Proposition 8.5.46FBA7BB8AC94

Proof. The first lemma gives $u\in W^{2,p}$ for every finite $p$ and a bounded complex Hessian. The matrices $A_ j=I+g^{-1/2}H(u_ j)g^{-1/2}$ converge weakly in $L^2$ to the analogous matrix $A$ for $u$. Mazur's lemma gives convex combinations of tails converging strongly in $L^2$, hence almost everywhere along a subsequence. The pointwise closed cone is convex, so $A\in C_{n,m}$ almost everywhere. The product lemma passes the equation to the limit as currents, since its right-hand side converges uniformly. Equality of the bounded densities then gives the equation almost everywhere.

The right-hand sides are bounded above and away from zero. The complex Hessian bound gives a uniform upper bound $L$ for $\operatorname {tr}A_ j$. Choose $f_0>0$ below all their normalized roots. Thus every $A_ j$, and almost every $A$, belongs to the compact set $K$ of the preceding lemma. Every $u_ j$ is a classical solution of

For completeness, if a smooth test $q$ touches $u$ from above at an interior point $x_0$, add a small positive quartic term to make the local contact strict. Local maximum points of $u_ j-q$ tend to $x_0$. At these points $D^2u_ j\leq D^2q$, and ellipticity gives $\mathcal G(x,D^2q)\geq r_ j(x)$. Pass to the limit, then remove the quartic term. A test from below gives the reverse inequality. The extension is defined on every real Hessian, so no admissibility of a lower test has been assumed. This proves the viscosity statement. $\square$