Theorem 14.19.450F1A655FC651

Proof. Set $a=u_1-\underline u$, $b=u_2-\underline u$ and $\theta =\chi +\mathrm{dd}^{\mathrm{c}}\underline u$. By proposition 14.19.44, $|a|,|b|\leq C_0$ for a fixed $C_0\geq 1$. It suffices to bound $s=\max _ X(b-a)$ when it is positive; the opposite difference is handled by interchanging the solutions.

Fix $r,R,\delta $ from lemma 14.19.43 and put $\alpha =r\omega $. For every smooth positive $\rho $ with

theorem 14.2.1 gives a smooth solution

At a maximum of $\psi $, $c\rho \leq 1$, so $c\leq 2$. Apply theorem 14.19.19 with $p=2$ and background $\alpha $. Its potential and lower-density bounds give fixed $c_0,C_\psi >0$ with

These constants depend only on the three bounds for $\rho $, and not on the cutoff function chosen below.

Choose $\kappa >0$ so that $2C_0\kappa \leq 1/4$, $\kappa C_0\leq 1/16$ and $\kappa C_\psi /2\leq 1/16$. Next choose $M\geq 1$ with $2^{-n}c_0Mr^ n>R^ n$, and then $\varepsilon _0>0$ with

If one of the inequalities

fails, then $s\leq C(\| h_1-h_2\| _\infty +\| u_1-u_2\| _{L^1})$, with $C$ depending only on $\kappa ,\delta ,\varepsilon _0$ and the volume of $X$. We assume both and derive a contradiction. Put $t=\kappa s$; since $s\leq 2C_0$, we have $t\leq 1/4$. On $\Omega =\{ (1-t)b-a>s/2\} $, we have $b-a>7s/16$, so by Chebyshev's inequality the relative volume of $\Omega $ is at most $16\varepsilon _0/7$.

Take a smooth $\zeta :\mathbb R\to [0,1]$, zero on $(-\infty ,1/2]$ and one on $[3/4,\infty )$, and set

The second term is supported in $\Omega $, so $1/2\leq b_0\leq 1$ and $\langle \rho ^2\rangle _\alpha \leq 2+2M^2(16\varepsilon _0/7)\leq 4$; here we use that averages with respect to $\alpha =r\omega $ and $\omega $ agree. Thus the auxiliary solution for this $\rho $ satisfies the bounds with the same constants $c_0,C_\psi $.

At a point where $b-a=s$, the function $\Phi =(1-t)b-a+t\psi /2$ is at least $7s/8$, by the choice of $\kappa $. At a maximum point $x_0$ of $\Phi $, therefore, $(1-t)b-a\geq 7s/8$ because $\psi \leq 0$, and so $\rho (x_0)\geq M$. The second derivative test gives, at $x_0$,

The identity

and concavity imply

Both forms in the convex combination are admissible, since $\mu -r\mathbf1\in \Gamma $ and $P>0$. By lemma 14.19.43, $P\leq R\omega $. On the other hand, $P\geq \frac12(\alpha +\mathrm{dd}^{\mathrm{c}}\psi )$, so the auxiliary equation gives

a contradiction. Hence one of the two smallness inequalities fails, which gives the required bound for $s$. Interchanging the solutions proves the theorem. $\square$