Pome

Proposition 14.19.50F13386AF12E0

Proposition 14.19.50.F13386AF12E0 There are constants $\eta ,C>0$, depending only on the fixed data and $H_1$, such that every smooth admissible $V$ with $\| U-V\| _\infty \leq \eta $ satisfies

\[ \| \partial U\| _\infty \leq C\bigl (1+\| \partial V\| _\infty ^2\bigr ). \]

The comparison radius $\eta $ is independent of the derivatives of $V$.

Proof. Let $a\geq 1$ and $\beta >1$ be constants, to be chosen below. Assuming $\| U-V\| _\infty \leq \beta ^{-1}$, put

\[ w=U-V-\beta ^{-1},\qquad \xi =a(\underline U-U)+\beta w^2,\qquad K_ V=1+\| \partial \underline U\| _\infty +\| \partial V\| _\infty . \]

Then $-2\beta ^{-1}\leq w\leq 0$ and

\[ \xi _ i=a(\underline U_ i-V_ i)+(-a+2\beta w)(U_ i-V_ i), \qquad |-a+2\beta w|\leq a+4. \]

Writing $D=|\partial (U-V)|_ B^2$, the Cauchy–Schwarz inequality gives

\begin{align*} \left|2\operatorname{Re}\sum _ iB_ i\lambda _ iU_ i\xi _{\bar i}\right| & \leq \frac18\sum _ iB_ i\lambda _ i^2+C_ aK_ V^4T+C_ a(Q+K_ V^2)D,\\ \left|2\operatorname{Re}\sum _{i,k}B_ i\chi _{i\bar k}U_ k\xi _{\bar i}\right| & \leq \tfrac 12QD+C_ aT+C_ aK_ VQ^{1/2}T. \end{align*}

For the first line, split $\xi _ i$ as above and use $\sum B_ i|U_ i|^2\leq 2D+2K_ V^2T$; then apply Young's inequality to the resulting three products, each absorbing at most $\frac1{24}\sum B_ i\lambda _ i^2$. For the second line, use $|\partial \xi |_ B\leq aK_ VT^{1/2}+(a+4)D^{1/2}$ and the bound for $\chi $. In both lines, $C_ a$ depends on $a$ but not on $\beta $ or on the derivatives of $V$.

By lemma 14.19.48 and $w\leq 0$,

\[ \mathcal L\xi \geq a\mathcal L(\underline U-U)-4H_0+2\beta D. \]

Choose $a$ so large that $(a\kappa -C_ g-2)\tau >4H_0+2H_1+1$, and then $\beta \geq 2C_ a+2$. Combining the preceding bounds with lemma 14.19.49, we obtain, at every point where $Q\geq C'(1+K_ V^4)$ for a suitable constant $C'$,

\begin{align*} \mathrm e^{-\xi }\mathcal L(\mathrm e^\xi Q) & \geq \frac18\sum _ iB_ i\lambda _ i^2\\ & \quad +Q\bigl [a\mathcal L(\underline U-U)-(C_ g+2)T +\beta D-(4H_0+2H_1)\bigr ]. \end{align*}

Indeed, the coefficient of $D$ before absorption is $(2\beta -C_ a-1/2)Q-C_ aK_ V^2$, which is at least $\beta Q$ once $C'$ is large enough. The other errors satisfy $C_ aK_ V^4\leq Q$ and $C_ a+C_ aK_ VQ^{1/2}\leq Q$, and they account for the additional term $-2T$ in the bracket.

Consider a maximum point of $\mathrm e^\xi Q$ at which $Q\geq C'(1+K_ V^4)$. In the first alternative of lemma 14.19.48, the bracket is positive by the choice of $a$, since $T\geq \tau $. In the second, $D\geq \kappa T|\partial (U-V)|^2$. For $Q\geq 4(1+K_ V^2)$, the triangle inequality gives $|\partial (U-V)|^2\geq c_*Q$, where $c_*=(\sqrt3-1)^2/4$. Since $\mathcal L(\underline U-U)\geq -H_0$, the bracket is then at least

\[ (\beta \kappa c_*Q-C_ g-2)T-(a+4)H_0-2H_1. \]

This is positive once $C'$ is large enough, because $T\geq \tau $. In both cases $\mathcal L(\mathrm e^\xi Q)>0$ at a maximum point, which contradicts the second derivative test. Thus $Q\leq C(1+K_ V^4)$ at the maximum point. The weight has uniformly bounded oscillation:

\[ \operatorname {osc}_ X\xi \leq a(\operatorname {osc}_ X\underline U+\operatorname {osc}_ XU)+4/\beta . \]

By the potential bound in proposition 14.19.44, this oscillation is bounded independently of $V$, so the estimate at the maximum point gives $Q\leq C(1+K_ V^4)$ everywhere. Taking square roots and setting $\eta =\beta ^{-1}$ proves the assertion. $\square$

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