[proofplan] We show that every bounded convex domain $\Omega \subset \mathbb{C}^n$ is pseudoconvex by proving that the log-distance function $\phi(z) = -\log d(z, \partial\Omega)$ is a plurisubharmonic exhaustion. The proof has three ingredients: (1) the distance function $d(\cdot, \partial\Omega)$ is concave on a convex domain, (2) $-\log$ composed with a concave positive function is convex, (3) convex functions on $\mathbb{R}^{2n}$ are subharmonic along every affine two-dimensional slice, hence in particular along every complex line, which gives plurisubharmonicity. [/proofplan] [step:Establish that $d(\cdot, \partial\Omega)$ is concave on $\Omega$] Let $z, w \in \Omega$ and $t \in [0,1]$. We claim \begin{align*} d((1-t)z + tw, \partial\Omega) \geq (1-t)\,d(z, \partial\Omega) + t\,d(w, \partial\Omega). \end{align*} Since $\Omega$ is convex, for any $p \in \Omega$ and any $r > 0$ with $B(p, r) \subset \Omega$, the open ball $B(p, r)$ is contained in $\Omega$. In particular, $B(z, d(z, \partial\Omega)) \subset \Omega$ and $B(w, d(w, \partial\Omega)) \subset \Omega$. The convex combination of these two balls is \begin{align*} (1-t)\,B(z, d(z, \partial\Omega)) + t\,B(w, d(w, \partial\Omega)) = B\!\big((1-t)z + tw,\; (1-t)\,d(z, \partial\Omega) + t\,d(w, \partial\Omega)\big), \end{align*} where the equality uses the fact that the Minkowski sum of two balls $B(a, r_1) + B(b, r_2) = B(a+b, r_1+r_2)$ after appropriate rescaling by $(1-t)$ and $t$. Since $\Omega$ is convex, the Minkowski combination $(1-t)A + tB \subset \Omega$ for any $A, B \subset \Omega$. Therefore \begin{align*} B\!\big((1-t)z + tw,\; (1-t)\,d(z, \partial\Omega) + t\,d(w, \partial\Omega)\big) \subset \Omega, \end{align*} which gives $d((1-t)z + tw, \partial\Omega) \geq (1-t)\,d(z, \partial\Omega) + t\,d(w, \partial\Omega)$. [guided] The concavity of the distance function on convex domains is a standard result in convex analysis. The geometric content is that if $z$ can "see" the boundary at distance $r_1$ and $w$ can "see" it at distance $r_2$, then the midpoint $(1-t)z + tw$ can see it at distance at least $(1-t)r_1 + tr_2$, because convexity of $\Omega$ prevents the boundary from "intruding" between $z$ and $w$. The Minkowski sum identity $(1-t)B(z, r_1) + tB(w, r_2) = B((1-t)z + tw, (1-t)r_1 + tr_2)$ holds because every point in the left-hand side has the form $(1-t)(z + r_1 u) + t(w + r_2 v) = (1-t)z + tw + (1-t)r_1 u + t r_2 v$ for $|u|, |v| < 1$, and the set $\{(1-t)r_1 u + t r_2 v : |u|, |v| < 1\}$ equals $B(0, (1-t)r_1 + tr_2)$ (by the triangle inequality and attainability of the bound when $u = v$). The convexity of $\Omega$ ensures $(1-t)A + tB \subset \Omega$ for subsets $A, B \subset \Omega$ because for any $a \in A \subset \Omega$ and $b \in B \subset \Omega$, the point $(1-t)a + tb \in \Omega$ by convexity. [/guided] [/step] [step:Show that $\phi = -\log d(\cdot, \partial\Omega)$ is convex on $\Omega$] Since $d(\cdot, \partial\Omega)$ is concave and positive on $\Omega$, and $-\log: (0, \infty) \to \mathbb{R}$ is a convex decreasing function, the composition $\phi = (-\log) \circ d(\cdot, \partial\Omega)$ is convex on $\Omega$. To verify this composition rule: let $g: \Omega \to (0, \infty)$ be concave and let $\chi: (0, \infty) \to \mathbb{R}$ be convex and decreasing. For $z, w \in \Omega$ and $t \in [0,1]$: \begin{align*} \chi(g((1-t)z + tw)) &\leq \chi((1-t)g(z) + tg(w)) & &\text{($\chi$ decreasing, $g$ concave)} \\ &\leq (1-t)\chi(g(z)) + t\chi(g(w)) & &\text{($\chi$ convex)}. \end{align*} The [first inequality](/theorems/2897) uses: $g((1-t)z + tw) \geq (1-t)g(z) + tg(w)$ by concavity of $g$, and $\chi$ decreasing reverses the inequality. The [second inequality](/theorems/2136) is [Jensen's inequality](/theorems/1977) for the convex function $\chi$. This gives $\phi((1-t)z + tw) \leq (1-t)\phi(z) + t\phi(w)$, confirming convexity of $\phi$. [/step] [step:Deduce that $\phi$ is plurisubharmonic by showing convex functions are subharmonic along complex lines] Fix $z \in \Omega$ and $w \in \mathbb{C}^n$. Define $\psi: D \to \mathbb{R}$ by $\psi(\zeta) = \phi(z + \zeta w)$, where $D = \{\zeta \in \mathbb{C} : z + \zeta w \in \Omega\}$. Since $\phi$ is convex on $\Omega$ and the map $\zeta \mapsto z + \zeta w$ is affine, $\psi$ is a convex function on the convex [open set](/page/Open%20Set) $D \subset \mathbb{C} \cong \mathbb{R}^2$. [claim:A convex function on an open convex subset of $\mathbb{R}^2$ is subharmonic] Let $\psi: D \to \mathbb{R}$ be convex on a convex [open set](/page/Open%20Set) $D \subset \mathbb{R}^2$. [proof] For any $\zeta_0 \in D$ and $r > 0$ with $\overline{B}(\zeta_0, r) \subset D$, we verify the sub-mean-value inequality. Each point $\zeta_0 + re^{i\theta}$ on the circle satisfies \begin{align*} \zeta_0 = \frac{1}{2}(\zeta_0 + re^{i\theta}) + \frac{1}{2}(\zeta_0 - re^{i\theta}), \end{align*} so by convexity of $\psi$: \begin{align*} \psi(\zeta_0) \leq \frac{1}{2}\psi(\zeta_0 + re^{i\theta}) + \frac{1}{2}\psi(\zeta_0 + re^{i(\theta + \pi)}). \end{align*} Integrating over $\theta \in [0, 2\pi]$ with respect to $\frac{1}{2\pi}\,d\mathcal{L}^1(\theta)$: \begin{align*} \psi(\zeta_0) &\leq \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{2}\psi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta) + \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{2}\psi(\zeta_0 + re^{i(\theta+\pi)})\,d\mathcal{L}^1(\theta). \end{align*} The substitution $\theta' = \theta + \pi$ in the second integral shows both integrals are equal to $\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{2}\psi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta)$. Therefore \begin{align*} \psi(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} \psi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta), \end{align*} which is the sub-mean-value inequality. Since $\psi$ is continuous (convex functions on open sets are continuous), $\psi$ is subharmonic. [/proof] [/claim] Since $\psi(\zeta) = \phi(z + \zeta w)$ is subharmonic in $\zeta$ for every $z \in \Omega$ and $w \in \mathbb{C}^n$, and $\phi$ is continuous (as the composition of the continuous functions $-\log$ and $d(\cdot, \partial\Omega)$) and upper semicontinuous, the function $\phi$ is plurisubharmonic on $\Omega$. [/step] [step:Verify that $\phi$ is a psh exhaustion of $\Omega$] For each $c \in \mathbb{R}$, the sublevel set is $\{\phi < c\} = \{z \in \Omega : d(z, \partial\Omega) > e^{-c}\}$. We show $\{\phi < c\} \subset\subset \Omega$. Since $\Omega$ is bounded, $\overline{\Omega}$ is compact, and $\{\phi < c\} \subset \Omega$ is bounded. The closure $\overline{\{\phi < c\}}$ satisfies $d(z, \partial\Omega) \geq e^{-c} > 0$ for every $z \in \overline{\{\phi < c\}}$, so $\overline{\{\phi < c\}} \cap \partial\Omega = \varnothing$. Therefore $\overline{\{\phi < c\}}$ is a closed bounded subset of $\Omega$, hence compact and compactly contained in $\Omega$. Moreover, $\phi(z) = -\log d(z, \partial\Omega) \to +\infty$ as $z \to \partial\Omega$ (since $d(z, \partial\Omega) \to 0$), confirming that $\phi$ is an exhaustion. Combined with plurisubharmonicity and continuity, $\phi$ is a continuous psh exhaustion of $\Omega$, so $\Omega$ is pseudoconvex. [/step]