[proofplan] We compute $\|\bar\partial u\|_{L^2}^2$ and $\|\bar\partial^* u\|_{L^2}^2$ in coordinates. A combinatorial expansion of the wedge product reduces $\|\bar\partial u\|_{L^2}^2$ to a "diagonal" piece $\sum_{J,j} \|\partial_{\bar z_j} u_J\|^2$ minus a "cross" piece $\sum_{K,j,k}(\partial_{\bar z_j} u_{kK}, \partial_{\bar z_k} u_{jK})$, while $\|\bar\partial^* u\|_{L^2}^2$ contributes a "divergence" piece $\sum_{K,j,k}(\partial_{z_j} u_{jK}, \partial_{z_k} u_{kK})$. Integrating by parts twice converts the difference of cross and divergence pieces into a pure boundary integral; the $\bar\partial$-Neumann boundary condition annihilates one of the resulting boundary terms, and tangential differentiation of the conjugate boundary condition converts the remaining boundary term into the complex Hessian (Levi form) of $\rho$ paired against the boundary values of $u$. [/proofplan] [step:Set up complex coordinates and the integration by parts identity] Write $z_j = x_j + i y_j$ and use the Wirtinger derivatives \begin{align*} \frac{\partial}{\partial z_j} = \tfrac{1}{2}\!\left(\frac{\partial}{\partial x_j} - i\frac{\partial}{\partial y_j}\right), \qquad \frac{\partial}{\partial \bar z_j} = \tfrac{1}{2}\!\left(\frac{\partial}{\partial x_j} + i\frac{\partial}{\partial y_j}\right). \end{align*} For $f, g \in C^\infty(\overline\Omega; \mathbb{C})$, applying the [Divergence Theorem](/theorems/3614) componentwise to the real and imaginary parts of $f\bar g$ and combining yields \begin{align*} (\partial_{z_j} f, g)_{L^2(\Omega)} &= -(f, \partial_{\bar z_j} g)_{L^2(\Omega)} + \int_{\partial\Omega} f\, \bar g \, \frac{\partial \rho}{\partial z_j} \, dS_\rho, \\ (\partial_{\bar z_j} f, g)_{L^2(\Omega)} &= -(f, \partial_{z_j} g)_{L^2(\Omega)} + \int_{\partial\Omega} f\, \bar g \, \frac{\partial \rho}{\partial \bar z_j} \, dS_\rho, \end{align*} where $(\phi, \psi)_{L^2(\Omega)} := \int_\Omega \phi \, \bar\psi \, d\mathcal{L}^{2n}$. We use these formulas repeatedly below. [guided] The Wirtinger calculus is essential: it lets us treat $z_j$ and $\bar z_j$ as independent. To verify the integration-by-parts formula, write \begin{align*} \partial_{z_j}(f \bar g) = (\partial_{z_j} f)\, \bar g + f\, (\partial_{z_j} \bar g) = (\partial_{z_j} f)\, \bar g + f\, \overline{\partial_{\bar z_j} g}, \end{align*} where the last equality uses $\partial_{z_j} \bar g = \overline{\partial_{\bar z_j} g}$. Integrating over $\Omega$ and applying the [Divergence Theorem](/theorems/3614) to the vector field whose Cartesian components recombine into $\partial_{z_j}(f\bar g)$ (specifically: $\partial_{z_j} = \frac{1}{2}(\partial_{x_j} - i \partial_{y_j})$ contributes the half-sum of the $x_j$ and $-i y_j$ divergences) gives \begin{align*} \int_\Omega \partial_{z_j}(f\bar g) \, d\mathcal{L}^{2n} = \int_{\partial\Omega} f \bar g \cdot \tfrac{1}{2}\!\left(\frac{\partial_{x_j}\rho}{|\nabla\rho|} - i \frac{\partial_{y_j}\rho}{|\nabla\rho|}\right) d\sigma = \int_{\partial\Omega} f\bar g \, \frac{\partial\rho}{\partial z_j} \, dS_\rho, \end{align*} since the outer unit normal at $x \in \partial\Omega$ is $\nabla \rho(x)/|\nabla\rho(x)|$ (because $\rho < 0$ inside $\Omega$ and $\rho = 0$ on $\partial\Omega$). Rearranging the product-rule identity yields the first formula; the second follows by complex conjugation. Throughout the proof we will need both formulas: $\partial_{z_j}$ and $\partial_{\bar z_j}$ are formal adjoints (up to a sign and boundary term) of $-\partial_{\bar z_j}$ and $-\partial_{z_j}$ respectively. [/guided] [/step] [step:Expand $\|\bar\partial u\|_{L^2}^2$ in coordinates] Since $\bar\partial u = \sum_{J}{}' \sum_j \partial_{\bar z_j} u_J \, d\bar z_j \wedge d\bar z_J$, we claim \begin{align*} \|\bar\partial u\|_{L^2(\Omega)}^2 = \sum_{|J|=q}{}' \sum_{j=1}^n \|\partial_{\bar z_j} u_J\|_{L^2(\Omega)}^2 \;-\; \sum_{|K|=q-1}{}' \sum_{j,k=1}^n (\partial_{\bar z_j} u_{kK},\, \partial_{\bar z_k} u_{jK})_{L^2(\Omega)}, \tag{$\ast$} \end{align*} where in the second sum the components $u_{jK}, u_{kK}$ are extended by total antisymmetry. [claim:Coordinate formula for $\|\bar\partial u\|^2$] Equation $(\ast)$ holds. [/claim] [proof] For $|L| = q+1$ and $j \in L$, let $\epsilon_{j,L} := \mathrm{sgn}(\pi)$ where $\pi$ is the permutation sending $(j, L \setminus j)$ (with $L \setminus j$ in increasing order) to $L$ in increasing order. Then \begin{align*} \bar\partial u = \sum_{|L|=q+1}{}' (\bar\partial u)_L \, d\bar z_L, \qquad (\bar\partial u)_L = \sum_{j \in L} \epsilon_{j,L}\, \partial_{\bar z_j} u_{L \setminus j}. \end{align*} Therefore \begin{align*} \|\bar\partial u\|_{L^2}^2 = \sum_{|L|=q+1}{}' \int_\Omega \left|\sum_{j \in L} \epsilon_{j,L}\, \partial_{\bar z_j} u_{L\setminus j}\right|^2 d\mathcal{L}^{2n} = \mathrm{D} + \mathrm{O}, \end{align*} where $\mathrm{D}$ and $\mathrm{O}$ are the diagonal ($j = k$) and off-diagonal ($j \neq k$) contributions of the expanded square. **Diagonal.** Reparametrising $L = \{j\} \cup J$ with $J = L \setminus j$ ordered increasingly and $j \notin J$, \begin{align*} \mathrm{D} = \sum_{|L|=q+1}{}' \sum_{j \in L} \|\partial_{\bar z_j} u_{L \setminus j}\|_{L^2}^2 = \sum_{|J|=q}{}' \sum_{j \notin J} \|\partial_{\bar z_j} u_J\|_{L^2}^2. \end{align*} **Off-diagonal.** For distinct $j, k \in L$, set $K := L \setminus \{j, k\}$ ordered increasingly, $|K| = q-1$. For $a \notin K$, define $\delta_{a,K} := \operatorname{sgn}((a,K) \to \{a\} \cup K)$, where $\{a\} \cup K$ is written in increasing order. The antisymmetric extension gives $u_{aK} = \delta_{a,K} u_{\{a\}\cup K}$. Since $L \setminus j = \{k\} \cup K$ and $L \setminus k = \{j\} \cup K$, \begin{align*} \partial_{\bar z_j}u_{L\setminus j} = \delta_{k,K}\,\partial_{\bar z_j}u_{kK}, \qquad \partial_{\bar z_k}u_{L\setminus k} = \delta_{j,K}\,\partial_{\bar z_k}u_{jK}. \end{align*} Moreover, \begin{align*} \epsilon_{j,L}\delta_{k,K} = \operatorname{sgn}((j,k,K) \to L), \qquad \epsilon_{k,L}\delta_{j,K} = \operatorname{sgn}((k,j,K) \to L). \end{align*} The ordered lists $(j,k,K)$ and $(k,j,K)$ differ by exactly one transposition, so \begin{align*} (\epsilon_{j,L}\delta_{k,K})(\epsilon_{k,L}\delta_{j,K}) = -1. \end{align*} Hence \begin{align*} \mathrm{O} = -\sum_{|K|=q-1}{}' \sum_{j \neq k} (\partial_{\bar z_j} u_{kK},\, \partial_{\bar z_k} u_{jK})_{L^2}. \end{align*} The antisymmetric extension makes $u_{jK} = 0$ whenever $j \in K$, so adding the $j = k$ diagonal of the second sum costs $\sum_{K}{}' \sum_j \|\partial_{\bar z_j} u_{jK}\|^2 = \sum_{J}{}'\sum_{j \in J}\|\partial_{\bar z_j} u_J\|^2$ (parametrising $J = \{j\} \cup K$). Combining, \begin{align*} \mathrm{D} + \mathrm{O} = \sum_{J}{}'\!\sum_{j \notin J}\|\partial_{\bar z_j} u_J\|^2 + \sum_{J}{}'\!\sum_{j \in J}\|\partial_{\bar z_j} u_J\|^2 - \sum_{K}{}'\sum_{j,k}(\partial_{\bar z_j} u_{kK}, \partial_{\bar z_k} u_{jK})_{L^2}, \end{align*} which is precisely $(\ast)$. [/proof] [/step] [step:Compute $\|\bar\partial^* u\|_{L^2}^2$ from the interior formula] By the interior formula defining $\bar\partial^* u$, \begin{align*} \|\bar\partial^* u\|_{L^2(\Omega)}^2 = \sum_{|K|=q-1}{}' \int_\Omega \left|\sum_{j=1}^n \partial_{z_j} u_{jK}\right|^2 d\mathcal{L}^{2n} = \sum_{|K|=q-1}{}' \sum_{j,k=1}^n (\partial_{z_j} u_{jK},\, \partial_{z_k} u_{kK})_{L^2(\Omega)}. \tag{$\ast\ast$} \end{align*} Combining $(\ast)$ and $(\ast\ast)$, \begin{align*} \|\bar\partial u\|_{L^2}^2 + \|\bar\partial^* u\|_{L^2}^2 = \sum_{|J|=q}{}'\sum_{j=1}^n \|\partial_{\bar z_j} u_J\|_{L^2}^2 + \Theta, \end{align*} where \begin{align*} \Theta := \sum_{|K|=q-1}{}' \sum_{j,k=1}^n \Big[(\partial_{z_j} u_{jK},\, \partial_{z_k} u_{kK})_{L^2} - (\partial_{\bar z_j} u_{kK},\, \partial_{\bar z_k} u_{jK})_{L^2}\Big]. \end{align*} It remains to show that $\Theta$ equals the boundary integral of the complex Hessian. [/step] [step:Convert $\Theta$ to a boundary integral via double integration by parts] Fix $K$ with $|K| = q-1$ and $j, k \in \{1, \dots, n\}$. Apply the first integration-by-parts identity of Step 1 to move $\partial_{z_j}$ off $u_{jK}$: \begin{align*} (\partial_{z_j} u_{jK}, \partial_{z_k} u_{kK})_{L^2} = -(u_{jK}, \partial_{\bar z_j} \partial_{z_k} u_{kK})_{L^2} + \int_{\partial\Omega} u_{jK}\, \overline{\partial_{z_k} u_{kK}}\, \frac{\partial \rho}{\partial z_j}\, dS_\rho. \end{align*} Since $u_{kK} \in C^\infty(\overline\Omega; \mathbb{C})$, the mixed partials commute: $\partial_{\bar z_j} \partial_{z_k} = \partial_{z_k} \partial_{\bar z_j}$. Now apply the formula $(f, \partial_{z_k} g)_{L^2} = -(\partial_{\bar z_k} f, g)_{L^2} + \int_{\partial\Omega} f \bar g \, \partial_{\bar z_k} \rho \, dS_\rho$ (obtained by complex-conjugating the second identity of Step 1) with $f = u_{jK}$ and $g = \partial_{\bar z_j} u_{kK}$: \begin{align*} (u_{jK}, \partial_{z_k} \partial_{\bar z_j} u_{kK})_{L^2} = -(\partial_{\bar z_k} u_{jK}, \partial_{\bar z_j} u_{kK})_{L^2} + \int_{\partial\Omega} u_{jK}\, \overline{\partial_{\bar z_j} u_{kK}}\, \frac{\partial \rho}{\partial \bar z_k}\, dS_\rho. \end{align*} Substituting, \begin{align*} (\partial_{z_j} u_{jK}, \partial_{z_k} u_{kK})_{L^2} = (\partial_{\bar z_k} u_{jK}, \partial_{\bar z_j} u_{kK})_{L^2} + B_{j,k,K}^{(1)} - B_{j,k,K}^{(2)}, \end{align*} where \begin{align*} B_{j,k,K}^{(1)} &:= \int_{\partial\Omega} u_{jK}\, \overline{\partial_{z_k} u_{kK}}\, \frac{\partial \rho}{\partial z_j}\, dS_\rho, \\ B_{j,k,K}^{(2)} &:= \int_{\partial\Omega} u_{jK}\, \overline{\partial_{\bar z_j} u_{kK}}\, \frac{\partial \rho}{\partial \bar z_k}\, dS_\rho. \end{align*} Under the swap $j \leftrightarrow k$, the inner product $(\partial_{\bar z_k} u_{jK}, \partial_{\bar z_j} u_{kK})_{L^2}$ becomes $(\partial_{\bar z_j} u_{kK}, \partial_{\bar z_k} u_{jK})_{L^2}$. Hence after summing over $j$ and $k$, \begin{align*} \sum_{j,k}(\partial_{\bar z_k} u_{jK}, \partial_{\bar z_j} u_{kK})_{L^2} = \sum_{j,k}(\partial_{\bar z_j} u_{kK}, \partial_{\bar z_k} u_{jK})_{L^2}, \end{align*} so the volume terms in $\Theta$ cancel and \begin{align*} \Theta = \sum_{|K|=q-1}{}' \sum_{j,k=1}^n \big[B_{j,k,K}^{(1)} - B_{j,k,K}^{(2)}\big]. \end{align*} [guided] Fix $K$ with $|K| = q-1$ and fix $j,k \in \{1,\dots,n\}$. We want to compare the divergence inner product \begin{align*} (\partial_{z_j} u_{jK}, \partial_{z_k} u_{kK})_{L^2(\Omega)} \end{align*} with the cross inner product \begin{align*} (\partial_{\bar z_j} u_{kK}, \partial_{\bar z_k} u_{jK})_{L^2(\Omega)}. \end{align*} The mechanism is two integrations by parts, with the commutation of mixed Wirtinger derivatives used between them. Apply the first integration-by-parts identity from Step 1 with $f = u_{jK}$ and $g = \partial_{z_k}u_{kK}$. The functions are smooth on $\overline\Omega$, so the identity applies. We obtain \begin{align*} (\partial_{z_j} u_{jK}, \partial_{z_k} u_{kK})_{L^2(\Omega)} &= -(u_{jK}, \partial_{\bar z_j}\partial_{z_k}u_{kK})_{L^2(\Omega)} \\ &\quad + \int_{\partial\Omega} u_{jK}\,\overline{\partial_{z_k}u_{kK}}\,\frac{\partial\rho}{\partial z_j}\,dS_\rho. \end{align*} Since $u_{kK} \in C^\infty(\overline\Omega;\mathbb{C})$, the real partial derivatives commute, and therefore the Wirtinger derivatives commute: \begin{align*} \partial_{\bar z_j}\partial_{z_k}u_{kK} = \partial_{z_k}\partial_{\bar z_j}u_{kK}. \end{align*} Thus the remaining volume term is \begin{align*} (u_{jK}, \partial_{z_k}\partial_{\bar z_j}u_{kK})_{L^2(\Omega)}. \end{align*} Now use the rearranged second integration-by-parts identity from Step 1, \begin{align*} (f,\partial_{z_k}g)_{L^2(\Omega)} = -(\partial_{\bar z_k}f,g)_{L^2(\Omega)} + \int_{\partial\Omega} f\,\bar g\,\frac{\partial\rho}{\partial\bar z_k}\,dS_\rho, \end{align*} with $f = u_{jK}$ and $g = \partial_{\bar z_j}u_{kK}$. This gives \begin{align*} (u_{jK}, \partial_{z_k}\partial_{\bar z_j}u_{kK})_{L^2(\Omega)} &= -(\partial_{\bar z_k}u_{jK}, \partial_{\bar z_j}u_{kK})_{L^2(\Omega)} \\ &\quad + \int_{\partial\Omega} u_{jK}\,\overline{\partial_{\bar z_j}u_{kK}}\,\frac{\partial\rho}{\partial\bar z_k}\,dS_\rho. \end{align*} Define the two boundary terms \begin{align*} B_{j,k,K}^{(1)} &:= \int_{\partial\Omega} u_{jK}\,\overline{\partial_{z_k}u_{kK}}\,\frac{\partial\rho}{\partial z_j}\,dS_\rho, \\ B_{j,k,K}^{(2)} &:= \int_{\partial\Omega} u_{jK}\,\overline{\partial_{\bar z_j}u_{kK}}\,\frac{\partial\rho}{\partial\bar z_k}\,dS_\rho. \end{align*} Substituting the second integration-by-parts formula into the first one yields \begin{align*} (\partial_{z_j} u_{jK}, \partial_{z_k} u_{kK})_{L^2(\Omega)} = (\partial_{\bar z_k}u_{jK}, \partial_{\bar z_j}u_{kK})_{L^2(\Omega)} + B_{j,k,K}^{(1)} - B_{j,k,K}^{(2)}. \end{align*} This is the exact point where the volume terms prepare to cancel: after summing over $j$ and $k$, the finite relabeling $(j,k) \mapsto (k,j)$ gives \begin{align*} \sum_{j,k=1}^n(\partial_{\bar z_k}u_{jK}, \partial_{\bar z_j}u_{kK})_{L^2(\Omega)} = \sum_{j,k=1}^n(\partial_{\bar z_j}u_{kK}, \partial_{\bar z_k}u_{jK})_{L^2(\Omega)}. \end{align*} Therefore, in the quantity \begin{align*} \Theta := \sum_{|K|=q-1}{}'\sum_{j,k=1}^n\Big[(\partial_{z_j}u_{jK},\partial_{z_k}u_{kK})_{L^2(\Omega)}-(\partial_{\bar z_j}u_{kK},\partial_{\bar z_k}u_{jK})_{L^2(\Omega)}\Big], \end{align*} the summed volume contribution cancels with the cross term. What remains is precisely the boundary contribution \begin{align*} \Theta = \sum_{|K|=q-1}{}'\sum_{j,k=1}^n\big[B_{j,k,K}^{(1)}-B_{j,k,K}^{(2)}\big]. \end{align*} This completes the conversion of $\Theta$ from an interior expression into boundary terms. [/guided] [/step] [step:Annihilate $B^{(1)}$ using the $\bar\partial$-Neumann boundary condition] For each $K$ with $|K| = q-1$, factor the $j$-sum out of $B^{(1)}_{j,k,K}$: \begin{align*} \sum_{j,k=1}^n B_{j,k,K}^{(1)} = \sum_{k=1}^n \int_{\partial\Omega} \left(\sum_{j=1}^n \frac{\partial \rho}{\partial z_j} u_{jK}\right) \overline{\partial_{z_k} u_{kK}} \, dS_\rho. \end{align*} By hypothesis, $\sum_{j} (\partial \rho / \partial z_j)\, u_{jK} = 0$ pointwise on $\partial\Omega$, so $\sum_{j,k} B^{(1)}_{j,k,K} = 0$ for every $K$. Therefore \begin{align*} \Theta = -\sum_{|K|=q-1}{}' \sum_{j,k=1}^n B_{j,k,K}^{(2)} = -\sum_{|K|=q-1}{}' \sum_{j,k=1}^n \int_{\partial\Omega} u_{jK}\, (\partial_{z_j} \overline{u_{kK}})\, \frac{\partial \rho}{\partial \bar z_k}\, dS_\rho, \end{align*} where we used $\overline{\partial_{\bar z_j} u_{kK}} = \partial_{z_j} \overline{u_{kK}}$. [/step] [step:Convert the residual boundary term to the Levi form by tangential differentiation] Take complex conjugates of the $\bar\partial$-Neumann condition: since $\rho$ is real, $\overline{\partial \rho / \partial z_j} = \partial \rho / \partial \bar z_j$, and so \begin{align*} \sum_{k=1}^n \frac{\partial \rho}{\partial \bar z_k}\, \overline{u_{kK}} = 0 \qquad \text{on } \partial\Omega \text{ for every } K \text{ with } |K| = q-1. \end{align*} A smooth $\mathbb{C}$-valued function on $\overline\Omega$ vanishing on $\partial\Omega$ is divisible by $\rho$ in $C^\infty(\overline\Omega)$ (Hadamard's lemma applied componentwise), so there exists $h_K \in C^\infty(\overline\Omega; \mathbb{C})$ with \begin{align*} \sum_{k=1}^n \frac{\partial \rho}{\partial \bar z_k}\, \overline{u_{kK}} = h_K\, \rho \qquad \text{on } \overline\Omega. \end{align*} Apply $\partial_{z_j}$ to both sides: \begin{align*} \sum_{k=1}^n \frac{\partial^2 \rho}{\partial z_j\, \partial \bar z_k}\, \overline{u_{kK}} + \sum_{k=1}^n \frac{\partial \rho}{\partial \bar z_k}\, \partial_{z_j} \overline{u_{kK}} = (\partial_{z_j} h_K)\, \rho + h_K\, \frac{\partial \rho}{\partial z_j}. \end{align*} Restricting to $\partial\Omega$, where $\rho = 0$, gives \begin{align*} \sum_{k=1}^n \frac{\partial \rho}{\partial \bar z_k}\, \partial_{z_j} \overline{u_{kK}} = -\sum_{k=1}^n \frac{\partial^2 \rho}{\partial z_j\, \partial \bar z_k}\, \overline{u_{kK}} + h_K\, \frac{\partial \rho}{\partial z_j} \qquad \text{on } \partial\Omega. \end{align*} Multiply by $u_{jK}$ and sum over $j$. The term $h_K \sum_j u_{jK} (\partial \rho/\partial z_j)$ vanishes on $\partial\Omega$ by the boundary condition, yielding the pointwise identity on $\partial\Omega$ \begin{align*} \sum_{j,k=1}^n u_{jK}\, \frac{\partial \rho}{\partial \bar z_k}\, \partial_{z_j} \overline{u_{kK}} = -\sum_{j,k=1}^n \frac{\partial^2 \rho}{\partial z_j\, \partial \bar z_k}\, u_{jK}\, \overline{u_{kK}}. \end{align*} Integrating against $dS_\rho$ and inserting into the formula for $\Theta$ from Step 5, \begin{align*} \Theta = -\sum_{|K|=q-1}{}' \int_{\partial\Omega} \sum_{j,k=1}^n u_{jK}\, \frac{\partial \rho}{\partial \bar z_k}\, \partial_{z_j} \overline{u_{kK}} \, dS_\rho = \sum_{|K|=q-1}{}' \int_{\partial\Omega} \sum_{j,k=1}^n \frac{\partial^2 \rho}{\partial z_j\, \partial \bar z_k}\, u_{jK}\, \overline{u_{kK}} \, dS_\rho. \end{align*} [guided] The key obstacle is that $\partial_{z_j}$ is *not* a tangential operator on $\partial\Omega$ in general — applying it directly to the boundary identity $\sum_k \rho_{\bar z_k} \overline{u_{kK}} = 0$ is not valid. Hadamard's lemma resolves this: any smooth function vanishing on $\partial\Omega$ equals $\rho$ times another smooth function, so the boundary identity *extends* to a global equation $\sum_k \rho_{\bar z_k} \overline{u_{kK}} = h_K \rho$. Differentiating the global equation by $\partial_{z_j}$ is now legitimate; the term $(\partial_{z_j} h_K)\rho$ then vanishes when we restrict back to $\partial\Omega$, leaving an honest boundary relation. The remaining nuisance is the unwanted $h_K \, \partial_{z_j}\rho$ term, which encodes the *normal* component of $\partial_{z_j}$. The boundary condition $\sum_j u_{jK}\, \partial_{z_j} \rho = 0$ is exactly the geometric statement that the vector $(u_{1K}, \dots, u_{nK})$ lies in the complex tangent space (the holomorphic tangent space) of $\partial\Omega$ at each point. Pairing the identity with this tangent vector kills the normal-direction nuisance and leaves the complex Hessian of $\rho$ — that is, the Levi form — paired against the boundary trace of $u$. This is precisely the "contraction condition" mentioned in the strategy: the projection onto the holomorphic tangent space is automatic once the boundary condition is consumed. [/guided] [/step] [step:Assemble the identity] Substituting the expression for $\Theta$ from Step 6 into the formula \begin{align*} \|\bar\partial u\|_{L^2}^2 + \|\bar\partial^* u\|_{L^2}^2 = \sum_{|J|=q}{}' \sum_{j=1}^n \|\partial_{\bar z_j} u_J\|_{L^2}^2 + \Theta \end{align*} of Step 3 yields \begin{align*} \|\bar\partial u\|_{L^2}^2 + \|\bar\partial^* u\|_{L^2}^2 = \sum_{|J|=q}{}' \sum_{j=1}^n \left\|\frac{\partial u_J}{\partial \bar z_j}\right\|_{L^2}^2 + \sum_{|K|=q-1}{}' \int_{\partial\Omega} \sum_{j,k=1}^n \frac{\partial^2 \rho}{\partial z_j\, \partial \bar z_k}\, u_{jK}\, \overline{u_{kK}}\, dS_\rho, \end{align*} which is the Morrey–Kohn–Hörmander formula. [/step]