Lemma 6.15.7365648DE5857

Proof. We write $\| \cdot \| _ m$ for the norm of $H_ m$. By the definition of $K_ m$, every $h\in H_ m$ satisfies $|h(\zeta )|^2\leq \| h\| _ m^2K_ m(\zeta )$ for all $\zeta \in \Omega $; if $\| h\| _ m=0$, then $h$ vanishes almost everywhere, hence identically.

Let $F\in H_{2m}$. We first extend $F$ from the diagonal to $\Omega \times \Omega $. By corollary 6.3.5, $\Omega \times \Omega $ is a bounded pseudoconvex open set in $\mathbb C^{2n}$, of diameter at most $\sqrt2L$, and the function $p(z,w)=mu(z)+mu(w)$ is plurisubharmonic on it (proposition 5.9.3, proposition 5.9.5). The diagonal $\Lambda =\{ (z,w)\in \mathbb C^{2n}:z=w\} $ is a complex linear subspace of codimension $n$, and the map $z\mapsto (z,z)$ multiplies Euclidean lengths by $\sqrt2$, so Lebesgue measure on $\Lambda $ corresponds to $2^ n$ times Lebesgue measure on $\mathbb C^ n$. On $\Lambda \cap (\Omega \times \Omega )$, identified with $\Omega $, the weight $p$ restricts to $2mu$, so

The extension statement of corollary 6.5.8, applied on $\Omega \times \Omega $ to the weight $p$ and the subspace $\Lambda $, gives $G\in \mathcal O(\Omega \times \Omega )$ with $G(z,z)=F(z)$ and $\int _{\Omega \times \Omega }|G|^2\mathrm e^{-p}\, \mathrm dV_{4n}\leq C^ n(\sqrt2L)^{2n}2^ nc_\varrho ^{-1}\| F\| _{2m}^2$, where $C$ is the constant of theorem 6.5.4. Hence

with $C_\Omega =4^ nC^ nL^{2n}C_\varrho ^2/c_\varrho $.

We now bound $G$ on the diagonal by $K_ m$. For $w\in \Omega $ put

By Tonelli's theorem, $N$ is measurable and $\int _\Omega N(w)\mathrm e^{-mu(w)}\varrho (w)\, \mathrm dV(w)=\| G\| ^2<\infty $; since $\mathrm e^{-mu}\varrho >0$ almost everywhere, $N(w)<\infty $ for almost every $w$. For such $w$, the function $G(\cdot ,w)$ belongs to $H_ m$ with $\| G(\cdot ,w)\| _ m^2=N(w)$, so $|G(z,w)|^2\leq N(w)K_ m(z)$ for every $z$. Fix $z\in \Omega $. Integrating this inequality in $w$, we obtain

so $G(z,\cdot )\in H_ m$ with $\| G(z,\cdot )\| _ m^2\leq K_ m(z)\| G\| ^2$, and therefore $|G(z,w)|^2\leq K_ m(z)K_ m(w)\| G\| ^2$ for every $w$. Taking $w=z$,

Taking the supremum over the unit ball of $H_{2m}$ proves the lemma. $\square$