[proofplan] We first record the weighted integration-by-parts formula for the complex derivatives $\partial_{z_j}$ and $\partial_{\bar z_j}$, where the weight contributes the zeroth-order operator $\delta_j^\phi=\partial_{z_j}-\phi_j$. The algebraic expansion of $\|\bar\partial u\|_\phi^2+\|\bar\partial_\phi^*u\|_\phi^2$ reduces the identity to commuting $\delta_j^\phi$ past $\partial_{\bar z_k}$ in each coefficient family $(u_{jK})_{j=1}^n$. The commutator produces the interior complex Hessian of $\phi$, while differentiating the boundary condition $\sum_j \rho_j u_{jK}=0$ along tangential complex directions produces the Levi-form boundary term. [/proofplan]

[step:Introduce the weighted adjoint operators and the boundary condition]Write \begin{align*} \phi_j := \frac{\partial \phi}{\partial z_j}, \qquad \phi_{j\bar k}:=\frac{\partial^2\phi}{\partial z_j\partial \bar z_k}, \qquad \rho_j := \frac{\partial \rho}{\partial z_j}, \qquad \rho_{\bar j}:=\frac{\partial \rho}{\partial \bar z_j}, \qquad \rho_{j\bar k}:=\frac{\partial^2\rho}{\partial z_j\partial \bar z_k}. \end{align*} For each $j \in \{1,\dots,n\}$ define the first-order differential operator \begin{align*} \delta_j^\phi:C^\infty(\overline{\Omega})&\to C^\infty(\overline{\Omega})\\ f&\mapsto \frac{\partial f}{\partial z_j}-\phi_j f. \end{align*} The weighted integration-by-parts formula is \begin{align*} \int_\Omega \frac{\partial f}{\partial \bar z_j}(z)\,\overline{g(z)}\,e^{-\phi(z)}\,d\mathcal L^{2n}(z) = -\int_\Omega f(z)\,\overline{\delta_j^\phi g(z)}\,e^{-\phi(z)}\,d\mathcal L^{2n}(z) +\int_{\partial\Omega} f(z)\,\overline{g(z)}\,\rho_{\bar j}(z)\,e^{-\phi(z)}\,dS_\rho(z) \end{align*} for $f,g\in C^\infty(\overline{\Omega})$. Indeed, applying the [divergence theorem](/theorems/2754) to the real and imaginary parts of $\partial_{\bar z_j}f\,\overline g\,e^{-\phi}$ gives the displayed boundary coefficient $\rho_{\bar j}$ with the chosen surface measure $dS_\rho$, while differentiating $e^{-\phi}$ produces the term $-\phi_jg$ inside $\delta_j^\phi g$. For a strictly increasing multi-index $K=(k_1,\dots,k_{q-1})$, define $u_{jK}$ by the insertion convention \begin{align*} u_{jK}:= \begin{cases} 0, & j\in K,\\ \operatorname{sgn}(j,K)u_J, & j\notin K, \end{cases} \end{align*} where $J$ is the increasing rearrangement of $(j,k_1,\dots,k_{q-1})$ and $\operatorname{sgn}(j,K)$ is the sign of the permutation carrying $(j,k_1,\dots,k_{q-1})$ to $J$. The boundary condition defining smooth elements of $\operatorname{Dom}(\bar\partial^*)$ is, for every strictly increasing multi-index $K$ of length $q-1$, \begin{align*} \sum_{j=1}^n \rho_j(z)u_{jK}(z)=0 \qquad z\in \partial\Omega. \end{align*} Because this condition annihilates the boundary term in the adjoint computation, the weighted adjoint is given coefficientwise by \begin{align*} (\bar\partial_\phi^*u)_K = -\sum_{j=1}^n \delta_j^\phi u_{jK}. \end{align*}[/step]

[guided]We need a precise replacement for the unweighted adjoint. The weight $e^{-\phi}$ changes the formal adjoint of $\partial_{\bar z_j}$ because differentiating the weight gives a zeroth-order term. Define \begin{align*} \delta_j^\phi:C^\infty(\overline{\Omega})&\to C^\infty(\overline{\Omega})\\ f&\mapsto \frac{\partial f}{\partial z_j}-\phi_j f, \end{align*} where \begin{align*} \phi_j=\frac{\partial \phi}{\partial z_j}. \end{align*} Then [integration by parts](/theorems/2098) gives \begin{align*} \int_\Omega \frac{\partial f}{\partial \bar z_j}\overline{g}\,e^{-\phi}\,d\mathcal L^{2n} = -\int_\Omega f\,\overline{\delta_j^\phi g}\,e^{-\phi}\,d\mathcal L^{2n} +\int_{\partial\Omega} f\,\overline{g}\,\rho_{\bar j}\,e^{-\phi}\,dS_\rho. \end{align*} Here the boundary coefficient is $\rho_{\bar j}$ because the outward complex normal component paired with $\partial_{\bar z_j}$ is represented by $\rho_{\bar j}$ for the chosen defining function and surface measure $dS_\rho$. The term $-\phi_j g$ in $\delta_j^\phi g$ is exactly the derivative of the weight. For a $(0,q)$-form \begin{align*} u=\sum_{|J|=q}'u_J\,d\bar z_J, \end{align*} the smooth $\bar\partial^*$ boundary condition says that, for every coefficient family indexed by $K$ with $|K|=q-1$, \begin{align*} \sum_{j=1}^n \rho_j u_{jK}=0 \qquad \text{on } \partial\Omega. \end{align*} This is the vanishing of the complex normal contraction of $u$. Therefore the boundary contribution in the adjoint computation cancels, and the weighted adjoint has coefficients \begin{align*} (\bar\partial_\phi^*u)_K = -\sum_{j=1}^n \delta_j^\phi u_{jK}. \end{align*}[/guided]

[step:Expand the two quadratic terms coefficient by coefficient]For each strictly increasing multi-index $K$ with $|K|=q-1$, set \begin{align*} v_j:=u_{jK}, \qquad j=1,\dots,n. \end{align*} The standard coefficient formula for $\bar\partial u$ gives \begin{align*} \|\bar\partial u\|_\phi^2 = \sum_{|J|=q}'\sum_{j=1}^n \left\|\frac{\partial u_J}{\partial \bar z_j}\right\|_\phi^2 - \sum_{|K|=q-1}'\sum_{j,k=1}^n \left( \frac{\partial u_{jK}}{\partial\bar z_k}, \frac{\partial u_{kK}}{\partial\bar z_j} \right)_\phi. \end{align*} Using the coefficient formula for $\bar\partial_\phi^*$, \begin{align*} \|\bar\partial_\phi^*u\|_\phi^2 = \sum_{|K|=q-1}'\sum_{j,k=1}^n \left( \delta_j^\phi u_{jK}, \delta_k^\phi u_{kK} \right)_\phi. \end{align*} Hence \begin{align*} \|\bar\partial u\|_\phi^2+\|\bar\partial_\phi^*u\|_\phi^2 = \sum_{|J|=q}'\sum_{j=1}^n \left\|\frac{\partial u_J}{\partial \bar z_j}\right\|_\phi^2 + \sum_{|K|=q-1}' A_K, \end{align*} where \begin{align*} A_K := \sum_{j,k=1}^n \left[ \left( \delta_j^\phi u_{jK}, \delta_k^\phi u_{kK} \right)_\phi - \left( \frac{\partial u_{jK}}{\partial\bar z_k}, \frac{\partial u_{kK}}{\partial\bar z_j} \right)_\phi \right]. \end{align*}[/step]

[guided]The identity is mostly algebra once the adjoint is known. For a fixed multi-index $K$ of length $q-1$, define the coefficient family \begin{align*} v_j:=u_{jK}, \qquad j=1,\dots,n. \end{align*} The coefficients of $\bar\partial u$ are alternating sums of the derivatives $\partial_{\bar z_j}u_J$. Squaring and summing over the alternating coefficients separates into the diagonal derivative terms and the cross terms: \begin{align*} \|\bar\partial u\|_\phi^2 = \sum_{|J|=q}'\sum_{j=1}^n \left\|\frac{\partial u_J}{\partial \bar z_j}\right\|_\phi^2 - \sum_{|K|=q-1}'\sum_{j,k=1}^n \left( \frac{\partial u_{jK}}{\partial\bar z_k}, \frac{\partial u_{kK}}{\partial\bar z_j} \right)_\phi. \end{align*} The minus sign comes from the antisymmetry of the wedge product. On the other hand, the previous step gives \begin{align*} (\bar\partial_\phi^*u)_K = -\sum_{j=1}^n \delta_j^\phi u_{jK}. \end{align*} Therefore \begin{align*} \|\bar\partial_\phi^*u\|_\phi^2 = \sum_{|K|=q-1}'\sum_{j,k=1}^n \left( \delta_j^\phi u_{jK}, \delta_k^\phi u_{kK} \right)_\phi. \end{align*} Adding the two formulas leaves the pure derivative square \begin{align*} \sum_{|J|=q}'\sum_{j=1}^n \left\|\frac{\partial u_J}{\partial \bar z_j}\right\|_\phi^2 \end{align*} plus, for each $K$, the difference \begin{align*} A_K = \sum_{j,k=1}^n \left[ \left( \delta_j^\phi u_{jK}, \delta_k^\phi u_{kK} \right)_\phi - \left( \frac{\partial u_{jK}}{\partial\bar z_k}, \frac{\partial u_{kK}}{\partial\bar z_j} \right)_\phi \right]. \end{align*} Thus the rest of the proof is the computation of $A_K$.[/guided]

[step:Compute the commutator and boundary term after summing over coefficient indices]Fix a strictly increasing multi-index $K$ with $|K|=q-1$ and define $v_j:=u_{jK}$ for $j\in\{1,\dots,n\}$. We compute only after summing over $j$ and $k$, since the boundary cancellation is a summed tangential identity. The boundary condition gives \begin{align*} \sum_{j=1}^n \rho_j(z)v_j(z)=0 \qquad z\in\partial\Omega. \end{align*} For $z\in\partial\Omega$, define the complex vector \begin{align*} V(z):=\sum_{j=1}^n v_j(z)\frac{\partial}{\partial z_j}. \end{align*} We call $V(z)$ complex tangential to $\partial\Omega$ when its derivative of the defining function vanishes, and this holds because $V(z)\rho=\sum_{j=1}^n \rho_j(z)v_j(z)=0$ on $\partial\Omega$. Apply the weighted integration-by-parts formula in the form \begin{align*} (\delta_j^\phi a,b)_\phi = -\left(a,\frac{\partial b}{\partial\bar z_j}\right)_\phi +\int_{\partial\Omega}a(z)\overline{b(z)}\rho_j(z)e^{-\phi(z)}\,dS_\rho(z) \end{align*} to $a=v_j$ and $b=\delta_k^\phi v_k$. After summing over $j,k$, the boundary term vanishes because $\sum_j \rho_jv_j=0$ on $\partial\Omega$. Thus \begin{align*} \sum_{j,k=1}^n(\delta_j^\phi v_j,\delta_k^\phi v_k)_\phi &=-\sum_{j,k=1}^n\left(v_j,\frac{\partial}{\partial\bar z_j}(\delta_k^\phi v_k)\right)_\phi. \end{align*} Using \begin{align*} \frac{\partial}{\partial\bar z_j}(\delta_k^\phi v_k) = \delta_k^\phi\left(\frac{\partial v_k}{\partial\bar z_j}\right) -[\delta_k^\phi,\partial_{\bar z_j}]v_k, \end{align*} and applying weighted [integration by parts](/theorems/210) once more to the first term gives \begin{align*} \sum_{j,k=1}^n(\delta_j^\phi v_j,\delta_k^\phi v_k)_\phi &=\sum_{j,k=1}^n\left(\frac{\partial v_j}{\partial\bar z_k},\frac{\partial v_k}{\partial\bar z_j}\right)_\phi \\ &\quad+\sum_{j,k=1}^n\left([\delta_k^\phi,\partial_{\bar z_j}]v_k,v_j\right)_\phi \\ &\quad-\int_{\partial\Omega}\sum_{j,k=1}^n v_j(z)\overline{\frac{\partial v_k}{\partial\bar z_j}(z)}\rho_{\bar k}(z)e^{-\phi(z)}\,dS_\rho(z). \end{align*} The commutator is computed directly: for $f\in C^\infty(\overline\Omega)$, \begin{align*} [\delta_k^\phi,\partial_{\bar z_j}]f &=\left(\frac{\partial}{\partial z_k}-\phi_k\right)\frac{\partial f}{\partial\bar z_j} - \frac{\partial}{\partial\bar z_j}\left(\frac{\partial f}{\partial z_k}-\phi_k f\right) \\ &=\phi_{k\bar j}f. \end{align*} Since $\phi$ is real-valued, $\phi_{k\bar j}=\overline{\phi_{j\bar k}}$, and the corresponding inner-product sum equals \begin{align*} \sum_{j,k=1}^n\left([\delta_k^\phi,\partial_{\bar z_j}]v_k,v_j\right)_\phi = \int_\Omega\sum_{j,k=1}^n\phi_{j\bar k}(z)v_j(z)\overline{v_k(z)}e^{-\phi(z)}\,d\mathcal L^{2n}(z). \end{align*} It remains to rewrite the boundary term. Because $u\in C^\infty_{(0,q)}(\overline\Omega)$, the functions $v_j|_{\partial\Omega}$ and $\rho_j|_{\partial\Omega}$ are smooth. Taking the complex conjugate of the boundary condition gives \begin{align*} \sum_{k=1}^n \rho_{\bar k}(z)\overline{v_k(z)}=0 \qquad z\in\partial\Omega. \end{align*} Differentiate this conjugate boundary condition along the smooth complex tangential field $V=\sum_{j=1}^n v_j\partial/\partial z_j$ on $\partial\Omega$. Since $V\rho=0$ on $\partial\Omega$, this tangential differentiation is intrinsic to the boundary, and it gives \begin{align*} 0 &=V\left(\sum_{k=1}^n\rho_{\bar k}\overline{v_k}\right) \\ &=\sum_{j,k=1}^n\rho_{j\bar k}v_j\overline{v_k} +\sum_{j,k=1}^n\rho_{\bar k}v_j\overline{\frac{\partial v_k}{\partial\bar z_j}}. \end{align*} Therefore \begin{align*} -\sum_{j,k=1}^n v_j\overline{\frac{\partial v_k}{\partial\bar z_j}}\rho_{\bar k} = \sum_{j,k=1}^n\rho_{j\bar k}v_j\overline{v_k} \qquad \text{on }\partial\Omega. \end{align*} Therefore \begin{align*} A_K = \int_\Omega \sum_{j,k=1}^n \phi_{j\bar k}(z)v_j(z)\overline{v_k(z)}e^{-\phi(z)}\,d\mathcal L^{2n}(z) + \int_{\partial\Omega} \sum_{j,k=1}^n \rho_{j\bar k}(z)v_j(z)\overline{v_k(z)}e^{-\phi(z)}\,dS_\rho(z). \end{align*}[/step]

[guided]We now compute $A_K$, but the computation must be done after the double sum over $j$ and $k$. A fixed pair $(j,k)$ does not separately satisfy the boundary cancellation; the boundary condition is the single summed identity \begin{align*} \sum_{j=1}^n \rho_j v_j=0 \qquad \text{on }\partial\Omega. \end{align*} This identity says that the vector \begin{align*} V(z):=\sum_{j=1}^n v_j(z)\frac{\partial}{\partial z_j} \end{align*} is complex tangential to $\partial\Omega$ in the following concrete sense: its derivative of the defining function vanishes, because $V(z)\rho=\sum_j\rho_j(z)v_j(z)=0$. First move $\delta_j^\phi$ off $v_j$ in the summed expression. The weighted integration-by-parts formula gives \begin{align*} (\delta_j^\phi a,b)_\phi = -\left(a,\frac{\partial b}{\partial\bar z_j}\right)_\phi +\int_{\partial\Omega}a\overline b\rho_j e^{-\phi}\,dS_\rho. \end{align*} With $a=v_j$ and $b=\delta_k^\phi v_k$, summing over $j,k$ gives \begin{align*} \sum_{j,k=1}^n(\delta_j^\phi v_j,\delta_k^\phi v_k)_\phi &=-\sum_{j,k=1}^n\left(v_j,\frac{\partial}{\partial\bar z_j}(\delta_k^\phi v_k)\right)_\phi \\ &\quad+\int_{\partial\Omega}\left(\sum_{j=1}^n\rho_jv_j\right)\overline{\left(\sum_{k=1}^n\delta_k^\phi v_k\right)}e^{-\phi}\,dS_\rho. \end{align*} The boundary integral is zero by the boundary condition. Next separate the derivative of $\delta_k^\phi v_k$ into a commutator term: \begin{align*} \frac{\partial}{\partial\bar z_j}(\delta_k^\phi v_k) = \delta_k^\phi\left(\frac{\partial v_k}{\partial\bar z_j}\right) -[\delta_k^\phi,\partial_{\bar z_j}]v_k. \end{align*} Apply weighted [integration by parts](/theorems/2098) again to the term containing $\delta_k^\phi(\partial v_k/\partial\bar z_j)$. This produces \begin{align*} \sum_{j,k=1}^n(\delta_j^\phi v_j,\delta_k^\phi v_k)_\phi &=\sum_{j,k=1}^n\left(\frac{\partial v_j}{\partial\bar z_k},\frac{\partial v_k}{\partial\bar z_j}\right)_\phi \\ &\quad+\sum_{j,k=1}^n\left([\delta_k^\phi,\partial_{\bar z_j}]v_k,v_j\right)_\phi \\ &\quad-\int_{\partial\Omega}\sum_{j,k=1}^n v_j\overline{\frac{\partial v_k}{\partial\bar z_j}}\rho_{\bar k} e^{-\phi}\,dS_\rho. \end{align*} The boundary coefficient is $\rho_{\bar k}$ because this second [integration by parts](/theorems/210) uses $(a,\delta_k^\phi b)_\phi=-(\partial_{\bar z_k}a,b)_\phi+\int_{\partial\Omega}a\overline b\rho_{\bar k}e^{-\phi}\,dS_\rho$. The first term cancels the mixed term in the definition of $A_K$. The commutator is local and direct. For $f\in C^\infty(\overline\Omega)$, \begin{align*} [\delta_k^\phi,\partial_{\bar z_j}]f &=\left(\frac{\partial}{\partial z_k}-\phi_k\right)\frac{\partial f}{\partial\bar z_j} - \frac{\partial}{\partial\bar z_j}\left(\frac{\partial f}{\partial z_k}-\phi_k f\right) \\ &=\frac{\partial^2f}{\partial z_k\partial\bar z_j} -\phi_k\frac{\partial f}{\partial\bar z_j} -\frac{\partial^2f}{\partial\bar z_j\partial z_k} +\phi_{k\bar j}f +\phi_k\frac{\partial f}{\partial\bar z_j} \\ &=\phi_{k\bar j}f. \end{align*} Since $\phi$ is real-valued, $\phi_{k\bar j}=\overline{\phi_{j\bar k}}$, so the summed commutator contributes \begin{align*} \int_\Omega\sum_{j,k=1}^n\phi_{j\bar k}v_j\overline{v_k}e^{-\phi}\,d\mathcal L^{2n}. \end{align*} Finally identify the boundary contribution. The smoothness $u\in C^\infty_{(0,q)}(\overline\Omega)$ and $\rho\in C^2$ allow differentiating boundary identities along smooth complex tangential fields. The remaining boundary integrand contains $\rho_{\bar k}\overline{\partial_{\bar z_j}v_k}$, so we use the conjugate boundary condition \begin{align*} \sum_{k=1}^n\rho_{\bar k}\overline{v_k}=0 \qquad \text{on }\partial\Omega. \end{align*} Differentiate it along $V=\sum_jv_j\partial/\partial z_j$. Since $V\rho=0$ on $\partial\Omega$, this is a tangential differentiation along the boundary. On $\partial\Omega$, \begin{align*} 0 &=V\left(\sum_{k=1}^n\rho_{\bar k}\overline{v_k}\right) \\ &=\sum_{j,k=1}^n\rho_{j\bar k}v_j\overline{v_k} +\sum_{j,k=1}^n\rho_{\bar k}v_j\overline{\frac{\partial v_k}{\partial\bar z_j}}. \end{align*} Thus \begin{align*} -\sum_{j,k=1}^n v_j\overline{\frac{\partial v_k}{\partial\bar z_j}}\rho_{\bar k} = \sum_{j,k=1}^n\rho_{j\bar k}v_j\overline{v_k}. \end{align*} This is exactly the Levi-form boundary term. Therefore \begin{align*} A_K = \int_\Omega \sum_{j,k=1}^n \phi_{j\bar k}v_j\overline{v_k}e^{-\phi}\,d\mathcal L^{2n} + \int_{\partial\Omega} \sum_{j,k=1}^n \rho_{j\bar k}v_j\overline{v_k}e^{-\phi}\,dS_\rho. \end{align*}[/guided]

[step:Sum over all coefficient families and obtain the identity]Substituting the expression for $A_K$ into the expansion from the previous step and recalling that $v_j=u_{jK}$ gives \begin{align*} \|\bar{\partial}u\|_{\phi}^2+\|\bar{\partial}_\phi^*u\|_{\phi}^2 &=\sum_{|J|=q}'\sum_{j=1}^n\left\|\frac{\partial u_J}{\partial\bar z_j}\right\|_{\phi}^2 \\ &\quad+\sum_{|K|=q-1}'\int_\Omega\sum_{j,k=1}^n \phi_{j\bar k}(z)u_{jK}(z)\overline{u_{kK}(z)}e^{-\phi(z)}\,d\mathcal L^{2n}(z) \\ &\quad+\sum_{|K|=q-1}'\int_{\partial\Omega}\sum_{j,k=1}^n \rho_{j\bar k}(z)u_{jK}(z)\overline{u_{kK}(z)}e^{-\phi(z)}\,dS_\rho(z). \end{align*} Since $\phi_{j\bar k}=\partial^2\phi/\partial z_j\partial\bar z_k$ and $\rho_{j\bar k}=\partial^2\rho/\partial z_j\partial\bar z_k$, this is precisely the claimed weighted Kohn-Morrey-Hörmander identity.[/step]

[guided]For each strictly increasing multi-index $K$ with $|K|=q-1$, the previous step proved that the leftover term $A_K$ equals the sum of the interior Hessian contribution and the Levi-form boundary contribution. Substituting that formula into the coefficient expansion gives \begin{align*} \|\bar{\partial}u\|_{\phi}^2+\|\bar{\partial}_\phi^*u\|_{\phi}^2 &=\sum_{|J|=q}'\sum_{j=1}^n\left\|\frac{\partial u_J}{\partial\bar z_j}\right\|_{\phi}^2 \\ &\quad+\sum_{|K|=q-1}'\int_\Omega\sum_{j,k=1}^n \phi_{j\bar k}(z)u_{jK}(z)\overline{u_{kK}(z)}e^{-\phi(z)}\,d\mathcal L^{2n}(z) \\ &\quad+\sum_{|K|=q-1}'\int_{\partial\Omega}\sum_{j,k=1}^n \rho_{j\bar k}(z)u_{jK}(z)\overline{u_{kK}(z)}e^{-\phi(z)}\,dS_\rho(z). \end{align*} The definitions introduced at the start identify $\phi_{j\bar k}$ with $\partial^2\phi/\partial z_j\partial\bar z_k$ and $\rho_{j\bar k}$ with $\partial^2\rho/\partial z_j\partial\bar z_k$. Therefore the displayed equality is exactly the weighted Kohn-Morrey-Hörmander identity stated in the theorem.[/guided]