[proofplan] We prove the Cartan--Thullen theorem: a domain $\Omega \subset \mathbb{C}^n$ is a domain of holomorphy if and only if it is holomorphically convex. The proof has two directions. For the forward direction (domain of holomorphy implies holomorphically convex), we show that if $\Omega$ is a domain of holomorphy, then for every compact $K \Subset \Omega$, the boundary distance function $d_\Omega(z) := d(z, \partial\Omega)$ satisfies $d_\Omega(z_0) \geq \inf_{w \in K} d_\Omega(w)$ for all $z_0 \in \widehat{K}_\Omega$, using the Cauchy inequalities on Taylor coefficients and the defining property of the holomorphic hull. This forces $\widehat{K}_\Omega \Subset \Omega$. For the reverse direction (holomorphically convex implies domain of holomorphy), we construct a [holomorphic function](/page/Holomorphic%20Function) on $\Omega$ that is unbounded near every boundary point by exploiting the compact exhaustion afforded by holomorphic convexity. [/proofplan] [step:Establish the distance formula $d(\widehat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega)$ for domains of holomorphy] Let $\Omega$ be a domain of holomorphy and $K \Subset \Omega$ a compact subset. Define $r := d(K, \partial\Omega) = \inf_{w \in K} d(w, \partial\Omega) > 0$ (since $K$ is compact and contained in the [open set](/page/Open%20Set) $\Omega$). Since $K \subset \widehat{K}_\Omega$, we have $d(\widehat{K}_\Omega, \partial\Omega) \leq d(K, \partial\Omega) = r$. For the reverse inequality, fix $z_0 \in \widehat{K}_\Omega$. We show $d(z_0, \partial\Omega) \geq r$. Fix any $w \in K$ and any $\rho$ with $0 < \rho < d(w, \partial\Omega)$. Since $B(w, d(w, \partial\Omega)) \subset \Omega$, in particular $\overline{B}(w, \rho) \subset \Omega$. For any $f \in \mathcal{O}(\Omega)$, the Cauchy inequalities for polydiscs (applied on the polydisc of polyradius $(\rho, \dots, \rho)$ centred at $w$, which lies in $B(w, \rho\sqrt{n}) \subset \Omega$ for $\rho$ sufficiently small, or more precisely, applying the [Cauchy integral formula](/theorems/345) on the distinguished boundary of such a polydisc) give \begin{align*} \left|\frac{\partial^\alpha f(w)}{\alpha!}\right| \leq \frac{\sup_{\overline{B}(w, \rho)} |f|}{\rho^{|\alpha|}}. \end{align*} The key observation is that for each multi-index $\alpha$, the function $\partial^\alpha f \in \mathcal{O}(\Omega)$. Since $z_0 \in \widehat{K}_\Omega$, the defining property of the holomorphic hull gives \begin{align*} \left|\frac{\partial^\alpha f(z_0)}{\alpha!}\right| \leq \sup_{w \in K} \left|\frac{\partial^\alpha f(w)}{\alpha!}\right|. \end{align*} Now consider the Taylor series of $f$ at $z_0$: \begin{align*} f(z) = \sum_{\alpha \in \mathbb{N}_0^n} \frac{\partial^\alpha f(z_0)}{\alpha!}(z - z_0)^\alpha. \end{align*} This series converges on the largest polydisc centred at $z_0$ on which the coefficients $\partial^\alpha f(z_0)/\alpha!$ decay sufficiently fast. Using the hull estimate above and the Cauchy inequality at each $w \in K$, for any $\rho < r = \inf_{w \in K} d(w, \partial\Omega)$: \begin{align*} \left|\frac{\partial^\alpha f(z_0)}{\alpha!}\right| \leq \sup_{w \in K} \left|\frac{\partial^\alpha f(w)}{\alpha!}\right| \leq \sup_{w \in K} \frac{\sup_{\overline{B}(w, \rho)} |f|}{\rho^{|\alpha|}} \leq \frac{M_{f,\rho}}{\rho^{|\alpha|}}, \end{align*} where $M_{f,\rho} := \sup_{w \in K} \sup_{\overline{B}(w,\rho)} |f| < \infty$ (since $\bigcup_{w \in K} \overline{B}(w, \rho) \Subset \Omega$ is compact when $\rho < r$, and $f$ is continuous on $\Omega$). This bound shows that the Taylor series of $f$ at $z_0$ converges absolutely for $|z - z_0| < \rho$. Since $\rho < r$ was arbitrary, the Taylor series converges on $B(z_0, r)$. Now suppose for contradiction that $d(z_0, \partial\Omega) < r$. Choose $b \in \partial\Omega$ with $|z_0 - b| = d(z_0, \partial\Omega) < r$. Since $\Omega$ is a domain of holomorphy, there exists $f \in \mathcal{O}(\Omega)$ that cannot be holomorphically continued past $b$. But the argument above shows the Taylor series of $f$ at $z_0$ converges on $B(z_0, r)$, and since $|z_0 - b| < r$, this ball contains $b$. On the intersection $B(z_0, r) \cap \Omega$, the Taylor series agrees with $f$ (by the identity theorem, since $B(z_0, r) \cap \Omega$ is a connected [open set](/page/Open%20Set) containing $z_0$). Therefore the Taylor series provides a holomorphic extension of $f$ to a neighbourhood of $b$ --- a contradiction. Hence $d(z_0, \partial\Omega) \geq r$ for every $z_0 \in \widehat{K}_\Omega$, giving $d(\widehat{K}_\Omega, \partial\Omega) \geq r = d(K, \partial\Omega)$. [guided] The distance formula $d(\widehat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega)$ is the quantitative core of the forward direction. The inequality $d(\widehat{K}_\Omega, \partial\Omega) \leq d(K, \partial\Omega)$ is immediate from $K \subset \widehat{K}_\Omega$. The substance is the reverse inequality: no point of the holomorphic hull can be closer to $\partial\Omega$ than $r := d(K, \partial\Omega)$. Fix $z_0 \in \widehat{K}_\Omega$. The strategy is to show that the Taylor series of any $f \in \mathcal{O}(\Omega)$ at $z_0$ converges on the ball $B(z_0, r)$. If $d(z_0, \partial\Omega) < r$, this convergence would extend $f$ past the boundary, contradicting the assumption that $\Omega$ is a domain of holomorphy. **Bounding the Taylor coefficients via the hull property.** For each multi-index $\alpha \in \mathbb{N}_0^n$, the function $h_\alpha := \partial^\alpha f / \alpha!$ is holomorphic on $\Omega$ (holomorphic functions have holomorphic partial derivatives of all orders). Since $z_0 \in \widehat{K}_\Omega$, the defining property of the holomorphic hull gives \begin{align*} |h_\alpha(z_0)| = \left|\frac{\partial^\alpha f(z_0)}{\alpha!}\right| \leq \sup_{w \in K} \left|\frac{\partial^\alpha f(w)}{\alpha!}\right| = \sup_{w \in K} |h_\alpha(w)|. \end{align*} Why is this useful? Because points $w \in K$ are at distance at least $r$ from $\partial\Omega$, so the Cauchy inequalities provide strong decay estimates on their Taylor coefficients. **Applying the Cauchy inequalities on $K$.** Fix $\rho$ with $0 < \rho < r$. For each $w \in K$, we have $d(w, \partial\Omega) \geq r > \rho$, so $\overline{B}(w, \rho) \subset \Omega$. The Cauchy inequalities for the Taylor coefficients of $f$ at $w$ (obtained by applying the [Cauchy integral formula on polydiscs](/theorems/3398) of polyradius $(\rho, \dots, \rho)$ contained in the ball) give \begin{align*} \left|\frac{\partial^\alpha f(w)}{\alpha!}\right| \leq \frac{\sup_{\overline{B}(w,\rho)} |f|}{\rho^{|\alpha|}}. \end{align*} Taking the supremum over $w \in K$ and combining with the hull estimate: \begin{align*} \left|\frac{\partial^\alpha f(z_0)}{\alpha!}\right| \leq \sup_{w \in K} \frac{\sup_{\overline{B}(w,\rho)} |f|}{\rho^{|\alpha|}} \leq \frac{M_{f,\rho}}{\rho^{|\alpha|}}, \end{align*} where $M_{f,\rho} := \sup_{w \in K} \sup_{\overline{B}(w,\rho)} |f|$. This quantity is finite because $\bigcup_{w \in K} \overline{B}(w, \rho)$ is compact (it is the closed $\rho$-neighbourhood of the compact set $K$, and $\rho < r$ ensures it lies in $\Omega$), and $f$ is continuous on $\Omega$. **Convergence of the Taylor series.** The Taylor series of $f$ at $z_0$ is \begin{align*} f(z) = \sum_{\alpha \in \mathbb{N}_0^n} \frac{\partial^\alpha f(z_0)}{\alpha!}(z - z_0)^\alpha. \end{align*} For any $z$ with $|z - z_0| < \rho$, the absolute value of the general term is bounded by \begin{align*} \left|\frac{\partial^\alpha f(z_0)}{\alpha!}\right| |z - z_0|^{|\alpha|} \leq M_{f,\rho} \left(\frac{|z - z_0|}{\rho}\right)^{|\alpha|}. \end{align*} Since $|z - z_0| / \rho < 1$, the multi-index series $\sum_\alpha (|z - z_0|/\rho)^{|\alpha|}$ converges (it equals $\prod_{j=1}^n (1 - |z_j - (z_0)_j|/\rho)^{-1}$ when each component is bounded appropriately; more precisely, grouping by total degree $|\alpha| = m$, the number of multi-indices with $|\alpha| = m$ is $\binom{m+n-1}{n-1}$, and $\sum_{m=0}^\infty \binom{m+n-1}{n-1} t^m = (1-t)^{-n}$ for $0 \leq t < 1$). Therefore the Taylor series converges absolutely on $B(z_0, \rho)$. Since $\rho < r$ was arbitrary, the series converges on $B(z_0, r)$. **The contradiction.** Suppose $d(z_0, \partial\Omega) < r$. Choose $b \in \partial\Omega$ with $|z_0 - b| = d(z_0, \partial\Omega) < r$. Since $\Omega$ is a domain of holomorphy, there exists $f \in \mathcal{O}(\Omega)$ that cannot be holomorphically continued past $b$. But we have just shown that the Taylor series of $f$ at $z_0$ converges on $B(z_0, r)$, which is an open ball containing $b$ (since $|z_0 - b| < r$). On $B(z_0, r) \cap \Omega$, the sum of the Taylor series agrees with $f$ by the identity theorem (since both are holomorphic on the connected [open set](/page/Open%20Set) $B(z_0, r) \cap \Omega$ and they agree near $z_0$). Therefore the Taylor series defines a holomorphic extension of $f$ to $B(z_0, r) \supset \{b\}$, contradicting the assumption that $f$ cannot be extended past $b$. We conclude $d(z_0, \partial\Omega) \geq r$ for all $z_0 \in \widehat{K}_\Omega$, so $d(\widehat{K}_\Omega, \partial\Omega) \geq r = d(K, \partial\Omega)$. Combined with the reverse inequality from $K \subset \widehat{K}_\Omega$, we obtain $d(\widehat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega)$. [/guided] [/step] [step:Conclude the forward direction: domain of holomorphy implies holomorphically convex] By the distance formula established above, for every compact $K \Subset \Omega$: \begin{align*} d(\widehat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega) > 0. \end{align*} We must show $\widehat{K}_\Omega$ is compact (i.e., $\widehat{K}_\Omega \Subset \Omega$). The set $\widehat{K}_\Omega$ is closed in $\Omega$: if $z_k \in \widehat{K}_\Omega$ with $z_k \to z_0 \in \Omega$, then for every $f \in \mathcal{O}(\Omega)$, $|f(z_0)| = \lim_k |f(z_k)| \leq \sup_K |f|$, so $z_0 \in \widehat{K}_\Omega$. The set $\widehat{K}_\Omega$ is bounded: for each coordinate function $z_j \in \mathcal{O}(\Omega)$, $|z_j(z)| \leq \sup_K |z_j|$ for all $z \in \widehat{K}_\Omega$, so $\widehat{K}_\Omega \subset \{z \in \mathbb{C}^n : |z_j| \leq \sup_K |z_j| \text{ for all } j\}$, which is bounded. Since $\widehat{K}_\Omega$ is bounded, closed in $\Omega$, and satisfies $d(\widehat{K}_\Omega, \partial\Omega) > 0$, it is a closed subset of $\mathbb{C}^n$ that does not approach $\partial\Omega$ and is bounded, hence compact. Therefore $\widehat{K}_\Omega \Subset \Omega$, and $\Omega$ is holomorphically convex. [/step] [step:Prove the reverse direction: holomorphically convex implies domain of holomorphy] Assume $\Omega$ is holomorphically convex. We construct $f \in \mathcal{O}(\Omega)$ that cannot be holomorphically extended beyond any boundary point. Choose a compact exhaustion $(K_j)_{j \geq 1}$ of $\Omega$: compact sets with $K_j \subset K_{j+1}^\circ$ and $\bigcup_j K_j = \Omega$. By holomorphic convexity, replace each $K_j$ with its holomorphic hull $\widehat{(K_j)}_\Omega$, which is still compact (and satisfies $\widehat{(K_j)}_\Omega \Subset \Omega$). Relabelling, assume $K_j = \widehat{(K_j)}_\Omega$. Choose a countable [dense subset](/page/Dense%20Subset) $\{b_k\}_{k \geq 1}$ of $\partial\Omega$. For each $k$, choose $w_k \in \Omega$ with $|w_k - b_k| < 1/k$ and $w_k \notin K_k$ (possible since $K_k \Subset \Omega$ and $b_k \in \partial\Omega$). Since $w_k \notin K_k = \widehat{(K_k)}_\Omega$, there exists $g_k \in \mathcal{O}(\Omega)$ with \begin{align*} |g_k(w_k)| > \sup_{K_k} |g_k|. \end{align*} Normalise: define \begin{align*} \varphi_k: \Omega &\to \mathbb{C} \\ z &\mapsto \frac{g_k(z)}{g_k(w_k)}, \end{align*} so that $\varphi_k(w_k) = 1$ and $\sup_{K_k} |\varphi_k| < 1$. [guided] The strategy is to combine the functions $\varphi_k$ into a single [holomorphic function](/page/Holomorphic%20Function) that blows up near every boundary point. The key property of holomorphic convexity is the ability to "separate" points outside the hull: for any point not in $\widehat{K}_\Omega$, some [holomorphic function](/page/Holomorphic%20Function) on $\Omega$ is larger at that point than on $K$. We exploit this to find, for each boundary point $b_k$, a [holomorphic function](/page/Holomorphic%20Function) $\varphi_k$ that is "large" near $b_k$ (at the nearby point $w_k$) and "small" on the compact set $K_k$. The construction then combines all these functions into a single $f$ that is unbounded near every boundary point. The normalisation ensures $\varphi_k(w_k) = 1$ and $\sup_{K_k} |\varphi_k| = \delta_k < 1$ for some $\delta_k$. Raising $\varphi_k$ to a high power $N_k$ makes $\sup_{K_k} |\varphi_k^{N_k}| = \delta_k^{N_k}$ as small as desired, while $|\varphi_k^{N_k}(w_k)| = 1$. [/guided] [/step] [step:Construct a holomorphic function unbounded near every boundary point] For each $k \geq 1$, set $\delta_k := \sup_{K_k} |\varphi_k| < 1$. Choose integers $N_k \geq 1$ such that \begin{align*} k \cdot \delta_k^{N_k} < \frac{1}{2^k}. \end{align*} This is possible since $\delta_k < 1$ implies $\delta_k^{N_k} \to 0$ as $N_k \to \infty$. Define \begin{align*} \psi_k: \Omega &\to \mathbb{C} \\ z &\mapsto k \cdot \varphi_k(z)^{N_k}, \end{align*} so $\psi_k \in \mathcal{O}(\Omega)$, $\psi_k(w_k) = k$, and $\sup_{K_k} |\psi_k| = k \cdot \delta_k^{N_k} < 2^{-k}$. Define \begin{align*} f: \Omega &\to \mathbb{C} \\ z &\mapsto \sum_{k=1}^{\infty} \psi_k(z). \end{align*} The series converges uniformly on each $K_j$: for $k \geq j$, we have $K_j \subset K_k$ (since $j \leq k$ and the exhaustion is increasing), so $\sup_{K_j} |\psi_k| \leq \sup_{K_k} |\psi_k| < 2^{-k}$. Hence \begin{align*} \sum_{k=j}^{\infty} \sup_{K_j} |\psi_k| \leq \sum_{k=j}^{\infty} \frac{1}{2^k} < \infty. \end{align*} The finitely many terms $\psi_1, \dots, \psi_{j-1}$ are each holomorphic on $\Omega$ and hence bounded on the compact set $K_j$. Therefore $\sum_{k=1}^\infty \psi_k$ converges uniformly on $K_j$. Since $K_j$ was an arbitrary member of a compact exhaustion and uniform limits of holomorphic functions are holomorphic, $f \in \mathcal{O}(\Omega)$. We claim $f$ cannot be extended past any boundary point. Fix $b \in \partial\Omega$ and a neighbourhood $V$ of $b$. Since $\{b_k\}$ is dense in $\partial\Omega$ and $|w_k - b_k| < 1/k$, there exist infinitely many $k$ with $w_k \in V$. We show $|f(w_k)| \to \infty$ along a subsequence of such $k$, which prevents holomorphic extension. For each $k$, write \begin{align*} f(w_k) = \psi_k(w_k) + \sum_{m \neq k} \psi_m(w_k) = k + \sum_{m \neq k} \psi_m(w_k). \end{align*} We bound the sum $\sum_{m \neq k} |\psi_m(w_k)|$ by splitting it into two parts. **Terms with $m > k$:** For $m > k$, we have $w_k \in \Omega$ and we need $w_k \in K_m$ to use the small-on-$K_m$ estimate. Since $w_k$ is a fixed point of $\Omega$ and $\bigcup_m K_m = \Omega$, there exists $j(k) \geq 1$ such that $w_k \in K_{j(k)}$. For all $m \geq j(k)$, we have $w_k \in K_{j(k)} \subset K_m$, so $|\psi_m(w_k)| \leq \sup_{K_m} |\psi_m| < 2^{-m}$. The finitely many terms with $k < m < j(k)$ are each bounded by some finite constant (each $\psi_m$ is continuous on $\Omega$). Therefore \begin{align*} \sum_{m=k+1}^{\infty} |\psi_m(w_k)| \leq \sum_{m=k+1}^{j(k)-1} |\psi_m(w_k)| + \sum_{m=j(k)}^{\infty} 2^{-m} < \infty. \end{align*} However, this bound may depend on $k$ through $j(k)$. We handle this more carefully. Since each $w_k$ satisfies $w_k \in K_{j(k)}$ for some $j(k)$, and $K_{j(k)} \subset K_m$ for $m \geq j(k)$, we have \begin{align*} \sum_{m > k} |\psi_m(w_k)| \leq \underbrace{\sum_{m=k+1}^{j(k)-1} |\psi_m(w_k)|}_{\text{finite, but may grow with } k} + \sum_{m=j(k)}^{\infty} 2^{-m} \leq \sum_{m=k+1}^{j(k)-1} |\psi_m(w_k)| + 1. \end{align*} To control the problematic middle terms, we refine the construction. After choosing $N_k$ to ensure $k \delta_k^{N_k} < 2^{-k}$, additionally require $N_k$ large enough so that $\sup_{K_j} |\psi_k| < 2^{-k}$ for all $j \leq k$ (which already holds since $K_j \subset K_k$ for $j \leq k$). The key estimate for the "tail past $k$" terms is already settled. **Terms with $m < k$:** For $m < k$, note that $w_k \to \partial\Omega$ as $k \to \infty$ (since $|w_k - b_k| < 1/k$ and $b_k \in \partial\Omega$). Each $\psi_m$ is a fixed [holomorphic function](/page/Holomorphic%20Function) on $\Omega$. We cannot bound $|\psi_m(w_k)|$ uniformly in $k$ because $w_k$ approaches the boundary. To avoid this difficulty, we use a refined construction. **Refined construction using partial sums.** Define $S_k := \sum_{m=1}^{k-1} \psi_m$, which is a finite sum of holomorphic functions and hence holomorphic on $\Omega$. In particular, $S_k$ is continuous on $\Omega$ and takes a definite value at $w_k$. We replace the scaling factor $k$ by a factor that dominates $|S_k(w_k)|$: choose $N_k$ (increasing it further if necessary) so that \begin{align*} k \cdot \delta_k^{N_k} < 2^{-k} \quad \text{and} \quad |\varphi_k(w_k)|^{N_k} = 1 \text{ (already ensured)}. \end{align*} Then define $c_k := k + |S_k(w_k)|$ and redefine $\psi_k := c_k \cdot \varphi_k^{N_k}$, choosing $N_k$ large enough that $c_k \cdot \delta_k^{N_k} < 2^{-k}$ (possible since $c_k$ is a fixed finite number for each $k$ and $\delta_k^{N_k} \to 0$). Note that $c_k$ is determined before $N_k$ is chosen: $S_k$ depends only on $\psi_1, \dots, \psi_{k-1}$, which have already been constructed. This inductive construction is valid. With this choice, $\psi_k(w_k) = c_k = k + |S_k(w_k)|$, and \begin{align*} |f(w_k)| &= \left|\psi_k(w_k) + S_k(w_k) + \sum_{m=k+1}^{\infty} \psi_m(w_k)\right| \\ &\geq |\psi_k(w_k)| - |S_k(w_k)| - \sum_{m=k+1}^{\infty} |\psi_m(w_k)|. \end{align*} The first term is $c_k = k + |S_k(w_k)|$. For the tail $\sum_{m > k} |\psi_m(w_k)|$: since $w_k \in K_{j(k)}$ for some $j(k)$, and $K_{j(k)} \subset K_m$ for $m \geq j(k)$, we have $|\psi_m(w_k)| \leq \sup_{K_m} |\psi_m| < 2^{-m}$ for $m \geq j(k)$. The terms with $k < m < j(k)$ are finite in number and each bounded. However, this intermediate sum could depend on $k$. To handle this cleanly, we make one further refinement: in the inductive step, after constructing $\psi_1, \dots, \psi_{k-1}$, the function $T_k := \sum_{m=1}^{k-1} \psi_m$ is holomorphic on $\Omega$. We set $c_k := k$ and choose $N_k$ large enough that both $k \delta_k^{N_k} < 2^{-k}$ and $\sup_{K_k} |k \varphi_k^{N_k}| < 2^{-k}$ hold. The tail estimate becomes: for $m > k$ and $w_k \in K_{j(k)}$, \begin{align*} \sum_{m=k+1}^{\infty} |\psi_m(w_k)| &= \sum_{\substack{m=k+1 \\ m < j(k)}}^{} |\psi_m(w_k)| + \sum_{m=j(k)}^{\infty} |\psi_m(w_k)| \\ &\leq \sum_{\substack{m=k+1 \\ m < j(k)}}^{} |\psi_m(w_k)| + \sum_{m=j(k)}^{\infty} 2^{-m}. \end{align*} The second sum is at most $2^{1-j(k)} \leq 1$. For the first sum: each $|\psi_m(w_k)| = k |\varphi_m(w_k)|^{N_m}$ is finite (a [holomorphic function](/page/Holomorphic%20Function) evaluated at a point of $\Omega$), but we have no uniform bound. The standard resolution is to use the following strengthened construction. Rather than attempting to bound the tail at the specific points $w_k$, we instead invoke the following classical lemma. [claim:A holomorphically convex domain admits a function unbounded on every sequence approaching the boundary] Let $\Omega \subset \mathbb{C}^n$ be holomorphically convex and $(w_k)_{k \geq 1}$ a sequence in $\Omega$ with no accumulation point in $\Omega$ (i.e., $w_k$ eventually leaves every compact subset of $\Omega$). Then there exists $f \in \mathcal{O}(\Omega)$ with $\sup_k |f(w_k)| = \infty$. [proof] We construct $f$ as a series $f = \sum_{k=1}^\infty \psi_k$ by induction, at each step ensuring two properties: (i) the series converges uniformly on compact subsets, and (ii) $|f(w_k)|$ is large. Since $(w_k)$ has no accumulation point in $\Omega$, for each $j$ the compact set $K_j$ contains only finitely many $w_k$. By passing to a subsequence (which suffices, since unboundedness on a subsequence implies unboundedness on the full sequence), assume $w_k \notin K_k$ for all $k$. **Inductive construction.** Set $\psi_0 := 0$ and $f_0 := 0$. At stage $k \geq 1$, the partial sum $f_{k-1} = \sum_{m=1}^{k-1} \psi_m$ is a [holomorphic function](/page/Holomorphic%20Function) on $\Omega$ already constructed. Since $w_k \notin K_k = \widehat{(K_k)}_\Omega$, there exists $\varphi_k \in \mathcal{O}(\Omega)$ with $\varphi_k(w_k) = 1$ and $\delta_k := \sup_{K_k} |\varphi_k| < 1$ (as constructed above). Choose the integer $N_k \geq 1$ large enough that \begin{align*} (k + |f_{k-1}(w_k)|) \cdot \delta_k^{N_k} < \frac{1}{2^k}. \end{align*} Define $\psi_k := (k + |f_{k-1}(w_k)|) \cdot \varphi_k^{N_k}$. Then: - $\sup_{K_k} |\psi_k| = (k + |f_{k-1}(w_k)|) \cdot \delta_k^{N_k} < 2^{-k}$, ensuring [uniform convergence](/page/Uniform%20Convergence) on compact subsets (since $K_j \subset K_k$ for $j \leq k$). - $|\psi_k(w_k)| = k + |f_{k-1}(w_k)|$. The series $f = \sum_{k=1}^\infty \psi_k$ converges uniformly on each $K_j$ (for $k \geq j$, $\sup_{K_j} |\psi_k| < 2^{-k}$), so $f \in \mathcal{O}(\Omega)$. At $w_k$: \begin{align*} |f(w_k)| &= \left|f_{k-1}(w_k) + \psi_k(w_k) + \sum_{m=k+1}^{\infty} \psi_m(w_k)\right| \\ &\geq |\psi_k(w_k)| - |f_{k-1}(w_k)| - \sum_{m=k+1}^{\infty} |\psi_m(w_k)|. \end{align*} For the tail: since $w_k \in K_{j(k)}$ for some $j(k)$ (as $w_k \in \Omega = \bigcup_j K_j$), for $m \geq j(k)$ we have $w_k \in K_m$ and $|\psi_m(w_k)| \leq \sup_{K_m} |\psi_m| < 2^{-m}$. For the finitely many terms $k < m < j(k)$, each $|\psi_m(w_k)|$ is finite. Define \begin{align*} R_k := \sum_{m=k+1}^{\infty} |\psi_m(w_k)| < \infty. \end{align*} Then $|f(w_k)| \geq (k + |f_{k-1}(w_k)|) - |f_{k-1}(w_k)| - R_k = k - R_k$. It remains to show $R_k$ does not grow as fast as $k$. We refine the choice of $N_k$ further: at stage $k$, after choosing $N_k$ as above, we additionally increase $N_k$ so that \begin{align*} (k + |f_{k-1}(w_k)|) \cdot |\varphi_k(w_m)|^{N_k} < \frac{1}{2^k} \quad \text{for all } m = 1, \dots, k-1 \end{align*} (this is possible since $|\varphi_k(w_m)|$ is a fixed finite number for each $m$, and if $|\varphi_k(w_m)| \leq 1$ the bound is immediate from the $K_k$ estimate when $w_m \in K_k$; if $|\varphi_k(w_m)| > 1$ we cannot ensure this, but we simply observe the following cleaner argument). **Cleaner approach:** We prove unboundedness without needing $R_k \to 0$. We have the recurrence $|f_k(w_k)| = |f_{k-1}(w_k) + \psi_k(w_k) + \sum_{m=k+1}^{\infty} \psi_m(w_k)|$. But we only need $\sup_k |f(w_k)| = \infty$, not $|f(w_k)| \to \infty$. Suppose for contradiction that $|f(w_k)| \leq M$ for all $k$. Then \begin{align*} |\psi_k(w_k)| \leq |f(w_k)| + |f_{k-1}(w_k)| + \sum_{m=k+1}^{\infty} |\psi_m(w_k)|. \end{align*} But $|\psi_k(w_k)| = k + |f_{k-1}(w_k)|$, so $k + |f_{k-1}(w_k)| \leq M + |f_{k-1}(w_k)| + R_k$, giving $k \leq M + R_k$. Since $R_k = \sum_{m > k} |\psi_m(w_k)| \to 0$ as $k \to \infty$ (for $k$ large enough that $j(k) = k+1$, i.e., $w_k \in K_{k+1}$, which happens for infinitely many $k$ since $w_k \in \Omega = \bigcup_j K_j$ and by passing to a further subsequence we may arrange $w_k \in K_{k+1}$), we get $k \leq M + 1$ for all large $k$, a contradiction. More precisely: by passing to a subsequence, assume $w_k \in K_{k+1} \setminus K_k$ for all $k$ (possible because $(w_k)$ eventually leaves each $K_j$, and by relabelling). Then for $m \geq k+1$, $w_k \in K_{k+1} \subset K_m$, so $|\psi_m(w_k)| \leq \sup_{K_m} |\psi_m| < 2^{-m}$. Hence $R_k = \sum_{m=k+1}^{\infty} |\psi_m(w_k)| < \sum_{m=k+1}^{\infty} 2^{-m} = 2^{-k}$, giving $k \leq M + 2^{-k} \leq M + 1$ for all $k$ --- a contradiction. [/proof] [/claim] We apply this claim to complete the proof. Since $\{b_k\}_{k \geq 1}$ is dense in $\partial\Omega$ and $|w_k - b_k| < 1/k$, the sequence $(w_k)$ has no accumulation point in $\Omega$ (any accumulation point would lie in $\partial\Omega$, which is disjoint from $\Omega$). By the claim, there exists $f \in \mathcal{O}(\Omega)$ with $\sup_k |f(w_k)| = \infty$. Now suppose $f$ could be holomorphically extended to an [open set](/page/Open%20Set) $U$ containing a boundary point $b \in \partial\Omega$. Then $f$ would be continuous (hence bounded) on some compact neighbourhood $\overline{B}(b, \varepsilon) \cap \overline{U}$. But since $\{b_k\}$ is dense in $\partial\Omega$, infinitely many $b_k$ lie in $B(b, \varepsilon/2)$, and for $k$ large enough that $1/k < \varepsilon/2$, the corresponding $w_k$ satisfy $|w_k - b| \leq |w_k - b_k| + |b_k - b| < 1/k + \varepsilon/2 < \varepsilon$, so $w_k \in B(b, \varepsilon) \cap \Omega \subset U$. Since $|f(w_k)|$ is unbounded along a subsequence, this contradicts the continuity of $f$ on $U$. Therefore $\Omega$ is a domain of holomorphy. [guided] The goal of this step is to produce a single [holomorphic function](/page/Holomorphic%20Function) $f \in \mathcal{O}(\Omega)$ that cannot be holomorphically extended past any boundary point. The key difficulty is ensuring that $f$ is genuinely unbounded near the boundary, not just "large at the principal term." Let us carry out the full construction with commentary explaining each choice. **Setting up the building blocks.** For each $k \geq 1$, recall from the previous step that we have $\varphi_k \in \mathcal{O}(\Omega)$ with $\varphi_k(w_k) = 1$ and $\delta_k := \sup_{K_k} |\varphi_k| < 1$. The number $\delta_k$ measures how much $\varphi_k$ is "suppressed" on $K_k$ compared to its value at $w_k$. Since $\delta_k < 1$, raising $\varphi_k$ to a large power $N_k$ drives $\sup_{K_k} |\varphi_k^{N_k}| = \delta_k^{N_k}$ to zero while keeping $|\varphi_k(w_k)|^{N_k} = 1$. Choose integers $N_k \geq 1$ such that \begin{align*} k \cdot \delta_k^{N_k} < \frac{1}{2^k}. \end{align*} This is possible since $\delta_k < 1$ implies $\delta_k^{N_k} \to 0$ as $N_k \to \infty$. Why this particular bound? We need the series $\sum_k \psi_k$ to converge uniformly on compact subsets, and the geometric decay $2^{-k}$ ensures this via the Weierstrass $M$-test. Define \begin{align*} \psi_k: \Omega &\to \mathbb{C} \\ z &\mapsto k \cdot \varphi_k(z)^{N_k}, \end{align*} so $\psi_k \in \mathcal{O}(\Omega)$ (a composition of holomorphic functions), $\psi_k(w_k) = k \cdot 1^{N_k} = k$, and $\sup_{K_k} |\psi_k| = k \cdot \delta_k^{N_k} < 2^{-k}$. **Defining the candidate function.** Define \begin{align*} f: \Omega &\to \mathbb{C} \\ z &\mapsto \sum_{k=1}^{\infty} \psi_k(z). \end{align*} We must verify that this series defines a [holomorphic function](/page/Holomorphic%20Function). The series converges uniformly on each $K_j$: for $k \geq j$, we have $K_j \subset K_k$ (since $j \leq k$ and the exhaustion is increasing), so $\sup_{K_j} |\psi_k| \leq \sup_{K_k} |\psi_k| < 2^{-k}$. Hence \begin{align*} \sum_{k=j}^{\infty} \sup_{K_j} |\psi_k| \leq \sum_{k=j}^{\infty} \frac{1}{2^k} < \infty. \end{align*} The finitely many terms $\psi_1, \dots, \psi_{j-1}$ are each holomorphic on $\Omega$ and hence bounded on the compact set $K_j$. Therefore $\sum_{k=1}^\infty \psi_k$ converges uniformly on $K_j$. Since $K_j$ was an arbitrary member of a compact exhaustion and uniform limits of holomorphic functions are holomorphic, $f \in \mathcal{O}(\Omega)$. Why does [uniform convergence](/page/Uniform%20Convergence) on compact subsets suffice? Because every point of $\Omega$ lies in some $K_j$, so the convergence is locally uniform, which is the standard criterion for the limit of holomorphic functions to be holomorphic. **Why a naive unboundedness argument fails.** We want to show $|f(w_k)| \to \infty$. The naive estimate is: write $f(w_k) = \psi_k(w_k) + \sum_{m \neq k} \psi_m(w_k) = k + \sum_{m \neq k} \psi_m(w_k)$ and try to show the error term $\sum_{m \neq k} |\psi_m(w_k)|$ is small. For the "tail" terms $m > k$, we could use the $\sup_{K_m}$ estimate if $w_k \in K_m$. But for the "past" terms $m < k$, the functions $\psi_m$ are fixed holomorphic functions, and as $w_k \to \partial\Omega$, the values $|\psi_m(w_k)|$ could be arbitrarily large. The [uniform convergence](/page/Uniform%20Convergence) on compact sets does not help here because $w_k$ eventually leaves every compact set. This is the central technical obstacle. **The resolution: an inductive construction with adaptive scaling.** The standard fix is to build the $\psi_k$ inductively, choosing the scaling factor at stage $k$ to dominate the contribution of all previously constructed terms. This is the content of the following claim. [claim:A holomorphically convex domain admits a function unbounded on every sequence approaching the boundary] Let $\Omega \subset \mathbb{C}^n$ be holomorphically convex and $(w_k)_{k \geq 1}$ a sequence in $\Omega$ with no accumulation point in $\Omega$ (i.e., $w_k$ eventually leaves every compact subset of $\Omega$). Then there exists $f \in \mathcal{O}(\Omega)$ with $\sup_k |f(w_k)| = \infty$. [proof] We construct $f$ as a series $f = \sum_{k=1}^\infty \psi_k$ by induction, at each step ensuring two properties: (i) the series converges uniformly on compact subsets, and (ii) $|f(w_k)|$ is large. Since $(w_k)$ has no accumulation point in $\Omega$, for each $j$ the compact set $K_j$ contains only finitely many $w_k$. Why? Because if infinitely many $w_k$ belonged to $K_j$, then by compactness of $K_j$ they would have an accumulation point in $K_j \subset \Omega$, contradicting the hypothesis. By passing to a subsequence (which suffices, since unboundedness on a subsequence implies unboundedness on the full sequence), assume $w_k \notin K_k$ for all $k$. **Inductive construction.** Set $\psi_0 := 0$ and $f_0 := 0$. At stage $k \geq 1$, the partial sum $f_{k-1} = \sum_{m=1}^{k-1} \psi_m$ is a [holomorphic function](/page/Holomorphic%20Function) on $\Omega$ already constructed. The value $f_{k-1}(w_k)$ is therefore a known complex number. Since $w_k \notin K_k = \widehat{(K_k)}_\Omega$, there exists $\varphi_k \in \mathcal{O}(\Omega)$ with $\varphi_k(w_k) = 1$ and $\delta_k := \sup_{K_k} |\varphi_k| < 1$ (as constructed in the previous step). The key idea of the inductive construction is to choose the scaling factor $c_k := k + |f_{k-1}(w_k)|$ so that it automatically dominates the contribution of all past terms at $w_k$. Choose the integer $N_k \geq 1$ large enough that \begin{align*} (k + |f_{k-1}(w_k)|) \cdot \delta_k^{N_k} < \frac{1}{2^k}. \end{align*} This is possible because $c_k = k + |f_{k-1}(w_k)|$ is a fixed finite number (determined before $N_k$ is chosen) and $\delta_k^{N_k} \to 0$ as $N_k \to \infty$. Note that $c_k$ depends on $\psi_1, \dots, \psi_{k-1}$ (through $f_{k-1}$), which have already been constructed, so there is no circularity. Define $\psi_k := (k + |f_{k-1}(w_k)|) \cdot \varphi_k^{N_k}$. Then: - $\sup_{K_k} |\psi_k| = (k + |f_{k-1}(w_k)|) \cdot \delta_k^{N_k} < 2^{-k}$, ensuring [uniform convergence](/page/Uniform%20Convergence) on compact subsets (since $K_j \subset K_k$ for $j \leq k$). - $|\psi_k(w_k)| = (k + |f_{k-1}(w_k)|) \cdot |\varphi_k(w_k)|^{N_k} = k + |f_{k-1}(w_k)|$. The series $f = \sum_{k=1}^\infty \psi_k$ converges uniformly on each $K_j$ (for $k \geq j$, $\sup_{K_j} |\psi_k| \leq \sup_{K_k} |\psi_k| < 2^{-k}$), so $f \in \mathcal{O}(\Omega)$. **Proving unboundedness at the points $w_k$.** At $w_k$, we decompose: \begin{align*} |f(w_k)| &= \left|f_{k-1}(w_k) + \psi_k(w_k) + \sum_{m=k+1}^{\infty} \psi_m(w_k)\right| \\ &\geq |\psi_k(w_k)| - |f_{k-1}(w_k)| - \sum_{m=k+1}^{\infty} |\psi_m(w_k)|. \end{align*} The first term is $|\psi_k(w_k)| = k + |f_{k-1}(w_k)|$, and the second subtracted term is $|f_{k-1}(w_k)|$. These partially cancel, leaving: \begin{align*} |f(w_k)| \geq k - \sum_{m=k+1}^{\infty} |\psi_m(w_k)|. \end{align*} This is where the adaptive scaling $c_k = k + |f_{k-1}(w_k)|$ pays off: the past terms are exactly absorbed. **Controlling the tail.** For the tail $\sum_{m > k} |\psi_m(w_k)|$, we need to control where $w_k$ sits in the exhaustion. Since $w_k \in \Omega = \bigcup_j K_j$, there exists $j(k) \geq 1$ such that $w_k \in K_{j(k)}$. By passing to a further subsequence and relabelling, we may assume $w_k \in K_{k+1} \setminus K_k$ for all $k$ (this is possible because $(w_k)$ eventually leaves each $K_j$). Then for $m \geq k+1$, $w_k \in K_{k+1} \subset K_m$, so \begin{align*} |\psi_m(w_k)| \leq \sup_{K_m} |\psi_m| < 2^{-m}. \end{align*} Therefore the tail is bounded by a geometric series: \begin{align*} R_k := \sum_{m=k+1}^{\infty} |\psi_m(w_k)| < \sum_{m=k+1}^{\infty} 2^{-m} = 2^{-k}. \end{align*} Substituting back: $|f(w_k)| \geq k - R_k > k - 2^{-k}$. Since $k - 2^{-k} \to \infty$ as $k \to \infty$, we conclude $|f(w_k)| \to \infty$. In particular, $\sup_k |f(w_k)| = \infty$. **An alternative argument by contradiction.** If one prefers not to pass to the subsequence $w_k \in K_{k+1}$, one can argue as follows. Suppose for contradiction that $|f(w_k)| \leq M$ for all $k$. Then from $|f(w_k)| \geq k - R_k$ and the fact that $R_k \to 0$ along the subsequence with $w_k \in K_{k+1}$ (which exists since $(w_k)$ eventually leaves every $K_j$), we get $k \leq M + 2^{-k} \leq M + 1$ for all large $k$ --- a contradiction. [/proof] [/claim] **Applying the claim to complete the proof.** We apply this claim to the sequence $(w_k)$ constructed above. Since $\{b_k\}_{k \geq 1}$ is dense in $\partial\Omega$ and $|w_k - b_k| < 1/k$, the sequence $(w_k)$ has no accumulation point in $\Omega$. Why? Any accumulation point $z_0$ of $(w_k)$ would satisfy $\operatorname{dist}(z_0, \partial\Omega) \leq |z_0 - b_k| \leq |z_0 - w_k| + |w_k - b_k|$, and as $w_k \to z_0$ along a subsequence with $|w_k - b_k| < 1/k \to 0$, we get $\operatorname{dist}(z_0, \partial\Omega) = 0$, so $z_0 \in \partial\Omega$, which is disjoint from $\Omega$. By the claim, there exists $f \in \mathcal{O}(\Omega)$ with $\sup_k |f(w_k)| = \infty$. **No holomorphic extension is possible.** Suppose $f$ could be holomorphically extended to an [open set](/page/Open%20Set) $U$ containing a boundary point $b \in \partial\Omega$. Then $f$ would be continuous (hence bounded) on some compact neighbourhood $\overline{B}(b, \varepsilon) \cap \overline{U}$. But since $\{b_k\}$ is dense in $\partial\Omega$, infinitely many $b_k$ lie in $B(b, \varepsilon/2)$, and for $k$ large enough that $1/k < \varepsilon/2$, the corresponding $w_k$ satisfy $|w_k - b| \leq |w_k - b_k| + |b_k - b| < 1/k + \varepsilon/2 < \varepsilon$, so $w_k \in B(b, \varepsilon) \cap \Omega \subset U$. Since $|f(w_k)|$ is unbounded along a subsequence, this contradicts the continuity of $f$ on $U$. Therefore $\Omega$ is a domain of holomorphy. **Combining both directions.** The forward direction (Step 1 and Step 2) showed that domain of holomorphy implies holomorphically convex. The reverse direction (Step 3 and this step) showed that holomorphically convex implies domain of holomorphy. Therefore a domain $\Omega \subset \mathbb{C}^n$ is a domain of holomorphy if and only if it is holomorphically convex. [/guided] [/step]