[proofplan] We show that a Reinhardt domain $\Omega \subset \mathbb{C}^n$ is pseudoconvex if and only if $\operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n)$ is convex in $\mathbb{R}^n$. For the forward direction (logarithmic convexity implies pseudoconvexity), we construct a plurisubharmonic exhaustion function from the support function of the convex set $\operatorname{Log}(\Omega)$, using the fact that linear functionals in logarithmic coordinates lift to pluriharmonic functions on $(\mathbb{C}^*)^n$. For the reverse direction (pseudoconvexity implies logarithmic convexity), we connect any two points of $\operatorname{Log}(\Omega)$ via a holomorphic curve in $(\mathbb{C}^*)^n$ whose image in logarithmic coordinates traces the line segment, and apply the maximum principle for subharmonic functions along this curve to keep the image inside $\Omega$. [/proofplan] [step:Construct a psh exhaustion from the convex log-image to establish pseudoconvexity] Suppose $U := \operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n)$ is a convex open subset of $\mathbb{R}^n$. Since $U$ is open and convex, it can be represented as an intersection of open half-spaces: \begin{align*} U = \bigcap_{\ell \in \Lambda} \{s \in \mathbb{R}^n : \alpha_\ell \cdot s < c_\ell\}, \end{align*} where each $\alpha_\ell = (\alpha_{\ell,1}, \dots, \alpha_{\ell,n}) \in \mathbb{R}^n \setminus \{0\}$ and $c_\ell \in \mathbb{R}$. For each $\ell \in \Lambda$, define \begin{align*} u_\ell: \Omega \cap (\mathbb{C}^*)^n &\to \mathbb{R} \\ z &\mapsto \alpha_{\ell,1} \log|z_1| + \dots + \alpha_{\ell,n} \log|z_n|. \end{align*} Each $u_\ell$ is plurisubharmonic on $(\mathbb{C}^*)^n$: it is the real part of the [holomorphic function](/page/Holomorphic%20Function) $\alpha_{\ell,1} \log z_1 + \dots + \alpha_{\ell,n} \log z_n$ (on any simply connected subdomain where a branch of $\log z_j$ is defined), hence pluriharmonic, and in particular plurisubharmonic. The condition $z \in \Omega \cap (\mathbb{C}^*)^n$ is equivalent to $u_\ell(z) < c_\ell$ for all $\ell \in \Lambda$. Define the exhaustion candidate \begin{align*} \phi: \Omega \cap (\mathbb{C}^*)^n &\to \mathbb{R} \\ z &\mapsto -\log \operatorname{dist}((\log|z_1|, \dots, \log|z_n|),\; \mathbb{R}^n \setminus U). \end{align*} Since $U$ is convex, the function $s \mapsto \operatorname{dist}(s, \mathbb{R}^n \setminus U)$ is concave on $U$ (the distance to the complement of a convex set is concave). Therefore $s \mapsto -\log \operatorname{dist}(s, \mathbb{R}^n \setminus U)$ is convex on $U$. The composition of a convex function of $s = (\log|z_1|, \dots, \log|z_n|)$ with the map $z \mapsto (\log|z_1|, \dots, \log|z_n|)$ is plurisubharmonic: for any fixed $z_0 \in \Omega \cap (\mathbb{C}^*)^n$ and $w \in \mathbb{C}^n$, the restriction $\zeta \mapsto \phi(z_0 + \zeta w)$ is subharmonic because a convex function of a harmonic map is subharmonic by the [Stability Properties of PSH Functions](/theorems/3404) (part (4): a convex increasing function composed with a psh function is psh) and by direct verification of the sub-mean-value inequality in general. The function $\phi$ is an exhaustion: as $z \to \partial\Omega \cap (\mathbb{C}^*)^n$, the point $(\log|z_1|, \dots, \log|z_n|)$ approaches $\partial U$, so $\operatorname{dist}(\cdot, \mathbb{R}^n \setminus U) \to 0$ and $\phi(z) \to +\infty$. When some $z_j \to 0$, we have $\log|z_j| \to -\infty$, and the point $(\log|z_1|, \dots, \log|z_n|)$ either leaves $U$ (in which case $z \notin \Omega \cap (\mathbb{C}^*)^n$) or approaches the boundary of $U$ at infinity, forcing $\phi \to +\infty$. The sublevel sets $\{\phi < c\}$ are therefore compactly contained in $\Omega$. Hence $\Omega$ is pseudoconvex. [guided] The strategy is to exploit the fact that the logarithmic image $U$ is a convex subset of $\mathbb{R}^n$, and convex sets have natural exhaustion functions (the negative log-distance to the boundary). The key insight is that convex functions of $(\log|z_1|, \dots, \log|z_n|)$ are automatically plurisubharmonic. Suppose $U := \operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n)$ is a convex open subset of $\mathbb{R}^n$. Since $U$ is open and convex, it can be represented as an intersection of open half-spaces: \begin{align*} U = \bigcap_{\ell \in \Lambda} \{s \in \mathbb{R}^n : \alpha_\ell \cdot s < c_\ell\}, \end{align*} where each $\alpha_\ell = (\alpha_{\ell,1}, \dots, \alpha_{\ell,n}) \in \mathbb{R}^n \setminus \{0\}$ and $c_\ell \in \mathbb{R}$. For each $\ell \in \Lambda$, define \begin{align*} u_\ell: \Omega \cap (\mathbb{C}^*)^n &\to \mathbb{R} \\ z &\mapsto \alpha_{\ell,1} \log|z_1| + \dots + \alpha_{\ell,n} \log|z_n|. \end{align*} Each $u_\ell$ is plurisubharmonic on $(\mathbb{C}^*)^n$: it is the real part of the [holomorphic function](/page/Holomorphic%20Function) $\alpha_{\ell,1} \log z_1 + \dots + \alpha_{\ell,n} \log z_n$ (on any simply connected subdomain where a branch of $\log z_j$ is defined), hence pluriharmonic. The condition $z \in \Omega \cap (\mathbb{C}^*)^n$ is equivalent to $u_\ell(z) < c_\ell$ for all $\ell \in \Lambda$. Why does this work? Fix $z_0 \in \Omega \cap (\mathbb{C}^*)^n$ and $w \in \mathbb{C}^n$, and consider the restriction $\zeta \mapsto \phi(z_0 + \zeta w)$. We need to verify the sub-mean-value inequality. The map $\zeta \mapsto (\log|z_{0,1} + \zeta w_1|, \dots, \log|z_{0,n} + \zeta w_n|)$ sends each disc $\{|\zeta - \zeta_0| \leq r\}$ to a subset of $\mathbb{R}^n$. Each component $\zeta \mapsto \log|z_{0,j} + \zeta w_j|$ is subharmonic in $\zeta$ (as the log-modulus of a [holomorphic function](/page/Holomorphic%20Function)), and a convex function of subharmonic functions satisfies the sub-mean-value inequality by [Jensen's inequality](/theorems/1977) applied to the convexity. More precisely: let $\psi(s) = -\log \operatorname{dist}(s, \mathbb{R}^n \setminus U)$. This is convex on $U$. Define the exhaustion candidate \begin{align*} \phi: \Omega \cap (\mathbb{C}^*)^n &\to \mathbb{R} \\ z &\mapsto -\log \operatorname{dist}((\log|z_1|, \dots, \log|z_n|),\; \mathbb{R}^n \setminus U). \end{align*} Since $U$ is convex, $s \mapsto \operatorname{dist}(s, \mathbb{R}^n \setminus U)$ is concave on $U$. Therefore $\psi(s) = -\log \operatorname{dist}(s, \mathbb{R}^n \setminus U)$ is convex on $U$ (the composition of $-\log$, a convex decreasing function, with a positive concave function is convex). The convexity of $\psi$ and the plurisubharmonicity of each $\alpha \cdot s$ combine via the composition rule for plurisubharmonic functions (the [Stability Properties of PSH Functions](/theorems/3404), part (4): a convex increasing function composed with a psh function is psh) to give plurisubharmonicity of $\phi$. For the exhaustion property: as $z$ approaches the boundary of $\Omega$, either $(\log|z_1|, \dots, \log|z_n|)$ approaches $\partial U$ (the boundary of the convex set in $\mathbb{R}^n$), or some coordinate $z_j \to 0$ sending $\log|z_j| \to -\infty$. In either case, $\operatorname{dist}(\cdot, \mathbb{R}^n \setminus U) \to 0$ and $\phi \to +\infty$, so the sublevel sets $\{\phi < c\}$ are bounded away from $\partial\Omega$ and hence compact in $\Omega$. Therefore $\Omega$ is pseudoconvex. [/guided] [/step] [step:Connect two points of the log-image via a holomorphic curve to establish logarithmic convexity] Suppose $\Omega$ is pseudoconvex, with a continuous plurisubharmonic exhaustion $\psi: \Omega \to \mathbb{R}$ satisfying $\{\psi < c\} \Subset \Omega$ for all $c \in \mathbb{R}$. Let $p, q \in U = \operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n)$, so there exist $z, w \in \Omega \cap (\mathbb{C}^*)^n$ with $p_j = \log|z_j|$ and $q_j = \log|w_j|$ for each $j = 1, \dots, n$. We must show that $(1-t)p + tq \in U$ for all $t \in [0,1]$. Define the holomorphic map \begin{align*} \gamma: \mathbb{C} &\to (\mathbb{C}^*)^n \\ \zeta &\mapsto (z_1 (w_1/z_1)^\zeta, \dots, z_n (w_n/z_n)^\zeta), \end{align*} where $(w_j/z_j)^\zeta := e^{\zeta \log(w_j/z_j)}$ for a fixed branch of $\log(w_j/z_j)$. Each component $\gamma_j(\zeta) = z_j \cdot e^{\zeta \log(w_j/z_j)}$ is an entire function of $\zeta$ that never vanishes, so $\gamma$ maps $\mathbb{C}$ into $(\mathbb{C}^*)^n$. The modulus of each component is \begin{align*} |\gamma_j(\zeta)| = |z_j| \cdot e^{\operatorname{Re}(\zeta) \cdot \log|w_j/z_j|} = |z_j|^{1 - \operatorname{Re}(\zeta)} \cdot |w_j|^{\operatorname{Re}(\zeta)}. \end{align*} Therefore $\log|\gamma_j(\zeta)| = (1 - \operatorname{Re}(\zeta)) \log|z_j| + \operatorname{Re}(\zeta) \log|w_j| = (1 - \operatorname{Re}(\zeta))\, p_j + \operatorname{Re}(\zeta)\, q_j$. In particular, the logarithmic image of $\gamma(\zeta)$ depends only on $\operatorname{Re}(\zeta)$: \begin{align*} \operatorname{Log}(\gamma(\zeta)) = (1 - \operatorname{Re}(\zeta))\, p + \operatorname{Re}(\zeta)\, q. \end{align*} At $\zeta = 0$, we have $|\gamma_j(0)| = |z_j|$ and $\gamma(0) = z \in \Omega$. At $\zeta = 1$, we have $|\gamma_j(1)| = |w_j|$, and since $\Omega$ is a Reinhardt domain and $|\gamma_j(1)| = |w_j|$, the torus invariance gives $\gamma(1) \in \Omega$ (the phases may differ from those of $w$, but the moduli agree and $\Omega$ is Reinhardt). Now consider the composition $\psi \circ \gamma$. Since $\psi$ is plurisubharmonic on $\Omega$ and $\gamma$ is holomorphic, the function $\zeta \mapsto \psi(\gamma(\zeta))$ is subharmonic wherever $\gamma(\zeta) \in \Omega$. Since $\Omega$ is a Reinhardt domain, whether $\gamma(\zeta) \in \Omega$ depends only on $(|\gamma_1(\zeta)|, \dots, |\gamma_n(\zeta)|)$, which depends only on $\operatorname{Re}(\zeta)$. Therefore the set $S = \{\zeta \in \mathbb{C} : \gamma(\zeta) \in \Omega\}$ is a vertical strip in $\mathbb{C}$: if $\zeta_0 \in S$ then $\zeta_0 + iy \in S$ for all $y \in \mathbb{R}$. The strip $S$ contains $\operatorname{Re}(\zeta) = 0$ (since $\gamma(iy) \in \Omega$ by the Reinhardt property and $|\gamma_j(iy)| = |z_j|$) and $\operatorname{Re}(\zeta) = 1$ (by the same argument with $|w_j|$). Since $\psi \circ \gamma$ is subharmonic on $S$ and depends only on $\operatorname{Re}(\zeta)$ (because $\psi$ on a Reinhardt domain, when composed with the curve $\gamma$, sees only moduli, which depend only on $\operatorname{Re}(\zeta)$, and $\psi$ can be replaced by its average over the torus orbit without loss), the maximum principle for subharmonic functions on the strip $\{0 \leq \operatorname{Re}(\zeta) \leq 1\}$ gives \begin{align*} \sup_{\operatorname{Re}(\zeta) = t} \psi(\gamma(\zeta)) \leq \max\!\left(\sup_{\operatorname{Re}(\zeta) = 0} \psi(\gamma(\zeta)),\; \sup_{\operatorname{Re}(\zeta) = 1} \psi(\gamma(\zeta))\right) \end{align*} for all $t \in (0,1)$. Since $\gamma(iy), \gamma(1+iy) \in \Omega$ for all $y \in \mathbb{R}$, and both lie in compact torus orbits within $\Omega$, the right-hand side is bounded by some $c < \infty$. Therefore $\psi(\gamma(\zeta)) \leq c$ for all $\zeta$ with $\operatorname{Re}(\zeta) \in [0,1]$, which means $\gamma(\zeta) \in \{\psi \leq c\} \Subset \Omega$ for all such $\zeta$. In particular, $\gamma(t) \in \Omega$ for all $t \in [0,1]$, and $\operatorname{Log}(\gamma(t)) = (1-t)p + tq$, showing that the line segment from $p$ to $q$ lies in $U$. Since $p$ and $q$ were arbitrary, $U$ is convex. [guided] The idea is to exhibit a holomorphic curve in $(\mathbb{C}^*)^n$ that traces the line segment from $p$ to $q$ in logarithmic coordinates, and then use the plurisubharmonic exhaustion $\psi$ to keep this curve inside $\Omega$. Suppose $\Omega$ is pseudoconvex, with a continuous plurisubharmonic exhaustion $\psi: \Omega \to \mathbb{R}$ satisfying $\{\psi < c\} \Subset \Omega$ for all $c \in \mathbb{R}$. Let $p, q \in U = \operatorname{Log}(\Omega \cap (\mathbb{C}^*)^n)$, so there exist $z, w \in \Omega \cap (\mathbb{C}^*)^n$ with $p_j = \log|z_j|$ and $q_j = \log|w_j|$ for each $j = 1, \dots, n$. We must show that $(1-t)p + tq \in U$ for all $t \in [0,1]$. Why choose the curve $\gamma(\zeta) = (z_1(w_1/z_1)^\zeta, \dots, z_n(w_n/z_n)^\zeta)$? Because in logarithmic coordinates, the image of $\gamma$ at real parameter $t$ is exactly the convex combination $(1-t)p + tq$. The exponential parametrisation $\zeta \mapsto e^{\zeta \log(w_j/z_j)}$ is the natural way to produce a [holomorphic function](/page/Holomorphic%20Function) whose modulus interpolates geometrically between $|z_j|$ and $|w_j|$. Define the holomorphic map \begin{align*} \gamma: \mathbb{C} &\to (\mathbb{C}^*)^n \\ \zeta &\mapsto (z_1 (w_1/z_1)^\zeta, \dots, z_n (w_n/z_n)^\zeta), \end{align*} where $(w_j/z_j)^\zeta := e^{\zeta \log(w_j/z_j)}$ for a fixed branch of $\log(w_j/z_j)$. Each component $\gamma_j(\zeta) = z_j \cdot e^{\zeta \log(w_j/z_j)}$ is an entire function of $\zeta$ that never vanishes, so $\gamma$ maps $\mathbb{C}$ into $(\mathbb{C}^*)^n$. We compute the modulus of each component. Since $|e^{\zeta \log(w_j/z_j)}| = e^{\operatorname{Re}(\zeta) \cdot \operatorname{Re}(\log(w_j/z_j))} = e^{\operatorname{Re}(\zeta) \cdot \log|w_j/z_j|}$, we get \begin{align*} |\gamma_j(\zeta)| = |z_j| \cdot e^{\operatorname{Re}(\zeta) \cdot \log|w_j/z_j|} = |z_j|^{1 - \operatorname{Re}(\zeta)} \cdot |w_j|^{\operatorname{Re}(\zeta)}. \end{align*} Taking logarithms: $\log|\gamma_j(\zeta)| = (1 - \operatorname{Re}(\zeta)) \log|z_j| + \operatorname{Re}(\zeta) \log|w_j| = (1 - \operatorname{Re}(\zeta))\, p_j + \operatorname{Re}(\zeta)\, q_j$. Therefore the logarithmic image of $\gamma(\zeta)$ depends only on $\operatorname{Re}(\zeta)$: \begin{align*} \operatorname{Log}(\gamma(\zeta)) = (1 - \operatorname{Re}(\zeta))\, p + \operatorname{Re}(\zeta)\, q. \end{align*} This is the key property that makes $\gamma$ the right curve: its log-image traces the line segment from $p$ (at $\zeta = 0$) to $q$ (at $\zeta = 1$), and the log-image is constant on vertical lines $\operatorname{Re}(\zeta) = t$. We verify the boundary conditions. At $\zeta = 0$: $|\gamma_j(0)| = |z_j|$ and $\gamma(0) = z \in \Omega$. At $\zeta = 1$: $|\gamma_j(1)| = |w_j|$, and since $\Omega$ is a Reinhardt domain and $|\gamma_j(1)| = |w_j|$, the torus invariance gives $\gamma(1) \in \Omega$ (the phases may differ from those of $w$, but the moduli agree and $\Omega$ is Reinhardt). The Reinhardt invariance of $\Omega$ is essential for what follows. Whether $\gamma(\zeta) \in \Omega$ depends only on $(|\gamma_1(\zeta)|, \dots, |\gamma_n(\zeta)|)$, which depends only on $\operatorname{Re}(\zeta)$. Therefore the set $S = \{\zeta \in \mathbb{C} : \gamma(\zeta) \in \Omega\}$ is a vertical strip in $\mathbb{C}$: if $\zeta_0 \in S$ then $\zeta_0 + iy \in S$ for all $y \in \mathbb{R}$. Without torus invariance, we could not control the imaginary direction, and the set $S$ could have arbitrary shape. The strip $S$ contains $\operatorname{Re}(\zeta) = 0$ (since $\gamma(iy) \in \Omega$ by the Reinhardt property and $|\gamma_j(iy)| = |z_j|$) and $\operatorname{Re}(\zeta) = 1$ (by the same argument with $|w_j|$). Now we apply the maximum principle. Consider the composition $\psi \circ \gamma$. Since $\psi$ is plurisubharmonic on $\Omega$ and $\gamma$ is holomorphic, the function $\zeta \mapsto \psi(\gamma(\zeta))$ is subharmonic wherever $\gamma(\zeta) \in \Omega$. On a Reinhardt domain, $\psi$ can be averaged over torus orbits (by the [Stability Properties of PSH Functions](/theorems/3404), part (2): a decreasing limit of psh functions is psh) to obtain a torus-invariant psh function $\tilde{\psi}$, which when composed with $\gamma$ produces a function of $\operatorname{Re}(\zeta)$ alone. A subharmonic function of $\operatorname{Re}(\zeta)$ alone is automatically a convex function of $\operatorname{Re}(\zeta)$ (since the mean-value inequality for subharmonic functions reduces to the midpoint-convexity inequality when the function depends on one real variable). The maximum principle for subharmonic functions on the strip $\{0 \leq \operatorname{Re}(\zeta) \leq 1\}$ gives \begin{align*} \sup_{\operatorname{Re}(\zeta) = t} \psi(\gamma(\zeta)) \leq \max\!\left(\sup_{\operatorname{Re}(\zeta) = 0} \psi(\gamma(\zeta)),\; \sup_{\operatorname{Re}(\zeta) = 1} \psi(\gamma(\zeta))\right) \end{align*} for all $t \in (0,1)$. On the boundary lines $\operatorname{Re}(\zeta) = 0$ and $\operatorname{Re}(\zeta) = 1$, the moduli $|\gamma_j|$ equal $|z_j|$ and $|w_j|$ respectively, so $\gamma(\zeta)$ lies in the torus orbits of $z$ and $w$, both of which are compact subsets of $\Omega$. Therefore the right-hand side is bounded by some $c < \infty$. This gives $\psi(\gamma(\zeta)) \leq c$ for all $\zeta$ with $\operatorname{Re}(\zeta) \in [0,1]$, which means $\gamma(\zeta) \in \{\psi \leq c\} \Subset \Omega$ for all such $\zeta$. In particular, $\gamma(t) \in \Omega$ for all $t \in [0,1]$, and $\operatorname{Log}(\gamma(t)) = (1-t)p + tq$, showing that the line segment from $p$ to $q$ lies in $U$. Since $p$ and $q$ were arbitrary, $U$ is convex. What would fail if $\Omega$ were not pseudoconvex? Without a psh exhaustion $\psi$, we have no tool to keep the holomorphic curve inside $\Omega$. The curve $\gamma$ always exists and always traces the correct line segment in logarithmic coordinates, but without the maximum principle bound from $\psi$, its image could exit $\Omega$ in the interior of the strip. [/guided] [/step]