Lemma 11.44.15FA5BBA96CCC3

Proof. The centered potential satisfies $\mathrm{MA}_\theta (w_\lambda )=c\mathrm{e}^{\lambda w_\lambda }f\, \mathrm{d}V$, while $\psi \leq 0$ gives $\mathrm{MA}_\theta (\psi )\geq c\mathrm{e}^{\lambda \psi }f\, \mathrm{d}V$. Part (2) of corollary 15.13.3 therefore gives $\psi \leq w_\lambda $, hence $k_\lambda \geq 0$. For each fixed $\lambda $, the density $\mathrm{e}^{\lambda w_\lambda }f$ lies in $L^ p$, and its critical scalar is $c$. Therefore

The moment estimate above gives $aw_\lambda \in L^\ell $, so proposition 11.44.9, with $z=w_\lambda $ and $t=\lambda $, gives the asserted integral inequalities. Fix one $\nu \in \mathcal P_ f$. Then

by (11.44.13), with $K$ independent of $\lambda $. The normalization $\sup _ Xv_\lambda =0$ gives $v_\lambda \leq V_\theta $, and subtracting $k_\lambda $ from the lower bound $\psi \leq w_\lambda $ proves (11.44.16).

The potentials $v_\lambda =u_\lambda -\sup _ Xu_\lambda $ form the normalized exponential family of proposition 15.17.3, which gives subsequences converging locally uniformly on $\Omega $ and in $L^1(X)$ to potentials of minimal type. Extracting a convergent subsequence of $k_\lambda \in [0,K]$ gives the same compactness for $w_\lambda =v_\lambda +k_\lambda $. The bound $|w_\lambda |\leq |\psi |+K$ also gives $L^1$ convergence along every subsequence converging locally uniformly on $\Omega $, since $X\setminus \Omega $ has zero volume. $\square$