# Asymptotic geometry and delta-points

Trond A. Abrahamsen  Department of Mathematics, University of Agder, Postboks
422, 4604 Kristiansand, Norway<EMAIL_ADDRESS>http://home.uia.no/trondaa/index.php3 ,  Vegard Lima  Department of
Engineering Sciences, University of Agder, Postboks 509, 4898 Grimstad,
Norway<EMAIL_ADDRESS>,  André Martiny  Department of Mathematics,
University of Agder, Postboks 422, 4604 Kristiansand, Norway.
<EMAIL_ADDRESS>and  Yoël Perreau  Laboratoire de Mathématiques de
Besançon, Université Bourgogne Franche-Comté, CNRS UMR- 6623, 16 route de
Gray, 25030 Besançon Cédex, Besançon, France<EMAIL_ADDRESS>
###### Abstract.

We study Daugavet- and $\Delta$-points in Banach spaces. A norm one element
$x$ is a Daugavet-point (respectively a $\Delta$-point) if in every slice of
the unit ball (respectively in every slice of the unit ball containing $x$)
you can find another element of distance as close to $2$ from $x$ as desired.

In this paper we look for criteria and properties ensuring that a norm one
element is not a Daugavet- or $\Delta$-point. We show that asymptotically
uniformly smooth spaces and reflexive asymptotically uniformly convex spaces
do not contain $\Delta$-points. We also show that the same conclusion holds
true for the James tree space as well as for its predual.

Finally we prove that there exists a superreflexive Banach space with a
Daugavet- or $\Delta$-point provided there exists such a space satisfying a
weaker condition.

###### Key words and phrases:

Delta-point, Daugavet-point, asymptotic uniform smoothness, asymptotic uniform
convexity, uniformly non-square norm

###### 2010 Mathematics Subject Classification:

Primary 46B20, 46B22, 46B04, 46B06

This work was supported by the AURORA mobility programme (project number:
309597) from the Norwegian Research Council and the $``$PHC Aurora$"$ program
(project number: 45391PF), funded by the French Ministry for Europe and
Foreign Affairs, the French Ministry for Higher Education, Research and
Innovation.

##  1\. Introduction

Daugavet- and $\Delta$-points first appeared in [1] as natural pointwise
versions of geometric characterizations of the Daugavet property [44, Lemma 2]
and of the so called spaces with bad projections [27, Theorem 1.4] (also known
as spaces with the diametral local diameter two property (DLD2P) [8]). We
refer to Section 2 for precise definitions and equivalent reformulations.

From their introduction on, Daugavet- and $\Delta$-points attracted a lot of
attention and were intensively studied in classical Banach spaces ([1, 2, 3,
23]). In particular a strong emphasis was put on finding linear or geometric
properties that would prevent norm one elements in a space to be Daugavet- or
$\Delta$-points. It soon appeared that even nice properties which on a global
level prevent the space to have the Daugavet property or the DLD2P do not
provide an obstruction to the existence of Daugavet- or $\Delta$-points in the
space. For example there exists a Banach space with a 1-unconditional basis
such that the set of Daugavet-points are weakly dense in the unit ball [3,
Theorem 4.7].

Another striking example illustrating this was obtained in the context of
Lipschitz-free spaces. The study of Daugavet- and $\Delta$-points in this
context started in [30] where a characterization of Daugavet-points in free-
spaces over compact metric spaces was discovered. It was observed that
Daugavet-points have to be at distance $2$ from every denting point of the
unit ball in any given Banach space, [30, Proposition 3.1], and it was proved
that the converse holds in every free-space in the compact setting. Extending
this result to the general setting, Veeorg was then able to provide in [46] a
surprising example of a metric space whose free-space has the Radon–Nikodým
property (RNP) and admits a Daugavet-point.

In the present paper we continue the investigation of Daugavet- and
$\Delta$-points in general Banach spaces by focusing on the interactions
between those points and the asymptotic geometry of the space. We provide new
examples of Banach spaces failing to contain $\Delta$-points and we introduce
weaker notions which can be viewed as a step forward in the direction of
constructing an example of a superreflexive space with a Daugavet- or a
$\Delta$-point.

Let us now describe the content of the paper and expose our main results. In
Section 2 we recall the notion of slices, give the definition of Daugavet- and
$\Delta$-points, and state a few simple geometric lemmata. As a warm up we
give a simple proof that uniformly non-square spaces do not admit
$\Delta$-points and explain why simple considerations on the diameter of
slices cannot rule out $\Delta$-points outside of this setting. We end the
section with the necessary background on asymptotic uniform properties of
Banach spaces.

In Section 3 we focus on asymptotic smoothness. Our main result there is a
condition on the modulus of asymptotic smoothness $\overline{\rho}_{X}(t,x)$
at some point $x$ which prevents the considered point to be a $\Delta$-point.
We will show in particular that no asymptotically smooth point can be a
$\Delta$-point. As a consequence we obtain that asymptotically uniformly
smooth spaces fail to contain $\Delta$-points. We then apply this theorem to
obtain new examples of classical spaces failing to contain $\Delta$-points,
specifically spaces with Kalton’s property $(M^{*})$ and the predual $JT_{*}$
of the James tree space.

In Section 4 we obtain partial pointwise results for asymptotic convexity. As
a consequence we show that spaces with the property $(\alpha)$ of Rolewicz and
in particular reflexive asymptotically uniformly convex spaces do not admit
$\Delta$-points. As a consequence we obtain that the Baernstein space $B$ as
well as its dual do not admit $\Delta$-points.

Motivated by the result from the preceding section, we look in Section 5 at
the existence of Daugavet- and $\Delta$-points in the James tree space $JT$.
Our main result there is that $JT$ does not admit $\Delta$-points. We also
have some partial results for the dual $JT^{*}$.

In the final section we introduce a natural weaker notion of $(2-\varepsilon)$
Daugavet- and $\Delta$-points and we look at Banach spaces in which such
points exits for every $\varepsilon>0$. We show that if there exists a
superreflexive space $X$ which has a $(2-\varepsilon)$ Daugavet-point
(respectively a $(2-\varepsilon)$ $\Delta$-point) for every $\varepsilon>0$,
then there exists a superreflexive space (an ultrapower of $X$) with a
Daugavet-point (respectively a $\Delta$-point).

##  2\. Preliminaries

Let $X$ be a Banach space, $B_{X}$ its unit ball, $S_{X}$ its unit sphere, and
$X^{*}$ its dual space. We consider real Banach spaces only.

For any $x^{*}\in S_{X^{*}}$ and $\delta>0$ we define a _slice_ of $B_{X}$ by

$S(x^{*},\delta):=\\{y\in B_{X}:x^{*}(y)>1-\delta\\}.$

The corresponding closed slice is denoted

$\overline{S}(x^{*},\delta):=\left\\{y\in B_{X}:x^{*}(y)\geq
1-\delta\right\\}.$

Slices and closed slices of $B_{X}$ can also be defined using non-zero
functionals in $X^{*}$ by replacing the $1$ above by the norm of the
corresponding functional. Now if $x^{*}$ is a non-zero functional we have
$S(x^{*},\delta)=S\left(\frac{x^{*}}{\left\|x^{*}\right\|},\frac{\delta}{\left\|x^{*}\right\|}\right)$
and we would like to point out that working with non-normalized functionals
can cause delicate computational problems (see e.g. the proof of Theorem 3.5).
We will avoid them as much as possible. We will also informally say that a
Banach space has _small slices of arbitrary large diameter_ , when the
diameter of $S(x^{*},\delta)$ is as close as we want to $2$ for small
$\delta$.

For every $x\in X$, let us write $D(x):=\\{x^{*}\in S_{X^{*}}:\
x^{*}(x)=\left\|x\right\|\\}$. We will make extensive use of the following
lemma.

######  Lemma 2.1.

Let $x\in S_{X}$. For every $n\geq 1$, $\delta>0$, and
$x_{1}^{*},\ldots,x_{n}^{*}\in D(x)$, we have:

$S\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}^{*},\frac{\delta}{n}\right)\subset\bigcap_{i=1}^{n}S(x_{i}^{*},\delta).$

###### Proof.

Let $x^{*}=\frac{1}{n}\sum_{i=1}^{n}x_{i}^{*}$. Since every $x_{i}^{*}$ is in
$D(x)$, we immediately obtain that $x^{*}$ is in $D(x)$ and in particular that
$\left\|x^{*}\right\|=1$. Now if $y\in S\left(x^{*},\frac{\delta}{n}\right)$,
we have, for every $1\leq i_{0}\leq n$,

$x_{i_{0}}^{*}(y)=nx^{*}(y)-\sum_{i=1\atop i\neq
i_{0}}^{n}x_{i}^{*}(y)>n-\frac{n\delta}{n}-(n-1)=1-\delta,$

and the conclusion follows. ∎

Note that the fact that every functional $x_{i}^{*}$ attains its norm at $x$
is crucial here to guarantee that the average of the $x_{i}^{*}$ remains a
norm one functional.

Let $X$ be a Banach space and $x\in S_{X}$. For $\varepsilon>0$ we define

$\Delta_{\varepsilon}(x):=\\{y\in B_{X}:\|x-y\|\geq 2-\varepsilon\\}.$

Following [1] we say that $x$ is a _Daugavet-point_ if
$B_{X}=\overline{\operatorname{conv}}\Delta_{\varepsilon}(x)$ for all
$\varepsilon>0$ and we say that $x$ is a _$\Delta$ -point_ if
$x\in\overline{\operatorname{conv}}\Delta_{\varepsilon}(x)$ for all
$\varepsilon>0$.

Using Hahn–Banach separation we get the following well known lemma that we
will use without reference.

######  Lemma 2.2.

Let $X$ be a Banach space and let $x\in S_{X}$.

  1. (a)

$x$ is a Daugavet-point if and only if for all $\varepsilon>0$, all
$\delta>0$, and all $x^{*}\in S_{X^{*}}$, there exists $y\in S(x^{*},\delta)$
such that $\|x-y\|>2-\varepsilon$.

  2. (b)

$x$ is a $\Delta$-point if and only if for all $\varepsilon>0$, all
$\delta>0$, and all $x^{*}\in S_{X^{*}}$ such that $x\in S(x^{*},\delta)$,
there exists $y\in S(x^{*},\delta)$ such that $\|x-y\|>2-\varepsilon$.

The above lemma appears in [1, Lemma 2.1], but note that there is a misprint
in the statement of (2) in [1, Lemma 2.1]. Following this we say that an
element $x^{*}\in S_{X^{*}}$ is a _weak ∗ Daugavet-point_ if for all
$\varepsilon>0$, all $\delta>0$, and all $x\in S_{X}$ (naturally identified
with an element of $X^{**}$) there exists $y^{*}\in S(x,\delta)$ such that
$\|x^{*}-y^{*}\|>2-\varepsilon$ , and we say that $x^{*}\in S_{X^{*}}$ is a
_weak ∗ $\Delta$-point_ if for all $\varepsilon>0$, all $\delta>0$, and all
$x\in S_{X}$ such that $x^{*}\in S(x,\delta)$, there exists $y^{*}\in
S(x,\delta)$ such that $\|x^{*}-y^{*}\|>2-\varepsilon$.

A Banach space is said to be _uniformly non-square_ if there exists
$\varepsilon>0$ such that for all $x,y\in B_{X}$ we have either
$\|\frac{1}{2}(x+y)\|\leq 1-\varepsilon$ or $\|\frac{1}{2}(x-y)\|\leq
1-\varepsilon$.

Uniformly non-square spaces were introduced by James [28]. It is well known
that uniformly non-square spaces are superreflexive and that uniformly convex
and uniformly smooth spaces are uniformly non-square (see e.g. [32,
Proposition 1]). The following simple observations characterizes uniformly
non-square spaces in terms of slices.

######  Proposition 2.3.

Let $X$ be a Banach space. The following are equivalent.

  1. (a)

$X$ is uniformly non-square.

  2. (b)

There exists $\varepsilon>0$ such that for all $x^{*}\in S_{X^{*}}$ the
diameter of $S(x^{*},\varepsilon)$ is less than $2-2\varepsilon$.

###### Proof.

(a) $\Rightarrow$ (b). Let $\varepsilon>0$ be such that for all $x,y\in B_{X}$
we have either $\|\frac{1}{2}(x+y)\|\leq 1-\varepsilon$ or
$\|\frac{1}{2}(x-y)\|\leq 1-\varepsilon$. If $x^{*}\in S_{X^{*}}$ and $x,y\in
S(x^{*},\varepsilon)$, then $\|x+y\|\geq x^{*}(x+y)\geq 2-2\varepsilon$, so
$\|x-y\|\leq 2-2\varepsilon$.

(b) $\Rightarrow$ (a). If $X$ is not uniformly non-square, then for every
$\varepsilon>0$ there exists $x,y\in B_{X}$ such that $\|x+y\|>2-\varepsilon$
and $\|x-y\|>2-\varepsilon$.

Find $x^{*}\in S_{X^{*}}$ such that $x^{*}(x+y)>2-\varepsilon$ and observe
that $x,y\in S(x^{*},\varepsilon)$ with $\|x-y\|>2-\varepsilon$. ∎

If $x$ is a $\Delta$-point then for all $\delta>0$ and $x^{*}\in S_{X^{*}}$
with $x^{*}(x)>1-\delta$ we have $\operatorname{diam}S(x^{*},\delta)=2$ so we
immediately get the following.

######  Corollary 2.4.

Let $X$ be a uniformly non-square Banach space. Then $X$ does not admit
$\Delta$-points.

If a Banach space $X$ is _not_ uniformly non-square then for every
$\varepsilon>0$ there is a slice $S(x^{*},\varepsilon)$, $x^{*}\in S_{X^{*}}$,
with diameter strictly greater than $2-2\varepsilon$, so $X$ admits small
slices of arbitrarily large diameter. Observe that if $X$ is a dual space,
then we may assume that the functional $x^{*}$ in the proof of (b)
$\Rightarrow$ (a) in Proposition 2.3 is weak∗ continuous. Hence a dual space
which is not uniformly non-square has a unit ball with small weak∗ slices of
diameter arbitrary close to $2$. These simple observations show that for
spaces that are not uniformly non-square there is no hope of ruling out
$\Delta$-points by some upper bound on the diameter of slices of the unit
ball. Note that any Banach space $X$ of dimension greater or equal to $2$ has
an equivalent norm $|\cdot|$ such that $(X,|\cdot|)$ is _not_ uniformly non-
square [32, Corollary 1]. In particular, Corollary 2.4 does not rule out
$\Delta$-points in superreflexive or finite dimensional spaces.

Let $X$ be a Banach space. Let $\operatorname{cof}(X)$ denote the set of all
subspaces of finite co-dimension of $X$. For $t>0$ and $x\in S_{X}$ consider

$\bar{\rho}_{X}(t,x)=\inf_{Y\in\operatorname{cof}(X)}\sup_{y\in
S_{Y}}\\{\|x+ty\|-1\\}$

and

$\bar{\delta}_{X}(t,x)=\sup_{Y\in\operatorname{cof}(X)}\inf_{y\in
S_{Y}}\\{\|x+ty\|-1\\}.$

The _modulus of asymptotic convexity_ of $X$ is given by

$\bar{\delta}_{X}(t)=\inf_{x\in S_{X}}\bar{\delta}_{X}(t,x).$

and the _modulus of asymptotic smoothness_ of $X$ is given by

$\bar{\rho}_{X}(t)=\sup_{x\in S_{X}}\bar{\rho}_{X}(t,x).$

The space $X$ is said to be _asymptotically uniformly smooth_ (AUS for short)
if $\lim_{t\to 0}t^{-1}\bar{\rho}_{X}(t)=0$ and it is _asymptotically
uniformly convex_ (AUC) if $\bar{\delta}_{X}(t)>0$ for all $t>0$. Similarly in
$X^{*}$ there is a weak∗-modulus of asymptotic uniform convexity defined by

$\bar{\delta}^{*}_{X}(t)=\inf_{x^{*}\in S_{X^{*}}}\sup_{E}\inf_{y^{*}\in
S_{E}}\\{\|x^{*}+ty^{*}\|-1\\},$

where $E$ runs through all weak∗-closed subspaces of $X^{*}$ of finite
codimension. We say that $X^{*}$ is _AUC_ ∗ if $\bar{\delta}^{*}_{X}(t)>0$ for
all $t>0$. Clearly, a dual space which is AUC∗ is also AUC. The converse does
not hold true since the James tree space $JT$ is known to have an AUC dual
space (see [20]) but to admit no equivalent norm whose dual norm is AUC∗.

It is well known (see e.g. [11, Corollay 2.4]) that $X$ is AUS if and only if
$X^{*}$ is AUC∗, and that if $X$ is reflexive then $X$ is AUS if and only if
$X^{*}$ is AUC. The space $\ell_{1}$ is an example of a non-reflexive AUC
space.

A Banach space $X$ is said to have the _uniform Kadec–Klee_ (UKK) property if
for every $\varepsilon>0$, there exists $\delta>0$ such that whenever
$(x_{n})\subset B_{X}$ with $\|x_{n}-x_{m}\|\geq\varepsilon$ for all $n\neq m$
and $x_{n}\to x$ weakly for some $x\in X$, then it follows that $\|x\|\leq
1-\delta$. This property was introduced by Huff [26]. The UKK property is a
close relative of asymptotic uniform convexity. It is not too difficult too
show that $X$ is AUC if and only if for all $\varepsilon>0$ there exists
$\delta>0$ such that if $x\in B_{X}$ and all weak neighborhoods $V$ of $x$
satisfy $\operatorname{diam}(V\cap B_{X})>\varepsilon$, then $\|x\|\leq
1-\delta$. The latter property is a generalization of UKK and it shows $X$ AUC
implies $X$ UKK. (This generalization was used in e.g. [33], and Lancien
actually used it as a definition of UKK.) It is known that if $X$ does not
contain a copy of $\ell_{1}$, then the reverse implication holds and $X$ is
AUC if and only if $X$ is UKK. In particular, AUC and UKK are equivalent for
reflexive spaces. The idea is that a Banach space $X$ failing this property
contains a point $x\in S_{X}$ and a net $(x_{V})\subset B_{X}$ converging
weakly to $x$ such that $\left\|x-x_{V}\right\|\geq\varepsilon$ for every weak
neighborhood $V$ of $x$ for some fixed $\varepsilon>0$. Now if $X$ does not
contain $\ell_{1}$, the non-separable version of Rosenthal’s result [42,
Theorem 3] proved in [21, Theorem 2.6] tells us that the set $\\{x_{V}\\}$ is
weakly sequentially dense in its weak-closure and we can thus extract a
sequence converging weakly to $x$ and providing an obstruction to the UKK
property (up to further extractions in order to obtain an
$\frac{\varepsilon}{2}$-separated sequence).

Let $X$ be a Banach space and $A$ a bounded subset of $X$. By $\alpha(A)$ we
denote the _Kuratowski measure of non-compactness_ of $A$ which is defined as
the infimum of $\varepsilon>0$ such that $A$ can be covered by a finite number
of sets with diameters less than $\varepsilon$, that is,

$\alpha(A):=\inf\left\\{\varepsilon>0:A\subset\bigcup_{i=1}^{n}A_{i},A_{i}\subset
X,\operatorname{diam}(A_{i})<\varepsilon,i=1,2,\ldots,n\right\\}.$

A Banach space $X$ has _Rolewicz’ property $(\alpha)$_ if for every $x^{*}\in
S_{X^{*}}$ and $\varepsilon>0$ there exists $\delta>0$ such that
$\alpha(S(x^{*},\delta))\leq\varepsilon$. We say that $X$ has _uniform
property $(\alpha)$_ if the same $\delta$ works for all $x^{*}\in S_{X^{*}}$.
These properties were introduced by Rolewicz in [41]. If $X$ has Rolewicz’
property $(\alpha)$, then it is reflexive (see e.g. [36]).

Implicit in Rolewicz [41, Theorem 3] is the result that $X$ is AUC and
reflexive if and only if $X$ has uniform property $(\alpha)$, Rolewicz uses
the term “$X$ is $\Delta$-uniformly convex” instead of $X$ is AUC and
reflexive. It is known $X$ is AUC and reflexive if and only if $X$ is nearly
uniformly convex (NUC) [26] if and only if $X$ is $\Delta$-uniformly convex.
The difference between these two types of uniform convexity is that they use
different (but equivalent) measures of non-compactness.

We also note that for dual spaces we have that if $X^{*}$ is AUC∗ then $X^{*}$
is has _weak ∗_ uniform property $(\alpha)$, that is, for all $\varepsilon>0$
there exists $\delta>0$ such that $\alpha(S(x,\delta))\leq\varepsilon$ for all
$x\in S_{X}$. Our Corollary 3.4 is a pointwise version of this result. Lennard
[34, Proposition 1.3] states that $X^{*}$ has the weak∗ version of the uniform
Kadec–Klee property if and only if $X^{*}$ has weak∗ uniform property
$(\alpha)$. Note that Lennard credits this result to Sims and he uses a
different measure of non-compactness.

##  3\. Asymptotic uniform smoothness

The goal of this section is to show that Banach spaces with an asymptotic
uniformly smooth norm do not admit $\Delta$-points.

Our first result connects the pointwise modulus of asymptotic smoothness
$\overline{\rho}_{X}(t,x)$ at $x\in S_{X}$ with the measure of non-compactness
of the slices defined by $x$. In his thesis Dutrieux gave a proof that a
separable Banach space is AUS if and only if its dual is weak∗ uniformly
Kadec–Klee [12, Proposition 36]. Our proof of the following proposition
follows closely part of the proof given by Dutrieux, but we do not assume
separability.

######  Proposition 3.1.

Let $X$ be a Banach space, and fix $x\in S_{X}$ and $\varepsilon>0$. If there
exists $t>0$ such that $\bar{\rho}_{X}(t,x)/t<\varepsilon$, then we can find
$\delta>0$ such that $\alpha(S(x,\delta))<2\varepsilon$.

To prove this proposition we will need two well known lemmas. The first lemma
is from [29, Lemma 2.13]. A nice different proof can be found in the thesis of
Dutrieux [12, Lemma 38].

######  Lemma 3.2.

Let $X$ be a Banach space. For all $Y\in\operatorname{cof}(X)$ and all
$\varepsilon>0$, there exists a compact set $K_{\varepsilon}$ such that
$B_{X}\subset K_{\varepsilon}+(2+\varepsilon)B_{Y}$.

The proof of the next lemma is simple using contradiction, so we skip it.

######  Lemma 3.3.

Let $X$ be a Banach space. Let $x^{*}\in B_{X^{*}}$ and $\varepsilon>0$.

If $\limsup_{\alpha}\|x^{*}-x^{*}_{\alpha}\|<\varepsilon$ whenever
$(x^{*}_{\alpha})\subseteq B_{X^{*}}$ and
$x^{*}_{\alpha}\overset{w^{*}}{\to}x^{*}$, then there exists a
weak∗-neighborhood $V$ of $x^{*}$ with $V\cap B_{X^{*}}\subseteq
B(x^{*},\varepsilon)$.

###### Proof of Proposition 3.1.

Let us assume that $\bar{\rho}_{X}(t,x)/t<\varepsilon$ and let us write
$\delta_{x}=\varepsilon t-\bar{\rho}_{X}(t,x)$. We then have the following.

Claim. Take any $\delta<\delta_{x}$. For every $x^{*}\in S(x,\delta)$ and
every net $(x_{\alpha}^{*})\subset B_{X^{*}}$ such that $x_{\alpha}^{*}\to
x^{*}$ weak∗ we have $\limsup_{\alpha}\|x^{*}-x^{*}_{\alpha}\|<2\varepsilon$.

Before proving the claim let us see how to finish the proof. Pick any
$\delta<\delta_{x}$ and take $\delta^{\prime}$ such that
$\delta<\delta^{\prime}<\delta_{x}$. By the claim and by Lemma 3.3 we can find
for every $x^{*}\in S(x,\delta^{\prime})$ a weak∗-neighborhood $V_{x^{*}}$ of
$x^{*}$ with $V_{x^{*}}\cap B_{X^{*}}\subseteq B(x^{*},2\varepsilon)$. Then
$(V_{x^{*}})_{x^{*}\in S(x,\delta^{\prime})}$ is an open cover of
$S(x,\delta^{\prime})$ and therefore an open cover of the weak∗ compact set
$\bar{S}(x,\delta)$. By compactness there is a finite subcover.

Proof of the claim. Fix some $\delta<\delta_{x}$ and let $x^{*}\in
S(x,\delta)$ and $(x_{\alpha}^{*})\subset B_{X^{*}}$ such that
$x_{\alpha}^{*}\to x^{*}$ weak∗. We will show that
$L=\limsup_{\alpha}\|x^{*}-x^{*}_{\alpha}\|<2\varepsilon$. Again pick any
$\delta^{\prime}$ such that $\delta<\delta^{\prime}<\delta_{x}$. By definition
of $\delta_{x}$ we then have $\bar{\rho}_{X}(t,x)<\varepsilon
t-\delta^{\prime}$ so there exists $Z\in\operatorname{cof}(X)$ such that

$\sup_{z\in S_{Z}}\|x+tz\|\leq 1+\varepsilon t-\delta^{\prime}.$

Since for any $s\in[0,1]$

$\|x+tsz\|=\|(1-s)x+s(x+tz)\|\leq 1+\varepsilon t-\delta^{\prime},$

we get

$\sup_{z\in B_{Z}}\|x+tz\|\leq 1+\varepsilon t-\delta^{\prime}.$

Now let $Y=Z\cap\ker(x^{*})$ and let $\eta>0$. By Lemma 3.2 there exists a
compact set $K$ in $X$ such that $B_{X}\subseteq K+(2+\eta)B_{Y}$.

By compactness of $K$ and boundedness of $(x^{*}_{\alpha})$ we have that
$x^{*}_{\alpha}\to x^{*}$ uniformly on $K$. We may therefore choose $\beta$
such that

  * •

$|\langle x^{*}_{\beta}-x^{*},k\rangle|<\eta$ for all $k\in K$;

  * •

$x^{*}_{\beta}(x)>1-\delta$;

  * •

$|\|x^{*}_{\beta}-x^{*}\|-L|<\eta$.

Choose $x_{\beta}\in S_{X}$ such that $\langle
x^{*}_{\beta}-x^{*},x_{\beta}\rangle>L-\eta$. Now write
$x_{\beta}=k_{\beta}+(2+\eta)y_{\beta}$ with $k_{\beta}\in K$ and
$y_{\beta}\in B_{Y}$. We get

$x^{*}_{\beta}(y_{\beta})=\langle
x^{*}_{\beta}-x^{*},y_{\beta}\rangle=\frac{\langle
x^{*}_{\beta}-x^{*},x_{\beta}\rangle-\langle
x^{*}_{\beta}-x^{*},k_{\beta}\rangle}{2+\eta}>\frac{L-2\eta}{2+\eta}$

since $y_{\beta}\in\ker(x^{*})$. Therefore

$1-\delta+\frac{t(L-2\eta)}{2+\eta}<\langle
x_{\beta}^{*},x\rangle+\frac{t(L-2\eta)}{2+\eta}<\langle
x^{*}_{\beta},x+ty_{\beta}\rangle\leq\|x+ty_{\beta}\|\leq 1+\varepsilon
t-\delta^{\prime}.$

Finally

$L-2\eta<(2+\eta)(\varepsilon-\theta)$

with $\theta=\frac{\delta^{\prime}-\delta}{t}>0$. Since $\eta>0$ was arbitrary
we get $L\leq 2(\varepsilon-\theta)<2\varepsilon$ as desired. ∎

As an immediate corollary we get.

######  Corollary 3.4.

Let $X$ be a Banach space and let $x\in S_{X}$ be an asymptotically smooth
point, that is a point for which $\lim_{t\to
0}\frac{\overline{\rho}_{X}(t,x)}{t}=0$. Then $\lim_{\delta\to
0}~\alpha(S(x,\delta))=0$.

We will now show that this condition on the Kuratowski index of the slices
$S(x,\delta)$ prevents the point $x$ to be a $\Delta$-point.

######  Theorem 3.5.

Let $X$ be a Banach space and let $x\in S_{X}$. If there exists a $\delta>0$
such that $\alpha(S(x,\delta))<\frac{2}{3}$, then $x$ is not a $\Delta$-point.

###### Proof.

Find $\delta>0$ and $\varepsilon>0$ such that
$\alpha\left(S(x,\delta)\right)<\varepsilon<\frac{2}{3}$. Since for every
$\delta^{\prime}\leq\delta$, we have $S(x,\delta^{\prime})\subset S(x,\delta)$
so
$\alpha\left(S(x,\delta^{\prime})\right)\leq\alpha\left(S(x,\delta)\right)<\varepsilon$,
we may, and do, assume that $\delta\leq\varepsilon$.

By assumption we can find $x_{1}^{*},\ldots,x_{n}^{*}\in B_{X^{*}}$ with
$n\geq 1$ such that
$S(x,\delta)\subset\bigcup_{i=1}^{n}B(x_{i}^{*},\varepsilon)$. In fact we have
the following.

Claim 1. For every $\delta^{\prime}\leq\delta$, we can find $1\leq m\leq n$
and $y_{1}^{*},\ldots,y_{m}^{*}\in S_{X^{*}}$ such that

  1. (a)

$y_{j}^{*}\in S(x,\delta^{\prime})$ for every $1\leq j\leq m$;

  2. (b)

$S_{X^{*}}\cap
S(x,\delta^{\prime})\subset\bigcup_{j=1}^{m}B(y_{j}^{*},2\varepsilon)$.

###### Proof of Claim 1.

Let us take $\delta^{\prime}\leq\delta$. As observed before, we have
$S(x,\delta^{\prime})\subset S(x,\delta)$, and thus $S_{X^{*}}\cap
S(x,\delta^{\prime})\subset
S(x,\delta)\subset\bigcup_{i=1}^{n}B(x_{i}^{*},\varepsilon)$.

Clearly the set $S_{X^{*}}\cap S(x,\delta^{\prime})$ is not empty since it
contains $D(x)$, so the set $J=\\{1\leq i\leq n:\ (S_{X^{*}}\cap
S(x,\delta^{\prime}))\cap B(x_{i}^{*},\varepsilon)\neq\emptyset\\}$ has
cardinality $\left|J\right|=m$ with $1\leq m\leq n$.

Since obviously $S_{X^{*}}\cap S(x,\delta^{\prime})\subset\bigcup_{j\in
J}B(x_{j}^{*},\varepsilon)$, the conclusion follows by picking, for every
$j\in J$, an element $y_{j}^{*}\in(S_{X^{*}}\cap S(x,\delta^{\prime}))\cap
B(x_{j}^{*},\varepsilon)$, and by observing that
$B(x_{j}^{*},\varepsilon)\subset B(y_{j}^{*},2\varepsilon)$. ∎

Now let us find $1\leq m\leq n$ and $y_{1}^{*},\ldots y_{m}^{*}\in S_{X^{*}}$
satisfying the properties of Claim $1$ for
$\delta^{\prime}=\frac{\delta}{2n}$, and let us define
$y^{*}=\frac{1}{m}\sum_{j=1}^{m}y_{j}^{*}$. We have the following.

Claim 2. The slice $S(y^{*},\delta^{\prime})=\\{y\in B_{X}:\
y^{*}(y)>\left\|y^{*}\right\|-\delta^{\prime}\\}$ satisfies

$x\in S(y^{*},\delta^{\prime})\subset\bigcap_{j=1}^{m}S(y_{j}^{*},\delta).$

###### Proof of Claim 2.

Since every $y_{j}^{*}$ is in $S_{X^{*}}$ and satisfies
$y_{j}^{*}(x)>1-\delta^{\prime}$, we clearly have $\left\|y^{*}\right\|\leq 1$
and $y^{*}(x)>1-\delta^{\prime}\geq\left\|y^{*}\right\|-\delta^{\prime}$ so
$x\in S(y^{*},\delta^{\prime})$ and $\left\|y^{*}\right\|\geq
1-\delta^{\prime}$.

Now for every $y\in S(y^{*},\delta^{\prime})$ we have
$y^{*}(y)>\left\|y^{*}\right\|-\delta^{\prime}\geq 1-2\delta^{\prime}$ so

$\frac{1}{m}y_{k}^{*}(y)>1-2\delta^{\prime}-\frac{1}{m}\sum_{j\neq
k}y_{j}^{*}(y)$

for every $1\leq k\leq m$ and thus
$y_{k}^{*}(y)>1-2m\delta^{\prime}=1-\frac{m\delta}{n}\geq 1-\delta$ since
$m\leq n$. ∎

Let $y\in S(y^{*},\delta^{\prime})$ and $z^{*}\in S_{X^{*}}$. If
$z^{*}(x)>1-\delta^{\prime}$, then

$z^{*}\in S_{X^{*}}\cap
S(x,\delta^{\prime})\subset\bigcup_{j=1}^{m}B(y_{j}^{*},2\varepsilon)$

so there exists $y^{*}_{j_{0}}$ with $\|z^{*}-y_{j_{0}}^{*}\|\leq
2\varepsilon$. Then

$z^{*}(y)\geq y_{j_{0}}^{*}(y)-2\varepsilon>1-\delta-2\varepsilon$

and hence

$z^{*}(x-y)\leq 1-z^{*}(y)<\delta+2\varepsilon\leq 3\varepsilon.$

While if $z^{*}(x)\leq 1-\delta^{\prime}$, then

$z^{*}(x-y)\leq 1-\delta^{\prime}+1=2-\delta^{\prime}.$

Hence $\|x-y\|\leq\max(2-\delta^{\prime},3\varepsilon)$. We have
$\delta^{\prime}>0$ and $\varepsilon<\frac{2}{3}$ so $x$ cannot be a
$\Delta$-point. ∎

Combining Theorem 3.5 and Proposition 3.1 we then get

######  Proposition 3.6.

Let $X$ be a Banach space and let $x\in S_{X}$ satisfy
$\overline{\rho}_{X}(t,x)<\frac{t}{3}$ for some $t>0$. Then $x$ is not a
$\Delta$-point.

In particular no asymptotically smooth point can be a $\Delta$-point and we
obtain

######  Theorem 3.7.

Let $X$ be an AUS Banach space. Then $X$ does not admit a $\Delta$-point.

Next let us collect some examples where the above corollary applies.

Recall that a Banach space has Kalton’s property $(M)$ if whenever $x,y\in X$
with $\|x\|=\|y\|$ and $(x_{\alpha})$ is a bounded weakly null net in $X$,
then

$\limsup_{\alpha}\|x+x_{\alpha}\|=\limsup_{\alpha}\|y+x_{\alpha}\|.$

Similarly $X$ has property $(M^{*})$ if whenever $x^{*},y^{*}\in X^{*}$ with
$\|x^{*}\|=\|y^{*}\|$ and $(x^{*}_{\alpha})$ is a bounded weak∗ null net in
$X^{*}$, then

$\limsup_{\alpha}\|x^{*}+x^{*}_{\alpha}\|=\limsup_{\alpha}\|y^{*}+x^{*}_{\alpha}\|.$

If $X$ has property $(M^{*})$, then $X$ has property $(M)$ and $X$ is an
$M$-ideal in $X^{**}$ (see e.g. [24, Proposition VI.4.15]). In particular, $X$
is an Asplund space (see e.g. [24, Theorem III.3.1]). It is well known that
property $(M^{*})$ is inherited by both subspaces and quotients (see e.g.
[38]).

By chasing references we find that the following proposition holds.

######  Proposition 3.8.

Assume a Banach space $X$ has property $(M)$. The following are equivalent:

  1. (a)

$X$ is AUS;

  2. (b)

$X$ contains no copy of $\ell_{1}$;

  3. (c)

$X$ has property $(M^{*})$.

###### Proof.

(a) $\Rightarrow$ (b). If $X$ is AUS, then $X$ is Asplund (see e.g. [29,
Proposition 2.4]). Hence $X$ contains no copy of $\ell_{1}$.

(b) $\Rightarrow$ (c). If $X$ contains no copy of $\ell_{1}$, then no
separable subspace of $X$ can contain $\ell_{1}$. Clearly every separable
closed subspace of $X$ has property $(M)$ (both net and sequential version,
see [37, Proposition 1]) and then they all have property $(M^{*})$ (both net
and sequential version) by Theorem 2.6 in [31]. Finally $X$ has property
$(M^{*})$ if every separable closed subspace does [38, Proposition 3.1].

(c) $\Rightarrow$ (a). Dutta and Godard [13] proved that if $X$ is a separable
Banach space with property $(M^{*})$, then $X$ is AUS. However, using property
($M^{*}$) and Proposition 2.2 in [16] one finds
$\bar{\rho}_{X}(t,x)=\bar{\rho}_{X}(t)$ for all $x\in S_{X}$ and their proof
also works in the non-separable case. ∎

Since $c_{0}$ has property $(M^{*})$ we have that all subspaces and quotients
of $c_{0}$ are AUS and they all fail to contain $\Delta$-points. All these
examples are $M$-ideals in their bidual, that is, they are $M$-embedded [24,
Chapter 3].

Note that there are $M$-embedded spaces which are not AUS. For example the
Schreier space $\mathcal{S}$ is not AUS since it does not have property
$(M^{*})$. Indeed, if a Banach space $X$ has property $(M^{*})$, then the
relative norm and weak∗ topologies on $S_{X^{*}}$ coincide (see e.g. [24,
Proposition VI.4.15]). But if $(e_{i})$ is the unit vector basis in
$\mathcal{S}$ and $(e_{i}^{*})$ the the biorthogonal functionals in the dual,
then $e_{2}^{*}+e_{i}^{*}\in S_{\mathcal{S}^{*}}$ and converges weak∗ to
$e_{2}^{*}$, but not in norm. Note however that $\mathcal{S}$ does admit
$\Delta$-point by Proposition 2.15 in [3].

Let $X$ be a Banach space with a normalized basis $(e_{i})$ (or more generally
an FDD $(E_{i})$). We say that _$(e_{i})$ admits block upper $\ell_{p}$
estimates_ for some $p\in(1,\infty)$ if there is a constant $C>0$ such that
for every finite blocks $x_{1},\ldots,x_{N}$ of $(e_{i})$ with consecutive
disjoint supports we have $\left\|\sum_{n=1}^{N}x_{n}\right\|^{p}\leq
C\sum_{n=1}^{N}\left\|x_{n}\right\|^{p}$. We say that $(e_{i})$ admits block
lower $\ell_{q}$ estimates for some $q\in(1,\infty)$ if there is a constant
$c>0$ such that for every finite blocks $x_{1},\ldots,x_{N}$ of $(e_{i})$ with
consecutive disjoint supports we have
$\left\|\sum_{n=1}^{N}x_{n}\right\|^{q}\geq
c\sum_{n=1}^{N}\left\|x_{n}\right\|^{q}$. It is well known that a basis
admitting upper $\ell_{p}$ estimates is shrinking while a basis admitting
lower $\ell_{q}$ estimates is boundedly complete. The latter can be proved by
using the following criterion, which is left as an exercise in [4, Exercise
3.8] and whose proof can be found in [10, Proposition 3.1]: a basis is
boundedly complete if and only if
$\sup_{N}\left\|\sum_{n=1}^{N}x_{n}\right\|=\infty$ for every block sequence
$(x_{n})$ of $(e_{i})$ that is bounded away from $0$. The former is then
obtained by duality. Applying [16, Corollary 2.4] we then have that any space
with a basis admitting block upper $\ell_{p}$ estimates is AUS (with power
type $p$) and that any space admitting a basis with block lower $\ell_{q}$
estimates is AUC∗ (with power type $q$) as the dual of the space of
$Y=[e_{i}^{*}]$. As a consequence, a Banach space with a basis admitting block
upper $\ell_{p}$ estimates does not admit $\Delta$-points and the predual of a
Banach space with a basis admitting lower $\ell_{q}$ estimates does not admit
$\Delta$-points. This applies in particular to the predual of the James tree
space (see Section 5) and to the Baernstein space $B$ (see at the end of
Section 4).

##  4\. Asymptotic uniform convexity

The main result of this section is that reflexive AUC spaces do not have
$\Delta$-points. The proof uses the characterization of reflexive AUC spaces
in terms of the measure of non-compactness of slices and relies on the
following result of Kuratowski (see e.g. [7, p. 151])

######  Lemma 4.1.

Let $(M,d)$ be complete metric space. If $(F_{n})$ is a decreasing sequence of
non-empty, closed, and bounded subsets of $M$ such that
$\lim_{n}\alpha(F_{n})=0$, then the intersection
$F_{\infty}=\bigcap_{n=1}^{\infty}F_{n}$ is a non-empty compact subset of $M$.

Let us first proof a pointwise version of the main result.

######  Theorem 4.2.

Let $X$ be a Banach space and $x\in S_{X}$. If there exists $x_{0}^{*}\in
D(x)$ such that $\lim_{\delta\to 0}\alpha(S(x_{0}^{*},\delta))=0$, then $x$ is
not a $\Delta$-point.

###### Proof.

Let $x\in S_{X}$ be such that there exists $x_{0}^{*}\in D(x)$ with
$\lim_{\delta\to 0}\alpha(S(x_{0}^{*},\delta))=0$ and let us assume for
contradiction that $x$ is a $\Delta$-point.

Define a set of functionals norming $x$ by

$D_{0}(x):=\left\\{\frac{x^{*}+x_{0}^{*}}{2}:x^{*}\in D(x)\right\\}.$

If $\delta>0$ and $f\in D_{0}(x)$, then by Lemma 2.1

$S\left(f,\frac{\delta}{2}\right)=S\left(\frac{x^{*}+x_{0}^{*}}{2},\frac{\delta}{2}\right)\subset
S(x^{*},\delta)\cap S(x_{0}^{*},\delta)$

and therefore $\lim_{\delta\to
0}\alpha(\overline{S}(f,\delta))=\lim_{\delta\to 0}\alpha(S(f,\delta))=0$ for
all $f\in D_{0}(x)$.

Given $f\in D_{0}(x)$ and a sequence $(\delta_{n})\subseteq(0,1)$ decreasing
to $0$ we define for each $n$

$F_{n}^{f}:=\overline{S}(f,\delta_{n})\cap\Delta_{\delta_{n}}(x).$

Each $F_{n}^{f}$ is a closed non-empty subset of $B_{X}$ with
$\alpha(F_{n}^{f})\to 0$ hence $F^{f}=\cap_{n}F_{n}^{f}$ is non-empty and
compact for every $f\in D_{0}(x)$ by Lemma 4.1. Using a compactness argument
we can then prove the following.

Claim.

$F:=\bigcap_{f\in D_{0}(x)}F^{f}\neq\emptyset.$

Before proving this claim let us see that it will give us the desired
conclusion. Indeed, if $z\in F$, then $\|x-z\|=2$ and for all $f\in D_{0}(x)$
we must have $f(z)=1$, in particular $x^{*}_{0}(z)=1$. But this is nonsense
since for any $x^{*}\in S_{X^{*}}$ with $x^{*}(x-z)=2$ we must have
$x^{*}(x)=1$ and $x^{*}(z)=-1$, so

$1=f_{x^{*}}(x)=\frac{x^{*}+x^{*}_{0}}{2}(z)=\frac{-1+1}{2}=0.$

This contradiction shows that $x$ is not a $\Delta$-point.

To finish the proof we only need to prove the claim that $F\neq\emptyset$. It
is enough to show that $(F^{f})_{f\in D_{0}(x)}$ has the finite intersection
property since all the sets $F^{f}$ are compact and non-empty.

Let $f_{1},\ldots,f_{k}\in D_{0}(x)$, which means
$f_{j}=\frac{x_{j}^{*}+x_{0}^{*}}{2}$ for $x^{*}_{j}\in D(x)$. Define

$f=\frac{1}{k}\sum_{j=1}^{k}f_{j}=\frac{x_{0}^{*}+\frac{1}{k}\sum_{j=1}^{k}x^{*}_{j}}{2}$

Clearly $f\in D_{0}(x)$. By Lemma 2.1 we have for all $\delta>0$

$S\left(f,\frac{\delta}{k}\right)\subset\bigcap_{j=1}^{k}S(f_{j},\delta)$

and hence

(1)
$S\left(f,\frac{\delta}{k}\right)\cap\Delta_{\frac{\delta}{k}}(x)\subset\bigcap_{j=1}^{k}S(f_{j},\delta)\cap\Delta_{\delta}(x).$

Since $(\delta_{n})$ is decreasing there must for any $n$ exist $m$ with
$\delta_{m}<\delta_{n}/k$. By (1) we get

$F_{m}^{f}\subset\bigcap_{j=1}^{k}F_{n}^{f_{j}}$

and hence by commutativity of intersections

$\emptyset\neq\bigcap_{n}F_{n}^{f}\subseteq\bigcap_{j=1}^{k}\bigcap_{n}F_{n}^{f_{j}}=\bigcap_{j=1}^{k}F^{f_{j}}$

and the claim is proved. ∎

From Theorem 4.2 we immediately get

######  Theorem 4.3.

If $X$ has Rolewicz’ property $(\alpha)$, then $X$ does not have
$\Delta$-points.

As we noted in Section 2 a Banach space $X$ is reflexive and AUC if and only
if it has uniform property $(\alpha)$. Also finite-dimensional spaces are
trivially AUC since for example $\alpha(S(x^{*},\delta))=0$ for slices of
$B_{X}$ in finite-dimensional spaces.

######  Theorem 4.4.

If $X$ is reflexive and AUC, then $X$ does not have $\Delta$-points.

In particular, if $X$ is finite-dimensional then $X$ does not have
$\Delta$-points.

######  Remark 4.5.

Let $X$ be a Banach space such that for every $x\in S_{X}$ there exists
$x^{*}\in D(x)$ with $\lim_{\delta\to 0}\alpha(S(x^{*},\delta))=0$. Then by
Theorem 4.2 $X$ does not admit a $\Delta$-point. Note that unlike Rolewicz’
property $(\alpha)$ (see [40]) this property does not imply reflexivity.

Indeed, every separable Banach space has an equivalent locally uniformly
rotund renorming and if (the norm of) $X$ is locally uniformly rotund, then
every $x\in S_{X}$ is strongly exposed by $x^{*}\in D(x)$ so that for all
$\varepsilon>0$ there exists $\delta>0$ such that $S(x^{*},\delta)$ has
diameter less than $\varepsilon$. In particular,
$\alpha(S(x^{*},\delta))<\varepsilon$.

Using the duality AUS/AUC in reflexive spaces we can in fact combine Theorem
3.7 and 4.4.

######  Corollary 4.6.

If $X$ is reflexive and AUC, then neither $X$ nor $X^{*}$ admit
$\Delta$-points.

In particular we can apply this result to the Baernstein’s space $B$ whose
construction and basic properties are given in the introductory Chapter 0 of
[9, Construction 0.9]. This space was originally introduced in [5] as an
example of a reflexive space failing the Banach–Saks property. It is known to
have a normalized (unconditional) basis with block lower $\ell_{2}$ estimates
and thus to be $2$-AUC (see [39, Theorem 3] with the NUC terminology). Also
the optimal modulus of near convexity of $B$ has been estimated in [6]. From
our preceding results the space $B$ and its dual space $B^{*}$ both fail to
have $\Delta$-points.

Here is a pointwise application of Theorem 4.2.

######  Corollary 4.7.

Let $X$ be a Banach space and let $x^{*}\in S_{X^{*}}$ be a norm one
functional which attains its norm at some $x\in S_{X}$. If $\lim_{t\to
0}\frac{\overline{\rho}(t,x)}{t}=0$, then $x^{*}$ is not a $\Delta$-point.

###### Proof.

By Corollary 3.4 we have $\lim_{\delta\to 0}\alpha(S(x,\delta))=0$ and the
conclusion follows directly from Theorem 4.2 since $x\in D(x^{*})$. ∎

Let $X$ be a Banach space. An element $x\in B_{X}$ is said to be a _quasi-
denting point_ if given $\varepsilon>0$, there exists $x^{*}\in S_{X^{*}}$ and
$\delta>0$ with $x\in S(x^{*},\delta)$ such that
$\alpha(S(x^{*},\delta))<\varepsilon$. Recall that $x$ is a _denting-point_ if
the above can be strengthened to $S(x^{*},\delta)\subset B(x,\varepsilon)$. In
dual spaces one can similarly define weak∗ (quasi-)denting points by requiring
$x^{*}$ to be weak∗ continuous. Giles and Moors [19] introduced quasi-denting
points (under the name $\alpha$-denting points) see e.g. [35] or [45].

Let $X$ be a Banach space such that the dual is AUC∗, then every $x^{*}\in
S_{X^{*}}$ is a quasi-denting point. This is essentially contained in e.g.
[22, Proposition 4.8], but we include the straightforward argument. The AUC∗
is equivalent to the following version of UKK∗: For every $\varepsilon>0$
there exists $\delta>0$ such that if $x^{*}\in B_{X^{*}}$ and all weak∗
neighborhoods $V$ of $x^{*}$ satisfy $\operatorname{diam}(V\cap
B_{X^{*}})>\varepsilon$, then $\|x^{*}\|\leq 1-\delta$. Now fix
$\varepsilon>0$ and choose $\delta>0$ as above. If $x^{*}\in S_{X^{*}}$, then
we can find $x\in S_{X}$ with $x^{*}(x)>1-\delta$. Let
$0<\delta^{\prime}<\delta$. If $y^{*}\in S(x,\delta)$, then
$\|y^{*}\|>1-\delta$ and the exits a weak∗ open neighborhood $V_{y^{*}}$ of
$y^{*}$ with $\operatorname{diam}(V_{y^{*}}\cap B_{X^{*}})\leq\varepsilon$. We
therefore have an open cover of the weak∗ compact set
$\bar{S}(x,\delta^{\prime})$ and by compactness we have a finite cover and
hence $x^{*}$ is a weak∗ quasi-denting point.

If the dual $X^{*}$ is AUC∗, then $X$ is AUS and by Corollary 4.7 no norm-
attaining $x^{*}\in S_{X^{*}}$ can be a $\Delta$-point, but we do not know if
a weak∗-quasi-denting point, or more generally a quasi-denting point, can be a
Daugavet-point or a $\Delta$-point. For non-reflexive AUC spaces, we do not
even know if every element of the unit sphere is quasi-denting.

##  5\. The James tree space

Let $T=\\{\emptyset\\}\cup\bigcup_{n\geq 1}\\{0,1\\}^{n}$ be the infinite
binary tree and let us denote by $\leq$ the natural ordering on $T$. A totally
ordered subset $S$ of $T$ is called a segment if it satisfies:

$\forall s,t\in S,\ [s,t]=\\{u\in T:s\leq u\leq t\\}\subset S.$

An infinite segment of $T$ is also called a branch of $T$. Let us denote by
$\mathcal{F}$ the set of all finite families of disjoint segments in $T$. The
James tree space $JT$ is define as follows:

$JT=\left\\{x=(x_{s})_{s\in T}\subset\mathbb{R}:\
\left\|x\right\|_{JT}^{2}=\sup_{F\in\mathcal{F}}\sum_{S\in
F}x_{S}^{2}<\infty\right\\}$

where $x_{S}=\sum_{s\in S}x_{s}$ for every non-empty segment $S$ and
$x_{\emptyset}=0$.

It is well known that $JT$ is a Banach space and that the set of canonical
unit vectors $\\{e_{t}\\}_{t\in T}$ of $c_{00}(T)$ forms, for the
lexicographic order, a normalized monotone boundedly complete basis of $JT$.
Moreover, it is also known that the closed linear span
$JT_{*}=[e_{t}^{*}]_{t\in T}$ of the set of biorthogonal functionals in
$JT^{*}$ is a unique isometric predual of $JT$. If $S$ is a segment of $T$ or
if $\beta$ is a branch of $T$ we will sometimes refer to the set
$\\{e_{s}\\}_{s\in S}$ as a segment of $JT$ and the set
$\\{e_{t}\\}_{t\in\beta}$ as a branch of $JT$. From the definition of the
norm, it is easy to show that
$\left\|x+y\right\|_{JT}^{2}\geq\left\|x\right\|_{JT}^{2}+\left\|y\right\|_{JT}^{2}$
whenever $x,y\in JT$ have totally disconnected supports (that is if
$\operatorname{conv}(\operatorname{supp}_{T}x)\cap\operatorname{conv}(\operatorname{supp}_{T}y)=\emptyset$).
Since we are working with the lexicographic order on $T$, this applies
whenever $\operatorname{supp}x<\operatorname{supp}y$ with respect to the
ordering of the basis, and $\\{e_{t}\\}_{t\in T}$ thus satisfies block lower
$\ell_{2}$ estimates. It follows that $JT$ is $2$-AUC∗ and Theorem 3.7 applied
to the $2$-AUS $JT_{*}$ yields the following result.

######  Theorem 5.1.

The predual $JT_{*}$ of the James tree space does not admit $\Delta$-points.

Let us also emphasize that
$\left\|x\right\|_{JT}\geq\left\|x\right\|_{\ell_{2}}$ for every $x\in JT$. It
is also worth mentioning that if one consider the equivalent norm
$\left\|x\right\|^{2}=\sup_{S_{1}<\dots<S_{n}}\sum_{i=1}^{n}x_{S_{i}}^{2}$ on
the James space $J$, where the $S_{i}$ are segments of $\mathbb{N}$, then one
obtains a Banach space isometric to the closed linear span
$[e_{t}]_{t\in\beta}$ of any branch of $JT$. In particular all the results we
will obtain for the space $JT$ will also apply to $(J,\left\|.\right\|)$.

As we see from Proposition 2.3 the unit ball of spaces that are not uniformly
non-square contain small slices of diameter arbitrary close to $2$. Let us
start by illustrating this by exhibiting slices of diameter $2$ in JT.

######  Proposition 5.2.

For every $\delta>0$ we can find some $x^{*}\in S_{JT^{*}}$ such that
$\operatorname{diam}S(x^{*},\delta)=2$ and the diameter is attained.

###### Proof.

For convenience let us work in the space $(J,\left\|.\right\|)$ introduced
above. Doing the same construction on any branch $(x_{t})_{t\in\beta}$ of $JT$
would do the work for $JT$.

Fix $\delta>0$, fix $n\geq 1$, and let

$x=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(-1)^{i+1}e_{2i-1}=\frac{1}{\sqrt{n}}(1,0,-1,0,\ldots,-1,0,0,0,\ldots)$

and

$y=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(-1)^{i}e_{2i}=\frac{1}{\sqrt{n}}(0,-1,0,1,\ldots,0,1,0,0,\ldots)$

so that

$x-y=\frac{1}{\sqrt{n}}(1,1,-1,-1,\ldots,-1,-1,0,0,\ldots)$

and

$x+y=\frac{1}{\sqrt{n}}(1,-1,-1,1,1,\ldots,-1,-1,1,0,0,\ldots).$

It is easy to check that $\left\|x\right\|=\left\|y\right\|=1$,
$\left\|x-y\right\|=2$, and $\left\|x+y\right\|=\sqrt{\frac{1+(n-1)\times
2^{2}+1}{n}}=\sqrt{4-\frac{2}{n}}$.

By assuming that $n$ was chosen large enough so that
$\sqrt{4-\frac{2}{n}}>2-\delta$ and by choosing a norming functional $x^{*}\in
S_{J^{*}}$ for $x+y$ we then have $x^{*}(x+y)=\left\|x+y\right\|>2-\delta$ and
this implies that $x,y\in S(x^{*},\delta)$. Since $\left\|x-y\right\|=2$ the
conclusion follows. ∎

We start by showing that the elements of the basis of $JT$ are weak∗ denting
(and even strongly exposed by an element of the predual) and then use this to
prove that $JT$ does not admit weak∗ Daugavet-points.

######  Lemma 5.3.

For each $t\in T$, $\lim_{\delta\to
0}\operatorname{diam}S(e_{t}^{*},\delta)=0$.

###### Proof.

Let us fix $t\in T$ and $\delta>0$, and let us take $y\in
S(e_{t}^{*},\delta)$. By the triangle inequality, we have

$\left\|y-e_{t}\right\|_{JT}\leq\left\|y-y_{t}e_{t}\right\|_{JT}+(1-y_{t})\leq\left\|y-y_{t}e_{t}\right\|_{JT}+\delta$

so we only need to estimate $\left\|y-y_{t}e_{t}\right\|_{JT}$.

If $t$ is not in the support of a family $F\in\mathcal{F}$, that is if
$t\notin\bigcup_{S\in F}S$, then the segment $\\{t\\}$ is disjoint from all
the segments in $F$ and we have

$y_{t}^{2}+\sum_{S\in F}(y-y_{t}e_{t})_{S}^{2}=y_{t}^{2}+\sum_{S\in
F}y_{S}^{2}\leq\left\|y\right\|_{JT}^{2}\leq 1$

so that

$\sum_{S\in F}(y-y_{t}e_{t})_{S}^{2}\leq 1-(1-\delta)^{2}\leq 2\delta.$

Now if we take any segment $S$ of $T$ containing $t$, we can write
$S=S^{-}\cup\\{t\\}\cup S^{+}$ with $S^{-}<\\{t\\}<S^{+}$ segments of $T$, and
by the preceding computations we have

$(y-y_{t}e_{t})_{S}^{2}\leq
y_{S^{-}}^{2}+y_{S^{+}}^{2}+2\sqrt{y_{S^{-}}^{2}y_{S^{+}}^{2}}\leq 6\delta.$

By combining the two observations we get $\sum_{S\in
F}(y-y_{t}e_{t})_{S}^{2}\leq 8\delta$ for every $F\in\mathcal{F}$, that is
$\left\|y-y_{t}e_{t}\right\|_{JT}^{2}\leq 8\delta$. The conclusion follows. ∎

######  Corollary 5.4.

The space $JT$ does not admit weak∗ Daugavet-points.

###### Proof.

Let us fix $x\in S_{JT}$. If $x=e_{t}$ for some $t\in T$ then $x$ is a weak∗
denting point by the preceding lemma and $x$ cannot be a weak∗ Daugavet-point.
So let us assume that $x$ is not of that form and let us take $t\in T$ such
that $x_{t}\neq 0$ and $x_{s}=0$ for every $s<t$.

Now let us write $x_{t}=\theta\alpha$ with $\alpha\in(0,1)$ and
$\theta\in\\{-1,1\\}$. Because of the choice of $t$ we clearly have

$\left\|x-x_{t}e_{t}\right\|_{JT}^{2}=\sup_{F\in\mathcal{F}_{t}}\sum_{S\in
F}^{n}x_{S}^{2}$

where $\mathcal{F}_{t}$ is the set of finite families of disjoint segments of
$T$ not intersecting $[\emptyset,t]$. Since

$x_{t}^{2}+\sum_{S\in F}x_{S}^{2}\leq\left\|x\right\|_{JT}^{2}$

for every such family we obtain

$\left\|x-x_{t}e_{t}\right\|_{JT}^{2}\leq 1-\alpha^{2}<1.$

Thus

$\left\|x-\theta
e_{t}\right\|_{JT}\leq\left\|x-x_{t}e_{t}\right\|_{JT}+(1-\alpha)\leq
2-\alpha$

so that $x$ is at distance strictly less than $2$ from a (weak∗) denting point
of $JT$. The conclusion follows from a weak∗ version of [30, Proposition 3.1].
∎

Now we will show that $JT$ does not admit $\Delta$-points. For this purpose
let us introduce some more notations. Let us write $\mathcal{F}_{\infty}$ for
the set of (finite or infinite) families of disjoint segments in $T$. Note
that every infinite $F\in\mathcal{F}_{\infty}$ has to be countable. For any
segment $S$ of $T$, let us write $\mathbbm{1}_{S}=\sum_{s\in S}e_{s}^{*}$. A
molecule in $JT^{*}$ is an element of the form $\sum_{i\geq
1}\lambda_{i}\mathbbm{1}_{S_{i}}$ where $\\{S_{i}\\}\in\mathcal{F}_{\infty}$
and $\lambda=(\lambda_{i})\in B_{\ell_{2}}$. The molecule is finite if the
family $\\{S_{i}\\}$ contains only finitely many non-empty segments, and we
will write $\mathcal{M}$ (resp. $\mathcal{M}_{\infty}$) for the set of finite
(resp. finite or infinite) molecules of $JT^{*}$.

Molecules play an important role in the study of the space $JT$ because they
turn computations in $JT$ into computations in $\ell_{2}$ and because of the
following result due to Schachermayer [43, Proposition 2.2].

######  Theorem 5.5.

The unit ball $B_{JT^{*}}$ of the dual of JT is the norm closed convex hull of
$\mathcal{M}$.

######  Remark 5.6.

It is also known that $\mathcal{M}_{\infty}$ is the weak∗ closure of
$\mathcal{M}$ in $B_{JT^{*}}$.

We will use some specific molecules to provide norming functionals for
elements of $S_{JT}$. For $x\in S_{JT}$ and $F\in\mathcal{F}_{\infty}$, let us
write $m_{x,F}=\sum_{S\in F}x_{S}\mathbbm{1}_{S}$. Since $\sum_{S\in
F}x_{S}^{2}\leq\left\|x\right\|_{JT}=1$, those elements are molecules in
$JT^{*}$. Moreover, we have the following result.

######  Lemma 5.7.

Let us assume that $\sum_{S\in F}x_{S}^{2}=\left\|x\right\|_{JT}=1$. Then the
molecule $m_{x,F}$ belongs to $D(x)$ (that is, it is a norm one functional and
it norms $x$). Moreover, if $y\in S(m_{x,F},\delta)$ for some $\delta>0$, then
we have

$\sum_{S\in F}(x_{S}-y_{S})^{2}\leq 2\delta.$

###### Proof.

The key observation here is that for every $y\in JT$, we have

$m_{x,F}(y)=\sum_{S\in F}x_{S}y_{S}=\langle(x_{S})_{S\in F},(y_{S})_{S\in
F}\rangle_{\ell_{2}}.$

In particular we have, by the Cauchy–Schwartz inequality,

$\left|m_{x,F}(y)\right|\leq\left\|(x_{S})_{S\in
F}\right\|_{\ell_{2}}\left\|(y_{S})_{S\in
F}\right\|_{\ell_{2}}\leq\left\|x\right\|_{JT}\left\|y\right\|_{JT},$

so $\left\|m_{x,F}\right\|_{JT^{*}}\leq 1$. By assumption
$m_{x,F}(x)=\left\|(x_{S})_{S\in F}\right\|_{\ell_{2}}^{2}=1$ and the first
part of the lemma follows.

Now $y\in S(m_{x,F},\delta)$ if and only if $\langle(x_{S})_{S\in
F},(y_{S})_{S\in F}\rangle_{\ell_{2}}>1-\delta$ and thus every $y$ in
$S(m_{x,F},\delta)$ satisfy

$\displaystyle\left\|(x_{S}-y_{S})_{S\in F}\right\|_{\ell_{2}}^{2}=$
$\displaystyle\left\|(x_{S})_{S\in
F}\right\|_{\ell_{2}}^{2}+\left\|(x_{S})_{S\in
F}\right\|_{\ell_{2}}^{2}-2\langle(x_{S})_{S\in F},(y_{S})_{S\in
F}\rangle_{\ell_{2}}$ $\displaystyle\leq 2-2(1-\delta)=2\delta.\qed$

Note that it is not obvious at first that such a norm attaining family exists.
We will first do a warm up with finitely supported elements in $JT$ for which
this is obvious, and then prove it in a lemma.

######  Proposition 5.8.

Let $x\in S_{JT}$ be an element of finite support. Then $x$ is not a weak∗
$\Delta$-point.

###### Proof.

Let $x\in S_{JT}$ be an element of finite support $\Sigma$ and let
$\mathcal{F}_{\Sigma}$ be the set of families of disjoint segments of the
convex hull of $\Sigma$ in $T$ (that is the smaller subset of $T$ containing
$[s,t]$ for every $s\leq t$ in $\Sigma$). Then $\mathcal{F}_{\Sigma}$ is
finite and we have

$\left\|x\right\|_{JT}^{2}=\max_{F\in\mathcal{F}_{\Sigma}}\sum_{S\in
F}x_{S}^{2}.$

Now let $\mathcal{D}=\\{F\in\mathcal{F}_{\Sigma}:\ \sum_{S\in
F}x_{S}^{2}=1\\}$. The set $\mathcal{D}$ is a non-empty subset of
$\mathcal{F}_{\Sigma}$ and since this set is finite we can find a constant
$\eta_{x}>0$ such that for every
$F\in\mathcal{F}_{\Sigma}\backslash\mathcal{D}$ we have

$\sum_{S\in F}x_{S}^{2}<(1-\eta_{x})^{2}.$

Let us introduce
$x^{*}=\frac{1}{\left|\mathcal{D}\right|}\sum_{F\in\mathcal{D}}m_{x,F}$ where
$m_{x,F}$ is the molecule associated to $x$ and $F$. For every
$F\in\mathcal{D}$, the molecule $m_{x,F}$ is in $D(x)$ by the preceding lemma,
and by Lemma 2.1 we know that $x^{*}$ is in $D(x)$ and that for every choice
of a $\delta>0$ we have

$S\left(x^{*},\frac{\delta}{n}\right)\subset\bigcap_{F\in\mathcal{D}}S(m_{x,F},\delta).$

To conclude choose any $\delta\in(0,1)$ and let us take $y\in
S\left(x^{*},\frac{\delta}{n}\right)$. If $F\in\mathcal{F}$ does not belong to
$\mathcal{D}$, then by Minkowski’s inequality we have

$\sum_{S\in F}(x_{S}-y_{S})^{2}\leq\left(\sqrt{\sum_{S\in
F}x_{S}^{2}}+\sqrt{\sum_{S\in F}y_{S}^{2}}\right)^{2}\leq(2-\eta_{x})^{2}.$

Now if $F\in\mathcal{D}$ we have $y\in S(m_{x,F},\delta)$ and by the preceding
lemma we get

$\sum_{S\in F}(x_{S}-y_{S})^{2}\leq 2\delta.$

Finally $\left\|x-y\right\|_{JT}\leq\max\\{2-\eta_{x},\sqrt{2\delta}\\}$ and
this quantity is strictly less than $2$ since $\delta<1$. To conclude, note
that $x^{*}\in JT_{*}$ since all the segments involved in the construction are
finite. ∎

To tackle the other elements of $JT$ we first need to ensure the existence of
norm attaining (possibly infinite) families.

######  Lemma 5.9.

Let $x\in S_{JT}$. Then there is an $F\in\mathcal{F}_{\infty}$ such that
$\sum_{S\in F}x_{S}^{2}=1$.

###### Proof.

Let $x\in S_{JT}.$ By Krein–Milman $x$ attains its norm on an extreme point
$x^{*}\in B_{JT*}.$ By Milman’s converse, Lemma 5.5 and Remark 5.6, $x^{*}\in
M_{\infty},$ so we can write $x^{*}=\sum_{i\geq
1}\lambda_{i}\mathbbm{1}_{S_{i}}$ for some $\lambda=(\lambda_{i})$ in
$B_{\ell_{2}}$ and $\\{S_{i}\\}$ in $\mathcal{F}_{\infty}$. To conclude,
consider $\mu=(x_{S_{i}})$ in $B_{\ell_{2}}$ and observe that

$\langle\lambda,\mu\rangle_{\ell_{2}}=x^{*}(x)=1,$

so $\lambda$ strongly exposes $\mu$ in $B_{\ell_{2}}.$ Thus $\lambda=\mu$ and
$\left\|\mu\right\|_{\ell_{2}}=1$ which is precisely the desired result. ∎

For every $x\in S_{JT}$ let us introduce
$\mathcal{D}(x)=\\{F\in\mathcal{F}_{\infty}:\ \sum_{S\in F}x^{2}_{S}=1\\}$. By
the preceding lemma $\mathcal{D}(x)$ is non-empty, but it does not need to be
finite. To get around this problem we will consider the restriction of
families in $\mathcal{D}(x)$ to a subtree of finite level. So for every $N\geq
1$ let us write
$T_{N}=\text{level}[0,N]=\\{\emptyset\\}\cup\bigcup_{n=1}^{N}\\{0,1\\}^{n}$
the binary tree of height $N$ and let us write $\mathcal{D}_{N}(x)=\\{F\cap
T_{N}:\ F\in\mathcal{D}(x)\\}$ where $F\cap T_{N}=\\{S\cap T_{N}:\ S\in F\\}$.
Then $\mathcal{D}_{N}(x)$ is a finite non-empty set and we have, similarly to
the finite support case, the following lemma.

######  Lemma 5.10.

For every $x\in S_{JT}$ and for every $N\geq 1$, there is a constant
$\eta_{x,N}>0$ such that $\sum_{S\in F}x_{S}^{2}<(1-\eta_{x,N})^{2}$ for every
$F$ in $\mathcal{F}$ for which $F\cap T_{N}$ does not belong to
$\mathcal{D}_{N}(x)$.

###### Proof.

If this was not true, then since $\mathcal{F}\cap T_{N}$ is also finite we
could find a family $F\in\mathcal{F}$ of segments of $T_{N}$ not belonging to
$\mathcal{D}_{N}(x)$ and a sequence $(F_{i})_{i\geq 1}\subset\mathcal{F}$ such
that $F_{i}\cap T_{N}=F$ and $\sum_{S\in
F_{i}}x_{S}^{2}>\left(1-\frac{1}{i}\right)^{2}$ for every $i\geq 1$. Using a
compactness argument (weak∗ extractions for the sequence of corresponding
molecules in $B_{JT^{*}}$ and metrizability of the weak∗ topology on
$B_{JT^{*}}$) we would then obtain a family $G\in\mathcal{F}_{\infty}$ for
which $\sum_{S\in G}x_{S}^{2}=1$ and such that $G\cap T_{N}=F$. But then this
would mean that $F\in\mathcal{D}_{N}(x)$ and we would get a contradiction. ∎

We will need a last easy fact which states that any $x\in S_{JT}$ has its norm
almost concentrated on a subtree of finite level. The proof is elementary and
comes from the definition of the norm of $JT$.

######  Lemma 5.11.

Let $x\in S_{JT}$ and let $\varepsilon>0$. There is an $N\geq 1$ such that for
every family $F\in\mathcal{F}_{\infty}$ of segments of $T$ which do not
intersect $T_{N}$ one has $\sum_{S\in F}x_{S}^{2}\leq\varepsilon^{2}$.

With those tools in hand we can now prove the main result of this section.

######  Theorem 5.12.

Let $x\in S_{JT}$. Then $x$ is not a $\Delta$-point.

###### Proof.

Let us fix some $x\in S_{JT}$ and let us assume that $x$ has infinite support.
Let us also fix some $\varepsilon>0$ and let us take some $N\geq 1$ for which
$x$ is almost concentrated on $T_{N}$ in the sense of the previous lemma.

For every $F$ in $\mathcal{D}_{N}(x)$ we pick a representative family
$F_{\mathcal{R}}$ in $\mathcal{D}(x)$ which satisfy $F_{\mathcal{R}}\cap
T_{N}=F$. This means that for every $S\in F$ there is (a unique)
$S_{\mathcal{R}}\in F_{\mathcal{R}}$ such that $S_{\mathcal{R}}\cap T_{N}=S$.
For every such $F$ we define $x_{F}^{*}=m_{x,F_{\mathcal{R}}}$ to be the
molecule associated to $x$ and $F_{\mathcal{R}}$ and we let $x^{*}$ be the
average of the $x_{F}^{*}s$ with $F\in\mathcal{D}_{N}(x)$. Since each
$x_{F}^{*}$ is in $D(x)$, Lemma 2.1 tells us that $x^{*}\in D(x)$ and that

$S\left(x^{*},\frac{\varepsilon^{2}}{2n}\right)\subset\bigcap_{F\in\mathcal{D}_{N}(x)}S\left(x_{F}^{*},\frac{\varepsilon^{2}}{2}\right)$

for $n=\left|\mathcal{D}_{N}(x)\right|$. We now want to get a uniform bound
for $\left\|x-y\right\|_{JT}$ on
$S\left(x^{*},\frac{\varepsilon^{2}}{2n}\right)$.

So let us take $y\in S\left(x^{*},\frac{\varepsilon^{2}}{2n}\right)$ and let
us fix some $G\in\mathcal{F}$. First observe that Lemma 5.10 allows us to get
rid of the case $G\cap T_{N}\notin\mathcal{D}_{N}(x)$ exactly as in the finite
support proof because it yields

$\sum_{S\in G}(x_{S}-y_{S})^{2}\leq(2-\eta_{x,N})^{2}.$

So let us assume that $F=G\cap T_{N}$ belongs to $\mathcal{D}_{N}(x)$. We will
split the family $G$ into $4$ disjoint subfamilies $G_{i}$, $1\leq i\leq 4$,
and we will estimate separately the sums $\sum_{S\in G_{i}}(x_{S}-y_{S})^{2}$.
For this we will use repeatedly the two following inequalities.

Claim $A$. Let $H$ be a subfamily of $F_{\mathcal{R}}$. Then

$\sum_{S\in H}(x_{S}-y_{S})^{2}\leq\varepsilon^{2}.$

###### Proof of claim $A$.

Since segments in $H$ are in $F_{\mathcal{R}}$ and since $y\in
S\left(x_{F}^{*},\frac{\varepsilon^{2}}{2}\right)$ with
$x_{F}^{*}=m_{x,F_{\mathcal{R}}}$ Lemma 5.7 yields

$\sum_{S\in H}(x_{S}-y_{S})^{2}\leq\sum_{S\in
F_{\mathcal{R}}}(x_{S}-y_{S})^{2}\leq\frac{2\varepsilon^{2}}{2}=\varepsilon^{2}.$

∎

Claim $B$. Let $H$ be a family of disjoint segments of $T$ which do not
intersect $T_{N}$. Then

$\sum_{S\in H}(x_{S}-y_{S})^{2}\leq\sum_{S\in
H}y_{S}^{2}+\varepsilon^{2}+2\varepsilon.$

###### Proof of claim $B$.

Since segments in $H$ do not intersect $T_{N}$ we have, by our initial choice
of $N$, $\sum_{S\in H}x_{S}^{2}\leq\varepsilon^{2}$ and so using Minkowski’s
inequality

$\displaystyle\sum_{S\in H}(x_{S}-y_{S})^{2}$ $\displaystyle\leq\sum_{S\in
H}x_{S}^{2}+\sum_{S\in H}y_{S}^{2}+2\sqrt{\sum_{S\in
H}x_{S}^{2}}\sqrt{\sum_{S\in H}y_{S}^{2}}$ $\displaystyle\leq\sum_{S\in
H}y_{S}^{2}+\varepsilon^{2}+2\varepsilon.$

∎

Now let us split the family $G$ and let us start the computations of the
corresponding sums.

Claim 1. Let $G_{1}=\\{S\in G:\ S\subset T_{N-1}\\}$. There is a
$\gamma_{1}=\gamma_{1}(\varepsilon)$ such that

$\sum_{S\in G_{1}}(x_{S}-y_{S})^{2}\leq\gamma_{1}.$

###### Proof of Claim 1.

Since segments in $G_{1}$ are contained in $T_{N-1}$, every segment $S$ in
$G_{1}$ has to be equal to its representative $S_{\mathcal{R}}$ because
$S_{\mathcal{R}}\cap T_{N}=S$. This means that $G_{1}$ is a subfamily of
$F_{\mathcal{R}}$ and the result follows directly from Claim $A$. ∎

Claim 2. Let $G_{2}=\\{S\in G:\ S\subset T\backslash T_{N}\\}$. There is a
$\gamma_{2}=\gamma_{2}(\varepsilon)$ such that

$\sum_{S\in G_{2}}(x_{S}-y_{S})^{2}\leq\sum_{S\in G_{2}}y_{S}^{2}+\gamma_{2}.$

###### Proof of Claim 2.

Since segments in $G_{2}$ do not intersect $T_{N}$ the result follows directly
from Claim $B$. ∎

The remaining segments of $G$ are those that intersect the $N^{\text{th}}$
level of $T$. For those, we will distinguish between segments $S$ such that
$S$ and its representative $S_{\mathcal{R}}$ are on the same branch of $T$ and
those for which $S$ and $S_{\mathcal{R}}$ split at some higher level. So let
$G_{3}$ be the subset of the remaining segments of $G$ satisfying the first
condition and let $G_{4}$ be the subset of those satisfying the second
condition.

Note that if $S$ is a segment in $G_{3}$, then since $S$ and $S_{\mathcal{R}}$
are on the same branch and are equal in $T_{N}$ we have either $S\subset
S_{\mathcal{R}}$ or $S_{\mathcal{R}}\subset S$ and the complements are in
either case segments which do not intersect $T_{N}$. Thus.

Claim 3. Let $G_{3,1}=\\{S\in G_{3}:\ S=S_{\mathcal{R}}\\}$, let
$G_{3,2}=\\{S\in G_{3}:\ S\subsetneqq S_{\mathcal{R}}\\}$ and let
$G_{3,3}=\\{S\in G_{3}:\ S_{\mathcal{R}}\subsetneqq S\\}$. There is a
$\gamma_{3}=\gamma_{3}(\varepsilon)$ such that

$\sum_{S\in G_{3}}(x_{S}-y_{S})^{2}\leq\sum_{S\in
G_{3,2}}y_{S_{\mathcal{R}}\backslash S}^{2}+\sum_{S\in G_{3,3}}y_{S\backslash
S_{\mathcal{R}}}^{2}+\gamma_{3}.$

###### Proof of Claim 3.

By definition $G_{3,1}$ is a subfamily of $F_{\mathcal{R}}$ so Claim $A$
yields

$\sum_{S\in G_{3,1}}(x_{S}-y_{S})^{2}\leq\varepsilon^{2}.$

Now let us deal with $G_{3,2}$. By Claim $A$ we know that

$\sum_{S\in
G_{3,2}}(x_{S_{\mathcal{R}}}-y_{S_{\mathcal{R}}})^{2}\leq\varepsilon^{2}.$

Moreover, for every $S\in G_{3,2}$, we know that the segment
$S_{\mathcal{R}}\backslash S$ does not intersect $T_{N}$ and thus Claim $B$
yields

$\sum_{S\in G_{3,2}}(x_{S_{\mathcal{R}}\backslash
S}-y_{S_{\mathcal{R}}\backslash S})^{2}\leq\sum_{S\in
G_{3,2}}y_{S_{\mathcal{R}}\backslash S}^{2}+\varepsilon^{2}+2\varepsilon.$

So finally we get

$\displaystyle\sum_{S\in G_{3,2}}(x_{S}-y_{S})^{2}$ $\displaystyle=\sum_{S\in
G_{3,2}}[(x_{S_{\mathcal{R}}}-y_{S_{\mathcal{R}}})-(x_{S_{\mathcal{R}}\backslash
S}-y_{S_{\mathcal{R}}\backslash S})]^{2}$ $\displaystyle\leq\sum_{S\in
G_{3,2}}(x_{S_{\mathcal{R}}}-y_{S_{\mathcal{R}}})^{2}+\sum_{S\in
G_{3,2}}(x_{S_{\mathcal{R}}\backslash S}-y_{S_{\mathcal{R}}\backslash S})^{2}$
$\displaystyle\quad\quad\quad+2\sqrt{\sum_{S\in
G_{3,2}}(x_{S_{\mathcal{R}}}-y_{S_{\mathcal{R}}})^{2}}\sqrt{\sum_{S\in
G_{3,2}}(x_{S_{\mathcal{R}}\backslash S}-y_{S_{\mathcal{R}}\backslash
S})^{2}}$ $\displaystyle\leq\sum_{S\in G_{3,2}}y_{S_{\mathcal{R}}\backslash
S}^{2}+\varepsilon^{2}+(\varepsilon^{2}+2\varepsilon)+4\varepsilon.$

Similar computations allow us to deal with $G_{3,3}$ and the conclusion
follows by combining the $3$ inequalities. ∎

Finally let us take $S\in G_{4}$. Since $S$ and its representative
$S_{\mathcal{R}}$ are equal on $T_{N}$ the segments $S^{-}=S\backslash(S\cap
S_{\mathcal{R}})$ and $S_{\mathcal{R}}^{-}=S_{\mathcal{R}}\backslash(S\cap
S_{\mathcal{R}})$ do not intersect $T_{N}$ (and are non-empty). Thus.

Claim 4. There is a $\gamma_{4}=\gamma_{4}(\varepsilon)$ such that

$\sum_{S\in G_{4}}(x_{S}-y_{S})^{2}\leq 1+\sum_{S\in
G_{4}}y_{S^{-}}^{2}+\sum_{S\in G_{4}}y_{S_{\mathcal{R}}^{-}}^{2}+\gamma_{4}.$

###### Proof of Claim 4.

Using again Claim $A$ and Claim $B$, we have

$a=\sum_{S\in
G_{4}}(x_{S_{\mathcal{R}}}-y_{S_{\mathcal{R}}})^{2}\leq\varepsilon^{2},$
$b=\sum_{S\in G_{4}}(x_{S^{-}}-y_{S^{-}})^{2}\leq\sum_{S\in
G_{4}}y_{S^{-}}^{2}+\varepsilon^{2}+2\varepsilon,$

and

$c=\sum_{S\in
G_{4}}(x_{S_{\mathcal{R}}^{-}}-y_{S_{\mathcal{R}}^{-}})^{2}\leq\sum_{S\in
G_{4}}y_{S_{\mathcal{R}}^{-}}^{2}+\varepsilon^{2}+2\varepsilon.$

Thus we get, using the $\sqrt{u+v}\leq\sqrt{u}+\sqrt{v}$ identity for the last
term:

$\displaystyle\sum_{S\in G_{4}}(x_{S}-y_{S})^{2}$ $\displaystyle=\sum_{S\in
G_{4}}[(x_{S_{\mathcal{R}}}-y_{S_{\mathcal{R}}})-(x_{S_{\mathcal{R}}^{-}}-y_{S_{\mathcal{R}}^{-}})+(x_{S^{-}}-y_{S^{-}})]^{2}$
$\displaystyle\leq a+b+c+2\sqrt{ab}+2\sqrt{ac}+2\sqrt{bc}$
$\displaystyle\leq\varepsilon^{2}+\sum_{S\in G_{4}}y_{S^{-}}^{2}+\sum_{S\in
G_{4}}y_{S_{\mathcal{R}}^{-}}^{2}+2(\varepsilon^{2}+2\varepsilon)+4\sqrt{\varepsilon^{2}(1+\varepsilon^{2}+2\varepsilon)}$
$\displaystyle\quad\quad\quad+2\sqrt{\sum_{S\in
G_{4}}y_{S^{-}}^{2}}\sqrt{\sum_{S\in
G_{4}}y_{S_{\mathcal{R}}^{-}}^{2}}+2\sqrt{2(\varepsilon^{2}+2\varepsilon)+(\varepsilon^{2}+2\varepsilon)^{2}}$

so it only remains to show that

$2\sqrt{\sum_{S\in G_{4}}y_{S^{-}}^{2}}\sqrt{\sum_{S\in
G_{4}}y_{S_{\mathcal{R}}^{-}}^{2}}\leq 1.$

To show this, observe that if you take two different segments $S$ and $T$ in
$G_{4}$, then you obviously have $S\cap T=\emptyset$ and $S_{\mathcal{R}}\cap
T_{\mathcal{R}}=\emptyset$ since we work with families of disjoint segments,
but you also have $S\cap T_{\mathcal{R}}=\emptyset$ and $T\cap
S_{\mathcal{R}}=\emptyset$ since $S$ and $T$ have disjoint starting parts in
$T_{N}$. Consequently:

$\sum_{S\in G_{4}}y_{S^{-}}^{2}+\sum_{S\in
G_{4}}y_{S_{\mathcal{R}}^{-}}^{2}\leq\left\|y\right\|_{JT}^{2}=1.$

The conclusion follows since the function $f(x)=x\sqrt{1-x^{2}}$ has maximum
$\frac{1}{2}$ on $[0,1]$ (attained at $\frac{1}{\sqrt{2}}$). ∎

Now letting $\gamma=\sum_{i=1}^{4}\gamma_{i}$ and combining the results from
the $4$ claims, we obtain:

$\displaystyle\sum_{S\in G}(x_{S}-y_{S})^{2}$ $\displaystyle\leq 1+\sum_{S\in
G_{2}}y_{S}^{2}+\sum_{S\in G_{3,3}}y_{S\backslash
S_{\mathcal{R}}}^{2}+\sum_{S\in G_{4}}y_{S^{-}}^{2}$
$\displaystyle\quad\quad\quad+\sum_{S\in G_{3,2}}y_{S_{\mathcal{R}}\backslash
S}^{2}+\sum_{S\in G_{4}}y_{S_{\mathcal{R}}^{-}}^{2}+\gamma,$

and since we work with families of disjoint segments we get

$\sum_{S\in G}(x_{S}-y_{S})^{2}\leq 3+\gamma.$

Finally, $\left\|x-y\right\|_{JT}\leq\max\\{2-\eta_{x,N}\sqrt{3+\gamma}\\}$
and this quantity is strictly less than $2$ if we take a small enough
$\varepsilon$ since $\gamma$ goes to $0$ as $\varepsilon$ goes to $0$. ∎

######  Remark 5.13.

Here the segments involved can be infinite so the functional $x^{*}$ we slice
with does not have to be in $JT_{*}$. This raises the following question.

######  Question 5.14.

Does $JT$ admit weak∗ $\Delta$-points?

From the work of [20] we know that the dual of the James tree space $JT^{*}$
is AUC. Now $JT$ does not admit an equivalent norm whose dual norm is AUC∗.
Indeed such norm would be AUS by the asymptotic duality. This is impossible
because $JT$ is not Asplund (it is separable while $JT^{*}$ is not), see [29,
Proposition 2.4]. In view of those observations $JT^{*}$ would be a natural
candidate for our study and the question of the existence of $\Delta$\- or
Daugavet-points there, as well as the question of the existence of non quasi-
denting points would be particularly relevant. Unfortunately it seems that
computations are still out of reach in this context, although the use of
molecules facilitates them, and we where only able to obtain a few partial
results even while trying to restrict ourselves to $J^{*}$.

######  Lemma 5.15.

Let $s,t\in T$ be two distinct points. For any $\varepsilon>0$ and $z^{*}\in
JT^{*}$ with $\\{s,t\\}\cap\operatorname{supp}(z^{*})=\emptyset$ we have

$\|e_{t}^{*}-\varepsilon e_{s}^{*}+z^{*}\|>1.$

###### Proof.

Let $x^{*}=e_{t}^{*}-\varepsilon e_{s}^{*}+z^{*}$ for some $\varepsilon>0$ and
$z^{*}$ supported in $T\backslash\\{s,t\\}$. Assume first that
$\varepsilon\geq 1$, and let $\alpha=1/\sqrt{2}$ and $x=\alpha e_{t}-\alpha
e_{s}$. We have $\|x\|=1$ and

$x^{*}(x)=\alpha+\varepsilon\alpha\geq 2/\sqrt{2}>1.$

Next if $\varepsilon<1$ we let $\alpha=\sqrt{1-\varepsilon^{2}}$ and $x=\alpha
e_{t}-\varepsilon e_{s}$. We have $\|x\|=1$ since
$\alpha^{2}+\varepsilon^{2}=1$ and $(\alpha-\varepsilon)^{2}\leq
1-2\alpha\varepsilon<1$. Now

$x^{*}(x)=\alpha+\varepsilon^{2}>\alpha^{2}+\varepsilon^{2}=1.$

In both cases, $\|x^{*}\|>1$. ∎

######  Lemma 5.16.

In $JT^{*}$ every basis vector $e_{t}^{*}$ is an extreme point of
$B_{JT^{*}}$, and therefore a weak∗ denting point.

###### Proof.

Assume that $x^{*}$ and $y^{*}$ are norm one elements such that
$e_{t}^{*}=\frac{x^{*}+y^{*}}{2}$. Then we have $x^{*}(e_{t})=y^{*}(e_{t})=1$
and $x^{*}(e_{s})=-y^{*}(e_{s})$ for all $s\neq t$. If we have
$x^{*}(e_{s})<0$ for some $s\neq t$, then with
$z^{*}=x^{*}-(e^{*}_{t}+x^{*}(e_{s})e^{*}_{s})$ we get

$\|x^{*}\|=\|e^{*}_{t}+x^{*}(e_{s})e^{*}_{s}+z^{*}\|>1$

by Lemma 5.15. Hence $x^{*}=y^{*}=e^{*}_{t}$.

By Proposition 3.d.19 in [15] we know that $e^{*}_{t}$ is a point of weak∗ to
norm continuity on the unit ball of $JT^{*}$ so the conclusion follows by
applying Choquet’s lemma (see for example [14, Lemma 3.69]) which tells that
the weak∗ slices form a neighborhood basis of $e_{t}^{*}$. Indeed the
continuity of the identity map at $e_{t}^{*}$ then ensures that any ball
around $e_{t}^{*}$ contains a weak∗ slice containing $e_{t}^{*}$. ∎

######  Corollary 5.17.

No molecule in $JT^{*}$ is a weak∗ Daugavet point.

###### Proof.

Let $x^{*}=\sum_{i\geq 1}\lambda_{i}\mathbbm{1}_{S_{i}}$ where $S_{i}$ are
disjoint segments of $T$ and $\sum_{i\geq 1}\lambda_{i}^{2}\leq 1$. If there
is an $i_{0}\geq 1$ such that $\lambda_{i_{0}}=1$, then
$x^{*}=\mathbbm{1}_{S_{i_{0}}}$. Since the biorthogonal functionals are
(weak∗) denting points we may assume that $S_{i_{0}}$ contains at least two
points of $T$. Now if we let $s_{i_{0}}$ be the starting point of this
segment, and if we let $T_{i_{0}}=S_{i_{0}}\backslash\\{s_{i_{0}}\\}$, we have
$\left\|x^{*}-e_{s_{i_{0}}}^{*}\right\|=\left\|\mathbbm{1}_{T_{i_{0}}}\right\|=1$
so $x^{*}$ is at distance strictly less than $2$ from a (weak∗) denting point
in $JT^{*}$ and thus cannot be a weak∗ Daugavet point.

Next let us assume that $\lambda_{i}\in(0,1)$ for every $i\geq 1$ and let us
fix any $i_{0}\geq 1$. We define as above $s_{i_{0}}$ and $T_{s_{i_{0}}}$
(which might eventually be empty) and we let $T_{i}=S_{i}$ for any $i\neq
i_{0}$. Since the $T_{i}$ are disjoint segments of $T$ we have
$\left\|x^{*}-e_{s_{i_{0}}}^{*}\right\|\leq
1-\left|\lambda_{i_{0}}\right|+\left\|x^{*}-\lambda_{s_{i_{0}}}e_{s_{i_{0}}}^{*}\right\|=1-\left|\lambda_{i_{0}}\right|+\left\|\sum_{i\geq
1}\lambda_{i}\mathbbm{1}_{T_{i}}\right\|\leq 2-\lambda_{s_{i_{0}}}<2$ and the
conclusion follows as above. ∎

In light of the above we ask:

######  Question 5.18.

Can molecules in $JT^{*}$ be $\Delta$-points?

Note that the observation from [30, Remark 2.4] applied to the set of
molecules in $JT^{*}$ which satisfies as mentioned
$\mathop{\overline{\mathrm{conv}}}\nolimits\mathcal{M}=B_{X^{*}}$ gives the
following simplification.

######  Lemma 5.19.

Let $x^{*}\in S_{JT^{*}}$. Then $x^{*}$ is a $\Delta$-point if and only if we
can find, for every $\varepsilon>0$ and for every slice $S$ of $B_{JT^{*}}$,
some $m\in\mathcal{M}\cap S$ such that $\left\|x-m\right\|\geq 2-\varepsilon$.

Although it is possible to do some computations in very specific cases (for
example when $x^{*}$ is a molecule supported on two segments) it seems to be
difficult to estimate the distance between two molecules in general even when
they are supported on the same branch of $T$. Moreover, it is not completely
trivial to distinguish norm one molecules in $\mathcal{M}$ and to find
suitable norming elements in $JT$.

Let us mention that with techniques similar to those for $JT,$ it is possible
to prove that also $J$ with the equivalent norms $\|\cdot\|_{\mathcal{J}}$ and
$\|\cdot\|_{0}$ as given on p. 62 in [4], fail to contain $\Delta$-points.

However, if we replace the binary tree $T$ by the countably branching tree
$T_{\infty}=\\{\emptyset\\}\cup\bigcup_{n\geq 1}\mathbb{N}^{n}$ in the
definition of the James tree space we obtain the Banach space $JT_{\infty}$
originally introduced in [17] which shares some of the basic properties of
$JT,$ but presents a few striking dissimilarities. Indeed it is proved in [17]
that the predual of $JT_{\infty}$ fails the PCP and as a consequence admits no
equivalent AUC norm (in fact
$\overline{\delta}_{\left|.\right|}(\frac{1}{2})=0$ for any equivalent norm
$\left|.\right|$ on $(JT_{\infty})_{*}$, see [20]) while it satisfy the so
called convex PCP, see [18]. For our study we have as before that
$JT_{\infty}$ is $2$-AUC∗ and admits no Daugavet-points, but our proof of non-
existence of $\Delta$-points fails for infinitely supported elements since
restricting the support of such element to a finite level of the tree does not
necessarily provide finitely many nodes of $T_{\infty}$ anymore. It is thus
natural to ask the following.

######  Question 5.20.

Does $JT_{\infty}$ admit $\Delta$-points?

##  6\. Almost Daugavet- and Delta-points

Looking at Corollary 2.4 it is natural to ask:

######  Question 6.1.

Do all superreflexive Banach spaces fail to contain $\Delta$-points?

This question is not trivial since, as we have seen, even superreflexive
spaces whose norm is not uniformly non-square have small slices of the unit
ball with diameter arbitrary close to $2$. However, for a superreflexive space
$X$ we know that the dual of an ultrapower $X^{\mathcal{U}}$ is isometric to
$(X^{*})^{\mathcal{U}}$, and this opens the door to using ultrafilter limits.
Since $X^{\mathcal{U}}$ is also superreflexive we do not leave the context of
superreflexive spaces when passing to ultrapowers. Motivated by this we
introduce the following definitions which are a weakening of the notions of
Daugavet- and $\Delta$-points, and we will see how those can help simplify the
problem from Question 6.1.

Let $X$ be a Banach space and let $\varepsilon>0$. We say that $x\in B_{X}$ is
a _$(2-\varepsilon)$ $Daugavet$-point_ if
$B_{X}\subset\mathop{\overline{\mathrm{conv}}}\nolimits{\Delta_{\varepsilon}(x)}$
or equivalently if we can find for every $\delta>0$ and for every $x^{*}\in
S_{X^{*}}$ an element $y\in S(x^{*},\delta)$ satisfying
$\left\|x-y\right\|\geq 2-\varepsilon$. We say that $x\in B_{X}$ is a
_$(2-\varepsilon)$ $\Delta$-point_ if
$x\in\mathop{\overline{\mathrm{conv}}}\nolimits{\Delta_{\varepsilon}(x)}$ or
equivalently if we can find for every $\delta>0$ and for every $x^{*}\in
S_{X^{*}}$ such that $x\in S(x^{*},\delta)$ an element $y\in S(x^{*},\delta)$
satisfying $\left\|x-y\right\|\geq 2-\varepsilon$.

We say that $X$ _admits almost Daugavet-points_ if it admits a
$(2-\varepsilon)$ Daugavet-point for every $\varepsilon>0$, and we say that
$X$ _admits almost $\Delta$-points_ if it admits a $(2-\varepsilon)$
$\Delta$-point for every $\varepsilon>0$.

We say that a Banach space $X$ _contains $\ell_{p}^{n}$’s uniformly_ ($1\leq
p\leq\infty$) if for all $\varepsilon>0$ and $n\in\mathbb{N}$ there exist
$x_{1},\ldots,x_{n}\in B_{X}$ such that for all sequences of scalars $(a_{k})$

$(1+\varepsilon)^{-1}\|(a_{n})\|_{p}\leq\left\|\sum_{k=1}^{n}a_{k}x_{k}\right\|\leq\|(a_{n})\|_{p}.$

We will start by studying the existence of almost Daugavet- and
$\Delta$-points in some classical spaces. We highlight the following
observation, which although obvious from the definition will provide an easy
way of constructing almost $\Delta$-points.

######  Observation 6.2.

Let $\varepsilon>0$. If $x\in\operatorname{conv}\Delta_{\varepsilon}(x)$, then
$x$ is a $(2-\varepsilon)$ $\Delta$-point.

Using this, we can prove the two following lemmas.

######  Lemma 6.3.

If a Banach space $X$ contains $\ell_{1}^{n}$’s uniformly, then $X$ admits
almost $\Delta$-points.

###### Proof.

Let $\varepsilon>0$ and $n\in\mathbb{N}$ and find $x_{1},\ldots,x_{n}\in
B_{X}$ from the definition of containing $\ell_{1}^{n}$’s uniformly. Let
$x=\frac{1}{n}\sum_{k=1}^{n}x_{k}$. Then $\|x\|\leq 1$ and for each $j$

$\|x-x_{j}\|=\left\|\sum_{k\neq
j}\frac{1}{n}x_{k}+(\frac{1}{n}-1)x_{j}\right\|\geq(1+\varepsilon)^{-1}\left(2-\frac{2}{n}\right)$

which can be made as close to $2$ as we like. ∎

######  Lemma 6.4.

If a Banach space $X$ contains $\ell_{\infty}^{n}$’s uniformly, then $X$
admits almost $\Delta$-points.

###### Proof.

Let $\varepsilon>0$ and $n\in\mathbb{N}$ and find $x_{1},\ldots,x_{n}\in
B_{X}$ from the definition of containing $\ell_{\infty}^{n}$’s uniformly. For
$i=1,\ldots,n$ we define

$y_{i}=-2x_{i}+\sum_{k=1}^{n}x_{k}$

and $x=\frac{1}{n}\sum_{i=1}^{n}y_{i}$, that is

$x=(1-\frac{2}{n})\sum_{k=1}^{n}x_{k}.$

Then $\|x\|\leq 1$ and for each $j$

$\|y_{j}-x\|=\left\|\sum_{k\neq
j}\frac{2}{n}x_{k}+\left(\frac{2}{n}-2\right)x_{j}\right\|\geq(1+\varepsilon)^{-1}\left(2-\frac{2}{n}\right)$

which can be made as close to $2$ as we like. ∎

The two lemmas above can be formulated as: If $X$ does not have finite co-type
or does not have non-trivial type, then $X$ admits almost $\Delta$-points. In
particular the spaces $c_{0}$ and $\ell_{1}$ both admits almost
$\Delta$-points. Let us recall that they do not admit $\Delta$-points (by e.g.
[3, Theorem 2.17]). For those spaces we can say more.

######  Lemma 6.5.

The space $c_{0}$ does not admit almost Daugavet-points.

###### Proof.

Let $x=(x_{i})\in S_{c_{0}}$. Given $\delta\in(0,1)$ there exists a finite
non-empty set $J\subset\mathbb{N}$ of cardinality $n\geq 1$ such that
$|x_{j}|\geq 1-\delta$ for every $j\in J$ and $|x_{i}|<1-\delta$ for every
$i\in\mathbb{N}\setminus J$. Now let $x^{*}=\frac{1}{n}\sum_{j\in
J}\operatorname{sign}(x_{j})e_{j}^{*}\in S_{\ell_{1}}$. By Lemma 2.1 we have
$S(x^{*},\frac{\delta}{n})\subset\bigcap_{j\in
J}S(\operatorname{sign}(x_{j})e_{j}^{*},\delta)$ so if $y=(y_{i})_{i\geq 1}$
is in $S(x^{*},\frac{\delta}{n})$, then it satisfies
$\left|x_{j}-y_{j}\right|\leq\delta$ for every $j\in J$, and thus
$\left\|x-y\right\|\leq\max\\{\delta,\ 2-\delta\\}$. The conclusion follows. ∎

######  Lemma 6.6.

The space $\ell_{1}$ admits almost Daugavet-points.

###### Proof.

Fix $n\geq 1$ and let $x=\frac{1}{n}\sum_{i=1}^{n}e_{i}\in S_{\ell_{1}}$ where
$(e_{i})$ is the unit vector basis of $\ell_{1}$. Then $x$ is a
$(2-\frac{2}{n})$ $\Delta$-point by construction and we will show that it is
in fact a $(2-\frac{2}{n})$ Daugavet-point. Indeed fix $x^{*}\in
S_{\ell_{\infty}}$ and $\delta>0$. Since $\sup_{i}\left|x^{*}(e_{i})\right|=1$
we can find some $i_{0}$ such that $\left|x^{*}(e_{i_{0}})\right|>1-\delta$
that is $e_{i_{0}}\in S(x^{*},\delta)$ or $-e_{i_{0}}\in S(x^{*},\delta)$. Now
it is easy to check that $\left\|x\pm e_{i_{0}}\right\|\geq 2-\frac{2}{n}$ so
we are done. ∎

######  Remark 6.7.

We have seen in Theorem 4.3 that if a Banach space $X$ has Rolewicz’ property
$(\alpha)$, then $X$ has no $\Delta$-points. But $X$ can contain almost
$\Delta$-points.

Indeed, let $T$ be the Tsirelson space. Even though $T$ fails Rolewicz’
property $(\alpha)$ there exists an equivalent norm $|\cdot|$, so that
$(T,|\cdot|)$ has Rolewicz’ property $(\alpha)$ by [36, Theorems 3 and 4].

Since $T$ contains $\ell_{1}^{n}$’s uniformly the same holds for $(T,|\cdot|)$
by James’ $\ell_{1}$-distortion theorem. Thus $(T,|\cdot|)$ admits almost
$\Delta$-points by Lemma 6.3.

The following result, whose proof is clear from Proposition 2.3, covers a lot
of classical norms, and in particular uniformly smooth and uniformly convex
ones.

######  Proposition 6.8.

If $X$ is uniformly non-square, then $X$ does not admit almost
$\Delta$-points.

It is clear from Proposition 2.3 that the unit ball of every non uniformly
non-square norm admits small slices of diameter arbitrarily close to $2$, so
ruling out almost Daugavet- and $\Delta$-points is non-trivial even in
superreflexive spaces.

The following result is the main reason for the introduction of the notions
almost Daugavet- and almost $\Delta$-points.

######  Proposition 6.9.

Let $X$ be a superreflexive Banach space and let $\mathcal{U}$ be some free
ultrafilter on $\mathbb{N}$.

If $X$ admits almost $\Delta$\- (resp. Daugavet-) points, then there exists
$x\in X^{\mathcal{U}}$ with $\|x\|=1$ such that for any slice $S$ of
$B_{X^{\mathcal{U}}}$ containing $x$ (resp. any slice $S$ of
$B_{X^{\mathcal{U}}}$) there exists $y\in S$ with $\|y\|=1$ and $\|x-y\|=2$.

In particular, if there exists a superreflexive Banach space which admits an
almost $\Delta$\- (resp. Daugavet-) point, then there exists a superreflexive
Banach space with a $\Delta$\- (resp. Daugavet-) point.

###### Proof.

Let $\mathcal{U}$ be a free ultrafilter on $\mathbb{N}$. Since $X$ is
superreflexive we have $(X^{\mathcal{U}})^{*}=(X^{*})^{\mathcal{U}}$ (see e.g.
[25, Proposition 7.1]). For each $n\in\mathbb{N}$ choose a $(2-\frac{1}{n})$
$\Delta$-point $x_{n}\in B_{X}$. Let $x=(x_{n})_{\mathcal{U}}$. We have
$\|x_{n}\|\geq 1-\frac{1}{n}$ hence $\|x\|=\lim_{\mathcal{U}}\|x_{n}\|=1$.

Let $x^{*}=(x_{n}^{*})_{\mathcal{U}}$ with $\|x^{*}\|=1$ and $\delta>0$.
Assume that $x\in S(x^{*},\delta)$. This means that there is an $\eta>0$ such
that $x^{*}(x)=\lim_{\mathcal{U}}x_{n}^{*}(x_{n})>1-\delta+\eta$ and thus
there is a set $A\in\mathcal{U}$ such that

$x_{n}^{*}(x_{n})>1-\delta+\eta=\|x_{n}^{*}\|-(\|x_{n}^{*}\|-(1-\delta+\eta)).$

for every $n\in A$. In particular we have
$\left\|x_{n}^{*}\right\|>1-\delta+\eta$ for these $n$ and

$x_{n}\in S_{n}:=\\{y\in
B_{X}:x_{n}^{*}(y)>\|x_{n}^{*}\|-(\|x_{n}^{*}\|-(1-\delta+\eta))\\}.$

We can then use the definition to find $y_{n}\in S_{n}$ with
$\|x_{n}-y_{n}\|\geq 2-\frac{1}{n}$. For $n\notin A$ we just let $y_{n}=0$.

Now $y=(y_{n})\in X^{\mathcal{U}}$ and

$\|y\|=\lim_{\mathcal{U}}\|y_{n}\|\geq\lim_{\mathcal{U}}(\|x_{n}-y_{n}\|-\|x_{n}\|)\geq\lim_{\mathcal{U}}(2-\frac{1}{n}-1)=1.$

Finally we check that $y$ is in the slice $S(x^{*},\delta)$

$x^{*}(y)=\lim_{\mathcal{U}}x_{n}^{*}(y_{n})=\lim_{\mathcal{U}}\|x_{n}^{*}\|\lim_{\mathcal{U}}\frac{x_{n}^{*}}{\|x_{n}^{*}\|}(y_{n})\geq
1-\delta+\eta>1-\delta.$

For the in particular part we just note that ultrapowers of superreflexive
spaces are superreflexive by finite representability of $X^{\mathcal{U}}$ in
$X$. ∎

######  Corollary 6.10.

If $X$ is finite dimensional, then it does not admit almost $\Delta$-points.

###### Proof.

This is an immediate consequence of previous proposition since finite
dimensional spaces do not admit $\Delta$-points and since any ultrapower of a
finite dimensional space is trivially identified with the space itself (under
the diagonal map). ∎

By Proposition 6.9 we can thus restate Question 6.1 in the following way.

######  Question 6.11.

Is it possible to find a superreflexive space admitting almost $\Delta$-points
or almost Daugavet-points?

In particular it would be interesting to investigate the following question.

######  Question 6.12.

Is it possible to find a renorming of $\ell_{2}$ with almost $\Delta$-points
or almost Daugavet points?

Also note that proving that superreflexive spaces do not admit $\Delta$\- (or
Daugavet-) points in general would immediately imply by the preceding
proposition that superreflexive spaces do not admit almost $\Delta$\- (or
Daugavet-) points.

## Acknowledgments

The majority of this work was done while Y. Perreau visited University of
Agder, Kristiansand, in October 2021, and while T. A. Abrahamsen and V. Lima
visited Université Bourgogne Franche-Comté, Besançon in November 2021. We
thank everyone who made these visits possible, in particular, the AURORA
travel programme supported by the Norwegian Research Council, the French
Ministry for Europe and Foreign Affairs, and the French Ministry for Higher
Education, Research and Innovation. We thank Gilles Lancien and Antonin
Prochazka from the university of Besançon for valuable discussions on the
subject and for their investment in the realization of the project. We also
thank Stanimir Troyanski for valuable discussions on the subject.

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