Proposition 5.10.8B979DF329B28

Proof. We prove the first estimate by induction on $n$. For $n=1$ it is lemma 5.10.5. Let $n\geq 2$, write $z=(z_1,z')$ with $z'\in \mathbb C^{n-1}$, and let $v$ be as in the statement with $v(0)>-\infty $. The function $w(z')=v(0,z')$ is the composition of $v$ with the holomorphic map $z'\mapsto (0,z')$, so by proposition 5.9.3 it is plurisubharmonic or identically $-\infty $ on $D^{n-1}$. Since $w(0)=v(0)>-\infty $, it is plurisubharmonic, hence locally integrable and finite outside a Lebesgue null set $E\subset D^{n-1}$. For $z'\in D^{n-1}\setminus E$, the function $z_1\mapsto v(z_1,z')$ is subharmonic and nonpositive on $D$, with the finite value $w(z')$ at the center, so lemma 5.10.5 gives

We integrate over $z'\in (\tfrac 12D)^{n-1}$ using Tonelli's theorem; the null set $E$, on which the inner integral may be infinite, does not contribute. The induction hypothesis applied to $w$ then gives

which is the first estimate with $C_{n,p}=C_ pC_{n-1,p}$.

For the second estimate, let $Q=(\tfrac 1{10}D)^ n$ and $|Q|$ its volume. If $\| v\| _{L^1(D^ n)}=0$, then $v=0$ almost everywhere, hence $v\equiv 0$ by theorem 5.9.1, and there is nothing to prove; we may also assume that $\| v\| _{L^1(D^ n)}<\infty $. Put $\lambda =2|Q|^{-1}\| v\| _{L^1(D^ n)}>0$. By Chebyshev's inequality, the set of points of $Q$ with $|v|>\lambda $ has measure at most $\lambda ^{-1}\int _ Q|v|\, \mathrm dV\leq |Q|/2$, so its complement in $Q$ has positive measure. Since $v$ is finite almost everywhere, there is a point $a\in Q$ with $v(a)>-\infty $ and

The polydisc of radius $4/5$ centered at $a$ lies in $D^ n$, and the polydisc of radius $2/5$ centered at $a$ contains $(\tfrac 14D)^ n$, since $\tfrac 14+\tfrac 1{10}<\tfrac 25$. We apply the first estimate to the plurisubharmonic function $\zeta \mapsto v(a+\tfrac 45\zeta )\leq 0$ on $D^ n$ and change variables:

This is (5.10.9). $\square$