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\title{A sharp entropy bound and the simplex inequality\\ for isotropic constants}
\author{}
\date{}

\begin{document}
\maketitle

\begin{abstract}
We prove the strong isotropic constant conjecture: in every dimension
$n$, among all convex bodies in $\R^n$ a simplex has the largest
isotropic constant, and it is the only maximizer. We also prove the
sharper, more general \emph{sharp entropy bound}
$h(f)\ge m+\tfrac12\log\det\Cov(f)$ for every log-concave density $f$
on $\R^m$, with equality if and only if $f$ is an invertible affine
image of a product of $m$ independent one-sided exponentials. The body
inequality then follows by a geometric reduction to one dimension higher.
\end{abstract}

\section{Introduction}

\subsection{The question}

A set $K\subset\R^n$ is a \emph{convex body} if it is closed, bounded,
convex, and has nonempty interior. The simplest convex bodies are the
\emph{simplices}: the convex hull of $n+1$ points not lying in a
hyperplane (a segment when $n=1$, a triangle when $n=2$, a tetrahedron
when $n=3$). Write $|K|$ for the $n$-dimensional volume of $K$ and
$b_K=|K|^{-1}\int_Kx\,\dd x$ for its centroid. The covariance matrix of
the uniform probability measure on $K$ is
\[
 \Sigma_K=\frac1{|K|}\int_K(x-b_K)(x-b_K)^{\mathsf T}\,\dd x,
\]
and the \emph{isotropic constant} of $K$ is
\begin{equation}\label{intro:isotropic-constant}
 L_K=\left(\frac{\det\Sigma_K}{|K|^2}\right)^{1/(2n)}.
\end{equation}
The exponent $1/(2n)$ is chosen so that $L_K$ is unchanged if $K$ is
replaced by an invertible affine image $A(K)$: the determinant of
$\Sigma_{A(K)}$ grows by $(\det A)^2$, the squared volume $|A(K)|^2$
grows by $(\det A)^2$, and the ratio is invariant. So $L_K$ measures
the shape of $K$, not its size or orientation.

\begin{definition}[The simplex constant]
Let $\Delta_n$ be a regular simplex in $\R^n$. A direct computation gives
\[
 L_{\Delta_n}=\frac{(n!)^{1/n}}{(n+1)^{(n+1)/(2n)}\sqrt{n+2}}.
\]
By affine invariance every simplex has this same value; call it $L_n^\star$.
\end{definition}

Bourgain, studying maximal functions associated with convex bodies,
asked whether $L_K$ is bounded independently of the dimension
\cite{Bourgain1986}. This is the \emph{slicing problem}. Klartag proved
$L_K\le Cn^{1/4}$ \cite{Klartag2006}; after a long line using Eldan's
stochastic localization \cite{Eldan2013,Chen2021,KlartagLehec2022,Klartag2023},
Klartag and Lehec proved a dimension-independent bound using Guan's
covariance estimate \cite{Guan2024,KlartagLehec2025}. That settles the
qualitative question but not the sharp value.

The \emph{strong isotropic constant conjecture} asks for the exact
extremal value and the extremizer: the simplex should maximize $L_K$
and be the unique maximizer. Our first theorem proves it.

\begin{theorem}[The simplex inequality]\label{intro:simplex}
For every integer $n\ge1$ and every convex body $K\subset\R^n$,
\begin{equation}\label{intro:simplex-bound}
 L_K\le\frac{(n!)^{1/n}}{(n+1)^{(n+1)/(2n)}\sqrt{n+2}}=L_n^\star,
\end{equation}
with equality if and only if $K$ is a simplex.
\end{theorem}

In the plane the bound was known, with uniqueness of the triangle
\cite{CampiColesantiGronchi1999,Saroglou2010}; partial restrictions had
been obtained for special bodies \cite{Rademacher2016,Kipp2026}.
Theorem~\ref{intro:simplex} settles all dimensions at once.

For a convex body $K$ with $0$ in its interior, the polar is
$K^\circ=\{y\in\R^n:\langle x,y\rangle\le1\text{ for all }x\in K\}$.
Klartag showed that the sharp simplex bound implies the nonsymmetric
\emph{Mahler inequality} \cite[Corollary~1.2]{Klartag2018}
\begin{equation}\label{intro:mahler}
 |K|\,|K^\circ|\ge\frac{(n+1)^{n+1}}{(n!)^2}.
\end{equation}
So Theorem~\ref{intro:simplex} also proves Mahler's inequality in every
dimension.

\subsection{The bridge: entropy of log-concave densities}

Optimizing $L_K$ directly over all convex bodies is hopeless. The key
idea (Fradelizi and Mar\'in Sola \cite{FradeliziMarinSola2026}) is to
pass to a larger, more flexible world, that of probability densities,
and to prove a sharp inequality there.

Let $f:\R^m\to[0,\infty)$ be a probability density. Its
\emph{differential entropy} and \emph{covariance} are
\[
 h(f)=-\int_{\R^m}f(x)\log f(x)\,\dd x,\qquad
 \Cov(f)=\int_{\R^m}(x-\bar x)(x-\bar x)^{\mathsf T}f(x)\,\dd x,
\]
with $\bar x=\int_{\R^m}xf(x)\,\dd x$. All logarithms are natural and
$0\log0=0$. A nonnegative function is \emph{log-concave} if it has the
form $f=e^{-V}$ with $V$ convex; up to a null set every such density has
a proper lower semicontinuous convex potential $V:\R^m\to(-\infty,+\infty]$,
where $V=+\infty$ encodes $f=0$.

\begin{theorem}[The sharp entropy bound]\label{intro:entropy}
Let $m\ge1$ and let $f$ be a log-concave probability density on $\R^m$.
Then
\begin{equation}\label{intro:entropy-bound}
 h(f)\ge m+\frac12\log\det\Cov(f).
\end{equation}
Equality holds if and only if, almost everywhere, $f$ is the density of
$M(E_1,\dots,E_m)^{\mathsf T}+b$, where $M$ is an invertible real
$m\times m$ matrix, $b\in\R^m$, and $E_1,\dots,E_m$ are independent
random variables with the one-sided exponential density
$e^{-u}\mathbf1_{\{u\ge0\}}$ (mean $1$, variance $1$).
\end{theorem}

The bound is sharp. The product of $m$ independent mean-one
exponentials has entropy $m$, covariance $I_m$, and $\log\det I_m=0$, so
equality holds. Under an invertible affine map $x\mapsto Ax+b$, both
$h(f)$ and $\tfrac12\log\det\Cov(f)$ increase by $\log|\det A|$, so the
gap
$\Dgap(f)=h(f)-m-\tfrac12\log\det\Cov(f)$ is affine invariant. Thus the
theorem says: among log-concave densities with covariance $I_m$ the
entropy is minimized by the independent-exponentials product, and only
by it.

The reduction to bodies is one clean construction
(Section~\ref{sec:geometry}): place $K$ at height one over the origin,
form the cone it generates in $\R^{n+1}$, and put the exponential
density on the cone. Applied in dimension $n+1$, the entropy bound
becomes exactly the simplex bound, and entropy equality becomes the
statement that the cone is an affine image of an orthant, which holds
if and only if $K$ is a simplex. This is the equivalence of Fradelizi
and Mar\'in Sola \cite[Proposition~5.2]{FradeliziMarinSola2026}, and it
is where the constant $L_n^\star$ comes from. In dimension one the
entropy bound is due to Melbourne, Nayar, and Roberto
\cite{MelbourneNayarRoberto2026}. Our Theorem~\ref{intro:entropy}
proves the inequality and classifies equality in every dimension.

\subsection{How the proof works, in words}

The route has five big steps. The single most important idea is that a
\emph{strict remainder} is kept the whole way, and it vanishes exactly
on the extremizers.

\paragraph{Step 1: transport from a Gaussian.} Push a standard Gaussian
forward to $f$ by the optimal (Brenier) map and let $J$ be its
Jacobian, a positive definite symmetric matrix at each point. Because
$J$ has positive eigenvalues, any continuous scalar function acts on
$J$ by spectral calculus. Define $A=q^{-1}(J)$, where
\[
 q(a)=\frac{\varphi(a)}{\Phi(-a)}
\]
is the derivative of $t(a)=-\log\Phi(-a)$, with $\varphi,\Phi$ the
standard Gaussian density and distribution function. The map $q$ is a
smooth increasing bijection $\R\to(0,\infty)$. An affine change of the
target makes $\E A=0$, the key normalization (not the usual covariance
normalization).

\paragraph{Step 2: first chaos versus the rest.} Gaussian measure has an
orthogonal decomposition into homogeneous polynomials of fixed degree
(chaoses). Since $\E A=0$, the degree-zero part of $A$ is gone, and
\[
 A=\sum_{r=1}^m x_rX_r+Y,\qquad X_r=\E[x_rA],\qquad B=\sum_{r=1}^mX_r^2,
\]
where $Y$ has only chaoses of degree $\ge2$. In the extremal
(independent-exponentials) case $B=I$. The proof makes both
$B\neq I$ and $Y\neq0$ cost a positive amount.

\paragraph{Step 3: express the gap.} A covariance identity involving
the Ornstein--Uhlenbeck operator rewrites $\Dgap$ in terms of $A$ and
its derivatives. Convexity of the target potential gives an energy
bound that keeps a nonnegative term $\mathcal C$ penalizing unequal
eigenvalues of $A$. Matrix functions of $A$ do not commute, so
$\mathcal C$ cannot be discarded.

\paragraph{Step 4: absorb, with slack.} Two scalar one-variable
inequalities (one for absorption, one for pairs of eigenvalues) feed two
matrix lemmas (a commutator square and a block Cauchy--Schwarz) to
reduce $\Dgap$ to a quadratic form nonpositive on every chaos degree
$\ge2$. The end is Theorem~\ref{mat:entropy-smooth}: a smooth uniformly
convex density satisfies
\[
 2h(\mu)-2m-\log\det\Cov(\mu)\ge
 \frac1{500}\E\tr(Y^2)+\frac1{1000}\sum_{i=1}^m\omega(\lambda_i(B)),
\]
where $\lambda_i(B)$ are the eigenvalues of $B$ and $\omega\ge0$ is an
explicit weight vanishing only at $1$. This strict remainder is stronger
than the inequality and is what classifies equality.

\paragraph{Step 5: remove smoothness, then classify equality.} An
arbitrary log-concave density is approximated by smooth uniformly convex
ones preserving entropy, covariance, and first moments
(Section~\ref{geo:section}, via the Moreau envelope), giving the
inequality for all $f$. For equality, approximate a sharp $f$ by smooth
ones and let the strict remainder shrink. It forces $Y\to0$ in Gaussian
$L^2$, $B\to I$, and the transport Jacobians to converge to $q(A)$ for
a linear field $A=\sum x_rX_r$ with $\sum X_r^2=I$. The mixed partial
derivatives of that limit commute, which
(Section~\ref{sec:rigidity}) forces the $X_r$ to commute, which
diagonalizes $A$ and yields independent one-dimensional exponential
transports. That is the equality case.

\paragraph{The geometry.} Section~\ref{sec:geometry} computes the entropy
of the exponential density on the cone over $K$ in terms of $|K|$ and
$\Sigma_K$. Plugging into Theorem~\ref{intro:entropy} in dimension $n+1$
gives the simplex bound; equality recycles the equality classification
to show the cone is an orthant, hence $K$ a simplex.

\subsection{Roadmap}
Section~\ref{tr:section} sets up the Gaussian transport, the affine
normalization, and the energy estimate with the surplus $\mathcal C$.
Section~\ref{mat:section} decomposes the entropy gap and proves the
smooth bound (Theorem~\ref{mat:entropy-smooth}) using two scalar
propositions. Section~\ref{sc:section} proves those two propositions,
and the Appendix~\ref{cert:verification} certifies every numerical
constant with exact rational arithmetic. Section~\ref{geo:section}
removes smoothness. Section~\ref{sec:rigidity} classifies equality in
the entropy bound. Section~\ref{sec:geometry} transfers everything to
convex bodies and completes Theorem~\ref{intro:simplex}. Throughout,
all finite decimals denote exact rational numbers and all matrices are
real; ``symmetric'' means real symmetric and affine maps are invertible.

\section{Gaussian transport and a matrix surplus}\label{tr:section}

This section builds the machinery. We start from a smooth, uniformly
convex density, push the standard Gaussian onto it by the optimal map,
reparametrize the Jacobian, choose affine coordinates so that the
reparametrized field has mean zero, and prove an energy estimate that
keeps track of eigenvalue separation.

\subsection{Standing smoothness assumptions}

For most of the proof we work with a probability density $e^{-V}$ on
$\R^m$ satisfying
\begin{equation}\label{tr:smooth-assumptions}
 V\in C^\infty(\R^m),\qquad 0<c_-I\le D^2V\le c_+I<\infty.
\end{equation}
Here $c_-,c_+$ may depend on the density but are finite and positive.
Write $C=\Cov(\mu)$ where $\dd\mu=e^{-V}\dd y$, and let $\gamma_m$ be
standard Gaussian measure. Throughout the proof, $\E$ denotes
integration against $\gamma_m$.

\subsection{The one-dimensional scalar model}

Let $\varphi$ and $\Phi$ be the standard Gaussian density and
distribution function. For $a\in\R$ define
\begin{equation}\label{tr:scalar-functions}
 t(a)=-\log\Phi(-a),\quad q(a)=t'(a)=\frac{\varphi(a)}{\Phi(-a)},\quad
 d(a)=q(a)-a,
\end{equation}
and then
\begin{equation}\label{tr:scalar-functions-2}
 k(a)=\frac{t(a)}{q(a)},\qquad b(a)=k'(a),\qquad l(a)=b'(a).
\end{equation}
There is a geometric meaning to $t$: if $G$ is standard Gaussian then
$\Phi(-G)$ is uniform on $(0,1)$, so $t$ is exactly the monotone
transport from a standard Gaussian to the exponential law of mean one
(the ``entropy benchmark'' with entropy $1$ and variance $1$). The
function $q=t'$ is therefore the derivative of that transport.

The identities
\begin{equation}\label{tr:scalar-identities}
 q'=qd,\qquad b=1-kd,\qquad l=k-q+ab
\end{equation}
follow by differentiating the definitions. In particular $q-k=ab-l$.
These recur constantly, so it is worth checking them once: $q'=t''$,
and differentiating $q=\varphi/\Phi(-\cdot)$ gives
$q'=q(q-a)=qd$. From $t=qk$ we get $t'=q'k+qk'=q$, i.e.
$qdk+qb=q$, i.e.\ $b=1-kd$. Then $l=b'=-k'd-kd'=-(1-kd)d-k(qd-1)$,
which after expanding equals $k-q+ab$. That last equality uses
$q'=qd$.

\begin{lemma}[Basic properties of $q$]\label{tr:q-properties}
The function $q$ is a smooth increasing bijection $\R\to(0,\infty)$,
and
\[
 d>0,\qquad 0<q'<1,\qquad q''\ge0,\qquad 0<1-qd\le d^2.
\]
\end{lemma}
\begin{proof}
Let $X_a$ be the random variable with the conditional law of $G-a$
given $G>a$, and write $\E_a$ for expectation under that law. Gaussian
integration by parts in $a$ gives
\[
 \E_aX_a=d(a),\qquad \Var(X_a)=1-q(a)d(a).
\]
Since $X_a>0$ almost surely, $d>0$, hence $q'=qd>0$: $q$ is
increasing. The positivity of the variance gives $q'<1$.

The variance is at most the square of the mean only for a
positive random variable of a special kind; here we use the survival
function
\[
 S_a(u)=\Pr(X_a>u)=\frac{\Phi(-a-u)}{\Phi(-a)},\qquad u\ge0.
\]
Because the hazard rate $q(a+u)$ is increasing,
$S_a(u+v)\le S_a(u)S_a(v)$. Integrating this inequality over
$u,v\ge0$ gives
\[
 \frac12\E_aX_a^2=\int_0^\infty uS_a(u)\dd u
 \le\Bigl(\int_0^\infty S_a(u)\dd u\Bigr)^2=d(a)^2,
\]
so $\Var(X_a)\le d(a)^2$, i.e.\ $1-qd\le d^2$. Differentiating $q'=qd$
gives $q''=q(d^2+qd-1)\ge0$. Finally $q(a)\to0$ as $a\to-\infty$, and
$q(a)>a$ for $a>0$, so $q$ is onto $(0,\infty)$ and hence a bijection.
\end{proof}

Two additional facts, proved in the Appendix~\ref{cert:verification}
with exact rational arithmetic, are the $b$-bounds and a growth rate of
a certain quantity. Put
\begin{equation}\label{tr:M-definition}
 D_0(a)=\frac1{d(a)},\qquad M(a)=D_0(a)^2(1-b(a)^2).
\end{equation}
The quantity $M$ will let us compare uniform averages in $q$ with
uniform averages in $\log q$. Define a nondecreasing step function
$m_0$ by
\begin{equation}\label{tr:bins}
\begin{array}{c|rrrrrr}
 &O&P_1&R_1&U_1&V_1&T_1\\\hline
 a&(-\infty,-3)&[-3,-2)&[-2,-1)&[-1,\tfrac12)&[\tfrac12,4)&[4,\infty)\\
 m_0(a)&0&.19&.38&.68&.68&.68
\end{array}
\end{equation}
where at a finite endpoint we take the right-hand value.

\begin{lemma}[Scalar bounds]\label{tr:scalar-basic}
On all of $\R$,
\begin{equation}\label{tr:scalar-basic-equations}
 0<b<\tfrac12,\qquad l>0,\qquad
 \frac{d\log M}{d\log q}:=\frac{M'}{Md}\ge m_0.
\end{equation}
\end{lemma}

\subsection{Spectral calculus and divided differences}

For a real symmetric matrix we interpret scalar functions by spectral
calculus: if $f$ is defined on the spectrum, $f(A)$ means
$U\operatorname{diag}(f(\lambda_i))U^{\mathsf T}$ for any orthonormal
eigenbasis $A=U\operatorname{diag}(\lambda_i)U^{\mathsf T}$. For
differentiable $f$ the \emph{divided difference} is
\begin{equation}\label{tr:divided-difference}
 f[u,v]=\begin{cases}\dfrac{f(u)-f(v)}{u-v},&u\neq v,\\[2mm]
 f'(u),&u=v.\end{cases}
\end{equation}
The Daleckii--Krein formula \cite{DaleckiiKrein1965} (see also
\cite[Section~2.4]{Noferini2017}) says: in an eigenbasis of a symmetric
$A$, the derivative of $f(A)$ in a symmetric direction $Z$ has entries
$f[a_i,a_j]Z_{ij}$. This is the tool that lets us handle matrix
functions whose derivatives do not commute.

\subsection{Transport, Jacobian, and the reparametrization}

Brenier's theorem \cite{Brenier1991,McCann1995} gives a convex potential
$\phi$ with $(\nabla\phi)_\#\gamma_m=\mu$, unique up to null sets. The
full-space regularity theorem of Cordero-Erausquin and Figalli
\cite[Theorem~1.1]{CorderoErausquinFigalli2019}, which extends
Caffarelli's interior theory \cite{Caffarelli1992} to unbounded domains,
makes $\phi$ smooth; for $m=1$ this follows directly from the two
distribution functions. Caffarelli's contraction theorem
\cite{Caffarelli2000} and its inverse bound then give the Jacobian
\begin{equation}\label{tr:Jacobian}
 J=D^2\phi.
\end{equation}
\begin{equation}\label{tr:Jacobian-bounds}
 c_+^{-1/2}I\le J\le c_-^{-1/2}I.
\end{equation}
The lower bound comes from applying contraction to the inverse
transport; quantitatively, source Hessian $\le aI$ and target Hessian
$\ge bI$ give a transport Lipschitz constant $\le\sqrt{a/b}$ (see also
\cite[Theorem~1]{ChewiPooladian2023}).

Because $J$ is symmetric positive definite, the symmetric matrix field
\begin{equation}\label{tr:A-definition}
 A=q^{-1}(J)
\end{equation}
is well defined. The spectral bounds on $J$ confine the spectrum of $A$
to a compact interval, and $q$ is smooth there with uniformly bounded
derivatives, so $A$ and its functions behave well.

The usual isotropic coordinates would normalize $\Cov(\mu)=I$. We
instead impose the weaker but better-suited normalization $\E A=0$.

\begin{proposition}[Affine normalization]\label{tr:normalization}
Let $\mu=e^{-V}\dd y$ satisfy \eqref{tr:smooth-assumptions}. There is a
positive definite symmetric matrix $P$ such that the Brenier map from
$\gamma_m$ to $P_\#\mu$ has Jacobian $J_P$ satisfying
$\E q^{-1}(J_P)=0$. The transformed density again satisfies
\eqref{tr:smooth-assumptions}, with possibly different positive
constants.
\end{proposition}
\begin{proof}
We use a homotopy from the Gaussian and show its zeros remain in a
compact subset of the positive definite cone, then invoke degree theory.

For $0\le\theta\le1$ put
\[
 W_\theta(y)=(1-\theta)\frac{|y|^2}2+\theta V(y),\qquad
 \dd\mu_\theta=Z_\theta^{-1}e^{-W_\theta(y)}\dd y.
\]
There are $0<a_-\le a_+<\infty$, independent of $\theta$, with
$a_-I\le D^2W_\theta\le a_+I$. The respective covariance matrices
$C_\theta$ are positive definite and depend continuously on $\theta$;
since the family is uniformly bounded above and below in the spectral
order, there are $c_0,C_0>0$ with
\begin{equation}\label{tr:homotopy-covariance}
 c_0I\le C_\theta\le C_0I.
\end{equation}

For a positive definite symmetric $P$, let $J_{\theta,P}$ be the
Jacobian of the transport to $P_\#\mu_\theta$, and set
\[
 F(\theta,P)=\E q^{-1}(J_{\theta,P}),
\]
a map into the vector space of symmetric matrices. Two-directional
contraction gives
\begin{equation}\label{tr:homotopy-spectral}
 \frac{\lambda_{\min}(P)}{\sqrt{a_+}}I\le J_{\theta,P}
 \le\frac{\|P\|_{\mathrm{op}}}{\sqrt{a_-}}I.
\end{equation}

\emph{Continuity.} Restrict $P$ to a compact subset of the positive
definite cone. The transports have common Lipschitz constants by
\eqref{tr:homotopy-spectral}; their values at zero are bounded because
means and second moments are bounded. Hence any sequence
$(\theta_j,P_j)$ with a limit inside the subset has locally uniformly
convergent transport maps; their limits push $\gamma_m$ to the limiting
target and are gradients of convex functions, so Brenier uniqueness
identifies the limit. Writing $J_j,J$ for the Jacobians, they converge
weak-$*$ locally with common positive spectral bounds. The
change-of-variables equation expresses $\log\det J_j$ locally uniformly
in terms of the target potential, so it converges to $\log\det J$. On
a fixed compact spectral interval, strict concavity of $\log\det$
provides $c>0$ with
\[
 \log\det J_j\le\log\det J+\tr\bigl(J^{-1}(J_j-J)\bigr)
 -c|J_j-J|_{\mathrm F}^2.
\]
Integrating on balls and using weak-$*$ convergence gives
$J_j\to J$ in local $L^2$, and functional calculus gives local $L^2$
convergence of $q^{-1}(J_j)$. Common spectral bounds control the
Gaussian tails, so $F$ is continuous.

\emph{Uniform boundedness of zeros.} Suppose $\E A=0$, where
$A=q^{-1}(J_{\theta,P})$. Convexity of $q$ and Jensen applied to the
expected spectral measure of each unit vector give
$\E J_{\theta,P}\ge q(0)I$. Gaussian integration by parts gives
$\E(x_r\partial_i\phi)=(\E J_{\theta,P})_{ir}$. Projecting the centered
components of $\nabla\phi$ onto $\mathrm{span}(x_1,\dots,x_m)$ yields
\[
 PC_\theta P\ge(\E J_{\theta,P})^2\ge q(0)^2I.
\]
Together with \eqref{tr:homotopy-covariance} this uniformly bounds
$P^{-1}$. The lower estimate in \eqref{tr:homotopy-spectral} then
gives $A\ge a_*I$ for some $a_*\le0$. Since $q'\le1$,
$q(a)\le q(a_*)+a-a_*$ for $a\ge a_*$, so
$\E\tr J_{\theta,P}\le m(q(a_*)-a_*)$. On the other hand Gaussian
Poincar\'e and \eqref{tr:homotopy-spectral} give
\[
 c_0\|P\|_{\mathrm{op}}^2\le\tr(PC_\theta P)
 \le\E\tr J_{\theta,P}^2
 \le\frac{\|P\|_{\mathrm{op}}}{\sqrt{a_-}}\E\tr J_{\theta,P},
\]
which also bounds $P$ uniformly.

Choose a bounded open set in the space of symmetric matrices whose
closure lies in the positive definite cone and whose interior contains
every zero just bounded. The map $F(\theta,\cdot)$ has no zeros on its
boundary. At $\theta=0$ the target is $N(0,P^2)$, so $J_{0,P}=P$ and
$F(0,P)=q^{-1}(P)$; its unique zero is $q(0)I$, and its derivative
there is multiplication by the positive scalar $1/q'(0)$, so the
Brouwer degree is nonzero. Homotopy invariance gives the required zero
at $\theta=1$.
\end{proof}

The quantity $m+\tfrac12\log\det\Cov(\mu)-h(\mu)$ is unchanged under
an invertible affine map: a linear map $P$ adds $\log|\det P|$ to
entropy and $2\log|\det P|$ to the covariance log determinant. So we
may impose the normalization above. From here on, $V,J,A,C$ refer to
this transformed density and transport and
\begin{equation}\label{tr:mean-zero}
 \E A=0.
\end{equation}

\subsection{The transport equation and its energy}

We will differentiate the Jacobian and integrate the resulting equation.
Set, for a matrix field $F$,
\[
 |F|_{\mathrm F}^2=\tr(F^{\mathsf T}F),\quad
 \|F\|^2=\E|F|_{\mathrm F}^2,\quad
 \|\partial F\|^2=\sum_{r=1}^m\|\partial_rF\|^2.
\]

\begin{lemma}[Jacobian equation and finite energy]\label{tr:energy}
Under \eqref{tr:smooth-assumptions}, put $J_r=\partial_rJ$ and
\[
 \mathcal L u=\tr(J^{-1}D^2u)-\nabla V(\nabla\phi)\cdot\nabla u.
\]
Then
\begin{equation}\label{tr:Jacobian-equation}
 \mathcal LJ_{ij}
 =-\delta_{ij}+(JD^2V(\nabla\phi)J)_{ij}
 +\tr(J^{-1}J_iJ^{-1}J_j).
\end{equation}
Moreover $\|\partial J\|<\infty$. The fields $A$ and $k(A)$ lie in
Gaussian $H^1$, and the subsequent integrations by parts with smooth
matrix functions of $J$ are justified by compact cutoffs.
\end{lemma}
\begin{proof}
The density equation (change of variables) is
\begin{equation}\label{tr:density-equation}
 \log\det J=V(\nabla\phi)-\frac{|x|^2}2-\frac m2\log(2\pi).
\end{equation}
Differentiating twice gives \eqref{tr:Jacobian-equation}: the divergence
of the cofactor matrix of a Hessian vanishes, and that identity applied
to $J$, together with one more derivative of \eqref{tr:density-equation},
turns the second derivatives into the displayed form. Then
\[
 \mathcal Lu=e^{|x|^2/2}\operatorname{div}\bigl(e^{-|x|^2/2}
 J^{-1}\nabla u\bigr),
\]
so for smooth $u$ and compactly supported smooth $v$,
\begin{equation}\label{tr:divergence-form}
 \E(v\mathcal Lu)=-\E\langle\nabla v,J^{-1}\nabla u\rangle.
\end{equation}

Fix $0<\lambda\le\Lambda<\infty$ with $\lambda I\le J\le\Lambda I$.
Taking the trace in \eqref{tr:Jacobian-equation} and using
$D^2V\ge0$ gives $\mathcal L(\tr J)\ge-m+\Lambda^{-2}\sum_r|J_r|^2$.
For a compactly supported $0\le\chi\le1$, integration by parts and
$|\nabla\tr J|\le\sqrt m\,|\partial J|$ give
\begin{align*}
 \Lambda^{-2}\E\chi^2|\partial J|^2
 &\le m+2\lambda^{-1}\sqrt m\,\E(\chi|\nabla\chi|\,|\partial J|)\\
 &\le m+\tfrac12\Lambda^{-2}\E\chi^2|\partial J|^2
 +2m\Lambda^2\lambda^{-2}\E|\nabla\chi|^2.
\end{align*}
Letting $\chi$ grow to one through standard radial cutoffs, Fatou gives
$\|\partial J\|<\infty$.

The Fr\'echet derivatives of $q^{-1}$ and of $k\circ q^{-1}$ are
bounded on the compact spectrum of $J$, so $A,k(A)\in H^1(\gamma_m)$;
the same holds for any smooth scalar function on that spectrum. For such
a matrix multiplier, insert a cutoff in \eqref{tr:divergence-form} and
let it tend to one: the gradient-of-cutoff terms vanish by
Cauchy--Schwarz and the energy bound, and the remaining differentiated
terms are bounded by $1+|\partial J|^2$. This licenses the integrations
by parts.
\end{proof}

\subsection{Keeping the eigenvalue separation}

Set $Z_r=\partial_rA$. At each point diagonalize $A$, writing $a_i$ for
its eigenvalues and $\lambda_i=q(a_i)$ for those of $J$. For real $a,b$
define
\begin{equation}\label{tr:surplus-weight}
 \delta(a,b)=|\log q(a)-\log q(b)|,\quad m_*(a,b)=m_0(\min(a,b)),
\end{equation}
and
\begin{equation}\label{tr:surplus-weight-2}
 r(a,b)=\frac{1+m_*(a,b)}{16}\Bigl(1+\frac{\delta(a,b)^2}{30}
 +\frac{\delta(a,b)^4}{1680}\Bigr).
\end{equation}
Write $\delta_{ij}=\delta(a_i,a_j)$, $r_{ij}=r(a_i,a_j)$, and
\begin{equation}\label{tr:surplus-definition}
 \mathcal C=\E\sum_{i,j,r}r_{ij}\delta_{ij}^2(Z_r)_{ij}^2.
\end{equation}
This does not depend on the orthonormal basis chosen within repeated
eigenspaces; the factor $\delta_{ij}^2$ vanishes when $a_i=a_j$.

\begin{proposition}[Matrix surplus]\label{tr:surplus}
For the transport satisfying \eqref{tr:smooth-assumptions}, with the
notation above,
\begin{equation}\label{tr:surplus-inequality}
 \E\tr k(A)^2\ge\|\partial A\|^2-\|\partial k(A)\|^2+\mathcal C.
\end{equation}
\end{proposition}
\begin{proof}
There are two parts: replace the transport equation by averages of the
scalar function $M$, then compare arithmetic and logarithmic averages
to produce $\mathcal C$. The identity $qk=t$ will later match the
entropy term in the covariance completion.

\emph{Step A.} Put $F(\lambda)=k(q^{-1}(\lambda))^2$. Multiply
\eqref{tr:Jacobian-equation} by $F(J)$ in trace and integrate by parts
(licit by Lemma~\ref{tr:energy}). At a fixed point choose an
orthonormal eigenbasis of $J$ and rotate all three indices of the
symmetric tensor $\Theta_{ijr}=\partial_rJ_{ij}=\partial_i\partial_j
\partial_r\phi$ into that basis. The quadratic term becomes
\[
 \sum_iF(\lambda_i)\tr(J^{-1}J_iJ^{-1}J_i)
 =\sum_{i,j,r}\frac{F(\lambda_i)}{\lambda_j\lambda_r}\Theta_{ijr}^2,
\]
using symmetry of $\Theta$, and the integration-by-parts term is
\[
 \sum_r\lambda_r^{-1}\tr\bigl((\partial_rF(J))J_r\bigr)
 =\sum_{i,j,r}\frac{F[\lambda_i,\lambda_j]}{\lambda_r}\Theta_{ijr}^2.
\]
The remaining term is $\tr(F(J)JD^2VJ)\ge0$. Discarding it,
\begin{equation}\label{tr:initial-multiplier}
 \E\tr k(A)^2\ge\E\sum_{i,j,r}\Theta_{ijr}^2
 \Bigl(\frac{F(\lambda_i)}{\lambda_j\lambda_r}
 +\frac{F[\lambda_i,\lambda_j]}{\lambda_r}\Bigr).
\end{equation}
Only pointwise changes of basis occur; no eigenvectors are
differentiated.

\emph{Step B.} Regard $M$ as a function of $\lambda=q(a)$. The
identities give
\begin{equation}\label{tr:lambdaF}
 \frac{d}{d\lambda}(\lambda F(\lambda))=k^2+\frac{2kb}d
 =\frac{1-b^2}{d^2}=M.
\end{equation}
For positive $u,v$ let $M^{\mathrm{ar}}_{uv}$ be the uniform average of
$M$ on the interval between $u$ and $v$ (continuous when $u=v$). With
weight $\Theta_{ijr}^2/(\lambda_i\lambda_j\lambda_r)$, symmetrizing in
$i,j$ changes the coefficient in \eqref{tr:initial-multiplier} to
$\frac{\lambda_i+\lambda_j}2M^{\mathrm{ar}}_{\lambda_i\lambda_j}$.
In the full sum we may replace this by
$\lambda_rM^{\mathrm{ar}}_{\lambda_i\lambda_j}$ (a decrease). To see it,
order any three eigenvalues $u\le v\le w$. By
Lemma~\ref{tr:scalar-basic}, $M$ is increasing, so
$M^{\mathrm{ar}}_{uv}\le M^{\mathrm{ar}}_{uw}\le M^{\mathrm{ar}}_{vw}$.
The coefficients $u+v-2w$, $u+w-2v$, $v+w-2u$ sum to zero, and their
scalar product with these three ordered averages is nonnegative by
rearrangement. Full symmetry of $\Theta$ allows exactly this
permutation averaging. Hence
\begin{equation}\label{tr:arithmetic-energy}
 \E\tr k(A)^2\ge\E\sum_{i,j,r}
 \frac{\Theta_{ijr}^2}{\lambda_i\lambda_j\lambda_r}\,
 \lambda_rM^{\mathrm{ar}}_{\lambda_i\lambda_j}.
\end{equation}

\emph{Step C.} Fix a pair of eigenvalues; bars denote uniform averages
in $\log\lambda$ over their interval. Set $x=\delta_{ij}/2$,
$w_0=x/\sinh x$ (with $w_0=1$ at $x=0$), and $m_*=m_*(a_i,a_j)$. Since
$\dd a/\dd\log\lambda=D_0$, the divided-difference formula shows the
coefficients of $(Z_r)_{ij}^2$ and $(\partial_rk(A))_{ij}^2$, written
with the same weight as in \eqref{tr:arithmetic-energy}, are
\begin{equation}\label{tr:derivative-coefficients}
 \lambda_rw_0^2\overline{D_0}^{\,2},
 \qquad\lambda_rw_0^2\overline{bD_0}^{\,2}.
\end{equation}
Both $D_0(1+b)$ and $D_0(1-b)=k$ are increasing ($d'=qd-1<0$,
$b'=l>0$, $k'=b>0$). The covariance inequality for two increasing
functions on an interval gives
\begin{equation}\label{tr:logarithmic-chebyshev}
 \overline M\ge M_1:=\overline{D_0}^{\,2}-\overline{bD_0}^{\,2}
 \ge\tfrac34\overline{D_0}^{\,2}.
\end{equation}
On the logarithmic interval, \eqref{tr:scalar-basic-equations} says
$M(\lambda)\lambda^{-m_*}$ is increasing. Exponential weighting and the
same covariance inequality give
\begin{equation}\label{tr:average-comparison}
 \frac{M^{\mathrm{ar}}_{\lambda_i\lambda_j}}{\overline M}
 \ge\frac{\operatorname{shc}((1+m_*)x)}
 {\operatorname{shc}(x)\operatorname{shc}(m_*x)},\qquad
 \operatorname{shc}(x)=\frac{\sinh x}{x}.
\end{equation}
Indeed, after centering the logarithmic interval, the left side is
$\E_0(e^sM)/(\E_0e^s\,\E_0M)$ for the uniform average $\E_0$; under the
measure proportional to $e^{m_*s}$ the increasing factor
$Me^{-m_*s}$ can only increase the mean of $e^s$, and taking
$M=e^{m_*s}$ gives the displayed ratio.

\emph{Step D.} A useful explicit lower bound. Define
$A_0(x)=x\coth x-1$ (and $A_0(0)=0$). The identity
\[
 x\coth x+\frac{x^2}{\sinh^2x}\ge2
\]
gives $A_0(x)\ge1-w_0^2$ and shows $A_0(x)/x^2$ is nonincreasing for
$x>0$. (To verify the identity, clear denominators and expand
$\frac x2\sinh(2x)+x^2-\cosh(2x)+1$ in powers of $x$; coefficients
vanish through degree four and are nonnegative thereafter.) Since
$0\le m_*\le1$, $A_0(m_*x)\ge m_*^2A_0(x)$, and the addition formula
for $\sinh$ gives
\begin{align}
 \frac{\operatorname{shc}((1+m_*)x)}
 {\operatorname{shc}(x)\operatorname{shc}(m_*x)}
 &=1+\frac{m_*A_0(x)+A_0(m_*x)}{1+m_*}\notag\\
 &\ge1+m_*(1-w_0^2).
 \label{tr:hyperbolic-comparison}
\end{align}
Everything extends continuously to $m_*=0$ or $x=0$.

\emph{Step E.} Combining
\eqref{tr:logarithmic-chebyshev}--\eqref{tr:hyperbolic-comparison}, the
amount by which the coefficient in \eqref{tr:arithmetic-energy} exceeds
the difference of the two coefficients in
\eqref{tr:derivative-coefficients} is at least
$\lambda_r(1+m_*)(1-w_0^2)M_1$. Relative to the coefficient of
$(Z_r)_{ij}^2$ this is at least
\[
 \tfrac34(1+m_*)(w_0^{-2}-1)\ge\frac{1+m_*}{16}
 \Bigl(\delta_{ij}^2+\frac{\delta_{ij}^4}{30}
 +\frac{\delta_{ij}^6}{1680}\Bigr),
\]
where the last inequality is the Taylor expansion of $(\sinh x/x)^2$
with positive omitted coefficients. Substituting into
\eqref{tr:arithmetic-energy} and summing proves
\eqref{tr:surplus-inequality}.
\end{proof}

We now have two inputs: the field $A$ has no constant Gaussian
component, and convexity gives \eqref{tr:surplus-inequality}. The next
section combines these with a Gaussian covariance identity and separates
the first chaos of $A$ from the rest.

\section{Gaussian decomposition of the entropy deficit}\label{mat:section}

The transport estimate of Proposition~\ref{tr:surplus} controls the
Gaussian energy of the reparametrized Jacobian. Its degree-one part,
however, has no Poincar\'e surplus. We use that part to choose a matrix
supporting plane for the log determinant; the remaining Gaussian chaoses
supply the coercivity needed for the entropy bound. A strict remainder
is kept throughout, and it will also classify the equality cases. In
this section $\E$ denotes integration against the standard Gaussian
measure $\gamma_m$.

Define a scalar weight on $[0,\infty)$ by
\begin{equation}\label{mat:rigidity-weight}
 \omega(\lambda)=
 \frac{3(\lambda-1)^2}{37+3\lambda}
 \Bigl(\frac3{20}+\frac{49}{100}\frac{37}{37+3\lambda}\Bigr)
 \min\Bigl\{\frac{13}{100},\frac{32}{25\sqrt\lambda}\Bigr\},
\end{equation}
with the minimum equal to $13/100$ at $\lambda=0$. The weight is
continuous and nonnegative, vanishes only at $1$, and satisfies
$\omega(\lambda)/\sqrt\lambda\to24/125$ as $\lambda\to\infty$.

\begin{theorem}[The smooth entropy bound with a remainder]
\label{mat:entropy-smooth}
Let $m\ge1$ and let $\mu=e^{-V}\dd y$ be a probability measure on
$\R^m$ with $V\in C^\infty(\R^m)$ and
$c_-\Id\le D^2V\le c_+\Id$ for some $0<c_-\le c_+<\infty$. Choose the
affine normalization of Proposition~\ref{tr:normalization}, and let
$A=q^{-1}(J)$ be the resulting Gaussian transport matrix field, so
$\E A=0$. Set
\[
 X_r=\E(x_rA),\qquad Y=A-\sum_{r=1}^m x_rX_r,\qquad
 B=\sum_{r=1}^mX_r^2.
\]
If $\lambda_1,\dots,\lambda_m$ are the eigenvalues of $B$, then
\begin{equation}\label{mat:strict-remainder}
 2h(\mu)-2m-\log\det\Cov(\mu)\ge\frac1{500}\E\tr(Y^2)
 +\frac1{1000}\sum_{i=1}^m\omega(\lambda_i).
\end{equation}
In particular $h(\mu)\ge m+\tfrac12\log\det\Cov(\mu)$.
\end{theorem}

We work in the normalization $\E A=0$ of
Proposition~\ref{tr:normalization}; it leaves
$h(\mu)-\tfrac12\log\det\Cov(\mu)$ unchanged. Scalar functions of $A$
are by spectral calculus; when arguments are omitted, $q,k,t,b,l$ and
the functions below are evaluated at $A$. For a matrix field $F$ put
\[
 \tau(F)=\E\tr F,\qquad\|F\|^2=\E\tr(F^{\mathsf T}F),\qquad
 \|F\|_S^2=\E\tr(F^{\mathsf T}SF),
\]
the last for a constant positive definite $S$. For a family indexed by
$r$, these norms include $\sum_r$. Lemma~\ref{tr:energy} supplies the
Sobolev regularity used below.

Write $C=\Cov(\mu)$ and $\mathscr D=2m+\log\det C-2h(\mu)$ (negative of
twice the entropy gap). The transport change of variables and
$\log q(a)=-\tfrac{a^2}2-\tfrac12\log(2\pi)+t(a)$ give
\begin{equation}\label{mat:deficit-exact}
 \mathscr D=\tau(A^2-2t)+\log\det C+m.
\end{equation}

\subsection{Covariance and the first Gaussian chaos}

Let $N=-\Delta+x\cdot\nabla$ be the nonnegative Ornstein--Uhlenbeck
operator and $R=(\Id+N)^{-1}$. The degree-$d$ Gaussian chaos is the
span of coordinatewise products of one-dimensional Hermite polynomials
with degrees summing to $d$; these spaces give an orthogonal
decomposition of $L^2(\gamma_m)$, on which $N$ and $R$ act by $d$ and
$(1+d)^{-1}$. Write $\partial=(\partial_1,\dots,\partial_m)$ and
$\partial^*v=\sum_r(x_rv_r-\partial_rv_r)$, so $N=\partial^*\partial$.
The Gaussian Sobolev norm is $\|F\|_{H^1}^2=\|F\|^2+\|\partial F\|^2$,
with dual norm
\[
 \|F\|_{-1}^2=\sum_{d\ge0}\frac{\|F_d\|^2}{1+d},
\]
where $F_d$ is the degree-$d$ chaos component (see
\cite[Section~2]{NourdinPeccati2009}). We use the Gaussian
Helffer--Sj\"ostrand covariance representation
\cite[Proposition~4.1(ii)]{DuerinckxOtto2020}, with a direct chaos proof
below. The term $-2\tau(t)$ in \eqref{mat:deficit-exact} determines its
completion: since $qk=t$, subtracting $(\Id+N)k$ from $q$ in the
covariance quadratic form produces exactly the cross term $-2t$.

\begin{lemma}[Covariance completion]\label{mat:covariance}
Define the symmetric $H^{-1}$ matrix field
$U=q(A)-(\Id+N)k(A)$ and let $W$ be the Gram matrix of its rows in
$H^{-1}$:
\[
 W_{ij}=\sum_{\ell=1}^m\langle U_{i\ell},U_{j\ell}\rangle_{-1}.
\]
Then
\begin{equation}\label{mat:covariance-completion}
 C=\E[J(RJ)]=W+\E\Bigl(2t-k^2-\sum_rK_r^2\Bigr),
 \qquad K_r=\partial_rk(A).
\end{equation}
Moreover, if $U=u-\partial^*v$ with $u$ and the family $v=(v_r)_r$
square-integrable, then for every constant $S>0$,
\begin{equation}\label{mat:dual-bound}
 \tr(SW)\le\|u\|_S^2+\|v\|_S^2.
\end{equation}
\end{lemma}
\begin{proof}
Put $F=\nabla\phi$. For two degree-$d$ components of $F$, Gaussian
integration by parts gives
$\sum_\ell\E[(\partial_\ell F_{i,d})(\partial_\ell F_{j,d})]
=d\,\E[F_{i,d}F_{j,d}]$. Differentiation lowers the degree by one, so
$R$ cancels the factor $d$, and summing over $d\ge1$ proves
$C_{ij}=\sum_\ell\E[J_{i\ell}RJ_{j\ell}]$, the first identity, using
symmetry of $J$. Expanding the row Gram matrix of $q-(\Id+N)k$ gives
cross terms $-2\E(qk)=-2\E t$; the remaining term is
$\E[k(\Id+N)k]=\E(k^2+\sum_rK_r^2)$ by entrywise integration by parts.
This is valid by the $H^1$--$H^{-1}$ pairing even when $Nk$ is not
square-integrable.

For a test field $F$, the pairing with $u-\partial^*v$ is
$\langle u,F\rangle-\langle v,\partial F\rangle$, bounded by
$(\|u\|^2+\|v\|^2)^{1/2}\|F\|_{H^1}$. Taking the dual norm gives
\eqref{mat:dual-bound} for $S=\Id$; applying the argument to $S^{1/2}U$
gives the general case.
\end{proof}

Separate the degree-one part of $A$:
\begin{equation}\label{mat:chaos}
 A=\sum_{r=1}^m x_rX_r+Y,\qquad X_r=\E(x_rA),\qquad B=\sum_rX_r^2.
\end{equation}
The matrices $X_r$ are constant and symmetric; $Y$ has only chaoses of
degree $\ge2$. We use
\begin{equation}\label{mat:gradient-fields}
 Z_r=\partial_rA=X_r+G_r,\qquad G_r=\partial_rY,\qquad
 g_r=\partial_rN^{-1}Y,\qquad V_0=\|G\|^2-\|Y\|^2,
\end{equation}
so $\partial^*g=Y$; all three derivative fields are symmetric.

The coordinatewise Gaussian-to-exponential transport has
$A(x)=\operatorname{diag}(x_1,\dots,x_m)$ and $B=\Id$. We therefore
choose the supporting plane for $\log\det C$ as a regularized inverse of
$B$ centered at $B=\Id$. Fix
\begin{equation}\label{mat:dual-constants}
 \alpha=\frac3{40},\qquad s=\frac1{1-\alpha},
\end{equation}
and set
\begin{equation}\label{mat:dual-matrices}
 S=(\Id/s+\alpha B)^{-1},\qquad T=S-\Id,\qquad H=-T(B-\Id).
\end{equation}
These constant matrices commute with one another (though generally not
with functions of $A$), and
\begin{equation}\label{mat:dual-relations}
 0<S\le s\Id,\quad T=-\alpha S(B-\Id),\quad
 H=\alpha S(B-\Id)^2\ge0,\quad T^2=\alpha HS.
\end{equation}
In particular $S=\Id$ when $B=\Id$, and deviations of the first chaos
from that value produce the nonnegative matrix $H$.

Concavity of the log determinant gives
$\log\det C+m\le\tr(SC)-\log\det S$. Combining this with
Lemma~\ref{mat:covariance} and Proposition~\ref{tr:surplus}, and using
$\|A\|^2-\|Z\|^2=-V_0$, gives
\begin{equation}\label{mat:deficit-reduction}
 \mathscr D\le-V_0-\mathcal C+\tr(T-\log S)+\tr(SW)
 +\tau\Bigl[T\Bigl(2t-k^2-\Id-\sum_rK_r^2\Bigr)\Bigr].
\end{equation}
Here $\mathcal C$ is the eigenvalue-separation surplus of
Proposition~\ref{tr:surplus}. It remains to estimate the last two terms
while keeping the negative contribution involving $H$.

\subsection{An exact expansion with noncommuting factors}

Recall the divided differences with $v[a,a]=v'(a)$, and for a smooth $v$
define the secant increment at the second endpoint by
$\Delta v_{ij}=v[a_i,a_j]-v'(a_j)$. We use the matrix derivative
formula \cite[Section~2.4]{Noferini2017}. At a fixed point choose an
orthonormal eigenbasis of $A$, with entrywise multiplication denoted
$\circ$, and define the derivative residuals
\begin{equation}\label{mat:residuals}
 D_r=K_r-X_rb=G_rb+Z_r\circ\Delta k,\qquad
 L_r=\partial_rb-X_rl=G_rl+Z_r\circ\Delta b.
\end{equation}
These need not be symmetric; the distinction between $D_r$ and $D_r^t$
is essential below.

The scalar identity $q-k=ab-l$ implies
\begin{equation}\label{mat:U-representation}
 U=Yb+\sum_rX_rL_r+(B-\Id)l-\partial^*D.
\end{equation}
Indeed $\partial^*(Xb)=(A-Y)b-Bl-\sum_rX_rL_r$. To divide the
higher-chaos terms between two squares fix
\begin{equation}\label{mat:constants}
 \beta=\frac7{25},\qquad c=\frac32,\qquad\gamma=\frac c2.
\end{equation}
Using $\partial^*g=Y$ in \eqref{mat:dual-bound},
\begin{equation}\label{mat:W-bound}
 \tr(SW)\le\left\|Y(b-\beta)+\sum_rX_rL_r+(B-\Id)l\right\|_S^2
 +\|D-\beta g\|_S^2.
\end{equation}
We group the expansion by its $H$ term and its derivative residuals. In
the remaining $Y$ coefficient we split off $ck$: the identity
$\partial^*g=Y$ transfers $c\tau(TYk)$ to derivatives, where completing
the square produces $D-\gamma g$. Introduce the scalar functions
\begin{equation}\label{mat:scalar-functions}
 f=k(1+b),\quad h_0=f-ck-\frac{2(b-\beta)l}{\alpha},\quad
 p=f'-b^2-\frac{l^2}{\alpha},\quad e=f'-2b^2-\frac{2l^2}{\alpha}.
\end{equation}
The identity $af-f'=2t-k^2-b^2-1$ follows from $q-k=ab-l$ and
$b=1-k(q-a)$. For each pair of eigenvalues of $A$ put
\[
 \psi_{ij}=\Delta f_{ij}-2b_j\Delta k_{ij}
 -\frac{2l_j}{\alpha}\Delta b_{ij},\qquad b_j=b(a_j),\ l_j=l(a_j),
\]
and define
\begin{equation}\label{mat:M-w}
 M_r=G_re+Z_r\circ\psi+cg_rb,\qquad w_r=D_r-\gamma g_r.
\end{equation}
Finally write $P_T=\tau(TbBb-TBb^2)$.

\begin{lemma}[Matrix expansion]\label{mat:expansion}
The sum of the last two terms in \eqref{mat:deficit-reduction} is at most
\begin{align}
 &-\tau(Hp)+\tau\Bigl[T\Bigl(Yh_0+\sum_rX_rM_r\Bigr)\Bigr]
 \label{mat:expanded-bound}\\
 &\quad+\Bigl\|Y(b-\beta)+\sum_rX_rL_r\Bigr\|_S^2
 +\|D-\beta g\|_S^2+\gamma^2\tau\Bigl(T\sum_rg_r^2\Bigr)\nonumber\\
 &\quad-P_T-2\tau\Bigl([T,b]\sum_rX_rw_r\Bigr)
 -\tau\Bigl(T\sum_rw_r^tw_r\Bigr).\nonumber
\end{align}
Except for the use of \eqref{mat:W-bound}, this expansion is exact.
\end{lemma}
\begin{proof}
Let $F_0=Y(b-\beta)+\sum_rX_rL_r$. Using \eqref{mat:dual-relations},
\[
 \|F_0+(B-\Id)l\|_S^2=\|F_0\|_S^2+\frac1\alpha\tau(Hl^2)
 -\frac2\alpha\tau(TF_0l),
\]
where the cross term is written using $\tr(F_0^tTl)=\tr(TF_0l)$.
Integration by parts on the linear part of $A$ gives
\[
 \tau(TAf)=\tau\Bigl[T\Bigl(Yf+\sum_rX_r\partial_rf\Bigr)\Bigr],
 \qquad\partial_rf=X_rf'+G_rf'+Z_r\circ\Delta f.
\]
Together with $af-f'=2t-k^2-b^2-1$ this accounts for the terms
involving $f'$ and $f$.

To expand the energy term use both $K_r=X_rb+D_r=bX_r+D_r^t$:
\begin{equation}\label{mat:K-square}
 \tau\Bigl(T\sum_rK_r^2\Bigr)=\tau(TbBb)
 +2\tau\Bigl(Tb\sum_rX_rD_r\Bigr)+\tau\Bigl(T\sum_rD_r^tD_r\Bigr).
\end{equation}
The two cross terms agree under transposition inside the trace. Moving
$b$ past $T$, rather than past $X_r$, gives
$\tau(TbX_rD_r)=\tau(TX_rD_rb)+\tau([T,b]X_rD_r)$. The terms containing
$B-\Id$ now combine as
\[
 \tau\bigl(T(B-\Id)f'\bigr)+\tau(Tb^2-TbBb)+\frac1\alpha\tau(Hl^2)
 =-\tau(Hp)-P_T.
\]
In the other terms, substituting \eqref{mat:residuals} produces
$G_re+Z_r\circ\psi$ and the $Y$ multiplier
$f-2(b-\beta)l/\alpha=h_0+ck$.

It remains to handle $c\tau(TYk)$. Since $\partial^*g=Y$,
$\tau(TYk)=\sum_r\tau(TK_rg_r)$; integration by parts first gives
$\tau(Tg_rK_r)$ and transposition makes them equal. Substituting
$K_r=bX_r+D_r^t$ and moving $b$ past $T$ introduces $cg_rb$ in $M_r$
and replaces $D_r$ by $D_r-\gamma g_r$ in the commutator term. The
remaining square is completed by
\[
 -\tau(TD_r^tD_r)+c\tau(TD_r^tg_r)
 =-\tau(Tw_r^tw_r)+\gamma^2\tau(Tg_r^2).
\]
This proves \eqref{mat:expanded-bound}.
\end{proof}

\subsection{The commutator and the weighted squares}

The commutator in Lemma~\ref{mat:expansion} has a compensating quadratic
term $P_T$. Keeping that term is what permits a dimension-free estimate
without assuming that the Jacobian and the first-chaos matrix commute.

\begin{lemma}[Commutator estimate]\label{mat:commutator}
For the matrices $X_r,B,S,T$ in \eqref{mat:chaos} and
\eqref{mat:dual-matrices}, any square-integrable symmetric matrix field
$b$, and arbitrary square-integrable matrix fields $w_r$,
\begin{equation}\label{mat:commutator-bound}
 -P_T-2\tau\Bigl([T,b]\sum_rX_rw_r\Bigr)-\tau\Bigl(T\sum_rw_r^tw_r\Bigr)
 \le\Bigl(1+\frac s8\Bigr)\|w\|^2.
\end{equation}
\end{lemma}
\begin{proof}
Work pointwise and diagonalize $S$. The relation
$B=\alpha^{-1}(S^{-1}-\Id/s)$ gives
\begin{equation}\label{mat:P-square}
 P_T=\frac1{2\alpha}\bigl\|S^{-1/2}[T,b]S^{-1/2}\bigr\|^2,
\end{equation}
with the norm including expectation; the contribution of $b_{ij}^2$ for
$i<j$ on either side is $(S_i-S_j)^2/(\alpha S_iS_j)$. Set
$F=\sum_rX_rw_r$ and $\operatorname{skew}M=(M-M^t)/2$. Completing the
square in the skew matrix $S^{-1/2}[T,b]S^{-1/2}$ bounds the first two
terms on the left by
$2\alpha\bigl\|\operatorname{skew}(S^{1/2}FS^{1/2})\bigr\|^2$.

For completeness, estimate this square together with the last term.
Write $S_i=su_i$, $0<u_i\le1$, so $\alpha S_iB_i=1-u_i$. The rows of
the block matrix $X=(X_1,\dots,X_m)$ are orthogonal since $XX^t=B$ is
diagonal. For column $j$ of the vertically stacked matrix $(w_r)_r$, let
$z_{ij}$ be its component along the normalized $i$th row of $X$ (zero
rows completed arbitrarily; they do not contribute to $F$). Then
$F_{ij}=\sqrt{B_i}z_{ij}$ and the remaining column components are
orthogonal to these directions. For $i<j$ the quadratic form in
$(z_{ij},z_{ji})$ has matrix
\[
 \begin{pmatrix}
 1-su_iu_j&-s\sqrt{u_iu_j(1-u_i)(1-u_j)}\\
 -s\sqrt{u_iu_j(1-u_i)(1-u_j)}&1-su_iu_j
 \end{pmatrix},
\]
whose largest eigenvalue is at most $1+s(z-2z^2)\le1+s/8$, where
$z=\sqrt{u_iu_j}$ and $(1-u_i)(1-u_j)\le(1-z)^2$. Diagonal and
orthogonal column components have coefficient at most $1$. Summation
and expectation prove \eqref{mat:commutator-bound}.
\end{proof}

We also need a block version of Cauchy--Schwarz. If
$X=(X_1,\dots,X_m)$ and $S^{-1}=\Id/s+\alpha XX^t$, then
\begin{equation}\label{mat:block-bound}
 \Bigl\|a+\sum_rX_rL_r\Bigr\|_S^2\le s\|a\|^2+\frac1\alpha\|L\|^2.
\end{equation}
This follows column by column because the block operator
$S^{1/2}(\Id/\sqrt s,\sqrt\alpha X)$ has product with its transpose
equal to $\Id$.

The scalar estimates in Lemma~\ref{sc:bounds} below give $p(a)>0$ for
all real $a$. To reserve part of the negative $H$ term for the cost of
$\log\det S$, fix
\begin{equation}\label{mat:completion-constants}
 \epsilon=\frac3{20},\qquad\kappa=\frac{49}{100},
\end{equation}
and put $\rho=\epsilon\Id+\kappa S/s$. Then $\Id-\rho>0$. A weighted
completion of the square gives
\begin{align}
 &\tau\Bigl[T\Bigl(Yh_0+\sum_rX_rM_r\Bigr)\Bigr]
 \le\tau\bigl(H(\Id-\rho)p\bigr)\notag\\
 &\qquad+\frac{\|Mp^{-1/2}\|^2}{4(1-\epsilon)}
 +\frac{\alpha s\|Yh_0p^{-1/2}\|^2}{4(1-\epsilon-\kappa)}.
 \label{mat:weighted-completion}
\end{align}
To see that the order of the weights is valid, write
$Q_0=H(\Id-\rho)$ and $F=Yh_0+\sum_rX_rM_r$, and apply
$\langle a,b\rangle\le\|a\|^2+\tfrac14\|b\|^2$ to $Q_0^{1/2}p^{1/2}$ and
$Q_0^{-1/2}TFp^{-1/2}$. The resulting second square has left weight
bounded above by $T^2Q_0^{-1}\le\alpha S(\Id-\rho)^{-1}$, with equality
on the range of $H$; inverses are pseudoinverses on the common nullspace
of $Q_0$ and $T$. Since
$(\Id-\rho)S^{-1}=\frac{1-\epsilon-\kappa}{s}\Id+(1-\epsilon)\alpha XX^t$,
the same block argument as for \eqref{mat:block-bound} gives
\eqref{mat:weighted-completion}. Thus $p$ need not commute with $S$.

\subsection{Two scalar estimates and the final chaos bound}

The next estimate combines $\tr(T-\log S)$ with the negative $H$ term
left by \eqref{mat:weighted-completion} and the terms quadratic in $Y$;
a second controls the derivative residuals, including their dependence
on $Z$, by $\mathcal C$ and the higher-chaos energy. The first is proved
in Section~\ref{sc:absorption-proof}; its strict remainder controls the
eigenvalues of $B$ through \eqref{mat:rigidity-weight}.

\begin{proposition}[Absorption with a first-chaos remainder]
\label{mat:scalar-absorption}
Let $A\in L^2(\gamma_m;\operatorname{Sym}_m)$ satisfy $\E A=0$, and
define $X_r,Y,B$ by \eqref{mat:chaos} and $S,T,H$ by
\eqref{mat:dual-matrices}. With \eqref{mat:dual-constants},
\eqref{mat:constants}, \eqref{mat:completion-constants}, the functions
\eqref{mat:scalar-functions}, and $\rho=\epsilon\Id+\kappa S/s$,
\begin{equation}\label{mat:absorption-inequality}
 \begin{split}
 &\tr(T-\log S)-\tau(H\rho p)
 +s\tau\Bigl[Y^2\Bigl((b-\beta)^2+\frac{\alpha h_0^2}
 {4(1-\epsilon-\kappa)p}\Bigr)\Bigr]\\
 &\hspace{15mm}\le\frac{69}{500}\|Y\|^2
 -\frac1{1000}\sum_{i=1}^m\omega(\lambda_i),
 \end{split}
\end{equation}
where $\lambda_1,\dots,\lambda_m$ are the eigenvalues of $B$ and scalar
factors in the traces are evaluated at $A$.
\end{proposition}

The remaining derivative terms are compared entrywise in an eigenbasis
of $A$; this comparison retains a multiple of the eigenvalue-separation
surplus $\mathcal C$ rather than discarding it.

\begin{proposition}[Two-eigenvalue estimate]\label{mat:scalar-pair}
Let $a_i,a_j\in\R$ and $G,g,Z\in\R$. Set
\[
 \delta=\bigl|\log(q(a_i)/q(a_j))\bigr|,\qquad
 r_{ij}=\frac{1+m_0(\min(a_i,a_j))}{16}
 \Bigl(1+\frac{\delta^2}{30}+\frac{\delta^4}{1680}\Bigr),
\]
where $m_0$ is the six-interval lower bound in \eqref{tr:bins}. Define
the endpoint residuals
\[
 \begin{array}{lll}
 D_j=Gb_j+Z\Delta k_{ij},&L_j=Gl_j+Z\Delta b_{ij},&
 M_j=Ge_j+Z\psi_{ij}+cgb_j,\\
 D_i=Gb_i+Z\Delta k_{ji},&L_i=Gl_i+Z\Delta b_{ji},&
 M_i=Ge_i+Z\psi_{ji}+cgb_i,
 \end{array}
\]
and let $Q(u,v)=\frac{43}{100}u^2+\frac{26}{100}uv+\frac{68}{25}v^2$.
Then
\begin{align}
 &\operatorname{avg}_{v=i,j}\Bigl[
 s(D_v-\beta g)^2+\Bigl(1+\frac s8\Bigr)(D_v-\gamma g)^2
 +\gamma^2(s-1)g^2+\frac{L_v^2}{\alpha}\notag\\
 &\qquad\qquad+\frac{M_v^2}{4(1-\epsilon)p_v}\Bigr]
 \le Q(G,g-G/2)+r_{ij}\delta^2Z^2,
 \label{mat:pair-inequality}
\end{align}
where $\operatorname{avg}_{v=i,j}$ is the arithmetic mean of the two
endpoint expressions (counted twice when $i=j$). Moreover $p_v=p(a_v)>0$
and all divided differences extend continuously to $a_i=a_j$.
\end{proposition}

The proof of Proposition~\ref{mat:scalar-pair} is in
Section~\ref{sc:pair-proof}. We now show that these two scalar estimates
close the matrix argument.

\begin{proof}[Proof of Theorem~\ref{mat:entropy-smooth}]
Apply Lemma~\ref{mat:expansion} in \eqref{mat:deficit-reduction}. Bound
its last line by Lemma~\ref{mat:commutator}, its first square by
\eqref{mat:block-bound}, and its linear $T$ term by
\eqref{mat:weighted-completion}. The $H$ terms combine to
$-\tau(H\rho p)$. Since $Y=Y^t$ and $h_0,p,b$ are commuting functions of
$A$, the two terms quadratic in $Y$ have exactly the form in
\eqref{mat:absorption-inequality} by cyclicity of the trace, with no
commutation with $Y$ required. Set
$\Omega_B=\sum_{i=1}^m\omega(\lambda_i(B))$. Then
\begin{align}
 \mathscr D\le{}&-V_0-\mathcal C+\frac{69}{500}\|Y\|^2
 -\frac{\Omega_B}{1000}+\|D-\beta g\|_S^2\notag\\
 &+\Bigl(1+\frac s8\Bigr)\|D-\gamma g\|^2+\frac1\alpha\|L\|^2
 +\gamma^2\tau\Bigl(T\sum_rg_r^2\Bigr)
 +\frac{\|Mp^{-1/2}\|^2}{4(1-\epsilon)}.
 \label{mat:pre-pair}
\end{align}
Use $S\le s\Id$ and $T\le(s-1)\Id$. In an eigenbasis of $A$ the entries
of $G_r,g_r,Z_r$ are symmetric in $i,j$; pairing the entries $(i,j)$ and
$(j,i)$ therefore gives the endpoint average in
\eqref{mat:pair-inequality}, and right multiplication by $p^{-1/2}$ gives
the denominator $p_j$ for entry $(i,j)$. Proposition~\ref{mat:scalar-pair}
now bounds the last five terms of \eqref{mat:pre-pair} by
\[
 \E\sum_{i,j,r}Q\bigl((G_r)_{ij},(g_r)_{ij}-(G_r)_{ij}/2\bigr)
 +\mathcal C.
\]
The separation surplus is the same $\mathcal C$ as in
Proposition~\ref{tr:surplus}: its coefficient at $(i,j)$ does not depend
on $r$, so summing over $r$ is unchanged by rotating that index. All
changes of eigenbasis are pointwise; no eigenvectors are
differentiated, and the surplus cancels. Because $Q$ has constant
coefficients, its summed quadratic form is invariant under orthonormal
changes of basis, so we may return to fixed coordinates and use the
Gaussian chaos decomposition.

For a degree-$d$ component of $Y$, $d\ge2$, the associated components
of $g$ and $G$ satisfy $g=G/d$ and $\|G\|^2=d\|Y\|^2$, and different
degrees stay orthogonal after differentiation. Degree by degree the
needed bound follows from
\begin{equation}\label{mat:final-polynomial}
 Q(1,u-1/2)+\frac7{50}u\le1-u,\qquad 0\le u\le\frac12,
\end{equation}
since the right side minus the left factors as
$\frac1{50}(1-2u)(1+68u)\ge0$. Set $u=1/d$, multiply by $\|G\|^2$, and
sum over the chaoses (the sums converge because $Y\in H^1(\gamma_m)$).
The summed $Q$ term is at most $V_0-\frac7{50}\|Y\|^2$, so substitution
into \eqref{mat:pre-pair} gives
\[
 \mathscr D\le-\frac1{500}\|Y\|^2-\frac{\Omega_B}{1000},
\]
which is \eqref{mat:strict-remainder}.
\end{proof}

\section{The two scalar estimates}\label{sc:section}

The matrix argument reduced the entropy inequality to two estimates.
Proposition~\ref{mat:scalar-absorption} controls the part depending on the
first Gaussian chaos; Proposition~\ref{mat:scalar-pair} controls the
remaining quadratic form, one pair of eigenvalues at a time. We prove
them here. The functions $q,d,k,b,l$ and the six intervals
$O,P_1,R_1,U_1,V_1,T_1$ are those of Section~\ref{tr:section}, and the
constants $\alpha,s,\beta,c,\gamma,\epsilon,\kappa$ and the functions
$p,e,h_0$ are those of Section~\ref{mat:section}. All finite decimals
below denote exact rational numbers.

We first state the one-variable bounds used in both proofs; their
verification is in Appendix~\ref{cert:verification}, and none of them is
inferred from numerical sampling. For the quadratic form in
Proposition~\ref{mat:scalar-pair} set
\begin{equation}\label{sc:auxiliary-functions}
 \begin{gathered}
 D_*=\frac{1}{4(1-\epsilon)},\qquad
 \mathsf D=\frac{D_*}{p},\qquad\mathsf L=\frac{l}{\alpha},\\
 \mathsf E=e+\gamma b=p+b(.75-b)-\frac{l^2}{\alpha},\\
 K_0=\mathsf D\mathsf E,\qquad J_0=\mathsf D cb,\\
 A_*=s+1+\frac s8,\qquad
 \eta_*=s\beta+\Bigl(1+\frac s8\Bigr)\gamma,\qquad
 \nu_*=s\beta^2+\frac{9s}{8}\gamma^2.
 \end{gathered}
\end{equation}
The symbols $\mathsf D,\mathsf L,\mathsf E$ are scalar functions, to be
distinguished from the matrix residuals $D_r,L_r,M_r$ above.

\begin{lemma}[One-variable bounds]\label{sc:bounds}
For every $a\in\R$,
\begin{equation}\label{sc:p-h-bounds}
 p(a)>0,\qquad|h_0(a)|\le.67\sqrt{p(a)},
\end{equation}
and
\[
 p(a)\ge\begin{cases}(7+a^2)^{-1},&a\le0,\\ .25,&a\ge0,\end{cases}
 \qquad p(a)\ge.46\quad(a\ge4).
\]
The interval bounds below hold (all rows use the bin
\eqref{tr:bins}):
\begin{equation}\label{sc:interval-table}
\begin{array}{c|cccccc}
 &b&\mathsf L\le&\mathsf D\le&K_0&J_0\le&\ell_1\\\hline
 O &[0,.09]&.60&.295(7+a^2)&[.37,.56]&.72&.13\\
 P_1 &[.083,.15]&1.01&4.3&[.29,.46]&.72&.13\\
 R_1 &[.14,.23]&1.19&3.00&[.29,.371]&.68&.13\\
 U_1 &[.22,.342]&1.19&1.72&[.29,.398]&.61&.123\\
 V_1 &[.337,.45]&.83&.87&[.31,.4]&.50&.060\\
 T_1 &[.445,.5]&.20&.640&[.32,.4]&.48&.022
\end{array}
\end{equation}
where $\ell_1$ is the positive number in the last column, and
\begin{equation}\label{sc:local-surplus}
 Q-
 \begin{pmatrix}
 A_*b^2-\eta_*b+\nu_*/4+l^2/\alpha&\nu_*/2-\eta_*b\\
 \nu_*/2-\eta_*b&\nu_*
 \end{pmatrix}
 -\mathsf D
 \begin{pmatrix}\mathsf E\\cb\end{pmatrix}
 \begin{pmatrix}\mathsf E&cb\end{pmatrix}
 \succeq\begin{pmatrix}\ell_1&0\\0&.75\end{pmatrix},
\end{equation}
with $Q=\left(\begin{smallmatrix}.43&.13\\.13&2.72\end{smallmatrix}\right)$
as in Proposition~\ref{mat:scalar-pair}. Finally, writing
\begin{equation}\label{sc:derivative-functions}
 z=\frac{l'}{l},\qquad h_*=1+3b+kz,\qquad\chi=\frac{l}{d},
\end{equation}
one has
\begin{equation}\label{sc:derivative-table}
\begin{array}{c|ccc}
 &z&h_*&\chi\le\\\hline
 a\le-1&[-.08,.70]&[.8,1.77]&.070\\
 a\ge-1&[-.475,0]&[.8,1.77]&.097
\end{array}
\end{equation}
At an endpoint shared by two intervals, either applicable row may be
used.
\end{lemma}

The first line of the lemma gives a lower bound for the mean of $p$ from
only a second moment, and the matrix surplus \eqref{sc:local-surplus}
then leaves room to absorb the secant errors in the second proposition.

\subsection{Absorption using a second moment}\label{sc:absorption-proof}

\begin{proof}[Proof of Proposition~\ref{mat:scalar-absorption}]
By $0<b<1/2$ and \eqref{sc:p-h-bounds},
\begin{equation}\label{sc:y-cost}
 s\Bigl((b-\beta)^2+\frac{\alpha h_0^2}{4(1-\epsilon-\kappa)p}\Bigr)
 \le s\Bigl(.28^2+\frac{.075\cdot.67^2}{4\cdot.36}\Bigr)<.111.
\end{equation}
It remains to bound the terms involving $T-\log S$ while keeping a
negative multiple of the weight $\omega$ in \eqref{mat:rigidity-weight}.

Fix a unit eigenvector $v$ of $B$ with eigenvalue $\lambda\ge0$. The
corresponding eigenvalues of $S,T,H,\rho$ are
\[
 S=(1/s+\alpha\lambda)^{-1},\quad T=S-1,\quad
 H=\alpha S(\lambda-1)^2,\quad\rho=\epsilon+\kappa S/s.
\]
Let $\nu_v$ be the probability measure obtained by averaging the
spectral measure of $A$ at $v$. The first-chaos decomposition of $A$
gives
\begin{equation}\label{sc:spectral-moments}
 \int a\,\dd\nu_v(a)=0,\qquad
 \int a^2\,\dd\nu_v(a)=\lambda+r,\qquad
 r=v^{\mathsf T}\E(Y^2)v\ge0,
\end{equation}
using orthogonality of the Gaussian chaoses, which gives
$\E A^2=B+\E Y^2$ including vanishing of both matrix cross terms.

For $0<u\le.13$ define
$n_u=u^{3/2}/2$ and $d_u=u^3/(16(.46-u))$. We claim the quadratic
minorant
\begin{equation}\label{sc:p-minorant}
 p(a)\ge u+n_ua-d_ua^2\qquad(a\in\R).
\end{equation}
For $a\le0$ put $x=-\sqrt u\,a$. If $x\ge2$ the right side is
nonpositive; if $0\le x\le2$ it is enough to use
$p(a)\ge u/(x^2+7u)$ together with
\[
 1-(1-x/2)(x^2+7u)\ge\frac{x(x-1)^2}{2}+.045(2-x)\ge0.
\]
For $0\le a\le4$ the right side of \eqref{sc:p-minorant} is at most
$.13+2(.13)^{3/2}<.25$; for $a\ge4$ its maximum over all real $a$ is
$u+n_u^2/(4d_u)=.46$. So \eqref{sc:p-minorant} follows from
\eqref{sc:p-h-bounds} in all three ranges.

Choose $u=\min(.13,1.28/\sqrt\lambda)$, with $u=.13$ when $\lambda=0$.
Since $\lambda u^2\le1.28^2$, integrating the minorant against
\eqref{sc:spectral-moments} gives
\begin{equation}\label{sc:p-mean}
 \int p\,\dd\nu_v\ge u-d_u(\lambda+r)\ge.689u-d_ur,
\end{equation}
where the second inequality uses
$1-1.28^2/[16(.46-.13)]>.689$. We also have
\begin{equation}\label{sc:rho-loss}
 H\rho d_u\le.027.
\end{equation}
To check it, note $\rho\le.64$ and $H\le\lambda$ for $\lambda\ge1$,
while $H\le\alpha s$ for $\lambda\le1$. In the former case
$H\rho d_u\le\frac{.64\cdot1.28^2\cdot.13}{16(.46-.13)}<.02582$; in the
latter, $\alpha s\cdot.64\cdot.13^3/[16(.46-.13)]<.027$.

For completeness, we verify the elementary logarithmic bound needed to
compare $T-\log S$ with the positive term in \eqref{sc:p-mean}. Set
\[
 F_0(S)=\frac{S(S-1-\log S)}{(1-S)^2},\qquad F_0(1)=\tfrac12.
\]
For $0<S\le s$ and $\lambda=(1/S-1/s)/\alpha$ we claim
\begin{equation}\label{sc:log-bound}
 \frac{\alpha F_0(S)}{\epsilon+\kappa S/s}
 \le\min\Bigl(.089,\frac{.88}{\sqrt\lambda}\Bigr)\le.688u,
\end{equation}
the middle bound being $+\infty$ at $\lambda=0$. First,
$F_0(S)=\int_0^1\frac{Sx}{1-x+xS}\dd x$ is concave. Its tangent at
$S=.28$ has intercept below $.16$ and slope below $.52$, so
$F_0(S)\le.16+.52S$. (Both strict inequalities can be checked from
$-\log(.28)=2\sum_{j\ge0}(9/16)^{2j+1}/(2j+1)$ with twelve terms and the
geometric remainder.) Coefficient comparison now gives
$.075(.16+.52S)<.089(.15+.45325S)$, proving the first bound of the
minimum.

For the second bound we use
\begin{equation}\label{sc:F0-sqrt}
 \frac{F_0(S)}{\sqrt S}\le.478+1.4S\qquad(S>0).
\end{equation}
Here is a direct verification, including $S=1$ by continuity. For $x>0$
let
\[
 J(x)=\frac{(.478+1.4x^2)(1-x^2)^2}{x}-(x^2-1-2\log x).
\]
Then
\[
 x^2J'(x)=P(x):=-.478+2x+.444x^2-2x^3-6.966x^4+7x^6.
\]
By Descartes' rule $P$ has at most three positive roots counting
multiplicity; the signs at $.24,.26,.6$ together with $P(1)=0$ and
$P'(1)>0$ give exactly three: one in $(.24,.26)$, one in $(.26,.6)$,
and $1$. So the only local minima of $J$ are at the first root and at
$1$. We have $J(1)=0$ and $J(.25)>.14$ (the latter from
$\log2<.694$); on $[.24,.26]$, $|P(.25)|<.01$ and $|P'|<4$ give
$|J'|<2$, so the first minimum is positive. Also $J(x)\to+\infty$ at
both ends of $(0,\infty)$. Hence $J\ge0$, which is \eqref{sc:F0-sqrt}
off $x=1$.

Using $\alpha\lambda=1/S-1/s$, inequality \eqref{sc:F0-sqrt} gives
\[
 \frac{\alpha F_0(S)\sqrt\lambda}{\rho}
 \le\sqrt\alpha\sqrt{1-S/s}\,\frac{.478+1.4S}{.15+.45325S}
 \le\frac{\sqrt{.075}\cdot.478}{.15}<.88,
\]
the last ratio decreasing in $S$ because
$1.4\cdot.15<.478\cdot.45325$. Finally $.089<.688\cdot.13$ and
$.88<.688\cdot1.28$ prove the last inequality of \eqref{sc:log-bound}.

For $\lambda\neq1$,
$\dfrac{T-\log S}{H\rho}=\dfrac{\alpha F_0(S)}{\rho}$. Combining
\eqref{sc:p-mean}--\eqref{sc:log-bound} and treating $\lambda=1$ by
continuity gives
\[
 T-\log S-H\rho\int p\,\dd\nu_v\le-.001H\rho u+.027r.
\]
By definition $H\rho u=\omega(\lambda)$. Summing over an orthonormal
eigenbasis of $B$ and adding \eqref{sc:y-cost} evaluates
$\tau(H\rho p(A))$ without requiring $p(A)$ to commute with $B$, and
integrates \eqref{sc:y-cost} against the positive semidefinite $Y^2$
without a commutation assumption. Since $\sum_vr=\|Y\|^2$ the result is
\[
 (.027+.111)\|Y\|^2-.001\sum_i\omega(\lambda_i)
 =\frac{69}{500}\|Y\|^2-\frac1{1000}\sum_i\omega(\lambda_i),
\]
as asserted.
\end{proof}

\subsection{A quadratic estimate for two endpoints}\label{sc:pair-proof}

\begin{proof}[Proof of Proposition~\ref{mat:scalar-pair}]
Fix two endpoints $a_-\le a_+$ and write
\[
 \delta=\log q(a_+)-\log q(a_-),\qquad m_*=m_0(a_-).
\]
When the endpoints coincide all secant increments vanish and
\eqref{sc:local-surplus} directly proves the assertion, so suppose
$\delta>0$.

\emph{One endpoint.} Use coordinates $(G,h)$ with $h=g-G/2$. With $Z=0$
the left side of the endpoint form in Proposition~\ref{mat:scalar-pair}
is exactly the two matrices subtracted from $Q$ in
\eqref{sc:local-surplus}. For an increment vector $(x,y,w)$ define
\begin{align}
 \mathcal N_j(x,y,w)^2={}&A_*x^2+\frac{y^2}{\alpha}+\mathsf D_jw^2\notag\\
 &+\frac{((A_*b_j-\eta_*/2)x+\mathsf L_jy+K_{0j}w)^2}{\ell_{1j}}
 +\frac{(-\eta_*x+J_{0j}w)^2}{.75}.
 \label{sc:increment-norm}
\end{align}
Completing squares against the surplus $\ell_{1j}G^2+.75h^2$ in
\eqref{sc:local-surplus} shows that the full endpoint form is bounded
above by
\begin{equation}\label{sc:one-endpoint}
 Q(G,h)+Z^2\mathcal N_j(\Delta k,\Delta b,\psi_j)^2.
\end{equation}
Indeed, the two coefficients of $2GZ$ and $2hZ$ are precisely the
numerators of the two squared terms in \eqref{sc:increment-norm}, and the
first line of that formula is the coefficient of $Z^2$ before the
completion of squares.

\emph{Secant as averaged derivative.} A secant slope is the uniform
average of the derivative over $[a_-,a_+]$. With $\xi=\log q(a)$,
differentiation gives the vector
\begin{equation}\label{sc:derivative-vector}
 v_j(a)=\chi(a)(1,z(a),W_j(a)),\qquad
 W_j(a)=h_*(a)-2b_j-2\mathsf L_jz(a).
\end{equation}
More explicitly, $(\Delta k,\Delta b,\psi_j)$ equals the uniform average
in $a$ of $\int_{\log q(a_j)}^{\log q(a)}v_j(q^{-1}(e^\xi))\dd\xi$,
which uses $b'=l$, $l'=zl$, $\dd(\log q)/\dd a=d$, and
$f''=l(1+3b+kz)=lh_*$. Put $F_j(a)=\mathcal N_j(v_j(a))^2$.

\emph{Pointwise bounds.} Lemma~\ref{sc:bounds} gives: if the endpoint
$a_j$ lies in $R_1,U_1,V_1,T_1$ respectively, then
\begin{equation}\label{sc:F-simple}
 F_j(a)\le.295,\quad.23,\quad.23,\quad.39\qquad(\text{all }a),
\end{equation}
with the $R_1$ bound improving to $.20$ for $a\le-1$. If $a_j\in P_1$
the bounds are $.42^2$ for $a\le-1$ and $.61^2$ for $a\ge-1$. If
$a_j\in O$, write $t_j=-a_j$; then
\begin{equation}\label{sc:F-negative}
 \frac{F_j(a)}{\chi(a)^2}\le
 \begin{cases}20+1.04(7+t_j^2),&a\le-1,\\
 12+1.63(7+t_j^2),&a\ge-1.\end{cases}
\end{equation}
The finite arithmetic behind these bounds is in the following table; its
three entries bound, in order,
$|W_j|$, $|A_*b_j-\eta_*/2+\mathsf L_jz+K_{0j}W_j|$, and
$|-\eta_*+J_{0j}W_j|$:
\begin{equation}\label{sc:substitution-table}
\begin{array}{c|cc}
 &a\le-1&a\ge-1\\\hline
 O&(1.867,.52,1.314)&(2.34,.55,1.155)\\
 P_1&(1.766,.49,1.813)&(2.564,.433,1.155)\\
 R_1&(1.681,.664,2.057)&(2.621,.420,1.155)\\
 U_1&(1.55,.847,2.101)&(2.461,.615,1.155)\\
 V_1&(1.263,.912,1.787)&(1.885,.770,1.205)\\
 T_1&(.913,.869,1.386)&(1.071,.840,1.251)
\end{array}
\end{equation}
To verify it, substitute the intervals of \eqref{sc:interval-table} and
\eqref{sc:derivative-table}; each expression is affine in each variable
separately, so its extremum is at a box corner. For the middle
expression, first write it as
\[
 (A_*-2K_{0j})b_j-\frac{\eta_*}2+K_{0j}h_*+(1-2K_{0j})\mathsf L_jz.
\]
Substituting these three bounds into \eqref{sc:increment-norm}, with the
bounds for $\mathsf D_j,\ell_{1j},z,\chi$, proves
\eqref{sc:F-simple}--\eqref{sc:F-negative}. All are finite rational
comparisons.

\emph{Comparing derivative cost with the separation surplus.} Let $u_-$
be the mean of $(\log q(a)-\log q(a_-))/\delta$ under the uniform
probability measure on $[a_-,a_+]$, and $u_+=1-u_-$. In the $\xi$
coordinate this measure has density proportional to $D_0=1/d$.
Lemma~\ref{tr:q-properties} gives
$0\le\dd\log D_0/\dd\xi=(1-qd)/d^2\le1$. The density $D_0$ is
increasing whereas $D_0e^{-\xi}$ is decreasing; the covariance of an
increasing function with an increasing function is nonnegative, and with
a decreasing function is nonpositive. Applying this to the coordinate
$\xi$ compares its mean first with constant density and then with
density proportional to $e^\xi$, giving
$u_-\le\tfrac12+\tfrac12\coth(\delta/2)-1/\delta$ and hence
\begin{equation}\label{sc:mean-distance}
 \tfrac12\le u_-\le\tfrac12+\frac\delta{12},\qquad
 u_-^2+u_+^2\le\tfrac12\Bigl(1+\frac{\delta^2}{36}\Bigr).
\end{equation}
The upper bound uses $\coth x-1/x\le x/3$, obtained by comparing power
series of $x\cosh x$ and $(1+x^2/3)\sinh x$.

Suppose $P_-,P_+$ satisfy: on every partial $\xi$-interval starting at
the indicated endpoint, the mean of $\mathcal N_j(v_j)$ is at most
$\sqrt{P_j}$ (pointwise bounds for $F_j$ suffice). Minkowski's
inequality, followed by the uniform average in $a$, gives
\[
 \mathcal N_-(\Delta k,\Delta b,\psi_-)\le\sqrt{P_-}\,\delta u_-,\qquad
 \mathcal N_+(\Delta k,\Delta b,\psi_+)\le\sqrt{P_+}\,\delta u_+.
\]
Since $u_-\ge u_+\ge0$,
$P_-u_-^2+P_+u_+^2\le\max(P_-,\tfrac{P_-+P_+}{2})(u_-^2+u_+^2)$.
Averaging \eqref{sc:one-endpoint} and applying \eqref{sc:mean-distance}
proves the desired estimate whenever
\begin{equation}\label{sc:threshold}
 \max\Bigl(P_-,\frac{P_-+P_+}{2}\Bigr)\le
 \Theta_{m_*}(\delta):=\frac{1+m_*}{4}\,
 \frac{1+\delta^2/30+\delta^4/1680}{1+\delta^2/36},
\end{equation}
the resulting coefficient of $Z^2$ being at most
$\delta^2(1+\delta^2/36)\Theta_{m_*}(\delta)/4=r_{-+}\delta^2$.

\emph{The threshold.} The function $\Theta_m$ increases in $\delta\ge0$:
differentiating with respect to $y=\delta^2$ leaves the positive
numerator $1/180+2y/1680+y^2/(1680\cdot36)$. We also use: if a smaller
bound holds on an initial segment of logarithmic length $L$, then in a
partial interval of length $h\le\delta$ its fraction is at least
$\min(1,L/h)\ge L/\delta$ when $\delta\ge L$.

If $a_-\in R_1$ take $P_-\le.295$, $P_+\le.39$; then
$(P_-+P_+)/2\le.3425<\Theta_{.38}(0)=.345$. If $a_-\in U_1$ or $V_1$,
$P_-\le.23$, $P_+\le.39$, and $\Theta_{.68}(0)=.42$. If $a_-\in T_1$,
both bounds are $.39<.42$.

Now $a_-\in P_1$. If $a_+\le-1$ both endpoint costs are at most
$.20<\Theta_{.19}(0)$, using the $R_1$ row at $a_+=-1$. Otherwise the
part of the logarithmic interval below $-1$ has length at least $1.62$
(from $q(-1)>.287$, $q(-2)<.056$), so every partial interval from the
lower endpoint has mean norm at most $.61-\frac{(.61-.42)1.62}{\delta}$,
i.e.\ $P_-=(.61-.3078/\delta)^2$. If $a_+\in T_1$ then $\delta>4$
because $q(4)>4$. For $\delta\le4$, $P_+\le.23$ and
$P_-\le.284143<\Theta_{.19}(0)=.2975$. For $4\le\delta\le5$,
$P_-<.300787$, $P_+\le.39$, and
$\max(.300787,(.300787+.39)/2)<.345394<\Theta_{.19}(4)$. For
$\delta\ge5$, $P_-\le.61^2=.3721$ and
$\max(.3721,(.3721+.39)/2)=.38105<\Theta_{.19}(5)$. This covers $P_1$.

Finally $a_-\in O$. For any endpoint $a_j\in O$ and any argument
$a\le-1$ between the endpoints,
\[
 t_j^2\le a^2+2\delta,\qquad|a|\chi(a)\le.09,
\]
the first by integrating $d\ge-a$ on the negative half-line, the second
from $l\le.075\cdot1.19<.09$ and $d\ge|a|$. With $\chi\le.07$ in
\eqref{sc:F-negative},
\begin{equation}\label{sc:O-low-cost}
 F_j(a)\le.144+.011\delta.
\end{equation}
If the whole interval is at or below $-1$, take
$P_-=.144+.011\delta$ and $P_+\le\max(P_-,.20)$ (the latter also covers
$P_1,R_1$ upper endpoints). For $0\le\delta\le4$ both are below $.25$.
For $\delta\ge4$ use
\begin{equation}\label{sc:theta-tail}
 \Theta_0(\delta)\ge.25+.00260\delta^2,
\end{equation}
which follows on clearing denominators and using $\delta^2\ge16$; the
quadratic $.106-.011\delta+.00260\delta^2$ is strictly positive, so
\eqref{sc:threshold} holds.

If $a_+>-1$, the first part of the logarithmic interval has length at
least $4$ (from $q(-3)<.005$, $q(-1)>.287$). For $a\ge-1$ integration
gives $t_-^2\le1+2\delta$; with $\chi\le.097$ in
\eqref{sc:F-negative}, $F_-(a)\le.236+.031\delta$. Jensen therefore
permits the uniform bound
\[
 P_-=.236+.031\delta-\frac4\delta(.092+.020\delta)
 =.156+.031\delta-.368/\delta,
\]
and the same bound holds on intervals below $-1$ by
\eqref{sc:O-low-cost} since $\delta\ge4$. By \eqref{sc:theta-tail},
\[
 \Theta_0(\delta)-P_-\ge.094-.031\delta+.00260\delta^2+.368/\delta>0,
\]
the quadratic having negative discriminant and positive leading
coefficient. If $a_+\notin T_1$ then $P_+\le.23<\Theta_0(\delta)$; if
$a_+\in T_1$ then $\delta>6.5$ (from $q(4)>4$, $q(-3)<.005$) and
$P_+\le.39<\Theta_0(6.5)<\Theta_0(\delta)$. This proves
\eqref{sc:threshold} in every case and completes the proof.
\end{proof}

\section{Approximation of log-concave densities}\label{geo:section}

Theorem~\ref{mat:entropy-smooth} assumes a potential with positive,
bounded Hessian. We now remove those restrictions while preserving
entropy, covariance, and first moments. The same approximation later
studies equality, by applying the strict remainder to smooth densities
whose entropy gaps tend to zero.

\subsection{Convex approximation with entropy convergence}

We use extended-valued convex potentials: $V:\R^m\to(-\infty,+\infty]$
is \emph{proper} if it is finite somewhere and never $-\infty$. The
convention $e^{-\infty}=0$ allows bounded supports. The next lemma
gives the common integrable bound used for the approximation, including
its entropy integrand.

\begin{lemma}\label{geo:coercivity}
Let $V:\R^m\to(-\infty,+\infty]$ be proper, lower semicontinuous, and
convex, with $0<\int_{\R^m}e^{-V(x)}\dd x<\infty$. Then there are $a>0$
and $b\ge0$ with
\begin{equation}\label{geo:linear-coercivity}
 V(x)\ge a|x|-b\qquad(x\in\R^m).
\end{equation}
In particular $e^{-V}$ has moments of every order and
$\int|V|e^{-V}<\infty$, with $|V|e^{-V}$ read as zero where $V=+\infty$.
\end{lemma}
\begin{proof}
The convex set $\{V<\infty\}$ has positive measure and hence nonempty
interior (otherwise its affine hull would be a proper affine subspace).
Choose a simplex with all vertices in this set and with nonempty
interior; convexity bounds $V$ above by the largest vertex value
throughout the simplex, so some ball $B(x_0,r)$ lies in a sublevel set
$\{V\le M\}$, $M\ge V(x_0)$.

The closed convex set $A=\{V\le M+1\}$ has finite volume, since
$|A|\le e^{M+1}\int e^{-V}$. It is bounded: the convex hull of
$B(x_0,r)$ and a point $z\in A$ contains a cone with base an
$(m-1)$-disk of radius $r$ through $x_0$, perpendicular to $z-x_0$, and
height $|z-x_0|$, whose volume tends to infinity with $|z-x_0|$ (in
dimension one, use the length of the interval from $x_0$ to $z$).

Choose $R>r$ so $A\subset B(x_0,R)$. If $d=|x-x_0|\ge R$ and
$V(x)<\infty$, put $y=x_0+(R/d)(x-x_0)$; then $y\notin A$ and convexity
gives
\[
 M+1<V(y)\le\Bigl(1-\frac Rd\Bigr)V(x_0)+\frac RdV(x),
\]
so $V(x)>V(x_0)+d/R$. The inequality is automatic if $V(x)=+\infty$. On
the compact ball $\overline B(x_0,R)$, lower semicontinuity and
properness give a finite lower bound (a sequence with values tending to
$-\infty$ would have a convergent subsequence and violate lower
semicontinuity). Combining the two bounds, after enlarging $b$, gives
\eqref{geo:linear-coercivity}.

For moments, $e^{-V(x)}\le e^be^{-a|x|}$. For entropy, the elementary
bound
\begin{equation}\label{geo:entropy-domination}
 |u|e^{-u}\le C_be^{-u/2}\qquad(u\ge-b)
\end{equation}
holds because $|u|e^{-u/2}$ is bounded on $[-b,\infty)$; substitute
$u=V(x)$ and use \eqref{geo:linear-coercivity}.
\end{proof}

\begin{lemma}\label{geo:approximation}
Let $f=e^{-V}$ be a log-concave probability density on $\R^m$,
represented by a proper lower semicontinuous convex potential $V$. There
are smooth convex potentials $V_j$ and probability densities
$f_j=Z_j^{-1}e^{-V_j}$, $Z_j=\int_{\R^m}e^{-V_j}$, such that
\begin{equation}\label{geo:hessian-bounds}
 \frac2j\Id\preceq D^2V_j\preceq\Bigl(j+\frac2j\Bigr)\Id
 \qquad(j\ge1),
\end{equation}
and
\begin{equation}\label{geo:approximation-limits}
 Z_j\to1,\quad\norm{f_j-f}_{L^1}\to0,\quad
 h(f_j)\to h(f),\quad\Cov(f_j)\to\Cov(f),
\end{equation}
the covariance limit entrywise, with the means of $f_j$ converging to
the mean of $f$; all these quantities are finite.
\end{lemma}
\begin{proof}
Fix $a,b$ from Lemma~\ref{geo:coercivity}. For a positive integer $j$
define the Moreau envelope \cite[Section~7]{Moreau1965}
\begin{equation}\label{geo:moreau}
 Q_j(x)=\inf_{y\in\R^m}\Bigl\{V(y)+\frac j2|x-y|^2\Bigr\}.
\end{equation}
The expression in braces is lower semicontinuous, strictly convex on its
effective domain, and tends to infinity as $|y|\to\infty$; it has a
unique minimizer $p_j(x)$ and $Q_j$ is finite everywhere. Moreover
\begin{equation}\label{geo:envelope-lower}
 Q_j(x)\ge a|x|-b-\frac{a^2}{2j},
\end{equation}
because $V(y)\ge a|x|-a|x-y|-b$ and minimizing $jr^2/2-ar$ over $r\ge0$
gives the bound.

For completeness, the regularity of $Q_j$ follows from the minimization.
The optimality condition is $j(x-p_j(x))\in\partial V(p_j(x))$, where
$\partial V$ is the convex subdifferential: compare the minimum at
$p=p_j(x)$ with $p+s(z-p)$, use convexity to bound $V(p+s(z-p))$ above,
divide by $s>0$, and let $s\downarrow0$, giving
$V(z)\ge V(p)+j\langle x-p,z-p\rangle$. Adding the two subgradient
inequalities at $p_j(x)$ and $p_j(x')$ gives
$\langle x-x',p_j(x)-p_j(x')\rangle\ge|p_j(x)-p_j(x')|^2$, so both
$p_j$ and $x-p_j(x)$ are $1$-Lipschitz. Evaluating the infimum at
$p_j(x)$ and, in reverse, at $p_j(x+h)$ shows $\nabla Q_j(x)=j(x-p_j(x))$
(an upper error $j|h|^2/2$ from the first comparison, a lower error of
order $j|h|^2$ from the second and the Lipschitz estimate). Thus $Q_j$
is continuously differentiable with $j$-Lipschitz gradient. It is
convex, since the minimized function is jointly convex in $(x,y)$.

We next check convergence before smoothing. The functions $Q_j$
increase with $j$, and $Q_j(x)\le V(x)$ where $V(x)$ is finite. If
$Q_j(x)$ stays bounded above as $j\to\infty$, the identity at the
minimizer and $V\ge-b$ give $|x-p_j(x)|^2\le2(Q_j(x)+b)/j\to0$, and
lower semicontinuity gives
$V(x)\le\liminf_jV(p_j(x))\le\lim_jQ_j(x)$. Hence
\begin{equation}\label{geo:envelope-limit}
 Q_j(x)\to V(x)\quad\text{for every }x,
\end{equation}
including the value $+\infty$.

Choose a nonnegative even smooth function $\eta$ supported in the unit
ball with integral one, and set
$\eta_\varepsilon(x)=\varepsilon^{-m}\eta(x/\varepsilon)$. Define
\begin{equation}\label{geo:smooth-potentials}
 V_j=Q_j*\eta_{1/j}+\frac{|x|^2}j.
\end{equation}
Convolution preserves convexity and the Lipschitz gradient bound, so
\eqref{geo:hessian-bounds} holds. Evenness gives $\int z\eta(z)\dd z=0$,
and Jensen plus \eqref{geo:envelope-lower} gives the common bound
\begin{equation}\label{geo:common-tail}
 V_j(x)\ge Q_j(x)\ge a|x|-B,\qquad B=b+\frac{a^2}2.
\end{equation}

Smoothing at this scale does not affect the limit. The $j$-Lipschitz
gradient bound gives
$Q_j(x-z)\le Q_j(x)-\langle\nabla Q_j(x),z\rangle+\frac j2|z|^2$;
integrating against $\eta_{1/j}$ cancels the linear term, so
\[
 0\le(Q_j*\eta_{1/j})(x)-Q_j(x)\le\frac1{2j}\qquad(x\in\R^m),
\]
and with \eqref{geo:envelope-limit} this proves $V_j(x)\to V(x)$
everywhere.

Now \eqref{geo:common-tail} dominates $(1+|x|^2)e^{-V_j(x)}$ by an
integrable function independent of $j$, and
\eqref{geo:entropy-domination} with $B$ in place of $b$ gives
$|V_j(x)|e^{-V_j(x)}\le Ce^{-a|x|/2}$. At points where $V=+\infty$ both
$e^{-V_j}$ and $V_je^{-V_j}$ tend to zero. Dominated convergence
therefore gives convergence of the normalizing constants, first and
second moments, and the integrals $\int V_je^{-V_j}$. Finally
\[
 h(f_j)=\frac1{Z_j}\int V_je^{-V_j}+\log Z_j,
\]
which proves all of \eqref{geo:approximation-limits}.
\end{proof}

\begin{proof}[Proof of the inequality in Theorem~\ref{intro:entropy}]
Apply Theorem~\ref{mat:entropy-smooth} to the normalized potentials
$V_j+\log Z_j$ of Lemma~\ref{geo:approximation}. Their Hessians satisfy
\eqref{geo:hessian-bounds}, so
\[
 h(f_j)\ge m+\frac12\log\det\Cov(f_j).
\]
The covariance of $f$ is positive definite: zero variance in a nonzero
direction would concentrate this Lebesgue density on an affine
hyperplane. Thus \eqref{geo:approximation-limits} lets us pass to the
limit on both sides, proving the inequality.
\end{proof}

\section{From vanishing slack to exponential products}\label{sec:rigidity}

Write $\Dgap(f)=h(f)-m-\tfrac12\log\det\Cov(f)$ for the nonnegative
entropy gap. Theorem~\ref{mat:entropy-smooth} controls the nonlinear part
of the normalized matrix field and the sum of squares of its linear
coefficients. We now show that this control determines every equality
case. The structural step is local: the first two nonconstant terms of
$q(A(x))$, for a linear symmetric field $A$, force its coefficients to
commute whenever $q(A)$ is a Jacobian.

\subsection{Compatibility of a nonlinear Jacobian}

\begin{lemma}[Rigidity of a linear matrix field]\label{lem:linear-jacobian}
Let $m\ge1$ and let $X_1,\dots,X_m$ be real symmetric $m\times m$
matrices with
\[
 \sum_{r=1}^mX_r^2=I,\qquad A(x)=\sum_{r=1}^m x_rX_r.
\]
Suppose $F\in W^{1,2}_{\mathrm{loc}}(\R^m;\R^m)$ has weak Jacobian
$DF=q(A)$ almost everywhere. Then there are an orthogonal matrix $U$,
signs $\sigma_1,\dots,\sigma_m\in\{-1,1\}$, and $w\in\R^m$ with
\begin{equation}\label{eq:rigid-map}
 U^{\mathsf T}F(Uy)=\bigl(\sigma_it(\sigma_iy_i)\bigr)_{i=1}^m+w
 \qquad\text{for almost every }y\in\R^m.
\end{equation}
In particular $F_\#\gamma_m$ is an affine image of the product of $m$
mean-one exponentials, with covariance $I$.
\end{lemma}
\begin{proof}
Write $e_1,\dots,e_m$ for the standard basis and $J=q(A)$.
Commutation of distributional derivatives of $F$ gives
\begin{equation}\label{eq:jacobian-compatibility}
 (\partial_rJ)e_s=(\partial_sJ)e_r\qquad(1\le r,s\le m).
\end{equation}
The function $q$ is real analytic near $0$, and $A(0)=0$, so its
convergent power series, in matrix multiplication, gives
\[
 J(x)=q(0)I+q'(0)A(x)+\tfrac12q''(0)A(x)^2+O(|x|^3)
\]
near zero, with differentiated remainder $O(|x|^2)$. The identity
$q'(a)=q(a)(q(a)-a)$ gives
\[
 q(0)=\sqrt{\frac2\pi},\qquad q'(0)=\frac2\pi>0,\qquad
 q''(0)=\sqrt{\frac2\pi}\Bigl(\frac4\pi-1\Bigr)>0.
\]
Thus \eqref{eq:jacobian-compatibility}, a pointwise identity between
analytic functions near zero, has constant term
\begin{equation}\label{eq:coefficient-symmetry}
 X_re_s=X_se_r.
\end{equation}
Equivalently $A(x)z=A(z)x$ for all $x,z\in\R^m$; in particular
$A(x)e_s=X_sx$. Its degree-one term is
\[
 (X_rA(x)+A(x)X_r)e_s=(X_sA(x)+A(x)X_s)e_r.
\]
Terms with leftmost factor $A(x)$ cancel by \eqref{eq:coefficient-symmetry},
leaving $(X_rX_s-X_sX_r)x=0$ near zero, hence
\begin{equation}\label{eq:coefficients-commute}
 X_rX_s=X_sX_r\qquad(1\le r,s\le m).
\end{equation}

Commuting real symmetric matrices admit a common orthonormal eigenbasis.
Let $U$ have those basis vectors as columns, and rotate input and output.
Then
\[
 \widetilde A(y)=U^{\mathsf T}A(Uy)U=\sum_{i=1}^m y_i\widetilde X_i,
 \qquad\widetilde X_i=U^{\mathsf T}\Bigl(\sum_{r=1}^mU_{ri}X_r\Bigr)U
\]
has diagonal coefficients. Orthogonality gives $\sum_i\widetilde X_i^2=I$,
and $A(x)z=A(z)x$ gives $\widetilde X_ie_j=\widetilde X_je_i$; for
$i\neq j$ the two vectors are multiples of different basis vectors, so
both vanish. Hence only the $i$th diagonal entry of $\widetilde X_i$ is
nonzero, and the sum-of-squares identity makes it a sign:
\[
 \widetilde X_i=\sigma_ie_ie_i^{\mathsf T},\qquad
 \widetilde A(y)=\operatorname{diag}(\sigma_1y_1,\dots,\sigma_my_m).
\]
Then
$D_y(U^{\mathsf T}F(Uy))=\operatorname{diag}(q(\sigma_1y_1),\dots,q(\sigma_my_m))$,
which is also the Jacobian of
$y\mapsto(\sigma_it(\sigma_iy_i))_{i=1}^m$ since $t'=q$ and $\sigma_i^2=1$.
The difference has zero weak gradient on the connected set $\R^m$ and is
therefore almost everywhere constant, proving \eqref{eq:rigid-map}.

For independent standard Gaussian $G_i$, the variables $t(\sigma_iG_i)$
are independent mean-one exponentials, with variance one, and signs and
orthogonal maps preserve covariance $I$.
\end{proof}

\subsection{Compactness and the proof of the equality classification}

We first record the functional-calculus estimate that turns convergence
of the normalized fields into convergence of transport Jacobians; the
Frobenius norm is essential to the elementary argument.

\begin{lemma}\label{lem:spectral-lipschitz}
For real symmetric matrices $M,N\in\R^{m\times m}$,
\begin{equation}\label{eq:spectral-lipschitz}
 |q(M)-q(N)|_{\mathrm F}\le|M-N|_{\mathrm F}.
\end{equation}
\end{lemma}
\begin{proof}
Let $(u_i)$ and $(v_k)$ be orthonormal eigenbases of $M$ and $N$ with
eigenvalues $\lambda_i$ and $\mu_k$. Then
$u_i^{\mathsf T}(q(M)-q(N))v_k=(q(\lambda_i)-q(\mu_k))u_i^{\mathsf T}v_k$,
while the corresponding entry of $M-N$ is
$(\lambda_i-\mu_k)u_i^{\mathsf T}v_k$. The scalar bound $0<q'<1$ bounds
each former entry in absolute value by the latter. Squaring and summing
over $i,k$ proves \eqref{eq:spectral-lipschitz}, the Frobenius norm being
unchanged by orthogonal changes of the row and column bases separately.
\end{proof}

\begin{proof}[Proof of the equality statement in Theorem~\ref{intro:entropy}]
Let $f$ be a log-concave probability density on $\R^m$ with $\Dgap(f)=0$.
Take the approximating densities $f_j$ of Lemma~\ref{geo:approximation},
with means $a_j,a$ and covariances $C_j,C$. That lemma gives
\begin{equation}\label{eq:equality-approximation}
 \norm{f_j-f}_{L^1(\R^m)}\to0,\quad a_j\to a,\quad C_j\to C,\quad
 \Dgap(f_j)\to0.
\end{equation}
Here $C$ is positive definite (as recalled in the introduction), so the
covariance convergence also gives convergence of log determinants.

For each $j$ apply Proposition~\ref{tr:normalization} to $f_j$. Let $P_j$
be its positive definite normalizing matrix and let $F_j$ transport
$\gamma_m$ to $(P_j)_\#(f_j(x)\dd x)$. In the notation of
Theorem~\ref{mat:entropy-smooth} set
\[
 J_j=DF_j,\quad A_j=q^{-1}(J_j),\quad X_{j,r}=\E[x_rA_j(x)],
\]
\[
 A_j(x)=\sum_{r=1}^m x_rX_{j,r}+Y_j(x),\qquad B_j=\sum_{r=1}^mX_{j,r}^2.
\]
The entropy gap is affine invariant, so Theorem~\ref{mat:entropy-smooth}
and \eqref{eq:equality-approximation} give
\begin{equation}\label{eq:vanishing-slack}
 \norm{Y_j}_{L^2(\gamma_m)}\to0,\qquad
 \sum_{i=1}^m\omega(\lambda_i(B_j))\to0.
\end{equation}
The function $\omega$ is continuous and nonnegative on $[0,\infty)$,
vanishes only at $1$, and tends to infinity at infinity; so for every
$\varepsilon>0$ it has a positive infimum on
$\{\lambda\ge0:|\lambda-1|\ge\varepsilon\}$. Hence every eigenvalue of
$B_j$ tends to $1$ and $B_j\to I$ in operator norm.

Since $\sum_r|X_{j,r}|_{\mathrm F}^2=\tr B_j$, the coefficients form a
bounded sequence in a finite-dimensional space. Pass to a subsequence
with $X_{j,r}\to X_r$ for every $r$. Each $X_r$ is symmetric, and
Gaussian orthogonality together with \eqref{eq:vanishing-slack} gives
\begin{equation}\label{eq:linear-field-limit}
 \sum_{r=1}^mX_r^2=I,\qquad
 A_j\to A(x):=\sum_{r=1}^mx_rX_r\quad\text{in }L^2(\gamma_m).
\end{equation}
Lemma~\ref{lem:spectral-lipschitz} then gives
\begin{equation}\label{eq:jacobian-limit}
 J_j=q(A_j)\to J:=q(A)\quad\text{in }L^2(\gamma_m).
\end{equation}

To obtain a limiting map, remove translations:
$\widehat F_j=F_j-\E F_j$. Each $\widehat F_j$ lies in
$H^1(\gamma_m;\R^m)$: its law has finite second moments and, for this
fixed $j$, its Jacobian obeys \eqref{tr:Jacobian-bounds} for the
transformed target. Gaussian Poincar\'e applied componentwise to the
centered difference gives
\begin{equation}\label{eq:map-cauchy}
 \norm{\widehat F_j-\widehat F_k}_{L^2(\gamma_m)}^2
 \le\norm{J_j-J_k}_{L^2(\gamma_m)}^2.
\end{equation}
Thus $\widehat F_j$ converges strongly in Gaussian $H^1$ to a centered
map $\widehat F$ with weak Jacobian $D\widehat F=J$: \eqref{eq:map-cauchy}
and \eqref{eq:jacobian-limit} give convergence of maps and first
derivatives, and passage to weak derivatives is valid on every bounded
set, where the Gaussian density has a positive lower bound. Hence
$\widehat F\in W^{1,2}_{\mathrm{loc}}(\R^m;\R^m)$, and
Lemma~\ref{lem:linear-jacobian} shows $\widehat F_\#\gamma_m$ is an affine
exponential product with covariance $I$.

It remains to return from normalized laws to $f$. By Cauchy--Schwarz the
strong $L^2(\gamma_m)$ convergence gives
\begin{equation}\label{eq:normalized-covariance-limit}
 D_j:=\Cov((\widehat F_j)_\#\gamma_m)=P_jC_jP_j^{\mathsf T}\to I.
\end{equation}
We can now bound the normalizing matrices uniformly, without uniform
Hessian bounds on the approximants. Choose $c,C_0,d,D>0$ with, for large
$j$,
\[
 cI\le C_j\le C_0I,\qquad dI\le D_j\le DI.
\]
Congruence by $P_j$ yields
\begin{equation}\label{eq:normalization-compactness}
 \frac d{C_0}I\le P_jP_j^{\mathsf T}\le\frac DcI,
\end{equation}
so $P_j$ and $P_j^{-1}$ are bounded. Pass to a further subsequence with
$P_j\to P$; the lower bound in \eqref{eq:normalization-compactness} makes
$P$ invertible.

The law of $\widehat F_j$ is the image of $f_j(x)\dd x$ under
$x\mapsto P_j(x-a_j)$. For bounded continuous $\psi$,
\begin{align*}
 &\Bigl|\int\psi(P_j(x-a_j))f_j(x)\dd x
 -\int\psi(P(x-a))f(x)\dd x\Bigr|\\
 &\le\norm{\psi}_\infty\norm{f_j-f}_{L^1}
 +\int\bigl|\psi(P_j(x-a_j))-\psi(P(x-a))\bigr|f(x)\dd x\to0,
\end{align*}
the first term by \eqref{eq:equality-approximation}, the second by
dominated convergence. Strong convergence also identifies the weak limit
of these laws as $\widehat F_\#\gamma_m$. So the image of $f(x)\dd x$
under $x\mapsto P(x-a)$ is the affine exponential product determined
above; since $P$ is invertible, $f$ itself has the asserted form up to
equality almost everywhere.

Conversely, the density of a product of $m$ independent mean-one
exponentials is log-concave, has entropy $m$, and covariance $I$; every
invertible affine image stays log-concave with the same entropy gap by
affine invariance. Thus every density in the stated class has
$\Dgap=0$.
\end{proof}

\section{The simplex inequality and its equality cases}\label{sec:geometry}

We transfer Theorem~\ref{intro:entropy} to convex bodies through the
exponential density on a cone. The construction appears in
\cite[Proposition~2(b)]{FradeliziMeyer2008}, its entropy reduction in
\cite[Lemma~5.2 and Proposition~5.2]{FradeliziMarinSola2026}, and the
moment formulas also in \cite[Lemma~2.5]{Klartag2018}. The computation
identifies equality in the body inequality with entropy equality in one
higher dimension; the support of the resulting exponential product then
determines the body.

\begin{proof}[Proof of Theorem~\ref{intro:simplex}]
Let $K\subset\R^n$ be a convex body. Translation preserves $L_K$ and
preserves being a simplex, so assume $K$ has centroid zero. Write
$\Sigma_K=\Cov(X)$ for $X$ uniform on $K$, and define
\begin{equation}\label{eq:body-cone}
 \mathcal C_K=\{(tx,t):x\in K,\ t\ge0\}\subset\R^{n+1}.
\end{equation}
Convexity of $K$ makes $\mathcal C_K$ convex. It is closed because $K$
is compact: a convergent sequence $(t_jx_j,t_j)$ with $t_j\to0$ has
$t_jx_j\to0$, and when the limiting height is positive one divides by
that height. The cone has nonempty interior and its only point at height
zero is the origin.

Define the log-concave density
\begin{equation}\label{eq:body-cone-density}
 f_K(y,t)=\frac{e^{-t}}{n!\,|K|}\,\mathbf1_{\mathcal C_K}(y,t).
\end{equation}
The change of variables $y=tx$ (Jacobian $t^n$ for $t>0$) gives
\[
 \int_{\mathcal C_K}e^{-t}\dd y\dd t
 =|K|\int_0^\infty t^ne^{-t}\dd t=n!\,|K|,
\]
so $f_K$ is a probability density. The same change of variables shows it
is the density of $(TX,T)$, where $T$ is independent of $X$ with density
$t^ne^{-t}/n!$ on $(0,\infty)$. Integration by parts gives
\[
 \E T=n+1,\qquad\E T^2=(n+1)(n+2),\qquad\Var(T)=n+1.
\]
Since $\E X=0$, independence makes the cross-covariance of $TX$ and $T$
vanish. Hence
\begin{align}
 h(f_K)&=n+1+\log(n!\,|K|),\label{eq:body-cone-entropy}\\
 \Cov(f_K)&=
 \begin{pmatrix}(n+1)(n+2)\Sigma_K&0\\0&n+1\end{pmatrix},
 \label{eq:body-cone-covariance}\\
 \det\Cov(f_K)&=(n+1)^{n+1}(n+2)^n\det\Sigma_K.
 \label{eq:body-cone-determinant}
\end{align}
Applying Theorem~\ref{intro:entropy} in dimension $n+1$ yields
\begin{equation}\label{eq:body-cone-bound}
 (n!)^2|K|^2\ge(n+1)^{n+1}(n+2)^n\det\Sigma_K,
\end{equation}
which is precisely
\[
 L_K\le\frac{(n!)^{1/n}}{(n+1)^{(n+1)/(2n)}\sqrt{n+2}}.
\]
Every step from the entropy inequality to \eqref{eq:body-cone-bound} is
reversible, so equality in the body bound holds if and only if $f_K$
has equality in Theorem~\ref{intro:entropy}.

Suppose equality holds. By that theorem the law with density $f_K$ is an
invertible affine image of a product of $n+1$ mean-one exponentials. The
closed support of $f_K$ is exactly $\mathcal C_K$: it has positive
density on the interior of the cone, and this interior is dense in the
cone. Taking closed supports gives
\begin{equation}\label{eq:body-cone-affine-orthant}
 \mathcal C_K=b+M[0,\infty)^{n+1}
\end{equation}
for an invertible matrix $M$ and a vector $b$.

The origin is the unique extreme point of $\mathcal C_K$: every nonzero
$z\in\mathcal C_K$ is the midpoint of $0$ and $2z$, while
$0=(u+v)/2$ with $u,v\in\mathcal C_K$ forces the nonnegative heights of
$u$ and $v$ to vanish, hence $u=v=0$. An orthant also has the origin as
its unique extreme point, and invertible affine maps preserve extreme
points, so \eqref{eq:body-cone-affine-orthant} forces $b=0$.

Choose positive multiples $w_0,\dots,w_n$ of the respective columns of
$M$, and let $a_i$ be the last coordinate of $w_i$. Each $w_i$ is a
nonzero point of $\mathcal C_K$, so $a_i>0$. Put $p_i=w_i/a_i$. The
section at height one is
\begin{equation}\label{eq:body-cone-section}
 K\times\{1\}=\Bigl\{\sum_{i=0}^nc_iw_i:c_i\ge0,\
 \sum_{i=0}^nc_ia_i=1\Bigr\}=\operatorname{conv}\{p_0,\dots,p_n\}.
\end{equation}
The $p_i$ are linearly independent (the $w_i$ are), hence affinely
independent in the height-one hyperplane, and $K$ is a simplex. (To
picture it: slice the cone by the hyperplane $t=1$; dividing each
generating ray $w_i$ by its positive height $a_i$ lands the endpoint on
that slice, and the slice is the convex hull of the $p_i$.)

Conversely, let $K=\operatorname{conv}\{v_0,\dots,v_n\}$ be a simplex,
and let $M$ have columns $(v_i,1)$. These are linearly independent and
$M[0,\infty)^{n+1}=\mathcal C_K$, $|\det M|=n!\,|K|$ (subtract the first
column from the others and expand along the last row). If $E_0,\dots,E_n$
are independent mean-one exponentials, the density of
$M(E_0,\dots,E_n)^{\mathsf T}$ at $(y,t)\in\mathcal C_K$ is
\[
 \frac1{|\det M|}\exp\Bigl(-\sum_{i=0}^n(M^{-1}(y,t))_i\Bigr)
 =\frac{e^{-t}}{n!\,|K|},
\]
because the last row of $M$ consists of ones. This is exactly $f_K$ up
to immaterial boundary values, so Theorem~\ref{intro:entropy} gives
equality in \eqref{eq:body-cone-bound}.
\end{proof}

For $n=1$ every convex body is an interval, hence a simplex, and the
constant is $1/\sqrt{12}$: an interval of length $\ell$ has variance
$\ell^2/12$, matching the formula.

\appendix
\section{Verification of the scalar estimates}\label{cert:verification}

This appendix proves the Gaussian estimates used in Sections~\ref{tr:section}
and \ref{sc:section}. All terminating decimals denote exact rational
numbers. The finite certificates below concern entire intervals: their
validity follows from coefficient bounds and polynomial positivity, with
no sampling assumption. In the negative tail, $k$ and its first three
derivatives are small and are approximated by Gaussian excess moments. In
the positive tail $k\sim a/2$, so we subtract this leading term and
control derivatives of the remaining logarithmic correction. On the
compact interval we propagate the differential equations in their stable
directions (backward for $q$ from $q(10)$, forward for $k$ from $k(-6)$)
and certify polynomial positivity by exact arithmetic.

\begin{proof}[Proof of Lemma~\ref{sc:bounds} and Lemma~\ref{tr:scalar-basic}]
We use the definitions of $q,k,d,b,l$ from Section~\ref{tr:section} and of
$p,h_0,\mathsf D,\mathsf L,\mathsf E,K_0,J_0,A_*,\eta_*,\nu_*$ from
Section~\ref{sc:section}. The differential identities needed below are
\begin{equation}\label{cert:identities}
 q'=qd,\qquad k'=b=1-kd,\qquad l=k-q+ab,\qquad l'=2b-qd+al.
\end{equation}
In particular,
\begin{equation}\label{cert:derived}
 \begin{gathered}
 p=b+kl-l^2/\alpha,\qquad
 \mathsf E=p+b(.75-b)-l^2/\alpha,\\
 h_0=k(b-.5)-2(b-.28)l/\alpha,\\
 \frac{\dd\log M}{\dd\log q}=\frac{2(1-qd)}{d^2}
 -\frac{2bl}{d(1-b^2)},\qquad M=d^{-2}(1-b^2).
 \end{gathered}
\end{equation}
We verify the estimates separately on $a\le-6$, $a\ge10$, and $[-6,10]$.
The central calculation is also specified in
\texttt{verification/verify\_scalar.py}, which uses rational arithmetic
only.

\subsection*{The negative tail}
Put $r=-a\ge6$, $u=r^{-2}$, and
\[
 I_j(r)=\int_0^\infty v^je^{-rv-v^2/2}\dd v\qquad(j\ge0).
\]
With $\varphi$ the standard Gaussian density, the lower tail is
$P=\Phi(-r)=\varphi(r)I_0(r)$, so
\[
 q=\frac{\varphi(r)}{1-P},\qquad
 k=\frac{(1-P)(-\log(1-P))}{\varphi(r)}.
\]
Integration by parts gives $I_1=1-rI_0$, $I_2=I_0-rI_1$, $I_3=2I_1-rI_2$.
The inequality $0\le1-(1-P)(-\log(1-P))/P\le2P$ gives $|k-I_0|\le2P/r$.
Since $P\le\varphi(r)/r$ and $\varphi(6)<7\cdot10^{-9}$, we have
$q\le1.01\varphi(r)$. Applying \eqref{cert:identities} successively gives
\[
 |b-I_1|\le4\varphi(r)/r,\quad|l-I_2|\le6\varphi(r),\quad
 |l'-I_3|\le9r\varphi(r).
\]
Each Gaussian error decreases faster than its required inverse power for
$r\ge6$ ($r^N\varphi(r)$ is decreasing there for $N\le7$). Checking at
$r=6$ yields, with $g_0=k,g_1=b,g_2=l,g_3=l'$,
\begin{equation}\label{cert:left-errors}
 |g_j-I_j|\le.01u\frac{j!}{r^{j+1}}\qquad(0\le j\le3).
\end{equation}
The bounds $1-v^2/2\le e^{-v^2/2}\le1$ also give
\begin{equation}\label{cert:integral-errors}
 1-\frac{(j+1)(j+2)}2u\le\frac{I_j}{j!/r^{j+1}}\le1.
\end{equation}
For $\zeta=1+.01u$ the consequences we use are
\[
 0<k\le\zeta/r,\quad u(1-3.01u)\le b\le u\zeta<.028,
\]
\[
 2r^{-3}(1-6.01u)\le l\le2\zeta r^{-3},\quad0<l'\le6\zeta r^{-4}.
\]
In particular $0<b<1/2$ and $l>0$. Moreover
\[
 p\ge u(1-3.01u)-4\zeta^2u^3/\alpha
 \ge u(1-4.54u)\ge\frac1{7+r^2},
\]
the last steps by inserting $u\le1/36$. The two summands of $h_0$ in
\eqref{cert:derived} have opposite signs and magnitudes at most $.501/r$
and $.43/r$, so
$h_0^2\le.501^2u<.67^2u(1-4.54u)\le.67^2p$. This gives the bounds on
$p,h_0$ in this tail.

The remaining substitutions, avoiding cancellation in $p$, use
\begin{equation}\label{cert:left-ratios}
 \frac{kl}{b}\le\frac{2u\zeta^2}{1-3.01u},\qquad
 \frac{l^2}{\alpha b}\le\frac{4u^2\zeta^2}{\alpha(1-3.01u)},\qquad
 .93<\frac bp<1.05.
\end{equation}
They imply the $O$ row of \eqref{sc:interval-table}: for example
$\mathsf D\le D_*(7+r^2)<.295(7+r^2)$, $\mathsf L\le2\zeta/(\alpha r^3)<.60$,
\[
 .37<K_0=D_*\Bigl(1+(.75-b)\frac bp-\frac{l^2}{\alpha b}\frac bp\Bigr)<.56,
 \qquad J_0=D_*c\frac bp<.72.
\]
The bounds for $z=l'/l$, $h_*=1+3b+kz$, and $\chi=l/d$ follow from
\[
 0<z\le\frac{3\zeta}{r(1-6.01u)}<.61,\quad
 1<h_*<1+3(.028)+\zeta(.61)/6<1.77,
\]
\[
 \chi\le2\zeta u^2<.070,
\]
with $d=r+q\ge r$. Since $d'<0$, both $d^{-1}(1+b)$ and
$d^{-1}(1-b)=k$ are increasing; their product is $M$, proving the
logarithmic slope bound $m_0=0$.

For the matrix inequality in this tail, write $N$ for the matrix
subtracted from $Q$ in \eqref{sc:local-surplus},
\begin{equation}\label{cert:N}
 N=\begin{pmatrix}
 A_*b^2-\eta_*b+\nu_*/4+l^2/\alpha&\nu_*/2-\eta_*b\\
 \nu_*/2-\eta_*b&\nu_*
 \end{pmatrix}
 +\mathsf D\binom{\mathsf E}{cb}\begin{pmatrix}\mathsf E&cb\end{pmatrix}.
\end{equation}
Equations \eqref{cert:left-ratios} give $\mathsf E<.05072$. Using
$b<.028$, $K_0<.56$, $J_0<.72$ gives
$N_{11}<.224$, $N_{22}<.802$, $.34<N_{12}<.386$. Since
$Q_{11}=.43$, $Q_{12}=.13$, $Q_{22}=2.72$, the leading minor of
$Q-N-\operatorname{diag}(.13,.75)$ is positive and its determinant is at
least $(.43-.224-.13)(2.72-.802-.75)-(.13-.386)^2=.023232>0$. This
proves the local matrix bound on $a\le-6$.

\subsection*{The positive tail}
For $r=a\ge10$ put $u=r^{-2}$ and $\theta=\log(r\sqrt{2\pi})$. Gaussian
tail integration gives
\[
 F(u)=\int_0^\infty e^{-v-uv^2/2}\dd v,\qquad
 \Phi(-r)=\frac{\varphi(r)}rF(u),\qquad
 k=\frac r2+\frac{C(u,\theta)}r,
\]
where
\begin{equation}\label{cert:C}
 C=F\theta+G,\qquad G=\frac{F-1}{2u}-F\log F.
\end{equation}
Let $\mathcal U=u\partial_u$ with $\theta$ held fixed and
$\mathcal D=r\,\dd/\dd r=\partial_\theta-2\mathcal U$. Define
$B=(1-\mathcal D)C$, $J=(2-\mathcal D)B$, $R=(3-\mathcal D)J$ (auxiliary
scalar functions used only here). Differentiating $k$ gives
\begin{equation}\label{cert:right-derivatives}
 b=.5-uB,\qquad l=J/r^3,\qquad l'=-R/r^4.
\end{equation}
We first establish uniform bounds
\begin{equation}\label{cert:right-errors}
 |B-(\theta-1.5)|<.04,\quad|J-(2\theta-4)|<.13,\quad
 |R-(6\theta-14)|<.75.
\end{equation}
Differentiation under the integral gives, for $0\le u\le.01$,
\[
 1-u\le F\le1,\quad|F'|\le1,\quad|F''|\le6,\quad|F'''|\le90,\quad
 |F^{(4)}|\le2520,
\]
the $j$th derivative bounded in absolute value by $(2j)!/2^j$. To control
the quotient in $G$, use $(F(u)-1)/u=\int_0^1F'(tu)\dd t$. Since $F\ge.99$
and $|1+\log F|\le1$, differentiating \eqref{cert:C} yields
\[
 |G''|\le15+6+1/.99<23,\qquad|G'''|\le315+90+18/.99+1/.99^2<435.
\]
The values $F(0)=1$, $F'(0)=-1$, $G(0)=-1/2$, $G'(0)=5/2$ give
$C=\theta-.5+(2.5-\theta)u+R_0$, with $g=11.5+3\theta$ and
$c_0=435+90\theta$, and
\begin{equation}\label{cert:Euler-remainder}
 |\mathcal U^jR_0|\le2^jgu^2+c_0u^3\mathbf1_{\{j=3\}}\qquad(0\le j\le3).
\end{equation}
The same bounds apply to $\partial_\theta R_0$ with $g=3,c_0=90$; higher
$\theta$ derivatives vanish. Expanding the three operators, the
linear-in-$u$ errors are $(8.5-3\theta)u$, $(37-12\theta)u$,
$(197-60\theta)u$, and those contributed by $R_0$ are bounded by
\begin{equation}\label{cert:operator-errors}
 (5g+3)u^2,\qquad(30g+33)u^2,\qquad(210g+321+8c_0u)u^2.
\end{equation}
For example the second operator is $2-3\mathcal D+\mathcal D^2$ and the
third is $6-11\mathcal D+6\mathcal D^2-\mathcal D^3$.

To check uniformity on the unbounded interval, write $x=\log(r/10)\ge0$,
so $u=.01e^{-2x}$ and $\theta=\theta_0+x$ with $3.221<\theta_0<3.222$.
The terms in \eqref{cert:operator-errors} decrease with $x$; for the
linear-in-$u$ terms, maximize a linear function times $e^{-2x}$ at an
endpoint or stationary point, using $e^{-t}\le1/(1+t)$ for $t\ge0$. Put
\[
 g_+=11.5+3(3.222),\quad c_+=435+90(3.222),\quad
\]
\[
 a_B=3(3.222)-8.5,\quad a_J=12(3.222)-37,\quad a_R=197-60(3.222).
\]
Upper bounds for the three total errors, giving larger rationals, are
\[
 \begin{array}{c|l|c}
 &\text{rational upper bound}&\text{larger rational}\\\hline
 B&\displaystyle\frac{.015}{2-2a_B/3}+\frac{5g_++3}{10^4}&.024\\[2mm]
 J&\displaystyle\frac{.06}{2-a_J/6}+\frac{30g_++33}{10^4}&.102\\[2mm]
 R&\displaystyle\max\Bigl\{\tfrac{197-60(3.221)}{100},
 \tfrac{.3}{2+a_R/30}\Bigr\}+\tfrac{210g_++321}{10^4}
 +\tfrac{8c_+}{10^6}&.624
 \end{array}
\]
which prove \eqref{cert:right-errors}. The bound on $\theta_0$ uses the
logarithm series $\log y=2\sum_{j=0}^N\frac{z^{2j+1}}{2j+1}+\mathcal R_N$,
$z=(y-1)/(y+1)$, $|\mathcal R_N|\le2|z|^{2N+3}/((2N+3)(1-z^2))$,
applied to $\theta_0=\tfrac12\log(200\pi)$, with Machin's identity
$\pi=16\arctan(1/5)-4\arctan(1/239)$; this gives
$3.221523626198<\theta_0<3.221523626199$.

Combining \eqref{cert:right-errors} with the Taylor bounds on $C$ gives
\begin{equation}\label{cert:right-consequences}
 1.7<R/J\le3,\quad.482<b<.5,\quad0<l<.0026,\quad
 .5<k/r<.528,\quad.493<p<.5.
\end{equation}
The ratio bounds come from $J>2\theta-4.13>0$, $R-1.7J>2.6\theta-8.171>0$,
and $3J-R>.86$. For $k$ and $p$: $C\ge.99\theta-.5>0$ by \eqref{cert:C},
while the Taylor expansion gives
$C-(\theta-.5)\le u\{2.5-\theta+.01(11.5+3\theta)\}<0$. Thus
$0<C<\theta-.5<\theta-.475$ and $0<uCJ<.075$, so
\[
 p=.5+u\{-B+(.5+uC)J\}-l^2/\alpha,\qquad-.605<-B+.5J<-.395,
\]
proving $.493<p<.5$. The inverse-power majorants take their maxima at
$r=10$. Similarly $B/r<.18$, so
\[
 |h_0|\le.528(.18)+2(.22)(.0026)/\alpha<.14<.67\sqrt p.
\]
These give the $T_1$ row of \eqref{sc:interval-table}: for example
$.367<K_0<.4$, $\mathsf D<D_*/.493<.640$,
$J_0<D_*c(.5)/.493<.48$, $\mathsf L<.0026/\alpha<.20$. They also give
$-.3<z<0$ and $.8<1+3(.482)-3(.528)\le h_*\le1+3(.5)-1.7(.5)<1.77$.

For the estimates involving $d$, let $V$ have density
$F(u)^{-1}e^{-v-uv^2/2}$ on $v>0$; the Gaussian excess above $r$ is $V/r$,
so $rd=\E V$, $1-qd=r^{-2}\Var V$. Taylor's lower bound and $F\le1$ give
$\E V\ge1-3u$ and $\E V^2\ge2(1-6u)$; monotone weighting gives $\E V\le1$.
Hence $d\ge.97/r$, $\Var V\ge.88$, and $\chi\le J/(.97r^2)<.027<.097$.
Equation \eqref{cert:derived} then gives the logarithmic slope
$\ge2(.88)-.027/(1-.5^2)>.68$.

For the matrix inequality, use $\mathsf E\le p+.482(.75-.482)$ in
\eqref{cert:N}. The first part of $N_{11}$ increases with $b$ on
$[.482,.5]$ and $(p+v)^2/p$ increases with $p$ when $0<v<p$, so
\[
 N_{11}\le A_*/4-\eta_*/2+\nu_*/4+.0026^2/\alpha
 +D_*\frac{(.5+.482(.75-.482))^2}{.5}<.403,
\]
\[
 N_{22}\le\nu_*+D_*\frac{(c/2)^2}{.493}<1.12,\qquad.080<N_{12}<.13.
\]
For the off-diagonal use $N_{12}=\nu_*/2+b(cK_0-\eta_*)$ with
$.367<K_0<.4$. Then $Q-N-\operatorname{diag}(.022,.75)$ has positive
leading minor and determinant at least
$(.43-.403-.022)(2.72-1.12-.75)-.05^2=.00175>0$. This finishes the
positive tail.

\subsection*{Polynomial enclosures on the central interval}
We now give all data and rules for the finite certificate on $[-6,10]$.
On each row of Table~\ref{cert:centers} write $a=w+rx$, $-1\le x\le1$;
the intervals are consecutive and cover $[-6,10]$. The tabulated
constants are rational seeds, not assumed Gaussian values. The
differential residuals and the two analytic anchors $q(10)$ and $k(-6)$
will certify the whole piecewise polynomial family, including all its
initial coefficients. Construct degree-$14$ polynomials $P,\widehat k$
from their tabulated constant terms by
\begin{equation}\label{cert:recurrence}
 D=P-a,\quad P_{i+1}=\frac r{i+1}(PD)_i,\quad
 \widehat k_{i+1}=\frac r{i+1}(1-\widehat k D)_i\quad(0\le i<14),
\end{equation}
where a subscript denotes a coefficient (the constant $1$ contributes
only at $i=0$, and only already-computed coefficients occur on the
right). Here $P,\widehat k,D$ denote polynomial approximations to
$q,k,d$.

\begin{table}[htbp]
\centering
\caption{Rational input data for the central polynomial certificate.}
\label{cert:centers}
\begin{tabular}{r r l l}
\hline
$w$&$r$&$P_0$&$\widehat k_0$\\\hline
$-5.25$&$.75$&$.0000004128471303$&$.1842076703035094$\\
$-3.75$&$.75$&$.0003526268606772$&$.2507500253088710$\\
$-2.5$&$.5$&$.0176378254869167$&$.3531628936050266$\\
$-1.5$&$.5$&$.1387897504588508$&$.4981884857033017$\\
$-.5$&$.5$&$.5091604338370335$&$.7246172144765372$\\
$.25$&$.25$&$.9635539794164037$&$.9475979387933716$\\
$1.25$&$.75$&$1.7288166273310539$&$1.3000948982566192$\\
$3$&$1$&$3.2830986549304364$&$2.0126493036044195$\\
$5$&$1$&$5.1865039671258417$&$2.9046537878842420$\\
$7$&$1$&$7.1375456132265036$&$3.8366560415481663$\\
$9$&$1$&$9.1085231050028685$&$4.7898159350739631$\\\hline
\end{tabular}
\end{table}

For a polynomial $A$ let $[A]$ be the sum of the absolute values of its
coefficients. Direct use of \eqref{cert:recurrence} gives
\begin{equation}\label{cert:residuals}
 r[(PD)_{\deg\ge14}]<
 \begin{cases}2\cdot10^{-9},&w=-3.75,\\7\cdot10^{-11},&\text{else},
 \end{cases}
 \quad r[(\widehat k D)_{\deg\ge14}]<2\cdot10^{-10}.
\end{equation}
For each adjacent pair the values of both $P$ and $\widehat k$ at the
common endpoint differ by less than $2\cdot10^{-10}$. Also
\begin{equation}\label{cert:additional-poly}
 D>.09,\qquad[\widehat k]<\begin{cases}2,&w\le.25,\\5.6,&w>.25,
 \end{cases}
\end{equation}
the first-row value $\widehat k(-1)$ differs from $.16237766$ by less
than $3\cdot10^{-9}$, and the last-row value $P(1)$ differs from
$10.098093234$ by less than $2\cdot10^{-10}$. These are rational
coefficient checks; for example the first residual is bounded by
$r\sum_{i=14}^{28}|\sum_{j=0}^{14}P_jD_{i-j}|$ with missing coefficients
zero, and the bound on $D$ follows from the quartic rule
\eqref{cert:Bernstein} below after charging discarded coefficients to the
remainder.

Analytic anchors for those checks: for any $r>0$, Taylor's formula under
the integral gives alternating bounds
\begin{equation}\label{cert:Mills}
 S_N(r)=\sum_{i=0}^{N}\frac{(-1)^i(2i-1)!!}{r^{2i+1}},\qquad
 S_{2j+1}(r)\le I_0(r)\le S_{2j}(r),\quad(-1)!!=1.
\end{equation}
At $r=10$ use $S_{11},S_{12}$ and $q(10)=1/I_0(10)$; at $r=6$ use
$S_{13},S_{14}$ and $|k(-6)-I_0(6)|\le2\varphi(6)/36$. Exact
substitution gives
\[
 |q(10)-10.098093234|<4\cdot10^{-10},\qquad
 |k(-6)-.16237766|<4\cdot10^{-8}.
\]
The rough bound $\varphi(6)<7\cdot10^{-9}$ follows, e.g., from
$\sqrt{2\pi}>12/5$ and $e^{18}>\sum_{i=0}^{59}18^i/i!$, whence
$(5/12)/\sum_{i=0}^{59}18^i/i!<7\cdot10^{-9}$.

These anchors and residuals imply uniform (not merely endpoint) errors
\begin{equation}\label{cert:global-errors}
 |q-P|\le e_q=10^{-8},\qquad|k-\widehat k|\le e_k=8\cdot10^{-7}.
\end{equation}
The equation for $q-P$ has multiplier $r(q+P-a)>0$, so backward
integration from $a=10$ is stable; its absolute error grows by at most
twice each residual and by each joining error, giving
$6\cdot10^{-10}+2(2\cdot10^{-9}+10\cdot7\cdot10^{-11})+10\cdot2\cdot10^{-10}
=8\cdot10^{-9}<e_q$. The equation for $k-\widehat k$ has multiplier
$-rd<0$, so forward integration from $a=-6$ is stable; the additional
forcing is bounded by $r[\widehat k]e_q$, and the portions with
$[\widehat k]<2$ and $[\widehat k]<5.6$ have total lengths $6.5$ and
$9.5$. Thus an upper bound for its accumulated error is
\[
 4.3\cdot10^{-8}+22\cdot10^{-10}\cdot2+20\cdot10^{-10}
 +10^{-8}(6.5\cdot2+9.5\cdot5.6)=7.114\cdot10^{-7}<e_k.
\]

Construct the following full polynomials before truncating any of them:
\begin{equation}\label{cert:aux-polys}
 B=1-\widehat k D,\qquad J=\widehat k-P+aB,\qquad Y=2B-PD+aJ,
\end{equation}
approximating $b,l,l'$. Their error bounds, from \eqref{cert:identities}
and \eqref{cert:global-errors}, are
\begin{equation}\label{cert:aux-errors}
 \begin{gathered}
 E_b=[\widehat k]e_q+e_k(e_q+[D]),\qquad
 E_l=e_q+e_k+[a]E_b,\\
 E_{l'}=2E_b+([P]+[D]+e_q)e_q+[a]E_l.
 \end{gathered}
\end{equation}
The order in \eqref{cert:aux-polys} matters: form $B,J,Y$ with the full
degree-$14$ inputs, and only then apply the quartic arithmetic below.

\subsection*{The rational positivity certificate}
An enclosure $(A,e)$ means the function differs from $A(x)$ by at most
$e$ on $[-1,1]$. Replace a polynomial by its terms of degree at most four
and add the sum of the absolute values of the discarded coefficients to
$e$. Addition and subtraction add error bounds; for a product use the
coefficient convolution and the error
\begin{equation}\label{cert:product-error}
 [A]e_B+[B]e_A+e_Ae_B,
\end{equation}
then truncate and charge discarded coefficients as before. Scalar
multiplication scales the error by the absolute value of the scalar.
Initialize with $(q,k,d,b,l,l')\leftrightarrow(P,\widehat k,D,B,J,Y)$ and
$(e_q,e_k,e_q,E_b,E_l,E_{l'})$, using the full polynomials of
\eqref{cert:aux-polys}; truncate each of these six enclosures to degree
four, charging discarded coefficients to its error, before evaluating the
groups. For a resulting quartic $A(x)=\sum_{j=0}^4A_jx^j$ a rational lower
bound is
\begin{equation}\label{cert:Bernstein}
 \min_{\substack{\sigma=\pm1\\0\le i\le4}}
 \sum_{j=0}^i\frac{\binom ij}{\binom4j}\sigma^jA_j-e.
\end{equation}
This is the Bernstein coefficient bound on each of $[0,1]$ and $[-1,0]$:
on $x=\sigma t$ the displayed coefficients are the degree-four Bernstein
coefficients, whose basis functions are nonnegative and sum to one.

We list every polynomial to which the rule is applied. In a given row,
use the corresponding bin of Lemma~\ref{sc:bounds} to define
$b_{\min},b_{\max},\mathsf L_{\max},\mathsf D_{\max},K_{\min},K_{\max},
J_{\max},\ell_1$; $m_0$ is from Lemma~\ref{tr:scalar-basic}; the triples
$(z_{\min},z_{\max},\chi_{\max})$ are $(-.08,.70,.070)$ left of $-1$ and
$(-.475,0,.097)$ right of it. Within the quartic arithmetic set
\[
 p=b+kl-\frac{l^2}{\alpha},\quad
 \mathsf E=p+b(.75-b)-\frac{l^2}{\alpha},\quad
 H=l+3bl+kl',\quad f_1=cb,
\]
\[
 h=.43-\ell_1-A_*b^2+\eta_*b-\frac{\nu_*}4-\frac{l^2}{\alpha},\quad
 v=1.97-\nu_*,\quad o=.13-\frac{\nu_*}2+\eta_*b.
\]
The four groups are
\begin{align*}
 \mathcal G_1:\quad& b-b_{\min},\ b_{\max}-b,\ l,\
 \alpha\mathsf L_{\max}-l,\\
 &p(7+a^2)-1\ \text{on rows }w<0,\quad
 p-.25\ \text{on }0<w<4,\quad
 p-.46\ \text{on }w>4,\\
 &.67^2p-\{k(b-.5)-2(b-.28)l/\alpha\}^2;\\
 \mathcal G_2:\quad& l'-z_{\min}l,\ z_{\max}l-l',\
 H-.8l,\ 1.77l-H,\ \chi_{\max}d-l,\\
 &2(1-qd)(1-b^2)-2bld-m_0d^2(1-b^2);\\
 \mathcal G_3:\quad& p-D_*/\mathsf D_{\max}\ (\text{omit for }O),\\
 &D_*\mathsf E-K_{\min}p,\ K_{\max}p-D_*\mathsf E,\
 J_{\max}p-D_*f_1;\\
 \mathcal G_4:\quad& hp-D_*\mathsf E^2,\\
 &p(hv-o^2)-D_*(\mathsf E^2v-2\mathsf E\,o\,f_1+h f_1f_1).
\end{align*}
Products are evaluated left to right, with repeated multiplication for
powers; in the last expression the middle product is $2\mathsf E\,o\,f_1$.
Each tabulated bound is the minimum over a group, so positivity proves
every inequality listed in it.

\begin{table}[htbp]
\centering
\caption{Lower bounds for the four polynomial groups, in units of $10^{-3}$.}
\label{cert:group-bounds}
\begin{tabular}{r r r r r}
\hline
$w$&$\mathcal G_1$&$\mathcal G_2$&$\mathcal G_3$&$\mathcal G_4$\\\hline
$-5.25$&$7$&$1$&$1$&$1$\\
$-3.75$&$1$&$1$&$2$&$2$\\
$-2.5$&$2$&$2$&$2$&$5$\\
$-1.5$&$2$&$2$&$2$&$8$\\
$-.5$&$2$&$3$&$5$&$14$\\
$.25$&$2$&$2$&$8$&$11$\\
$1.25$&$2$&$.8$&$8$&$20$\\
$3$&$2$&$.8$&$8$&$7$\\
$5$&$1$&$.8$&$10$&$15$\\
$7$&$3$&$.6$&$11$&$11$\\
$9$&$2$&$.35$&$12$&$9$\\\hline
\end{tabular}
\end{table}

Table~\ref{cert:group-bounds} is obtained by the rational operations
\eqref{cert:recurrence}, \eqref{cert:aux-polys}, \eqref{cert:aux-errors},
\eqref{cert:product-error}, \eqref{cert:Bernstein}, with no additional
function evaluations. The accompanying checker reproduces all $44$ bounds,
all $22$ differential residual bounds, the $20$ joining bounds, the anchor
checks, and the coefficient norms used for \eqref{cert:global-errors}. The
same endpoint enclosures give $q(-1)>.287$, $q(-2)<.056$, $q(-3)<.005$,
$q(4)>4$, as used in the secant argument, and the logarithm series
certifies
\[
 \log\frac{q(-1)}{q(-2)}>1.62,\quad
 \log\frac{q(-1)}{q(-3)}>4,\quad
 \log\frac{q(4)}{q(-2)}>4,\quad
 \log\frac{q(4)}{q(-3)}>6.5.
\]

Why do these finite checks finish the proof? Group $\mathcal G_1$ gives
$0<b<1/2$, $l>0$, the bounds for $p,h_0$, and those for $b,\mathsf L$; its
strict positive margins give $p>0$. Group $\mathcal G_2$ gives the bounds
for $z,h_*,\chi$ after division by $l$ or $d$, and the logarithmic slope
bound by division by $d^2(1-b^2)$. Group $\mathcal G_3$ gives the
remaining coefficient bounds after division by $p$; the $O$ bound for
$\mathsf D$ instead follows from $D_*<.295$ and the lower bound on $p$.
Finally
\[
 Q-N-\operatorname{diag}(\ell_1,.75)
 =\begin{pmatrix}h&o\\o&v\end{pmatrix}
 -\frac{D_*}p\binom{\mathsf E}{f_1}(\mathsf E\ \ f_1),
\]
whose leading minor and determinant, multiplied by $p$, are exactly the
two expressions in $\mathcal G_4$; their positivity proves the matrix
inequality. Continuity supplies every bin endpoint, including the
stronger of the two adjacent slope bounds. Together with the two tails,
this proves both lemmas on the whole real line.
\end{proof}

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\end{document}