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On strong orthogonality and strictly convex normed linear spaces
Journal of Inequalities and Applications volume 2013, Article number: 242 (2013)
Abstract
We introduce the notion of a strongly orthogonal set relative to an element in the sense of BirkhoffJames in a normed linear space to find a necessary and sufficient condition for an element x of the unit sphere {S}_{X} to be an exposed point of the unit ball {B}_{X}. 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 \lambda \in {S}_{K}.
MSC:46B20, 47A30.
1 Introduction
Suppose (X,\parallel \cdot \parallel ) 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\in X:\parallel x\parallel =1\} is an extreme point of the unit ball {B}_{X}=\{x\in X:\parallel x\parallel \le 1\}. There are many equivalent characterizations of the strict convexity of a normed space, some of them given in [1, 2] are as follows.

(i)
If x,y\in {S}_{X}, then we have \parallel x+y\parallel <2.

(ii)
Every nonzero continuous linear functional attains a maximum on at most one point of the unit sphere.

(iii)
If \parallel x+y\parallel =\parallel x\parallel +\parallel y\parallel, x\ne 0, then y=cx for some c\ge 0.
The notion of strict convexity plays an important role in the studies of the geometry of Banach spaces. One may go through [1–11] for more information related to strictly convex spaces.
An element x is said to be orthogonal to y in X in the sense of BirkhoffJames [1, 8, 12], written as, x{\perp}_{B}y, iff
If X is an inner product space, then x{\perp}_{B}y implies \parallel x\parallel <\parallel x+\lambda y\parallel for all scalars \lambda \ne 0. Motivated by this fact, we here introduce the notion of strong orthogonality as follows.
Strongly orthogonal in the sense of BirkhoffJames: In a normed linear space X, an element x is said to be strongly orthogonal to another element y in the sense of BirkhoffJames, written as x{\perp}_{SB}y, iff
If x{\perp}_{SB}y, then x{\perp}_{B}y, but the converse is not true. In {l}_{\mathrm{\infty}}({R}^{2}) the element (1,0) is orthogonal to (0,1) in the sense of BirkhoffJames, but not strongly orthogonal.
Strongly orthogonal set relative to an element: A finite set of elements S=\{{x}_{1},{x}_{2},\dots ,{x}_{n}\} is said to be a strongly orthogonal set relative to an element {x}_{{i}_{0}} contained in S in the sense of BirkhoffJames iff
whenever not all {\lambda}_{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 BirkhoffJames iff every finite subset containing that element is strongly orthogonal relative to that element in the sense of BirkhoffJames.
Strongly orthogonal set: A finite set of elements \{{x}_{1},{x}_{2},\dots ,{x}_{n}\} is said to be a strongly orthogonal set in the sense of BirkhoffJames iff for each i\in \{1,2,\dots ,n\}
whenever not all {\lambda}_{j}’s are 0.
An infinite set of elements is said to be a strongly orthogonal set in the sense of BirkhoffJames iff every finite subset of the set is a strongly orthogonal set in the sense of BirkhoffJames.
Clearly if a set is strongly orthogonal in the sense of BirkhoffJames, then it is strongly orthogonal relative to every element of the set in the sense of BirkhoffJames. If X has a Hamel basis which is strongly orthogonal in the sense of BirkhoffJames, then we call the Hamel basis a strongly orthogonal Hamel basis in the sense of BirkhoffJames, and if X has a Hamel basis which is strongly orthogonal relative to an element of the basis in the sense of BirkhoffJames, then we call the Hamel basis a strongly orthogonal Hamel basis relative to that element of the basis in the sense of BirkhoffJames. 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,\dots ,0),(0,1,0,\dots ,0),\dots ,(0,0,\dots ,1)\} is a strongly orthonormal Hamel basis in the sense of BirkhoffJames in {l}_{1}({R}^{n}), but not in {l}_{\mathrm{\infty}}({R}^{n}).
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 BirkhoffJames, but not relative to (0,1,1).
In this paper we give another characterization of strictly convex normed linear spaces by using the HahnBanach theorem and the notion of a strongly orthogonal Hamel basis relative to an element in the sense of BirkhoffJames. More precisely, we explore the relation between the existence of a strongly orthogonal Hamel basis relative to an element with the unit norm in the sense of BirkhoffJames 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 \lambda \in {S}_{K}.
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 {x}_{0}\in {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 BirkhoffJames, then {x}_{0} is an extreme point of {B}_{X}.
Proof Let D=\{{x}_{0},{x}_{\alpha}:\alpha \in \mathrm{\Lambda}\} be a strongly orthonormal Hamel basis relative to {x}_{0} in the sense of BirkhoffJames.
If possible, suppose that {x}_{0} is not an extreme point of {B}_{X}, then {x}_{0}=t{z}_{1}+(1t){z}_{2} where 0<t<1 and \parallel {z}_{1}\parallel =\parallel {z}_{2}\parallel =1.
So, there exists {\alpha}_{1},{\alpha}_{2},\dots ,{\alpha}_{n} in Λ such that
for some scalars {\beta}_{j}, {\gamma}_{j} (j=0,1,2,\dots ,n).
If {\beta}_{0}=0 and {\gamma}_{0}=0, then {x}_{0}=t{z}_{1}+(1t){z}_{2} implies that
which contradicts the fact that every finite subset of D is linearly independent. So, the case {\beta}_{0}=0 and {\gamma}_{0}=0 is ruled out.
If {\beta}_{0}\ne 0, {\gamma}_{0}=0, then as \{{x}_{0},{x}_{{\alpha}_{1}},{x}_{{\alpha}_{2}},\dots ,{x}_{{\alpha}_{n}}\} is a strongly orthonormal set relative to {x}_{0} in the sense of BirkhoffJames, so we get
Now
and so t{\beta}_{0}=1, which is not possible as {\beta}_{0}\le 1 and 0<t<1.
Similarly {\beta}_{0}=0, {\gamma}_{0}\ne 0 is also ruled out.
Thus we have {\beta}_{0}\ne 0 and {\gamma}_{0}\ne 0.
Our claim is that at least one of {\beta}_{0}, {\gamma}_{0} must be less than 1.
If possible, suppose that {\beta}_{0}>1. Then
This contradicts \parallel {z}_{1}\parallel =1. Thus {\beta}_{0}\le 1. Similarly {\gamma}_{0}\le 1. We next show that {\beta}_{0}=1 and {\gamma}_{0}=1 cannot hold simultaneously.
Case 1. X is a real normed linear space.
Then {\beta}_{0}=1 implies that
unless {\beta}_{i}=0 \mathrm{\forall}i=1,2,\dots ,n.
Thus {\beta}_{0}=1\Rightarrow {z}_{1}={\beta}_{0}{x}_{0}\Rightarrow {z}_{1}=\pm {x}_{0}\Rightarrow {x}_{0}={z}_{1}={z}_{2} or t=0, which is not possible. Thus {\beta}_{0}\ne 1. Similarly {\gamma}_{0}\ne 1.
Case 2. X is a complex normed linear space.
Then {\beta}_{0}=1 implies that
unless {\beta}_{i}=0 \mathrm{\forall}i=1,2,\dots ,n.
Thus {\beta}_{0}=1\Rightarrow {z}_{1}={\beta}_{0}{x}_{0}\Rightarrow {z}_{1}={e}^{i\theta}{x}_{0}, similarly {\gamma}_{0}=1\Rightarrow {z}_{2}={e}^{i\varphi}{x}_{0}. Then {x}_{0}=t{e}^{i\theta}{x}_{0}+(1t){e}^{i\varphi}{x}_{0}\Rightarrow {x}_{0}={z}_{1}={z}_{2}, which is not possible. Thus {\beta}_{0}=1 and {\gamma}_{0}=1 cannot hold simultaneously.
So, at least one of {\beta}_{0}, {\gamma}_{0} is less than 1.
Now {x}_{0}=t{z}_{1}+(1t){z}_{2} implies
But {\beta}_{0}<1 or {\gamma}_{0}<1 implies
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 BirkhoffJames.
Example 2.2 (i) Consider ({R}^{2},\parallel \cdot \parallel ) where the unit sphere S is given by S = {(x,y)\in {R}^{2}:x=\pm 1 and 1\le y\le 1} ∪ {(x,y)\in {R}^{2}:{x}^{2}2y+{y}^{2}=0 and y\ge 1} ∪ {(x,y)\in {R}^{2}:{x}^{2}+2y+{y}^{2}=0 and y\le 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 BirkhoffJames.

(ii)
Consider ({R}^{2},\parallel \cdot \parallel ) where the unit sphere S is given by S = {(x,y)\in {R}^{2}:x=\pm 1 and 1\le y\le 1} ∪ {(x,y)\in {R}^{2}:{x}^{2}+2y3=0 and y\ge 1} ∪ {(x,y)\in {R}^{2}:{x}^{2}2y3=0 and y\le 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 BirkhoffJames.

(iii)
In {l}_{\mathrm{\infty}}({R}^{3}) the extreme points of the unit ball are of the form (\pm 1,\pm 1,\pm 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 BirkhoffJames 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 neighborhood of (1,1) contains both extreme as well as nonextreme 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\cap S=\{x\}. The notion of exposed points can be found in [5, 13–15]. 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 BirkhoffJames.
Theorem 2.3 Let X be a normed linear space and {x}_{0}\in {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 BirkhoffJames.
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\cap {B}_{X}=\{{x}_{0}\}. Then we can find a linear functional f on X such that H=\{x\in X:f(x)=1\}. Let {H}_{0}=\{x\in X:f(x)=0\}. Then {H}_{0} is a subspace of X. Let D=\{{x}_{\alpha}:\alpha \in \mathrm{\Lambda}\} be a Hamel basis of {H}_{0} with \parallel {x}_{\alpha}\parallel =1. Clearly \{{x}_{0}\}\cup D is a Hamel basis of X. We claim that \{{x}_{0}\}\cup D is a strongly orthonormal set relative to {x}_{0} in the sense of BirkhoffJames.
Consider a finite subset \{{x}_{{\alpha}_{1}},{x}_{{\alpha}_{2}},\dots ,{x}_{{\alpha}_{n1}}\} of D and let ({\lambda}_{1},{\lambda}_{2},\dots ,{\lambda}_{n1})\ne (0,0,\dots ,0). Now if z={x}_{0}+{\sum}_{j=1}^{n1}{\lambda}_{j}{x}_{{\alpha}_{j}}, then
So \parallel {x}_{0}+{\sum}_{j=1}^{n1}{\lambda}_{j}{x}_{{\alpha}_{j}}\parallel >1=\parallel {x}_{0}\parallel. Thus \{{x}_{0}\}\cup D is a Hamel basis containing {x}_{0} which is strongly orthonormal relative to {x}_{0} in the sense of BirkhoffJames.
This completes the proof. □
We next prove the following theorem.
Theorem 2.4 Let X be a normed linear space and {x}_{0}\in {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 BirkhoffJames, then there exists a bounded invertible linear operator A on X such that \parallel A\parallel =\parallel A{x}_{0}\parallel >\parallel Ay\parallel for all y in {S}_{X} with y\ne \lambda {x}_{0}, \lambda \in {S}_{K}.
Proof Let \{{x}_{0},{x}_{\alpha}:\alpha \in \mathrm{\Lambda}\} be a Hamel basis of X which is strongly orthonormal relative to {x}_{0} in the sense of BirkhoffJames.
Define a linear operator A on X by A({x}_{0})={x}_{0} and A({x}_{\alpha})=\frac{1}{2}{x}_{\alpha} \mathrm{\forall}\alpha \in \mathrm{\Lambda}.
Clearly A is invertible. Take any z\in X such that \parallel z\parallel =1. Then z={\lambda}_{0}{x}_{0}+{\sum}_{j=1}^{n1}{\lambda}_{j}{x}_{{\alpha}_{j}} for some scalars {\lambda}_{j}’s and {\lambda}_{0}.
If {\lambda}_{0}=0, then Az=\frac{1}{2}z and so
If {\lambda}_{0}\ne 0, then as \{{x}_{0},{x}_{\alpha}:\alpha \in \mathrm{\Lambda}\} is a strongly orthonormal Hamel basis relative to {x}_{0} in the sense of BirkhoffJames, so we get
Hence we get
This proves that \parallel A\parallel \le 1. Also \parallel Az\parallel =1 iff {\lambda}_{0}=1 and {\lambda}_{j}=0 \mathrm{\forall}j=1,2,\dots ,n1.
Thus \parallel Az\parallel =1 iff z={\lambda}_{0}{x}_{0} with {\lambda}_{0}\in {S}_{K}. This completes the proof. □
We now prove the following theorem.
Theorem 2.5 Let X be a normed linear space and {x}_{0}\in {S}_{X}. If there exists a bounded linear operator A:X\to X which attains its norm only at the points of the form \lambda {x}_{0} with \lambda \in {S}_{K}, then {x}_{0} is an exposed point of {B}_{X}.
Proof Assume, without loss of generality, that \parallel A\parallel =1 and by the HahnBanach theorem, there exists f\in {S}_{{X}^{\ast}} such that f(A({x}_{0}))=1. Clearly \parallel foA\parallel =1 as f\in {S}_{{X}^{\ast}} and \parallel A\parallel =\parallel A{x}_{0}\parallel =1. If y\in {S}_{X} is such that foA(y)=1, then \parallel Ay\parallel =1.
Now \parallel A\parallel =1 and A attains its norm only at the points of the form \lambda {x}_{0} with \lambda \in {S}_{K}, so y\in \{\lambda {x}_{0}:\lambda \in {S}_{K}\}.
Thus foA attains its norm only at the points of the form \lambda {x}_{0} with \lambda \in {S}_{K}. Considering the hyperplane H=\{x\in X:foA(x)=1\}, it is easy to verify that H\cap {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\in {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 BirkhoffJames.

3.
There exists a bounded linear operator A on X which attains its norm only at the points of the form λx with \lambda \in {S}_{K}.
We next give a characterization of a 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\in {S}_{X}, there exists a Hamel basis of X containing x which is strongly orthonormal relative to x in the sense of BirkhoffJames.

3.
For each x\in {S}_{X}, there exists a bounded linear operator A on X which attains its norm only at the points of the form λx with \lambda \in {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 BirkhoffJames orthogonality and BirkhoffJames orthogonality coincide in a Hilbert space, they do not characterize Hilbert spaces as ({R}^{n},{\parallel \cdot \parallel}_{p}) (1<p<\mathrm{\infty}, p\ne 2) is not a Hilbert space, but the notions of strong BirkhoffJames orthogonality and BirkhoffJames orthogonality coincide there.
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Acknowledgements
We would like to thank the referee for their invaluable suggestion. We would also like to thank Professor TK 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 DSTPURSE project and the second author would like to thank UGC, Govt. of India for the financial support.
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Paul, K., Sain, D. & Jha, K. On strong orthogonality and strictly convex normed linear spaces. J Inequal Appl 2013, 242 (2013). https://doi.org/10.1186/1029242X2013242
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DOI: https://doi.org/10.1186/1029242X2013242
Keywords
 orthogonality
 strict convexity
 extreme point