Proposition 16.28.51DDC2E6D4F75F

Proof. By the preceding lemma, every neighborhood of a point of $K$ contains a nonempty open subset of $K$, namely a product of open disks. This proves the first assertion, which applies in particular to each $K\cap \overline B(a,r)$, $r>0$.

Put $b(\zeta )=\log |\zeta -2|-\log 3$. Since $|\zeta -2|\leq 3$ on $E$, we have $b\leq 0$ there, and $b(2)=-\infty $. For $\varepsilon >0$, the plurisubharmonic function

belongs to $\mathcal L_1$ and satisfies $u_\varepsilon \leq 0$ at every point of $K$, including the points with $z_1=2$, where it equals $-\infty $. Thus $u_\varepsilon \leq F_{1,K}$. Letting $\varepsilon \downarrow 0$ gives $v(z_1)\leq F_{1,K}(z)$ off the hyperplane $\{ z_1=2\} $, and lemma 5.24.3 extends this inequality everywhere. The value $v(2)=4\log 2/9$ gives the lower bound for $\delta =1$. Since $0$ is a competitor, $V_{\delta ,K}|_ K=0$, and scaling of competitors gives $F_{\delta ,K}=\delta F_{1,K}$. Replacing $K$ by a smaller compact set only increases the envelope, which proves the local assertion.

Finally, let $f$ be plurisubharmonic near $a_0$. If $f(a_0)=-\infty $, (16.28.53) is automatic. Otherwise, restrict $f$ to the complex line through $a_0$ in which only the second coordinate varies. Small circles in this line centered at $a_0$ lie in $K\setminus \{ a_0\} $, and the submean inequality on these circles gives (16.28.53). $\square$