% The Mahler Conjecture for General Convex Bodies
% A self-contained, unified proof. Every statement needed for the main
% theorem is proved; nothing below relies on external files.
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\title{The Mahler Conjecture for General Convex Bodies}
\author{}
\date{}
\begin{document}
\maketitle
\begin{abstract}
We prove the Mahler conjecture for general convex bodies, in every
dimension and with no symmetry or regularity hypothesis. For every
convex body $K\subset\R^n$ and its Santal\'o point $s(K)$,
\[
|K|\,|(K-s(K))^\circ|\ \ge\ \frac{(n+1)^{n+1}}{(n!)^2},
\]
with equality if and only if $K$ is a simplex. Two consequences are
given: a sharp functional (log-concave) Mahler inequality, and an
entropy--transport inequality.
\end{abstract}
\tableofcontents
% =====================================================================
\section{Introduction and overview}\label{sec:intro}
A \emph{convex body} in $\R^n$ is a compact convex set with nonempty
interior. Write $|A|$ for $n$-dimensional Lebesgue volume and
$\langle\cdot,\cdot\rangle$ for the Euclidean inner product. A simplex
is the convex hull of $n+1$ affinely independent points.
\paragraph{Polarity and the Santal\'o point.}
For $z\in\operatorname{int}K$ let
\[
(K-z)^\circ=\{y\in\R^n:\ \langle y,x-z\rangle\le 1\ \text{for all }x\in K\}
\]
be the polar body of $K$ re-centered at $z$. The function
$z\mapsto |(K-z)^\circ|$ is strictly convex on $\operatorname{int}K$ and
tends to $+\infty$ at the boundary; hence it has a unique interior
minimizer, the \emph{Santal\'o point} $s(K)$. (We prove this in
Section~\ref{sec:cones}.) The \emph{volume product} is
\[
P(K)=|K|\,|(K-s(K))^\circ|
=\inf_{z\in\operatorname{int}K}|K|\,|(K-z)^\circ|.
\]
It is invariant under invertible affine maps.
\begin{theorem}[Mahler conjecture]\label{thm:mahler}
For every $n\ge1$ and every convex body $K\subset\R^n$,
\[
P(K)\ \ge\ \frac{(n+1)^{n+1}}{(n!)^2}.
\]
Equality holds if and only if $K$ is a simplex.
\end{theorem}
Two remarks. First, because $P(K)$ is the minimum over translation
points, Theorem~\ref{thm:mahler} also gives the displayed lower bound
for $|K|\,|(K-z)^\circ|$ at \emph{every} interior point $z$.
Second, the \emph{symmetric} Mahler conjecture asks for the stronger
constant $4^n/n!$ on the restricted class of centrally symmetric bodies.
That sharper statement is different and is proved separately in the
companion paper~\cite{OpenAISymmetricMahler2026}; the argument here does
not use symmetry.
\paragraph{Context.}
The problem originates with Mahler, who settled the planar case and its
triangle equality among polygons \cite{Mahler1938}, and who formulated
the higher-dimensional symmetric version \cite{Mahler1939}. Meyer gave
a new proof of the planar general inequality with a complete equality
characterization (triangles are the only minimizers)
\cite{Meyer1991}. The affine center used here belongs to Santal\'o's
theory \cite{Santalo1949}. Asymptotic lower bounds (inverse
Santal\'o / Bourgain--Milman type) go back to
\cite{BourgainMilman1987, Kuperberg2008, Nazarov2012,
MastrantonisRubinstein2024}; these give exponential constants but not
the sharp simplex value. Sharp results for special classes include the
polytope case with at most $n+3$ vertices via shadow systems
\cite{MeyerReisner2006}, strict local minimality at every simplex
\cite{KimReisner2011}, the three-dimensional symmetric case
\cite{IriyehShibata2020}, and the general three-dimensional case
\cite{ChenLiXiXu2026}. The present argument works for arbitrary bodies
in all dimensions and also pins down the global equality case.
\paragraph{How to read the proof.}
The proof has a single spine:
\begin{enumerate}[label=\arabic*.]
\item \emph{Cone reduction} (Section~\ref{sec:cones}): translate $K$
and lift it to a cone $C\subset\R^{n+1}$. The volume product equals,
up to an explicit constant, a product of two Laplace integrals on the
cone and its positive dual $D$. It suffices to show that product is
$\ge1$.
\item \emph{Gaussian projections} (Section~\ref{sec:proj}): build two
maps of a Gaussian into the cones using orthogonal projections, with
carefully chosen biases and covariance. Their Jacobians turn the
Laplace product into an entropy-type lower bound.
\item \emph{Entropy inequality} (Section~\ref{sec:entropy}): change of
variables plus Jensen reduces the cone inequality to $\mathcal E\le0$
for an explicit error functional $\mathcal E$.
\item \emph{Matrix comparison} (Sections~\ref{sec:quad}--\ref{sec:layers}):
compare the whole field of projection derivatives to spectral
thresholds of a linear Gaussian matrix $L=\sum_i G_iM_i$. Hermite
calculus separates the linear part; the leftover ``layer'' terms are
bounded below by zero.
\item \emph{Linear field and equality}
(Section~\ref{sec:linear}): express the remaining terms as averages
over eigenvalue pairs of $L$, absorb them with a strict one-variable
inequality, and show equality forces the projection to be
coordinatewise, hence $K$ a simplex.
\end{enumerate}
The dimension-independent scalar inequalities used throughout are
collected once (Proposition~\ref{prop:scalar-input}) and proved in the
two appendices, Sections~\ref{sec:scalar} and~\ref{sec:segments}.
\paragraph{Functional and transport consequences.}
The geometric theorem implies a sharp functional (log-concave) Mahler
inequality for arbitrary convex potentials with no centering
assumption:
\begin{corollary}[Functional Mahler]\label{cor:functional-mahler}
For every $n\ge1$ and every proper lower-semicontinuous convex
$\varphi:\R^n\to\R\cup\{+\infty\}$ with
$0<\int e^{-\varphi}<\infty$,
\[
\Big(\int_{\R^n}e^{-\varphi(x)}\,dx\Big)
\Big(\int_{\R^n}e^{-\varphi^*(y)}\,dy\Big)\ \ge\ e^n,
\]
where $\varphi^*(y)=\sup_x\{\langle x,y\rangle-\varphi(x)\}$.
The second integral may be infinite, in which case the inequality is
trivial.
\end{corollary}
This is sharp ($e^n$ is attained). Via Gozlan's equivalence it yields
an entropy--transport inequality (Corollary~\ref{cor:entropy-transport}).
Both are proved in Section~\ref{sec:consequences} and are \emph{not}
inputs to the geometric argument.
% =====================================================================
\section{Polarity, the cone reduction, and the scalar input}\label{sec:cones}
\subsection{The minimizing center}
The support function is $h_K(u)=\sup_{x\in K}\langle u,x\rangle$. For
$z\in\operatorname{int}K$ and $n\ge2$, polar coordinates give
\[
|(K-z)^\circ|=\frac1n\int_{S^{n-1}}\big(h_K(u)-\langle z,u\rangle\big)^{-n}\,d\sigma(u),
\]
where $\sigma$ is surface measure on the unit sphere. As a function of
$z$ this is strictly convex: $t\mapsto t^{-n}$ is strictly convex for
$t>0$, and two distinct centers have inner products differing on a set
of positive surface measure. It blows up at $\partial K$: at a limiting
boundary point with supporting normal $u_0$, the quantity
$\varepsilon=h_K(u_0)-\langle z,u_0\rangle\to0$; the integrands share a
Lipschitz constant in $z\in K$, are $O(\varepsilon)$ on a spherical cap
of radius $\varepsilon$, and that cap has measure $\gtrsim
\varepsilon^{n-1}$, so its contribution is $\gtrsim\varepsilon^{-1}$.
Hence a unique interior minimizer exists. In dimension one, with
$K=[a,b]$, the same conclusion follows from
$(z-a)^{-1}+(b-z)^{-1}$.
For an invertible affine map $x\mapsto Bx+a$ we have
$(BK+a-(Bz+a))^\circ=B^{-\mathsf T}(K-z)^\circ$, the two volume factors
change by reciprocal determinants, and $s(BK+a)=Bs(K)+a$. Thus $P$ is
affine-invariant and the Santal\'o point is affine-equivariant.
\subsection{The cone reduction}
Put $m=n+1$. Fix $z_0\in\operatorname{int}K$ and define the cone
\[
C=\{(y,t):\ t\ge0,\ y\in t(K-z_0)\},\qquad
D=C^*=\{w:\ \langle w,x\rangle\ge0\ \text{for all }x\in C\}.
\]
Here the star is \emph{positive} duality. Both cones are closed, convex,
solid (nonempty interior) and pointed (contain no line). For interior
vectors $V\in D$ and $U\in C$ write the Laplace integrals
\[
\chi_C(V)=\int_C e^{-\langle V,x\rangle}\,dx,\qquad
\chi_D(U)=\int_D e^{-\langle U,x\rangle}\,dx
\]
in $m$-dimensional Lebesgue measure.
\begin{lemma}[Cone reduction]\label{lem:cone-reduction}
For $U=(0,m)$ and $V=(0,1)$,
\[
\chi_C(V)\,\chi_D(U)=\frac{(n!)^2}{m^m}\,|K|\,|(K-z_0)^\circ|.
\]
For \emph{any} closed convex solid pointed cone $C$ with dual $D=C^*$
and interior $U\in C$, $V\in D$, both Laplace integrals are finite, and
their product and the inner product $\langle U,V\rangle$ are preserved
under $(C,D,U,V)\mapsto(BC,B^{-\mathsf T}D,BU,B^{-\mathsf T}V)$ for
every invertible linear map $B$.
\end{lemma}
\begin{proof}
A point $(w,t)$ lies in $D$ iff $t\ge\sup_{x\in K-z_0}\langle -w,x\rangle$.
Since $0$ is interior to $K-z_0$ this forces $t\ge0$, and the slice at
height $t$ is $-t(K-z_0)^\circ$. Integrating slices,
\[
\chi_C((0,1))=n!\,|K|,\qquad
\chi_D((0,m))=n!\,m^{-m}\,|(K-z_0)^\circ|,
\]
giving the product formula. Interiority of $V$ gives
$\langle V,x\rangle\ge c|x|$ on $C$, and interiority of $U$ the analogue
on $D$, so both integrals are finite. Under the linear rescaling the
two change-of-variables factors are $|\det B|$ and $|\det B|^{-1}$,
while inner products in the exponentials are unchanged.
\end{proof}
By Lemma~\ref{lem:cone-reduction} it suffices to prove
\begin{equation}\label{eq:goal}
\chi_C(V)\,\chi_D(U)\ \ge\ 1
\end{equation}
whenever $C\subset\R^m$ is closed, convex, solid, pointed, $D=C^*$,
$U\in\operatorname{int}C$, $V\in\operatorname{int}D$, and
$\langle U,V\rangle=m$. These are the \emph{standing cone
assumptions}; the constant $m$ in the inner product is what makes the
Gaussian normalization below come out exactly. We will additionally
see that equality forces $C$ to be a linear image of an orthant, which
by Section~\ref{sec:linear} is equivalent to $K$ being a simplex.
\subsection{Scalar profiles and the analytic input}
The argument uses a fixed family of one-variable functions, independent
of $C$ and of the dimension. Let
$\phi(x)=(2\pi)^{-1/2}e^{-x^2/2}$, $p(x)=\int_{-\infty}^x\phi$, and
$\gamma f=\int f\,\phi$. Put $\mathcal N=x\partial_x-\partial_x^2$.
Temporarily write
\[
X=\frac{p}{\phi},\qquad Y=\frac{1-p}{\phi},\qquad
f_0=-(1-p)-\log p,\qquad j_0=-p-\log(1-p),
\]
and define the three even profiles
\begin{equation}\label{eq:profiles}
d=(1+xX)^2 f_0+(1-xY)^2 j_0,\qquad
g=\tfrac12\big(1-X^2f_0-Y^2j_0\big),\qquad v=\log(XY).
\end{equation}
The functions $d,g,v$ are smooth, even, and $v'=-\log\frac{p}{1-p}$ up
to an additive constant.
\begin{lemma}[Profile identities]\label{lem:profile-ids}
With $d,g,v$ as above,
\begin{equation}\label{eq:profile-ids}
(\mathcal N+2)g=d,\qquad d'=2xd-v'-2g',\qquad
\mathcal N d-v''=-(\mathcal N+2)(d+g'').
\end{equation}
\end{lemma}
\begin{proof}
One checks $X'=1+xX$, $Y'=xY-1$, $f_0'=-\phi(1-p)/p$, $j_0'=\phi p/(1-p)$.
Direct substitution gives
$X^2f_0'+Y^2j_0'=0$, $XX'f_0'+YY'j_0'=-1$,
$(X')^2f_0'+(Y')^2j_0'=-v'$. Also $X''=xX'+X$ and $Y''=xY'+Y$.
Differentiating $g'=-(XX'f_0+YY'j_0)$ once more gives
$(\mathcal N+2)g=d$. Differentiating $d=(X')^2f_0+(Y')^2j_0$ gives the
second identity, and differentiating the first identity twice and
combining with the second gives the third.
\end{proof}
\paragraph{Fixed parameters.}
All finite decimals below denote \emph{exact} rational numbers. Set
\begin{equation}\label{eq:params}
\begin{gathered}
b=.602,\quad c=.365,\quad k=\sqrt{b-c},\quad r=.22,\quad w=4.6,\quad s=3.6,\\
\eta=.022,\quad \kappa=1.16,\quad a_R=7.5,\quad \lambda=.65,\\
t_-=-.022,\quad t_+=.064,\quad t_c=.021,\quad t_r=.043,\quad m_*=.352.
\end{gathered}
\end{equation}
The vector profile and the auxiliary scalar profile are
\[
\pi(x)=r(\cos(w\arsinh(x/s)),\ \sin(w\arsinh(x/s))),\quad
q(x)=k-\sqrt{b-r^2-d(x)},\quad u=(\pi,q).
\]
Appendices~\ref{sec:scalar}--\ref{sec:segments} establish, for these
exact constants and profiles, the consequences stated next. They are
\emph{purely one-variable} facts; the rest of the paper is finite
linear algebra and Gaussian analysis.
\begin{proposition}[Scalar input]\label{prop:scalar-input}
The following hold.
\begin{enumerate}[label=(\roman*)]
\item (Bounds) $b-r^2-d>0$, $|v'(x)-x|\le.319$, and
$d\to1/2$ as $|x|\to\infty$; moreover
$\mathsf C:=-d+(a_R-1)g''\le -.341$ and $S_q:=(2+\mathcal N)q>0$,
with $d,g,v,\mathsf K,\mathsf C,\pi,q$ smooth with polynomially
bounded derivatives, where $\mathsf K=d-2g$.
\item (Covariance eigenvalue bounds) for $T\in[t_-I,t_+I]$,
\[
\lambda t_-^2+t_-\mathsf C+\mathsf K>0,\qquad
\lambda t_+^2+t_+\mathsf C+\mathsf K<0,
\]
and $2(-.37)\mathsf C-(.37)^2-4\lambda\mathsf K>m_*^2$.
\item (Pointwise layer) with $b_m=\frac{(1+t_+)(b-3\eta)}{3(3(b+\eta)-4c)}$,
$e_0=.256$, $\alpha=2.26$, $h_s=.5$, and
\[
H(z,x)=4c+4kq(x)+2(k-q(x))S_q(z)-2S_\pi(z)\cdot\pi(x),
\quad H_0=H+v''(x)-2\kappa,
\]
that
\begin{align}
H+t_cH_0-\tfrac12t_r(h_s+H_0^2/h_s)
&>2(b+\eta+(b+\eta-\kappa)t_-)+2b_mS_q(z)S_q(x)+2e_0(2t_r)^2+H_0^2/\alpha,\label{eq:pt}\\
4c+2k(S_q(z)+S_q(x))-S_q(z)S_q(x)-S_\pi(z)\cdot S_\pi(x)&>0. \label{eq:pt2}
\end{align}
\item (Segment) for endpoints $x_-<x_+$ (subscript $-\sigma$ the
opposite endpoint, bar the uniform average on the segment),
\[
e_K+\Gamma+\tfrac1{8\lambda}\textstyle\sum_\sigma\big(\overline{\mathsf C}-\mathsf C_{-\sigma}-\overline{|u-u_\sigma|^2}\big)^2
+\tfrac{t_+}{2}\textstyle\sum_\sigma|\bar u-u_\sigma|^2<\overline{|u|^2}-|\bar u|^2,
\]
where $e_K=\overline{\mathsf K}-\tfrac12(\mathsf K_++\mathsf K_-)$ and
\[
\Gamma=\frac{.80}{m_*^2}\big(1.25(\mathsf K_+-\mathsf K_-+t_c(\mathsf C_+-\mathsf C_-))^2+5t_r^2(\mathsf C_+-\mathsf C_-)^2\big),
\]
with equality (both sides zero) iff the endpoints coincide.
\end{enumerate}
\end{proposition}
The bound $d\to1/2$ makes the radicand in $q$ tend to
$b-r^2-\tfrac12=.0536>0$, so $q$ is globally smooth. That
$S_q>0$ and $4k-2S_q>0$, $2(k-q)>0$ ensure the monotonicities used
later; these are consequences of the appendix bounds
$S_q\le.514<2k$ and $q<k$. We take them as part of the scalar input.
The rest of Sections~\ref{sec:proj}--\ref{sec:linear} use only
Proposition~\ref{prop:scalar-input} together with elementary Gaussian and
matrix facts. The appendices prove the proposition once and for all.
% =====================================================================
\section{Biased Gaussian projections and simultaneous normalization}
\label{sec:proj}
We now attach to the cone data a family of Gaussian projections whose
means move along a fixed curve in $C$. Their derivative field supplies
both the Jacobians in the entropy estimate and the matrix fields in the
later quadratic analysis. Throughout, fix cone data satisfying the
standing assumptions and a symmetric $T$ with
\[
t_-I\le T\le t_+I,\qquad \Sigma=I+T,\qquad Z=\Sigma^{1/2}G,
\]
where $G$ is standard Gaussian in $\R^m$. Thus every eigenvalue of
$\Sigma$ lies in $[.978,1.064]$. Expectation is in $G$ unless stated.
\paragraph{Matrix trace conventions.}
For matrix fields, set
\[
\tau B=\tfrac1m\E\operatorname{tr}B,\quad
\tau_\Lambda B=\tau(\Lambda B),\quad
\langle B,E\rangle_\Lambda=\tau_\Lambda(B\circ E),\quad
B\circ E=\tfrac12(BE+EB),
\]
for symmetric $B,E$ and deterministic symmetric weight $\Lambda$;
subscript $I$ is omitted. Transposition and cyclicity give
$\operatorname{tr}(\Lambda BE)=\operatorname{tr}(\Lambda EB)$ without
assuming commutation, so $\langle\cdot,\cdot\rangle_\Sigma$ is a
positive inner product (as $\Sigma>0$). We write
$\|B\|_2^2=\tau(B^{\mathsf T}B)$ and $[B,E]=BE-EB$.
\subsection{The biased projection field}
Let $\Pi_C,\Pi_D$ be Euclidean nearest-point projections onto $C,D$, and
put $a(z)=\phi(z)+zp(z)$, so $a>0$, $a'=p$.
\begin{lemma}[Biased projections]\label{lem:biased}
For each $z\in\R$ there is a unique vector $\xi_z$ such that
\[
X_z=\Pi_C(Z+\xi_z),\quad Y_z=\Pi_D(-Z-\xi_z),\quad
\E X_z=a(z)U. \tag{5}\label{eq:mean}
\]
The map $z\mapsto\xi_z$ is smooth; the a.e. derivative
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