Circle-Uniqueness of Pythagorean Orthogonality in Normed Linear Spaces

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 𝑋=(𝑋,‖⋅‖) a real normed linear space whose dimension is at least 2. The origin, unit ball, and unit sphere of 𝑋 are denoted by 𝑜, 𝐵_𝑋, and 𝑆_𝑋, respectively. When 𝑋 is two-dimensional, it is called a Minkowski plane. Its unit sphere 𝑆_𝑋 is then called the unit circle of 𝑋, and each homothetic copy of 𝑆_𝑋 is a circle. For two distinct points (or vectors) 𝑥 and 𝑦 in 𝑋, we denote by ⟨𝑥,𝑦⟩ the line passing through 𝑥 and 𝑦, by [𝑥,𝑦⟩ the ray starting from 𝑥 and passing through 𝑦, and by [𝑥,𝑦] the (nondegenerate) segment connecting 𝑥 and 𝑦. Moreover, 𝑋 is said to be strictly convex if 𝑆_𝑋 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 𝑥 and 𝑦 be two vectors in a real normed linear space. If 𝑥−𝑦²=‖𝑥‖²+𝑦², (1) then 𝑥 and 𝑦 are said to be Pythagorean orthogonal to each other (denoted by 𝑥⟂_𝑃𝑦). James showed that the following facts are equivalent:(1)𝑥,𝑦∈𝑋, 𝛼∈ℝ, 𝑥⟂_𝑃𝑦⇒𝑥⟂_𝑃𝛼𝑦;(2)𝑋 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 𝑥 and 𝑦 in 𝑋, there exists a number 𝛼 such that 𝑥⟂_𝑃𝛼𝑥+𝑦. That is, James proved that in each line parallel to the line ⟨−𝑥,𝑥⟩ there exists a vector that is Pythagorean orthogonal to 𝑥.

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 𝑋, where a binary relation ⟂ on 𝑋 is said to be line-unique if and only if for each 𝑥≠𝑜 and 𝑦∈𝑋 there exists a unique real number 𝛼 such that 𝑥⟂𝛼𝑥+𝑦. 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 𝑋 is strictly convex, which updates the knowledge about uniqueness of Pythagorean orthogonality.

For each 𝑥∈𝑋, we denote by 𝑃(𝑥) the set of points that are Pythagorean orthogonal to 𝑥; that is, 𝑃(𝑥)∶=𝑧∈𝑋∶𝑧⟂_𝑃𝑥=𝑧∶‖𝑧−𝑥‖²=‖𝑧‖²+‖𝑥‖². (2)

For two linearly independent vectors 𝑥 and 𝑦 we denote by 𝐿_𝑥,𝑦 the two-dimensional subspace of 𝑋 spanned by 𝑥 and 𝑦 and by 𝐻_𝑥,𝑦 the closed halfplane of 𝐿_𝑥,𝑦 bounded by the line ⟨−𝑥,𝑥⟩ and containing 𝑦.

Definition 1. Pythagorean orthogonality on 𝑋 is said to be circle-unique if, for each pair of linearly independent vectors 𝑥 and 𝑦 and each nonnegative real number 𝛼, there exists a unique vector 𝑧 in 𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝑃(𝑥).

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 𝐶₁∶=𝛾₁𝑆_𝑋+𝑐₁ and 𝐶₂∶=𝛾₂𝑆_𝑋+𝑐₂ be two circles in a Minkowski plane 𝑋, where 𝑐₁ and 𝑐₂ are two distinct points, and let 𝑝 and 𝑞 be the points of intersection of ⟨𝑐₁,𝑐₂⟩ and 𝐶₁. Then the set 𝐶₁∩𝐶₂ has one of the following forms: (1)𝐶₁∩𝐶₂=∅;(2)𝐶₁∩𝐶₂ is the union of two closed, disjoint segments (one or both of them may degenerate to a singleton) lying on the opposite sides of ⟨𝑐₁,𝑐₂⟩;(3)𝐶₁∩𝐶₂ is the union of two segments (one or both of them may degenerate to a singleton) with common point 𝑝 or 𝑞.

One can easily verify the following proposition.

Proposition 3. Let 𝑥 and 𝑦 be two points in 𝑋. Then 𝑥⟂_𝑃𝑦 if and only if 𝑦∈‖𝑥‖²+𝑦²𝑆_𝑋+𝑥. (3)

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 𝑥 and 𝑦 and each number 𝛼≥0, the set 𝑃∶=𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝑃(𝑥) (4) is a nonempty segment that may degenerate to a singleton.

Proof. We only consider the nontrivial case 𝛼>0. Clearly, 𝑃^′∶=𝐿_𝑥,𝑦∩𝛼𝑆_𝑋∩𝑃(𝑥)=𝐿_𝑥,𝑦∩𝛼𝑆_𝑋∩‖𝑥‖²+𝛼²𝑆_𝑋=𝐿+𝑥_𝑥,𝑦∩𝛼𝑆_𝑋∩𝐿_𝑥,𝑦∩‖𝑥‖²+𝛼²𝑆_𝑋+𝑥. (5) Since ‖𝑥‖²+𝛼²−𝛼<‖𝑥‖<‖𝑥‖²+𝛼²+𝛼, (6)𝑃^′ is not empty. It is also clear that 𝑃^′∩⟨−𝑥,𝑥⟩=∅. Thus, by Lemma 2, 𝑃^′ is the union of two closed, disjoint segments contained in 𝐿_𝑥,𝑦, one or both of which may degenerate to a singleton, lying in opposite halfplanes with respect to the line ⟨−𝑥,𝑥⟩. This completes the proof.

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

Lemma 5. Let 𝑋 be a Minkowski plane, 𝑥≠𝑜 a vector in 𝑋, and 𝛼,𝛽>0 two numbers such that 0<𝛽−𝛼<‖𝑥‖. (7) Then there are two common supporting lines of 𝛼𝐵_𝑋 and 𝛽𝐵_𝑋+𝑥 passing through the point 𝑝=(𝛼/(𝛼−𝛽))𝑥.

Proof. By the hypothesis of the lemma, 𝑝 is exterior to 𝛼𝐵_𝑋. Thus two supporting lines 𝑙₁ and 𝑙₂ of 𝛼𝐵_𝑋 can be drawn through 𝑝. In the following we show that these two lines are two common supporting lines of 𝛼𝐵_𝑋 and 𝛽𝐵_𝑋+𝑥.
Clearly, there exists a point 𝑢∈𝑆_𝑋 such that 𝑙₁ supports 𝛼𝐵_𝑋 at 𝛼𝑢. Put 𝜆₀=(𝛼−𝛽)/𝛼. Then 𝜆₀𝑝+1−𝜆₀𝛼𝑢=𝛼−𝛽𝛼⋅𝛼𝛽𝛼−𝛽𝑥+𝛼𝛼𝑢=𝛽𝑢+𝑥∈𝛽𝑆_𝑋+𝑥, (8) which implies that 𝑙₁ intersects 𝛽𝑆_𝑋+𝑥 in 𝛽𝑢+𝑥. Moreover, inf_𝜆∈ℝ𝑥−𝜆𝑝+(1−𝜆)𝛼𝑢=inf_𝜆∈ℝ𝑥−𝜆₀𝑝−1−𝜆₀𝛼𝑢+𝜆₀𝑝+1−𝜆₀𝛼𝑢−𝜆𝑝+(1−𝜆)𝛼𝑢=inf_𝜆∈ℝ𝛽𝑢−𝜆₀=𝛽−𝜆𝑝−𝛼𝑢𝛼inf_𝜆∈ℝ𝛼𝛽𝜆−𝜆₀𝛼𝑝+𝛽𝛽𝛼−𝜆−𝜆₀‖𝑢‖;𝛼𝑢=𝛽 (9) that is, the distance from 𝑥 to 𝑙₁ is 𝛽‖𝑢‖=𝛽. Thus 𝑙₁ is a common supporting line of 𝛼𝐵_𝑋 and 𝛽𝐵_𝑋+𝑥. In a similar way we can show that 𝑙₂ is also a common supporting line of these two discs.

Theorem 6. Let 𝑥 and 𝑦 be two linearly independent vectors, and let 𝛼 be a positive number, 𝛽=𝛼²+‖𝑥‖², and 𝑝=(1/(𝛼−𝛽))𝑥. Then 𝑃∶=𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝑃(𝑥) (10) is a nondegenerate segment if and only if there exist two unit vectors 𝑢 and 𝑣 in 𝐻_𝑥,𝑦 such that (1)[𝑢,𝑣] is a nondegenerate maximal segment contained in 𝑆_𝑋∩𝐻_𝑥,𝑦;(2)[𝑣,𝑢⟩  intersects  ⟨−𝑥,𝑥⟩ at 𝑝;(3)‖𝑢−𝑣‖/‖𝑢−𝑝‖>𝛽/𝛼−1 or, equivalently, ‖𝛼𝑢−𝛼𝑣‖>‖𝛽𝑢+𝑥−𝛼𝑢‖.

Proof. It is clear that 𝑝 is exterior to 𝐵_𝑋 and 𝛼𝑝 is exterior to 𝛼𝐵_𝑋.
First suppose that 𝑃 is a segment [𝑚,𝑛]. Then [𝑚,𝑛]⊂𝛼𝑆_𝑋∩𝛽𝑆_𝑋+𝑥, (11) which implies that ⟨𝑚,𝑛⟩ is one of the two common supporting lines of 𝛼𝐵_𝑋 and 𝛽𝐵_𝑋+𝑥. Lemma 5 shows that ⟨𝑚,𝑛⟩ intersects ⟨−𝑥,𝑥⟩ at 𝛼𝑝. Then there exist two unit vectors 𝑢 and 𝑣 such that (1)[𝑢,𝑣] is a maximal segment contained in 𝑆_𝑋∩𝐻_𝑥,𝑦;(2)[(1/𝛼)𝑚,(1/𝛼)𝑛]⊆[𝑢,𝑣];(3)𝑝∈[𝑣,𝑢⟩∩⟨−𝑥,𝑥⟩.
Thus there exists a number 𝜂∈(0,1) such that 𝑢=𝜂𝑝+(1−𝜂)𝑣. Since [𝛽𝑢+𝑥,𝛽𝑣+𝑥] is the unique maximal segment contained in (𝛽𝑆_𝑋+𝑥)∩𝐻_𝑥,𝑦 and parallel to [𝑚,𝑛], the lines ⟨𝑚,𝑛⟩=⟨𝛼𝑢,𝛼𝑣⟩ and ⟨𝛽𝑢+𝑥,𝛽𝑣+𝑥⟩ coincide.
From [𝑚,𝑛]⊆[𝛼𝑢,𝛼𝑣]⊂𝛼𝑆_𝑋,[𝑚,𝑛]⊆𝛽𝑢+𝑥,𝛽𝑣+𝑥⊂𝛽𝑆_𝑋,+𝑥,𝛽𝑢+𝑥−𝛼𝑝=𝛽𝜂𝑝+𝛽1−𝜂𝑣+𝑥−𝛼𝑝=𝛽𝜂𝑝+𝛽1−𝜂𝑣−𝛽𝑝=𝛽1−𝜂𝑣−𝑝𝛼𝑢−𝛼𝑝=𝛼1−𝜂𝑣−𝑝 (12) it follows that 𝛽𝑢+𝑥∈[𝛼𝑢,𝛼𝑣]⧵{𝛼𝑢,𝛼𝑣}. Thus 𝛼‖‖=‖‖>=.𝑢−𝑣𝛼𝑢−𝛼𝑣𝛽𝑢+𝑥−𝛼𝑢𝛽−𝛼𝑢−𝑝 (13) Therefore, 𝑢 and 𝑣 are two unit vectors having the desired properties.
Conversely, suppose that 𝑢 and 𝑣 are two unit vectors having these properties. Clearly, [𝛼𝑢,𝛼𝑣]⊂𝛼𝑆_𝑋∩𝐻_𝑥,𝑦,𝛽𝑢+𝑥,𝛽𝑣+𝑥⊂𝛽𝑆_𝑋+𝑥∩𝐻_𝑥,𝑦. (14) Next we show that the lines ⟨𝛼𝑢,𝛼𝑣⟩ and ⟨𝛽𝑢+𝑥,𝛽𝑣+𝑥⟩ coincide. Since these two lines are parallel, we only need to show that they intersect. Clearly, there exists a number 𝜂∈(0,1) such that 𝑢=𝜂𝑝+(1−𝜂)𝑣 or, equivalently, 𝛼𝑢=𝜂𝛼𝑝+(1−𝜂)𝛼𝑣. It follows that +.𝛽𝑢+𝑥=𝛽𝜂𝑝+𝛽1−𝜂𝑣+𝑥=𝛽𝜂𝑝+1−𝜂𝛽𝑣+𝑥+𝜂𝑥=𝜂𝛽𝑝+𝑥1−𝜂𝛽𝑣+𝑥=𝜂𝛼𝑝+1−𝜂𝛽𝑣+𝑥 (15) Thus 𝛼𝑝∈⟨𝛼𝑢,𝛼𝑣⟩∩⟨𝛽𝑢+𝑥,𝛽𝑣+𝑥⟩. (16) In the rest of the proof we show that the intersection of the segments [𝛼𝑢,𝛼𝑣] and [𝛽𝑢+𝑥,𝛽𝑣+𝑥] is a nontrivial segment, which forces the set 𝑃 to be a nondegenerate segment. It suffices to show that 𝛽𝑢+𝑥 is a relatively interior point of the segment [𝛼𝑢,𝛼𝑣].
On the one hand, we have ,.𝛽𝑢+𝑥−𝛼𝑝=1−𝜂𝛽𝑣+𝑥−𝛼𝑝=𝛽1−𝜂𝑣−𝑝𝛼𝑢−𝛼𝑝=𝛼1−𝜂𝑣−𝑝 (17) Thus, 𝛽𝑢+𝑥 lies in the set [𝛼𝑢,𝛼𝑣⟩⧵{𝛼𝑢}. On the other hand, we have ‖‖‖‖>=.𝛼𝑢−𝛼𝑣=𝛼𝑢−𝑣𝛽−𝛼𝑢−𝑝𝛽𝑢+𝑥−𝛼𝑢 (18) It follows that 𝛽𝑢+𝑥 is from the relative interior of [𝛼𝑢,𝛼𝑣].

Corollary 7. Let 𝑥 and 𝑦 be two linearly independent vectors and 𝛼 be a positive number, 𝛽=𝛼²+‖𝑥‖², and 𝑝=(1/(𝛼−𝛽))𝑥. If 𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝑃(𝑥) is a non-degenerate segment [𝑚,𝑛], then there exist two unit vectors 𝑢 and 𝑣 such that [𝑢,𝑣] is a maximal segment contained in 𝑆_𝑋∩𝐻_𝑥,𝑦 and containing [(1/𝛼)𝑚,(1/𝛼)𝑛], [𝑣,𝑢⟩ intersects ⟨−𝑥,𝑥⟩ at 𝑝, and ‖‖>‖𝑥‖𝑢−𝑣𝛼𝛽+1−𝛼. (19)

Proof. Let 𝑢 and 𝑣 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 ‖‖>=1𝛼𝑣−𝛼𝑢𝛽𝑢+𝑥−𝛼𝑢𝛽−𝛼𝑢+𝑥≥‖𝑥‖𝛽−𝛼𝛽−𝛼1−=‖𝑥‖𝛽−𝛼𝛽−𝛼=‖𝑥‖𝛽−𝛼−1+𝛼−𝛼²+‖𝑥‖², (20) from which (19) follows.

Corollary 8. Let 𝑥 and 𝑦 be two linearly independent vectors, 𝛼 be a positive number, 𝛽=𝛼²+‖𝑥‖², and 𝑝=(1/(𝛼−𝛽))𝑥. If 𝑆_𝑋∩𝐻_𝑥,𝑦 contains a segment [𝑢^′,𝑣^′] such that the ray [𝑣^′,𝑢^′⟩ intersects the line ⟨−𝑥,𝑥⟩ at 𝑝 and that 𝑢^′−𝑣^′>𝛽𝛼‖𝑥‖−1+𝛼, (21) then the set 𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝑃(𝑥) is a nondegenerate segment.

Proof. Let 𝑢 and 𝑣 be two unit vectors such that [𝑢^′,𝑣^′]⊆[𝑢,𝑣], [𝑢,𝑣] is a maximal segment contained in 𝑆_𝑋∩𝐻_𝑥,𝑦, and 𝑝∈[𝑣,𝑢⟩∩⟨−𝑥,𝑥⟩. By Theorem 6, we only need to show ‖𝛼𝑢−𝛼𝑣‖>‖𝛽𝑢+𝑥−𝛼𝑢‖. In fact, ‖𝑥‖𝛽𝛽𝑢+𝑥−𝛼𝑢≤𝛽−𝛼+=𝛼𝛼‖𝑥‖−1+𝛼<𝛼𝑢^′−𝑣^′‖‖=‖‖.≤𝛼𝑢−𝑣𝛼𝑢−𝛼𝑣 (22) The proof is complete.

Now we have sufficient tools to prove the following theorem.

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

Proof. If 𝑋 is strictly convex, then Corollary 7 shows that Pythagorean orthogonality is circle-unique.
Conversely, suppose that Pythagorean orthogonality is circle-unique. If 𝑋 is not strictly convex, then there exist two distinct unit vectors 𝑢 and 𝑣 in 𝑋 such that [𝑢,𝑣]⊂𝑆_𝑋. Let 𝛼>0 be a number such that ‖‖>√𝑢−𝑣𝛼²+1𝛼1−1+𝛼. (23) Put 𝛽=√𝛼²+1. Since 1=1𝛽−𝛼√𝛼²=√+1−𝛼𝛼²+1+𝛼>1, (24) the line ⟨𝑢,𝑣⟩ intersects 1/(𝛽−𝛼)𝑆_𝑋 in a point 𝑝. By interchanging 𝑢 and 𝑣 if necessary, we may assume that 𝑝∈[𝑣,𝑢⟩. Put 𝑥=(𝛼−𝛽)𝑝. Then Corollary 8 implies that Pythagorean orthogonality on 𝑋 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]: 𝑥 and 𝑦 are said to be isosceles orthogonal to each other if ‖𝑥+𝑦‖=‖𝑥−𝑦‖. 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 𝑥 and 𝑦 are two linearly independent vectors and 𝐼(𝑥) is the set of vectors isosceles orthogonal to 𝑥, then the property whether 𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝐼(𝑥) is a singleton is determined by the length of the segment (possibly degenerated to a point) contained in 𝑆_𝑋∩𝐿_𝑥,𝑦 and parallel to ⟨−𝑥,𝑥⟩; see [8]. As we have shown, if 𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝑃(𝑥) is not a singleton, then its structure is determined by a segment contained in 𝑆_𝑋 which is not parallel to ⟨−𝑥,𝑥⟩. Moreover, for different values of 𝛼, the segment determining the structure of 𝛼𝑆_𝑋∩𝐻_𝑥,𝑦∩𝑃(𝑥) 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.