[proofplan] We show that the holomorphic hull $\hat{K}_\Omega$ of a torus-invariant compact set $K$ in a logarithmically convex Reinhardt domain $\Omega$ is the Reinhardt domain whose logarithmic image is the convex hull of $\operatorname{Log}(K \cap (\mathbb{C}^*)^n)$ intersected with $\operatorname{Log}(\Omega)$. The proof proceeds in two parts: first, we show that the monomials $z^\alpha$ for $\alpha \in \mathbb{Z}^n$ already impose the constraint that $\operatorname{Log}(\hat{K}_\Omega)$ lies in the convex hull; second, we show that monomials suffice to determine the hull, since every [holomorphic function](/page/Holomorphic%20Function) on $\Omega$ is a limit of Laurent polynomials on compact subsets. [/proofplan] [step:Show that monomial inequalities force the log-image of $\hat{K}_\Omega$ into the convex hull] Let $L := \operatorname{Log}(K \cap (\mathbb{C}^*)^n) \subset \mathbb{R}^n$ denote the logarithmic image of $K$ restricted to $(\mathbb{C}^*)^n$, and let $\operatorname{conv}(L)$ denote its convex hull. We claim that if $z \in \hat{K}_\Omega \cap (\mathbb{C}^*)^n$, then $s := (\log|z_1|, \dots, \log|z_n|) \in \operatorname{conv}(L) \cap \operatorname{Log}(\Omega)$. For each $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{Z}^n$, the monomial \begin{align*} f_\alpha: \Omega \cap (\mathbb{C}^*)^n &\to \mathbb{C} \\ z &\mapsto z^\alpha = z_1^{\alpha_1} \cdots z_n^{\alpha_n} \end{align*} is holomorphic on $\Omega \cap (\mathbb{C}^*)^n$ (and on all of $\Omega$ when $\alpha \in \mathbb{N}_0^n$). Since $z \in \hat{K}_\Omega$, the definition of the holomorphic hull gives $|f_\alpha(z)| \leq \sup_K |f_\alpha|$, i.e., \begin{align*} |z^\alpha| \leq \sup_{\zeta \in K} |\zeta^\alpha|. \end{align*} Taking logarithms and using $|z^\alpha| = |z_1|^{\alpha_1} \cdots |z_n|^{\alpha_n} = e^{\alpha \cdot s}$ where $s = (\log|z_1|, \dots, \log|z_n|)$: \begin{align*} \alpha \cdot s \leq \sup_{p \in L} \alpha \cdot p. \end{align*} This holds for every $\alpha \in \mathbb{Z}^n$. By the hyperplane separation theorem for convex sets, a point $s \in \mathbb{R}^n$ belongs to the closed convex hull $\overline{\operatorname{conv}(L)}$ if and only if $\alpha \cdot s \leq \sup_{p \in L} \alpha \cdot p$ for every $\alpha \in \mathbb{R}^n$. Since $\mathbb{Z}^n$ is dense in $\mathbb{R}^n$ in the sense that for every $\alpha \in \mathbb{R}^n$ and every $\varepsilon > 0$, the functional $\alpha \cdot s$ can be approximated by $(\alpha'/m) \cdot s$ for $\alpha' \in \mathbb{Z}^n$ and $m \in \mathbb{N}$ (using rational approximation of each component of $\alpha$), the constraint $\alpha \cdot s \leq \sup_L \alpha \cdot p$ for all $\alpha \in \mathbb{Z}^n$ implies the same for all $\alpha \in \mathbb{R}^n$. Therefore $s \in \overline{\operatorname{conv}(L)}$. Since $\hat{K}_\Omega \subset \Omega$ (the holomorphic hull is contained in the domain by definition), we also have $s \in \operatorname{Log}(\Omega)$. As $\operatorname{Log}(\Omega)$ is open and $L \subset \operatorname{Log}(\Omega)$, the intersection $\operatorname{conv}(L) \cap \operatorname{Log}(\Omega)$ is open, and $s$ lies in this intersection. [guided] The key idea is that monomials $z^\alpha$ act as "linear functionals in disguise." In logarithmic coordinates, the inequality $|z^\alpha| \leq \sup_K |z^\alpha|$ becomes the linear inequality $\alpha \cdot s \leq \sup_{p \in L} \alpha \cdot p$. This is precisely the condition that $s$ lies in the half-space $\{s : \alpha \cdot s \leq h_L(\alpha)\}$, where $h_L(\alpha) := \sup_{p \in L} \alpha \cdot p$ is the support function of $L$. Let $L := \operatorname{Log}(K \cap (\mathbb{C}^*)^n) \subset \mathbb{R}^n$ and $\operatorname{conv}(L)$ its convex hull. For each $\alpha \in \mathbb{Z}^n$, the monomial $f_\alpha(z) = z^\alpha = z_1^{\alpha_1} \cdots z_n^{\alpha_n}$ is holomorphic on $\Omega$. Since $z \in \hat{K}_\Omega$, the definition of the holomorphic hull gives $|z^\alpha| \leq \sup_K |z^\alpha|$. Taking logarithms with $s = (\log|z_1|, \dots, \log|z_n|)$: \begin{align*} \alpha \cdot s \leq \sup_{p \in L} \alpha \cdot p =: h_L(\alpha). \end{align*} The convex hull of a set $L \subset \mathbb{R}^n$ is characterised by support functions: a point $s$ belongs to $\overline{\operatorname{conv}(L)}$ if and only if $\alpha \cdot s \leq h_L(\alpha)$ for every direction $\alpha \in \mathbb{R}^n$. We have this inequality for all $\alpha \in \mathbb{Z}^n$ (from the monomial constraints), and we need it for all $\alpha \in \mathbb{R}^n$. Why does $\mathbb{Z}^n$ suffice? For any $\alpha \in \mathbb{R}^n$, choose a sequence $\alpha^{(k)} \in \mathbb{Q}^n$ with $\alpha^{(k)} \to \alpha$. Writing $\alpha^{(k)} = \beta^{(k)}/m_k$ with $\beta^{(k)} \in \mathbb{Z}^n$ and $m_k \in \mathbb{N}$, the monomial constraint gives $\beta^{(k)} \cdot s \leq \sup_L \beta^{(k)} \cdot p$. Dividing by $m_k$: $\alpha^{(k)} \cdot s \leq \sup_L \alpha^{(k)} \cdot p$. Taking $k \to \infty$: $\alpha \cdot s \leq \sup_L \alpha \cdot p = h_L(\alpha)$. The passage to the limit is valid because both sides are continuous in $\alpha$ (the left side $\alpha \cdot s$ is linear in $\alpha$, and the right side $h_L(\alpha) = \sup_{p \in L} \alpha \cdot p$ is continuous on $\mathbb{R}^n$ when $L$ is compact, being the supremum of a family of continuous linear functions). Therefore $s \in \overline{\operatorname{conv}(L)}$, and since $s \in \operatorname{Log}(\Omega)$ (an [open set](/page/Open%20Set)), we get $s \in \operatorname{conv}(L) \cap \operatorname{Log}(\Omega)$. [/guided] [/step] [step:Show that monomials suffice by approximation via Laurent polynomials] We now show the reverse inclusion: every point $z \in \Omega$ whose logarithmic image $s = (\log|z_1|, \dots, \log|z_n|)$ lies in $\operatorname{conv}(L) \cap \operatorname{Log}(\Omega)$ belongs to $\hat{K}_\Omega$. It suffices to show that for every $f \in \mathcal{O}(\Omega)$, the inequality $|f(z)| \leq \sup_K |f|$ holds. Since $\Omega$ is a logarithmically convex Reinhardt domain, by the [Domains of Holomorphy Among Reinhardt Domains](/theorems/3394) it is a domain of holomorphy, and by the [Laurent Series in Complete Reinhardt Domains](/theorems/3419) every $f \in \mathcal{O}(\Omega)$ has a [Laurent series expansion](/theorems/350) \begin{align*} f(z) = \sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha \end{align*} converging absolutely and uniformly on compact subsets of $\Omega \cap (\mathbb{C}^*)^n$. The partial sums \begin{align*} S_N(z) = \sum_{|\alpha| \leq N} a_\alpha z^\alpha \end{align*} are Laurent polynomials converging uniformly to $f$ on $K$ (since $K$ is compact in $\Omega$). For each Laurent polynomial $S_N$, the inequality $|S_N(z)| \leq \sup_K |S_N|$ holds for all $z$ with $s \in \operatorname{conv}(L)$. This follows from the torus invariance of $K$: since $K$ is torus-invariant, the [maximum modulus principle](/page/Maximum%20Modulus%20Principle) on torus orbits gives $|S_N(z)| \leq \sup_K |S_N|$ directly for each $N$. (On each torus orbit $\{(\zeta_1, \dots, \zeta_n) : |\zeta_j| = |z_j|\}$, the Laurent polynomial $S_N$ restricted to the torus attains its maximum modulus on the boundary of $K$'s intersection with that torus orbit, and since $K$ is torus-invariant, this maximum is $\sup_K |S_N|$.) Taking $N \to \infty$: [uniform convergence](/page/Uniform%20Convergence) $S_N \to f$ on $K$ gives $\sup_K |S_N| \to \sup_K |f|$, and $S_N(z) \to f(z)$ pointwise. Since $|S_N(z)| \leq \sup_K |S_N|$ for each $N$, passing to the limit yields $|f(z)| \leq \sup_K |f|$. Therefore $z \in \hat{K}_\Omega$, and the holomorphic hull is exactly the Reinhardt domain whose logarithmic image is $\operatorname{conv}(L) \cap \operatorname{Log}(\Omega)$. [guided] The previous step showed that monomial constraints force $\operatorname{Log}(\hat{K}_\Omega) \subset \operatorname{conv}(L)$. This step proves the reverse inclusion: every point $z \in \Omega$ whose logarithmic image $s = (\log|z_1|, \dots, \log|z_n|)$ lies in $\operatorname{conv}(L) \cap \operatorname{Log}(\Omega)$ belongs to $\hat{K}_\Omega$. To show $z \in \hat{K}_\Omega$, we must verify that $|f(z)| \leq \sup_K |f|$ for every $f \in \mathcal{O}(\Omega)$. Why are monomials sufficient to determine the hull? Because on a logarithmically convex Reinhardt domain, the Laurent expansion expresses every [holomorphic function](/page/Holomorphic%20Function) as a convergent sum of monomials, and the monomial constraints from the previous step pass to arbitrary holomorphic functions by approximation. Since $\Omega$ is a logarithmically convex Reinhardt domain, by the [Domains of Holomorphy Among Reinhardt Domains](/theorems/3394) it is a domain of holomorphy, and by the [Laurent Series in Complete Reinhardt Domains](/theorems/3419) every $f \in \mathcal{O}(\Omega)$ has a [Laurent series](/page/Laurent%20Series) expansion \begin{align*} f(z) = \sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha \end{align*} converging absolutely and uniformly on compact subsets of $\Omega \cap (\mathbb{C}^*)^n$. The partial sums \begin{align*} S_N(z) = \sum_{|\alpha| \leq N} a_\alpha z^\alpha \end{align*} are Laurent polynomials. Since $K$ is compact in $\Omega$, the convergence $S_N \to f$ is uniform on $K$, so $\sup_K |S_N| \to \sup_K |f|$ and $S_N(z) \to f(z)$ pointwise. Now we show that the monomial constraints pass to Laurent polynomials. Each $S_N$ is a finite linear combination of monomials $z^\alpha$. For each such monomial, the previous step established $|z^\alpha| \leq \sup_K |z^\alpha|$ (equivalently $\alpha \cdot s \leq \sup_{p \in L} \alpha \cdot p$) whenever $s \in \operatorname{conv}(L)$. The sharper argument uses the torus invariance of $K$: since $K$ is torus-invariant, the [maximum modulus principle](/theorems/491) on torus orbits gives $|S_N(z)| \leq \sup_K |S_N|$ directly for each $N$. On each torus orbit $\{(\zeta_1, \dots, \zeta_n) : |\zeta_j| = |z_j|\}$, the Laurent polynomial $S_N$ restricted to the torus attains its maximum modulus on the boundary of $K$'s intersection with that torus orbit, and since $K$ is torus-invariant, this maximum is $\sup_K |S_N|$. Taking $N \to \infty$: the [uniform convergence](/page/Uniform%20Convergence) $S_N \to f$ on $K$ gives $\sup_K |S_N| \to \sup_K |f|$, and pointwise convergence gives $S_N(z) \to f(z)$. Since $|S_N(z)| \leq \sup_K |S_N|$ for each $N$, passing to the limit yields $|f(z)| \leq \sup_K |f|$. Therefore $z \in \hat{K}_\Omega$, and the holomorphic hull is exactly the Reinhardt domain whose logarithmic image is $\operatorname{conv}(L) \cap \operatorname{Log}(\Omega)$. This is an instance of a general principle: when a domain has a large symmetry group (here, the torus $\mathbb{T}^n$), the holomorphic hull can be computed using a smaller class of "equivariant" functions (here, monomials). The torus symmetry decomposes $\mathcal{O}(\Omega)$ into weight spaces indexed by $\alpha \in \mathbb{Z}^n$, each one-dimensional (spanned by $z^\alpha$). The analogy with convex geometry is exact: the holomorphic hull of $K$ in $\Omega$ corresponds to the convex hull of $L$ in $\operatorname{Log}(\Omega)$, with monomials playing the role of linear functionals and the [maximum modulus principle](/page/Maximum%20Modulus%20Principle) playing the role of the hyperplane separation theorem. [/guided] [/step]