6.5 Interpolation and extension across a hyperplane6FAB73FA6CE4

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 $C(n,r,L)=2+2c_0^2L^{2n}\mathrm e^{L^2}r^{-2n-2}$. $\square$

Proof. The idea is that, if $u$ is the minimal solution in $L^2(\Omega ,\mathrm e^{-p})$, then $u\mathrm e^\psi $ is the minimal solution of a twisted equation in $L^2(\Omega ,\mathrm e^{-p-\psi })$, to which we apply Hörmander's estimate. Put $m=\sup _\Omega |\psi |<\infty $. Since $\mathrm e^{-m}\le \mathrm e^{\pm \psi }\le \mathrm e^ m$, the spaces $L^2(\Omega ,\mathrm e^{-p})$ and $L^2(\Omega ,\mathrm e^{-p-\psi })$ consist of the same functions, with equivalent norms, and so contain the same holomorphic functions. By the discussion preceding theorem 6.4.7, applied with the weight $p-\psi $, which is bounded above on compact sets, $\alpha \in L^2_{\mathrm{loc},(0,1)}(\Omega )$; so $\bar\partial \alpha $ is defined. Since $\mathrm e^{-p-\psi }\le \mathrm e^{2m}\mathrm e^{\psi -p}$, we have $\int _\Omega |\alpha |_{A^{-1}}^2\mathrm e^{-p-\psi }\, \mathrm dV\le \mathrm e^{2m}E<\infty $. Hence theorem 6.4.9, applied with the plurisubharmonic function $p$ and the smooth function $q=\psi $, whose complex Hessian matrix $A$ is positive semidefinite, shows that the equation $\bar\partial u=\alpha $ has a solution in $L^2(\Omega ,\mathrm e^{-p-\psi })=L^2(\Omega ,\mathrm e^{-p})$. Let $u$ be the minimal solution in $L^2(\Omega ,\mathrm e^{-p})$, given by lemma 6.4.10(2).

The function $v=u\mathrm e^\psi $ belongs to $L^2(\Omega ,\mathrm e^{-p-\psi })$ and solves

For every holomorphic $g$ in $L^2(\Omega ,\mathrm e^{-p-\psi })$ we have $\int v\bar g\, \mathrm e^{-p-\psi }\, \mathrm dV=\int u\bar g\, \mathrm e^{-p}\, \mathrm dV=0$, so $v$ is the minimal solution of $\bar\partial v=\beta $ in $L^2(\Omega ,\mathrm e^{-p-\psi })$, by lemma 6.4.10(2).

We now estimate $v$ by theorem 6.4.9, applied again with $p$ and $q=\psi $. The form $\beta =\bar\partial v$ is $\bar\partial $-closed, and it lies in $L^2_{\mathrm{loc},(0,1)}(\Omega )$ because $\alpha $ and $u$ are locally square integrable and $\psi $ is smooth. Since $E<\infty $ and $h<1$, the dual energies of $\alpha $ and $\bar\partial \psi $ are finite almost everywhere, and at such points both vectors lie in the subspace on which the dual energy is a Hermitian quadratic form. The triangle inequality $|\alpha +u\, \bar\partial \psi |_{A^{-1}}\le |\alpha |_{A^{-1}}+|u|h^{1/2}$ therefore gives

So theorem 6.4.9 yields a solution of $\bar\partial w=\beta $ whose squared norm in $L^2(\Omega ,\mathrm e^{-p-\psi })$ is at most this integral, and the minimal solution $v$ has no larger norm. As $\int |v|^2\mathrm e^{-p-\psi }\, \mathrm dV=\int |u|^2\mathrm e^{\psi -p}\, \mathrm dV$, we obtain

At almost every point, let $\langle \cdot ,\cdot \rangle _{A^{-1}}$ be the Hermitian inner product associated with the dual energy on the subspace where it is finite; by the polarization identity $\langle x,y\rangle _{A^{-1}}=\frac14\sum _{k=0}^3\mathrm i^ k|x+\mathrm i^ ky|_{A^{-1}}^2$, the function $\langle \alpha ,\bar\partial \psi \rangle _{A^{-1}}$ is measurable. We expand

the second inequality being the Cauchy–Schwarz inequality. The function $h|u|^2\mathrm e^{\psi -p}$ is integrable, since $h<1$ almost everywhere and $u\in L^2(\Omega ,\mathrm e^{-p})$; moving its integral to the left, we find that $J=\int (1-h)|u|^2\mathrm e^{\psi -p}\, \mathrm dV$, which is finite, satisfies

If $a=0$, then $h=0$ almost everywhere on $\{ \alpha \ne 0\} $, the last integral vanishes, and $J\le E$, as claimed. Let $a>0$. The last integrand vanishes outside $\{ \alpha \ne 0\} $, where $h\le a$ and hence $h/(1-h)\le a/(1-a)$. For $t>1$, the inequality $2xy\le t^{-1}x^2+ty^2$ with $x=(1-h)^{1/2}|u|$ and $y=(h/(1-h))^{1/2}|\alpha |_{A^{-1}}$ gives there

Integrating, we obtain

Since $J$ is finite, this gives $J\le \frac t{t-1}\bigl (1+\frac{ta}{1-a}\bigr )E$. Choose $t=1+a^{-1/2}$. Then $1+ta/(1-a)=(1+\sqrt a)/(1-a)=1/(1-\sqrt a)$ and $t/(t-1)=1+\sqrt a$, so $J\le (1+\sqrt a)E/(1-\sqrt a)$. $\square$

Proof. Since $|\zeta |^2+\varepsilon ^2<1/8$, we have $s>\log 8>2$, so $\eta >s>2$ and $\psi _\varepsilon <0$ is well defined. For fixed $\varepsilon $, $s$ ranges in a bounded interval, so $\psi _\varepsilon $ is bounded; and $\psi _\varepsilon =G(|\zeta |^2)$ for a function $G$ that is smooth near $[0,1/16)$, so $\psi _\varepsilon $ is smooth.

For $\zeta \ne 0$ put $x=|\zeta |^2$, $t=\log x$ and $b=x/(x+\varepsilon ^2)\in (0,1)$, and denote derivatives with respect to $t$ by primes. Then $s'=-b$ and $b'=b(1-b)$, so $s''=-b(1-b)$. With $D=1+1/s$ we have $\eta '=s'+s'/s=-bD$ and $D'=-s'/s^2=b/s^2$. Hence $\psi _\varepsilon '=-\eta '/\eta =bD/\eta $ and $\psi _\varepsilon ''=(b'D+bD')/\eta -bD\eta '/\eta ^2$, that is,

For a function $g$ of $t=\log |\zeta |^2$ we have $\partial _{\bar\zeta }g=g'/\bar\zeta $ and $\partial _\zeta \partial _{\bar\zeta }g=g''/|\zeta |^2$. Hence $(\psi _\varepsilon )_{\zeta \bar\zeta }=\psi _\varepsilon ''/x>0$ and $h=(\psi _\varepsilon ')^2/\psi _\varepsilon ''=A/(A+B)<1$ for $\zeta \ne 0$. In general $\partial _{\bar\zeta }\psi _\varepsilon =G'(|\zeta |^2)\zeta $ and $(\psi _\varepsilon )_{\zeta \bar\zeta }=G''(|\zeta |^2)|\zeta |^2+G'(|\zeta |^2)$, so at $\zeta =0$ we have $\partial _{\bar\zeta }\psi _\varepsilon =0$ and $(\psi _\varepsilon )_{\zeta \bar\zeta }=G'(0)$, where $G'(x)=\psi _\varepsilon '/x=D/(\eta (x+\varepsilon ^2))>0$; so $(\psi _\varepsilon )_{\zeta \bar\zeta }>0$ and $h=0$ there. Thus $\psi _\varepsilon $ is strictly subharmonic, and $h<1$ everywhere.

For the first estimate let $\zeta \ne 0$. Since $B\ge b^2/(\eta s^2)$, we have $A/B\le D^2s^2/\eta =(s+1)^2/\eta $, and therefore

because $\eta \le 2s\le 2s^2$ and $(s+1)^2\le 4s^2$ for $s\ge 1$. At $\zeta =0$ we have $(1-h)\mathrm e^{\psi _\varepsilon }=1/\eta \ge 1/(6s^2)$ as well. For the second estimate let $0<x\le \varepsilon ^2$. Then $b\le 1/2$, and $B\ge b(1-b)D/\eta $ gives

since $s+1\le 2s\le 2\eta $; moreover $s\ge -\log (2\varepsilon ^2)$, because $x+\varepsilon ^2\le 2\varepsilon ^2$. Together with $h(0)=0$, this proves the second estimate. For the third, let $\varepsilon /2\le |\zeta |\le \varepsilon $. Then $1/5\le b\le 1/2$, so $\eta \psi _\varepsilon ''\ge \eta B\ge b(1-b)D\ge b(1-b)\ge 4/25$, and $\mathrm e^{\psi _\varepsilon }/(|\zeta |^2(\psi _\varepsilon )_{\zeta \bar\zeta })=1/(\eta \psi _\varepsilon '')\le 25/4$. $\square$

Proof. After a normalization, Step 1 treats smooth weights on relatively compact pseudoconvex subsets, by correcting a cutoff of $f$ with lemma 6.5.2, lemma 6.5.3, and Step 2 regularizes $p$ and passes to the limit.

Normalization. The substitution $z_ n=4R\zeta $ maps $\mathbb C^{n-1}\times \Delta _{1/4}$ onto $\mathbb C^{n-1}\times \Delta _ R$, preserves the hyperplane and the integral $I$, and multiplies Lebesgue measure on $\mathbb C^ n$ by $16R^2$. Hence it suffices to show the following: if $\Omega \subset \mathbb C^{n-1}\times \Delta _{1/4}$, then $f$ has an extension $F\in \mathcal O(\Omega )$ with $\int _\Omega |F|^2\mathrm e^{-p}\, \mathrm dV\le C_0I$ for an absolute constant $C_0$; then $C=16C_0$ works. From now on we write the points of $\Omega $ as $(z',\zeta )$ with $z'\in \mathbb C^{n-1}$ and $|\zeta |<1/4$; for $n=1$ the variable $z'$ is absent. We may assume $\Omega _0\ne \varnothing $.

Step 1: smooth weights on a relatively compact set. Let $D\Subset \Omega $ be a pseudoconvex open set, let $U\subset \Omega $ be an open neighborhood of $\overline D$, and let $\varphi $ be a smooth plurisubharmonic function on $U$. We show that there are an open set $P\Subset \Omega _0$ with $\overline P\times \{ 0\} \subset U$ and a function $F\in \mathcal O(D)$ with $F(z',0)=f(z')$ whenever $(z',0)\in D$, such that

where $C_0$ is an absolute constant; we write $I_ P$ for the integral over $P$ on the right.

The set $K_0=\{ z':(z',0)\in \overline D\} $ is a compact subset of the open set $U_0=\{ z':(z',0)\in U\} \subset \Omega _0$; choose an open set $P$ with $K_0\subset P\Subset U_0$. There is $\varepsilon _0\in (0,1/4)$ such that every $(z',\zeta )\in D$ with $|\zeta |\le \varepsilon _0$ satisfies $z'\in P$: otherwise there would be points $(z'_ k,\zeta _ k)\in D$ with $\zeta _ k\to 0$ and $z'_ k\notin P$, and a subsequence would converge, $D$ being bounded, to a point $(z',0)\in \overline D$ with $z'\notin P$, which is impossible. Decreasing $\varepsilon _0$, we may also assume that $\overline P\times \overline\Delta _{\varepsilon _0}\subset U$ and, by the uniform continuity of $\varphi $ on this compact set, that

Fix a smooth function $\chi $ on $\mathbb R$ with $0\le \chi \le 1$, $\chi =1$ on $(-\infty ,1/4]$ and $\chi =0$ on $[1,\infty )$, for instance $\chi (x)=\kappa _0((4x-1)/3)$ with the function $\kappa _0$ fixed at the beginning of section 6.4, and put $c_\chi =\sup |\chi '|^2$. For $0<\varepsilon \le \varepsilon _0$ define $g_\varepsilon $ on $D$ by $g_\varepsilon (z',\zeta )=f(z')\chi (|\zeta |^2/\varepsilon ^2)$ if $z'\in P$, and $g_\varepsilon =0$ otherwise. The open sets $D\cap (P\times \mathbb C)$ and $D\cap \{ |\zeta |>\varepsilon \} $ cover $D$ by the choice of $\varepsilon _0$, and $g_\varepsilon $ is smooth on each of them (on the second it vanishes); hence $g_\varepsilon $ is smooth on $D$, and $g_\varepsilon (z',0)=f(z')$ for $(z',0)\in D$. Since $f$ is holomorphic,

is a $\bar\partial $-closed smooth form supported in $\{ \varepsilon /2\le |\zeta |\le \varepsilon \} $.

We solve $\bar\partial u=\alpha _\varepsilon $ on $D$ by lemma 6.5.2, with the function $\widetilde\varphi =\varphi +\log |\zeta |^2$ in place of $p$, which is plurisubharmonic on $D$ by proposition 5.10.3, proposition 5.9.5, and with $\psi =\psi _\varepsilon (\zeta )$ from lemma 6.5.3, a smooth bounded function on $D$ which is plurisubharmonic by proposition 5.9.3, applied to the projection $(z',\zeta )\mapsto \zeta $. The complex Hessian matrix $A=H_\psi $ has the single nonzero entry $(\psi _\varepsilon )_{\zeta \bar\zeta }$, so $|\bar\partial \psi |_{A^{-1}}^2$ is the function $h$ of lemma 6.5.3; thus $h<1$, and $h\le a_\varepsilon =2/(-\log (2\varepsilon ^2))$ on $\{ \alpha _\varepsilon \ne 0\} \subset \{ |\zeta |\le \varepsilon \} $. The dual energy of $\alpha _\varepsilon $ is $|f|^2|\chi '|^2|\zeta |^2\varepsilon ^{-4}/(\psi _\varepsilon )_{\zeta \bar\zeta }$, so the third estimate of lemma 6.5.3 and $|\zeta |\le \varepsilon $ give, on its support,

By (6.5.6) and Fubini's theorem, since every point of the support has $z'\in P$,

If $\varepsilon $ is so small that $a_\varepsilon \le 1/4$, then $(1+\sqrt{a_\varepsilon })/(1-\sqrt{a_\varepsilon })\le 3$, and lemma 6.5.2 gives $u_\varepsilon $ with $\bar\partial u_\varepsilon =\alpha _\varepsilon $ and $\int _ D(1-h)|u_\varepsilon |^2\mathrm e^{\psi _\varepsilon -\widetilde\varphi }\, \mathrm dV\le 3E$. By the first estimate of lemma 6.5.3, $(1-h)\mathrm e^{\psi _\varepsilon }\ge 1/(6\log ^2(|\zeta |^2+\varepsilon ^2))$, and $\mathrm e^{-\widetilde\varphi }=\mathrm e^{-\varphi }|\zeta |^{-2}$. Hence, with $C_1=225\pi c_\chi $,

On $D\cap \{ |\zeta |<\varepsilon /2\} $ we have $\bar\partial u_\varepsilon =0$, so $u_\varepsilon $ is holomorphic there, by lemma 6.4.10(1). Suppose that $u_\varepsilon (z'_0,0)\ne 0$ for some $(z'_0,0)\in D$. On a small polydisc around this point, $|u_\varepsilon |$ is bounded below by a positive constant, $\varphi $ is bounded above, and $\log ^2(|\zeta |^2+\varepsilon ^2)\le \log ^2(\varepsilon ^2)$; so the integrand in (6.5.7) is at least a positive constant times $|\zeta |^{-2}$, whose integral over a disc around $\zeta =0$ diverges. This contradicts (6.5.7). Hence $u_\varepsilon =0$ on $D\cap \{ \zeta =0\} $, and $F_\varepsilon =g_\varepsilon -u_\varepsilon $ is holomorphic on $D$, by the same lemma, because $\bar\partial F_\varepsilon =0$; moreover $F_\varepsilon (z',0)=f(z')$.

It remains to estimate $F_\varepsilon $. For $0<x=|\zeta |^2<1/16$ we have $x<x+\varepsilon ^2<1/8$, so $x\log ^2(x+\varepsilon ^2)\le x\log ^2x\le 4\mathrm e^{-2}<1$. Thus (6.5.7) gives $\int _ D|u_\varepsilon |^2\mathrm e^{-\varphi }\, \mathrm dV\le C_1I_ P$. By (6.5.6) and Fubini's theorem, $\int _ D|g_\varepsilon |^2\mathrm e^{-\varphi }\, \mathrm dV\le 2\pi \varepsilon ^2I_ P\le I_ P$. Therefore $\int _ D|F_\varepsilon |^2\mathrm e^{-\varphi }\, \mathrm dV\le 2(1+C_1)I_ P$, which is (6.5.5) with $C_0=2(1+C_1)$.

Step 2: the general case. By corollary 6.3.5 there are pseudoconvex open sets $D_1\subset D_2\subset \cdots $ with $D_ j\Subset \Omega $ and $\bigcup _ jD_ j=\Omega $. Choose $\varepsilon _ j\downarrow 0$ with $\varepsilon _ j<\operatorname {dist}(\overline D_ j,\mathbb C^ n\setminus \Omega )$, and let $\rho _\varepsilon $ be the radial mollifier of section 2.5; then, by proposition 5.9.2(2), $p_ j=p*\rho _{\varepsilon _ j}$ is a smooth plurisubharmonic function on the open neighborhood $U_ j=\{ z\in \Omega :\operatorname {dist}(z,\mathbb C^ n\setminus \Omega )>\varepsilon _ j\} $ of $\overline D_ j$, and $p_ j\downarrow p$ pointwise. Step 1, applied to $D_ j$, $U_ j$ and $\varphi =p_ j$, gives $F_ j\in \mathcal O(D_ j)$ with $F_ j(z',0)=f(z')$ for $(z',0)\in D_ j$ and

because $p_ j\ge p$. (On a component of $\Omega _0$ on which $p(\cdot ,0)\equiv -\infty $ we have $f=0$, and the last inequality holds with the convention $0\cdot \infty =0$.)

We show that a subsequence of $(F_ j)$ converges locally uniformly on $\Omega $. Let $K\subset \Omega $ be compact. Then $K\subset D_ k$ for some $k$, and for $j\ge k$ we have $p_ j\le p_ k$ on $K$, so $\int _ K|F_ j|^2\, \mathrm dV\le \mathrm e^{\max _ Kp_ k}C_0I$. It follows that $(F_ j)$ is uniformly bounded on compact subsets of $\Omega $. Indeed, for a holomorphic function $G$ the complex Hessian matrix of $|G|^2$ is $\bigl (\partial _{z_\mu }G\, \overline{\partial _{z_\nu }G}\bigr )_{\mu ,\nu }$, which is positive semidefinite; so $|G|^2$ is plurisubharmonic by proposition 5.9.2(1), and hence subharmonic by theorem 5.9.1. If $K\subset \Omega $ is compact, $\delta >0$ and $K_\delta =\{ z:\operatorname {dist}(z,K)\le \delta \} \subset \Omega $, then the mean value inequality over balls in theorem 3.4.5, applied to $|F_ j|^2$, gives $|F_ j(z)|^2\le |B(z,\delta )|^{-1}\int _{K_\delta }|F_ j|^2\, \mathrm dV$ for $z\in K$ and large $j$, and the right side is bounded independently of $j$ by the preceding estimate applied to $K_\delta $.

Next, let $K\subset \Omega $ be compact, and choose $r>0$ such that the compact set $K'$ of points at distance at most $2\sqrt n\, r$ from $K$ lies in $\Omega $. For $z\in K'_0=\{ z:\operatorname {dist}(z,K)\le \sqrt n\, r\} $, the closed polydisc of radius $r$ centered at $z$ lies in $K'$, and the coefficient bound in (5.2.1) with $a=z$ gives $|\partial _{z_ l}F_ j(z)|\le r^{-1}\max _{K'}|F_ j|$ for $1\le l\le n$ and all large $j$ (those with $K'\subset D_ j$). Since $\mathrm dF_ j=\sum _ l\partial _{z_ l}F_ j\, \mathrm dz_ l$, it follows that these $F_ j$ are Lipschitz with a common constant on every ball of radius $\sqrt n\, r$ centered at a point of $K$, so the sequence $(F_ j)$ is equicontinuous on $K$ from some index on. By proposition 2.7.4 on a compact exhaustion of $\Omega $ and a diagonal argument, a subsequence converges locally uniformly on $\Omega $ to a continuous function $F$. Since $\bar\partial F_ j=0$ and $F_ j\to F$ in the sense of distributions, $\bar\partial F=0$, and $F$ is holomorphic by lemma 6.4.10(1). Moreover $F(z',0)=\lim _ jF_ j(z',0)=f(z')$ for $z'\in \Omega _0$.

Finally, let $K\subset \Omega $ be compact, and let $k$ be so large that $K\subset D_ k$. For $j\ge k$ we have $p_ j\le p_ k$ on $K$, so locally uniform convergence gives

This holds for all large $k$, and $\mathrm e^{-p_ k}$ increases to $\mathrm e^{-p}$ as $k\to \infty $; so the monotone convergence theorem gives $\int _ K|F|^2\mathrm e^{-p}\, \mathrm dV\le C_0I$, and exhausting $\Omega $ by compact sets gives $\int _\Omega |F|^2\mathrm e^{-p}\, \mathrm dV\le C_0I$. $\square$

Proof. We prove the second assertion; the first is the case $V=\{ a\} $, $r=n$ and $f=1$, whose slice norm with respect to the counting measure $\mathrm dV_0$ is $\mathrm e^{-p(a)}$. If $r=0$ we take $F=f$, and if $\Omega \cap V=\varnothing $ we take $F=0$.

Assume now that $r\ge 1$ and $\Omega \cap V\ne \varnothing $. Choose $b\in \Omega \cap V$ and unitary affine coordinates $w=(w_1,\ldots ,w_ n)$ on $\mathbb C^ n$ centered at $b$ with $V=\{ w_1=\cdots =w_ r=0\} $, and put $\Omega _ k=\Omega \cap \{ w_1=\cdots =w_ k=0\} $ for $0\le k\le r$, an open subset of a space isomorphic to $\mathbb C^{n-k}$. Unitary coordinates preserve Lebesgue measure. Each $\Omega _ k$ with $k<n$ is pseudoconvex: the restriction of a continuous plurisubharmonic exhaustion function of $\Omega $ to the closed subset $\Omega _ k$ of $\Omega $ is continuous and plurisubharmonic by proposition 5.9.3, applied to the linear embedding of $\{ w_1=\cdots =w_ k=0\} $, and its sublevel sets are compact. By the discussion following definition 6.3.1, every connected component of $\Omega _ k$ is pseudoconvex as well. Since $\Omega $ is open and has diameter at most $L$, we have $|z-b|<L$ for $z\in \Omega $ (a point $z\in \Omega $ with $|z-b|=L$ could be moved slightly away from $b$ inside $\Omega $), so the coordinate $w_ k$ takes its values in $\Delta _ L$ on $\Omega _{k-1}$.

Put $f_ r=f$, let $1\le k\le r$, and assume that $f_ k\in \mathcal O(\Omega _ k)$ has finite norm $\int _{\Omega _ k}|f_ k|^2\mathrm e^{-p}\, \mathrm dV_{2n-2k}$. Let $W$ be a connected component of $\Omega _{k-1}$; ordering the coordinates as $(w_{k+1},\ldots ,w_ n,w_ k)$, we regard $W$ as an open subset of $\mathbb C^{n-k}\times \Delta _ L$ whose slice by $\{ w_ k=0\} $ is the open subset $W\cap \Omega _ k$ of $\Omega _ k$. By proposition 5.9.3, the restriction of $p$ to $W$ is plurisubharmonic or identically $-\infty $. In the second case $\mathrm e^{-p}=\infty $ on $W\cap \Omega _ k$, so the finiteness of the norm of $f_ k$ forces $f_ k=0$ almost everywhere on $W\cap \Omega _ k$, hence everywhere there, since $f_ k$ is continuous, and we put $f_{k-1}=0$ on $W$. In the first case, theorem 6.5.4, applied to $W$ with normal coordinate $w_ k$, gives a holomorphic function $f_{k-1}$ on $W$ extending the restriction of $f_ k$ to $W\cap \Omega _ k$, with

The components $W$ are open, disjoint and cover $\Omega _{k-1}$, and the sets $W\cap \Omega _ k$ are disjoint and cover $\Omega _ k$. Hence these functions define $f_{k-1}\in \mathcal O(\Omega _{k-1})$ extending $f_ k$, and summing over $W$ gives

Proceeding in this way for $k=r,r-1,\ldots ,1$ and multiplying the $r$ estimates, we see that $F=f_0$ has the required properties. $\square$

Proof. Let $k=\dim Y$ and $n=\dim X$ near $a$. By the definition of a complex submanifold (definition 5.2.3), there are holomorphic coordinates $w=(w',w'')\in \mathbb C^ k\times \mathbb C^{n-k}$ centered at $a$ in which $Y=\{ w''=0\} $; by proposition 5.9.3, $p$ is plurisubharmonic near $0$ in these coordinates. On a compact neighborhood of $a$, the densities of smooth positive volume forms with respect to coordinate Lebesgue measure are bounded above and below by positive constants. Hence we may test integrability with Lebesgue measure $\mathrm dV_{2k}$ on $Y=\mathbb C^ k\times \{ 0\} $ and $\mathrm dV_{2n}$ in the coordinates $w$.

Choose a ball $B$ around $0$ so small that $p$ is plurisubharmonic on a neighborhood of $\overline B$, that $f$ is defined on $B\cap Y$, and that $\int _{B\cap Y}|f|^2\mathrm e^{-p}\, \mathrm dV_{2k}<\infty $. The ball $B$ is a bounded pseudoconvex domain, by corollary 6.3.5 applied to the exhaustion function $|w|^2$ of $\mathbb C^ n$, and $B\cap Y$ is its intersection with the linear subspace $\{ w''=0\} $ of codimension $n-k$. By corollary 6.5.8 there is $F\in \mathcal O(B)$ with $F|_{B\cap Y}=f$ and $\int _ B|F|^2\mathrm e^{-p}\, \mathrm dV_{2n}<\infty $. (If $k=0$, the integral over $Y$ is the weighted value $|f(a)|^2\mathrm e^{-p(a)}$; if $p|_ Y$ is identically $-\infty $ near $a$, then $f=0$ and $F=0$, in accordance with the conventions of theorem 6.5.4.) $\square$