On strong orthogonality and strictly convex normed linear spaces

Abstract

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 $S_X$ 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 non-zero 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 Birkhoff-James [1, 8, 12], written as, $x{\perp }_{B}y$, iff

$$\parallel x\parallel \le \parallel x+\lambda y\parallel \text{for all scalars }\lambda .$$

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 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 $x{\perp }_{SB}y$, iff

$$\parallel x\parallel <\parallel x+\lambda y\parallel \text{for all scalars }\lambda \ne 0.$$

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 Birkhoff-James, 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 Birkhoff-James iff

$$\parallel x_{i_0}\parallel <\parallel x_{i_0}+\sum _{j=1,j\ne i_0}^n{\lambda }_j{x}_j\parallel$$

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 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 $\{x_1,x_2,\dots ,x_n\}$ is said to be a strongly orthogonal set in the sense of Birkhoff-James iff for each $i\in \{1,2,\dots ,n\}$

$$\parallel x_i\parallel <\parallel x_i+\sum _{j=1,j\ne i}^n{\lambda }_j{x}_j\parallel$$

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 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 $\{(1,0,\dots ,0),(0,1,0,\dots ,0),\dots ,(0,0,\dots ,1)\}$ is a strongly orthonormal Hamel basis in the sense of Birkhoff-James 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 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 a strongly orthogonal Hamel basis relative to an element in the sense of Birkhoff-James. 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 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 $\lambda \in S_K$.

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 Birkhoff-James, 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 Birkhoff-James.

If possible, suppose that $x_0$ is not an extreme point of $B_X$, then $x_0=t{z}_1+(1-t)z_2$ 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

$$z_1={\beta }_0{x}_0+\sum _{j=1}^n{\beta }_j{x}_{{\alpha }_j}\text{and}{z}_2={\gamma }_0{x}_0+\sum _{j=1}^n{\gamma }_j{x}_{{\alpha }_j}$$

for some scalars ${\beta }_j$, ${\gamma }_j$ ($j=0,1,2,\dots ,n$).

If ${\beta }_0=0$ and ${\gamma }_0=0$, then $x_0=t{z}_1+(1-t)z_2$ implies that

$$x_0=\sum _{j=1}^n(t{\beta }_j+(1-t){\gamma }_j)x_{{\alpha }_j},$$

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 Birkhoff-James, so we get

$$1=\parallel z_1\parallel =|{\beta }_0|\parallel x_0+\sum _{j=1}^n\frac{{\beta }_j}{{\beta }_0}{x}_{{\alpha }_j}\parallel \ge |{\beta }_0|.$$

Now

$$x_0=t{\beta }_0{x}_0+\sum _{j=1}^n(t{\beta }_j+(1-t){\gamma }_j)x_{{\alpha }_j}$$

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

$$\parallel {\beta }_0{x}_0+\sum _{j=1}^n{\beta }_j{x}_{{\alpha }_j}\parallel =|{\beta }_0|\parallel x_0+\sum _{j=1}^n\frac{{\beta }_j}{{\beta }_0}{x}_{{\alpha }_j}\parallel \ge |{\beta }_0|>1.$$

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

$$1=\parallel z_1\parallel =|{\beta }_0|\parallel x_0+\sum _{j=1}^n\frac{{\beta }_j}{{\beta }_0}{x}_{{\alpha }_j}\parallel >\parallel x_0\parallel ,$$

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

$$1=\parallel z_1\parallel =|{\beta }_0|\parallel x_0+\sum _{j=1}^n\frac{{\beta }_j}{{\beta }_0}{x}_{{\alpha }_j}\parallel >\parallel x_0\parallel ,$$

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 $x_0=t{e}^{i\theta }{x}_0+(1-t)e^{i\varphi }{x}_0\Rightarrow x_0=z_1=z_2$, 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 $x_0=t{z}_1+(1-t)z_2$ implies

$$t{\beta }_0+(1-t){\gamma }_0=1,t{\beta }_j+(1-t){\gamma }_j=0\mathrm{\forall }j=1,2,\dots ,n.$$

But $|{\beta }_0|<1$ or $|{\gamma }_0|<1$ implies

$$1=|t{\beta }_0+(1-t){\gamma }_0|\le t|{\beta }_0|+(1-t)|{\gamma }_0|<1,$$

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.

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 Birkhoff-James.

• (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 $-1\le y\le 1$} ∪ {$(x,y)\in R^2:x^2+2y-3=0$ and $y\ge 1$} ∪ {$(x,y)\in R^2:x^2-2y-3=0$ 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 Birkhoff-James.

• (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 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 neighborhood 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 $H\cap S=\{x\}$. 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 Birkhoff-James.

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 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 $H\cap B_X=\{x_0\}$. 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 Birkhoff-James.

Consider a finite subset $\{x_{{\alpha }_1},x_{{\alpha }_2},\dots ,x_{{\alpha }_{n-1}}\}$ of D and let $({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{n-1})\ne (0,0,\dots ,0)$. Now if $z=x_0+{\sum }_{j=1}^{n-1}{\lambda }_j{x}_{{\alpha }_j}$, then

So $\parallel x_0+{\sum }_{j=1}^{n-1}{\lambda }_j{x}_{{\alpha }_j}\parallel >1=\parallel x_0\parallel$. Thus $\{x_0\}\cup D$ is a Hamel basis containing $x_0$ which is strongly orthonormal relative to $x_0$ in the sense of Birkhoff-James.

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 Birkhoff-James, 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 Birkhoff-James.

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 $z={\lambda }_0{x}_0+{\sum }_{j=1}^{n-1}{\lambda }_j{x}_{{\alpha }_j}$ for some scalars ${\lambda }_j$’s and ${\lambda }_0$.

If ${\lambda }_0=0$, then $Az=\frac{1}{2}z$ and so

$$\parallel A{x}_0\parallel =1>\frac{1}{2}=\parallel Az\parallel .$$

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 Birkhoff-James, so we get

$$1=\parallel z\parallel =\parallel {\lambda }_0{x}_0+\sum _{j=1}^{n-1}{\lambda }_j{x}_{{\alpha }_j}\parallel \ge |{\lambda }_0|.$$

Hence we get

$$\begin{array}{rcl}\parallel Az\parallel & =& \parallel {\lambda }_0{x}_0+\frac{1}{2}\sum _{j=1}^{n-1}{\lambda }_j{x}_{{\alpha }_j}\parallel \\ =& \parallel \frac{1}{2}({\lambda }_0{x}_0+\sum _{j=1}^{n-1}{\lambda }_j{x}_{{\alpha }_j})+\frac{1}{2}{\lambda }_0{x}_0\parallel \\ \le & \frac{1}{2}\parallel z\parallel +\frac{1}{2}|{\lambda }_0|\\ \le & 1=\parallel A{x}_0\parallel .\end{array}$$

This proves that $\parallel A\parallel \le 1$. Also $\parallel Az\parallel =1$ iff $|{\lambda }_0|=1$ and ${\lambda }_j=0$ $\mathrm{\forall }j=1,2,\dots ,n-1$.

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 Hahn-Banach 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 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 $\lambda \in S_K$.

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 Birkhoff-James.

• 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 Birkhoff-James orthogonality and Birkhoff-James 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 Birkhoff-James orthogonality and Birkhoff-James orthogonality coincide there.