Proposition 12.10.37.A98696DE3546 Under the hypotheses and notation of proposition 12.10.20, put
\[ A_\lambda =\omega +\mathrm{dd}^{\mathrm{c}}v_\lambda ,\qquad A_0=\omega +\mathrm{dd}^{\mathrm{c}}u_0. \]
There is a constant $C$ such that
\[ \| A_\lambda \wedge T-A_0\wedge T\| _{\mathrm{TV}} \leq C\lambda \qquad (\lambda >0). \]
For each compact set $K\Subset X\setminus Z$, there is a constant $C_ K$ such that
\[ \| A_\lambda -A_0\| _{\mathrm{TV}(K)} \leq C_ K\sqrt\lambda \qquad (0<\lambda \leq 1). \]
Proof.
Put $p=f^{1/n}$, use the probability measure $\nu =A_0\wedge T=b^{1/n}p k_ T\, \mathrm{d}V$ from the proof of proposition 12.10.20, where the equality is (12.10.18), and set
\[ q_\lambda =(c_\lambda /b)^{1/n}\mathrm{e}^{\lambda v_\lambda /n}, \qquad R_\lambda =A_\lambda \wedge T-q_\lambda \nu \geq 0. \]
Here $R_\lambda \geq 0$ by the trace inequality used in that proof, which also gives $c_\lambda \geq b$ and $\int _ X(-v_\lambda )\, \mathrm{d}\nu \leq C_0$ for a constant $C_0$. Write $s_+=\max (s,0)$ for the positive part of a real number $s$. Since $q_\lambda \geq \mathrm{e}^{\lambda v_\lambda /n}$ and $1-\mathrm{e}^{x}\leq -x$, we obtain
\[ (1-q_\lambda )_+\leq 1-\mathrm{e}^{\lambda v_\lambda /n}, \qquad \int _ X(1-q_\lambda )_+\, \mathrm{d}\nu \leq C_0\lambda /n. \]
Since $A_\lambda \wedge T$ and $\nu $ both have mass one,
\[ \| A_\lambda \wedge T-\nu \| _{\mathrm{TV}} \leq R_\lambda (X)+\int _ X|q_\lambda -1|\, \mathrm{d}\nu =2\int _ X(1-q_\lambda )_+\, \mathrm{d}\nu . \]
This proves the first estimate, with $C=2C_0/n$, and also gives $R_\lambda (X)\leq C_0\lambda /n$ and $\int _ X|q_\lambda -1|\, \mathrm{d}\nu \leq 2C_0\lambda /n$.
On $U=X\setminus Z$, write $T=\gamma ^{n-1}$ and $A_0=a\gamma $ as in the proof of theorem 12.10.17, and put $\alpha _\lambda =q_\lambda a$. Decompose the positive current $A_\lambda $ into its absolutely continuous part, with matrix density $B_\lambda $ relative to $\gamma $, and its positive singular part. Since $\mathrm{MA}^{\mathrm{ext}}_\omega (v_\lambda )=c_\lambda \mathrm{e}^{\lambda v_\lambda }f\, \mathrm{d}V$, the trace inequalities for all Hermitian forms imply, as in the proof of corollary 12.10.12,
\[ (\det B_\lambda )^{1/n}\geq \alpha _\lambda \quad \text{almost everywhere on }U. \]
This is a statement about determinants of matrices; no nonlinear product of coefficient measures is involved. Put $s_\lambda =\operatorname {tr}B_\lambda /n$ and $\delta _\lambda =s_\lambda -\alpha _\lambda \geq 0$. On $U$, the absolutely continuous part of the trace error $R_\lambda $ is $\delta _\lambda \gamma ^ n$, and its singular part is the trace measure of the singular part of $A_\lambda $. Thus
\[ \int _ U s_\lambda \gamma ^ n\leq 1,\qquad \int _ U\delta _\lambda \gamma ^ n\leq C_0\lambda /n, \]
and the singular trace mass on $U$ has the same upper bound.
For nonnegative numbers $\ell _1,\ldots ,\ell _ n$ with geometric mean at least $\alpha \geq 0$ and arithmetic mean $s$, the arithmetic–geometric mean inequality and the Cauchy–Schwarz inequality give
\[ \sum _ i(\sqrt{\ell _ i}-\sqrt\alpha )^2\leq n(s-\alpha ), \qquad \sum _ i|\ell _ i-\alpha |\leq 2n\sqrt{s(s-\alpha )}. \]
Indeed $\sum _ i\sqrt{\ell _ i}\geq n\sqrt\alpha $, while $\sum _ i(\sqrt{\ell _ i}+\sqrt\alpha )^2\leq 4ns$. Applying this to the eigenvalues of $B_\lambda $ relative to $\gamma $ and integrating, with the Cauchy–Schwarz inequality once more, we see that the absolutely continuous part of $A_\lambda -\alpha _\lambda \gamma $ has integrated trace norm at most $2n\sqrt{C_0\lambda /n}$ on $U$. Its singular part has norm at most $C_0\lambda $ in the same metric. Finally
\[ \int _ U|\alpha _\lambda -a|\gamma ^ n =\int _ U|q_\lambda -1|\, \mathrm{d}\nu \leq 2C_0\lambda /n. \]
Since $\gamma $ is comparable to the fixed metric on the compact set $K\subset U$, these estimates and $\lambda \leq \sqrt\lambda $ for $0<\lambda \leq 1$ give the second conclusion.
$\square$
Pome
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