The Ptolemy constant of absolute normalized norms on ℝ²

Abstract

We determine and estimate the Ptolemy constant of absolute normalized norms on ℝ² 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 [3–8]. 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

$$C_p\left(X\right):=\text{sup}\left\{\frac{∥x-y∥∥z∥}{∥x-z∥∥y∥+∥z-y∥∥x∥}:x,y,z\in X\backslash \left\{0\right\},x\ne y\ne z\ne x\right\}$$

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 = (ℝ², ||.||_∞), it follows that C _p (X) = 2 whenever X contains an isometric copy of (ℝ², ||.||_∞).

Recall that a norm on ℝ² is called absolute if ||(z, w)|| = ||(|z|, |w|)|| for all z, w ∈ ℝ and normalized if ||(1, 0)|| = ||(0, 1)|| = 1. Let N _α denotes the family of all absolute normalized norms on ℝ², 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

$${∥\left(z,\omega \right)∥}_{\psi }:=\left\{\begin{array}{cc}\hfill \left(\left|z\right|+\left|\omega \right|\right)\psi \left(\frac{\left|\omega \right|}{\left|z\right|+\left|\omega \right|}\right),\hfill & \hfill \left(z,\omega \right)\ne \left(0,0\right);\hfill \\ \hfill 0,\hfill & \hfill \left(z,\omega \right)=\left(0,0\right).\hfill \end{array}\right.$$

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:

$${\psi }_p\left(t\right)=\left\{\begin{array}{c}\hfill {\left\{{\left(1-t\right)}^p+t^p\right\}}^{1/p},1\le p<\infty ,\hfill \\ \hfill \text{max}\left\{1-t,t\right\},p=\infty .\hfill \end{array}\right.$$

Also, the above correspondence enable us to get many non-l _p norms on ℝ². One of the properties of these norms is stated in the following result.

Proposition 1.3 Let ψ, φ ∈ Ψ and φ ≤ ψ. Put $M={\max }_{0\le t\le 1}\frac{\psi (t)}{\phi (t)}$, then

$${∥.∥}_{\phi }\le {∥.∥}_{\psi }\le M{∥.∥}_{\phi }.$$

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

$$ce{s}_p=\left\{x\in l:∥x∥=∥\left(x\left(i\right)\right)∥={\left(\sum _{n=1}^{\infty }{\left(\frac{1}{n}\sum _{i=1}^n\left|x\left(i\right)\right|\right)}^p\right)}^{1/p}<\infty \right\}$$

The geometry of Cesà ro sequence spaces have been extensively studied in [13–21]. Let us restrict ourselves to the 2D Cesà ro sequence space $ce{s}_p^{\left(2\right)}$ which is just ℝ² equipped with the norm defined by norm defined by

$$∥\left(x,y\right)∥={\left({\left|x\right|}^p+{\left(\frac{\left|x\right|+\left|y\right|}{2}\right)}^p\right)}^{1/p}$$

2. Main results

In this section, we give a simple method to determine and estimate the Ptolemy constant of absolute normalized norms on ℝ². Moreover, the exact values were calculated in some concrete Banach spaces. For a norm ||.|| on ℝ², we write C _p (||.||) for C _p (ℝ²,||.||).

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) ∈ ℝ² and a ≠ 0, b ≠ 0, put $\tilde{x}=\left(b,a\right)$. Then

$${∥x∥}_{\phi }=\left(\left|a\right|+\left|b\right|\right)\phi \left(\frac{\left|b\right|}{\left|a\right|+\left|b\right|}\right)=\left(\left|b\right|+\left|a\right|\right)\psi \left(\frac{\left|a\right|}{\left|a\right|+\left|b\right|}\right)={∥\tilde{x}∥}_{\psi }.$$

Consequently, we have

$$\begin{array}{ll}\hfill C_p\left({∥.∥}_{\phi }\right)& =\text{sup}\left\{\frac{{∥x-y∥}_{\phi }{∥z∥}_{\phi }}{{∥x-z∥}_{\phi }{∥y∥}_{\phi }+{∥z-y∥}_{\phi }{∥x∥}_{\phi }}:x,y,z\in X\backslash \left\{0\right\},x\ne y\ne z\ne x\right\}\\ =\text{sup}\left\{\frac{{∥\tilde{x}-ỹ∥}_{\psi }{∥\tilde{z}∥}_{\psi }}{{∥\tilde{x}-\tilde{z}∥}_{\psi }{∥ỹ∥}_{\psi }+{∥\tilde{z}-ỹ∥}_{\psi }{∥\tilde{x}∥}_{\psi }}:\tilde{x},ỹ,\tilde{z}\in X\backslash \left\{0\right\},\tilde{x}\ne ỹ\ne \tilde{z}\ne \tilde{x}\right\}\\ =C_p\left({∥.∥}_{\psi }\right).\end{array}$$

We now consider the Ptolemy constant of a class of absolute normalized norms on ℝ². Now let us put

$$M_1=\underset{0\le t\le 1}{\text{max}}\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}and{M}_2=\underset{0\le t\le 1}{\text{max}}\frac{\psi \left(t\right)}{{\psi }_2\left(t\right)}$$

Theorem 2.2 Let ψ ∈ Ψ and ψ ≤ ψ₂, if the function $\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}$ attains its maximum at t = 1/2, then

$$C_p\left({∥.∥}_{\psi }\right)=\frac{1}{2{\psi }^2\left(1/2\right)}.$$

Proof. By Proposition 1.3, we have ||.|| _ψ ≤ ||.||₂ ≤ M₁||.|| _ψ . Let x, y ∈ X, (x, y) ≠ (0, 0), where X = ℝ². Then

$$\begin{array}{ll}\hfill \frac{{∥x-y∥}_{\psi }{∥z∥}_{\psi }}{{∥x-z∥}_{\psi }{∥y∥}_{\psi }+{∥z-y∥}_{\psi }{∥x∥}_{\psi }}& \le \frac{{∥x-y∥}_2{∥z∥}_2}{\left(1/M_1^2\right){∥x-z∥}_2{∥y∥}_2+{∥z-y∥}_2{∥x∥}_2}\\ \le M_1^2{C}_p\left({∥.∥}_2\right)\\ \le M_1^2\end{array}$$

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

$$C_p\left({∥.∥}_{\psi }\right)\le M_1^2=\underset{0\le t\le 1}{\text{max}}\frac{{\psi }_2^2\left(t\right)}{{\psi }^2\left(t\right)}.$$

(1)

On the other hand, note that the function $\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}$ attains its maximum at t = 1/2, i.e., $M_1=\frac{{\psi }_2\left(1/2\right)}{\psi \left(1/2\right)}$. Let us put x = (1/2, 1/2), y = (1/2, -1/2), z = (1, 0), then

$$\begin{array}{ll}\hfill \frac{{∥x-y∥}_{\psi }{∥z∥}_{\psi }}{{∥x-z∥}_{\psi }{∥y∥}_{\psi }+{∥z-y∥}_{\psi }{∥x∥}_{\psi }}& =\frac{{∥\left(0,1\right)∥}_{\psi }{∥\left(1,0\right)∥}_{\psi }}{{∥\left(-1/2,1/2\right)∥}_{\psi }{∥\left(1/2,-1/2\right)∥}_{\psi }+{∥\left(1/2,1/2\right)∥}_{\psi }{∥\left(1/2,1/2\right)∥}_{\psi }}\\ =\frac{1}{2{\psi }^2\left(1/2\right)}\\ =\frac{2\times 1/2\times \left(1-1/2\right)}{{\psi }^2\left(1/2\right)}=\frac{{\left(1/2\right)}^2+{\left(1-1/2\right)}^2}{{\psi }^2\left(1/2\right)}\\ =\frac{{\psi }_2^2\left(1/2\right)}{{\psi }^2\left(1/2\right)}=M_1^2.\end{array}$$

From (1) and the above equality, we have

$$C_p\left({∥.∥}_{\psi }\right)=M_1^2=\frac{1}{2{\psi }^2\left(1/2\right)}.$$

Theorem 2.3 Let ψ ∈ Ψ and ψ ≥ ψ₂, if the function $\frac{\psi \left(t\right)}{{\psi }_2\left(t\right)}$ attains its maximum at t = 1/2, then

$$C_p\left({∥.∥}_{\psi }\right)=2{\psi }^2\left(1/2\right).$$

Proof. By Proposition 1.3, we have ||.||₂ ≤ ||.|| _ψ ≤ M₂||.||₂. Let x, y ∈ X, (x, y) ≠ (0, 0), where X = ℝ². Then

$$\begin{array}{ll}\hfill \frac{{∥x-y∥}_{\psi }{∥z∥}_{\psi }}{{∥x-z∥}_{\psi }{∥y∥}_{\psi }+{∥z-y∥}_{\psi }{∥x∥}_{\psi }}& \le \frac{M_2^2{∥x-y∥}_2{∥z∥}_2}{{∥x-z∥}_2{∥y∥}_2+{∥z-y∥}_2{∥x∥}_2}\\ \le M_2^2{C}_p\left({∥.∥}_2\right)\\ \le M_2^2\end{array}$$

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

$$C_p\left({∥.∥}_{\psi }\right)\le M_2^2=\underset{0\le t\le 1}{\text{max}}\frac{{\psi }^2\left(t\right)}{{\psi }_2^2\left(t\right)}.$$

(2)

On the other hand, note that the function $\frac{\psi \left(t\right)}{{\psi }_2\left(t\right)}$ attains its maximum at t = 1/2, i.e., $M_2=\frac{\psi \left(1/2\right)}{{\psi }_2\left(1/2\right)}$. Let us put x = (1/2, 0), y = (0, 1/2), z = (1/2, 1/2), then

$$\begin{array}{ll}\hfill \frac{{∥x-y∥}_{\psi }{∥z∥}_{\psi }}{{∥x-z∥}_{\psi }{∥y∥}_{\psi }+{∥z-y∥}_{\psi }{∥x∥}_{\psi }}& =\frac{{∥\left(1/2,-1/2\right)∥}_{\psi }{∥\left(1/2,1/2\right)∥}_{\psi }}{{∥\left(0,-1/2\right)∥}_{\psi }{∥\left(0,1/2\right)∥}_{\psi }+{∥\left(1/2,0\right)∥}_{\psi }{∥\left(1/2,0\right)∥}_{\psi }}\\ =2{\psi }^2\left(1/2\right)\\ =\frac{{\psi }^2\left(1/2\right)}{{\left(1/2\right)}^2+{\left(1-1/2\right)}^2}\\ =\frac{{\psi }^2\left(1/2\right)}{{\psi }_2^2\left(1/2\right)}=M_2^2.\end{array}$$

From (2) and the above equality, we have

$$C_p\left({∥.∥}_{\psi }\right)=M_2^2=2{\psi }^2\left(1/2\right).$$

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

$$C_p\left({∥.∥}_p\right)=\text{max}\left\{2^{2/p-1},2^{2/q-1}\right\}.$$

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

$${\psi }_2\left(t\right)\le {\psi }_p\left(t\right)\le 2^{1/p-1/2}{\psi }_2\left(t\right)\left(0\le t\le 1\right),$$

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

$$\frac{{\psi }_p\left(1/2\right)}{{\psi }_2\left(1/2\right)}=\frac{{\left({\left(1-1/2\right)}^p+{\left(1/2\right)}^p\right)}^{1/p}}{{\left({\left(1-1/2\right)}^2+{\left(1/2\right)}^2\right)}^{1/2}}=2^{1/p-1/2}$$

Therefore, by Theorem 2.3, we have

$$C_p\left({∥.∥}_p\right)=2{\psi }_p^2\left(1/2\right)=2^{2/p-1}.$$

(3)

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

$$C_p\left({∥.∥}_p\right)=\frac{1}{2{\psi }_p^2\left(1/2\right)}=2^{2/q-1}.$$

(4)

From (3) and (4), we have

$$C_p\left({∥.∥}_p\right)=\text{max}\left\{2^{2/p-1},2^{2/q-1}\right\}.$$

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

$$\frac{a^2{C}_p\left(\left|.\right|\right)}{b^2}\le C_p\left(∥.∥\right)\le \frac{b^2{C}_p\left(\left|.\right|\right)}{a^2}$$

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

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

$$\begin{array}{ll}\hfill C_p\left(∥.∥\right)& =\text{sup}\left\{\frac{∥x-y∥∥z∥}{∥x-z∥∥y∥+∥z-y∥∥x∥}:x,y,z\in X\backslash \left\{0\right\},x\ne y\ne z\ne x\right\}\\ \le \text{sup}\left\{\frac{b^2\left|x-y\right|\left|z\right|}{a^2\left|x-z\right|\left|y\right|+\left|z-y\right|\left|x\right|}:x,y,z\in X\backslash \left\{0\right\},x\ne y\ne z\ne x\right\}\\ =\frac{b^2}{a^2}\text{sup}\left\{\frac{\left|x-y\right|\left|z\right|}{\left|x-z\right|\left|y\right|+\left|z-y\right|\left|x\right|}:x,y,z\in X\backslash \left\{0\right\},x\ne y\ne z\ne x\right\}\\ \le \frac{b^2}{a^2}{C}_p\left(\left|.\right|\right).\end{array}$$

Similarly, we also have

$$\frac{a^2{C}_p\left(\left|.\right|\right)}{b^2}\le C_p\left(∥.∥\right).$$

Example 2.6 Let X = ℝ² with the norm

$$∥x∥=\text{max}\left\{{∥x∥}_2,\lambda {∥x∥}_1\right\}\left(1/\sqrt{2}\le \lambda \le 1\right).$$

Then

$$C_p\left(∥.∥\right)=2{\lambda }^2.$$

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

$$\psi \left(t\right)=∥\left(1-t,t\right)∥=\text{max}\left\{{\psi }_2\left(t\right),\lambda \right\}\ge {\psi }_2\left(t\right).$$

Therefore

$$\frac{\psi \left(t\right)}{{\psi }_2\left(t\right)}=\text{max}\left\{1,\frac{\lambda }{{\psi }_2\left(t\right)}\right\}.$$

Since ψ₂(t) attains minimum at t = 1/2 and hence $\frac{\psi \left(t\right)}{{\psi }_2\left(t\right)}$ attains maximum at t = 1/2. Therefore, from Theorem 2.3, we have

$$C_p\left(∥.∥\right)=2{\psi }^2\left(1/2\right)=2{\lambda }^2$$

Example 2.7 Let X = ℝ² with the norm

$$∥x∥=\text{max}\left\{{∥x∥}_2,\lambda {∥x∥}_{\infty }\right\}\left(1\le \lambda \le \sqrt{2}\right).$$

Then

$$C_p\left(∥.∥\right)={\lambda }^2.$$

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

$$\left|.\right|=\frac{∥.∥}{\lambda }=\text{max}\left\{\frac{{∥.∥}_2}{\lambda },{∥.∥}_{\infty }\right\}$$

Then |.| ∈ ℕ _α and its corresponding function is

$$\psi \left(t\right)=∥\left(1-t,t\right)∥=\text{max}\left\{\frac{{\psi }_2\left(t\right)}{\lambda },{\psi }_{\infty }\left(t\right)\right\}\le {\psi }_2\left(t\right).$$

Thus

$$\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}=\text{min}\left\{\lambda ,\frac{{\psi }_2\left(t\right)}{{\psi }_{\infty }\left(t\right)}\right\}.$$

Consider the increasing continuous function $g\left(t\right)=\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}\left(0\le t\le 1/2\right)$. Because g(0) = 1 and $g\left(1/2\right)=\sqrt{2}$, 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

$$g\left(t\right)=\left\{\begin{array}{cc}\hfill \frac{{\psi }_2\left(t\right)}{\psi \left(t\right)},\hfill & \hfill t\in \left[0,a\right]\cup \left[1-a,a\right];\hfill \\ \hfill \lambda ,\hfill & \hfill t\in \left[a,1-a\right]\hfill \end{array}\right.$$

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

$$C_p\left(∥.∥\right)=C_p\left(\left|.\right|\right)=\frac{1}{2{\psi }^2\left(1/2\right)}={\lambda }^2.$$

Example 2.8 Let X = ℝ² with the norm

$$∥x∥=\left({∥x∥}_2^2+\lambda {∥x∥}_{\infty }^2\right)\left(\lambda \ge 0\right)$$

Then

$$C_p\left(∥.∥\right)=2\left(1+\lambda \right)/\lambda +2.$$

Proof. It is obvious to check that the norm $∥x∥=\left({∥x∥}_2^2+\lambda {∥x∥}_{\infty }^2\right)$ is absolute, but not normalized, since ||(1, 0)|| = ||(0, 1)|| = (1 + λ)^1/2. Let us put

$$\left|.\right|=\frac{∥.∥}{\sqrt{1+\lambda }}.$$

Therefore |.| ∈ ℕ _α and its corresponding function is

$$\psi \left(t\right)=∥\left(1-t,t\right)∥=\left\{\begin{array}{c}\hfill {\left[{\left(1-t\right)}^2+t^2/\left(1+\lambda \right)\right]}^{1/2},t\in \left[0,1/2\right],\hfill \\ \hfill {\left[t^2+{\left(1-t\right)}^2/\left(1+\lambda \right)\right]}^{1/2},t\in \left[1/2,1\right].\hfill \end{array}\right.$$

Obvious ψ(t) ≤ ψ₂(t). Since $\lambda \ge 0,\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}$ is symmetric with respect to t = 1/2, it suffices to consider $\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}$ for t ∈ [0, 1/2]. Note that, for any t ∈ [0, 1/2], put $g\left(t\right)=\frac{{\psi }_2{\left(t\right)}^2}{\psi {\left(t\right)}^2}$. Taking derivative of the function g(t), then we have

$$g^{\prime }\left(t\right)=\frac{2\lambda }{1+\lambda }\times \frac{t\left(1-t\right)}{{\left[{\left(1-t\right)}^2+t^2/\left(1+\lambda \right)\right]}^2}.$$

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 $\frac{{\psi }_2\left(t\right)}{\psi \left(t\right)}$ attains its maximum at t = 1/2, by Theorem 2.2 and Lemma 2.5, we have

$$C_p\left(∥.∥\right)=C_p\left(\left|.\right|\right)=\frac{1}{2{\psi }^2\left(1/2\right)}=2\left(1+\lambda \right)/\lambda +2.$$

Example 2.9 (Lorentz sequence spaces) Let 0 < a < 1. Two-dimensional Lorentz sequence space, i.e., ℝ² with the norm

$${∥\left(z,\omega \right)∥}_{a,2}={\left({\left(x_1^{*}\right)}^2+a{\left(x_2^{*}\right)}^2\right)}^{1/2},$$

where $\left(x_1^{*},x_2^{*}\right)$ is the rearrangement of (|z|, |ω|) satisfying $\left(x_1^{*}\ge x_2^{*}\right)$, then

$$C_p\left({∥\left(z,\omega \right)∥}_{a,2}\right)=\frac{2}{a+1}.$$

Proof. Indeed, ||(z, ω)||_a,2∈ ℕ _α , and the corresponding convex function is given by

$${\psi }_{a,2}\left(t\right)={∥\left(1-t,t\right)∥}_{a,2}=\left\{\begin{array}{c}\hfill {\left[{\left(1-t\right)}^2+a{t}^2\right]}^{1/2},t\in \left[0,1/2\right],\hfill \\ \hfill {\left[t^2+a{\left(1-t\right)}^2\right]}^{1/2},t\in \left[1/2,1\right].\hfill \end{array}\right.$$

Obvious ψ_a,2(t) ≤ ψ₂(t). Repeating the arguments in the proof of Example 2.8, we can easily get the conclusion that $\frac{{\psi }_2\left(t\right)}{{\psi }_{a,2}\left(t\right)}$ attains its maximum at t = 1/2. By Theorem 2.2, we have

$$C_p\left({∥\left(z,\omega \right)∥}_{a,2}\right)=\frac{2}{a+1}.$$

Example 2.10 Let X be a 2D Cesà ro space $ce{s}_2^{\left(2\right)}$, then

$$C_p\left(ce{s}_2^{\left(2\right)}\right)=1+\frac{1}{\sqrt{5}}.$$

Proof. We first define

$$\left|x,y\right|={∥\left(\frac{2x}{\sqrt{5}},2y\right)∥}_{ce{s}_2^{\left(2\right)}}$$

for (x, y) ∈ ℝ². It follows that $ce{s}_2^{\left(2\right)}$ is isometrically isomorphic to (ℝ²,|.|) and |.| is absolute and normalized norm, and the corresponding convex function is given by

$$\psi \left(t\right)={\left[\frac{4{\left(1-t\right)}^2}{5}+{\left(\frac{1-t}{\sqrt{5}}+t\right)}^2\right]}^{\frac{1}{2}}$$

Indeed, $T:ce{s}_2^{\left(2\right)}\to \left({ℝ}^2,\left|.\right|\right)$ defined by $T\left(x,y\right)=\left(\frac{x}{\sqrt{5}},2y\right)$ is an isometric isomorphism. We prove that ψ(t) ≥ ψ₂(t). Note that

$${\left(\frac{1-t}{\sqrt{5}}+t\right)}^2\ge {\left(\frac{1-t}{\sqrt{5}}\right)}^2+t^2$$

Consequently,

$$\psi \left(t\right)\ge {\left({\left(1-t\right)}^2+t^2\right)}^{1/2}={\psi }_2\left(t\right)$$

Some elementary computation shows that $\frac{\psi \left(t\right)}{{\psi }_2\left(t\right)}$ attains its maximum at t = 1/2. Therefore, from Theorem 2.3, we have

$$C_p\left(ce{s}_2^{\left(2\right)}\right)=2{\psi }^2\left(1/2\right)=1+\frac{1}{\sqrt{5}}.$$