Proposition 8.4.6.23966DC8CDBF Fix $M,B\geq 1$ and a smooth real $h(x,t)$ on a neighborhood of $X\times [-B-1,B+1]$. There is $0<\varepsilon \leq 1$, depending only on $M,B,h,\omega ,n,m$, with the following property. For every smooth real $\varphi $ with $\| \varphi \| _\infty \leq B$, there is $C<\infty $, depending also on its $C^2$ norm, such that if smooth real $u,v$ satisfy
\begin{gather*} W\in \Gamma _ m,\qquad I+v_{i\bar j}\in \overline{\Gamma _ m},\\ \| u\| _\infty \leq M,\qquad \| v-\varphi \| _\infty \leq \varepsilon ,\\ F(W)=\exp \bigl ((u+h(x,v))/m\bigr ), \end{gather*}
then
\[ \operatorname {tr}_\omega (\omega +\mathrm{dd}^{\mathrm{c}}u) \leq C\bigl (1+\sup _ X|\partial u|_ g^2\bigr ). \]
The cone condition on $v$ is understood in a unitary frame. The tolerance $\varepsilon $ uses no derivatives of $\varphi $; the constant $C$ uses no derivatives of $v$.
Proof.
For $\varepsilon \leq 1$ the density has fixed bounds $0<f_0\leq f\leq f_1$. Its first derivative is
\[ f_ k=\frac fm(u_ k+h_ k(x,v)+h_ t(x,v)v_ k). \]
The nonnegative $u_ k$ contribution can be dropped from the gradient lemma. At a point with $\lambda _1>1$, differentiating once more gives
\[ f_{1\bar1}\geq -C(1+|\partial v|^2+|1+v_{1\bar1}|). \]
Here $u_{1\bar1}=\lambda _1-1>0$, the square from differentiating the exponential is nonnegative, and all derivatives of $h$ are bounded on the fixed interval.
Set
\begin{align*} \Phi (s)& =-\tfrac 12\log (1-s/(2K)),\qquad \Psi (u)=-2Au+(A\tau /2)u^2,\\ \chi (t)& =-\tfrac 1{12}\log (2\varepsilon -t),\qquad \tau =(M+1)^{-1}. \end{align*}
Then $(4K)^{-1}\leq \Phi '\leq (2K)^{-1}$, $\Phi ''=2(\Phi ')^2$, $-3A\leq \Psi '\leq -A$, $\Psi ''=A\tau $, and $q=\chi '$ lies between $1/(36\varepsilon )$ and $1/(12\varepsilon )$, with $\chi ''=12q^2$. Take a maximum of $G=\log \lambda _1+\Phi (s)+\Psi (u)+\chi (v-\varphi )$ and use the smooth lower support from the eigenvalue lemma. At this point put
\[ a_ i=v_ i,\quad b_ i=\varphi _ i,\quad P_ v=I+v_{i\bar j},\quad W_ v=\sum _ iF_ i(P_ v)_{i\bar i},\quad x_ i=\Phi 's_ i,\quad y_ i=\Psi 'u_ i. \]
The first derivative condition and the torsion identity give
\[ H_{1\bar1;i}/\lambda _1=-x_ i-y_ i-q(a_ i-b_ i),\qquad Z_ i=-x_ i-y_ i-qa_ i+e_ i,\quad e_ i=qb_ i+(1-\lambda _1^{-1})T^1_{i1}. \]
Thus $|e_ i|\leq E_0$, with $E_0$ independent of $A$ and of derivatives of $v$.
We make the initial maximum inequality explicit. The gradient lemma and the identity $u_{i\bar j}=(\lambda _ i-1)\delta _{ij}$ give, after multiplication by $\Phi '$,
\[ \Phi 'Ls\geq \frac1{8K}\sum _{i,k}F_ i(|u_{ki}|^2+|u_{k\bar i}|^2) +\frac{3}{64K}Q-C\mathcal F -\frac{C_ G'}S\sum _ i|a_ i|^2. \]
To justify the constants, retain half of the positive quadratic term, use $(\lambda _ i-1)^2\geq \lambda _ i^2/2-1$, and choose the torsion parameter to cost at most $Q/(128K)$. The remaining density-gradient error is $C|a|/\sqrt K$. Young's inequality bounds it by $C_ G'|a|^2/S+cS/K$; choose $C_ G'$ large enough that the quadratic coercivity lemma absorbs $cS/K$ by $Q/(128K)$, with a fixed constant error. This leaves $3Q/(64K)$. The linear $H_{1\bar1;i}$ error in the eigenvalue lemma is estimated using the first derivative condition. Its $x_ i$ term costs at most half of the retained quadratic sum and $C\mathcal F$. Its $y_ i$ term costs at most $\Psi ''\sum _ iF_ i|u_ i|^2/4+\mathcal F$ whenever $\lambda _1\geq L_0(A)$, where $L_0(A)$ is fixed independently of $K,v$. Its $qa_ i$ term costs at most $q\sum _ iF_ i|a_ i|$ when $\lambda _1$ exceeds a fixed metric constant; its $qb_ i$ term costs $C\mathcal F$. If either eigenvalue threshold fails, the required bound follows directly. Otherwise, with fixed $C_ H,C_ G,C_0$ independent of $A$, the second derivative condition gives
\begin{align*} 0\geq {}& N-\sum _ iF_ i|Z_ i|^2+2\sum _ iF_ i|x_ i|^2 +\tfrac 34\Psi ''\sum _ iF_ i|u_ i|^2+\frac{Q}{32K} +\Psi '(f-\mathcal F)-C_0\mathcal F\\ & +q(W_ v-\mathcal F)-\frac{C_ H}{\lambda _1}|(P_ v)_{1\bar1}| -\frac{C_ G}S\sum _ i|a_ i|^2 +\sum _ iF_ i\bigl (12q^2|a_ i-b_ i|^2-q|a_ i|\bigr ). \end{align*}
The fixed $qL\varphi $ term is included in $C_0$. Choose $\varepsilon $ so the lower bound for $q$ meets all thresholds of the derivative-absorption lemma. These thresholds depend on $M,B,h$ and the metric, and not on derivatives of $\varphi $. This choice precedes that of $A$.
First suppose $\lambda _ n<-\delta \lambda _1$. Concavity gives $N\geq 0$, and
\[ -|Z_ i|^2+2|x_ i|^2 \geq -6|y_ i|^2-6q^2|a_ i|^2-6E_0^2. \]
The derivative-absorption lemma applies after decreasing the right side further by $(3q^2+2q)|a_ i|^2$. Drop its nonnegative reserves and $\Psi ''$ term. Since $\Psi 'f\geq -3Af_1$, this gives
\[ 0\geq \frac{Q}{32K}-(54A^2+C_*)K\mathcal F-3Af_1, \]
where $C_*$ is fixed independently of $A,\delta $.
Now suppose $\lambda _ n\geq -\delta \lambda _1$, and let $I=\{ i:F_ i>\delta ^{-1}F_1\} $. For $i\in I$, $\lambda _ i<\lambda _1$ and
\[ \frac{F_ i-F_1}{\lambda _1-\lambda _ i} \geq \frac{(1-\delta )F_ i}{(1+\delta )\lambda _1} \geq \frac{(1-2\delta )F_ i}{\lambda _1}. \]
The spectral formula for $D^2F$, including its nonpositive diagonal part, therefore gives $N\geq (1-2\delta )\sum _{i\in I}F_ i|Z_ i|^2$. This uses the original eigenvalues; repeated eigenvalues do not occur in $I$. Using $x_ i=-Z_ i-y_ i-qa_ i+e_ i$ and $|z+w|^2\geq \delta |z|^2-\delta |w|^2/(1-\delta )$, one obtains
\begin{align*} N-\sum _{i\in I}F_ i|Z_ i|^2+2\sum _{i\in I}F_ i|x_ i|^2 & \geq -\frac{4\delta (\Psi ')^2}{1-\delta }\sum _{i\in I}F_ i|u_ i|^2\\ & \quad -r q^2\sum _{i\in I}F_ i|a_ i|^2-rE_0^2\mathcal F, \qquad r=\frac{8\delta }{1-\delta }. \end{align*}
Choose $\delta \leq \min \{ 1/17,\tau /(72A+\tau )\} $. Then the first coefficient is at most $\Psi ''/2$, and $r\leq 1/2$. On the complementary indices the preceding elementary bound gives
\[ -\sum _{i\notin I}F_ i|Z_ i|^2+2\sum _{i\notin I}F_ i|x_ i|^2 \geq -\frac{54A^2+6nE_0^2}{\delta }KF_1 -6q^2\sum _{i\notin I}F_ i|a_ i|^2. \]
Apply the derivative-absorption lemma, with penalty $r$ on $I$ and zero outside it; adding the unused negative $(9q^2+2q)|a_ i|^2$ on $I$, and $(3q^2+2q)|a_ i|^2$ off $I$, only decreases its input. The constants in its fixed-comparison error and the high-index $rE_0^2$ error are bounded independently of $A,\delta $. Enlarge $C_*$ to include them. We obtain
\[ 0\geq \frac{Q}{32K}-\frac{54A^2+6nE_0^2}{\delta }KF_1 +(A-C_*)\mathcal F-3Af_1. \]
Take $A\geq \max \{ 1,C_*+1\} $, then fix the displayed $\delta $ and the threshold $L_0(A)$. Both case inequalities now satisfy the hypotheses of the conditional eigenvalue proposition, with fixed data. Hence $\lambda _1\leq CK$ at the maximum point. Finally $\Phi ,\Psi ,\chi $ have oscillations bounded independently of $K$ and of derivatives of $v$. Comparing $G$ at any other point with its maximum gives the same bound for $\lambda _1$ everywhere. Since $S\leq n\lambda _1$, this proves the trace estimate.
$\square$
Pome
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