64c3814
==== TABLE OF CONTENTS ====
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//approximation.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//certificate.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//geometry.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//introduction.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//main.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//matrix.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//rigidity.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//scalar.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//transport.tex
A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//references.bib
===========================
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//approximation.tex ====
\section{Approximation of log-concave densities}\label{geo:section}
The smooth estimate of Theorem~\ref{mat:entropy-smooth} requires a
potential with positive, bounded Hessian. We remove those restrictions
while preserving the entropy, covariance, and first moments. The same
approximation will let us study equality by applying the strict
remainder estimate 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
proper if it is finite somewhere and never takes the value $-\infty$.
The convention $e^{-\infty}=0$ allows bounded supports.
The elementary tail estimate below supplies a common integrable bound 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 $b\geq0$ such that
\begin{equation}\label{geo:linear-coercivity}
V(x)\geq a|x|-b\qquad(x\in\R^m).
\end{equation}
In particular, $e^{-V}$ has moments of every order, and
$\int |V|e^{-V}<\infty$, where $|V|e^{-V}$ is interpreted as zero when
$V=+\infty$.
\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 of its values at the vertices
throughout the simplex. Consequently, some ball $B(x_0,r)$ lies in a
sublevel set $\{V\leq M\}$, where $M\geq V(x_0)$.
The closed convex set $A=\{V\leq M+1\}$ has finite volume, since
\[
|A|\leq e^{M+1}\int e^{-V}.
\]
It is bounded. Indeed, the convex hull of $B(x_0,r)$ and a point $z\in A$
contains a cone with base an $(m-1)$-dimensional disk of radius $r$ through
$x_0$, perpendicular to $z-x_0$, and height $|z-x_0|$. Its volume tends to
infinity with $|z-x_0|$. In dimension one the same assertion follows from
the length of the interval joining $x_0$ to $z$.
Choose $R>r$ so large that $A\subset B(x_0,R)$. If
$d=|x-x_0|\geq R$ and $V(x)<\infty$, put
$y=x_0+(R/d)(x-x_0)$. Then $y\notin A$, and convexity gives
\[
M+1<V(y)\leq \left(1-\frac Rd\right)V(x_0)+\frac RdV(x).
\]
Thus $V(x)>V(x_0)+d/R$. 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 these two bounds proves \eqref{geo:linear-coercivity}, after
enlarging $b$.
The moment assertion follows from $e^{-V(x)}\leq e^b e^{-a|x|}$.
For the entropy assertion, the elementary bound
\begin{equation}\label{geo:entropy-domination}
|u|e^{-u}\leq C_b e^{-u/2}\qquad(u\geq-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
\[
f_j=Z_j^{-1}e^{-V_j},\qquad Z_j=\int_{\R^m}e^{-V_j},
\]
such that
\begin{equation}\label{geo:hessian-bounds}
\frac2j\Id\preceq D^2V_j\preceq\left(j+\frac2j\right)\Id
\qquad(j\geq1),
\end{equation}
and
\begin{equation}\label{geo:approximation-limits}
Z_j\longrightarrow1,\qquad \norm{f_j-f}_{L^1}\longrightarrow0,\qquad
h(f_j)\longrightarrow h(f),\qquad
\Cov(f_j)\longrightarrow\Cov(f).
\end{equation}
The covariance limit is entrywise, the means of $f_j$ converge to the mean
of $f$, and all these quantities are finite.
\end{lemma}
\begin{proof}
Fix $a,b$ from Lemma~\ref{geo:coercivity}. For each positive integer $j$,
define the Moreau envelope \cite[Section~7]{Moreau1965}
\begin{equation}\label{geo:moreau}
Q_j(x)=\inf_{y\in\R^m}\left\{V(y)+\frac j2|x-y|^2\right\}.
\end{equation}
The expression in braces is lower semicontinuous, strictly convex on its
effective domain, and tends to infinity as $|y|\to\infty$. It therefore
has a unique minimizer $p_j(x)$, and $Q_j$ is finite everywhere. Moreover,
\begin{equation}\label{geo:envelope-lower}
Q_j(x)\geq a|x|-b-\frac{a^2}{2j}.
\end{equation}
Indeed, $V(y)\geq a|x|-a|x-y|-b$, and minimizing
$jr^2/2-ar$ over $r\geq0$ proves the bound.
For completeness, the needed regularity of $Q_j$ follows directly from
the minimization. The optimality condition is
$j(x-p_j(x))\in\partial V(p_j(x))$, where $\partial V$ denotes the convex
subdifferential. To obtain it, 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 $s\downarrow0$. The resulting inequality is
$V(z)\geq V(p)+j\langle x-p,z-p\rangle$.
Adding the two subgradient inequalities at $p_j(x)$ and
$p_j(x')$ gives
\[
\langle x-x',p_j(x)-p_j(x')\rangle
\geq |p_j(x)-p_j(x')|^2.
\]
Consequently both $p_j$ and $x\mapsto x-p_j(x)$ are $1$-Lipschitz.
Evaluating the infimum at $p_j(x)$ and, in the reverse direction, at
$p_j(x+h)$ shows that
\[
\nabla Q_j(x)=j(x-p_j(x)).
\]
For example, the first comparison gives an upper error $j|h|^2/2$
after subtracting $j\langle x-p_j(x),h\rangle$, and the second gives
a lower error of order $j|h|^2$ by the Lipschitz estimate just proved.
Thus $Q_j$ is continuously differentiable with $j$-Lipschitz gradient.
It is convex, since the function minimized in \eqref{geo:moreau} is jointly
convex in $(x,y)$.
We next check convergence before smoothing. The functions $Q_j$ increase
with $j$, and $Q_j(x)\leq V(x)$ when $V(x)$ is finite. If $Q_j(x)$ stays
bounded above as $j\to\infty$, the identity at the minimizer and $V\geq-b$
imply
\[
|x-p_j(x)|^2\leq\frac{2(Q_j(x)+b)}j\longrightarrow0.
\]
Lower semicontinuity then gives
$V(x)\leq\liminf_j V(p_j(x))\leq\lim_jQ_j(x)$.
This proves
\begin{equation}\label{geo:envelope-limit}
Q_j(x)\longrightarrow V(x)\quad\hbox{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 bound on the gradient;
hence \eqref{geo:hessian-bounds} holds. Evenness gives
$\int z\eta(z)\,\dd z=0$, so Jensen's inequality and
\eqref{geo:envelope-lower} yield the common bound
\begin{equation}\label{geo:common-tail}
V_j(x)\geq Q_j(x)\geq a|x|-B,
\qquad B=b+\frac{a^2}2.
\end{equation}
Smoothing at this scale does not affect the limit. The $j$-Lipschitz
bound on the gradient gives
\[
Q_j(x-z)\leq Q_j(x)-\langle\nabla Q_j(x),z\rangle+\frac j2|z|^2.
\]
Integrating against $\eta_{1/j}$ cancels the linear term, so
\[
0\leq (Q_j*\eta_{1/j})(x)-Q_j(x)\leq\frac1{2j}
\qquad(x\in\R^m).
\]
Together 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 $j$. Applying
\eqref{geo:entropy-domination} with $B$ in place of $b$ also gives
\[
|V_j(x)|e^{-V_j(x)}\leq C e^{-a|x|/2}.
\]
At points where $V=+\infty$, both $e^{-V_j}$ and $V_je^{-V_j}$ tend to
zero. Dominated convergence therefore proves convergence of the
normalizing constants, first moments, 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
\[
h(f_j)\geq m+\frac12\log\det\Cov(f_j).
\]
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} permits passage to the limit in both
sides and proves the inequality.
\end{proof}
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//approximation.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//certificate.tex ====
\section{Verification of the scalar estimates}\label{cert:verification}
This appendix proves the Gaussian estimates used in the transport and scalar
arguments. All terminating decimals in this appendix 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
can be 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
from $q(10)$ and forward from $k(-6)$---and then 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 the transport section and of
$p,h_0,\mathsf D,\mathsf L,\mathsf E,K_0,J_0,A_*,\eta_*,\nu_*$
from the scalar 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{split}
p&=b+kl-l^2/\alpha,\\
\mathsf E&=p+b(.75-b)-l^2/\alpha,\\
h_0&=k(b-.5)-2(b-.28)l/\alpha,\\
\frac{d\log M}{d\log q}
&=\frac{2(1-qd)}{d^2}
-\frac{2bl}{d(1-b^2)},\qquad M=d^{-2}(1-b^2).
\end{split}
\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}; it uses rational arithmetic only.
\subsection*{The negative tail}
Put $r=-a\ge6$, $u=r^{-2}$, and
\[
I_j(r)=\int_0^\infty v^j e^{-rv-v^2/2}\,dv\qquad(j\ge0).
\]
Writing $\varphi$ for the standard Gaussian density, the lower Gaussian tail
is $P=\Phi(-r)=\varphi(r)I_0(r)$. Thus
\[
q=\frac{\varphi(r)}{1-P},\qquad
k=\frac{(1-P)(-\log(1-P))}{\varphi(r)}.
\]
Integration by parts gives
\[
I_1=1-rI_0,\qquad I_2=I_0-rI_1,\qquad I_3=2I_1-rI_2.
\]
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 therefore 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
$r\ge6$: indeed $r^N\varphi(r)$ 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 elementary 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 $\zeta=1+.01u$, consequences we shall use are
\[
\begin{array}{c}
0<k\le\zeta/r,\quad
u(1-3.01u)\le b\le u\zeta<.028,\\
2r^{-3}(1-6.01u)\le l\le2\zeta r^{-3},\quad
0<l'\le6\zeta r^{-4}.
\end{array}
\]
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 two inequalities follow by inserting $u\le1/36$; after dividing
by positive factors they reduce to
$3.01+4\zeta^2u/\alpha<4.54$ and
$(1-4.54u)(1+7u)\ge1$.
The two summands of $h_0$ in \eqref{cert:derived} have opposite signs
and magnitudes at most $.501/r$ and $.43/r$. Hence
\[
h_0^2\le .501^2u<.67^2u(1-4.54u)\le .67^2p.
\]
This establishes the required bounds on $p,h_0$ in this tail.
For clarity, the remaining substitutions can be made with the following
ratios, avoiding cancellation in $p$:
\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 the coefficient table in Lemma~\ref{sc:bounds}.
For example, $\mathsf D\le D_*(7+r^2)<.295(7+r^2)$,
$\mathsf L\le2\zeta/(\alpha r^3)<.60$,
\[
.37<K_0=D_*\left(1+( .75-b)\frac bp-
\frac{l^2}{\alpha b}\frac bp\right)<.56,
\qquad J_0=D_*c\frac bp<.72.
\]
The bounds for $z=l'/l$, $h_*=1+3b+kz$, and $\chi=l/d$ follow from
\[
0<z\le\frac{3\zeta}{r(1-6.01u)}<.61,\qquad
1<h_*<1+3(.028)+\zeta(.61)/6<1.77,
\qquad \chi\le2\zeta u^2<.070,
\]
where $d=r+q\ge r$. Since $d'<0$, both $d^{-1}(1+b)$ and
$d^{-1}(1-b)=k$ are increasing. Their product is $M$, proving the
required logarithmic slope bound $m_0=0$.
It remains to check the matrix inequality in this tail. Write $N$ for the
matrix subtracted from $Q$ in that inequality, so that
\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 $b<.028$, $K_0<.56$, and $J_0<.72$ in \eqref{cert:N} gives
\[
N_{11}<.224,\qquad N_{22}<.802,\qquad .34<N_{12}<.386.
\]
Since $Q_{11}=.43$, $Q_{12}=.13$, and $Q_{22}=2.72$, 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
\[
F(u)=\int_0^\infty e^{-v-uv^2/2}\,dv,\qquad
\Phi(-r)=\frac{\varphi(r)}r F(u),\qquad
k=\frac r2+\frac{C(u,\theta)}r,
\]
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 let
$\mathcal D=r\,d/dr=\partial_\theta-2\mathcal U$. Define
\[
B=(1-\mathcal D)C,\qquad J=(2-\mathcal D)B,\qquad
R=(3-\mathcal D)J.
\]
Here $B,J,R$ are auxiliary scalar functions used only in this appendix.
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 for these three derivatives:
\begin{equation}\label{cert:right-errors}
|B-(\theta-1.5)|<.04,\qquad
|J-(2\theta-4)|<.13,\qquad
|R-(6\theta-14)|<.75.
\end{equation}
Differentiation under the integral gives, for $0\le u\le.01$,
\[
1-u\le F\le1,\qquad |F'|\le1,\qquad
|F''|\le6,\qquad |F'''|\le90,\qquad |F^{(4)}|\le2520.
\]
Indeed the $j$th derivative is bounded in absolute value by
$(2j)!/2^j$. To control the apparent quotient in $G$, use
$(F(u)-1)/u=\int_0^1F'(tu)\,dt$. 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 $F(0)=1$, $F'(0)=-1$, $G(0)=-1/2$, and $G'(0)=5/2$
therefore give
\[
C=\theta-.5+(2.5-\theta)u+R_0,\qquad
g=11.5+3\theta,\qquad c_0=435+90\theta,
\]
with
\begin{equation}\label{cert:Euler-remainder}
|\mathcal U^jR_0|\le2^jg u^2+c_0u^3\mathbf1_{\{j=3\}}
\qquad(0\le j\le3).
\end{equation}
The same bounds apply to $\partial_\theta R_0$ with $g=3,c_0=90$;
all higher $\theta$ derivatives vanish. For example,
$\mathcal U^2R_0=uR_0'+u^2R_0''$ and
$\mathcal U^3R_0=uR_0'+3u^2R_0''+u^3R_0'''$, so these statements
follow directly from Taylor's theorem and the derivative bounds just given.
Expanding the three operators in \eqref{cert:right-derivatives}, the
linear-in-$u$ errors are respectively
\[
(8.5-3\theta)u,\qquad (37-12\theta)u,\qquad
(197-60\theta)u.
\]
The errors contributed by $R_0$ are bounded respectively 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
$6-11\mathcal D+6\mathcal D^2-\mathcal D^3$; substituting
$\mathcal D=\partial_\theta-2\mathcal U$ into
\eqref{cert:Euler-remainder} gives exactly the displayed constants.
Here is a rational way to check uniformity on the unbounded interval.
Write $x=\log(r/10)\ge0$, so $u=.01e^{-2x}$ and
$\theta=\theta_0+x$, where $3.221<\theta_0<3.222$.
The terms in \eqref{cert:operator-errors} decrease with $x$. For the
linear-in-$u$ terms, maximize a linear function times $e^{-2x}$ at
its endpoint or stationary point, using $e^{-t}\le1/(1+t)$ for $t\ge0$.
More explicitly, set
\[
\begin{aligned}
g_+&=11.5+3(3.222),& c_+&=435+90(3.222),\\
a_B&=3(3.222)-8.5,& a_J&=12(3.222)-37,\\
a_R&=197-60(3.222).
\end{aligned}
\]
Upper bounds for the total errors are
\[
\begin{array}{c|l|c}
&\text{rational upper bound}&\text{a larger rational}\\\hline
B&\displaystyle\frac{.015}{2-2a_B/3}
+\frac{5g_++3}{10^4}&.024\\[2mm]
J&\displaystyle\frac{.06}{2-a_J/6}
+\frac{30g_++33}{10^4}&.102\\[2mm]
R&\displaystyle\max\left\{\frac{197-60(3.221)}{100},
\frac{.3}{2+a_R/30}\right\}
+\frac{210g_++321}{10^4}+\frac{8c_+}{10^6}&.624
\end{array}
\]
These prove \eqref{cert:right-errors}. The bound on $\theta_0$ itself
can be checked with the logarithm series
\[
\log y=2\sum_{j=0}^{N}\frac{z^{2j+1}}{2j+1}+\mathcal R_N,\qquad
z=\frac{y-1}{y+1},\qquad
|\mathcal R_N|\le
\frac{2|z|^{2N+3}}{(2N+3)(1-z^2)},
\]
applied to $\theta_0=\tfrac12\log(200\pi)$. For a completely
specified enclosure, use Machin's identity
$\pi=16\arctan(1/5)-4\arctan(1/239)$ and the alternating arctangent
sums through degrees $47$ and $49$. In the logarithm series first divide
the argument by a power of two to place it in $[1,2)$, and take $N=35$
for that argument and for $\log2$. These rational operations give
$3.221523626198<\theta_0<3.221523626199$.
Combining \eqref{cert:right-errors} with the preceding Taylor bounds
on $C$ gives the following estimates needed in the lemmas:
\begin{equation}\label{cert:right-consequences}
1.7<R/J\le3,\quad .482<b<.5,\quad 0<l<.0026,\quad
.5<k/r<.528,\quad .493<p<.5.
\end{equation}
To see the ratio bounds directly, $J>2\theta-4.13>0$,
$R-1.7J>2.6\theta-8.171>0$, and $3J-R>.86$.
For the bounds involving $k$ and $p$, we have
$C\ge .99\theta-.5>0$ by \eqref{cert:C}, whereas its 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 $0<uCJ<.075$. In particular,
\[
p=.5+u\{-B+(.5+uC)J\}-l^2/\alpha,\qquad
-.605<-B+.5J<-.395,
\]
which proves $.493<p<.5$. The inverse-power majorants used for these consequences take their maxima at $r=10$,
as follows by differentiating the linear or quadratic polynomials in
$\log(r/10)$ times the indicated inverse power of $r$. For example,
$u(\theta-.475)(2\theta-3.87)$ decreases because
$1/(\theta-.475)+2/(2\theta-3.87)<2$ at $\theta=3.221$ and hence
for every larger $\theta$. Similarly $B/r<.18$, and consequently
\[
|h_0|\le .528(.18)+2(.22)(.0026)/\alpha<.14<.67\sqrt p.
\]
These bounds prove the claimed estimates on $p,h_0$ and the $T_1$ row
of the coefficient table; for example
\[
.367<K_0<.4,\qquad \mathsf D<D_*/.493<.640,\qquad
J_0<D_*c(.5)/.493<.48,\qquad \mathsf L<.0026/\alpha<.20.
\]
They also give $-.3<z<0$ and
\[
.8<1+3(.482)-3(.528)\le h_*
\le1+3(.5)-1.7(.5)<1.77.
\]
To check the remaining 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
\[
rd=\mathbb EV,\qquad 1-qd=r^{-2}\operatorname{Var}V.
\]
Taylor's lower bound in the numerator and $F\le1$ give
$\mathbb EV\ge1-3u$ and $\mathbb EV^2\ge2(1-6u)$.
Monotone weighting of the exponential density gives
$\mathbb EV\le1$. Hence $d\ge.97/r$,
$\operatorname{Var}V\ge.88$, and
\[
\chi\le\frac{J}{.97r^2}<.027<.097.
\]
Equation \eqref{cert:derived} now implies
\[
\frac{d\log M}{d\log q}
\ge2(.88)-\frac{.027}{1-.5^2}>.68.
\]
This verifies the logarithmic slope required by Lemma~\ref{tr:scalar-basic}.
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
$0<v<p$. Therefore
\[
\begin{split}
N_{11}&\le A_*/4-\eta_*/2+\nu_*/4+.0026^2/\alpha
+D_*\frac{(.5+.482(.75-.482))^2}{.5}<.403,\\
N_{22}&\le\nu_*+D_*\frac{(c/2)^2}{.493}<1.12,
\qquad .080<N_{12}<.13.
\end{split}
\]
For the off-diagonal estimate use
$N_{12}=\nu_*/2+b(cK_0-\eta_*)$ and $.367<K_0<.4$.
Thus $Q-N-\operatorname{diag}(.022,.75)$ 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 next provide 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 corresponding closed intervals are consecutive and cover $[-6,10]$.
The tabulated constants are rational seeds; we do not assume that they are
Gaussian function values. The differential residuals and the two analytic
anchors $q(10)$ and $k(-6)$ will certify the entire piecewise polynomial
family, including all its initial coefficients.
Construct polynomials $P,\widehat k$ of degree $14$ from their tabulated constant
terms by the recurrences
\begin{equation}\label{cert:recurrence}
D=P-a,\qquad
P_{i+1}=\frac r{i+1}(PD)_i,\qquad
\widehat k_{i+1}=\frac r{i+1}(1-\widehat k D)_i\quad(0\le i<14).
\end{equation}
The subscript denotes a coefficient, so the constant $1$ contributes only
when $i=0$. At each step only coefficients already computed occur on the
right. The letters $P,\widehat k,D$ denote polynomial approximations to $q,k,d$,
respectively, throughout this central calculation.
\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 $[A]$ denote 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{otherwise},\end{cases}
\qquad r[(\widehat k D)_{\deg\ge14}]<2\cdot10^{-10}.
\end{equation}
For each adjacent pair, the values of both $P$ and $\widehat k$ at their 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}
and the first-row value $\widehat k(-1)$ differs from $.16237766$ by less than
$3\cdot10^{-9}$, while 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
\[
r\sum_{i=14}^{28}\left|\sum_{j=0}^{14}P_jD_{i-j}\right|,
\]
with missing coefficients taken as zero. The bound on $D$ follows from
the quartic lower-bound rule \eqref{cert:Bernstein} below, after charging
discarded coefficients to the remainder.
Here are analytic anchors for those coefficient checks. For any $r>0$,
Taylor's formula under the integral gives alternating upper and lower
bounds for $I_0(r)$ by
\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),
\end{equation}
where $(-1)!!=1$. 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 $\varphi(6)<7\cdot10^{-9}$ used here follows, for example,
from $\sqrt{2\pi}>12/5$ and
$e^{18}>\sum_{i=0}^{59}18^i/i!$: division gives
$(5/12)/\sum_{i=0}^{59}18^i/i!<7\cdot10^{-9}$.
These anchors and residuals imply the following
uniform, rather than just 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}
Indeed the equation for $q-P$ has multiplier $r(q+P-a)>0$, so backward
integration from $a=10$ is stable. Its absolute error increases by at most
twice each bound in \eqref{cert:residuals} and by each joining error.
The resulting bound is
\[
6\cdot10^{-10}+2(2\cdot10^{-9}+10\cdot7\cdot10^{-11})
+10\cdot2\cdot10^{-10}=8\cdot10^{-9}<e_q.
\]
The equation for $k-\widehat k$ has multiplier $-rd<0$, so forward integration
from $a=-6$ is stable. The additional forcing is bounded by $r[\widehat k]e_q$.
The portions with $[\widehat k]<2$ and $[\widehat k]<5.6$ have total lengths $6.5$ and
$9.5$, respectively. 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 \emph{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}
They approximate $b,l,l'$. Their error bounds, obtained directly from
\eqref{cert:identities} and \eqref{cert:global-errors}, are
\begin{equation}\label{cert:aux-errors}
\begin{split}
E_b&=[\widehat k]e_q+e_k(e_q+[D]),\\
E_l&=e_q+e_k+[a]E_b,\\
E_{l'}&=2E_b+([P]+[D]+e_q)e_q+[a]E_l.
\end{split}
\end{equation}
The order in \eqref{cert:aux-polys} matters for the strength of the
certificate: form $B,J,Y$ with the full degree-$14$ inputs, and only then
apply the quartic arithmetic described next.
\subsection*{The rational positivity certificate}
An enclosure $(A,e)$ means that the function in question 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 all 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 the discarded coefficients as before. Scalar
multiplication scales the error by the absolute value of the scalar.
Initialize this arithmetic with
\[
(q,k,d,b,l,l')\longleftrightarrow
(P,\widehat k,D,B,J,Y),\qquad
(e_q,e_k,e_q,E_b,E_l,E_{l'}),
\]
using the full polynomials in \eqref{cert:aux-polys}. Truncate each of
these six initial enclosures to degree four, charging discarded coefficients
to its error, before evaluating the groups below.
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 precisely the coefficients
in the degree-four Bernstein basis, whose basis functions are nonnegative
and sum to one.
We list every polynomial to which this rule is applied. In a given row,
use the corresponding bin in Lemma~\ref{sc:bounds} to define the bounds
\[
b_{\min},\ b_{\max},\ \mathsf L_{\max},\ \mathsf D_{\max},\qquad
K_{\min},\ K_{\max},\ J_{\max},\ \ell_1.
\]
The constant $m_0$ is from Lemma~\ref{tr:scalar-basic}.
The triples $(z_{\min},z_{\max},\chi_{\max})$ are
$(-.08,.70,.070)$ for rows to the left of $-1$, and
$(-.475,0,.097)$ for rows to the right. Within the quartic arithmetic set
\[
\begin{aligned}
p&=b+kl-l^2/\alpha,&
\mathsf E&=p+b(.75-b)-l^2/\alpha,\\
H&=l+3bl+kl',& f_1&=cb,
\end{aligned}
\]
\[
\begin{aligned}
h&=.43-\ell_1-A_*b^2+\eta_*b-\nu_*/4-l^2/\alpha,\\
v&=1.97-\nu_*,\qquad o=.13-\nu_*/2+\eta_*b.
\end{aligned}
\]
The four groups are as follows:
\begin{align*}
\mathcal G_1:\quad& b-b_{\min},\ b_{\max}-b,\ l,\
\alpha\mathsf L_{\max}-l,\\
&p(7+a^2)-1\quad\text{on rows with }w<0,\\
&p-.25\quad\text{on rows with }0<w<4,\qquad
p-.46\quad\text{on rows with }w>4,\\
&.67^2p-\{k(b-.5)-2(b-.28)l/\alpha\}^2;\\[1mm]
\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);\\[1mm]
\mathcal G_3:\quad& p-D_*/\mathsf D_{\max}\quad
\text{(omitted for bin }O\text{)},\\
&D_*\mathsf E-K_{\min}p,\ K_{\max}p-D_*\mathsf E,\
J_{\max}p-D_*f_1;\\[1mm]
\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_1 f_1).
\end{align*}
Products are evaluated from left to right, with repeated multiplication
for powers. In the last expression the product in the middle is
$2\mathsf E\,o\,f_1$. The bounds below are the minimum over all members
of 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}, and
\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 norm bounds used to obtain
\eqref{cert:global-errors}. The same endpoint enclosures give
$q(-1)>.287$, $q(-2)<.056$, $q(-3)<.005$, and $q(4)>4$, as used
in the secant argument. The logarithm series above also certifies
\[
\log\frac{q(-1)}{q(-2)}>1.62,\quad
\log\frac{q(-1)}{q(-3)}>4,\quad
\log\frac{q(4)}{q(-2)}>4,\quad
\log\frac{q(4)}{q(-3)}>6.5.
\]
We explain why these finite checks finish the proof. Group $\mathcal G_1$
gives $0<b<1/2$, $l>0$, the bounds for $p,h_0$, and the bounds 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
\[
Q-N-\operatorname{diag}(\ell_1,.75)
=\begin{pmatrix}h&o\\o&v\end{pmatrix}
-\frac{D_*}{p}\binom{\mathsf E}{f_1}
\begin{pmatrix}\mathsf E&f_1\end{pmatrix}.
\]
Its 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 logarithmic slope bounds. Together with the two tails,
this proves both lemmas on the whole real line.
\end{proof}
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//certificate.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//geometry.tex ====
\section{The simplex inequality and its equality cases}\label{sec:geometry}
We transfer Theorem~\ref{intro:entropy} to convex bodies using the exponential
density on a cone. This construction appears in
\cite[Proposition~2(b)]{FradeliziMeyer2008}; its entropy reduction is
\cite[Lemma~5.2 and Proposition~5.2]{FradeliziMarinSola2026}.
The moment formulas also appear in \cite[Lemma~2.5]{Klartag2018}.
The calculation 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 both $L_K$
and the property of being a simplex, so assume that $K$ 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\geq0\}\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 can divide
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|}\,
\mathbf 1_{\mathcal C_K}(y,t).
\end{equation}
The change of variables $y=tx$, with 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
that it is the density of $(TX,T)$, where $T$ is independent of $X$
and has density $t^ne^{-t}/n!$ on $(0,\infty)$. Integration by parts
therefore 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 the entropy inequality of Theorem~\ref{intro:entropy} in
dimension $n+1$ yields
\begin{equation}\label{eq:body-cone-bound}
(n!)^2|K|^2\geq
(n+1)^{n+1}(n+2)^n\det\Sigma_K.
\end{equation}
This is precisely
\[
L_K\leq
\frac{(n!)^{1/n}}{(n+1)^{(n+1)/(2n)}\sqrt{n+2}}.
\]
Every step from the entropy inequality to \eqref{eq:body-cone-bound}
is reversible. Thus equality in this 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 the product of $n+1$ mean-one
exponential laws. 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 $\mathcal C_K$. Indeed,
every nonzero $z\in\mathcal C_K$ is the midpoint of $0$ and $2z$,
whereas $0=(u+v)/2$ with $u,v\in\mathcal C_K$ forces the
nonnegative heights of $u$ and $v$ to vanish, and hence $u=v=0$.
The orthant also has the origin as its unique extreme point.
Invertible affine maps preserve extreme points, so
\eqref{eq:body-cone-affine-orthant} forces $b=0$.
Choose positive multiples $w_0,\ldots,w_n$ 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\}
=\left\{\sum_{i=0}^n c_iw_i:
c_i\geq0,\ \sum_{i=0}^n c_i a_i=1\right\}
=\operatorname{conv}\{p_0,\ldots,p_n\}.
\end{equation}
The vectors $p_i$ are linearly independent, since the vectors $w_i$
are. They are consequently affinely independent in the height-one
hyperplane. Thus $K$ is a simplex. Figure~\ref{fig:cone-section}
illustrates the normalization of the generating rays in this step.
Conversely, let $K=\operatorname{conv}\{v_0,\ldots,v_n\}$ be a
simplex, and let $M$ have columns $(v_i,1)$. These columns are
linearly independent and
\[
M[0,\infty)^{n+1}=\mathcal C_K,
\qquad |\det M|=n!\,|K|.
\]
For the determinant identity, subtract the first column from the
others and expand along the last row; the remaining determinant is
$n!$ times the volume of $K$, up to sign. If $E_0,\ldots,E_n$ are
independent mean-one exponential variables, the density of
$M(E_0,\ldots,E_n)^{\mathsf T}$ at $(y,t)\in\mathcal C_K$ is
\[
\frac{1}{|\det M|}\exp\left(-\sum_{i=0}^n
(M^{-1}(y,t))_i\right)
=\frac{e^{-t}}{n!\,|K|},
\]
because the last row of $M$ consists of ones. This is exactly
$f_K$, up to immaterial boundary values. Theorem~\ref{intro:entropy}
therefore 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 $1/\sqrt{12}$. Indeed, an interval of length $\ell$ has
variance $\ell^2/12$, in agreement with the formula above.
\begin{figure}[t]
\centering
\begin{tikzpicture}[scale=1.25,
x={(1cm,0cm)},y={(0.45cm,0.35cm)},z={(0cm,1.5cm)},
line cap=round,line join=round,font=\small]
\coordinate (O) at (0,0,0);
\coordinate (P0) at (-1,-0.65,1);
\coordinate (P1) at (1,-0.65,1);
\coordinate (P2) at (0,1,1);
\coordinate (W0) at (-1.4,-0.91,1.4);
\fill[gray!8] (-1.4,-1,1)--(1.4,-1,1)--(1.4,1.3,1)
--(-1.4,1.3,1)--cycle;
\draw[gray!55] (-1.4,-1,1)--(1.4,-1,1)--(1.4,1.3,1)
--(-1.4,1.3,1)--cycle;
\node[anchor=west] at (1.4,1.3,1) {$t=1$};
\draw[gray!70] (O)--(P2);
\fill[blue!12] (P0)--(P1)--(P2)--cycle;
\draw[blue!65!black,thick] (P0)--(P1)--(P2)--cycle;
\draw[thick] (O)--(P0) (O)--(P1);
\draw[thick,dashed,->] (P0)--(-1.65,-1.0725,1.65);
\draw[thick,dashed,->] (P1)--(1.5,-0.975,1.5);
\draw[thick,dashed,->] (P2)--(0,1.5,1.5);
\fill (W0) circle[radius=1.3pt]
node[above right] {$w_0$} node[left] {$t=a_0>1$};
\fill (O) circle[radius=1.3pt] node[below] {$0$};
\fill (P0) circle[radius=1.3pt] node[left] {$p_0$};
\fill (P1) circle[radius=1.3pt] node[right] {$p_1$};
\fill (P2) circle[radius=1.3pt] node[above left] {$p_2$};
\node at (0,0,1) {$K\times\{1\}$};
\node[align=left,anchor=west] at (2.3,-1,0.65)
{$p_i=w_i/a_i$\\$a_i=\text{height of }w_i>0$};
\end{tikzpicture}
\caption{The height-one section of a cone generated by three independent
rays. Dividing each generator $w_i$ by its positive height $a_i$
produces the vertices $p_i$ of the section; one generator above the
section is shown explicitly. Formula
\eqref{eq:body-cone-section} gives the same correspondence in every
dimension.}
\label{fig:cone-section}
\end{figure}
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//geometry.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//introduction.tex ====
\section{Introduction}
The isotropic constant compares the covariance of a uniform distribution
with the volume of its support. For a convex body $K\subset\R^n$,
meaning a compact convex set with nonempty interior, write $|K|$ for its
Lebesgue volume, $b_K=|K|^{-1}\int_Kx\,\dd x$ for its centroid, and
\[
\Sigma_K=\frac1{|K|}\int_K(x-b_K)(x-b_K)^{\mathsf T}\,\dd x.
\]
Its isotropic constant is
\begin{equation}\label{intro:isotropic-constant}
L_K=\left(\frac{\det\Sigma_K}{|K|^2}\right)^{1/(2n)}.
\end{equation}
This quantity is unchanged by invertible affine maps. The slicing
problem, arising in Bourgain's work on convex bodies
\cite{Bourgain1986}, asks whether it admits an upper bound independent
of dimension. Klartag proved an $O(n^{1/4})$ bound
\cite{Klartag2006}. Subsequent progress used stochastic localization,
introduced by Eldan \cite{Eldan2013}, leading through subpolynomial
and polylogarithmic estimates \cite{Chen2021,KlartagLehec2022,Klartag2023}
to Klartag and Lehec's dimension-independent bound using Guan's
covariance estimate \cite{Guan2024,KlartagLehec2025}.
The sharp form, also called the strong isotropic constant conjecture,
asks, in each dimension, whether a simplex maximizes $L_K$.
The distinction is substantial: a universal bound does not
identify the extremal value or an extremizing body.
In the plane, Campi, Colesanti, and Gronchi proved the simplex bound,
and Saroglou established uniqueness of the maximizing body
\cite{CampiColesantiGronchi1999,Saroglou2010}.
In higher dimensions, geometric variation arguments have imposed
restrictions on possible maximizers. Rademacher proved that a
simplicial polytope maximizing the isotropic constant among all convex
bodies must be a simplex \cite{Rademacher2016}. More recent restrictions on local
maximizers in terms of decomposability appear in \cite{Kipp2026}.
We prove the sharp inequality for arbitrary convex bodies and classify
all equality cases.
\begin{theorem}[The simplex inequality]\label{intro:simplex}
For every integer $n\ge1$, every convex body $K\subset\R^n$ satisfies
the following bound, with equality if and only if $K$ is a simplex:
\begin{equation}\label{intro:simplex-bound}
L_K\le
\frac{(n!)^{1/n}}{(n+1)^{(n+1)/(2n)}\sqrt{n+2}}.
\end{equation}
\end{theorem}
The theorem also has a volume-product consequence. For a convex body
$K$ with $0$ in its interior, define its polar by
$K^\circ=\{y\in\R^n:\langle x,y\rangle\le1\text{ for every }x\in K\}$.
Using exponential measures on convex cones, Klartag proved that the
sharp simplex bound implies
\begin{equation}\label{intro:mahler}
|K|\,|K^\circ|\ge\frac{(n+1)^{n+1}}{(n!)^2},
\end{equation}
the nonsymmetric Mahler inequality
\cite[Corollary~1.2]{Klartag2018}. Thus Theorem~\ref{intro:simplex}
also gives this inequality in every dimension.
The proof proceeds through differential entropy. For a probability
density $f$ on $\R^m$, define, when the integrals are finite,
\[
h(f)=-\int_{\R^m}f(x)\log f(x)\,\dd x,
\qquad
\Cov(f)=\int_{\R^m}(x-\bar x)(x-\bar x)^{\mathsf T}f(x)\,\dd x,
\quad \bar x=\int_{\R^m}xf(x)\,\dd x.
\]
All logarithms are natural, and $0\log0=0$. A log-concave density can
be represented, up to a null set, as $f=e^{-V}$ for a proper lower
semicontinuous convex function $V:\R^m\to(-\infty,+\infty]$.
The value $+\infty$ permits a density to vanish. Such a probability
density has finite entropy and moments, and its covariance is positive
definite; the needed tail bound is proved in Lemma~\ref{geo:coercivity}.
\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 $f$ is, almost everywhere, the density
of $M(E_1,\ldots,E_m)^{\mathsf T}+b$, where $M$ is an invertible real
$m\times m$ matrix, $b\in\R^m$, and the $E_i$ are independent random
variables with density $e^{-u}\mathbf1_{\{u\ge0\}}$.
\end{theorem}
Sharpness follows from a product of $m$ independent exponential variables
of mean one: its entropy is $m$ and its covariance is the identity.
Both sides of \eqref{intro:entropy-bound} increase by $\log|\det A|$
under an invertible affine change $x\mapsto Ax+b$. Thus the theorem
asserts that a log-concave density with identity covariance has entropy
at least $m$.
Bobkov and Madiman developed the relation between entropy bounds for
log-concave measures and the slicing problem \cite{BobkovMadiman2011}.
Fradelizi and Mar\'in Sola formulated the sharp entropy statement
\eqref{intro:entropy-bound} and proved its equivalence, across dimensions,
with the simplex bound \cite[Conjecture~3(iii) and Proposition~5.2]{FradeliziMarinSola2026}.
In dimension one, Melbourne, Nayar, and Roberto proved the sharp bound,
attained by a one-sided exponential distribution
\cite[Theorem~1.1]{MelbourneNayarRoberto2026}.
Theorem~\ref{intro:entropy} establishes the inequality and identifies
all equality cases in every dimension.
The cone reduction used here follows the preceding geometric and
entropic work; we include it both to derive the dimensional constant
and to transfer the equality classification to convex bodies.
\subsection{Ideas of the proof}
We first assume that $f=e^{-V}$ is smooth and that the Hessian of $V$ is
bounded above and below by positive multiples of the identity.
Let $\nabla\phi$ transport standard Gaussian measure to $f$, and put
$J=D^2\phi$. Brenier's theorem and Caffarelli's regularity and
contraction theory supply this positive definite matrix field and its
uniform bounds
\cite{Brenier1991,McCann1995,Caffarelli2000,CorderoErausquinFigalli2019}.
The scalar model is the transport from a Gaussian variable to a
mean-one exponential variable,
\[
t(a)=-\log\Phi(-a),\qquad q(a)=t'(a),
\]
where $\Phi$ is the standard Gaussian distribution function. Since $q$
is an increasing bijection onto $(0,\infty)$, spectral calculus defines
the symmetric matrix field $A=q^{-1}(J)$.
Two features of this parametrization drive the proof. First, an affine
change of the target can be chosen so that $\E A=0$, where expectation
is Gaussian. This nonlinear normalization is proved by a homotopy and
a compactness argument; ordinary covariance normalization does not
give the required identity. Second, differentiating the transport
equation and using convexity of $V$ gives an energy estimate for $A$.
The estimate retains a nonnegative term involving pairs of distinct
eigenvalues of $A$. Discarding that term would lose the control needed
for matrix functions whose derivatives do not commute.
The zero mean permits an orthogonal decomposition
\[
A(x)=\sum_{r=1}^m x_rX_r+Y(x),
\qquad X_r=\E[x_rA(x)],
\]
where $Y$ has Gaussian chaos degrees at least two. The first term lies
in the equality space of the Gaussian Poincar\'e inequality and hence
has no energy surplus. We instead use its constant matrices $X_r$ to
choose a supporting plane for the covariance log determinant.
The higher-degree part supplies a strict spectral gap. A covariance
identity involving the resolvent of the Ornstein--Uhlenbeck operator
then expresses the signed deficit
$m+\tfrac12\log\det\Cov(f)-h(f)$ in terms of these two parts.
The resulting estimate has both noncommuting matrix terms and scalar
divided differences. A commutator square controls the former.
One scalar inequality absorbs the cost of the supporting plane and
the terms without derivatives. A second controls the divided
differences, charging their off-diagonal error to the
eigenvalue-separation term retained earlier.
After this cancellation, the entropy deficit is bounded above by a
quadratic expression that is nonpositive on every chaos of degree at
least two. Retaining strict slack also controls $Y$ and forces
$\sum_rX_r^2$ towards the identity when the entropy gap tends to zero. All numerical constants in the scalar inequalities are
exact rationals. Their verification uses polynomial bounds on a compact
interval and separate analytic estimates on both tails.
For equality, it is essential that this remainder applies to nearly
sharp densities. The expected extremizers have restricted support and
affine potentials, so equality cannot be studied solely within the
smooth, uniformly convex class. We approximate an arbitrary equality
density while preserving entropy and covariance. The remainder then
forces the reparametrized Jacobians to converge in Gaussian $L^2$ to
a linear field $A(x)=\sum_rx_rX_r$, with $\sum_rX_r^2=I$.
Since $q$ is Lipschitz in Frobenius norm, the original Jacobians
converge to $q(A)$, and Gaussian Poincar\'e gives convergence of the
centered maps.
The local compatibility of this limiting Jacobian is decisive.
The constant and linear terms of its mixed-derivative identities
force the symmetric matrices $X_r$ to commute. Simultaneous
orthogonal diagonalization then gives independent one-dimensional
exponential transports. Only the nonvanishing of $q'(0)$ and $q''(0)$
is needed for the commutation argument, isolating a potentially
reusable algebraic step. No uniform Hessian bounds on the smooth
approximants are required. Their affine normalizations are controlled
only after the limiting law and its covariance have been identified.
Section~\ref{tr:section} constructs the normalization and proves the
transport energy estimate. Section~\ref{mat:section} decomposes the
entropy deficit and proves the smooth case using two scalar
propositions. Section~\ref{sc:section} proves those propositions;
Appendix~\ref{cert:verification} supplies the quantitative one-variable
bounds. Section~\ref{geo:section} removes the smoothness assumptions.
Section~\ref{sec:rigidity} classifies equality in the entropy bound.
Section~\ref{sec:geometry} applies that classification to the exponential
density on the cone over $K$, completing Theorem~\ref{intro:simplex}.
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//introduction.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//main.tex ====
\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amssymb,amsthm,mathtools}
\usepackage{booktabs,array,longtable}
\usepackage{tikz}
\usepackage{microtype}
\usepackage{xcolor}
\usepackage[numbers,sort&compress]{natbib}
\usepackage[colorlinks=true,linkcolor=blue!55!black,citecolor=blue!55!black,urlcolor=blue!55!black]{hyperref}
\hypersetup{pdftitle={A sharp entropy bound and the simplex inequality for isotropic constants},pdfauthor={OpenAI}}
\newcommand{\R}{\mathbb R}
\newcommand{\N}{\mathbb N}
\newcommand{\Dgap}{\delta}
\newcommand{\mathscr}[1]{\mathcal{#1}}
\newcommand{\E}{\mathbb E}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\Cov}{Cov}
\DeclareMathOperator{\Var}{Var}
\DeclareMathOperator{\Sym}{Sym}
\newcommand{\Id}{I}
\newcommand{\dd}{\,\mathrm d}
\newcommand{\norm}[1]{\lVert #1\rVert}
\newcommand{\abs}[1]{\lvert #1\rvert}
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\newtheorem{remark}[theorem]{Remark}
\numberwithin{equation}{section}
\allowdisplaybreaks[1]
\title{A sharp entropy bound and the simplex inequality\protect\\ for isotropic constants}
\author{OpenAI}
\date{October 5, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove the strong isotropic constant conjecture: in each dimension,
simplices are the unique maximizers of the isotropic constant among
convex bodies. We also prove the sharp entropy bound
$h(f)\ge m+\tfrac12\log\det\Cov(f)$ for every log-concave probability
density on $\R^m$, with equality precisely for invertible affine
images of products of one-sided exponential laws.
\end{abstract}
\input{introduction}
\input{transport}
\input{matrix}
\input{scalar}
\input{approximation}
\input{rigidity}
\input{geometry}
\appendix
\input{certificate}
\bibliographystyle{plainnat}
\bibliography{references}
\end{document}
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//main.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//matrix.tex ====
\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 then supply the coercivity needed to prove the entropy bound.
We retain a strict remainder that will also identify the equality cases.
Throughout 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}
\left(\frac3{20}+\frac{49}{100}\frac{37}{37+3\lambda}\right)
\min\left\{\frac{13}{100},\frac{32}{25\sqrt\lambda}\right\},
\end{equation}
where the minimum is $13/100$ at $\lambda=0$.
This 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}\label{prop:slack}
Let $m\ge1$ be an integer and let $\mu=e^{-V}\,\dd y$ be a
probability measure on $\R^m$, where
$V\in C^\infty(\R^m)$ and
$c_-\Id\le D^2V\le c_+\Id$ for constants $0<c_-\le c_+<\infty$.
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 $\lambda_1,\ldots,\lambda_m$ 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,
\[
h(\mu)\ge m+\frac12\log\det\Cov(\mu).
\]
\end{theorem}
We use the affine normalization of
Proposition~\ref{tr:normalization}: the Brenier Jacobian $J$ and the
symmetric matrix field $A=q^{-1}(J)$ satisfy $\E A=0$. This normalization
does not change $h(\mu)-\frac12\log\det\Cov(\mu)$. For the proof,
we replace $\mu$ by this affine image and retain the notation $\mu$.
Scalar functions of $A$
are interpreted by spectral calculus; when their arguments are omitted,
$q,k,t,b,l$ and the functions introduced below are evaluated at $A$. For a matrix field $F$, put
\[
\tau(F)=\E\tr F,\qquad
\|F\|^2=\E\tr(F^tF),\qquad
\|F\|_S^2=\E\tr(F^tSF)
\]
when $S$ is a constant positive definite matrix. For a family indexed
by $r$, these squared norms include summation over $r$. Lemma~\ref{tr:energy}
provides the Sobolev regularity used below.
Write $C=\Cov(\mu)$ and let
\[
\mathscr D=2m+\log\det C-2h(\mu)
\]
be the negative of twice the entropy gap. The transport change of variables and
$\log q(a)=-a^2/2-\frac12\log(2\pi)+t(a)$ 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 let $R=(\Id+N)^{-1}$. The degree-$d$ Gaussian chaos is
the span of coordinatewise products of one-dimensional Hermite polynomials whose
degrees sum to $d$. These spaces give an orthogonal decomposition of
$L^2(\gamma_m)$; on the degree-$d$ space, $N$ and $R$ act by $d$ and
$(1+d)^{-1}$, respectively.
We write $\partial=(\partial_1,\ldots,\partial_m)$ and
$\partial^*v=\sum_r(x_rv_r-\partial_rv_r)$, so $N=\partial^*\partial$.
The usual Gaussian Sobolev norm is
$\|F\|_{H^1}^2=\|F\|^2+\|\partial F\|^2$; its dual norm is
\[
\|F\|_{-1}^2=\sum_{d\ge0}\frac{\|F_d\|^2}{1+d},
\]
initially for square-integrable fields, and then by completion.
Here $F_d$ denotes the degree-$d$ chaos component. These spectral
conventions are standard; see \cite[Section~2]{NourdinPeccati2009}.
We use the Gaussian Helffer--Sj\"ostrand covariance representation;
see \cite[Proposition~4.1(ii)]{DuerinckxOtto2020}. We include 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\bigl[J(RJ)\bigr]
=W+\E\left(2t-k^2-\sum_r K_r^2\right),
\qquad K_r=\partial_r k(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, and $R$ therefore cancels the
factor $d$. Summing over $d\ge1$ proves
$C_{ij}=\sum_\ell\E[J_{i\ell}R J_{j\ell}]$. Symmetry of $J$ gives
the first matrix identity in \eqref{mat:covariance-completion}.
Expanding the row Gram matrix of $q-(\Id+N)k$ gives cross terms
$-2\E(qk)=-2\E t$. The remaining term is
\[
\E\bigl[k(\Id+N)k\bigr]
=\E\left(k^2+\sum_r K_r^2\right),
\]
by entrywise integration by parts. This proves the second identity.
The calculation is valid by the $H^1$--$H^{-1}$ pairing, even if $Nk$
is not square-integrable.
For a test field $F$, the pairing with $u-\partial^*v$ is
$\langle u,F\rangle-\langle v,\partial F\rangle$. Cauchy--Schwarz
bounds it by
$(\|u\|^2+\|v\|^2)^{1/2}\|F\|_{H^1}$. Taking the dual norm proves
\eqref{mat:dual-bound} for $S=\Id$, and applying this argument to
$S^{1/2}U$ proves the general case.
\end{proof}
Separate the degree-one part of $A$ by writing
\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; $Y$ contains only chaoses
of degree at least two. We shall 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}
Thus $\partial^*g=Y$. All three derivative fields are symmetric.
For comparison, the coordinatewise Gaussian-to-exponential transport
satisfies
\[
A(x)=\operatorname{diag}(x_1,\ldots,x_m),\qquad B=\Id.
\]
We therefore choose the supporting plane for $\log\det C$ as a
regularized inverse of $B$ centered at this value. Fix
\begin{equation}\label{mat:dual-constants}
\alpha=\frac{3}{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 $A$. Directly from their definition,
\begin{equation}\label{mat:dual-relations}
0<S\le s\Id,\qquad T=-\alpha S(B-\Id),\qquad
H=\alpha S(B-\Id)^2\ge0,\qquad T^2=\alpha HS.
\end{equation}
In particular, $S=\Id$ when $B=\Id$. Deviations of the first chaos from
this value produce the nonnegative matrix $H$.
Concavity of the log determinant yields
$\log\det C+m\le\tr(SC)-\log\det S$. 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\left[T\left(2t-k^2-\Id-\sum_rK_r^2\right)\right].
\end{equation}
Here $\mathcal C$ is the eigenvalue-separation surplus in
Proposition~\ref{tr:surplus}. The remaining task is to estimate the last
two terms while retaining the negative contribution involving $H$.
\subsection{An exact expansion with noncommuting factors}
Recall the divided differences from Section~\ref{tr:section}, with
$v[a,a]=v'(a)$. For a smooth scalar function $v$, define its secant
increment at the second endpoint by
\[
\Delta v_{ij}=v[a_i,a_j]-v'(a_j).
\]
We use the matrix derivative formula in
\cite[Section~2.4]{Noferini2017}.
At a fixed point, choose an orthonormal eigenbasis of $A$. Entrywise
multiplication is denoted by $\circ$. 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 matrices need not be symmetric. The distinction between $D_r$
and $D_r^t$ will be 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 the 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}, we obtain
\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 according to its $H$ term and its derivative
residuals. In the remaining $Y$ coefficient, we
split off $ck$: the identity $\partial^*g=Y$ will transfer
$c\tau(TYk)$ to derivatives, where completing the square produces
$D-\gamma g$. Introduce the scalar functions
\begin{equation}\label{mat:scalar-functions}
\begin{aligned}
f&=k(1+b),&
h_0&=f-ck-\frac{2(b-\beta)l}{\alpha},\\
p&=f'-b^2-\frac{l^2}{\alpha},&
e&=f'-2b^2-\frac{2l^2}{\alpha}.
\end{aligned}
\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
\[
\psi_{ij}=\Delta f_{ij}-2b_j\Delta k_{ij}
-\frac{2l_j}{\alpha}\Delta b_{ij},
\qquad b_j=b(a_j),\quad l_j=l(a_j),
\]
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
\begin{equation}\label{mat:P-definition}
P_T=\tau(TbBb-TBb^2).
\end{equation}
\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\left[T\left(Yh_0+\sum_rX_rM_r\right)\right]
\label{mat:expanded-bound}\\
&\quad+\left\|Y(b-\beta)+\sum_rX_rL_r\right\|_S^2
+\|D-\beta g\|_S^2+\gamma^2\tau\left(T\sum_rg_r^2\right)\nonumber\\
&\quad-P_T-2\tau\left([T,b]\sum_rX_rw_r\right)
-\tau\left(T\sum_rw_r^tw_r\right).\nonumber
\end{align}
Except for the use of \eqref{mat:W-bound}, this expansion is exact.
\end{lemma}
\begin{proof}
Let $F_0=Y(b-\beta)+\sum_rX_rL_r$. The relations
\eqref{mat:dual-relations} give
\[
\|F_0+(B-\Id)l\|_S^2
=\|F_0\|_S^2+\frac1\alpha\tau(Hl^2)
-\frac2\alpha\tau(TF_0l).
\]
For the cross term, the order of the factors follows from
$\tr(F_0^tTl)=\tr(TF_0l)$, by transposition and cyclicity.
Integration by parts on the linear part of $A$ yields
\[
\tau(TAf)=\tau\left[T\left(Yf+\sum_rX_r\partial_rf\right)\right],
\qquad
\partial_rf=X_rf'+G_rf'+Z_r\circ\Delta f.
\]
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 versions of
$K_r=X_rb+D_r=bX_r+D_r^t$:
\begin{equation}\label{mat:K-square}
\tau\left(T\sum_rK_r^2\right)
=\tau(TbBb)+2\tau\left(Tb\sum_rX_rD_r\right)
+\tau\left(T\sum_rD_r^tD_r\right).
\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 $\partial^*g=Y$,
\[
\tau(TYk)=\sum_r\tau(TK_rg_r).
\]
Here integration by parts first gives $\tau(Tg_rK_r)$; transposition
makes the two expressions 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\left([T,b]\sum_rX_rw_r\right)
-\tau\left(T\sum_rw_r^tw_r\right)
\le\left(1+\frac s8\right)\|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}
Indeed, the contribution of $b_{ij}^2$ for $i<j$ on either side is
$(S_i-S_j)^2/(\alpha S_iS_j)$. The norm in
\eqref{mat:P-square} includes expectation.
Set $F=\sum_rX_rw_r$ and let $\operatorname{skew}M=(M-M^t)/2$.
Completing the square in the skew-symmetric matrix
$S^{-1/2}[T,b]S^{-1/2}$ bounds the first two terms on the left of
\eqref{mat:commutator-bound} by
\[
2\alpha\bigl\|\operatorname{skew}(S^{1/2}FS^{1/2})\bigr\|^2.
\]
For completeness, we estimate this square together with the last term.
Write $S_i=su_i$, where $0<u_i\le1$, so
$\alpha S_iB_i=1-u_i$. The rows of the block matrix
$X=(X_1,\ldots,X_m)$ 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 can
be completed to orthonormal directions, and cause no contribution 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
\[
\begin{pmatrix}
1-su_iu_j&-s\sqrt{u_iu_j(1-u_i)(1-u_j)}\\
-s\sqrt{u_iu_j(1-u_i)(1-u_j)}&1-su_iu_j
\end{pmatrix}.
\]
Its largest eigenvalue is at most
$1+s(z-2z^2)\le1+s/8$, where $z=\sqrt{u_iu_j}$; here
$(1-u_i)(1-u_j)\le(1-z)^2$.
Diagonal and orthogonal column components have coefficient at most
$1$. Summation and expectation prove \eqref{mat:commutator-bound}.
\end{proof}
We also use a block version of Cauchy--Schwarz. If
$X=(X_1,\ldots,X_m)$ and $S^{-1}=\Id/s+\alpha XX^t$, then
\begin{equation}\label{mat:block-bound}
\left\|a+\sum_rX_rL_r\right\|_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} give $p(a)>0$ for every
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\left[T\left(Yh_0+\sum_rX_rM_r\right)\right]
&\le\tau\bigl(H(\Id-\rho)p\bigr)
+\frac{\|M p^{-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 verify that the order of the weights is valid, write
$Q_0=H(\Id-\rho)$ and $F=Yh_0+\sum_rX_rM_r$.
Apply $\langle a,b\rangle\le\|a\|^2+\frac14\|b\|^2$ 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 $H$. Here inverses are interpreted as
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 in \eqref{mat:block-bound} proves
\eqref{mat:weighted-completion}. Thus this step does not require $p$
to commute with $S$.
\subsection{Two scalar estimates and the final chaos bound}
The following estimate combines the cost $\tr(T-\log S)$ with the
negative $H$ term left by \eqref{mat:weighted-completion} and the
terms quadratic in $Y$. A second estimate will control the derivative
residuals, including their dependence on $Z$, by $\mathcal C$ and the
higher-chaos energy. The proof of the first estimate is given in
Section~\ref{sc:absorption-proof}. Its strict remainder controls the
eigenvalues of $B$ through the weight \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 the constants
\eqref{mat:dual-constants}, \eqref{mat:constants}, and
\eqref{mat:completion-constants}, the functions \eqref{mat:scalar-functions}, and
$\rho=\epsilon\Id+\kappa S/s$, one has
\begin{equation}\label{mat:absorption-inequality}
\begin{split}
&\tr(T-\log S)-\tau(H\rho p)
+s\tau\left[Y^2\left((b-\beta)^2+
\frac{\alpha h_0^2}{4(1-\epsilon-\kappa)p}\right)\right]\\
&\hspace{15mm}\le\frac{69}{500}\|Y\|^2
-\frac1{1000}\sum_{i=1}^m\omega(\lambda_i),
\end{split}
\end{equation}
where $\lambda_1,\ldots,\lambda_m$ are the eigenvalues of $B$.
All scalar factors in the expected 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
\[
\delta=\bigl|\log(q(a_i)/q(a_j))\bigr|,\qquad
r_{ij}=\frac{1+m_0(\min(a_i,a_j))}{16}
\left(1+\frac{\delta^2}{30}+\frac{\delta^4}{1680}\right),
\]
where $m_0$ is the six-interval lower bound in \eqref{tr:bins}. Define the endpoint residuals
\[
\begin{array}{lll}
D_j=Gb_j+Z\Delta k_{ij},&L_j=Gl_j+Z\Delta b_{ij},&
M_j=Ge_j+Z\psi_{ij}+cgb_j,\\
D_i=Gb_i+Z\Delta k_{ji},&L_i=Gl_i+Z\Delta b_{ji},&
M_i=Ge_i+Z\psi_{ji}+cgb_i.
\end{array}
\]
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}\left[
s(D_v-\beta g)^2+\left(1+\frac s8\right)(D_v-\gamma g)^2
+\gamma^2(s-1)g^2+\frac{L_v^2}{\alpha}
+\frac{M_v^2}{4(1-\epsilon)p_v}\right]\notag\\
&\hspace{15mm}\le Q(G,g-G/2)+r_{ij}\delta^2Z^2.
\label{mat:pair-inequality}
\end{align}
Here $\operatorname{avg}_{v=i,j}$ is the arithmetic mean of the two
endpoint expressions, counting both also 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} occupies
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 appearing
in \eqref{mat:absorption-inequality}, by cyclicity of the trace;
no commutation with $Y$ is required. Set
$\Omega_B=\sum_{i=1}^m\omega(\lambda_i)$. Consequently,
\begin{align}
\mathscr D\le {}&-V_0-\mathcal C+\frac{69}{500}\|Y\|^2
-\frac{\Omega_B}{1000}
+\|D-\beta g\|_S^2
+\left(1+\frac s8\right)\|D-\gamma g\|^2
+\frac1\alpha\|L\|^2\nonumber\\
&\hspace{22mm}+\gamma^2\tau\left(T\sum_rg_r^2\right)
+\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 precisely the endpoint average in
\eqref{mat:pair-inequality}; 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 $(i,j)$ is independent
of the derivative index $r$, so summing over $r$ is unchanged by rotating
that index. All changes of eigenbasis are pointwise; no eigenvectors are
differentiated. The separation surplus cancels.
Because $Q$ has constant coefficients,
its summed quadratic form is invariant under orthonormal changes of
basis. We may therefore return to fixed coordinates and apply the
Gaussian chaos decomposition.
For a degree-$d$ component of $Y$, where $d\ge2$, the associated
components of $g$ and $G$ satisfy $g=G/d$, and
$\|G\|^2=d\|Y\|^2$. Different degrees remain orthogonal after
differentiation. The required bound follows, degree by degree, 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}
Indeed, the right side minus the left side 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
$V_0-\frac7{50}\|Y\|^2$. Substitution into
\eqref{mat:pre-pair} therefore gives
\[
\mathscr D\le-\frac1{500}\|Y\|^2-\frac{\Omega_B}{1000},
\]
which is \eqref{mat:strict-remainder}.
\end{proof}
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//matrix.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//rigidity.tex ====
\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 $q(A(x))$ force the
coefficients of a linear symmetric matrix field $A$ 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 $X_1,\ldots,X_m$ be real symmetric $m\times m$ matrices satisfying
\[
\sum_{r=1}^m X_r^2=I,
\qquad A(x)=\sum_{r=1}^m x_rX_r.
\]
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 $\sigma_1,\ldots,\sigma_m\in\{-1,1\}$, and $w\in\R^m$ such that
\begin{equation}\label{eq:rigid-map}
U^{\mathsf T}F(Uy)
=\bigl(\sigma_i t(\sigma_i y_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 exponential laws, and its covariance matrix is $I$.
\end{lemma}
\begin{proof}
Write $e_1,\ldots,e_m$ 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 scalar function $q$ is real analytic near zero. Since $A(0)=0$, its
convergent power series, interpreted by matrix multiplication, gives
\[
J(x)=q(0)I+q'(0)A(x)+\tfrac12q''(0)A(x)^2+O(|x|^3)
\]
near zero; the differentiated remainder is $O(|x|^2)$.
The identity $q'(a)=q(a)(q(a)-a)$ yields
\[
q(0)=\sqrt{\frac2\pi},\qquad
q'(0)=\frac2\pi>0,\qquad
q''(0)=\sqrt{\frac2\pi}\left(\frac4\pi-1\right)>0.
\]
Thus \eqref{eq:jacobian-compatibility}, which is 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
\[
(X_rA(x)+A(x)X_r)e_s
=(X_sA(x)+A(x)X_s)e_r.
\]
The terms with leftmost factor $A(x)$ cancel by
\eqref{eq:coefficient-symmetry}. The remaining equality is
$(X_rX_s-X_sX_r)x=0$ near zero, and 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.
Choose the orthogonal matrix $U$ with those basis vectors as columns,
and rotate both the input and the output. Then
\[
\widetilde A(y)=U^{\mathsf T}A(Uy)U
=\sum_{i=1}^m y_i\widetilde X_i,
\qquad
\widetilde X_i=U^{\mathsf T}
\left(\sum_{r=1}^m U_{ri}X_r\right)U
\]
has diagonal coefficients. Orthogonality of $U$ gives
$\sum_i\widetilde X_i^2=I$, and the identity $A(x)z=A(z)x$ gives
$\widetilde X_i e_j=\widetilde X_j e_i$. If $i\ne j$, these two vectors
are multiples of different basis vectors, so both are zero.
Consequently, only the $i$th diagonal entry of $\widetilde X_i$ can be
nonzero. The sum-of-squares identity makes that entry a sign:
\[
\widetilde X_i=\sigma_i e_ie_i^{\mathsf T},
\qquad
\widetilde A(y)=\operatorname{diag}(\sigma_1y_1,\ldots,\sigma_my_m).
\]
It follows that
\[
D_y\bigl(U^{\mathsf T}F(Uy)\bigr)
=\operatorname{diag}\bigl(q(\sigma_1y_1),\ldots,q(\sigma_my_m)\bigr).
\]
Since $t'=q$ and $\sigma_i^2=1$, the right side is also the Jacobian of
$y\mapsto(\sigma_i t(\sigma_i y_i))_{i=1}^m$.
Their difference has zero weak gradient on the connected set $\R^m$
and is therefore almost everywhere constant. This proves
\eqref{eq:rigid-map}.
For independent standard Gaussian coordinates $G_i$, the variables
$t(\sigma_iG_i)$ are independent mean-one exponentials. Their variances
are one, and the signs and the orthogonal matrix preserve covariance
$I$. This proves the final assertion.
\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 this 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
\[
u_i^{\mathsf T}(q(M)-q(N))v_k
=(q(\lambda_i)-q(\mu_k))\,u_i^{\mathsf T}v_k,
\]
whereas the corresponding entry of $M-N$ is
$(\lambda_i-\mu_k)u_i^{\mathsf T}v_k$. The scalar bound $0<q'<1$
therefore bounds the absolute value of each former entry by the absolute
value of the latter.
Squaring and summing over $i,k$ proves \eqref{eq:spectral-lipschitz},
because the Frobenius norm is 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
$\Dgap(f)=0$. Choose the approximating densities $f_j$ from
Lemma~\ref{geo:approximation}. Write $a_j,a$ for their means and the mean of
$f$, and $C_j,C$ for their covariance matrices. That lemma gives
\begin{equation}\label{eq:equality-approximation}
\norm{f_j-f}_{L^1(\R^m)}\longrightarrow0,\qquad
a_j\longrightarrow a,\qquad C_j\longrightarrow C,\qquad
\Dgap(f_j)\longrightarrow0.
\end{equation}
Here $C$ is positive definite, as recalled in the introduction, so the
covariance convergence also implies convergence of the 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}, put
\[
\begin{gathered}
J_j=DF_j,\qquad A_j=q^{-1}(J_j),\qquad
X_{j,r}=\E[x_rA_j(x)],\\
A_j(x)=\sum_{r=1}^m x_rX_{j,r}+Y_j(x),\qquad
B_j=\sum_{r=1}^m X_{j,r}^2.
\end{gathered}
\]
The entropy gap is affine invariant. Theorem~\ref{mat:entropy-smooth} and
\eqref{eq:equality-approximation} therefore imply
\begin{equation}\label{eq:vanishing-slack}
\norm{Y_j}_{L^2(\gamma_m)}\longrightarrow0,
\qquad
\sum_{i=1}^m \omega(\lambda_i(B_j))\longrightarrow0.
\end{equation}
The function $\omega$ is continuous and nonnegative on $[0,\infty)$,
vanishes only at $1$, and tends to infinity at infinity.
For every $\varepsilon>0$, it consequently has a positive infimum on
$\{\lambda\ge0:|\lambda-1|\ge\varepsilon\}$. Thus 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
such that $X_{j,r}\to X_r$ 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\longrightarrow A(x):=\sum_{r=1}^m x_rX_r
\quad\text{in }L^2(\gamma_m).
\end{equation}
Lemma~\ref{lem:spectral-lipschitz} now gives
\begin{equation}\label{eq:jacobian-limit}
J_j=q(A_j)\longrightarrow J:=q(A)
\quad\text{in }L^2(\gamma_m).
\end{equation}
To obtain a limiting map, remove the irrelevant 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 has the bounds \eqref{tr:Jacobian-bounds}
applied to 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}
Consequently $\widehat F_j$ converges strongly in Gaussian $H^1$ to a
centered map $\widehat F$ with weak Jacobian $D\widehat F=J$.
Indeed, \eqref{eq:map-cauchy} and \eqref{eq:jacobian-limit} give
convergence of both maps and first derivatives; passage to weak
derivatives is also valid on every bounded set, where the Gaussian
density has a positive lower bound. In particular,
$\widehat F\in W^{1,2}_{\mathrm{loc}}(\R^m;\R^m)$.
Lemma~\ref{lem:linear-jacobian} applies to \eqref{eq:linear-field-limit}
and shows that $\widehat F_\#\gamma_m$ is an affine exponential product
with covariance $I$.
It remains to return from the normalized laws to $f$.
By Cauchy--Schwarz, the strong $L^2(\gamma_m)$ convergence of the
centered maps gives
\begin{equation}\label{eq:normalized-covariance-limit}
D_j:=\Cov((\widehat F_j)_\#\gamma_m)
=P_jC_jP_j^{\mathsf T}\longrightarrow I.
\end{equation}
We can now bound the normalizing matrices uniformly, without any
uniform Hessian bounds on the approximating potentials.
Choose positive constants $c,C_0,d,D$ such that, for all sufficiently
large $j$,
\[
cI\le C_j\le C_0I,\qquad dI\le D_j\le DI.
\]
Congruence by $P_j$ then yields
\begin{equation}\label{eq:normalization-compactness}
\frac d{C_0}I\le P_jP_j^{\mathsf T}\le\frac Dc I.
\end{equation}
Hence $P_j$ 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 a bounded continuous function $\psi$,
\begin{align*}
&\left|\int\psi(P_j(x-a_j))f_j(x)\,\dd x
-\int\psi(P(x-a))f(x)\,\dd x\right|\\
&\quad\le\norm{\psi}_{\infty}\norm{f_j-f}_{L^1}
+\int\left|\psi(P_j(x-a_j))-\psi(P(x-a))\right|f(x)\,\dd x
\longrightarrow0.
\end{align*}
The first term tends to zero by \eqref{eq:equality-approximation};
the second does so by dominated convergence. Strong convergence of
$\widehat F_j$ also identifies the weak limit of these laws as
$\widehat F_\#\gamma_m$. It follows that the image of $f(x)\,\dd x$
under $x\mapsto P(x-a)$ is the affine exponential product already
determined above. Since $P$ is invertible, $f$ itself has the form
asserted in Theorem~\ref{intro:entropy}, up to equality almost everywhere.
Conversely, the density of a product of $m$ independent mean-one
exponentials is log-concave, has entropy $m$, and has covariance $I$.
Every invertible affine image remains log-concave and has the same
entropy gap by affine invariance. Thus every density in the stated
class satisfies $\Dgap=0$.
\end{proof}
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//rigidity.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//scalar.tex ====
\section{The two scalar estimates}\label{sc:scalar}\label{sc:section}
The matrix argument has 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 the transport section, and the constants
$\alpha,s,\beta,c,\gamma,\epsilon,\kappa$ and functions $p,e,h_0$ are those
of the matrix section. All finite decimals below denote exact rational
numbers.
We first state the one-variable bounds used in both proofs. Their verification
is given in Appendix~\ref{cert:verification}; in particular, none of the
bounds in this section 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},\quad \mathsf L=\frac{l}{\alpha},\quad
\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+s/8,\quad
\eta_*=s\beta+(1+s/8)\gamma,\quad
\nu_*=s\beta^2+(9s/8)\gamma^2.
\end{gathered}
\end{equation}
The symbols $\mathsf D,\mathsf L,\mathsf E$ denote scalar functions, to
be distinguished from the matrix residuals $D_r,L_r,M_r$ of the preceding
section.
\begin{lemma}[One-variable bounds]\label{sc:bounds}
For every $a\in\R$,
\begin{equation}\label{sc:p-h-bounds}
\begin{gathered}
p(a)>0,\qquad |h_0(a)|\le .67\sqrt{p(a)},\\
p(a)\ge\begin{cases}(7+a^2)^{-1},&a\le0,\\ .25,&a\ge0,\end{cases}
\qquad p(a)\ge .46\quad(a\ge4).
\end{gathered}
\end{equation}
The following interval bounds hold:
\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}
Here $\ell_1$ 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}
Here $Q=\left(\begin{smallmatrix}.43&.13\\.13&2.72\end{smallmatrix}\right)$,
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}
we have
\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 supplies a lower bound for the mean of $p$ from
only a second moment. The matrix surplus in~\eqref{sc:local-surplus}
then provides room to absorb the secant errors in the second proposition.
We treat these two uses separately.
\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\left((b-\beta)^2+
\frac{\alpha h_0^2}{4(1-\epsilon-\kappa)p}\right)
\le s\left(.28^2+\frac{.075\cdot .67^2}{4\cdot .36}\right)<.111.
\end{equation}
It remains to bound the terms involving $T-\log S$ while retaining 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 decomposition of $A$ into its first chaos and
$Y$ gives
\begin{equation}\label{sc:spectral-moments}
\int a\,d\nu_v(a)=0,\qquad
\int a^2\,d\nu_v(a)=\lambda+r,
\qquad r=v^{\mathsf T}\E(Y^2)v\ge0.
\end{equation}
Indeed, orthogonality of the Gaussian chaoses gives
$\E A^2=B+\E Y^2$, including the vanishing of both matrix cross terms.
For $0<u\le.13$, define
\[
n_u=\frac{u^{3/2}}{2},\qquad
d_u=\frac{u^3}{16(.46-u)}.
\]
We claim the following 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
$p(a)\ge u/(x^2+7u)$ and
\[
1-(1-x/2)(x^2+7u)
\ge \frac{x(x-1)^2}{2}+.045(2-x)\ge0.
\]
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
$u+n_u^2/(4d_u)=.46$. Thus~\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 $\lambda u^2\le1.28^2$, integration of the
minorant against~\eqref{sc:spectral-moments} yields
\begin{equation}\label{sc:p-mean}
\int p\,d\nu_v\ge u-d_u(\lambda+r)
\ge .689u-d_ur.
\end{equation}
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 this, note that $\rho\le.64$ and that $H\le\lambda$ for $\lambda\ge1$,
whereas $H\le\alpha s$ 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 next 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(.089,.88/\sqrt\lambda)\le .688u,
\end{equation}
where the second bound in the minimum is interpreted as $+\infty$ at
$\lambda=0$. First,
\[
F_0(S)=\int_0^1 \frac{Sx}{1-x+xS}\,dx
\]
is concave. Its tangent at $S=.28$ has intercept below $.16$ and slope
below $.52$, so $F_0(S)\le .16+.52S$. These two strict inequalities
can, for example, be checked from
\[
-\log(.28)=2\sum_{j=0}^{\infty}\frac{(9/16)^{2j+1}}{2j+1},
\]
using twelve terms and the geometric bound for the remainder.
Coefficient comparison now gives
\[
.075(.16+.52S)<.089(.15+.45325S),
\]
which proves the first bound in 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 the point $S=1$ 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
\[
x^2J'(x)=P(x):=-.478+2x+.444x^2-2x^3-6.966x^4+7x^6.
\]
Descartes' rule of signs gives at most three positive roots of $P$,
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 $1$. Thus the only local minima of $J$ occur at the
first root and at $1$. We have $J(1)=0$ and $J(.25)>.14$; the latter
follows already from $\log2<.694$. On $[.24,.26]$,
$|P(.25)|<.01$ and $|P'|<4$, hence $|J'|<2$.
The first minimum is therefore positive. Also $J(x)\to+\infty$ at
both ends of $(0,\infty)$. This proves $J\ge0$, which is equivalent to
\eqref{sc:F0-sqrt} away from $x=1$.
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 decreases in $S$ because
$1.4\cdot.15<.478\cdot.45325$. Finally,
$.089<.688\cdot.13$ and $.88<.688\cdot1.28$ prove the last inequality
in~\eqref{sc:log-bound}.
For $\lambda\ne1$ we have
\[
\frac{T-\log S}{H\rho}=\frac{\alpha F_0(S)}{\rho}.
\]
Combining~\eqref{sc:p-mean}--\eqref{sc:log-bound}, and treating $\lambda=1$
by continuity, gives
\[
T-\log S-H\rho\int p\,d\nu_v
\le -.001H\rho u+.027r.
\]
The definitions give $H\rho u=\omega(\lambda)$. Sum over an orthonormal
eigenbasis of the matrix $B$, and add~\eqref{sc:y-cost}.
This summation evaluates $\tau(H\rho p(A))$ without requiring $p(A)$
to commute with $B$. Likewise, \eqref{sc:y-cost} may be integrated
against the positive semidefinite matrix $Y^2$ without a commutation
assumption. Since $\sum_v r=\norm{Y}^2$, the result is
\[
(.027+.111)\norm{Y}^2-.001\sum_i\omega(\lambda_i)
=\frac{69}{500}\norm{Y}^2-\frac1{1000}\sum_i\omega(\lambda_i),
\]
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} immediately proves the assertion.
We henceforth suppose $\delta>0$.
We first isolate the cost of the secant increments at one endpoint $a_j$.
Use coordinates $(G,h)$ with $h=g-G/2$. With $Z=0$, the left side of
the endpoint quadratic 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+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}.
The first line of that formula is the coefficient of $Z^2$ before
completion of squares.
The next step estimates these increment norms by integrating derivatives
between the endpoints. Recall that a secant slope is the uniform average
of the derivative over $[a_-,a_+]$. With $\xi=\log q(a)$, differentiation
therefore 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, $(\Delta k,\Delta b,\psi_j)$ is the uniform average in
$a$ of $\int_{\log q(a_j)}^{\log q(a)}v_j(q^{-1}(e^\xi))\,d\xi$.
This follows from $b'=l$, $l'=zl$, $d(\log q)/da=d$, and
$f''=l(1+3b+kz)=lh_*$.
Put
\[
F_j(a)=\mathcal N_j(v_j(a))^2.
\]
We record the bounds furnished by Lemma~\ref{sc:bounds}. If the endpoint
$a_j$ belongs to $R_1,U_1,V_1,T_1$, respectively, then
\begin{equation}\label{sc:F-simple}
F_j(a)\le .295,\quad .23,\quad .23,\quad .39
\qquad\hbox{for all }a.
\end{equation}
For $a\le-1$, the $R_1$ bound improves to $.20$.
For $a_j\in P_1$, the bounds are $.42^2$ when $a\le-1$ and $.61^2$
when $a\ge-1$. 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}
For clarity, the complete finite arithmetic behind these bounds is given
by the following table. Its three entries bound, in order,
\[
|W_j|,\qquad
|A_*b_j-\eta_*/2+\mathsf L_jz+K_{0j}W_j|,\qquad
|-\eta_*+J_{0j}W_j|.
\]
\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 in
\eqref{sc:interval-table} and~\eqref{sc:derivative-table}; each expression
is affine in each of its variables separately, so its extrema occur at
corners of the corresponding box. For the middle expression, first write
it as
\[
(A_*-2K_{0j})b_j-\eta_*/2+K_{0j}h_*+
(1-2K_{0j})\mathsf L_jz.
\]
Substituting these three bounds into~\eqref{sc:increment-norm}, together
with the bounds for $\mathsf D_j,\ell_{1j},z,\chi$, proves
\eqref{sc:F-simple}--\eqref{sc:F-negative}. All these are finite rational
comparisons; the exact verifier accompanying Appendix~\ref{cert:verification}
also checks them.
It remains to compare the integrated derivative cost with the logarithmic
separation surplus. Let $u_-$ be the mean of
$(\log q(a)-\log q(a_-))/\delta$ under the uniform probability measure
on $[a_-,a_+]$, and put $u_+=1-u_-$. In the $\xi$ coordinate this measure
has density proportional to $D_0=1/d$. The excess-variance bound in
Lemma~\ref{tr:q-properties} gives
\[
0\le \frac{d\log D_0}{d\xi}=\frac{1-qd}{d^2}\le1.
\]
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
observation to the coordinate $\xi$ compares its mean first with constant
density and then with density proportional to $e^\xi$. The latter comparison
gives
$u_-\le\tfrac12+\tfrac12\coth(\delta/2)-1/\delta$, and hence
\begin{equation}\label{sc:mean-distance}
\tfrac12\le u_-\le\tfrac12+\delta/12,
\qquad u_-^2+u_+^2\le\tfrac12(1+\delta^2/36).
\end{equation}
For the upper bound one uses $\coth x-1/x\le x/3$, obtained by comparing
the power series of $x\cosh x$ and $(1+x^2/3)\sinh x$.
Suppose $P_-$ and $P_+$ have the following property: 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 $F_j$
are one way to ensure this property. 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 $u_-\ge u_+\ge0$,
\[
P_-u_-^2+P_+u_+^2
\le\max\left(P_-,\frac{P_-+P_+}{2}\right)(u_-^2+u_+^2).
\]
Averaging~\eqref{sc:one-endpoint} and applying
\eqref{sc:mean-distance} therefore proves the desired estimate whenever
\begin{equation}\label{sc:threshold}
\max\left(P_-,\frac{P_-+P_+}{2}\right)\le
\Theta_{m_*}(\delta):=
\frac{1+m_*}{4}\,
\frac{1+\delta^2/30+\delta^4/1680}{1+\delta^2/36}.
\end{equation}
Indeed, the resulting coefficient of $Z^2$ is at most
$\delta^2(1+\delta^2/36)\Theta_{m_*}(\delta)/4
=r_{-+}\delta^2$.
We verify~\eqref{sc:threshold} for all endpoint positions. The function
$\Theta_m$ increases in $\delta\ge0$: differentiation with respect to
$y=\delta^2$ leaves a positive numerator
$1/180+2y/1680+y^2/(1680\cdot36)$.
We will also use the following observation about partial intervals:
if a smaller bound holds on an initial segment of logarithmic length $L$,
then its fraction in a partial interval of length $h\le\delta$ is at least
$\min(1,L/h)\ge L/\delta$, provided $\delta\ge L$.
If $a_-\in R_1$, take $P_-\le.295$ and $P_+\le.39$;
then $(P_-+P_+)/2\le.3425<\Theta_{.38}(0)=.345$.
For $a_-\in U_1$ or $V_1$, use $P_-\le.23$, $P_+\le.39$ and
$\Theta_{.68}(0)=.42$. For $a_-\in T_1$, both bounds are $.39<.42$.
Now suppose $a_-\in P_1$. 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 $1.62$. Hence every partial
interval from the lower endpoint has mean norm at most
\[
.61-\frac{(.61-.42)1.62}{\delta},\qquad
P_-=(.61-.3078/\delta)^2.
\]
The length assertion follows from $q(-1)>.287$, $q(-2)<.056$.
If the upper endpoint lies in $T_1$, then $\delta>4$ because $q(4)>4$.
For $\delta\le4$, therefore, $P_+\le.23$ and
$P_-\le.284143<\Theta_{.19}(0)=.2975$.
For $4\le\delta\le5$, use $P_-<.300787$, $P_+\le.39$, and
\[
\max(.300787,(.300787+.39)/2)<.345394
<\Theta_{.19}(4).
\]
For $\delta\ge5$, use $P_-\le.61^2=.3721$ and
\[
\max(.3721,(.3721+.39)/2)=.38105<\Theta_{.19}(5).
\]
This covers $P_1$.
Finally suppose $a_-\in O$. 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 follows by integrating $d\ge-a$ on the negative half-line;
the second follows from $l\le .075\cdot1.19<.09$ and $d\ge|a|$.
Using $\chi\le.07$ in~\eqref{sc:F-negative} gives
\begin{equation}\label{sc:O-low-cost}
F_j(a)\le .144+.011\delta.
\end{equation}
If the whole interval lies at or below $-1$, take
$P_-=.144+.011\delta$ and
$P_+\le\max(P_-,.20)$; the latter bound also covers upper endpoints
in $P_1$ and $R_1$.
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}
This follows on clearing denominators and using $\delta^2\ge16$.
The quadratic $.106-.011\delta+.00260\delta^2$ is strictly positive,
so~\eqref{sc:threshold} follows in this case too.
If $a_+>-1$, the first part of the logarithmic interval has length at
least $4$, since $q(-3)<.005$ and $q(-1)>.287$.
For $a\ge-1$, the same integration gives $t_-^2\le1+2\delta$;
using $\chi\le.097$ in~\eqref{sc:F-negative} yields
$F_-(a)\le .236+.031\delta$.
On every partial interval from the lower endpoint, Jensen's inequality
therefore permits the uniform bound
\[
P_-=.236+.031\delta-\frac{4}{\delta}(.092+.020\delta)
=.156+.031\delta-.368/\delta.
\]
For partial intervals contained below $-1$, the same bound follows from
\eqref{sc:O-low-cost}, because $\delta\ge4$.
By~\eqref{sc:theta-tail},
\[
\Theta_0(\delta)-P_-
\ge .094-.031\delta+.00260\delta^2+.368/\delta>0;
\]
the quadratic has 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 $q(4)>4$ and $q(-3)<.005$,
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}
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//scalar.tex ====
==== BEGIN A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//transport.tex ====
\section{Gaussian transport and a matrix surplus}\label{tr:section}
We first work with a probability density $e^{-V}$ on $\mathbb R^m$ satisfying
\begin{equation}\label{tr:smooth-assumptions}
V\in C^\infty(\mathbb R^m),\qquad
0<c_- I\le D^2V\le c_+I<\infty.
\end{equation}
The constants $c_-,c_+$ may depend on the density. Write
$C=\operatorname{Cov}(\mu)$, where $d\mu=e^{-V}\,dy$, and let
$\gamma_m$ be standard Gaussian measure. Throughout this section,
$\mathbb E$ denotes integration against $\gamma_m$.
There are two objectives. We choose affine coordinates in which a
reparametrization of the transport Jacobian has mean zero. We then use
convexity of $V$ to control derivatives of that reparametrization. The
resulting estimate contains an additional nonnegative term measuring
separation between eigenvalues; retaining this term is essential in the
matrix argument that follows.
\subsection{The Gaussian parametrization}
Let $\Phi$ and $\varphi$ denote the distribution function and density of a
standard one-dimensional Gaussian. Define, for $a\in\mathbb R$,
\begin{equation}\label{tr:scalar-functions}
\begin{gathered}
t(a)=-\log\Phi(-a),\qquad q(a)=t'(a)=\frac{\varphi(a)}{\Phi(-a)},
\qquad d(a)=q(a)-a,\\
k(a)=\frac{t(a)}{q(a)},\qquad b(a)=k'(a),\qquad l(a)=b'(a).
\end{gathered}
\end{equation}
The map $t$ transports a standard Gaussian to the exponential law of
mean one: $\Phi(-G)$ is uniform on $(0,1)$ when $G$ is standard Gaussian.
Thus $q$ is the derivative of the scalar transport associated with the
entropy benchmark $h=1$, $\operatorname{Var}=1$.
The identities
\begin{equation}\label{tr:scalar-identities}
q'=qd,\qquad b=1-kd,\qquad l=k-q+ab
\end{equation}
follow by differentiation. In particular, $q-k=ab-l$. We record the
elementary properties that permit us to use $q$ as a change of variable.
\begin{lemma}\label{tr:q-properties}
The function $q$ is a smooth increasing bijection from $\mathbb R$ to
$(0,\infty)$. Moreover,
\[
d>0,\qquad 0<q'<1,\qquad q''\ge0,
\qquad 0<1-qd\le d^2.
\]
\end{lemma}
\begin{proof}
Let $X_a$ have the conditional law of $G-a$ given $G>a$, and write
$\mathbb E_a$ for expectation under this law. Gaussian integration
by parts gives
\[
\mathbb E_a X_a=d(a),\qquad
\operatorname{Var}(X_a)=1-q(a)d(a).
\]
Since $X_a>0$ almost surely, $d>0$ and hence $q'=qd>0$.
Its strictly positive variance gives $q'<1$.
For completeness, this variance is at most the square of its mean.
The survival function
\[
S_a(u)=\frac{\Phi(-a-u)}{\Phi(-a)},\qquad u\ge0,
\]
satisfies $S_a(u+v)\le S_a(u)S_a(v)$, since its hazard rate $q(a+u)$
is increasing. Integrating over $u,v\ge0$ yields
\[
\frac12\mathbb E_a X_a^2
=\int_0^\infty uS_a(u)\,du
\le\left(\int_0^\infty S_a(u)\,du\right)^2=d(a)^2.
\]
Consequently $1-qd=\operatorname{Var}(X_a)\le d^2$. Differentiating
$q'=qd$ now gives
$q''=q(d^2+qd-1)\ge0$. Finally, $q(a)\to0$ as $a\to-\infty$,
whereas $q(a)>a$ for $a>0$. These limits prove surjectivity.
\end{proof}
We shall also need two positivity bounds and a quantitative growth estimate.
Their proof, including
the rational bounds on the compact interval and estimates on both tails,
is given in Appendix~\ref{cert:verification}.
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}
A lower bound for the growth of $M$ in $\log q$ will let us compare
uniform averages in $q$ with uniform averages in $\log q$ in the
surplus estimate.
Define a nondecreasing step function $m_0$ by the following table. At a
finite endpoint we take the value in the interval to its right.
\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}
All terminating decimals in this paper denote exact rational numbers.
\begin{lemma}[Scalar bounds]\label{tr:scalar-basic}
The functions in \eqref{tr:scalar-functions} satisfy, on $\mathbb R$,
\begin{equation}\label{tr:scalar-basic-equations}
0<b<\frac12,\qquad l>0,\qquad
\frac{d\log M}{d\log q}
:=\frac{M'}{Md}\ge m_0.
\end{equation}
\end{lemma}
For a real symmetric matrix, scalar functions will always be interpreted
by spectral functional calculus. If $f$ is differentiable, its divided
difference is
\begin{equation}\label{tr:divided-difference}
f[u,v]=\begin{cases}(f(u)-f(v))/(u-v),&u\ne v,\\f'(u),&u=v.
\end{cases}
\end{equation}
The classical Daleckii--Krein formula
\cite{DaleckiiKrein1965} states that, in an eigenbasis of a symmetric
matrix $A$, the derivative of $f(A)$ in a symmetric direction $Z$ has
entries $f[a_i,a_j]Z_{ij}$; see also
\cite[Section~2.4]{Noferini2017}. This formula
will allow us to keep the off-diagonal entries of matrix derivatives.
\subsection{Transport and the choice of coordinates}
Brenier's theorem gives a convex potential $\phi$ such that
$(\nabla\phi)_\#\gamma_m=\mu$; its gradient is unique up to null sets
\cite{Brenier1991,McCann1995}. For $m\ge2$ we use the full-space
regularity theorem of Cordero-Erausquin and Figalli
\cite[Theorem~1.1]{CorderoErausquinFigalli2019}, which extends
Caffarelli's interior theory \cite{Caffarelli1992} to the unbounded
domains considered here. It applies with both domains equal to $\R^m$, because both densities
are smooth and locally bounded away from zero. For $m=1$, the same
smoothness follows directly by expressing monotone transport through
the two distribution functions. Thus $\phi$ is smooth in every dimension.
Caffarelli's contraction theorem \cite{Caffarelli2000} and its inverse
bound then imply that the Jacobian
\begin{equation}\label{tr:Jacobian}
J=D^2\phi
\end{equation}
satisfies
\begin{equation}\label{tr:Jacobian-bounds}
c_+^{-1/2}I\le J\le c_-^{-1/2}I.
\end{equation}
For the lower bound, apply the contraction theorem to the inverse
transport. We use its quantitative form: a source potential with
Hessian at most $aI$ and a target potential with Hessian at least $bI$
give a transport Lipschitz constant at most $\sqrt{a/b}$; see also
\cite[Theorem~1]{ChewiPooladian2023}.
Define the symmetric matrix field
\begin{equation}\label{tr:A-definition}
A=q^{-1}(J).
\end{equation}
The bounds \eqref{tr:Jacobian-bounds} place the spectrum of $A$ in a
compact interval. Our preferred coordinates are characterized by
$\mathbb EA=0$. They need not be the usual isotropic coordinates.
\begin{proposition}[Affine normalization]\label{tr:normalization}
Let $\mu=e^{-V}dy$ satisfy \eqref{tr:smooth-assumptions}. There exists a
positive definite symmetric matrix $P$ such that the Brenier map from
$\gamma_m$ to $P_\#\mu$ has Jacobian $J_P$ satisfying
\[
\mathbb E q^{-1}(J_P)=0.
\]
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 that its zeros stay in a
compact subset of the positive definite cone. For $0\le\theta\le1$,
define
\[
W_\theta(y)=(1-\theta)|y|^2/2+\theta V(y),\qquad
d\mu_\theta=Z_\theta^{-1}e^{-W_\theta(y)}\,dy,
\]
where $Z_\theta$ is the normalizing integral.
There are constants $a_-,a_+>0$, independent of $\theta$, with
$a_-I\le D^2W_\theta\le a_+I$. These densities have uniformly
Gaussian tails. Their covariance matrices $C_\theta$ depend
continuously on $\theta$, are positive definite, and therefore satisfy
\begin{equation}\label{tr:homotopy-covariance}
c_0I\le C_\theta\le C_0I
\end{equation}
for fixed $c_0,C_0>0$.
For a positive definite symmetric $P$, let $J_{\theta,P}$ be the
Jacobian of the transport to $P_\#\mu_\theta$. Define a map with
values in the vector space of symmetric matrices by
\[
F(\theta,P)=\mathbb E q^{-1}(J_{\theta,P}).
\]
Contraction in both directions 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}
We first verify continuity of $F$. Restrict $P$ to any 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 their means and second moments are bounded. Every
convergent sequence $(\theta_j,P_j)$ thus has locally uniformly
convergent subsequences of transport maps. Their limits push forward
$\gamma_m$ to the limiting target and remain gradients of convex
functions, so Brenier uniqueness identifies the limit. Write $J_j,J$
for the corresponding Jacobians. They converge weak-* locally and
obey common positive spectral bounds.
The change-of-variables equation expresses $\log\det J_j$ in terms
of the target potential composed with its transport map. It therefore
converges locally uniformly to $\log\det J$. On a fixed compact
spectral interval, strict concavity of $\log\det$ gives a constant
$c>0$ such that
\[
\log\det J_j\le\log\det J+
\operatorname{tr}\bigl(J^{-1}(J_j-J)\bigr)-c|J_j-J|_{\mathrm F}^2.
\]
Integration on any ball, followed by weak-* convergence, proves
$J_j\to J$ in local $L^2$. Functional calculus then gives local
$L^2$ convergence of $q^{-1}(J_j)$. Their common spectral bounds
control the Gaussian tails, proving continuity of $F$.
We next bound all zeros, uniformly in $\theta$. Suppose
$\mathbb EA=0$, where $A=q^{-1}(J_{\theta,P})$. Convexity of $q$
and Jensen's inequality on the expected spectral measure of each unit
vector give
\[
\mathbb E J_{\theta,P}\ge q(0)I.
\]
Gaussian integration by parts gives
$\mathbb E(x_r\partial_i\phi)=(\mathbb E J_{\theta,P})_{ir}$.
Orthogonal projection of the centered components of $\nabla\phi$
onto the span of $x_1,\ldots,x_m$ consequently yields
\[
P C_\theta P\ge(\mathbb E J_{\theta,P})^2\ge q(0)^2I.
\]
Together with \eqref{tr:homotopy-covariance}, this bounds $P^{-1}$
uniformly. The lower estimate in \eqref{tr:homotopy-spectral} now
gives a uniform lower bound $A\ge a_*I$, where we may take $a_*\le0$.
Since $q'\le1$,
\[
q(a)\le q(a_*)+a-a_*\quad(a\ge a_*),
\qquad
\mathbb E\operatorname{tr}J_{\theta,P}
\le m\bigl(q(a_*)-a_*\bigr).
\]
On the other hand, Gaussian Poincar\'e and
\eqref{tr:homotopy-spectral} imply
\[
c_0\|P\|_{\mathrm{op}}^2
\le\operatorname{tr}(P C_\theta P)
\le\mathbb E\operatorname{tr}J_{\theta,P}^2
\le\frac{\|P\|_{\mathrm{op}}}{\sqrt{a_-}}
\mathbb E\operatorname{tr}J_{\theta,P}.
\]
This 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 $q(0)I$;
its derivative there is multiplication by the positive scalar
$1/q'(0)$. Its Brouwer degree is therefore nonzero. Homotopy
invariance of degree gives a zero at $\theta=1$, as required.
\end{proof}
The quantity
$m+\tfrac12\log\det\operatorname{Cov}(\mu)-h(\mu)$ is unchanged
by an invertible affine transformation. Indeed, a linear map $P$
adds $\log|\det P|$ to entropy and $2\log|\det P|$ to the covariance
log determinant. We may therefore impose the normalization from
Proposition~\ref{tr:normalization}. From now on $V,J,A,C$ refer to
this transformed density and transport, and
\begin{equation}\label{tr:mean-zero}
\mathbb EA=0.
\end{equation}
\subsection{The transport equation and its energy}
We shall differentiate the Jacobian and integrate the resulting equation.
The twice-differentiated change-of-variables equation is part of
Caffarelli's transport method \cite[Section~5]{Caffarelli2000}.
The next lemma establishes the finite energy needed to do this without
additional decay assumptions on higher derivatives of $V$.
For a matrix field $F$, set
\[
|F|_{\mathrm F}^2=\operatorname{tr}(F^{\mathsf T}F),\qquad
\|F\|^2=\mathbb E|F|_{\mathrm F}^2,\qquad
\|\partial F\|^2=\sum_{r=1}^m\|\partial_rF\|^2.
\]
The same convention sums over any displayed derivative index of a
collection of matrix fields.
\begin{lemma}[Jacobian equation and finite energy]\label{tr:energy}
Under \eqref{tr:smooth-assumptions}, put $J_r=\partial_rJ$ and
\[
\mathcal L u=\operatorname{tr}(J^{-1}D^2u)
-\nabla V(\nabla\phi)\cdot\nabla u.
\]
Then
\begin{equation}\label{tr:Jacobian-equation}
\mathcal L J_{ij}
=-\delta_{ij}+(JD^2V(\nabla\phi)J)_{ij}
+\operatorname{tr}(J^{-1}J_iJ^{-1}J_j).
\end{equation}
Moreover $\|\partial J\|<\infty$. The matrix fields $A$ and $k(A)$
belong to Gaussian $H^1$, and the integrations by parts below with
smooth matrix functions of $J$ are justified by compact cutoffs.
\end{lemma}
\begin{proof}
The density equation is
\begin{equation}\label{tr:density-equation}
\log\det J=V(\nabla\phi)-|x|^2/2-(m/2)\log(2\pi).
\end{equation}
Differentiating twice gives \eqref{tr:Jacobian-equation}. The divergence
of the cofactor matrix of a Hessian vanishes. Applying this identity
to $J$, and then differentiating \eqref{tr:density-equation} once,
shows that
\[
\mathcal L u=e^{|x|^2/2}\operatorname{div}
\bigl(e^{-|x|^2/2}J^{-1}\nabla u\bigr).
\]
Thus, for smooth $u$ and compactly supported smooth $v$,
\begin{equation}\label{tr:divergence-form}
\mathbb E(v\mathcal Lu)
=-\mathbb E\langle\nabla v,J^{-1}\nabla u\rangle.
\end{equation}
Choose $0<\lambda\le\Lambda<\infty$ such that
$\lambda I\le J\le\Lambda I$. Taking the trace in
\eqref{tr:Jacobian-equation}, and using $D^2V\ge0$, gives
\[
\mathcal L(\operatorname{tr}J)
\ge-m+\Lambda^{-2}\sum_r|J_r|_{\mathrm F}^2.
\]
For a compactly supported cutoff $0\le\chi\le1$, integration by parts
and $|\nabla\operatorname{tr}J|\le\sqrt m\,|\partial J|$ imply
\begin{align*}
\Lambda^{-2}\mathbb E\chi^2|\partial J|^2
&\le m+2\lambda^{-1}\sqrt m\,
\mathbb E\bigl(\chi|\nabla\chi|\,|\partial J|\bigr)\\
&\le m+\tfrac12\Lambda^{-2}\mathbb E\chi^2|\partial J|^2
+2m\Lambda^2\lambda^{-2}\mathbb E|\nabla\chi|^2.
\end{align*}
Let $\chi$ tend to one through standard radial cutoffs with gradients
bounded by a constant divided by their radii. Fatou's lemma proves
$\|\partial J\|<\infty$.
All eigenvalues of $J$ lie in $[\lambda,\Lambda]$. The Fr\'echet
derivatives of $q^{-1}$ and of $k\circ q^{-1}$ are bounded on that
spectral interval, so $A,k(A)\in H^1(\gamma_m)$. The same applies to
any smooth scalar function on this interval. For such a matrix
multiplier, insert a cutoff into \eqref{tr:divergence-form} and let it
tend to one. The terms containing its gradient vanish by
Cauchy--Schwarz and the energy bound; the remaining differentiated
terms are bounded by a constant times $1+|\partial J|^2$.
This justifies the asserted integrations by parts.
\end{proof}
\subsection{Keeping the eigenvalue separation}
Set $Z_r=\partial_r A$. At each point diagonalize $A$, and write
$a_i$ for its eigenvalues and $\lambda_i=q(a_i)$ for those of $J$.
For real $a,b$, define
\begin{equation}\label{tr:surplus-weight}
\begin{gathered}
\delta(a,b)=|\log q(a)-\log q(b)|,\qquad
m_*(a,b)=m_0(\min(a,b)),\\
r(a,b)=\frac{1+m_*(a,b)}{16}
\left(1+\frac{\delta(a,b)^2}{30}
+\frac{\delta(a,b)^4}{1680}\right).
\end{gathered}
\end{equation}
We use the abbreviations $\delta_{ij}=\delta(a_i,a_j)$ and
$r_{ij}=r(a_i,a_j)$. The quantity
\begin{equation}\label{tr:surplus-definition}
\mathcal C=\mathbb E\sum_{i,j,r}
r_{ij}\delta_{ij}^2(Z_r)_{ij}^2
\end{equation}
is independent of the choice of orthonormal basis within repeated
eigenspaces. In particular, 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}
\mathbb E\operatorname{tr}k(A)^2
\ge\|\partial A\|^2-\|\partial k(A)\|^2+\mathcal C.
\end{equation}
\end{proposition}
\begin{proof}
The proof has two parts. Symmetry of the third derivatives of $\phi$
first replaces the transport equation by averages of the scalar
function $M$. Comparing arithmetic and logarithmic averages then
produces the term $\mathcal C$.
The identity $qk=t$ will match the entropy term in the covariance
completion of Section~\ref{mat:section}; we first estimate the derivative
cost of $k(A)$. Put $F(\lambda)=k(q^{-1}(\lambda))^2$. Multiply
\eqref{tr:Jacobian-equation} by $F(J)$ in trace and integrate by parts.
Lemma~\ref{tr:energy} justifies this operation. 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 on the right of the Jacobian
equation has trace contraction
\[
\sum_i F(\lambda_i)\operatorname{tr}(J^{-1}J_iJ^{-1}J_i)
=\sum_{i,j,r}\frac{F(\lambda_i)}{\lambda_j\lambda_r}\Theta_{ijr}^2,
\]
where we used the symmetry of $\Theta$. The term obtained by integration
by parts is
\[
\sum_r\lambda_r^{-1}\operatorname{tr}\bigl((\partial_rF(J))J_r\bigr)
=\sum_{i,j,r}\frac{F[\lambda_i,\lambda_j]}{\lambda_r}\Theta_{ijr}^2.
\]
The remaining term is
$\operatorname{tr}(F(J)JD^2VJ)\ge0$.
Discarding it gives
\begin{equation}\label{tr:initial-multiplier}
\mathbb E\operatorname{tr}k(A)^2
\ge\mathbb E\sum_{i,j,r}\Theta_{ijr}^2
\left(\frac{F(\lambda_i)}{\lambda_j\lambda_r}
+\frac{F[\lambda_i,\lambda_j]}{\lambda_r}\right).
\end{equation}
Only pointwise changes of basis are made here; no eigenvectors are
differentiated.
Regard $M$ as a function of $\lambda=q(a)$. The identities
\eqref{tr:scalar-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 two positive numbers $u,v$, let $M^{\mathrm{ar}}_{uv}$ be the
uniform average of $M$ on the interval between $u$ and $v$, with its
continuous interpretation when $u=v$. With the weight
$\Theta_{ijr}^2/(\lambda_i\lambda_j\lambda_r)$, symmetrization in
$i,j$ changes the coefficient in \eqref{tr:initial-multiplier} to
\[
\frac{\lambda_i+\lambda_j}{2}\,
M^{\mathrm{ar}}_{\lambda_i\lambda_j}.
\]
In the full sum we can replace this coefficient by
$\lambda_rM^{\mathrm{ar}}_{\lambda_i\lambda_j}$, decreasing its
value. To see this, order three eigenvalues as $u\le v\le w$.
Lemma~\ref{tr:scalar-basic} makes $M$ increasing, so
\[
M^{\mathrm{ar}}_{uv}\le M^{\mathrm{ar}}_{uw}
\le M^{\mathrm{ar}}_{vw}.
\]
The ordered coefficients $u+v-2w$, $u+w-2v$, and $v+w-2u$ sum to
zero. Their scalar product with these three ordered averages is
nonnegative, by the rearrangement inequality. Full symmetry of
$\Theta$ permits precisely this averaging over permutations of the
three indices. We have proved
\begin{equation}\label{tr:arithmetic-energy}
\mathbb E\operatorname{tr}k(A)^2
\ge\mathbb E\sum_{i,j,r}
\frac{\Theta_{ijr}^2}{\lambda_i\lambda_j\lambda_r}\,
\lambda_r M^{\mathrm{ar}}_{\lambda_i\lambda_j}.
\end{equation}
We now fix a pair of eigenvalues. Bars in the rest of the proof denote
uniform averages in $\log\lambda$ over their interval. Set
\[
x=\delta_{ij}/2,\qquad w_0=\frac{x}{\sinh x},\qquad m_*=m_*(a_i,a_j),
\]
with $w_0=1$ at $x=0$. Since $da/d\log\lambda=D_0$,
the divided-difference formula shows that the coefficients of
$(Z_r)_{ij}^2$ and $(\partial_r k(A))_{ij}^2$, respectively, when
written with the same weight as in \eqref{tr:arithmetic-energy}, are
\begin{equation}\label{tr:derivative-coefficients}
\lambda_r w_0^2\overline{D_0}^{\,2},\qquad
\lambda_r w_0^2\overline{bD_0}^{\,2}.
\end{equation}
Both $D_0(1+b)$ and $D_0(1-b)=k$ are increasing: $d'=qd-1<0$,
and $b'=l>0$, $k'=b>0$. The covariance inequality for two increasing
functions on an interval therefore yields
\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}
We also need a quantitative comparison of the two averaging measures.
On the logarithmic interval, \eqref{tr:scalar-basic-equations} says
that $M(\lambda)\lambda^{-m_*}$ 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
$\mathbb E_0(e^sM)/[\mathbb E_0e^s\,\mathbb E_0M]$, where
$\mathbb E_0$ is its uniform average. Under the measure proportional
to $e^{m_*s}$, the increasing factor $Me^{-m_*s}$ can only increase
the mean of $e^s$. Taking $M=e^{m_*s}$ gives the displayed ratio.
Here is a convenient explicit lower bound for this ratio. Define
$A_0(x)=x\coth x-1$, with $A_0(0)=0$. The identity
\[
x\coth x+\frac{x^2}{\sinh^2x}\ge2
\]
implies both $A_0(x)\ge1-w_0^2$ and that $A_0(x)/x^2$ is
nonincreasing for $x>0$. To verify the identity, clear the denominator
and expand
$\tfrac{x}{2}\sinh(2x)+x^2-\cosh(2x)+1$ in powers of $x$;
the coefficients vanish through degree four and are nonnegative
thereafter. Since $0\le m_*\le1$, we obtain
$A_0(m_*x)\ge m_*^2A_0(x)$. The addition formula for $\sinh$ now 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}
Every formula here extends continuously to $m_*=0$ or $x=0$.
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
\[
\frac34(1+m_*)(w_0^{-2}-1)
\ge\frac{1+m_*}{16}
\left(\delta_{ij}^2+\frac{\delta_{ij}^4}{30}
+\frac{\delta_{ij}^6}{1680}\right).
\]
The last inequality is the Taylor expansion of
$(\sinh x/x)^2$, whose omitted coefficients are positive.
Substituting in \eqref{tr:arithmetic-energy} and summing proves
\eqref{tr:surplus-inequality}.
\end{proof}
We have obtained two inputs for the entropy argument: the matrix field
$A$ has no constant Gaussian component, and convexity supplies
\eqref{tr:surplus-inequality}. The next section combines these facts
with a Gaussian covariance identity and separates the first chaos of
$A$ from its higher chaoses.
==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//transport.tex ====
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==== END A-sharp-entropy-bound-and-the-simplex-inequality-for-isotropic-constants//references.bib ====