
The constant \(f(z_0)\) and the linear function \(f'(z_0)(z - z_0)\) have primitives (e.g. \(f'(z_0)(z - z_0)^2/2\)), and \(\int_{T_n} dz = \int_{T_n} z\, dz = 0\) for a closed contour.
2.7. \(\abs{\int_{T_n} f} = \abs{\int_{T_n} (z - z_0)\varepsilon(z)\, dz} \leq \sup_{z \in T_n} \abs{\varepsilon(z)} \cdot \operatorname{diam}(T_n) \cdot \operatorname{perim}(T_n)\).
Standard estimate: modulus of a line integral is bounded by the maximum of the integrand times the length of the contour.
2.8. Contradiction.
\(\operatorname{diam}(T_n) = 2^{-n} d\) and \(\operatorname{perim}(T_n) = 2^{-n} p\) with \(d, p\) fixed, so the bound in 7 is \(\sup\abs\varepsilon \cdot dp/4^n\); but \(\sup_{z \in T_n}\abs{\varepsilon(z)} \to 0\) (<2>5), so \(\abs{\int_{T_n} f} < M/4^n\) for large \(n\), contradicting 3.
3. By Morera’s theorem, \(f\) is holomorphic.
Morera: if \(f\) is continuous on \(\Omega\) (<1>1) and \(\int_T f = 0\) for every triangle \(T \subset \Omega\) (<1>2), then \(f\) is holomorphic.
4. Q.E.D.
3 is the claim.