Proposition 5.9.5.7EBEE004A7FA Let $X$ be a complex manifold. Nonnegative multiples, finite sums and maxima of nonempty finite families of plurisubharmonic functions on $X$ are plurisubharmonic; here the zero multiple of a function is the zero function. Moreover, let $u\in \mathrm{PSH}(X)$, let $D\subset X$ be open, and let $v\in \mathrm{PSH}(D)$. If
\[ \varlimsup _{D\ni y\to x}v(y)\le u(x)\qquad \text{for every }x\in \partial D, \]
where $\partial D$ is the boundary of $D$ in $X$, then the function equal to $\max \{ u,v\} $ on $D$ and to $u$ on $X\setminus D$ is plurisubharmonic on $X$. This boundary inequality holds in particular if
\[ \varliminf _{\substack {D\ni y\to x\\ v(y)>-\infty }}(u(y)-v(y))\ge 0 \qquad \text{for every }x\in \partial D. \]
Proof.
In local coordinates, upper semicontinuity and the submean inequality on complex lines are preserved by nonnegative multiples, sums and maxima; for sums we use that circle means are additive on functions with values in $[-\infty ,\infty )$. It remains to check that the resulting functions are not identically $-\infty $ on any component. For $cu$ with $c>0$ this holds because $cu=-\infty $ exactly where $u=-\infty $; a sum of plurisubharmonic functions is locally integrable by theorem 5.9.1, and a maximum dominates each of its members. This proves the first assertion.
Let $w$ denote the glued function. On $D$ it equals $\max \{ u,v\} $, which is plurisubharmonic by the first part, and on the open set $X\setminus \overline D$ it equals $u$. At $x\in \partial D$ we have $w(x)=u(x)$, and the upper semicontinuity of $u$ together with the boundary inequality gives
\[ \varlimsup _{y\to x}w(y)\leq \max \Bigl \{ \varlimsup _{y\to x}u(y),\ \varlimsup _{D\ni y\to x}v(y)\Bigr \} \leq u(x)=w(x). \]
Hence $w$ is upper semicontinuous. To verify the line condition, work in a chart and restrict $w$ to a connected component of an affine complex-line section. At points of $D$ and of $X\setminus \overline D$, the restriction satisfies the submean inequality on small circles, because $w$ is plurisubharmonic near such points. At a point $x\in \partial D$ we have $w\geq u$ everywhere and $w(x)=u(x)$, so each small circle mean of $w$ is at least the corresponding circle mean of $u$, which is at least $u(x)=w(x)$; if the restriction of $u$ is identically $-\infty $, then $w(x)=u(x)=-\infty $ and there is nothing to prove. Thus the restriction of $w$ is upper semicontinuous and satisfies the submean inequality on small circles at every point; that is, it is subharmonic. Finally $w\geq u$, so $w$ is not identically $-\infty $ in any chart, and $w$ is plurisubharmonic.
For the sufficient condition, fix $x\in \partial D$ and $\varepsilon >0$. By hypothesis, $x$ has a neighborhood $W$ such that $v(y)<u(y)+\varepsilon $ for all $y\in W\cap D$ with $v(y)>-\infty $. Hence $v\leq u+\varepsilon $ on $W\cap D$, the inequality being trivial where $v=-\infty $, and the upper semicontinuity of $u$ gives
\[ \varlimsup _{D\ni y\to x}v(y)\leq \varlimsup _{y\to x}u(y)+\varepsilon \leq u(x)+\varepsilon . \]
Letting $\varepsilon \downarrow 0$ gives the boundary inequality.
$\square$
Pome
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