On strong orthogonality and strictly convex normed linear spaces

We introduce the notion of a strongly orthogonal set relative to an element in the sense of Birkhoff-James in a normed linear space to find a necessary and sufficient condition for an element x of the unit sphere SX to be an exposed point of the unit ball BX. We then prove that a normed linear space is strictly convex iff for each element x of the unit sphere, there exists a bounded linear operator A on X which attains its norm only at the points of the form λx with λ∈SK. MSC:46B20, 47A30.


Introduction.
Suppose (X, .) is a normed linear space over the field K, real or complex.X is said to be strictly convex iff every element of the unit sphere S X = {x ∈ X : x = 1} is an extreme point of the unit ball B X = {x ∈ X : x ≤ 1}.There are many equivalent characterizations of the strict convexity of a normed space, some of them given in [8,12] are (i) If x, y ∈ S X then we have x + y < 2, (ii) Every non-zero continuous linear functional attains a maximum on atmost one point of the unit sphere, (ii) If x + y = x + y , x = 0 then y = cx for some c ≥ 0. The notion of strict convexity plays an important role in the studies of the geometry of Banach Spaces.One may go through [1,[3][4][5][6][7][8][11][12][13][14] for more information related to strictly convex spaces.An element x is said to be orthogonal to y in X in the sense of Birkhoff-James [2,7,8], written as, x⊥ B y, iff x ≤ x + λy for all scalars λ.
If X is an inner product space then x⊥ B y implies x < x + λy for all scalars λ = 0. Motivated by this fact we here introduce the notion of strong orthogonality as follows: Strongly orthogonal in the sense of Birkhoff-James: In a normed linear space X an element x is said to be strongly orthogonal to another element y in the sense of Birkhoff-James, written as x ⊥ SB y, iff x < x + λy for all scalars λ = 0.If x ⊥ SB y then x ⊥ B y but the converse is not true.In l ∞ (R 2 ) the element (1, 0) is orthogonal to (0, 1) in the sense of Birkhoff-James but not strongly orthogonal.Strongly orthogonal set relative to an element : A finite set of elements S = {x 1 , x 2 , . . .x n } is said to be a strongly orthogonal set relative to an element x i0 contained in S in the sense of Birkhoff-James iff whenever not all λ j 's are 0.An infinite set of elements is said to be a strongly orthogonal set relative to an element contained in the set in the sense of Birkhoff-James iff every finite subset containing that element is strongly orthogonal relative to that element in the sense of Birkhoff-James.

Strongly orthogonal Set:
A finite set of elements {x 1 , x 2 , . . .x n } is said to be a strongly orthogonal set in the sense of Birkhoff-James iff for each i ∈ {1, 2, . . .n} whenever not all λ j 's are 0.An infinite set of elements is said to be a strongly orthogonal set in the sense of Birkhoff-James iff every finite subset of the set is a strongly orthogonal set in the sense of Birkhoff-James.Clearly if a set is strongly orthogonal in the sense of Birkhoff-James then it is strongly orthogonal relative to every element of the set in the sense of Birkhoff-James.If X has a Hamel basis which is strongly orthogonal in the sense of Birkhoff-James then we call the Hamel basis a strongly orthogonal Hamel basis in the sense of Birkhoff-James and if X has a Hamel basis which is strongly orthogonal relative to an element of the basis in the sense of Birkhoff-James then we call the Hamel basis a strongly orthogonal Hamel basis relative to that element of the basis in the sense of Birkhoff-James.If in addition the norm of each element of a strongly orthogonal set is 1 then accordingly we call them orthonormal.As for example the set {(1, 0, . . ., 0), (0, 1, 0 . . ., 0), . . ., (0, 0, . . ., 1)} is a strongly orthonormal Hamel basis in the sense of Birkhoff-James in In l 2 (R 3 ) the set {(1, 0, 0), (0, 1, 0), (0, 1, 1)} is strongly orthogonal relative to (1,0,0) in the sense of Birkhoff-James but not relative to(0,1,1).In this paper we give another characterization of strictly convex normed linear spaces by using the Hahn-Banach theorem and the notion of strongly orthogonal Hamel basis relative to an element in the sense of Birkhoff-James, more precisely we explore the relation between the existence of strongly orthogonal Hamel basis relative to an element with unit norm in the sense of Birkhoff-James in a normed space and that of an extreme point of the unit ball in the space.We also prove that a normed linear space is strictly convex iff for each point x of the unit sphere there exists a bounded linear operator A on X which attains its norm only at the points of the form λx with λ ∈ S K .

Main Results
We first obtain a sufficient condition for an element in the unit sphere to be an extreme point of the unit ball in an arbitrary normed linear space.
Theorem 2.1.Let X be a normed linear space and x 0 ∈ S X .If there exists a Hamel basis of X containing x 0 which is strongly orthonormal relative to x 0 in the sense of Birkhoff-James then x 0 is an extreme point of B X .
If β 0 = 0 and γ 0 = 0 then which contradicts the fact that every finite subset of D is linearly independent.So the case β 0 = 0 and γ 0 = 0 is ruled out.
Thus we have β 0 = 0 and γ 0 = 0. Our claim is that at least one of Then Case2.X is a complex normed linear space.
which is not possible.Thus x 0 is an extreme point of B X .This completes the proof.The converse of the above Theorem is however not always true.If x 0 is an extreme point of B X then there may or may not exist a strongly orthonormal Hamel basis relative to x 0 in the sense of Birkhoff-James.
In the first two examples the extreme point (1,1) is such that every neighbourhood of (1,1) contains both extreme as well as non-extreme points whereas in the third case the extreme point (1,1,1) is an isolated extreme point.An element x in the boundary of a convex set S is called an exposed point of S iff there exists a hyperplane of support H to S through x such that H ∩ S = {x}.The notion of exposed points can be found in [4,9,10,15].We next prove that if the extreme point x 0 is an exposed point of B X then there exists a Hamel basis of X containing x 0 which is strongly orthonormal relative to x 0 in the sense of Birkhoff-James.
Theorem 2.3.Let X be a normed linear space and x 0 ∈ S X be an exposed point of B X .Then there exists a Hamel basis of X containing x 0 which is strongly orthonormal relative to x 0 in the sense of Birkhoff-James.
Proof.As x 0 is an exposed point of B X so there exists a hyperplane of support H to B X through x 0 such that H ∩ B X = {x 0 }.Then we can find a linear functional f on X such that . Then H 0 is a subspace of X.Let D = {x α : α ∈ Λ} be a Hamel basis of H 0 with x α = 1.Clearly {x 0 } ∪ D is a Hamel basis of X.We claim that {x 0 } ∪ D is a strongly orthonormal set relative to x 0 in the sense of Birkhoff-James.Consider a finite subset {x α1 , x α2 , . . ., x αn−1 } of D and let (λ Thus {x 0 } ∪ D is a Hamel basis containing x 0 which is strongly orthonormal relative to x 0 in the sense of Birkhoff-James.This completes the proof.We next prove that Theorem 2.4.Let X be a normed linear space and x 0 ∈ S X .If there exists a Hamel basis of X containing x 0 which is strongly orthonormal relative to x 0 in the sense of Birkhoff-James then there exists a bounded invertible linear operator A on X such that A = Ax 0 > Ay for all y in S X with y = λx 0 , λ ∈ S K . Proof.Let {x 0 , x α : α ∈ Λ} be a Hamel basis of X which is strongly orthonormal relative to x 0 in the sense of Birkhoff-James.Define a linear operator A on X by A(x 0 ) = x 0 and A(x α ) = 1 2 x α ∀α ∈ Λ. Clearly A is invertible.Take any z ∈ X such that z = 1.Then z = λ 0 x 0 + n−1 j=1 λ j x αj for some scalars λ j 's and λ 0 .If λ 0 = 0 then Az = 1 2 z and so If λ 0 = 0 then as {x 0 , x α : α ∈ Λ} is a strongly orthonormal Hamel basis relative to x 0 in the sense of Birkhoff-James so we get
Thus Az = 1 iff z = λ 0 x 0 with λ 0 ∈ S K .This completes the proof.We now prove that Theorem 2.5.Let X be a normed linear space and x 0 ∈ S X .If there exists a bounded linear operator A : X → X which attains its norm only at the points of the form λx 0 with λ ∈ S K then x 0 is an exposed point of B X .
Proof.Assume without loss of generality that A = 1 and by the Hahn-Banach theorem there exists Now A = 1 and A attains its norm only at the points of the form λx 0 with λ ∈ S K , so y ∈ {λx 0 : λ ∈ S K }.Thus f oA attains its norm only at the points of the form λx 0 with λ ∈ S K .Considering the hyperplane H = {x ∈ X : f oA(x) = 1} it is easy to verify that H ∩ B X = {x 0 } and so x 0 is an exposed point of B X .
Thus we obtained complete characterizations of exposed points which is stated clearly in the following theorem Theorem 2.6.For a normed linear space X and a point x ∈ S X the following are equivalent : 1.
x is an exposed point of B X .
2. There exists a Hamel basis of X containing x which is strongly orthonormal relative to x in the sense of Birkhoff-James.
3. There exists a bounded linear operator A on X which attains its norm only at the points of the form λx with λ ∈ S K .
We next give a characterization of strictly convex space as follows : Theorem 2.7.For a normed linear space X the following are equivalent : 1. X is strictly convex.2. For each x ∈ S X there exists a Hamel basis of X containing x which is strongly orthonormal relative to x in the sense of Birkhoff-James.
3. For each x ∈ S X there exists a bounded linear operator A on X which attains its norm only at the points of the form λx with λ ∈ S K .
Proof.The proof follows from previous theorem and the fact that a normed linear space X is strictly convex iff every element of S X is an exposed point of B X .
Remark 2.8.Even though the notions of strong Birkhoff-James orthogonality and Birkhoff-James orthogonality coincide in Hilbert space they do not characterize Hilbert spaces as (R n , .p )(1 < p < ∞, p = 2) is not a Hilbert space but the notions of strong Birkhoff-James orthogonality and Birkhoff-James orthogonality coincide there.