5.29 Hermitian capacityDD310AD8B258

Proof. Cover $X$ by finitely many coordinate balls $B'_ j$, with larger concentric balls $B_ j$, $B'_ j\Subset B_ j$. A local capacity test $0\leq v\leq 1$ on $B_ j$ can be turned into a global test as follows. Choose a smooth function $\chi _ j$ with $-2\leq \chi _ j\leq 0$, equal to zero near $\overline{B'_ j}$ and to $-2$ near $\partial B_ j$. Choose $\delta >0$ so small that

The function

is $\omega $-plurisubharmonic on $X$, because near $\partial B_ j$ the constant branch is the larger one, and it takes values between zero and one. On $B'_ j$ we have $\omega _ V\geq \delta \mathrm{dd}^{\mathrm{c}}v$. By the positivity of bounded products,

The local volume–capacity estimate, summed over the finite cover, proves (5.29.3).

For the second estimate, choose on each $B_ j$ a smooth negative quadratic function $r_ j$ with $\mathrm{dd}^{\mathrm{c}}r_ j\geq \omega $. For every global test $0\leq v\leq 1$, the functions $u+r_ j\leq 0$ and $v+r_ j$ are plurisubharmonic on $B_ j$. Thus

Here the second inequality uses $\omega _ v\leq \mathrm{dd}^{\mathrm{c}}(v+r_ j)$, and the third is the weighted Chern–Levine–Nirenberg inequality, applied to local test functions with a common bound. The $L^1$ bound for normalized potentials in theorem 5.25.2 completes the proof after taking the supremum over $v$. $\square$

Proof. By Hölder's inequality and (5.29.3), the integral is at most

The function $x\mapsto x^{-2}\exp (-ax^{-1/n}/p')$ is bounded on $0<x\leq \operatorname {Cap}(X)<\infty $, and sets of capacity zero have zero volume. $\square$

Proof. Choose intermediate concentric balls and a smooth cutoff $0\leq \chi \leq 1$ supported in the innermost of them and equal to one on $B'$. All integrations below take place away from $\partial B$. Let $p_\varepsilon $ be a nonnegative radial convolution of $p$ on a larger intermediate ball. It is plurisubharmonic, and, for small $\varepsilon >0$,

Both references have a common bounded range. Set $f=|a-b|=2\max (a,b)-a-b$, and put

This is a closed positive current. Factorization and integration by parts give

We check that the last signed measure has variation at most $C(1+M)^2$, uniformly in $\varepsilon $. The term $\chi \mathrm{dd}^{\mathrm{c}}f\wedge T$ is a linear combination of the positive measures formed with $\max (a,b),a,b$, whose masses are at most $CM$ by local Chern–Levine–Nirenberg estimates. The term $f\mathrm{dd}^{\mathrm{c}}\chi \wedge T$ has variation at most $CM$.

For the two terms involving $\mathrm df$, use a cutoff $\eta $ equal to one near $\operatorname {supp}\chi $. If $q$ is any of $a,b,\max (a,b)$, translate it so that $0\leq q\leq 2M$. The bounded chain rule and closedness of $T$ give

The last term is nonnegative before subtraction, and the first is bounded using the local mass of $T$. Since $f=2\max (a,b)-a-b$, its gradient energy is also bounded by $CM^2$. Cauchy–Schwarz against the positive current $T$ now bounds the variation of $\mathrm d\chi \wedge \mathrm d^{\mathrm c}f\wedge T$ and $\mathrm d^{\mathrm c}\chi \wedge \mathrm df\wedge T$ by $CM$. These are all the terms of $\mathrm{dd}^{\mathrm{c}}(\chi f)\wedge T$.

The bounded formulas follow first for local smooth regularizations. To pass to bounded potentials, regularize $a,b,p$ decreasingly on the intermediate balls, write $f$ as the displayed linear combination, and apply lemma 5.27.2 to the expanded identities. The energy bounds are uniform; thus the integration by parts and its estimate survive this limit.

Writing $L=\| a-b\| _{L^1(B)}$, we have proved

For small positive $L$, take $\varepsilon =L^{1/(\alpha +2n)}$. A fixed admissible radius and the bounded local mass handle the remaining $L$. For $L=0$, let $\varepsilon \downarrow 0$ in the same inequality. $\square$

Proof. On each of finitely many coordinate balls, choose a fixed smooth strictly plurisubharmonic quadratic $Q$ such that $a+Q,b+Q,u+Q$ are plurisubharmonic and $\mathrm{dd}^{\mathrm{c}}Q\geq \omega $. There $0\leq \omega _ u\leq \mathrm{dd}^{\mathrm{c}}(u+Q)$, so positivity of bounded products gives $\omega _ u^ n\leq (\mathrm{dd}^{\mathrm{c}}(u+Q))^ n$. Apply lemma 5.29.7 on smaller balls covering $X$. The quadratic cancels in the difference, and the local volumes are equivalent to $\mathrm dV$. Summing proves the assertion. $\square$

Proof. Choose a smooth quadratic $Q$ with $\mathrm{dd}^{\mathrm{c}}Q\geq \Omega $ on $B$ and a fixed constant $D$ such that $w=v+Q-D\leq 0$. For integers $m\geq 1$, set

These are bounded plurisubharmonic truncations; $0\leq f_ m\leq 1$, it vanishes on $\{ w\geq -m\} $, and it equals one on $\{ w<-m-1\} $. By lemma 5.26.2 on $(Y,\Omega )$ and the bounded correction $Q-D$,

for a fixed $a>0$. Use the proof of lemma 5.29.7, with $M=m+1$, keeping its estimate (5.29.8) before optimization. Taking $\varepsilon =\mathrm e^{-am/(\alpha +2n)}$ gives

for some $c>0$. Large $m$ ensure that the convolution radius is admissible; the common local mass handles the other integers. The bounded difference between $w$ and $v$, and then choosing an integer just below the given height, prove the estimate for every real $t\geq 0$. $\square$

Proof. The pluripolar case follows from the recalled theorem and the characterization of capacity-zero sets. Otherwise let $M=\sup V_ K^*<\infty $. Subtracting $\sup _ Kv$ from a normalized competitor, and conversely normalizing competitors for $V_ K$, shows directly that $T(K)=\mathrm e^{-M}$.

If $M\geq 1$, the function $q=V_ K^*/M$ satisfies $0\leq q\leq 1$ and is an admissible capacity test. Since the background is closed,

Here $\omega _{V_ K^*}\leq M\omega _ q$ follows from $M\geq 1$. Thus $M\geq \operatorname {Cap}_{\omega _{\mathrm{FS}}}(K)^{-1/n}$. If $M<1$, $V_ K^*$ itself is a capacity test, so the same supported probability gives $\operatorname {Cap}_{\omega _{\mathrm{FS}}}(K)=1$. Since $T(K)\leq 1$, the stated factor $\mathrm e$ covers this case. $\square$

Proof. Fix nested coordinate balls $B'\Subset B$ and a smooth quadratic $Q$ with $\mathrm{dd}^{\mathrm{c}}Q\geq \omega $ on $B$. The function $p=u+Q$ is plurisubharmonic with a common Hölder bound, and $\omega _ u^ n\leq (\mathrm{dd}^{\mathrm{c}}p)^ n$ there. Regard $B$ as a ball in the affine chart $\mathbb C^ n\subset \mathbb P^ n$ and restrict this local measure to $B'$, extending it by zero. By lemma 5.29.10 on $(\mathbb P^ n,\omega _{\mathrm{FS}})$, it has a uniform exponential tail on every normalized global Fubini–Study potential. Its total mass is uniformly bounded by local Chern–Levine–Nirenberg estimates.

For a non-pluripolar compact $K\subset B'$, choose a normalized potential $v$ with $\sup _ Kv\leq \log T(K)+1$. For small $T(K)$ the tail estimate gives

The common mass bound handles the remaining $T(K)$, and pluripolar sets have zero local product mass. A bounded local potential for $\omega _{\mathrm{FS}}$ turns every global test $0\leq v\leq 1$ into a plurisubharmonic function of a fixed bounded range on $B$. Hence

where the last inequality is the gluing construction of lemma 5.29.2 on the original manifold. These inequalities have the required direction after taking negative inverse powers. Inner regularity of the finite local product extends the bound from compact to Borel subsets of $B'$. Summing over a fixed finite cover proves the first assertion. The second follows by bounding $x^{-s}\exp (-ax^{-1/n})$ on $0<x\leq \operatorname {Cap}_\omega (X)$. $\square$

Proof. Choose open sets $O_ j$ of capacity at most $2^{-j}$ such that the restriction of $f$ to $X\setminus O_ j$ is continuous, and set $G_ j=\bigcup _{k\geq j}O_ k$. Then the sets $G_ j$ decrease, their capacities tend to zero, and $f$ is continuous on the closed complement of each $G_ j$. Choose $M\geq \sup _ Xf$ and define

At a boundary point of $G_ j$, the values of $f_ j$ on the closed complement tend to the value of $f$ there, while the values inside $G_ j$ are at least that value. Thus $f_ j$ is lower semicontinuous. The functions $f_ j$ decrease to $f$ outside $\bigcap _ jG_ j$, a set of capacity zero. Since $f_ j-f$ is supported in $G_ j$, they also converge to $f$ in capacity. $\square$

Proof. Translate the $v_ i$ so that $0\leq v_ i\leq 2A$, and put $q=(v_1+\cdots +v_ k)/B$, where $B\geq \max \{ 2k,2kA,1\} $. Then $0\leq q\leq 1$ and

where all coefficients are positive. Expanding $\omega _ q^ n$ shows that the given mixed product is bounded by a constant multiple of $\omega _ q^ n$, and the definition of capacity proves the first claim. The case $k=0$ follows by testing with $v=0$. For a bounded function $z$, split its integral into the parts over $\{ |z|>t\} $ and over its complement; first use convergence in capacity for fixed $t>0$, and then let $t\downarrow 0$. $\square$

Proof. On a relatively compact coordinate ball, choose a fixed quadratic function $q$ such that the functions $v_ j+q$ are plurisubharmonic; they still have common bounds. Expand

The absolute value of the $\ell $-th summand is bounded by a fixed constant times

where $\kappa _0$ is a fixed positive Euclidean form. This positive mixed measure is bounded, uniformly in $j$, by a constant times the local capacity: add a fixed strictly plurisubharmonic quadratic function to $v_ j+q$, expand the top power of its Hessian, and rescale its uniformly bounded range into $[0,1]$ to apply the definition of capacity. This also covers $\ell =0$. Summing over a finite cover gives the asserted control of the total variations. The ordinary decreasing continuity of Bedford–Taylor products gives weak convergence term by term. For a bounded quasi-continuous coefficient, remove an open set of small capacity and extend the restriction of the coefficient to the compact complement to a continuous function on $X$. The variation bound controls the resulting error uniformly for the sequence and its limit, and letting the capacity of the removed set tend to zero proves the assertion. $\square$

Proof. If the first assertion failed, there would be $\epsilon >0$ and Borel sets $E_ j$ of capacity at most $2^{-j}$ with $\mu (E_ j)\geq \epsilon $. By countable subadditivity of capacity,

Hence $\limsup _ jE_ j$ is pluripolar and has $\mu $-measure zero. On the other hand, since $\mu $ is finite, continuity from above applied to the sets $\bigcup _{k\geq j}E_ k$ shows that $\limsup _ jE_ j$ has measure at least $\epsilon $, a contradiction. If $u_ j\to u$ in capacity, applying the first assertion to the sets $\{ |u_ j-u|>\delta \} $ gives the consequence. $\square$

Proof. In the decreasing case $u$ is a common lower bound for the sequence and its limit; in the increasing case $u_1$ is one. Choose a Hermitian form $\Omega \geq \theta $ and truncate all functions at $-T$. The truncations are bounded, monotone and $\Omega $-plurisubharmonic, so the convergence in capacity of locally bounded monotone sequences, recalled in section 5.27, applies to them on the balls of a finite cover. By quasi-continuity of the common lower bound, there is an open set of arbitrarily small capacity containing its pole set such that the restriction of the lower bound to the compact complement is continuous. There it is finite and hence bounded below, so the capacity of the sublevel set of the lower bound at height $-T$ tends to zero as $T\to \infty $. Since the functions agree with their truncations outside this sublevel set, letting $T\to \infty $ proves the lemma. $\square$

Proof. For $n=1$, Stokes' theorem gives $\int _ X\theta _ v=\int _ X\theta $, so assume $n\geq 2$. Choose a fixed Hermitian form $\Omega \geq \theta $. All currents on the left-hand side are positive, and $\Omega _\psi \geq \theta _\psi $. Hence Stokes' theorem for bounded potentials gives

The first term on the right is bounded independently of $v$. To estimate the second, differentiate:

where the second term is absent for $n=2$. Since $\mathrm{dd}^{\mathrm{c}}\Omega $ and $\mathrm d\Omega \wedge \mathrm d^{\mathrm c}\Omega $ are smooth forms, the absolute values of these terms are bounded by fixed multiples of the positive measures $\Omega ^2\wedge \Omega _\psi ^{n-2}$ and $\Omega ^3\wedge \Omega _\psi ^{n-3}$. The weighted local Chern–Levine–Nirenberg inequality of section 5.27 bounds the integrals of $|v|$ against these measures. Indeed, on each larger coordinate ball, add fixed smooth corrections to $v$ and $\psi $, shift the corrected $v$ so that it becomes negative, expand in the corrected positive Hessian of $\psi $, and apply that inequality on a smaller ball. The resulting bound is

By theorem 5.25.2, normalized $\theta $-plurisubharmonic functions have uniformly bounded $L^1$ norms, and summing the local bounds proves the assertion. $\square$

Proof. The smooth form $\sigma -A\eta $ has integral zero, so the Poisson equation on the compact connected surface $Y$ has a smooth solution $h$, unique up to an additive constant. If $u\in \mathrm{PSH}(Y,\sigma )$, positivity and Stokes' theorem give

Conversely, $\sigma +\mathrm{dd}^{\mathrm{c}}(-h)=A\eta $, so $-h\in \mathrm{PSH}(Y,\sigma )$ when $A\geq 0$. If $A=0$, every positive curvature measure has total mass zero and therefore vanishes. Thus $\mathrm{dd}^{\mathrm{c}}(u+h)=0$ for every $u\in \mathrm{PSH}(Y,\sigma )$, and elliptic regularity and the uniqueness of harmonic functions on $Y$ show that $u+h$ is constant. Normalizing the supremum gives the stated formula. $\square$

Proof. Choose $C>0$ with $\sigma _ j\leq C\eta $ for every $j$. Every sequence of normalized $\sigma _ j$-plurisubharmonic functions then lies in the compact family of normalized $C\eta $-plurisubharmonic functions of theorem 5.25.2. Every $L^1$ limit $u$ of a subsequence is normalized and, by passing to the limit in the sense of distributions, satisfies $\sigma _0+\mathrm{dd}^{\mathrm{c}}u\geq 0$. The preceding lemma gives $u=u_0$. If the first uniform limit failed, a sequence violating it would have a subsequence converging to this same limit, a contradiction.

For a potential $u$ in the $j$-th normalized family, put $T=\sigma _ j+\mathrm{dd}^{\mathrm{c}}u$. Positivity and Stokes' theorem give $\| T\| _{\mathrm{TV}}=A_ j$, and $A_ j\to \int _ Y\sigma _0=0$. Since $\sigma _0+\mathrm{dd}^{\mathrm{c}}u_0=0$,

uniformly over that family. Apply lemma 5.27.11 on a fixed finite cover of $Y$ by concentric coordinate disks. The first limit and the bound above make each local capacity tend to zero uniformly, and the comparison between local and global capacity at the beginning of this section proves the second limit. The identity $\| T\| _{\mathrm{TV}}=A_ j$ proves the final assertion. $\square$