[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*}
[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]
[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]
