[proofplan] We prove the obstruction by expanding $\bar{\partial}$ in coefficients, rather than merely citing $\bar{\partial}^2=0$. In the smooth case, applying $\bar{\partial}$ twice gives paired terms whose mixed derivatives are equal and whose wedge factors differ by a minus sign. In the distributional case, the same cancellation is valid because distributional $\bar z$-derivatives commute when tested against compactly supported smooth functions. Thus every $\bar{\partial}$-exact datum is automatically $\bar{\partial}$-closed. [/proofplan] [step:Fix coefficient and primed-sum notation] For each integer $r$ with $0\le r\le n$, let $\mathcal I_r$ be the set of strictly increasing $r$-tuples \begin{align*} I=(i_1,\dots,i_r), \qquad 1\le i_1<\cdots<i_r\le n. \end{align*} For $I\in\mathcal I_p$ and $J=(j_1,\dots,j_s)\in\mathcal I_s$, define \begin{align*} dz_I&=dz_{i_1}\wedge\cdots\wedge dz_{i_p},\\ d\bar z_J&=d\bar z_{j_1}\wedge\cdots\wedge d\bar z_{j_s}. \end{align*} The prime in a sum means summation over strictly increasing multi-indices. Thus a smooth $(p,q-1)$-form $u$ has the expansion \begin{align*} u=\sum_{I\in\mathcal I_p,\ J\in\mathcal I_{q-1}}' u_{I,J}\,dz_I\wedge d\bar z_J, \end{align*} with $u_{I,J}\in C^\infty(\Omega)$. The same notation is used for distribution-valued coefficients when $u$ is only locally $L^2$. [/step] [step:Compute the smooth cancellation coefficient by coefficient] Assume first that $u\in\Omega^{p,q-1}(\Omega)$ and $f=\bar{\partial}u$. The coefficient formula for $\bar{\partial}$ is \begin{align*} \bar{\partial}u = \sum_{I,J}'\sum_{k=1}^n \frac{\partial u_{I,J}}{\partial\bar z_k}\, d\bar z_k\wedge dz_I\wedge d\bar z_J. \end{align*} Applying $\bar{\partial}$ again gives \begin{align*} \bar{\partial}(\bar{\partial}u) = \sum_{I,J}'\sum_{k=1}^n\sum_{\ell=1}^n \frac{\partial^2 u_{I,J}}{\partial\bar z_\ell\,\partial\bar z_k}\, d\bar z_\ell\wedge d\bar z_k\wedge dz_I\wedge d\bar z_J. \end{align*} For the terms with $k=\ell$, the exterior factor contains \begin{align*} d\bar z_k\wedge d\bar z_k=0, \end{align*} so those terms vanish. For distinct indices $k\ne\ell$, group the ordered pair $(k,\ell)$ with the ordered pair $(\ell,k)$. Since $u_{I,J}$ is smooth, mixed Wirtinger derivatives commute: \begin{align*} \frac{\partial^2 u_{I,J}}{\partial\bar z_\ell\,\partial\bar z_k} = \frac{\partial^2 u_{I,J}}{\partial\bar z_k\,\partial\bar z_\ell}. \end{align*} The wedge factors satisfy \begin{align*} d\bar z_\ell\wedge d\bar z_k =-d\bar z_k\wedge d\bar z_\ell. \end{align*} Therefore each paired contribution is \begin{align*} & \frac{\partial^2 u_{I,J}}{\partial\bar z_\ell\,\partial\bar z_k}\, d\bar z_\ell\wedge d\bar z_k\wedge dz_I\wedge d\bar z_J + \frac{\partial^2 u_{I,J}}{\partial\bar z_k\,\partial\bar z_\ell}\, d\bar z_k\wedge d\bar z_\ell\wedge dz_I\wedge d\bar z_J\\ &= \frac{\partial^2 u_{I,J}}{\partial\bar z_\ell\,\partial\bar z_k} \bigl(d\bar z_\ell\wedge d\bar z_k+d\bar z_k\wedge d\bar z_\ell\bigr) \wedge dz_I\wedge d\bar z_J\\ &=0. \end{align*} Summing over $I$, $J$, $k$, and $\ell$ gives \begin{align*} \bar{\partial}f = \bar{\partial}(\bar{\partial}u) =0. \end{align*} This proves the smooth assertion. [/step] [step:Verify commutation of distributional $\bar z$-derivatives by testing] Let $U\in\mathcal D'(\Omega)$ be a scalar distribution and let $k,\ell\in\{1,\dots,n\}$. For a [test function](/page/Test%20Function) $\psi\in C_c^\infty(\Omega)$, the distributional Wirtinger derivative is defined by \begin{align*} \left(\frac{\partial U}{\partial\bar z_k}\right)(\psi) = -U\left(\frac{\partial\psi}{\partial\bar z_k}\right). \end{align*} Applying this definition twice gives \begin{align*} \left(\frac{\partial^2 U}{\partial\bar z_\ell\,\partial\bar z_k}\right)(\psi) &= -\left(\frac{\partial U}{\partial\bar z_k}\right) \left(\frac{\partial\psi}{\partial\bar z_\ell}\right)\\ &= U\left(\frac{\partial^2\psi}{\partial\bar z_k\,\partial\bar z_\ell}\right). \end{align*} Interchanging $k$ and $\ell$ gives \begin{align*} \left(\frac{\partial^2 U}{\partial\bar z_k\,\partial\bar z_\ell}\right)(\psi) = U\left(\frac{\partial^2\psi}{\partial\bar z_\ell\,\partial\bar z_k}\right). \end{align*} The [test function](/page/Test%20Function) $\psi$ is smooth, so its mixed Wirtinger derivatives commute: \begin{align*} \frac{\partial^2\psi}{\partial\bar z_k\,\partial\bar z_\ell} = \frac{\partial^2\psi}{\partial\bar z_\ell\,\partial\bar z_k}. \end{align*} Hence \begin{align*} \frac{\partial^2 U}{\partial\bar z_\ell\,\partial\bar z_k} = \frac{\partial^2 U}{\partial\bar z_k\,\partial\bar z_\ell} \end{align*} as distributions on $\Omega$. [/step] [step:Apply the same cancellation to distribution-valued coefficients] Assume now that $u\in L^2_{\mathrm{loc}}(\Omega;\Lambda^{p,q-1})$ and $\bar{\partial}u=f$ in $\mathcal D'(\Omega;\Lambda^{p,q})$. The coefficients $u_{I,J}$ define regular scalar distributions because they are locally $L^2$ and hence locally $L^1$. The coefficient formula for $\bar{\partial}u$ is the same as in the smooth case, with distributional derivatives replacing classical derivatives: \begin{align*} \bar{\partial}u = \sum_{I,J}'\sum_{k=1}^n \frac{\partial u_{I,J}}{\partial\bar z_k}\, d\bar z_k\wedge dz_I\wedge d\bar z_J. \end{align*} Applying $\bar{\partial}$ once more gives the distribution-valued form \begin{align*} \bar{\partial}(\bar{\partial}u) = \sum_{I,J}'\sum_{k=1}^n\sum_{\ell=1}^n \frac{\partial^2 u_{I,J}}{\partial\bar z_\ell\,\partial\bar z_k}\, d\bar z_\ell\wedge d\bar z_k\wedge dz_I\wedge d\bar z_J. \end{align*} The terms with $k=\ell$ vanish because $d\bar z_k\wedge d\bar z_k=0$. For $k\ne\ell$, the test-function calculation from the previous step gives equality of the two mixed distributional derivatives, while exterior antisymmetry gives \begin{align*} d\bar z_\ell\wedge d\bar z_k =-d\bar z_k\wedge d\bar z_\ell. \end{align*} Thus the paired ordered terms cancel exactly as in the smooth computation. Therefore \begin{align*} \bar{\partial}(\bar{\partial}u)=0 \end{align*} in $\mathcal D'(\Omega;\Lambda^{p,q+1})$. [/step] [step:Conclude closedness of the datum] Since $\bar{\partial}u=f$ in $\mathcal D'(\Omega;\Lambda^{p,q})$, applying $\bar{\partial}$ to both sides gives \begin{align*} \bar{\partial}f &=\bar{\partial}(\bar{\partial}u)\\ &=0. \end{align*} This proves the distributional assertion, and the convention $\Lambda^{p,n+1}=\{0\}$ covers the top-degree case $q=n$. [/step]