64c3814
==== TABLE OF CONTENTS ====
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//01_introduction.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//02_gauges.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//03_spherical.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//04_norm.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//05_position.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//06_conclusion.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//07_consequences.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//bibliography.tex
Pettys-projection-volume-conjecture-in-dimensions-at-least-four//main.tex
===========================
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//01_introduction.tex ====
\section{Introduction}
\label{sec:introduction}
For a convex body $K\subset\R^n$, its projection body $\Pi K$ is the
origin-symmetric convex body whose support function is
\[
h_{\Pi K}(u)=\operatorname{vol}_{n-1}
\bigl(\operatorname{proj}_{u^\perp}K\bigr),
\qquad u\in S^{n-1}.
\]
Here a convex body is compact, convex, and has nonempty interior, and
$h_L(x)=\max_{y\in L}\ip{x}{y}$ denotes its support function.
Thus $\Pi K$ records the volumes of the orthogonal shadows of $K$ in
all directions. Write $|L|=\operatorname{vol}_n(L)$ and let $B_2^m$
be the Euclidean unit ball in $\R^m$, with volume $\kappa_m$.
The normalized projection volume is
\[
R_n(K)=\frac{|\Pi K|}{|K|^{n-1}}.
\]
It is invariant under translations and invertible linear maps: the
classical contravariance identity is
$\Pi(TK)=|\det T|T^{-t}\Pi K$; see
\cite[Equation~(1)]{Ludwig2002}. We verify this identity directly in
Section~\ref{sec:conclusion}.
Petty's projection-volume conjecture asserts, for $n\ge3$, that
$R_n$ is minimized precisely by ellipsoids \cite{Petty1971}.
The ball satisfies $\Pi B_2^n=\kappa_{n-1}B_2^n$, so the proposed
minimum is fixed by the Euclidean case. This is an affine isoperimetric
problem in which directional projection data are assembled into a
second body before its volume is measured. The classical Petty
inequality for the \emph{polar} of the projection body concerns a
different functional; see \cite[Section~18.6]{Gardner2002}.
We prove the unpolarized conjecture in every dimension at least four.
\begin{theorem}\label{thm:main}
For every integer $n\ge4$ and every convex body $K\subset\R^n$,
\begin{equation}\label{eq:main}
\frac{|\Pi K|}{|K|^{n-1}}
\ge \kappa_{n-1}^{\,n}\kappa_n^{\,2-n}.
\end{equation}
Equality holds if and only if $K=a+TB_2^n$ for some $a\in\R^n$ and some
invertible linear map $T$.
\end{theorem}
The theorem includes bodies without symmetry or boundary regularity.
Its equality assertion is global and imposes no proximity to an
ellipsoid.
\subsection*{Previous results and methods}
One approach studies a body together with its second projection body.
The class-reduction inequality $R_n(\Pi K)\le R_n(K)$, together with
its equality condition, reduces the study of minimizers to bodies
homothetic to $\Pi^2K$; see \cite[Sections~1--2]{Saroglou2015}.
Saroglou and Zvavitch proved a local minimum theorem under the
hypothesis that the density of the surface-area measure, after an
invertible linear change of variables, is sufficiently close to that
of the ball in $L^\infty$ \cite[Theorem~1.3]{SaroglouZvavitch2017}.
Ivaki established local rigidity for $C^2$ solutions of
$\Pi^2K=cK$ near the ball after an invertible linear change of
variables \cite{Ivaki2018}. These results use the harmonic structure
of the projection operator and the special role of the directions
arising from linear changes of coordinates.
Chen, Feng, Li, Xi, and Xu proved the unrestricted three-dimensional
inequality, including ellipsoid equality
\cite[Theorem~1.1]{ChenEtAl2026}. Mielke-Sulz proved the conjecture
for bodies of revolution in every dimension $n\ge3$
\cite{MielkeSulz2026}. Theorem~\ref{thm:main} treats arbitrary convex
bodies in the remaining dimensions. Combining it with the separately
proved three-dimensional theorem of Chen et al. establishes the
conjecture for every $n\ge3$; that external theorem is not used in our
proof for $n\ge4$.
Lutwak reformulated the conjecture as a lower bound for a two-body
integral with kernel $|u\cdot v|$ against surface-area measures
\cite{LutwakPetty1990}; see the explicit formulation in
\cite[Introduction, item~(i)]{Saroglou2015}.
Our reduction uses the same pairing for a broader class of positive
measures satisfying a lower bound on integrals of support functions.
Our proof uses the classical first-variation and spherical-transform
frameworks, but requires estimates and an affine normalization that
apply to arbitrary bodies. The Wulff variations are closely related
to the Aleksandrov variations for dual curvature measures developed
by Huang, Lutwak, Yang, and Zhang
\cite{HuangLutwakYangZhang2016}. The variational representation of a
support functional belongs to the $L_p$ dual Minkowski framework;
compare Huang and Zhao \cite{HuangZhao2018}. We prove the needed
representation directly, including its logarithmic endpoint.
The cosine and Funk spectra are classical
\cite{OlafssonPasqualeRubin2013}; we derive their multipliers in our
normalization. The further tasks are a strict estimate for all norms
and a continuous choice of variational minimizers through changes of
matrix rank. The latter permits a global choice of affine coordinates,
where proximity to the ball is unavailable.
\subsection*{The argument}
The proof passes from a body to a positive measure on vectors.
After translating and normalizing $|K|=\kappa_n$, the first variation
of volume gives a bounded, spanning probability measure $\eta_K$ on
$\R^n$ such that
\[
\int h_M\,d\eta_K\ge (|M|/\kappa_n)^{1/n}
\]
for every origin-symmetric convex body $M$.
The same measure represents $h_{\Pi K}(y)$ as a fixed multiple of
$\int|x\cdot y|\,d\eta_K(x)$.
Theorem~\ref{thm:bilinear} bounds the pairing
\[
\iint|x\cdot y|\,d\eta(x)d\zeta(y)
\]
for any two bounded, spanning positive measures satisfying this
support-functional inequality. Applying the result to the measures
of $K$ and its volume-normalized projection body yields the desired
lower bound. Sharpness in the pairing returns equality to the
homothety case of the first variation of volume.
Two ingredients have statements independent of projection bodies.
Theorem~\ref{thm:norm-estimate} bounds the higher spherical harmonics
of a power of an arbitrary norm by a difference of its spherical
means. The estimate is strict except for the Euclidean norm, and no
smoothness is needed. Proposition~\ref{prop:representation} represents
a positive support functional using the gradient of a variationally
selected norm. A spherical sign test bounds the pairing below by a
bilinear form involving the cosine and Funk transforms. The strict norm estimate
controls the contribution of harmonics of degree at least four.
The degree-two component must be removed by choosing affine
coordinates; it is not a zero eigenspace of the sign-test operator.
Proposition~\ref{prop:position} makes this choice by extending the
family of variational minimizers continuously to singular matrices.
At that boundary the gauges become seminorms. An explicit calculation
in their missing directions gives an inward-pointing matrix field,
and a fixed-point argument yields an invertible matrix at which the
degree-two component vanishes. This position is needed for only one
of the two measures. The other is transformed contragrediently, which
preserves their pairing. These constructions and the strict norm
estimate also supply the equality case.
Combining Theorem~\ref{thm:main} with the separate three-dimensional
result of Chen et al. gives several sharp inequalities in dimensions
$n\ge3$. Lutwak's implication yields Euclidean-ball lower bounds for
lower-degree projection-body ratios, in which intrinsic volumes replace
the volumes of shadows. Haddad's $p=1$ implication bounds the integral of
the absolute determinant of gradients below by the product of critical
Lebesgue norms. Its
diagonal case strengthens
the Sobolev--Zhang affine $L^1$ inequality. The theorem also gives the
sharp Holmes--Thompson isoperimetric inequality in normed spaces, where
equality requires an ellipsoidal norm ball and a body homothetic to it.
The full statements below distinguish the hypotheses and equality
conclusions of these consequences.
Section~\ref{sec:gauges} establishes the volume and gauge facts used in
the reduction. Section~\ref{sec:spherical} derives the sign test and
its harmonic multipliers. Section~\ref{sec:norm} proves the strict norm
estimate, including the scalar inequalities in every dimension.
Section~\ref{sec:position} constructs the variational representation
and affine position. Section~\ref{sec:conclusion} proves the bilinear
inequality and deduces Theorem~\ref{thm:main}.
Section~\ref{sec:consequences} develops its consequences. We use the classical
Brunn--Minkowski inequality with its homothety equality case and
Brouwer's fixed-point theorem in their stated forms; the first-variation
reduction, harmonic estimates, variational construction, and equality
argument are proved below.
\subsection*{Notation}
Throughout Sections~\ref{sec:gauges}--\ref{sec:conclusion}, $n\ge4$,
$p=n-1$, and $\sigma$ is uniform probability on
$S^{n-1}$. The symbol $\E$ denotes integration against $\sigma$, and
\[
b_n=\E|u_1|.
\]
Repeated spherical integrals use independent directions.
A symmetric body always means an origin-symmetric body.
When a function on the sphere is used as a degree-one homogeneous test
on vectors, it is extended by positive homogeneity and given value zero
at the origin. Euclidean gradients are denoted by $\nabla$; the
tangential gradient and Laplacian are $\gradS$ and $\lapS$.
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//01_introduction.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//02_gauges.tex ====
\section{Gauges and projection measures}\label{sec:gauges}
We first associate a measure to an arbitrary convex body. A volume
variation will give both the lower bound satisfied by this measure and
its relation to the projection body. No symmetry of the original body
is needed.
\subsection{Differentiating a gauge}
For a convex body $L$ containing the origin in its interior, its
\emph{gauge} is
\[
g_L(x)=\inf\{a>0:x\in aL\}.
\]
It is convex and positively homogeneous, and $L=\{x:g_L(x)\le1\}$.
When $L$ is origin-symmetric, its gauge is a norm. A continuous function
$\phi$ on $\Sph$, when evaluated at a vector, will mean its positively
homogeneous extension
\[
\phi(x)=|x|\phi(x/|x|)\quad(x\ne0),\qquad \phi(0)=0.
\]
This convention does not impose evenness on $\phi$.
Varying the supporting constraints in this way is a Wulff variation.
The almost-everywhere radial differentiation below is the gauge form
of the Aleksandrov variation used in dual Brunn--Minkowski theory;
compare \cite[Lemmas~4.1--4.3 and Corollary~4.9]{HuangLutwakYangZhang2016}. We give the
argument directly because the later minimization uses arbitrary
continuous constraints of both signs, without boundary regularity.
\begin{lemma}[Gauge variation]\label{lem:gauge-variation}
Let $s:\Sph\to(0,\infty)$ and $\phi:\Sph\to\R$ be continuous. For
sufficiently small $|t|$, define
\[
W_t=\bigcap_{v\in\Sph}\{x\in\R^n:x\cdot v\le s(v)+t\phi(v)\},
\qquad g_t=g_{W_t}.
\]
Then $W_t$ is a convex body containing the origin in its interior, and
\begin{equation}\label{eq:gauge-formulas}
g_t(x)=\max_{v\in\Sph}\frac{x\cdot v}{s(v)+t\phi(v)},
\qquad
\frac{|W_t|}{\kappa_n}=\E g_t(u)^{-n}.
\end{equation}
For almost every $u\in\Sph$, the maximizing vector $v/s(v)$ at $t=0$
is unique and equals the Euclidean gradient $\nabla g_0(u)$. At these
directions,
\begin{equation}\label{eq:gauge-derivative}
\left.\frac{d}{dt}\right|_{t=0}g_t(u)
=-g_0(u)\phi\bigl(\nabla g_0(u)\bigr).
\end{equation}
On the sphere the gauges are uniformly bounded above and away from
zero for small $|t|$, and the difference quotients
$(g_t-g_0)/t$ are uniformly bounded. In particular,
\begin{equation}\label{eq:constraint-volume-derivative}
\left.\frac{d}{dt}\right|_{t=0}|W_t|
=n\kappa_n\E\!\left[
g_0(u)^{-n}\phi\bigl(\nabla g_0(u)\bigr)\right].
\end{equation}
For $s=h_L$, one has $W_0=L$.
\end{lemma}
\begin{proof}
Choose $0<m\le s(v)\le M$. For small $|t|$ the constraints
$s+t\phi$ lie between $m/2$ and some fixed finite constant $M'$.
Consequently
\[
(m/2)B_2^n\subset W_t\subset M'B_2^n.
\]
The defining inequalities show that $x\in aW_t$ precisely when
$x\cdot v\le a(s(v)+t\phi(v))$ for every $v$. Taking the least
such $a$ gives the maximum formula in \eqref{eq:gauge-formulas}.
The radial function of $W_t$ is $1/g_t(u)$, so polar integration,
with spherical area $n\kappa_n$, gives the volume formula. If
$s=h_L$, separation identifies the defining intersection with $L$.
For completeness, the maximum formula also gives the differentiability
needed here without any regularity assumption on the boundary of
$W_0$. The function $g_0$ is convex and Lipschitz: it is the maximum
of linear functions whose coefficient vectors $v/s(v)$ form a compact
set. On every coordinate line, a finite convex function has a
derivative except on a set of one-dimensional measure zero. Slicing
therefore shows that all coordinate partial derivatives of $g_0$
exist at almost every point of $\R^n$. If a vector $w=v/s(v)$ attains
the maximum at such a point $x$, then
\[
g_0(x+he_i)\ge g_0(x)+h w_i\qquad(h\in\R).
\]
The two signs of $h$ imply $w_i=\partial_i g_0(x)$ for every $i$.
Thus the maximizing vector is unique.
Uniqueness implies differentiability. Indeed, if $w_z$ maximizes at
$x+z$ and $w$ maximizes at $x$, compactness and continuity imply
$w_z\to w$ as $z\to0$. The maximum formula gives
\[
0\le g_0(x+z)-g_0(x)-w\cdot z
\le (w_z-w)\cdot z=o(|z|).
\]
The nondifferentiability set away from the origin is invariant under
positive dilations, by homogeneity. Since it has Lebesgue measure
zero, polar integration shows that its intersection with $\Sph$ has
$\sigma$-measure zero.
Fix a differentiability direction $u$, and let $v_0$ maximize at
$t=0$. Uniqueness of $v_0/s(v_0)$ implies uniqueness of $v_0$
itself, since $s$ is positive. Every maximizing direction $v_t$ for
$g_t(u)$ tends to $v_0$ as $t\to0$. The derivatives with respect
to $t$ of the functions
\[
(u,v,t)\longmapsto\frac{u\cdot v}{s(v)+t\phi(v)}
\]
are continuous on the relevant compact set. Comparing the maxima at
$t$ and $0$, using $v_t$ and $v_0$, therefore gives
\[
\left.\frac{d}{dt}\right|_{0}g_t(u)
=-\frac{(u\cdot v_0)\phi(v_0)}{s(v_0)^2}
=-g_0(u)\phi\!\left(\frac{v_0}{s(v_0)}\right).
\]
This is \eqref{eq:gauge-derivative}. Notice that the argument uses
neither convexity nor evenness of the function $s+t\phi$.
The same uniform bound on the derivatives of the displayed quotients
bounds $(g_t-g_0)/t$ uniformly in $u$. The ball inclusions bound
$g_t$ above and away from zero on $\Sph$. Dominated convergence
now differentiates the volume formula and proves
\eqref{eq:constraint-volume-derivative}.
\end{proof}
\subsection{A volume inequality with its equality case}
We use the classical Brunn--Minkowski inequality, including its equality
case, in the following form; see \cite[Sections~3 and~5]{Gardner2002}.
The first-variation statement and its equality consequence are then
proved in the gauge coordinates needed for our measures.
\begin{theorem}[Brunn--Minkowski inequality]\label{thm:brunn-minkowski}
For every integer $d\ge1$ and convex bodies $A,B\subset\R^d$, write $|\cdot|$ for
$d$-dimensional volume. For $0\le t\le1$,
\begin{equation}\label{eq:brunn-minkowski}
|(1-t)A+tB|^{1/d}
\ge (1-t)|A|^{1/d}+t|B|^{1/d}.
\end{equation}
If $0<t<1$, equality holds if and only if $B=a+cA$ for some
$a\in\R^d$ and $c>0$.
\end{theorem}
\begin{corollary}[First variation of volume]\label{cor:volume-first-variation}
Let $K,M\subset\R^n$ be convex bodies, with the origin interior to
$K$. Then
\begin{align}
\left.\frac{d}{dt}\right|_{0+}|K+tM|
&=n\kappa_n\E\!\left[g_K(u)^{-n}
h_M\bigl(\nabla g_K(u)\bigr)\right]
\label{eq:minkowski-volume-derivative}\\
&\ge n|K|^{(n-1)/n}|M|^{1/n}.\nonumber
\end{align}
Equality in the inequality holds if and only if $K$ and $M$ are
homothetic up to translation. The derivative identity remains valid
when $M$ is an arbitrary nonempty compact convex set.
\end{corollary}
\begin{proof}
Support functions add under Minkowski addition, so
Lemma~\ref{lem:gauge-variation}, with $s=h_K$ and $\phi=h_M$,
gives the derivative identity. This applies equally to a compact
convex set $M$ of lower dimension. For full-dimensional $M$,
Theorem~\ref{thm:brunn-minkowski} and homogeneity give
\[
F(t):=|K+tM|^{1/n}\ge |K|^{1/n}+t|M|^{1/n}
\qquad(t\ge0).
\]
In addition, $F$ is concave, by applying that theorem to $K+sM$
and $K+tM$. Its right derivative at zero therefore satisfies
$F'(0+)\ge |M|^{1/n}$, which is the asserted lower bound.
If equality holds in this derivative bound, concavity gives
\[
F(t)\le F(0)+tF'(0+)=|K|^{1/n}+t|M|^{1/n}.
\]
Thus the preceding volume inequality is an equality for every $t>0$.
The equality case of Theorem~\ref{thm:brunn-minkowski}, applied for
example to $K+M$, forces homothety. Conversely, if $M=a+cK$
with $c>0$, then $K+tM=ta+(1+tc)K$, which verifies equality in
the derivative bound.
\end{proof}
\subsection{The measure associated with a convex body}
The construction below is a gauge-coordinate form of the classical
surface-area and mixed-volume description of projection bodies;
see \cite[Section~6]{Gardner2002} and \cite[Section~1]{Saroglou2015}.
The local segment calculation will fix its normalization and retain
the equality information from the first variation.
For a finite positive Borel measure $\eta$ on $\R^n$ with bounded
support, write
\[
\tau_\eta(M)=\int h_M(x)\,d\eta(x).
\]
\begin{definition}\label{def:admissible-measure}
We call $\eta$ \emph{admissible} if its support spans $\R^n$ and
\begin{equation}\label{eq:functional-condition}
\tau_\eta(M)\ge\left(\frac{|M|}{\kappa_n}\right)^{1/n}
\quad\text{for every origin-symmetric convex body }M\subset\R^n.
\end{equation}
The measure itself is not required to be symmetric.
\end{definition}
\begin{proposition}[Projection measure]\label{prop:projection-measure}
Translate and dilate a convex body $K$ so that $0\in\operatorname{int}K$
and $|K|=\kappa_n$. Define $\eta_K$ by
\begin{equation}\label{eq:eta-definition}
\int \psi(x)\,d\eta_K(x)
=\E\!\left[g_K(u)^{-n}\psi\bigl(\nabla g_K(u)\bigr)\right]
\end{equation}
for nonnegative Borel functions $\psi$; the choice of the gradient on
its null exceptional set is immaterial. Then $\eta_K$ is an
admissible probability measure of bounded support. Equality in
\eqref{eq:functional-condition} for $\eta_K$ and a symmetric body
$M$ holds if and only if $K$ is a translate of a positive dilate of
$M$. Moreover,
\begin{equation}\label{eq:projection-measure}
h_{\Pi K}(y)=\frac{n\kappa_n}{2}
\int |x\cdot y|\,d\eta_K(x)
\qquad(y\in\R^n),
\end{equation}
and
\begin{equation}\label{eq:eta-normalization}
\int h_K(x)\,d\eta_K(x)=1.
\end{equation}
For the unit ball, $\eta_{B_2^n}=\sigma$, and consequently
\begin{equation}\label{eq:ball-constant}
\frac{n\kappa_n b_n}{2}=\kappa_{n-1}.
\end{equation}
\end{proposition}
\begin{proof}
By \eqref{eq:gauge-formulas}, $\E g_K^{-n}=1$, so
\eqref{eq:eta-definition} defines a probability measure. The active
vectors $v/h_K(v)$ lie in a bounded set because $h_K$ is positive
on the sphere. Hence $\eta_K$ has bounded support. Applying
Corollary~\ref{cor:volume-first-variation} gives, for every symmetric
convex body $M$,
\[
\tau_{\eta_K}(M)
=\frac1{n\kappa_n}
\left.\frac{d}{dt}\right|_{0+}|K+tM|
\ge\left(\frac{|M|}{\kappa_n}\right)^{1/n},
\]
with precisely the stated homothety equality condition. The same
derivative identity with $M=K$, using
$|K+tK|=(1+t)^n\kappa_n$, proves
\eqref{eq:eta-normalization}; this identity does not require $K$
to be symmetric.
Now let $y\ne0$ and take $M=[-y,y]$. Every nonempty fiber of
$K$ parallel to $y$ is a compact interval. Adding $t[-y,y]$
increases its length by $2t|y|$, without changing its projection
onto $y^\perp$. Fubini's theorem therefore gives
\[
|K+t[-y,y]|=|K|+
2t|y|\operatorname{vol}_{n-1}
(\operatorname{proj}_{y^\perp}K)
=|K|+2t h_{\Pi K}(y).
\]
Since the support function of the segment is $x\mapsto|x\cdot y|$,
the derivative identity yields \eqref{eq:projection-measure}.
The formula is also true at $y=0$. The projection of a
full-dimensional convex body has positive $(n-1)$-volume in every
direction. If $\supp\eta_K$ failed to span $\R^n$, a nonzero
vector perpendicular to its span would make the integral in
\eqref{eq:projection-measure} vanish. This contradiction proves
the spanning property and hence admissibility.
Finally, for $K=B_2^n$ one has $g_K(u)=1$ and
$\nabla g_K(u)=u$ on the sphere, so $\eta_K=\sigma$.
Its projection volume in every unit direction is $\kappa_{n-1}$.
Substituting this into \eqref{eq:projection-measure} proves
\eqref{eq:ball-constant}.
\end{proof}
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//02_gauges.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//03_spherical.tex ====
\section{A spherical sign estimate}\label{sec:spherical}
The estimate in this section separates the constant part of a spherical
function from its harmonics of degree at least four. The degree-two part
will later be removed by a linear change of coordinates. Throughout,
$p=n-1$, surface measure $\sigma$ has total mass one, and inner products
and norms of functions are those of $L^2(\sigma)$.
\subsection{Harmonics and the spectral gap}
Let $\mathcal H_l$ be the restrictions to $\Sph$ of homogeneous harmonic
polynomials of degree $l$, and put
\[
\lambda_l=l(l+n-2)=l(l+p-1).
\]
For an even function, let $Q$ denote orthogonal projection onto the
closed sum of $\mathcal H_4,\mathcal H_6,\ldots$.
\begin{lemma}\label{lem:spherical-harmonics}
The spaces $\mathcal H_l$ form an orthogonal decomposition of
$L^2(\sigma)$, and $\lapS Y=-\lambda_lY$ for $Y\in\mathcal H_l$.
Every even $H\in C^1(\Sph)$ satisfies
\begin{equation}\label{eq:spectral-gap}
\lambda_2\Var(H)+(\lambda_4-\lambda_2)\norm{QH}_2^2
\le \E\abs{\gradS H}^2,
\qquad \lambda_2=2n,\quad \lambda_4=4(n+2).
\end{equation}
\end{lemma}
\begin{proof}
The polar-coordinate formula for the Euclidean Laplacian gives
\[
\Delta_{\R^n}\big(r^lY(u)\big)
=r^{l-2}\big(l(l+n-2)Y(u)+\lapS Y(u)\big).
\]
This proves the eigenvalue formula. Integration by parts on the sphere
then proves orthogonality of spaces with distinct degrees.
We give the completeness argument explicitly. If $Y_l$ is homogeneous
and harmonic, direct differentiation yields
\[
\Delta_{\R^n}\big(\abs{x}^{2j}Y_l(x)\big)
=2j(2l+2j+n-2)\abs{x}^{2j-2}Y_l(x).
\]
Induction on the degree now decomposes every homogeneous polynomial
into terms $\abs{x}^{2j}Y_l(x)$. Indeed, decompose its Laplacian by the
induction hypothesis and use the displayed identity to find a sum of
such terms with the same Laplacian. The difference is harmonic.
Restricting to the unit sphere shows that every polynomial restriction
belongs to the span of the $\mathcal H_l$.
Polynomial restrictions uniformly approximate continuous functions on
$\Sph$. To see this without an approximation theorem, set
\[
K_j(u,v)=
\frac{((1+u\cdot v)/2)^j}
{\displaystyle\int_{\Sph}((1+u\cdot w)/2)^j\,d\sigma(w)}.
\]
The denominator is positive and independent of $u$, so
$T_jH(u)=\int K_j(u,v)H(v)\,d\sigma(v)$ is a polynomial in $u$.
For $0<\delta<2$, let $m_\delta>0$ be the measure of the cap
$\{v:\abs{u-v}<\delta/2\}$. Since
$(1+u\cdot v)/2=1-\abs{u-v}^2/4$,
\[
\int_{\abs{u-v}\ge\delta}K_j(u,v)\,d\sigma(v)
\le \frac1{m_\delta}
\left(\frac{1-\delta^2/4}{1-\delta^2/16}\right)^j.
\]
Uniform continuity of $H$ therefore gives $T_jH\to H$ uniformly.
Continuous functions are dense in $L^2(\sigma)$: approximate the sets
in a simple function from inside by compact sets and from outside by
open sets, then interpolate their indicators continuously using
distance to these sets. The squared error is bounded by the measure
of the intervening sets. This proves completeness.
Choose real orthonormal bases $Y_{l,a}$ of the spaces $\mathcal H_l$,
with $Y_{0,1}=1$. Integration by parts shows that the tangent fields
$\gradS Y_{l,a}/\sqrt{\lambda_l}$, for $l\ge1$, are orthonormal.
Bessel's inequality, which follows by expanding the squared distance
from any finite orthogonal sum, consequently gives
\[
\E\abs{\gradS H}^2
\ge \sum_{l\ge1,a}
\left|\E\,\gradS H\cdot
\frac{\gradS Y_{l,a}}{\sqrt{\lambda_l}}\right|^2
=\sum_{l\ge1,a}\lambda_l\left|\E(HY_{l,a})\right|^2.
\]
No differentiation of an infinite series is needed. If $H$ is even,
its odd-degree coefficients vanish. Completeness identifies its
variance with the sum of the squared nonconstant coefficients and
$\norm{QH}_2^2$ with the corresponding sum over even $l\ge4$.
Using $\lambda_l\ge\lambda_2$ for even $l\ge2$ and
$\lambda_l\ge\lambda_4$ for $l\ge4$ proves
\eqref{eq:spectral-gap}.
\end{proof}
\subsection{The sign-test operator}
For $k>0$ and an even differentiable function $f$ on $\Sph$, define
\begin{equation}\label{eq:vector-field}
X_f(u)=f(u)u+k^{-1}\gradS f(u).
\end{equation}
For even $C^1$ functions $f_1,f_2$, set
\begin{equation}\label{eq:sign-form}
\mathcal B(f_1,f_2)
=\frac1{b_n}\iint_{\Sph\times\Sph}
\sgn(u\cdot v)\,X_{f_1}(u)\cdot X_{f_2}(v)
\,d\sigma(u)d\sigma(v).
\end{equation}
The value assigned to $\sgn(0)$ is immaterial. Since
$|a|\ge\sgn(u\cdot v)a$ for every real $a$, the form gives the lower bound
\[
\iint |X_{f_1}(u)\cdot X_{f_2}(v)|\,d\sigma(u)d\sigma(v)
\ge b_n\mathcal B(f_1,f_2).
\]
We estimate this lower bound using two spherical transforms.
Write $\sigma_{u^\perp}$
for uniform probability on the equator $u^\perp\cap\Sph$, and define
\[
(Cf)(u)=\frac1{b_n}\int_{\Sph}\abs{u\cdot v}f(v)\,d\sigma(v),
\qquad
(Pf)(u)=\int_{u^\perp\cap\Sph}f(v)\,d\sigma_{u^\perp}(v).
\]
These are the cosine and Funk transforms, here normalized to preserve
constants; compare \cite[Section~2, especially (2.3), (2.5)--(2.6)]
{OlafssonPasqualeRubin2013}. The local multiplier calculation below
uses this probability-measure convention throughout.
Both are self-adjoint contractions on $L^2(\sigma)$ and preserve
constants. For $C$ this follows from symmetry of its nonnegative
kernel and Jensen's inequality. For $P$, average first over the band
$\abs{u\cdot v}<\epsilon$, dividing by its measure. The resulting
kernel is symmetric and preserves constants. Spherical coordinates
show that, for continuous $f$, these band averages tend to $Pf$;
Jensen's inequality and passage to the limit give the same conclusions
for $P$, followed by extension to $L^2(\sigma)$.
\begin{lemma}\label{lem:sign-operator}
For even $C^1$ functions,
\begin{equation}\label{eq:sign-operator}
\mathcal B(f_1,f_2)
=\left\langle f_1,
\left[C+\left(\frac{2p}{k}+\frac{p^2}{k^2}\right)(C-P)\right]
f_2\right\rangle.
\end{equation}
\end{lemma}
\begin{proof}
Let $d_0$ be the density at zero of $u\cdot v$ for uniform $u$ and fixed
$v\in\Sph$. Equivalently, $d_0$ is the ratio of the surface areas of
$S^{n-2}$ and $S^{n-1}$. For a fixed ambient vector $w$, its tangential
projection has divergence $-p\,u\cdot w$. Integrating separately on
the two hemispheres gives
\begin{align}
&\int_{\Sph}\sgn(u\cdot v)\,\gradS f(u)\cdot w\,d\sigma(u)
\notag\\
&\quad=p\int_{\Sph}\sgn(u\cdot v)f(u)\,u\cdot w\,d\sigma(u)
-2d_0\int_{v^\perp\cap\Sph}f(u)\,v\cdot w
\,d\sigma_{v^\perp}(u).
\label{eq:hemisphere-ibp}
\end{align}
The outward conormals on the equator are $-v$ and $v$, respectively;
the sign change between the hemispheres makes their contributions
add with the displayed negative sign. Taking $f=1$ and $w=v$ yields
\[
2d_0=p b_n.
\]
The radial--radial term in \eqref{eq:sign-form}, before division by
$b_n$, is $b_n\langle f_1,Cf_2\rangle$. Apply
\eqref{eq:hemisphere-ibp} with $w=v$ and then integrate against
$f_2(v)$. Each of the two mixed radial--gradient terms, before its
factor $k^{-1}$, equals
\[
p b_n\langle f_1,(C-P)f_2\rangle.
\]
For the gradient--gradient term, first integrate in $u$ with
$w=\gradS f_2(v)$. Its equator term vanishes because
$v\cdot\gradS f_2(v)=0$. What remains is
\[
p\iint f_1(u)\sgn(u\cdot v)\,
u\cdot\gradS f_2(v)\,d\sigma(u)d\sigma(v).
\]
A second application of \eqref{eq:hemisphere-ibp}, now in $v$ with
$w=u$, makes this $p^2b_n\langle f_1,(C-P)f_2\rangle$.
Adding the four terms proves \eqref{eq:sign-operator}.
\end{proof}
\subsection{Multipliers and passage to norms}
\begin{lemma}\label{lem:spherical-multipliers}
On $\mathcal H_{2j}$ the operators $P$ and $C$ are scalar multipliers,
with
\begin{equation}\label{eq:transform-multipliers}
P_{2j}=(-1)^j
\frac{1\cdot3\cdots(2j-1)}{p(p+2)\cdots(p+2j-2)},
\qquad
C_{2j}=\frac{pP_{2j}}{p-\lambda_{2j}},
\end{equation}
where empty products equal one. In particular,
\begin{equation}\label{eq:cosine-recurrence}
\begin{gathered}
C_0=1,\qquad C_2=\frac1{p+2},\qquad
C_4=-\frac1{(p+2)(p+4)},\\[3pt]
\frac{C_{l+2}}{C_l}=-\frac{l-1}{l+p+2}\quad(l\ge2\text{ even}).
\end{gathered}
\end{equation}
\end{lemma}
\begin{proof}
Fix $u\in\Sph$ and a homogeneous harmonic polynomial $Y_{2j}$.
Let $Z$ be a standard Gaussian in the $p$-dimensional space $u^\perp$.
Its direction is uniform on the equator and independent of its radius,
so polar integration gives
\[
\mathbb E Y_{2j}(Z)
=p(p+2)\cdots(p+2j-2)\,(PY_{2j})(u).
\]
The radial moment here follows by successive integration by parts in
$\int_0^\infty r^{p+2j-1}e^{-r^2/2}\,dr$.
The one-dimensional Gaussian moment recurrence
$\mathbb E Z_1^{2a}=(2a-1)\mathbb E Z_1^{2a-2}$, together with
independence of coordinates, also gives
\[
\mathbb E Y_{2j}(Z)
=\frac{(\Delta_{u^\perp})^jY_{2j}(0)}{2^j j!}.
\]
This identity follows term by term on monomials; monomials with an odd
exponent have both sides zero, and the other coefficients are exactly
the multinomial coefficients in the power of the Laplacian.
Harmonicity and commutation of constant-coefficient derivatives imply
\[
(\Delta_{u^\perp})^jY_{2j}
=(-1)^j\partial_u^{2j}Y_{2j},
\qquad
\partial_u^{2j}Y_{2j}(0)=(2j)!Y_{2j}(u).
\]
Since $(2j)!/(2^j j!)=1\cdot3\cdots(2j-1)$, the formula for $P_{2j}$
follows.
To obtain $C$, integrate by parts on the two hemispheres in the
variable $v$. Off the equator,
$\lapS\abs{u\cdot v}=-p\abs{u\cdot v}$; the jump of its normal
derivative contributes $2d_0$ times equator averaging. Thus, for any
smooth $Y$,
\[
\int_{\Sph}\abs{u\cdot v}\lapS Y(v)\,d\sigma(v)
=-p\int_{\Sph}\abs{u\cdot v}Y(v)\,d\sigma(v)
+2d_0(PY)(u).
\]
For $Y\in\mathcal H_l$, division by $b_n$ gives
$-\lambda_l CY=p(PY-CY)$. The denominator
$p-\lambda_l=-(l-1)(l+p)$ is nonzero for every even $l$, proving the
second formula in \eqref{eq:transform-multipliers}. Its first three
values give those in \eqref{eq:cosine-recurrence}; division of successive
values, using $P_{l+2}/P_l=-(l+1)/(l+p)$, gives the recurrence.
\end{proof}
Set
\begin{equation}\label{eq:tail-constant}
c_{n,k}=
\frac{1+\left(2/k+(n-1)/k^2\right)4(n+2)}{(n+1)(n+3)}.
\end{equation}
\begin{proposition}\label{prop:sign-estimate}
Let $k>0$ and let $f_1,f_2$ be even $C^1$ functions on $\Sph$.
If $f_1$ has zero projection onto $\mathcal H_2$, then
\begin{equation}\label{eq:sign-bound}
\mathcal B(f_1,f_2)
\ge (\E f_1)(\E f_2)
-c_{n,k}\norm{Qf_1}_2\norm{Qf_2}_2.
\end{equation}
The operator identity \eqref{eq:sign-operator} and this estimate also
hold when $f_i=g_i^k$ for arbitrary norms $g_i$ on $\R^n$, with the
vector fields defined almost everywhere. In that case
\begin{equation}\label{eq:norm-vector-field}
X_{g_i^k}(u)=g_i(u)^{k-1}\nabla g_i(u).
\end{equation}
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:spherical-multipliers}, the operator in
\eqref{eq:sign-operator} has multiplier one on constants and, on degree
$l$, multiplier
\[
\mu_l=C_l\left[1+\left(\frac2k+\frac p{k^2}\right)\lambda_l\right].
\]
Indeed, $C_l-P_l=(\lambda_l/p)C_l$.
Put $a=2/k+p/k^2>0$. For even $l\ge4$,
\[
(l+p+2)\lambda_l-(l-1)\lambda_{l+2}
=(p-1)l^2+(p^2-1)l+2(p+1)>0.
\]
Since also $l-1<l+p+2$, it follows that
\[
\frac{\abs{\mu_{l+2}}}{\abs{\mu_l}}
=\frac{l-1}{l+p+2}\,
\frac{1+a\lambda_{l+2}}{1+a\lambda_l}\le1.
\]
Consequently $\abs{\mu_l}\le\abs{\mu_4}=c_{n,k}$ on all even
$l\ge4$. The constant term of the bilinear form is
$(\E f_1)(\E f_2)$, and its degree-two term vanishes by hypothesis.
Cauchy--Schwarz on the remaining harmonic coefficients proves
\eqref{eq:sign-bound}. This expansion is justified by the completeness
in Lemma~\ref{lem:spherical-harmonics} and the boundedness of $C$ and
$P$.
We finish by proving the passage to norms, retaining the details
needed when their unit balls have corners. Let $g$ be a norm and let
$g_\epsilon=g*\rho_\epsilon$ on $\R^n$, where $\rho_\epsilon$ is a
smooth, even, nonnegative function of integral one, supported in the
ball of radius $\epsilon$. The functions $g_\epsilon$ are smooth,
even and convex. If $L$ is the Euclidean Lipschitz constant of $g$,
then
\[
\abs{g_\epsilon-g}\le L\epsilon,
\qquad \abs{\nabla g_\epsilon}\le L.
\]
At a differentiability point $x$ of $g$, these gradients converge to
$\nabla g(x)$. Indeed, every convergent subsequence of the bounded
gradients has a limit $a$, and convexity gives, on passing to the
limit,
\[
g(y)\ge g(x)+a\cdot(y-x)\qquad(y\in\R^n).
\]
Taking $y=x\pm th$ and then $t\downarrow0$ forces
$a\cdot h=\nabla g(x)\cdot h$ for every $h$, so the only possible
limit is $\nabla g(x)$.
Lemma~\ref{lem:gauge-variation} supplies differentiability of $g$ at
almost every spherical direction. Equivalently, one may use its
almost-everywhere differentiability in $\R^n$ and homogeneity: the
exceptional set is a union of rays, so polar integration makes its
intersection with the sphere null. On the sphere $g$ is bounded away
from zero. Thus the restrictions
$f_\epsilon=g_\epsilon^k|_{\Sph}$ converge uniformly to $g^k$, and
\[
\gradS f_\epsilon(u)
=k g_\epsilon(u)^{k-1}
(\Id-uu^t)\nabla g_\epsilon(u)
\longrightarrow
k g(u)^{k-1}(\Id-uu^t)\nabla g(u)
\]
almost everywhere and in $L^2(\sigma)$, by bounded convergence.
Euler's identity $u\cdot\nabla g(u)=g(u)$, obtained by differentiating
$g(tu)=tg(u)$, identifies the limiting vector field with
\eqref{eq:norm-vector-field}.
The approximants $g_\epsilon$ need not be homogeneous; only their
restrictions and tangential gradients are used here. The fields
$X_{f_\epsilon}$ converge in $L^2$, so the bounded sign kernel in
\eqref{eq:sign-form} permits passage to the limit. The right side of
\eqref{eq:sign-operator} also converges, because $C$ and $P$ are
bounded on $L^2$. This proves the operator identity for norm powers.
Finally apply its harmonic multiplier estimate directly to the
limiting functions. The approximants themselves need not have
vanishing degree-two part: that hypothesis is used only for the
limiting $f_1$. This proves \eqref{eq:sign-bound} in the stated
nonsmooth setting.
\end{proof}
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//03_spherical.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//04_norm.tex ====
\section{A strict inequality for norms}\label{sec:norm}
We now bound the higher spherical harmonics of a power of a norm by two
of its means. Convexity supplies the differential estimate; the remaining
step consists of scalar inequalities. Throughout this section set
\begin{equation}\label{eq:k-choice}
k=\begin{cases}1,&n=4,\\2,&n\ge5,\end{cases}
\qquad c=c_{n,k},\qquad d_* =\lambda_4-\lambda_2=2n+8.
\end{equation}
Thus the first power is used in dimension four and the square in higher
dimensions. The scalar estimates below establish these two choices with
the same strict conclusion.
\begin{theorem}\label{thm:norm-estimate}
Let $g$ be a norm on $\R^n$, where $n\ge4$, normalized by
$\E g^{-n}=1$ on $\Sph$. Then
\begin{equation}\label{eq:norm-estimate}
c_{n,k}\|Q(g^k)\|_2^2
\le (\E g^k)^2-(\E g^{k-1})^2.
\end{equation}
The inequality is strict unless $g=1$ on $\Sph$.
\end{theorem}
\subsection{Convexity and a scalar decomposition}
For $z>0$ define
\begin{align}
e(z)&=\frac{z^{-n}-1}{n},&
M(z)&=z^k-1+k e(z),&
W(z)&=z^k-z^{k-1}+e(z).\label{eq:scalar-basic}
\end{align}
Since $M'(z)=k(z^{k-1}-z^{-n-1})$, the function $M$ decreases up to
$1$, increases thereafter, and satisfies $M(1)=M'(1)=0$. In particular,
$M\ge0$. For a norm with $\E g^{-n}=1$, the correction $e(g)$
has mean zero. We will split $g^k-1$ into a term controlled by
convexity and a remainder controlled by its variance. Choose the constants
\begin{equation}\label{eq:scalar-constants}
\begin{array}{c|cc}
n&\beta&w\\ \hline
4&1/2&3/14\\
5,6&2/3&1/3\\
n\ge7&2/3&1/2
\end{array}
\end{equation}
and split
\begin{equation}\label{eq:scalar-split}
G(z)=\beta M(z)\mathbf 1_{\{z>1\}},\qquad
D(z)=z^k-1-G(z).
\end{equation}
Thus $G$ vanishes for $z\le1$, while $D$ retains the remaining part
of $z^k-1$. To estimate $G(g)$ by convexity,
let $I(z)=0$ for $0<z\le1$, and, for $z>1$, let
\begin{align}
I(z)&=\int_1^z M'(a)^2\,da\notag\\
&=k^2\left(\frac{z^{2k-1}-1}{2k-1}
-\frac{2(1-z^{k-1-n})}{n+1-k}
+\frac{1-z^{-2n-1}}{2n+1}\right).
\label{eq:scalar-I}
\end{align}
Both $G$ and $I$ are continuously differentiable across $1$;
$I\ge0$ and $G'^2=\beta^2 I'$.
\begin{lemma}\label{lem:curvature}
For every norm $g$ on $\R^n$,
\begin{equation}\label{eq:norm-curvature}
2n\Var(G(g))+d_*\|QG(g)\|_2^2
\le p\beta^2\E[gI(g)],\qquad p=n-1.
\end{equation}
\end{lemma}
\begin{proof}
Suppose first that $g$ is smooth away from the origin. The trace of its
Euclidean Hessian on the tangent space at $u\in\Sph$ is
$\lapS g(u)+p g(u)$, and it is nonnegative by convexity. Multiplying by
$I(g)\ge0$ and integrating by parts gives
\begin{align*}
\E|\gradS G(g)|^2
&=\beta^2\E[I'(g)|\gradS g|^2]
=-\beta^2\E[I(g)\lapS g]
\le p\beta^2\E[gI(g)].
\end{align*}
The spectral gap estimate \eqref{eq:spectral-gap} proves
\eqref{eq:norm-curvature} in this case.
Here is a passage to arbitrary norms that does not assume the
approximants are homogeneous. Convolve the ambient norm with smooth,
even, nonnegative mollifiers of mass one and shrinking compact support,
and write the resulting smooth convex functions as $g_\varepsilon$.
On the unit sphere their tangent Hessian trace is
\[
\lapS g_\varepsilon+p\partial_r g_\varepsilon\ge0.
\]
The preceding argument and \eqref{eq:spectral-gap} therefore give
\[
2n\Var(G(g_\varepsilon))+d_*\|QG(g_\varepsilon)\|_2^2
\le p\beta^2\E[I(g_\varepsilon)\partial_r g_\varepsilon].
\]
The functions $g_\varepsilon$ converge uniformly to $g$ on compact
sets, are uniformly Lipschitz there, and their gradients converge to
$\nabla g$ at every differentiability point of $g$. To see the last
assertion, every limit of their bounded gradients satisfies the
subgradient inequality for $g$ by convexity and uniform convergence;
differentiability makes that subgradient unique. The exceptional
directions have spherical measure zero, as in
Lemma~\ref{lem:gauge-variation}. Homogeneity of $g$ then gives
$\partial_r g_\varepsilon(u)\to\nabla g(u)\cdot u=g(u)$ almost
everywhere. On the sphere the approximants stay in a fixed positive
compact interval, while their radial derivatives are bounded. Uniform
convergence on the left and dominated convergence on the right prove
\eqref{eq:norm-curvature}.
\end{proof}
\subsection{From scalar bounds to the norm estimate}\label{subsec:norm-deduction}
Define, for $z>0$,
\begin{equation}\label{eq:scalar-L}
\mathcal L(z)=c\left(
\frac{p\beta^2 zI(z)-2nG(z)^2}{wd_*}
+\frac{D(z)^2}{1-w}\right).
\end{equation}
\begin{lemma}\label{lem:scalar-estimates}
The constants and functions above satisfy
\begin{equation}\label{eq:scalar-mean-bound}
c\left(\frac{2n\beta^2}{wd_*}
-\frac{(1-\beta)^2}{1-w}\right)
\le\frac{2k-1}{k^2}
\end{equation}
and
\begin{equation}\label{eq:scalar-pointwise}
\mathcal L(z)\le2W(z)\qquad(z>0).
\end{equation}
Equality in \eqref{eq:scalar-pointwise} holds only at $z=1$.
\end{lemma}
We first deduce the norm estimate from the two scalar bounds. Their
verification, including the strict pointwise inequality in every
dimension, occupies the remainder of the section.
\begin{proof}[Proof of Theorem~\ref{thm:norm-estimate}]
In the following expectations, $G,D,I$ denote their compositions with
$g$. Set
\[
m=\E g^k-1,\qquad \ell=\E g^{k-1}-1,\qquad t=\E G.
\]
The normalization gives $\E e(g)=0$, so $m=\E M(g)\ge0$,
$\E D=m-t$, and $\E W(g)=m-\ell$. Since
$0\le G\le\beta M$, we have $0\le t\le\beta m$.
For every $a>0$, Jensen's inequality applied to the convex function
$y\mapsto y^{-a/n}$ yields
$\E g^a\ge(\E g^{-n})^{-a/n}=1$.
Together with
\[
z^{k-1}\le\frac{(k-1)z^k+1}{k},
\]
this gives
\begin{equation}\label{eq:norm-mean-range}
0\le\ell\le\frac{k-1}{k}\,m.
\end{equation}
For $k=1$ this simply says $\ell=0$.
Since $Q$ annihilates constants,
$Q(g^k)=QG+QD$. The Hilbert-space inequality
\[
\|a+b\|^2\le\frac{\|a\|^2}{w}
+\frac{\|b\|^2}{1-w}
\qquad(0<w<1)
\]
and $\|QD\|_2^2\le\Var(D)$, followed by
Lemma~\ref{lem:curvature}, give
\begin{align}
c\|Q(g^k)\|_2^2
&\le\frac{c}{wd_*}
\left(p\beta^2\E[gI(g)]-2n\Var(G)\right)
+\frac{c}{1-w}\Var(D)\notag\\
&=\E\mathcal L(g)+c\left(
\frac{2n}{wd_*}t^2-\frac{(m-t)^2}{1-w}\right).
\label{eq:norm-reduction}
\end{align}
The quadratic expression in parentheses is increasing for
$0\le t\le m$: its derivative is
\[
\frac{4n}{wd_*}t+\frac{2(m-t)}{1-w}\ge0.
\]
It is therefore no larger than its value at $t=\beta m$.
Using \eqref{eq:scalar-mean-bound} and
\eqref{eq:norm-mean-range}, we conclude that the final term of
\eqref{eq:norm-reduction} is at most
\[
\frac{2k-1}{k^2}m^2\le m^2-\ell^2.
\]
Now \eqref{eq:scalar-pointwise} yields
\begin{align*}
c\|Q(g^k)\|_2^2
&\le 2\E W(g)+m^2-\ell^2\\
&=2(m-\ell)+m^2-\ell^2
=(\E g^k)^2-(\E g^{k-1})^2,
\end{align*}
as claimed. If $g$ is not identically $1$ on the sphere, continuity
gives a set of positive spherical measure on which $g\ne1$.
Lemma~\ref{lem:scalar-estimates} then gives
$\E\mathcal L(g)<2\E W(g)$, so the final inequality is strict.
The only constant norm on the sphere with $\E g^{-n}=1$ is $g=1$;
in this case both sides of \eqref{eq:norm-estimate} vanish.
\end{proof}
\subsection{Proof of the scalar bounds}
\begin{proof}[Proof of Lemma~\ref{lem:scalar-estimates}]
We first record the values of the spectral constant:
\begin{equation}\label{eq:scalar-c}
c=\frac{121}{35}\quad(n=4),\qquad
c=\frac{n^2+5n+7}{(n+1)(n+3)}\quad(n\ge5).
\end{equation}
For $n=4,5,6$ the left side of \eqref{eq:scalar-mean-bound} is,
respectively,
\[
\frac{11}{12},\qquad \frac{589}{864},\qquad \frac{1387}{1890}.
\]
These are smaller than $1,3/4,3/4$, the respective right sides.
For $n\ge7$ the left side is
\[
\frac{2(3n-4)(n^2+5n+7)}{9(n+1)(n+3)(n+4)}.
\]
Its difference from $3/4$, with the latter first, is
\[
\frac{3n^3+128n^2+505n+548}{36(n+1)(n+3)(n+4)}>0.
\]
This proves \eqref{eq:scalar-mean-bound} for every dimension.
\medskip
\noindent\emph{The interval $0<z\le1$.}
Here $G=I=0$ and $|D(z)|=|z^k-1|\le k|z-1|$.
Also $W(1)=W'(1)=0$ and
\[
W''(z)=\begin{cases}
(n+1)z^{-n-2},&k=1,\\
2+(n+1)z^{-n-2},&k=2.
\end{cases}
\]
Thus $2W(z)\ge(n+2k-1)(z-1)^2$. The positive margins
\begin{equation}\label{eq:scalar-local-margin}
n+2k-1-\frac{ck^2}{1-w}
=\begin{cases}
3/5,&n=4,\\
7/8,&n=5,\\
43/21,&n=6,\\
n+3-8c,&n\ge7
\end{cases}
\end{equation}
give the desired strict inequality for $z<1$. In the last case the
margin is $9/10$ at $n=7$ and increases thereafter: regarding $n$ as a
real variable in \eqref{eq:scalar-c},
\[
c'(n)=-\frac{n^2+8n+13}{(n+1)^2(n+3)^2}<0.
\]
At $z=1$ both sides of \eqref{eq:scalar-pointwise} vanish.
\medskip
\noindent\emph{The interval $z>1$ in dimensions $4,5,6$.}
Set $x=z-1>0$. Multiplication by a positive factor clears the negative
powers of $z$: if
\[
P_n(z)=s_nz^{2n}\bigl(2W(z)-\mathcal L(z)\bigr),
\qquad (s_4,s_5,s_6)=(2880,712800,221130),
\]
then substitution of \eqref{eq:scalar-basic}--\eqref{eq:scalar-L}
and collection of equal powers give
\begin{align}
P_4(z)={}&462z^{10}-983z^9+93z^8
+3399z^5-3378z^4+407,\notag\\
P_5(z)={}&151525z^{14}-273570z^{12}-680800z^{11}
+938157z^{10}\notag\\
&\quad+225720z^7-462264z^5+101232,\notag\\
P_6(z)={}&48399z^{16}-75894z^{14}-218004z^{13}
+285961z^{12}\notag\\
&\quad+56940z^8-127478z^6+30076.
\label{eq:small-polynomials}
\end{align}
For clarity, the coefficients needed from $P_n(1+x)$ are listed below;
the constant and linear coefficients are zero.
\[
\begin{array}{c|rrr}
[x^j]&n=4&n=5&n=6\\ \hline
j=2&1728&623700&452790\\
j=3&-1446&-1505900&679380\\
j=4&-6711&-5802775&-132249\\
j=5&1173&12849430&7800780\\
j=6&17052&86314305&51269452\\
j=7&20796&191506920&146058120
\end{array}
\]
All remaining coefficients are nonnegative, as can be seen without
listing them. Let $a_nz^{d_n}$ be the leading term in
\eqref{eq:small-polynomials}. At $j=8$, its contribution
$a_n\binom{d_n}{8}$ minus the contributions of all negative terms is
\[
11943,\qquad 207280425,\qquad 114414300
\quad\text{for } n=4,5,6,
\]
respectively. For every $i<d_n$, the ratio
$\binom{i}{j}/\binom{d_n}{j}$ is nonincreasing in $j$; while it is
nonzero, its ratio at successive indices is $(i-j)/(d_n-j)\le1$.
Once $j>i$ it is zero. Thus the same leading term covers all negative
contributions for every $8\le j\le d_n$.
It follows that $P_n(1+x)/x^2$ is bounded below by, respectively,
\begin{align*}
U_4(x)&=1728-1446x-6711x^2+17052x^4,\\
U_5(x)&=623700-1505900x-5802775x^2+86314305x^4,\\
U_6(x)&=452790-132249x^2+51269452x^4.
\end{align*}
These lower bounds are positive. Indeed
\begin{align*}
1446x&\le400+1400x^2,&
8111x^2&\le1000+17052x^4,\\
1505900x&\le200000+2900000x^2,&
8702775x^2&\le230000+86314305x^4.
\end{align*}
Each follows from $dt\le a+bt^2$ when $d^2\le4ab$; the corresponding
values of $4ab-d^2$ are
\[
149084,\quad 2419679,\quad 52265190000,\quad 3670867899375.
\]
The first row gives $U_4(x)\ge328$ and the second gives
$U_5(x)\ge193700$. Finally $U_6$, regarded as a quadratic in $x^2$,
has positive leading coefficient and negative discriminant, since
\[
4\cdot452790\cdot51269452-132249^2=92839690886319>0.
\]
Therefore $P_n(z)>0$ for $z>1$ in all three dimensions.
\medskip
\noindent\emph{The interval $z>1$ in all dimensions $n\ge7$.}
Put $H=18(n+4)/c$. Direct collection of powers in
\eqref{eq:scalar-L} now gives
\begin{equation}\label{eq:large-polynomial}
\frac{9(n+4)}{c}z^{2n}(2W-\mathcal L)
=z^{2n}(A_4z^4+A_2z^2+A_1z+A_0)
+z^n(B_2z^2+B_0)+T_0,
\end{equation}
where
\begin{align}
A_4&=\frac{2n-8}{3},&
A_2&=H-12n-32-\frac{64}{n},\notag\\
A_1&=-H+\frac{16(n-1)}3+32-\frac{16(n-1)}{2n+1},&&\notag\\
A_0&=-\frac Hn+6n+40+\frac{64}{n}-\frac{128}{n^2},&&\notag\\
B_2&=16+\frac{64}{n},&
B_0&=\frac Hn-48-\frac{64}{n}+\frac{256}{n^2},\notag\\
T_0&=\frac{16(n-1)}{2n+1}-\frac{128}{n^2}.&&
\label{eq:large-coefficients}
\end{align}
We prove that the polynomial on the right of
\eqref{eq:large-polynomial}, expressed in $x=z-1$, has nonnegative
coefficients and a positive quadratic coefficient.
Write its two parentheses as $\sum_{i=0}^4q_ix^i$ and
$\sum_{i=0}^2r_ix^i$. Thus
\begin{align*}
q_0&=A_4+A_2+A_1+A_0,&q_1&=4A_4+2A_2+A_1,\\
q_2&=6A_4+A_2,&q_3&=4A_4,\qquad q_4=A_4,\\
r_0&=B_2+B_0,&r_1&=2B_2,\qquad r_2=B_2.
\end{align*}
In particular $q_3,q_4,r_1,r_2>0$. The constant and linear
coefficients of the full polynomial vanish. Its quadratic coefficient
is
\begin{equation}\label{eq:large-quadratic}
\frac{9(n+4)}c\,(n+3-8c)>0.
\end{equation}
One can obtain these three coefficients either from
\eqref{eq:large-coefficients} or directly from the order of vanishing at
$z=1$: $G=O(x^2)$, $I=O(x^3)$, and $D=2x+O(x^2)$, so that
$2W-\mathcal L=(n+3-8c)x^2+O(x^3)$. Positivity is
\eqref{eq:scalar-local-margin}.
The remaining signs admit explicit polynomial certificates. Set
\begin{align*}
V_0(n)&=34n^5-95n^4-1666n^3-5701n^2-8052n-2688,\\
V_1(n)&=2n^5+9n^4-162n^3-845n^2-1220n-448,\\
V_2(n)&=5n^3+8n^2-41n-56,\\
R(n)&=7n^4+8n^3-187n^2-748n-896.
\end{align*}
Substituting $H=18(n+4)(n+1)(n+3)/(n^2+5n+7)$ gives
\begin{align}
q_0+\frac{r_0}{8}
&=\frac{V_0(n)}{4n^2(2n+1)(n^2+5n+7)},\notag\\
q_1&=\frac{2V_1(n)}{n(2n+1)(n^2+5n+7)},\notag\\
q_2&=\frac{2(n+4)V_2(n)}{n(n^2+5n+7)},&
r_0&=-\frac{2R(n)}{n^2(n^2+5n+7)}.
\label{eq:large-signs}
\end{align}
The required signs hold throughout their stated ranges because, for
$t\ge0$,
\begin{align}
V_0(10+t)={}&34t^5+1605t^4+28534t^3
+227319t^2\notag\\
&\quad+698128t+130692,\notag\\
V_1(10+t)={}&2t^5+109t^4+2198t^3+19695t^2
+69280t+30852,\notag\\
V_2(7+t)={}&5t^3+113t^2+806t+1764,\notag\\
R(7+t)={}&7t^4+204t^3+2039t^2+7414t+4256.
\label{eq:shift-certificates}
\end{align}
Consequently $q_2>0$ and $r_0<0$ for every $n\ge7$, while
$q_0+r_0/8>0$ and $q_1>0$ for every $n\ge10$.
Use the convention that $\binom mj=0$ when $j<0$ or $j>m$. For
$3\le j\le2n$, comparison of the factors in the binomial products gives
\begin{equation}\label{eq:binomial-comparison}
\binom nj\le 2^{-j}\binom{2n}{j}.
\end{equation}
Indeed $(n-a)/(2n-a)\le1/2$ for $0\le a<n$; if $j>n$ the left
side is zero. For $n\ge10$, the sum of the $q_0$ and $r_0$
contributions to the coefficient of $x^j$ is therefore at least
\[
\left(q_0+\frac{r_0}{2^j}\right)\binom{2n}{j}
\ge\left(q_0+\frac{r_0}{8}\right)\binom{2n}{j}>0.
\]
All other contributions are nonnegative. For $j>2n$, the $q_0$ and
$r_0$ contributions vanish and every remaining term is nonnegative.
Together with \eqref{eq:large-quadratic}, this proves coefficient
positivity for every $n\ge10$.
It remains to treat $n=7,8,9$, where $q_1$ can be negative.
For $3\le j\le2n$, divide the coefficient of $x^j$ by
$\binom{2n}{j}$, discard the nonnegative $q_3,q_4,r_2$ contributions,
and use \eqref{eq:binomial-comparison} for the negative $r_0$ term.
The resulting lower bound is
\begin{equation}\label{eq:small-large-dimension-bound}
\begin{split}
q_0+\frac{r_0}{2^j}
+\frac{j}{2n-j+1}
\left(q_1+q_2\frac{j-1}{2n-j+2}\right)
+r_1\frac{\binom n{j-1}}{\binom{2n}{j}}.
\end{split}
\end{equation}
Substitution in \eqref{eq:large-coefficients} gives the componentwise
lower bounds
\[
\begin{array}{c|rrrrr}
n&q_0&q_1&q_2&r_0&r_1\\ \hline
7&-2&-11&60&-2&50\\
8&-1&-6&72&-6&46\\
9&-1&-6&72&-6&46
\end{array}
\]
All weights in \eqref{eq:small-large-dimension-bound} are nonnegative,
so these bounds may be substituted directly. They give the following
positive numbers at $j=3,4$; at $j=5$ we have already omitted the final
$r_1$ term:
\[
\begin{array}{c|ccc}
n&j=3&j=4&j=5\ \text{without the final term}\\ \hline
7&5/26&1233/1144&589/176\\
8&37/28&10707/3640&1153/208\\
9&101/136&3851/2040&395/112
\end{array}
\]
These finite substitutions can also be checked from the exact tuples
$(q_0,q_1,q_2,r_0,r_1)$:
\begin{align*}
n=7:&\quad
\left(-\frac{5984}{3185},-\frac{4832}{455},
\frac{792}{13},-\frac{1216}{637},\frac{352}{7}\right),\\
n=8:&\quad
\left(-\frac{421}{629},-\frac{3736}{629},
\frac{2688}{37},-\frac{145}{37},48\right),\\
n=9:&\quad
\left(\frac{2416}{10773},-\frac{2192}{1197},
\frac{100568}{1197},-\frac{57968}{10773},\frac{416}{9}\right).
\end{align*}
For $5\le j\le2n$, the parenthesis
$q_1+q_2(j-1)/(2n-j+2)$ is positive and increasing in $j$.
In fact the displayed componentwise bounds give at $j=5$ the positive
lower bounds $119/11$, $210/13$, and $66/5$, respectively.
The prefactor $j/(2n-j+1)$ is positive and increasing too, while
$r_0/2^j$ increases because $r_0<0$. The expression
\eqref{eq:small-large-dimension-bound} with its last term omitted is
therefore increasing throughout this range. Positivity at $j=5$
proves positivity at every subsequent index through $2n$.
For $j>2n$ all $r_i$ contributions vanish, since $n+2<2n+1$.
The only possibly negative term is the $q_1$ contribution at
$j=2n+1$. At this index the $q_1$ and $q_2$ contributions sum to
$q_1+2nq_2$, which the same componentwise bounds make at least
$829,1146,1290$ for $n=7,8,9$, respectively. For $j\ge2n+2$ only
the nonnegative $q_2,q_3,q_4$ terms remain. This handles every
coefficient, including the endpoints.
We have proved that the polynomial in \eqref{eq:large-polynomial}
has nonnegative coefficients and a strictly positive coefficient of
$x^2$ for every $n\ge7$. It is consequently positive for every
$x>0$. Its multiplier $9(n+4)z^{2n}/c$ is positive, so
$2W(z)-\mathcal L(z)>0$ for $z>1$. This finishes the proof of
\eqref{eq:scalar-pointwise} and its strictness in every dimension.
\end{proof}
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//04_norm.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//05_position.tex ====
\section{Variational representation and affine position}\label{sec:position}
We next represent a support functional by a vector field of the form
\eqref{eq:vector-field}, and choose coordinates in which the associated
norm power has no degree-two component. Throughout this section, $k$ has the value in
\eqref{eq:k-choice}, and $\eta$ is a finite positive Borel measure with
bounded support spanning $\R^n$. Neither symmetry of $\eta$ nor the
functional condition \eqref{eq:functional-condition} is needed here.
Let $A$ be a nonzero symmetric positive semidefinite matrix, and write
$V=\range(A)$. For an origin-symmetric convex body $L$ in $V$, with
interior relative to $V$, put
\[
\tau_A(L)=\int h_L(Ax)\,d\eta(x).
\]
We extend its gauge to $\R^n$ by $g=g_L\circ P_V$, where $P_V$ denotes
orthogonal projection onto $V$. This extension is a seminorm, with
kernel $V^\perp$. Define
\begin{equation}\label{eq:variational-objective}
J(g)=
\begin{cases}
\E\log g,&k=1,\\
\E g,&k=2.
\end{cases}
\end{equation}
The logarithm is integrated outside the kernel, which is a spherical
null set.
When $A$ is positive definite, these objectives are chosen so that their
Wulff first variations involve $g^{k-1}\nabla g=X_{g^k}$, the vector
field in \eqref{eq:norm-vector-field}.
This variational construction belongs to the framework for dual
curvature measures. The optimizer-to-measure argument corresponds to
the $L_1$ dual Minkowski problem with dual parameter $q=1-k$:
for $k=2$, compare \cite[Section~3.1, Lemmas~3.1--3.3]{HuangZhao2018};
for $k=1$ and full-dimensional bodies, minimization under $\tau_A(L)=1$
is equivalent, by polarity, to the $p=1$ entropy maximization in
\cite[Section~5, Lemmas~5.1--5.3]{HuangLutwakYangZhang2018}.
Wang and Zhou established uniqueness and continuity for the relevant
full-dimensional dual Minkowski problems
\cite[arXiv version, Theorems~1.1--1.3]{WangZhou2026}.
Here we prove the representation in our normalization, together with
uniqueness and continuity through changes of rank, where the gauges
become seminorms. This continuity is required to select affine
coordinates. The measures may have atoms, and all
variations remain valid for nonsmooth minimizing bodies.
\begin{lemma}\label{lem:minimizer}
For every nonzero symmetric positive semidefinite $A$, there is a unique
origin-symmetric convex body $L_A$ in $V=\range(A)$ that minimizes
\eqref{eq:variational-objective} subject to $\tau_A(L_A)=1$.
Denote its extended gauge by $g_A$.
\end{lemma}
\begin{proof}
The spanning hypothesis gives a norm on $V$:
\[
q_A(y)=\int |y\cdot Ax|\,d\eta(x).
\]
Indeed, $q_A(y)=0$ implies that $Ay$ is orthogonal to $\supp\eta$,
and hence $Ay=0$; because $y\in V$, this forces $y=0$.
For any normalized competitor $L$ and $y\in L$, symmetry gives
$q_A(y)\le\tau_A(L)=1$. Thus all normalized competitors have a common
outer radius in $V$.
We also need a bound on their gauges along a minimizing sequence.
For any nonzero seminorm $g$, let $R=\max_{\Sph}g$. By separation,
$g$ is the support function of the compact symmetric set
$\{z:z\cdot y\le g(y)\text{ for every }y\in\R^n\}$.
This set has maximal radius $R$: if $Re$ is a point of maximal norm
in it, then
\begin{equation}\label{eq:seminorm-coordinate-bound}
g(u)\ge R|u\cdot e|\qquad(u\in\Sph).
\end{equation}
Consequently,
\[
J(g)\ge
\begin{cases}
\log R+\E\log|u_1|,&k=1,\\
Rb_n,&k=2.
\end{cases}
\]
Here $\E|\log|u_1||^2<\infty$: the first coordinate has a density
proportional to $(1-t^2)^{(n-3)/2}$ on $(-1,1)$, and
$\int_0^{1/2}|\log t|^2\,dt<\infty$. In particular every nonzero
seminorm has a finite logarithmic objective.
Set $b_A=\int|Ax|\,d\eta(x)>0$. The ball of radius $1/b_A$ in $V$
is a normalized competitor. Comparison with this ball bounds $J$
above along a minimizing sequence, so the preceding inequalities bound
$R$ above. The bodies therefore contain a common inner ball in $V$,
as well as lying in a common outer ball. Their gauges are uniformly
bounded and equicontinuous on the unit sphere of $V$. A subsequence
converges uniformly to the gauge of a body with the same two ball
bounds. The support functions converge uniformly as well, so the
normalization passes to the limit. The objective also passes to the
limit: in the logarithmic case the extended gauges are uniformly
comparable to $|P_Vu|$, whose logarithm is integrable by
\eqref{eq:seminorm-coordinate-bound}. This proves existence.
For uniqueness, let $L_0,L_1$ be two minimizers and
$L_t=(1-t)L_0+tL_1$, where $0<t<1$. Its normalization remains one.
For $P_Vu\ne0$, write $\rho_i(u)=1/g_{L_i}(P_Vu)$.
Taking points on the ray through $P_Vu$ gives
\[
\frac{1}{g_{L_t}(P_Vu)}
\ge (1-t)\rho_0(u)+t\rho_1(u).
\]
Both $-\log\rho$ and $1/\rho$ are strictly convex decreasing
functions of $\rho>0$. Thus the objective of $L_t$ is at most the
interpolated objectives, and the inequality is strict wherever
$\rho_0\ne\rho_1$. Distinct bodies have different radial functions
on an open set of directions in $V$, which gives a set of positive
spherical measure after projection. They would therefore give a
strictly smaller objective, a contradiction.
\end{proof}
\begin{proposition}[Variational representation]\label{prop:representation}
Suppose $A$ is positive definite, and let $L$ be any positive dilate
of $L_A$, with gauge $g=g_L$. For every continuous even function
$\phi:\R^n\to\R$ that is positively homogeneous of degree one,
\begin{equation}\label{eq:representation}
\int\phi(Ax)\,d\eta(x)
=s\,\E\phi\bigl(X_{g^k}(u)\bigr),
\qquad
s=\frac{\tau_A(L)}{\E g^{k-1}}.
\end{equation}
Here, almost everywhere on the sphere,
\begin{equation}\label{eq:norm-power-field}
X_{g^k}=g^ku+g^{k-1}\gradS g=g^{k-1}\nabla g.
\end{equation}
\end{proposition}
\begin{proof}
Normalization turns the minimization problem into minimizing
$J(\tau_A(L)g_L)$ over all symmetric bodies. Put
$\tau=\tau_A(L)$ and $I=\int\phi(Ax)\,d\eta(x)$. For small
positive or negative $t$, define the body
\[
W_t=\bigl\{y\in\R^n:
y\cdot v\le h_L(v)+t\phi(v)\text{ for all }v\in\Sph\bigr\},
\qquad g_t=g_{W_t}.
\]
The constraint is positive for sufficiently small $|t|$, so $W_t$ is
a symmetric convex body. Although that constraint need not be a
support function, it bounds the actual support function above.
Therefore
\[
\tau_A(W_t)\le\tau+tI.
\]
Both quantities are positive for small $|t|$. Since $J$ increases
under pointwise increase of a positive gauge, minimality gives
\begin{equation}\label{eq:wulff-minimum}
J\bigl((\tau+tI)g_t\bigr)
\ge J\bigl(\tau_A(W_t)g_t\bigr)
\ge J(\tau g),
\end{equation}
with equality at $t=0$.
Lemma~\ref{lem:gauge-variation} gives
\[
\left.\frac{d}{dt}\right|_{t=0}g_t(u)
=-g(u)\phi(\nabla g(u))
\]
almost everywhere, with uniform bounds that permit differentiation
of the objective. The left side of \eqref{eq:wulff-minimum} has a
two-sided minimum at zero. Its derivative is consequently zero:
\[
\frac{I}{\tau}=\E\phi(\nabla g)\quad(k=1),
\qquad
I\E g=\tau\E\bigl[g\phi(\nabla g)\bigr]\quad(k=2).
\]
Equivalently,
\[
\frac{I}{\tau}
=\frac{\E\bigl[g^{k-1}\phi(\nabla g)\bigr]}{\E g^{k-1}}.
\]
Euler's identity for the homogeneous gauge gives
\eqref{eq:norm-power-field}, and positive homogeneity of $\phi$
then gives \eqref{eq:representation}. The argument applies to both
signs of $t$ without imposing convexity on the varied constraint.
\end{proof}
To select a position we allow the matrix to approach the boundary of
the compact convex set
\begin{equation}\label{eq:matrix-domain}
\mathcal A=\{A=A^t\ge0:\tr A=1\}.
\end{equation}
The minimizers extend continuously to this boundary, even though
their ambient unit sets may become cylinders.
\begin{lemma}\label{lem:matrix-continuity}
The map $A\mapsto g_A|_{\Sph}$ is continuous on $\mathcal A$ in
the uniform norm.
\end{lemma}
\begin{proof}
Let $A_j\to A$ in $\mathcal A$, and write
$V_j=\range(A_j)$, $g_j=g_{A_j}$, and
$b_j=\int|A_jx|\,d\eta(x)$. Then
$b_j\to b=\int|Ax|\,d\eta(x)>0$. The normalized ball in $V_j$
has extended gauge $b_j|P_{V_j}u|$, and hence objective at most
$\log b_j$ if $k=1$, and at most $b_j$ if $k=2$.
By \eqref{eq:seminorm-coordinate-bound}, this bounds
$R_j=\max_{\Sph}g_j$ uniformly above. The minimizing body contains
the ball of radius $1/R_j$ in $V_j$, so its normalization gives
$1\ge b_j/R_j$. Thus $R_j\ge b_j$, bounded away from zero.
The seminorms $g_j$ have uniformly bounded Lipschitz constants.
Every subsequence therefore has a further subsequence converging
uniformly on $\Sph$ to a nonzero seminorm $g$. Along such a
subsequence,
\begin{equation}\label{eq:objective-limit}
J(g_j)\longrightarrow J(g).
\end{equation}
For $k=2$ this is immediate. For $k=1$, choose unit vectors $e_j$
with $g_j(u)\ge R_j|u\cdot e_j|$. The positive parts of $\log g_j$
are uniformly bounded, and their negative parts have uniformly
bounded squared integrals, by rotation invariance and
$\E|\log|u_1||^2<\infty$. These logarithms are therefore uniformly
integrable. They converge almost everywhere to $\log g$, since
the kernel of a nonzero seminorm is a proper subspace. Truncating
the logarithms below at $-M$ and then sending $M$ to infinity proves
\eqref{eq:objective-limit}; the discarded integrals are uniformly
$O(M^{-1})$ by the squared-integral bound.
We identify this subsequential limit. Set
$Z=\{g\le1\}$ and $Z_j=\{g_j\le1\}$, allowing infinite values
for their support functions. If $g(y)<1$, then $y\in Z_j$ for all
sufficiently large $j$. Testing the support functions against $y$
and then taking the supremum gives, for every $x$,
\begin{equation}\label{eq:cylinder-support-limit}
h_Z(Ax)\le\liminf_j h_{Z_j}(A_jx).
\end{equation}
The supremum over $g(y)<1$ equals $h_Z$, by scaling towards zero.
Since $Z_j=L_{A_j}+V_j^\perp$ and $A_jx\in V_j$, its support
function at $A_jx$ equals that of $L_{A_j}$. All these support
functions are nonnegative, so Fatou's lemma yields
\begin{equation}\label{eq:cylinder-normalization}
\int h_Z(Ax)\,d\eta(x)\le1.
\end{equation}
Let $V=\range(A)$ and $L_0=\overline{P_VZ}$. This projection is
bounded despite the possible unboundedness of $Z$. Indeed, for
every $z\in Z$, symmetry and \eqref{eq:cylinder-normalization} give
\[
q_A(P_Vz)=\int|z\cdot Ax|\,d\eta(x)
\le\int h_Z(Ax)\,d\eta(x)\le1.
\]
The norm $q_A$ bounds the radius in $V$. Moreover $Z$ contains a
Euclidean ball, so $L_0$ has interior in $V$. Its extended gauge
$\bar g=g_{L_0}\circ P_V$ satisfies $\bar g\le g$, and
\begin{equation}\label{eq:projected-normalization}
0<c_0:=\tau_A(L_0)=\int h_Z(Ax)\,d\eta(x)\le1.
\end{equation}
For the opposite comparison, let $N$ be any full-dimensional
symmetric convex body in $\R^n$. The extended gauge of its
projection to $V_j$ is at most $g_N$, and this projected body's
normalization is $\tau_{A_j}(N)$. Minimality and monotonicity give
\[
J(g_j)\le J\bigl(\tau_{A_j}(N)g_N\bigr).
\]
Passing to the limit, using \eqref{eq:objective-limit} and
continuity of $\tau_{A_j}(N)$, gives
\begin{equation}\label{eq:full-body-comparison}
J(g)\le J\bigl(\tau_A(N)g_N\bigr).
\end{equation}
Take $N=N_R=L_A+RB_{V^\perp}$, the orthogonal product of the
minimizing body in $V$ and a ball of radius $R$ in $V^\perp$.
Then $\tau_A(N_R)=1$ and
\[
g_{N_R}(u)=\max\{g_A(u),|P_{V^\perp}u|/R\}\downarrow g_A(u).
\]
For $k=1$, the integrable function $\log g_A$ is a lower bound
for these logarithms, and their positive parts are uniformly
bounded for $R\ge1$; hence the objectives converge. The case
$k=2$ is immediate. Thus \eqref{eq:full-body-comparison} implies
$J(g)\le J(g_A)$.
Finally, $c_0\bar g$ is the extended gauge of the normalized
body $L_0/c_0$. Since $c_0\bar g\le g$, we have
\[
J(g_A)\le J(c_0\bar g)\le J(g)\le J(g_A).
\]
Strict increase of the objective integrands forces
$c_0\bar g=g$ almost everywhere, and then everywhere by continuity.
Uniqueness in Lemma~\ref{lem:minimizer} gives
$c_0\bar g=g_A$. Thus every subsequential limit is $g_A$, which
proves the asserted continuity.
\end{proof}
We use Brouwer's fixed-point theorem to complete the choice of
position. We record the compact-convex formulation and its short
reduction to the simplex theorem \cite[Satz~4, p.~115]{Brouwer1911};
the nearest-point inequality in the reduction will also be used below.
\begin{lemma}[A finite-dimensional fixed-point principle]\label{lem:fixed-point}
Every continuous self-map of a nonempty compact convex subset of a
finite-dimensional Euclidean space has a fixed point.
\end{lemma}
\begin{proof}
Let the convex set be $D$, and work in its affine hull, of dimension
$d$. The case $d=0$ is immediate. The nearest-point projection
$\pi$ onto $D$ exists by compactness and is unique by strict
convexity of squared distance along a segment between distinct
minimizers. Minimizing along segments gives
\[
\langle x-\pi(x),z-\pi(x)\rangle\le0\qquad(z\in D).
\]
Applying this twice yields
\[
|\pi(x)-\pi(y)|^2
\le\langle x-y,\pi(x)-\pi(y)\rangle,
\]
so $\pi$ is continuous. If $T:D\to D$ is continuous, choose a
$d$-simplex $\Delta$ containing $D$ and set
$G=T\circ\pi:\Delta\to\Delta$.
Brouwer's theorem for a simplex gives $v=G(v)$.
Since $G(\Delta)\subset D$, this point lies in $D$, hence
$\pi(v)=v$ and $T(v)=v$.
\end{proof}
\begin{proposition}[Affine position]\label{prop:position}
There is a symmetric positive definite matrix $A$ with $\det A=1$
such that $g_A^k$ has zero component in $\mathcal H_2$.
\end{proposition}
\begin{proof}
On $\mathcal A$ define the symmetric traceless matrix
\begin{equation}\label{eq:position-field}
\mathcal F(A)
=\frac{\E[g_A(u)^k uu^t]}{\E g_A^k}-\frac1n\Id.
\end{equation}
It is continuous by Lemma~\ref{lem:matrix-continuity}; its
denominator is positive because $g_A$ is a nonzero seminorm.
The direction $-\mathcal F(A)$ points strictly into the positive
cone along every missing direction of a singular matrix. To compute this, let
$A$ be singular and $e$ a unit vector in $\Ker A$. For a standard
Gaussian $Y$ in $\R^n$, $g_A(Y)$ depends only on the projection
of $Y$ to $\range(A)$, and is independent of $Y\cdot e$.
Separating the Gaussian radial and angular variables therefore gives
\begin{align*}
\frac{\E[g_A(u)^k(u\cdot e)^2]}{\E g_A^k}
&=\frac{\mathbb E[g_A(Y)^k(Y\cdot e)^2]}
{\mathbb E g_A(Y)^k}
\frac{\mathbb E|Y|^k}{\mathbb E|Y|^{k+2}}\\
&=\frac{1}{n+k}.
\end{align*}
The last radial moment ratio follows by integration by parts in
$\int_0^\infty r^{n+k-1}e^{-r^2/2}\,dr$.
It follows, by polarization on $\Ker A$, that
\begin{equation}\label{eq:kernel-field}
\left.\mathcal F(A)\right|_{\Ker A}
=-\frac{k}{n(n+k)}I_{\Ker A}.
\end{equation}
The range--kernel block of $\mathcal F(A)$ is zero: reflecting
all kernel coordinates preserves $g_A$ and reverses each mixed
product. Consequently $A-\epsilon\mathcal F(A)$ is positive
definite for sufficiently small $\epsilon>0$. On the kernel this
follows from \eqref{eq:kernel-field}; on the range it follows from
the positive definiteness of the restriction of $A$. Its trace is
one. The same perturbation is allowed at a positive definite $A$
for sufficiently small $\epsilon$.
Use the Euclidean inner product $\langle U,W\rangle=\tr(UW)$ on
symmetric matrices, and consider the continuous map
\[
A\longmapsto
\operatorname{proj}_{\mathcal A}\bigl(A-\mathcal F(A)\bigr).
\]
Lemma~\ref{lem:fixed-point} provides a fixed point $A$. The
nearest-point condition there is
\[
\langle\mathcal F(A),B-A\rangle\ge0
\qquad(B\in\mathcal A).
\]
We may take $B=A-\epsilon\mathcal F(A)$, as just shown. This
gives $-\epsilon\tr(\mathcal F(A)^2)\ge0$, hence
$\mathcal F(A)=0$. Formula~\eqref{eq:kernel-field} rules out
singularity, so $A$ is positive definite.
For every symmetric traceless matrix $H$, the vanishing field gives
\[
\E\bigl[g_A(u)^k u^tHu\bigr]=0.
\]
These quadratics are exactly $\mathcal H_2$, since the Euclidean
Laplacian of $x^tHx$ is $2\tr H$. Thus the required degree-two
component vanishes.
Finally, for $a>0$ the correspondence $L\mapsto L/a$ between
normalized competitors for $A$ and $aA$ gives
\begin{equation}\label{eq:minimizer-scaling}
g_{aA}=a g_A.
\end{equation}
Indeed, $\tau_{aA}(L/a)=\tau_A(L)$, while multiplying a gauge by
$a$ adds $\log a$ to the logarithmic objective and multiplies
the linear objective by $a$. The minimizers therefore correspond.
Taking $a=(\det A)^{-1/n}$ makes the determinant one and preserves
the vanishing degree-two component.
\end{proof}
Only one measure needs the position of Proposition~\ref{prop:position}.
For a second measure, Proposition~\ref{prop:representation} provides
the required vector field in the resulting coordinates without a
simultaneous position condition.
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//05_position.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//06_conclusion.tex ====
\section{The bilinear inequality and equality cases}
\label{sec:conclusion}
We now combine the spherical and variational estimates. Recall that an
admissible measure satisfies the support-functional inequality
\eqref{eq:functional-condition}, and need not itself be symmetric.
\begin{theorem}\label{thm:bilinear}
Let $\eta,\zeta$ be finite positive measures on $\R^n$ with bounded,
spanning support. Suppose that both satisfy
\eqref{eq:functional-condition}. Then
\begin{equation}\label{eq:bilinear}
\iint |x\cdot y|\,d\eta(x)d\zeta(y)\ge b_n.
\end{equation}
If equality holds, there is an origin-centered ellipsoid $E_0$ such that
\[
\int h_{E_0}\,d\eta=(|E_0|/\kappa_n)^{1/n}.
\]
\end{theorem}
\begin{proof}
Choose $k$ as in \eqref{eq:k-choice} and write $c=c_{n,k}$.
Apply Proposition~\ref{prop:position} to $\eta$, obtaining a symmetric
positive definite matrix $A$ with determinant one.
Replace $\eta$ by its image under $A$ and $\zeta$ by its image under
$A^{-1}$. Their pairing is unchanged, since
$(Ax)\cdot(A^{-1}y)=x\cdot y$.
Their support-functional inequalities also persist: for any linear map
$T$,
\[
h_M(Tx)=h_{T^tM}(x),
\]
and $|T^tM|=|M|$ when $|\det T|=1$.
For the first transformed measure, the support functional at the
identity is precisely the original $\tau_A$. Its unique minimizing
body is therefore $L_A$, with the degree-two cancellation supplied by
Proposition~\ref{prop:position}.
For each transformed measure choose its minimizing body as in
Proposition~\ref{prop:representation}, and rescale that body to volume
$\kappa_n$. Let the resulting gauges be $g_1,g_2$, and put
\[
f_i=g_i^k,\qquad t_i=\E g_i^{k-1},\qquad
s_i=\frac{\tau_i(L_i)}{t_i},\qquad
a_i=s_i\E f_i\quad(i=1,2),
\]
where $\tau_i$ is the functional of the corresponding transformed
measure. Thus
\begin{equation}\label{eq:normalized-parameters}
\E g_i^{-n}=1,\qquad s_it_i\ge1,\qquad a_i\ge1.
\end{equation}
For the last inequality, if $k=1$ then $\E g_i\ge1=t_i$ by the
negative-moment normalization. If $k=2$, the same normalization gives
$\E g_i\ge1$, and $\E g_i^2\ge(\E g_i)^2\ge\E g_i=t_i$.
The first function $f_1$ has no degree-two harmonic component; scalar
rescaling does not alter this property.
Apply \eqref{eq:representation} to $|x\cdot y|$ successively in both
variables. These are even continuous homogeneous tests, so the formula
applies despite any asymmetry of the measures. It gives
\begin{align}
\iint |x\cdot y|\,d\eta(x)d\zeta(y)
&=s_1s_2\iint
|X_{f_1}(u)\cdot X_{f_2}(v)|\,d\sigma(u)d\sigma(v)
\notag\\
&\ge b_ns_1s_2\mathcal B(f_1,f_2).
\label{eq:pairing-sign}
\end{align}
The pointwise inequality in the last line uses the test
$\sgn(u\cdot v)$. Proposition~\ref{prop:sign-estimate} now gives
\[
b_n^{-1}\iint |x\cdot y|\,d\eta(x)d\zeta(y)
\ge a_1a_2-s_1s_2c\norm{Qf_1}_2\norm{Qf_2}_2.
\]
By Theorem~\ref{thm:norm-estimate} and
\eqref{eq:normalized-parameters},
\begin{equation}\label{eq:error-absorption}
s_i^2c\norm{Qf_i}_2^2
\le a_i^2-s_i^2t_i^2\le a_i^2-1.
\end{equation}
Consequently the last lower bound is at least
\[
a_1a_2-\sqrt{(a_1^2-1)(a_2^2-1)}\ge1.
\]
Indeed $a_1a_2-1\ge0$ and
\[
(a_1a_2-1)^2-(a_1^2-1)(a_2^2-1)=(a_1-a_2)^2\ge0.
\]
This proves \eqref{eq:bilinear}.
Suppose equality holds. If $a_i>1$, the inequality
$s_i^2c\norm{Qf_i}_2^2\le a_i^2-1$ is strict: this follows from
Theorem~\ref{thm:norm-estimate} when $g_i$ is nonconstant, and from its
zero left side when $g_i$ is constant. If both $a_i$ exceed one, the
product error in \eqref{eq:error-absorption} is therefore strictly
smaller than the displayed square root. If exactly one exceeds one,
the error product vanishes and $a_1a_2>1$.
Both alternatives contradict equality, so $a_1=a_2=1$.
In particular, $\E g_1^k\le\E g_1^{k-1}$.
For $k=1$, this forces $\E g_1=1$ and equality in the negative-moment
power-mean bound; hence $g_1=1$.
For $k=2$, the chain
\[
1\le\E g_1\le(\E g_1)^2\le\E g_1^2\le\E g_1
\]
again forces constancy and then $g_1=1$.
Thus $s_1=1$, and the support-functional condition of the transformed
first measure is sharp on $B_2^n$. Undoing the transformation makes the
original condition sharp on $A^tB_2^n$, which is an origin-centered
ellipsoid of volume $\kappa_n$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Translate $K$ so that the origin is in its interior, and first rescale
it to satisfy $|K|=\kappa_n$.
Proposition~\ref{prop:projection-measure} supplies the admissible
measure $\eta_K$. Set
\[
r=\left(\frac{|\Pi K|}{\kappa_n}\right)^{1/n},
\qquad D=\frac1r\Pi K.
\]
The symmetric body $D$ has volume $\kappa_n$, so it too has an
admissible measure $\eta_D$. Its own support-functional integral is
one: the volume derivative under $D+tD=(1+t)D$ gives
\[
\int h_D\,d\eta_D
=\frac1{n\kappa_n}\left.\frac{d}{dt}\right|_{0+}|(1+t)D|=1.
\]
Using \eqref{eq:projection-measure} and Theorem~\ref{thm:bilinear},
\[
1=\frac{n\kappa_n}{2r}
\iint|x\cdot y|\,d\eta_K(x)d\eta_D(y)
\ge\frac{n\kappa_nb_n}{2r}.
\]
Equation~\eqref{eq:ball-constant} yields $r\ge\kappa_{n-1}$, or
$|\Pi K|\ge\kappa_n\kappa_{n-1}^n$ under the current normalization.
The projection body scales as $\Pi(aK)=a^{n-1}\Pi K$ for $a>0$.
Undoing the volume normalization proves \eqref{eq:main}.
If equality holds, the pairing of $\eta_K$ and $\eta_D$ equals $b_n$.
Theorem~\ref{thm:bilinear} gives an ellipsoid on which the
support-functional bound for $\eta_K$ is sharp.
The equality assertion of Proposition~\ref{prop:projection-measure}
then makes $K$ a translate of a positive dilate of that ellipsoid.
For the converse, the quotient is invariant under translations and
invertible linear maps. To verify the latter directly, the segment
volume derivative gives, for every vector $y$ and every invertible $T$,
\begin{align*}
2h_{\Pi(TK)}(y)
&=\left.\frac{d}{dt}\right|_{0+}|TK+t[-y,y]|\\
&=|\det T|\left.\frac{d}{dt}\right|_{0+}
|K+t[-T^{-1}y,T^{-1}y]|\\
&=2|\det T|h_{\Pi K}(T^{-1}y).
\end{align*}
Hence $\Pi(TK)=|\det T|T^{-t}\Pi K$ and
$|\Pi(TK)|=|\det T|^{n-1}|\Pi K|$, exactly matching the factor in
$|TK|^{n-1}$. Finally, $\Pi B_2^n=\kappa_{n-1}B_2^n$ gives equality
for the ball, and therefore for every ellipsoid.
\end{proof}
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//06_conclusion.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//07_consequences.tex ====
\section{Consequences}\label{sec:consequences}
Theorem~\ref{thm:main}, together with the separate three-dimensional
theorem of Chen, Feng, Li, Xi, and Xu \cite[Theorem~1.1]{ChenEtAl2026},
gives Petty's projection-volume inequality and its ellipsoid equality
case in every dimension $n\ge3$. We now apply this combined conclusion
to lower-degree projection bodies, affine functional inequalities, and
isoperimetry in normed spaces. The implications and equality assertions
are specified separately in each setting.
\subsection{Lower-degree projection bodies}
For a convex body $L\subset\R^m$, use the standard intrinsic volumes
$V_j(L)$, normalized by the Steiner formula
\[
\operatorname{vol}_m(L+tB_2^m)
=\sum_{j=0}^m\kappa_{m-j}V_j(L)t^{m-j},
\qquad t\ge0,\qquad \kappa_0=1.
\]
This is the normalization in
\cite[author version, Section~2, p.~6]{OrtegaMorenoSchuster2021};
on a hyperplane we use its induced Euclidean structure. For
$1\le i\le n-1$, the degree-$i$ projection body $\Pi_iK$ is defined by
\[
h_{\Pi_iK}(u)=V_i(K|u^\perp),\qquad u\in S^{n-1},
\]
where $K|u^\perp=\operatorname{proj}_{u^\perp}K$. In particular,
$\Pi_{n-1}K=\Pi K$ because $V_{n-1}$ on $u^\perp$ is its
$(n-1)$-dimensional volume.
\begin{corollary}[Lutwak--Petty lower-degree inequalities]
\label{cor:lutwak-petty}
Let $n\ge3$ and $1\le i\le n-1$ be integers, and let
$K\subset\R^n$ be a convex body with nonempty interior. Then
\begin{equation}\label{eq:lutwak-petty}
\begin{aligned}
\frac{V_{i+1}(\Pi_iK)}{V_{i+1}(K)^i}
&\ge
\frac{V_{i+1}(\Pi_iB_2^n)}{V_{i+1}(B_2^n)^i}\\
&=
\frac{\left(\displaystyle\binom{n-1}{i}
\frac{\kappa_{n-1}}{\kappa_{n-i-1}}\right)^{i+1}}
{\left(\displaystyle\binom{n}{i+1}
\frac{\kappa_n}{\kappa_{n-i-1}}\right)^{i-1}}.
\end{aligned}
\end{equation}
The lower bound is sharp, since $K=B_2^n$ attains it.
\end{corollary}
\begin{proof}
Theorem~\ref{thm:main} and the separately proved three-dimensional
result \cite[Theorem~1.1]{ChenEtAl2026} establish Petty's conjecture for
every $n\ge3$. Lutwak's conditional result \cite{Lutwak1990}, in the
explicit formulation of Ortega-Moreno and Schuster
\cite[author version, Introduction, p.~2]{OrtegaMorenoSchuster2021},
states that Petty's conjecture implies the Euclidean-ball lower bound
for these ratios for every $1\le i\le n-1$. Applying that implication
gives the inequality. For its explicit value, the Steiner normalization
gives
\[
V_j(B_2^m)=\binom{m}{j}\frac{\kappa_m}{\kappa_{m-j}},
\qquad
\Pi_iB_2^n=V_i(B_2^{n-1})B_2^n.
\]
The $(i+1)$-homogeneity of $V_{i+1}$ now gives the displayed ball
ratio. Only the inequality conclusion of the conditional result is used.
\end{proof}
At $i=n-1$, the displayed constant is
$\kappa_{n-1}^{\,n}\kappa_n^{\,2-n}$, exactly the constant in
Equation~\eqref{eq:main}. The case $i=1$ was already established by
Lutwak, as recalled in
\cite[author version, Introduction, p.~2]{OrtegaMorenoSchuster2021}.
Thus the additional range furnished here is $2\le i\le n-2$ for
$n\ge4$. No classification of equality cases is asserted for
Corollary~\ref{cor:lutwak-petty}.
\subsection{A sharp affine \texorpdfstring{$L^1$}{L1} Sobolev consequence}
For functions $f_1,\ldots,f_n\in C_c^1(\R^n)$, define the mixed
determinant-gradient integral
\[
\mathcal D_n(f_1,\ldots,f_n)
=\int_{(\R^n)^n}
\left|\det\bigl(\nabla f_1(x_1),\ldots,\nabla f_n(x_n)\bigr)\right|
\,dx_1\cdots dx_n.
\]
The determinant is the volume of the spanned parallelepiped, with no
factor $1/n!$. Under $f_i(x)\mapsto f_i(Ax+b)$ with $A$ invertible,
this integral and the product of the $L^{n/(n-1)}$ norms below both
scale by $|\det A|^{1-n}$.
\begin{corollary}[Mixed determinant-gradient and affine Sobolev inequalities]
\label{cor:affine-sobolev}
Let $n\ge3$ be an integer, put $q=n/(n-1)$, and let
$f_1,\ldots,f_n\in C_c^1(\R^n)$ be nonnegative. Then
\begin{equation}\label{eq:mixed-affine-sobolev}
\mathcal D_n(f_1,\ldots,f_n)
\ge n!\,\kappa_{n-1}^{\,n}\kappa_n^{\,2-n}
\prod_{i=1}^n\norm{f_i}_q.
\end{equation}
In particular, every nonnegative $f\in C_c^1(\R^n)$ satisfies the
affine $L^1$ Sobolev inequality
\begin{equation}\label{eq:affine-sobolev}
\mathcal D_n(f,\ldots,f)^{1/n}
\ge (n!)^{1/n}\kappa_{n-1}\kappa_n^{(2-n)/n}\norm{f}_q.
\end{equation}
Both constants are sharp in this class, through smooth approximation
of ellipsoid indicators in $BV(\R^n)$; no equality assertion for nonzero
$C_c^1$ functions is made.
\end{corollary}
\begin{proof}
Write $c_n=\kappa_{n-1}^{\,n}\kappa_n^{\,2-n}$.
Haddad's geometric functional
$\widetilde I_1(K_1,\ldots,K_n)$ integrates the absolute determinant
against the surface-area measures of the convex bodies $K_i$.
He uses
\[
h_{\Pi K}(u)=\frac12\int_{S^{n-1}}|\ip{u}{v}|\,dS_K(v),
\]
which is the projection-volume normalization above by Cauchy's
projection formula. His mixed-volume identity and the
Aleksandrov--Fenchel inequality give
\[
\frac1{n!}\widetilde I_1(K_1,\ldots,K_n)
=V(\Pi K_1,\ldots,\Pi K_n)
\ge\prod_{i=1}^n|\Pi K_i|^{1/n}.
\]
Here $V$ denotes mixed volume, and Haddad's $\omega_m$ is our
unit-ball volume $\kappa_m$; see
\cite[arXiv version, Introduction, p.~5]{Haddad2020}.
Theorem~\ref{thm:main} for $n\ge4$ and the separately proved
three-dimensional case \cite[Theorem~1.1]{ChenEtAl2026} therefore give
\[
\widetilde I_1(K_1,\ldots,K_n)
\ge n!c_n\prod_{i=1}^n|K_i|^{(n-1)/n}.
\]
Haddad's Lemma~5.1 and the equivalence of Equations~(18)--(21)
\cite[arXiv version, Section~5, pp.~16--17]{Haddad2020}
transfer this $p=1$ geometric inequality to the functional inequality
for $\mathcal D_n$, with critical exponent $p^*=n/(n-1)=q$.
At this endpoint his normalized weak profile for $K$ is $\mathbf1_K$,
with the same surface-area measure as
$K$ and $\norm{\mathbf1_K}_q=|K|^{(n-1)/n}$; see
\cite[arXiv version, Equation~(9) and Definition~2.2, pp.~8--9]{Haddad2020}.
Consequently the functional and geometric sharp constants agree, as
also follows from the constant relations after his Equation~(21).
This proves Equation~\eqref{eq:mixed-affine-sobolev}; taking all
$f_i=f$ gives Equation~\eqref{eq:affine-sobolev}, the $p=1$ case of
his Equation~(22).
For sharpness, take nonnegative compactly supported smooth
mollifications $f_j$ of the indicator of an ellipsoid $E$. They converge to
$\mathbf1_E$ strictly in $BV(\R^n)$ and in $L^q$. Define finite measures
$\nu_j$ on the sphere by
\[
\int\psi\,d\nu_j
=\int_{\{\nabla f_j\ne0\}}
\psi\!\left(-\frac{\nabla f_j}{|\nabla f_j|}\right)
|\nabla f_j|\,dx,
\qquad \psi\in C(S^{n-1}).
\]
Reshetnyak's continuity theorem
\cite[author version, Theorem~1.3, p.~3]{Spector2011} gives
$\nu_j\rightharpoonup S_E$; the minus sign selects the outward normal
in $D\mathbf1_E$. The product measures also converge weakly on the
compact product sphere. Since the absolute determinant is continuous
and homogeneous in each variable, this gives
$\mathcal D_n(f_j,\ldots,f_j)\to
\widetilde I_1(E,\ldots,E)=n!|\Pi E|$. Since
$\norm{\mathbf1_E}_q^n=|E|^{n-1}$, ellipsoid equality in Petty's
inequality gives the stated constants in the limit. This is the weak
$p=1$ interpretation of sharpness.
\end{proof}
To compare the diagonal statement with Zhang's affine Sobolev
inequality, let $f\ne0$ be as in Corollary~\ref{cor:affine-sobolev}
and define the origin-symmetric convex body $Z_f$ by
\[
h_{Z_f}(u)=\frac12\int_{\R^n}|u\cdot\nabla f(x)|\,dx.
\]
This integral is positive for every unit vector $u$: otherwise $f$
would be constant on every line parallel to $u$, contradicting compact
support and $f\ne0$. Thus $Z_f$ has nonempty interior. Haddad's
zonoid identity gives $|Z_f|=\mathcal D_n(f,\ldots,f)/n!$, and his
Blaschke--Santal\'o comparison
\cite[arXiv version, Equations~(24)--(25), p.~17]{Haddad2020} yields
\[
|Z_f^\circ|^{-1/n}
\ge \left(\frac{\mathcal D_n(f,\ldots,f)}{n!\kappa_n^2}\right)^{1/n}
\ge \frac{\kappa_{n-1}}{\kappa_n}\norm f_q.
\]
The resulting bound on $|Z_f^\circ|$ is Zhang's sharp affine Sobolev
inequality \cite{Zhang1999}. Thus Equation~\eqref{eq:affine-sobolev}
is a stronger affine $L^1$ inequality; the case $f=0$ is immediate.
This application uses only $p=1$ and gives no conclusion for
Haddad's $p>1$ problems.
\subsection{Holmes--Thompson isoperimetry}
\begin{corollary}[Holmes--Thompson isoperimetry]
\label{cor:holmes-thompson}
Let $n\ge3$, let $B=-B\subset\R^n$ be the unit ball of a real normed
space, and let $C\subset\R^n$ be a convex body. Write $B^\circ$ for the
polar unit ball and put $I_B=\Pi(B^\circ)/\kappa_{n-1}$. Normalize the
Holmes--Thompson volume and boundary area by
\[
\mathcal V_B(C)=\frac{|B^\circ|}{\kappa_n}|C|,
\qquad
\mathcal A_B(\partial C)=nV(C[n-1],I_B),
\]
where $V(C[n-1],I_B)$ is mixed volume with $n-1$ copies of $C$,
normalized by $V(C[n])=|C|$. Then
\begin{equation}\label{eq:holmes-thompson}
\mathcal A_B(\partial C)^n
\ge n^n\kappa_n\,\mathcal V_B(C)^{n-1}.
\end{equation}
Equality holds if and only if $B$ is an ellipsoid and $C=x+tB$ for some
$x\in\R^n$ and $t>0$.
\end{corollary}
\begin{proof}
The fixed-norm isoperimetrix formula and the equivalence with symmetric
unpolarized Petty are classical; see Martini--Mustafaev
\cite[Definition~1(i), Equation~(2), Section~5, and Theorem~12]{MartiniMustafaev2008}.
Minkowski's mixed-volume inequality gives
\[
\frac{\mathcal A_B(\partial C)^n}{\mathcal V_B(C)^{n-1}}
\ge n^n\frac{\kappa_n^{n-1}}{\kappa_{n-1}^{n}}R_n(B^\circ)
\ge n^n\kappa_n.
\]
The last step applies Theorem~\ref{thm:main} to $B^\circ$ for $n\ge4$,
and the separate three-dimensional result
\cite[Theorem~1.1]{ChenEtAl2026} for $n=3$. Equality in the first step
means that $C$ is homothetic to $I_B$, while equality in the last makes
$B^\circ$, hence $B$, an ellipsoid. Then $I_B$ is homothetic to $B$,
giving exactly the stated equality cases.
\end{proof}
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//07_consequences.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//bibliography.tex ====
\begingroup
\small
\begin{thebibliography}{99}
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\endgroup
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//bibliography.tex ====
==== BEGIN Pettys-projection-volume-conjecture-in-dimensions-at-least-four//main.tex ====
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\hypersetup{pdftitle={Petty's projection-volume conjecture in dimensions at least four},
pdfauthor={OpenAI},
pdfsubject={The projection-volume inequality and its equality cases in every dimension at least four}}
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\title{Petty's projection-volume conjecture\\
in dimensions at least four}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove that ellipsoids uniquely minimize the volume of the projection body
among convex bodies of fixed volume in every dimension at least four.
This proves Petty's projection-volume conjecture in these dimensions.
\end{abstract}
\input{sections/01_introduction}
\input{sections/02_gauges}
\input{sections/03_spherical}
\input{sections/04_norm}
\input{sections/05_position}
\input{sections/06_conclusion}
\input{sections/07_consequences}
\input{bibliography}
\end{document}
==== END Pettys-projection-volume-conjecture-in-dimensions-at-least-four//main.tex ====