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The Birkhoff-James orthogonality of operators on infinite dimensional Banach spaces
Linear Algebra and its Applications 582 (2019) 440–451
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Linear Algebra and its Applications www.elsevier.com/locate/laa
The Birkhoff-James orthogonality of operators on infinite dimensional Banach spaces ✩ Sun Kwang Kim a , Han Ju Lee b,∗ a
Department of Mathematics, Chungbuk National University, 1 Chungdae-ro, Seowon-gu, Cheongju, Chungbuk 28644, Republic of Korea b Department of Mathematics Education, Dongguk University - Seoul, 30 Pildong-ro 1-gil, Jung-gu, Seoul 04620, Republic of Korea
a r t i c l e
i n f o
Article history: Received 24 February 2019 Accepted 13 August 2019 Available online 16 August 2019 Submitted by P. Semrl MSC: primary 46B20 secondary 46B04, 46B22, 46B28
a b s t r a c t We study the Birkhoff-James orthogonality on operator spaces in terms of the Bhatia-Šemrl property. We first prove that every functional on a Banach space X has the BhatiaŠemrl property if and only if X is reflexive. We also find some geometric conditions of Banach space which ensure the denseness of operators with Bhatia-Šemrl property. Finally, we investigate operators with the Bhatia-Šemrl property when a domain space is L1 [0, 1] or C[0, 1]. © 2019 Elsevier Inc. All rights reserved.
Keywords: Birkhoff-James orthogonality The Bhatia-Šemrl property Banach space
✩ The first author was supported by the research grant of the Chungbuk National University in 2018. The second author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF2016R1D1A1B03934771). * Corresponding author. E-mail addresses: [email protected] (S.K. Kim), [email protected] (H.J. Lee).
https://doi.org/10.1016/j.laa.2019.08.012 0024-3795/© 2019 Elsevier Inc. All rights reserved.
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1. Introduction The famous Birkhoff-James orthogonality was considered first in 1935 by G. Birkhoff in order to discuss orthogonality on linear metric spaces [3]. This concept allows many authors to consider not only orthogonality of operators but also the geometry of the space the operators are defined. Especially, Bhatia and Šemrl [2] found some characterization of operators T and S on finite dimensional Hilbert space such that T is Birkhoff-James orthogonal to S in terms of norm attaining points of T , and they asked a question whether the same result can be proved on other finite dimensional Banach spaces. The answer of this question is known to be negative by Li and Schneider [8]. Since then, many authors focused on the characterization (called the Bhatia-Šemrl property or, in short, BŠ property), and very recently it was proved that there are some Banach spaces such that sufficiently many operators satisfy the Bhatia-Šemrl property [7]. The main aim of the present paper is to continue this study for other Banach spaces. For more details, we give necessary notions and backgrounds. Throughout this paper, we only consider real Banach spaces, and X, Y are reserved for Banach spaces if there is no explanation. We write L(X, Y ) for the Banach space of all bounded linear operators from X into Y , equipped with the norm T = sup{T (x) : x ∈ X and x ≤ 1} for T ∈ L(X, Y ). Especially if Y is the real field R, L(X, R) is the dual space of X and it is denoted by X ∗ . An operator T ∈ L(X, Y ) is said to attain its norm at x ∈ X (x is a norm attaining point of T ) if x = 1 and T (x) = T . For subsets C and D of X, the canonical metric d is defined by d(C, D) = inf{x − y : x ∈ C, y ∈ D}. For two vectors x, y ∈ X, we say that x is Birkhoff-James orthogonal to y (in short, x⊥B y) if x ≤ x + λy for every λ ∈ R [3]. The aforementioned result of Bhatia and Šemrl is the following. Theorem 1.1. [2] Let H be a finite dimensional Hilbert space. Then for arbitrary T, S ∈ L(H, H), T is Birkhoff-James orthogonal to S if and only if there exists x ∈ H such that T attains its norm at x and T x⊥B Sx. The question of Bhatia and Šemrl is whether it is possible to deduce the same result when H is replaced by other finite dimensional spaces. The first counterexample was given by Li and Schneider [8]. Moreover, it had been proved that it holds only on Hilbert space [1]. Recall that an operator T ∈ L(X, Y ) said to have the Bhatia-Šemrl property (in short, BŠ property) if for arbitrary S ∈ L(X, Y ) with T ⊥B S there exists x ∈ X such that T attains its norm at x and T x⊥B Sx. Very recently, there were many efforts to study operator T ∈ L(X, Y ) with this property (see [7,11–14]). Among all the results, we present the following which shows that if an operator attains its norm ‘strongly’ at norm attaining points then it has the BŠ property. We will use it later for our main result.
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Theorem 1.2. [12, Theorem 2.2] Suppose that X, Y are Banach spaces and T ∈ L(X, Y ) satisfies that the set MT = {x ∈ X : T x = T , x = 1} is the union of D and −D for some nonempty compact connected set D. If sup{T x : x ∈ C} < T for all closed subset C ⊂ {x ∈ X : x = 1} with d(MT , C) > 0, then T has the BŠ property. Very recently, Sain, Paul and Hait [14] proved that the operators with the BŠ property are dense in L(X, X) for finite dimensional strictly convex spaces X and it is proved that the same result holds if X has the Radon-Nikodým property. That is, the set of operators with the BŠ property is dense in L(X, Y ) for arbitrary Y [7]. In this paper, we investigate how many operators have the BŠ property. In Section 2, we deal with a ‘Bhatia-Šemrl type result’ which means that every operator has the BŠ property. We use a simple characterization of the Birkhoff-James orthogonality for vectors in terms of norm attaining functionals, and prove that every functional has the BŠ property if and only if the domain is reflexive. We also prove that a one-dimensional space is a unique domain space (resp. range space) in the sense that every operator in L(X, Y ) has the BŠ property for arbitrary space Y (resp. arbitrary reflexive spaces X). In Section 3, we study a ‘denseness result’ which means the cases that the set of operators with the BŠ property is dense. We find many pairs of Banach spaces such that the denseness holds, for example, the case is either that X has the property α or that X is locally uniformly convex and Y has the property β. We also deduce some related results using renorming of Banach spaces. More precisely, we prove that every weakly compactly generated space can be renormed so that the set of operators with the BŠ property is dense. Finally, we also provide some negative results on L1 [0, 1] and C[0, 1]. 2. Bhatia-Šemrl type results We consider the cases that a ‘Bhatia-Šemrl type’ result holds for a pair (X, Y ) which means that the set of operators with the BŠ property is equal to the whole space of operators. For the convenience, we use BŠ(X, Y ) and NA(X, Y ) for the set of operators from X to Y with the BŠ property and the set of norm attaining operators, respectively. It is clear that BŠ(X, Y ) is contained in NA(X, Y ) by the definition. As we already mentioned before, it is proved [2] that a finite dimensional Banach space X satisfies BŠ(X, X) = L(X, X) if and only if X is isometric to a Hilbert space. In this section, we deal with the natural question whether the similar results can be obtained for L(X, Y ) when X and Y are different. First, we consider the case that the range space is one-dimensional. For this, we present the following charaterization of Birkhoff-James orthogonality for vectors. This must be well known fact, but we give details for the completeness. Proposition 2.1. For any x, y ∈ X, x⊥B y if and only if there exists a unit functional x∗ ∈ X ∗ such that x∗ (x) = x and x∗ (y) = 0.
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Proof. For the proof of ‘if’ part, we see x + λy ≥ |x∗ (x + λy)| = x for any λ ∈ R. For ‘only if’ part, we may assume that x, y = 0 since it is clear otherwise. Also, it is x enough to show only for the case x = 1 since x⊥B y if and only if x ⊥B y. Let Z = span{x, y}, and then it is clear that dimension of Z is 2 and x⊥B y on Z. From the well known separation theorem, choose a unit functional z ∗ ∈ Z ∗ whose kernel is span{y} with z ∗ (x) > 0. We now show that z ∗ (x) = 1. Otherwise, 0 < z ∗ (x) < 1 and there is z ∈ Z so that z ∗ (z) = 1 = z. Since z ∗ (x − z ∗ (x)z) = 0, we see that x − z ∗ (x)z = αy for some α ∈ R \ {0}. This shows that x − αy = z ∗ (x)z = |z ∗ (x)| < 1 which contradicts to x⊥B y. Finally we take the Hahn-Banach extension x∗ of z ∗ on X to get the desired functional. 2 According to Theorem 2.1, we see that if x∗ , y ∗ ∈ X ∗ satisfy x∗ ⊥B y ∗ , then there exists a unit vector x∗∗ ∈ X ∗∗ such that x∗∗ (x∗ ) = x∗ and x∗∗ (y ∗ ) = 0. If X is reflexive, the bidual element x∗∗ can be identified with an element x in X. This shows that x∗ has the BŠ property. On the otherhand, it is well known that X is reflexive if and only if every functional on X attains its norm [6]. Hence, we get the following corollary. Corollary 2.2. Let X be a Banach space. Then X is reflexive if and only if BŠ(X, R) = X ∗ . We noticed above that x∗ ∈ X ∗ has the BŠ property if and only if whenever x∗ ⊥B y ∗ , there is a unit vector x in X satisfying x∗ (x) = x and y ∗ (x) = 0. The general statement for operators is as follows. Corollary 2.3. An operator T ∈ L(X, Y ) has the BŠ property if and only if for each S ∈ L(X, Y ) such that T ⊥B S there exist unit elements x ∈ X and y ∗ ∈ Y ∗ such that y ∗ (T x) = T and y ∗ (Sx) = 0. We proved that if Y is one-dimensional, then a Bhatia-Šemrl type result holds for (X, Y ) for arbitrary reflexive space X. We now prove that one-dimensional space is the unique space having this property. Proposition 2.4. If BŠ(X, Y ) = L(X, Y ) for every reflexive X, then Y is onedimensional. Proof. Let X = n1 be an n dimensional space with 1 norm with a canonical basis {ei }ni=1 and Y be a space whose dimension is greater than 1. Then it is easy to see that there exist linearly independent unit vectors u, v ∈ Y so that, for scalars α and β with |α| + |β| = 1, αu + βv = 1 if and only if α = 0 or β = 0. We now define T ∈ L(X, Y ) by n n T ( i=1 αi ei ) = α1 u + α2 v and S ∈ L(X, Y ) by S( i=1 αi ei ) = α1 u − α2 v for arbitary
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sequence of scalars (αi )ni=1 . Then, it is clear that T ⊥B S and T only attains its norm at ±e1 and ±e2 . However, T e1 and T e2 are not orthogonal to Se1 and Se2 , respectively. This shows that T does not have the BŠ property. 2 We now consider a similar problem on domain space whether there exists X such that a Bhatia-Šemrl type result holds for (X, Y ) for arbitrary space Y , and show that one-dimensional space is the unique space. To prove this, we use a well known result of Stegall that if X has the Radon-Nikodým property, then for arbitrary Y , the set of operators in L(X, Y ) which only attain their norm at two points x and −x for some x ∈ X is dense [15]. Proposition 2.5. For a Banach space X, the equality BŠ(X, Y ) = L(X, Y ) holds for every Y if and only if X is one-dimensional. Proof. From Corollary 2.3, we see that BŠ(R, Y ) = L(R, Y ). For the converse, we first observe that X is reflexive by the theorem of James [6] since the equality BŠ(X, Y ) = L(X, Y ) for some nonzero Y implies BŠ(X, R) = X ∗ . Let 2∞ be a 2-dimensional space with ∞ norm with a canonical basis {e1 , e2 }, and we claim that BŠ(X, 2∞ ) = L(X, 2∞ ). To define an operator without the BŠ property, we use the fact that reflexivity implies the Radon-Nikodým property. From the result of Stegall, first take a unit functional x∗ ∈ X ∗ which attains its norm only at two points, say x, −x ∈ X. We next consider unit functionals z ∗ , y ∗ ∈ X ∗ such that z ∗ (x) = 0, y ∗ − z ∗ < 1/2 and y ∗ attains its norm only at some two points y, −y ∈ X. Note that x and y are linearly independent. Let T, S ∈ L(X, Y ) be operators given by T (·) = (x∗ (·), y ∗ (·)) and S(·) = (x∗ (·), −y ∗ (·)), then T ⊥B S and M = {±x, ±y} is the set of vectors where T attains its norm. However, T z is not orthogonal to Sz for arbitrary z ∈ M . Indeed, for example, let T z = (1, α) and Sz = (1, −α) for some |α| < 1. Choose λ > 0 so that (1 + λ)|α| < 1, T x − λSx = (1 − λ, (1 + λ)α) < 1. The rest cases can be proved in a similar way. 2 3. Denseness results The denseness of operators with the BŠ property was first considered in [14]. The authors showed that if a finite dimensional space X is not Hilbertian, then there exist operators in L(X, X) without the BŠ property. However, operators with the BŠ property are dense in L(X, X) for every finite dimensional strictly convex space X. In [7], it had been proved that the denseness holds not only on arbitrary finite dimensional space, but also on infinite dimensional spaces with the Radon-Nikodým property. The main aim of this section is also to find other infinite dimensional spaces such that the denseness holds.
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To do so, recall that a Banach space X is said to have the property α with constant 0 ≤ ρ < 1 if there exist sets {xi ∈ X : xi = 1, i ∈ Λ} and {x∗i ∈ X ∗ : x∗i = 1, i ∈ Λ} for some index set Λ such that (i) x∗i (xi ) = 1 for all i ∈ Λ, (ii) |x∗i (xj )| ≤ ρ < 1 for all i, j ∈ Λ with i = j, (iii) BX = aconv{xi : i ∈ Λ}, where aconv(A) is the convex hull of {±x : x ∈ A} for A ⊂ X. Notice that 1 is the most typical infinite dimensional Banach space with the property α. This property is a well known criterion for the (Lindenstrauss) property A, that is, N A(X, Y ) = L(X, Y ) holds for every Y [16]. We have an analogous result for denseness of operators with the BŠ property. Theorem 3.1. If X has the property α, then BŠ(X, Y ) = L(X, Y ) for every Y . Proof. We follow the argument in [16, Proposition 1.3], but we give details for the completeness. Let X have the property α with {xi : i ∈ Λ} ⊂ X, {x∗i : i ∈ Λ} ⊂ X ∗ and ρ > 0. Since the zero operator always has the BŠ property, we start with a nonzero operator T ∈ L(X, Y ). For arbitrary ε > 0, our aim is to find a norm attaining operator S with the BŠ property such that T − S < ε. We first pick 0 < ε < ε and λ ∈ Λ so that T xλ > T
1 + ε ρ . 1 + ε
Define an operator S by Sz = T z + ε x∗λ (z)T xλ , then it is clear that T − S < ε. It is easy to check that Sxλ > T (1 + ε ρ) and Sxi ≤ T (1 + ε ρ) for i = λ. We claim that S only attains its norm at ±xλ and it has the BŠ property. For this we show the conditions in Theorem 1.2 for {±xλ }. Indeed, fix 0 < γ < 2 and a unit vector x ∈ X satisfying d({x}, {±xλ }) > γ > 0. Take z = j∈F θj xj ∈ aconv{xi : i ∈ Λ} such that F ⊂ Λ is finite, i∈F |θi | ≤ 1, λ ∈ F and x − z <
γ(S − T (1 + ε ρ)) and d({z}, {±xλ }) > γ > 0. 4S
Then, we see that |θλ | ≤ 1 − γ/2. Otherwise, d({z}, {±xλ }) < γ. Hence,
j∈F \{λ}
|θj | < γ/2 which shows
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Sz ≤ |θλ |Sxλ +
|θj |Sxj
j∈F \{λ}
≤ |θλ |S + (1 − |θλ |)T (1 + ε ρ) ≤ S − (1 − |θλ |)(S − T (1 + ε ρ)) ≤ S −
γ(S − T (1 + ε ρ)) 2
and so Sx ≤ Sz + Sz − Sx < S −
γ(S − T (1 + ε ρ)) . 4
Therefore, we see that {±xλ } is the set of norm attaining points of S and sup{Sx : x ∈ C} < S for all closed subset C of unit vectors in X with d({±xλ }, C) > 0 which means S has the BŠ property. 2 According to the renorming results on [16], we see that there exists (1 + ε) renorming of c0 which has the property α. Moreover, every weakly compactly generated spaces can be renormed to have the property α. Since it is known that the set of operators from c0 to c0 with the BŠ property is not dense [7], we have the following observation. Corollary 3.2. Every weakly compactly generated Banach space X is isomorphic to a ˜ such that BŠ(X, ˜ Y ) = L(X, ˜ Y ) for every Y . In particular, there exist Banach space X ˜ Y ) = L(X, ˜ Y ) for Banach spaces X and Y such that BŠ(X, Y ) = L(X, Y ) but BŠ(X, ˜ some renorming X of X. In the next, we deal with a natural question whether there exists a space X such that ˜ of X, the set of norm attaining operators with the BŠ property for arbitrary renorming X ˜ Y ) for every Y . We give its characterization with the Radon-Nikodým is dense in L(X, property. Theorem 3.3. Let X be a Banach space X. Then the following are equivalent. ˜ is isomorphic to X, then for every Y (1) If X ˜ Y ) = L(X, Y ). BŠ(X, (2) X has the Radon-Nikodým property. Proof. It is well known that the Radon-Nikodým property is isomorphiclly invariant, ˜ has it [4]. Hence, from [7, Theorem 2.4], we see that (2) implies (1). and X ˜ Y ) = L(X, Y ) for every Banach space X ˜ isomorphic In [4], it is shown that if NA(X, to X and arbitrary Banach space Y , then X has the Radon-Nikodým property. Since ˜ Y ) is a subset of NA(X, ˜ Y ), we see that (1) implies (2). 2 BŠ(X,
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Now, we focus on range spaces. In [9], a geometric property called the property β was introduced to prove the existence of Y so that the set of norm attaining operators from X into Y is dense for arbitrary X. Recall that a Banach space Y is said to have the property β if there exist 0 ≤ ρ < 1, index set Λ, {yi ∈ Y : yi = 1, i ∈ Λ} and {yi∗ ∈ Y ∗ : yi∗ = 1, i ∈ Λ} such that (i) yi∗ (yi ) = 1 for all i ∈ Λ, (ii) |yi∗ (yj )| ≤ ρ < 1 for all i, j ∈ Λ with i = j, (iii) y = supi∈Λ |yi∗ (y)| for all y ∈ Y . The most typical example with the property β is c0 , and this is a kind of a dual of the property α. Surprisingly, it is known that every Banach space can be renormed to have the property β [10]. In the next result, using a similar argument in [9], we prove that if the set of functionals on X with the BŠ property is dense in X ∗ , then the denseness also holds for operators whose range is a space with the property β. Theorem 3.4. If Y has the property β and the set of functionals on X with the BŠ property is dense in X ∗ , then BŠ(X, Y ) = L(X, Y ). Proof. Let ε > 0 be given and ρ > 0 be the number in the definition of the property β. We will show that for an arbitrary nonzero operator T ∈ L(X, Y ) we find S ∈ L(X, Y ) with the BŠ property so that T − S < ε. ε Let T˜ = T and fix 0 < ξ < ε so that 1 + ρ +ξ < 1+ ε (1 − ξ). T
4T
4T
4T
From the definition of the property β, we see that T˜ = supi∈Λ T˜∗ yi∗ , and so there exists an index λ ∈ Λ such that T˜ ∗ yλ∗ > 1 − ξ. From we find a unit functional x∗1 ∈ X ∗ with the BŠ property such the assumption, ∗ T˜∗ yλ∗ that x1 − T˜∗ y∗ < ξ. This gives that x∗1 − T˜∗ yλ∗ < 2ξ. λ Define S˜ : X −→ Y by ˜ := T˜(·) + S(·)
1+
ε 4T
x∗1 (·) − T˜∗ yλ∗ (·) yλ ,
ε ε ˜∗ ∗ we have S˜ − T˜ < 4T + 2ξ < T . Moreover, we have S yλ = 1 + ∗ ∗ S˜ y = 1 + ε . On the other hand, we deduce that λ
ε 4T
x∗1 , and so
4T
∗ ∗ S˜ y ≤ 1 + ρ i
ε +ξ 4T
<
1+
ε 4T
(1 − ξ) < 1 +
ε 4T
for i ∈ Λ \ {λ}. Therefore we see that the sets of norm attaining points of S˜ and S˜∗ yλ∗ coincide and they are non-empty.
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Now we prove that S˜ has the BŠ property. To do so, take arbitrary A ∈ L(X, Y ) such ˜ B A and we claim that S˜∗ y ∗ ⊥B A∗ y ∗ . We may assume that A = 0. Since S⊥ ˜ B A, that S⊥ λ λ ∗ ∗ ∗ ˜ ˜ we have S + rA ≥ S for arbitrary r ∈ R. Notice that, for each r ∈ R, sup S˜∗ yi∗ + rA∗ yi∗ = S˜∗ + rA∗ . i∈Λ
If 0 < |r| <
1 A
1+
ε 4T
− 1+
ε 4T
(1 − ξ) , then
sup S˜∗ yi∗ + rA∗ yi∗ i∈Λ\{λ}
sup S˜∗ yi∗ + rA
≤
i∈Λ\{λ}
≤1+ρ < So, for 0 < |r| <
1 A
ε +ξ 4T
ε ε + 1+ − 1+ (1 − ξ) 4T 4T
ε + ξ. 4T 1+
ε 4T
− 1+
ε 4T
(1 − ξ) , we have
S˜∗ + rA∗ = S˜∗ yλ∗ + rA∗ yλ∗ ≥ S˜∗ ≥ S˜∗ yλ∗ . ∗ ∗ ˜ for all r ∈ R and S˜∗ y ∗ ⊥B A∗ y ∗ . ˜ + rA∗ ≥ Sy From the convexity of norm, Sy λ λ λ λ λ ∗ ˜ has the BŠ property, using Corollary 2.3, we conclude that there is a unit Since Sy λ ˜ and y ∗ A(x0 ) = 0. Hence, S = T S˜ is the desired ˜ 0 = S vector x0 ∈ X such that yλ∗ Sx λ ˜ operator with S − T < ε. 2 Recall that a Banach space X is said to be locally uniformly convex if for every unit vector x ∈ X and for every sequence of unit vectors {xn } in X with limn x +xn = 2, the sequence xn converges to x. We see that if X is a locally uniformly convex space, then the set of functionals with the BŠ property is dense. To prove this we first show that arbitrary nonzero norm attaining functional x∗ ∈ X ∗ attains its norm strongly only at two points. Indeed, for a norm attaining point x0 ∈ X of x∗ we have sup{|x∗ (x)| : x ∈ C} < x∗ for every closed subset C ⊂ X of unit vectors with d({±x0 }, C) > 0. Otherwise there exists a sequence xn ∈ C so that |x∗ (xn )| converges to x∗ . Hence, there exists a subsequence (xnk ) such that either limn x0 + xnk = 2 or limn x0 − xnk = 2, and from the assumption, xnk converges to x0 or −x0 which is the contradiction. Now, we use Theorem 1.2 to see that x∗ has the BŠ property, and the celebrated Bishop-Phelps Theorem shows that the set of norm-attaining functionals is dense in X ∗ . We summarize this observation as the following. Corollary 3.5. If X is locally uniformly convex and Y has the property β, then BŠ(X, Y ) = L(X, Y ).
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Finally, we present some results on classical spaces L1 [0, 1] and C[0, 1], the space of Lebesgue integrable functions and that of continuous functions on [0, 1]. In [7], it is proved that the set of norm attaining operators without the BŠ property is dense in L(L1 [0, 1], Y ) whenever Y has the Radon-Nikodým property. In the following, we show that zero operator is the only one with the BŠ property on L1 [0, 1] for arbitrary range. Proposition 3.6. For every Y , BŠ(L1 [0, 1], Y ) = {0}. Proof. Let T ∈ L(L1 [0, 1], Y ) be a nonzero norm attaining operator and T attains its norm at f ∈ L1 [0, 1]. Let A = {t ∈ [0, 1] : f (t) = 0}. We choose an infinite sequence of disjoint non-empty measurable sets Ai ⊂ A such that A = ∪i Ai . For a measurable set A ⊂ [0, 1], let χA be the characteristic function on A, and define fχ i a sequence of measurable functions fi = f χA . Then, we see that T fi = T since Ai f = i f χAi fi and i f χAi = 1. 1 1 c We define an operator S by Sg = T i i gχAi + 2 gχA , then T is orthogonal to S in the sense of Birkhoff-James. Indeed, for λ ∈ R, T + λS ≥ T fi + λSfi =
1 1+λ i
T fi −→ T .
On the other hand, for arbitrary unit norm attaining point h ∈ L1 [0, 1] of T , i − 1 1 hχAi + hχAc < T h = T h. T h − Sh = T i 2 i Therefore, T does not have the BŠ property. 2 It is proved that BŠ(c0 , Y ) is not dense in L(c0 , Y ) if Y is strictly convex [7]. We get the same result when the domain is C[0, 1] and the range Y is reflexive and strictly convex. To get this, we use the representation theorem of weakly compact operators in terms of vector measures [5]. Proposition 3.7. For every reflexive strictly convex Y , BŠ(C[0, 1], Y ) ⊂ {T ∈ L(C[0, 1], Y ) : MT = {±χ[0,1] }} ∪ {0} where χA denotes the characteristic function on A ⊂ [0, 1] and MT is the set of norm attaining points of T . Proof. Let T ∈ L(C[0, 1], Y ) be a nonzero norm attaining operator such that there exists f ∈ MT \ {±χ[0,1] }. Since Y is reflexive, T is weakly compact, and so there exists a Y -valued regular countably additive representing Borel measure μ of T (see [5, Theorem 1 in P.152, Theorem 3 in P153, Corollary 14 in P. 159]). Note that the measure
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μ is regular means for arbitrary measurable subset A of [0, 1] and ε > 0, there exists a compact subset K and an open subset U of [0, 1] so that K ⊂ A ⊂ U and μ(U \K) < ε where μ is the total variation of μ. Let A = {t ∈ [0, 1] : |f (t)| = 1} and Ac = [0, 1] \ A. So Ac is a nonempty open subset of [0, 1]. We claim that μ(Ac ) = 0, which is equivalent to supg∈BC[0,1] Ac gdμ = 0. Fix arbitrary ε > 0 and take a nonempty compact subset K ⊂ Ac such that μ(Ac \ K) < ε from regularity of μ. We consider an open set GK ⊂ Ac so that K ⊂ GK ⊂ GK ⊂ Ac and take a function gK ∈ C[0, 1] whose support is a subset of GK so that gK (t) = gK = min{1 − |f (s)| : s ∈ GK } > 0 for every t ∈ K by Urysohn’s lemma. For arbitrary g ∈ C[0, 1] with g ≤ 1, we see that f + gK g, f − gK g ≤ 1 and T (f + gK g) + T (f − gK g) Tf = ∈ Y. 2T T So, we have we have
g gdμ [0,1] K
= T (gK g) = 0 because of the strict convexity of Y . Therefore,
g g K K gdμ ≤ 1− gdμ gK gdμ + gK c c c A
A
A
gK gK 1− = gdμ + gdμ gK gK Ac \K [0,1] < 2ε.
Hence we prove the claim. Since Ac is a non-empty open subset of [0, 1], there exists an open interval (α, β) ⊂ Ac . Then, we define an operator S by constructing its representing measure ν as follows. For the Lebesgue measure σ on [0, 1], we set
ν = −μ +
1 (β − α)
χα, α+β σ − χ α+β ,β σ T f. 2
2
It is easy to see that for arbitrary λ ∈ R, T + λS = μ + λν
1 = (1 − λ)μ + λ χ α, α+β σ − χ α+β ,β σ T f (β − α) 2 2 = μ = T .
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Moreover, for arbitrary norm attaining point h of T and 0 < λ < 1, T h + λSh = hdμ + λ hdν [0,1] [0,1] 1 = (1 − λ)μ + λμ χα, α+β σ − χ α+β ,β σ hd (β − α) 2 2 [0,1] < μ = T h. 2 Corollary 3.8. If Y is reflexive strictly convex, then BŠ(C[0, 1], Y ) is not dense in L(C[0, 1], Y ). Proof. Take an arbitrary unit functional x∗ ∈ C[0, 1]∗ such that whose kernel contains χ[0,1] and fix a unit vector y ∈ Y . Define T by T (·) = x∗ (·)y, then it is clear that S ∈ L(C[0, 1], Y ) with S − T < 1/2 attains its norm at neither χ[0,1] nor −χ[0,1] . 2 Declaration of competing interest There is no competing interest. References [1] C. Benítez, M. Fernández, M.L. Soriano, Orthogonality of matrices, Linear Algebra Appl. 422 (2007) 155–163. [2] R. Bhatia, P. Šemrl, Orthogonality of matrices and some distance problems, Linear Algebra Appl. 287 (1–3) (1999) 77–85. [3] G. Birkhoff, Orthogonality in linear metric spaces, Duke Math. J. 1 (2) (1935) 169–172. [4] J. Bourgain, On dentability and the Bishop-Phelps property, Israel J. Math. 28 (4) (1977) 265–271. [5] J. Diestel, J.J. Uhl Jr, Vector Measures, Math. Surveys, vol. 15, Amer. Math. Soc., 1977. [6] R.C. James, Weak compactness and reflexivity, Israel J. Math. 2 (1964) 101–119. [7] S.K. Kim, Quantity of operators with the Bhatia-Šemrl property, Linear Algebra Appl. 537 (2018) 22–37. [8] C.K. Li, H. Schneider, Orthogonality of matrices, Linear Algebra Appl. 347 (2002) 115–122. [9] J. Lindenstrauss, On operators which attain their norm, Israel J. Math. 1 (1963) 139–148. [10] J.R. Partington, Norm attaining operators, Israel J. Math. 43 (1982) 273–276. [11] K. Paul, D. Sain, Orthogonality of operators on (Rn , ∞ ), Novi Sad J. Math. 43 (1) (2013) 121–129. [12] K. Paul, D. Sain, P. Ghosh, Birkhoff-James orthogonality and smoothness of bounded linear operators, Linear Algebra Appl. 506 (2016) 551–563. [13] D. Sain, K. Paul, Operator norm attainment and inner product spaces, Linear Algebra Appl. 439 (8) (2013) 2448–2452. [14] D. Sain, K. Paul, S. Hait, Operator norm attainment and Birkhoff-James orthogonality, Linear Algebra Appl. 476 (2015) 85–97. [15] C. Stegall, Optimization of functions on certain subsets of Banach spaces, Math. Ann. 236 (2) (1978) 171–176. [16] W. Schachermayer, Norm attaining operators and renormings of Banach spaces, Israel J. Math. 44 (3) (1983) 201–212.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
The Birkhoff-James orthogonality of operators on infinite dimensional Banach spaces ✩ Sun Kwang Kim a , Han Ju Lee b,∗ a
Department of Mathematics, Chungbuk National University, 1 Chungdae-ro, Seowon-gu, Cheongju, Chungbuk 28644, Republic of Korea b Department of Mathematics Education, Dongguk University - Seoul, 30 Pildong-ro 1-gil, Jung-gu, Seoul 04620, Republic of Korea
a r t i c l e
i n f o
Article history: Received 24 February 2019 Accepted 13 August 2019 Available online 16 August 2019 Submitted by P. Semrl MSC: primary 46B20 secondary 46B04, 46B22, 46B28
a b s t r a c t We study the Birkhoff-James orthogonality on operator spaces in terms of the Bhatia-Šemrl property. We first prove that every functional on a Banach space X has the BhatiaŠemrl property if and only if X is reflexive. We also find some geometric conditions of Banach space which ensure the denseness of operators with Bhatia-Šemrl property. Finally, we investigate operators with the Bhatia-Šemrl property when a domain space is L1 [0, 1] or C[0, 1]. © 2019 Elsevier Inc. All rights reserved.
Keywords: Birkhoff-James orthogonality The Bhatia-Šemrl property Banach space
✩ The first author was supported by the research grant of the Chungbuk National University in 2018. The second author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF2016R1D1A1B03934771). * Corresponding author. E-mail addresses: [email protected] (S.K. Kim), [email protected] (H.J. Lee).
https://doi.org/10.1016/j.laa.2019.08.012 0024-3795/© 2019 Elsevier Inc. All rights reserved.
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1. Introduction The famous Birkhoff-James orthogonality was considered first in 1935 by G. Birkhoff in order to discuss orthogonality on linear metric spaces [3]. This concept allows many authors to consider not only orthogonality of operators but also the geometry of the space the operators are defined. Especially, Bhatia and Šemrl [2] found some characterization of operators T and S on finite dimensional Hilbert space such that T is Birkhoff-James orthogonal to S in terms of norm attaining points of T , and they asked a question whether the same result can be proved on other finite dimensional Banach spaces. The answer of this question is known to be negative by Li and Schneider [8]. Since then, many authors focused on the characterization (called the Bhatia-Šemrl property or, in short, BŠ property), and very recently it was proved that there are some Banach spaces such that sufficiently many operators satisfy the Bhatia-Šemrl property [7]. The main aim of the present paper is to continue this study for other Banach spaces. For more details, we give necessary notions and backgrounds. Throughout this paper, we only consider real Banach spaces, and X, Y are reserved for Banach spaces if there is no explanation. We write L(X, Y ) for the Banach space of all bounded linear operators from X into Y , equipped with the norm T = sup{T (x) : x ∈ X and x ≤ 1} for T ∈ L(X, Y ). Especially if Y is the real field R, L(X, R) is the dual space of X and it is denoted by X ∗ . An operator T ∈ L(X, Y ) is said to attain its norm at x ∈ X (x is a norm attaining point of T ) if x = 1 and T (x) = T . For subsets C and D of X, the canonical metric d is defined by d(C, D) = inf{x − y : x ∈ C, y ∈ D}. For two vectors x, y ∈ X, we say that x is Birkhoff-James orthogonal to y (in short, x⊥B y) if x ≤ x + λy for every λ ∈ R [3]. The aforementioned result of Bhatia and Šemrl is the following. Theorem 1.1. [2] Let H be a finite dimensional Hilbert space. Then for arbitrary T, S ∈ L(H, H), T is Birkhoff-James orthogonal to S if and only if there exists x ∈ H such that T attains its norm at x and T x⊥B Sx. The question of Bhatia and Šemrl is whether it is possible to deduce the same result when H is replaced by other finite dimensional spaces. The first counterexample was given by Li and Schneider [8]. Moreover, it had been proved that it holds only on Hilbert space [1]. Recall that an operator T ∈ L(X, Y ) said to have the Bhatia-Šemrl property (in short, BŠ property) if for arbitrary S ∈ L(X, Y ) with T ⊥B S there exists x ∈ X such that T attains its norm at x and T x⊥B Sx. Very recently, there were many efforts to study operator T ∈ L(X, Y ) with this property (see [7,11–14]). Among all the results, we present the following which shows that if an operator attains its norm ‘strongly’ at norm attaining points then it has the BŠ property. We will use it later for our main result.
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Theorem 1.2. [12, Theorem 2.2] Suppose that X, Y are Banach spaces and T ∈ L(X, Y ) satisfies that the set MT = {x ∈ X : T x = T , x = 1} is the union of D and −D for some nonempty compact connected set D. If sup{T x : x ∈ C} < T for all closed subset C ⊂ {x ∈ X : x = 1} with d(MT , C) > 0, then T has the BŠ property. Very recently, Sain, Paul and Hait [14] proved that the operators with the BŠ property are dense in L(X, X) for finite dimensional strictly convex spaces X and it is proved that the same result holds if X has the Radon-Nikodým property. That is, the set of operators with the BŠ property is dense in L(X, Y ) for arbitrary Y [7]. In this paper, we investigate how many operators have the BŠ property. In Section 2, we deal with a ‘Bhatia-Šemrl type result’ which means that every operator has the BŠ property. We use a simple characterization of the Birkhoff-James orthogonality for vectors in terms of norm attaining functionals, and prove that every functional has the BŠ property if and only if the domain is reflexive. We also prove that a one-dimensional space is a unique domain space (resp. range space) in the sense that every operator in L(X, Y ) has the BŠ property for arbitrary space Y (resp. arbitrary reflexive spaces X). In Section 3, we study a ‘denseness result’ which means the cases that the set of operators with the BŠ property is dense. We find many pairs of Banach spaces such that the denseness holds, for example, the case is either that X has the property α or that X is locally uniformly convex and Y has the property β. We also deduce some related results using renorming of Banach spaces. More precisely, we prove that every weakly compactly generated space can be renormed so that the set of operators with the BŠ property is dense. Finally, we also provide some negative results on L1 [0, 1] and C[0, 1]. 2. Bhatia-Šemrl type results We consider the cases that a ‘Bhatia-Šemrl type’ result holds for a pair (X, Y ) which means that the set of operators with the BŠ property is equal to the whole space of operators. For the convenience, we use BŠ(X, Y ) and NA(X, Y ) for the set of operators from X to Y with the BŠ property and the set of norm attaining operators, respectively. It is clear that BŠ(X, Y ) is contained in NA(X, Y ) by the definition. As we already mentioned before, it is proved [2] that a finite dimensional Banach space X satisfies BŠ(X, X) = L(X, X) if and only if X is isometric to a Hilbert space. In this section, we deal with the natural question whether the similar results can be obtained for L(X, Y ) when X and Y are different. First, we consider the case that the range space is one-dimensional. For this, we present the following charaterization of Birkhoff-James orthogonality for vectors. This must be well known fact, but we give details for the completeness. Proposition 2.1. For any x, y ∈ X, x⊥B y if and only if there exists a unit functional x∗ ∈ X ∗ such that x∗ (x) = x and x∗ (y) = 0.
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Proof. For the proof of ‘if’ part, we see x + λy ≥ |x∗ (x + λy)| = x for any λ ∈ R. For ‘only if’ part, we may assume that x, y = 0 since it is clear otherwise. Also, it is x enough to show only for the case x = 1 since x⊥B y if and only if x ⊥B y. Let Z = span{x, y}, and then it is clear that dimension of Z is 2 and x⊥B y on Z. From the well known separation theorem, choose a unit functional z ∗ ∈ Z ∗ whose kernel is span{y} with z ∗ (x) > 0. We now show that z ∗ (x) = 1. Otherwise, 0 < z ∗ (x) < 1 and there is z ∈ Z so that z ∗ (z) = 1 = z. Since z ∗ (x − z ∗ (x)z) = 0, we see that x − z ∗ (x)z = αy for some α ∈ R \ {0}. This shows that x − αy = z ∗ (x)z = |z ∗ (x)| < 1 which contradicts to x⊥B y. Finally we take the Hahn-Banach extension x∗ of z ∗ on X to get the desired functional. 2 According to Theorem 2.1, we see that if x∗ , y ∗ ∈ X ∗ satisfy x∗ ⊥B y ∗ , then there exists a unit vector x∗∗ ∈ X ∗∗ such that x∗∗ (x∗ ) = x∗ and x∗∗ (y ∗ ) = 0. If X is reflexive, the bidual element x∗∗ can be identified with an element x in X. This shows that x∗ has the BŠ property. On the otherhand, it is well known that X is reflexive if and only if every functional on X attains its norm [6]. Hence, we get the following corollary. Corollary 2.2. Let X be a Banach space. Then X is reflexive if and only if BŠ(X, R) = X ∗ . We noticed above that x∗ ∈ X ∗ has the BŠ property if and only if whenever x∗ ⊥B y ∗ , there is a unit vector x in X satisfying x∗ (x) = x and y ∗ (x) = 0. The general statement for operators is as follows. Corollary 2.3. An operator T ∈ L(X, Y ) has the BŠ property if and only if for each S ∈ L(X, Y ) such that T ⊥B S there exist unit elements x ∈ X and y ∗ ∈ Y ∗ such that y ∗ (T x) = T and y ∗ (Sx) = 0. We proved that if Y is one-dimensional, then a Bhatia-Šemrl type result holds for (X, Y ) for arbitrary reflexive space X. We now prove that one-dimensional space is the unique space having this property. Proposition 2.4. If BŠ(X, Y ) = L(X, Y ) for every reflexive X, then Y is onedimensional. Proof. Let X = n1 be an n dimensional space with 1 norm with a canonical basis {ei }ni=1 and Y be a space whose dimension is greater than 1. Then it is easy to see that there exist linearly independent unit vectors u, v ∈ Y so that, for scalars α and β with |α| + |β| = 1, αu + βv = 1 if and only if α = 0 or β = 0. We now define T ∈ L(X, Y ) by n n T ( i=1 αi ei ) = α1 u + α2 v and S ∈ L(X, Y ) by S( i=1 αi ei ) = α1 u − α2 v for arbitary
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sequence of scalars (αi )ni=1 . Then, it is clear that T ⊥B S and T only attains its norm at ±e1 and ±e2 . However, T e1 and T e2 are not orthogonal to Se1 and Se2 , respectively. This shows that T does not have the BŠ property. 2 We now consider a similar problem on domain space whether there exists X such that a Bhatia-Šemrl type result holds for (X, Y ) for arbitrary space Y , and show that one-dimensional space is the unique space. To prove this, we use a well known result of Stegall that if X has the Radon-Nikodým property, then for arbitrary Y , the set of operators in L(X, Y ) which only attain their norm at two points x and −x for some x ∈ X is dense [15]. Proposition 2.5. For a Banach space X, the equality BŠ(X, Y ) = L(X, Y ) holds for every Y if and only if X is one-dimensional. Proof. From Corollary 2.3, we see that BŠ(R, Y ) = L(R, Y ). For the converse, we first observe that X is reflexive by the theorem of James [6] since the equality BŠ(X, Y ) = L(X, Y ) for some nonzero Y implies BŠ(X, R) = X ∗ . Let 2∞ be a 2-dimensional space with ∞ norm with a canonical basis {e1 , e2 }, and we claim that BŠ(X, 2∞ ) = L(X, 2∞ ). To define an operator without the BŠ property, we use the fact that reflexivity implies the Radon-Nikodým property. From the result of Stegall, first take a unit functional x∗ ∈ X ∗ which attains its norm only at two points, say x, −x ∈ X. We next consider unit functionals z ∗ , y ∗ ∈ X ∗ such that z ∗ (x) = 0, y ∗ − z ∗ < 1/2 and y ∗ attains its norm only at some two points y, −y ∈ X. Note that x and y are linearly independent. Let T, S ∈ L(X, Y ) be operators given by T (·) = (x∗ (·), y ∗ (·)) and S(·) = (x∗ (·), −y ∗ (·)), then T ⊥B S and M = {±x, ±y} is the set of vectors where T attains its norm. However, T z is not orthogonal to Sz for arbitrary z ∈ M . Indeed, for example, let T z = (1, α) and Sz = (1, −α) for some |α| < 1. Choose λ > 0 so that (1 + λ)|α| < 1, T x − λSx = (1 − λ, (1 + λ)α) < 1. The rest cases can be proved in a similar way. 2 3. Denseness results The denseness of operators with the BŠ property was first considered in [14]. The authors showed that if a finite dimensional space X is not Hilbertian, then there exist operators in L(X, X) without the BŠ property. However, operators with the BŠ property are dense in L(X, X) for every finite dimensional strictly convex space X. In [7], it had been proved that the denseness holds not only on arbitrary finite dimensional space, but also on infinite dimensional spaces with the Radon-Nikodým property. The main aim of this section is also to find other infinite dimensional spaces such that the denseness holds.
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To do so, recall that a Banach space X is said to have the property α with constant 0 ≤ ρ < 1 if there exist sets {xi ∈ X : xi = 1, i ∈ Λ} and {x∗i ∈ X ∗ : x∗i = 1, i ∈ Λ} for some index set Λ such that (i) x∗i (xi ) = 1 for all i ∈ Λ, (ii) |x∗i (xj )| ≤ ρ < 1 for all i, j ∈ Λ with i = j, (iii) BX = aconv{xi : i ∈ Λ}, where aconv(A) is the convex hull of {±x : x ∈ A} for A ⊂ X. Notice that 1 is the most typical infinite dimensional Banach space with the property α. This property is a well known criterion for the (Lindenstrauss) property A, that is, N A(X, Y ) = L(X, Y ) holds for every Y [16]. We have an analogous result for denseness of operators with the BŠ property. Theorem 3.1. If X has the property α, then BŠ(X, Y ) = L(X, Y ) for every Y . Proof. We follow the argument in [16, Proposition 1.3], but we give details for the completeness. Let X have the property α with {xi : i ∈ Λ} ⊂ X, {x∗i : i ∈ Λ} ⊂ X ∗ and ρ > 0. Since the zero operator always has the BŠ property, we start with a nonzero operator T ∈ L(X, Y ). For arbitrary ε > 0, our aim is to find a norm attaining operator S with the BŠ property such that T − S < ε. We first pick 0 < ε < ε and λ ∈ Λ so that T xλ > T
1 + ε ρ . 1 + ε
Define an operator S by Sz = T z + ε x∗λ (z)T xλ , then it is clear that T − S < ε. It is easy to check that Sxλ > T (1 + ε ρ) and Sxi ≤ T (1 + ε ρ) for i = λ. We claim that S only attains its norm at ±xλ and it has the BŠ property. For this we show the conditions in Theorem 1.2 for {±xλ }. Indeed, fix 0 < γ < 2 and a unit vector x ∈ X satisfying d({x}, {±xλ }) > γ > 0. Take z = j∈F θj xj ∈ aconv{xi : i ∈ Λ} such that F ⊂ Λ is finite, i∈F |θi | ≤ 1, λ ∈ F and x − z <
γ(S − T (1 + ε ρ)) and d({z}, {±xλ }) > γ > 0. 4S
Then, we see that |θλ | ≤ 1 − γ/2. Otherwise, d({z}, {±xλ }) < γ. Hence,
j∈F \{λ}
|θj | < γ/2 which shows
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Sz ≤ |θλ |Sxλ +
|θj |Sxj
j∈F \{λ}
≤ |θλ |S + (1 − |θλ |)T (1 + ε ρ) ≤ S − (1 − |θλ |)(S − T (1 + ε ρ)) ≤ S −
γ(S − T (1 + ε ρ)) 2
and so Sx ≤ Sz + Sz − Sx < S −
γ(S − T (1 + ε ρ)) . 4
Therefore, we see that {±xλ } is the set of norm attaining points of S and sup{Sx : x ∈ C} < S for all closed subset C of unit vectors in X with d({±xλ }, C) > 0 which means S has the BŠ property. 2 According to the renorming results on [16], we see that there exists (1 + ε) renorming of c0 which has the property α. Moreover, every weakly compactly generated spaces can be renormed to have the property α. Since it is known that the set of operators from c0 to c0 with the BŠ property is not dense [7], we have the following observation. Corollary 3.2. Every weakly compactly generated Banach space X is isomorphic to a ˜ such that BŠ(X, ˜ Y ) = L(X, ˜ Y ) for every Y . In particular, there exist Banach space X ˜ Y ) = L(X, ˜ Y ) for Banach spaces X and Y such that BŠ(X, Y ) = L(X, Y ) but BŠ(X, ˜ some renorming X of X. In the next, we deal with a natural question whether there exists a space X such that ˜ of X, the set of norm attaining operators with the BŠ property for arbitrary renorming X ˜ Y ) for every Y . We give its characterization with the Radon-Nikodým is dense in L(X, property. Theorem 3.3. Let X be a Banach space X. Then the following are equivalent. ˜ is isomorphic to X, then for every Y (1) If X ˜ Y ) = L(X, Y ). BŠ(X, (2) X has the Radon-Nikodým property. Proof. It is well known that the Radon-Nikodým property is isomorphiclly invariant, ˜ has it [4]. Hence, from [7, Theorem 2.4], we see that (2) implies (1). and X ˜ Y ) = L(X, Y ) for every Banach space X ˜ isomorphic In [4], it is shown that if NA(X, to X and arbitrary Banach space Y , then X has the Radon-Nikodým property. Since ˜ Y ) is a subset of NA(X, ˜ Y ), we see that (1) implies (2). 2 BŠ(X,
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Now, we focus on range spaces. In [9], a geometric property called the property β was introduced to prove the existence of Y so that the set of norm attaining operators from X into Y is dense for arbitrary X. Recall that a Banach space Y is said to have the property β if there exist 0 ≤ ρ < 1, index set Λ, {yi ∈ Y : yi = 1, i ∈ Λ} and {yi∗ ∈ Y ∗ : yi∗ = 1, i ∈ Λ} such that (i) yi∗ (yi ) = 1 for all i ∈ Λ, (ii) |yi∗ (yj )| ≤ ρ < 1 for all i, j ∈ Λ with i = j, (iii) y = supi∈Λ |yi∗ (y)| for all y ∈ Y . The most typical example with the property β is c0 , and this is a kind of a dual of the property α. Surprisingly, it is known that every Banach space can be renormed to have the property β [10]. In the next result, using a similar argument in [9], we prove that if the set of functionals on X with the BŠ property is dense in X ∗ , then the denseness also holds for operators whose range is a space with the property β. Theorem 3.4. If Y has the property β and the set of functionals on X with the BŠ property is dense in X ∗ , then BŠ(X, Y ) = L(X, Y ). Proof. Let ε > 0 be given and ρ > 0 be the number in the definition of the property β. We will show that for an arbitrary nonzero operator T ∈ L(X, Y ) we find S ∈ L(X, Y ) with the BŠ property so that T − S < ε. ε Let T˜ = T and fix 0 < ξ < ε so that 1 + ρ +ξ < 1+ ε (1 − ξ). T
4T
4T
4T
From the definition of the property β, we see that T˜ = supi∈Λ T˜∗ yi∗ , and so there exists an index λ ∈ Λ such that T˜ ∗ yλ∗ > 1 − ξ. From we find a unit functional x∗1 ∈ X ∗ with the BŠ property such the assumption, ∗ T˜∗ yλ∗ that x1 − T˜∗ y∗ < ξ. This gives that x∗1 − T˜∗ yλ∗ < 2ξ. λ Define S˜ : X −→ Y by ˜ := T˜(·) + S(·)
1+
ε 4T
x∗1 (·) − T˜∗ yλ∗ (·) yλ ,
ε ε ˜∗ ∗ we have S˜ − T˜ < 4T + 2ξ < T . Moreover, we have S yλ = 1 + ∗ ∗ S˜ y = 1 + ε . On the other hand, we deduce that λ
ε 4T
x∗1 , and so
4T
∗ ∗ S˜ y ≤ 1 + ρ i
ε +ξ 4T
<
1+
ε 4T
(1 − ξ) < 1 +
ε 4T
for i ∈ Λ \ {λ}. Therefore we see that the sets of norm attaining points of S˜ and S˜∗ yλ∗ coincide and they are non-empty.
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Now we prove that S˜ has the BŠ property. To do so, take arbitrary A ∈ L(X, Y ) such ˜ B A and we claim that S˜∗ y ∗ ⊥B A∗ y ∗ . We may assume that A = 0. Since S⊥ ˜ B A, that S⊥ λ λ ∗ ∗ ∗ ˜ ˜ we have S + rA ≥ S for arbitrary r ∈ R. Notice that, for each r ∈ R, sup S˜∗ yi∗ + rA∗ yi∗ = S˜∗ + rA∗ . i∈Λ
If 0 < |r| <
1 A
1+
ε 4T
− 1+
ε 4T
(1 − ξ) , then
sup S˜∗ yi∗ + rA∗ yi∗ i∈Λ\{λ}
sup S˜∗ yi∗ + rA
≤
i∈Λ\{λ}
≤1+ρ < So, for 0 < |r| <
1 A
ε +ξ 4T
ε ε + 1+ − 1+ (1 − ξ) 4T 4T
ε + ξ. 4T 1+
ε 4T
− 1+
ε 4T
(1 − ξ) , we have
S˜∗ + rA∗ = S˜∗ yλ∗ + rA∗ yλ∗ ≥ S˜∗ ≥ S˜∗ yλ∗ . ∗ ∗ ˜ for all r ∈ R and S˜∗ y ∗ ⊥B A∗ y ∗ . ˜ + rA∗ ≥ Sy From the convexity of norm, Sy λ λ λ λ λ ∗ ˜ has the BŠ property, using Corollary 2.3, we conclude that there is a unit Since Sy λ ˜ and y ∗ A(x0 ) = 0. Hence, S = T S˜ is the desired ˜ 0 = S vector x0 ∈ X such that yλ∗ Sx λ ˜ operator with S − T < ε. 2 Recall that a Banach space X is said to be locally uniformly convex if for every unit vector x ∈ X and for every sequence of unit vectors {xn } in X with limn x +xn = 2, the sequence xn converges to x. We see that if X is a locally uniformly convex space, then the set of functionals with the BŠ property is dense. To prove this we first show that arbitrary nonzero norm attaining functional x∗ ∈ X ∗ attains its norm strongly only at two points. Indeed, for a norm attaining point x0 ∈ X of x∗ we have sup{|x∗ (x)| : x ∈ C} < x∗ for every closed subset C ⊂ X of unit vectors with d({±x0 }, C) > 0. Otherwise there exists a sequence xn ∈ C so that |x∗ (xn )| converges to x∗ . Hence, there exists a subsequence (xnk ) such that either limn x0 + xnk = 2 or limn x0 − xnk = 2, and from the assumption, xnk converges to x0 or −x0 which is the contradiction. Now, we use Theorem 1.2 to see that x∗ has the BŠ property, and the celebrated Bishop-Phelps Theorem shows that the set of norm-attaining functionals is dense in X ∗ . We summarize this observation as the following. Corollary 3.5. If X is locally uniformly convex and Y has the property β, then BŠ(X, Y ) = L(X, Y ).
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Finally, we present some results on classical spaces L1 [0, 1] and C[0, 1], the space of Lebesgue integrable functions and that of continuous functions on [0, 1]. In [7], it is proved that the set of norm attaining operators without the BŠ property is dense in L(L1 [0, 1], Y ) whenever Y has the Radon-Nikodým property. In the following, we show that zero operator is the only one with the BŠ property on L1 [0, 1] for arbitrary range. Proposition 3.6. For every Y , BŠ(L1 [0, 1], Y ) = {0}. Proof. Let T ∈ L(L1 [0, 1], Y ) be a nonzero norm attaining operator and T attains its norm at f ∈ L1 [0, 1]. Let A = {t ∈ [0, 1] : f (t) = 0}. We choose an infinite sequence of disjoint non-empty measurable sets Ai ⊂ A such that A = ∪i Ai . For a measurable set A ⊂ [0, 1], let χA be the characteristic function on A, and define fχ i a sequence of measurable functions fi = f χA . Then, we see that T fi = T since Ai f = i f χAi fi and i f χAi = 1. 1 1 c We define an operator S by Sg = T i i gχAi + 2 gχA , then T is orthogonal to S in the sense of Birkhoff-James. Indeed, for λ ∈ R, T + λS ≥ T fi + λSfi =
1 1+λ i
T fi −→ T .
On the other hand, for arbitrary unit norm attaining point h ∈ L1 [0, 1] of T , i − 1 1 hχAi + hχAc < T h = T h. T h − Sh = T i 2 i Therefore, T does not have the BŠ property. 2 It is proved that BŠ(c0 , Y ) is not dense in L(c0 , Y ) if Y is strictly convex [7]. We get the same result when the domain is C[0, 1] and the range Y is reflexive and strictly convex. To get this, we use the representation theorem of weakly compact operators in terms of vector measures [5]. Proposition 3.7. For every reflexive strictly convex Y , BŠ(C[0, 1], Y ) ⊂ {T ∈ L(C[0, 1], Y ) : MT = {±χ[0,1] }} ∪ {0} where χA denotes the characteristic function on A ⊂ [0, 1] and MT is the set of norm attaining points of T . Proof. Let T ∈ L(C[0, 1], Y ) be a nonzero norm attaining operator such that there exists f ∈ MT \ {±χ[0,1] }. Since Y is reflexive, T is weakly compact, and so there exists a Y -valued regular countably additive representing Borel measure μ of T (see [5, Theorem 1 in P.152, Theorem 3 in P153, Corollary 14 in P. 159]). Note that the measure
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μ is regular means for arbitrary measurable subset A of [0, 1] and ε > 0, there exists a compact subset K and an open subset U of [0, 1] so that K ⊂ A ⊂ U and μ(U \K) < ε where μ is the total variation of μ. Let A = {t ∈ [0, 1] : |f (t)| = 1} and Ac = [0, 1] \ A. So Ac is a nonempty open subset of [0, 1]. We claim that μ(Ac ) = 0, which is equivalent to supg∈BC[0,1] Ac gdμ = 0. Fix arbitrary ε > 0 and take a nonempty compact subset K ⊂ Ac such that μ(Ac \ K) < ε from regularity of μ. We consider an open set GK ⊂ Ac so that K ⊂ GK ⊂ GK ⊂ Ac and take a function gK ∈ C[0, 1] whose support is a subset of GK so that gK (t) = gK = min{1 − |f (s)| : s ∈ GK } > 0 for every t ∈ K by Urysohn’s lemma. For arbitrary g ∈ C[0, 1] with g ≤ 1, we see that f + gK g, f − gK g ≤ 1 and T (f + gK g) + T (f − gK g) Tf = ∈ Y. 2T T So, we have we have
g gdμ [0,1] K
= T (gK g) = 0 because of the strict convexity of Y . Therefore,
g g K K gdμ ≤ 1− gdμ gK gdμ + gK c c c A
A
A
gK gK 1− = gdμ + gdμ gK gK Ac \K [0,1] < 2ε.
Hence we prove the claim. Since Ac is a non-empty open subset of [0, 1], there exists an open interval (α, β) ⊂ Ac . Then, we define an operator S by constructing its representing measure ν as follows. For the Lebesgue measure σ on [0, 1], we set
ν = −μ +
1 (β − α)
χα, α+β σ − χ α+β ,β σ T f. 2
2
It is easy to see that for arbitrary λ ∈ R, T + λS = μ + λν
1 = (1 − λ)μ + λ χ α, α+β σ − χ α+β ,β σ T f (β − α) 2 2 = μ = T .
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Moreover, for arbitrary norm attaining point h of T and 0 < λ < 1, T h + λSh = hdμ + λ hdν [0,1] [0,1] 1 = (1 − λ)μ + λμ χα, α+β σ − χ α+β ,β σ hd (β − α) 2 2 [0,1] < μ = T h. 2 Corollary 3.8. If Y is reflexive strictly convex, then BŠ(C[0, 1], Y ) is not dense in L(C[0, 1], Y ). Proof. Take an arbitrary unit functional x∗ ∈ C[0, 1]∗ such that whose kernel contains χ[0,1] and fix a unit vector y ∈ Y . Define T by T (·) = x∗ (·)y, then it is clear that S ∈ L(C[0, 1], Y ) with S − T < 1/2 attains its norm at neither χ[0,1] nor −χ[0,1] . 2 Declaration of competing interest There is no competing interest. References [1] C. Benítez, M. Fernández, M.L. Soriano, Orthogonality of matrices, Linear Algebra Appl. 422 (2007) 155–163. [2] R. Bhatia, P. Šemrl, Orthogonality of matrices and some distance problems, Linear Algebra Appl. 287 (1–3) (1999) 77–85. [3] G. Birkhoff, Orthogonality in linear metric spaces, Duke Math. J. 1 (2) (1935) 169–172. [4] J. Bourgain, On dentability and the Bishop-Phelps property, Israel J. Math. 28 (4) (1977) 265–271. [5] J. Diestel, J.J. Uhl Jr, Vector Measures, Math. Surveys, vol. 15, Amer. Math. Soc., 1977. [6] R.C. James, Weak compactness and reflexivity, Israel J. Math. 2 (1964) 101–119. [7] S.K. Kim, Quantity of operators with the Bhatia-Šemrl property, Linear Algebra Appl. 537 (2018) 22–37. [8] C.K. Li, H. Schneider, Orthogonality of matrices, Linear Algebra Appl. 347 (2002) 115–122. [9] J. Lindenstrauss, On operators which attain their norm, Israel J. Math. 1 (1963) 139–148. [10] J.R. Partington, Norm attaining operators, Israel J. Math. 43 (1982) 273–276. [11] K. Paul, D. Sain, Orthogonality of operators on (Rn , ∞ ), Novi Sad J. Math. 43 (1) (2013) 121–129. [12] K. Paul, D. Sain, P. Ghosh, Birkhoff-James orthogonality and smoothness of bounded linear operators, Linear Algebra Appl. 506 (2016) 551–563. [13] D. Sain, K. Paul, Operator norm attainment and inner product spaces, Linear Algebra Appl. 439 (8) (2013) 2448–2452. [14] D. Sain, K. Paul, S. Hait, Operator norm attainment and Birkhoff-James orthogonality, Linear Algebra Appl. 476 (2015) 85–97. [15] C. Stegall, Optimization of functions on certain subsets of Banach spaces, Math. Ann. 236 (2) (1978) 171–176. [16] W. Schachermayer, Norm attaining operators and renormings of Banach spaces, Israel J. Math. 44 (3) (1983) 201–212.