Lemma 3.1
For any number
\(0\leq a\leq2\), \(0\leq b\leq2\)
we have
$$ \biggl(\frac{a+b}{2C}-\frac{1}{C} \biggr)^{2}+ \frac{1}{C^{2}}\geq\frac{a^{2}+b^{2}}{4C^{2}} . $$
(3.1)
Proof
$$\begin{aligned} \biggl(\frac{a+b}{2C}-\frac{1}{C} \biggr)^{2}+ \frac{1}{C^{2}} =& \frac {a^{2}+b^{2}+2ab}{4C^{2}}+\frac{1}{C^{2}}-\frac{(a+b)}{C^{2}}+ \frac{1}{C^{2}} \\ =& \frac{a^{2}+b^{2}}{4C^{2}}+\frac{ab}{2C^{2}}+\frac{1}{C^{2}}-\frac {(a+b)}{C^{2}}+ \frac{1}{C^{2}} \\ =& \frac{a^{2}+b^{2}}{4C^{2}}+\frac{1}{C^{2}} \biggl[\frac{ab}{2}+2-(a+b) \biggr] \\ \geq& \frac{a^{2}+b^{2}}{4C^{2}}. \end{aligned}$$
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Lemma 3.2
Let
a
be a real number and let
\(b>0\), then
$$\frac{at^{2}+bt}{1+t^{2}}\leq\frac{a+\sqrt{a^{2}+b^{2}}}{2}, \quad \forall t\geq0. $$
The first theorem is a relation between \(\mathcal{C}_{NJ}(\mathcal{B})\) and the modulus of uniform smoothness.
Theorem 3.3
Let
\(\mathcal{B}\)
be a quasi-Banach space, then
$$ \mathcal{C}_{NJ}(\mathcal{B})\leq1+C\rho_{\mathcal{B}}(1) \bigl[ \sqrt{\bigl\{ 1-C\rho_{\mathcal{B}}(1)\bigr\} ^{2}+1}-\bigl\{ 1-C \rho_{\mathcal{B}}(1)\bigr\} \bigr] . $$
(3.2)
Proof
We know that
$$ \mathcal{C}_{NJ}(\mathcal{B})=\sup \biggl\{ \frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})}: \forall x_{1},x_{2}\in S_{\mathcal{B}} \text{ with } (x_{1},x_{2})\neq (0,0), 0\leq t\leq1 \biggr\} . $$
By using Lemma 3.1
$$ \begin{aligned} &\frac{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}}{4C^{2}}\leq \biggl\{ \biggl(\frac{{\Vert x_{1}+x_{2} \Vert }+{\Vert x_{1}-x_{2}\Vert }}{2C}-\frac{1}{C} \biggr)^{2}+\frac {1}{C^{2}} \biggr\} , \\ &{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}\leq 4C^{2} \biggl\{ \rho_{\mathcal{B}}^{2}(1)+\frac{1}{C^{2}} \biggr\} . \end{aligned} $$
(3.3)
Also we have
$$ {\Vert x_{1}+x_{2} \Vert }+{\Vert x_{1}-x_{2}\Vert }\leq2C \biggl\{ \rho_{\mathcal{B}}(1)+ \frac{1}{C} \biggr\} . $$
(3.4)
Since
$$\begin{aligned}& {\Vert x_{1}+tx_{2} \Vert }={\bigl\Vert t(x_{1}+x_{2})+(1-t)x_{1} \bigr\Vert }\leq C \bigl(t{\Vert x_{1}+x_{2} \Vert }+(1-t) \bigr), \\& {\Vert x_{1}-tx_{2} \Vert }={\bigl\Vert t(x_{1}-x_{2})+(1-t)x_{1} \bigr\Vert }\leq C \bigl(t{\Vert x_{1}-x_{2} \Vert }+(1-t) \bigr), \end{aligned}$$
we have
$$\begin{aligned}& {\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2} \leq C^{2} \bigl[ \bigl(t{\Vert x_{1}+x_{2} \Vert }+(1-t) \bigr)^{2}+ \bigl(t{\Vert x_{1}-x_{2} \Vert }+(1-t) \bigr)^{2} \bigr] \\& \hphantom{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}} = C^{2}\bigl[ \bigl({\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2} \Vert }^{2} \bigr)t^{2} \\& \hphantom{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}=} {}+ \bigl({\Vert x_{1}+x_{2} \Vert }+{\Vert x_{1}-x_{2} \Vert } \bigr)2t(1-t)+2(1-t)^{2} \bigr] \\& \hphantom{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}} \leq C^{2}\biggl[ 4C^{2} \biggl\{ \rho_{\mathcal{B}}^{2}(1)+ \frac{1}{C^{2}} \biggr\} t^{2} \\& \hphantom{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}=} {}+ 2\biggl(2C\biggl(\frac{1}{C}+\rho_{\mathcal{B}}(1)\biggr)t(1-t)+2(1-t)^{2} \biggr)\biggr] \\& \hphantom{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}= C^{2}\bigl[4Ct^{2}\rho_{\mathcal{B}}(1) \bigl(C \rho_{\mathcal{B}}(1)-1\bigr)+4Ct\rho _{\mathcal{B}}(1)+\bigl(1+t^{2} \bigr)\bigr], \\& \frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})} \leq \frac{2Ct^{2}\rho_{\mathcal{B}}(1)(C\rho _{\mathcal{B}}(1)-1)+2Ct\rho_{\mathcal{B}}(1)+2(1+t^{2})}{1+t^{2}} \\& \hphantom{\frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})}} \leq C\rho_{\mathcal{B}}(1)\bigl\{ C\rho_{\mathcal{B}}(1)-1\bigr\} \\& \hphantom{\frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})}=}{} + \sqrt {C^{2}\rho^{2}_{\mathcal{B}}(1)\bigl\{ C \rho_{\mathcal{B}}(1)-1\bigr\} ^{2}+C^{2} \rho^{2}_{\mathcal{B}}(1)}+1 \\& \hphantom{\frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})}=} (\text{by using Lemma~3.2}) \\& \hphantom{\frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})}} = 1+C\rho_{\mathcal{B}}(1)\sqrt{\bigl\{ C\rho_{\mathcal{B}}(1)-1\bigr\} ^{2}+1}-\bigl\{ 1-C\rho_{\mathcal{B}}(1)\bigr\} . \end{aligned}$$
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The next result is the relation between the James constant and the modulus of convexity.
Theorem 3.4
Let
\(\mathcal{B}\)
be a quasi-Banach space then
$$ J(\mathcal{B})=\sup \biggl\{ \varepsilon: \delta(\varepsilon)\leq1- \frac {\varepsilon}{2} \biggr\} . $$
Proof
Let
$$\alpha=\sup \biggl\{ \varepsilon: \delta(\varepsilon)\leq1-\frac {\varepsilon}{2} \biggr\} . $$
We shall show that \(J(\mathcal{B})\leq\alpha\). For this purpose, if \(\alpha=2\), then there is nothing to prove. So, we may assume that \(\alpha< 2\). For any \(\beta> \alpha\), we have for any \(x_{1},x_{2}\in S_{\mathcal{B}}\) and \(\frac{\Vert x_{1}-x_{2}\Vert }{C}\geq\beta\)
$$\delta(\beta)>1-\frac{\beta}{2}. $$
From the definition of δ, we have
$$1- \frac{\Vert x_{1}+x_{2}\Vert }{2C}> 1-\frac{\beta}{2}, $$
which implies that
$$\frac{\Vert x_{1}+x_{2}\Vert }{C}< \beta; $$
therefore
$$\min \biggl( \frac{\Vert x_{1}+x_{2}\Vert }{C}, \frac{\Vert x_{1}-x_{2}\Vert }{C} \biggr)\leq\beta. $$
As \(J(\mathcal{B})\leq\beta\) and since β was arbitrary we have \(J(\mathcal{B})\leq\alpha\).
For the reverse we use the definition of δ, so \(\forall \gamma>0\) there exist \(x_{1},x_{2}\in S_{\mathcal{B}}\) such that
$$\frac{\Vert x_{1}-x_{2}\Vert }{C}\geq\varepsilon $$
and
$$1-\frac{\Vert x_{1}+x_{2}\Vert }{2C}\leq\delta(\varepsilon)+\gamma, $$
where we have \(\varepsilon=\alpha-\gamma\), so
$$\frac{\Vert x_{1}+x_{2}\Vert }{C}> 2-2\delta(\varepsilon)-2\gamma; $$
therefore
$$\begin{aligned} J(\mathcal{B}) \geq& \min \biggl( \frac{\Vert x_{1}+x_{2}\Vert }{C}, \frac{\Vert x_{1}-x_{2}\Vert }{C} \biggr) \\ \geq&\min \bigl( 2\bigl(1-\delta(\varepsilon)-\gamma\bigr),\varepsilon \bigr) \\ \geq&\min(\varepsilon-2\gamma, \varepsilon) \\ =& \varepsilon-2\gamma \\ =&\alpha-3\gamma, \end{aligned}$$
where γ was arbitrary, so we have \(J(\mathcal{B})\geq\alpha\). □
Corollary 3.5
For any quasi-Banach space
\(\mathcal{B}\), we have
$$J(\mathcal{B})\geq\sqrt{2}. $$
Corollary 3.6
Let
\(\mathcal{B}\)
be any quasi-Banach space and
\(J(\mathcal{B})\leq2\), then
$$\delta\bigl(J(\mathcal{B})\bigr)=1-\frac{J(\mathcal{B})}{2}. $$
Now, we are going to give an equivalent formula of the generalized von Neumann-Jordan constant.
Theorem 3.7
We have
$$ \mathcal{C}_{NJ}(\mathcal{B})=\sup \biggl\{ \frac{{\Vert x_{1}+x_{2} \Vert }^{p}+{\Vert x_{1}-x_{2}\Vert }^{p}}{2^{p-1}C^{p}({\Vert x_{1}\Vert }^{p}+{\Vert x_{2}\Vert }^{p})}: x_{1},x_{2}\in\mathcal{B} \textit{ with } \Vert x_{1}\Vert =1, \Vert x_{2}\Vert \leq1 \biggr\} , $$
(3.5)
where
\(1\leq p< \infty\).
Proof
If \(0\neq \Vert x_{1}\Vert \geq \Vert x_{2}\Vert \)
$$\begin{aligned}& {\Vert x_{1}+x_{2}\Vert }^{p}={\Vert x_{1}\Vert }^{p}{\biggl\Vert \frac{x_{1}}{\Vert x_{1}\Vert }+ \frac{x_{2}}{\Vert x_{1}\Vert }\biggr\Vert }^{p}, \\& {\Vert x_{1}-x_{2}\Vert }^{p}={\Vert x_{1}\Vert }^{p}{\biggl\Vert \frac{x_{1}}{\Vert x_{1}\Vert }- \frac{x_{2}}{\Vert x_{1}\Vert }\biggr\Vert }^{p}, \\& {\Vert x_{1}+x_{2}\Vert }^{p}+{\Vert x_{1}-x_{2}\Vert }^{p}={\Vert x_{1} \Vert }^{p} \biggl[{\biggl\Vert \frac{x_{1}}{\Vert x_{1}\Vert }+ \frac{x_{2}}{\Vert x_{1}\Vert }\biggr\Vert }^{p}+{\biggl\Vert \frac{x_{1}}{\Vert x_{1}\Vert }- \frac{x_{2}}{\Vert x_{1}\Vert }\biggr\Vert }^{p} \biggr], \\& \frac{{\Vert x_{1}+x_{2} \Vert }^{p}+{\Vert x_{1}-x_{2}\Vert }^{p}}{2^{p-1}C^{p}({\Vert x_{1}\Vert }^{p}+{\Vert x_{2}\Vert }^{p})}=\frac{{\Vert \frac{x_{1}}{\Vert x_{1}\Vert }+\frac{x_{2}}{\Vert x_{1}\Vert } \Vert }^{p}+{\Vert \frac{x_{1}}{\Vert x_{1}\Vert }-\frac{x_{2}}{\Vert x_{1}\Vert }\Vert }^{p}}{2^{p-1}C^{p}{ (1+ (\frac{x_{2}}{\Vert x_{1}\Vert } )^{p} )}}. \end{aligned}$$
This shows that
$$\mathcal{C}_{NJ}(\mathcal{B})=\sup \biggl\{ \frac{{\Vert x_{1}+x_{2} \Vert }^{p}+{\Vert x_{1}-x_{2}\Vert }^{p}}{2^{p-1}C^{p}({\Vert x_{1}\Vert }^{p}+{\Vert x_{2}\Vert }^{p})}: x_{1},x_{2}\in\mathcal{B} \text{ with } \Vert x_{1}\Vert =1, \Vert x_{2}\Vert \leq1 \biggr\} . $$
□
The next theorems show the relation between the generalized von Neumann-Jordan and James constants.
Theorem 3.8
For any quasi-Banach space
\(\mathcal{B}\), we have
$$ \frac{1}{2}J(\mathcal{B})^{2}\leq\mathcal{C}_{NJ}( \mathcal{B})\leq\frac {J^{2}(\mathcal{B})}{(J(\mathcal{B})-1)^{2}+1}. $$
(3.6)
Proof
For any \(x_{1},x_{2}\in S_{\mathcal{B}}\), we have
$$\begin{aligned}& 2\bigl(\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \bigr)^{2} \leq 2 \biggl(\frac{{\Vert x_{1}+x_{2}\Vert }+{\Vert x_{1}-x_{2}\Vert }}{2} \biggr)^{2} \\& \hphantom{2\bigl(\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \bigr)^{2}} \leq 2 \biggl(\frac{{\Vert x_{1}+x_{2}\Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}}{2} \biggr) \\& \hphantom{2\bigl(\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \bigr)^{2}}= \frac{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}}{2C^{2}({\Vert x_{1}\Vert }^{2}+{\Vert x_{2}\Vert }^{2})}\cdot {2C^{2}\bigl({\Vert x_{1}\Vert }^{2}+{\Vert x_{2}\Vert }^{2}}\bigr) \\& \hphantom{2\bigl(\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \bigr)^{2}} \leq 2C^{2}\mathcal{C}_{NJ}(\mathcal{B}) \bigl({\Vert x_{1}\Vert }^{2}+{\Vert x_{2}\Vert }^{2}\bigr) \\& \hphantom{2\bigl(\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \bigr)^{2}} = 4C^{2}\mathcal{C}_{NJ}(\mathcal{B}), \\& \bigl(\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \bigr)^{2} \leq 2 C^{2}\mathcal{C}_{NJ}(\mathcal{B}), \\& \frac{1}{C}\min\bigl\{ \Vert x_{1}+x_{2}\Vert , \Vert x_{1}-x_{2} \Vert )\bigr\} \leq 2\sqrt{ \mathcal{C}_{NJ}(\mathcal{B})}. \end{aligned}$$
Hence
$$\sup \biggl(\frac{1}{C}\min\bigl\{ \Vert x_{1}+x_{2} \Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \biggr)\leq2 \sqrt{\mathcal{C}_{NJ}(\mathcal{B})}. $$
Therefore
$$\frac{1}{2}J^{2}(\mathcal{B})\leq\mathcal{C}_{NJ}( \mathcal{B}). $$
To prove the right hand side we use Theorem 3.7, so we only take \(\Vert x_{1}\Vert =1\) and \(\Vert x_{2}\Vert \leq1\).
Case 1: If \(\Vert x_{2}\Vert =t\geq J(\mathcal{B})-1\), then
$$\begin{aligned}& {\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2} \leq \bigl[C\bigl( \Vert x_{1}\Vert +\Vert x_{2}\Vert \bigr) \bigr]^{2}+ \biggl(\frac{C}{C}\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \biggr)^{2} \\& \hphantom{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2} }= C^{2} \biggl[\bigl(\Vert x_{1}\Vert +\Vert x_{2}\Vert \bigr)^{2}+ \biggl(\frac{1}{C}\min\bigl\{ \Vert x_{1}+x_{2}\Vert ,\Vert x_{1}-x_{2} \Vert \bigr\} \biggr)^{2} \biggr] \\& {\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2} \leq C^{2} \bigl[\bigl(\Vert x_{1}\Vert +\Vert x_{2}\Vert \bigr)^{2}+J^{2}(\mathcal{B}) \bigr], \\& \frac{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}}{2C^{2}(\Vert x_{1}\Vert ^{2}+\Vert x_{2}\Vert ^{2})} \leq \frac{(\Vert x_{1}\Vert +\Vert x_{2}\Vert )^{2}+J^{2}(\mathcal{B})}{2(\Vert x_{1}\Vert ^{2}+\Vert x_{2}\Vert ^{2})} \\& \hphantom{\frac{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}}{2C^{2}(\Vert x_{1}\Vert ^{2}+\Vert x_{2}\Vert ^{2})}} = \frac{(1+t)^{2}+J^{2}(\mathcal{B})}{2(1+t^{2})}. \end{aligned}$$
The function
$$f(t)=\frac{(1+t)^{2}+J^{2}(\mathcal{B})}{2(1+t^{2})} $$
is increasing on \((0,\mu)\) and decreasing on \((\mu,1)\) where
$$\mu=\frac{-J^{2}(\mathcal{B})+\sqrt{J^{4}(\mathcal{B})+4}}{2}. $$
Since \(J(\mathcal{B})-1\geq\mu\) and \(J(\mathcal{B})-1\leq t\), we have \(f(t)\leq f(J(\mathcal{B})-1)\)
$$\frac{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}}{2C^{2}(\Vert x_{1}\Vert ^{2}+\Vert x_{2}\Vert ^{2})}\leq\frac{J^{2}(\mathcal {B})}{1+(J(\mathcal{B})-1)^{2}}, $$
taking the supremum, we get
$$ \mathcal{C}_{NJ}(\mathcal{B})\leq\frac{J^{2}(\mathcal{B})}{1+(J(\mathcal {B})-1)^{2}}. $$
(3.7)
Case 2: If \(\Vert x_{2}\Vert =t\leq J(\mathcal{B})-1\), then
$$\begin{aligned}& {\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2} \leq 2C^{2} \bigl(\Vert x_{1}\Vert +\Vert x_{2}\Vert \bigr)^{2} \\& \hphantom{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}} = 2C^{2}(1+t)^{2}, \\& \frac{{\Vert x_{1}+x_{2} \Vert }^{2}+{\Vert x_{1}-x_{2}\Vert }^{2}}{2C^{2}(\Vert x_{1}\Vert ^{2}+\Vert x_{2}\Vert ^{2})} \leq \frac{(1+t)^{2}}{(1+t^{2})}. \end{aligned}$$
Let
$$g(t)=\frac{(1+t)^{2}}{(1+t^{2})}, $$
since g is increasing on \((0,1]\)
$$g(t)\leq g\bigl(J(\mathcal{B})-1\bigr), $$
hence
$$ \mathcal{C}_{NJ}(\mathcal{B})\leq\frac{J^{2}(\mathcal{B})}{1+(J(\mathcal {B})-1)^{2}}. $$
(3.8)
□
Corollary 3.9
If
\(\mathcal{B}\)
is not uniformly non-square then
$$J(\mathcal{B})=\mathcal{C}_{NJ}(\mathcal{B})=2. $$
Theorem 3.10
For any quasi-Banach space
\(\mathcal{B}\), we have
$$ \mathcal{C}_{NJ}(\mathcal{B})\leq1+ \biggl(\frac{CJ(\mathcal {B})}{2} \biggr)^{2}. $$
(3.9)
Proof
Since \(\mathcal{C}_{NJ}(\mathcal{B})=\sup\{\mathcal{C}_{NJ}(t,\mathcal {B}): t\in[0,1]\}\) where
$$\mathcal{C}_{NJ}(t,\mathcal{B})=\sup \biggl\{ \frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})}: \forall x_{1},x_{2}\in S_{\mathcal{B}} \biggr\} . $$
First we prove that
$$ \mathcal{C}_{NJ}(t,\mathcal{B})\leq1+\frac{C^{2}t^{2}J^{2}(\mathcal {B})+2Ct(1-t)J(\mathcal{B})}{2(1+t^{2})}. $$
(3.10)
For this purpose
$$\begin{aligned}& {\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2} \leq \bigl[C(1+t) \bigr]^{2}+ \bigl(\min\bigl\{ {\Vert x_{1}+tx_{2} \Vert },{\Vert x_{1}-tx_{2}\Vert }\bigr\} \bigr)^{2} \\& \hphantom{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}} \leq C^{2}\bigl[(1+t)^{2}+J^{2}(t,\mathcal{B}) \bigr], \\& \frac{{\Vert x_{1}+tx_{2} \Vert }^{2}+{\Vert x_{1}-tx_{2}\Vert }^{2}}{2C^{2}(1+t^{2})} \leq \frac{(1+t)^{2}+J^{2}(t,\mathcal{B})}{2(1+t^{2})}. \end{aligned}$$
Taking the supremum we get
$$ \mathcal{C}_{NJ}(t,\mathcal{B})\leq\frac{ (1+t)^{2}+J^{2}(t,\mathcal {B})}{2(1+t^{2})}. $$
(3.11)
Also note that
$$ \begin{aligned} &\min\bigl\{ {\Vert x_{1}+tx_{2} \Vert }+{\Vert x_{1}-tx_{2}\Vert }\bigr\} \leq C\min\bigl\{ t{\Vert x_{1}+x_{2} \Vert }+(1-t),t{\Vert x_{1}-x_{2} \Vert }+(1-t)\bigr\} , \\ & J(t,\mathcal{B})\leq tCJ(\mathcal{B})+(1-t). \end{aligned} $$
(3.12)
Using (3.11) and (3.12), we get
$$\begin{aligned} \mathcal{C}_{NJ}(t,\mathcal{B}) \leq& \frac{[CtJ(\mathcal {B})+(1-t)]^{2}+(1+t)^{2}}{2(1+t^{2})} \end{aligned}$$
(3.13)
$$\begin{aligned} =& 1+\frac{C^{2}t^{2}J^{2}(\mathcal{B})+2t(1-t)J(\mathcal{B})}{2(1+t^{2})}. \end{aligned}$$
(3.14)
Taking the supremum over t, we get
$$ \mathcal{C}_{NJ}(\mathcal{B})\leq1+ \biggl(\frac{CJ(\mathcal {B})}{2} \biggr)^{2}. $$
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