A characterization of Banach spaces containing \ell_1(κ) via ball-covering properties
aa r X i v : . [ m a t h . F A ] F e b A CHARACTERIZATION OF BANACH SPACESCONTAINING ℓ ( κ ) VIA BALL-COVERING PROPERTIES
STEFANO CIACI, JOHANN LANGEMETS, ALEKSEI LISSITSIN
Abstract.
In 1989, G. Godefroy proved that a Banach space containsan isomorphic copy of ℓ if and only if it can be equivalently renormed tobe octahedral. It is known that octahedral norms can be characterizedby means of covering the unit sphere by a finite number of balls. Thisobservation allows us to connect the theory of octahedral norms withball-covering properties of Banach spaces introduced by L. Cheng in2006. Following this idea, we extend G. Godefroy’s result to highercardinalities. We prove that, for an infinite cardinal κ , a Banach space X contains an isomorphic copy of ℓ ( κ + ) if and only if it can be equivalentlyrenormed in such a way that its unit sphere cannot be covered by κ manyopen balls not containing αB X , where α ∈ (0 , Introduction
The existence of isomorphic copies of ℓ in a Banach space has beenthe interest of many mathematicians. Probably one of the most knownresults in this direction is the one given by H. Rosenthal in [24], which isindependent of the norm considered in the space. There are also purelygeometrical characterizations in terms of the considered norm in the space.For this reason octahedral norms were introduced in an unpublished paperby G. Godefroy and B. Maurey (see [12] and [14]).We recall that the norm k · k on a normed space X (or X ) is called octahedral if, for every finite-dimensional subspace E of X and ε >
0, thereis y ∈ S X such that k x + y k > (1 − ε ) (cid:0) k x k + k y k (cid:1) for all x ∈ E .If X is a separable Banach space, then G. Godefroy and N. Kalton (see[13, Lemma 9.1]) proved that the octahedrality of X is equivalent to theexistence of some x ∗∗ ∈ X ∗∗ \ { } such that k x + x ∗∗ k = k x k + k x ∗∗ k for all x ∈ X. Mathematics Subject Classification.
Primary 46B20; Secondary 46B03, 46B04,46B26.
Key words and phrases.
Octahedral norm, Ball-covering, ℓ -subspace, Renorming.This work was supported by the Estonian Research Council grants (PSG487) and(PRG877). Therefore, it is clear that ℓ (also L [0 ,
1] and C ([0 , Theorem 1.1 (see [12, Theorem II.4]) . Let X be a Banach space. Then X contains an isomorphic copy of ℓ if and only if there exists an equivalentnorm ||| · ||| in X such that ( X, ||| · ||| ) is octahedral. It is known that a Banach space X is octahedral whenever X ∗∗ is. Theconverse is not true in general. The natural norm of C ([0 , X is a separable Banach spacecontaining ℓ , then there always exists an equivalent norm on X such thatthe bidual X ∗∗ is octahedral (see [21]). The non-separable case remainsunknown as far as the authors know.G. Godefroy also remarked (without a proof) in [12, p. 12] that the normon a Banach space X is octahedral if and only if every finite covering of B X ,the closed unit ball of X , with closed balls has at least one member of thecovering that itself contains B X . A proof of this fact can be found in [17].Thus, octahedral norms are closely connected to coverings of the unitball or the unit sphere in a Banach space, which is a frequent topic in theliterature (see [23] for a nice survey). In this direction L. Cheng introducedthe ball-covering property (see [4]). We recall that a normed space is saidto have the ball-covering property (BCP, for short) if its unit sphere canbe covered by the union of countably many closed balls not containing theorigin. It is known that separable Banach spaces have the BCP, but theconverse is not true, because ℓ ∞ has the BCP. In recent years the BCP ofBanach spaces has been studied by various authors (see e.g., [7], [8], [10],[15] and [22]).Following [6] and clarifying the state-of-the-art on the BCP, A. J. Guirao,A. Lissitsin and V. Montesinos (see [15]) extended the original BCP to the α -BCP, where α ∈ [ − , X is said to have the α -BCP ,for α ∈ [0 , αB X , while, for α ∈ ( − , αB X . In this language the original BCP corresponds to the 0-BCP.In this paper we consider a slight generalisation of the α -BCP, namelythe α -BCP κ (see Definition 2.3), in which we substitute the condition ofthe covering from being countable to being of cardinality κ , where κ issome infinite cardinal. In addition, we consider the property α -BCP <ω (see Definition 2.4), in which we require the covering to be finite. Ourstarting point is the observation that a Banach space fails the α -BCP <ω forall α ∈ ( − ,
1) if and only if it is octahedral (see Proposition 2.5). Therefore,G. Godefroy’s result (see Theorem 1.1) can be restated in terms of failing the α -BCP <ω (see Remark 2.6). Our main result in this paper is an extensionof Theorem 1.1 to infinite cardinals. CHARACTERIZATION OF BANACH SPACES CONTAINING ℓ ( κ ) 3 Theorem.
Let X be a Banach space and κ an infinite cardinal. Then X contains an isomorphic copy of ℓ ( κ + ) if and only if it can be equivalentlyrenormed to fail the α -BCP κ for every ( or some ) α ∈ ( − , . Let us now describe the organization of the paper. In Section 2 we intro-duce some notation for coverings and collect some elementary properties ofthem, which will be used throughout the paper. Also, we define the α -BCP κ and show that a Banach space fails the α -BCP <ω for all α ∈ ( − ,
1) if andonly if it is octahedral (see Proposition 2.5). Section 3 is purely devotedto proving the promised characterization of Banach spaces containing ℓ ( κ )(see Theorem 3.1). In Section 4 we investigate the stability of coverings indirect sums of Banach spaces. On one hand these results complement theexisting ones from the literature. On the other hand, we will use them inSection 5 to construct various spaces with and without the α -BCP κ . Webegin Section 5 by proving that in Lipschitz spaces the notions of octahedralnorm and failure of the ( − α -BCP κ and octahedral norms (see Figure 1). Namely, we prove thatthere exists a Banach space, which fails the α -BCP κ for every α ∈ ( − , − κ -octahedral norm (see Definition 5.3) and show that thereexists a Banach space which is κ -octahedral, but that has the α -BCP for all α ∈ [ − , X , X ∗ denotes its topological dual and B ( x, r )the open ball centered in x of radius r . By B X we denote the closed unitball and by S X the unit sphere of X . By dens( E ) we represent the densityof a topological space E and by supp( f ) the support of a function f . For acardinal κ , by cf( κ ) we denote its cofinality and by κ + its successor cardinal.Given a metric space M with a designated origin 0, we will denote byLip ( M ) the Banach space of all real-valued Lipschitz functions on M whichvanish at 0. 2. Preliminaries
Let X be a normed space, A ⊂ X , κ a cardinal and α, β ∈ [ − , C ( α, β, A ) := { x ∈ X : ∃ a ∈ A ( k x − a k < α k x k + β k a k ) } and C ( α, β, κ ) := { B ⊂ C ( α, β, A ) : for some set A of cardinality κ } . Notice that this notation can represent coverings of the sphere since thecondition S X ∈ C ( α, β, κ ) holds if and only if there exists a set A ⊂ X of STEFANO CIACI, JOHANN LANGEMETS, ALEKSEI LISSITSIN cardinality κ such that S X ⊂ [ a ∈ A B ( a, α + β k a k ) . Now we list some elementary but useful properties regarding the set C ( α, β, A ). Remark . Let X be a normed space, α, β ∈ [ − ,
1] and
A, B ⊂ X . Then( a ) C ( α, β, A ) = C ( α, β, A ).( b ) If B ⊂ C ( α, β, A ), then tB ⊂ C ( α, β, tA ) for all t > c ) If S X ⊂ C ( α, β, A ), then X \ { } = C ( α, β, R + A ).( d ) Let { t a : a ∈ A } ⊂ [1 , ∞ ). If B ⊂ C ( α, , A ), then B ⊂ C ( α, , { t a a : a ∈ A } ) . Proposition 2.2.
Let X be a normed space, κ an infinite cardinal and α, β ∈ [ − , . Then the following are equivalent: ( i ) S X / ∈ C ( α, β, κ ) . ( ii ) X \ { } / ∈ C ( α, β, κ ) . ( iii ) For all subspaces Y ⊂ X with dens( Y ) ≤ κ there exists x ∈ S X suchthat for all λ ∈ R and y ∈ Y we have k λx + y k ≥ α | λ | + β k y k .Proof. ( ii ) = ⇒ ( i ) follows from Remark 2.1(a) and (c).( i ) = ⇒ ( iii ). Assume that Y = A for some set A ⊂ X of cardinality κ . Remark 2.1(a) implies that S X C ( α, β, A ) = C ( α, β, Y ). Thereforethere exists x ∈ S X such that k x + y k ≥ α + β k y k for all y ∈ Y , thus for all λ ∈ R \ { } and y ∈ Y k λx + y k = | λ |k x + y/λ k ≥ | λ | ( α + β k y/λ k ) = α | λ | + β k y k . ( iii ) = ⇒ ( ii ) is trivial. (cid:3) We are now ready to extend the α -BCP to bigger cardinals. Definition 2.3.
Let X be a normed space, κ an infinite cardinal and α ∈ [ − , X has the α -BCP κ if S X ∈ C ( − α, , κ ).In this notation the usual α -BCP corresponds to the α -BCP ω . By Propo-sition 2.2, a normed space X has the α -BCP κ if and only if X \ { } ∈ C ( − α, , κ ). Moreover, as already noted earlier, X has the α -BCP κ if andonly if there exists a set A ⊂ X of cardinality κ such that S X ⊂ [ a ∈ A B ( a, k a k − α ) . In addition, let us now consider the following finite version of the BCP.
Definition 2.4.
Let X be a normed space and α ∈ [ − , X has the α -BCP <ω if there exists a finite set A ⊂ X such that S X ∈ C ( − α, , A ).We observe now that octahedral norms can be characterized in terms offailing the α -BCP <ω . CHARACTERIZATION OF BANACH SPACES CONTAINING ℓ ( κ ) 5 Proposition 2.5.
Let X be a normed space. Then the following are equiv-alent: ( i ) X is octahedral. ( ii ) For all n ∈ N , a , . . . , a n ∈ S X and ε > there exists x ∈ S X suchthat k x − a j k ≥ − ε for all j = 1 , . . . , n . ( iii ) X fails the α -BCP <ω for all α ∈ ( − , .Proof. ( i ) ⇐⇒ ( ii ) is known from [18, Proposition 2.2].( i ) = ⇒ ( iii ). Let A ⊂ X be a finite set and α ∈ ( − , β ∈ (0 , β ≥ ( − α + k a k ) / (1 + k a k ) for all a ∈ A . Since X has an octahedralnorm, then there exists x ∈ S X such that k x − a k ≥ β (1 + k a k ) ≥ − α + k a k for all a ∈ A . Thus, X fails the α -BCP <ω for all α ∈ ( − , iii ) = ⇒ ( ii ) is obvious. (cid:3) Remark . Using Proposition 2.5, we can now rephrase Theorem 1.1 asfollows.Let X be a Banach space. Then X contains an isomorphic copy of ℓ if and only if it can be equivalently renormed to fail the α -BCP <ω for all α ∈ ( − , Characterizing Banach spaces containing ℓ ( κ )We can now state and prove the main result of this paper. Theorem 3.1.
Let X be a Banach space and κ an infinite cardinal. Then X contains an isomorphic copy of ℓ ( κ + ) if and only if it can be equivalentlyrenormed to fail the α -BCP κ for every ( or some ) α ∈ ( − , . The proof follows immediately from Propositions 3.2 and 3.3 combinedwith the fact that cf( κ + ) = κ + for an infinite cardinal κ .In [15, Proposition 23] it was shown that if X fails the ( − X contains ℓ (Γ) for some uncountable set Γ. Using essentially the samescheme, we generalize this result to arbitrary infinite cardinals, while (strictly)weakening the assumption on X . Indeed, in Theorem 5.11, we will prove, forall infinite cardinals κ , the existence of a Banach space failing the α -BCP κ for all α ∈ ( − , − Proposition 3.2.
Let X be a Banach space, α ∈ [ − , and κ an infinitecardinal. If X fails the α -BCP κ , then it contains an isomorphic copy of ℓ ( κ + ) .Proof. We will follow the ideas of [15, Proposition 23]. It suffices to showthat there is a set A ⊂ S X of cardinality κ + such that k n X i =1 λ i a i k ≥ − α n X i =1 | λ i | STEFANO CIACI, JOHANN LANGEMETS, ALEKSEI LISSITSIN for all n ∈ N , { a , . . . , a n } ⊂ A and { λ , . . . , λ n } ⊂ R .Call any set (not necessarily of cardinality κ + ) satisfying this property good and let P be the partially ordered set of all good sets in S X .Notice that P 6 = ∅ since { x } ∈ P for any x ∈ S X . Moreover, for any chain( A η ) ⊂ P , we have that S η A η ∈ P is an upper bound. By Zorn’s Lemma P contains a maximal element A .We claim that | A | > κ + . Assume by contradiction otherwise. Since X fails the α -BCP κ , by Proposition 2.2, we can find x ∈ S X such that k λx + a k ≥ − α | λ | + k a k for all a ∈ span( A ) and λ ∈ R . In particular k λx + n X i =1 λ i a i k ≥ − α (cid:0) | λ | + n X i =1 | λ i | (cid:1) holds for all n ∈ N , { a , . . . , a n } ⊂ A and { λ, λ , . . . , λ n } ⊂ R . This meansthat A ∪ { x } ∈ P , which contradicts the maximality of A . (cid:3) The second ingredient in the proof of Theorem 3.1 is the following result.The cornerstone of its proof is the norm construction technique from [19,Section 4].
Proposition 3.3.
Let X be a Banach space and κ an infinite cardinal. If ℓ ( κ ) ⊂ X isomorphically, then there exists an equivalent norm ||| · ||| on X such that ( X, ||| · ||| ) fails the α -BCP η for all α ∈ ( − , and η < cf ( κ ) . The proof of Proposition 3.3 makes use of the following lemma.
Lemma 3.4.
Let X be a normed space and Y a closed subspace of X . If p is a seminorm on X dominated by the norm of X and equivalent to it on Y , then ||| x ||| := p ( x ) + k x + Y k X/Y defines an equivalent norm on X .Furthermore, if κ is an infinite cardinal, α ∈ [ − , and for every set A ⊂ X of cardinality κ there exists y ∈ Y \ { } such that p ( y − a ) ≥ − αp ( y ) + p ( a ) for all a ∈ A ,then ( X, ||| · ||| ) fails the α -BCP κ .Proof. Assume p C k·k on X and p > c k·k on Y for some C, c >
0. Onone hand ||| · ||| ( C + 1) k·k . On the other hand we claim that ||| · ||| ≥ ˜ c k · k ,where ˜ c := c/ (1 + c + C ).Fix x ∈ X . Without loss of generality we can assume that k x + Y k < ˜ c k x k , otherwise the claim is trivial. There is y ∈ Y such that k x − y k < ˜ c k x k and hence k y k > k x k − k x − y k > (1 − ˜ c ) k x k . Therefore, ||| x ||| > p ( x ) > p ( y ) − p ( x − y ) > c k y k − C k x − y k >> c (1 − ˜ c ) k x k − C ˜ c k x k = ˜ c k x k . CHARACTERIZATION OF BANACH SPACES CONTAINING ℓ ( κ ) 7 Thus the claim is proved. For the furthermore part fix a set A ⊂ X ofcardinality κ . There is y ∈ Y \ { } such that p ( y − a ) ≥ − αp ( y ) + p ( a ) forall a ∈ A . Thus, ||| y − a ||| = p ( y − a ) + k a + Y k X/Y ≥ − αp ( y ) + ||| a ||| = − α ||| y ||| + ||| a ||| . (cid:3) We are now ready to prove Proposition 3.3.
Proof of Proposition 3.3. If ℓ ( κ ) ⊂ X isomorphically, then we can renorm X such that Y := ℓ ( κ ) ⊂ X isometrically. Let { e η : η < κ } ⊂ S Y satisfy k n X j =1 λ j e η j k = n X j =1 | λ j | for all n ∈ N , { λ , . . . , λ n } ⊂ R and { η , . . . , η n } ⊂ κ . Let P be the familyof all seminorms on X dominated by its norm such that they coincide withit on Y . Notice that P 6 = ∅ since k · k ∈ P . Let p := inf P and observe that p ∈ P is a well defined seminorm. Applying the claim in the proof of [3,Theorem 1.3] to ultrafilters in κ , we obtain thatlim η p ( x + e η ) = p ( x ) + 1 for all x ∈ X. Now fix α ∈ ( − ,
1) and a set A ⊂ X of cardinality strictly smaller thencf( κ ). For all a ∈ A there exists η a < κ such that for all η a < η < κp ( a + e η ) ≥ p ( a ) − α. Since | A | < cf( κ ), there exists some θ < κ such that for all a ∈ A we havethat η a < θ and of course p ( a + e θ ) ≥ p ( a ) − α. Finally, we apply Lemma 3.4 to obtain that ( X, ||| · ||| ) fails the α -BCP η when-ever η < cf( κ ). (cid:3) Ball-covering properties in direct sums
In [15, Proposition 8] the authors investigated the stability of the α -BCPin ℓ p -sums of normed spaces. They proved that if X and Y have the α -BCP,then so does X ⊕ p Y for any 1 p ∞ . An inspection of their proof yieldsthat a similar statement also holds for the α -BCP κ . Our first aim in thissection is to investigate the converses to the aforementioned results and forthat we will use the notion of an absolute sum.Recall that a norm N on R is called absolute (see [2]) if N ( a, b ) = N ( | a | , | b | ) for all ( a, b ) ∈ R and normalized if N (1 ,
0) = N (0 ,
1) = 1.For example, the ℓ p -norm on R is absolute and normalized for every p ∈ [1 , ∞ ]. If N is an absolute norm on R , then N ( a, b ) N ( c, d ) whenever | a | | c | and | b | | d | . STEFANO CIACI, JOHANN LANGEMETS, ALEKSEI LISSITSIN If X and Y are normed spaces and N is an absolute normalized norm on R , then we denote by X ⊕ N Y the product space X × Y endowed with thenorm k ( x, y ) k N := N ( k x k , k y k ). Proposition 4.1.
Let X and Y be normed spaces, N an absolute normalizednorm on R , α ∈ [ − , and β ∈ [ − , . If S X ⊕ N Y ⊂ C ( α, β, A ) for someset A ⊂ X ⊕ N Y , then S X ⊂ C ( α, β, π X ( A )) , where π X is the projectionfrom X ⊕ N Y onto X . Moreover, if N is the ℓ norm on R , then thestatement holds also for α ∈ (0 , .Proof. Fix x ∈ S X and consider the element ( x, ∈ S X ⊕ N Y . By assumptionthere is ( a X , a Y ) ∈ A such that N ( k x − a X k , k a Y k ) < α + βN ( k a X k , k a Y k ) . Assume by contradiction that k x − a X k ≥ α + β k a X k . Therefore, α + βN ( k a X k , k a Y k ) > N ( k x − a X k , k a Y k ) ≥ N ( α + β k a X k , k a Y k ) ≥≥ − N ( α,
0) + N ( β k a X k , k a Y k ) = −| α | + N ( β k a X k , k a Y k ) . Since α ∈ [ − ,
0] we have that −| α | = α . In addition notice that N ( β k a X k , k a Y k ) ≥ βN ( k a X k , k a Y k ) , which leads to a contradiction.For the moreover part, if N is the ℓ norm on R , we only need to adjustthe previous inequality to α + β ( k a X k + k a Y k ) > k x − a X k + k a Y k ≥ α + β k a X k + k a Y k to get a contradiction. (cid:3) Corollary 4.2.
Let X and Y be normed spaces, N an absolute normalizednorm in R , κ an infinite cardinal and α ∈ [0 , . If X ⊕ N Y has the α -BCP κ , then X and Y have the α -BCP κ . We now turn our attention to the behaviour of the α -BCP κ in infinitedirect sums of normed spaces. Firstly, recall that for a sequence ( X n ) ofnormed spaces and 1 p ∞ , the normed space ℓ p ( X n ) has the BCP ifand only if each X n has the BCP (see [22, Corollary 2.7 and Theorem 2.8]).Our goal in the remaining part of this section is to investigate the α -BCP κ in infinite ℓ and ℓ ∞ -sums in more detail. Proposition 4.3.
Let ( X η ) be a family of normed spaces and for each η let A η be a subset of X η . For every θ consider the map ˜( · ) : X θ → ℓ ∞ ( X η ) defined by ˜ x = ( . . . , , x, , . . . ) for all x ∈ X θ . ( a ) If α ∈ [0 , and S X η ⊂ C ( α, , A η ) for all η , then S ℓ ∞ ( X η ) ⊂ C ( α, , R + S η ˜ A η ) . ( b ) If α ∈ ( − , and S X η ⊂ C ( α, , A η ) for all η , then S ℓ ∞ ( X η ) ⊂ C ( α + ε, , R + S η ˜ A η ) for every ε > . CHARACTERIZATION OF BANACH SPACES CONTAINING ℓ ( κ ) 9 Proof.
Assume that S X η ⊂ C ( α, , A η ) for all η . Then, by Remark 2.1(c),one has that X η \ { } = C ( α, , R + A η ). Fix x ∈ S ℓ ∞ ( X η ) .( a ) If α ∈ [0 , η such that k x ( η ) k > b ) If α ∈ ( − , ε ∈ (0 , − α ) and find η such that k x ( η ) k ≥ ε/α .By our assumption there is some a ∈ R + A η \ { } such that k x ( η ) − a k <α k x ( η ) k + k a k . Hence, k x ( η ) − ta k < α k x ( η ) k + k ta k for any t ≥ max { / k a k , } by Remark 2.1(d). Notice that with this choiceof t we have that k ta k ≥
2, therefore k x − t ˜ a k ∞ = k x ( η ) − ta k . We deducethat k x − t ˜ a k ∞ = k x ( η ) − ta k < α k x ( η ) k + k ta k = α k x ( η ) k + k t ˜ a k ∞ . ( a ) If α ∈ [0 , α k x ( η ) k ≤ α . This proves that S ℓ ∞ ( X η ) ⊂ C ( α, , R + S ˜ A η )( b ) If α ∈ ( − , α k x ( η ) k ≤ α (1 + ε/α ) = α + ε . This shows that S ℓ ∞ ( X η ) ⊂ C ( α + ε, , R + S ˜ A η ). (cid:3) Corollary 4.4.
Let κ be an infinite cardinal and { X η : η < κ } a family ofnormed spaces. ( a ) If α ∈ [ − , and each X η satisfies the α -BCP κ , then ℓ ∞ ( X η ) hasthe α -BCP κ . ( b ) If α ∈ (0 , and each X η satisfies the α -BCP κ , then, for every ε > , ℓ ∞ ( X η ) has the ( α − ε ) -BCP κ . Notice that the converse of Corollary 4.4, for α ∈ [0 , Proposition 4.5.
Let ( X n ) be a sequence of normed spaces, for each n let A n be a subset of X n and let β ∈ [ − , . ( a ) Let α ∈ [0 , . If X n \ { } = C ( α, β, A n ) for all n ∈ N , then ℓ ( X n ) \{ } = C ( α + ε, β, c ( A n ∪ { } )) for all ε > . ( b ) Let α ∈ [ − , . If each A n is a subspace and there exists some c > such that S X n ⊂ C ( α, β, cB A n ) for all n ∈ N , then S ℓ ( X n ) ⊂ C ( α, β, cB ℓ ( A n ) ) .Proof. ( a ). Fix x ∈ ℓ ( X n ) \ { } and ε >
0. Find some n ∈ N suchthat P ∞ j = n +1 k x ( j ) k ≤ ε k x k . There are a ∈ A , . . . , a n ∈ A n such that k x ( j ) − a j k < α k x ( j ) k + β k a j k for every j ≤ n that satisfies x ( j ) = 0.Define a ∈ c ( A n ∪ { } ) by a ( j ) := ( a j if j n and x ( j ) = 0 , . Therefore, k x − a k = n X j =1 k x ( j ) − a j k + ∞ X j = n +1 k x ( j ) k < ( α + ε ) k x k + β k a k . ( b ). Fix x ∈ S ℓ ( X n ) . For all n ∈ N with x ( n ) = 0, we can find y n ∈ c k x ( n ) k · B A n satisfying k x ( n ) − y n k < α k x ( n ) k + β k y n k , whose existenceis guaranteed by Remark 2.1(b). Define y ∈ cB ℓ ( A n ) by y ( n ) := ( y n if n ∈ N is such that x ( n ) = 0 , . Thus, k x − y k < α k x k + β k y k . (cid:3) We end this section by spelling out two useful consequences of the afore-mentioned results.
Corollary 4.6.
Let ( X n ) be a sequence of normed spaces, κ an infinitecardinal and β ∈ [ − , . ( a ) Let α ∈ [ − , . If there is a m ∈ N such that S X m / ∈ C ( α, β, κ ) ,then S ℓ ( X n ) / ∈ C ( α, β, κ ) . ( b ) Let α ∈ [0 , . If S ℓ ( X n ) / ∈ C ( α, β, κ ) , then, for every ε > , thereexists n ∈ N such that S X n / ∈ C ( α − ε, β, κ ) .Proof. ( a ) follows by Proposition 4.1.( b ) is an easy consequence of Proposition 4.5(a). (cid:3) Examples and counterexamples of spaces with the α -BCP κ We begin this section by showing that in a big class of Banach spaces,namely in Lipschitz spaces, the notions of having an octahedral norm andfailing the ( − X is said to have the Daugavet property if every rank-oneoperator T : X → X satisfies the norm equality k I + T k = 1 + k T k , where I : X → X denotes the identity operator. In 1963, I. K. Daugavet (see[9]) was the first to observe that the space C [0 ,
1] has the Daugavet property.By now this property has attracted a lot of attention and many exampleshave emerged. Indeed, C ( K ) spaces for a compact Hausdorff space K with-out isolated points, L ( µ ) and L ∞ ( µ ) spaces for some atomless measure µ ,and Lipschitz spaces Lip ( M ) whenever M is a length space – all have theDaugavet property (see [11] and [25]). Lemma 5.1 (see [20, Lemma 2.12]) . If X has the Daugavet property, then X ∗ fails the ( − -BCP. CHARACTERIZATION OF BANACH SPACES CONTAINING ℓ ( κ ) 11 Note that in [22, Theorem 3.5] it was proved that L ∞ [0 ,
1] fails the BCP.Since L [0 ,
1] has the Daugavet property and ( L [0 , ∗ = L ∞ [0 , L ∞ [0 ,
1] even fails the ( − L ∞ [0 ,
1] = Lip ([0 , Proposition 5.2.
Let M be a complete metric space. Then the followingare equivalent: ( i ) Lip ( M ) has the Daugavet property. ( ii ) Lip ( M ) is octahedral. ( iii ) Lip ( M ) fails the ( − -BCP.Proof. The equivalence ( i ) ⇐⇒ ( ii ) follows from [1, Theorem 1.5].( iii ) = ⇒ ( ii ) is clear from definition.( i ) = ⇒ ( iii ). Assume that Lip ( M ) has the Daugavet property. It isknown that Lip ( M ) is a dual space, whose predual is the Lipschitz-freespace F ( M ). Since the Daugavet property passes from the dual space to itspredual (see [25, Lemmata 2.2 and 2.4]), then F ( M ) also has the Daugavetproperty. Therefore, by Lemma 5.1, Lip ( M ) fails the ( − (cid:3) As noted in Remark 2.6, Theorem 1.1 can be expressed also by meansof the failure of the α -BCP <ω . Nevertheless, it is natural to ask if we canrestate Theorem 3.1 in terms of some infinite version of octahedrality. Forthis reason we introduce the following definition. Definition 5.3.
Let X be a normed space and κ an infinite cardinal. Wesay that X is κ -octahedral if for every subspace Y ⊂ X with dens( Y ) ≤ κ and ε >
0, there exists x ∈ S X such that for all λ ∈ R and y ∈ Y we have k λx + y k ≥ (1 − ε )( | λ | + k y k ).Note that, by Proposition 2 .
2, a normed space X is κ -octahedral if andonly if S X / ∈ C ( α, α, κ ) (or equivalently X \ { } / ∈ C ( α, α, κ )) for all α ∈ (0 , ℓ ( κ + )can be characterized in terms of κ -octahedrality, but we can compare the α -BCP κ with κ -octahedrality. Namely, we finish this section by showingthat the following diagram holds. Figure 1. X fails the ( − κ X fails the α -BCP κ for all α ∈ ( − , X is κ -octahedral ℓ ( κ + ) ⊂ X isomorphically X is octahedral Thm. 5.11 Thm. 5.13 Ex. 5.4Ex. 5.4Ex. 5.5Ex. 5.5 Prop. 3.2 // / ///
The unlabelled arrows in Figure 1 follow easily from the correspondingdefinitions. We begin with some elementary examples.
Example 5.4.
The Banach space ℓ is clearly octahedral, but for everyinfinite cardinal κ , it doesn’t contain ℓ ( κ + ) and it is not ω -octahedral.Indeed, S ℓ ⊂ C ( α, α, ℓ ) for every α ∈ (0 , S ℓ ∈ C ( α, α, ω )thanks to Remark 2.1(a). In general, notice that there is no normed space X which is dens( X )-octahedral. Example 5.5.
Fix an infinite cardinal κ . On one hand, by [16, Fact 7.26],we have that ℓ ( k + ) ⊂ ℓ (2 κ ) ⊂ ℓ ∞ ( κ ). On the other hand, fix any x ∈ S ℓ ∞ ( κ ) . • If | x (1) | = 1, then k x − sgn( x (2)) e k ∞ = 1. • If | x (1) | 6 = 1, then k x − sgn( x (1)) e k ∞ = 1.This shows that ℓ ∞ ( κ ) is not octahedral since S ℓ ∞ ( κ ) is contained in C (2 / , / , {∓ e , ∓ e } ). In fact for all x ∈ S ℓ ∞ ( κ ) there exists a ∈ {∓ e , ∓ e } such that k x − a k ∞ = 1 < / / · ( k x k ∞ + k a k ∞ ) . Moreover, it also shows that ℓ ∞ ( κ ) has the α -BCP <ω for all α ∈ [ − ,
0) since S ℓ ∞ ( κ ) ⊂ C ( α, , {∓ e , ∓ e } ). In fact we only need to change the previousinequality to k x − a k ∞ = 1 < α + k a k ∞ . Our next aim is to show that there exists a Banach space, which fails the α -BCP κ for every α ∈ ( − , − κ = ω building on the construction from [5]. Remark . The main idea of the necessity proof of [5, Theorem 2.1] actu-ally gives that the quotient space ℓ ∞ /c fails the ( − Proposition 5.7.
Let X and Y be Banach spaces, let T ∈ L ( X, Y ) beonto and let Y fail the α -BCP κ for some α ∈ [ − , and some infinitecardinality κ . Then for every ε > there exists λ > such that ( X, k·k λ ) CHARACTERIZATION OF BANACH SPACES CONTAINING ℓ ( κ ) 13 fails the ( α + ε ) -BCP κ , where k x k λ = λ k x k + k T x k is an equivalent normon X .Proof. By the open mapping theorem there exists r > rB Y ⊂ T ( B X ). Take a set A ⊂ X with | A | κ . Find an element y ∈ Y such that k y − T a k > k T a k − α k y k for all a ∈ A and find x such that T x = y and r k x k k T x k . Note that k x k λ > ( λ + r ) k x k > r k x k . Then k x − a k λ = λ k x − a k + k T x − T a k > λ ( k a k − k x k ) + k T a k − α k T x k = k a k λ − α k x k λ − λ (1 − α ) k x k > k a k λ − α k x k λ − λ (1 − α ) r k x k λ , so that ( X, k·k λ ) fails the (cid:16) α + λ (1 − α ) r (cid:17) -BCP κ . (cid:3) Remark . Note that in the proposition above, a set A ⊂ ker T satisfies S X ⊂ C (1 , , A ) in the new norm if and only if it does so in the originalnorm. Example 5.9.
Applying the above to the quotient map T : ℓ ∞ → ℓ ∞ /c ,we get, for any ε >
0, a renorming X ε of ℓ ∞ , failing the ( − ε )-BCP, butsatisfying S X ε ⊂ C (1 , , B c ) (in particular, enjoying the ( − ℓ ( X /n ), [15, Proposition 7] (or its more general ver-sion, Corollary 4.6.(a)) say that it fails the ( − ε )-BCP for all ε >
0. Onthe other hand, ℓ ( X /n ) has the ( − κ is not necessarily ω , let us start with adifferent building block. Lemma 5.10.
Let κ be an infinite cardinal and α ∈ ( − , . Let X α,κ :=( ℓ ( κ + ) , k · k α ) , where k · k α := − α k · k + (1 + α ) k · k ∞ . Then X α,κ fails the α -BCP κ , but S X α,κ ⊂ C (1 , , {∓ e } ) .Proof. Fix a set A ⊂ X α,κ of cardinality κ and find θ ∈ κ + \ S a ∈ A supp( a ).Then for all a ∈ A k a − e θ k α = − α ( k a k + 1) + (1 + α ) max {k a k ∞ , } ≥≥ − α ( k a k + 1) + (1 + α ) k a k ∞ = k a k α − α. This proves that X α,κ fails the α -BCP κ .Finally, we show that S X α,κ ⊂ C (1 , , {∓ e } ). Let x ∈ S X α,κ . If x (1) = 0,then k x + e k ∞ = 1. Otherwise, say x (1) <
0, note that k x + e k < x (1) >
0, then use − e instead). In either case we conclude that for all x ∈ X α,κ there exists e ∈ {∓ e } such that k x − e k α < (cid:3) Theorem 5.11.
For every infinite cardinal κ there exists a Banach space X which fails the α -BCP κ for every α ∈ ( − , , but that has the ( − -BCP. Proof.
Let κ be an infinite cardinal. Fix a sequence ( α n ) ⊂ ( − ,
0) suchthat inf n α n = −
1. For every n ∈ N define a Banach space X n := X α n ,κ asin Lemma 5.10 and set X := ℓ ( X n ). Now, Corollary 4.6(a) implies that X fails the α -BCP κ for every α ∈ ( − , X has the ( − (cid:3) We end this section by proving the existence of a Banach space which is κ -octahedral, but that has the α -BCP for all α ∈ [ − , Lemma 5.12.
For all α ∈ (1 / , and infinite cardinal κ there exists aBanach space X α,κ that has the β -BCP for all β ∈ [ − , and such that S X α,κ / ∈ C ( α, α, κ ) .Proof. Set p := (1 + log ( α )) − ∈ (1 , ∞ ) and X α,κ := ℓ p ( κ + ). Fix a set A ⊂ X α,κ of cardinality κ . Find some θ ∈ κ + \ S a ∈ A supp( a ). For all a ∈ A ,by Jensen’s inequality, we have that k e θ − a k p = (cid:18) X | a ( η ) | p (cid:19) p ≥
12 2 p + 12 (cid:18) X | a ( η ) | p (cid:19) p == 2 p − (1 + k a k p ) = α (1 + k a k p ) . This proves that S X α,κ / ∈ C ( α, α, A ). On the other hand X α,κ is reflexive,hence it does not contain ℓ . Therefore, by Proposition 3.2, X α,κ has the β -BCP for all β ∈ [ − , (cid:3) Theorem 5.13.
For every infinite cardinal κ there exists a Banach spacewhich is κ -octahedral, but that has the α -BCP for all α ∈ [ − , .Proof. Fix a sequence ( α n ) ⊂ (1 / ,
1) such that sup n α n = 1 and for all n ∈ N pick a Banach space X n := X α n ,κ as in Lemma 5.12. Now, Corollary 4.6(b)implies that X := ℓ ( X n ) has the α -BCP for all α ∈ [ − , X is κ -octahedral. (cid:3) Remarks and open questions
Let us end the paper with some questions that are suggested by thecurrent work.Recall that Theorem 3.1 shows that if a Banach space contains ℓ ( κ + ),then it can be equivalently renormed to fail the α -BCP κ for all α ∈ ( − , − κ . Note that Theorem 5.11 suggests that the answer tothe following question could be negative. Question 6.1.
Let κ be an infinite cardinal. Can every Banach space con-taining an isomorphic copy of ℓ ( κ + ) be equivalently renormed so that it failsthe ( − -BCP κ ? As already mentioned in Section 5, it is natural to ask whether we canrephrase Theorem 3.1 in terms of κ -octahedrality in order to obtain a resultcloser to Theorem 1.1. Notice that Proposition 3.3 provides one direction CHARACTERIZATION OF BANACH SPACES CONTAINING ℓ ( κ ) 15 since the failure of the α -BCP κ for all α ∈ ( − ,
1) implies κ -octahedrality,but Theorem 5.13 suggests that the converse might not hold. Question 6.2.
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