[proofplan] We prove the compactly supported formulation by constructing a top-degree $(0,n)$-form whose integral against the standard holomorphic volume form is nonzero. If this form were $\bar\partial$ of a compactly supported smooth $(0,n-1)$-form, wedging with the holomorphic volume form would make that nonzero top-degree integral equal to the integral of an [exterior derivative](/theorems/1525) of a compactly supported form. A coordinate Stokes computation shows every such integral is zero, giving the contradiction. [/proofplan]

[step:Choose a compactly supported top degree form with nonzero integral] Write $z_j=x_j+iy_j$ for the standard complex coordinates on $\mathbb C^n$, and let $\mathcal L^{2n}$ denote Lebesgue measure on $\mathbb C^n\cong\mathbb R^{2n}$. Let \begin{align*} dV\in C^\infty(\mathbb C^n;\Lambda^{n,n}) \end{align*} denote the real orientation form \begin{align*} dV_z=dx_1\wedge dy_1\wedge\cdots\wedge dx_n\wedge dy_n. \end{align*} Define the standard holomorphic and antiholomorphic volume forms \begin{align*} \omega\in C^\infty(\mathbb C^n;\Lambda^{n,0}),\qquad \omega_z&=dz_1\wedge\cdots\wedge dz_n,\\ \theta\in C^\infty(\mathbb C^n;\Lambda^{0,n}),\qquad \theta_z&=d\bar z_1\wedge\cdots\wedge d\bar z_n. \end{align*} Set \begin{align*} c_n:=(-1)^{n(n-1)/2}(2i)^n. \end{align*} Reordering the factors in $\theta\wedge\omega$ and using $d\bar z_j\wedge dz_j=2i\,dx_j\wedge dy_j$ gives \begin{align*} \theta\wedge\omega=c_n dV. \end{align*} Choose a map $\psi:\Omega\to[0,\infty)$ with $\psi\in C_c^\infty(\Omega)$, $\operatorname{supp}\psi\subset B((1/2,0,\dots,0),1/4)$, and $\psi$ not identically zero. Define \begin{align*} A:=\int_\Omega \psi(z)\,d\mathcal L^{2n}(z)>0 \end{align*} and define the normalized bump function \begin{align*} \rho:\Omega&\to\mathbb R,\\ z&\mapsto A^{-1}\psi(z). \end{align*} Then $\rho\in C_c^\infty(\Omega)$ and \begin{align*} \int_\Omega \rho(z)\,d\mathcal L^{2n}(z)=1. \end{align*} For any $\alpha\in C^\infty(\Omega;\Lambda^{n,n})$, let $\operatorname{coef}_{dV}(\alpha):\Omega\to\mathbb C$ denote the unique smooth function satisfying $\alpha=\operatorname{coef}_{dV}(\alpha)dV$. Define \begin{align*} f:\Omega&\to\Lambda^{0,n},\\ z&\mapsto c_n^{-1}\rho(z)\theta_z. \end{align*} Then $f\in C_c^\infty(\Omega;\Lambda^{0,n})$, and the preceding normalization gives \begin{align*} \int_\Omega \operatorname{coef}_{dV}(f\wedge\omega)(z)\,d\mathcal L^{2n}(z)=\int_\Omega \rho(z)\,d\mathcal L^{2n}(z)=1. \end{align*} Finally, $\bar\partial f=0$ because $f$ has top antiholomorphic degree. In coordinates, \begin{align*} \bar\partial f=c_n^{-1}\sum_{j=1}^n \frac{\partial \rho}{\partial \bar z_j}\,d\bar z_j\wedge d\bar z_1\wedge\cdots\wedge d\bar z_n=0, \end{align*} since every summand contains a repeated factor $d\bar z_j\wedge d\bar z_j$. [/step]

[step:Derive the contradiction forced by a compactly supported primitive] Assume, toward a contradiction, that there exists a form $u\in C_c^\infty(\Omega;\Lambda^{0,n-1})$ satisfying $\bar\partial u=f$. Define \begin{align*} \beta\in C_c^\infty(\Omega;\Lambda^{n,n-1}),\qquad \beta:=u\wedge\omega. \end{align*} Because $\operatorname{supp}u$ is compact in $\Omega$, the zero extension $\widetilde\beta\in C_c^\infty(\mathbb C^n;\Lambda^{2n-1})$ is smooth and compactly supported. [claim:The integral of a compactly supported exterior derivative on $\mathbb C^n$ vanishes] For every $\gamma\in C_c^\infty(\mathbb C^n;\Lambda^{2n-1})$, \begin{align*} \int_{\mathbb C^n}\operatorname{coef}_{dV}(d\gamma)(z)\,d\mathcal L^{2n}(z)=0. \end{align*} [/claim] [proof] Let $X=(X_1,\dots,X_{2n})=(x_1,y_1,\dots,x_n,y_n)$ be the real coordinate system on $\mathbb C^n$. There are coefficient functions $\gamma_j:\mathbb C^n\to\mathbb C$, $1\le j\le 2n$, with $\gamma_j\in C_c^\infty(\mathbb C^n)$ such that \begin{align*} \gamma=\sum_{j=1}^{2n}(-1)^{j-1}\gamma_j\,dX_1\wedge\cdots\wedge\widehat{dX_j}\wedge\cdots\wedge dX_{2n}. \end{align*} Taking the [exterior derivative](/theorems/1525) gives \begin{align*} d\gamma=\left(\sum_{j=1}^{2n}\frac{\partial\gamma_j}{\partial X_j}\right)dV. \end{align*} For each $j$, let $X_{\widehat j}\in\mathbb R^{2n-1}$ denote the list of all real coordinates except $X_j$. Since $\gamma_j$ has compact support, [Fubini's theorem](/theorems/2961) and the one-dimensional [fundamental theorem of calculus](/theorems/632) give \begin{align*} \int_{\mathbb C^n}\frac{\partial\gamma_j}{\partial X_j}(X)\,d\mathcal L^{2n}(X) &=\int_{\mathbb R^{2n-1}}\left(\int_{\mathbb R}\frac{\partial\gamma_j}{\partial X_j}(X_j,X_{\widehat j})\,d\mathcal L^1(X_j)\right)d\mathcal L^{2n-1}(X_{\widehat j})\\ &=0. \end{align*} Summing this identity over $1\le j\le 2n$ proves the claim. [/proof] Apply the claim to $\gamma=\widetilde\beta$. On $\Omega$, the form $\beta$ has holomorphic degree $n$, so $\partial\beta=0$. Also $\bar\partial\omega=0$, because $\omega$ has constant coefficients. Using the graded Leibniz rule for $\bar\partial$ and the equation $\bar\partial u=f$, we obtain \begin{align*} d\beta=\bar\partial\beta=\bar\partial(u\wedge\omega)=\bar\partial u\wedge\omega+(-1)^{n-1}u\wedge\bar\partial\omega=f\wedge\omega. \end{align*} Since $\widetilde\beta$ agrees with $\beta$ on $\Omega$ and vanishes outside a compact subset of $\Omega$, the claim gives \begin{align*} 0&=\int_{\mathbb C^n}\operatorname{coef}_{dV}(d\widetilde\beta)(z)\,d\mathcal L^{2n}(z)\\ &=\int_\Omega\operatorname{coef}_{dV}(d\beta)(z)\,d\mathcal L^{2n}(z)\\ &=\int_\Omega\operatorname{coef}_{dV}(f\wedge\omega)(z)\,d\mathcal L^{2n}(z)\\ &=1, \end{align*} which is impossible. [/step]

[step:Conclude compactly supported non-solvability] The constructed form $f$ belongs to $C_c^\infty(\Omega;\Lambda^{0,n})$ and satisfies $\bar\partial f=0$. The contradiction above shows that no $u\in C_c^\infty(\Omega;\Lambda^{0,n-1})$ can satisfy $\bar\partial u=f$. This proves the compactly supported non-solvability statement. [/step]