Corollary 16.27.10.9C70131AD005 Under the hypotheses of proposition 16.27.8, let $p>1$, $0\leq f\in L^ p(X,\omega ^ n)$ and $r>0$. There is $B_ r>0$, depending only on $X$, $\theta $, $\omega $, $p$ and $r$, such that every Borel set $E$ satisfies
8CF36AE13615
16.27.11
\begin{equation} \int _ E f\omega ^ n\leq B_ r\| f\| _{L^ p} \left(\frac{\operatorname {Cap}_{\theta ,\phi }(E)}m\right)^ r. \end{equation}
In particular, for every $r>1$ the measure $f\omega ^ n$ is dominated by a constant multiple of $\operatorname {Cap}_{\theta ,\phi }^ r$.
Proof.
If $\operatorname {Cap}_{\theta ,\phi }(E)=0$, then $E$ has zero volume by proposition 16.27.8, and both sides vanish. Otherwise let $q=p/(p-1)$. By Hölder's inequality and (16.27.9), the integral is at most
\[ \| f\| _{L^ p}C^{1/q}\exp \left[-\frac\alpha q \left(\frac{m}{\operatorname {Cap}_{\theta ,\phi }(E)}\right)^{1/n}\right]. \]
Put $t=(m/\operatorname {Cap}_{\theta ,\phi }(E))^{1/n}$; then $t\geq 1$, since the capacity is at most $\int _ X\langle \theta _\phi ^ n\rangle =m$. The function $t^{nr}\mathrm e^{-\alpha t/q}$ is bounded on $[1,\infty )$: its logarithmic derivative is $nr/t-\alpha /q$, so its maximum is attained at $1$ or at $nrq/\alpha $. Hence $\mathrm e^{-\alpha t/q}\leq A\, t^{-nr}$ for a constant $A$ depending only on $n,r,q,\alpha $, and this proves (16.27.11).
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.