Theorem 14.5.5A61645907A4D

Proof. Use $h=g+(u_{i\bar j})$, $S=\operatorname {tr}_ g h$, and $B=\operatorname {tr}_ h g$ as in lemma 14.5.1. Write $s=u-\inf _ Xu+1$, so that $1\leq s\leq D+1$, and consider the maximum of

on $X$, where $A\geq 1$ will be fixed below. Let $x$ be a maximum point. If $S(x)<1$, an upper bound for $S$ on $X$ already follows from the oscillation bound and the definition of $Q$, as at the end of the proof. We may therefore assume $S(x)\geq 1$ and use lemma 14.5.1.

Since the eigenvalues of $h$ relative to $g$ are positive, $SB\geq n^2$, and hence $\Delta _ gG/S\geq -KB/n^2$. Moreover, $\partial Q=0$ at $x$ gives

Because $|\tau |_ h^2\leq C_ gB$, the absolute value of the torsion term is bounded by

If $S(x)^2\leq (A+1)^2(D+1)^3$, then $S(x)$ is bounded once $A$ is fixed, and again so is $S$ on $X$. In the remaining case the coefficient of $B$ in the last term is at most $C_ g$.

On the other hand, direct computation gives

The trace inequality and (14.5.6) imply at $x$

after a fixed enlargement of $C_ g$. Choosing $A>C_ g+K/n^2+1$, we conclude that $B(x)$ is bounded. If $\lambda _1,\ldots ,\lambda _ n$ are the eigenvalues of $h$ relative to $g$, then $\prod _ i\lambda _ i=\mathrm{e}^ G$, and for $n\geq 2$

For $n=1$, directly $S=\mathrm{e}^ G$. Thus $S(x)$ is bounded in all cases. For every $y\in X$, the maximality of $Q$ at $x$ gives

This proves $S\leq C$ on $X$. The lower bound for the determinant then gives $\lambda _ i\geq \mathrm{e}^{-K}C^{-(n-1)}$, which proves the uniform equivalence. $\square$