Proposition 6.5.166BCDC65B27D

Proof. Fix a smooth function $\kappa $ on $\mathbb R$ with $0\le \kappa \le 1$, $\kappa =1$ on $(-\infty ,1]$ and $\kappa =0$ on $[3/2,\infty )$, for instance $\kappa (x)=\kappa _0(2x-2)$ with the function $\kappa _0$ fixed at the beginning of section 6.4, and put $\chi (z)=\kappa (|z-a|/r)$. Then $\chi $ is smooth, because $\kappa $ is constant near $0$; it has compact support in $B(a,2r)$, equals one on $B(a,r)$, and satisfies $|\bar\partial \chi |\le c_0/r$ with $c_0=\sup |\kappa '|/2$, because $|\bar\partial |z-a||=1/2$ for $z\ne a$. Extend $\chi h$ by zero to $\Omega $. Then $\alpha =\bar\partial (\chi h)=h\, \bar\partial \chi $ is a smooth $\bar\partial $-closed form supported in the shell $r\le |z-a|\le 2r$.

The function $p+n\log |z-a|^2$ is plurisubharmonic on $\Omega $ by proposition 5.10.3, proposition 5.9.5. We apply theorem 6.4.9 with this function in place of $p$ and with $q=|z-a|^2$, for which $A=I$. Since $\mathrm e^{-|z-a|^2}\le 1$ and $|z-a|^{-2n}\le r^{-2n}$ on the support of $\alpha $, we obtain a solution of $\bar\partial u=\alpha $ with

On $B(a,r)$ we have $\bar\partial u=\alpha =0$, so $u$ is holomorphic there, by lemma 6.4.10(1). Since $p$ is bounded above on $B(a,r)$, the factor $\mathrm e^{-p-|z-a|^2}$ is bounded below there by a positive constant. If $u(a)$ were nonzero, then $|u|$ would be bounded below by a positive constant near $a$, and the integral on the left would diverge, because $\int _{B(a,\rho )}|z-a|^{-2n}\, \mathrm dV$ is a positive multiple of $\int _0^\rho \tau ^{-2n}\tau ^{2n-1}\, \mathrm d\tau =\infty $ by integration in polar coordinates (3.1.2). Thus $u(a)=0$. The function $F=\chi h-u$ satisfies $\bar\partial F=0$, so it is holomorphic by the same lemma, and $F(a)=h(a)$. Since $|z-a|<L$ on $\Omega $, we have $|z-a|^{2n}\mathrm e^{|z-a|^2}\le L^{2n}\mathrm e^{L^2}$, and therefore

Finally, $|F|^2\le 2|\chi h|^2+2|u|^2$ and $|\chi h|\le |h|\mathbf1_{B(a,2r)}$ give the assertion with $C(n,r,L)=2+2c_0^2L^{2n}\mathrm e^{L^2}r^{-2n-2}$. $\square$