15.33 Power capacity bounds and threshold equations119B0BE69A7D

Proof. Step 1: the coefficient and root tails. Put $g_ j=\min (g,j)$ and let $\phi _ j$ be normalized continuous solutions of

given by theorem 14.19.2. The definition (15.28.1) gives $c_ j\downarrow b_ h(g)$. Their measures satisfy a power capacity bound with constant $c_ jA$.

Take competitor zero, $H=1$ and $\varepsilon =1/2$ in lemma 14.19.16, lemma 14.19.14. Choose $D>0$ fixed with $2D<\delta _{1/2}$ and put $U_ D=\{ \phi _ j<\inf _ X\phi _ j+D\} $. These estimates give

Combining them yields

Taking the limit proves the lower bound for $b_ h(g)$. The identity (15.29.2), with its full family of Gauduchon tests, gives the root estimate. Apply it to $g=f\mathbf1_{\{ f>K^ n\} }$ whenever this density is nonzero. Its mass tends to zero; the zero density has root norm zero. Thus the tails of $f^{1/n}$ tend to zero in both $L^ n$ and $\rho _ h$, and proposition 15.32.9 gives $f^{1/n}\in \mathcal D_ h^0$.

Step 2: uniform convergence of a subsequence. Now use $f_ j=\min (f,j)$, $c_ j=c_ h(f_ j)$ and normalized continuous solutions $\phi _ j$. Step 1 gives $b_ h(f)>0$, while $c_ j\leq c_1<\infty $. Hence

The first part of the proof of proposition 14.19.23, using the capacity tail estimate, gives one $M\geq 1$ with $-M\leq \phi _ j\leq 0$ for all $j$.

By theorem 5.25.2, extract a subsequence converging in $L^1(\mathrm dV)$ and almost everywhere. Since $f\, \mathrm dV$ is a fixed finite absolutely continuous measure and the potentials have a common bound, dominated convergence gives

along this subsequence as both indices tend to infinity. Apply proposition 14.19.23 in both orders, with $H=M$. The subsequence is uniformly Cauchy. Its limit $\phi $ is continuous, $h$-plurisubharmonic by distributional positivity, and has supremum zero.

Step 3: passage of the products. On a coordinate ball choose a fixed smooth $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq h$, and put $S_ j=\mathrm{dd}^{\mathrm{c}}(\phi _ j+q)$ and $S=\mathrm{dd}^{\mathrm{c}}(\phi +q)$. For every $\ell \geq 1$, the calculus of bounded products gives

The local Chern–Levine–Nirenberg estimate bounds its pairing with a compactly supported smooth test form by $C\| \phi _ j-\phi \| _\infty $. Expand the Hermitian products as in proposition 5.28.2:

Every term converges, with its smooth correction factor, to the corresponding term for $\phi $. On the other hand, $c_ jf_ j\to b_ h(f)f$ in $L^1(\mathrm dV)$. The two limits prove (15.33.4); its prescribed mass is $b_ h(f)\int _ Xf\, \mathrm dV$. $\square$

Proof. Choose a holomorphic ball with coordinates centred at $x_0$ and $R=|z|<1$, and a smooth cutoff $0\leq \chi \leq 1$ supported in it and equal to one on a smaller ball. Put

Set $\psi (x_0)=0$ and extend $\psi =1$ outside the ball. In the smaller ball, $q'=\alpha (1-s)^{-\alpha -1}>0$ and $q''=\alpha (\alpha +1)(1-s)^{-\alpha -2}>0$, so $\psi $ is plurisubharmonic there, including its continuous extension at $x_0$. The transition is smooth and $\mathrm{dd}^{\mathrm{c}}\psi \geq -K_0h$ globally for a fixed $K_0$. For sufficiently small $\delta >0$, the normalized potential $u=\delta \psi -\sup _ X\delta \psi $ satisfies $h+\mathrm{dd}^{\mathrm{c}}u\geq h/2$.

The bounded product gives zero mass to $x_0$. Indeed, after a smooth local correction, its expansion is dominated in total variation by a constant times the top product of a bounded plurisubharmonic function plus a strictly plurisubharmonic quadratic function. Rescaling this sum into $[0,1]$ bounds that top product by local capacity, and a pluripolar point has capacity zero. Off $x_0$ the potential is smooth. Thus the measure is absolutely continuous; write it as $f\, \mathrm dV$. The strict lower bound implies $f\geq c>0$ almost everywhere.

In the smaller ball the tangential and radial eigenvalues of the Hessian of $q(\log R^2)$ are $q'/R^2$ and $q''/R^2$. In the operator convention of the book this gives

where $\mathrm d\lambda $ is Euclidean volume.

Integration over the ball gives

For $k<n$, the degree-$k$ mixed terms with the smooth positive form $h$ have density bounded by $CR^{-2k}(1-\log R^2)^{-k(\alpha +1)}$ and mass $O(r^2)$ on $B_ r$. The expansion has positive terms in this ball. Hence $f\in L^1$. For $p>1$, the pure top term gives a positive multiple of the divergent integral

as a lower bound for $\int _{B_ r}f^ p\, \mathrm dV$, since smooth volume densities are comparable. Thus $f\notin L^ p$. For all sufficiently small $r$,

By (5.29.3),

Let $s=\operatorname {Cap}_ h(E)>0$ be small and choose $r=\exp [-as^{-1/n}/(4n)]$. Splitting $E$ inside and outside $B_ r$ and applying (15.33.9), we obtain

For larger $s$ use the finite total mass; for $s=0$ the volume estimate gives $\mathrm dV(E)=0$. This proves (15.33.2) for every Borel set.

By theorem 15.33.1, there is a normalized continuous $\phi $ with $\mathrm{MA}_ h(\phi )=b_ h(f)f\, \mathrm dV$. The explicit potential $u$ is a subsolution with coefficient one, so theorem 15.28.4 gives $b_ h(f)\geq 1$. Suppose $b_ h(f)>1$, and choose $0<\varepsilon <1$ with $(1-\varepsilon )^ nb_ h(f)>1$. Positive multilinearity gives

Apply theorem 14.19.6 to $u,\phi $ on

For sufficiently small $s>0$ its reverse factor $Q_\varepsilon (s)$ is defined, tends to one as $s\downarrow 0$, and satisfies

Choose $s$ so that $Q_\varepsilon (s)<(1-\varepsilon )^ nb_ h(f)$. The set $U_ s$ is open and nonempty, and $f\geq c>0$ almost everywhere, so its $f\, \mathrm dV$ measure is positive. This contradiction proves $b_ h(f)=1$.

Finally take $r_ K=K^{-1/(4n)}$. The outside bound in (15.33.9) gives $f\leq CK^{1/2}<K$ almost everywhere on $X\setminus B_{r_ K}$ for large $K$. Therefore

Apply (15.33.3) to $g=f\mathbf1_{\{ f>K\} }$ to get (15.33.7). $\square$