Every Banach space admitting an infinite dimensional separable quotient can be alently renormed to make the set of norm-attaining functionals lineable.. However, they appeared again in 2
Preliminaries
We begin by recalling the following relatively new concepts related to the “size” of subsets of Banach spaces.
Definition 1.1.1 (Gurariy, 1991) A subset M of a Banach space is said to be
1 n-lineable ifM ∪ {0} contains an n-dimensional vector subspace;
2 lineableif M∪ {0} contains an infinite dimensional vector subspace;
3 dense-lineable if M∪ {0} contains an infinite dimensional dense vector subspace;
4 spaceable if M∪ {0} contains an infinite dimensional closed vector subspace.
In order to have a better perspective of these new concepts, we refer the reader to the papers in [7], [8], and [25], where it is proved that several pathological properties occur more often than one might expect in the sense described in the definitions above.
The main questions that we want to attack in this chapter are the following:
Problem 1.1.2 (Aron/Gurariy, 2004) Is the set of norm-attaining functionals on an infinite dimensional Banach space always lineable?
The very first results relative to this question and trying to answer it are presented now. All of them appeared in [9].
Remark 1.1.3 (Bandyopadhyay/Godefroy, 2005) Let X be a Banach space Then,
X⊆NA(X ∗ ) and hence NA(X ∗ ) is spaceable ifX is infinite dimensional Observe thatX is reflexive if and only if X=NA(X ∗ ).
Theorem 1.1.4 (Bandyopadhyay/Godefroy, 2005) LetXbe a Banach space such that
BX ∗ is ω ∗ -sequentially compact The following conditions are equivalent:
1 There exists an equivalent norm on X that makes NA(X) spaceable.
2 There exists an infinite dimensional quotient ofX which is isomorphic to a dual space.
Theorem 1.1.5 (Bandyopadhyay/Godefroy, 2005) LetXbe an Asplund Banach space with the Dunford-Pettis property Then, the closed vector subspaces of NA(X) are finite di- mensional In particular, X cannot be equivalently renormed to make NA(X) spaceable.
Theorem 1.1.6 (Bandyopadhyay/Godefroy, 2005) Let X be a Banach space with the Radon-Nikodym property or an almost locally uniformly rotund norm Then, we have that span (NA(X)) =X ∗ In particular, if X is not reflexive but has the Radon-Nikodym prop- erty then X cannot be equivalently renormed to make NA(X) be a vector space.
Now we will present several results on concrete spaces giving partially solutions to the previous question.
We begin with spaces of continuous functions, but first we need the following two lemmas involving compact Hausdorff spaces and subspaces of `1.
Lemma 1.1.7 LetK be an infinite compact Hausdorff topological space Then, there exists a sequence (lj) j∈
N ⊆ K such that li 6= lj if i 6= j, and verifying one of the following conditions:
2 The set {l j :j∈N} does not contain isolated points.
Proof Firstly, note that ifK is scattered then we deduce (see [32]) thatK is sequentially compact Therefore, we can assume that K is not scattered It is known (see [32]) that
K = P ∪D where P is closed and perfect (the perfect kernel of K) and D is open and scattered (the scattered kernel of K.) (Note that this decomposition holds not only for compact Hausdorff spaces, but for any topological space.) Now, there exists a sequence
N ⊆ P such that l i 6= l j for i 6= j and the set {l j :j ∈N} does not contain isolated points.
Lemma 1.1.8 There exists an infinite dimensional vector subspaceM of`1 such that every (αi) ∈ N ∈M\ {0} verifies one of the following conditions:
1 The set{j∈N: Re (αj)>0}is non-empty and finite and the set{j∈N: Re (αj)0} is infinite and the set {j∈N: Re (α j ) 0 Now, consider the infinite dimensional closed subspace
N∈c 0 Observe thatMis a closed subspace ofL∞(à) linearly isometric toc0 Now, if (αn) n∈ N ∈c0 then there existsn∈Nsuch that|α n |(αn) n∈ N kP∞ n=1αnχA nk, and if t∈An then|P∞ n=1αnχA n(t)|=|α n |.
Let us second show the spaceability of L1(à) ∗ \ NA(L1(à)) Since L1(à) is infinite dimensional, there exists a countable family (An) n∈ N of pairwise disjoint measurable sets such that, for each n∈N, à(A n ) >0 and A n =S∞ k=1B n k , where B k n k∈ N is a sequence of pairwise disjoint measurable sets of positive measure Next, consider the vector subspace of L∞(à) defined as
) , where α := (α(k)) k∈ N is a fixed strictly increasing sequence of real positive numbers such thatkαk ∞ = 1 If x ∗ =P∞ n=1x(n)
P∞ k=1α(k)χ B k n is an element ofM, then kx ∗ k ∞ = sup{kx ∗ χ A n k ∞ :n∈N}
For 06=x ∗ ∈M and t∈Ω, we clearly have that x ∗ (t) = 0 for all t /∈S n∈ NAn If t∈An for some n ∈ N, then there is k ∈ N such that x ∈ B n k and so we have that |x ∗ (t)| |x(n)α(k)|< kxk ∞ = kx ∗ k ∞ Therefore for every t ∈Ω, we have that |x ∗ (t)| 0 We will show that the ball centered at x ∗ of radiusr contains a functional that does not attain its norm We can clearly assume that x ∗ attains its norm and x ∗ 6= 0, and hence there is a measurable set A ⊂Ω with à(A) >0 such that |x ∗ (t)|=kx ∗ k ∞ for all t∈ A Since à is atomless, we can write A =S n∈ NAn, where (An) n∈ N is a sequence of pairwise disjoint measurable sets such that à(An)>0 for eachn∈N Next, we choose a strictly increasing convergent sequence (rn) n∈ N of real numbers such thatrn≥1 for alln∈Nand whose limit l satisfiesl≤1 + kx ∗ r k ∞ We will denote byy ∗ the element of L∞(à) given by y ∗ ∞
X n=1 r n χ A n x ∗ + (1−χ A )x ∗ , where the above convergence is pointwise Then ky ∗ k ∞ = max{ky ∗ χAk ∞ ,k(1−χA)y ∗ k ∞ }
In fact, rnkx ∗ k ∞ = kr n χ A n x ∗ k ∞ ≤ ky ∗ k ∞ for all n ∈ N, so lkx ∗ k ∞ ≤ ky ∗ k ∞ , and we have that ky ∗ k =lkx ∗ k ∞ Since the sequence (rn) n∈ N is strictly increasing and since the essential supremum is not attained at any measurable set with positive measure, y ∗ does not attain its norm at L1(à) Also, ky ∗ −x ∗ k ∞
As a consequence, we have checked thaty ∗ ∈x ∗ +rB L 1 (à) and y ∗ does not attain its norm.
Remark 1.1.14 If the measure space (Ω,Σ, à) is σ-finite and has an atom A of finite measure, then for all x ∗ ∈ L∞(à) such that kx ∗ −χAk∞ < 1 2 , we have that x ∗ is norm- attaining As a consequence, L 1 (à) ∗ \NA(L 1 (à)) is not dense.
Note that these previous results motivate the following question.
Question 1.1.15 Is the set of non-norm-attaining functionals on a non-reflexive Banach space always lineable?
Lineability of NA (X)
We will begin by providing positive results, in other words, sufficient conditions for the set NA(X) to be lineable (sometimes under renormings.)
In the first place, we have the following result.
Proposition 1.2.1 LetXbe an infinite dimensional Banach space Then, for everyn∈N,
X can be equivalently renormed so that NA(X) is n-lineable.
Proof Observe that X is isomorphic to a Banach space of the form Y =H⊕ 2 M where
H a Hilbert space of dimension n and M is a Banach space Next, Y ∗ = H ∗ ⊕ 2 M ∗ and
Theorem 1.2.2 Let X be a separable Banach space with a Markushevich basis (e n ) n∈ N Then:
1 If (en) n∈ N is monotone, then NA(X) is lineable.
Nis a monotone and shrinking Schauder basis, thenNA(X) is dense-lineable.
1 We will show that every vector in
) is norm-attaining, where (e ∗ n ) n∈ N is the sequence of biorthogonal functionals associated to (en) n∈ N Notice that, since the basis is monotone, n
, and hence every element of M is norm-attaining.
2 Note that, if (en) n∈ N is a monotone and shrinking Schauder basis, then the sequence of biorthogonal functionals (e ∗ n ) n∈ N in X ∗ is a Schauder basis for the dual X ∗ , and henceM is dense in X ∗
Observe that the previous theorem says that any example of an infinite dimensional separable Banach space such that the set of norm-attaining functionals on it is not lineable is an example of a separable space admitting no monotone Markushevich basis On the other hand, following the line proposed by Theorem 1.2.2, we can try to see what happens for Banach spaces admitting an infinite dimensional separable quotient.
Theorem 1.2.3 Let X be a Banach space Assume that M is a closed subspace of X. Then:
1 If M is 1-complemented and NA(M) is lineable (spaceable) then NA(X) is lineable (spaceable.)
2 If M is proximinal and NA(X/M) is lineable (spaceable) then NA(X) is lineable (spaceable.)
1 Let p : X −→ X be a linear projection of norm 1 so that p(X) = M If Y ⊆
NA(M) is an infinite dimensional vector space thenZ :={f◦p:f ∈Y}is an infinite dimensional vector space contained inNA(X) Indeed, sincekpk= 1 and everyf ∈Y is norm-attaining, we have that kfk =kf◦pk and f ◦p is norm-attaining for every f ∈Y (at the same point at whichf is.) Finally, if Y is closed thenZ is closed too.
Indeed, if (fn◦p) n∈ N is a sequence converging to g, then (fn) n∈ N is Cauchy (because kf n k=kf n ◦pk,) therefore it converges to some f ∈Y, and henceg=f ◦p.
2 LetY ⊆NA(X/M) be an infinite dimensional vector space Then we have thatZ :{f ◦p:f ∈Y} is an infinite dimensional vector space contained in NA(X), where p denotes the quotient map fromXontoX/M Indeed, sinceM is proximinal and every f ∈Y is norm-attaining, we have that kfk=kf◦pk and f ◦p is norm-attaining for every f ∈ Y Finally, if Y is closed then Z is closed too Indeed, if (fn◦p) n∈ N is a sequence converging tog, then (fn) n∈ N is Cauchy (becausekf n k=kf n ◦pk,) therefore it converges to some f ∈Y, and hence g=f◦p.
Corollary 1.2.4 Let X be a Banach space If X admits an infinite dimensional separable quotient, then X can be equivalently renormed so that NA(X) is lineable.
Proof According to [35], if X admits an infinite dimensional separable quotient, then X admits an infinite dimensional quotientX/M with a Schauder basis Now, according to [22], X/M can be endowed with an equivalent norm |ã|so that X/M has a monotone Schauder basis Next, according to [9, Lemma 2.4], there exists an equivalent normk|ã|kon X which coincides with the original norm onM, whose quotient norm onX/M is a positive multiple of |ã|, and which makes M proximinal Finally, X with the norm k|ã|k has a proximinal subspace M so thatNA(X/M) is lineable (by Theorem 1.2.2,) and so, by Theorem 1.2.3,
Now, we will try to approach the converse of the previous corollary To do this, the following will be helpful (see [9].)
Lemma 1.2.5 (Bandyopadhyay/Godefroy, 2005) Let X be a Banach space Then:
1 If Y ⊆NA(X) is a closed separable subspace not containing`1 thenX has a quotient isomorphic toY ∗
2 If B X ∗ is sequentially ω ∗ -compact then every closed separable subspace Y ⊆ NA(X) does not contain `1.
Theorem 1.2.6 Let X be a Banach space such that BX ∗ is sequentially ω ∗ -compact If
NA(X) contains an infinite dimensional closed subspace with an unconditional basic se- quence thenX admits an infinite dimensional separable quotient.
Proof Let Y ⊆ NA(X) be a closed separable subspace with an unconditional basic se- quence Then, we have three possibilities forY:
1 c0 ⊆Y In this case, by Lemma 1.2.5, we have that X has a quotient isomorphic to
2 ` 1 ⊆Y This is impossible if we take into account Lemma 1.2.5.
3 There exists an infinite dimensional reflexive Banach spaceR contained in Y Then, take any closed separable subspace W of R Then, W is reflexive and it does not contain ` 1 Again by Lemma 1.2.5 we have that X has a quotient isomorphic to W ∗ which is separable.
At this moment, we will concentrate on providing negative results, in other words, on the non-lineability of NA(X) As we will see, the hypothesis of smoothness will be crucial for the development of this section, and everything here will be based upon the following fact.
Remark 1.2.7 Let X be a smooth Banach space If x ∗ 6= y ∗ ∈ X ∗ \ {0} are so that kx ∗ k= ky ∗ k x ∗ +y ∗ 2
, then x ∗ +y ∗ cannot be norm-attaining As a consequence, every element of NA(X)∩S X ∗ is an extreme point of B X ∗
A direct consequence of this remark is the next sufficient condition.
Theorem 1.2.8 LetX be a smooth Banach space IfX ∗ does not contain rotund subspaces thenNA(X) is not even 2-lineable.
Remark 1.2.9 Observe that the Banach space c 0 almost verifies the conditions of the pre- vious theorem, because according to the Mazur Theorem (see [33]) we have that smo (B c 0 ) is a G δ dense subset of S c 0 Besides, c 0 is also closed to verify the thesis of Theorem 1.2.8 since rot (B ` 1 ) =∅.
Lineability of X ∗ \ NA (X)
In this section we will show that the answer to Question 1.1.15 is negative Nevertheless, we will begin by presenting some positive results, in other words, by showing conditions for
Remark 1.3.1 Let X be a smooth Banach space Ifx ∗ ∈NA(X)∩SX ∗ thenx ∗ is not only an extreme point of B X ∗ but a (ω ∗ -strongly) exposed point.
A direct consequence of this remark is the next “hint.”
Proposition 1.3.2 Let X be a smooth Banach space If Y is a vector subspace of X ∗ so thatY ∩exp (BX ∗ ) =∅, then Y ⊆X ∗ \NA(X)∪ {0}.
Lemma 1.3.3 Let X be a Banach space with a Schauder basis (en) n∈ N Then, the set
Proof It can be checked that span{e 1 +e 2 , e 3 +e 4 , e 5 +e 6 , } ⊆X\ {±e n :n∈N}.
Lemma 1.3.4 LetK be an infinite Hausdorff compact topological space LetXbe a rotund Banach space Then, C(K, X)\ext BC(K,X) is spaceable.
Proof Let us fix an arbitrary s∈K Then, consider the continuous linear operator δ s : C(K, X) −→ X f 7−→ f(s). Sinceδs is a non-zero operator andC(K, X) is infinite dimensional, we deduce that ker (δs) is an infinite dimensional closed subspace ofC(K, X) Now, according to [15], ext B C(K,X)
={f ∈ C(K, X) :kf(t)k= 1 for all t∈K}, therefore ker (δs)⊆ C(K, X)\ext B C(K,X)
Theorem 1.3.5 LetX be a smooth Banach space Assume that either one of the following conditions hold:
2 There exists an infinite compact Hausdorff topological space K and a rotund Banach space Y such thatC(K, Y)⊆X ∗
1 In the first place, observe that, if (e n ) n∈
Ndenotes the` 1 -basis, then` 1 \{±e n :n∈N}∩ exp (B X ∗ ) =∅ The result follows now from Proposition 1.3.2 along Lemma 1.3.3.
2 In the second place, in accordance with Lemma 1.3.4, C(K, Y) \ ext BC(K,Y ) is spaceable Finally, by Proposition 1.3.2 we have thatX ∗ \NA(X)∪ {0}is spaceable.
Now, it is time to take care of the negative result, which actually solves Question 1.1.15.
Theorem 1.3.6 Let X be a Banach space such that X is a maximal subspace of X ∗∗ Then, X ∗∗ \NA(X ∗ ) is not even 2-lineable.
ProofIfY is a vector space contained inX ∗∗ \NA(X ∗ )∪ {0}, thenX∩Y ={0}, therefore
In virtue of [34, Theorem 4.5.7] and [34, Theorem 4.5.9] we have the following remark.
Remark 1.3.7 The James Space J verifies that J is of codimension 1 in its bidual J ∗∗ , and is isometrically isomorphic toJ ∗∗ In particular, by Theorem 1.3.6, J ∗∗∗ \NA(J ∗∗ )is not2-lineable, since J ∗ is of codimension1 inJ ∗∗∗ , therefore J ∗ \NA(J) is not2-lineable either.
We want to finish this section by presenting a generalization of Theorem 1.3.6.
Theorem 1.3.8 LetX be a Banach space such thatXis a maximal subspace ofX ∗∗ Then
X ∗ cannot be equivalently dually renormed to makeX ∗∗ \NA(X ∗ ) be 2-lineable.
Proof Ifkãkis an equivalent dual norm onX ∗ then there exists an equivalent norm|ã|on
X so that |ã| ∗ =kãk Now, (X,|ã|) verifies the conditions of Theorem 1.3.6.
Density of X ∗ \ NA (X)
In this section we pretend to take advantage of some geometrical techniques to keep studing the set of non-norm-attaining functionals in a topological way In particular, we will begin by finding some sufficient conditions for the setX ∗ \NA(X) to be dense.
Lemma 1.4.1 Let X be a Banach space Then,
[{C ⊆S X :C is a face of B X and diam (C)>0} ⊆cl (S X \ext (B X )).
Proof LetC be a non-trivial face of B X and x∈C.Then there isy∈C\ {x}and so the segment betweenx and y lies inC⊂S X To conclude the proof, it suffices to observe that for everyt∈(0,1), tx+ (1−t)y is not an extreme point ofB X
Theorem 1.4.2 Let X be a smooth Banach space Then,
[{C ⊆S X ∗ :C is a face of B X ∗ and diam (C)>0} ⊆cl (S X ∗ \NA(X)).
In particular, if cl (SX ∗ \NA(X))contains all rotund points of BX ∗ , then SX ∗ \NA(X) is dense in SX ∗
Proof According to Remark 1.2.7 we have that SX ∗ \ NA(X) ⊇ SX ∗ \ext (BX ∗ ), and hence cl (SX ∗ \NA(X)) ⊇ cl (SX ∗ \ext (BX ∗ )) In view of Lemma 1.4.1, we have that cl (S X ∗ \ext (B X ∗ )) contains all the elements in S X ∗ which are not rotund points So the stated result follows from the assumption.
Corollary 1.4.3 Let X be a smooth Banach space Ifrot (B X ∗ ) =∅then X ∗ \NA(X) is dense.
Next remark shows that the assumption of smoothness in the above result cannot be dropped In fact, there are spaces with a dense set of smooth points in the unit sphere and satisfying the second assumption that we imposed, but not the conclusion.
Remark 1.4.4 The unit ball of `∞ has no rotund points and S` 1 possesses a dense set of smooth points However, as we already remarked, NẰ 1 ) has non-empty interior and so
S ` ∞ \NẰ 1 ) is not dense in S ` ∞
Now, what we will do in the final part of this section is give another proof of the fact that every real Banach space can be equivalently renormed so that the set of non-norm-attaining functionals is non-dense (see [1] and [30].) We want to indicate that we did not know of the existence of these preceding proofs when we obtained ours.
Theorem 1.4.5 Let X be a Banach space Then, cl (SX ∗ \NA(X))⊆[ bdS X ∗(C) :C is a face of BX ∗
In particular, we have the following:
1 If B X ∗ possesses a smooth face, then S X ∗ \NA(X) cannot be dense in S X ∗
2 If the union of the smooth faces ofB X ∗ is dense inS X ∗ , thenS X ∗ \NA(X)is nowhere dense in SX ∗
ProofIn the first place, letx ∗ ∈SX ∗ \NA(X) and assume that there exists a smooth faceC ofB X ∗ such thatx ∗ ∈int S X ∗(C) By the Bishop-Phelps Theorem, there is a norm-attaining functional y ∗ ∈int S X ∗(C) Letx ∈S X so that y ∗ (x) = 1 Now, int S X ∗(C) is contained in the set of all smooth points of B X ∗ Since int S X ∗(C) is a convex set contained inS X ∗ , we deduce thatx(x ∗ ) = 1; that is, x ∗ (x) = 1 which is a contradiction In the second place, we have that given any two distinct facesCandDofB X ∗ , int S X ∗(C)∩int S X ∗(D) =∅, and on the other hand, given any face C ofBX ∗ with int S X ∗(C) 6=∅, cl int S X ∗(C)
=C There- fore, the sets S bdS X ∗(C) :C is a face of BX ∗ and S intS X ∗(C) :C is a face of BX ∗ are disjoint Since their union is SX ∗ , we have that S bdS X ∗(C) :C is a face of BX ∗ is closed.
[ bdS X ∗(C) :C is a face of BX ∗ 6=SX ∗
2 By hypothesis, int S X ∗(cl (S X ∗ \NA(X))) ⊆ int S X ∗
Theorem 1.4.5 leads us to the idea that, for non-density of the set X ∗ \NA(X), it is sufficient to find an equivalent norm on X so that BX ∗ has a smooth face For this, the following result, which explains the relation between the faces of B X L
∞ Y in terms of the faces ofB X and B Y ,will be helpful.
Theorem 1.4.6 Let X andY be Banach spaces Then:
1 If C is a maximal face of B X ⊕ ∞ Y, then C is either of the form B X ×D, where D is a maximal face of B Y , or of the form E×B Y , where E is a maximal face of B X Conversely, for Dand E maximal faces of B Y and B X respectively, both B X ×Dand
E×BY are maximal faces of BX⊕ ∞ Y.
2 If C is a smooth face of BX⊕ ∞ Y, then C is either of the form B X ×D, where D is a smooth face of B Y , or of the form E ×B Y , where E is a smooth face of B X Conversely, for D and E smooth faces ofB Y and B X respectively, bothB X ×D and
E×BY are smooth faces ofBX ⊕ ∞ Y.
1 In the first place, assume thatCis a maximal face ofBX ⊕ ∞ Y The setπX(C)×π Y (C) is a face ofBX ⊕ ∞ Y as well asπX(C) andπY (C) are faces ofBX andBY, respectively. Let us see thatπX(C)×π Y (C)6=BX ⊕ ∞ Y Otherwise, keeping in mind thatBX⊕ ∞ Y B X ×B Y , we have that 0 ∈π X (C) therefore there exists y ∈B Y so that (0, y)∈ C. Now, y ∈ π Y (C) so −y ∈ π Y (C) therefore there exists x ∈ B X so that (x,−y) ∈
C Since C is convex, we deduce that 1 2 x,0
∞ ≤ 1 2 butC ⊂S X ⊕ ∞ Y As a consequence,π X (C)×π Y (C)6=BX⊕ ∞ Y This means that πX(C)×πY (C) is a proper face ofBX⊕ ∞ Y Since C⊆πX(C)×πY (C) and C is maximal, we deduce that C = πX(C) ×πY (C) By using again that πX(C)×πY (C) 6= BX ⊕ ∞ Y, we deduce that either πX(C) 6= BX or πY (C) 6= BY. Without loss, we can assume thatπX(C)6=BX Then,πX(C) is a proper face ofBX, so there exists a maximal face E of B X containing π X (C) Now, E×B Y is a proper face ofBX⊕ ∞ Y containingC By maximality, they have to be equal Conversely, both
B X ×D and E ×B Y are proper faces of BX⊕ ∞ Y, and according to what is proved right above they have to be maximal.
2 As above, let us start by assuming that C is a smooth face of BX ⊕ ∞ Y Since C is maximal, C is either of the form BX ×D, where D is a maximal face of BY, or of the formE×BY, whereE is a maximal face of BX Since the projections are open, we conclude the result Conversely, both BX ×D and E×BY are maximal faces of
BX⊕ ∞ Y Moreover, int S X ⊕∞ Y (B X ×D) = U X ×int S Y (D) and int S X ⊕∞ Y (E×B Y ) int S X (E)×U Y , so they are smooth faces.
Theorem 1.4.7 Let X be a real Banach space Then, X can be equivalently renormed so that its new unit ball possesses a smooth face.
Proof Consider x ∗ ∈SX ∗ to be a norm-attaining functional and consider a pointa∈SX so thatx ∗ (a) = 1 Now, it suffices to consider the equivalent norm onX given by kxk ∞ max{|x ∗ (x)|,kx−x ∗ (x)ak} for every x ∈ X Notice that (X,kãk ∞ ) = ker (x ∗ )⊕ ∞ Ra. Finally, we apply Theorem 1.4.6.
Corollary 1.4.8 Let X be a real Banach space Then X can be equivalently renormed so that the set of non-norm-attaining functionals is not dense.
Proof Let a∈SX and a ∗ ∈SX ∗ with a ∗ (a) = 1 and consider the equivalent norm on X given bykxk 1 =|a ∗ (x)|+kx−a ∗ (x)akfor everyx∈X Notice that (X,kãk 1 ) = ker (a ∗ )⊕ 1
Ra The dual norm of kãk 1 is kãk ∗ 1 =kãk ∞ , where kx ∗ k ∞ = max{|a(x ∗ )|,kx ∗ −a(x ∗ )a ∗ k} for everyx ∗ ∈X ∗ Again, (X ∗ ,kãk ∞ ) = ker (a)⊕ ∞ Ra ∗ Finally, by applying first Theorem 1.4.7 and then Theorem 1.4.5, we deduce the result.
Preliminaries
In this chapter everything comes up to the following result (see [28].)
Theorem 2.1.1 (James, 1964) Let X be a Banach space Then, X is reflexive if and only if every functional on X is norm-attaining.
James was asked for the possibility of removing the completeness hypothesis, and like this he came up with the following counterexample (see [29].)
Remark 2.1.2 (James, 1971) There exists a non-complete real normed space on which every functional is norm-attaining.
Later, Blatter realized the following (see [17].)
Remark 2.1.3 (Blatter, 1976) LetX be a normed space Iff ∈X ∗ \{0}andα >0then dist 0, f −1 (α)
=α/kfk, thus f is norm-attaining if and only if f −1 (α) has a minimum- norm element.
And then he came up with the following nice result (see [17].)
Theorem 2.1.4 (Blatter, 1976) LetX be a normed space Then, X is complete if every closed convex set has a minimum-norm element In particular, X is reflexive if and only if every closed convex set has a minimum-norm element.
With this theorem the minimum-norm elements appeared in the literature And the next theorem improved Blatter’s result (see [4].)
Theorem 2.1.5 (Aizpuru/Garc´ıa-Pacheco, 2005) LetX be a normed space Then,X is complete if every bounded closed convex set with non-empty interior has a minimum-norm element In particular, X is reflexive if and only if every bounded closed convex set with non-empty interior has a minimum-norm element.
So, the point is to characterize the non-complete normed spaces on which every func- tional is norm-attaining As shown above (Theorem 2.1.4,) we cannot use the existence of minimum-norm elements, but maybe we can use the existence of translations to obtain them Following this line, here we present the conjecture on which we have been working.
Conjecture 2.1.6 Given a normed space, then:
1 If every functional is norm-attaining, then every closed convex set with non-empty boundary can be translated to have a non-zero minimum-norm element.
2 If every bounded closed convex set with non-empty interior can be translated to have a non-zero minimum-norm element, then every functional on X is norm-attaining.
Later on we will see more precisely the motivation to state this conjecture, which can be found in Theorem 2.2.2 Another problem related to this comes from the following result of James (see [29].)
Theorem 2.1.7 (James, 1971) LetXa non-complete normed space on which every func- tional is norm-attaining Then, the completion Y of X is reflexive but not rotund.
This result motivates the following question.
Question 2.1.8 Given any non-rotund reflexive space, does it contain a proper dense sub- space on which every functional is norm-attaining?
Before starting with the main results, let us see some basic ones related to translations and minimum-norm elements.
Theorem 2.1.9 Let X be a normed space and consider M to be a closed convex subset of X such that M has a maximum-norm element m Then, we can find a∈ X such that m+a6= 0 and m+a is a minimum-norm element ofM+a.
Proof We can start by assuming thatm6= 0 Let us take a=− m kmk −m.
Ifx∈M then kx+ak x−1 +kmk kmk m
≥ 1 kmk|kmk kxk −(1 +kmk)kmk|
That is, m+ais a minimum-norm element of M+a.
The following theorems show that the converse to Theorem 2.1.9 does not hold, that is, the existence of non-zero minimum-norm elements does not imply that we can find translations which map these non-zero minimum-norm elements into maximum-norm ele- ments Here, we can see the relations between translations, minimum-norm elements, and norm-attaining functionals.
Lemma 2.1.10 Let X be an infinite dimensional normed space on which every functional is norm-attaining Then, X can be equivalently renormed to be locally uniformly rotund (and hence to have the Radon-Riesz property.) As a consequence, X contains a ω-null sequence of elements of norm 1.
Proof In the first place, remember (see [34]) that a Banach space has the Radon-Riesz property if on its unit sphere everyω-convergent sequence is convergent Now, the comple- tion ofX is a reflexive Banach space, and every reflexive Banach space can be equivalently renormed to be locally uniformly rotund (see [20].) Finally, every reflexive Banach space has aω-null sequence of norm 1 (see [31] and [36],) therefore due to the density of X in its completion we deduce the existence of aω-null sequence in SX.
Theorem 2.1.11 LetX be an infinite dimensional normed space on which every functional is norm-attaining Assume that X has the Radon-Riesz property There exists a bounded closed convex subset M with a non-zero minimum-norm element x ∈ M such that no translation exists which maps x into a maximum-norm element.
Proof In accordance with Lemma 2.1.10, we can pick a sequence (yn) n∈ N in S X which is ω-convergent to 0 By taking a subsequence if it is necessary, we can assume without loss of generality that there existsf ∈SX ∗ such that Ref(yn)≥0 for every n∈N Letx∈SX withf(x) = 1 For everyn∈N, let us denoteyn+xbyxn Then, (xn) n∈ N isω-convergent to x but none of its subsequences converge to x Let us see that x is a minimum-norm element of
M = co ({x n :n∈N} ∪ {x}). Take an element λx+λ1xn 1 +ã ã ã+λ k xn k ∈co ({x n :n∈N} ∪ {x}) Then kλx+λ 1 x n 1 +ã ã ã+λ k x n k k ≥ Ref(λx+λ 1 x n 1 +ã ã ã+λ k x n k )
Next, suppose we could find a ∈X so that x+ais a maximum-norm element of M +a.
N ω-converges to x+a and there exists a subsequence of (kx n +ak) n∈
N which converges to kx+ak Therefore, sinceX has the Radon-Riesz property, there exists a subsequence of (x n +a) n∈
Nconverging tox+a, in other words, there exists a subsequence of (xn) n∈ N converging to x, which is a contradiction.
The following theorems are more precise than the previous one (which, by the way, requires thatX be infinite dimensional.) To show them, we require the following lemma.
Lemma 2.1.12 Let X be a real normed space Assume thatx∈X andf ∈X ∗ are so that δ:=f(x)>0 andt:=kxk>0 Then:
1 If there exists r > 0 such that B X (x, r)∩f −1 (δ) ⊆ S X (0, t), then there exists 0 < r 0 ≤r so that BX(x, r 0 )∩SX(0, t)⊆f −1 (δ).
2 If there existsr >0such thatB X (x, r)∩S X (0, t)⊆f −1 (δ), then there exists 0< r 0 ≤ r so that B X (x, r 0 )∩f −1 ([0, δ])⊆B X (0, t) In this situation, B X (x, r 0 )∩f −1 (δ)⊆
S X (0, t) Furthermore, r 0 can be chosen to diverge to∞ ifr diverges to ∞.
1 In the first place, let us see thatkfk= δ t Obviously, fx t
= δ t thereforekfk ≥ δ t Ify∈BX and f(y)> δ t then we can take 0< α 0 and δ α(ty) + (1−α)x f(α(ty) + (1−α)x) ∈BX(x, r)∩f −1 (δ), which means that t δ α(ty) + (1−α)x f(α(ty) + (1−α)x)
≤ δ f(α(ty) + (1−α)x)t, and δ < f(α(ty) + (1−α)x)≤δ, which is impossible In the second place, choose
With this choice ofr 0 we have that, if y∈B X (x, r 0 ) then f(y)≥f(x)− |f(y)−f(x)| ≥δ−δ tr 0 ≥ δ
Therefore, δy f(y) ∈BX(x, r)∩f −1 (δ)⊆SX(0, t), in other words, t δy f(y)
2 In the first place, let us see thatkfk= δ t Obviously, f x t
= δ t thereforekfk ≥ δ t Ify∈BX and f(y)> δ t then we can take 0< α t kα(ty) + (1−α)xkδ, and t t Then ty kyk −x
≤ t kykky−xk+ 1 kyk|kxk − kyk|t
Therefore, ty kyk ∈B X (x, r)∩S X (0, t)⊆f −1 (δ), in other words, δ =f ty kyk
≤ t kykδ < δ, which is a contradiction In the third and last place, since kfk = δ t , we have that
Theorem 2.1.13 Let X be a real normed space The following conditions are equivalent:
1 There exists a norm-attaining x ∗ ∈SX ∗ such that (x ∗ ) −1 (1)∩BX has empty interior relative to S X
2 There exists a bounded closed convex subsetM with non-empty interior so thatM has a non-zero minimum-norm element x ∈M and cannot be translated mapping x into a maximum-norm element.
Proof Assume that 1 holds We can pick x ∗ ∈S X ∗ and x∈S X such that x ∗ (x) = 1 and (x ∗ ) −1 (1)∩BX has empty interior relative to SX Let M = BX(x,1)∩(x ∗ ) −1 ([1,∞)). Let us set C := BX(x,1)∩(x ∗ ) −1 (1) Suppose that there is a∈ X such that x+a is a maximum-norm element of M+a Then, kx+ak 6= 0 andM +a⊆ BX(0,kx+ak) Let us show that C+a ⊂ S X (0,kx+ak) If c+a ∈ C+a with c ∈ C, then, by assuming that c 6= x, we can find d ∈ C such that x ∈ (c, d) Now, x+a ∈ (c+a, d+a) which means that kc+ak = kx+ak = kd+ak, because x+a is a maximum-norm element of
M+a On the other hand,C+a=B X (x+a,1)∩(x ∗ ) −1 (1 +x ∗ (a)) Next, let us see that
1 +x ∗ (a)1 Now,ky+ak ≥x ∗ (y+a)>
1 +x ∗ (a) =kx+ak, which contradicts the fact that x+ais a maximum-norm element of
B X (x+a,1)∩(−x ∗ ) −1 (−1−x ∗ (a))⊂S X (0,kx+ak), we deduce, according to paragraph 1 of Lemma 2.1.12, the existence of a positive number
∩S X (0,kx+ak)⊂(−x ∗ ) −1 (−1−x ∗ (a)), which is impossible since (x ∗ ) −1 (1)∩BX has empty interior relative toSX.
Assume that 2 holds and consider M to be a bounded closed convex subset with non- empty interior so that M has a non-zero minimum-norm element x ∈ M and cannot be translated mappingxinto a maximum-norm element On the contrary, let us suppose that f −1 (1)∩B X has non-empty interior relative to S X for all norm-attaining f ∈ S X ∗ Let x ∗ ∈S X ∗ verify that x ∗ (u)< x ∗ (m) for allu∈U X (0,dist (0, M)) and allm∈M Clearly, x ∗ (x) = dist (0, M) = kxk, that is, x ∗ is norm-attaining Since M is bounded we can consider a number K > 0 such that M ⊆ BX(0, K) Now, by assumption (−x ∗ ) −1 (K)∩
BX(0, K) has non-empty interior relative to SX(0, K), so we can take an interior point z ∈ B X (z, r)∩S X (0, K) ⊆ (−x ∗ ) −1 (K) By paragraph 2 of Lemma 2.1.12 there exists
0 < r 0 ≤ r such that BX(z, r 0 )∩(−x ∗ ) −1 ([0, K]) ⊆ BX(0, K) Observe that we may assume thatr 0 >diam (M) by taking K large enough Finally, consider the translated set
M+ (z−x) Ifm∈M then m+ (z−x)∈BX(z,diam (M))⊆BX 0, r 0
(−x ∗ ) (m+ (z−x)) =−x ∗ (m)−x ∗ (z) +x ∗ (x)≤ −dist (0, M) +K+ dist (0, M) =K, and hence m + (z−x) ∈ B X (0, K) And kx+ (z−x)k = kzk = K, which means that x+ (z−x) is a maximum-norm element of M+ (z−x) reaching a contradiction.
Corollary 2.1.14 Let X be a complex normed space Then, there exists a bounded closed convex subsetM with non-empty interior so thatM has a non-zero minimum-norm element x∈M and cannot be translated mapping x into a maximum-norm element.
Proof Let us consider X R , that is, the underlining real space of X By Lemma 1.4.11 the unit sphere ofX R is free of convex sets with non-empty interior relative to it, therefore Theorem 2.1.13 allows us to conclude the result.
Lemma 2.1.15 LetX be a real Banach space withdim (X)>1 Assume thatf −1 (1)∩BX has non-empty interior relative to SX for everyf ∈NA(X)∩SX ∗ Then:
1 Letx∈SX We will show thatxis a smooth point ofBX Otherwise, letf 6=g∈SX ∗ such thatf(x) =g(x) = 1 We have that f+g 2 ∈S X ∗ and f+g
(1)∩B X cannot have non-empty interior with respect toS X
2 If x ∈ exp (BX) then there exists f ∈ SX ∗ such that {x} = f −1 (1)∩ BX This contradicts the fact thatf −1 (1)∩BX has non-empty interior with respect toSX.
3 On the one hand, char (X) = char (SX) On the other hand,
Finally, the map f ∈NA(X)∩S X ∗ 7−→int S X f −1 (1)∩B X is a bijection.
4 Assume that X is separable Then we have that NA(X)∩S X ∗ is countable If Y is a 2-dimensional subspace of X then SY ∗ is uncountable By the Hahn-Banach Theorem, we can extend all the elements in SY ∗ to an uncountable set contained in
NA(X)∩SX ∗ , which is impossible.
Corollary 2.1.16 Let X be a real normed space withdim (X)>1 Then:
Minimum-norm elements and norm-attaining functionals
In this section we will provide a positive approach to Conjecture 2.1.6 The first theorem we present in this section, a primary version of which appears for the first time in [4] as a “key lemma,” will clarify why we try to characterize normed spaces having only norm- attaining functionals in terms of translations to obtain minimum-norm elements.
Definition 2.2.1 Let X be a normed space Let H be a convex subset of X A subset C of H is said to be a norm-attaining exposed face ofH if C is an exposed face ofH and the functional that determines C is norm-attaining.
Theorem 2.2.2 Let X be a normed space Consider a closed convex subset M of X and a point x∈bd (M) The following are equivalent:
1 There exists a∈X such that x+a is a non-zero minimum-norm element ofM+a.
2 There exists a norm-attaining exposed face C of M containing x.
3 There exists a closed ball B such that x∈M∩B and M∩int (B) =∅.
Proof Let us see that 3 implies 2 The Hahn-Banach Theorem allows us to deduce the existence of an element f ∈ S X ∗ such that Ref(u) > Ref(m) for every u ∈ int (B) and every m ∈ M Since the interior of B is dense in B, we have that f defines a non-trivial exposed face in M Let us see that f is norm-attaining Let b be the center of B Since x∈bd (M), we have that the radius ofB is kx−bk We will show that
For everyz∈UX(0,kx−bk), we have that Re (−f) (z+b)0 so thatsup (Ref(M+a))≥δ for allf ∈SX ∗ Indeed, it suffices to take ato be the opposite of the center of a closed ball contained in M and δ the radius of this ball.
Lemma 2.4.1 together with Theorem 2.2.2 afford the following result.
Theorem 2.4.3 Let X be a normed space Assume that every functional on X is norm- attaining Let M be a closed convex subset with non-empty boundary that verifies either one of the following conditions:
1 There are a∈X and δ >0 so thatsup (Ref(M+a))≥δ for allf ∈SX ∗
2 There are a∈X and f ∈S X ∗ such that Ref(M+a) ={0}.
Then M can be translated to have a non-zero minimum-norm element.
Proof It is sufficient to take into account that Lemma 2.4.1 may be applied to the com- pletionY ofX to deduce that everyx∈bd (M) is contained in an exposed face Theorem 2.2.2 allows us to assure that these exposed faces are indeed norm-attaining exposed faces.
In the previous section we have seen that the completion of a non-complete real normed space on which every functional is norm-attaining cannot be rotund Following this line, we present the following theorem.
Theorem 2.4.4 Let X be a non-complete normed space If every bounded closed convex subset of X with non-empty interior can be translated to have a non-zero minimum-norm element, then the completion Y of X cannot be rotund.
ProofAssume thatY is rotund Lety∈S Y \S X and considerM =B Y (y,1/2)∩X There isa∈X such thatM+ahas a non-zero minimum-norm elementm+awithm∈M We have thatM+a=BY (y+a,1/2)∩X and, since 0∈/ M+a, we deduce thatky+ak>1/2. Next, M+ais dense inBY (y+a,1/2) Thus dist (0, M +a) = dist (0,BY (y+a,1/2)).
Now, the rotundity of Y allows us to deduce that B Y (y+a,1/2) has a unique element of minimum-norm Therefore m+a= ky+ak −1/2 ky+ak (y+a),which implies that y∈X, and this is a contradiction.
Preliminaries
Let us recall the definition of transitivity for Banach spaces.
Definition 3.1.1 (Banach, 1932) Let X be a Banach space Then:
1 The space X is said to be transitive if for every x, y ∈ SX there exists a surjective linear isometry T :X−→X such thatT(x) =y.
2 The space X is said to be almost transitive if for every ε >0 and for every x, y∈S X there exists a surjective linear isometry T :X −→X such that kT(x)−yk ≤ε.
3 The space X ∗ is said to be ω ∗ -transitive if for every x, y∈ S X ∗ there exists a ω ∗ -ω ∗ continuous surjective linear isometry T :X ∗ −→X ∗ such that T(x) =y.
4 The space X ∗ is said to be almost ω ∗ -transitive if for every ε > 0 and for every x, y∈ S X ∗ there exists a ω ∗ -ω ∗ continuous surjective linear isometry T :X ∗ −→ X ∗ such that kT(x)−yk ≤ε.
In 1932 Mazur and Banach had a conversation in a bar, and Mazur asked Banach for an example of a transitive and separable Banach space different from `2 This was the beginning of the famous Banach-Mazur conjecture for rotations.
Problem 3.1.2 (Banach/Mazur, 1932) Is every transitive and separable Banach space a Hilbert space?
The main results trying to answer this question are given now The first one is given by Mazur and appears in [33].
Theorem 3.1.3 (Mazur, 1933) LetXbe a Banach space IfXis separable thensmo (BX) is a G δ dense subset of SX As a consequence, if X is transitive and separable, then X is smooth andX ∗ is almost ω ∗ -transitive.
Fifty years later Finet came up with this important result (see [23]).
Theorem 3.1.4 (Finet, 1986) Let X be a Banach space If X is almost transitive and super-reflexive, then both X and X ∗ are uniformly convex.
This theorem motivated Cabello to relax its hypothesis to keep obtaining the same thesis (see [18].)
Theorem 3.1.5 (Cabello, 1999) Let X be an almost transitive Banach space If X is Asplund or has the Radon-Nikodym property, then X is super-reflexive.
Finally, the following result improved the two previous ones (see [12].)
Theorem 3.1.6 (Becerra-Guerrero/Rodr´ıguez-Palacios, 1999) Let X be a Banach space The following conditions are equivalent:
1 The space X is almost transitive and super-reflexive.
2 There exists x∈SX such that co{T(x) :T :X −→X surjective linear isometry}=B X and the norm of X is Frech´et differentiable at x.
According to the Mazur Theorem (see [33]) in order to assure the existence of smooth points it suffices to consider separable Banach spaces Unfortunately, this hypothesis is not enough to assure the existence of rotund points as remarked as follows.
Remark 3.1.7 The space ` 2 ∞ is separable and its unit ball is free of rotund points.
Note that the Banach space in the previous remark is not smooth As we will see in this section, smoothness is crucial to obtain rotund points A primary example is given by the following theorem, which shows that the unit balls of 2-dimensional real smooth Banach spaces have uncountably many rotund points.
Theorem 3.1.8 LetX be a2-dimensional real Banach space IfXis smooth, thenrot (B X ) is uncountable.
Proof Let C denote the set of all exposed faces of B X Note that, in this situation, the rotund points are exactly the exposed faces with null diameter Due to the smoothness of X, the exposed faces are pairwise disjoint, and thus there is a bijection between C and
SX ∗ Now, SX is separable and every exposed face with diameter greater than zero has non-empty interior with respect to SX Therefore there must be countably many exposed faces with diameter greater than zero This leaves uncountably many rotund points.
Let us continue by reviewing some geometrical concepts related to smoothness, which will be fundamental to study some topological properties of the set of rotund points, and can be widely consulted in [19].
Remark 3.1.9 Let X stand for a Banach space Then a point x of the unit sphere of X is said to be
1 a Y-smooth point ofB X (where Y is a closed subspace ofX ∗ ) if everyf, g∈S Y ∗ such thatf(x) =g(x) = 1 verify f =g;
2 a smooth-A point of B X (where A is a subset of S X ) if every f, g ∈ S X ∗ such that f(x) =g(x) = 1 verify f| A =g| A
The sets of Y-smooth points of BX and smooth-A points of BX will be denoted, respec- tively, by smo Y (B X ) and smo A (B X ) Furthermore, if A is a ω ∗ -dense subset of S X ∗ , then smo A (B X ∗ ) ⊆ smo X (B X ∗ ); and if A is a ω-dense subset of S X , then smo A (B X ) smo (B X ) Finally, in 1966 (see [38]) Phelps obtained (by using the Baire Category Theo- rem) the following results for a Banach space X:
1 For every a ∈ S X and every m ∈ N, the set P X (a, m) of all x ∈ S X such that
|f(a)−g(a)|