8.13.9 Domination and proper equationsDBEDA039CF16

Proof. Add the same constant to $u,v$ so that $v\geq 1$. This preserves the measure hypothesis and the desired conclusion. Choose $0<\alpha <1$ with $\alpha ^ m>c$ and put $V=\alpha v$, $\eta =1-\alpha $. Suppose $S=\inf _ X(u-V)<0$. For $0<s<-S$, the set $E_ s=\{ u<V+S+s\} $ is contained in $\{ u<V\} $ and hence in $\{ u<v\} $. It has positive volume. Indeed, if its volume were zero, then $u\geq V+S+s$ almost everywhere. Both functions are bounded admissible with their canonical representatives, so 8.2.10 would give this order everywhere, contradicting the definition of $S$. The argument does not require $E_ s$ to be open or the infimum to be attained.

Choose also $s<\eta ^3/C$, where $C$ is from 8.13.8. By 8.12.3 and the measure hypothesis,

If $c>0$, take $s$ still smaller so that $c(1+Cs\eta ^{-m})<\alpha ^ m$. If $c=0$, this strict inequality already holds. All measures are finite and positive, so the display forces the positive-volume term to be zero, a contradiction. Hence $S\geq 0$ and $u\geq \alpha v$. Let $\alpha $ increase to one, with $\alpha ^ m>c$. Since $v$ is bounded, $\alpha v\to v$ uniformly, and therefore $u\geq v$. The final assertion is $c=0$. $\square$

Proof. For $\delta >0$, put $E_\delta =\{ u<v-\delta \} $. Multiplying the assumed measure inequality by the positive bounded Borel function $e^{au}$ gives on $E_\delta $

The last identity uses invariance under constants. Since $e^{-a\delta }<1$, 8.13.10 applied to $u,v-\delta $ gives $u\geq v-\delta $ everywhere. Letting $\delta \downarrow 0$ gives $u\geq v$. Two solutions of the stated equation have equal weighted measures $e^{-au}H_ m(u)=\nu $. Applying the order assertion in both directions proves uniqueness. $\square$

Proof. Put $K=\| h_1-h_2\| _{L^\infty (\nu )}$ and fix $s>K/\kappa $. On $E_ s=\{ v+s<u\} $, the actual values $u,v$ lie in $[A,B]$, so $f(x,u)-f(x,v)\geq \kappa s$. The equations and the essential bound on $h_2-h_1$ therefore give, as restricted measures,

The constant is smaller than one. Applying 8.13.10 to $v+s,u$ gives $u\leq v+s$ pointwise. Let $s\downarrow K/\kappa $ and exchange $u,v$ to obtain the norm bound. The exponent is evaluated only at $u,v$; the shifted comparison potential need not take its values in $[A,B]$. $\square$

Proof. For a continuous admissible potential, smooth decreasing admissible approximations converge uniformly on compact $X$ by Dini's theorem. Thus 8.11.12, 8.7.1 define the same Hessian measure on the continuous class. The smooth solution of 8.8.3 therefore solves the bounded equation as well. Apply 8.13.12 with $\nu =\omega ^ n$ and $h_1=h_2=0$ to identify any other bounded solution in $[A,B]$ with it. For the final assertion, smooth existence is already 8.6.1, and 8.13.11 with $\nu =e^ H\omega ^ n$ gives uniqueness among all bounded solutions. $\square$