Generalized von Neumann-Jordan constant and its relationship to the fixed point property

Abstract

We introduce a new geometric constant $C_{NJ}^{(p)}(X)$ for a Banach space X, called a generalized von Neumann-Jordan constant. Next, it is shown that $1\leq C_{NJ}^{(p)}(X)\leq2$ for any Banach space X and that the right hand side inequality is sharp if and only if X is uniformly non-square. Moreover, a relationship between the James constant $J(X)$ and $C_{NJ}^{(p)}(X)$ is presented. Finally, the generalized von Neumann-Jordan constant of the Lebesgue space $L_{r}([0,1])$ is calculated and a relationship between $C_{NJ}^{(p)}(X)$ and the fixed point property is found.

1 Introduction

Recently many geometric constants for a Banach space X have been investigated. In particular, the von Neumann-Jordan constant $C_{NJ}(X)$ and the James constant $J(X)$ are widely treated. We introduce a new geometric constant, called the generalized von Neumann-Jordan constant $C_{NJ}^{(p)}(X)$, which is related to the von Neumann-Jordan constant of a Banach space X and can be used for much better characterization of a Banach space X.

In connection with the famous work [1] (see also [2]) of Jordan and von Neumann concerning inner products, the von Neumann-Jordan constant $C_{NJ}(X)$ for a Banach space X was introduced by Clarkson [3] as the smallest constant C, for which the estimates

$$\frac{1}{C}\leq\frac{\|x+y\|^{2}+\|x-y\|^{2}}{2(\|x\|^{2}+\|y\|^{2})}\leq C $$

hold for all $x,y\in X$ with $(x,y)\neq(0,0)$. Equivalently,

$${C_{NJ}(X)}:=\sup \biggl\{ \frac{\|x+y\|^{2}+\|x-y\|^{2}}{2(\|x\|^{2}+\|y\|^{2})} : x,y\in X\mbox{ with } (x,y)\neq(0,0) \biggr\} . $$

The classical von Neumann-Jordan constant $C_{NJ}(X)$ was investigated in many papers (see for instance [4–7]).

A Banach space X is said to be uniformly non-square in the sense of James if there exists a positive number $\delta<2$ such that for any $x,y\in S_{X}:= \{ x\in X\colon\|x\| = 1 \}$, we have

$$\operatorname{min}\bigl(\|x+y\|,\|x-y\|\bigr)\leq\delta. $$

The James constant $J(X)$ of a Banach space X is defined by

$$J(X):=\sup \bigl\{ \operatorname{min}\bigl(\|x+y\|,\|x-y\|\bigr): x,y\in S_{X} \bigr\} . $$

It is obvious that X is uniformly non-square if and only if $J(X)<2$.

In this paper we introduce a new constant $C_{NJ}^{(p)}(X)$, generalizing the von Neumann-Jordan constant $C_{NJ}(X)$. By the definition of $C_{NJ}^{(p)}(X)$, we will get a relationship between $C_{NJ}^{(p)}(X)$ and $J(X)$, as well as we will estimate the value of $C_{NJ}^{(p)}(X)$. Furthermore, the constant $C_{NJ}^{(p)}(X)$ enable us to establish some new equivalent conditions for the uniform non-squareness of a Banach space X. Since any uniformly non-square Banach space X has the fixed point property (see [8]), our constant $C_{NJ}^{(p)}(X)$ is related to the fixed point theory. Moreover, the value of the generalized von Neumann-Jordan constant for the space $L_{r}[0,1]$ will be calculated. Finally, we will find a relationship between the constant $C_{NJ}^{(p)}(X)$ and normal structure of X, and in such a way we have again its relationship to the fixed point theory.

2 Preliminaries

Let $X = (X, \|\cdot\|)$ be a real Banach space. Geometrical properties of a Banach space X are determined by its unit sphere $S_{X}$ or its unit ball $B(X)$.

Definition 1

The generalized von Neumann-Jordan constant $C_{NJ}^{(p)}(X)$ is defined by

$$C_{NJ}^{(p)}(X):=\sup \biggl\{ \frac{\|x+y\|^{p}+\|x-y\|^{p}}{2^{p-1}\bigl(\|x\| ^{p}+\|y\|^{p}\bigr)}: x,y\in X , (x,y)\neq(0,0) \biggr\} , $$

where $1\leq p<\infty$.

We will also use the following parametrized formula for the constant $C_{NJ}^{(p)}(X)$ (see [9] and [7] in the case of the classical von Neumann-Jordan constant):

$$C_{NJ}^{(p)}(X)=\sup \biggl\{ \frac{\|x+ty\|^{p}+\|x-ty\|^{p}}{2^{p-1}(1+t^{p})}: x,y\in S_{X} ,0\leq t\leq1 \biggr\} , $$

where $1\leq p<\infty$. By taking $t=1$ and $x=y$, we obtain the estimate

$$C_{NJ}^{(p)}(X)\geq\frac{\|2x\|^{p}}{2^{p-1}(1+1)}=\frac {2^{p}}{2^{p-1}\cdot2}=1. $$

Definition 2

(see [10])

The modulus of uniform smoothness of X is defined as

$$\rho_{X}(t):=\sup \biggl\{ \frac{\|x+ty\|+\|x-ty\|}{2}-1: x,y\in S_{X}, t>0 \biggr\} . $$

It is clear that $\rho_{X}(t)$ is a convex function on the interval $[0,\infty)$ satisfying $\rho_{X}(0)=0$, whence it follows that $\rho_{X}$ is nondecreasing on $[0,\infty)$. It is also easy to show that $\max \{0,t-1 \}\leq\rho_{X}(t)\leq t$.

Definition 3

(see [11])

A Banach space X is said to be uniformly smooth if $(\rho_{X})'_{+}(0):=\lim_{t \to0^{+}}\frac{\rho_{X}(t)}{t}=0$.

Definition 4

(see [12] or [13])

A Banach space X is said to be q-uniformly smooth ($1< q\leq2$) if there exists a constant $K>0$ such that $\rho_{X}(t)\leq K t^{q}$ for all $t>0$.

Definition 5

(see [13])

Given any Banach space X and a number $p\in[1,\infty)$, another function $J_{X,p}(t)$ is defined by

$$J_{X,p}(t):=\sup \biggl\{ \biggl( \frac{\|x+ty\|^{p}+\|x-ty\|^{p}}{2} \biggr)^{\frac{1}{p}} : x,y\in S_{X} \biggr\} $$

on the interval $[0,\infty)$.

By the inequality

$$\frac{\Vert x+ty\Vert ^{p}+\Vert x-ty\Vert ^{p}}{2}\geq \biggl(\frac {\Vert x+ty\Vert +\Vert x-ty\Vert }{2} \biggr)^{p}, $$

which follows by convexity of the function $f(u)=u^{p}$ on $[0,\infty)$, we get $J_{X,p}(t)\geq\rho_{X}(t)+1$ when $1\leq p<\infty$. For $p=1$ and $p=2$, we have the equalities $J_{X,1}(t)=\rho_{X}(t)+1$ and $2J_{X,2}^{2}(t)=E(t,X)$, respectively, where the constant $E(t,X)$ was introduced by Gao [14] in 2005, and it is defined by the formula

$$E(t,X)=\sup \bigl\{ \|x+ty\|^{2}+\|x-ty\|^{2}: x,y\in S_{X} \bigr\} . $$

Definition 6

(see [15])

For any Banach space X, we define

$$\begin{aligned} \mu(X) :=&\inf \Bigl\{ r>0\colon\limsup_{n\rightarrow\infty} \Vert x+x_{n}\Vert \leq r\limsup_{n\rightarrow\infty} \Vert x-x_{n}\Vert ,\mbox{for any }(x_{n})\subset X\\ &{}\mbox{with }x_{n}\overset{\mathrm{w}}{\rightarrow} 0 \mbox{ and any } x\in X \Bigr\} . \end{aligned}$$

Definition 7

A Banach space X is said to have normal (resp. weak normal) structure if X contains no bounded and closed (resp. weakly compact) convex subset C with more than one point which is diametral in the sense that, for all $x\in C$,

$$\sup\bigl\{ \|y-x\|\colon y\in C\bigr\} =\operatorname{diam{C}}:= \sup\bigl\{ \|y-z\| \colon y,z \in C\bigr\} . $$

Recall that the weak normal structure (so the normal structure as well) of a Banach space X implies the weak fixed point property for X (see [16, 17]).

Remark 2.1

(see [18])

A sufficient condition for normal structure of a Banach space X is the following: there exists $\varepsilon\in(0,2)$ such that

$$\frac{1}{\mu(X)}>\max \biggl\{ \frac{\varepsilon}{2},1-\delta_{x}( \varepsilon ) \biggr\} , $$

where $\delta_{x}:[0,2]\rightarrow[0,1]$ is the classical modulus of convexity of X defined as

$$\delta_{x}(\varepsilon)=\inf\biggl\{ 1-\frac{1}{2}\|x+y\|:x,y \in B_{X},\|x-y\|\geq \varepsilon\biggr\} . $$

Lemma 2.2

(see [13])

For any Banach space X and any $1\leq p<\infty$ the following statements are true:

• (1) $J_{X,p}(\cdot)$ is nondecreasing on $(0,\infty)$.

• (2) $J_{X,p}(\cdot)$ is convex on $(0,\infty)$.

• (3) $J_{X,p}(\cdot)$ is continuous on $(0,\infty)$.

• (4) $\frac{J_{X,p}(\cdot)-1}{t}$ is nondecreasing on $(0,\infty)$.

The proof of this lemma can be found in [13].

Lemma 2.3

For any $1\leq p<\infty$ a Banach space X is uniformly smooth if and only if $\lim_{t \to0^{+}}\frac{J_{X,p}(t)-1}{t}=0$.

Proof

Since $J_{X,p}(t)\geq\rho_{X}(t)+1$ for any $t>0$ and $1\leq p<\infty$, the sufficiency is obvious. Now we will prove the necessity. Assume, to derive a contradiction, that $\lim_{t \to0^{+}}\frac {J_{X,p}(t)-1}{t}>0$. By Lemma 2.2(4), there exists $0< c<1$ such that $\lim_{t \to0^{+}}\frac {J_{X,p}(t)-1}{t}\geq c$. In particular, we can choose $0< t<1$ and x, y in X with $\|x\|=1$, $\|y\|=t$ satisfying

$$ \|x+y\|^{p}+\|x-y\|^{p}\geq2(1+ct)^{p}. $$

(2.1)

We can assume without loss of generality that $\min\{\|x+y\|,\|x-y\|\} =\|x-y\|$. Then, denoting $\|x-y\|=h$, we have $h\in[1-t,1+t]$, which follows from the inequalities $\vert \|x\|-\|y\|\vert \leq\|x-y\|\leq\|x\|+\|y\|$. By inequality (2.1), we obtain

$$\|x+y\|+\|x-y\|\geq h+\bigl(2(1+ct)^{p}-h^{p} \bigr)^{\frac{1}{p}}=: f(h). $$

Since

$$f'(h)=1-\frac{h^{p-1}}{ (2(1+ct)^{p}-h^{p} )^{\frac{p-1}{p}}}, $$

it is easy to see that f is an increasing function with respect to h on the interval $[1-t,1+ct]$ and decreasing on the interval $[1+ct,1+t]$. Hence the minimum value of the function $f(h)$ can be attained either at $h=1-t$ or at $h=1+t$. In the case when the minimum value is attained at the point $1-t$, we have by the definition of the modulus of uniform smoothness that

$$\frac{\rho_{X}(t)}{t}\geq\frac{f(1-t)-2}{2t}=\frac {1-t+(2(1+ct)^{p}-(1-t)^{p})^{\frac{1}{p}}-2}{2t}. $$

In the second case, we have

$$\frac{\rho_{X}(t)}{t}\geq\frac{f(1+t)-2}{2t}=\frac {1+t+(2(1+ct)^{p}-(1+t)^{p})^{\frac{1}{p}}-2}{2t}. $$

In both cases, letting $t\rightarrow0^{+}$ and using the L’Hôpital rule, we easily obtain $\lim _{t \to0^{+}}\frac{\rho _{X}(t)}{t}\geq c>0$. Obviously, this contradicts the definition of uniform smoothness of X, and thus we completed the proof. □

Lemma 2.4

(see [12])

Let $1\leq p< \infty$ and $1< q \leq2$. A Banach space X is q-uniformly smooth if and only if there exists a constant $K\geq1$ such that

$$\frac{\|x+y\|^{p}+\|x-y\|^{p}}{2}\leq\|x\|^{q}+\|Ky\|^{q} , \quad\forall x,y\in X. $$

Therefore, according to Lemma 2.4 and the definition of $J_{X,p}(\cdot)$, the following lemma holds.

Lemma 2.5

Let $1\leq p< \infty$ and $1< q \leq2$. The following statements are equivalent:

• (1) X is q-uniformly smooth.

• (2) There exists a constant $K\geq1$ such that the inequality $J_{X,p}(t)\leq(1+Kt^{q})^{\frac{1}{q}}$ is satisfied for any $t>0$.

3 Main results

Theorem 3.1

For any Banach space X and any $1\leq p<\infty$ the generalized von Neumann-Jordan constant $C_{NJ}^{(p)}(X)$ satisfies the inequality $C_{NJ}^{(p)}(X)\leq2$.

Proof

We will use in the proof the following parametrized formula for the generalized von Neumann-Jordan constant $C_{NJ}^{(p)}(X)$, where $1\leq p<\infty$:

$$C_{NJ}^{(p)}(X)=\sup \biggl\{ \frac{\|x+ty\|^{p}+\|x-ty\| ^{p}}{2^{p-1}(1+t^{p})}: x,y\in S_{X} ,0\leq t\leq1 \biggr\} . $$

Since

$$\begin{aligned} \|x+ty\|^{p}+\|x-ty\|^{p} \leq& \bigl(\|x\|+t\|y \|\bigr)^{p}+\bigl(\|x\|+t\|y\|\bigr)^{p}\\ =& 2\bigl(\|x\|+t\|y\|\bigr)^{p}\\ =& 2(1+t)^{p}, \end{aligned}$$

so

$$\begin{aligned} \frac{\|x+ty\|^{p}+\|x-ty\|^{p} }{2^{p-1}(1+t^{p})} \leq\frac{2(1+t)^{p}}{{2^{p-1}}(1+t^{p})}. \end{aligned}$$

(3.1)

Applying convexity of the function $\varphi(u)=|u|^{p}$, we get

$$(1+t)^{p}= \biggl(2\cdot\frac{1+t}{2} \biggr)^{p}=2^{p} \biggl(\frac{1+t}{2} \biggr)^{p}\leq2^{p}\cdot \frac{1+t^{p}}{2}=2^{p-1}\bigl(1+t^{p}\bigr). $$

Combining this estimate with inequality (3.1), we get

$$\begin{aligned} \frac{\|x+ty\|^{p}+\|x-ty\|^{p} }{2^{p-1}(1+t^{p})} \leq&\frac{2(1+t)^{p}}{{2^{p-1}}(1+t^{p})} \leq\frac{1}{2^{p-2}}\cdot2^{p-1} =2. \end{aligned}$$

Hence

$$C_{NJ}^{(p)}(X)=\sup \biggl\{ \frac{\|x+ty\|^{p}+\|x-ty\| ^{p}}{2^{p-1}(1+t^{p})}: x,y\in S_{X} ,0\leq t\leq1 \biggr\} \leq2, $$

and the proof is completed. □

Lemma 3.2

(see [6])

Let $1< p<\infty$. A Banach space X is uniformly non-square if and only if there exists $\delta\in(0,1)$ such that for any $x,y\in X$, we have

$$\biggl\Vert \frac{x+y}{2}\biggr\Vert ^{p}+\biggl\Vert \frac{x-y}{2}\biggr\Vert ^{p}\leq (2-\delta)\frac{\Vert x\Vert ^{p}+\Vert y\Vert ^{p}}{2}. $$

According to Lemma 3.2, we directly obtain the following theorem.

Theorem 3.3

Let $1\leq p<\infty$. A Banach space X is uniformly non-square if and only if $C_{NJ}^{(p)}(X)<2$.

Now let us present the following theorem indicating the relationship between constants $J(X)$ and $C_{NJ}^{(p)}(X)$.

Theorem 3.4

For any $1< p<\infty$ and any Banach space X, the following inequality holds:

$$J(X)\leq2^{\frac{p-1}{p}}\sqrt[p]{C_{NJ}^{(p)}(X)}. $$

Proof

Indeed, if $1< p<\infty$, then for any $x,y\in S_{X}$, we have

$$\begin{aligned} 2\bigl(\min \bigl\{ \|x+y\|,\|x-y\| \bigr\} \bigr)^{p} \leq&\|x+y\|^{p}+ \|x-y\|^{p} \\ \leq& 2^{p-1}\bigl(\|x\|^{p}+\|y\|^{p} \bigr)C_{NJ}^{(p)}(X) \\ =&2^{p-1}\cdot2C_{NJ}^{(p)}(X), \end{aligned}$$

so

$$\begin{aligned} \min \bigl\{ \|x+y\|,\|x-y\| \bigr\} \leq2^{\frac{p-1}{p}}\sqrt [p]{C_{NJ}^{(p)}(X)}, \end{aligned}$$

and the proof is completed. □

By Theorem 3.4, we obtain the following corollary.

Corollary 3.5

For any Banach space X and any $1\leq p<\infty$ the inequalities $C_{NJ}^{(p)}(X)<2$ and $J(X)<2$ are equivalent. Moreover, if X is a Banach space with $C_{NJ}^{(p)}(X)<2$, then X has the fixed point property.

Proof

It is well known that $J(X)<2$ if and only if a Banach space X is uniformly non-square. However, by Theorem 3.3, we know that a Banach space X is uniformly non-square if and only if $C_{NJ}^{(p)}(X)<2$. Hence, $J(X)<2$ if and only if $C_{NJ}^{(p)}(X)<2$. Moreover, every uniformly non-square Banach space have the fixed point property (see [8]), so if X is a Banach space with $C_{NJ}^{(p)}(X)<2$, then X has the fixed point property. □

Now we will calculate the generalized von Neumann-Jordan constant for the space $L_{r}[0,1]$.

Theorem 3.6

Let X be the Banach space $L_{r}[0,1]$. Let $1< r\leq2$ and $\frac {1}{r}+\frac{1}{r^{\prime}}=1$. Then

• (1) if $1< p\leq r$ then $C_{NJ}^{(p)}(L_{r}[0,1])=2^{2-p}$ and if $r< p\leq r^{\prime}$ then $C_{NJ}^{(p)}(L_{r}[0,1])=2^{\frac{p}{r}-p+1}$;

• (2) if $r^{\prime}< p<\infty$ then $C_{NJ}^{(p)}(L_{r}[0,1])=1$.

Proof

Let us note that $r\leq2\leq r'$ and

(1) for any $x,y\in S_{X}$ and any $0\leq t\leq1$, if $1< p\leq r'$, then in virtue of Remark 2.3 from [19], we have

$$\bigl(\|x+ty\|_{r}^{p}+\|x-ty\|_{r}^{p} \bigr)^{\frac{1}{p}}\leq2^{\frac {1}{p}}\bigl(\|x\|_{r}^{r}+ \|ty\|_{r}^{r}\bigr)^{\frac{1}{r}}=2^{\frac{1}{p}} \bigl(1+t^{r}\bigr)^{\frac{1}{r}}, $$

which is equivalent to

$$\|x+ty\|_{r}^{p}+\|x-ty\|_{r}^{p}\leq2 \bigl(1+t^{r}\bigr)^{\frac{p}{r}}. $$

Consequently,

$$\frac{\|x+ty\|_{r}^{p}+\|x-ty\|_{r}^{p}}{2^{p-1}(1+t^{p})}\leq\frac {2(1+t^{r})^{\frac{p}{r}}}{2^{p-1}(1+t^{p})}, $$

whence

$$\sup \biggl\{ \frac{\|x+ty\|_{r}^{p}+\|x-ty\|_{r}^{p}}{2^{p-1}(1+t^{p})}: x,y\in S_{X} \biggr\} \leq \frac{2(1+t^{r})^{\frac{p}{r}}}{2^{p-1}(1+t^{p})}, $$

and from the definition of $C_{NJ}^{(p)}(L_{r}[0,1])$, we have

$$C_{NJ}^{(p)}\bigl(L_{r}[0,1]\bigr)\leq\sup \biggl\{ \frac{2(1+t^{r})^{\frac{p}{r}}}{2^{p-1}(1+t^{p})}: 0\leq t\leq1 \biggr\} . $$

Defining $f(t)=\frac{(1+t^{r})^{\frac{p}{r}}}{1+t^{p}}$, we get $(f(t))^{r}=\frac{(1+t^{r})^{p}}{(1+t^{p})^{r}}=:G(t)$. Obviously, both functions $f(t)$ and $G(t)$ are continuous and

$$G'(t)= \frac {p(1+t^{r})^{p-1}rt^{r-1}(1+t^{p})^{r}-r(1+t^{p})^{r-1}pt^{p-1}(1+t^{r})^{p}}{(1+t^{p})^{2r}}, $$

whence it follows that $G'(t)=0$ if and only if

$$p\bigl(1+t^{r}\bigr)^{p-1}rt^{r-1} \bigl(1+t^{p}\bigr)^{r}-r\bigl(1+t^{p} \bigr)^{r-1}pt^{p-1}\bigl(1+t^{r}\bigr)^{p}=0, $$

i.e. $t^{r}(1+t^{p})-t^{p}(1+t^{r})=0$, which means that $t^{r}=t^{p}$. Let us observe that if $p=r$, then $G(t)=1$ for any $t\in[0,1]$, so $G'(t)=0$ on the whole interval $[0,1]$.

Notice also that if $1< p\neq r$, then there is no interior point of the interval $[0,1]$ at which the derivative $G'(t)$ vanishes. Therefore, the function $f(t)$ can reach its biggest value on the interval $[0,1]$ either at the point 0 ($f(0)=1$) or at the point 1 ($f(1)=2^{\frac {p}{r}-1}$), depending on the relationship between p and r. Namely:

• if $1< p\leq r$, then $2^{\frac{p}{r}-1}\leq1$, so $C_{NJ}^{(p)}(L_{r}[0,1])\leq\frac{2}{2^{p-1}}\cdot1=2^{2-p}$;

• if $r< p\leq r'$, then $2^{\frac{p}{r}-1}>1$, so $C_{NJ}^{(p)}(L_{r}[0,1])\leq\frac{2}{2^{p-1}}\cdot2^{\frac {p}{r}-1}=2^{\frac{p}{r}-p+1}$.

On the other hand, notice that the space $L_{r}[0,1]$ is r-uniformly smooth if $1< r\leq2$, and the following Clarkson inequality is satisfied:

$$\biggl(\frac{\|x+ty\|^{r'}+\|x-ty\|^{r'}}{2} \biggr)^{\frac{1}{r'}}\leq\bigl( \| x\|^{r}+\|y\|^{r}\bigr)^{\frac{1}{r}}. $$

If $1< p\leq r'$, the thesis in Lemma 2.4 holds with $K=1$. Therefore, we have the inequality $J_{X,p}(t)\leq(1+t^{r})^{\frac{1}{r}}$ for any $t\geq0$. Take x and y from the space $L_{r}[0,1]$, satisfying $\int_{0}^{b}|x(s)|^{r} \,ds=1$ and $\int_{b}^{1}|y(s)|^{r} \,ds=1$ with some $b\in(0,1)$ and let

$$\begin{aligned} x_{1}(s)=\left \{ \begin{array}{@{}l@{\quad}l} x(s),& 0\leq s< b,\\ 0, & b\leq s\leq1, \end{array} \right .\qquad y_{1}(s)=\left \{ \begin{array}{@{}l@{\quad}l} 0, & 0\leq s<b,\\ y(s),& b\leq s\leq1. \end{array} \right . \end{aligned}$$

Then $\|x_{1}(s)\|_{r}=\|y_{1}(s)\|_{r}=1$, and if $1< p<r'$, we have

$$\begin{aligned} \biggl(\frac{\|x_{1}(s)+ty_{1}(s)\|_{r}^{p}+\|x_{1}(s)-ty_{1}(s)\|_{r}^{p}}{2} \biggr)^{\frac{1}{p}}=\bigl(1+t^{r} \bigr)^{\frac{1}{r}}. \end{aligned}$$

Thus

$$\begin{aligned} \frac{\|x_{1}(s)+ty_{1}(s)\|_{r}^{p}+\|x_{1}(s)-ty_{1}(s)\| _{r}^{p}}{2^{p-1}(1+t^{p})}=\frac{2(1+t^{r})^{\frac{p}{r}}}{2^{p-1}(1+t^{p})}, \end{aligned}$$

which means that if $1< p\leq r'$. Therefore

$$\begin{aligned} C_{NJ}^{(p)}\bigl(L_{r}[0,1]\bigr)\geq \frac{2(1+t^{r})^{\frac{p}{r}}}{2^{p-1}(1+t^{p})}\quad \bigl(\forall t\in[0,1]\bigr). \end{aligned}$$

Taking $t=1$, we get $C_{NJ}^{(p)}(L_{r}[0,1])\geq2^{\frac{p}{r}-p+1}$, while taking $t=0$, we obtain $C_{NJ}^{(p)}(L_{r}[0,1])\geq2^{2-p}$. Therefore:

• if $1< p\leq r$ then $2^{2-p}\geq2^{\frac{p}{r}-p+1}$ and $C_{NJ}^{(p)}(L_{r}[0,1])\geq2^{2-p}$;

• if $r< p\leq r'$ then $2^{\frac{p}{r}-p+1}> 2^{2-p}$ and $C_{NJ}^{(p)}(L_{r}[0,1])\geq2^{\frac{p}{r}-p+1}$.

From what has been discussed above, the results from the thesis (1) of the theorem follow immediately.

(2) In the case when $r'< p<\infty$, in virtue of Remark 2.3 from [19] we know that for any $x,y\in S_{X}$ and any $0\leq t\leq1$, we have

$$\begin{aligned} \bigl(\|x+ty\|_{r}^{p}+\|x-ty\|_{r}^{p} \bigr)^{\frac{1}{p}}\leq2^{\frac{1}{r'}}\bigl(\|x\|_{r}^{r}+t \|y\|_{r}^{r}\bigr)^{\frac{1}{r}}=2^{\frac{1}{r'}} \bigl(1+t^{r}\bigr)^{\frac{1}{r}}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \|x+ty\|_{r}^{p}+\|x-ty\|_{r}^{p} \leq2^{\frac{p}{r'}}\bigl(1+t^{r}\bigr)^{\frac{p}{r}}. \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{\|x+ty\|_{r}^{p}+\|x-ty\|_{r}^{p}}{2^{p-1}(1+t^{p})}\leq\frac{2^{\frac {p}{r'}}(1+t^{r})^{\frac{p}{r}}}{2^{p-1}(1+t^{p})}=2^{\frac{p}{r'}-p+1}\cdot \frac{(1+t^{r})^{\frac{p}{r}}}{1+t^{p}}. \end{aligned}$$

By the proof of thesis (1), if $r< p$ then the supremum of the function f is equal to $2^{\frac{p}{r}-1}$, so we have

$$\begin{aligned} C_{NJ}^{(p)}\bigl(L_{r}[0,1]\bigr) \leq2^{\frac{p}{r'}-p+1}\cdot2^{\frac{p}{r}-1}=1. \end{aligned}$$

By the observation just after Definition 1 of $C_{NJ}^{(p)}(X)$, we have $C_{NJ}^{(p)}(X)\geq1$, so thesis (2) is proved and the proof of the theorem is completed. □

The following theorem gives a relationship between the constant $C_{NJ}^{(p)}(X)$ and the normal structure of X. It is a generalization of a similar result from [20] concerning only the case $p=2$.

Theorem 3.7

If $1\leq p<\infty$ and X is a Banach space with $C_{NJ}^{(p)}(X)<\frac{1}{2^{p-1}}(1+\frac{1}{\mu(X)})^{p}$, then X has normal structure.

Proof

Let us observe that by the inequality $\mu(X)\geq1$, we have $C_{NJ}^{(p)}(X)<2$. We know that if $J(X)<2$, then X is reflexive (see [21]). Therefore, by Corollary 3.5, $C_{NJ}^{(p)}(X)<2$, and so X is reflexive and it has normal structure if and only if it has weak normal structure.

Looking for a contradiction, suppose that X fails to have weak normal structure. Then it is well known (see [17]) that there exists a bounded sequence $(x_{n})$ in X satisfying the following statements:

• (i) $(x_{n})$ is weakly convergent to 0 in X,

• (ii) $\operatorname{diam}(\{x_{n}:n=1,2,\ldots\})=1$,

• (iii) for all $x\in\overline{\operatorname{conv}} (\{x_{n}:n=1,2,\ldots\})$, we have

  $$\begin{aligned} \lim_{n\rightarrow\infty}\| x-x_{n}\|=\operatorname {diam} \bigl(\{x_{n}:n=1,2,\ldots\}\bigr)=1. \end{aligned}$$

Let us fix $\varepsilon>0$ as small as needed. Then, using the above properties of $(x_{n})$ and the definition of $\mu:=\mu(X)$, we can find two positive integers n, m, with $m>n$, such that

• (1) $\| x_{n}\|\geq1-\varepsilon$,

• (2) $\| x_{m}-x_{n}\|\leq1$,

• (3) $\|x_{m}+x_{n}\|\leq\mu+\varepsilon$,

• (4) $\|(1+\frac{1}{\mu+\varepsilon})x_{m}-(1-\frac{1}{\mu +\varepsilon})x_{n}\|\geq(1+\frac{1}{\mu+\varepsilon})(1-\varepsilon)$,

• (5) $\|(1-\frac{1}{\mu+\varepsilon})x_{m}-(1+\frac{1}{\mu +\varepsilon})x_{n}\|\geq(1+\frac{1}{\mu+\varepsilon})\|x_{n}\|-\varepsilon$.

Since

$$\begin{aligned} \limsup _{n\rightarrow\infty}\| x_{m}+x_{n}\|\leq \mu \limsup _{n\rightarrow\infty}\| x_{m}-x_{n}\|, \end{aligned}$$

by condition (2), when m is big enough, we get

$$\begin{aligned} \| x_{m}+x_{n}\|\leq\mu+\varepsilon, \end{aligned}$$

and condition (3) is proved. We just need to prove conditions (4) and (5).

Let us fix $n\in\mathbb{N}$ and define again $\mu:=\mu(X)$. Notice that we can easily get from the Mazur theorem

$$ \biggl[ \biggl(1-\frac{1}{\mu+\varepsilon} \biggr)\Big/ \biggl(1+ \frac{1}{\mu +\varepsilon} \biggr) \biggr]x_{n}\in\overline{\operatorname{conv}}\bigl( \{x_{k}:k\in \mathbb{N}\}\bigr) $$

(3.2)

for any $n\in\mathbb{N}$. Indeed, since $x_{n}\rightarrow0$ weakly as $n\rightarrow\infty$, then by the Mazur theorem $0\in\overline{\operatorname {conv}}(\{x_{k}:k\in\mathbb{N}\})$, whence (3.2) follows immediately. Since (3.2) holds, so by the assumption that X fails to have weak normal structure, for some $m>n$, we have

$$\begin{aligned} \biggl\Vert x_{m}-\frac{1-\frac{1}{\mu+\varepsilon}}{1+\frac{1}{\mu+\varepsilon }}x_{n}\biggr\Vert \geq1-\varepsilon, \end{aligned}$$

and condition (4) follows. In the same way, we can get condition (5).

Next, put $x=x_{m}-x_{n}$, $y=(\mu+\varepsilon)^{-1}(x_{m}+x_{n})$ and use the previous estimates to obtain $\|x\|\leq1$, $\|y\|\leq1$, and

$$\begin{aligned}& \begin{aligned}[b] \| x+y\|&=\biggl\Vert \biggl(1+\frac{1}{\mu+\varepsilon}\biggr)x_{m}- \biggl(1-\frac{1}{\mu +\varepsilon}\biggr)x_{n}\biggr\Vert \\ &\geq\biggl(1+\frac{1}{\mu+\varepsilon}\biggr) (1-\varepsilon), \end{aligned}\\& \begin{aligned}[b] \| x-y\|&=\biggl\Vert \biggl(1-\frac{1}{\mu+\varepsilon}\biggr)x_{m}- \biggl(1+\frac{1}{\mu +\varepsilon}\biggr)x_{n}\biggr\Vert \\ &\geq\biggl(1+\frac{1}{\mu+\varepsilon}\biggr)\|x_{n}\|-\varepsilon \\ &\geq\biggl(1+\frac{1}{\mu+\varepsilon}\biggr) (1-\varepsilon)-\varepsilon. \end{aligned} \end{aligned}$$

By the definition of $C_{NJ}^{(p)}(X)$, we get the estimate

$$\begin{aligned} C_{NJ}^{(p)}(X) \geq&\frac{\|x+y\|^{p}+\|x-y\|^{p}}{2^{p-1}(\|x\|^{p}+\|y\| ^{p})} \\ \geq&\frac{(1+\frac{1}{\mu+\varepsilon})^{p}(1-\varepsilon)^{p}+ [(1+\frac{1}{\mu+\varepsilon})(1-\varepsilon)-\varepsilon]^{p}}{ 2^{p-1}(1+1)}. \end{aligned}$$

Finally, letting $\varepsilon\rightarrow0^{+}$, we obtain

$$\begin{aligned} C_{NJ}^{(p)}(X)\geq\frac{1}{2^{p-1}}\biggl(1+ \frac{1}{\mu}\biggr)^{p}, \end{aligned}$$

which contradicts the hypothesis. This contradiction finishes the proof of the theorem. □