Lemma 15.18.8.3EBB652C42D7 Let $B$ be a coordinate ball, let $\vartheta $ and $F>0$ be smooth near $\overline B$, and let $q$ be smooth with $R=\vartheta +\mathrm{dd}^{\mathrm{c}}q\geq \varepsilon \beta $ for a fixed $\varepsilon >0$. Fix a finite $L\geq 0$. If $u\in C^\infty (B)$ satisfies
\[ T=\vartheta +\mathrm{dd}^{\mathrm{c}}u>0, \qquad T^ n=F\beta ^ n, \qquad |u-q|\leq L, \]
put $S=\operatorname {tr}_\beta T$ and $v=u-q$. For $A>0$, define
\[ Q_ A=\log S-Av+\mathrm{e}^{-v}. \]
There are $A_0,C$, depending only on the fixed data and $L$, such that at every interior local maximum of $Q_ A$, with $A\geq A_0$,
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15.18.9
\begin{equation} S\leq C\sqrt A. \end{equation}
The constants do not depend on any bound for the gradient of $u$.
Proof.
This is the maximum-principle computation behind the classical second-order estimate. The non-closedness of $\vartheta $ produces a first-order error term, which is absorbed by the convex term $\mathrm{e}^{-v}$.
At the point in question make a constant unitary change of coordinates so that the matrix of $T$ is diagonal, with eigenvalues $\lambda _ i$. Write
\[ \Delta _ T=\sum _ i\lambda _ i^{-1}\partial _ i\partial _{\bar i}, \qquad H=\operatorname {tr}_ T\beta . \]
Differentiating $\log \det T=\log F$ and commuting derivatives gives
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15.18.10
\begin{equation} \Delta _ TS=\Delta _\beta \log F+P+E_0, \quad P=\sum _{k,i,j}\frac{|\partial _ kT_{i\bar j}|^2}{\lambda _ i\lambda _ j}, \quad |E_0|\leq C_0H. \end{equation}
The non-closedness of $\vartheta $ enters through the bounded vector
\[ E_ p=\sum _ k(\partial _ p\vartheta _{k\bar k}-\partial _ k\vartheta _{p\bar k}). \]
Since $\partial _ pS-E_ p=\sum _ k\partial _ kT_{p\bar k}$, weighted Cauchy–Schwarz gives $|\partial S-E|_ T^2\leq SP$. From (15.18.10),
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15.18.11
\begin{equation} \Delta _ T\log S \geq -\frac{C_0(1+H)}S- \frac2S\operatorname {Re}\langle \partial \log S,E\rangle _ T. \end{equation}
Set $g_ A(s)=-As+\mathrm{e}^{-s}$. At a local maximum of $Q_ A$, $\partial \log S=-g_ A'(v)\partial v$ and $\Delta _ Tv=n-\operatorname {tr}_ TR$. Combining these identities with (15.18.11), then completing the square in $\partial v$, yields
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15.18.12
\begin{equation} 0\geq \left[\varepsilon A-\frac{C_2}{S}-\frac{C_2(A+1)^2}{S^2}\right]H -C_3(A+1)-\frac{C_2}{S}. \end{equation}
Here $|E|_ T^2\leq CH$ and the bounds for $v$ have been used. If $S\geq K\sqrt A$ and $K$ is large, the bracket is at least $\varepsilon A/2$. Thus $H\leq C$ independently of $A$ and $u$. The determinant upper bound then gives $S\leq C'$. Enlarging $K$ makes this incompatible with $S\geq K\sqrt A$, proving (15.18.9).
$\square$
Pome
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