[proofplan] The proof is the compact-support form of the Morrey-Kohn-Hörmander identity. We first record the weighted adjoint of $\partial_{\bar z_j}$ and compute the exact identity relating $\|\bar\partial v\|_\varphi^2+\|\bar\partial_\varphi^*v\|_\varphi^2$ to a nonnegative first-derivative term plus the curvature term of $\varphi$. The lower bound on the complex Hessian then gives the desired estimate after integration. [/proofplan]

[step:Define the weighted inner product and adjoint differential operators] For scalar functions $f,g\in C_c^\infty(\Omega)$, define the weighted inner product \begin{align*} (f,g)_\varphi := \int_\Omega f(z)\overline{g(z)}e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z). \end{align*} For each $j\in\{1,\dots,n\}$, define the first-order operator \begin{align*} \delta_j:C_c^\infty(\Omega)&\to C_c^\infty(\Omega)\\ f&\mapsto \partial_{z_j}f-(\partial_{z_j}\varphi)f. \end{align*} Since all functions under consideration have compact support in $\Omega$, [integration by parts](/theorems/2098) has no boundary term. Hence, for $f,g\in C_c^\infty(\Omega)$, \begin{align*} (\partial_{\bar z_j}f,g)_\varphi = -(f,\delta_j g)_\varphi. \end{align*} Therefore, for \begin{align*} v=\sum_{j=1}^n v_j\,d\bar z_j, \end{align*} the weighted adjoint is \begin{align*} \bar\partial_\varphi^*v = -\sum_{j=1}^n \delta_j v_j = \sum_{j=1}^n\left((\partial_{z_j}\varphi)v_j-\partial_{z_j}v_j\right). \end{align*} [/step]

[step:Expand the Morrey-Kohn-Hörmander identity for compactly supported forms]For $j,k\in\{1,\dots,n\}$, define \begin{align*} A_{jk}:\Omega&\to\mathbb{C}\\ z&\mapsto \partial_{\bar z_j}v_k(z). \end{align*} The weighted norm of $\bar\partial v$ is \begin{align*} \|\bar\partial v\|_\varphi^2 = \sum_{1\le j<k\le n} \int_\Omega |A_{jk}(z)-A_{kj}(z)|^2 e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z). \end{align*} Expanding the square gives \begin{align*} \|\bar\partial v\|_\varphi^2 = \sum_{j,k=1}^n (A_{jk},A_{jk})_\varphi - \sum_{j,k=1}^n (A_{jk},A_{kj})_\varphi. \end{align*} Next, \begin{align*} \|\bar\partial_\varphi^*v\|_\varphi^2 = \left(\sum_{j=1}^n \delta_jv_j,\sum_{k=1}^n \delta_kv_k\right)_\varphi = \sum_{j,k=1}^n(\delta_jv_j,\delta_kv_k)_\varphi. \end{align*} Using the adjoint relation from the previous step, \begin{align*} (\delta_jv_j,\delta_kv_k)_\varphi = (v_j,-\partial_{\bar z_j}\delta_kv_k)_\varphi. \end{align*} The commutator identity \begin{align*} \partial_{\bar z_j}\delta_k f = \delta_k\partial_{\bar z_j}f-\varphi_{k\bar j}f \end{align*} holds for every $f\in C_c^\infty(\Omega)$, by differentiating $\delta_k f=\partial_{z_k}f-(\partial_{z_k}\varphi)f$. Therefore \begin{align*} (\delta_jv_j,\delta_kv_k)_\varphi = (v_j,-\delta_kA_{jk})_\varphi + (v_j,\varphi_{k\bar j}v_k)_\varphi. \end{align*} Applying the adjoint relation again to the first term gives \begin{align*} (v_j,-\delta_kA_{jk})_\varphi = (A_{kj},A_{jk})_\varphi. \end{align*} Thus \begin{align*} \|\bar\partial_\varphi^*v\|_\varphi^2 = \sum_{j,k=1}^n(A_{kj},A_{jk})_\varphi + \int_\Omega \sum_{j,k=1}^n \varphi_{k\bar j}(z)v_j(z)\overline{v_k(z)} e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z). \end{align*} Adding the two expansions cancels the mixed derivative terms and yields \begin{align*} \|\bar\partial v\|_\varphi^2+\|\bar\partial_\varphi^*v\|_\varphi^2 = \sum_{j,k=1}^n \int_\Omega |\partial_{\bar z_j}v_k(z)|^2e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z) + \int_\Omega \sum_{j,k=1}^n \varphi_{j\bar k}(z)v_j(z)\overline{v_k(z)} e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z). \end{align*}[/step]

[guided]The point of introducing $\delta_j$ is that it is exactly the weighted adjoint partner of $\partial_{\bar z_j}$. The weight $e^{-\varphi}$ changes the ordinary adjoint by the first-order correction $-(\partial_{z_j}\varphi)f$, so the correct operator is \begin{align*} \delta_j f=\partial_{z_j}f-(\partial_{z_j}\varphi)f. \end{align*} Because each $v_j$ has compact support in $\Omega$, [integration by parts](/theorems/210) produces no boundary term. Now define \begin{align*} A_{jk}(z)=\partial_{\bar z_j}v_k(z). \end{align*} The coefficient of $d\bar z_j\wedge d\bar z_k$ in $\bar\partial v$ is $A_{jk}-A_{kj}$ for $j<k$, so \begin{align*} \|\bar\partial v\|_\varphi^2 = \sum_{1\le j<k\le n} \int_\Omega |A_{jk}(z)-A_{kj}(z)|^2e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z). \end{align*} Expanding the square and writing the result symmetrically over all ordered pairs gives \begin{align*} \|\bar\partial v\|_\varphi^2 = \sum_{j,k=1}^n(A_{jk},A_{jk})_\varphi - \sum_{j,k=1}^n(A_{jk},A_{kj})_\varphi. \end{align*} For the adjoint term, use \begin{align*} \bar\partial_\varphi^*v=-\sum_{j=1}^n\delta_jv_j. \end{align*} The sign disappears after taking the norm, so \begin{align*} \|\bar\partial_\varphi^*v\|_\varphi^2 = \sum_{j,k=1}^n(\delta_jv_j,\delta_kv_k)_\varphi. \end{align*} Move $\delta_j$ from the first factor to the second factor using the weighted adjoint relation: \begin{align*} (\delta_jv_j,\delta_kv_k)_\varphi = (v_j,-\partial_{\bar z_j}\delta_kv_k)_\varphi. \end{align*} The only noncommuting part is the derivative of the coefficient $\partial_{z_k}\varphi$. Direct differentiation gives \begin{align*} \partial_{\bar z_j}\delta_kv_k = \delta_k\partial_{\bar z_j}v_k-\varphi_{k\bar j}v_k. \end{align*} Therefore \begin{align*} (\delta_jv_j,\delta_kv_k)_\varphi = (v_j,-\delta_kA_{jk})_\varphi + (v_j,\varphi_{k\bar j}v_k)_\varphi. \end{align*} Move $\delta_k$ back to the first factor: \begin{align*} (v_j,-\delta_kA_{jk})_\varphi = (A_{kj},A_{jk})_\varphi. \end{align*} Hence \begin{align*} \|\bar\partial_\varphi^*v\|_\varphi^2 = \sum_{j,k=1}^n(A_{kj},A_{jk})_\varphi + \int_\Omega \sum_{j,k=1}^n\varphi_{k\bar j}(z)v_j(z)\overline{v_k(z)} e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z). \end{align*} When this is added to the expansion of $\|\bar\partial v\|_\varphi^2$, the mixed terms cancel exactly. Since the complex Hessian matrix $(\varphi_{j\bar k})$ is Hermitian, relabelling indices gives \begin{align*} \|\bar\partial v\|_\varphi^2+\|\bar\partial_\varphi^*v\|_\varphi^2 = \sum_{j,k=1}^n \int_\Omega |\partial_{\bar z_j}v_k(z)|^2e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z) + \int_\Omega \sum_{j,k=1}^n \varphi_{j\bar k}(z)v_j(z)\overline{v_k(z)} e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z). \end{align*}[/guided]

[step:Use the curvature lower bound to obtain the estimate] For each $z\in\Omega$, define the coefficient vector \begin{align*} v(z):=(v_1(z),\dots,v_n(z))\in\mathbb{C}^n. \end{align*} The hypothesis on the complex Hessian gives the pointwise inequality \begin{align*} \sum_{j,k=1}^n\varphi_{j\bar k}(z)v_j(z)\overline{v_k(z)} \ge c\sum_{j=1}^n |v_j(z)|^2. \end{align*} Using this in the identity from the previous step and discarding the nonnegative derivative term, we obtain \begin{align*} \|\bar\partial v\|_\varphi^2+\|\bar\partial_\varphi^*v\|_\varphi^2 &\ge \int_\Omega \sum_{j,k=1}^n \varphi_{j\bar k}(z)v_j(z)\overline{v_k(z)} e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z)\\ &\ge c\int_\Omega \sum_{j=1}^n |v_j(z)|^2e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z)\\ &= c\|v\|_\varphi^2. \end{align*} This is precisely the asserted Hörmander $L^2$ estimate for the compactly supported $(0,1)$-form $v$. [/step]