Theorem 9.11.71.127BF0018B1A Let $f\in H^ s(Y)$ be strictly positive and set
\[ \nu =\pi _*(f\Omega ),\qquad c=\frac{M}{\int _ Y f\Omega }. \]
If $\| cf-f_\varepsilon \| _{H^ s}\leq \delta $, there is a unique bounded $u\in \mathrm{PSH}(X,h)$ with $\sup _ Xu=0$ and
\[ \mathrm{MA}^{\mathrm{ext}}_{X,h}(u)=c\nu . \]
The function $u$ is continuous and locally Hölder continuous of exponent $1/3$ in ambient coordinates. The measure $\nu $ is non-pluripolar and gives the singular locus zero mass. On the regular locus its density with respect to $h^ n$ is $f/J$, where $J=H^ n/\Omega $; near the cusp this density is $\lambda f/\rho $. The equation holds on every resolution and every common smooth refinement. Every bounded solution of an equation $\mathrm{MA}^{\mathrm{ext}}_{X,h}(u')=c'\nu $ with $c'>0$ has $c'=c$.
Proof.
The function $g=cf-f_\varepsilon $ has zero mean. Let $w$ be the solution given by theorem 9.11.66. The potential $v_0+w$ is bounded and $H$-plurisubharmonic. Descent through the resolution $\pi $ (theorem 9.11.5) gives
\[ u=(v_0+w)\circ \pi ^{-1}-\sup _ Y(v_0+w), \]
which satisfies the disk condition with the background $h$ itself. Since the normalization is a homeomorphism, $u$ is continuous. The definition of the extended measure and theorem 9.11.26 give the equation. Any other bounded solution pulls back to a bounded $H$-plurisubharmonic potential. Since $\pi $ is a Borel homeomorphism, equality of the pushforward measures means equality on $Y$. Subtracting $v_0$ and applying proposition 9.11.69 gives uniqueness. The mass identity of lemma 9.11.68 forces the scalar of every bounded solution to be $M/\nu (X)=c$.
The singular locus pulls back to the divisor $\{ p\} \times E_2\times \cdots \times E_ n$, which has zero smooth volume. If a set is locally pluripolar, a local plurisubharmonic function equal to $-\infty $ on it pulls back to a plurisubharmonic function, not identically $-\infty $, away from that divisor; its pole set has zero smooth volume. Thus $\nu $ is non-pluripolar. On the regular locus, the change of variables formula gives the density $f/J$. Near the cusp $\alpha =0$, so
\[ H^ n=n!\rho \, \kappa _1\wedge \cdots \wedge \kappa _ n, \qquad \Omega =n!\lambda \, \kappa _1\wedge \cdots \wedge \kappa _ n. \]
The ratio gives the density $\lambda f/\rho $; the generic degree of $\pi $ and both factorials have been accounted for.
For the Hölder assertion, put $d=z-z'$ and $a=z+z'$. Then
\[ z^2-z'^2=da,\qquad z^3-z'^3=d(3a^2+d^2)/4. \]
If $|a|\geq |d|/4$, the first modulus is at least $|d|^2/4$, which controls $|d|^3$ on a bounded chart. Otherwise the second modulus is at least $13|d|^3/64$. Hence $|z-z'|\leq C(|z^2-z'^2|+|z^3-z'^3|)^{1/3}$. The $C^1$ regularity of $v_0+w$ in $z$ and in the other product coordinates gives the asserted ambient estimate.
Every resolution $\sigma :Z\to X$ factors through $Y$: in the cusp chart, the quotient $y/x$ is bounded off the exceptional set and extends holomorphically to the smooth source, giving the normalization coordinate. Write $\sigma =\pi \circ \mu $. The map $\mu $ is a proper modification. The bounded pullback products agree off its exceptional locus, and they give this analytic set zero mass by proposition 5.28.2. Hence their pushforwards agree on all of $Y$. This proves the equation on every resolution, and the same argument applies to further common refinements. The full potential is exactly $u\circ \sigma $, which is constant on every fiber, and surjectivity preserves all ordinary comparisons.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.