# How to prove that a holomorphic function on an open simply connected space has a primitive?

I want to prove:

Let $$\Omega$$ be an open simply connected subspace of $$\mathbb C$$. Let $$f:\Omega\to\mathbb C$$ be holomorphic. Then $$f$$ has a primitive (antiderivative) on $$\Omega$$.

I don't want to use Cauchy's integral theorem, because I need the above theorem to prove Cauchy's integral theorem. In my complex analysis book, the above theorem and Cauchy's integral theorem is proved only for open discs or some special domains, but not general open simply connected domains. So I want to know the proof for the general situation. Here is my attempt. I know the following facts.

Theorem 1. Let $$\Omega$$ be an open connected subsace of $$\mathbb C$$ and let $$x,y\in\Omega$$. Then $$x$$ and $$y$$ can be joined by a finite number of straight line segments. More precisely, there is $$z:[0,1]\to\Omega$$ and $$0=a_0 ($$n$$ is a nonzero natural number) such that $$z(0)=x,z(1)=y$$ and that for all $$0\leq i and $$t\in[a_i,a_{i+1}]$$, $$$$z(t)=\frac{(a_{i+1}-t)z(a_i)+(t-a_i)z(a_{i+1})}{a_{i+1}-a_i}\text{.}$$$$ Call such $$z$$ a piecewise-straight path from $$x$$ to $$y$$.

Theorem 2 (Goursat's Theorem). If $$\Omega$$ is an open set in $$\mathbb C$$, and $$T\subseteq\Omega$$ a triangle whose interior (bounded component) is also contained in $$\Omega$$, then $$$$\int_Tf(z)dz=0$$$$ whenever $$f$$ is holomorphic in $$\Omega$$.

Goursat's theorem citation: Stein, Elias M.; Shakarchi, Rami, Complex analysis, Princeton Lectures in Analysis. 2. Princeton, NJ: Princeton University Press. xvii, 379 p. (2003). ZBL1020.30001.

Now let $$\Omega$$ be a nonempty open simply connected subspace of $$\mathbb C$$ and let $$f:\Omega\to\mathbb C$$ be holomorphic. Choose $$p\in\Omega$$. For each $$z\in\Omega$$ define $$$$F(z)=\int_{\gamma_z}f(w)dw$$$$ where $$\gamma_z$$ is a curve parametrized by a piecewise-straight path (defined in theorem 1) from $$p$$ to $$z$$ ($$\gamma_z$$ may not be unique for each $$z\in\Omega$$, but just choose one for each). I want to show that $$F$$ is a primitive of $$f$$. Fix $$z_0\in\Omega$$ and $$z\in D$$, where $$D\subseteq\Omega$$ is an open disc centered at $$z_0$$. To show that $$(F(z)-F(z_0))/(z-z_0)\to f(z)$$ as $$z\to z_0$$, I want to compute $$F(z)-F(z_0)$$. I think I should show that $$F(z)-F(z_0)=\int_\eta f(w)dw$$, where $$\eta$$ is the straight line segment from $$z_0$$ to $$z$$. Let $$\gamma_{z_0}\ast\eta$$ be the curve from $$p$$ to $$z$$ defined by joining $$\gamma_{z_0}$$ and $$\eta$$. Since $$\Omega$$ is simply connected, there is a path homotopy $$H:I^2\to\Omega$$ from $$\gamma_{z_0}\ast\eta$$ to $$\gamma_z$$, where $$I=[0,1]$$. If $$H_s:I\to\Omega,t\mapsto H(s,t)$$ is a straight line segment for each $$s\in I$$, then I think I can use Goursat's theorem to show that $$F(z)-F(z_0)=\int_\eta f(w)dw$$. But can we choose $$H$$ to satisfy this? Or maybe my attempt is wrong.

You forgot to check $$F$$ is well-defined since it is apparently dependent on the path you choose. First of all cover $$\Omega$$ by open balls. We shall show there exists a primitive on each ball.

Let $$B(z_0,R)$$ be a ball. For $$z\in B(z_0,R)$$, choose the radial path from $$z_0$$ to $$z$$, call that $$\gamma_z$$.

Next define $$F(z)=\int_{\gamma_z}f(\xi)d\xi$$

Then we observe that for $$h$$ small, $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_{L(z,z+h)}f(\xi)d\xi$$by Gousrat Theorem where $$L(z,z+h)$$ is the straight line joining $$z$$ to $$z+h$$ Then we get $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_0^1f(z+\theta h)hd\theta=\int_0^1f(z+\theta h)d\theta \rightarrow f(z)$$as $$h\rightarrow 0$$ by your favourite convergence theorem.

Thus we have showed the existence of an anti-derivative on an open ball.

This in particular shows that integral of a holomorphic function on a closed curve in any open ball is 0.

Now let $$H$$ be a fixed end point homotopy between two paths $$\gamma_0,\gamma_1$$ in a region $$\Omega$$ Say $$H: I^2\rightarrow \Omega$$ Choose a partition of $$I^2$$ into a grid $$\{G_{ij}\}$$ so that any small tile $$G_{ij}$$ falls into an open ball in $$\Omega$$ via $$H$$ using continuity and compactness. Join the corners of the tiles in $$\Omega$$ by straight lines and for the sake of simplicity call them $$G_{ij}$$ as well.

Then one can write $$\int_{\gamma_0}f(\xi)d\xi -\int_{\gamma_1}f(\xi)d\xi =\sum_{i,j}\int_{\partial G_{ij}} f(\xi) d\xi$$ Each term in the last sum is $$0$$ as it is the integral of a holomorphic function on an open ball by our choice of the partition.

This shows integral of a holomorphic function on 2 fixed end-point homotopic curves is the same. In particular this shows integral of a holomorphic function on a closed curve in any simply-connected domain is 0.

Then you can proceed as you were doing.

If you are interested in a purely algebraic topology approach, here's one way to proceed.

We have solved the primitive problem locally an open cover say balls $$\mathcal B=\{ B_i\}$$. Any $$2$$ such solutions on a ball differ by a constant. Say we fix a local solution $$\{f_i \}_i$$ on the local cover $$B_i$$

Then on the intersection $$B_i\cap B_j$$ we get a complex number $$c_{ij}$$ such $$f_i-f_j=c_{ij}$$. Thus we get a co-cycle in the Cech cohomology group $$\hat {H^1}(\Omega ;\mathcal B)$$

The cover we chose was a Leray cover as intersections of convex sets are convex and hence all contractible. So the obstruction is an element of $$\hat{H^1}(\Omega; \mathbb C)\cong {H_{dR}^1}(\Omega; \mathbb C)$$

For simply-connected smooth manifolds, the $$1$$st de-Rham cohomology group is $$0$$ and hence we get the obstruction we got isn't an obstruction at all.

• Thanks! I got most of the first part (up to where the algebraic topology approach begins) of your answer. But I don't know why $\partial G_{ij}$ can be considered piecewise-smooth. It has to be piecewise-smooth in order for $\int_{\partial G_{ij}}f(\xi)d\xi$ to make sense. Should I add some restriction on $H$ so that $\partial G_{ij}$ becomes piecewise-smooth? I only know the undergraduate algebraic topology (2nd part of Munkres's "Topology"). I don't know manifold or homology. If the proof seems too hard for me, just giving me the relevant theorem would be great. – zxcv May 24 '20 at 4:37
• You have chosen the tiling of $I^2$ so that each small tile falls in an open ball. In that ball join the corner points of the tiles using straight lines. Then you get a region in $\Omega$. That is my new $G_{ij}$. If you want you can call it $\tilde G_{ij}$ and then do the same thing with $\tilde G_{ij}$ in place of $G_{ij}$ – Baidehi May 24 '20 at 10:09
• As for the algebraic topology, I used some ideas of sheaf cohomology. If you want a concrete reference, I'd suggest you look up the book on Riemann Surfaces by Jurgen Jost. See the proposition of gluing together maps on simply connected manifolds in the first chapter. That will solve your problem. – Baidehi May 24 '20 at 10:12
• Oh, you redefined $G_{ij}$ with straight lines. I thought I might need some advanced math to prove that $\partial G_{ij}$ is piecewise-smooth. That's why I said I don't know manifold and homology. Anyway, thank you so much! – zxcv May 24 '20 at 11:39