Proof.
The extension is unique thanks to proposition 5.14.6. We shall prove the existence.
2 Thanks to the uniqueness of the extension, the problem is local, so we may assume that $X$ is a domain in $\mathbb {C}^ n$ with $n>0$ and there is a non-zero holomorphic function $f$ on $X$ vanishing on $Z$. For each $\epsilon >0$, we claim that the function $\varphi _{\epsilon }$ defined by
\[ \varphi _{\epsilon }(x)\coloneqq \left\{ \begin{aligned} \varphi (x)+\epsilon \log |f(x)|^2,& \quad x\in X\setminus Z;\\ -\infty ,& \quad x\in Z \end{aligned} \right. \]
is plurisubharmonic on $X$. By definition 5.8.1, it suffices to verify the case $n=1$. In this case, we may assume that $Z=\{ 0\} $. It is clear that $ \varphi _{\epsilon }\in \mathrm{SH}(X\setminus Z)$. It suffices to verify the sub-mean value inequality at $0$, which is immediate.
Next observe that the net $\varphi _{\epsilon }$ is increasing as $\epsilon \searrow 0$ and $\varphi _{\epsilon }$ is locally uniformly bounded from above. It follows from proposition 5.14.2 that $\tilde{\varphi }\coloneqq \operatorname *{sup}_{\epsilon >0}{}\! {}^{*}\varphi _{\epsilon }\in \mathrm{PSH}(X)$. Moreover, $\tilde{\varphi }$ clearly extends $\varphi $. Note that (5.14.11) follows from the construction.
1 We invite the reader to consult 5.14.12 for the notations used in the proof.
Thanks to 2, it suffices to verify that $\varphi $ is locally bounded from above near each point of $Z$. The problem is local, so we may assume that $X$ is a domain in $\mathbb {C}^ n$ with $n\geq 2$.
Suppose the assertion fails. Then there exist $z\in Z$ and a sequence $(x_ j)_{j\geq 1}$ in $X\setminus Z$ converging to $z$ such that
CD940E826F95
5.14.13
\begin{equation} \lim _{j\to \infty } \varphi (x_ j)=\infty . \end{equation}
Since $Z$ has codimension at least $2$, we may choose a complex line $L$ through $z$ which meets $Z$ in a discrete set; after shrinking $X$ we may further assume that
\[ L\cap Z=\{ z\} . \]
Pick an open ball $B_ n(z,r)\Subset X$. After adding a constant to $\varphi $, we may arrange $\varphi <0$ on $L\cap \partial B_ n(z,r)$. Since $\varphi $ is upper semicontinuous, there is an open neighborhood $U\subseteq X\setminus Z$ of $L\cap \partial B_ n(z,r)$ such that
\[ \varphi |_ U<0. \]
For each $j\geq 1$ we now choose a complex line $L_ j$ through $x_ j$, disjoint from $Z\cap B_ n(z,r)$, with $L_ j\to L$ as $j\to \infty $ (the convergence being in the Grassmannian of affine complex lines in $\mathbb {C}^ n$). The existence of such $L_ j$ is precisely where the assumption $\operatorname {codim} Z\geq 2$ enters: after replacing $B_ n(z,r)$ by a slightly smaller ball if necessary, the closure of $Z\cap B_ n(z,r)$ is analytic in a neighborhood of the closed ball. The projection to the space of directions is proper on this compact analytic set, so by Remmert's theorem its image is contained in a proper analytic subset of $\mathbb {P}^{n-1}$. Hence its complement contains directions arbitrarily close to that of $L$.
For $j$ large enough we now have
\[ L_ j\cap \partial B_ n(z,r)\subseteq U, \]
so applying the maximum principle to $\varphi |_{L_ j\cap X}$ at $x_ j$ yields $\varphi (x_ j)<0$, contradicting (5.14.13).
$\square$
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