# Complex Analysis on Holomorphic Anti-derivatives

Question: Let $$U_{1} \subseteq U_2 \subseteq U_{3} \subseteq ... \subseteq \mathbb{C}$$ be connected open sets and let $$U = \cup_{i = 1}^{\infty} U_i$$. Let $$f$$ be holomorphic on $$U$$. Suppose for each $$U_i$$, $$f |_{U_i}$$ has a holomorphic anti-derivative on $$U_i$$. Prove that $$f$$ has a holomorphic anti-derivative on all of $$U$$.

Answer: Since $$f_i=f |_{U_i}$$ has a holomorphic anti-derivative on $$U_i$$ and for some index j $$\in$$ $$\mathbb{N}$$, $$U_{i} \subseteq U_j$$, $$\frac{df_i}{dz} = f = \frac{df_j}{dz}$$, this implies that $$f_j - f_i = C_i$$ which $$C_i$$ is a constant. We also know that $$\cap_{i =1} ^{\infty} U_i = U_1$$ is nonempty. This is true for each pair of open sets $$U_i \subseteq U_j$$. I have an idea of how to define the function $$H(z)$$ on $$U$$. I don't know how to type in a piecewise function on latex. Am I on the right track?

You have the right idea. Since we have an inclusion chain, if $$z \in U_i$$ then for all $$j \ge i$$ we have $$z \in U_j.$$ Let $$F_i$$ be the antiderivative of $$f$$ on $$U_i.$$ As you note, if an antiderivative exists, it is unique up to a constant, so for all $$z \in U_i,$$ we have $$F_j(z) = F_i(z) + c_j$$ for all $$j \ge i.$$ We can now start gluing together an antiderivative.

On $$U_1$$ we have an antiderivative $$F_1$$. So far, so good. For notational convenience, let $$F'_1 = F_1.$$

On $$U_2$$ we define $$F'_2 = F_2-c_2.$$ Then $$F'_2$$ is an antiderivative satisfying $$F'_2 = F_1$$ wherever both are defined.

On $$U_3$$ we define $$F'_3 = F_3-c_3-c_2.$$ On $$U_2$$ we have $$F'_3 = F_2 - c_2 = F'_2$$ as desired.

In general, on $$U_i$$ we define $$F'_i = F_i - \sum_{2 \le k \le i} c_k,$$ so that on $$U_{i-1}$$ we have $$F'_i = F_{i-1}- \sum_{2 \le k \le i-1} = F'_{i-1}.$$

Finally, given $$z \in U,$$ we know $$z \in U_i$$ for some (minimal if desired) index $$i,$$ so we can let $$F(x) = F'_i(z).$$

By the work above, we know $$F(z)$$ will be holomorphic in a neighborhood of $$z$$ (we merely shifted by constants, so local holomorphicity was preserved) and satisfies $$F'(z) = f(z).$$ Since these conditions hold for all $$z \in U,$$ we conclude $$F$$ is holomorphic in $$U$$ since it is locally holomorphic at each point, and that $$F$$ is an antiderivative of $$f$$ on $$U.$$

Note: proving $$F$$ is holomorphic is not required, since if $$F$$ is an antiderivative of $$f$$ then $$F$$ is by definition complex differentiable and is thus holomorphic.

• That helps. I have another question. If we let $(a,b) \in \mathbb{R}^2$ be fixed and we are given a holomorphic function on an open rectangle, $U_{i} := \{ (x,y) \in \mathbb{R}^2| |x -a | < \delta, |y -b| < \epsilon \}$, how can we find a holomorphic anti-derivative such that $F(a,b) =0$? Could we use secant planes?? – user586464 Dec 11 '18 at 17:00
• @Overachiever find any antiderivative $F$, subtract $F(a,b)$ from it. Will still be an antiderivative. If you have a seperate question, ask it seperately. Be aware though, this is not a homework solving site. – Brevan Ellefsen Dec 12 '18 at 23:11