Let $X$ be a complex manifold and $\omega$ a Kahler form on $X$. A smooth function $\rho$ is called a potential of $\omega$ if $i\partial\bar\partial\rho=\omega$. By intuition, it seems that $\rho$ can never be bounded if $\omega$ is a complete Kahler metric. 


**Is the following claim true: the potential of a complete Kahler metric can never be bounded?**


We can consider an example that $X=\Delta=\{z\in \mathbb C; |z|<1\}$. Assume $\rho$ is a bounded smooth funtion on $\Delta$ such that $i\partial\bar\partial\rho$ gives a complete metric on $\Delta$. Replace $\rho$ by the averaging of it w.r.t rotations, it still induces a complete metric. So we can assume $\rho$ depends only on $r=|z|$. Then up to a constant $i\partial\bar\partial\rho=(\rho''(r)+\rho'(r)/r)dx\wedge dy$. The completeness of $i\partial\bar\partial\rho$ implies $\int^1\_0\sqrt{\rho''(r)+\rho'(r)/r}dr=\infty$. 


So the above question can more or less be reduced to the following one: deos there exit a bounded smooth function $\rho$ on $[0,1)$ such that $\rho''(r)+\rho'(r)/r>0$ and $\int^1\_0\sqrt{\rho''(r)+\rho'(r)/r}dr=\infty$?
I believe the answer is no, but I can not give a proof. 


Thanks a lot!



Can the potential of a complete Kahler metric be bounded?
**The claim is false.** The potential $\psi$ of the complete Kahler metric on the disk given in Theorem 1.1 of <http://arxiv.org/pdf/math/0603530v7.pdf> is bounded.
Indeed, such potential $\psi$ is the pullback of the potential of $P := |z\_1|^2 + |z\_2|^2 + |z\_3|^2$ by the holomorphic map $X$. Since the image of $X$ is bounded it follows that $ \psi = X^\* P$ is bounded.


