14.19.30 Smoothness of bounded solutions6CE6BC329E93

Proof. The extrema of $u$ give the first bound, as in theorem 14.7.2. Put $k=g+(v_{i\bar j})$, $S_ v=\operatorname {tr}_ gk$, and $D=2\| G\| _{C^0}$. The chain rule and Cauchy–Schwarz give, for $H=u+G(x,v)$,

Indeed $\Delta _ gu=S-n\geq -n$, $\Delta _ gv=S_ v-n$, and the mixed derivatives of $G(x,v)$ are bounded below by $-C_0(1+|\partial v|_ g^2)$. Here and below constants depend only on the declared data.

Set $r=v+M+1$, $R=2M+1$ and $s=u-\inf _ Xu+1$, so $1\leq r\leq R$ and $1\leq s\leq D+1$. For constants $A,L>0$ to be chosen, maximize

on $X$. If $S<1$ at the maximum, the bounded ranges of $r,u,s$ already bound $S$ everywhere. Otherwise lemma 14.5.1 applies there. Positivity gives

The middle inequality follows by diagonalizing $h$ relative to $g$: every eigenvalue of $h$ is at most $S$, and the diagonal entries of $k$ are positive. Consequently,

At the maximum,

Since $|\tau |_ h^2\leq C_ gB$, Young's inequality bounds the torsion term by

Also

Combining these inequalities with (14.5.2) yields

Choose $A>C_0+1$ and then $L>2AR+C_2+2$. If the fraction in the coefficient of $B$ is at least one, $S^2$ at the maximum is bounded by its numerator. Otherwise the displayed inequality gives $B\leq n(L+1)$ there. For $n\geq 2$, the bounded determinant gives

For $n=1$, $S=\mathrm e^{u+G(x,v)}$ directly. Thus $S$ is bounded at the maximum in every case. Maximality of $Q$ and the bounded ranges of $r,u,s$ give the same bound globally. The positive lower determinant bound then bounds every eigenvalue of $h$ away from zero. $\square$

Proof. The density $\mathrm e^{F(x,u)}$ is bounded and strictly positive, so corollary 14.19.29 makes $u$ continuous. Choose smooth strictly $\omega $-plurisubharmonic $v_ j\downarrow u$ by theorem 5.32.1. They have a common bound $M$ and converge uniformly by Dini's theorem. Define

By theorem 14.7.2, there is a smooth strictly $\omega $-plurisubharmonic $a_ j$ with

Put $\delta _ j=\| G(x,v_ j)-G(x,u)\| _{C^0}\to 0$. The given $u$ solves the proper equation with coefficient $G(x,u)$. The weighted measures of $a_ j-\delta _ j$ and $a_ j+\delta _ j$ bracket its coefficient, so corollary 14.19.12 gives

Hence $a_ j\to u$ uniformly. By lemma 14.19.31, $C^{-1}\omega \leq \omega _{a_ j}\leq C\omega $ uniformly. In particular $\Delta _ ga_ j$ is uniformly bounded. Fixed-operator elliptic estimates give uniform $W^{2,q}$ bounds for every finite $q$, so $u\in C^{1,\alpha }(X)$ for every $0<\alpha <1$.

We explain the uniformly elliptic passage to higher regularity, using the concave extension from corollary 14.5.8. On a relatively compact coordinate chart, let $P$ send a real Hessian to its complex Hermitian part, with the coordinate normalization of $\mathrm{dd}^{\mathrm{c}}$. All matrices $g(x)+P(D^2a_ j)$ lie in a fixed compact interval $\mathcal K=\{ \lambda I\leq A\leq \Lambda I\} $ of positive matrices. Define

The tangent inequality for the concave logarithmic determinant shows that this is the original operator whenever $g(x)+P(N)\in \mathcal K$. It is concave and uniformly elliptic on real symmetric matrices: for $N'\geq 0$, $P(N')\geq 0$ and its trace is a fixed positive multiple of $\operatorname {tr}N'$, while $A^{\pm 1}$ have uniform bounds. Its dependence on $x$ is uniformly Lipschitz on this chart. Thus the $a_ j$ are classical, and hence viscosity, solutions of

Both sides converge uniformly except for the Hessian variable. Stability of viscosity solutions gives $\mathcal F(x,D^2u)=u+G(x,u)$ in the viscosity sense. The right-hand side is Hölder because $u\in C^{1,\alpha }$. The local Evans–Krylov estimate therefore gives $u\in C^{2,\beta }$ for some $\beta >0$. The metric bounds pass to the limit, so the original equation is now uniformly elliptic with $\omega +\mathrm{dd}^{\mathrm{c}}u>0$. Differentiating it and applying Schauder estimates repeatedly proves $u\in C^\infty (X)$. $\square$