If every holomorphic function $f$ defined on an open subset $\Omega$ of $\mathbb C$ has a primitive. Is it true that $\Omega$ is simply connected?

I saw this post: If every harmonic function on $\Omega$ has a harmonic conjugate on $\Omega$, then $\Omega$ is simply connected. Then I am curious whether it is true for assuming the existence of primitive.

I think we can decompose an open set into the union of connected components. Then suppose there is a connected component that is not simply connected. By the definition of simply connected, there exists a point $z_0\notin\Omega$ encompassed by a Jordan curve $\gamma$ lying in a small circle which is completely contained in the connected component. Consider the holomorphic function $f=\frac 1{z-z_0}$, it should be clear that $f$ does not have a primitive in the small neighborhood. Contradiction!

Is my proof correct?

Edit: I should have claimed that every connected component of $\Omega$ is simply connected.


It is necessary to assume that your open set is connected. If you consider the union of two disjoint open disks then any analytic function on it has a primitive but the set is not even connected.

Assuming connectedness, you are making things too complicated. Rudin's RCA has a theorem giving various ways of characterizing simply connected regions and this is one of them. If you know that a region is simply connected iff the integral of any analytic function over any closed path in it is $0$ then this result is obvious because the integral of a derivative over a close path is $0$ by the very definition of the integral.

  • $\begingroup$ I didn't know that theorem, but thank you! $\endgroup$ – Bach Jun 6 at 10:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.