[proofplan] We prove that $d(\hat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega)$ for every compact $K \subset \Omega$, where $\Omega$ is a domain of holomorphy in $\mathbb{C}^n$. The inequality $d(\hat{K}_\Omega, \partial\Omega) \leq d(K, \partial\Omega)$ is immediate from $K \subset \hat{K}_\Omega$. For the reverse, we fix $z_0 \in \hat{K}_\Omega$, derive Cauchy coefficient bounds at points of $K$, transfer them to $z_0$ via the holomorphic hull property, and conclude that the Taylor series at $z_0$ converges on polydiscs of polyradius approaching $r\mathbf{1}$ (where $r = d(K, \partial\Omega)$). Since $\Omega$ is a domain of holomorphy -- equivalently, holomorphically convex by the [Cartan-Thullen Theorem](/theorems/3385) -- these polydiscs must lie inside $\Omega$, giving $d(z_0, \partial\Omega) \geq r$. [/proofplan]

[step:Establish the inequality $d(\hat{K}_\Omega, \partial\Omega) \leq d(K, \partial\Omega)$] Since $K \subset \hat{K}_\Omega$, every point of $K$ is also a point of $\hat{K}_\Omega$. Therefore \begin{align*} d(\hat{K}_\Omega, \partial\Omega) = \inf_{z \in \hat{K}_\Omega} d(z, \partial\Omega) \leq \inf_{z \in K} d(z, \partial\Omega) = d(K, \partial\Omega). \end{align*} [/step]

[step:Bound the Taylor coefficients at $w \in K$ using the Cauchy inequalities on polydiscs]Let $r := d(K, \partial\Omega) > 0$ (since $K$ is compact and $K \subset \Omega$). Fix any $f \in \mathcal{O}(\Omega)$ and any $w \in K$. For each $0 < \rho < r$, the closed polydisc $\overline{\mathbb{D}^n(w, \rho\mathbf{1})}$ (where $\rho\mathbf{1} = (\rho, \dots, \rho) \in \mathbb{R}^n_{+}$) satisfies $\overline{\mathbb{D}^n(w, \rho\mathbf{1})} \subset \Omega$, so $f$ is holomorphic on a neighbourhood of $\overline{\mathbb{D}^n(w, \rho\mathbf{1})}$ and admits the convergent Taylor expansion \begin{align*} f(z) = \sum_{\alpha \in \mathbb{N}_0^n} c_\alpha(w) (z - w)^\alpha, \qquad c_\alpha(w) = \frac{1}{\alpha!} \partial^\alpha f(w), \qquad z \in \mathbb{D}^n(w, \rho\mathbf{1}), \end{align*} where $\mathbb{N}_0 = \{0, 1, 2, \dots\}$. By the [Cauchy Integral Formula on Polydiscs](/theorems/3398), applied to $f$ on $\mathbb{D}^n(w, \rho\mathbf{1})$, the Cauchy inequalities give \begin{align*} |c_\alpha(w)| \leq \frac{M_f(w, \rho)}{\rho^{|\alpha|}}, \end{align*} where $M_f(w, \rho) := \sup_{\zeta \in \overline{\mathbb{D}^n(w, \rho\mathbf{1})}} |f(\zeta)|$.[/step]

[guided]The strategy is classical: we want to show that the [power series](/page/Power%20Series) of $f$ centred at some point $w \in K$ converges not just in a polydisc inside $\Omega$, but at the point $z_0 \in \hat{K}_\Omega$ as well. The Cauchy inequalities are the tool that converts a bound on $|f|$ over a polydisc into bounds on the Taylor coefficients $c_\alpha(w)$. The key will be that the definition of $\hat{K}_\Omega$ lets us transfer these coefficient bounds from $w$ to $z_0$. Fix any $f \in \mathcal{O}(\Omega)$ and $w \in K$. Let $r = d(K, \partial\Omega) > 0$. For any $0 < \rho < r$, the closed polydisc $\overline{\mathbb{D}^n(w, \rho\mathbf{1})}$ is contained in $\Omega$, so $f$ is holomorphic on a neighbourhood of $\overline{\mathbb{D}^n(w, \rho\mathbf{1})}$ and has the convergent Taylor expansion \begin{align*} f(z) = \sum_{\alpha \in \mathbb{N}_0^n} c_\alpha(w)(z - w)^\alpha, \qquad c_\alpha(w) = \frac{1}{\alpha!}\partial^\alpha f(w), \qquad z \in \mathbb{D}^n(w, \rho\mathbf{1}). \end{align*} The [Cauchy Integral Formula on Polydiscs](/theorems/3398) expresses $c_\alpha(w)$ as an integral over the distinguished boundary (the torus $\{|\zeta_j - w_j| = \rho : j = 1, \dots, n\}$), and bounding the integrand pointwise gives the Cauchy inequalities: \begin{align*} |c_\alpha(w)| \leq \frac{M_f(w, \rho)}{\rho^{|\alpha|}}, \end{align*} where $M_f(w, \rho) = \sup_{\zeta \in \overline{\mathbb{D}^n(w, \rho\mathbf{1})}} |f(\zeta)|$.[/guided]

[step:Transfer the coefficient bounds from $K$ to $z_0$ via the holomorphic hull property]Fix $z_0 \in \hat{K}_\Omega$. For each multi-index $\alpha \in \mathbb{N}_0^n$, define the [holomorphic function](/page/Holomorphic%20Function) \begin{align*} g_\alpha := \frac{1}{\alpha!}\partial^\alpha f \in \mathcal{O}(\Omega). \end{align*} By definition of the holomorphic hull, $z_0 \in \hat{K}_\Omega$ means $|g(z_0)| \leq \sup_K |g|$ for every $g \in \mathcal{O}(\Omega)$. Applying this to $g_\alpha$: \begin{align*} |c_\alpha(z_0)| = |g_\alpha(z_0)| \leq \sup_{w \in K} |g_\alpha(w)| = \sup_{w \in K} |c_\alpha(w)|. \end{align*} Combining with the Cauchy inequality from the previous step at each $w \in K$: \begin{align*} |c_\alpha(z_0)| \leq \sup_{w \in K} |c_\alpha(w)| \leq \sup_{w \in K} \frac{M_f(w, \rho)}{\rho^{|\alpha|}} = \frac{M}{\rho^{|\alpha|}}, \end{align*} where $M := \sup_{w \in K} M_f(w, \rho) = \sup_{w \in K}\, \sup_{\zeta \in \overline{\mathbb{D}^n(w, \rho\mathbf{1})}} |f(\zeta)|$. Since $K$ is compact and $\rho < r = d(K, \partial\Omega)$, the set $K_\rho := \bigcup_{w \in K} \overline{\mathbb{D}^n(w, \rho\mathbf{1})}$ is a compact subset of $\Omega$, so $M = \sup_{K_\rho} |f| < \infty$ (as $f$ is continuous on the compact set $K_\rho$).[/step]

[guided]The crucial idea: the definition of $\hat{K}_\Omega$ gives us control over $|g(z_0)|$ for any $g \in \mathcal{O}(\Omega)$, but only in terms of $\sup_K |g|$. The Taylor coefficients $c_\alpha(z_0) = \frac{1}{\alpha!}\partial^\alpha f(z_0)$ are values of holomorphic functions $g_\alpha = \frac{1}{\alpha!}\partial^\alpha f$ at $z_0$, so the hull property applies to each coefficient individually. For each $\alpha$, $g_\alpha \in \mathcal{O}(\Omega)$. Since $z_0 \in \hat{K}_\Omega$: \begin{align*} |c_\alpha(z_0)| = |g_\alpha(z_0)| \leq \sup_{w \in K} |g_\alpha(w)| = \sup_{w \in K} |c_\alpha(w)|. \end{align*} Now at each $w \in K$, we bound $|c_\alpha(w)|$ using the Cauchy inequalities on the polydisc of radius $\rho < r$ centred at $w$ (which is compactly contained in $\Omega$): \begin{align*} |c_\alpha(w)| \leq \frac{M_f(w, \rho)}{\rho^{|\alpha|}}. \end{align*} Combining and taking the supremum over $w \in K$: \begin{align*} |c_\alpha(z_0)| \leq \frac{M}{\rho^{|\alpha|}}, \qquad M = \sup_{w \in K} M_f(w, \rho). \end{align*} Since $K$ is compact and $\rho < r$, the $\rho$-neighbourhood $K_\rho = \bigcup_{w \in K}\overline{\mathbb{D}^n(w, \rho\mathbf{1})}$ is a compact subset of $\Omega$, so $M = \sup_{K_\rho} |f| < \infty$ (as $f$ is continuous on the compact set $K_\rho$). This uniform bound on the Taylor coefficients at $z_0$ is the ingredient for showing convergence in the next step.[/guided]

[step:Show that the Taylor series at $z_0$ converges on $\mathbb{D}^n(z_0, \rho\mathbf{1})$ for every $\rho < r$, and conclude $d(z_0, \partial\Omega) \geq r$]Consider the Taylor expansion of $f$ at $z_0$. For any $z \in \mathbb{C}^n$ and any multi-index $\alpha$, we have \begin{align*} |(z - z_0)^\alpha| = \prod_{j=1}^n |z_j - z_{0,j}|^{\alpha_j} \leq \|z - z_0\|_\infty^{|\alpha|}, \end{align*} where $\|z - z_0\|_\infty := \max_{1 \leq j \leq n} |z_j - z_{0,j}|$ is the sup-norm on $\mathbb{C}^n$. Using the coefficient bound $|c_\alpha(z_0)| \leq M / \rho^{|\alpha|}$ from the previous step: \begin{align*} \sum_{\alpha \in \mathbb{N}_0^n} |c_\alpha(z_0)(z - z_0)^\alpha| \leq M \sum_{\alpha \in \mathbb{N}_0^n} \left(\frac{\|z - z_0\|_\infty}{\rho}\right)^{|\alpha|} = M \left(\frac{1}{1 - \|z - z_0\|_\infty / \rho}\right)^n, \end{align*} provided $\|z - z_0\|_\infty < \rho$, where the last equality uses the identity $\sum_{\alpha \in \mathbb{N}_0^n} t^{|\alpha|} = (1 - t)^{-n}$ for $0 \leq t < 1$ (grouping by $|\alpha| = k$ and counting the $\binom{n + k - 1}{k}$ multi-indices of total degree $k$). The Taylor series of $f$ at $z_0$ therefore converges absolutely on the polydisc $\mathbb{D}^n(z_0, \rho\mathbf{1}) = \{z \in \mathbb{C}^n : \|z - z_0\|_\infty < \rho\}$. Since $\rho < r$ is arbitrary, the Taylor series at $z_0$ converges absolutely on $\mathbb{D}^n(z_0, \rho\mathbf{1})$ for every $\rho < r$, and the sum defines a [holomorphic function](/page/Holomorphic%20Function) there that agrees with $f$ on $\mathbb{D}^n(z_0, \rho\mathbf{1}) \cap \Omega$. By the [Cartan-Thullen Theorem](/theorems/3385), $\Omega$ is a domain of holomorphy if and only if $\Omega$ is holomorphically convex, and in particular $f$ cannot be holomorphically continued past any boundary point of $\Omega$. If $\mathbb{D}^n(z_0, \rho\mathbf{1}) \not\subset \Omega$ for some $\rho < r$, then the convergent Taylor series would provide a holomorphic extension of $f$ beyond $\Omega$, contradicting the domain-of-holomorphy hypothesis. Therefore $\mathbb{D}^n(z_0, \rho\mathbf{1}) \subset \Omega$ for every $\rho < r$, which gives \begin{align*} d(z_0, \partial\Omega) \geq r = d(K, \partial\Omega). \end{align*}[/step]

[guided]**Convergence.** From the previous step, $|c_\alpha(z_0)| \leq M/\rho^{|\alpha|}$ for all $\alpha$, where $M < \infty$ depends on $f$, $K$, and $\rho$ but not on $\alpha$. To estimate the Taylor series at $z_0$, we need to bound $|(z - z_0)^\alpha| = \prod_j |z_j - z_{0,j}|^{\alpha_j}$. The natural norm to use is the sup-norm $\|z - z_0\|_\infty = \max_j |z_j - z_{0,j}|$, because \begin{align*} \prod_{j=1}^n |z_j - z_{0,j}|^{\alpha_j} \leq \left(\max_j |z_j - z_{0,j}|\right)^{\alpha_1 + \cdots + \alpha_n} = \|z - z_0\|_\infty^{|\alpha|}. \end{align*} Why the sup-norm and not the Euclidean norm $|z - z_0|$? Because the factorisation $|(z - z_0)^\alpha| = \prod_j |z_j - z_{0,j}|^{\alpha_j}$ involves coordinate-wise absolute values, and each satisfies $|z_j - z_{0,j}| \leq \|z - z_0\|_\infty$ by definition. The same bound with the Euclidean norm would be false: $\prod_j |z_j - z_{0,j}|^{\alpha_j}$ does not equal $|z - z_0|^{|\alpha|}$ in general (consider $z - z_0 = (1, 1)$ with $\alpha = (1, 1)$: then $\prod_j |z_j - z_{0,j}|^{\alpha_j} = 1$ but $|z - z_0|^{|\alpha|} = (\sqrt{2})^2 = 2$). With this bound: \begin{align*} \sum_\alpha |c_\alpha(z_0)(z - z_0)^\alpha| \leq M \sum_\alpha \left(\frac{\|z - z_0\|_\infty}{\rho}\right)^{|\alpha|} = M \left(\frac{1}{1 - \|z - z_0\|_\infty/\rho}\right)^n \end{align*} provided $\|z - z_0\|_\infty < \rho$. The last equality uses the identity $\sum_{\alpha \in \mathbb{N}_0^n} t^{|\alpha|} = (1 - t)^{-n}$ for $0 \leq t < 1$, which follows by grouping: there are $\binom{n + k - 1}{k}$ multi-indices with $|\alpha| = k$, and $\sum_{k=0}^\infty \binom{n+k-1}{k} t^k = (1-t)^{-n}$ is the generating function for multisets. Since $\rho$ can be taken arbitrarily close to $r$, the series converges whenever $\|z - z_0\|_\infty < r$, i.e., whenever $z \in \mathbb{D}^n(z_0, r\mathbf{1})$. **Distance conclusion.** The convergent Taylor series defines a [holomorphic function](/page/Holomorphic%20Function) on $\mathbb{D}^n(z_0, \rho\mathbf{1})$ that agrees with $f$ on $\mathbb{D}^n(z_0, \rho\mathbf{1}) \cap \Omega$ (by the identity theorem, since both are holomorphic and agree on a non-empty open subset). If $\mathbb{D}^n(z_0, \rho\mathbf{1}) \not\subset \Omega$ for some $\rho < r$, we would have a holomorphic extension of $f$ past $\partial\Omega$. But $\Omega$ is a domain of holomorphy, which by the [Cartan-Thullen Theorem](/theorems/3385) means no $f \in \mathcal{O}(\Omega)$ extends holomorphically past any boundary point. This contradicts the existence of such an extension, so $\mathbb{D}^n(z_0, \rho\mathbf{1}) \subset \Omega$ for every $\rho < r$. Taking $\rho \to r$, we conclude $d(z_0, \partial\Omega) \geq r = d(K, \partial\Omega)$.[/guided]

[step:Combine both inequalities to conclude $d(\hat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega)$] The first step gives $d(\hat{K}_\Omega, \partial\Omega) \leq d(K, \partial\Omega)$. The argument in the preceding steps shows that for every $z_0 \in \hat{K}_\Omega$, $d(z_0, \partial\Omega) \geq d(K, \partial\Omega)$. Taking the infimum over all $z_0 \in \hat{K}_\Omega$: \begin{align*} d(\hat{K}_\Omega, \partial\Omega) = \inf_{z_0 \in \hat{K}_\Omega} d(z_0, \partial\Omega) \geq d(K, \partial\Omega). \end{align*} Therefore $d(\hat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega)$. [/step]