Proposition 16.28.2295F6C861B41

Proof. The solutions exist by theorem 14.7.2. Choose $A\geq 0$ with $\mathrm{dd}^{\mathrm{c}}f\leq A\omega $. At a maximum point of $u_\beta -f$, the second derivative test gives

Evaluating the equation at this point, we obtain

Thus $u_\beta -C_ f/\beta $ is a competitor for $P_\omega (f)$, and $u_\beta \leq p+C_ f/\beta $.

For the lower bound, note that $p$ is continuous by proposition 5.40.3, and choose smooth strictly $\omega $-plurisubharmonic functions $v_ j\downarrow p$ by theorem 5.32.1; by Dini's theorem the convergence is uniform. For any $\delta >0$, choose $j$ so that

Then $v_ j-\delta \leq f$, since $p\leq f$. Since $\omega _{v_ j}>0$ and $X$ is compact, there is $c_ j>0$ such that $\omega _{v_ j}^ n\geq c_ j\omega ^ n$. Put $C_ j=\max (0,-\log c_ j)$. The function

satisfies

Exponential comparison, corollary 14.19.12 with $\lambda =\beta $, gives $w_{j,\beta }\leq u_\beta $. Consequently

Letting first $\beta \to \infty $ with $j$ fixed, so that no lower bound for $c_ j$ uniform in $j$ is needed, and then $\delta \downarrow 0$, we obtain uniform convergence. Bounded Bedford–Taylor continuity then gives weak convergence of the measures.

On a compact subset of the open set $\{ p<f\} $, uniform convergence gives $u_\beta -f\leq -\epsilon $ for some $\epsilon >0$ and all large $\beta $. There the measures $\omega _{u_\beta }^ n$ have densities at most $\mathrm{e}^{-\beta \epsilon }$ with respect to $\omega ^ n$. Testing against continuous functions with compact support in $\{ p<f\} $ and passing to the limit gives $\omega _ p^ n=0$ on this open set. $\square$