15.12.5 Existence for a positive exponent3EFAC1693CBD

of theorem 15.12.1, part (1). Step 1: a maximizing sequence. Fix $\lambda >0$. By lemma 15.12.4, there is $w\sim \varphi $ such that

Consider all potentials $z\sim \varphi $ for which

and let $B$ be the supremum of their weighted masses $\nu _ z(X)$. This family is nonempty, since it contains $w$, and $B\leq \mu (X)<+\infty $. Choose $u_ j$ in the family with $\nu _{u_ j}(X)\to B$. Since $\nu _0\leq \nu _{u_ j}$, corollary 15.2.3 gives $u_ j\leq w$.

Step 2: uniform bounds. We need a lower bound for the suprema of the $u_ j$. Put $\varphi _\varepsilon =P_{\varepsilon \theta }(\varphi )\leq 0$, and let $\psi $ be the bounded solution of

given by theorem 15.3.3. For $h=\psi +\varphi _\varepsilon $, positivity of the splitting $\theta _ h=(\varepsilon \theta +\mathrm{dd}^{\mathrm{c}}\varphi _\varepsilon )+((1-\varepsilon )\theta +\mathrm{dd}^{\mathrm{c}}\psi )$ gives

Here $h\preceq u_ j$, because $h\preceq \varphi $ and $u_ j\sim \varphi $, and each $u_ j$ admits a smaller-background envelope because $u_ j\sim \varphi $. Corollary 15.2.3 therefore gives $u_ j\geq h$. Consequently

and the lower bound is finite.

We now apply theorem 15.12.3 to $v_ j=u_ j-M_ j$. Their measures satisfy

so $\varphi -C\leq v_ j\leq \varphi $. Together with the bounds for $M_ j$, this shows that all $u_ j$ lie within a fixed bounded distance of $\varphi $.

Step 3: the limit. Extract an $L^1$ limit $u$. By corollary 15.9.4, applied with the dominating measure $\mathrm{e}^{\lambda \sup _ Xw}\mu $, the convergence holds in capacity, $u\sim \varphi $, and

After a further extraction we may also assume $\nu _{u_ j}\rightharpoonup \nu $. Since $\nu _0\leq \nu _{u_ j}\leq \mu $, the limit satisfies the same inequalities. We claim that

Indeed, $\mathrm{e}^{\lambda u_ j}\to \mathrm{e}^{\lambda u}$ in $L^1(\mu )$ by lemma 5.44.2, and all the measures $\nu _{u_ j}$ are at most $\mu $; this controls the error caused by changing the weight. The fixed bounded weight $\mathrm{e}^{\lambda u}$ can be approximated in $L^1(\mu )$ by continuous functions, and since both $\nu _{u_ j}$ and $\nu $ are dominated by $\mu $, this approximation is uniform in the integrals. The weak convergence $\nu _{u_ j}\rightharpoonup \nu $ then gives the claim, and the same argument applies after multiplication by any continuous test function.

Since $\mathrm{e}^{\lambda u_ j}\nu _{u_ j}=\mathrm{MA}^{\mathrm{ext}}_\theta (u_ j)$, it follows that

The second identity holds off the pole set of $u$, which has zero $\mu $-measure. Hence $u$ belongs to the family and attains the supremum $B$. Write $\nu =g\, \mathrm{d}V$, with $0\leq g\leq f$.

Step 4: no deficit. Suppose that $g\neq f$ in $L^1(\mathrm{d}V)$. Applying lemma 15.12.4 to $f-g$, we find $v\sim \varphi $ with

This measure is nonzero. In fact, no potential $v$ admitting a smaller-background envelope has zero Monge–Ampère measure, since theorem 15.2.2, applied with coefficient zero to $v$ and $v+1$, would give $v\geq v+1$. Thus $\nu _ v(X)>0$.

The sum $\nu +\nu _ v$ is finite, so theorem 15.11.1 gives $a\sim \varphi $ with

This measure still lies between $\nu _0$ and $\mu $, but its mass is $B+\nu _ v(X)>B$, a contradiction. Therefore $g=f$, and $u$ solves the required equation. Uniqueness in the prescribed ordinary type follows from corollary 15.2.3. $\square$