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i am working for some time on a problem and at some point i cant go further. here the critical part: Let $U \subset \mathbb{C}^{n}$ be a open set and consider $c : U \rightarrow \mathbb{R}$ a smooth positive function. Does there exist any holomorphic function $f : U \rightarrow \mathbb{C}$ such that $f(z) \overline{f(z)} = c(z)$ ??? it would be very nice if somebody could help me.


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Sorry, but I think this question is off-topic for this site. The FAQ describes what this website is for, and the section gives some links to sites where this question would be a better fit. (No, not all smooth functions are real analytic.) – Jonas Meyer Aug 13 '11 at 9:19
For $U$ simply connected, the condition that $c^{-1}\partial c$ be a closed (hence exact) holomorphic 1-form seems to be necessary and sufficient: if $l$ is a primitive for $c^{-1}\partial c$, we can set $f(z)=\sqrt{c(z_0)}\exp(l(z)-l(z_0))$ for some fixed $z_0\in U$. For $n=1$, the condition amounts to $c\cdot\Delta c=c_x^2+c_y^2$. – user2035 Aug 13 '11 at 12:51
Miriam? Andrei? – Yemon Choi Aug 13 '11 at 21:07

Obviously $c$ would have to be real analytic. The Laplacian of $f(z)\overline{f(z)}$ must be positive, so $c$ would have to be a positive function with positive Laplacian. Clearly every positive power of the Laplacian on $c$ would also have to be a positive function. I don't know if that is sufficient.

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