[proofplan] We use the analytic form of pseudoconvexity: $\Omega$ admits a continuous plurisubharmonic exhaustion. Given any sequence escaping compact subsets of $\Omega$, we choose a locally finite family of small disjoint balls around a subsequence and solve carefully weighted $\bar\partial$-problems with logarithmic poles at the chosen centers. The logarithmic pole forces each correction term to vanish at its center, while the exhaustion weight lets us make the resulting holomorphic functions small on prescribed compact subsets. A rapidly convergent series then produces one [holomorphic function](/page/Holomorphic%20Function) whose values become unbounded along the chosen escaping subsequence; this proves both the boundary blow-up statement and holomorphic convexity. [/proofplan]

[step:Choose an escaping subsequence with separated coordinate balls]Let $\rho:\Omega \to \mathbb{R}$ be a continuous [plurisubharmonic exhaustion](/page/Plurisubharmonic%20Exhaustion) of $\Omega$, meaning that for each $c \in \mathbb{R}$ the set \begin{align*} K_c := \{z \in \Omega : \rho(z) \leq c\} \end{align*} is compact in $\Omega$, after increasing $\rho$ by a constant if necessary. This uses the [analytic characterization of pseudoconvex domains](/page/Pseudoconvex%20Domain) by continuous plurisubharmonic exhaustions. Let $(z_j)_{j=1}^{\infty}$ be any sequence in $\Omega$ with no accumulation point in $\Omega$. Since $\rho$ is an exhaustion, $\rho(z_j) \to \infty$ along a subsequence. Passing to that subsequence and relabelling, choose points $a_j \in \Omega$ such that \begin{align*} \rho(a_j) \geq j^2 \end{align*} for every $j \in \mathbb{N}$. For each $j$, choose $r_j > 0$ so that the Euclidean ball \begin{align*} B_j := B(a_j, r_j) = \{z \in \mathbb{C}^n : |z-a_j| < r_j\} \end{align*} satisfies $\overline{B}(a_j, r_j) \subset \Omega$. By passing to a further subsequence and shrinking the radii, we may also assume that the closed balls $\overline{B}(a_j,r_j)$ are pairwise disjoint and locally finite in $\Omega$. For each $j$, choose a smooth cutoff function \begin{align*} \chi_j:\Omega &\to [0,1] \end{align*} such that $\chi_j=1$ on $B(a_j,r_j/3)$ and $\operatorname{supp}\chi_j \subset B(a_j,2r_j/3)$.[/step]

[guided]The exhaustion $\rho$ is the device that records escape to the boundary, including escape to infinity when $\Omega$ is unbounded. Since each sublevel set $K_c=\{\rho\le c\}$ is compact in $\Omega$, a sequence with no compactly convergent subsequence must eventually leave every $K_c$. Thus, after passing to a subsequence, we can force $\rho(a_j)$ to grow as fast as we like; the concrete lower bound $\rho(a_j)\ge j^2$ is chosen only to make later convergence estimates simple. We next isolate the points $a_j$ in pairwise disjoint coordinate balls. Because each $a_j$ lies in the [open set](/page/Open%20Set) $\Omega$, there is a positive Euclidean distance from $a_j$ to $\mathbb{C}^n\setminus\Omega$ on a sufficiently small scale, so we can choose $r_j>0$ with $\overline{B}(a_j,r_j)\subset\Omega$. Since the sequence has no accumulation point in $\Omega$, a subsequence can be chosen so that these balls are pairwise disjoint and locally finite. Local finiteness means that each compact subset of $\Omega$ intersects only finitely many of the balls. Finally, the smooth cutoff \begin{align*} \chi_j:\Omega &\to [0,1] \end{align*} is chosen to equal $1$ near $a_j$ and to vanish outside a slightly larger ball. The annulus on which $\bar\partial\chi_j$ can be nonzero is separated from $a_j$; this separation is essential because the weight used in the $\bar\partial$ equation will have a logarithmic pole at $a_j$.[/guided]

[step:Construct holomorphic functions with prescribed large values and small compact norms]We prove the following interpolation estimate. [claim:Weighted interpolation near an escaping point] For every $j \in \mathbb{N}$, every number $M_j>0$, and every finite set $P_j\subset\Omega\setminus\operatorname{supp}\chi_j$ with $a_j\notin P_j$, there exists $G_j \in \mathcal{O}(\Omega)$ such that \begin{align*} G_j(a_j)=M_j \end{align*} and \begin{align*} G_j(p)=0 \end{align*} for every $p\in P_j$. Define the annulus associated to the cutoff $\chi_j$ by \begin{align*} A_j:=\{z\in\Omega : r_j/3 \le |z-a_j|\le 2r_j/3\}. \end{align*} Moreover, let $L\subset\Omega$ be compact, suppose $L\cap\operatorname{supp}\chi_j=\varnothing$, and suppose there is a number $\gamma>0$ such that \begin{align*} \inf_{w\in A_j}\rho(w)\geq \sup_{z\in L}\rho(z)+\gamma. \end{align*} Then, in the construction below, one may choose the weight coefficient $N_j>0$ sufficiently large so that $G_j$ satisfies any prescribed bound \begin{align*} \sup_{z\in L}|G_j(z)|\le \varepsilon_j \end{align*} with $\varepsilon_j>0$. [/claim] [proof] Fix $j$. Let $\mathcal{L}^{2n}$ denote Lebesgue measure on $\mathbb{C}^n\cong\mathbb{R}^{2n}$, and let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $(0,\infty)$. Define the smooth compactly supported $(0,1)$-form \begin{align*} \alpha_j := M_j\,\bar\partial\chi_j . \end{align*} Since $\chi_j=1$ on $B(a_j,r_j/3)$ and $\operatorname{supp}\chi_j\subset B(a_j,2r_j/3)$, the support of $\alpha_j$ is contained in the annulus \begin{align*} A_j:=\{z\in\Omega : r_j/3 \le |z-a_j|\le 2r_j/3\}. \end{align*} In particular, $\alpha_j$ vanishes in a neighbourhood of $a_j$. Let $P_j=\{p_{j,1},\dots,p_{j,m_j}\}$, where $m_j\in\mathbb{N}\cup\{0\}$ is the cardinality of $P_j$. For $N_j>0$, define the plurisubharmonic weight \begin{align*} \varphi_j:\Omega &\to [-\infty,\infty)\\ z &\mapsto N_j\rho(z)+2(n+1)\log|z-a_j|+2(n+1)\sum_{q=1}^{m_j}\log|z-p_{j,q}|. \end{align*} The functions $z\mapsto \log|z-a_j|$ and $z\mapsto \log|z-p_{j,q}|$ are plurisubharmonic on $\Omega$, and therefore $\varphi_j$ is plurisubharmonic. We use [Hörmander's $L^2$ estimate for the $\bar\partial$-equation](/page/Hormander%20L2%20Estimate) in its plurisubharmonic-weight form: if $\psi:\Omega\to[-\infty,\infty)$ is plurisubharmonic, possibly non-smooth and with logarithmic poles, and $\beta$ is a smooth compactly supported $\bar\partial$-closed $(0,1)$-form with finite weighted norm, then there is a measurable function $u:\Omega\to\mathbb{C}$ with $\bar\partial u=\beta$ in the sense of distributions and finite weighted $L^2$ norm. The constant in this form is the universal constant supplied by the theorem after the standard approximation of plurisubharmonic weights from above; in particular, it does not depend on adding the coefficient $N_j$ to the exhaustion term except through the displayed weighted norm of $\beta$. Applying the theorem to $\psi=\varphi_j$ and $\beta=\alpha_j$ is legitimate because $\Omega$ is pseudoconvex by assumption, $\varphi_j$ is plurisubharmonic, $\alpha_j$ is smooth, compactly supported, and $\bar\partial$-closed, and all logarithmic singularities lie outside $A_j\supset\operatorname{supp}\alpha_j$. On the fixed compact annulus $A_j$, all logarithmic factors in $\varphi_j$ are bounded above and below. Hence the weighted norm of $\alpha_j$ is bounded by a constant times $M_j^2\exp(-N_j\inf_{w\in A_j}\rho(w))$, where the constant depends on $j$, $P_j$, $\chi_j$, and $A_j$, but not on $N_j$. Hörmander's estimate gives a measurable function \begin{align*} u_j:\Omega &\to \mathbb{C} \end{align*} such that \begin{align*} \bar\partial u_j=\alpha_j \end{align*} in the sense of distributions and a constant $C_j>0$, independent of $N_j$, with \begin{align*} \int_{\Omega}|u_j(z)|^2 e^{-\varphi_j(z)}\,d\mathcal{L}^{2n}(z) \leq C_jM_j^2\exp\left(-N_j\inf_{w\in A_j}\rho(w)\right). \end{align*} Here $C_j$ depends only on $j$, the finite set $P_j$, the cutoff $\chi_j$, the annulus $A_j$, and the bounded logarithmic factors on $A_j$; it is independent of $N_j$ because the only $N_j$-dependence on $A_j$ is the explicit factor $e^{-N_j\rho}$ already displayed. Set \begin{align*} G_j:\Omega &\to \mathbb{C}\\ z &\mapsto M_j\chi_j(z)-u_j(z). \end{align*} Then \begin{align*} \bar\partial G_j = M_j\bar\partial\chi_j-\bar\partial u_j = \alpha_j-\alpha_j = 0, \end{align*} so $G_j\in\mathcal{O}(\Omega)$. Since $\alpha_j=0$ on $B(a_j,r_j/3)$, the equation $\bar\partial u_j=0$ holds there, and hence $u_j$ is holomorphic on $B(a_j,r_j/3)$. The finite weighted norm forces $u_j(a_j)=0$. Indeed, if $u_j(a_j)\ne0$, then by continuity of the [holomorphic function](/page/Holomorphic%20Function) $u_j$ there are constants $\delta>0$ and $c>0$ such that $|u_j(z)|\ge c$ for $z\in B(a_j,\delta)$. On this ball, \begin{align*} e^{-\varphi_j(z)} = e^{-N_j\rho(z)}|z-a_j|^{-2(n+1)}. \end{align*} Since $\rho$ is continuous, $e^{-N_j\rho}$ is bounded below by a positive constant on $\overline{B}(a_j,\delta)$. Thus \begin{align*} \int_{B(a_j,\delta)} |u_j(z)|^2 e^{-\varphi_j(z)}\,d\mathcal{L}^{2n}(z) \ge C\int_{B(a_j,\delta)} |z-a_j|^{-2(n+1)}\,d\mathcal{L}^{2n}(z) = \infty, \end{align*} because in real dimension $2n$ the radial integral contains \begin{align*} \int_0^\delta r^{2n-1}r^{-2(n+1)}\,d\mathcal{L}^1(r) = \int_0^\delta r^{-3}\,d\mathcal{L}^1(r). \end{align*} This contradicts the finite weighted norm, so $u_j(a_j)=0$. Since $\chi_j(a_j)=1$, we obtain \begin{align*} G_j(a_j)=M_j. \end{align*} The same argument at each point $p\in P_j$ gives $u_j(p)=0$, because $\alpha_j$ vanishes near $p$ and the weight contains the factor $|z-p|^{-2(n+1)}$. Since $p\notin\operatorname{supp}\chi_j$, we have $\chi_j(p)=0$, and therefore \begin{align*} G_j(p)=0 \end{align*} for every $p\in P_j$. It remains to record the compact smallness under the separation hypothesis in the claim. Let $L\subset\Omega$ be compact and assume that there is $\gamma>0$ with \begin{align*} \inf_{w\in A_j}\rho(w)\geq \sup_{z\in L}\rho(z)+\gamma. \end{align*} Together with the hypothesis $L\cap\operatorname{supp}\chi_j=\varnothing$, this permits us to cover $L$ by finitely many balls $B(\zeta_m,s_m)$ compactly contained in $\Omega\setminus\operatorname{supp}\chi_j$. On each such ball, $G_j=-u_j$ is holomorphic. Let $V\subset\Omega\setminus\operatorname{supp}\chi_j$ be the finite union of slightly larger balls used for this cover, chosen so that $L\Subset V$ and $\overline V\cap\operatorname{supp}\chi_j=\varnothing$. Define \begin{align*} R_L:=\sup_{z\in V}\rho(z). \end{align*} By shrinking $V$ if necessary and using the separation inequality, we may arrange $R_L\leq \sup_{z\in L}\rho(z)+\gamma/2$. Since $\overline V$ is compact and avoids $a_j$, the logarithmic factor $|z-a_j|^{-2(n+1)}$ is bounded above and below on $V$ by positive constants depending only on $j$ and $L$. For points $p\in P_j\cap V$, the factors $|z-p|^{-2(n+1)}$ in $e^{-\varphi_j}$ only strengthen the weighted integral near $p$; equivalently, when passing from the weighted estimate to an unweighted estimate on $V$, we need only an upper bound for the corresponding factors $|z-p|^{2(n+1)}$ in $e^{\varphi_j}$, and this upper bound holds on the compact set $\overline V$. Hence there is a constant $D_{j,L}>0$ independent of $N_j$ such that \begin{align*} \int_V |u_j(z)|^2\,d\mathcal{L}^{2n}(z) \leq D_{j,L}\exp(N_jR_L) \int_{\Omega}|u_j(z)|^2e^{-\varphi_j(z)}\,d\mathcal{L}^{2n}(z). \end{align*} Combining this with the weighted Hörmander estimate gives \begin{align*} \int_V |u_j(z)|^2\,d\mathcal{L}^{2n}(z) \leq D_{j,L}C_jM_j^2\exp\left(-N_j\left(\inf_{w\in A_j}\rho(w)-R_L\right)\right). \end{align*} The finite-cover mean-value inequality for holomorphic functions on the balls covering $L$ gives another constant $E_{j,L}>0$, independent of $N_j$, such that \begin{align*} \sup_{z\in L}|G_j(z)| = \sup_{z\in L}|u_j(z)| \leq E_{j,L}M_j\exp\left(-\frac{N_j\gamma}{4}\right). \end{align*} The right-hand side tends to $0$ as $N_j\to\infty$. Given $\varepsilon_j>0$, choose $N_j$ large enough to make \begin{align*} \sup_{z\in L}|G_j(z)|\le \varepsilon_j. \end{align*} This proves the claim. [/proof][/step]

[guided]The goal of this step is to build a [holomorphic function](/page/Holomorphic%20Function) that behaves like a prescribed constant $M_j$ at $a_j$ but is very small on compact sets already fixed far inside $\Omega$. Start with the local seed $M_j\chi_j$. It equals $M_j$ near $a_j$, but it is not holomorphic because the cutoff is not holomorphic. Its failure to be holomorphic is exactly \begin{align*} \alpha_j := M_j\bar\partial\chi_j. \end{align*} The form $\alpha_j$ is supported in the annulus \begin{align*} A_j=\{z\in\Omega:r_j/3\le |z-a_j|\le 2r_j/3\}, \end{align*} so it is separated from the pole at $a_j$. Write the finite interpolation set as \begin{align*} P_j=\{p_{j,1},\dots,p_{j,m_j}\}, \end{align*} where $m_j\in\mathbb{N}\cup\{0\}$ is the cardinality of $P_j$. We solve \begin{align*} \bar\partial u_j=\alpha_j \end{align*} using the plurisubharmonic weight \begin{align*} \varphi_j:\Omega&\to[-\infty,\infty)\\ z&\mapsto N_j\rho(z)+2(n+1)\log|z-a_j|+2(n+1)\sum_{q=1}^{m_j}\log|z-p_{j,q}|. \end{align*} The exhaustion term $N_j\rho$ is used to make the solution small deep inside the domain. The logarithmic pole at $a_j$ forces any finite-weight solution to vanish at $a_j$, and the additional logarithmic poles at the points of $P_j$ force the prescribed zeroes. [Hörmander's $L^2$ estimate for the $\bar\partial$-equation](/page/Hormander%20L2%20Estimate) applies in its plurisubharmonic-weight form. Its hypotheses are satisfied here: $\Omega$ is pseudoconvex, $\varphi_j$ is plurisubharmonic although it has logarithmic poles, and $\alpha_j$ is a smooth $\bar\partial$-closed $(0,1)$-form with compact support away from every logarithmic singularity of the weight. If the theorem is first proved for smooth strictly plurisubharmonic weights, this is exactly the standard decreasing-approximation extension to arbitrary plurisubharmonic weights, and the resulting estimate has the same universal constant. On the annulus supporting $\alpha_j$, those logarithmic factors are bounded, so the only varying weight contribution is the displayed factor $e^{-N_j\rho}$. Thus we obtain \begin{align*} u_j:\Omega\to\mathbb{C} \end{align*} with $\bar\partial u_j=\alpha_j$ and \begin{align*} \int_{\Omega}|u_j(z)|^2e^{-\varphi_j(z)}\,d\mathcal{L}^{2n}(z)<\infty. \end{align*} Now define \begin{align*} G_j(z)=M_j\chi_j(z)-u_j(z). \end{align*} Then \begin{align*} \bar\partial G_j=M_j\bar\partial\chi_j-\bar\partial u_j=\alpha_j-\alpha_j=0, \end{align*} so $G_j$ is holomorphic on $\Omega$. Why does $G_j(a_j)$ still equal $M_j$? Near $a_j$, the form $\alpha_j$ is zero, so $u_j$ is holomorphic near $a_j$. If $u_j(a_j)\ne0$, then $|u_j|$ is bounded below by a positive constant on a smaller ball around $a_j$. But the weight contains the factor \begin{align*} |z-a_j|^{-2(n+1)}. \end{align*} In real dimension $2n$, this produces the divergent radial integral \begin{align*} \int_0^\delta r^{2n-1}r^{-2(n+1)}\,d\mathcal{L}^1(r) = \int_0^\delta r^{-3}\,d\mathcal{L}^1(r) = \infty. \end{align*} This contradicts the finite weighted $L^2$ norm. Hence $u_j(a_j)=0$, and because $\chi_j(a_j)=1$, \begin{align*} G_j(a_j)=M_j. \end{align*} The same logarithmic-pole argument applies at every prescribed point $p\in P_j$: the equation is holomorphic near $p$, the finite weighted norm forces $u_j(p)=0$, and $p\notin\operatorname{supp}\chi_j$ gives $G_j(p)=0$. Finally, we explain why $G_j$ can be made small on a compact set $L$ satisfying $L\cap\operatorname{supp}\chi_j=\varnothing$ and the separation condition \begin{align*} \inf_{w\in A_j}\rho(w)>\sup_{z\in L}\rho(z). \end{align*} The weighted estimate gives \begin{align*} \int_{\Omega}|u_j(z)|^2e^{-\varphi_j(z)}\,d\mathcal{L}^{2n}(z) \leq C_jM_j^2\exp\left(-N_j\inf_{w\in A_j}\rho(w)\right), \end{align*} with $C_j$ independent of $N_j$. On a fixed neighbourhood $V$ of $L$ whose closure avoids $A_j$ and lies below the annulus in the exhaustion, the factor $e^{N_j\rho}$ is bounded above by $\exp(N_j\sup_V\rho)$. Hence \begin{align*} \int_V |u_j(z)|^2\,d\mathcal{L}^{2n}(z) \leq D_{j,L}C_jM_j^2\exp\left(-N_j\left(\inf_{w\in A_j}\rho(w)-\sup_{z\in V}\rho(z)\right)\right), \end{align*} where $D_{j,L}>0$ is independent of $N_j$. Choosing $V$ so that $\sup_V\rho$ is still strictly smaller than $\inf_{A_j}\rho$, the exponent is negative with a fixed gap. Since $G_j=-u_j$ on $L$ and is holomorphic near $L$, the finite-cover mean-value inequality converts this $L^2$ bound on $V$ into \begin{align*} \sup_{z\in L}|G_j(z)|\leq E_{j,L}M_j e^{-c_{j,L}N_j} \end{align*} for constants $E_{j,L}>0$ and $c_{j,L}>0$ independent of $N_j$. Therefore increasing $N_j$ makes the supremum on $L$ smaller than any prescribed $\varepsilon_j>0$.[/guided]

[step:Sum the interpolating functions to obtain one function unbounded along the sequence]Choose a strictly increasing sequence $(c_m)_{m=1}^{\infty}$ in $\mathbb{R}$ with $c_m\to\infty$, and define compact exhaustion sublevels \begin{align*} L_m:=\{z\in\Omega:\rho(z)\leq c_m\}. \end{align*} The exhaustion property of $\rho$ implies that each $L_m$ is compact in $\Omega$ and that every compact subset of $\Omega$ is contained in some $L_m$ after increasing $m$. Before constructing $G_j$, pass to a further subsequence, still denoted $(a_j)$, such that \begin{align*} \rho(a_j)\geq c_j+2 \end{align*} for every $j\in\mathbb{N}$. Then shrink $r_j$ so that \begin{align*} \inf_{w\in A_j}\rho(w)\geq c_j+1 \end{align*} for every $j\in\mathbb{N}$, and so that $\operatorname{supp}\chi_j$ is disjoint from $L_j\cup\{a_1,\dots,a_{j-1}\}$. This is possible because $\rho(a_j)>c_j+1$, $\rho$ is continuous at $a_j$, $L_j\subset\{z\in\Omega:\rho(z)\leq c_j\}$, and the points $a_1,\dots,a_{j-1}$ are distinct from $a_j$, after discarding finitely many initial indices and relabelling. Hence $L_j$ is separated from $A_j$ in the exhaustion sense required by the weighted interpolation claim, while the earlier points are handled by the finite zero interpolation in that claim. We construct $G_j\in\mathcal{O}(\Omega)$ inductively so that \begin{align*} G_j(a_j)=M_j, \end{align*} \begin{align*} G_j(a_k)=0\quad\text{for }1\leq k<j, \end{align*} and \begin{align*} \sup_{z\in L_j}|G_j(z)|\le 2^{-j}. \end{align*} This is possible by the weighted interpolation claim applied with $L=L_j$ and $P_j=\{a_1,\dots,a_{j-1}\}$, after choosing $N_j$ sufficiently large. We also choose $M_j>0$ recursively so large that \begin{align*} M_j> j+\sum_{k=1}^{j-1}|G_k(a_j)|+1. \end{align*} Define \begin{align*} F:\Omega &\to \mathbb{C}\\ z &\mapsto \sum_{j=1}^{\infty}G_j(z). \end{align*} Fix $m\in\mathbb{N}$. The finite partial sum $\sum_{j=1}^{m-1}G_j$ is holomorphic on $\Omega$, so it does not affect [uniform convergence](/page/Uniform%20Convergence) of the tail on $L_m$. For the tail, since $L_m\subset L_j$ whenever $j\ge m$, we have \begin{align*} \sum_{j=m}^{\infty}\sup_{z\in L_m}|G_j(z)| \le \sum_{j=m}^{\infty}\sup_{z\in L_j}|G_j(z)| \le \sum_{j=m}^{\infty}2^{-j} <\infty. \end{align*} Thus the Weierstrass $M$-test gives [uniform convergence](/page/Uniform%20Convergence) on each compact sublevel $L_m$. Since every compact subset of $\Omega$ is contained in some $L_m$, the series converges uniformly on compact subsets of $\Omega$. By the [Weierstrass theorem for holomorphic functions](/page/Weierstrass%20Theorem), $F\in\mathcal{O}(\Omega)$. At the point $a_j$, \begin{align*} |F(a_j)| &\ge |G_j(a_j)| - \sum_{k=1}^{j-1}|G_k(a_j)| - \sum_{k=j+1}^{\infty}|G_k(a_j)|\\ &\ge M_j - \sum_{k=1}^{j-1}|G_k(a_j)| - \sum_{k=j+1}^{\infty}0. \end{align*} For every $k>j$, the point $a_j$ belongs to the finite interpolation set $P_k=\{a_1,\dots,a_{k-1}\}$, so $G_k(a_j)=0$. By the choice of $M_j$, \begin{align*} |F(a_j)|\ge j+1>j. \end{align*} Thus $F$ is unbounded along the escaping subsequence $(a_j)$.[/step]

[guided]We now combine the individual functions $G_j$ into a single [holomorphic function](/page/Holomorphic%20Function). The issue is convergence: each $G_j$ is designed to be large at $a_j$, so we must also force it to be small on compact sets that have already been fixed. Let $(L_m)$ be the compact exhaustion by sublevels of $\rho$ constructed above. At stage $j$, we apply the weighted interpolation claim to the compact set $L_j$ and to the finite set of earlier points \begin{align*} P_j=\{a_1,\dots,a_{j-1}\}. \end{align*} The radii have been chosen so that $L_j$ is disjoint from $\operatorname{supp}\chi_j$ and is separated from $A_j$ in the exhaustion, while each point of $P_j$ is outside $\operatorname{supp}\chi_j$. Thus we may prescribe \begin{align*} \sup_{z\in L_j}|G_j(z)|\le 2^{-j} \end{align*} and also impose $G_j(a_k)=0$ for $1\leq k<j$. The summable bound guarantees compact convergence after discarding only a finite initial segment. Fix $m\in\mathbb{N}$. The sum of the first $m-1$ functions is already a finite holomorphic sum, so convergence on $L_m$ is controlled by the tail $j\ge m$. For such $j$, we have $L_m\subset L_j$, and hence \begin{align*} \sup_{z\in L_m}|G_j(z)|\le \sup_{z\in L_j}|G_j(z)|\le 2^{-j}. \end{align*} Therefore \begin{align*} \sum_{j=m}^{\infty}\sup_{z\in L_m}|G_j(z)| \le \sum_{j=m}^{\infty}2^{-j} <\infty. \end{align*} Since every compact subset of $\Omega$ is contained in some $L_m$, the Weierstrass theorem for locally uniform limits of holomorphic functions then gives a holomorphic limit \begin{align*} F(z)=\sum_{j=1}^{\infty}G_j(z). \end{align*} The remaining point is to ensure that the value at $a_j$ is not cancelled by the other terms. The earlier terms $G_1,\dots,G_{j-1}$ are already fixed when $M_j$ is chosen, so their values at $a_j$ form a finite number: \begin{align*} \sum_{k=1}^{j-1}|G_k(a_j)|. \end{align*} We choose $M_j$ larger than this finite quantity by at least $j+1$. The later terms do not contribute at $a_j$ because, for every $k>j$, the point $a_j$ is one of the prescribed zeroes in $P_k$. Hence \begin{align*} G_k(a_j)=0. \end{align*} Consequently \begin{align*} |F(a_j)| \ge M_j-\sum_{k=1}^{j-1}|G_k(a_j)| \ge j+1>j. \end{align*} Thus the single [holomorphic function](/page/Holomorphic%20Function) $F$ becomes unbounded along the escaping sequence.[/guided]

[step:Produce boundary blow-up at every boundary point]Fix $a\in\partial\Omega$ and an open neighbourhood $U\subset\mathbb{C}^n$ of $a$. Choose a sequence $(z_j)_{j=1}^{\infty}$ in $U\cap\Omega$ such that $z_j\to a$ in $\mathbb{C}^n$. This sequence has no accumulation point in $\Omega$, because its only Euclidean limit is the boundary point $a$. Applying the preceding construction to this sequence gives $F\in\mathcal{O}(\Omega)$ and a subsequence $(a_j)$ of $(z_j)$ such that \begin{align*} |F(a_j)|\to\infty. \end{align*} Since $a_j\in U\cap\Omega$ for every $j$, the function $F$ is unbounded on $U\cap\Omega$. This proves the stated boundary blow-up property. By the [boundary blow-up characterization of domains of holomorphy](/page/Domain%20of%20Holomorphy), a domain $\Omega\subset\mathbb{C}^n$ for which every boundary point $a\in\partial\Omega$ and every neighbourhood $U$ of $a$ admit some $F\in\mathcal{O}(\Omega)$ unbounded on $U\cap\Omega$ is a domain of holomorphy. The hypotheses of this characterization are exactly the property proved above, so $\Omega$ is a domain of holomorphy.[/step]

[guided]Fix a boundary point $a\in\partial\Omega$ and an open neighbourhood $U\subset\mathbb{C}^n$ of $a$. Because $a$ lies in the boundary, every neighbourhood of $a$ meets $\Omega$, so we may choose a sequence \begin{align*} z_j\in U\cap\Omega \end{align*} with $z_j\to a$ in $\mathbb{C}^n$. This sequence has no accumulation point in $\Omega$: any Euclidean accumulation point of the sequence is $a$, and $a\notin\Omega$. The escaping-sequence construction applies precisely to sequences in $\Omega$ with no accumulation point in $\Omega$. Applying it to $(z_j)_{j=1}^{\infty}$ gives a [holomorphic function](/page/Holomorphic%20Function) $F\in\mathcal{O}(\Omega)$ and a subsequence $(a_j)_{j=1}^{\infty}$ of $(z_j)_{j=1}^{\infty}$ such that \begin{align*} |F(a_j)|\to\infty. \end{align*} Since each $a_j$ still lies in $U\cap\Omega$, this proves that $F$ is unbounded on $U\cap\Omega$. The [boundary blow-up characterization of domains of holomorphy](/page/Domain%20of%20Holomorphy) says that this local boundary unboundedness condition implies that no boundary point admits a holomorphic extension neighbourhood common to all functions in $\mathcal{O}(\Omega)$. Hence $\Omega$ is a domain of holomorphy.[/guided]

[step:Deduce holomorphic convexity from escaping-sequence separation]Let $K\subset\Omega$ be compact. Suppose, for contradiction, that its [holomorphic hull](/page/Holomorphic%20Hull) \begin{align*} \widehat{K}_{\Omega} = \{z\in\Omega: |f(z)|\le \sup_{w\in K}|f(w)| \text{ for every } f\in\mathcal{O}(\Omega)\} \end{align*} is not compact in $\Omega$. Then there exists a sequence $(z_j)_{j=1}^{\infty}$ in $\widehat{K}_{\Omega}$ with no accumulation point in $\Omega$. By the escaping-sequence construction, there exists $F\in\mathcal{O}(\Omega)$ and a subsequence $(z_{j_\ell})_{\ell=1}^{\infty}$ such that \begin{align*} |F(z_{j_\ell})|\to\infty. \end{align*} However, since each $z_{j_\ell}\in\widehat{K}_{\Omega}$, the defining property of the hull gives \begin{align*} |F(z_{j_\ell})|\le \sup_{w\in K}|F(w)| \end{align*} for every $\ell\in\mathbb{N}$. The right-hand side is finite because $F$ is continuous and $K$ is compact. This contradicts $|F(z_{j_\ell})|\to\infty$. Therefore $\widehat{K}_{\Omega}$ is compact in $\Omega$ for every compact $K\subset\Omega$. Hence $\Omega$ is [holomorphically convex](/page/Holomorphic%20Convexity). Together with the boundary blow-up property proved above, this completes the proof.[/step]

[guided]To prove holomorphic convexity, we must show that the holomorphic hull of every compact set stays compactly inside $\Omega$. Fix a compact set $K\subset\Omega$ and define \begin{align*} \widehat{K}_{\Omega} = \{z\in\Omega: |f(z)|\le \sup_{w\in K}|f(w)| \text{ for every } f\in\mathcal{O}(\Omega)\}. \end{align*} Assume for contradiction that $\widehat{K}_{\Omega}$ is not compact in $\Omega$. Then we can choose a sequence $(z_j)_{j=1}^{\infty}$ in $\widehat{K}_{\Omega}$ with no accumulation point in $\Omega$. The escaping-sequence construction applies to this sequence because its only required hypothesis is that the sequence lie in $\Omega$ and have no accumulation point in $\Omega$, which is exactly how $(z_j)_{j=1}^{\infty}$ was chosen. It gives a [holomorphic function](/page/Holomorphic%20Function) $F\in\mathcal{O}(\Omega)$ and a subsequence $(z_{j_\ell})_{\ell=1}^{\infty}$ such that \begin{align*} |F(z_{j_\ell})|\to\infty. \end{align*} But every point $z_{j_\ell}$ lies in the hull $\widehat{K}_{\Omega}$. Applying the hull inequality to this particular [holomorphic function](/page/Holomorphic%20Function) $F$ gives \begin{align*} |F(z_{j_\ell})|\le \sup_{w\in K}|F(w)| \end{align*} for every $\ell\in\mathbb{N}$. Since $F$ is continuous and $K$ is compact, the supremum on the right is finite. This contradicts the divergence of $|F(z_{j_\ell})|$. Therefore the hull $\widehat{K}_{\Omega}$ is compact in $\Omega$ for every compact $K\subset\Omega$, which is exactly holomorphic convexity. Combining this with the boundary blow-up property proves the theorem.[/guided]