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18.28.38 Bernstein–Markov measuresB2D64A3DB08E

Let $K\subseteq X$ be a closed non-pluripolar subset, and $v\colon K\rightarrow [-\infty ,\infty )$ be a measurable function.

Definition 18.28.39.E365D575C322 Let $\mu $ be a Borel probability measure on $K$. We introduce the following functions on $\mathrm{H}^0(X,L^ k \otimes T)$ ($k\geq 1$), with values possibly equal to $\infty $:

46ED22785DA5 18.28.40
\begin{equation} \begin{aligned} N^ k_{v,\mu }(s) \coloneqq & \left(\int _ K h^ k \otimes h_ T(s,s) \mathrm{e}^{-kv} \, \mathrm{d}\mu \right)^{1/2},\\ N^ k_{v,K}(s) \coloneqq & \sup _{K\setminus \{ v=-\infty \} } \left( h^ k \otimes h_ T(s,s)\mathrm{e}^{-kv}\right)^{1/2}. \end{aligned} \end{equation}
Observe that if $\mu $ puts no mass on $\{ v=-\infty \} $ then we always have
F1111E54880E 18.28.41
\begin{equation} N^ k_{v,\mu }(s)\leq N^ k_{v,K}(s). \end{equation}
We start with the following elementary observation:

Lemma 18.28.42.E97173FEFAFD Let $v_1,v_2\colon K\rightarrow [-\infty ,\infty )$ be two measurable functions such that

  1. 7AB5A208122C

    $v_1\leq v_2$;

  2. 60A51A0A77E1

    $\{ v_1=-\infty \} =\{ v_2=-\infty \} $.

Then for any $s\in \mathrm{H}^0(X,L^ k\otimes T)$ ($k\geq 1$), we have

\[ N_{v_1,K}^ k(s)\geq N_{v_2,K}^ k(s). \]

Proof. This follows immediately from the definition (18.28.40). $\square$

Definition 18.28.43.48A0DBADBA86 Let $(K,v)$ be a weighted subset of $X$. A Borel probability measure $\mu $ on $K$ is Bernstein–Markov with respect to $(K,v)$ if for each $\epsilon >0$, there is a constant $C_\epsilon >0$ such that

FE92486453EB 18.28.44
\begin{equation} N^ k_{v,K}(s) \leq C_\epsilon \mathrm{e}^{\epsilon k} N^ k_{v,\mu }(s) \end{equation}

for any $s \in \mathrm{H}^0(X,L^ k \otimes T)$ and any $k\in \mathbb {Z}_{>0}$. We write $\mathrm{BM}(K,v)$ for the set of Bernstein–Markov measures with respect to $(K,v)$.

As pointed out in [BBWN11], any volume form on $X$ is Bernstein–Markov with respect to $(X,v)$, with $v \in C^\infty (X)$.

Proposition 18.28.45.A86573651304 Assume that $(K,v)$ is a weighted subset of $X$. Then

  1. A757D6ACA763

    $N^ k_{v,K}$ is a norm on $\mathrm{H}^0(X,L^ k\otimes T)$.

  2. 6EBD970292B6

    For any $\mu \in \mathrm{BM}(K,v)$, $N^ k_{v,\mu }$ is a norm on $\mathrm{H}^0(X,L^ k\otimes T)$.

Proof. 1 As $v$ is bounded, $N^ k_{v,K}$ is clearly finite on $\mathrm{H}^0(X,L^ k\otimes T)$. In order to show that it is a norm, it suffices to show that for any $s\in \mathrm{H}^0(X,L^ k\otimes T)$, $N^ k_{v,K}(s)=0$ implies that $s=0$. In fact, we have $s|_ K=0$, hence $s=0$ by the connectedness of $X$ and the assumption that $K$ is non-pluripolar.

2 As $v$ is bounded, clearly $N^ k_{v,\mu }$ is finite and satisfies the triangle inequality. Non-degeneracy follows from the fact that $N^ k_{v,K}$ is a norm and (18.28.44). $\square$

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