Lemma 5.27.14B54F79E95767

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 $f\leq 0$,

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 $\Delta f(K)\geq 0$. $\square$