A bounded density against a Hölder reference measure gives an exact homogeneous equation. The coefficient is obtained with the solution. Continuous approximation of the density is needed before passing to smooth reference measures, since weak convergence does not preserve integrals of a fixed Borel function.
Theorem 15.25.1.8CE230CD94D4 Let $h$ be a Hermitian form on a connected compact complex manifold $X$. Suppose $u\in \mathrm{PSH}(X,h)\cap C^{0,\alpha }(X)$ for some $0<\alpha \leq 1$, and let $\nu $ be a finite nonzero positive measure with
\[ \nu \leq A\mathrm{MA}_ h(u),\qquad A>0. \]
There are a constant $c>0$ and a Hölder continuous $w\in \mathrm{PSH}(X,h)$ such that
\[ \sup _ Xw=0,\qquad \mathrm{MA}_ h(w)=c\nu . \]
In particular, its mass is finite and positive. No mass bound over all bounded $h$-plurisubharmonic functions is assumed.
Proof.
Step 1: approximate the density and reference. Write $\sigma =\mathrm{MA}_ h(u)$ and $\nu =A g\sigma $, with $0\leq g\leq 1$. Choose continuous $g_ l\in [0,1]$ tending to $g$ in $L^1(\sigma )$. Regularity of the finite measure supplies this approximation. The inequality $|s^{1/n}-t^{1/n}|^ n\leq |s-t|$ on $[0,1]$, followed by Hölder's inequality, gives $g_ l^{1/n}\to g^{1/n}$ in $L^1(\sigma )$.
By lemma 5.42.7, there are smooth $v_ j\in \mathrm{PSH}(X,h)$ tending uniformly to $u$ with a common Hölder norm. Their bounded products $\sigma _ j=h_{v_ j}^ n$ tend weakly to $\sigma $. Choose a dense countable family of continuous tests, including the constant one. For each $l$, choose $j(l)\geq l$ so that integration of $g_ l^{1/n}$, and of $g_ l$ multiplied by the first $l$ tests, against $\sigma _{j(l)}-\sigma $ has absolute value at most $1/l$. Write $\widehat v_ l=v_{j(l)}$ and
\[ \nu _ l=A g_ l h_{\widehat v_ l}^ n. \]
The common bounded-reference mass and $0\leq g_ l\leq 1$ give uniformly bounded masses. The diagonal and density approximation imply
\[ \nu _ l\rightharpoonup \nu ,\qquad \lim _ l\int _ Xg_ l^{1/n}h_{\widehat v_ l}^ n =\int _ Xg^{1/n}\, \mathrm d\sigma >0. \]
Discard finitely many terms if needed. Each $\nu _ l$ has a continuous density with respect to smooth volume and positive mass. Thus theorem 14.19.2 gives continuous normalized $h$-plurisubharmonic $\varphi _ l$ and $c_ l>0$ with $h_{\varphi _ l}^ n=c_ l\nu _ l$.
Step 2: control the coefficients and potentials. Since both right-hand sides have smooth-volume densities, lemma 5.28.7 gives
\[ h_{\varphi _ l}\wedge h_{\widehat v_ l}^{n-1} \geq (c_ lA)^{1/n}g_ l^{1/n}h_{\widehat v_ l}^ n. \]
The trace bound in lemma 5.29.20 is uniform for references with a fixed supremum norm, as its proof shows. Integrating therefore bounds $c_ l$ above. By proposition 5.29.13, the solution measures $\mu _ l=h_{\varphi _ l}^ n\leq C h_{\widehat v_ l}^ n$ have a common square-capacity bound.
Fix $\varepsilon =1/2$, the zero competitor, and $D>0$ with $2D<\delta _{1/2}$. The backwards estimate of lemma 14.19.16 gives
\[ \operatorname {Cap}_ h\{ \varphi _ l<\inf _ X\varphi _ l+D\} \geq a_0>0. \]
The normalized tail estimate (5.29.4) then bounds $-\inf _ X\varphi _ l$ uniformly. Applying lemma 14.19.14 at $s=t=D$ also gives
\[ D^ na_0\leq C\mu _ l(X)=Cc_ l\nu _ l(X). \]
Since the masses of $\nu _ l$ are uniformly bounded above, $c_ l$ is bounded away from zero.
Step 3: take a uniform limit. Extract a subsequence with $c_ l\to c>0$ and $\varphi _ l$ converging in $L^1(\mathrm dV)$, using theorem 5.25.2. All potentials now have common upper and lower bounds. The common Hölder norms of $\widehat v_ l$ and corollary 5.29.9 give, uniformly in $j,k$,
\[ \int _ X|\varphi _ j-\varphi _ k|\, \mathrm d\mu _ j \leq C\| \varphi _ j-\varphi _ k\| _{L^1(\mathrm dV)}^{\alpha /(\alpha +2n)} \longrightarrow 0. \]
Use proposition 14.19.23 in both orders to see that this subsequence is uniformly Cauchy. Its continuous limit $w$ is $h$-plurisubharmonic and has supremum zero. Uniform bounded-product continuity and $\nu _ l\rightharpoonup \nu $ yield $\mathrm{MA}_ h(w)=c\nu $. Finally theorem 15.5.4, with reference $u$ and domination constant $cA$, gives the Hölder assertion. Its mass is $c\nu (X)$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.