Lemma 14.3.2D570CDA4FA55

Proof. For a fixed real $p\geq 1$, define $M_ k$ for $0\leq k\leq n$ and $J_ k$ for $0\leq k\leq n-1$ by

All these numbers are nonnegative. The proof first bounds $pJ$ in terms of the masses $M_ k$, and then bounds these masses by a recurrence. Let $T=\sum _{k=0}^{n-1}\eta ^ k\wedge \omega ^{n-k-1}$, so that $\eta ^ n-\omega ^ n=\mathrm{dd}^{\mathrm{c}}u\wedge T$. Since $\mathrm{d}\eta =\mathrm{d}\omega $, differentiation gives

The sum is empty when $n=1$. Stokes' theorem gives the exact identity

The density assumption bounds the left-hand side above by $(\Lambda +1)M_0$. Apply lemma 14.3.1 to the last integral, with $\delta =K/p$ for a sufficiently large fixed $K$. The contribution proportional to $J/\delta $ is then at most $pJ/2$, and after absorbing it we obtain

This estimate alone does not suffice, because the mixed weighted masses on the right have not yet been bounded.

For $1\leq k\leq n-2$, a second integration by parts gives

The derivative in $R_ k$ has two terms, with coefficients $k-1$ and $n-k$. Applying the pointwise estimate with $\delta =1/p$ to each of them, and omitting terms with negative indices, we obtain

Consequently, for $p\geq 1$,

Induction over the finitely many values of $k$ proves

For $n=2$ this is immediate, and for $n=1$ the sum is empty. Inserting this bound in (14.3.3) gives $pJ\leq CM_0+C'J$. For $p\geq 2C'$ the last term can be absorbed, and $pJ\leq CM_0$.

Finally,

This proves the claimed estimate for all $p\geq p_0$. $\square$