Pome

Lemma 14.19.3155D3763AB355

Lemma 14.19.31.55D3763AB355 Fix $M\geq 0$ and a smooth real function $G$ on $X\times [-M,M]$. There is $C$, depending only on $\omega $, $M$ and $\| G\| _{C^2}$, with the following property. If $v,u$ are smooth, $\omega _ v,\omega _ u>0$, $|v|\leq M$, and

\[ \omega _ u^ n=\mathrm e^{u+G(x,v)}\omega ^ n, \]

then

\[ |u|\leq \| G\| _{C^0},\qquad C^{-1}\omega \leq \omega _ u\leq C\omega . \]

Proof. The extrema of $u$ give the first bound, as in theorem 14.7.2. Put $k=g+(v_{i\bar j})$, $S_ v=\operatorname {tr}_ gk$, and $D=2\| G\| _{C^0}$. The chain rule and Cauchy–Schwarz give, for $H=u+G(x,v)$,

\[ \Delta _ gH\geq -C_0\bigl (1+S_ v+|\partial v|_ g^2\bigr ). \]

Indeed $\Delta _ gu=S-n\geq -n$, $\Delta _ gv=S_ v-n$, and the mixed derivatives of $G(x,v)$ are bounded below by $-C_0(1+|\partial v|_ g^2)$. Here and below constants depend only on the declared data.

Set $r=v+M+1$, $R=2M+1$ and $s=u-\inf _ Xu+1$, so $1\leq r\leq R$ and $1\leq s\leq D+1$. For constants $A,L>0$ to be chosen, maximize

\[ Q=\log S+Ar^2-Lu+s^{-1} \]

on $X$. If $S<1$ at the maximum, the bounded ranges of $r,u,s$ already bound $S$ everywhere. Otherwise lemma 14.5.1 applies there. Positivity gives

\[ SB\geq n^2,\qquad S_ v\leq S\operatorname {tr}_ hk, \qquad |\partial v|_ g^2\leq S|\partial v|_ h^2. \]

The middle inequality follows by diagonalizing $h$ relative to $g$: every eigenvalue of $h$ is at most $S$, and the diagonal entries of $k$ are positive. Consequently,

\[ \frac{\Delta _ gH}{S} \geq -C_1 B-C_0\operatorname {tr}_ hk-C_0|\partial v|_ h^2. \]

At the maximum,

\[ \frac{\partial S}{S}=(L+s^{-2})\partial u-2Ar\partial v. \]

Since $|\tau |_ h^2\leq C_ gB$, Young's inequality bounds the torsion term by

\[ \left|\frac2{S^2}\operatorname {Re}(h^{i\bar j}\tau _ iS_{\bar j})\right| \leq s^{-3}|\partial u|_ h^2+A|\partial v|_ h^2 +\frac{C_ g\bigl ((L+1)^2(D+1)^3+AR^2\bigr )}{S^2}B. \]

Also

\begin{align*} \Delta _ h(Ar^2)& \geq 2A\operatorname {tr}_ hk-2AR B+2A|\partial v|_ h^2,\\ \Delta _ h(-Lu+s^{-1})& =(L+s^{-2})(B-n)+2s^{-3}|\partial u|_ h^2. \end{align*}

Combining these inequalities with (14.5.2) yields

\[ \begin{split} 0\geq \Delta _ hQ\geq {}& \left(L-2AR-C_2- \frac{C_ g((L+1)^2(D+1)^3+AR^2)}{S^2}\right)B-n(L+1)\\ & +(2A-C_0)\operatorname {tr}_ hk+(A-C_0)|\partial v|_ h^2 +s^{-3}|\partial u|_ h^2. \end{split} \]

Choose $A>C_0+1$ and then $L>2AR+C_2+2$. If the fraction in the coefficient of $B$ is at least one, $S^2$ at the maximum is bounded by its numerator. Otherwise the displayed inequality gives $B\leq n(L+1)$ there. For $n\geq 2$, the bounded determinant gives

\[ S\leq n\mathrm e^{2\| G\| _{C^0}}B^{n-1}. \]

For $n=1$, $S=\mathrm e^{u+G(x,v)}$ directly. Thus $S$ is bounded at the maximum in every case. Maximality of $Q$ and the bounded ranges of $r,u,s$ give the same bound globally. The positive lower determinant bound then bounds every eigenvalue of $h$ away from zero. $\square$

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