[proofplan] The argument is an integration-by-parts computation followed by a duality dichotomy. Pairing $\bar{\partial} v$ against $u$ in $L^2$ coefficient by coefficient and applying the [Divergence Theorem](/theorems/3614) to each Wirtinger derivative produces two contributions: an interior term that equals $(v, \vartheta u)_{L^2}$, where $\vartheta u$ is the formal adjoint expression, and a boundary integral whose density is $\sum_j (\partial\rho/\partial z_j)\, u_{jK}$ paired against the boundary trace of $v_K$. Because smooth $(0,q-1)$-forms with prescribed boundary values are dense in $\operatorname{Dom}(\bar{\partial})$ in graph norm, the boundary integral must vanish identically for $u \in \operatorname{Dom}(\bar{\partial}^*)$; conversely, when it vanishes, the interior identity gives $\bar{\partial}^* u = \vartheta u$. The independence of the boundary trace of $v$ converts the integral vanishing into the pointwise contraction condition via the [Fundamental Lemma of Calculus of Variations](/theorems/45). [/proofplan]

[step:Set up the inner product and reduce to a coefficient-by-coefficient computation]We work in the [Hilbert space](/page/Hilbert%20Space) $L^2_{(0,q)}(\Omega)$ with inner product \begin{align*} (u, w)_{L^2_{(0,q)}} &= \sum_{|J|=q}{}' \int_\Omega u_J(z)\, \overline{w_J(z)} \, d\mathcal{L}^{2n}(z), \end{align*} where $\sum'$ denotes summation over strictly increasing multi-indices. For $v \in C^\infty_{(0,q-1)}(\overline{\Omega})$ with $v = \sum'_{|K|=q-1} v_K\, d\bar{z}_K$, the Dolbeault differential is \begin{align*} \bar{\partial} v &= \sum_{|K|=q-1}{}' \sum_{j=1}^n \frac{\partial v_K}{\partial \bar{z}_j}\, d\bar{z}_j \wedge d\bar{z}_K. \end{align*} Expanding $d\bar{z}_j \wedge d\bar{z}_K$ in the basis $\{d\bar{z}_J : |J|=q\}$ via the antisymmetric extension and taking the inner product against $u = \sum'_{|J|=q} u_J\, d\bar{z}_J$, the standard combinatorial identity for the wedge–coefficient pairing gives \begin{align*} (\bar{\partial} v, u)_{L^2_{(0,q)}} &= \sum_{|K|=q-1}{}' \sum_{j=1}^n \int_\Omega \frac{\partial v_K}{\partial \bar{z}_j}(z)\, \overline{u_{jK}(z)} \, d\mathcal{L}^{2n}(z). \tag{$\ast$} \end{align*} This identity is purely combinatorial: it follows from $u_{jK} = 0$ whenever $j \in K$ and from $u_{\sigma J} = \operatorname{sgn}(\sigma)\, u_J$ for permutations $\sigma$.[/step]

[guided]We start by unpacking the pairing $(\bar\partial v, u)_{L^2}$ in coefficients, since the strategy is to integrate by parts inside each Wirtinger derivative $\partial/\partial \bar z_j$. The [Hilbert space](/page/Hilbert%20Space) $L^2_{(0,q)}(\Omega)$ has inner product \begin{align*} (u, w)_{L^2_{(0,q)}} &= \sum_{|J|=q}{}' \int_\Omega u_J(z)\, \overline{w_J(z)} \, d\mathcal{L}^{2n}(z), \end{align*} where the prime restricts the sum to strictly increasing multi-indices to avoid double-counting (each $(0,q)$-form has a unique representation with increasing indices). The Dolbeault differential of $v = \sum'_K v_K\, d\bar z_K$ is \begin{align*} \bar{\partial} v &= \sum_{|K|=q-1}{}' \sum_{j=1}^n \frac{\partial v_K}{\partial \bar{z}_j}\, d\bar{z}_j \wedge d\bar{z}_K. \end{align*} Here $j$ ranges over $\{1,\dots,n\}$ without restriction — when $j \in K$, the wedge $d\bar z_j \wedge d\bar z_K$ vanishes automatically. Why expand $\bar\partial v$ this way rather than reindexing to increasing multi-indices? Because then the antisymmetric extension $u_{jK}$ absorbs the sign work: the pairing of $d\bar z_j \wedge d\bar z_K$ against $d\bar z_J$ (where $J$ is the unique increasing rearrangement of $\{j\} \cup K$) introduces $\operatorname{sgn}(\sigma)$, and this is exactly the sign convention defining $u_{jK}$. Computing the inner product term by term and collecting the antisymmetric coefficients, one obtains \begin{align*} (\bar{\partial} v, u)_{L^2_{(0,q)}} &= \sum_{|K|=q-1}{}' \sum_{j=1}^n \int_\Omega \frac{\partial v_K}{\partial \bar{z}_j}(z)\, \overline{u_{jK}(z)} \, d\mathcal{L}^{2n}(z). \tag{$\ast$} \end{align*} The identity $(\ast)$ is the key bookkeeping step: it converts the wedge product on the left into an ordinary scalar sum over $(j, K)$ on the right. The proof from here on is just [integration by parts](/theorems/2098) applied to each scalar integral.[/guided]

[step:Integrate each Wirtinger derivative by parts using the Divergence Theorem]Fix indices $j$ and $K$, and consider the scalar integral $\int_\Omega \frac{\partial v_K}{\partial \bar{z}_j}\, \overline{u_{jK}} \, d\mathcal{L}^{2n}$. Recall $\frac{\partial}{\partial \bar{z}_j} = \frac{1}{2}\!\left(\frac{\partial}{\partial x_j} + i \frac{\partial}{\partial y_j}\right)$ where $z_j = x_j + i y_j$. Applying the product rule, \begin{align*} \frac{\partial}{\partial \bar{z}_j}\!\left(v_K \, \overline{u_{jK}}\right) &= \frac{\partial v_K}{\partial \bar{z}_j}\, \overline{u_{jK}} + v_K \cdot \frac{\partial \overline{u_{jK}}}{\partial \bar{z}_j} = \frac{\partial v_K}{\partial \bar{z}_j}\, \overline{u_{jK}} + v_K \cdot \overline{\frac{\partial u_{jK}}{\partial z_j}}, \end{align*} where the second equality uses the conjugation identity $\partial_{\bar{z}_j} \bar{h} = \overline{\partial_{z_j} h}$ for any $C^1$ function $h$. Integrate this identity over $\Omega$. Writing $\partial/\partial \bar{z}_j = \frac{1}{2}(\partial_{x_j} + i\partial_{y_j})$ as a real linear combination of partial derivatives, the [Divergence Theorem](/theorems/3614) applied to the real and imaginary parts of the smooth $C^1$ function $v_K \overline{u_{jK}}$ yields \begin{align*} \int_\Omega \frac{\partial}{\partial \bar{z}_j}\!\left(v_K\, \overline{u_{jK}}\right) d\mathcal{L}^{2n} &= \int_{\partial\Omega} v_K \, \overline{u_{jK}} \cdot \nu^{(\bar{z}_j)} \, d\mathcal{H}^{2n-1}, \end{align*} where $\nu^{(\bar{z}_j)} := \frac{1}{2}(\nu_{x_j} + i\, \nu_{y_j})$ is the $\bar{z}_j$-component of the outward unit normal $\nu = \nabla \rho$ on $\partial\Omega$ (using $|\nabla\rho|=1$). Since $\rho$ is real-valued, \begin{align*} \nu^{(\bar{z}_j)} &= \frac{1}{2}\!\left(\frac{\partial \rho}{\partial x_j} + i\frac{\partial \rho}{\partial y_j}\right) = \frac{\partial \rho}{\partial \bar{z}_j} = \overline{\frac{\partial \rho}{\partial z_j}}. \end{align*} Substituting and rearranging: \begin{align*} \int_\Omega \frac{\partial v_K}{\partial \bar{z}_j}\, \overline{u_{jK}} \, d\mathcal{L}^{2n} &= - \int_\Omega v_K\, \overline{\frac{\partial u_{jK}}{\partial z_j}} \, d\mathcal{L}^{2n} + \int_{\partial \Omega} v_K\, \overline{u_{jK}}\, \overline{\frac{\partial \rho}{\partial z_j}} \, d\mathcal{H}^{2n-1}. \end{align*}[/step]

[guided]We want to move the derivative $\partial_{\bar z_j}$ off $v_K$ and onto $\overline{u_{jK}}$. The standard tool is the [Divergence Theorem](/theorems/3614). Since both $v_K$ and $u_{jK}$ are in $C^\infty(\overline\Omega)$, no regularity issue arises. First the product rule: \begin{align*} \frac{\partial}{\partial \bar{z}_j}\!\left(v_K \, \overline{u_{jK}}\right) &= \frac{\partial v_K}{\partial \bar{z}_j}\, \overline{u_{jK}} + v_K \cdot \frac{\partial \overline{u_{jK}}}{\partial \bar{z}_j}. \end{align*} The conjugation identity $\partial_{\bar z_j} \bar h = \overline{\partial_{z_j} h}$ (immediate from $\partial_{\bar z_j} = \frac{1}{2}(\partial_{x_j} + i \partial_{y_j})$ and conjugating both Wirtinger pieces) lets us rewrite the second term as $v_K \overline{\partial_{z_j} u_{jK}}$. Now integrate over $\Omega$. The Wirtinger derivative is a complex linear combination of real partial derivatives, so the [Divergence Theorem](/theorems/2754) applies coordinate by coordinate: \begin{align*} \int_\Omega \frac{\partial}{\partial \bar{z}_j}\!\left(v_K\, \overline{u_{jK}}\right) d\mathcal{L}^{2n} &= \int_{\partial\Omega} v_K \, \overline{u_{jK}} \cdot \nu^{(\bar{z}_j)} \, d\mathcal{H}^{2n-1}. \end{align*} The boundary "Wirtinger normal" $\nu^{(\bar{z}_j)} = \frac{1}{2}(\nu_{x_j} + i \nu_{y_j})$ is, since $\nu = \nabla \rho$ (using $|\nabla \rho|=1$) and $\rho$ is real, \begin{align*} \nu^{(\bar{z}_j)} &= \frac{1}{2}\!\left(\frac{\partial \rho}{\partial x_j} + i\frac{\partial \rho}{\partial y_j}\right) = \frac{\partial \rho}{\partial \bar{z}_j} = \overline{\frac{\partial \rho}{\partial z_j}}. \end{align*} The last equality is again $\bar\rho = \rho$. This is the *only* place where the realness of $\rho$ matters: it lets us write the boundary density as $\overline{\partial \rho/\partial z_j}$, which is the conjugate of the $z_j$-derivative used in the statement. Combining and isolating the term of interest: \begin{align*} \int_\Omega \frac{\partial v_K}{\partial \bar{z}_j}\, \overline{u_{jK}} \, d\mathcal{L}^{2n} &= - \int_\Omega v_K\, \overline{\frac{\partial u_{jK}}{\partial z_j}} \, d\mathcal{L}^{2n} + \int_{\partial \Omega} v_K\, \overline{u_{jK}}\, \overline{\frac{\partial \rho}{\partial z_j}} \, d\mathcal{H}^{2n-1}. \end{align*} We have produced exactly the two contributions the proof needs: an interior term that we will recognise as $(v, \vartheta u)$, and a boundary term that we will recognise as the contraction $\sum_j (\partial \rho/\partial z_j) u_{jK}$ paired against $v_K$.[/guided]

[step:Assemble the interior and boundary contributions] Summing the previous identity over $j \in \{1,\dots,n\}$ and over increasing $K$ with $|K|=q-1$, and substituting into $(\ast)$: \begin{align*} (\bar{\partial} v, u)_{L^2_{(0,q)}} = - \sum_{|K|=q-1}{}' \int_\Omega v_K \, \overline{\sum_{j=1}^n \frac{\partial u_{jK}}{\partial z_j}} \, d\mathcal{L}^{2n} + \sum_{|K|=q-1}{}' \int_{\partial \Omega} v_K \, \overline{\sigma_K(u)} \, d\mathcal{H}^{2n-1}, \end{align*} where for each increasing $K$ with $|K|=q-1$ we have introduced \begin{align*} \sigma_K(u): \partial\Omega &\to \mathbb{C}, \\ z &\mapsto \sum_{j=1}^n \frac{\partial \rho}{\partial z_j}(z)\, u_{jK}(z). \end{align*} Define the smooth $(0,q-1)$-form \begin{align*} \vartheta u &:= -\sum_{|K|=q-1}{}' \sum_{j=1}^n \frac{\partial u_{jK}}{\partial z_j}\, d\bar{z}_K \;\in\; C^\infty_{(0,q-1)}(\overline{\Omega}). \end{align*} Then the previous identity reads \begin{align*} (\bar{\partial} v, u)_{L^2_{(0,q)}} &= (v, \vartheta u)_{L^2_{(0,q-1)}} + \sum_{|K|=q-1}{}' \int_{\partial \Omega} v_K \, \overline{\sigma_K(u)} \, d\mathcal{H}^{2n-1}, \tag{$\dagger$} \end{align*} valid for every $v \in C^\infty_{(0,q-1)}(\overline{\Omega})$. [/step]

[step:Prove sufficiency: vanishing of $\sigma_K(u)$ on $\partial\Omega$ implies $u \in \operatorname{Dom}(\bar{\partial}^*)$] Suppose $\sigma_K(u) \equiv 0$ on $\partial\Omega$ for every increasing $K$ with $|K|=q-1$. Then $(\dagger)$ reduces to \begin{align*} (\bar{\partial} v, u)_{L^2_{(0,q)}} &= (v, \vartheta u)_{L^2_{(0,q-1)}} \qquad \text{for all } v \in C^\infty_{(0,q-1)}(\overline{\Omega}). \tag{$\ddagger$} \end{align*} Because $\vartheta u \in C^\infty_{(0,q-1)}(\overline{\Omega})$ and $\Omega$ is bounded, $\vartheta u \in L^2_{(0,q-1)}(\Omega)$, so by Cauchy–Schwarz \begin{align*} \left|(\bar{\partial} v, u)_{L^2_{(0,q)}}\right| &\le \|\vartheta u\|_{L^2_{(0,q-1)}(\Omega)}\, \|v\|_{L^2_{(0,q-1)}(\Omega)} \qquad \text{for all } v \in C^\infty_{(0,q-1)}(\overline{\Omega}). \end{align*} The space $C^\infty_{(0,q-1)}(\overline{\Omega})$ is a core for $\bar{\partial}$ on bounded smooth domains: every $w \in \operatorname{Dom}(\bar{\partial}) \subset L^2_{(0,q-1)}(\Omega)$ can be approximated by a sequence $v^{(k)} \in C^\infty_{(0,q-1)}(\overline{\Omega})$ with $v^{(k)} \to w$ in $L^2$ and $\bar{\partial} v^{(k)} \to \bar{\partial} w$ in $L^2$. This is the Friedrichs density lemma for the maximal $\bar{\partial}$, established via convolution with a Friedrichs mollifier (citing a result not yet in the wiki: *Friedrichs density of smooth forms in $\operatorname{Dom}(\bar\partial)$ on smoothly bounded domains*). Passing to the limit in $(\ddagger)$ along this approximation, \begin{align*} (\bar{\partial} w, u)_{L^2_{(0,q)}} &= (w, \vartheta u)_{L^2_{(0,q-1)}} \qquad \text{for all } w \in \operatorname{Dom}(\bar{\partial}). \end{align*} By the definition of the [Hilbert space](/page/Hilbert%20Space) adjoint, this says $u \in \operatorname{Dom}(\bar{\partial}^*)$ and $\bar{\partial}^* u = \vartheta u$, establishing both the sufficiency direction and the formula for $\bar{\partial}^* u$. [/step]

[step:Prove necessity: $u \in \operatorname{Dom}(\bar{\partial}^*)$ forces $\sigma_K(u) \equiv 0$ on $\partial\Omega$]Assume $u \in \operatorname{Dom}(\bar{\partial}^*)$, so $\bar{\partial}^* u \in L^2_{(0,q-1)}(\Omega)$ exists and \begin{align*} (\bar{\partial} v, u)_{L^2_{(0,q)}} &= (v, \bar{\partial}^* u)_{L^2_{(0,q-1)}} \qquad \text{for all } v \in \operatorname{Dom}(\bar{\partial}). \end{align*} In particular, this holds for every $v \in C^\infty_{c,(0,q-1)}(\Omega)$, the space of smooth compactly supported $(0,q-1)$-forms. For such $v$, the boundary term in $(\dagger)$ vanishes (boundary traces are zero), so $(v, \vartheta u)_{L^2} = (v, \bar{\partial}^* u)_{L^2}$ for every $v \in C^\infty_{c,(0,q-1)}(\Omega)$. Since $C^\infty_c$ is dense in $L^2$ and $\vartheta u, \bar{\partial}^* u \in L^2_{(0,q-1)}(\Omega)$, this gives $\vartheta u = \bar{\partial}^* u$ as elements of $L^2_{(0,q-1)}(\Omega)$. Substituting back into $(\dagger)$ and using $(\bar\partial v, u) = (v, \bar\partial^* u) = (v, \vartheta u)$ for every $v \in C^\infty_{(0,q-1)}(\overline{\Omega})$ (which lies in $\operatorname{Dom}(\bar\partial)$), the boundary integral must satisfy \begin{align*} \sum_{|K|=q-1}{}' \int_{\partial \Omega} v_K\, \overline{\sigma_K(u)} \, d\mathcal{H}^{2n-1} &= 0 \qquad \text{for all } v \in C^\infty_{(0,q-1)}(\overline{\Omega}). \tag{$\sharp$} \end{align*} We now show $(\sharp)$ forces $\sigma_K(u) \equiv 0$ on $\partial\Omega$ for each increasing $K$. Fix an increasing multi-index $K_0$ with $|K_0|=q-1$ and an arbitrary $\phi \in C^\infty(\partial\Omega; \mathbb{C})$. Extend $\phi$ to a function $\tilde{\phi} \in C^\infty(\overline{\Omega}; \mathbb{C})$ via the boundary trace [extension theorem](/theorems/59) ([Trace Theorem](/theorems/60); since $\phi \in C^\infty(\partial\Omega)$, an extension can be constructed directly using a tubular neighbourhood of $\partial\Omega$ and a cutoff). Define the test form $v := \tilde{\phi}\, d\bar{z}_{K_0} \in C^\infty_{(0,q-1)}(\overline{\Omega})$. Then $v_{K_0} = \tilde\phi$ and $v_K = 0$ for every increasing $K \ne K_0$. Applying $(\sharp)$: \begin{align*} \int_{\partial \Omega} \phi \, \overline{\sigma_{K_0}(u)} \, d\mathcal{H}^{2n-1} &= 0. \end{align*} Since $\phi \in C^\infty(\partial\Omega)$ was arbitrary, the [Fundamental Lemma of Calculus of Variations](/theorems/45) applied on the smooth compact manifold $\partial\Omega$ with surface measure $\mathcal{H}^{2n-1}$ forces $\overline{\sigma_{K_0}(u)} = 0$, hence $\sigma_{K_0}(u) \equiv 0$ on $\partial\Omega$. Since $K_0$ was an arbitrary increasing $(q-1)$-index, the boundary condition holds.[/step]

[guided]The boundary integral in $(\dagger)$ now becomes the focus. We have already shown that $u$ being in $\operatorname{Dom}(\bar\partial^*)$ forces $\bar\partial^* u = \vartheta u$ as an $L^2$ identity — this came from testing against compactly supported $v$, which kills the boundary term. Substituting back into $(\dagger)$, the identity $(\bar\partial v, u) = (v, \vartheta u)$ now holds for *all* $v \in C^\infty_{(0,q-1)}(\overline\Omega)$, and reading off the boundary term yields $(\sharp)$. The remaining content is the following dichotomy: $(\sharp)$ is one integral equation for each test form $v$, but the boundary traces $v_K|_{\partial\Omega}$ can be chosen freely and independently across the different increasing multi-indices $K$. Why? Because $C^\infty(\overline{\Omega})$ surjects onto $C^\infty(\partial\Omega)$ (any smooth function on $\partial\Omega$ extends — explicitly, use a tubular neighbourhood $\partial\Omega \times (-\delta, \delta)$ parametrised by signed distance, set $\tilde\phi(z',t) := \phi(z')\, \chi(t)$ with $\chi \in C^\infty_c((-\delta, \delta))$ equal to $1$ near $t=0$, and transport back; this is a classical construction underlying the [Trace Theorem](/theorems/60)). And different multi-indices $K$ correspond to different basis $1$-forms $d\bar z_K$, so we can isolate one $v_{K_0}$ by setting all other $v_K = 0$. Concretely, fix $K_0$ and $\phi \in C^\infty(\partial\Omega)$. Choose $\tilde\phi \in C^\infty(\overline\Omega)$ with $\tilde\phi|_{\partial\Omega} = \phi$, and set $v := \tilde\phi\, d\bar z_{K_0}$. Plugging into $(\sharp)$: \begin{align*} \int_{\partial \Omega} \phi\, \overline{\sigma_{K_0}(u)} \, d\mathcal{H}^{2n-1} &= 0. \end{align*} This holds for *every* smooth $\phi$ on $\partial\Omega$. By the [Fundamental Lemma of Calculus of Variations](/theorems/45) on the smooth compact $(2n-1)$-manifold $\partial\Omega$ — the hypotheses are met since $\sigma_{K_0}(u)$ is continuous on $\partial\Omega$ (as the boundary trace of a $C^\infty$ function), so the lemma forces it to vanish pointwise — we get $\sigma_{K_0}(u) \equiv 0$ on $\partial\Omega$. Repeating for each increasing $K_0$ with $|K_0|=q-1$ completes the proof.[/guided]

[step:Verify the conclusion of the theorem] Combining the two directions: when $\sigma_K(u) \equiv 0$ on $\partial\Omega$ for every increasing $K$ with $|K|=q-1$, the previous steps show $u \in \operatorname{Dom}(\bar{\partial}^*)$ with $\bar{\partial}^* u = \vartheta u$, which is the displayed formula. Conversely, when $u \in \operatorname{Dom}(\bar{\partial}^*)$, the necessity step shows $\sigma_K(u) \equiv 0$ on $\partial\Omega$ for every such $K$. The biconditional and the formula for $\bar{\partial}^* u$ are therefore both established, completing the proof. Finally, we note that the boundary condition $\sum_j (\partial \rho/\partial z_j)\, u_{jK} = 0$ on $\partial\Omega$ is invariant under the normalisation choice: if $\tilde\rho = h\, \rho$ with $h \in C^\infty(\mathbb{C}^n; (0,\infty))$, then on $\partial\Omega = \{\rho = 0\}$ we have $\partial \tilde\rho/\partial z_j = h\, \partial \rho/\partial z_j$, so $\sum_j (\partial \tilde\rho/\partial z_j) u_{jK} = h \sum_j (\partial \rho/\partial z_j) u_{jK}$, which vanishes on $\partial\Omega$ iff $\sum_j (\partial \rho/\partial z_j) u_{jK}$ does. Hence the normalisation $|\nabla \rho|=1$ used in the proof entails no loss of generality. [/step]