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

Journal of Inequalities and Applications20152015:79

https://doi.org/10.1186/s13660-015-0598-3

• Accepted: 13 February 2015
• Published:

The Erratum to this article has been published in Journal of Inequalities and Applications 2015 2015:145

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.

Keywords

• James type constant
• $$l_{p}-l_{1}$$ space
• von Neuman-Jordan type constant

• 46B20
• 47H10

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 ).
The von Neumann-Jordan constant, introduced by Clarkson in , 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 , Wang in  and Yang-Li in  almost at the same time.

Recently, as a generalization of the James constant and the von Neumann-Jordan constant, Takahashi in  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 , Baronti-Casini-Papini’s constant $$A_{2}(X)$$ in , Gao’s constant $$E(X)$$ in  and Yang-Wang’s modulus $$\gamma_{X}(t)$$ in . 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, 1114].
1. (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. (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. (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. (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. (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. (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 , 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
1. (a)

$$2\tau^{p}+p-2-p\tau^{2}\geq0$$;

2. (b)

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

3. (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 ). 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.
1. (a)

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

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

Authors’ Affiliations

(1)
College of Mathematics and Information Science, 46 East of Construction Road, Xinxiang, Henan, 453007, P.R. China

References 