Ebook norm derivatives and characterizations of inner product spaces part 2

Chapter 4 Perpendicular Bisectors in Normed Spaces The center of the circumscribed circumference of a triangle is the intersec tion point of the three perpendicular bisectors of its three sides This i[.]

Chapter 4

Perpendicular Bisectors in Normed

Spaces

The center of the circumscribed circumference of a triangle is the intersec-tion point of the three perpendicular bisectors of its three sides This is

a classical situation in an i.p.s What happens in a general normed linear space where various notions of orthogonality lead to alternative expressions for perpendicular bisectors? We explore such questions in this chapter

In a real i.p.s (X, h·, ·i) with dimension greater than or equal to 2, given two linearly independent vectors x, y ∈ X, we define the perpendicular bisector of the linear segment

[x, y] = {αx + (1 − α)y : 0 ≤ α ≤ 1}

as the set

 x + y

 ,

where u 6= 0 is a vector in the subspace generated by x and y (denoted by span(x, y)) and orthogonal to x − y

If u := αx + βy, then by the requirement hx − y, ui = 0, we have

− hx, yi), and we may take

103

y u

x+y

2 + λu

Figure 4.1.1

In a real normed linear space (X, k·k), we can replace the inner product

for all x, y in X, and define



It follows from the above that if x and y are linearly dependent, e.g y = λx for some λ ∈ R, then

We call the lines (4.1.2) perpendicular bisectors of the linear segment [x, y], and later we justify their associated orthogonal relations

Furthermore, in the real normed linear space (X, k·k), we can also define the perpendicular bisector of the linear segment [x, y] as the set of metric nature:

but

product spaces

dim X ≥ 2 Then X is an inner product space if and only if for all x, y in X

For the proof of this theorem, we need the following lemma based on the James orthogonality

for all vectors x and y in X, we have



kw − (x − y)k = 2kz − xk = 2kz − yk = kw + (x − y)k



independent vectors x, y ∈ span (v, ς) and such that ς = x − y Then by the previous lemma

x + y

1



∩ span (v, ς)

= {z ∈ X : kz − xk = kz − yk} ∩ span (x, y)

 x + y



Therefore,

i.e., the James orthogonality is homogeneous, which is a characterization of inner product space [James (1945); Amir (1986)] The converse implication

In an inner product space, the perpendicular bisectors of the linear seg-ments determined by the sides of a rhombus are its diagonals Indeed, this property characterizes inner products spaces

dim X ≥ 2 Then X is an inner product space if and only if for all x, y in

X with kxk = kyk one has

so ρ′

Since in an inner product space, two vectors x and y are orthogonal

if and only if they are orthogonal in the sense of James, if we take the rhombus with sides x + y and x − y, then by the last theorem

Based on this property, we define a new orthogonality relation

we consider the perpendicular bisector orthogonality

+

The next lemma contains some basic properties of this orthogonality relation

the following properties for all x, y in X and for all a, t in R:

equivalent to the usual orthogonality hx, yi = 0

Proposition 4.2.2 If (X, k · k) is a real normed linear space with dim X ≥ 2, then for all pairs x, y in X\{0} there exist two unique real

equivalent to

kyk2t2+ (ρ′sgn(t)(y, x) − ρ′sgn(t)(x, y))t − kxk2= 0 (4.2.2)

So taking

′

and

′

dim X ≥ 2 Then X is an inner product space if and only if for all x, y in X,

i.e.,

kxk = tkyk

on the account of Theorem 2.1.1 means that X is an inner product space



+ = ρ′

Proof Assume that ⊥M-orthogonality is symmetric For each pair of

satisfied, i.e.,

ρ′

In general, the perpendicular bisector orthogonality is not symmetric;

per-pendicular bisectors, we obtain a new geometrical characterizations of inner product spaces

dim X ≥ 2 The following conditions are equivalent:

(i) X is an inner product space;

and

then

Note that (ii) gives a characterization of the perpendicular bisectors of the rectangle sides (see Figure 4.2.1), (iii) states that in a rhombus the diagonals are orthogonal, and (iv) establishes that if the diagonals of a parallelogram are orthogonal then it is a rhombus

M (x, x + y)

Figure 4.2.1

x

i.e.,

then

and therefore,

Finally, from last two inequalities, we deduce

and by Theorem 1.4.4, we infer that X is an inner product space [Amir (1986); Ben´ıtez and del Rio (1984)]

If we assume condition (iii), then for all u, v in SX if u + v ⊥M u − v,

′

be the unique positive root of (4.2.2) such that x + αy ⊥ x − αy Then, by (4.2.4)

and consequently,

so ρ′

dim X ≥ 2 Then X is an inner product space if and only if for all

orthogonalities

If (X, h·, ·i) is an i.p.s and x, y are two vectors in X, we can consider the

to x − y Now we show how this property characterizes i.p.s., whenever we consider in a real normed space (X, k · k) the classical orthogonal relations

dim X ≥ 2 Then the following conditions are equivalent:

(i) X is an inner product space;

(vi) x − y ⊥Bu(x, y) for all x, y in X.

properties (ii), (iii), (iv), (v) and (vii) implies (i)

If we assume condition (ii), i.e., for all x, y in X

replacing y by x − z and using Theorem 2.1.1, we obtain for all x, z in X

However,

lim

So, for λ > 0 in a neighborhood of zero, F (z, λx) > 0, and replacing x

by λx and using Theorem 2.1.1 in (4.3.2), we have

Taking limit when λ tends to zero in the last equality, by using Proposition

+= ρ′

−

On the other hand, by the substitutions z := x − y, y := y in (4.3.1),

Now take linearly independent unit vectors w, v in X, and substitute

z := w, y := v in the above equality:

(4.3.4)

By interchanging the roles of w and v in (4.3.1), using Theorem 2.1.1,

+= ρ′

convex and w, v are linearly independent) and dividing by the common factor, we obtain

account of Theorem 1.4.3 gives a characterization of an i.p.s

To prove that (iii) implies (i), we first use the same argument as that

+= ρ′

not necessary) The rest follows from the previous implication

If condition (iv) holds, then for all x, y in X

we have

by 2γ and taking limit when γ tends to zero, finally using the definition of

ρ′

If we assume condition (v), i.e., for all x, y in X

ku(x, y) − x + yk = ku(x, y) + x − yk, replacing x by tx and y by ty for all t > 0,

and for all γ > 0

± we obtain,

By the substitutions, z := x − y, x := x in (4.3.6) and using (v) from Theorem 2.1.1

where F (z, x) is defined in (4.3.2) Replacing x by λx, λ > 0, dividing by λ, taking limit when λ tends to zero, and using (4.3.3) and Proposition 2.1.6,

+ = ρ′

have condition (ii) and X is an i.p.s

Finally, if condition (vi) holds, then by Proposition 2.1.7 for all x, y in X

If we first make the substitution z := x−y, x := x and then z := −ty, t > 0,

+(x, y) > ρ′

In this last case, by using Proposition 2.1.6 and Corollary 2.1.1, we get lim

by (4.3.8) and Theorem 2.1.1,

Now, for all t > 0, if we consider x and x + ty using (iii) and (v) from Theorem 2.1.1, we have

If ρ′

+(x, y) > ρ′

and taking limit when t tends to zero, we obtain the contradiction kxk = 0

+= ρ′

Now we generalize another property of perpendicular bisectors

we have kx − zk = ky − zk

Proof By hypothesis for all λ > 0

x + y



triangle

If (X, h·, ·i) is an i.p.s and x, y are two linearly independent vectors in X,

in the triangle of sides x, y and x − y, the radius R of the circumscribed circumference is given by the formula

kxkkykkx − yk

where s = (kxk + kyk + kx − yk) /2 is the semiperimeter of the triangle Moreover, another equivalent expression for R is

which we get by finding R as kαx+βyk, and determining α and β by means

must be orthogonal to y

Then, in a real normed space (X, k · k), given two linearly independent vectors x and y, we can define the radius of the circumscribed circumference

in the triangle of sides x, y and x − y as:

This definition is only possible if

ρ′−(x, y)2+ ρ′−(y, x)2< 2kxk2kyk2, i.e.,

ρ′

ρ′

< kxkkyk For this reason we assume that X is strictly convex

Then X is an i.p.s if and only if for all linearly independent vectors x, y

in X,

where s = (kxk + kyk + kx − yk) /2

we obtain

=



lim

t→0 +

= kxkkzk

 lim

t→0 +

tkx − tzk

t→0 +

tkx − tzk

= kxkkzk

s

kxk

s

kxk

and therefore,

x

′

−

 z

x kxk

 z

−



x kxk, z kzk

−



z kzk, x kxk

2

r

−



x kxk,kzkz 2

z

Analogous definitions of R(x, y) can be given by replacing the role of

ρ′

− by ρ′

example, assume that the radius of the circumscribed circumference is given by

ˆ

(which is equal to R(x, y) in an i.p.s.) Then a strictly convex space X with dim X ≥ 2 is an i.p.s if and only if

ˆ

The proof is immediate and uses the fact that the right-hand side

ρ′

Let (X, k·k) be a real normed linear space with dim X ≥ 2 In the following, when speaking about points we will always refer to vectors with the initial point at the origin Then, consider the triangle ∆xyz with sides x − y,

y − z, x − z, where x, y, z belong to X, x, y are linearly independent and z

is in the plane determined by x and y (see Figure 4.5.1)

△xyz

Figure 4.5.1

In an i.p.s., the perpendicular bisectors of the three sides of a triangle all pass through the circumcenter, which is the center of the circumscribed circle (see [Coxeter (1969)], p 12-13 and [Puig-Adam (1986)], p 92) Then,

in a natural way we have the following

Definition 4.5.1 The triangle ∆xyz has a circumcenter C if and only if

First we consider the class of triangles ∆xy(x + y), where we can show the following result:

Then X is an i.p.s if and only if there exists the circumcenter of ∆xy(x+y) for all linearly independent vectors x, y in X

y, it is a straightforward computation to prove that the three straight lines

following equality holds

+,

in the right part of last equality; using the equality

and finally by taking limit when λ tends to zero, by Proposition 2.1.6,

ρ′

+= ρ′

ư Moreover, by substituting x := u, y := vưu in (4.5.1), using the equality

ρ′

+ = ρ′

way, we get

+= ρ′

ư and

we have

or

hypothesis, X is strictly convex and we obtain a contradiction (cf Theorem 2.1.1 (vii))

Then, for all u, v in X

+= ρ′

−,

Then for all t > 0, the function defined by

2t

+(u, v), ρ′

lim

Note that Theorem 4.5.1 covers a general case because ∆xy(x + y) is equivalent to the triangle determined by x and y (i.e., with sides x, y and

x − y) (see Figure 4.5.2)

△xy(x + y)

→x +→y

Figure 4.5.2

Then, with a similar proof to that of the last theorem, we have the following

dim X ≥ 2 such that there exists the circumcenter of ∆xy(αx + y) for all linearly independent vectors x, y in X and α in R fixed Then (X, k · k)

is smooth

In a real normed linear space (X, k · k) we consider the triangle ∆xyz with sides x − y, y − z and x − z, where x and y are linearly independent

Then by a straightforward computation, we can prove that the

give us three points















(4.5.4)

Theorem 4.5.1, in a real normed space, the three points are, in general, different

circumcenter points of ∆xyz

We consider two classical properties concerning the circumcenter of a triangle in the Euclidean plane, and we translate these properties into a

new characterizations of inner product spaces

dim X ≥ 2 and let x, y, z be vectors in X with kxk = kyk = kzk Then X

is the origin

Proof The direct part is a well-known result Reciprocally, if for all x, y, z

then in particular it is true for unit vectors such that x and y are linearly

using (4.5.4) and the linear independence of x and y, we infer that the system

where A, B, C and D are defined after (4.5.4) gives C = D, and therefore

ρ′

of ∆xy(λx + y) is

whenever x, y are linearly independent vectors in X with kxk = kyk and λ belongs to R

1 + λ

By the last equalities, with a long and tedious computation we can prove that

Now, we claim that |ρ′

Let x and y be two unit linearly independent vectors in X

If λ = 1, the result is evident Assume that λ 6= 1 If





ρ′

λx+y kλx+yk λx+y kλx+yk

and repeat the process

possibilities

+(x, bn)

1 − λ

For n = 1 we have proved that it is true If it is true for n − 1, we wish

to prove that it is true for n

+



kb n−1 k



=

λ +ρ′+ (x,b n−1 )

1−λ

new characterizations of inner product spaces

dim X ≥ and let x, y, z be vectors in X with kxk... dim X ≥ is an i.p.s if and only if

ˆ

The proof is immediate and uses the fact that the right-hand side

ρ′

Let (X, k·k) be a real normed linear space with... data-page="19">

Proof The direct part is a well-known result Reciprocally, if for all x, y, z

then in particular it is true for unit vectors such that x and y are linearly

using (4.5.4) and the