==== TABLE OF CONTENTS ====
Diamond-distortion-from-midpoint-and-tree-energies//d-applications.tex
Diamond-distortion-from-midpoint-and-tree-energies//diamond-replacement.tex
Diamond-distortion-from-midpoint-and-tree-energies//direct.tex
Diamond-distortion-from-midpoint-and-tree-energies//f-applications.tex
Diamond-distortion-from-midpoint-and-tree-energies//introduction.tex
Diamond-distortion-from-midpoint-and-tree-energies//main.tex
Diamond-distortion-from-midpoint-and-tree-energies//metrics.tex
Diamond-distortion-from-midpoint-and-tree-energies//preamble.tex
Diamond-distortion-from-midpoint-and-tree-energies//segment-transfer-variants.tex
Diamond-distortion-from-midpoint-and-tree-energies//transfer-extensions.tex
Diamond-distortion-from-midpoint-and-tree-energies//references.bib
===========================
==== BEGIN Diamond-distortion-from-midpoint-and-tree-energies//d-applications.tex ====
\section{Graph consequences of path and segment costs}\label{met:d-section}
The next four models change what a tree representation costs: a sum of
Hilbert and path costs, their Euclidean combination, a charge at each
segment's start, or a restriction to incomparable segments. These are
different norms. Within one norm, several estimates can also give
different bounds. We first extract their squared losses, then retain
the actual-child or qualitative conclusions when the method supplies
different information. All intrinsic inputs come from the companion on
bounded tree potentials and path costs; the graph deductions use the
transfers proved above.
\subsection{Linear path cost}\label{met:linear-section}
Recall the linear path-cost space $X_{\rm lin}$ from
Section~\ref{dir:layers}. Its component norm is the infimum of
$\|h\|_2+\|\mu\|_1$ over decompositions into a Hilbert vector and
root paths, including the component root. For later comparison write
$J_n\mu=\sum_s\mu(s)b_s$ for the root-path sum on
$T_n=\mathbb N^{\le n}$.
The finite ancestor-closed coordinate projections are contractive,
finite supports are dense, and this space and its dual are separable and
reflexive. The squared estimates below and the weakly null estimate in
Section~\ref{met:weak-linear} give lower bounds for this same norm;
the latter will give the strongest coefficient displayed here.
The positive-flow argument gives, for $Pp=p$, $Py=0$
\cite[Theorem~5.7]{TreePotentials},
\begin{equation}\label{met:linear-flow-input}
\|y\|^2\le C^2(\|p+y\|^2+\|p-y\|^2-2\|p\|^2),
\qquad C=6(\sqrt2+1).
\end{equation}
A separate trimming argument \cite[Theorem~5.10]{TreePotentials}
gives the same inequality with $C^2$
replaced by $242$. Since each endpoint is at most $R$, these are
\eqref{met:square-input} with $a=1/(2C^2)$ and $a=1/484$,
respectively. Proposition~\ref{met:tail-to-lens} and
Theorem~\ref{met:squared-transfer} therefore give
\begin{equation}\label{met:linear-flow-graph}
D^2\ge1+\frac{k}{2C^2},\qquad
D^2\ge1+\frac{k}{484},
\end{equation}
using the corresponding proof in each case. In upper-one normalization,
the first argument says $r_j^2\le r_{j+1}^2-1/(2C^2D^2)$ and hence
also retains $k\le2C^2D^2$.
The inward-clipping argument \cite[Theorem~5.4]{TreePotentials}
gives the different parameterized estimate
\begin{equation}\label{met:linear-clipping-input}
\max_\pm\|p\pm y\|
\ge\frac{\|p\|+c\|y\|/2}{\sqrt{1+4c^2}}
\qquad(c>0, Pp=p, Py=0).
\end{equation}
Here is its direct actual-child application, retaining the error budget.
For a parent $x$ in the noncontracting normalization set
\[
c=\frac1{8D},\quad Q=\sqrt{1+4c^2},\quad
\gamma=\frac c2-(Q-1)D,\quad
\eta=\frac{\gamma}{2(1+c/2+2Q)}.
\]
Since $Q-1\le2c^2$, we have $\gamma\ge c/4>0$.
Choose a finite head $P$ with $\|x-Px\|<\eta$, and choose the pair in
Lemma~\ref{met:finite-pair}; write $w=(z_i-z_j)/2$ and $y=w-Pw$.
The greatest of the four actual child norms is at least
\begin{align*}
\max_\pm\|x\pm w\|
&\ge \frac{\|Px\|+c\|y\|/2}{Q}-2\eta\\
&\ge \frac{\|x\|-\eta+c(1-\eta)/2}{Q}-2\eta
\ge\|x\|+\frac{\gamma}{2Q}.
\end{align*}
Selecting such a child through all $k$ generations gives
\begin{equation}\label{met:linear-clipping-graph}
1+\frac{k\gamma}{2Q}\le D.
\end{equation}
The additive increment is attached to the clipping estimate, independently
of the positive-flow and trimming proofs.
\subsection{Weakly null pairs for the linear norm}\label{met:weak-linear}
A further form of inward clipping \cite[Theorem~5.5]{TreePotentials}
states that for every $x\in X_{\rm lin}$
and every bounded weakly null sequence $(y_j)$,
\begin{equation}\label{met:weak-input}
\liminf_j\max_\pm\|x\pm y_j\|
\ge\sqrt{\|x\|^2+\frac1{16}(\liminf_j\|y_j\|)^2}.
\end{equation}
This input retains a larger squared loss when we pair deviations before
applying it:
\begin{equation}\label{met:linear-weak-graph}
D^2\ge1+k/16.
\end{equation}
For a parent with child bound $R$ in the noncontracting normalization,
the bounded sequence of deviations $z_i$ has a weakly Cauchy
subsequence. Indeed, diagonalize its evaluations on a countable
norm-dense subset of $X_{\rm lin}^*$, then use boundedness to
approximate every other functional. Relabel this subsequence and set
$w_j=(z_{2j}-z_{2j-1})/2$. It is weakly null, and sibling separation
gives $\|w_j\|\ge1$. Crossed convex averages give
\[
x+w_j=\tfrac12((x+z_{2j})+(x-z_{2j-1})),\qquad
x-w_j=\tfrac12((x-z_{2j})+(x+z_{2j-1})),
\]
so both norms are at most $R$. Equation~\eqref{met:weak-input} yields
$\|x\|^2+1/16\le R^2$. Taking the stage suprema and summing proves
\eqref{met:linear-weak-graph}. Alternatively, reflexivity supplies a
weak limit $z$ and one can use $(z_i-z)/2$. Separation then gives only
$\liminf\|(z_i-z)/2\|\ge1/2$, yielding the weaker
$D^2\ge1+k/64$; pairing two deviations avoids that loss.
The qualitative bottom-up mass-removal argument
\cite[Theorem~5.9]{TreePotentials} can also be transferred
without replacing it by~\eqref{met:weak-input}. It states that pure-tail
perturbations tend uniformly to zero as finite-head centers approach
the unit sphere, allowing the heads to vary. Thus it satisfies the
hypothesis of Lemma~\ref{met:bridge:sequential-lens}, which supplies
a uniform separated-lens gap. Proposition~\ref{met:backward} gives
$D^{-1}\le\rho_D^k$ for some $\rho_D<1$. This route uses the
qualitative input alone and requires no weak-limit extraction or
numerical modulus.
\subsection{Quadratic path cost}\label{met:quadratic-section}
On the same trees define
\[
Q_n(z)=\inf_{z=h+J_n\mu}(\|h\|_2^2+\|\mu\|_1^2)^{1/2},\qquad
X_{\rm quad}=\left(\bigoplus_{n\ge1}(\ell_2(T_n),Q_n)\right)_2.
\]
This changes the norm: the root singleton has quadratic-cost norm
$1/\sqrt2$, whereas its linear-cost norm is one. The two cancellation
estimates \cite[Theorem~7.6 and Proposition~7.10]{TreePotentials}
have the same arbitrary-displacement form, with constants $8$ and $7$:
\begin{equation}\label{met:quadratic-input}
\|(I-P)y\|\le K\sqrt{R^2-\|p\|^2},
\qquad Pp=p,\quad \|p\pm y\|\le R,\qquad K\in\{8,7\}.
\end{equation}
Finite ancestral heads have bounded finite-rank projections, and finite
coordinate vectors are dense. These facts permit the same packing
argument for either constant.
For an infinite $\varepsilon$-separated $R$-lens at arbitrary $x$, choose
a finite-support $p$ with $\|p-x\|<\eta$. Apply
\eqref{met:quadratic-input} to every deviation at radius $R+\eta$,
using the same finite-rank head $P$. Two head projections are arbitrarily
close. Hence
\[
\varepsilon\le2K\sqrt{(R+\eta)^2-\|p\|^2}.
\]
Letting $\eta$ decrease to zero gives
$\|x\|^2\le R^2-\varepsilon^2/(4K^2)$. Thus
Theorem~\ref{met:squared-transfer} gives $D^2\ge1+k/K^2$.
In particular,
\begin{equation}\label{met:quadratic-graph}
D^2\ge1+k/49.
\end{equation}
The coefficient-$8$ proof independently gives $D^2\ge1+k/64$.
Only boundedness and finite rank of $P$ entered the packing step.
The separate stopped-flow energy estimate
\cite[Proposition~7.8]{TreePotentials} has input
\[
\|p\|^2+\|y\|^2/144
\le\tfrac12(\|p+y\|^2+\|p-y\|^2),
\qquad Pp=p,\quad Py=0.
\]
It gives $r_j^2\le r_{j+1}^2-1/(144D^2)$ in upper-one normalization,
and hence the quantitative necessary condition $k\le144D^2$.
The finite-head pairing, the $R+2\eta$ endpoint error, and the limiting
step are exactly those in Proposition~\ref{met:tail-to-lens}.
\paragraph{Packing at a fixed radius.}
The same arbitrary-tail input also allows a direct finite-error packing
argument. Its coefficient-$7$ version gives, for $\alpha=1/D$ and
$\delta=\alpha^2/2000$, the backward factor $1-\delta$. Indeed, if a
unit-ball lens had center norm greater than $1-\delta$, approximate it
within $\delta$ by $p$. Then $\|p\|>1-2\delta$ and the endpoint
radius is $1+\delta$. Its common tails have norm at most
\[
7\sqrt{(1+\delta)^2-(1-2\delta)^2}
\le7\sqrt{6\delta}<\alpha/2.
\]
Two head projections within $\alpha$ contradict separation $2\alpha$.
For the pruning estimate with coefficient $9$
\cite[Proposition~7.12]{TreePotentials}, choose instead
\begin{equation}\label{met:pruning-parameter}
0<\gamma<1/2,\qquad
9\sqrt{(1+\gamma)^2-(1-2\gamma)^2}<\frac{2\alpha}{3}.
\end{equation}
The same approximation bounds every tail by $2\alpha/3$; choose two
head projections within $2\alpha/3$. Their full distance is less than
$2\alpha$, a contradiction. Thus the backward factor is $1-\gamma$.
These choices retain the bounds supplied by their respective proofs.
The threshold-removal argument \cite[Proposition~7.14]{TreePotentials}
gives a different relative-radius input:
for every nonzero $x$ and $0<\delta\le1$, a single finite
head bounds all deviations satisfying
$\|x\pm y\|\le(1+\delta)\|x\|$ by
\[
\|(I-P)y\|\le72\sqrt\delta\,\|x\|.
\]
Take $\delta=(200D)^{-2}$. If every child of a noncontracting parent
had norm at most $(1+\delta)\|x\|$, their deviations would have a
common tail bound $72\sqrt\delta D$. Two head projections can be
chosen within one, giving full distance less than
$1+144\sqrt\delta D<2$. This contradicts sibling separation. An
actual child therefore grows by the factor $1+\delta$, and
$D\ge(1+\delta)^k$.
\paragraph{A finite-codimensional input.}
The component-to-sum argument
\cite[Lemma~7.15, Proposition~7.16 and Corollary~7.17]{TreePotentials}
for the quadratic norm gives, at arbitrary unit centers, a finite-codimensional subspace
on which
\[
\max_\pm\|x\pm y\|\le1+e
\quad\Longrightarrow\quad \|y\|\le17(2e)^{1/6},
\qquad00$, choose $p$ and the paired tail
$y$ by Lemma~\ref{met:finite-pair}, with $\|y\|\ge1-\eta$.
Then
\[
M\ge\sqrt{\|p\|^2+(1-\eta)^2/64}-2\eta.
\]
On letting $\eta$ decrease to zero, $M^2\ge\|x\|^2+1/64$.
Apply this to every parent before taking the stage suprema. Summing
their squared losses gives
\begin{equation}\label{met:segment-graph}
D^2\ge1+k/64.
\end{equation}
If one instead wants to follow actual children, some child has squared
norm greater than
$\|x\|^2+1/128$: otherwise the supremum would be at most this
strictly smaller number. Selecting an actual child at each generation
retains the separate bound $D^2\ge1+k/128$.
No contractivity of $I-P_C$ was used, and no child was presumed to
attain the supremum.
\subsection{Incomparable segments}\label{met:incomparable-section}
Let $Y_I$ be the completion on the disjoint union of $T_n$ for the
supremum of Euclidean sums of segment evaluations, where the admissible
segments are pairwise incomparable: every node of one segment is
incomparable with every node of another. Put $X_I=Y_I^*$. This
restriction differs from requiring the segments merely to be disjoint.
The intrinsic estimate for $X_I$ \cite[Theorem~9.5]{TreePotentials}
is
\[
\|(I-P_E)y\|\le K\sqrt{1-\|p\|},\qquad
P_Ep=p,\quad \|p\pm y\|\le1,\quad K=\sqrt2+\sqrt6,
\]
where $E$ is finite and ancestral and $y$ may have head coordinates.
Finite ancestral projections are contractive and finite coordinate
vectors are dense. For a pure tail, homogeneity gives
$\|y\|^2\le K^2R(R-\|p\|)$ at radius $R$.
Proposition~\ref{met:tail-to-lens} and
Theorem~\ref{met:squared-transfer} therefore give
\begin{equation}\label{met:incomparable-additive}
k\le K^2D(D-1).
\end{equation}
At radius one the same lens estimate reads, for an infinite
$2\alpha$-separated family,
\[
\|x\|\le1-(\alpha/K)^2.
\]
Under lower factor $\alpha=1/D$ and upper factor one, normalize each
parent lens by its child bound $R\le1$. Its separation is at least
$2\alpha/R\ge2\alpha$, so every parent bound contracts by
$\rho=1-(\alpha/K)^2$. Hence $\alpha\le\rho^k$ by
Proposition~\ref{met:backward}. This is an application to the
incomparable-segment norm itself, regardless of the equality of its
displayed tail constant with one obtained for the disjoint-segment norm.
==== END Diamond-distortion-from-midpoint-and-tree-energies//d-applications.tex ====
==== BEGIN Diamond-distortion-from-midpoint-and-tree-energies//diamond-replacement.tex ====
\begin{figure}[!bp]
\centering
\begin{tikzpicture}[scale=0.85,every node/.style={font=\small}]
\draw[thick] (0,0)--node[above] {$\ell$}(2,0);
\fill (0,0) circle (2pt) node[left] {$u$};
\fill (2,0) circle (2pt) node[right] {$v$};
\draw[-{Stealth},thick] (2.8,0)--(4,0);
\coordinate (u) at (5,0); \coordinate (v) at (9,0);
\foreach \h/\lab in {1.4/1,0.45/2,-1.4/j}{
\draw[thick] (u)--(7,\h)--(v);
\fill (7,\h) circle (2pt) node[above] {$w_{\lab}$};
}
\node at (5.75,1.0) {$\ell/2$};
\node at (8.25,1.0) {$\ell/2$};
\node at (7,-0.5) {$\vdots$};
\fill (u) circle (2pt) node[left] {$u$};
\fill (v) circle (2pt) node[right] {$v$};
\node at (1,-1.9) {one old edge};
\node at (7,-1.9) {countably many two-edge paths};
\end{tikzpicture}
\caption{One normalized diamond replacement, with only three of the
countably many paths shown. Each new edge has length $\ell/2$.
The old endpoints remain at distance $\ell$, and any two distinct new
middle vertices also have distance $\ell$. The drawing is schematic.}
\label{fig:diamond}
\end{figure}
==== END Diamond-distortion-from-midpoint-and-tree-energies//diamond-replacement.tex ====
==== BEGIN Diamond-distortion-from-midpoint-and-tree-energies//direct.tex ====
\section{The segment forest and its atoms}\label{dir:forest-definition}
The direct graph arguments below use the same finite-height segment forest,
with James's disjoint-segment norm \cite[p.~739]{James1974}.
We first give its norm and the two forms of its atomic unit ball that the
arguments need.
Let
\[
T=\coprod_{n\ge1} T_n,\qquad T_n=\mathbb N^{\le n},
\]
where each component has its own root and is ordered by extension. A segment
is a set $[a,b]=\{t:a\le t\le b\}$ with $a\le b$ in one component.
Disjoint segments need only be disjoint as sets. A set is ancestral if it contains every ancestor of each of its vertices. On real finitely supported
arrays define
\begin{equation}\label{dir:atomic:forest-norm}
\rho(y)=
\sup_{\mathcal I}
\left(\sum_{I\in\mathcal I}\left|\sum_{t\in I}y(t)\right|^2\right)^{1/2},
\end{equation}
where the supremum runs over finite disjoint families of segments. Let $J$
be its completion and $X=J^*$. Write $e_t^*$ for the coordinate functional
and $s_I^*=\sum_{t\in I}e_t^*$, with $s_\varnothing^*=0$. We recall the elementary facts that the
argument needs, including the exact form of the atomic representation.
\begin{lemma}\label{dir:atomic:atoms}
The spaces $J$ and $X$ are reflexive, and finite coordinate arrays are dense
in both. If a set $B\subset T$ meets each segment in either a segment or the
empty set, its coordinate projection $P_B$ is contractive on both spaces.
For an antichain $V$, the coordinate space supported on $V$ has its
$\ell_2(V)$ norm in both spaces. In particular,
\begin{equation}\label{dir:atomic:atom-bound}
\left\|\sum_{I\in\mathcal I}\lambda_I s_I^*\right\|_X
\le\left(\sum_{I\in\mathcal I}\lambda_I^2\right)^{1/2}
\end{equation}
for every disjoint segment family.
Call a vector on the left with $\sum_I\lambda_I^2\le1$ an atom. The unit
ball of $X$ is the norm-closed convex hull of the atoms. If $S$ is finite
and ancestral, the unit ball of $P_SX$ is the convex hull of atoms
whose segments lie in $S$; every point has a finite convex representation.
\end{lemma}
\begin{proof}
Singleton tests give the lower estimate in
\[
\|y\|_2\le\rho(y)\le\sqrt{n+1}\,\|y\|_2
\qquad(\operatorname{supp}y\subset T_n).
\]
For the upper estimate apply Cauchy--Schwarz on every segment and use its
length at most $n+1$, followed by disjointness. Norm squares add over the
components: one inequality follows by restricting each segment family to
each component, and the reverse inequality follows by combining families
that approximate the component suprema. Thus $J$ is the $\ell_2$-sum of
spaces isomorphic to $\ell_2(T_n)$. The dual of an $\ell_2$-sum is the
$\ell_2$-sum of the duals: Cauchy--Schwarz gives the upper norm bound and
finite coordinate tests give the reverse bound. Applying this description
to the bidual proves reflexivity, and truncation first in components and
then in coordinates proves density in $J$ and $X$.
Intersecting a disjoint segment family with $B$ proves contractivity on
$J$, hence on $X$ by duality. This applies to ancestral sets, their
complements, differences of nested ancestral sets, and antichains. Each
segment meets an antichain in at most one point, so singleton tests identify
its $\rho$ norm with $\ell_2$; duality and the contractive projection give the
same identification in $X$. Equation~\eqref{dir:atomic:atom-bound} follows
directly from Cauchy--Schwarz and \eqref{dir:atomic:forest-norm}.
The supremum of pairing with the atoms equals the right side of
\eqref{dir:atomic:forest-norm}. If their norm-closed convex hull omitted a
point of the unit ball, Hahn--Banach separation would give a separating
functional in $X^*=J$, contradicting this equality. Here reflexivity is
used to identify the separating functional with an element of $J$. For
finite ancestral $S$, the same argument is finite-dimensional: projection
clips atoms to atoms in $S$, and the family of atoms in $S$ is compact,
being a finite union of continuous images of finite-dimensional Euclidean
unit balls. Its convex hull is compact and equals the coordinate unit
ball. Every point therefore has a finite convex representation.
\end{proof}
For ancestral $S$ put $Q_S=I-P_S$. Finite-support density and
contractivity imply that, for a finite collection of vectors and any
$\eta>0$, a finite ancestral $S$ makes every $Q_S$-tail smaller than
$\eta$. Indeed, take finite approximations and add all ancestors of their
supports; each vertex has only finitely many ancestors.
\section{Direct descent with primal tests}\label{dir:accumulated}
The next two arguments follow actual edges of a diamond and construct a
test vector at each step. Earlier diamond obstructions include
Lee--Naor's telescoping squared-distance inequality
\cite[Lemma~2.2]{LeeNaor2004} and Johnson--Schechtman's iterated
modulus-of-convexity estimate \cite[Section~4]{JohnsonSchechtman2009}.
Here the accumulated evaluation is compared with the cost of its primal
test. The first argument pays a bounded squared-norm cost for
each increase in its evaluation. The second balances prefix sums of
comparable sizes and controls the final test all at once.
\subsection{The graph slopes}\label{dir:graph-slopes}
For an oriented parent edge of represented length $\ell$, use the
slope $x$ and deviations $z_j$ of Section~\ref{met:graph-definition}:
\begin{equation}\label{dir:splitting}
x=\frac{f(b)-f(a)}\ell,\qquad
z_j=\frac{2f(m_j)-f(a)-f(b)}\ell,\qquad
\text{child slopes }x+z_j,\ x-z_j.
\end{equation}
In the noncontracting normalization all slopes have norm at most $D$
and the deviations are $2$-separated. After division by $D$, these
bounds are one and $2/D$. The symmetric splitting is the finite-level
version of the sprawling-tree construction in
\cite[Definition~3.7 and Proposition~6.4]{BassetLancienProchazka2026}.
The selections below use the same compactness step as
Lemma~\ref{met:finite-pair}: two bounded projections onto a fixed finite
head can be made arbitrarily close. Their separated full vectors then
have a large tail difference, so at least one has a large tail. This
choice requires the infinitely many middle vertices. Once a middle
vertex has been selected, either sign in~\eqref{dir:splitting} follows
an actual child edge.
\subsection{A correction at the first fresh roots}
The correction controls the extra squared cost of segments crossing
the head boundary. Its value is placed at each first exterior vertex.
For crossings with a large head sum, the proof either makes the cross
term nonpositive or controls it by splitting the head portion.
\begin{lemma}\label{dir:root-correction}
Suppose $v,u\in c_{00}(T)$, $v=P_Av$, and $u=Q_Au$, where $A$ is finite
and ancestral. There is $c=Q_Ac\in c_{00}(T)$ such that
\begin{equation}\label{dir:correction-cost}
\rho(c)\le\rho(u),\qquad
\rho(v+c+r u)^2\le\rho(v)^2+16\rho(u)^2
\quad(r\in\{-1,1\}).
\end{equation}
\end{lemma}
\begin{proof}
The complement of $A$ is the disjoint union of subtrees $U_d$ whose
roots are its first vertices. Put $u_d=u|_{U_d}$ and $a_d=\rho(u_d)$.
Independent segment tests in these subtrees give
$\sum_d a_d^2=\rho(u)^2$; only finitely many terms are nonzero.
If $d$ has parent $p\in A$, examine the suffix sums
$V_d(h)=\sum_{q\in[h,p]}v(q)$. If some suffix has absolute sum greater
than $2a_d$, choose the shortest such suffix, with initial vertex $h_d$, and set
\[
c_d=-\operatorname{sign}(V_d(h_d))a_d.
\]
Otherwise set $c_d=0$, also when $d$ has no parent in $A$. Let
$c=\sum_dc_de_d$. The roots $d$ are incomparable, so
$\rho(c)^2=\sum_dc_d^2\le\rho(u)^2$.
Fix a disjoint segment family testing $v+c+ru$. Segments wholly outside
$A$ cost at most $4\sum_da_d^2$ in squared sum, since
$\rho(c_de_d+ru_d)\le2a_d$. A crossing into $U_d$ has head sum $P$ and
tail sum $Q$, with $|Q|\le2a_d$. We claim that
\[
(P+Q)^2\le E+12a_d^2,
\]
where $E$ is the squared-sum test of one or two disjoint subsegments of
the original head portion. If $|P|\le2a_d$, use $E=P^2$, since
$2|PQ|+Q^2\le12a_d^2$. Otherwise the preferred suffix exists. If its
sum has the same sign as $P$, the choice of $c_d$ gives $PQ\le0$, and
again $E=P^2$ suffices. In the remaining case let $P'=V_d(h_d)$.
The preferred suffix lies properly inside the head portion: the latter
qualifies, and the former is shortest and has the opposite sign.
Splitting there gives
\[
E=(P-P')^2+(P')^2\ge P^2+4|P|a_d\ge P^2+2PQ.
\]
Adding $Q^2\le4a_d^2$ proves the claim in this case as well.
All the replacement head segments remain disjoint. Their total squared
test, including the unchanged segments in $A$, is at most $\rho(v)^2$.
At most one crossing uses each fresh root $d$. Combining the crossing
cost $12\sum_da_d^2$ and the wholly exterior cost proves
\eqref{dir:correction-cost}.
\end{proof}
\begin{proposition}\label{dir:deterministic-descent}
Every embedding of $\mathcal D_k$ into $X$ with distortion at most $D$
satisfies $k\le256D^2$.
\end{proposition}
\begin{proof}
Rescale to upper Lipschitz constant one and lower constant $1/D$.
Start at the initial edge with test $v=0$. At a current edge of slope
$x$, enlarge a finite ancestral $A$ to contain the support of $v$ and
make $\|Q_Ax\|<1/(4D)$. In \eqref{dir:splitting}, choose two deviations
whose head projections are less than $1/(2D)$ apart. Since the
deviations are $2/D$-separated, at least one of them, denoted $z$, has
$\|Q_Az\|>1/(2D)$. Finite-support density in $J$ and contractivity of
$Q_A$ supply $u=Q_Au\in c_{00}(T)$ with
$\rho(u)\le1$ and $z(u)>1/(2D)$.
Apply Lemma~\ref{dir:root-correction}. Because $\rho(c)\le1$,
$|x(c)|<1/(4D)$. Averaging the two matching evaluations gives
\[
\frac12\sum_{r=\pm1}(x+rz)(v+c+ru)
=x(v)+x(c)+z(u)>x(v)+\frac1{4D}.
\]
Choose a sign attaining at least this average, follow that actual child
edge, and replace $v$ by $v+c+ru$. Each step increases $\rho(v)^2$ by
at most $16$. After $k$ steps the evaluation is at least $k/(4D)$,
whereas the final slope has norm at most one and the final test has
norm at most $4\sqrt{k}$. Hence $k\le256D^2$.
\end{proof}
\subsection{Balancing dyadic sizes along a random descent}
The next construction does not control the increase of the test norm
at an individual step. It controls prefix sums separately at every
dyadic size, then bounds any segment by charging its largest size to a
root through which that segment passes.
\begin{proposition}\label{dir:bin-descent}
Every noncontracting $D$-Lipschitz embedding of $\mathcal D_k$ into $X$
satisfies $D>\sqrt{k}/144$ for $k\ge1$.
\end{proposition}
\begin{proof}
We descend at random, choosing a fair sign after selecting a middle
vertex. Write $x$ for the current slope and $G$ for the accumulated
test, initially zero. Along a fixed history, the preceding tests lie
in finite blocks $Q\setminus P$ with ancestral endpoints, and every
new $P$ contains the previous $Q$. Enlarge $P$ until
$\|Q_Px\|<1/32$. By the separation in \eqref{dir:splitting} and
finite-dimensional total boundedness, choose a deviation $z$ with
$\|Q_Pz\|\ge3/4$. There is a finite test $g$ supported in a block
$Q\setminus P$ such that
\begin{equation}\label{dir:bin-witness}
\rho(g)\le1,\qquad z(g)>1/2.
\end{equation}
Here $Q$ is a finite ancestral enlargement containing the support of
$g$; both choices are made before tossing the next sign.
Decompose the block into its connected tree components $C$. If $r_C$
is the root of $C$, define
\[
p_C(v)=\sum_{w\in[r_C,v]}g(w),\qquad
A_C=\max_{v\in C}|p_C(v)|.
\]
Ignore components with $A_C=0$. Choose a power of two $B_C$ with
$A_C\le B_C<2A_C$. Disjoint root-prefix tests give
$\sum_CA_C^2\le1$. Differencing the potentials
$\max(p_C,0)$ and $\min(p_C,0)$ gives arrays $g_C^+,g_C^-$ with
$g|_C=g_C^++g_C^-$. On every subsegment, either array has sum in
absolute value at most the absolute value of the corresponding sum of
$g$, since both scalar truncations are 1-Lipschitz.
To state the balancing rule precisely, call a component earlier if it
was selected before the current update. Label each earlier component by
its bin size and, for every vertex $v$ and dyadic size $B$, put
\[
S_B(v)=\sum_{\substack{t\preceq v:\ t\text{ lies in an earlier}\\
\text{component of bin size }B}}G(t).
\]
The invariant is $|S_B(v)|\le B$ for every $v$. Before adding a new
component $C$, the sum of earlier bin-$B_C$ contributions above its root
$r_C$ is fixed; choose a weak sign opposite to this sum, with either
choice at zero. Allow the new test on $C$ to be zero or either of
$g_C^+,g_C^-$ oriented so that all its prefix sums have this chosen sign.
Each such prefix has magnitude at most $B_C$.
This preserves the invariant at every vertex. All earlier supports lie
in $P$, so their prefix contribution is constant through $C$. Adding
an opposite-sign quantity of magnitude at most $B_C$ keeps its absolute
value at most $B_C$. In a gap there is no new contribution; an ignored
component has $g|_C=0$. Beyond $Q$, the new prefix is the one at its last
ancestor in $Q$, or zero when there is no such ancestor. Thus the same
bound holds after every update, for every ancestor path.
The sign restriction is fixed before the next fair sign $s\in\{-1,1\}$.
For each $C$, choose an allowed option maximizing $s z$ on that
component; let $w$ be the resulting increment. The choice depends on
$s$. Nevertheless $\rho(w)\le1$: every tested segment meets this block
in a segment in one component, and the preceding contraction estimate
applies simultaneously to disjoint tests. If $a,b$ are the evaluations
under $z$ of the two oriented offers in a component, then
\[
\mathbb E_s\max(0,sa,sb)
\ge\frac{|a|+|b|}{4}
=\frac{|z(g_C^+)|+|z(g_C^-)|}{4}
\ge\frac{z(g|_C)}4.
\]
Sum over $C$ and use \eqref{dir:bin-witness} to obtain
$\mathbb E_s[s z(w)]>1/8$. Also $|x(w)|<1/32$ and
$\mathbb E_s[s z(G)]=0$. Thus the next actual slope $x+sz$ satisfies
\[
\mathbb E_s[(x+sz)(G+w)]>x(G)+1/8-1/32>x(G)+1/16.
\]
After $k$ steps the expected evaluation exceeds $k/16$. The entire
probability space has only $2^k$ outcomes.
It remains to bound $\rho(G)$ for every final outcome. Fix any finite
disjoint segment family. On a segment, first separate the portion
lying in the component that contains its initial vertex, if this is
a nonignored component. Across all segments these initial portions
have squared-sum total at most $k$: grouping them by construction step
gives disjoint tests of an increment of norm at most one.
The remainder of a segment enters every nonignored component it meets
at its root, since all blocks have ancestral endpoints and are
successively outside previous blocks. For a bin of size $B$, its sum
on this remainder is a difference of two values of $S_B$, hence
has magnitude at most $2B$. A remainder meeting no nonignored component
contributes zero. Otherwise, if $B_*$ is the largest encountered bin,
geometric summation gives a total magnitude at most $4B_*$. Charge the
square of this bound to an encountered component with size $B_*$.
Two disjoint segments cannot charge the same component: both would
contain its root. Since
\[
\sum_{\text{all components}}B_C^2
\le4\sum_{\text{steps}}\sum_CA_C^2\le4k,
\]
the remaining portions have squared-sum total at most $64k$.
The Euclidean triangle inequality for the vector of segment sums now
gives $\rho(G)\le\sqrt{k}+8\sqrt{k}=9\sqrt{k}$.
Every final slope has norm at most $D$, so every final evaluation is at
most $9D\sqrt{k}$. Comparison with its expectation gives
$k/16<9D\sqrt{k}$, as claimed.
\end{proof}
\section{Direct graph obstructions from atomic prediction energy}
\label{dir:atomic:section}
We give two direct proofs that the countably branching diamonds do not embed
with uniformly bounded distortion into the finite-height segment-forest dual.
Both proofs first select finitely many edge slopes from the infinitely
branching graph. An atomic representation then turns these slopes into
conditional means. The first proof counts segment endpoints and predicts the
coefficients of segments crossing a long interval of layers. The second
counts segment starts and follows each segment until it terminates. Their
energy estimates concern these finite martingales; no midpoint-modulus
estimate is needed.
\begin{theorem}\label{dir:atomic:obstruction}
For the real space $X$ defined by \eqref{dir:atomic:forest-norm}, no finite
$D$ has the following property: for every $k\ge1$ there are a map
$f_k:\mathcal D_k\to X$ and a scalar $a_k>0$ such that
\[
a_k d_k(u,v)\le\|f_k(u)-f_k(v)\|_X\le D a_k d_k(u,v)
\qquad(u,v\in\mathcal D_k).
\]
\end{theorem}
Rescaling by $D a_k$ reduces the theorem to excluding maps satisfying
\begin{equation}\label{dir:atomic:normalization}
D^{-1}d_k(u,v)\le\|f(u)-f(v)\|_X\le d_k(u,v).
\end{equation}
Necessarily $D\ge1$. The graph normalization and old-vertex distance
calculation in Section~\ref{met:graph-definition} apply throughout.
Orient the edges consistently and write their normalized slopes as in
\eqref{met:slopes}. They have norm at most one, and each retained parent
has the symmetric children
\begin{equation}\label{dir:atomic:splittings}
x+d_j,\quad x-d_j,\qquad
\|d_j-d_l\|_X\ge 2/D\quad(j\ne l).
\end{equation}
At every retained edge choose one replacement path, retain both its
slopes, and then descend by fair binary choices. The resulting random
slopes $M_0,\ldots,M_k$ form a martingale, because the two children average
to their parent. Let $\mathcal F_i$ record the first $i$ binary choices.
Every history at level $i$ has probability $2^{-i}$, and each history has
two children of conditional probability $1/2$. Infinite branching is
used only to select a suitable replacement path before this finite
probability space is formed.
\begin{lemma}[A finite slope martingale and its exact atomic law]
\label{dir:atomic:preparation}
Suppose a depth-$k$ embedding satisfies~\eqref{dir:atomic:normalization}.
For every $0<\eta<1/(4D)$ one can select a fair binary slope martingale
$(M_i)_{i=0}^k$, nested finite ancestral sets $S_0\subseteq\cdots\subseteq S_k$,
and a finitely valued random atom $Z$ supported in $S_k$, such that, with
$E_i=\mathbb E[\,\cdot\mid\mathcal F_i]$ and $U_i=E_iZ=P_{S_k}M_i$,
\begin{equation}\label{dir:atomic:preparation-bounds}
\|Q_{S_{i-1}}U_{i-1}\|<\eta,\qquad
\|P_{S_i\setminus S_{i-1}}U_i\|\ge1/D-4\eta
\quad(1\le i\le k).
\end{equation}
The unprojected slopes also satisfy $\|Q_{S_i}M_i\|<\eta$ for
$0\le i\le k$. Every $M_i$ and $U_i$ has norm at most one. The additional terminal
sampling used to produce $Z$ is excluded from each $\mathcal F_i$,
including $\mathcal F_k$.
\end{lemma}
\begin{proof}
Choose a finite ancestral $S_0$ with $\|Q_{S_0}M_0\|<\eta$.
Suppose that the selected level-$(i-1)$ slopes have tails smaller than
$\eta$ outside a common finite ancestral $S_{i-1}$. For each such parent
$x$, the first-child slopes $x+d_j$ in \eqref{dir:atomic:splittings} have
bounded projections to the finite-dimensional space $P_{S_{i-1}}X$.
Two distinct projections are within $2\eta$. Their tail difference has
norm at least $2/D-2\eta$, so one first-child tail has norm at least
$1/D-\eta$. Retain that path. Since its two slopes add to $2x$, the
mate's tail has norm at least $1/D-3\eta$.
Only finitely many slopes have now been selected. Enlarge $S_{i-1}$ to a
finite ancestral $S_i$ making all their tails smaller than $\eta$.
Writing $B_i=S_i\setminus S_{i-1}$, both children satisfy
\[
\|P_{B_i}M_i\|\ge1/D-4\eta.
\]
Repeat through level $k$. All sets are common to an entire level, not
chosen after its final random history.
The finite-support martingale $P_{S_k}M_i$ has norm at most one.
By Lemma~\ref{dir:atomic:atoms}, at every terminal binary history its value
is a finite convex combination of atoms supported in $S_k$. Enlarge the
finite sample space by selecting one atom from that combination and call
it
\[
Z=\sum_{I\in\mathcal I}a_I s_I^*,\qquad \sum_{I\in\mathcal I}a_I^2\le1.
\]
The extra terminal sampling is excluded from every $\mathcal F_i$, even
$\mathcal F_k$. With $E_i=\mathbb E[\,\cdot\mid\mathcal F_i]$, the tower
property gives $E_iZ=P_{S_k}M_i$. The projections commute, and
$S_i\subseteq S_k$, so the displayed
layer lower bounds also hold for $U_i$. Moreover,
$Q_{S_{i-1}}U_{i-1}=P_{S_k}Q_{S_{i-1}}M_{i-1}$; contractivity gives
the first bound. This proves~\eqref{dir:atomic:preparation-bounds}.
\end{proof}
The lemma supplies the same finite martingale for two different energy
arguments. In either argument a large child layer is decomposed into
three terms: segments charged to the chosen interval, coefficients
already predicted from the parent, and the change in their prediction.
The first term has a counting budget, the second is controlled by the
small parent tail, and scalar $L^2$ orthogonality bounds the third.
\subsection{Endpoint intervals and crossing predictions}
\label{dir:atomic:endpoint-route}
We prove Theorem~\ref{dir:atomic:obstruction} first by charging the two
endpoints of every terminal atom. Assume that \eqref{dir:atomic:normalization}
is available for every $k$, with $D$ fixed, and set
\begin{equation}\label{dir:atomic:endpoint-constants}
\eta=\frac1{32D},\qquad c=\frac1{2D}.
\end{equation}
Apply Lemma~\ref{dir:atomic:preparation} with this $\eta$, write
$B_i=S_i\setminus S_{i-1}$, and write its random atom as
\[
Z=\sum_{I\in\mathcal I}w_I s_I^*,\qquad \sum_Iw_I^2\le1.
\]
Since $1/D-4\eta>c$, we have on every binary history
\begin{equation}\label{dir:atomic:endpoint-layer}
\|Q_{S_{i-1}}E_{i-1}Z\|<\eta,\qquad
\|P_{B_i}E_iZ\|\ge c
\quad(1\le i\le k).
\end{equation}
Take $k=ML$ and partition the layers into $M$ consecutive intervals of
length $L$. A segment has at most two endpoints, so the sum, over these
intervals, of its squared coefficient charged to an interval containing
an endpoint is at most $2w_I^2$. Consequently some interval $p\le i\le q$,
$q-p+1=L$, has
\[
\mathbb E\sum_{\substack{I:\text{an endpoint of }I\text{ lies in}
\,S_q\setminus S_{p-1}}}w_I^2\le\frac2M.
\]
Call these segments endpoint segments and let $Z^{\rm end}$ be their
sum. Put $b_0=\sqrt{2/M}$. By \eqref{dir:atomic:atom-bound},
\begin{equation}\label{dir:atomic:endpoint-error}
\|Z^{\rm end}\|_{L^2(X)}\le b_0.
\end{equation}
Here $\|V\|_{L^2(X)}=(\mathbb E\|V\|_X^2)^{1/2}$; contractive
projections and conditional expectations preserve upper bounds in this
norm, by Jensen's inequality.
Every other segment meeting a layer in this interval begins in
$S_{p-1}$ and ends outside $S_q$: heredity puts its first endpoint in
$S_q$, and the absence of either endpoint in $S_q\setminus S_{p-1}$
then forces these two alternatives. Call such segments crossing
segments and write $Z^{\rm cr}$ for their sum. On every relevant layer
only $Z^{\rm end}$ and $Z^{\rm cr}$ contribute.
For $p\le i\le q+1$ let $A_i$ be the minimal vertices of
$S_k\setminus S_{i-1}$, and put $A_i'=A_i\cap S_i$ for $i\le q$.
These are finite antichains. Each crossing segment meets every $A_i$
in exactly one vertex, since it begins in $S_{p-1}$ and ends outside
$S_q$. Label it by its vertex $r\in A_p$. Disjointness allows at most
one crossing segment with a given label. Let $W_r$ be its coefficient,
and $d_i^r$ its vertex in $A_i$; in the absence of such a segment put
$W_r=0$ and $d_i^r=\dagger$ for every $i$.
A later location determines an earlier one by taking its unique
$A_i$-ancestor, with $\dagger$ mapping to $\dagger$. For each fixed
$r$, the sigma-algebras
\[
\mathcal G_{i-1}^r=\mathcal F_{i-1}\vee\sigma(d_i^r)
\qquad(p\le i\le q+1)
\]
therefore increase with $i$. Define $H_{i-1}^r=\mathbb E[W_r\mid
\mathcal G_{i-1}^r]$. Scalar conditional expectation is orthogonal
projection in $L^2$, whence
\begin{equation}\label{dir:atomic:crossing-energy}
\sum_{i=p}^q h_i^2\le1,\qquad
h_i^2:=\sum_{r\in A_p}\mathbb E|H_i^r-H_{i-1}^r|^2.
\end{equation}
Indeed the sum for each $r$ telescopes to at most $\mathbb E W_r^2$,
and the sum of those quantities is at most one. Different labels may
use different sigma-algebras; no joint refinement across labels is
being asserted.
Let $J_i^r$ be the part of the crossing segment with label $r$ in $B_i$,
empty if the label is absent. It is determined by $d_{i+1}^r$: the
previous location is recovered from it, and the portion runs from
$d_i^r$ up to but excluding $d_{i+1}^r$. If these locations agree the
portion is empty. Therefore
\[
E_i\sum_r W_r s_{J_i^r}^*=E_i\sum_r H_i^r s_{J_i^r}^*.
\]
Replacing $H_i^r$ by $H_{i-1}^r$ in this expression costs at most $h_i$
in $L^2(X)$, by disjointness, \eqref{dir:atomic:atom-bound}, and
conditional Jensen. Together with \eqref{dir:atomic:endpoint-error},
this reduces the layer estimate to bounding
\[
V_i=E_i\sum_r H_{i-1}^r s_{J_i^r}^*.
\]
Fix a history through time $i-1$. For $v\in A_i'$, let $r(v)$ be its
unique $A_p$-ancestor and let $m(v)$ be the mean of $W_{r(v)}$
conditional on this history and $d_i^{r(v)}=v$. Set it to zero on an
empty conditional event. Write
\[
p_j(v)=\Pr(d_i^{r(v)}=v\mid\mathcal F_j),\qquad j=i-1,i.
\]
On either child history, $p_i(v)\le2p_{i-1}(v)$, because that history
has half its parent's probability. The contribution from $v$ to $V_i$
is $m(v)p_i(v)$ times an average of segment functionals starting at $v$.
The descendant subtrees at different $v$ are disjoint. Independently
sampling one segment from each of these averages and applying
\eqref{dir:atomic:atom-bound} before averaging proves
\[
\|V_i\|_X\le
\left(\sum_{v\in A_i'}|m(v)p_i(v)|^2\right)^{1/2}
\le2\left(\sum_{v\in A_i'}|m(v)p_{i-1}(v)|^2\right)^{1/2}.
\]
The products in the final sum are precisely the coordinates of
$E_{i-1}Z^{\rm cr}$ on $A_i'$. The antichain norm, the parent-tail
bound in \eqref{dir:atomic:endpoint-layer}, and
\eqref{dir:atomic:endpoint-error} now give
\[
\|V_i\|_{L^2(X)}\le2(\eta+b_0).
\]
Combining the three contributions yields, for every $p\le i\le q$,
\begin{equation}\label{dir:atomic:endpoint-final}
c\le2\eta+3\sqrt{2/M}+h_i.
\end{equation}
By \eqref{dir:atomic:crossing-energy}, some $h_i\le L^{-1/2}$.
Since $2\eta=c/80,\quad3\eta<\frac{c_0}{2},\quad4\eta<\frac\delta4.
\end{equation}
Apply Lemma~\ref{dir:atomic:preparation}, put $H_i=S_i$ and
$J_i=H_{i+1}\setminus H_i$, and write
\[
Z=\sum_{I\in\mathcal I}\lambda_I s_I^*,\qquad
\sum_I\lambda_I^2\le1,\qquad U_i=E_iZ=P_{S_k}M_i.
\]
The common preparation gives the exact conditional-mean identity.
Since $1/D=2c_0$ and $\eta\delta.
\end{equation}
We keep this slack in the estimates below. The distinction from the
endpoint argument will be that segments may terminate inside the chosen
interval; their last vertices must therefore remain in the prediction
information.
Take $k=NL$ and divide its steps into intervals $r\le i4/\delta^2$ first. Then choose $N$ large enough that
$3/\sqrt N<\delta/4$, and take $k=NL$. Some interval satisfies
$b_r\le1/N$, since $\sum_r b_r\le1$. For this interval,
\eqref{dir:atomic:status-constants} and \eqref{dir:atomic:status-final}
force $\sqrt{g_i}>\delta/2$ at every step. This contradicts
\eqref{dir:atomic:status-energy}, because
\[
\sum_{i=r}^{r+L-1}g_i>L\delta^2/4>1.
\]
This completes the second proof of
Theorem~\ref{dir:atomic:obstruction}.
The two estimates retain different information. In the endpoint proof,
the selected interval has endpoint energy at most $2/M$, all surviving
segments cross the entire interval, and the coefficient predictions are
scalar martingales for each boundary label. In the start proof, total
start energy is at most one, segments may terminate inside the interval,
and prediction uses their refining active or inactive statuses on a
finite measure space. These are direct obstruction arguments with
their respective parameter choices; neither calculation above asserts
an asymptotic midpoint modulus.
\section{Fresh coordinate layers along diamond slopes}\label{dir:layers}
We now use the linear path-cost space. Its intrinsic estimates control
either a fresh part of a finite martingale or the squared-norm increase
of a symmetric split. The graph arguments below supply precisely those
inputs, with their separate support and measurability conditions.
For $T_n=\mathbb N^{\le n}$, including its root, let
$b_t=\sum_{s\preceq t}e_s$ and
\[
L_n(z)=\inf_{z=h+\sum_t\alpha_tb_t}
\bigl(\|h\|_{\ell_2(T_n)}+\|\alpha\|_{\ell_1(T_n)}\bigr),
\qquad
X_{\rm lin}=\left(\bigoplus_{n\ge1}(\ell_2(T_n),L_n)\right)_2.
\]
The infimum ranges over real $h\in\ell_2(T_n)$ and
$\alpha\in\ell_1(T_n)$; it need not be attained. The intrinsic estimates
below allow arbitrarily close decompositions. This definition is the
linear path-cost model of
\cite[Propositions~5.1 and~5.2]{TreePotentials}. In particular, that
paper proves the equivalence with the bounded global-test norm on each
finite-height component. No infinite-height completion is substituted
for this Hilbert sum.
For a finite ancestral set $A$ in the forest, $P_A$ is contractive on
$X_{\rm lin}$: it restricts a Hilbert vector and truncates every $b_t$ to another
root path or zero. Finite coordinate vectors are dense because
$\|z\|_2/\sqrt{n+1}\le L_n(z)\le\|z\|_2$ on each component.
Thus the $P_A$ approximate the identity strongly. For $A\subseteq B$,
the layer projection is $P_{B\setminus A}=P_B-P_A$ and has norm at most
two. The tail projection $I-P_A$ also has norm at most two.
We use three forms of coordinate selection. Exact graph slopes can be
paired with windows chosen at the current history, or with one head
fixed across a whole level. Estimates requiring exact support instead
use finite approximations: the approximating slopes retain the exact
parent--child averaging identity, and the next support is fixed before
the next sign. We give each intrinsic estimate immediately before the
construction that meets its hypotheses.
Throughout, the probability space is a uniform finite sign cube,
$\mathcal F_i$ records its first $i$ signs, and a dyadic martingale
satisfies $W_i=\mathbb E[W_m\mid\mathcal F_i]$. The estimates are the
separate sign-prediction, variation, and stopping arguments proved in
\cite{TreePotentials}; no constant is transferred between these proofs.
\subsection{Exact slopes with adaptive windows}\label{dir:adaptive-windows}
The weighted sign-prediction estimate
\cite[Theorem~6.4]{TreePotentials} supplies an absolute constant
$C_{\rm sign}$ such that, for finite ancestral windows
$A_i\subseteq B_i$ measurable with respect to $\mathcal F_i$, with
$B_i\subseteq A_{i+1}$ whenever $i+11/4
\]
on either outcome. For the parent,
$P_{B_i\setminus A_i}W_i=P_{B_i}(I-P_{A_i})W_i$, so its layer norm is
at most $2^{-k}$. All windows are known before the next sign and are
disjoint along each complete history. Applying
\eqref{dir:sign-input} therefore gives
\begin{equation}\label{dir:adaptive-conclusion}
\frac{k}{16}\le C_{\rm sign}(D^2+k2^{-2k})
\le C_{\rm sign}(D^2+1).
\end{equation}
The constant remains independent of $k$ and $D$.
\subsection{Exact slopes with a common head at each level}
\label{dir:deterministic-heads}
For fixed finite ancestral $A_0\subseteq\cdots\subseteq A_m$, with
$E_i=A_i\setminus A_{i-1}$, the deterministic-head leakage estimate
\cite[Theorem~6.5]{TreePotentials} is
\begin{equation}\label{dir:leakage-input}
\sum_{i=1}^m\mathbb E\|P_{E_i}W_i\|^2
\le1024\left(\mathbb E\|W_m\|^2+
\sum_{i=1}^m\mathbb E\|P_{E_i}W_{i-1}\|^2\right).
\end{equation}
It imposes no support restriction on the martingale.
Return to the noncontracting normalization and keep the true slopes
$W_i$, so $\|W_i\|\le D$. Put
$\varepsilon=\min\{1/8,1/\sqrt{k}\}$. Once the finitely many slopes at
level $i$ have been selected, choose a single finite ancestral $A_i$
containing $A_{i-1}$ when $i\ge1$ and satisfying
\[
\|(I-P_{A_i})W_i\|\le\varepsilon
\]
for every history at that level. At each current edge the first-child
slopes are pairwise $2$-separated. Two have head projections less than
$1/2$ apart; one therefore has tail norm at least $3/4$. Keep that
middle vertex and its two half-edges as the next fair choice. After all
choices at this level, enlarge the common head to approximate all next
slopes as well.
For $E_{i+1}=A_{i+1}\setminus A_i$, the first-child outcome satisfies
$\|P_{E_{i+1}}W_{i+1}\|\ge3/4-\varepsilon\ge1/2$. This outcome has
conditional probability one half. On all outcomes the parent layer
norm is at most $\varepsilon$, because
$P_{E_{i+1}}W_i=P_{A_{i+1}}(I-P_{A_i})W_i$.
The sets $A_i$ are fixed across each level, exactly as required by
\eqref{dir:leakage-input}. It follows that
\begin{equation}\label{dir:leakage-conclusion}
\frac{k}{8}\le1024(D^2+k\varepsilon^2)
\le1024(D^2+1).
\end{equation}
\subsection{Supported approximations with predictable heads}
\label{dir:predictable-application}
The predictable-layer estimate \cite[Theorem~6.1]{TreePotentials}
applies when $P_0$ is a fixed finite ancestral set,
$P_i$ is finite ancestral and $\mathcal F_{i-1}$-measurable for $i\ge1$,
$P_{i-1}\subseteq P_i$, and $W_i=P_{P_i}W_i$. If
$\|W_m\|\le C$ on every outcome, it gives
\begin{equation}\label{dir:predictable-input}
\sum_{i=1}^m\mathbb E\|P_{P_i\setminus P_{i-1}}W_i\|^2\le495C^2.
\end{equation}
For the estimate with constant $495$, keep the noncontracting
normalization. Set $\alpha=1/[8(k+1)]$. Approximate the initial true
slope within $\alpha$ by a finite vector $W_0$ supported on a finite
ancestral $P_0$. At a history of length $i-1$, choose from the true
edge a deviation $z$ with
\[
\|(I-P_{P_{i-1}})z\|\ge3/4.
\]
To do so, take two of the $2$-separated deviations with projected
difference less than $1/2$, and use the triangle inequality for their
tails. Approximate $z$ within $\alpha$ by a finite vector $v_i$ and
choose a finite ancestral $P_i\supseteq P_{i-1}$ containing its support.
For the next fair sign put $W_i=W_{i-1}\pm v_i$ and follow the
corresponding actual half-edge.
The sets $P_i$ and vectors $v_i$ are fixed before this sign. Consequently
the $W_i$ form an exact dyadic martingale, are supported in $P_i$, and
differ from their true slopes by less than $(i+1)\alpha$. In particular,
$\|W_k\|\le D+1$. Since the parent has no coordinates outside
$P_{i-1}$, on either sign
\[
\|P_{P_i\setminus P_{i-1}}W_i\|
=\|(I-P_{P_{i-1}})v_i\|
\ge3/4-2\alpha\ge1/2.
\]
The factor two uses only the bound for $I-P_{P_{i-1}}$.
Every hypothesis of \eqref{dir:predictable-input} now holds, giving
\begin{equation}\label{dir:495-conclusion}
\frac{k}{4}\le495(D+1)^2.
\end{equation}
Under the same predictable support hypotheses, the separate
variation-and-mass-removal estimate
\cite[Theorem~6.3]{TreePotentials} gives
\begin{equation}\label{dir:variation-input}
\sum_{i=1}^m\mathbb E\|P_{P_i\setminus P_{i-1}}W_i\|^2\le252C^2.
\end{equation}
In particular its bound is $1008$ when $C=2$.
The variation estimate has a useful different graph normalization.
Rescale the graph to upper constant one and lower constant $1/D$, and
put $\varepsilon=\min\{1/(4D),1/(k+1)\}$. Approximate the initial
slope within $\varepsilon$ as before. Given a finite head space
$E=\operatorname{ran}P_{P_{i-1}}$, some deviation has distance from $E$
greater than $1/(2D)$. Otherwise all deviations would have bounded
approximants in $E$ with errors less than $3/(4D)$, and two close
approximants would contradict their separation $2/D$.
Choose this deviation, approximate it by a finite $v_i$ with error less
than $\varepsilon$, and update $W_i=W_{i-1}\pm v_i$ using a predictable
ancestral enlargement $P_i$. The cumulative error is less than
$(i+1)\varepsilon$, and hence $\|W_k\|\le2$. Moreover,
\[
\|P_{P_i\setminus P_{i-1}}W_i\|
=\|v_i-P_{P_{i-1}}v_i\|
\ge\operatorname{dist}(v_i,E)>\frac1{4D}.
\]
Using \eqref{dir:variation-input} with $C=2$ proves
\begin{equation}\label{dir:1008-conclusion}
\frac{k}{16D^2}\le1008.
\end{equation}
The distance-to-head argument avoids requiring the tail projection to
be contractive.
\subsection{A telescoping squared-gap argument}\label{dir:gap-application}
For finite ancestral $A\subseteq A'$, $x=P_Ax$, and $y=P_{A'}y$, the
fresh-layer squared-gap estimate
\cite[Theorem~5.11]{TreePotentials} is
\begin{equation}\label{dir:gap-input}
\|P_{A'\setminus A}y\|^2\le
400\left(\frac{\|x+y\|^2+\|x-y\|^2}{2}-\|x\|^2\right).
\end{equation}
The increment $y$ may have a nonzero part in $A$. Averaging a convexity
gain and telescoping is the martingale method for uniformly convex norms
\cite[Theorem~4.6 and its proof]{Pisier2011}. Here the estimate controls
only the part of the increment in the new coordinate layer.
Normalize the graph to upper constant one and lower constant $1/D$.
Let $a=2/D$ and choose
$0<\gamma<\min\{a/8,1/(k+1)\}$. At the root use a finite ancestral
projection $W_0=P_{A_0}x_0$ of the true slope, with error at most
$\gamma$. Work successively by levels. Given a current true slope
and $A_{i-1}$, its $a$-separated deviations have one member $z$ with
$\|(I-P_{A_{i-1}})z\|\ge a/4$, by the same finite-head compactness
argument. Choose one at each of the finitely many histories. Enlarge
to a common finite ancestral $A_i$ so that
$\|z-P_{A_i}z\|\le\gamma$ for all these choices. Set
$v_i=P_{A_i}z$ and define $W_i=W_{i-1}\pm v_i$, associated with the
actual half-edges. The parent is supported in $A_{i-1}$, the increment
is supported in $A_i$, and every approximation error is at most
$(i+1)\gamma$. Thus $\|W_i\|\le2$.
For $E_i=A_i\setminus A_{i-1}$, nesting of the projections gives
\[
P_{E_i}v_i=(P_{A_i}-P_{A_{i-1}})z,
\qquad \|P_{E_i}v_i\|\ge a/4-\gamma>a/8.
\]
Here the error is exactly $z-P_{A_i}z$, whose norm is at most
$\gamma$; no layer-projection bound enters this comparison.
Apply \eqref{dir:gap-input} at every parent with $x=W_{i-1}$ and
$y=v_i$. Averaging over a level and then summing the squared-norm
increases yields
\begin{equation}\label{dir:400-conclusion}
4\ge\mathbb E\|W_k\|^2
\ge\|W_0\|^2+\frac{k(a/8)^2}{400}
\ge\frac{k(a/8)^2}{400}.
\end{equation}
Thus this direct argument also excludes a uniform distortion bound.
Only a finite binary tree was averaged after the coordinate choices;
the availability of infinitely many middle vertices was used at each
selection step.
==== END Diamond-distortion-from-midpoint-and-tree-energies//direct.tex ====
==== BEGIN Diamond-distortion-from-midpoint-and-tree-energies//f-applications.tex ====
\section{Graph consequences of recursive upper-array norms}
\label{met:f:applications}
Recursive upper arrays give three further target norms. The first uses
a height-dependent child exponent and an outer Hilbert sum; its sixth-
and cubic-power estimates yield additive losses in squared slope norm.
The other two use Euclidean child aggregation on a joined tree and
differ at the joining root. We retain the qualitative propagation
argument for the free-root norm and an explicit finite-error cubic
argument for the fixed-zero-root norm. The constants below belong to
these stated inputs and deductions.
Let $T_h=\bigcup_{j=0}^h\mathbb N^j$ and $p_h=1+1/h$.
For real $v\in\ell_{p_h}(T_h)$, let $P(v),Q(v)$ be the least nonnegative
arrays whose difference is $v$ and which satisfy
\[
P_s(v)\ge\Big(\sum_{t\in C(s)}P_t(v)^{p_h}\Big)^{1/p_h},
\qquad
Q_s(v)\ge\Big(\sum_{t\in C(s)}Q_t(v)^{p_h}\Big)^{1/p_h}.
\]
Here $C(s)$ is the set of children of $s$ and the sums at leaves are
zero. More explicitly, if the two child norms are $a_s,b_s$, then
$P_s(v)=\max\{a_s,v_s+b_s\}$ and
$Q_s(v)=\max\{a_s-v_s,b_s\}$. Write
\[
\nu_h(v)=P_\emptyset(v)+Q_\emptyset(v),\qquad
X_{\mathrm v}=\Big(\bigoplus_{h\ge1}(\ell_{p_h}(T_h),\nu_h)
\Big)_{\ell_2}.
\]
The intrinsic upper-array results supply completeness, reflexivity,
separability of $X_{\mathrm v}^*$, density of finite coordinate vectors,
and the estimates used below; see \cite[Section~7]{RecursivePotentials}.
We use the slopes from~\eqref{met:slopes}. To discuss paired children,
write
\begin{equation}\label{met:f:slopes}
y_j=x+z_j,\qquad \bar y_j=x-z_j.
\end{equation}
In noncontracting normalization the $y_j$, and also the $z_j$, are
$2$-separated. In upper-one normalization the separation is $2/D$.
\subsection{The sixth-power estimate}
A finite head in $X_{\mathrm v}$ means a finite ancestor-closed set in
each of finitely many components, containing the component root whenever
that component is used. Its coordinate projection is a bounded
finite-rank operator. No estimate on its operator norm is needed in
the following argument.
The finite-head estimate in \cite[Theorem~7.4]{RecursivePotentials}
allows arbitrary completed tails.
\begin{proposition}\label{met:f:sixth}
For $X_{\mathrm v}$ one has
\begin{equation}\label{met:f:sixth-input}
S-2\|x\|_{\mathrm v}^2
\ge c_6S^{-2}\|y\|_{\mathrm v}^6,
\qquad
S=\|x+y\|_{\mathrm v}^2+\|x-y\|_{\mathrm v}^2,
\qquad c_6=2\exp(-(\mathrm e-1))4^{-6},
\end{equation}
whenever $x$ is supported on a finite head and $y$ vanishes there.
The right side is interpreted as zero when $S=0$.
Every distortion-$D$ embedding of $\mathcal D_k$ into this norm satisfies
\begin{equation}\label{met:f:sixth-power-depth}
\frac{c_6 k}{8}\le D^6-D^4.
\end{equation}
For $0<\varepsilon\le1$, put
\begin{equation}\label{met:f:sixth-q}
q_6(\varepsilon)=
\left(1-\frac{c_6\varepsilon^6}{512}\right)^{1/2}.
\end{equation}
Every infinite family satisfying
\[
\|x+u_j\|_{\mathrm v},\ \|x-u_j\|_{\mathrm v}\le1,
\qquad \|u_j-u_l\|_{\mathrm v}\ge\varepsilon\quad(j\ne l)
\]
has $\|x\|_{\mathrm v}\le q_6(\varepsilon)$.
Keeping a fixed unit-lens contraction gives the additional form
\begin{equation}\label{met:f:sixth-depth}
D^{-1}\le q_6(D^{-1})^k.
\end{equation}
\end{proposition}
\begin{proof}
If $\|x\pm y\|_{\mathrm v}\le R$ with $R>0$, then $S\le2R^2$.
For $S>0$, \eqref{met:f:sixth-input} implies
\[
R^2\ge\|x\|_{\mathrm v}^2+
\frac{c_6}{8R^4}\|y\|_{\mathrm v}^6.
\]
It is trivial when $S=0$. Theorem~\ref{met:power-transfer}, with
power $6$ and coefficient $c_6/8$, gives
\eqref{met:f:sixth-power-depth}.
For the unit-lens statement, Lemma~\ref{met:finite-pair} supplies
pure head--tail pairs $x',y'$ with
$\|x'-x\|_{\mathrm v}<\tau$,
$\|y'\|_{\mathrm v}\ge\varepsilon/2-\tau$, and endpoints of
norm at most $1+2\tau$. Apply the displayed homogeneous estimate
and let $\tau\downarrow0$. It gives
$\|x\|_{\mathrm v}^2+c_6\varepsilon^6/512\le1$.
In upper-one graph normalization use the fixed separation parameter
$\varepsilon=1/D$; actual sibling separation is at least $2/D$.
Proposition~\ref{met:backward} proves \eqref{met:f:sixth-depth}.
\end{proof}
The power bound keeps the child radius in the loss at every level.
The geometric form instead uses one contraction factor chosen at the
start from $D$.
\subsection{The cubic estimate and weakly null paired differences}
The intrinsic cubic inequality and its weakly null consequence are
\cite[Theorem~7.6 and Proposition~7.7]{RecursivePotentials}.
The first gives another additive power bound. The second exhibits a
different selection: pairing a weakly Cauchy subsequence. The paired
differences and crossed convex averages also occur in the approximate
midpoint argument of \cite[proof of Corollary~2.3]{DKRRZ2016}; here
dual separability supplies the subsequence by diagonal extraction.
\begin{proposition}\label{met:f:cubic-weak}
The dual $X_{\mathrm v}^*$ is separable, and under the same
finite-head support conditions as above one has
\begin{equation}\label{met:f:cubic-input}
\|x\pm y\|_{\mathrm v}\le R
\quad\Longrightarrow\quad
R^2\ge\|x\|_{\mathrm v}^2+
\frac{c_3}{2\sqrt2\,R}\|y\|_{\mathrm v}^3,
\qquad R>0,\quad c_3=(6\mathrm e)^{-3}.
\end{equation}
Consequently every distortion-$D$ embedding of $\mathcal D_k$ satisfies
\begin{equation}\label{met:f:cubic-power-depth}
\frac{c_3 k}{2\sqrt2}\le D^3-D.
\end{equation}
For $0<\varepsilon\le1$, define
\begin{equation}\label{met:f:cubic-theta}
\theta(\varepsilon)=
\left(1-\frac{c_3\varepsilon^3}{2\sqrt2}\right)^{1/2}.
\end{equation}
If $(h_j)$ is weakly null, $\|h_j\|_{\mathrm v}\ge\varepsilon$,
and $\|x\pm h_j\|_{\mathrm v}\le1$, then
$\|x\|_{\mathrm v}\le\theta(\varepsilon)$.
Every distortion-$D$ embedding of the depth-$k$ countably branching
diamond into $X_{\mathrm v}$ satisfies
\begin{equation}\label{met:f:cubic-depth}
1\le D\theta(D^{-1})^k.
\end{equation}
\end{proposition}
\begin{proof}
Theorem~\ref{met:power-transfer}, applied directly to
\eqref{met:f:cubic-input}, gives \eqref{met:f:cubic-power-depth}.
The weakly null implication is the cited Proposition~7.7.
To use it in the graph, normalize to lower factor one and upper
factor $D$, and let $R\le D$ bound the children of one parent $x$.
Dual separability gives a weakly Cauchy subsequence of $(y_j)$.
The diagonal argument is the one in Section~\ref{met:weak-linear}.
Take successive disjoint pairs $y_{j_{2r}},y_{j_{2r+1}}$ from this
subsequence and put
\[
h_r=\frac{y_{j_{2r}}-y_{j_{2r+1}}}{2R}.
\]
The sequence $(h_r)$ is weakly null. Sibling separation gives
$\|h_r\|_{\mathrm v}\ge1/R\ge1/D$. The identities
\[
\frac{x}{R}+h_r
=\frac{y_{j_{2r}}+\bar y_{j_{2r+1}}}{2R},
\qquad
\frac{x}{R}-h_r
=\frac{y_{j_{2r+1}}+\bar y_{j_{2r}}}{2R}
\]
show that both points belong to the unit ball. The weakly null
conclusion, with $\varepsilon=1/D$, gives $\|x\|_{\mathrm v}
\le R\theta(D^{-1})$. Taking the parent suprema and iterating
this contraction proves \eqref{met:f:cubic-depth}.
\end{proof}
\subsection{A qualitative contraction for the free-root sum norm}
We next change the norm. Join disjoint copies of $T_h$ at a new root
$\rho$ whose children are their roots. At every node use the least
nonnegative arrays $p,q$ with difference equal to the coordinate and
with each array dominating the $\ell_2$ norm of its child values.
For a finitely supported vector with an unrestricted root coordinate,
put
\[
N_J(x)=p_\rho(x)+q_\rho(x),
\]
and let $X_J$ be the completion. In this subsection
$X_J$ refers to the tree formed by joining these finite-height components,
with a free root coordinate. In particular, the norm of an indicator of an initial
path starting at a component root is two.
Restriction $P_H$ to a finite ancestral set containing $\rho$ is
contractive. On finitely supported arrays, restriction removes child
values outside $H$; the recursion is monotone in their nonnegative
norms, so neither root field can increase. This proves contractivity
on the dense finite arrays and hence on the completion. Applying
$I-P_H$ to finite approximations also proves density of finitely
supported vectors in $\ker P_H$.
The intrinsic probabilistic propagation estimate
\cite[Lemmas~6.1 and~6.2]{RecursivePotentials} supplies, for every
$0<\zeta\le2$, a number $\gamma(2,\zeta/4)>0$ such that
\begin{equation}\label{met:f:sum-input}
\frac{N_J(x+y)+N_J(x-y)}2
\ge N_J(x)+2\gamma(2,\zeta/4)
\end{equation}
whenever $x,y$ are finitely supported, $x$ is supported on a finite
ancestor-closed set $H$ containing $\rho$, $y$ vanishes on $H$,
$N_J(x\pm y)\le2$, and $N_J(y)\ge\zeta$.
The number $\gamma$ is uniform in the finite tree and its height.
The estimate comes from the logarithmic-martingale propagation of the
two upper arrays; only its positivity is used here.
The resulting separated-lens statement is also
\cite[Theorem~6.3]{RecursivePotentials}.
\begin{proposition}\label{met:f:sum}
For every $0<\varepsilon\le2$ there is
$\eta(\varepsilon)\in(0,1)$ such that
\[
N_J(x\pm z_j)\le1,
\qquad N_J(z_j-z_l)\ge\varepsilon\quad(j\ne l)
\quad\Longrightarrow\quad
N_J(x)\le1-\eta(\varepsilon).
\]
For $\varepsilon>2$ the hypotheses are impossible, so any
$\eta(\varepsilon)\in(0,1)$ is admissible.
An embedding of the depth-$k$ countably branching diamond into
$X_J$ with distortion at most $D$ therefore satisfies
\begin{equation}\label{met:f:sum-depth}
D^{-1}\le\bigl(1-\eta(2/D)\bigr)^k.
\end{equation}
\end{proposition}
\begin{proof}
Fix $\varepsilon$ and put $\zeta=\varepsilon/4$.
For each sufficiently small $\delta>0$, take a finite coordinate
vector $x'$ with $N_J(x-x')<\delta$ and a finite
ancestor-closed head $H$ containing its support and $\rho$.
The sequence $(z_j)$ is bounded. Finite-dimensional total boundedness
therefore supplies a pair with
\[
u=(z_j-z_l)/2,
\qquad N_J(u)\ge\varepsilon/2,
\qquad N_J(P_Hu)<\delta.
\]
Also $N_J(x\pm u)\le1$ by convexity. Approximate
$(I-P_H)u$ by a finitely supported vector $y$ vanishing on $H$ so
that $N_J(y-u)<2\delta$. This is possible by first using
density and then applying the bounded operator $I-P_H$; no
contractivity is assumed. For small enough $\delta$ we have
\[
N_J(y)\ge\zeta,
\qquad N_J(x'\pm y)\le1+3\delta\le2.
\]
Now \eqref{met:f:sum-input} gives
\[
1+3\delta\ge N_J(x')+2\gamma(2,\varepsilon/16)
\ge N_J(x)-\delta+2\gamma(2,\varepsilon/16).
\]
Let $\delta\downarrow0$. For example,
$\eta(\varepsilon)=\min\{\gamma(2,\varepsilon/16),1/2\}$
has the required property. This choice records positivity, not a
numerical power law in $\varepsilon$.
In upper-one normalization use the fixed separation $2/D$ and
apply Proposition~\ref{met:backward} with
$\rho=1-\eta(2/D)$. This proves \eqref{met:f:sum-depth}.
\end{proof}
\subsection{An additive cubic loss for the fixed-zero-root norm}
On the same joined tree, now require $x_\rho=0$ and define
\[
N_0(x)=p_\rho(x)=q_\rho(x)
=\max\left\{
\Big(\sum_h p_{r_h}(x)^2\Big)^{1/2},
\Big(\sum_h q_{r_h}(x)^2\Big)^{1/2}\right\},
\]
where $r_h$ is the root of the $h$th component. Denote its
completion by $X_0$. The root coordinate is fixed at zero and the
cost is the common root value. On the root-zero subspace the
preceding sum norm is $2N_0$; keeping the actual normalization is
essential for the constants below.
The same ancestral projections preserve this subspace and are
contractive for $N_0$; finitely supported tails are dense here as well.
The deterministic upper-array estimate for $X_0$ is
\cite[Theorem~2.1, equation~(2.4)]{RecursivePotentials}:
\begin{equation}\label{met:f:zero-input}
\frac{N_0(x+y)+N_0(x-y)}2
\ge N_0(x)+\frac{N_0(y)^3}{8(2N_0(x)+N_0(y))^2},
\end{equation}
for every finite ancestor-closed head containing $\rho$, every $x$
supported on it, and every nonzero completed tail $y$ vanishing
there. Finite support is required of $x$, but not of $y$.
\begin{proposition}\label{met:f:zero}
If the depth-$k$ countably branching diamond admits an embedding into
$X_0$ with distortion at most $D\ge1$, then
\begin{equation}\label{met:f:zero-depth}
k\le4608D^3(1-D^{-1}).
\end{equation}
\end{proposition}
\begin{proof}
Normalize to lower constant $A=1/D$ and upper constant one.
For each level $i$, let $L_i$ be the supremum of the $N_0$-norms
of all its historical edge slopes. Thus $A\le L_i\le1$.
Fix any parent slope $x$ at level $i0$ satisfies
\[
s d_k(u,v)\le\|f(u)-f(v)\|\le Ds d_k(u,v)
\qquad(u,v\in\mathcal D_k).
\]
The scale $s$ may depend on the map and on $k$. Its removal by
multiplying $f$ by a scalar does not change $D$.
Our main example applies James's disjoint-segment norm
\cite[p.~739]{James1974} to a forest of finite heights. Let
$\mathcal T=\coprod_{h\ge1}\mathbb N^{\le h}$, with the prefix order
within each rooted tree. A segment is the set of consecutive vertices
between two comparable nodes. Complete the finitely supported real
arrays on $\mathcal T$ in the norm
\[
\rho(a)=\sup\left\{
\left(\sum_{j=1}^m\left|\sum_{s\in S_j}a_s\right|^2\right)^{1/2}:
S_1,\ldots,S_m\text{ are pairwise disjoint segments}\right\},
\]
and call the resulting space $J$. We consider $X=J^*$ with its given
dual norm. On the infinite tree $T_\infty=\mathbb N^{<\omega}$,
including its root, let $J_\infty$ be the completion of the finitely
supported real arrays for the same disjoint-segment formula. Define
\[
B_\infty=\overline{\operatorname{span}}\{e_t^*:t\in T_\infty\}
\subset J_\infty^*,
\]
where $e_t^*$ is coordinate evaluation and the closure uses the dual
norm. The companion on midpoint lenses proves the midpoint estimates
and the finite-coordinate approximation facts for these two spaces
\cite{SegmentLenses}. Section~\ref{dir:forest-definition} also gives
the forest foundations needed for our direct proofs.
James's original construction uses the infinite binary tree.
Girardi proved asymptotic uniform convexity for the coordinate predual
and the dual of that binary-tree space
\cite[Theorems~3 and~5]{Girardi2001}. The countably branching spaces
defined here use their own segment estimates.
\begin{theorem}\label{met:intro-main}\label{met:joint-distortion}
For every integer $k\ge0$, every distortion-$D$ embedding of
$\mathcal D_k$ into either $X$ or $B_\infty$ satisfies
\[
D^2\ge1+k/4.
\]
\end{theorem}
In both cases the input is the following tail inequality. A finite set
$H$ is ancestral if it contains every ancestor of each of its vertices;
$P_H$ retains its coordinates and $Q_H=I-P_H$. If $P_Hp=p$, then
\[
\|p\pm y\|\le R
\quad\Longrightarrow\quad
\|Q_Hy\|\le2\sqrt{R^2-\|p\|^2}.
\]
It holds for arbitrary displacements $y$. In a symmetric lens
$\{z:\|x\pm z\|\le R\}$ containing infinitely many separated deviations,
two projected deviations can be made arbitrarily close, and their
half-difference remains in the same lens. This turns the tail inequality
into a loss in squared edge stretch. Section~\ref{met:joint-section}
gives the complete deduction. The coefficient $1/4$ results from retaining the full quadratic budget
in this calculation.
\subsection{Midpoint methods and earlier work}
Diamond graphs connect metric embeddings with convexity of Banach-space
norms. For binary diamonds, Lee and Naor obtain
quantitative lower bounds in $L_p$, $10$ there is $\delta>0$
such that every unit center $x$ admits a finite-codimensional subspace
on whose unit directions $y$ one has
$\max\{\|x+ty\|,\|x-ty\|\}\ge1+\delta$. Replacing the maximum
by $\|x+ty\|$ gives asymptotic uniform convexity.
Baudier, Causey, Dilworth, Kutzarova, Randrianarivony, Schlumprecht, and Zhang turn this property into a countably branching
diamond obstruction \cite[Section~4]{BaudierEtAl2017}. Their
Theorem~4.1 includes a quantitative conclusion: a midpoint modulus bounded
below by a positive multiple of $t^p$ for $0j$.
Distances between old vertices are unchanged at a replacement. Indeed,
an old path can be replaced by paths of the same total length. Conversely,
a path between old vertices consists of two-edge portions through new
vertices; each portion either returns to the same old vertex or traverses
the endpoints of one old edge. Compressing these portions gives an old
walk of no greater length. It follows that a stage-$j$ edge has endpoint
distance $\ell_j=2^{-j}$ in $\mathcal D_k$. Its distinct replacement
midpoints have mutual distance $\ell_j$, and distance $\ell_j/2$ from
either endpoint. At their introduction the assertion is immediate because
two distinct middle vertices are nonadjacent, and every incident edge has
length $\ell_j/2$; subsequent replacements preserve it.
\input{figures/diamond-replacement}
The normalized metric $d_k$ defined above differs by the global factor
$2^{-k}$ from the convention
in which every edge of the final graph has length one. In that convention
old-vertex distances double at each replacement and a stage-$j$ edge has
represented length $2^{k-j}$. Both conventions give exactly the same
embedding distortion.
Suppose $f:\mathcal D_k\to X$, $s>0$ and $D\ge1$ satisfy
\begin{equation}\label{met:all-pairs}
s d_k(u,v)\le\|f(u)-f(v)\|\le Ds d_k(u,v)
\qquad(u,v\in\mathcal D_k).
\end{equation}
The lower inequality is required for all pairs, including distinct
replacement midpoints. Dividing $f$ by $s$ gives the noncontracting
normalization $d_k\le\|f(u)-f(v)\|\le Dd_k$. Dividing instead by $Ds$
gives lower factor $\alpha=1/D$ and upper factor one. We use both
normalizations and state which one is in force.
Orient the original edge and orient every replacement path consistently.
For a parent edge $[a,b]$ of represented length $\ell$, define
\begin{equation}\label{met:slopes}
x=\frac{f(b)-f(a)}{\ell},\qquad
z_i=\frac{2}{\ell}\left(f(m_i)-\frac{f(a)+f(b)}2\right).
\end{equation}
Its child slopes are $x+z_i$ and $x-z_i$. Under lower factor $\alpha$,
\begin{equation}\label{met:sibling-separation}
\|z_i-z_j\|\ge2\alpha\qquad(i\ne j).
\end{equation}
Every slope lies between $\alpha$ and the upper embedding factor. The
sequence $(z_i)$ is bounded, since each $z_i$ is half the difference of
two child slopes.
\subsection{The finite-dimensional pair selection}\label{met:pairing-section}
The next observation records the approximation errors that will be used
throughout. It requires bounded finite-rank projections, rather than
contractivity of the complementary projections.
\begin{lemma}\label{met:finite-pair}
Let $X$ be a Banach space, $R>0$, and let $(z_i)_{i\in\mathbb N}$ satisfy
$\|x\pm z_i\|\le R$ with $\|z_i-z_j\|\ge\varepsilon>0$ for $i\ne j$.
Every half-difference $w=(z_i-z_j)/2$ with $i\ne j$ satisfies
$\|w\|\ge\varepsilon/2$ and $\|x\pm w\|\le R$.
Fix $\eta>0$. The following two selections are available.
\begin{enumerate}
\item Suppose $p$ has finite support, $\|p-x\|<\eta$, and $P$ is a bounded
finite-rank coordinate projection satisfying $Pp=p$. There are distinct
$i,j$ for which, with $w=(z_i-z_j)/2$ and $y=w-Pw$,
\begin{equation}\label{met:pair-errors}
\|Pw\|<\eta,\quad
\|y\|\ge\varepsilon/2-\eta,\quad
\|p\|\ge\|x\|-\eta,\quad
\|p\pm y\|\le R+2\eta.
\end{equation}
\item If $F$ is a closed finite-codimensional subspace, there are distinct
$i,j$ whose half-difference satisfies
$\operatorname{dist}(w,F)<\eta$.
This selection requires no coordinate structure on $X$.
\end{enumerate}
\end{lemma}
\begin{proof}
For the common endpoint bound use
\[
x+w=\tfrac12((x+z_i)+(x-z_j)),\qquad
x-w=\tfrac12((x-z_i)+(x+z_j)).
\]
In the first selection, the bounded sequence $(Pz_i)$ has two distinct
members at distance less than $2\eta$, because bounded subsets of a
finite-dimensional space are totally bounded. This gives the first
inequality in~\eqref{met:pair-errors} and hence the second.
Replacing $x$ by $p$ and $w$ by $w-Pw$ costs less than $2\eta$.
The last assertion follows by applying the same argument to the bounded
images of $z_i$ in the finite-dimensional quotient $X/F$.
\end{proof}
\subsection{The quadratic loss from the pooled atomic estimate}\label{met:joint-section}
For the segment forest space, the pooled atomic estimate has the form
\begin{equation}\label{met:joint-input}
\|Q_Hy\|\le2\sqrt{R^2-\|p\|^2},
\qquad P_Hp=p,\quad \|p\pm y\|\le R.
\end{equation}
Here $H$ is finite and ancestral, $Q_H=I-P_H$, and the displacement
$y$ is arbitrary. The estimate is supplied by the joint-budget
theorem in the companion on midpoint lenses
\cite[Theorem~1.1]{SegmentLenses}. Its target is the
projected tail $Q_Hy$; an individual endpoint-law estimate is not being
used with an arbitrary head.
The same companion establishes, for both the finite-forest dual and
the infinite-height closed coordinate span, density of finite coordinate
vectors and contractivity of $P_H$ and $Q_H$ for ancestral $H$.
Thus $P_Hx$ can approximate any center $x$: take a finite coordinate
approximation and let $H$ contain its support and all its ancestors.
These approximation properties refer to each space with its own norm.
\begin{lemma}\label{met:joint-lens-lemma}
In the finite-height segment forest dual, every infinite
$\varepsilon$-separated symmetric $R$-lens satisfies
\begin{equation}\label{met:joint-lens}
\|x\|^2+\varepsilon^2/16\le R^2.
\end{equation}
The same lens conclusion holds for the infinite-height closed coordinate
span with its own segment-dual norm and finite-head approximation theorem.
\end{lemma}
\begin{proof}
Fix $0<\eta<\varepsilon/2$ and choose $H$ so that
$p=P_Hx$ has $\|p-x\|<\eta$. Choose two deviations with
$w=(z_i-z_j)/2$ satisfying $\|P_Hw\|<\eta$.
Then $\|Q_Hw\|\ge\varepsilon/2-\eta$ and
$\|p\pm w\|\le R+\eta$, because $\|x\pm w\|\le R$.
Apply~\eqref{met:joint-input} to the arbitrary displacement $w$:
\[
(\varepsilon/2-\eta)^2
\le4\bigl((R+\eta)^2-\|p\|^2\bigr).
\]
Letting $\eta\downarrow0$ proves~\eqref{met:joint-lens}. The same
argument applies with the infinite-height coordinate span's own version
of~\eqref{met:joint-input}.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{met:intro-main}]
Normalize to lower factor one and upper factor $D$.
Let $R_j$ be the supremum of stage-$j$ slope norms. Every parent
slope is the center of a $2$-separated lens whose endpoints have norms
at most $R_{j+1}$. Thus $\|x_e\|^2+1/4\le R_{j+1}^2$ for every
parent. Only now take the supremum: $R_j^2+1/4\le R_{j+1}^2$.
Since $R_0\ge1$ and $R_k\le D$, summing gives $D^2\ge1+k/4$.
No supremum need be attained. The infinite-height statement uses exactly
the same scalar argument with that space's proved version of
\eqref{met:joint-input}; no identification with the full infinite-height
dual is required.
\end{proof}
The $1/4$ coefficient comes from retaining the joint quadratic budget
in~\eqref{met:joint-input}. The direct graph-energy arguments follow next. General transfer rules
and alternative forest estimates appear in Section~\ref{met:transfer-section}
and Appendix~\ref{met:forest-variants}.
==== END Diamond-distortion-from-midpoint-and-tree-energies//metrics.tex ====
==== BEGIN Diamond-distortion-from-midpoint-and-tree-energies//preamble.tex ====
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\date{September 27, 2026}
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==== BEGIN Diamond-distortion-from-midpoint-and-tree-energies//segment-transfer-variants.tex ====
\section{Alternative midpoint estimates for the segment forest}
\label{met:forest-variants}
The joint quadratic estimate gives the strongest forest bound in this
paper. The following inputs come from different intrinsic arguments and
illustrate the transfer rules of Section~\ref{met:transfer-section}.
They control radial deficit, parameterized test gains, or tails selected
by sequential compactness and a common primal norming test.
\subsection{Radial deficit}
\begin{corollary}[The atomic forest bound]\label{met:atomic-distortion}
For the segment forest space $X$, every map satisfying~\eqref{met:all-pairs}
has $D^2\ge1+k/16$.
\end{corollary}
\begin{proof}
For the forest atomic estimate, the constant
$C_*=\sqrt2+\sqrt6<4$ in the companion's arbitrary-displacement radial estimate
\cite[Proposition~12.1]{SegmentLenses} gives
\eqref{met:radial-input}. In particular,
\[
\varepsilon^2/64\le R(R-\|x\|)\le R^2-\|x\|^2,
\qquad D^2\ge1+k/16.
\]
This radial input gives a separate proof from the joint quadratic estimate.
\end{proof}
Keeping the
exact constant in the same calculation also gives
$k\le C_*^2D(D-1)$. Different atomic derivations with their own radial
constant $C$ give \eqref{met:additive-bound} with that constant; replacing
one proof's $C$ by another proof's value would discard its quantitative
content.
\subsection{Parameterized clipping tests}
Apply Proposition~\ref{met:parametric} with $a=1/2$ and $b=3$, as in the clipping input of
\cite[Section~10.1, Equation~(10.9)]{SegmentLenses}. Take $\lambda=1/(6D)$;
$Q-1\le3\lambda^2/2$ gives $\Delta\ge1/(24D)$.
Thus the full parameterized consequence is
\begin{equation}\label{met:parametric-twentyfour}
1+\frac{k}{24D}\le D.
\end{equation}
For an optimized input
$\max_\pm\|p\pm y\|\ge\sqrt{\|p\|^2+a\|y\|^2}$,
the same limiting selection gives a child supremum at least
$\sqrt{\|x\|^2+a}$. Since
$\sqrt{r^2+a}-r$ decreases with $r>0$, some actual child has norm
at least $\|x\|+g/2$, where $g=\sqrt{D^2+a}-D>0$.
With $a=1/12$, as in
\cite[Proposition~10.4 and Equation~(10.10)]{SegmentLenses}, this gives
$1+kg/2\le D$. For this value of $a$, one
may make the finite selection before passing to a limit by taking approximation error
$\eta=g/8$: the norm comparison costs at most $4\eta$, leaving
$g/2$. Both descriptions retain the same actual-child loss.
Likewise, a radial input with constant $C$ gives a child supremum at
least $\|x\|+1/(C^2D)$. Selecting an actual child above the smaller
threshold $\|x\|+1/(2C^2D)$ gives
$1+k/(2C^2D)0$ such that
\begin{equation}\label{met:bridge:fixed-bound-input}
\begin{gathered}
1\le\|x\|\le D,\qquad \|z_i\|\le2D,\qquad
\|z_i-z_j\|\ge1\quad(i\ne j)\\
\Longrightarrow\quad
\max_\pm\|x\pm z_i\|>\|x\|+c_D
\quad\text{for some }i.
\end{gathered}
\end{equation}
For fixed $D$, the same $c_D$ works for every center and every family
satisfying the displayed assumptions. The selected deviation need not
annihilate the head.
\begin{proposition}\label{met:bridge:fixed-bound}
If a target satisfies \eqref{met:bridge:fixed-bound-input}, every
noncontracting embedding of $\mathcal D_k$ with upper factor $D$ obeys
\[
1+kc_D\le D.
\]
\end{proposition}
\begin{proof}
For every parent slope $x$ and its deviations $z_i$ from
\eqref{met:slopes}, one has $1\le\|x\|\le D$,
$\|z_i\|\le D\le2D$, and $\|z_i-z_j\|\ge2\ge1$.
The input therefore selects an actual child slope with norm greater
than $\|x\|+c_D$. Starting at the original slope and making this
choice through $k$ replacements gives a final slope greater than
$1+kc_D$ when $k\ge1$, and every final slope is at most $D$.
For $k=0$ the conclusion is $1\le D$.
\end{proof}
\subsection{Coordinate-null and sequential estimates}
Use the upper-one normalization with $\alpha=1/D$. At a parent with
child bound $00$ there is $0<\rho<1$ such that an infinite
$\varepsilon$-separated symmetric unit lens in the segment forest
space has center norm at most $\rho$.
\end{proposition}
\begin{proof}
Otherwise choose centers $x_n\in B_X$ with $\|x_n\|\to1$ and
infinite $\varepsilon$-separated families $(v_{n,i})_i$ satisfying
$\|x_n\pm v_{n,i}\|\le1$. First choose a finitely supported primal
near-normer $f_n$ with
\[
\rho(f_n)\le1,\qquad x_n(f_n)>\|x_n\|-1/n.
\]
Next choose a finite ancestral $H_n$ containing
$\operatorname{supp}f_n$ and satisfying $\|Q_{H_n}x_n\|<1/n$.
This is possible by density of finite coordinate vectors in $X$ and
contractivity of $Q_{H_n}$: include the support of an approximation
to $x_n$ before taking its finite ancestral closure.
Only now choose two members of the $n$th lens whose projections
onto $H_n$ differ by less than $1/n$, and set
$u_n=(v_{n,i}-v_{n,j})/2$. Finite-dimensional total boundedness
justifies this choice, and crossed convex averages give
\begin{equation}\label{met:bridge:normer-selection}
\|x_n\pm u_n\|\le1,\qquad
\|P_{H_n}u_n\|<\frac1{2n},\qquad
\|Q_{H_n}u_n\|\ge\frac\varepsilon2-\frac1{2n}.
\end{equation}
Apply \eqref{met:bridge:weighted-input} with the already chosen
$f_n,H_n$ and with $z_\pm=x_n\pm u_n$. Here
$d_n=1-x_n(f_n)\to0$ and $a_n=\|Q_{H_n}x_n\|\to0$.
Consequently both endpoint tails tend to zero. Their half-difference
$Q_{H_n}u_n$ also tends to zero, contradicting
\eqref{met:bridge:normer-selection}. No head has been chosen from
the selected pair; the order is near-normer, head, and then pair.
\end{proof}
For the original upper-one graph normalization, retain the separation
parameter $\varepsilon=1/D$. The normalized sibling deviations have
separation at least $2/(DL)\ge1/D$ when the child bound is
$00$:
\begin{align}
\|y\|^2&\le C^2 R(R-\|p\|)
&&\text{if }\|p\pm y\|\le R, \label{met:radial-input}\\
\|p\|^2+a\|y\|^2&\le R^2
&&\text{if }\|p\pm y\|\le R. \label{met:square-input}
\end{align}
Here $C>0$, respectively $a>0$, is uniform over $\mathcal P$, the
vectors, and $R$. No uniform bound on $\|P\|$ or $\|I-P\|$ is assumed.
For $R>0$ and $\varepsilon>0$, an infinite $\varepsilon$-separated family in the symmetric $R$-lens at
$x$ then satisfies, respectively,
\begin{align}
\varepsilon^2/4&\le C^2R(R-\|x\|),\label{met:radial-lens}\\
\|x\|^2+a\varepsilon^2/4&\le R^2.\label{met:square-lens}
\end{align}
\end{proposition}
\begin{proof}
Choose $p$ and $P$ as in Lemma~\ref{met:finite-pair}, and apply the
assumed estimate to $p,y$ at radius $R+2\eta$. In the first case,
\[
(\varepsilon/2-\eta)^2
\le C^2(R+2\eta)(R+2\eta-\|p\|)
\le C^2(R+2\eta)(R+3\eta-\|x\|).
\]
In the second case,
$\max\{\|x\|-\eta,0\}^2+a(\varepsilon/2-\eta)^2\le(R+2\eta)^2$.
Take $0<\eta<\varepsilon/2$ and let $\eta$ decrease to zero.
\end{proof}
\subsection{Squared losses and additive losses}\label{met:squared-section}
For the noncontracting normalization define
\[
R_j=\sup\{\|x_e\|: e\text{ is a stage-$j$ edge}\},
\qquad 1\le R_j\le D.
\]
The countable supremum need not be attained. Our estimates apply to each
parent edge before the supremum is taken.
The squared-lens hypothesis below is the $p=2$ Banach-space form of
the metric diamond inequality in
\cite[Equation~(2) and Theorem~3.7 of arXiv v2]{Malthaner2025}:
its endpoints are $-x,x$ and its intermediate points are the deviations
$z_i$. We retain the coefficient by summing the squared losses directly.
\begin{theorem}\label{met:squared-transfer}
If every infinite $\varepsilon$-separated symmetric $R$-lens satisfies
$\|x\|^2\le R^2-b\varepsilon^2$ for some $b>0$, then every embedding
\eqref{met:all-pairs} satisfies $D^2\ge1+4bk$.
If instead \eqref{met:radial-lens} holds, then
\begin{equation}\label{met:additive-bound}
k\le C^2D(D-1).
\end{equation}
\end{theorem}
\begin{proof}
For each stage-$j$ parent, its children lie in the lens of radius
$R_{j+1}$ and its deviations are $2$-separated. The first hypothesis
gives $\|x\|^2\le R_{j+1}^2-4b$. Taking suprema gives
$R_j^2\le R_{j+1}^2-4b$, and summing yields
$1\le R_0^2\le D^2-4bk$.
For the radial estimate,
\[
1\le C^2R_{j+1}(R_{j+1}-\|x\|),\qquad
R_j\le R_{j+1}-\frac1{C^2R_{j+1}}
\le R_{j+1}-\frac1{C^2D}.
\]
Summing proves \eqref{met:additive-bound}. Empty sums cover $k=0$.
\end{proof}
\subsection{Retaining a homogeneous power loss}
The radius dependence of a higher-power input can be accumulated directly.
Replacing it by a fixed contraction factor would discard this information.
\begin{theorem}\label{met:power-transfer}
Assume the density and projection-family hypotheses of
Proposition~\ref{met:tail-to-lens}. Let $q\ge2$ and $c>0$. Suppose that
for every admissible projection, every pure head--tail pair $p,y$, and
every $R>0$,
\[
\|p\pm y\|\le R
\quad\Longrightarrow\quad
\|p\|^2+cR^{2-q}\|y\|^q\le R^2.
\]
Then every distortion-$D$ embedding of $\mathcal D_k$ satisfies
\[
ck\le D^{q-2}(D^2-1).
\]
\end{theorem}
\begin{proof}
Use the noncontracting normalization. Fix a parent $x$ whose child
slopes have norm at most $R$. Lemma~\ref{met:finite-pair}, followed
by $\eta\downarrow0$, gives
\[
\|x\|^2+cR^{2-q}\le R^2.
\]
Here the paired tail has limiting lower norm one. Set $R=R_{j+1}$
and take the supremum over stage-$j$ parents. Since $1\le R_{j+1}\le D$
and $q\ge2$,
\[
R_{j+1}^2-R_j^2\ge cR_{j+1}^{2-q}\ge cD^{2-q}.
\]
Summing between $R_0\ge1$ and $R_k\le D$ gives
$D^2-1\ge kcD^{2-q}$, as required.
\end{proof}
\subsection{Keeping the parameterized test losses}\label{met:parametric-section}
Some primal tests give a linear gain before optimization. The graph
argument can preserve that parameter instead of replacing the input by a
single modulus.
\begin{proposition}\label{met:parametric}
Assume the density and projection-family hypotheses of
Proposition~\ref{met:tail-to-lens}. Fix constants $a,b>0$. Suppose that
for every admissible projection and every pure head--tail pair $p,y$,
the following estimate holds for all $\lambda>0$:
\[
\max_\pm\|p\pm y\|\ge
\frac{\|p\|+a\lambda\|y\|}{\sqrt{1+b\lambda^2}}.
\]
Put $Q=\sqrt{1+b\lambda^2}$. For noncontracting diamond slopes
bounded above by $D$, the stage suprema satisfy
\[
R_j\le Q R_{j+1}-a\lambda
\le R_{j+1}-\bigl(a\lambda-(Q-1)D\bigr).
\]
In particular, whenever $\Delta=a\lambda-(Q-1)D>0$, one has
$1+k\Delta\le D$.
\end{proposition}
\begin{proof}
For each parent $x$, choose a finite-support approximation $p$ and a
paired tail $y$ as in Lemma~\ref{met:finite-pair}. The half-difference
has norm at least one, so $\|y\|\ge1-\eta$. If the child bound is
$R$, then $R+2\eta\ge(\|p\|+a\lambda(1-\eta))/Q$.
Let $\eta\downarrow0$ to obtain $\|x\|\le QR-a\lambda$,
and then take suprema. Summation proves the last assertion.
\end{proof}
\subsection{Backward contraction}\label{met:backward-section}
An exact power-law lens estimate is useful but unnecessary for a
qualitative obstruction. The following form also handles estimates whose
constants are selected only after the separation is specified.
\begin{proposition}\label{met:backward}
Fix $0<\alpha\le1$. Suppose some $0\le\rho<1$ has the following
property: whenever $\|x\pm z_i\|\le1$ and
$\|z_i-z_j\|\ge2\alpha$ for $i\ne j$, one has $\|x\|\le\rho$.
An embedding with lower factor $\alpha$ and upper factor one must satisfy
\begin{equation}\label{met:backward-bound}
\alpha\le\rho^k,
\end{equation}
where the empty product $\rho^0$ is one.
\end{proposition}
\begin{proof}
Let $r_j$ be the supremum of stage-$j$ slope norms. Initially $r_k\le1$,
and every $r_j\ge\alpha>0$. If all child slopes are at most $r_{j+1}$,
normalize \eqref{met:slopes} by this positive number. The resulting
unit lens has separation at least $2\alpha/r_{j+1}\ge2\alpha$.
The hypothesis gives $\|x\|\le\rho r_{j+1}$ for every parent, hence
$r_j\le\rho r_{j+1}$. Induction proves the assertion. If $\rho=0$,
the existence of any replacement already contradicts $r_j\ge\alpha$.
\end{proof}
For example, \eqref{met:radial-lens} gives
$\rho=1-\alpha^2/C^2$ when this number is positive. If it is
nonpositive, a parent lens with center norm at least $\alpha>0$
is already impossible. With upper factor one it also gives the stronger
intermediate inequality
\[
\|x\|\le r-\frac{\alpha^2}{C^2r}
\le r-\frac{\alpha^2}{C^2},\qquad 00$. If $(1+\delta)^k>D$, every
noncontracting embedding with upper factor $D$ has a parent edge all of whose
child slopes have norm at most $(1+\delta)$ times its own slope norm.
If, for every unit $u$, the enlarged lens
\[
M(u,\delta)=\{z:\max_\pm\|u\pm z\|\le1+\delta\}
\]
contains no infinite $2/D$-separated family, then every such embedding
satisfies
\begin{equation}\label{met:bridge:enlarged-depth}
1\le \frac{D}{(1+\delta)^k}.
\end{equation}
\end{proposition}
\begin{proof}
If no stable parent existed, choose at each stage an actual child whose
slope norm is greater than $1+\delta$ times its parent's norm. Starting
with norm at least one would contradict the terminal upper bound $D$.
At a stable parent of norm $r\le D$, division by $r$ puts all deviations
in $M(x/r,\delta)$ with separation at least $2/r\ge2/D$.
This proves the second assertion by contradiction.
The same exclusion also gives a backward proof. If all children of a
parent of norm $r$ have norm at most $L\le D$, then
$r\ge L/(1+\delta)$ would put the normalized deviations in the forbidden
lens. Thus every parent has norm less than $L/(1+\delta)$; taking
suprema gives the non-strict bound, and iteration from $L=D$ proves
\eqref{met:bridge:enlarged-depth}. Neither proof assumes attainment of a
child supremum.
\end{proof}
The stable parent may instead be divided by its child bound
$(1+\delta)r$. Its unit-ball lens has center norm $1/(1+\delta)$
and separation at least $2/((1+\delta)D)$. This is useful when the
intrinsic estimate is stated for unit-ball lenses rather than enlarged
unit-center lenses.
\subsection{Uniform finite-head bounds and their parameters}\label{met:packing-section}
Many estimates above control all lens vectors using one head selected
from the center. This order of quantifiers is precisely what packing
requires.
\begin{lemma}\label{met:packing}
Let $P$ be bounded and finite rank, let $A\subset X$ be bounded,
and let $b\ge0$ and $\varepsilon>0$.
If $\|(I-P)v\|\le b$ for all $v\in A$, then $A$ contains no infinite
$\varepsilon$-separated family when $2b<\varepsilon$.
Consequently the following two inputs have different graph conclusions.
\begin{enumerate}
\item Let $0<\delta<1$. For each $\|x\|>1-\delta$, suppose one such $P$ bounds by
$b(\delta)$ the tails of every $v$ with $\|x\pm v\|\le1$.
If $2b(\delta)<2/D$, backward contraction gives
$D^{-1}\le(1-\delta)^k$.
\item Let $\delta>0$. For every unit $x$, suppose one such $P$ bounds by
$b(\delta)$ the tails of every $v$ with $\|x\pm v\|\le1+\delta$.
If $2b(\delta)<2/D$, actual-child selection gives
$D\ge(1+\delta)^k$.
\end{enumerate}
The same conclusions hold when the bound is on the first endpoint of
every pair with $(u+v)/2=x$: take $u,v\in B_X$ in case (1), and
$u,v\in(1+\delta)B_X$ in case (2).
\end{lemma}
\begin{proof}
If $(v_i)$ were such a separated family, choose distinct $i,j$ with
$\|Pv_i-Pv_j\|<\varepsilon-2b$. The triangle inequality contradicts
$\|v_i-v_j\|\ge\varepsilon$. The graph conclusions follow from
Propositions~\ref{met:backward} and~\ref{met:stable-parent}. For first
endpoints, translation by their common center preserves separation and
the same finite-dimensional argument applies directly to $(u_i)$.
\end{proof}
Under the projection-family hypotheses of Proposition~\ref{met:tail-to-lens},
a bound only on pure tails uses Lemma~\ref{met:finite-pair} for a
second form of compactness. For example, suppose unit-ball
head--tail pairs obey $\|y\|\le g(1-\|p\|)$, where
$g:[0,1]\to[0,\infty)$ is nondecreasing and $g(r)\to0$ as
$r\downarrow0$. For $t>0$, an infinite
$2t$-separated unit lens cannot have center norm greater than $1-\Delta$
if
\begin{equation}\label{met:g-parameters}
0<\Delta<1/4,\qquad (1+2\Delta)g(4\Delta)+\Delta0$, there is $0<\rho<1$
such that every infinite $\varepsilon$-separated symmetric unit lens
has center norm at most $\rho$.
\end{lemma}
\begin{proof}
If no such $\rho$ existed, choose centers $x_n$ with
$\|x_n\|\to1$ and infinite $\varepsilon$-separated families
$(z_{n,i})_i$ with $\|x_n\pm z_{n,i}\|\le1$.
Choose $\eta_n\downarrow0$, finitely supported $p_n$ with
$\|p_n-x_n\|<\eta_n$, and admissible heads $P_n$ satisfying
$P_np_n=p_n$. Lemma~\ref{met:finite-pair} gives a half-difference
$w_n$ and the tail $y_n=w_n-P_nw_n$ with
\[
\|y_n\|\ge\varepsilon/2-\eta_n,\qquad
\|p_n\|\ge\|x_n\|-\eta_n,\qquad
\|p_n\pm y_n\|\le1+2\eta_n.
\]
Divide both vectors by $1+2\eta_n$. The resulting pure head--tail
pairs have endpoints in the unit ball and centers whose norms tend
to one. Their tail norms have lower limit at least
$\varepsilon/2>0$, contradicting the assumed sequential statement.
This contradiction gives one $\rho$ valid for all centers and heads.
\end{proof}
In the upper-one graph normalization, take $\varepsilon=2/D$.
Dividing a parent configuration by any child bound $00$ and $g>0$, every unit vector $u\in X$ admits
a closed finite-codimensional subspace $F_u$ such that
\[
y\in F_u,\quad\|y\|\ge t
\quad\Longrightarrow\quad
\max\{\|u+y\|,\|u-y\|\}\ge1+g.
\]
If $D\ge1$ and $t\le1/(2D)$, every noncontracting embedding of
$\mathcal D_k$ with upper factor $D$ satisfies $D\ge(1+g/2)^k$.
\end{proposition}
\begin{proof}
For a parent slope $x$, put $r=\|x\|\in[1,D]$ and use the subspace
$F_{x/r}$. Apply the quotient part of Lemma~\ref{met:finite-pair} to
choose $w=(z_i-z_j)/2$, with $\|w\|\ge1$, and $w'\in F_{x/r}$
such that $\|w-w'\|<\min\{1/2,rg/2\}$. Then
$\|w'/r\|\ge1/(2D)\ge t$, and
\[
\max_\pm\|x\pm w\|\ge r(1+g)-\|w-w'\|>r(1+g/2).
\]
Each $x\pm w$ is the average of two of the four selected child slopes.
One of these four actual slopes is therefore greater than $r(1+g/2)$.
Choose it and repeat through $k$ generations. This argument makes no
claim that a supremum over all children is attained.
\end{proof}
A bound initially stated only for $\|y\|=t$ extends to $\|y\|\ge t$:
for a fixed direction $v$, the function
$h(r)=\max\{\|u+rv\|,\|u-rv\|\}$ is even and convex, and hence
nondecreasing on $[0,\infty)$. This observation applies likewise to the
average of the two norms.
\begin{proposition}[The full asymptotic gain in a backward step]
\label{met:full-gain}
Assume the finite-codimensional hypothesis of
Proposition~\ref{met:quotient-transfer}. Consider a noncontracting embedding
with upper factor $D\ge1$. If $t\le1/D$ and all child
slopes of one parent have norm at most $L$, then
$\|x\|\le L/(1+g)$. Consequently every depth-$k$ embedding with
upper factor $D$ satisfies $1\le D/(1+g)^k$.
\end{proposition}
\begin{proof}
Put $r=\|x\|\in[1,D]$. Quotient compactness supplies half-differences
$w$ with $\|w\|\ge1$ and vectors $v\in F_{x/r}$ with
$\|v-w\|<\eta$, where $0<\eta<1/2$ is arbitrarily small.
Set $w'=\|w\|v/\|v\|$. Then $w'\in F_{x/r}$,
$\|w'\|/r=\|w\|/r\ge1/D\ge t$, and
\[
\|w'-w\|\le \bigl|\|w\|-\|v\|\bigr|+\|v-w\|<2\eta.
\]
The intrinsic gain and the crossed-average child bound imply
$(1+g)r\le\max_\pm\|x\pm w'\|\le L+2\eta$.
Let $\eta\downarrow0$ before taking suprema over parents, then
iterate backward. Without the norm-preserving rescaling, the same
proof works under the smaller threshold $t\le1/(2D)$.
\end{proof}
\begin{lemma}[The local midpoint estimate gives an enlarged lens gap]
\label{met:bridge:local-amuc}
Let $R>0$ and put $t=R/4$. Suppose $\gamma>0$ is such that every
unit $u$ admits a closed finite-codimensional subspace $F_u$ with
\begin{equation}\label{met:bridge:local-amuc-input}
\max_\pm\|u\pm t y\|\ge1+\gamma
\qquad(y\in F_u,\ \|y\|=1).
\end{equation}
Then $\beta=\gamma/2$ excludes infinite $R$-separated families in
$M(u,\beta)$, uniformly over unit $u$.
\end{lemma}
\begin{proof}
If $(z_i)\subset M(u,\beta)$ were such a family, symmetry and
convexity would place every half-difference $(z_i-z_j)/2$ in the
same lens. The family is bounded, so its image in the
finite-dimensional quotient $X/F_u$ has arbitrarily close distinct
pairs. Hence there are $i\ne j$ and $w\in F_u$ such that, with
$v=(z_i-z_j)/2$,
\[
\|v\|\ge R/2,\qquad
\|v-w\|<\min\{R/4,\gamma/2\}.
\]
In particular $\|w\|>R/4=t$. For a fixed unit direction $y$, the
function $s\mapsto\max_\pm\|u\pm s y\|$ is even and convex,
hence nondecreasing for $s\ge0$. Apply
\eqref{met:bridge:local-amuc-input} to $y=w/\|w\|$ to obtain
$\max_\pm\|u\pm w\|\ge1+\gamma$. On the other hand,
$v\in M(u,\beta)$ and the strict approximation error give
\[
\max_\pm\|u\pm w\|
\le1+\beta+\|w-v\|<1+\gamma,
\]
a contradiction. For $R=2/D$, Proposition~\ref{met:bridge:enlarged-backward}
now gives \eqref{met:bridge:enlarged-depth} with $\delta=\beta=\gamma/2$.
\end{proof}
\subsection{The obstruction for an equivalent midpoint-convex norm}
\label{met:bridge:renorming-section}
The following obstruction is the countably branching diamond case of
\cite[Theorem~4.1]{BaudierEtAl2017}. We include the short deduction
from the quotient transfer to specify the norm convention and the
effect of an equivalent norm.
\begin{theorem}\label{met:bridge:equivalent-amuc}
If a Banach space admits an equivalent asymptotically midpoint
uniformly convex norm, then the family $(\mathcal D_k)_{k\ge0}$
does not embed into it with uniformly bounded distortion.
Write $\operatorname{cof}(X)$ for the closed finite-codimensional
linear subspaces of $X$. In infinite dimension, midpoint convexity may be expressed by
positivity of the maximum modulus
\begin{equation}\label{met:bridge:max-modulus}
\delta_N^{\max}(t)=
\inf_{N(u)=1}\
\sup_{F\in\operatorname{cof}(X)}\
\inf_{\substack{y\in F\\N(y)=1}}
\bigl(\max_\pm N(u\pm t y)-1\bigr)>0
\qquad(00$. Embeddings with common distortion $D_0$ for the original
norm, and arbitrary positive scales, would give embeddings with common
distortion $D=(b/a)D_0$ for $N$: divide each map by its original
lower scale times $a$.
Fix $t=1/(2D)$ and $0