Proposition 5.10.12219E06AF4230

Proof. Put $A_ n=\max \{ 1,2/|\frac18B|\} $. The mean of $|v|$ over $\frac18B$ is at most $1/|\frac18B|<A_ n$. Hence the set of points of $\frac18B$ with $|v|\leq A_ n$ has positive measure, since otherwise $|v|>A_ n$ almost everywhere on $\frac18B$ and the mean would be at least $A_ n$; so there is a point $a\in \frac18B$ with $|v(a)|\leq A_ n$. For each unit vector $e\in \mathbb C^ n$, the disc $\{ a+\tfrac 34\zeta e:|\zeta |<1\} $ lies in $B$, so the function

is subharmonic and nonpositive on $D$, with value at least $-1$ at $0$. Set $\alpha _ n=\beta /A_ n$, with $\beta $ as in lemma 5.10.11. That lemma and the change of variables $\zeta \mapsto \frac34\zeta $ give a bound, independent of $e$ and $v$, for

It remains to pass from these integrals over discs in complex lines to an integral over a ball. Let $\tau $ and $s_{2n-1}$ be as in (5.8.3). By that identity, applied with center $a$, and since $|\zeta |^{2n-2}\leq r^{2n-2}$ for $|\zeta |<r$, every Borel function $F\geq 0$, possibly taking the value $+\infty $, satisfies

for every $r>0$. We apply this with $F=\mathrm e^{-\alpha _ n v}$ and $r=3/8$. Since $\frac14B\subset B(a,3/8)$, the uniform bound for the line integrals proves (5.10.13). Finally, if $\| v\| _{L^1(B)}\leq M$ with $M>0$, then $v/M$ is a nonpositive plurisubharmonic function with $\| v/M\| _{L^1(B)}\leq 1$, and (5.10.13) for $v/M$ is the last assertion. $\square$