More precisely, given a holomorphic function $f$ on a simply connected domain $\Omega \subset \mathbb{C}$ where for any $z \in \Omega$, $z+1$ is also in $\Omega$, does there exist a holomorphic function $g(z)$ such that $g(z+1) = f(z) + g(z)$? Obviously, it won't be unique, but I'm wondering if existence is guaranteed. It seems like it should be but I don't know how to prove it.

I am aware of the Euler-Maclaurin formula which gives an asymptotic expression for $g(z)$ but it doesn't seem obvious to me how this would give a holomorphic function here.

  • 2
    $\begingroup$ Find some information here en.wikipedia.org/wiki/Indefinite_sum $\endgroup$
    – GEdgar
    Jun 17, 2018 at 16:35
  • $\begingroup$ Is $\Omega$ simply connected? $\endgroup$
    – Szeto
    Jun 27, 2018 at 4:32
  • $\begingroup$ Yes, I should clarify that $\endgroup$ Jun 27, 2018 at 4:43
  • $\begingroup$ @GEdgar I don't see any connection between the article you refer to and this question. $\endgroup$ Jun 27, 2018 at 5:54

1 Answer 1


Here is a simple proof when $\Omega = \mathbb{C}$. Here is a sophisticated proof for any $\Omega$. More precisely, the latter link proves the result whenever $\Omega/\mathbb{Z}$ is Stein but, since the map $z \mapsto \exp(2 \pi i z)$ identifies $\Omega/\mathbb{Z}$ with an open subset of $\mathbb{C}^{\ast}$, it will always be Stein.

I'm thinking about if I can remove the sophistication.

  • $\begingroup$ Sophistication is fine but that is using some material that I haven't learned yet. I'll leave the question open for a bit longer to see if anyone can find a more elementary proof. $\endgroup$ Jul 17, 2018 at 2:27

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