Spear operators between Banach spaces
Vladimir Kadets, Miguel Martin, Javier Meri, and Antonio Perez

TL;DR
This paper introduces and develops the theory of spear operators between Banach spaces, exploring their properties, examples, and relations to other geometric properties of Banach spaces, including Lipschitz variants.
Contribution
It defines spear operators, introduces related properties like the alternative Daugavet property and lushness, and establishes their theoretical framework and examples, including the lushness of the Fourier transform on L1.
Findings
Lushness of the Fourier transform on L1.
$ ext{ell}_1$ contained in duals of operators with infinite rank.
Lush operators are Lipschitz spear operators.
Abstract
The aim of this manuscript is to study \emph{spear operators}: bounded linear operators between Banach spaces and satisfying that for every other bounded linear operator there exists a modulus-one scalar such that To this end, we introduce two related properties, one weaker called the alternative Daugavet property (if rank-one operators satisfy the requirements), and one stronger called lushness, and we develop a complete theory about the relations between these three properties. To do this, the concepts of spear vector and spear set play an important role. Further, we provide with many examples among classical spaces, being one of them the lushness of the Fourier transform on . We also study the relation of these properties with the Radon-Nikod\'ym property, with Asplund spaces, with the duality, and…
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Key words and phrases:
Banach space, bounded linear operator, Daugavet center, numerical range of operator; alternative Daugavet property; lush spaces; spear set; Lipschitz operator.
2010 Mathematics Subject Classification:
Primary 46B04. Secondary 46B20, 46B22, 46B25, 46J10, 47A12, 47A30, 47A99.
The research of the first author is done in frames of Ukrainian Ministry of Science and Education Research Program 0115U000481, and it was partially done during his stay in Murcia under the support of MINECO/FEDER projects MTM2014-57838-C2-1-P and partially during his visits to the University of Granada which were supported by the Spanish MINECO/FEDER project MTM2015-65020-P. The second and the third authors were partially supported by Spanish MINECO/FEDER project MTM2015-65020-P, and by Junta de Andalucía and FEDER grant FQM-185. The fourth author was partially supported by the MINECO/FEDER project MTM2014-57838-C2-1-P and a Ph.D. fellowship of “La Caixa Foundation”.
Abstract
The aim of this manuscript is to study spear operators: bounded linear operators between Banach spaces and satisfying that for every other bounded linear operator there exists a modulus-one scalar such that
[TABLE]
To this end, we introduce two related properties, one weaker called the alternative Daugavet property (if rank-one operators satisfy the requirements), and one stronger called lushness, and we develop a complete theory about the relations between these three properties. To do this, the concepts of spear vector and spear set play an important role. Further, we provide with many examples among classical spaces, being one of them the lushness of the Fourier transform on . We also study the relation of these properties with the Radon-Nikodým property, with Asplund spaces, with the duality, and we provide some stability results. Further, we present some isometric and isomorphic consequences of these properties as, for instance, that is contained in the dual of the domain of every real operator with infinite rank and the alternative Daugavet property, and that these three concepts behave badly with smoothness and rotundity. Finally, we study Lipschitz spear operators (that is, those Lipschitz operators satisfying the Lipschitz version of the equation above) and prove that (linear) lush operators are Lipschitz spear operators.
Contents
Chapter 1 Introduction
1.1. Motivation
Let and be Banach spaces. The main goal of this manuscript is to study bounded linear operators such that for every other bounded linear operator there is a modulus-one scalar such that the norm equality
[TABLE]
holds. In this case, we say that is a spear operator. When and is the identity operator, the study of this equation goes back to the 1970 paper [27]. It was proved there that for bounded and linear, the existence of a modulus-one scalar such that the norm equality
[TABLE]
holds, is equivalent to the equality between the numerical radius of and its norm. The list of spaces for which the identity is a spear operator (they are called spaces with numerical index one) contains all spaces and spaces, as well as some spaces of analytic functions and vector-valued functions, which motivated the intensive study of this class of spaces in the past decades. Let us say that, as we will see here, the extension to general operators produces other important (and very different from the identity) examples of operators which are spear. One of the most striking one is the Fourier transform on the space on a locally compact Abelian group (see subsection 4.1).
The concepts of numerical range and numerical radius of operators, and the one of numerical index of Banach spaces, played an important role in operator theory, particularly in the study and classification of operator algebras (see the fundamental Bonsall and Duncan books [7, 8], the survey paper [42], and the sections §2.1 and §2.9 of the recent book [14]). For general operators , a concept of numerical range of operators with respect to has been recently introduced [3], and there is a relation between (1.1) and numerical ranges, analogous to the case of (1.2). For instance, a spear operator is geometrically unitary in the strongest possible form. These concepts provide also a natural motivation for the study of spear operators; we will give a short account on this in section 1.4.
The property of being a spear operator is formulated in terms of all bounded linear operators between two Banach spaces, which leads to several difficulties for its study in abstract spaces, and also in some concrete spaces. It would be much more convenient to have a geometric definition of this property (in terms of ), but unfortunately until now a description of this property in pure geometrical terms has not been discovered, even for the case when . In order to manage this difficulty for this case, two other Banach space properties were introduced: the alternative Daugavet Property (aDP for short) and lushness (see [56] and [13], respectively). These two properties are of geometric nature, the aDP is weaker and lushness is stronger than the fact that the identity is a spear operator. On the other hand, in some classes of Banach spaces these properties are equivalent (say, in Asplund spaces, in spaces with the Radon-Nikodým property and, more generally, in SCD spaces introduced in [4]). The study of these two properties has been crucial in the development of the theory in the case when . We will give a short account on this in section 1.3, where we also present the main examples and results about SCD spaces.
So, now naturally appears the task of re-constructing the theory of the aDP and lushness in such a way that it could be applied to spear operators. We will do so here. On this way, we not only transfer the known results to the new setting, but in fact make much more. Namely, we introduce a unified approach to a huge number of previously known results, substantially simplify the system of notations and, in many cases, present the general results for operators in a more clear way than it was done earlier for the identity operator. To do so it has been crucial to study the new concept of target operator (see section 3.3), which plays the same role here that the concept of strong Daugavet operator plays for the study of the Daugavet property [46] and of Daugavet centers [9]. Let us also mention that the aDP, to be a target operator and lushness are separably determined properties, while we do not know whether the concept of spear operator is. This separable determination allows us to use the full power of the theory of SCD sets and operators, a task which will be crucial in the development of the subject.
Finally, another motivation to do this job was the potential applicability of the extended theory to the study of non-linear Lipschitz maps, which would allow to extend some recent results from [67] and [41]. In particular, this will allow to reprove the results of the latter papers in a more reasonable (i.e. linear) way. Namely, the standard technique of Lipschitz-free spaces (see [32] or [68]) reduces equation (1.2) for a non-linear Lipschitz map to an analogous equation for the linearization of , but this linearization acts from the Lipschitz-free space to . Hence, in order to use this technique, we are in need of studying equation (1.1) instead of (1.2), that is, to study spear operators between two different spaces. See section 7 for more details.
1.2. Notation, terminology, and preliminaries
By we denote the scalar field ( or ), and we use the standard notation for its unit sphere. We use the letters , , for Banach spaces over and by subspace we always mean closed subspace. In some cases, we have to distinguish between the real and the complex case, but for most results this difference is insignificant. The closed unit ball and unit sphere of are denoted respectively by and . We denote the Banach space of all bounded linear operators from to by , and write for . The identity operator is denoted by , or if it is necessary to precise the space. The dual space of is denoted by , and denotes the natural isometric inclusion of into its bidual . For and , we write to denote the rank-one operator given by for every . We write and to mean that is, respectively, the -sum and the -sum of and . In the first case, we say that is an -summand of ; in the second case, we say that is an -summand of .
For a subset we write if is bounded and if it is unbounded. Observe that this function has the following properties:
[TABLE]
for every and every subsets , , of . The diameter of a (bounded) set can be calculated as . We will also use the notation
[TABLE]
For a subset and for we write
[TABLE]
A subset of is said to be rounded if .
Given , we denote by the convex hull of , and by its absolutely convex hull, i.e. . We say that is norming for if for every we have that or, equivalently, .
Given and , we put
[TABLE]
and we say that this is a slice of . A face of is a (non-empty) subset of the form
[TABLE]
where is such that its real part attains its supremum on . If and the functional defining the slice or the face is taken in the predual, i.e. , then is called a -slice of and is called a -face of .
Given , and , we define
[TABLE]
and we call it a generalized slice of (observe that it is a union of slices when non-empty). We also define
[TABLE]
and call it a generalized face of . If , then we will simply write (which is a slice of when non-empty). The following easy results about generalized slices will be frequently used.
Remark 1.1*.*
Let be a Banach space, let be a rounded set, let be a set, and let . Then:
- (a)
; 2. (b)
if attains its supremum on , then the set
[TABLE]
coincides with .
Given , we denote by the set of extreme points of . When is convex and it is compact in a locally convex topology, then the set of its extreme points has many good topological properties. Here, we will repeatedly use the following ones.
Lemma 1.2**.**
Let be a Banach space and let convex and weak∗-compact.
- (a)
(Choquet’s Lemma) If , then for every weak∗-neighborhood of in there is a weak∗-slice of such that . In other words, the weak∗-slices of containing form a base of the relative weak∗-neighborhoods of in . 2. (b)
(Milman’s Theorem) If satisfies that , then . 3. (c)
* is a Baire space, so the intersection of every sequence of Gδ dense subsets of is again (Gδ) dense.*
Assertion (a) can be found in [21, p. 107]; (b) is an immediate consequence of (a) and Hahn-Banach separation Theorem; (c) appears in [21, p. 146, Theorem 27.9].
Recall that a Banach space is said to be strictly convex if , and smooth if the mapping is Gâteaux differentiable at every point of (equivalently, for each there is a unique with ). If moreover, the mapping is Fréchet differentiable at every point of , then is said to be Fréchet smooth. A point is said to be strongly extreme if given a sequence in such that , we have that . A point is denting if it belongs to slices of of arbitrarily small diameter. If is a dual space and the slices can be taken to be weak∗-open, then the point is called weak∗-denting. Observe that denting points are strongly extreme and strongly extreme points are extreme points, and none of the implications reverses in general (see [48], for instance). Finally, we recall some common notation for spaces of vector-valued function spaces. Given a compact Hausdorff topological space and a Banach space , is the Banach space of all continuous functions from into endowed with the supremum norm. Given a positive measure space and a Banach space , is the Banach space of all (clases of) measurable functions from into which are essentially bounded, endowed with the essential supremum norm; is the Banach space of all (clases of) Bochner-integrable functions from into , endowed with the integral norm.
1.3. A short account on the results for the identity
Our goal here is to briefly present some results about Banach spaces with numerical index one and the related properties aDP and lushness. For more information and background, we refer the reader to [4, 12, 13, 18, 27, 37, 42, 45, 56] and references therein. Let us start with the main definitions.
Definition 1.3**.**
Let be a Banach space.
- (a)
has numerical index one [27] if is a spear operator, i.e. for every . 2. (b)
has the Daugavet property (DPr in short) [45] if for every rank-one . 3. (c)
has the alternative Daugavet property (aDP in short) [56] if the equality holds for every rank-one . 4. (d)
is lush [13] if for every , every , and every , there exists such that \operatorname{dist}\bigl{(}x_{0},\operatorname{aconv}(\operatorname{gSlice}(S_{X},x^{\ast},\varepsilon))\bigr{)}<\varepsilon.
These four properties are related as follows:
[TABLE]
The first implication appears in [13, Proposition 2.2], while the second and the third ones are obvious. None of them reverses in general (see [40, Remarks 4.2.(a)], [56, Remark 2.4], and [56, Example 3.2]), and the DPr and numerical index one are not related. Examples of lush spaces include, among others, spaces and their isometric preduals (so, in particular, spaces), uniform algebras, and finite codimensional subspaces of (see [12, 18, 42]). They also are the main examples of Banach spaces with numerical index one. On the other hand, and have the DPr (and so the aDP) no matter the range space [45], while they have numerical index one if and only does (see e.g. [42, §1]). This provides many examples of spaces having the aDP but failing to have numerical index one as, for instance, and . Finally, is an example of a space with the aDP, but failing the DPr and not having numerical index one.
One of the milestones of the theory was reached in the 2010 paper [4], where a general condition was given to make properties (a), (c), and (d) equivalent, the SCD property.
Definition 1.4** ([4]).**
Let , be Banach spaces. A bounded subset is said to be slicely countably determined (SCD in short) if there exists a countable family of slices of such that whenever for every . The space is said to be SCD if every convex bounded subset of is SCD. Finally, a bounded linear operator is an SCD operator if is an SCD subset of .
While the definition of SCD set is valid for arbitrary bounded sets, it was only introduced for convex bounded sets in [4] since it is mainly used for the image of the unit ball by bounded linear operators. The next result contains the main examples of (convex) SCD sets, spaces, and operators.
Example 1.5** ([4]).**
Let , be Banach spaces.
- (a)
A convex bounded subset of is SCD provided:
- (a.1)
is separable and has the convex point of continuity property; in particular, is separable and has the Radon-Nikodým property. 2. (a.2)
is separable and it does not contain -sequences; in particular, is separable and Asplund. 2. (b)
The following conditions on imply that every separable subspace of is SCD:
- (b.1)
has the convex point of continuity property; in particular, has the Radon-Nikodým Property. 2. (b.2)
does not contain copies of ; in particular, is Asplund. 3. (c)
If a Banach space has the DPr, then its unit ball is not an SCD set. Actually it is shown in [4, Example 2.13] that given and a sequence of slices of we can find for each so that does not belong to the closed linear hull of . In particular, the unit balls of and are not SCD sets. 4. (d)
The following conditions on an operator guarantee that its restriction to every separable subspace of is SCD:
- (d.1)
does not fix any copy of . 2. (d.2)
has the convex point of continuity property; in particular, has the Radon-Nikodým Property.
The main applications of the SCD property in our context are the following.
Proposition 1.6** ([4, §4]).**
Let be a Banach space. If has the aDP and is SCD for every separable subspace of , then is lush. In particular, if has the aDP and it has the convex point of continuity property, the Radon-Nikodým Property, or it does not contain copies of , then is lush.
Proposition 1.7** ([4, §5]).**
Let be a Banach space with the aDP. For every such that is SCD for every separable subspace of , one has In particular, this happens if has the convex point of continuity property, or the Radon-Nikodým Property, or if does not fix copies of .
Observe that SCD sets are separable, and we do not know whether numerical index one is separably determined. However, the aDP and lushness are, and this is crucial in the way the results above were proved.
Let us finally say that lushness is surprisingly related to the study of Tingley’s problem about extensions of surjective isometries between unit spheres of Banach spaces [65] and to the study of norm attaining operators [20, 47].
1.4. A brushstroke on numerical ranges and numerical indices
Our aim here is to give a short account on numerical ranges which will be very useful for the motivation and better understanding of the concepts of spear vector and spear operator. The study of the numerical range of an operator started with O. Toeplitz’s field of values of a matrix of 1918, a concept which quickly extended to bounded linear operators on Hilbert spaces. An extension of it to elements of unital Banach algebras was used in the 1950’s to relate geometrical and algebraic properties of the unit (that the unit is a vertex of the unit ball of every complex unital Banach algebra), and in the developing of Vidav’s characterization of -algebras. Later on, in the 1960’s, F. Bauer and G. Lumer gave independent but related extensions of Toeplitz’s numerical range to bounded linear operators on arbitrary Banach spaces, which do not use the algebraic structure of the space of operators. All these notions are essential to define and study when an operator on a general Banach space is hermitian, skew-hermitian, dissipative…We refer the reader to the monographs by F. Bonsall and J. Duncan [7, 8] for background. In the 1985 paper [60], an abstract notion of numerical range, which had already appeared implicitly in the 1950’s paper [6], was developed. We refer to sections §2.1 and §2.9 of the very recent book [14] for more information and background.
Given a Banach space and a distinguished element , we define the numerical range and the numerical radius of with respect to as, respectively,
[TABLE]
Then, the numerical index of is
[TABLE]
With this notation, is a vertex if for every (that is, separates the points of ), is a geometrically unitary element if the linear hull of equals the whole space or, equivalently, if (see e.g. [14, Theorem 2.1.17]). By Definition 2.1, is a spear vector if for every , and this is equivalent, by Hahn-Banach Theorem, to . So, spear vectors are geometrically unitary elements (in the strongest possible way!), geometrically unitary elements are vertices, and vertices are extreme points. None of this implications reverses (see [14, §2.1]). Let us also comment that the celebrated Bohnenblust-Karlin Theorem [6] states that algebraic unitary elements of a unital complex Banach algebra (i.e. elements such that and have norm one) are geometrically unitary, see [61] for a detailed account on this.
We then have an easy way to define the numerical range of an operator: given a Banach space and , the algebra numerical range or intrinsic numerical range of is just . As in order to study this concept we have to deal with the (wild) dual of , there are other concepts of numerical range which simplify such a task. The (Bauer) spatial numerical range of is defined as
[TABLE]
It is a classical result that the two numerical ranges are related as follows:
[TABLE]
(see e.g. [14, Proposition 2.1.31]), so they produce the same numerical radius of operators.
Let now , be Banach spaces and let us deal with numerical ranges with respect to with . First, the intrinsic numerical range of with respect to is easy to define: just consider
[TABLE]
and so we have the corresponding numerical radius and numerical index . But this definition forces us to deal with the dual of , which is not a nice task. On the other hand, the possible extension of the definition of spatial numerical range has many problems as, for instance, it is empty if does not attain its norm; moreover, even in the case when is an isometric embedding, it does not have a good behaviour, see [55]. Very recently, a new notion has appeared [3]: the approximated spatial numerical range of with respect to is defined by
[TABLE]
We then have the corresponding numerical radius and numerical index:
[TABLE]
The relationship between these two numerical ranges is analogous to the one for the identity operator [54, Theorem 2.1]:
[TABLE]
for every norm-one and every . Therefore, both concepts produce the same numerical radius of operators and so, the same numerical index of , the same concepts of vertex and geometrically unitary elements. In particular, both numerical ranges produce the same concept of spear operator: is a spear operator if and only if N\bigl{(}L(X,Y),G\bigr{)}=1 if and only if .
1.5. The structure of the manuscript
The main part of this manuscript is divided in seven chapters. In chapter 2 we recall the concept of spear vector and introduce the new concept of spear set. These concepts are used here as “leitmotiv” to give a unified presentation of the concepts of spear operator, lush operator, alternative Daugavet property, and other notions that we will introduce here for operators. We collect some properties of spear sets and vectors, together with some examples of spear vectors.
Chapter 3 includes the main definitions of the manuscript for operators: spearness, the alternative Daugavet property and lushness. We start presenting some preliminary results and easy examples of spear operators in section 3.1. Next, in section 3.2 we study operators with the alternative Daugavet Property (aDP). These are operators satisfying for every rank-one operator . This definition is a generalization of the aDP, which is analogous to the generalization of the DPr given by Daugavet centers [10]: is a Daugavet center if for every rank-one operator . Of course, Daugavet centers have the aDP, but the converse result is not true. There is a clear parallelism between the study of the aDP and of Daugavet centers. We give several characterizations of operators with the aDP (some of them in terms of spear sets) and prove that this is a separably determined property. Section 3.3 starts with the definition of target operator for . This property guarantees that an operator that has it satisfies . Interestingly, if has the aDP and the operator is SCD, then is a target for , and this will be frequently used to deduce important results. Our new concept of target operator naturally plays an analogous role that the one played by strong Daugavet operators in the study of the DPr [46] and in the study of Daugavet centers [9]. Let us say that even for the case , this concept is new and provides with non trivial new results. We characterize target operators for a given operator , show that this property is separably determined, and prove that if has the aDP, then every operator whose restriction to separable subspaces is SCD is a target for . In section 3.4, we introduce the notion of lush operator, which generalizes the concept of lush space. This generalization is closely connected with target operators from the previous section, which, on the one hand, reduces some results about lush spaces to results from the previous section, and on the other hand, gives more motivation for the study of target operators. We give several characterizations of lush operators, prove that this property is separably determined, show that the aDP and lushness are equivalent when every separable subspace of the domain space is SCD (so, for instance, when the domain is Asplund, has the Radon-Nikodým Property, or does not contain copies of ), and present some sufficient conditions for lushness which will be used in the chapter about examples and applications. Besides, we prove that lush operators with separable domain fulfill a stronger version of lushness which has to do with spear functionals.
Chapter 4 is devoted to present some examples in classical Banach spaces. Among other results, we show that the Fourier transform is lush, we characterize operators from spaces which have the aDP, and we study lushness, spearness and the aDP for operators which arrive to spaces of continuous functions. In particular, we show that every uniform algebra isometrically embeds by a lush operator into the space of bounded continuous functions on a completely regular Hausdorff topological space (for unital algebras, this space is just its Choquet boundary).
Next, we devote chapter 5 to provide further results on our properties. We characterize lush operators when the domain space has the Radon-Nikodým Property or the codomain space is Asplund, and we get better results when the domain or the codomain is finite-dimensional or when the operator has rank-one. Further, we study the behaviour of lushness, spearness and the aDP with respect to the operation of taking adjoint operators; in particular, we show that these properties pass from an operator to its adjoint if the domain has the Radon-Nikodým Property or the codomain is -embedded; we also show that the aDP and spearness pass from an operator to its adjoint when the codomain is -embedded.
In chapter 6 we provide with some isomorphic and isometric consequences of the properties as, among others, that the dual of the domain of an operator with the aDP and infinite rank contains in the real case. Many results showing that the aDP, spearness and lushness do not combine well with rotundity or smoothness properties are also presented.
We study Lipschitz spear operators in chapter 7. These are just the spear vectors of the space of Lipschitz operators between two Banach spaces endowed with the Lipschitz norm. The main result here is that every (linear) lush operator is a Lipschitz spear operator, a result which can be applied, for instance, to the Fourier transform. We also provide with an analogous result for aDP operators and for Daugavet centers.
Finally, a collection of stability results for our properties is given in chapter 8. We include results for various operations like absolute sums, vector-valued function spaces, and ultraproducts. The results we got are in most cases extensions of previously known results for the case of the identity.
We complement the manuscript with a collection of open problems in chapter 9.
We finish the introduction with a diagram about the relationships between the properties of a norm-one operator that we have presented in this introduction:
[TABLE]
None of the implications above reverses, and Daugavet centers and spear operators do not imply each other.
Chapter 2 Spear vectors and spear sets
The following definition will be crucial in our further discussion.
Definition 2.1** ([3, Definition 4.1]).**
Let be a Banach space. An element is a spear (or spear vector) if for every . We write to denote the set of all elements of a Banach space which are spear.
As we commented in section 1.4, this is equivalent to the fact that . In particular, the definition was motivated in [3] by the fact that is a spear element of if and only if has numerical index one. Let us also comment that the concept of spear vector appeared, without name, in the paper [49] by Å. Lima about intersection properties of balls. It had also appeared tangentially in the monograph [50] by J. Lindenstrauss about extension of compact operators.
Remark 2.2*.*
Observe that using a standard convexity argument, is a spear vector if and only if for every .
The next notion extends the definition of spear from vectors to sets.
Definition 2.3**.**
Let be a Banach space. is called a spear set if for every .
Observe that if is a spear set, then every subset of containing is also a spear set. In particular, if a subset of contains a spear vector, then is a spear set. On the other hand, it is not true that every spear set contains a spear vector: in every Banach space , is obviously a spear set, but there are Banach spaces containing no spear vectors at all (for instance, a two-dimensional Hilbert space, see Example 2.11.(h)).
We start the exposition with the following fundamental result.
Theorem 2.4**.**
Let be a Banach space, let be a subset of and let with . The following statements are equivalent:
- (i)
* for every .* 2. (ii)
* for every .* 3. (iii)
* for every .* 4. (iv)
* is a dense subset of .*
If is a dual Banach space, this is also equivalent to
- (v)
* for every .*
Moreover, remark that the set is Gδ which makes item (iv) more applicable.
* Proof (ii): Given , we just have to check that every -slice of intersects . We can assume that for some and . Using (i) and the condition on , we can find and such that*
[TABLE]
In particular, and , so .
(ii) (i): Given and , the hypothesis allows us to find such that . Also, by definition of , there is such that . Now,
[TABLE]
and the arbitrariness of gives the result.
The equivalence between (i) and (iii) is just a particular case of the already proved equivalence between (i) and (ii) since satisfies the condition above by the Krein-Milman Theorem.
(iii) (iv): For each , is a relatively weak∗-open subset of , as it can be written as union of weak∗-slices. Moreover, condition (iv) together with Milman’s Theorem (see Lemma 1.2.(b)) yields that the set is weak∗-dense in . Using that the set is a Baire space (see Lemma 1.2.(c)), we conclude that
[TABLE]
satisfies the properties above.
(iv) (iii): Given , since we deduce that this last set is also dense in , and so using the Krein-Milman Theorem we conclude that
[TABLE]
Finally, if , (v) is a particular case of (ii) with by Goldstine’s Theorem.
*The “moreover” part follows from equation (2.1). *
Now we may present a characterization of spear sets which is an easy consequence of the above theorem and the fact that for every set and every vector .
Corollary 2.5**.**
Let be a Banach space and let with . For , the following assertions are equivalent:
- (i)
* is a spear set, i.e. for each .* 2. (ii)
* for every .* 3. (iii)
* is a dense Gδ subset of .*
If is a dual Banach space, this is also equivalent to
- (iv)
* for every .*
The following result is of interest in the complex case.
Proposition 2.6**.**
Let be a Banach space. If is a spear set, then for every .
[* Proof]Let and . Using Corollary 2.5.(ii), we get that*
[TABLE]
The case in which a spear set is a singleton coincides, of course, with the concept of spear vector of Definition 2.1. Most of the assertions of the next corollary follow from Corollary 2.5 and the fact that is -closed. The other ones are consequences of the general theory of numerical range spaces (see section 1.4) and can be found in [14, §2.1].
Corollary 2.7**.**
Let be a Banach space and let with . The following assertions are equivalent for :
- (i)
* (that is, for every ).* 2. (ii)
* for each .* 3. (iii)R
If is a real space, B_{X^{\ast}}=\operatorname{conv}\bigl{(}\operatorname{Face}(S_{X^{\ast}},z)\cup-\operatorname{Face}(S_{X^{\ast}},z)\bigr{)}. 4. (iii)C
If is a complex space, \operatorname{int}(B_{X^{\ast}})\subset\operatorname{aconv}\bigl{(}\operatorname{Face}(S_{X^{\ast}},z)\bigr{)} so, in particular, B_{X^{\ast}}=\overline{\operatorname{aconv}}\bigl{(}\operatorname{Face}(S_{X^{\ast}},z)\bigr{)}. 5. (iv)
* for every .*
If is a dual Banach space and , this is also equivalent to:
- (v)
B_{Y}=\overline{\operatorname{aconv}}{\bigl{(}\operatorname{Slice}(S_{Y},y^{\ast},\varepsilon)\bigr{)}}* for every .*
[* Proof]The equivalence between (i), (ii) and (iv) is just particular case of Corollary 2.5, as it is the equivalence with (v) when is a dual space.*
(i) (iii) is contained in [14, Theorem 2.1.17] (both in the real and in the complex case), but we give the easy argument here. Recall that (i) is equivalent to the fact that for every or, equivalently, is norming for or, equivalently,
[TABLE]
In the real case, we have that the set
[TABLE]
is weak∗-compact as so is , and the result follows from (2.2). In the complex case, for we take such that , where are the th roots of in . Then we have
[TABLE]
Since is weak∗-compact and is contained in the set , it follows from (2.2) that , and this gives the result moving .
*The implication (iii) (i) follows immediately from (ii) (i) for the particular case . *
The next surprising result about spear vectors of a dual space appeared literally in [2, Corollary 3.5] and it is also consequence of the earlier [31, Theorem 2.3] using Corollary 2.7.(iii). In both cases, the main tool is the use of norm-to-weak upper semicontinuity of the duality and pre-duality mappings. We include here an adaptation of the proof of [31, Theorem 2.3] to our particular situation which avoids the use of semicontinuities (which, on the other hand, are automatic in our context, see [14, Fact 2.9.3 and Theorem 2.9.18]).
Theorem 2.8** **([31, Theorem 2.3],
[2, Corollary 3.5]).
Let be a Banach space and let . Then, if and only if .
[* Proof]The “if” part follows immediately from Corollary 2.7.(v), so we just have to prove the “only if” part. To simplify, we will denote and .*
Claim 1*. If satisfies , then . Indeed, let and with . Since by Corollary 2.7.(iii), we can find and with , and such that*
[TABLE]
Therefore,
[TABLE]
Since was arbitrary, the claim is proved.
Claim 2*. Given with we have that the set satisfies . Indeed, let with . Using the Principle of Local Reflexivity [28, Theorem 6.3], we have that for each we can find an element such that , , and .*
Claim 3*. (in particular, ) and if satisfies , then . Indeed, let and fix . By Claim 1, we have that , so taking with we have that satisfies . Repeating the same process with , we can find such that satisfies . Iterating this process, we will have a Cauchy sequence whose limit satisfies that and that , so .*
Claim 4*. . Indeed, given a slice of , Corollary 2.7.(v) shows that , so Claim 3 provides that . It is now routine to show that for every slice of . *
The following proposition collects all the properties of spear vectors we know. In order to prove it we need the following technical lemma.
Lemma 2.9**.**
Let be a Banach space and let be a decreasing sequence of spear sets of such that tends to zero. If , then is a spear element.
[* Proof]For every we can write*
[TABLE]
*But the hypothesis implies that . *
Proposition 2.10**.**
Let be a Banach space. Then:
- (a)
* for each and every .* 2. (b)
Every is a strongly extreme point of . In particular, . 3. (c)
J_{X}\bigl{(}\operatorname{Spear}(X)\bigr{)}\subset\operatorname{Spear}(X^{\ast\ast}). In particular, J_{X}\bigl{(}\operatorname{Spear}(X)\bigr{)}\subset\operatorname{ext}B_{X^{\ast\ast}}. 4. (d)
* is norm-closed.* 5. (e)
If , then is nowhere-dense in . 6. (f)
If then . 7. (g)
If , then is a spear element if and only if is a strongly extreme point of if and only if . 8. (h)
If is strictly convex and , then . 9. (i)
If is smooth and , then .
If is a real space, we can add:
- (j)
If is infinite, then contains a copy of or . 2. (k)
If and is norm-attaining with , then is norm-attaining and . 3. (l)
If is smooth and , then .
[* Proof of Proposition 2.10] Statement (a) is given by Proposition 2.6, and (b) is an obvious consequence of it.*
(c). Fixed , we have that B_{X^{\ast}}=\overline{\operatorname{aconv}}^{\|\cdot\|}\bigl{(}\operatorname{Face}(S_{X^{\ast}},z)\bigr{)} by Corollary 2.7.(iii) so, using Goldstine’s Theorem for , we obtain that
[TABLE]
and, a fortiori,
[TABLE]
Now, Milman’s Theorem (see Lemma 1.2.(b)) gives that
[TABLE]
Then, Corollary 2.7.(iv) gives that . An alternative proof is the following: for we have that for every by Corollary 2.7.(iv), and so for every by [63, Proposition 3.5], so by using again Corollary 2.7.(iv).
(d). Given a norm-convergent sequence in , apply Lemma 2.9 to the family of sets .
(e). Fixed , take an element . Given , there exists such that and so, by convexity, for every . Then, for each , belongs to , satisfies , and , so is not a spear by Corollary 2.7.(iv).
(f). Fix and let . The hypothesis implies that for every we can find with . Since (as is a spear) we conclude that . This gives that and the equality follows from (b).
(g). If we assume that , then for each by Corollary 2.7.(iv). But Milman’s theorem (see Lemma 1.2.(c)) applied to our assumption gives that is weak∗-dense in , so we conclude that for every , which implies that is a spear by Corollary 2.7.(iv).
(h). If , we have that for every . If is strictly convex, this implies that and so . Next, suppose that there exists , so by Theorem 2.8. If then contains at least two points, so is not strictly convex.
(i). If is smooth, the set is a singleton for every . If there is , then the above observation and Corollary 2.7.(iii) imply that is one-dimensional, so is one-dimensional as well.
(j). Is just a reformulation of [52, Proposition 2], but we include the short argument for completeness. Suppose that does not contain . Then, by Rosenthal’s -Theorem [25, Chapter XI] there is a weakly Cauchy sequence of distinct members of . Write for the closed linear span of and observe that, obviously, for every . Therefore, by Corollary 2.7.(iv), the fact that the sequence is weakly Cauchy and that we are in the real case, we have that \operatorname{ext}(B_{Y^{\ast}})=\bigcup\nolimits_{n\in\mathbb{N}}\bigl{(}E_{n}\cup-E_{n}\bigr{)} where
[TABLE]
As separates the points of , each is finite, so must be countable. Fonf’s Theorem [29] gives us that , finishing the proof.
The proof of (k) is based on ideas of [49]. Let . By Corollary 2.7.(iii), we have that . Let attain its norm at , i.e. , and suppose that . We can write for some and . If we assume that , then necessarily, which is not possible as . Therefore, and we get that , so .
*(l). Suppose that there exists . By (j), we can take a norm-attaining functional with , such that is norm-attaining and . Hence there is with , which means that is not smooth. *
Below we list the known examples of spear vectors from [3] and some easy-to-check generalizations of those examples.
Example 2.11**.**
- (a)
[3, p. 170] If is an arbitrary set, then
[TABLE]
where is the function on with value one at and zero on the rest. The proof is straightforward. 2. (b)
[3, p. 170] Let be a measure space. The spear vectors of the space are functions of the form where and is an atom. That is, spear vectors coincide with extreme points of the unit ball. The proof of this result is straightforward. 3. (c)
[3, p. 170] If is a Hausdorff compact space, then
[TABLE]
That is, again spear vectors coincide with extreme points of the unit ball. Again, the proof is elementary. 4. (d)
With the same ideas of the example above, one can even show that if is a Banach space, then for one has that
[TABLE] 5. (e)
Let be a measure space and let . Then,
[TABLE]
Indeed, suppose that and there exists a measurable subset with such that for every . Then we have that
[TABLE]
so there exists such that the measurable set has positive measure. Now,
[TABLE]
and thus is not a spear vector, a contradiction. Conversely, suppose that for -almost every . For and , there is with such that for every . By the hypothesis, there is , , such that for every . Now, using the compactness of we can give a lower bound for . Indeed, fixed an -net of we can find an element and a subset of with positive measure such that for every . Therefore, we can write
[TABLE]
and the arbitrariness of gives . 6. (f)
It will be proved in Corollary 4.23 that the vector valued case of (e) is also valid. Let be a measure space, let be a Banach space and let . Then if and only if for -almost every . Actually, the proof of the “if” part is just an straightforward adaptation of the corresponding one for (e). 7. (g)
It is straightforward to show that if the linear span of is an -summand of , then . Observe that this is what happens in examples (a) and (b) above. On the other hand, the converse result does not hold. Indeed, just note that , where is the Cantor set, contains no proper -summand by Behrends - Theorem (see [33, Theorem I.1.8] for instance). 8. (h)
The space contains no spear vector if and (use Proposition 2.10.(h), for instance). 9. (i)
Let , be Banach spaces and let . Then, if, and only if, and . The proof is straightforward. 10. (j)
Let , be Banach spaces and let . Then, if, and only if, either and or and . The proof is again straightforward.
As an application of examples (b) and (e) above and Theorem 2.8, we get easily the following well-known old result (see [58] for an exposition which also covers the complex case).
Corollary 2.12**.**
Let be a Banach space such that either or is isometrically isomorphic to an space. Then, for every .
Chapter 3 Three definitions for operators: spearness, the alternative Daugavet property, and lushness
This is the main chapter of our manuscript, as we introduce and deeply study the main definitions: the one of spear operator, the weaker of operator with the alternative Daugavet property and the stronger lush operator.
3.1. A first contact with spear operators
Even though it has been given in the introduction, we formally state the definition of spear operator as it is the main concept of the manuscript.
Definition 3.1**.**
Let , be Banach spaces and let be a norm-one operator. We say that is a spear operator if the norm equality
[TABLE]
holds for every , that is, if G\in\operatorname{Spear}\bigl{(}L(X,Y)\bigr{)}.
We would like now to list some of the equivalent reformulations of the concept of spear operator which one can get particularizing the results of the previous chapter. We will also include a characterization in terms of numerical ranges that comes from section 1.4.
Proposition 3.2**.**
Let , be Banach spaces and let be a norm-one operator. The following assertions are equivalent:
- (i)
* is a spear operator, i.e. for every .* 2. (ii)
* for every \zeta\in\operatorname{ext}\bigl{(}B_{L(X,Y)^{\ast}}\bigr{)}.* 3. (iii)
Given ,
[TABLE]
for every .
* Proof and (ii) are equivalent by Corollary 2.7, and (iii) (i) is immediate, so only (i) (iii) needs an explanation. For and , write*
[TABLE]
Fix now and observe that
[TABLE]
*On the other hand, is a spear operator if and only if or, equivalently, if , that is, if for every , and so all the inequalities above become equalities. *
We next present some examples of spear operators which may help to better understand the definition and see how far from the Identity a spear operator can be. The first family appeared in [3, Theorem 4.2].
Proposition 3.3**.**
Let be an arbitrary set, let , be Banach spaces, and let be the canonical basis of (as defined in Example 2.11.(a)).
- (a)
* is a spear operator if and only if for each .* 2. (b)
* is a spear operator if and only if for every .*
[* Proof]As indicated, (a) appears in [3, Theorem 4.2] and the proof is elementary. Let us prove (b). The sufficiency of the condition is given by the obvious fact that is a spear operator when is (as taking adjoint preserves the norm) and the result in (a). For the necessity, suppose that there is such that and find such that . We then consider the norm-one operator given by and if for every , and observe that and for . Therefore,*
[TABLE]
*so is not a spear operator. *
This result will be improved in Example 5.5. More involved examples of spear operators, lush operators, operators with the aDP…will appear in chapters 4, 5, and 7.
The following observations follow straightforwardly from the definition of spear operator.
Remark 3.4*.*
Let be Banach spaces and let .
- (i)
Composing with isometric isomorphisms preserves spearness: Let , be Banach spaces, and let and be isometric isomorphisms. Then G\in\operatorname{Spear}\bigl{(}L(X,Y)\bigr{)} if and only if \Phi_{2}G\Phi_{1}\in\operatorname{Spear}\bigl{(}L(X_{1},Y_{1})\bigr{)}. 2. (ii)
We may restrict the codomain of a spear operator keeping the property of being spear operator: If is a spear operator and is a subspace of containing , then is a spear operator. On the other hand, the extension of the codomain does not preserve spears: the map is not a spear operator. 3. (iii)
As an easy consequence of (i) and (ii), we get that the following assertions are equivalent: (a) has numerical index one (i.e. is a spear), (b) there exists a Banach space and an isometric isomorphism which is a spear in or , (c) there exists a Banach space and an isometric embedding of into which is a spear operator.
3.2. Alternative Daugavet Property
We start presenting the definition of the alternative Daugavet property for an operator, which extends the analogous definition for a Banach space (through the Identity).
Definition 3.5**.**
Let , be Banach spaces. We say that has the alternative Daugavet property (aDP in short), if the norm equality
[TABLE]
holds for every rank-one operator .
Substituting in (aDE) we deduce that if has the aDP then .
The following fundamental result characterizes the aDP of an operator in terms of the behaviour of the operator with respect to slices, spear sets…
Theorem 3.6**.**
Let be a norm-one operator between two Banach spaces , , let with and let with . The following assertions are equivalent:
- (i)
* has the aDP.* 2. (ii)
* is a spear set for every slice of .* 3. (ii*∗*)
* is a spear set for every weak**∗**-slice of .* 4. (iii)
For every and
[TABLE] 5. (iv)
For every , the set
[TABLE]
is a dense Gδ set in .
* Proof (ii): Let be a slice of where and . Given any consider the rank-one operator (i.e. for every ) which satisfies that . Since , for every there exists such that*
[TABLE]
Making a rotation of if necessary, we can assume that . Using the hypothesis on , we deduce from (3.1) the existence of some satisfying
[TABLE]
Since , we deduce that . Hence and so
[TABLE]
(ii) (iii): Given , and a slice of , since is a spear set, we can find with , which means that every slice of intersects
[TABLE]
Therefore
[TABLE]
(iii) (i): Let be a rank-one operator. By a convexity argument, we may and do suppose that . Then, it is of the form for some and . Given , the hypothesis implies that intersects , so there exists and such that . Hence
[TABLE]
(ii) (iv): We may and do suppose that . Given , the set G\bigl{(}\operatorname{Slice}(\mathcal{B},x_{0}^{\ast},\varepsilon)\bigr{)} is a spear of by hypothesis, which by Corollary 2.5.(iii) means that the set \operatorname{gFace}\bigl{(}\operatorname{ext}B_{Y^{\ast}},\mathbb{T}G\bigl{(}\operatorname{Slice}(\mathcal{B},x_{0}^{\ast},\varepsilon)\bigr{)}\bigr{)} is a dense Gδ set in the Baire space . Hence,
[TABLE]
is also dense in , and it is easy to check that
[TABLE]
(iv) (ii∗): Fix . If is any weak∗-slice of , then is a non-empty open subset of . The hypothesis implies that the slice contains an element of , so .
(ii∗) (i): Let , where and , be an arbitrary rank-one operator. Given any put . Notice that for every ,
[TABLE]
so using that is a spear we deduce that
[TABLE]
The next result shows that the aDP is separably determined, and will be very useful in the next section where we deal with SCD operators.
Proposition 3.7**.**
Let , be Banach spaces and let . Then, has the aDP if and only if for every separable subspaces and , there exist separable subspaces , satisfying and and such that and has the aDP.
[* Proof]Suppose first that has the aDP. Pick a sequence of with and consider and , both separable subspaces. By Theorem 3.6.(iii) we have that*
[TABLE]
But since and are separable, it is easy to deduce the existence of a countable set such that
[TABLE]
Define then and , which are again separable. Repeating the same process as above, we can construct an increasing sequence of closed separable subspaces and such that
[TABLE]
This implies that and , satisfy that
[TABLE]
which means that has the aDP by using again Theorem 3.6.(iii).
Conversely, take a non-null rank-one operator , consider a separable subspace such that and , and write . By hypothesis, there are separable subspaces and such that has norm one and has the aDP. As is rank-one and , it follows that and . Then we may apply that has the aDP to get that . But, clearly,
[TABLE]
*and the reverse inequality is always true, so has the aDP. *
As we did for spear operators, we may directly deduce from the definition the following three elementary results about operators with the aDP.
Remark 3.8*.*
Let be Banach spaces and .
- (i)
The composition with isometric isomorphisms preserves the aDP: If , are Banach spaces and , are isometric isomorphisms, then, has the aDP if and only if has the aDP. 2. (ii)
If has the aDP and is a subspace of containing , then has the aDP. However, the property of aDP is not preserved by extending the codomain of the operator, as the same example of Remark 3.4 shows. 3. (iii)
As an easy consequence of (i) and (ii), we have that the following statements are equivalent: (a) has the aDP, (b) there exist a Banach space and an isometric isomorphism in or in which has the aDP, (d) there exist a Banach space and an isometric embedding which has the aDP.
3.3. Target operators
Our goal in this section is to present and study the concept of target operator, which will be the key in the next section to relate the aDP and lushness so, in particular, to relate the aDP and spear operators. As far as we know, this is a new concept even in the particular case in which is the identity operator of a Banach space.
Definition 3.9**.**
Let , , be Banach spaces and let be a norm-one operator. We say that is a target for if each has the following property:
[TABLE]
Remark that if satisfies ( ‣ 3.9) then there is a finite subset of satisfying the same condition.
At the end of the section we will include a result characterizing spear vectors in terms of target operators which will allow to better understand this definition, see Proposition 3.22. Next, we provide with several characterizations of this kind of operators which will be very useful in the sequel.
Proposition 3.10**.**
Let , , be Banach spaces, let with , let , and let with . Given , the following assertions are equivalent:
- (i)
* satisfies ( ‣ 3.9).* 2. (ii)
For every and there is such that
[TABLE] 3. (iii)
For every , the set
[TABLE]
intersects every -slice of . 4. (iv)
The set
[TABLE]
is a dense (Gδ) subset of .
* Proof (ii): Let and . Fixed such that , by ( ‣ 3.9) in Definition 3.9, we can find , with , and such that*
[TABLE]
Let be such that
[TABLE]
Then and, moreover, satisfies
[TABLE]
Hence, we get that
[TABLE]
Using the right-hand side inequality of (3.2) and (3.3), we can write
[TABLE]
(ii) (iii): statement (ii) claims that intersects for every and every . Since for every , contains for every , we conclude the result.
(iii) (iv): Using Lemma 1.2.(a), we have that intersects every -open subset of . In other words, is a dense subset of . Since
[TABLE]
we just have to show that is open and then apply Lemma 1.2 (c). Indeed, notice that for every we can find a finite subset of such that . The set is a relatively weak*∗**-open subset of , which contains by definition. Also , since for every . So, is open as desired.*
(iv) (i): Given and , intersects . Taking an element in such intersection, by definition of , we can find a finite set in such that
[TABLE]
But the condition yields that every satisfies
[TABLE]
One of the main applications of target operators is the following result.
Proposition 3.11**.**
Let , be Banach spaces and let be a norm-one operator. If is a target for , then
[TABLE]
[* Proof]We can assume that . For , take with and write . By Proposition 3.10, there exists with*
[TABLE]
We can then find and elements , , for such that
[TABLE]
Since , a standard convexity argument leads to the existence of some with . Therefore
[TABLE]
The following observations can be proved directly from the definition of target, but they follow easier from Proposition 3.10.
Remark 3.12*.*
Let , , , , be Banach spaces and let be a norm-one operator.
- (a)
If is a target for , then is a target for for every . 2. (b)
Let and be operators such that for every . If is a target for , then so is .
Target operators are separably determined, and this fact will be crucial in the study of the relationship with the aDP and SCD.
Theorem 3.13**.**
Let , , be Banach spaces, let be a norm-one operator and consider . Then, is a target for if and only if for every separable subspaces and , there exist separable subspaces , satisfying and and such that has norm one and is a target for .
[* Proof]Let us assume first that is a target for .*
Claim:* given separable subspaces and , we can find a countable set with the following property (P): given , and , there exists with and .*
Indeed, fixing and countable dense subsets of and respectively, we can apply the definition of target operator to construct a countable set satisfying the property (P) for all , and . But using the density of and , it turns out that has the same property for each , and .
Now we prove the theorem: We can assume that . Put and , both separable Banach spaces. Using the claim, we deduce the existence of a countable set with the property (P) for and . Define and . Repeating this process inductively, we construct increasing sequences of closed separable subspaces and such that has the property (P) above for and . Taking and , we conclude the result, as and satisfy the definition of target operator by construction.
*Let us check the converse implication. Given and we can find separable subspaces and with the properties above, so applying the definition of target for , and the previous elements we get the result. *
We need one more ingredient to be able to present the main result about the relationship between target operators and SCD operators.
Proposition 3.14**.**
Let , , be Banach spaces and let be a norm-one operator. Let be an operator such that the set
[TABLE]
is dense in . Then, is a target for , and in the case of , we have .
[* Proof]In the notation of Proposition 3.10, the inclusion holds for every , so by Proposition 3.10.(iv), this means that is a target for . If , an application of Proposition 3.11 implies that . *
The converse of the above result holds for operators with separable image.
Proposition 3.15**.**
Let , , be Banach spaces and let be a norm-one operator. Suppose that is a target for such that is separable. Then,
[TABLE]
is a dense Gδ subset of .
[* Proof]Since is a target for , for every we have by Proposition 3.10 that the set*
[TABLE]
*is a dense Gδ subset of . If we choose a sequence in so that is dense in , then . Since all are dense subsets of , so is (see Lemma 1.2.(c)). *
Theorem 3.16**.**
Let , , , be Banach spaces and let be an operator with the aDP. If for there is an SCD operator such that for every , then is a target for . In the case of , we have .
[* Proof]By Remark 3.12 we may assume that is an SCD operator and that . Let \bigl{\{}\widehat{S}_{n}\colon n\in\mathbb{N}\bigr{\}} be a determining family of slices for . Then, is a slice of for each . Since has the aDP, Theorem 3.6.(ii) tells us that is a spear set for every , which implies that is a dense set in by Corollary 2.5.(iii). As is a Baire space (see Lemma 1.2.(c)), we deduce that the intersection is weak∗-dense in . Observe that, by Proposition 3.14, it suffices to show that this intersection is contained in . Given belonging to this intersection, we have that for every and , . Therefore, , and so . Using that the family \bigl{\{}\widehat{S}_{n}\colon n\in\mathbb{N}\bigr{\}} is determining for , we conclude that*
[TABLE]
*and, therefore, . *
As a consequence of the previous results we may present a class of operators which is a two-sided operator ideal consisting of operators satisfying the condition whenever has the aDP. Let us recall the needed definitions which we borrow from [4] and [44]. Let , be Banach spaces, an operator is hereditarily SCD if is a hereditarily SCD set, that is, if every convex subset of is SCD. Obviously, hereditarily SCD operators are SCD. The operator is HSCD-majorized if there is a Banach space and a hereditarily SCD operator such that for every . It is shown in [44, Theorem 3.1] that the class of HSCD-majorized operators is a two sided operator ideal. By Example 1.5, this ideal contains those operators with separable range such that the image of the unit ball has the Radon-Nikodým Property, or the convex point of continuity property, or it is an Asplund set, and those operators with separable rank which do not fix copies of .
Corollary 3.17**.**
Let , , be Banach spaces and let be an operator with the aDP. If is a HSCD-majorized operator then is a target for . In the case of , we have .
Corollary 3.18**.**
Let , be Banach spaces and let be an operator with the aDP. The class of operators satisfying contains the component in of the two-sided operator ideal formed by HSCD-majorized operators.
Even in the case when , the result above was unknown.
We can extend Theorem 3.16 to the non-separable setting in the following way.
Proposition 3.19**.**
Let , , be Banach spaces and let . If has the aDP and satisfies that is an SCD set for every separable subspace of , then is a target for . Therefore, if then .
[* Proof]Our aim is to use Theorem 3.13 to deduce that is a target for . To do so let us fix separable subspaces and . By Proposition 3.7 we can find separable subspaces and such that and has norm one and the aDP. Now, as is SCD, Theorem 3.16 tells us that is a target for and we can apply Theorem 3.13 to get that is a target for . *
Using the known results about SCD sets (which were commented in section 1.3), we may provide with the following consequence.
Corollary 3.20**.**
Let , , be Banach spaces and let . Suppose that has the aDP and satisfies that has one of the following properties: Radon-Nikodým Property, Asplund Property, convex point of continuity property or absence of -sequences. Then, is a target for . Therefore, if then .
[* Proof]In [4, §5] (see Examples 1.5) it is shown that any of the previous properties implies that the requirements of Proposition 3.19 are satisfied. *
To finish the discussion about which operators are targets for a given aDP operator, we may also extend Corollary 3.18 to operators with non separable range as follows. Given two Banach spaces and , consider the class of those operators such that for every separable subspace of , is HSCD-majorized. As a consequence of the cited result [44, Theorem 3.1], this class is a two sided operator ideal. Therefore, extending straightforwardly the proof of Proposition 3.19, we get the following result.
Corollary 3.21**.**
Let , be Banach spaces and let be an operator with the aDP. The class of operators satisfying contains the component in of the two-sided operator ideal of those operators such that their restrictions to separable subspaces are HSCD-majorized. Moreover, this ideal contains those operators for which the image of the unit ball has one of the following properties: Radon-Nikodým Property, Asplund Property, convex point of continuity property, or absence of -sequences.
We finish this section with two results concerning elements satisfying property ( ‣ 3.9) in Definition 3.9.
Proposition 3.22**.**
Let be a Banach space and let . Then, is a spear vector if and only if belongs to and satisfies ( ‣ 3.9) with .
[* Proof]Using Proposition 3.10, has property ( ‣ 3.9) for if and only if*
[TABLE]
*is dense in . If is a spear, then by Proposition 2.10.(b). Moreover, the definition of spear yields that for each , and so . Let us see the converse. If then for each and , we have that is an extreme point of \overline{\operatorname{aconv}}^{\sigma(X^{\ast\ast},X^{\ast})}\bigl{(}\operatorname{gSlice}(S_{X},x^{\ast},\varepsilon)\bigr{)}. By Milman’s Theorem (see Lemma 1.2.(b)), we deduce that , and so . Since is arbitrary and is weak∗-dense in , we conclude that for each , which means that is a spear by Corollary 2.7.(iv). *
Proposition 3.23**.**
Let , , be Banach spaces and let be a norm-one operator. Given , the set of points satisfying ( ‣ 3.9) in Definition 3.9 is absolutely convex and closed.
[* Proof]Denote by the set of points satisfying ( ‣ 3.9) for and . Fixed , ,and , there is a finite subset with . By Proposition 3.10.(iv), we have that is a dense Gδ subset of . Take any . Then, belongs to \overline{T\bigl{(}\operatorname{aconv}{\operatorname{gSlice}(S_{X},G^{\ast}y^{\ast},\varepsilon)}\bigr{)}} for each , and so*
[TABLE]
*A straightforward normalization gives that satisfies ( ‣ 3.9) for and , so , as desired. *
3.4. Lush operators
We start with the definition of lush operator, which generalizes the concept of lush space when applied to the Identity.
Definition 3.24**.**
Let , be Banach spaces and let be a norm-one operator. We say that is lush if is a target for .
From the definition of target (or better from Remark 3.12.(b)), it follows immediately the following observation.
Remark 3.25*.*
is lush if and only if every operator whose domain is is a target for . In particular, every lush is a spear operator, that is,
[TABLE]
for every .
Let us summarize the results of the previous section when applied to lushness.
Proposition 3.26**.**
Let , be Banach spaces, let with and let with . Then the following assertions are equivalent for a norm-one operator :
- (i)
* is lush.* 2. (ii)
For every , and there is such that
[TABLE] 3. (iii)
For every , and there exists such that
[TABLE] 4. (iv)
For every , the set
[TABLE]
is a dense (Gδ) subset of . 5. (v)
For every , every and every , there exists such that
[TABLE] 6. (vi)
For every separable subspaces and , we can find separable subspaces and such that , and is lush.
[* Proof]The equivalences are consequence of Theorem 3.13, together with Proposition 3.23 to pass from to . *
Next, we get from the previous sections some conditions for an operator having the aDP to be lush. The main result in this line is the next one, which follows from Theorem 3.16 applied to .
Theorem 3.27**.**
Let , be Banach spaces and let be a norm-one operator. Suppose that is SCD. Then, has the aDP if and only if is lush.
As all the properties involved in the above result are separably determined, we have the following generalization.
Corollary 3.28**.**
Let , be Banach spaces and let be a norm-one operator. Suppose that is SCD for every separable subspace . Then, has the aDP if and only if is lush.
[* Proof]Since every lush operator is a spear, it has in particular the aDP. The converse is consequence of Proposition 3.19 applied to . *
The most interesting particular cases of the above results are summarized in the next corollary, which uses the examples of SCD spaces provided in Example 1.5.
Corollary 3.29**.**
Let , be Banach spaces and let be a norm-one operator. Suppose that has one of the following properties: Radon-Nikodým Property, Asplund Property, convex point of continuity property or absence of isomorphic copies of . Then, has the aDP if and only if is lush.
A result of this kind for the codomain space will be given in Proposition 5.3: if has the aDP and is Asplund, then is lush.
Our next aim is to provide the following sufficient conditions for an operator to be lush which will be used in the next chapters.
Proposition 3.30**.**
Let , be Banach spaces and let be a norm-one operator. Then, each of the following conditions ensures to be lush.
- (a)
The set \bigl{\{}y^{\ast}\in B_{Y^{\ast}}\colon G^{\ast}y^{\ast}\in\operatorname{Spear}(X^{\ast})\bigr{\}} is norming for . 2. (b)
The set \bigl{\{}y^{\ast}\in\operatorname{ext}B_{Y^{\ast}}\colon G^{\ast}y^{\ast}\in\operatorname{Spear}(X^{\ast})\bigr{\}} is dense in . 3. (c)
B_{X}=\overline{\operatorname{conv}}\bigl{\{}x\in B_{X}\colon Gx\in\operatorname{Spear}(Y)\bigr{\}}.
[* Proof]The fact that (a) implies lushness follows from Proposition 3.26.(v), as Theorem 2.8 gives that B_{X}=\overline{\operatorname{aconv}}\bigl{(}\operatorname{Face}(S_{X},G^{\ast}y^{\ast})\bigr{)} for every such that . Condition (b) is a particular case of condition (a). Finally, by using Corollary 2.7.(iv), condition (c) implies that every satisfies , so is a spear vector by (the easy part of) Theorem 2.8. *
We do not know whether the conditions (a) or (b) above are necessary for lushness in general, but they are when the domain space is separable as the following deep result shows. We will see later that they are also necessary when the codomain is an Asplund space (see Proposition 5.3).
Theorem 3.31**.**
Let be a separable Banach space and let be a Banach space. If is lush, then the set \Omega=\bigl{\{}y^{\ast}\in\operatorname{ext}B_{Y^{\ast}}\colon G^{\ast}y^{\ast}\in\operatorname{Spear}(X^{\ast})\bigr{\}} is a Gδ dense subset of . In other words, if is lush, there exists a Gδ dense subset of such that
[TABLE]
for every .
[* Proof]This is consequence of Proposition 3.15 and the characterization of spear vectors given in Corollary 2.7. The last part is a consequence of Theorem 2.8. *
On the other hand, condition (c) of Proposition 3.30 is not in general necessary for lushness: consider and , which is lush as is lush, but is empty as contains no extreme points. We will see later that condition (c) is necessary when the domain space has the Radon-Nikodým Property (see Proposition 5.2).
We finish the section with some elementary observations analogous to the ones given for spear operators and for operators with the aDP.
Remark 3.32*.*
Let be Banach spaces and .
- (i)
The composition with isometric isomorphisms preserves lushness: If , are Banach spaces and , are isometric isomorphisms, then, is lush if and only if is lush. 2. (ii)
If is lush and is a subspace of containing , then is lush. However, lushness is not preserved by extending the codomain of the operator, as the same example of Remark 3.4 shows. 3. (iii)
As an easy consequence of (i) and (ii), we have that the following statements are equivalent: (a) is lush, (b) there exist a Banach space and an isometric isomorphism in or in which is lush, (d) there exist a Banach space and an isometric embedding which is lush.
Chapter 4 Some examples in classical Banach spaces
Our aim here is to present examples of operators which are lush, spear, or have the aDP, defined in some classical Banach spaces. One of the most intriguing examples is the Fourier transform on , which we prove that is lush. Next, we study a number of examples of operators arriving to spaces of continuous functions. In particular, it is shown that every uniform algebra is lush-embedded into a space of bounded continuous functions. Finally, examples of operators acting from spaces of integrable functions are studied.
4.1. Fourier transform
Let be a locally compact Abelian group and let be the Haar measure on . The dual group of is the set of all continuous homomorphisms endowed with a topology that makes it a locally compact group (see [62, §1.2] for the details). If is the space of -integrable functions over , and is the space of continuous functions on which vanish at infinity, then the Fourier transform is defined as
[TABLE]
Theorem 4.1**.**
Let be a locally compact Abelian group and let be its dual group. Then, the Fourier transform is lush. In particular, is a spear operator, that is,
[TABLE]
for every .
[* Proof]For each , corresponds to the function given by for every . Hence, for every , and so is a spear of by Example 2.11.(e). As is the set of extreme points of , Proposition 3.30.(b) shows that is lush. *
4.2. Operators arriving to sup-normed spaces
Our goal here is to study various families of operators arriving to spaces of continuous functions. We start with a general result.
Proposition 4.2**.**
Let be a Banach space, let be a locally compact Hausdorff topological space and let be a norm-one operator. Consider the following statements:
- (i)
The set \bigl{\{}t\in L\colon G^{\ast}\delta_{t}\in\operatorname{Spear}(X^{\ast})\bigr{\}} is dense in . 2. (ii)
* is lush.* 3. (iii)
* is a spear operator.* 4. (iv)
* has the aDP.* 5. (v)
* is a spear set of for every open subset .*
Then . Besides, we have the following:
- (a)
If is scattered, then all of the statements are equivalent. 2. (b)
If is separable, then (i) (ii).
[* Proof]The implications (ii) (iii) (iv) are clear. Using the fact that we can identify homeomorphically and the fact that , we conclude easily (i) (ii) from Proposition 3.30, while the equivalence (iv) (v) follows from Theorem 3.6.*
(a). is scattered if and only if is Asplund (see [1, Comment after Corollary 2.6], for instance). Notice that is weak∗-weak∗-continuous, so we just have to prove that given an open set , there exists such that is a spear. Since is weak∗-fragmentable (see [28, Theorem 11.8] for the relation and defintion between Asplundness and fragmentability), for each there exists a weak∗-open set satisfying that has diameter less than . Now, since is weak∗-continuous, we can find an open set with . Because of the local compactness of we can select in such a way that is compact. In particular is a closed spear set with diameter less than . It is clear that we can iterate this process to construct a decreasing sequence \bigl{(}W_{n}\bigr{)}_{n\in\mathbb{N}} of open subsets of such that \operatorname{diam}\bigl{(}G^{\ast}(\overline{W_{n}})\bigr{)} tends to zero. Since is a spear set by (v), it follows then from Lemma 2.9 that an element must satisfy that is a spear.
*(b). If is separable, the result follows from Theorem 3.31. *
Let us mention that (a) is also a particular case of Proposition 5.3 in the next chapter, using the stated above equivalence between being scattered and being Asplund. In the more restrictive case in which has the discrete topology (which is trivially scattered), the result will be also proved, with a different approach, in Example 5.5.
The next result characterizes lush spaces. We need some notation. Let be a completely regular Hausdorff topological space and denote by the Banach space of all scalar bounded and continuous functions on endowed with the supremum norm.
Theorem 4.3**.**
Let be a Banach space. Then, is lush if and only if the canonical inclusion is lush.
Recall that for a completely regular Hausdorff topological space the corresponding can be canonically seen as a space by taking , the Stone-Cech compactification of . Since is a dense subset of , we have that the set is dense in .
[* Proof]The space endowed with the weak∗-topology is completely regular. As we mentioned before, the set*
[TABLE]
is dense in being moreover a Baire space with the induced topology (Lemma 1.2.(c)). By Proposition 3.10, is lush if and only if for every and the set intersects every -slice of . Actually, the properties of given in Lemma 1.2 allow us to repeat the arguments in the proof of Proposition 3.10 to deduce that is lush if and only if given an arbitrary the set
[TABLE]
is dense in . But using the natural homeomorphism between and , we can reformulate the previous condition as
[TABLE]
*being dense in for every , which, by Proposition 3.26, is equivalent to say that is lush, that is, is lush. *
The following very general result will allow us to deduce many other interesting examples.
Theorem 4.4**.**
Let be a non empty set and let . Let be Banach spaces satisfying the following properties:
- (i)
For every , and there are and such that ,
[TABLE] 2. (ii)
For each there is such that
[TABLE]
Then, the inclusion is lush.
[* Proof]Fix , and . Since , we can find and such that*
[TABLE]
Again by (i) we can find such that , and with such that
[TABLE]
By (ii), there is , with , , and . Let us fix with .
We claim that for each with one has that
[TABLE]
Indeed, the conditions , and imply that . We distinguish two cases: if then which gives , so we get that
[TABLE]
On the other hand, if then
[TABLE]
This finishes the proof of the claim.
Observe that [math] belongs to the convex hull of the set as . We can then find and such that . Take with , pick satisfying \theta_{1}\big{(}x(t_{0})+\gamma_{1}bx_{0}(t_{0})\big{)}\geqslant 0, \theta_{2}\big{(}x(t_{0})+\gamma_{2}bx_{0}(t_{0})\big{)}\geqslant 0 and . Define
[TABLE]
By the claim above, we have that . Besides, we can write
[TABLE]
Moreover, and for every we have that
[TABLE]
Therefore, we can estimate as follows
[TABLE]
and, moreover,
[TABLE]
*This shows that is lush by (ii) of Proposition 3.26 with . *
The following notion for subspaces of was introduced in [19, Definition 2.3], generalizing the analogous concept for subspaces of introduced in [13, Definition 2.3] (for perfect compact spaces it was considered earlier in [43] and was studied extensively in frames of the Daugavet Property theory).
Definition 4.5**.**
Let be a completely regular Hausdorff topological space. A closed subspace is called C-rich if for every and every open subset there exists a norm-one function in such that and .
It follows from Urysohn’s Lemma that is C-rich in itself for every completely regular space .
The main tool in the rest of the section will be the following.
Theorem 4.6**.**
Let be a completely regular Hausdorff topological space. If is C-rich, then the inclusion is lush. In particular, we have for every .
[* Proof]We just have to check that the hypothesis of Theorem 4.4 are satisfied for and taking as the family of all open subsets of . Hypothesis (i) satisfied just using the continuity, while (ii) is consequence of the C-richness of . *
Remark 4.7*.*
Let us observe that there are natural inclusions which are lush without being a C-rich subspace of . For instance, using Theorem 4.3 we deduce that the inclusion
[TABLE]
is lush. However, is not C-rich in . Indeed, we argue by contradiction. Let , consider the open set , and suppose that and satisfy that
[TABLE]
Taking supremum over all , we deduce that
[TABLE]
While implies that
[TABLE]
Taking small enough, (4.2) and (4.1) contradict each other.
Let us present some applications of Theorem 4.6. First, it was shown in [43, Proposition 1.2] that if is a perfect compact Hausdorff topological space, then every finite-codimensional subspace of is C-rich, but this is not always the case when has isolated points. Actually, finite-codimensional subspaces of general spaces were characterized in [13, Proposition 2.5] in terms of the supports of the functionals defining the subspace. We recall that the support of an element (represented by the regular measure ) is . Then, for , the subspace is C-rich in if and only if does not intersect the set of isolated points of .
Corollary 4.8**.**
Let be a compact Hausdorff topological space, consider functionals and let . If does not intersect the set of isolated points of , then the natural inclusion is lush. In particular, if is perfect, then for every finite-codimensional subspace of , the inclusion is lush.
For we may even go to smaller subspaces, using a result of [43]: if is a subspace of such that does not contain isomorphic copies of , then is C-rich in [43, Proposition 1.2 and Definition 2.1].
Corollary 4.9**.**
Let be a subspace of such that does not contain isomorphic copies of (in particular, if is reflexive). Then, the inclusion is lush.
Let us now go to present the main part of this section. Recall that a uniform algebra (on a compact Hausdorff topological space ) is a closed subalgebra that separates the points of . We refer to [22] for background.
Theorem 4.10**.**
Let be a compact Hausdorff topological space and let be a uniform algebra on . Then, there exists subset such that (isometrically) is C-rich, and so the inclusion is lush. Moreover, if is unital, then is just its Choquet boundary.
[* Proof]If is a unital uniform algebra, then consider being the Choquet boundary of . Given and with , take small enough so that every with satisfies that . By [15, Lemma 2.5], there exists and such that , for each and*
[TABLE]
Put and . These are disjoint compact subsets of , so there exists continuous such that and . The element belongs to and satisfies . We just have to check that . Indeed, if then , so and so ; on the other hand, if then . The restriction satisfies the definition of C-rich for the given and .
*If is not unital, then we can repeat the same argument that above but now using [15, Lemma 2.7] and taking as the set that appears in the referenced lemma. *
The Choquet boundary of the disk algebra is (see [22, Proposition 4.3.13], for instance), so by the previous result we have the following consequence.
Corollary 4.11**.**
The natural inclusion is lush.
Another family of interesting C-rich subspaces is the following. Let be an infinite compact Abelian group, let be the Haar measure on , let be the dual group of , let be the space of all regular Borel measures on and, finally, let be the Fourier-Stieltjes transform, which is the natural extension of the classical Fourier transform of section 4.1 (see [62, §1.3]). For , the space of -spectral continuous functions is defined by
[TABLE]
and similarly it is defined the space of -spectral measures . These spaces are known to be precisely the closed translation invariant subspaces of and , respectively. A subset of is said to be a semi-Riesz set [70, p. 126] if all elements of are diffuse (i.e. if they map singletons to [math]). Semi-Riesz sets include Riesz sets, defined as those such that ; the chief example of a Riesz subset of the dual group of is . We refer to [33, §IV.4] for background. It is shown in [53, Theorem 4.13] that is a semi-Riesz set if and only if is C-rich in . Therefore, we have the following consequence of Theorem 4.6.
Corollary 4.12**.**
Let be an infinite compact Abelian group and let be a subset of the dual group . If is a semi-Riesz set, then the inclusion is lush.
Remark 4.13*.*
It is proved in [70, Theorem 3.7] that if is a semi-Riesz set, then is nicely embedded into , that is, the isometric embedding satisfies that for every , and the linear span of is an -summand in (this is actually a straightforward consequence of the definition of semi-Riesz set). Then, it follows immediately from Example 2.11.(g) that for every , so is lush by Proposition 3.30.(a). This is thus an alternative elementary proof of Corollary 4.12 which does not need the more complicated [53, Theorem 4.13].
Let us also comment that it was proved in [70, Proof of Theorem 3.3] that unital function algebras are nicely embedded into , where is the Choquet boundary of the algebra, so Theorem 4.10, in the unital case, and Corollary 4.11, can be also proved by using the argument above.
We next provide with more applications of Theorem 4.4. The following definition appears in [38, Definition 3.2] for vector-valued spaces of continuous functions.
Definition 4.14**.**
Let be a compact and Hausdorff space. We say that a closed subspace is a -superspace if it contains and for each , every open subset and each , there are an open subset and an element such that
[TABLE]
The result for -superspaces is the following.
Corollary 4.15**.**
If is a -superspace, then the inclusion is lush.
[* Proof]Using Theorem 4.4 for the inclusions with being the set of open subsets of , we have that (ii) is satisfied by Urysohn’s Lemma, while (i) is just the definition of -superspace. *
An interesting application is given by the next example.
Example 4.16**.**
Let be the space of bounded functions on which are right-continuous, have left limits everywhere and are continuous at . It is shown in [38, Proposition 3.3] that is a -superspace (this is because is the closure in of the span of the step functions and ). Therefore, the inclusion is lush.
4.3. Operators acting from spaces of integrable functions
Our aim here is to describe operators from spaces which have the aDP. For commodity, we only deal with probability spaces, but this is not a mayor restriction as -spaces associated to -finite measures are (up to an isometric isomorphism) spaces associated to probability measures (see [17, Proposition 1.6.1], for instance). We introduce some notation. Let be a probability space and let be a Banach space. We write \Sigma^{+}:=\bigl{\{}B\in\Sigma\colon\mu(B)>0\bigr{\}} and for we consider
[TABLE]
Recall that an operator is (Riesz) representable if there exists (i.e. a strongly measurable and essentially bounded function) such that
[TABLE]
Rank-one operators are representable by the classical Riesz representation theorem assuring that , and then so are all finite-rank operators. Actually, compact operators [26, p. 68, Theorem 2] and even weakly compact operators [26, p. 65, Theorem 12] are representable, but the converse result is not true [26, p. 79, Example 22]. Finally, let us say that the set of representable operators can be isometrically identified with [26, p. 62, Lemma 4]. We refer the reader to chapter III of [26] for more information and background on representable operators.
Here is the characterization of aDP operators which is the main result of this section.
Theorem 4.17**.**
Let be a probability space, let be a Banach space and let be a norm-one operator. The following assertions are equivalent:
- (i)
* has the aDP.* 2. (ii)
* is a spear set for every .* 3. (iii)
* for every representable.*
Let us recall the following exhaustion argument that we briefly prove here.
Observation 4.18**.**
Let be a finite measure space. If for each there is satisfying a certain property , then we can find a countable family of disjoint sets such that every satisfies property (P) and is -null.
Indeed, this follows from a simple argument: using Zorn’s lemma we can take a maximal family of disjoint sets in satisfying property (P), which must be countable as is finite. To see the last condition, notice that if had positive measure, then we could use the hypothesis to find a subset satisfying (P), and hence would contradict the maximality of .
We need a preliminary result.
Lemma 4.19**.**
Let be a probability space and let be a Banach space. For every one has that
[TABLE]
[* Proof]The inequalities are clear in both cases, so we just have to see that*
[TABLE]
is greater than or equal to . Let be a simple function, where is a finite partition of into elements of , so . Given , we have that for each there is such that . Using Observation 4.18 in each set , we can find a countable partition of such that every of positive measure is contained in some element of and satisfies . If we write whenever , then
[TABLE]
*Since was arbitrary, we conclude that . As runs on all simple functions, it follows that . *
[* Proof of Theorem 4.17] (i) (ii): Fix , , and . Consider the rank-one operator given by*
[TABLE]
Then, . By Lemma 4.19 we can find such that
[TABLE]
It follows that , consequently , so the set satisfies that
[TABLE]
The combination of the above two inequalities leads to
[TABLE]
(ii) (iii): Let and . By Lemma 4.19, we can find satisfying . If is representable, then there exists such that \operatorname{diam}\bigl{(}T(\Gamma_{B})\bigr{)}<\varepsilon (see [26, p. 62, Lemma 4 and p. 135, Lemma 6]), so taking any and using that is a spear set we obtain
[TABLE]
*(iii) (i) is obvious as rank-one operators are representable. *
Remark 4.20*.*
A direct way to prove (i) (iii) in Theorem 4.17 is the following: every representable operator factorizes through , i.e. there are operators and such that . But then, is an SCD operator satisfying for each , and so Corollary 3.17 implies that is a target for . However, item (ii) in Theorem 4.17 gives an intrinsic characterization of aDP operators acting from an space which has its own interest.
As an obvious consequence of Theorem 4.17, if for a Banach space all bounded linear operators from to are representable, then the aDP is equivalent to be spear for every . This is the case when has the Radon-Nikodým Property with respect to [26, p. 63, Theorem 5]. Therefore, the following result follows.
Corollary 4.21**.**
Let be a probability space, let be a Banach space which has the Radon-Nikodým Property with respect to , and let be a norm-one operator. Then has the aDP if and only if is a spear operator.
Let us observe that if in the above corollary has the Radon-Nikodým Property, then the result also follows from Corollary 3.20. On the other hand, it is easy to see that every Banach space has the Radon-Nikodým Property with respect to the counting measure on (use [26, Theorem 8, p. 66], for instance) and so has the aDP if and only if is a spear operator. But, in this case, we will show that actually has the aDP if and only if is lush (see Example 5.5).
We next present an example showing that spearness and lushness are not equivalent for an operator having an space as domain,
Example 4.22**.**
There is a Banach space such that the quotient operator is a spear operator but it is not lush.
Indeed, let be the space introduced in [40, Theorem 4.1]: this is defined as , where is a -closed subspace with no modulus-one (a.e.) functions. If were lush, then its dual operator would send “many” extreme points to spear vectors by Theorem 3.31. However, is an isometric embedding and does not contain any modulus-one function, while the spears of are characterized that way, see Example 2.11.(e). This is a contradiction. On the other hand, is constructed to be C-rich in (viewed as a space), and so is lush by Theorem 4.6. This, in particular, implies that is a spear operator.
We finish this section showing that a representable operator has the aDP if and only if it is represented by a spear vector of . As a consequence, we will describe the spear vectors of as those functions which take spear values almost everywhere, extending Example 2.11.(e) to the vector-valued case.
Corollary 4.23**.**
Let be a norm-one operator which is representable by . Then, the following are equivalent:
- (i)
* has the aDP.* 2. (ii)
* for a.e. .* 3. (iii)
g\in\operatorname{Spear}\bigl{(}L_{\infty}(\mu,Y)\bigr{)}.
[* Proof]The implication (ii) (iii) is an easy adaptation of the scalar case proved in Example 2.11.(e). Indeed, for and , there is such that for every . By (ii), there is with such that for every . Now, fixed an -net of we can find an element and a subset of with positive measure such that for every . Therefore, we can write*
[TABLE]
and the arbitrarines of gives the result. (iii) (i) is consequence of the fact that every rank-one operator is representable, so for every rank-one there is which represents and so
[TABLE]
Let us prove (i) (ii). Fix . Since is strongly measurable, given any there exists such that (see [16, Proposition 2.2], for instance). By Observation 4.18 we can take a countable family of disjoint sets with the property that for each and is -null. Given , it must belong to some , and since (see [26, p. 48, Corollary 8]), we deduce that . Using that is a spear set by Theorem 4.17, it follows that for every ,
[TABLE]
*Finally, if we take now a decreasing sequence of positive numbers converging to zero and consider the correspondent for each , then every satisfies for each , i.e. is a spear. *
Chapter 5 Further results
Our goal here is to complement the previous chapter with some interesting results. We characterize lush operators when the domain space has the Radon-Nikodým Property or the codomain space is Asplund, and we get better results when the domain or the codomain is finite-dimensional or when the operator has rank one. Further, we study the behaviour of lushness, spearness and the aDP with respect to the operation of taking adjoint operators. Finally, we collect some stability results.
5.1. Radon-Nikodým Property in the domain or Asplund codomain
We first provide a result about the relationship of an operator with the aDP and spear vectors of the unit ball of the range space and the spear vectors of the dual ball of the domain space.
Proposition 5.1**.**
Let , be Banach spaces and let be an operator with the aDP, then
- (a)
* for every denting point of .* 2. (b)
* for every weak*∗-denting point of .
[* Proof]We only illustrate the proof of (a), since the other one is completely analogous. If is denting, then we can find a decreasing sequence of slices of containing and such that tends to zero. Since has the aDP, Theorem 3.6.(iii) gives that \bigl{(}G(S_{n})\bigr{)}_{n\in\mathbb{N}} is a decreasing sequence of spear sets whose diameters tend to zero, so is a spear vector by Lemma 2.9. *
We now characterize spear operators acting from a Banach space with the Radon-Nikodým Property.
Proposition 5.2**.**
Let be a Banach space with the Radon-Nikodým Property, let be a Banach space and let be a norm-one operator. Then, the following assertions are equivalent:
- (i)
* is lush.* 2. (ii)
* is a spear operator.* 3. (iii)
* has the aDP.* 4. (iv)
* for every and every denting point of .* 5. (v)
B_{X}=\overline{\operatorname{conv}}\bigl{\{}x\in B_{X}\colon Gx\in\operatorname{Spear}(Y)\bigr{\}}* or, equivalently,*
[TABLE]
* Proof (ii) (iii) are clear. (iii) (iv) follows from Proposition 5.1 and Corollary 2.7.(iv). (iv) (v) is consequence of the fact that is the closed convex hull of its denting points since has the Radon-Nikodým Property (see [11, §2] for instance), and the equivalent reformulation is a consequence of Theorem 2.8. Finally, (v) (i) follows from Proposition 3.30.(c). *
For Asplund spaces, we have the following characterization.
Proposition 5.3**.**
Let be a Banach space, let be an Asplund space and let be a norm-one operator. Then, the following assertions are equivalent:
- (i)
* is lush.* 2. (ii)
* is a spear operator.* 3. (iii)
* has the aDP* 4. (iv)
* for every and every weak*∗-denting point of . 5. (v)
The set \bigl{\{}y^{\ast}\in\operatorname{ext}B_{Y^{\ast}}\colon G^{\ast}y^{\ast}\in\operatorname{Spear}(X^{\ast})\bigr{\}} is dense in or, equivalently, there is a dense subset of such that
[TABLE]
for every . 6. (vi)
B_{Y^{\ast}}=\overline{\operatorname{conv}}^{w^{\ast}}{\bigl{\{}y^{\ast}\in B_{Y^{\ast}}\colon G^{\ast}y^{\ast}\in\operatorname{Spear}(X^{\ast})\bigr{\}}}.
* Proof (ii) (iii) are clear. (iii) (iv) follows from Proposition 5.1 and Corollary 2.7.(iv). (iv) (v): the set contains all weak∗-denting points of by Proposition 5.1, so it is weak∗ dense since for Asplund spaces, weak∗-denting points are weak∗ dense in the set of extreme points of the dual ball (see [11, §2] for instance). The equivalent reformulation is consequence of Theorem 2.8. Finally, (v) (vi) is clear and (vi) (i) follows from Proposition 3.30.(b). *
We do not know whether the above result extends to the case when is SCD. What is easily true, using Theorem 3.16, is that aDP and spearness are equivalent in this case.
Remark 5.4*.*
Let be a Banach space, let be an SCD Banach space, and let be a norm-one operator. Then, has the aDP if and only if is a spear operator.
As a consequence of the results above, we may improve Proposition 3.3.
Example 5.5**.**
Let be an arbitrary set, let , be Banach spaces and let be the canonical basis of (as defined in Example 2.11.(a)).
- (a)
For the following are equivalent: is lush, is a spear operator, has the aDP, for every , for every and every . 2. (b)
For the following are equivalent: is lush, is a spear operator, has the aDP, for every , B_{X}=\overline{\operatorname{aconv}}\bigl{(}\operatorname{Face}(S_{X},G^{\ast}e_{\gamma})\bigr{)} for every .
Part of assertion (a) above also follows from Corollary 4.21; the whole assertion (b) also follows from Proposition 4.2.
5.2. Finite-dimensional domain or codomain
Our goal now is to discuss the situation about spear operators when the domain or the codomain is finite-dimensional. We start with the case in which the domain is finite-dimensional, where the result is just an improvement of Proposition 5.2. To get it, we only have to recall that for finite-dimensional spaces, the concepts of denting point and extreme point coincide thanks to the compactness of the unit ball and Choquet’s Lemma (Lemma 1.2.(a)).
Proposition 5.6**.**
Let be a finite-dimensional space, let be a Banach space and let be a norm-one operator. Then, the following are equivalent:
- (i)
* is lush.* 2. (ii)
* is a spear operator.* 3. (iii)
* has the aDP.* 4. (iv)
* for every and every .* 5. (v)
* for every .* 6. (vi)
.
The next example shows that even in the finite-dimensional case, bijective lush operators can be very far away from being isometries and that their domain and codomain are not necessarily spaces with numerical index one.
Example 5.7**.**
There exists a bijective lush operator such that neither its domain nor its codomain has the aDP.
Indeed, let be the real four-dimensional space whose unit ball is given by
[TABLE]
Let be the real space , let and, finally, let . Consider the operator given by for every and consider . Finally, calling and , the operator we are looking for is given by for every .
We start showing that is lush. To this end, by Proposition 5.6, all we have to do is to check that carries extreme points of to spear vectors of . By Example 2.11.(i), this is equivalent to show that both and carry extreme points to spear elements. This is evident for and it is also straightforward to show for (alternatively, the first assertion gives that is lush by Proposition 5.6, so is also lush by Corollary 5.22 in the next section, so carries extreme points of to spear elements in by using again Proposition 5.6).
Finally, let us show that does not have the aDP (i.e. that does not have the aDP). By Proposition 5.6, it is enough to find an extreme point of which is not a spear vector of . By Example 2.11.(i), it is enough to find an extreme point of which is not a spear vector of . Let us show that this happens for . On the one hand, is clearly an extreme point of by construction. On the other hand, if were a spear vector, we would have for every by Corollary 2.7.(iv), so we would get a contradiction if we show that the functional is an extreme point of . Let us show this last assertion. First, belongs to since for every we have that
[TABLE]
Next, consider such that both and lie in . This, together with the fact that
[TABLE]
implies that
[TABLE]
so since it vanishes on a basis of . This gives that is an extreme point, as desired.
When the codomain is finite-dimensional, we can improve Proposition 5.3 as follows, just taking into account that weak∗-denting points and extreme points of the dual ball are the same for a finite-dimensional space.
Proposition 5.8**.**
Let be a Banach space, let be a finite-dimensional space and let be a norm-one operator. Then, the following assertions are equivalent:
- (i)
* is lush.* 2. (ii)
* is a spear operator.* 3. (iii)
* has the aDP* 4. (iv)
* for every and every .* 5. (v)
* for every .* 6. (vi)
B_{X}=\overline{\operatorname{aconv}}\bigl{(}\operatorname{Face}(S_{X},G^{\ast}y^{\ast})\bigr{)}* for every .* 7. (vii)
B_{Y^{\ast}}=\operatorname{conv}{\bigl{\{}y^{\ast}\in B_{Y^{\ast}}\colon G^{\ast}y^{\ast}\in\operatorname{Spear}(X^{\ast})\bigr{\}}}.
We do not know whether this result, or part of it, is also true when just the range of the operator is finite-dimensional. But we can provide with the following result for rank-one operators.
Corollary 5.9**.**
Let , be Banach spaces and let be a norm-one rank-one operator, and write for suitable and . Then, the following assertions are equivalent:
- (i)
* is lush.* 2. (ii)
* is a spear operator.* 3. (iii)
* has the aDP* 4. (iv)
* and .*
* Proof (ii) (iii) are clear. Let us prove (iii) (iv). First, for every , consider the rank-one operator and observe that*
[TABLE]
so . Next, we have that , also has the aDP (use Remark 3.8) and we may use Proposition 5.1 to get that .
*(iv) (i). Observe that for every . Now, for each we have that by Corollary 2.7.(iv) (as ), so . Now, Proposition 3.30.(b) gives us that is lush. *
5.3. Adjoint Operators
We would like to discuss here the relationship of the aDP, spearness and lushness with the operation of taking the adjoint.
As the norm of an operator and the one of its adjoint coincide, the following observation is immediate.
Remark 5.10*.*
Let , be Banach spaces and let be a norm-one operator. If is a spear operator, then is a spear operator. If has the aDP, then has the aDP.
With respect to lushness, the above result is not true, even for equal to the Identity, as the following example shows.
Example 5.11** ([40, Theorem 4.1]).**
There is a (separable) Banach space such that is lush, but is not lush. Therefore, is not lush, while is lush. Actually, where is the space defined in Example 4.22.
This example has some more properties which are interesting.
Remark 5.12*.*
Let be the Banach space of Example 5.11 and consider the operator .
- (a)
is a spear operator but it is not lush (use Remark 5.10). 2. (b)
([40, Remarks 4.2.(c)]). Therefore:
- •
Theorem 3.31 is far from being true for spear operators;
- •
Proposition 5.2 is far from being true for spaces without the Radon-Nikodým Property;
- •
there is no lush operator whose domain is .
We may give a positive result in this line: if the second adjoint of an operator is lush, then the operator itself is lush. This will be given in Corollary 5.16, but we need some preliminary work to get the result. We start with a general result which allows to restrict the domain of a lush operator.
Proposition 5.13**.**
Let , be Banach spaces and let be a weak∗-weak∗ continuous norm-one operator. If is lush, then is lush.
For the sake of clearness, we include the most technical part of the proof of this result in the following lemma.
Lemma 5.14**.**
Let , , be Banach spaces and let and be norm-one operators. Suppose that there is a subset such that satisfies the following property
[TABLE]
Suppose also that is lush and there is a subset with such that . Then is lush.
[* Proof]Fix , and . Let . Since is lush, applying Proposition 3.26.(iii) with , we can find such that*
[TABLE]
Then, there are , , for each , with and
[TABLE]
Notice that for each we have that , and
[TABLE]
This implies that
[TABLE]
Hence, every slice of containing satisfies that
[TABLE]
As , the hypothesis (P1) yields that
[TABLE]
and, therefore,
[TABLE]
Since was arbitrary, we conclude that
[TABLE]
and so
[TABLE]
*As , we get that . This, together with the first part of (5.1), gives that is lush by using again Proposition 3.26.(iii). *
[* Proof of Proposition 5.13] We will use the above lemma with , , , and . To this end, we first show that satisfies condition (P1) of Lemma 5.14 with . Indeed, let us fix a slice of the form of , and , and suppose that*
[TABLE]
Since
[TABLE]
we actually have that
[TABLE]
and so, a fortiori,
[TABLE]
But it then follows that
[TABLE]
This clearly implies that J_{X}\bigl{(}\operatorname{Slice}(B_{X},x_{1}^{\ast},\delta)\bigr{)}\cap\operatorname{gSlice}(B_{X^{\ast\ast}},\mathbb{T}\,J_{X^{\ast}}(x^{\ast}),\varepsilon)\neq\emptyset, as desired.
Now, let and which is norming for . As is weak∗-weak∗ continuous, we have that (indeed, let such that and observe that for every ).
*Therefore, all the requirements of Lemma 5.14 are satisfied, so is lush. *
We get a couple of corollaries of this result. The first one deals with the natural inclusion of a lush Banach space into its bidual. It is an immediate consequence of the result above applied to .
Corollary 5.15**.**
Let be a Banach space. If is lush, then the canonical inclusion is lush.
The next consequence is the promised result saying that lushness passes from the biadjoint operator to the operator.
Corollary 5.16**.**
Let , be Banach spaces and let be a norm-one operator. If is lush, then is lush.
[* Proof]Apply Proposition 5.13 to , which is weak∗-weak∗ continuous, to get that is lush. But, clearly, and then, restricting the codomain and considering that and are isometrically isomorphic, Remark 3.32 gives us that is lush. *
These two corollaries improve [40, Proposition 4.3] where it is proved that a Banach space is lush whenever is lush.
Remark 5.17*.*
The technical hypothesis in Lemma 5.14 is fundamental to get the result. Indeed, consider the inclusion and the projection . Notice that satisfies the condition (P1) of Lemma 5.14 with (this is shown in the proof of Proposition 5.13). On the other hand is lush since it carries every spear vector of into a spear vector of . This can be easily seen using the canonical (isometric) identifications and , so that is just the restriction operator. On the other hand, , which clearly is not lush. The technical hypothesis of the lemma is not satisfied, since every with must be zero.
The same example also shows the need of the operator in Proposition 5.13 to be weak∗-weak∗-continuous: indeed, just take and observe that .
Let us now discuss the more complicated direction: when lushness, spearness or the aDP passes from an operator to its adjoint. It is easy to provide examples of operators with the aDP whose adjoint do not share the property: for instance this is the case of the Identity operator on the space (indeed, this space has the aDP by [45, Example in p. 858], while its dual contains as -summand and so it fails the aDP by [56, Proposition 3.1]). But the same question for spearness of the Identity (that is, whether the numerical index one passes from a Banach space to its dual) was one of the main open questions of the theory of numerical index until 2007, when it was solved in the negative [13]. The counterexample given there is actually lush, and its dual even fails the aDP. Let us state all of this here.
Example 5.18** ([13, §3] ).**
Consider the Banach space
[TABLE]
and write . Then, is lush ( is actually a C-rich subspace of ), but does not even have the aDP.
Some remarks about this example are interesting. Recall that a James boundary for a Banach space is a subset of such that . As a consequence of the Hanh-Banach and the Krein-Milman theorems, the set is a James boundary for .
Remark 5.19*.*
Let be the space of Example 5.18 and . Then the set is norming for , but it is not a James boundary for so, in particular, it does not coincide with [13, Example 3.4]. Therefore:
- •
Theorem 3.31 cannot be improved to get that the set is the whole set of extreme points, nor a James boundary for ;
- •
the Gδ dense set in Proposition 5.3.(v) does not always coincide with the set of all extreme points of the dual ball, nor is always a James boundary for .
Our next goal is to provide sufficient conditions which allow to pass the properties of an operator to its adjoint. The first of these conditions is that the domain space has the Radon-Nikodým Property.
Proposition 5.20**.**
Let be a Banach space with the Radon-Nikodým Property, let be a Banach space and let be a norm-one operator. If has the aDP, then is lush. Therefore, the following six assertions are equivalent: has the aDP, is a spear operator, is lush, has the aDP, is a spear operator, is lush.
[* Proof]Write for the set of denting points of . If has the aDP, Proposition 5.1 gives that for every , and then Proposition 2.10.(c) gives that for every . Therefore, the set*
[TABLE]
contains which is norming for as has the Radon-Nikodým Property (see [11, §2] for instance). Then, Proposition 3.30.(a) gives that is lush.
*Finally, let us comment the proof of the last part. The three first assertions are equivalent by Proposition 5.2 since has the Radon-Nikodým Property; lush \ \Rightarrow\ spear \ \Rightarrow\ has the aDP \ \Rightarrow\ has the aDP by Remark 5.10. The remaining implication is just what we have proved above. *
Another result in this line is the following. We recall that a Banach space is -embedded if is an -summand in (which is actually equivalent to the fact that the Dixmier projection on is an -projection). We refer the reader to the monograph [33] for more information and background. Examples of -embedded spaces are reflexive spaces (trivial), and all of its closed subspaces, (the space of compact operators on a Hilbert space ), , the little Bloch space , among others (see [33, Examples III.1.4]).
Proposition 5.21**.**
Let be a Banach space, let be an -embedded Banach space, and let be a norm-one operator. If has the aDP, then is lush. Therefore, the following nine assertions are equivalent: has the aDP, is a spear operator, is lush, has the aDP, is a spear operator, is lush, has the aDP, is a spear operator, is lush.
[* Proof]We use Proposition 5.1 to get that the set \bigl{\{}y^{\ast}\in B_{Y^{\ast}}\colon G^{\ast}y^{\ast}\in\operatorname{Spear}(X^{\ast})\bigr{\}} contains the set of those weak∗-denting points of . By [33, Corollary III.3.2], we have that so, a fortiori,*
[TABLE]
Then, Proposition 3.30.(c) gives that is lush.
*Finally, for the last part, the three first assertions are equivalent by Proposition 5.3 since is Asplund [33, Theorem III.3.2]. The middle three assertions are equivalent by Proposition 5.2 since has the Radon-Nikodým Property [33, Theorem III.3.2]. If has the aDP, so does (Remark 5.10) and this implies that is lush by the above. As has the Radon-Nikodým Property, if has the aDP, then is lush by Proposition 5.20, and this gives the equivalence with the last three assertions. *
Even though part of what we have used in the proof above is Asplundness of -embedded spaces, just this hypothesis on is not enough to get the result as Example 5.18 shows.
A consequence of the two results above is that lushness passes from an operator with finite-dimensional domain or codomain to all of its successive adjoint operators.
Corollary 5.22**.**
Let , be Banach spaces such that at least one of them is finite-dimensional space, and let be a norm-one operator. If has the aDP, then all the successive adjoint operators of are lush.
[* Proof]If is finite-dimensional, then it has the Radon-Nikodým Property, so Proposition 5.20 gives that is lush. If is finite-dimensional, then it is clearly -embedded, so Proposition 5.21 gives us that is lush. For the successive adjoint operators, one of the above two arguments applies. *
We do not know whether the above result can be extended to finite-rank operators. We may do when the operator has actually rank one.
Proposition 5.23**.**
Let , be Banach spaces, and let be a rank-one norm-one operator. If has the aDP, then is lush. Therefore, all the successive adjoints of are lush.
[* Proof]If has the aDP, by Corollary 5.9 we have that with and . Observe that . Since by Proposition 2.10.(c), we get that is lush by using again Corollary 5.9. *
The last result deals with -embedded spaces. Recall that a Banach space is -embedded if for suitable closed subspace of . We refer to the monograph [33] for background. Examples of -embedded spaces are reflexive spaces (trival), predual of von Neumann algebras so, in particular, spaces, the Lorentz spaces and , the Hardy space , the dual of the disk algebra , among others (see [33, Examples IV.1.1 and III.1.4]).
Proposition 5.24**.**
Let be a Banach space, let be an -embedded space, and let be a norm-one operator.
- (a)
If is a spear operator, then is a spear operator. 2. (b)
If has the aDP, then has the aDP.
* Proof. Write for the projection associated to the decomposition . We fix and consider the operators*
[TABLE]
and observe that . Given , since is dense in by Goldstine’s Theorem and is weak∗-weak∗-continuous, we may find such that
[TABLE]
Now, we may find and such that
[TABLE]
We define by for every , and observe that . As is a spear operator, we have that , so we may find , , and such that
[TABLE]
Finally, consider which has norm-one (here we use the -embeddedness hypothesis) and observe that
[TABLE]
Moving , we get that is a spear operator, as desired.
*(b). If just has the aDP, we may repeat the above argument for rank-one operators , and everything works fine as the operator constructed there has finite rank, so by Theorem 3.16 (as, clearly, finite-rank operators are SCD). *
Chapter 6 Isometric and isomorphic consequences
Our goal here is to present consequences on the Banach spaces and of the fact that there is which is a spear operator, is lush or has the aDP.
We first start with a deep structural consequence which generalizes [4, Corollary 4.10] where it was proved for real infinite-dimensional Banach spaces with the aDP.
Theorem 6.1**.**
Let , be real Banach spaces and let . If has the aDP and has infinite rank, then contains a copy of .
[* Proof]Using Proposition 3.7, we can find separable subspaces and such that has the aDP, and still it has infinite rank. By Remark 3.8, we may and do suppose that . It is enough to show that contains a copy of since, in this case, also contains such a copy by the lifting property of (see [51, Proposition 2.f.7] or [66, p. 11]). We have two possibilities. If contains a copy of , then contains a quotient isomorphic to and so contains a copy of again by the lifting property of . If does not contain copies of then is an SCD set by [4, Theorem 2.22] (see Example 1.5), so Theorem 3.27 gives that is lush. Then, by Theorem 3.31, the set \bigl{\{}y^{\ast}\in\operatorname{ext}B_{Y_{\infty}^{\ast}}\colon G_{\infty}^{\ast}(y^{\ast})\in\operatorname{Spear}(X_{\infty}^{\ast})\bigr{\}} is weak∗-dense in . As has dense range, is injective, and since is infinite-dimensional, it follows that the set must be infinite. Now, Proposition 2.10.(i) gives us that contains a copy of or . But a dual space contains a copy of whenever it contains a copy of [51, Proposition 2.e.8]. *
Another result in this line is the following.
Proposition 6.2**.**
Let be a real Banach space with the Radon-Nikodým Property, let be a real Banach space, and let . If has the aDP and has infinite rank, then or .
[* Proof]By Proposition 5.2 we have that B_{X}=\overline{\operatorname{conv}}\bigl{\{}x\in B_{X}\colon Gx\in\operatorname{Spear}(Y)\bigr{\}}, so*
[TABLE]
*Now, if has infinite rank, has to be infinite and so Proposition 2.10.(j) gives the result. *
Remark 6.3*.*
Let us observe that both possibilities in the result above may happen. On the one hand, is lush by Example 5.5. On the other hand, the operator given by if and if is also lush by Example 5.5.
We next deal with isometric consequences of the existence of operators with the aDP. The following result generalizes [39, Theorem 2.1] where it was proved for . Let us remark that the proof given there relied on a non-trivial result of the theory of numerical range: that the set of operators whose adjoint attain its numerical radius is norm dense in the space of operators.
Proposition 6.4**.**
Let , be Banach spaces and let be an operator with the aDP. Then
- (a)
If is strictly convex, then . 2. (b)
If is smooth, then . 3. (c)
If is strictly convex, then .
* Proof. We start showing that has rank one. Using Theorem 3.6.(iv) we can find with . By the same result, there is a -dense subset of whose elements satisfy that*
[TABLE]
It follows from the definition of strict convexity that for every such , and we deduce by the Krein-Milman Theorem and the weak∗ continuity of , that is contained in . Hence, has rank one. Therefore, has rank one and is non empty by Corollary 5.9. Finally, is one-dimensional by Proposition 2.10.(h).
(b). Given arbitrary elements we use use Theorem 3.6.(iv) and the fact that is a Baire space (see Lemma 1.2.(c)) to deduce the existence of some with for . Take now with for each . If is smooth, then and hence . So every element of is a spear and then Proposition 2.10.(e) tells us that is one-dimensional.
The proof of (c) follows the lines of the one of (a). Indeed, arguing like in (a), we find and a weak∗-dense subset of whose elements satisfy (6.1) so, a fortiori, they satisfy that
[TABLE]
*Being strictly convex, we get that for every such , but this implies that , and so , is one-dimensional by the Krein-Milman Theorem. *
The following result generalizes [39, Proposition 2.5].
Proposition 6.5**.**
Let , be Banach spaces and let be an operator with the aDP.
- (a)
If the norm of is Fréchet smooth, then . 2. (b)
If and are real spaces and the norm of is Fréchet smooth, then .
* Proof. By Proposition 5.1 we have that for every weak∗-strongly exposed point of . Since the norm of is Fréchet smooth, every functional in attaining its norm is a weak∗-strongly exposed point of (see [24, Corollary I.1.5] for instance). As norm-one norm attaining functionals are dense in by the Bishop-Phelps Theorem and is norm closed by Proposition 2.10.(d), we get in fact that for every . So, given arbitrary elements we can write*
[TABLE]
which gives that every element in is a spear. Therefore, is one-dimensional by Proposition 2.10.(e).
(b). Fixed and arbitrary separable subspaces we can use Proposition 3.7 to find separable subspaces and such that and has norm one and the aDP. Next, we fix a countable dense subset and we consider given by
[TABLE]
which is countable since is countable and is smooth. Therefore, we can use the fact that is a Baire space (see Lemma 1.2.(c)) and Theorem 3.6.(iv) to deduce the existence of some with for every . We will show that . To do so, fix attaining its norm at and recall that strongly exposes as is Fréchet smooth. Let be a sequence in converging to and let satisfying for every . Then we have that
[TABLE]
so converges in norm to and, therefore,
[TABLE]
*which gives . Since norm-attaining norm-one functionals are dense in by the Bishop-Phelps Theorem, we deduce that is a spear in . Finally, Proposition 2.10.(k) tells us that , and thus , is one-dimensional as it is smooth. The arbitrariness of implies that is one-dimensional. *
The next result deals with WLUR points. Given a Banach space , a point is said to be LUR (respectively WLUR) if for every sequence in such that one has that in norm (respectively weakly). It is clear that LUR points are WLUR, but the converse result is known to be false [64].
Proposition 6.6**.**
Let , be Banach spaces and let be an operator with the aDP. Then
- (a)
If contains a WLUR point, then . 2. (b)
If contains a WLUR point, then .
* Proof. Let be a WLUR point of . We start showing that . To do so, take with and use Theorem 3.6.(iv) to find such that . Therefore, there is a sequence in satisfying*
[TABLE]
which clearly implies |y^{\ast}(Gx_{n})|=\bigl{|}[G^{\ast}y^{\ast}](x_{n})\bigr{|}\longrightarrow 1 and . Hence, there is a sequence in such that and so
[TABLE]
Now since is a WLUR point we get that converges weakly to . Therefore, converges weakly to , and the fact that tells us that .
Suppose that is not one-dimensional, then there is with . Consider the operator which satisfies . We have that since has the aDP, so there are sequences in and in such that
[TABLE]
which implies and . Hence, we may find a sequence in such that and so
[TABLE]
Since is a WLUR point we get that converges weakly to . This, together with , tells us that which is a contradiction.
(b). Let be a WLUR point of . Since has the aDP, Theorem 3.6.(iv) provides us with a dense Gδ set in such that for every . We claim that for every . Indeed, fixed , consider the rank-one operator which satisfies . So there are sequences in and in such that
[TABLE]
This implies that and that there is a sequence in such that
[TABLE]
Being a WLUR point, we deduce that converges weakly to and, therefore, we get , finishing the proof of the claim.
To finish the proof, fix and observe that
[TABLE]
*This, together with being a WLUR point, gives that . Therefore, is one-dimensional as desired. *
Our next result improves Proposition 6.4 but only for lush operators. We do not know whether it is also true for operators with the aDP.
Proposition 6.7**.**
Let , be Banach spaces and let be a norm-one operator which is lush. Then:
- (a)
If is strictly convex then . 2. (b)
In the real case, if is smooth then . 3. (c)
If is strictly convex then .
[* Proof]Given arbitrary separable subspaces and , we can use Proposition 3.26.(vi) to get the existence of separable subspaces and such that , , and is lush. Now Theorem 3.31 tells us that there exists a Gδ dense subset of such that or, equivalently, that*
[TABLE]
for every .
(a). If is strictly convex so is , and then Proposition 2.10.(h) tells us that is one-dimensional as is non-empty. Thus, is one-dimensional and its arbitrariness gives that is one-dimensional.
(b). If is smooth so is . Using this time Proposition 2.10.(l) we get that is one-dimensional as is non-empty. Therefore, is one-dimensional and its arbitrariness tells us that is one-dimensional.
*(c). In this case we have that is strictly convex. Observe that, fixed , every element in the set satisfies that , so by the strict convexity of the set G_{\infty}\bigl{(}\operatorname{Face}(S_{X_{\infty}},G_{\infty}^{\ast}y^{\ast})\bigr{)} must consist of one point. This, together with (6.2), implies that has rank one. Therefore, is non-empty by Corollary 5.9 and so (and thus ) is one-dimensional by Proposition 2.10.(h). The arbitrariness of tells us that is one-dimensional. *
Our last result in this chapter is an extension of Theorem 2.8 to arbitrary lush operators: every lush operator attains its norm (i.e. the supremum defining its norm is actually a maximum).
Proposition 6.8**.**
Let , be Banach spaces and let be a norm-one operator. If is lush, then it is norm-attaining. Actually,
[TABLE]
[* Proof]Fix an arbitrary . By Proposition 3.26, there are separable Banach spaces and satisfying that is lush. Using Theorem 3.31, there exists such that is a spear, so Theorem 2.8 gives us that*
[TABLE]
We will see in Example 8.14 that the aDP is not enough to get norm-attainment.
Chapter 7 Lipschitz spear operators
Let , be Banach spaces. We denote by the set of all Lipschitz mappings such that . This is a Banach space when endowed with the norm
[TABLE]
Observe that, clearly, with equality of norms.
Our aim in this chapter is to study those elements of which are spears. First, let us give a name for this.
Definition 7.1**.**
Let , be Banach spaces. A norm-one operator is a Lipschitz spear operator if for every .
We will prove here that every (linear) lush operator is a Lipschitz spear operator and present similar results for Daugavet centers and for operators with the aDP. To do so, we will use the technique of the Lipschitz-free space. We need some definitions and preliminary results. Let be a Banach space. Observe that we can associate to each an element which is just the evaluation map for every . The Lipschitz-free space over is the Banach space
[TABLE]
It turns out that is an isometric predual of (which has been very recently shown to be the unique predual [69]). The map establishes an isometric non-linear embedding since for all . The name Lipschitz-free space appeared for the first time in the paper [32] by G. Godefroy and N. Kalton, but the concept was studied much earlier and it is also known as the Arens-Ells space of (see [68, §2.2]). The main features of the Lipschitz-free space which we are going to use here are contained in the following result. The first four assertions are nowadays considered folklore in the the theory of Lipschitz operators, and may be found in the cited paper [32] (written for the real case, but also working in the complex case), section 2.2 of the book [68] by N. Weaver, and Lemma 1.1 of [36]. The fifth assertion was proved in [41, Lemma 2.4]. For background on Lipschitz-free spaces we refer the reader to the already cited [32, 36, 68] and the very recent survey [30] by G. Godefroy.
Lemma 7.2**.**
Let , be Banach spaces.
- (a)
For every , there exists a unique linear operator such that and . Moreover, the application is an isometric isomorphism from onto . 2. (b)
There exists a norm-one -linear quotient map which is a left inverse of , that is, . It is called the barycenter map in **[32*]**, and is given by the formula *
[TABLE] 3. (c)
From the uniqueness in item (a), it follows that for every . 4. (d)
The set
[TABLE]
is norming for , i.e. . 5. (e)
Given and a slice of ,
[TABLE]
A comment on item (e) above could be clarifying. Let be a Banach space. As and , a slice of has the form
[TABLE]
where has norm one and is a positive real number. Then, we have that
[TABLE]
is what is called in [41] a Lipschitz slice of . Then, item (e) above means that if a Lipschitz slice of does not intersect a subset , then it does not intersect either. This was proved in [41, Lemma 2.4] with a completely elemental proof. Let us also say that assertion (e) is equivalent to the following fact [5, Lemma 2.3]: given a Lipschitz slice of and a point , there is a linear slice of such that (indeed, one direction is obvious and for the non trivial one, let which clearly satisfies that ; then, and so the Hahn-Banach theorem gives the result). The proof of this last result given in [5] is independent of the above one and uses generalized derivatives and the Fundamental Theorem of Calculus for them.
The next one is the main result of this chapter. It is an application of our theory to Lipschitz-free spaces from which we will deduce the commented result about Lipschitz spear operators.
Theorem 7.3**.**
Let , be Banach spaces and let be a norm-one operator. If is lush, then is lush.
We need the following general technical result.
Lemma 7.4**.**
Let , , be Banach spaces and let and be norm-one operators. Suppose that there is a subset norming for (i.e. ) such that satisfies the following property
[TABLE]
If is lush, then is lush.
[* Proof]Fix , , and . As is lush, by Proposition 3.26.(v) we may find such that*
[TABLE]
Therefore, for every slice of containing we have that
[TABLE]
and so (P2) gives us that
[TABLE]
Therefore, we have that
[TABLE]
This has been proved for every slice of containing , but it is a fortiori also true for every slice of containing , so it follows that
[TABLE]
*As is norming for , Proposition 3.26.(v) gives us the result. *
[* Proof of Theorem 7.3] By Lemma 7.2.(e), it follows that satisfies condition (P2) of Lemma 7.4. As is lush, it follows from this lemma that is lush. But by Lemma 7.2.(c). *
The identification of with given in Lemma 7.2.(a) allows to deduce the promised result about Lipschitz spear operators from Theorem 7.3.
Corollary 7.5**.**
Let , be Banach spaces and let be a norm-one operator. If is lush, then is a Lipschitz spear operator, i.e., for every .
A first particular case of this result follows when we consider a lush Banach space and . This result appeared previously in [41, 67]
Corollary 7.6** ([41, Theorem 4.1] and [67, Theorem 2.6]).**
Let be a lush Banach space. Then, is a Lipschitz spear operator, i.e. for every .
As we commented, this result is already known, as it is contained in [67, Theorem 2.6] and [41, Theorem 4.1]. But to get it from those references, the concept of Lipschitz numerical index of a Banach space is needed. Let be a Banach space. For , the Lipschitz numerical range of [67] is
[TABLE]
the Lipschitz numerical radius of is just w_{L}(F):=\sup\bigl{\{}|\lambda|\colon\lambda\in W_{L}(F)\bigr{\}}, and the Lipschitz numerical index of is
[TABLE]
It is shown in [67, Corollary 2.3] that is a Lipschitz spear operator if and only if . With this in mind, Corollary 7.6 is just [67, Theorem 2.6] in the real case and [41, Theorem 4.1] in the complex case. Let us comment that the main difficulty of the proofs in [67] and [41] is to deal with Lipschitz operators. With our approach using the Lipschitz-free spaces, we avoid this.
Theorem 7.3 and Corollary 7.5 apply to all the lush operators presented in this manuscript. We would like to emphasise the following two particular ones, which follow from Theorem 4.1 and Corollary 4.11, respectively.
Example 7.7**.**
Let be a locally compact Abelian group and let be its dual group. Then, the Fourier transform is a Lipschitz spear operator, that is,
[TABLE]
for every .
Example 7.8**.**
The inclusion is a Lipschitz spear operator, that is,
[TABLE]
for every .
The last consequence of Theorem 7.3 (actually, of Corollary 7.5) we would like to present here is the following.
Corollary 7.9**.**
Let be a Banach space. Then,
[TABLE]
That is, every spear functional is actually a Lipschitz spear functional.
[* Proof]Let . It follows from Corollary 5.8 that is lush. Then, Corollary 7.5 gives that for every , that is, as desired. *
We would like next to deal with operators with the aDP. The main result here is that we may extend the aDP of a (linear) operator to its linearization to the Lipschitz-free space.
Theorem 7.10**.**
Let , be Banach spaces and let be a norm-one operator. If has the aDP, then has the aDP.
[* Proof]We fix and . As has the aDP, we have that*
[TABLE]
by Theorem 3.6.(iii). Then, if is an arbitrary slice of , we obviously have that
[TABLE]
and so Lemma 7.2.(e) gives us that
[TABLE]
Therefore, S\cap\bigl{\{}\xi\in S_{\mathcal{F}(X)}\colon\bigl{\|}\bigl{[}G\circ\beta_{X}\bigr{]}(\xi)+\mathbb{T}\,y_{0}\bigr{\|}>2-\varepsilon\bigr{\}}\neq\emptyset, that is, using that by Lemma 7.2.(c),
[TABLE]
The arbitrariness of gives then that
[TABLE]
As , we actually have that
[TABLE]
*and Theorem 3.6.(iii) gives that has the aDP, as desired. *
The identification of with given in Lemma 7.2.(a) allows to write Theorem 7.10 in terms of the Lipschitz norm of Lipschitz operators. We need some preliminary work to write the results only in terms of the Lipschitz operators. Let , be Banach spaces. For , we define the slope of [41] as the set
[TABLE]
Observe that if , then . On the other hand, it is clear that \operatorname{slope}(F)=\widehat{F}\bigl{(}\mathcal{B}_{X}\bigr{)} and, in particular,
[TABLE]
With this in mind, we get that if has the aDP and satisfies that is SCD, then by Theorems 7.10 and 3.16. But, actually, we can go further and avoid to use the absolutely closed convex hull in the assumption.
Corollary 7.11**.**
Let , be Banach spaces and let be a norm-one operator with the aDP. If satisfies that is SCD, then .
The result will follow from (7.1) in the proof of Theorem 7.10 and the following general result.
Lemma 7.12**.**
Let , , be Banach spaces, let such that , and let be a norm-one operator such that is a spear set for every slice of . Then, if satisfies that is SCD, then is a target for . In the case of , we have .
The proof of this lemma is an easy adaptation of the one of Theorem 3.16.
[* Proof of Corollary 7.11] If has the aDP, it follows from (7.1) in the proof of Theorem 7.10 that satisfies the hypothesis of the above lemma with . Now, if is SCD, we have that , that is, as desired. *
If we apply this result in the particular case when and , we get the following result which already appeared in [41].
Corollary 7.13** ([41, Theorem 3.7]).**
Let be a Banach space with the aDP. Then for every such that is SCD.
Our next result extends Corollary 7.11 to the non-separable case.
Corollary 7.14**.**
Let , be Banach spaces and let be a norm-one operator with the aDP. If satisfies that for every separable subspace of , is SCD, then .
The result follows immediately from Corollary 7.11 and the following lemma.
Lemma 7.15**.**
Let , be Banach spaces and let be an operator with the aDP. Given , there are separable subspaces of and of such that has the aDP, and .
[* Proof]Consider two sequences and such that for every and , let be the closed linear span in of the elements of the two sequences and let be the closed linear span of . By Proposition 3.7, there are separable subspaces of and of such that has the aDP. By construction, we have that . Now, we may apply again Proposition 3.7 starting with and the closed linear span of to get separable subspaces , such that has the aDP, , and . Repeating the process, it is straightforward to check that the separable subspaces of and of work. *
The main particular cases in which Corollary 7.14 apply are the following.
Corollary 7.16**.**
Let , be Banach spaces and let be an operator with the aDP. If satisfies that has the Radon-Nikodým property, the convex point of continuity property or it is an Asplund set, or that does not contain -sequences, then .
[* Proof]If has the Radon-Nikodým property, or the convex point of continuity property or it is an Asplund set, then is SCD for every separable subspace of (use Example 1.5). By [41, Lemma 3.1], it follows that is SCD for every separable subspace of , so Corollary 7.14 gives the result. If does not contain -sequences, neither does for every separable subspace of . This gives that is SCD for every separable subspace of (the proof of [4, Theorem 2.22] also works for not convex subsets) and then Corollary 7.14 gives the result. *
It is immediate that the above result applies to the recently introduced Lipschitz compact and Lipschitz weakly compact operators [36, Definition 2.1]: in is Lipschitz compact (respectively, Lipschitz weakly compact) if is relatively compact (respectively, relatively weakly compact).
The last aim in this chapter is to give for Daugavet centers analogous results to the ones we have for the aDP. In this case, we have to deal with the real version of the Lipschitz-free space. Given a (real or complex) Banach space , we write to denote the Lipschitz-free space of in the sense of the real scalars, that is, it is the canonical predual of . If and are real Banach spaces, nothing changes, but if they are complex spaces, we are considering only their real structure and so Lemma 7.2 is only valid for real scalars.
The main result for Daugavet centers is the following one.
Theorem 7.17**.**
Let , be Banach spaces and let be a norm-one operator. If is a Daugavet center, then is a Daugavet center.
We need the following characterization of Daugavet centers which follows immediately from [10] and which will play the role of our Theorem 3.6.(iii). In particular, it follows from it that to be a Daugavet center only depends on the real structure of the Banach spaces involved.
Lemma 7.18** (see [10, Theorem 2.1]).**
Let , be Banach spaces and let be a norm-one operator. Then is a Daugavet center if and only if
[TABLE]
for every and every .
[* Proof of Theorem 7.17] We just have to adapt mutatis mutandis the proof of Theorem 7.10, using the above lemma instead of Theorem 3.6.(iii). But observe that the game played in (7.2) and (7.3) is not valid here as the set is not rounded. At this point is where we have to go to the real version of the Lipschitz-free space, since the set is clearly -rounded and then we actually have . With this in mind, everything works. *
Our next aim is to get consequences of Theorem 7.17 just in terms of the Lipschitz norm and the slope of Lipschitz operators.
Corollary 7.19**.**
Let , be Banach spaces and let be a Daugavet center. If satisfies that is SCD, then .
This result immediately follows from Theorem 7.17 and the next proposition which allows to pass the SCD from a set to its convex hull and which may have its own interest.
Proposition 7.20**.**
Let be a Banach space and let be a bounded set. If is SCD, then is SCD.
We need a lemma which relates the slices of a set and the slices of its convex hull.
Lemma 7.21**.**
Let be a Banach space, let be a bounded set, let , and let . Then
[TABLE]
[* Proof]Every element of is a (finite) convex combination of the form where for every and*
[TABLE]
Put . Then
[TABLE]
which allows to deduce that . Since for each , the result follows from the following estimation:
[TABLE]
[* Proof of Proposition 7.20] First, by [4, Remark 2.7], it is enough to show that is SCD. Suppose that is a family of slices determining for and observe that we may and do assume that for every . We consider for the following (countable) family of slices:*
[TABLE]
Given any and any , we will show that contains one element of , showing then that is determining. Indeed, for we know that there is such that . Taking big enough, we can assure, using the previous lemma, that
[TABLE]
We can then conclude that
[TABLE]
*This shows that is SCD, as desired. *
[* Proof of Corollary 7.19] If is a Daugavet center, then is also a Daugavet center by Theorem 7.17. Now, if satisfies that is SCD, so is . Therefore, by [9, Corollary 1]. Finally, this is equivalent to by Lemma 7.2. *
We may extend the result to the non-separable case as we did for the aDP.
Corollary 7.22**.**
Let , be Banach spaces and let be a Daugavet center. If satisfies that for every separable subspace of , is SCD, then .
The result follows immediately from Corollary 7.19 and the following lemma which allows a reduction to the separable case and is completely analogous to Lemma 7.15. Its proof follows from [35, Theorem 1] in the same manner that the proof of Lemma 7.15 follows from Proposition 3.7.
Lemma 7.23**.**
Let , be Banach spaces and let be a Daugavet center. Given , there are separable subspaces of and of such that is a Daugavet center, and .
The most interesting particular cases of Corollary 7.22 are summarized in the following result, whose proof is completely analogous to the one of Corollary 7.16.
Corollary 7.24**.**
Let , be Banach spaces and let be a Daugavet center. If satisfies that has the Radon-Nikodým property, the convex point of continuity property or it is an Asplund set, or that does not contain -sequences, then .
Chapter 8 Some stability results
Our aim here is to provide several results on the stability of our properties by several operations like absolute sums, vector-valued function spaces, and ultraproducts.
8.1. Elementary results
The first result shows that we may produce an injective operator with the aDP from any operator with the aDP.
Proposition 8.1**.**
Let , be Banach spaces and let be an operator with the aDP and let be the quotient map. Then, the quotient operator \widetilde{G}\in L\big{(}X/\ker G,Y\big{)} satisfying has the aDP.
[* Proof]By Theorem 3.6 it suffices to show that is a spear set in for every slice of So, we fix an arbitrary slice of and find and such that*
[TABLE]
Since , the set S=\bigl{\{}x\in B_{X}\colon\operatorname{Re}[P^{\ast}z^{\ast}](x)>1-\alpha\bigr{\}} is a slice of . Observe that if then and which give that . Therefore, we have that and so
[TABLE]
*Now, as is a spear set by Theorem 3.6, so is a fortiori , as desired. *
The reciprocal result is not true: consider given by and observe that is clearly lush, while does not even have the aDP (use Proposition 5.6 for instance).
Our next aim is to provide a way to extend the domain and the codomain keeping the properties of being spear, lush, or the aDP. The first result deals with extending the domain.
Proposition 8.2**.**
Let , , be Banach spaces, let be a norm-one operator, and consider the norm-one operator given by for every . Then:
- (a)
if is a spear operator, so is ; 2. (b)
if has the aDP, so does ; 3. (c)
if is lush, so is .
* Proof. Fix with and . Take and satisfying . Now pick so that and define the operator by*
[TABLE]
which satisfies and so . Now we can estimate as follows
[TABLE]
The arbitrariness of gives .
(b). Just observe that if is a rank-one operator in the argument above, then also is rank-one.
(c). Consider , and . As is lush, Proposition 3.26.(iii) allows to find such that
[TABLE]
Now, observe that and it is then immediate that
[TABLE]
*Now, Proposition 3.26.(iii) gives that is lush. *
We can get an analogous result to extend the codomain space.
Proposition 8.3**.**
Let , , be Banach spaces, let be a norm-one operator, and consider the norm-one operator given by for every . Then:
- (a)
if is a spear operator, so is ; 2. (b)
if has the aDP, so does ; 3. (c)
if is lush, so is .
* Proof. Fix with , , and such that . Denote by and the respective projections from to and . Take satisfying and pick so that . Now define by*
[TABLE]
which satisfies
[TABLE]
Finally, using the triangle inequality and the fact that is a spear operator, we can estimate as follows:
[TABLE]
The arbitrariness of finishes the proof.
(b). Observe that if is a rank-one operator in the argument above, then also is rank-one.
(c). Fix , and . As is lush, we may find such that
[TABLE]
Indeed, if , apply Proposition 3.26.(iii) to ; if , we may apply that proposition to any vector in . Now, pick such that and consider . Observe that, on the one hand,
[TABLE]
and, on the other hand, , so we have that
[TABLE]
*We then get that is lush by Proposition 3.26.(iii). *
8.2. Absolute sums
We show in this section the stability of our properties by , , and sums of Banach spaces. The following result borrows the ideas from [58, Proposition 1] and [12, §5].
Proposition 8.4**.**
Let , be two families of Banach spaces and let be a norm-one operator for every . Let be one of the Banach spaces , , or , let and , and define the operator by G\bigl{[}(x_{\lambda})_{\lambda\in\Lambda}\bigr{]}=(G_{\lambda}x_{\lambda})_{\lambda\in\Lambda} for every . Then
- (a)
* is a spear operator if and only if is a spear operator for every ;* 2. (b)
* has the aDP if and only if has the aDP for every ;* 3. (c)
* is lush if and only if is lush for every .*
* Proof. We suppose first that is a spear operator and, fixed , we have to show that is a spear operator. Observe that calling and , we can write and when is or and and when is . Given a non-zero operator , define by which obviously satisfies . Let and denote the projections from onto and respectively. When is or we can write*
[TABLE]
Since , it follows that and so is a spear operator. When is equality can be continued as follows
[TABLE]
We prove now the sufficiency when is or . Given an operator and fixed , we find such that and write where . Since is the convex hull of we may find and such that
[TABLE]
Now fix with and define the operator given by
[TABLE]
which satisfies . Observe finally that
[TABLE]
So, the arbitrariness of gives that , finishing the proof for .
Suppose now that . Fix an operator and observe that it may be seen as a family of operators where for every , and . Given , find such that , and write , , and where and . Now we choose such that
[TABLE]
we find , satisfying
[TABLE]
and define the operator by
[TABLE]
Then
[TABLE]
Moreover, since is a spear operator, fixed we may find and such that
[TABLE]
Now take and , and observe that
[TABLE]
So, the arbitrariness of gives that , finishing the proof for .
(b). For the aDP, the arguments above apply just taking into account that when one starts with rank-one operators, the constructed operators are also rank-one.
(c). We assume first that is lush. Fixed , , , we consider the elements and given by
[TABLE]
Now Proposition 3.26.(iii) provides with such that
[TABLE]
From this point we have to distinguish two cases depending on the space . Suppose first that or and observe that . In this case, given , it follows that which, together with (8.1), allows us to deduce that
[TABLE]
We consider now the more bulky case in which . Using (8.1) we can find scalars with and elements such that
[TABLE]
Since , we deduce that so Lemma 8.13 tells us that the set satisfies that . Hence
[TABLE]
But every satisfies that
[TABLE]
from where it follows that
[TABLE]
for each . This, together with (8.2), tells us that
[TABLE]
finishing the proof of the necessity for .
Let us prove the sufficiency when or . Fixed , , and , there is such that . Using that is lush we may find satisfying
[TABLE]
Defining by for every , we clearly have . Observe that, fixed and , the element given by for and belongs to . Using this it is easy to deduce that
[TABLE]
which tells us that is lush by Proposition 3.26.(iii).
Suppose that . We take the set \mathcal{B}=\bigl{\{}(x_{\lambda})_{\lambda\in\Lambda}\in B_{X}\colon\#\operatorname{supp}(x_{\lambda})=1\bigr{\}} which is norming for . Fixed , there is so that and for . Given and , we may and do assume that and, since is lush, we may use Proposition 3.26.(iii) for and to find such that
[TABLE]
For each we take satisfying and we define by
[TABLE]
Then it obviously follows that and, thanks to the shape of , one can easily deduce that
[TABLE]
*This finishes the proof by using Proposition 3.26.(iii) since . *
8.3. Vector-valued function spaces
Our next aim is to present several results concerning the behaviour of our properties for vector-valued function spaces. We start analysing the situation for spaces of continuous functions.
Theorem 8.5**.**
Let be Banach spaces, let be a compact Hausdorff topological space and let be a norm-one operator. Consider the norm-one composition operator given by for every . Then:
- (a)
* is a spear operator if and only if is a spear operator.* 2. (b)
* is lush if and only if is lush.* 3. (c)
If contains isolated points, then has the aDP if and only if does. 4. (d)
If is perfect, then has the aDP if and only if is a spear set.
Remark 8.6*.*
All the information given in the above result was previously known for the case of the identity (see [40, 56, 57]).
[* Proof of Theorem 8.5.(a)] This is an easy adaptation of [57, Theorem 5]. Suppose first that is a spear operator. Fixed with and , find with and such that*
[TABLE]
Define and find a continuous function such that and if . Now write with , , and consider the functions
[TABLE]
Then meaning that
[TABLE]
and, using (8.3), we must have
[TABLE]
By making the right choice of or we get such that
[TABLE]
Next, we fix with , denote
[TABLE]
and consider the operator given by
[TABLE]
which, by (8.4), obviously satisfies . Now, we use that is a spear operator to find satisfying , and observe that
[TABLE]
The arbitrariness of gives that and so is a spear operator.
Suppose conversely that is a spear operator. Fix , and define the operator T\in L\big{(}C(K,X),C(K,Y)\big{)} by
[TABLE]
which satisfies . Since is a spear operator we may find and such that
[TABLE]
and we can write
[TABLE]
*The arbitrariness of tells us that is a spear operator. *
We next deal with lushness for spaces of vector-valued continuous functions.
[* Proof of Theorem 8.5.(b)] Suppose that is lush and let us show that is lush. This part of the proof is an easy adaptation of [40, Proposition 5.1]. Let , , and be fixed. Then, we take with and, using that is lush together with Proposition 3.26.(iii), we find such that*
[TABLE]
So, there are , with , and such that
[TABLE]
Next, we take an open set such that and
[TABLE]
and we fix a continuous function with and . Now we consider the functional given by for which clearly satisfies . Finally, for each , we define by
[TABLE]
and we observe that
[TABLE]
Therefore, we deduce that for every . On the other hand, for an arbitrary we have that
[TABLE]
So, if , then and, therefore,
[TABLE]
If, otherwise, , then and thus . All this tells us that
[TABLE]
and shows that is lush.
Suppose now that is lush and let us show that is lush. To do so, we will use Proposition 3.26.(iii) with the set
[TABLE]
where for . Observe that is norming and rounded, so . Fixed , , and , we consider and given respectively by
[TABLE]
Now we use that is lush and Proposition 3.26.(iii) to find such that
[TABLE]
Therefore, as , we clearly get that . Moreover, using that if then , we easily deduce that
[TABLE]
*which gives that is lush. *
[* Proof of Theorem 8.5.(c)] We start showing that has the aDP when does. Indeed, observe that in the first part of the proof of Theorem 8.5.(a), if the operator has rank one then so does the operator constructed there.*
To prove the reversed implication, fix an isolated point and observe that we can identify and . Now, if for we write
[TABLE]
and we consider the operator given by
[TABLE]
then, we can write
[TABLE]
*Therefore, as has the aDP, we may use Proposition 8.4 to deduce that has aDP. *
[* Proof of Theorem 8.5.(d)] We prove first the sufficiency. We will use Theorem 3.6.(iii) to show that has the aDP. So, fixed and , we write*
[TABLE]
and we have to show that . The argument follows the lines of [71, p. 81]: let be the open set and pick, given , open pairwise disjoint non-void subsets and points . Next, we use the hypothesis to find and such that . Now, fixed , we may choose functions such that in and . Indeed, take Urysohn functions such that
[TABLE]
and define
[TABLE]
On the one hand, observe that :
[TABLE]
On the other hand, for we have that
[TABLE]
and, for , it follows that . This proves that and so has the aDP.
Suppose now that has the aDP. Fixed and a non-zero , we take the constant function given by and we use Theorem 3.6.(iii) to find such that . So, there is satisfying and, therefore, is the element in we are looking for:
[TABLE]
We next deal with spaces of essentially bounded measurable functions.
Theorem 8.7**.**
Let be a Banach spaces, let be a -finite measure space and let be a norm-one operator. Consider the norm-one composition operator given by for every . Then:
- (a)
* is a spear operator if and only if is a spear operator.* 2. (b)
* is lush if and only if is lush.* 3. (c)
If has an atom, then has the aDP if and only if does. 4. (d)
If is nonatomic, then has the aDP if and only if is a spear set.
Remark 8.8*.*
The results in items (a), (c), and (d) of the above theorem were known for the case of the identity (see [56, 59]). The content of (b) is completely new even for the identity.
Corollary 8.9**.**
Let be a Banach space and let be a -finite measure space. Then, is lush if and only if is lush.
We will use the following notation during the proof of the theorem:
[TABLE]
Observe that when the measure is finite, this notation is consistent with the one given in section 4.3. Moreover, we will also use the subset of given by
[TABLE]
which clearly satisfies that
[TABLE]
[* Proof of Theorem 8.7.(a)] This is an easy adaptation of [59, Theorem 2.3]. Suppose first that is a spear operator. We fix T\in L\big{(}L_{\infty}(\mu,X),L_{\infty}(\mu,Y)\big{)} with . Given we may follow the first part of the proof of [59, Theorem 2.3] to find , , and with , such that*
[TABLE]
Now we fix with , we write
[TABLE]
and we define the operator given by
[TABLE]
which, by (8.5), satisfies . Next, we use that is a spear operator to find such that , so we can take satisfying
[TABLE]
Finally, define the functional by
[TABLE]
and observe that
[TABLE]
The arbitrariness of gives that is a spear operator.
Assume now that is a spear operator. Fix and define the operator T\in L\big{(}L_{\infty}(\mu,X),L_{\infty}(\mu,Y)\big{)} by
[TABLE]
which clearly satisfies . As we mentioned at the beginning of the proof, we can find , , and with , such that and
[TABLE]
Therefore, we can write
[TABLE]
*Thus, we get and so is a spear operator. *
We next deal with lushness for .
[* Proof of Theorem 8.7.(b)] Assume first that is lush. To prove that so is , we will check that Proposition 3.26.(iii) is satisfied. Let , and . By density, we can assume that can be written as*
[TABLE]
where is a countable partition of and , for each . Since , we can assume without loss of generality that there is with . Using that is lush, we can find such that
[TABLE]
and elements , , for () with satisfying that
[TABLE]
Consider now , and for each let
[TABLE]
Then, by (8.6) we have that
[TABLE]
A similar argument shows that
[TABLE]
for every , since . Moreover, using (8.7) we immediately conclude that
[TABLE]
Let us see the converse: assume that is lush, fix an element , and let , and . Then, defining and , we can use the hypothesis to find such that
[TABLE]
Since we can write for some and , condition can be rewritten as
[TABLE]
By (8.8) we find elements , and for satisfying and
[TABLE]
Next, for each , we consider the element
[TABLE]
which clearly satisfies that
[TABLE]
Moreover, making use of (8.9) we get that
[TABLE]
and, combining this with (8.10), we conclude that
[TABLE]
*which finishes the proof. *
[* Proof of Theorem 8.7.(c)] We start showing that when has the aDP so does . Indeed, observe that in the first part of the proof of Theorem 8.7.(a), if the operator has rank one then the operator constructed there also has rank one.*
To prove the reversed implication, fix which is an atom for and observe that we can identify
[TABLE]
and
[TABLE]
Now, if for we write
[TABLE]
and we consider the operator given by
[TABLE]
then we can write
[TABLE]
*Therefore, as has the aDP, we may use Proposition 8.4 to deduce that has the aDP. *
[* Proof of Theorem 8.7.(d)] We prove first the sufficiency. We will use Theorem 3.6.(iii) to show that has the aDP. So, fixed and , we write*
[TABLE]
and we have to show that . Using Lemma 2.2 in [59], we may find and such that
[TABLE]
As is atomless, for every we may and do pick pairwise disjoint sets with positive measure such that for every . Next, we use that is a spear set to find and such that . Now, fixed , we define for . On the one hand, observe that for every we get the following estimation
[TABLE]
so which implies that . On the other hand, for we have that
[TABLE]
and, for , it follows that . This proves that and so has the aDP.
Conversely, suppose now that has the aDP. Fixed and a non-zero , we take the constant function given by and we use Theorem 3.6.(iii) to find such that . So, there is satisfying that and, therefore,
[TABLE]
*This shows that is a spear set, concluding thus the proof. *
Finally, we would like to work with spaces of integrable functions.
Theorem 8.10**.**
Let be Banach spaces, let be a -finite measure space, and let be a norm-one operator. Consider the norm-one composition operator given by for every . Then:
- (a)
* is a spear operator if and only if is a spear operator.* 2. (b)
* is lush if and only if is lush.* 3. (c)
If has an atom, then has the aDP if and only if has the aDP. 4. (d)
If is nonatomic, then has the aDP if and only if
[TABLE]
Remark 8.11*.*
The results in items (a), (c), and (d) of the above theorem were known for the case of the identity (see [56, 57]). The content of (b) is completely new even for the identity.
Corollary 8.12**.**
Let be a Banach space and let be a -finite measure space. Then, is lush if and only if is lush.
We claim that for the proof of Theorem 8.10 we can assume without loss of generality that is a probability space. This is clear if is finite by normalizing the measure. If is -finite but not finite, we can find a countable partition of . For each we have a finite measure space where
[TABLE]
Moreover, we have a canonical isometry
[TABLE]
(see [23, pp. 501], for instance). Then, the composition operator of Theorem 8.10, can be seen as where is the composition operator correspondant to . An application of Proposition 8.4 gives that is a spear operator (resp. lush, aDP) if and only if is a spear operator (resp. lush, aDP) for each . Therefore, the claim is valid for the proofs of (a) and (b). In the case of (c) and (d) it is also valid taking into account that has an atom if and only if some does, and that if has the aDP then it satisfies
[TABLE]
Hence, the claim is established.
In order to prove Theorem 8.10 we need to introduce some notation. If is a probability space, the set
[TABLE]
satisfies that
[TABLE]
since is rounded and it is clearly norming for the simple functions of . On the other hand, we will write
[TABLE]
which satisfies that
[TABLE]
Indeed, it is enough to notice that every simple function in belongs to the convex hull of : such an can be written as , where is a finite family of pairwise disjoint sets of and for each . Then
[TABLE]
and hence
[TABLE]
[* Proof of Theorem 8.10.(a)] Suppose first that is a spear operator. Fix and consider \widetilde{T}\in L\big{(}L_{1}(\mu,X),L_{1}(\mu,Y)\big{)} given by , which satisfies . Given , we can find and such that*
[TABLE]
But notice that
[TABLE]
This, together with (8.13) and the arbitrariness of , tells us that is a spear operator.
Assume now that is a spear operator. Fixed T\in L\big{(}L_{1}(\mu,X),L_{1}(\mu,Y)\big{)} with and , we may find by (8.12) elements and such that
[TABLE]
Using now (8.11), there exists , where is a finite partition of into sets of and for each , satisfying that
[TABLE]
Then, we can write
[TABLE]
so by a standard convexity argument we can assume that there is such that and (8.14) is still satisfied. By the density of norm-attaining functionals, we can and do assume that every is norm-ataining, so there is such that . Define the operator by
[TABLE]
It is easy to check that , and moreover since as a consequence of (8.14) we obtain that
[TABLE]
By hypothesis, we can then find and such that . We claim that
[TABLE]
Indeed, the left-hand side expression of (8.15) is equal to
[TABLE]
Our next aim is to deal with lushness. To this end, we will make use of the following immediate numerical result which we prove for the sake of completeness.
Lemma 8.13**.**
Let , , and let for . Suppose that are such that for and satisfy . Then,
[TABLE]
In particular, if , then
[TABLE]
[* Proof]Calling it suffices to observe that*
[TABLE]
*from where it easily follows that . The last claim is clear. *
[* Proof of Theorem 8.10.(b)] Assume that is lush. To check that is lush, we just have to show that Proposition 3.26.(iii) is satisfied. Fix , and of the form for some and . By density, we can assume that*
[TABLE]
where is a finite partition of and satisfy that . By Proposition 3.26.(iii), the lushness of lets us find for each an element such that and
[TABLE]
Let , which satisfies that as
[TABLE]
Our aim is to prove now that
[TABLE]
which will finish the proof. First notice that for each with we have that
[TABLE]
since every satisfies
[TABLE]
In particular, if for each we take an element satisfying , which exists by (8.16), then the inclusion in (8.18) yields that
[TABLE]
Moreover we have that
[TABLE]
which shows that (8.17) holds.
Let us see the converse: To check that is lush we will show that condition (iii) of Proposition 3.26 is satisfied, for which let , and . The mentioned condition applied to the lush operator for , and provides , functions
[TABLE]
and scalars , with satisfying that
[TABLE]
By density, we can assume that the functions are simple and moreover that there is a finite partition of such that
[TABLE]
where and for every . Then, conditions (8.19) and (8.20) can be rewritten as
[TABLE]
[TABLE]
and
[TABLE]
Applying Lemma 8.13 to (8.21) and (8.22), we obtain respectively that
[TABLE]
and
[TABLE]
Using that , the last inequality yields in particular that
[TABLE]
Combining the relations of (8.23) in a convex sum with the ’s as coefficients we obtain that
[TABLE]
Actually, from the right-hand equality of the previous expression we get that
[TABLE]
which, together with (8.26), implies that the number
[TABLE]
satisfies the relation
[TABLE]
A simple computation shows that necessarily , and so
[TABLE]
On the other hand, an application of Lemma 8.13 to the left-hand side part of (8.27) gives that
[TABLE]
Using that combined with (8.24), (8.25), (8.28), and (8.29), we deduce the existence of some satisfying simultaneaously
[TABLE]
[TABLE]
[TABLE]
and
[TABLE]
If we denote
[TABLE]
then again we can apply Lemma 8.13 to (8.32) with
[TABLE]
to get that
[TABLE]
This, together with (8.31), yields that
[TABLE]
Consider now the elements
[TABLE]
which satisfy
[TABLE]
Finally, using (8.30) and (8.33) we conclude that
[TABLE]
*which finishes the proof. *
[* Proof of Theorem 8.10.(c)] Let us fix an atom . Assume that has the aDP. We will show that satisfies condition (iii) of Theorem 3.6: Let , and . By hypothesis, we have that*
[TABLE]
Then, for each we can find a finite family , elements satisfying
[TABLE]
and scalars with such that
[TABLE]
But is either [math] or for each as is an atom. Then, if we just integrate in (8.35) over the atom we will get that
[TABLE]
In particular, this yields that
[TABLE]
Combining (8.36) and (8.37), we deduce that
[TABLE]
Since was arbitrary, we deduce from (8.38) that belongs to the closed absolute convex hull of the set of all for which there is satisfying and
[TABLE]
But then such elements and satisfy in particular that
[TABLE]
and hence
[TABLE]
We then conclude that
[TABLE]
Let us prove now the converse of (c). We remark here that this implication does not use that is atomic. Assuming that has the aDP, we will now check that satisfies Theorem 3.6.(ii). For this, it is enough to prove that given a simple function of the form
[TABLE]
where is a finite partition of and (), we have that
[TABLE]
Let and . Then
[TABLE]
and so in order to show that
[TABLE]
we can assume without loss of generality, by using a standard convexity argument, that is contained in some . Since has the aDP, using Theorem 3.6.(iii), for each there is a finite set
[TABLE]
such that . In particular, this implies that
[TABLE]
Finally, notice that each satisfies that
[TABLE]
Therefore, (8.39) leads to
[TABLE]
*for arbitrary . *
[* Proof of Theorem 8.10.(d)] Assuming hat has no atoms, we claim that given a simple function , for every we can write as*
[TABLE]
where is a finite partition of and the coefficients for each . Let us check this: of course, we can write as in (8.40) for a partition and elements with . But since has no atoms, we can find for each a partition of into elements satisfying . If is the collection of all such subsets, then this is a finer partition than and
[TABLE]
and the proof of the claim is over.
Let us then prove now that if
[TABLE]
for every , then has the aDP. By Theorem 3.6.(iii), it is enough to show that given a simple function as in (8.40) and , we have that
[TABLE]
for every . Let , and . Let and () as in the claim above for the given . We can moreover assume that every is either contained in or in . Using (8.41), we can find , with and such that and
[TABLE]
Then, it is easy to check that
[TABLE]
This shows that is -approximated by an absolutely convex sum of elements of the form for some and . Finally, notice that every such element satisfies that
[TABLE]
Therefore,
[TABLE]
Using that was arbitrary, we conclude the result.
Conversely, suppose that has the aDP and let , and . By Theorem 3.6.(iii), we have that
[TABLE]
Therefore, given any there is a finite set and elements , () such that satisfying that
[TABLE]
and
[TABLE]
It easily follows from (8.44) that
[TABLE]
On the other hand, (8.45) yields that
[TABLE]
Therefore,
[TABLE]
*and since and were arbitrary, we conclude that (8.41) holds. *
We may use Theorem 8.10.(d) to produce an example of an operator with the aDP which does not attain its norm. Recall that it was proved in Proposition 6.8 that this cannot happen if the operator is actually lush.
Example 8.14**.**
Let be the operator given by
[TABLE]
The operator defined by for has the aDP and it does not attain its norm.
[* Proof]It is clear that and that it does not attain its norm. Moreover, for each it easy to check that*
[TABLE]
Indeed, fixed , , and , we take large enough to satisfy
[TABLE]
and we consider given by
[TABLE]
Then, it is clear that and
[TABLE]
Now the arbitrariness of gives that .
Finally, has the aDP thanks to Theorem 8.10.(d) as (8.46) holds for every . Besides, for each non-zero we have that
[TABLE]
*since does not attain its norm. So we deduce that and, therefore, does not attain its norm. *
8.4. Target operators, lushness and ultraproducts
Now, we will prove the stability of target operators and lush operators with respect to the operation of taking ultraproducts. These results extend Corollaries 4.4 and 4.5 of [12] about stability of lush spaces with respect to ultraproducts.
Let us recall the basic definitions, taken from [34]. Let be a free ultrafilter on . The limit of a sequence with respect to the ultrafilter is denoted by , or , if it is necessary to stress that the limit is taken with respect to the variable . Let be a sequence of Banach spaces. We can consider the -sum of the family, , together with its closed subspace
[TABLE]
The quotient space is called the ultraproduct of the family relative to the ultrafilter . Let stand for the element of containing a given representative . It is easy to check that
[TABLE]
Moreover, every can be represented as in such a way that for all .
If all the are equal to the same Banach space , the ultraproduct of the family is called the -ultrapower of . We denote this ultrapower by .
Let , be two sequences of Banach spaces and let be a norm-bounded sequence of operators where for every . We denote the operator that acts from to as follows: . Evidently,
[TABLE]
Now, we state our main result about ultraproducts.
Theorem 8.15**.**
Let be a free ultrafilter on , , , be sequences of Banach spaces and let , be norm bounded sequences of operators such that and for every . If each is a target for the corresponding for every , then is a target for .
We need the following easy remark about the absolutely convex hull of a convex set. In fact, this idea already appeared implicitly in the proof of implication (i) (iii) of Corollary 2.7.
Proposition 8.16**.**
Let be a convex set. If is a real space, then
[TABLE]
If is a complex space, then for every and every there are , and such that
[TABLE]
[* Proof]We demonstrate only the more complicated complex case. As there are , with , and satisfying*
[TABLE]
Taking into account that the points form a -net of we can represent the set of indices as a disjoint union of sets , in such a way that
[TABLE]
Let us show that
[TABLE]
and
[TABLE]
fulfill the desired condition (8.47). Indeed, it is clear that and . Now,
[TABLE]
[* Proof of Theorem 8.15] We demonstrate the theorem only for the more complicated complex case. Also, we may and do suppose that for every .*
Let , and be fixed. Evidently, the “coordinates” can be selected in such a way that and for every . For each applying ( ‣ 3.9) in Definition 3.9 for , and we obtain the corresponding satisfying
[TABLE]
Without loss of generality, we assume that is convex, otherwise we just substitute by its convex hull. Our choice means that there is such that
[TABLE]
Select such that . Using Proposition 8.16, we can find for each corresponding , and such that
[TABLE]
and, consequently,
[TABLE]
For each denote and . Also, denote . Because , and is convex, by (8.48) we have
[TABLE]
Also, since is fixed, (8.49) implies
[TABLE]
that is
[TABLE]
*Consequently, satisfies ( ‣ 3.9) for , and , i.e., is the set we are looking for. *
In the case of ultrapowers the converse result is also true.
Theorem 8.17**.**
Let be a free ultrafilter on , let , , be Banach spaces, and let and be operators. If is a target for , then is a target for .
[* Proof]For given , and , we apply the definition of target to and . By ( ‣ 3.9) in Definition 3.9, we can find , with , and such that*
[TABLE]
and
[TABLE]
Write and denote by the set of those for which
[TABLE]
We claim that . Indeed, if this is not so, then . For every choose with such that
[TABLE]
Then, for we have
[TABLE]
which contradicts (8.50).
Now, since , according to (8.51) there is an such that
[TABLE]
*The corresponding fulfills ( ‣ 3.9) in Definition 3.9. *
Since lushness of an operator reduces to the fact that the identity operator is a target for it, we obtain the following two corollaries.
Corollary 8.18**.**
Let be a free ultrafilter on , , be sequences of Banach spaces, be a sequence of lush operators where for every . Then is lush.
Corollary 8.19**.**
Let be a free ultrafilter on , , be Banach spaces, . If is lush, then is lush.
Chapter 9 Open problems
Corresponding to Spear sets and spear vectors:
Problem 9.1**.**
Let be a complex Banach space. If is not compact, does contain a copy of or ?
Problem 9.2**.**
If is a complex smooth Banach space and , can we deduce that ?
Corresponding to Lush operators:
Problem 9.3**.**
Are items (a) and (b) in Proposition 3.30 necessary for to be lush? If there is a counterexample, notice that the domain must be non separable.
Corresponding to Examples in classical Banach spaces:
Problem 9.4**.**
Is lush the dual of the Fourier transform on ? Is lush the Fourier-Stieltjes transform?
Problem 9.5**.**
Is Proposition 4.2 always an equivalence?
Problem 9.6**.**
Is there a characterization of lush or spear operators acting from an space analogous to the one given in Theorem 4.17 for the aDP?
Corresponding to Further results:
Problem 9.7**.**
Are spearness and lushness equivalent when the codomain space is SCD? The aDP and spearness are, see Remark 5.4.
Problem 9.8**.**
Are the aDP and lushness equivalent when the image of the operator is Asplund? They are equivalent when the codomain is Asplund, see Proposition 5.3.
Problem 9.9**.**
Can the results about rank-one operators be extended to finite-rank operators? They are Corollary 5.9 and Proposition 5.23.
Problem 9.10**.**
If is lush and is -embedded, is lush? This is true for spearness and the aDP (see Proposition 5.24).
Corresponding to Isomorphic and isometric consequences:
Problem 9.11**.**
Is Theorem 6.1 valid in the complex case? That is, does the dual of the domain of a complex operator with the aDP always contain ?
Problem 9.12**.**
Let be an operator with the aDP. Does if is strictly convex or smooth? Does if is strictly convex or smooth?
Problem 9.13**.**
Does every spear operator attain its norm? This is true for lush operators, see Proposition 6.8, but it is not true for operators with the aDP, see Example 8.14.
Corresponding to Stability results:
Problem 9.14**.**
Are there results about the relationship between spear and lush operators with quotients by the kernel of the operator analogous to the one given in Proposition 8.1 for the aDP?
Index
- §3.3
- §4.3
- §4.3
- §4.3
- §4.3
- absolutely convex hull §1.2
- aDP §1.3
- aDP for an operator §3.2
- algebra numerical range §1.4
- alternative Daugavet property §1.3
- alternative Daugavet property for an operator §3.2
- approximated spatial numerical range with respect to an operator §1.4
- barycenter map item (b)
- C-rich §4.2
- -superspace Definition 4.14
- §1.2
- convex hull §1.2
- Daugavet center §1.5
- Daugavet property §1.3
- denting §1.2
- DPr §1.3
- §1.2
- face §1.2
- §1.2
- Fourier transform §4.1
- Fréchet smooth §1.2
- generalized face §1.2
- generalized slice §1.2
- geometrically unitary §1.4
- hereditarily SCD §3.3
- HSCD-majorized §3.3
- intrinsic numerical range §1.4
- intrinsic numerical range with repect to an operator §1.4
- §1.2
- -sum §1.2
- -embedded §5.3
- -sum §1.2
- §1.2
- Lipschitz numerical index Chapter 7
- Lipschitz numerical range Chapter 7
- Lipschitz slice Chapter 7
- Lipschitz spear operator Chapter 7
- Lipschitz-free space Chapter 7
- -summand §1.2
- LUR Chapter 6
- lush §1.3
- lush operator §3.4
- -embedded §5.3
- -summand §1.2
- nicely embedded Remark 4.13
- norming §1.2
- numerical index §1.4
- numerical index one §1.3
- numerical radius §1.4
- numerical range §1.4
- Riesz representable §4.3
- Riesz set §4.2
- rounded §1.2
- SCD §1.3
- SCD operator §1.3
- SCD space §1.3
- semi-Riesz set §4.2
- slice §1.2
- §1.2
- slicely countably determined §1.3
- slope Chapter 7
- smooth §1.2
- spatial numerical range §1.4
- spear Chapter 2
- spear operator §3.1
- spear set Chapter 2
- spear vector Chapter 2
- Chapter 2
- strictly convex §1.2
- strongly extreme §1.2
- supp §4.2
- target §3.3
- ultrapower §8.4
- ultraproduct §8.4
- uniform algebra §4.2
- §1.4
- §1.4
- vertex §1.4
- §1.4
- weak∗-denting §1.2
- WLUR Chapter 6
- -face §1.2
- -slice §1.2
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