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A New Approximate Birkhoff Orthogonality Type

Chuanjiang Zhou, Qi Liu, Yongjin Li∗
Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, P. R. China

1090871744@qq.com, liuq325@mail2.sysu.edu.cn, stslyj@mail.sysu.edu.cn
∗Correspondence: stslyj@mail.sysu.edu.cn

Abstract. In this note, we introduce a new approximate Birkhoff orthogonality type and give a char-acterization for inner product spaces using the approximate orthogonality. We show some generalproperties of the approximate Birkhoff orthogonality type as well as applications. In particular, westudy the relationship between the new approximate Birkhoff orthogonality type and other approx-imate orthogonality types that have been defined before. Furthermore we study the approximatepreserving mapping and give some properties.

1. Introduction
One of the important ideas playing a fundamental role in geometry of normed spaces is the con-cept of orthogonality. Many mathematicians have introduced different types of orthogonality for thenormed linear spaces, cf. [2, 20, 24]. In 1934 [23], the first orthogonality type:Roberts orthogonalitywas introduced by Roberts. After that in 1935 [5], Birkhoff introduced one of the most importantorthogonality types: x is said to be Birhoff orthogonal to y (x ⊥B y) if ‖x +ty‖≥‖x‖ for all t ∈R.Then James in 1945 [15] introduced the Pythagorean orthogonality and isosceles orthogonality:

x is said to be isosceles orthogonal to y (x ⊥I y) if ‖x + y‖ = ‖x − y‖. There are also otherorthognality types related to norm limit such as ρ-orthogonality and g-orthogonality [10, 18].Let X be inner product spaces (X,〈·|·〉), all the orthogonality types are equivalent to x ⊥ y orequivalently, 〈x|y〉 = 0. In inner product spaces a natural way to generalize orthogonality is todefine the approximate orthogonality by: x ⊥� y if and only if |〈x|y〉| ≤ �‖x‖‖y‖, x,y ∈ X [9, 26].Inspired by the approximate orthogonality, Dragomir [13] gave the definition of the approximateBirkhoff orthogonality x� ⊥B y : ‖x + ty‖ ≥ (1 − �)‖x‖ for all t ∈ R. It is easy to see thatthis type of approximate orthogonality is equivalent to ⊥� in inner product spaces [13]. After thatJacek Chmieliński [21] introduced the approximate Birkhoff orthogonality x ⊥�B y : ‖x + ty‖2 ≥
‖x‖2− 2�‖x‖‖ty‖ for all t ∈R, the approximate isosceles orthogonality [11] x ⊥�I y : |‖x + y‖2−

Received: 20 Feb 2022.
Key words and phrases. Birkhoff orthogonality; isosceles-orthogonality; approximate orthogonality; orthogonalitypreserving mappings. 1

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 2
‖x − y‖2| ≤ 4�‖x‖‖y‖ for all t ∈ R, and x� ⊥I y : |‖x + y‖−‖x − y‖| ≤ �‖x + y‖‖x − y‖ forall t ∈R. Many meaningful results have been found about approximate orthogonality through thetireless efforts of mathematicians, see [12, 14].In this paper we will introduce a new approximate Birkhoff othogonality type and investigateits properties and its relationship with other approximate orthogonality types. Moreover we give acharacterization of inner product spaces by approximate orthogonality types and some propertiesabout approximately orthogonality preserving mapping.Throughout the paper we will only consider normed spaces with dimX ≥ 2, we use 〈·|·〉 denotingthe inner product and (·|·) denoting the angle between x and y , i,e, in inner product spaces
(x,y) =

‖x+y‖2−‖x‖2−‖y‖2
2‖x‖‖y‖ .

2. Approximate Birkhoff Orthogonality ⊥B�
Let � ∈ [0, 1) and x,y be elements of inner product spaces X, we have the vertical relationship:

x ⊥ y ⇐⇒ |〈x|y〉| = 0. To generalize the orthogonality, it is natural to consider the approximateorthogonality (�-orthogonality: x ⊥� y) defined by:
x ⊥� y ⇐⇒ |〈x|y〉| ≤ �‖x‖‖y‖ ⇐⇒ |cos(x,y)| ≤ �.

Now we consider normed spaces, many mathematicians have introduced different types of orthogo-nality to represent orthogonality such as Birkhoff orthogonality [5] and Isosceles orthogonality [15].As an extension for the orthogonality, approximately orthogonality such as approximate Birkhofforthogonalty [13, 21]:
x� ⊥B y ⇐⇒ ‖x + ty‖≥ (1 − �)‖x‖ t ∈R.

x ⊥�B y ⇐⇒ ‖x + ty‖
2 ≥‖x‖2 − 2�‖x‖‖ty‖ t ∈R,

and approximate isosceles orthogonality [11]:
x ⊥�I y ⇐⇒ |‖x + y‖

2 −‖x −y‖2| ≤ 4�‖x‖‖y‖ t ∈R.

x� ⊥I y ⇐⇒ |‖x + y‖−‖x −y‖|≤ �‖x + y‖‖x −y‖ t ∈R,have been defined and studied. Notice that the definition of � ⊥B is quadratic while the definitionof ⊥�B is of first order, we give a new approximate Birkhoff orthogonality type:
x ⊥B� y ⇐⇒ ‖x + ty‖≥‖x‖− �‖ty‖,

which is also of first order but different from � ⊥B. It is easy to see that the inequality is alwayscorrect if t ≥ ‖x‖
�‖y‖.

Example 2.1. Let X = (R2,‖ · ‖1), assume that x = (1, 0), y = (z, 1 − z), z ∈ [0, 1). If we
want x� ⊥B y, then the inequality ‖x + ty‖≥ (1 −�)‖x‖ should hold for all t ∈R, thus we have:

‖(1 + tz,t(1 −z))‖≥ 1 − � t ∈R.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 3
If t ≥ 0, the inequality is always correct. For t < 0, if 1 + tz ≥ 0, we have:

1 + tz − t + tz ≥ 1 − �.

thus z ≤ 1
2
− �
2t
. By 1 + tz ≥ 0, we get z ≤ 1

2−�. If 1 + tz < 0, similarly we need t ≤ �−2. Since
z ≤ 1

2−�, from 1 + tz < 0, we get t ≤ �− 2. Thus x
� ⊥B y iff z ≤ 12−�.

On the other hand, if we want x ⊥B� y, the inequality ‖x + ty‖≥‖x‖− �‖ty‖ should hold for
all t ∈R, thus we have:

‖(1 + tz,t(1 −z))‖≥ 1 − �|t| t ∈R.

If t ≥ 0, the inequality is also always correct. For t < 0, if 1 + tz ≥ 0, we have:

1 + tz − t + tz ≥ 1 + �t.

Thus z ≤ 1+�
2
. Similarly we can get ‖(1 + tz,t(1 −z))‖≥ 1 − �|t| for 1 + tz < 0 if z ≤ 1+�

2
.

Thus x ⊥B� y iff z ≤ 1+�2 . We have the result that ⊥B� is not always equivalent to
� ⊥B in X.

Since the definition of approximate Birkhoff orthogonality comes from the notion of approximateorthogonality ⊥� in inner product spaces, it is natural to require the equivalence: x ⊥B� y ⇐⇒
x ⊥� y in inner product spaces. Now we give some basic properties about ⊥B� before prove theequivalence.

Proposition 2.2. Let X be normed spaces, then ⊥B� is homogeneous., this is

x ⊥B� y implies αx ⊥B� βy (x,y ∈ X,α,β ∈R).

Proof. Since x ⊥B� y , we have ‖x + ty‖ ≥ ‖x‖− �‖ty‖ for any t ∈ R. If α = 0 ,αx ⊥B� βy is
always correct; if α 6= 0, we have

‖αx + tβy‖ = |α|‖x +
β

α
ty‖≥ |α|{‖x‖− �‖

β

α
ty‖} = ‖ax‖− �‖tβy‖.

Thus αx ⊥B� βy . �

Recall that the limits [16] :
N±(x; y) = lim

n→±∞
‖nx + y‖−|nx‖ = lim

h→0±
‖x + hy‖−‖x‖

h
,

exist and satisfy the weakened linearity condition [4]. x,y are said to be Gateaux differentiable [1]at 0 if N−(x,y) = N+(x,y). Moreover we have [16]:
N±(x; rx + sy) = r‖x‖ + s ·N±(x; y), f or s ≥ 0 and all r.

We then give a characterization of x ⊥B� y using the definition of N±(x,y).

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 4
Proposition 2.3. Let X be normed spaces, then

x ⊥B� y if and only if N+(x,y) + �‖y‖≥ 0 ≥ N−(x,y) − �‖y‖.

Proof. Let x ⊥B� y and t ∈R\{0} then
‖x + ty‖−‖x‖

|t|
≥ �‖y‖.

Let t → 0+, we have N+(x,y) ≥−�‖y‖. Similarly, let t → 0−, we have N−(x,y) ≤ �‖y‖. To sum
up, N+(x,y) + �‖y‖≥ 0 ≥ N−(x,y) − �‖y‖.

Conversely, if N+(x,y) ≥−�‖y‖, for ∀η > 0, there ∃δ such that if 0 < t ≤ δ, we have:
‖x + ty‖−‖x‖

t
≥−(� + η)‖y‖,

or equivalently
‖x + ty‖−‖x‖≥−t(� + η)‖y‖ f or t ∈ (0,δ].

Because of the convexity of ‖x + ty‖, we have

‖x + ty‖−‖x‖≥−t(� + η)‖y‖ f or t > 0.

Let δ → 0, we have
‖x + ty‖−‖x‖≥−�‖ty‖ f or t > 0.

Similarly, using N−(x,y) ≤ �‖y‖, we have:‖x + ty‖−‖x‖≥ t(�)‖y‖ f or t < 0.
If t = 0,‖x + ty‖−‖x‖≥ �‖ty‖ is obvious. To conclude, we have:

‖x + ty‖−‖x‖≥ �‖ty‖ f or t ∈R.

Thus x ⊥B� y . �

To verify the validity of the new approximate Birkhoff orthogonality, we have the followingproposition:
Proposition 2.4. Let X be normed spaces, we have:

x ⊥B� y if and only if x ⊥� y.

Proof. Since x ⊥B� y , for 0 < t ≤
‖x‖
�‖y‖ we have

‖x + ty‖≥‖x‖− �‖ty‖.

Square both sides we get

‖x‖2 + t2‖y‖2 + 2t(x,y) ≥‖x‖2 + �2t2‖y‖2 − 2�‖x‖‖ty‖.

Thus (1−�2)t‖y‖≥−2‖x‖(� + cos(x,y)) When t tends to 0, (1−�2)t‖y‖ tends to 0, so we have

� + cos(x,y) ≥ 0 ⇐⇒ cos(x,y) ≥−�.

Similarly for 0 > t ≥− ‖x‖
�‖y‖ , we have cos(x,y) ≤ �, thus x ⊥

� y .

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 5
Conversely, if |cos(x,y)| ≤ �, we have

‖x + ty‖−‖x‖≥ �‖ty‖ f or |t| ∈ [0,
‖x‖
�‖y‖

].

On the other hand, ‖x + ty‖−‖x‖≥ �‖ty‖ is always correct for |t| ≥ ‖x‖
�‖y‖ . To conclude,

‖x + ty‖−‖x‖≥ �‖ty‖ f or t ∈R.

Thus x ⊥B� y . �

From N±(x; rx + sy) = r‖x‖ + s ·N±(x; y), f or s ≥ 0 and all r, we have the following:
Proposition 2.5. In the normed space X, if x ⊥B� y , then we have x ⊥B� rx + sy for s ≥ 0, r

satisfying

�(‖rx + sy‖− s‖y‖) ≥ r‖x‖≥ �(s‖y‖−‖rx + sy‖).

Proof. Since x ⊥B� y, we have N+(x,y) ≥−�‖y‖ and N−(x,y) ≤ �‖y‖, so

N+(x,rx + sy) ≥ r‖x‖ + (−s�‖y‖) ≥−�‖rx + sy‖.

Similarly we have N−(x,rx + sy) ≤ �‖rx + sy‖, thus x ⊥B� rx + sy. �

Let X be normed spaces, it is known that [16] for any x,y ∈ X there exists a real number asuch that x ⊥B ax + y, moreover, such a number satisfies |a| ≤ ‖y‖‖x‖. On this basis, Chmieliński [9]discovered that x ⊥�B y if and only if there exists a real number |a| ≤ ‖y‖‖x‖� such that x ⊥B ax + y.In fact, in inner product spaces, it is easy to see that x ⊥� y if and only if there exists |a| ≤ ‖y‖‖x‖�such that x ⊥B ax + y by taking a = −〈x|y〉‖x‖2 x + y for x 6= 0.In the following we will prove that it is also true for ⊥B�, that is, in normed spaces,
x ⊥B� y if and only if there exists |a| ≤ ‖y‖‖x‖� such that x ⊥B ax + y.

Before the proof, we need some lemma.
Lemma 2.6. [16] Let X be normed spaces,

N−(x,y) ≤ N+(x,y).

Lemma 2.7. [16] Let X be normed spaces, a ≤ b, a, b ∈ X, if x ⊥B ax + y,x ⊥B bx + y , then

x ⊥B cx + y f or c ∈ [a,b].

Lemma 2.8. [16] Let X be normed spaces,

x ⊥B ax + y ⇐⇒ N−(x,y) ≤−a‖x‖≤ N+(x,y).

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 6
Theorem 2.9. Let X be normed spaces, x ⊥B� y if and only if there exists |a| ≤

‖y‖
‖x‖� such that

x ⊥B ax + y.
Proof. If x ⊥B� y , first we have

N−(x,y) ≤ �‖y‖ , N+(x,y) ≥−�‖y‖.

On the other hand, from Lemma 2.8, we have if

−
N+(x,y)

‖x‖
≤ a ≤−

N−(x,y)

‖x‖
,

then x ⊥B ax + y. If there exists no |a| ≤
‖y‖
‖x‖� such that x ⊥B ax + y , then

−
N+(x,y)

‖x‖
>
‖y‖
‖x‖

� or −
N−(x,y)

‖x‖
< −
‖y‖
‖x‖

�.

Thus

N+(x,y) < ‖y‖ or N(x,y) > ‖y‖�.

Contradict to x ⊥B� y , so there must exists |a| ≤
‖y‖
‖x‖�, such that x ⊥B ax + y.

Conversely, if there exists |a| ≤ �‖y‖‖x‖ such that x ⊥B ax + y , we have:

x ⊥B ax + y =⇒ −N+(x,y) ≤ a‖x‖≤−N−(x,y)

=⇒ N+(x,y) + �‖y‖≥ 0 ≥ N−(x,y) − �‖y‖

=⇒ x ⊥B� y.

To conclude, in normed spaces, x ⊥B� y if and only if there exists |a| ≤
‖y‖
‖x‖� such that x ⊥B ax +y.

�

Since both x ⊥B� y and x ⊥B� y are equivalent to there exists |a| ≤ ‖y‖‖x‖� such that x ⊥B ax +y.we have x ⊥B� y if and only if x ⊥�B y in normed spaces. Now we give a direct proof for this.
Theorem 2.10. Let X be normed spaces,

x ⊥B� y if and only if x ⊥�B y.

Proof.We can assume that |t| ≤ ‖x‖
�‖y‖ and x 6= 0. If x ⊥B� y , we have:

‖x + ty‖≥‖x‖− �‖ty‖≥ 0.

Take square on both sides, we get

‖x + ty‖2 ≥‖x‖2 − �2‖ty‖2 − 2�‖x‖‖ty‖.

Thus ‖x + ty‖2 ≥‖x‖2 − 2�‖x‖‖ty‖, which means that x ⊥�B y.
Conversely, if x ⊥�B y , we have

‖x + ty‖2 ≥‖x‖2 − 2�‖x‖‖ty‖≥ 0.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 7
Let both sides be divided by ‖x + ty‖ + ‖x‖, we get:

‖x + ty‖−‖x‖≥
−2�‖x‖‖ty‖
‖x + ty‖ + ‖x‖

.

Since ‖x + ty‖ tends to ‖x‖ when t → 0, for every 2‖x‖ > δ > 0, we can find η > 0, such that

‖x + ty‖ + ‖x‖≥ 2‖x‖−δ if |t| ≤ η.

We then have

‖x + ty‖−‖x‖≥
−2‖x‖‖ty‖

2‖x‖−δ
.

Let δ → 0 we have

‖x + ty‖−‖x‖≥
−2‖x‖‖ty‖

2‖x‖
when t → 0.

Thus

N+(x,y) + �‖y‖≥ 0 ≥ N−(x,y) − �‖y‖ =⇒ x ⊥B� y.

�

Recall that Dragomir gave the following definition about approximate Birkhoff orthogonality:
x� ⊥B y ⇐⇒ ‖x + ty‖≥ (1 − �)‖x‖.

It is known that [19] in normed spaces, x ⊥�B y implies xδ ⊥B, where δ = 1 −√1 − 4�. Now wegive a more accurate estimate of δ as an application of the above Proposition.
Proposition 2.11. Let X be normed spaces, let x,y ∈ X, then:

x ⊥�B y =⇒ x
δ ⊥B y where δ = 2�.

Proof. Let f (t) = ‖x + ty‖ and assume that f (t) attains its minimum at t0, hence

‖x + t0y + ty‖≥‖x + t0y‖ f or all t ∈R.

Choose t = −t0 we have

‖x‖≥‖x + t0y‖≥ |‖x‖−|t0|‖y‖|,

thus we get |t0| ≤
2‖x‖
‖y‖ , then

‖x + ty‖≥‖x + t0y‖≥‖x‖− �|t0|‖y‖≥ (1 − 2�)‖x‖ f or all t ∈R.

Thus xδ ⊥B y , where δ = 2�. By the equivalence between ⊥B� and ⊥�B, we have the result that
x ⊥�B y implies x

δ ⊥B y. Since 2� ≤ 1 −
√

1 − 4�, 2� can be seen as a more accurate estimate. �

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 8
3. Approximate Isosceles Orthogonality And Approximate Birkhoff Orthogonality

In the following we will use the notion of approximate isosceles orthogonality [11], recall thatthe approximate isosceles orthogonality is defined by:
x ⊥�I y : |‖x + y‖

2 −‖x −y‖2| ≤ 4�‖x‖‖y‖.

x� ⊥I y : |‖x + y‖−‖x −y‖|≤ �(‖x + y‖ + ‖x −y‖).

It is easy to see that in inner product spaces we have:
x ⊥�I y ⇐⇒ |cos(x,y)| ≤ � ⇐⇒ x ⊥B� y,

and [11]
x� ⊥I y ⇐⇒ |cos(x,y)| ≤

�

1 + �2
(‖x‖2 + ‖y‖2).

In the following we give some simple properties about approxiamte isosceles orthogonality.
Proposition 3.1. Let X be normed spaces, if there exists |a| ≤ ‖y‖‖x‖� such that x ⊥I ax + y , then

x� ⊥I y .
Proof. Since x ⊥I ax + y we have ‖x + ax + y‖ = ‖x −ax −y‖, then

|‖x + y‖−‖x −y‖| = |‖x + ax + y −ax‖−‖x −ax −y + ax‖|.

On the other hand, by trigonometric inequality we have:

‖x + ax + y‖−‖ax‖− (‖x −ax −y‖ + ‖ax‖)

≤‖x + ax + y −ax‖−‖x −ax −y + ax‖,

and

‖x + ax + y −ax‖−‖x −ax −y + ax‖

≤‖x + ax + y‖ + ‖ax‖− (‖x −ax −y‖−‖ax‖).

Thus

|‖x + ax + y −ax‖−‖x −ax −y + ax‖|≤ 2‖ax‖,

then

|‖x + y‖−‖x −y‖|≤ 2‖ax‖≤ �(‖x + y‖ + ‖x −y‖),

thus x� ⊥I y .

Proposition 3.2. Let X be normed spaces, if for every ‖x‖ = ‖y‖ = 1, there is no 0 ≤ � < 1
such that x ⊥�I y , then X is a strictly convex space.
Proof. For any ‖x‖ = ‖y‖ = ‖x+y‖

2
= 1, if x 6= y ,we have

|‖x + y‖2 −‖x −y‖2| = |4 −‖x −y‖| < 4,

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 9
thus there must exist a 0 ≤ � < 1 such that |4 −‖x − y‖2| ≤ 4� which means that x ⊥�I y ,
contradict to the condition. so there must be x = y. From the equivalent characterization of strictly
convex space [23]. we get the result that X must be a strictly convex space.

It is known that in inner product spaces, different orthogonality types such as isosceles,pythagorean, and Birkhoff orthogonality is equivalent [3]. Using the notions of orthogonality innormed linear spaces it is possible to give different characterizations for inner product spaces. Forinstance [17], if x ⊥I y =⇒ x ⊥B y in a normed space X, then X must be inner product spaces.Inspired by this, now we give a characterization for inner product spaces using approximate or-thogonality.
Theorem 3.3. Let X be normed spaces, then X is inner product spaces iff the following two

conditions are satisfied.
(1) If there exists |a| ≤ ‖y‖‖x‖� such that x ⊥I ax + y, then x ⊥

�
I y.

(2) x ⊥�I y impliesx ⊥B� y.
Proof. If X is an inner product space, we have

x ⊥I y ⇐⇒ x ⊥B y and x ⊥�I y ⇐⇒ x ⊥B� y.

Thus (2) is satisfied. If there exists |a| ≤ ‖y‖‖x‖� such that x ⊥I ax + y , we have:

x ⊥B ax + y, |a| ≤
‖y‖
‖x‖

�.

Thus x ⊥B� y which implies x ⊥�I y. Thus both (1) and (2) are satisfied.
Conversely, assume that both (1) and (2) are satisfied, let x ⊥I y, x 6= 0. If |a| ≤ �

‖ax+y‖
‖x‖ , let

b = −a, then |b| ≤ �‖ax+y‖‖x‖ and x ⊥I bx + ax + y , thus form (1) we have x ⊥
�
I ax + y. To conclude

we have:
x ⊥�I ax + y if |a| ≤ �

‖ax + y‖
‖x‖

.

Now define: f (t) = ‖tx+y‖‖x‖ �, we have

f (t) =
‖tx + y‖
‖x‖

� ≤
‖tx‖ + ‖y‖
‖x‖

� = �|t| +
‖y‖
‖x‖

�.

Since 0 ≤ � < 1, when |t| tends to infinite, f (t) < |t|.
When t = 0, f (0) = ‖y‖‖x‖� > 0. By the convexity of f (t) we have

x ⊥�I t1x + y, t1 < 0,
‖t1x + y‖
‖x‖

� = −t1.

x ⊥�I t2x + y, t2 > 0,
‖t2x + y‖
‖x‖

� = t2.

From (2) we have x ⊥B� t1x + y and x ⊥B� t2x + y . from proposition 2.9,there must exist

|a1| ≤
‖t1x + y‖
‖x‖

� = −t1 and |a2| ≤
‖t2x + y‖
‖x‖

� = t2

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 10
such that x ⊥B a1 + t1 + y, x ⊥B a2 + t2 + y .
By a1 + t1 ≤ 0 , a2 + t2 ≥ 0 and lemma 2.7, we have x ⊥B y . Thus we have

x ⊥I y =⇒ x ⊥B y,

which means that X is inner product spaces.

Example 3.4. Let X = (R2,‖ · ‖∞), that is, ‖(x1,x2)‖ = max(|x1|, |x2|), assume that x =
(1, 0),y = (z, 1), |z| < 1. In order to satisfy x ⊥B� y or equivalently ‖x + ty‖≥‖x‖− �‖ty‖, the
following inequality shuold hold for all t ∈R:

‖(1 + tc,t)‖≥ 1 − �|t|.

Since ‖(1 + tc,t)‖ ≥ ‖t‖, we know the above inequality is always correct if |t| ≥ 1
1+�

, then we
may assume that |t| < 1

1+�
≤ 1.

when 1 > t ≥ 0, If 1 + tc ≥ t which means that t ≤ 1
1−c , we have

1 + tc ≥ 1 − �t

which implies that c ≥ −�. If 1 + tc < t or equivalently t > 1
1−c , we have t ≥ 1 − �t that is

t ≥ 1
1+�

. so there must be
1

1 −c
≥

1

1 + �

which implies that c ≥ −�. Similarly when −1 < t ≤ 0, we can get c ≤ �. To conclude we have:
(1, 0) ⊥B� (c, 1) if |c| ≤ �.
In order to satisfy x ⊥I� y or equivalently |‖x +y‖2−‖x−y‖2| ≤ 4�‖x‖‖y‖, the following inequality
shuold hold:

|‖(1 + c, 1)‖2 −‖(1 −c,−1)‖2| ≤ 4�.

If c ≥ 0, we have (1 + c)2 − 1 ≤ 4� =⇒ 0 ≤ c < −1 +
√

1 + 4�.
If c < 0, we have (1 −c)2 − 1 ≤ 4� =⇒ 1 −

√
1 + 4� ≤ c < 0.

To conclude we have (1, 0) ⊥�I (c, 1) if |c| ≤ −1 +
√

1 + 4�. Since � ≤ −1 +
√

1 + 4�, it can be
seen as an example that x ⊥�I y does not imply x ⊥B� y.

Example 3.5. Let X = (R2,‖ ·‖∞), we assume that x = (1, 1),y = (−1 −
√
2
2
�, 1 −

√
2
2
�),

z = (−1, 1). We have x ⊥I z, and

z = −
�
√

2
x + y, � ≤

‖y‖
‖x‖

�.

Thus |− �√
2
| ≤ ‖y‖‖x‖�, which implies that there exists |a| ≤

‖y‖
‖x‖� such that x ⊥I ax + y.

On the other hand, since

‖x + y‖ = ‖(−
√

2

2
�, 2 −

√
2

2
�)‖ = 2 −

√
2

2
�.

‖x −y‖ = ‖(2 +
√

2

2
�,

√
2

2
�)‖ = 2 +

√
2

2
�.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 11
we have |‖x + y‖2−‖x −y‖2| = 4

√
2� ≥ 4

√
2. so x 6⊥�I y , thus it can be seen as an example that

condition (1) is not satisfied.

A mapping T : H→K which satisfies the condtion
x ⊥ y =⇒ T (x) ⊥ T (y).

is called orthogonality preserving(o.p.) [7, 8], and T is said to be an isometry maping [22] if ‖Tx‖ =
‖x‖. To promote the concept, Jacek Chmielinski [6] introduced the notion of approximately orthogo-nality preserving (a.o.p.) mapping and have studied the properties of mapping that is approximarelyisosceles orthogonality preserving(T: x ⊥I y =⇒ T (x) ⊥�I T (y)). After that many mathematicianshave show great interest in the a.o.p mapping [25], and Aleksej Turnšek [19] studied the mappingthat is approximately Birkhoff orthogonality preserving(T : x ⊥B y =⇒ T (x) ⊥�B T (x))innormed spaces. Now we try to study the approximarely orthogonality preserving mapping types:

T : x ⊥I y =⇒ Tx ⊥�B Ty,

and
T : x ⊥I y =⇒ Tx �x ⊥B Ty.

Proposition 3.6. Let T : X → Y be a nontrivial linear mapping satisfying

x ⊥I y =⇒ Tx ⊥�B Ty, x,y ∈ X.

Then T is a bounded and bounded from below, ‖Tx‖≥ (1−�)
2

3−�2+2
√
2−�2

.

Proof. Take two arbitrary unit vectors x and y and note that x+y
2
⊥I x−y2 , it follows that

T (x + y) ⊥�B T (x −y),

hence for all λ ∈R we have

‖T (x + y) + λT (x −y)‖2 ≥‖T (x + y)‖2 − 2�‖T (x + y)‖‖λT (x −y)‖,

by the triangle inequality and the linearity of T it follows that

‖T (x + y)‖2 ≤‖(1 + λ)Tx + (1 −λ)Ty‖2 + 2�|λ|‖Tx + Ty‖2.

On the other hand we have

‖T (x + y)‖2 ≥ (‖Tx‖−‖Ty‖)2,

thus we get:

‖Tx‖2+‖Ty‖2−2‖Tx‖‖Ty‖≤ (1+λ)2‖Tx‖2+(1−λ)2‖Ty‖2+2(1−λ2)‖Tx‖‖Ty‖+2�|λ|‖Tx+Ty‖2.

If Tx = 0, then let y ∈ X such that Ty 6= 0, substitute x,y into the above formula,we have

0 ≤ λ2 − 2λ + 2�|λ| f or all λ ∈R.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 12
It is impossible cause 0 ≤ � < 0, so we can divide both sides of the inequality by ‖Tx‖2 and
denote z = ‖Ty‖‖Tx‖ . we get:

(z − 1)2λ2 +
(

2 − 2z2 + 2�(1 + z)2
)
λ + 4z ≥ 0 f or all λ ≥ 0.

the inequality is satisfied when − b
2a
≤ 0 or ∆ = b2 − 4ac ≤ 0.

∆ = 4
(

(1 −z2)2 + �2(1 + z)4 + 2�(1 + z)2(1 −z2) − (4z)(z − 1)2
)
.

(1 −z2)2 + �2(1 + z)4 + 2�(1 + z)2(1 −z2) − (4z)(z + 1)2 ≤ 0 =⇒ ∆ ≤ 0.

Thus we get z ≤ 3−�
2+2
√
2−�2

(1−�)2 which implies ‖Ty‖ ≤
3−�2+2

√
2−�2

(1−�)2 ‖Tx‖. Since x,y are arbitrary,
T is bounded and

‖Tx‖≥
(1 − �)2

3 − �2 + 2
√

2 − �2
‖T‖‖x‖.

Theorem 3.7. Let T : X → Y be a nontrivial linear mapping satisfying

x ⊥I y =⇒ Tx � ⊥B Ty, x,y ∈ X.

Then T is a scalar mutiple of a isometric mapping, i.e., for some γ > 0, ‖Tx‖ = γ‖x‖.
Proof. Take two arbitrary unit vectors x and y and note that x+y

2
⊥I x−y2 , it follows that

T (x + y)� ⊥B T (x −y).

Hence for all λ ∈R we have

‖T (x + y) + λT (x −y)‖≥ (1 − �)‖T (x + y)‖,

thus

(
‖(1 + λ)Tx‖ + ‖(1 −λ)Ty‖

)2 ≥ (1 − �)2‖Tx + Ty‖2 ≥ (1 − �)2(‖Tx‖−‖Ty‖)2.
If Tx = 0, then let y ∈ X such that Ty 6= 0, substitute x,y into the above formula, we have

‖(1 −λ)Ty‖2 ≥ (1 − �)2‖Ty‖2 =⇒ (1 −λ)2 ≥ (1 − �)2 f or all λ ∈R,

it is impossible, then we can divide both sides by ‖Tx‖2 like before and denote z = ‖Ty‖‖Tx‖ , we get

(1 −z)2λ2 + 2(1 −z2)λ + (z + 1)2 − (1 − �)2(1 −z)2 ≥ 0 f or all λ ∈R.

If z 6= 1, then ∆ = 4 · (1 − z)4(1 − �)2 > 0. Thus the inequality is satisfied only when z = 1, so
z = 1 which means that ‖Tx‖ = ‖Ty‖, thus T must be a scalar mutiple of a isometric mapping.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.16 13
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https://doi.org/10.1016/j.jmaa.2007.03.016
https://doi.org/10.1016/j.jmaa.2007.03.016
https://doi.org/10.1007/s00010-013-0233-7
https://doi.org/10.1007/s00010-013-0233-7

	1. Introduction
	2. Approximate Birkhoff Orthogonality B
	3. Approximate Isosceles Orthogonality And Approximate Birkhoff Orthogonality
	References

