14.14.12 Cancellation when the smallest eigenvalue is controlled.DA6F05F8F203

Fix $0<\delta <1/2$, to be chosen after $D_1,D_2$. First suppose that $\lambda _ n\geq -\delta \lambda $. Put $I=\{ k:f_ k>\delta ^{-1}f_1\} $, so that $1\notin I$. Equation (14.14.4) and $\phi ''=2(\phi ')^2$ give

Choose $\delta $ small enough that $4\delta (\psi ')^2\leq \psi ''/2$ throughout $[-M,0]$. By (14.13.8),

Here we used $(1-\delta )/(1+\delta )\geq 1-2\delta $.

We need the following comparison of the two derivatives $\nabla _1g_{k\bar1}$ and $a_ k=\nabla _ kg_{1\bar1}$. For every $\varepsilon >0$, provided $\lambda /K$ is sufficiently large depending on the fixed function $\psi $,

To prove this, let $U=\sum _{p>1,q}|v_{pq}|$. Commuting partial derivatives in $g$ gives, for $k>1$,

The second inequality uses the absence of $v_1$ in $Z_{k\bar1}$, so that differentiation in the direction $1$ introduces no pure derivative $v_{11}$. All mixed second derivatives are bounded by $C\lambda $, and every other pure second derivative occurs in $U$.

Use $|a+e|^2-|a|^2\geq -2|a||e|$, and then (14.14.4). For $k>1$, the pure second derivatives in $V_ k$ all occur in $U$. With $H_ k=\sum _ p|v_{p\bar k}|$, the resulting negative terms, divided by $\lambda ^2$, are bounded by a constant times

Here we used $H_ k\leq C\lambda $ and $\lambda \geq K$ to simplify the cross terms. The weight comparison (14.13.6), the symmetry of pure derivatives and the Cauchy–Schwarz inequality give $\mathcal F U^2\leq C E$. Consequently, after multiplication by $f_ k$ and summation, the first two terms are bounded by any fixed small fraction of $E/K$ plus $C\mathcal F$. The third is at most $CE/(\lambda \sqrt K)$, which is absorbed by another such fraction when $\lambda /K$ is large. For the last two terms we use

The contribution of $U^2$ is at most $C|\psi '|E/\lambda $, which can again be absorbed when $\lambda /K$ is large. Choosing the fractions with sum $1/12$ proves (14.14.13).

Insert this cancellation and the preceding inequalities into (14.14.11). The remaining derivative error satisfies, by (14.14.4),

after Young's inequality and an adjustment of constants. The torsion contraction also satisfies

The first inequality uses again the missing gradient index in $Z_{1\bar1}$. Since $\psi '<0$ and $v_{k\bar k}=\lambda _ k-\chi _{k\bar k}-W_{k\bar k}$, we conclude, after decreasing the remaining coefficient of $E$ if necessary, that

For the first term we used $|v_{1\bar1}|^2\geq \lambda ^2/2-CK$, which follows from $\chi +W=O(\sqrt K)$, and $E\geq f_1|v_{1\bar1}|^2$.

For large $\lambda $, lemma 14.13.5 bounds the last sum below by $\tau \mathcal F/2$. Choose first $\varepsilon $ with $\varepsilon C_0<\tau /4$. Next choose $D_2$ so that $\psi ''/2>C_\varepsilon |\psi '|$, and then choose $D_1$ so that $|\psi '|\tau /4>C_0$ on $[-M,0]$. Finally choose $\delta $ as above. All these choices depend only on the fixed data. Inequality (14.14.14) then forces $\lambda ^2\leq 80\, \delta ^{-1}\sup _{[-M,0]}|\psi '|^2K^2$, which proves the desired bound in the first case.