14.19 Continuous solutions and normalization080A6052D542

Proof. Write $v_\varepsilon =(1-\varepsilon )v$, $U=U_{\varepsilon ,s}$ and

Set $I_{-1}=I_{-2}=0$. Then $0\leq \phi -u\leq s$, this difference vanishes outside $U$, and on $U$ plurifine locality gives $\omega _\phi =\omega _{v_\varepsilon }\geq \varepsilon \omega $ in all bounded products.

By lemma 5.28.3, for every fixed positive mixed product $W$ of $n-1$ bounded-potential factors,

For $0\leq k\leq n-1$, apply this with $W=\omega _ u^ k\wedge \omega _\phi ^{n-k-1}$. Since $u=\phi $ on $X\setminus U$, the difference $T_ k-T_{k+1}=(\omega _\phi -\omega _ u)\wedge W$ is nonnegative there. Consequently

This integration by parts is valid for bounded potentials. To see this, approximate $u$ and $\phi $ by decreasing sequences of smooth, uniformly bounded $\omega $-plurisubharmonic functions, using theorem 5.32.1. For smooth functions the identity is ordinary integration by parts. The form $\mathrm{dd}^{\mathrm{c}}W$ expands as below into bounded Hessian products and smooth forms, and bounded Bedford–Taylor convergence, also against bounded quasi-continuous coefficients converging in capacity, passes each term to the limit.

Set $r=n-k-1$. Since $\mathrm{d}\omega _ u=\mathrm{d}\omega _\phi =\mathrm{d}\omega $, expansion gives

Terms containing a negative power have coefficient zero and are omitted. On $U$ we may replace each $\omega ^ j$ in an upper bound by $\varepsilon ^{-j}\omega _\phi ^ j$, since $\omega _\phi \geq \varepsilon \omega $ there. Using $\phi -u\leq s$ and $\varepsilon ^{-2}\leq \varepsilon ^{-3}$ in (14.19.8), we obtain

Each coefficient is at most $n^2$, and $a=B_0n^2s\varepsilon ^{-3}$. Hence

Put $q=1-4a$; the assumption on $s$ gives $a<1/32$, so $q>7/8$. Suppose inductively that $qI_ j\leq I_{j+1}$ for $j<k$. Then $I_{k-1}\leq q^{-1}I_ k$ and $I_{k-2}\leq q^{-2}I_ k$ whenever the indices exist. Since

(14.19.9) gives $qI_ k\leq I_{k+1}$. The case $k=0$ follows from the same formula with the two absent terms set to zero. By induction, $q^ n I_0\leq I_ n$. Since $T_0=\omega _{v_\varepsilon }^ n$ on $U$ and $T_ n=\omega _ u^ n$, this is the theorem. In dimension one all torsion terms vanish and the same argument applies. $\square$

Proof. Suppose $m=\inf _ X(u-v)<0$. Choose $\varepsilon >0$ so small that

omitting the second restriction if $v=0$. Let $m_\varepsilon $ and $U_{\varepsilon ,s}$ be defined as in theorem 14.19.6 for this pair $u,v$. Since $|m_\varepsilon -m|\leq \varepsilon \| v\| _\infty $, for all sufficiently small $s>0$ the set $U=U_{\varepsilon ,s}$ is contained in $\{ u<v\} $: there

The set $U$ is nonempty by the definition of the infimum and has positive volume by lemma 5.24.4.

Decrease $s$ further so that theorem 14.19.6 applies and $(1-4B_0n^2s\varepsilon ^{-3})^ n(1-\varepsilon )^ n>c$. Write the first factor as $q^ n$. Positivity of every mixed term gives

Modified comparison and the assumption would therefore imply

All integrals are finite, so the choice $q^ n(1-\varepsilon )^ n>c$ and the positive volume of $U$ make this impossible. Hence $m\geq 0$. $\square$

Proof. Let $u$ be a bounded $\omega $-plurisubharmonic function. If $\omega _ u^ n=0$, then theorem 14.19.10, applied with $c=0$ and the constant potential $v=\sup _ Xu+1$, would give $u\geq v$, a contradiction. $\square$

Proof. For every $t>0$, the assumption gives $\omega _ u^ n\leq \mathrm{e}^{-\lambda t}\omega _ v^ n$ on $\{ u<v-t\} $. Since $\omega _{v-t}=\omega _ v$, theorem 14.19.10, applied to $u$ and $v-t$, gives $u\geq v-t$. Letting $t\downarrow 0$ proves $v\leq u$. $\square$

Proof. Let $B_0$ be the constant of theorem 14.19.6, and set

For a capacity test $0\leq \rho \leq 1$, put

Then $0<\tau <1$ and $0\leq (1-\varepsilon )(w-v)\leq t/2$, so $m-t/2\leq m'\leq m$. Consequently

We apply modified comparison to $u,w$ at height $s+t/2<\delta _\varepsilon $; by the choice of $\delta _\varepsilon $, its inverse factor is at most $(8/7)^ n$. Moreover,

Integrating over $U_ s\subset V$, applying modified comparison on $V$, and taking the supremum over $\rho $, we obtain (14.19.15) with $C_ H=[2(H+1)]^ n(8/7)^ n$. Finally, each $U_ s$ is nonempty by the definition of the infimum, so it has positive smooth volume by lemma 5.24.4, and hence positive capacity, as the constant test function shows. $\square$

Proof. Put $a(s)=\operatorname {Cap}_\omega (U_ s)^{1/n}>0$ and $K=(C_ HA)^{1/n}$. The thin-sublevel estimate gives

Suppose $C_\alpha K a(D)^\alpha <D$. Starting with $s_0=D$, set

If $a(s_ i)\leq 2^{-i}a(D)$ for $0\leq i\leq j$, then

Thus $s_{j+1}>0$, and the sublevel estimate, applied with $s=s_{j+1}$ and $t=t_ j$, gives $a(s_{j+1})\leq a(s_ j)/2$. Induction justifies every step. The decreasing sequence $s_ j$ has a limit

Monotonicity now gives $0<a(s_\infty )\leq 2^{-j}a(D)$ for every $j$, a contradiction. This proves (14.19.17), including the case in which $U_ D=X$. $\square$

Proof. By corollary 14.19.11, $F$ is not almost everywhere zero. Apply lemma 14.19.16 with $v=0$, $H=1$, $\varepsilon =1/2$, and $A=C\| F\| _ p>0$, where $C$ is the constant of corollary 5.29.5. Fix $D=\delta _{1/2}/2$, and put $M=-\inf _ Xu$. Since $U_ D=\{ u<-M+D\} $, the resulting inequality reads

If $M>D+1$, (5.29.4) bounds this capacity above by $C/(M-D)$, and the first assertion follows; the case $M\leq D+1$ is immediate. Bounding the same capacity above by $\operatorname {Cap}(X)$ gives the second assertion. $\square$

Proof. By theorem 14.19.19, there is a common $H\geq 1$ with $-H\leq u_ j\leq 0$. Suppose that the sequence is not uniformly Cauchy. After exchanging indices if necessary, we find $a>0$ and arbitrarily large $j,k$ with

Choose $0<\varepsilon <1$ such that $\varepsilon H<a/4$, and choose a fixed $D>0$ with $2D<\delta _\varepsilon $ and $2D<a/4$. Apply the preceding sublevel lemmas with $u=u_ k$ and $v=u_ j$. Since $v\leq 0$, $m=\inf (u-(1-\varepsilon )v)\leq S$. Thus

The capacity lower bound (14.19.18), with a constant $A$ independent of $j,k$ given by corollary 5.29.5, yields $\operatorname {Cap}(U_ D)\geq c>0$. On the other hand, (14.19.15) and the inequalities of Hölder and Chebyshev give

This contradiction proves that the sequence is uniformly Cauchy. For the last assertion, the normalized compactness theorem, theorem 5.25.2, gives an $L^1$-Cauchy subsequence of any such family. Its uniform limit is continuous, and distributional convergence preserves $\omega $-plurisubharmonicity. $\square$

Proof. Write $L=\int _ X(v-u)_+\, \mathrm d\mu $ and $S=\sup _ X(v-u)_+$. Apply lemma 14.19.16 with competitor zero, $\varepsilon =1/2$ and one fixed $0<D<\delta _{1/2}$. It bounds $\operatorname {Cap}_\omega \{ u<\inf _ Xu+D\} $ below by a positive constant. The tail estimate (5.29.4) therefore bounds $-\inf _ Xu$ in terms of $(X,\omega ),A,\alpha $. Consequently $S$ has a fixed upper bound.

For $0<\varepsilon \leq 1/2$, choose a fixed small $c>0$ and put $D=c\varepsilon ^3$. The formula for $\delta _\varepsilon $ in lemma 14.19.14 allows $c$ to be chosen so that $2D<\delta _\varepsilon $ and $4D\leq \varepsilon $. With

we have $m\leq -S$ when $S>0$, since $v\leq 0$. On $U_{2D}$,

If $S>2\varepsilon H+4D$, this is greater than $S/2$. The backwards and thin-sublevel estimates give

where $c_0>0$ depends only on the stated data. Hence $S\leq CL\varepsilon ^{-3n(1+1/\alpha )}$ in this case. In the other case, $S\leq (2H+1)\varepsilon $.

For sufficiently small $L>0$, choose $\varepsilon =L^{\sigma _\alpha }$. The two alternatives give (14.19.24). The remaining positive values of $L$ are covered by the fixed upper bound for $S$. If $L=0$, the first alternative is impossible for every small $\varepsilon $, and the second gives $S=0$ on letting $\varepsilon \downarrow 0$. $\square$

of theorem 14.19.2. Choose smooth strictly positive functions $f_ j\to f$ in $L^ p(X,\mathrm{d}V)$. By theorem 14.2.1, there are smooth $u_ j$ with $\omega _{u_ j}>0$ and constants $c_ j$ such that

We first bound the constants. For $n\geq 2$, fix a Gauduchon metric $\gamma $. By proposition 11.5.4,

Since $|x^{1/n}-y^{1/n}|\leq |x-y|^{1/n}$ for $x,y\geq 0$, Hölder's inequality gives $A_ j\to A>0$, where $A=\int _ X(f\mathrm{d}V/\gamma ^ n)^{1/n}\gamma ^ n$. Thus $c_ j$ is bounded above. In dimension one, Stokes' theorem gives the same estimate, with $A_ j=\int _ Xf_ j\, \mathrm{d}V$.

The lower bound for the density norm in theorem 14.19.19 gives $c_ j\| f_ j\| _ p\geq c_ p$, so the constants are bounded away from zero. The densities $c_ jf_ j$ are therefore uniformly bounded in $L^ p$, and the same theorem gives a common bound for the $u_ j$. By theorem 5.25.2, we may pass to a subsequence with $c_ j\to c>0$ and $u_ j$ Cauchy in $L^1$. By theorem 14.19.21, $u_ j\to u$ uniformly, with $u$ continuous and $\sup _ Xu=0$.

Bounded Bedford–Taylor continuity gives $\omega _{u_ j}^ n\rightharpoonup \omega _ u^ n$, while $c_ jf_ j\to cf$ in $L^ p$. Hence $\omega _ u^ n=cf\, \mathrm{d}V$, as required. $\square$

Proof. Absorbing the fixed smooth positive ratio $\mathrm{d}V/\omega ^ n$ into $f_ j$ and $f$ preserves their $L^ p$ convergence, so we may assume $\mathrm{d}V=\omega ^ n$. Choose $\rho >0$ smooth with $\mathrm{dd}^{\mathrm{c}}(\rho \omega ^{n-1})=0$, using theorem 11.4.2; for $n=1$, take $\rho =1$. Put

Set $A=\int _ X f^{1/n}\rho \omega ^ n$. The inequality for $n$th roots used in the preceding existence proof gives $A_ j\to A>0$. Since $\mathrm{dd}^{\mathrm{c}}(\rho \omega ^{n-1})=0$, the trace $\int _ X\omega _{u_ j}\wedge \rho \omega ^{n-1}$ equals $\alpha $; the arithmetic–geometric mean inequality therefore implies

Applying Jensen's inequality to the probability measure $A_ j^{-1}f_ j^{1/n}\rho \omega ^ n$ and taking logarithms, we obtain

By lemma 5.26.2, the normalized potentials $\psi _ j$ have uniform $L^ q$ bounds for every finite $q$. With $q=np/(np-1)$, Hölder's inequality gives a uniform bound for the last integral, since $\| f_ j^{1/n}\| _{np}=\| f_ j\| _ p^{1/n}$. This proves the upper bound for $M_ j$.

Finally, the density $F_ j=\mathrm{e}^{\lambda u_ j}f_ j$ satisfies $\| F_ j\| _ p\leq \mathrm{e}^{\lambda M_ j}\| f_ j\| _ p$. The lower-density bound of theorem 14.19.19, applied to $\psi _ j$, gives $c_ p\leq \mathrm{e}^{\lambda M_ j}\| f_ j\| _ p$, which bounds $M_ j$ below. $\square$

of theorem 14.19.3. Choose smooth strictly positive functions $f_ j\to f$ in $L^ p$. By theorem 14.7.2, there are smooth $u_ j$ with $\omega _{u_ j}>0$ and $\omega _{u_ j}^ n=\mathrm{e}^{\lambda u_ j}f_ j\, \mathrm{d}V$. The preceding lemma bounds $M_ j=\sup _ Xu_ j$, and consequently bounds $F_ j=\mathrm{e}^{\lambda u_ j}f_ j$ in $L^ p$. The normalized potentials $\psi _ j=u_ j-M_ j$ are uniformly bounded by theorem 14.19.19. We may pass to a subsequence with $M_ j\to M$ and $\psi _ j$ Cauchy in $L^1$. Theorem 14.19.21 gives uniform convergence $u_ j\to u\in \mathrm{PSH}(X,\omega )\cap C(X)$.

Uniform convergence and $f_ j\to f$ in $L^ p$ give $\mathrm{e}^{\lambda u_ j}f_ j\to \mathrm{e}^{\lambda u}f$ in $L^ p$, and continuity of bounded products therefore gives $\omega _ u^ n=\mathrm{e}^{\lambda u}f\, \mathrm{d}V$. Uniqueness, even among all bounded solutions, follows from bounded exponential comparison, corollary 14.19.12. $\square$

Proof. Since $f_ a\leq f_ b$, the two equations give $\mathrm{e}^{-\lambda b}\omega _ b^ n\geq \mathrm{e}^{-\lambda a}\omega _ a^ n$. Hence $b\leq a$ by corollary 14.19.12, applied with $u=a$ and $v=b$. $\square$

Proof. Since $u$ is bounded, the density $\mathrm{e}^{-u}f$ is nonnegative, nonzero and belongs to $L^ p$. By theorem 14.19.3, applied with $\lambda =1$ to this density, there is a continuous solution $r$ of

The bounded potential $u$ solves the same exponential equation. Applying lemma 14.19.28 with $f_ a=f_ b=\mathrm{e}^{-u}f$ in both orders gives $u=r$, so $u$ is continuous. $\square$

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 $u\in C^\infty (X)$. $\square$

Proof. Set $a=u+\lambda ^{-1}\log c-M$ and $b=u+\lambda ^{-1}\log c-m$. These translates of $u$ have the same Monge–Ampère measure as $u$, and since $m\leq u\leq M$,

Thus $\mathrm{e}^{-\lambda a}\omega _ a^ n\geq f\, \mathrm{d}V=\mathrm{e}^{-\lambda u_\lambda }\omega _{u_\lambda }^ n\geq \mathrm{e}^{-\lambda b}\omega _ b^ n$. Applying corollary 14.19.12 first to the pair $(u_\lambda ,a)$ and then to the pair $(b,u_\lambda )$ gives $a\leq u_\lambda \leq b$, which is (14.19.35). Multiplying these inequalities by $\lambda $ and using $m\leq u\leq M$ gives the norm bound, and the uniform convergence follows.

If $(v,c')$ is another such pair, the same argument applied to $v$ gives $\lambda u_\lambda \to \log c'$, because $u_\lambda $ depends only on $f$ and $\lambda $. Hence $c'=c$. $\square$

Proof. Fix $\varepsilon >0$. At a maximum point of $u-(1+\varepsilon )v$, the complex Hessian of this function is nonpositive, so that

At this point $B>0$, because $A+\varepsilon \omega >0$. Monotonicity of the determinant and $A^ n=cB^ n$ give $c\leq (1+\varepsilon )^ n$. Letting $\varepsilon \downarrow 0$ gives $c\leq 1$, and interchanging the two potentials gives $c^{-1}\leq 1$. $\square$

Proof. Integrating the equation gives $\int _ X\omega _ u^3=6c_ f I_ f$. Both the inequality $c_ fI_ f\geq 1$ and (14.19.39) therefore follow from (23.1.39), whose left-hand side is nonnegative.

Suppose $c_ f I_ f=1$. By proposition 23.1.38, $u=\pi ^*q$ for a bounded $\omega _{\mathrm{FS}}$-plurisubharmonic function $q$; local holomorphic sections of $\pi $ show that $q$ is continuous. Thus $u\circ \tau _ a=u$ for every $a\in E$. Since bounded products are natural under biholomorphisms and $\omega $ is invariant under the translations, it follows that

As $f$ is defined only almost everywhere, we use this invariance as an identity of measures.

By Fubini's theorem, the fiber average $Af$ is defined almost everywhere. It equals $\pi ^*g$ for a measurable function $g$ on the base, because local trivializations differ by translations of the Haar variable. Jensen's inequality gives

For every continuous real test function $\chi $ on $X$, the invariance of the measure gives $\int _ X\chi f\, \mathrm{d}\nu =\int _ X(\chi \circ \tau _ a)f\, \mathrm{d}\nu $. Integrating in $a$, and using Fubini's theorem together with the invariance of Haar measure under $a\mapsto -a$, we obtain $\int _ X\chi f\, \mathrm{d}\nu =\int _ X\chi Af\, \mathrm{d}\nu $. Hence $f=Af=\pi ^*g$ almost everywhere, with $g\geq 0$.

Conversely, suppose $f=\pi ^*g$ almost everywhere. Disintegration gives $I_ f=\int _{\mathbb P^2}g\omega _{\mathrm{FS}}^2$. Applying theorem 14.19.2 on the base, we find a continuous $\omega _{\mathrm{FS}}$-plurisubharmonic function $q$ and a constant $k>0$ with $(\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}q)^2=kg\, \omega _{\mathrm{FS}}^2$. Bounded decreasing regularization and Stokes' theorem give $\int _{\mathbb P^2}(\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}q)^2=1$, so $k=I_ f^{-1}$.

Put $u=\pi ^*q$. To compute its bounded product, take smooth strictly $\omega _{\mathrm{FS}}$-plurisubharmonic functions $q_ j\downarrow q$ by theorem 5.32.1. Their pullbacks decrease to $u$ and are strictly $\omega $-plurisubharmonic. Since $\alpha ^3=\beta ^2=0$, the smooth cube identity gives, for each $j$,

In a local trivialization, the right-hand side is $6\mathrm{d}A_ E$ times the base measure $(\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}q_ j)^2$: the horizontal components of $\beta $ vanish against the base top-degree form, and its vertical restriction is $2\mathrm{d}A_ E$. Continuity of bounded products on the base and on $X$, tested in these trivializations and patched by a partition of unity, therefore gives

Uniqueness of the constant implies $c_ f=I_ f^{-1}$. Every continuous solution then has mass $6$, so the mass characterization in proposition 23.1.38 makes it a pullback from the base, and the descended potential is continuous by local sections, as above. $\square$

Proof. Since the constants $c_{f_ j}$ converge, the densities $c_{f_ j}f_ j$ have a common $L^ p$ bound. By theorem 14.19.21, every subsequence has a further subsequence on which $u_ j\to u$ uniformly. Since $L^ p$ is reflexive, a further extraction gives $f_ j\rightharpoonup f$ weakly in $L^ p$. Positivity, the unit integral and the bound $\| f\| _ p\leq M$ pass to the limit, and continuity of bounded products under uniform convergence gives

Thus $c_ f=1$. Since also $I_ f=1$, proposition 14.19.38 gives $u=\pi ^*q$ and $f=\pi ^*g$. The bounded cube identity proved there, tested against pullbacks of continuous functions on the base, gives the asserted base equation. In particular, both limits are fixed by $A$.

Haar averaging is a contraction on $C^0(X)$ and on $L^ p(X,\nu )$. It is also self-adjoint with respect to $\nu $: Fubini's theorem and the invariance of $\nu $ under the fiber translations give $\int \chi Af\, \mathrm{d}\nu =\int A\chi f\, \mathrm{d}\nu $ for $\chi \in L^{p'}$. Along the extracted subsequence, therefore, $u_ j-Au_ j\to 0$ uniformly and $f_ j-Af_ j\rightharpoonup 0$ in $L^ p$. If either convergence failed for the original sequence, some subsequence would stay away from the limit (for the weak convergence, when tested against one fixed function in $L^{p'}$), and a further subsequence of the kind just constructed would give a contradiction. This proves both conclusions for the whole sequence. $\square$

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. $\square$

Proof. If $\lambda (P)\in \Gamma $, then $\mathcal A_ U+tP$ is admissible for $t\geq 0$. The function $t\mapsto F(\mathcal A_ U+tP)$ is positive and concave on $[0,\infty )$, so its derivative at $t=0$ cannot be negative; thus $\operatorname {tr}(BP)\geq 0$. Since $F$ is concave and $F(0)=0$, the tangent inequality at $\mathcal A_ U$, evaluated at the origin, gives $\operatorname {tr}(B\mathcal A_ U)\leq h$. Choose $a_*>0$ with $F(a_*\operatorname {Id})>H_0+1$. Concavity gives

Thus $T\geq a_*^{-1}$. Since $\mathcal L(U-V)=\operatorname {tr}(B\mathcal A_ U)-\operatorname {tr}(B\mathcal A_ V)$ and $\mathcal L(\underline U-U)=\operatorname {tr}(B\mathcal A_{\underline U})-\operatorname {tr}(B\mathcal A_ U)$, the remaining two bounds follow from the nonnegativity of $\operatorname {tr}(B\mathcal A_ V)$ and $\operatorname {tr}(B\mathcal A_{\underline U})$.

We derive the alternative from the strict gap. Fix $r,H,\delta $ from lemma 14.19.43, and put $\mu =\lambda (\mathcal A_{\underline U})$. Compactness and the gap give one $S>r$ such that

everywhere. We first prove a scalar alternative. At a point of $X$, let $\lambda \in \Gamma $ satisfy $f(\lambda )=\sigma $, where $\sigma =h$ at that point, and consider the set

It is contained in $\Gamma $ because $\mu -r\mathbf1\in \Gamma $, it is closed, and the matrix bound of lemma 14.19.43, applied to diagonal matrices, makes it bounded; thus $K$ is compact. When nonempty, it contains $\mu -r\mathbf1$ and is star-shaped about that point: moving along a segment toward it decreases each coordinate, hence decreases $f$. Thus $K$ is connected. Let $N=\nabla f(\lambda )$, whose entries are positive, and let $\Pi $ be the tangent hyperplane $N\cdot (v-\lambda )=0$.

If $f(\mu -r\mathbf1)>\sigma $, concavity gives $N\cdot (\mu -\lambda )>r\sum _ iN_ i$. Otherwise $\mu -r\mathbf1\in K$. The diagonal ray $\mu -r\mathbf1+t\mathbf1$, $t\geq 0$, meets $\{ f=\sigma \} $, since $f$ diverges along it; at the meeting point $v$, the tangent inequality gives $N\cdot (v-\lambda )\geq 0$. If $\Pi $ misses $K$, connectedness makes $N\cdot (v-\lambda )$ strictly positive throughout $K$, and in particular at $\mu -r\mathbf1$; this again gives $N\cdot (\mu -\lambda )>r\sum _ iN_ i$.

If $v\in \Pi \cap K$, put $z_ i=v-r\mathbf1+Se_ i$. Then $z_ i\geq \mu -2r\mathbf1+Se_ i$, so $f(z_ i)>\sigma $. The tangent inequality at $\lambda $ gives

Thus $N_ i>(r/S)\sum _ jN_ j$ for every $i$. We have proved that $N\cdot (\mu -\lambda )>r\sum _ iN_ i$ or $N_ i>(r/S)\sum _ jN_ j$ for all $i$, at every positive level $\sigma $, with constants independent of $\sigma $.

To pass to matrices, take $\lambda =\lambda (\mathcal A_ U)$, and order $\lambda $ and $\mu $ decreasingly. By concavity and symmetry, the numbers $f_ i(\lambda )$ are then ordered increasingly. In the eigenframe of $\mathcal A_ U$, the diagonal of $\mathcal A_{\underline U}$ is a doubly stochastic average of permutations of $\mu $, so the rearrangement inequality gives

Since $N=(f_ i(\lambda ))_ i=(B_ i)_ i$, taking $\kappa $ smaller than $\min \{ r,r/S\} $ proves the matrix alternative. $\square$

Proof. Write $\mathscr T^ p_{ik}$ and $R^ p_{i\bar jk}$ for the torsion and curvature of the Chern connection. The commutation formula and the differentiated equation read

The product rule for $|\partial U|^2$ gives both squares $\sum B_ i|U_{ki}|^2$ and $\sum B_ i|U_{k\bar i}|^2$, the term $2\operatorname{Re}\sum _ kh_ kU_{\bar k}$, a curvature error bounded by $CT|\partial U|^2$, and the errors from $\nabla \chi $ and torsion. The latter are bounded by

Absorb the last term into half of the mixed square. Since $U_{i\bar j}=\lambda _ i\delta _{ij}-\chi _{i\bar j}$, the remaining half is at least $\frac14\sum _ iB_ i\lambda _ i^2-CT$. Hence

Finally,

In the product rule for $\mathrm e^\xi Q$, combine the cross term coming from the first sum with $|\partial U|^2|\partial \xi |_ B^2$ and with the square $\sum B_ i|U_{ki}|^2$. This gives $\sum B_ i|U_{ki}+U_ k\xi _ i|^2$, while the remaining weight term is $|\partial \xi |_ B^2$. $\square$

Proof. Let $a\geq 1$ and $\beta >1$ be constants, to be chosen below. Assuming $\| U-V\| _\infty \leq \beta ^{-1}$, put

Then $-2\beta ^{-1}\leq w\leq 0$ and

Writing $D=|\partial (U-V)|_ B^2$, the Cauchy–Schwarz inequality gives

For the first line, split $\xi _ i$ as above and use $\sum B_ i|U_ i|^2\leq 2D+2K_ V^2T$; then apply Young's inequality to the resulting three products, each absorbing at most $\frac1{24}\sum B_ i\lambda _ i^2$. For the second line, use $|\partial \xi |_ B\leq aK_ VT^{1/2}+(a+4)D^{1/2}$ and the bound for $\chi $. In both lines, $C_ a$ depends on $a$ but not on $\beta $ or on the derivatives of $V$.

By lemma 14.19.48 and $w\leq 0$,

Choose $a$ so large that $(a\kappa -C_ g-2)\tau >4H_0+2H_1+1$, and then $\beta \geq 2C_ a+2$. Combining the preceding bounds with lemma 14.19.49, we obtain, at every point where $Q\geq C'(1+K_ V^4)$ for a suitable constant $C'$,

Indeed, the coefficient of $D$ before absorption is $(2\beta -C_ a-1/2)Q-C_ aK_ V^2$, which is at least $\beta Q$ once $C'$ is large enough. The other errors satisfy $C_ aK_ V^4\leq Q$ and $C_ a+C_ aK_ VQ^{1/2}\leq Q$, and they account for the additional term $-2T$ in the bracket.

Consider a maximum point of $\mathrm e^\xi Q$ at which $Q\geq C'(1+K_ V^4)$. In the first alternative of lemma 14.19.48, the bracket is positive by the choice of $a$, since $T\geq \tau $. In the second, $D\geq \kappa T|\partial (U-V)|^2$. For $Q\geq 4(1+K_ V^2)$, the triangle inequality gives $|\partial (U-V)|^2\geq c_*Q$, where $c_*=(\sqrt3-1)^2/4$. Since $\mathcal L(\underline U-U)\geq -H_0$, the bracket is then at least

This is positive once $C'$ is large enough, because $T\geq \tau $. In both cases $\mathcal L(\mathrm e^\xi Q)>0$ at a maximum point, which contradicts the second derivative test. Thus $Q\leq C(1+K_ V^4)$ at the maximum point. The weight has uniformly bounded oscillation:

By the potential bound in proposition 14.19.44, this oscillation is bounded independently of $V$, so the estimate at the maximum point gives $Q\leq C(1+K_ V^4)$ everywhere. Taking square roots and setting $\eta =\beta ^{-1}$ proves the assertion. $\square$

Proof. The right-hand sides in the family with $\| \partial h\| _\infty \leq H_1$ form an equicontinuous set. By corollary 14.19.46, the corresponding functions $U$ form a precompact subset of $C^0(X)$. Choose a finite $\eta $-net in this set, consisting of smooth admissible functions $V_1,\ldots ,V_ N$. Each solution is within $\eta $ of some $V_ j$, so proposition 14.19.50 bounds its gradient by $C(1+\max _ j\| \partial V_ j\| _\infty ^2)$. The set of solutions, and hence the chosen finite net, depends only on the fixed data and $H_1$. Finally, $u=2\pi U$, so the same conclusion holds for $\mathrm du$. $\square$