Corollary 17.4.585CFF04E2CEC

Proof. For $u$ in the first set, lemma 16.22.2 gives $u\leq \phi +\sup _ Xu$, so monotonicity and the shift formula give $-B\leq E_\theta ^\phi (u)\leq m\sup _ Xu$. Also $\sup _ Xu\leq \sup _ X\phi =0$. Hence the set lies in $\{ u\in \mathrm{PSH}(X,\theta ):-B/m\leq \sup _ Xu\leq 0\} $, which is compact in $L^1(X)$ by proposition 10.3.1. The set is closed in $L^1(X)$: the inequality $u\leq \phi $ passes to $L^1$ limits by lemma 17.4.3, since $\phi $ is upper semicontinuous, and the energy bound passes to limits by upper semicontinuity (proposition 17.4.4). It is therefore compact. Its members have finite energy, so they lie in $\mathcal E^1(X,\theta ,\phi )$ by theorem 17.3.14.

The set $K_\phi $ is contained in $\{ u\in \mathrm{PSH}(X,\theta ):\sup _ Xu=0\} $, which is compact in $L^1(X)$ by proposition 10.3.1, and it is closed in $L^1(X)$, because the supremum is continuous for $L^1$ convergence of potentials (by the same proposition) and the inequality $u\leq \phi $ passes to $L^1$ limits by lemma 17.4.3. Hence $K_\phi $ is compact. $\square$