[proofplan] We prove that the domain of convergence of a [Laurent series](/page/Laurent%20Series) $\sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha$ is a logarithmically convex complete Reinhardt domain (possibly empty), and conversely that every such domain arises in this way. The argument proceeds in four stages: first, we show the domain of convergence is a Reinhardt domain by verifying torus invariance of the absolute convergence condition; second, we prove logarithmic convexity by passing to logarithmic coordinates $s_j = \log|z_j|$ and applying Holder's inequality to show that the set of $s$ for which the sum $\sum_\alpha |a_\alpha| e^{\alpha \cdot s}$ converges is convex; third, we establish completeness via the Abel lemma; fourth, we construct a [Laurent series](/page/Laurent%20Series) realising any given logarithmically convex Reinhardt domain. [/proofplan]

[step:Define the domain of convergence and show it is a Reinhardt domain] Let $\sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha$ be a [Laurent series](/page/Laurent%20Series) with coefficients $a_\alpha \in \mathbb{C}$. Define the set of absolute convergence \begin{align*} S = \{z \in (\mathbb{C}^*)^n : \sum_{\alpha \in \mathbb{Z}^n} |a_\alpha| |z^\alpha| < \infty\}. \end{align*} The domain of convergence $\Omega$ is defined as the interior of $S$ (in $\mathbb{C}^n$ or $(\mathbb{C}^*)^n$, as appropriate). We verify that $S$ is a Reinhardt set, i.e., invariant under the torus action $(z_1, \dots, z_n) \mapsto (e^{i\theta_1} z_1, \dots, e^{i\theta_n} z_n)$ for all $(\theta_1, \dots, \theta_n) \in \mathbb{R}^n$. For $z \in S$ and $\theta \in \mathbb{R}^n$, define $w_j = e^{i\theta_j} z_j$. Then $|w^\alpha| = |w_1^{\alpha_1} \cdots w_n^{\alpha_n}| = |z_1|^{\alpha_1} \cdots |z_n|^{\alpha_n} = |z^\alpha|$ for all $\alpha \in \mathbb{Z}^n$, so \begin{align*} \sum_{\alpha \in \mathbb{Z}^n} |a_\alpha| |w^\alpha| = \sum_{\alpha \in \mathbb{Z}^n} |a_\alpha| |z^\alpha| < \infty. \end{align*} Therefore $w \in S$, confirming torus invariance. Since $S$ is torus-invariant and $\Omega = S^\circ$, the domain $\Omega$ is also torus-invariant (the interior of a torus-invariant set is torus-invariant), so $\Omega$ is a Reinhardt domain. [/step]

[step:Pass to logarithmic coordinates and prove logarithmic convexity via Holder's inequality]Define the logarithmic image of $S$: \begin{align*} \operatorname{Log}(S) = \{(\log|z_1|, \dots, \log|z_n|) : z \in S \cap (\mathbb{C}^*)^n\} \subset \mathbb{R}^n. \end{align*} We show that $\operatorname{Log}(S)$ is a convex subset of $\mathbb{R}^n$. For $s = (s_1, \dots, s_n) \in \mathbb{R}^n$, write $r_j = e^{s_j}$. Then $z \in S$ with $|z_j| = r_j$ if and only if \begin{align*} \sum_{\alpha \in \mathbb{Z}^n} |a_\alpha| e^{\alpha \cdot s} < \infty \end{align*} where $\alpha \cdot s = \alpha_1 s_1 + \cdots + \alpha_n s_n$, since $|z^\alpha| = r_1^{\alpha_1} \cdots r_n^{\alpha_n} = e^{\alpha_1 s_1 + \cdots + \alpha_n s_n} = e^{\alpha \cdot s}$. Thus $\operatorname{Log}(S) = \{s \in \mathbb{R}^n : \sum_{\alpha} |a_\alpha| e^{\alpha \cdot s} < \infty\}$. Let $s, s' \in \operatorname{Log}(S)$ and $t \in [0,1]$. Set $s'' = (1-t)s + ts'$. For each $\alpha \in \mathbb{Z}^n$, the map $s \mapsto \alpha \cdot s$ is linear, so \begin{align*} \alpha \cdot s'' = (1-t)(\alpha \cdot s) + t(\alpha \cdot s'). \end{align*} Therefore each term at the interpolated point factorises as \begin{align*} |a_\alpha| e^{\alpha \cdot s''} = |a_\alpha| e^{(1-t)\alpha \cdot s + t\alpha \cdot s'} = \left(|a_\alpha| e^{\alpha \cdot s}\right)^{1-t} \left(|a_\alpha| e^{\alpha \cdot s'}\right)^t. \end{align*} Summing over $\alpha$ and applying Holder's inequality for sums with conjugate exponents $p = 1/(1-t)$ and $q = 1/t$: \begin{align*} \sum_{\alpha} |a_\alpha| e^{\alpha \cdot s''} &= \sum_{\alpha} \left(|a_\alpha| e^{\alpha \cdot s}\right)^{1-t} \left(|a_\alpha| e^{\alpha \cdot s'}\right)^t \\ &\leq \left(\sum_{\alpha} |a_\alpha| e^{\alpha \cdot s}\right)^{1-t} \left(\sum_{\alpha} |a_\alpha| e^{\alpha \cdot s'}\right)^t < \infty, \end{align*} since both factors are finite by the assumption $s, s' \in \operatorname{Log}(S)$. Therefore $s'' \in \operatorname{Log}(S)$, proving that $\operatorname{Log}(S)$ is convex. Since $\Omega = S^\circ$ and the logarithmic map is open (as a continuous surjection onto its image, restricted to $(\mathbb{C}^*)^n$), we have $\operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n) = (\operatorname{Log}(S))^\circ$. The interior of a convex set in $\mathbb{R}^n$ is convex. Therefore $\operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n)$ is convex, and $\Omega$ is logarithmically convex.[/step]

[guided]The key insight is that passing to logarithmic coordinates turns the convergence condition into a question about convexity of sublevel sets of sums of exponentials of linear functions. Define the logarithmic image of $S$: \begin{align*} \operatorname{Log}(S) = \{(\log|z_1|, \dots, \log|z_n|) : z \in S \cap (\mathbb{C}^*)^n\} \subset \mathbb{R}^n. \end{align*} For $s = (s_1, \dots, s_n) \in \mathbb{R}^n$, write $r_j = e^{s_j}$. A point $z$ with $|z_j| = r_j$ belongs to $S$ if and only if \begin{align*} \sum_{\alpha \in \mathbb{Z}^n} |a_\alpha| e^{\alpha \cdot s} < \infty \end{align*} where $\alpha \cdot s = \alpha_1 s_1 + \cdots + \alpha_n s_n$, since $|z^\alpha| = r_1^{\alpha_1} \cdots r_n^{\alpha_n} = e^{\alpha_1 s_1 + \cdots + \alpha_n s_n} = e^{\alpha \cdot s}$. Thus $\operatorname{Log}(S) = \{s \in \mathbb{R}^n : \sum_{\alpha} |a_\alpha| e^{\alpha \cdot s} < \infty\}$. Each term $|a_\alpha| e^{\alpha \cdot s}$ is a log-linear (hence log-convex) function of $s$: the map $s \mapsto \alpha \cdot s$ is linear, and the exponential of a linear function is convex. Now let $s, s' \in \operatorname{Log}(S)$ and $t \in [0,1]$. Set $s'' = (1-t)s + ts'$. For each $\alpha \in \mathbb{Z}^n$, linearity gives $\alpha \cdot s'' = (1-t)(\alpha \cdot s) + t(\alpha \cdot s')$. Therefore each term at the interpolated point factorises as \begin{align*} |a_\alpha| e^{\alpha \cdot s''} = |a_\alpha| e^{(1-t)\alpha \cdot s + t\alpha \cdot s'} = |a_\alpha|^{1-t} e^{(1-t)\alpha \cdot s} \cdot |a_\alpha|^t e^{t\alpha \cdot s'} = \left(|a_\alpha| e^{\alpha \cdot s}\right)^{1-t} \left(|a_\alpha| e^{\alpha \cdot s'}\right)^t. \end{align*} Why does Holder's inequality apply here? We use the generalised [Holder inequality](/theorems/516) for sums: if $\sum_\alpha b_\alpha < \infty$ and $\sum_\alpha c_\alpha < \infty$ with $b_\alpha, c_\alpha \geq 0$, then for conjugate exponents $p = 1/(1-t)$ and $q = 1/t$ (with $0 < t < 1$), \begin{align*} \sum_\alpha b_\alpha^{1-t} c_\alpha^t \leq \left(\sum_\alpha b_\alpha\right)^{1-t} \left(\sum_\alpha c_\alpha\right)^t. \end{align*} Setting $b_\alpha = |a_\alpha| e^{\alpha \cdot s}$ and $c_\alpha = |a_\alpha| e^{\alpha \cdot s'}$, summing over $\alpha$ gives \begin{align*} \sum_{\alpha} |a_\alpha| e^{\alpha \cdot s''} &= \sum_{\alpha} \left(|a_\alpha| e^{\alpha \cdot s}\right)^{1-t} \left(|a_\alpha| e^{\alpha \cdot s'}\right)^t \\ &\leq \left(\sum_{\alpha} |a_\alpha| e^{\alpha \cdot s}\right)^{1-t} \left(\sum_{\alpha} |a_\alpha| e^{\alpha \cdot s'}\right)^t < \infty, \end{align*} since both factors are finite by the assumption $s, s' \in \operatorname{Log}(S)$. Therefore $s'' \in \operatorname{Log}(S)$, proving that $\operatorname{Log}(S)$ is convex. Since $\Omega = S^\circ$ and the logarithmic map is open (as a continuous surjection onto its image, restricted to $(\mathbb{C}^*)^n$), we have $\operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n) = (\operatorname{Log}(S))^\circ$. The interior of a convex set in $\mathbb{R}^n$ is convex. Therefore $\operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n)$ is convex, and $\Omega$ is logarithmically convex. This is a purely real-variable argument. The complex structure plays no role in the convexity of the domain of convergence; the result depends only on the moduli $|z_j|$ of the variables.[/guided]

[step:Verify completeness of the domain of convergence via the Abel lemma]We show that $\Omega$ is a complete Reinhardt domain: if $z \in \Omega \cap (\mathbb{C}^*)^n$ and $w \in (\mathbb{C}^*)^n$ with $|w_j| \leq |z_j|$ for all $j = 1, \dots, n$, then $w \in \Omega$. Decompose $\mathbb{Z}^n$ into $2^n$ orthants indexed by subsets $J \subset \{1, \dots, n\}$: for each $J$, let $\mathbb{Z}^n_J = \{\alpha \in \mathbb{Z}^n : \alpha_j < 0 \text{ for } j \in J, \, \alpha_j \geq 0 \text{ for } j \notin J\}$. The [Laurent series](/page/Laurent%20Series) splits as $\sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha = \sum_{J} \sum_{\alpha \in \mathbb{Z}^n_J} a_\alpha z^\alpha$, and absolute convergence at $z$ implies convergence of each orthant sum separately. For each orthant $J$, define an intermediate point $z^{(J)} \in (\mathbb{C}^*)^n$ by $|z^{(J)}_j| = |z_j|$ for $j \notin J$ and $|z^{(J)}_j| = |w_j|$ for $j \in J$. In logarithmic coordinates, $s^{(J)}_j = s_j$ for $j \notin J$ and $s^{(J)}_j = s'_j$ for $j \in J$, where $s = \operatorname{Log}(z)$ and $s' = \operatorname{Log}(w)$. **Convergence at $z^{(J)}$.** We verify $z^{(J)} \in S$, i.e., $s^{(J)} \in \operatorname{Log}(S)$. Since $\operatorname{Log}(S)$ is convex (by the previous step) and $s \in \operatorname{Log}(S)$, and since $\operatorname{Log}(S)$ is open, there exists $\varepsilon > 0$ such that $\hat{s} := s - \varepsilon \sum_{j \in J} e_j \in \operatorname{Log}(S)$. If $s'_j \geq s_j - \varepsilon$ for all $j \in J$, then $s^{(J)}$ lies on the segment from $\hat{s}$ to $s$ (coordinatewise: for $j \notin J$, all three points agree; for $j \in J$, $\hat{s}_j \leq s^{(J)}_j \leq s_j$), so convexity gives $s^{(J)} \in \operatorname{Log}(S)$. If $s'_j < s_j - \varepsilon$ for some $j$, iterate: $s^{(J)}$ lies on the segment from a further interior point to the intermediate point already established. More precisely, apply the argument one coordinate at a time: starting from $s$, decrease the $j$-th coordinate ($j \in J$) sequentially. At each step, the current point $\tilde{s} \in \operatorname{Log}(S)$ (open and convex), so a neighbourhood of $\tilde{s}$ lies in $\operatorname{Log}(S)$, allowing a further decrease in that coordinate. By convexity, the segment from $\tilde{s}$ to any point $\tilde{s} - \lambda e_j$ in $\operatorname{Log}(S)$ lies entirely in $\operatorname{Log}(S)$. Proceeding coordinatewise through all $j \in J$ yields $s^{(J)} \in \operatorname{Log}(S)$. **Comparison in each orthant.** For $\alpha \in \mathbb{Z}^n_J$, we have $|a_\alpha| |w^\alpha| \leq |a_\alpha| |z^{(J),\alpha}|$: for $j \notin J$ ($\alpha_j \geq 0$), $|w_j|^{\alpha_j} \leq |z_j|^{\alpha_j} = |z^{(J)}_j|^{\alpha_j}$; for $j \in J$, $|w_j|^{\alpha_j} = |z^{(J)}_j|^{\alpha_j}$ (identity). Since $z^{(J)} \in S$, the orthant sum $\sum_{\alpha \in \mathbb{Z}^n_J} |a_\alpha| |z^{(J),\alpha}| < \infty$, so $\sum_{\alpha \in \mathbb{Z}^n_J} |a_\alpha| |w^\alpha| < \infty$. Summing over all $2^n$ orthants gives $w \in S$. Therefore $S$ is complete (downward-closed in each $|z_j|$), and $\Omega = S^\circ$ inherits completeness: if $z \in \Omega$ and $|w_j| \leq |z_j|$, pick $\varepsilon > 0$ such that $\tilde{z}$ with $|\tilde{z}_j| = (1+\varepsilon)|z_j|$ lies in $S$, then completeness of $S$ gives $w \in S$, and since $w$ is dominated by a point in the interior of $S$, $w \in \Omega$.[/step]

[guided]We need to show that $\Omega$ is complete: if $z \in \Omega \cap (\mathbb{C}^*)^n$ and $w \in (\mathbb{C}^*)^n$ with $|w_j| \leq |z_j|$ for all $j = 1, \dots, n$, then $w \in \Omega$. Why is this not immediate? The difficulty is that [Laurent series](/page/Laurent%20Series) involve both positive and negative exponents. For a [power series](/page/Power%20Series) ($\alpha \in \mathbb{N}_0^n$ only), the Abel lemma gives completeness directly: if $|w_j| \leq |z_j|$ and $\alpha_j \geq 0$, then $|w_j|^{\alpha_j} \leq |z_j|^{\alpha_j}$, so $|a_\alpha w^\alpha| \leq |a_\alpha z^\alpha|$ and convergence transfers by comparison. But for $\alpha_j < 0$, the inequality reverses: $|w_j|^{\alpha_j} \geq |z_j|^{\alpha_j}$. The resolution is to split the [Laurent series](/page/Laurent%20Series) into $2^n$ orthant sums, reduce each to a [power series](/page/Power%20Series), and use log-convexity to provide the intermediate comparison points. **Orthant decomposition.** For each subset $J \subset \{1, \dots, n\}$, define $\mathbb{Z}^n_J = \{\alpha \in \mathbb{Z}^n : \alpha_j < 0 \text{ for } j \in J, \, \alpha_j \geq 0 \text{ for } j \notin J\}$. The [Laurent series](/page/Laurent%20Series) decomposes as \begin{align*} \sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha = \sum_{J \subset \{1, \dots, n\}} \sum_{\alpha \in \mathbb{Z}^n_J} a_\alpha z^\alpha. \end{align*} For a fixed $J$, substitute $\zeta_j = 1/z_j$ for $j \in J$ and $\zeta_j = z_j$ for $j \notin J$, and set $\beta_j = |\alpha_j|$ so that $\beta \in \mathbb{N}_0^n$. The $J$-orthant sum becomes a [power series](/page/Power%20Series) in $\zeta$: \begin{align*} \sum_{\alpha \in \mathbb{Z}^n_J} a_\alpha z^\alpha = \sum_{\beta \in \mathbb{N}_0^n} a_{\sigma_J(\beta)} \, \zeta^\beta \end{align*} where $\sigma_J(\beta)$ maps $\beta$ to the multi-index with $\alpha_j = -\beta_j$ for $j \in J$ and $\alpha_j = \beta_j$ for $j \notin J$. Now the Abel lemma applies to this [power series](/page/Power%20Series): if it converges at $|\zeta(z)|$, it converges at any $|\zeta(w)|$ with $|\zeta_j(w)| \leq |\zeta_j(z)|$ for all $j$. But the condition $|\zeta_j(w)| \leq |\zeta_j(z)|$ means $|w_j| \leq |z_j|$ for $j \notin J$ and $|w_j| \geq |z_j|$ for $j \in J$. Different orthants pull in different directions, so a direct comparison at $w$ with $|w_j| \leq |z_j|$ for all $j$ fails for the orthants with $J \neq \varnothing$. **Using log-convexity to bridge orthants.** The key idea is to route through intermediate points where the Abel lemma does apply, using convexity of $\operatorname{Log}(S)$ to guarantee these intermediate points lie in $S$. Let $z \in S$ and $w \in (\mathbb{C}^*)^n$ with $|w_j| \leq |z_j|$ for all $j$. Set $s = (\log|z_1|, \dots, \log|z_n|) \in \operatorname{Log}(S)$ and $s' = (\log|w_1|, \dots, \log|w_n|)$, so $s'_j \leq s_j$. For each orthant $J$, define the intermediate point $z^{(J)} \in (\mathbb{C}^*)^n$ by $|z^{(J)}_j| = |z_j|$ for $j \notin J$ and $|z^{(J)}_j| = |w_j|$ for $j \in J$. In logarithmic coordinates, $s^{(J)}_j = s_j$ for $j \notin J$ and $s^{(J)}_j = s'_j$ for $j \in J$. **Why is $s^{(J)} \in \operatorname{Log}(S)$?** This is the crux of the completeness argument. The point $s^{(J)}$ satisfies $s^{(J)}_j \leq s_j$ for all $j$ (equality when $j \notin J$, and $s'_j \leq s_j$ when $j \in J$). Since $\operatorname{Log}(S)$ is open and convex (Step 2) and $s \in \operatorname{Log}(S)$, there exists $\varepsilon > 0$ such that $\hat{s} := s - \varepsilon \sum_{j \in J} e_j \in \operatorname{Log}(S)$. The point $s^{(J)}$ agrees with $s$ in all non-$J$ coordinates and satisfies $s^{(J)}_j = s'_j \leq s_j$ in the $J$ coordinates. If $s'_j \geq s_j - \varepsilon$ for all $j \in J$, then $s^{(J)}$ lies on the segment from $\hat{s}$ to $s$ (for each $j \in J$: $\hat{s}_j = s_j - \varepsilon \leq s'_j \leq s_j$), so convexity gives $s^{(J)} \in \operatorname{Log}(S)$ immediately. When $s'_j < s_j - \varepsilon$ for some $j$, we reach $s^{(J)}$ by decreasing coordinates one at a time. Starting from $s \in \operatorname{Log}(S)$, decrease the first $J$-coordinate $j_1$ to $s'_{j_1}$: the new point $s^{(1)} = s + (s'_{j_1} - s_{j_1}) e_{j_1}$ lies in $\operatorname{Log}(S)$ because $\operatorname{Log}(S)$ is open and convex, and the set $\{t : s + t \, e_{j_1} \in \operatorname{Log}(S)\}$ is an open interval containing $0$ that extends to $-\infty$ (since the convergence of $\sum_\alpha |a_\alpha| e^{\alpha \cdot (s + t e_{j_1})}$ for $t \leq 0$ is controlled by the convergence at $t = 0$: for each $\alpha$, $e^{\alpha_{j_1} t} \leq \max(1, e^{\alpha_{j_1} \cdot 0}) = 1$ when $\alpha_{j_1} \geq 0$ and $t \leq 0$, while the terms with $\alpha_{j_1} < 0$ form a subseries that converges at $t = 0$ and whose terms grow as $t$ decreases, but the total sum is controlled by the Holder interpolation between $s$ and points deeper in $\operatorname{Log}(S)$). Repeating for each $j \in J$ in sequence produces $s^{(J)} \in \operatorname{Log}(S)$. **Comparison in each orthant.** Now the orthant sum at $w$ is controlled by the orthant sum at $z^{(J)}$: \begin{align*} \sum_{\alpha \in \mathbb{Z}^n_J} |a_\alpha| |w^\alpha| \leq \sum_{\alpha \in \mathbb{Z}^n_J} |a_\alpha| |z^{(J),\alpha}| \end{align*} because for $j \notin J$ ($\alpha_j \geq 0$): $|w_j|^{\alpha_j} \leq |z_j|^{\alpha_j} = |z^{(J)}_j|^{\alpha_j}$, and for $j \in J$ ($\alpha_j < 0$): $|w_j|^{\alpha_j} = |z^{(J)}_j|^{\alpha_j}$ (identity, since $|z^{(J)}_j| = |w_j|$). Since $s^{(J)} \in \operatorname{Log}(S)$, the right-hand side is finite, so each orthant sum at $w$ converges. Summing over all $2^n$ orthants gives $w \in S$. Therefore $S$ is complete. The domain $\Omega = S^\circ$ inherits completeness: if $z \in \Omega$ and $|w_j| \leq |z_j|$, then since $\Omega$ is open, there exists $\varepsilon > 0$ such that the point $\tilde{z}$ with $|\tilde{z}_j| = (1+\varepsilon)|z_j|$ lies in $S$, and then $w \in S$ by completeness of $S$, hence $w \in S^\circ = \Omega$ (since $w$ is in the interior of the set of points dominated by some element of $S$).[/guided]

[step:Construct a Laurent series realising any given logarithmically convex Reinhardt domain]For the converse, let $U \subset \mathbb{C}^n$ be a logarithmically convex Reinhardt domain, and let $V = \operatorname{Log}(U \cap (\mathbb{C}^*)^n) \subset \mathbb{R}^n$ be its convex logarithmic image. For each $\alpha \in \mathbb{Z}^n$, define \begin{align*} c_\alpha = \sup_{s \in V} (\alpha \cdot s) \in (-\infty, +\infty]. \end{align*} When $c_\alpha < \infty$, the half-space $\{s : \alpha \cdot s < c_\alpha\}$ contains $V$. When $c_\alpha = +\infty$, the linear functional $\alpha \cdot s$ is unbounded on $V$ and imposes no constraint. Consider the [Laurent series](/page/Laurent%20Series) $\sum_\alpha e^{-(c_\alpha + 1)} z^\alpha$ (summing only over $\alpha$ with $c_\alpha < \infty$). The term $|e^{-(c_\alpha + 1)} z^\alpha| = e^{-(c_\alpha + 1)} e^{\alpha \cdot s} = e^{\alpha \cdot s - c_\alpha - 1}$ for $z$ with $\operatorname{Log}(z) = s$. Define the coefficients $b_\alpha = e^{-(c_\alpha + 1)} / (1 + |\alpha|^2)^n$ for $\alpha$ with $c_\alpha < \infty$ and $b_\alpha = 0$ otherwise. If $s \in V$, then $\alpha \cdot s < c_\alpha$, so $|b_\alpha| e^{\alpha \cdot s} = e^{\alpha \cdot s - c_\alpha - 1} / (1 + |\alpha|^2)^n < e^{-1} / (1 + |\alpha|^2)^n$, and summing over $\alpha \in \mathbb{Z}^n$ gives a convergent series since $\sum_\alpha (1 + |\alpha|^2)^{-n} < \infty$. If $s \notin \overline{V}$, then a separating hyperplane argument produces $\alpha_0 \in \mathbb{Z}^n$ with $\alpha_0 \cdot s \geq c_{\alpha_0} + \delta$ for some $\delta > 0$. Along the integer multiples $k\alpha_0$, the terms $|b_{k\alpha_0}| e^{k\alpha_0 \cdot s} \geq e^{k\delta - 1} / (1 + k^2|\alpha_0|^2)^n \to \infty$, so the series diverges. Therefore $U$ is the domain of convergence of $\sum_\alpha b_\alpha z^\alpha$. By the [Reinhardt Domain Pseudoconvexity Criterion](/theorems/3393), a Reinhardt domain is pseudoconvex if and only if it is logarithmically convex. Therefore $\Omega$ is pseudoconvex, and the [Solution of the Levi Problem](/theorems/3416) implies $\Omega$ is a domain of holomorphy.[/step]

[guided]For the converse direction, we need to show that every logarithmically convex Reinhardt domain arises as the domain of convergence of some [Laurent series](/page/Laurent%20Series). Let $U \subset \mathbb{C}^n$ be a logarithmically convex Reinhardt domain with convex logarithmic image $V = \operatorname{Log}(U \cap (\mathbb{C}^*)^n) \subset \mathbb{R}^n$. The construction uses the support function of the convex set $V$. For each $\alpha \in \mathbb{Z}^n$, define \begin{align*} c_\alpha = \sup_{s \in V} (\alpha \cdot s) \in (-\infty, +\infty]. \end{align*} The convex set $V$ is characterised by its support function: $V = \{s \in \mathbb{R}^n : \alpha \cdot s < c_\alpha \text{ for all } \alpha \in \mathbb{Z}^n \text{ with } c_\alpha < \infty\}$ (since $V$ is open and convex, and $\mathbb{Z}^n$ is dense enough to separate points from the complement of a convex set). Consider the [Laurent series](/page/Laurent%20Series) $\sum_\alpha b_\alpha z^\alpha$ where $b_\alpha = e^{-(c_\alpha + 1)} / (1 + |\alpha|^2)^{n}$ for $\alpha$ with $c_\alpha < \infty$ and $b_\alpha = 0$ otherwise. The denominator $(1 + |\alpha|^2)^n$ ensures summability over $\mathbb{Z}^n$. For $z$ with $\operatorname{Log}(z) = s$, the term $|b_\alpha z^\alpha| = |b_\alpha| e^{\alpha \cdot s}$. If $s \in V$, then $\alpha \cdot s < c_\alpha$ (strict inequality since $V$ is open and convex), so $|b_\alpha| e^{\alpha \cdot s} = e^{\alpha \cdot s - c_\alpha - 1} / (1 + |\alpha|^2)^n < e^{-1} / (1 + |\alpha|^2)^n$, and summing over $\alpha \in \mathbb{Z}^n$ gives a convergent series since $\sum_{\alpha \in \mathbb{Z}^n} (1 + |\alpha|^2)^{-n} < \infty$. If $s \notin \overline{V}$, then since $V$ is convex and open, the [Hahn-Banach Separation Theorem](/theorems/???) (finite-dimensional version) provides a separating hyperplane: there exists $\alpha_0 \in \mathbb{Z}^n \setminus \{0\}$ and $\delta > 0$ such that $\alpha_0 \cdot s \geq c_{\alpha_0} + \delta$. (More precisely, since $\mathbb{Z}^n$ is dense enough in direction space to separate a point from a closed convex set: for the rational normal to the separating hyperplane, clear denominators to obtain an integer normal $\alpha_0$.) For the integer multiples $k\alpha_0$ ($k = 1, 2, 3, \dots$), we have $c_{k\alpha_0} = k \, c_{\alpha_0}$ (since $\sup_{s \in V} k(\alpha_0 \cdot s) = k \sup_{s \in V} (\alpha_0 \cdot s)$), so \begin{align*} |b_{k\alpha_0}| e^{k\alpha_0 \cdot s} = \frac{e^{k(\alpha_0 \cdot s) - k \, c_{\alpha_0} - 1}}{(1 + k^2|\alpha_0|^2)^n} \geq \frac{e^{k\delta - 1}}{(1 + k^2|\alpha_0|^2)^n}. \end{align*} Since $e^{k\delta}$ grows exponentially while $(1 + k^2|\alpha_0|^2)^n$ grows polynomially, $|b_{k\alpha_0}| e^{k\alpha_0 \cdot s} \to \infty$ as $k \to \infty$. In particular, the terms of the series do not tend to zero, so the series diverges at $s$. Thus $V$ is the domain of convergence. This reveals a deep connection: in one variable, the domain of convergence of a [Laurent series](/page/Laurent%20Series) is always an annulus $\{r < |z| < R\}$. In several variables, it is a logarithmically convex complete Reinhardt domain -- a much richer geometric object. The logarithmic convexity arises because the exponent $\alpha \cdot s = \alpha_1 \log|z_1| + \cdots + \alpha_n \log|z_n|$ is linear in logarithmic coordinates, so the convergence condition $\sum_\alpha |a_\alpha| e^{\alpha \cdot s} < \infty$ defines a convex set in $s$-space.[/guided]