# Circle-Uniqueness of Pythagorean Orthogonality in Normed Linear Spaces

Authors: Senlin Wu; Xinjian Dong; Dan Wang
Journal: Journal of Function Spaces
 (2014)

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## 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.


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## Body

## 1. Introduction

We denote byX=(X,∥·∥) a real normed linear space whose dimension is at least 2. Theorigin,unit ball, andunit sphere of X are denoted by o, BX, and SX, respectively. When X is two-dimensional, it is called aMinkowski plane. Its unit sphere SX is then called theunit circle of X, and each homothetic copy of SX is acircle. For two distinct points (or vectors) x and y in X, we denote by 〈x,y〉 theline passing through x and y, by [x,y〉 theray starting from x and passing through y, and by [x,y] the(nondegenerate) segment connecting x and y. Moreover, X is said to bestrictly convex if SX 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
(1)∥x-y∥2=∥x∥2+∥y∥2,
then x and y are said to bePythagorean 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 nothomogeneous in general normed linear spaces. Among other things, James proved theline-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 beline-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 eachx∈X, we denote by P(x) the set of points that are Pythagorean orthogonal to x; that is,
(2)P(x)≔{z∈X:z⊥Px}={z:∥z-x∥2=∥z∥2+∥x∥2}.For two linearly independent vectorsx and y we denote by Lx,y the two-dimensional subspace of X spanned by x and y and by Hx,y the closed halfplane of Lx,y bounded by the line 〈-x,x〉 and containing y.Definition 1.
Pythagorean orthogonality onX is said to becircle-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 αSX∩Hx,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]).
LetC1:=γ1SX+c1 and C2:=γ2SX+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.
Letx and y be two points in X. Then x⊥Py if and only if
(3)y∈(∥x∥2+∥y∥2SX+x).First we show that Pythagorean orthogonality has thecircle-existence property. More precisely, we show the following proposition.Proposition 4.
For each pair of linearly independent vectorsx and y and each number α≥0, the set
(4)P:=αSX∩Hx,y∩P(x)
is a nonempty segment that may degenerate to a singleton.Proof.
We only consider the nontrivial caseα>0. Clearly,
(5)P′:=Lx,y∩(αSX)∩P(x)=Lx,y∩(αSX)∩(∥x∥2+α2SX+x)=(Lx,y∩(αSX))∩(Lx,y∩(∥x∥2+α2SX+x)).
Since
(6)∥x∥2+α2-α<∥x∥<∥x∥2+α2+α,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 Lx,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.
LetX be a Minkowski plane, x≠o a vector in X, and α,β>0 two numbers such that
(7)0<β-α<∥x∥.
Then there are two common supporting lines of αBX and βBX+x passing through the point p=(α/(α-β))x.Proof.
By the hypothesis of the lemma,p is exterior to αBX. Thus two supporting lines l1 and l2 of αBX can be drawn through p. In the following we show that these two lines are two common supporting lines of αBX and βBX+x.
Clearly, there exists a pointu∈SX such that l1 supports αBX at αu. Put λ0=(α-β)/α. Then
(8)λ0p+(1-λ0)αu=α-βα·αα-βx+βααu=βu+x∈βSX+x,
which implies that l1 intersects βSX+x in βu+x. Moreover,
(9)inf⁡λ∈ℝ∥x-(λp+(1-λ)αu)∥=inf⁡λ∈ℝ∥x-λ0p-(1-λ0)αu+λ0p+(1-λ0)αu-(λp+(1-λ)αu)∥=inf⁡λ∈ℝ∥βu-(λ0-λ)(p-αu)∥=βαinf⁡λ∈ℝ∥αβ(λ-λ0)p+αβ(βα-(λ-λ0))αu∥=β∥u∥;
that is, the distance from x to l1 is β∥u∥=β. Thus l1 is a common supporting line of αBX and βBX+x. In a similar way we can show that l2 is also a common supporting line of these two discs.Theorem 6.
Letx 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
(10)P:=αSX∩Hx,y∩P(x)
is a nondegenerate segment if and only if there exist two unit vectors u and v in Hx,y such that (1)
[u,v] is a nondegenerate maximal segment contained in SX∩Hx,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 thatp is exterior to BX and αp is exterior to αBX.
First suppose thatP is a segment [m,n]. Then
(11)[m,n]⊂αSX∩(βSX+x),
which implies that 〈m,n〉 is one of the two common supporting lines of αBX and βBX+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 SX∩Hx,y;(2)
[(1/α)m,(1/α)n]⊆[u,v];(3)
p∈[v,u〉∩〈-x,x〉.
Thus there exists a numberη∈(0,1) such that u=ηp+(1-η)v. Since [βu+x,βv+x] is the unique maximal segment contained in (βSX+x)∩Hx,y and parallel to [m,n], the lines 〈m,n〉=〈αu,αv〉 and 〈βu+x,βv+x〉 coincide.
From(12)[m,n]⊆[αu,αv]⊂αSX,[m,n]⊆[βu+x,βv+x]⊂βSX+x,βu+x-αp=βηp+β(1-η)v+x-αp=βηp+β(1-η)v-βp=β(1-η)(v-p),αu-αp=α(1-η)(v-p)
it follows that βu+x∈[αu,αv]∖{αu,αv}. Thus
(13)α∥u-v∥=∥αu-αv∥>∥βu+x-αu∥=(β-α)∥u-p∥.
Therefore, u and v are two unit vectors having the desired properties.
Conversely, suppose thatu and v are two unit vectors having these properties. Clearly,
(14)[αu,αv]⊂αSX∩Hx,y,[βu+x,βv+x]⊂(βSX+x)∩Hx,y.
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
(15)βu+x=βηp+β(1-η)v+x=βηp+(1-η)(βv+x)+ηx=η(βp+x)+(1-η)(βv+x)=ηαp+(1-η)(βv+x).
Thus
(16)αp∈〈αu,αv〉∩〈βu+x,βv+x〉.
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(17)βu+x-αp=(1-η)(βv+x-αp)=β(1-η)(v-p),αu-αp=α(1-η)(v-p).
Thus, βu+x lies in the set [αu,αv〉∖{αu}. On the other hand, we have
(18)∥αu-αv∥=α∥u-v∥>(β-α)∥u-p∥=∥βu+x-αu∥.
It follows that βu+x is from the relative interior of [αu,αv].Corollary 7.
Letx and y be two linearly independent vectors and α be a positive number, β=α2+∥x∥2, and p=(1/(α-β))x. If αSX∩Hx,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 SX∩Hx,y and containing [(1/α)m,(1/α)n], [v,u〉 intersects 〈-x,x〉 at p, and
(19)∥u-v∥>∥x∥α+1-βα.Proof.
Letu 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
(20)∥αv-αu∥>∥βu+x-αu∥=(β-α)∥u+1β-αx∥≥(β-α)|1-∥x∥β-α|=(β-α)(∥x∥β-α-1)=∥x∥+α-α2+∥x∥2,
from which (19) follows.Corollary 8.
Letx and y be two linearly independent vectors, α be a positive number, β=α2+∥x∥2, and p=(1/(α-β))x. If SX∩Hx,y contains a segment [u′,v′] such that the ray [v′,u′〉 intersects the line 〈-x,x〉 at p and that
(21)∥u′-v′∥>βα-1+∥x∥α,
then the set αSX∩Hx,y∩P(x) is a nondegenerate segment.Proof.
Letu and v be two unit vectors such that [u′,v′]⊆[u,v], [u,v] is a maximal segment contained in SX∩Hx,y, and p∈[v,u〉∩〈-x,x〉. By Theorem 6, we only need to show ∥αu-αv∥>∥βu+x-αu∥. In fact,
(22)∥βu+x-αu∥≤β-α+∥x∥=α(βα-1+∥x∥α)<α∥u′-v′∥≤α∥u-v∥=∥αu-αv∥.
The proof is complete.Now we have sufficient tools to prove the following theorem.Theorem 9.
Pythagorean orthogonality onX is circle-unique if and only if X is strictly convex.Proof.
IfX is strictly convex, then Corollary 7 shows that Pythagorean orthogonality is circle-unique.
Conversely, suppose that Pythagorean orthogonality is circle-unique. IfX is not strictly convex, then there exist two distinct unit vectors u and v in X such that [u,v]⊂SX. Let α>0 be a number such that
(23)∥u-v∥>α2+1α-1+1α.
Put β=α2+1. Since
(24)1β-α=1α2+1-α=α2+1+α>1,
the line 〈u,v〉 intersects 1/(β-α)SX 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 beisosceles 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 αSX∩Hx,y∩I(x) is a singleton is determined by the length of the segment (possibly degenerated to a point) contained in SX∩Lx,y and parallel to 〈-x,x〉; see [8]. As we have shown, if αSX∩Hx,y∩P(x) is not a singleton, then its structure is determined by a segment contained in SX which is not parallel to 〈-x,x〉. Moreover, for different values of α, the segment determining the structure of αSX∩Hx,y∩P(x) might be different.



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