Proposition 10.31.15.3464170481EE Set $N=2n+1$ and $r=1+(4N)^{-1}$. For $u\in \mathrm{PSH}(X,\theta )$ with $u\sim V$ and $\sup _ Xu=0$, put $B=\int _ X(V-u)\, \mu _ u$. There is a constant $C$, depending only on $n,\ell _ V,M_ V$, such that, for every non-pluripolar probability measure $\nu $ with $A=A_\theta (\nu )<\infty $,
\[ \int _ X(V-u)^ r\, \mathrm{d}\nu \leq C(1+A)(1+B)^{1+1/N} . \]
In particular, normalized potentials $u\sim V$ with uniformly bounded $B$ are uniformly integrable with respect to $\nu $.
Proof.
From the distribution of $V-u$ under $\mu _ u$ we build a concave weight adapted to $u$, apply the preceding lemma, and use $A$ to bound the resulting envelope against $\nu $. Throughout, $c$ and $C$ denote positive constants depending only on $n,\ell _ V,M_ V$.
The normalization gives $u\leq V$. Put $x=V-u\geq 0$ and
\[ T=\sup \{ t\geq 0:\mu _ u(\{ x>t\} )>0\} , \]
with $T=0$ if this set is empty. Since $u\sim V$, $T$ is finite. Zero domination, applied to $u$ and $V-t$ for $t>T$, shows that $u\geq V-T$; thus $x\leq T$ off the pole set. If $T\leq 2$ the claim is immediate. We may therefore suppose $T>2$; in particular $B>0$.
Fix a real number $s$ with $0<s<T-2$. For $t\geq 0$, define
\[ g_ s(t)= \begin{cases} \bigl ((1+t)^{3/2}\mu _ u(\{ x>t\} )\bigr )^{-1},& 0\leq t\leq T-s,\\ (1+t)^{-3/2},& t>T-s, \end{cases} \quad F_ s(t)=1+\int _0^ t g_ s(a)\, \mathrm{d}a, \]
and set
\[ W_ s(t)=\int _0^ t F_ s(a)^{1/N}\, \mathrm{d}a, \qquad \chi _ s(-t)=-W_ s(t). \]
The tail $\mu _ u(\{ x>t\} )$ in the first denominator has a positive lower bound on $[0,T-s]$. Hence $F_ s$ is locally Lipschitz, and $\chi _ s$ has the regularity and concavity required in the lemma.
Tonelli's theorem gives
9761605EDA18
10.31.16
\begin{equation} \int _ X F_ s(x)\, \mu _ u =\mu _ u(X)+\int _0^ Tg_ s(t)\mu _ u(\{ x>t\} )\, \mathrm{d}t \leq 3\max (1,M_ V), \end{equation}
since $\int _0^\infty (1+t)^{-3/2}\, \mathrm{d}t=2$ and the tail is at most $M_ V$ on the second interval. Chebyshev's inequality gives $\mu _ u(\{ x>t\} )\leq B/t$. Integrating first for $F_ s$ and then for $W_ s$, we obtain
430EB5E01BC8
10.31.17
\begin{equation} W_ s(t)\geq c(1+B)^{-1/N}t^{\beta }, \qquad 1\leq t\leq T-s, \qquad \beta =1+\frac1{2N}. \end{equation}
Indeed, $g_ s(t)\geq cB^{-1}t^{-1/2}$ for $t\geq 1$ in this interval, so $F_ s(t)\geq c(1+B)^{-1}t^{1/2}$. For $t\geq 2$ integrate this estimate over $[t/2,t]$; for $1\leq t\leq 2$ use $W_ s(t)\geq t$.
Let $p_ s=P_\theta (V-W_ s(x))$, the envelope of the preceding lemma for $\chi =\chi _ s$. The lemma, the bound $W_ s(t)\leq tF_ s(t)^{1/N}$ and the Cauchy–Schwarz inequality give, since $N=2n+1$,
\[ \begin{split} \int _ X(V-p_ s)^{1/2}\, \mu _{p_ s} & \leq \int _ X x^{1/2}F_ s(x)^{(n+1/2)/N}\, \mu _ u\\ & \leq B^{1/2}\left(\int _ XF_ s(x)\, \mu _ u\right)^{1/2} \leq C B^{1/2}. \end{split} \]
Write $a_ s=\sup _ Xp_ s\leq 0$. By the definition of $V$, $p_ s-a_ s\leq V$, so $V-p_ s\geq -a_ s$. Since $\mu _{p_ s}(X)\geq \ell _ V$, the last estimate yields $\ell _ V\sqrt{-a_ s}\leq C\sqrt B$, and hence $-a_ s\leq CB$.
The definition of $A$ now gives
\[ \int _ XW_ s(x)\, \mathrm{d}\nu \leq \int _ X(V-p_ s)\, \mathrm{d}\nu \leq A-a_ s\leq A+CB. \]
Combining this with (10.31.17), and then letting $s\downarrow 0$ for each fixed $1\leq t<T$, we obtain
\[ \nu (\{ x>t\} )\leq C(1+B)^{1/N}(A+B)t^{-\beta }. \]
There is no tail for $t\geq T$. The layer-cake formula therefore gives
\[ \int _ Xx^ r\, \mathrm{d}\nu =r\int _0^\infty t^{r-1}\nu (\{ x>t\} )\, \mathrm{d}t \leq 1+C(1+B)^{1/N}(A+B), \]
because $1<r<\beta $. This implies the stated bound.
Finally $-u=-V+(V-u)$, where $-V\in L^1(\nu )$ because $A<\infty $. The first summand is fixed and integrable, and the second has a uniform $L^ r(\nu )$ bound. This proves uniform integrability; no $L^ r$ bound for $V$ itself is claimed.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.