Last updated: October 13, 2026
The following note is based on ongoing conversations between me and Claude about OpenAI's recent claimed proof of the symmetric Mahler conjecture. While the main ideas are taken from the original proof, some of the details are being rewritten to facilitate my understanding, and some figures and interactive applets are being added for visualization. This is a work in progress. I do not claim to have verified the proof completely.
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For background on the Mahler conjecture, see the survey of Fradelizi, Meyer and Zvavitch1. Complex-analytic methods were used earlier by Nazarov2 and by Berndtsson3,4 to prove the Bourgain–Milman inequality.
A convex body is a compact convex set \(K\subset\mathbb{R}^d\) with nonempty interior; it is symmetric if \(K=-K\). Recall that the polar body \(K^\circ\) of a convex body \(K\) containing the origin in its interior is defined by \[K^{\circ}:=\{y\in\mathbb{R}^d\colon\langle x,y\rangle\le1\ \ \forall\,x\in K\}.\]
For every symmetric convex body \(K\subset \mathbb{R}^{d}\),
\[ \operatorname{Vol}(K)\operatorname{Vol}(K^{\circ}) \ge \frac{4^{d}}{d!}.\tag{1} \]Equality holds if and only if \(K\) is a linear image of a Hanner polytope.
For the definition of Hanner polytopes, see The equality case.
We first assume that \(K\) is a polytope; the general case of the inequality follows by a standard approximation argument, see Lemma 7. Let \(b_1,\dots, b_{m}\) be nonzero, pairwise nonproportional vectors such that \[ K=\{ x \in \mathbb{R}^{d}\,\colon\,|\langle b_{j},x \rangle| \leq 1\quad\forall\, 1\leq j\leq m \}. \] We will construct, for every \(x \in \operatorname{int} K\), a star-shaped set \(\Sigma_{x}\subset K^{\circ}\), and prove that \[ \int_{K}\operatorname{Vol}(\Sigma_{x})\,d\operatorname{Vol}(x) \geq \frac{4^{d}}{d!}.\tag{2} \] Since \(\operatorname{Vol}(\Sigma_{x})\leq \operatorname{Vol}(K^{\circ})\), this will imply (1).
Define \(\lambda:[-1,1]\to[0,\infty)\) by \[ \lambda(t):=\frac{2}{\pi}\int_{|t|}^{1}\operatorname{arctanh}\left(\sin\frac{\pi s}{2}\right)ds, \] and let \[ D:=\{ w\in\mathbb{C}\,\,\colon\,\,|\mathrm{Re}(w)|\lt 1,\,\,|\mathrm{Im}(w)|\lt\lambda(\mathrm{Re}(w)) \}. \]
Figure 1: the domain \(D\)
Since \(\lambda\) is concave and bounded, \(D\) is a bounded convex domain, so the map \(F\) in Lemma 0 exists and is unique by the Riemann mapping theorem, and extends to \(\overline{\mathbb D}\) by Carathéodory’s theorem. Lemma 0 is proved in Appendix: Properties of the function \(F\).
Let \[ \Omega :=\{ z\in \mathbb{C}^{d}\,\,\colon\,\, \langle b_{j},z \rangle \in D \quad \forall\,1\leq j\leq m\}. \] Since \(D\cap \mathbb{R}=(-1,1)\), we have \[ \Omega \cap \mathbb{R}^{d}=\operatorname{int} K. \] For every \(x \in \operatorname{int} K\), denote by \(L_x\) the fiber of \(\overline{\Omega}\) over \(x\): \[ L_{x}:=\{ y\in \mathbb{R}^{d}\,\,\colon\,\,x+iy\in \overline{\Omega} \}, \qquad x \in \operatorname{int} K. \] Since \(\overline{D}=\{ w\colon|\mathrm{Re}(w)|\le1,\,|\mathrm{Im}(w)|\le\lambda(\mathrm{Re}(w)) \}\), \(L_{x}\) is the symmetric polytope \[ L_{x}=\{ y\in \mathbb{R}^{d}\,\,\colon\,\, |\langle b_{j},y \rangle|\leq \lambda\left(\langle b_{j},x \rangle\right)\quad \forall\,1\leq j\leq m \}, \qquad x \in \operatorname{int} K. \] Note that \[ L_x=\{y\in\mathbb{R}^{d}\,\colon\,\langle h,y\rangle\le\lambda(\langle h,x\rangle)\quad\forall\,h\in K^{\circ}\}.\tag{3} \] Indeed, for fixed \(y\in\mathbb{R}^d\), the function \(h\mapsto\lambda(\langle h,x\rangle)-\langle h,y\rangle\) is concave, since \(\lambda\) is. Its minimum over the polytope \(K^{\circ}=\operatorname{conv}\{\pm b_l\}\) is therefore attained at one of the vertices \(\pm b_l\). As \(\lambda\) is even, the inequalities \(\langle\pm b_l,y\rangle\le\lambda(\langle\pm b_l,x\rangle)\) are precisely the inequalities defining \(L_x\), and (3) follows.
For \(J\subset[m]\), define a linear map \(B_J:\mathbb{R}^d\to \mathbb{R}^{J}\) by \[ (B_{J}x)_{j}=\langle b_{j},x \rangle ,\qquad j\in J, \qquad x \in \mathbb{R}^d, \] and let \[ \mathcal{J}:=\left\{ J\in{[m]\choose d}\,\,\colon\,\,B_J\text{ is invertible} \right\} \] (note that \(\mathbb{R}^d\) and \(\mathbb{R}^J\) have the same dimension).
Extend \(\lambda\) to all of \(\mathbb{R}\) by setting \(\lambda(t):=0\) for \(|t|>1\). For \(J\in\mathcal{J}\), \(\varepsilon \in \{ -1,1 \}^{J}\) and \(x\in\mathbb{R}^d\), let \(y_{J,\varepsilon}(x)\) be the unique solution to \[ \langle b_j,y_{J,\varepsilon}(x)\rangle=\varepsilon_j\lambda(\langle b_j,x\rangle),\qquad j\in J. \] For \(x\in\operatorname{int} K\), the point \(y_{J,\varepsilon}(x)\) satisfies \(d\) of the inequalities defining \(L_x\) with equality, and the vectors \(\{ \varepsilon_jb_j\}_{j\in J}\) are linearly independent. Hence either \(y_{J,\varepsilon}(x)\notin L_x\), or \(y_{J,\varepsilon}(x)\) is a vertex of \(L_x\).
Define \[ \Sigma_x:=\bigcup_{y\in \partial L_x}\operatorname{conv}\big(\{0\}\cup\{h\in K^{\circ}\,\colon\,\langle h,y\rangle=\lambda(\langle h,x\rangle)\}\big)\subseteq K^{\circ}.\tag{4} \]
For \(J\in \mathcal{J}\) and \(\varepsilon \in \{ -1,1 \}^{J}\), write \(\Delta_{J,\varepsilon}\) for the simplex \[ \Delta_{J,\varepsilon}:=\operatorname{conv} \left( \{0\}\cup \{\varepsilon_{j}b_{j}\}_{j\in J}\right)\subset K^{\circ}, \] and let \[ \mathcal{E}_{J}(x):=\left\{ \varepsilon \in \{ -1,1 \}^{J}\,\, \colon\,\, y_{J,\varepsilon}(x)\in L_{x} \right\}. \] Then for every \(x\in\operatorname{int} K\), \[ \Sigma_{x}=\bigcup\limits_{J\in \mathcal{J}}\bigcup\limits_{\varepsilon \in \mathcal{E}_{J}(x)}\Delta_{J,\varepsilon}.\tag{5} \] We shall only need the inclusion \(\supseteq\), which is immediate: if \(\varepsilon\in\mathcal{E}_J(x)\), then \(y=y_{J,\varepsilon}(x)\) is a vertex of \(L_x\) and \[ \langle\varepsilon_jb_j,y\rangle=\lambda(\langle b_j,x\rangle)=\lambda(\langle\varepsilon_jb_{j},x\rangle)\qquad \text{for all $j\in J$,} \] so \(\Delta_{J,\varepsilon}\subseteq\Sigma_x\). The inclusion \(\subseteq\) is proved in Appendix: \(\Sigma_x\) as a union of simplices. In Figure 2 below, each vertex \(y_{J,\varepsilon}(x)\) of \(L_x\) with \(\varepsilon\in\mathcal{E}_J(x)\) (left) and the simplex \(\Delta_{J,\varepsilon}\) (right) are drawn in the same colour.
Figure 2: the set \(\Sigma_x\)
For almost every \(x \in \operatorname{int} K\), the simplices \(\{\Delta_{J,\varepsilon}\}_{J\in \mathcal{J},\,\varepsilon \in \mathcal{E}_J(x)}\) have disjoint interiors.
Proof. Let \(\varepsilon\in\mathcal{E}_J(x)\) and let \(\xi\in\operatorname{int}\Delta_{J,\varepsilon}\). Then there exist \(\alpha_{j}>0\) such that \(\sum\limits_{j\in J}\alpha_{j}\lt 1\) and \[ \xi=\sum_{j\in J}\alpha_j\varepsilon_jb_j. \] For \(y\in L_x\), \[ \langle \xi,y\rangle\le\sum_{j\in J}\alpha_j\lambda(\langle b_j,x\rangle), \] with equality if and only if \(y=y_{J,\varepsilon}(x)\). So \(y_{J,\varepsilon}(x)\) is the unique maximizer of the functional \(\xi\) on \(L_x\), and so it suffices to show that for almost every \(x\), the points \((y_{J,\varepsilon}(x))_{J\in\mathcal{J},\varepsilon\in\mathcal{E}_J(x)}\) are distinct.
For fixed \(J\), since \(\lambda>0\) on \((-1,1)\), the defining equations of \(y_{J,\varepsilon}(x)\) have distinct solutions for different \(\varepsilon\) so the points \(y_{J,\varepsilon}(x)\) are distinct. Thus it suffices to consider pairs \((J,\varepsilon), (J',\varepsilon')\) with \(J\neq J'\); fix such a pair and fix \(k\in J'\setminus J\). If \(y_{J,\varepsilon}(x)=y_{J',\varepsilon'}(x)\), then \[ \langle b_k,y_{J,\varepsilon}(x)\rangle=\varepsilon'_k\lambda(\langle b_k,x\rangle).\tag{6} \] Write \(b_k=\sum_{j\in J}c_jb_j\) and use the coordinates \(q_j=\langle b_j,x\rangle\), \(j\in J\), on \(\operatorname{int} K\). Since \(\langle b_j,y_{J,\varepsilon}(x)\rangle=\varepsilon_j\lambda(q_j)\), equation (6) reads \[ G(q):=\sum_{j\in J}c_j\varepsilon_j\lambda(q_j)-\varepsilon'_k\lambda\Big(\sum_{j\in J}c_jq_j\Big)=0. \] Since \(b_k\) is not proportional to \(b_j\) for any \(j\neq k\), at least two coefficients \(c_{j_1},c_{j_2}\) are nonzero. Hence, by Lemma 0, \[ \frac{\partial^{2}G}{\partial q_{j_1}\,\partial q_{j_2}}=-\varepsilon'_k c_{j_1}c_{j_2}\lambda''\Big(\sum_{j\in J}c_jq_j\Big)\neq0 \] everywhere. By the implicit function theorem, both \(\{G=0,\ \nabla G\neq0\}\) and \(\{\partial G/\partial q_{j_1}=0\}\supseteq\{\nabla G=0\}\) are locally \(C^1\) hypersurfaces, so the level set \(\{G=0\}\) has measure zero. Since there are finitely many pairs \((J,\varepsilon),(J',\varepsilon')\), this proves the lemma. \(\square\)
By (5) and Lemma 1, we get
\[ \operatorname{Vol}(\Sigma_{x})\ge\frac{1}{d!}\sum\limits_{J\in \mathcal{J}}|\det B_{J}|\cdot|\mathcal{E}_J(x)| \qquad \text{for almost every \(x\in\operatorname{int} K\).} \]Thus, in order to prove (2), it suffices to prove that \[ \begin{aligned} 4^{d} & \leq \int_{K}\sum\limits_{J\in \mathcal{J}}|\det B_{J}|\cdot|\mathcal{E}_J(x)|\,d\operatorname{Vol}(x) \\ & = \sum\limits_{J\in \mathcal{J}}\sum\limits_{\varepsilon \in \{ -1,1 \}^{J}}|\det B_{J}|\cdot \operatorname{Vol}\left(\{x \in K\,\,\colon\,\,\varepsilon \in \mathcal{E}_{J}(x)\}\right), \end{aligned} \] or equivalently, using the definitions of \(\mathcal{E}_J\), \(y_{J,\varepsilon}(x)\) and \(L_x\), \[ 4^{d}\leq\sum\limits_{J\in \mathcal{J}}\sum\limits_{\varepsilon \in \{ -1,1 \}^{J}}|\det B_{J}|\cdot \operatorname{Vol}\left(\{x \in K\,\,\colon\,\,\left| \langle b_{k},y_{J,\varepsilon}(x) \rangle \right| \leq \lambda(\langle b_{k},x \rangle )\quad\forall k\not\in J\}\right).\tag{7} \]
Fix \(J\in \mathcal{J}\), \(q\in[-1,1]^J\) and \(\varepsilon\in\{-1,1\}^J\), and write \[ z_J(q,\varepsilon):=B_J^{-1}q+iy_{J,\varepsilon}(B_J^{-1}q). \] Since \(\langle b_j,B_J^{-1}q\rangle=q_j\) for \(j\in J\), by the definition of \(y_{J,\varepsilon}\) \[ \langle b_j,z_J(q,\varepsilon)\rangle=q_j+i\varepsilon_j\lambda(q_j)\in\partial D, \qquad j\in J. \] Let \(x=B_J^{-1}q\) and \(y=y_{J,\varepsilon}(x)\), so that \(z_J(q,\varepsilon)=x+iy\). Since \(\overline\Omega=\{z\colon\langle b_l,z\rangle\in\overline D\ \forall l\}\) and \(\overline D=\{w\colon|\mathrm{Re}\,w|\le1,\ |\mathrm{Im}\,w|\le\lambda(\mathrm{Re}\,w)\}\), \[ x+iy\in\overline\Omega\iff|\langle b_l,x\rangle|\le1\ \text{ and }\ |\langle b_l,y\rangle|\le\lambda(\langle b_l,x\rangle)\qquad\text{for all }1\le l\le m. \] The conditions \(|\langle b_l,x\rangle|\le1\), \(1\le l\le m\), are equivalent to \(x\in K\); and among the conditions \(|\langle b_l,y\rangle|\le\lambda(\langle b_l,x\rangle)\), those with \(l\in J\) hold with equality by the definition of \(y_{J,\varepsilon}\). Hence \[ z_J(q,\varepsilon)\in\overline\Omega\iff B_J^{-1}q\in K\ \text{ and }\ |\langle b_k,y_{J,\varepsilon}(B_J^{-1}q)\rangle|\le\lambda(\langle b_k,B_J^{-1}q\rangle)\quad\forall k\notin J. \] Since \(q\) and \(\varepsilon\) were arbitrary, and since \[ K\subseteq\{x\colon|\langle b_j,x\rangle|\le1\ \forall j\in J\}=B_J^{-1}([-1,1]^J), \]the change of variables \(q=B_Jx\) gives \[ \begin{aligned} &|\det B_J|\operatorname{Vol}\left(\{x \in K\,\,\colon\,\,\left| \langle b_{k},y_{J,\varepsilon}(x) \rangle \right| \leq \lambda(\langle b_{k},x \rangle )\quad\forall k\not\in J\}\right)\\ &\qquad=\operatorname{Vol}\left(\{q\in[-1,1]^J\,\,\colon\,\,z_{J}(q,\varepsilon)\in \overline\Omega\}\right). \end{aligned} \tag{8} \]
By Lemma 0, \(F\) is a homeomorphism of \(\overline{\mathbb D}\) onto \(\overline D\). Define \(\Phi_J:[-1,1]^J\times\{-1,1\}^J\to(\partial\mathbb D)^J\) by \[ \Phi_J(q,\varepsilon)_j:=F^{-1}\big(q_{j}+i\varepsilon_{j}\lambda(q_{j})\big),\qquad j\in J. \] Extend \(B_J\) (and its inverse) complex-linearly to a map from \(\mathbb{C}^d\) to \(\mathbb{C}^J\); then, by the formula for \(y_{J,\varepsilon}(B_J^{-1}q)\), \[ z_J(q,\varepsilon)=B_J^{-1}\,\mathrm{Re}\,F(\Phi_J(q,\varepsilon))+i\,B_J^{-1}\,\mathrm{Im}\,F(\Phi_J(q,\varepsilon))=B_J^{-1}F(\Phi_J(q,\varepsilon)). \]
Let \(\mu_J\) be the Haar probability measure on \((\partial\mathbb D)^J\). By our choice of \(F\), the function \(\Phi_J\) pushes the product of Lebesgue measure on \([-1,1]^J\) and counting measure on \(\{-1,1\}^J\) forward to \(4^d\mu_J\). Hence, summing (8) over \(\varepsilon\) gives \[ \begin{aligned} &\sum_{\varepsilon\in\{-1,1\}^J}|\det B_J|\operatorname{Vol}\left(\{x \in K\,\,\colon\,\,\left| \langle b_{k},y_{J,\varepsilon}(x) \rangle \right| \leq \lambda(\langle b_{k},x \rangle )\quad\forall k\not\in J\}\right)\\ &\qquad=4^{d}\mu_J\left(\{\omega\in(\partial\mathbb D)^J\,\,\colon\,\,B_J^{-1}F(\omega)\in \overline\Omega\}\right). \end{aligned} \]
Since \(\overline D=F(\overline{\mathbb D})\) and \(\langle b_j,B_J^{-1}F(\omega)\rangle=F(\omega_j)\in F(\overline{\mathbb D})\) for \(j\in J\), (7) becomes \[ \sum_{J\in\mathcal{J}}\mu_J\left(\left\{\omega \in (\partial\mathbb D)^J\,\,\colon\,\,\langle b_k,B_J^{-1}F(\omega)\rangle\in F(\overline{\mathbb D})\quad\forall\, k\notin J\right\}\right)\geq 1. \] For \(J\in\mathcal{J}\) let \[ A_J:=F^{-1}\big(B_J(\overline\Omega)\big)\subseteq\overline{\mathbb D}^J, \] with \(F\) applied coordinatewise; then the set \(A_J\) is closed, since \(B_J(\overline\Omega)\) is closed and \(F\) is continuous. For \(\omega\in(\partial\mathbb D)^J\), \(\omega\in A_J\) if and only if \(B_J^{-1}F(\omega)\in\overline\Omega\), so the inequality to prove is \[ \sum_{J\in\mathcal{J}}\mu_J(A_J)\geq 1. \]
Figure 3: from \(K\) to the cube to the polycircle
Recap:
and our goal is to prove that \[ \sum_{J\in\mathcal{J}}\mu_J(A_J)\geq 1, \] where \(\mu_{J}\) is the Haar probability measure on \((\partial\mathbb D)^{J}\).
Let \(B:=B_{[m]}:\mathbb{C}^d\to\mathbb{C}^m\), so that \(B_J=\pi_J\circ B\) for the coordinate projection \(\pi_J:\mathbb{C}^m\to\mathbb{C}^J\).
For \(k\ge1\) let \(p_k\) be the coordinatewise \(k\)-th power, \(p_k(v)=(v_j^k)_j\), on \(\mathbb{C}^m\) and on each \(\mathbb{C}^J\); then \(\pi_J\circ p_k=p_k\circ\pi_J\). Let \[ dV_J:=\bigwedge_{j\in J}\frac{i}{2}\,dv_j\wedge d\bar v_j \] denote the volume form of \(\mathbb{C}^J\), let \(dV:=dV_{[d]}\) denote the volume form of \(\mathbb{C}^d\), and let \(M_{k}\) be the \(d\)-dimensional complex submanifold \[ M_k:=p_k^{-1}(\mathbb B_m)\cap F^{-1}\big(B(\mathbb{C}^d)\big)\subseteq\mathbb D^m, \] where \(\mathbb B_m\) denotes the open Euclidean unit ball in \(\mathbb{C}^m\).
The map \[ \phi:=F^{-1}\circ B \] is a biholomorphism of \(\Omega\) onto \(F^{-1}(B(\mathbb{C}^d))\), so \(M_k=\phi(\Omega_k)\) where \[ \Omega_k:=\{z\in\Omega\colon p_k(\phi(z))\in\mathbb B_m\}. \] For \(u\in M_k\) and \(J\in\mathcal{J}\), \[ F(u)=B\,B_J^{-1}F(\pi_Ju), \] so \(\pi_J\) is injective on \(M_k\) and \(\pi_J(M_k)\subseteq A_J\).
Define a \(2d\)-form \(\sigma_{k}\) on \(M_{k}\) by \[ \sigma_k:=\sum_{J\in\mathcal{J}}(\pi_J|_{M_k})^*\,p_k^*\,dV_J. \]
Each term of \(\sigma_k\) is a nonnegative \(2d\)-form on \(M_k\), and the integral \[ \int_{M_k}\sigma_k \] is the \(2d\)-dimensional volume of \(p_k(M_k)\), counted with multiplicity.
Proof. Let \(f_{k}:=p_k\circ\phi\), let \(P:=Df_k(z)\), and for \(J \in {[m]\choose d}\) denote by \(P_J\) the corresponding minor of \(P\). Since \(f_k=(p_k\circ F^{-1})\circ B\) with \(p_k\circ F^{-1}\) acting coordinatewise, \(P_J\) is \(B_J\) multiplied on the left by a diagonal matrix, so \(\det P_J=0\) for \(J\notin\mathcal{J}\). Then \[ \phi^*\sigma_k=\sum_{J\in\mathcal{J}}(\pi_J\circ f_k)^*dV_J=\sum_{J\in\mathcal{J}}|\det P_J|^2\,dV=\sum_{J\in{[m]\choose d}}|\det P_J|^2\,dV, \] where the first equality follows from \(\pi_J\circ p_k=p_k\circ\pi_J\), the second holds because a complex-linear map \(T\colon\mathbb{C}^d\to\mathbb{C}^J\) has real Jacobian \(|\det T|^2\), and the third holds since \(\det P_J=0\) for all \(J\notin\mathcal{J}\). By Cauchy–Binet, \(\sum_{J\in{[m]\choose d}}|\det P_J|^2\) is the \(2d\)-dimensional Jacobian of \(f_k\) at \(z\), and the lemma follows. \(\square\)
For all \(J \in \mathcal{J}\),
\[ \limsup_{k \to \infty} \frac{1}{k^{d}}\int_{M_{k}}(\pi_J|_{M_k})^*p_k^*dV_J\leq \frac{\pi^{d}}{d!}\mu_{J}(A_{J}). \]Proof. Write \[ R_{J}:=\{\rho\in\mathbb{R}^J\,\,\colon\,\,\sum_j\rho_j^2\lt 1,\,\,\rho_{j}>0\,\,\forall j\in J\} \] and \(\rho^{1/k}\omega:=(\rho_j^{1/k}\omega_j)_{j\in J}\). Then \[ \begin{align*} \limsup_k\frac{1}{k^d}\int_{M_k}(\pi_J|_{M_k})^*p_k^*dV_J &\le\limsup_k\frac{1}{k^d}\int_{A_J\cap p_k^{-1}(\mathbb B_J)}p_k^*dV_J\\ &=\limsup_k\,(2\pi)^d\int_{R_{J}\times(\partial\mathbb D)^J}\mathbf 1_{A_J}(\rho^{1/k}\omega)\prod_j\rho_j\,d\rho\,d\mu_J(\omega)\\ &\le(2\pi)^d\int_{R_{J}\times(\partial\mathbb D)^J}\mathbf 1_{A_J}(\omega)\prod_j\rho_j\,d\rho\,d\mu_J(\omega)\\ &=\frac{\pi^d}{d!}\,\mu_J(A_J). \end{align*} \] The first inequality holds because \(\pi_J\) is injective on \(M_k\) and \(\pi_J(M_k)\subseteq A_J\cap p_k^{-1}(\mathbb B_J)\). In the second passage we applied the substitution \(v_j=\rho_j^{1/k}\omega_j\) and used \(p_k^*dV_J=k^d(2\pi)^d\prod_j\rho_j\,d\rho\otimes\mu_J\). The second inequality is Fatou's lemma, since \(A_J\) is closed and \(\rho^{1/k}\omega\to\omega\), so \(\limsup_k\mathbf 1_{A_J}(\rho^{1/k}\omega)\le\mathbf 1_{A_J}(\omega)\). \(\square\)
Summing Lemma 3 over \(J\in \mathcal{J}\) we get \[ \limsup_{k \to \infty} \frac{1}{(\pi k)^{d}}\int_{M_{k}}\sigma_{k}\leq \frac{1}{d!}\sum\limits_{J\in \mathcal{J}}\mu_{J}(A_{J}). \] The proof will thus be finished once we prove that \[ \frac{1}{(\pi k)^{d}}\int_{M_{k}}\sigma_{k}\geq \frac{1}{d!}. \]
Recall that \[ M_k:=p_k^{-1}(\mathbb B_m)\cap F^{-1}\big(B(\mathbb{C}^d)\big), \] and that, by Lemma 2, the integral \(\int_{M_k}\sigma_{k}\) is the \(2d\)-dimensional volume of the image \(p_{k}(M_{k})\), counted with multiplicity. Thus it suffices to apply the following key lemma:
Let \(d\geq 1\) and \(m\geq d\). Let \(U\subset \mathbb{C}^{d}\) be an open set containing the origin, and let \(f:U\to \mathbb{C}^{m}\) be a holomorphic function satisfying
Then the \(2d\)-dimensional volume of \(f(U_{1})\), counted with multiplicity, is at least \(\frac{(\pi k)^{d}}{d!}\).
Let us verify that the hypotheses of Lemma 4 are satisfied with \(f=f_k=p_{k}\circ F^{-1}\circ B\). The set \(\Omega\) is open and contains \(0\), and \(f_k\) is holomorphic on \(\Omega\) because \(B(\Omega)\subseteq D^m\) and \(F^{-1}\) acts coordinatewise. In the notation of the lemma, we have \(U_1=\{ |f_{k}|\lt 1 \}=\Omega_k\), so \(f_k(U_1)=p_k(\phi(\Omega_k))=p_k(M_k)\). Since \(F(0)=0\) and \(F\) is injective, \(f_k(z)=0\) if and only if \(Bz=0\), that is, \(z=0\), because \(b_1,\dots,b_m\) span \(\mathbb{R}^d\).
Since \(F^{-1}\) is holomorphic near \(0\) and \(F^{-1}(0)=0\), there are \(\delta>0\) and \(C>0\) with \(|F^{-1}(w)|\le C|w|\) for \(|w|\lt\delta\). Since \(|f_k(z)|^{2}=\sum_{j}\big|F^{-1}(\langle b_j,z\rangle)\big|^{2k}\), for \(z\) near \(0\) we have \[ |f_k(z)|\le\sqrt m\,C^{k}\max_j|\langle b_j,z\rangle|^{k}. \] Since \(B\) is injective, \(\max_j|\langle b_j,\cdot\rangle|\) is a norm on \(\mathbb{C}^d\), so \(|f_k(z)|\le C'|z|^{k}\) for some \(C'>0\) near the origin.
Let \(0\lt s \lt1\). For \(z\in U_s\) and every \(j\), \[|F^{-1}(\langle b_j,z\rangle)|^{k}\le|f_k(z)|\lt s,\] so \(\langle b_j,z\rangle\in F(s^{1/k}\overline{\mathbb D})\). Hence \[ \overline{U_s}\subseteq\{z\in\mathbb{C}^d\colon\langle b_j,z\rangle\in F(s^{1/k}\overline{\mathbb D})\ \ \forall\,1\le j\le m\}=B^{-1}\big(F(s^{1/k}\overline{\mathbb D})^{m}\big). \] The set \(F(s^{1/k}\overline{\mathbb D})\) is a compact subset of \(D\) and \(B\) is injective, so the right-hand side is a compact subset of \(\Omega\). Thus \(U_s\) is relatively compact in \(\Omega\).
The hypotheses of Lemma 4 are satisfied; hence the theorem (modulo the equality cases) would follow once we prove Lemma 5.
Proof of Lemma 5. We use the normalization \[ d := \partial + \bar\partial, \qquad d^{c}:=\frac{i}{4}(\bar\partial-\partial),\qquad dd^{c}=\frac{i}{2}\partial\bar\partial, \] so that \(dd^{c}|z|^{2}=\sum_{j}dx_{j}\wedge dy_{j}\) and \((dd^{c}|z|^{2})^{d}=d!\,dV\).
On \(\mathbb{C}^m\), \((dd^{c}|w|^{2})^{d}=d!\sum_{J\in{[m]\choose d}}dV_J\), so the computation in the proof of Lemma 2, with \(Df\) in place of \(P\), gives \((dd^{c}|f|^{2})^{d}=d!\sqrt{\det(Df_\mathbb{R}^TDf_\mathbb{R})}\,dV\).
Hence it suffices to prove that \[ \int_{U_{1}}(dd^{c}|f|^{2})^{d}\geq(\pi k)^{d}. \] Let \(s \in(0,1)\) be a regular value of \(|f|\). By Stokes' theorem, \[ \int_{U_{s}}(dd^{c}|f|^{2})^{d}=\int_{\partial U_{s}}d^{c}|f|^{2}\wedge(dd^{c}|f|^{2})^{d-1}. \] Write \[ \gamma:=d^{c}\log|f|^{2}=|f|^{-2}d^{c}|f|^{2}, \] and note that \[ d\gamma=dd^{c}\log|f|^{2}=|f|^{-2}dd^{c}|f|^{2}-|f|^{-4}d|f|^{2}\wedge d^{c}|f|^{2}. \] Since \(|f|\equiv s\) on \(\partial U_{s}\), we have \[ \gamma=s^{-2}d^{c}|f|^{2}\qquad \text{and}\qquad d\gamma=s^{-2}dd^{c}|f|^{2}\qquad \text{on $\partial U_s$.} \] Hence \[ \int_{U_{s}}(dd^{c}|f|^{2})^{d}=s^{2d}\int_{\partial U_{s}}\gamma \wedge(d\gamma)^{d-1}. \] The function \(\log|f|^{2}\) is smooth and plurisubharmonic on \(U_{s}\setminus\{0\}\) by the first assumption.
For \(0 \lt \eta\lt 1\) set \[ g:=\log\frac{|f|^{2}}{s^{2}}\qquad \text{and}\qquad h:=(1-\eta)k\log\frac{|z|^{2}}{\max_{\overline{U_{s}}}|z|^{2}}-2 . \] On \(\overline{U_s}\) we have \(h\le-2\), while \(g=0\) on \(\partial U_s\). Hence \(g>h+1\) near \(\partial U_s\).
By the second assumption \(g\le k\log|z|^{2}+O(1)\) near the origin, so \[ h-g\ge-\eta k\log|z|^{2}+O(1)\to\infty \qquad \text{as $z\to 0$.} \] Let \(r>0\) be small enough that \(r\overline{\mathbb B}_d\subset U_{s}\) and \(h>g+1\) near \(\partial(r\mathbb B_d)\), where \(\mathbb B_d\) is the open unit ball of \(\mathbb{C}^d\). Let \(\widetilde{\max}:\mathbb{R}^{2}\to\mathbb{R}\) be smooth, convex and nondecreasing in each variable, with \(\widetilde{\max}(x,y)=\max(x,y)\) whenever \(|x-y|\geq 1\). Then the function \(\widetilde{\max}(g,h)\) is smooth and plurisubharmonic on \(U_{s}\setminus r\overline{\mathbb B}_d\), and equals \(g\) near \(\partial U_{s}\) and \(h\) near \(\partial(r\mathbb B_d)\). Hence, since \(d^{c}g=\gamma\) and \((dd^{c}\,\widetilde{\max}(g,h))^{d}\ge0\) on \(U_{s}\setminus r\overline{\mathbb B}_d\), Stokes' theorem on \(U_{s}\setminus r\overline{\mathbb B}_d\) gives \[ \int_{\partial U_{s}}\gamma\wedge(d\gamma)^{d-1}\geq\int_{\partial(r\mathbb B_d)}d^{c}h\wedge(dd^{c}h)^{d-1}=\frac{((1-\eta)k)^{d}}{r^{2d}}\int_{r\mathbb B_d}(dd^{c}|z|^{2})^{d}=((1-\eta)\pi k)^{d}. \] Combining with the previous identity and letting \(\eta\to 0\), we get \[ \int_{U_{s}}(dd^{c}|f|^{2})^{d}\geq s^{2d}(\pi k)^{d}. \] Letting \(s\uparrow 1\) along regular values and using monotone convergence gives the claim.5 \(\square\)
For a symmetric convex body \(A\), write \(\|x\|_A:=\min\{t\ge0\colon x\in tA\}\) for the norm with unit ball \(A\). For symmetric convex bodies \(A\subset\mathbb{R}^{k}\) and \(A'\subset\mathbb{R}^{l}\), define the symmetric convex bodies \[ \begin{aligned} A\oplus_\infty A'&:=A\times A'=\{(x,x')\in\mathbb{R}^{k+l}\colon\max\{\|x\|_A,\|x'\|_{A'}\}\le1\},\\ A\oplus_1A'&:=\operatorname{conv}\big((A\times\{0\})\cup(\{0\}\times A')\big)=\{(x,x')\in\mathbb{R}^{k+l}\colon\|x\|_A+\|x'\|_{A'}\le1\}. \end{aligned} \] Then \((A\oplus_1A')^{\circ}=A^{\circ}\oplus_\infty A'^{\circ}\) and \((A\oplus_\infty A')^{\circ}=A^{\circ}\oplus_1 A'^{\circ}\). Hanner polytopes form the smallest class of convex bodies that contains the segments \([-a,a]\subset\mathbb{R}\), \(a>0\), and is closed under \(\oplus_1\) and \(\oplus_\infty\).
Plan. The proof above gives \(\frac{4^{d}}{d!}\le\int_K\operatorname{Vol}(\Sigma_x)\,dx\le\operatorname{Vol}(K)\operatorname{Vol}(K^{\circ})\), where the first inequality follows from Lemma 6 and the second holds true since \(\Sigma_x\subseteq K^{\circ}\). Equality in the second inequality forces \(\Sigma_x=K^{\circ}\) for almost every \(x \in K\). Lemma 6 shows that this is then true for all \(x \in \operatorname{int} K\). Lemma 6 for \(\widehat K=K\oplus_1[-1,1]\) at points \((\delta u,1-\delta)\), \(\delta\downarrow0\), together with Lemma 0, gives that every triple of points of \((\mathbb{R}^d,\|\cdot\|_{K^{\circ}})\) has a metric median (Lemma 7), and Theorem 8, which is due to Hansen and Lima, identifies the norm.
We extend the definitions of \(L_{x}\) and \(\Sigma_{x}\) to arbitrary convex bodies as follows: for \(x\in\operatorname{int} K\) define \(L_x\) by the right-hand side of (3), and \(\Sigma_x\) by (4). Thus \(\xi\in\Sigma_x\) iff \(\xi=\sum_i\alpha_ih_i\) for finitely many \(h_i\in K^{\circ}\) with \(\langle h_i,y\rangle=\lambda(\langle h_i,x\rangle)\) for one \(y\in \partial L_x\) and all \(i\), where \(\alpha_i\ge0\) and \(\sum_i\alpha_i\le1\); by Carathéodory's theorem, \(d+1\) terms suffice. Since \(\lambda\le\lambda(0)\), every \(y\in L_x\) satisfies \(\langle h,y\rangle\le\lambda(0)\) for all \(h\in K^{\circ}\), that is, \[ L_x\subseteq\lambda(0)K .\tag{9} \]
Let \(K^{(j)}\supseteq K\) be symmetric convex bodies with \(K^{(j)}\subseteq t_j^{-1}K\) and \(t_j\to1\), and let \(L^{(j)}_x\) and \(\Sigma^{(j)}_x\) be defined as \(L_x\) in (3) and \(\Sigma_x\) in (4), with \(K\) replaced by \(K^{(j)}\). If \(x_j\to x\in\operatorname{int} K\), \(\xi_j\to\xi\) and \(\xi_j\in\Sigma^{(j)}_{x_j}\), then \(\xi\in\Sigma_x\).
Proof. Write \(\xi_j=\sum_{i=0}^{d}\alpha_{ij}h_{ij}\) with \(h_{ij}\in (K^{(j)})^{\circ}\), \(\langle h_{ij},y_j\rangle=\lambda(\langle h_{ij},x_j\rangle)\) for some \(y_j\in L^{(j)}_{x_j}\), \(\alpha_{ij}\ge0\), \(\sum_i\alpha_{ij}\le1\). By (9) for \(K^{(j)}\), \(y_j\in\lambda(0)K^{(j)}\subseteq\lambda(0)t_j^{-1}K\), so all data range in a compact set. Pass to a subsequence with \(y_j\to y\), \(h_{ij}\to h_i\in K^{\circ}\), \(\alpha_{ij}\to\alpha_i\). For \(h\in K^{\circ}\) we have \(t_jh\in (K^{(j)})^{\circ}\), so \(\langle t_jh,y_j\rangle\le\lambda(\langle t_jh,x_j\rangle)\); continuity of \(\lambda\) gives \(y\in L_x\). The same continuity turns \(\langle h_{ij},y_j\rangle=\lambda(\langle h_{ij},x_j\rangle)\) into \(\langle h_i,y\rangle=\lambda(\langle h_i,x\rangle)\). If all \(\alpha_i\) vanish then \(\xi=0\in\Sigma_x\). Otherwise some \(h_i\) has \(\alpha_i>0\), and then \(h_i\neq0\) since \(\lambda(0)>0\); as \(y+\varepsilon h_i\) violates (3) for every \(\varepsilon>0\), \(y\in\partial L_x\). So \(\xi=\sum\alpha_ih_i\in\Sigma_x\). \(\square\)
For every symmetric convex body \(K\), \[ \int_K\operatorname{Vol}(\Sigma_x)\,dx\ge\frac{4^{d}}{d!}. \] If \(\operatorname{Vol}(K)\operatorname{Vol}(K^{\circ})=4^{d}/d!\), then \(\Sigma_x=K^{\circ}\) for every \(x\in\operatorname{int} K\). In particular, for every \(x\in\operatorname{int} K\) and every \(\xi\in\partial K^{\circ}\) there exist \(y\in L_x\) and \(h_i\in K^{\circ}\) such that \(\xi\in\operatorname{conv}\{h_i\}\) and \(\langle h_i,y\rangle=\lambda(\langle h_i,x\rangle)\) for all \(i\).
Proof. Put \[ E_K:=\{(x,\xi)\,\colon\,x\in\operatorname{int} K,\ \xi\in\Sigma_x\}\subseteq K\times K^{\circ}. \] Take \(K^{(j)}=K\) and \(t_j=1\) in Lemma 5: if \((x_j,\xi_j)\in E_K\) and \((x_j,\xi_j)\to(x,\xi)\) with \(x\in\operatorname{int} K\), then \((x,\xi)\in E_K\). So \(E_K\) is relatively closed in \(\operatorname{int} K\times\mathbb{R}^{d}\), and each fiber \(\Sigma_x\) is closed; since \(\Sigma_x\subseteq K^{\circ}\), it is compact. By Fubini, \(\operatorname{Vol}_{2d}(E_K)=\int_K\operatorname{Vol}(\Sigma_x)\,dx\).
Now choose \(t_j\in(0,1)\) with \(t_j\to1\), and symmetric polytopes \(K^{(j)}\) with \[ K\subseteq K^{(j)}\subseteq t_j^{-1}K\qquad\text{for all }j. \] Since \(K^{(j)}\) is a polytope, (2) applies to it, so \[ \operatorname{Vol}_{2d}\big(E_{K^{(j)}}\big)=\int_{K^{(j)}}\operatorname{Vol}\big(\Sigma^{(j)}_x\big)\,dx\ge\frac{4^{d}}{d!}\qquad\text{for all }j. \] Put \[ E_j:=E_{K^{(j)}}\cap\big(\operatorname{int} K\times\mathbb{R}^{d}\big)=\big\{(x,\xi)\,\colon\,x\in\operatorname{int}K,\ \xi\in\Sigma^{(j)}_x\big\}. \] Since \(K\subseteq K^{(j)}\), we have \(\Sigma^{(j)}_x\subseteq(K^{(j)})^{\circ}\subseteq K^{\circ}\) for every \(x\). Hence \(E_j\subseteq K\times K^{\circ}\), and \(E_{K^{(j)}}\setminus E_j\subseteq(K^{(j)}\setminus\operatorname{int}K)\times K^{\circ}\), so \[ \operatorname{Vol}_{2d}(E_j)\ge\frac{4^{d}}{d!}-\operatorname{Vol}\big(K^{(j)}\setminus K\big)\operatorname{Vol}(K^{\circ})\ge\frac{4^{d}}{d!}-\big(t_j^{-d}-1\big)\operatorname{Vol}(K)\operatorname{Vol}(K^{\circ}). \] The last term tends to \(0\) as \(j\to\infty\). The set \(E_j\) is contained in \(K\times K^{\circ}\) for all $j$, so by the reverse Fatou lemma \[ \operatorname{Vol}_{2d}\Big(\limsup_{j\to\infty}E_j\Big)\ge\limsup_{j\to\infty}\operatorname{Vol}_{2d}(E_j)\ge\frac{4^{d}}{d!}. \] Finally, \(\limsup_jE_j\subseteq E_K\). Indeed, let \((x,\xi)\in E_j\) for infinitely many \(j\), say for \(j\) in an infinite set \(J_0\). Then \(x\in\operatorname{int}K\) and \(\xi\in\Sigma^{(j)}_x\) for all \(j\in J_0\), and Lemma 5, applied along \(J_0\) with the constant sequences \(x_j=x\) and \(\xi_j=\xi\), gives \(\xi\in\Sigma_x\), i.e., \((x,\xi)\in E_K\). Therefore \[ \int_K\operatorname{Vol}(\Sigma_x)\,dx=\operatorname{Vol}_{2d}(E_K)\ge\operatorname{Vol}_{2d}\Big(\limsup_{j\to\infty}E_j\Big)\ge\frac{4^{d}}{d!}, \] which proves the inequality.
Finally, assume \(\operatorname{Vol}(K)\operatorname{Vol}(K^{\circ})=4^{d}/d!\). Then \(E_K\subseteq K\times K^{\circ}\) and \(\operatorname{Vol}(E_K)\ge\operatorname{Vol}(K\times K^{\circ})\), so \((\operatorname{int} K\times\operatorname{int} K^{\circ})\setminus E_K\) is null. It is open by Lemma 5, hence empty. So \(\Sigma_x\supseteq\operatorname{int} K^{\circ}\), and \(\Sigma_x\) is closed. For \(\xi\in\partial K^{\circ}\), write \(\xi=\sum\alpha_ih_i\) as in the definition of \(\Sigma_x\); the weights \(\alpha_i\) sum to \(1\) since \[1=\|\xi\|_{K^{\circ}}\le\sum\alpha_i\|h_i\|_{K^{\circ}}\le\sum\alpha_i\le1,\] so \(\xi\in\operatorname{conv}\{h_i\}\). \(\square\)
Let \((V,\|\cdot\|)\) be a normed space. A median of a triple \(x_1,x_2,x_3\in V\) is a point \(m\) with \(\|x_i-m\|+\|m-x_j\|=\|x_i-x_j\|\) for all \(i\neq j\).
If \(\operatorname{Vol}(K)\operatorname{Vol}(K^{\circ})=4^{d}/d!\) then every triple of points has a median in \((\mathbb{R}^{d},\|\cdot\|_{K^{\circ}})\).
Proof. Write \(\|\cdot\|=\|\cdot\|_{K^{\circ}}\). It suffices to consider triples of the form \(0,v,u\). We may also assume that the three points are distinct (otherwise any repeated point is a median). The set \[ W:=\{w \in \mathbb{R}^{d}:\ \|v-w\|+\|u-w\|=\|v-u\|\} \] is compact and convex; indeed by the triangle inequality, it is the set of minimizers of the convex function \(\|v-\cdot\|+\|u-\cdot\|\), which tends to \(\infty\) at \(\infty\).
We claim that
\[ \min_{w\in W}\langle w,x\rangle\le\tfrac12\big(\|v\|+\|u\|-\|v-u\|\big)\qquad\text{for every }x\in K.\tag{10} \]First let us deduce the lemma from (10). Since \(\|w\|=\max_{x\in K}\langle w,x\rangle\), the minimax theorem for the bilinear pairing on the compact convex sets \(W\) and \(K\) gives \[ \min_{w\in W}\|w\|=\max_{x\in K}\min_{w\in W}\langle w,x\rangle\le\tfrac12\big(\|v\|+\|u\|-\|v-u\|\big). \] Pick a minimizer \(w_{0}\); then by the triangle inequality \[ \begin{align*} & 2\big(\|w_{0}\|+\|v-w_{0}\|+\|u-w_{0}\|\big)\\ &\qquad\qquad \le\|v\|+\|u\|+\|v-u\|\\ &\qquad\qquad \le\big(\|w_{0}\|+\|v-w_{0}\|\big)+\big(\|w_{0}\|+\|u-w_{0}\|\big)+\big(\|v-w_{0}\|+\|u-w_{0}\|\big), \end{align*} \] so the three triangle inequalities in the second step are equalities and \(w_{0}\) is a median.
It remains to prove (10). By continuity, we may assume that \(x\in\operatorname{int} K\), so that \(\|x\|_K\lt 1\). Let \[ \widehat K:=K\oplus_1[-1,1]=\{(z,s):\ \|z\|_K+|s|\le1\},\qquad \widehat K^{\circ}=K^{\circ}\times[-1,1]. \] The section of \(\widehat K\) at height \(s\) is \((1-|s|)K\), so \(\operatorname{Vol}(\widehat K)=\frac2{d+1}\operatorname{Vol}(K)\), while \(\operatorname{Vol}(\widehat K^{\circ})=2\operatorname{Vol}(K^{\circ})\). Hence \(\widehat K\) is also an equality case, in dimension \(d+1\). For \(\hat x\in\operatorname{int}\widehat K\), we write \(\hat L_{\hat x}\) for the set \(L_{\hat x}\) with the body \(K\) replaced by \(\widehat K\). Fix \(\delta\in(0,1)\) and set \[\hat x:=(\delta x,1-\delta)\in\operatorname{int}\widehat K. \] Let \(\hat v:=(v,\|v\|)\) and \(\hat u:=(u,\|u\|)\) be the lifts of \(v\) and \(u\) to the boundary of the cone \(\{(z,t)\colon\|z\|\le t\}\). Since \(\|(h,s)\|_{\widehat K^{\circ}}=\max\{\|h\|,|s|\}\), we have
\[ \|\hat v-\hat u\|_{\widehat K^{\circ}}=\max\big\{\|v-u\|,\,\big|\|v\|-\|u\|\big|\big\}=\|v-u\|,\qquad \|\hat v\|_{\widehat K^{\circ}}=\|v\|,\qquad \|\hat u\|_{\widehat K^{\circ}}=\|u\|. \]By Lemma 6 for \(\widehat K\) at \(\hat x\), applied to \((\hat v-\hat u)/\|v-u\|\in\partial\widehat K^{\circ}\), there exist \(\hat y\in \hat L_{\hat x}\), finitely many points \((h_i,s_i)\in\widehat K^{\circ}\) and weights \(a_i\ge0\) with \(\sum_ia_i=\|v-u\|\) and \(\langle(h_i,s_i),\hat y\rangle=\lambda\big(\langle(h_i,s_i),\hat x\rangle\big)\) for all \(i\), such that \[ \hat v-\hat u=\sum_ia_i(h_i,s_i). \] For \((h,s)\in\widehat K^{\circ}\), concavity and evenness of \(\lambda\) give \[ \begin{aligned} \lambda\big(\langle(h,s),\hat x\rangle\big)&=\lambda\Big(\frac{1+s}2\big(1-\delta(1-\langle h,x\rangle)\big)-\frac{1-s}2\big(1-\delta(1+\langle h,x\rangle)\big)\Big)\\ &\ge\frac{1+s}2\,\lambda\big(1-\delta(1-\langle h,x\rangle)\big)+\frac{1-s}2\,\lambda\big(1-\delta(1+\langle h,x\rangle)\big). \end{aligned}\tag{11} \] Apply (11) to each \((h_i,s_i)\), then pair with \(\hat y\), and finally use \(\hat y\in \hat L_{\hat x}\) for \(\hat v/\|v\|\) and \(-\hat u/\|u\|\), which lie in \(\widehat K^{\circ}\), together with evenness of \(\lambda\) and the definition of \(\hat x\): \[ \begin{aligned} &\sum_ia_i\Big(\frac{1+s_i}2\,\lambda\big(1-\delta(1-\langle h_i,x\rangle)\big)+\frac{1-s_i}2\,\lambda\big(1-\delta(1+\langle h_i,x\rangle)\big)\Big)\\ &\qquad\le\sum_ia_i\lambda\big(\langle(h_i,s_i),\hat x\rangle\big)=\langle\hat v-\hat u,\hat y\rangle\\ &\qquad\le\|v\|\,\lambda\Big(1-\delta\Big(1-\frac{\langle v,x\rangle}{\|v\|}\Big)\Big)+\|u\|\,\lambda\Big(1-\delta\Big(1-\frac{\langle u,x\rangle}{\|u\|}\Big)\Big). \end{aligned}\tag{12} \] Now divide (12) by \(\lambda(1-\delta)\) and let \(\delta\searrow0\). Every argument of \(\lambda\) in (12) is of the form \(1-\delta c\) with \(c\in[1-\|x\|_K,1+\|x\|_K]\), so by Lemma 0 each ratio \(\lambda(1-\delta c)/\lambda(1-\delta)\) may be replaced by \(c\), with a total error of \(o(1)\): \[ \begin{aligned} \sum_ia_i\big(1-s_i\langle h_i,x\rangle\big)&=\sum_ia_i\Big(\frac{1+s_i}2\big(1-\langle h_i,x\rangle\big)+\frac{1-s_i}2\big(1+\langle h_i,x\rangle\big)\Big)\\ &\le\|v\|\Big(1-\frac{\langle v,x\rangle}{\|v\|}\Big)+\|u\|\Big(1-\frac{\langle u,x\rangle}{\|u\|}\Big)+o(1)\\ &=\|v\|+\|u\|-\langle v+u,x\rangle+o(1). \end{aligned}\tag{13} \] Now put \[w:=v-\sum_ia_i\frac{1+s_i}2h_i. \] Since \(\sum_ia_ih_i=v-u\), we have \(u-w=-\sum_ia_i\frac{1-s_i}2h_i\), and since \(\|h_i\|\le1\), \[ \|v-w\|+\|u-w\|\le\sum_ia_i\Big(\frac{1+s_i}2+\frac{1-s_i}2\Big)=\|v-u\|, \] so \(w\in W\).
Figure 4: proof of Lemma 7
Adding the expressions for \(v-w\) and \(u-w\) gives \(v+u-2w=\sum_ia_is_ih_i\), so the left side of (13) is \(\|v-u\|-\langle v+u-2w,x\rangle\), and (13) reads \[ \langle w,x\rangle\le\tfrac12\big(\|v\|+\|u\|-\|v-u\|\big)+o(1)\qquad\text{as }\delta\searrow0. \] This proves (10), since \(w\in W\). \(\square\)
Let \(\|\cdot\|\) be a norm on \(\mathbb{R}^{d}\). Suppose that every triple of points has a median in \((\mathbb{R}^{d},\|\cdot\|)\). Then the unit ball of \(\|\cdot\|\) is a linear image of a Hanner polytope.
Proof. A normed space has the 3.2 intersection property if any three pairwise intersecting closed balls, of arbitrary radii, have a common point.8 Let the closed balls of radii \(\rho_i\) centered at \(x_i\), \(i=1,2,3\), intersect pairwise. Let \(m\) be a median of the centers and \(r_i=\|m-x_i\|\), so \(r_i+r_j=\|x_i-x_j\|\le\rho_i+\rho_j\). If every \(r_i\le\rho_i\), then \(m\) is a common point. Otherwise say \(t:=r_1-\rho_1>0\), so \(r_j+t\le\rho_j\) for \(j=2,3\). The point \(z=m+\frac t{r_1}(x_1-m)\) has \(\|z-x_1\|=\rho_1\) and \(\|z-x_j\|\le t+r_j\le\rho_j\). Hence the space has the 3.2 intersection property, and the conclusion is Corollary 7.4 of Hansen–Lima.9 \(\square\)
Suppose \(\operatorname{Vol}(K)\operatorname{Vol}(K^{\circ})=4^{d}/d!\). By Lemma 7 and Theorem 8, the unit ball \(K^{\circ}\) of \(\|\cdot\|_{K^{\circ}}\) is a linear image of a Hanner polytope. Polarity swaps \(\oplus_1\) and \(\oplus_\infty\) and commutes with linear maps up to transpose-inverse, so \(K=(K^{\circ})^{\circ}\) is one too.
Conversely, for symmetric bodies \(A\subset\mathbb{R}^{k}\), \(A'\subset\mathbb{R}^{l}\), the fiber of \(A\oplus_1A'\) over \(x\in A\) is \((1-\|x\|_A)A'\), so \[ \operatorname{Vol}(A\oplus_1A')=k\operatorname{Vol}(A)\operatorname{Vol}(A')\int_0^{1}t^{k-1}(1-t)^{l}\,dt=\frac{k!\,l!}{(k+l)!}\operatorname{Vol}(A)\operatorname{Vol}(A') \] and \[ \operatorname{Vol}(A\oplus_\infty A')=\operatorname{Vol}(A)\operatorname{Vol}(A'). \] Since polarity swaps the two sums, both satisfy \(\operatorname{Vol}(A\oplus A')\operatorname{Vol}\big((A\oplus A')^{\circ}\big)=\operatorname{Vol}(A)\operatorname{Vol}(A^{\circ})\cdot\operatorname{Vol}(A')\operatorname{Vol}(A'^{\circ})\big/\binom{k+l}{k}\). A segment gives \(4\), so induction gives \(4^{d}/d!\) for every Hanner polytope, and linear maps do not change the volume product. \(\square\)
Define an analytic function \(F\) on the unit disc \(\mathbb{D}\subset \mathbb{C}\) by \[ F(z):=\frac{8}{\pi^{2}}\sum\limits_{j=0}^{\infty} \frac{z^{2j+1}}{(2j+1)^{2}},\qquad z\in\mathbb{D}. \]
Proof of Lemma 0. The integrand \(\operatorname{arctanh}\big(\sin\frac{\pi s}{2}\big)\) is positive on \((0,1)\) and \(O\big(\log\frac{1}{1-s}\big)\) as \(s\uparrow1\), so \(\lambda\) is continuous on \([-1,1]\), \(\lambda(\pm1)=0\), and \(\lambda>0\) on \((-1,1)\). For \(0\lt |t| \lt 1\), since \(\lambda\) is even, \[ \lambda'(t)=-\frac{2}{\pi}\operatorname{arctanh}\Big(\sin\frac{\pi t}{2}\Big), \] and this also holds at \(t=0\), because the right-hand side is continuous and vanishes there. Differentiating again, \[ \lambda''(t)=-\frac{2}{\pi}\cdot\frac{\frac{\pi}{2}\cos\frac{\pi t}{2}}{1-\sin^{2}\frac{\pi t}{2}}=-\sec\frac{\pi t}{2}\lt0,\qquad -1\lt t \lt 1, \] so \(\lambda\) is strictly concave on \([-1,1]\); being also even, it attains its maximum at \(0\). It follows that \(D\) is a bounded convex domain.
The series converges absolutely on \(\overline{\mathbb D}\), so \(F\) is continuous there, \(F(0)=0\), \(F'(0)=8/\pi^{2}\gt0\), \(F(\pm1)=\pm\frac{8}{\pi^{2}}\sum_j(2j+1)^{-2}=\pm1\), and \(F(\bar z)=\overline{F(z)}\). Differentiating the series and multiplying by \(z\), we get \[ zF'(z)=\frac{8}{\pi^{2}}\operatorname{arctanh}z, \] which extends analytically across \(\partial\mathbb D\setminus\{\pm1\}\). For \(0 \lt \theta\lt\pi\), \(\operatorname{arctanh}e^{i\theta}=\frac12\log\big(i\cot\frac\theta2\big)\), so \[ \frac{d}{d\theta}F(e^{i\theta})=\frac{8i}{\pi^{2}}\operatorname{arctanh}e^{i\theta}=-\frac{2}{\pi}+\frac{4i}{\pi^{2}}\log\cot\frac{\theta}{2}. \] Hence \(\mathrm{Re}\,F(e^{i\theta})=1-2\theta/\pi=:t\), and, by \(\log\cot\frac{x}{2}=\operatorname{arctanh}(\cos x)\), \(\frac{d}{dt}\mathrm{Im}\,F=-\frac{2}{\pi}\operatorname{arctanh}\big(\sin\frac{\pi t}{2}\big)=\lambda'(t)\) with \(\mathrm{Im}\,F=0\) at \(t=\pm1\).
So \(\mathrm{Im}\,F(e^{i\theta})=\lambda(t)\), and since \(F(\bar z)=\overline{F(z)}\), \[ F(e^{\pm i\theta})=t\pm i\lambda(t),\qquad t=1-\frac{2\theta}{\pi},\quad 0\le\theta\le\pi. \] Thus \(F\) maps the upper and the lower semicircle bijectively onto the upper and the lower half of \(\partial D\), and \(F|_{\partial\mathbb D}\) is a homeomorphism onto the Jordan curve \(\partial D\), which it traverses counterclockwise.
Let \(w\in\mathbb{C}\setminus\partial D\). As \(r\uparrow1\), the curves \(\theta\mapsto F(re^{i\theta})\) converge uniformly to \(\theta\mapsto F(e^{i\theta})\), so for \(r\) close to \(1\) their winding number around \(w\) is \(1\) if \(w\in D\) and \(0\) if \(w\notin\overline D\). By the argument principle, \(F\) takes every value in \(D\) exactly once in \(\mathbb D\) and no value outside \(\overline D\). Since \(F(\mathbb D)\) is open, it does not meet \(\partial D\). Hence \(F\) is a biholomorphism of \(\mathbb D\) onto \(D\), and, with the boundary correspondence, a continuous bijection of the compact set \(\overline{\mathbb D}\) onto \(\overline D\), hence a homeomorphism.
Let \(\zeta\) be uniform on \(\partial\mathbb D\), and write \(\zeta=e^{i\varepsilon\theta}\) with \(\varepsilon\in\{-1,1\}\) and \(\theta\in(0,\pi)\), which is possible almost surely. Then \(\varepsilon\) is a uniform sign, and \(\theta\) is uniform on \((0,\pi)\) and independent of \(\varepsilon\). So \(T:=1-2\theta/\pi\) is uniform on \([-1,1]\) and independent of \(\varepsilon\), and \(F(\zeta)=T+i\varepsilon\lambda(T)\).
It remains to prove the endpoint asymptotics. Substituting \(s=1-\tau\) in the definition of \(\lambda\) and using \(\operatorname{arctanh}(\cos u)=\log\cot\frac u2\), \[ \lambda(1-\eta)=\frac2\pi\int_0^{\eta}\log\cot\frac{\pi\tau}{4}\,d\tau=\frac2\pi\,\eta\Big(\log\frac1\eta+1+\log\frac4\pi\Big)+O(\eta^{3}). \] The ratio at \(\eta=\delta c\) and \(\eta=\delta\) is \(c\,\big(1+O(1/\log\tfrac1\delta)\big)\), uniformly for \(c\in[a,b]\). \(\square\)
We prove the inclusion \(\subseteq\) in the equality \(\Sigma_{x}=\bigcup_{J\in \mathcal{J}}\bigcup_{\varepsilon \in \mathcal{E}_{J}(x)}\Delta_{J,\varepsilon}\) stated after (4); it is not used elsewhere in this note. Call \(h\in K^{\circ}\) active at \(y\in L_x\) if \(\langle h,y\rangle=\lambda(\langle h,x\rangle)\), so that (4) reads \[ \Sigma_x=\bigcup_{y\in \partial L_x}\operatorname{conv}\big(\{0\}\cup\{h\in K^{\circ}\,\colon\,h\text{ is active at }y\}\big). \]
Fix \(y\in\partial L_x\) and let \[\xi\in\operatorname{conv}\big(\{0\}\cup\{h\in K^{\circ}\colon h\text{ is active at }y\}\big).\]
Let \(h\in K^{\circ}\) be active at \(y\). Since \(K^{\circ}=\operatorname{conv}\{\pm b_l\}\), \(h\) is a convex combination, with positive weights, of some of the vectors \(\pm b_l\). The function \(\lambda(\langle \cdot,x\rangle)-\langle \cdot,y\rangle\) is concave, nonnegative on \(K^{\circ}\) by (3), and vanishes at \(h\), so it vanishes at each of these vectors, that is, they are active at \(y\). Consequently, \[ \xi=\sum_v\beta_v\,v,\qquad\beta_v\ge0,\quad\sum_v\beta_v\le1, \] where \(v\) ranges over the vectors \(\pm b_l\) active at \(y\). Among all such representations of \(\xi\), choose one with the fewest nonzero coefficients. Then the vectors \(v\) with \(\beta_v>0\) are linearly independent.
The points of \(L_x\) at which every \(v\) with \(\beta_v>0\) is active form a face of \(L_x\) containing \(y\); let \(y'\) be a vertex of this face. Then \(y'\) is a vertex of \(L_x\), so the vectors active at \(y'\) span \(\mathbb{R}^d\). Moreover, \(b_l\) and \(-b_l\) are never both active at \(y'\), since this would give \(\lambda(\langle b_l,x\rangle)=-\lambda(\langle b_l,x\rangle)\), whereas \(\lambda(\langle b_l,x\rangle)>0\) because \(|\langle b_l,x\rangle|\lt 1\). Hence the linearly independent vectors \(\{v\colon\beta_v>0\}\) can be completed, by vectors active at \(y'\), to a basis of \(\mathbb{R}^d\) of the form \(\{\varepsilon_jb_j\}_{j\in J}\) with \(J\in\mathcal{J}\) and \(\varepsilon\in\{-1,1\}^J\). Since every \(\varepsilon_jb_j\) is active at \(y'\) and \(\lambda\) is even, \[ \langle b_j,y'\rangle=\varepsilon_j\lambda(\langle b_j,x\rangle),\qquad j\in J, \] that is, \(y'=y_{J,\varepsilon}(x)\). As \(y'\in L_x\), we get \(\varepsilon\in\mathcal{E}_J(x)\), and \(\xi=\sum_v\beta_vv\in\Delta_{J,\varepsilon}\). \(\square\)
The tables below match the notation and numbered statements of this note with those of the original preprint (version of September 22, 2026). Equation numbers of the preprint are written as [8], [26], etc., to distinguish them from equation numbers (1)–(13) of this note.
The main difference in conventions is that the preprint's lens is the domain \(D\) of this note rotated by a quarter turn: the preprint uses \(F_{\mathrm{pre}}(z)=\frac{8}{\pi^2}\sum_j\frac{(-1)^jz^{2j+1}}{(2j+1)^2}\), so that \(F(z)=-iF_{\mathrm{pre}}(iz)\) and \(D=-iD_{\mathrm{pre}}\). Accordingly, the preprint's lens is \(\{v+it\colon|t|\le1,\ |v|\le\lambda(t)\}\), and a point of \(\mathbb{C}^d\) is written there as \(Y+iX\) rather than \(x+iy\).
Notation
| This note | Preprint | Remarks |
|---|---|---|
| \(d\) | \(n\) or \(d\) | The preprint writes \(n\) for the dimension of a general body \(K\) and \(d\) for that of the polytope \(A\); in Sections 7–8, \(d=n+1\). |
| \(\operatorname{Vol}(K)\) | \(|K|\) | The volume product \(|K|\,|K^\circ|\) is denoted \(P(K)\). |
| \(\langle b_j,x\rangle\) | \(b_jX\) | The \(b_j\) are row vectors acting on \(X\). |
| \(K=\{|\langle b_j,x\rangle|\le1\}\) | \(A\) | [21]; the polar is \(C=A^\circ\). |
| \(x,\ y\) | \(X,\ Y\) | \(x+iy\) corresponds to \(Y+iX\). |
| \(F,\ D\) | \(F,\ D\) rotated | [8], [9]; see above. \(\lambda\) is the same function. |
| \(F^{-1}\) | \(g\) | |
| \(\Omega\) | \(\Omega\) rotated | Defined in the proof of Lemma 5.1. |
| \(\phi=F^{-1}\circ B\) | \((h_1,\dots,h_m)\) | \(h_j(z)=g(b_jz)\). |
| \(J\in\mathcal{J}\) | \(I\in\mathcal{B}\) | Sets of \(d\) independent rows. |
| \(B_J,\ |\det B_J|\) | \(B_I,\ D_I\) | |
| \(\varepsilon\) | \(\epsilon\) | Sign vectors. |
| \(L_x\) | \(L_X\) | [25]. |
| \(y_{J,\varepsilon}(x)\) | \(Y_{I,\epsilon}(X)\) | |
| \(\varepsilon\in\mathcal{E}_J(x)\) | \((I,\epsilon)\) feasible at \(X\) | |
| \(\Delta_{J,\varepsilon}\) | \(T_{I,\epsilon}\) | [26]. |
| \(\Sigma_x\) | \(\Sigma_X\) | The preprint defines \(\Sigma_X\) as the union of feasible simplices [26]; the intrinsic definition (4) is not in the preprint. |
| \(\mu_J(A_J)\) | \(P_I\) | [22]; \(\sum_J\mu_J(A_J)\) is \(S\). |
| \(f_k=p_k\circ\phi\) | \(f_k=(h_1^k,\dots,h_m^k)\) | \(|f_k|^2=\tau_k\), and \(\Omega_k=\{\tau_k\lt1\}\). |
| \(\int_{M_k}(\pi_J|_{M_k})^*p_k^*dV_J\) | \(M_{I,k}\) | [23]. The submanifolds \(M_k\) and the form \(\sigma_k\) are not used in the preprint. |
| \(m,\ U,\ U_s\) | \(N,\ \mathcal U,\ E_{s^2}\) | In Lemma 4.1, \(E_s=\{|f|^2\lt s\}\). |
| \(\|\cdot\|_A\) | \(p_A\) | Gauge; \(h_A\) is the support function. |
| \(K^\circ\) | \(Q\) | In Sections 6.1, 7 and 8. |
| \(\widehat K,\ \widehat K^\circ\) | \(B,\ C\) | [32]. |
| \(\hat x=(\delta x,1-\delta)\) | \(X_\delta=(\delta u,1-\delta)\) | The point \(x\in\operatorname{int}K\) of Lemma 7 is \(u\) in the preprint. |
| \(v,\ u\) | \(x,\ y\) | The triple \(0,v,u\) in Lemma 7 is \(0,x,y\) in Proposition 8.1. |
| \((h_i,s_i),\ a_i\) | \((b_i,\sigma_i),\ \alpha_i\) | In the preprint \(\sigma_i\in\{-1,1\}\), while here \(s_i\in[-1,1]\). |
| \(W\) | \(D_0\) | [43]. Not the same set: the median is sought as \(w\in W\) here, and as \(m=\frac12(x+y-e)\) with \(e\in D_0\) and \(x+y-e\in(s-p)Q\) there. |
| median | metric median | |
| linear image of a Hanner polytope | linear Hanner body | |
| 3.2 intersection property | three-ball intersection property | The preprint uses both names. |
Statements
| This note | Preprint | Remarks |
|---|---|---|
| Theorem, (1) | Theorem 1.1, [1] | The equality case is Proposition 8.4 with Lemma 2.2. |
| Lemma 0 | Lemmas 3.1, 3.2 | Up to the rotation above; the endpoint asymptotics are Lemma 3.2, [14]. The proofs of Lemma 3.1 differ: here via \(zF'(z)=\frac{8}{\pi^2}\operatorname{arctanh}z\) and the argument principle, there via convexity of the curves \(F(re^{i\theta})\). |
| (3), Appendix: \(\Sigma_x\) as a union of simplices | — | The preprint takes the union of feasible simplices as the definition of \(\Sigma_X\). |
| Lemma 1 | Proof of Proposition 6.1, Lemma 5.2 | The preprint shows that extra tight constraints have probability zero on the complex side (Lemma 5.2); here this is done directly in \(x\), using \(\lambda''\neq0\). |
| Formula for \(\operatorname{Vol}(\Sigma_x)\) | [30] | |
| (7), (8) and the reduction to \(\sum_J\mu_J(A_J)\ge1\) | [27], [29] | Proposition 6.1: \(S=\frac{d!}{4^d}\int_A|\Sigma_X|\,dX\). |
| Lemma 2 | Proofs of Lemmas 4.1 and 5.1 | Cauchy–Binet, and vanishing of the minors of dependent rows. |
| Lemma 3 | [24] and the end of the proof of Lemma 5.1 | Radial substitution and Fatou's lemma. |
| Lemma 4 | Lemma 4.1, [17] | The preprint assumes a nondegenerate homogeneous leading term of degree \(k\), where this note assumes \(|f|\le C|z|^k\). The proofs differ: here by gluing with a regularized maximum (as in Demailly), there by computing the flux through a small sphere. |
| Verification of the hypotheses of Lemma 4 | Proof of Lemma 5.1 | Lemma 5.1 holds for any biholomorphism \(F\) of \(\mathbb D\) onto a bounded domain with \(F(0)=0\). |
| Lemma 6, inequality | Corollary 6.2 | The preprint approximates \(K^\circ\) from inside by polytopes and applies Proposition 6.1. |
| Lemma 5, Lemma 6, equality | Proposition 7.1, [33]–[35] | The preprint works only with the lifted body \(B\), and obtains a common witness \(Y\) as a limit of feasible pairs [39]–[41], using the vanishing deficit [38]. |
| Definition (median) | Beginning of Section 8 | |
| Lemma 7 | Proposition 8.1 | (10) corresponds to [45], and (11)–(13) to [44] and the endpoint comparison. |
| Theorem 8 | Lemma 8.2, Theorem 8.3, Proposition 8.4 | Medians imply the 3.2 intersection property (Lemma 8.2), then Hansen–Lima (Theorem 8.3). |
| Volume product of Hanner polytopes | Lemmas 2.1, 2.2 | |
| — | Corollaries 1.2, 1.3 | Functional and entropy–transport consequences; not covered in this note. |
If you use material from this note, please cite it as: Rotem Assouline, Notes on OpenAI's proof of the symmetric Mahler conjecture, https://rotemassouline.github.io/symmetric-mahler/.
The text of this note is licensed under CC BY 4.0, and the code for the figures under the MIT license; see the LICENSE file. The original preprint is distributed under the Apache 2.0 license.