12.9 Integrability of Gauduchon determinant measuresD02094988A03

Proof. Fix $p>1$ and a smooth strictly positive function $h$. By theorem 14.2.1 there are a smooth form $\Omega _ h=\omega +\mathrm{dd}^{\mathrm{c}}u_ h>0$ and a constant $c_ h>0$ such that

The lower-density estimate of theorem 14.19.19 gives a constant $c_ p>0$, independent of $h$, such that

The pointwise matrix arithmetic–geometric mean inequality, applied to $\Omega _ h$ and the coefficient density of $T$, gives the inequality of measures

Integrating and using $\mathrm{dd}^{\mathrm{c}}T=0$ yields

Approximation by smooth strictly positive functions extends this bound to every continuous nonnegative $h$, and applied to $|h|$ it bounds the integral of every continuous function against $\Gamma _0(T)$ by its $L^{np}$ norm. Since continuous functions are dense in $L^{np}$, the duality between $L^{np}$ and $L^{np/(np-1)}$ shows that $\Gamma _0(T)$ has a density in $L^{np/(np-1)}$, with norm at most $c_ p^{-1/n}$. As $p$ ranges over $(1,+\infty )$, the exponent $np/(np-1)$ ranges over $(1,n/(n-1))$. This proves the theorem. $\square$

Proof. Suppose first that $v_-(\omega )>0$. For the smooth equation in the preceding proof, the definition of $v_-$ gives

The same pointwise inequality and Stokes' theorem therefore give

By duality, $\| k_ T\| _{L^{n/(n-1)}}\leq v_-(\omega )^{-1/n}$ for every $T\in \mathscr G_\omega $.

For the reverse bound, let $g$ be any smooth positive probability density relative to $\mathrm{d}V$, and solve

Let $T_ g$ be the trace-normalized $(n-1)$st power of the Gauduchon metric conformal to $\Omega _ g$. Equality in the pointwise matrix inequality gives

Since $\| g^{1/n}\| _{L^ n}=1$, Hölder's inequality gives $\| k_{T_ g}\| _{L^{n/(n-1)}}\geq c(g)^{-1/n}$. By theorem 12.15.4, theorem 12.2.3,

Taking the supremum over these smooth currents $T_ g$ proves the matching lower bound; if $v_-(\omega )=0$, the same argument makes this supremum infinite. Since the currents $T_ g$ are smooth, this also proves that allowing nonsmooth currents does not change the supremum. $\square$

Proof. Recall that $M(F)=\exp (-B_0(F)/n)$, and represent $\nu $ by a maximizing current $T\in \mathscr M(F)$, using theorem 12.8.2. Since $\Gamma _ F(T)=\mathrm{e}^{F/n}\Gamma _0(T)$ and $B_0$ is $1$-Lipschitz,

Theorem 12.9.1 therefore gives the subcritical bound with constant $\mathrm{e}^{2R/n}M(0)^{-1}C_ q$. When $v_-(\omega )>0$, the same argument with the endpoint constant $v_-(\omega )^{-1/n}$ of theorem 12.9.2 gives the case $q=n/(n-1)$. $\square$

Proof. For $\| v\| _{C^{4,\alpha }}\leq 1/(2D)$ put

The logarithm series starts at degree two. If the endomorphism norms of $\omega ^{-1}\mathrm{dd}^{\mathrm{c}}v$ and $\omega ^{-1}\mathrm{dd}^{\mathrm{c}}w$ are at most $\rho \leq 1/2$, telescoping their powers bounds the difference series by $C_{\mathrm{tr}}\rho /(1-\rho )$ times the endomorphism difference. Hence

On the closed radius-$r$ ball of normalized pairs $(v,c)$, the map

has image norm at most $K\varepsilon +KQr^2\leq 3r/4$ and Lipschitz constant at most $2KQr\leq 1/2$. Its fixed point $(u,c)$ satisfies

The estimate $Dr\leq 1/2$ gives the form comparison. The scalar is $b=\log V+c$. Differentiating the local equation gives a uniformly elliptic equation for each first derivative of $u$, with coefficients initially in $C^{2,\alpha }$. Interior Schauder estimates improve the regularity one order at a time, so $u$ is smooth. Uniqueness in theorem 14.2.1 identifies it with $u_ F$ and gives $b=B_0(F)$. No uniform higher-derivative bounds on $F$ enter the estimates above.

Put $\Delta A=A_{\Omega _ F}-A_\omega $. Stokes' theorem makes its image lie in $Z_0$. On the complete space $E_\chi $, the affine map

has Lipschitz constant $\kappa =GC_ ADr\leq m/(4C_0)\leq 1/4$. Its fixed point obeys

Since $A_\omega \chi _0=0$ and $A_\omega \mathcal R$ is the identity on $Z_0$, the function $\chi _ F=\chi _0+\eta $ satisfies $A_{\Omega _ F}\chi _ F=0$, and it has the stated normalization and bounds. Elliptic regularity makes it smooth. Every Gauduchon factor of $\Omega _ F$ with the same normalization satisfies the same affine fixed-point equation, so this factor is the one given by theorem 11.4.2. $\square$

Proof. Set $\Omega =\Omega _ F$, $\chi =\chi _ F$ and

The form comparison and $\int _ X\chi \omega ^ n=1$ give

Since $\mathrm{dd}^{\mathrm{c}}(\chi \Omega ^{n-1})=0$, Stokes' theorem also gives $Z=\int _ X\chi \Omega ^ n$. The form $T=(\chi /Z)\Omega ^{n-1}$ is normalized and is a constant multiple of the $(n-1)$st power of the Gauduchon metric conformal to $\Omega $. By theorem 12.7.3, it is the unique maximizing current, so $T=T_ F$. The equality case in the proof of that theorem gives $\nu _ F=\Omega \wedge T_ F$, and hence

The last identity uses the positive root $\gamma =(\chi /Z)^{1/(n-1)}\Omega $, for which $T_ F=\gamma ^{n-1}$. Uniqueness of the supporting probability follows from theorem 12.8.2.

Now $\chi /Z\leq (3a/2)2^{n-1}$ and $\Omega ^ n/\mathrm{d}V\leq (3/2)^ nH_{\max }$, and multiplying gives exactly $q_ F\leq C_\nu $ and $k_{T_ F}\leq C_ k$. Since $\int _ Xq_ F\, \mathrm{d}V=1$, we have $\int _ Xq_ F^ p\, \mathrm{d}V\leq C_\nu ^{p-1}$, which gives the sharper $L^ p$ estimate for $q_ F$. The $L^ p$ estimate for $k_{T_ F}$ follows from $\int _ X\mathrm{d}V=1$. $\square$

Proof. For a small smooth function $f$, put

If $F(f)=F(g)$, then $f-g$ is constant, so the map $f\mapsto F(f)$ is injective on functions of zero $\mathrm{d}V$-mean. By its continuity in $C^{2,\alpha }$, it maps a sufficiently small neighborhood of zero in any fixed finite-dimensional space of such smooth functions into the ball, and these spaces can have arbitrary dimension.

For the path, choose a smooth real $\psi $ of zero $\mathrm{d}V$-mean whose complex Hessian at some point $x_0$ has, in an $\omega $-unitary frame, a matrix $Q_0$ that is not a multiple of the identity. A cutoff of $|z_1|^2-|z_2|^2$ in local coordinates, minus its mean, is such a function. For small real $t$ put

These normalized data tend to $F_*$ in $C^{2,\alpha }$ as $t\to 0$, and by uniqueness in theorem 14.2.1 their solution forms are the forms $\Omega _ t$. If $F_ s=F_ t$, then $\log (\Omega _ s^ n/\Omega _ t^ n)$ is the constant $\log (V_ s/V_ t)$. The signs of the complex Hessian of $(s-t)\psi $ at a maximum and at a minimum force this constant to be zero. Integrating the derivative of $\log \det $ along the segment between $\Omega _ t$ and $\Omega _ s$ then gives a uniformly elliptic equation for $(s-t)\psi $ without zeroth-order term, and the maximum principle forces $(s-t)\psi $ to be constant. Since $\psi $ is not constant, $s=t$.

If the maximizing currents for $F_ s$ and $F_ t$ agree, (12.9.9) makes $\Omega _ s^{n-1}$ and $\Omega _ t^{n-1}$ proportional at $x_0$. Positive cofactor matrices determine positive forms up to scale, so $I+sQ_0$ and $I+tQ_0$ are proportional. Since $I$ and $Q_0$ are linearly independent, comparing coefficients gives $s=t$. $\square$

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 $. $\square$