12.10 Integrable densities in the compact Gauduchon problem5CB8CFC32A19

Proof. For a smooth strictly positive $p$, we have $D_ p=\mathfrak D_{n\log p}$ and $\mathrm{e}^{-B_0(n\log p)/n}=\rho _\omega (p)$, so theorem 12.7.3, applied to $F=n\log p$, proves (12.10.2). By Hölder's inequality and theorem 12.9.1, for every real $r>n$ there is a constant $C_ r$ such that, for every $q\in L^ r$,

Next, approximate a bounded nonnegative $p$ in $L^{2n}$ by smooth strictly positive functions. This estimate and (15.32.8) give uniform convergence on $\mathscr G_\omega $ of the corresponding functionals $D$ and convergence of the root norms, so the supremum formula holds for bounded $p$. For such $p$, the functional $D_ p$ is also upper semicontinuous and concave, by lemma 12.7.2 and uniform convergence. For general $p\in L^ n$, we use the truncations $\min (p,K)$ at positive integer levels $K$. Monotone convergence and the definition of $b_\omega $ through truncations give

This step only interchanges two suprema and gives no attainment.

Now suppose $p\in \mathcal D_\omega ^0$. The supremum formula applied to the tail gives

uniformly in $T$, by proposition 15.32.9. Hence $D_ p$ is finite and is a uniform limit of upper semicontinuous concave functionals, and compactness of $\mathscr G_\omega $ gives a maximizer. Finally, fix $1<q<n/(n-1)$. By Hölder's inequality and theorem 12.9.1, for every measurable set $E$,

Choosing first $K$ large and then the volume of $E$ small proves uniform absolute continuity. Since the total masses are bounded by $\rho _\omega (p)$, the densities are uniformly integrable. $\square$

Proof. The first condition implies the second by theorem 12.10.1. Conversely, $\mathrm{d}V(\{ p>K\} )\to 0$ as $K\to \infty $, so uniform integrability and (12.10.2) give

This is the first condition by proposition 15.32.9.

If $p\in \mathcal D_\omega ^0$ and $|F_ j|\leq A$, then $p(\mathrm{e}^{F_ j/n}-1)$ tends to zero in measure and its absolute value is at most $(\mathrm{e}^{A/n}+1)p$. The dominated-convergence criterion of proposition 15.32.9 makes its root norm tend to zero, and the triangle inequality gives $\rho _\omega (p\mathrm{e}^{F_ j/n})\to \rho _\omega (p)>0$. Taking negative $n$th powers proves the third condition, also for bounded measurable $F_ j$.

Suppose now that the tails do not tend to zero, and put $R=\rho _\omega (p)>0$. The formula of theorem 15.29.1, which uses only smooth tests, yields $\delta >0$, numbers $K_ j\to \infty $ and smooth strictly positive $T_ j\in \mathscr G_\omega $ such that

Since the finite measure $p k_{T_ j}\, \mathrm{d}V$ is regular, we can choose a compact set $C_ j\subset \{ p>K_ j\} $ of mass greater than $\delta /2$ for this measure. By outer regularity of $\mathrm{d}V$, choose an open set $O_ j\supset \{ p>K_ j\} $ with $\mathrm{d}V(O_ j)\leq \mathrm{d}V(\{ p>K_ j\} )+1/j$, and a smooth cutoff $0\leq \chi _ j\leq 1$ that equals one on $C_ j$ and has support in $O_ j$. Fix $A>0$ with $\mathrm{e}^{A/n}\delta /2>2R$, and set $F_ j=A\chi _ j$. Then $F_ j\to 0$ in measure, while

Since $b_\omega (\mathrm{e}^{F_ j}p^ n)=\rho _\omega (p\mathrm{e}^{F_ j/n})^{-n}$, this gives the asserted uniform gap. Finally, $|\mathrm{e}^{F_ j}-1|p^ n\leq (\mathrm{e}^ A-1)p^ n\mathbf1_{O_ j}$, whose integral tends to zero by absolute continuity of the integral. This proves the last assertion and the converse to the third condition. $\square$

Proof. For positive integers $K$, put $p_ K=\min (p,K)$. Since $p$ is finite almost everywhere, uniform integrability gives

For every nonnegative continuous function $h$, the weight $hp_ K$ is bounded and measurable, so the upper semicontinuity of $D_{hp_ K}$ proved in theorem 12.10.1 gives

Now let $K\to \infty $. For $h=1$ this yields $R\leq D_ p(T)$, and the supremum formula (12.10.2) gives the reverse inequality. If $\Lambda $ is a weak cluster limit of the measures $p k_{T_ j}\, \mathrm{d}V$, the same estimate for all nonnegative continuous $h$ gives $\Lambda \leq p k_ T\, \mathrm{d}V$. Both measures have mass $R$, so they agree. Since these measures have bounded masses, they form a weakly relatively compact family, and every cluster limit has now been identified. The final assertion follows from the compactness of $\mathscr G_\omega $. $\square$

Proof. For each $T$, Hölder's inequality makes $F\mapsto \int _ Xp \mathrm{e}^{F/n}k_ T\, \mathrm{d}V$ logarithmically convex, and taking suprema preserves the inequality defining logarithmic convexity. Thus $M_ f$ is logarithmically convex and $B_ f=-n\log M_ f$ is concave. Monotonicity and homogeneity of $\rho _\omega $ give the translation rule and, by comparison with constant translates, the Lipschitz estimate.

For fixed $F$, the weight $p\mathrm{e}^{F/n}$ belongs to $\mathcal D_\omega ^0$, so the densities $p\mathrm{e}^{F/n}k_ T$, $T\in \mathscr G_\omega $, are uniformly integrable by theorem 12.10.1. If $T_ j\to T$ and $D_{p\mathrm{e}^{F/n}}(T_ j)\to M_ f(F)$, proposition 12.10.6 makes $T$ a maximizer and gives convergence of the weighted measures. For $T_1,T_2\in \mathscr G_\omega $ and real $0\leq t\leq 1$, matrix determinant concavity gives

For two maximizing currents, the current on the left also maximizes, and equal total masses turn this inequality into an equality. The set $\mathscr N_ f(F)=\{ \nu _{T,F}:T\in \mathscr M_ f(F)\} $ is therefore nonempty, compact and convex.

A maximizing $T$ gives

Taking $-n$ times the logarithm and applying Jensen's inequality proves (12.10.8), so $\mathscr N_ f(F)\subset \mathcal A_ f(F)$. For the reverse inclusion, a Taylor expansion of the exponential gives

where the error is uniform in $T$ since $D_{p \mathrm{e}^{F/n}}(T)\leq M_ f(F)$. Maximizers at $F+tH$ are near-maximizers at $F$, by the multiplicative bounds $\mathrm{e}^{-t\| H\| _\infty /n}\leq M_ f(F+tH)/M_ f(F)\leq \mathrm{e}^{t\| H\| _\infty /n}$. The convergence of their weighted measures proved above and this uniform expansion are exactly the two inputs of the one-sided differentiation argument in the proof of theorem 12.8.2, which gives

The separation argument of that proof then identifies all supporting probabilities with $\mathscr N_ f(F)$, and replacing $H$ by $-H$ gives the left derivative.

Finally, $M_ f(F)\geq \mathrm{e}^{-R/n}\rho _\omega (p)$ when $\| F\| _\infty \leq R$, so the densities of the probabilities $\nu _{T,F}$ are at most $\mathrm{e}^{2R/n}\rho _\omega (p)^{-1}p k_ T$. The uniform integrability of the densities $pk_ T$ in theorem 12.10.1 proves the remaining assertions. $\square$

Proof. A smooth strictly positive $T\in \mathscr G_\omega $ is $\gamma ^{n-1}$ for a Gauduchon metric $\gamma $ normalized by $\int _ X\omega \wedge \gamma ^{n-1}=1$. Lemma 13.14.12 proves positivity of $R_{u,T}$, and Stokes' theorem for currents gives $((\omega +\mathrm{dd}^{\mathrm{c}}u)\wedge T)(X)=\int _ X\omega \wedge T=1$. This proves the mass formula. Since $\rho _\omega (p)=b^{-1/n}$, the nonnegative mass tends to zero along a maximizing sequence; it equals the total variation norm and does not depend on $u$.

Under the tail hypothesis, theorem 12.10.1 gives uniform integrability of the densities $p k_{T_ j}$. Thus proposition 12.10.6 makes $T$ a maximizer and gives $p k_{T_ j}\, \mathrm{d}V\rightharpoonup p k_ T\, \mathrm{d}V$. Adding the error measures $R_{u,T_ j}$, which tend to zero in total variation, proves the limit formula. $\square$

Proof. Let $u$ be any threshold subsolution; one exists by theorem 15.28.4. Write $T=\gamma ^{n-1}$, where $\gamma $ is the smooth Gauduchon metric given by the positive $(n-1)$st root of $T$, and set

Put $A=\omega +\mathrm{dd}^{\mathrm{c}}u\geq 0$. Since $T$ maximizes, the positive measure $R_{u,T}$ of proposition 12.10.11 has mass zero, which gives the equality

The coefficient measures of the positive current $A$ are controlled by this trace measure, which is absolutely continuous. Thus $A$ has a matrix density that is positive semidefinite almost everywhere, with coefficients in $L^ n$ because $a\in L^ n$.

For every smooth Hermitian form $\beta $, lemma 13.14.12 also gives

The resulting pointwise matrix inequalities hold almost everywhere, simultaneously for all positive definite matrices: on each coordinate chart, first use a countable dense family of constant positive matrices, realized on smaller charts by global smooth Hermitian forms, and then use continuity in the matrix. The variational formula for $(\det P)^{1/n}$ in the proof of lemma 12.7.2 therefore yields

Here $A^ n$ denotes only the determinant of the measurable coefficient matrix of $A$; no nonlinear product of currents is used. The trace equality gives the reverse inequality, by the arithmetic–geometric mean inequality. Equality then forces all eigenvalues of $A$ relative to $\gamma $ to equal $a$, also where $a=0$. Consequently $A=a\gamma $ as currents.

By lemma 5.22.4, applied with $\theta =\omega $, both $a$ and $u$ are smooth, and $f^{1/n}=b^{-1/n}a(\gamma ^ n/\mathrm{d}V)^{1/n}$ has a smooth nonnegative representative. Since $u$ is smooth, we may now take the top power: $(\omega +\mathrm{dd}^{\mathrm{c}}u)^ n=a^ n\gamma ^ n=bf\, \mathrm{d}V$. Finally, the function $a$ was defined from $f$, $b$ and $\gamma $ alone, so two threshold subsolutions $u_1,u_2$ satisfy $\omega +\mathrm{dd}^{\mathrm{c}}u_1=a\gamma =\omega +\mathrm{dd}^{\mathrm{c}}u_2$. Their smooth difference is therefore pluriharmonic, hence constant on the connected compact manifold $X$. $\square$

Proof. By corollary 12.10.12, $f$ agrees almost everywhere with a smooth function, so it belongs to $L^ p$ for every finite $p$. By proposition 15.28.7, the constant $c$ for which $\mathrm{MA}_\omega (v)=cf\, \mathrm{d}V$ has a bounded solution is $b_\omega (f)$. Thus proposition 15.20.1, with the trivial symmetry group, shows that the normalized family $u_\lambda -\sup _ Xu_\lambda $ is uniformly precompact as $\lambda \downarrow 0$ and that every cluster limit is a normalized solution with this constant. Each cluster limit is therefore a normalized threshold subsolution and equals $u$ by corollary 12.10.12. If the whole family did not converge uniformly, a sequence staying at a fixed positive uniform distance from $u$ would have a uniformly convergent subsequence, a contradiction. $\square$

Proof. The Kähler formula in proposition 15.28.7 gives $b_\omega (f)>0$. If a smooth strictly positive maximizer exists, the proof of corollary 12.10.12 gives

with $a,u$ smooth and $a\not\equiv 0$. Since $\mathrm{d}\omega =0$, the first-order equation in lemma 5.22.4 becomes $\mathrm{d}a=-aC$ for a smooth one-form $C$. Along any smooth path $\sigma $, this is the linear scalar equation

whose solution is the initial value multiplied by a positive exponential. Since $X$ is connected and $a\not\equiv 0$, it follows that $a>0$ everywhere, and $f=b_\omega (f)^{-1}a^ n\gamma ^ n/\mathrm{d}V$ is smooth and strictly positive. Conversely, if $f$ is smooth and strictly positive, the smooth equation and theorem 12.2.3 provide the maximizing Gauduchon form. $\square$

Proof. Fix a Gauduchon metric $\gamma $ and a point $x$. Choose a smooth real function $\psi $ such that $\mathrm{dd}^{\mathrm{c}}\psi \leq -\gamma $ on a coordinate neighborhood $U$ of $x$; in holomorphic coordinates $z=(z_1,\ldots ,z_ n)$ centered at $x$, such a function is obtained by multiplying $-M|z|^2$, with $M$ large enough, by a cutoff that equals one near $x$, and then shrinking $U$.

Choose a smooth function $\chi $ with $0\leq \chi \leq 1$ that vanishes on a smaller neighborhood of $x$ and equals one outside $U$. Put

Since $X$ is compact, $\omega $ is strictly positive for sufficiently large $L>0$: on $U$ the term $-\mathrm{dd}^{\mathrm{c}}\psi $ is at least $\gamma $, and outside $U$ the term $L\gamma $ dominates $\mathrm{dd}^{\mathrm{c}}\psi $. Set $f=a^ n\gamma ^ n/\mathrm{d}V$. The smooth function $\psi $ satisfies $\omega +\mathrm{dd}^{\mathrm{c}}\psi =a\gamma \geq 0$ and $\mathrm{MA}_\omega (\psi )=f\, \mathrm{d}V$, so $b_\omega (f)\geq 1$.

Stokes' theorem and $\mathrm{dd}^{\mathrm{c}}\gamma ^{n-1}=0$ give

Testing (15.29.3) with $\gamma $ therefore gives the opposite inequality $b_\omega (f)\leq 1$. Hence $b_\omega (f)=1$, the metric $\gamma $ attains the infimum, and the normalized current $\gamma ^{n-1}/\int _ X\omega \wedge \gamma ^{n-1}$ is a smooth strictly positive maximizer of $D_{f^{1/n}}$. If $\omega $ were closed, corollary 12.10.14 would force $f$ to be strictly positive, contrary to its vanishing on an open set. Thus $\omega $ is nonclosed. $\square$

Proof. Put $p=f^{1/n}$ and $A=\omega +\mathrm{dd}^{\mathrm{c}}u$, where $u$ is any threshold subsolution; one exists by theorem 15.28.4. Fix a smooth strictly positive $T_0\in \mathscr G_\omega $, and for real $\varepsilon >0$ set $T_\varepsilon =(T+\varepsilon T_0)/(1+\varepsilon )$. Each $T_\varepsilon $ is a smooth strictly positive member of $\mathscr G_\omega $, so proposition 12.10.11 gives

For a smooth semipositive form $T$ with coefficient matrix $P'$ relative to $\omega ^ n$, (12.7.1) gives $k_ T\, \mathrm{d}V=f_0(\det P')^{1/n}\omega ^ n$, so that $k_ T=f_0^{1-n}(\det P')^{1/n}$ is a continuous function of that matrix. Thus $k_{T_\varepsilon }\to k_ T$ uniformly on the compact manifold $X$. Since $p\in L^1$, passing to the limit gives

The left side has mass one by Stokes' theorem and $\mathrm{dd}^{\mathrm{c}}T=0$, and the right side has mass $b^{1/n}D_ p(T)=1$ because $T$ maximizes. Their difference is positive and has mass zero, hence

On $U=X\setminus Z$ write $T=\gamma ^{n-1}$, where $\gamma $ is smooth and strictly positive, and set $a=(bf\, \mathrm{d}V/\gamma ^ n)^{1/n}$. The trace equality becomes $A\wedge \gamma ^{n-1}=a\gamma ^ n$ on $U$. On relatively compact subsets of $U$, it controls every coefficient measure of $A$ by an $L^ n$ density. The trace inequalities for all Hermitian forms and the matrix arithmetic–geometric mean inequality, used exactly as in the proof of corollary 12.10.12, force $A=a\gamma $ on $U$. This step uses only the positivity of $\gamma $ on $U$, since the Hermitian test forms needed there can be taken to be global forms obtained by extension from smaller coordinate charts.

The local regularity lemma lemma 5.22.4 makes $u$ and $a$ smooth on $U$. Taking the top power proves the equation on $U$, and $f^{1/n}=b^{-1/n}a(\gamma ^ n/\mathrm{d}V)^{1/n}$ is smooth and nonnegative there. Both $\mathrm{MA}^{\mathrm{ext}}_\omega (u)$ and $f\, \mathrm{d}V$ put no mass on the pluripolar set $Z$. Hence the equality holds on $X$, and its total mass is $b\int _ Xf\, \mathrm{d}V<+\infty $.

It remains to prove uniqueness. Let $v$ be another threshold subsolution and put $w=\max (u,v)$. By proposition 13.14.6,

Here the pole sets carry no mass for any of the measures involved. Thus $w$ is also a threshold subsolution, and the preceding argument gives the same smooth current $\omega +\mathrm{dd}^{\mathrm{c}}u=\omega +\mathrm{dd}^{\mathrm{c}}v=\omega +\mathrm{dd}^{\mathrm{c}}w=a\gamma $ on $U$.

We apply lemma 5.38.2 to $u\leq w$ and to $v\leq w$: their complex Hessians agree on $X\setminus Z$, so $w-u$ and $w-v$ are constant. Hence $u-v$ is constant. $\square$

Proof. Write $p=f^{1/n}$ and introduce the probability measure

It has mass one because $T$ maximizes. Its density belongs to $L^ n$, since $p\in L^ n$ and the smooth semipositive form $T$ has bounded determinant density $k_ T$. By lemma 5.26.2, $\omega $-plurisubharmonic functions with supremum zero have uniformly bounded moments of every finite order. Hölder's inequality with the exponents $n$ and $n/(n-1)$, which is finite because $n\geq 2$, therefore gives

Since $\mathrm{MA}^{\mathrm{ext}}_\omega (v_\lambda )=c_\lambda \mathrm{e}^{\lambda v_\lambda }f\, \mathrm{d}V$, the trace inequality of lemma 13.14.12, applied with the positive regularizations of $T$ and passed to the limit as in the preceding proof, gives $(\omega +\mathrm{dd}^{\mathrm{c}}v_\lambda )\wedge T\geq (c_\lambda /b)^{1/n}\mathrm{e}^{\lambda v_\lambda /n}\, \nu $. Integrating, we obtain

where the last step is Jensen's inequality. This proves the upper bound for $c_\lambda $. The lower bound and the comparison

follow from corollary 15.35.3 applied with $v=u_0$, and together with the upper bound they give the common lower barrier $u_0-C$.

Take any sequence $\lambda _ j\downarrow 0$. By theorem 5.25.2, a subsequence of $(v_{\lambda _ j})$ converges in $L^1$ to some $v\in \mathrm{PSH}(X,\omega )$ with supremum zero, and proposition 15.28.6 makes $v$ a threshold subsolution. Uniqueness in theorem 12.10.17 therefore gives $v=u_0$. Since every sequence tending to zero has a subsequence with this same $L^1$ limit, the whole family converges. $\square$

Proof. Fix a sequence $\lambda _ j\downarrow 0$. Proposition 12.10.20 gives $u_0-C\leq v_{\lambda _ j}\leq 0$ and $v_{\lambda _ j}\to u_0$ in $L^1$. For all $j$ with $\lambda _ j\leq 1$,

The bounded functions $|\mathrm{e}^{v_{\lambda _ j}}-\mathrm{e}^{u_0}|$ tend to zero in $\mathrm{d}V$-measure. Since $f\in L^1$, their integrals against $f\, \mathrm{d}V$ tend to zero, so the error term in (15.9.2) tends to zero. We apply proposition 15.9.3 with $\theta =\omega $ and $\varphi =u_0$; its additional hypothesis that the $L^1$ limit $u$ satisfies $u\preceq \varphi $, in the notation of definition 16.7.1, holds trivially, since $u=u_0$. This proves convergence in capacity along every sequence, and hence for the family. Finally, if $u_0$ is bounded, a constant below it is a competitor for $P_{\delta \omega }(u_0)$, so this envelope is not identically $-\infty $. $\square$

Proof. By theorem 11.1.7, every bounded $\omega $-plurisubharmonic function has mass $V$, and truncation therefore bounds every extended mass by $V$. For $f_ k=\min (f,k)$, integration of the bounded equation gives $c_\omega (f_ k)=V/\int _ Xf_ k\, \mathrm{d}V$, so the definition (15.28.1) gives $b_\omega (f)=V/M$. A threshold subsolution has mass at least $b_\omega (f)M=V$. Its mass is consequently $V$, and the nonnegative difference between its measure and $b_\omega (f)f\, \mathrm{d}V$ has mass zero.

Now let $u\in \mathrm{PSH}(X,\omega )$ have extended mass $V$. Normalize $u\leq 0$, put $\mu _ u=\mathrm{MA}^{\mathrm{ext}}_\omega (u)$, and fix $a>1$. For $u_ j=\max (u,-j)$, let $w_ j=P_\omega (au_ j)$. Comparison with constants gives $-aj\leq w_ j\leq 0$. The bounded contact theorem theorem 16.28.4 shows that $\mathrm{MA}_\omega (w_ j)$ is concentrated on the contact set $C_ j=\{ w_ j=au_ j\} $. Since the two functions are canonical representatives and $w_ j\leq au_ j$ almost everywhere, this inequality holds everywhere. Both functions are $a\omega $-plurisubharmonic, so positivity and proposition 13.14.6 give

For integers $j\geq N$, locality of the products on $\{ u>-N\} $ and the mass identity for bounded functions yield

Choose $N$ with $a^ n\mu _ u(\{ u\leq -N\} )<V$; this is possible because $\mu _ u$ is finite and puts no mass on the pole set of $u$. If $\sup _ Xw_ j<-aN$, the contact set $C_ j$ lies in $\{ u_ j<-N\} $, and the preceding inequalities contradict $\int _ X\mathrm{MA}_\omega (w_ j)=V$. Hence $\sup _ Xw_ j\geq -aN$ for $j\geq N$, and for smaller $j$ the bound $\sup _ Xw_ j\geq -aj$ suffices.

The envelopes $w_ j$ decrease with $j$, and by compactness of normalized quasi-plurisubharmonic functions this uniform lower bound for their suprema prevents their limit from being identically $-\infty $. Their limit $w\in \mathrm{PSH}(X,\omega )$ satisfies $w\leq au$, also at the poles of $u$. With $\delta =1/a$, the function $\delta w\in \mathrm{PSH}(X,\delta \omega )$ is therefore a competitor for $P_{\delta \omega }(u)$, so this envelope is not identically $-\infty $. Since $a>1$ was arbitrary, this holds for every $0<\delta <1$, and the Lelong estimate preceding the theorem completes the proof. The normalization $u\leq 0$ does not affect any conclusion. $\square$

Proof. Choose $\alpha <\beta <1$ and $\varepsilon >0$ with $\varepsilon \beta C\leq \delta $. The number

is finite, and $v_\delta =-\varepsilon (1-\psi )^\beta -K\leq u$. The function $t\mapsto -\varepsilon (1-t)^\beta $ is increasing and convex on $t\leq 0$, with derivative at most $\varepsilon \beta $. Applying the chain inequality for quasi-plurisubharmonic functions to $\max (\psi ,-j)$ and passing to the decreasing limit as $j\to \infty $, we obtain $\mathrm{dd}^{\mathrm{c}}v_\delta \geq -\varepsilon \beta C\omega \geq -\delta \omega $. The limit is not identically $-\infty $ because $\psi $ is finite almost everywhere; at the poles of $\psi $ it takes the value $-\infty $, where the comparison with $u$ still holds. Thus $v_\delta \in \mathrm{PSH}(X,\delta \omega )$ is a competitor for $P_{\delta \omega }(u)$ for each $\delta $, and the Lelong estimate preceding theorem 12.10.23 proves the last assertion. $\square$

Proof. Put $U=X\setminus Z$ and $A=\omega +\mathrm{dd}^{\mathrm{c}}\Phi _ f$. By theorem 12.10.17, $A$ is smooth on $U$ and $\Phi _ f$ satisfies the equation on $X$; the root $\gamma $ is positive on $U$. Stokes' theorem gives $\int _ XA\wedge T=1$, while attainment gives

The difference $A\wedge \gamma ^{n-1}-(A^ n/\gamma ^ n)^{1/n}\gamma ^ n$ is nonnegative and continuous on $U$ by the arithmetic–geometric mean inequality, and its integral vanishes. The equality case gives $A=r\gamma $ on $U$ with the smooth function $r=\operatorname {tr}_\gamma A/n\geq 0$; where $A^ n=0$, equality forces $A=0$ and $r=0$.

Since $\mathrm{dd}^{\mathrm{c}}\Phi _ f$ is closed,

For a semipositive form $\gamma $ of rank at least two, the real-linear map $W_\gamma :\xi \mapsto \xi \wedge \gamma $ from real one-forms to three-forms is injective at every point. Indeed, write $\gamma =\sum _ k\lambda _ k\, \mathrm dx_ k\wedge \mathrm dy_ k$ at the point in suitable real coordinates $z_ k=x_ k+\mathrm iy_ k$, with $\lambda _ k\geq 0$ and at least two $\lambda _ k$ positive. For each index $j$ choose $k\neq j$ with $\lambda _ k>0$. The coefficient of $\mathrm dx_ j\wedge \mathrm dx_ k\wedge \mathrm dy_ k$ in $\xi \wedge \gamma $ is $\lambda _ k$ times the coefficient of $\mathrm dx_ j$ in $\xi $, and similarly for $\mathrm dy_ j$. Hence $\xi \wedge \gamma =0$ forces $\xi =0$.

Choose an auxiliary Riemannian metric. The left inverse $L_\gamma =(W_\gamma ^*W_\gamma )^{-1}W_\gamma ^*$ is smooth and uniformly bounded on compact $X$. Thus

with smooth one-forms on $X$. In particular $|\mathrm d\log (1+r)|\leq C$ on $U$.

Points of $U$ can be joined by short paths avoiding $Z$. In a coordinate ball, stratify $Z$ into countably many smooth strata of real dimension at most $2n-2$. For a fixed point outside $Z$, the endpoints of segments from it that meet one stratum lie in a cone of dimension at most $2n-1$, which is Lebesgue-null. Given two nearby points outside $Z$, choose a generic intermediate point in a ball of radius comparable to their distance; both connecting segments then avoid $Z$, and their combined length is bounded by a fixed multiple of that distance. Using a finite connected cover by coordinate balls, with intermediate points chosen in the overlaps, we also obtain a uniform bound for the lengths of connecting paths in $U$.

The bound on $\mathrm d\log (1+r)$ now bounds $r$ on $U$, and (12.10.27) bounds its derivative. The short paths therefore give a unique locally Lipschitz extension of $r$ across $Z$. Since $Z$ is Lebesgue-null, the weak derivative of the extension satisfies (12.10.27) on $X$. The right side is continuous, so $r$ is $C^1$, and iterating makes it smooth.

Set $S=r\gamma $. It is smooth and semipositive, and $\mathrm dS=\mathrm d\omega $ on $X$ by density of $U$. The current $R=A-S$ is therefore closed, and it is supported on $Z$. It is also positive: since $A=S$ off $Z$ and $S$ puts no mass on $Z$, the current $R$ is the restriction of the positive current $A$ to $Z$. The support theorem recalled in section 5.38 expresses $R$ as a nonnegative combination of currents of integration along the components of $Z$ of codimension one. Locally $S-\omega $ is smooth and closed, so it has a smooth local $\mathrm{dd}^{\mathrm{c}}$-potential, and the coefficients of $R$ are the generic Lelong numbers of $\Phi _ f$ along these components. All of them vanish, so $\mathrm{dd}^{\mathrm{c}}\Phi _ f=S-\omega $ is smooth. Taking a Euclidean trace and applying elliptic regularity gives a smooth function equal to $\Phi _ f$ almost everywhere, and hence everywhere, since $\Phi _ f$ is its canonical representative. $\square$

Proof. The form is invariant under $z\mapsto 2z$, so it descends to $X$, and it is positive. For $s=|z|^2$ and $F(s)=s^{-1}$, the identity $sF''+2F'=0$ gives $\mathrm{dd}^{\mathrm{c}}\omega =\mathrm{dd}^{\mathrm{c}}(s^{-1})\wedge \mathrm{dd}^{\mathrm{c}}s=0$, while $\omega $ is not closed. Choose a nonconstant smooth real $q$ and small nonzero $\varepsilon $ such that $A=\omega +\mathrm{dd}^{\mathrm{c}}(\varepsilon q)>0$. Put $V=\int _ X\omega ^2$, $\mathrm{d}V=\omega ^2/V$, $f=A^2/\mathrm{d}V$ and $u=\varepsilon q-\sup _ X\varepsilon q$. Invariance of the mass gives $\int _ Xf\, \mathrm{d}V=\int _ XA^2=V$, hence $b_\omega (f)=1$ by theorem 12.10.23.

The form $T=A/V$ satisfies $\mathrm{dd}^{\mathrm{c}}T=0$, $\int _ X\omega \wedge T=1$ and $D_{\sqrt f}(T)=1$. For every smooth strictly positive $Q\in \mathscr G_\omega $, the arithmetic–geometric mean inequality and Stokes' theorem give $D_{\sqrt f}(Q)\leq \int _ XA\wedge Q=1$. Thus $T$ is a smooth strictly positive maximizer, and since $u$ is a threshold solution with supremum zero, corollary 12.10.12 gives $\Phi _ f=u$. This is a bounded nonconstant example. $\square$

Proof. Let

The quadratic form is nonsingular, and $F$ is finite of degree two, with smooth ramification divisor $Z=\{ w=0\} $. Local branch coordinates show that $\mathrm dF$ has rank $n-1$ on $Z$ and $n$ elsewhere. Let $\omega $ be the restriction to $X$ of the normalized Fubini–Study form of $\mathbb P^{n+1}$, and let $A=F^*\omega _{\mathrm{FS}}$, where $\omega _{\mathrm{FS}}$ is the normalized Fubini–Study form of $\mathbb P^ n$. Then

is a smooth function on $X$ with $-\log 2\leq u\leq 0$, $\sup _ Xu=0$ and $A=\omega +\mathrm{dd}^{\mathrm{c}}u$. The change of variables for the branched cover gives $\int _ XA^ n=2$, and Stokes' theorem gives $V=\int _ X\omega ^ n=2$.

Set $\mathrm{d}V=\omega ^ n/V$, $f=A^ n/\mathrm{d}V$, $T=A^{n-1}/V$ and $\gamma =A/V^{1/(n-1)}$. Since $\omega $ is closed, theorem 12.10.23 gives $b_\omega (f)=V/\int _ Xf\, \mathrm{d}V=1$. Stokes' theorem gives $\mathrm{dd}^{\mathrm{c}}T=0$ and $\int _ X\omega \wedge T=1$, and computing the determinant density gives $D_{f^{1/n}}(T)=1$. The arithmetic–geometric mean inequality, which remains valid for semipositive $A$, bounds $D_{f^{1/n}}(Q)$ by $\int _ XA\wedge Q=1$ for every smooth strictly positive $Q\in \mathscr G_\omega $. Thus $T$ is a maximizer, strictly positive off the proper analytic set $Z$, and theorem 12.10.17 gives $\Phi _ f=u$. Its root $\gamma $ has rank $n-1\geq 2$ on $Z$ and is positive elsewhere, as required. $\square$

Proof. By proposition 15.28.7, the constant $c$ for which $\mathrm{MA}_\omega (v)=cf\, \mathrm{d}V$ has a continuous solution is $b_\omega (f)$. The continuous normalized solution is therefore a threshold solution, and it equals $u_0$ by the uniqueness in theorem 12.10.17. For $L^ p$ data the greatest exponential solution coincides with the bounded solution, as explained after theorem 15.35.2. By proposition 15.20.1, applied with the trivial group, the normalized family is uniformly precompact, and every cluster limit solves the threshold equation. Uniqueness identifies every cluster limit with $u_0$, which proves uniform convergence of the whole family. $\square$

Proof. The numerator in the definition of $\psi $ vanishes only at $o$, and locally $\omega +\mathrm{dd}^{\mathrm{c}}\psi =\mathrm{dd}^{\mathrm{c}}\log \sum _{i=1}^ n|z_ i|^2\geq 0$. The function $\chi (t)=1-(1-t)^\alpha $ is increasing and convex on $t\leq 0$, with $0\leq \chi '\leq \alpha $. Applying the chain inequality to bounded truncations of $\psi $ and taking the decreasing limit, we obtain $u=\chi \circ \psi \in \mathrm{PSH}(X,\omega )$, with

The function $u$ is smooth off $o$, and it attains its supremum zero on $\{ z_0=0\} $. Since $u=-(1-\psi )^\alpha +1$, proposition 12.10.24 shows that all the envelopes $P_{\delta \omega }(u)$ are not identically $-\infty $.

To compute the mass, use the chart $z_0=1$ and write $s=\sum _{i=1}^ n|z_ i|^2$. A local potential $F$ for $A$, as a function of $s$, and the function $q=sF'$ are

The eigenvalues of $A$ are $q/s$ in the tangential directions and $q'$ in the radial direction. Polar integration gives the radial top-degree measure $nq^{n-1}q'\, \mathrm{d}s$, where the angular normalization follows from the Fubini–Study case $q=s/(1+s)$, whose radial density is $ns^{n-1}(1+s)^{-n-1}\, \mathrm{d}s$ and whose total mass is one. Consequently, for $0<s_1<s_2$,

Here $q(s)\to 0$ as $s\downarrow 0$ and $q(s)\to 1$ as $s\to \infty $. On each compact annulus the extended measure equals the smooth top power, and it puts no mass on $o$ or on the hyperplane at infinity, since both are pluripolar. Exhausting the affine chart by annuli therefore gives extended mass exactly one. (This computation does not use the vanishing of the Lelong numbers.)

Near $o$ write $L=1-\log s$. Differentiating (12.10.34) gives

The smooth volume form, written in terms of $s$, is comparable to $s^{n-1}\, \mathrm{d}s$, so the contribution of a neighborhood of $o$ to $\int _ Xf\, \mathrm{d}V$ is comparable to $\int ^\infty L^{-1-n(1-\alpha )}\, \mathrm{d}L$, which is finite. For every $p>1$, the corresponding integral of $f^ p$ contains an exponentially growing factor $\exp (n(p-1)L)$ and diverges. The measure is smooth off $o$ and puts no mass on $o$, so it is $f\, \mathrm{d}V$ on all of $X$. Since its total mass is one and $\omega $ is closed, theorem 12.10.23 gives $b_\omega (f)=1$, and hence $\rho _\omega (f^{1/n})=b_\omega (f)^{-1/n}=1$.

Suppose that a smooth semipositive $T\in \mathscr G_\omega $ attained this value. The positive current $A$ satisfies $\int _ XA\wedge T=1$ by Stokes' theorem for currents. On $X\setminus \{ o\} $, the matrix arithmetic–geometric mean inequality gives

The right side has total mass one by attainment, so the nonnegative difference of the two sides vanishes off $o$. Since $A$ is smooth and strictly positive there, the equality case gives $T=tA^{n-1}$ on $X\setminus \{ o\} $, also at points where $T=0$: applying the arithmetic–geometric mean inequality to the positive semidefinite coefficient matrix of $T$ conjugated by $A^{1/2}$ shows that all its eigenvalues agree, possibly at zero. The function $t=(T\wedge A)/A^ n$ is smooth and nonnegative on $X\setminus \{ o\} $.

Since $A$ is closed, the equation $\mathrm{dd}^{\mathrm{c}}T=0$ becomes $\mathrm{dd}^{\mathrm{c}}t\wedge A^{n-1}=0$, a locally strictly elliptic scalar equation on $X\setminus \{ o\} $. All eigenvalues of $A$ tend to infinity at $o$, and since $n\geq 2$, so do all eigenvalues of $A^{n-1}$, which are products of $n-1$ eigenvalues of $A$. Since the coefficient matrix of the smooth form $T$ is bounded, it follows that $t\to 0$ at $o$.

Set $t(o)=0$. The resulting continuous nonnegative function on compact $X$ is not identically zero, by the normalization of $T$, so it attains a positive maximum away from $o$. The strong maximum principle on the connected manifold $X\setminus \{ o\} $ forces it to be constant there, contrary to its zero limit at $o$. This excludes every smooth semipositive maximizing test; no assumption on the degeneracy set of $T$ was needed. $\square$

Proof. Only a neighborhood of $o$ can contribute a divergence. There the preceding estimates give $f\, \mathrm{d}V\asymp L^{-1-n(1-\alpha )}\, \mathrm{d}L$, $\log f\sim nL$ and $-u\asymp L^\alpha $. Hence the entropy integral converges exactly when $n(1-\alpha )>1$, and the weighted integral exactly when $\alpha r<n(1-\alpha )$. In the equality cases both integrals are comparable to $\int ^\infty L^{-1}\, \mathrm{d}L$ and diverge. The negative part of $f\log f$ is bounded, hence integrable for the probability volume, so the entropy criterion is unambiguous. $\square$

Proof. Put $p=f^{1/n}$, use the probability measure $\nu =A_0\wedge T=b^{1/n}p k_ T\, \mathrm{d}V$ from the proof of proposition 12.10.20, where the equality is (12.10.18), and set

Here $R_\lambda \geq 0$ by the trace inequality used in that proof, which also gives $c_\lambda \geq b$ and $\int _ X(-v_\lambda )\, \mathrm{d}\nu \leq C_0$ for a constant $C_0$. Write $s_+=\max (s,0)$ for the positive part of a real number $s$. Since $q_\lambda \geq \mathrm{e}^{\lambda v_\lambda /n}$ and $1-\mathrm{e}^{x}\leq -x$, we obtain

Since $A_\lambda \wedge T$ and $\nu $ both have mass one,

This proves the first estimate, with $C=2C_0/n$, and also gives $R_\lambda (X)\leq C_0\lambda /n$ and $\int _ X|q_\lambda -1|\, \mathrm{d}\nu \leq 2C_0\lambda /n$.

On $U=X\setminus Z$, write $T=\gamma ^{n-1}$ and $A_0=a\gamma $ as in the proof of theorem 12.10.17, and put $\alpha _\lambda =q_\lambda a$. Decompose the positive current $A_\lambda $ into its absolutely continuous part, with matrix density $B_\lambda $ relative to $\gamma $, and its positive singular part. Since $\mathrm{MA}^{\mathrm{ext}}_\omega (v_\lambda )=c_\lambda \mathrm{e}^{\lambda v_\lambda }f\, \mathrm{d}V$, the trace inequalities for all Hermitian forms imply, as in the proof of corollary 12.10.12,

This is a statement about determinants of matrices; no nonlinear product of coefficient measures is involved. Put $s_\lambda =\operatorname {tr}B_\lambda /n$ and $\delta _\lambda =s_\lambda -\alpha _\lambda \geq 0$. On $U$, the absolutely continuous part of the trace error $R_\lambda $ is $\delta _\lambda \gamma ^ n$, and its singular part is the trace measure of the singular part of $A_\lambda $. Thus

and the singular trace mass on $U$ has the same upper bound.

For nonnegative numbers $\ell _1,\ldots ,\ell _ n$ with geometric mean at least $\alpha \geq 0$ and arithmetic mean $s$, the arithmetic–geometric mean inequality and the Cauchy–Schwarz inequality give

Indeed $\sum _ i\sqrt{\ell _ i}\geq n\sqrt\alpha $, while $\sum _ i(\sqrt{\ell _ i}+\sqrt\alpha )^2\leq 4ns$. Applying this to the eigenvalues of $B_\lambda $ relative to $\gamma $ and integrating, with the Cauchy–Schwarz inequality once more, we see that the absolutely continuous part of $A_\lambda -\alpha _\lambda \gamma $ has integrated trace norm at most $2n\sqrt{C_0\lambda /n}$ on $U$. Its singular part has norm at most $C_0\lambda $ in the same metric. Finally

Since $\gamma $ is comparable to the fixed metric on the compact set $K\subset U$, these estimates and $\lambda \leq \sqrt\lambda $ for $0<\lambda \leq 1$ give the second conclusion. $\square$

Proof. Since $u$ is smooth off $Z$, these divisors are among the finitely many components of $Z$ of codimension one. Put $a_ j=\nu (u,D_ j)>0$. By lemma 5.38.1,

Suppose that real numbers $c_1,\ldots ,c_ k$ satisfy $\sum _ j c_ j\{ D_ j\} =0$. Fix a Gauduchon metric $\eta $ and apply lemma 20.4.20 to the currents of integration $[D_ j]$, with the integrable potentials $v_\eta ([D_ j])$ given there. Since the smooth representative in that lemma depends linearly on the Bott–Chern class, we obtain

Choose $t\ne 0$ sufficiently small that $a_ j+tc_ j\geq 0$ for every $j$. The distribution $u_ t=u+th$ satisfies

so it is represented by a function $u_ t\in \mathrm{PSH}(X,\omega )$. On $X\setminus Z$, the function $h$ is smooth and pluriharmonic, so $u_ t$ has the same Monge–Ampère measure as $u$ there. Since neither extended measure charges the pluripolar set $Z$, $u_ t$ is another threshold solution. Uniqueness makes $th=u_ t-u$ constant. Hence $\sum _ jc_ j[D_ j]=0$, and testing near a generic point of each of the distinct divisors $D_ j$ gives $c_ j=0$. $\square$

Proof. Define the holomorphic map $F:\mathbb P^ n\to \mathbb P^ n$ and the semipositive form $A$ by

The homogeneous formula

defines a smooth function on projective space, and $A=\omega +\mathrm{dd}^{\mathrm{c}}u$. The form $A$ is semipositive, and it is strictly positive exactly where every homogeneous coordinate is nonzero. Define

Since $\omega $ and $A$ are closed and cohomologous, Stokes' theorem gives $\int _ X f\omega ^ n=\int _ X\omega \wedge T=1$, and $\mathrm{dd}^{\mathrm{c}}T=0$; thus $T\in \mathscr G_\omega $. Every $v\in \mathrm{PSH}(X,\omega )$ has extended mass at most one: each bounded truncation $\max (v,-k)$ has mass one, and restricting its Monge–Ampère measure to $\{ v>-k\} $ can only decrease that mass. Consequently an inequality $\mathrm{MA}^{\mathrm{ext}}_\omega (v)\geq c f\omega ^ n$ forces $c\leq 1$. The smooth solution $u$ gives the reverse bound, so $b_\omega (f)=1$.

The determinant formula for smooth semipositive forms gives

By (12.10.2) and $\rho _\omega (f^{1/n})=b_\omega (f)^{-1/n}=1$, the form $T$ is a maximizer. It is strictly positive away from the coordinate hyperplanes because $A$ is; on these hyperplanes $A$ has deficient rank, and hence so does the coefficient matrix of $T=A^{n-1}$.

For the density root, use the chart $Z_0=1$ with coordinates $z_1,\ldots ,z_ n$. Pulling back the Fubini–Study volume by $(z_1,\ldots ,z_ n)\mapsto (z_1^2,\ldots ,z_ n^2)$ and including the factor $2^{-n}$ gives

Near a point of $\{ z_1=0\} $ at which all the other coordinates are nonzero, $f^{1/n}$ is $|z_1|^{2/n}$ times a smooth strictly positive function. Restricting to a real line transverse to that hyperplane proves nondifferentiability, also in the case $n=2$, where $|z_1|^{2/n}=|z_1|$. By symmetry of the coordinates, the same holds on every coordinate hyperplane. The absence of a smooth strictly positive maximizer follows from corollary 12.10.14, since $f$ has zeros. Finally, $f$ is bounded, so corollary 12.10.31 gives the uniform convergence of the normalized exponential solutions. $\square$

Proof. Since $\mathrm{MA}^{\mathrm{ext}}_\omega (v_\lambda )=c_\lambda \mathrm{e}^{\lambda v_\lambda }f\, \mathrm{d}V$, the trace inequality of lemma 13.14.12 and Stokes' theorem give, for each $j$ and $\lambda >0$,

The second inequality follows from Jensen's inequality and (12.10.44). Letting $j\to \infty $ and using $R=b^{-1/n}$, we obtain the upper bound for $c_\lambda $. The lower bound and the comparison with every normalized threshold subsolution were proved in corollary 15.35.3; applying that comparison to $\Phi _ f$, we obtain

Now let $v_{\lambda _ k}\to v$ in $L^1$. By proposition 15.28.6, the limit $v$ lies in $\mathcal S_ f$, hence below $\Phi _ f$. The lower bound passes to the limit, so $\Phi _ f-C\leq v\leq \Phi _ f$; in particular $v\sim \Phi _ f$ in the sense of definition 16.7.1. For all $k$ with $\lambda _ k\leq 1$,

The functions $|\mathrm{e}^{v_{\lambda _ k}}-\mathrm{e}^ v|$ are uniformly bounded and tend to zero in $\mathrm{d}V$-measure, so their integrals against the fixed $L^1$ density $f$ tend to zero. Hence the error term in (15.9.2) tends to zero, as required by proposition 15.9.3 with $\varphi =\Phi _ f$. The additional hypothesis $v\preceq \Phi _ f$ of that proposition holds since $v\leq \Phi _ f$, and the condition on $P_{\delta \omega }(\Phi _ f)$ is assumed. That proposition therefore gives convergence in capacity.

The continuity of damped measures under convergence in capacity, theorem 13.15.9, now identifies the equation without a mass-conservation argument. Write $g_ k=c_{\lambda _ k}\mathrm{e}^{\lambda _ k v_{\lambda _ k}}f$. We know $g_ k\to bf$ in $L^1$ and $0\leq g_ k\leq b\mathrm{e}^ Cf$ for large $k$. Hence

in total variation, while theorem 13.15.9 gives the weak limit $\mathrm{e}^ v\mathrm{MA}^{\mathrm{ext}}_\omega (v)$. Dividing by $\mathrm{e}^ v$ off the pole set of $v$, which neither measure charges, gives the equation and the finiteness of the mass. Finally, compactness of normalized $\omega $-plurisubharmonic functions provides an $L^1$ cluster limit along any sequence $\lambda _ k\downarrow 0$, which proves existence. $\square$