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The Ptolemy constant of absolute normalized norms on 2

Abstract

We determine and estimate the Ptolemy constant of absolute normalized norms on 2 by means of their corresponding continuous convex functions on [0, 1]. Moreover, the exact values were calculated in some concrete Banach spaces.

2000 Mathematics Subject Classification: 46B20.

1. Introduction and preliminaries

There are several constants defined on Banach spaces such as the Gao [1] and von Neumann-Jordan constants [2]. It has been shown that these constants are very useful in geometric theory of Banach spaces, which enable us to classify several important concepts of Banach spaces such as uniformly non-squareness and uniform normal structure [38]. On the other hand, calculation of the constant for some concrete spaces is also of some interest [5, 6, 9].

Throughout this article, we assume that X is a real Banach space. By S X and B X we denote the unit sphere and the unit ball of a Banach space X, respectively. The notion of the Ptolemy constant of Banach spaces was introduced in [10] and recently it has been studied by Llorens-Fuster in [9].

Definition 1.1 For a normed space (X, ||.||) the real number

C p ( X ) : = sup x - y z x - z y + z - y x : x , y , z X \ { 0 } , x y z x

is called the Ptolemy constant of (X, ||.||).

As we have already mentioned [10], 1 ≤ C p (X) ≤ 2 for all normed spaces X. The Ptolemy inequality shows that C p (H) = 1 whenever (H, ||.||) is an inner product space. It is obvious that if Y is a subspace of (X, ||.||), then C p (Y) ≤ C p (X). Since C p (Y) = 2 for Y = (2, ||.||), it follows that C p (X) = 2 whenever X contains an isometric copy of (2, ||.||).

Recall that a norm on 2 is called absolute if ||(z, w)|| = ||(|z|, |w|)|| for all z, w and normalized if ||(1, 0)|| = ||(0, 1)|| = 1. Let N α denotes the family of all absolute normalized norms on 2, and let Ψ denotes the family of all continuous convex functions on [0, 1] such that ψ(1) = ψ(0) = 1 and max{1 - t, t} ≤ ψ(t) ≤ 1(0 ≤ t ≤ 1). It has been shown that N α and Ψ are a one-to-one correspondence in view of the following proposition in [11].

Proposition 1.2 If ||.|| N α , then ψ(t) = ||(1 - t, t)|| Ψ. On the other hand, if ψ(t) Ψ, defining the norm ||.|| ψ as

( z , ω ) ψ : = ( z + ω ) ψ ω z + ω , ( z , ω ) ( 0 , 0 ) ; 0 , ( z , ω ) = ( 0 , 0 ) .

then the norm ||.|| ψ N α .

A simple example of absolute normalized norm is usual l p (1 ≤ p ≤ ∞) norm. From Proposition 1.2, one can easily get the corresponding function of the l p norm:

ψ p ( t ) = { ( 1 - t ) p + t p } 1 / p , 1 p < , max { 1 - t , t } , p = .

Also, the above correspondence enable us to get many non-l p norms on 2. One of the properties of these norms is stated in the following result.

Proposition 1.3 Let ψ, φ Ψ and φψ. Put M = max 0 t 1 ψ ( t ) φ ( t ) , then

. φ . ψ M . φ .

The Cesà ro sequence space was defined by Shue [12]. It is very useful in the theory of matrix operators and others. Let l be the space of real sequences. For 1 < p < ∞, the Cesà ro sequence space ces p is defined by

c e s p = x l : x = ( x ( i ) ) = n = 1 1 n i = 1 n x ( i ) p 1 / p <

The geometry of Cesà ro sequence spaces have been extensively studied in [1321]. Let us restrict ourselves to the 2D Cesà ro sequence space ce s p ( 2 ) which is just 2 equipped with the norm defined by norm defined by

( x , y ) = x p + x + y 2 p 1 / p

2. Main results

In this section, we give a simple method to determine and estimate the Ptolemy constant of absolute normalized norms on 2. Moreover, the exact values were calculated in some concrete Banach spaces. For a norm ||.|| on 2, we write C p (||.||) for C p (2,||.||).

Proposition 2.1 Let φ Ψ and ψ(t) = φ(1 - t). Then C p (||.|| φ ) = C p (||.|| ψ )

Proof. For any x = (a, b) 2 and a ≠ 0, b ≠ 0, put x ̃ = ( b , a ) . Then

x φ = ( a + b ) φ b a + b = ( b + a ) ψ a a + b = x ̃ ψ .

Consequently, we have

C p ( . φ ) = sup x - y φ z φ x - z φ y φ + z - y φ x φ : x , y , z X \ { 0 } , x y z x = sup x ̃ - ψ z ̃ ψ x ̃ - z ̃ ψ ψ + z ̃ - ψ x ̃ ψ : x ̃ , , z ̃ X \ { 0 } , x ̃ z ̃ x ̃ = C p ( . ψ ) .

We now consider the Ptolemy constant of a class of absolute normalized norms on 2. Now let us put

M 1 = max 0 t 1 ψ 2 ( t ) ψ ( t ) a n d M 2 = max 0 t 1 ψ ( t ) ψ 2 ( t )

Theorem 2.2 Let ψ Ψ and ψψ2, if the function ψ 2 ( t ) ψ ( t ) attains its maximum at t = 1/2, then

C p ( . ψ ) = 1 2 ψ 2 ( 1 / 2 ) .

Proof. By Proposition 1.3, we have ||.|| ψ ≤ ||.||2M1||.|| ψ . Let x, y X, (x, y) ≠ (0, 0), where X = 2. Then

x - y ψ z ψ x - z ψ y ψ + z - y ψ x ψ x - y 2 z 2 ( 1 / M 1 2 ) x - z 2 y 2 + z - y 2 x 2 M 1 2 C p ( . 2 ) M 1 2

from the definition of C p (X), implies that

C p ( . ψ ) M 1 2 = max 0 t 1 ψ 2 2 ( t ) ψ 2 ( t ) .
(1)

On the other hand, note that the function ψ 2 ( t ) ψ ( t ) attains its maximum at t = 1/2, i.e., M 1 = ψ 2 ( 1 / 2 ) ψ ( 1 / 2 ) . Let us put x = (1/2, 1/2), y = (1/2, -1/2), z = (1, 0), then

x - y ψ z ψ x - z ψ y ψ + z - y ψ x ψ = ( 0 , 1 ) ψ ( 1 , 0 ) ψ ( - 1 / 2 , 1 / 2 ) ψ ( 1 / 2 , - 1 / 2 ) ψ + ( 1 / 2 , 1 / 2 ) ψ ( 1 / 2 , 1 / 2 ) ψ = 1 2 ψ 2 ( 1 / 2 ) = 2 × 1 / 2 × ( 1 - 1 / 2 ) ψ 2 ( 1 / 2 ) = ( 1 / 2 ) 2 + ( 1 - 1 / 2 ) 2 ψ 2 ( 1 / 2 ) = ψ 2 2 ( 1 / 2 ) ψ 2 ( 1 / 2 ) = M 1 2 .

From (1) and the above equality, we have

C p ( . ψ ) = M 1 2 = 1 2 ψ 2 ( 1 / 2 ) .

Theorem 2.3 Let ψ Ψ and ψψ2, if the function ψ ( t ) ψ 2 ( t ) attains its maximum at t = 1/2, then

C p ( . ψ ) = 2 ψ 2 ( 1 / 2 ) .

Proof. By Proposition 1.3, we have ||.||2 ≤ ||.|| ψ M2||.||2. Let x, y X, (x, y) ≠ (0, 0), where X = 2. Then

x - y ψ z ψ x - z ψ y ψ + z - y ψ x ψ M 2 2 x - y 2 z 2 x - z 2 y 2 + z - y 2 x 2 M 2 2 C p ( . 2 ) M 2 2

from the definition of C p (X), implies that

C p ( . ψ ) M 2 2 = max 0 t 1 ψ 2 ( t ) ψ 2 2 ( t ) .
(2)

On the other hand, note that the function ψ ( t ) ψ 2 ( t ) attains its maximum at t = 1/2, i.e., M 2 = ψ ( 1 / 2 ) ψ 2 ( 1 / 2 ) . Let us put x = (1/2, 0), y = (0, 1/2), z = (1/2, 1/2), then

x - y ψ z ψ x - z ψ y ψ + z - y ψ x ψ = ( 1 / 2 , - 1 / 2 ) ψ ( 1 / 2 , 1 / 2 ) ψ ( 0 , - 1 / 2 ) ψ ( 0 , 1 / 2 ) ψ + ( 1 / 2 , 0 ) ψ ( 1 / 2 , 0 ) ψ = 2 ψ 2 ( 1 / 2 ) = ψ 2 ( 1 / 2 ) ( 1 / 2 ) 2 + ( 1 - 1 / 2 ) 2 = ψ 2 ( 1 / 2 ) ψ 2 2 ( 1 / 2 ) = M 2 2 .

From (2) and the above equality, we have

C p ( . ψ ) = M 2 2 = 2 ψ 2 ( 1 / 2 ) .

Theorem 2.4 If X is the l p (1 ≤ p ≤ ∞) space, then

C p ( . p ) = max { 2 2 / p - 1 , 2 2 / q - 1 } .

In particular, C p (||.||1) = C p (||.||) = 2.

Proof. Let 1 ≤ p ≤ 2, then we have ψ p (t) ≥ ψ2(t) and ψ p (t)/ψ2(t) attains s maximum at t = 1/2. Since

ψ 2 ( t ) ψ p ( t ) 2 1 / p - 1 / 2 ψ 2 ( t ) ( 0 t 1 ) ,

where the constant 21/p-1/2is the best possible. On the other hand, for t = 1/2, we have

ψ p ( 1 / 2 ) ψ 2 ( 1 / 2 ) = ( ( 1 - 1 / 2 ) p + ( 1 / 2 ) p ) 1 / p ( ( 1 - 1 / 2 ) 2 + ( 1 / 2 ) 2 ) 1 / 2 = 2 1 / p - 1 / 2

Therefore, by Theorem 2.3, we have

C p ( . p ) = 2 ψ p 2 ( 1 / 2 ) = 2 2 / p - 1 .
(3)

Similarly, for 2 < p < ∞, then we have 1 < q < 2 and ψ p (t) ≤ ψ2(t). By Theorem 2.2, we have

C p ( . p ) = 1 2 ψ p 2 ( 1 / 2 ) = 2 2 / q - 1 .
(4)

From (3) and (4), we have

C p ( . p ) = max { 2 2 / p - 1 , 2 2 / q - 1 } .

Lemma 2.5 Let ||.|| and |.| be two equivalent norms on a Banach space. If a|.| ≤ ||.|| ≤ b|.|(ba > 0), then

a 2 C p ( . ) b 2 C p ( . ) b 2 C p ( . ) a 2

Moreover, if ||x|| = a|x|, then C p (||.||) = C p (|.|).

Proof. From the definition of C p (X), we have

C p ( . ) = sup x - y z x - z y + z - y x : x , y , z X \ { 0 } , x y z x sup b 2 x - y z a 2 x - z y + z - y x : x , y , z X \ { 0 } , x y z x = b 2 a 2 sup x - y z x - z y + z - y x : x , y , z X \ { 0 } , x y z x b 2 a 2 C p ( . ) .

Similarly, we also have

a 2 C p ( . ) b 2 C p ( . ) .

Example 2.6 Let X = 2 with the norm

x = max { x 2 , λ x 1 } ( 1 / 2 λ 1 ) .

Then

C p ( . ) = 2 λ 2 .

Proof. It is very easy to check that ||x|| = max{||x||2, λ||x||1} α and its corresponding function is

ψ ( t ) = ( 1 - t , t ) = max { ψ 2 ( t ) , λ } ψ 2 ( t ) .

Therefore

ψ ( t ) ψ 2 ( t ) = max { 1 , λ ψ 2 ( t ) } .

Since ψ2(t) attains minimum at t = 1/2 and hence ψ ( t ) ψ 2 ( t ) attains maximum at t = 1/2. Therefore, from Theorem 2.3, we have

C p ( . ) = 2 ψ 2 ( 1 / 2 ) = 2 λ 2

Example 2.7 Let X = 2 with the norm

x = max { x 2 , λ x } ( 1 λ 2 ) .

Then

C p ( . ) = λ 2 .

Proof. It is obvious to check that the norm ||x|| = max{||x||2, λ||x||} is absolute, but not normalized, since ||(1, 0)|| = ||(0, 1)|| = λ. Let us put

. = . λ = max . 2 λ , .

Then |.| α and its corresponding function is

ψ ( t ) = ( 1 - t , t ) = max ψ 2 ( t ) λ , ψ ( t ) ψ 2 ( t ) .

Thus

ψ 2 ( t ) ψ ( t ) = min λ , ψ 2 ( t ) ψ ( t ) .

Consider the increasing continuous function g ( t ) = ψ 2 ( t ) ψ ( t ) ( 0 t 1 / 2 ) . Because g(0) = 1 and g ( 1 / 2 ) = 2 , hence, there exists a unique 0 ≤ a ≤ 1 such that g(a) = λ. In fact g(t) is symmetric with respect to t = 1/2, then we have

g ( t ) = ψ 2 ( t ) ψ ( t ) , t [ 0 , a ] [ 1 - a , a ] ; λ , t [ a , 1 - a ]

Obvious, g(t) attains its maximum at t = 1/2. Hence, from Theorem 2.2 and Lemma 2.5, we have

C p ( . ) = C p ( . ) = 1 2 ψ 2 ( 1 / 2 ) = λ 2 .

Example 2.8 Let X = 2 with the norm

x = ( x 2 2 + λ x 2 ) ( λ 0 )

Then

C p ( . ) = 2 ( 1 + λ ) / λ + 2 .

Proof. It is obvious to check that the norm x = ( x 2 2 + λ x 2 ) is absolute, but not normalized, since ||(1, 0)|| = ||(0, 1)|| = (1 + λ)1/2. Let us put

. = . 1 + λ .

Therefore |.| α and its corresponding function is

ψ ( t ) = ( 1 - t , t ) = [ ( 1 - t ) 2 + t 2 / ( 1 + λ ) ] 1 / 2 , t [ 0 , 1 / 2 ] , [ t 2 + ( 1 - t ) 2 / ( 1 + λ ) ] 1 / 2 , t [ 1 / 2 , 1 ] .

Obvious ψ(t) ≤ ψ2(t). Since λ0, ψ 2 ( t ) ψ ( t ) is symmetric with respect to t = 1/2, it suffices to consider ψ 2 ( t ) ψ ( t ) for t [0, 1/2]. Note that, for any t [0, 1/2], put g ( t ) = ψ 2 ( t ) 2 ψ ( t ) 2 . Taking derivative of the function g(t), then we have

g ( t ) = 2 λ 1 + λ × t ( 1 - t ) [ ( 1 - t ) 2 + t 2 / ( 1 + λ ) ] 2 .

We always have g'(t) ≥ 0 for 0 ≤ t ≤ 1/2, this implies that the function g(t) is increased for 0 ≤ t ≤ 1/2. Therefore, the function ψ 2 ( t ) ψ ( t ) attains its maximum at t = 1/2, by Theorem 2.2 and Lemma 2.5, we have

C p ( . ) = C p ( . ) = 1 2 ψ 2 ( 1 / 2 ) = 2 ( 1 + λ ) / λ + 2 .

Example 2.9 (Lorentz sequence spaces) Let 0 < a < 1. Two-dimensional Lorentz sequence space, i.e., 2 with the norm

( z , ω ) a , 2 = ( ( x 1 * ) 2 + a ( x 2 * ) 2 ) 1 / 2 ,

where ( x 1 * , x 2 * ) is the rearrangement of (|z|, |ω|) satisfying ( x 1 * x 2 * ) , then

C p ( ( z , ω ) a , 2 ) = 2 a + 1 .

Proof. Indeed, ||(z, ω)||a,2 α , and the corresponding convex function is given by

ψ a , 2 ( t ) = ( 1 - t , t ) a , 2 = [ ( 1 - t ) 2 + a t 2 ] 1 / 2 , t [ 0 , 1 / 2 ] , [ t 2 + a ( 1 - t ) 2 ] 1 / 2 , t [ 1 / 2 , 1 ] .

Obvious ψa,2(t) ≤ ψ2(t). Repeating the arguments in the proof of Example 2.8, we can easily get the conclusion that ψ 2 ( t ) ψ a , 2 ( t ) attains its maximum at t = 1/2. By Theorem 2.2, we have

C p ( ( z , ω ) a , 2 ) = 2 a + 1 .

Example 2.10 Let X be a 2D Cesà ro space ce s 2 ( 2 ) , then

C p ( c e s 2 ( 2 ) ) = 1 + 1 5 .

Proof. We first define

x , y = 2 x 5 , 2 y c e s 2 ( 2 )

for (x, y) 2. It follows that ce s 2 ( 2 ) is isometrically isomorphic to (2,|.|) and |.| is absolute and normalized norm, and the corresponding convex function is given by

ψ ( t ) = 4 ( 1 - t ) 2 5 + 1 - t 5 + t 2 1 2

Indeed, T:ce s 2 ( 2 ) ( 2 , . ) defined by T ( x , y ) = x 5 , 2 y is an isometric isomorphism. We prove that ψ(t) ≥ ψ2(t). Note that

1 - t 5 + t 2 1 - t 5 2 + t 2

Consequently,

ψ ( t ) ( ( 1 - t ) 2 + t 2 ) 1 / 2 = ψ 2 ( t )

Some elementary computation shows that ψ ( t ) ψ 2 ( t ) attains its maximum at t = 1/2. Therefore, from Theorem 2.3, we have

C p ( c e s 2 ( 2 ) ) = 2 ψ 2 ( 1 / 2 ) = 1 + 1 5 .

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Acknowledgements

This research was supported by the fund of Scientific research in Southeast University (the support project of fundamental research) and NSF of CHINA, Grant No. 11126329.

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Zuo, Z. The Ptolemy constant of absolute normalized norms on 2. J Inequal Appl 2012, 107 (2012). https://doi.org/10.1186/1029-242X-2012-107

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Keywords

  • Ptolemy constant
  • absolute normalized norm
  • Lorentz sequence space
  • the Cesà ro sequence space