Pome

Theorem 15.27.3C7AA7377C144

Theorem 15.27.3.C7AA7377C144 Suppose $v_-(h)=0$. There is a probability measure $\mu =f\, \mathrm{d}V$, with $f\in L^1(\mathrm{d}V)$ and $f\geq m$ almost everywhere for a constant $m>0$, such that no $u\in \mathrm{PSH}(X,h)$ and $a>0$ satisfy

8D5EB0B8CCD7 15.27.4
\begin{equation} \mathrm{MA}^{\mathrm{ext}}_ h(u)\geq a\mu . \end{equation}

In particular, the equation $\mathrm{MA}^{\mathrm{ext}}_ h(u)=c\mu $ has no solution $u\in \mathrm{PSH}(X,h)$ for any $c>0$; neither a singularity type nor finiteness of the mass of $u$ is imposed.

Proof. By theorem 12.15.4, theorem 12.2.3, $v_-(h)$ is the infimum of the normalization scalars $c_ h(\nu )$ over smooth strictly positive probability measures $\nu $. Since $v_-(h)=0$, we can choose such measures $\nu _ j=f_ j\, \mathrm{d}V$ with

\[ c_ h(\nu _ j)\leq 2^{-2j}\qquad (j\geq 1). \]

Set

\[ \mu =\sum _{j=1}^{\infty }2^{-j}\nu _ j, \qquad f=\sum _{j=1}^{\infty }2^{-j}f_ j. \]

By monotone convergence, $\int _ Xf\, \mathrm{d}V=1$, so $f\in L^1$. Moreover $f\geq \tfrac 12\min _ X f_1=:m>0$, and $\mu $ is non-pluripolar, being absolutely continuous with respect to $\mathrm{d}V$.

Suppose that (15.27.4) held for some $u$ and $a>0$. Since $\mu \geq 2^{-j}\nu _ j$, we would have

\[ \mathrm{MA}^{\mathrm{ext}}_ h(u)\geq a2^{-j}\nu _ j \]

for every $j$, and lemma 15.27.1 would give $a2^{-j}\leq c_ h(\nu _ j)\leq 2^{-2j}$. Thus $a\leq 2^{-j}$ for every $j$, contradicting $a>0$. $\square$

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