Generating operators between Banach spaces

We introduce and study the notion of generating operators as those norm-one operators $G\colon X\longrightarrow Y$ such that for every $0<\delta<1$, the set $\{x\in X\colon \|x\|\leq 1,\ \|Gx\|>1-\delta\}$ generates the unit ball of $X$ by closed convex hull. This class of operators includes isometric embeddings, spear operators (actually, operators with the alternative Daugavet property), and other examples like the natural inclusions of $\ell_1$ into $c_0$ and of $L_\infty[0,1]$ into $L_1[0,1]$. We first present a characterization in terms of the adjoint operator, make a discussion on the behaviour of diagonal generating operators on $c_0$-, $\ell_1$-, and $\ell_\infty$-sums, and present examples in some classical Banach spaces. Even though rank-one generating operators always attain their norm, there are generating operators, even of rank-two, which do not attain their norm. We discuss when a Banach space can be the domain of a generating operator which does not attain its norm in terms of the behaviour of some spear sets of the dual space. Finally, we study when the set of all generating operators between two Banach spaces $X$ and $Y$ generates all non-expansive operators by closed convex hull. We show that this is the case when $X=L_1(\mu)$ and $Y$ has the Radon-Nikod\'ym property with respect to $\mu$. Therefore, when $X=\ell_1(\Gamma)$, this is the case for every target space $Y$. Conversely, we also show that a real finite-dimensional space $X$ satisfies that generating operators from $X$ to $Y$ generate all non-expansive operators by closed convex hull only in the case that $X$ is an $\ell_1$-space.


Introduction
Let X and Y be Banach spaces over the field K (K = R or K = C).We denote by L(X, Y ) the space of all bounded linear operators from X to Y and write X * = L(X, K) to denote the dual space.By B X and S X we denote the closed unit ball and the unit sphere of X, respectively, and we write T for the set of modulus one scalars.Some more notation and definitions (which are standard) are included in Subsection 1.1 at the end of this introduction.
The concept of spear operator was introduced in [1] and deeply studied in the book [7].A norm-one operator G ∈ L(X, Y ) is said to be an spear operator if the norm equality max θ∈T G + θT = 1 + T holds for all T ∈ L(X, Y ).This concept extends the properties of the identity operator in those Banach spaces having numerical index one and it is satisfied, for instance, by the Fourier transform on L 1 .There are isometric and isomorphic consequences on the domain and range spaces of a spear operator as, for instance, in the real case, the dual of the domain of a spear operator with infinite rank has to contain a copy of ℓ 1 .For more information and background, we refer the interested reader to the already cited book [7].Even though the definition of spear operator given above does not need numerical ranges, it is well known that spear operators are exactly those operators such that the numerical radius with respect to them coincides with the operator norm.Let us introduce the relevant definitions.Fixed a norm-one operator G ∈ L(X, Y ), the numerical radius with respect to G is the seminorm defined as v G (T ) := sup{|φ(T )| : φ ∈ L(X, Y ) * , φ(G) = 1} = inf δ>0 sup{|y * (T x)| : y * ∈ S Y * , x ∈ S X , Re y * (Gx) > 1 − δ} for every T ∈ L(X, Y ) (the equality above was proved in [14,Theorem 2.1]).Observe that v G (•) is a seminorm in L(X, Y ) which clearly satisfies Then, G is a spear operator if and only if v G (T ) = T for every T ∈ L(X, Y ) (see [7,Proposition 3.2]).
Our discussion here starts with the observation that it is possible to introduce a natural seminorm between v G (T ) and T in Eq. ( 1): the (semi-)norm relative to G. Let us introduce the needed notation and definitions.Let X, Y , Z be Banach spaces and let G ∈ L(X, Y ) be a norm-one operator.For δ > 0, we write att(G, δ) to denote the δ-attainment set of G, that is, att(G, δ) := {x ∈ S X : Gx > 1 − δ}.
If there exists x ∈ S X such that Gx = 1, we say that G attains its norm and we denote by att(G) the attainment set of G: att(G) := {x ∈ S X : Gx = 1}.We consider the parametric family of norms on L(X, Z) defined by which are equivalent to the usual norm on L(X, Z) (this is so since att(G, δ) has nonempty interior).We are interested in the (semi-)norm obtained taking infimum on this parametric family.
Definition 1.1.Let X, Y and Z be Banach spaces and let G ∈ L(X, Y ) be a norm-one operator.For T ∈ L(X, Z), we define the (semi-)norm of T relative to G by When Z = Y , we clearly have that and so this • G is the promised seminorm to extend Eq. ( 1).We may study the possible equality between v G (•) and • G and between • G and the usual operator norm.We left the first relation for a subsequent paper which is still in process [9].The main aim in this manuscript is to study when the norm equality T G = T (2) holds true.Definition 1.2.Let X, Y be Banach spaces.We say that G ∈ L(X, Y ) with norm-one is generating (or a generating operator ) if equality (2) holds true for all T ∈ L(X, Y ).We denote by Gen(X, Y ) the set of all generating operators from X to Y .
Observe that both • G and the operator norm can be defined for operators with domain X and arbitrary range, so one may wonder if there are different definitions of generating requiring that Eq. ( 2) holds replacing Y for other range spaces.This is not the case, as we will show in Section 2 that a generating operator G satisfies that T G = T for every T ∈ L(X, Z) and every Banach space Z (see Corollary 2.3).This is so thanks to a characterization of generating operators in terms of the sets att(G, δ): G is generating (if and) only if conv(att(G, δ)) = B X for every δ > 0, see Corollary 2.3 again.When the dimension of X is finite, this is clearly equivalent to the fact that conv(att(G)) = B X (actually, the same happens for compact operators defined on reflexive spaces, see Proposition 2.5).For some infinite-dimensional X, there are generating operators from X which do not attain their norm, even of rank-two (see Example 3.2); but there are even generating operators attaining the norm such that conv(att(G)) has empty interior (see Example 3.4).
There is another characterization which involves the geometry of the dual space.We need some definitions.A subset F of the unit ball of a Banach space Z is said to be a If z ∈ S Z satisfies that F = {z} is a spear set, we just say that z is a spear vector and we write Spear(Z) for the set of spear vectors of Z.We refer the reader to [7, Chapter 2] for more information and background.We will show that a norm-one operator G ∈ L(X, Y ) is generating if and only if G * (B Y * ) is a spear set of X * , see Corollary 2.17.These characterizations appear in Section 2, together with a discussion on the behaviour of diagonal generating operators on c 0 -, ℓ 1 -, and ℓ ∞ -sums, and examples in some classical Banach spaces.
We next discuss in Section 3 the relationship between generating operators and norm attainment.On the one hand, we show that rank-one generating operators attain their norm (see Corollary 3.1) and, clearly, the same happens with isometric embeddings (which are generating), or with generating operators whose domain has the RNP (see Corollary 2.12), as every generating operator attains its norm on denting points (see Lemma 2.8).But, on the other hand, there are generating operators, even of rank two, which do not attain their norm (see Example 3.2).We further discuss the possibility for a Banach space X to be the domain of a generating operator which does not attain its norm in terms of the behaviour of some spear sets of X * (see Theorem 3.5).
Finally, Section 4 is devoted to the study of the set Gen(X, Y ).We show that it is closed (see Proposition 4.1), and show that for every Banach space Y , there is a Banach space X such that Gen(X, Y ) = ∅ (see Proposition 4.2), but this result is not true for Y = C[0, 1] if we restrict the space X to be separable (Example 4.5).We next study properties of Gen(X, Y ) when X is fixed.We first show that Gen(X, Y ) = ∅ for every Y if and only if Spear(X * ) = ∅ (see Corollary 4.6) and that the only case in which there is Y such that Gen(X, Y ) = S L(X,Y ) is when X is one-dimensional (see Corollary 4.7).We then study the possibility that the set Gen(X, Y ) generates the unit ball of L(X, Y ) by closed convex hull, showing first that this is the case when X = L 1 (µ) and Y has the RNP (Theorem 4.10) and when X = ℓ 1 (Γ) and Y is arbitrary (see Proposition 4.12) and that this is the only possibility for real finite-dimensional spaces (see Proposition 4.14).

1.1.
A bit of notation.Let X, Y be Banach spaces.We write J X : X −→ X * * to denote the natural inclusion of X into its bidual space.
where f ∈ X * and α > 0, and observe that every slice of C is of the above form.
For A ⊂ X, conv(A) and aconv(A) are, respectively, the convex hull and the absolutely convex hull of A; conv(A) and aconv(A) are, respectively, the closures of these sets.For B ⊂ X convex, ext(B) denotes the set of extreme points of B.

Characterizations, first results, and some examples
Our first result gives different characterizations for the equivalence of • and • G on L(X, Z).As one may have expected, this does not depend on the range space Z.
Proposition 2.1.Let X, Y be Banach spaces, let G ∈ L(X, Y ) be a norm-one operator, and let r ∈ (0, 1].Then, the following are equivalent: (i) T G r T for every Banach space Z and every T ∈ L(X, Z). (ii) There is a (non null) Banach space Z such that T G r T for every T ∈ L(X, Z). (iii) There is a (non null) Banach space Z such that T G r T for every rank-one operator The remaining implication (v) ⇒ (vi) follows from the Bipolar theorem.Indeed, for δ > 0, take x ∈ rB X , we have to prove that J X (x) belongs to att(G, δ) where the second inequality follows from (iv) and the last one from the fact that x * ∈ att(G, δ) • .Therefore J X (x) ∈ att(G, δ) •• = conv w * (att(G, δ)).
Observe that item (vi) in the previous result just means that, for every δ ∈ (0, 1), the set att(G, δ) is r-norming for X * .This leads to the following concept which extends the one of generating operator.Definition 2.2.Let X, Y be Banach spaces, let G ∈ L(X, Y ) be a norm-one operator and let r ∈ (0, 1].We say that G is r-generating if conv(att(G, δ)) ⊇ rB X for every δ > 0.
Of course, the case r = 1 coincides with the generating operators introduced in the introduction.For them, the following characterization deserves to be emphasized.
Corollary 2.3.Let X, Y be Banach spaces, let G ∈ L(X, Y ) be a norm-one operator.Then, the following are equivalent: (ii) T G = T for every T ∈ L(X, Z) and every Banach space Z. (iii) There is a (non null) Banach space Z such that T G = T for every rank-one operator T ∈ L(X, Z). (iv) B X = conv(att(G, δ)) for every δ > 0.
In particular, if there exists A ⊆ B X which satisfies aconv(A) = B X and A ⊆ att(G, δ) for every δ > 0, then G is generating.
In the next list we give the first easy examples of generating operators.(1) The identity operator on every Banach space is generating.
(3) Spear operators are generating since, in this case, v G (T ) = T for every T ∈ L(X, Y ).(4) Actually, operators with the alternative Daugavet property (i.e.those G ∈ L(X, Y ) such that v G (T ) = T for every T ∈ L(X, Y ) with rank-one, cf.[7,Section 3.2]) are also generating by using Corollary 2.3 with Z = Y in item (iii).( 5) The natural embedding G of ℓ 1 into c 0 is a generating operator.
Indeed, for every δ > 0, we have that Indeed, for every δ > 0, notice that ) (this should be well known, but in any case it follows from Lemma 4.11 which includes the vector-valued case).Observe then that, for every We will provide some more examples in classical Banach spaces in Subsection 2.2.
The next result deals with compact operators defined on a reflexive Banach space.Proposition 2.5.Let X be a reflexive Banach space, let Y be a Banach space, and let G ∈ L(X, Y ) be a compact operator with G = 1.Then, Consequently, G is r-generating if and only if rB X ⊆ conv(att(G)).
Clearly, the previous result applies when X is finite-dimensional.
Corollary 2.6.Let X be a finite-dimensional space, let Y be a Banach space, and let G ∈ L(X, Y ) with G = 1.Then, Consequently, G is r-generating if and only if rB X ⊆ conv(att(G)).
The next result characterizes those operators acting from a finite-dimensional space which are rgenerating for some 0 < r 1.
Proposition 2.7.Let X be a Banach space with dim(X) = n, let Y be a Banach space, and let G ∈ L(X, Y ) with G = 1.The following are equivalent: (i) G is r-generating for some r ∈ (0, 1].(ii) The set att(G) contains n linearly independent elements.
Proof.(i) ⇒ (ii).By Corollary 2.6, we have that rB X ⊆ conv(att(G)).Therefore, att(G) contains n linearly independent elements.(ii) ⇒ (i).We start proving that the set conv(att(G)) is absorbing.Indeed, let {x 1 , . . ., x n } be a linearly independent subset of att(G).Then, fixed 0 = x ∈ X, there are λ 1 , . . ., λ n ∈ K such that x = n j=1 λ j x j .Calling 0 < ρ = n j=1 |λ j | we can write where we used that λ j |λ j | x j ∈ att(G) as this set is balanced.Hence, the set conv(att(G)) is absorbing.Besides, conv(att(G)) is clearly balanced, convex, and compact.So its Minkowski functional defines a norm on X which must be equivalent to the original one.Then, there is r > 0 such that rB X ⊆ conv(att(G)) and, therefore, G is r-generating by Corollary 2.6.
We next would like to present the relationship of generating operators with denting points (and so with the Radon-Nikodým property, RNP in short).We need some notation.Let A be a bounded closed convex set.Recall that x 0 ∈ A is a denting point if for every δ > 0 x 0 / ∈ conv(A \ B(x 0 , δ)) or, equivalently, if x 0 belongs to slices of B X of arbitrarily small diameter.We write dent(A) to denote the set of denting points of A. A closed convex subset C of X has the Radon-Nikodým property (RNP in short), if all of its closed convex bounded subsets contain denting points or, equivalently, if all of its closed convex bounded subsets are equal to the closed convex hull of their denting points.In particular, the whole space X may also have this property.
The following result tells us that generating operators must attain their norms on every denting point.
Lemma 2.8.Let X, Y be Banach spaces and let G ∈ L(X, Y ) be a (norm-one) generating operator.
The above result can be slightly improved by using the following definition.Definition 2.9.Let x 0 ∈ S X .We say that x 0 is a point of sliced fragmentability if for every δ > 0 there is a slice Observe that this notion is weaker than that of denting point (for instance, points in the closure of the set of denting points are of sliced fragmentability but they do not need to be denting, even in the finite-dimensional case).
Lemma 2.10.Let X, Y be Banach spaces, let G ∈ S L(X,Y ) be a generating operator, and let x 0 ∈ S X be a point of sliced fragmentability, then Gx 0 = 1.
Proof.Fixed δ > 0, by our assumption, conv(att(G, δ)) = B X for every δ > 0. This implies that, fixed δ > 0, the set att(G, δ) intersects every slice of B X .Applying this to the slice S δ from Definition 2.9, we obtain that there is a point and the arbitrariness of δ finishes the proof.
We do not know if Lemma 2.10 is a characterization, but in Proposition 3.6 we will characterize those points on which every generating operator attains its norm.
Proposition 2.11.Let X, Y be Banach spaces and let G ∈ L(X, Y ) be a norm-one operator.Suppose that B X = conv(dent(B X )).Then, G is generating if and only if Gx = 1 for every x ∈ dent(B X ).
Corollary 2.12.Let X, Y be Banach spaces and let G ∈ L(X, Y ) be a norm-one operator.Suppose that X has the Radon-Nikodým property.Then, G is generating if and only if Gx = 1 for every x ∈ dent(B X ).
In the finite-dimensional case, the RNP is for free and denting points and extreme points coincide.Therefore, the following particular case holds.Corollary 2.13.Let X be a finite-dimensional space, let Y be a Banach space, and let G ∈ L(X, Y ) be a norm-one operator.Then, G is generating if and only if Gx = 1 for every x ∈ ext(B X ).
The following particular case of Corollary 2.12 is especially interesting.
Example 2.14.Let Y be a Banach space and let G ∈ L(ℓ 1 , Y ) be a norm-one operator.Then, G is generating if and only if Ge n = 1 for every n ∈ N.
When every point of the unit sphere of the domain is a denting point, Proposition 2.11 tells us that generating operators are isometric embeddings.Spaces with such property of the unit sphere are average locally uniformly rotund (ALUR for short) spaces.They were introduced in [20] and it can be deduced from [13,Theorem] that a Banach space is ALUR if and only if every point of the unit sphere is a denting point.
Corollary 2.15.Let X, Y be Banach spaces and suppose that X is ALUR.Then, every generating operator G ∈ L(X, Y ) is an isometric embedding.
The next result gives another useful characterization of r-generating operators.
Theorem 2.16.Let X, Y be Banach spaces, let G ∈ L(X, Y ) be a norm-one operator, let r ∈ (0, 1], and let Proof.If G is r-generating, fixed x * ∈ X * and δ > 0, we can write where the last inequality holds by Proposition 2.1.The arbitrariness of δ gives the desired inequality.
To prove the converse, fixed x * ∈ S X * and δ > 0, it suffices to show that x * G,δ r by Proposition 2.1.We use the hypothesis for δ 2 x * to get that max θ∈T sup So, given 0 < ε < δ 2 , there are y * ∈ A, θ ∈ T, and x ∈ B X such that The arbitrariness of ε gives x * G,δ r as desired.
Of course, one can always use A = B Y * in Theorem 2.16 if no other interesting choice for A is available and still one obtains a useful characterization of r-generating operators.
In the case of generating operators, we emphasize the following result.
Corollary 2.17.Let X, Y be Banach spaces, let A ⊂ B Y * be one-norming for Y , and let G ∈ L(X, Y ) with G = 1.Then, the following are equivalent: Only item (iv) is new, and follows immediately from the following remark.Indeed, to prove the sufficiency, fixed 0 = z 1 ∈ X, observe that max , the triangle inequality allows to write What we have shown is that it suffices to use elements x * ∈ S X * in Theorem 2.16 when r = 1.However, the following example shows that this is not the case for any other value of 0 < r < 1.
Example 2.19.Let 0 < r < 1 be fixed, let X be the real two-dimensional Hilbert space, {e 1 , e 2 } be its orthonormal basis with {e * 1 , e * 2 } being the corresponding coordinate functionals.The norm-one operator G ∈ L(X) given by G = r Id +(1 − r)e * 1 ⊗ e 1 is not r-generating but satisfies Observe that G attains its norm only at ±e 1 so Proposition 2.7 tells us that G is not r-generating (in fact, it is not s-generating for any 0 < s 1).
If we are able to guarantee that G * (B Y * ) is a spear set of X * , Corollary 2.17 shows that G is generating.The most naive way to do so is to require This obviously means that G is a rank one operator; in this case, G * (B Y * ) is a spear set of X * if and only if x * 0 is a spear vector of X * .In this particular case, Corollary 2.17 reads as follows.Corollary 2.20.Let X, Y be Banach spaces, x * 0 ∈ S X * , and y 0 ∈ S Y .Then, the rank-one operator G = x * 0 ⊗ y 0 is generating if and only if x * 0 ∈ Spear(X * ).
Observe the similarity with [7, Corollary 5.9] which states that G = x * 0 ⊗ y 0 is spear if and only if x * 0 is a spear functional and y 0 is a spear vector.Here the condition is easier to satisfy, of course.
2.1.Some stability results.The following result shows that the property of being generating is stable by c 0 -, ℓ 1 -, and ℓ ∞ -sums of Banach spaces.
Proof.Suppose first that G is generating and, fixed κ ∈ Λ, let us show that G κ is generating.Observe To prove the sufficiency when , we have that B X = conv (S Xκ × S W ) and so we may find x 0 ∈ S Xκ and w 0 ∈ S W such that Take x * 0 ∈ S Xκ * with x * 0 (x 0 ) = 1 and define the operator S ∈ L(X κ , Y κ ) by which satisfies S Sx 0 = P κ T (x 0 , w 0 ) > T − ε and S Gκ = S since G κ is generating.Moreover, fixed δ > 0, S Gκ = S > T − ε and the arbitrariness of ε gives that T G T as desired.
In the case when E = ℓ 1 , fixed δ > 0, consider the set where in the last equality we have used Corollary 2.3.iv as G λ is generating for every λ ∈ Λ.Therefore, B X = conv λ∈Λ {x ∈ X : ) and the arbitrariness of δ gives that G is generating by Corollary 2.3.iv.
We next discuss the relationship of being generating with the operation of taking the adjoint.We show next that if the second adjoint is r-generating then the operator itself is r-generating.
r since ε > 0 was arbitrary.So G is r-generating by Proposition 2.1.
We do not know if the converse of the above result holds in general or even for r = 1.On the other hand, the following example shows that there is no good behaviour of the property of being generating with respect to taking one adjoint, as the property does not pass from an operator to its adjoint, nor the other way around.For any x ∈ S c 0 with x(1) ∈ T we have that G(x) = 1 and, consequently, x ∈ att(G, δ) for every δ > 0. Since such elements are enough to recover the whole unit ball of c 0 by taking closed convex hull, G is generating by Corollary 2.3.iv.
• The adjoint operator G * : is not generating by Example 2.14 since is again generating following an analogous argument to the one used for G, using this time elements x ∈ S ℓ∞ with x(1) ∈ T.

2.2.
Some examples in classical Banach spaces.Our aim here is to provide some characterizations of generating operators when the domain space is L 1 (µ) or the range space is C 0 (L) by making use of Corollary 2.17.
The question of which operators acting from L 1 (µ) are generating leads to study the spear sets in L ∞ (µ).We do so in the next result which is valid for arbitrary measures.Proposition 2.24 (Spear sets in B L∞(µ) ).Let (Ω, Σ, µ) be a positive measure space and let F ⊂ B L∞(µ) .Then, the following are equivalent: Proof.Suppose first that F is a spear set.Given A ∈ Σ with µ(A) = 0 and ε > 0, since max there exists f 0 ∈ F and θ 0 ∈ T such that f 0 + θ 0 ½ A ∞ > 2 − ε and thus, there exists B ⊂ A with µ(B) = 0 such that |f (t)| > 1 − ε for every t ∈ B. To prove the converse implication, given x ∈ L ∞ (µ) and ε > 0, there is A ∈ Σ with µ(A) = 0 such that |x(t)| x ∞ − ε for every t ∈ A. By the hypothesis, there is a subset B of A with µ(B) = 0 and f ∈ F such that |f 0 (t)| > 1 − ε for every t ∈ B. Now, thanks to the compactness of T we can fix an ε-net T ε of T, then we may find θ 0 ∈ T ε and and the arbitrariness of ε gives max As an immediate consequence we get the following characterization of generating representable operators acting on L 1 (µ).
Remark 2.26.The restriction on the measure µ being finite in Corollary 2.25 can be relaxed to being σ-finite.
Indeed, given a σ-finite measure µ, there is a suitable probability measure ν such that 2 ) given by G(f ) = 2.2.2.Operators arriving to C 0 (L).Let L be a Hausdorff locally compact topological space.It is immediate from the definition of the norm, that the set A = {δ t : t ∈ L} ⊂ C 0 (L) * is one-norming for C 0 (L).Hence, Corollary 2.17 reads in this case as follows.
Proposition 2.27.Let X be a Banach space, let L be a Hausdorff locally compact topological space, and let G ∈ L(X, C 0 (L)) be a norm-one operator.Then, the following are equivalent: We would like to compare the result above with [7,Proposition 4.2] where it is proved that G ∈ L(X, C 0 (L)) has the alternative Daugavet property if and only if {G * (δ t ) : t ∈ U } is a spear set of X * for every open subset U ⊂ L. It is then easy to construct examples of generating operators arriving to C 0 (L) spaces which do not have the alternative Daugavet property.For instance, consider G ∈ L(c 0 , c 0 ) given by

Generating operators and norm-attainment
We discuss here when generating operators are norm-attaining.On the one hand, it is shown in [7, Theorem 2.9] that every spear x * ∈ X * attains its norm.So rank-one generating operators also attain their norm by Corollary 2.20.
Corollary 3.1.Let X, Y be Banach spaces and G ∈ Gen(X, Y ) of rank-one.Then, G attains its norm.
Besides, if B X contains denting points, all generating operators with domain X are norm attaining by Lemma 2.8.
On the other hand, operators with the alternative Daugavet property are generating (see Example 2.4.(4)), and there are operators with the alternative Daugavet property which do not attain their norm (see [7,Example 8.7]).The construction of the cited example in [7] is not easy at all, but we may construct easier examples of generating operators which do not attain their norm, even with rank two.
Proof.Observe that G is generating by Corollary 2.25 as g(t) = 1 for every t ∈ [0, 1].To prove that G does not attain its norm, recall that for an integrable complex-valued function f the equality |dt holds if and only if there is λ ∈ T such that f = λ|f | except for a set of zero measure.Suppose, to find a contradiction, that there is a non-zero x ∈ L 1 [0, 1] satisfying Gx = x .Then, as xg can be seen as a complex-valued function and we can identify the norm on ℓ 2 2 with the modulus in C, we have that Therefore, there is λ ∈ T such that xg = λ|xg| = λ|x| except for a set of zero measure.But this is impossible since x takes real values and g covers a non-trivial arc of the unit circumference.Moreover, if the previous assertions hold, we have that ) with g ∞ = 1 and g(t) = 1 almost everywhere by Corollary 2.25.Since S Y is a finite or countable union of segments, we may find a partition π of [0, 1] in measurable subsets of positive measure such that g(A) is contained in a segment of S Y almost everywhere for every A ∈ π.Then, for every ∆ ∈ π and every measurable subset A ⊂ ∆ of positive measure, consider , where |A| denotes the Lebesgue measure of A, and let us show that G attains its norm at x A .Indeed, as g(A) is contained in a segment of S Y a.e., there exists y * ∈ S Y * such that y * (g(t)) = 1 a.e. in A, thus and so G(x A ) = 1 as desired.Moreover, for this π To prove (ii) ⇒ (i), suppose that S Y cannot be written as a finite or countable union of segments and let us construct a generating operator G ∈ L(L 1 [0, 1], Y ) not attaining its norm.Observe that the number of open maximal segments in S Y is finite or countable as S Y is a curve on a two-dimensional space with finite length.Let ∆ n , n ∈ N, be the open maximal segments in S Y and denote D = S Y \ (∪ n∈N ∆ n ).Clearly, D is an uncountable metric compact subset of S Y , hence it contains a homeomorphic copy of the Cantor set K [11, Chapter I] and so there exists an injective continuous function ϕ : K −→ D. Now, let us construct an injection from [0, 1] to K. To do so, recall that the Cantor set is the set of numbers of [0, 1] that have a triadic representation consisting purely of 0's and 2's, that is, where α k (t) ∈ {0, 1}.This representation is unique except for a countable subset of [0, 1] consisting of those numbers with finite dyadic representation.Consider φ : [0, 1] −→ K given by where α k (t) ∈ {0, 1} are the coefficients in the dyadic representation of t.The function φ is welldefined almost everywhere on [0, 1], injective, measurable, and its image lies on K.Then, the function and it is injective.Consider the operator G : G is generating by Corollary 2.25 as g(t) = 1 almost everywhere but it does not attain its norm.Indeed, suppose on the contrary that there is a non-zero We may find y * 0 ∈ S Y * such that This equality implies the existence of a measurable subset A of [0, 1] with positive measure such that |x(t)| = x(t)y * 0 (g(t)) for every t ∈ A, thus y * 0 (g(t)) ∈ {1, −1} for every t ∈ A. Note that g(A) ⊆ y ∈ D : y * 0 (y) ∈ {1, −1} .However, this leads to a contradiction.On the one hand, the latter set has at most four elements as D does not contain open segments of S Y .On the other hand, since g is injective and A has positive measure, g(A) has infinitely many elements.Thus, G cannot attain its norm.
The next example shows that, even in the case of norm-attaining operators, the set att(G) cannot be used to characterize when G is generating outside the case when X is reflexive and G is compact covered by Proposition 2.5.
Example 3.4.Let G ∈ L(X, Y ) be a generating operator between two Banach spaces X and Y such that it does not attain its norm.Then, the operator G : is generating by Proposition 2.21 and attains its norm, but conv(att The following result characterizes the possibility to construct a generating operator not attaining its norm acting from a given Banach space which somehow extend Example 3.2.Theorem 3.5.Let X be a Banach space, the following are equivalent: (i) There exists a Banach space Y and a norm-one operator G ∈ L(X, Y ) such that G is generating but att(G) = ∅.(ii) There exists a spear set B ⊆ B X * such that sup ), since G is generating, we can use Corollary 2.17 to deduce that B is a spear set.Besides, as G does not attain its norm, we have that On the one hand, for x ∈ S X , we have that On the other hand, using that B is a spear set, for every ε > 0 we may find x * ∈ B with x * > 1 − ε and so Therefore, G = 1 but the norm is not attained.
To show that G is generating, we start claiming that, for every g ∈ ℓ 1 (B) ⊂ ℓ ∞ (B) * , we have Indeed, given g ∈ ℓ 1 (B), observe that Therefore, by the arbitrariness of x * 0 ∈ B, we get G * (B ℓ∞(B) * ) ⊃ B, so G * (B ℓ∞(B) * ) is a spear set and G is generating by Corollary 2.17.
The above proof, when read pointwise, allows to give a characterization of those points at which every generating operator attains its norm.Proposition 3.6.Let X be a Banach space and x 0 ∈ S X .Then, the following are equivalent: (i) ⇒ (ii) Suppose that (ii) does not hold.Then, there is a spear set B ⊆ B X * such that sup is generating (as shown in the proof of Theorem 3.5) and satisfies Therefore, (i) does not hold.

The set of all generating operators
Our aim here is to study the set Gen(X, Y ) of all generating operators between the Banach spaces X and Y .Recall, on the one hand, that Id X ∈ Gen(X, X) for every Banach space X, so Gen(X, X) = ∅ for every Banach space X.On the other hand, recall that Corollary 2.20 shows that Gen(X, K) = Spear(X * ), so Gen(X, K) is empty for many Banach spaces X: those for which Spear(X * ) = ∅ as uniformly smooth spaces, strictly convex spaces, or real smooth spaces with dimension at least two (see [7,Proposition 2.11]).We will be interested in finding conditions to ensure that Gen(X, Y ) is non-empty and, in those cases, to study how big the set Gen(X, Y ) can be.We start with an easy observation on Gen(X, Y ).
, where the last inequality holds by Corollary 2.17 since G n is generating.Now, it follows again from Corollary 2.17 that G 0 ∈ Gen(X, Y ).
Next, we study the problem of finding out whether Gen(X, Y ) is empty or not for the Banach spaces X and Y from two points of view: fixing the space Y and fixing the space X.
4.1.Gen(X, Y ) when Y is fixed.We will show that for every Banach space Y there is another Banach space X such that Gen(X, Y ) = ∅.Proposition 4.2.For every Banach space Y there is a Banach space X such that Gen(X, Y ) = ∅.
We need the following obstructive result for the existence of generating operators that will serve to our purpose.We are now able to provide the pending proof.For a Banach space X let dens(X) denote its density character.
The above argument is based on the possibility of considering Banach spaces in the domain with a very big density character.It is then natural to raise the following question.This question is easily solvable for separable spaces.Indeed, the space Y = C[0, 1] contains isometrically every separable Banach space.Since isometric embeddings are generating, we get the following example.The question of whether the same trick works for all density characters is involved and depends on the Axiomatic Set Theory.On the one hand, assuming CH, ℓ ∞ /c 0 is isometrically universal for all Banach spaces of density character the continuum [17] but, on the other hand, it is consistent that no such a universal space exists [19], even a isomorphically universal space, see [4].
For the moreover part, it is enough to see that extm(B X * ) is actually a James boundary for X and so B X * = conv(extm(B X * )) by [6,Theorem III.1].
Our next aim is to show that the set Gen(L 1 (µ), Y ) is quite big for every finite measure µ and many Banach spaces Y , and that in some cases it allows to recover the unit ball of L(L 1 (µ), Y ) by taking closed convex hull.Given a finite measure space (Ω, Σ, µ) and a Banach space Y we write Theorem 4.10.Let (Ω, Σ, µ) be a finite measure space and let Y be a Banach space.Then, As a consequence, if Y has the RNP, then Observe that the restriction on the measure µ to be finite can be relaxed to be σ-finite as in Remark 2.26.
The proof of the theorem follows immediately using Corollary 2.25 and the next lemma, which we do not know whether it is already known.Step one.Let f ∈ S L∞(µ,Y ) and suppose that there are N ∈ N, numbers α , and pairwise disjoint subsets B k ⊂ Ω with µ(B k ) = 0 for k = 1, . . ., N such that N k=1 B k = Ω and f (t) = α k for every t ∈ B k and every k = 1, . . ., N (observe that α N = 1 as f = 1).Then, f can be written as a convex combination of 2 N −1 functions in B.
Indeed, we proceed by induction on N : for N = 1, the function f belongs to B. The case N = 2 gives the flavour of the proof.In this case we have that f (t) = α 1 for every t ∈ B 1 and f If otherwise α 1 = 0, fix y 0 ∈ S Y , and define g 1 (t) = y 0 and g 2 (t) = −y 0 for every t ∈ B 1 .It is clear that in any case we have f = λ 1 g 1 + λ 2 g 2 and that g 1 , g 2 ∈ B. Suppose now that the result is true for N 2 and let us prove it for N + 1.So, let f ∈ S L∞(µ,Y ) and suppose that there are numbers α 1 < • • • < α N +1 ∈ [0, 1] with α N +1 = 1, and pairwise disjoint subsets B k ⊂ Ω with µ(B k ) = 0 for k = 1, . . ., N + 1 such that N +1 k=1 B k = Ω and f (t) = α k for every t ∈ B k and every k = 1, . . ., N + 1. Observe that, as N 2, we have that α N > 0.Then, we . So, we can apply the induction step for f 1 and f 2 to write Therefore, the convex combination we are looking for is which finishes the induction process.
Step two.Every function f ∈ S L∞(µ,Y ) can be approximated by functions of the class described in the first step.
Let us now discuss the case of purely atomic measures.When µ is purely atomic and σ-finite (so L 1 (µ) can be easily viewed as L 1 (ν) for a suitable purely atomic and finite measure ν, see [3, Proposition 1.6.1]for instance), every operator in L(L 1 (µ), Y ) is representable for every Banach space Y (see [5, p. 62], for instance).So, Theorem 4.10 gives that B L(ℓ 1 (Γ),Y ) = conv(Gen(ℓ 1 (Γ), Y )) for every Banach space Y and every countable set Γ. Actually, the restriction of countability for the set Γ can be remove and the proof in this case is much more direct.For finite-dimensional ℓ 1 -spaces, we get a better result.The next result shows that the only finite-dimensional real spaces with this property are ℓ n 1 for n ∈ N. Proposition 4.14.Let X be a real Banach space with dim(X) = n and such that B L(X,Y ) = conv(Gen(X, Y )) for every Banach space Y .Then, X = ℓ n 1 .Proof.Proposition 4.8 tells us that X * is an almost CL-space so n(X * ) = n(X) = 1.Therefore, as X is real, the set ext(B X ) is finite by [16,Theorem 3.2].Our goal is to show that ext(B X ) contains exactly 2n elements as this clearly implies that X is isometrically isomorphic to the real space ℓ n 1 .We suppose that ext(B X ) has more than 2n elements and we show that, in such a case, there is a Banach space Y (= X with a new norm) such that B L(X,Y ) = conv(Gen(X, Y )).Since dim(X) = n and ext(B X ) has more than 2n elements, we may find {e 1 , . . ., e n } ⊂ ext(B X ) linearly independent and e n+1 ∈ ext(B X ) satisfying e n+1 / ∈ {±e j : j = 1, . . ., n}.For each j = 1, . . ., n, as ext(B X ) is finite, we can pick f j ∈ X * such that 1 = f j (e j ) > c j = max {f j (x) : x ∈ ext(B X ) \ {e j }} .

Remark 2 . 18 .
Let Z be a Banach space and F ⊂ B Z .Then, F is a spear set if and only if max θ∈T sup z∈F z + θz 0 = 2 for every z 0 ∈ S Z .

Example 3 .
2 can be generalized for other two-dimensional spaces Y , but we need some assumptions on the shape of S Y .If S Y can be expressed as a finite or countable union of segments, then every generating operator G ∈ L(L 1 [0, 1], Y ) attains its norm, leading to a complete characterization.Proposition 3.3.Let Y be a real two-dimensional space.Then, the following are equivalent:(i) S Y is a finite or countable union of segments.(ii) Every generating operator G ∈ L(L 1 [0, 1], Y ) attains its norm.

Question 4 . 4 .
Does there exist a Banach space Y with dens(Y ) = Γ such that Gen(X, Y ) = ∅ for every Banach space X satisfying dens(X) Γ?
see [3, Proposition 1.6.1]for instance).Compare Corollary 2.25 the above result with [7, Corollary 4.22] which says that G ∈ L(L 1 (µ), Y ) of norm-one which is representable by g ∈ L ∞ (µ, Y ) is a spear operator if and only if it has the alternative Daugavet property if and only if g(t) ∈ Spear(Y ) for a.e.t ∈ Ω.It is then easy to construct generating operators from L 1 (µ) which do not have the alternative Daugavet property: for instance By Corollary 2.17, G * (B Y * ) is a spear set, so we can find a sequence {x* n } in G * (B Y * ) and a sequence {θ n } in T such that θ n x * n + x * 0 → 2. Therefore, there is a sequence {x n } in S X satisfying Re x * 0 (x n ) → 1 and |x * n (x n )| → 1.Since the norm of X * is Fréchet differentiable at x * 0 ∈ S X *, by the Šmulyian's test, we have that x n − x → 0. Thus, we get |x * n (x)| → 1 which contradicts (4).