PomeLemma 5.27.14.B54F79E95767 Let $B\subset \mathbb C$ be a disk and let $f\in L^1_{\mathrm{loc}}(B)$ have a distributional Laplacian that is a signed Radon measure. If $f\leq 0$ almost everywhere, then $(\Delta f)^{\mathrm{pol}}\geq 0$.
5.27.13 Polar parts on curves1953F5546AC6
Every locally finite signed measure $\nu $ on a Riemann surface splits uniquely as the sum of its polar part $\nu ^{\mathrm{pol}}$, which is carried by a polar Borel set, and a measure that charges no polar set. To construct the splitting locally, choose countably many polar Borel sets whose $|\nu |$-masses approach the supremum over all polar Borel sets, and restrict $\nu $ to their union, which is again polar. Uniqueness allows these local decompositions to be patched together and shows that $\nu \mapsto \nu ^{\mathrm{pol}}$ is linear.
For a locally quasi-subharmonic function $u$, the polar curvature of $u$ is the positive measure $\mu _ u=(\mathrm{dd}^{\mathrm{c}}u)^{\mathrm{pol}}=\mathbf1_{\{ u=-\infty \} }\mathrm{dd}^{\mathrm{c}}u$. The second equality holds because smooth corrections do not change polar parts, while bounded truncations and plurifine locality show that $\mathbf1_{\{ u>-j\} }\mathrm{dd}^{\mathrm{c}}u$ charges no polar set. The following inverse maximum principle shows that an upper bound for the difference of two potentials orders their polar curvatures.
Proof. It suffices to prove $\Delta f(K)\geq 0$ for every compact polar set $K\subset B$; restriction to smaller disks and inner regularity then give the assertion. Choose open neighborhoods $O_ j$ of $K$ with smooth boundaries such that $\overline O_{j+1}\subset O_ j\Subset B$ and $\bigcap _ j\overline O_ j=K$. Let $h_ j$ be the function that equals one on $\overline O_ j$, is harmonic on $B\setminus \overline O_ j$, and vanishes on $\partial B$. The Dirichlet problem on this smooth domain shows that $h_ j$ is continuous with $0\leq h_ j\leq 1$. Its outward normal derivative on $\partial O_ j$ is nonpositive, so the jump of the derivative across $\partial O_ j$ gives $\Delta h_ j\leq 0$ on $B$ in the sense of distributions. The maximum principle gives $h_{j+1}\leq h_ j$.
Since $0\leq h_ j\leq 1$, the sequence converges in $L^1_{\mathrm{loc}}(B)$. Let $h$ be the superharmonic representative of this limit; it may differ from the pointwise limit on $K$. It is harmonic off $K$, so its Laplacian is supported on the polar set $K$. On the other hand, $-h$ is bounded and subharmonic, so its Laplacian charges no polar set, by theorem 5.27.1. Hence $h$ is harmonic on $B$. Since $0\leq h\leq h_1$ and $h_1$ tends to zero at $\partial B$, the maximum principle gives $h=0$. It follows that $h_ j\to 0$ smoothly on compact subsets of $B\setminus K$, while $h_ j=1$ on $K$.
Choose $0\leq \chi \in C_ c^\infty (B)$ equal to one near $K$. For large $j$, regularize $h_ j$ by convolution on a neighborhood of $\operatorname {supp}\chi $, obtaining smooth functions $h_{j,\epsilon }$ with $\Delta h_{j,\epsilon }\leq 0$. Since $f\leq 0$,
\[ \begin{split} \int _ B\chi h_{j,\epsilon }\, \mathrm{d}(\Delta f) & =\int _ B f\, \Delta (\chi h_{j,\epsilon })\, \mathrm{d}x\, \mathrm{d}y\\ & \geq \int _ B f\bigl (2\nabla \chi \cdot \nabla h_{j,\epsilon } +h_{j,\epsilon }\Delta \chi \bigr )\, \mathrm{d}x\, \mathrm{d}y. \end{split} \]
For fixed $j$, the convolutions converge uniformly to $h_ j$ on $\operatorname {supp}\chi $, because $h_ j$ is continuous. On the support of the derivatives of $\chi $, the function $h_ j$ is harmonic for all large $j$. We may therefore first let $\epsilon \downarrow 0$ and then let $j\to \infty $. The smooth convergence away from $K$ makes the right-hand side tend to zero, and dominated convergence for the signed measure $\Delta f$ makes the left-hand side tend to $\Delta f(K)$. This proves $\Delta f(K)\geq 0$. $\square$
Corollary 5.27.15.F9517F96057F Let $u,v$ be locally quasi-subharmonic functions on a Riemann surface. If $u\leq v+C$ on an open set for some constant $C$, then $\mu _ u\geq \mu _ v$ there. Moreover,
where the right-hand side is the greatest measure dominated by both $\mu _ u$ and $\mu _ v$.
Proof. Apply lemma 5.27.14 to $u-v-C$ in local coordinates. Since smooth functions have zero polar part, linearity of the polar part gives the first assertion. Applying it to $u-\max (u,v)$ and to $v-\max (u,v)$ gives the inequality $\leq $ in (5.27.16).
For the reverse inequality, add a common smooth correction to make $u$ and $v$ subharmonic on a coordinate disk. On a smaller disk, let $g$ be the Green potential of the restriction of $\lambda =\mu _ u\wedge \mu _ v$ to that disk, so that $\mathrm{dd}^{\mathrm{c}}g=\lambda $. The distributions $u-g$ and $v-g$ have positive Hessians and hence subharmonic representatives $a$ and $b$. Since $\max (u,v)=g+\max (a,b)$ almost everywhere, the Hessian of $\max (u,v)$ dominates $\lambda $. Restricting this inequality to a polar Borel set carrying $\lambda $ proves the reverse inequality on the smaller disk, and these local statements give it on the whole surface. $\square$
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