Pome

Theorem 15.35.98FF7FBD67552

Theorem 15.35.9.8FF7FBD67552 Let $f\geq 0$ be a nonzero integrable function on $X$, and put $F=f/R$.

  1. If $\lambda >0$ and $F\leq AB_\lambda $ almost everywhere for a constant $A>0$, then

    \[ s_\lambda =u_\alpha +\lambda ^{-1}\log (CV_0/A) \]

    is an exponential subsolution at exponent $\lambda $, of mass $CV_0$, and the equation $\mathrm{MA}^{\mathrm{ext}}_ H(u)=\mathrm{e}^{\lambda u}f\, \mathrm{d}V$ has a greatest solution, whose mass is finite and positive.

  2. Suppose that $F$ is locally essentially bounded on $X\setminus (X_0\times \{ o\} )$ and

    \[ \lim _{T\to +\infty }\operatorname *{ess\, sup}_{x\in X_0,\ L(y)>T} \frac{\log ^+(F(x,y)/q_\alpha (y))}{L(y)^\alpha }=0, \]

    where $\log ^+r=\log \max (1,r)$ for $r\geq 0$. Then the hypothesis of (1), and hence its conclusion, holds for every $\lambda >0$.

Proof. (1) By theorem 15.30.3, $\mathrm{MA}^{\mathrm{ext}}_ H(u_\alpha )=CV_0q_\alpha \mu $. Off the pole fiber $X_0\times \{ o\} $ we have $\mathrm e^{\lambda u_\alpha }B_\lambda =q_\alpha $ and $f\, \mathrm{d}V=F\mu \leq AB_\lambda \mu $, hence

\[ \mathrm e^{\lambda s_\lambda }f\, \mathrm dV \leq (CV_0/A)\mathrm e^{\lambda u_\alpha }AB_\lambda \mu =CV_0q_\alpha \mu =\mathrm{MA}^{\mathrm{ext}}_ H(s_\lambda ), \]

and both measures vanish on the pole fiber. Thus $s_\lambda $ is an exponential subsolution, and its mass is $CV_0$, since adding a constant does not change the Monge–Ampère measure. Theorem 15.35.2 now gives the greatest solution $u$ and the finiteness of its mass. The mass $\int _ X\mathrm{e}^{\lambda u}f\, \mathrm{d}V$ is positive because $u$ is finite almost everywhere and $f$ has positive integral.

(2) Fix $\lambda >0$. By hypothesis, there is $T$ such that $F/q_\alpha \leq \mathrm e^{(\lambda /2)L^\alpha }$ almost everywhere on $\{ L>T\} $. Since $\mathrm e^{(\lambda /2)L^\alpha }\leq \mathrm e^{\lambda }\mathrm e^{\lambda (L^\alpha -1)}$ for $L\geq 1$, we have $F\leq \mathrm e^{\lambda }B_\lambda $ there. On the complement $\{ L\leq T\} $, which is compact and disjoint from the pole fiber, $F$ is essentially bounded, while $B_\lambda \geq q_\alpha \geq b_0>0$. Increasing the constant, we obtain $F\leq AB_\lambda $ almost everywhere on $X$. $\square$

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