PomeF5E70269636CCorollary 12.10.12.F5E70269636C Let $0\leq f\in L^1(\mathrm{d}V)$ be nonzero with $b=b_\omega (f)>0$. If $D_{f^{1/n}}$ has a smooth strictly positive maximizer $T\in \mathscr G_\omega $, then $f^{1/n}$ agrees almost everywhere with a smooth nonnegative function, and every threshold subsolution $u$ is smooth and satisfies
In particular a smooth threshold solution exists, and any two threshold subsolutions differ by a constant.
Proof. Let $u$ be any threshold subsolution; one exists by theorem 15.28.4. Write $T=\gamma ^{n-1}$, where $\gamma $ is the smooth Gauduchon metric given by the positive $(n-1)$st root of $T$, and set
Put $A=\omega +\mathrm{dd}^{\mathrm{c}}u\geq 0$. Since $T$ maximizes, the positive measure $R_{u,T}$ of proposition 12.10.11 has mass zero, which gives the equality
The coefficient measures of the positive current $A$ are controlled by this trace measure, which is absolutely continuous. Thus $A$ has a matrix density that is positive semidefinite almost everywhere, with coefficients in $L^ n$ because $a\in L^ n$.
For every smooth Hermitian form $\beta $, lemma 13.14.12 also gives
The resulting pointwise matrix inequalities hold almost everywhere, simultaneously for all positive definite matrices: on each coordinate chart, first use a countable dense family of constant positive matrices, realized on smaller charts by global smooth Hermitian forms, and then use continuity in the matrix. The variational formula for $(\det P)^{1/n}$ in the proof of lemma 12.7.2 therefore yields
Here $A^ n$ denotes only the determinant of the measurable coefficient matrix of $A$; no nonlinear product of currents is used. The trace equality gives the reverse inequality, by the arithmetic–geometric mean inequality. Equality then forces all eigenvalues of $A$ relative to $\gamma $ to equal $a$, also where $a=0$. Consequently $A=a\gamma $ as currents.
By lemma 5.22.4, applied with $\theta =\omega $, both $a$ and $u$ are smooth, and $f^{1/n}=b^{-1/n}a(\gamma ^ n/\mathrm{d}V)^{1/n}$ has a smooth nonnegative representative. Since $u$ is smooth, we may now take the top power: $(\omega +\mathrm{dd}^{\mathrm{c}}u)^ n=a^ n\gamma ^ n=bf\, \mathrm{d}V$. Finally, the function $a$ was defined from $f$, $b$ and $\gamma $ alone, so two threshold subsolutions $u_1,u_2$ satisfy $\omega +\mathrm{dd}^{\mathrm{c}}u_1=a\gamma =\omega +\mathrm{dd}^{\mathrm{c}}u_2$. Their smooth difference is therefore pluriharmonic, hence constant on the connected compact manifold $X$. $\square$
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