Theorem 6.15.45AEABFD276A6

Proof. Let $\Omega _0$ be a connected component of $\Omega $, and choose $a\in \Omega _0$ with $u(a)>-\infty $. By corollary 6.5.8, applied on $\Omega $ to the weight $mu$, some element of $H_ m$ takes the value $1$ at $a$, so $H_ m$ contains a function that does not vanish identically on $\Omega _0$. For every $F$ in the unit ball of $H_ m$ that is not identically zero on $\Omega _0$, $\log |F|^2$ is plurisubharmonic on $\Omega _0$ by proposition 5.10.3, and $\log K_ m$ is the supremum of these functions on $\Omega _0$. Since $K_ m$ is continuous by lemma 6.15.3, this family is locally uniformly bounded above, and $\log K_ m$ is upper semicontinuous with values in $[-\infty ,\infty )$, so it equals its upper semicontinuous regularization. Hence theorem 5.10.2 shows that $u_ m$ is plurisubharmonic. The comparison in lemma 6.15.3 gives $u_ m=m^{-1}\log \sum _ i|g_ i|^2+O(1)$ locally, with generators $g_ i$ of $\mathcal I(mu)$; this is the asserted type.

For the lower bound in (6.15.5), let $u(z)>-\infty $; at points of the pole set of $u$ there is nothing to prove. By corollary 6.5.8, applied on $\Omega $ to the weight $mu$, there is $F\in H_ m$ with $F(z)=1$ and $\| F\| _ m^2\leq C_1\mathrm e^{-mu(z)}$, where $C_1$ depends only on $n$ and $L$. Hence $K_ m(z)\geq |F(z)|^2/\| F\| _ m^2\geq C_1^{-1}\mathrm e^{mu(z)}$. For the upper bound, let $\| F\| _ m\leq 1$. The mean-value inequality (6.8.7) of lemma 6.8.6 for $|F|^2$ on $B(z,r)$, where $|B(z,r)|=\pi ^ nr^{2n}/n!$, gives

with $C_2=n!/\pi ^ n$, and taking the supremum over $F$ proves the upper bound.

Now fix $z$ and $r$, and let $m\to \infty $ in (6.15.5): we obtain $u(z)\leq \varliminf _ mu_ m(z)\leq \varlimsup _ mu_ m(z)\leq \sup _{B(z,r)}u$. Letting $r\downarrow 0$ and using the upper semicontinuity of $u$ gives $u_ m(z)\to u(z)$, also at points of the pole set. On a compact set $K\subset \Omega $, a fixed $r<\operatorname {dist}(K,\partial \Omega )$ in the upper bound gives a uniform upper bound for all $u_ m$, while $u_ m\geq u-|\log C_1|$. Since $u\in L^1(K)$, dominated convergence gives $u_ m\to u$ in $L^1(K)$.

For the Lelong numbers, let $g_1,\ldots ,g_ s$ generate $\mathcal I(mu)$ near $z$. The stalk $\mathcal I(mu)_ z$ is not zero: by the first paragraph, applied to the component of $\Omega $ containing $z$, it contains the germ of an element of $H_ m$ that does not vanish identically on that component, and this germ is not zero by the identity theorem. Since $u_ m=m^{-1}\log \sum _ i|g_ i|^2+O(1)$ near $z$, lemma 6.14.7 gives $\nu (u_ m,z)=m^{-1}\operatorname {ord}_ z\mathcal I(mu)$, and (6.15.6) follows from lemma 6.9.1.

Finally, near each point $K_ m\leq C\sum _ i|g_ i|^2$ with $g_ i\in H_ m$, by lemma 6.15.3, and each $|g_ i|^2\mathrm e^{-mu}$ is integrable on $\Omega $. Hence $\exp (m(u_ m-u))=K_ m\mathrm e^{-mu}$ is locally integrable.

Now replace Lebesgue measure by $\varrho \, \mathrm dV$, and let $K^\varrho _ m$ be the corresponding kernel. The space of holomorphic functions is the same, and the two norms satisfy $c_\varrho \| F\| _ m^2\leq \| F\| _{m,\varrho }^2\leq C_\varrho \| F\| _ m^2$. If $\| F\| _ m\leq 1$, then $F/C_\varrho ^{1/2}$ lies in the unit ball for $\varrho $, and if $\| F\| _{m,\varrho }\leq 1$, then $c_\varrho ^{1/2}F$ lies in the unit ball for Lebesgue measure; hence

The proof of plurisubharmonicity applies verbatim to $m^{-1}\log K^\varrho _ m$, with the unit ball taken for $\varrho $ and the continuity of $K^\varrho _ m$ given by lemma 6.15.3. The lower and upper bounds follow from the last display, with the constants $C_\varrho C_1$ and $C_2/c_\varrho $. The convergence and the Lelong numbers then follow as before, since lemma 6.15.3 applies to $K^\varrho _ m$. Finally, $K^\varrho _ m\mathrm e^{-mu}\leq c_\varrho ^{-1}K_ m\mathrm e^{-mu}$ is locally integrable. $\square$