Banach spaces which always produce octahedral spaces of operators
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Universidad de Granada/CBUA
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Collectanea Mathematica https://doi.org/10.1007/s13348-023-00394-9 Banach spaces which always produce octahedral spaces of operators Abraham Rueda Zoca1 Received: 18 July 2022 / Accepted: 10 January 2023 © The Author(s) 2023 Abstract We characterise those Banach spaces Xwhich satisfy that L(Y,X)is octahedral for every non-zero Banach space Y. They are those satisfying that, for every finite dimensional subspace Z,∞can be finitely-representable in a part of Xkind of 1-orthogonal to Z.Wealsoprove that L(Y,X)is octahedral for every Yif, and only if, L(n p,X)is octahedral for every n∈N and 1 <p<∞. Finally, we find examples of Banach spaces satisfying the above conditions like Lip0(M)spaces with octahedral norms or L1-preduals with the Daugavet property. Keywords Spaces of operators ·Universally octahedral ·Finite representability Mathematics Subject Classification 46B06 ·46B20 ·46B28 1 Introduction Accordingto[10, Remark II.5.2], the norm of a Banach space Xis octahedral if, for every finite dimensional subspace Eof Xand every ε>0, there exists y∈SXsuch that x+λy≥(1−ε)(x+|λ|)for every x∈Eand every λ∈R. Octahedral norms were studied at the end of the eighties in succesive papers [10,11] because it turns out that such norms characterise the containment of 1.Indeed,in[10, Theorem II.4] it is proved that a Banach space Xcontains an isomorphic copy of 1if, and only if, Xcan be equivalently renormed with an octahedral norm. However, octahedral norms have received much more attention in the recent years because in [6, Theorem 2.1] it is proved that Xis octahedral if, and only if, every convex combination The research of Abraham Rueda Zoca was supported by MCIN/AEI/10.13039/501100011033: Grants PID2021-122126NB-C31 and PID2021-122126NB-C32; and by Junta de Andalucía: Grants FQM-0185 and PY20_00255. BAbraham Rueda Zoca [email protected] https://arzenglish.wordpress.com 1Departamento de Análisis Matemático, Facultad de Ciencias, Universidad de Granada, 18071 Granada, Spain 123
A. R. Zoca of w∗-slices of BX∗has diameter two. Since then, octahedral norms and variations of such norms have been studied in many different contexts (see. e.g. [4,8,12,18,21]). One of the areas where octahedrality has been intensively studied is in spaces of operators, that is, it has been analysed when the space of bounded operators L(X,Y)between two Banach spaces Xand Yis octahedral. The motivation for this interest comes from [1, Question (b)], where the authors asked when the projective tensor product X ⊗πYsatisfies that all the convex combination of slices of its unit ball have diameter two. Thanks to the duality (X ⊗πY)∗=L(X,Y∗)and the above mentioned [6, Theorem 2.1], the above question is equivalent to determining when the norm of the space of operators is octahedral. In [7, Theorem 3.5] it is proved that if Y∗and Xare octahedral then His octahedral for any subspace H⊆L(Y,X)containing finite-rank operators F(Y,X). More examples of octahedral spaces of operators were given in [14]. It was shown, however, that the above is not the case if we remove octahedrality on Y∗, and in fact octahedrality of L(Y,X)is connected with finite-representability of Yin X. Indeed, in [17, Lemma 3.7] it is proved that if some subspace Hof L(Y,X)is octahedral and Yis uniformly convex then Yis finitely representable in X. The connection between finite-representability and octahedrality of spaces of operators have shown to be much deeper. Indeed, a kind of converse is established in [19, Theorem 3.2] where it is proved that if Xis a Banach space which is finitely representable in 1and with the metric approximation property, then L(X,Y)is octahedral if Yis octahedral. In this note we will focus on the following problem: which Banach spaces Xsatisfies that L(Y,X)is octahedral for every Banach space Y? We will refer to these spaces as universally octahedral (see Definition 3.1). Observe that, in order to solve a problem about octahedrality of spaces of vector-valued Lipschitz functions, it is proved in [19, Theorem 3.1] that Lip0(M), the space of Lipschitz functions over M, is universally octahedral whenever Lip0(M)is octahedral. Anyway, in view of [17, Lemma 3.7] one should think that if Xis universally octahedral then it should be an octahedral space such that every uniformly convex Banach space is finitely representable in it. This intuition is confirmed in Lemma 3.2, where we observe that a necessary condition for universal octahedrality is that, roughly speaking, given any finite dimensional subspace Zof Xand any ε>0, then any finite dimensional uniformly convex Banach space can be (1+ε)-embedded in the space of “ε-orthogonal vectors to Z”. In a further step, making use of approximations in Banach-Mazur distance, we obtain in Theorem 3.4 that we can replace in the above statement uniformly convex Banach spaces with n ∞for every n.Moreprecisely,weprovethatifXis universally octahedral then, given any finite dimensional subspace Zof X,anyn∈Nand any ε>0, we can find a norm-one operator :n ∞−→ Xsuch that z+(y)≥(1−ε)(z+y) holds for every z∈Zand every y∈∞. The converse, making use of the finite-representability of every Banach space in c0together with the celebrated characterisation of L1-preduals due to J. Lindenstrauss [20, Theorem 6.1], is proved in Theorem 3.6. As a consequence we obtain, in Theorem 3.7, that a Banach space Xis universally octahedral if, and only if, L(n p,X)is octahedral for every 1 ≤p≤∞and every n∈N, which is in turn equivalent to the condition that, given any finite dimensional subspace Zof X,anyn∈Nand any ε>0, we can find a norm-one operator :n ∞−→ X such that z+(y)≥(1−ε)(z+y) 123
Banach spaces which always produce octahedral… holds for every z∈Zand every y∈n ∞. Observe that the above condition is strictly stronger than the mere finite-representability of ∞in X. Indeed, in Example 3.8 we construct an example of a octahedral Banach space which contains an isomorphic copy of ∞but failing the universal octahedrality. This shows that, in order to obtain universal octahedrality, the requierement that the copies of n ∞can be found in the orthogonal part of any finite dimensional subspace can not be relaxed. Another relevant example is given in Example 3.9, where it is shown that a universally octahedral space does not have to contain c0isomorphically. In Sect.4we aim to find new examples of Banach spaces Xwhich are universally octahedral. We begin by observing that a sufficient condition for universal octahedrality of a X is the following: for every finite dimensional subspace Zof Xand every ε>0, there exists a subspace Yof Xwhich is isometrically isomorphic to c0and such that z+y≥(1−ε)(z+y) holds for every z∈Zand every y∈Y(we define this property in Definition 4.1 as c0octahedral). The reason to introduce this definition is double. The first one is to recover the technique followed in [19, Theorem 3.1], where it is proved that Lip0(M)is universally octahedral when it is octahedral, but whose proof is based on [19, Lemma 3.3], where it is preciselly proved that Lip0(M)is c0-octahedral. On the other hand, in spite of the fact that Example 3.9 shows that universal octahedrality does not imply c0-octahedrality, there is a strong connection through ultrapower spaces. In fact, in Proposition 4.4 it is proved that Xis universally octahedral if, and only if, XUis c0-octahedral for every free ultrafilter U over N. We end the paper with Theorem 4.6, where we prove that every L1-predual which is octahedral is indeed universally octahedral, using recent tools developed in [22]. 2 Notation and preliminary results We will consider real Banach spaces. Given a Banach space X, we will denote the closed unit ball and the unit sphere of Xby BXand SXrespectively. We will also denote by X∗the topological dual of X. Given two Banach spaces Xand Ydenote by L(X,Y)(respectively F(X,Y)) the space of linear bounded operators (respectively the finite-rank operators) from Xto Y. Accordingto[2, Definition 11.1.1], given two Banach spaces Xand Y,wesaythatXis finitely representable in Y if, given any finite dimensional subspace Eof Xand any ε>0, there exist a subspace Fof Yand a linear continuous bijection T:E−→ Fsuch that TT−1≤1+ε. This notion encodes the idea that Ycontains all the finite dimensional structure of X. Observe that every Banach space is finitely-representable in c0[2,Example 11.1.2]. Moreover, as a consequence of the Principle of Local Reflexivity (c.f. e.g. [9, Lemma 9.15]), for every Banach space Xit follows that X∗∗ is finitely representable in X.Werefer the interested reader to [2, Chapter 11] and references therein for background about finite representability of Banach spaces. Let us include here, for easy reference, the following lemma, which is extracted from [2, Lemma 11.1.11]. Lemma 2.1 Let E be a finite dimensional Banach space and let {xj:1≤j≤N}⊆SXbe an ε-net of SE.LetT :E−→ X be a linear mapping such that (1−ε) ≤T(xj)≤(1+ε) 123
A. R. Zoca holds for every 1≤j≤N. Then, for every e ∈E, we have 1−3ε 1−εe≤T(e)≤1+ε 1−εe. Given a sequence of Banach spaces {Xn:n∈N}we denote ∞(N,Xn):= f:N−→ n∈N Xn:f(n)∈Xn∀nand sup n∈Nf(n)<∞. Given a non-principal ultrafilter Uover N, consider c0,U(N,Xn):= {f∈∞(N,Xn): limUf(n)=0}.Theultrapower of {Xn:n∈N}with respect to Uis the Banach space (Xn)U:= ∞(N,Xn)/c0,U(N,Xn). We will naturally identify a bounded function f:N−→ n∈N Xnwith the element (f(n))n∈N. In this way, we denote by (xn)Uor simply by (xn), if no confusion is possible, the coset in (Xn)Ugiven by (xn)n∈N+c0,U(N,(Xn)). From the definition of the quotient norm, it is not difficult to prove that (xn)U= limUxnholds for every (xn)∈(Xn)U. When Xn=Xholds for every n∈N, the definition of the norm on XUyields a canonical inclusion j:X−→ XUgiven by the equation j(x):= (x)U. This inclusion is an into linear isometry, so Xcan be isometrically embedded in XU.Moreover, XUis finitely representable in X[2, Proposition 11.1.12]. Given a Banach space Xwe say that Xis an L1-predual if X∗=L1(μ) isometrically for some measure μ. Let us include here for easy reference in the text the following result. Theorem 2.2 [20, Theorem 6.1] Let X be a Banach space. The following assertions are equivalent: (1) XisanL 1-predual. (2) Every compact operator T :Y−→ X has, for every ε>0and every Banach space Z containing Y , an extension ˆ T:Z−→ X such that ˆ T≤(1+ε)T. Strongly related to L1(μ)-spaces are the L-summands. A projection P:X−→ Xon a Banach space Xis said to be an L-projection if x=Px+x−Pxfor every x∈X. The range of an L-projection is called an L-summand. We refer the reader to [15]foravast background about L-summands. 3 Characterisation of universally octahedral spaces Let us start with the main definition of the paper. Definition 3.1 Let Xbe a Banach space. We will say that Xis universally octahedral if L(Y,X)is octahedral for every non-zero Banach space Y. The aim of this section is to provide a characterisation of universally octahedral Banach spaces. In order to do so, let us start with the following preliminary lemma, which is a 123
Banach spaces which always produce octahedral… strengthening of [17, Lemma 3.7]. Recall that a Banach space Xis said to be uniformly convex if, for every ε>0, there exists δ(ε) > 0 such that x,y∈BX x+y>2−δ(ε)⇒x−y<ε. Examples of uniformly convex Banach spaces are Lp(μ) for 1 <p<∞thanks to Clarkson inequality (see [9, Chapter 9] for background about uniform convexity). Lemma 3.2 Let X be a Banach space and Y be a finite dimensional uniformly convex Banach space. Assume that L(Y,X)is octahedral. Then, for every ε>0and for every finite dimensional subspace Z of X, there exists an element T ∈BL(Y,X)such that z+T(y)≥(1−ε)2(z+y) holds for every y ∈Y and every z ∈Z. Observe that the mapping Tis a (1+ε)-isometry (just take z=0) and that Xis octahedral. Proof Since Yis uniformly convex there exists a mapping δ:R+−→ R+such that limε→0δ(ε) =0 and with the property that, given η>0, if x,y∈BYsatisfy x+y> 2−δ(η) then x−y<η. Take ε>0andη>0 small enough such that δ(η) +4η<ε. Pick a finite dimensional subspace Zof X.Take{y1,...,yn}aη-net of SYand take {z1,...,zp}aη-net of SZ.For every i∈{1,...,n}take fi∈SY∗such that fi(yi)=1. Define Tij := fi⊗zj∈L(Y,X)by Tij(x):= fi(x)zj, and note that Tij is a norm-one element. By the assumption that L(Y,X)is octahedral we can find an operator T∈SL(Y,X) such that Tij +T>2−δ(η) holds for every 1 ≤i≤nand 1 ≤j≤p. Fix 1 ≤i≤nand 1 ≤j≤p. By the definition of the operator norm we can find yij ∈SYsuch that 2 −δ(η) < Tij(yij)+T(yij). By the Hahn-Banach theorem we can find x∗ ij ∈SX∗such that 2−δ(η) < x∗ ij(fi(yij)zj+T(yij)). Up to a change of sign we can assume with no loss of generality that fi(yij)≥0. Since all the elements in the above inequality are norm-one elements we get that fi(yij)>1−δ(η). Moreover, since fi(yi)=1 we get that yi+yij>2−δ(η), and the uniform convexity implies that yi−yij<η. This implies that 2−δ(η) < x∗ ij(fi(yij)zj+T(yij)) ≤x∗ ij(fi(yi)zj+T(yi)) +2yi−yij ≤zj+T(yi)+2η. Since {y1,...,yn}is a η-net in SYand {z1,...,zp}is a η-net in SZwe conclude that z+T(y)>2−δ(η) −4η>2−ε holds for every z∈SZand every y∈SY. Let us conclude from here the desired result. To this end, take arbitrary z∈SZand y∈SY, and take t1,t2∈[0,1]such that t1+t2=1. Let us estimate t1z+t2T(y). Assume with 123
A. R. Zoca no loss of generality that t1≥t2(the other case runs similar). Then t1z+t2T(y)=t1(z+T(y)) +(t2−t1)T(y)≥t1z+T(y)−|t2−t1|T(y) ≥t1(2−ε) +t2−t1=t1+t2−t1ε≥1−ε. Observe that this proves in particular that T(y)≥1−εholds for every y∈SYand, in consequence, T(y)≥(1−ε)yfor every y= 0. Now, given z∈Z\{0}and y∈Y\{0},wegetthat z+T(y) z+T(y)=z z+T(y) z z+T(y) z+T(y) T(y) T(y)>1−ε, from where z+T(y)>(1−ε)(z+T(y))>(1−ε)(z+(1−ε)y)) > (1−ε)2(z+y), and the lemma is proved. Remark 3.3 Observe that, from the last part of the above proof, the following holds true: Given two Banach spaces Xand Ywith Yfinite dimensional, the following assertions are equivalent: (1) For every finite dimensional subspace Zof Xand every ε>0 we can find a norm-one operator T:Y−→ Xsuch that z+T(y)≥(1−ε)(z+y) holds for every y∈Yand every z∈Z. (2) For every finite subsets {z1,...,zn}⊆SXand {y1,...,ym}⊆SYand every ε>0there exists a norm-one operator T:Y−→ Xsuch that zi+T(yj)>2−ε holds for every 1 ≤i≤nand 1 ≤j≤m. We will use this remark throughout the text. Now we are ready to prove the following necessary condition for a Banach space being universally octahedral. Theorem 3.4 Let X be a universally octahedral Banach space. Then, for every ε>0,for every finite dimensional subspace Z of X and for every n ∈N, there exists an operator T:n ∞−→ X such that T≤1and such that z+T(y)≥(1−ε)(z+y) holds for every y ∈n ∞and every z ∈Z. Proof Observe that, given x∈Rn,wehavethatx∞≤xp≤n1 px∞,son ∞is, for every ε>0, (1+ε)-isometric to a uniformly convex Banach space. The proof is simple from now. Take ε>0andn∈N,andtakep∈Nsuch that a suitable scalling of the formal identity φ:n ∞−→ n psatisfies √1−εx≤φ(x)≤x∀x∈X. Now set Z⊆Xbe a finite dimensional subspace. Applying Lemma 3.2 we can find a bounded operator T:n p−→ Xwith T≤1 and such that z+T(y)≥√1−ε(z+y) holds for every z∈Zand every y∈Y.NowT◦φis the desired operator. 123
Banach spaces which always produce octahedral… Remark 3.5 A couple of remarks are pertinent. (1) In the above proof we have only used that L(n p,X)is octahedral for every n∈Nand every 1 <p<∞. (2) Observe that this in particular implies that c0is finitely representable in X[2, Lemma 11.1.6], which implies that every Banach space is finitely representable in X.Inparticular ∞is finitely representable in X, which implies that Xhas a trivial cotype [2, Theorem 11.1.14]. Now it is time to prove that the converse holds true. Theorem 3.6 Let X be a Banach space. Assume that, for every ε>0, for every finite dimensional subspace Z of X and for every n ∈N, there exists an operator T :n ∞−→ X such that T≤1and such that z+T(y)≥(1−ε)(z+y) holds for every y ∈n ∞and every z ∈Z. Then, for every Banach space Y and for every subspace H of L(Y,X)containing the finite rank operators, the norm of H is octahedral. Proof Let Ybe a non-zero Banach space and H⊆L(Y,X)as in the hypothesis. In order to prove that His octahedral pick T1,...,Tn∈SHand ε>0, and let us find an element ∈SHsuch that Ti+>2−εholds for every 1 ≤i≤n. This is enough by [13, Proposition 2.1]. In order to do so find, for every i∈{1,...,n},anelementyi∈SYsuch that Ti(yi)>1−ε.SetZ:= span{T(yi):1≤i≤n}. Set also V:= span{y1,...,yn}⊆Y. Since every Banach space is finitely representable in c0we can find an operator φ:V−→ c0with φ(yi)>1−εfor every iandsuchthat φ<1. This operator can be extended by Theorem 2.2 to an operator Q:Y−→ c0which satisfies that Q(yi)>1−εfor every iand still Q<1. By the definition of the c0norm we can find nlarge enough such that, if we define P:c0−→ n ∞the natural projection, we get P(Q(yi))>1−εfor every i. Now, by the hypothesis, we can find an operator T:n ∞−→ Xsuch that T≤1and such that z+T(y)≥(1−ε)(z+y) holds for every y∈n ∞and every z∈Z. Now the desired operator is := T◦P◦Q:Y−→ X, which belongs to F(Y,X)⊆H. Observe that ≤1. Moreover, given 1 ≤i≤n,weget Ti+≥Ti(yi)+T(Q(P(yi)))≥(1−ε)(Ti(yi)+Q(P(yi))) ≥(1−ε)(1−ε+1−ε) =2(1−ε)2. Since εwas arbitrary we conclude the result. As a consequence we get the following result. Theorem 3.7 Let X be a Banach space. The following are equivalent: (1) For every Banach space Y and every H ⊆L(Y,X)containing the finite-rank operators, the space H is octahedral. (2) X is universally octahedral. 123
A. R. Zoca (3) For every finite dimensional Banach space Y , the space L(Y,X)is octahedral. (4) For every finite dimensional uniformly convex Banach space Y , the space L(Y,X)is octahedral. (5) For every 1<p<∞and every n ∈Nthe space L(n p,X)is octahedral. (6) For every ε>0, for every finite dimensional subspace Z of X and for every n ∈N,there exists an element T :n ∞−→ X with T≤1and such that z+T(y)≥(1−ε)(z+y) holds for every y ∈n ∞and every z ∈Z. Proof (1)⇒(2)⇒(3)⇒(4)⇒(5) are immediate, and (5)⇒(6) follows by Remark 3.5. Finally, (6)⇒(1) is Theorem 3.6. Observe that condition (6) requires not only that ∞is finitely representable in X,butalso that ∞must be finitely representable in a part of Xwhich kind of 1-orthogonal to Zfor any finite-dimensional subspace Zof X. This fact will become more clear in the following example. Example 3.8 Let X:= ∞⊕11.Xis octahedral [13, Proposition 3.10] and clearly contains ∞isometrically. However, we claim that Xis not universally octahedral. Indeed, assume by contradiction that Xis universally octahedral. Let Z:= span{(e1,0)}⊆ X=∞⊕11.Fixn∈Nand ε>0. By the above characterisation we can find φ:n ∞−→ X with φ≤1 and such that (e1,0)±φ(y)>(1−ε)(1+y) holds for every y∈n ∞. Take Q:X=∞⊕11−→ 1the natural projection. Let us prove that Q◦φ:n ∞−→ 1 satisfies that (1−2ε)y≤Q(φ(y))≤y, from where the arbitrariness of nand εwill imply that ∞is finitely-representable in 1, which is a contradiction because 1has cotype 2 and the finite representability of ∞in a Banach space Zimplies that Zfails to have cotype qfor every q<∞[2, Theorem 11.1.14]. So take x∈Sn ∞, call φ(x):= (a,b)∈∞⊕11and notice that a∞+b1≤1. Note that 2(1−ε) < (e1,0)±φ(x)=(e1±a)∞+b1. It is direct computation that either e1+a∞≤1ore1−a∞≤1. Assume without loss of generality that e1+a∞≤1. Now 2(1−ε) ≤e1+a∞+b1≤1+b1, which implies b1≥1−2ε. This implies that Q(φ(x))≥1−2ε. The arbitrariness of x∈Sn ∞forces that Q(φ(x))≥1−2εholds for every x∈Sn ∞, and a homogeneity argument yields that, for every x∈n ∞,weget (1−2ε)x≤Q(φ(x))≤Qφx≤x, as desired. 123
Banach spaces which always produce octahedral… Another exotic example is the following. Example 3.9 Let X:= (⊕∞ i=1i ∞)1. It is immediate by the main characterisation that Xis universally octahedral. Indeed, given n∈N,ε>0and{z1,...,zk}⊆SX, it is enough by Remark 3.3 to find an operator T:n ∞−→ Xwith zi+T(y)>(1−ε)(1+y) for every 1 ≤i≤kand every y∈Sn ∞. Up to a density argument we can assume that zi∈(⊕∞ i=1i ∞)have finite support (say contained in ⊕p i=1i ∞). Take q>max{p,n}.Now take the canonical inclusion operator φ:n ∞→q ∞. Take the canonical inclusion operator j:q ∞→⊕ ∞ i=1i ∞by j(x)(q)=xif i=qand 0 otherwise. Now T=j◦φ:n ∞−→ X satisfies the desired requirement. In fact, given any y∈n ∞observe that, since the support of ziand T(y)are disjoint by construction, we obtain that zi+T(y)=zi+T(y)=1+y since Tis an isometry. This proves that Xis universally octahedral. However Xdoes not contain c0isomorphically. Indeed, X=(c0(N, n 1))∗is a dual space, so if Xcontained c0then it would indeed contain ∞by [9, Theorem 6.39], which is impossible since Xis clearly separable. Let us end with an observation concerning the existence of L-orthogonal elements. Remark 3.10 Let Xbe a Banach space. Following the notation of [21], we say that an element u∈X∗∗ is an L-orthogonal element if it satisfies x+u=x+u holds for every x∈X. The existence of non-zero L-orthogonal elements is strongly connected with octahedral norms. It is a consequence of the Principle of Local Reflexivity (and explicitly mentioned in [11, Lemma 9.1]) that if Xhas a non-zero L-orthogonal element then the norm of Xis octahedral. Moreover, the converse is true if Xis separable [11, Lemma 9.1]. The question whether octahedrality implies the existence of non-zero L-orthogonals has remained open until the recent work [21], where many examples of octahedral spaces without any non-zero L-orthogonal element is exhibited. A natural question at this point is whether or not there exists a Banach space Xsatisfying that, for every non-zero Banach space Yand for every H⊆L(Y,X)containing F(Y,X), the space Hhas non-zero L-orthogonal elements. The answer is no. Indeed, given any Banach space X,takingY=2(I)and Hto be the space of compact operators from Yto X(denoted by K(Y,X)), if Hhas a non-zero Lorthogonal element, then Yis isometrically isomorphic to a subspace of X∗∗ by [21, Lemma 3.1], so it remains to take Ibig enough so that there is no injective mapping φ:I−→ X∗∗ to conclude that K(2(I), X)does not have any non-zero L-orthogonal element. 4 Examples In this section we will analyse examples of universally octahedral Banach spaces. Observe that the condition in Theorem 3.6 is very difficult to check in a particular example. So, in order to provide examples where universal octahedrality holds, we give a criterion which implies it. In order to save notation, let us make the following definition. 123