5.24 Maxima and the maximum principleAA6F3BC0AD9F

Proof. Adding a common local smooth correction to $u$ and $v$ shows that $\max (u,v)$ is quasi-plurisubharmonic. To obtain the sharp lower bound for its Hessian, fix a positive constant $(1,1)$-form $\kappa $ on a coordinate ball. For each $\varepsilon >0$, every point of the ball has a neighborhood on which $\theta $ is close to a constant form $\theta '$, in the sense that $-\varepsilon \kappa \leq \theta -\theta '\leq \varepsilon \kappa $. Write $\theta '+\varepsilon \kappa =\mathrm{dd}^{\mathrm{c}}r$ with a quadratic polynomial $r$. Then $\mathrm{dd}^{\mathrm{c}}(u+r)\geq \theta +\mathrm{dd}^{\mathrm{c}}u\geq 0$, so $u+r$ and, likewise, $v+r$ are plurisubharmonic; hence so is their maximum. It follows that

on each such neighborhood, and letting $\varepsilon \to 0$ gives $\theta +\mathrm{dd}^{\mathrm{c}}\max (u,v)\geq 0$. The last assertion follows because constants belong to $\mathrm{PSH}(X,\theta )$ when $\theta \geq 0$. $\square$

Proof. Denote the glued function by $w$; it is upper semicontinuous by the boundary inequality. Freeze $\beta $ on a small coordinate ball as in the proof of lemma 5.24.1, and choose a quadratic correction $r$ such that $u+r$ and $v+r$ are plurisubharmonic on their respective domains. Off $\partial D$, the function $w+r$ is plurisubharmonic, being locally either $u+r$ or the maximum of two plurisubharmonic functions. At a point of $\partial D$, it equals $u+r$ and dominates $u+r$ on the ball, so it satisfies the submean inequality on every complex line through that point. Thus $w+r$ is plurisubharmonic on the whole ball. Letting the error of the frozen coefficients tend to zero gives $\beta +\mathrm{dd}^{\mathrm{c}}w\geq 0$. $\square$

Proof. On a coordinate ball, add the same smooth function $q$ to both functions so that $a+q$ and $b+q$ are plurisubharmonic. By the submean inequality and upper semicontinuity, a plurisubharmonic function is, at each point, the limit of its averages over balls centered there as the radius tends to zero. The almost-everywhere inequality orders these averages, so passing to the limit gives $a+q\leq b+q$. Subtracting $q$ proves the claim. $\square$

Proof. Choose a number $c$ with $a(x)<c<b(x)$. By upper semicontinuity, $a<c$ on a neighborhood $V$ of $x$. If $\{ b>c\} \cap V$ had zero volume, then $b\leq c$ almost everywhere on $V$, and lemma 5.24.3 would give $b(x)\leq c$, a contradiction. Thus $\{ b>c\} \cap V$, which is contained in $\{ a<b\} $, has positive volume. $\square$

Proof. First let $f$ be a smooth subharmonic function on $\mathbb C$ with $f\leq M$. Write $\sigma =(\Delta f)\, \mathrm{d}A$ for its nonnegative Laplacian measure, where $\Delta $ and $\mathrm{d}A$ are Euclidean. Green's formula for circular means gives, for $R>0$,

For fixed $r_0>0$ and $R>r_0$, the right-hand side is at least $\sigma (B(0,r_0))\log (R/r_0)/(2\pi )$, while the left-hand side is at most $M-f(0)$. Letting $R\to \infty $ gives $\sigma (B(0,r_0))=0$. Thus $f$ is harmonic, and the Liouville theorem for the positive harmonic function $M+1-f$ shows that $f$ is constant.

Now let $p$ be plurisubharmonic on $\mathbb C^ m$ and bounded above. Its convolutions with a smooth nonnegative radial probability kernel are smooth plurisubharmonic functions with the same upper bound. By the planar case, their restrictions to every affine complex line are constant, so each convolution is constant on $\mathbb C^ m$. Since the convolutions converge to $p$ in $L^1_{\mathrm{loc}}$, the function $p$ is almost everywhere equal to a constant, and hence constant by lemma 5.24.3. $\square$

Proof. By the growth bound, the real translates $h(\cdot +s)$, $s\in [0,1]$, form a family of subharmonic functions that is locally uniformly bounded above. The local upper-envelope theorem therefore shows that

is subharmonic. By periodicity, $H$ is invariant under every real translation, so it depends only on $\operatorname {Im}\zeta $, and its subharmonicity means that it is a convex function of $\operatorname {Im}\zeta $. On the imaginary axis it satisfies $H(\mathrm i y)\leq A\log (2+|y|)+B$. Since a nonconstant convex function on $\mathbb R$ grows at least linearly at one end, this bound shows that $H$ is constant. Thus $h\leq H$ is bounded above, and lemma 5.24.5 shows that $h$ is constant. $\square$

Proof. First suppose that $u$ is plurisubharmonic. Near a point of $Z$, choose a plurisubharmonic function $\psi \leq 0$ with $\psi =-\infty $ on $Z$. For $t>0$, extend $u+t\psi $ by $-\infty $ on $Z$. The local upper bound for $u$ makes this extension upper semicontinuous. It is plurisubharmonic off $Z$, and its submean inequality on complex lines is automatic at points of $Z$, where its value is $-\infty $. It is therefore plurisubharmonic on the whole neighborhood. As $t\downarrow 0$, these functions increase, with a common local upper bound. By the local upper-envelope theorem, the regularization of their limit is a plurisubharmonic function that agrees with $u$ almost everywhere off $Z$, hence everywhere off $Z$ by lemma 5.24.3. By upper semicontinuity and the submean inequality, its value at a point of $Z$ is the upper limit of $u$ off $Z$, so it is the upper-limit extension of $u$.

For a smooth background, fix a positive constant form $\kappa $ on a coordinate chart and let $\epsilon >0$. On a sufficiently small ball, freeze $\theta $ to a constant form $\theta '$ with $-\epsilon \kappa \leq \theta -\theta '\leq \epsilon \kappa $, and choose a quadratic potential $q$ for $\theta '+\epsilon \kappa $. Applying the preceding argument to the plurisubharmonic function $u+q$ shows that the upper-limit extension satisfies $\theta +\mathrm{dd}^{\mathrm{c}}u\geq -2\epsilon \kappa $ on the ball. Since the extension is defined by upper limits, it does not depend on the ball, and letting $\epsilon \downarrow 0$ gives $\theta +\mathrm{dd}^{\mathrm{c}}u\geq 0$. Finally, analytic subsets are locally pluripolar: locally, a proper analytic subset is contained in the $-\infty $ locus of the logarithm of the sum of the squared moduli of local defining functions. $\square$

Proof. Choose coordinates centered at $p$ and write $\Delta _ R=\{ z\in \mathbb C:|z|<R\} $. On the coordinate polydisk $\Delta _ R^ n$, add a smooth quadratic function to the given function to obtain a plurisubharmonic function $h$ on $\Delta _ R^ n\setminus \{ 0\} $. Write the coordinates as $(z_1,z_2,z')$, with $z'$ absent when $n=2$, and choose $0<r<R$. The compact set

avoids the puncture, so $h\leq M$ on $K$ for some finite $M$. Consider a point of the punctured closed polydisk of radius $r$. If $z_1\ne 0$, the disk in the $z_2$-direction through the point avoids the puncture and has its boundary in $K$, so the maximum principle gives $h\leq M$ at the point. If $z_1=0$ and $(z_2,z')\ne 0$, the disk in the $z_1$-direction through the point avoids the puncture, and its boundary consists of points with $z_1\neq 0$, where the bound has already been proved; the maximum principle again gives $h\leq M$. Subtracting the smooth correction proves the local upper bound.

By lemma 5.24.7, applied with $Z=\{ p\} $, the upper-limit extension belongs to $\mathrm{PSH}(X,\theta )$. It is unique by lemma 5.24.3. If $X$ is compact, the extension is an upper semicontinuous function on a compact space and is therefore bounded above. $\square$

Proof. In coordinates centered at $x$, let $q$ be a quadratic polynomial with $\mathrm{dd}^{\mathrm{c}}q=\theta (x)$, and let $\varepsilon >0$. On a sufficiently small ball about $x$, we have $\theta \leq \theta (x)+\varepsilon \mathrm{dd}^{\mathrm{c}}|z|^2$, so $u+q+\varepsilon |z|^2$ is plurisubharmonic, and $a+q+\varepsilon |z|^2$ touches it from above at $x$. By the submean inequality on complex lines through $x$, a $C^2$ function touching a plurisubharmonic function from above at $x$ has nonnegative complex Hessian at $x$. Hence $\theta (x)+\mathrm{dd}^{\mathrm{c}}a(x)+\varepsilon \mathrm{dd}^{\mathrm{c}}|z|^2\geq 0$, and letting $\varepsilon \to 0$ proves the lemma. $\square$