[proofplan] We first make the coefficient conventions explicit for forms and fix the Euclidean Hermitian inner product. Next, we verify the finite-dimensional adjoint relation between raising the antiholomorphic degree by one and contracting in the corresponding coordinate direction; the sign is checked directly on basis forms. The analytic part is a scalar compact-support [integration by parts](/theorems/2098) identity in real coordinates. Combining the algebraic and scalar identities coefficient by coefficient gives the stated formula for the formal adjoint. [/proofplan] [step:Fix the form basis and coefficient conventions] For each integer $r$ with $0\le r\le n$, let $\mathcal I_r$ denote 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 $J=(j_1,\dots,j_q)\in\mathcal I_q$, 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_q},\\ e_{I,J}&=dz_I\wedge d\bar z_J. \end{align*} The [vector space](/page/Vector%20Space) $\Lambda^{p,q}$ is the complex span of the basis vectors $e_{I,J}$ with $I\in\mathcal I_p$ and $J\in\mathcal I_q$. The Euclidean Hermitian inner product on $\Lambda^{p,q}$ is the one for which this basis is orthonormal: \begin{align*} (e_{I,J},e_{I',J'})_{\Lambda^{p,q}} = \begin{cases} 1, & I=I'\text{ and }J=J',\\ 0, & \text{otherwise}. \end{cases} \end{align*} It is linear in the first argument and conjugate-linear in the second. The prime on a sum means summation over strictly increasing multi-indices. Thus a form $u\in\Omega_c^{p,q}(\Omega)$ is written \begin{align*} u=\sum_{I\in\mathcal I_p,\ J\in\mathcal I_q}' u_{I,J}\,dz_I\wedge d\bar z_J, \end{align*} where each coefficient $u_{I,J}$ belongs to $C_c^\infty(\Omega)$. [/step] [step:Prove the exterior-algebra adjoint identity on basis forms] For $1\le j\le n$, define exterior multiplication \begin{align*} \varepsilon_j:\Lambda^{p,q-1}&\to\Lambda^{p,q}\\ \alpha&\mapsto d\bar z_j\wedge\alpha, \end{align*} and contraction \begin{align*} \iota_j:\Lambda^{p,q}&\to\Lambda^{p,q-1}\\ \beta&\mapsto \iota_{\partial/\partial\bar z_j}\beta. \end{align*} The contraction $\iota_j$ is the graded derivation of degree $-1$ determined on coordinate one-forms, for $1\le \ell\le n$, by \begin{align*} \iota_j(dz_\ell)&=0,\\ \iota_j(d\bar z_\ell)&= \begin{cases} 1, & \ell=j,\\ 0, & \ell\ne j, \end{cases} \end{align*} and, for homogeneous forms $\gamma$ and $\delta$ with total exterior degree $|\gamma|$, by \begin{align*} \iota_j(\gamma\wedge\delta) = \iota_j\gamma\wedge\delta +(-1)^{|\gamma|}\gamma\wedge\iota_j\delta. \end{align*} We prove that $\iota_j$ is the adjoint of $\varepsilon_j$ with respect to the Euclidean Hermitian inner products: \begin{align*} (\varepsilon_j\alpha,\beta)_{\Lambda^{p,q}} = (\alpha,\iota_j\beta)_{\Lambda^{p,q-1}}. \end{align*} It suffices to check basis vectors. Fix $I,I'\in\mathcal I_p$, $J\in\mathcal I_{q-1}$, and $K\in\mathcal I_q$. If $j\in J$, then $d\bar z_j\wedge d\bar z_J=0$, so $\varepsilon_j e_{I,J}=0$. Also $K$ cannot be the increasing rearrangement of $\{j\}\cup J$ because this set has a repeated element, and hence the coefficient of $e_{I,J}$ in $\iota_j e_{I',K}$ is zero. Both inner products vanish. Assume now that $j\notin J$. Let $K(j,J)\in\mathcal I_q$ be the increasing rearrangement of the set $\{j,j_1,\dots,j_{q-1}\}$. Define $\sigma(j,J)\in\{1,-1\}$ by \begin{align*} d\bar z_j\wedge d\bar z_J=\sigma(j,J)d\bar z_{K(j,J)}. \end{align*} Since $d\bar z_j$ must first pass across the $p$ holomorphic factors in $dz_I$, set \begin{align*} \tau(j,J)=(-1)^p\sigma(j,J). \end{align*} Then \begin{align*} \varepsilon_j e_{I,J} &=d\bar z_j\wedge dz_I\wedge d\bar z_J\\ &=\tau(j,J)e_{I,K(j,J)}. \end{align*} By the defining rule for contraction with $\partial/\partial\bar z_j$, \begin{align*} \iota_j e_{I',K} = \begin{cases} \tau(j,J)e_{I',J}, & K=K(j,J),\\ 0, & K\ne K(j,J). \end{cases} \end{align*} Therefore \begin{align*} (\varepsilon_j e_{I,J},e_{I',K})_{\Lambda^{p,q}} &= \begin{cases} \tau(j,J), & I=I'\text{ and }K=K(j,J),\\ 0, & \text{otherwise}, \end{cases}\\ (e_{I,J},\iota_j e_{I',K})_{\Lambda^{p,q-1}} &= \begin{cases} \tau(j,J), & I=I'\text{ and }K=K(j,J),\\ 0, & \text{otherwise}. \end{cases} \end{align*} The two expressions agree on basis vectors, and sesquilinearity gives the identity for all $\alpha\in\Lambda^{p,q-1}$ and $\beta\in\Lambda^{p,q}$. [/step] [step:Prove the scalar compact-support integration by parts identity] Let $a,b\in C_c^\infty(\Omega)$ and fix $j\in\{1,\dots,n\}$. Write $z_j=x_j+iy_j$, so \begin{align*} \frac{\partial}{\partial \bar z_j} &=\frac{1}{2}\left(\frac{\partial}{\partial x_j} +i\frac{\partial}{\partial y_j}\right),\\ \frac{\partial}{\partial z_j} &=\frac{1}{2}\left(\frac{\partial}{\partial x_j} -i\frac{\partial}{\partial y_j}\right). \end{align*} Because $a$ and $b$ have compact support in $\Omega$, their zero extensions to $\mathbb R^{2n}$ are smooth compactly supported functions. The integral over $\Omega$ equals the integral over $\mathbb R^{2n}$ of the zero extensions. Applying the one-dimensional integration-by-parts formula in the $x_j$ and $y_j$ variables, with no boundary term because the zero extensions have compact support, gives \begin{align*} \int_\Omega \frac{\partial a}{\partial\bar z_j}\,\overline{b}\, d\mathcal L^{2n}(z) &= \frac{1}{2}\int_\Omega \left(\frac{\partial a}{\partial x_j} +i\frac{\partial a}{\partial y_j}\right)\overline{b}\, d\mathcal L^{2n}(z)\\ &= -\frac{1}{2}\int_\Omega a\left(\frac{\partial\overline b}{\partial x_j} +i\frac{\partial\overline b}{\partial y_j}\right) d\mathcal L^{2n}(z)\\ &= -\int_\Omega a\,\overline{\frac{\partial b}{\partial z_j}}\, d\mathcal L^{2n}(z). \end{align*} Thus the scalar formal adjoint of $\partial/\partial\bar z_j$ on compactly supported functions is $-\partial/\partial z_j$. [/step] [step:Apply the scalar identity coefficientwise] For $v\in\Omega_c^{p,q-1}(\Omega)$, the Dolbeault operator can be written as \begin{align*} \bar{\partial}v =\sum_{j=1}^n \varepsilon_j\left(\frac{\partial v}{\partial\bar z_j}\right), \end{align*} where $\partial v/\partial\bar z_j$ denotes coefficientwise differentiation. Let $u\in\Omega_c^{p,q}(\Omega)$. Since the sums over form coefficients are finite, we may combine the exterior-algebra adjoint identity and the scalar integration-by-parts identity term by term: \begin{align*} (\bar{\partial}v,u)_{L^2} &= \sum_{j=1}^n \int_\Omega \left( \varepsilon_j\frac{\partial v}{\partial\bar z_j}(z), u(z) \right)_{\Lambda^{p,q}} d\mathcal L^{2n}(z)\\ &= \sum_{j=1}^n \int_\Omega \left( \frac{\partial v}{\partial\bar z_j}(z), \iota_j u(z) \right)_{\Lambda^{p,q-1}} d\mathcal L^{2n}(z)\\ &= \sum_{j=1}^n \int_\Omega \left( v(z), -\frac{\partial}{\partial z_j}(\iota_j u)(z) \right)_{\Lambda^{p,q-1}} d\mathcal L^{2n}(z). \end{align*} The last line uses the scalar identity on every coefficient of $v$ and $\iota_j u$; the coefficients of $\iota_j u$ are compactly supported because $u$ is compactly supported. By the stated coordinate definition of the Euclidean formal adjoint, \begin{align*} \bar{\partial}^*u =-\sum_{j=1}^n \frac{\partial}{\partial z_j}\bigl(\iota_j u\bigr). \end{align*} Substitution gives \begin{align*} (\bar{\partial}v,u)_{L^2}=(v,\bar{\partial}^*u)_{L^2}. \end{align*} This proves the integration-by-parts formula. [/step]