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.

  • $\begingroup$ 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?? $\endgroup$ – user586464 Dec 11 '18 at 17:00
  • $\begingroup$ @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. $\endgroup$ – Brevan Ellefsen Dec 12 '18 at 23:11

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