%------------------------------------------------------------------------------ % Here please write the date of submission of paper or its revisions: %------------------------------------------------------------------------------ % \documentclass[12pt, reqno]{amsart} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks, plainpages]{hyperref} \textheight 22.5truecm \textwidth 14.5truecm \setlength{\oddsidemargin}{0.35in}\setlength{\evensidemargin}{0.35in} \setlength{\topmargin}{-.5cm} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{summary}[theorem]{Summary} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \input{mathrsfs.sty} \begin{document} \setcounter{page}{22} \noindent\parbox{2.95cm}{\includegraphics*[keepaspectratio=true,scale=0.125]{AFA.jpg}} \noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent \qquad Ann. Funct. Anal. 2 (2011), no. 2, 22--33\\ {\footnotesize \qquad \textsc{\textbf{$\mathscr{A}$}nnals of \textbf{$\mathscr{F}$}unctional \textbf{$\mathscr{A}$}nalysis}\\ \qquad ISSN: 2008-8752 (electronic)\\ \qquad URL: \textcolor[rgb]{0.00,0.00,0.99}{www.emis.de/journals/AFA/} }\\[.5in]} \title[Geometric constants of $\Bbb R^2$]{Some geometric constants of absolute normalized norms on $\Bbb R^2$} \author[H. Mizuguchi, K.-S. Saito]{Hiroyasu Mizuguchi and Kichi-Suke Saito$^{*}$} \address{Department of Mathematical Sciences, Graduate School of Science and Technology, Niigata University, Niigata, 950-2181 Japan.} \email{\textcolor[rgb]{0.00,0.00,0.84}{mizuguchi@m.sc.niigata-u.ac.jp}} \email{\textcolor[rgb]{0.00,0.00,0.84}{saito@math.sc.niigata-u.ac.jp}} \dedicatory{{\rm Communicated by J. I. Fujii}} \subjclass[2010]{Primary 46B20; Secondary 46B25.} \keywords{ Zb{\rm \u{a}}ganu constant, absolute norm, von Neumann-Jordan constant.} \date{Received: 31 March 2011; Accepted: 14 June 2011. \newline \indent $^{*}$ Corresponding author} \begin{abstract} We consider the Banach space $X=(\Bbb R^2, \|\cdot\|)$ with a normalized, absolute norm. Our aim in this paper is to calculate the modified Neumann-Jordan constant $C'_{NJ}(X)$ and the Zb{\rm \u{a}}ganu constant $C_Z(X)$. \end{abstract} \maketitle \section{Introduction and preliminaries} \noindent Let $X$ be a Banach space with the unit ball $B_X=\{x\in X: \|x\|\leq 1\}$ and the unit sphere $S_X=\{x\in X: \|x\| = 1\}$. Many geometric constants for a Banach space $X$ have been investigated. In this paper we shall consider the following constants; \[C_{NJ}(X)=\sup \left\{ \frac{\| x+y\| ^2+\| x-y\| ^2}{2(\| x\| ^2+\| y\| ^2)}~\biggl|~ (x,y)\not = (0,0)\biggr. \right\},\] \[C'_{NJ}(X)=\sup \left\{ \frac{\| x+y\| ^2+\| x-y\| ^2}{4}~\biggl|~ x,y\in S_{X}\biggr. \right\}, \] \[ C_{Z}(X)=\sup \left\{ \frac{\| x+y\| \| x-y\| }{\| x\| ^2+\| y\| ^2}~\biggl|~ x,y \in X,~(x,y)\not = (0,0)\biggr. \right\}. \] The constant $C_{NJ}(X)$, called the von Neumann-Jordan constant (hereafter referred to as NJ constant) have been considered in many papers (\cite{C,V,MT,T} and so on). The constant $C'_{NJ}(X)$, called the modified von Neumann-Jordan constant (shortly, modified NJ constant) was introduced by Gao in \cite{G} and does not necessarily coincide with $C_{NJ}(X)$ (cf. \cite{A, G2, GS2}). The constant $C_Z(X)$ was introduced by Zb{\rm \u{a}}ganu (\cite{Z}) and was conjectured that $C_Z(X)$ coincides with the von Neumann-Jordan constant $C_{NJ}(X)$, but Alonso and Martin \cite{J} gave an example that $C_{NJ}(X) \not= C_Z(X)$ (cf.\cite{GS, L}). A norm $\|\cdot \|$ on $\mathbb{R}^2$ is said to be absolute if $\| (a,b) \|=\| (|a|,|b|) \| $ for any $(a,b) \in \mathbb{R}^2$, and normalized if $ \|(1,0)\|=\|(0,1)\|=1$. Let $AN_2$ denote the family of all absolute normalized norm on $\mathbb{R}^2$, and $\Psi _2$ denote the family of all continuous convex function $\psi$ on $[0,1]$ such that $\psi(0)=\psi(1)=1$ and $\max\{1-t,t\}\leq\psi(t) \leq 1$ for all $0\leq t\leq 1$. As in \cite{S}, it is well known that $AN_2$ and $\Psi _2$ are in a one-to-one correspondence under the equation $\psi(t)=\|(1-t,t)\|$ $(0\leq t\leq 1)$. Denote $\| \cdot \|_{\psi}$ be an absolute normalized norm associated with a convex function $\psi \in \Psi _2$. For $\psi, \varphi \in \Psi _2$, we denote $\psi \leq \varphi$ if $\psi(t) \leq \varphi(t)$ for any t in $[0,1]$. Let \[M_{1}=\max_{0\leq t\leq 1}\frac{\psi(t)}{\psi_{2}(t)} ~~~{\rm and} ~~~M_{2}=\max_{0\leq t\leq 1}\frac{\psi_{2}(t)}{\psi(t)},\] where $\psi_{2}(t)=\|(1-t,t)\|_{2}=\sqrt{(1-t)^2+t^2}$ corresponds to the $l_2$-norm. In \cite{S}, Saito, Kato and Takahashi proved that, if $\psi \geq \psi_2$ (resp. $\psi \leq \psi_2$), then $C_{NJ}\left(\mathbb{C}^2,\|\cdot\|_{\psi} \right)=M_1^2$ (resp. $M_2^2$). We put $X = (\Bbb R^2, \|\cdot\|_{\psi})$ for $\psi \in \Psi_2$. Our aim in this paper is to consider the conditions of $\psi$ that $C_{NJ}(X) = C_Z(X)$ or $C_{NJ}(X) = C'_{NJ}(X)$. In \S 2, we consider the modified von Neumann-Jordan constant. We prove that if $\psi \leq \psi_2$, then $C'_{NJ}(X) =C_{NJ}(X) = M_2^2$. If $\psi \geq \psi_2$, then we present the necessarily and sufficient condition that $C'_{NJ} \left( \mathbb{R}^2, \| \cdot \|_{\psi} \right) =C_{NJ} \left( \mathbb{R}^2, \| \cdot \|_{\psi} \right) =M_1^2$. Further, we consider the conditions that $C'_{NJ}(\Bbb R^2, \| \cdot \|_{\psi}) =C_{NJ}(\Bbb R^2, \| \cdot \|_{\psi}) =M_1^2M_2^2$. In \S 3, we study the Zb{\rm \u{a}}ganu constant. First, we show that, if $\psi \geq \psi_2$, then $C_Z \left( \mathbb{R}^2, \| \cdot \|_{\psi} \right)=C_{NJ} \left( \mathbb{R}^2, \| \cdot \|_{\psi} \right) = M_1^2$. If $\psi \leq \psi_2$, then we give the necessarily and sufficient condition for that $C_Z \left( \mathbb{R}^2, \| \cdot \|_{\psi} \right) =C_{NJ} \left( \mathbb{R}^2, \| \cdot \|_{\psi} \right) = M_2^2$. Further we study the conditions that $C_Z(\Bbb R^2, \| \cdot \|_{\psi}) =C_{NJ}(\Bbb R^2, \| \cdot \|_{\psi}) =M_1^2M_2^2$. In \S 4, we calculate the modified NJ-constant $C'_{NJ}(X)$ and the Zb{\rm \u{a}}ganu constant $C_Z(X)$ for some normed liner spaces. \section{The modified NJ constant of R$^2$} In this section, we consider the Banach space $X= ( \mathbb{R}^2, \| \cdot \|_{\psi}).$ From the definition of the modified NJ constant, it is clear that $C'_{NJ}(X) \leq C_{NJ}(X)$. In this section, we consider the condition that $C'_{NJ}(X)=C_{NJ}(X)$. \begin{proposition} Let $\psi \in \Psi_2$. If $\psi\leq \psi_{2}$, then $C'_{NJ}(X)=C_{NJ}(X)=M_{2}^2$. \end{proposition} \begin{proof} For any $x,y \in S_X$, by \cite[Lemma 3]{S}, \begin{align*} \|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2 & \leq \|x+y\|_2^2+\|x-y\|_2^2\\ &=2\left( \|x\|_2^2+\|y\|_2^2\right)\\ & \leq 2M_2^2\left( \|x\|_{\psi}^2+\|y\|_{\psi}^2\right)=4M_2^2. \end{align*} Now let $\psi_2/\psi$ attain the maximum at $t=t_0$ ($0\leq t_0\leq 1$), and put \[x=\frac{1}{\psi(t_0)}(1-t_0,~t_0), ~y=\frac{1}{\psi(t_0)}(1-t_0,~-t_0).\] \\ Then $x,y \in S_X$ and \begin{align*}\|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2 &=\frac{4(1-t_0)^2+4t_0^2}{\psi(t_0)^2}\\ &=4\frac{\psi_2(t_0)^2}{\psi(t_0)^2}=4M_2^2, \end{align*} which implies that $C'_{NJ}(X)=M_{2}^2$. By \cite[Theorem 1]{S}, we have this proposition. \end{proof} If $\psi\geq \psi_{2}$, by \cite[Theorem 1]{S}, then $C_{NJ}(X)=M_1^2$. We now give the necessarily and sufficient condition of $C'_{NJ}(X)=M_1^2$. \begin{theorem} Let $\psi \in \Psi_2$ such that $\psi\geq \psi_2$. Then $C'_{NJ}(X)=M_1^2$ if and only if there exist $s,t\in [0,1]~~(s < t)$ satisfying one of the following conditions: {\rm (1)} $\psi(s)=\psi_2(s)$, $\psi(t)=\psi_2(t)$ and, if we put $r=\frac{\psi(s)t+\psi(t)s}{\psi(s)+\psi(t)}$, then $\frac{\psi(r)}{\psi_2(r)}=\frac{\psi(1-r)}{\psi_2(1-r)}=M_1$. {\rm (2)} $\psi(s)=\psi_2(s)$, $\psi(t)=\psi_2(t)$ and, if we put $r = \frac{\psi(t)s+\psi(s)t}{\psi(t)+\psi(s)(2t-1)}$, then $\frac{\psi(r)}{\psi_2(r)}=\frac{\psi(1-r)}{\psi_2(1-r)}=M_1$. \end{theorem} \begin{proof} ($\Longrightarrow$) Suppose that $C'_{NJ}(X)=M_1^2$. First, for any $x,y \in S_X$, by \cite[Lemma 3]{S}, we have \begin{align*} \|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2 & \leq M_1^2(\|x+y\|_2^2+\|x-y\|_2^2)\\ &=2M_1^2\left( \|x\|_2^2+\|y\|_2^2\right)\\ & \leq 2M_1^2\left( \|x\|_{\psi}^2+\|y\|_{\psi}^2\right)=4M_1^2. \end{align*} Since $X=\left( \mathbb{R}^2, \| \cdot \|_{\psi} \right)$ is finite dimensional, \[C'_{NJ}(X)=\max \left\{ \frac{\| x+y\|_{\psi} ^2+\| x-y\|_{\psi} ^2}{4}~\biggl|~ x,y\in S_{X}\biggr. \right\}. \] Therefore, $C'_{NJ}(X)=M_1^2$ if and only if there exist $x,y \in S_X~~(x \neq y)$ such that \[\| x+y\|_{\psi} ^2+\| x-y\|_{\psi} ^2= 4 M_1^2.\] From the above inequality, the elements $x,y \in S_X~~(x \neq y)$ satisfy $\|x\|_{\psi}=\|x\|_2=1$, $\|y\|_{\psi}=\|y\|_2=1$ and \[\frac{\|x+y\|_{\psi}}{\|x+y\|_2}=\frac{\|x-y\|_{\psi}}{\|x-y\|_2}=M_1.\] Since $\|\cdot\|_{\psi}$ is absolute and $x,y \in S_X~~(x \neq y)$ satisfy $\|x\|_2=\|y\|_2=1$, it is sufficient to consider the following three cases: (i) There exist $s,t\in [0,1]~~(s\neq t)$ satisfying $x=\frac{1}{\psi_2(s)}(1-s,s)$ and $y=\frac{1}{\psi_2(t)}(1-t,t).$ (ii) There exist $s,t\in [0,1]~~(s < t)$ satisfying $x=\frac{1}{\psi_2(s)}(1-s,s)$ and $y=\frac{1}{\psi_2(t)}(-1+t,t).$ (iii) There exist $s,t\in [0,1]~~(s > t)$ satisfying $x=\frac{1}{\psi_2(s)}(1-s,s)$ and $y=\frac{1}{\psi_2(t)}(-1+t,t).$ \noindent Case (i). We may suppose that $s t)$ satisfying $x=\frac{1}{\psi_2(s)}(1-s,s)$ and $y=\frac{1}{\psi_2(t)}(-1+t,t).$ Then, we put $s_0 = t$ and $t_0 =s$. We define $x_0,~~y_0$ in $S_X$ by $$ x_0 = \frac{1}{\psi(s_0)}(1-s_0,s_0), ~~~y_0 = \frac{1}{\psi(t_0)}(-1+t_0,t_0). $$ Then we can reduce Case (ii). \noindent ($\Longleftarrow$). If we suppose (1) (resp. (2)), then we put $x=\frac{1}{\psi_2(s)}(1-s,s)$ (resp. $x=\frac{1}{\psi_2(s)}(1-s,s)$) and $y=\frac{1}{\psi_2(t)}(1-t,t)$ (resp. $y=\frac{1}{\psi_2(t)}(-1+t,t)$). Then we have $||x||_{\psi} = ||x||_2 =1$, $||y||_{\psi} = ||y||_2 =1$, $||x+y||_{\psi} = M_1||x+y||_2$ and $||x-y||_{\psi} = M_1||x-y||_2$. Hence it is clear to prove that $C'_{NJ}(X) = M_1^2$. This completes the proof. \end{proof} We next study the modified NJ constant in the general case. If $\psi \in \Psi$, then by \cite[Therem 3]{S}, we have $$\max\{M_1^2, M_2^2\} \leq C_{NJ}(X) \leq M_1^2M_2^2.$$ However, by Theorem 2.2, there exist many $\psi \in \Psi$ satisfying $\psi \geq \psi_2$ such that $$C'_{NJ}(X) < \max\{M_1^2, M_2^2\} = C_{NJ}(X).$$ From \cite[Theorem 3]{S}, $C_{NJ}(X)=M_1^2M_2^2$ if either $\psi/\psi_2$ or $\psi_2/\psi$ attains a maximum at $t=1/2$. Then, we have the following \begin{proposition} Let $\psi \in \Psi_2$ and let $\psi(t)=\psi(1-t)$ for all $t\in[0,1]$. If $\psi/\psi_2$ attains a maximum at $t=1/2$, then $C'_{NJ}(X)=C_{NJ}(X)=M_1^2M_2^2.$ \end{proposition} \begin{proof} Suppose first $M_1 = \psi(1/2)/\psi_2(1/2)$. Take an arbitrary $t\in[0,1]$ and put \[x=\frac{1}{\psi(t)}(t,1-t)~,~y=\frac{1}{\psi(t)}(1-t,t).\] Then $x,y\in S_X$ and \[\|x+y\|_{\psi}=\frac{2}{\psi(t)}\psi(\frac{1}{2}) ~,~\|x-y\|_{\psi}=\frac{2|2t-1|}{\psi(t)}\psi(\frac{1}{2}).\] Therefore we have \begin{align*} \frac{\|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2}{4} &=\left\{(2t-1)^2+1\right\}\frac{\psi(1/2)^2}{\psi(t)^2}\\ &=2\psi_{2}(t)^2\frac{\psi(1/2)^2}{\psi(t)^2}\\ &=\frac{\psi_{2}(t)^2}{\psi(t)^2} \frac{\psi(1/2)^2}{\psi_{2}(1/2)^2} =M_1^2 \frac{\psi_{2}(t)^2}{\psi(t)^2}. \end{align*} Since $t$ is arbitrary, we have $C'_{NJ}(X)\geq M_1^2M_2^2$ which prove that $C'_{NJ}(X) = M_1^2M_2^2$. \end{proof} In the case that $M_2= \psi_2(1/2)/\psi(1/2)$, $C'_{NJ}(X)$ does not necessarily coincide with $M_1^2M_2^2$. However, we have the following \begin{theorem} Let $\psi \in \Psi_2$ and let $\psi (t)=\psi (1-t)$ for all $t\in[0,1]$. Assume that $M_2= \psi_2(1/2)/\psi(1/2)$ and $M_1 >1$. Then $C'_{NJ}(X)=M_1^2M_2^2$ if and only if there exist $s,t\in [0,1]~~(s < t)$ satisfying one of the following conditions: {\rm (1)} $\psi_2(s)=M_2\psi(s)$, $\psi_2(t)=M_2\psi(t)$ and, if we put $r=\frac{\psi(s)t+\psi(t)s}{\psi(s)+\psi(t)}$, then $\psi(r)=M_1\psi_2(r)$. {\rm (2)} $\psi_2(s)=M_2\psi(s)$, $\psi_2(t)=M_2\psi(t)$ and, if we put $r = \frac{\psi(t)s+\psi(s)t}{\psi(t)+\psi(s)(2t-1)}$, then $\psi(r)=M_1\psi_2(r)$. \end{theorem} \begin{proof} ($\Longrightarrow$). For all $x,y \in S_X$, we have \begin{align*} \|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2&\leq M_1^2\left(\|x+y\|_2^2+\|x-y\|_2^2 \right)\\ &=2M_1^2\left(\|x\|_2^2+\|y\|_2^2 \right)\\ &\leq 2M_1^2M_2^2\left(\|x\|_{\psi}^2+\|y\|_{\psi}^2 \right) =4M_1^2M_2^2. \end{align*} From this inequality, $C'_{NJ}(X)=M_{1}^2M_2^2~$ if and only if there exist $x,y \in S_X~~(x \neq y)$ such that \[\| x+y\|_{\psi} ^2+\| x-y\|_{\psi} ^2= 4 M_1^2M_2^2.\] Suppose that $C'_{NJ}(X)=M_1^2M_2^2$. Then, the elements $x,y \in S_X~~(x \neq y)$ satisfy \[ \|x\|_2=\|y\|_2=M_2,~\|x+y\|_{\psi}=M_1\|x+y\|_2,~\|x-y\|_{\psi}=M_1\|x-y\|_2 . \] Since $\|\cdot\|_{\psi}$ is absolute, it is sufficient to consider the following three cases: (i) There exist $s,t\in [0,1]~~(s\neq t)$ satisfying $x=\frac{1}{\psi(s)}(1-s,s)$ and $y=\frac{1}{\psi(t)}(1-t,t).$ (ii) There exist $s,t\in [0,1]~~(s < t)$ satisfying $x=\frac{1}{\psi(s)}(1-s,s)$ and $y=\frac{1}{\psi(t)}(-1+t,t).$ (iii) There exist $s,t\in [0,1]~~(s > t)$ satisfying $x=\frac{1}{\psi(s)}(1-s,s)$ and $y=\frac{1}{\psi(t)}(-1+t,t).$ As in the proof of Theorem 2.2, we can prove this theorem. This completes the proof. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section {The Zb\u{a}ganu constant of $\Bbb R^2$ } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The Zb{\rm \u{a}}ganu constant $C_Z(X)$ in \cite{Z} is defined by \[C_Z(X)=\sup \left\{ \frac{\| x+y\| \| x-y\| }{\| x\| ^2+\| y\| ^2}~\biggl|~ x,y \in X,~(x,y)\not = (0,0)\biggr. \right\} .\] Then it is clear that $C_Z(X) \leq C_{NJ}(X)$ for any Banach space $X$. In this section, we consider the condition that $C_Z(X)=C_{NJ}(X)$ for $X=\left( \mathbb{R}^2, \| \cdot \|_{\psi} \right)$. Then, we have the following \begin{proposition} Let $\psi \in \Psi_2$. If $\psi\geq \psi_{2}$, then $C_Z(X)= C_{NJ}(X) = M_{1}^2$. \end{proposition} \begin{proof} For any $x,y \in X$, \begin{align*} 2\|x+y\|_{\psi}\|x-y\|_{\psi}&\leq \|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2\\ &\leq M_1^2\left( \|x+y\|_2^2+\|x-y\|_2^2 \right)\\ &=2M_1^2\left( \|x\|_2^2+\|y\|_2^2 \right)\\ &\leq 2M_1^2\left( \|x\|_{\psi}^2+\|y\|_{\psi}^2 \right). \end{align*} Since $\psi/\psi_2$ attains the maximum at $t=t_0$ ($0\leq t_0\leq 1$), we put $x=(1-t_0,0)$ and $y=(0,t_0)$, respectively. Then we have \begin{align*} \| x+y \|_{\psi}^2+\| x-y \|_{\psi}^2 &= 2\psi(t_0)^2\\ &= 2M_1^2\psi_2(t_0)^2\\ &= 2M_1^2 \left(\|x\|_{\psi}^2+\|y\|_{\psi}^2 \right). \end{align*} Since $\|x+y\|_{\psi}=\psi(t_0)=\|x-y\|_{\psi}$, we have \begin{align*} 2\|x+y\|_{\psi}\|x-y\|_{\psi} &=\|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2\\ &=2M_1^2\left (\|x\|_{\psi}^2+\|y\|_{\psi}^2 \right). \end{align*} Therefore we have \[\frac{\|x+y\|_{\psi}\|x-y\|_{\psi}}{\|x\|_{\psi}^2+\|y\|_{\psi}^2}=M_1^2,\] which implies that $C_Z(X)=M_1^2$. \end{proof} We next consider the case that $\psi \leq \psi_2$. We remark that the Zb{\rm \u{a}}ganu constant $C_Z(X)$ is in the following form; \[C_{Z}(X) =\sup \left\{ \frac{4\| x\| \| y\| }{\|x+y\| ^2+\| x-y\| ^2}~\biggl|~ x,y \in X,~(x,y)\not = (0,0)\biggr. \right\}.\] Then we have the following \begin{theorem}Let $\psi \in \Psi_2$. Assume that $\psi\leq \psi_2$. Then $C_Z(X)=M_2^2$ if and only if there exist $s,t\in [0,1]~~(s < t)$ satisfying one of the following conditions: {\rm (1)} $\psi(s)=\psi_2(s)$, $\psi(t)=\psi_2(t)$ and, if we put $r=\frac{\psi(s)t+\psi(t)s}{\psi(s)+\psi(t)}$, then $\frac{\psi_2(r)}{\psi(r)}=\frac{\psi(1-r)}{\psi_2(1-r)}=M_2$. {\rm (2)} $\psi(s)=\psi_2(s)$, $\psi(t)=\psi_2(t)$ and, if we put $r = \frac{\psi(t)s+\psi(s)t}{\psi(t)+\psi(s)(2t-1)}$, then $\frac{\psi_2(r)}{\psi(r)}=\frac{\psi(1-r)}{\psi_2(1-r)}=M_2$. \end{theorem} \begin{proof} For any $x,y \in X$, \begin{align*} 4\|x\|_{\psi}\|y\|_{\psi}&\leq 2\left(\|x\|_{\psi}^2+\|y\|_{\psi}^2\right)\\ &\leq 2\left( \|x\|_2^2+\|y\|_2^2 \right)\\ &= \|x+y\|_2^2+\|x-y\|_2^2 \\ &\leq M_2^2\left( \|x+y\|_{\psi}^2+\|x-y\|_{\psi}^2 \right). \end{align*} Since $X=\left( \mathbb{R}^2, \| \cdot \|_{\psi} \right)$ is finite dimensional, \[C_{Z}(X) =\max \left\{ \frac{4\| x\|_{\psi} \| y\|_{\psi} }{\|x+y\|_{\psi} ^2+\| x-y\|_{\psi} ^2}~\biggl|~ x,y \in X,~(x,y)\not = (0,0)\biggr. \right\}.\] Then $C_Z(X)=M_2^2$ if and only if there exist $x,y \in S_X~~(x \neq y)$ such that \[\frac{4\| x\|_{\psi} \| y\|_{\psi} }{\|x+y\|_{\psi} ^2+\|x-y\|_{\psi} ^2}=M_2^2.\] From the above inequality, $\|x\|_2=\|x\|_{\psi}=\|y\|_{\psi}=\|y\|_2$ and \[\frac{\|x+y\|_2}{\|x+y\|_{\psi}}=\frac{\|x-y\|_2}{\|x-y\|_{\psi}}=M_2^2.\] Hence we may assume that \[\|x\|_2=\|x\|_{\psi}=\|y\|_{\psi}=\|y\|_2=1. \] As in the proof of Theorem 2.2, it is sufficient to consider the following three cases: (i) There exist $s,t\in [0,1]~~(s\neq t)$ satisfying $x=\frac{1}{\psi_2(s)}(1-s,s)$ and $y=\frac{1}{\psi_2(t)}(1-t,t).$ (ii) There exist $s,t\in [0,1]~~(s < t)$ satisfying $x=\frac{1}{\psi_2(s)}(1-s,s)$ and $y=\frac{1}{\psi_2(t)}(-1+t,t).$ (iii) There exist $s,t\in [0,1]~~(s > t)$ satisfying $x=\frac{1}{\psi_2(s)}(1-s,s)$ and $y=\frac{1}{\psi_2(t)}(-1+t,t).$ As in the proof of Theorem 2.2, we can similarly prove this theorem. \end{proof} We next study the Zb{\rm \u{a}}ganu constant $C_Z(X)$ in general case. If $\psi \in \Psi$, by \cite[Theorem 3]{S}, then we have $$\max\{M_1^2, M_2^2\} \leq C_Z(X) \leq C_{NJ}(X) \leq M_1^2M_2^2.$$ However, by Theorem 3.2, there exist many $\psi \in \Psi$ satisfying $\psi \geq \psi_2$ such that $$C_{Z}(X) < C_{NJ}(X) \leq \max\{M_1^2, M_2^2\}.$$ From \cite[Theorem 3]{S}, $C_{NJ}(X)=M_1^2M_2^2$ if either $\psi/\psi_2$ or $\psi_2/\psi$ attains a maximum at $t=1/2$. Then, we have the following \begin{proposition} Let $\psi \in \Psi_2$ and let $\psi(t)=\psi(1-t)$ for all $t\in[0,1]$. If $M_2= \frac{\psi_2(1/2)}{\psi(1/2)}$, then $C_Z(X)=C_{NJ}(X)=M_1^2M_2^2.$ \end{proposition} \begin{proof} From the definition, we have $C_Z(X)\leq C_{NJ}(X) = M_1^2M_2^2$. Take an arbitrary $t\in[0,1]$ and put $x=(t,1-t)$ and $y=(1-t,t)$. Then $\|x\|_{\psi}=\|y\|_{\psi}=\psi(t)$ and $\|x+y\|_{\psi}=\|(1,1)\|_{\psi}=2\psi(1/2)$, $\|x-y\|_{\psi}=\|(2t-1,1-2t)\|_{\psi}=2|2t-1|\psi(1/2)$. Hence we have \begin{align*} \frac{4\| x\|_{\psi} \| y\|_{\psi} }{\|x+y\|_{\psi} ^2+\| x-y\|_{\psi} ^2} &=\frac{2\left(\|x\|_{\psi}^2 +\|y\|_{\psi}^2\right) }{\|x+y\|_{\psi} ^2+\| x-y\|_{\psi} ^2}\\ &=\frac{\psi(t)^2}{\left(1+(2t-1)^2\right)\psi(1/2)^2}\\ &=\frac{\psi(t)^2}{2\psi_2(t)^2\psi(1/2)^2}\\ &=\frac{\psi(t)^2}{\psi_2(t)^2}\frac{\psi_2(1/2)^2}{\psi(1/2)^2}=M_2^2\frac{\psi(t)^2}{\psi_2(t)^2} \end{align*} Since $t$ is arbitrary, we have $C_Z(X)\geq M_1^2M_2^2$. Therefore we have $C_Z(X)= M_1^2M_2^2$. This completes the proof. \end{proof} In case that $M_1= \psi(1/2)/\psi_2(1/2)$, we have the following theorem as in the proof of Theorem 2.2 and so omit the proof. \begin{theorem} Let $\psi \in \Psi_2$ and let $\psi (t)=\psi (1-t)$ for all $t\in[0,1]$. If $M_1= \frac{\psi(1/2)}{\psi_2(1/2)}$ and $M_2 > 1$, then $C_Z(X)=M_1^2M_2^2$ if and only if there exist $s,t\in [0,1]~~(s < t)$ satisfying one of the following conditions: {\rm (1)} $\psi_2(s)=M_2\psi(s)$, $\psi_2(t)=M_2\psi(t)$ and, if we put $r=\frac{\psi(s)t+\psi(t)s}{\psi(s)+\psi(t)}$, then $\psi(r)=M_1\psi_2(r)$. {\rm (2)} $\psi_2(s)=M_2\psi(s)$, $\psi_2(t)=M_2\psi(t)$ and, if we put $r = \frac{\psi(t)s+\psi(s)t}{\psi(t)+\psi(s)(2t-1)}$, then $\psi(r)=M_1\psi_2(r)$. \end{theorem} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section {Examples} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In this section, we calculate $C'_{NJ}(X)$ and $C_Z(X)$ of some Banach spaces $X = (\Bbb R^2,\|\cdot\|_{\psi})$, where $\psi \in \Psi$. First, we consider the case that $\psi = \psi_{p}$. \begin{example} Let $1 \leq p \leq \infty$ and $1/p +1/q = 1$. We put $t=\min (p,q)$. Then $C'_{NJ}(\Bbb R^2, \|\cdot\|_p)= C_Z(\Bbb R^2,||\cdot||_p)= C_{NJ}(\Bbb R^2, \|\cdot\|_p)= 2^{\frac{2}{t}-1}$. Suppose that $1 \leq p \leq 2$. Since $\psi_{p} \geq \psi_{2}$, we have $C_Z(\Bbb R^2,||\cdot||_p) = 2^{\frac{2}{p}-1}$ by Proposition 3.1. On the other hand, as in Theorem 2.2, we take $s=0$ and $t=1$. Since $r = \frac{\psi(0)\cdot 1+\psi(1) \cdot 0}{\psi(0)+\psi(1)}=\frac{1}{2}$ and $M_1 = \psi_p(1/2)/\psi_2(1/2) = 2^{\frac{1}{p}-\frac{1}{2}}$, by Theorem 2.2, we have $C'_{NJ}(\Bbb R^2, \|\cdot\|_p) = M_1^2 = 2^{\frac{2}{p}-1}$. If $2\leq p \leq \infty$, then we similarly have, by Proposition 2.1 and Theorem 3.2, $C'_{NJ}(\Bbb R^2, \|\cdot\|_p)= C_Z(\Bbb R^2,||\cdot||_p)= C_{NJ}(\Bbb R^2, \|\cdot\|_p)= 2^{\frac{2}{p}-1}$. In \cite[Example]{YL}, C. Yang and H. Li calculated the modified NJ constant of the following normed linear space. From our theorems, we have \end{example} \begin{example} Let $\lambda > 0$ and $X_{\lambda} = \Bbb R^2$ endowed with norm $$ ||(x,y)||_{\lambda} = (||(x,y)||_p^2 + \lambda ||(x,y)||_q^2)^{1/2}.$$ (i) If $2 \leq p \leq q \leq \infty$, then $C_{NJ}(X_{\lambda}) = C'_{NJ}(X_{\lambda}) = C_Z(X_{\lambda}) = \frac{2(\lambda +1)}{2^{2/p} + \lambda 2^{2/q}}.$ \noindent (ii) If $1 \leq p \leq q \leq 2$, then $C_{NJ}(X_{\lambda}) = C'_{NJ}(X_{\lambda}) = C_Z(X_{\lambda}) = \frac{2^{2/p} + \lambda 2^{2/q}}{2(\lambda +1)}.$ To see this, first, we remark that $(p,q)$ is not necessarily a H\"{o}lder pair. We define the normalized norm $||\cdot||_{\lambda}^0$ by $$ ||(x,y)||_{\lambda}^0 = \frac{||(x,y)||_{\lambda}}{\sqrt{1+ \lambda}}.$$ Then $||\cdot||_{\lambda}^0$ is absolute and so put the corresponding function $\psi_{\lambda}(t) = ||(1-t,t)||_{\lambda}^0$. \noindent (i) Suppose that $2 \leq p \leq q \leq \infty$. Since $\psi_{\lambda} \leq \psi_2$, by Proposition 2.1, we have $C_{NJ}(X_{\lambda}) = C'_{NJ}(X_{\lambda}) = M_2^2 = \frac{2(\lambda +1)}{2^{2/p} + \lambda 2^{2/q}}.$ On the other hand, in Theorem 3.2, we take $s=0$ and $t=1$. Then we have $r=1/2$ and $\frac{\psi_2(1/2)}{\psi_{\lambda}(1/2)}=M_2$. Thus we have $C_Z(X_{\lambda}) = M_2^2 = \frac{2^{2/p} + \lambda 2^{2/q}}{2(\lambda +1)}.$ \noindent (ii) Suppose that $1 \leq p \leq q \leq 2$. Since $\psi_{\lambda} \geq \psi_2$, by Theorem 2.2 and Proposition 3.1, we similarly have (ii). \end{example} \begin{example} Put \[ \psi(t)=\left\{ \begin{array}{cc} \psi_2(t) & (0\leq t \leq 1/2), \\[1em] \left( 2-\sqrt{2} \right) t+ \sqrt{2}-1& (1/2\leq t\leq 1). \\ \end{array} \right. \] Then $C'_{NJ}(\Bbb R^2, \|\cdot\|_{\psi} )< C_Z(\Bbb R^2, \|\cdot\|_{\psi})= C_{NJ}(\Bbb R^2, \|\cdot\|_{\psi})=2\sqrt{2}(\sqrt{2}-1)$. In fact, $\psi \in \Psi_2$ and the norm of $\|\cdot\|_{\psi}$ is \[ \| (a,b) \|_{\psi}=\left\{ \begin{array}{cc} \sqrt{|a|^2+|b|^2} & \left(|a|\geq |b| \right) \\[1em] \left(\sqrt{2}-1\right)|a|+|b| & \left(|a| \leq |b| \right). \\ \end{array} \right. \] Since $\psi\geq \psi_2$, by Proposition 3.1, we have $C_Z\left( \mathbb{R}^2, \|\cdot\|_{\psi} \right )=M_1^2=2\sqrt{2}\left(\sqrt{2}-1\right)$. We assume that $C'_{NJ}(\Bbb R^2, \|\cdot\|_{\psi} )= M_1^2$. By Theorem 2.2, we can choose $r \in [0,1]$ such that $\frac{\psi(r)}{\psi_2(r)}=\frac{\psi(1-r)}{\psi_2(1-r)}=M_1$. This is impossible by the definition of $\psi$. Therefore we have $C'_{NJ}(\Bbb R^2, \|\cdot\|_{\psi} ) < M_1^2$. \end{example} \begin{example} Let $1/2 \leq \beta \leq 1$. We define a convex function $\psi_{\beta} \in \Psi_2$ by $$\psi_{\beta}(t) = \max\{1-t, t, \beta \}.$$ By \cite[Example 4]{S}, we have \begin{align*} C_{NJ}(\Bbb R^2,\|\cdot\|_{\psi_{\beta}}) = \begin{cases} \dfrac{\beta^2 + (1-\beta)^2}{\beta^2} &(\beta \in [\frac 12,\frac{1}{\sqrt 2}])\\ 2(\beta^2 + (1-\beta)^2) &(\beta \in (\frac{1}{\sqrt 2}, 1]). \end{cases} \end{align*} Indeed, \begin{align*} M_1 = \begin{cases} 1 &(\beta \in [\frac 12,\frac{1}{\sqrt 2}])\\ \frac{\psi_{\beta}(1/2)}{\psi_2(1/2)} = \frac{\beta}{1/\sqrt{2}} = \sqrt{2}\beta &(\beta \in (\frac{1}{\sqrt 2}, 1]) \end{cases} \end{align*} and $$ M_2 = \frac{\psi_2(\beta)}{\psi_{\beta}(\beta)} = \frac{1}{\beta}\{(1-\beta)^2+\beta^2\}^{1/2}.$$ If $1/2 \leq \beta \leq 1/\sqrt{2}$, then $\psi_{\beta} \leq \psi_2$ and so, by Proposition 2.1, we have $$C'_{NJ} (\Bbb R^2,\|\cdot\|_{\psi_{\beta}}) = M_2^2=\dfrac{\beta^2 + (1-\beta)^2}{\beta^2}.$$ By Theorem 3.2, we have $C_Z(\Bbb R^2,\|\cdot\|_{\psi_{\beta}}) < M_2^2$. Assume that $1/\sqrt{2} < \beta \leq 1$. Since $M_1 = \frac{\psi_{\beta}(1/2)}{\psi_2(1/2)}$, we have, by Proposition 2.3, $$C'_{NJ} (\Bbb R^2,\|\cdot\|_{\psi_{\beta}}) = M_1^2M_2^2 = 2(\beta^2 + (1-\beta)^2).$$ On the other hand, we take $s=\beta$ and $t=1-\beta$ in Theorem 3.4. Then we have $r = \frac{\psi(\beta)(1-\beta)+\psi(1-\beta)\beta}{\psi(\beta)+\psi(1-\beta)} = 1/2$. By Theorem 3.4, we have $$C_Z(\Bbb R^2,\|\cdot\|_{\psi_{\beta}})=M_1^2M_2^2=2(\beta^2 + (1-\beta)^2).$$ \end{example} \begin{example} We consider $\psi_{\beta}$ in Example 4.4 in case of $\beta = 1/\sqrt{2}$. Then we have $$C'_{NJ}( \mathbb{R}^2, \|\cdot\|_{\psi_{\beta}} )=C_{NJ}( \mathbb{R}^2, \|\cdot\|_{\psi_{\beta}})=M_2^2 =2\sqrt{2}(\sqrt{2}-1).$$ On the other hand, we have $C_Z( \Bbb R^2, \|\cdot\|_{\psi_{\beta}})=M_2^2 =2\sqrt{2} (\sqrt{2}-1)$. For this $\psi_{\beta}$, define a convex function $\varphi \in \Psi_2$ by \[ \varphi(t)=\left\{ \begin{array}{cc} \psi_{\beta}(t) & (0\leq t \leq 1/2), \\[1.0em] \psi_2(t) &(1/2 \leq t \leq 1). \end{array} \right. \] As in Example 4.2, we similarly have $$C_Z(\Bbb R^2, \|\cdot\|_{\varphi })