Lemma 16.15.4F24675566E64

Proof. At every point the eigenvalues of $g$ relative to $h$ are positive, their sum is $H$ and the sum of their inverses is $\operatorname {tr}_ gh$, so the Cauchy–Schwarz inequality gives $H\operatorname {tr}_ gh\geq n^2$.

Fix finitely many charts and compact subsets of their domains whose union is $X$, and let $x\in X$. By lemma 16.15.3, applied to $h$ in one of these charts, followed by a unitary change of coordinates that diagonalizes $g$ at $x$, there are coordinates as in the statement. In them, the derivatives of order at most two of $h$ at $x$ are bounded by a constant depending only on $X$ and $\omega $. All the following computations are at $x$, in these coordinates. Locally $\kappa $ and $\kappa +\mathrm{dd}^{\mathrm{c}}a$ have Kähler potentials (lemma 6.14.2): $g_{i\bar j}=\partial _ i\partial _{\bar j}\varphi $ with $\varphi $ smooth. Hence the derivatives $\partial _ pg_{i\bar q}$ are symmetric in $p$ and $i$, and the second derivatives $\partial _ p\partial _{\bar l}g_{i\bar q}$ are symmetric in $p,i$ and in $\bar l,\bar q$. Differentiating $H=h^{i\bar q}g_{i\bar q}$ twice, and using that the first derivatives of $h$, hence of $h^{-1}$, vanish at $x$, we get

Since $g$ is diagonal with positive eigenvalues, $|E|\leq C_ EH\operatorname {tr}_ gh$, where $C_ E$ bounds the second derivatives of $h^{-1}$ at $x$. As $h=I$ and $\mathrm dh=0$ at $x$, the formula $\partial (h^{-1})=-h^{-1}(\partial h)h^{-1}$ shows that these are the second derivatives of $-h$, up to transposition; so $C_ E$ depends only on $X$ and $\omega $.

On the other hand, differentiating the identity $\partial _ p\log \det g=g^{i\bar q}\partial _ pg_{i\bar q}$ and using $\partial (g^{-1})=-g^{-1}(\partial g)g^{-1}$ gives

By the symmetry of the second derivatives, $\partial _ p\partial _{\bar p}g_{i\bar i}=\partial _ i\partial _{\bar i}g_{p\bar p}$. Summing over $p$, comparing with the formula for $D_ gH$, and noting that $\sum _ p\partial _ p\partial _{\bar p}=D_ h$ at $x$, we obtain the identity (16.15.5), with the bound for $E$ just proved.

For the inequality for $\log H$, note that $\partial _ pH=\sum _ i\partial _ pg_{i\bar i}=\sum _ i\partial _ ig_{p\bar i}$ by the symmetry of the first derivatives. The Cauchy–Schwarz inequality with weights $\lambda _ i$ therefore gives

the last sum containing the previous one (take $p=q$ in the term of index $(i,p,q)$ and rename). Since $D_ g\log H=H^{-1}D_ gH-H^{-2}|\partial H|_ g^2$, the quadratic term in $D_ gH$ absorbs the negative term, and (16.15.6) follows. $\square$