Proposition 16.15.21E43C0935B545

Proof. At each point, the eigenvalues of $k$ relative to $h$ are positive, and the arithmetic–geometric mean inequality gives $\operatorname {tr}_ hk\geq n(\det k/\det h)^{1/n}\geq nd_0^{1/n}$. Since $\sup _ XD_ hf\geq 0$ by (16.15.1), it follows that $B\geq nd_0^{1/n}$, that is, $K_0\geq 1$.

The upper bound. Let $\beta >0$, write $a=a_\beta $ and $g=k+(a_{i\bar j})$, and let $x_0$ be a maximum point of $a-f$. There the complex Hessian of $a-f$ is negative semidefinite, so $0<g\leq k+(f_{i\bar j})$. In particular $k+(f_{i\bar j})$ is positive definite at $x_0$, and the arithmetic–geometric mean inequality for its eigenvalues relative to $h$ gives

at $x_0$. Hence $a-f\leq \log K_0/\beta $ on $X$. The function $a-\log K_0/\beta $ is therefore a smooth $\kappa $-psh function below $f$, that is, a competitor for $P_\kappa (f)$, and so $a\leq P_\kappa (f)+\log K_0/\beta $.

The lower bound. Let $\beta \geq 1$ and put $v=P_\kappa (f)$; as noted at the beginning of this subsection, $\min _ Xf\leq v\leq f$. For $0<t\leq 1$ put

This is a bounded $\kappa $-psh function, and we show that $w\leq a$.

For a bounded $\kappa $-psh function $b$ write $\mu _ b=\langle (\kappa +\mathrm{dd}^{\mathrm{c}}b)^ n\rangle $, the non-pluripolar Monge–Ampère measure, which is the Bedford–Taylor measure since $b$ is bounded (proposition 13.4.3, applied with $q=p=n$ to the sums of $b$ with smooth local potentials of $\kappa $), and equals the smooth form $(\kappa +\mathrm{dd}^{\mathrm{c}}b)^ n$ if $b$ is smooth (proposition 7.2.2(1)). Locally $\kappa =\mathrm{dd}^{\mathrm{c}}\rho $ with $\rho $ smooth, and $\kappa +\mathrm{dd}^{\mathrm{c}}w=\mathrm{dd}^{\mathrm{c}}\bigl ((1-t)(\rho +v)+t\rho \bigr )$. Expanding the $n$th power by multilinearity and positive homogeneity (proposition 13.2.6), and discarding the mixed terms, which are positive measures, we get $\mu _ w\geq t^ n\kappa ^ n$. On the other hand $w\leq f-n\log (1/t)/\beta $, so $\mathrm e^{\beta (w-f)}\leq t^ n$.

Let $E=\{ a<w\} $, a Borel set. Since $a$ and $w$ are bounded, $a\geq w-O(1)$, so proposition 13.9.5 applies without the correction terms; together with the two inequalities just proved and the equation for $a$ it gives

On $E$ the first integrand is strictly larger than the last one, so $E$ has volume zero. Thus $w\leq a$ almost everywhere, and hence everywhere by lemma 6.14.4(1), since both functions are quasi-plurisubharmonic.

Now take $t=1/\beta $, which is allowed since $\beta \geq 1$. Since $\min _ Xf-v\geq \min _ Xf-\max _ Xf=-\operatorname {osc}_ Xf$, we obtain

The Laplacian bound. Let $\beta _0$ be the constant of lemma 16.15.7. For $\beta >\beta _0$ that lemma, together with $D_ ha_\beta =H-\operatorname {tr}_ hk$, $H>0$ and $\operatorname {osc}_ Xf\leq 2\| f\| _\infty $, gives

Since $a_\beta \to v$ uniformly, $\langle D_ hv,\chi \rangle =\int _ Xv\, D_ h\chi \, \mathrm dV=\lim _{\beta \to \infty }\int _ X(D_ ha_\beta )\chi \, \mathrm dV$ for smooth $\chi $, which proves (16.15.22). By the remark following (16.15.2), $Lv=-4\pi D_ hv$ is therefore a bounded function, with a bound depending only on $X$, $\omega $, $\kappa $ and upper bounds for $\| f\| _\infty $ and $\sup _ XD_ hf$. Finally $v$ is continuous, as a uniform limit of continuous functions, so by proposition 16.14.9 it belongs to $C^{1,\alpha }(X)$ for every $0<\alpha <1$. $\square$