arXiv:1211.6489v1 [math.FA] 28 Nov 2012 omdlna pc ofidancsayadsffiin conditi sufficient and necessary a find to space linear normed a Abstract lmn fteui peeteeeit one ierope linear bounded a form exists the of there points sphere unit the of x element ups ( Suppose Introduction. 1 Classification Subject Mathematical Keywords nepsdpito h ntball unit the of point exposed an X ffeeyeeeto h ntsphere unit the of element every iff i If (i) are [8,12] in given them of i)Eeynnzr otnoslna ucinlatisamxmmo maximum If a (ii) attains functional linear continuous non-zero Every (ii) 1 nsrn rhgnlt n titycne omdlinear Paul Kallol normed convex strictly and spaces orthogonality strong On mi:Klo [email protected];Dbay Sai Debmalya [email protected]; - Paul Kallol Email: ∗ 2 eateto ahmtc,Jdvu nvriy okt - Kolkata University, Jadavpur Mathematics, of Department eateto ahmtclSine,Sho fSine Ka Science, of School Sciences, Mathematical of Department orsodn author Corresponding : eitouetento fsrnl rhgnlstrelative set orthogonal strongly of notion the introduce We k x ,y x, k ≤ k x + ∈ X, 1 y rhgnlt,Src ovxt,Eteepoint. Extreme convexity, Strict Orthogonality, : S } k k 1 . X . = hr r ayeuvln hrceiain ftesrc convex strict the of characterizations equivalent many are There k emlaSain Debmalya , hnw have we then sanre iersaeoe h edK elo ope.Xi adt b to said is X complex. or real K, field the over space linear normed a is ) k x λx k + with k y k λ x , ∈ k then 0 6= S x K + B 1 . X y n ahiaJha Kanhaiya and S k . X < ete rv htanre iersaei titycne iff convex strictly is space linear normed a that prove then We = y 2 , { = rmr 62,Scnay47A30. Secondary 46B20, Primary : x cx ∈ X o some for anemlagalcm ahiaJha Kanhaiya [email protected]; - n : k 002 INDIA 700032, had nvriy Oo ubr65,Ktmnu NEPAL Kathmandu, 6250, Number POBox University, thmandu x 1 k 1 = ∗ 2 c ≥ } 0 nfra element an for on sa xrm on fteui ball unit the of point extreme an is ao nXwihatisisnr nya the at only its attains which X on A rator oa lmn ntesneo iko-ae in Birkhoff-James of sense the in element an to . totoepito h ntsphere, unit the of point one atmost n x t fanre pc,some space, normed a of ity fteui sphere unit the of ∗ [email protected]; - tityconvex strictly e B X S o each for X = obe to { x ∈ The notion of strict convexity plays an important role in the studies of the geometry of Banach Spaces. One may go through [1,3-5,6-8,11-14] 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⊥By, iff

kxk≤kx + λyk for all scalars λ.

If X is an then x⊥By implies kxk < kx + λyk for all scalars λ 6= 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

kxk < kx + λyk for all scalars λ 6=0.

2 If x ⊥SB y then x ⊥B y but the converse is not true. In l∞(R ) 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 = {x1, x2,...xn} is said

to be a strongly orthogonal set relative to an element xi0 contained in S in the sense of Birkhoff-James iff

n

kxi0 k < kxi0 + X λj xj k j=1,j6=i0

whenever not all λj ’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 {x1, x2,...xn} is said to be a strongly orthogonal set in the sense of Birkhoff-James iff for each i ∈{1, 2,...n}

n kxik < kxi + X λj xj k j=1,j6=i

whenever not all λj ’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

2 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

n n in the sense of Birkhoff-James in l1(R ) but not in l∞(R ). 3 In l2(R ) 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 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 ∈ SK.

2 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 x0 ∈ SX . If there exists a Hamel basis of X containing x0 which is strongly orthonormal relative to x0 in the sense of Birkhoff-James then x0 is an extreme point of BX .

Proof. Let D = {x0, xα : α ∈ Λ} be a strongly orthonormal Hamel basis relative to x0 in the sense of Birkhoff-James.

If possible suppose x0 is not an extreme point of BX , then x0 = tz1 + (1 − t)z2 where 0

So there exists α1, α2,...αn in Λ such that

n n

z1 = β0x0 + X βjxαj and z2 = γ0x0 + X γj xαj j=1 j=1

3 for some scalars βj ,γj (j =0, 1, 2,...n).

If β0 = 0 and γ0 = 0 then x0 = tz1 + (1 − t)z2 implies that

n

x0 = X(tβj + (1 − t)γj )xαj , j=1 which contradicts the fact that every finite subset of D is linearly independent. So the case β0 = 0 and

γ0 = 0 is ruled out.

If β0 6=0, γ0 = 0 then as {x0, xα1 , xα2 ,...,xαn } is a strongly orthonormal set relative to x0 in the sense of Birkhoff-James so we get n βj 1= kz1k = |β0|kx0 + X xαj k ≥ |β0|. β0 j=1 Now n

x0 = tβ0x0 + X(tβj + (1 − t)γj )xαj j=1 and so tβ0 = 1 which is not possible as |β0|≤ 1 and 0

Similarly β0 =0, γ0 6= 0 is also ruled out.

Thus we have β0 6= 0 and γ0 6=0.

Our claim is that at least one of | β0 |, | γ0 | must be less than 1.

If possible suppose | β0 |> 1. Then

n n βj kβ0x0 + X βj xαj k =| β0 |kx0 + X xαj k ≥| β0 |> 1. β0 j=1 j=1

This contradicts that kz1k =1. Thus | β0 |≤ 1. Similarly | γ0 |≤ 1. We next show that | β0 |= 1 and | γ0 |=1 can not hold simultaneously. Case1. X is a real normed linear space.

Then | β0 |= 1 implies that n βj 1= kz1k =| β0 |kx0 + X xαj k > kx0k, β0 j=1 unless βi =0 ∀i =1, 2, . . . n.

Thus | β0 |= 1 ⇒ z1 = β0x0 ⇒ z1 = ±x0 ⇒ x0 = z1 = z2 or t = 0, which is not possible. Thus | β0 |6= 1.

Similarly | γ0 |6= 1. Case2. X is a complex normed linear space.

Then | β0 |= 1 implies that n βj 1= kz1k =| β0 |kx0 + X xαj k > kx0k, β0 j=1

4 unless βi =0 ∀i =1, 2, . . . n.

iθ iφ iθ Thus | β0 |= 1 ⇒ z1 = β0x0 ⇒ z1 = e x0, similarly | γ0 |= 1 ⇒ z2 = e x0. Then x0 = te x0 + (1 −

iφ t)e x0 ⇒ x0 = z1 = z2, which is not possible. Thus | β0 |= 1 and | γ0 |= 1 can not hold simultaneously.

So at least one of | β0 |, | γ0 | is less than 1.

Now x0 = tz1 + (1 − t)z2 implies

tβ0 + (1 − t)γ0 =1, tβj + (1 − t)γj =0 ∀ j =1, 2, . . . n.

But | β0 |< 1 or | γ0 |< 1 implies

1=| tβ0 + (1 − t)γ0 |≤ t | β0 | +(1 − t) | γ0 |< 1, which is not possible.

Thus x0 is an extreme point of BX . This completes the proof.

The converse of the above Theorem is however not always true. If x0 is an extreme point of BX then there may or may not exist a strongly orthonormal Hamel basis relative to x0 in the sense of Birkhoff-James.

Example 2.2 (i) Consider (R2, k.k) where the unit sphere S is given by S = {(x, y) ∈ R2 : x = ±1 and −1 ≤ y ≤ 1}∪{(x, y) ∈ R2 : x2 − 2y + y2 = 0 and y ≥ 1}∪{(x, y) ∈ R2 : x2 +2y + y2 = 0 and y ≤ −1}. Then (1, 1) is an extreme point of the unit ball but there exists no strongly orthonormal Hamel basis relative to (1, 1) in the sense of Birkhoff-James. (ii) Consider (R2, k.k) where the unit sphere S is given by S = {(x, y) ∈ R2 : x = ±1 and − 1 ≤ y ≤ 1}∪{(x, y) ∈ R2 : x2 +2y − 3=0 and y ≥ 1}∪{(x, y) ∈ R2 : x2 − 2y − 3=0 and y ≤−1}. Then (1, 1) is an extreme point of the unit ball and {(1, 1), (−1, 1)} is a strongly orthonormal basis relative to (1,1) in the sense of Birkhoff-James. 3 (iii) In l∞(R ) the extreme points of the unit ball are of the form (±1, ±1, ±1) and for the extreme point (1,1,1) we can find a strongly orthonormal basis relative to (1,1,1) in the sense of Birkhoff-James which is {(1, 1, 1), (1, 0, −1), (0, 1, −1)}.

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 S is called an exposed point of S iff there exists a hyperplane

5 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 x0 is an exposed point of BX then there exists a Hamel basis of X containing x0 which is strongly orthonormal relative to x0 in the sense of Birkhoff-James.

Theorem 2.3 Let X be a normed linear space and x0 ∈ SX be an exposed point of BX . Then there exists a

Hamel basis of X containing x0 which is strongly orthonormal relative to x0 in the sense of Birkhoff-James.

Proof. As x0 is an exposed point of BX so there exists a hyperplane of support H to BX through x0 such that H ∩ BX = {x0}. Then we can find a linear functional f on X such that H = {x ∈ X : f(x)=1}. Let

H0 = {x ∈ X : f(x)=0}. Then H0 is a subspace of X. Let D = {xα : α ∈ Λ} be a Hamel basis of H0 with kxαk = 1. Clearly {x0} ∪ D is a Hamel basis of X. We claim that {x0} ∪ D is a strongly orthonormal set relative to x0 in the sense of Birkhoff-James.

Consider a finite subset {xα1 , xα2 ,...,xαn−1 } of D and let (λ1, λ2,...,λn−1) 6= (0, 0,..., 0). Now if z = n−1 0 x + Pj=1 λj xαj then n−1

f(z)= f(x0 + X λj xαj )= f(x0)=1 j=1 ⇒ z ∈ H

⇒ z∈ / BX , as H ∩ BX = {x0}

n−1 0 0 0 0 So kx + Pj=1 λj xαj k > 1 = kx k. Thus {x } ∪ D is a Hamel basis containing x which is strongly orthonormal relative to x0 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 x0 ∈ SX . If there exists a Hamel basis of X containing x0 which is strongly orthonormal relative to x0 in the sense of Birkhoff-James then there exists a bounded invertible linear operator A on X such that kAk = kAx0k > kAyk for all y in SX with y 6= λx0, λ ∈ SK .

Proof. Let {x0, xα : α ∈ Λ} be a Hamel basis of X which is strongly orthonormal relative to x0 in the sense of Birkhoff-James. 1 Define a linear operator A on X by A(x0)= x0 and A(xα)= 2 xα ∀α ∈ Λ. n−1 0 0 Clearly A is invertible. Take any z ∈ X such that kzk =1. Then z = λ x + Pj=1 λj xαj for some scalars

λj ’s and λ0. 1 If λ0 = 0 then Az = 2 z and so 1 kAx0k =1 > = kAzk. 2

6 If λ0 6= 0 then as {x0, xα : α ∈ Λ} is a strongly orthonormal Hamel basis relative to x0 in the sense of Birkhoff-James so we get n−1

1= kzk = kλ0x0 + X λj xαj k ≥ |λ0|. j=1 Hence we get

n−1 1 kAzk = kλ0x0 + λ x k 2 X j αj j=1 n−1 1 1 = k (λ0x0 + λ x )+ λ0x0k 2 X j αj 2 j=1 1 1 ≤ kzk + |λ0| 2 2

≤ 1= kAx0k.

This proves that kAk≤ 1. Also kAzk = 1 iff | λ0 |= 1 and λj =0 ∀j =1, 2,...n − 1.

Thus kAzk = 1 iff z = λ0x0 with λ0 ∈ SK . This completes the proof. We now prove that

Theorem 2.5 Let X be a normed linear space and x0 ∈ SX . If there exists a bounded linear operator

A : X → X which attains its norm only at the points of the form λx0 with λ ∈ SK then x0 is an exposed point of BX .

Proof. Assume without loss of generality that kAk = 1 and by the Hahn-Banach theorem there exists f ∈ SX∗ such that f(A(x0)) = 1. Clearly kfoAk =1 as f ∈ SX∗ and kAk = kAx0k = 1. If y ∈ SX is such that | foA(y) |=1, then kAyk =1.

Now kAk = 1 and A attains its norm only at the points of the form λx0 with λ ∈ SK , so y ∈{λx0 : λ ∈ SK }.

Thus foA attains its norm only at the points of the form λx0 with λ ∈ SK . Considering the hyperplane

H = {x ∈ X : foA(x)=1} it is easy to verify that H ∩ BX = {x0} and so x0 is an exposed point of BX .

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 ∈ SX the following are equivalent :

1. x is an exposed point of BX . 2. There exists a Hamel basis of X containing x which is strongly orthonormal relative to x in the sense of Birkhoff-James.

7 3. There exists a bounded linear operator A on X which attains its norm only at the points of the form λx

with λ ∈ SK .

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 ∈ SX 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 ∈ SX there exists a bounded linear operator A on X which attains its norm only at the points

of the form λx with λ ∈ SK .

Proof. The proof follows from previous theorem and the fact that a normed linear space X is strictly convex

iff every element of SX is an exposed point of BX .

Remark 2.8 Even though the notions of strong Birkhoff-James orthogonality and Birkhoff-James orthogo-

n nality coincide in they do not characterize Hilbert spaces as (R , k.kp)(1

Competing interests

The authors declare that they have no competing interests.

Acknowledgements

We would like to thank the referee for their invaluable suggestion. We would also like to thank Professor T. K. Mukherjee for his invaluable suggestion while preparing this paper. The first author would like to thank Jadavpur University and DST, Govt. of India for the partial financial support provided through DST-PURSE project and the second author would like to thank UGC, Govt. of India for the financial support.

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