On the James type constant of $l_{p}-l_{1}$

Abstract

For any $\tau\geq0$, $t\geq1$ and $p\geq1$, the exact value of the James type constant $J_{X,t}(\tau)$ of the $l_{p}-l_{1}$ space is investigated. As an application, the exact value of the von Neuman-Jordan type constant of the $l_{p}-l_{1}$ space can also be obtained.

1 Introduction and preliminaries

Throughout this paper, we shall assume that X stands for a nontrivial Banach space, i.e., $\dim X\geq2$. We will use $S_{X}$ and $B_{X}$ to denote the unit sphere and unit ball of X, respectively.

A Banach space X is called uniformly non-square in the sense of James if there exists a positive number $\delta<1$ such that $\frac{\|x+y\| }{2}\leq\delta$ or $\frac{\|x-y\|}{2}\leq\delta$, whenever $x,y \in S_{X}$. The non-square or James constant is defined by

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

Obviously, X is uniformly non-square in the sense of James if and only if $J(X)< 2$ (see [1]).

The von Neumann-Jordan constant, introduced by Clarkson in [2], is defined as follows:

$$C_{\mathrm{NJ}}(X)=\sup \biggl\{ \frac{\|x+y\|^{2}+\|x-y\|^{2}}{2(\|x\|^{2}+\|y\|^{2})}: x\in S_{X}, y\in B_{X} \biggr\} . $$

It is well known that the von Neumann-Jordan constant is not larger than the James constant. This result $C_{\mathrm{NJ}}(X)\leq J(X)$ was obtained by Takahashi-Kato in [3], Wang in [4] and Yang-Li in [5] almost at the same time.

Recently, as a generalization of the James constant and the von Neumann-Jordan constant, Takahashi in [6] introduced the James type constant $J_{X,t}(\tau)$ and the von Neumann-Jordan type constant $C_{t}(X)$, respectively, as follows:

$$J_{X,t}(\tau)=\sup \bigl\{ \mu_{t}\bigl(\Vert x+\tau y \Vert ,\Vert x-\tau y\Vert \bigr):x,y\in S_{X} \bigr\} , $$

where $\tau\geq0$, $-\infty\leq t < +\infty$. Here, we denote $\mu _{t}(a,b)=(\frac{a^{t}+b^{t}}{2})^{\frac{1}{t}}$ ($t\neq0$) and $\mu_{0}(a,b)=\lim_{t\rightarrow0}\mu_{t}(a,b)=\sqrt{ab}$ for two positive numbers a and b. It is well known that $\mu_{t}(a,b)$ is nondecreasing and $\mu_{-\infty}(a,b)=\lim_{t\rightarrow-\infty}\mu_{t}(a,b)=\min(a,b)$. Therefore, $J(X)=J_{X,-\infty}(1)$,

$$C_{t}(X)=\sup \biggl\{ \frac{J_{X,t}(\tau)^{2}}{1+\tau^{2}}: 0\leq\tau\leq1 \biggr\} . $$

It is obvious that $C_{2}(X)=C_{\mathrm{NJ}}(X)$ and the James type constants include some known constants such as Alonso-Llorens-Fuster’s constant $T(X)$ in [7], Baronti-Casini-Papini’s constant $A_{2}(X)$ in [8], Gao’s constant $E(X)$ in [9] and Yang-Wang’s modulus $\gamma_{X}(t)$ in [10]. These constants are defined by $T(X)=J_{X,0}(1)$, $A_{2}(X)=J_{X,1}(1)$, $E(X)=2J_{X,2}^{2}(1)$ and $\gamma_{X}(t)=J_{X,2}^{2}(t)$.

Now let us list some known results of the constant $J_{X,t}(\tau)$; for more details, see [6, 11–14].

• (1) If $-\infty\leq t_{1}\leq t_{2}<\infty$, then $J_{X,t_{1}}(\tau)\leq J_{X,t_{2}}(\tau)$ for any $\tau\geq0$.

• (2) Let $t\geq1$, $\tau\geq0$ and $X=l_{1}-l_{2}$, then

  $$ J_{X,t}(\tau)= \biggl(\frac{(1+\tau^{2})^{\frac{t}{2}}+(1+\tau)^{t}}{2} \biggr)^{\frac{1}{t}}. $$

  (1.1)

• (3) Let X be an $l_{\infty}-l_{1}$ space. If $0\leq\tau\leq1$, then

  $$J_{X,t}(\tau)= \left \{ \begin{array}{l@{\quad}l} (\frac{1+(1+\tau)^{t}}{2})^{\frac{1}{t}},& t\geq1, \\ 1+\frac{\tau}{2},& t\leq1. \end{array} \right . $$

• (4) Let $1\leq t\leq p\leq\infty$, $2\leq p$ and $0\leq\tau\leq1$. Then

  $$J_{X,t}(\tau)=1+2^{-\frac{1}{p}}\tau, $$

  where X is an $l_{\infty}-l_{p}$ space.

• (5) Let $t_{2}\geq t_{1}\geq1$ and $0\leq\tau\leq1$. Then, for any Banach space X,

  $$ J_{X,t_{1}}^{t_{2}}(\tau)\leq J_{X,t_{2}}^{t_{2}}(\tau) \leq \frac{(1+\tau)^{t_{2}}+ \{2J_{X,t_{1}}^{t_{1}}(\tau)-(1+\tau)^{t_{1}} \} ^{\frac{t_{2}}{t_{1}}}}{2}. $$

  (1.2)

• (6) $J_{X,t_{1}}(\tau)=1+\tau$ if and only if $J_{X,t_{2}}(\tau)=1+\tau$.

For $p\geq1$, the $l_{p}-l_{1}$ space is defined by $X= \mathbf{R}^{2}$ with the norm

$$\|x\|=\bigl\Vert (x_{1},x_{2})\bigr\Vert = \left \{ \begin{array}{l@{\quad}l} \|x\|_{p},& x_{1}x_{2}\geq0, \\ \|x\|_{1},& x_{1}x_{2}\leq0. \end{array} \right . $$

For any $\tau\geq0$ and $p\geq1$, we have calculated the exact value of the James type constant $J_{l_{p}-l_{1},t}(\tau)$ for $t\geq1$. As an application, we also give the exact value of the von Neumann-Jordan type constant $C_{t}(l_{p}-l_{1})$ for $1\leq t\leq2$. In [11], for $1< p\leq2$, it is known that $C_{\mathrm{NJ}}(l_{p}-l_{1})=1+2^{\frac{2}{p}-2}$ was given. In this paper, for $p\geq2$, $(p-2)2^{\frac{2}{p}-2}\leq1$ and $p>2$, $(p-2)2^{\frac{2}{p}-2}\geq1$, the exact value of the von Neumann-Jordan constant $C_{\mathrm{NJ}}(l_{p}-l_{1})$ is obtained.

2 Main results and their proofs

To give the value of $J_{X,t}(\tau)$ for $X=l_{p}-l_{1}$, we need the following lemmas.

Lemma 2.1

Let $x_{1}, x_{2}, y_{1}, y_{2}\geq0$ and $p\geq1$ such that

$$x_{1}^{p}+x_{2}^{p}=1\quad \textit{and} \quad y_{1}^{p}+y_{2}^{p}=1. $$

If $0\le \tau\le1$, $0\leq\tau y_{1}\leq x_{1}$ and $0\leq x_{2}\leq\tau y_{2}$, then

$$\bigl[(x_{1}+\tau y_{1})^{p}+(x_{2}+ \tau y_{2})^{p}\bigr]^{\frac{1}{p}}+x_{1}-\tau y_{1}+\tau y_{2}-x_{2}\leq1+\tau+\bigl(1+ \tau^{p}\bigr)^{\frac{1}{p}}. $$

Proof

It is readily seen that $0\le x_{1}-\tau y_{1}+\tau y_{2}-x_{2}\le1+\tau$. Let us now consider two possible cases.

Case 1. $0\le x_{1}-\tau y_{1}+\tau y_{2}-x_{2}\leq(1+\tau ^{p})^{1/p}$. Hence

$$\begin{aligned}& \bigl[(x_{1}+\tau y_{1})^{p}+(x_{2}+ \tau y_{2})^{p}\bigr]^{\frac{1}{p}}+x_{1}-\tau y_{1}+\tau y_{2}-x_{2} \\& \quad \le \bigl[\bigl(x_{1}^{p}+x_{2}^{p} \bigr)^{1/p}+\bigl(\tau^{p} y_{1}^{p}+ \tau^{p} y_{2}^{p}\bigr)^{1/p}\bigr]+ \bigl(1+\tau^{p}\bigr)^{\frac{1}{p}} \\& \quad = 1+\tau+\bigl(1+\tau^{p}\bigr)^{\frac{1}{p}}. \end{aligned}$$

Case 2. $(1+\tau^{p})^{1/p}\le x_{1}-\tau y_{1}+\tau y_{2}-x_{2}\le 1+\tau$. By Minkowski’s inequality,

$$\begin{aligned}& \bigl[(x_{1}+\tau y_{1})^{p}+(x_{2}+ \tau y_{2})^{p}\bigr]^{1/p}+x_{1}-\tau y_{1}+\tau y_{2}-x_{2} \\& \quad \leq \bigl(x_{1}^{p}+\tau^{p} y_{2}^{p}\bigr)^{1/p}+\bigl(\tau^{p} y_{1}^{p}+x_{2}^{p}\bigr)^{1/p}+x_{1}- \tau y_{1}+\tau y_{2}-x_{2} \\& \quad \leq \bigl(x_{1}^{p}+\tau^{p} y_{2}^{p}\bigr)^{1/p}+\tau y_{1}+x_{2}+x_{1}- \tau y_{1}+\tau y_{2}-x_{2} \\& \quad \leq (1+\tau)+\bigl(1+\tau^{p}\bigr)^{1/p}, \end{aligned}$$

where the second inequality follows from the fact $\|\cdot\|_{p}\le\|\cdot\|_{1}$. Consequently, the proof is complete. □

Lemma 2.2

Let $\tau\in(0,1)$, $t\in[1,2]$ and $p\geq2$. Then

• (a) $2\tau^{p}+p-2-p\tau^{2}\geq0$;

• (b) $1-\tau^{2p-2}-(p-1)(\tau^{p-2}-\tau^{p})\geq0$;

• (c) the function

  $$f(\tau)=\frac{\tau-\tau^{p-1}}{(1-\tau)(1+\tau)^{t-1}}\bigl(1+\tau^{p}\bigr)^{\frac{t}{p}-1} $$

  is nondecreasing; moreover, $0\leq f(\tau)\leq(p-2)2^{\frac{t}{p}-t}$.

Proof

(a) Letting $h(\tau)=2\tau^{p}+(p-2)-p\tau^{2}$, we have $h'(\tau)=2p(\tau^{p-1}-\tau)\leq0$, and $h(\tau)\geq h(1)=0$.

(b) Letting $g(\tau)=1-\tau^{2p-2}-(p-1)(\tau^{p-2}-\tau^{p})$, we have

$$g'(\tau)=-(p-1)\tau^{p-3}\bigl(2\tau^{p}+p-2-p \tau^{2}\bigr). $$

Hence, $g'(\tau)\leq0$ by (a) and $g(\tau)\geq g(1)=0$.

(c) By a basic calculation, then by use of (b), we have

$$\begin{aligned} f'(\tau) =&\frac{1}{[(1-\tau)(1+\tau)^{t-1}]^{2}}\bigl\{ (1-\tau) (1+ \tau)^{t-1}\bigl[ \bigl(1-(p-1)\tau^{p-2}\bigr) \bigl(1+ \tau^{p}\bigr)^{\frac{t}{p}-1} \\ &{}+\bigl(\tau-\tau^{p-1}\bigr) (t-p)\tau^{p-1}\bigl(1+ \tau^{p}\bigr)^{\frac{t}{p}-2}\bigr] \\ &{}-\bigl(\tau-\tau^{p-1}\bigr) \bigl(1+\tau^{p} \bigr)^{\frac{t}{p}-1}\bigl[-(1+\tau)^{t-1}+(1-\tau ) (t-1) (1+ \tau)^{t-2}\bigr]\bigr\} \\ =&\frac{(1+\tau^{p})^{\frac{t}{p}-2}(1+\tau)^{t-2}}{[(1-\tau)(1+\tau )^{t-1}]^{2}}\bigl\{ (1+\tau) \bigl(1+\tau^{p}\bigr) \bigl[1-(p-1)\tau^{p-2} -\tau+(p-1)\tau^{p-1} \\ &{}+\tau-\tau^{p-1}\bigr]+(1-\tau) \bigl(\tau-\tau^{p-1} \bigr)\bigl[(t-p) (1+\tau)\tau^{p-1} -\bigl(1+\tau^{p}\bigr) (t-1)\bigr]\bigr\} \\ =&\frac{(1+\tau^{p})^{\frac{t}{p}-2}(1+\tau)^{t-2}}{[(1-\tau)(1+\tau )^{t-1}]^{2}}\bigl\{ \bigl(1+\tau^{2}\bigr)\bigl[1- \tau^{2p-2}-(p-1)\tau^{p-2}\bigl(1-\tau^{2}\bigr)\bigr] \\ &{}+(2-t) (1-\tau) \bigl(\tau-\tau^{p-1}\bigr) \bigl(1- \tau^{p-1}\bigr)\bigr\} \geq 0. \end{aligned}$$

Now from $\lim_{\tau\rightarrow1^{-}}f(\tau)=(p-2)2^{\frac{t}{p}-t}$, we have $0\leq f(\tau)\leq(p-2)2^{\frac{t}{p}-t}$. □

Theorem 2.3

Let $t\geq1$, $p\geq1$, $\tau\geq0$ and $X=l_{p}-l_{1}$ space. Then

$$ J_{X,t}(\tau)= \biggl(\frac{(1+\tau^{p})^{\frac{t}{p}}+(1+\tau)^{t}}{2} \biggr)^{\frac{1}{t}}. $$

(2.1)

Proof

As $J_{X,t}(\tau)=\tau J_{X,t}(\frac{1}{\tau})$ is valid for any $\tau>0$, we only consider the case $0\leq\tau\leq1$. We claim that the following inequality is valid for any $x,y\in S_{l_{p}-l_{1}}$:

$$ \|x+\tau y\|+\|x-\tau y\|\leq\bigl(1+\tau^{p} \bigr)^{\frac{1}{p}}+1+\tau. $$

(2.2)

In fact, by the convexity of norm, we only need to show that this inequality is valid for any $x,y\in \operatorname{ext}(S_{l_{p}-l_{1}})$, where $\operatorname{ext}(S_{l_{p}-l_{1}})$ denotes the set of extreme points of $S_{l_{p}-l_{1}}$. From $\operatorname{ext}(S_{l_{p}-l_{1}})=\{(x_{1},x_{2}):x_{1}^{p}+x_{2}^{p}=1, x_{1}x_{2}\geq0\}$, we may assume that $x=(a,b)$, $y=(c,d)$, where $a,b,c,d\geq0$ with $a^{p}+b^{p}=c^{p}+d^{p}=1$.

(I) If $(a-c\tau)(b-d\tau)\geq0$,

$$\begin{aligned} \|x+\tau y\|+\|x-\tau y\| =&\|x+\tau y\|_{p}+\|x-\tau y \|_{p} \\ \leq&1+\tau+\bigl[\vert a-c\tau \vert ^{p}+|b-d\tau|^{p} \bigr]^{\frac{1}{p}} \\ \leq&1+\tau+\max\bigl\{ \bigl[a^{p}+b^{p} \bigr]^{\frac{1}{p}},\bigl[(c\tau)^{p}+(d\tau)^{p} \bigr]^{\frac {1}{p}}\bigr\} \\ \leq&2+\tau \\ \leq&\bigl(1+\tau^{p}\bigr)^{\frac{1}{p}}+1+\tau. \end{aligned}$$

(II) If $(a-c\tau)(b-d\tau)\leq0$.

We may assume that $a-c\tau>0$ and $b-d\tau\leq0$. Then, by use of Lemma 2.1, we also have

$$\|x+\tau y\|+\|x-\tau y\|=\|x+\tau y\|_{p}+\|x-\tau y\|_{1} \leq\bigl(1+\tau^{p}\bigr)^{\frac{1}{p}}+1+\tau. $$

Thus (2.2) is valid.

Now, by taking $x=(1,0)$ and $y=(0,1)$, we have $2J_{l_{p}-l_{1},1}(\tau)=(1+\tau^{p})^{\frac{1}{p}}+1+\tau$. Therefore by (1.2) we have

$$J_{X,t}^{t}(\tau)\leq\frac{(1+\tau)^{t}+[2J_{X,1}(\tau)-(1+\tau)]^{t}}{2} =\frac{(1+\tau)^{t}+(1+\tau^{p})^{\frac{t}{p}}}{2}. $$

On the other hand, by taking $x=(1,0)$, $y=(0,1)$, we have

$$\|x+\tau y\|=\bigl(1+\tau^{p}\bigr)^{\frac{1}{p}},\qquad \|x-\tau y \|=1+\tau, $$

so

$$J_{X,t}^{t}(\tau)\geq\frac{(1+\tau)^{t}+(1+\tau^{p})^{\frac{t}{p}}}{2}. $$

Therefore, (2.1) is valid for $t\geq1$. □

Theorem 2.4

Let $p=2$, $t\geq1$ or $p>2$, $t\in[1,2]$, and X be an $l_{p}-l_{1}$ space.

For p and t such that $(p-2)2^{\frac{t}{p}-t}\leq1$, then

$$ C_{t}(X)= \biggl(\frac{2^{\frac{t}{p}-\frac{t}{2}}+2^{\frac{t}{2}}}{2} \biggr)^{\frac{2}{t}}. $$

(2.3)

For p and t such that $(p-2)2^{\frac{t}{p}-t}>1$, then

$$C_{t}(X)=\frac{1}{1+\tau_{0}^{2}} \biggl(\frac{(1+\tau_{0})^{t}+(1+\tau_{0}^{p})^{\frac {t}{p}}}{2} \biggr)^{\frac{2}{t}}, $$

where $\tau_{0}$ is the unique solution of the equation

$$ \frac{(\tau-\tau^{p-1})(1+\tau^{p})^{\frac {t}{p}-1}}{(1-\tau)(1+\tau)^{t-1}}=1. $$

(2.4)

Proof

By (2.1), we have

$$C_{t}(X)=\bigl[\sup\bigl\{ h(\tau):0\leq\tau\leq1\bigr\} \bigr]^{\frac{2}{t}},\quad \mbox{where } h(\tau)=\frac{(1+\tau)^{t}+(1+\tau^{p})^{\frac{t}{p}}}{2(1+\tau^{2})^{\frac{t}{2}}}. $$

A simple computation yields

$$h'(\tau)=\frac{t(1-\tau)(1+\tau)^{t-1}}{2(1+\tau^{2})^{\frac{t}{2}+1}} \biggl[1-\frac{(\tau-\tau^{p-1})(1+\tau^{p})^{\frac{t}{p}-1}}{(1-\tau )(1+\tau)^{t-1}}\biggr]. $$

If $p=2$, $t\geq1$ or $p>2$, $t\in[1,2]$ such that $(p-2)2^{\frac{t}{p}-t}\leq1$, Lemma 2.2 implies $h'(\tau)\geq0$, so that h is nondecreasing. Hence

$$C_{t}(X)=h(1)^{\frac{2}{t}}= \biggl(\frac{2^{\frac{t}{p}-\frac {t}{2}}+2^{\frac{t}{2}}}{2} \biggr)^{\frac{2}{t}}. $$

Otherwise, let $\tau_{0}\in(0,1)$ be the unique solution to equation (2.4). It then follows from Lemma 2.2 that $h'(\tau)\geq0$ for $\tau\in[0,\tau_{0}]$ and $h'(\tau)\leq0$ for $\tau\in[\tau_{0},1]$. In other words, h attains its maximum at $\tau_{0}$. Hence

$$C_{t}(X)=\frac{1}{1+\tau_{0}^{2}} \biggl(\frac{(1+\tau_{0})^{t}+(1+\tau_{0}^{p})^{\frac {t}{p}}}{2} \biggr)^{\frac{2}{t}}. $$

 □

For $1< p\leq2$, $C_{\mathrm{NJ}}(l_{p}-l_{1})=1+2^{\frac{2}{p}-2}$ (see [11]). Now, by taking $t=2$ in Theorem 2.3, as a generalization, we can obtain the following corollary on the von Neumann-Jordan constant of $l_{p}-l_{1}$ space.

Corollary 2.5

Let X be the $l_{p}-l_{1}$ space.

• (a) If $p\geq2$ and $(p-2)2^{\frac{2}{p}-2}\leq1$, then $C_{\mathrm{NJ}}(X)=1+2^{\frac{2}{p}-2}$.

• (b) If $p>2$ and $(p-2)2^{\frac{2}{p}-2}\geq1$, then

  $$C_{\mathrm{NJ}}(X)= \frac{1}{2}+\frac{1-\tau_{0}^{p}}{2(\tau_{0}-\tau_{0}^{p-1})}, $$

  where $\tau_{0}\in(0,1)$ is the unique solution to the equation

  $$\frac{(\tau-\tau^{p-1})(1+\tau^{p})^{\frac{2}{p}-1}}{1-\tau^{2}}=1. $$