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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

  • Received: 8 October 2014
  • 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

MSC

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

Notes

Declarations

Acknowledgements

The research was supported by the National Natural Science Foundation of China (Nos. 11271112; 11201127) and IRTSTHN (14IRTSTHN023).

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

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

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© Yang and Li; licensee Springer. 2015

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