12.9.11 Uniformly controlled complex Hessians5603574CC3A6

Proof. We first estimate positive and negative powers, then compare their small moments by an estimate for $\log u$.

For $p\ne 0,1$, test the weak equation by $\eta ^2u^{p-1}$, where $\eta $ is a smooth cutoff. The principal term is

The other terms contain respectively $2\eta u^{p-1}A\nabla u\cdot \nabla \eta $, $(p-1)\eta ^2u^{p-1}B\cdot \nabla u$, and $2\eta u^ pB\cdot \nabla \eta $. Taking absolute values after moving the principal term to one side and applying Young's inequality absorbs half of its energy. Thus, for either sign of $p$,

with $C_1$ depending only on the ellipticity bounds. Each test is legitimate since this individual smooth positive solution is bounded above and away from zero on the cutoff support.

Put $\beta =m/(m-2)>1$ and choose $p_0=\beta ^{-k-1/2}$, with $k$ to be fixed later. Apply the Euclidean Sobolev inequality to $\eta u^{p/2}$ on nested cubes shrinking from $Q$ to $Q/2$. Choose the cutoff gaps proportional to $2^{-j}$ and the powers $p_ j=p_0\beta ^ j$. The positive powers never equal one: the closest two are $\beta ^{-1/2}$ and $\beta ^{1/2}$. Consequently the constants in (12.9.13), together with the cutoff terms from $\nabla (\eta u^{p/2})$, grow at most polynomially in $p_ j$ and exponentially in $j$. Since $\sum _ j(j+1)\beta ^{-j}<\infty $, their iteration product converges. Negative powers give the same calculation for $u^{-1}$. We obtain

Here $C_2$ also depends on $p_0$ and the fixed cube size.

Testing instead by $\eta ^2/u$ gives

For every subcube $S\subset Q$ of side length $\ell $, a cutoff supported in $2S\subset 3Q$, equal to one on $S$, has derivative bounded by $C/\ell $. Poincaré's inequality and Cauchy–Schwarz therefore give

where $K_0\geq 1$ is uniform and $(\log u)_ S$ denotes the mean. We spell out the exponential consequence of this mean oscillation bound.

Write $f=\log u$ and fix $a>2$. Select the maximal dyadic subcubes $S$ of $Q$ on which $\fint _ S|f-f_ Q|>aK_0$. Their total measure is at most $|Q|/a$. The parent of each selected cube has average at most $aK_0$, hence $|f_ S-f_ Q|\leq 2^ maK_0$. Outside their union, differentiation of dyadic averages gives $|f-f_ Q|\leq aK_0$ almost everywhere. Repeat this selection inside each chosen cube, subtracting its own mean. The union at generation $j$ has measure at most $a^{-j}|Q|$, and outside generation $j+1$ the accumulated mean changes and last oscillation give

The intersection of all generations is null. Choosing $c>0$ so that $\exp (c2^ ma)<a$ and summing over the generations yields

with dimensional constants. Now fix $k$ large enough that $p_0<c/K_0$. Multiplying the estimates for the positive and negative exponentials cancels $f_ Q$ and gives

Combining this with (12.9.14) proves the result. Every constant comes from the stated coefficient bounds and the fixed cube. $\square$

Proof. Existence and uniqueness of $\rho _ v$ follow from theorem 11.4.2. Fix a real reference metric on $X$, and put $Q_ v=\Omega _ v^{n-1}$ and $J_ v=\Omega _ v^ n/\mathrm{d}V$. Since $d\Omega _ v=d\omega $,

Both $Q_ v$ and $dQ_ v$ have uniformly bounded coefficients in fixed charts.

Define a real tensor $A_ v$ and vector field $b_ v$ by

Only the $(1,1)$ part of $d\psi \wedge \mathrm d^{\mathrm c}\phi $ contributes to the first identity; it is

Thus $A_ v$ is symmetric and uniformly positive definite, with a uniform upper bound. The second identity and (12.9.16) give a uniform bound for $b_ v$.

For $L_{\Omega _ v}\phi =n\mathrm{dd}^{\mathrm{c}}\phi \wedge Q_ v/\Omega _ v^ n$, Stokes' theorem gives

In particular, the adjoint equation for $\rho _ v$ is

In each coordinate chart, multiplying the coefficients by the fixed smooth volume density converts this to Euclidean divergence form while preserving uniform ellipticity and a uniform bound for the vector coefficient. Lemma 12.9.12 applies. A fixed finite cover by inner cubes has a connected overlap graph, since $X$ is connected. Chaining its local estimates gives $\sup _ X\rho _ v\leq C_ H\inf _ X\rho _ v$. The normalization and $\mathrm{d}V(X)=1$ imply the claimed two-sided bound. $\square$

Proof. Use the normalization of proposition 12.9.15 and put

The factor equation and $\mathrm{dd}^{\mathrm{c}}$-Stokes give

Hence $T_ v\in \mathscr G_\omega $. It is the $(n-1)$st power of the Gauduchon metric $\gamma _ v=(\rho _ v/N_ v)^{1/(n-1)}\Omega _ v$. The equation $\Omega _ v^ n=c_ v\mathrm e^{F_ v}\mathrm{d}V$ gives $B_0(F_ v)=\log c_ v$. Thus theorem 12.7.3 identifies $T_ v$ as the unique exact maximizer, and $M(F_ v)=c_ v^{-1/n}$. The determinant and probability normalizations are therefore

Here $j_-\leq J_ v\leq j_+$ and, since $\int _ X\rho _ v\, \mathrm{d}V=1$, also $j_-\leq N_ v\leq j_+$. The factor bound gives the first pair of inequalities in (12.9.19); the identity $k_{T_ v}=J_ v^{-1/n}r_ v$ gives the second. The critical norms are bounded by the respective upper constants because $\mathrm{d}V$ is a probability volume.

The inequalities

give the bound for $F_ v-F_0$ by subtracting logarithms. To prove nontriviality, choose a nonconstant smooth real $\phi $. For small $t$, the potential $t\phi -\sup _ X(t\phi )$ belongs to $\mathcal V_\varepsilon $ and gives the same density as $t\phi $. Differentiation at zero gives

This function is nonconstant: otherwise $L_\omega \phi $ would be constant, its values at a maximum and a minimum would force that constant to be zero, and the strong maximum principle would make $\phi $ constant. Choose two points where the displayed derivative differs. The difference of the densities at these points is strictly monotone for $t$ in a small interval. These give infinitely many distinct densities, with at most one parameter having equal values at the chosen points. Discarding that parameter gives the nonconstant family. $\square$

Proof. Choose holomorphic coordinates centered at a point $p$, put $x=\operatorname {Re}z_1$, and take a smooth real cutoff $\zeta $ supported in the chart and equal to one near $p$. We may assume $0\leq \zeta \leq 1$. On the chart set

This is a nonzero semipositive form of rank one. For a constant $a>0$ to be chosen, extend

by zero outside the chart, and put $v_ j=w_ j-\sup _ Xw_ j$. The product rule gives

All fixed forms in this expression have compact support after multiplication by the indicated cutoff or its derivatives. Thus a fixed $C>0$ bounds the operator norm of $\mathrm{dd}^{\mathrm{c}}w_ j$ relative to $\omega $ by $aC$ for every $j\geq 1$. Choose $a$ so small that $aC\leq \varepsilon /2$. Then $v_ j\in \mathcal V_\varepsilon $, and $\| v_ j\| _{C^0}\leq 2a/j^2$.

Write $s=\operatorname {tr}_{\omega (p)}\beta >0$. Since $x(p)=0$ and $\zeta =1$ near $p$, we have $\Omega _{v_ j}(p)=\omega (p)$ and

Differentiating the logarithmic determinant, and observing that its normalizing integral is constant in the spatial variable, gives the exact identity

These derivatives are pairwise different and their absolute values tend to infinity. Hence the densities are pairwise distinct and their $C^1$ norms are unbounded. Their uniform $C^0$ bound is part of theorem 12.9.18.

To check failure of equicontinuity, take points $p_ j^\pm $ with $x=\pm \pi /(2j)$ and all other real coordinates zero. For large $j$ they lie where $\zeta =1$, and $\Omega _{v_ j}(p_ j^\pm )=\omega (p_ j^\pm )\mp a\beta $. The integral normalizing $F_{v_ j}$ cancels from the difference, so

Indeed, $\beta ^2=0$ gives $(\omega (p)\pm a\beta )^ n=(1\pm as)\omega (p)^ n$, and $as<1$ by the chosen form comparison. Both points tend to $p$, which proves the failure of equicontinuity. The exact-adjoint bounds still apply because the entire sequence lies in $\mathcal V_\varepsilon $. $\square$