There is no operatorwise version of the Bishop-Phelps-Bollobás property
Sheldon Dantas, Vladimir Kadets, Sun Kwang Kim, Han Ju Lee, Miguel Martín
TTHERE IS NO OPERATORWISE VERSION OF THEBISHOP–PHELPS–BOLLOB ´AS PROPERTY
SHELDON DANTAS, VLADIMIR KADETS, SUN KWANG KIM, HAN JU LEE, AND MIGUEL MART´IN
Dedicated to the memory of Bernardo Cascales
Abstract.
Given two real Banach spaces X and Y with dimensions greater than one, it is shown thatthere is a sequence { T n } n ∈ N of norm attaining norm-one operators from X to Y and a point x ∈ X with (cid:107) x (cid:107) = 1, such that (cid:107) T n ( x ) (cid:107) −→ n ∈ N (cid:8) dist (cid:0) x , { x ∈ X : (cid:107) T n ( x ) (cid:107) = (cid:107) x (cid:107) = 1 } (cid:1)(cid:9) > . This shows that a version of the Bishop–Phelps–Bollob´as property in which the operator is not changedis possible only if one of the involved Banach spaces is one-dimensional. Introduction
Let X be a Banach space. We denote by X ∗ , S X , and B X the topological dual, the unit sphere,and the closed unit ball of X , respectively. We say that x ∗ ∈ X ∗ attains its norm (or that x ∗ is a norm attaining functional ) if there exists x ∈ S X such that | x ∗ ( x ) | = (cid:107) x ∗ (cid:107) = sup x ∈ S X | x ∗ ( x ) | . It iswell-known that the set of all norm attaining functionals NA( X ) is always norm-dense in X ∗ . This is thefamous 1961 Bishop–Phelps theorem [6]. Shortly after this result was established, Bollob´as [7] sharpedit in the following way: given 0 < ε < / x ∈ B X , and x ∗ ∈ S X ∗ satisfying that | − x ∗ ( x ) | < ε / x ∈ S X and x ∗ ∈ S X ∗ such that x ∗ ( x ) = 1, (cid:107) x − x (cid:107) < ε and (cid:107) x ∗ − x ∗ (cid:107) < ε (we are givingthe statement in a little bit improved form, which can be found in [8] or [10]).If X , Y are Banach spaces, we denote by L ( X, Y ) the space of all bounded linear operators from X to Y and we say that T ∈ L ( X, Y ) attains its norm (or that T is norm attaining ) if there is x ∈ S X such that (cid:107) T ( x ) (cid:107) = (cid:107) T (cid:107) = sup x ∈ S X (cid:107) T ( x ) (cid:107) . Lindenstrauss [19] was the first one who studied thepossible validity of the Bishop–Phelps theorem for operators, i.e., the density of the set of norm attainingoperators between two Banach spaces. He showed that such density is not always true and also gavesome conditions on the involved Banach spaces X and Y to get the density of the set of norm attainingoperators. We refer to the survey paper [1] for an account of the results on this area. In 2008, M. Acosta,R. Aron, D. Garc´ıa, and M. Maestre [2] introduced the so-called Bishop–Phelps–Bollob´as property tocheck when we can get a Bollob´as’ type theorem for bounded linear operators. More precisely, a pair( X, Y ) of Banach spaces has the
Bishop–Phelps–Bollob´as property ( BPBp for short) if, given ε >
0, thereis η ( ε ) > T ∈ L ( X, Y ) with (cid:107) T (cid:107) = 1 and x ∈ S X satisfy (cid:107) T ( x ) (cid:107) > − η ( ε ), thereare S ∈ L ( X, Y ) with (cid:107) S (cid:107) = 1 and x ∈ S X such that (cid:107) S ( x ) (cid:107) = 1, (cid:107) x − x (cid:107) < ε , and (cid:107) S − T (cid:107) < ε . Date : June 25th, 2018.2010
Mathematics Subject Classification.
Primary 46B04; Secondary 46B20.
Key words and phrases.
Banach space; norm attaining operators; Bishop–Phelps–Bollob´as property.The first author was supported by Pohang Mathematics Institute (PMI), POSTECH, Korea and Basic Science ResearchProgram through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science andTechnology (NRF-2015R1D1A1A 09059788). The research of the second author is done in frames of Ukrainian Ministryof Science and Education Research Program 0118U002036, and it was partially supported by Spanish MINECO/FEDERprojects MTM2015-65020-P and MTM2017-83262-C2-2-P. Third author was partially supported by Basic Science ResearchProgram through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science andTechnology (NRF-2017R1C1B1002928). Fourth author was partially supported by Basic Science Research Program throughthe National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF-2016R1D1A1B03934771). Fifth author partially supported by Spanish MINECO/FEDER grant MTM2015-65020-P. a r X i v : . [ m a t h . F A ] O c t DANTAS, KADETS, KIM, LEE, AND MART´IN
Among other results, they showed that any pair of finite dimensional Banach spaces have the BPBp andcharacterized the pairs ( (cid:96) , Y ) to satisfy it via a geometric property on Y . After 2008, a lot of attentionwas given to this topic and there is a vast literature about the Bishop–Phelps–Bollob´as property. Werefer the reader to the very recent papers [3, 9, 11, 13] and references therein. It is important to remarkthat the Bishop–Phelps–Bollob´as property has geometric consequences on the involved Banach spaces.For instance, if X is a finite-dimensional Banach space, then all operators from X into any other Banachspace Y attain their norm but, unless the dimension of X is equal to one, it is possible to construct arenorming (cid:101) X of X and to find a Banach space Y such that the pair ( (cid:101) X, Y ) fails the BPBp [5, Theorem3.1].In the last years, some variations of the BPBp have appeared in the literature. For instance, there isa property, stronger than the BPBp, in which only the operator moves: a pair (
X, Y ) of Banach spaceshas the pointwise Bishop–Phelps–Bollob´as property [14, 15] if given ε >
0, there is η ( ε ) > T ∈ L ( X, Y ) with (cid:107) T (cid:107) = 1 and x ∈ S X satisfy (cid:107) T ( x ) (cid:107) > − η ( ε ), there is S ∈ L ( X, Y ) with (cid:107) S (cid:107) = 1 such that (cid:107) S ( x ) (cid:107) = 1 and (cid:107) S − T (cid:107) < ε . That is, the new operator S attains its norm at thesame point at which T almost attains its norm. This property has deep consequences on the structure ofthe involved spaces as, for instance, if a pair ( X, Y ) has the pointwise Bishop–Phelps–Bollob´as property,then X has to be uniformly smooth [15, Proposition 2.3] (actually, if Y is equal to the base field, thischaracterizes uniform smoothness). If ( X, Y ) has the pointwise Bishop–Phelps–Bollob´as property forevery Banach space Y , then the space X also has to be uniformly convex with a power type [14, Theorem3.1].Thinking on an “operatorwise” version of the above property, the following definition appeared in [12](with the name of “property 2”), where it is shown that many pairs of (even finite-dimensional) Banachspaces fail it. Definition ([12, Definition 2.8]) . Let X , Y be Banach spaces. The pair ( X, Y ) has property (P2) if given ε >
0, there exists ¯ η ( ε ) > T ∈ L ( X, Y ) with (cid:107) T (cid:107) = 1 and x ∈ S X satisfy that (cid:107) T ( x ) (cid:107) > − ¯ η ( ε ) , then there is x ∈ S X such that (cid:107) T ( x ) (cid:107) = 1 and (cid:107) x − x (cid:107) < ε. For the case when Y is the base field, this property had appeared earlier in [18], where it is proved thata Banach space X is uniformly convex if and only if the pair ( X, K ) has property (P2) [18, Theorem 2.1].On the other hand, it is immediate that the pairs of the form ( K , Y ) have property (P2) for every Banachspace Y . Our aim in this paper is to prove that, for real Banach spaces, these are the only possible casesin which property (P2) can be satisfied: if the real Banach spaces X and Y have dimension greater thanor equal to two, then the pair ( X, Y ) fails property (P2).Let us finally comment that there is a property weaker than property (P2) also introduced in [12](with the name of property 1) where the function ε (cid:55)−→ η ( ε ) depends on the operator T . This property issatisfied, for instance, by the pairs ( (cid:96) p , (cid:96) q ) for 1 (cid:54) q < p < ∞ [12] and it has some geometric consequencesas it has been pointed out in [21].We would like to dedicate this paper to the memory of our dear friend Bernardo Cascales , who passedaway last April, 2018. Bernardo was an enormous mathematician who in the last years worked, amongmany other topics, on the Bishop–Phelps–Bollob´as property. His deep knowledge of functional analysis,his enthusiasm, and his nice way to explain mathematics, have had an huge impact both on the BPBpand on the people working on it. We would like to highlight the following references [4, 8, 9] containinghis contributions to this field.
HERE IS NO OPERATORWISE VERSION OF THE BISHOP–PHELPS–BOLLOB´AS PROPERTY 3 The Result
Let us state the main result of the paper.
Theorem 1.
Let X and Y be real Banach spaces of dimension greater than or equal to . Then the pair ( X, Y ) fails property (P2). In other words, one may find a sequence { T n } n ∈ N of norm attaining norm-oneelements of L ( X, Y ) and a point x ∈ S X , such that (cid:107) T n ( x ) (cid:107) −→ and inf n ∈ N (cid:8) dist (cid:0) x , { x ∈ S X : (cid:107) T n ( x ) (cid:107) = 1 } (cid:1)(cid:9) > . The proof of this result is rather involved, so we will present it divided into several steps. We startwith the reduction to the case of X and Y being two-dimensional Banach spaces. Proposition 2.
Let X and Y be Banach spaces of dimension greater than or equal to . Suppose thatthe pair ( X, Y ) has property (P2). If Y (cid:54) Y and X (cid:54) X are such that dim( Y ) = dim( X/X ) = 2 ,then the pair ( X/X , Y ) has property (P2).Proof. Let ε >
X, Y ) has property (P2) with some function ¯ η ( ε ) > (cid:101) T : X/X −→ Y with (cid:107) (cid:101) T (cid:107) = 1 and [ x ] ∈ S X/X be such that (cid:107) (cid:101) T ([ x ]) (cid:107) > − ¯ η ( ε/ . Pick a sequence { x n } n ∈ N ⊂ X with (cid:107) x n (cid:107) −→ x n ] = [ x ] for every n ∈ N . Consider the quotientmapping Q : X −→ X/X , define the operator T := (cid:101) T ◦ Q , and observe that (cid:107) T (cid:107) = 1 as Q is a quotientmap. Then (cid:107) T ( x n ) (cid:107) = (cid:107) (cid:101) T ( Q ( x n )) (cid:107) = (cid:107) (cid:101) T ([ x ]) (cid:107) > − ¯ η ( ε/ . Therefore, we may find n ∈ N such that (cid:13)(cid:13)(cid:13)(cid:13) T (cid:18) x n (cid:107) x n (cid:107) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) > − ¯ η ( ε/
2) and (cid:12)(cid:12) − (cid:107) x n (cid:107) (cid:12)(cid:12) < ε/ . The hypothesis provides us with y ∈ S X such that (cid:107) T ( y ) (cid:107) = 1 and (cid:13)(cid:13)(cid:13)(cid:13) y − x n (cid:107) x n (cid:107) (cid:13)(cid:13)(cid:13)(cid:13) < ε/ . Then, (cid:107) x n − y (cid:107) < ε and so (cid:107) [ x ] − Q ( y ) (cid:107) (cid:54) (cid:107) x n − y (cid:107) < ε. On the other hand, 1 = (cid:107) T ( y ) (cid:107) = (cid:107) (cid:101) T ( Q ( y )) (cid:107) (cid:54) (cid:107) Q ( y ) (cid:107) (cid:54) (cid:107) Q ( y ) (cid:107) = 1 = (cid:107) (cid:101) T ( Q ( y )) (cid:107) . (cid:3) Therefore, the proof of Theorem 1 finishes if we are able to prove it for two-dimensional spaces X and Y . This is what we will do in Proposition 5, but we need some preliminary work.Let X be a 2-dimensional real Banach space. We assume that X = R and we consider the standardunit basis vectors e = (1 ,
0) and e = (0 ,
1) of X . The unit sphere S X of X can be represented by thecontinuous curve γ defined as follows: γ ( θ ) := cos θe + sin θe (cid:107) cos θe + sin θe (cid:107) ( θ ∈ [0 , π ]) . Given a point x = γ ( θ ) ∈ S X for some θ ∈ R , we call the curve γ x defined by γ x ( θ ) := γ ( θ + θ ) (0 (cid:54) θ (cid:54) π )as the half arc starting at x . For x ∗ ∈ S X ∗ , we define F ( x ∗ ) to be the face F ( x ∗ ) := { x ∈ S X : x ∗ ( x ) = 1 } .We note that for a given x ∗ ∈ S X ∗ , if γ ( θ ) and γ ( θ ) with 0 (cid:54) θ − θ (cid:54) π are in the face F ( x ∗ ), then γ ( θ ) ∈ F ( x ∗ ) for all θ (cid:54) θ (cid:54) θ . Indeed, the line segment [0 , γ ( θ )] from 0 to γ ( θ ) intersects the linesegment [ γ ( θ ) , γ ( θ )] from γ ( θ ) to γ ( θ ) whenever θ (cid:54) θ (cid:54) θ with 0 (cid:54) θ − θ (cid:54) π (see Figure 1). DANTAS, KADETS, KIM, LEE, AND MART´IN
We will use this observation in the following result. More in general, we have for any x ∗ ∈ X ∗ that if x ∗ ( γ ( θ )) (cid:62) x ∗ ( γ ( θ )) (cid:62)
1, then x ∗ ( γ ( θ )) (cid:62) θ (cid:54) θ (cid:54) θ . Figure 1.
The half arc starting at x Proposition 3.
Let X and Y be two-dimensional real Banach spaces and consider T ∈ L ( X, Y ) with (cid:107) T (cid:107) = 1 . Let γ be the half arc starting at x . Suppose that for (cid:54) θ (cid:54) π , the image T ( γ ( θ )) intersectsthe unit sphere S Y in three points at , θ c and π for some θ c . Also, suppose that there are θ , θ with (cid:54) θ (cid:54) θ c and θ c (cid:54) θ (cid:54) π such that T ( γ ( θ )) and T ( γ ( θ )) are in the interior of B Y (see Figure 2).Then T ( γ ( θ c )) does not belong to F ( y ∗ ) ∪ F ( − y ∗ ) for any y ∗ ∈ S Y ∗ with y ∗ ( T ( x )) = 1 . Figure 2.
Proof.
Suppose that there exists some y ∗ ∈ S Y ∗ such that y ∗ ( T ( x )) = 1 and y ∗ ( T ( γ ( θ c ))) = 1. Then[ T ∗ y ∗ ]( x ) = 1 = [ T ∗ y ∗ ]( γ ( θ c )) . Note that T ∗ y ∗ ∈ S X ∗ . So the points x = γ (0) and γ ( θ c ) are both in the face F ( T ∗ y ∗ ). By the observationjust before this proposition, we get that γ ( θ ) ∈ F ( T ∗ y ∗ ) for all 0 (cid:54) θ (cid:54) θ c . This implies that1 = [ T ∗ y ∗ ]( γ ( θ )) = y ∗ ( T ( γ ( θ ))) (cid:54) (cid:107) T ( γ ( θ )) (cid:107) (cid:54) (cid:54) θ (cid:54) θ c . This shows that (cid:107) T ( γ ( θ )) (cid:107) = 1 for all 0 (cid:54) θ (cid:54) θ c which contradicts the hypothesis on θ . If we have T ( γ ( θ c )) ∈ F ( − y ∗ ), then we can use the same arguments as before to get a contradictionwith the hypothesis on θ . (cid:3) HERE IS NO OPERATORWISE VERSION OF THE BISHOP–PHELPS–BOLLOB´AS PROPERTY 5
The most intriguing part of the proof of Theorem 1 for a pair of two-dimensional real spaces is containedin the following proposition which may have its own interest.
Proposition 4.
Let X and Y be -dimensional real Banach spaces. Then there exists T ∈ L ( X, Y ) with (cid:107) T (cid:107) = 1 such that (i) T ( B X ) ⊂ B Y and (ii) T ( B X ) ∩ S Y contains two points y and y such that for some y ∗ ∈ S Y ∗ with y ∗ ( y ) = 1 we have dist ( y , F ( y ∗ ) ∪ F ( − y ∗ )) > .Proof. We divide the proof in two cases.
Case 1: we assume that X is a Hilbert space. Since Y is finite-dimensional, by using John’s theorem(see [22, Corollary 15.2, p. 121] for example), there is a unique ellipsoid E of maximal volume such that E ⊂ B Y . Since X is a Hilbert space, there is T ∈ L ( X, Y ) with (cid:107) T (cid:107) = 1 such that T ( B X ) = E ⊂ B Y .Now using [22, Theorem 15.3], since Y is 2-dimensional, there are at least two linearly independent points y , y ∈ T ( B X ) ∩ S Y . Let y ∗ ∈ S Y ∗ be such that y ∗ ( y ) = 1. Since the boundary of E does not containline segments, we get that y (cid:54)∈ F ( y ∗ ) ∪ F ( y ∗ ). Figure 3.
Before consider Case 2 in which X is not a Hilbert space, we review the proof of [20, Theorem]. Let Z be any 2-dimensional Banach space and let γ be a parametrization of S Z . If two unit vectors z and z are rotated around S Z while their difference z − z has constantly norm equal to ε , the vector ( z + z )describes a curve Γ ε . Let r ( θ ) = (cid:107) γ ( θ ) (cid:107) where (cid:107) · (cid:107) denotes the Euclidean norm in R . Let z θ be thepoint where the segment [0 , γ ( θ )] intersects Γ ε (see Figure 3) and let ∆( ε, θ ) = 1 − (cid:107) z θ (cid:107) = (cid:107) γ ( θ ) − z θ (cid:107) .So (cid:107) z θ (cid:107) = 1 − ∆( ε, θ ) and z θ = (cid:107) z θ (cid:107) γ ( θ ). So (cid:107) z θ (cid:107) = (cid:107) z θ (cid:107)(cid:107) γ ( θ ) (cid:107) = (1 − ∆( ε, θ )) (cid:107) γ ( θ ) (cid:107) = (1 − ∆( ε, θ )) r ( θ ) . Using this and denoting B Z ε the region inside Γ ε , we have thatArea( B Z ε ) = 12 (cid:90) π (cid:107) z θ (cid:107) dθ = 12 (cid:90) π (1 − ∆( ε, θ )) r ( θ ) dθ. DANTAS, KADETS, KIM, LEE, AND MART´IN
Also, Area( B Z ) = 12 (cid:90) π (cid:107) γ ( θ ) (cid:107) dθ = 12 (cid:90) π r ( θ ) dθ. On the other hand, [20, Lemma] says thatArea( B Z ε ) = (cid:18) − ε (cid:19) Area( B Z ) . And then(1) (cid:90) π (cid:20) (1 − ∆( ε, θ )) − (cid:18) − ε (cid:19)(cid:21) r ( θ ) dθ = 0 . Case 2:
Now we assume that X is not a Hilbert space. By the Day-Nordlander theorems (see [17, p. 60]or [16, Theorem 4.1] and [20, Theorem], respectively), there is some ε > δ X ( ε ) is strictly lessthan the modulus of convexity of a Hilbert space δ H ( ε ) = 1 − (cid:113) − ε . So by (1), there is θ such that(1 − ∆( ε, θ )) − (cid:18) − ε (cid:19) < . as well as θ such that (1 − ∆( ε, θ )) − (cid:18) − ε (cid:19) > . It means that there are x , x ∈ S X such that (cid:107) x − x (cid:107) = ε and (cid:107) x + x (cid:107) < √ − ε . By moving oneof the points x or x on S X a little, we may assume that those points satisfy (cid:107) x + x (cid:107) < √ − ε and (cid:107) x − x (cid:107) < ε .Now for the Banach space Y , using the continuity of ∆, we can find θ such that(1 − ∆( ε, θ )) − (cid:18) − ε (cid:19) = 0 . So there are y , y ∈ S Y such that (cid:107) y − y (cid:107) = ε and (cid:107) y + y (cid:107) = √ − ε .Define the operator S : X −→ Y to be such that S ( x ) = y and S ( x ) = y . So (cid:107) S ( x ) (cid:107) = (cid:107) S ( x ) (cid:107) = 1, S ( x − x ) = y − y and S ( x + x ) = y + y . Moreover, (cid:13)(cid:13)(cid:13)(cid:13) S (cid:18) x − x (cid:107) x − x (cid:107) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) = (cid:107) y − y (cid:107)(cid:107) x − x (cid:107) > (cid:13)(cid:13)(cid:13)(cid:13) S (cid:18) x + x (cid:107) x + x (cid:107) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) = (cid:107) y + y (cid:107)(cid:107) x + x (cid:107) > S by 1 − δ for some small δ >
0, we may assume that (cid:107) S ( x ) (cid:107) < , (cid:107) S ( x ) (cid:107) < , (cid:13)(cid:13)(cid:13)(cid:13) S (cid:18) x − x (cid:107) x − x (cid:107) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) > (cid:13)(cid:13)(cid:13)(cid:13) S (cid:18) x + x (cid:107) x + x (cid:107) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) > . Consider γ to be the half arc starting at x + x (cid:107) x + x (cid:107) (see Figure 4). Then there are 0 (cid:54) t < t < t (cid:54) π such that (cid:107) S ( γ ( t )) (cid:107) < , (cid:107) S ( γ ( t )) (cid:107) > (cid:107) S ( γ ( t )) (cid:107) < . Let a := max (cid:8) (cid:107) S ( γ ( t )) (cid:107) : 0 (cid:54) t (cid:54) t , t (cid:54) t (cid:54) π (cid:9) and b := max (cid:8) (cid:107) S ( γ ( t )) (cid:107) : t (cid:54) t (cid:54) t (cid:9) . We may assume that a (cid:54) b . Otherwise, we consider the half arc γ starting at x − x (cid:107) x − x (cid:107) instead of γ .Now we consider two cases. Subcase 1:
We assume that a = b and we consider the operator T := a S ∈ S L ( X,Y ) . So (cid:107) T ( γ ( t )) (cid:107) = 1 a (cid:107) S ( γ ( t )) (cid:107) (cid:54) (cid:107) S ( γ ( t )) (cid:107) < . HERE IS NO OPERATORWISE VERSION OF THE BISHOP–PHELPS–BOLLOB´AS PROPERTY 7
Figure 4.
Analogously, (cid:107) T ( γ ( t )) (cid:107) <
1. Also, by the definition of a and b , there are s , s such that s (cid:54) t Subcase 2:
Now we assume that a < b . Let s ∈ [0 , t ] ∪ [ t , π ] be such that a = (cid:107) S ( γ ( s )) (cid:107) . Define theoperator T := a S . Then (cid:107) T ( γ ( s )) (cid:107) = a (cid:107) S ( γ ( s )) (cid:107) = 1. Let γ be the half arc starting at γ ( s ). So(see Figure 5) there are 0 < s < s < s < π such that (cid:107) T ( γ (0)) (cid:107) = (cid:107) T ( γ ( π )) (cid:107) = 1 , (cid:107) T ( γ ( t )) (cid:107) (cid:54) t ∈ [0 , s ] ∪ [ s , π ]as well as (cid:107) T ( γ ( s )) (cid:107) < , (cid:107) T ( γ ( s )) (cid:107) < (cid:107) T ( γ ( s )) (cid:107) > . Let y ∗ ∈ S Y ∗ be such that y ∗ ( T ( γ (0))) = 1 and define P : Y −→ Y by P ( y ) := y ∗ ( y ) T ( γ (0)) ( y ∈ Y ) . Note that P is a projection with (cid:107) P (cid:107) = 1. For all λ ∈ [0 , P λ : Y −→ Y by P λ := λ Id Y +(1 − λ ) P. So (cid:107) P λ (cid:107) (cid:54)
1. Let T λ := P λ T ∈ L ( X, Y ) and define ϕ : [0 , −→ R by ϕ ( λ ) := max (cid:8) (cid:107) P λ T ( γ ( t )) (cid:107) : s (cid:54) t (cid:54) s (cid:9) ( λ ∈ [0 , . DANTAS, KADETS, KIM, LEE, AND MART´IN
Figure 5.
Then ϕ is continuous, ϕ (0) = max (cid:8) | y ∗ ( T ( γ ( t ))) | : s (cid:54) t (cid:54) s (cid:9) and ϕ (1) = max (cid:8) (cid:107) T ( γ ( t )) (cid:107) : s (cid:54) t (cid:54) s (cid:9) . We note that | y ∗ ( T ( γ ( t ))) | < s (cid:54) t (cid:54) s . Indeed, otherwise there is some s < (cid:101) t < s suchthat y ∗ ( T ( γ ( (cid:101) t ))) = 1 or −
1. We assume that y ∗ ( T ( γ ( (cid:101) t ))) = 1, so[ T ∗ y ∗ ]( γ ( (cid:101) t )) = 1 = [ T ∗ y ∗ ]( γ (0)) . Hence [ T ∗ y ∗ ]( γ ( t )) (cid:62) (cid:54) t (cid:54) (cid:101) t and this is a contradiction with the fact that y ∗ ( T ( γ ( s ))) (cid:54) (cid:107) T ( γ ( s )) (cid:107) < . Therefore, ϕ (0) <
1. Since ϕ (1) (cid:62) (cid:107) T ( γ ( s )) (cid:107) >
1, there exists λ ∈ (0 ,
1) such that ϕ ( λ ) = 1.Consider T := T λ = P λ T ∈ L ( X, Y ). Then (cid:107) T (cid:107) (cid:54)
1. Also, (cid:107) T ( γ (0)) (cid:107) = (cid:107) T ( γ ( π )) (cid:107) = 1 , (cid:107) T ( γ ( s )) (cid:107) < (cid:107) T ( γ ( s )) (cid:107) < . Also, by the definition of ϕ ( λ ), there is (cid:101) s ∈ [ s , s ] such that (cid:107) T ( γ ( (cid:101) s )) (cid:107) = 1. So taking y := T ( γ (0)), y := T ( γ ( (cid:101) s )) and y ∗ ∈ S Y ∗ to be such that y ∗ ( y ) = 1, one has dist ( y , F ( y ∗ ) ∪ F ( − y ∗ )) > (cid:3) We are now ready to prove that a pair (
X, Y ) with dim( X ) = dim( Y ) = 2 cannot satisfy property(P2). As announced, this, together with Proposition 2, provide the proof of Theorem 1. Proposition 5.
Let X and Y be -dimensional real Banach spaces. Then there are δ > , T n ∈ L ( X, Y ) with (cid:107) T n (cid:107) = 1 for every n ∈ N , and x ∈ S X , such that (cid:107) T n ( x ) (cid:107) −→ but dist (cid:0) x , (cid:8) x ∈ S X : (cid:107) T n ( x ) (cid:107) = 1 (cid:9)(cid:1) > δ for every n ∈ N .Proof. By Proposition 4, there exists an operator T ∈ L ( X, Y ) with (cid:107) T (cid:107) = 1 so that T ( B X ) ∩ S Y containstwo points y and y in such a way that for some y ∗ ∈ S Y ∗ with y ∗ ( y ) = 1 we have δ := dist ( y , F ( y ∗ ) ∪ F ( − y ∗ )) > . HERE IS NO OPERATORWISE VERSION OF THE BISHOP–PHELPS–BOLLOB´AS PROPERTY 9
Let P : Y −→ Y be the projection defined by P ( y ) := y ∗ ( y ) y ( y ∈ Y ) . For all λ ∈ [0 , P λ := λ Id Y +(1 − λ ) P ∈ B L ( Y,Y ) . Since (cid:107) T (cid:107) = 1 and y ∈ T ( B X ) ∩ S Y , there exists x ∈ S X such that T ( x ) = y . Note that if y ∈ B Y \ ( F ( y ∗ ) ∪ F ( − y ∗ )), then (cid:107) P ( y ) (cid:107) < P λ ( y ) is in the interior of B Y for all 0 (cid:54) λ < λ → (cid:107) P λ ( T ( x )) (cid:107) = lim λ → (cid:107) P λ ( y ) (cid:107) = lim λ → (cid:107) λy + (1 − λ ) P ( y ) (cid:107) = 1 . Let T λ := P λ T ∈ L ( X, Y ). If x ∈ S X is such that (cid:107) T λ ( x ) (cid:107) = 1, we have that1 = (cid:107) T λ ( x ) (cid:107) = (cid:107) P λ ( T ( x )) (cid:107) (cid:54) λ (cid:107) T ( x ) (cid:107) + (1 − λ ) | y ∗ ( T ( x )) |(cid:107) y (cid:107) (cid:54) | y ∗ ( T ( x )) | = 1 and so T ( x ) ∈ F ( y ∗ ) ∪ F ( − y ∗ ). So (cid:107) x − x (cid:107) (cid:62) (cid:107) T ( x ) − T ( x ) (cid:107) = (cid:107) T ( x ) − y (cid:107) > δ. (cid:3) References [1]
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Departament of Mathematics, POSTECH, 790-784, (Pohang), Republic of KoreaORCID:
E-mail address : [email protected] (Kadets) School of Mathematics and Computer Sciences, V. N. Karazin Kharkiv National University, pl. Svo-body 4, 61022 Kharkiv, UkraineORCID:
E-mail address : [email protected] (Kim) Department of Mathematics, Chungbuk National University, 1 Chungdae-ro, Seowon-Gu, Cheongju,Chungbuk 28644, Republic of KoreaORCID:
E-mail address : [email protected] (Lee) Department of Mathematics Education, Dongguk University – Seoul, 04620 (Seoul), Republic of KoreaORCID:
E-mail address : [email protected] (Mart´ın) Departamento de An´alisis Matem´atico, Facultad de Ciencias, Universidad de Granada, 18071 Granada,SpainORCID:
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