14.19.42 Stability for concave equationsA73917A5BC62

Proof. Each $f_{\infty ,i}$ is an increasing limit of positive concave functions, hence concave with values in $(0,+\infty ]$. Suppose it is infinite at one point and finite at another point of $\Gamma $. Extend the segment joining them slightly beyond the second point, inside the open cone. The second point is then an interior point of the longer segment, so concavity forces the value there to be infinite, a contradiction. Thus $f_{\infty ,i}$ is either finite everywhere, and then continuous, or infinite everywhere. By symmetry, the same alternative holds for every $i$.

Take $\delta =\frac18\min \{ \delta _0,\min _ Xf(\mu )\} >0$. In the finite case set $H=\min \{ H_0,f_\infty (\mu )-4\delta \} $; otherwise set $H=H_0$. Since $f_\infty (\mu )\geq f(\mu )\geq 8\delta $, this function is positive. The conditions on the family give $h\leq H$ and $f_\infty (\mu )\geq H+4\delta $. Since $\mu (X)$ is a compact subset of $\Gamma $ and the functions $f_{\infty ,i}$ are continuous (or identically infinite), a sufficiently small $r>0$ gives $\mu -2r\mathbf1\in \Gamma $ and $f_\infty (\mu -2r\mathbf1)\geq H+3\delta $ on $X$.

If the matrix bound failed, there would be points $x_ j\to x_*$ and $P_ j\geq 0$ with unbounded norms and $F(A_{\underline u}(x_ j)-r\operatorname {Id}+P_ j)\leq H(x_ j)+2\delta $. By monotonicity of eigenvalues, the ordered eigenvalues $\lambda ^{(j)}$ of $A_{\underline u}(x_ j)-r\operatorname {Id}+P_ j$ satisfy $\lambda _ i^{(j)}\geq \mu _ i(x_ j)-r$. Their sum tends to infinity, so some component $k$ tends to infinity along a subsequence. For every fixed $t>0$, eventually

Since $f$ is increasing in each variable, letting $t\to \infty $ shows that the lower limit of $F(A_{\underline u}(x_ j)-r\operatorname {Id}+P_ j)$ is at least $f_\infty (\mu (x_*)-2r\mathbf1)\geq H(x_*)+3\delta $, which contradicts the upper bound $H(x_ j)+2\delta $. Adding positive semidefinite matrices preserves the cone, so all these comparisons take place in $\Gamma $. $\square$

Proof. Put $\theta =\chi +\mathrm{dd}^{\mathrm{c}}\underline u$ and $w=u-\underline u-\sup _ X(u-\underline u)\leq 0$. Since $\Gamma \subseteq \Gamma _1$, we have $\operatorname {tr}_\omega \mathrm{dd}^{\mathrm{c}}w\geq -\operatorname {tr}_\omega \theta \geq -C$. The Green-kernel argument in the proof of proposition 14.11.1, applied to the operator $\operatorname {tr}_\omega \mathrm{dd}^{\mathrm{c}}$, therefore bounds $\| w\| _{L^1}$ uniformly.

Center a holomorphic coordinate ball of fixed size at a minimum point of $w$, and rescale it to the real unit ball. For a fixed sufficiently small $\varepsilon >0$, set $q=w+\varepsilon |x|^2$ and $m=\min _ Xw$. The real contact-set construction of (14.11.2) gives a compact set $E$ on which

On $E$, $\mathrm{dd}^{\mathrm{c}}w\geq -C\varepsilon \omega $. Choose $\varepsilon $ so that $C\varepsilon <r$. Then $P=\omega ^{-1}(\mathrm{dd}^{\mathrm{c}}w+r\omega )\geq 0$ and

Lemma 14.19.43 bounds $P$, and hence the complex trace of the Hessian of $q$. The nonnegative real Hessian of $q$ on $E$ therefore has bounded trace, and hence bounded determinant, so the contact inequality implies $|E|\geq c>0$. Since $w\leq m+\varepsilon /2$ on $E$, the $L^1$ bound gives $c(-m-\varepsilon /2)\leq C$. This bounds the oscillation of $w$, hence that of $u$; with the normalization $\inf _ Xu=0$, this is the assertion. $\square$

Proof. Set $a=u_1-\underline u$, $b=u_2-\underline u$ and $\theta =\chi +\mathrm{dd}^{\mathrm{c}}\underline u$. By proposition 14.19.44, $|a|,|b|\leq C_0$ for a fixed $C_0\geq 1$. It suffices to bound $s=\max _ X(b-a)$ when it is positive; the opposite difference is handled by interchanging the solutions.

Fix $r,R,\delta $ from lemma 14.19.43 and put $\alpha =r\omega $. For every smooth positive $\rho $ with

theorem 14.2.1 gives a smooth solution

At a maximum of $\psi $, $c\rho \leq 1$, so $c\leq 2$. Apply theorem 14.19.19 with $p=2$ and background $\alpha $. Its potential and lower-density bounds give fixed $c_0,C_\psi >0$ with

These constants depend only on the three bounds for $\rho $, and not on the cutoff function chosen below.

Choose $\kappa >0$ so that $2C_0\kappa \leq 1/4$, $\kappa C_0\leq 1/16$ and $\kappa C_\psi /2\leq 1/16$. Next choose $M\geq 1$ with $2^{-n}c_0Mr^ n>R^ n$, and then $\varepsilon _0>0$ with

If one of the inequalities

fails, then $s\leq C(\| h_1-h_2\| _\infty +\| u_1-u_2\| _{L^1})$, with $C$ depending only on $\kappa ,\delta ,\varepsilon _0$ and the volume of $X$. We assume both and derive a contradiction. Put $t=\kappa s$; since $s\leq 2C_0$, we have $t\leq 1/4$. On $\Omega =\{ (1-t)b-a>s/2\} $, we have $b-a>7s/16$, so by Chebyshev's inequality the relative volume of $\Omega $ is at most $16\varepsilon _0/7$.

Take a smooth $\zeta :\mathbb R\to [0,1]$, zero on $(-\infty ,1/2]$ and one on $[3/4,\infty )$, and set

The second term is supported in $\Omega $, so $1/2\leq b_0\leq 1$ and $\langle \rho ^2\rangle _\alpha \leq 2+2M^2(16\varepsilon _0/7)\leq 4$; here we use that averages with respect to $\alpha =r\omega $ and $\omega $ agree. Thus the auxiliary solution for this $\rho $ satisfies the bounds with the same constants $c_0,C_\psi $.

At a point where $b-a=s$, the function $\Phi =(1-t)b-a+t\psi /2$ is at least $7s/8$, by the choice of $\kappa $. At a maximum point $x_0$ of $\Phi $, therefore, $(1-t)b-a\geq 7s/8$ because $\psi \leq 0$, and so $\rho (x_0)\geq M$. The second derivative test gives, at $x_0$,

The identity

and concavity imply

Both forms in the convex combination are admissible, since $\mu -r\mathbf1\in \Gamma $ and $P>0$. By lemma 14.19.43, $P\leq R\omega $. On the other hand, $P\geq \frac12(\alpha +\mathrm{dd}^{\mathrm{c}}\psi )$, so the auxiliary equation gives

a contradiction. Hence one of the two smallness inequalities fails, which gives the required bound for $s$. Interchanging the solutions proves the theorem. $\square$

Proof. Admissibility and $\Gamma \subseteq \Gamma _1$ give $\operatorname {tr}_\omega \mathrm{dd}^{\mathrm{c}}u\geq -C$. By theorem 11.4.2, there is a smooth function $\rho _0>0$ with $\mathrm{dd}^{\mathrm{c}}(\rho _0\omega ^{n-1})=0$, so that $\int _ X\mathrm{dd}^{\mathrm{c}}\varphi \wedge \rho _0\omega ^{n-1}=0$ for every smooth $\varphi $. Integrating the product rule $\mathrm{dd}^{\mathrm{c}}(u^2)=2u\mathrm{dd}^{\mathrm{c}}u+2\mathrm du\wedge \mathrm d^{\mathrm c}u$ against $\rho _0\omega ^{n-1}$, and using $0\leq u\leq C$ from proposition 14.19.44, we obtain a uniform bound for $\int _ X|\mathrm du|_\omega ^2\omega ^ n$, because $\rho _0$ is bounded above and below. Rellich's theorem therefore gives precompactness in $L^1$.

Given a sequence of such solutions, the Arzelà–Ascoli theorem gives a subsequence whose right-hand sides converge uniformly, since they are equicontinuous and bounded by $H_0$. A further subsequence of the solutions is Cauchy in $L^1$, and theorem 14.19.45 makes it uniformly Cauchy. This proves precompactness in $C^0(X)$. A common modulus of continuity follows from a finite uniform $\varepsilon /3$-net, since each of its finitely many members is uniformly continuous. $\square$