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 εisometry. Thus, whenever εisometries 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: In a normed space we introduce an exact and approximate orthogonality relation connected with “norm derivatives” r¢±{\rho^{\prime}_{\pm}} . We also consider classes of linear mappings preserving (exactly and approximately) this kind of orthogonality. Mathematics Subject Classification (2010)Primary 46B2046C50Secondary 39B82 KeywordsOrthogonalityapproximate orthogonalityorthogonality preserving mappingsnorm derivativeAequationes Mathematicae 80(1):4555. · 0.42 Impact Factor  [Show abstract] [Hide abstract]
ABSTRACT: We survey mainly recent results on the two most important orthogonality types in normed linear spaces, namely on Birkhoff orthogonality and on isosceles (or James) orthogonality. We lay special emphasis on their fundamental properties, on their differences and connections, and on geometric results and problems inspired by the respective theoretical framework. At the beginning we also present other interesting types of orthogonality. This survey can also be taken as an update of existing related representations.Aequationes Mathematicae 83(12). · 0.42 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 εisometry. Thus, whenever εisometries 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
εorthogonal (x⊥εy) whenever ?x,y? ≤ ε?x??y?. A mapping T : H → K is an εAOP 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 εAOP mapping, then there is an isometry U such that ?T − ?T?U? ≤
?
see [2] for isoscelesorthogonality, where it was shown that linear AOP mappings are scalar multiples of εisometries. (We
adopt a more general definition of εisometries. We say that a mapping T is an εisometry if (1 − δ1(ε))?x? ≤ ?Tx? ≤
(1+δ2(ε))?x?, where δ1(ε),δ2(ε) → 0 as ε → 0.) Thus, whenever εisometries 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.
Email addresses: blaz.mojskerc@ef.unilj.si (B. Mojškerc), aleksej.turnsek@fmf.unilj.si (A. Turnšek).
0362546X/$ – 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
εisometry. 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?. Thenf(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− 2ελ0?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)arescalarmultiplesofεisometries.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 νna.e. on Rn.
We will also need the following lemma, see [6, Lemma 2.4].
Lemma 3.4. Let ?·?beanynormonK2andlet D ⊆ K2beasetofallnonsmoothpoints.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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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)?
νa.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)? νa.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 twodimensional real normed space spanned by x and ζ(y − x) (since
ζ ?= 0, the space is twodimensional). 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 twodimensional real normed space spanned by x+ζ(y−x) and y (since ζ ?= 1 the space
is twodimensional). 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.
ThefollowingpropositionshowsthatascalarmultipleofanεisometryisAOPmappinginsense(II).Recallthatamapping
T is an εisometry 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 εisometry. Then
x ⊥ y ?⇒ Tx⊥δ(ε)
where δ(ε) =δ1(ε)+δ2(ε)
Proof. Let x ⊥ y and T = cU, where U is an εisometry and c ∈ K. Then
?Tx + λTy? = ?T(x + λy)? = ?cU(x + λy)? ≥ c(1 − δ1(ε))?x + λy?
≥ c(1 − δ1(ε))?x? ≥ c1 − 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 finitedimensional spaces, with respect to a small change of the norm. In other words, εisometries 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? ≤ x1cosε + x2sinε + x1sinε + x2cosε
= (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 εisometry. 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. εisometries, 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 εisometry. In some cases the converse also holds, that is εisometries 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 εisometry, 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 finitedimensional 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
εnAOP 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 εisometry? 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 εisometry, 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 λcone 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 λncone 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 theλncones at±ento the unit ball of?·?. Note thatλ−1
that λns are chosen so ±xns do not ‘‘see’’ each other, i.e., there is no line in S between two distinct ±xns. 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 λncones. 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
imagemustbecontainedinλ1coneandtheircommonendpointmustbe±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 λ2cone. But if they were contained in λ1cone, 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 εisometry, 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
1x. 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
λiei = 1, hence
T − S ≥ λ−1
iei+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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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
1Tytanα
?????≤ ?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
iλi− 1 + tanα
????
x
?x?−
x
x−
+
x?
?x??
x?
?x??
????
x
??x?
x
????
?
????
x
x−
?
x?
?x??
????
?
1Ty
?
,
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 onetoone correspondence between support functionals of xnand hyperplanes, tangent to the unit sphere at xn. On the
other hand, there is a onetoone 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.
1Ty
?
tanα
2i+ tanα
2i+ λ−1
i1 − 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
1Tytan
2i= ?Ty?tanα
2i≤ λ−1
1Tytanα
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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