EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 3, 2018, 793-802
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On the Continuity of Orthogonal Sets in the Sense of
Operator Orthogonality

M. Iranmanesh1,∗, M. Saeedi Khojasteh1, M. K. Anwary2

1 Department of Pure Mathematics, Shahrood University of Technology, Iran
2 Department of Pure Mathematics, Ferdowsi University of Mashhad, Iran

Abstract. In this paper, we introduce the operator approach for orthogonality in linear spaces.
In particular, we represent the concept of orthogonal vectors using an operator associated with
them, in normed spaces. Moreover, we investigate some of continuity properties of this kind of
orthogonality. More precisely, we show that the set valued function

F (x; y) = {µ : µ ∈ C, p(x− µy, y) = 1}

is upper and lower semi continuous, where

p(x, y) = sup{pz1,...,zn−2
(x, y) : z1, . . . , zn−2 ∈ X}

and
pz1,...,zn−2

(x, y) = ‖Px,z1,...,zn−2,y‖
−1

where Px,z1,...,zn−2,y denotes the projection parallel to y from X to the subspace generated by
{x, z1, . . . , zn−2}. This can be considered as an alternative definition for numerical range in linear
spaces.

Key Words and Phrases: Birkhoff orthogonality, Minkowski plane, set valued function, upper
semi continuous, lower semi continuous

1. Introduction

Orthogonality, is one of the important concepts in mathematical and numerical anal-
ysis. Perhaps, it is the main property in linear spaces, normed spaces and inner product
spaces. There are some various kinds of orthogonality. In fact, it has been defined different
kinds in mathematical spaces.

In inner product spaces, it is easily said that two vectors x, y are orthogonal if

〈x, y〉 = 0.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3237

Email addresses: m.iranmanesh2012@gmail.com (M. Iranmanesh),
m.saeedi64@gmail.com (M. Saeedi Khojasteh), abdh1248@gmail.com (M. K. Anwary)

http://www.ejpam.com 793 c© 2018 EJPAM All rights reserved.



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 794

But, in normed spaces, there is no simple tool for define orthogonal vectors. However,
there are some good suggestions. One of them, is the Birkhoff James orthogonality [2].

Let X be a real normed space, and x, y be in X. We say that x is Birkhoff orthogonal
to y if for every constant a,

‖x‖ 6 ‖x+ ay‖. (1)

It is not difficult to show that this definition is the same in inner product spaces [6].
In 1993, Milicic [7] introduced g-orthogonality in normed spaces via Gateaux deriva-

tives. In fact, one has the notion of g-angle related to g-orthogonality.
In this paper, the authors define a new type of orthogonality in a linear space by using

projection operators.
Let X be a Minkowski plane. Denote by ‖.‖ the norm of X. Fix a basis {e1, e2} of

X. Then we can write each x ∈ X as x = (x1, x2) under this basis, where x1, x2 ∈ R.
Moreover, {δe1 , δe2} is a basis of the dual space X∗, where δei for i = 1, 2 is a bounded
linear function on X with

δei(ej) =

{
0 i 6= j;

1 i = j.

Denote by L(X) the set of all bounded linear operators from X to X. For T ∈ L(X),
the operator T ∗ ∈ L(X∗) is said to be the Banach conjugate operator of T if for any z ∈ X
and any z∗ ∈ X∗, there must be (T ∗z∗)(z) = z∗(Tz). Note that if we use the following
notation

f(x) = 〈x, f〉

then the property of conjugate can be rewritten as the following way

〈x, T ∗f〉 = 〈Tx, f〉

as usual in inner product spaces.
Recall that an operator P is an orthogonal projection if it is idempotent and self-

adjoint, i.e. P 2 = P and P ∗ = P . In an inner product space it is equivalent to

〈Px, x〉 = 〈Px, Px〉 = 〈x, Px〉.

Suppose that x = (x1, x2)T and y = (y1, y2)T are two linearly independent vectors in
X under the basis {e1, e2}. Put

Dxy =

[
x1 y1

x2 y2

]
notice

|Dxy| = x1y2 − x2y1 6= 0

since x and y are linearly independent. Define by Pxy the projection parallel to y from X
to the subspace {λx;λ ∈ R}. Then Pxy depends only on the vectors x and y, and has the
following presentation under the basis {e1, e2}:

Pxy = Dxy .

[
1 0
0 0

]
. Dxy

−1 =
1

|Dxy|

[
x1y2 −x1y1

x2y2 −x2y1

]
.



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 795

It is clear for any two linearly independent vectors x and y in X,

1 ≤ ‖Pxy‖ < +∞.

Note that if x, y are orthogonal, in the sense of inner product space, then Pxy is an
orthogonal projection.

Furthermore, denote

p(x, y) =

{
0 x and y are linearly dependent;

‖Pxy‖−1 x and y are linearly independent.

For any x, y ∈ X, the p-angle between x and y is defined by

Ap(x, y) = arcsin(p(x, y)).

In an inner product space (X, 〈., .〉), obviously

p(x, y) =
〈x, y〉
‖x‖‖y‖

,

and consequently, the p-angle is identical with the usual angle.
Let (X, ‖.‖) Be a complex Banach space, (X∗, ‖.‖) be its dual space, and B(X) be the

algebra of all bounded linear operators acting on X. Define the set of normalized states
Ω = {ω ∈ B(X)∗ : ω(I) = ‖ω‖ = 1}, where I denotes the identity operator. For any
operator A ∈ B(X), the (algebraic) numerical range (also known as field of values) of A
is defined by

F (A) = {ω(A) : ω ∈ Ω}.

In the finite-dimensional case (X, ‖.‖) = (Cn, ‖.‖2), where ‖.‖2 is the spectral norm,
the numerical range of a square matrix A ∈ Cn×n is also written

F (A) = {x∗Ax ∈ C : x ∈ Cn, x∗x = 1}.

The suggested references on numerical ranges of operators and matrices are [3] and [5].
We recall that for two compact subsets Ω1 and Ω2 of a metric space (X, ρ), the Haus-

dorff distance between Ω1 and Ω2 is defined by

dH(Ω1,Ω2) = max{max
x1∈Ω1

min
x2∈Ω2

ρ(x1, x2), max
x2∈Ω2

min
x1∈Ω1

ρ(x1, x2)}

For any x0 ∈ X and δ > 0, we define the closed ball B(x0, δ) = {x ∈ X : ρ(x0, x) ≤ δ}.

Definition 1. [1] Suppose (X, ρX) is a metric space and (Y, ρY ) is a complete metric
space. Consider a multi-valued mapping F : X → Y , and let x0 ∈ X.

(i) F is called upper semi-continuous at x0 if for every neighborhood N(F (x0)) ⊂ Y of
the set F (x0), there is a neighborhood N(x0) ⊂ X of x0 such that

F (x) ⊂ N(F (x0)), ∀x ∈ N(x0).



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 796

(ii) F is called lower semi-continuous at x0 if for every y0 ∈ F (x0) and every neighbor-
hood N(y0) ⊂ Y of y0, there exist a neighborhood N(x0) ⊂ X of x0 such that

F (x) ∩N(y0) 6= ∅, ∀x ∈ N(x0).

(iii) F is said to be semi continuous at x0 if it is upper and lower semi-continuous.

The following example is showing some of the common behavior of upper semi contin-
uous functions.

Example 1. The following functions is upper semi continuous

f(α) = {x : x ∈ X, ‖x‖ ≤ |α|}.

Note that this function is increasing in the mean that if |α| ≤ |β| then f(α) is contained
in f(β).

We will investigate the upper semi continuity at α0 = 1. Investigating other points are
similar.

Assume that N(f(1)) is an open set containing f(1). Since f(1) is closed, there is a
scalar β such that

f(1) ⊆ f(β) = {x : x ∈ X, ‖x‖ ≤ |β|} ⊆ N(f(1)).

It is clear that 1 < |β|. Now consider the following open set

N(1) = {α : α ∈ C, |α| < |β|},

It is clear that for any α in N(1), f(α) is contained in f(β), since f is increasing. So
f(α) ⊆ N(f(1)).

2. Main Results

Definition 2. Let X be a linear space with dimension n. Suppose that

xk = (xk1 , . . . , xkn)T , k = 1, . . . , n

are n linearly independent vectors in X. Put

Dx1,...,x2 =

x11 . . . xn1
...

. . .
...

x1n . . . xnn


since x1, . . . , x2 are linearly independent, we have

|Dx1,...,x2 | 6= 0.



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 797

Definition 3. Let X be a linear space with dimension n. Suppose that

x = (x1, . . . , xn)T , y = (y1, . . . , yn)T

are two linearly independent vectors in X. Extend x, y to a basis for X by adding n − 2
vector as

zk = (zk1 , . . . , zkn)T , k = 1, . . . , n− 2.

Denote by Px,z1,...,zn−2,y the projection parallel to y from X to the subspace generated
by x, z1, . . . , zn−2. Since the vectors x, z1, . . . , zn−2, y are the eigenvectors of Px,z1,...,zn−2,y,
it turn implies that Px,z1,...,zn−2,y is similar to the following

1 . . . 0 0
...

. . .
...

...
0 . . . 1 0
0 . . . 0 0

 .
In fact, Px,z1,...,zn−2,y has a representation as follows

Px,z1,...,zn−2,y = Dx,z1,...,zn−2,y .


1 . . . 0 0
...

. . .
...

...
0 . . . 1 0
0 . . . 0 0

 . D−1
x,z1,...,zn−2,y.

Proposition 1. For any two linearly independent vectors x and y in X,

1 ≤ ‖Px,z1,...,zn−2,y‖ < +∞

in other words, Px,z1,...,zn−2,y is a bounded operator.

Furthermore, denote

pz1,...,zn−2(x, y) = ‖Px,z1,...,zn−2,y‖
−1

and let
p(x, y) = sup{pz1,...,zn−2(x, y) : z1, . . . , zn−2 ∈ X}.

It is obvious that

p(x, y) = max{‖Px,z1,...,zn−2,y‖
−1 : z1, . . . , zn−2 ∈ X , ‖z1‖ = 1, . . . , ‖zn−2‖ = 1}.

Definition 4. For any linearly independent x, y in X , the p-angle between x, y is defined
by

Ap(x, y) = arcsin(p(x, y)).

Note that p-angle is not depending on selected vectors z1, . . . , zn−2.



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 798

Definition 5. For linearly independent vectors x, y in X, we say that x is p-orthogonal
to y if

Ap(x, y) =
π

2
.

It is clear that x, y are p-orthogonal if there exist suitable vectors z1, . . . , zn−2 such
that

‖Px,z1,...,zn−2,y‖ = 1.

Theorem 1. The concept of p-orthogonality is compatible with the usual orthogonality in
the inner product spaces.

Proof. Let X be an inner product space. First, assume that x, y are orthogonal. We
shall show that

‖Px,z1,...,zn−2,y‖ = 1

for suitable choice of z1, . . . , zn−2 . To this end, extending x, y to a basis as {x, z1, . . . , zn−2, y}
to an orthogonal basis for X, we show that

‖P‖ = 1

where P = Px,z1,...,zn−2,y is the orthogonal projection associated with the subspace gener-
ated by {x, z1, . . . , zn−2, y}.

Since
y ∈ [span{x, z1, . . . , zn−2}]⊥

and for any z in X, we have

Pz ∈ span{x, z1, . . . , zn−2}

we conclude that
y ⊥ Pz

and we have
‖z‖2 = ‖z − Pz + Pz‖2 = ‖z − Pz‖2 + ‖Pz‖2 ≥ ‖Pz‖2

therefore
‖p‖ ≤ 1,

now, taking z = x, we have
Px = x

so
‖Px‖ = ‖x‖

hence
‖P‖ = 1.



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 799

Next, assume that x, y are not orthogonal. We shall show that

‖Px,z1,...,zn−2,y‖ > 1

for all choices of z1, . . . , zn−2 . Since

{y}⊥ 6= span{x, z1, . . . , zn−2}

there exists a nonzero vector z in {y}⊥ that does not belong to
span{x, z1, . . . , zn−2}. For this z we have

Pz − z ⊥ z.

We conclude that

‖Pz‖2 = ‖Pz − z + z‖2 = ‖Pz − z‖2 + ‖z‖2 > ‖z‖2

therefore
‖P‖ > 1

as claimed.

For a complex linear space, we have already defined the operator orthogonality. We
denote this kind of orthogonality by notation ⊥p.

Let x, y be two vectors in X. Similar to [? ], we consider the following set in C as the
orthogonality set of x with respect to y:

F (x; y) = {µ : µ ∈ C, (x− µy) ⊥p y}

or equivalently
F (x; y) = {µ : µ ∈ C, ‖Px−µy,z1,...,zn−2,y‖ = 1}.

Moreover, we can involve an other parameter α for more benefits:

F (x; y;α) = {µ : µ ∈ C, (αx− µy) ⊥p y}

or equivalently
F (x; y;α) = {µ : µ ∈ C, ‖Pαx−µy,z1,...,zn−2,y‖ = 1}.

Lemma 1. For any non zero α, we have the following

F (x; y;α) = αF (x; y; 1).

Proof. By definition, we have

F (x; y;α) = {µ : µ ∈ C, ‖Pαx−µy,z1,...,zn−2,y‖ = 1}
= {µ : µ ∈ C, ‖Pα(x− µ

α
y),z1,...,zn−2,y‖ = 1}

= {µ : µ ∈ C, ‖P(x− µ
α
y),z1,...,zn−2,y‖ = 1}



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 800

since P is homogenized. It turn implies that if µ ∈ F (x; y;α), then µ
α ∈ F (x; y; 1); or

µ ∈ αF (x; y; 1); it completes the proof.

In the following theorems we will see upper semi continuity and lower semi continuity
of F (x; y;α) in α, using lemma 1.

Theorem 2. The set valued function which maps α to F (x; y;α), is upper semi continu-
ous.

Proof. With out loss of generality, we will show upper continuity at α0 = 1. Continuity
at other points are similar.

For more simplicity, fix x, y and let f(α) = F (x; y;α).
Assume that N(f(1)) is an open set containing f(1). The following scalars are well

defined

β1 = inf{β : f(β) is contained in N(f(1))}
β2 = sup{β : f(β) is contained in N(f(1))}.

On the other hand, since f(1) is closed, we have β1 < 1 < β2. Now consider the open
set around 1,

N(1) = {α : α ∈ C, β1 < |α| < β2}.

It is clear that for any α in N(1), f(α) is contained in N(f(1)).

Theorem 3. The set valued function which maps α to F (x; y;α) , is lower semi continuous

Proof. With out loss of generality, we will show lower continuity at α0 = 1. Continuity
at other points are similar.

For more simplicity, fix x, y and let f(α) = F (x; y;α).
Assume that β0 ∈ f(1) is an arbitrary point. Moreover, assume that N(β0) is an open

neighborhood of β0. The following scalars are well defined

β1 = inf{β : f(β) intersects N(β0)}
β2 = sup{β : f(β) intersects N(β0)}.

On the other hand, since f(1) is closed, we have β1 < 1 < β2. Now consider the open
set around 1,

N(1) = {α : α ∈ C, β1 < |α| < β2}.

It is clear that for any α in N(1), f(α) intersects N(β0).

Corollary 1. The set valued function which maps α to F (x; y;α) , is semi continuous

As we proved, this functions is both upper and lower semi continuous, the corollary is
hold.



M. Iranmanesh, M. Saeedi Khojasteh, M. K. Anwary / Eur. J. Pure Appl. Math, 11 (3) (2018), 793-802 801

As we saw in theorem 1 , the concept of operator orthogonal vectors in the inner
product spaces, is the same of usual orthogonal vectors, i.e. for vector x, y in an inner
product space X, we have

p(x, y) = 1

if and only if
〈x, y〉 = 0

This leads us to a simple computation for the set F (x; y;α) in an inner product space.

Example 2. In an inner product space we have

F (x; y;α) = {µ : µ ∈ C, 〈αx− µy, y〉 = 0}.

It turn implies that

F (x; y;α) = {µ : µ ∈ C, α〈x, y〉 = µ〈y, y〉}.

or equivalently

F (x; y;α) = {α〈x, y〉
〈y, y〉

}.

This means that in inner product spaces, F (x; y;α) is a singleton set. Therefore the
concept of semi continuity of this set valued function is the same of its usual continuity.

Example 3. Let 1 ≤ r ≤ ∞. For any two vectors x = (x1, x2) and y = (y1, y2) in l2
r, it

has been shown in [8] that the operator angle between x, y is the following

arcsin (
|x1y2 − x2y1|
‖x‖r‖y‖ r−1

r

).

It implies that x, y are operator orthogonal if

‖x‖r‖y‖ r−1
r

= |x1y2 − x2y1|.

Therefore, in this case we have

F (x; y;α) = {µ : µ ∈ C, (αx− µy) ⊥p y}
= {µ : µ ∈ C, ‖αx− µy‖r‖y‖ r−1

r
= |(αx1 − µy1)y2 − (αx2 − µy2)y1|}.

So by definition of ‖x‖r, we have

F (x; y;α) = {µ : µ ∈ C,(|αx1 − µy1|r + |αx2 − µy2|r)
1
r (|y1|

r−1
r + |y2|

r−1
r )

r
r−1 =

|(αx1 − µy1)y2 − (αx2 − µy2)y1|}.

Specially, for r = 2,

F (x; y;α) = {µ : µ ∈ C,(|αx1 − µy1|2 + |αx2 − µy2|2)
1
2 (|y1|

1
2 + |y2|

1
2 )

2
=

|(αx1 − µy1)y2 − (αx2 − µy2)y1|}.



REFERENCES 802

Giving x1, x2, y1, y2, α, this is an equation on µ; In fact we have

A((αx1 − µy1)2 + (αx2 − µy2)2) = ((αx1 − µy1)y2 − (αx2 − µy2)y1)2

where

A = (|y1|
1
2 + |y2|

1
2 )

4
.

It leads to the following

A((y1
2 + y2

2)µ2 − 2α(x1y1 + x2y2)µ+ α2(x1
2 + x2

2))

= α2(x1
2y2

2 − 2x1x2y1y2 + x2
2y1

2),

and µ is obtained from this equation.

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