Proposition 10.31.5436BCEF7C4255

Proof. As in proposition 10.31.15, we build a concave weight from the distribution of $\varphi -u$ and apply the preceding lemma. The moment bound on $\mu $ now yields a differential inequality for this weight, which bounds the depth of $\varphi -u$ under $\mu _ u$.

The envelope identity gives $u\leq \varphi \leq 0$. Put $x=\varphi -u$, defined off the poles, and $N=2n+1$. Since $x\leq -u$ and $\mu _ u\leq \mu $, Hölder's inequality gives the uniform first-moment bound

Let $H(t)=\mu _ u(\{ x>t\} )$ and $T=\sup \{ t\geq 0:H(t)>0\} $, with $T=0$ if this set is empty; at this point $T$ may be infinite. If $T<\infty $, zero domination applied to $u$ and $\varphi -t$ for $t>T$ gives $u\geq \varphi -T$. It is therefore enough to bound $T$. We may assume $T>1$, and we choose any $S$ with $1<S<T$.

Define, for $t\geq 0$,

and set $W(t)=\int _0^ t F(s)^{1/N}\, \mathrm{d}s$ and $\chi (-t)=-W(t)$. Since $H(S)>0$, $F$ is bounded on the whole half-line, although its bound may depend on $S$ and $u$. It is locally Lipschitz and nondecreasing. Hence $\chi $ has bounded derivative, is increasing and concave, and satisfies the hypotheses of the preceding lemma. Also $W(t)\geq t$ and $W(t)\leq tF(t)^{1/N}$.

Since $H(t)\leq M$, Tonelli's theorem gives a bound independent of $S$ and $u$:

Let $p=P_\theta (\varphi -W(x))\in \mathcal E_\varphi $, the envelope of the preceding lemma for this $\chi $. On its contact set the lemma bounds its measure by $F(x)^{n/N}\mu _ u$. Hence, using $N=2n+1$ and the Cauchy–Schwarz inequality,

The identity $P_\theta [p]=\varphi $ implies $p-\sup _ Xp\leq \varphi $. Since $p\leq \varphi \leq 0$ and $\mu _ p(X)\geq \ell $, the last integral is at least $\ell \sqrt{-\sup _ Xp}$. Consequently

Using $W(x)\leq \varphi -p\leq -p$, and applying the moment hypothesis to $p-\sup _ Xp$, we obtain

where we choose $C_0=1+2^{m-1}(A_ m+D^ m\mu (X))$. This constant is independent of $S$ and $u$.

For $0<t\leq S$, Chebyshev's inequality and $\mu _ u\leq \mu $ give

Since $W'=F^{1/N}$, the derivative of $(W')^{N+1}$ is at least

All functions being differentiated are absolutely continuous on compact intervals. Integrating from $0$ to $t$ and using $W(0)=0$ gives

In particular

with $c_0>0$ independent of $S$ and $u$. Integrating the derivative of $W^{1-\kappa }$ over $[1,S]$ yields

because $W(1)\geq 1$. Since the integral on the left diverges as $S\to \infty $, every $S<T$ is bounded by a constant independent of $S$ and $u$, and hence so is $T$. The zero-domination argument at the beginning of the proof now gives $u\geq \varphi -C$. The case $T\leq 1$, which includes $B=0$, was covered there as well. $\square$