Theorem 14.3.5.EEF74F1DEBB8 For each $\Lambda >0$ there is $C_\Lambda $ such that every smooth $u$ with
satisfies $\| u\| _\infty \leq C_\Lambda $.
EEF74F1DEBB8Theorem 14.3.5.EEF74F1DEBB8 For each $\Lambda >0$ there is $C_\Lambda $ such that every smooth $u$ with
satisfies $\| u\| _\infty \leq C_\Lambda $.
Proof. Let $\mathrm{d}\mu =\omega ^ n/\int _ X\omega ^ n$. The Sobolev inequality for the Riemannian metric of $\omega $ gives, for some $\beta >1$ and every smooth real function $w$,
We may take $\beta =n/(n-1)$ if $n\geq 2$, and for example $\beta =2$ if $n=1$. Choose $p_0$ as in lemma 14.3.2. Applying the Sobolev inequality to $w=\mathrm{e}^{-pu/2}$ and using that lemma, we obtain, with $Z=\mathrm{e}^{-u}$,
Iterating with $p=p_0\beta ^ j$, $j=0,1,2,\ldots $, we obtain a bound for $\sup _ XZ$; the product of the constants is finite because $\sum _ j(1+\log (p_0\beta ^ j))/(p_0\beta ^ j)<\infty $. Hence, for a uniform $C_1\geq 1$ and $m=\inf _ Xu$,
0F99936A9B52
14.3.6
The right-hand side still involves the unknown potential. Divide by $\mathrm{e}^{-p_0m}$ and set $A=p_0^{-1}\log (2C_1)$. The integrand $\mathrm{e}^{-p_0(u-m)}$ is at most one everywhere and at most $(2C_1)^{-1}$ outside $S=\{ u\leq m+A\} $. Therefore (14.3.6) implies
Since $\sup _ Xu=0$, theorem 5.25.2 gives $\int _ X(-u)\, \mathrm{d}\mu \leq C_0$. On $S$ we have $-u\geq -m-A$, so
when $-m\geq A$; otherwise the desired bound is immediate. Thus $-m\leq A+C_0/\delta $, proving the theorem. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.