[proofplan]
We view $\bar\partial$ as a closed densely defined operator between weighted Hilbert spaces of forms and use the [Hilbert space](/page/Hilbert%20Space) duality criterion for solving $Tu=f$. The strict plurisubharmonicity hypothesis gives the higher-degree Hörmander-Kodaira estimate, which controls closed $(0,q)$-forms by their weighted $\bar\partial$-adjoint with constant $qc$. This estimate bounds the functional induced by $f$ on the range of the adjoint; Hahn-Banach and Riesz then produce the desired form $u$, and the adjoint identity identifies $\bar\partial u$ with $f$.
[/proofplan]
[step:Set up the weighted $\bar\partial$ Hilbert complex]
Let $\mathcal{L}^{2n}$ denote Lebesgue measure on $\mathbb{C}^n$ after the standard identification $\mathbb{C}^n\cong\mathbb{R}^{2n}$. For each integer $r$ with $0 \le r \le n$, define the weighted [Hilbert space](/page/Hilbert%20Space)
\begin{align*}
H_r := L^2_{(0,r)}(\Omega,e^{-\varphi})
\end{align*}
with inner product
\begin{align*}
(\alpha,\beta)_{H_r}
:=
\int_{\Omega}
\langle \alpha(z),\beta(z)\rangle_{\Lambda^{0,r}}\,
e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z),
\end{align*}
where $\langle \cdot,\cdot\rangle_{\Lambda^{0,r}}$ is the pointwise Hermitian inner product induced by the Euclidean metric on $\mathbb{C}^n$. Also set
\begin{align*}
H_{n+1}:=\{0\}
\end{align*}
with its unique [Hilbert space](/page/Hilbert%20Space) structure. This convention is used only for the top-degree case $q=n$.
Let
\begin{align*}
T:\operatorname{Dom}(T)\subset H_{q-1}\to H_q
\end{align*}
be the maximal distributional $\bar\partial$ operator on $(0,q-1)$-forms, and let
\begin{align*}
S:\operatorname{Dom}(S)\subset H_q\to H_{q+1}
\end{align*}
be the maximal distributional $\bar\partial$ operator on $(0,q)$-forms. Thus $T u=\bar\partial u$ and $S v=\bar\partial v$ whenever these distributional derivatives lie in the corresponding weighted $L^2$ spaces. Since $\bar\partial^2=0$ in the distributional sense, $S T=0$.
The hypothesis $\bar\partial f=0$ says exactly that $f\in \operatorname{Dom}(S)$ and $S f=0$.
[guided]
We first translate the theorem into [Hilbert space](/page/Hilbert%20Space) language. Let $\mathcal{L}^{2n}$ denote Lebesgue measure on $\mathbb{C}^n$ after identifying $\mathbb{C}^n$ with $\mathbb{R}^{2n}$. For each degree $r$, the space
\begin{align*}
H_r := L^2_{(0,r)}(\Omega,e^{-\varphi})
\end{align*}
consists of measurable $(0,r)$-forms whose squared pointwise norm is integrable against the measure $e^{-\varphi}\,d\mathcal{L}^{2n}$. Its inner product is
\begin{align*}
(\alpha,\beta)_{H_r}
:=
\int_{\Omega}
\langle \alpha(z),\beta(z)\rangle_{\Lambda^{0,r}}\,
e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z).
\end{align*}
The measure is written explicitly because all weighted $L^2$ estimates in this theorem are with respect to $e^{-\varphi}\,d\mathcal{L}^{2n}$.
We now define two closed operators. The first is
\begin{align*}
T:\operatorname{Dom}(T)\subset H_{q-1}\to H_q,
\end{align*}
where $T u=\bar\partial u$ in the distributional sense and $\operatorname{Dom}(T)$ consists of those $u\in H_{q-1}$ for which $\bar\partial u\in H_q$. The second is
\begin{align*}
S:\operatorname{Dom}(S)\subset H_q\to H_{q+1}.
\end{align*}
Here we use the convention $H_{n+1}=\{0\}$. If $q<n$, then $S v=\bar\partial v$ and $\operatorname{Dom}(S)$ consists of those $v\in H_q$ for which $\bar\partial v\in H_{q+1}$. If $q=n$, then $\operatorname{Dom}(S)=H_n$ and $S$ is the zero map $H_n\to\{0\}$, reflecting the fact that there are no nonzero $(0,n+1)$-forms. The identity $\bar\partial^2=0$ for distributions gives $S T=0$. Therefore the image of $T$ lies in the kernel of $S$, which is the Hilbert-complex structure needed for the duality argument.
Finally, the assumption $\bar\partial f=0$ means precisely that $f\in \operatorname{Dom}(S)$ and $S f=0$; in the top-degree case $q=n$, this condition is automatic with the convention just stated.
[/guided]
[/step]
[step:Apply the higher-degree Hörmander-Kodaira estimate on closed forms]
Let
\begin{align*}
T_\varphi^*:\operatorname{Dom}(T_\varphi^*)\subset H_q\to H_{q-1}
\end{align*}
denote the [Hilbert space](/page/Hilbert%20Space) adjoint of $T$ with respect to the weighted inner products $ (\cdot,\cdot)_{H_{q-1}}$ and $(\cdot,\cdot)_{H_q}$. The subscript $\varphi$ records that the adjoint depends on the weight $e^{-\varphi}$. We use the same Euclidean normalization as in the estimate: $\omega$ is the standard Kähler form on $\mathbb{C}^n$, and $\langle\cdot,\cdot\rangle_{\Lambda^{0,r}}$ is the induced Euclidean pointwise form inner product. The [higher-degree Hörmander-Kodaira $L^2$ estimate](/page/Higher-Degree%20Hormander-Kodaira%20Estimate) on pseudoconvex domains states that, under the hypotheses $i\partial\bar\partial\varphi\ge c\omega$ and $1\le q\le n$,
\begin{align*}
\|S v\|_{H_{q+1}}^2+\|T_\varphi^*v\|_{H_{q-1}}^2
\ge
q c\,\|v\|_{H_q}^2
\end{align*}
for every $v\in \operatorname{Dom}(S)\cap \operatorname{Dom}(T_\varphi^*)$.
In particular, if $v\in \operatorname{Dom}(T_\varphi^*)\cap \ker S$, then $S v=0$, and hence
\begin{align*}
q c\,\|v\|_{H_q}^2
\le
\|T_\varphi^*v\|_{H_{q-1}}^2.
\end{align*}
[guided]
The analytic input is the [higher-degree Hörmander-Kodaira $L^2$ estimate](/page/Higher-Degree%20Hormander-Kodaira%20Estimate). We are using its Euclidean normalization: $\omega$ is the standard Kähler form on $\mathbb{C}^n$, and the pointwise form norm is the one induced by the Euclidean Hermitian metric, which is exactly the norm used in the definition of $H_r$. It applies here because $\Omega$ is pseudoconvex, $\varphi\in C^2(\Omega;\mathbb{R})$, and the curvature lower bound
\begin{align*}
i\partial\bar\partial\varphi\ge c\omega
\end{align*}
holds with $c>0$. For every form
\begin{align*}
v\in \operatorname{Dom}(S)\cap \operatorname{Dom}(T_\varphi^*),
\end{align*}
the estimate gives
\begin{align*}
\|S v\|_{H_{q+1}}^2+\|T_\varphi^*v\|_{H_{q-1}}^2
\ge
q c\,\|v\|_{H_q}^2.
\end{align*}
This is the point at which the form degree matters: for $(0,q)$-forms the curvature contribution is bounded below by $qc$.
We will use the estimate only on the closed subspace $\ker S$. If
\begin{align*}
v\in \operatorname{Dom}(T_\varphi^*)\cap \ker S,
\end{align*}
then $S v=0$, so the first term on the left vanishes. Therefore
\begin{align*}
q c\,\|v\|_{H_q}^2
\le
\|T_\varphi^*v\|_{H_{q-1}}^2.
\end{align*}
This converts control of the adjoint derivative $T_\varphi^*v$ into control of the whole form $v$, which is exactly the estimate needed for the [Hilbert space](/page/Hilbert%20Space) solvability argument.
[/guided]
[/step]
[step:Bound the dual functional on the adjoint range]
Define the closed subspace
\begin{align*}
K:=\ker S\subset H_q.
\end{align*}
Regard $T$ as an operator
\begin{align*}
T_K:\operatorname{Dom}(T)\subset H_{q-1}\to K,
\end{align*}
which is legitimate because $S T=0$. Its adjoint is the restriction of $T_\varphi^*$ to $\operatorname{Dom}(T_\varphi^*)\cap K$.
Define a linear functional
\begin{align*}
\Lambda:T_K^*(\operatorname{Dom}(T_K^*))\to \mathbb{C}
\end{align*}
by
\begin{align*}
\Lambda(T_K^*v):=(v,f)_{H_q}.
\end{align*}
This is well-defined: if $T_K^*v=0$, then the estimate from the previous step gives $\|v\|_{H_q}=0$, so $(v,f)_{H_q}=0$.
For every $v\in \operatorname{Dom}(T_K^*)$, the [Cauchy-Schwarz inequality](/theorems/432) in $H_q$ and the adjoint estimate give
\begin{align*}
|\Lambda(T_K^*v)|
=
|(v,f)_{H_q}|
\le
\|v\|_{H_q}\|f\|_{H_q}
\le
\frac{1}{\sqrt{qc}}\,
\|T_K^*v\|_{H_{q-1}}\,
\|f\|_{H_q}.
\end{align*}
Thus $\Lambda$ is bounded on its domain with operator norm at most $(qc)^{-1/2}\|f\|_{H_q}$.
[guided]
Now we turn the estimate into a solvability statement. Let
\begin{align*}
K:=\ker S\subset H_q.
\end{align*}
Since $S$ is closed, $\ker S$ is a closed Hilbert subspace of $H_q$. Because $S T=0$, every form $T u$ lies in $K$, so we may regard $T$ as an operator
\begin{align*}
T_K:\operatorname{Dom}(T)\subset H_{q-1}\to K.
\end{align*}
The goal is to solve $T_Ku=f$. [Hilbert space](/page/Hilbert%20Space) duality says to test $f$ against elements in the domain of the adjoint. The adjoint of $T_K$ is the restriction of the weighted adjoint $T_\varphi^*$ to the closed-form subspace $K$:
\begin{align*}
T_K^*v=T_\varphi^*v,
\qquad
v\in \operatorname{Dom}(T_\varphi^*)\cap K.
\end{align*}
Define
\begin{align*}
\Lambda:T_K^*(\operatorname{Dom}(T_K^*))\to \mathbb{C}
\end{align*}
by
\begin{align*}
\Lambda(T_K^*v):=(v,f)_{H_q}.
\end{align*}
We must first check that this definition does not depend on the choice of $v$. Suppose $T_K^*v=0$. Since $v\in K=\ker S$, the estimate from the previous step gives
\begin{align*}
q c\,\|v\|_{H_q}^2
\le
\|T_K^*v\|_{H_{q-1}}^2
=
0.
\end{align*}
Hence $\|v\|_{H_q}=0$, so $v=0$ in $H_q$ and therefore $(v,f)_{H_q}=0$. Thus $\Lambda$ is well-defined.
Next we prove boundedness. By the [Cauchy-Schwarz inequality](/theorems/432) in the [Hilbert space](/page/Hilbert%20Space) $H_q$,
\begin{align*}
|\Lambda(T_K^*v)|
=
|(v,f)_{H_q}|
\le
\|v\|_{H_q}\|f\|_{H_q}.
\end{align*}
The closed-form Hörmander estimate gives
\begin{align*}
\|v\|_{H_q}
\le
\frac{1}{\sqrt{qc}}\,
\|T_K^*v\|_{H_{q-1}}.
\end{align*}
Combining these two inequalities yields
\begin{align*}
|\Lambda(T_K^*v)|
\le
\frac{1}{\sqrt{qc}}\,
\|T_K^*v\|_{H_{q-1}}\,
\|f\|_{H_q}.
\end{align*}
So $\Lambda$ is a bounded linear functional on the subspace $T_K^*(\operatorname{Dom}(T_K^*))\subset H_{q-1}$, with norm at most $(qc)^{-1/2}\|f\|_{H_q}$.
[/guided]
[/step]
[step:Use Hahn-Banach and Riesz to construct the solution]
By the [Hahn-Banach extension theorem](/page/Hahn-Banach%20Theorem), $\Lambda$ extends to a bounded linear functional
\begin{align*}
\widetilde{\Lambda}:H_{q-1}\to\mathbb{C}
\end{align*}
with
\begin{align*}
\|\widetilde{\Lambda}\|_{H_{q-1}^*}
\le
\frac{1}{\sqrt{qc}}\|f\|_{H_q}.
\end{align*}
By the [Riesz representation theorem for Hilbert spaces](/page/Riesz%20Representation%20Theorem), there exists $u\in H_{q-1}$ such that
\begin{align*}
\widetilde{\Lambda}(h)=(h,u)_{H_{q-1}}
\end{align*}
for every $h\in H_{q-1}$, and
\begin{align*}
\|u\|_{H_{q-1}}
=
\|\widetilde{\Lambda}\|_{H_{q-1}^*}
\le
\frac{1}{\sqrt{qc}}\|f\|_{H_q}.
\end{align*} Therefore
\begin{align*}
\|u\|_{H_{q-1}}^2
\le
\frac{1}{qc}\|f\|_{H_q}^2.
\end{align*}
For every $v\in \operatorname{Dom}(T_K^*)$,
\begin{align*}
(T_K^*v,u)_{H_{q-1}}
=
\widetilde{\Lambda}(T_K^*v)
=
\Lambda(T_K^*v)
=
(v,f)_{H_q}.
\end{align*}
Taking complex conjugates, using the Hermitian symmetry of the inner product, gives
\begin{align*}
(u,T_K^*v)_{H_{q-1}}
=
(f,v)_{H_q}.
\end{align*}
[guided]
The functional $\Lambda$ is currently defined only on the subspace
\begin{align*}
T_K^*(\operatorname{Dom}(T_K^*))\subset H_{q-1}.
\end{align*}
The [Hahn-Banach extension theorem](/page/Hahn-Banach%20Theorem) extends it to a bounded linear functional
\begin{align*}
\widetilde{\Lambda}:H_{q-1}\to\mathbb{C}
\end{align*}
without increasing its norm. Thus
\begin{align*}
\|\widetilde{\Lambda}\|_{H_{q-1}^*}
\le
\frac{1}{\sqrt{qc}}\|f\|_{H_q}.
\end{align*}
The [Riesz representation theorem for Hilbert spaces](/page/Riesz%20Representation%20Theorem) then converts this functional into an element of the [Hilbert space](/page/Hilbert%20Space). Since the inner product is linear in the first argument, there exists $u\in H_{q-1}$ such that
\begin{align*}
\widetilde{\Lambda}(h)=(h,u)_{H_{q-1}}
\end{align*}
for every $h\in H_{q-1}$, and
\begin{align*}
\|u\|_{H_{q-1}}
=
\|\widetilde{\Lambda}\|_{H_{q-1}^*}.
\end{align*}
Consequently,
\begin{align*}
\|u\|_{H_{q-1}}
\le
\frac{1}{\sqrt{qc}}\|f\|_{H_q},
\end{align*}
and squaring gives
\begin{align*}
\|u\|_{H_{q-1}}^2
\le
\frac{1}{qc}\|f\|_{H_q}^2.
\end{align*}
It remains to record the identity that $u$ satisfies. If $v\in \operatorname{Dom}(T_K^*)$, then $T_K^*v$ lies in the original domain of $\Lambda$, so
\begin{align*}
(T_K^*v,u)_{H_{q-1}}
=
\widetilde{\Lambda}(T_K^*v)
=
\Lambda(T_K^*v)
=
(v,f)_{H_q}.
\end{align*}
By Hermitian symmetry,
\begin{align*}
(u,T_K^*v)_{H_{q-1}}
=
(f,v)_{H_q}.
\end{align*}
This is the weak adjoint identity that will identify $u$ as a solution. More explicitly, $T_K$ is densely defined because $C_c^\infty$ $(0,q-1)$-forms are dense in $H_{q-1}$, and $T_K$ is closed as an operator into $K$ because convergence in the closed subspace $K\subset H_q$ is the same as convergence in $H_q$ and the maximal distributional $\bar\partial$ operator is closed. Also $f\in K$ since $S f=0$. The standard double-adjoint theorem for closed densely defined operators gives $T_K^{**}=T_K$. Therefore the identity above says precisely that $u\in\operatorname{Dom}(T_K^{**})=\operatorname{Dom}(T_K)$ and $T_Ku=f$.
[/guided]
[/step]
[step:Identify the Hilbert space solution with the distributional equation]
The identity
\begin{align*}
(u,T_K^*v)_{H_{q-1}}=(f,v)_{H_q}
\end{align*}
for every $v\in \operatorname{Dom}(T_K^*)$ is an identity against the adjoint domain. Let $C_c^\infty(\Omega;\Lambda^{0,q-1})$ denote the space of smooth compactly supported $(0,q-1)$-forms on $\Omega$. The operator $T_K$ is densely defined because $C_c^\infty(\Omega;\Lambda^{0,q-1})$ is dense in $H_{q-1}$. It is closed as an operator into $K$ because $K$ is a closed subspace of $H_q$ and the maximal distributional $\bar\partial$ operator $T:H_{q-1}\supset\operatorname{Dom}(T)\to H_q$ is closed. Since $f\in K$, the defining criterion for the double adjoint says that the displayed identity is exactly the assertion that $u\in \operatorname{Dom}(T_K^{**})$ and $T_K^{**}u=f$. By the [double-adjoint theorem for closed densely defined operators](/page/Double-Adjoint%20Theorem), $T_K^{**}=T_K$. Hence $u\in \operatorname{Dom}(T_K)$ and $T_Ku=f$.
Since $T_K$ is $T$ with codomain restricted to $\ker S$, this says $u\in\operatorname{Dom}(T)$ and
\begin{align*}
T u=f.
\end{align*}
By the definition of $T$ as the maximal distributional $\bar\partial$ operator,
\begin{align*}
\bar\partial u=f
\end{align*}
in the distributional sense. Rewriting the Hilbert norms gives
\begin{align*}
\|u\|_{L^2(\Omega,e^{-\varphi})}^2
\le
\frac{1}{qc}
\|f\|_{L^2(\Omega,e^{-\varphi})}^2.
\end{align*}
This is the claimed solution and estimate.
[/step]
