Research Article | Open Access

# Circle-Uniqueness of Pythagorean Orthogonality in Normed Linear Spaces

**Academic Editor:**Yongqiang Fu

#### Abstract

We introduce the circle-uniqueness of Pythagorean orthogonality in normed linear spaces and show that Pythagorean orthogonality is circle-unique if and only if the underlying space is strictly convex. Further related results providing more detailed relations between circle-uniqueness of Pythagorean orthogonality and the shape of the unit sphere are also presented.

#### 1. Introduction

We denote by a real normed linear space whose dimension is at least 2. The* origin*,* unit ball*, and* unit sphere* of are denoted by , , and , respectively. When is two-dimensional, it is called a* Minkowski plane*. Its unit sphere is then called the* unit circle* of , and each homothetic copy of is a* circle*. For two distinct points (or vectors) and in , we denote by the* line* passing through and , by the* ray* starting from and passing through , and by the* (nondegenerate) segment* connecting and . Moreover, is said to be* strictly convex* if does not contain a nondegenerate segment.

Pythagorean orthogonality, which was introduced by James in [1], is one of the most natural extensions of orthogonality in inner product spaces to normed linear spaces (for other orthogonality types in normed linear spaces, we refer to [2–4] and the references therein). Let and be two vectors in a real normed linear space. If
then and are said to be* Pythagorean orthogonal* to each other (denoted by ). James showed that the following facts are equivalent:(1), , ;(2) is an inner product space.

In other words, Pythagorean orthogonality is not* homogeneous* in general normed linear spaces. Among other things, James proved the* line-existence* of Pythagorean orthogonality: for each pair of vectors and in , there exists a number such that . That is, James proved that in each line parallel to the line there exists a vector that is Pythagorean orthogonal to .

However, James did not obtain any essential result on the uniqueness of this orthogonality type. Kapoor and Prasad [5] fixed this gap by proving that Pythagorean orthogonality is line-unique in each normed linear space , where a binary relation on is said to be* line-unique* if and only if for each and there exists a unique real number such that . It appears that the uniqueness of Pythagorean orthogonality has nothing to do with the shape of the unit ball. By introducing the circle-uniqueness (see Definition 1 next) of Pythagorean orthogonality, we show that this is not true. Our main result shows that Pythagorean orthogonality is circle-unique if and only if is strictly convex, which updates the knowledge about uniqueness of Pythagorean orthogonality.

For each , we denote by the set of points that are Pythagorean orthogonal to ; that is,

For two linearly independent vectors and we denote by the two-dimensional subspace of spanned by and and by the closed halfplane of bounded by the line and containing .

*Definition 1. *Pythagorean orthogonality on is said to be* circle-unique* if, for each pair of linearly independent vectors and and each nonnegative real number , there exists a unique vector in .

#### 2. Results and Proofs

The following lemma concerning the intersection of two circles in a Minkowski plane is one of our main tools.

Lemma 2 (Theorem 2.4 in [6]). *Let and be two circles in a Minkowski plane , where and are two distinct points, and let and be the points of intersection of and . Then the set has one of the following forms: *(1)*;*(2)* is the union of two closed, disjoint segments (one or both of them may degenerate to a singleton) lying on the opposite sides of ;*(3)* is the union of two segments (one or both of them may degenerate to a singleton) with common point or .*

One can easily verify the following proposition.

Proposition 3. *Let and be two points in . Then if and only if
*

First we show that Pythagorean orthogonality has the* circle-existence* property. More precisely, we show the following proposition.

Proposition 4. *For each pair of linearly independent vectors and and each number , the set
**
is a nonempty segment that may degenerate to a singleton.*

*Proof. *We only consider the nontrivial case . Clearly,
Since
is not empty. It is also clear that . Thus, by Lemma 2, is the union of two closed, disjoint segments contained in , one or both of which may degenerate to a singleton, lying in opposite halfplanes with respect to the line . This completes the proof.

Next we state a simple result on common supporting lines of two circles.

Lemma 5. *Let be a Minkowski plane, a vector in , and two numbers such that
**
Then there are two common supporting lines of and passing through the point .*

*Proof. *By the hypothesis of the lemma, is exterior to . Thus two supporting lines and of can be drawn through . In the following we show that these two lines are two common supporting lines of and .

Clearly, there exists a point such that supports at . Put . Then
which implies that intersects in . Moreover,
that is, the distance from to is . Thus is a common supporting line of and . In a similar way we can show that is also a common supporting line of these two discs.

Theorem 6. *Let and be two linearly independent vectors, and let be a positive number, , and . Then
**
is a nondegenerate segment if and only if there exist two unit vectors and in such that *(1)* is a nondegenerate maximal segment contained in ;*(2)* intersects at ;*(3)* or, equivalently, .*

*Proof. *It is clear that is exterior to and is exterior to .

First suppose that is a segment . Then
which implies that is one of the two common supporting lines of and . Lemma 5 shows that intersects at . Then there exist two unit vectors and such that (1) is a maximal segment contained in ;(2);(3).

Thus there exists a number such that . Since is the unique maximal segment contained in and parallel to , the lines and coincide.

From
it follows that . Thus
Therefore, and are two unit vectors having the desired properties.

Conversely, suppose that and are two unit vectors having these properties. Clearly,
Next we show that the lines and coincide. Since these two lines are parallel, we only need to show that they intersect. Clearly, there exists a number such that or, equivalently, . It follows that
Thus
In the rest of the proof we show that the intersection of the segments and is a nontrivial segment, which forces the set to be a nondegenerate segment. It suffices to show that is a relatively interior point of the segment .

On the one hand, we have
Thus, lies in the set . On the other hand, we have
It follows that is from the relative interior of .

Corollary 7. *Let and be two linearly independent vectors and be a positive number, , and . If is a non-degenerate segment , then there exist two unit vectors and such that is a maximal segment contained in and containing , intersects at , and
*

*Proof. *Let and be defined as in the first part of the proof of Theorem 6. Then we only need to show (19). By the first part of the proof of Theorem 6 and the triangle inequality, we have
from which (19) follows.

Corollary 8. *Let and be two linearly independent vectors, be a positive number, , and . If contains a segment such that the ray intersects the line at and that
**
then the set is a nondegenerate segment.*

*Proof. *Let and be two unit vectors such that , is a maximal segment contained in , and . By Theorem 6, we only need to show . In fact,
The proof is complete.

Now we have sufficient tools to prove the following theorem.

Theorem 9. *Pythagorean orthogonality on is circle-unique if and only if is strictly convex.*

*Proof. *If is strictly convex, then Corollary 7 shows that Pythagorean orthogonality is circle-unique.

Conversely, suppose that Pythagorean orthogonality is circle-unique. If is not strictly convex, then there exist two distinct unit vectors and in such that . Let be a number such that
Put . Since
the line intersects in a point . By interchanging and if necessary, we may assume that . Put . Then Corollary 8 implies that Pythagorean orthogonality on is not circle-unique, a contradiction.

In the end of this section we mention some result on the uniqueness of isosceles orthogonality, which was introduced by James in [1]: and are said to be* isosceles orthogonal* to each other if . This orthogonality is not homogeneous in general normed linear spaces. The line-existence, line-uniqueness, circle-existence, and circle-uniqueness for isosceles orthogonality can be defined in a similar way. The uniqueness of isosceles orthogonality attracted much attention; see [5, 7, 8]. It has been shown that line-uniqueness and circle-uniqueness of isosceles orthogonality are equivalent to strict convexity of the underlying space. If and are two linearly independent vectors and is the set of vectors isosceles orthogonal to , then the property whether is a singleton is determined by the length of the segment (possibly degenerated to a point) contained in and parallel to ; see [8]. As we have shown, if is not a singleton, then its structure is determined by a segment contained in which is not parallel to . Moreover, for different values of , the segment determining the structure of might be different.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The first named author is supported by the 973 program (Grant no. 2013CB834201), a foundation from the Ministry of Education of Heilongjiang Province (Grant no. 1251H013), the National Nature Science Foundation of China (Grant nos. 11371114 and 11171082), China Postdoctoral Science Foundation (Grant nos. 2012M520097 and 2013T60019), and the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

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#### Copyright

Copyright © 2014 Senlin Wu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.