# Can the potential of a complete Kahler metric be bounded?

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!

• Potential of Kahler-Einstein metric is bounded, more generally, potential of canonical metric is bounded
– user21574
Jan 27, 2017 at 3:03

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.