Theorem 19.8.11.1E6D5AA611CC In this situation $0\leq d\leq \ell $, and
Moreover,
1E6D5AA611CCTheorem 19.8.11.1E6D5AA611CC In this situation $0\leq d\leq \ell $, and
Moreover,
Proof. Step 1: the section spaces. A germ of a section of $L^ m$ vanishing to order $k$ at $p_ i$ has weighted squared norm comparable to $|z_ i|^{2(k-mc_ i)}$. As in example 5.33.3, it is locally integrable if and only if $k-mc_ i>-1$, that is, $k\geq \lfloor mc_ i\rfloor $; it is bounded if and only if $k\geq mc_ i$, that is, $k\geq \lceil mc_ i\rceil $. Away from the points $p_ i$ the weight is locally bounded and imposes no condition. Since $X$ is compact, this gives the displayed global identifications.
Step 2: the mass. The positive measure $\beta +\mathrm{dd}^{\mathrm{c}}u$ has total mass $\ell $, and it has an atom of mass $c_ i$ at $p_ i$. Subtracting these atoms leaves a positive measure, which proves $d\leq \ell $. This remaining measure does not charge polar sets: after subtraction of the logarithmic singularity, its local potential has a bounded quasi-plurisubharmonic representative, and by proposition 5.28.2 its curvature gives polar sets zero mass. Thus the remaining measure is exactly the extended measure, and its mass is $\ell -d$.
Step 3: growth. If $\ell >d$, the degrees of both line bundles in the display equal $m(\ell -d)+O(1)$ and tend to infinity. Let $g$ be the genus of $X$. By Riemann–Roch and Serre duality, $h^0=\deg +1-g$ once the degree exceeds $2g-2$. Dividing by $m$ proves both limits.
If $\ell =d$, both degrees remain bounded. A line bundle of negative degree has no nonzero section, and one of nonnegative degree $k$ has at most $k+1$ independent sections: the sections vanishing at $k+1$ given distinct points are sections of a line bundle of negative degree, hence zero. Therefore both dimensions are uniformly bounded, and both limits vanish. These arguments include the case $r=0$. $\square$
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