[proofplan] The smooth solutions $u_j$ obey weighted bounds with respect to the smooth weights $\varphi_j$. Because the $\varphi_j$ decrease and are smooth, on every compact subset of $\Omega$ the weights $e^{-\varphi_j}$ are bounded below by a positive constant, which converts the uniform weighted bounds into uniform local $L^2$ bounds for the $u_j$. A diagonal application of [weak sequential compactness](/theorems/214) in the separable Hilbert spaces $L^2(\overline{\Omega_m})$ extracts a subsequence converging weakly on every compact set to a limit $u \in L^2_{\mathrm{loc}}(\Omega)$. Testing against compactly supported smooth functions transfers the equation $\bar\partial u_j = f$ to the limit in $\mathcal{D}'(\Omega)$. Finally, weighted [lower semicontinuity of the norm](/theorems/215) against each fixed weight $e^{-\varphi_\ell}$ gives the bound $C$ with that weight, and two applications of the [Monotone Convergence Theorem](/theorems/509) — first letting $\ell \to \infty$ so that $e^{-\varphi_\ell} \uparrow e^{-\varphi}$, then exhausting $\Omega$ — upgrade this to the desired estimate against $e^{-\varphi}$. [/proofplan] [step:Fix a pseudoconvex exhaustion and record the uniform weighted bound] We use the exhaustion $\Omega_1 \Subset \Omega_2 \Subset \cdots$ supplied by the hypotheses, and set \begin{align*} K_m := \overline{\Omega_m}, \qquad m \ge 1. \end{align*} Each $K_m$ is compact since $\Omega_m \Subset \Omega$, and $K_m \subset \Omega_{m+1} \subset K_{m+1}$, so $(K_m)_{m \ge 1}$ is an increasing sequence of compact sets with $\bigcup_{m=1}^\infty K_m = \bigcup_{m=1}^\infty \Omega_m = \Omega$. The functions $u_j \in L^2_{\mathrm{loc}}(\Omega_j)$ are those produced by [Hörmander's $L^2$ Theorem](/theorems/3500) applied on the pseudoconvex subdomain $\Omega_j$ with the smooth strictly plurisubharmonic weight $\varphi_j$; we take as given the uniform bound \begin{align*} \int_{\Omega_j} |u_j|^2\, e^{-\varphi_j}\, d\mathcal{L}^{2n}(z) \le C \qquad \text{for all } j \ge 1, \tag{1} \end{align*} with $C = \int_\Omega |f|^2_{\mathcal{L}_\varphi^{-1}}\, e^{-\varphi}\, d\mathcal{L}^{2n} < \infty$ independent of $j$. The whole argument concerns the passage to the limit $j \to \infty$. [/step] [step:Convert the uniform weighted bounds into uniform local $L^2$ bounds] Fix $m \ge 1$. For every index $j \ge m+1$ we have $K_m = \overline{\Omega_m} \subset \Omega_{m+1} \subset \Omega_j$, so $u_j$ is defined on $K_m$. Because the weights decrease, $\varphi_j \le \varphi_{m+1}$ on $K_m$ for all $j \ge m+1$; and since $\varphi_{m+1}$ is smooth on a neighbourhood of the compact set $K_m$, the quantity \begin{align*} M_m := \sup_{z \in K_m} \varphi_{m+1}(z) \end{align*} is finite. Hence $e^{-\varphi_j} \ge e^{-M_m}$ on $K_m$ for all $j \ge m+1$, and using (1), \begin{align*} \int_{K_m} |u_j|^2\, d\mathcal{L}^{2n}(z) &\le e^{M_m} \int_{K_m} |u_j|^2\, e^{-\varphi_j}\, d\mathcal{L}^{2n}(z) \\ &\le e^{M_m} \int_{\Omega_j} |u_j|^2\, e^{-\varphi_j}\, d\mathcal{L}^{2n}(z) \le e^{M_m} C. \end{align*} Therefore \begin{align*} \sup_{j \ge m+1} \|u_j\|_{L^2(K_m)}^2 \le e^{M_m} C < \infty. \tag{2} \end{align*} [guided] The given bound (1) is stated with respect to the moving weight $e^{-\varphi_j}$, which is awkward for extracting a limit: both the integrand and the weight change with $j$. To apply weak compactness we want a uniform bound in a fixed, $j$-independent [Hilbert space](/page/Hilbert%20Space). The natural choice on a compact set $K_m$ is unweighted $L^2(K_m)$. How do we pass from the weighted bound to an unweighted one? We need a lower bound on the weight $e^{-\varphi_j}$ over $K_m$, i.e. an upper bound on $\varphi_j$ over $K_m$. This is exactly where the monotonicity $\varphi_1 \ge \varphi_2 \ge \cdots$ pays off: for every $j \ge m+1$ we have $\varphi_j \le \varphi_{m+1}$ on $K_m$, so a single function $\varphi_{m+1}$ dominates the entire tail of the sequence. Since $\varphi_{m+1}$ is smooth, it is bounded on the compact set $K_m$, giving the finite constant $M_m = \sup_{K_m}\varphi_{m+1}$. We restrict to indices $j \ge m+1$ for two reasons at once: this guarantees both that $K_m$ lies inside the domain $\Omega_j$ of $u_j$ (so the integral makes sense) and that $\varphi_j \le \varphi_{m+1}$ there. Then on $K_m$, \begin{align*} e^{-\varphi_j} \ge e^{-\varphi_{m+1}} \ge e^{-M_m} > 0, \end{align*} and we may divide out the weight at the cost of the constant $e^{M_m}$: \begin{align*} \int_{K_m} |u_j|^2\, d\mathcal{L}^{2n}(z) \le e^{M_m}\int_{K_m} |u_j|^2 e^{-\varphi_j}\, d\mathcal{L}^{2n}(z). \end{align*} Enlarging the integration domain on the right-hand side from $K_m$ to the full $\Omega_j$ only increases the (nonnegative) integral, and the latter is at most $C$ by (1). The constant $e^{M_m}C$ depends on $m$ but not on $j$, which is all we need: for each fixed level $m$, the tail $(u_j)_{j \ge m+1}$ is bounded in $L^2(K_m)$. [/guided] [/step] [step:Extract a diagonal subsequence converging weakly in $L^2$ on every compact set] We first record the abstract tool. For each $m$, the space $L^2(K_m)$ is a [Hilbert space](/page/Hilbert%20Space). The [measure space](/page/Measure%20Space) $(K_m, \mathcal{B}(K_m), \mathcal{L}^{2n})$ is $\sigma$-finite — indeed $\mathcal{L}^{2n}(K_m) < \infty$ since $K_m$ is compact — and $\mathcal{B}(K_m)$ is countably generated; with $p = 2 \in [1, \infty)$, the [Separability of $L^p$ Spaces](/theorems/548) shows $L^2(K_m)$ is separable. By the [Riesz Representation Theorem](/theorems/218), $L^2(K_m)$ is isometrically isomorphic to its own dual, so weak and weak* convergence coincide on it; applying the [Sequential Banach–Alaoglu Theorem](/theorems/496) to the [separable space](/page/Separable%20Space) $L^2(K_m)$ then shows that every norm-bounded sequence in $L^2(K_m)$ has a weakly convergent subsequence. Now build the limit by diagonalisation. By (2), the tail $(u_j)_{j \ge 2}$ is bounded in $L^2(K_1)$, so there is a subsequence $(u_{j})_{j \in S_1}$ (with $S_1 \subseteq \{2,3,\dots\}$ infinite) converging weakly in $L^2(K_1)$ to some $u^{(1)} \in L^2(K_1)$. By (2) again, $(u_j)_{j \in S_1,\, j \ge 3}$ is bounded in $L^2(K_2)$; pass to a further infinite subset $S_2 \subseteq S_1$ along which $u_j \rightharpoonup u^{(2)}$ in $L^2(K_2)$. Continuing, we obtain nested infinite index sets $S_1 \supseteq S_2 \supseteq \cdots$ and limits $u^{(m)} \in L^2(K_m)$ with $u_j \rightharpoonup u^{(m)}$ in $L^2(K_m)$ along $S_m$. For $m' < m$, the restriction map $L^2(K_m) \to L^2(K_{m'})$ is bounded linear, hence weakly continuous, so $u_j \rightharpoonup u^{(m)}\big|_{K_{m'}}$ in $L^2(K_{m'})$ along $S_m \subseteq S_{m'}$; uniqueness of weak limits gives $u^{(m)}\big|_{K_{m'}} = u^{(m')}$ $\mathcal{L}^{2n}$-a.e. The functions $u^{(m)}$ therefore patch to a single $u \in L^2_{\mathrm{loc}}(\Omega)$ with $u\big|_{K_m} = u^{(m)}$. Choosing the diagonal indices $j_k := \min(S_k \setminus \{j_1,\dots,j_{k-1}\})$, the sequence $(j_k)_{k \ge 1}$ is strictly increasing and, for each fixed $m$, eventually lies in $S_m$; hence \begin{align*} u_{j_k} \rightharpoonup u \quad \text{in } L^2(K_m) \text{ for every } m \ge 1. \tag{3} \end{align*} [guided] We have, for each compact $K_m$, a uniformly bounded tail of $(u_j)$ in the [Hilbert space](/page/Hilbert%20Space) $L^2(K_m)$. Bounded sequences in a [Hilbert space](/page/Hilbert%20Space) need not converge strongly, but in a separable [Hilbert space](/page/Hilbert%20Space) they always have weakly convergent subsequences. Let us make precise why. A separable [Hilbert space](/page/Hilbert%20Space) $H$ is reflexive, and the [Riesz Representation Theorem](/theorems/218) identifies $H$ isometrically with $H^*$ via $h \mapsto (\,\cdot\,, h)_H$; under this identification a sequence $h_j \in H$ corresponds to functionals on $H$, and weak* convergence of those functionals — meaning $(x, h_j)_H \to (x, h)_H$ for all $x \in H$ — is exactly [weak convergence](/page/Weak%20Convergence) $h_j \rightharpoonup h$ in $H$. The [Sequential Banach–Alaoglu Theorem](/theorems/496) requires the predual to be separable: here the predual is $H = L^2(K_m)$ itself, which is separable by the [Separability of $L^p$ Spaces](/theorems/548) (we checked $\sigma$-finiteness from compactness of $K_m$, that the Borel $\sigma$-algebra is countably generated, and that $p = 2 < \infty$). A norm-bounded sequence in $H^* \cong H$ thus has a weak* — equivalently weakly — convergent subsequence. The subtlety is that each $u_j$ lives in many different spaces $L^2(K_m)$ simultaneously, and we want a single subsequence that works for all $m$. This is the classical diagonal argument. We extract a subsequence on $K_1$, then thin it further on $K_2$, then on $K_3$, and so on, producing nested index sets $S_1 \supseteq S_2 \supseteq \cdots$. The diagonal sequence $j_1 < j_2 < \cdots$, with $j_k$ chosen from $S_k$, is eventually contained in each $S_m$ (because $S_m \supseteq S_{m'}$ for $m \le m'$ and the $j_k$ are drawn from progressively smaller sets), so it inherits [weak convergence](/page/Weak%20Convergence) on every $K_m$. Why do the limits on different compacta agree? Because restriction $L^2(K_m) \to L^2(K_{m'})$ for $m' < m$ is a bounded [linear map](/page/Linear%20Map), and bounded linear maps are continuous for the weak topologies; so the weak limit on $K_m$, restricted to $K_{m'}$, must coincide with the weak limit on $K_{m'}$. Uniqueness of weak limits (a consequence of the Hahn–Banach separation of points) forces $u^{(m)}\big|_{K_{m'}} = u^{(m')}$ almost everywhere. The pieces therefore glue to a well-defined function $u$ on $\Omega = \bigcup_m K_m$, square-integrable on each compact set, i.e. $u \in L^2_{\mathrm{loc}}(\Omega)$. [/guided] [/step] [step:Pass the equation $\bar\partial u_j = f$ to the limit in the sense of distributions] Fix a [test function](/page/Test%20Function) $\psi \in C_c^\infty(\Omega)$ and an index $\kappa \in \{1,\dots,n\}$. Choose $m$ so large that $\operatorname{supp}\psi \subset \Omega_m \subset K_m$. For every $j_k \ge m+1$ we have $\operatorname{supp}\psi \subset \Omega_{j_k}$, and the equation $\bar\partial u_{j_k} = f$ on $\Omega_{j_k}$ reads, in its $\kappa$-th component and in the distributional sense $\partial_{\bar z_\kappa} u_{j_k} = f_\kappa$, \begin{align*} \int_\Omega u_{j_k}\, \partial_{\bar z_\kappa}\psi\, d\mathcal{L}^{2n}(z) = -\int_\Omega f_\kappa\, \psi\, d\mathcal{L}^{2n}(z). \tag{4} \end{align*} The function $h := \partial_{\bar z_\kappa}\psi$ lies in $C_c^\infty(\Omega)$ with support in $K_m$, so $h \in L^2(K_m)$, and the map \begin{align*} \Lambda : L^2(K_m) &\to \mathbb{C}, \\ v &\mapsto \int_{K_m} v\, h\, d\mathcal{L}^{2n}(z) \end{align*} is a bounded linear functional by the Cauchy–Schwarz inequality, $|\Lambda(v)| \le \|v\|_{L^2(K_m)}\|h\|_{L^2(K_m)}$. The [weak convergence](/page/Weak%20Convergence) (3) gives $\Lambda(u_{j_k}) \to \Lambda(u)$, that is, the left-hand side of (4) converges to $\int_\Omega u\, \partial_{\bar z_\kappa}\psi\, d\mathcal{L}^{2n}$. The right-hand side of (4) is independent of $k$. Passing to the limit, \begin{align*} \int_\Omega u\, \partial_{\bar z_\kappa}\psi\, d\mathcal{L}^{2n}(z) = -\int_\Omega f_\kappa\, \psi\, d\mathcal{L}^{2n}(z). \end{align*} As $\psi \in C_c^\infty(\Omega)$ and $\kappa$ were arbitrary, $\partial_{\bar z_\kappa} u = f_\kappa$ in $\mathcal{D}'(\Omega)$ for each $\kappa$, i.e. $\bar\partial u = f$ in $\mathcal{D}'(\Omega)$. [guided] The relation $\bar\partial u_{j_k} = f$ holds in the distributional sense, meaning that for every smooth compactly supported $\psi$ the integration-by-parts identity (4) holds — the minus sign is the defining feature of the [distributional derivative](/page/Distributional%20Derivative) $\partial_{\bar z_\kappa}$, which moves the derivative onto the [test function](/page/Test%20Function). We want to take $k \to \infty$ inside this identity. The obstruction to passing to a limit is that $u_{j_k}$ converges only weakly, not strongly, so we cannot pass the limit through an arbitrary nonlinear expression. But (4) is *linear* in $u_{j_k}$, and it pairs $u_{j_k}$ against the fixed function $h = \partial_{\bar z_\kappa}\psi$. This is precisely the setting in which [weak convergence](/page/Weak%20Convergence) is designed to operate: [weak convergence](/page/Weak%20Convergence) in $L^2(K_m)$ means exactly that $\int_{K_m} u_{j_k}\, g\, d\mathcal{L}^{2n} \to \int_{K_m} u\, g\, d\mathcal{L}^{2n}$ for every $g \in L^2(K_m)$ — equivalently, against every bounded linear functional. Why is the relevant functional $\Lambda(v) = \int_{K_m} v\, h\, d\mathcal{L}^{2n}$ bounded on $L^2(K_m)$? Because $h = \partial_{\bar z_\kappa}\psi$ is smooth with compact support in $K_m$, hence square-integrable there, and Cauchy–Schwarz gives $|\Lambda(v)| \le \|v\|_{L^2(K_m)}\|h\|_{L^2(K_m)}$. We must also ensure the equation (4) is available for our subsequence: it holds whenever $\operatorname{supp}\psi \subset \Omega_{j_k}$, which is guaranteed for all $j_k \ge m+1$ once $m$ is chosen with $\operatorname{supp}\psi \subset \Omega_m$. Since only the tail $j_k \ge m+1$ is needed and the right-hand side of (4) does not depend on $k$ at all, taking $k \to \infty$ yields the identity for $u$. Because this holds for every [test function](/page/Test%20Function) and every component $\kappa$, the limit $u$ satisfies $\bar\partial u = f$ distributionally — no regularity of $u$ beyond $L^2_{\mathrm{loc}}$ is used or needed. [/guided] [/step] [step:Recover the weighted estimate by weighted lower semicontinuity and monotone convergence] Fix $m \ge 1$ and an index $\ell \ge m+1$. Then $\varphi_\ell$ is smooth on a neighbourhood of the compact set $K_m$, so there are constants \begin{align*} 0 < a_{m,\ell} := e^{-\sup_{K_m}\varphi_\ell} \le e^{-\varphi_\ell} \le e^{-\inf_{K_m}\varphi_\ell} =: b_{m,\ell} < \infty \quad \text{on } K_m. \end{align*} Equip $K_m$ with the weighted [Hilbert space](/page/Hilbert%20Space) $H_{m,\ell} := L^2\big(K_m,\, e^{-\varphi_\ell}\, d\mathcal{L}^{2n}\big)$, whose norm satisfies $a_{m,\ell}^{1/2}\|v\|_{L^2(K_m)} \le \|v\|_{H_{m,\ell}} \le b_{m,\ell}^{1/2}\|v\|_{L^2(K_m)}$. These equivalent norms induce the same bounded linear functionals on $K_m$, hence the same notion of [weak convergence](/page/Weak%20Convergence); so by (3), $u_{j_k} \rightharpoonup u$ in $H_{m,\ell}$. The [Lower Semicontinuity of the Norm](/theorems/215), applied in the [Banach space](/page/Banach%20Space) $H_{m,\ell}$, gives \begin{align*} \int_{K_m} |u|^2\, e^{-\varphi_\ell}\, d\mathcal{L}^{2n}(z) = \|u\|_{H_{m,\ell}}^2 \le \liminf_{k \to \infty} \|u_{j_k}\|_{H_{m,\ell}}^2 = \liminf_{k \to \infty} \int_{K_m} |u_{j_k}|^2\, e^{-\varphi_\ell}\, d\mathcal{L}^{2n}(z). \tag{5} \end{align*} For every $j_k \ge \ell$ the monotonicity $\varphi_{j_k} \le \varphi_\ell$ on $K_m$ gives $e^{-\varphi_\ell} \le e^{-\varphi_{j_k}}$, whence \begin{align*} \int_{K_m} |u_{j_k}|^2\, e^{-\varphi_\ell}\, d\mathcal{L}^{2n}(z) \le \int_{K_m} |u_{j_k}|^2\, e^{-\varphi_{j_k}}\, d\mathcal{L}^{2n}(z) \le \int_{\Omega_{j_k}} |u_{j_k}|^2\, e^{-\varphi_{j_k}}\, d\mathcal{L}^{2n}(z) \le C, \end{align*} using (1) in the last step. Since this holds for all sufficiently large $k$ (those with $j_k \ge \ell$), the $\liminf$ in (5) is at most $C$, and therefore \begin{align*} \int_{K_m} |u|^2\, e^{-\varphi_\ell}\, d\mathcal{L}^{2n}(z) \le C \qquad \text{for all } \ell \ge m+1. \tag{6} \end{align*} Now let $\ell \to \infty$ with $m$ fixed. Since $\varphi_\ell(z) \downarrow \varphi(z)$ for every $z$, the sequence $-\varphi_\ell$ increases to $-\varphi$, so the nonnegative [measurable functions](/page/Measurable%20Functions) $|u|^2 e^{-\varphi_\ell}$ increase pointwise to $|u|^2 e^{-\varphi}$ on $K_m$. The [Monotone Convergence Theorem](/theorems/509) and (6) yield \begin{align*} \int_{K_m} |u|^2\, e^{-\varphi}\, d\mathcal{L}^{2n}(z) = \lim_{\ell \to \infty} \int_{K_m} |u|^2\, e^{-\varphi_\ell}\, d\mathcal{L}^{2n}(z) \le C. \end{align*} Finally let $m \to \infty$. The functions $\mathbb{1}_{K_m}\, |u|^2 e^{-\varphi}$ increase pointwise to $\mathbb{1}_\Omega\, |u|^2 e^{-\varphi}$ because $K_m \uparrow \Omega$, so a second application of the [Monotone Convergence Theorem](/theorems/509) gives \begin{align*} \int_\Omega |u|^2\, e^{-\varphi}\, d\mathcal{L}^{2n}(z) = \lim_{m \to \infty} \int_{K_m} |u|^2\, e^{-\varphi}\, d\mathcal{L}^{2n}(z) \le C. \end{align*} In particular $u \in L^2(\Omega, e^{-\varphi}) \subseteq L^2_{\mathrm{loc}}(\Omega, e^{-\varphi})$. Combined with $\bar\partial u = f$ in $\mathcal{D}'(\Omega)$ from the previous step, this is exactly the assertion of the theorem. [guided] We now have a candidate solution $u$ and must show it inherits the weighted bound $C$ — but with respect to the *limit* weight $e^{-\varphi}$, whereas every uniform bound we possess is with respect to a *smooth* weight $e^{-\varphi_j}$. There are two gaps to bridge: [weak convergence](/page/Weak%20Convergence) does not preserve norms (only bounds them from below), and the target weight $e^{-\varphi}$ is neither smooth nor bounded below. We close the first gap with lower semicontinuity and the second with monotone convergence. First fix a smooth weight $e^{-\varphi_\ell}$ on a compact $K_m$. Why is this a legitimate Hilbert-space weight? Because $\varphi_\ell$ is smooth on a neighbourhood of the compact $K_m$ (this is why we take $\ell \ge m+1$), the weight is pinched between two positive constants $a_{m,\ell}$ and $b_{m,\ell}$. Consequently the weighted norm $\|\cdot\|_{H_{m,\ell}}$ is equivalent to the plain $L^2(K_m)$ norm. Equivalent norms have the same continuous dual, hence the same [weak topology](/page/Weak%20Topology), so the [weak convergence](/page/Weak%20Convergence) $u_{j_k} \rightharpoonup u$ established in $L^2(K_m)$ automatically holds in $H_{m,\ell}$ as well. Now we invoke the key inequality: the norm is weakly lower semicontinuous. Under [weak convergence](/page/Weak%20Convergence) the norm can only drop in the limit, so $\|u\|_{H_{m,\ell}}^2 \le \liminf_k \|u_{j_k}\|_{H_{m,\ell}}^2$. This is the direction we want — it lets us estimate the limit's weighted energy by the (uniformly bounded) energies of the approximants. To bound those, note that for $j_k \ge \ell$ the weight $e^{-\varphi_\ell}$ is *smaller* than $e^{-\varphi_{j_k}}$ (because $\varphi_\ell \ge \varphi_{j_k}$ when $\ell \le j_k$, the weights decreasing in the index), so replacing $e^{-\varphi_\ell}$ by the larger $e^{-\varphi_{j_k}}$ and enlarging the domain to $\Omega_{j_k}$ only increases the integral, which is then $\le C$ by the standing hypothesis (1). The tail $j_k \ge \ell$ is all the $\liminf$ sees, so we obtain (6): $\int_{K_m}|u|^2 e^{-\varphi_\ell} \le C$ for every admissible $\ell$. The estimate (6) holds for the smooth weights but not yet for the singular limit weight. Here monotonicity is decisive: since $\varphi_\ell \downarrow \varphi$, we have $-\varphi_\ell \uparrow -\varphi$ and therefore $e^{-\varphi_\ell} \uparrow e^{-\varphi}$, an increasing sequence of nonnegative functions. This is exactly the hypothesis of the [Monotone Convergence Theorem](/theorems/509): the integrals $\int_{K_m}|u|^2 e^{-\varphi_\ell}$ converge up to $\int_{K_m}|u|^2 e^{-\varphi}$, and since each is $\le C$, so is the limit. (Monotone convergence is essential and Fatou alone would also suffice here; what we may not do is dominate, since $e^{-\varphi}$ can be unbounded where $\varphi = -\infty$.) The final step exhausts $\Omega$: as $m \to \infty$ the indicators $\mathbb{1}_{K_m}$ increase to $\mathbb{1}_\Omega$, so $\mathbb{1}_{K_m}|u|^2 e^{-\varphi} \uparrow \mathbb{1}_\Omega |u|^2 e^{-\varphi}$, and one more application of the [Monotone Convergence Theorem](/theorems/509) delivers the global bound $\int_\Omega |u|^2 e^{-\varphi} \le C$. This shows $u$ has finite weighted energy — in particular $u \in L^2_{\mathrm{loc}}(\Omega, e^{-\varphi})$ — and together with the distributional identity $\bar\partial u = f$ from the previous step, the proof is complete. [/guided] [/step]