[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 $f\in\ker S$.[/step]

[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: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 $H_1$.[/step]

[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: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*}[/step]

[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: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.[/step]

[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: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]