[proofplan]
We formulate the $\bar\partial$-equation as a Hilbert complex between weighted $L^2$ spaces of $(0,q)$-forms. The key analytic input is the Hörmander-Kodaira basic estimate, which uses pseudoconvexity of $\Omega$ and the lower curvature bound $i\partial\bar\partial\varphi \geq c\omega$ to control closed $(0,1)$-forms by their weighted adjoint derivative. This estimate makes the functional $T_\varphi^*v \mapsto (v,f)_{e^{-\varphi}}$ bounded on the range of the adjoint. Hahn-Banach and the [Riesz representation theorem](/theorems/221) then produce a function $u$, and the defining identity for $u$ is exactly the weak formulation of $\bar\partial u=f$.
[/proofplan]
[step:Declare the weighted Hilbert complex for $\bar\partial$]
Define the weighted scalar [Hilbert space](/page/Hilbert%20Space)
\begin{align*}
H_0
:=
L^2(\Omega,e^{-\varphi})
=
\left\{
u:\Omega\to\mathbb{C}\ \text{measurable}:
\int_\Omega |u|^2 e^{-\varphi}\,d\mathcal{L}^{2n}<\infty
\right\},
\end{align*}
with inner product
\begin{align*}
(u,w)_{H_0}
:=
\int_\Omega u\overline{w}\,e^{-\varphi}\,d\mathcal{L}^{2n}.
\end{align*}
For $q\in\{1,2\}$, define $H_q:=L^2_{(0,q)}(\Omega,e^{-\varphi})$, the [Hilbert space](/page/Hilbert%20Space) of measurable $(0,q)$-forms whose coefficient vector is square-integrable with respect to $e^{-\varphi}d\mathcal{L}^{2n}$. If
\begin{align*}
\alpha
=
\sum_{|J|=q}\alpha_J\,d\bar z_J,
\qquad
\beta
=
\sum_{|J|=q}\beta_J\,d\bar z_J,
\end{align*}
define
\begin{align*}
(\alpha,\beta)_{H_q}
:=
\int_\Omega
\sum_{|J|=q}\alpha_J\overline{\beta_J}\,
e^{-\varphi}\,d\mathcal{L}^{2n}.
\end{align*}
Let
\begin{align*}
T:\operatorname{Dom}(T)\subset H_0&\to H_1,\\
u&\mapsto \bar\partial u
\end{align*}
be the maximal distributional $\bar\partial$ operator, where
\begin{align*}
\operatorname{Dom}(T)
:=
\{u\in H_0:\bar\partial u\in H_1\ \text{in the distributional sense}\}.
\end{align*}
Let
\begin{align*}
S:\operatorname{Dom}(S)\subset H_1&\to H_2,\\
v&\mapsto \bar\partial v
\end{align*}
be the corresponding maximal distributional $\bar\partial$ operator on $(0,1)$-forms. Since $\bar\partial^2=0$ in the distributional sense, $S\circ T=0$ on $\operatorname{Dom}(T)$. Denote by
\begin{align*}
T_\varphi^*:\operatorname{Dom}(T_\varphi^*)\subset H_1\to H_0
\end{align*}
the Hilbert-space adjoint of $T$ with respect to the weighted inner products.
The hypothesis $\bar\partial f=0$ says precisely that $f\in\ker S$.
[guided]
We first translate the theorem into operator language. The unknown $u$ is a scalar function, so it belongs to the weighted [Hilbert space](/page/Hilbert%20Space)
\begin{align*}
H_0
=
L^2(\Omega,e^{-\varphi})
=
\left\{
u:\Omega\to\mathbb{C}\ \text{measurable}:
\int_\Omega |u|^2e^{-\varphi}\,d\mathcal{L}^{2n}<\infty
\right\}.
\end{align*}
The data $f$ is a $(0,1)$-form, so it belongs to
\begin{align*}
H_1=L^2_{(0,1)}(\Omega,e^{-\varphi}).
\end{align*}
For a $(0,q)$-form
\begin{align*}
\alpha=\sum_{|J|=q}\alpha_J\,d\bar z_J,
\end{align*}
the weighted norm is computed coefficientwise:
\begin{align*}
\|\alpha\|_{H_q}^2
=
\int_\Omega
\sum_{|J|=q}|\alpha_J|^2e^{-\varphi}\,d\mathcal{L}^{2n}.
\end{align*}
The measure is explicitly $d\mathcal{L}^{2n}$ because $\Omega\subset\mathbb{C}^n$ is identified with an open subset of $\mathbb{R}^{2n}$.
Now define the maximal distributional operator
\begin{align*}
T:\operatorname{Dom}(T)\subset H_0&\to H_1,\\
u&\mapsto\bar\partial u,
\end{align*}
where
\begin{align*}
\operatorname{Dom}(T)
=
\{u\in H_0:\bar\partial u\in H_1\ \text{in the distributional sense}\}.
\end{align*}
Similarly, define
\begin{align*}
S:\operatorname{Dom}(S)\subset H_1&\to H_2,\\
v&\mapsto\bar\partial v.
\end{align*}
The identity $\bar\partial^2=0$ gives $S\circ T=0$. Thus the range of $T$ lies in $\ker S$, which is why the condition $\bar\partial f=0$ is the natural compatibility condition for solving $Tu=f$.
Finally, $T_\varphi^*$ denotes the Hilbert-space adjoint of $T$ with respect to the weighted inner products. Concretely, $v\in\operatorname{Dom}(T_\varphi^*)$ if there exists $g\in H_0$ such that
\begin{align*}
(Tu,v)_{H_1}=(u,g)_{H_0}
\end{align*}
for every $u\in\operatorname{Dom}(T)$, and then $T_\varphi^*v:=g$.
[/guided]
[/step]
[step:Use the Hörmander-Kodaira estimate on closed $(0,1)$-forms]
We use the following standard estimate: by the Hörmander-Kodaira $L^2$ estimate on pseudoconvex domains with weight curvature bounded below by $c\omega$ (citing a result not yet in the wiki: Hörmander-Kodaira Basic Estimate), every form
\begin{align*}
v\in \operatorname{Dom}(T_\varphi^*)\cap\operatorname{Dom}(S)\cap\ker S
\end{align*}
satisfies
\begin{align*}
c\|v\|_{H_1}^2
\leq
\|T_\varphi^*v\|_{H_0}^2.
\end{align*}
In particular, if such a form $v$ satisfies $T_\varphi^*v=0$, then $v=0$ in $H_1$.
[guided]
The analytic heart of the proof is the weighted $\bar\partial$ estimate. The lower curvature bound
\begin{align*}
i\partial\bar\partial\varphi\geq c\omega
\end{align*}
means that the complex Hessian of $\varphi$ dominates $c$ times the Euclidean Hermitian metric. Together with pseudoconvexity of $\Omega$, the Hörmander-Kodaira basic estimate gives the coercive inequality
\begin{align*}
c\|v\|_{H_1}^2
\leq
\|T_\varphi^*v\|_{H_0}^2
\end{align*}
for every
\begin{align*}
v\in \operatorname{Dom}(T_\varphi^*)\cap\operatorname{Dom}(S)\cap\ker S.
\end{align*}
Here the condition $v\in\ker S$ means $\bar\partial v=0$ in the distributional sense. The full Bochner-Kodaira estimate normally contains an additional term $\|Sv\|_{H_2}^2$ on the right-hand side; for closed forms this term is zero.
This estimate has an important algebraic consequence. If $v$ is closed and satisfies $T_\varphi^*v=0$, then
\begin{align*}
c\|v\|_{H_1}^2
\leq
0.
\end{align*}
Since $c>0$, this implies $\|v\|_{H_1}=0$, hence $v=0$ as an element of $H_1$. This injectivity is what makes the functional in the next step well-defined.
[/guided]
[/step]
[step:Define a bounded functional on the adjoint range]
Let
\begin{align*}
\mathcal{D}
:=
\operatorname{Dom}(T_\varphi^*)\cap\operatorname{Dom}(S)\cap\ker S
\end{align*}
and define
\begin{align*}
R
:=
T_\varphi^*(\mathcal{D})
\subset H_0.
\end{align*}
Define a map
\begin{align*}
\Lambda:R&\to\mathbb{C},\\
T_\varphi^*v&\mapsto (v,f)_{H_1}.
\end{align*}
This map is well-defined: if $T_\varphi^*v_1=T_\varphi^*v_2$ for $v_1,v_2\in\mathcal{D}$, then $v_1-v_2\in\mathcal{D}$ and $T_\varphi^*(v_1-v_2)=0$, hence $v_1-v_2=0$ by the estimate above.
For $v\in\mathcal{D}$, the [Cauchy-Schwarz inequality](/theorems/432) in $H_1$ gives
\begin{align*}
|\Lambda(T_\varphi^*v)|
=
|(v,f)_{H_1}|
\leq
\|v\|_{H_1}\|f\|_{H_1}.
\end{align*}
Using the Hörmander-Kodaira estimate,
\begin{align*}
\|v\|_{H_1}
\leq
\frac{1}{\sqrt c}\|T_\varphi^*v\|_{H_0}.
\end{align*}
Therefore
\begin{align*}
|\Lambda(T_\varphi^*v)|
\leq
\frac{1}{\sqrt c}\|f\|_{H_1}\|T_\varphi^*v\|_{H_0}.
\end{align*}
Thus $\Lambda$ is a bounded linear functional on the subspace $R\subset H_0$ with operator norm satisfying
\begin{align*}
\|\Lambda\|_{R^*}
\leq
\frac{1}{\sqrt c}\|f\|_{H_1}.
\end{align*}
[guided]
The estimate tells us how to control $v$ by $T_\varphi^*v$. This suggests defining a functional whose input is $T_\varphi^*v$ and whose value is the pairing of $v$ with the given closed form $f$.
Set
\begin{align*}
\mathcal{D}
=
\operatorname{Dom}(T_\varphi^*)\cap\operatorname{Dom}(S)\cap\ker S.
\end{align*}
This is the class of closed $(0,1)$-forms on which the coercive estimate applies. Define
\begin{align*}
R=T_\varphi^*(\mathcal{D})\subset H_0.
\end{align*}
For an element of $R$, write it as $T_\varphi^*v$ with $v\in\mathcal{D}$ and set
\begin{align*}
\Lambda(T_\varphi^*v)=(v,f)_{H_1}.
\end{align*}
We must first check that this does not depend on the chosen representative $v$. Suppose
\begin{align*}
T_\varphi^*v_1=T_\varphi^*v_2
\end{align*}
for $v_1,v_2\in\mathcal{D}$. Then $v_1-v_2\in\mathcal{D}$ and
\begin{align*}
T_\varphi^*(v_1-v_2)=0.
\end{align*}
The Hörmander-Kodaira estimate gives
\begin{align*}
c\|v_1-v_2\|_{H_1}^2
\leq
\|T_\varphi^*(v_1-v_2)\|_{H_0}^2
=
0.
\end{align*}
Since $c>0$, $v_1=v_2$ in $H_1$. Hence
\begin{align*}
(v_1,f)_{H_1}=(v_2,f)_{H_1},
\end{align*}
so $\Lambda$ is well-defined.
Next we prove boundedness. The [Cauchy-Schwarz inequality](/theorems/432) in the [Hilbert space](/page/Hilbert%20Space) $H_1$ gives
\begin{align*}
|\Lambda(T_\varphi^*v)|
=
|(v,f)_{H_1}|
\leq
\|v\|_{H_1}\|f\|_{H_1}.
\end{align*}
The estimate gives
\begin{align*}
\|v\|_{H_1}
\leq
\frac{1}{\sqrt c}\|T_\varphi^*v\|_{H_0}.
\end{align*}
Substituting this into the Cauchy-Schwarz bound yields
\begin{align*}
|\Lambda(T_\varphi^*v)|
\leq
\frac{1}{\sqrt c}\|f\|_{H_1}\|T_\varphi^*v\|_{H_0}.
\end{align*}
Thus the operator norm of $\Lambda$ on $R$ is at most $\frac{1}{\sqrt c}\|f\|_{H_1}$.
[/guided]
[/step]
[step:Extend the functional and represent it by a weighted $L^2$ function]
By the [Hahn-Banach theorem](/page/Hahn-Banach%20Theorem) for normed complex vector spaces (citing a result not yet in the wiki: [Hahn-Banach Theorem](/theorems/879)), $\Lambda$ extends to a bounded linear functional
\begin{align*}
\widetilde{\Lambda}:H_0\to\mathbb{C}
\end{align*}
such that
\begin{align*}
\|\widetilde{\Lambda}\|_{H_0^*}
=
\|\Lambda\|_{R^*}
\leq
\frac{1}{\sqrt c}\|f\|_{H_1}.
\end{align*}
By the [Riesz representation theorem](/theorems/218) for Hilbert spaces (citing a result not yet in the wiki: [Riesz Representation](/theorems/67) Theorem for Hilbert Spaces), there exists a unique $u\in H_0$ such that
\begin{align*}
\widetilde{\Lambda}(g)=(g,u)_{H_0}
\end{align*}
for every $g\in H_0$, and
\begin{align*}
\|u\|_{H_0}
=
\|\widetilde{\Lambda}\|_{H_0^*}
\leq
\frac{1}{\sqrt c}\|f\|_{H_1}.
\end{align*}
Squaring this inequality gives
\begin{align*}
\|u\|_{H_0}^2
\leq
\frac{1}{c}\|f\|_{H_1}^2.
\end{align*}
[/step]
[step:Extend the adjoint identity from closed test forms to all adjoint test forms]
For every $v\in\mathcal{D}$, since $\widetilde{\Lambda}$ extends $\Lambda$,
\begin{align*}
(T_\varphi^*v,u)_{H_0}
=
\widetilde{\Lambda}(T_\varphi^*v)
=
\Lambda(T_\varphi^*v)
=
(v,f)_{H_1}.
\end{align*}
Taking complex conjugates gives
\begin{align*}
(u,T_\varphi^*v)_{H_0}
=
(f,v)_{H_1}.
\end{align*}
Let $K:=\ker S\subset H_1$, where the kernel is taken with respect to the closed maximal operator $S$. Since $S$ is closed, $K$ is a closed linear subspace of $H_1$. Let
\begin{align*}
P_K:H_1&\to K
\end{align*}
denote the [orthogonal projection](/theorems/437) onto $K$. Because $S\circ T=0$, one has $\operatorname{Range}(T)\subset K$, hence
\begin{align*}
K^\perp\subset \operatorname{Range}(T)^\perp=\ker T_\varphi^*.
\end{align*}
Now fix $v\in\operatorname{Dom}(T_\varphi^*)$. Define
\begin{align*}
v_0:=P_Kv,
\qquad
v_1:=v-P_Kv.
\end{align*}
Then $v_0\in K=\operatorname{Dom}(S)\cap\ker S$, while $v_1\in K^\perp\subset\ker T_\varphi^*$. In particular $v_1\in\operatorname{Dom}(T_\varphi^*)$ and $T_\varphi^*v_1=0$. Therefore $v_0=v-v_1\in\operatorname{Dom}(T_\varphi^*)$, so $v_0\in\mathcal{D}$, and
\begin{align*}
T_\varphi^*v_0=T_\varphi^*v.
\end{align*}
Since $f\in K$ and $v_1\in K^\perp$,
\begin{align*}
(f,v)_{H_1}
=
(f,v_0)_{H_1}+(f,v_1)_{H_1}
=
(f,v_0)_{H_1}.
\end{align*}
Applying the identity already proved for $v_0\in\mathcal{D}$ gives
\begin{align*}
(u,T_\varphi^*v)_{H_0}
=
(u,T_\varphi^*v_0)_{H_0}
=
(f,v_0)_{H_1}
=
(f,v)_{H_1}.
\end{align*}
Thus the adjoint identity holds for every $v\in\operatorname{Dom}(T_\varphi^*)$.
By the definition of the Hilbert-space adjoint, this says exactly that $u\in\operatorname{Dom}(T)$ and
\begin{align*}
Tu=f.
\end{align*}
Equivalently,
\begin{align*}
\bar\partial u=f
\end{align*}
in the distributional sense.
[guided]
The Riesz representative $u$ was chosen so that
\begin{align*}
\widetilde{\Lambda}(g)=(g,u)_{H_0}
\end{align*}
for all $g\in H_0$. In particular, if $g=T_\varphi^*v$ with $v\in\mathcal{D}$, then
\begin{align*}
(T_\varphi^*v,u)_{H_0}
=
\widetilde{\Lambda}(T_\varphi^*v)
=
\Lambda(T_\varphi^*v)
=
(v,f)_{H_1}.
\end{align*}
Conjugate symmetry of the Hilbert inner products gives
\begin{align*}
(u,T_\varphi^*v)_{H_0}
=
(f,v)_{H_1}
\end{align*}
for every closed adjoint test form $v\in\mathcal{D}$.
We now explain why this restricted identity is enough. Let $K:=\ker S\subset H_1$, where $S$ is the closed maximal distributional $\bar\partial$ operator from $H_1$ to $H_2$. Since $S$ is closed, its kernel $K$ is a closed linear subspace of the [Hilbert space](/page/Hilbert%20Space) $H_1$, so the [orthogonal projection](/theorems/437)
\begin{align*}
P_K:H_1&\to K
\end{align*}
exists. The relation $S\circ T=0$ means $\operatorname{Range}(T)\subset K$. Taking orthogonal complements in $H_1$ gives
\begin{align*}
K^\perp\subset \operatorname{Range}(T)^\perp.
\end{align*}
By the definition of the Hilbert-space adjoint, $w\in\operatorname{Range}(T)^\perp$ is exactly the same as $w\in\operatorname{Dom}(T_\varphi^*)$ and $T_\varphi^*w=0$; indeed, for every $a\in\operatorname{Dom}(T)$,
\begin{align*}
(Ta,w)_{H_1}=0=(a,0)_{H_0}.
\end{align*}
Thus
\begin{align*}
K^\perp\subset\ker T_\varphi^*.
\end{align*}
Now take an arbitrary $v\in\operatorname{Dom}(T_\varphi^*)$. Decompose it orthogonally by defining
\begin{align*}
v_0:=P_Kv,
\qquad
v_1:=v-P_Kv.
\end{align*}
Then $v_0\in K=\operatorname{Dom}(S)\cap\ker S$ and $v_1\in K^\perp$. From $v_1\in K^\perp\subset\ker T_\varphi^*$ we get $v_1\in\operatorname{Dom}(T_\varphi^*)$ and $T_\varphi^*v_1=0$. Since $v\in\operatorname{Dom}(T_\varphi^*)$ as well, subtraction in the linear domain gives $v_0=v-v_1\in\operatorname{Dom}(T_\varphi^*)$. Therefore
\begin{align*}
v_0\in\operatorname{Dom}(T_\varphi^*)\cap\operatorname{Dom}(S)\cap\ker S=\mathcal{D},
\end{align*}
and
\begin{align*}
T_\varphi^*v_0=T_\varphi^*v-T_\varphi^*v_1=T_\varphi^*v.
\end{align*}
The closedness of the datum $f$ now matters. The hypothesis $Sf=0$ says $f\in K$. Since $v_1\in K^\perp$, orthogonality gives
\begin{align*}
(f,v_1)_{H_1}=0.
\end{align*}
Hence
\begin{align*}
(f,v)_{H_1}
=
(f,v_0)_{H_1}+(f,v_1)_{H_1}
=
(f,v_0)_{H_1}.
\end{align*}
Applying the identity already proved for $v_0\in\mathcal{D}$, and using $T_\varphi^*v_0=T_\varphi^*v$, yields
\begin{align*}
(u,T_\varphi^*v)_{H_0}
=
(u,T_\varphi^*v_0)_{H_0}
=
(f,v_0)_{H_1}
=
(f,v)_{H_1}.
\end{align*}
This proves the adjoint identity for every $v\in\operatorname{Dom}(T_\varphi^*)$, not only for closed test forms.
Finally recall the definition of the adjoint. A function $u\in H_0$ belongs to $\operatorname{Dom}(T)$ and satisfies $Tu=f$ precisely when
\begin{align*}
(u,T_\varphi^*v)_{H_0}
=
(f,v)_{H_1}
\end{align*}
for all $v\in\operatorname{Dom}(T_\varphi^*)$. Hence $Tu=f$, which is the same as
\begin{align*}
\bar\partial u=f
\end{align*}
in the distributional sense.
[/guided]
[/step]
[step:Translate the Hilbert norm estimate into the stated integral estimate]
The norm estimate obtained from the extension and representation steps is
\begin{align*}
\|u\|_{H_0}^2
\leq
\frac{1}{c}\|f\|_{H_1}^2.
\end{align*}
By the definitions of the weighted norms,
\begin{align*}
\|u\|_{H_0}^2
=
\int_\Omega |u|^2e^{-\varphi}\,d\mathcal{L}^{2n}
\end{align*}
and, writing $f=\sum_{j=1}^n f_j\,d\bar z_j$,
\begin{align*}
\|f\|_{H_1}^2
=
\int_\Omega
\sum_{j=1}^n |f_j|^2e^{-\varphi}\,d\mathcal{L}^{2n}
=
\int_\Omega |f|^2e^{-\varphi}\,d\mathcal{L}^{2n}.
\end{align*}
Therefore
\begin{align*}
\int_\Omega |u|^2e^{-\varphi}\,d\mathcal{L}^{2n}
\leq
\frac{1}{c}
\int_\Omega |f|^2e^{-\varphi}\,d\mathcal{L}^{2n}.
\end{align*}
This is the asserted estimate, and the proof is complete.
[/step]
