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\noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
Banach J. Math. Anal. 2 (2008), no. 2, 31--41\\
$\frac{\rule{4.55in}{0.05in}}{{}}$\\
{\footnotesize
\textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach
\textbf{\large{J}}ournal of \textbf{\large{M}}athematical
\textbf{\large{A}}nalysis}}\\
ISSN: 1735-8787 (electronic)\\
\textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\
$\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.5in]}


\title[Triangle inequality]{Some remarks on the triangle inequality for norms}

\author[L. Maligranda]{Lech Maligranda}

\address{Department of Mathematics, Lule{\aa} University of Technology, SE--971 87 Lule{\aa}, Sweden.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{lech@sm.luth.se}}


\dedicatory{Dedicated to Professor Josip Pe\v{c}ari\'{c}\\
\vspace{.5cm}{\rm Submitted by D. E. Alspach}}

\subjclass[2000]{Primary 46B20, 46B99; Secondary 51M16.}

\keywords{Inequalities, normed space, norm inequality, triangle inequality,
reversed triangle inequality, angular distance, Fischer--Musz\'ely equality.}

\date{Received: 11 April 2008; Accepted 24 April 2008.}

\begin{abstract}
Remarks about strengthening of the triangle inequality
and its reverse inequality in normed spaces for two and more elements
are collected. There is also a discussion on Fischer--Musz\'ely
equality for $n$-elements in a normed space. Some other estimates
which follow from the triangle inequality are also presented.
\end{abstract} \maketitle

\section{Introduction and preliminaries}
\noindent We present three results on refinements of the triangle
inequality and some related estimates of independent interest.

\medskip

{\bf A)} The following strengthening of the triangle inequality and
its reverse inequality in normed spaces were observed already in
2003. The paper \cite{MALI} was sent to AMM in 2003 and published in
2006.

\medskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem} For any nonzero elements $x$ and $y$ in a normed space
$X = (X, \|\cdot\|)$ we have
\begin{eqnarray}\label{1}
\|x + y \| \leq \|x\| + \|y\| - \left( 2 - \| \frac{x}{\|x\|} +
\frac{y}{\|y\|} \| \right) \min \left\{ \|x\|, \|y\| \right\}
\end{eqnarray}
and
\begin{eqnarray}\label{2}
\|x + y \| \geq \|x\| + \|y\| - \left( 2 - \| \frac{x}{\|x\|} +
\frac{y}{\|y\|} \| \right) \max \left\{ \|x\|, \|y\| \right\}.
\end{eqnarray}
Moreover, if either $\|x\| = \|y\|$ or $y = c x$ with $c > 0$, then
equality holds in both $\eqref{1}$ and \eqref{2}.
\end{theorem}
\begin{proof}(\cite{MALI}, p. 257) \rm Without loss of generality we may
assume that $\|x\| \leq \|y\|$. Then, by the triangle inequality
\begin{eqnarray*}
\|x + y\|
&=&
\| \frac{\|x\|}{\|x\|} x +  \frac{\|x\|}{\|y\|} y +
(1 -  \frac{\|x\|}{\|y\|} ) y \| \\
&\leq&
\|x\| ~\|  \frac{x}{\|x\|} +  \frac{y}{\|y\|} \| + \|y\| - \|x\| \\
&=&
\|y\| + ( \| \frac{x}{\|x\|} +  \frac{y}{\|y\|} \| - 1 ) \|x\| \\
&=&
\|x\| + \|y\| + ( \| \frac{x}{\|x\|} +  \frac{y}{\|y\|} \| - 2 ) \|x\|,
\end{eqnarray*}
which establishes the estimate \eqref{1}. Similarly, the computation
\begin{eqnarray*}
\|x + y\|
&=&
\| \frac{\|y\|}{\|y\|} y +  \frac{\|y\|}{\|x\|} x +
(1 -  \frac{\|y\|}{\|x\|} ) x \| \\
&\geq &
\|y\| \|  \frac{y}{\|y\|} +  \frac{x}{\|x\|} \| -
\left| ~\|x\| - \|y\| ~\right| \\
&=&
\|y\| \| \frac{y}{\|y\|} +  \frac{x}{\|x\|} \| -
\|y\| + \|x\| \\
&=&
\|x\| + \|y\| - ( 2 - \| \frac{x}{\|x\|} +  \frac{y}{\|y\|}
\| ) \|y\|
\end{eqnarray*}
gives the inequality \eqref{2}.
\end{proof}

Estimates \eqref{1} and \eqref{2} were explicitly stated and proved in
\cite[Theorem 1]{MALI}. Estimate \eqref{1} appeared also in \cite[Lemma
1.1]{HU-L} and implicitly they appeared in \cite[Lemma 2]{K-S-T}.
We can put  estimates \eqref{1} and \eqref{2} together as
$$
\|x + y \| + \left( 2 - \| \frac{x}{\|x\|} +
\frac{y}{\|y\|} \| \right) \min \left\{ \|x\|, \|y\| \right\} \leq
\|x\| + \|y\|
$$
$$\leq \|x + y \| + \left( 2 - \| \frac{x}{\|x\|} +
\frac{y}{\|y\|} \| \right) \max \left\{ \|x\|, \|y\| \right\}.
$$
Moreover, we can rewrite them as the estimates for the
so-called {\it norm-angular distance} (called also the {\it Clarkson distance}
since he defined it in \cite{CL}) between nonzero $x$ and $y$ as
$d[x, y] = \|\frac{x}{\|x\|} - \frac{y}{\|y\|} \|$ (cf.
\cite[Remark 3]{MALI}):

\medskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{corollary} For any nonzero elements $x$ and $y$ in a normed space
$X = (X, \|\cdot\|)$ we have
\begin{eqnarray}\label{3}
\| \frac{x}{\|x\|} -
\frac{y}{\|y\|} \| \leq \frac{ \|x - y \| + \left| ~\|x\| - \|y\|
~\right|}{\max \{ \|x\|, \|y\|\}}
\end{eqnarray}
and
\begin{eqnarray}\label{4}
\| \frac{x}{\|x\|} -
\frac{y}{\|y\|} \| \geq \frac{ \|x - y \| - \left|~\|x\| - \|y\|
~\right|}{\min \{ \|x\|, \|y\|\}}.
\end{eqnarray}
\end{corollary}
\begin{proof} \rm Estimate \eqref{3} follows directly from \eqref{2} and estimate \eqref{4} directly
from \eqref{1}. In fact, inequality \eqref{2} implies that
\begin{eqnarray*}
\|\frac{x}{\|x\|} - \frac{y}{\|y\|} \| \max \{ \|x\|, \|y\|\}
&\leq &
\|x - y\| + 2 \max \{ \|x\|, \|y\|\} - \|x\| - \|y\| \\
&=&
\|x - y\| + |~\|x - y\|~|,
\end{eqnarray*}
and inequality \eqref{1} gives that
\begin{eqnarray*}
\|\frac{x}{\|x\|} - \frac{y}{\|y\|} \| \max \{ \|x\|, \|y\|\}
&\geq &
\|x - y\| + 2 \min \{ \|x\|, \|y\|\} - \|x\| - \|y\| \\
&=&
\|x - y\| - |~\|x - y\|~|.
\end{eqnarray*}
\end{proof}

\medskip
Estimates \eqref{1} and \eqref{2} mean for the norm-angular distance that
$$
\left( 2 - d[x, -y]\right) \min\{\|x\|, \|y\|\} \leq \|x \| + \|y\| -
\| x + y\|
$$
$$
\leq \left( 2 - d[x, -y]\right) \max \{\|x\|, \|y\|\}
$$
and they were mentioned in the book by E. Deza and M.-M. Deza \cite[p. 52]{DE-D} as
a result from \cite{MALI}. Estimates \eqref{3} and \eqref{4} mean that
$$
\frac{ \|x - y \| - \left|~\|x\| - \|y\|
~\right|}{\min \{ \|x\|, \|y\|\}} \leq d[x, y] \leq
\frac{ \|x - y \| + \left| ~\|x\| - \|y\|
~\right|}{\max \{ \|x\|, \|y\|\}}.
$$
Estimate \eqref{3} is a refinement of the Massera--Sch\"affer
inequality proved in 1958 (see \cite[Lemma 5.1]{M-S}; see also
\cite[Theorem 1]{GUR} and \cite[p. 516]{M-P-F}): for nonzero vectors $x$ and
$y$ in $X$ we have that $d[x, y] \leq \frac{2 \|x - y\|}{\max \left( \|x\|,
\|y\| \right)}$, which is stronger than the Dunkl--Williams
inequality $d[x, y] \leq \frac{4 \|x - y\|}{\|x\| +
\|y\|}$ proved in \cite{DU-W}. Estimates \eqref{2} and \eqref{4} can be seen
as the reverse inequalities of \eqref{1} and \eqref{3}, respectively. Another
proof of estimate \eqref{4} appeared recently in \cite[Theorem 1]{ME}. By
the way, Mercer \cite{ME} is using the name {\it Maligranda inequality} for \eqref{3}
and {\it reverse Maligranda inequality} for \eqref{4}, but
Pe\v{c}ari\'{c}-Raji\'{c} \cite{P-R} called them {\it Maligranda--Mercer
inequalities}.

\medskip
Note that the Dunkl--Williams inequality holds with constant $2$ if
and only if $X$ is an inner product space (\cite{KI-S}; cf. also \cite[p. 31]{AM})
but one cannot replace the constant $2$ by $1$ in the Massera--Sch\"affer
inequality even for an inner product space. Conditions for equality are
proved in \cite{KI-S} and, consequently, we can also ask for the equality
conditions in \eqref{3} and \eqref{4}.
The Dunkl--Williams estimate was used in the proof of the fact that the Lipschitz
norm of the radial projection $k(X)$ is smaller than 2 (see \cite{DF-K},
\cite{SC}, \cite{TH}, \cite[p. 142]{AM}; see also \cite{MA}, where the
radial projection was used in the proof of the Dugunji theorem on
a failure of the Brouwer fixed point theorem in arbitrary infinite
dimensional Banach space, and \cite[Theorem 1]{DE}, where Desbiens showed
that the Sch\"affer constant $s(X):=
\limsup_{\alpha[x,y] \rightarrow 0^{+}} \frac{\alpha[x,y] \max\{
\|x\|, \|y\| \}}{\|x - y\|}$ is equal to $k(X)$ in every
finite-dimensional Banach space $X$). Inequality \eqref{1} was also used in the
proof of a certain estimate of the Jordan--von Neumann constant $C_{JN}(X)$
by the James constant $J(X)$ (see \cite[Lemma 1]{MAL} and \cite[Lemma 1]{M-N-P-Z}).

Recently, Betiuk--Pilarska and Prus \cite{BP-P} used the inequalities \eqref{1}
and \eqref{2} to find estimate of the James constant of a direct sum of
the spaces $X_{s}$ by its James constants:
$J( (\sum_{s\in S} X_{s})_{Z} ) \leq 2 - \frac{1}{2} [ 2 - \sup_{s \in S}
J(X_{s})] [ 2 - J(Z) ]$. Moreover, Jim{\'e}nez--Melado, Llorens--Fuster and
Mazu{\~n}{\'a}n--Nararro \cite{J-L-M} introduced the notion of {\it
Dunkl--Williams constant}
$$DW(X) = \sup \left\{ \frac{\| x\| + \| y\|}{\|x - y \|}: ~x, y \in
X, x \neq 0, y \neq 0, x \neq y\right \}$$ and collected its connection
with some other constants.

\medskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{remark} The right-hand sides of \eqref{1} and \eqref{2} can be
written also in another way since
\vspace{-5mm}
\begin{eqnarray*}
\|x\| + \|y\|
&-&
\left( 2 - \|\frac{x}{\|x\|} +
\frac{y}{\|y\|} \| \right) \min \left\{ \|x\|, \|y\| \right\} \\
&=&
\min \left\{ \|x\|, \|y\| \right\} \| \frac{x}{\|x\|} +
\frac{y}{\|y\|} \| + \left|~\|x\| - \|y\|~\right|
\end{eqnarray*}
\vspace{-5mm}
and
\begin{eqnarray*}
\|x\| + \|y\|
&-&
\left( 2 - \| \frac{x}{\|x\|} +
\frac{y}{\|y\|} \| \right) \max \left\{ \|x\|, \|y\| \right\} \\
&=&
\max \left\{ \|x\|, \|y\| \right\} \| \frac{x}{\|x\|} +
\frac{y}{\|y\|} \| - \left|~\|x\| - \|y\|~\right|.
\end{eqnarray*}
\end{remark}

Kato, Saito and Tamura (\cite{KA-S-T}, Theorem 1) generalized
inequalities \eqref{1} and \eqref{2} to $n$-elements with $n \geq
2$: {\it For any nonzero elements $x_{1}, x_{2}, \ldots, x_{n}$ in a
normed space $X = (X, \|\cdot\|)$ we have
\begin{eqnarray}\label{5}
\| \sum_{k=1}^{n} x_{k} \| \leq \sum_{k=1}^{n} \|x_{k}\| - \left( n -
\| \sum_{k=1}^{n}\frac{x_{k}}{\|x_{k}\|} \| \right)
~\min_{k=1, 2,\ldots,n} ~\|x_{k}\|
\end{eqnarray}
and
\begin{eqnarray}\label{6}
\| \sum_{k=1}^{n} x_{k} \| \geq \sum_{k=1}^{n} \|x_{k}\| - \left( n -
\| \sum_{k=1}^{n}\frac{x_{k}}{\|x_{k}\|} \| \right)
~\max_{k=1, 2,\ldots,n} ~\|x_{k}\|.
\end{eqnarray}
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{remark} If either $\|x_{1}\| = \|x_{2}\| = \ldots = \|x_{n} \|$
or for a fixed $i \in I = \{1, 2, \ldots, n\}$ we have that $0 \neq
x_{i} \in X$ and $x_{k} = c_{k} x_{i}$ with $c_{k} >
0$ for $k \in I \setminus \{i\}$, then equality holds in both \eqref{5} and \eqref{6}.
\end{remark}

\medskip
\noindent
Immediately from inequalities \eqref{5} and \eqref{6} we have the following equivalence.

\medskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{corollary}\label{c2}
 For nonzero vectors $x_{1}, x_{2}, \ldots, x_{n}$
in a normed space $X = (X, \|\cdot\|)$ we have equality
$\|\sum_{k=1}^{n} x_{k}\| = \sum_{k=1}^{n} \| x_{k}\|$
if and only if $\|\sum_{k=1}^{n} \frac{x_{k}}{\|x_{k}\|} \| = n$.
\end{corollary}

\medskip
Let us compare the above considerations with the Fischer--Musz\'ely
equality for $n$-elements in a normed space. The statement was proved
for two-elements ($n = 2$) by Fischer--Musz\'ely \cite{FI-M} (see also
\cite[Lemma 1]{BA}, \cite[Lemma 2.1]{AB-A-B} and \cite[Lemma 11.4]{AB-A})
and for three-elements ($n = 3$) by Lin \cite[Lemma 1]{LIN}. An
induction proof for all $n \geq 2$ can be found in \cite[pp.
335-336]{A-AL} and \cite[p. 463]{AB-A}. We present a direct proof
following the ideas of Baker \cite{BA} and Lin \cite {LIN}. Some results,
without knowledge of the Fisher--Musz{\'e}ly theorem, were presented
also by Kato, Saito and Tamura (cf. \cite{KA-S-T}, Lemma 1 and Theorems
3, 4).


\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem}\label{t2} {\bf (a)}. If $x_{1}, x_{2}, \ldots, x_{n}$
are elements in a normed space $X = (X, \|\cdot\|)$, then the equality
\begin{eqnarray}\label{7}
\| \sum_{k=1}^{n} x_{k} \| = \sum_{k=1}^{n} \|x_{k}\|
\end{eqnarray}
holds if and only we have equality
\begin{eqnarray}\label{8}
\| \sum_{k=1}^{n} a_{k} x_{k} \| = \sum_{k=1}^{n} a_{k} \|x_{k}\|
\end{eqnarray}
for any positive numbers $a_{1}, a_{2}, \ldots, a_{n}$.

{\bf (b).}  If, in addition, $X$ is a strictly convex normed space, that is,
its sphere does not contain any segment, then
the equalities \eqref{7} and \eqref{8} for nonzero $x_{1}, x_{2}, \ldots, x_{n}
\in X$ are equivalent to the equalities
\begin{eqnarray}\label{9}
\frac{x_{1}}{\|x_{1}\|} = \frac{x_{2}}{\|x_{2}\|} = \ldots
= \frac{x_{n}}{\|x_{n}\|}.
\end{eqnarray}
\end{theorem}
\begin{proof} (a) Of course, it is sufficient to prove the implication
\eqref{7}$\Longrightarrow$ \eqref{8}. Without loss of generality we can assume
that $a_{1} = \max_{k = 1, 2, \ldots, n} a_{k}$. Then, by \eqref{7}, we
obtain
\begin{eqnarray*}
\| \sum_{k=1}^{n} a_{k} x_{k} \|
&=&
\| a_{1} \sum_{k=1}^{n} x_{k} - \sum_{k=1}^{n} (a_{1} - a_{k}) x_{k} \| \\
&\geq&
a_{1} \| \sum_{k=1}^{n} x_{k}\| - \|\sum_{k=1}^{n} (a_{1} - a_{k}) x_{k} \| \\
&\geq&
a_{1} \| \sum_{k=1}^{n} x_{k}\| - \sum_{k=1}^{n} (a_{1} - a_{k}) \|x_{k} \| \\
&=&
\sum_{k=1}^{n} a_{k} \|x_{k}\|.
\end{eqnarray*}
The reverse inequality follows from the triangle inequality and thus we
obtain the equality \eqref{8}.

(b) It is well-known that a normed space $X$ is strictly convex if
and only if the equality $\|x + y \| = \|x\| + \|y\|$ for nonzero $x,
y$ implies that $x = c y$ for some $c > 0$.

If we assume that \eqref{7} holds, then for any $1 < j \leq n$
\begin{eqnarray*}
\| x_{1} + x_{j} \|
&\geq&
\| \sum_{k=1}^{n} x_{k} \| - \| \sum_{k\neq 1, j} x_{k} \| \geq
\| \sum_{k=1}^{n} x_{k} \| - \sum_{k\neq 1, j} \|x_{k} \| \\
&=&
\sum_{k=1}^{n} \| x_{k} \| - \sum_{k\neq 1, j} \|x_{k} \| =
\|x_{1}\| + \|x_{j}\|,
\end{eqnarray*}
and whence $\| x_{1} + x_{j} \| = \|x_{1}\| + \|x_{j}\|$, which by the
strict convexity of $X$ gives that $x_{1} = c_{j} x_{j}$ with $c_{j} > 0,
1 < j \leq n$. Consequently, $c_{j} = \frac{\|x_{1}\|}{\|x_{j}\|}$
or $\frac{x_{1}}{\|x_{1}\|} = \frac{x_{j}}{\|x_{j}\|}$ for any
$1 < j \leq n$, and \eqref{9} is proved.

If \eqref{9} holds, then for any positive numbers $a_{1}, a_{2}, \ldots, a_{n}$
\begin{eqnarray*}
\| \sum_{k=1}^{n} a_{k} x_{k} \|
&=&
\| \sum_{k=1}^{n} a_{k} \|x_{k}\| \frac{x_{k}}{\|x_{k}\|} \| =
\| \sum_{k=1}^{n} a_{k} \|x_{k}\| \frac{x_{i}}{\|x_{i}\|} \| \\
&=&
\sum_{k=1}^{n} a_{k} \|x_{k}\| ~\| \frac{x_{i}}{\|x_{i}\|} \| =
\sum_{k=1}^{n} a_{k} \|x_{k}\|,
\end{eqnarray*}
and the theorem is proved.
\end{proof}
%%%%%%%%%%%%%%%%%
\begin{remark} If \eqref{9} holds, then we have equalities \eqref{7} and
\eqref{8} without any restriction on a normed space $X$. It will be
interesting to characterize equalities in \eqref{5} and \eqref{6}.
\end{remark}

\medskip
Some other sharpenings of \eqref{3} and \eqref{4} in the case $n
\geq 3$ (for $n = 2$ they are just estimates \eqref{1} and \eqref{2}
was proved by Mitani, Saito, Kato and Tamura \cite[Theorem
1]{M-S-K-T}: {\it For any nonzero elements $x_{1}, x_{2}, \ldots,
x_{n}$ in a normed space $X = (X, \|\cdot\|)$ we have
$$
\| \sum_{k=1}^{n} x_{k} \| + \sum_{k=2}^{n}  \left( k - \| \sum_{i=1}^{k} \frac{x_{i}^{*}} {\|x_{i}^{*}\| } \| \right)
\left(\|x_{k}^{*}\|  - \|x_{k+1}^{*}\| \right)  \leq
\sum_{k=1}^{n} \| x_{k} \|
$$
$$
\leq \| \sum_{k=1}^{n} x_{k} \| -
\sum_{k=2}^{n}  \left( k - \| \sum_{i=n-(k-1)}^{n} \frac{x_{i}^{*}} {\|x_{i}\|} \right)
\left(\|x_{n-k}^{*}\|  - \|x_{n-(k-1)}^{*}\| \right),
$$
where $x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}$ are the
rearrangements\,~~ of\,~~ $\|x_{1}\|, \|x_{2}\|, \ldots,
\|x_{n}\|$\,\, satisfying~~~ \, $\|x_{1}^{*} \| \geq \|x_{2}^{*}\|
\geq \ldots \geq \|x_{n}^{*}\|$ and $x_{0}^{*} = x_{n+1}^{*}= 0.$ }

\medskip
Pe\v{c}ari\'{c} and Raji\'{c} \cite{P-R} generalized inequalities \eqref{3}
and \eqref{4} to $n$-elements with $n \geq 2$: {\it if nonzero $x_{1}, x_{2},
\ldots, x_{n} \in X$, then
$$
\max_{1 \leq i \leq n} \{ \frac{S - D_{i}}{\|x_{i}\|} \}
\leq \| \sum_{k=1}^{n}\frac{x_{k}}{\|x_{k}\|}\| \leq
\min_{1 \leq i \leq n} \{ \frac{S + D_{i}}{\|x_{i}\|} \},
$$
where $S = \| \sum_{k=1}^{n} x_{k} \|$ and
$D_{i} = \sum_{k=1}^{n} \left| \| x_{k} \| - \|x_{i}\| \right|, i = 1, 2, \ldots, n.$}
They also observed (cf. \cite{P-R}, Corollary 2.3) that from these
estimates we can obtain the estimates \eqref{5} and \eqref{6} in the form
$$
\sum_{k=1}^{n} \|x_{k}\| \leq \| \sum_{k=1}^{n} x_{k} \| + \left( n -
\| \sum_{k=1}^{n}\frac{x_{k}}{\|x_{k}\|} \| \right)
~\max_{k=1, 2,\ldots,n} ~\|x_{k}\|
$$
and
$$
\sum_{k=1}^{n} \|x_{k}\| \geq\| \sum_{k=1}^{n} x_{k} \| + \left( n -
\| \sum_{k=1}^{n}\frac{x_{k}}{\|x_{k}\|} \| \right)
~\min_{k=1, 2,\ldots,n} ~\|x_{k}\|.
$$
These last inequalities were also obtained, in a more general form and even
for convex functions, by Dragomir \cite{DR}:
$$
\sum_{k=1}^{n} \|x_{k}\|^{p} - n^{1-p} \|\sum_{k=1}^{n} x_{k}\|^{p}
\leq \left( \sum_{k=1}^{n} \|x_{k}\|^{p-1} - \|
\sum_{k=1}^{n}\frac{x_{k}}{\|x_{k}\|} \|^{p}\right) ~\max_{k=1, 2,\ldots,n} ~\|x_{k}\|
$$
and
$$
\sum_{k=1}^{n} \|x_{k}\|^{p} - n^{1-p} \|\sum_{k=1}^{n} x_{k}\|^{p}
\geq \left( \sum_{k=1}^{n} \|x_{k}\|^{p-1} - \|
\sum_{k=1}^{n}\frac{x_{k}}{\|x_{k}\|} \|^{p}\right) ~\min_{k=1, 2,\ldots,n} ~\|x_{k}\|,
$$
where $p \geq 1$ and $n \geq 2$.

\bigskip
{\bf B)} In 1930 Alfred Tarski posed in \cite{TA} the question to prove that
$$
\left |~ | x| - | y| ~\right | = | x + y| + | x - y| - | x| - |y|
$$
holds for all real numbers $x, y$, and in 1931 appeared his nice solution
\cite{TA}. Of course, this equality is not true for complex numbers
(as a counterexample it is enough to take $x = 1$ and $y = i$).

\noindent
Note also that since for all $x, y \in {\Bbb R}$ we have the equality
$\max\{|x + y|, |x - y|\} = |x| + |y|$ and also
\begin{eqnarray*}
|x + y| + |x - y| - |x| - |y|
&=&
|x + y| + |x - y| - \max\{|x + y|, |x - y|\}\\
&=&
\min\{|x + y|, |x - y|\},
\end{eqnarray*}
it yields that for all real $x, y$:
\begin{eqnarray*}
\left |~ | x| - | y| ~\right | = | x + y| + | x - y| - | x| - |y| =
\min\{|x + y|, |x - y|\}.
\end{eqnarray*}
Next we state a corresponding result in normed spaces.

\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem} \label{t3}
For any elements $x, y$ in a normed space $X =
(X, \|\cdot\|)$ we have that
\begin{eqnarray}\label{11}
\left |~\| x\| - \| y\| ~\right | \leq \| x + y\| + \| x - y\| - \| x\| - \|y\|
\leq \min \{ \|x + y\|, \| x - y\|\}
\end{eqnarray}
and
\begin{eqnarray}\label{12}
\left |~\| x\| - \| y\| ~\right | \leq \|x\| + \|y\| - \left |~\| x +
y\| - \| x - y\| ~\right |.
\end{eqnarray}
\end{theorem}
\begin{proof} It is clear that for every $x, y \in X$
$$
\| x\| + \|y\| + \left |~\| x\| - \| y\| ~\right | = 2 \max \{ \|x\|,
\|y\|\}
$$
and
$$
\| x\| + \|y\| - \left |~\| x\| - \| y\| ~\right | = 2 \min \{ \|x\|,
\|y\|\}.
$$
By the triangle inequality
$$
\| x + y\| + \| x - y\| \geq \| x + y \pm (x - y)\|  = 2
\|x\| ~~{\rm or} ~= 2 \|y\|.
$$
Thus,
$$
\| x + y\| + \| x - y\| \geq 2 \max \{ \|x\|, \|y\|\},
$$
and by combining these facts we obtain the first estimate in \eqref{11}. The second
estimate in \eqref{11} follows directly from the triangle inequality. \\
By the triangle inequality
$$
\left |~\| x + y\| - \| x - y\| ~\right | \leq \|x + y \pm(x - y)\| = 2
\|x\| ~~{\rm or} ~= 2 \|y\|,
$$
so that
$$
\left |~\| x + y\| - \| x - y\| ~\right | \leq 2 \min \{ \|x\|, \|y\|\}
$$
and also the estimates in \eqref{12} are proved.
\end{proof}

\bigskip
{\bf C)} In a normed space $X = (X, \|\cdot\|)$ for a fixed $u \in X$ and $p
\geq 1$ we consider new norms $\|\cdot\|_{u, p}$ defined by
\begin{eqnarray}\label{13}
\|x\|_{u, p} = \left( \| x + \|x\| u \|^{p} + \| x - \|x\| u \|^{p} \right)^{1/p}.
\end{eqnarray}
The norm $\|\cdot\|_{u, 1}$ was considered by Odell and Schlumprecht
\cite[p. 178]{O-S} to produce a strictly convex norm in every separable
Banach space (cf. also \cite[p. 118]{PK}).

\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem} \label{t4} The functionals $\|\cdot\|_{u, p}$ defined by \eqref{13} are norms in
$X$ which are equivalent to the norm $\|\cdot\|$.
\end{theorem}

\bigskip
\noindent
To prove this theorem we will need the following lemma of independent
interest.

\medskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma} \label{l1} For fixed $x, y$ in a normed space $X$ and $p \geq
1$ consider the function $f_{p}: {\Bbb R} \rightarrow [0, \infty)$
defined by
$$
f_{p}(t) = \left( \| x + t y \|^{p} + \| x - t y \|^{p}
\right)^{1/p}, ~~t \in {\Bbb R}.
$$
Then $f_{p}$ is an even convex function and so is increasing on $[0,
\infty)$.
\end{lemma}
\begin{proof} Let $0 \leq \alpha, \beta \leq 1$ be such that
$\alpha + \beta = 1$ and $s, t \in {\Bbb R}$. Then, by the triangle
inequality, homogeneouity of the norm $\|\cdot\|$ and the Minkowski
inequality for two-dimensional $l^{p}_{2}$-norm, that is, for $a, b,
c, d \geq 0$ it yields that
$$
\|(a+c, b + d)\|_{p} \leq \|(a, b)\|_{p} +,\|(c, d)\|_{p}
$$
or, equivalently,
$$
[(a+c)^{p} + (b + d)^{p}]^{1/p} \leq (a^{p} + b^{p})^{1/p} + (c^{p} + d^{p})^{1/p},
$$
we obtain the convexity of $f_{p}$:
\begin{eqnarray*}
f_{p}( \alpha s + \beta t)
&=&
\left( \| x + (\alpha s + \beta t) y) \|^{p} + \| x - (\alpha s +
\beta t) y) \|^{p} \right)^{1/p}\\
&=&
\left( \| \alpha (x + s y) + \beta (x + t y) \|^{p} + \| \alpha (x - s
y) + \beta (x - t y) \|^{p} \right)^{1/p}\\
&\leq&
\left( [ \alpha \| x + s y\| + \beta \|x + t y \|]^{p} + [ \alpha \|x - s
y\| + \beta \|x - t y \|]^{p} \right)^{1/p}\\
&\leq&
\left( [\alpha \| x + s y\|]^{p} + [\alpha \|x - s
y\|]^{p} \right)^{1/p} \\
&+&
\left( [\alpha \| x + t y\|]^{p} + [\alpha \|x -
t y\|]^{p} \right)^{1/p} = \alpha f_{p}(s) + \beta f_{p}(t).
\end{eqnarray*}
Moreover, by the triangle inequality and the concavity of $u^{1/p}$, we get, for
$t \in {\Bbb R}$, that
\begin{eqnarray*}
f_{p}(0)
&=&
2^{1/p} \|x\| \leq 2^{1/p} \frac{ \| x + t y \| + \| x - t y \|}{2}\\
&\leq&
2^{1/p} ( \frac{\| x + t y \|^{p} + \| x - t y \|^{p}}{2} )^{1/p} =
f_{p}(t).
\end{eqnarray*}
In particular, if $0 \leq s < t$, then
\begin{eqnarray*}
f_{p}(s)
&=&
f_{p} \left (\frac{s}{t} t + (1-\frac{s}{t}) 0 \right ) \leq
\frac{s}{t} f_{p}(t) + (1-\frac{s}{t}) f_{p}(0) \\
&\leq &
\frac{s}{t} f_{p}(t) + (1-\frac{s}{t}) f_{p}(t) = f_{p}(t).
\end{eqnarray*}
Note also that $|f_{p}(s) - f_{p}(t)| \leq 2^{1/p} |s - t| \|y\|$ for
all $s, t \in {\Bbb R}$.
\end{proof}

\medskip
We are now ready to prove Theorem \ref{t4}.
\begin{proof} We need to show the triangle inequality for
$\|\cdot\|_{u, p}$. For any $x, y \in X$, by using the monotonicity from
Lemma \ref{l1}, twice the triangle inequality of the norm $\|\cdot\|$ and the
Minkowski inequality for the two-dimensional $l^{p}_{2}$-norm, we obtain that
\begin{eqnarray*}
\| x + y \|_{u, p}
&=&
\left( \| x + y + \|x + y\| u \|^{p} + \| x + y - \|x + y \| u \|^{p}
\right)^{1/p}\\
&\leq&
\left( \| x + y + (\|x\| + \|y\|) u \|^{p} + \| x + y - (\|x\| +
\|y\| ) u \|^{p} \right)^{1/p}\\
&=&
\left( \| (x + \|x\| u) + (y + \|y\| u) \|^{p} + \| (x - \|x\| u) +
(y - \|y\| u) \|^{p} \right)^{1/p}\\
&\leq&
\left( [\| x + \|x\| u\| + \| y + \|y\| u\|]^{p} + [\| x - \|x\| u\| +
\|y - \|y\| u \|]^{p} \right)^{1/p}\\
&\leq&
\left( \| x + \|x\| u \|^{p} + \| x - \|x\| u \|^{p} \right)^{1/p}\\
&+&
\left( \| y + \|y\| u \|^{p} + \|y - \|y\| u \|^{p} \right)^{1/p} =
\| x \|_{u, p} + \| y \|_{u, p}.
\end{eqnarray*}
Moreover, by the convexity of $u^{p}$ and the triangle inequality,
$$
\| x \|_{u, p} \geq 2^{1/p-1} (\| x + \|x\| u \| + \| x - \|x\| u \|)
\geq 2^{1/p} \| x\|,
$$
and also, by the triangle inequality,
$$
\| x \|_{u, p} \leq 2^{1/p} (\|x\| + \|x\| \|u\|).
$$
Thus
$$
2^{1/p} \| x\| \leq \| x \|_{u, p} \leq 2^{1/p} (1 + \|u\|) \|x\|.
$$
\end{proof}


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\end{document}


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