Lemma 16.15.7EA811ACFD26E

Proof. Step 1: the value bounds. Since $(\kappa +\mathrm{dd}^{\mathrm{c}}a)^ n/\kappa ^ n=\det g/\det k$, the equation says that $\det g=\mathrm e^{\beta (a-f)}\det k$. At a maximum point of $a$ the complex Hessian $(a_{i\bar j})$ is negative semidefinite, so $0<g\leq k$ and therefore $\det g\leq \det k$ there, the eigenvalues of $k^{-1/2}gk^{-1/2}$ lying in $(0,1]$. Hence $a\leq f\leq \max _ Xf$ at that point, and therefore everywhere. At a minimum point of $a$ the inequalities are reversed, and $a\geq \min _ Xf$.

Step 2: a differential inequality for $\log H$. Let $x\in X$, and take the coordinates of lemma 16.15.4 at $x$. By that lemma, (16.15.6) holds at $x$, with $|E|\leq C_ EH\operatorname {tr}_ gh$, and $H\operatorname {tr}_ gh\geq n^2$. Now we use the equation in the form $\log \det g=\log \det k+\beta (a-f)$, valid in every chart. Since $D_ ha=H-\operatorname {tr}_ hk$,

The function $D_ h\log \det k$ is a smooth function on $X$, so it is bounded by a constant depending only on $X$, $\omega $ and $\kappa $. Dividing by $H$ and using $H^{-1}\leq \operatorname {tr}_ gh/n^2$ and $|E|/H\leq C_ E\operatorname {tr}_ gh$, we arrive at

where $C_0$ depends only on $X$, $\omega $ and $\kappa $. Both sides are independent of the coordinates, so (16.15.9) holds everywhere on $X$.

Step 3: the maximum principle. By (16.15.1) we have $B>0$. Let $B'\geq B$ be a constant, so that $\operatorname {tr}_ hk+D_ hf\leq B'$ on $X$. For a constant $A>0$, to be chosen, let $x_0$ be a maximum point of $Q=\log H-Aa$. Since

inequality (16.15.9) gives at $x_0$, where the complex Hessian of $Q$ is negative semidefinite,

We first prove the uniform equivalence of $g$ and $h$. Let $\beta $ be fixed, let $M\geq \| f\| _\infty $ and $N\geq \sup _ XD_ hf$ be given, take $B'=\sup _ X\operatorname {tr}_ hk+N$, and $A=(C_0+\beta B'/n^2+1)/\gamma $. Since $1/H\leq \operatorname {tr}_ gh/n^2$, the right side of (16.15.10) is at least $\beta +\operatorname {tr}_ gh-An$, so $\operatorname {tr}_ gh\leq An$ at $x_0$. Thus every eigenvalue of $g$ relative to $h$ is at least $(An)^{-1}$ at $x_0$. By Step 1, $|a-f|\leq \operatorname {osc}_ Xf\leq 2M$, and therefore, by the equation, the product of the eigenvalues satisfies

everywhere on $X$. Together with the lower bound for the eigenvalues at $x_0$, the upper bound for their product bounds each eigenvalue above at $x_0$, and hence bounds $H(x_0)$. For every $x\in X$, since $Q(x)\leq Q(x_0)$ and $\min _ Xf\leq a\leq \max _ Xf$,

Thus $H$ is bounded on $X$, so every eigenvalue of $g$ relative to $h$ is bounded above on $X$, and the lower bound for their product then bounds every eigenvalue below. All the constants depend only on $X$, $\omega $, $\kappa $, $\beta $, $M$ and $N$, which proves the uniform equivalence of $g$ and $h$.

Next we prove the bound for $H$ that is uniform in $\beta $. Take $B'=B$, $A=(C_0+1)/\gamma $ and $\beta _0=2An$; both $A$ and $\beta _0$ depend only on $X$, $\omega $ and $\kappa $. Suppose that $\beta >\beta _0$, and let $x_0$ be a maximum point of $Q=\log H-Aa$ for this $A$. If $H(x_0)>2B$, then $1-B/H>1/2$ at $x_0$, and the right side of (16.15.10) is at least $\beta /2+\operatorname {tr}_ gh-An>0$, a contradiction. Hence $H(x_0)\leq 2B$, and the same argument as above gives $H\leq 2B\, \mathrm e^{A\operatorname {osc}_ Xf}$ on $X$. Since $\operatorname {osc}_ Xf\leq 2\| f\| _\infty $ and $B\leq \sup _ X\operatorname {tr}_ hk+N$, this proves the last assertion. $\square$