[proofplan] We work in the weighted Hilbert complex associated to the maximal distributional Cauchy-Riemann operator. The weighted Bochner-Kodaira-Morrey estimate controls the curvature norm of every closed test form by the norm of the corresponding Hilbert-space adjoint. The inverse-curvature assumption then makes the natural pairing with the datum bounded on the closed-form adjoint domain. Applying Hahn-Banach and Riesz to this conjugate-linear functional produces $u$, and the double-adjoint theorem identifies the resulting weak identity with the desired distributional equation. [/proofplan] [step:Set up the weighted Hilbert complex] Define the weighted Hilbert spaces \begin{align*} H_0&=L^2(\Omega,e^{-\varphi}),\\ H_1&=L^2_{0,1}(\Omega,e^{-\varphi}),\\ H_2&=L^2_{0,2}(\Omega,e^{-\varphi}), \end{align*} with inner products linear in the first argument. For functions $a,b\in H_0$ and $(0,1)$-forms $\alpha=\sum_{j=1}^n\alpha_j\,d\bar z_j$, $\beta=\sum_{j=1}^n\beta_j\,d\bar z_j$ in $H_1$, these inner products are \begin{align*} (a,b)_\varphi &=\int_\Omega a(z)\overline{b(z)}e^{-\varphi(z)}\,d\mathcal L^{2n}(z),\\ (\alpha,\beta)_\varphi &=\int_\Omega \sum_{j=1}^n \alpha_j(z)\overline{\beta_j(z)}e^{-\varphi(z)}\,d\mathcal L^{2n}(z). \end{align*} Let \begin{align*} T=\bar{\partial}: \operatorname{Dom}(T)\subset H_0&\to H_1,\\ S=\bar{\partial}: \operatorname{Dom}(S)\subset H_1&\to H_2 \end{align*} be the maximal distributional $\bar{\partial}$ operators. Thus $a\in\operatorname{Dom}(T)$ precisely when the distributional $(0,1)$-form $\bar{\partial}a$ is represented by an element of $H_1$, and similarly for $S$ on $(0,1)$-forms. Compactly supported smooth forms are dense in the weighted $L^2$ spaces, so $T$ and $S$ are densely defined. They are closed because if $a_m\to a$ in weighted $L^2$ and $\bar{\partial}a_m\to g$ in weighted $L^2$, then the same convergence holds locally in ordinary $L^2$ because $e^{-\varphi}$ is bounded above and below on compact subsets of $\Omega$; testing against compactly supported smooth forms gives $\bar{\partial}a=g$ distributionally. Set \begin{align*} K=\ker S=\{w\in\operatorname{Dom}(S)\mid Sw=0\}\subset H_1. \end{align*} For every $a\in\operatorname{Dom}(T)$, the element $Ta\in H_1$ is the $L^2$ representative of the distributional form $\bar{\partial}a$. Since $\bar{\partial}(\bar{\partial}a)=0$ as a distribution, the form $Ta$ belongs to $\operatorname{Dom}(S)$ and satisfies \begin{align*} S(Ta)=0. \end{align*} Thus $ST=0$ as an operator identity on $\operatorname{Dom}(T)$, and \begin{align*} T(\operatorname{Dom}(T))\subset K. \end{align*} Because $S$ is closed, its kernel is closed in $H_1$: if $w_m\in K$, $w_m\to w$ in $H_1$, then $Sw_m=0\to 0$ in $H_2$, so closedness of $S$ gives $w\in\operatorname{Dom}(S)$ and $Sw=0$. Hence $K$ is a closed Hilbert subspace of $H_1$ with the inherited inner product. The Hilbert-space adjoint of $T$ is denoted \begin{align*} T^*=\bar{\partial}_\varphi^*:\operatorname{Dom}(T^*)\subset H_1\to H_0. \end{align*} Since $f\in H_1$ and $\bar{\partial}f=0$ distributionally, $f$ belongs to $K$. [/step] [step:State the weighted Bochner-Kodaira-Morrey estimate on the graph domain] For $v=\sum_{j=1}^n v_j\,d\bar z_j\in\operatorname{Dom}(T^*)\cap\operatorname{Dom}(S)$, define its curvature norm by \begin{align*} \|v\|_{M_\varphi,\varphi}^2 = \int_\Omega \sum_{i,j=1}^n \frac{\partial^2\varphi}{\partial z_i\partial\bar z_j}(z) v_i(z)\overline{v_j(z)} e^{-\varphi(z)}\,d\mathcal L^{2n}(z). \end{align*} Strict plurisubharmonicity means that $M_\varphi(z)$ is a positive definite Hermitian matrix for every $z\in\Omega$, so the integrand is nonnegative. The weighted Bochner-Kodaira-Morrey estimate used here is the following precise graph-domain statement: if $\Omega$ is pseudoconvex and $\varphi\in C^2(\Omega)$ is plurisubharmonic, then every $v\in\operatorname{Dom}(T^*)\cap\operatorname{Dom}(S)$ satisfies \begin{align*} \|v\|_{M_\varphi,\varphi}^2 \le \|T^*v\|_{\varphi}^2+\|Sv\|_{\varphi}^2. \end{align*} The hypotheses are satisfied because $\Omega$ is pseudoconvex and $\varphi\in C^2(\Omega)$. The passage from compactly supported smooth forms to the maximal graph domain is part of the estimate: it is obtained by graph-norm approximation on a pseudoconvex exhaustion, and the Levi boundary term is nonnegative on the exhausting domains. If $v\in\operatorname{Dom}(T^*)\cap K$, then $Sv=0$, and the estimate becomes \begin{align*} \|v\|_{M_\varphi,\varphi}^2\le \|T^*v\|_{\varphi}^2. \end{align*} [/step] [step:Bound the closed datum by the inverse-curvature norm] Let $v=\sum_{j=1}^n v_j\,d\bar z_j\in\operatorname{Dom}(T^*)\cap K$. For each $z\in\Omega$, the [Cauchy-Schwarz inequality](/theorems/432) for the positive definite Hermitian matrix $M_\varphi(z)$ gives \begin{align*} \left|\sum_{j=1}^n f_j(z)\overline{v_j(z)}\right|^2 &\le \left(\sum_{i,j=1}^n (M_\varphi(z)^{-1})_{ij} f_i(z)\overline{f_j(z)}\right) \left(\sum_{i,j=1}^n (M_\varphi(z))_{ij} v_i(z)\overline{v_j(z)}\right). \end{align*} Let $\mathcal B(\Omega)$ denote the Borel $\sigma$-algebra on $\Omega$. Applying the [Cauchy-Schwarz inequality](/theorems/432) in the [measure space](/page/Measure%20Space) $(\Omega,\mathcal B(\Omega),e^{-\varphi}\mathcal L^{2n})$ yields \begin{align*} |(f,v)_\varphi|^2 &= \left| \int_\Omega \sum_{j=1}^n f_j(z)\overline{v_j(z)} e^{-\varphi(z)}\,d\mathcal L^{2n}(z) \right|^2\\ &\le A\,\|v\|_{M_\varphi,\varphi}^2\\ &\le A\,\|T^*v\|_{\varphi}^2. \end{align*} The last inequality is the closed-form Bochner-Kodaira-Morrey estimate from the previous step. This is the only estimate on the datum used in the duality argument; the constant is exactly $A^{1/2}$. [/step] [step:Apply Hahn-Banach and Riesz with the linear-first convention] Regard $K$ as a [Hilbert space](/page/Hilbert%20Space) with inner product inherited from $H_1$. Since the previous step showed $Ta\in K$ for every $a\in\operatorname{Dom}(T)$, define the restricted-codomain operator \begin{align*} T_K:\operatorname{Dom}(T)\subset H_0&\to K,\\ a&\mapsto Ta. \end{align*} Here $\operatorname{Dom}(T_K)=\operatorname{Dom}(T)$ is a dense subspace of $H_0$, because $T$ is densely defined as an operator from $H_0$ to $H_1$. The operator $T_K$ is closed as an operator from $H_0$ to $K$. Indeed, suppose $a_m\in\operatorname{Dom}(T_K)$, $a_m\to a$ in $H_0$, and $T_Ka_m\to k$ in $K$. The norm on $K$ is the restriction of the norm on $H_1$, so $Ta_m\to k$ in $H_1$. Since $T$ is closed as an operator from $H_0$ to $H_1$, it follows that $a\in\operatorname{Dom}(T)$ and $Ta=k$. As $k\in K$, this says $a\in\operatorname{Dom}(T_K)$ and $T_Ka=k$. We now identify the adjoint of $T_K$. By definition, a vector $v\in K$ belongs to $\operatorname{Dom}(T_K^*)$ exactly when there exists $h\in H_0$ such that \begin{align*} (T_Ka,v)_K=(a,h)_\varphi \end{align*} for every $a\in\operatorname{Dom}(T_K)$. The inner product on $K$ is inherited from $H_1$, and $\operatorname{Dom}(T_K)=\operatorname{Dom}(T)$, so this condition is equivalent to \begin{align*} (Ta,v)_\varphi=(a,h)_\varphi \end{align*} for every $a\in\operatorname{Dom}(T)$. This is precisely the defining condition for $v\in\operatorname{Dom}(T^*)$ with $T^*v=h$. Conversely, if $v\in\operatorname{Dom}(T^*)\cap K$, the same identity shows that $v\in\operatorname{Dom}(T_K^*)$ and $T_K^*v=T^*v$. Therefore \begin{align*} \operatorname{Dom}(T_K^*)=\operatorname{Dom}(T^*)\cap K, \qquad T_K^*v=T^*v, \end{align*} where the domain on the left is understood as a subspace of the [Hilbert space](/page/Hilbert%20Space) $K$. Define \begin{align*} \mathcal R=T_K^*(\operatorname{Dom}(T_K^*))\subset H_0. \end{align*} Define a conjugate-linear functional $\Lambda:\mathcal R\to\mathbb C$ by \begin{align*} \Lambda(T_K^*v)=(f,v)_\varphi, \qquad v\in\operatorname{Dom}(T_K^*). \end{align*} The functional is well-defined. If $T_K^*v=0$, the estimate from the previous step gives \begin{align*} |\Lambda(T_K^*v)|^2=|(f,v)_\varphi|^2\le A\|T_K^*v\|_\varphi^2=0. \end{align*} The same estimate gives \begin{align*} |\Lambda(T_K^*v)|\le A^{1/2}\|T_K^*v\|_\varphi. \end{align*} The complex [Hahn-Banach theorem](/page/Hahn-Banach%20Theorem) for conjugate-linear functionals extends $\Lambda$ to a bounded conjugate-linear functional on all of $H_0$ with norm at most $A^{1/2}$. The [Riesz representation theorem](/theorems/221) for complex Hilbert spaces with inner product linear in the first argument says that there exists $u\in H_0$ such that \begin{align*} \Lambda(g)=(u,g)_\varphi \end{align*} for every $g\in H_0$, and \begin{align*} \|u\|_{\varphi}\le A^{1/2}. \end{align*} Thus, for every $v\in\operatorname{Dom}(T_K^*)$, \begin{align*} (u,T_K^*v)_\varphi=(f,v)_\varphi. \end{align*} [/step] [step:Identify the represented vector as a solution] The preceding identity gives, after conjugate symmetry of the weighted Hilbert inner products, \begin{align*} (T_K^*v,u)_\varphi=(v,f)_\varphi \end{align*} for every $v\in\operatorname{Dom}(T_K^*)$. This is the defining condition for \begin{align*} u\in\operatorname{Dom}\bigl((T_K^*)^*\bigr), \qquad (T_K^*)^*u=f \end{align*} as an equality in the [Hilbert space](/page/Hilbert%20Space) $K$, because $f\in K$ and the inner product on $K$ is inherited from $H_1$. The previous step proved that $T_K:\operatorname{Dom}(T)\subset H_0\to K$ is densely defined and closed. The double-adjoint theorem for densely defined closed operators states that if $A$ is densely defined and closed, then $(A^*)^*=A$. Applying this theorem to $A=T_K$ gives \begin{align*} (T_K^*)^*=T_K. \end{align*} Therefore $u\in\operatorname{Dom}(T)$ and \begin{align*} \bar{\partial}u=Tu=T_Ku=f \end{align*} as an equality in $H_1$. Equality in $H_1$ implies the same equality in the distributional sense. Finally, the norm estimate from [Riesz representation](/theorems/67) gives \begin{align*} \int_\Omega |u|^2e^{-\varphi}\,d\mathcal L^{2n}(z) &=\|u\|_\varphi^2\\ &\le A\\ &= \int_\Omega \sum_{i,j=1}^n (M_\varphi(z)^{-1})_{ij} f_i(z)\overline{f_j(z)} e^{-\varphi(z)}\,d\mathcal L^{2n}(z). \end{align*} This is the asserted Hörmander estimate. [/step]