PomeProposition 18.31.1.5E124FAEA311 Let $v\in C(X)$. Every smooth positive probability measure on $X$ is Bernstein–Markov for $(X,v)$.
18.31 A local Bernstein–Markov estimate71CAA1E8F43F
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A Bernstein–Markov estimate compares the supremum norm of sections on a weighted compact set with an integral norm, up to an exponential loss in the degree whose rate can be made arbitrarily small. It is what keeps partial Fubini–Study potentials below the weight in the limit. For a smooth positive volume and a continuous weight on the whole manifold, it follows from the submean inequality.
For a compact set $K\subseteq X$ and $w\in C(K)$, write
\[ \| s\| _{m,w,K}=\sup _ K |s|_{h^ m}\mathrm{e}^{-mw/2}. \]
A probability measure $\mu $ on $K$ is Bernstein–Markov for $(K,w)$ if, for every $\epsilon >0$, there is a constant $C_\epsilon $ such that
\[ \| s\| _{m,w,K}\leq C_\epsilon \mathrm{e}^{\epsilon m} \left(\int _ K|s|_{h^ m}^2\mathrm{e}^{-mw}\, \mathrm{d}\mu \right)^{1/2} \]
for all $m\geq 1$ and all $s\in H^0(X,L^ m)$. If $K$ is non-pluripolar, the supremum norm is a norm: the zero set of a nonzero section is an analytic hypersurface, hence pluripolar, and cannot contain $K$. The Bernstein–Markov inequality then shows that the integral
\[ \| s\| _{m,w,\mu }^2=\int _ K|s|_{h^ m}^2\mathrm{e}^{-mw}\, \mathrm{d}\mu \]
is also a norm on $H^0(X,L^ m)$, and in particular on every selected space $V_ m$. The definitions of section 18.30 therefore apply with $(X,v,\nu )$ replaced by $(K,w,\mu )$. We write $B_{m,w,\varphi ,\mu }$ for the resulting partial Bergman kernel on $K$ and $\beta _ m^{(r)}=(r!/m^ r)B_{m,w,\varphi ,\mu }\, \mu $ for the partial kernel measure; the extremal formula and the total mass $r!N_ m/m^ r$ hold as before.
Proof. Let $\nu $ be a smooth positive probability measure, and fix $\epsilon >0$. Cover $X$ by finitely many coordinate charts on which $L$ has holomorphic frames, and choose relatively compact open subsets of these charts that still cover $X$. In a frame $e$ write $|e|_ h^2=\mathrm{e}^{-\chi }$ and $s=f e^{\otimes m}$, so that $|s|_{h^ m}^2\mathrm{e}^{-mv}=|f|^2\mathrm{e}^{-m(\chi +v)}$. By compactness and uniform continuity of the finitely many functions $\chi +v$, there is a radius $\delta >0$ with the following property: every point $x$ lies in one of the smaller sets, the coordinate ball $B(x,\delta )$ lies in the corresponding chart, and the oscillation of $\chi +v$ on $B(x,\delta )$ is at most $2\epsilon $. On these balls the density of $\nu $ with respect to coordinate Lebesgue measure is bounded below by a positive constant independent of $x$.
The submean inequality for the holomorphic function $f$ on $B(x,\delta )$ therefore gives
\[ \begin{split} |s(x)|_{h^ m}^2\mathrm{e}^{-mv(x)} & =|f(x)|^2\mathrm{e}^{-m(\chi +v)(x)}\\ & \leq C_\delta \mathrm{e}^{2\epsilon m} \int _{B(x,\delta )}|f|^2\mathrm{e}^{-m(\chi +v)}\, \mathrm{d}\nu \\ & \leq C_\delta \mathrm{e}^{2\epsilon m}\| s\| _{m,v,\nu }^2, \end{split} \]
with $C_\delta $ independent of $x$ and $m$. Taking the supremum over $x$ and then the square root gives the Bernstein–Markov inequality with $C_\epsilon =C_\delta ^{1/2}$. $\square$
For a general compact set $K$ and weight $w\in C(K)$, the Bernstein–Markov property is not automatic, and we assume it explicitly whenever it is needed.
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