I've been thinking about the following problem. If $f$ is continuous on the entire complex plane, is it necessarily true that the function $G_R(w)= \int_0^w f(z)dz$ is holomorphic on the closed disk of radius $R$ centered at 0? The real-valued analogue holds, but complex is alittle more delicate it seems. Does anyone have any suggestions?
2026-03-27 19:29:57.1774639797
Complex-valued continuous is holomorphic as an integral.
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I assume that the integral is taken over the line segment from $0$ to $w$.
Let $f(z) = \bar z$. Then $f$ is continuous on the complex plane, but $$ G(w) = \int_0^w \bar z\,dz = \int_0^1 \overline{t w} \, w \, dt = \frac12 |w|^2 $$ which is not holomorphic on any disc. (I used the parametrization $t \mapsto t\,w$).