[proofplan] We prove the theorem by reducing the general bounded pseudoconvex domain and possibly singular plurisubharmonic weight to the smooth bounded case. On each smoothly bounded pseudoconvex exhaustion subdomain, we apply the uniform geometric form of the smooth weighted Ohsawa-Takegoshi estimate obtained from the Hörmander $L^2$ estimate for $\bar\partial$ with the singular normal weight. The constants are uniform because that smooth theorem gives a constant depending only on $\operatorname{diam}(\Omega)$ and $\sup_{z\in\Omega}|z_n|$, and the exhaustion does not increase either quantity. A normal-family and Fatou argument then passes to the limit and preserves both holomorphicity and the weighted $L^2$ bound. [/proofplan]

[step:Fix the geometric scales and choose a pseudoconvex exhaustion]Define \begin{align*} D &:= \operatorname{diam}(\Omega), & R &:= \sup_{z \in \Omega} |z_n|. \end{align*} Since $\Omega$ is bounded, both $D$ and $R$ are finite. We use the following precise exhaustion form: by the [smooth strictly plurisubharmonic exhaustion characterization of pseudoconvex domains](/page/Smooth%20Strictly%20Plurisubharmonic%20Exhaustion), there is a smooth strictly plurisubharmonic exhaustion map $\rho: \Omega \to \mathbb{R}$ whose regular sublevel sets are relatively compact smoothly bounded strongly pseudoconvex domains. Choose an increasing sequence of regular values $(a_j)_{j=1}^{\infty}$ with $a_j \to \infty$ and define \begin{align*} \Omega_j := \{z \in \Omega : \rho(z) < a_j\}. \end{align*} Then $(\Omega_j)_{j=1}^{\infty}$ is an increasing sequence of smoothly bounded pseudoconvex domains satisfying \begin{align*} \overline{\Omega_j} \subset \Omega_{j+1}, \qquad \bigcup_{j=1}^{\infty} \Omega_j = \Omega. \end{align*} For each $j \in \mathbb{N}$ define the hyperplane slice \begin{align*} H_j := \Omega_j \cap \{z_n = 0\}. \end{align*} Then $H_j \subset H$, $\operatorname{diam}(\Omega_j) \le D$, and $\sup_{z \in \Omega_j} |z_n| \le R$. For an open complex submanifold $G \subset \mathbb{C}^m$ and a real-valued measurable weight $\psi: G \to [-\infty,\infty)$, define \begin{align*} A^2(G,e^{-\psi}) := \left\{g: G \to \mathbb{C} \mid g \text{ is holomorphic and } \int_G |g|^2 e^{-\psi}\,d\mathcal{L}^{2m} < \infty\right\}. \end{align*} When $G \subset H$ has complex dimension $n-1$, the measure in this definition is the induced Lebesgue measure $\mathcal{L}^{2n-2}$ on the hyperplane $\{z_n=0\}$. We use the standard convention that a plurisubharmonic function is upper semicontinuous and is not identically $-\infty$ on any connected component of its domain; in particular, it is locally bounded above on compact subsets.[/step]

[guided]The first task is to isolate the geometric quantities on which the final constant is allowed to depend. Define \begin{align*} D &:= \operatorname{diam}(\Omega), & R &:= \sup_{z \in \Omega} |z_n|. \end{align*} The boundedness hypothesis on $\Omega$ gives $D < \infty$ and $R < \infty$. These two numbers control the size of the exhaustion domains and the scale in the normal direction to the hypersurface $\{z_n = 0\}$. Because $\Omega \subset \mathbb{C}^n$ is bounded and pseudoconvex, the [smooth strictly plurisubharmonic exhaustion characterization of pseudoconvex domains](/page/Smooth%20Strictly%20Plurisubharmonic%20Exhaustion) supplies a smooth strictly plurisubharmonic exhaustion map $\rho: \Omega \to \mathbb{R}$. Choose regular values $a_j \to \infty$ and set \begin{align*} \Omega_j := \{z \in \Omega : \rho(z) < a_j\}. \end{align*} Because each $a_j$ is a regular value, $\Omega_j$ has smooth boundary; because $\rho$ is plurisubharmonic, each $\Omega_j$ is pseudoconvex; because $\rho$ is an exhaustion, $\overline{\Omega_j} \subset \Omega_{j+1}$ after passing to a strictly increasing subsequence of regular values and $\bigcup_{j=1}^{\infty}\Omega_j=\Omega$. Thus the precise exhaustion properties used below follow from the cited theorem rather than from a bare smoothing of the original boundary. For each exhaustion domain we set \begin{align*} H_j := \Omega_j \cap \{z_n = 0\}. \end{align*} This is the part of the hypersurface visible inside $\Omega_j$. Since $\Omega_j \subset \Omega$, we have $H_j \subset H$, and the two geometric bounds are inherited: \begin{align*} \operatorname{diam}(\Omega_j) \le D, \qquad \sup_{z \in \Omega_j} |z_n| \le R. \end{align*} These inequalities are the reason the constants in the smooth estimates below do not depend on $j$.[/guided]

[step:Regularise the plurisubharmonic weight from above on each exhaustion domain]For each $j \in \mathbb{N}$ choose a sequence $(\varphi_{j,k})_{k=1}^{\infty}$ of smooth plurisubharmonic functions on a neighbourhood of $\overline{\Omega_j}$ such that \begin{align*} \varphi_{j,k} \downarrow \varphi \quad \text{pointwise on } \Omega_j. \end{align*} This regularisation follows from the [regularisation theorem for plurisubharmonic functions](/page/Regularization%20of%20Plurisubharmonic%20Functions), applied on the relatively compact subdomain $\Omega_j \Subset \Omega$. Its hypotheses hold because $\Omega$ is pseudoconvex, $\Omega_j$ has compact closure in $\Omega$, and $\varphi$ is plurisubharmonic on $\Omega$. Since $\varphi_{j,k} \ge \varphi$ on $\Omega_j$, the weights satisfy \begin{align*} e^{-\varphi_{j,k}} \le e^{-\varphi} \quad \text{on } \Omega_j. \end{align*} Hence, for every $j,k \in \mathbb{N}$, \begin{align*} \int_{H_j} |f|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n-2} \le \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2} < \infty. \end{align*}[/step]

[guided]We need smooth weights because the next step invokes a smooth [extension theorem](/theorems/59). For a fixed $j$, the closure $\overline{\Omega_j}$ is compactly contained in $\Omega$, and $\Omega$ is pseudoconvex. The [regularisation theorem for plurisubharmonic functions](/page/Regularization%20of%20Plurisubharmonic%20Functions) therefore applies to the plurisubharmonic function $\varphi$ on a neighbourhood of $\overline{\Omega_j}$. It gives smooth plurisubharmonic functions $(\varphi_{j,k})_{k=1}^{\infty}$, each defined on a neighbourhood of $\overline{\Omega_j}$, such that \begin{align*} \varphi_{j,k} \downarrow \varphi \quad \text{pointwise on } \Omega_j. \end{align*} The monotone decrease means $\varphi_{j,k} \ge \varphi$ for every $k$. Exponentiating reverses the sign in the exponent but preserves the inequality in the needed direction: \begin{align*} e^{-\varphi_{j,k}} \le e^{-\varphi} \quad \text{on } \Omega_j. \end{align*} Restricting this inequality to $H_j \subset H$ and using the assumption $f \in A^2(H,e^{-\varphi})$ gives \begin{align*} \int_{H_j} |f|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n-2} \le \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2} < \infty. \end{align*} Thus the same boundary datum is square-integrable for each smooth approximating weight.[/guided]

[step:Apply the smooth weighted extension estimate with constants independent of the exhaustion]Apply the [uniform geometric smooth Ohsawa-Takegoshi extension estimate for coordinate hyperplane sections](/page/Smooth%20Ohsawa-Takegoshi%20Extension%20Theorem) to the smoothly bounded pseudoconvex domain $\Omega_j$, the coordinate hyperplane slice $H_j=\Omega_j\cap\{z_n=0\}$, the smooth plurisubharmonic weight $\varphi_{j,k}$, and the [holomorphic function](/page/Holomorphic%20Function) $f|_{H_j}: H_j \to \mathbb{C}$. We use the formulation of the smooth theorem in which the datum is prescribed on the analytic set $\Omega_j\cap\{z_n=0\}$ with the induced Lebesgue measure on the hyperplane; no separate smooth-boundary hypothesis on $H_j$ is required. Its hypotheses hold by construction: $\Omega_j$ is smoothly bounded and pseudoconvex, $H_j$ is exactly the coordinate hyperplane section allowed by the theorem, $\varphi_{j,k}$ is smooth plurisubharmonic near $\overline{\Omega_j}$, and the preceding estimate gives $f|_{H_j} \in A^2(H_j,e^{-\varphi_{j,k}})$. Therefore there exists a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} F_{j,k}: \Omega_j &\to \mathbb{C} \end{align*} such that $F_{j,k}|_{H_j}=f|_{H_j}$ and \begin{align*} \int_{\Omega_j} |F_{j,k}|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n} \le C(D,R) \int_{H_j} |f|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n-2}. \end{align*} Here $C(D,R)>0$ is the constant supplied by the uniform geometric smooth estimate. The precise constant statement used is: for every smoothly bounded pseudoconvex $U\subset\mathbb{C}^n$, every smooth plurisubharmonic weight $\psi$ near $\overline U$, and every datum on $U\cap\{z_n=0\}$, the extension constant is bounded by a function of $\operatorname{diam}(U)$ and $\sup_{z\in U}|z_n|$ alone. Applying this with $U=\Omega_j$ and $\psi=\varphi_{j,k}$ gives a constant depending only on the two bounds $\operatorname{diam}(\Omega_j) \le D$ and $\sup_{z \in \Omega_j}|z_n| \le R$, and not on $j$, $k$, $f$, or the particular regularised weight. Using $e^{-\varphi_{j,k}} \le e^{-\varphi}$ on $H_j$ gives \begin{align*} \int_{\Omega_j} |F_{j,k}|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*}[/step]

[guided]This is the only place where the deep analytic input enters. The cited smooth [extension theorem](/theorems/59) is the uniform geometric version proved from the Hörmander $L^2$ estimate for the $\bar\partial$-operator with a logarithmic pole in the normal coordinate $z_n$. Its formulation applies to the analytic coordinate hyperplane section $U\cap\{z_n=0\}$ of a smoothly bounded pseudoconvex domain $U$ and integrates the datum with respect to the induced Lebesgue measure on that hyperplane; it does not require the slice itself to have smooth boundary as a domain in $\mathbb{C}^{n-1}$. We apply it on $\Omega_j$ with weight $\varphi_{j,k}$. Let us check the hypotheses one by one. The domain $\Omega_j$ is smoothly bounded and pseudoconvex by the exhaustion chosen above. The slice $H_j=\Omega_j\cap\{z_n=0\}$ is exactly the coordinate hyperplane section allowed by the theorem. The weight $\varphi_{j,k}$ is smooth and plurisubharmonic on a neighbourhood of $\overline{\Omega_j}$ by construction. The boundary datum is the map \begin{align*} f|_{H_j}: H_j &\to \mathbb{C}, \end{align*} which is holomorphic because $f$ is holomorphic on $H$ and $H_j \subset H$. Its weighted square integrability follows from \begin{align*} \int_{H_j} |f|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n-2} \le \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2} < \infty. \end{align*} Thus the smooth theorem supplies a holomorphic map \begin{align*} F_{j,k}: \Omega_j &\to \mathbb{C} \end{align*} with $F_{j,k}|_{H_j}=f|_{H_j}$ and \begin{align*} \int_{\Omega_j} |F_{j,k}|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n} \le C(D,R) \int_{H_j} |f|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n-2}. \end{align*} The constant is uniform in $j$ and $k$ by the precise constant statement in the smooth theorem: for a smoothly bounded pseudoconvex domain $U$, a smooth plurisubharmonic weight $\psi$ near $\overline U$, and a datum on $U\cap\{z_n=0\}$, the extension constant is bounded by a function only of $\operatorname{diam}(U)$ and $\sup_{z\in U}|z_n|$. For every exhaustion domain these quantities are bounded by $D$ and $R$, so the same number $C(D,R)$ works for all $j$ and $k$. Combining this with $e^{-\varphi_{j,k}} \le e^{-\varphi}$ on $H_j$ yields \begin{align*} \int_{\Omega_j} |F_{j,k}|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*}[/guided]

[step:Pass from smooth weights to the original weight on each fixed exhaustion domain]Fix $j \in \mathbb{N}$. For every compact set $K \subset \Omega_j$, the smooth function $\varphi_{j,1}: \Omega_j \to \mathbb{R}$ is bounded above on $K$; choose \begin{align*} M_{K,j} := \sup_{z \in K}\varphi_{j,1}(z) < \infty. \end{align*} Because $\varphi_{j,k} \downarrow \varphi$, we have $\varphi_{j,k} \le \varphi_{j,1} \le M_{K,j}$ on $K$ for every $k \in \mathbb{N}$. Therefore $1 \le e^{M_{K,j}}e^{-\varphi_{j,k}}$ on $K$, and \begin{align*} \int_K |F_{j,k}|^2\,d\mathcal{L}^{2n} \le e^{M_{K,j}}\int_{\Omega_j} |F_{j,k}|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n}. \end{align*} The right-hand side is bounded independently of $k$. The [local $L^2$ mean-value estimate for holomorphic functions](/page/Local%20L2%20Mean%20Value%20Estimate) therefore makes $(F_{j,k})_{k=1}^{\infty}$ locally uniformly bounded on $\Omega_j$. By [Montel's theorem](/page/Montel%27s%20Theorem), after passing to a subsequence $(k_\ell)_{\ell=1}^{\infty}$, the functions $F_{j,k_\ell}$ converge locally uniformly on $\Omega_j$ to a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} F_j: \Omega_j &\to \mathbb{C}. \end{align*} The convergence is locally uniform on $H_j$, so $F_j|_{H_j}=f|_{H_j}$. Since $e^{-\varphi_{j,k_\ell}} \uparrow e^{-\varphi}$ pointwise as $\ell\to\infty$ and $F_{j,k_\ell} \to F_j$ pointwise, [Fatou's lemma](/page/Fatou%27s%20Lemma) applied to the non-negative functions $|F_{j,k_\ell}|^2e^{-\varphi_{j,k_\ell}}$ gives \begin{align*} \int_{\Omega_j} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le \liminf_{\ell\to\infty}\int_{\Omega_j}|F_{j,k_\ell}|^2e^{-\varphi_{j,k_\ell}}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*}[/step]

[guided]Fix $j \in \mathbb{N}$. We first need compact-uniform control of the holomorphic functions $F_{j,k}$ while the smooth weights decrease to the original possibly singular weight. Let $K \subset \Omega_j$ be compact. Since $\varphi_{j,1}$ is smooth on a neighbourhood of $\overline{\Omega_j}$, it is bounded above on $K$; define \begin{align*} M_{K,j} := \sup_{z \in K}\varphi_{j,1}(z) < \infty. \end{align*} The monotonicity $\varphi_{j,k} \downarrow \varphi$ gives $\varphi_{j,k} \le \varphi_{j,1}$ for every $k$, hence $\varphi_{j,k} \le M_{K,j}$ on $K$. This is the needed uniform upper bound in $k$. It implies $1 \le e^{M_{K,j}}e^{-\varphi_{j,k}}$ on $K$, so \begin{align*} \int_K |F_{j,k}|^2\,d\mathcal{L}^{2n} \le e^{M_{K,j}}\int_K |F_{j,k}|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n} \le e^{M_{K,j}}\int_{\Omega_j} |F_{j,k}|^2 e^{-\varphi_{j,k}}\,d\mathcal{L}^{2n}. \end{align*} The last integral is bounded independently of $k$ by the estimate from the previous step. The [local $L^2$ mean-value estimate for holomorphic functions](/page/Local%20L2%20Mean%20Value%20Estimate) applies because each $F_{j,k}$ is holomorphic on $\Omega_j$ and gives sup-norm bounds on compact subsets strictly inside $\Omega_j$. Hence the family $(F_{j,k})_{k=1}^{\infty}$ is locally uniformly bounded. By [Montel's theorem](/page/Montel%27s%20Theorem), a locally uniformly bounded family of holomorphic functions on a domain has a subsequence converging locally uniformly to a [holomorphic function](/page/Holomorphic%20Function). Passing to such a subsequence, write it as $(F_{j,k_\ell})_{\ell=1}^{\infty}$. We obtain a holomorphic map \begin{align*} F_j: \Omega_j &\to \mathbb{C} \end{align*} with $F_{j,k_\ell} \to F_j$ locally uniformly on $\Omega_j$. Since $H_j \subset \Omega_j$ and $F_{j,k_\ell}|_{H_j}=f|_{H_j}$ for every $\ell$, locally [uniform convergence](/page/Uniform%20Convergence) restricted to $H_j$ gives $F_j|_{H_j}=f|_{H_j}$. It remains to preserve the weighted estimate. Pointwise on $\Omega_j$, the convergence $F_{j,k_\ell} \to F_j$ and the monotone convergence $e^{-\varphi_{j,k_\ell}} \uparrow e^{-\varphi}$ as $\ell\to\infty$ imply \begin{align*} |F_{j,k_\ell}|^2e^{-\varphi_{j,k_\ell}} \to |F_j|^2e^{-\varphi}. \end{align*} These functions are non-negative and measurable, so [Fatou's lemma](/page/Fatou%27s%20Lemma) gives \begin{align*} \int_{\Omega_j} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le \liminf_{\ell \to \infty}\int_{\Omega_j} |F_{j,k_\ell}|^2 e^{-\varphi_{j,k_\ell}}\,d\mathcal{L}^{2n}. \end{align*} Using the uniform estimate for the right-hand side yields \begin{align*} \int_{\Omega_j} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*}[/guided]

[step:Extract a global holomorphic limit and preserve the trace]The estimates for $F_j$ are uniform in $j$: \begin{align*} \int_{\Omega_j} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*} Let $K \subset \Omega$ be compact. Choose $j_K \in \mathbb{N}$ such that $K \subset \Omega_j$ for all $j \ge j_K$. By the plurisubharmonic convention stated at the start, $\varphi$ is locally bounded above on compact subsets; combining this local upper bound on $K$ with the local mean-value estimate gives local uniform boundedness of $(F_j)_{j \ge j_K}$ on $K$. A diagonal application of Montel's theorem gives a subsequence, still denoted $(F_j)$, converging locally uniformly on $\Omega$ to a holomorphic map \begin{align*} F: \Omega &\to \mathbb{C}. \end{align*} For every compact set $K_H \subset H$, choose $j_H \in \mathbb{N}$ such that $K_H \subset H_j$ for $j \ge j_H$. Since $F_j|_{H_j}=f|_{H_j}$ and $F_j \to F$ locally uniformly on $\Omega$, the restriction satisfies $F|_{K_H}=f|_{K_H}$. As $K_H$ was arbitrary, $F|_H=f$.[/step]

[guided]We now remove the exhaustion. The estimate \begin{align*} \int_{\Omega_j} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2} \end{align*} is independent of $j$. To obtain a holomorphic limit on all of $\Omega$, fix a compact set $K \subset \Omega$. Since $(\Omega_j)$ exhausts $\Omega$, there is $j_K \in \mathbb{N}$ such that $K \subset \Omega_j$ whenever $j \ge j_K$. By the convention stated in the first step, the plurisubharmonic function $\varphi$ is upper semicontinuous and not identically $-\infty$ on any component; hence it is locally bounded above. Therefore choose $M_K \in \mathbb{R}$ with $\varphi \le M_K$ on $K$. Then \begin{align*} \int_K |F_j|^2\,d\mathcal{L}^{2n} \le e^{M_K}\int_{\Omega_j} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \end{align*} for all $j \ge j_K$. The right-hand side is uniformly bounded. The [local $L^2$ mean-value estimate for holomorphic functions](/page/Local%20L2%20Mean%20Value%20Estimate) converts this local $L^2$ bound into a local sup-norm bound on smaller compact subsets. [Montel's theorem](/page/Montel%27s%20Theorem) then gives a locally uniformly convergent subsequence on $K$. Applying this on a countable compact exhaustion of $\Omega$ and diagonalising gives a subsequence, relabeled as $(F_j)$, and a holomorphic map \begin{align*} F: \Omega &\to \mathbb{C} \end{align*} such that $F_j \to F$ locally uniformly on $\Omega$ along a subsequence. It remains to check that the limit has the prescribed trace. Let $K_H \subset H$ be compact. Since $H_j$ exhausts $H$, there is $j_H \in \mathbb{N}$ such that $K_H \subset H_j$ for all $j \ge j_H$. For those $j$, the construction gives $F_j|_{K_H}=f|_{K_H}$. Passing to the locally uniform limit on a neighbourhood of $K_H$ yields $F|_{K_H}=f|_{K_H}$. Because every point of $H$ lies in such a compact neighbourhood inside $H$, we conclude $F|_H=f$.[/guided]

[step:Use Fatou's lemma to obtain the global weighted estimate]Choose an increasing compact exhaustion $(K_m)_{m=1}^{\infty}$ of $\Omega$ with $K_m \subset \Omega_m$. We use the subsequence obtained in the previous step and relabeled as $(F_j)$. For each fixed $m$, local [uniform convergence](/page/Uniform%20Convergence) gives pointwise convergence $|F_j|^2 e^{-\varphi} \to |F|^2 e^{-\varphi}$ on $K_m$. Let $\mathcal{B}(K_m)$ denote the Borel $\sigma$-algebra of the compact [metric space](/page/Metric%20Space) $K_m$. [Fatou's lemma](/page/Fatou%27s%20Lemma) for the [measure space](/page/Measure%20Space) $(K_m,\mathcal{B}(K_m),\mathcal{L}^{2n})$ gives \begin{align*} \int_{K_m} |F|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le \liminf_{j \to \infty} \int_{K_m} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n}. \end{align*} For $j$ large enough that $K_m \subset \Omega_j$, the last integral is bounded by the uniform estimate on $\Omega_j$, hence \begin{align*} \int_{K_m} |F|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*} Letting $m \to \infty$ and applying the [monotone convergence theorem](/page/Monotone%20Convergence%20Theorem) to the increasing sequence of non-negative functions $\mathbb{1}_{K_m}|F|^2 e^{-\varphi}$ yields \begin{align*} \int_{\Omega} |F|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*} Thus $F \in A^2(\Omega,e^{-\varphi})$, $F|_H=f$, and the theorem holds with $C_\Omega := C(D,R)$.[/step]

[guided]The final step is to pass the uniform estimate to the limit. Choose compact sets $(K_m)_{m=1}^{\infty}$ with \begin{align*} K_m \subset K_{m+1}, \qquad \bigcup_{m=1}^{\infty} K_m = \Omega, \qquad K_m \subset \Omega_m. \end{align*} For a fixed $m$, the locally [uniform convergence](/page/Uniform%20Convergence) $F_j \to F$ implies pointwise convergence on $K_m$: \begin{align*} |F_j|^2 e^{-\varphi} \to |F|^2 e^{-\varphi}. \end{align*} Let $\mathcal{B}(K_m)$ denote the Borel $\sigma$-algebra of $K_m$. These functions are non-negative and measurable with respect to $\mathcal{B}(K_m)$, so [Fatou's lemma](/page/Fatou%27s%20Lemma) on $(K_m,\mathcal{B}(K_m),\mathcal{L}^{2n})$ applies and gives \begin{align*} \int_{K_m} |F|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le \liminf_{j \to \infty} \int_{K_m} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n}. \end{align*} When $j$ is large enough, $K_m \subset \Omega_j$, so enlarging the domain of integration from $K_m$ to $\Omega_j$ is valid because the integrand is non-negative. Therefore \begin{align*} \int_{K_m} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le \int_{\Omega_j} |F_j|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*} Combining the two inequalities gives \begin{align*} \int_{K_m} |F|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*} Finally, the functions $\mathbb{1}_{K_m}|F|^2e^{-\varphi}$ increase pointwise to $|F|^2e^{-\varphi}$ on $\Omega$. The [monotone convergence theorem](/page/Monotone%20Convergence%20Theorem) gives \begin{align*} \int_{\Omega} |F|^2 e^{-\varphi}\,d\mathcal{L}^{2n} \le C(D,R) \int_H |f|^2 e^{-\varphi}\,d\mathcal{L}^{2n-2}. \end{align*} Thus $F$ is holomorphic on $\Omega$, belongs to $A^2(\Omega,e^{-\varphi})$, restricts to $f$ on $H$, and satisfies the required estimate. Setting $C_\Omega := C(D,R)$ gives the stated dependence of the constant and completes the proof.[/guided]