Lemma 5.27.11C56C5CBA0E0D

Proof. Put $f=u-v$ and write the Jordan decomposition on $B$ as $\mathrm{dd}^{\mathrm{c}}f=\alpha ^+-\alpha ^-$. Let $g^\pm \leq 0$ be the Green potentials of $\alpha ^\pm $ on $B$ with zero boundary values. The $L^1(B)$ norm of the Green kernel of the disk is bounded uniformly in its pole, so

Indeed, on the unit disk the kernel is $\log |(z-\zeta )/(1-\overline\zeta z)|^2$; its logarithmic singularity is uniformly integrable, and $|1-\overline\zeta z|\leq 2$. Scaling gives the bound on $B$.

The distribution $h=f-g^++g^-$ is harmonic, and the interior mean-value estimate gives

The identity $u+g^-=v+g^++h$ holds almost everywhere, hence everywhere by lemma 5.24.3. Thus $|u-v|\leq -g^+-g^-+|h|$ off the polar sets of $u$ and $v$. Let $0\leq a\leq 1$ be a subharmonic test function on $B$. Since the measure $\mathrm{dd}^{\mathrm{c}}a$ charges no polar set, this pointwise bound, the weighted Chern–Levine–Nirenberg estimate and (5.27.3) yield

Restricting the integral to $B'\cap \{ |u-v|>t\} $, dividing by $t$ and taking the supremum over $a$ proves (5.27.12). $\square$