==== TABLE OF CONTENTS ====
Uniform-Bi-Holder-Transport-from-Weak-MTW//main.tex
Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/conclusion.tex
Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/geometry.tex
Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/introduction.tex
Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/jets.tex
Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/scales.tex
Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/spectral.tex
Uniform-Bi-Holder-Transport-from-Weak-MTW//references.bib
===========================

==== BEGIN Uniform-Bi-Holder-Transport-from-Weak-MTW//main.tex ====
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\hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black,
 pdftitle={Uniform Bi-H\"older Transport from Weak MTW},pdfauthor={OpenAI}}
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\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
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\newcommand{\R}{\mathbb R}
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\DeclareMathOperator{\Hess}{Hess}
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\title{Uniform Bi-H\"older Transport from Weak MTW}
\author{OpenAI}
\date{September 25, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove that the weak Ma--Trudinger--Wang condition on a fixed smooth
connected compact boundaryless Riemannian manifold of dimension at least
two implies a common H\"older estimate for optimal transport maps and their
inverses over the entire class of probability densities with fixed positive
upper and lower bounds. The maps have homeomorphic representatives, conjugate
cut points are allowed, and no density regularity is assumed.
\end{abstract}
\tableofcontents
\input{sections/introduction}
\input{sections/geometry}
\input{sections/spectral}
\input{sections/scales}
\input{sections/jets}
\input{sections/conclusion}
\bibliographystyle{plain}
\bibliography{references}
\end{document}
==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//main.tex ====

==== BEGIN Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/conclusion.tex ====
\section{Completion of the uniform estimate}
\label{sec:conclusion}

Proposition~\ref{jet:contradiction} was applied to the section-oscillation
supremum over the entire class of densities with the fixed bounds
$\lambda,\Lambda$ on $(M,g)$. It therefore gives constants
$C_0<\infty$, $\beta>0$, and $r_0>0$, common to that class, such that
\[
 F(r)\le C_0r^\beta\qquad(0<r<r_0).
\]
Lemma~\ref{sc:power} now gives a common H\"older estimate with
$\alpha=\tfrac12\min\{\beta,1\}$. In particular, every pole has a
single contact endpoint. Applying the same estimate to the reversed
density pair gives singleton contacts on the other side, with the same
constants. The two contact maps are continuous inverses and agree
almost everywhere with the optimal maps. Enlarging the H\"older constant
to cover distances outside the small-scale range proves
Theorem~\ref{thm:main}.

\begin{remark}
The proof produces a positive exponent by contradiction and compactness.
It gives no explicit exponent or estimate uniform over varying metrics
or density bounds. The inverse map comes from applying the result to the
reversed positive-density pair; no time-one injectivity is asserted for
an arbitrary cost potential.
\end{remark}
==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/conclusion.tex ====

==== BEGIN Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/geometry.tex ====
\section{The density-free foundation}
\label{sec:geometry}

We record the precise geometric input from
\cite{Foundation2026}. Every statement in this section is
density-free and uses only the fixed compact manifold and weak
MTW. The proofs, including the treatment of conjugate cut
endpoints, are in that paper.

A potential is a function
$u(x)=\max_y\{-c(x,y)-w(y)\}$ with a continuous datum $w$;
replacing $w$ by the cost transform of $u$ gives a dual pair.
Put $D_g=\operatorname{diam}M$.
The total minimizing-vector set $\{(x,p):p\in G_x\}$ is compact.
For $p\in G_x$ and $y_s=\exp_x(sp)$, $0<s<1$, divided
action gives
\begin{equation}\label{geo:split-support}
 c(z,\exp_xp)\leq c_s(z,y_s)+\frac{1-s}{2}|p|^2,
 \qquad c_s(x,y_s)=\frac{s}{2}|p|^2.
\end{equation}
The right side is a smooth upper support near $x$.

\begin{lemma}[Uniform supports and active logs]\label{geo:prelim}
There is a geometric constant $C_g$ such that the cost is
uniformly semiconcave in either variable, including at cut
points. Every potential is $D_g$-Lipschitz and uniformly
semiconvex. For a continuous representing datum $w$, define
\[
 \mathcal V_u(x)=\{p\in G_x:
              u(x)+c(x,\exp_xp)+w(\exp_xp)=0\}.
\]
Then $\mathcal V_u(x)$ is compact and
$\partial u(x)=\operatorname{conv}\mathcal V_u(x)$.
Every $p\in\partial u(x)$ has a positive representation using at most
$n+1$ affinely independent active logs.
The set $\mathcal V_u(x)$ includes every minimizing log to
each endpoint active for the specified datum $w$.
\end{lemma}

This is \cite[Lemma~4.1]{Foundation2026}. In particular, the smooth upper
supports obtained by splitting at time $1/2$ have a common
neighborhood size and Hessian bound. The representing datum
need not itself be a cost potential.

Write $H_s(p)=A_x(sp)/s$. At a specified nonconjugate
minimizing branch, $A_x$ will also denote the smooth branch
Hessian.

\begin{lemma}[Radial divided action]\label{geo:radial-index}
Let $p\in G_x$, and let $S_p(r)$ be the Jacobi tensor along
$\exp_x(rp)$ with $S_p(0)=0$, $D_rS_p(0)=\Id$, in parallel
orthonormal frames. For $0<t<s<1$,
\[
 \frac{d}{dr}H_r(p)=-S_p(r)^{-1}S_p(r)^{-T},
 \qquad H_t(p)-H_s(p)\geq\kappa_g(s-t)\Id,
\]
where $\kappa_g>0$ depends only on the manifold. On every
smooth minimizing branch, $A_x(w)w=w$.
\end{lemma}

\begin{lemma}[Transverse slack]\label{geo:slack}
Let $p_i\in G_x$, $m_i>0$, $\sum_i m_i=1$, and set
$p=\sum_i m_ip_i$, $E=\Span\{p_i-p\}$. If
$t\operatorname{conv}\{p_i\}\subset I(x)$ and $0<t<s<1$,
then
\[
 \left[\frac{A_x(tp)}t-\sum_i m_i\frac{A_x(sp_i)}s\right]
       \bigg|_{E^\perp}\geq\kappa_g(s-t)\Id.
\]
The conclusion also holds at a limiting time $t$ if the
radially contracted hulls lie in $I(x)$ and the displayed
Hessians have nonconjugate branch limits. No minimizing
assumption about $sp$ is needed in this limiting assertion.
\end{lemma}

These are \cite[Lemmas~4.2--4.3]{Foundation2026}. The latter applies to
every positive finite representation, not merely a selected
Carath\'eodory representation. We reconstruct the Jacobi
identities and the additional uniform spectral bounds in
Section~\ref{sec:spectral}.

\begin{lemma}[Short time for a semiconvex datum]\label{geo:short-time}
Let $f$ be Lipschitz and semiconvex. For sufficiently small
$t>0$, depending only on these two bounds and the manifold,
\[
 F_t:\Gamma_f\to M,\qquad
 \Gamma_f=\{(x,p):p\in\partial f(x)\},\qquad
 F_t(x,p)=\exp_x(tp)
\]
is a homeomorphism with locally Lipschitz inverse into the
tangent bundle. The minimizing pole $x_t(z)$ of $Q_tf$ is
unique, $Q_tf\in C^{1,1}$, and
$\grad Q_tf(z)=-t^{-1}\log_zx_t(z)$.
The graph $\Gamma_f$ is an $n$-dimensional Lipschitz manifold.
\end{lemma}

This is \cite[Lemma~4.4]{Foundation2026}. The local mechanism is the curved
version of proximal regularization and quadratic
inf-convolution; compare \cite[Sections~3--5 and~7]{Moreau1965}.
It is needed below for increasing scalar modifications of
potentials, which need not themselves be cost potentials.

\begin{theorem}[Global supporting property]\label{geo:support}
For every potential $u$, every $x\in M$, and every
$p\in\partial u(x)$, one has $p\in G_x$ and
\[
 u(x')\geq u(x)+c(x,\exp_xp)-c(x',\exp_xp)
 \qquad(x'\in M).
\]
For each $0<t<1$, $F_t:\Gamma_u\to M$ is a
homeomorphism, $tp\in I(x)$, and its poles are the unique
global minimizers of $Q_tu$.
\end{theorem}

\begin{corollary}[Convex injectivity domains]\label{geo:convexity}
Both $G_x$ and $I(x)$ are convex for every $x\in M$.
\end{corollary}

\begin{corollary}[Convex lifted gap sections]\label{geo:sections}
For a dual pair $(u,v)$, each set
\[
 \mathcal S_x(r)=\left\{p\in G_x:
      u(x)+\frac{|p|^2}{2}+v(\exp_xp)\leq r\right\}
\]
is compact and convex, including ordinary and conjugate cut
endpoints. If it is nonempty and $r\geq0$, then
\[
 \operatorname{diam}\mathcal S_x(r)^2
 \leq8\left(r+\osc_{p\in\mathcal S_x(r)}v(\exp_xp)\right).
\]
\end{corollary}

These are \cite[Theorem~4.8 and Corollaries~4.9--4.10]{Foundation2026}.
Convexity of
injectivity domains alone is not the whole input: the
ordinary-subgradient supporting assertion is also used for
modified envelopes and local lower tests.

\begin{proposition}[Intermediate regularity and pole control]
\label{geo:intermediate}
Let $(u,v)$ be a dual pair and $0<t<1$. Then
$Q_tu=-Q_{1-t}v\in C^{1,1}(M)$, with almost-everywhere bounds
\[
 -\frac{C_g}{1-t}g\leq\Hess Q_tu\leq\frac{C_g}{t}g.
\]
If $(x_t(z),p_t(z))=F_t^{-1}(z)$ and $\Phi_s$ is geodesic
flow on $TM$, then
\[
 (x_t(z),p_t(z))=\Phi_{-t}(z,\grad Q_tu(z)).
\]
In particular, the pole and momentum maps are Lipschitz
with constants depending only on $M,g,t$.
There are also $r_t,b_t>0$, depending only on $M,g,t$,
such that every $z=\exp_x(tp)$, $p\in\partial u(x)$,
satisfies
\[
 u(x')+c_t(x',z)\geq
 u(x)+c_t(x,z)+b_t d(x,x')^2
 \qquad\text{if }d(x,x')<r_t.
\]
All these constants are uniform when $t$ ranges in a
compact subinterval of $(0,1)$.
\end{proposition}

This is \cite[Proposition~4.11]{Foundation2026}. Its proof uses the two-sided
supports from the split contact geodesic and smooth
backward geodesic flow, not inversion of the full
endpoint exponential map.

\begin{remark}\label{geo:time-one}
No time-one injectivity is claimed for an arbitrary
potential. For example, $u(x)=-c(x,y_0)$ can have many
poles with endpoint $y_0$. All inverse projections and
$C^{1,1}$ regularizations in this paper are used at
times strictly between zero and one.
\end{remark}
==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/geometry.tex ====

==== BEGIN Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/introduction.tex ====
\section{Introduction}
\label{sec:introduction}

For the squared-distance cost on a compact Riemannian manifold,
positive upper and lower density bounds give an almost-everywhere
optimal map, but continuity depends on the geometry of the cost.
The weak Ma--Trudinger--Wang condition controls that geometry where
the cost is smooth. We prove that it gives a common H\"older estimate
for the map and its inverse throughout each fixed density-bounded
class, even when transport graphs approach conjugate pairs.

Let $(M,g)$ be a smooth connected compact Riemannian manifold without
boundary, of dimension $n\geq2$. Write $d$ and $\vol$ for its distance
and volume measure, and set
\[
 c(x,y)=\frac12d(x,y)^2,\qquad
 I(x)=\{p\in T_xM:\ d(x,\exp_x(ap))=a|p|
                         \text{ for some }a>1\}.
\]
We assume the weak Ma--Trudinger--Wang condition: for every
$p\in I(x)$ and $\xi,\eta\in T_xM$ with $\xi\perp\eta$,
\begin{equation}\label{eq:weak-mtw}
 -\frac32\left.\frac{\partial^4}{\partial s^2\partial t^2}
 c\bigl(\exp_x(t\xi),\exp_x(p+s\eta)\bigr)\right|_{s=t=0}\geq0.
\end{equation}
The cost is smooth near the pair in this formula. No curvature
inequality at a nonsmooth cost pair is assumed.
For probability densities $\rho_0,\rho_1$, an optimal map is a
measurable map $T$ with $T_\#(\rho_0\vol)=\rho_1\vol$ that minimizes
$\int_M c(x,T(x))\rho_0(x)\,d\vol(x)$ among maps with that pushforward.

\begin{theorem}[Uniform bi-H\"older transport]\label{thm:main}
Fix $(M,g)$ as above and $0<\lambda\leq\Lambda<\infty$ for which the
following density class is nonempty. There are
$\alpha\in(0,1]$ and $C<\infty$, depending only on
$(M,g),\lambda,\Lambda$, such that the following holds for every pair
of measurable densities satisfying
\[
 \int_M\rho_i\,d\vol=1,\qquad
 \lambda\leq\rho_i\leq\Lambda\quad\vol\text{-almost everywhere},
 \qquad i=0,1.
\]
The almost-everywhere unique optimal map from $\rho_0\vol$ to
$\rho_1\vol$ for the cost $c$ has a homeomorphic representative
$\widetilde T:M\to M$, and
\[
 d(\widetilde T(x),\widetilde T(x'))\leq C d(x,x')^\alpha,
 \qquad
 d(\widetilde T^{-1}(y),\widetilde T^{-1}(y'))
       \leq C d(y,y')^\alpha
\]
for all $x,x',y,y'\in M$.
\end{theorem}

The constants are common to the full density class on the fixed
manifold. They do not depend on individual densities or their
derivatives. The theorem includes conjugate cut points, and does not
assume nonfocality, strict MTW, or cut avoidance. We obtain a positive
exponent by contradiction; no explicit exponent or uniformity over
varying metrics is asserted.

\subsection{Context and geometric input}

McCann's polar factorization theorem supplies the almost-everywhere
unique optimal map and its cost potential \cite{McCann2001}.
The curvature condition originates in the work of
Ma--Trudinger--Wang \cite{MTW2005}; Loeper established its connection
with supporting cost functions and continuity \cite{Loeper2009}.
Sections and contact-volume comparison are central tools in
Caffarelli's theory \cite{Caffarelli1992} and in the weak-MTW
regularity theory of Figalli--Kim--McCann
\cite[Theorem~2.1 and Corollary~9.6]{FKM2013Holder}. These results already treat rough densities.
Their smooth-cost H\"older estimates require the appropriate
cost-convexity and nondegeneracy hypotheses. A density-controlled
exponent after localization does not by itself provide a common
localization scale or H\"older constant for a family whose transport
graphs approach conjugate pairs.

Uniform control over a density-bounded class is known in important
geometric settings. Loeper--Villani obtain such estimates under
strict MTW and nonfocality, with a further extension under their
uniform-regularity hypothesis and positivity of the minimizing-fiber
diameter $\delta(M)$
\cite[Corollaries~8.3--8.4]{LoeperVillani2010}.
Figalli--Kim--McCann prove uniform cut separation on products of
round spheres \cite[Theorem~5.1 and Corollary~5.2]{FKM2013Spheres}.
The issue addressed here is the general fixed weak-MTW manifold,
without imposing such separation. A different quantitative stability
result of Warren \cite[Theorem~1]{Warren2026} assumes common
$C^2$ norm bounds on the logarithms of smooth densities and sufficiently small
Wasserstein distance; those assumptions are not part of the
measurable-density class considered here.

The density-free input is proved in the companion paper
\emph{Global Support and Convex Injectivity Domains under Weak MTW}
\cite{Foundation2026}. It derives convexity of open and closed
injectivity domains, global support for every ordinary subgradient,
and uniform intermediate regularity from \eqref{eq:weak-mtw}.
We state the complete interface in Section~\ref{sec:geometry}.
The global supporting conclusion has an important antecedent:
Figalli--Rifford--Villani \cite[Lemma~3.1]{FRV2011} prove the
global double-mountain principle assuming convex injectivity
domains as well as weak MTW. The companion supplies the geometric
hypothesis under weak MTW alone, together with the intermediate
statements needed below. Figalli--Gallou\"et--Rifford
\cite[Theorem~1.7]{FGR2015} previously proved the convexity
implication under nonfocality. On compact surfaces,
Figalli--Rifford--Villani \cite[Theorem~1.3]{FRV2011} proved that
transport continuity is equivalent to the conjunction of convex
injectivity domains and weak MTW. Their introduction poses the
equivalence with weak MTW alone; the separate convexity and continuity
questions also appear in \cite[Questions~8.6--8.7]{LPZ2019}.

Figalli--Rifford--Villani's transport continuity property concerns
all density pairs bounded above and away from zero. They prove that
this property implies convex injectivity domains and weak MTW
\cite[Definition~1.1 and Theorem~1.2(i)]{FRV2011}.
Theorem~\ref{thm:main} implies that property by choosing bounds for
each pair. Combined with their necessity result, it yields the
weak-MTW--transport-continuity equivalence, with the forward
direction strengthened to the class-uniform estimates above.

\subsection{Conventions}

We use dual potentials with
\[
 u(x)=\sup_y\{-c(x,y)-v(y)\},\qquad
 v(y)=\sup_x\{-c(x,y)-u(x)\}.
\]
Thus $u(x)+c(x,y)+v(y)\geq0$, with equality at a contact.
The symbol $\partial u(x)$ denotes the ordinary subdifferential
of the semiconvex function, with covectors identified with tangent
vectors by the metric. At a differentiable contact its gradient
is the initial minimizing log. We write
\[
 G_x=\{p\in T_xM:d(x,\exp_xp)=|p|\}=\overline{I(x)},\qquad
 c_t=c/t,\qquad Q_tf(z)=\min_x\{f(x)+c_t(x,z)\},
\]
and, for $p\in I(x)$,
\[
 A_x(p)=D^2_{xx}c(x,\exp_xp),\qquad
 \sigma_x(p)=|\det(d\exp_x)_p|.
\]
The function $\sigma_x$ extends continuously to $G_x$. At an
ordinary cut point, a specified nonconjugate minimizing branch
has a smooth upper cost branch; we state explicitly when such a
branch is used. A star denotes the adjoint. Constants denoted by
$c,C$ may change between formulas. Constants in the final
comparison depend only on the fixed geometry and density bounds;
constants used solely to pass limits at fixed scales may also
depend on the already fixed templates.

\subsection{Proof structure}

For a dual pair $(u,v)$, the lifted gap section at $x$ and
height $r\geq0$ is
\[
 \mathcal S_x(r)=\left\{p\in G_x:
       u(x)+\frac{|p|^2}{2}+v(\exp_xp)\leq r\right\}.
\]
We study the oscillation of $v$ on its endpoint image
$\exp_x\mathcal S_x(r)$. Convexity of the lifted section
turns a power bound on that oscillation into a power bound
on the diameter of the endpoint image. Taking the oscillation
supremum over all admissible density pairs makes this a single
scalar estimate for the entire class. Once it is proved, every
zero-height section has one endpoint, and comparison of nearby
poles gives the H\"older estimate in both transport directions.

The foundation theorem applies to all ordinary subgradients, including
convex combinations of all minimizing logs at active cut endpoints.
The projection $(x,p)\mapsto\exp_x(tp)$ of the full subgradient graph
is a homeomorphism for $0<t<1$; only the supporting inequality is
asserted at time one for an arbitrary potential. It also gives compact
convex lifted gap sections and $C^{1,1}$ intermediate potentials.
The minimizing point in $Q_tu(z)$, called its pole, depends
Lipschitzly on $z$.

Section~\ref{sec:spectral} develops the quantitative estimates needed
when a minimizing endpoint approaches conjugacy. A fixed joining
point factors the full Jacobi matrix into nonsingular piece
matrices and a positive semidefinite broken-action Hessian.
Radial shortening adds a uniformly positive form. Transverse MTW,
the radial identity, and a common separating logarithmic scale
then compare ordered singular values along an extended affine
segment.

Section~\ref{sc:section} takes a supremum of section oscillations
over the entire density class. Failure of every power bound gives
two separated gap scales on which this supremum is almost constant.
Two scalar modifications produce a positive comparison maximum.
One modifies the value at a center endpoint; the other supplies
the active endpoints whose minimizing logs have the center log
as a convex combination.
Varying their amplitude gives an inequality for every positive
active barycentric representation, with constants fixed before
the curvature of the outer modification is chosen.

Section~\ref{sec:jets} keeps the order of the limiting argument
explicit. An exact pole-matrix identity transfers lower jets to
active endpoints. The original contact-volume bound first excludes
conjugacy at a selected outer endpoint and bounds its determinant.
This bounds the center Hessian. Only then does the short-time
minimizer have nonsingular derivative, allowing the original
almost-everywhere density equation to be used at the sampled
center endpoints. An oblique-projection determinant gain and the
spectral comparison give the contradiction. Finally,
Section~\ref{sec:conclusion} converts the resulting power bound
into the common estimate for the map and its inverse.
==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/introduction.tex ====

==== BEGIN Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/jets.tex ====
\section{Lower jets and the determinant contradiction}
\label{sec:jets}

The comparison maxima of Section~\ref{sc:section} relate a center
endpoint to finitely many active outer endpoints. We seek determinant
bounds for lower tests of the original potential at these endpoints.
The main obstacle is that the center selected by a maximum need not
belong to the full-measure set where the original transport Jacobian
equation holds. Even though the short-time minimizing map is Lipschitz,
it could send a set of positive volume into that exceptional set.

We first transfer lower jets from the prefix regularization to its
active outer endpoints. The original contact-volume bound applies to
these lower tests and bounds the selected outer determinant. That bound
controls the regularized center Hessian, which in turn makes the
short-time minimizing map nonsingular at the sampled points. Only then
can we use the original Jacobian equation at the centers. The resulting
matrix inequalities give a determinant gain proportional to the outer
curvature parameter~$K$, contradicting the geometric comparison from
Section~\ref{sec:spectral}.

Throughout the argument the manifold and density bounds are fixed.
In the local calculations, $F,G$ and their indexed versions denote
regularized potentials; the scalar oscillation function $F$ from
Section~\ref{sc:section} returns in Proposition~\ref{jet:contradiction}.

Throughout this section, a \emph{lower Taylor jet} $(\zeta,B)$ of a function $f$ at the
origin means
\[
 f(h)\ge f(0)+\zeta\cdot h+\tfrac12 B(h,h)+o(|h|^2).
\]
Replacing $B$ by $B-\eps\Id$, for any $\eps>0$, gives a genuine quadratic
lower test on a sufficiently small neighborhood. All the tests to which
we apply a determinant inequality will have a strictly positive relative
Hessian, so this replacement and the subsequent limit $\eps\downarrow0$
are harmless.

\subsection{The original contact measure and lower tests}

\begin{lemma}[Contact-volume comparison]\label{jet:contact-volume}
Let $u,v$ be the dual potentials for an admissible pair of densities,
and let
\[
 \mathcal C=\{(x,y):u(x)+c(x,y)+v(y)=0\}.
\]
For a Borel set $A$ on either side, its contact image satisfies
\begin{equation}\label{jet:contact-volume-eq}
 \frac{\lambda}{\Lambda}\vol(A)
 \le \vol(\mathcal C(A))
 \le \frac{\Lambda}{\lambda}\vol(A).
\end{equation}
Projections are interpreted in the completed volume measures.
\end{lemma}

\begin{proof}
The contact relation is closed. Its projections of Borel sets are
analytic, hence measurable for the completed measures. The optimal
coupling has full mass on $\mathcal C$. At every differentiability point
of either potential its contact endpoint is unique: all minimizing
first-variation logs must equal its gradient, and their exponential
endpoints therefore coincide. Since both marginals are absolutely
continuous, almost every contact pair lies in the sets where both
potentials are differentiable. The two contact maps on these sets are
inverse to each other.

Write $\mu=\rho_0\vol$ and $\nu=\rho_1\vol$. Consequently
$\nu(\mathcal C(A))=\mu(A)$ for sets on the first side, and
$\mu(\mathcal C(A))=\nu(A)$ for sets on the second side: any additional
contact image outside the almost-everywhere map image consists, up to a
null set, of points where the opposite potential is not differentiable.
The density bounds give \eqref{jet:contact-volume-eq}.
\end{proof}

\begin{lemma}[Upper determinant bound for a lower test]
\label{jet:viscosity}
Let $v$ be one of the original dual potentials. Suppose a $C^2$
function $\phi$ touches $v$ from below at $y$, its gradient
$r=\grad\phi(y)$ belongs to $I(y)$, and
\[
 W=\Hess\phi(y)+A_y(r)>0.
\]
Then
\begin{equation}\label{jet:viscosity-eq}
 \sigma_y(r)\det W\le \Lambda/\lambda.
\end{equation}
The same conclusion holds for a lower Taylor jet with these
properties, by approximation of its quadratic part.
\end{lemma}

\begin{proof}
It suffices first to prove the result after decreasing the Hessian by a
small positive multiple of the metric. Thus we may arrange strict
touching, with
$v-\phi\ge\eps d(\cdot,y)^2$ in a small coordinate ball. The map
\[
 S(z)=\exp_z\grad\phi(z)
\]
is a local diffeomorphism near $y$. Indeed its defining equation is
$D_zc(z,S(z))+\grad\phi(z)=0$, and differentiation gives
\begin{equation}\label{jet:smooth-jacobian}
 |\det DS(z)|
 =\sigma_z(\grad\phi(z))
   \det\{\Hess\phi(z)+A_z(\grad\phi(z))\}.
\end{equation}
The relative Hessian is positive throughout a sufficiently small ball,
and the logs remain before cut.

For $h>0$, let $\Omega_h$ be the connected component containing $y$ of
$\{\phi+h>v\}$ inside that ball. It is relatively compact and shrinks to
$y$. If $z\in\Omega_h$ and $x=S(z)$, the cost mountain
\[
 m_x(w)=\phi(z)+c(z,x)-c(w,x)
\]
is tangent to $\phi$ at $z$. On a fixed smaller coordinate ball,
$\phi-m_x$ has positive Hessian, uniformly for $x$ close to $S(y)$;
thus $\phi\ge m_x$ there. On $\partial\Omega_h$ we have
$v-m_x\ge h$, whereas at $z$ we have $v(z)-m_x(z)<h$.
The minimum of $v-m_x$ on $\overline{\Omega_h}$ is therefore attained
at an interior point $z'$. Thus $\grad m_x(z')\in\partial v(z')$.
By Theorem~\ref{geo:support}, the corresponding local contact is
a global cost contact. Its endpoint is $x$. This proves
\[
 S(\Omega_h)\subset\mathcal C(\Omega_h).
\]
Use the area formula for the diffeomorphism $S$
\cite[Chapter~2, Theorem~3.3]{Simon2018}, followed by
Lemma~\ref{jet:contact-volume}, to obtain
\[
 \inf_{\Omega_h}|\det DS|\,\vol(\Omega_h)
 \le \vol(S(\Omega_h))
 \le(\Lambda/\lambda)\vol(\Omega_h).
\]
Let $h\downarrow0$ and then restore the original Hessian.
\end{proof}

Only the original correspondence occurs in these two Lemmas.
No density bound is claimed for a modified potential or its envelope.
For later use, the almost-everywhere counterpart of
\eqref{jet:smooth-jacobian} is the following consequence of
\cite[Proposition~4.1 and Theorem~4.2]{CEMS2001}. There is a
volume-null Borel set $N\subset M$ such that, at every $y\notin N$,
$v$ has an Alexandrov Hessian, its true log $r=\grad v(y)$ is before
cut, and
\begin{equation}\label{jet:ae-jacobian}
 W_v(y):=\Hess v(y)+A_y(r)>0,\qquad
 \frac{\lambda}{\Lambda}
 \le \sigma_y(r)\det W_v(y)
 \le\frac{\Lambda}{\lambda}.
\end{equation}
We enlarge $N$ to include the exceptional sets in the contact-map
differential of \cite[Proposition~4.1(b)]{CEMS2001}. The Jacobian
equation in \cite[Theorem~4.2]{CEMS2001} assumes only absolute
continuity with $L^1$ densities. The lower density bound makes its
full-source-measure set a full-volume set; differentiability of a
density is not required.

\subsection{An exact endpoint-jet calculation}

Fix a continuous datum $w_o$ and put
\[
 g(x)=\max_y\{-c(x,y)-w_o(y)\},\qquad F=Q_tg,\quad 0<t<1.
\]
By Proposition~\ref{geo:intermediate}, $F$ is locally $C^{1,1}$ and its
minimizing pole $x(z)$ is unique and locally Lipschitz. The following
calculation concerns the first minimum alone.

\begin{lemma}[The pole matrix and endpoint lower jets]
\label{jet:transfer}
Suppose $z_0$ is a point where $F$ has a second-order expansion and
$x(z)$ is differentiable. Write $x=x(z_0)$,
$z_0=\exp_x(tp)$, and choose active logs
\[
 p=\sum_{j=1}^Nm_jp_j,\qquad m_j>0,\quad
 \sum_jm_j=1,\qquad N\le n+1,
\]
where $y_j=\exp_xp_j$ and
$g(x)+c(x,y_j)+w_o(y_j)=0$. Set
$E=\Span\{p_j-p:1\le j\le N\}$. In the fixed-pole coordinates
$z(q)=\exp_x(tq)$, define
\begin{equation}\label{jet:L-definition}
 L=D_q^2\{c_t(x,z(q))-F(z(q))\}\big|_{q=p}.
\end{equation}
Then
\begin{equation}\label{jet:L-properties}
 L\ge0,\qquad LE=0.
\end{equation}
For every $j$, the function
\[
 \Psi_j(d)=w_o(\exp_x(p_j+d))+\tfrac12|p_j+d|^2
\]
has at $d=0$ the lower Taylor jet $(0,m_jL)$, after subtracting
its value at zero. If $p_j$ is nonconjugate, this is an endpoint
lower test in the smooth chosen cost branch.

More precisely, for $t<s<1$ put
\[
 Q_s=\frac1tA_x(tp)-\frac1s\sum_jm_j A_x(sp_j).
\]
Then $Q_s\ge0$ on $E^\perp$, and
$m_j(L+LQ_sL)$ is a lower Taylor Hessian for $\Psi_j$.
\end{lemma}

\begin{proof}
Use normal coordinates $x'=\exp_xa$ and set
\[
 B(a,q)=c_t(\exp_xa,z(q)),\quad
 f(q)=F(z(q)),\quad a(q)=\log_xx(z(q)).
\]
At the reference point $a(p)=0$. First variation gives the exact
identities
\[
 B(0,q)=\tfrac t2|q|^2,\qquad B_a(0,q)=-q.
\]
Thus $B_{qq}=t\Id$, $B_{qa}=-\Id$ at $(0,p)$.
The envelope identity $Df(q)=B_q(a(q),q)$ and differentiability of the
pole imply
\begin{equation}\label{jet:pole-identity}
 D^2f=t\Id-R,\qquad R=D_qa(p)=L.
\end{equation}
In particular $R$ is symmetric. Since
$f(q)\le g(x)+B(0,q)$ with equality at $p$, we have $L\ge0$.

Choose a common $s\in(t,1)$. For each $j$ define
\[
 y_j(d)=\exp_x(p_j+d),\qquad
 z_j(d)=\exp_x(s(p_j+d)),
\]
and the split upper cost
\begin{equation}\label{jet:split-cost}
 C_j(a,d)=
 \frac{c(\exp_xa,z_j(d))}{s}
 +\frac{c(z_j(d),y_j(d))}{1-s}.
\end{equation}
Both cost pieces are smooth near $(0,0)$ because they are proper
subsegments of a minimizing ray. Energy splitting gives
$c(\exp_xa,y_j(d))\le C_j(a,d)$. At $a=0$, the chosen geodesic
remains minimizing on both pieces for small $d$, and consequently
\begin{equation}\label{jet:split-derivatives}
 C_j(0,d)=\tfrac12|p_j+d|^2,\quad
 (C_j)_a(0,d)=-(p_j+d),\quad
 (C_j)_{ad}=-\Id,\quad
 (C_j)_{aa}=\frac1sA_x(sp_j).
\end{equation}
The use of the fixed reference pole in $z_j(d)$ is what makes the
mixed derivative exactly $-\Id$.

The nonnegative function
\[
 g(x(z(q)))+C_j(a(q),0)+w_o(y_j)
\]
vanishes at $p$. It is differentiable there, since
$g(x(z(q)))=f(q)-B(a(q),q)$, and its derivative equals
$(p-p_j)^*R$. This derivative is zero. Symmetry in
\eqref{jet:pole-identity} proves $LE=0$.

On $E^\perp$, MTW concavity on the convex hull of the $sp_j$, followed
by divided-action monotonicity, gives
\[
 Q_s\ge \frac1tA_x(tp)-\frac1s A_x(sp)>0.
\]
All points used in this inequality are before cut by
Theorem~\ref{geo:support}; equivalently one may invoke
Lemma~\ref{geo:slack}. Notice that unsplit endpoints may still be
conjugate.

Fix $i$ and write $m=m_i$. Given $d$, choose any linear map
$U:T_xM\to E$, and set $k=m(d+Ud)$ and $a=a(p+k)$.
In the lower bound defining $g(\exp_xa)$, move endpoint $i$ to
$y_i(d)$, keep all other endpoints fixed, replace the costs by
\eqref{jet:split-cost}, and average with the old weights. Write
$\widehat\Psi_i(d)=\Psi_i(d)-\Psi_i(0)$. The resulting inequality is
\[
 f(p+k)\ge
 B(a,p+k)-\sum_jm_j\{C_j(a,d_j)+w_o(y_j(d_j))\},
\]
where $d_i=d$ and $d_j=0$ for $j\ne i$. Subtract the equality at
$(a,k,d)=(0,0,0)$ and use
\eqref{jet:pole-identity}--\eqref{jet:split-derivatives}. The linear
terms cancel: the pole terms cancel because $\sum m_jp_j=p$, and
the prefix terms cancel because $Df(p)=tp$. The remaining inequality is
\begin{equation}\label{jet:quadratic-transfer}
 m\widehat\Psi_i(d)\ge
 \tfrac12L(k,k)+\tfrac12Q_s(a,a)
 +a\cdot(md-k)+o(|d|^2).
\end{equation}
Here $a=Lk+o(|d|)$. Since $md-k=-mUd$ and
$\operatorname{range}L\subset E^\perp$, the mixed term is
$o(|d|^2)$. Hence the lower Hessian is
\[
 m(\Id+U)^*(L+LQ_sL)(\Id+U).
\]
In fact this equals $m(L+LQ_sL)$ because $LU=0$. The latter
dominates $mL$, proving the asserted jet. The linear term is exactly
zero independently of any second function or maximum condition.

The map $d\mapsto y_i(d)$ need not be invertible for the preceding
pullback calculation. When $p_i$ is nonconjugate it is a local
diffeomorphism, and $\tfrac12|p_i+d|^2$ is the smooth chosen
upper branch of $c(x,y_i(d))$. This converts the pullback jet into
an endpoint jet.
\end{proof}

Let $G$ be another locally $C^{1,1}$ function, and define
\begin{equation}\label{jet:H-definition}
 H=D_q^2\{c_t(x,z(q))+G(z(q))\}\big|_{q=p}.
\end{equation}
The exact identity
\begin{equation}\label{jet:H-minus-L}
 H-L=D_q^2\{(F+G)(z(q))\}\big|_{q=p}
\end{equation}
will distinguish the first-minimum calculation from the maximum
argument. At a twice differentiable maximum, it gives $H\le L$ and
$H|_E\le0$. At a Jensen point with gradient and upper Hessian errors
tending to zero in a fixed smooth chart, it gives
\begin{equation}\label{jet:Jensen-error}
 H\le L+\eta\Id,\qquad \eta\longrightarrow0.
\end{equation}
The coordinate-change term uses the gradient error and also tends to
zero. Lemma~\ref{jet:transfer} is exact at such a point. For every
linear $U$ with image in $E$, it therefore supplies a lower jet
$H_i=m_iL$ satisfying
\[
 H_i\ge m_i(\Id+U)^*H(\Id+U)
       -m_i\eta(\Id+U)^*(\Id+U).
\]
In particular no approximate stationarity term occurs in the
linear coefficient of the endpoint test.

\subsection{The center semibound without a density equation}

We use the scales and templates of the preceding section. To distinguish
a scalar argument from a point of the manifold, write
\[
 \varphi_c(h)=h+bB_c((h-a)/D),\quad
 \varphi_o(h)=h+bB_o((h-a)/D),\quad
 w_c=\varphi_c\circ v,\quad w_o=\varphi_o\circ v.
\]
For each fixed outer template the scales are sufficiently small that
both scalar maps have derivative between $1/2$ and $2$. On the fixed
compact interval of possible center levels, $B_c'$ is negative and
bounded away from zero. Its derivative bounds are fixed independently
of the outer curvature parameter $K$.

\begin{lemma}[A lower semibound before the center Jacobian]
\label{jet:semibound}
Fix one set of scales, one original potential $v$, and the two
templates. Let $t_j\uparrow1$ and $b_j\in[b_-,b_+]$. Put
\[
 F_j=Q_{t_j}g_{b_j},\qquad G_j=Q_{1-t_j}w_{c,b_j}.
\]
Let $z_j$ be simultaneous differentiability points for the objects in
Lemma~\ref{jet:transfer}, with second-order expansions of $G_j$, whose center levels lie in the
fixed compact interval above. Suppose
\[
 |\grad(F_j+G_j)(z_j)|\longrightarrow0.
\]
Let $x_j$ be the prefix pole and
$z_j=\exp_{x_j}(t_jp_j)$. The matrix $H_j$ of
\eqref{jet:H-definition} satisfies
\begin{equation}\label{jet:semibound-eq}
 H_j\ge c\frac{b_j}{D}\Id
       -C\frac{b_j}{D^2}p_jp_j^*-\eps_j\Id,
 \qquad \eps_j\longrightarrow0.
\end{equation}
The constants $c,C$ depend on the geometry and the fixed center
template, but not on $K$. The errors tend to zero with the prefix and
gradient errors at the fixed scales. No upper bound on $H_j$, and no
Alexandrov Hessian of $v$ at the center, is assumed.
\end{lemma}

\begin{proof}
Write $\tau_j=1-t_j$, and let $y_j$ be the unique final minimizer.
Set
\[
 r_j=\tau_j^{-1}\log_{y_j}z_j,\qquad
 l_j=\varphi_c'(v(y_j))<1.
\]
The short minimum gives a smooth local lower test for $w_c$ with
gradient $r_j$. Inverting $\varphi_c$ gives a local lower test for
$v$ with gradient $r_j/l_j$. Theorem~\ref{geo:support} therefore
implies that
\[
 \zeta_j=\exp_{y_j}(r_j/l_j)
\]
is a true contact endpoint of $v$, joined by the indicated minimizing
ray. In particular $\widehat x_j=\exp_{y_j}r_j$ is strictly before
cut along that ray.

For the moment omit subscripts. Put
$\theta=(1+l^{-1})/2$ and $\xi=\exp_y(\theta r)$.
Split the true support at $\xi$, using the fraction $l\theta$ of its
unit duration. The resulting upper cost support is
\[
 C(h)=\frac{c(h,\xi)}{l\theta}
       +\frac{c(\xi,\zeta)}{1-l\theta}.
\]
It agrees with $c(h,\zeta)$ at $h=y$, and $v(h)\ge
v(y)+c(y,\zeta)-C(h)$. Thus
\[
 \psi(h)=\varphi_c\{v(y)+c(y,\zeta)-C(h)\}
\]
is a smooth local lower support for $w_c$, with
\begin{equation}\label{jet:center-support-jet}
 \grad\psi(y)=r,\qquad
 \Hess\psi(y)
 =-\theta^{-1}A_y(\theta r)
   +\frac{\varphi_c''(v(y))}{l^2}rr^*.
\end{equation}
The original true endpoint $\zeta$ may be conjugate. The point $\xi$
is halfway into the available extension and is before cut, which is
all that this construction uses.

At the fixed scales, $1-l$ has a positive lower bound and the
parameters $(y,r,l,v(y))$ range over a compact set. The first cost
piece in $C$ is uniformly smooth on a fixed neighborhood of $y$;
the second piece is constant in $h$. Consequently the supports
$\psi$ have uniform smooth bounds and a uniform supporting radius.
For sufficiently small $\tau$, the local infimum $Q_\tau\psi$ is
smooth near $z$, has minimizer $y$ at $z$, and supports
$Q_\tau w_c$ from below there. To justify the last assertion, all
minimizers of the latter lie within $O(\tau)$ of $z$, by the
Lipschitz bound on $w_c$, and so lie in the supporting neighborhood.
For $\psi$ the local Hessian of the minimizing expression is
$\tau^{-1}\Id+O(1)>0$. Equality and stationary first variation hold
at $y$. The implicit-function theorem and the Schur formula for a
minimum give, after parallel identification,
\begin{equation}\label{jet:smooth-short-limit}
 \Hess Q_\tau\psi(z)=\Hess\psi(y)+O(\tau).
\end{equation}
All constants in this statement are uniform for the fixed-scale
family just described.

Let $\widehat p=\log_{\widehat x}y$ and
$\widehat L=(d\exp_{\widehat x})_{\widehat p}$.
Lemma~\ref{spec:pullback}, applied to the minimizing ray through
duration $1/l$, gives
\[
 \widehat L^*
 \{A_y(r)-\theta^{-1}A_y(\theta r)\}\widehat L
 \ge c(1-l)\Id.
\]
Gauss's Lemma gives $\widehat L^*r=-\widehat p$. Combining this
with \eqref{jet:center-support-jet} yields
\begin{equation}\label{jet:center-support-pullback}
 \widehat L^*\{\Hess\psi(y)+A_y(r)\}\widehat L
 \ge c(1-l)\Id
       +\frac{\varphi_c''(v(y))}{l^2}
          \widehat p\widehat p^*.
\end{equation}
Here $1-l\ge c b/D$ and
$|\varphi_c''(v(y))|/l^2\le Cb/D^2$.

We finally compare the reference pole with $\widehat x_j$.
The envelope identities give
\[
 x_j=\exp_{z_j}(-t_j\grad F_j(z_j)),\qquad
 \widehat x_j=\exp_{z_j}(t_j\grad G_j(z_j)).
\]
They are at distance tending to zero. The pairs
$(\widehat x_j,y_j)$ have a uniform fixed-scale extension, so their
cost charts and inverse exponential charts are uniformly smooth.
Omit subscripts again and put $U=Q_\tau\psi$, $Z(q)=\exp_x(tq)$,
$M=D_qZ(p)$, and $R=\nabla_q^2Z(p)$.
Since $G-U$ has a local minimum at $z$, its first derivative
vanishes and its second-order form is positive semidefinite.
Consequently
\[
 H\ge H_\psi
 :=t\Id+M^*\Hess U(z)M+\langle\grad U(z),R\rangle.
\]
Only the smooth support $U$ occurs on the right. Its gradient
and Hessian are uniformly bounded at the fixed scales, and
\eqref{jet:smooth-short-limit} identifies its Hessian with that
of $\psi$, up to $O(\tau)$ after parallel transport.

Write $\delta=|\grad(F+G)(z)|$ and identify nearby pole
tangent spaces by parallel transport. The explicit pole formulas
give $d(x,\widehat x)=O(\delta)$.
Moreover $p=t^{-1}\log_xz$ and
$\widehat p=t^{-1}\log_{\widehat x}z$ differ by
$O_{\mathrm{fixed}}(\delta)$, since the latter ray has a
uniform fixed-scale extension. Thus replacing $x,p$ by
$\widehat x,\widehat p$ in $M,R$ changes their entries by
$O_{\mathrm{fixed}}(\delta)$. Changing $t$ to $1$ then
changes them by $O(\tau)$. Every resulting error in $H_\psi$
multiplies a bounded derivative of $U$.

For the full endpoint map $Y(q)=\exp_{\widehat x}q$, let
$B=DY(\widehat p)$. The identity
$c(\widehat x,Y(q))=|q|^2/2$ on the chosen branch gives
\[
 \Id=B^*A_y(r)B-\langle r,\nabla_q^2Y(\widehat p)\rangle.
\]
Since $\grad\psi(y)=r$, the preceding expansions yield
\[
 H\ge H_\psi
 =B^*\{\Hess\psi(y)+A_y(r)\}B
       +O_{\mathrm{fixed}}(\tau+\delta).
\]
Combining this with \eqref{jet:center-support-pullback}
proves \eqref{jet:semibound-eq}. No coordinate error
multiplies the unknown Hessian of $G$.
\end{proof}

The true and modified endpoints occur in opposite orders along the
center and outer reverse rays. In the center calculation, $l<1$ makes $\widehat x=\exp_y r$ a proper intermediate
point on the true ray to $\zeta=\exp_y(r/l)$. At an active outer
endpoint $y_i$, a slope $l_i=\varphi_o'(v(y_i))>1$ instead makes the true endpoint
$\zeta_i=\exp_{y_i}(r_i/l_i)$ an intermediate point on the modified
ray to $x=\exp_{y_i}r_i$, where $r_i$ is its minimizing reverse log.
Figure~\ref{fig:reverse-rays} displays this distinction. We next use it
to exclude conjugacy at the full modified outer endpoint.

\begin{figure}[htbp]
\centering
\setlength{\unitlength}{1mm}
\begin{picture}(128,57)
 \put(0,50){\makebox(128,5)[l]{Center modification: $l<1$}}
 \put(9,38){\vector(1,0){110}}
 \put(13,38){\circle*{1.2}}
 \put(72,38){\circle*{1.2}}
 \put(107,38){\circle*{1.2}}
 \put(8,40){\makebox(10,5){$y$}}
 \put(67,40){\makebox(10,5){$\widehat x$}}
 \put(102,40){\makebox(10,5){$\zeta$}}
 \put(8,31){\makebox(10,5){$0$}}
 \put(67,31){\makebox(10,5){$1$}}
 \put(97,31){\makebox(20,5){$1/l>1$}}
 \put(0,22){\makebox(128,5)[l]{Selected outer modification: $l_i>1$}}
 \put(9,10){\vector(1,0){110}}
 \put(13,10){\circle*{1.2}}
 \put(58,10){\circle*{1.2}}
 \put(107,10){\circle*{1.2}}
 \put(8,12){\makebox(10,5){$y_i$}}
 \put(53,12){\makebox(10,5){$\zeta_i$}}
 \put(102,12){\makebox(10,5){$x$}}
 \put(8,3){\makebox(10,5){$0$}}
 \put(43,3){\makebox(30,5){$1/l_i<1$}}
 \put(102,3){\makebox(10,5){$1$}}
\end{picture}
\caption{Two independent reverse rays, shown schematically with their
duration parameters. The modified center endpoint $\widehat x$ and
the true outer endpoint $\zeta_i$ are strictly before cut. The full
endpoints $\zeta$ and $x$ may still be conjugate at this stage of the
argument. The two rows are not drawn to a common scale.}
\label{fig:reverse-rays}
\end{figure}

\subsection{The outward endpoint and conjugacy}

\begin{lemma}[Conjugacy is excluded by an active increasing modification]
\label{jet:outer-conjugacy}
Fix $b>0$ and an active outer pair
\[
 g_b(x)+c(x,y)+\varphi_o(v(y))=0.
\]
Choose any minimizing reverse log $r$ from $y$ to $x$. Suppose
\[
 l=\varphi_o'(v(y))>1,\qquad \varphi_o''(v(y))<0.
\]
Then the chosen modified ray is nonconjugate at its endpoint.
The true log $r/l$ is before cut.
\end{lemma}

\begin{proof}
The active inequality reads
$\varphi_o(v(h))\ge-c(x,h)-g_b(x)$.
For $1/l<s<1$, put $\xi_s=\exp_y(sr)$ and form the smooth upper
support
\[
 C_s(h)=c(h,\xi_s)/s+c(\xi_s,x)/(1-s).
\]
It has value $c(y,x)$ and gradient $-r$ at $y$. Since
$\varphi_o$ is increasing, $\varphi_o^{-1}(-C_s-g_b(x))$
is a smooth lower test for $v$ with gradient $r/l$.
That log is strictly before cut because it shortens a minimizing
ray. It also gives a global contact by
Theorem~\ref{geo:support}, but its before-cut property does not
depend on a density equation.

The scaled true relative Hessian of this test is
\begin{equation}\label{jet:outer-truncated}
 W_s=lA_y(r/l)-s^{-1}A_y(sr)
       -\frac{\varphi_o''(v(y))}{l^2}rr^*.
\end{equation}
Fix $s_0\in(1/l,1)$. For $s\ge s_0$, divided-action monotonicity
gives a positive definite lower bound for the difference of the
first two terms, independent of $s$. The last term is positive
semidefinite. If the endpoint at duration one were conjugate,
the index-form characterization of conjugacy, equivalently
Lemma~\ref{spec:radial}, would force an eigenvalue of
$s^{-1}A_y(sr)$ to tend to $-\infty$ as $s\uparrow1$.
Thus $\det W_s\to\infty$, while $W_s$ retains a positive
definite baseline. Lemma~\ref{jet:viscosity}, at the fixed true
log $r/l$, instead gives
\[
 \det W_s\le l^n(\Lambda/\lambda)/\sigma_y(r/l)<\infty.
\]
This is a contradiction.
\end{proof}

\begin{lemma}[Undoing the outer modification]
\label{jet:outer-test}
At a pair as in Lemma~\ref{jet:outer-conjugacy}, let
$p=\log_xy$ be the corresponding forward log,
$B=(d\exp_x)_p$, and let $H_o\ge0$ be a lower Taylor Hessian,
with zero linear part, of
$d\mapsto w_o(\exp_x(p+d))+\tfrac12|p+d|^2$.
Define
\begin{align}
 V_o={}&H_o+B^*\{lA_y(r/l)-A_y(r)\}B
             -\frac{\varphi_o''(v(y))}{l^2}pp^* .
 \label{jet:outer-form}
\end{align}
Then $V_o>0$ and
\begin{equation}\label{jet:outer-det}
 \det V_o\le
 l^n(\Lambda/\lambda)\,
       \frac{\sigma_x(p)^2}{\sigma_y(r/l)}.
\end{equation}
The middle term in \eqref{jet:outer-form} is positive definite.
On a compact family of active pairs where $b>0$ is bounded away
from zero, $\varphi_o'>1$ and $\varphi_o''<0$ hold with fixed
strict margins, this term has a uniform positive definite
baseline and the chosen exponential branches are uniformly
nonsingular.
\end{lemma}

\begin{proof}
Lemma~\ref{jet:outer-conjugacy} makes the endpoint coordinate map
locally invertible. Its smooth cost branch has endpoint Hessian
$A_y(r)$, and the zero linear coefficient fixes the gradient of
the induced lower test of $w_o$ to be $r$. Inverting
$\varphi_o$ and using the chain rule shows that
\eqref{jet:outer-form} is $l$ times the true relative lower-test
Hessian, pulled back by $B$. Gauss's Lemma gives the rank-one
term displayed there.

The divided-action difference between durations $1/l$ and $1$
is strictly positive definite, so $V_o>0$. Decrease the lower
Taylor Hessian by $\eps\Id$ to obtain a genuine lower test,
apply Lemma~\ref{jet:viscosity}, and let $\eps\downarrow0$.
Taking the determinant of the scaling and pullback gives
\eqref{jet:outer-det}.

For the uniformity assertion, include the minimizing log among
the parameters. The family is compact and active contact is a
closed condition. Lemma~\ref{jet:outer-conjugacy} excludes a
conjugate limit anywhere in that family; radial shortening
excludes true cut limits. The smooth branch matrices and the
positive divided-action difference therefore have the claimed
uniform bounds by compactness. Multiple ordinary branches
cause no difficulty: each specified nonconjugate minimizing
branch is a smooth upper branch and has been included among
the parameters.
\end{proof}

\subsection{The order of the limiting argument}

We now select points at which the preceding lower tests and the original
Jacobian equation can be combined. The following elementary fact will
justify differentiation through a short-time minimizer once its endpoint
is an Alexandrov point: neighboring endpoints need not be differentiable
points of the datum.

\begin{lemma}[Subgradients at an Alexandrov point]\label{jet:subgradient-differential}
If $f$ is semiconvex near $0$ and has the Alexandrov expansion
\[
 f(h)=f(0)+p\cdot h+\tfrac12 A(h,h)+o(|h|^2),
\]
then every nearby subgradient satisfies
\[
 \zeta=p+Ah+o(|h|),\qquad \zeta\in\partial f(h),
\]
uniformly over the choice of $\zeta$.
\end{lemma}

\begin{proof}
Adding a fixed quadratic reduces the assertion to a
convex function, and subtracting it later restores the displayed
formula. In the convex case write the remainder as $R(h)$, where
$|R(h)|\le\omega(|h|)|h|^2$ for a nondecreasing modulus
$\omega(r)\to0$. For $r=|h|>0$, a unit vector $e$, and
$0<\alpha\le1$, the subgradient inequalities at
$h\pm\alpha r e$ give
\[
 |(\zeta-p-Ah)\cdot e|
 \le \tfrac12\|A\|\alpha r
        +5\omega(2r)r/\alpha.
\]
For small $r$, choose
$\alpha=\sqrt{\max\{\omega(2r),r^2\}}$ and take the supremum
over unit $e$. This proves the uniform assertion.
\end{proof}

The selection has two stages. At each fixed prefix time we first pass
from nearby twice differentiable points to an exact maximum, where the
parameter inequality holds for every active representation. After
selecting an outer endpoint in that limit, we return to the nearby
points to obtain the Hessian bound and the original center Jacobian.
This order preserves the parameter inequality without assuming that a
prescribed active vertex persists under perturbation.

\begin{proposition}[The limiting matrix configuration]
\label{jet:determinants}
Fix the center template and the constants of
Lemmas~\ref{sc:excess}, \ref{sc:maximum}, \ref{sc:parameter}, and
\ref{sc:outward}. Choose $K$, then an outer template with curvature
less than $-K$ in the prescribed capped region. For all sufficiently
small scales in the sequence of Lemma~\ref{sc:plateau}, there are
a parameter $b\in[b_-,b_+]$, a pole $x$, center and active logs
$p,p_i\in G_x$, weights $m_i>0$ summing to one, and symmetric
matrices $H,L,V,V_i$ on $T_xM$, with the following properties.
For one selected index $i$,
\begin{gather}
 p=\sum_jm_jp_j,\qquad
 E=\Span\{p_j-p\}=\R e,\qquad |e|=1,\quad
 q=p\cdot e>0,\quad q^2\ge cD, \label{jet:limiting-geometry}\\
 m_i\ge m_*>0,\qquad p_i=p+d_ie,\quad d_i>0, \label{jet:limiting-outward}\\
 L\ge0,\qquad LE=0,\qquad H\le L,\qquad
 H\ge V+c\frac bD\Id-C\frac b{D^2}pp^*,\qquad V>0,
 \label{jet:limiting-center}\\
 V_i\ge m_iL+cK\frac b{D^2}p_ip_i^*,\qquad V_i>0.
 \label{jet:limiting-outer}
\end{gather}
The center modified log and the selected outer modified log are
nonconjugate. Let $r$ and $r_i$ denote the reverse logs of these
specified rays, let $y=\exp_xp$, $y_i=\exp_xp_i$, and set
$l=\varphi_c'(v(y))$, $l_i=\varphi_o'(v(y_i))$. The chosen true
limiting branches are also nonconjugate, and
\begin{equation}\label{jet:limiting-determinants}
 \det V\ge c\,\frac{\sigma_x(p)^2}{\sigma_y(r/l)},
 \qquad
 \det V_i\le C\,\frac{\sigma_x(p_i)^2}{\sigma_{y_i}(r_i/l_i)}.
\end{equation}
Here $c,C,m_*$ are independent of $K$ and the outer template.
The level and norm-increment conclusions of
Lemma~\ref{sc:outward} hold as well.
\end{proposition}

\begin{proof}
\emph{Step 1: fixed scales and exact maxima.}
Throughout the proof the original potential $v$, the scales, and
both templates are fixed. Constants used only to pass limits may
depend on all of them. Choose $t_k\uparrow1$ and interior maxima
$(z_k^0,b_k)$ supplied by Lemma~\ref{sc:maximum}, for
\[
 F_k(z)=Q_{t_k}g_{b_k}(z),\qquad
 G_k(z)=Q_{1-t_k}w_{c,b_k}(z).
\]
The functions are $C^{1,1}$ in space:
Proposition~\ref{geo:intermediate} applies to the first, and
Lemma~\ref{geo:short-time} to the second. At an exact maximum the
velocities align. If $x_k^0$ and $y_k^0$ are its two minimizers
and $p_k^0=t_k^{-1}\log_{x_k^0}z_k^0$, then
$y_k^0=\exp_{x_k^0}p_k^0$ along the concatenated minimizing ray.
Every finite positive active representation of $p_k^0$ satisfies
the parameter inequality of Lemma~\ref{sc:parameter}. The level
relations and common section bounds of Lemma~\ref{sc:levels}
hold for these representations.

\emph{Step 2: null sets are removed before Jensen sampling.}
Let $N$ be the null set for the original potential in
\eqref{jet:ae-jacobian}. For this fixed $k$, the final minimizer
map $Y_k(z)$ is locally Lipschitz and hence differentiable
almost everywhere \cite[Chapter~2, Theorem~1.4]{Simon2018}.
In each coordinate chart the
locally Lipschitz area formula
\cite[Chapter~2, Theorem~3.3]{Simon2018} implies that
\begin{equation}\label{jet:null-preimage}
 Z_k=\{z:DY_k(z)\text{ exists},\ \det DY_k(z)\ne0,\
                     Y_k(z)\in N\}
\end{equation}
is a null set. Indeed the integral of $|\det DY_k|$ on
$Y_k^{-1}(N)$ is zero. This argument does not require us to know
in advance where $DY_k$ is nonsingular.

Apply the semiconvex maximum Lemma
\cite[Appendix, Lemmas~A.3--A.4]{CIL1992} to $F_k+G_k$ at
$z_k^0$, using a small fourth-order strictness penalty if needed.
Its positive-measure conclusion allows us to choose a sequence
$z_{k,\nu}\to z_k^0$ off $Z_k$ and off all the null sets where
$F_k,G_k$, the prefix pole map, or $Y_k$ lack the derivatives
used below. We have
\begin{equation}\label{jet:Jensen-sequence}
 \grad(F_k+G_k)(z_{k,\nu})\longrightarrow0,\qquad
 D^2(F_k+G_k)(z_{k,\nu})\le \eta_{k,\nu}\Id,\quad
 \eta_{k,\nu}\longrightarrow0
\end{equation}
in the fixed chart.
At every such point choose a representation with at most $n+1$
active logs, padding by zero weights only for compactness.
Lemma~\ref{jet:transfer} applies at each point. Its matrices
$L_{k,\nu}$ are positive semidefinite and annihilate the active
differences, exactly; its endpoint lower jets are $m_{i,k,\nu}
L_{k,\nu}$. The matrices $H_{k,\nu}$ satisfy
\eqref{jet:Jensen-error}.

\emph{Step 3: take the first, fixed-prefix limits.}
For each fixed $k$, the local $C^{1,1}$ bounds are finite.
After taking a subsequence in $\nu$, all configuration parameters,
weights, and the two matrices $H_{k,\nu},L_{k,\nu}$ converge.
Denote the limits by $p_k,p_{i,k},m_{i,k},H_k,L_k$ and the
corresponding base points. Discard zero limiting weights.
The limiting endpoints are active at the exact maximum, and
their weighted sum is $p_k$. In particular the \emph{universal}
parameter inequality applies to this limiting representation;
we have not asserted forward persistence of any prescribed
active vertex. We retain
\begin{equation}\label{jet:fixed-prefix-limits}
 L_k\ge0,\qquad L_kE_k=0,\qquad H_k\le L_k,
 \quad E_k=\Span\{p_{i,k}-p_k:m_{i,k}>0\}.
\end{equation}
The quantitative proof of Lemma~\ref{jet:semibound} has an error
bounded by a fixed-scale constant times
$1-t_k+|\grad(F_k+G_k)(z_{k,\nu})|$. Passing $\nu\to\infty$
therefore gives
\[
 H_k\ge c\frac{b_k}{D}\Id
           -C\frac{b_k}{D^2}p_kp_k^*
           -C_{\mathrm{fixed}}(1-t_k)\Id.
\]
Since $b_k\ge b_->0$, for sufficiently large $k$ this implies,
after reducing $c$,
\begin{equation}\label{jet:absorbed-semibound}
 H_k\ge c\frac{b_k}{D}\Id
           -C\frac{b_k}{D^2}p_kp_k^*.
\end{equation}

If $\dim E_k\ge2$, a nonzero vector in $E_k\cap p_k^\perp$
contradicts \eqref{jet:fixed-prefix-limits} and
\eqref{jet:absorbed-semibound}. If $E_k=0$, all active logs in
the representation equal $p_k$ and all endpoints equal the
center. But $B_c\le B_o$ at every level, contradicting the
strict parameter inequality. Thus $E_k=\R e_k$. Orient $e_k$
so $q_k=p_k\cdot e_k>0$; the same two inequalities give
$q_k^2\ge cD$. Lemma~\ref{sc:outward} now supplies an index
with $m_{i,k}\ge m_*$, a strict outward displacement, and a
bounded template value. We choose an index carrying a fixed
share of the outward moment, as permitted by that Lemma, so
its displacement has a positive lower bound at the fixed
scales. Relabel it as index $1$.

\emph{Step 4: fix the outer branch before bounding any center Hessian.}
Pass to a subsequence in $k$ so that the selected configurations
and their weights converge. Their centers and selected
endpoints remain in the compact level ranges of
Lemma~\ref{sc:outward}. The selected limiting weight is at
least $m_*$, its displacement is strictly outward, and its
outer template value is below the prescribed cap. By imposing
the template conditions under a slightly larger fixed cap,
the limiting slope and curvature conditions have strict
margins:
$B_o'>0$ and $B_o''<-K$. The limiting selected pair is active.
Lemma~\ref{jet:outer-conjugacy} excludes conjugacy of its
modified branch, using only this activity and the original
contact-volume bound. Its shortened true log is before cut.
Lemma~\ref{jet:outer-test} consequently supplies a neighborhood
of this configuration with constants $\delta>0$ and $M<\infty$
such that its divided-action gain is at least $\delta\Id$
and its determinant upper bound is at most $M$.
These constants may depend on the fixed scales and templates.

For every remaining $k$, return to its already chosen
Jensen subsequence and choose $\nu=\nu(k)$ sufficiently large.
We require the configuration and matrices to be within $1/k$
of their fixed-prefix limits, the selected weight to be at
least $m_*/2$, and the selected endpoint to lie in that
neighborhood. We also require both errors in
\eqref{jet:Jensen-sequence}, including their transformed
versions in \eqref{jet:Jensen-error}, to be at most
$(1-t_k)^3/k$. All these requirements are compatible because
the limit in $\nu$ is taken with $k$ fixed.
For the rest of the proof, a subscript $k$ denotes this
diagonal sequence of actual Jensen points.

\emph{Step 5: the outer test bounds the exact pole matrix.}
At each diagonal point, Lemma~\ref{jet:transfer} gives the
actual lower Taylor Hessian
$H_{o,k}=m_{1,k}L_k\ge0$ at the selected endpoint. Use
Lemma~\ref{jet:outer-test} to undo the modification and call
the resulting form $V_{1,k}$. The positive outer curvature
term and the uniform divided-action baseline give
\begin{equation}\label{jet:bootstrap}
 V_{1,k}\ge m_{1,k}L_k+\delta\Id,\qquad
 \det V_{1,k}\le M.
\end{equation}
Every eigenvalue of $V_{1,k}$ is at least $\delta$, so its
largest eigenvalue is at most $M\delta^{1-n}$. Since
$m_{1,k}\ge m_*/2$, this bounds $L_k$ from above.
The comparison $H_k\le L_k+o(1)\Id$ and
Lemma~\ref{jet:semibound} now bound $H_k$ from both sides.
There has still been no use of a center density equation.

The center extension gives uniformly nonsingular
fixed-scale prefix-log coordinate maps, as in the proof of
Lemma~\ref{jet:semibound}. Their derivatives and second
derivatives are bounded; the gradients of $G_k$ are bounded
by the Lipschitz bound of the datum. Thus the coordinate
change in \eqref{jet:H-definition} converts the bound on
$H_k$ into
\begin{equation}\label{jet:G-bound}
 \|\Hess G_k(z_k)\|\le C_{\mathrm{fixed}}.
\end{equation}

\emph{Step 6: the chosen centers are good original endpoints.}
Write $\tau_k=1-t_k$, $y_k=Y_k(z_k)$, and
$K_k(z,y)=c(z,y)/\tau_k$. The envelope identity
$\grad G_k=(K_k)_z(z,Y_k(z))$ and differentiability of $Y_k$
give
\[
 \Hess G_k=(K_k)_{zz}+(K_k)_{zy}DY_k.
\]
In parallel orthonormal frames, the distance between $z_k$
and $y_k$ is $O(\tau_k)$, and the smooth short-cost expansions
are
\[
 (K_k)_{zz}=\tau_k^{-1}\Id+O(\tau_k),\quad
 (K_k)_{yy}=\tau_k^{-1}\Id+O(\tau_k),\quad
 (K_k)_{zy}=-\tau_k^{-1}\Pi+O(\tau_k),
\]
where $\Pi$ is the appropriate parallel isometry. Solving
the preceding identity, using \eqref{jet:G-bound}, gives
\begin{equation}\label{jet:Y-identity}
 DY_k=\Pi^{-1}\{\Id-\tau_k\Hess G_k\}+O(\tau_k^2).
\end{equation}
It is nonsingular for large $k$. The Jensen points were
chosen off the set \eqref{jet:null-preimage}; hence
$y_k\notin N$. This is the promised justification for
using the original Alexandrov and Jacobian statements at
the centers.

In the following formulas the scalar modifications are evaluated
with $b=b_k$. At these points $w_c=\varphi_c\circ v$ has an Alexandrov
Hessian. The minimizing property supplies a proximal subgradient
\[
 \xi(z)=-(K_k)_y(z,Y_k(z))\in\partial w_c(Y_k(z))
\]
at every nearby point, even if $w_c$ is not differentiable there.
Apply Lemma~\ref{jet:subgradient-differential} at the Alexandrov point
$y_k=Y_k(z_k)$. Since $Y_k$ is Lipschitz and differentiable at
$z_k$, it gives, in normal coordinates at $z_k$ and $y_k$,
\[
 \xi(z_k+h)=\grad w_c(y_k)
       +\Hess w_c(y_k)DY_k(z_k)h+o(|h|).
\]
Expanding the smooth function $(K_k)_y$ in its two variables
therefore gives
\[
 \{\Hess w_c(y_k)+(K_k)_{yy}\}DY_k+(K_k)_{yz}=0.
\]
Thus stationary differentiation uses the subgradient expansion even
when neighboring endpoints are nonsmooth.
Solving it with \eqref{jet:Y-identity} gives, in parallel
frames,
\begin{equation}\label{jet:undo-short}
 \Hess w_c(y_k)=\Hess G_k(z_k)+O(\tau_k).
\end{equation}
In particular these endpoint Hessians are uniformly bounded.
The error statement follows directly from the displayed
short-cost expansions and the bounded inverse in
\eqref{jet:Y-identity}; it does not assume a bound for
$\Hess w_c$ in advance.

\emph{Step 7: insert the true center form and take matrix limits.}
Set
\[
 r_k=\tau_k^{-1}\log_{y_k}z_k,\quad
 l_k=\varphi_c'(v(y_k)),\quad
 \widehat x_k=\exp_{y_k}r_k,\quad
 \widehat p_k=\log_{\widehat x_k}y_k.
\]
The true center endpoint is
$\zeta_k=\exp_{y_k}(r_k/l_k)$, and it is before cut because
$y_k\notin N$. Identify $T_{x_k}M$ with
$T_{\widehat x_k}M$ by parallel transport over their short
joining segment, and let $B_k:T_{x_k}M\to T_{y_k}M$ be the
exponential differential at $\widehat p_k$ after that
identification. Define
\begin{equation}\label{jet:V-sequence}
 V_k=l_k B_k^*
       \{\Hess v(y_k)+A_{y_k}(r_k/l_k)\}B_k>0.
\end{equation}
The exact chain rule for the scalar modification gives
\begin{align}
 B_k^*\{\Hess w_c(y_k)+A_{y_k}(r_k)\}B_k
 ={}& V_k
 +B_k^*\{A_{y_k}(r_k)-l_kA_{y_k}(r_k/l_k)\}B_k
 \nonumber\\
 &+\frac{\varphi_c''(v(y_k))}{l_k^2}
      \widehat p_k\widehat p_k^* .
 \label{jet:center-exact}
\end{align}
All vectors on the last line are understood after the
same isometric identification.
Lemma~\ref{spec:pullback} bounds the middle term below by
$c(1-l_k)\Id$. Equation~\eqref{jet:undo-short} and the
vanishing pole mismatch show that the left side differs
from $H_k$ by $o(1)\Id$. The verification is the same smooth
coordinate comparison as in
Lemma~\ref{jet:semibound}, now using the bounded
unregularized Hessians in place of the smooth support
Hessians. This proves
\[
 H_k\ge V_k+c\frac{b_k}{D}\Id
       -C\frac{b_k}{D^2}p_kp_k^*-o(1)\Id.
\]
It also bounds $V_k$ from above, because $V_k>0$ and $H_k$
is bounded.

Taking the determinant in \eqref{jet:V-sequence} and using
\eqref{jet:ae-jacobian} gives the exact factorization
\begin{equation}\label{jet:V-jacobian}
 \det V_k
 =l_k^n\sigma_{\widehat x_k}(\widehat p_k)^2
       \frac{|\det D(\exp_{\cdot}\grad v)(y_k)|}
            {\sigma_{y_k}(r_k/l_k)}
 \ge c\,\frac{\sigma_{\widehat x_k}(\widehat p_k)^2}
                 {\sigma_{y_k}(r_k/l_k)}.
\end{equation}
The squared factor is the determinant of the Hessian
pullback. All scalar derivatives lie between $1/2$ and $2$,
so the constants in this density inequality are independent
of $K$.

There is no assertion that the limiting endpoint is an
Alexandrov point of $v$. We take limits of matrices at the good
endpoints just selected. The center modified rays have a uniform
extension at the fixed scales: $1-l_k\ge c b_-/D>0$, and the
ray through $\widehat x_k$ continues to the true endpoint through
duration $1/l_k$. Compactness of this family of proper minimizing
subsegments therefore gives
\[
 \sigma_{\widehat x_k}(\widehat p_k)\ge c_{\mathrm{fixed}}>0.
\]
Since $0<V_k\le C_{\mathrm{fixed}}\Id$, equation
\eqref{jet:V-jacobian} now implies
\[
 \sigma_{y_k}(r_k/l_k)
 \ge \frac{c\,\sigma_{\widehat x_k}(\widehat p_k)^2}
             {\det V_k}
 \ge c'_{\mathrm{fixed}}>0.
\]
Continuity of the exponential differential excludes a conjugate
limit of the true center rays. An ordinary cut limit may remain;
the specified nonsingular branch suffices.

Finally take a subsequence so that $b_k,H_k,L_k,V_k,V_{1,k}$
and all configuration parameters converge, in a smooth
orthonormal trivialization. Discard any additional zero
limiting weights. The selected weight remains at least $m_*$,
and its outward displacement retains the positive fixed-scale
lower bound from Step~3. The barycenter identity then gives
another surviving vertex, so the surviving difference span
is still the same nonzero line. The selected modified and true
outer branches were already uniformly nonsingular in
Step~4; the center modified branch has its fixed-scale
extension. Equation~\eqref{jet:V-jacobian} gives a positive
lower determinant in the limit, so $V>0$. Equations
\eqref{jet:L-properties}, \eqref{jet:Jensen-error}, and
\eqref{jet:center-exact} give
\eqref{jet:limiting-center}.

At each selected outer endpoint in the diagonal sequence, formula
\eqref{jet:outer-form}, with $H_{o,k}=m_{1,k}L_k$, has a positive
divided-action term. Its scalar curvature coefficient satisfies
\[
 -\frac{\varphi_o''(v(y_{1,k}))}{l_{1,k}^2}
 =-\frac{b_kB_o''((v(y_{1,k})-a)/D)}{D^2l_{1,k}^2}
 \ge cK b_k/D^2.
 \]
Passing this inequality and \eqref{jet:outer-det} along
$V_{1,k}\to V_i$ gives \eqref{jet:limiting-outer} and the other
inequality in \eqref{jet:limiting-determinants}. No lower Taylor
jet is asserted at the limiting endpoint. The weights, directions,
and section-level estimates pass through the same chosen
subsequence and give
\eqref{jet:limiting-geometry}--\eqref{jet:limiting-outward}.
Their constants and the cap were fixed before $K$.
Only $\delta,M$, the rate of the prefix limit, and the
Jensen radii depended on the fixed outer template.
\end{proof}

\subsection{The determinant gain}

\begin{lemma}[Oblique projection and the missing direction]
\label{jet:matrix}
Suppose the matrices and vectors satisfy
\eqref{jet:limiting-geometry}--\eqref{jet:limiting-outer}.
Then
\begin{equation}\label{jet:matrix-gain}
 \det V_i\ge cK\det V,
\end{equation}
where $c>0$ depends on the fixed center-template constants,
the dimension, and $m_*$, and is independent of $K$.
\end{lemma}

\begin{proof}
Put $\beta=Cb/D^2$, increasing $C$ if necessary, and
$\gamma=cKb/D^2$. Define
\[
 P=\Id-e p^*/q.
\]
This is the projection onto $p^\perp$ along $e$.
Since $LE=0$, $P^*LP=L$. On $p^\perp$ the center inequality
gives $H\ge V$, and $L\ge H$. Thus
\begin{equation}\label{jet:projected-matrix}
 V_i\ge m_iP^*VP+\gamma p_ip_i^*.
\end{equation}
Also $H(e,e)\le L(e,e)=0$ implies
$V(e,e)\le\beta q^2$.

Let $Q$ have orthonormal columns spanning $p^\perp$, and
use the nonsingular basis matrix $[Q,e]$. In this basis write
\[
 [Q,e]^*V[Q,e]
 =\begin{pmatrix} A&z\\z^*&h\end{pmatrix},\qquad
 S=h-z^*A^{-1}z>0.
\]
Then $S\le h=V(e,e)\le\beta q^2$.
The determinant of the right side of
\eqref{jet:projected-matrix}, in the same basis, is
$m_i^{n-1}\det A\,\gamma(p_i\cdot e)^2$.
Taking the ratio cancels the determinant of the
nonorthogonal change of basis and gives
\[
 \frac{\det V_i}{\det V}
 \ge m_i^{n-1}\frac{\gamma(p_i\cdot e)^2}{S}
 \ge m_i^{n-1}\frac{\gamma}{\beta}
        \frac{(p_i\cdot e)^2}{q^2}.
\]
Since $p_i=p+d_ie$ with $d_i>0$ and $q>0$, we have
$p_i\cdot e\ge q$. Now use $m_i\ge m_*$ and
$\gamma/\beta\ge cK$.
\end{proof}

\begin{proposition}[Contradiction to a non-power oscillation]
\label{jet:contradiction}
The section-oscillation function of Lemma~\ref{sc:plateau}
has a positive power upper bound at zero.
\end{proposition}

\begin{proof}
Suppose otherwise, and take the sequence of scales supplied
by that Lemma. Fix all center and weight constants, then
choose $K$ and its outer template, and then take sufficiently
small scales to apply Proposition~\ref{jet:determinants}.
Lemma~\ref{jet:matrix} and
\eqref{jet:limiting-determinants} imply
\[
 cK\,\frac{\sigma_x(p)^2}{\sigma_y(r/l)}
 \le \det V_i
 \le C\,\frac{\sigma_x(p_i)^2}{\sigma_{y_i}(r_i/l_i)}.
\]
The radial comparison of Corollary~\ref{spec:radial-determinants} gives
\[
 \sigma_y(r/l)\le C\sigma_x(p),\qquad
 \sigma_{y_i}(r_i/l_i)\ge c\sigma_x(p_i).
\]
For the selected outward endpoint,
\eqref{jet:limiting-geometry}--\eqref{jet:limiting-outward} give
$m_i\ge m_*$ and $q^2\ge cD$. Its norm increment satisfies
$|p_i|^2-|p|^2\le3D$ by Lemma~\ref{sc:outward}.
Corollary~\ref{spec:outward} therefore applies with
$\mu=m_*$, $\kappa=c$, and $L_0=3$, and gives
$\sigma_x(p)\ge c'\sigma_x(p_i)$. These parameters were fixed
before $K$. Hence
\[
 cK\sigma_x(p)\le C\sigma_x(p_i)\le C'\sigma_x(p).
\]
The center modified branch is nonconjugate, so
$\sigma_x(p)>0$. This bounds $K$ by a constant fixed before
it was chosen. Choosing $K$ larger gives a contradiction.
\end{proof}
==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/jets.tex ====

==== BEGIN Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/scales.tex ====
\section{Sections, scales, and a quantitative separation}
\label{sc:section}

Throughout this section the manifold and the density bounds are fixed.
We use the negative-cost convention
\[
 f^c(x)=\max_{y\in M}\{-c(x,y)-f(y)\}.
\]
Thus a cost-dual pair satisfies $u=v^c$, $v=u^c$, and
$u(x)+c(x,y)+v(y)\geq0$.
Let $\mathcal C$ denote the class of these pairs for all admissible density
pairs, with any fixed additive normalization.
All members of $\mathcal C$ have a common Lipschitz bound, and their
oscillations are uniformly bounded.
None of the arguments below uses derivatives of the densities.

\subsection{A scalar quantity controlling both transport directions}

For $(u,v)\in\mathcal C$ and $r\geq0$, put
\[
 S_r(x;u,v)=\{y\in M:u(x)+c(x,y)+v(y)\leq r\},
 \qquad
 F(r)=\sup_{(u,v)\in\mathcal C,\ x\in M}
       \osc_{S_r(x;u,v)}v.
\]
The sets $S_r$ are nonempty and compact.  The function $F$ is bounded
and nondecreasing.

\begin{lemma}[The power reduction]\label{sc:power}
If there are $\beta>0$, $C_0<\infty$, and $r_0>0$ such that
$F(r)\leq C_0r^\beta$ for $0<r<r_0$, then all the optimal maps in the
fixed density class have homeomorphic representatives with a common
H\"older estimate in both directions.
One may take any exponent no larger than
$\frac12\min\{\beta,1\}$.
\end{lemma}

\begin{proof}
Write
\[
 h_x(p)=u(x)+\frac12|p|^2+v(\exp_xp),\qquad p\in G_x.
\]
By Corollary~\ref{geo:sections}, the lifted section
$\{p\in G_x:h_x(p)\leq r\}$ is convex.
If $p,q$ belong to it and $m=(p+q)/2$, the squared-norm identity gives
\begin{align}
 \frac18|p-q|^2
 &=v(\exp_xm)-\frac{v(\exp_xp)+v(\exp_xq)}2
   +\frac{h_x(p)+h_x(q)}2-h_x(m) \notag\\
 &\leq F(r)+r.                                      \label{sc:diameter}
\end{align}
The exponential map has a common Lipschitz bound on the compact bundle
of minimizing velocities.  Hence the target diameter of a section is
at most $C\sqrt{F(r)+r}$.

Let $(x,y)$ and $(x',y')$ be contacts of the same dual pair.  The common
Lipschitz bounds for $u$ and $c$ imply
\[
 0\leq u(x)+c(x,y')+v(y')\leq C d(x,x').
\]
Thus $y,y'$ belong to a section at $x$ of that height.  For sufficiently
small $d(x,x')$, \eqref{sc:diameter} gives
\[
 d(y,y')\leq C d(x,x')^{\frac12\min\{\beta,1\}}.
\]
In particular, letting the section height tend to zero shows that
every pole has exactly one contact endpoint.  The bound proves that
this endpoint depends continuously on the pole. The class also contains
the reversed pair $(v,u)$ after renormalization, which does not affect
its section oscillations. The same reasoning therefore gives a unique
continuous contact in the reverse direction. The dual identities give a contact above every point
of either copy of $M$, and the two resulting maps are inverse.
They agree almost everywhere with the optimal maps by optimality and
uniqueness.  Finally, compactness absorbs the remaining, larger
distances into the same uniform constant.
\end{proof}

We shall contradict failure of a power bound for $F$.
The following elementary selection makes the required quantifiers
explicit.

\begin{lemma}[Logarithmic plateaus]\label{sc:plateau}
Suppose that $F$ has no power upper bound near zero.
There are $r_j<R_j\to0$, pairs $(u_j,v_j)\in\mathcal C$, poles $x_j$,
and numbers $a_j,D_j$ with $D_j>0$ such that
\begin{gather}
 \min_{S_{r_j}(x_j;u_j,v_j)}v_j=a_j,\qquad
 \max_{S_{r_j}(x_j;u_j,v_j)}v_j=a_j+D_j,               \label{sc:endlevels}\\
 \frac{r_j}{R_j}\longrightarrow0,\qquad
 F(R_j)\leq(1+\delta_j)D_j,\qquad \delta_j\longrightarrow0,\qquad
 \frac{|\log D_j|}{|\log R_j|}\longrightarrow0.       \label{sc:plateau-data}
\end{gather}
Consequently $R_j/D_j^k\to0$ for every fixed $k>0$.
\end{lemma}

\begin{proof}
If $F$ vanished at a positive scale, monotonicity would already give a
power bound below that scale.  Thus $F(r)>0$.
Fix $C_F\geq\max\{1,\sup_r F(r)\}$ and set
\[
 f(s)=\log\frac{C_F}{F(e^{-s})}.
\]
This function is nonnegative and nondecreasing.  Failure of every
power bound implies $\liminf_{s\to\infty}f(s)/s=0$; otherwise
$f(s)\geq\beta s$ eventually for some $\beta>0$.
Choose $N_j\to\infty$ with $f(N_j)=o(N_j)$, and divide
$[N_j/2,N_j]$ into
\[
 m_j=\left\lceil\sqrt{N_j\max\{1,f(N_j)\}}\right\rceil
\]
equal intervals.  Their length $N_j/(2m_j)$ tends to infinity.
Since their total $f$-increment is at most $f(N_j)$, some interval
$[A_j,B_j]$ has increment at most $f(N_j)/m_j=o(1)$.
Put $R_j=e^{-A_j}$ and $r_j=e^{-B_j}$.  Then
\[
 \log(R_j/r_j)=B_j-A_j\longrightarrow\infty,\qquad
 \frac{F(R_j)}{F(r_j)}=e^{f(B_j)-f(A_j)}\longrightarrow1.
\]
Choose an actual section with oscillation
$D_j\geq(1-1/j)F(r_j)$, discarding the first few indices.
Its extrema are attained, which gives \eqref{sc:endlevels}.
Moreover $A_j\geq N_j/2$ and $f(B_j)\leq f(N_j)=o(N_j)$,
so $|\log D_j|=o(A_j)$.
This proves \eqref{sc:plateau-data}.  Finally
\[
 \log(R_j/D_j^k)=-A_j-k\log D_j=-A_j+o(A_j)\longrightarrow-\infty.
\]
\end{proof}

We suppress the index $j$ from now on.  All asymptotic statements refer
to the sequence in Lemma~\ref{sc:plateau}, after the templates below
have been fixed.  It is not necessary that $D$ tend to zero.
Set
\begin{equation}
 b_-=r(R/r)^{1/3},\qquad b_+=r(R/r)^{2/3}.
 \label{sc:binterval}
\end{equation}
Then $r\ll b_-\ll b_+\ll R$.  Uniformly for $b\in[b_-,b_+]$ and
$0\leq\tau\leq8b/D$,
\begin{equation}
 \frac rb\to0,\quad \frac bR\to0,\quad
 \frac b{D^k}\to0\ (k>0\text{ fixed}),\quad
 \frac\tau D\to0,\quad \frac{\tau^2}{b}\to0.
 \label{sc:smallness}
\end{equation}
For example, $\tau^2/b\leq64b/D^2$.
These estimates, rather than an additional restriction on $R/r$,
will control the Taylor errors.

\subsection{Templates and the order of choices}

At the normalized level $s=(v-a)/D$, we will use the two
modifications $v+bB_c(s)$ and $v+bB_o(s)$.
The first is the center datum; we compare it with the double
cost transform of the second. A positive excess in this
comparison will produce an interior maximum. We first fix
the center profile and a cap on the outer values, then choose
the outer curvature parameter $K$ and its profile, and only
afterward take the gap scales small. The construction below
makes that order explicit.

We use the fixed numerical choices
\begin{equation}
 \eps_0=2^{-10},\qquad c_0=2^{-10},\qquad c_1=c_0/4.
 \label{sc:numerical}
\end{equation}
The constants are deliberately smaller than needed.

The value constraints below serve the envelope comparison: the center
profile stays near one on the lower half of the selected oscillation,
whereas the outer profile vanishes at level one and places a barrier
on intermediate levels. We also prescribe strong outer concavity
where the outer value is bounded by a fixed cap. That derivative
condition will enter the endpoint lower tests in
Section~\ref{sec:jets}; the height bound below keeps the constants in
the switch and excess bounds independent of $K$.

\begin{lemma}[Smooth templates]\label{sc:templates}
There is a fixed smooth function $B_c:\R\to(-1,1)$ with bounded
derivatives such that
\[
 B_c'<0\text{ on }\R,\qquad
 B_c(s)\geq1-\eps_0\quad(s\leq1/2),\qquad
 B_c(s)<0\quad(s\geq3/4).
\]
Given a cap $M_0\geq2$ and $K\geq1$, there are
$0<\eta\leq1/128$ and a smooth bounded function $B_o\geq0$, with
bounded derivatives, satisfying
\begin{align}
 &B_o(s)\geq1\quad(s\leq1-\eta),\qquad
 B_o(1)=0,\qquad 1\leq B_o(0)\leq1+\eps_0,       \label{sc:outer-values}\\
 &B_o(s)\geq B_o(0)+1
       \quad(1/16\leq s\leq1-\eta),            \label{sc:barrier}\\
 &B_o'(s)>0,\quad B_o''(s)\leq-K
       \quad\text{if }-2\eta\leq s\leq1-\eta
                      \text{ and }B_o(s)\leq M_0.     \label{sc:curvature}
\end{align}
One can arrange $\|B_o\|_\infty\leq2M_0+8$, independently of $K$,
and make $\eta$ arbitrarily smaller if desired.
\end{lemma}

\begin{proof}
Choose $\gamma$ so large that $2/(1+e^{\gamma/8})\leq\eps_0$, and set
\[
 B_c(s)=1-\frac{2}{1+e^{-\gamma(s-5/8)}}.
\]
This has all the asserted properties.  In particular, $|B_c'|$ has a
positive minimum on every fixed compact interval.

For the outer function let
\[
 \beta(z)=\begin{cases}e^{-1/z},&z>0,\\0,&z\leq0,\end{cases}
 \qquad
 \chi(z)=\frac{\beta(z)}{\beta(z)+\beta(1-z)}.
\]
Thus $\chi$ is a smooth step, equal to zero on $(-\infty,0]$ and to
one on $[1,\infty)$.  Put
\[
 \kappa=K+1,\qquad A=M_0+3,\qquad
 L=128A\kappa,\qquad h_0=\frac1{128\kappa},\qquad
 0<\eta\leq\min\left\{\frac1{128},\frac{\eps_0}{24L}\right\}.
\]
Define
\begin{align*}
 w(s)&=(L-\kappa s)
       \chi\!\left(\frac{s+3\eta}{\eta}\right)
       \left[1-\chi\!\left(\frac{s-h_0}{h_0}\right)\right],\\
 J(s)&=1+\frac{\eps_0}2+\int_0^s w(q)\,dq,\\
 B_o(s)&=J(s)
       \left[1-\chi\!\left(\frac{s-(1-\eta)}{\eta}\right)\right].
\end{align*}
The function $w$ is nonnegative and supported in $[-3\eta,2h_0]$.
Also
\[
 \int_{-3\eta}^0w\leq3\eta(L+3\kappa\eta)
       \leq6\eta L\leq\eps_0/4,\qquad
 \int_0^{2h_0}w\leq2Lh_0=2A.
\]
It follows that
\[
 1+\eps_0/4\leq J(s)\leq1+\eps_0/2+2A.
\]
On $[-2\eta,h_0]$, one has $J'=L-\kappa s>0$ and
$J''=-\kappa$.  Moreover
\[
 J(h_0)-J(0)=A-\frac1{32768\kappa}>M_0+1.
\]
Since $J$ is nondecreasing, the condition $B_o(s)\leq M_0$ on
$[-2\eta,1-\eta]$ forces $s<h_0$; here $B_o=J$, proving
\eqref{sc:curvature}.  The function $J$ is constant beyond
$2h_0\leq1/64$, so \eqref{sc:barrier} also follows.
The final factor is identically one up to $1-\eta$, and decreases to
zero only in $(1-\eta,1)$.  This proves the remaining assertions,
including smoothness at all joins and the uniform height bound.
\end{proof}

The cap $M_0$ will be fixed independently of $K$ by
Lemma~\ref{sc:outward}; the choice
$M_0=24(n+1)/c_1$ suffices.  The center template is already fixed.
After fixing $M_0$, choose $K$ and the outer template, and only then
take the scales small.  Besides \eqref{sc:smallness}, this order gives
\begin{equation}
 \delta_j=o(\eta),\qquad
 \frac bD\max\{\|B_o'\|_\infty,\|B_c'\|_\infty\}\to0,\qquad
 \frac b{D^2}\max\{\|B_o''\|_\infty,\|B_c''\|_\infty\}\to0.
 \label{sc:template-smallness}
\end{equation}
In particular, both scalar modifications used below are increasing.
Constants in intermediate local estimates may depend on a fixed
template, but the constants $c_0,c_1$ and the cap do not depend on $K$.

\subsection{The first switch}

Fix a selected section with extreme endpoints $y_0,y_1$, so that
$v(y_0)=a$ and $v(y_1)=a+D$.  Choose minimizing logs
$p_0,p_1\in G_{x_0}$ to these endpoints, and put $P=|p_0|^2$.
For every target write
\[
 s_y=\frac{v(y)-a}{D},\qquad
 w_{o,b}(y)=v(y)+bB_o(s_y),\qquad
 w_{c,b}(y)=v(y)+bB_c(s_y),\qquad
 g_b=w_{o,b}^c.
\]
Let $H_o=\|B_o\|_\infty$.  Since $B_o\geq0$,
\begin{equation}
 u-bH_o\leq g_b\leq u.
 \label{sc:g-bounds}
\end{equation}

We will move the pole toward $y_0$. Initially the outer modification
lowers the level-zero mountain by about $b$, while leaving $y_1$
unpenalized. As long as mountains with $s_y\geq1-\eta$ dominate, their
minimizing logs are shorter than the log to $y_0$, so they grow more
slowly along this motion. The barrier excludes intermediate-level
winners, forcing a switch with both a low and a high active endpoint.
Their two logs will then provide a supported intermediate log with a
prescribed squared norm; the next subsection converts this construction
into a positive excess.

We record the uniform distance estimate used in the proof.
Splitting a minimizing geodesic at its midpoint gives a smooth upper
support for the squared-distance cost with a uniformly bounded
Hessian.  Consequently, for any minimizing log $q$ from $x$ to $y$,
\begin{equation}
 c(\exp_x z,y)\leq c(x,y)-\langle q,z\rangle+C_g|z|^2
 \quad\text{for }|z|\text{ sufficiently small},
 \label{sc:cost-upper}
\end{equation}
with one geometric constant and radius.  Along a smooth curve, the
one-sided derivative of $-c(\cdot,y)$ is the maximum of the pairings
with its minimizing logs.  In particular, whenever this restriction
is differentiable,
\begin{equation}
 \frac d{d\tau}[-c(x_\tau,y)]
       \leq d(x_\tau,y)|\dot x_\tau|.
 \label{sc:advance}
\end{equation}
These assertions hold also at a cut endpoint.  The same upper
derivative bound applies to an active mountain of a maximum: at a
time when the maximum is differentiable, a semiconvex active mountain
has the same derivative.  Indeed its one-sided derivatives straddle
that derivative in the opposite order to their semiconvex ordering,
and hence coincide.  The restrictions in question are Lipschitz, so
these differential inequalities can be integrated.

\begin{lemma}[First switch with absolute level control]\label{sc:switch}
For all sufficiently small selected scales, uniformly for
$b\in[b_-,b_+]$, there is
\begin{equation}
 \frac b{32D}\leq\tau_b\leq\frac{8b}D
 \label{sc:switch-time}
\end{equation}
such that, at $x_b=\exp_{x_0}(\tau_b p_0)$, the function $g_b$ has
active endpoints $y_-$ and $y_+$ with
\[
 -2\eta\leq s_{y_-}<1/16,\qquad s_{y_+}\geq1-\eta.
\]
The original endpoints $y_0,y_1$ and every active endpoint at this
pole lie in one original section $S_{C b}(x_b;u,v)$, where $C$ depends
only on the geometry and $H_o$.  In particular every active level is
in $[-\delta_j,1+\delta_j]$.
For every minimizing log $q_i$ to an active endpoint,
\begin{equation}
 |q_i|^2=P-2D s_{y_i}+O_{H_o}(\tau_b+b)
        =P-2D s_{y_i}+o(D).
 \label{sc:switch-norms}
\end{equation}
The constants in \eqref{sc:switch-time} are independent of $K$ and
of the height and derivatives of the outer template.
\end{lemma}

\begin{proof}
Set $x_\tau=\exp_{x_0}(\tau p_0)$ for
$0\leq\tau\leq T:=8b/D$, and let $\Pi_\tau$ be parallel transport
along this geodesic.  For small scales $T<1/2$.  The left cost is exact:
\[
 f_y(\tau):=-c(x_\tau,y)-v(y),\qquad
 f_0(\tau)=-\frac12(1-\tau)^2P-a,\qquad
 q_0(\tau)=(1-\tau)\Pi_\tau p_0.
\]
The initial gaps at $y_0,y_1$ lie in $[0,r]$, whence
\begin{equation}
 P-|p_1|^2=2D+O(r),\qquad
 |p_0-p_1|^2\leq8(D+r),\qquad
 P-\langle p_0,p_1\rangle\leq6D
 \label{sc:initial-logs}
\end{equation}
for small scales; the second inequality is \eqref{sc:diameter}
applied using the actual oscillation $D$ of this section.
In particular $P\geq D$.

First exclude the middle levels.  If $1/16\leq s_y\leq1-\eta$, put
\[
 h_y(\tau)=f_y(\tau)-bB_o(s_y)
                -f_0(\tau)+bB_o(0).
\]
Initially $h_y(0)\leq r-b$.  Wherever $h_y\geq-b/2$, the identity
relating values and squared distances gives, for every minimizing
log $q_y$ at the current pole,
\[
 |q_y|^2-|q_0|^2
   =-2Ds_y-2b(B_o(s_y)-B_o(0))-2h_y
   \leq-D/8-b.
\]
Equations~\eqref{sc:advance} and $\dot x_\tau=\Pi_\tau p_0$ imply
$h_y'\leq-D/16$ at almost every such time.
The positive part of $h_y+b/2$ is therefore nonincreasing and starts
at zero.  Thus $h_y\leq-b/2$ throughout $[0,T]$.
No middle-level mountain can be active.

Let $G_L$ and $G_R$ be the maxima of $f_y-bB_o(s_y)$ over the compact
sets $s_y\leq1/16$ and $s_y\geq1-\eta$, respectively.
Both are continuous, and the preceding paragraph shows
$g_b(x_\tau)=\max\{G_L(\tau),G_R(\tau)\}$.
Initially $G_L(0)\leq u(x_0)-b$ and
$G_R(0)\geq f_1(0)\geq u(x_0)-r$.
While $G_R>G_L$, every active endpoint is right and comparison with
the left modified mountain gives
\[
 |q_y|^2\leq|q_0|^2-2D(1-\eta)+2bB_o(0)
             \leq|q_0|^2-D.
\]
By rationalizing the difference of the square roots in
\eqref{sc:advance},
\[
 \frac d{d\tau}\{g_b(x_\tau)-f_0(\tau)\}\leq-D/2
 \quad\text{almost everywhere while }G_R>G_L.
\]
This difference starts at most $r$ and is always at least
$-bB_o(0)\geq-2b$.  It cannot obey the displayed inequality throughout
$[0,T]$.  There is therefore a first equality time $\tau_b\leq T$.
At that time both maxima are attained, and the middle-level exclusion
gives $s_{y_-}<1/16$.
Throughout the preceding interval, including its endpoint,
\begin{equation}
 -bB_o(0)\leq g_b(x_\tau)-f_0(\tau)\leq r.
 \label{sc:left-anchor}
\end{equation}

We next anchor the absolute levels.  By \eqref{sc:g-bounds} and
\eqref{sc:left-anchor},
$u(x_\tau)-f_0(\tau)\leq bH_o+r$.
The initial-cost upper support and \eqref{sc:initial-logs} give
\begin{equation}
 f_1(\tau)\geq
 f_0(0)-r+(P-6D)\tau-C_g\tau^2=:\mathcal L(\tau).
 \label{sc:benchmark}
\end{equation}
Since $f_0(\tau)=f_0(0)+P\tau-P\tau^2/2$, it follows that
$f_0-f_1\leq r+6D\tau+C\tau^2=O(b)$ on this interval.
Thus both $y_0,y_1$ lie in an original section of height $Cb$.
For any active endpoint $y$,
\[
 u(x_\tau)+c(x_\tau,y)+v(y)
      =u(x_\tau)-g_b(x_\tau)-bB_o(s_y)\leq bH_o.
\]
All these endpoints belong to the same section of height $Cb<R$.
Its oscillation is at most $(1+\delta_j)D$ and it contains levels
$a,a+D$.  Hence every active level lies in
$[-\delta_j,1+\delta_j]$, proving in particular the required lower
bound for $s_{y_-}$.

To prove the lower time estimate, keep the winning left endpoint
$y_-$ fixed, and put
\[
 k(\tau)=f_{y_-}(\tau)-bB_o(s_{y_-})-\mathcal L(\tau).
\]
One has $k(0)\leq2r-b<-b/2$, while
$k(\tau_b)\geq0$, since $g_b(x_{\tau_b})\geq f_1(\tau_b)$.
Let $\theta$ be the last time before $\tau_b$ at which $k=-b/2$.
Then $k\geq-b/2$ on $[\theta,\tau_b]$.
On this interval, using $s_{y_-}\geq-2\eta$ and $B_o\geq0$, the
inequality defining this last strip gives
\begin{align*}
 |q_{y_-}|^2-|q_0|^2
 &\leq4\eta D+2r+12D\tau+2C_g\tau^2+b\\
 &\leq D
\end{align*}
for all sufficiently small scales.  This uses only
\eqref{sc:smallness}; in particular $\tau/D\to0$.
Rationalizing once more gives
$f_{y_-}'\leq(1-\tau)P+2D$ almost everywhere in the strip.
Since $\mathcal L'=P-6D-2C_g\tau$, we have $k'\leq10D$ there
for small scales.  Therefore
\[
 b/2\leq k(\tau_b)-k(\theta)\leq10D(\tau_b-\theta),
\]
which implies the stated weaker lower bound $\tau_b\geq b/(32D)$.
No upper bound for the template or its derivatives occurs in this
last estimate.

Finally, if $y_i$ is active at the switch, the exact equality of its
modified value with $g_b$ gives
\[
 |q_i|^2=(1-\tau_b)^2P-2Ds_{y_i}
         -2bB_o(s_{y_i})-2\{g_b(x_b)-f_0(\tau_b)\}.
\]
Equation~\eqref{sc:left-anchor} and \eqref{sc:smallness} prove
\eqref{sc:switch-norms}.
\end{proof}

\subsection{Positive excess and an interior maximum}

Write $\widehat w_b=g_b^c=w_{o,b}^{cc}$ and
\[
 E_b(y)=w_{c,b}(y)-\widehat w_b(y),\qquad
 E(b)=\max_{y\in M}E_b(y).
\]
Order reversal of one cost transform and the double-transform
inequality give
\begin{equation}
 v\leq\widehat w_b\leq w_{o,b},\qquad
 E_b(y)\leq bB_c(s_y)\leq b.
 \label{sc:excess-upper}
\end{equation}

\begin{lemma}[Uniform positive excess]\label{sc:excess}
For all sufficiently small selected scales and all
$b\in[b_-,b_+]$,
\begin{equation}
 c_0b\leq E(b)\leq b,
 \label{sc:excess-bounds}
\end{equation}
with the constant in \eqref{sc:numerical}, independently of $K$.
\end{lemma}

\begin{proof}
Use the two active logs at the switch supplied by
Lemma~\ref{sc:switch}.  Equation~\eqref{sc:switch-norms} and their
level bounds show that their squared norms lie on opposite sides of
$P-D/2$.  By continuity along their segment, choose
$p\in\partial g_b(x_b)$ with
\[
 |p|^2=P-D/2.
\]
Theorem~\ref{geo:support} says that $p$ is minimizing and that
$y=\exp_{x_b}p$ is a contact endpoint for $g_b$.  Thus
\[
 E_b(y)=v(y)+bB_c(s_y)+c(x_b,y)+g_b(x_b).
\]
Minorizing $g_b$ by the left modified mountain gives the direct bound
\begin{equation}
 E_b(y)\geq D(s_y-1/4)+P\tau_b-P\tau_b^2/2
                  +b\{B_c(s_y)-B_o(0)\}.
 \label{sc:direct-excess}
\end{equation}
If $s_y>1/2$, the right side is at least $D/4-3b$, which is larger
than $c_0b$ for small scales.

It remains to consider $s_y\leq1/2$.  The original support inequality
at $x_0$ yields
\[
 v(y)-a\geq c(x_0,y_0)-c(x_0,y)-r.
\]
Apply \eqref{sc:cost-upper} at $x_b$ with backward displacement
$-\tau_b\Pi_{\tau_b}p_0$ to get
\[
 c(x_b,y)-c(x_0,y)
       \geq-\tau_b\langle p,\Pi_{\tau_b}p_0\rangle-C\tau_b^2.
\]
The corresponding cost difference for $y_0$ is exactly
$-P\tau_b+P\tau_b^2/2$.  Therefore
\begin{equation}
 \frac{E_b(y)}b\geq B_c(s_y)-B_o(0)
   +\bigl(P-\langle p,\Pi_{\tau_b}p_0\rangle\bigr)\frac{\tau_b}b
   -\frac rb-C\frac{\tau_b^2}b.
 \label{sc:geometric-excess}
\end{equation}
The norm choice gives
\[
 P-\langle p,\Pi_{\tau_b}p_0\rangle
   =\frac{P-|p|^2+|p-\Pi_{\tau_b}p_0|^2}{2}\geq D/4.
\]
The middle term in \eqref{sc:geometric-excess} is consequently at
least $1/128$.  The template difference is at least $-2\eps_0$,
and the last two terms tend to zero uniformly by
\eqref{sc:smallness}.  The numerical choices
\eqref{sc:numerical} give the asserted lower bound.
The upper bound is \eqref{sc:excess-upper}.
\end{proof}

To obtain an interior maximum, we now subtract a penalty from the
excess. A small linear term preserves its positive lower bound,
and a quadratic term makes the upper amplitude boundary unfavorable.
Define
\begin{equation}
 \pi(b)=\frac{c_0}4b+\frac{2b^2}{b_+}.
 \label{sc:penalty}
\end{equation}
In particular, $\pi'(b)\geq c_1$.

\begin{lemma}[Regularized interior maxima]\label{sc:maximum}
Fix a sufficiently small selected scale.  For every $t<1$ sufficiently
close to one, the continuous function
\[
 \Phi_t(b,z)=Q_tg_b(z)+Q_{1-t}w_{c,b}(z)-\pi(b),\qquad
 (b,z)\in[b_-,b_+]\times M,
\]
has a positive global maximum with $b$ in the interior of its interval.
Both spatial terms are $C^{1,1}$, and both minimizing points are unique.
If $x$ and $y$ are these minimizing points at a maximum, there is
$p\in\partial g_b(x)\cap G_x$ such that
\begin{equation}
 z=\exp_x(tp),\qquad y=\exp_xp,\qquad
 Q_tg_b(z)+Q_{1-t}w_{c,b}(z)=E_b(y)>0.
 \label{sc:aligned}
\end{equation}
Moreover $(x,y)$ is before cut.
\end{lemma}

\begin{proof}
At $b=b_+$, \eqref{sc:excess-bounds} makes $E(b)-\pi(b)$ negative.
At $b=b_-$, it is at most $b_-$.  At
$b_*=c_0b_+/8$, which belongs to the interval for small scales,
\[
 E(b_*)-\pi(b_*)\geq c_0^2b_+/16.
\]
Since $b_-/b_+\to0$, these inequalities give a strictly positive
interior maximum with a positive separation from both boundary values.

For a fixed scale and $b$ in its compact interval,
$Q_tg_b\to Q_1g_b=-\widehat w_b$ uniformly as $t\uparrow1$.
This follows directly from the uniform bound for
$|c/t-c|$.  The functions $w_{c,b}$ have a common Lipschitz constant
$L_*$ on that interval, and
\[
 w_{c,b}-\tfrac12L_*^2(1-t)\leq Q_{1-t}w_{c,b}\leq w_{c,b}.
\]
Thus $\Phi_t\to E_b-\pi$ uniformly in $(b,z)$, preserving the
positive interior maximum when $t$ is sufficiently close to one.

Proposition~\ref{geo:intermediate} gives the asserted regularity and unique
pole for $Q_tg_b$.  To check the other term, the scalar function
$q\mapsto q+bB_c((q-a)/D)$ has positive derivative, and its first two
derivatives are bounded at the fixed scales.  Composing a Lipschitz
semiconvex function with this function preserves semiconvexity: the
lower Taylor support for $v$, its Lipschitz bound, and the ordinary
second-order Taylor inequality for the scalar function give a common
quadratic lower support for $w_{c,b}$.
For completeness, a minimizer in $Q_\epsilon w_{c,b}(z)$ has
$d(z,y)\leq2L_*\epsilon$, by comparison with $y=z$.
For small $\epsilon$, all minimizers therefore lie in one convex
normal ball, where the Hessian of $c(z,\cdot)/\epsilon$ dominates
the negative semiconvexity bound.  Strict convexity gives uniqueness.
In one fixed coordinate chart, let $K_*$ be the semiconvexity constant
and let $\mu/\epsilon$ be a lower Hessian bound for the smooth cost
term.  The subgradient inequalities at the minimizers $y,y'$ for
nearby $z,z'$ and the mixed cost derivative bound give
\[
  (\mu/\epsilon-K_*)|y-y'|^2
       \leq (C/\epsilon)|z-z'|\,|y-y'|.
\]
For small $\epsilon$ this proves local Lipschitz dependence on $z$.
The envelope gradient formula then gives $C^{1,1}$ regularity.
This is uniform in $b$ on the fixed interval.

At a maximum, the two spatial gradients sum to zero.  Their envelope
formulas show that the geodesic from $x$ to $z$, continued for the
remaining time $1-t$ with the same velocity, reaches $y$.
The pole minimum gives $p\in\partial g_b(x)$.
By Theorem~\ref{geo:support}, this full vector is minimizing and its
endpoint is a support for $g_b$.  Splitting its energy at time $t$
proves \eqref{sc:aligned} exactly.

Finally put $l=1+bB_c'(s_y)/D\in(0,1)$.
The short final minimization gives its backward velocity in
$\partial w_{c,b}(y)=l\partial v(y)$; the equality follows from the
increasing scalar chain rule, which follows in both directions by
Taylor expansion and inversion of that scalar function.
Dividing by $l$ gives a minimizing velocity by
Theorem~\ref{geo:support}.  The backward velocity to $x$ is a strict
radial shortening of this vector, and is therefore before cut.
\end{proof}

\subsection{The parameter inequality for every representation}

\begin{lemma}[The parameter derivative]\label{sc:parameter}
At any maximum in Lemma~\ref{sc:maximum}, let
\[
 p=\sum_{i=1}^N m_i p_i,\qquad m_i>0,\qquad\sum_i m_i=1,
\]
be any finite representation by minimizing logs active for the displayed
representation $g_b=w_{o,b}^c$ at its minimizing pole $x$.
Thus $g_b(x)+c(x,\exp_xp_i)+w_{o,b}(\exp_xp_i)=0$.
Write $y_i=\exp_xp_i$ and
$s_i=(v(y_i)-a)/D$, and let $s=(v(y)-a)/D$ be the center level.
Then
\begin{equation}
 B_c(s)-\sum_i m_iB_o(s_i)\geq \pi'(b)\geq c_1.
 \label{sc:parameter-inequality}
\end{equation}
\end{lemma}

\begin{proof}
Fix the prefix time and the spatial maximum point $z$, and abbreviate
$A(b)=Q_tg_b(z)$ and $C(b)=Q_{1-t}w_{c,b}(z)$.
Decrease $b$ by $h>0$, and let $x_h$ be the new pole for $A(b-h)$.
Because $0\leq g_{b-h}-g_b\leq hH_o$, compactness and uniqueness of
the old pole imply $x_h\to x$.
The strict local quadratic growth in
Proposition~\ref{geo:intermediate} gives, in normal coordinates
$\Delta_h=\exp_x^{-1}x_h$,
\begin{equation}
 |\Delta_h|=O(\sqrt h),\qquad 0\leq A(b-h)-A(b)\leq hH_o.
 \label{sc:pole-movement}
\end{equation}

Choose one duration $q\in(t,1)$ and split each active endpoint cost
at $z_i=\exp_x(qp_i)$.  The energy inequality
\[
 c(x',y_i)\leq c_q(x',z_i)+c_{1-q}(z_i,y_i)
\]
is an equality at $x'=x$.  Its first term is smooth near $x$, because
$qp_i$ is before cut.  Its gradient is $-p_i$ and its Hessian is
$A_x(qp_i)/q$.  Combining this upper support with the smooth prefix
cost gives
\begin{align}
 A(b-h)&\geq A(b)+hB_o(s_i)
       +\langle p_i-p,\Delta_h\rangle
       +\frac12 Q_i(\Delta_h,\Delta_h)+o(|\Delta_h|^2), \label{sc:individual-pole}\\
 Q_i&=\frac1t A_x(tp)-\frac1q A_x(qp_i).\notag
\end{align}
All expansions here concern fixed scales, a fixed time, and a fixed
finite representation.
Equations~\eqref{sc:pole-movement}--\eqref{sc:individual-pole} bound
each $\langle p_i-p,\Delta_h\rangle$ above by $O(h)$.
Their positive weighted sum is zero, so each is also bounded below by
$-O(h)$.  Hence, for $E=\Span\{p_i-p\}$,
\begin{equation}
 |\operatorname{proj}_E\Delta_h|=O(h).
 \label{sc:linear-movement}
\end{equation}

On $E^\perp$, MTW concavity on the simplex of the $qp_i$ gives
\[
 A_x(qp)\geq\sum_i m_iA_x(qp_i).
\]
Radial monotonicity of divided action, equivalently
Lemma~\ref{geo:slack}, then gives
\begin{equation}
 \sum_i m_iQ_i
 =\frac1t A_x(tp)-\sum_i\frac{m_i}qA_x(qp_i)\geq0
 \quad\text{on }E^\perp.
 \label{sc:transverse-pole}
\end{equation}
Average \eqref{sc:individual-pole}.  The linear terms cancel.
By \eqref{sc:pole-movement}, \eqref{sc:linear-movement}, and
\eqref{sc:transverse-pole}, its quadratic term is bounded below by
$-o(h)$: the transverse part is nonnegative, mixed terms are
$O(h^{3/2})$, and the remaining part is $O(h^2)$.
The Taylor remainders are $o(h)$ as well.  Thus
\[
 A(b-h)-A(b)\geq h\sum_i m_iB_o(s_i)+o(h).
\]
The unique center minimizer and the elementary envelope argument
(compare the old and new minimizers and use their convergence) give
\[
 C(b-h)-C(b)=-hB_c(s)+o(h).
\]
Since $b$ is an interior maximum of $A+C-\pi$, division of the
resulting one-sided comparison by $h$ and passage to the limit give
\eqref{sc:parameter-inequality}.  The argument applies to the
arbitrary representation fixed at the start, proving the asserted
universal quantifier.
\end{proof}

\begin{lemma}[Levels at a regularized maximum]\label{sc:levels}
At a maximum in Lemma~\ref{sc:maximum}, the center and all active
endpoints belong to $S_{(H_o+2)b}(x;u,v)$.
For every active log $p_i$,
\begin{equation}
 v(y_i)-v(y)=\frac{|p|^2-|p_i|^2}{2}
          +b\{B_c(s)-B_o(s_i)\}-E_b(y).
 \label{sc:level-identity}
\end{equation}
In every positive barycentric representation the total weight on
levels $s_i\geq1-\eta$ is at least $c_1$.
Moreover
\begin{equation}
 s<3/4,\qquad
 \max\{s,s_i\}-\min\{s,s_i\}\leq1+\delta_j,\qquad
 \min\{s,s_i\}\geq-2\eta
 \label{sc:level-bounds}
\end{equation}
for small scales.  These conclusions hold at each exact maximum,
before taking a prefix-time limit.
\end{lemma}

\begin{proof}
An active endpoint has original gap
$u-g_b-bB_o(s_i)\in[0,bH_o]$.
By \eqref{sc:aligned}, the center gap is
$u-g_b+E_b(y)-bB_c(s)$, which is at most $(H_o+2)b$ by
\eqref{sc:excess-upper}.  This proves the common section assertion
and its oscillation bound.  Subtracting the active contact equality
from \eqref{sc:aligned} gives \eqref{sc:level-identity}.
Since $E_b(y)>0$ and $E_b(y)\leq bB_c(s)$, one has $B_c(s)>0$,
and therefore $s<3/4$.

Let $m_R$ be the total right weight.  Nonnegativity of $B_o$ and its
lower bound by one outside the right region give
\[
 1-m_R\leq\sum_i m_iB_o(s_i)\leq B_c(s)-c_1\leq1-c_1.
\]
Thus $m_R\geq c_1$.  In particular a right endpoint is present.
The common range bound then puts every level above
$(1-\eta)-(1+\delta_j)=-\eta-\delta_j\geq-2\eta$.
\end{proof}

\subsection{A quantitative outward endpoint}

The next Lemma selects an endpoint once the active logs lie on a line
and the center log has a nonzero component along that line.
The center Hessian estimate in Section~\ref{sec:jets} will establish
these two properties; the selection itself is finite-dimensional.

\begin{lemma}[Weight, cap, and norm increment]\label{sc:outward}
Suppose $p=\sum_{i=1}^N m_ip_i$ in an $n$-dimensional Euclidean
space, with $N\leq n+1$, $m_i>0$, and $\sum_i m_i=1$.
Suppose $E=\Span\{p_i-p\}=\R e$, where $|e|=1$, and
$q=p\cdot e>0$.
Assume the template and level inequalities
\eqref{sc:parameter-inequality}, \eqref{sc:level-bounds}, and
\[
 \left|\ |p_i|^2-|p|^2+2D(s_i-s)\ \right|\leq\zeta D,\qquad
 0\leq\delta_j\leq\zeta\leq1/16.
\]
There is an outward index $i$ with $p_i=p+d_i e$, $d_i>0$, such that
\begin{gather}
 m_i\geq m_*:=\frac{c_1}{12(n+1)},\qquad
 m_id_i\geq\frac{c_1D}{8(n+1)q},\qquad
 d_i\leq\frac{3D}{2q},                         \label{sc:outward-weight}\\
 0<|p_i|^2-|p|^2\leq3D,                       \label{sc:outward-norm}\\
 -2\eta\leq s_i<1-\eta,\qquad B_o(s_i)\leq1/m_*.
                                                        \label{sc:outward-cap}
\end{gather}
In particular the cap $M_0=2/m_*$ can be fixed before $K$.
All these constants are independent of $K$.
\end{lemma}

\begin{proof}
Write $p_i=p+d_i e$.  The level inequality gives total right mass
at least $c_1$, as in Lemma~\ref{sc:levels}.
A right point has $s_i-s>1/4-\eta$, hence
\[
 |p_i|^2-|p|^2=2qd_i+d_i^2\leq-D/4.
\]
It follows that $d_i<0$ and $|d_i|\geq D/(8q)$.
Every outward point satisfies, by the common level range,
\[
 0<2qd_i+d_i^2\leq2D(1+\delta_j)+\zeta D\leq3D,
\]
and so $d_i\leq3D/(2q)$.
The barycenter identity $\sum_i m_id_i=0$ therefore implies
\[
 \sum_{d_i>0}m_id_i\geq\frac{c_1D}{8q}.
\]
Choose an outward index contributing at least $1/N$ of this moment.
Its upper displacement bound gives $m_i\geq c_1/(12N)\geq m_*$,
proving \eqref{sc:outward-weight}.

The parameter inequality and nonnegativity of the outer template give
$m_iB_o(s_i)\leq1$, proving its cap bound.
The lower level bound is already known.  The positive norm increment
implies $s_i\leq s+\zeta/2<3/4+1/32<1-\eta$.
Thus the selected endpoint lies in the region where
\eqref{sc:curvature} applies when $M_0=2/m_*$.
\end{proof}

For a convergent sequence of active representations, an endpoint
selected by Lemma~\ref{sc:outward} in the limit carries a definite
positive moment as well as a definite weight.  Its strict outward
displacement and its strict separation from the right region persist
for the corresponding approximating vertices.  The remaining analytic
task is to produce the
one-dimensional difference space and a bound $q^2\geq\kappa D$ with
$\kappa>0$ independent of $K$. Together with
\eqref{sc:outward-norm}, these are the hypotheses of
Corollary~\ref{spec:outward}, which will compare the center and selected
outer Jacobians.
==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/scales.tex ====

==== BEGIN Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/spectral.tex ====
\section{Jacobi fields and uniform comparison of exponential Jacobians}
\label{sec:spectral}

This section proves two estimates that remain uniform near conjugate
endpoints: a positive divided-Hessian difference after pullback by an
exponential differential, and a comparison of exponential Jacobians
along an extended affine segment. The radial estimates use only the
fixed compact Riemannian manifold. Weak MTW enters when we compare
different rays on the segment. None of these constants depends on the
scalar perturbations introduced in Section~\ref{sc:section}.
The pullback estimate is Lemma~\ref{spec:pullback}; it supplies the
positive term in the center lower bound of Lemma~\ref{jet:semibound}.
The affine-segment estimate is Lemma~\ref{spec:comparison}.
Corollary~\ref{spec:outward} applies it to a positive barycentric
representation. A fixed positive vertex weight supplies the opposite
extension required for the affine comparison; radial and norm bounds
control its remaining hypotheses. This is the geometric comparison
needed in the final determinant argument.

For a linear map between tangent spaces, singular values and adjoints are
defined using the Riemannian inner products.  We write
\[
  s_1(x,v)\le\cdots\le s_n(x,v)
\]
for the singular values of $(d\exp_x)_v$, including zero singular values
when $v\in G_x$ is conjugate.  Thus
\[
  \sigma_x(v)=\prod_{j=1}^n s_j(x,v).
\]
For a positive definite symmetric matrix $Q$, the notation
$\lambda_j^{\downarrow}(Q)$ lists its eigenvalues in decreasing order.
All matrix identities below are written in parallel orthonormal frames;
their statements are independent of that choice.

\subsection{The divided-action identities}

On a minimizing ray, the divided Hessian $S(T)$ below is $H_T(v)$
from Lemma~\ref{geo:radial-index}. We record its Jacobi form together
with reversal, and distinguish the local action from the global cost
when a branch passes through cut.

\begin{lemma}[Jacobi matrices and reversal]\label{spec:identities}
Let $\gamma(s)=\exp_x(sv)$ be a geodesic and identify its tangent spaces
by parallel transport.  Put $\mathcal R(s)w=R(w,\dot\gamma(s))\dot\gamma(s)$
and let $C,J$ solve
\[
  U''+\mathcal R U=0,
  \qquad C(0)=\Id,\quad C'(0)=0,\quad
  J(0)=0,\quad J'(0)=\Id.
\]
Then
\begin{equation}\label{spec:exp-jacobi}
  (d\exp_x)_{Tv}=\frac{J(T)}{T}
\end{equation}
for $T>0$. Whenever the geodesic to $\gamma(T)$ is nonconjugate,
let $c_{\gamma,T}$ be its smooth local action: one half the squared
length of the geodesic branch obtained by varying the two endpoints.
The source Hessian of its divided action is
\begin{equation}\label{spec:divided-hessian}
  S(T):=D^2_{xx}\frac{c_{\gamma,T}(x,\gamma(T))}{T}
       =J(T)^{-1}C(T),
  \qquad
  S'(T)=-J(T)^{-1}J(T)^{-*}.
\end{equation}
The endpoint is held fixed when taking $D^2_{xx}$; the derivative in $T$
then follows the displayed geodesic.  The identities apply to a smooth
action branch whether or not it minimizes. Before cut, this action is
the cost $c$; at a nonconjugate minimizing cut vector, it is the chosen
smooth upper branch of $c$.

If $y=\exp_xv$ and $\bar v=-P_{x\to y}v$ is the reverse initial velocity,
then
\begin{equation}\label{spec:reversal}
  (d\exp_y)_{\bar v}=\bigl((d\exp_x)_v\bigr)^*,
  \qquad \sigma_y(\bar v)=\sigma_x(v),
\end{equation}
with the indicated source and target tangent spaces.  Finally, for
$v\in I(x)$,
\begin{equation}\label{spec:radial-identity}
  A_x(v)v=v.
\end{equation}
\end{lemma}

\begin{proof}
Varying the initial velocity in the geodesic equation gives
\eqref{spec:exp-jacobi}.  Since $\mathcal R$ is symmetric, the fundamental
matrix
\[
  \Phi(T)=
  \begin{pmatrix}C(T)&J(T)\\ C'(T)&J'(T)\end{pmatrix}
\]
is symplectic.  In particular,
\begin{equation}\label{spec:wronskians}
  J^*J'=J'^*J,
  \qquad C^*J'-C'^*J=\Id,
  \qquad CJ^*=JC^*.
\end{equation}
These identities can also be obtained by differentiating their left
sides and using the Jacobi equation; their values at zero give the
displayed constants.

For a variation of the initial point by $a$ with terminal variation zero,
the variational field is $C(s)a+J(s)b$, where
$b=-J(T)^{-1}C(T)a$.  The initial gradient of the divided action is the
negative initial velocity.  Differentiating it in a normal frame at $x$
therefore gives $S(T)=J(T)^{-1}C(T)$.  The last identity in
\eqref{spec:wronskians} shows that this matrix is symmetric.  Moreover,
the first two identities give
\[
  J^*\bigl(C'-J'J^{-1}C\bigr)
    =J^*C'-J'^*C=-\Id.
\]
Consequently
\[
  \frac{d}{dT}(J^{-1}C)
    =J^{-1}\bigl(C'-J'J^{-1}C\bigr)
    =-J^{-1}J^{-*},
\]
including the order of the two inverse factors in
\eqref{spec:divided-hessian}.

For completeness, symplecticity also gives
\[
  \Phi(T)^{-1}=
  \begin{pmatrix}J'(T)^*&-J(T)^*\\-C'(T)^*&C(T)^*\end{pmatrix}.
\]
A Jacobi field for the reversed geodesic with initial position zero and
initial derivative $a$ has, in the forward parametrization, terminal
data $(0,-a)$.  Multiplication by this inverse matrix shows that its
value at the other endpoint is $J(T)^*a$.  The reversed Jacobi matrix
at time $T$ is thus $J(T)^*$.  Dividing both matrices by $T$ as in
\eqref{spec:exp-jacobi} proves \eqref{spec:reversal}; this argument does
not require $J(T)$ to be invertible.

In the radial direction the Jacobi equation has the explicit solutions
$J(s)v=sv$ and $C(s)v=v$. Thus $S(T)v=v/T$. For $v\in I(x)$,
taking $T=1$ gives \eqref{spec:radial-identity}; the case $v=0$
is immediate.
\end{proof}

\subsection{Radial shortening at a conjugate endpoint}

\begin{lemma}[Uniform radial transversality]\label{spec:radial}
There are constants $c_{\rm r}>0$ and $C_{\rm r}<\infty$, depending only
on $(M,g)$, such that for $x\in M$, $v\in G_x$, $0<\delta\le1/2$, and
$1\le j\le n$,
\begin{equation}\label{spec:singular-shortening}
  c_{\rm r}\bigl(s_j(x,v)+\delta\bigr)
   \le s_j(x,(1-\delta)v)
   \le C_{\rm r}\bigl(s_j(x,v)+\delta\bigr).
\end{equation}
There is also a positive definite form $Q_{x,v,\delta}$ on $T_xM$ such
that, on putting
\[
  N_{x,\delta}(v)
       =-\delta\bigl(A_x((1-\delta)v)-\Id\bigr),
\]
one has
\begin{align}
  &\|N_{x,\delta}(v)-Q_{x,v,\delta}\|\le C_{\rm r}\delta,
       \label{spec:bounded-background}\\
  &c_{\rm r}\frac{\delta}{s_j(x,v)+\delta}
     \le\lambda_j^{\downarrow}(Q_{x,v,\delta})
     \le C_{\rm r}\frac{\delta}{s_j(x,v)+\delta}.
       \label{spec:hessian-spectrum}
\end{align}
In particular, uniformly on this set,
\begin{equation}\label{spec:bounded-N}
  \|N_{x,\delta}(v)\|\le C_{\rm r},\qquad
  N_{x,\delta}(v)\succeq-C_{\rm r}\delta\Id,
  \qquad N_{x,\delta}(v)v=0.
\end{equation}
No condition on the multiplicity of conjugacy is required.
\end{lemma}

\begin{proof}
The bundle
\[
  \mathcal G=\{(x,v):x\in M,\ v\in G_x\}
\]
is compact: its defining distance equality is closed, and $|v|$ is
bounded by $\operatorname{diam}(M)$.  For each such vector consider
$\gamma(s)=\exp_x(sv)$ on $[0,1]$.  Fix the splitting time $h=1/4$ and
allow the final time $T$ to vary in $[1/2,1]$.
The prefix $hv$ belongs to $I(x)$, since it extends by the factor
$1/h=4$ to the original minimizing ray. For the suffix, reverse the
ray at $y_T=\gamma(T)$ and put
\[
 w_T=-(T-h)\dot\gamma(T),\qquad
 a_T=\frac{T}{T-h}\ge\frac43.
\]
Then $\exp_{y_T}w_T=\gamma(h)$ and
$d(y_T,\exp_{y_T}(a_Tw_T))=a_T|w_T|$, so $w_T\in I(y_T)$.
The reversed suffix is therefore before cut; reversal preserves
uniqueness and, by Lemma~\ref{spec:identities}, nonconjugacy.
The parameter family is compact and its image lies in the open
injectivity domain. Thus both piece differentials and their inverses
have uniform bounds. Their time lengths are bounded below by $1/4$.

Let $C,J$ be the matrices from Lemma~\ref{spec:identities}.  Starting
at time $h$, let $C_h(s),J_h(s)$ have the same initial data and solve
the Jacobi equation along $\gamma(h+s)$.  Smooth dependence on
$(x,v,T)$ and the preceding nonconjugacy show that
\begin{equation}\label{spec:piece-bounds}
  J(h),\ J(h)^{-1},\ J_h(T-h),\ J_h(T-h)^{-1},\ C(h),\ J'(h)
\end{equation}
all have uniformly bounded norms.  This compactness assertion also
includes $v=0$, for which $J(s)=s\Id$.

Write $z=\gamma(h)$ and $y_T=\gamma(T)$.  The Hessian in the joining
point of the broken action
\[
  z'\longmapsto \frac{c(x,z')}{h}
                     +\frac{c(z',y_T)}{T-h}
\]
at $z$ is
\begin{equation}\label{spec:middle-Hessian}
  D(T)=J'(h)J(h)^{-1}
            +J_h(T-h)^{-1}C_h(T-h).
\end{equation}
The first term is the terminal Hessian of the first piece; the second
is the source Hessian of the second piece.  At $T=1$, the broken
action has a minimum at $z$.  To see this directly, the triangle
inequality and Cauchy--Schwarz give
\[
  \frac{d(x,z')^2}{2h}+\frac{d(z',y_1)^2}{2(1-h)}
    \ge\frac{(d(x,z')+d(z',y_1))^2}{2}
    \ge\frac{d(x,y_1)^2}{2},
\]
with equality at $z$.  Hence $D(1)\succeq0$.

Only the second term in \eqref{spec:middle-Hessian} depends on $T$.
Applying \eqref{spec:divided-hessian} with initial point $z$ yields
\begin{equation}\label{spec:middle-crossing}
\begin{split}
  D(1-\delta)&=D(1)+\delta G_\delta,\\
  G_\delta&=\frac1\delta\int_{1-\delta}^{1}
       J_h(s-h)^{-1}J_h(s-h)^{-*}\,ds,
       \qquad c\Id\preceq G_\delta\preceq C\Id.
\end{split}
\end{equation}
The bounds on $G_\delta$ are precisely the bounds in
\eqref{spec:piece-bounds}: the smallest eigenvalue of the integrand
is $\|J_h(s-h)\|^{-2}$, and its largest eigenvalue is
$\|J_h(s-h)^{-1}\|^2$.  Thus the crossing is positive on the whole
space, including any null space of $D(1)$.

Jacobi propagation through the joining point gives the exact
factorization
\begin{equation}\label{spec:Jacobi-factorization}
\begin{split}
  J(T)
    &=C_h(T-h)J(h)+J_h(T-h)J'(h)\\
    &=J_h(T-h)D(T)J(h).
\end{split}
\end{equation}
This formula is also valid at $T=1$ when either of the two full
matrices is singular.  The splitting point is kept fixed while $T$
varies; differentiating a moving midpoint is unnecessary.  Taking
$h=1/2$ gives the usual midpoint factorization on $T\in[3/4,1]$.
The choice $h=1/4$ covers all the parameters in the statement with
one set of bounds.

Write $\lambda_1(D)\le\cdots\le\lambda_n(D)$ for the increasing
eigenvalues of a positive semidefinite matrix $D$.  The min--max
characterization applied to \eqref{spec:middle-crossing} gives
\[
  \lambda_j(D(1))+c\delta
     \le\lambda_j(D(1-\delta))
     \le\lambda_j(D(1))+C\delta.
\]
Multiplication of a matrix on either side by a uniformly bounded
invertible matrix changes each ordered singular value by at most a
uniform multiplicative factor.  Apply this fact to
\eqref{spec:Jacobi-factorization}, use
$(d\exp_x)_{Tv}=J(T)/T$, and note $1/2\le T\le1$.
At $T=1$, the singular values of $J(1)$ are comparable to the
eigenvalues of $D(1)$; at $T=1-\delta$, they are comparable to the
eigenvalues of $D(1-\delta)$.  These observations prove
\eqref{spec:singular-shortening} with uniform constants.

We give the Hessian calculation as well, because it identifies the
bounded term that must not be discarded at a radial eigenvector.
Set
\[
  P=J(h)^{-1}C(h),\qquad E=J(h)^{-*}.
\]
The source--join mixed Hessian of the first divided action is
$-J(h)^{-1}=-E^*$.  For $T<1$, eliminating the joining-point
variation in the Hessian of the broken action gives
\begin{equation}\label{spec:Schur-action}
  \frac{A_x(Tv)}{T}=P-E^*D(T)^{-1}E.
\end{equation}
To verify the elimination, write the quadratic form for source
variation $a$ and joining variation $b$ as
$a^*Pa-2a^*E^*b+b^*D(T)b$.  Its unique stationary value in $b$ is
at $b=D(T)^{-1}Ea$ and has value
$a^*(P-E^*D(T)^{-1}E)a$.  This is the Hessian of the stationary
composed action.  Positivity of $D(T)$ follows from
\eqref{spec:middle-crossing}; the composed action is the smooth
cost branch of the shortened minimizing ray.

For $T=1-\delta$, formula \eqref{spec:Schur-action} gives
\begin{equation}\label{spec:Q-definition}
\begin{split}
  N_{x,\delta}(v)&=Q_{x,v,\delta}
                         +\delta(\Id-TP),\\
  Q_{x,v,\delta}&=T\delta E^*D(T)^{-1}E.
\end{split}
\end{equation}
The second form is positive definite.  The bounds on $P,E,E^{-1}$,
the min--max inequalities just proved, and inversion of $D(T)$ give
\eqref{spec:bounded-background} and \eqref{spec:hessian-spectrum}.
They imply the norm and lower bounds in \eqref{spec:bounded-N}.
The remaining identity follows from
\eqref{spec:radial-identity}.  In particular, for $v\ne0$ the radial
eigenvalue of $N_{x,\delta}(v)$ is exactly zero, while the radial
singular value of $(d\exp_x)_v$ is one.  These two facts are consistent
because the comparison with $Q_{x,v,\delta}$ includes the bounded
background $\delta(\Id-TP)$.
\end{proof}

\begin{corollary}[Radial comparison of exponential determinants]
\label{spec:radial-determinants}
With a constant depending only on $(M,g)$, if $v\in G_x$ and
$1/2\le\theta\le1$, then
\begin{equation}\label{spec:radial-det}
  \sigma_x(\theta v)\ge c\sigma_x(v).
\end{equation}
Suppose $y=\exp_xp$, $p\in G_x$, and
$\bar p=-P_{x\to y}p$.  If $l\in[1/2,1)$ and $\bar p/l\in G_y$,
then
\begin{equation}\label{spec:center-true-det}
  \sigma_y(\bar p/l)\le C\sigma_x(p).
\end{equation}
If $l_+\in[1,2]$, then
\begin{equation}\label{spec:outer-true-det}
  \sigma_y(\bar p/l_+)\ge c\sigma_x(p).
\end{equation}
Comparison~\eqref{spec:center-true-det} permits the longer reversed
vector $\bar p/l$ to have a conjugate endpoint.
\end{corollary}

\begin{proof}
For $\theta<1$, multiply the lower bounds in
\eqref{spec:singular-shortening} with $\delta=1-\theta$ and discard
the nonnegative terms $\delta$.  The case $\theta=1$ is immediate.
For \eqref{spec:center-true-det}, apply \eqref{spec:radial-det} to
the longer vector $\bar p/l$ with shortening factor $l$, and then
use \eqref{spec:reversal}.  For \eqref{spec:outer-true-det}, apply
it to $\bar p$ with shortening factor $1/l_+$ and use the same
reversal identity.  Thus the direction of the two inequalities
reflects, respectively, a longer and a shorter true reversed ray.
\end{proof}

\subsection{A uniform gain from an extra minimizing extension}

\begin{lemma}[Isotropic gain after pullback]\label{spec:pullback}
There is $c_{\rm p}>0$, depending only on $(M,g)$, with the following
property.  Let $y\in M$, $r\in T_yM$, and $1<T_*\le2$ satisfy
$T_*r\in G_y$.  Put
\[
  \gamma(s)=\exp_y(sr),\quad x=\gamma(1),\quad
  p=-P_{y\to x}r,\quad L=(d\exp_x)_p,
  \qquad \tau=1+\frac{T_*-1}{2}.
\]
Then all costs appearing in
\begin{equation}\label{spec:half-extension-gain}
  L^*\left[A_y(r)-\frac1\tau A_y(\tau r)\right]L
       \succeq c_{\rm p}(T_*-1)\Id
\end{equation}
are smooth.  If the terminal ray at time $T_*$ is nonconjugate,
the same inequality holds with $\tau$ replaced by $T_*$ and its
chosen smooth branch.  In particular, for $T_*=1/l$ this latter
inequality reads
\begin{equation}\label{spec:full-extension-gain}
  L^*\left[A_y(r)-l A_y(r/l)\right]L
       \succeq c_{\rm p}(1-l)\Id,
       \qquad 1/2\le l<1.
\end{equation}
Without any uniformity assertion, the source Hessian of divided
action decreases strictly in the positive definite order between
any two nonconjugate times on a minimizing ray.
\end{lemma}

\begin{proof}
Every time less than $T_*$ lies strictly before the cut time of this
ray, so $r$ and $\tau r$ are in $I(y)$, and the reverse vector $p$
is in $I(x)$.  Let $J(s)$ be the Jacobi matrix based at $y$ with
unit initial derivative.  By reversal,
\begin{equation}\label{spec:pullback-transpose}
  L=J(1)^*.
\end{equation}
Apply Lemma~\ref{spec:radial} to the minimizing vector $T_*r$,
with shortening factor $1/T_*$.  Since
$\delta=(T_*-1)/T_*\ge(T_*-1)/2$, its lower bound gives
\begin{equation}\label{spec:J-one-inverse}
  s_1(J(1))=s_1(y,r)\ge c(T_*-1),
  \qquad \|J(1)^{-1}\|\le\frac{C}{T_*-1}.
\end{equation}
There is also a uniform bound on $\|J'(s)\|$ for $0\le s\le T_*$.
Indeed, $|T_*r|\le\operatorname{diam}(M)$, $T_*\le2$, and smooth
Jacobi evolution on this compact set of initial data bounds its
fundamental matrices.  Therefore, for $1\le s\le\tau$,
\begin{equation}\label{spec:noncommuting-ratio}
\begin{split}
  \|J(s)J(1)^{-1}\|
   &\le 1+\|J(s)-J(1)\|\,\|J(1)^{-1}\|\\
   &\le 1+C\frac{s-1}{T_*-1}\le C.
\end{split}
\end{equation}

Write $S(s)=A_y(sr)/s$.  Lemma~\ref{spec:identities} and
\eqref{spec:pullback-transpose} give
\begin{equation}\label{spec:noncommuting-positive}
  L^*[-S'(s)]L
    =J(1)J(s)^{-1}J(s)^{-*}J(1)^*.
\end{equation}
The matrix on the right is $B_sB_s^*$ with
$B_s=J(1)J(s)^{-1}$.  Its smallest eigenvalue equals
$\|B_s^{-1}\|^{-2}=\|J(s)J(1)^{-1}\|^{-2}$ and is thus bounded
below by a uniform positive constant by
\eqref{spec:noncommuting-ratio}.  This uses the displayed order of
the factors and does not commute any Jacobi matrices.  Integration
over $[1,\tau]$, whose length is $(T_*-1)/2$, proves
\eqref{spec:half-extension-gain}.

If the terminal branch is nonconjugate, $S$ is defined through time
$T_*$ and $-S'(s)$ is positive definite on the remaining interval.
Adding its integral over $[\tau,T_*]$ preserves the lower bound.
Since $T_*-1=(1-l)/l\ge1-l$, this gives
\eqref{spec:full-extension-gain}.  For arbitrary two nonconjugate
times on a minimizing ray, integration of
$-S'(s)=J(s)^{-1}J(s)^{-*}\succ0$ proves strict decrease.  When the
true terminal point is conjugate, only the half-extension formula
is used; no undefined terminal Hessian has been differentiated.
\end{proof}

\subsection{From transverse MTW to comparison along an affine segment}

\begin{lemma}[Uniform comparison along an extended segment]
\label{spec:comparison}
Assume weak MTW.  Fix $a_0>0$ and $B<\infty$.  There is
$c_{\rm s}>0$, depending only on $(M,g),a_0,B$, such that the
following holds.  Let $x\in M$, $p,q\in T_xM$, $q\ne0$, satisfy
\begin{equation}\label{spec:segment-hypotheses}
  p+tq\in G_x\quad(-a_0\le t\le1),\qquad
  p\cdot q>0,\qquad
  \frac{|q|^2}{p\cdot q}\le B.
\end{equation}
Then, for every $1\le j\le n$,
\begin{equation}\label{spec:ordered-comparison}
  s_j(x,p)\ge c_{\rm s}s_j(x,p+q).
\end{equation}
In particular,
\begin{equation}\label{spec:product-comparison}
  \sigma_x(p)\ge c_{\rm s}^n\sigma_x(p+q).
\end{equation}
\end{lemma}

\begin{proof}
If there is a configuration in the statement then $B>0$.  Set
\[
  a=\min\{a_0,(2B)^{-1}\}.
\]
We use only the segment $[-a,1]$.  First suppose that all its
points lie in $I(x)$.  Write
\[
  p(t)=p+tq,\qquad \beta=\frac{|q|^2}{p\cdot q},\qquad
  L(t)=\Id+t\frac{qq^*}{p\cdot q}.
\]
The eigenvalues of $L(t)$ are $1$ on $q^\perp$ and $1+t\beta$
on $\R q$.  Hence
\begin{equation}\label{spec:L-bounds}
  \|L(t)\|\le1+B,\qquad \|L(t)^{-1}\|\le2
       \qquad(-a\le t\le1).
\end{equation}
For $0<\delta\le1/2$, put
\begin{equation}\label{spec:F-definition}
  N_\delta(t)=-\delta\bigl(A_x((1-\delta)p(t))-\Id\bigr),
  \qquad F_\delta(t)=L(t)^*N_\delta(t)L(t).
\end{equation}
These matrices are uniformly bounded by Lemma~\ref{spec:radial}
and \eqref{spec:L-bounds}, and
$F_\delta(t)\succeq-C\delta\Id$.

We next prove full matrix convexity of $F_\delta$.  For any fixed
$z\in T_xM$, the vector
\[
  \xi=z-\frac{z\cdot q}{p\cdot q}p
\]
is perpendicular to $q$.  Moreover,
\[
  \xi-L(t)z=-\frac{z\cdot q}{p\cdot q}p(t).
\]
The last vector belongs to the kernel of $N_\delta(t)$ by the
radial identity.  Symmetry consequently gives the exact equality
\begin{equation}\label{spec:transverse-conjugation}
  z^*F_\delta(t)z=\xi^*N_\delta(t)\xi.
\end{equation}
Weak MTW means that
$t\mapsto A_x((1-\delta)p(t))(\xi,\xi)$ is concave whenever
$\xi\perp q$.  The negative sign in the definition of $N_\delta$
therefore makes the right side of
\eqref{spec:transverse-conjugation} convex.  Since this holds for
every $z$, $F_\delta$ is convex in the Loewner order.

This argument remains uniform when $p\cdot q$ is small.  The
fixed test vector $\xi$ itself might be large, but its value
modulo the radial kernel is $L(t)z$.  Equations
\eqref{spec:L-bounds} and \eqref{spec:transverse-conjugation}
bound the scalar convex function by $C|z|^2$, independently of
the size of $\xi$.  For any compact interval
$K\subset(-a,1)$, a real convex function bounded in absolute
value by $C|z|^2$ has slopes on $K$ bounded by
$2C|z|^2/\operatorname{dist}(K,\{-a,1\})$: sandwich each
secant slope between secants to the two endpoints.  Applying
this estimate to basis vectors and their sums and differences
shows by polarization that every matrix entry of $F_\delta$
has a uniform Lipschitz bound on $K$.

To turn this bounded matrix convexity into a spectral comparison,
we argue by contradiction. The spectral profile
$\delta/(s_i+\delta)$ from \eqref{spec:hessian-spectrum}
detects singular values below the radial scale: it stays positive
when $s_i\ll\delta$ and tends to zero when $s_i\gg\delta$.
At a scale separated from the singular values, the rank of a
limiting matrix therefore counts those below the scale.
A failing ordered singular-value ratio
will provide a radial scale between the center and endpoint values.
We choose it to separate asymptotically from every singular value
at each fixed point of a countable dense set of segment parameters.
The resulting positive semidefinite
matrix limit has a uniform gap between zero and its positive
eigenvalues, forcing constant interior rank. Convexity then prevents
rank loss at the endpoint, whereas the failing ratio will force
precisely such a loss.

Suppose that no uniform constant in
\eqref{spec:ordered-comparison} exists.  There are configurations
$(x_k,p_k,q_k)$ satisfying the same $a_0,B$, an index $j$ fixed
after passing to a subsequence, and
\begin{equation}\label{spec:failed-ratio}
  \frac{s_j(x_k,p_k)}{s_j(x_k,p_k+q_k)}\longrightarrow0.
\end{equation}
For now all the segments are in their open injectivity domains,
so the singular values are positive.  Choose an orthonormal
identification of each $T_{x_k}M$ with $\R^n$.  No smooth choice
in $k$ is needed.  Abbreviate
\[
  s_{i,k}(t)=s_i(x_k,p_k+tq_k),\qquad
  R_k=\log\frac{s_{j,k}(1)}{s_{j,k}(0)}\longrightarrow\infty.
\]
Let $\{t_1,t_2,\ldots\}$ be a fixed countable dense subset of
$(-a,1)$, enlarged to include $0$ and $1$.  We now choose one
radial scale that separates every fixed member of this set.
Take $\log\delta_k$ in the middle half of the interval
\[
  [\log s_{j,k}(0),\log s_{j,k}(1)].
\]
Exclude the intervals of radius $R_k^{1/3}$ centered at
\[
  \log s_{i,k}(t_m),\qquad
  1\le i\le n,\quad 1\le m\le\lfloor R_k^{1/3}\rfloor.
\]
Their total length is at most $2nR_k^{2/3}$, while the available
middle interval has length $R_k/2$.  For all sufficiently large
$k$ there is a remaining point, which we choose as
$\log\delta_k$.  This construction ensures
\begin{align}
  &s_{j,k}(0)/\delta_k\longrightarrow0,
       \qquad s_{j,k}(1)/\delta_k\longrightarrow\infty,
       \label{spec:separating-scale}\\
  &\left|\log\frac{s_{i,k}(t_m)}{\delta_k}\right|
       \longrightarrow\infty
       \quad\text{for every fixed }i,m.
       \label{spec:dense-separation}
\end{align}
All singular values on $\mathcal G$ are uniformly bounded above.
Since $\delta_k\le s_{j,k}(1)e^{-R_k/4}$, we also have
$\delta_k\to0$.  In particular these scales are eventually
admissible in Lemma~\ref{spec:radial}.

Form the matrices $F_k(t)$ in
\eqref{spec:F-definition} using the $k$th segment and scale
$\delta_k$.  Their uniform bounds and interior Lipschitz bounds
give, by successive extraction on compact subintervals, a locally
uniform limit $F(t)$ on $(-a,1)$.  Extract also a limit
$E=\lim_kF_k(1)$.  Both $F(t)$ and $E$ are positive semidefinite,
because their approximants are bounded below by
$-C\delta_k\Id$.  The limit $F$ is matrix convex and bounded.

To identify its possible ranks, use
\eqref{spec:bounded-background} to write
\[
  F_k(t)=L_k(t)^*Q_{x_k,p_k+tq_k,\delta_k}L_k(t)
                     +\mathcal E_k(t),
  \qquad \|\mathcal E_k(t)\|\le C\delta_k.
\]
The positive form in this expression has its $i$th decreasing
eigenvalue between uniform multiples of
\begin{equation}\label{spec:rank-spectral-profile}
  \frac{\delta_k}{s_{i,k}(t)+\delta_k},
\end{equation}
by \eqref{spec:hessian-spectrum} and \eqref{spec:L-bounds}.
At a fixed dense point $t_m$, put $\eta_k=\exp(-R_k^{1/3})$.
For all sufficiently large $k$, depending on $m$,
\eqref{spec:dense-separation} says that
$s_{i,k}(t_m)/\delta_k$ is at most $\eta_k$ or at least
$\eta_k^{-1}$. Weyl's eigenvalue inequality therefore gives
one of the two alternatives
\[
 \lambda_i^\downarrow(F_k(t_m))
       \ge \frac{c_0}{1+\eta_k}-C\delta_k,
 \qquad\text{or}\qquad
 -C\delta_k\le\lambda_i^\downarrow(F_k(t_m))
       \le C_0\eta_k+C\delta_k,
\]
with $c_0,C_0>0$ uniform in $i,m,k$.
Since $F_k(t_m)$ already converges, its ordered eigenvalues converge.
Their limits are therefore either zero or at least $c_0$, even if
the alternatives initially alternate. No simultaneous rate at all
dense points is required. Set $c_*=c_0/2$.

The ordered eigenvalues of a symmetric matrix are continuous in
its entries.  Density of the set $\{t_m\}$ and continuity of
$F(t)$ therefore give
\begin{equation}\label{spec:uniform-limit-gap}
  \operatorname{spec}F(t)\subset\{0\}\cup[c_*,C_*]
        \qquad(-a<t<1)
\end{equation}
after decreasing $c_*$ once if necessary.  Indeed, an eigenvalue
strictly between zero and $c_*$ at an interior point would remain
in that interval in a neighborhood and contradict the dense-set
conclusion.  The rank is now locally constant, because a
continuous eigenvalue cannot pass from zero to a positive value
without entering that gap.  It is consequently constant on the
connected interval $(-a,1)$.

The endpoint also has to be treated; constancy of the interior
rank alone would not suffice.  For every fixed vector $z$, the
bounded scalar convex function $z^*F(t)z$ has a finite limit as
$t\uparrow1$.  Polarization yields a matrix limit $F(1-)$.
The gap \eqref{spec:uniform-limit-gap} persists in this limit,
so its rank equals the interior rank.  For fixed $s<t<1$,
matrix convexity of $F_k$ says
\[
  F_k(t)\preceq\frac{1-t}{1-s}F_k(s)
                         +\frac{t-s}{1-s}F_k(1).
\]
First let $k\to\infty$, then let $t\uparrow1$; this gives
\begin{equation}\label{spec:endpoint-order}
  F(1-)\preceq E.
\end{equation}
Positive semidefinite order implies
$\operatorname{rank}E\ge\operatorname{rank}F(1-)$, since the
kernel of $E$ is contained in the kernel of $F(1-)$.

On the other hand, the first inequality in
\eqref{spec:separating-scale} and the ordering of singular values
make the first $j$ eigenvalues in
\eqref{spec:rank-spectral-profile} at $t=0$ bounded away from
zero.  Thus $\operatorname{rank}F(0)\ge j$.  The second inequality
makes all eigenvalues with decreasing index $i\ge j$ tend to
zero at $t=1$, so $\operatorname{rank}E\le j-1$.  This contradicts
the interior rank constancy and \eqref{spec:endpoint-order}.
The uniform ordered comparison is proved for open segments.

Finally, replace a closed segment in \eqref{spec:segment-hypotheses}
by $(1-\eps)(p+tq)$, with $0<\eps<1$.  It lies in $I(x)$ by
radial shortening.  Its extension interval and the ratio
$|q|^2/(p\cdot q)$ are unchanged, so the same constant applies.
Letting $\eps\downarrow0$ and using continuity of the singular
values proves \eqref{spec:ordered-comparison} on $G_x$.
Multiplication over $j$ proves
\eqref{spec:product-comparison}.  Every bound in the contradiction
argument depended only on $(M,g),a_0,B$, which establishes the
stated uniformity.
\end{proof}

\subsection{Jacobian comparison for a barycentric configuration}

\begin{corollary}[Barycentric configurations]\label{spec:outward}
Assume weak MTW and the convexity of $G_x$ established in
Corollary~\ref{geo:convexity}.  Fix $\mu\in(0,1)$ and positive constants
$\kappa,L_0$.  Suppose
\[
  p=\sum_{i=1}^N m_i p_i,\qquad
  p_i\in G_x,\quad m_i>0,\quad\sum_i m_i=1,
\]
and for one index $i$ there are $D>0$, a unit vector $e$, and
$t_i>0$ such that
\begin{equation}\label{spec:outward-data}
  p_i=p+t_i e,\quad h:=p\cdot e>0,\quad
  h^2\ge\kappa D,\quad
  |p_i|^2-|p|^2\le L_0D,\quad m_i\ge\mu.
\end{equation}
Then
\begin{equation}\label{spec:outward-det}
  \sigma_x(p)\ge c\sigma_x(p_i),
\end{equation}
where $c>0$ depends only on $(M,g),\mu,\kappa,L_0$.
\end{corollary}

\begin{proof}
Set $q=t_ie$.  Expansion of the squared norms in
\eqref{spec:outward-data} gives
\[
  2ht_i+t_i^2\le L_0D,
  \qquad
  \frac{|q|^2}{p\cdot q}=\frac{t_i}{h}
          \le\frac{L_0D}{2h^2}\le\frac{L_0}{2\kappa}.
\]
Because $q\ne0$, $m_i<1$.  The remaining barycenter belongs to
$G_x$ by its convexity and is
\[
  \frac1{1-m_i}\sum_{j\ne i}m_jp_j
       =p-\frac{m_i}{1-m_i}q.
\]
The coefficient $m_i/(1-m_i)$ is at least $m_i\ge\mu$.
Convexity therefore puts $p+s q$ in $G_x$ for every
$s\in[-\mu,1]$.  Lemma~\ref{spec:comparison} applies with
$a_0=\mu$ and $B=L_0/(2\kappa)$ and proves the result.
No upper bound on $m_i/(1-m_i)$ is needed: a shorter common
opposite extension is enough.
\end{proof}
==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//sections/spectral.tex ====

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==== END Uniform-Bi-Holder-Transport-from-Weak-MTW//references.bib ====