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On the James type constant of \(l_{p}-l_{1}\)

  • 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.

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.

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

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

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Acknowledgements

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

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Correspondence to Changsen Yang.

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The authors completed the paper, and read and approved the final manuscript.

An erratum to this article is available at http://dx.doi.org/10.1186/s13660-015-0663-y.

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Yang, C., Li, H. On the James type constant of \(l_{p}-l_{1}\) . J Inequal Appl 2015, 79 (2015). https://doi.org/10.1186/s13660-015-0598-3

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MSC

  • 46B20
  • 47H10

Keywords

  • James type constant
  • \(l_{p}-l_{1}\) space
  • von Neuman-Jordan type constant