Lemma 6.16.282376665E5F7

Proof. Fix $j,l$ with $\overline{U'_ j}\cap \overline{U'_ l}\neq \varnothing $, and work in the coordinates of the ball $\mathcal B_{jl}$. By lemma 6.14.2 choose a smooth function $g$ on $\mathcal B_{jl}$ with $\mathrm{dd}^{\mathrm{c}}g=\theta $. The functions $h_ j-g$ and $h_ l-g$ are pluriharmonic on the coordinate balls $B_ j^+$ and $B_ l^+$, so by the same lemma $h_ j-g=\operatorname{Re}b_ j$ and $h_ l-g=\operatorname{Re}b_ l$ with $b_ j,b_ l$ holomorphic on these balls (the constants are absorbed into $b_ j,b_ l$). Put $u=\varphi +g$, a plurisubharmonic function on $\mathcal B_{jl}$. Since $|\mathrm e^{-mb_ j/2}|^2=\mathrm e^{-m\operatorname{Re}b_ j}$, multiplication by $\mathrm e^{-mb_ j/2}$ is an isometry from the weighted space defining $K_{j,m}$ onto the space of holomorphic functions on $B_ j$ with weight $mu$ and volume form $\omega ^ n/n!$. If $K^{mu}_{B_ j}$ denotes the Bergman kernel of the latter space, then $K^{mu}_{B_ j}=\mathrm e^{-m\operatorname{Re}b_ j}K_{j,m}$, and therefore

Hence it suffices to prove $K^{mu}_{B_ j}\leq C_0^2K^{mu}_{B_ l}$ on $\overline{U'_ j}\cap \overline{U'_ l}$ with $C_0\geq 1$ depending only on the fixed geometry.

Choose a smooth function $\rho \colon X\to [0,1]$ with compact support in $B_ j\cap B_ l$, equal to one on a neighborhood of the compact set $\overline{U'_ j}\cap \overline{U'_ l}$ (proposition 5.2.6); this choice belongs to the fixed geometry. Let $r>0$ be the distance, in the coordinates of $\mathcal B_{jl}$, from $\overline{U'_ j}\cap \overline{U'_ l}$ to the support of $\bar\partial \rho $, and let $c>0$ and $C''$ be lower and upper bounds for the density of $\omega ^ n/n!$ with respect to Lebesgue measure on $B_ j\cup B_ l$ in these coordinates. The ball $B_ l$ is pseudoconvex also in the coordinates of $\mathcal B_{jl}$, since a biholomorphism transports a continuous plurisubharmonic exhaustion function (proposition 5.9.3).

To compare the kernels at a point $x\in \overline{U'_ j}\cap \overline{U'_ l}$, we correct $\rho f$, for $f$ holomorphic on $B_ j$, to a holomorphic function on $B_ l$ with the same value at $x$. Let $f$ be holomorphic on $B_ j$ with norm at most one in the weighted space, and extend $\rho f$ by zero to a smooth function on $B_ l$. We apply theorem 6.4.9 on $B_ l$ to the smooth $\bar\partial $-closed form $f\bar\partial \rho $, with the plurisubharmonic function $p=mu+n\log |z-x|^2$ and the smooth function $q=|z|^2$, whose complex Hessian matrix is the identity. On the support of $\bar\partial \rho $ we have $|z-x|^{-2n}\leq r^{-2n}$, so

Hence there is a solution of $\bar\partial w=f\bar\partial \rho $ on $B_ l$ with

and $C$ depends only on the fixed geometry. The function $|z-x|^{2n}\mathrm e^{|z|^2}$ is bounded on the bounded set $B_ l$ by a constant of the same kind, and the density of $\omega ^ n/n!$ is at most $C''$. Hence the squared weighted norm of $w$, for the weight $mu$ and the volume form $\omega ^ n/n!$, is at most a constant $C'$ depending only on the fixed geometry.

Near $x$ we have $\bar\partial w=f\bar\partial \rho =0$, because $\rho =1$ there; hence $\Delta w=0$ near $x$, and by theorem 3.1.11 $w$ is smooth, hence holomorphic, near $x$. Suppose that $w(x)\neq 0$. Then $|w|^2\geq |w(x)|^2/4$ on a ball $B(x,\delta )$ on which $u\leq M$ for some constant $M$, since $u$ is upper semicontinuous, and the last integral is at least

because $\int _{B(x,\delta )}|z-x|^{-2n}\, \mathrm dV=s_{2n-1}\int _0^\delta t^{-1}\, \mathrm dt$ in polar coordinates (3.1.2). Hence $w(x)=0$. The function $F=\rho f-w$ satisfies $\bar\partial F=0$ on $B_ l$, so it is holomorphic by theorem 3.1.11, and $F(x)=f(x)$. Since $0\leq \rho \leq 1$ and $\rho $ vanishes outside $B_ j$, the weighted norm of $F$ on $B_ l$ is at most $1+C'^{1/2}=C_0$. By the definition of the kernel, $|f(x)|^2=|F(x)|^2\leq C_0^2K^{mu}_{B_ l}(x)$, and taking the supremum over $f$ gives $K^{mu}_{B_ j}(x)\leq C_0^2K^{mu}_{B_ l}(x)$.

No bound on $u$, $\theta $ or $m$ enters these constants. Taking $C_ B$ to be the maximum of $2\log C_0$ over the finitely many pairs $(j,l)$, with the roles of $j$ and $l$ also exchanged, proves the lemma. $\square$