Uniform approximation of Nakayama-Zariski decomposition

Let $X$ be a complex smooth projective manifold, $L$ be a pseudo-effective $\mathbb{Q}$-divisor on $X$, and $A$ be an ample divisor. Is it true that, for any $\varepsilon>0$, there exists $\delta$ such that $\sigma_{E}(\|L\|)-\mathrm{ord}_{E}(\|L+\delta A\|)<\varepsilon$ for any prime divisor $E$ on $X$?

Moreover, let $\pi:Y \longrightarrow X$ be a birational projective morphism from a smooth manifold $Y$. Does the inequality $\sigma_{E} (\|f^{\ast}L\|)-ord_{E}(\|f^{\ast}L+\delta A\|)<\varepsilon (1+ ord_{E}(K_{Y/X})$ hold?

Notations: $ord_{E}(\|L\|)=\inf \frac{1}{m}ord_{E} Fix|mL|$, $\sigma_{E}(\|L\|)=\lim_{\delta \rightarrow 0} ord_{E}(\|L+\delta A\|)$.

edited Dec 7 2011 at 12:35

asked Dec 7 2011 at 11:31

Zhengyu Hu
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could you please explain a little bit better your notations, please? – diverietti Dec 7 2011 at 12:14

I add some explanations for the notations... if I don't explain clearly enough, let me know about it... – Zhengyu Hu Dec 7 2011 at 12:36

and $||L||$ is the asymptotic base locus (seen as a scheme)? – diverietti Dec 7 2011 at 16:00

Aren't there only finitely many $E$ for which $\sigma_E(\lVert L \rVert) > 0$, and only for those $E$ can there exist a $\delta$ so $\sigma_E(\lVert L+\delta A\rVert)> 0$? So you only have to worry about a finite set of $E$, and so the first statement at least should be true. – John L. Dec 7 2011 at 21:46