15.12.6 The normalized equation21A9095B1925

of theorem 15.12.1, part (2). For each $j\geq 1$, part (1) gives $u_ j\sim \varphi $ with

Write $M_ j=\sup _ Xu_ j$, $v_ j=u_ j-M_ j$ and $c_ j=\mathrm{e}^{M_ j/j}$. Let $h$ be a bounded normalized solution of $\mathrm{MA}_\theta (h)=\alpha f\, \mathrm{d}V$, with $\alpha >0$, given by theorem 15.3.5.

For $H_ j=h+\| h\| _\infty +j\log \alpha $ we have

The bounded function $H_ j$ admits a smaller-background envelope, and $u_ j\preceq H_ j$, so corollary 15.2.3 gives $u_ j\leq H_ j$. In particular,

The lower density bound of lemma 15.12.2, applied to $v_ j$ with $q=p$, gives

so the $c_ j$ are also bounded away from zero.

Since $\sup _ Xv_ j=0$ and $\mathrm{MA}^{\mathrm{ext}}_\theta (v_ j)=c_ j\mathrm{e}^{v_ j/j}f\, \mathrm{d}V\leq C f\, \mathrm{d}V$, theorem 15.12.3 gives $\varphi -C'\leq v_ j\leq \varphi $. Extract a subsequence with $v_ j\to v$ in $L^1(X)$ and $c_ j\to c>0$. By corollary 15.9.4, the convergence holds in capacity, $v\sim \varphi $, and the Monge–Ampère measures converge weakly. Meanwhile, lemma 5.44.5 gives

Thus the right-hand sides converge in total variation to $cf\, \mathrm{d}V$, and $\mathrm{MA}^{\mathrm{ext}}_\theta (v)=cf\, \mathrm{d}V$.

For the uniqueness of the constant, suppose that $a,b\sim \varphi $ solve the normalized equation with constants $c_ a<c_ b$. For every real $t$,

so theorem 15.2.2 gives $a\geq b+t$. This is impossible as $t\to +\infty $, at any point where both potentials are finite. Interchanging the roles of the two constants proves that $c$ is unique. $\square$