Daugavet- and delta-points in Banach spaces with unconditional bases

We study the existence of Daugavet- and delta-points in the unit sphere of Banach spaces with a $1$-unconditional basis. A norm one element $x$ in a Banach space is a Daugavet-point (resp. delta-point) if every element in the unit ball (resp. $x$ itself) is in the closed convex hull of unit ball elements that are almost at distance $2$ from $x$. A Banach space has the Daugavet property (resp. diametral local diameter two property) if and only if every norm one element is a Daugavet-point (resp. delta-point). It is well-known that a Banach space with the Daugavet property does not have an unconditional basis. Similarly spaces with the diametral local diameter two property do not have an unconditional basis with suppression unconditional constant strictly less than $2$. We show that no Banach space with a subsymmetric basis can have delta-points. In contrast we construct a Banach space with a $1$-unconditional basis with delta-points, but with no Daugavet-points, and a Banach space with a $1$-unconditional basis with a unit ball in which the Daugavet-points are weakly dense.


Introduction
Let X be a Banach space with unit ball B X , unit sphere S X , and topological dual X * . For x ∈ S X and ε > 0 let ∆ ε (x) = {y ∈ B X : x − y ≥ 2 − ε}. We say that X has the (i) Daugavet property if for every x ∈ S X and every ε > 0 we have B X = conv∆ ε (x); (ii) diametral local diameter two property if for every x ∈ S X and every ε > 0 we have x ∈ conv∆ ε (x). In [Kad96, Corollary 2.3] Kadets proved that any Banach space with the Daugavet property fails to have an unconditional basis (see also [Wer01,Proposition 3.1]). These arguments are probably the easiest known proofs of the absence of unconditional bases in the classical Banach spaces C[0, 1] and L 1 [0, 1]. The diametral local diameter two property was named and studied in [BGLPRZ18], but it was first introduced in [IK04] under the name space with bad projections. (See the references in [IK04] for previous unnamed appearances of this property.) Using the characterizations in [IK04] we see that if a Banach space with the diametral local diameter two property has an unconditional basis, then the unconditional suppression basis constant is at least 2. But note that we do not know of any Banach space with an unconditional basis and the diametral local diameter two property.
In the present paper we study pointwise versions of the Daugavet property and the diametral local diameter two property in spaces with 1-unconditional bases.
Definition 1.1. Let X be a Banach space and let x ∈ S X . We say that x is (i) a Daugavet-point if for every ε > 0 we have B X = conv∆ ε (x); (ii) a delta-point if for every ε > 0 we have x ∈ conv∆ ε (x).
Daugavet-points and delta-points were introduced in [AHLP20]. For the spaces L 1 (µ), for preduals of such spaces, and for Müntz spaces these notions are the same [AHLP20, Theorems 3.1, 3.7, and 3.13]. However, C[0, 1] ⊕ 2 C[0, 1] is an example of a space with the diametral local diameter two property, but with no Daugavet-points [AHLP20, Example 4.7]. Stability results for Daugavet-and delta-points in absolute sums of Banach spaces was further studied in [HPV20].
In Section 2 we consider Banach spaces with 1-unconditional bases and study a family of subsets of the support of a vector x. We find properties of these subsets that are intimately linked to x not being a delta-point. Quite general results are obtained in this direction. We apply these results to show that Banach spaces with subsymmetric bases (these include separable Lorentz and Orlicz sequence spaces) always fail to contain delta-points.
In Section 3 we construct a Banach space with a 1-unconditional basis which contains a delta-point, but contain no Daugavet-points. The example is a Banach space of the type h A,1 generated by an adequate family of subsets of a binary tree. The norm of the space is the supremum of the ℓ 1 -sum of branches in the binary tree.
In Section 4 we modify slightly the binary tree from Section 3 and the associated adequate family, to obtain an h A,1 space with some remarkable properties: It has Daugavet-points; the Daugavet-points are even weakly dense in the unit ball; the diameter of every slice of the unit ball is two, but is has relatively weakly open subsets of the unit ball of arbitrary small diameter.
Finally, let us also remark that the examples in both Section 3 and Section 4 contain isometric copies of c 0 and ℓ 1 . Both the ℓ 1 -ness of the branches and c 0 -ness of antichains in the binary tree play an important role in our construction of Daugavet-and delta-points in these spaces (see e.g. Theorems 3.1 and 4.2, and Corollary 4.3).

1-unconditional bases and the sets M(x)
The main goal of this section is to prove that Banach spaces with a subsymmetric basis fail to have delta-points. Before we start this mission, let us point out some results and concepts that we will need. First some characterizations of Daugavet-and delta-points that we will frequently use throughout the paper.
Recall that a slice of the unit ball B X of a Banach space X is a subset of the form where x * ∈ X * and ε > 0. (ii) for every slice S of B X with x ∈ S and for every ε > 0 there exists y ∈ S X such that x − y ≥ 2 − ε.
Let X be a Banach space. Recall that a Schauder basis (e i ) i∈N of X is called unconditional if for every x ∈ X its expansion x = i∈N x i e i converges unconditionally. If, moreover, i∈N θ i x i e i = i∈N x i e i for any x = i∈N x i e i ∈ X and any sequence of signs (θ i ) i∈N , then (e i ) i∈N is called 1-unconditional. A Schauder basis is called subsymmetric, or 1subsymmetric, if it is unconditional and i∈N θ i x i e k i = i∈N x i e i for any x = i∈N x i e i ∈ X, any sequence of signs (θ i ) i∈N , and any infinite increasing sequence of naturals (k i ) i∈N . Trivially a subsymmetric basis is 1-unconditional. In the following we will assume that the basis (e i ) i∈N is normalized, i.e. e i = 1 for all i ∈ N. With (e * i ) i∈N we denote the conjugate in X * to the basis (e i ) i∈N . Clearly (e * i ) i∈N is a 1-unconditional basic sequence whenever (e i ) i∈N is. When studying Daugavet-points or delta-points in a Banach space X with 1-unconditional basis (e i ) i∈N we can restrict our investigation to the positive cone K X generated by the basis, where The reason for this is that for every sequence of signs θ = (θ i ) i∈N the operator T θ : X → X defined by T θ ( i∈N x i e i ) = i∈N θ i x i e i is a linear isometry. Hence x = i∈N x i e i is a Daugavet-point (resp. delta-point) if and only if |x| = i∈N |x i |e i is.
The following result is well-known.
Moreover P A = 1 where, for A ⊂ N, P A is the projection defined by From this we immediately get a fact that will be applied several times throughout the paper.
Fact 2.4. Let X be a Banach space with a 1-unconditional basis (e i ) i∈N and let x, y ∈ X and E ⊂ N. Then the following holds.
• If |x i | ≤ |y i | and sgn x i = sgn y i for all i ∈ E, then y − P E x ≤ y .
The upshot of Fact 2.4 is that it can be used to find an upper bound for the distance between x ∈ S X and elements in a given subset of the unit ball. Indeed, suppose we can find E ⊆ N, η > 0 and a subset S of the unit ball such that x − P E x < 1 − η and the assumption in Fact 2.4 holds for any y ∈ S. Then If such a set S is a slice (resp. a slice containing x), then x cannot be a Daugavet-point (resp. delta-point). We will see in Theorem 2.17 that any unit sphere element in a space with a subsymmetric basis, is contained in a slice of the above type. Our tool to investigate the existence of slices of this type in a Banach space with a 1-unconditional basis, are certain families of subsets of the support of the elements in the space.
Remark 2.5. If only the moreover part of Proposition 2.3 holds, then the basis is called 1-suppression unconditional. In this case the conclusion of Proposition 2.3 still holds if sgn a i = sgn b i , for all i. This is all that is needed in Fact 2.4. Similarly, one can check that all the results about 1-unconditional bases in the rest of this section also holds for a Banach space X with a 1-suppression unconditional basis.
Definition 2.6. For any Banach space X with 1-unconditional basis (e i ) i∈N and for x ∈ X, define We can think of M(x) as a collection of minimal "norm-giving" subsets of the support of x. If for example X = c 0 and x ∈ c 0 , then Our first observation about the families M(x) is that they are always non-empty.
Lemma 2.7. Let X be a Banach space with 1-unconditional basis Suppose we have found n 1 < · · · < n k−1 such that A k−1 = A k−2 \{n k−1 } satisfies P A k−1 x = x and P A k−1 x − x j e j < x for all j ∈ A k−1 ∩ {1, . . . , n k−1 − 1}. Then either A k−1 ∈ M(x) or there exists a smallest integer n k greater than n k−1 such that Either this process terminates and A k ∈ M(x), or we get a set N = If j ∈ A, find k such that j < n k , then by 1-unconditionality Our next goal is to prove that certain classes of subsets of M F (x) and M ∞ (x) are finite (see Lemma 2.10 below). We will use the next result as a stepping stone. In the proof, and throughout the paper, we will assume that the sets A = {a 1 , a 2 , . . .} ∈ M(x) are ordered so that a 1 < a 2 < · · · < a n < · · · , and we will use A(n) to denote the set {a 1 , . . . , a n }.
Lemma 2.8. Let X be a Banach space with 1-unconditional basis (e i ) i∈N . If x ∈ X, then for every n ∈ N, In particular, Proof. Let us prove (i) inductively. For k ∈ N, let R k = I − P N k , where N k = {1, . . . , k}. For n = 1 the result follows from R k x → 0. Now assume that |{A(n − 1) : A ∈ M(x), |A| > n − 1}| < ∞, and let s n−1 := max P A(n−1) x : In order to find the sets E ⊆ N mentioned in the remarks following Fact 2.4 we need the following families of subsets of M(x).
If it is clear from the context what element x we are considering, we will simply denote these sets by F n , G n , and E n .
It is pertinent with a couple of comments about these families of sets. Trivially, if M ∞ (x) = ∅, then G n = F n = M(x) for all n ∈ N. We can think of the elements of E n as essential for the norm of x, i.e. x − P E x < x for all E ∈ E n . According to Lemma 2.11 below the drop in norm is also uniformly bounded away from 0. The main reason for this is that F n and E n are finite for all n ∈ N. We will prove this now.
Assume for contradiction that |F n | = ∞. Then there exists a sequence (A k ) ⊂ F n such that |A k | ≥ k. By compactness of {0, 1} N and passing to a subsequence if necessary, we may assume that A k → A ∈ N pointwise and A ∩ {1, . . . , N} = A k ∩ {1, . . . , N} for all k. In par- Finally, (ii) follows from (i) and Lemma 2.8.
With the knowledge that the cardinality of E n is finite for every n ∈ N, we now obtain the following result.
Let X be a Banach space and x ∈ S X . If x is a delta-point, then for every slice S with x ∈ S, we have that x is at one end of a line segment in S with length as close to 2 as we want. Suppose we replace the slice S with a non-empty relatively weakly open subset W of B X with x ∈ W . If X has the Daugavet property, then x is at one end of a line segment in W with length as close to 2 as we want ([Shv00, Lemma 3]). Next we show that this is never the case if X has a 1-unconditional basis.
Proposition 2.12. Let X be a Banach space with 1-unconditional basis (e i ) i∈N . If x ∈ S X , then there exist δ > 0 and a relatively weakly open For any y ∈ W , we get that and we are done.
Let us remark a fun application of the above proposition.
Remark 2.13. Let K be an infinite compact Hausdorff space. Then C(K) does not have a 1-unconditional (or a 1-suppression unconditional) basis. Let f be a function which attains its norm on a limit point of K. Arguing similarly as in [AHLP20, Theorem 3.4] we may find a sequence of norm one functions g k with distance as close to 2 as we want from f that converge pointwise, and thus weakly, to f . The conclusion follows from Proposition 2.12.
The next result is the key ingredient in our proof that there are no delta-points in Banach spaces with subsymmetric bases. Its proof draws heavily upon Lemma 2.11.
Lemma 2.14. Let X be a Banach space with 1-unconditional basis (e i ) i∈N and let x ∈ S X . Assume that there exists a slice S(x * , δ), an n ∈ N and some η > 0 such that Then x is not a delta-point.
For any y ∈ S(z * , z * − 1 + δ |Fn|+1 ), we get that Solving for x * (y) we get that For any y ∈ S(z * , z * − 1 + δ |Fn|+1 ) we now get from Lemma 2.11 that If x ∈ S X with M ∞ (x) = ∅ in the above lemma, then any slice S(x * , δ) containing x trivially satisfies Lemma 2.14 (ii). We record this in the following proposition.
By definition of the sets M(x) and a convexity argument, the next result should be clear.
Finally it is time to cash in some dividends and prove the main result of this section.
Theorem 2.17. If X has subsymmetric basis (e i ) i∈N , then X has no delta-points.
Proof. Assume x ∈ S X ∩ K X . By Proposition 2.15 we may assume that M ∞ (x) = ∅. Let s := max{n : For contradiction assume that there exists A = {a 1 , a 2 , . . .} ∈ M ∞ (x) with s ∈ A. Let a 0 = 0 and j ∈ N be such that a j−1 < s < a j . Let t > 0 such that x s = x a j + t and let A s be A with a j replaced by s. Using that (e i ) i∈N is subsymmetric and Lemma 2.16 we get If we let n = s, then s ∈ D(n) for all D ∈ M ∞ (x), and the slice S(e * s , 1 − xs 2 ) and η = 1 2 satisfies the criteria in Lemma 2.14 and we are done.
In the proof above we saw that if X has a subsymmetric basis, then for any x ∈ S X either M ∞ (x) = ∅ or all A ∈ M ∞ (x) has a common element. In the case X has a 1-symmetric basis we can say a lot about the sets M(x) for any given x ∈ S X .
Recall that a Schauder basis (e i ) i∈N is called 1-symmetric if it is unconditional and i∈N θ i x i e π(i) = i∈N x i e i for any x = i∈N x i e i ∈ X, any sequence of signs (θ i ) i∈N , and any permutation π of N. A 1symmetric basis is subsymmetric [LT77, Proposition 3.a.3].
Proposition 2.18. Let X be a Banach space with 1-symmetric basis (e i ) i∈N and let x ∈ S X .
Since |A| = ∞, there exists k ∈ A and t > 0 with x k + t = x l . Using that (e i ) i∈N is 1symmetric and Lemma 2.16 we get (ii). Suppose that x is not constant on A△B and let k, l ∈ A△B with x k = x l , say k ∈ A, l ∈ B, and x k < x l . Then argue as in (i) to get a contradiction, so x is constant on A△B. As x is constant on A△B, we cannot have |A| < |B| since then a subset of B would be in M(x) contradicting the definition of M(x).

A space with 1-unconditional basis and delta-points
In this section we will prove the following theorem. Before giving a proof of the theorem we will need some notation. By definition, a tree is a partially ordered set (T , ) with the property that, for every t ∈ T , the set {s ∈ T : s t} is well ordered by . In any tree we use normal interval notation, so that for instance a segment is [s, t] = {r ∈ T : s r t}. If a tree has only one minimal member, it is said to be rooted and the minimal member is called the root of the tree and is denoted ∅. We have ∅ t for all t ∈ T . We say that t is an immediate successor of s if s ≺ t and the set {r ∈ T : s ≺ r ≺ t} is empty. The set of immediate successors of s we denote with s + . A sequence B = {t n } ∞ n=0 is a branch of T if t n ∈ T for all n, t 0 = ∅ and t n+1 ∈ t + n for all n ≥ 0. If s, t ∈ B are nodes such that neither s t nor t s, then s and t are incomparable. An antichain in a tree is a collection of elements which are pairwise incomparable.
We consider the infinite binary tree, B = ∞ n=0 {0, 1} n , that is, finite sequences of zeros and ones. The order on B is defined as follows: If s = {s 1 , s 2 , . . . , s k } ∈ {0, 1} k ⊂ B and t = {t 1 , t 2 , . . . , t l } ∈ {0, 1} l ⊂ B, then s t if and only if k ≤ l and s i = t i , 1 ≤ i ≤ k. As usual we denote with |s| the cardinality of s, i.e. |s| = k. The concatenation of s and t is s ⌢ t = {s 1 , s 2 , . . . , s k , t 1 , t 2  . Since A is compact we get that for every x ∈ h A,1 there exists A ∈ A such that There is a bijection between B and N where the natural order on N corresponds to the lexicographical order on B (see [AT04,p. 69]). The family A of all subsets of N corresponding to the branches of B and their subsets is an adequate family. We get that X B := h A,1 is a Banach space with 1-unconditional basis (e t ) t∈B . Note that the span of the basis vectors corresponding to any infinite antichain in X B is isometric to c 0 , and that the span of the basis vectors corresponding to any branch in X B is isometric to ℓ 1 .

Proof of Theorem 3.1 (i). Consider
Summing over branches we find that x = 1. We will show that x is a delta-point. Define z ∅ = 0 and then for t 0 ∈ B z t ⌢ 0 0 = z t 0 + e t ⌢ 0 1 and z t ⌢ 0 1 = z t 0 + e t ⌢ 0 0 . Here is a picture of z (0,0) and z (0,1) : By definition of z t 0 we have {u ∈ B : u t 0 } ∩ supp(z t 0 ) = ∅ hence z t 0 − e tn ∈ S(x * , δ). Summing over a branch containing t n we get Next is the proof that X B does not have Daugavet-points. We first need a general lemma about Daugavet-points.
Let (e i ) i∈N be a 1-unconditional basis in a Banach spaces X. Define Proof. Assume x ∈ S X ∩ K X and that there exists η > 0 and E ∈ E X such that , then it follows that 1 − γ < e * i (y) for all i ∈ E and so x is not a Daugavet-point.

Proof of Theorem 3.1 (ii)
. Assume x ∈ S X B ∩K X B . Let E = A∈M (x) A(1). From Lemma 2.8 we see that |E| is finite. Note that E is an antichain. Indeed, assume t 0 , t 1 ∈ E with t 0 t 1 where A(1) = {t 1 } for some A ∈ M(x). Then since x ∈ K X B and we must have t 0 = t 1 .
We have x − P E x < 1 by Lemma 2.11 (i). From Lemma 3.2 we get that x is not a Daugavet-point since E ∈ E X B .
Let us end this section with a remark about the proof of Theorem 3.1 (i). In order to prove that X B has a delta-point we could have used dyadic trees. Recall that a dyadic tree in a Banach space is a sequence (x t ) t∈B , such that x t = 1 2 (x t ⌢ 0 + x t ⌢ 1 ). In fact, x = |t|>0 2 −|t| e t is the root of a dyadic tree. In order to show this one uses the same z t 's as in the above proof, but attach a copy of x to the node t. Finally, we have the following result about dyadic trees and delta-points.
Proof. Let ε > 0 and find n with x ∅ − x t ≥ 2 − ε for all t with |t| = n. This means that x t ∈ ∆ ε (x ∅ ). By definition of a dyadic tree so we have x ∅ ∈ conv ∆ ε (x ∅ ).

A space with 1-unconditional basis and daugavet-points
In this section we will cut of the root of the binary tree and modify the norm from the example in the previous section to allow the space to have Daugavet-points.
Let M = ∞ n=1 {0, 1} n be the binary tree with the root removed. Note Using the lexicographical order ≤ on M we have a bijective correspondence to N with the natural order. Let A be the adequate family of subsets of N corresponding to subsets of branches and subsets of λ-segments. Using this adequate family we get a Banach space X M := h A,1 with 1-unconditional basis (e t ) t∈M . We call X M the modified binary tree space. Note that X M contains isometric copies of c 0 and ℓ 1 just like X B .
As we saw in the proof of Theorem 3.1 (ii) the antichains in the tree play an important role for the existence of Daugavet-points. Define The set E X M from Section 3 can be described as the set of all non-void finite antichains E of M such that |A ∩ E| ≤ 1 for all A ∈ A. Clearly supp(z) ∈ E X M for every z ∈ F \ {0} and every z with supp(z) ∈ E X M and z(M) ⊂ {0, ±1} belongs to F. It is also clear that for every E ∈ E X M there exists a branch B such that B ∩ E = ∅. We will see in Lemma 4.1 and Theorem 4.2 that the sets E X M and F will play an essential role in characterizing the Daugavet-points of X M .
If M is a finite subset of M, then we will use the notation K M = { t∈M a t e t : a t ≥ 0} and F M = {z ∈ F : supp(z) ⊂ M}.
First we prove a lemma which says that convex combinations of elements in F are dense in the unit ball of X M .
Assume the induction hypothesis holds for n ∈ N. Let x ∈ K M 2(n+1) ∩ B X M . Let t ∈ M be the node such that t ⌢ 0 corresponds to 2n + 1 and t ⌢ 1 to 2n + 2. Define For k ∈ I we let z k,0 := z k + e t ⌢ 0 and z k,1 := z k + e t ⌢ 1 .
Thus, by definition of the norm we have, which is a convex combination of elements in K supp(x) ∩ F.
With the above lemma in hand we are able to characterize Daugavetpoints in X M in terms of E X M . This will give us an easy way to identify and give examples of Daugavet-points.
Theorem 4.2. Let x ∈ S X M , then the following are equivalent (iii) for any z ∈ F, either x − z = 2 or for all ε > 0 there exists s ∈ M such that z ± e s ∈ F and x − z ± e s > 2 − ε.
Proof. As usual we will assume that x ∈ K X M throughout.
. If there, for some A ∈ M(x − P E x), exists t ∈ E and s 0 ∈ A such that t s 0 , or t ∈ E such that s t for all s ∈ A, then we are done since e * t (x) = 0 and So from now on we assume that no such A exists. Assume that there exists A ∈ M(x−P E x) that is a subset of a branch B. By definition of the norm, we have e * t (x) = 0 for t ∈ B \ A, and by the assumption above, we also have B ∩ E = ∅. Since |e * t (x)| → 0 as |t| → ∞ for t ∈ B we can find s ∈ B with |e * s (x)| < ε/2 and hence This concludes the case where A is a subset of a branch. Suppose for contradiction that no A ∈ M(x − P E x) is a subset of a branch, then every B ∈ M(x − P E x) is a subset of a λ-segment. By Lemma 2.10 we must have |M(x − P E x)| < ∞.
Choose any B ∈ M(x − P E x) and write Otherwise, by definition of the norm, we get the contradiction is a subset of a branch. We can use the argument above a finite number of times until we are left with E m ∈ E X M with x − P Em x = 1 and M(x − P Em x) = ∅ which contradicts Lemma 2.7.
Finally, (iii) ⇒ (i). Choose ε > 0. Let y ∈ B X M with finite support. Then by Lemma 4.1, we can write y = n k=1 λ k z k , with z k ∈ F, λ k ≥ 0 and n k=1 λ k = 1. Let D 1 = {k ∈ {1, . . . , n} : x − z k = 2} and D 2 = {1, . . . , n} \ D 1 . We can, by assumption, for each k ∈ D 2 find s k ∈ M such that z k ± e s k ∈ F with x − z k ± e s k > 2 − ε. Then y ∈ conv ∆ ε (x) since The set of all such y is dense in B X M , hence B X M = conv∆ ε (x) so x is a Daugavet-point.
With a characterization of Daugavet-points in hand we can now prove the main result of this section. Proof. Let x = t∈M 2 −|t| e t . We have that x is a Daugavet-point by Corollary 4.3. The next part of the proof is similar to the proof of Theorem 3.1 (i). We will show that a shifted version of x is a delta-point which is not a Daugavet-point. Define an operator on the modified binary tree: where t = ∅ when |t| = 0.
Define w = L(x). Let x * ∈ S X * M and δ > 0 such that w ∈ S(x * , δ). Just as in the proof of Theorem 3.1 (i) we can find z t 0 ∈ S X M whose support is an antichain (i.e. z t 0 ∈ F) and we can find e tn such that z t 0 − e tn ∈ S(x * , δ). Summing over a branch containing t n we get w − (z t 0 − e tn ) = 2.
In [AHLP20], the property that the unit ball of a Banach space is the closed convex hull of its delta-points was studied. We will next show that X M satisfies something much stronger, the unit ball is the closed convex hull of a subset of its Daugavet-points.
If D is the set of all Daugavet-points in X M define The proof of Theorem 4.4 shows that D B is non-empty. For t 0 ∈ M, let S t 0 be the shift operator on X M that shifts the root to t 0 , that is It is clear that S t 0 is an isometry on X M .
Proof. Let y ∈ B X M . We may assume that y has finite support, since such y are dense in B X M . By Lemma 4.1, we can write y = n k=1 λ k z k where z k ∈ F, λ k ≥ 0 and n k=1 λ k = 1. Fix z ∈ F. Let m := max{|t| : t ∈ supp(z)}.
Choose any x 0 ∈ D B and use the shift operator in (2) to define Observe that z ± x takes its norm along every branch, so by Corollary 4.3 both z ± x ∈ D B . Repeat this construction for z k to create x k for k ∈ {1, . . . , n}. Then is a convex combination of Daugavet-points in D B .
Our next result is that X M has the remarkable property that the Daugavet-points are weakly dense in the unit ball. So in a sense there are lots of Daugavet-points, but of course not enough of them in order for X M to have the Daugavet property. First we need a lemma. For t ∈ M, S t denotes the shift operator defined in (2) above.
Lemma 4.6. Let x * ∈ S X * M and s ∈ B. For any x ∈ S X M and ε > 0 there exist some infinite antichain Proof. Pick any x * ∈ S X * M , s ∈ B and x ∈ S X M . It is not difficult to find an infinite antichain E = {t i } ∞ i=1 satisfying (i) and (ii). Since E is an antichain we have n i=1 S t i (x) = 1 for all n ∈ N. Hence lim i→∞ x * (S t i x) = 0, and then we can find n ∈ N such that |x * (S t i x)| < ε for all i ≥ n. Now By construction x ∈ S X M and we have x ∈ W since Using Theorem 4.2 we will show that x is a Daugavet-point. Indeed, let E ∈ E X M . Then there exists a branch A with A ∩ E = ∅. Let t ∈ A with |t| = m. If t / ∈ N , then If t ∈ N , then since S bt (g) is a Daugavet-point, there exists a branch B with t ∈ B such that S bt (g) − P E S bt (g) = s∈B |S bt (g) s | = 1. Thus and we are done.
Question 4.8. How "massive" does the set of Daugavet-points in S X have to be in order to ensure that a Banach space X fails to have an unconditional basis?
If S is a slice of the unit ball of X M , then the above proposition tells us that S contains a Daugavet-point x. Then by definition of Daugavetpoints there exists for any ε > 0 a y ∈ S with x − y ≥ 2 − ε. Thus the diameter of every slice of the unit ball of X M is 2, that is X M has the local diameter two property.
The next natural question is whether the diameter of every nonempty relatively weakly open neighborhood in B X M equals 2, that is, does X M have the diameter two property? The answer is no, in fact, every Daugavet-point in D B has a weak neighborhood of arbitrary small diameter. Let us remark that the first example of a Banach space with the local diameter two property, but failing the diameter two property was given in [BGLPRZ15].
Proposition 4.9. In X M every x ∈ D B is a point of weak-to normcontinuity for the identity map on X M . In particular, X M fails the diameter two property.
Proof. Let ε > 0 and x ∈ D B . Let n ∈ N be such that |t|>n x t e t < ε 8 . Consider the weak neighborhood W of x W = {y ∈ B X M : |e * t (x − y)| < ε 2 |t|+3 , |t| ≤ n}. We want to show that the diameter of W is less than ε. Let y = t∈M y t e t ∈ W . Let A be a subset of a branch or of a λ-segment in M.
Since |x t − y t | < ε2 −|t|−3 for |t| ≤ n, |t|>n x t e t < ε 8 , and x attains its norm along every branch of M, we have t∈A |t|≤n From this it follows that the diameter of W is less than ε.
Recall from [ALL16] that a Banach space X is locally almost square if for every x ∈ S X and ε > 0 there exists y ∈ S X such that x±y ≤ 1+ε.

4
> x ± y = max ± |x s ± y s | + t∈A t =s |y t | = |x s | + |y s | + 1 − |y s | and we get the contradiction |x s | < 1 4 . Recall from [HLP15] that a Banach space X is locally octahedral if for every x ∈ S X and ε > 0, there exists y ∈ S X such that x ± y ≥ 2 − ε.
It is known that every Banach space with the Daugavet property is octahedral. Even though the modified binary tree space have lots of Daugavet-points, as seen in Proposition 4.5, it is not even locally octahedral.
Proposition 4.11. X M is not locally octahedral.
Proof. Consider x = 1 2 (e (0) + e (1) ) ∈ S X M . We want to show that for all y ∈ S X M we have min x ± y ≤ 3 2 . Let y = t∈M y t e t ∈ S X M . Let A be a subset of a branch or a λ-segment. If A = {(0), (1)}, then t∈A |x t ± y t | ≤ If A = {(0), (1)}, then, since |y (0) | + |y (1) | ≤ 1 and a convex function attains its maximum at the extreme points, we get Hence min x ± y ≤ 3 2 .