Lemma 10.9.1529371C8618C2

Proof. Put $\rho _ i=r/2+ir/(2n)$ and $B_ i=B(a,\rho _ i)$ for $0\leq i\leq n$, so that $B_0=B(a,r/2)$ and $B_ n=B(a,r)$. For $0\leq i<n$ fix a smooth function $\eta _ i\colon \mathbb R\to [0,1]$ with $\eta _ i=1$ on $(-\infty ,(\rho _ i/r)^2]$ and $\eta _ i=0$ on $[((\rho _ i+\rho _{i+1})/(2r))^2,\infty )$; the numbers $\rho _ i/r$ depend only on $n$ and $i$, so $\eta _ i$ can be chosen depending only on $n$ and $i$. By (10.9.14), the functions $\chi _ i(z)=\eta _ i(|z-a|^2/r^2)$ satisfy $\chi _ i\in C^\infty _ c(B_{i+1})$, $0\leq \chi _ i\leq 1$, $\chi _ i=1$ on $B_ i$ and $-C'r^{-2}\beta \leq \mathrm{dd}^{\mathrm{c}}\chi _ i\leq C'r^{-2}\beta $, where $C'=\max _ iC_{\eta _ i}$ depends only on $n$. For integers $j,m\geq 0$ with $j+m\leq n$ and $0\leq i\leq n$, put

Let $m\geq 1$, $i<n$, and let $R=A^ j\wedge D^{m-1}\wedge \beta ^{n-j-m}$, a closed positive current by proposition 7.2.2. By the definition and the symmetry of Bedford–Taylor products, $D\wedge R=A^ j\wedge D^ m\wedge \beta ^{n-j-m}$ and $D\wedge R-A\wedge R=\mathrm{dd}^{\mathrm{c}}((V-U)R)$, so

Since $R$ is positive, the forms $C'r^{-2}\beta \pm \mathrm{dd}^{\mathrm{c}}\chi _ i$ give positive measures $R\wedge (C'r^{-2}\beta \pm \mathrm{dd}^{\mathrm{c}}\chi _ i)$; as $|V-U|\leq \kappa r^2$ and $\operatorname{Supp}\chi _ i\subset B_{i+1}$, the last integral is at most $C'\kappa \int _{B_{i+1}}R\wedge \beta $ in absolute value. Using $\mathbf1_{B_ i}\leq \chi _ i\leq \mathbf1_{B_{i+1}}$ and the positivity of the currents involved, we obtain

Both terms on the right have one factor $D$ fewer. Applying this inequality $m$ times, starting from $i=0$, we find

Take $m=k-j$ and sum over $0\leq j\leq k$. With $l=j+t$ we have $m-t=k-l$, so every term with $A^ l$ carries the factor $\kappa ^{k-l}$, and the coefficients depend only on $n$. This proves the lemma; for $l=k$ the factor is $\kappa ^0=1$. $\square$