Theorem 15.7.2B0FDBEE3B071

Proof. Step 1: boundary barriers. We first take $\varphi $ smooth and $0<\varepsilon \leq 1$, and let $v_\varepsilon $ be the solution of lemma 15.7.1 with the constant density $F\equiv \varepsilon $. The strict subsolution $\underline v$ constructed there for $F\equiv 1$ lies below every $v_\varepsilon $, by smooth comparison. Choose a smooth function $q$ with $q=\varphi $ on the boundary and $\Delta q=-C$, where $C$ is so large that $\theta +\mathrm{dd}^{\mathrm{c}}v_\varepsilon \geq 0$ implies $\Delta v_\varepsilon \geq -C$. The maximum principle for the Laplacian gives $v_\varepsilon \leq q$. Both barriers are smooth and equal $\widetilde\varphi $ on the boundary. Consequently

with $C_1$ independent of $\varepsilon $.

Step 2: a Lipschitz bound independent of $\varepsilon $. Let $h\in \mathbb C^ n$ be small, and set $D_ h=\{ z\in B:z+h\in B\} $. Since $\theta $ is smooth, there is $C_2>0$, independent of the translation $h$, such that

For a constant $C_3>0$ to be fixed, define on $D_ h$

Then $\theta +\mathrm{dd}^{\mathrm{c}}w\geq \theta (z+h)+\mathrm{dd}^{\mathrm{c}}v_\varepsilon (z+h)>0$, and the determinant of $\theta +\mathrm{dd}^{\mathrm{c}}w$ is at least $\varepsilon \beta ^ n$, because the Euclidean volume form is translation invariant. On $\partial D_ h$, one of the points $z$ and $z+h$ lies on $\partial B$, and the other has distance at most $|h|$ from $\partial B$. Estimate (15.7.3) and the Lipschitz bound for $\widetilde\varphi $ therefore give $w\leq v_\varepsilon $ on $\partial D_ h$ if $C_3$ is chosen sufficiently large, and smooth comparison extends this inequality to all of $D_ h$. Together with the same argument for $-h$, this gives

with constants independent of $\varepsilon $. The same bound for arbitrary pairs of points follows by subdividing the segment joining them inside the convex ball, and it extends to $\overline B$ because each $v_\varepsilon $ is continuous there.

Step 3: the limit $\varepsilon \to 0$. Smooth comparison also shows that $v_\varepsilon $ increases as $\varepsilon $ decreases. By the barriers and the common Lipschitz bound, the limit $H_\varphi $ is Lipschitz and the convergence is uniform on $\overline B$. The inequality $\theta +\mathrm{dd}^{\mathrm{c}}v_\varepsilon >0$ of currents passes to the limit, and continuity of local bounded products under uniform convergence gives $(\theta +\mathrm{dd}^{\mathrm{c}}H_\varphi )^ n=0$. For smooth boundary values $\varphi ,\psi $, denote the solutions of lemma 15.7.1 with $F\equiv \varepsilon $ by $v_{\varepsilon ,\varphi }$ and $v_{\varepsilon ,\psi }$. Smooth comparison gives $\| v_{\varepsilon ,\varphi }-v_{\varepsilon ,\psi }\| _\infty \leq \| \varphi -\psi \| _\infty $, and passing to the limit proves the contraction for smooth data.

Step 4: continuous boundary values. For continuous $\varphi $, choose smooth $\varphi _ j\to \varphi $ uniformly on $\partial B$. By the contraction, $H_{\varphi _ j}$ is uniformly Cauchy. Its limit $H_\varphi $ is continuous and $\theta $-plurisubharmonic, has the prescribed boundary values, and satisfies the homogeneous equation, again by continuity of bounded products under uniform convergence. Passing to the limit also proves the contraction for all continuous boundary data.

Step 5: domination. It remains to prove the domination property, which does not follow from a comparison of total masses. First take $\varphi $ smooth and $p$ as in the statement. For $\delta >0$ put $p_\delta =p+\delta (|z|^2-R^2)$, so that $\theta +\mathrm{dd}^{\mathrm{c}}p_\delta \geq \delta \beta $. By lemma 5.24.9, every $C^2$ function $a$ touching $p_\delta $ from above at an interior point $z_0$ satisfies

Suppose that $p_\delta -v_\varepsilon $ had a positive maximum. Since this difference is nonpositive on the boundary, the maximum would be attained at an interior point, where $v_\varepsilon $ plus the maximum touches $p_\delta $ from above. The determinant of $\theta +\mathrm{dd}^{\mathrm{c}}v_\varepsilon $ at that point would then be at least $\delta ^ n\beta ^ n$, contradicting the equation for $v_\varepsilon $ when $\varepsilon <\delta ^ n$. Hence $p_\delta \leq v_\varepsilon $ for small $\varepsilon $. Letting first $\varepsilon $ and then $\delta $ decrease to zero, we obtain $p\leq H_\varphi $.

Finally, for continuous $\varphi $, take the smooth approximants $\varphi _ j$ above and set $a_ j=\| \varphi _ j-\varphi \| _\infty $. The boundary values of $p-a_ j$ are at most $\varphi _ j$, so the result for smooth data gives $p-a_ j\leq H_{\varphi _ j}$. Uniform convergence proves $p\leq H_\varphi $. $\square$