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Quantity of operators with Bhatia–Šemrl property
Linear Algebra and its Applications 537 (2018) 22–37
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Linear Algebra and its Applications www.elsevier.com/locate/laa
Quantity of operators with Bhatia–Šemrl property Sun Kwang Kim 1 Department of Mathematics, Kyonggi University, Suwon 16227, Republic of Korea
a r t i c l e
i n f o
Article history: Received 28 May 2017 Accepted 19 September 2017 Available online 21 September 2017 Submitted by R. Bhatia MSC: primary 46B20 secondary 46B04, 46B22, 46B2
a b s t r a c t We study the Bhatia–Šemrl property which is related to Birkhoff–James orthogonality on operator spaces. Recently, many authors obtained some conditions for operators to have Bhatia–Šemrl property. Using them, we explore the denseness of operators with Bhatia–Šemrl property or without Bhatia– Šemrl property on some classes of Banach spaces. © 2017 Elsevier Inc. All rights reserved.
Keywords: Birkhoff–James orthogonality Bhatia–Šemrl property Banach space
1. Introduction In 1935, G. Birkhoff considered a notion of orthogonality in linear metric spaces which we call Birkhoff–James orthogonality [3]. This concept has been played a very important role for studying geometry of Banach space. There has been many interesting works about orthogonality on operator space. Among them, on finite dimensional Hilbert space, Bhatia and Šemrl [2] found necessary and sufficient condition for operators A and B to be that A is orthogonal to B in terms of specific vectors. They also asked whether the
1
E-mail address: [email protected]. This work was supported by Kyonggi University Research Grant 2016.
https://doi.org/10.1016/j.laa.2017.09.021 0024-3795/© 2017 Elsevier Inc. All rights reserved.
S.K. Kim / Linear Algebra and its Applications 537 (2018) 22–37
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same characterization holds for arbitrary finite dimensional space, and later a negative answer was given by Li and Schneider [7]. In this paper, we investigate the condition of Bhatia and Šemrl. Especially, we focus on ‘quantity of operators having the condition’. Let us now restart the introduction. This time we give necessary notions and background material. Let X and Y be Banach spaces. We will use the common notation SX , BX , X ∗ for the unit sphere, the closed unit ball and the dual space of X respectively, L(X, Y ) for the Banach space of every bounded linear operators from X to Y . For the convenience we write L(X) to denote L(X, X). We say that an operator T ∈ L(X, Y ) attains its norm if there exists x0 ∈ SX such that T (x0 ) = T = supx∈BX T (x). We use MT for the set of points where T attains its norm. For any x, y ∈ X, x is said to be orthogonal to y in the sense of Birkhoff–James (x⊥B y) if x ≤ x + λy for every λ ∈ R [3]. On the operator space L(X, Y ), it is clear that for T and A in L(X, Y ) if there exists x ∈ SX such that T (x) = T and T (x)⊥B A(x), then T ⊥B A. When H is finite dimensional Hilbert space, the converse for L(H) was proved by Bhatia and Šemrl [2]. Indeed, for a finite dimensional Hilbert space H and T, A ∈ H, T ⊥B A ⇔ ∃x ∈ MT such that T x⊥B Ax. Later, this is generalized by Benítez, Fernández and Soriano to inner product space [1]. Motivated by above results, Bhatia–Šemrl property is defined as follows. Definition 1.1. [11] For Banach spaces X and Y , an operator T ∈ L(X, Y ) said to have Bhatia–Šemrl property (BŠ property) if for each A with T ⊥B A there exists x ∈ MT such that T x⊥B Ax. After it is known that there exists some finite dimensional normed space X and an operator T ∈ L(X) without BŠ property [7], many authors tried find sufficient condition for T to have BŠ property (see [8–11]). Especially, Sain, Paul and Hait [11] proved that there are sufficiently many operators with BŠ property in L(X) when X is finite dimensional strictly convex real Banach space. More precisely, they showed that the set of operators having BŠ property is dense in L(X) for such X. Moreover, they remarked that strict convexity is not necessary by giving example which satisfies such situation. Our aim in this paper is to get the generalization of the result of Sain, Paul and Hait. Especially, we deal with not only finite dimensional space but also some infinite dimensional Banach spaces, and just in case there exists an operator which does not attains its norm. Hence, we will consider only norm attaining operators for BŠ property in the remaining part of this paper. In section 2, we show that for a Banach space X with Radon–Nikodým property, the set of norm attaining operators having BŠ property is dense in L(X, Y ). It is worth to note that the set of norm attaining operators is dense in L(X, Y ) when X has Radon– Nikodým property due to a result of Bourgain [4]. There are many classes of Banach
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S.K. Kim / Linear Algebra and its Applications 537 (2018) 22–37
spaces having Radon–Nikodým property, for example, reflexive Banach spaces and 1 space. We also deal with c0 space which is a well known example without Radon–Nikodým property. In section 3, we study about denseness of norm attaining operator without BŠ property. The first object is L1 ([0, 1]) space. We show that the set of norm attaining operators without BŠ property is dense in L(L1 ([0, 1]), X) when X has Radon–Nikodým property. This is somewhat strange result since we see that the set of norm attaining operators with BŠ property is dense in L(1 , X) for arbitrary X because 1 has Radon–Nikodým property, and the set of norm attaining operators without BŠ property is not dense in L(1 , X) for arbitrary X. We show also that the set of norm attaining operators without BŠ property is dense in L(c0 ). In section 4, we study stability of denseness of norm attaining operators having BŠ property or without BŠ property under some direct sums. More precisely, let {Xi} and {Yi } be sequences of Banach spaces and X = i∈I Xi 1 and Y = j∈J Yj ∞ or Y = j∈J Yj c0 . The set of norm attaining operators with BŠ property (without BŠ property) is dense in L(Xi , Yj ) for each i, j ∈ N if and only if the set of norm attaining operators with BŠ property (without BŠ property) is dense in L(X, Y ). 2. The set of norm attaining operators with BŠ property Before we start this section, we introduce some definitions. Let A be a subset of a Banach space X. A real valued function f defined on A strongly exposes A if there is an x0 in A such that f attains its supremum on A at x0 and if {xi } is any sequence in A such that f (xi ) → f (x0 ) then xi − x0 → 0. A set B ⊂ X is said to be a set having Radon–Nikodým property (RNP set) if it is a convex, closed and bounded subset such that for any probability space (Ω, Σ, μ) every absolutely μ continuous vector measure F : Σ −→ X of bounded variation having average range {μ(E)−1 F(E) : E ∈ Σ, μ(E) > 0} ⊂ B is representable by a Bochner integrable function. A Banach space X is said to have the Radon–Nikodým property (RNP) if for any finite measure space (Ω, Σ, μ) every μ continuous vector measure F : Σ −→ X of bounded variation is representable by a Bochner integrable function. By the definition, we see that BX is an RNP set when ever X has RNP. For more information, we give a reference [5]. There has been many efforts to study RNP set. Among them, we use the following result of Stegall about strongly exposing functions. Theorem 2.1. [12, p. 174 Theorem] Let D be an RNP set and f : D → R that is upper semi continuous (u.s.c.) and bounded above. Then, for ε > 0, there exists x∗ ∈ X ∗ , x∗ < ε such that f + x∗ and f + |x∗ | strongly expose D.
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Remark 2.2. It is worth to remark that the statement ‘f + |x∗ | strongly expose D’ is a little bit confusing since when D and f is symmetric then the number of points that f + |x∗ | attains its supremum D can be more than 1. According to proof of Theorem 2.1, the statement can be written as follows. Let D be an RNP set and f : D → R that is u.s.c. and bounded above. Then, for ε > 0, there exists x∗ ∈ X ∗ , x∗ < ε such that f + x∗ strongly expose D and supx∈D f (x) + x∗ (x) = supx∈D f (x) + |x∗ (x)|. Recently, Paul, Sain, and Ghosh obtained the following useful characterization of orthogonal operators. With these two results we deduce our main results of this section. Theorem 2.3. [9, Theorem 2.2] Let X be a Banach space, T ∈ L(X, Y ), MT = D ∪ (−D) (D is a non-empty compact connected subset of SX ). When sup{T x : x ∈ C} < T for all closed subset C of SX with d(MT , C) > 0, for any A ∈ L(X, Y ), T ⊥B A if and only if there exists z ∈ MT such that T z⊥B Az. Theorem 2.4. For a Banach space X, if the closed unit ball BX is an RNP set, then the set of norm attaining operators satisfying the BŠ property is dense in L(X, Y ) for every Banach space Y . Proof. Fix arbitrary T ∈ L(X, Y ) which is not 0 and arbitrary 0 < ε < T . Our aim is to choose S ∈ L(X, Y ) with BŠ property such that T − S < ε. For f : BX −→ R defined by f (x) = T (x), there exists x∗ ∈ X ∗ such that x∗ < ε, supx∈BX f (x) + x∗ (x) = supx∈BX f (x) + |x∗ (x)| and that f + x∗ strongly expose BX by Theorem 2.1. Let f + |x∗ | strongly exposes BX at x0 ∈ BX . Note that T = supx∈BX f (x) ≤ supx∈BX f (x) + |x∗ (x)| = f (x0 ) + |x∗ (x0 )|, and so T x0 = f (x0 ) > 0. Define S ∈ L(X, Y ) by Sx = T x + x∗ (x)
T x0 . T x0
Since Sx ≤ T x + |x∗ (x)| = f (x) + |x∗ (x)|, we have that S ≤ f (x0 ) + |x∗ (x0 )| = T (±x0 ) + |x∗ (±x0 )| T (±x0 ) ∗ T (±x ) + |x (±x )| = 0 0 T x0 = S(±x0 ) ≤ S. For δ > 0, consider Dδ = SX \(B(x0 , δ)∪B(−x0 , δ)). We see that supx∈Dδ Sx < S. Indeed, if supx∈Dδ Sx = S, there exists xi ∈ Dδ such that S(xi ) converges to S. By taking zi = xi if x∗ (xi ) = |x∗ (xi )| and zi = −xi if −x∗ (xi ) = |x∗ (xi )|, we construct zi ∈ Dδ so that x∗ (zi ) = |x∗ (zi )| for every i ∈ N and S(zi ) converges to S. Since
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S(zi ) ≤ T zi + |x∗ (zi )| = f (zi ) + x∗ (zi ) ≤ S, zi converges to x0 . This does not happen since xi ∈ Dδ . Therefore, we have MS = {±x0 } which is a compact set, we use Theorem 2.3 to see that S⊥B A if and only if Sx0 ⊥B Ax0 . Hence S satisfies BŠ property. 2 There are many known examples of Banach spaces having RNP (see for example, p. 218 of [5]). The most typical ones are reflexive space and 1 space. Hence, we have the following corollary. Corollary 2.5. If a Banach space X is one of the followings, then the set of norm attaining operators satisfying the BŠ property is dense in L(X). (1) Finite dimensional space. (2) Reflexive space (for example Lp (μ) for 1 < p < ∞). (3) p for 1 ≤ p < ∞. Now, we find a pair of Banach spaces such that the set of norm attaining operators having the BŠ property is not dense even though the set of norm attaining operators is dense. In order to do this, we consider c0 space which is a typical example that the unit ball is not an RNP set. From the following Theorem 2.6, when Y is uniformly convex, the set of norm attaining operators having the BŠ property is not dense in L(c0 , Y ), but it is known that the set of norm attaining ones is dense (for example, [6]). Theorem 2.6. If Y is strictly convex, then the set of norm attaining operators having the BŠ property is not dense in L(c0 , Y ). Proof. Let (ei ) be the canonical basis of c0 . It is enough to show that arbitrary norm attaining operator S ∈ SL(c0 ,Y ) does not have BŠ property. Let z ∈ Sc0 be a norm attaining point of S, i.e. S(z) = S. We see that there exists a finite subset A ⊂ N such that SPA = S where PA : c0 → c0 be a canonical projection to ∞ (A). Since z = (zi )∞ i=1 ∈ Sc0 , a set A = {i ∈ N : |zi | = 1} is finite. Because S(z + (1 − |zj |)ej ) + S(z − (1 − |zj |)ej ) = S(2z) = 2 for any j ∈ Ac , we have S(ej ) = 0 from the strict convexity of Y . To show that S does not have the BŠ property, we define an operator U such that 1 S⊥B U by U (ei ) = 0 for i < min A, U (ei ) = −S(ei ) for i ∈ A and U (ei ) = 2i−max A Sz for i > max A. Then for arbitrary λ ∈ R and m > maxA, we have (S + λU ) PA (z) +
m i=max A+1
m−max A 1 ei = (1 − λ)S(z) + λ S(z) 2i i=1
which converges to S(z) = 1 as m goes to infinity. This shows that S + λU ≥ S and so S⊥B U .
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On the other hand, for any norm attaining point x ∈ Bc0 of S and 0 < λ < 1, we deduce S(x) + λU (x) = (1 − λ)SPA (x) + λU (I − PA )(x) < 1 = S(x) which implies S(x) is not orthogonal to U x in the sense of Birkhoff–James. 2 3. The set of norm attaining operators without BŠ property In section 1, we obtained a result for denseness of the set of norm attaining operators with BŠ property when the domain space has RNP. In this section, we deal with spaces without RNP. The most typical spaces are L1 ([0, 1]) with Lebesgue measure and a sequence space c0 . On L1 ([0, 1]) space, we will show that the set of norm attaining linear operators without the BŠ property is dense in L(L1 ([0, 1]), Y ) whenever Y has RNP. This is interesting since we already proved that the set of norm attaining operators satisfying the BŠ property is dense in L(1 , Y ) for every Y . Moreover, we will show that the set of norm attaining operators without the BŠ property is not dense in L(1 , Y ) for arbitrary Y . These observation show that the quantity of operators having BŠ property (or without BŠ property) is related to the measure of domain space. Theorem 3.1. The set of norm attaining operators without the BŠ property is not dense in L(1 , Y ) for arbitrary Y . Proof. For arbitrary y ∈ SY , we define an operator T ∈ L(1 , Y ) by T (x) = yx1 for x = (xi ) ∈ 1 . Let S ∈ L(1 , Y ) satisfy S − T < 1/3. Then, S attains its norm only at e1 where (ei ) is the canonical base of 1 . Indeed, we see that S(e1 ) ≥ T (e1 ) − T (e1 ) − S(e1 ) > 2/3 and S(ei ) ≤ T (ei ) + T (ei ) − S(ei ) < 1/3 < S(e1 ) − 1/3 for i = 1. Since S(x) = S(ei )xi for x = (xi ) ∈ 1 , S(x) = S if and only if x = e1 . Let x = (xi ) ∈ Dδ = SX \ (B(e1 , δ) ∪ B(−e1 , δ)) (we may assume that δ < 2). We see that δ < e1 − x = |1 − x1 | + i=k |xi | = |1 − xk | + (1 − |x1 |). Similarly, we have δ < − e1 − x = |1 + x1 | + (1 − |x1 |). Hence we have δ/2 < 1 − |x1 | ≤ 1 and so S(x) ≤ S(e1 )|x1 | +
S(ei )|xi |
i=1
< S|x1 | + (S − 1/3)
|xi |
i=1
= S|x1 | + (S − 1/3)(1 − |x1 |) = S − (1 − |x1 |)/3 < S − δ/6. Therefore, we see that supx∈Dδ Sx < S and so we deduce S(e1 )⊥B A(e1 ) for every A ∈ L(1 , Y ) such that S⊥B A from Theorem 2.3. 2
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Theorem 3.2. Let X be a Banach space with RNP. Then the set of norm attaining operators without the BŠ property is dense in L(L1 ([0, 1]), X). Proof. In this proof, we denote the Lebesgue measure on [0, 1] by μ and the set of Lebesgue measurable set by Σ. Fix a non-zero operator T ∈ L(L1 ([0, 1]), X) and ε > 0. Our aim is to find a norm attaining operator S ∈ L(L1 ([0, 1]), X) without BŠ property such that T − S < ε. Using [5, Theorem 5, p. 63], we represent T by a μ-measurable function h : [0, 1] → Y
such that T (f ) = [0,1] hf dμ for all f ∈ L1 ([0, 1]) and h∞ = T . Since it is possible to approximate h with countably valued μ-measurable functions [5, Corollary 3, p. 42], there exists a set {yi : i = 1, ...∞} ⊂ Y and a mutually disjoint partition {Ai : i = 1, ...∞} ⊂ Σ of [0, 1] such that h − h1 ∞ < min{ε/2, h∞ } where ∞ h1 = i=1 yi χAi . Since it is obvious that h1 ∞ = sup{yi : μ(Ai ) = 0} > 0, we take k such that yk > h1 ∞ − min{ε/2, h1 ∞ }. We define a μ-measurable function h2 : [0, 1] → Y by h2 = 1 −
ε 2h1 ∞
i=k
yi χAi +
h1 ∞ yk χAk . yk
Then ε h 1 ∞ yk χAk h1 − h2 = yi χAi + 1 − yk 2h1 ∞ i=k < max{ε/2, min{ε/2, h1 ∞ }} ≤ ε/2.
Let S be the operator represented by h2 , i.e. S(f ) = [0,1] h2 f dμ for all f ∈ L1 ([0, 1]). We see that T − S = h − h2 ≤ h − h1 + h1 − h2 < ε. Now we show that S is a norm attaining operator without BŠ property from the following 2 steps. First step. We see that S attains its norm only at MS = D ∪ (−D) where D = {f ∈ SL1 ([0,1]) : f = 0 on Ack a.e. , f ≥ 0 on Ak a.e. }. h1 ∞ 1 ∞ Moreover, we deduce that S(f ) = h yk yk for every f ∈ D and S(f ) = − yk yk for every f ∈ (−D). These are direct consequences of the following computations. For arbitrary f ∈ L1 ([0, 1]),
S(f ) = h2 f dμ [0,1]
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ε h1 ∞ yk f χAk dμ = 1− yi f χAi + 2h1 ∞ yk i=k [0,1]
h1 ∞ ε yi f χAi dμ + yk f χAk dμ = 1− 2h1 ∞ yk i=k [0,1] [0,1] ε h1 ∞ ≤ 1− yi f χAi dμ + f χAk dμ yk 2h1 ∞ yk i=k [0,1] [0,1] ε ≤ h1 ∞ − f χAi dμ + h1 ∞ f χAk dμ . 2 i=k [0,1] [0,1] For arbitrary f ∈ D,
S(f ) =
h2 f dμ [0,1]
1−
= [0,1]
=
h1 ∞ yk yk
ε 2h1 ∞
yi f χAi +
i=k
f χAk dμ =
h1 ∞ yk f χAk dμ yk
h1 ∞ yk yk
[0,1]
Second step. We construct an operator G ∈ L(L1 ([0, 1]), X) such that S⊥B G but Sf ⊥B Gf fails for every f ∈ MS . Since μ(Ak ) > 0, there exists a countably infinite partition {Bi : i = 1, ..., ∞} of Ak such that each Bi having positive measure.
Consider G ∈ L(L1 ([0, 1]), X) defined by G(f ) = [0,1] gf dμ for every f ∈ L([0, 1]) where g : [0, 1] → Y is a μ-measurable function defined by ∞
g=
h1 ∞ 1 yk χB . yk i i i=1
For arbitrary λ, we see that (S + λG)(χBi /μ(Bi )) =
h1 ∞ 1 h1 ∞ yk + λ yk . yk i yk
Since i is arbitrarily, S + λG ≥ h1 ∞ = S which means S⊥B G. For f ∈ MS = D ∪ (−D), we see that
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∞ h h1 ∞ 1 1 ∞ yk yk (S + λG)(f ) = f χAk dμ + λ f χBi dμ yk yk i i=1 [0,1] [0,1] = h1 ∞ |(1 + λ)|
∞ 1 i=1
i
|f χBi |dμ
[0,1]
< h1 ∞ |(1 + λ)| = S|(1 + λ)| = Sf |(1 + λ)|. Hence for each −1 < λ < 0, we see that (S + λG)(f ) < Sf and this shows that Sf is not orthogonal to Gf in the sense of Birkhoff–James. 2 Remark 3.3. It is worth to note that the operator S which fails to have BŠ property in the proof of Theorem 3.2 satisfies that sup{Sx : x ∈ C} < S for all closed subset C of SX with d(MS , C) > 0. Indeed, for 0 < δ < 1, assume g ∈ SL1 ([0,1]) satisfy that
d(g, MS ) > δ. Then Ac |g|dμ + Ak |g − f |dμ = [0,1] |g − f |dμ > δ for every f ∈ MS . If k
|g|dμ > δ/2, then we have Ac k
ε S(g) ≤ h1 ∞ − g χAi dμ + h1 ∞ g χAk dμ 2 i=k [0,1] [0,1] ⎛ ⎞
ε ⎜ ⎟ ≤ h1 ∞ − |g|dμ + h1 ∞ ⎝1 − |g|dμ⎠ 2
< h1 ∞ −
ε 2
Ack
Ack
|g|dμ < S −
δε . 4
Ack
|g|dμ ≤ δ/2, then 1 ≥
|g|dμ ≥ 1 − δ/2, and so take f0 ∈ MS so that f0 =
Since f0 ≥ |g| on Ak and Ak |g − f0 |dμ ≥ δ/2, we see that Ak |f0 − Ak
g|dμ = Ak (f0 − g)dμ and so Ak gdμ = Ak f0 dμ − Ak (f0 − g)dμ ≤ 1 − δ/2. Similarly,
gdμ = Ak −f0 dμ − Ak (g − f0 )dμ ≥ −1 + δ/2. Ak Therefore, we have If
Ack |g| χ . |g|dμ Ak
Ak
ε S(g) ≤ h1 ∞ − g χAi dμ + h1 ∞ g χAk dμ 2 i=k [0,1] [0,1] ⎞ ⎛ ε⎝ ≤ h1 ∞ − 1 − gdμ⎠ + h1 ∞ gdμ 2
Ak
Ak
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⎞ ⎛ δε ε⎝ < h1 ∞ − 1 − gdμ⎠ < S − . 2 4 Ak
This is an example which shows that the condition ‘compactness of D’ is needed for T to have BŠ property in Theorem 2.3. We now prove the similar result for c0 space as in the case of L1 ([0, 1]). In the proof of Theorem 2.6, for a strict convex space Y , we see that every norm attaining operators in L(c0 , Y ) does not have the BŠ property. Hence, if the set of norm attaining operators is dense in L(c0 , Y ), then the set of norm attaining operators without the BŠ property is dense and a space having uniform convexity is such Y . In the next theorem we study L(c0 ). Theorem 3.4. The set of norm attaining operators without the BŠ property is dense in L(c0 ). Proof. Before we start the proof, we note a fact that T ∈ L(X ∗ ) is w∗ − w∗ continuous if and only if there exists S ∈ L(X) such that S ∗ = T . In this case, it is also known that T = S. For given T0 ∈ L(c0 ) which is not 0 and ε > 0, we choose k ∈ N such that T0∗ (ek ) > T0 − min{ε/3, T0 }. Take z = (zi ) ∈ 1 having some finite support Λ ⊂ N so that z − T0∗ (ek ) < ε/3 and z = T0 . Define S1 ∈ L(1 ) by S1 (x) = zxk + (1 − ε/3)
T0∗ (ei )xi
i=k
where x = (xi ) ∈ 1 and (ei ) is the canonical base of 1 . By the construction, S1 is w∗ − w∗ continuous, and so there exists T1 ∈ L(c0 ) so that ∗ T1 = S1 . Moreover, it is easy to see that T1 = S1 = T0 = z and S1 − T0∗ = T1 − T0 < ε/3. We now claim that T1 does not have the BŠ property. To show this, we first get MT1 . Since for every x ∈ Bc0 T1 (x) = sup |ei (T1 (x))| = sup |(S1 (ei ))(x)| ≤ max{(1 − ε/3)T ∗ , |(S1 (ek ))(x)|}, i
i
T1 attains its norm at x0 ∈ Sc0 if and only if |S1 (ek )(x0 )| = S1 (ek ) = z. Hence MT1 = D ∪ (−D) where D = {x = (xi ) ∈ Sc0 | xi = S1 (ek )(ei )/|S1 (ek )(ei )| for every i ∈ Λ} and (ei ) is the canonical base of c0 . We see that S1 (ek )(ei )/|S1 (ek )(ei )| is the sign of i-th coordinate of z (for the convenience we write sign zi ).
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In order to construct an operator A ∈ L(c0 ) such that T1 ⊥B A, we define an element f = (fi ) ∈ 1 by fi = −(sign zi )/ max Λ for each i ∈ Λ, fi = 1/2i−(max Λ) for each i > max Λ and fi = 0 otherwise. This element satisfies that i∈Λ |fi | = i∈N\Λ |fi | = 1. Define A by A(x) = f (x)ek for x ∈ c0 . And for each m ∈ N, we fix y max Λ+m ∈ c0 satisfying ei (y max Λ+m ) = sign zi for i ∈ Λ, ei (y max Λ+m ) = 1 for max Λ < i ≤ max Λ + m and ei (y max Λ+m ) = 0 for others. Since A∗ (ei ) = f if i = k and A∗ (ei ) = 0 otherwise, for λ ∈ R we see that T1 + λA = S1 + λA∗ ≥ S1 (ek ) + λA∗ (ek ) ≥ |(S1 (ek ) + λA∗ (ek ))(y max Λ+m )| max Λ+m i−(max Λ) = (|zi |) + λ −1 + 1/2 . i∈Λ
i=max Λ+1
Since the last part converges to z = T1 as m goes to infinity, we see that T1 ⊥B A. Finally, for arbitrary y = (yi ) ∈ MT1 we will see T1 (y) is not orthogonal to A(y) in the sense of Birkhoff–James. In order to show this, we need to see T1 (y) > (T1 +λA)(y) = supi |ei (T1 + λA)(y)| for some λ ∈ R. If i = k, |ei (T1 + λA)(y)| = |S1 (ei )(y) + λA∗ (ei )(y)| ≤ (1 − ε/3)T0∗ = (1 − ε/3)T1 (y). If i = k and 0 < λ < z, |ei (T1 + λA)(y)| = |S1 (ei )(y) + λA∗ (ei )(y)| = |z(y) + λf (y)| < z = T1 (y). The reason why the last inequality holds is that z(y) = z and −2 < f (y) < 0 for y ∈ D, and z(y) = −z and 0 < f (y) < 2 for y ∈ (−D). 2 4. Direct sum and BŠ property In this section, we study stability of denseness of norm attaining operators having BŠ property (or without BŠ property) under some direct sums. Theorem 4.1. Let {Xi } and {Yi } be sequences of Banach spaces and X = i∈I Xi 1 and Y = j∈J Yj ∞ or Y = j∈J Yj c0 . The set of norm attaining operators with BŠ property is dense in L(Xi , Yj ) for each i, j ∈ N if and only if the set of norm attaining operators with BŠ property is dense in L(X, Y ).
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Proof. Let Ei1 and Ej2 be natural embeddings from Xi and Yj to X and Y , respectively, and let Pi1 and Pj2 denote the natural coordinate maps from X and Y to Xi and Yj , respectively. Before we prove the statement, we observe that for arbitrary T ∈ L(X, Y ) T = supi,j Pj T Ei as follows. T = sup T (x) ≤ sup sup Pj2 T (x) ≤ sup Pj2 T ≤ T . x∈BX
x∈BX
j
j
2 Pj2 T Ek1 Pi1 (x) Pj T = sup Pj2 T (x) = sup x∈BX
≤ sup
x∈BX
x∈BX
i
sup{Pj2 T Ei1 Pi1 (x)} i
≤ sup Pj2 T Ei1 Pi1 ≤ Pj2 T . i
Proof of ‘if’ part. Fix ε > 0 and a non-zero operator Ti,j : Xi −→ Yj . Define T : X −→ Y by T (x) = Ej2 Ti,j Pi1 (x) for each x ∈ X. Then, we have that T = Ti,j . From the assumption we choose a norm attaining operator with BŠ property S : X −→ Y such that S − T < min{ε, Ti,j /2} and S = T . Let Si,j = Pj2 SEi1 . Since we see that Ti,j −Si,j = Pj2 T Ei1 −Pj2 SEi1 ≤ T −S < ε, we only need to show that Si,j is a norm attaining operator with BŠ property. Assume that S(z) = S at an element z ∈ SX . Since for each k = j Pk2 T = 0, we T deduce Pk2 S = Pk2 T − Pk2 S < 2i,j , and so we see that S(z) = Pj2 S(z) from T
S > 2i,j . We also see that Ei1 Pi1 (z) = z which implies S = Si,j = Si,j Pi1 (z). If not, we have contradiction as follows. 1 1 1 1 1 1 Pj S(z) = Pj SEk Pk (z) ≤ Pj SEi Pi (z) + Pj SEk Pk (z) k
k=i
1 1 1 1 1 ≤ S Pk (z) + Pj T Ek Pk (z) + Pj (S − T )Ek Pk (z) k=i k=i Ti,j 1 < Ti,j = S. P (z) < Ti,j Pk1 (z) + k 2 k=i
Take arbitrary Ui,j ∈ L(Xi , Yj ) so that Si,j ⊥B Ui,j . For U = Ej2 Ui,j Pi1 ∈ L(X, Y ), we see that S⊥B U . Indeed, S + λU ≥ Pj2 SEi1 + λPj2 U Ei1 = Si,j + λUi,j ≥ Si,j = S for every λ ∈ R.
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Since S has BŠ property, there exists z0 ∈ SX such that S(z0 ) = S, and S(z0 )⊥B U (z0 ). T
For λ ∈ R, if k = j, then Pk2 S(z0 ) + λPk2 U (z0 ) = Pk2 S(z0 ) < i,j = S 2 2 . This 2 2 shows that S(z0 ) ≤ S(z0 ) + λU (z0 ) = Pj S(z0 ) + λPj U (z0 ). Hence, from the fact that Ei1 Pi1 (z0 ) = z0 , Si,j Pi1 (z0 ) = Pj2 SEi1 Pi1 (z0 ) = S(z0 ) ≤ Pj2 S(z0 ) + λPj2 U (z0 ) = Pj2 SEi1 Pi1 (z0 ) + λPj2 U Ei1 Pi1 (z0 ) = Si,j Pi1 (z0 ) + λUi,j Pi1 (z0 ) . We finally conclude that Si,j Pi1 (z) = Si,j and Si,j Pi1 (z)⊥B Ui,j Pi1 (z). Proof of ‘only if’ part. Fix ε > 0 and T ∈ L(X, Y ) \ {0}. Let δ = min {T , ε}. Take i and j so that Pj2 T Ei1 > T − 4δ . Choose a norm attaining operator with BŠ property Si,j ∈ L(Xi , Yj ) such that Si,j = Pj2 T Ei1 and Si,j − Pj2 T Ei1 < ε. We define an operator S so that Pk2 S = 1 −
δ Pk2 T for k = j and 2T δ 2 1 Pj2 T Ek1 Pk1 (x). Pj S(x) = Si,j Pi (x) + 1 − 2T k=i
We see that S = Si,j . Indeed, we have
δ δ ≤ 1− Pk2 T ≤ T − for k = j, 2T 2 δ δ Pj2 SEk1 ≤ 1 − Pj2 T Ek1 Pk1 ≤ T − for k = i and 2T 2 Pk2 S
δ Pj2 SEi1 = Si,j > T − . 4 Moreover, this shows that S attains norm only at Ei1 (z) for z ∈ SXi with Si,j (z) = Si,j . On the other hand,
2 T − S = max Pj T − Pj2 S , sup Pk2 T − Pk2 S k=j
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2 2 δ 1 2 1 1 2 1 ≤ max Pj T Ei − Pj SEi , sup Pj T Ek − Pj SEk , 2 k=i δ ≤ max Pj2 T Ei1 − Si,j , 2 < ε. To show S has BŠ property, take U ∈ L(X, Y ) such that S⊥B U and we prove that Pj2 SEi1 ⊥B Pj2 U Ei1 . We assume that U = 0, otherwise it is clear. 2 1 2 1 2 < Pj SEi1 . This implies 0 = 0 so that Pj SEi + λ0 Pj U Ei If not, there exists λ 2 Pj SEi1 + |λλ00 | λPj2 U Ei1 < Pj2 SEi1 for any positive number λ < |λ0 |. λ0 δ Pick any 0 < α < min 4U , |λ0 | , and define λ1 = |λ0 | α. We see that for k = j, 2 Pk S + λ1 Pk2 U ≤ Pk2 S + λ1 Pk2 U < T − δ < Pj2 SEi1 and 4 for k = i, 2 Pj SEk1 + λ1 Pj2 U Ek1 ≤ Pj2 SEk1 + λ1 Pj2 U Ek1 < T − δ < Pj2 SEi1 . 4 2 Hence, we get S + λ1 U = supk,t Pk SEt1 + λ1 Pk2 U Et1 < Pj2 SEi1 ≤ S which is a contradiction. Hence, we see that Si,j ⊥B Pj2 U Ei1 and so Si,j (z0 )⊥B Pj2 U Ei1 (z0 ) for some z0 ∈ SXi such that Si,j (z0 ) = Si,j . Since this implies SEi1 (z0 )⊥B U Ei1 (z0 ), we finish the proof. 2 Theorem 4.2. Let {Xi } and {Yi } be sequences of Banach spaces and X = i∈I Xi 1 and Y = j∈J Yj ∞ or Y = j∈J Yj c0 . The set of norm attaining operators without BŠ property is dense in L(Xi , Yj ) for each i, j ∈ N if and only if the set of norm attaining operators without BŠ property is dense in L(X, Y ). Proof. The proof is similar to that of Theorem 4.1, but we give it for a sake of completeness. We use the same notions for embeddings and coordinate maps. Proof of ‘if’ part. For ε > 0 and a non-zero operator Ti,j : Xi −→ Yj , we define T : X −→ Y by T (x) = Ej2 Ti,j Pi1 (x) for each x ∈ X. We choose a norm attaining operator without BŠ property S : X −→ Y such that S − T < min{ε, Ti,j /2} and S = T = Ti,j . Define Si,j = Pj2 SEi1 . As we checked in the proof of ‘if’ part of Theorem 4.1, if S(z) = S for an element z ∈ SX , then S(z) = Pj2 S(z), Ei1 Pi1 (z) = z, Si,j = S and Ti,j − Si,j < ε. Now we show that Si,j does not have BŠ property. Take U ∈ L(X, Y ) so that S⊥B U but S(z) is not orthogonal to U (z) in the sense of Birkhoff–James for every z ∈ SX with S(z) = S. It is clear that U = 0 and
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2 1 more over we see Si,j ⊥B Ui,j where Ui,j = Pj U Ei. If not, there exists λ0 = 0 so that Si,j + λ0 Ui,j < Si,j . Hence, Si,j + |λλ00 | λUi,j < Si,j for any positive number λ < |λ0 |. Ti,j Pick any 0 < α < min 3U , |λ | , and define λ1 = |λλ00 | α. We see that 0 for k = j,
2 Pk S + λ1 Pk2 U ≤ Pk2 (S − T ) + λ1 Pk2 U < 5 S and 6 for k = i, 2 Pj SEk1 + λ1 Pj2 U Ek1 ≤ Pj2 (S − T )Ek1 + λ1 Pj2 U Ek1 < 5 S . 6 To finish our proof, we assume that Si,j (xi )⊥B Ui,j (xi ) for some xi ∈ SXi with Si,j (xi ) = Si,j , then we get the desired contradiction that for every λ ∈ R 1 SE (xi ) + λU E 1 (xi ) ≥ Si,j (xi ) + λUi,j (xi ) ≥ Si,j = S. Proof of ‘onlyif’ part.Fix ε > 0 and T ∈ L(X, Y ) \ {0}. Let δ = min {T , ε}. Take i and j so that Pj2 T Ei1 > T − 4δ . Choose a norm attaining operator without BŠ property Si,j ∈ L(Xi , Yj ) such that Si,j = Pj2 T Ei1 and Si,j − Pj2 T Ei1 < ε. We define an operator S using the same construction in the proof of ‘only if’ part of Theorem 4.1, then we get that T − S < ε, S = Si,j and that S attains norm only at Ei1 (z) for z ∈ SXi with Si,j (z) = Si,j . Take Ui,j ∈ L(Xi , Yj ) such that Si,j ⊥B Ui,j but Si,j (zi ) is not orthogonal to Ui,j (zi ) in the sense of Birkhoff–James for every zi ∈ SXi with Si,j (zi ) = Si,j . It is clear that Ui,j = 0. To show S does not have BŠ property, we define an operator U ∈ L(X, Y ) by U = 2 Ej Ui,j Pi1 . We observe that S⊥B U . Indeed, for arbitrary λ ∈ R S + λU ≥ Si,j + λUi,j ≥ Si,j = S . Assume that there exists an element z0 ∈ SX such that S(z0 ) = S and δ S(z0 )⊥B U (z0 ). Let λ be a number satisfying |λ| < 4U , then for k = j, 2 Pk S(z0 ) + λ1 Pk2 U (z0 ) ≤ Pk2 S(z0 ) + λ1 Pk2 U (z0 ) < S(z0 ) − δ 4 for k = i, 2 Pj SEk1 (z0 ) + λ1 Pj2 U Ek1 (z0 ) ≤ Pj2 SEk1 (z0 ) + λ1 Pj2 U Ek1 (z0 ) < S(z0 ) − δ 4
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Since S(z0 ) + λU (z0 ) ≥ S(z0 ), we see the desired contradiction that Si,j Ei1 (z0 ) + λUi,j Ei1 (z0 ) ≥ S(z0 ) = Si,j Ei1 (z0 ).
2
Finally, we summarize this section as follows. Corollary 4.3. (1) (2) (3) (4)
The The The The
set set set set
of of of of
norm norm norm norm
attaining attaining attaining attaining
operator operator operator operator
with BŠ property is dense in L(1 ). without BŠ property is not dense in L(1 ). with BŠ property is not dense in L(c0 ). without BŠ property is dense in L(c0 ).
Proof. It is obvious that for arbitrary Y the set of norm attaining operator with BŠ property is dense L(X, Y ) whenever X is one-dimensional. Hence, statement (1) and (2) follows from Theorem 4.1. The statement (3) follows from Theorem 2.6 and 4.2, and (4) is Theorem 3.4. 2 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] S.K. Kim, The Bishop–Phelps–Bollobás theorem for operators from c0 to uniformly convex spaces, Israel J. Math. 197 (2013) 425–435. [7] C.K. Li, H. Schneider, Orthogonality of matrices, Linear Algebra Appl. 347 (2002) 115–122. [8] K. Paul, D. Sain, Orthogonality of operators on (Rn , ∞ ), Novi Sad J. Math. 43 (1) (2013) 121–129. [9] K. Paul, D. Sain, P. Ghosh, Birkhoff–James orthogonality and smoothness of bounded linear operators, Linear Algebra Appl. 506 (2016) 551–563. [10] D. Sain, K. Paul, Operator norm attainment and inner product spaces, Linear Algebra Appl. 439 (8) (2013) 2448–2452. [11] D. Sain, K. Paul, S. Hait, Operator norm attainment and Birkhoff–James orthogonality, Linear Algebra Appl. 476 (2015) 85–97. [12] C. Stegall, Optimization of functions on certain subsets of Banach spaces, Math. Ann. 236 (2) (1978) 171–176.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Quantity of operators with Bhatia–Šemrl property Sun Kwang Kim 1 Department of Mathematics, Kyonggi University, Suwon 16227, Republic of Korea
a r t i c l e
i n f o
Article history: Received 28 May 2017 Accepted 19 September 2017 Available online 21 September 2017 Submitted by R. Bhatia MSC: primary 46B20 secondary 46B04, 46B22, 46B2
a b s t r a c t We study the Bhatia–Šemrl property which is related to Birkhoff–James orthogonality on operator spaces. Recently, many authors obtained some conditions for operators to have Bhatia–Šemrl property. Using them, we explore the denseness of operators with Bhatia–Šemrl property or without Bhatia– Šemrl property on some classes of Banach spaces. © 2017 Elsevier Inc. All rights reserved.
Keywords: Birkhoff–James orthogonality Bhatia–Šemrl property Banach space
1. Introduction In 1935, G. Birkhoff considered a notion of orthogonality in linear metric spaces which we call Birkhoff–James orthogonality [3]. This concept has been played a very important role for studying geometry of Banach space. There has been many interesting works about orthogonality on operator space. Among them, on finite dimensional Hilbert space, Bhatia and Šemrl [2] found necessary and sufficient condition for operators A and B to be that A is orthogonal to B in terms of specific vectors. They also asked whether the
1
E-mail address: [email protected]. This work was supported by Kyonggi University Research Grant 2016.
https://doi.org/10.1016/j.laa.2017.09.021 0024-3795/© 2017 Elsevier Inc. All rights reserved.
S.K. Kim / Linear Algebra and its Applications 537 (2018) 22–37
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same characterization holds for arbitrary finite dimensional space, and later a negative answer was given by Li and Schneider [7]. In this paper, we investigate the condition of Bhatia and Šemrl. Especially, we focus on ‘quantity of operators having the condition’. Let us now restart the introduction. This time we give necessary notions and background material. Let X and Y be Banach spaces. We will use the common notation SX , BX , X ∗ for the unit sphere, the closed unit ball and the dual space of X respectively, L(X, Y ) for the Banach space of every bounded linear operators from X to Y . For the convenience we write L(X) to denote L(X, X). We say that an operator T ∈ L(X, Y ) attains its norm if there exists x0 ∈ SX such that T (x0 ) = T = supx∈BX T (x). We use MT for the set of points where T attains its norm. For any x, y ∈ X, x is said to be orthogonal to y in the sense of Birkhoff–James (x⊥B y) if x ≤ x + λy for every λ ∈ R [3]. On the operator space L(X, Y ), it is clear that for T and A in L(X, Y ) if there exists x ∈ SX such that T (x) = T and T (x)⊥B A(x), then T ⊥B A. When H is finite dimensional Hilbert space, the converse for L(H) was proved by Bhatia and Šemrl [2]. Indeed, for a finite dimensional Hilbert space H and T, A ∈ H, T ⊥B A ⇔ ∃x ∈ MT such that T x⊥B Ax. Later, this is generalized by Benítez, Fernández and Soriano to inner product space [1]. Motivated by above results, Bhatia–Šemrl property is defined as follows. Definition 1.1. [11] For Banach spaces X and Y , an operator T ∈ L(X, Y ) said to have Bhatia–Šemrl property (BŠ property) if for each A with T ⊥B A there exists x ∈ MT such that T x⊥B Ax. After it is known that there exists some finite dimensional normed space X and an operator T ∈ L(X) without BŠ property [7], many authors tried find sufficient condition for T to have BŠ property (see [8–11]). Especially, Sain, Paul and Hait [11] proved that there are sufficiently many operators with BŠ property in L(X) when X is finite dimensional strictly convex real Banach space. More precisely, they showed that the set of operators having BŠ property is dense in L(X) for such X. Moreover, they remarked that strict convexity is not necessary by giving example which satisfies such situation. Our aim in this paper is to get the generalization of the result of Sain, Paul and Hait. Especially, we deal with not only finite dimensional space but also some infinite dimensional Banach spaces, and just in case there exists an operator which does not attains its norm. Hence, we will consider only norm attaining operators for BŠ property in the remaining part of this paper. In section 2, we show that for a Banach space X with Radon–Nikodým property, the set of norm attaining operators having BŠ property is dense in L(X, Y ). It is worth to note that the set of norm attaining operators is dense in L(X, Y ) when X has Radon– Nikodým property due to a result of Bourgain [4]. There are many classes of Banach
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spaces having Radon–Nikodým property, for example, reflexive Banach spaces and 1 space. We also deal with c0 space which is a well known example without Radon–Nikodým property. In section 3, we study about denseness of norm attaining operator without BŠ property. The first object is L1 ([0, 1]) space. We show that the set of norm attaining operators without BŠ property is dense in L(L1 ([0, 1]), X) when X has Radon–Nikodým property. This is somewhat strange result since we see that the set of norm attaining operators with BŠ property is dense in L(1 , X) for arbitrary X because 1 has Radon–Nikodým property, and the set of norm attaining operators without BŠ property is not dense in L(1 , X) for arbitrary X. We show also that the set of norm attaining operators without BŠ property is dense in L(c0 ). In section 4, we study stability of denseness of norm attaining operators having BŠ property or without BŠ property under some direct sums. More precisely, let {Xi} and {Yi } be sequences of Banach spaces and X = i∈I Xi 1 and Y = j∈J Yj ∞ or Y = j∈J Yj c0 . The set of norm attaining operators with BŠ property (without BŠ property) is dense in L(Xi , Yj ) for each i, j ∈ N if and only if the set of norm attaining operators with BŠ property (without BŠ property) is dense in L(X, Y ). 2. The set of norm attaining operators with BŠ property Before we start this section, we introduce some definitions. Let A be a subset of a Banach space X. A real valued function f defined on A strongly exposes A if there is an x0 in A such that f attains its supremum on A at x0 and if {xi } is any sequence in A such that f (xi ) → f (x0 ) then xi − x0 → 0. A set B ⊂ X is said to be a set having Radon–Nikodým property (RNP set) if it is a convex, closed and bounded subset such that for any probability space (Ω, Σ, μ) every absolutely μ continuous vector measure F : Σ −→ X of bounded variation having average range {μ(E)−1 F(E) : E ∈ Σ, μ(E) > 0} ⊂ B is representable by a Bochner integrable function. A Banach space X is said to have the Radon–Nikodým property (RNP) if for any finite measure space (Ω, Σ, μ) every μ continuous vector measure F : Σ −→ X of bounded variation is representable by a Bochner integrable function. By the definition, we see that BX is an RNP set when ever X has RNP. For more information, we give a reference [5]. There has been many efforts to study RNP set. Among them, we use the following result of Stegall about strongly exposing functions. Theorem 2.1. [12, p. 174 Theorem] Let D be an RNP set and f : D → R that is upper semi continuous (u.s.c.) and bounded above. Then, for ε > 0, there exists x∗ ∈ X ∗ , x∗ < ε such that f + x∗ and f + |x∗ | strongly expose D.
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Remark 2.2. It is worth to remark that the statement ‘f + |x∗ | strongly expose D’ is a little bit confusing since when D and f is symmetric then the number of points that f + |x∗ | attains its supremum D can be more than 1. According to proof of Theorem 2.1, the statement can be written as follows. Let D be an RNP set and f : D → R that is u.s.c. and bounded above. Then, for ε > 0, there exists x∗ ∈ X ∗ , x∗ < ε such that f + x∗ strongly expose D and supx∈D f (x) + x∗ (x) = supx∈D f (x) + |x∗ (x)|. Recently, Paul, Sain, and Ghosh obtained the following useful characterization of orthogonal operators. With these two results we deduce our main results of this section. Theorem 2.3. [9, Theorem 2.2] Let X be a Banach space, T ∈ L(X, Y ), MT = D ∪ (−D) (D is a non-empty compact connected subset of SX ). When sup{T x : x ∈ C} < T for all closed subset C of SX with d(MT , C) > 0, for any A ∈ L(X, Y ), T ⊥B A if and only if there exists z ∈ MT such that T z⊥B Az. Theorem 2.4. For a Banach space X, if the closed unit ball BX is an RNP set, then the set of norm attaining operators satisfying the BŠ property is dense in L(X, Y ) for every Banach space Y . Proof. Fix arbitrary T ∈ L(X, Y ) which is not 0 and arbitrary 0 < ε < T . Our aim is to choose S ∈ L(X, Y ) with BŠ property such that T − S < ε. For f : BX −→ R defined by f (x) = T (x), there exists x∗ ∈ X ∗ such that x∗ < ε, supx∈BX f (x) + x∗ (x) = supx∈BX f (x) + |x∗ (x)| and that f + x∗ strongly expose BX by Theorem 2.1. Let f + |x∗ | strongly exposes BX at x0 ∈ BX . Note that T = supx∈BX f (x) ≤ supx∈BX f (x) + |x∗ (x)| = f (x0 ) + |x∗ (x0 )|, and so T x0 = f (x0 ) > 0. Define S ∈ L(X, Y ) by Sx = T x + x∗ (x)
T x0 . T x0
Since Sx ≤ T x + |x∗ (x)| = f (x) + |x∗ (x)|, we have that S ≤ f (x0 ) + |x∗ (x0 )| = T (±x0 ) + |x∗ (±x0 )| T (±x0 ) ∗ T (±x ) + |x (±x )| = 0 0 T x0 = S(±x0 ) ≤ S. For δ > 0, consider Dδ = SX \(B(x0 , δ)∪B(−x0 , δ)). We see that supx∈Dδ Sx < S. Indeed, if supx∈Dδ Sx = S, there exists xi ∈ Dδ such that S(xi ) converges to S. By taking zi = xi if x∗ (xi ) = |x∗ (xi )| and zi = −xi if −x∗ (xi ) = |x∗ (xi )|, we construct zi ∈ Dδ so that x∗ (zi ) = |x∗ (zi )| for every i ∈ N and S(zi ) converges to S. Since
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S(zi ) ≤ T zi + |x∗ (zi )| = f (zi ) + x∗ (zi ) ≤ S, zi converges to x0 . This does not happen since xi ∈ Dδ . Therefore, we have MS = {±x0 } which is a compact set, we use Theorem 2.3 to see that S⊥B A if and only if Sx0 ⊥B Ax0 . Hence S satisfies BŠ property. 2 There are many known examples of Banach spaces having RNP (see for example, p. 218 of [5]). The most typical ones are reflexive space and 1 space. Hence, we have the following corollary. Corollary 2.5. If a Banach space X is one of the followings, then the set of norm attaining operators satisfying the BŠ property is dense in L(X). (1) Finite dimensional space. (2) Reflexive space (for example Lp (μ) for 1 < p < ∞). (3) p for 1 ≤ p < ∞. Now, we find a pair of Banach spaces such that the set of norm attaining operators having the BŠ property is not dense even though the set of norm attaining operators is dense. In order to do this, we consider c0 space which is a typical example that the unit ball is not an RNP set. From the following Theorem 2.6, when Y is uniformly convex, the set of norm attaining operators having the BŠ property is not dense in L(c0 , Y ), but it is known that the set of norm attaining ones is dense (for example, [6]). Theorem 2.6. If Y is strictly convex, then the set of norm attaining operators having the BŠ property is not dense in L(c0 , Y ). Proof. Let (ei ) be the canonical basis of c0 . It is enough to show that arbitrary norm attaining operator S ∈ SL(c0 ,Y ) does not have BŠ property. Let z ∈ Sc0 be a norm attaining point of S, i.e. S(z) = S. We see that there exists a finite subset A ⊂ N such that SPA = S where PA : c0 → c0 be a canonical projection to ∞ (A). Since z = (zi )∞ i=1 ∈ Sc0 , a set A = {i ∈ N : |zi | = 1} is finite. Because S(z + (1 − |zj |)ej ) + S(z − (1 − |zj |)ej ) = S(2z) = 2 for any j ∈ Ac , we have S(ej ) = 0 from the strict convexity of Y . To show that S does not have the BŠ property, we define an operator U such that 1 S⊥B U by U (ei ) = 0 for i < min A, U (ei ) = −S(ei ) for i ∈ A and U (ei ) = 2i−max A Sz for i > max A. Then for arbitrary λ ∈ R and m > maxA, we have (S + λU ) PA (z) +
m i=max A+1
m−max A 1 ei = (1 − λ)S(z) + λ S(z) 2i i=1
which converges to S(z) = 1 as m goes to infinity. This shows that S + λU ≥ S and so S⊥B U .
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On the other hand, for any norm attaining point x ∈ Bc0 of S and 0 < λ < 1, we deduce S(x) + λU (x) = (1 − λ)SPA (x) + λU (I − PA )(x) < 1 = S(x) which implies S(x) is not orthogonal to U x in the sense of Birkhoff–James. 2 3. The set of norm attaining operators without BŠ property In section 1, we obtained a result for denseness of the set of norm attaining operators with BŠ property when the domain space has RNP. In this section, we deal with spaces without RNP. The most typical spaces are L1 ([0, 1]) with Lebesgue measure and a sequence space c0 . On L1 ([0, 1]) space, we will show that the set of norm attaining linear operators without the BŠ property is dense in L(L1 ([0, 1]), Y ) whenever Y has RNP. This is interesting since we already proved that the set of norm attaining operators satisfying the BŠ property is dense in L(1 , Y ) for every Y . Moreover, we will show that the set of norm attaining operators without the BŠ property is not dense in L(1 , Y ) for arbitrary Y . These observation show that the quantity of operators having BŠ property (or without BŠ property) is related to the measure of domain space. Theorem 3.1. The set of norm attaining operators without the BŠ property is not dense in L(1 , Y ) for arbitrary Y . Proof. For arbitrary y ∈ SY , we define an operator T ∈ L(1 , Y ) by T (x) = yx1 for x = (xi ) ∈ 1 . Let S ∈ L(1 , Y ) satisfy S − T < 1/3. Then, S attains its norm only at e1 where (ei ) is the canonical base of 1 . Indeed, we see that S(e1 ) ≥ T (e1 ) − T (e1 ) − S(e1 ) > 2/3 and S(ei ) ≤ T (ei ) + T (ei ) − S(ei ) < 1/3 < S(e1 ) − 1/3 for i = 1. Since S(x) = S(ei )xi for x = (xi ) ∈ 1 , S(x) = S if and only if x = e1 . Let x = (xi ) ∈ Dδ = SX \ (B(e1 , δ) ∪ B(−e1 , δ)) (we may assume that δ < 2). We see that δ < e1 − x = |1 − x1 | + i=k |xi | = |1 − xk | + (1 − |x1 |). Similarly, we have δ < − e1 − x = |1 + x1 | + (1 − |x1 |). Hence we have δ/2 < 1 − |x1 | ≤ 1 and so S(x) ≤ S(e1 )|x1 | +
S(ei )|xi |
i=1
< S|x1 | + (S − 1/3)
|xi |
i=1
= S|x1 | + (S − 1/3)(1 − |x1 |) = S − (1 − |x1 |)/3 < S − δ/6. Therefore, we see that supx∈Dδ Sx < S and so we deduce S(e1 )⊥B A(e1 ) for every A ∈ L(1 , Y ) such that S⊥B A from Theorem 2.3. 2
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Theorem 3.2. Let X be a Banach space with RNP. Then the set of norm attaining operators without the BŠ property is dense in L(L1 ([0, 1]), X). Proof. In this proof, we denote the Lebesgue measure on [0, 1] by μ and the set of Lebesgue measurable set by Σ. Fix a non-zero operator T ∈ L(L1 ([0, 1]), X) and ε > 0. Our aim is to find a norm attaining operator S ∈ L(L1 ([0, 1]), X) without BŠ property such that T − S < ε. Using [5, Theorem 5, p. 63], we represent T by a μ-measurable function h : [0, 1] → Y
such that T (f ) = [0,1] hf dμ for all f ∈ L1 ([0, 1]) and h∞ = T . Since it is possible to approximate h with countably valued μ-measurable functions [5, Corollary 3, p. 42], there exists a set {yi : i = 1, ...∞} ⊂ Y and a mutually disjoint partition {Ai : i = 1, ...∞} ⊂ Σ of [0, 1] such that h − h1 ∞ < min{ε/2, h∞ } where ∞ h1 = i=1 yi χAi . Since it is obvious that h1 ∞ = sup{yi : μ(Ai ) = 0} > 0, we take k such that yk > h1 ∞ − min{ε/2, h1 ∞ }. We define a μ-measurable function h2 : [0, 1] → Y by h2 = 1 −
ε 2h1 ∞
i=k
yi χAi +
h1 ∞ yk χAk . yk
Then ε h 1 ∞ yk χAk h1 − h2 = yi χAi + 1 − yk 2h1 ∞ i=k < max{ε/2, min{ε/2, h1 ∞ }} ≤ ε/2.
Let S be the operator represented by h2 , i.e. S(f ) = [0,1] h2 f dμ for all f ∈ L1 ([0, 1]). We see that T − S = h − h2 ≤ h − h1 + h1 − h2 < ε. Now we show that S is a norm attaining operator without BŠ property from the following 2 steps. First step. We see that S attains its norm only at MS = D ∪ (−D) where D = {f ∈ SL1 ([0,1]) : f = 0 on Ack a.e. , f ≥ 0 on Ak a.e. }. h1 ∞ 1 ∞ Moreover, we deduce that S(f ) = h yk yk for every f ∈ D and S(f ) = − yk yk for every f ∈ (−D). These are direct consequences of the following computations. For arbitrary f ∈ L1 ([0, 1]),
S(f ) = h2 f dμ [0,1]
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ε h1 ∞ yk f χAk dμ = 1− yi f χAi + 2h1 ∞ yk i=k [0,1]
h1 ∞ ε yi f χAi dμ + yk f χAk dμ = 1− 2h1 ∞ yk i=k [0,1] [0,1] ε h1 ∞ ≤ 1− yi f χAi dμ + f χAk dμ yk 2h1 ∞ yk i=k [0,1] [0,1] ε ≤ h1 ∞ − f χAi dμ + h1 ∞ f χAk dμ . 2 i=k [0,1] [0,1] For arbitrary f ∈ D,
S(f ) =
h2 f dμ [0,1]
1−
= [0,1]
=
h1 ∞ yk yk
ε 2h1 ∞
yi f χAi +
i=k
f χAk dμ =
h1 ∞ yk f χAk dμ yk
h1 ∞ yk yk
[0,1]
Second step. We construct an operator G ∈ L(L1 ([0, 1]), X) such that S⊥B G but Sf ⊥B Gf fails for every f ∈ MS . Since μ(Ak ) > 0, there exists a countably infinite partition {Bi : i = 1, ..., ∞} of Ak such that each Bi having positive measure.
Consider G ∈ L(L1 ([0, 1]), X) defined by G(f ) = [0,1] gf dμ for every f ∈ L([0, 1]) where g : [0, 1] → Y is a μ-measurable function defined by ∞
g=
h1 ∞ 1 yk χB . yk i i i=1
For arbitrary λ, we see that (S + λG)(χBi /μ(Bi )) =
h1 ∞ 1 h1 ∞ yk + λ yk . yk i yk
Since i is arbitrarily, S + λG ≥ h1 ∞ = S which means S⊥B G. For f ∈ MS = D ∪ (−D), we see that
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∞ h h1 ∞ 1 1 ∞ yk yk (S + λG)(f ) = f χAk dμ + λ f χBi dμ yk yk i i=1 [0,1] [0,1] = h1 ∞ |(1 + λ)|
∞ 1 i=1
i
|f χBi |dμ
[0,1]
< h1 ∞ |(1 + λ)| = S|(1 + λ)| = Sf |(1 + λ)|. Hence for each −1 < λ < 0, we see that (S + λG)(f ) < Sf and this shows that Sf is not orthogonal to Gf in the sense of Birkhoff–James. 2 Remark 3.3. It is worth to note that the operator S which fails to have BŠ property in the proof of Theorem 3.2 satisfies that sup{Sx : x ∈ C} < S for all closed subset C of SX with d(MS , C) > 0. Indeed, for 0 < δ < 1, assume g ∈ SL1 ([0,1]) satisfy that
d(g, MS ) > δ. Then Ac |g|dμ + Ak |g − f |dμ = [0,1] |g − f |dμ > δ for every f ∈ MS . If k
|g|dμ > δ/2, then we have Ac k
ε S(g) ≤ h1 ∞ − g χAi dμ + h1 ∞ g χAk dμ 2 i=k [0,1] [0,1] ⎛ ⎞
ε ⎜ ⎟ ≤ h1 ∞ − |g|dμ + h1 ∞ ⎝1 − |g|dμ⎠ 2
< h1 ∞ −
ε 2
Ack
Ack
|g|dμ < S −
δε . 4
Ack
|g|dμ ≤ δ/2, then 1 ≥
|g|dμ ≥ 1 − δ/2, and so take f0 ∈ MS so that f0 =
Since f0 ≥ |g| on Ak and Ak |g − f0 |dμ ≥ δ/2, we see that Ak |f0 − Ak
g|dμ = Ak (f0 − g)dμ and so Ak gdμ = Ak f0 dμ − Ak (f0 − g)dμ ≤ 1 − δ/2. Similarly,
gdμ = Ak −f0 dμ − Ak (g − f0 )dμ ≥ −1 + δ/2. Ak Therefore, we have If
Ack |g| χ . |g|dμ Ak
Ak
ε S(g) ≤ h1 ∞ − g χAi dμ + h1 ∞ g χAk dμ 2 i=k [0,1] [0,1] ⎞ ⎛ ε⎝ ≤ h1 ∞ − 1 − gdμ⎠ + h1 ∞ gdμ 2
Ak
Ak
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⎞ ⎛ δε ε⎝ < h1 ∞ − 1 − gdμ⎠ < S − . 2 4 Ak
This is an example which shows that the condition ‘compactness of D’ is needed for T to have BŠ property in Theorem 2.3. We now prove the similar result for c0 space as in the case of L1 ([0, 1]). In the proof of Theorem 2.6, for a strict convex space Y , we see that every norm attaining operators in L(c0 , Y ) does not have the BŠ property. Hence, if the set of norm attaining operators is dense in L(c0 , Y ), then the set of norm attaining operators without the BŠ property is dense and a space having uniform convexity is such Y . In the next theorem we study L(c0 ). Theorem 3.4. The set of norm attaining operators without the BŠ property is dense in L(c0 ). Proof. Before we start the proof, we note a fact that T ∈ L(X ∗ ) is w∗ − w∗ continuous if and only if there exists S ∈ L(X) such that S ∗ = T . In this case, it is also known that T = S. For given T0 ∈ L(c0 ) which is not 0 and ε > 0, we choose k ∈ N such that T0∗ (ek ) > T0 − min{ε/3, T0 }. Take z = (zi ) ∈ 1 having some finite support Λ ⊂ N so that z − T0∗ (ek ) < ε/3 and z = T0 . Define S1 ∈ L(1 ) by S1 (x) = zxk + (1 − ε/3)
T0∗ (ei )xi
i=k
where x = (xi ) ∈ 1 and (ei ) is the canonical base of 1 . By the construction, S1 is w∗ − w∗ continuous, and so there exists T1 ∈ L(c0 ) so that ∗ T1 = S1 . Moreover, it is easy to see that T1 = S1 = T0 = z and S1 − T0∗ = T1 − T0 < ε/3. We now claim that T1 does not have the BŠ property. To show this, we first get MT1 . Since for every x ∈ Bc0 T1 (x) = sup |ei (T1 (x))| = sup |(S1 (ei ))(x)| ≤ max{(1 − ε/3)T ∗ , |(S1 (ek ))(x)|}, i
i
T1 attains its norm at x0 ∈ Sc0 if and only if |S1 (ek )(x0 )| = S1 (ek ) = z. Hence MT1 = D ∪ (−D) where D = {x = (xi ) ∈ Sc0 | xi = S1 (ek )(ei )/|S1 (ek )(ei )| for every i ∈ Λ} and (ei ) is the canonical base of c0 . We see that S1 (ek )(ei )/|S1 (ek )(ei )| is the sign of i-th coordinate of z (for the convenience we write sign zi ).
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In order to construct an operator A ∈ L(c0 ) such that T1 ⊥B A, we define an element f = (fi ) ∈ 1 by fi = −(sign zi )/ max Λ for each i ∈ Λ, fi = 1/2i−(max Λ) for each i > max Λ and fi = 0 otherwise. This element satisfies that i∈Λ |fi | = i∈N\Λ |fi | = 1. Define A by A(x) = f (x)ek for x ∈ c0 . And for each m ∈ N, we fix y max Λ+m ∈ c0 satisfying ei (y max Λ+m ) = sign zi for i ∈ Λ, ei (y max Λ+m ) = 1 for max Λ < i ≤ max Λ + m and ei (y max Λ+m ) = 0 for others. Since A∗ (ei ) = f if i = k and A∗ (ei ) = 0 otherwise, for λ ∈ R we see that T1 + λA = S1 + λA∗ ≥ S1 (ek ) + λA∗ (ek ) ≥ |(S1 (ek ) + λA∗ (ek ))(y max Λ+m )| max Λ+m i−(max Λ) = (|zi |) + λ −1 + 1/2 . i∈Λ
i=max Λ+1
Since the last part converges to z = T1 as m goes to infinity, we see that T1 ⊥B A. Finally, for arbitrary y = (yi ) ∈ MT1 we will see T1 (y) is not orthogonal to A(y) in the sense of Birkhoff–James. In order to show this, we need to see T1 (y) > (T1 +λA)(y) = supi |ei (T1 + λA)(y)| for some λ ∈ R. If i = k, |ei (T1 + λA)(y)| = |S1 (ei )(y) + λA∗ (ei )(y)| ≤ (1 − ε/3)T0∗ = (1 − ε/3)T1 (y). If i = k and 0 < λ < z, |ei (T1 + λA)(y)| = |S1 (ei )(y) + λA∗ (ei )(y)| = |z(y) + λf (y)| < z = T1 (y). The reason why the last inequality holds is that z(y) = z and −2 < f (y) < 0 for y ∈ D, and z(y) = −z and 0 < f (y) < 2 for y ∈ (−D). 2 4. Direct sum and BŠ property In this section, we study stability of denseness of norm attaining operators having BŠ property (or without BŠ property) under some direct sums. Theorem 4.1. Let {Xi } and {Yi } be sequences of Banach spaces and X = i∈I Xi 1 and Y = j∈J Yj ∞ or Y = j∈J Yj c0 . The set of norm attaining operators with BŠ property is dense in L(Xi , Yj ) for each i, j ∈ N if and only if the set of norm attaining operators with BŠ property is dense in L(X, Y ).
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Proof. Let Ei1 and Ej2 be natural embeddings from Xi and Yj to X and Y , respectively, and let Pi1 and Pj2 denote the natural coordinate maps from X and Y to Xi and Yj , respectively. Before we prove the statement, we observe that for arbitrary T ∈ L(X, Y ) T = supi,j Pj T Ei as follows. T = sup T (x) ≤ sup sup Pj2 T (x) ≤ sup Pj2 T ≤ T . x∈BX
x∈BX
j
j
2 Pj2 T Ek1 Pi1 (x) Pj T = sup Pj2 T (x) = sup x∈BX
≤ sup
x∈BX
x∈BX
i
sup{Pj2 T Ei1 Pi1 (x)} i
≤ sup Pj2 T Ei1 Pi1 ≤ Pj2 T . i
Proof of ‘if’ part. Fix ε > 0 and a non-zero operator Ti,j : Xi −→ Yj . Define T : X −→ Y by T (x) = Ej2 Ti,j Pi1 (x) for each x ∈ X. Then, we have that T = Ti,j . From the assumption we choose a norm attaining operator with BŠ property S : X −→ Y such that S − T < min{ε, Ti,j /2} and S = T . Let Si,j = Pj2 SEi1 . Since we see that Ti,j −Si,j = Pj2 T Ei1 −Pj2 SEi1 ≤ T −S < ε, we only need to show that Si,j is a norm attaining operator with BŠ property. Assume that S(z) = S at an element z ∈ SX . Since for each k = j Pk2 T = 0, we T deduce Pk2 S = Pk2 T − Pk2 S < 2i,j , and so we see that S(z) = Pj2 S(z) from T
S > 2i,j . We also see that Ei1 Pi1 (z) = z which implies S = Si,j = Si,j Pi1 (z). If not, we have contradiction as follows. 1 1 1 1 1 1 Pj S(z) = Pj SEk Pk (z) ≤ Pj SEi Pi (z) + Pj SEk Pk (z) k
k=i
1 1 1 1 1 ≤ S Pk (z) + Pj T Ek Pk (z) + Pj (S − T )Ek Pk (z) k=i k=i Ti,j 1 < Ti,j = S. P (z) < Ti,j Pk1 (z) + k 2 k=i
Take arbitrary Ui,j ∈ L(Xi , Yj ) so that Si,j ⊥B Ui,j . For U = Ej2 Ui,j Pi1 ∈ L(X, Y ), we see that S⊥B U . Indeed, S + λU ≥ Pj2 SEi1 + λPj2 U Ei1 = Si,j + λUi,j ≥ Si,j = S for every λ ∈ R.
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Since S has BŠ property, there exists z0 ∈ SX such that S(z0 ) = S, and S(z0 )⊥B U (z0 ). T
For λ ∈ R, if k = j, then Pk2 S(z0 ) + λPk2 U (z0 ) = Pk2 S(z0 ) < i,j = S 2 2 . This 2 2 shows that S(z0 ) ≤ S(z0 ) + λU (z0 ) = Pj S(z0 ) + λPj U (z0 ). Hence, from the fact that Ei1 Pi1 (z0 ) = z0 , Si,j Pi1 (z0 ) = Pj2 SEi1 Pi1 (z0 ) = S(z0 ) ≤ Pj2 S(z0 ) + λPj2 U (z0 ) = Pj2 SEi1 Pi1 (z0 ) + λPj2 U Ei1 Pi1 (z0 ) = Si,j Pi1 (z0 ) + λUi,j Pi1 (z0 ) . We finally conclude that Si,j Pi1 (z) = Si,j and Si,j Pi1 (z)⊥B Ui,j Pi1 (z). Proof of ‘only if’ part. Fix ε > 0 and T ∈ L(X, Y ) \ {0}. Let δ = min {T , ε}. Take i and j so that Pj2 T Ei1 > T − 4δ . Choose a norm attaining operator with BŠ property Si,j ∈ L(Xi , Yj ) such that Si,j = Pj2 T Ei1 and Si,j − Pj2 T Ei1 < ε. We define an operator S so that Pk2 S = 1 −
δ Pk2 T for k = j and 2T δ 2 1 Pj2 T Ek1 Pk1 (x). Pj S(x) = Si,j Pi (x) + 1 − 2T k=i
We see that S = Si,j . Indeed, we have
δ δ ≤ 1− Pk2 T ≤ T − for k = j, 2T 2 δ δ Pj2 SEk1 ≤ 1 − Pj2 T Ek1 Pk1 ≤ T − for k = i and 2T 2 Pk2 S
δ Pj2 SEi1 = Si,j > T − . 4 Moreover, this shows that S attains norm only at Ei1 (z) for z ∈ SXi with Si,j (z) = Si,j . On the other hand,
2 T − S = max Pj T − Pj2 S , sup Pk2 T − Pk2 S k=j
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2 2 δ 1 2 1 1 2 1 ≤ max Pj T Ei − Pj SEi , sup Pj T Ek − Pj SEk , 2 k=i δ ≤ max Pj2 T Ei1 − Si,j , 2 < ε. To show S has BŠ property, take U ∈ L(X, Y ) such that S⊥B U and we prove that Pj2 SEi1 ⊥B Pj2 U Ei1 . We assume that U = 0, otherwise it is clear. 2 1 2 1 2 < Pj SEi1 . This implies 0 = 0 so that Pj SEi + λ0 Pj U Ei If not, there exists λ 2 Pj SEi1 + |λλ00 | λPj2 U Ei1 < Pj2 SEi1 for any positive number λ < |λ0 |. λ0 δ Pick any 0 < α < min 4U , |λ0 | , and define λ1 = |λ0 | α. We see that for k = j, 2 Pk S + λ1 Pk2 U ≤ Pk2 S + λ1 Pk2 U < T − δ < Pj2 SEi1 and 4 for k = i, 2 Pj SEk1 + λ1 Pj2 U Ek1 ≤ Pj2 SEk1 + λ1 Pj2 U Ek1 < T − δ < Pj2 SEi1 . 4 2 Hence, we get S + λ1 U = supk,t Pk SEt1 + λ1 Pk2 U Et1 < Pj2 SEi1 ≤ S which is a contradiction. Hence, we see that Si,j ⊥B Pj2 U Ei1 and so Si,j (z0 )⊥B Pj2 U Ei1 (z0 ) for some z0 ∈ SXi such that Si,j (z0 ) = Si,j . Since this implies SEi1 (z0 )⊥B U Ei1 (z0 ), we finish the proof. 2 Theorem 4.2. Let {Xi } and {Yi } be sequences of Banach spaces and X = i∈I Xi 1 and Y = j∈J Yj ∞ or Y = j∈J Yj c0 . The set of norm attaining operators without BŠ property is dense in L(Xi , Yj ) for each i, j ∈ N if and only if the set of norm attaining operators without BŠ property is dense in L(X, Y ). Proof. The proof is similar to that of Theorem 4.1, but we give it for a sake of completeness. We use the same notions for embeddings and coordinate maps. Proof of ‘if’ part. For ε > 0 and a non-zero operator Ti,j : Xi −→ Yj , we define T : X −→ Y by T (x) = Ej2 Ti,j Pi1 (x) for each x ∈ X. We choose a norm attaining operator without BŠ property S : X −→ Y such that S − T < min{ε, Ti,j /2} and S = T = Ti,j . Define Si,j = Pj2 SEi1 . As we checked in the proof of ‘if’ part of Theorem 4.1, if S(z) = S for an element z ∈ SX , then S(z) = Pj2 S(z), Ei1 Pi1 (z) = z, Si,j = S and Ti,j − Si,j < ε. Now we show that Si,j does not have BŠ property. Take U ∈ L(X, Y ) so that S⊥B U but S(z) is not orthogonal to U (z) in the sense of Birkhoff–James for every z ∈ SX with S(z) = S. It is clear that U = 0 and
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2 1 more over we see Si,j ⊥B Ui,j where Ui,j = Pj U Ei. If not, there exists λ0 = 0 so that Si,j + λ0 Ui,j < Si,j . Hence, Si,j + |λλ00 | λUi,j < Si,j for any positive number λ < |λ0 |. Ti,j Pick any 0 < α < min 3U , |λ | , and define λ1 = |λλ00 | α. We see that 0 for k = j,
2 Pk S + λ1 Pk2 U ≤ Pk2 (S − T ) + λ1 Pk2 U < 5 S and 6 for k = i, 2 Pj SEk1 + λ1 Pj2 U Ek1 ≤ Pj2 (S − T )Ek1 + λ1 Pj2 U Ek1 < 5 S . 6 To finish our proof, we assume that Si,j (xi )⊥B Ui,j (xi ) for some xi ∈ SXi with Si,j (xi ) = Si,j , then we get the desired contradiction that for every λ ∈ R 1 SE (xi ) + λU E 1 (xi ) ≥ Si,j (xi ) + λUi,j (xi ) ≥ Si,j = S. Proof of ‘onlyif’ part.Fix ε > 0 and T ∈ L(X, Y ) \ {0}. Let δ = min {T , ε}. Take i and j so that Pj2 T Ei1 > T − 4δ . Choose a norm attaining operator without BŠ property Si,j ∈ L(Xi , Yj ) such that Si,j = Pj2 T Ei1 and Si,j − Pj2 T Ei1 < ε. We define an operator S using the same construction in the proof of ‘only if’ part of Theorem 4.1, then we get that T − S < ε, S = Si,j and that S attains norm only at Ei1 (z) for z ∈ SXi with Si,j (z) = Si,j . Take Ui,j ∈ L(Xi , Yj ) such that Si,j ⊥B Ui,j but Si,j (zi ) is not orthogonal to Ui,j (zi ) in the sense of Birkhoff–James for every zi ∈ SXi with Si,j (zi ) = Si,j . It is clear that Ui,j = 0. To show S does not have BŠ property, we define an operator U ∈ L(X, Y ) by U = 2 Ej Ui,j Pi1 . We observe that S⊥B U . Indeed, for arbitrary λ ∈ R S + λU ≥ Si,j + λUi,j ≥ Si,j = S . Assume that there exists an element z0 ∈ SX such that S(z0 ) = S and δ S(z0 )⊥B U (z0 ). Let λ be a number satisfying |λ| < 4U , then for k = j, 2 Pk S(z0 ) + λ1 Pk2 U (z0 ) ≤ Pk2 S(z0 ) + λ1 Pk2 U (z0 ) < S(z0 ) − δ 4 for k = i, 2 Pj SEk1 (z0 ) + λ1 Pj2 U Ek1 (z0 ) ≤ Pj2 SEk1 (z0 ) + λ1 Pj2 U Ek1 (z0 ) < S(z0 ) − δ 4
S.K. Kim / Linear Algebra and its Applications 537 (2018) 22–37
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Since S(z0 ) + λU (z0 ) ≥ S(z0 ), we see the desired contradiction that Si,j Ei1 (z0 ) + λUi,j Ei1 (z0 ) ≥ S(z0 ) = Si,j Ei1 (z0 ).
2
Finally, we summarize this section as follows. Corollary 4.3. (1) (2) (3) (4)
The The The The
set set set set
of of of of
norm norm norm norm
attaining attaining attaining attaining
operator operator operator operator
with BŠ property is dense in L(1 ). without BŠ property is not dense in L(1 ). with BŠ property is not dense in L(c0 ). without BŠ property is dense in L(c0 ).
Proof. It is obvious that for arbitrary Y the set of norm attaining operator with BŠ property is dense L(X, Y ) whenever X is one-dimensional. Hence, statement (1) and (2) follows from Theorem 4.1. The statement (3) follows from Theorem 2.6 and 4.2, and (4) is Theorem 3.4. 2 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] S.K. Kim, The Bishop–Phelps–Bollobás theorem for operators from c0 to uniformly convex spaces, Israel J. Math. 197 (2013) 425–435. [7] C.K. Li, H. Schneider, Orthogonality of matrices, Linear Algebra Appl. 347 (2002) 115–122. [8] K. Paul, D. Sain, Orthogonality of operators on (Rn , ∞ ), Novi Sad J. Math. 43 (1) (2013) 121–129. [9] K. Paul, D. Sain, P. Ghosh, Birkhoff–James orthogonality and smoothness of bounded linear operators, Linear Algebra Appl. 506 (2016) 551–563. [10] D. Sain, K. Paul, Operator norm attainment and inner product spaces, Linear Algebra Appl. 439 (8) (2013) 2448–2452. [11] D. Sain, K. Paul, S. Hait, Operator norm attainment and Birkhoff–James orthogonality, Linear Algebra Appl. 476 (2015) 85–97. [12] C. Stegall, Optimization of functions on certain subsets of Banach spaces, Math. Ann. 236 (2) (1978) 171–176.