# Mappings approximately preserving orthogonality in normed spaces

**ABSTRACT** We answer many open questions regarding approximately orthogonality preserving mappings (in Birkhoff–James sense) in normed spaces. In particular, we show that every approximately orthogonality preserving linear mapping (in Chmieliński sense) is necessarily a scalar multiple of an ε-isometry. Thus, whenever ε-isometries are close to isometries we obtain stability. An example is given showing that approximately orthogonality preserving mappings are in general far from scalar multiples of isometries, that is, stability does not hold.

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**ABSTRACT:**We discuss the problem of stability of relations preserving property for finite-dimensional normed spaces. Moreover, we obtain similar results for other orthogonality relations. Next, we show a counterexample for infinite-dimensional normed spaces.Linear Algebra and its Applications 11/2014; 460:125–135. · 0.98 Impact Factor -
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**ABSTRACT:**We assume that H,KH,K are the complex Hilbert spaces and a linear operator T∈B(H;K)T∈B(H;K) is given. The object of this paper is to define and consider the certain basis of H connected with the operator T∈B(H;K)T∈B(H;K). This basis will play an important role in describing the approximately orthogonality preserving operators. Next, we will give some application of our results. We give new results and more general ones. Namely, we answer a question, whether a linear operator which approximately preserves orthogonality must be close to an orthogonality preserving one.Journal of Mathematical Analysis and Applications 10/2014; · 1.12 Impact Factor

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Nonlinear Analysis 73 (2010) 3821–3831

Contents lists available at ScienceDirect

Nonlinear Analysis

journal homepage: www.elsevier.com/locate/na

Mappings approximately preserving orthogonality in normed spaces

Blaž Mojškerca, Aleksej Turnšekb,c,∗

aFaculty of Economics, University of Ljubljana, Kardeljeva ploščad 17, 1000 Ljubljana, Slovenia

bFaculty of Maritime Studies and Transport, University of Ljubljana, Pot pomorščakov 4, 6320 Portorož, Slovenia

cInstitute of Mathematics, Physics and Mechanics, Jadranska 19, 1000 Ljubljana, Slovenia

a r t i c l ei n f o

Article history:

Received 8 June 2010

Accepted 7 August 2010

MSC:

46B20

47B99

39B82

Keywords:

Birkhoff–James orthogonality

Approximate orthogonality

Orthogonality preserving mapping

Stability

Isometry

a b s t r a c t

We answer many open questions regarding approximately orthogonality preserving

mappings (in Birkhoff–James sense) in normed spaces. In particular, we show that

every approximately orthogonality preserving linear mapping (in Chmieliński sense) is

necessarily a scalar multiple of an ε-isometry. Thus, whenever ε-isometries are close

to isometries we obtain stability. An example is given showing that approximately

orthogonality preserving mappings are in general far from scalar multiples of isometries,

that is, stability does not hold.

© 2010 Elsevier Ltd. All rights reserved.

1. Introduction

Orthogonality preserving property was first introduced in the setting of inner product spaces. If H and K are real or

complex inner product spaces, then a mapping T : H → K is called orthogonality preserving, OP in short, if Tx and Ty

are orthogonal whenever x and y are orthogonal vectors. If the mapping does not preserve orthogonality exactly we are

dealing with approximately orthogonality preserving mapping, AOP in short. Recall that for ε ∈ [0,1), vectors x and y are

ε-orthogonal (x⊥εy) whenever |?x,y?| ≤ ε?x??y?. A mapping T : H → K is an ε-AOP mapping if

x ⊥ y ⇒ Tx⊥εTy.

Since OP linear mappings are precisely scalar multiples of isometries, the natural question is whether AOP mappings are

close to OP mappings, that is, to scalar multiples of isometries. The answer to this question is positive, that is, we have

stability in Ulam’s sense. More precisely, if T is an ε-AOP mapping, then there is an isometry U such that ?T − ?T?U? ≤

?

see [2] for isosceles-orthogonality, where it was shown that linear AOP mappings are scalar multiples of ε-isometries. (We

adopt a more general definition of ε-isometries. We say that a mapping T is an ε-isometry if (1 − δ1(ε))?x? ≤ ?Tx? ≤

(1+δ2(ε))?x?, where δ1(ε),δ2(ε) → 0 as ε → 0.) Thus, whenever ε-isometries are close to isometries we obtain stability.

However this is not always the case and an example of a normed space and of an AOP mapping were given such that

1 −

Theabovequestionscanbeaskedalsointhesettingofnormedspaceswithrespecttovariousdefinitionsoforthogonality;

?

1−ε

1+ε

?

?T?, and this estimate is sharp; see [1].

∗Corresponding author at: Faculty of Maritime Studies and Transport, University of Ljubljana, Pot pomorščakov 4, 6320 Portorož, Slovenia.

E-mail addresses: blaz.mojskerc@ef.uni-lj.si (B. Mojškerc), aleksej.turnsek@fmf.uni-lj.si (A. Turnšek).

0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved.

doi:10.1016/j.na.2010.08.007

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the mapping in question is far from all multiples of isometries. In the same paper the authors noted that the analogous

questions for the most important Birkhoff–James orthogonality remain open. The main purpose of this note is to answer

these questions.

This paper is organized as follows. In the next section we recall some basic facts about Birkhoff–James orthogonality,

smooth points, and differentiability in normed spaces. We explain two different definitions of approximate orthogonality

given by Chmieliński and Dragomir. In Section 3 we investigate how the two notions of approximate orthogonality relate to

each other and prove the main result of the note: every AOP linear mapping in Chmieliński sense is a scalar multiple of an

ε-isometry. In the last section we give an example of a normed space and of an AOP mapping such that the stability fails.

2. Some preliminaries

Recall that a vector x in a normed space X is orthogonal in a Birkhoff–James sense [3,4] to a vector y in X (x ⊥ y) if

for every scalar λ ∈ K, where K is either R or C, we have ?x + λy? ≥ ?x?. For inner product spaces this last definition is

equivalent to the usual notion of orthogonality. However in general normed spaces Birkhoff–James orthogonality is neither

symmetric nor additive but it is homogeneous. It is known that a linear mapping T between normed spaces X and Y is OP

if and only if it is a scalar multiple of an isometry; see [5] for the case of real normed spaces and [6] for the complex case.

Therefore OP mappings between normed spaces are of the same form as OP mappings between inner product spaces. So the

naturalquestioniswhetherAOPmappingsareclosetoOPmappingsasisthecaseininnerproductspaces.Howeverwehave

yet to define what approximately orthogonal means. There are two proposed ways, see [7,8], but first we need some facts

about smoothness and differentiability in normed spaces. Recall that a support functional Fxat a nonzero x ∈ X is a norm

one functional such that Fx(x) = ?x?. Recall also that 0 ?= x ∈ X is a smooth point if there is a unique support functional at

x. It is well known that x ?= 0 is a smooth point if and only if the norm is Gateaux differentiable at x, that is, the following

limit

?x + ty? − ?x?

t

exists for all y ∈ X. When this limit exists it is equal to the real part of the (unique) support functional at x, see [9]. It is also

easy to see that Fx(y) = 0 implies x ⊥ y. Indeed, this follows from ?x + λy? ≥ |Fx(x + λy)| = Fx(x) = ?x?. Conversely, if

x ⊥ y then there is a support functional Fxat x such that Fx(y) = 0. To see this define a linear functional f on a linear span of

x and y by f(αx+βy) = α?x?. Then|f(αx+βy)| = ?αx? ≤ ?αx+βy?, hence?f? = 1. Now use the Hahn–Banach theorem

to obtain the desired support functional. Moreover, these last two observations show that if x is smooth then Fx(y) = 0 is

equivalent to x ⊥ y.

In an inner product space support functional Fxis given by Fx(y) =

of approximate orthogonality can be rewritten as |Fx(y)| ≤ ε?y?. This fact motivated Chmieliński to define approximate

orthogonality for smooth points in normed spaces in the same way, i.e., if x ∈ X is smooth, then x⊥εy if |Fx(y)| ≤ ε?y?.

However, if x is smooth, the condition |Fx(y)| ≤ ε?y? is equivalent to ?x+λy?2≥ ?x?2− 2ε?x??λy? for all λ ∈ K, see [7].

Thus it is natural to define approximate orthogonality for all points as follows. Let ε ∈ [0,1). Then

x⊥εy

On the other hand it is easy to see that in an inner product space the condition |?x,y?| ≤ ε?x??y? is equivalent to

?x + λy? ≥

orthogonality. Let ε ∈ [0,1). Then

x⊥ε

Therefore we can study AOP mappings T : X → Y satisfying

x ⊥ y ⇒ Tx⊥εTy

or

lim

t→0,t∈R

(1)

1

?x??y,x?. Therefore the definition |?x,y?| ≤ ε?x??y?

if ?x + λy?2≥ ?x?2− 2ε?x??λy? (λ ∈ K).

√

1 − ε2?x? for all λ ∈ K. This motivated Dragomir, see [8], to give the following definition of approximate

(DI)

ny

if ?x + λy? ≥ (1 − ε)?x?(λ ∈ K).

(DII)

(I)

x ⊥ y ⇒ Tx⊥ε

nTy.

(II)

3. Main results

The next proposition shows that approximate orthogonality of (DI) implies that of (DII). (Of course not with the same ε.)

ny, where δ = 1 −√

Proof. We may assume that y ?= 0. The function λ ?→ ?x + λy? attains its minimum at, say, λ0(there may be of course

many such points). Then x + λ0y ⊥ y, hence ?x + λ0y + λy? ≥ ?x + λ0y? for all λ ∈ K. Choose λ = −λ0to get

?x? ≥ ?x + λ0y? ≥ |?x? − |λ0|?y?|.

Proposition 3.1. Let x,y ∈ X. Then x⊥εy ⇒ x⊥δ

1 − 4ε.

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This implies |λ0| ≤

?x + λy?2≥ ?x + λ0y?2≥ ?x?2− 2ε|λ0|?x??y? ≥ (1 − 4ε)?x?2

Thus x⊥δ

The following lemma shows that linear AOP mappings are automatically bounded and bounded from below. We give

the proof for AOP mappings (I). Exactly the same proof works for mappings (II), just a slightly different bound is obtained,

namely ?Tx? ≥

Lemma 3.2. Let X and Y be normed spaces, ε ∈?

Proof. Let us first show that T is bounded. Clearly we can assume T ?= 0. Take two arbitrary unit vectors x and y and note

that 3x + y ?⊥ x. Let α =

the kernel of F3x+y, so

3x + y ⊥ 3x − α(3x + y).

Write F3x+y(y) = θ?y?eiφ,θ ∈ [0,1], and note that ?3x + y? ≥ |?3x? − ?y?| = 2. Then

|F3x+y(3x)| = |F3x+y(3x + y − y)| = |?3x + y? − θ?y?eiφ|

≥ |?3x + y? − θ?y?| = ?3x + y? − θ?y? ≥ 1.

So 1 ≤ |F3x+y(3x)| ≤ 3 and since 2 ≤ ?3x + y? ≤ 4, we conclude that

1

4≤ |α| ≤

From (2) it follows that

2?x?

?y?. Using this last estimate, it follows from x⊥εy that

(λ ∈ K).

ny, where δ = 1 −√

1 − 4ε and the proof is completed.

?

1−ε

15−3ε?T??x?.

0,1

2

?

, and suppose that a linear mapping T : X → Y satisfies x ⊥ y ⇒

1−2ε

Tx⊥εTy. Then T is bounded and bounded from below, ?Tx? ≥

15+18ε?T??x?.

F3x+y(3x)

?3x+y?, where F3x+yis a support functional at the point 3x + y. Then 3x − α(3x + y) belongs to

(2)

3

2.

(3)

3Tx + Ty⊥ε3Tx − α(3Tx + Ty),

hence for all λ ∈ K

?(3 + 3λ − 3αλ)Tx + (1 − αλ)Ty?2≥ ?3Tx + Ty?2− 2ε?3Tx + Ty??λ(3Tx − α(3Tx + Ty))?.

Take λ =

????

????

Since ?3Tx + Ty? ≥ |3?Tx? − ?Ty?|, we get

9?Tx?2− 6?Tx??Ty? + ?Ty?2≤ 144?Tx?2+ 2ε(45?Tx?2+ 18?Tx??Ty? + ?Ty?2).

If Tx = 0, then choose y ∈ X such that Ty ?= 0. But then (4) implies ?Ty?2≤ 2ε?Ty?2, which is a contradiction because

ε <

Divide both sides of the inequality (4) by ?Tx?2and denote z =?Ty?

(1 − 2ε)z2− (6 + 36ε)z − (135 + 90ε) ≤ 0.

Then z ≤ z1, where z1is bigger of the two roots of the preceding quadratic equation. Thus

3 + 18ε + 12

1 − 2ε

Since ε2−1

3 + 18ε + 12

1 − 2ε

1

αto obtain

????

3

αTx

2

≥ ?3Tx + Ty?2− 2ε?3Tx + Ty?

????

????

3

αTx − 3Tx − Ty

????,

????

hence by the triangle inequality and (3) it follows that

?3Tx + Ty?2≤

3

αTx

????

2

+ 2ε?3Tx + Ty?

3

αTx − 3Tx − Ty

≤ 144?Tx?2+ 2ε(3?Tx? + ?Ty?)(15?Tx? + ?Ty?).

(4)

1

2.

?Tx?. Rewrite (4) as

z ≤

?

ε2−ε

2+ 1

.

2ε < 0, we get the estimate

z ≤=

15 + 18ε

1 − 2ε

.

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From this we obtain

?Ty? ≤

15 + 18ε

1 − 2ε

?Tx?.

Unit vectors x and y are arbitrary, so we conclude that T is bounded. Furthermore, on the left side take the supremum over

all unit vectors to obtain ?Tx? ≥

Thenexttheoremshowsthatmappingssatisfying(I)arescalarmultiplesofε-isometries.Itsproofisbasedon[6,Theorem

3.1]. We will also need the following facts about the smooth points in normed spaces.

Let νnbe the Lebesgue measure on Rn. We recall the following consequence of Rademacher’s theorem on differentiable

points of Lipschitz functions, see [10, Theorem 3.2.5].

1−2ε

15+18ε?T?, that is, T is bounded from below.

?

Proposition 3.3. Every norm on Rnis Gateaux differentiable νn-a.e. on Rn.

We will also need the following lemma, see [6, Lemma 2.4].

Lemma 3.4. Let ?·?beanynormonK2andlet D ⊆ K2beasetofallnon-smoothpoints.Thenthereexistsapathγ : [0,2] → K2

of the form:

?(1,tζ),

for some ζ ∈ K,ζ ?= 0,1, so that ν{t : γ(t) ∈ D} = 0.

Theorem 3.5. Let X,Y be normed spaces, ε ∈?

(1 − 16ε)?T??x? ≤ ?Tx? ≤ ?T??x?.

γ(t) :=

t ∈ [0,1];

t ∈ [1,2],(1,(2 − t)ζ + (t − 1)),

0,1

2

?

and T : X → Y a linear mapping satisfying x ⊥ y ⇒ Tx⊥εTy. Then

Proof. By Lemma 3.2 T is bounded, so we may and do assume that ?T? = 1. Let x,y ∈ X be unit vectors. We will show

that |?Tx? − ?Ty?| ≤ 16ε. If x and y are linearly dependent this is true. Let x and y be linearly independent and let M be the

complex linear span of x and y. On M we define a second norm by ?u?T = ?Tu?. By Lemma 3.2 T is injective, so ? · ?Tis

indeed a norm.

Let ∆ be the set of all those points in M at which at least one of the norms ? · ? or ? · ?Tis not Gateaux differentiable.

Now take 0 ?= u ∈ M \ ∆ and denote by Fuand Gusupport functionals at u for the norms ? · ? and ? · ?T, respectively. Let

w ∈ M be arbitrary and writew = γu+v,γ ∈ K andv ∈ kerFu. Then Fu(w) = γ?u? and Gu(w) = γ?u?T+Gu(v). Hence

Gu(w) − λ(u)Fu(w) = Gu(v),

where λ(u) =?u?T

Sincev ∈ kerFuit follows that u ⊥ v, hence Tu⊥εTv. By assumption u is smooth for?·?T, hence the condition Tu⊥εTv

means |Gu(v)| ≤ ε?v?T. Thus

|Gu(w) − λ(u)Fu(w)| = |Gu(v)| ≤ ε?v?T= ε?Tv? ≤ ε?v?

= ε?w − γu? ≤ ε(?w? + ?γu?) ≤ 2ε?w?,

where in the last inequality we have used the fact that ?γu? ≤ ?γu + v? = ?w? because of u ⊥ v. Since λ(u) is real, it

follows that

?u?.

|gu(w) − λ(u)fu(w)| ≤ 2ε?w?,

where fuand guare the real parts of Fuand Gu, respectively.

Now consider the linear isomorphism L : C2→ M, (α,β) ?→ αx + β(y − x) and set D = L−1(∆). Then D is the set of

those points (α,β) ∈ C2at which at least one of the functions (α,β) ?→ ?L(α,β)? or (α,β) ?→ ?L(α,β)?Tis not Gateaux

differentiable. Both these functions are norms in C2= R4, hence by Proposition 3.3 ν4(D) = 0. Let γ : [0,2] → C2be the

path obtained in Lemma 3.4. Then Γ : [0,2] → M defined by

Γ(t) =

u ∈ M \ ∆,w ∈ M,

(5)

Lγ(t)

?Lγ(t)?,

t ∈ [0,2],

is a path from x to y such that ?Γ(t)? = 1 for all t ∈ [0,2] and ν{t : Γ(t) ∈ ∆} = 0. It is also easy to see that t ?→ ?Lγ(t)?

and t ?→ ?Lγ(t)?Tare Lipschitz functions and, therefore, absolutely continuous. Then, since ?Lγ(t)?Tand ?Lγ(t)? are

bounded and bounded from below,

?Γ(t)?T=?Lγ(t)?T

?Lγ(t)?

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3825

is also absolutely continuous and

ν{t : Γ?(t) does not exist} = ν{t : ?Lγ(t)??does not exist} = 0.

Since fΓ(t)(Γ?(t)) =

|gΓ(t)(Γ?(t))| ≤ 2ε?Γ?(t)?

ν-a.e. on [0,2], thus??d

0

0

and, since t ?→ ?Γ(t)?Tis absolutely continuous, it follows that

d

dt?Γ(t)? = 0 (because t ?→ ?Γ(t)? is a constant function), it follows from (5) that

dt?Γ(t)?T

????≤

??≤ 2ε?Γ?(t)? ν-a.e. on [0,2]. Integrating both sides we get

dt?Γ(t)?T

0

????

?2

d

dt?Γ(t)?Tdt

?2

????

d

????dt ≤ 2ε

?2

?Γ?(t)?dt,

|?Tx? − ?Ty?| = |?Γ(2)?T− ?Γ(0)?T| ≤ 2ε

?2

0

?Γ?(t)?dt = 2ε · l(Γ),

(6)

where l(Γ) is the length of the curve Γ (see [10, Theorem 3.4.6] for more details). To estimate the length of Γ we

proceed as follows. Write Γ = Γ1∪ Γ2, where Γ1(t) =

γ2(t) = (1,(2 − t)ζ + (t − 1)), t ∈ [1,2], see Lemma 3.4. Then t ?→ Lγ1(t) = x + tζ(y − x) is a line segment joining

x and x + ζ(y − x) and this line segment lies in the two-dimensional real normed space spanned by x and ζ(y − x) (since

ζ ?= 0, the space is two-dimensional). Hence the length of Γ1is at most half of the circumference of the unit sphere (circle)

which is by [11, Theorem 4I] at most 4. Similarly, t ?→ Lγ2(t) = (2 − t)(x + ζ(y − x)) + (t − 1)y is a line segment joining

x+ζ(y−x) and y and lies in the two-dimensional real normed space spanned by x+ζ(y−x) and y (since ζ ?= 1 the space

is two-dimensional). Thus the length of Γ2is also at most 4, and so the length of Γ is at most 8. From (6) it follows that

Lγ1(t)

?Lγ1(t)?, γ1(t) = (1,tζ), t ∈ [0,1] and Γ2(t) =

Lγ2(t)

?Lγ2(t)?,

|?Tx? − ?Ty?| ≤ 16ε.

This yields ?Ty? − 16ε ≤ ?Tx?. Taking the supremum of the left side over all unit vectors y, we get 1 − 16ε ≤ ?Tx? for any

unit vector x ∈ X. Hence

(1 − 16ε)?x? ≤ ?Tx? ≤ ?x?

for any vector and the proof is completed.

?

Remark 3.1. If K = R we can obtain a better result

(1 − 8ε)?T??x? ≤ ?Tx? ≤ ?T??x?.

The reason is that in the real case for a path we can choose just one line segment. Indeed, let γ : [0,1] → R2be a

line segment joining (1,0) and (0,1) given by γ(t) = (1 − t,t) and let L : R2→ M, (α,β) ?→ αx + βy. Then

γ(t) ∈ R2is smooth if and only if Lγ(t) ∈ M is smooth. Suppose that ν{t ∈ [0,1] : γ(t) is not smooth} = a ?= 0.

Then ν2{λγ(t) : λ ∈ [0,1],γ(t) is not smooth} =

we obtain the estimate

a

2, a contradiction with Proposition 3.3. As in the proof of Theorem 3.5

|?Tx? − ?Ty?| ≤ 2ε · l(Γ)

and l(Γ) ≤ 4. The rest is clear.

Thefollowingpropositionshowsthatascalarmultipleofanε-isometryisAOPmappinginsense(II).Recallthatamapping

T is an ε-isometry if (1 − δ1(ε))?x? ≤ ?Tx? ≤ (1 + δ2(ε))?x?, where δ1(ε),δ2(ε) → 0 as ε → 0.

Proposition 3.6. Let X,Y be normed spaces and T : X → Y a scalar multiple of a linear ε-isometry. Then

x ⊥ y ?⇒ Tx⊥δ(ε)

where δ(ε) =δ1(ε)+δ2(ε)

Proof. Let x ⊥ y and T = cU, where U is an ε-isometry and c ∈ K. Then

?Tx + λTy? = ?T(x + λy)? = ?cU(x + λy)? ≥ |c|(1 − δ1(ε))?x + λy?

≥ |c|(1 − δ1(ε))?x? ≥ |c|1 − δ1(ε)

n

Ty,

1+δ2(ε)

.

1 + δ2(ε)?Ux? =

→ 0 as ε → 0, the proof is completed.

1 − δ1(ε)

1 + δ2(ε)?Tx?.

Since

1−δ1(ε)

1+δ2(ε)= 1 −δ1(ε)+δ2(ε)

1+δ2(ε)

andδ1(ε)+δ2(ε)

1+δ2(ε)

?

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Remark 3.2. ThelatterpropositionshowsinfactthatDragomir’snotionofapproximateorthogonalityisstablewithrespect

toasmallchangeofthenorm.However,asshowninthenextexample,Chmieliński’sapproximateorthogonalityisnotstable,

even in finite-dimensional spaces, with respect to a small change of the norm. In other words, ε-isometries in general do

not satisfy (I). Therefore the two definitions of approximate orthogonality are not equivalent in general.

Example 3.7. LetX = (R2,?·?1), that is,?(x1,x2)? = |x1|+|x2|, and let T : X → X be the rotation for an angleε ∈ (0,1)

in the anticlockwise direction, so Tx = (x1cosε − x2sinε,x1sinε + x2cosε). Then

?Tx? ≤ |x1|cosε + |x2|sinε + |x1|sinε + |x2|cosε

= (|x1| + |x2|)(sinε + cosε)

=

SinceT−1x = (x1cosε+x2sinε,−x1sinε+x2cosε),in thesamewayasbefore weget?T−1x? ≤ (1+ε)?x?,which implies

?Tx? ≥ (1 + ε)−1?x? ≥ (1 − ε)?x?. Thus T satisfies

(1 − ε)?x? ≤ ?Tx? ≤ (1 + ε)?x?,

hence T is an ε-isometry. We will show that T does not satisfy (I). Let x = (1,0) and y = (0,1). Then x ⊥ y and

Tx = (cosε,sinε) and Ty = (−sinε,cosε). Since both cosε and sinε are nonzero, Tx is a smooth point and the support

functional at Tx is given by FTx(u1,u2) = u1+ u2. Then

?FTx(Ty)

≥

Thus |FTx(Ty)| ≥ (1 − 2ε)?Ty?.

In some cases with additional assumptions we obtain the equivalence of the two definitions of approximate

orthogonality. Recall that a normed space X is called uniformly smooth if limit (1) exists uniformly in the set {(x,y) : ?x? =

?y? = 1}. For a normed space X ?= {0} a modulus of convexity or modulus of rotundity is a function δX: [0,2] → [0,1]

defined by the formula

?

The space X is uniformly convex or uniformly rotund if δX(ε) > 0 whenever 0 < ε ≤ 2. The characteristic or coefficient of

convexity of a normed space X is the number

ε0= ε0(X) = sup{ε ≥ 0 : δX(ε) = 0}.

Proposition 3.8. Let X be uniformly smooth and let ε ∈ [0,2δX∗(1)). Then x⊥ε

Proof. Recall first that X is uniformly smooth if and only if X∗is uniformly convex, see [9, p. 207], and that the modulus of

convexity is continuous on[0,2) and strictly increasing on[ε0,2], see [12, Lemma 5.1], so in our caseδX∗ is continuous and

strictly increasing on [0,2). Take unit vectors x,y ∈ X such that x⊥ε

λ. Let minλ?x + λy? = ?x + λ0y?. Then x + λ0y ⊥ y, or equivalently, Fx+λ0y(y) = 0, where Fx+λ0yis the support functional

at x + λ0y. Then

Fx+λ0y(x) = Fx+λ0y(x + λ0y) = ?x + λ0y? ≥ 1 − ε.

Hence

?Fx+λ0y+ Fx? ≥ |Fx+λ0y(x) + Fx(x)| ≥ 1 − ε + 1 = 2 − ε.

Since X∗is uniformly convex, ?Fx+λ0y− Fx? ≤ δ−1

because of the condition ε < 2δX∗(1)), which ends the proof.

By Propositions 3.1 and 3.8 the following corollary is immediate.

√

1 + sin2ε?x? ≤√

1 + 2ε?x? ≤ (1 + ε)?x?.

?Ty?

?2

=(cosε − sinε)2

1 − 2ε

(cosε + sinε)2=

1 + 2ε≥ (1 − 2ε)2.

1 − sin2ε

1 + sin2ε

δX(ε) = inf1 − ?1

2(x + y)? : ?x? = ?y? = 1,?x − y? ≥ ε

?

.

ny ⇒ x⊥ηy, where η = δ−1

X∗

?ε

2

?

.

ny, which is equivalent to ?x + λy? ≥ 1 − ε for every

X∗

?ε

2

?

, hence |Fx(y)| = |Fx+λ0y(y) − Fx(y)| ≤ δ−1

?

X∗

?ε

2

?, (δ−1

X∗

?ε

2

?

< 1

Corollary 3.9. Let X be a uniformly smooth normed space. Then (DI) and (DII) are equivalent.

Proposition 3.10. Let X,Y be normed spaces, Y uniformly smooth, ε ∈

satisfying x ⊥ y ⇒ Tx⊥ε

?

Proof. From Proposition 3.8 we get Tx⊥ηTy, where η = δ−1

?

0,2δY∗?1

2

??

and T : X → Y a linear mapping

nTy. Then

??

1 − 16δ−1

Y∗

?ε

2

?T??x? ≤ ?Tx? ≤ ?T??x?.

Y∗

?ε

2

?<

1

2. Use Theorem 3.5 to complete the proof.

?

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4. ε-isometries, stability and coned unit ball

If a linear mapping T : X → Y is close to an isometry, say ?T − U? ≤ ε, where U is an isometry, then from

|?Tx? − ?x?| = |?Tx? − ?Ux?| ≤ ?Tx − Ux? ≤ ε?x?

it follows that T is an ε-isometry. In some cases the converse also holds, that is ε-isometries are close to isometries. More

precisely, let us say that a pair (X,Y) of normed spaces has the stability of linear isometries (SLI) property if there exists a

function δ : [0,1) → R+satisfying limε→0δ(ε) = 0 such that whenever T is an ε-isometry, then there exists an isometry

U such that ?T − U? ≤ δ(ε). The function δ depends on X and Y but not on T, that is, the approximation is uniform.

This is known to be true for example if X and Y are finite-dimensional normed spaces, Hilbert spaces, see [13], Lpspaces,

1 < p < ∞, see [14], spaces of continuous functions if the domain space is metrizable, see [15]. In general the converse

does not hold, see [16] or Example 4.5. The class of all pairs of normed spaces with the (SLI) property is exactly the class A

from [2, section 4]. (A proposed new name is just an attempt to be more descriptive). For a more detailed discussion of class

A (and B) please see [2, section 4].

Turning to the context of linear AOP mappings the question about stability is: If T : X → Y is a linear AOP mapping, is

there a mapping U which is a multiple of an isometry such that

?T − U? ≤ δ(ε)?T?,

FromTheorem3.5andtheaboveconsiderationsweobtainthefollowingtheoremshowingthatthe(SLI)propertyimplies

stability in sense (I).

δ(ε) → 0 if ε → 0.

(7)

Theorem 4.1. Suppose that a pair of normed spaces (X,Y) has the (SLI) property and let T : X → Y be a linear mapping

approximately preserving orthogonality in sense (I). Then T is close to a multiple of an isometry, that is, (7) holds.

Proposition 3.6 implies the following partial converse of the preceding theorem.

Proposition 4.2. Let XandY benormedspacessuchthatthestabilityoftheorthogonalitypreservingpropertyinsense(II)holds.

Then a pair (X,Y) has the (SLI) property.

If the spaceY is uniformly smooth, we obtain the equivalence of the (SLI) property and stability in both senses (I) and (II).

Theorem 4.3. Let X and Y be normed spaces and let Y be uniformly smooth. Then the following assertions are equivalent:

(i) A pair (X,Y) has the (SLI) property.

(ii) Stability of the orthogonality preserving property in sense (I) holds.

(iii) Stability of the orthogonality preserving property in sense (II) holds.

Proof. The implication (i)⇒ (ii) follows from Theorem 4.1 and (iii)⇒ (i) follows from Proposition 4.2. Finally, (ii)⇒ (iii) is

a consequence of Proposition 3.8.

?

It is worth mentioning another special case not covered by the above results.

Proposition 4.4. Let dimX,dimY < ∞. If T : X → Y approximately preserves orthogonality in sense (II), it is close to a

multiple of an isometry, that is, (7) holds.

Proof. The proof is by contradiction. Suppose that there is a sequence of positive numbers εn → 0 and a sequence of

εn-AOP mappings in sense (II) Tn : X → Y, none of which is close to CIso, the set of scalar multiples of isometries,

i.e., d(Tn,CIso) > δ0?Tn? for each n. Then d(Tn/?Tn?,CIso) > δ0> 0. Since {Tn/?Tn?} is a bounded sequence in a finite-

dimensionalspace,ithasaconvergentsubsequence,sayTin/?Tin? → U.ButU preservesorthogonality,henceby[6,Theorem

3.1] U is a scalar multiple of an isometry which contradicts d(U,CIso) ≥ δ0.

Finally, we are tempted to pose the following questions:

Questions: Let X and Y be normed spaces and T : X → Y a linear AOP mapping in sense (II). Is it true that T is a scalar

multiple of an ε-isometry? Does stability in sense (I) always imply the (SLI) property?

We finish this paper with an example of a separable Banach space X, the idea comes from [17], and a linear mapping

T : X → X which is an ε-isometry, AOP mapping in both senses (I) and (II) but far from all isometries, hence stability in

general does not hold.

?

Example 4.5. Let ? · ? be the usual norm on a real ?2and let {en : n ∈ N} be the standard basis for this space. Roughly

speaking, we need a ‘‘nice’’ new norm on ?2in order to have explicit expressions for support functionals, but on the other

hand we do not want many isometries so that the (SLI) property will fail. The idea is to obtain a new norm on ?2by creating

a modified unit ball by ‘‘adding’’ cones at the points ±ento the standard unit ball. In general we add a cone at an arbitrary

point as follows. Take x0∈ ?2with ?x0? = 1 and 0 < λ < 1 to denote xλ= λ−1x0. Now add ±xλto the standard unit ball

and convexify this union. The new norm ? · ?λis called a λ-cone at x0.

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The unit sphere of ? · ?λ,S = {y : ?y?λ = 1}, contains line segments, i.e., there are points y,z ∈ S so that

?ty + (1 − t)z?λ = 1 for each 0 ≤ t ≤ 1. The points y and z are said to be endpoints of a maximal line segment in the

unit sphere of ? · ?λ, if

L = {ty + (1 − t)z : 0 ≤ t ≤ 1} ⊂ S

and if L ⊂ L?= {su + (1 − s)v : 0 ≤ s ≤ 1} ⊂ S then L = L?.

For 0 < α <π

6we define

αn=α

λn= cosαn,

xn= λ−1

Consider the λn-cone at en. Observe that if ?z? is different from ?z?λn, then the angle between z and one of ±enis less than

αn. Also note that the vector w is such that xnand xn+ w are the endpoints of a maximal line segment in the unit sphere of

this cone, if and only if ?w? = tanαnand the angle between w and enis 90° − αn.

Let|||·||| be the equivalent norm obtained by adding all theλn-cones at±ento the unit ball of?·?. Note thatλ−1

that λn-s are chosen so ±xn-s do not ‘‘see’’ each other, i.e., there is no line in S between two distinct ±xn-s. It is easy to see

that |||x||| ≤ ?x? ≤ λ−1

||| · |||, then |||w||| ≤ ?w? = tanαn. Furthermore, w can be chosen so that this maximal line segment has a maximal possible

length, that is |||w||| = ?w?. Indeed,

w = tanαn(sinαnen+ cosαnen+1)

is an example of such a vector, with |||w||| = ?w? = tanαn. Because of symmetry diametrically opposed (with respect to

the cone) vector (line segment) has the same maximal length. These facts will be needed in showing that isometries map xn

to ±xn.

Let S be an isometry in ||| · |||. Note that the only proper lines in S are contained in λn-cones. Choose two diametrically

opposed maximal line segments (with respect to the cone) with maximal possible length having one endpoint in x1. Since S

is a linear isometry, it maps lines inS into lines inS. These lines share one endpoint, so their images must share an endpoint

due to connectedness. Additionally, images of these two lines are not contained in a single maximal line segment since their

underlying vectors are linearly independent and S is injective. The two lines have the maximal length in S, therefore their

imagemustbecontainedinλ1-coneandtheircommonendpointmustbe±x1,henceSx1= ±x1.Nowtaketwodiametrically

opposed maximal line segments with maximal possible length having one endpoint in x2. Due to the length of these lines,

their images may only be contained in λ1- or λ2-cone. But if they were contained in λ1-cone, that would be a contradiction

withinjectivityofS,hencewiththesamereasoningasbeforewegetSx2= ±x2.ContinuinginthiswayweobtainSxn= ±xn

for all n. (Incidentally this shows that all isometries in this space are surjective).

Now take i ∈ N and define a linear mapping T : ?2→ ?2by

Tei= ei+1,

Tei+1= ei,

Tej= ej

We will show that T approximately preserves orthogonality, hence it is an ε-isometry, and that there is no isometry close

to T.

The last part of the statement is easy to see. Take any isometry S in ||| · |||. Then (T − S)ei= ei+1± ei,

?ei± ei+1? =λ1

λi

2n

nen.

n

↓ 1 and

1|||x|||. If w is such that xnand xn+w are the endpoints of a maximal line segment in the unit sphere of

for j ?= i,i + 1.

1

λi|||ei||| = 1, hence

|||T − S||| ≥ λ−1

i|||ei+1± ei||| ≥λ1

λi

√

2.

Now denote with C+

n= {x ∈ S : ?(x,−en) <

x. Note that when x ?∈ Cnfor n ∈ N,Fxcan be written as Fx= ?·,

Fx? =

Case 1: x ?∈ Ci∪ Ci+1,Tx ?∈ Ci∪ Ci+1.

If x ⊥ y, then Fx(y) =

and i + 1), hence Tx ⊥ Ty.

Case 2: x ?∈ Ci∪ Ci+1,Tx ∈ Ci∪ Ci+1.

n

= {x ∈ S : ?(x,en) <

α

2n}. Define Cn = R+C+

α

2n}, where ?(x,en) is the angle between x and en, and similarly,

n∪ R+C−

x

?x??. On the other hand, if x ∈ Cn\ R+{±xn}, then Fxequals

C−

n. With Fxwe denote a support linear functional at a point

?

·,

x?

?x??

?

, where x?is some endpoint of a maximal line segment containing x. We consider the following cases:

?

y,

x

?x?

?

= 0. But FTx(Ty) =

?

Ty,

Tx

?Tx?

?

=

?

y,

x

?x?

?

= 0 (since T acts as a transposition of indices i

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3829

By definition of T note that Tx ?∈ R+{±ei,±ei+1}. Take y so that x ⊥ y. Then Fx(y) =

of a maximal line segment containing Tx. Then by using ?Ty,Tx? = ?y,x? = 0, we get

?

Estimate

????

hence Tx⊥εTy.

Case 3: x ∈ (Ci∪ Ci+1) \ R+{±ei,±ei+1},Tx ?∈ Ci∪ Ci+1.

Since x ∈ Ci∪ Ci+1, there exists x?which is an endpoint of a maximal line segment containing x. Let x ⊥ y. Then

Fx(y) = Fx?(y) =

This implies

?

=

?x??

Estimate

????

= ?Ty?

?x?

≤ ?Ty?

|||x|||

≤ λ−1

2i

due to x ∈ Ci∪ Ci+1, so Tx⊥εTy, since |λi− 1| can be arbitrarily small.

Case 4: x,Tx ∈ (Ci∪ Ci+1) \ R+{±ei,±ei+1}.

Take x ⊥ y. There exists x?such that x?is the endpoint of a maximal line segment containing x. Then Fx(y) = Fx?(y) =

?

?

=

?x???

Estimate

????

????

≤

|||Tx|||

≤ tanα

≤ tanα

|||Tx?|||

?

y,

x

?x?

?

= 0. Let x?be an endpoint

FTx(Ty) = Fx?(Ty) =

Ty,

x?

?x??

?

=

?

Ty,

x?

?x??−

Tx

|||Tx|||

?

.

|FTx(Ty)| =

?

Ty,

x?

?x??−

2i≤ λ−1

Tx

|||Tx|||

1|||Ty|||tanα

?????≤ ?Ty?

????

x?

?x??−

Tx

|||Tx|||

????

≤ ?Ty?tanα

2i,

?

y,

x?

?x??

?

= 0. The fact Tx ?∈ Ci∪ Ci+1implies that Tx??∈ Ci∪ Ci+1. Hence FTx?(Ty) =

?

Ty,

Tx?

?Tx??

?

= 0.

FTx(Ty) =

Ty,

Tx

?Tx?

x

?x?−

?

=

x?

?

y,

?

x

?x?

?

−

?

y,

x?

?x??

?

?

y,

.

|FTx(Ty)| =

?

y,

x

?x?−

????

x?

?x??

x

|||x|||−

?|||x|||

λ−1

?????≤ ?y?

|||x|||+

?x?− 1

i|λi− 1| + tanα

????

x

?x?−

x

|||x|||−

+

x?

?x??

x?

?x??

????

|||x|||

??x?

x

????

?

????

x

|||x|||−

?

x?

?x??

????

?

1|||Ty|||

?

,

y,

FTx= Fx??. Then

FTx(Ty) = Fx??(Ty) =

x?

?x??

?

= 0. The same goes for Tx, there exists x??which is the endpoint of a maximal line segment containing Tx, so

Ty,

x??

?x???

Ty,

?

Tx?

?Tx??

=

?

?

Ty,

x??

?x???

?

?

−

?

y,

x?

?x??

Tx?

?Tx??

?

?

Ty,

x??

?

−

?

=

Ty,

x??

?x???−

?

.

|FTx(Ty)| ≤ ?Ty?

Rewrite

x??

?x???−

x??

?x???−

Tx?

?Tx??

????.

Tx Tx?

?Tx??

????=

????

x??

?x???−

x??

?x???−

2i+

|||Tx|||+

Tx

Tx

|||Tx|||−

????+

|||Tx|||−

Tx

|||Tx|||−

Tx?

?Tx??

????

????????

|||Tx?|||+

Tx?

Tx

|||Tx|||−

Tx?

Tx?

?Tx??

Tx?

|||Tx?|||−|||Tx?|||

????+?Tx??

????

????

Tx

?Tx??

????1 −|||Tx?|||

Tx?

|||Tx?|||

????

2i+

????

|||Tx?|||· ?Tx??

????.

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Combining these estimates we get

|FTx(Ty)| ≤ λ−1

so Tx⊥εTy.

Case 5: x ∈ {±xi,±xi+1},Tx ∈ R+{±ei,±ei+1}.

First note that the unit sphere of ||| · ||| is not smooth at points {±xn}∞

a one-to-one correspondence between support functionals of xnand hyperplanes, tangent to the unit sphere at xn. On the

other hand, there is a one-to-one correspondence between tangent hyperplanes at xnand the points, say z, of a unit sphere

in ? · ? with ?z? ?= |||z|||. Indeed, take the hyperplane that is tangent on the unit sphere of ||| · ||| at xnand take the line which

is orthogonal to the hyperplane and that goes through the origin. The intersection between the line and the unit sphere in

?·? is the required point z = (zi). Observe that zn?= 0 and that at every such point we can find a unique support functional

in ? · ?, since the unit sphere of ? · ? is smooth. Every such linear functional is of the form Fz= ?·,z?. In order for Fzto be a

support functional at xnin |||·|||, we only need to multiply Fzwith an appropriate constant α so that αFz(xn) = α?xn,z? = 1.

Since xn= λ−1

Now we need the following general observation.

Observation: Let X be any normed space and let Fxbe a support functional at a not necessarily smooth point x ∈ X. Then

|Fx(y)| ≤ ε?y? implies x⊥εy. To see this, write

?x? = |Fx(x)| = |Fx(x + λy) − Fx(λy)| ≤ |Fx(x + λy)| + |Fx(λy)|

≤ ?x + λy? + ε?λy?,

hence ?x + λy? ≥ ?x? − ε?λy?. Consider two cases:

If ?x? − ε?λy? ≥ 0, square both sides to obtain

?x + λy?2≥ ?x?2− 2ε?x??λy? + ε2?λy?2≥ ?x?2− 2ε?x??λy?.

If ?x? − ε?λy? < 0, then ?x? − 2ε?λy? < 0, hence

?x?(?x? − 2ε?λy?) = ?x?2− 2ε?x??λy? < 0.

But then ?x + λy?2≥ ?x?2− 2ε?x??λy?.

Now we can proceed with the case 5. Let x = xiand Tx =

functional Gi,zfor xiso that Gi,z(y) = 0. Now take an arbitrary support functional at xi+1, say Gi+1,z?, and use the inequality

|z?

????

≤ λ−1

To estimate |yi| remember that ?y,z? = 0 and write

|yi| = |?y,ei? − ?y,z? + ?y,z?| ≤ |?y,z?| + |?y,ei− z?|

≤ ?y?tanα

Similarly we consider the other possible cases to see that T is AOP.

Case 6: x ∈ Cn\ R+{xn,−xn},n ?∈ {i,i + 1}.

Take x so that |||x||| = 1 and let x ⊥ y. We may assume that x is contained in a maximal line segment with endpoints xn

and x?. Again, Fx(y) = Fx?(y) = ?y,x?? = 0. Note that Txn= xnand that because of symmetry Tx?is an endpoint of a maximal

line segment containing Tx. Hence FTx(Ty) = FTx?(Ty) = ?Ty,Tx?? = ?y,x?? = 0, which shows that T preserves orthogonality.

Case 7: x ∈ {xn,−xn},n ?∈ {i,i + 1}.

Consider the case x = xnand take y so that x ⊥ y. Then there exists a support functional Fxat x such that Fx(y) = 0. By

the arguments from Case 5 there exists z ∈ Cnwith ?z? = 1 such that Fx=λn

symmetry Tz ∈ Cn, hence the functional G = α?·,Tz? is also a support functional at x(=Tx) for some appropriate constant

α. But then G(Ty) = α?Ty,Tz? = α?y,z? = 0, hence Tx ⊥ Ty and T preserves orthogonality.

1|||Ty|||

?

tanα

2i+ tanα

2i+ λ−1

i|1 − λi|

?

,

n=1. By definition of a support functional there is

nen, we get α =λn

zn. Therefore every support functional at xnin ||| · ||| can be written as Gn,z=λn

zn?·,z?.

λi+1

λixi+1. Take y such that xi⊥ y. Then there exists a support

i+1| ≥ λi+1to estimate

|Gi+1,z?(Ty)| =

λi+1

z?

i+1

?Ty,z??

????≤???Ty,z?? − ?Ty,ei+1? + ?Ty,ei+1???

α

2i+1+ |yi|.

≤?|?Ty,z?− ei+1?| + |yi|?≤ ?Ty??z?− ei+1? + |yi|

1|||Ty|||tan

2i= ?Ty?tanα

2i≤ λ−1

1|||Ty|||tanα

2i.

zn?·,z?. This yields ?y,z? = 0. Because of the

Acknowledgements

We thank Professor Bojan Magajna and an anonymous referee for their helpful suggestions. This research was supported

in part by the Ministry of Science and Education of Slovenia.

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