# Holomorphic functions as sums

Are there any holomorphic functions on a connected domain in $\mathbb C$ that can not be written as a sum of two univalent (holomorphic and injective) functions? What about as a sum of finitely many univalent functions? Or even infinitely many?

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If the domain is all of $\mathbb C$ then the answer is trivially yes, since all univalent functions are linear. – user31373 Sep 29 '12 at 16:15
Yes, that seems pretty clear, but I am really interested in more general domains. Given how there can be no entire functions into any bound domain, the picture might well be very different. – user29124 Sep 29 '12 at 16:27
Any such function must have unbounded derivative; if $|f'(z)| < B$ on the domain, then $g(z)=f(z)+Bz$ is univalent. It also follows that any function on a bounded domain with at worst simple poles on its boundary is the sum of finitely many univalent functions, and if your domain is convex you can immediately extend that to "at worst double poles." – Micah Sep 29 '12 at 16:37
Thanks, but you did not have to accept my incomplete answer... it's still interesting to know if every holomorphic function on $\mathbb D$ can be written as an infinite sum of univalent functions (uniformly convergent on open subsets). – user31373 Sep 30 '12 at 1:40
Well it sufficed for the application I had in mind, so while you didn't answer the exact question I posed, your answer was enough, that is why I accepted the answer. That said, I do agree that it would be interesting to know the full answer. – user29124 Sep 30 '12 at 19:09
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## 1 Answer

There is a growth obstruction for finite sum representation. Indeed, a theorem of Prawitz (1927) says that every univalent function on the unit disk belongs to the Hardy space $H^p$ for all $p<1/2$. Consequently, $f(z)=(1-z)^{-q}$ is not a finite sum of univalent functions when $q>2$.

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Wow! What a nice result. – Ewan Delanoy Oct 1 '12 at 19:36