# The extension of a plurisubharmonic Functions

I am reading the paper "Two Theorems on Extensions of Holomorphic Mappings" by Phillip A. Griffiths. Proposition 2.9 of the paper is: If $$\Psi$$ is a plurisubharmonic on the punctured ball $$B_n^{*}$$ and $$n\ge 2$$, then $$\Psi$$ extends to a plurisubharmonic function on the whole ball $$B_n$$. Then, the paper gives a proof in the special case for $$n=2$$, where I am stuck.

By defining $$\Psi(0)=\lim\sup \Psi(z),$$ he attempts to show that $$\Psi(0)<+\infty$$. When $$z_1\not= 0$$, we have the estimate $$\Psi(z_1,z_2)\le \frac{1}{2\pi}\int_{0}^{2\pi}\Psi(z_1,z_2+\epsilon e^{i\theta})d\theta$$ for $$\epsilon>0$$ small enough so that the integrand does not pass through $$z=0$$.

My question is: in the paper, it follows immediately that $$\Psi(0)<\infty$$. I do not know why it follows, since it seems that we do not have a uniform bound through this estimate.

Any hint is welcome. Thanks in advance!

Fix $$\epsilon>0$$. We have $$\Phi(z_1,z_2)\leq M<\infty$$ when $$|z_1|=|z_2|=\epsilon$$. Then by Maximum Principle, applied to $$1$$-dimensional disks $$D_{z_1}=\{(z_1,z_2):|z_2|\leq\epsilon\},\quad 0<|z_1|<\epsilon,$$ we obtain $$\Phi(z_1,z_2)\leq M$$ for all $$0<|z_1|\leq\epsilon$$, $$|z_2|\leq\epsilon$$. Similarly, the same holds or all $$|z_1|\leq\epsilon, \; 0<|z_2|\leq\epsilon$$. So we have a uniform estimate everywhere in the polydisk, except its center, and the result holds by the one-dimensional removable singularity theorem for subharmonic functions.