Hartogs extension by a compact-support dbar correction

If $K=\mathrm{\varnothing }$, then $G=\mathrm{\Omega }$ and take $F=f$. Any other holomorphic extension agrees with $f$ on a nonempty open subset of $\mathrm{\Omega }$ by [F13], so the identity theorem [F14] gives uniqueness.

Suppose $K\mathrm{\ne }\mathrm{\varnothing }$. The function $z\mapsto \parallel z\parallel$ is continuous: the triangle inequality and $\parallel -z\parallel =\parallel z\parallel$ from [F16] give $\mathrm{\mid }\parallel z\parallel -\parallel w\parallel \mathrm{\mid }\le \parallel z-w\parallel$, and [F9] identifies this norm distance with the metric. By [F10], there are $z_0\in K$ and $r\ge 0$ with $\parallel z_0\parallel =r={\max }_{z\in K}\parallel z\parallel$. Choose $\epsilon >0$ with $\{z:\parallel z-z_0\parallel <\epsilon \}\subseteq \mathrm{\Omega }$. If $r>0$, put $w=(1+\epsilon \mathrm{/}(2r))z_0$; if $r=0$, then $z_0=0$ and put $w=(\epsilon \mathrm{/}2)e_1$, where $e_1=(1,0,\dots ,0)$. In either case $\parallel w-z_0\parallel =\epsilon \mathrm{/}2$, so $w\in \mathrm{\Omega }$. With $R:=r+\epsilon \mathrm{/}4$, we have $\parallel w\parallel >R$ and $K\subset \{z:\parallel z\parallel <R\}$.

Apply [F4] to $K\subseteq W:=\mathrm{\Omega }\cap \{z:\parallel z\parallel <R\}$ to obtain $\chi \in C^{\mathrm{\infty }}({\mathbb{C}}^n,[0,1])$ equal to one near $K$ and satisfying $\mathrm{supp}\chi \subseteq W$. Define $f_0=(1-\chi )f$ on $G$ and $f_0=0$ on $K$. This is smooth on $\mathrm{\Omega }$: on a neighborhood of $K$ it is identically zero, and off $K$ it is a product of smooth functions by [F6].

On $\mathrm{\Omega }$ set $g=\bar{\mathrm{\partial }}{f}_0$, and define $g=0$ on ${\mathbb{C}}^n\setminus \mathrm{\Omega }$. On $G$, the product rule and $\bar{\mathrm{\partial }}f=0$ from [F7] give $g=-f\bar{\mathrm{\partial }}\chi$; near $K$, $g=0$. Thus the global nonzero locus lies in the closed set $\mathrm{supp}\chi \subseteq \mathrm{\Omega }\cap \{z:\parallel z\parallel <R\}$. Since $\mathrm{supp}\chi$ is a closed subset of the open set $\mathrm{\Omega }$, every point outside $\mathrm{\Omega }$ has a neighborhood disjoint from $\mathrm{supp}\chi$; inside $\mathrm{\Omega }$ there $\chi =0$ and $f_0=f$ with $\bar{\mathrm{\partial }}f=0$, while outside $\mathrm{\Omega }$ we set $g=0$. Hence the zero extension is smooth. Its support is a closed subset of $\mathrm{supp}\chi$, so it lies in $\mathrm{\Omega }$ and is bounded; [F15] makes it compact. On $\mathrm{\Omega }$, $\bar{\mathrm{\partial }}g={\bar{\mathrm{\partial }}}^2{f}_0=0$ by [F3], and around the complement of $\mathrm{\Omega }$ the extended form is zero, so globally $g$ is $\bar{\mathrm{\partial }}$-closed.

Invoke [F1] under the stated full AC hypothesis [F17] to obtain the unique smooth compactly supported $u$ with $\bar{\mathrm{\partial }}u=g$, vanishing on the unique unbounded connected component of ${\mathbb{C}}^n\setminus \mathrm{supp}g$. If $f=0$, then $f_0=g=0$ and uniqueness in [F1] gives $u=0$, so this construction also covers the zero function.

Let $E_R:=\{z:\parallel z\parallel >R\}$. It is disjoint from $\mathrm{supp}g$ by step 3.1 and is unbounded. It is path-connected: for $x,y\in E_R$, choose $L>\max (R,\parallel x\parallel ,\parallel y\parallel )$, move each point radially to the sphere of radius $L$, and join the resulting directions by a path on $S^{2n-1}$, rescaled by $L$. The sphere path exists by [F11], since $2n\ge 2$; all three paths stay in $E_R$. Thus $E_R$ is connected by [F11]. Its component containing $(R+1)e_1$, with $e_1=(1,0,\dots ,0)$, contains all of $E_R$ by [F12], so that component is unbounded and therefore is the unique unbounded component in [F1]. Hence $u=0$ on $E_R$. Also $\chi =0$ there by [F4] and [F5], since $E_R$ is disjoint from $\mathrm{supp}\chi$.

The set $G$ is open by [F8] and nonempty because the point $w$ from step 1.2 lies in $\mathrm{\Omega }\cap E_R\subseteq G$. On $G$ define $h:=u+\chi f$. It is smooth by [F6] and step 4.1, and $\bar{\mathrm{\partial }}h=g+f\bar{\mathrm{\partial }}\chi =0$ by steps 3.1 and 4.1. Therefore [F7], with library coordinate $k=j-1$, makes $h$ holomorphic on $G$. Step 4.2 gives $h=0$ on the nonempty open set $\mathrm{\Omega }\cap E_R$; because $G$ is connected, [F14] yields $h=0$ throughout $G$.

Set $F:=f_0-u$ on $\mathrm{\Omega }$. It is smooth, and $\bar{\mathrm{\partial }}F=g-g=0$, so [F7], with library coordinate $k=j-1$, makes $F$ holomorphic. On $G$, $F=(1-\chi )f-u=f-h=f$ by step 5.1. Thus $F$ is an extension of $f$ to $\mathrm{\Omega }$ in the sense of [F13].

If $\tilde{F}$ is any other holomorphic extension to $\mathrm{\Omega }$, [F13] gives a nonempty open $V\subseteq G$ on which $\tilde{F}=f$. By step 6.1, $F=f$ on all of $G$, so $\tilde{F}-F$ vanishes on $V$. The identity theorem [F14] on the connected domain $\mathrm{\Omega }$ gives $\tilde{F}=F$.