Proposition 7.4.5 (Błocki)EB80820D2146

Proof. First suppose that $u$, $h$ and $v$ are smooth; then $\mathrm{dd}^{\mathrm{c}}u$, $\mathrm{dd}^{\mathrm{c}}h$ and $\mathrm{dd}^{\mathrm{c}}v$ are positive $(1,1)$-forms, by proposition 5.9.2. Subtracting $\sup _ Bv$ from $v$, we may assume $-M\le v\le 0$ on $B$. Put $w=h-u\ge 0$; it vanishes on a collar of $\partial B$. Let $p\ge 2$, and let $R$ be a smooth closed strongly positive form of bidegree $(n-1,n-1)$; the exterior product of $R$ with a positive $(1,1)$-form is then a nonnegative multiple of the volume form, by proposition 5.3.2, since positive $(1,1)$-forms are strongly positive. Since $\mathrm{d}w\wedge \mathrm{d}^{\mathrm c}w$ is a positive $(1,1)$-form by (7.3.1) and $\mathrm{dd}^{\mathrm{c}}h\wedge R\ge 0$, the chain rule $\mathrm{dd}^{\mathrm{c}}(w^ p)=pw^{p-1}\mathrm{dd}^{\mathrm{c}}w+p(p-1)w^{p-2}\mathrm dw\wedge \mathrm d^{\mathrm c}w$ gives

Since $w^ p$ has compact support in $B$ and $R$ is closed, (7.2.3) gives $\int _ Bw^ p\mathrm{dd}^{\mathrm{c}}v\wedge R=(\mathrm{dd}^{\mathrm{c}}(vR))(w^ p)=(vR)(\mathrm{dd}^{\mathrm{c}}(w^ p))$. As $v\le 0$ and $-v\le M$, the preceding inequality therefore yields

For $k=1,\ldots ,n$, apply this with $p=n+2-k$ and $R=(\mathrm{dd}^{\mathrm{c}}v)^{n-k}\wedge (\mathrm{dd}^{\mathrm{c}}u)^{k-1}$, a product of positive $(1,1)$-forms, hence strongly positive (proposition 5.3.2) and closed. This gives

and chaining these $n$ inequalities from $k=1$ to $k=n$ gives the asserted inequality, since $(n+1)n\cdots 2=(n+1)!$.

In general, write $B=B(a,r)$ and let $V$ be a collar of $\partial B$ on which $u=h$. Since $\partial B$ is compact, there is $r''<r$ with $\{ z:r''\le |z-a|\le r\} \subset V$; choose $r'\in (r'',r)$ and put $B'=B(a,r')$. Then $h-u$ vanishes on $B\setminus \overline{B(a,r'')}$. Convolve $u$, $h$ and $v$ with the same radial kernel $\rho _\varepsilon $, with $\varepsilon <\min \{ (r'-r'')/2,r-r'\} $ so small that the regularizations are defined, smooth and plurisubharmonic on a neighborhood of $\overline{B'}$ (proposition 5.9.2). At points of $\overline{B'}$ they are averages over balls of radius $\varepsilon $ contained in $B$, where $u\le h$; hence $u_\varepsilon \le h_\varepsilon $ on $\overline{B'}$. Moreover $h_\varepsilon -u_\varepsilon $ vanishes on the collar $\{ z:r''+\varepsilon <|z-a|<r-\varepsilon \} $ of $\partial B'$ and outside $\overline{B(a,r''+\varepsilon )}$ in $B'$, hence outside the compact set $\overline{B(a,(r'+r'')/2)}\subset B'$. The values of $v_\varepsilon $ on $B'$ are averages of values of $v$ on $B$, so the oscillation of $v_\varepsilon $ on $B'$ is at most $M$.

Apply the smooth case on $B'$ and let $\varepsilon =\varepsilon _ k\downarrow 0$. The weights $(h_\varepsilon -u_\varepsilon )^{n+1}$ and $h_\varepsilon -u_\varepsilon $ are polynomials in decreasing sequences of bounded plurisubharmonic functions, and they, as well as their limits $(h-u)^{n+1}$ and $h-u$, vanish in $B'$ outside the fixed compact set $\overline{B(a,(r'+r'')/2)}$. Testing against a function $\chi \in C^\infty _ c(B')$ equal to $1$ on this set, corollary 7.3.8 lets both sides pass to the limit. Since $h-u$ vanishes on $B\setminus B'$, the limiting inequality on $B'$ is the asserted inequality on $B$. $\square$