I was wondering, suppose I have a non-compact Kähler manifold $M$ and suppose that outside some compact subset $A\subset M$, there exists a smooth function $f:M\backslash A\longrightarrow\mathbb{R}$ such that $i\partial\bar{\partial}f>0$. Is it always possible for me to find a smooth function $h:M\longrightarrow\mathbb{R}$ such that $h|\_{M\backslash A}=f$ and $i\partial\bar{\partial}h>0$ on the whole of $M$? If not in general, are there sufficient conditions on $M$ that will allow me to do this? Many thanks!



Extension of strictly plurisubharmonic functions on a Kähler manifold
It is instructive to conisder the case of Kahler metrics invariant under torus action. In this case your question becomes a certain (nontivial) question on convex functions.


Recall first, that Kahler metrics on $(\mathbb C^\*)^n$ invariant under the action of $(S^1)^n$ have global potential that is given by a convex function $F$ on $\mathbb R^n$. Here $\mathbb R^n$ is identified with the quotient 


$(\mathbb C^\*)^n/(S^1)^n$ and we take coordinates $log|z\_i|$ on $\mathbb R^n$.
So we can translate your original question as follows


QESTION. Suppose you have a smooth convex function $F$, defined on $\mathbb R^n$ outside compact $\Omega$. Is it possible to extend $F$ to a smooth convex function on the whole $\mathbb R^n$?


It easy to construct an example of a non-convex $\Omega$ on $\mathbb R^2$, with convex $F$ defined on $\mathbb R^2\setminus \Omega$, so that $F$ can not be extended. For the moment I don't see how to make such an example when $\Omega$ is the unite disk, but it sounds plausible that such examples exist. 


