[proofplan] We first solve $\bar\partial u=f$ without support control, using smooth global solvability for the Dolbeault-Neumann problem on smooth bounded strictly pseudoconvex domains. Since $f$ is compactly supported in the interior, this preliminary solution is holomorphic on a collar of the boundary. The Hartogs-Bochner [extension theorem](/theorems/59) in dimension $n\ge 2$ extends that collar [holomorphic function](/page/Holomorphic%20Function) across the compact interior region. Subtracting the extension kills the preliminary solution near the boundary while preserving its $\bar\partial$-equation, giving a compactly supported smooth solution. [/proofplan]

[step:Choose a connected boundary collar outside which $f$ vanishes] Let \begin{align*} K_0:=\operatorname{supp} f\subset\Omega. \end{align*} Since $f\in C_c^\infty(\Omega;\Lambda^{0,1}T^*\Omega)$, the set $K_0$ is compact in $\Omega$. Choose a smooth defining function \begin{align*} \rho:U\to\mathbb R \end{align*} for $\partial\Omega$ on an open neighbourhood $U\subset\mathbb C^n$ of $\partial\Omega$, with $\Omega\cap U=\{z\in U:\rho(z)<0\}$ and $d\rho_p\ne 0$ for every $p\in\partial\Omega$. Because $K_0$ is compactly contained in $\Omega$, choose $\varepsilon>0$ small enough that \begin{align*} W:=\{z\in\Omega:-\varepsilon<\rho(z)<0\} \end{align*} is a connected open collar neighbourhood of $\partial\Omega$ in $\Omega$ and satisfies $W\cap K_0=\varnothing$. The connectedness follows from the collar diffeomorphism between $W$ and $\partial\Omega\times(-\varepsilon,0)$, together with the hypothesis that $\partial\Omega$ is connected. Set \begin{align*} G:=\{z\in\Omega:\rho(z)<-\varepsilon\}. \end{align*} After decreasing $\varepsilon$ if necessary, $G$ has smooth boundary, $K_0\subset G$, $\overline G\subset\Omega$, and $W=\Omega\setminus\overline G$. Since $W\cap K_0=\varnothing$, we have $f=0$ on $W$. [/step]

[step:Solve $\bar\partial u=f$ smoothly up to the boundary]We use the standard global regularity and solvability theorem for the Dolbeault-Neumann problem on smooth bounded strictly pseudoconvex domains, in the smooth solvability consequence of [citetheorem:9221]. Its hypotheses apply because $\Omega\subset\mathbb C^n$ is bounded, has $C^\infty$ strictly pseudoconvex boundary, and $f\in C_c^\infty(\Omega;\Lambda^{0,1}T^*\Omega)$ is smooth and satisfies $\bar\partial f=0$. Therefore there exists a function \begin{align*} u:\overline\Omega\to\mathbb C \end{align*} with \begin{align*} u\in C^\infty(\overline\Omega;\mathbb C) \end{align*} such that \begin{align*} \bar\partial u=f \end{align*} on $\Omega$.[/step]

[guided]The first goal is only to solve the equation, not yet to control the support of the solution. The appropriate input is the global regularity and solvability theorem for the Dolbeault-Neumann problem on smooth bounded strictly pseudoconvex domains, using the smooth solvability consequence of [citetheorem:9221]. That theorem says, in the degree relevant here, that a smooth $\bar\partial$-closed $(0,1)$-form on such a domain has a smooth up-to-the-boundary solution to the $\bar\partial$-equation. We verify the hypotheses. The domain $\Omega\subset\mathbb C^n$ is bounded and has $C^\infty$ strictly pseudoconvex boundary by assumption. The form \begin{align*} f\in C_c^\infty(\Omega;\Lambda^{0,1}T^*\Omega) \end{align*} is in particular smooth on $\Omega$ and extends by zero to a smooth form near $\partial\Omega$, because its support is compactly contained in $\Omega$. The compatibility condition required for solving the equation in degree $(0,1)$ is exactly \begin{align*} \bar\partial f=0, \end{align*} which is one of the hypotheses. Thus the solvability theorem gives a function \begin{align*} u:\overline\Omega\to\mathbb C \end{align*} with \begin{align*} u\in C^\infty(\overline\Omega;\mathbb C) \end{align*} and \begin{align*} \bar\partial u=f \end{align*} on $\Omega$. The smoothness up to $\partial\Omega$ matters because the next step restricts $u$ to a boundary collar and then applies a Hartogs-type extension theorem to that collar holomorphic function.[/guided]

[step:Observe that the preliminary solution is holomorphic on the boundary collar] Since $f=0$ on $W$ and $\bar\partial u=f$ on $\Omega$, we have \begin{align*} \bar\partial u=0 \end{align*} on $W$. Hence the restriction \begin{align*} u|_W:W\to\mathbb C \end{align*} is holomorphic. [/step]

[step:Extend the collar holomorphic function across the compact interior]We apply the Hartogs-Bochner extension theorem in the following form: if $n\ge 2$, $\Omega\subset\mathbb C^n$ is a bounded domain with connected $C^\infty$ boundary, and $W\subset\Omega$ is a connected one-sided collar neighbourhood of $\partial\Omega$, then every holomorphic function $g:W\to\mathbb C$ extends to a holomorphic function $H:\Omega\to\mathbb C$ agreeing with $g$ on a possibly smaller connected collar $W_0\subset W$ of $\partial\Omega$. Its hypotheses apply because $n\ge 2$, $\Omega$ is bounded, $\partial\Omega$ is connected and $C^\infty$, and the preceding step constructed $W$ as a connected boundary collar. Applying this theorem to the holomorphic function $u|_W:W\to\mathbb C$ gives a holomorphic function $h:\Omega\to\mathbb C$ such that $h=u$ on some connected open collar $W_0\subset W$ of $\partial\Omega$ inside $\Omega$. Since $h$ is holomorphic on $\Omega$, it is smooth on $\Omega$.[/step]

[guided]The delicate point is that the holomorphic data near the boundary must produce one function on all of $\Omega$, not separate local extensions on unrelated collar pieces. This is why the statement now assumes $\partial\Omega$ is connected and why the first step chose $W$ to be a connected one-sided collar. The Hartogs-Bochner extension theorem applies precisely to such a collar in complex dimension $n\ge 2$: a holomorphic function on a connected neighbourhood of the boundary inside a bounded domain extends holomorphically through the compact set left inside the domain. We verify the hypotheses one by one. The dimension condition is $n\ge 2$, which is part of the theorem statement. The domain $\Omega\subset\mathbb C^n$ is bounded and has connected $C^\infty$ boundary by hypothesis. The set $W=\{z\in\Omega:-\varepsilon<\rho(z)<0\}$ is a connected one-sided collar of $\partial\Omega$ by construction. Finally, the map $u|_W:W\to\mathbb C$ is holomorphic because the previous step proved $\bar\partial u=0$ on $W$. Therefore the Hartogs-Bochner extension theorem gives a holomorphic function $h:\Omega\to\mathbb C$ and a smaller connected open collar $W_0\subset W$ of $\partial\Omega$ such that $h=u$ on $W_0$. The agreement on an open collar is the exact support-control input needed later: after subtracting $h$ from $u$, the difference vanishes near the boundary rather than merely having zero boundary trace. Since holomorphic functions on open subsets of $\mathbb C^n$ are smooth, $h$ is smooth on $\Omega$.[/guided]

[step:Subtract the holomorphic extension to obtain compact support] Define $v:\Omega\to\mathbb C$ by $v(z):=u(z)-h(z)$. Since $u\in C^\infty(\overline\Omega;\mathbb C)$ and $h:\Omega\to\mathbb C$ is holomorphic, we have $v\in C^\infty(\Omega;\mathbb C)$. Using linearity of $\bar\partial$ and holomorphicity of $h$, we get $\bar\partial v=\bar\partial u-\bar\partial h=f-0=f$. Moreover, $h=u$ on $W_0$, so $v=0$ on $W_0$. Therefore $\operatorname{supp} v\subset \Omega\setminus W_0$. Because $W_0$ is an open neighbourhood of $\partial\Omega$ in $\Omega$ and $\Omega$ is bounded, the [closed set](/page/Closed%20Set) $\Omega\setminus W_0$ is compactly contained in $\Omega$. Hence $v\in C_c^\infty(\Omega;\mathbb C)$. This proves the asserted compactly supported solvability. [/step]

[step:Identify the solvability statement with vanishing compactly supported cohomology] Let \begin{align*} Z_c^{0,1}(\Omega):=\ker(\bar\partial:C_c^\infty(\Omega;\Lambda^{0,1}T^*\Omega)\to C_c^\infty(\Omega;\Lambda^{0,2}T^*\Omega)) \end{align*} and let \begin{align*} B_c^{0,1}(\Omega):=\operatorname{im}(\bar\partial:C_c^\infty(\Omega;\mathbb C)\to C_c^\infty(\Omega;\Lambda^{0,1}T^*\Omega)). \end{align*} The compactly supported solvability just proved says that every element of $Z_c^{0,1}(\Omega)$ belongs to $B_c^{0,1}(\Omega)$. The reverse inclusion follows from nilpotence of the Dolbeault operator, since $\bar\partial^2=0$. Therefore \begin{align*} Z_c^{0,1}(\Omega)=B_c^{0,1}(\Omega). \end{align*} Consequently, \begin{align*} H_c^{0,1}(\Omega)=Z_c^{0,1}(\Omega)/B_c^{0,1}(\Omega)=0. \end{align*} [/step]