Unit balls of Polyhedral Banach spaces with many extreme points

Let $E$ be a $(\mathrm{IV})$-polyhedral Banach space. We show that, for each $\epsilon>0$, $E$ admits an $\epsilon$-equivalent $\mathrm{(V)}$-polyhedral norm such that the corresponding closed unit ball is the closed convex hull of its extreme points. In particular, we obtain that every separable isomorphically polyhedral Banach space, for each $\epsilon>0$, admits an $\epsilon$-equivalent $(\mathrm{V})$-polyhedral norm such that the corresponding closed unit ball is the closed convex hull of its extreme points.


Introduction
Let E be a real infinite-dimensional Banach space with topological dual E * .We say that E is polyhedral if the unit ball of each of its finite-dimensional subspaces is a polytope.Infinite-dimensional polyhedral Banach spaces were introduced by Victor Klee in [12], and in the same paper it was proved that c 0 is polyhedral.Polyhedral Banach spaces were studied by various authors, we refer the reader to [11] and the references therein for the main basic results on the subject.
The set ext(B E * ) consisting of all extreme points of the dual unit ball B E * plays a fundamental role in the study of polyhedral Banach spaces.Indeed, different notions of polyhedrality known in the literature are defined by means of geometric and topological properties of ext(B E * ) (for the different notions of polyhedrality see Definition 2.4 below or [9]).On the other hand, the results about polyhedrality and involving the set ext(B E ), consisting of all extreme points of the unit ball B E , are much more sporadic in the literature.This is due to the fact that extreme points of the unit ball of a polyhedral Banach space are in some sense quite rare, and certain stronger notions of polyhedrality exclude existence of extreme points of B E .For example, we have the following couple of facts: • if E is (IV)-polyhedral, then ext(B E ) = ∅ (see [9,Theorem 3.6]) (a wellknown example of this kind is c 0 ); • if E is (VI)-polyhedral, then each point in ext(B E ) is • -isolated (see Observation 6.9), and hence card ext(B E ) ≤ dens(E).Existence of a polyhedral Banach space whose unit ball contains infinitely many extreme points was already observed in the famous paper [14], in which Joram Lindenstrauss proved that every infinite-dimensional Banach space has a two-dimensional quotient whose unit ball is not a polygon, and hence no infinite-dimensional dual Banach space is polyhedral.For other examples of polyhedral Banach spaces whose unit ball contains infinitely many extreme points we refer the reader to [9].
A natural question about ext(B E ), posed by Lindenstrauss himself in [14], is whether there exist a polyhedral infinite-dimensional Banach space whose unit ball is the closed convex hull of its extreme points.This problem was solved in the affirmative in the paper [4], by considering a suitable renorming of c 0 .After that, other examples of polyhedral Banach spaces with the same property were provided in [1].We point out that the examples contained both in [4] and in [1] are separable and satisfy a stronger form of polyhedrality, namely (V)-polyhedrality.The results contained in our paper concern the following natural question: to what extent, given a polyhedral Banach space, is it possible to find a polyhedral renorming whose unit ball is the closed convex hull of its extreme points?Let us briefly describe the structure and the main results of our paper.
After same notation and preliminaries, contained in Section 2, in Section 3 we consider a polyhedral Banach space E, satisfying a certain additional condition, called by us property ( * * ) and implied by (V)-polyhedrality.Then, given a positive number ε, we define a certain ε-equivalent renorming of E, denoted by X.The main ingredients used in our construction are: (i) the well-known structural theorem by V.P. Fonf, asserting that the unit sphere of an infinite-dimensional polyhedral Banach space E is covered by Γ's many true faces (see Theorem 2.8 below), where Γ = dens(E); (ii) Lemma 3.1, which, roughly speaking, asserts that if we do not remove too much true faces from the unit sphere of E, then the unit ball B E coincides with the closed convex hull of the remaining true faces.For the sake of clearness we divide the construction in two cases: the separable one and the non-separable one.Even if the two cases are formally almost identical, we think that the idea of the construction can be more easily understood by the former.Despite some unavoidable technicalities, the geometrical argument used in our construction is quite simple and it is in part inspired by the results contained in [4] and in [14, Section 5, (e)].
In Section 4, we prove that X is a polyhedral Banach space.In order to do that, we introduce a result, namely Proposition 4.2, asserting that B X can be seen as the union of B E and Γ's many sets, each of them obtained as the convex hull of a true face of B E and a singleton.Moreover, at the end of the section, we show that B X is the closed convex hull of its extreme points.
In Section 5, we show that if in addition E (the Banach space taken as starting point of our construction) is (IV)-polyhedral then X is (V)-polyhedral.Finally, in Section 6 we presents some easy consequences of our results and some open problems.For example, it remains open whether each polyhedral Banach space admits an equivalent polyhedral renorming such that the corresponding unit ball is the closed convex hull of its extreme points (cf.Problem 6.3).

Notation and preliminaries
Throughout all this paper, we consider real Banach spaces.If not differently stated, we assume that the Banach spaces considered are infinite-dimensional.Given a Banach space E, with topological dual E * , we denote by B E , B 0 E and S E the closed unit ball, the open unit ball and the unit sphere of E, respectively.Given an ordinal number Γ, it will be often identified with the corresponding interval of ordinal numbers [0, Γ), endowed with the usual order.A set For x, y ∈ E, [x, y] denotes the closed segment in E with endpoints x and y, and (x, y) = [x, y] \ {x, y} is the corresponding "open" segment.Even if the same notation may be used also for intervals in R and for intervals of ordinals, the meaning should be clear from the context.We shall need the following elementary fact, the proof of which is left to the reader.In the sequel, if B ⊂ E * , we denote by B the set of all w * -cluster points of B. Let us recall that a set B ⊂ B E * is called 1-norming if for each x ∈ E we have x = sup x(B); using the Hahn-Banach theorem, it is easy to see that this is equivalent to say that conv Let us recall that the duality map D E : Definition 2.2.Let K be a nonempty closed convex subset of E.
(i) An element x ∈ K is said to be an extreme point of K if it does not lie in any "open" segment contained in K. (ii) By ext(K) we denote the set of all extreme points of K. (iii) A slice of K is a set of the form where α > 0 and x ∈ E * \ {0} is bounded above on K. (iv) An element x ∈ K is said to be a strongly exposed point of K if there exists x * ∈ E * \{0} such that x * (x) = sup x * (K) and for each norm neighbourhood V of x there exists α > 0 such that S(K, x * , α) ⊂ V .In this case, we say that x is strongly exposed by x * .(v) We denote by str exp (K) the set of all strongly exposed points of K.
We shall need the following easy-to-prove fact.
Fact 2.3.Let C be a bounded subset of a normed space E. Let x ∈ E and f ∈ E * be such that f (x) > sup f (C).Then x is a strongly exposed point of the set conv (C ∪ {x}) and it is strongly exposed by f .
A closed convex and bounded set P ⊂ E is called polytope if every finite-dimensional section of P is a finite-dimensional polytope, i.e., the convex hull of finitely many points (equivalently, if every finite-dimensional section of P is a finite-dimensional polyhedron, i.e., the intersection of finitely many halfspaces).We say that the Banach space E is polyhedral if B E is a polytope.It is well-known [13] that E is polyhedral iff the unit ball of each of its 2-dimensional subspaces is a polytope.In [9] and [6], the authors classified the principal known notions of infinite-dimensional polyhedrality, let us recall the main definitions.(I) (extB For j ∈ {I, II, III, IV, V, VI, VII}, E is called (j)-polyhedral if it satisfies property (j).Moreover, E is said isomorphically (j)-polyhedral if it admits an equivalent norm satisfying property (j).Definition 2.5.We say that E satisfies property (∆) if, for each x ∈ S X , the set ext D E (x) is finite.
It is well known (see [9] and the references therein) that and that none of the implications above can be reversed.Moreover, (IV) implies (∆).Definition 2.6.Let E be a Banach space and B ⊂ B E * .We say that B satisfies property (IV) iff we have, (1) sup{f (x); f ∈ B } < 1, whenever x ∈ S E .
Remark 2.7.Proceeding as in [9, Section 2.4], it is immediate to prove that condition (2) above is equivalent to the following condition: We shall say that a set F ⊂ S E is a true face of B E if there exists f ∈ S X * such that, if H = f −1 (1), then F = H ∩ B E and int H F (the relative topological interior of F in H) is nonempty.The set int H F will be simply called the interior of the true face F .We have the following important structural result for polyhedral Banach spaces (see [7,8,15]).
Theorem 2.8.Let E be a polyhedral Banach space.Then the sphere S E is covered by the true faces of B E .Hence the set is a boundary for E.Moreover, B E * = conv B 0 and card(B 0 ) = dens(E) = dens(E * ), where dens(E) and dens(E * ) denote the density character w.r.t. the norm topology of E and E * , respectively.
Observe that the boundary B 0 from Theorem 2.8 is minimal, in the sense that each boundary of E contains B 0 .In the sequel we shall simply say that B 0 is the minimal boundary of E.Moreover, Theorem 2.8 implies that polyhedral spaces are in particular Asplund spaces.The following result holds.Proposition 2.9 ([15, Lemma 2]).Let E be a polyhedral Banach space and x 0 ∈ S E .Then the following assertions are equivalent.
(i) x 0 is in the interior of a true face of In the sequel, we shall need the following consequence of Theorem 2.8.
Corollary 2.10.Suppose that D 0 is a polytope with nonempty interior in a Banach space E and suppose that there exist families Then card(I) ≥ dens(E).
Proof.Without any loss of generality, we can suppose that 0 ∈ intD 0 and that a i = 1, whenever i ∈ I. Without any loss of generality, we can also suppose that the norm of E is polyhedral and that the closed unit ball B E coincides with the set D 0 ∩ (−D 0 ).Let us denote by B 0 the minimal boundary of E, we claim that B 0 ⊂ {±g i } i∈I .In order to prove the claim, let Hence g i = f and the claim is proved.By our claim and Theorem 2.8, we have card The following lemma is an analogue of [10, Remark 1.4], stated here for both properties (IV) and (V).For the proof concerning property (V) see also [4,Lemma 3.4].
Lemma 2.11.Let E be a Banach space.Then the following conditions are equivalent. (i (iii) E is polyhedral and the minimal boundary B 0 satisfies property (IV) [respectively, property (V)].
In the sequel, we shall need the following notion of polyhedrality.
Definition 2.12 (Property ( * * )).Let E be a polyhedral space and let B 0 its minimal boundary.We shall say that E satisfies property ( * * ) if where FD(S E ) denotes the set of all elements in S E that are Fréchet differentiability points of B E .

The main construction
In this section, we suppose that ε > 0 and that E is an infinite-dimensional polyhedral Banach space satisfying property ( * * ).The norm of E will be denoted by • .For the sake of clearness we divide our construction in several subsections.
3.1.Auxiliary results.By Theorem 2.8, the unit sphere S E is covered by the true faces of B E , let B 0 be the corresponding minimal boundary.Let Γ = card(B 0 ) and let us represent Γ as the interval of ordinals [0, Γ).Clearly, B 0 can be represented as the disjoint union of two families For each γ ∈ [0, Γ), let us denote By our hypotheses and by Theorem 2.8, we have Γ = dens(E) = dens(E * ).
Roughly speaking, the next lemma asserts that, if we do not remove too much true faces from the unit sphere S E , then the unit ball B E coincides with the closed convex hull of the remaining true faces.
In particular, for each z ∈ B E and each θ > 0, there exist A, a finite nonempty subset of [0, Γ) \ C, and G, a finite nonempty set of such that z ∈ conv (G) + θB E and such that, for each w ∈ A, the sets Proof.Let us prove (4).Suppose on the contrary that then, by the Hahn-Banach theorem, there exists g ∈ X * such that Let us consider the set D 0 = {x ∈ B E ; g(x) ≥ 1}.Then D 0 is a closed convex set with nonempty interior.Moreover, D 0 is clearly a polyhedron since it is the intersection between B E , which is a polyhedron, and a closed half-space.We claim that and that, for each d ∈ ∂D 0 , at least one of the following condition holds: Proof of the claim.The containment D 0 ⊂ F is clear.For the other containment it is clearly sufficient to prove that F ⊂ B E .Suppose on the contrary that there exists x ∈ F \ B E .Take any x 0 ∈ intD 0 and observe that g(x 0 ) > 1 and Now, since y ∈ S E , there exists w ∈ [0, Γ) such that f + w (y) = 1 or f − w (y) = 1, and necessarily by (i) we have that w ∈ C. Hence, y ∈ D and, by the fact that sup g(D) < 1, we get a contradiction by (ii).The proof of D 0 = F is concluded.
For the second part of the claim, suppose that d ∈ ∂D 0 and that g(d) > 1, then clearly d ∈ S E and hence there exists w ∈ [0, Γ) such that f + w (d) = 1 or f − w (d) = 1.By the definition of D and since sup g(D) < 1, we have that w ∈ C. The proof of the claim is concluded.Now, by our claim and Corollary 2.10, we have that card(C) ≥ dens(E) = Γ, a contradiction and ( 4) is proved.
For the latter part, since there exists A, a finite nonempty subset of [0, Γ) \ C, and G, a finite nonempty set of such that z ∈ conv (G) + θB E .Since, for each w ∈ A, the sets int H + w (S + w ) and int H − w (S − w ) are convex, we can suppose without any loss of generality that, for each w ∈ A, the sets G ∩ int H + w (S + w ) and G ∩ int H − w (S − w ) contains at most an element.If necessary, we can add finitely many elements to the set G in such a way that the thesis holds.
We are grateful to L. Vesely for the idea of the proof of the present form of Lemma 3.1, indeed the first version of the proof of this lemma required the additional hypothesis dens(E) = w * -dens(E * ).
In the sequel we shall also need the following fact, for the sake of completeness we include a sketch of a proof.Fact 3.2.Let ε and Γ be as above.Then there exists a function θ : [0, Γ) → (0, ε) such that (5) inf Sketch of the proof.Let us denote by [Γ] n the set of all n-elements subsets of Γ, and by [Γ] <∞ the set of all finite subsets of Γ. Observe that, whenever n ∈ N, we have In particular, there exists a one-to-one correspondence p : [0, Γ) → Γ <∞ .If A ∈ Γ <∞ , let us denote by #A the number of its elements.By (6), it is easy to see that the map θ : [0, Γ) → (0, ε), defined by and such that, for each w ∈ A n 0 , the sets are singletons.Applying Lemma 3.1 finitely many times, we can easily define, for k < H n , finite sets G n k and A n k such that (a), (b), and (c) in (P n 4 ) are satisfied.Then define Clearly (P n 2 ) holds, properties (P n 1 ) and (P n 4 ) follow by our construction.

Construction of the equivalent norm | • |.
We are going to define an equivalent norm | • | on E, by means of the sets A γ and E γ defined in the previous subsections.
We will describe the details of the construction of | • | and its main properties (namely, conditions (A)-(E) and Proposition 3.3 below) only in the non-separable case, the corresponding definition and results in the separable case are formally identical.In the remaining part of this section, we assume that E is a non-separable polyhedral Banach space satisfying condition ( * * ).
(B) For each γ < Γ, we have (C) Since E satisfies property ( * * ), by (7), we have that (D) For each γ ∈ [0, Γ), we can take a number θ γ ∈ (0, ε) such that and observe that B is a symmetric closed convex set contained in (1 + ε)B E .Moreover, observe that For γ ∈ [0, Γ), let the set W γ be defined as in the construction of the sets A γ and E γ , contained in the Subsection 3.3.In the proof of the proposition we shall need the following lemma.
Proof.Since the sets B, conv (W γ ), and B E are convex, we can suppose, without any loss of generality, that y ∈ W η .By (P η 4 ), there exists η ∈ [0, η) and a finite set and hence, if we define α = w∈A η η α w and β = w∈A η η β w , we have Proof of Proposition 3.3.It is clearly sufficient to prove that, for each t > 0 and γ 0 ∈ [0, Γ), we have Hence, there exist b 1 ∈ B and w 1 ∈ conv (W γ 2 ) such that Repeating (n − 1)-times the same argument yields existence of elements 2 and α i + β i = 1, whenever i = 2, . . ., (n − 1), and such that, if we denote then we have In particular, since b ∈ B (this is easy to see by the definition of b and since 0 ∈ B) and since W γn ⊂ B E , we have , and the proof is concluded.
In particular, B is the unit ball of an equivalent norm on E and we have From now on, we denote by X the Banach space E endowed by the equivalent norm Let X be defined as in the previous section and let us denote by • the norm of E. The aim of this section is to prove that X is a polyhedral Banach space.The main ingredient of our proof is Proposition 4.2, called by us Pyramidal decomposition of B X and asserting that B X can be seen as the union of B E and Γ's many sets (pyramids), each of them obtained as the convex hull of a true face of B E and a singleton (see Figure 1).Moreover, at the end of the section, we show that B X is the closed convex hull of its strongly exposed points (and hence of its extreme points).Let us start with the following lemma.Lemma 4.1.We have Proof.By Fact 2.1 and symmetry, it is sufficient to prove that, if γ 1 , γ 2 , ∈ [0, Γ), γ 1 = γ 2 , and if we denote Suppose that this is not the case, then by the Hahn-Banach theorem there exists Since B E * = conv w * (B 0 ), we can suppose without any loss of generality that where n ∈ N\{1, 2}, λ 1 , . . ., λ n are non-negative numbers such that λ 1 +. ..+λ n = 1, and A similar argument yields λ 2 > 1 2 , a contradiction and the proof is concluded.Proposition 4.2 (Pyramidal decomposition of B X ).We have Proof.Let us prove the first equality.The containment ⊃ is trivial since (ii) For each γ ∈ [0, Γ), we have that eventually (as . Since B 0 is a boundary (and hence a 1-norming set) for E, we have that x ∈ B E .In the case (i) is satisfied, there exists γ ∈ [0, Γ) such that In both cases, we have For the second equality, observe that the containment ⊃ is trivial.For the other containment, it is sufficient to observe that if γ ∈ [0, Γ), x = (1 + θ γ )e γ , and b ∈ B E , then the segment [b, x] intersects H + γ in a unique point x (indeed, f + γ (x) > 1 and Lemma 2.7]).Let P be a polytope in a Banach space Z and W ⊂ Z a finite set.Then K := conv (P ∪ W ) is a polytope.Proof.Let Y be a finite-dimensional subspace of X.Since B E ∩ Y is a polytope in Y and hence it has finitely many true faces, the set Suppose on the contrary that this is not the case, by Proposition 4.2 there exists γ ∈ [0, Γ) \ Γ 0 , b ∈ S + γ , and λ ∈ (0, 1] such that So, we have , a contradiction since γ ∈ [0, Γ) \ Γ 0 and the claim is proved.By Lemma 4.3 and our claim, the set B X ∩ Y is a polytope in Y ; the proof is concluded by the arbitrariness of Y .

Stronger polyhedrality properties: X is (V)-polyhedral
The aim of this section is to prove that, under the assumption that E is (IV)polyhedral, the Banach space X is (V)-polyhedral.In the first part of the section we construct a boundary B of X, then in Theorem 5.4 we prove that if E is (IV)polyhedral then B satisfies property (V). Let For γ and ξ as above, let us denote .Proof.By definition of g + ξ,γ and since θ γ + µ γ + µ γ θ γ < 1, we have Hence, λ + ξ,γ > 1 2 and similarly we have λ − ξ,γ > 1 2 .
Theorem 5.4.Suppose that E is (IV)-polyhedral, then B satisfies property (V ), and hence X is a (V)-polyhedral Banach space.
Proof.Let us prove that, for each x ∈ S X , we have sup{f (x); f ∈ B \ D X (x)} < 1.By Remark 2.7, this is equivalent to prove that, for each x ∈ S X , we have Fix x ∈ S X , and observe that, proceeding as in the final part of the proof of Proposition 5.3, we can suppose without any loss of generality, that there exists . We consider three different cases.Case ϕ = 0.In this case x ∈ ∂ H + γ S + γ ⊂ S E .Suppose on the contrary that there exists a net {g α } α∈I ⊂ B \ D X (x) such that g α →g ∈ D X (x) in the w * -topology.Observe that since each element of B is a convex combination of two elements of B 0 , there exist µ 1 α , µ 2 α ∈ [0, 1] and h 1 α , h 2 α ∈ B 0 such that g α = µ 1 α h 1 α + µ 2 α h 2 α , whenever α ∈ I.Moreover, by w * -compactness and since {g α } α∈I ⊂ B \ D X (x), we can suppose without any loss of generality that: ], and µ 1 + µ 2 = 1.We claim that µ 1 = 0 (and hence that µ 2 = 1).Indeed, suppose that this is not case and observe that, since 1 = g(x) = µ 1 h 1 (x) + µ 2 h 2 (x), we should have 1 = h 1 (x), a contradiction by the fact that h 1 α ∈ D E (x), whenever α ∈ I, and by the fact that E is (IV )-polyhedral.The same argument shows that, by passing to a subnet if necessary, we can suppose that h 2 α ∈ D E (x), whenever α ∈ I. Since E is (IV )polyhedral, it has property (∆) (see Definition 2.5), and hence D E (x) ∩ B 0 is finite; so, by passing to a subnet if necessary, we can suppose that there exists γ ∈ [0, Γ) such that h Case ϕ ∈ (0, 1).Let us observe that in this case we have: where the last inequality holds since, by Lemma 5.1, λ + ξ,η > 1 2 .Since by definition µ γ + θ γ µ γ + θ γ < 1, we have g + ξ,η (x) < 1. Similarly g − ξ,η (x) < 1.By the previous observation, we have Proceeding as in the observation at the beginning of this case we get (8) sup Now, suppose on the contrary that there exists a net {g α } α∈I ⊂ B \ D X (x) such that g α →g ∈ D X (x) in the w * -topology.By (8) and by passing to a subnet if necessary, we can suppose that {g α } α∈I ⊂ B 1 .Since g(x) = 1, we necessarily have that g(b) = 1.Moreover, since {g α } α∈I ⊂ B 1 , we have that g α (b) < 1, whenever α ∈ I. Proceeding as in the previous case, we get a contradiction.Case ϕ = 1.In this case x = (1 + θ γ )e γ , let us observe that we have: • if ξ ∈ Γ \ {γ}, then by definition g + ξ,γ (x) = 1 and g − ξ,γ (x) = 1; • if η = γ and ξ ∈ Γ \ {η}, then proceeding as in the previous case, we have: 1+θγ +µγ +θγ µγ 2 .
By the previous observation, we have that , and hence, proceeding as in the previous cases, we have sup{f The proof is concluded.

Conclusion, final remarks, and open problems
Let us resume the results obtained in the previous sections.Definition 6.1.Let E be a Banach space, ε > 0. The space E endowed with an equivalent norm, with unit ball B, is called ε-equivalent renorming of E, if Theorem 6.2.Let E be a polyhedral Banach space satisfying property ( * * ) and ε > 0. Then the Banach space X, defined as in Section 3, is an ε-equivalent polyhedral renorming of E satisfying Whether the hypothesis concerning property ( * * ) can be omitted in the first part of the previous result remains an open question.Problem 6.3.Suppose that E is a polyhedral Banach space.Does E admits an equivalent polyhedral renorming X satisfying B X = conv ext(B X ) ?
Let us point out that the problem above has an affirmative answer in the case E is separable.Moreover, in this case, it is also possible to prove that the equivalent norms satisfying (9) are dense w.r.t. the Banach-Mazur distance (see Subsection 6.1 below for the details).Problem 6.3 has an affirmative answer also in the case E is a Lindenstrauss space, indeed in this case property ( * * ) is automatically satisfied (see Subsection 6.2). 6.1.Approximation by polyhedral norms such that the corresponding closed unit ball is the closed convex hull of its extreme points.The following result holds (see [9,Theorem 3.3] and [5, Theorem 1.1]).Theorem 6.4.For a separable Banach space E the following statements are equivalent.
Whether the previous theorem holds in the non-separable case is not known.However, the results contained in [2] imply that this is the case for the Banach space c 0 (Γ).By combining, the previous results we get the following corollary.
Corollary 6.5.Suppose that we are in one of the following cases: • E is isomorphic to c 0 (Γ) for some nonempty set Γ; • E is a separable isomorphically polyhedral Banach space.Then, for each ε > 0, E admits an ε-equivalent (V)-polyhedral renorming such that the corresponding closed unit ball is the closed convex hull of its extreme points.
6.2.Lindenstrauss spaces.A Banach space E is called an L 1 -predual space or a Lindenstrauss space if its dual is isometric to L 1 (µ) for some measure µ.The following result about polyhedral Lindenstrauss spaces holds.Theorem 6.6 ([3, Theorem 4.3] ).Let E be a Lindenstrauss space.The following properties are equivalent: (i) E is a polyhedral space; (ii) E does not contain an isometric copy of c; (iii) E is (V)-polyhedral.
The theorem above implies that if E is a polyhedral Lindenstrauss space then condition ( * * ) is satisfied, hence we have the following corollary.Corollary 6.7.Let E be a polyhedral Lindenstrauss space.Then, for each ε > 0, E admits an ε-equivalent polyhedral renorming such that the corresponding closed unit ball is the closed convex hull of its extreme points.
In view of the previous result, the following problem arises.Problem 6.8.Does there exist a polyhedral Lindenstrauss space E satisfying B E = conv ext(B E ) ? 6.3.Cardinality of ext(B E ).As pointed out above, the unit ball of a (IV)-polyhedral Banach space does not admit extreme points.However, the results presented above and those contained in [4,1] show that it is conceivable that the unit ball of a polyhedral Banach space has many extreme points, in the sense that it is the closed convex hull of its extreme points.
Another related problem could be the following: how many extreme points, in the sense of cardinality, the unit ball of a polyhedral Banach space could have?Let us point out that the separable polyhedral Banach spaces constructed in [4,1] are such that the corresponding unit ball contains countably many extreme points.This fact can be directly verified or, alternatively, one can apply the following easy-to-prove observation, taking into account that these spaces are (V)-polyhedral (and hence (VI)-polyhedral).Observation 6.9.Let E be a (VI)-polyhedral Banach space, then each point in ext(B E ) is • -isolated.This leads to the following problem, which, by the observation above, has obviously an affirmative answer in the case in which the space considered is (VI)polyhedral.Problem 6.10.Does E separable polyhedral imply ext(B E ) is countable?More in general, does card(ext(B E )) ≤ dens(E) hold, whenever E is a polyhedral Banach space?

Fact 2 . 1 .
Let C, D be subset of E and suppose that, for every x, y ∈ D with x = y, the segment [x, y] intersects C. Then conv (D) ⊂ x∈D conv (C ∪ {x}).

Definition 2 . 4 ([ 9 ,
Definition 1.1]).Let E be a Banach space and let us consider the following properties of E.

Fact 5 . 2 .
Let H be a hyperplane in a Banach space X, w ∈ X \ H, and let S ⊂ H be a convex set such that int H S = ∅.Then the set conv (S ∪ {w}) has nonempty interior in X.More precisely, we haveint[conv (S ∪ {w})] = λ∈(0,1) [λ int H (S) + (1 − λ)w].Proposition 5.3.B is contained in B X * and it is a boundary for X.Proof.Let γ, η ∈ [0, Γ), and ξ ∈ [0, Γ) \ {γ}, then the following assertions hold.(i) If η = γ, we have g + ξ,γ (1 + θ η )e η = 1 = g − ξ,γ (1 + θ η )e η ; finite and we can write W n = {z n k } k<Hn ; (P n 4 ) there exist {A n k } k<Hn , a finite family of finite pairwise disjoint subsets of [0, ω), and {G n k } k<Hn , a finite family of finite subsets of B E , such that E n = k<Hn G n k , A n = k<Hn A n k , and such that, for each k < H n , the following conditions are satisfied: Let us show that this is possible.Let n ∈ (0, ω) and suppose that E m , A m , and H m , satisfying properties (P m 1 )-(P m 4 ), are already defined, whenever m < n.Observe that, for m < n, the sets E m and A m are finite and hence also the set m<n E m is finite.Therefore we can find H n and {z n k } k<Hn such that (P n 3 ) is satisfied.Now, observe that, by Lemma 3.1, there exist A n 0 , a finite nonempty subset of [0, ω) \ m<n A m , and G n 0 , a finite nonempty set of and since, by Proposition 3.3, B E ⊂ B X .For the other containment: let x ∈ B X and let x n ∈ conv γ<Γ {±(1 + θ γ )e γ } (n ∈ N) be such that x n → x in the norm-topology.By Lemma 4.1, for each n ∈ N, there exists γ n ∈ [0, Γ) such that x n ∈ conv B E ∪ {±(1 + θ γn )e γn } .Now, exactly one of the following conditions is satisfied.(i)The sequence {γ n } n admits a constant subsequence.