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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!

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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.

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