Take $f:D\subseteq\mathbb{C}\rightarrow \mathbb{C}$ to be holomorphic on some path connected $D$. Suppose there exists a function $F:D\rightarrow \mathbb{C}$, such that for all $z,z'\in D$ and any path $\gamma:[0;1]\rightarrow D$ with $\gamma(0)=z$ and $\gamma(1)=z'$ we have: $$\int_\gamma f(z)\, dz = F(z')-F(z)$$ Can we show that $F$ is differentiable and $F'=f$ on $D$ (so $F$ is a complex antiderivative of $f$) without using Cauchy's integral formula or equivalence between the notions holomorphic and analytic?
I already tried fixing one point of the path and then splitting it up (or alternatively the whole integral) into real and imaginary part to be able to use some real analysis but I didn't find something useful.Any ideas?
Regards
Yes, we can (assuming that $D$ is an open set). Let $z\in D$ and take $\varepsilon>0$. I will prove that there is a $\delta>0$ such that$$\lvert h\rvert<\delta\implies\left\lvert\frac{F(z+h)-F(z)}h-f(z)\right\rvert<\varepsilon.$$Let $r>0$ be such that $\overline{B(z,r)}\subset D$. Take $\delta>0$ such that $\delta<r$ and that $\lvert w-z\rvert<\delta\implies\bigl\lvert f(w)-f(z)\bigr\rvert<\varepsilon$. Then, for each $h$ such that $\lvert h\rvert<\delta$, consider the path $\gamma\colon[0,1]\longrightarrow\mathbb C$ defined $\gamma(t)=z+th$. Then\begin{align}\left\lvert\frac{F(z+h)-F(z)}h-f(z)\right\rvert&=\left\lvert\frac{\int_\gamma f(w)\,\mathrm dw-hf(z)}h\right\rvert\\&=\left\lvert\frac{\int_\gamma f(w)-f(z)\,\mathrm dw}h\right\rvert\\&<\frac{\varepsilon\lvert h\rvert}{\lvert h\rvert}\\&<\varepsilon.\end{align}