Theorem 12.9.1.BF52D98B114F For every $T\in \mathscr G_\omega $, the measure $\Gamma _0(T)$ is absolutely continuous with respect to $\mathrm{d}V$; we write $\Gamma _0(T)=k_ T\, \mathrm{d}V$ and call $k_ T\geq 0$ the determinant density of $T$. For every $1<q<n/(n-1)$ there is $C_ q>0$ such that
Theorem 12.9.1BF52D98B114F
Proof. Fix $p>1$ and a smooth strictly positive function $h$. By theorem 14.2.1 there are a smooth form $\Omega _ h=\omega +\mathrm{dd}^{\mathrm{c}}u_ h>0$ and a constant $c_ h>0$ such that
The lower-density estimate of theorem 14.19.19 gives a constant $c_ p>0$, independent of $h$, such that
The pointwise matrix arithmetic–geometric mean inequality, applied to $\Omega _ h$ and the coefficient density of $T$, gives the inequality of measures
Integrating and using $\mathrm{dd}^{\mathrm{c}}T=0$ yields
Approximation by smooth strictly positive functions extends this bound to every continuous nonnegative $h$, and applied to $|h|$ it bounds the integral of every continuous function against $\Gamma _0(T)$ by its $L^{np}$ norm. Since continuous functions are dense in $L^{np}$, the duality between $L^{np}$ and $L^{np/(np-1)}$ shows that $\Gamma _0(T)$ has a density in $L^{np/(np-1)}$, with norm at most $c_ p^{-1/n}$. As $p$ ranges over $(1,+\infty )$, the exponent $np/(np-1)$ ranges over $(1,n/(n-1))$. This proves the theorem. $\square$
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