Let $\mathcal{A}$ be the family of open disks and open squares in the complex plane. Let $f:D_1 \cup D_2 \to \mathbb{C}$ be a holomorphic function where $D_1, D_2 \in \mathcal{A}$. Prove that $f$ has a primitive in $D_1 \cup D_2$. The problem here is that all the theorems we were taught that guarantee the existence of primitive require at least an open and connected set. But $D_1 \cup D_2$ is not connected unless $D_1 \cap D_2 \neq \emptyset$.

Now the proof I came up with (but I'm not really sure about) is the following:

  • If $D_1 \cap D_2 \neq \emptyset$ then $D_1 \cup D_2$ is open, connected and star-shaped. Since $f$ is holomorphic we know that it has a primitive (this is a known theorem)

  • If $D_1 \cap D_2 = \emptyset$ then $f|_{D_1}$ is holomorphic and $D_1$ is open connected and star-shaped so $f$ has a primitive on $D_1$. The same holds on $D_2$ and we can construct a primitive on $D_1 \cup D_2$ by "combining" the primitives on $D_1$ and $D_2$.

Is this correct? If not, how can I prove it? Thank you.

  • $\begingroup$ Is the union of two disks star-shaped? Your second argument is fine but could do with some justification about why "combination" works (since you don't seem sure). You can certainly define a primitive candidate $F$ piecewise by using the primitives on each disk. You can then always compute $F'(z)$ by restricting to a sufficiently small open nbhd of $z$ (lying entirely inside one of the disks). $\endgroup$ Jun 15, 2021 at 21:34
  • $\begingroup$ @preferred_anon You're right. The union of open discs is not always star-shaped! Is there a fix for that? $\endgroup$ Jun 15, 2021 at 21:40
  • $\begingroup$ Of course there is - but you should think about it first :-) $\endgroup$ Jun 15, 2021 at 21:41
  • $\begingroup$ @preferred_anon I could use a similar combining argument for $D_1 \setminus (D_1\cap D_2)$, $D_1 \cap D_2$ and $D_2 \setminus (D_1 \cap D_2)$ since all of these sets are star-shaped. Apart form $D_1 \cap D_2$ though I don't know for sure that the others are open. $\endgroup$ Jun 15, 2021 at 21:48
  • $\begingroup$ Let $D_1$, $D_2$ be the circles of radius a little more than 1 centred at $1$ and $-1$. Choose $f(z) = 0$. On $D_1$ my primitive is $F_1(z) = 1$. On $D_2$ my primitive is $F_2(z) = 2$. What value does my primitive take at $0$? Alternatively, on $D_1 \cap D_2$ I choose the primitive $F_3(z) = 3$. Is this holomorphic? $\endgroup$ Jun 15, 2021 at 21:51

1 Answer 1


Here is a writeup of the discussion in the comments.

  1. $D_1 \cap D_2 = \emptyset$. Let $F_1: D_1 \to \mathbb{C}$, $F_2: D_2 \to \mathbb{C}$ be the two given primitives of $f$. Claim: A primitive of $f$ on $D_1 \cup D_2$ is

$$F(z) = \cases{F_1(z) & $z \in D_1$ \\F_2(z) & $z \in D_2$ }.$$

Proof: $F$ is well-defined since $D_1 \cap D_2 = \emptyset$. Let $z \in D_1 \cup D_2$. $z \in D_1$ without loss of generality. We compute $F'(z)$: Since $D_1$ is open, there is some $\varepsilon > 0$ such that $B_{z}(\varepsilon) \subset D_1$. Then $\forall w \in B_z(\varepsilon)$, $F(w) = F_1(w)$. Therefore $F'(z) = F_1'(z) = f(z)$. Similarly for $D_2$.

  1. $D_1 \cap D_2$. In this case $D_1 \cup D_2$ is simply connected so it should come as no surprise that $f$ has an antiderivative. Let $F_1, F_2$ be as above. On the connected open set $D_1 \cap D_2$, $F_1'(z) = F_2'(z)$. Therefore $F_1(z) - F_2(z) = C$ throughout $D_1 \cap D_2$. Claim: A primitive of $f$ on $D_1 \cup D_2$ is

$$F(z) = \cases{F_1(z) - C & $z \in D_1$ \\F_2(z) & $z \in D_2$ }.$$ Proof: $F$ is well defined by the previous argument. Just like before, at any point of $D_1 \cup D_2$, we can restrict $F$ to a small enough ball that $F$ is equal to one of the branches. Then $F'(z) = f(z)$ as required.


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