text:
\begin{document}

\begin{frontmatter}

\title{Endpoints of  multi-valued weak  contractions on the  metric space valued in partially ordered groups}

\author{Congdian Cheng}

\address{College of Mathematics and Systems
Science, Shenyang Normal University,
 Shenyang,  110034,  P.R. China
{\bf Tel:}024-86170136; {\bf Fax:} 024-86171095 {\bf E-mail:}
 zhiyang918@163.com}

\begin{abstract}

We introduce the  metric space valued in partially ordered groups,

and define the convergence of sequences

 and
the multi-valued weak contractions, etc. , on the space. We then
establish  endpoint theorems for the defined maps. Our contributions
extend the theory of cone metric space constructed by Huang and
Zhang (2007) and some recent results on  the fixed point and
 endpoint theory, such as  the endpoint theorem given by Amini-Harandi
(2010).\end{abstract}

\begin{keyword}Fixed point\sep endpoint\sep metric space valued in partially ordered group
\sep topological structure\sep weak
 contraction

\MSC 47H10 \sep  54C60

\end{keyword}

\end{frontmatter}

\section{Introduction}
\label{1}Let $(X, d)$ be a complete metric space.  Denote by $CB(X)$
the class of all nonempty closed and bounded subsets of $X$.  Denote
by $H(A,B)$ the Hausdorff metric of $A$ and $B$ with respect to $d$,
that is,
$$H(A,B)=\max\{\sup \limits_{x\in A}d(x,B), \sup \limits_{y\in B}d(y,A)\},$$
for all $A,B\in CB(X)$, where $d(x,B)=\inf\limits_{y\in B}d(x,y)$.
Further let $T : X\rightarrow 2^X$  be a multi-valued / set-valued
map. A point $x$ is called a fixed point of $T$ if $x\in Tx$. Define
$Fix(T) = \{x\in X : x \in Tx\}$.

A point $x$ is called an endpoint / a stationary point of a
multi-valued map $T$ if $Tx = \{x\}$. We denote the set of all
endpoints of $T$ by $End(T)$.

The investigation of endpoint of multi-valued mappings is an
important extending  of the study of  fixed point, which

was made as early as 30 years ago, and has received great attention
in recent years, see e.g. [1-2] and the references therein.

In particular, Amini-Harandi [1]

 (2010)  proved the Theorem 1.1 below.\\

 \noindent{\bf
Theorem 1.1} ([1, Theorem 2.1]).\textit{ Let $(X, d)$ be a complete
metric space and $T: X\rightarrow CB(X)$ be a set-valued map that
satisfies
$$H(Tx, Ty)
\leq \psi(d(x, y)),\eqno (1.1)$$ for each $x, y \in X$ , where $\psi
: [0,+\infty) \rightarrow [0,+\infty)$ is upper semicontinuous
(u.s.c.),  $\psi(t) < t$ for each $t
> 0$  and  satisfies $\liminf \limits_{
t\rightarrow +\infty }(t - \psi(t)) > 0$. Then $T$ has a unique
endpoint if and only if T has the approximate endpoint property. (
i.e. $\inf\limits_{x\in X}\sup\limits_{y\in Tx}d(x,y)=0$.
)}\\

Huang and Zhang [3]

(2007) introduced the concept of cone metric space, and  established
some fixed point theorems for contractive type maps in a normal cone
metric space. Subsequently, some other authors gave many results
about the fixed point theory in cone metric spaces. For example,
Rezapour and Hamlbarani [4] (2008)
 generalized some results of [3].

  Raja and Vaezpour [5]
(2008) presented some extensions of Banach$'$s Contraction Principle
in complete cone metric spaces. Aage and Salunke [6] (2011) proved
some fixed point theorems for the expansion onto mappings on
complete cone metric spaces. Also, many common fixed point theorems
were proved for maps on cone metric spaces in some literatures, for
example, see Ili\'{c} and Rako\v{c}evi\'{c} [7] (2008); Arshad, Azam
and Vetro [8] (2009), whose results generalized and unified many
fixed point theorems. Rezapour and Haghi [9] (2009), as well as
Haghi and Rezapour [10] (2010) studied fixed points of
multifunctions (i.e. multi-valued mappings) on normal cone metric
spaces and on regular cone metric spaces, respectively. Moreover,
Wardowski [11] (2009) introduced a kind of set-valued contractions
in cone metric spaces and established endpoint and fixed point
theorems for his contractions.

 In addition, Rezapour and Haghi [9]
(2009) introduced the concept of cone topology on cone metric space.
Lakshmikantham and \'{C}iri\'{c} [12] (2009) introduced the concept
of a mixed g-monotone mapping and prove coupled coincidence and
coupled common fixed point theorems for such nonlinear contractive
mappings in partially ordered complete metric spaces. Harjani and
Sadarangani [13] (2009) present some fixed point theorems for weakly
contractive maps in a complete metric space endowed with a partial
order. And Zhang [14] (2010)  proved some new fixed point and
coupled fixed point theorems for multivalued monotone mappings in
ordered metric spaces. Finally, Amini-Harandi [15] (2011) studied
fixed point theorems for a kind of generalized quasicontraction maps
in so called the vector modular spaces.

Motivated by the contributions stated above, the present work
introduces the  metric space valued in a partially ordered group
endowed with a topological structure and the metric space valued in
a partially ordered module endowed with a topological structure, and
establishes some fundamental concepts of analysis on the introduced
spaces, such as the convergence of sequences, which extends the
theory of cone metric space.  It also defines multi-valued weak
contractions, etc., on the introduced spaces. And then it focus on
addressing the endpoint theory of the multi-valued weak
contractions.

\section{Preliminaries}
\label{1} This section provides necessary preliminaries for our
 discussions.

We first make the following explanations. For a partial order
$\preceq$ of a set, we write $a\prec b$ to indicate that $a\preceq
b$ but $a\neq b$, where $a$ and $b$ are elements of the set. And for
a group $G$ with partial order $\preceq$, we write $G_+$ and $G^+$
to  indicate respectively the sets $\{a\in G:a\succeq \theta\}$ and
$\{a\in G:a\succ\theta\}$,
where $\theta$ indicates the identity element of $G$.\\

\noindent{\bf Definition 2.1.} Let $G$ be an abelian/a commutative
group with partial order $\preceq$. We call $G$  a
$\preceq$-partially ordered group, a partially ordered group for
simplicity,
 if $\prec$ satisfies the  law (g1)
$a\prec b \Rightarrow a+c\prec b+c, \forall a,b,c\in G$. Let further
$G$ be an $R$-module and the integral ring $R$ be a $\leq$-partially
ordered group.  Assume that the partial order $<$

satisfies  the  law (r1): $1>0$, where $1$ and $0$ are the unit
element and the identity element of $R$, respectively.

 Assume also that the partial orders $<$ and $\prec$ satisfy  the  law
 (m1): $a\prec b \Rightarrow ra\prec rb, \forall a,b\in G$ and
$\forall r\in R^+$ (i.e. $r>0$).

Then we call $G$  an $(R,\leq,\preceq)$-partially ordered module, a
partially ordered module for
simplicity.\\

\noindent{\bf Remark 2.2.} (i)For convenience, we focus our
attention to study under the assumption that there exist
non-identity elements in group $G$ below. (ii) Note that each
element of a group has an inverse element. From (g1),

we can easily obtain  the order relation: (g1)$'$ $a\preceq b
\Leftrightarrow a+c\preceq b+c, \forall a,b,c\in G$.

  In
addition, from (m1), we can easily obtain  the order relation:
(m1)$'$ $a\preceq b \Rightarrow ra\preceq rb, \forall a,b\in G$ and
$\forall r\in R_+$. (iii) From (m1), we can also obtain the order
relations: (m2)  $r< s\Rightarrow ra\prec sa, \forall r,s\in R$ and
$\forall a\in G^+$; and (m2)$'$  $r\leq s\Rightarrow ra\preceq sa,
\forall r,s\in R$ and $\forall a\in G_+$. In fact, let $r< s$, then,
by (g1), we have $0< s-r$. Let also $a\in G^+$, i.e. $\theta\prec
a$. Then, by (m1), we have $(s-r)\theta\prec (s-r)a\Rightarrow
\theta\prec (s-r)a$. From (g1), this leads to $ra\prec ra+(s-r)a=
sa$. So we have (m2). Finally, from (m2), it is obvious that we have
(m2)$'$.

 (iv) It is
obvious that the partially ordered
module is a special kind of the partially ordered group.\\

\noindent{\bf Example 2.3.} Let $E$ be a Banach space over the  real
field $\mathbb{R}$ and $P$ be a subset of $E$. $P$ is called a cone
if and only if: (i) $P$ is closed, nonempty, and $P\neq\{\theta\}$;
(ii) $ r, s \in{\mathbb{R}_+}, a,b \in P\Rightarrow ra + sb \in P $;
(iii) $ a \in P$ and $-a \in P \Rightarrow a = \theta$. Here
${\mathbb{R}}_+$ denotes all the non-negative real numbers. For a
given cone $P$ of $E$, define the partial order $\preceq$ on $E$ by
$x\preceq y $ if and only if $y-x\in P$, see [3]. Then it can be
easily verified that $E$ is an $(\mathbb{R},\leq,\preceq)$-partially
ordered module, and therefore, of course, is a $\preceq$-partially
ordered group. Here $\leq$ is the usual order of $\mathbb{R}$.\\

In the following part of this section $G$ is supposed either of a
$\preceq$-partially ordered group and

an $(R,\leq,\preceq)$-partially ordered module unless
otherwise specified.\\

\noindent{\bf Definition 2.4.}   Let $\ll$  be a non-empty relation
of $G$.

$\ll$ is called an analytic  topological structure of partially
ordered group $G$ if it satisfies:

 (t1) $a\ll b\Rightarrow a\prec b, \forall a, b
\in G$; (t2) $a\preceq b, b\ll c\Rightarrow a\ll c$; (t3)  $a\ll b
\Rightarrow a+c\ll b+c, \forall a,b,c\in G$; (t4) $\theta \preceq
a\ll \varepsilon, \forall \varepsilon\gg \theta \Rightarrow
a=\theta$; and (t5) $\forall \varepsilon\gg\theta$, there exists
$\eta\gg\theta$ such that $\eta\ll\varepsilon$. $\ll$ is called an
analytic topological structure of partially ordered module $G$ if it
also satisfies: (t6) $a\ll b \Rightarrow ra\ll rb, \forall
a,b\in G$ and $\forall r\in R^+$.\\

\noindent{\bf Remark 2.5.}

In the definition above, for $\ll$  is non-empty,  there are
actually infinite elements $\varepsilon$ such that
$\varepsilon\gg\theta$ in $G$. In fact, since $\ll$  is a non-empty
relation, there exist at least two elements $a$ and $b$ such that
$a\ll b$. By (t3),  we have $\theta\ll b-a$. Thus,
according to (t5), the result holds.  \\

\noindent{\bf Example 2.6.} For the partially ordered module $E$ of
Example 2.3, define the relation $\ll$ by $x\ll y$ if and only if
$y-x\in intP$, where $intP$ denotes the interior of $P$,

see [3] and [4].

Then we can  verify that $\ll$ is an analytic topological structure
of $E$. In fact, it is obvious that $\ll$ satisfies (t1), (t3), (t5)
and (t6).  To prove (t2), let $a\preceq b$ and $b\ll c$. Then, from
$b\ll c$, we have $\theta\ll (c-b)$. So there is an $r\in
  {\mathbb{R}}^+$ such that $N((c-b), r)\subset P$, where $N((c-b), r)
  =\{x\in E: \|x-(c-b)\|<r\}$ and $\|x\|$ indicates the norm of $x$. Consider $N((c-a),
  r)$.
Let $u\in N((c-a),
  r)$. Then $\|u-(b-a)-(c-b)\|=\|u-(c-a)\|<r$. This implies
  $u-(b-a)\in N((c-b), r)\subset
  P$. On the other hand, $(b-a)\in P$ for $a\preceq b$. So
  $u=u-(b-a)+(b-a)\in P$. Namely  $N((c-a),
  r)\subset P$. Hence $\theta\ll c-a$, e.g. $a\ll c$, that is, (t2)
holds. To prove (t4), assume $\theta \preceq a\ll \varepsilon,
\forall \varepsilon\gg \theta$. Let $c\gg\theta$. Then
$\frac{c}{n}\gg\theta$. By regarding $\frac{c}{n}$ as $\varepsilon$,
we have  $\frac{c}{n}\gg a\Rightarrow\frac{c}{n}-a\gg \theta
\Rightarrow\frac{c}{n}- a \in P$ for all $n\in \mathbb{N}$, where
$\mathbb{N}$ represents all the natural numbers. This leads to $- a
\in P$ because $\frac{c}{n}\rightarrow\theta$ (in norm) and $P$ is
closed. So, by $ a \in P$, we have $a=\theta$. That is, (t4) holds.
Therefore, $\ll$ is an analytic topological structure of $E$.\\

\noindent{\bf Definition 2.7.} Let $\ll$ be an analytic topological
structure of $G$ and $a\in G_{+}$. A sequence $\{a_{n}\}$ of $G_{+}$
is said to be convergent to $a$ (in $\ll$) if $\forall
\varepsilon\gg \theta$, there is a natural number $N$ such that
$\theta\preceq a_{n}-a\ll \varepsilon$ for all $n>N$, denoted
by $a_{n}\rightarrow a$ or $\lim\limits_{n\rightarrow\infty}a_n = a$.\\

\noindent{\bf Remark 2.8.} (i) Let $\ll$ be an analytic topological
structure of $G$, which is different from the $\prec$. Suppose
$\prec$ is also an analytic topological structure of $G$, and

sequence $\{a_{n}\}$ of $G_{+}$ converges to $\theta$ in $\prec$.

 Then we can easily know that $a_{n}$ converges to $\theta$ in $\ll$.

(In fact, let $\varepsilon\gg\theta$. Then, from (t5), there exists
$\eta\gg\theta$ such that $\eta\ll\varepsilon$. For the $\eta$,
since $a_{n}\rightarrow \theta$ in $\prec$, there is a natural
number $N$ such that $a_{n}\prec\eta$ for all $n>N$. By (t2) and
$\eta\ll\varepsilon$, this leads to $a_{n}\ll\varepsilon$ for all
$n>N$. Hence $a_{n}\rightarrow \theta$ in $\ll$.) That is, the
convergence in $\prec$ is stronger than in $\ll$. So, in the case,
the convergence in $\ll$ can be regarded as a kind of weak
convergence. (ii) For the analytic topological structure $\ll$ of
the partially ordered module $E$ in Example 2.6, it can be easily
verified that $\ll$ is different from the $\prec$ if $E$ is a
two-dimensional Euclidean space and $P=\{(x,y): x\geq 0, y\geq 0\}$.
(iii) It can be easily verified that for an analytic topological
structure $\ll$ of $G$, $a_{n}\rightarrow a$ $\Leftrightarrow$
$\forall b\gg a\succeq\theta$, there is a natural number $N$ such
that $a\preceq a_{n}\ll b$ for all $n>N$. In fact, assume
$a_{n}\rightarrow a$. $\forall b\gg a$, let $\varepsilon= b-a$.
Then, by (t3), we have $\varepsilon\gg\theta$. So, there is a
natural number $N$ such that $\theta\preceq a_{n}-a\ll \varepsilon$
for all $n>N$. From (g1)$'$ and (t3), this leads to $a\preceq
a_{n}\ll b$ for all $n>N$. Conversely,
$\forall\varepsilon\gg\theta$, let $ b=a+\varepsilon$. Then
 $ b\gg a\succeq\theta$. So, there is a natural number $N$ such that
$a\preceq a_{n}\ll b\Rightarrow\theta\preceq a_{n}-a\ll \varepsilon$
for all $n>N$. Note that $a\succeq\theta$.
This shows  $a_{n}\rightarrow a$.\\

\noindent{\bf Remark 2.9.} For  the $E$ and the  analytic
topological structure $\ll$ of Example 2.6, let $\{a_{n}\}$ be a
sequence in $E_+$. Assume
 $a_{n}\rightarrow \theta$ in norm. Then,
$\forall\varepsilon\gg\theta$, there exists $r\in \mathbb{R}^+$ such
that $N(\varepsilon,r)\subset P$. Due to  $a_{n}\rightarrow \theta$
in norm, there exists also a natural number $N$ such that
$\|a_{n}\|<r$ for all $n>N$. Therefore,
$\|(\varepsilon-a_{n})-\varepsilon\|<r\Rightarrow(\varepsilon-a_{n})\in
N(\varepsilon,r)\Rightarrow (\varepsilon-a_{n})\in intP$, that is,
$a_{n}\ll \varepsilon$, for all $n>N$. This implies that
$a_{n}\rightarrow \theta$ in $\ll$ if $a_{n}\rightarrow \theta$ in
norm.\\

$G$ always associates with an analytic topological structure $\ll$
and the  convergence of the sequences of $G_{+}$
  is in $\ll$ are assumed below.\\

\noindent{\bf Lemma 2.10.} \textit{Let $\{a_{n}\}$ and $\{b_{n}\}$
be two sequences of $G_{+}$. We have the three conclusions as
follows. (i) If $a_{n}\rightarrow\theta$, then
$\lim\limits_{n\rightarrow\infty}a_n $ is unique. (ii) If
$a_{n}\rightarrow\theta$ and $b_{n}\rightarrow\theta$, then
$a_{n}+b_{n}\rightarrow\theta$. (iii) If $b_{n}\succeq a_{n}
\succeq a\succeq\theta$ for all $n\in {\mathbb{N}}$ and
$b_{n}\rightarrow a$, then $(b_{n}-a_{n})\rightarrow\theta$.}\\

\noindent{\bf Proof}. Proving (i). Let
$\lim\limits_{n\rightarrow\infty}a_n =a$. Then there is a natural
number $N_{1}$ such that $a_{n}\succeq a$ for all $n>N_{1}$. On the
other hand, $\forall\varepsilon\gg\theta$, since
$a_{n}\rightarrow\theta$, there is a natural number $N_{2}$ such
that $\varepsilon\gg a_{n}\succeq\theta$ for all $n>N_{2}$. Let
$n=\max\{N_{1}, N_{2}\}+1$. Then, $\varepsilon\gg a_n$ and $a_n
\succeq a\succeq\theta$. From (t2), this leads to   $\varepsilon \gg
a\succeq\theta$.

 By virture of (t4), we
have $a=\theta$. Hence (i) holds.

Proving (ii).  Let $\varepsilon\gg\theta$. By (t5), there exists
$\eta\gg\theta$ such that $\varepsilon-\eta\gg\theta$. For
$a_{n}\rightarrow\theta$ and $b_{n}\rightarrow\theta$, there are
natural numbers $N_{1}$ and $N_{2}$ such that $a_n\ll \eta, \forall
n>N_{1}$ and $b_n\ll\varepsilon-\eta, \forall n>N_{2}$. Put
$N=\max\{N_{1},N_{2}\}$. We have: $a_n+b_n\ll
\eta+\varepsilon-\eta=\varepsilon, \forall n>N$. Hence,
$a_{n}+b_{n}\rightarrow\theta$. That is (ii) holds.

Proving (iii). Arguing by contradiction, assume
$\lim\limits_{n\rightarrow\infty}(b_{n}-a_{n})\neq\theta$.  Then
there exists a $\delta\gg\theta$ and a subsequence
$\{b_{n_i}-a_{n_i}\}$ such that $(b_{n_i}-a_{n_i})\ll\delta$ does
not hold for all $i\in {\mathbb{N}}$. From $a_{n_i}\succeq a$, by
(g1)$'$, we  have $b_{n_i}-a_{n_i}\preceq b_{n_i}-a$. This implies
that
 $b_{n_i}-a\ll\delta$ does not hold for all $i\in {\mathbb{N}}$. (In
 fact, if for some $i\in {\mathbb{N}}$,
 $b_{n_i}-a\ll\delta$, then, from  (t2) and  $b_{n_i}-a_{n_i}\preceq b_{n_i}-a$,
 we have $b_{n_i}-a_{n_i}\ll\delta$,
which contradicts that $(b_{n_i}-a_{n_i})\ll\delta$ does not hold.)
Hence $\lim\limits_{n\rightarrow\infty}b_{n}\neq a$. The
contradiction shows (iii) holds.

$\square$\\

\noindent{\bf Definition 2.11.}  $G$ is called regular if   every
decreasing sequence $\{a_n\}$ of $G_+$ is convergent. That is, if a
sequence $\{a_n\}$ of $G_+$  satisfies $a_n\preceq a_{n+1}$ for all
$n\in {\mathbb{N}}$, then

 exists a $a\in G_+$ such that

$a_n$ converges to $a$.\\

\noindent{\bf Remark 2.12.} (i)

Let $A\subseteq G_+, A\neq\emptyset$ and $a\in G_+$.  $a$ is called
the infimum of $A$ if and only if $a$ is a lower bound of $A$,  and
$c\preceq a$ for each lower bound $c$ of $A$, denoted by $inf A$.
(ii) It is obvious that there is

at most one infimum for each subset of $G_+$. In fact, for any
$A\subseteq G_+$, let $a$ and $b$ be two infimums of $A$. Then both
$a\preceq b$ and $b\preceq a$ hold. Hence $a=b$. This shows that $A$
has at most one infimum. (iii) In particular, $G$ is regular if for
each non-empty subset $A$ of $G_+$,
 $inf A$ exists and there exists a sequence $\{a_n\}$ of $A$ such that
$a_n$ converges to $inf A$. Actually, let $\{a_n\}$ be a decreasing sequence
 of $G_+$. Then, in the case,

   $inf \{a_n\}$ exists and there exists a subsequence $\{a_{n_i}\}$ of
  $\{a_n\}$ such that $\{a_{n_i}\}$ converges to $inf \{a_n\}$. Since $\{a_{n_i}\}$ converges to $inf \{a_n\}$,
  $\forall \varepsilon\gg\theta$, there is a natural number $I$
   such that $\varepsilon\gg a_{n_i}-inf \{a_n\}\succeq\theta$ for all $i\geq I$.
  For $\{a_n\}$ decreasing, this leads to
  $\varepsilon\gg a_{n}-inf \{a_n\}\succeq\theta$
 for all $n\geq n_I$. That is,  $\{a_{n}\}$ converges
 to $inf \{a_n\}$. Hence $G$ is regular.\\

\noindent{\bf Definition 2.13.} Let X be a non-empty
set. Suppose the mapping $d : X \times X \rightarrow G$ satisfies\\
(d1) $d(x, y)\succeq\theta$ for all $x, y \in X$ and $d(x, y) =
\theta$ if and
only if $x = y$,\\ (d2) $d(x, y) =d(y, x)$ for all $x, y \in X$,\\
(d3) $d(x, y)\preceq d(x, z)+d(z, y)$ for all $x, y, z \in X$.\\
Then d is called a metric ( on $X$) valued in partially ordered
group $G$, and $(X,d)$ is called a metric space valued in partially
ordered group $G$, when $G$ is a partially ordered group; a metric
of group and a metric space of group for simplicity, respectively.
(Then d is called a metric valued in partially ordered module $G$,
and $(X,d)$ is called a metric space valued in partially ordered
module $G$, when $G$ is a
partially ordered module.)\\

In the rest of this section we always assume that $(X,d)$ is either
of a metric space valued in partially ordered group $G$ and a metric
space
valued in partially ordered module $G$.\\

\noindent{\bf Definition 2.14.} For given $(X,d)$, let $x \in X$ and
$\{x_n\}$ be a sequence in $X$.
\\(i) We call that $\{x_n\}$ converges to $x$ if and only if
 $d(x_n,x)\rightarrow \theta$, denoted by
$\lim\limits_{n\rightarrow\infty}x_n = x$ or $x_n\rightarrow x$.
\\(ii) $\{x_n\}$ is a Cauchy sequence if and only if $d(x_n,
x_m)\rightarrow \theta$, that is, $\forall \varepsilon \gg\theta$,
 there is a natural number $N$ such that $d(x_n,
x_m)\ll\varepsilon$ for all $n,m \geq N$. \\(iii) $(X,d)$ is
complete
 if and only if every Cauchy
sequence is convergent. \\(iv) $(X,d)$ is regular
 if and only if $G$ is regular.\\

\noindent{\bf Remark 2.15.}

The relation between the regular space and the
complete space is an  interesting question for further research.\\

\noindent{\bf Definition 2.16.} Given $(X,d)$, let $T : X
\rightarrow (2^X-\emptyset)$ be
 a multi-valued mapping and $\varphi: X
\times X \rightarrow G_+$ be a  mapping with $\varphi(x,y)\prec
d(x,y)$ for all $d(x,y)\succ\theta$. $T$ is called  a multi-valued
($\varphi$-)weak contraction on $(X,d)$ if, for all different
$x,y\in X$, $\forall x'\in Tx$, there exists $\bar{y}\in Ty$ such
that
$$d(x',\bar{y})\preceq \varphi (x,y).\eqno (2.1)$$
$T$ is called a global multi-valued ($\varphi$-)weak contraction on
$(X,d)$ if, for all different $x,y\in X$,  we have
$$d(x',y')\preceq \varphi(x,y), \forall x'\in Tx, \forall y'\in
Ty.\eqno (2.2)$$ The weak contraction $T$ is called to satisfy
C-condition (convergence condition) if $d(x_n, y_n) - \varphi(x_n,
y_n)\rightarrow\theta$,

then $d(x_n, y_n)\rightarrow\theta$, where $x_n$ and $y_n$ are two sequences of $X$.\\

\noindent{\bf Remark 2.17.}  (i) It is obvious that a global
multi-valued weak contraction is a  multi-valued weak contraction.
(ii) The weak contraction $T$ is called to satisfy C$'$-condition if
$d(x_n, x_m) - \varphi(x_n, x_m)\rightarrow\theta (n\neq m)$,
 that is, $\forall \varepsilon\gg\theta$,
there is a $N$ such that $d(x_n, x_m)-\varphi(x_n,
x_m)\ll\varepsilon$ for all $n,m>N$ and $n\neq m$, then $d(x_n,
x_m)\rightarrow\theta$. (iii) If $T$ satisfies C-condition, then it
also satisfies C$'$-condition. In fact, for the set $\{(n,m): n,
m\in \mathbb{N}, n\neq m\}$ is countable, it can be rewritten as the
sequence $\{(x'_{i}, y'_{i})\}$. Assume $d(x_n, x_m) - \varphi(x_n,
x_m)\rightarrow\theta (n\neq m)$. Let $\varepsilon\gg\theta$. Then
there exists a natural number $N$ such that $d(x_{n},
x_{m})-\varphi(x_{n}, x_{m})\ll\varepsilon$ whenever $n,m>N$ and
$n\neq m$. Because the set $\{(n,m): n, m\leq N\}$ is finite, there
is a natural number $I$ such that if $i>I$ and $(x'_{i},
y'_{i})=(x_{n}, x_{m})$, then $n, m> N$. This implies $\forall i>I$,
we have $d(x'_{i}, y'_{i})-\varphi(x'_{i}, y'_{i})\ll\varepsilon$.
Hence  $d(x'_{i}, y'_{i})-\varphi(x'_{i}, y'_{i})\rightarrow\theta$.
For $T$ satisfies C-condition, we have $d(x'_{i},
y'_{i})\rightarrow\theta$. Further, due to $d(x'_{i},
y'_{i})\rightarrow\theta$, $\forall \varepsilon\gg\theta$, there is
 a natural number $I'$ such that $d(x'_{i},
y'_{i})\ll\varepsilon$ whenever $i>I'$. Since the set $\{i: i\leq
I'\}$ is finite, there is a natural number $N'$ such that if
$n,m>N'$, $n\neq m$ and $(x_{n}, x_{m})=(x'_{i}, y'_{i})$, then
$i>I'$. That is,  $\forall n,m>N'$, $n\neq m$, we have $d(x_n,
x_m)\ll\varepsilon$. Note that $d(x_n, x_m)=\theta$ when $n= m$.
This leads to $d(x_n,
x_m)\rightarrow\theta$.\\

\noindent{\bf Definition 2.18.} A map $T : X \rightarrow
(2^X-\emptyset)$ on $(X,d)$ is said to have approximate endpoint
property if there exist a sequence $\{x_n\}$ of $X$ and a sequence
$\{a_n\}$ of $G_+$ with $a_n\rightarrow\theta$ such that
$$d(x_n,x'_n)\preceq a_n, \forall x'_n\in Tx_n, \eqno (2.3)$$
for all $n\in {\mathbb{N}}$.\\

\noindent{\bf Remark 2.19.} When $(X,d)$ is the usual complete
metric space, it can be easily verified that $T$ has the approximate
endpoint property in Theorem 1.1 and $T$ has the approximate
endpoint property  defined in Definition 2.18 are equivalent. (In
fact, if $T$ has the approximate endpoint property in Theorem 1.1,
that is, $\inf\limits_{x\in X}\sup\limits_{y\in Tx}d(x,y)=0$, then
there is a sequence $\{x_n\}$ of $X$ such that
 $\sup\limits_{y\in Tx_n}d(x_n,y)\rightarrow0$. Let
  $a_n=\sup\limits_{y\in Tx_n}d(x_n,y)$. Then   $d(x_n,x'_n)\leq a_n, \forall x'_n\in Tx_n$ and
   $a_n\rightarrow0$. This shows that $T$ has the approximate
endpoint property  defined in Definition 2.18. On the other hand, if
$T$ has the approximate endpoint property  defined in Definition
2.18, that is, there exist a sequence $\{x_n\}$ of $X$ and a
sequence $\{a_n\}$ of ${\mathbb{R}}_+$ with $a_n\rightarrow 0$ such
that $d(x_n,x'_n)\leq a_n, \forall x'_n\in Tx_n$, then
$\sup\limits_{y\in Tx_n}d(x_n,y)\leq a_n$
 for all $n\in {\mathbb{N}}$. This implies that $\inf\limits_{x\in X}\sup\limits_{y\in Tx}d(x,y)=0$, namely,
 $T$ has the approximate endpoint property in Theorem 1.1. Hence we have the equivalence stated above.) \\

 \noindent{\bf Lemma 2.20.} \textit{
 Let $T$ be a multi-valued weak
  contraction on $(X,d)$. Then we have the following two conclusions.
  (i) $T$ has approximate
 endpoint property if $T$ has endpoints. (ii) $T$ has one endpoint at most.
 (i.e. $|End(T)|\leq 1$. Here $|End(T)|$
denotes the cardinal number of $End(T)$.)}\\

\noindent{\bf Proof}.  (i) is obvious. In fact, let $x$ be an
endpoint of $T$. Put $x_n=x$ and $a_n=\theta$ for all $n\in
{\mathbb{N}}$. Then $a_n\rightarrow\theta$ and (2.3) holds for all
$n\in {\mathbb{N}}$. Hence $T$ has the approximate endpoint
property. To prove (ii), assume $|End(T)|>1$. Then, there exist
$x,y\in End(T)$ such that $x\neq y$. From (2.1), we have
$d(x,y)\preceq\varphi(x,y).$

Note that $\varphi(x,y)\prec d(x,y)$ for any $d(x,y)\succ\theta$.
This implies  $d(x,y)=\theta$. Hence, from (d1), we have $x=y$. This
contradicts
$x\neq y$. So $|End(T)|\leq 1$, that is, (ii) holds. $\square$\\

\section{Main results}

In this section, we always assume that $(X,d)$ is a metric space
valued in partially ordered group $G$.

Now we are ready to  prove our main results. We first present the
following Theorem 3.1,  which extends Theorem 1.1 (Amini-Harandi [1,
Theorem 2.1]) to the
case of the metric space of group.\\

\noindent{\bf Theorem 3.1.} \textit{Let
 $T$ be a multi-valued weak
contraction on complete $(X,d)$ and

satisfy   C-condition. Then $T$ has a unique endpoint if and only if
it has the
approximate endpoint property.}\\

\noindent{\bf Proof.} The necessity is clear from the (i) of Lemma
2.20. Next we prove the sufficiency.

Since $T$ has the approximate endpoint
 property, there exist sequences $\{x_n\}$ of $X$ and $\{a_n\}$ of $G_+$ satisfying
 (2.3) and $a_n\rightarrow \theta$. If there exists a subsequence $\{x_{n_i}\}$
 such that $x_{n_i}$ being the same point $x$ of $X$ for all $i\in{\mathbb{N}}$, then we can easily know
 that $x$ is an endpoint of $T$ from (2.3). (In fact, for any given $x'\in
 Tx$, we have $d(x, x')\preceq a_{n_i}$ for all $i\in {\mathbb{N}}$. Since  $a_n\rightarrow
 \theta$, we have $a_{n_i}\rightarrow
 \theta(i\rightarrow\infty)$. So we have $\theta\preceq d(x, x')\ll \varepsilon$ for all
  $\varepsilon\gg\theta$. This implies $d(x, x')=\theta$ from (t4), that is,  $x=x'$.  Hence $x$ is an endpoint of
 $T$.)

 Otherwise, without loss of generality, we can assume $x_n\neq x_m$ whenever
 $n\neq m$ and continue to prove as follows.

For any different $n,m\in\mathbb{N}$, let $x'_n\in Tx_n$. Then
$$d(x_n,x_m)\preceq d(x_n,x'_n)+d(x'_n,x_m). \eqno(3.1)$$
Since $x_n\neq x_m$, according to  (2.1), there exists $\bar{x}_m\in
Tx_m$ such that
$$d(x'_n,\bar{x}_m)\preceq \varphi(x_n,x_m).$$
Using this and (3.1), we further obtain
$$\begin{array}{rcl}d(x_n,x_m)&\preceq & d(x_n,x'_n)+d(x'_n,\bar{x}_m)+d(\bar{x}_m,x_m)
\\&\preceq& d(x_n,x'_n)+\varphi(x_n,x_m)+d(\bar{x}_m,x_m).\end{array}\eqno(3.2)$$
By (2.3), we have $d(x_n,x'_n)\preceq a_n$ and
$d(\bar{x}_m,x_m)\preceq a_m$. From (3.2), this leads to
$$\begin{array}{rcl}& &d(x_n,x_m)\preceq a_n+\varphi(x_n,x_m)+a_m\\&\Rightarrow&
d(x_n,x_m)-\varphi(x_n,x_m)\preceq a_n+a_m, n\neq m.
\end{array}\eqno(3.3)$$

For $x_n\neq x_m$, we have $d(x_n,x_m)\succ\theta$. So,
$d(x_n,x_m)-\varphi(x_n,x_m)\succeq\theta$. On the other hand,
noting that $a_n\rightarrow\theta$, following the proof on the (ii)
of Lemma 2.10, we can easily know that $a_n+a_m\rightarrow \theta$.
Thus we can obtain $d(x_n,x_m)-\varphi(x_n,x_m)\rightarrow \theta
(x_n\neq x_m)$ from (3.3). This implies $d(x_n,x_m)\rightarrow
\theta$ for $T$ satisfies the C-condition, which leads to $\varphi$
satisfies the C$'$-condition, see the (ii) and (iii) of Remark 2.17.

Hence, $\{x_n\}$ is a Cauchy sequence. Since $(X, d)$ is  complete,
there is a $x\in X$ such that $x_n\rightarrow x$.

We show $x$ is an endpoint of $T$ below.

Since $x_n\neq x_m$ whenever $n\neq m$, without loss of generality,
we can assume $x_n\neq x$ for any $n\in{\mathbb{N}}$. Let $x'\in
Tx$. Then, for all $n\in{\mathbb{N}}$, we have
$$d(x',x)\preceq d(x',x_n)+d(x_n, x). \eqno(3.4)$$
For $x_n\neq x$, by (2.1), there exists $\bar{x}_n\in Tx_n$ such
that
$$d(x',\bar{x}_n)\preceq \varphi(x,x_n)\prec d(x,x_n). \eqno(3.5)$$
In terms of  (3.4), (3.5)  and  (2.3),  we obtain
$$\begin{array}{rcl}d(x',x)&\preceq& d(x',\bar{x}_n)+d(\bar{x}_n,x_n)+d(x_n,x)\\
&\preceq&\varphi(x,x_n)+a_n+d(x, x_n)\\&\preceq&d(x, x_n)+a_n+d(x,
x_n).\end{array}$$ Since $d(x, x_n)\rightarrow\theta$ and
$a_n\rightarrow\theta$, by the (ii) of Lemma 2.10, we obtain $d(x,
x_n)+d(x, x_n)+a_n\rightarrow \theta$. So,  $d(x,x')\ll\varepsilon$
for all $\varepsilon\gg\theta$. Note also that
$d(x,x')\succeq\theta$. From (t4), we have $d(x,x')=\theta$. Hence
$x=x'$. That is, $x\in End(T)$.

Finally, the uniqueness is directly obtained from the (ii) of Lemma
2.20. The
proof completes. $\square$\\

\noindent{\bf Remark 3.2.} Here we make a simple explanation for
Theorem 3.1  extending Theorem 1.1. Firstly, it is obvious that for
the usual order $\leq$ of the real field $\mathbb{R}$, $\mathbb{R}$
is a

$\leq$-partially ordered group with analytic topological structure
$>$. Further, due that

$(X,d)$ in Theorem 1.1 is a complete metric space, it is a complete
metric space of the group $\mathbb{R}$ with analytic topological
structure $>$.
 That is, $(X,d)$ satisfies
the requirement of Theorem 3.1. Secondly, for the $\psi(t)$ in
Theorem 1.1, let $\varphi(x,y)=\psi(d(x,y))$, then $\varphi(x,y)$ is
a mapping from $X \times X$ to $\mathbb{R}_+$ and
$\varphi(x,y)<d(x,y)$ for all $d(x,y)>\theta$.
For the mapping $T$
in Theorem 1.1 and the $\varphi$ defined above, we have $H(Tx,
Ty)\leq\varphi(x,y)$ for all different $x, y\in X$, that is,

$$\max\{\sup \limits_{x'\in Tx}d(x',Ty), \sup \limits_{y'\in
Ty}d(Tx,y')\}\leq\varphi(x,y).$$ This leads to $\sup \limits_{x'\in
Tx}d(x',Ty)\leq\varphi(x,y)\Rightarrow d(x',Ty)\leq \varphi(x,y),
\forall x'\in Tx$. Thus, for $Ty$ is  closed and bounded, $\forall
x'\in Tx$, there is a $\overline{y}\in Ty$ such that
$d(x',\overline{y})\leq \varphi(x,y)$.

This shows that

$T$  is  a multi-valued weak contraction on the space $(X,d)$.
Thirdly, if $d(x_n, y_n)-\varphi(x_n, y_n)\rightarrow0$, then
$d(x_n, y_n)\rightarrow 0$, that is, $T$ satisfies the C-condition.
In fact, if $d(x_n, y_n)$ does not converge to $0$, then there exist
a $\delta>0$ and a subsequence $\{d(x_{n_i}, y_{n_i})\}$ such that
$d(x_{n_i}, y_{n_i})-\psi(d(x_{n_i}, y_{n_i}))>\delta$ for all
$i\in{\mathbb{N}}$. $\lceil$ We show the fact is true as follows.
Let $d_n=d(x_n, y_n)$ and $d_n$ do not converge to $0$. If $\{d_n\}$
is unbounded, without loss of generality, we can assume that
$\{d_n\}$ increases and converges to $+\infty$. For $\liminf
\limits_{ t\rightarrow +\infty }(t - \psi(t))=\lim \limits_{
t\rightarrow +\infty }[\inf\{(s - \psi(s)): s>t\}]> 0$ and
$\inf\{(d_k - \psi(d_k)): k>n\}\geq\inf\{(s - \psi(s)): s>d_n\}$, we
have $\liminf \limits_{ n\rightarrow \infty }(d_n - \psi(d_n))=\lim
\limits_{ n\rightarrow \infty }[\inf\{(d_k - \psi(d_k)):
k>n\}]\geq\lim \limits_{ n\rightarrow \infty }[\inf\{(s - \psi(s)):
s>d_n\}]=\lim \limits_{ t\rightarrow +\infty }[\inf\{(s - \psi(s)):
s>t\}]> 0$.

Hence there exist a $\delta>0$ and a subsequence $\{d_{n_i}\}$ such
that $d_{n_i}-\psi(d_{n_i})>\delta$  for all $i\in{\mathbb{N}}$,
that is, the fact is true.

If $\{d_n\}$ is bounded, without loss of generality,  we  assume
that $\{d_n\}$ increases and converges to a point $t'>0$. Then, for
$\psi$ is u.s.c. at $t'$, i.e. $\limsup\limits_{ t\rightarrow t'
}\psi(t)\leq\psi(t')$,
 and $\psi(t') < t'$, we have $\liminf\limits_{ t\rightarrow t'
}(t-\psi(t))=t'-\limsup\limits_{ t\rightarrow t' }\psi(t)\geq
t'-\psi(t')>0$. Note that $\{d_n\}$ increases and $\lim
\limits_{\Delta t\rightarrow 0}[\inf\{(t - \psi(t)): 0<|t-t'|<\Delta
t\}]=\liminf\limits_{ t\rightarrow t' }(t-\psi(t))$. We have
$\liminf \limits_{ n\rightarrow \infty }(d_n - \psi(d_n))=\lim
\limits_{ n\rightarrow \infty }[\inf\{(d_k - \psi(d_k)):
k>n\}]\geq\lim \limits_{ n\rightarrow \infty }[\inf\{(t - \psi(t)):
d_n<t<t'\}]=\lim \limits_{\Delta t\rightarrow 0}[\inf\{(t -
\psi(t)): 0<t-t'<\Delta t\}]\geq\lim \limits_{\Delta t\rightarrow
0}[\inf\{(t - \psi(t)): 0<|t-t'|<\Delta t\}]=\liminf\limits_{
t\rightarrow t' }(t-\psi(t))>0$. Hence the fact is also
true.$\rceil$

 That is, $d(x_{n_i}, y_{n_i})-\varphi(x_{n_i}, y_{n_i})>\delta$ for all $i\in{\mathbb{N}}$.

This contradicts $d(x_n, y_n)-\varphi(x_n, y_n)$ converges to $0$.
Hence $T$ satisfies the C-condition.

Finally, for the $T$ has the approximate endpoint property of
Theorem 1.1, from Remark 2.19,

it has the approximate endpoint property (defined in Definition
2.18).

Hence we can directly obtain
Theorem 1.1  from Theorem 3.1.\\

Next we further present the following Theorem 3.3, which shows, in
the setting that $(X,d)$ is  complete and  regular, if the global
multi-valued weak contraction satisfies C-condition, then it
has the approximate endpoint property, so has a unique endpoint from Theorem 3.1.\\

\noindent{\bf Theorem 3.3.} \textit{Let $(X,d)$ be complete and
regular, $T$ be a global multi-valued weak contraction on $(X,d)$
and satisfy  C-condition.

Then $T$ has a unique endpoint.}\\

\noindent{\bf Proof}. We first prove the existence of  endpoints.

Arguing by contradiction, assume $T$ has no endpoint. Then for any
$x\in X$, there is at least one $y\in Tx$ such that $y\neq x$. Hence
 there must be a sequence $\{y_n\}$ of $X$ such that
$y_{n+1}\in Ty_n$ and $y_{n+1}\neq y_n$ for all $n\in \mathbb{N}$.
Note $T$ is a global multi-valued weak contraction.  In terms of
$y_{n+1}\in Ty_n$, $y_{n+1}\neq y_n$, (2.2) and $\varphi(x,y)\prec
d(x,y)$ for $d(x,y)\succ \theta$, we have
$$d(y_{n+1}, y_{n+2})\preceq \varphi(y_{n}, y_{n+1})\prec d(y_{n}, y_{n+1})\eqno(3.6)$$
for all $n\in\mathbb{N}$. Hence the sequence

  $\{d(y_{n}, y_{n+1})\}$ is decreasing. So, for $G$ is
regular, there exists $a\in G_+$ such that $d(y_{n},
y_{n+1})\rightarrow a$. Hence, from (3.6), we have $a\preceq
\varphi(y_{n}, y_{n+1})\prec d(y_{n}, y_{n+1})$. Further, according
to the (iii) of Lemma 2.10, we  obtain $d(y_{n},
y_{n+1})-\varphi(y_{n}, y_{n+1})\rightarrow \theta$.  For $T$
satisfies  C-condition, this leads to $d(y_{n}, y_{n+1})\rightarrow
\theta$. And by (3.6) we further have  $\varphi(y_{n},
y_{n+1})\rightarrow \theta$. Now let $x_{n}=y_{n+1}$ and
$a_{n}=\varphi(y_{n}, y_{n+1})$. Then we have $d(x_{n},
x'_{n})\preceq a_{n}, \forall x'_{n}\in Tx_{n}$ and
$a_{n}\rightarrow \theta$. That is, $T$ has the approximate endpoint
property. Thus, by Theorem 3.1, $T$ has endpoints. This contradicts
our assumption.  Hence the existence of endpoints is true.

Finally, the uniqueness follows directly from the (ii) of Lemma
2.20. This ends the proof.
$\square$\\

For the single-valued weak contraction can be regarded as a kind of
specific global multi-valued weak contraction, from Theorem 3.3, we
can immediately derive the Corollary 3.4 below, which generalizes
Lemma 2.4 and Corollary 2.5 of [1].\\

\noindent{\bf Corollary 3.4.} \textit{Let $(X,d)$ be complete and
regular,
 $f$ be a single-valued weak
contraction on $(X,d)$ and satisfy  C-condition. Then $T$ has a
unique fixed point.}\\

\noindent{\bf Remark 3.5.}  The multi-valued weak contraction can
not have the approximate endpoint property, even in the usual metric
space, for instance, see the Example 2.3 of [1].

\section{Endpoint theory for the metric space of module}

Note that a metric space of module is  a special metric space of
group. As applications of the  results proved above,  this section
 discusses the endpoint theory for the metric space of module. We
always assume  $(X,d)$ is a metric space of module

$G$ in this section.

By regarding $\alpha(x,y)d(x,y)$ as $\varphi(x,y)$,
we can easily derive the next Theorem 4.1 from Theorem 3.1 and Theorem 3.3.\\

\noindent{\bf Theorem 4.1.} \textit{Let $(X,d)$ be complete, $T : X
\rightarrow (2^X-\emptyset)$ be a multi-valued mapping. Let also
$\alpha: X\times X\rightarrow [0, 1) (=\{r\in R: 0\leq r< 1\})$
 be a  mapping, which satisfies:

 for any two
  sequences $\{x_n\}$ and $\{y_n\}$ of $X$ with $[1-\alpha(x_n,y_n)]d(x_n,y_n)\rightarrow\theta$,
  there is an $\alpha\in [0, 1)$
 such that $\alpha(x_n,y_n)\leq \alpha$ for all $n\in{\mathbb{N}}$,
 as well as
  $(1-\alpha)$ has multiplicative inverse
 $(1-\alpha)^{-1}$  and $(1-\alpha)^{-1}>0$.
 Then we have the following two conclusions.\\ (i) Suppose  for
all different $x,y\in X$, $\forall x'\in Tx$, there exists
$\bar{y}\in
 Ty$ such that
$d(x',\bar{y})\preceq \alpha(x,y)d(x,y)$.

 Then $T$ has a
unique endpoint if and only if $T$ has the approximate endpoint
 property.\\
(ii) Let $(X,d)$ be  regular. Suppose for all different $x,y\in X$,
we have $d(x', y')\preceq \alpha(x,y)d(x,y)$, $\forall x'\in Tx,
\forall y'\in Ty$.  Then $T$ has a
unique endpoint.}\\

\noindent{\bf Proof.}

Let  $\varphi(x,y)=
\alpha(x,y)d(x,y)$. Then

$\varphi: X\times X \rightarrow G_+$ is a mapping. Since
$\alpha(x,y)\prec 1$, by (m2), see the (iii) of remark 2.2, we have
 $\varphi(x,y)=\alpha(x,y)d(x,y)\prec d(x,y)$ for all  $d(x,y)\succ\theta$. We  prove the statement that
 if
$d(x_n, y_n)-\varphi(x_n, y_n)\rightarrow\theta$, then
$d(x_n,y_n)\rightarrow\theta$ below.

Let $\{x_n\}$ and $\{y_n\}$ be two sequences of $X$. Then $d(x_n,
y_n)-\varphi(x_n, y_n)=[1-\alpha(x_n,y_n)]d(x_n,y_n)$. So, if
$d(x_n, y_n)-\varphi(x_n, y_n)\rightarrow\theta$, then
$$[1-\alpha(x_n,y_n)]d(x_n,y_n)\rightarrow\theta. \eqno(4.1)$$
Hence there exists an $\alpha\in [0, 1)$   such that
$\alpha(x_n,y_n)\leq\alpha, \forall n\in{\mathbb{N}}$. By (g1)$'$,
this further leads to $(1-\alpha)\leq [1-\alpha(x_n,y_n)]$. Since
also $d(x_n,y_n)\succeq\theta$, by (m2)$'$, we have
$$(1-\alpha)d(x_n,y_n)\preceq[1-\alpha(x_n,y_n)]d(x_n,y_n), \forall n\in{\mathbb{N}}. \eqno(4.2)$$
In terms of  (4.1) and (4.2), we obtain
$(1-\alpha)d(x_n,y_n)\rightarrow\theta$. For $\alpha<1$, we have
$1-\alpha>0$. Let $\varepsilon\gg\theta$. By (t6), we have
$(1-\alpha)\varepsilon\gg\theta$. Hence,
 there exists a natural $N$ such that
$(1-\alpha)d(x_n,y_n)\ll(1-\alpha)\varepsilon$ for all $n>N$. For
$(1-\alpha)^{-1}>0$, from (t6), we obtain also
$d(x_n,y_n)\ll\varepsilon$ for all $n>N$. This implies
$d(x_n,y_n)\rightarrow\theta$.

For conclusion (i), by $\varphi(x,y)=\alpha(x,y)d(x,y)\prec d(x,y)$
for all $d(x,y)\succ\theta$ and the statement proved above, it is
obvious that $T$ is a multi-valued weak contraction on the space
$(X,d)$ and satisfies  C-condition. Hence we can immediately know
that the conclusion is true from Theorem 3.1. For conclusion (ii),
$T$ is clearly a global multi-valued weak contraction and satisfies
C-condition. Note that $(X,d)$ is regular. We can immediately know
that the conclusion is true from Theorem 3.3.

This completes the proof. $\square$\\

Replacing, in  Theorem 4.1, $\alpha(x,y)$ by $\alpha$, we
 directly obtain the following Corollary 4.2.\\

\noindent{\bf Corollary  4.2.} \textit{Let $(X,d)$ be complete,  $T
: X \rightarrow (2^X-\emptyset)$ be a multi-valued mapping. Let also
$\alpha\in [0, 1)$ with $(1-\alpha)^{-1}$ and $(1-\alpha)^{-1}>0$.
(i) Suppose $T$ satisfies for all different $x,y\in X$, $\forall
x'\in Tx$, there exists $\bar{y}\in
 Ty$ such that
$d(x',\bar{y})\preceq \alpha d(x,y)$. Then $T$ has a unique endpoint
if and only if $T$ has the approximate endpoint
 property. (ii) Let  $(X,d)$ be  regular. Suppose $T$ satisfies for all
  different $x,y\in X$,
$d(x',y')\preceq \alpha d(x,y), \forall x'\in Tx, \forall y'\in Ty$.
Then $T$ has a unique endpoint.}\\

\noindent{\bf Proof}. Let $\alpha(x,y)=\alpha$. And then applying
Theorem 4.1, we obtain  the Corollary instantly. $\square$\\

Finally,  in the (ii) of Corollary 4.2,  replacing also multi-valued
mapping by single-valued mapping, we obtain Corollary 4.3 below.\\

\noindent{\bf Corollary  4.3.} \textit{Let $(X,d)$ be complete and
 regular, $T : X \rightarrow X$ be a single-valued mapping,
$\alpha\in [0, 1)$ with $(1-\alpha)^{-1}$ and $(1-\alpha)^{-1}>0$.
Suppose $T$ satisfies for all different $x,y\in X$, $d(Tx,
Ty)\preceq \alpha d(x,y)$. Then $T$ has a unique fixed point.}\\

\noindent{\bf Remark 4.4.} In particular,   when $(X,d)$ is the
usual complete metric space, Corollary 4.3 is just the famous Banach
fixed point theorem.

\noindent{\bf Acknowledgements}

The author cordially thanks the anonymous referees for their
valuable comments which lead to the improvement of this paper.

\bibliographystyle{model1-num-names}
\bibliography{<your-bib-database>}


TITLE: On analytic function with $f(e^z)=z$
QUESTION [3 upvotes]: Show that there does not exist any function $f$ which is analytic in $\mathbb C$ minus a (denumerable) set of isolated singularities and such that $f(e^z)=z$ wherever $f(e^z)$ is defined.

My attempt:
Suppose $w_0\ne 0$ is an isolated singularity of $f$, then for every sequence $\{z_n\}$ in the domain of $f$, such that $e^{z_n}\to w_0$ ($e^{z_n}$ can take any value in the nbhd of $w_0$ since we can choose an analytic branch of $\ln w_0$ for each $w_0\ne 0$) and $f(e^{z_n})=z_n$, $\lim_{n\to\infty}z_n=w$ and we conclude that $f$ can only have an isolated singularity at $z=0$.
Now note that $f(e^1)=1$ and $f(e^{2\pi i})=2\pi i$ but $e^1=e^{2\pi i}$ which is a contradiction.
Is that correct?

Edit:
I think the main idea of this question is that we have to take a branch cut along $[0,\infty]$ to make $f$ well-defined, otherwise it is nonsense to claim $f(e^z)=z$ as was pointed out in my attempt above.
On the other hand, $f$ is forced to exist other than a set with isolated singularities which is unrealistic, since $[0,\infty]$ does not consist of isolated points.

REPLY [2 votes]: Lemma: Suppose $C\subset \mathbb R$ is countable. Let $E=\mathbb R \setminus C.$ Then there exists $t_1,t_2\in E$ such that $t_2=t_1+2\pi.$
Proof: If not, then the uncountable set $\{t+2\pi: t\in E\}$ is a subset of the countable set $C,$ contradiction.
Now to the problem: Suppose we have such an $f.$ Let $D$ be the denumerable set of singularities of $f.$ Write $g(z)=e^z.$ Then $g^{-1}(D)$ is countable, so we can write this set as $\{x_1+iy_1,x_2+iy_2,\dots\}.$
Let $C=\{y_1,y_2,\dots\}$ and $E=\mathbb R \setminus C.$ The lemma then gives $t_1,t_2\in E$ such that $t_2=t_1+2\pi.$ This implies the horizontal lines $\{s+it_1:s\in \mathbb R\}$ and $\{s+i(t_1+2\pi):s\in \mathbb R\}$ both miss $g^{-1}(D).$ Thus $f\circ g$ is defined on both of these lines. We then have
$$s+it_1 =(f\circ g)(s+it_1)$$ $$ = (f\circ g)(s+i(t_1+2\pi))=s+i(t_1+2\pi)$$
for all $s,$ contradicition.


text:
{\bf Problem.} Round 3967149.1587234 to the nearest integer.
{\bf Level.} Level 2
{\bf Type.} Prealgebra
{\bf Solution.} Because 3967149.1587234 is between 3967149 and 3967150, rounding will give us either 3967149 or 3967150. Since 0.1587234 is less than 0.5, we get that 3967149.1587234 is closer to $\boxed{3967149}$.

set_name: MATH


TITLE: Continuous conditional distribution
QUESTION [1 upvotes]: This is about continuous conditional probability distributions. Why is it that they are allowed to take on single values, while this is a no-no with non conditional continuous distributions(to my knowledge) .
What makes these the exception?
For an example, let X and Y be continuous random variables, it seems to be allowed to say P[ Y | X = 4] can be displayed as:
$f(x,y)/f_{x}(4) $
Thanks

REPLY [1 votes]: Conditional probability is classically defined as:
$$\mathsf P(Y\in \mathcal B\mid X\in\mathcal A) ~=~ \dfrac{\mathsf P(X\in\mathcal A, Y\in\mathcal B)}{\mathsf P(X\in\mathcal A)}$$
This is okay as long as the event $\{X\in\mathcal A\}$ has non-zero probability mass.   In particular when the interval $\mathcal A$ contains discrete value(s) but the random variable is continuous, this definition is a "no-no", as you say ; eg: when $\mathsf P(X=x)=0$ it can not be done.
However that is probability mass.   We can work analogously with probability densities; if we take care.   If $X,Y$ are both continuous then conditional probability density is the ratio of the joint density to the marginal density (as long as that is non-zero at the point in question).
$$f_{Y\mid X}(y\mid x) ~=~ \dfrac{f_{X,Y}(x,y)}{f_X(x)}$$
And if $\mathcal B$ is a continuous interval.
$$\mathsf P(Y\in\mathcal B\mid X=x) ~=~ \dfrac{\int_\mathcal B f_{X,Y}(x,y)\operatorname d y}{f_X(x)}$$
We can also work with mixtures.   If $Y$ is a discrete random variable and $X$ a continuous one then we can say:
$$\mathsf P(Y=y\mid X=x) = \dfrac{f_{X\mid Y}(x\mid y)~\mathsf P(Y=y)}{f_X(x)}$$

Well, basically.   The modern approach is to use Measure Theory to tie this all together under a more abstract but rigorous umbrella.


TITLE: primitive roots of composite numbers
QUESTION [1 upvotes]: I am looking for a way to find primitive roots of composite numbers by primitive roots of its prime factors.im looking for a analytic way no algebraic.
I meant a way without meanings of abstract algebra like groups.
it is a great help if you help me with your links or files.

REPLY [1 votes]: The only numbers that admit a primitive root are those of the form
$p^k$  and $2p^k$ with $p$ an odd prime, and $2$ and $4$.
Let's ignore the simple explicit cases $2$ and $4$.
Then we have the following paraphrased from the Wikipedia page:

If $g$ is a primitive root modulo $p$, then $g$ is a primitive root modulo all powers $p^k$,  unless $g^{p – 1} \equiv 1 \pmod{p^2}$. In case $g^{p – 1} \equiv 1 \pmod{p^2}$ we have $g + p$ is primitive root.
If $g$ is a primitive root modulo $p^k$, then $g$ or $g + p^k$ (whichever one is odd) is a primitive root modulo $2p^k$.

Thus, the problem is easily reduced to the one for primes.


TITLE: Why do we not define vector multiplication and division?
QUESTION [2 upvotes]: Consider, two planar vectors:
$$V=  a \hat{x} + b \hat{y}$$
And
$$ U = a' \hat{x} + b' \hat{y}$$
These are analogous to the complex numbers:
$$ v = a + bi$$
and,
$$u= a' + b' i$$
Now, there are clear rules for multiply $ v \cdot u$ and also the expression: $ \frac{v}{u}$ , also there exists a geometric interpretation (if you view in polar form). Then, why is it that we don't bring up these products when speaking vectors and only discuss dot and cross product?

REPLY [0 votes]: For planar vectors we can indeed multiply and divide.
First identify $V = a \hat{x} + b \hat{y}$ with $(a, b)$ and
$U = c \hat{x} + d \hat{y} $ with $(c, d)$
Now define $(a, b) \times (c, d) := (ac - bd, ad + bc)$.  You can check that this definition of multiplication is well-defined.
Also, $\frac{(a,b)}{(c,d)} = \frac{1}{c^2 + d^2}(ac + bd, ad - bc)$ is well defined.
These formulas may look foreign at first site but a closer inspection reveals that they are just multiplication and division of complex numbers.


TITLE: Prove that if $\phi: G\to H$ is a homomorphism and $G_{1}\leq G$ is cyclic, then $\phi (G_{1})$ is cyclic.
QUESTION [2 upvotes]: Would someone tell me whether my proof is correct or is it missing anything important, please?

Prove that if $\phi: G\to H$ is a homomorphism and $G_{1}\leq G$ is cyclic, then $\phi (G_{1})$ is cyclic.

Let $G_{1}$ be cyclic. Then $G_{1}=\left \langle g \right \rangle$ for some $g\in G_{1}$. Since $\phi(G_{1})=\left \{ \phi(g_{1}):g_{1}\in G_{1} \right \}$ and $g_{1}=g^{k}$ for all $g_{1}\in G_{1}$ and $k\in \mathbb{Z}$, by definition of homomorphisms, $\phi(g_{1})=\phi(g^{k})=(\phi(g))^{k}$ for all $\phi(g_{1})\in\phi(G_{1})$. Then $\phi(G_{1})=\left \langle \phi(g) \right \rangle$ hence cyclic.

REPLY [1 votes]: Your proof is correct. Well done.
The phrasing at the end, though, is a little off; try a new sentence reading "Hence $\phi(G_1)$ is cyclic." It's just a minor suggestion.


text:
\begin{document}

\keywords{Approximation scheme, approximation error, Bernstein's Lethargy, Haar space, de La Vallée-Poussin Theorem, smoothness}
\subjclass[2000]{41A25, 41A65, 41A27}
 \baselineskip=16pt

\numberwithin{equation}{section}

\maketitle \markboth{Lethargy result for real analytic functions}{J. M. Almira}

\begin{abstract}
We prove that, if $(C[a,b],\{A_n\})$ is an approximation scheme and $(A_n)$ satisfies de La Vallée-Poussin Theorem, there are instances of continuous functions on $[a,b]$,
real analytic on $(a,b]$, which are ``poorly approximable'' by the elements of  $\{A_n\}$. This i\-llus\-tra\-tes the thesis
that the smoothness conditions guaranteeing that a function is ``well approximable''
must be ``global''. The failure of smoothness at endpoints may result in an
arbitrarily slow rate of approximation.
\end{abstract}

\section{Motivation} After the work by Bernstein  \cite{bernsteininverso} and  Jackson \cite{jackson} in Approximation Theory it is accepted that there is a strong relationship between the smoothness properties of a function and the rate of decay to zero of the sequences of errors of best uniform polynomial approximation $\{E(f,\Pi_n)\}$. Indeed, this connection between smoothness and the order of convergence to zero of a sequence of best approximation errors holds for many other approximation schemes such as rational approximation, spline approximation, etc. There are many papers and monographs devoted to show this claim in its many distinct cases (see, for example, \cite[Chapter 7]{devore} and \cite{pinkus}, just to mention two beautiful references where this philosophy is neatly shown) In particular, if the approximation scheme $(X,\{A_n\})$ satisfies both Jackson's inequality $E(f,A_n)\leq Cn^{-r}|f|_Y$, $(f\in Y)$, and Bernstein's inequality $|a_n|_Y\leq Cn^r\|a_n\|_X$, $(a_n\in A_n)$, for $n=0,1,\cdots $, with respect to a  quasi-semi-normed proper subspace $Y$ of $X$, the approximation space $A_q^{\alpha}(X,\{A_n\})=\{x\in X:\{(n+1)^{\alpha-\frac{1}{q}}E(x,A_n)\}\in \ell_q\}$ coincides with the interpolation space $(X,Y)_{\alpha/r,q}$ for $0<\alpha<r$ and $0<q\leq\infty$, which is usually viewed as a smoothness space (see \cite[Chapter 7, Theorem 9.1]{devore} for a proof of this result). Another well known result which is usually understood as a proof of the strong relationship that exists between smoothness of functions and the order of decay to zero of best approximation errors is Bernstein's characterization of the continuous functions on $[-1,1]$ which result from the restriction of an analytic function on the interior of the ellipse
$$E_{\rho}=\{(\frac{1}{2}(\rho+\frac{1}{\rho})\cos\theta, \frac{1}{2}(\rho-\frac{1}{\rho})\sin\theta):\theta\in [0,2\pi]\},$$
as those functions $f\in C[-1,1]$ such that $\lim\sup_{n\to\infty}E(f,\Pi_n)^{\frac{1}{n}}=\frac{1}{\rho}$.

In this short note we prove that, if $(C[a,b],\{A_n\})$ is an approximation scheme and $(A_n)$ satisfies de La Vallée-Poussin Theorem, there are instances of continuous functions on $[a,b]$,
real analytic on $(a,b]$, which are ``poorly approximable'' by the elements of the approximation scheme $(A_n)$. This illustrates the thesis
that the smoothness conditions guaranteeing that a function is ``well approximable''
must be, at least in these cases, ``global''. The failure of smoothness at endpoints may result in an
arbitrarily slow rate of approximation.  A result of this kind,  which is highly nonconstructive, based on different arguments, and applicable to different approximation schemes, was  recently proved by Almira and Oikhberg \cite{almira_oikhberg}.

\section{The main result}
Let us start by recalling what an approximation scheme is.
\begin{definition}[See, for example, \cite{almira}, \cite{almiraluther2}]
Let $(X,\|\cdot\|)$ be a quasi-Banach space, and let
$\{0\}=A_0\subset A_1\subset\cdots \subset A_n\subset\cdots \subset X$
be an infinite chain of subsets of $X$, where all inclusions are
strict. We say that $(X,\{A_n\})$ is an {\it approximation scheme} (or that $(A_n)$ is an approximation scheme in $X$) if:
\begin{itemize}
\item[$(i)$] There exists a map $K:\mathbb{N}\to\mathbb{N}$ such that $K(n)\geq n$ and $A_n+A_n\subseteq A_{K(n)}$ for all $n\in\mathbb{N}$.

\item[$(ii)$] $\lambda A_n\subset A_n$ for all $n\in\mathbb{N}$ and all scalars $\lambda$.

\item[$(iii)$] $\bigcup_{n\in\mathbb{N}}A_n$ is a dense subset of $X$
\end{itemize}
A particular example is a {\it linear approximation scheme}, arising when the
sets $A_n$ are linear subspaces of $X$. In this setting, we can take $K(n) = n$.
An approximation scheme is called {\it non-trivial} if $X \neq \cup_n \overline{A_n}$.
\end{definition}

\begin{definition} The vector space $A \subset C[a,b]$ is named Haar on $[a,b]$ if $\dim A=n$ and the only element from $A$ which has more than $n-1$ zeroes is the null function.
\end{definition}

\begin{definition} Let $(C[a,b],\{A_n\})$ be an approximation scheme. We say that $(A_n)$ satisfies de La Vallée-Poussin Theorem if there exists a function $\phi_{\{A_n\}}:\mathbb{N}\to\mathbb{N}$, $\phi_{\{A_n\}}(n)=m_n$, such that, if $f\in C[a,b]$ and $a\leq t_{n,0}<t_{n,1}<\cdots <t_{n,m_n}\leq b$ satisfy  $\mathbf{sign}(f(t_{n,k})f(t_{n,k+1}))=-1$ for all $k$ and $\min_{0\leq k<m_{n}}|f(t_{n,k})|>\varepsilon_n$, then $E(f,A_n)\geq \varepsilon_n$.  \end{definition}

Classical de La Vallée-Poussin Theorem \cite{poussin} (see also \cite[page 74, Theorem 5.2]{devore}) appears when $(C[a,b],\{A_n\})$ is a linear approximation scheme, $m_n=\dim A_n$,  and each $A_n$ is Haar. More examples of approximation schemes satisfying de La Vallée-Poussin Theorem appear in connection with rational approximation. Indeed, $\{n_k\},\{m_k\}\to +\infty$ are a pair of increasing sequences of natural numbers, and $R_{n,m}$ denotes the set of rational functions $r(t)=p(t)/q(t)$ with poles outside $[a,b]$ and $\deg(p)\leq n$, $\deg(q)\leq m$, the approximation scheme $(C[a,b],\{R_{n_k,m_k}\}_{k=0}^\infty)$ satisfies de La Vallée-Poussin Theorem \cite[Theorem 98]{meinardus}. Finally, it is an easy exercise to check that if the approximation scheme $(C[a,b],\{A_n\})$ satisfies de La Vallée-Poussin Theorem with $\phi_{\{A_n\}}(n)=m_n$ and we denote by $\Sigma_n(A_n)$ the set of splines with $n$ free knots constructed with the elements of $A_n$ then $(C[a,b],\{\Sigma_n(A_n)\})$  satisfies de La Vallée-Poussin Theorem with $\phi_{\{\Sigma_n(A_n)\}}(n)=(n+1)^2m_n$. Recall that  $f\in \Sigma_n(A_n)$ if and only if there exists a partition $a=t_0<t_1<\cdots<t_n<b=t_{n+1}$ such that  $f(t)=\sum_{k=0}^{n-1} a_k(t)\chi_{[t_k,t_{k+1})}(t)+a_k(t)\chi_{[t_n,t_{n+1}]}(t)$ for some $a_0,a_1,\cdots a_n\in A_n$. Here, $\chi_{I}$ denotes the characteristic function associated to the interval $I$.

\begin{theorem}\label{mainresult} Let us assume that $0<\alpha<\beta$ and let $[a,b]=[0,1]$ or $[a,b]=[\alpha,\beta]$. Let $(C[a,b],\{A_n\})$ be an approximation scheme such that  $(A_n)$ satisfies de La Vallée-Poussin Theorem with $\phi_{\{A_n\}}(n)=m_n$. Let $\{\varepsilon_n\}_{n=0}^{\infty}\in c_0$ be a non-increasing sequence of positive numbers converging to zero. Then there exists $f\in C[a,b]$ such that $f$ is real analytic on $(a,b]$ and $E(f,A_n)\geq \varepsilon_n$ for all $n=0,1,2,\cdots$.
\end{theorem}
\noindent \textbf{Proof.} Let us first consider the case $[a,b]=[0,1]$.  The result is trivial if $\{\varepsilon_n\}$ is stationary, so that we can assume $\varepsilon_n>0$ for all $n\in\mathbb{N}$. Let $p:\mathbb{R}\to\mathbb{R}$ be the continuous polygonal line with vertices $\{(n,p(n))\}_{n\in\mathbb{Z}}$ defined by
  \begin{itemize}
  \item[$(i)$] $p(t)$ is an even function, $p(0)=p(1)=\cdots=p(m_0-1)=3\varepsilon_0$.
  \item[$(ii)$] $p(m_0+m_1+\cdots+m_n+k)=3\varepsilon_{n+1}$ for all $n\in\mathbb{N}$ and $k=0,1,\cdots,m_{n+1}-1$.
  \end{itemize}
It follows from Carleman's Theorem \cite{carleman} (see also \cite[Chapter 1, Theorem 4.3]{makovoz}) that there exists an entire function $e(t)$ such that $e(\mathbb{R})\subset \mathbb{R}$ and $\sup_{t\in\mathbb{R}}|p(t)-e(t)|\leq \frac{p(t)}{3}$. Hence $e([0,m_0-1]) \subseteq (\varepsilon_0,5\varepsilon_0)$ and, for $n\in\mathbb{N}$, $e([m_0+\cdots+m_n,m_0+\cdots+m_{n+1}-1))\subseteq (\varepsilon_{n+1},5\varepsilon_{n+1})$. Let us set $f(0)=0$ and, for $t\in (0,1]$, $f(t)=e(\frac{1}{t})\cos(\frac{2\pi}{t})$. This is our function. Obviously, $f$ is analytic on $(0,1]$ and continuous on $[0,1]$. To prove that $E(f,A_n)\geq \varepsilon_n$ it is enough to take into account that $A_n$ satisfies de La Valleé-Poussin Theorem with $\phi_{\{A_n\}}(n)=m_n$ and, by construction, there exists an ordered set of points
$\frac{1}{m_0+m_1+\cdots+m_{n-1}}\leq t_{n,0}<t_{n,1}<\cdots <t_{n,m_n}\leq \frac{1}{m_0+m_1+\cdots+m_{n}-1}$ such that $\mathbf{sign}(f(t_{n,k})f(t_{n,k+1}))=-1$ for all $k$ and $\min_{0\leq k<m_{n}}|f(t_{n,k})|>\varepsilon_n$. This ends the proof for $[a,b]=[0,1]$. If $0<a<b$, the arguments above show that $E(g,A_n)\geq \varepsilon_n$ for all $n$, where $g(t)=f(\frac{t-a}{b-a})$. {\hfill $\Box$}

\begin{remark} The same kind of phenomenon also holds in different situations for higher dimensions. To prove this claim, let $(C([-1,1]^s),\{A_n\}_{n=0}^{\infty})$ be an approximation scheme such that, for each $k\in\{1,\cdots,s\}$ and $a=(a_1,\cdots,a_s)\in [-1,1]^s$, the sets $B_{k,n}^a=\{g(x)=f(a_1,\cdots,a_{k-1},x,a_{k+1},\cdots,a_s):f\in A_n\}$ form an approximation scheme in $C[0,1]$ and $(C[0,1],\{B_{k,n}^a\}_{n=0}^\infty)$ satisfies de La Vallée-Poussin Theorem with $\phi_{\{B_{k,n}^a\}}(n)=m_{n}$. These approximations schemes are easy to find. For example, for $s=2$ we can take $$A_n=\Pi_{n}[x,y]:=\{P(x,y)=\sum_{0\leq i,j\leq i+j\leq n}a_{ij}x^iy^j:a_{ij}\in\mathbb{R} \text{ for all }i,j\}$$ or $$A_n=\{P(x,y)/Q(x,y):P,Q\in \Pi_n[x,y] \text{ and } Q(a,b)\neq 0 \text{ for all } (a,b)\in [-1,1]^2\}.$$ Then, given a non-increasing sequence $\{\varepsilon_n\}\in c_0$ we choose $f\in C[0,1]$ as in Theorem \ref{mainresult} and define $g(x_1,\cdots,x_s)=f(x_1^2+x_2^2+\cdots+x_s^2)$. This function is continuous on $[-1,1]^s$ and real analytic on $\Omega=[-1,1]^s\setminus\{\mathbf{0}\}$, where $\mathbf{0}=(0,0,\cdots,0)\in \mathbb{R}^s$ (see \cite[Proposition 2.2.8]{krantz}). Moreover, $h(x)=f(x^2)$ also satisfies the conclusion of Theorem \ref{mainresult} since the change of $x$ by the new variable $x^2$ does not eliminates the oscillations of $f$. Hence, for each $n\in\mathbb{N}$ and $k\in\{1,\cdots,s\}$, we have that $E(g,A_n)_{C([-1,1]^s)}\geq E(g(0,\cdots,0,x_k,0,\cdots,0),B_{k,n}^{\mathbf{0}})_{C[0,1]}=E(h,B_{k,n}^{\mathbf{0}})_{C[0,1]}\geq \varepsilon_n$. \end{remark}

config: arxiv


TITLE: Can I choose my universal set to be convenient for the given problem?
QUESTION [1 upvotes]: I'm attempting to prove this fact from Abbot's Understanding Analysis

Now what I would like to do is assume that $a + t \not\in \mathbb{I}$ or $at \not\in \mathbb{I}$ with the assumption that my universal set $U$ is $\mathbb{R}$, so $a + t \in \mathbb{Q}$ and $at \in \mathbb{Q}$ because $\mathbb{I}^C = \mathbb{Q}$ and then derive a contradiction.
This would work just fine if I assume that $\mathbb{R}$ is my universal set. But there's the obvious possibility of choosing $U = \mathbb{C}$, which would break my argument. Is doing this reasonable?

REPLY [1 votes]: This has nothing to do with a universal set. $\mathbb I$ is unambiguously just $\mathbb R\setminus \mathbb Q$.
So since $\mathbb R$ is a field, both $a+t$ and $at$ are in $\mathbb R$. You need to show they are, in fact, both in $\mathbb R\setminus \mathbb Q$.


TITLE: Prove two definitions of $\limsup$ are equivalent
QUESTION [2 upvotes]: Let ($x_n$) be a real sequence. Prove the following:
If both
$\displaystyle\lim_{n\to\infty}\Big(\sup_{m\geq n}x_m\Big)$
 and
$\displaystyle\inf_{n\geq 0}\,\sup_{m\geq n}x_m$
 exist, then
$\displaystyle\lim_{n\to\infty}\Big(\sup_{m\geq n}x_m\Big) = \inf_{n\geq 0}\,\sup_{m\geq n}x_m$

REPLY [2 votes]: Hint: Note that the suprema are over smaller and smaller sets, since they are over shrinking tails of the underlying sequence. Do you see that you are therefore dealing with the limit of a monotone sequence?


TITLE: Proof: For all real numbers x and y if x + floor of (x) = y + floor of (y) then x = y
QUESTION [0 upvotes]: I am trying to determine if this statement is true or false:
I think that it is true, if i let x = 2.5 then the left side is 4.5 and if i let y be anything but 2.5 then x + floor of (x) cant not equal y + floor of (y).
Since n<= x < n + 1 any number y between n and n + 1 will have the same floor as x but if x is not y the statement cant be true.
How would i start to prove this?

REPLY [1 votes]: Recall that $\lfloor t+n\rfloor=\lfloor t\rfloor +n$ for all $t\in\mathbb{R}$ and $n\in\mathbb{Z}$.  If $x$ and $y$ are real numbers such that $$x+\lfloor x\rfloor= y+\lfloor y\rfloor\,,$$ then taking the floor function on both sides, we get $$2\,\lfloor x\rfloor=\big\lfloor x+\lfloor x\rfloor\big\rfloor=\big\lfloor y+\lfloor y\rfloor\big\rfloor=2\,\lfloor y\rfloor\,.$$
Thus, $\lfloor x\rfloor=\lfloor y\rfloor$.  The rest should be trivial.


TITLE: Fixed point subspaces $V^B$ and $V^G$ for a Borel subgroup $B\subset G$ coincide
QUESTION [3 upvotes]: Assume that $G$ is a linear algebraic group, and let $B \subset G$ a Borel subgroup of it. Let $(V,\rho)$ a rational $G$-module. Define $$V^G := \{ v \in V \mid g \cdot v = v \quad \forall g \in G \}$$ prove that $V^B = V^G$.
My attempt Obviously $V^G \subset V^B$ because $B \subset G$. My strategy is to find a way to prove (if it is true!) that the action $G \to GL_n(V)$ can be restarted to an action $G\to GL_n(V^B)$. Assuming that true, then by hypothesis the morphism factor through the quotient $G/B$ which is a projective variety and whose image is still a closed subset of $GL_n(V^B)$, hence affine and hence the image is a finite set of points (because is proper) and hence by connectedness it's the identity. Hence $V^B$ is $G$-stable and hence $V^B \subset V^G$. The point is that I'm not able to prove that I can restrict the action to $V^B$. I hope the rest of the reasoning is clear and correct.
Thanks in advance!

REPLY [2 votes]: As far as I can see, your reasoning is fine, and you don't even need an action of $G$ on $V^B$: It suffices to pick some $v\in V^B$ and to consider the map $G\to V$ given by $g\mapsto g.v$. Since $v\in V^B$, this map factors through $G/B\to V$, and your arguments show that it must be constant.


TITLE: the induced homomorphism between fundemental groups
QUESTION [2 upvotes]: Suppose $X,Y$ are topological spaces, $i:X\to Y$ is the inclusion map,$i$ induces a homomorphism $i_{*}:\pi_1(X,x_0)\to \pi_1(Y,x_0)$ such that $i_{*}([f])=[i\circ f]$.
According to the definition ,we have $i_{*}([f])=[i\circ f].$ Since $i$ is the inclusion map,we have $i\circ f(t)=f(t)$ for all $t\in [0,1]$,then we have $i_{*}([f])=[i\circ f]=[f]$.We can conclude that $i_{*}$ is injective without the condition that $Y$ retracts to $X$ .
Can anyone points out the wrong place.Thanks.

REPLY [4 votes]: The point is that, even if two classes $[f]$ and $[g]$ are different in $\pi_1(X,x_0)$ (that is, $f$ and $g$ are not homotopic in $X$), it can be the case that $[f]=[g]$ in $\pi_1(Y,x_0)$ (when $f$ and $g$ are homotopic in $Y$).
Indeed, when you write "$i_*([f])=[f]$" (which is true!), the "$[f]$" on the left side is different from the "$[f]$" on the right side, even if you write them with the same symbol: the one on the left side is the class of the loops (on $x_0$) homotopic (relatively to $x_0$) to $f$ in $X$, while the one on the right side is the class of the loops homotopic to $f$ in $Y$.
For instance, consider the inclusion $i$ of $D\setminus \{p\}$ into $D$, being $D$ a disk and $p$ one of its points. Let $f$ be a loop on a fixed $x_0 \in D\setminus \{p\}$ such that $f$ goes around $p$. Then clearly $[f]\neq[\sigma_{x_0}]$ in $\pi_1(D\setminus \{p\})$, but of course $[f]=[\sigma_{x_0}]$ in $\pi_1(D)$. Here $\sigma_{x_0}$ denotes the constant loop on $x_0$. Hence in this case $i_*$ is not injective. And indeed $D\setminus \{p\}$ is not a retract of $D$!
Edit. I rewrite everything with the clarifying notation.
The point is that, even if two classes $[f]_X$ and $[g]_X$ are different in $\pi_1(X,x_0)$ (that is, $f$ and $g$ are not homotopic in $X$), it can be the case that $[f]_Y=[g]_Y$ in $\pi_1(Y,x_0)$ (when $f$ and $g$ are homotopic in $Y$).
Indeed, when you write "$i_*([f]_X)=[f]_Y$" (which is true!), the "$[f]_X$" on the left side is different from the "$[f]_Y$" on the right side: $[f]_X$ the one on the left side is the class of the loops (on $x_0$) homotopic (relatively to $x_0$) to $f$ in $X$, while $[f]_Y$ is the class of the loops homotopic to $f$ in $Y$.
For instance, consider the inclusion $i$ of $D\setminus \{p\}$ into $D$, being $D$ a disk and $p$ one of its points. Let $f$ be a loop on a fixed $x_0 \in D\setminus \{p\}$ such that $f$ goes around $p$. Then clearly $[f]_{D\setminus\{p\}}\neq[\sigma_{x_0}]_{D\setminus\{p\}}$ in $\pi_1(D\setminus \{p\})$, but of course $[f]_D=[\sigma_{x_0}]_D$ in $\pi_1(D)$. Here $\sigma_{x_0}$ denotes the constant loop on $x_0$. Hence in this case $i_*$ is not injective. And indeed $D\setminus \{p\}$ is not a retract of $D$!


TITLE: Does Energy change sign when time is reversed?
QUESTION [1 upvotes]: In classical physics if one reverses time then energy does not change sign. For example in the formula for kinetic energy one has:
$$E = \frac{1}{2}m v^2$$
If you reverse time the velocity $v$ changes to $-v$ but $E$ stays the same.
What happens in quantum mechanics?
In QM one has the formula:
$$E = h \nu = \frac{h}{\Delta t} $$
If $\Delta t$ changes to $-\Delta t$ then in QM the energy should change sign if the time is reversed.
Why is there a difference between classical and QM?

REPLY [5 votes]: Energy, frequency and period are all scalars. It's correct to write:
$$ E = \frac{h}{\tau} $$
where $\tau$ is the period, but the sign of $\tau$ does not change if you run time backwards so the sign of energy doesn't change in QM either.

REPLY [5 votes]: As John Rennie wrote in his answer, what one should consider is not a generic $\Delta t$ but the period $\tau$ which is a positive number. However positivity also arises form coherence with other relations. In particular, in relativistic quantum mechanics $E= \sqrt{\hbar^2 \vec{k}^2 + m^2c^4}$. For $m=0$ you have $E= \hbar |\vec{k}| = \hbar \omega = h\nu$ where $\omega$ and $\nu =1/\tau$ are therefore supposed to be non negative. These numbers are trivially invariant under time reversal.


TITLE: How to find a number $n$ when $n$ mod $n_1$ and $n$ mod $n_2$ is given?
QUESTION [3 upvotes]: For example:
suppose we need to find x given that x mod 7 = 5 and x mod 13 = 8.
x = 47 is a solution but needs hit and trial.
Is there any shortcut to calculate such number?

REPLY [0 votes]: $x=\frac{5*13+8*7}{7+13}=121/20=121*41=47\mod{(7*13)}$
$x=m_1\mod{n_1}$
$x=m_2\mod{n_2}$
$x=m_3\mod{n_3}$
...
$x=m_n\mod{n_n}$
$$x=\frac{m_1/n_1+m_2/n_2+m_3/n_3+...m_n/n_n}{1/n_1+1/n_2+1/n_3+...1/n_n}\mod{(n_1*n_2*n_3...n_n)}$$


TITLE: Prove by strong induction the Jacobstahl Series
QUESTION [1 upvotes]: I am just beginning to learn about strong induction, I can't figure out how to complete this homework question:
Jacobstahl numbers are defined as
$$J_1 = 1$$
$$J_2 = 1$$
$$J_n = J_{n-1} + 2J_{n-2}$$
Prove:
$$J_n = 2^{n-1}-J_{n-1}$$
for n >= 2.
Here is what I have done so far:
Assume:
$$J_k = 2^{k-1} + J_{k-1}$$
then
$$J_{k+1} = 2^{k+0} - J_{k+0}$$
Then from the definition:
$$J_{k+1} = 2^{k+0} - J_{k+0} = 2^{k-1}-J_{k-1}+2(2^{k-2}-J_{k-2})$$
$$2^{k+0} - J_{k+0} = 2^{k-1}-j_{k-1}+2^1*2^{k-2} - 2^1*J_{k-2}$$
$$= 2(2^{k-1})-J_{k-1}-2J_{k-2}$$
$$= 2^{k+0} - J_{k-1}-2J_{k-2}$$
So I have gotten part of the way there, by finding the $2^{k+0}$ term, but I'm not sure what to do with the rest of it.  I think I've probably gone wrong in setting up the assumption, but I can't work out how.

REPLY [1 votes]: Your assumption for strong induction should be that $J_k=2^{k-1}-J_{k-1}$ for $k=2,\ldots,n$; from that assumption you want to prove that $J_{n+1}=2^n-J_n$. If you then attempt the usual kind of argument, your calculation of $J_{n+1}$ will start like this:
$$\begin{align*}
J_{n+1}&=J_n+2J_{n-1}\\
&=\left(2^{n-1}-J_{n-1}\right)+2J_{n-1}\\
&=2^{n-1}+J_{n-1}\\
&=2^{n-1}+2^{n-2}-J_{n-2}\,.
\end{align*}$$
It’s pretty clear that if you continue in the most straightforward fashion, replacing $J_{n-2}$ by $2^{n-3}-J_{n-3}$, you’ll still have a $J_k$ term of some kind: you’re not obviously reaching what you want.
It actually is possible to get somewhere this way, but for a simple induction proof you’re better off looking at the desired relationship in a different way. Notice that $J_k=2^{k-1}-J_{k-1}$ is equivalent to $J_k+J_{k-1}=2^{k-1}$. Thus, we could just as well assume as our induction hypothesis that $J_k+J_{k-1}=2^{k-1}$ for $k=2,\ldots,n$ and try to prove that $J_{n+1}+J_n=2^n$. Then
$$J_{n+1}+J_n=J_n+2J_{n-1}+J_n\,;$$
now just simplify that and apply the induction hypothesis. (You’ll find that you don’t actually need strong induction here: ordinary induction is enough.)


TITLE: Proving: M finitely generated as an R-module and $R/\text{Ann} _R(M)$ noetherian implies M noetherian
QUESTION [1 upvotes]: Let $R$ be a commutative Ring, $M$ an $R$-module, we define $\text{Ann} _R(M) := \{ r \in R : rm = 0 \text{ }\forall m \in M\} $. As an exercise we have to show:
"$M$ noetherian $\iff$ $M$ finitely generated, $R/\text{Ann}_R(M)$ noetherian "
I was able to prove the "=>" implication, however i am struggling with the other direction. I do not really know how I can use the condition "$R/\text{Ann}_R(M)$ noetherian ". What does this tell me about the submodules of $M$? I do not see the connection. My first guess was to consider finitely generated $R/\text{Ann}_R(M)$-modules, which would automatically be noetherian, maybe there could exist such a module that is isomorphic to $M$ (or a submodule of $M$?), however, my attempt didn't succeed.
My second guess was to use induction over the number of M-generating elements. Here, I was unable to go back to the inductive hypothesis. What else could I try here? Does somebody have some advice for me? Thanks in advance!

REPLY [2 votes]: Let us put $R_M=R/\mathrm{Ann}_R(M)$. Note that

$M$ is canonically an $R_M$-module and
$\Big\{R$-submodules of $M\Big\}=\Big\{R_M$-submodules of $M\Big\}$.

The lattices of $R$-submodules of $M$ and that of $R_M$-submodules of $M$ are equal. This is relevant because the noetherian property you are after is a property of the lattice of submodules: $M$ noetherian $R$-module means that every nonempty collection of submodules has a maximal element. Thus $M$ is noetherian as an $R$-module iff it is noetherian as an $R_M$-module.
Note furthermore that:

A subset of $M$ is $R$-generating iff it is $R_M$-generating.

In particular, $M$ is finitely generated over $R$ if and only if it is finitely generated over $R_M$.

Proof of $(\!\!\impliedby\!\!)$ Suppose $M$ is finitely generated over $R$. By the preceding point it is finitely generated over $R_M$. If we furthermore assume $R_M$ to be noetherian, then so is the finitely generated $R_M$-module $M$. By the point in bold, $M$ is noetherian over $R$.


TITLE: Why do we 'invert and multiply' when dividing fractions?
QUESTION [0 upvotes]: I read in a textbook that the explanation to this rule lies in the fact that division is the reverse operation to multiplication. Unfortunately, the author did not elaborate on this point. Based on this, can someone help me to understand why
$\frac{a}{b}$ / $\frac{c}{d}$ = $\frac{a}{b}$ * $\frac{d}{c}$?

REPLY [1 votes]: If you accept that you can multiply the numerator and denominator by the same nonzero number without changing the value of the fraction, then just multiply the numerator and denominator by $d/c$:
$$
\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{\frac{a}{b}\cdot \frac{d}{c}}{\frac{c}{d}\cdot \frac{d}{c}} = \frac{\frac{ad}{bc}}{1} = \frac{ad}{bc}.
$$


TITLE: Average cost and the level of output
QUESTION [0 upvotes]: A firm's average cost function is: $$AC = x^2 - 2x + 50 , \ 0 \leq x$$
a) Find the level of output to minimalize average cost (AC) function
b) Find the level of output when average cost equals marginal cost (AC=MC)
Could you please check my calculations ? I have done a) and a little b):
a)$$AC = x^2 - 2x + 50$$
$$f'(x)=2x-2=0$$
$$x-1=0$$ $$x=1$$
$$f(1) = 1 - 2 + 50 = 49$$
b) This one I'm totally not sure
$$TC=AC·x$$
$$TC=x^3 - 2x^2 + 50x$$
$$MC=TC'=3x^2-4x+50$$
$$AC=MC$$
$$x^2 - 2x + 50=3x+2-4x+50$$
$$x_1=1$$$$x_2=0$$

REPLY [1 votes]: The average cost function is the parabola for $x\ge 0$
$$\operatorname{AC}(x)=(x-1)^2+49$$
which has the minimum at the point $P=(1,49)$.
The total cost function is $\operatorname{TC}(x)=\operatorname{AC}(x)\cdot x$ and then differentiating
$$
\underbrace{\operatorname{TC}'(x)}_{\operatorname{MC}(x)}=\operatorname{AC}'(x)\cdot x+\operatorname{AC}(x)
$$
and using $\operatorname{MC}(x)=\operatorname{AC}(x)$ and $\operatorname{AC}'(x)=2(x-1)$ we have $$
0=\operatorname{AC}'(x)\cdot x=2(x-1)x
$$
that is $x=0$ or $x=1$.


TITLE: Vertices, Edges and Line Segment intersection points
QUESTION [0 upvotes]: So, I have a bunch of graph edges defined by start and end vertices
i.e. edge = (startVertex, EndVertex).

No coordinates i.e x or y points provided.
How do I find the corresponding line segments and then discover if the resulting segments intersect or not?

REPLY [1 votes]: You can't.  Abstractly, it doesn't make sense to ask about which edges interesect, or even to ask if any of them do intersect.  You can really only ask this for a geometric realization of the graph, such as a drawing in the plane, or three space.  But for an abstract ( finite ) graph, you can always embed it in threespace with no interesections.  it's not always possible to embed a graph in the plane with no intersections, but it is always possible to avoid interesecting any given pair of edges, so even if you know the graph is embedded in the plane, without knowing something about the embedding you can't say if two edges intersect.