# On stability of metric spaces and Kalton’s property $Q$

F. Baudier  Department of Mathematics, Texas A&M University, College Station,
TX 77843, USA<EMAIL_ADDRESS>,  Th. Schlumprecht  Department of
Mathematics, Texas A&M University, College Station, TX 77843, USA and Faculty
of Electrical Engineering, Czech Technical University in Prague, Zikova 4, 166
27, Prague, Czech Republic<EMAIL_ADDRESS>and  A. Zsák  Peterhouse,
Cambridge CB2 1RD and Department of Pure Mathematics and Mathematical
Statistics, Centre for Mathematical Sciences, University of Cambridge,
Wilberforce Road, Cambridge CB3 0WB, United Kingdom<EMAIL_ADDRESS>
###### Abstract.

The first named author introduced the notion of upper stability for metric
spaces in [baudier:22] as a relaxation of stability. The motivation was a
search for a new invariant to distinguish the class of reflexive Banach spaces
from stable metric spaces in the coarse and uniform category. In this paper we
show that property $Q$ does in fact imply upper stability. We also provide a
direct proof of the fact that reflexive spaces are upper stable by relating
the latter notion to the asymptotic structure of Banach spaces.

###### Key words and phrases:

stable spaces, property Q, coarse embeddings

###### 2020 Mathematics Subject Classification:

46B85, 46B20, 51F30, 05C63

F. Baudier was supported by the National Science Foundation under Grant Number
DMS-2055604. Th. Schlumprecht was supported by the National Science Foundation
under Grant Number DMS-2054443. A. Zsák was supported by the Workshop in
Analysis and Probability at Texas A&M University in 2023.

##  1\. Introduction

In [kalton:07], Kalton proved two landmark results about the coarse and
uniform geometry of infinite-dimensional Banach spaces. The first result was a
nonlinear embedding result that states that every stable metric space admits a
coarse and uniform embedding into a reflexive Banach space.

While uniform embeddings have long been considered in nonlinear Banach space
geometry (see [ben-lin:00] for an extensive account), the notion of coarse
embedding was introduced more recently by Gromov (under a different
terminology) in [gromov:93] as a notion of metric faithfulness that turned out
to be of paramount importance in geometric group theory and noncommutative
geometry. We refer to the monograph [now-yu:12] for a further discussion of
coarse embeddings. Kalton’s embedding result was quantitatively refined in
[bau-lan:15], where it was observed that the embeddings possess even better
metric faithfulness properties: they can be arbitrarily close to isometric
embeddings.

Krivine and Maurey introduced the notion of stability for separable Banach
spaces [kri-mau:81] in order to extend a result of Aldous [aldous:81] by
showing that every stable Banach space admits an isomorphic copy of $\ell_{p}$
for some $p\in[1,\infty)$. Garling [garling:82] then studied this isometric
Banach space notion in the more general context of arbitrary metric spaces.

######  Definition.

A metric space $(M,d_{M})$ is _stable_ if for any pair
$\mathcal{U},\mathcal{V}$ of free ultrafilters on $\mathbb{N}$ and for any two
bounded sequences $(x_{n}),(y_{n})$ in $M$ we have

$\lim_{m,\mathcal{U}}\lim_{n,\mathcal{V}}d_{M}(x_{m},y_{n})=\lim_{n,\mathcal{V}}\lim_{m,\mathcal{U}}d_{M}(x_{m},y_{n})\
.$

A close connection between stability and reflexivity in the category of
metrizable topological spaces and continuous maps was known since the original
work of Krivine and Maurey, and the new insight of Kalton was to establish
another connection in the category of metric spaces and Lipschitz maps. In the
latter connection, certain spaces of Lipschitz functions play a decisive role.
Both connections rely on an important factorization result for weakly compact
operators via reflexive spaces [dav-fig-joh-pel:74]. Kalton raised a few
natural problems.

######  Problem A ([kalton:07]*Problem 6.1).

Does every (separable) reflexive Banach space coarsely (or uniformly) embed
into a stable metric space? Does every reflexive Banach space admit an
embedding, which is simultaneously coarse and uniform, into a stable metric
space?

Kalton explicitly suspected that Problem A might have a negative solution, but
that Problem B below could have a positive solution.

######  Problem B ([kalton:07]*Problem 6.2).

If $X$ is a separable reflexive Banach space, does $B_{X}$ embed uniformly
into a stable metric space?

We will now mainly focus on the coarse case of Problem A (as the discussion of
the uniform case is similar). Recall that a metric space $(M,d_{M})$ _coarsely
embeds_ into a metric space $(N,d_{N})$ if there exist a map $f\colon M\to N$
and non-decreasing functions $\omega,\rho\colon[0,\infty)\to[0,\infty)$ with
$\lim_{t\to\infty}\rho(t)=\infty$ such that for all $x,y\in M$ we have

(1) $\rho\big{(}d_{M}(x,y)\big{)}\leqslant
d_{N}\big{(}f(x),f(y)\big{)}\leqslant\omega\big{(}d_{M}(x,y)\big{)}\ .$

A time-tested method to provide a negative answer to Problem A is to find a
coarse invariant, i.e., a property that is preserved under coarse embeddings,
that is possessed by every stable metric space but fails for at least one
separable reflexive Banach space. In [kalton:07], Kalton introduced several
variants of a concentration property for Lipschitz maps defined on certain
infinite graphs, namely the _interlacing graphs_. Given an infinite subset $L$
of $\mathbb{N}$ and an integer $k\in\mathbb{N}$, we consider the set
${[L]}^{k}$ of all subsets of $L$ of size $k$ as the vertex set of a graph
equipped with the interlacing adjacency relation, where two vertices
$\accentset{\rule{5.60002pt}{0.6pt}}{m}=\\{m_{1},\dots,m_{k}\\}$ and
$\accentset{\rule{2.79996pt}{0.6pt}}{n}=\\{n_{1},\dots,n_{k}\\}$ in
${[L]}^{k}$ with $m_{1}<\dots<m_{k}$ and $n_{1}<\dots<n_{k}$ are adjacent if
and only if they _interlace_ , i.e.,
$\accentset{\rule{5.60002pt}{0.6pt}}{m}\neq\accentset{\rule{2.79996pt}{0.6pt}}{n}$
and either $m_{1}\leqslant n_{1}\leqslant m_{2}\leqslant
n_{2}\leqslant\dots\leqslant m_{k}\leqslant n_{k}$ or $n_{1}\leqslant
m_{1}\leqslant n_{2}\leqslant m_{2}\leqslant\dots\leqslant n_{k}\leqslant
m_{k}$. The graph metric induced by the interlacing adjacency relation will be
denoted by $d_{I}$.

######  Definition.

A metric space $(M,d_{M})$ is said to have Kalton’s _property $Q$ _if there is
a constant $C\geqslant 0$ such that for all $k\in\mathbb{N}$ and for all
bounded functions $f\colon({[\mathbb{N}]}^{k},d_{I})\to(M,d_{M})$, there
exists $L\in{[\mathbb{N}]}^{\omega}$ such that

$\sup_{\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[L]}^{k}}d_{M}\big{(}f(\accentset{\rule{5.60002pt}{0.6pt}}{m}),f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\big{)}\leqslant
C\operatorname{\mathrm{Lip}}(f)\ .$

The infimum of all constants $C$ satisfying the above will be denoted $Q_{M}$
and called the _property $Q$ constant of $M$_.

Here ${[\mathbb{N}]}^{\omega}$ denotes the set of all infinite subsets of
$\mathbb{N}$ and $\operatorname{\mathrm{Lip}}(f)$ is the Lipschitz constant of
$f$. Kalton’s original definition of property $Q$ for Banach spaces was given
in terms of a modulus [kalton:07]*pages 403–404, but eventually took the form
above in [kalton:13]*page 1057 (see also [kalton:12]*page 1283).

The study of concentration inequalities for Lipschitz maps defined on infinite
graphs with countable degree, which began with property $Q$ in [kalton:07] for
interlacing graphs and continued shortly thereafter in [kal-ran:08] for
Hamming graphs, was a genuine breakthrough towards unlocking many mysteries
regarding the nonlinear geometry of Banach spaces in terms of their infinite-
dimensional and asymptotic properties (see [bau-kal-
lan:10][kalton:11][bcdkrsz:17][lan-raj:18][bau-lan-sch:18][netillard:18][bra-
lan-pet-pro:23][lan-pet-pro:20][bau-lan-mot-sch:21a][bau-lan-mot-sch:21b][bau-
gar:21][fovelle:23]).

Kalton observed [kalton:07]*Proposition 3.1 that every stable metric space has
property $Q$, and that if a Banach space $X$ admits a coarse embedding into a
metric space with property $Q$, then $X$ has property $Q$. The interlacing
graph metric can be seen as being induced by the summing basis of
$\mathrm{c}_{0}$, and thus every Banach space containing an isomorphic copy of
$\mathrm{c}_{0}$ contains bi-Lipschitz copies of the sequence
$\big{(}{[\mathbb{N}]}^{k},d_{I}\big{)}_{k=1}^{\infty}$ of interlacing graphs
with uniformly bounded distortion. It follows that $\mathrm{c}_{0}$ does not
embed coarsely into any metric space with property $Q$. The second remarkable
result of Kalton from [kalton:07] is that reflexive Banach spaces also have
property $Q$. An important consequence is that $\mathrm{c}_{0}$ does not
coarsely embed into any reflexive Banach space, thereby solving a longstanding
open question in the coarse geometry of Banach spaces. Kalton also showed that
property $Q$ is a uniform as well as coarse invariant for Banach spaces: if a
Banach space $X$, or its closed unit ball $B_{X}$, admits a uniform embedding
into a metric space with property $Q$, then $X$ has property $Q$. It follows
that $\mathrm{c}_{0}$, or indeed the closed unit ball of $\mathrm{c}_{0}$,
does not uniformly embed into any reflexive Banach space, solving yet another
old open problem in the nonlinear theory of Banach spaces.

The fact that reflexive spaces have property $Q$ rules out property $Q$ as a
potential coarse invariant to resolve negatively Problem A. In an attempt to
attack Problem A, the first author introduced in [baudier:22] another
invariant called upper stability. This bi-Lipschitz invariant is a natural
relaxation of the notion of stability in light of a characterization of
stability due to Raynaud [raynaud:83] (see Theorem 1 below). Upper stability
was shown to imply property $Q$ and to be preserved under coarse embeddings
(for Banach space domains and metric space targets akin to property $Q$).

######  Definition.

Let $(M,d_{M})$ be a metric space and $C$ be a non-negative real number. We
say that $M$ is _$C$ -upper stable _if for all integers $1\leqslant l<k$,
bounded functions $f\colon{[\mathbb{N}]}^{l}\to M$ and
$g\colon{[\mathbb{N}]}^{k-l}\to M$, infinite sets $L\subset\mathbb{N}$ and
permutations $\pi\in S_{k}$ preserving the order on $\\{1,\dots,l\\}$ and on
$\\{l+1,\dots,k\\}$,

$\inf_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[L]}^{k}}d_{M}\big{(}f(n_{1},\dots,n_{l}),g(n_{l+1},\dots,n_{k})\big{)}\leqslant\\\
C\cdot\sup_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[L]}^{k}}d_{M}\big{(}f(n_{\pi(1)},\dots,n_{\pi(l)}),g(n_{\pi(l+1)},\dots,n_{\pi(k)})\big{)}\
.$

We say that $M$ is _upper stable_ if it is $C$-upper stable for some $C$.

There are good reasons to think that an approach to solving Problem A via
upper stability would work. Indeed, if $X$ is a Banach space containing no
copies of $\ell_{p}$ for $1\leqslant p<\infty$, then $X$ cannot isomorphically
embed into a stable Banach space because of the result of Krivine and Maurey
mentioned above. If in addition $X$ is separable, then $X$ is not even bi-
Lipschitzly embeddable into a reflexive stable Banach space due to a result of
Heinrich and Mankiewicz [hei-man:82]*Theorem 3.5. Thus, any such space $X$
(Tsirelson’s space, Schlumprecht’s space, Gowers-Maurey space, etc.) could be
a candidate for a reflexive space failing upper stability. In this paper, we
dash this hope in two different ways, and at the same time, we further our
understanding of the connection between stability, reflexivity and property
$Q$.

In Section 3, we prove that every reflexive Banach space is upper stable.
Since upper stability implies property $Q$ (see [baudier:22]), we thus recover
the fact, originally proved by Kalton in [kalton:07], that reflexive spaces
have property $Q$. Kalton’s original proof relies on some clever “calculus”
for certain limiting operators in reflexive spaces. In order to prove the a
priori stronger property of upper stability, we take a more conceptual
approach as we relate upper stability with the asymptotic structure of Banach
spaces in the sense of Maurey, Milman and Tomczak-Jaegermann [mau-mil-tom:95].

First, however, we prove in Section 2 the more general result that property
$Q$ and upper stability are equivalent notions (up to the value of the
constants involved) in the category of metric spaces.

##  2\. Upper stability and property $Q$

We begin by stating the following characterization of stability, which
combines work of Krivine and Maurey [kri-mau:81] and Raynaud [raynaud:83] (see
also [baudier:22]).

######  Theorem 1.

The following are equivalent for a metric space $(M,d_{M})$.

  * (i)

$M$ is stable.

  * (ii)

For all $1\leqslant l<k$, bounded functions $f\colon{[\mathbb{N}]}^{l}\to M$
and $g\colon{[\mathbb{N}]}^{k-l}\to M$, free ultrafilters
$\mathcal{U}_{1},\dots,\mathcal{U}_{k}$ and permutations $\pi\in S_{k}$
preserving the order on $\\{1,\dots,l\\}$ and on $\\{l+1,\dots,k\\}$,

$\lim_{n_{1},\mathcal{U}_{1}}\dots\lim_{n_{k},\mathcal{U}_{k}}d_{M}\big{(}f(n_{1},\dots,n_{l}),g(n_{l+1},\dots,n_{k})\big{)}=\\\
\lim_{n_{\pi^{-1}(1)},\mathcal{U}_{\pi^{-1}(1)}}\dots\lim_{n_{\pi^{-1}(k)},\mathcal{U}_{\pi^{-1}(k)}}d_{M}\big{(}f(n_{1},\dots,n_{l}),g(n_{l+1},\dots,n_{k})\big{)}\
.$

  * (iii)

For all $1\leqslant l<k$, bounded functions $f\colon{[\mathbb{N}]}^{l}\to M$
and $g\colon{[\mathbb{N}]}^{k-l}\to M$, free ultrafilters $\mathcal{U}$ and
permutations $\pi\in S_{k}$ preserving the order on $\\{1,\dots,l\\}$ and on
$\\{l+1,\dots,k\\}$,

$\lim_{n_{1},\mathcal{U}}\dots\lim_{n_{k},\mathcal{U}}d_{M}\big{(}f(n_{1},\dots,n_{l}),g(n_{l+1},\dots,n_{k})\big{)}=\\\
\lim_{n_{1},\mathcal{U}}\dots\lim_{n_{k},\mathcal{U}}d_{M}\big{(}f(n_{\pi(1)},\dots,n_{\pi(l)}),g(n_{\pi(l+1)},\dots,n_{\pi(k)})\big{)}\
.$

  * (iv)

For all $1\leqslant l<k$, bounded functions $f\colon{[\mathbb{N}]}^{l}\to M$
and $g\colon{[\mathbb{N}]}^{k-l}\to M$, infinite sets $L\subset\mathbb{N}$ and
permutations $\pi\in S_{k}$ preserving the order on $\\{1,\dots,l\\}$ and on
$\\{l+1,\dots,k\\}$,

$\inf_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[L]}^{k}}d_{M}\big{(}f(n_{1},\dots,n_{l}),g(n_{l+1},\dots,n_{k})\big{)}\leqslant\\\
\sup_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[L]}^{k}}d_{M}\big{(}f(n_{\pi(1)},\dots,n_{\pi(l)}),g(n_{\pi(l+1)},\dots,n_{\pi(k)})\big{)}\
.$

In light of the above characterization, the definition of upper stability
given in the Introduction is a very natural relaxation of stability. There is
a certain asymmetry in part (iv) above, and hence in the definition of upper
stability, in the sense that the permutation is applied only to the right-hand
side of the defining inequality. We will now introduce an a priori stronger
version of upper stability by “symmetrizing” this inequality. We will also
introduce a different formulation in terms of cuts instead of permutations.

Fix $k,l,m\in\mathbb{N}$ with $k=l+m$. By a _cut of $\\{1,\dots,k\\}$ of size
$l$ _we mean simply a subset $P$ of $\\{1,\dots,k\\}$ of size $l$. If
$P=\\{p_{1},\dots,p_{l}\\}$ with $p_{1}<\dots<p_{l}$, then for
$n_{1}<\dots<n_{k}$ in $\mathbb{N}$, we write $(n_{i}:\,i\in P)$ for
$(n_{p_{1}},\dots,n_{p_{l}})$. The proof of Raynaud’s characterization of
stability shows the following.

######  Proposition 2.

For a metric space $(M,d_{M})$ and a constant $C\geqslant 1$, the following
are equivalent.

  * (i)

For all $k,l,m\in\mathbb{N}$ with $k=l+m$, free ultrafilters $\mathcal{U}$,
bounded functions $f\colon{[\mathbb{N}]}^{l}\to M$ and
$g\colon{[\mathbb{N}]}^{m}\to M$ and cuts $P,Q$ of $\\{1,\dots,k\\}$ of size
$l$, we have

$\lim_{n_{1},\mathcal{U}}\dots\lim_{n_{k},\mathcal{U}}d_{M}\big{(}f(n_{i}:\,i\in
P),g(n_{i}:\,i\in P^{c})\big{)}\leqslant\\\
C\cdot\lim_{n_{1},\mathcal{U}}\dots\lim_{n_{k},\mathcal{U}}d_{M}\big{(}f(n_{i}:\,i\in
Q),g(n_{i}:\,i\in Q^{c})\big{)}\ .$

  * (ii)

For all $k,l,m\in\mathbb{N}$ with $k=l+m$, bounded functions
$f\colon{[\mathbb{N}]}^{l}\to M$ and $g\colon{[\mathbb{N}]}^{m}\to M$,
infinite sets $L\subset\mathbb{N}$ and cuts $P,Q$ of $\\{1,\dots,k\\}$ of size
$l$, we have

$\inf_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[L]}^{k}}d_{M}\big{(}f(n_{i}:\,i\in
P),g(n_{i}:\,i\in P^{c})\big{)}\leqslant\\\
C\cdot\sup_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[L]}^{k}}d_{M}\big{(}f(n_{i}:\,i\in
Q),g(n_{i}:\,i\in Q^{c})\big{)}\ .$

######  Definition.

We say a metric space $(M,d_{M})$ is _$C$ -cut stable _if it satisfies one of
the two equivalent conditions in Proposition 2. We say $M$ is _cut stable_ if
it is $C$-cut stable for some $C\geqslant 1$.

######  Remarks.

We recover the notion of upper stability by taking $P$ to be the trivial cut
$\\{1,2,\dots,l\\}$ above. We will later see that every upper stable metric
space is cut stable, although this is not entirely trivial. It is clear that
upper stability is a Lipschitz invariant, and hence in particular an
isomorphic invariant for Banach spaces. In contrast, stability is an isometric
invariant. It is easy to find equivalent norms on the stable Banach space
$\ell_{1}$, for instance, which is not stable.

As mentioned in the Introduction, Kalton’s original definition of property $Q$
was given in terms of a modulus, which we now recall.

######  Definition (Kalton [kalton:07]).

Let $(M,d_{M})$ be a metric space. Given $\varepsilon>0$ and $\delta\geqslant
0$, we say that $M$ has the _$Q(\varepsilon,\delta)$ -property _if for all
$k\in\mathbb{N}$ and for all functions
$f\colon\big{(}{[\mathbb{N}]}^{k},d_{I}\big{)}\to M$ with
$\operatorname{\mathrm{Lip}}(f)\leqslant\delta$, there exists
$L\in{[\mathbb{N}]}^{\omega}$ such that

$d_{M}(f(\accentset{\rule{5.60002pt}{0.6pt}}{m}),f(\accentset{\rule{2.79996pt}{0.6pt}}{n}))<\varepsilon\qquad\text{for
all
}\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}\text{
in }{[L]}^{k}\ .$

For $\varepsilon>0$ we define

$\Delta_{M}(\varepsilon)=\sup\\{\delta\geqslant 0:\,M\text{ has the
$Q(\varepsilon,\delta)$-property}\\}\ ,$

and set $q_{M}$ to be the supremum of all constants $c\geqslant 0$ for which
$\Delta_{M}(\varepsilon)\geqslant c\varepsilon$ for all $\varepsilon>0$.

######  Remark.

It is immediate from above that $M$ has property $Q$ if and only if $q_{M}>0$
in which case $q_{M}^{-1}\leqslant Q_{M}\leqslant 2q_{M}^{-1}$. Kalton proved
in [kalton:07] that for a stable metric space $M$, $q_{M}\geqslant 1$, and for
a reflexive Banach space $X$, $q_{X}\geqslant 1/2$ (and $Q_{X}\leqslant 2$).
Note that for any infinite-dimensional Banach space $X$, $q_{X}\leqslant 1$.
Indeed, for any $\delta>1$ there exists $f\colon\mathbb{N}\to X$ with
$1<\lVert f(m)-f(n)\rVert<\delta$ for all $m<n$, and thus $X$ does not have
the $Q(1,\delta)$-property.

######  Remark.

Let $(e_{i})$ and $(s_{i})$ denote the unit vector basis and, respectively,
the summing basis of $\mathrm{c}_{0}$. Thus, $s_{n}=\sum_{i=1}^{n}e_{i}$,
$n\in\mathbb{N}$. For $k\in\mathbb{N}$ define
$f\colon({[\mathbb{N}]}^{k},d_{I})\to\mathrm{c}_{0}$ by
$f(\accentset{\rule{2.79996pt}{0.6pt}}{n})=\sum_{i=1}^{k}s_{n_{i}}$. Note that
$\lVert
f(\accentset{\rule{5.60002pt}{0.6pt}}{m})-f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert=d_{I}(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n})=1$
for adjacent vertices, and hence $\operatorname{\mathrm{Lip}}(f)=1$. Moreover,
$\lVert
f(\accentset{\rule{5.60002pt}{0.6pt}}{m})-f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert=k$
for all
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
in ${[\mathbb{N}]}^{k}$. Thus $\mathrm{c}_{0}$ has the worst possible
behaviour for property $Q$. Indeed, this behaviour of the summing basis
motivated the proof by Kalton that $\mathrm{c}_{0}$ does not uniformly or
coarsely embed into a reflexive space. It was also the motivation behind the
interlacing graph. Indeed,

$\lVert
f(\accentset{\rule{5.60002pt}{0.6pt}}{m})-f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert\leqslant
d_{I}(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n})\leqslant
2\lVert
f(\accentset{\rule{5.60002pt}{0.6pt}}{m})-f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert\qquad\text{for
all
}\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[\mathbb{N}]}^{k}$

and for every $k\in\mathbb{N}$.

######  Remark.

If follows easily by scaling that for a Banach space $X$, the function
$\Delta_{X}$ is linear, and thus $\Delta_{X}(\varepsilon)=q_{X}\varepsilon$
for all $\varepsilon>0$. Another observation of Kalton seems slightly less
obvious, so we provide a proof next.

######  Proposition 3.

For a Banach space $X$, $\Delta_{X}(\varepsilon)=\Delta_{B_{X}}(\varepsilon)$
for all $0<\varepsilon<1$.

###### Proof.

Clearly, $\Delta_{X}(\varepsilon)\leqslant\Delta_{B_{X}}(\varepsilon)$ for any
$\varepsilon>0$. For the converse, we will first show that $\Delta_{B_{X}}$ is
“linear” on $(0,1)$. Fix $b$ with $0<b<1$. Given

$0<\varepsilon_{1}<\varepsilon_{2}\leqslant b\ ,$

write $\varepsilon_{2}=a\varepsilon_{1}$ for suitable $a>1$. We show that
$\Delta_{B_{X}}(\varepsilon_{2})=a\Delta_{B_{X}}(\varepsilon_{1})$. This will
show linearity on $(0,b]$, and hence on $(0,1)$. To show that
$\Delta_{B_{X}}(\varepsilon_{2})=a\Delta_{B_{X}}(\varepsilon_{1})$, we may
clearly assume that $a\leqslant b^{-1}$ as the general case will then follow
by repeated applications of the special case. The inequality
$\Delta_{B_{X}}(\varepsilon_{2})\geqslant a\Delta_{B_{X}}(\varepsilon_{1})$
follows by simple scaling. Conversely, assume that $B_{X}$ has the
$Q(\varepsilon_{2},\delta)$-property. Let $k\in\mathbb{N}$ and
$f\colon\big{(}{[\mathbb{N}]}^{k},d_{I}\big{)}\to B_{X}$ be a function with
$\operatorname{\mathrm{Lip}}(f)\leqslant\delta/a$. Then in particular,
$\operatorname{\mathrm{Lip}}(f)\leqslant\delta$, and hence there exists
$L\in{[\mathbb{N}]}^{\omega}$ such that $\lVert
f(\accentset{\rule{5.60002pt}{0.6pt}}{m})-f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert<\varepsilon_{2}$
for all
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
in ${[L]}^{k}$. Enumerate $L$ as $l_{1}<l_{2}<\dots$ and set
$\accentset{\rule{2.79996pt}{0.6pt}}{l}=\\{l_{1},\dots,l_{k}\\}$. Define
$g\colon\big{(}{[\mathbb{N}]}^{k},d_{I}\big{)}\to X$ by

$g(n_{1},\dots,n_{k})=\big{(}f(l_{k+n_{1}},\dots,l_{k+n_{k}})-f(\accentset{\rule{2.79996pt}{0.6pt}}{l})\big{)}\cdot
a\ .$

Since $a\varepsilon_{2}\leqslant ab\leqslant 1$, it follows that $g$ takes
values in $B_{X}$. Moreover, $\operatorname{\mathrm{Lip}}(g)\leqslant
a\operatorname{\mathrm{Lip}}(f)\leqslant\delta$, and hence there exists
$N\in{[\mathbb{N}]}^{\omega}$ such that $\lVert
g(\accentset{\rule{5.60002pt}{0.6pt}}{m})-g(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert<\varepsilon_{2}$
for all
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
in ${[N]}^{k}$. It follows that $\lVert
f(\accentset{\rule{5.60002pt}{0.6pt}}{m})-f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert<\varepsilon_{1}$
for all
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
in ${[P]}^{k}$, where $P=\\{l_{k+n}:\,n\in N\\}$. We have proved that $B_{X}$
has the $Q(\varepsilon_{1},\delta/a)$-property, and hence
$a\cdot\Delta_{B_{X}}(\varepsilon_{1})\geqslant\Delta_{B_{X}}(\varepsilon_{2})$,
as required.

Having shown linearity of $\Delta_{B_{X}}$ on $(0,1)$, we now finish the proof
by showing that $\Delta_{B_{X}}(\varepsilon)\leqslant\Delta_{X}(\varepsilon)$
for any $0<\varepsilon<1$. Assume that
$0\leqslant\delta<\Delta_{B_{X}}(\varepsilon)$ for some $\delta$ and
$0<\varepsilon<1$. Let $k\in\mathbb{N}$ and
$f\colon\big{(}{[\mathbb{N}]}^{k},d_{I}\big{)}\to X$ be a function with
$\operatorname{\mathrm{Lip}}(f)\leqslant\delta$. Then for some
$n\in\mathbb{N}$, the function $n^{-1}f$ takes values in $B_{X}$, and moreover
$\operatorname{\mathrm{Lip}}(n^{-1}f)\leqslant{}n^{-1}\delta<n^{-1}\Delta_{B_{X}}(\varepsilon)=\Delta_{B_{X}}(n^{-1}\varepsilon)$.
Hence there exists $L\in{[\mathbb{N}]}^{\omega}$ such that $\lVert
n^{-1}f(\accentset{\rule{5.60002pt}{0.6pt}}{m})-n^{-1}f(\accentset{\rule{2.79996pt}{0.6pt}}{n})\rVert<n^{-1}\varepsilon$
for all
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
in ${[L]}^{k}$. It follows that $X$ has the $Q(\varepsilon,\delta)$-property,
and thus $\Delta_{X}(\varepsilon)\geqslant\Delta_{B_{X}}(\varepsilon)$ as
required. ∎

The following result shows that property $Q$ is preserved under coarse and
uniform embeddings with Banach space domain and metric space target. This is a
trivial extension of [kalton:07]*Theorem 4.2. We give the short proof for
convenience. Let us first recall that for a function
$f\colon(M,d_{M})\to(N,d_{N})$ between metric spaces, its _modulus of
continuity $\omega_{f}$ _and _compression modulus $\rho_{f}$ _are defined,
respectively, by

$\omega_{f}(t)=\sup\\{d_{N}(f(x),f(y)):\,d_{M}(x,y)\leqslant t\\}$

and

$\rho_{f}(t)=\inf\\{d_{N}(f(x),f(y)):\,d_{M}(x,y)\geqslant t\\}$

for $t\in[0,\infty)$. Note that $f$ is a uniform embedding if and only if
$\lim_{t\to 0^{+}}\omega_{f}(t)=0$ and $\rho_{f}(t)>0$ for all $t>0$, and that
$f$ is a coarse embedding if and only if $\lim_{t\to\infty}\rho_{f}(t)=\infty$
and $\omega_{f}(t)<\infty$ for all $t>0$. Note also that if $M$ is a graph and
$d_{M}$ is the graph distance, then
$\operatorname{\mathrm{Lip}}(f)=\omega_{f}(1)$.

######  Proposition 4.

Let $h\colon(M,d_{M})\to(N,d_{N})$ be a function between metric spaces, and
assume that $N$ has property $Q$. Then for all $\varepsilon>0$,

$\Delta_{M}(\varepsilon)\geqslant\sup\\{\delta\geqslant
0:\,q_{N}\cdot\rho_{h}(\varepsilon)>\omega_{h}(\delta)\\}\ .$

This has the following consequences.

  * (i)

If $h$ is a uniform embedding, then $\Delta_{M}(\varepsilon)>0$ for all
$\varepsilon>0$.

  * (ii)

If $h$ is a coarse embedding, then $\Delta_{M}(\varepsilon)\to\infty$ as
$\varepsilon\to\infty$.

In particular, if a Banach space $X$ embeds coarsely into a metric space with
property $Q$ or if the unit ball $B_{X}$ embeds uniformly into a metric space
with property $Q$, then $X$ has property $Q$.

###### Proof.

Let $\delta\geqslant 0$ and $\varepsilon>0$ be such that $M$ fails the
$Q(\varepsilon,\delta)$-property. Then there exist $k\in\mathbb{N}$ and a
function $f\colon{[\mathbb{N}]}^{k}\to M$ such that
$\omega_{f}(1)\leqslant\delta$ and for all $L\in{[\mathbb{N}]}^{\omega}$ there
exist
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
in ${[L]}^{k}$ with
$d_{M}(f(\accentset{\rule{5.60002pt}{0.6pt}}{m}),f(\accentset{\rule{2.79996pt}{0.6pt}}{n}))\geqslant\varepsilon$.
Then $\omega_{hf}(1)\leqslant\omega_{h}(\delta)$ and for all
$L\in{[\mathbb{N}]}^{\omega}$ there exist
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
in ${[L]}^{k}$ with
$d_{N}(hf(\accentset{\rule{5.60002pt}{0.6pt}}{m}),hf(\accentset{\rule{2.79996pt}{0.6pt}}{n}))\geqslant\rho_{h}(\varepsilon)$.
It follows that $N$ does not have the
$Q(\rho_{h}(\varepsilon),\omega_{h}(\delta))$-property, and hence

$\omega_{h}(\delta)\geqslant\Delta_{N}(\rho_{h}(\varepsilon))\geqslant
q_{N}\cdot\rho_{h}(\varepsilon)\ .$

The result follows. ∎

Recall that for $k\in\mathbb{N}$, the interlacing graph on the vertex set
${[\mathbb{N}]}^{k}$ joins two vertices
$\accentset{\rule{5.60002pt}{0.6pt}}{m}=\\{m_{1},\dots,m_{k}\\}$ and
$\accentset{\rule{2.79996pt}{0.6pt}}{n}=\\{n_{1},\dots,n_{k}\\}$ if and only
if $\accentset{\rule{5.60002pt}{0.6pt}}{m}$ and
$\accentset{\rule{2.79996pt}{0.6pt}}{n}$ are interlacing, i.e., either

  * (i)

$m_{1}\leqslant n_{1}\leqslant m_{2}\leqslant n_{2}\leqslant\dots\leqslant
m_{k}\leqslant n_{k}$ or

  * (ii)

$n_{1}\leqslant m_{1}\leqslant n_{2}\leqslant m_{2}\leqslant\dots\leqslant
n_{k}\leqslant m_{k}$.

######  Remark.

Using the above notation, let $m(x)=\lvert\\{i:\,m_{i}\leqslant x\\}\rvert$,
$x\in\mathbb{N}$. The function $x\mapsto m(x)-n(x)$ measures the discrepancy
between $\accentset{\rule{5.60002pt}{0.6pt}}{m}$ and
$\accentset{\rule{2.79996pt}{0.6pt}}{n}$, which in turn measures the graph
distance between $\accentset{\rule{5.60002pt}{0.6pt}}{m}$ and
$\accentset{\rule{2.79996pt}{0.6pt}}{n}$:

$d_{I}(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n})=\max_{x\in\mathbb{N}}\big{(}m(x)-n(x)\big{)}-\min_{x\in\mathbb{N}}\big{(}m(x)-n(x)\big{)}\
.$

It follows that the diameter of the interlacing graph on ${[\mathbb{N}]}^{k}$
is $k$, and moreover,
$d_{I}(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n})=k$
if and only if either $\accentset{\rule{5.60002pt}{0.6pt}}{m}$ lies strictly
between two consecutive elements of $\accentset{\rule{2.79996pt}{0.6pt}}{n}$
including the degenerate cases
$\accentset{\rule{5.60002pt}{0.6pt}}{m}<\accentset{\rule{2.79996pt}{0.6pt}}{n}$
(which means that $m_{k}<n_{1}$) and
$\accentset{\rule{2.79996pt}{0.6pt}}{n}<\accentset{\rule{5.60002pt}{0.6pt}}{m}$,
or vice versa. As noted earlier, the summing basis of $\mathrm{c}_{0}$
provides another description of the graph distance $d_{I}$.

We will need the following stronger version of the interlacing adjacency
relation.

######  Definition.

Given $k\in\mathbb{N}$ and
$\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[\mathbb{N}]}^{k}$,
we say that the pair
$(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n})$
is _strongly interlacing_ if either

  * (i)

$m_{1}\leqslant n_{1}<m_{2}\leqslant n_{2}<\dots<m_{k}\leqslant n_{k}$ or

  * (ii)

$n_{1}\leqslant m_{1}<n_{2}\leqslant m_{2}<\dots<n_{k}\leqslant m_{k}$.

######  Lemma 5.

Given $k\in\mathbb{N}$ and an interlacing pair
$(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n})$
in ${[2\mathbb{N}]}^{k}$, there exists
$\accentset{\rule{2.79996pt}{0.6pt}}{p}\in{[\mathbb{N}]}^{k}$ such that the
pairs
$(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{p})$
and
$(\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{2.79996pt}{0.6pt}}{p})$
are strongly interlacing.

###### Proof.

Without loss of generality, we may assume that

$m_{1}\leqslant n_{1}\leqslant m_{2}\leqslant n_{2}\leqslant\dots\leqslant
m_{k}\leqslant n_{k}\ .$

Define $\accentset{\rule{2.79996pt}{0.6pt}}{p}=\\{p_{1},\dots,p_{k}\\}$ in
${[\mathbb{N}]}^{k}$ by

$p_{i}=\begin{cases}n_{i}-1&\text{if }n_{i}=m_{i+1}\ ,\\\ n_{i}&\text{if
}n_{i}<m_{i+1}\ .\end{cases}$

It is straightforward to verify that this works. ∎

We are now ready to state and prove the main result of this section.

######  Theorem 6.

A metric space with property $Q$ is cut stable.

###### Proof.

Let $(M,d_{M})$ be a metric space with property $Q$. We will show that $M$ is
$C$-cut stable with $C=8Q_{M}+1$. Fix integers $1\leqslant l<k$, bounded
functions $f\colon{[\mathbb{N}]}^{l}\to M$, $g\colon{[\mathbb{N}]}^{k-l}\to M$
and cuts $P,Q$ of $\\{1,\dots,k\\}$ of size $l$. Set

$\displaystyle A$
$\displaystyle=\inf_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[\mathbb{N}]}^{k}}d_{M}\big{(}f(n_{i}:\,i\in
P),g(n_{i}:\,i\in P^{c})\big{)}$ $\displaystyle B$
$\displaystyle=\sup_{\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[\mathbb{N}]}^{k}}d_{M}\big{(}f(n_{i}:\,i\in
Q),g(n_{i}:\,i\in Q^{c})\big{)}\ .$

Put $L=\\{2nk:\,n\in\mathbb{N}\\}$ and let $L_{f}$ and $L_{g}$ be the
Lipschitz constants of $f\\!\\!\restriction_{{[L]}^{l}}$ and of
$g\\!\\!\restriction_{{[L]}^{k-l}}$, respectively. We first show that
$L_{f}\leqslant 4B$ and $L_{g}\leqslant 4B$. To see this, set
$L^{\prime}=\\{nk:\,n\in\mathbb{N}\\}$ and let
$(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n})$
be an interlacing pair in ${[L]}^{l}$. By Lemma 5 there exists
$\accentset{\rule{2.79996pt}{0.6pt}}{p}\in{[L^{\prime}]}^{l}$ such that the
pairs
$(\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{p})$
and
$(\accentset{\rule{2.79996pt}{0.6pt}}{p},\accentset{\rule{2.79996pt}{0.6pt}}{n})$
are strongly interlacing. We can then choose $m_{1}<\dots<m_{k}$ and
$p_{1}<\dots<p_{k}$ in $L^{\prime}$ such that
$\accentset{\rule{5.60002pt}{0.6pt}}{m}=\\{m_{i}:\,i\in Q\\}$,
$\accentset{\rule{2.79996pt}{0.6pt}}{p}=\\{p_{i}:\,i\in Q\\}$ and
$m_{i}=p_{i}$ for all $i\in Q^{c}$. It follows that

$d_{M}(f(\accentset{\rule{5.60002pt}{0.6pt}}{m}),f(\accentset{\rule{2.79996pt}{0.6pt}}{p}))\leqslant
d_{M}\big{(}f(m_{i}:\,i\in Q),g(m_{i}:\,i\in Q^{c})\big{)}\\\
+d_{M}\big{(}f(p_{i}:\,i\in Q),g(p_{i}:\,i\in Q^{c})\big{)}\leqslant 2B\ .$

Similarly, we get
$d_{M}(f(\accentset{\rule{2.79996pt}{0.6pt}}{p}),f(\accentset{\rule{2.79996pt}{0.6pt}}{n}))\leqslant
2B$, and hence
$d_{M}(f(\accentset{\rule{5.60002pt}{0.6pt}}{m}),f(\accentset{\rule{2.79996pt}{0.6pt}}{n}))\leqslant
4B$, as required. A similar argument shows that $L_{g}\leqslant 4B$.

Now fix $C>Q_{M}$. Since $M$ has property $Q$, there exists
$N\in{[L]}^{\omega}$ such that
$d_{M}(f(\accentset{\rule{5.60002pt}{0.6pt}}{m}),f(\accentset{\rule{2.79996pt}{0.6pt}}{n}))\leqslant
CL_{f}$ for all
$\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[N]}^{l}$
and
$d_{M}(g(\accentset{\rule{5.60002pt}{0.6pt}}{m}),g(\accentset{\rule{2.79996pt}{0.6pt}}{n}))\leqslant
CL_{g}$ for all
$\accentset{\rule{5.60002pt}{0.6pt}}{m},\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[N]}^{k-l}$.
Now fix any $n_{1}<\dots<n_{k}$ in $N$. We get

$\displaystyle A$ $\displaystyle\leqslant$ $\displaystyle
d_{M}\big{(}f(n_{i}:\,i\in P),g(n_{i}:\,i\in P^{c})\big{)}$
$\displaystyle\leqslant$ $\displaystyle d_{M}\big{(}f(n_{i}:i\in
P),f(n_{i}:i\in Q)\big{)}$ $\displaystyle+d_{M}\big{(}f(n_{i}:i\in
Q),g(n_{i}:i\in Q^{c})\big{)}$ $\displaystyle+d_{M}\big{(}g(n_{i}:i\in
Q^{c}),g(n_{i}:i\in P^{c})\big{)}\leqslant(8C+1)B\ .$

The result follows. ∎

The implication that an upper stable metric space has property $Q$ was shown
in [baudier:22]. Combining this with Theorem 6, we obtain

######  Corollary 7.

The following are equivalent for a metric space $(M,d_{M})$.

  * (i)

$M$ has property $Q$.

  * (ii)

$M$ is cut stable.

  * (iii)

$M$ is upper stable.

Combining this with Proposition 4, we obtain the following.

######  Corollary 8.

If a Banach space $X$ embeds coarsely into an upper stable metric space, or if
the unit ball $B_{X}$ embeds uniformly into an upper stable metric space, then
$X$ is upper stable.

##  3\. Reflexive spaces, upper stability and asymptotic structure

Since reflexive spaces have property $Q$ (Kalton [kalton:07]), the following
is an immediate consequence of Theorem 6.

######  Corollary 9.

Every reflexive Banach space is upper stable.

In this section, we give a direct proof of Corollary 9 without using property
$Q$. Instead, we shall relate stability and asymptotic structure. We begin by
recalling the notion of asymptotic structure introduced by Maurey, Milman and
Tomczak-Jaegermann [mau-mil-tom:95].

For each $k\in\mathbb{N}$, let $\mathcal{E}_{k}$ denote the set of norms on
$\mathbb{R}^{k}$ with respect to which the standard basis $(e_{i})_{i=1}^{k}$
is a normalized, monotone basis. Note that $\mathcal{E}_{k}$ is a compact
subset of $C\big{(}[-1,1]^{k}\big{)}$ by the Arzelà–Ascoli theorem. Fix an
infinite-dimensional Banach space $X$ with norm $\lVert\cdot\rVert$, and
consider the following two-player game between $\boldsymbol{\mathrm{S}}$
(subspace chooser) and $\boldsymbol{\mathrm{P}}$ (point chooser). At each move
of the game $\boldsymbol{\mathrm{S}}$ first picks a finite-codimensional
subspace $Y$ of $X$ and $\boldsymbol{\mathrm{P}}$ responds with a vector $x\in
Y$ with $\lVert x\rVert=1$. The game finishes after exactly $k$ moves
producing a sequence $Y_{1},x_{1},Y_{2},x_{2},\dots,Y_{k},x_{k}$. Given a norm
$N$ on $\mathbb{R}^{k}$ and $\delta>0$, we say that _$\boldsymbol{\mathrm{P}}$
wins the $(N,\delta)$-game _if

(2)
$\Big{\lvert}N\big{(}\textstyle\sum_{i=1}^{k}a_{i}e_{i}\big{)}-\big{\lVert}\textstyle\sum_{i=1}^{k}a_{i}x_{i}\big{\rVert}\Big{\rvert}<\delta\qquad\text{for
all }(a_{i})_{i=1}^{k}\in[-1,1]^{k}\ .$

The _$k^{\text{th}}$ asymptotic structure of $X$ _is the set $\\{X\\}_{k}$ of
all norms $N$ on $\mathbb{R}^{k}$ for which $\boldsymbol{\mathrm{P}}$ has a
winning strategy for the $(N,\delta)$-game for all $\delta>0$. One of the
results in [mau-mil-tom:95] states that $\\{X\\}_{k}$ is a non-empty closed
subset of $\mathcal{E}_{k}$, and moreover it is the smallest closed subset
$\mathcal{F}$ of $\mathcal{E}_{k}$ such that for every $\delta>0$,
$\boldsymbol{\mathrm{S}}$ has a winning strategy for forcing $\mathcal{P}$ to
choose vectors $x_{1},\dots,x_{k}$ such that (2) holds for some
$N\in\mathcal{F}$.

Winning strategies in the game described above naturally correspond to trees.
For $k\in\mathbb{N}$ and $L\subset\mathbb{N}$ we denote by ${[L]}^{\leqslant
k}$ the set of all subsets of $L$ of size at most $k$. By a _(countably
branching) tree in $X$ of height $k$ _we mean a function
$t\colon{[\mathbb{N}]}^{\leqslant k}\to X$ or
$t\colon{[\mathbb{N}]}^{\leqslant k}\setminus\\{\emptyset\\}\to X$; in the
former case $t$ is a _rooted_ tree with root $t(\emptyset)$, and in the latter
case $t$ is an _unrooted_ tree. We say $t$ is _normalized_ if it takes values
in the unit sphere $S_{X}$ of $X$. Given a sequence
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}=(\mathcal{U}_{1},\dots,\mathcal{U}_{k})$
of free ultrafilters, we say that $t$ is
_$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}$ -weakly null _if for each
$1\leqslant i\leqslant k$ we have

$\forall\,_{\mathcal{U}_{1}}n_{1}\ \forall\,_{\mathcal{U}_{2}}n_{2}\ \dots\
\forall\,_{\mathcal{U}_{i-1}}n_{i-1}\quad\operatorname*{w\mathrm{-}\lim}_{n,\mathcal{U}_{i}}t(n_{1},\dots,n_{i-1},n)=0\
.$

In the unrooted case, given
$\accentset{\rule{2.79996pt}{0.6pt}}{n}=\\{n_{1},\dots,n_{k}\\}\in{[\mathbb{N}]}^{k}$
with $n_{1}<\dots<n_{k}$, the sequence
$\big{(}t(n_{1},\dots,n_{i})\big{)}_{i=1}^{k}$ is called a _branch_ of $t$.

Since every weakly null sequence is eventually almost contained in any finite-
codimensional subspace, it follows that for every $k\in\mathbb{N}$ and
$\delta>0$, every normalized, weakly null (with respect to some sequence of
free ultrafilters) unrooted tree in $X$ of height $k$ has a branch
$(x_{1},\dots,x_{k})$ satisfying (2) for some $N\in\\{X\\}_{k}$. We shall need
a somewhat stronger and more precise formulation of this statement. Let us fix
$k\in\mathbb{N}$, a sequence
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}=(\mathcal{U}_{1},\dots,\mathcal{U}_{k})$
of free ultrafilters and a normalized,
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}$-weakly null tree $t$ in
$X$. Define

$\displaystyle N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\colon
X\oplus\mathbb{R}^{k}$ $\displaystyle\to$ $\displaystyle\mathbb{R}$
$\displaystyle
N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{(}x\oplus\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}$
$\displaystyle=$
$\displaystyle\lim_{\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{\lVert}x+\textstyle\sum_{i=1}^{k}a_{i}t(n_{1},\dots,n_{i})\Big{\rVert}$

where
$\lim_{\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}$
is an abbreviation for the iterated limit
$\lim_{n_{1},\mathcal{U}_{1}}\dots\lim_{n_{k},\mathcal{U}_{k}}$. It is clear
that $N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}$ is a seminorm
on $X\oplus\mathbb{R}^{k}$.

######  Lemma 10.

In the situation described above, we have

$N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{(}x\oplus\textstyle\sum_{i=1}^{j}a_{i}e_{i}\Big{)}\leqslant
N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{(}x\oplus\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}$

for all $0\leqslant j\leqslant k$, $x\in X$ and
$(a_{i})_{i=1}^{k}\in\mathbb{R}^{k}$. It follows that

$N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{(}0\oplus\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}\leqslant
2N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{(}x\oplus\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}$

for all $x\in X$ and $(a_{i})_{i=1}^{k}\in\mathbb{R}^{k}$.

###### Proof.

The second inequality follows from the first one by triangle inequality. To
see the first inequality, fix $x\in X$, $(a_{i})_{i=1}^{k}\in\mathbb{R}^{k}$
and $0\leqslant j\leqslant k$. For every
$\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[\mathbb{N}]}^{j}$, fix a norming
functional $z^{*}(\accentset{\rule{2.79996pt}{0.6pt}}{n})$ for
$x+\sum_{i=1}^{j}a_{i}t(n_{1},\dots,n_{i})$. It follows that

$\displaystyle
N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{(}x\oplus\textstyle\sum_{i=1}^{j}a_{i}e_{i}\Big{)}$
$\displaystyle=\lim_{n_{1},\mathcal{U}_{1}}\dots\lim_{n_{j},\mathcal{U}_{j}}\Big{\lVert}x+\textstyle\sum_{i=1}^{j}a_{i}t(n_{1},\dots,n_{i})\Big{\rVert}$
$\displaystyle=\lim_{n_{1},\mathcal{U}_{1}}\dots\lim_{n_{j},\mathcal{U}_{j}}z^{*}(n_{1},\dots,n_{j})\Big{(}x+\textstyle\sum_{i=1}^{j}a_{i}t(n_{1},\dots,n_{i})\Big{)}$
$\displaystyle=\lim_{n_{1},\mathcal{U}_{1}}\dots\lim_{n_{k},\mathcal{U}_{k}}z^{*}(n_{1},\dots,n_{j})\Big{(}x+\textstyle\sum_{i=1}^{k}a_{i}t(n_{1},\dots,n_{i})\Big{)}$
$\displaystyle\leqslant\lim_{n_{1},\mathcal{U}_{1}}\dots\lim_{n_{k},\mathcal{U}_{k}}\Big{\lVert}x+\textstyle\sum_{i=1}^{k}a_{i}t(n_{1},\dots,n_{i})\Big{\rVert}$
$\displaystyle=N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{(}x\oplus\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}\
.$

∎

######  Remark.

Lemma 10 shows that $N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}$
is in fact a norm on $X\oplus\mathbb{R}^{k}$ extending the norm of $X$, and
the unit vector basis of $\mathbb{R}^{k}$ is a normalized, monotone basis with
respect to $N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}$. By an
earlier remark, the restriction of
$N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}$ to $\mathbb{R}^{k}$
is in the $k^{\text{th}}$ asymptotic structure of $X$. It is not very hard to
show that if $X$ is separable and reflexive, then the converse also holds:
every member of $\\{X\\}_{k}$ is of the form
$N_{t}^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}$ for some
normalized, weakly null tree $t$ in $X$ with
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}$ being arbitrary.

Our next result approximates bounded functions, like those appearing in
Raynaud’s characterization of stability, by weakly null trees.

######  Lemma 11.

Assume $X$ is reflexive. Let $k\in\mathbb{N}$,
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}=(\mathcal{U}_{1},\dots,\mathcal{U}_{k})$
be a sequence of free ultrafilters and $f\colon{[\mathbb{N}]}^{k}\to X$ a
bounded function. Then there exist a normalized,
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}$-weakly null rooted tree $t$
in $X$ and non-negative scalars $a_{0},a_{1},\dots,a_{k}$ such that

$\lim_{\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{\lVert}f(n_{1},\dots,n_{k})-\textstyle\sum_{i=0}^{k}a_{i}t(n_{1},\dots,n_{i})\Big{\rVert}=0\
.$

###### Proof.

Define for each $1\leqslant j\leqslant k$ the function

$\displaystyle\partial^{j}f\colon{[\mathbb{N}]}^{k-j}$ $\displaystyle\to$
$\displaystyle X$ $\displaystyle\partial^{j}f(n_{1},\dots,n_{k-j})$
$\displaystyle=$
$\displaystyle\operatorname*{w\mathrm{-}\lim}_{n_{k-j+1},\mathcal{U}_{k-j+1}}\dots\operatorname*{w\mathrm{-}\lim}_{n_{k},\mathcal{U}_{k}}f(n_{1},\dots,n_{k})\
.$

We will identify the function $\partial^{k}f$ on
${[\mathbb{N}]}^{0}=\\{\emptyset\\}$ with its image in $X$. Note that for each
$\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[\mathbb{N}]}^{k}$ we have

(3)
$f(n_{1},\dots,n_{k})=\partial^{k}f+\sum_{i=1}^{k}\big{(}\partial^{k-i}f(n_{1},\dots,n_{i})-\partial^{k-i+1}f(n_{1},\dots,n_{i-1})\big{)}\
.$

Define an unrooted tree $a$ in $\mathbb{R}$ of height $k$ by setting

$a(n_{1},\dots,n_{i})=\big{\lVert}\partial^{k-i}f(n_{1},\dots,n_{i})-\partial^{k-i+1}f(n_{1},\dots,n_{i-1})\big{\rVert}$

for $1\leqslant i\leqslant k$ and $n_{1}<\dots<n_{i}$ in $\mathbb{N}$. We next
define $t(\emptyset)\in S_{X}$ and non-negative scalars
$a_{0},a_{1},\dots,a_{k}$ by

$\displaystyle\partial^{k}f$ $\displaystyle=a_{0}t(\emptyset)$ $\displaystyle
a_{i}$
$\displaystyle=\lim_{n_{1},\mathcal{U}_{1}}\dots\lim_{n_{i},\mathcal{U}_{i}}a(n_{1},\dots,n_{i})\qquad\text{for
}1\leqslant i\leqslant k\ .$

We now define our rooted tree $t$ in $S_{X}$ as follows. We have already
defined $t(\emptyset)$. Fix $1\leqslant i\leqslant k$. If $a_{i}=0$, then for
any $n_{1}<\dots<n_{i-1}$ in $\mathbb{N}$, we choose
$\big{(}t(n_{1},\dots,n_{i-1},n)\big{)}_{n=n_{i-1}+1}^{\infty}$ to be an
arbitrary normalized weakly null sequence. If $a_{i}\neq 0$, then for
$n_{1}<\dots<n_{i}$ we set

$t(n_{1},\dots,n_{i})=\begin{cases}\frac{\partial^{k-i}f(n_{1},\dots,n_{i})-\partial^{k-i+1}f(n_{1},\dots,n_{i-1})}{a(n_{1},\dots,n_{i})}&\text{if
}a(n_{1},\dots,n_{i})\neq 0\ ,\\\ \text{an arbitrary norm-$1$
vector}&\text{otherwise.}\end{cases}$

If $a_{i}\neq 0$, then
$\forall\,_{\mathcal{U}_{1}}n_{1}\dots\forall\,_{\mathcal{U}_{i}}n_{i}$
$a(n_{1},\dots,n_{i})>a_{i}/2$, from which it follows that $t$ is
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}$-weakly null. Now observe
that for $n_{1}<\dots<n_{k}$ we have

$f(n_{1},\dots,n_{k})-\sum_{i=0}^{k}a_{i}t(n_{1},\dots,n_{i})\\\
=\sum_{i=1}^{k}\Big{[}\big{(}\partial^{k-i}f(n_{1},\dots,n_{i})-\partial^{k-i+1}f(n_{1},\dots,n_{i-1})\big{)}-a(n_{1},\dots,n_{i})t(n_{1},\dots,n_{i})\Big{]}\\\
+\sum_{i=1}^{k}\big{(}a(n_{1},\dots,n_{i})-a_{i}\big{)}t(n_{1},\dots,n_{i})\
,$

where we used (3). It follows immediately that

$\lim_{\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\Big{\lVert}f(n_{1},\dots,n_{k})-\textstyle\sum_{i=0}^{k}a_{i}t(n_{1},\dots,n_{i})\Big{\rVert}=0\
,$

as required. ∎

An important tool in showing that reflexive spaces are cut stable will be the
following operation on trees.

######  Definition.

Let $k,l,m\in\mathbb{N}$ with $k=l+m$. Let $s$ and $t$ be trees in $X$ of
height $l$ and $m$, respectively. For a cut $P$ of $\\{1,\dots,k\\}$ of size
$l$, the _$P$ -intertwining of $s$ and $t$ _is the unrooted tree
$s\text{ⓩ}_{P}t$ in $X$ of height $k$ defined by

$(s\text{ⓩ}_{P}t)(n_{1},\dots,n_{j})=\begin{cases}s(n_{i}:\,i\in P,\
i\leqslant j)&\text{if }j\in P\ ,\\\ t(n_{i}:\,i\in P^{c},\ i\leqslant
j)&\text{if }j\in P^{c}\end{cases}$

for $1\leqslant j\leqslant k$ and $n_{1}<\dots<n_{j}$ in $\mathbb{N}$.

Note that if $P=\\{p_{1},\dots,p_{l}\\}$ with $p_{1}<\dots<p_{l}$ and
$P^{c}=\\{q_{1},\dots,q_{m}\\}$ with $q_{1}<\dots<q_{m}$, then

$\displaystyle(s\text{ⓩ}_{P}t)(n_{1},n_{2},\dots,n_{p_{i}})$
$\displaystyle=s(n_{p_{1}},\dots,n_{p_{i}})$ $\displaystyle\text{for
}1\leqslant i\leqslant l\ ,$
$\displaystyle(s\text{ⓩ}_{P}t)(n_{1},n_{2},\dots,n_{q_{j}})$
$\displaystyle=t(n_{q_{1}},\dots,n_{q_{j}})$ $\displaystyle\text{for
}1\leqslant j\leqslant m\ .$

Given a sequence
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}=(\mathcal{U}_{1},\dots,\mathcal{U}_{k})$
of free ultrafilters, let
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}=(\mathcal{U}_{p_{1}},\dots,\mathcal{U}_{p_{l}})$
and
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P^{c}}=(\mathcal{U}_{q_{1}},\dots,\mathcal{U}_{q_{m}})$.
Note that if $s$ is
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}$-weakly null and $t$ is
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P^{c}}$-weakly null, then
$s\text{ⓩ}_{P}t$ is $\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}$-weakly
null.

Although the standard basis of $\mathbb{R}^{k}$ is not in general
unconditional with respect to a norm in the $k^{\text{th}}$ asymptotic
structure of a Banach space, we have the following result.

######  Proposition 12.

Assume $X$ is reflexive. Let $k,l,m\in\mathbb{N}$ with $k=l+m$ and $P$ be a
cut of $\\{1,\dots,k\\}$ of size $l$. Let
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}=(\mathcal{U}_{1},\dots,\mathcal{U}_{k})$
be a sequence of free ultrafilters. Let $s$ and $t$ be normalized trees in $X$
of height $l$ and $m$, respectively, such that $s$ is
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}$-weakly null and $t$ is
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P^{c}}$-weakly null. Then

(4)
$N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{s\text{ⓩ}_{P}t}\Big{(}\textstyle\sum_{i\in
P}a_{i}e_{i}\Big{)}\leqslant
N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{s\text{ⓩ}_{P}t}\Big{(}\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}$

and

(5)
$N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{s\text{ⓩ}_{P}t}\Big{(}\textstyle\sum_{i\in
P^{c}}a_{i}e_{i}\Big{)}\leqslant
N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{s\text{ⓩ}_{P}t}\Big{(}\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}$

for all $(a_{i})_{i=1}^{k}\in\mathbb{R}^{k}$.

###### Proof.

Write $P$ as $P=\\{p_{1},\dots,p_{l}\\}$ with $p_{1}<\dots<p_{l}$ and $P^{c}$
as $P^{c}=\\{q_{1},\dots,q_{m}\\}$ with $q_{1}<\dots<q_{m}$. Put
$u=s\text{ⓩ}_{P}t$. We claim that

(6)
$N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\textstyle\sum_{i=1}^{l}a_{p_{i}}e_{p_{i}}+\textstyle\sum_{j=1}^{m-1}a_{q_{j}}e_{q_{j}}\Big{)}\leqslant
N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}\
.$

for all $(a_{i})_{i=1}^{k}\in\mathbb{R}^{k}$. Using the proof of this claim,
we can repeatedly eliminate coefficients to deduce (4). The proof of (5) is
similar.

To see the claim, fix $\varepsilon>0$, $n_{1}<\dots<n_{q_{m}-1}$ in
$\mathbb{N}$ and $L\in\mathcal{U}_{q_{m}}$ such that

$0\in\overline{\\{u(n_{1},n_{2},\dots,n_{q_{m}}):\,n_{q_{m}}\in L\\}}^{w}$

and for all $n_{q_{m}}\in L$ and for all $(a_{i})_{i=1}^{k}\in[-1,1]^{k}$, we
have

$\bigg{\lvert}N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}-\\\
\lim_{n_{q_{m}+1},\mathcal{U}_{q_{m}+1}}\dots\lim_{n_{k},\mathcal{U}_{k}}\Big{\lVert}\sum_{i=1}^{k}a_{i}u(n_{1},\dots,n_{i})\Big{\rVert}\bigg{\rvert}<\varepsilon\
.$

Note that
$\forall\,_{\mathcal{U}_{1}}n_{1}\dots\forall\,_{\mathcal{U}_{q_{m}-1}}n_{q_{m}-1}$
such $L\in\mathcal{U}_{q_{m}}$ exists. By Mazur’s theorem there is a finite
set $F\subset L$ and a convex combination

$\sum_{n\in F}\tau_{n}u(n_{1},n_{2},\dots,n_{q_{m}-1},n)$

with norm

$\Big{\lVert}\textstyle\sum_{n\in
F}\tau_{n}u(n_{1},n_{2},\dots,n_{q_{m}-1},n)\Big{\rVert}<\varepsilon\ .$

Since $F$ is finite, there exist $\max F<n_{q_{m}+1}<\dots<n_{k}$ such that

$\bigg{\lvert}N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}-\\\
\Big{\lVert}\sum_{i=1}^{k}a_{i}u(n_{1},\dots,n_{i})\Big{\rVert}\bigg{\rvert}<2\varepsilon$

for all $n_{q_{m}}\in F$ and for all $(a_{i})_{i=1}^{k}\in[-1,1]^{k}$. Now fix
$(a_{i})_{i=1}^{k}\in[-1,1]^{k}$ and observe that

$\displaystyle
N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\textstyle\sum_{i=1}^{l}a_{p_{i}}e_{p_{i}}$
$\displaystyle+\textstyle\sum_{j=1}^{m-1}a_{q_{j}}e_{q_{j}}\Big{)}$
$\displaystyle\leqslant\Big{\lVert}\textstyle\sum_{i=1}^{q_{m}-1}a_{i}u(n_{1},\dots,n_{i})+\textstyle\sum_{i=q_{m}+1}^{k}a_{i}u(n_{1},\dots,n_{i})\Big{\rVert}+2\varepsilon$
$\displaystyle\leqslant\Big{\lVert}\textstyle\sum_{n_{q_{m}}\in
F}\tau_{n_{q_{m}}}\textstyle\sum_{i=1}^{k}a_{i}u(n_{1},\dots,n_{i})\Big{\rVert}+3\varepsilon$
$\displaystyle\leqslant
N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\textstyle\sum_{i=1}^{k}a_{i}e_{i}\Big{)}+5\varepsilon\
.$

Here it is crucial that for $i>q_{m}$, $u(n_{1},\dots,n_{i})$ does not depend
on $n_{q_{m}}$. This completes the proof of the claim. ∎

We are now ready to prove the main result of this section.

######  Theorem 13.

Every reflexive Banach space is $5$-cut stable, and thus, in particular, upper
stable.

###### Proof.

Fix $k,l,m\in\mathbb{N}$ with $k=l+m$, a sequence
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}=(\mathcal{U}_{1},\dots,\mathcal{U}_{k})$
of free ultrafilters and cuts $P,Q$ of $\\{1,\dots,k\\}$ each of size $l$. Let
$X$ be an infinite-dimensional, reflexive Banach space and
$f\colon{[\mathbb{N}]}^{l}\to X$ and $g\colon{[\mathbb{N}]}^{m}\to X$ be
bounded functions. We will show that

(7)
$\lim_{\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\lVert
f(n_{i}:\,i\in P)+g(n_{i}:\,i\in P^{c})\rVert\\\ \leqslant
5\lim_{\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\lVert
f(n_{i}:\,i\in Q)+g(n_{i}:\,i\in Q^{c})\rVert$

in the special case when the ultrafilters $\mathcal{U}_{i}$ are all the same.
This will then complete the proof.

Write $P=\\{p_{1}\,\dots,p_{l}\\}$ with $p_{1}<\dots<p_{l}$ and
$P^{c}=\\{q_{1},\dots,q_{m}\\}$ with $q_{1}<\dots<q_{m}$. By Lemma 11 we may
assume that there exist normalized, rooted trees $s$ and $t$ in $X$ of height
$l$ and $m$, respectively, such that $s$ is
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}$-weakly null, $t$ is
$\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P^{c}}$-weakly null, and
there exist non-negative scalars $a_{0},a_{1},\dots,a_{l}$ and
$b_{0},b_{1},\dots,b_{m}$ such that

$f(n_{1},\dots,n_{l})=a_{0}s(\emptyset)+\sum_{i=1}^{l}a_{i}s(n_{1},\dots,n_{i})\qquad\text{for
}n_{1}<\dots<n_{l}$

and

$g(n_{1},\dots,n_{m})=b_{0}t(\emptyset)+\sum_{j=1}^{m}b_{j}t(n_{1},\dots,n_{j})\qquad\text{for
}n_{1}<\dots<n_{m}\ .$

Let $u=s\text{ⓩ}_{P}t$. Set $c_{p_{i}}=a_{i}$ for $1\leqslant i\leqslant l$
and $c_{q_{j}}=b_{j}$ for $1\leqslant j\leqslant m$. We shall frequently use
the fact that

$N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\textstyle\sum_{i\in
P}c_{i}e_{i}\Big{)}=N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}}_{s}\Big{(}\textstyle\sum_{i=1}^{l}a_{i}e_{i}\Big{)}\
.$

Now observe that for
$\accentset{\rule{2.79996pt}{0.6pt}}{n}\in{[\mathbb{N}]}^{k}$ we have

$f(n_{i}:\,i\in P)+g(n_{i}:\,i\in
P^{c})=a_{0}s(\emptyset)+b_{0}t(\emptyset)+\sum_{i=1}^{k}c_{i}u(n_{1},\dots,n_{i})\
.$

It follows that

(8)
$\lim_{\accentset{\rule{2.79996pt}{0.6pt}}{n},\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}\lVert
f(n_{i}:\,i\in P)+g(n_{i}:\,i\in
P^{c})\rVert=N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\big{(}a_{0}s(\emptyset)+b_{0}t(\emptyset)\big{)}\oplus\textstyle\sum_{i=1}^{k}c_{i}e_{i}\Big{)}\
.$

By the triangle inequality, we have

(9)
$N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\big{(}a_{0}s(\emptyset)+b_{0}t(\emptyset)\big{)}\oplus\textstyle\sum_{i=1}^{k}c_{i}e_{i}\Big{)}\\\
\leqslant\lVert
a_{0}s(\emptyset)+b_{0}t(\emptyset)\rVert+N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}}_{s}\Big{(}\textstyle\sum_{i=1}^{l}a_{i}e_{i}\Big{)}+N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P^{c}}}_{t}\Big{(}\textstyle\sum_{j=1}^{m}b_{j}e_{j}\Big{)}$

On the other hand, it follows from Lemma 10 and Proposition 12 that

(10) $\lVert
a_{0}s(\emptyset)+b_{0}t(\emptyset)\rVert+N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}}_{s}\Big{(}\textstyle\sum_{i=1}^{l}a_{i}e_{i}\Big{)}+N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P^{c}}}_{t}\Big{(}\textstyle\sum_{j=1}^{m}b_{j}e_{j}\Big{)}\\\
\leqslant 5\cdot
N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}}_{u}\Big{(}\big{(}a_{0}s(\emptyset)+b_{0}t(\emptyset)\big{)}\oplus\textstyle\sum_{i=1}^{k}c_{i}e_{i}\Big{)}\
.$

When the ultrafilters $\mathcal{U}_{i}$ are all identical, then

$N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P}}_{s}\Big{(}\textstyle\sum_{i=1}^{l}a_{i}e_{i}\Big{)}=N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{Q}}_{s}\Big{(}\textstyle\sum_{i=1}^{l}a_{i}e_{i}\Big{)}$

and

$N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{P^{c}}}_{t}\Big{(}\textstyle\sum_{j=1}^{m}b_{j}e_{j}\Big{)}=N^{\accentset{\rule{4.20004pt}{0.6pt}}{\mathcal{U}}_{Q^{c}}}_{t}\Big{(}\textstyle\sum_{j=1}^{m}b_{j}e_{j}\Big{)}\
.$

Hence (7) follows immediately from (8), (9) and (10) applied to
$u=s\text{ⓩ}_{P}t$ or $u=s\text{ⓩ}_{Q}t$ as appropriate. ∎

## References