[proofplan] Write $v = \sum_j v_j \, d\bar z_j$ and expand both $\|\bar\partial v\|^2$ and $\|\bar\partial^*_\varphi v\|^2$ into sums of weighted $L^2$ inner products of the partial derivatives of the components, isolating the cross-terms $(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k)_\varphi$ and $(\delta_j v_j, \delta_k v_k)_\varphi$, where $\delta_j = \partial_{z_j} - \varphi_j$ is the weighted formal adjoint of $-\partial_{\bar z_j}$. A single integration-by-parts identity, obtained from the Gauss–Green theorem applied in $\mathbb{R}^{2n}$, lets us move $\delta_j$ across $\partial_{\bar z_k}$; the obstruction to commuting them is exactly the commutator $[\partial_{\bar z_k}, \delta_j] = -\varphi_{j\bar k}$, which produces the Hessian term. Two integrations by parts convert the adjoint cross-terms into the $\bar\partial$ cross-terms plus the Hessian term plus boundary integrals; the boundary condition $\sum_j v_j \rho_j = 0$ kills one boundary integral, and differentiating that same condition tangentially turns the other into the Levi form $\sum_{j,k} \rho_{j\bar k} v_j \overline{v_k}$. Finally, pseudoconvexity says the Levi form is positive semidefinite on the complex tangent space, where the boundary values of $v$ live, so $B_{\partial\Omega}(v) \geq 0$. [/proofplan]

[step:Set up the weighted formal adjoints and the boundary characterisation of $\mathrm{Dom}(\bar\partial^*_\varphi)$]For $1 \le j \le n$ write $\partial_{z_j} = \partial/\partial z_j$ and $\partial_{\bar z_j} = \partial/\partial \bar z_j$, and define the first-order operator \begin{align*} \delta_j : C^\infty(\overline{\Omega}) &\to C^\infty(\overline{\Omega}) \\ f &\mapsto \frac{\partial f}{\partial z_j} - \varphi_j \, f . \end{align*} Since $\varphi$ is real-valued, $\varphi_{\bar k} := \partial\varphi/\partial\bar z_k = \overline{\varphi_k}$ and $\rho_{\bar k} := \partial\rho/\partial\bar z_k = \overline{\rho_k}$. The operator $\bar\partial$ acts on a function $f$ by $\bar\partial f = \sum_k (\partial_{\bar z_k} f) \, d\bar z_k$, and on the $(0,1)$-form $v = \sum_j v_j \, d\bar z_j$ by \begin{align*} \bar\partial v = \sum_{k < j} \left(\partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k\right) d\bar z_k \wedge d\bar z_j . \end{align*} We first record that for $v \in C^\infty_{(0,1)}(\overline{\Omega})$, \begin{align*} v \in \mathrm{Dom}(\bar\partial^*_\varphi) \iff \sum_{j=1}^n v_j \, \rho_j = 0 \ \text{ on } \partial\Omega, \quad\text{and then}\quad \bar\partial^*_\varphi v = -\sum_{j=1}^n \delta_j v_j . \end{align*} This is established in Step 3 once the integration-by-parts identity is in hand; we use it from Step 5 onward.[/step]

[guided]The whole proof is an exercise in integrating by parts with the weight $e^{-\varphi}$, so the first task is to identify the formal adjoints of the basic differential operators in the weighted inner product $(f,g)_\varphi = \int_\Omega f \, \overline{g} \, e^{-\varphi} \, d\mathcal{L}^{2n}$. The natural object is $\delta_j = \partial_{z_j} - \varphi_j$. Why this combination? Because $e^{\varphi} \partial_{z_j}(e^{-\varphi} f) = \partial_{z_j} f - \varphi_j f = \delta_j f$, i.e. $\delta_j$ is $\partial_{z_j}$ conjugated by the weight. This is precisely the operator that will turn out to be the weighted formal adjoint of $-\partial_{\bar z_j}$; we verify this in Step 3. Because $\varphi$ and $\rho$ are real-valued, complex conjugation interchanges holomorphic and antiholomorphic first derivatives: $\overline{\varphi_k} = \varphi_{\bar k}$ and $\overline{\rho_k} = \rho_{\bar k}$. We will use these identities repeatedly when conjugating integration-by-parts formulas. Finally, the two facts about $\mathrm{Dom}(\bar\partial^*_\varphi)$ recorded above are the only place where the "domain of $\bar\partial^*_\varphi$" hypothesis enters. The boundary condition $\sum_j v_j \rho_j = 0$ on $\partial\Omega$ says geometrically that the coefficient vector $(v_1, \dots, v_n)$ is annihilated by $\partial\rho$, i.e. it lies in the complex tangent space $T_p^{1,0}(\partial\Omega) = \ker(\partial\rho|_p)$ at each boundary point $p$. This is exactly the subspace on which pseudoconvexity controls the Levi form, which is why this condition is the bridge between the algebra and the final positivity statement.[/guided]

[step:Decompose the two norms into component inner products]For the antiholomorphic norm, the pointwise norm of the $(0,2)$-form $\bar\partial v = \sum_{k<j}(\partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k)\, d\bar z_k \wedge d\bar z_j$ is $\sum_{k<j}|\partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k|^2$. Doubling the strict-inequality sum to a full sum over $(j,k)$ and expanding the square, \begin{align*} \|\bar\partial v\|_{e^{-\varphi}}^2 &= \frac{1}{2} \sum_{j,k=1}^n \big\| \partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k \big\|_{e^{-\varphi}}^2 \\ &= \sum_{j,k=1}^n \big\| \partial_{\bar z_k} v_j \big\|_{e^{-\varphi}}^2 - \sum_{j,k=1}^n \big( \partial_{\bar z_k} v_j, \, \partial_{\bar z_j} v_k \big)_\varphi . \tag{$\star$} \end{align*} For the adjoint, using $\bar\partial^*_\varphi v = -\sum_j \delta_j v_j$ (Step 3), \begin{align*} \|\bar\partial^*_\varphi v\|_{e^{-\varphi}}^2 = \Big( \sum_{j} \delta_j v_j, \ \sum_{k} \delta_k v_k \Big)_\varphi = \sum_{j,k=1}^n \big( \delta_j v_j, \, \delta_k v_k \big)_\varphi . \tag{$\star\star$} \end{align*} Adding $(\star)$ and $(\star\star)$, \begin{align*} \|\bar\partial v\|_{e^{-\varphi}}^2 + \|\bar\partial^*_\varphi v\|_{e^{-\varphi}}^2 = \sum_{j,k=1}^n \big\| \partial_{\bar z_k} v_j \big\|_{e^{-\varphi}}^2 + \sum_{j,k=1}^n \Big[ \big( \delta_j v_j, \delta_k v_k \big)_\varphi - \big( \partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k \big)_\varphi \Big]. \tag{$\dagger$} \end{align*} The first sum on the right of $(\dagger)$ is already the square term in the statement, since $\|\partial_{\bar z_k} v_j\|_{e^{-\varphi}}^2 = \int_\Omega |\partial v_j/\partial\bar z_k|^2 e^{-\varphi}\, d\mathcal{L}^{2n}$. It remains to evaluate the bracketed cross-term sum.[/step]

[guided]The strategy of the entire proof is now visible in $(\dagger)$: the right-hand side splits into a manifestly nonnegative "good" term $\sum_{j,k}\|\partial_{\bar z_k} v_j\|^2$ and a bracketed remainder of cross-terms. The theorem will follow once we show the remainder equals the Hessian term plus the boundary term. Let us justify the expansion $(\star)$ in detail. The $(0,2)$-form $\bar\partial v$ has, in the orthonormal frame $\{d\bar z_k \wedge d\bar z_j\}_{k<j}$, the coefficients $\partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k$. Its squared pointwise norm is therefore $\sum_{k<j}|\partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k|^2$. Each ordered pair $(k,j)$ with $k<j$ corresponds to the same antisymmetric quantity as $(j,k)$, so summing over all ordered pairs double-counts: \begin{align*} \sum_{k<j}|\partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k|^2 = \tfrac12 \sum_{j,k} |\partial_{\bar z_k} v_j - \partial_{\bar z_j} v_k|^2 . \end{align*} Expanding the modulus squared gives $|\partial_{\bar z_k} v_j|^2 + |\partial_{\bar z_j} v_k|^2 - 2\,\mathrm{Re}\big(\partial_{\bar z_k} v_j \, \overline{\partial_{\bar z_j} v_k}\big)$. Upon summing over all $(j,k)$ and integrating against $e^{-\varphi}\,d\mathcal{L}^{2n}$, the first two terms each give $\sum_{j,k}\|\partial_{\bar z_k} v_j\|^2_{e^{-\varphi}}$ (relabel $j\leftrightarrow k$ in the second), while the cross sum $\sum_{j,k}(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k)_\varphi$ is invariant under $j \leftrightarrow k$ and hence real, so the $\mathrm{Re}$ may be dropped. This yields $(\star)$. The identity $(\star\star)$ is immediate from $\bar\partial^*_\varphi v = -\sum_j \delta_j v_j$ and the sesquilinearity of $(\cdot,\cdot)_\varphi$. Note the minus sign squares away.[/guided]

[step:Record the weighted integration-by-parts identity and identify $\bar\partial^*_\varphi$]Identify $\mathbb{C}^n \cong \mathbb{R}^{2n}$ via $z_k = x_k + i y_k$, so $\partial_{\bar z_k} = \tfrac12(\partial_{x_k} + i \partial_{y_k})$. Applying the [Gauss–Green Theorem](/theorems/28) in $\mathbb{R}^{2n}$ to the coordinates $x_k$ and $y_k$ and combining, we get for any $u \in C^1(\overline{\Omega})$ \begin{align*} \int_\Omega \partial_{\bar z_k} u \, d\mathcal{L}^{2n} = \int_{\partial\Omega} u \, \rho_{\bar k} \, d\mathcal{H}^{2n-1}, \end{align*} where the outward unit normal is $\nu = \nabla\rho$ (a unit vector because $|\nabla\rho| = 1$ on $\partial\Omega$) and $\rho_{\bar k} = \tfrac12(\rho_{x_k} + i \rho_{y_k}) = \tfrac12(\nu^{x_k} + i \nu^{y_k})$. Apply this to $u = f \, \overline{g} \, e^{-\varphi}$ with $f, g \in C^1(\overline{\Omega})$. Since \begin{align*} \partial_{\bar z_k}\!\big(\overline{g}\, e^{-\varphi}\big) = \big(\partial_{\bar z_k} \overline{g} - \varphi_{\bar k}\, \overline{g}\big) e^{-\varphi} = \overline{\big(\partial_{z_k} g - \varphi_k g\big)}\, e^{-\varphi} = \overline{\delta_k g}\, e^{-\varphi}, \end{align*} the product rule gives the master identity \begin{align*} \big(\partial_{\bar z_k} f, \, g\big)_\varphi + \big(f, \, \delta_k g\big)_\varphi = \int_{\partial\Omega} f \, \overline{g} \, \rho_{\bar k} \, e^{-\varphi} \, d\mathcal{H}^{2n-1}. \tag{IBP} \end{align*} Taking complex conjugates of (IBP) with the roles of $f,g$ exchanged and using $\overline{\rho_{\bar j}} = \rho_j$ yields the companion identity \begin{align*} \big(\delta_j a, \, b\big)_\varphi = -\big(a, \, \partial_{\bar z_j} b\big)_\varphi + \int_{\partial\Omega} a \, \overline{b} \, \rho_j \, e^{-\varphi} \, d\mathcal{H}^{2n-1}. \tag{IBP$'$} \end{align*} **Identification of $\bar\partial^*_\varphi$.** For $f \in C^\infty(\overline{\Omega})$ and $v = \sum_k v_k \, d\bar z_k \in C^\infty_{(0,1)}(\overline{\Omega})$, summing (IBP) over $k$ with $g = v_k$ gives \begin{align*} \big(\bar\partial f, v\big)_\varphi = \sum_k \big(\partial_{\bar z_k} f, v_k\big)_\varphi = \Big(f, -\sum_k \delta_k v_k\Big)_\varphi + \int_{\partial\Omega} f \, \overline{\Big(\sum_k v_k \rho_k\Big)} \, e^{-\varphi} \, d\mathcal{H}^{2n-1}, \end{align*} using $\overline{v_k}\,\rho_{\bar k} = \overline{v_k \rho_k}$. The form $v$ lies in $\mathrm{Dom}(\bar\partial^*_\varphi)$ precisely when the map $f \mapsto (\bar\partial f, v)_\varphi$ is represented by an $L^2_\varphi$ function with no boundary contribution for all such $f$; since $f|_{\partial\Omega}$ is arbitrary, this holds iff $\sum_k v_k \rho_k = 0$ on $\partial\Omega$, and then $\bar\partial^*_\varphi v = -\sum_k \delta_k v_k$. This proves the claim deferred in Step 1.[/step]

[guided]We need one computational engine — a weighted integration-by-parts formula — and we extract everything from it. Start from the complex form of the [divergence theorem](/theorems/2754). The [Gauss–Green Theorem](/theorems/28) is stated for $C^1$ domains in $\mathbb{R}^N$; here $N = 2n$ and $\Omega$ has smooth boundary, so its hypotheses hold. Writing $z_k = x_k + i y_k$ and $\partial_{\bar z_k} = \tfrac12(\partial_{x_k} + i\partial_{y_k})$, we apply Gauss–Green to the real and imaginary directions and add. The boundary normal is $\nu = \nabla\rho$, which is a genuine unit normal because we normalised $|\nabla\rho| = 1$ on $\partial\Omega$; this normalisation is exactly what removes a factor of $1/|\nabla\rho|$ that would otherwise clutter every boundary integral. The result is $\int_\Omega \partial_{\bar z_k} u\, d\mathcal{L}^{2n} = \int_{\partial\Omega} u\, \rho_{\bar k}\, d\mathcal{H}^{2n-1}$. Now feed in $u = f\,\overline{g}\,e^{-\varphi}$ and differentiate by the product rule. The key computation is $\partial_{\bar z_k}(\overline{g}\, e^{-\varphi}) = \overline{\delta_k g}\, e^{-\varphi}$. Let us verify it: $\partial_{\bar z_k}\overline{g} = \overline{\partial_{z_k} g}$ (conjugation swaps $\partial_{\bar z_k}$ and $\partial_{z_k}$), and $\partial_{\bar z_k} e^{-\varphi} = -\varphi_{\bar k} e^{-\varphi} = -\overline{\varphi_k}\, e^{-\varphi}$. Hence $\partial_{\bar z_k}(\overline{g}\, e^{-\varphi}) = (\overline{\partial_{z_k} g} - \overline{\varphi_k}\,\overline{g})e^{-\varphi} = \overline{(\partial_{z_k} g - \varphi_k g)}\,e^{-\varphi} = \overline{\delta_k g}\, e^{-\varphi}$. This is the precise sense in which **$\delta_k$ is the weighted formal adjoint of $-\partial_{\bar z_k}$**: the weight $e^{-\varphi}$, when differentiated, manufactures exactly the zeroth-order term $-\varphi_k$ that distinguishes $\delta_k$ from $\partial_{z_k}$. Substituting into the divergence formula gives (IBP). Identity (IBP$'$) is just (IBP) read backwards: conjugate (IBP) (with $f,g$ swapped) and use that $\rho$ is real so $\overline{\rho_{\bar j}} = \rho_j$. Having both forms lets us move a derivative off of either slot of the inner product. Why does this immediately pin down $\bar\partial^*_\varphi$? By definition the adjoint satisfies $(\bar\partial f, v)_\varphi = (f, \bar\partial^*_\varphi v)_\varphi$ for all $f$ in the domain of $\bar\partial$, with no leftover boundary integral. Summing (IBP) over $k$ produces the candidate interior adjoint $-\sum_k \delta_k v_k$ plus a boundary integral $\int_{\partial\Omega} f \, \overline{\sum_k v_k \rho_k}\, e^{-\varphi}\,d\mathcal{H}^{2n-1}$. Because $f$ may be any smooth function and its boundary restriction is unconstrained, the boundary integral vanishes for all $f$ if and only if its kernel vanishes, i.e. $\sum_k v_k \rho_k = 0$ on $\partial\Omega$. What would fail without this condition? The map $f \mapsto (\bar\partial f, v)_\varphi$ would carry a boundary distribution that is not of the form $(f, h)_\varphi$ for any $L^2_\varphi$ function $h$, so $v$ would not be in the domain of the adjoint at all.[/guided]

[step:Compute the commutator $[\partial_{\bar z_k}, \delta_j]= -\varphi_{j\bar k}$] For $f \in C^2(\overline{\Omega})$, \begin{align*} \partial_{\bar z_k}\big(\delta_j f\big) &= \partial_{\bar z_k}\big(\partial_{z_j} f - \varphi_j f\big) = \partial_{z_j}\partial_{\bar z_k} f - \big(\partial_{\bar z_k}\varphi_j\big) f - \varphi_j \, \partial_{\bar z_k} f, \\ \delta_j\big(\partial_{\bar z_k} f\big) &= \partial_{z_j}\partial_{\bar z_k} f - \varphi_j \, \partial_{\bar z_k} f . \end{align*} Subtracting and using $\partial_{\bar z_k}\varphi_j = \dfrac{\partial^2\varphi}{\partial z_j \, \partial \bar z_k} = \varphi_{j\bar k}$, \begin{align*} \big[\partial_{\bar z_k}, \delta_j\big] f = \partial_{\bar z_k}\delta_j f - \delta_j \partial_{\bar z_k} f = -\varphi_{j\bar k}\, f . \end{align*} Equivalently, $\partial_{\bar z_k}\delta_j = \delta_j \partial_{\bar z_k} - \varphi_{j\bar k}$ as operators on $C^2(\overline{\Omega})$.[/step]

[guided]This short computation is the algebraic heart of the identity: it is where the complex Hessian $\varphi_{j\bar k}$ is born. We want to know by how much $\partial_{\bar z_k}$ and $\delta_j$ fail to commute. The mixed second derivatives $\partial_{z_j}\partial_{\bar z_k} f$ and $\partial_{\bar z_k}\partial_{z_j} f$ agree (Clairaut/Schwarz, valid since $f \in C^2$), so those cancel. The term $-\varphi_j\,\partial_{\bar z_k} f$ also appears in both expansions and cancels. The only survivor is the term where $\partial_{\bar z_k}$ lands on the coefficient $\varphi_j$ rather than on $f$: that is $-(\partial_{\bar z_k}\varphi_j) f = -\varphi_{j\bar k} f$. So $[\partial_{\bar z_k}, \delta_j] = -\varphi_{j\bar k}$, multiplication by minus the $(j,k)$ entry of the complex Hessian of $\varphi$. This is precisely why the weight contributes its Hessian to the final identity: when, in the next step, we slide $\delta_j$ past $\partial_{\bar z_k}$ to match the $\bar\partial$ cross-term against the $\bar\partial^*_\varphi$ cross-term, the price of commuting is one factor of $\varphi_{j\bar k}$ per term. A flat weight $\varphi \equiv \mathrm{const}$ would make this commutator vanish, and the Hessian term in the theorem would disappear — which is the unweighted Kohn–Morrey case.[/guided]

[step:Convert each adjoint cross-term by integrating by parts twice]Fix $j, k$. Apply (IBP) with $f = \delta_j v_j$ and $g = v_k$: \begin{align*} \big(\delta_j v_j, \delta_k v_k\big)_\varphi = -\big(\partial_{\bar z_k}\delta_j v_j, \, v_k\big)_\varphi + \int_{\partial\Omega} \big(\delta_j v_j\big)\overline{v_k}\, \rho_{\bar k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1}. \end{align*} Use the commutator identity $\partial_{\bar z_k}\delta_j = \delta_j \partial_{\bar z_k} - \varphi_{j\bar k}$ from Step 4: \begin{align*} -\big(\partial_{\bar z_k}\delta_j v_j, v_k\big)_\varphi = -\big(\delta_j \partial_{\bar z_k} v_j, \, v_k\big)_\varphi + \big(\varphi_{j\bar k} v_j, \, v_k\big)_\varphi . \end{align*} Now apply (IBP$'$) with $a = \partial_{\bar z_k} v_j$ and $b = v_k$: \begin{align*} \big(\delta_j \partial_{\bar z_k} v_j, \, v_k\big)_\varphi = -\big(\partial_{\bar z_k} v_j, \, \partial_{\bar z_j} v_k\big)_\varphi + \int_{\partial\Omega} \big(\partial_{\bar z_k} v_j\big)\overline{v_k}\, \rho_j\, e^{-\varphi}\, d\mathcal{H}^{2n-1}. \end{align*} Combining the three displays and cancelling $(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k)_\varphi$ from both sides yields, for each $(j,k)$, \begin{align*} \big(\delta_j v_j, \delta_k v_k\big)_\varphi - \big(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k\big)_\varphi = \big(\varphi_{j\bar k} v_j, v_k\big)_\varphi + \int_{\partial\Omega} \Big[ \big(\delta_j v_j\big)\rho_{\bar k} - \big(\partial_{\bar z_k} v_j\big)\rho_j \Big] \overline{v_k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1}. \tag{$\ddagger$} \end{align*}[/step]

[guided]Our goal is to evaluate the bracketed remainder in $(\dagger)$, namely $\sum_{j,k}\big[(\delta_j v_j, \delta_k v_k)_\varphi - (\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k)_\varphi\big]$. The two pieces look unrelated — one built from $\delta = \partial_z - \varphi_z$, the other from $\partial_{\bar z}$ — so the plan is to integrate by parts in the $\delta$-term until it matches the $\partial_{\bar z}$-term, recording every commutator and boundary contribution along the way. **First [integration by parts](/theorems/2098).** In $(\delta_j v_j, \delta_k v_k)_\varphi$ we move the $\delta_k$ off the second slot. Recall from (IBP) that the weighted adjoint of $\delta_k$ is $-\partial_{\bar z_k}$ up to a boundary term; taking $f = \delta_j v_j$, $g = v_k$ produces $-(\partial_{\bar z_k}\delta_j v_j, v_k)_\varphi$ plus the boundary integral $\int_{\partial\Omega}(\delta_j v_j)\overline{v_k}\rho_{\bar k}e^{-\varphi}$. **Commute.** We now have $\partial_{\bar z_k}\delta_j$ acting on $v_j$, but we want $\delta_j \partial_{\bar z_k}$ so that the next [integration by parts](/theorems/210) will create $\partial_{\bar z_j} v_k$ and match the $\bar\partial$ cross-term. Step 4 tells us the cost of swapping is exactly $-\varphi_{j\bar k}$: $\partial_{\bar z_k}\delta_j v_j = \delta_j \partial_{\bar z_k} v_j - \varphi_{j\bar k} v_j$. This is where the Hessian term $(\varphi_{j\bar k} v_j, v_k)_\varphi$ enters. **Second [integration by parts](/theorems/2098).** Apply (IBP$'$) to $(\delta_j \partial_{\bar z_k} v_j, v_k)_\varphi$ with $a = \partial_{\bar z_k} v_j$, $b = v_k$. This moves $\delta_j$ off the first slot, turning it into $-\partial_{\bar z_j}$ acting on $v_k$ and producing $-(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k)_\varphi$ — exactly the $\bar\partial$ cross-term we want to cancel — plus a second boundary integral $\int_{\partial\Omega}(\partial_{\bar z_k} v_j)\overline{v_k}\rho_j e^{-\varphi}$. Assembling the three displays, the interior term $(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k)_\varphi$ produced by the second IBP cancels the one subtracted on the left, leaving the clean per-index identity $(\ddagger)$: the Hessian inner product plus two boundary integrals. The two boundary integrals are what Steps 6 and 7 now dispose of.[/guided]

[step:Sum over indices and kill the first boundary integral via the domain condition]Sum $(\ddagger)$ over $j,k = 1, \dots, n$. The first boundary integral factors: \begin{align*} \sum_{j,k} \int_{\partial\Omega} \big(\delta_j v_j\big)\rho_{\bar k}\,\overline{v_k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1} = \int_{\partial\Omega} \Big(\sum_j \delta_j v_j\Big) \, \overline{\Big(\sum_k v_k \rho_k\Big)} \, e^{-\varphi}\, d\mathcal{H}^{2n-1}, \end{align*} where we used $\rho_{\bar k}\,\overline{v_k} = \overline{v_k \rho_k}$. By the domain condition $\sum_k v_k \rho_k = 0$ on $\partial\Omega$ (Step 3), this integral vanishes. Hence \begin{align*} \sum_{j,k}\Big[\big(\delta_j v_j, \delta_k v_k\big)_\varphi - \big(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k\big)_\varphi\Big] = \sum_{j,k}\big(\varphi_{j\bar k} v_j, v_k\big)_\varphi \;-\; \underbrace{\sum_{j,k}\int_{\partial\Omega} \big(\partial_{\bar z_k} v_j\big)\rho_j\,\overline{v_k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1}}_{=: -\,B} . \tag{$\sharp$} \end{align*} We have written the surviving boundary contribution as $B$, to be identified in Step 7.[/step]

[guided]With the per-index identity $(\ddagger)$ in hand, we sum over $j$ and $k$ and watch the boundary integrals. The first boundary integral is the easy one. Its integrand contains $(\delta_j v_j)\rho_{\bar k}\overline{v_k}$; the index $j$ appears only through $\delta_j v_j$ and the index $k$ only through $\rho_{\bar k}\overline{v_k}$, so the double sum factors as a product of two single sums. The $k$-sum is $\sum_k \rho_{\bar k}\overline{v_k}$, which equals $\overline{\sum_k \rho_k v_k}$ because $\rho$ is real ($\rho_{\bar k} = \overline{\rho_k}$). But $\sum_k \rho_k v_k = 0$ on $\partial\Omega$ — this is exactly the domain condition characterising $\mathrm{Dom}(\bar\partial^*_\varphi)$ from Step 3. So the entire first boundary integral collapses to zero. This is the second and last time the domain hypothesis is used: once to identify $\bar\partial^*_\varphi$, now to discard the "normal" boundary integral. What remains is the interior Hessian term and a single boundary integral $B = \sum_{j,k}\int_{\partial\Omega}(\partial_{\bar z_k} v_j)\rho_j \overline{v_k} e^{-\varphi}$. We have written $(\sharp)$ with a $-B$ on the right; equivalently the surviving contribution to the right-hand side is $+B$ after moving it across, but the bookkeeping is cleaner if we first identify $B$ as the negative of the Levi term, which is the content of the next step.[/guided]

[step:Identify the Levi boundary term by differentiating the domain condition tangentially]Consider the function $h := \sum_j v_j \rho_j \in C^\infty(\overline{\Omega})$, which vanishes on $\partial\Omega$ by the domain condition. Since $\partial\Omega$ is smooth and $h$ vanishes there, $h = \lambda \rho$ near $\partial\Omega$ for some $\lambda \in C^\infty$ (with $\rho$ the smooth defining function). Differentiating, \begin{align*} \partial_{\bar z_k} h = \lambda \, \rho_{\bar k} + \big(\partial_{\bar z_k}\lambda\big)\rho, \qquad\text{so on } \partial\Omega \ (\rho = 0): \quad \partial_{\bar z_k} h = \lambda \, \rho_{\bar k}. \end{align*} On the other hand, by the product rule and $\partial_{\bar z_k}\rho_j = \rho_{j\bar k}$, \begin{align*} \partial_{\bar z_k} h = \sum_j \big(\partial_{\bar z_k} v_j\big)\rho_j + \sum_j v_j \, \rho_{j\bar k}. \end{align*} Equating the two expressions on $\partial\Omega$ gives, for each $k$, \begin{align*} \sum_j \big(\partial_{\bar z_k} v_j\big)\rho_j = \lambda\, \rho_{\bar k} - \sum_j v_j \, \rho_{j\bar k} \qquad \text{on } \partial\Omega . \end{align*} Substitute into $B = \sum_{j,k}\int_{\partial\Omega}(\partial_{\bar z_k} v_j)\rho_j \,\overline{v_k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1}$: \begin{align*} B = \int_{\partial\Omega} \lambda \, \overline{\Big(\sum_k v_k \rho_k\Big)} \, e^{-\varphi}\, d\mathcal{H}^{2n-1} - \int_{\partial\Omega} \sum_{j,k} \rho_{j\bar k}\, v_j \, \overline{v_k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1}, \end{align*} where in the first integral $\sum_k \rho_{\bar k}\overline{v_k} = \overline{\sum_k \rho_k v_k} = 0$ on $\partial\Omega$. Hence the first integral vanishes and \begin{align*} B = -\int_{\partial\Omega} \sum_{j,k} \rho_{j\bar k}\, v_j\, \overline{v_k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1} = -\,B_{\partial\Omega}(v) . \end{align*}[/step]

[guided]We must turn the leftover boundary integral $B$ into the Levi form $\sum_{j,k}\rho_{j\bar k} v_j \overline{v_k}$. The obstacle is that $B$ contains a derivative $\partial_{\bar z_k} v_j$ of the unknown components, not the components themselves. The resolution is to use the only information we have about derivatives on the boundary: the domain condition $\sum_j v_j \rho_j = 0$ holds *along all of* $\partial\Omega$, so it may be differentiated in directions tangent to $\partial\Omega$. Concretely, set $h = \sum_j v_j \rho_j$. This smooth function vanishes identically on the zero set of the smooth defining function $\rho$, so it is divisible by $\rho$: $h = \lambda\rho$ for a smooth $\lambda$ near $\partial\Omega$. Differentiating with $\partial_{\bar z_k}$ and restricting to $\partial\Omega$ (where $\rho = 0$) leaves only $\partial_{\bar z_k} h = \lambda \rho_{\bar k}$. The geometric content: the gradient of a function constant on $\partial\Omega$ is normal to $\partial\Omega$, i.e. proportional to $\nabla\rho$; the factor $\lambda$ is the proportionality constant and the term $\lambda\rho_{\bar k}$ is the "normal" part, which we will show is harmless. Computing $\partial_{\bar z_k} h$ directly by the product rule splits it as $\sum_j (\partial_{\bar z_k} v_j)\rho_j$ (the quantity appearing in $B$) plus $\sum_j v_j \rho_{j\bar k}$ (the Levi form, since $\partial_{\bar z_k}\rho_j = \rho_{j\bar k}$). Equating the two expressions for $\partial_{\bar z_k} h$ on $\partial\Omega$ solves for the unknown derivative combination: \begin{align*} \sum_j (\partial_{\bar z_k} v_j)\rho_j = \lambda \rho_{\bar k} - \sum_j v_j \rho_{j\bar k}. \end{align*} Now substitute into $B$ and sum against $\overline{v_k}$. The $\lambda\rho_{\bar k}$ piece carries the factor $\sum_k \rho_{\bar k}\overline{v_k} = \overline{\sum_k \rho_k v_k} = 0$ — the domain condition strikes again — so the normal part contributes nothing. What is left is exactly minus the Levi boundary term $B_{\partial\Omega}(v)$. The single minus sign is what makes $B = -B_{\partial\Omega}(v)$, so that $-B$ in $(\sharp)$ becomes $+B_{\partial\Omega}(v)$ in the final identity.[/guided]

[step:Assemble the identity and conclude nonnegativity from pseudoconvexity]Insert $B = -B_{\partial\Omega}(v)$ into $(\sharp)$: \begin{align*} \sum_{j,k}\Big[\big(\delta_j v_j, \delta_k v_k\big)_\varphi - \big(\partial_{\bar z_k} v_j, \partial_{\bar z_j} v_k\big)_\varphi\Big] = \sum_{j,k}\big(\varphi_{j\bar k} v_j, v_k\big)_\varphi + B_{\partial\Omega}(v) . \end{align*} Substituting this bracketed sum into $(\dagger)$ from Step 2 gives \begin{align*} \|\bar\partial v\|_{e^{-\varphi}}^2 + \|\bar\partial^*_\varphi v\|_{e^{-\varphi}}^2 = \sum_{j,k} \int_\Omega \varphi_{j\bar k}\, v_j\, \overline{v_k}\, e^{-\varphi}\, d\mathcal{L}^{2n} + \sum_{j,k} \int_\Omega \left|\frac{\partial v_j}{\partial \bar z_k}\right|^2 e^{-\varphi}\, d\mathcal{L}^{2n} + B_{\partial\Omega}(v), \end{align*} since $(\varphi_{j\bar k} v_j, v_k)_\varphi = \int_\Omega \varphi_{j\bar k} v_j \overline{v_k} e^{-\varphi}\, d\mathcal{L}^{2n}$ and $\|\partial_{\bar z_k} v_j\|_{e^{-\varphi}}^2 = \int_\Omega |\partial v_j/\partial\bar z_k|^2 e^{-\varphi}\, d\mathcal{L}^{2n}$. This is the asserted identity. It remains to show $B_{\partial\Omega}(v) \geq 0$. Fix $p \in \partial\Omega$. The domain condition gives $\sum_j \rho_j(p)\, v_j(p) = 0$, i.e. the vector $\xi := (v_1(p), \dots, v_n(p))$ satisfies $\xi \in \ker(\partial\rho|_p) = T_p^{1,0}(\partial\Omega)$, the complex tangent space at $p$. By [The Levi Condition Is a Boundary Test](/theorems/3392), pseudoconvexity of $\Omega$ is equivalent to the Levi form \begin{align*} \mathcal{L}_\rho(p)(\xi, \xi) = \sum_{j,k} \rho_{j\bar k}(p)\, \xi_j \, \overline{\xi_k} \end{align*} being positive semidefinite on $T_p^{1,0}(\partial\Omega)$ at every $p \in \partial\Omega$. Therefore $\sum_{j,k} \rho_{j\bar k}(p)\, v_j(p)\,\overline{v_k(p)} \geq 0$ for every $p \in \partial\Omega$. Integrating the nonnegative integrand against the positive measure $e^{-\varphi}\, d\mathcal{H}^{2n-1}$ gives \begin{align*} B_{\partial\Omega}(v) = \int_{\partial\Omega} \sum_{j,k} \rho_{j\bar k}\, v_j\, \overline{v_k}\, e^{-\varphi}\, d\mathcal{H}^{2n-1} \geq 0, \end{align*} which completes the proof.[/step]

[guided]We now collect the pieces. Step 6 reduced the bracketed remainder in $(\dagger)$ to the Hessian inner products plus the surviving boundary integral $B$, and Step 7 identified $B = -B_{\partial\Omega}(v)$. Substituting back, the bracketed remainder equals $\sum_{j,k}(\varphi_{j\bar k} v_j, v_k)_\varphi + B_{\partial\Omega}(v)$, and feeding this into $(\dagger)$ — whose "good" term $\sum_{j,k}\|\partial_{\bar z_k} v_j\|^2$ is the displayed square term — produces the Morrey–Kohn–Hörmander identity verbatim. The final task is the inequality $B_{\partial\Omega}(v) \geq 0$, and this is the one place pseudoconvexity is used. The mechanism is the matching of two subspaces. On the analytic side, the domain condition forces the boundary value $\xi = (v_1(p), \dots, v_n(p))$ to be annihilated by $\partial\rho$ at each boundary point $p$; that is exactly the defining equation of the complex tangent space $T_p^{1,0}(\partial\Omega) = \ker(\partial\rho|_p)$. On the geometric side, [The Levi Condition Is a Boundary Test](/theorems/3392) states that pseudoconvexity of a smoothly bounded domain is equivalent to positive semidefiniteness of the Levi form $\sum_{j,k}\rho_{j\bar k}\xi_j\overline{\xi_k}$ *restricted to that same complex tangent space*. The two subspaces coincide, so the boundary values of $v$ live exactly where pseudoconvexity guarantees the Levi form is $\geq 0$. What would go wrong without pseudoconvexity? The identity itself would still hold — it is a pure integration-by-parts computation valid for any smoothly bounded domain — but $B_{\partial\Omega}(v)$ could be negative, and the right-hand side would no longer be a sum of manifestly controllable terms. The positivity of $B_{\partial\Omega}(v)$ is precisely what upgrades the identity into the basic a priori estimate underpinning the solution of the $\bar\partial$-Neumann problem on pseudoconvex domains.[/guided]