Lipschitz isomorphism and fixed point theorem for normed groups

Abstract

This paper aims to propose normed structures for groups and to establish the Lipschitz mapping of a normed group $G$ to itself. We also investigate some conjugate and isomorphic Lipschitz mappings to determine the equivalent norm and inverse Lipschitz mappings. Specifically, in the main result, we present a fixed point theorem for self-mappings satisfying certain contraction principles on a complete normed group.

Keywords:

• Banach group
• generalized Lipschitz conditions
• fixed point theorems
• Klee property
• expansive mapping

PUBLIC INTEREST STATEMENT

In mathematical analysis, Lipschitz continuity (named after Rudolf Lipschitz) is the Picard–Lindelöf theorem’s central condition. A particular type of Lipschitz continuity, called contraction, is used in the Banach fixed-point theorem (known as the contraction mapping theorem) is an essential tool that ensures the existence and uniqueness of fixed points of certain self-mappings and provides a constructive method to find those fixed points. From previous studies, it has been shown that Banach fixed-point theorem is a useful mechanism for numerous branches of mathematical analysis, general topology, functional analysis, and economics. One of the most known applications of Banach’s fixed point theorem for economists is Bellman’s functional equations.

This article mainly focused on understanding the general properties of Lipschitz self-mappings and its connection with the existence of isomorphic group-norm. We have investigated Banach fixed point theorem for normed groups to determine the existence of unique fixed points following specific conditions.

1. Introduction and preliminaries

Normed groups are just groups associated with a right-invariant metric. The right-invariant metric performs an important role in the construction of group-norm. The fundamental metrization theorem for groups presented by Birkhoff (Birkhoff, 1936) and Kakutani (Kakutani, 1936) is usually stated that a first-countable Hausdorff group has a right-invariant metric. Klee (Klee, 1952) introduced invariant metrics in groups, which provided a different way to define a group-norm associated with the invariant metrics.

When $G$ is an abelian group and also topologically complete, then it admits a metric that is both right and left-invariant (bi-invariant). See, for instance, (Bingham & Ostaszewski, 2010; Klee, 1952). Bingham et al. (Bingham & Ostaszewski, 2010) also made contributions for the equivalent representation of a group-norm as a right $(d_R)$ and left $(d_L)$ invariant metric. The reader may also consult (Batagelj, 1995; Bökamp, 1994; Nourouzi & Pourmoslemi, 2017; Sarfraz & Li, 2019; Zorzitto, 1985) to understand the notion of $group-norm$ and its association with right and left-invariant metric. This metric enhances the group structure, and accordingly, the group becomes a metric space with a norm-like function.

The subsequent proposition is well recognized (see (Bingham & Ostaszewski, 2010)).

Proposition 1.1 If $\parallel \cdot \parallel$ is a group-norm, then $d(g,h):=\parallel g{h}^{-1}\parallel$ is a right-invariant metric; equivalently, $d_1(g^{-1},h^{-1}):=\parallel g^{-1}h\parallel$ is the conjugate left-invariant metric on the group. Conversely, if d is a right-invariant metric, then $\parallel g\parallel :=d(e,g)=d_1(e,g)$ is a group-norm.

We say that group $G$ is normed if it has a group-norm as defined below.

Definition 1.2 (Batagelj, 1995; Bingham & Ostaszewski, 2010; Sarfraz & Li, 2019) Let $(G,e,\ast )$ be a group under binary operation $\ast$ with identity element $e$, then its norm $\parallel \cdot \parallel :G\to [0,+\mathrm{\infty })$ is said to be a group-norm if the subsequent properties hold for every $g_1,g_2\in G$,

(1) $\parallel g_1\ast g_2\parallel \le \parallel g_1\parallel +\parallel g_2\parallel$;

(2) $\parallel g_1\parallel \ge 0$, with $\parallel g_1\parallel =0$ if and only if $g_1=e$;

(3) $\parallel g_1^{-1}\parallel =\parallel g_1\parallel$.

If property (1) holds, then it is said to be a semi-norm; if (1) and (2) satisfies with $\parallel e\parallel =0$ then one speaks of a pseudo-norm; if properties (1) and (2) holds, it is known as a pre-norm. A normed group having group-norm $\parallel \cdot \parallel$ and group $(G,e,\ast )$ with identity element $e$ is denoted by $(G,\parallel \cdot \parallel ,e,\ast )$. Moreover, if $G$ is complete for group-norm $\parallel \cdot \parallel$, then we say that $(G,\parallel \cdot \parallel )$ is a complete normed group.

It follows from the following theorem that the cardinal norm of a generated group $G$ (generated by set $S$) induces a metric given by $d_c(g_1,g_2)=\parallel g_1^{-1}{g}_2\parallel$ for every $g_1,g_2\in G$, called the cardinal metric, and so $G$ becomes a metric space. See, for instance, (Suksumran, 2019).

Theorem 1.3. (see (Suksumran, 2019), Theorem 2.1) Assume that $(G,S)$ be a generated group. The cardinal norm induced by $S$ is a group-norm; that is, it satisfies the group-norm properties exhibited in Definition 1.2.

Fixed point theory serves as an essential mechanism for numerous branches of mathematical analysis, general topology, functional analysis, and its applications. Loosely speaking, there are three main approaches in this theory: the topological, the metric, and the order-theoretic approach, where illustrative examples of these are Brouwer’s, Banach’s (Banach, 1922) and Tarski’s theorems, respectively.

The Banach fixed point theorem is also recognized as contraction mapping theorem, and it is a very useful to show the existence of a unique solution to functional and differential equations. One of the most known applications of Banach’s fixed point theorem for economists is Bellman’s functional equations. For comprehensive analyses and studies concerning Banach contraction principle, readers see (Jain et al., 2012; Jleli and Samet ((2014)); Shatanawi & Nashine, 2012).

In this paper, the fixed point theorem plays a significant role to determine the fixed points of $\eta$ in Banach group $G$. This study’s main purpose is to investigate the existence of a fixed point for self-mappings $\eta$ defined on a complete normed group (Banach group) $G$ that satisfies certain conditions. We recommend the readers to see (Debnath, 2020; Debnath et al., 2016, 2020; Debnath & Sen, 2020; Debnath & Srivastava, 2020, 2020; Merryfield et al., 2002; Neog et al., 2019; Pata, 2011) and the references cited therein to obtain comprehensive results relating to the fixed points. Fixed point theorems concern maps $\eta$ of a set $G$ into itself that, under certain conditions, admit a fixed point $g^{\ast }\in G$ such that $\eta (g^{\ast })=g^{\ast }$.

Apart from this introduction, the paper is arranged into two sections. In Section 2, we determine the Lipschitz mapping $\eta$ of a normed group $G$ (may or may not be complete) to itself and also introduce some isomorphic and continuous inverse Lipschitz mappings. This section also describes an equivalent norm and the construction of a conjugate Lipschitz map from a normed group $G$ to itself.

Section 3 deals with employing the fixed point theorem to determine the unique fixed points of $\eta$ from a Banach group $G$ to itself. Moreover, we also find the fixed points of $\eta$ from a normed cyclic group $G$ to itself whenever $\eta$ is an expansive mapping.

In our considerations the following definition will play an important role.

Definition 1.4. (Debnath & Sen, 2020; Hussain et al., 2015) Assume that G be a normed group defined over a norm $\parallel \cdot \parallel$. A mapping $\eta :G\to G$ is said to be a Lipschitz continuous if there exists $\gamma \ge 0$ such that

$\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel \le \gamma \parallel g_1{g}_2^{-1}\parallel forany{\mathrm{g}}_1,{\mathrm{g}}_2\in \mathrm{G},$

whenever $g_1\ne g_2$. The smallest value of $\gamma$ for which the above inequality holds is said to be a Lipschitz constant of $\eta$. If $\gamma =1$, then $\eta$ is termed as non-expansive mapping; if $\gamma <1$, then $\eta$ is said to be a contraction.

Definition 1.5. (Bingham & Ostaszewski, 2010; Sarfraz & Li, 2019) A group-norm $\parallel \cdot \parallel :G\to [0,+\mathrm{\infty })$ is stated as the abelian group-norm if

$\parallel g_1{g}_2\parallel =\parallel g_2{g}_1\parallel$ for all $g_1,g_2\in G$.

Thus the metric is bi-invariant if and only if the group-norm is abelian.

Subadditivity intimates that $\parallel e\parallel \ge 0$, and this collectively with symmetry implies that $\parallel g\parallel \ge 0$, since $\parallel e\parallel =\parallel g{g}^{-1}\parallel \le 2\parallel g\parallel$. Thus, $\parallel g\parallel \ge 0$ for any $g\in G$. Subadditivity also refers that $\parallel g^n\parallel \le n\parallel g\parallel$ for every $n\in \mathbb{N}$.

2. Lipschitz mapping and Banach groups

Assume that $(G,\parallel \cdot \parallel )$ be a normed group, we say it a complete normed group or Banach group if for any $g\in G$ with $\parallel g_m{g}_n^{-1}\parallel \to 0$ when $m,n\to \mathrm{\infty }$, then there exists $g_0\in G$ such that $\parallel g_n{g}_0^{-1}\parallel \to 0$.

Theorem 2.1. Suppose that $(G,\parallel \cdot \parallel )$ be a normed finite group, then $(G,\parallel \cdot \parallel )$ is a Banach group.

Proof. Let $\{g_n\}$ be a Cauchy sequence in $G$, noted $\epsilon =min\{\parallel g_1{g}_2^{-1}\parallel |g_1,g_2\in G,g_1\ne g_2\}$, since $G$ is a finite group, we can get $\epsilon >0$ and also take ${\epsilon }_0=\frac{\epsilon }{2}>0$, then there exists $N$ such that for any $m>n>N$, we have $\parallel g_m{g}_n^{-1}\parallel <{\epsilon }_0$. Thus $\parallel g_m{g}_n^{-1}\parallel =0$ for any $m>n>N$. Hence $\{g_n\}$ converges to $g_N$, therefore, $(G,\parallel \cdot \parallel )$ is a complete normed group.

The following corollary concerning the finite subgroup can be easily proved:

Corollary 2.2. Let $(G,\parallel \cdot \parallel )$ be a normed group and $H$ is a finite subgroup of $G$, then $(H,\parallel \cdot \parallel )$ is a closed subgroup of $(G,\parallel \cdot \parallel )$.

When we consider non-trivial cyclic or non-cyclic normed subgroup $G$, then we have following theorem:

Theorem 2.3. Assume that $(G,\parallel \cdot \parallel ,+)$ be a non-trivial normed subgroup of an additive normed group $(\mathbb{R},\parallel \cdot \parallel ,+)$. Then

(a) $G$ is non-cyclic and dense in $\mathbb{R}$ if and only if $G$ does not contain minimal positive element.

(b) $G$ is cyclic and closed in $\mathbb{R}$ if and only if $G$ contains minimal positive element.

Proof. Define a norm $\parallel \cdot \parallel :\mathbb{R}\to [0,+\mathrm{\infty })$ by

$\parallel x\parallel =|x|,(x\in \mathbb{R}).$

Then, it can be seen that $\mathbb{R}$ is an additive normed group. Assume that $G$ be a non-trivial normed subgroup of $\mathbb{R}$.

(a). Define $B=\{\parallel g\parallel :g\in G\mathrm{\setminus }\{0\}\}$ and $\beta =inf\{B\}$. Suppose that $G$ does not contain minimal positive element. Therefore $\beta =0$. Let $x\in \mathbb{R}$ and . Since $\beta =0$, therefore, there exists an element $g\in (0,\in )\cap G$. Also, assume that $x\ge 0$ and $n\in \mathbb{N}$, then

$ng\le x<(n+1)g,$

yielding that $ng\in G$ and $ng>0$, therefore

$0\le x-ng<(n+1)g-ng$

$0\le x-ng<g.$

Also, which implies that

$0\le x-ng<\in$.

$∥x-ng∥<\in and\in >0$.

which also infers that $x$ is within $\in$ distance from some elements $ng\in G$. Since this result holds for every $\in >0$ and also for all $x$; therefore, we can conclude $\bar{G}=\mathbb{R}$. Conversely, it is easily seen that the condition $\beta =inf\{B\}=0$ provided that $\bar{G}=\mathbb{R}$.

(b). Assume that $G$ contains minimal positive element $\beta$ under the assumption that $\beta \ne 0$. So, by division algorithm, for any $g\in G$ and for any positive integer $n$, we have

$\parallel g\parallel =n\beta +\alpha ,0\le \alpha <\beta .$

Since $G$ is a subgroup, then $\alpha =[\parallel g\parallel -n\beta ]\in G$. But $\alpha <\beta$ and $\beta$ is minimum positive element,so $\alpha =0$. Which concludes that $\parallel g\parallel =n\beta$. Hence $G=\beta \mathbb{Z}$

If possible, let $\alpha >0$ and utilize the definition of infimum, choose for every $\in >0$ there exist an element $g_1\in G$ such that $-ng\in (g_1,g_1+\in )$.Take $\in =\beta +\alpha$

$-\beta =\alpha -\in <n\beta +g_1+\alpha$.

$-\beta =\parallel g\parallel +g_1+\alpha .$

Since $(\parallel g\parallel +g_1)<0$ and $(\parallel g\parallel +g_1)\in G$, therefore,

$-\beta =\parallel g\parallel +g_1<\alpha <\beta$

$\parallel \parallel g\parallel +g_1\parallel <\beta ,$

which is a contradiction because $\beta$ is the minimal positive element. Thus, $\alpha =0$ and $G=\beta \mathbb{Z}$ is closed and cyclic. Conversely, it is also obvious that when $G$ is cyclic and closed, then there exists positive minimal element $\beta \ne 0$.

Example 2.4. The group of real numbers $\mathbb{R}$ (complex numbers) is a complete normed group with respect to the group-norm $|\cdot |$.

The notion of bi-Lipschitz equivalent inspires us to derive the following result:

Theorem 2.5. Assume that $(G,\parallel .\parallel )$ be a normed group and

(a) $\eta :G\to G$ is bi-Lipschitz equivalent, then

(i) $\eta$ is bijective mapping.

(ii) ${\eta }^{-1}$ is also Lipschitz mapping.

(b) Every isomorphic mapping $\eta :G\to G$ is the bi-Lipschitz equivalent, but the converse may or may not be true.

Proof. (a). Suppose $\eta$ is bi-Lipschitz equivalent, then for any ${\gamma }_1,{\gamma }_2\in {\mathbb{R}}_{+}$ and $g_1,g_2\in G$ we have

${\gamma }_1\parallel g_1{g}_2^{-1}\parallel \le \parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel \le {\gamma }_2\parallel g_1{g}_2^{-1}\parallel .$

Let $\eta (g_1)=\eta (g_2)$ then we can evaluate

$\begin{array}{cc}\hspace{0.17em}& {\gamma }_1\parallel g_1{g}_2^{-1}\parallel \le \parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \\ \iff & {\gamma }_1\parallel g_1{g}_2^{-1}\parallel =0\\ \iff & g_1=g_2,\end{array}$

which concludes that $\eta$ is injective. Also, $\eta$ is surjective. Therefore, ${\eta }^{-1}$ exists. Consequently, for any $g_1,g_2\in G$, the following computation shows that ${\eta }^{-1}$ is $\frac{1}{{\gamma }_1}$-Lipschitz:

$\begin{array}{rl}& {\gamma }_1\parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2){]}^{-1}\parallel \le \parallel \eta ({\eta }^{-1}(g_1))[\eta ({\eta }^{-1}(g_2)){]}^{-1}\parallel \\ & {\gamma }_1\parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2){]}^{-1}\parallel \le {\gamma }_2\parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2){]}^{-1}\parallel \\ & {\gamma }_1\parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2){]}^{-1}\parallel \le \parallel g_1{g}_2^{-1}\parallel \\ & \parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2){]}^{-1}\parallel \le \frac{1}{{\gamma }_1}\parallel g_1{g}_2^{-1}\parallel .\end{array}$

(b). Since $\eta$ is isomorphic, then for every $g_1\in G$, we have $\parallel g_1\parallel \le \parallel \eta (g_1)\parallel \le \parallel g_1\parallel$. Then for any $g_1,g_2\in G$, we can determine a bi-Lipschitz equivalent mapping as

$\begin{array}{cc}\parallel g_1{g}_2^{-1}\parallel & \le \parallel \eta (g_1{g}_2^{-1})\parallel \le \parallel g_1{g}_2^{-1}\parallel \\ \parallel g_1{g}_2^{-1}\parallel & \le \parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \le \parallel g_1{g}_2^{-1}\parallel ,\end{array}$

where ${\gamma }_1={\gamma }_2=1$. Conversely, bi-Lipschitz mapping $\eta$ can only be isomorphic for particular values ${\gamma }_1={\gamma }_2=1$ and can not be isomorphic for ${\gamma }_1,{\gamma }_2>1$.

The following corollary, which admittedly is unmotivated here, will be needed later.

Corollary 2.6. Let $(G,\parallel .\parallel )$ be a normed group, then every isomorphic mapping $\eta :G\to G$ is 1-Lipschitz.

Definition 2.7. (Bingham & Ostaszewski, 2010) Let $(G,\parallel .\parallel )$ be a normed group. For some $g\in G$, there is self-mapping ${\eta }_g:G\to G$; then g-conjugate norm is defined as

$\parallel x{\parallel }_g=\parallel {\eta }_g(x)\parallel =\parallel gx{g}^{-1}\parallel forallx\in G.$

Moreover, Left-shift norm ($L_g-norm$) and Right-shift norm ($R_g-norm$) can also be defined as

$\parallel L_g(x)\parallel =\parallel gx\parallel ,\parallel R_g(x)\parallel =\parallel xg\parallel forallx\in G.$

Definition 2.8. (Klee, 1952) Let $(G,\parallel .\parallel )$ be a normed group, then Klee property of a normed group $G$ is defined as

$\parallel g_{1}xy{g}_2\parallel \le \parallel g_1{g}_2\parallel +\parallel xy\parallel forall{g}_1,g_2,x,y\in G.$

If a normed group $G$ is abelian, then we can see that it satisfies the Klee property.

Definition 2.9. (Bingham & Ostaszewski, 2010) Let $(G,\parallel .\parallel )$ be a normed group and for some $g\in G$, Lipschitz constant for g-conjugacy (${\gamma }_g$), Left-shift ($L({\gamma }_g)$) and Right-shift ($R({\gamma }_g)$) of a normed group $G$, can be defined as

${\gamma }_g=\underset{x\ne e}{sup}\frac{\parallel {\eta }_g(x)\parallel }{\parallel x\parallel },li{p}_L({\gamma }_g)=\underset{x\ne e}{sup}\frac{\parallel gx\parallel }{\parallel x\parallel },li{p}_R({\gamma }_g)=\underset{x\ne e}{sup}\frac{\parallel xg\parallel }{\parallel x\parallel }.$

The following theorem gives a complete description of the conjugate mapping and Klee property (Klee, 1952) on abelian normed group.

Theorem 2.10. Let $(G,\parallel \cdot \parallel )$ be a normed group. A conjugacy self-mapping ${\eta }_g:G\to G$ is 1-Lipschitz if and only if $G$ is an abelian normed group.

Proof. For some $g\in G$, let us define g-conjugacy self-mapping as ${\eta }_g(x)=gx{g}^{-1}$, for all $x\in G$. Assume that G is an abelian normed group. Since $G$ is abelian then $Klee$ property holds, then g-conjugate norm yields that

$\begin{array}{cc}\parallel {\eta }_g(x)\parallel & =\parallel gx{g}^{-1}\parallel =\parallel gxe{g}^{-1}\parallel \\ \parallel {\eta }_g(x)\parallel & \le \parallel g{g}^{-1}\parallel +\parallel xe\parallel \\ \parallel {\eta }_g(x)\parallel & \le \parallel x\parallel .\end{array}$

Hence ${\eta }_g$ is a 1-Lipschitz.

Conversely, suppose that ${\eta }_g$ is a 1-Lipschitz, which concludes that ${\gamma }_g=1$ for any $g\in G$, then we can compute

(1) $\parallel gx{g}^{-1}\parallel \le \parallel x\parallel .$(1)

Since ${\eta }_g$ is a 1-Lipschitz, so applying (1), we have

$\parallel gx\parallel =\parallel g(xg)g^{-1}\parallel \le \parallel xg\parallel$

(2) $\parallel gx\parallel \le \parallel xg\parallel .$(2)

Since ${\eta }_{g^{-1}}$ also exists and is 1-Lipschitz, which intimates that ${\gamma }_{g^{-1}}=1$ for any $g\in G$, therefore, we can deduce

$\parallel xg\parallel =\parallel g^{-1}(gx)g\parallel \le \parallel gx\parallel$

(3) $\parallel xg\parallel \le \parallel gx\parallel .$(3)

Inequalities (2) and (3) yields that $\parallel xg\parallel =\parallel gx\parallel$ for any $x,g\in G$. Hence, $G$ is an abelian normed group.

In the proof of Theorem 2.13, we use the following technical result about conjugate ${\gamma }_g$-Lipschitz and equivalent group-norm satisfying certain conditions.

Theorem 2.11. Let $(G,\parallel .\parallel )$ be a normed group and ${\eta }_g:G\to G$ be a conjugacy ${\gamma }_g$-Lipschitz, then

(a) ${\eta }_g^{-1}={\eta }_{g^{-1}}$ also exists and is ${\gamma }_{g^{-1}}$-Lipschitz.

(b) For ${\gamma }_{g^{-1}}={\gamma }_g=1$, ${\eta }_g:G\to G$ has a isomorphic norm.

(c) ${\gamma }_e=1$ and ${\gamma }_g\ge 1$ for any $g\in G$.

(d) ${\gamma }_{g_1{g}_2}\le {\gamma }_{g_1}{\gamma }_{g_2}$ for $g_1,g_2\in G$.

(e) Conjugate norm is an equivalent as $\frac{1}{{\gamma }_{g^{-1}}}\parallel x\parallel \le \parallel {\eta }_g(x)\parallel \le {\gamma }_g\parallel x\parallel$.

Proof. (a). From definition of conjugacy norm and also ${\eta }_g$ is a ${\gamma }_g$-Lipschitz, we can deduce $\parallel {\eta }_g(x)\parallel =\parallel gx{g}^{-1}\parallel \le {\gamma }_g\parallel x\parallel$. It is obvious that the mapping is surjective evenly for every $x\in G$, there exists $g^{-1}xg\in G$ such that

${\eta }_g(g^{-1}xg)=g{g}^{-1}xg{g}^{-1}=x.$

Then there exists an inverse mapping of ${\eta }_g$ indicated as ${\eta }_g^{-1}={\eta }_{g^{-1}}$ such that ${\eta }_{g^{-1}}(x)=g^{-1}xg$, so we have

$\parallel {\eta }_{g^{-1}}(x_1)[{\eta }_{g^{-1}}(x_2){]}^{-1}\parallel =\parallel g^{-1}{x}_{1}g[g^{-1}{x}_{2}g{]}^{-1}\parallel$

$\parallel {\eta }_{g^{-1}}(x_1)[{\eta }_{g^{-1}}(x_2){]}^{-1}\parallel =\parallel g^{-1}{x}_1{x}_2^{-1}g\parallel$

$\parallel {\eta }_{g^{-1}}(x_1)[{\eta }_{g^{-1}}(x_2){]}^{-1}\parallel \le {\gamma }_{g^{-1}}\parallel x_1{x}_2^{-1}\parallel ,$

which signifies that inverse mapping is also ${\gamma }_{g^{-1}}$-Lipschitz.

(b). Suppose that ${\gamma }_g={\gamma }_{g^{-1}}=1$ for every $g\in G$. when ${\gamma }_g=1$, then we can obtain

(4) $\parallel gx{g}^{-1}\parallel \le \parallel x\parallel \mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{y}g,x\in G.$(4)

Also ${\gamma }_{g^{-1}}=1$, then we can compute

$\parallel x\parallel =\parallel g^{-1}(gx{g}^{-1})g\parallel$

$\le {\gamma }_{g^{-1}}\parallel gx{g}^{-1}\parallel$

(5) $\parallel x\parallel \le \parallel gx{g}^{-1}\parallel .$(5)

Inequalities (4) and (5) yields that $\parallel gx{g}^{-1}\parallel =\parallel x\parallel$, which implies that $\parallel {\eta }_g(x)\parallel =\parallel x\parallel$ for all $x\in G$. Moreover, mapping ${\eta }_g$ is also homomorphic and bijective; hence, a normed group G has an isomorphic group norm.

(c). It is obvious that ${\gamma }_e=1$ whenever $g=e$. Let $g\ne e$, then we can get

${\eta }_g(g)=gg{g}^{-1}=g.$

$\parallel g\parallel =\parallel {\eta }_g(g)\parallel \le {\gamma }_g\parallel g\parallel .$

Then $\parallel g\parallel \le {\gamma }_g\parallel g\parallel$, which implies that ${\gamma }_g\ge 1$.

(d). For $g_1,g_2\in G$, we compute that

$\parallel {\eta }_{g_1{g}_2}(x)\parallel =\parallel g_1{g}_{2}x(g_1{g}_2{)}^{-1}\parallel =\parallel g_1(g_{2}x{g}_2^{-1})g_1^{-1}\parallel$

$\parallel {\eta }_{g_1{g}_2}(x)\parallel \le {\gamma }_{g_1}\parallel g_{2}x{g}_2^{-1}\parallel$

$\parallel {\eta }_{g_1{g}_2}(x)\parallel \le {\gamma }_{g_1}\parallel g_{2}x{g}_2^{-1}\parallel$

$\parallel {\eta }_{g_1{g}_2}(x)\parallel \le {\gamma }_{g_1}{\gamma }_{g_2}\parallel x\parallel .$

Since ${\gamma }_{g_1},{\gamma }_{g_2}\ge 1$ whenever $x\ne e$, then we can get ${\gamma }_{g_1{g}_2}\le {\gamma }_{g_1}{\gamma }_{g_2}$ for any $g_1,g_2\in G$.

(e). As we know that

(6) $\parallel {\eta }_g(x)\parallel \le {\gamma }_g\parallel x\parallel ,$(6)

then from part (a) ${\eta }_{g^{-1}}$ also exists and is ${\gamma }_{g^{-1}}$-Lipschitz. Then, for all $x\in G$, we have

$\parallel {\eta }_{g^{-1}}(x)\parallel \le {\gamma }_{g^{-1}}\parallel x\parallel .$

Also,

$\parallel {\eta }_{g^{-1}}({\eta }_g(x))\parallel \le {\gamma }_{g^{-1}}\parallel {\eta }_g(x)\parallel$

$\parallel x\parallel \le {\gamma }_{g^{-1}}\parallel {\eta }_g(x)\parallel$

(7) $\frac{1}{{\gamma }_{g^{-1}}}\parallel x\parallel \le \parallel {\eta }_g(x)\parallel .$(7)

From inequalities (6) and (7), we conclude that conjugate norm is an equivalent because $\frac{1}{{\gamma }_{g^{-1}}}\parallel x\parallel \le \parallel {\eta }_g(x)\parallel \le {\gamma }_g\parallel x\parallel$.

Corollary 2.12. A group $G$ is an abelian normed group if and only if ${\gamma }_g=1$ for all $g\in G$.

In the following results, left-shift $(L_g)$ and left-shift $(R_g)$ mappings plays an important role in the existence of equivalent norms and 1-Lipschitz mappings.

Theorem 2.13. Let $(G,\parallel .\parallel )$ be a normed group. Also $L_g$, $R_g$, and ${\eta }_g$ be self-mappings from $G$ to $G$ and also $li{p}_L({\gamma }_g)$, $li{p}_R({\gamma }_g)$, ${\gamma }_g$-Lipschitz respectively, then

(a) $L_g$ under left-invariant metric $d_L$ is 1-Lipschitz and is ${\gamma }_g$-Lipschitz concerning $d_R$.

(b) $R_g$ under right-invariant metric $d_R$ is 1-Lipschitz and is ${\gamma }_{g^{-1}}$-Lipschitz concerning $d_L$.

(c) $li{p}_L({\gamma }_e)=li{p}_R({\gamma }_e)=1$ and $li{p}_L({\gamma }_g)li{p}_L({\gamma }_{g^{-1}})\ge 1$, $li{p}_R({\gamma }_g)li{p}_R({\gamma }_{g^{-1}})\ge 1$ for any $g\in G$.

(d) $li{p}_L({\gamma }_{g_1{g}_2})\le li{p}_L({\gamma }_{g_1})li{p}_L({\gamma }_{g_2})$ and $li{p}_R({\gamma }_{g_1{g}_2})\le li{p}_R({\gamma }_{g_1})li{p}_R({\gamma }_{g_2})$ for all $g_1,g_2\in G$.

(e) If $li{p}_L({\gamma }_g)=li{p}_R({\gamma }_g)=1$ for all $g\in G$, then $G$ is an abelian normed group.

(f) $L_g$ and $R_g$ have equivalent norms.

Proof. (a). By left-invariant metric $d_L(x_1,x_2)=\parallel x_1^{-1}{x}_2\parallel$ for $x_1,x_2\in G$, then proof of (a) consists of the following simple computation:

$\parallel [L_g(x_1){]}^{-1}{L}_g(x_2)\parallel =d_L(L_g(x_1),L_g(x_2))$

$=\parallel (g{x}_1{)}^{-1}g{x}_2\parallel$

$=\parallel x_1^{-1}{g}^{-1}g{x}_2\parallel$

$=\parallel x_1^{-1}{x}_2\parallel$

$\parallel [L_g(x_1){]}^{-1}{L}_g(x_2)\parallel =\parallel x_1^{-1}{x}_2\parallel .$

Also, by right-invariant metric $d_R(x_1,x_2)=\parallel x_1{x}_2^{-1}\parallel$ for $x_1,x_2\in G$, we can get

$\parallel L_g(x_1)[L_g(x_2){]}^{-1}\parallel =d_R(L_g(x_1),L_g(x_2))$

$=\parallel (g{x}_1)(g{x}_2{)}^{-1}\parallel$

$=\parallel g{x}_1{x}_2^{-1}{g}^{-1}\parallel$

$\le {\gamma }_g\parallel x_1{x}_2^{-1}\parallel$

$\parallel L_g(x_1)[L_g(x_2){]}^{-1}\parallel \le {\gamma }_g\parallel x_1{x}_2^{-1}\parallel ,$

which is required proof.

(b). Assume that $x_1,x_2\in G$, then required proof consists of the following simple computation:

$\parallel R_g(x_1)[R_g(x_2){]}^{-1}\parallel =d_R(R_g(x_1),R_g(x_2))$

$=\parallel (x_{1}g)(x_{2}g{)}^{-1}\parallel$

$=\parallel x_{1}g{g}^{-1}{x}_2^{-1}\parallel$

$=\parallel x_1{x}_2^{-1}\parallel$

$\parallel R_g(x_1)[R_g(x_2){]}^{-1}\parallel =\parallel x_1{x}_2^{-1}\parallel .$

Also,

$\parallel [R_g(x_1){]}^{-1}{R}_g(x_2)\parallel =d_L(R_g(x_1),R_g(x_2))$

$=\parallel (x_{1}g{)}^{-1}(x_{2}g)\parallel$

$=\parallel g^{-1}{x}_1^{-1}{x}_{2}g\parallel$

$\le {\gamma }_{g^{-1}}\parallel x_1^{-1}{x}_2\parallel$

$\parallel [R_g(x_1){]}^{-1}{R}_g(x_2)\parallel \le {\gamma }_{g^{-1}}\parallel x_1^{-1}{x}_2\parallel ,$

as required.

(c). Let $g=e$, then we can conclude that $li{p}_L({\gamma }_e)=1=li{p}_R({\gamma }_e)$. Assume that $g\ne e$, then we can compute

$\parallel x\parallel =\parallel g(g^{-1}x)\parallel$

$\le li{p}_L({\gamma }_g)\parallel g^{-1}x\parallel$

$\parallel x\parallel \le li{p}_L({\gamma }_g)li{p}_L({\gamma }_{g^{-1}})\parallel x\parallel .$

Then $\parallel x\parallel \le li{p}_L({\gamma }_g)li{p}_L({\gamma }_{g^{-1}})\parallel x\parallel$, which implies that $li{p}_L({\gamma }_g)li{p}_L({\gamma }_{g^{-1}})\ge 1$. Similarly, for right shift ($L_g$) mapping we can also calculate $li{p}_R({\gamma }_g)li{p}_R({\gamma }_{g^{-1}})\ge 1$.

(d). Let $g_1,g_2,x\in G$, then

$\parallel L_{g_1{g}_2}(x)\parallel =\parallel g_1(g_{2}x)\parallel$

$\parallel L_{g_1{g}_2}(x)\parallel \le li{p}_L({\gamma }_{g_1})\parallel g_{2}x\parallel$

$\parallel L_{g_1{g}_2}(x)\parallel \le li{p}_L({\gamma }_{g_1})li{p}_L({\gamma }_{g_2})\parallel x\parallel .$

By taking suprimum whenever $x\ne e$, we can obtain that $li{p}_L({\gamma }_{g_1{g}_2})\le li{p}_L({\gamma }_{g_1})li{p}_L({\gamma }_{g_2})$ for $g_1,g_2\in G$. The conclusion is easily seen to hold $li{p}_R({\gamma }_{g_1{g}_2})\le li{p}_R({\gamma }_{g_1})li{p}_R({\gamma }_{g_2})$ for right shift mapping ($R_g$).

(e). Assume that $li{p}_L({\gamma }_g)=li{p}_R({\gamma }_g)=1$ for all $g\in G$. Then, it is obvious that $L_g$ and $R_g$ are 1-Lipschitz. Consider $\parallel gx{g}^{-1}\parallel$ and using $L_g$ and $R_g$ as a 1-Lipschitz, we have

$\parallel gx{g}^{-1}\parallel \le li{p}_L({\gamma }_g)\parallel x{g}^{-1}\parallel$

$\le li{p}_L({\gamma }_g)li{p}_R({\gamma }_{g^{-1}})\parallel x\parallel$

$\parallel gx{g}^{-1}\parallel \le \parallel x\parallel ,$

which holds for all $g\in G$, and implies that ${\eta }_g$ is a 1-Lipschitz. Using the theorem that a conjugacy self-mapping ${\eta }_g:G\to G$ is 1-Lipschitz, then $G$ is an abelian normed group. Hence, $G$ is an abelian normed group.

(f). As we know that for some fix element $g\in G$ and for every $x\in G$, we have

(8) $\parallel L_g(x)\parallel \le li{p}_L({\gamma }_g)\parallel x\parallel .$(8)

Also, it can be seen that left shit $L_g$ is surjective and inverse mapping $L_g^{-1}=L_{g^{-1}}$ also exists, and is $li{p}_L({\gamma }_{g^{-1}})$-Lipschitz, then we have

$\parallel L_{g^{-1}}(x)\parallel \le li{p}_L({\gamma }_{g^{-1}})\parallel x\parallel .$

Also,

$\parallel L_{g^{-1}}(L_g(x))\parallel \le li{p}_L({\gamma }_{g^{-1}})\parallel L_g(x)\parallel$

$\parallel x\parallel \le li{p}_L({\gamma }_{g^{-1}})\parallel L_g(x)\parallel$

(9) $\frac{1}{li{p}_L({\gamma }_{g^{-1}})}\parallel x\parallel \le \parallel L_g(x)\parallel .$(9)

Inequalities (8) and (9) implies that $L_g$ norm is an equivalent because $\frac{1}{li{p}_L({\gamma }_{g^{-1}})}\parallel x\parallel \le \parallel {\eta }_g(x)\parallel \le li{p}_L({\gamma }_g)\parallel x\parallel$. Similarly, the $R_g$ norm also has an equivalent norm.

The following example gives a complete description of the conjugate mappings on additive normed group.

Example 2.14. Let $(G,\parallel \cdot \parallel ,+)$ be a normed group of Complex or Real numbers. Then for some $g\in G$ all mappings ${\eta }_g:G\to G$ are 1-Lipschitz.

Using continuous mappings, we prove the following theorem for isomorphic norm.

Theorem 2.15. A continuous self-mapping on a normed finite group $G$ has an isomorphic norm.

Proof. Assume that $\eta :G\to G$ be a continuous mapping. Since $\eta$ is continuous and $G$ is a normed finite group, then there exists a finite $n\in \mathbb{N}$ such that ${\eta }^n=$identity mapping, as ${\eta }^n(g)=g$ for all $g\in G$. So, we have an automorphism mapping ${\eta }^n:G\to G$ satisfies ${\eta }^n(g)=g$ for all $g\in G$. Therefore, $\parallel {\eta }^n(g)\parallel =\parallel g\parallel$ holds for all $g\in G$, and hence group norm is isomorphic.

Corollary 2.16. For a normed group $G$, isomorphic self-mapping is 1-Lipschitz.

Proof. Assume that $\eta :G\to G$ is an isomorphic self-mapping, then $\parallel \eta (g)\parallel =\parallel g\parallel$ holds for all $g\in G$. Consequently, $\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel =\parallel \eta (g_1)\eta (g_2^{-1})\parallel =\parallel \eta (g_1{g}_2^{-1})\parallel =\parallel g_1{g}_2^{-1}\parallel$. Hence this self-mapping is 1-Lipschitz.

3. The fixed point theorem in Banach groups

A contraction mapping theorem is an important tool in the theory of normed groups; it guarantees the existence and uniqueness of fixed points of certain self-maps and provides a constructive method to find those fixed points. The following example gives the completion of a normed group $G$.

Example 3.1. Let $G$ be a Banach space (it can be seen that a Banach space is an additive commutative group). Then $\parallel g\parallel =d(e,g)$ defines a group-norm on $G$ and is also complete normed group.

Our concern is to find sufficient conditions on normed group $G$ and self-mapping $\eta$ to ensure a fixed point of $\eta$ in G. We are also interested in uniqueness and procedures for the calculation of fixed points.

For this purpose, the first main result is giving by the following theorem.

Theorem 3.2. Let $\eta$ be a contraction on a complete normed group $G$. Then $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

Proof. Notice first that if $g_1,g_2\in G$ are fixed points of $\eta$, then

$\parallel g_1{g}_2^{-1}\parallel =\parallel \eta (g_1)\eta (g_2{)}^{-1}\parallel \le \gamma \parallel g_1{g}_2^{-1}\parallel ,$

which implies that $g_1=g_2$.

Choose now any $g_1\in G$, and define the iterate sequence $g_{n+1}=\eta (g_n)$. By induction on $n>1$,

$\parallel g_{n+1}{g}_n^{-1}\parallel \le {\gamma }^n\parallel \eta (g_1)g_1^{-1}\parallel .$

If $n\in N$ and $m\ge 1$,

$\parallel g_{m+n}{g}_n^{-1}\parallel =\parallel g_{m+n}{g}_{m+n-1}^{-1}{g}_{m+n-1}{g}_{m+n-2}^{-1}{g}_{m+n-2}\cdots g_{n+1}{g}_n^{-1}\parallel$

$\le \parallel g_{m+n}{g}_{m+n-1}^{-1}\parallel +\cdots +\parallel g_{n+1}{g}_n^{-1}\parallel$

$\le ({\gamma }^{m+n}+{\gamma }^{m+n-1}+\cdots +{\gamma }^n)\parallel \eta (g_1)g_1^{-1}\parallel$

$\parallel g_{m+n}{g}_n^{-1}\parallel \le \frac{{\gamma }^n}{1-\gamma }\parallel \eta (g_1)g_1^{-1}\parallel .$

Hence $\{g_n\}$ is a Cauchy sequence and admits a limit $g_0\in G$ because $G$ is complete. Since $\eta$ is continuous, we have $\eta (g_0)=lim\eta (g_n)=lim{g}_{n+1}=g_0$. Hence $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

The following corollary is easily seen to hold if normed group $G$ is finite.

Corollary 3.3. Let $\eta$ be a contraction mapping of finite order from a normed finite group $(G,\parallel .\parallel )$ into itself. Then $\eta$ has a fixed point in $G$.

Let $\eta$ be a homomorphic mapping of a normed group $(G,\parallel .\parallel )$ into itself. Then, $\eta$ must have a fixed point $e\in G$ such that $\eta (e)=e$.

The following theorem provides sufficient conditions for the existence of inverse Lipschitz mapping and isomorphic mapping whenever we have contraction map.

Theorem 3.4. Assume that $(G,\parallel .\parallel )$ is an abelian Banach group and $\eta$ be a homomorphic ${\gamma }_{\eta }-Lipschitz$ mapping of a Banach group $G$ into itself such that ${\gamma }_{\eta }<1$. Then

(a) There exists an isomorphic $\gamma -Lipschitz$ mapping $p:G\to G$,

(b) $p^{-1}$ also exists and is $(1-{\gamma }_{\eta }{)}^{-1}$-Lipschitz,

where

${\gamma }_{\eta }=min\{\gamma \in {\mathbb{R}}_{+}:\eta is\gamma -Lipschitz\}.$

Proof. (a). For any $u\in G$, a self-mapping ${\eta }_u:G\to G$ is defined by ${\eta }_u(g)=u\eta (g)$. We proceed in a few steps. In the first step, we show that the defined mapping ${\eta }_u$ is also contraction mapping concerning $\eta$. Since $G$ is abelian and $\eta$ be a homomorphic ${\gamma }_{\eta }-Lipschitz$ then for every $g_1,g_2\in G$, we have

$\parallel {\eta }_u(g_1)[{\eta }_u(g_2){]}^{-1}\parallel =\parallel u\eta (g_1)[u\eta (g_2){]}^{-1}\parallel$

$=\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel$

$\parallel {\eta }_u(g_1)[{\eta }_u(g_2){]}^{-1}\parallel \le {\gamma }_{\eta }\parallel g_1{g}_2^{-1}\parallel ,$

which infers that ${\eta }_u$ is also a contraction mapping.

Since $G$ is a Banach group and ${\eta }_u$ is a contraction mapping, according to the contraction mapping theorem, there exists a unique element $g^{\ast }\in G$ as follows

${\eta }_u(g^{\ast })=u\eta (g^{\ast }).$

Define a mapping $p:G\to G$ by

$p(g)=g[\eta (g){]}^{-1}.$

Thus, ${\eta }_u(g^{\ast })=g^{\ast }$ if and only if $p(g^{\ast })=u$. Assume that $g_1,g_2\in G$ such that

$p(g_1)=p(g_2)$

$g_1[\eta (g_1){]}^{-1}=g_2[\eta (g_2){]}^{-1}$

$g_1{g}_2^{-1}=\eta (g_1)[\eta (g_2){]}^{-1}$

$g_1{g}_2^{-1}=\eta (g_1{g}_2^{-1}).$

Since $\eta$ is also contraction mapping, it is possible whenever $g_1{g}_2^{-1}=e$, so $g_1=g_2$. Consequently, $p$ is an injective mapping. Also, ${\eta }_u(g^{\ast })=g^{\ast }$ if and only if $p(g^{\ast })=u$, therefore, $p$ is also surjective. Next, we determine that $p$ is also a homomorphism.

$p(g_1{g}_2)=g_1{g}_2[\eta (g_1{g}_2){]}^{-1}$

$p(g_1{g}_2)=g_1{g}_2[\eta (g_2){]}^{-1}[\eta (g_1){]}^{-1}$

$p(g_1{g}_2)=g_1[\eta (g_1){]}^{-1}{g}_2[\eta (g_2){]}^{-1}$

$p(g_1{g}_2)=p(g_1)p(g_2),$

which indicates that $p$ is a homomorphism. Lastly, to check the existence of Lipschitz mapping, let $g_1,g_2\in G$ such that

$\begin{array}{cc}\parallel p(g_1)[p(g_2{)]}^{-1}\parallel & =\parallel g_1{[\eta (g_1)]}^{-1}{[g_2{[\eta (g_2)]}^{-1}]}^{-1}\parallel \\ \hspace{0.17em}& =\parallel g_1{[\eta (g_1)]}^{-1}\eta (g_2)g_2^{-1}\parallel \\ \hspace{0.17em}& =\parallel g_1{g}_2^{-1}{[\eta (g_1)]}^{-1}\eta (g_2)\parallel \\ \hspace{0.17em}& \le \parallel g_1{g}_2^{-1}\parallel +\parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \\ \hspace{0.17em}& \le \parallel g_1{g}_2^{-1}\parallel +{\gamma }_{\eta }\parallel g_1{g}_2^{-1}\parallel \\ \hspace{0.17em}& \le (1+{\gamma }_{\eta })\parallel g_1{g}_2^{-1}\parallel \\ \hspace{0.17em}& =\gamma \parallel g_1{g}_2^{-1}\parallel \\ \parallel p(g_1)[p(g_2{)]}^{-1}\parallel & \le \gamma \parallel g_1{g}_2^{-1}\parallel ,\\ \hspace{0.17em}& \hspace{0.17em}\end{array}$

where $1+{\gamma }_{\eta }=\gamma$, and so the mapping $p$ is also $\gamma -Lipschitz$. Hence, $p$ is an isomorphic $\gamma -Lipschitz$.

(b). Since $p$ is surjective, therefore, inverse mapping $p^{-1}$ also exist. Now we prove that the inverse mapping is also $\gamma -Lipschitz$. So for $u_1,u_2\in G$, we have

$\begin{array}{cc}\parallel p^{-1}(u_1)[p^{-1}(u_2{)]}^{-1}\parallel & =\parallel u_1{u}_2^{-1}\parallel \\ \hspace{0.17em}& =\parallel {\eta }_{u_1}(g_1)[{\eta }_{u_2}(g_2{)]}^{-1}\parallel \\ \hspace{0.17em}& =\parallel u_1\eta (g_1)[u_2\eta (g_2{)]}^{-1}\parallel \\ \hspace{0.17em}& =\parallel u_1\eta (g_1)[\eta (g_2{)]}^{-1}{u}_2^{-1}\parallel \\ \hspace{0.17em}& \le \parallel u_1{u}_2^{-1}\parallel +\parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \\ \hspace{0.17em}& \le \parallel u_1{u}_2^{-1}\parallel +{\gamma }_{\eta }\parallel g_1{g}_2^{-1}\parallel \\ \hspace{0.17em}& \le \parallel u_1{u}_2^{-1}\parallel +{\gamma }_{\eta }\parallel p^{-1}(u_1)[p^{-1}(u_2{)]}^{-1}\parallel \\ (1-{\gamma }_{\eta })\parallel p^{-1}(u_1)[p^{-1}(u_2{)]}^{-1}\parallel & \le \parallel u_1{u}_2^{-1}\parallel \\ \parallel p^{-1}(u_1)[p^{-1}(u_2{)]}^{-1}\parallel & \le \frac{1}{1-{\gamma }_{\eta }}\parallel u_1{u}_2^{-1}\parallel \\ \parallel p^{-1}(u_1)[p^{-1}(u_2{)]}^{-1}\parallel & \le {(1-{\gamma }_{\eta })}^{-1}\parallel u_1{u}_2^{-1}\parallel .\\ \hspace{0.17em}& \hspace{0.17em}\end{array}$

Hence the inverse of $p$ is also $(1-{\gamma }_{\eta }{)}^{-1}$-Lipschitz.

The following theorem shows that there exists singleton set consisting a unique fixed point.

Theorem 3.5. Let $\eta$ be a contraction mapping from a Banach group $G$ to itself. Then there exists a function $h:G\to {\mathbb{R}}_{\ge 0}$ such that $h^{-1}(\{0\})$ is a singleton and $h(\eta (g))\le \gamma h(g)$ for every $g\in G$ and $0<\gamma <1$.

Proof. Since $\eta :G\to G$ be a contraction mapping, therefore, for every $g_1,g_2\in G$ and $0<\gamma <1$, we have

$\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel \le \gamma \parallel g_1{g}_2^{-1}\parallel .$

By contraction principle, $\eta$ has a unique fixed element $g^{\ast }\in G$ such that $\eta (g^{\ast })=g^{\ast }$. Define a mapping $h:G\to {\mathbb{R}}_{\ge 0}$ such that

$h(g)=\parallel g(g^{\ast }{)}^{-1}\parallel ,g\in G.$

Then $h(g)=0$ if and only if $g=g^{\ast }$, which implies that $h^{-1}(\{0\})=\{g^{\ast }\}$. Also

$\begin{array}{cc}h(\eta (g))& =\parallel \eta (g)[\eta (g^{\ast }{)]}^{-1}\parallel \\ \hspace{0.17em}& \le \gamma \parallel g{(g^{\ast })}^{-1}\parallel \\ h(\eta (g))& \le \gamma h(g),\\ \hspace{0.17em}& \hspace{0.17em}\end{array}$

which is required proof.

Let $G^{\ast }=[0,1)$ and its norm $\parallel \cdot \parallel :G^{\ast }\to [0,+\mathrm{\infty })$ defined by $\parallel r\parallel =min\{r,1-r\}$ for all $r\in G^{\ast }$. Then, $(G^{\ast },\parallel \cdot \parallel ,+)$ is a normed group. Consequently, we have the following fixed point theorems concerning the Banach group using normed group $G^{\ast }$.

Theorem 3.6. Assume that $(G,\parallel .\parallel )$ be a Banach group and $\eta :G\to G$ is a mapping satisfies the condition

$\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel \le p\parallel g_1\eta (g_1{)}^{-1}\parallel +q\parallel g_2\eta (g_2{)}^{-1}\parallel +r\parallel g_1{g}_2^{-1}\parallel$

for all $g_1,g_2\in G$, where $p,q,r\in G^{\ast }$ such that $\{t(q+r)+p\}\in G^{\ast }$ for $t\ge 1$. Then $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

Proof. Let $g_0\in G$ and $\{g_n\}$ be a sequence of elements in $G$ such that we can define the iterate sequence $g_n=\eta (g_{n-1})$, then we can obtain

$\begin{array}{cc}\parallel g_{n+1}\eta {(g_n)}^{-1}\parallel & =\parallel \eta (g_n)[\eta (g_{n-1}{)]}^{-1}\parallel \\ \hspace{0.17em}& \le p\parallel g_n\eta {(g_n)}^{-1}\parallel +q\parallel g_{n-1}\eta {(g_{n-1})}^{-1}\parallel +r\parallel g_n{g}_{n-1}^{-1}\parallel \\ \parallel g_{n+1}\eta {(g_n)}^{-1}\parallel & \le p\parallel g_n{g}_{n+1}^{-1}\parallel +q\parallel g_{n-1}{g}_n^{-1}\parallel +r\parallel g_n{g}_{n-1}^{-1}\parallel \\ (1-p)\parallel g_{n+1}\eta {(g_n)}^{-1}\parallel & \le q\parallel g_n{g}_{n-1}^{-1}\parallel +r\parallel g_n{g}_{n-1}^{-1}\parallel \\ (1-p)\parallel g_{n+1}\eta {(g_n)}^{-1}\parallel & \le (q+r)\parallel g_n{g}_{n-1}^{-1}\parallel \\ \parallel g_{n+1}\eta {(g_n)}^{-1}\parallel & \le \frac{q+r}{1-p}\parallel g_n{g}_{n-1}^{-1}\parallel \\ \parallel g_{n+1}\eta {(g_n)}^{-1}\parallel & \le \gamma \parallel g_n{g}_{n-1}^{-1}\parallel ,\end{array}$

where $\gamma =\frac{q+r}{1-p}\in G^{\ast }$. Continuing this process we can get

$\parallel g_{n+1}\eta (g_n{)}^{-1}\parallel \le {\gamma }^n\parallel g_1{g}_0^{-1}\parallel .$

As $\{t(q+r)+p\}\in G^{\ast }$, then $\gamma =\frac{q+r}{1-p}<\frac{1}{t}$, which refers that $\gamma \in G^{\ast }$ whenever $t\ge 1$, therefore mapping $\eta$ is a contraction mapping. Further, we will demonstrate that $\{g_n\}$ is a Cauchy sequence in $G$. Assume that $m>n$ for $m,n>0$, then using triangle inequality, we can ascertain that

$\begin{array}{cc}\parallel g_n{g}_m^{-1}\parallel & \le \parallel g_n{g}_{n+1}^{-1}\parallel +\parallel g_{n+1}{g}_m^{-1}\parallel \\ \hspace{0.17em}& \le \parallel g_n{g}_{n+1}^{-1}\parallel +\parallel g_{n+1}{g}_{n+2}^{-1}\parallel +\cdots +\parallel g_{m-1}{g}_m^{-1}\parallel \\ \hspace{0.17em}& \le {\gamma }^n\parallel g_0{g}_1^{-1}\parallel +{\gamma }^{n+1}\parallel g_0{g}_1^{-1}\parallel +{\gamma }^{n+2}\parallel g_0{g}_1^{-1}\parallel +\cdots +{\gamma }^{m-1}\parallel g_0{g}_1^{-1}\parallel \\ \hspace{0.17em}& \le {\gamma }^n\parallel g_0{g}_1^{-1}\parallel [1+\gamma +{\gamma }^2+\cdots ]\\ \parallel g_n{g}_m^{-1}\parallel & \le \frac{{\gamma }^n}{1-\gamma }\parallel g_0{g}_1^{-1}\parallel .\end{array}$

Taking limit $n\to \mathrm{\infty }$ when $\gamma \in G^{\ast }$, then we have $\underset{n\to \mathrm{\infty }}{lim}\parallel g_n{g}_m^{-1}\parallel =0$. Since $G$ is a complete normed group, then sequence $g_n$ converges to an element $g_0\in G$. Subsequently, we will demonstrate that $g_0$ is a fixed point of $\eta$. Consider

$\begin{array}{cc}\parallel \eta (g_0)g_0^{-1}\parallel & \le \parallel \eta (g_0)g_n^{-1}{g}_n{g}_0^{-1}\parallel \\ \hspace{0.17em}& \le \parallel \eta (g_0)g_n^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel \\ \hspace{0.17em}& =\parallel \eta (g_{n-1})\eta {(g_0)}^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel \\ \hspace{0.17em}& \le p\parallel g_{n-1}\eta {(g_{n-1})}^{-1}\parallel +q\parallel g_0\eta {(g_0)}^{-1}\parallel +r\parallel g_{n-1}{g}_0^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel \\ (1-q)\parallel \eta (g_0)g_0^{-1}\parallel & \le p{\gamma }^n\parallel g_0{g}_1^{-1}\parallel +r\parallel g_{n-1}{g}_0^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel ,\\ \hspace{0.17em}& \hspace{0.17em}\end{array}$

then taking limit $n\to \mathrm{\infty }$ when $\gamma \in G^{\ast }$, we can obtain

$\underset{n\to \mathrm{\infty }}{lim}\parallel \eta (g_0)g_0^{-1}\parallel =0,$

then $\eta (g_0)g_0^{-1}=e$, which intimates that $\eta (g_0)=g_0$. Therefore $g_0$ is a fixed point of $\eta$ in $G$. To prove the fixed point’s uniqueness, assume that $g_1$ and $g_2$ be two fixed points of $\eta$ such that $\eta (g_1)=g_1$ and $\eta (g_2)=g_2$, then given condition yields that

$\begin{array}{cc}\parallel g_1{g}_2^{-1}\parallel & =\parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \\ \hspace{0.17em}& \le p\parallel g_1\eta {(g_1)}^{-1}\parallel +q\parallel g_2\eta {(g_2)}^{-1}\parallel +r\parallel g_1{g}_2^{-1}\parallel \\ \hspace{0.17em}& \le p\parallel g_1{g}_1^{-1}\parallel +q\parallel g_2{g}_2^{-1}\parallel +r\parallel g_1{g}_2^{-1}\parallel \\ \parallel g_1{g}_2^{-1}\parallel & \le r\parallel g_1{g}_2^{-1}\parallel ,\end{array}$

which is a contradiction because $r\notin G^{\ast }$, therefore $g_1=g_2$. Hence $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

Example 3.7. Let $G=\mathbb{Q}$, then it is a normed group with the $p-adic$ norm $|\cdot {|}_p$ for prime $p\ge 2$. Taking the sequence $g_n=1+p+p^2+\cdots +p^{n-1}$, then $g_n$ is a Cauchy sequence and this sequence has a limit with respect to group-norm $|\cdot {|}_p$ such that ${lim}_{n\to \mathrm{\infty }}^{(p)}{g}_n=\frac{1}{1-p}\in \mathbb{Q}$, therefore $(\mathbb{Q},|\cdot {|}_p)$ is a complete normed group.

Before proving our next theorem concerning fixed point theorem, we need the following lemma:

Lemma 3.8. Let $\eta$ be a mapping from a Banach group $G$ to itself and satisfies the following condition

$\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel \le a_1\parallel g_1\eta (g_1{)}^{-1}\parallel +a_2\parallel g_2\eta (g_2{)}^{-1}\parallel +a_3\parallel g_1\eta (g_2{)}^{-1}\parallel +a_4\parallel g_2\eta (g_1{)}^{-1}\parallel +a_5\parallel g_1{g}_2^{-1}\parallel ,$

where $\sum _{i=1}^5{a}_i=u\in G^{\ast }$ and for each $i$, $a_i\in G^{\ast }$. Then for every $u\in G^{\ast }$ there exists $v\in G^{\ast }$ such that

$\parallel \eta (g)[{\eta }^2(g){]}^{-1}\parallel \le v\parallel g[\eta (g){]}^{-1}\parallel .$

Proof. Let $g_2=\eta (g_1)$, then according to given condition we have,

$\begin{array}{cc}\parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel & \le a_1\parallel g_1\eta {(g_1)}^{-1}\parallel +a_2\parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel +a_3\parallel g_1{[{\eta }^2(g_1)]}^{-1}\parallel \\ \hspace{0.17em}& +a_4\parallel \eta (g_1)\eta {(g_1)}^{-1}\parallel +a_5\parallel g_1{g}_1^{-1}\parallel \\ (1-a_2)\parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel & \le (a_1+a_5)\parallel g_1\eta {(g_1)}^{-1}\parallel +a_3\parallel g_1{[{\eta }^2(g_1)]}^{-1}\parallel \\ \parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel & \le \frac{a_1+a_5}{1-a_2}\parallel g_1\eta {(g_1)}^{-1}\parallel +\frac{a_3}{1-a_2}\parallel g_1{[{\eta }^2(g_1)]}^{-1}\parallel ,\end{array}$

then using triangle inequality, we can derive

$\begin{array}{cc}\parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel & \le \frac{a_1+a_5}{1-a_2}\parallel g_1\eta {(g_1)}^{-1}\parallel +\frac{a_3}{1-a_2}\parallel g_1{[\eta (g_1)]}^{-1}\parallel \\ \hspace{0.17em}& +\frac{a_3}{1-a_2}\parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel \\ \frac{1-a_2-a_3}{1-a_2}\parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel & \le \frac{a_1+a_3+a_5}{1-a_2}\parallel g_1\eta {(g_1)}^{-1}\parallel \\ \parallel \eta (g_1)[{\eta }^2(g_1{)]}^{-1}\parallel & \le \frac{a_1+a_3+a_5}{1-a_2-a_3}\parallel g_1\eta {(g_1)}^{-1}\parallel .\end{array}$

By the symmetry of given condition in the theorem, write $a_2$ instead of $a_1$ and $a_4$ instead of $a_3$, then we can conclude that

$\parallel \eta (g_1)[{\eta }^2(g_1){]}^{-1}\parallel \le \frac{a_2+a_4+a_5}{1-a_1-a_4}\parallel g_1\eta (g_1{)}^{-1}\parallel .$

Assume that $v=min[\frac{a_1+a_3+a_5}{1-a_2-a_3},\frac{a_2+a_4+a_5}{1-a_1-a_4}]$, then we obtain that $v\in G^{\ast }$ such that

$\parallel \eta (g_1)[{\eta }^2(g_1){]}^{-1}\parallel \le v\parallel g_1\eta (g_1{)}^{-1}\parallel .$

Using Lemma 3.8, the following theorem provides sufficient conditions for the existence and uniqueness of fixed point under specified condition.

Theorem 3.9. Let $\eta$ be a mapping from a Banach group $G$ to itself and satisfies the condition

$\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel \le a_1\parallel g_1\eta (g_1{)}^{-1}\parallel +a_2\parallel g_2\eta (g_2{)}^{-1}\parallel +a_3\parallel g_1\eta (g_2{)}^{-1}\parallel +a_4\parallel g_2\eta (g_1{)}^{-1}\parallel +a_5\parallel g_1{g}_2^{-1}\parallel$

for all $g_1,g_2\in G$, where ${\sum }_{i=1}^5{a}_i=u\in G^{\ast }$ and for each $i$, $a_i\in G^{\ast }$. Then $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

Proof. Let $g_0\in G$ and $\{g_n\}$ be a sequence of elements in $G$ such that we can define the iterate sequence $g_n=\eta (g_{n-1})$. Also, by Lemma 3.8, it follows that for every $u\in G^{\ast }$ there exists $\gamma \in G^{\ast }$ such that

$\parallel g_{n+1}{g}_n^{-1}\parallel \le {\gamma }^n\parallel g_1{g}_0^{-1}\parallel .$

Further, we will demonstrate that $\{g_n\}$ is a Cauchy sequence in $G$. Assume that $m>n$ for $m,n>0$, then using triangle inequality, we can obtain that

$\begin{array}{cc}\parallel g_n{g}_m^{-1}\parallel & \le \parallel g_n{g}_{n+1}^{-1}\parallel +\parallel g_{n+1}{g}_m^{-1}\parallel \\ \hspace{0.17em}& \le \parallel g_n{g}_{n+1}^{-1}\parallel +\parallel g_{n+1}{g}_{n+2}^{-1}\parallel +\cdots +\parallel g_{m-1}{g}_m^{-1}\parallel \\ \hspace{0.17em}& \le {\gamma }^n\parallel g_0{g}_1^{-1}\parallel +{\gamma }^{n+1}\parallel g_0{g}_1^{-1}\parallel +{\gamma }^{n+2}\parallel g_0{g}_1^{-1}\parallel +\cdots +{\gamma }^{m-1}\parallel g_0{g}_1^{-1}\parallel \\ \hspace{0.17em}& \le {\gamma }^n\parallel g_0{g}_1^{-1}\parallel [1+\gamma +{\gamma }^2+\cdots +{\gamma }^{m-1}]\\ \hspace{0.17em}& \le {\gamma }^n\parallel g_0{g}_1^{-1}\parallel \sum _{i=0}^{\mathrm{\infty }}{\gamma }^i\\ \parallel g_n{g}_m^{-1}\parallel & \le \frac{{\gamma }^n}{1-\gamma }\parallel g_0{g}_1^{-1}\parallel .\end{array}$

Taking limit $n\to \mathrm{\infty }$ when $\gamma \in G^{\ast }$, then we conclude $\underset{n\to \mathrm{\infty }}{lim}\parallel g_n{g}_m^{-1}\parallel =0$. Since $G$ is a Banach group, therefore, the sequence $g_n$ converges to an element $g_0\in G$. Subsequently, we will demonstrate that $g_0$ is a fixed point of $\eta$. For this purpose consider

$\begin{array}{cc}\parallel \eta (g_0)g_0^{-1}\parallel & \le \parallel \eta (g_0)g_n^{-1}{g}_n{g}_0^{-1}\parallel \\ \hspace{0.17em}& \le \parallel \eta (g_0)g_n^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel \\ \hspace{0.17em}& =\parallel \eta (g_{n-1})\eta {(g_0)}^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel \\ \hspace{0.17em}& \le a_1\parallel g_{n-1}\eta {(g_{n-1})}^{-1}\parallel +a_2\parallel g_0\eta {(g_0)}^{-1}\parallel +a_3\parallel g_{n-1}\eta {(g_0)}^{-1}\parallel \\ \hspace{0.17em}& +a_4\parallel g_0\eta {(g_{n-1})}^{-1}\parallel +a_5\parallel g_{n-1}{g}_0^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel \\ (1-a_2)\parallel \eta (g_0)g_0^{-1}\parallel & \le a_1{\gamma }^{n-1}\parallel g_0{g}_1^{-1}\parallel +a_3\parallel g_{n-1}\eta {(g_0)}^{-1}\parallel +a_4\parallel g_0\eta {(g_{n-1})}^{-1}\parallel \\ \hspace{0.17em}& +a_5\parallel g_{n-1}{g}_0^{-1}\parallel +\parallel g_n{g}_0^{-1}\parallel ,\end{array}$

and taking limit $n\to \mathrm{\infty }$ whenever $\gamma \in G^{\ast }$ then we get

$\parallel \eta (g_0)g_0^{-1}\parallel \le (a_2+a_3)\parallel \eta (g_0)g_0^{-1}\parallel .$

Since $\sum _{i=1}^5{a}_i\in G^{\ast }$, therefore, $(a_2+a_3)\in G^{\ast }$. So, which is possible only if $\eta (g_0)g_0^{-1}=e$ which implies that $\eta (g_0)=g_0$, therefore $g_0$ is a fixed point of $\eta$ in $G$. To prove the fixed point’s uniqueness, assume that $g_1$ and $g_2$ be two fixed points of $\eta$ such that $\eta (g_1)=g_1$ and $\eta (g_2)=g_2$, then given condition yields that

$\begin{array}{cc}\parallel g_1{g}_2^{-1}\parallel & =\parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \\ \hspace{0.17em}& \le a_1\parallel g_1\eta {(g_1)}^{-1}\parallel +a_2\parallel g_2\eta {(g_2)}^{-1}\parallel +a_3\parallel g_1\eta {(g_2)}^{-1}\parallel \\ \hspace{0.17em}& +a_4\parallel g_2\eta {(g_1)}^{-1}\parallel +a_5\parallel g_1{g}_2^{-1}\parallel \\ \hspace{0.17em}& \le a_3\parallel g_1\eta {(g_2)}^{-1}\parallel +a_4\parallel g_2\eta {(g_1)}^{-1}\parallel +a_5\parallel g_1{g}_2^{-1}\parallel \\ \parallel g_1{g}_2^{-1}\parallel & \le (a_3+a_4+a_5)\parallel g_1{g}_2^{-1}\parallel ,\end{array}$

which is a contradiction because $(a_3+a_4+a_5)\in G^{\ast }$, therefore $g_1=g_2$. Hence $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

The following corollary gives an important characterization of Banach group $G$ that possess continuous mappings having fixed point (may not be unique).

Corollary 3.10. Let $\eta$ be a continuous mapping of a Banach group $G$ into itself and also holds the condition

$\parallel \eta (g)[{\eta }^2(g){]}^{-1}\parallel \le v\parallel g[\eta (g){]}^{-1}\parallel$

for any $g\in G$ such that $\parallel g[\eta (g){]}^{-1}\parallel <\mathrm{\infty }$ and $v\in G^{\ast }$. Then there exists $n\ge 0$ such that $\parallel {\eta }^n(g)[{\eta }^{n+1}(g){]}^{-1}\parallel <\mathrm{\infty }$. And also there exists a fixed point $w\in G$ such that $\eta (w)=w$.

Remarks 3.11. The mapping $\eta$ satisfies the condition of Corollary 3.10 may have more than one fixed points when we consider the mapping $\eta (g)=g$ for any $g\in G$.

The following corollary about normed finite group gives the existence of unique fixed point.

Corollary 3.12. Let $\eta$ be a contraction mapping of a normed finite group $G$ into itself. Then $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$. In fact, for any $g\in G,g_1=\eta (g),\cdots ,g_{n+1}=\eta (g_n)$, there exist some integer $N$ such that for any $n\ge N$, we have $g_n=g_N$.

Definition 3.13. (Hussain et al., 2015) Let $(G,\parallel \cdot \parallel )$ be a normed group. A mapping $\eta$ of a normed group $G$ into itself is said to be expansive if there is $\gamma >1$ such that

$\parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel \ge \gamma \parallel g_1{g}_2^{-1}\parallel$

for any $g_1,g_2\in G$ and $g_1\ne g_2$.

It is easy to see that any expansive mapping must be injective; therefore, we have the following theorem to determine a fixed point for expansive mappings.

Theorem 3.14. Let $\eta$ be an expansive and surjective mapping of a Banach group $G$ into itself. Then $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

Proof. Since any expansive mapping is injective, therefore $\eta$ is bijective. Hence ${\eta }^{-1}$ exists and satisfies the following;

$\parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2){]}^{-1}\parallel \le \frac{1}{\gamma }\parallel g_1{g}_2^{-1}\parallel ,$

where $\frac{1}{\gamma }<1$, thus ${\eta }^{-1}$ is a contraction mapping. According to the fixed point theorem, there exists $g_0\in G$ such that ${\eta }^{-1}(g_0)=g_0$. Hence $g_0$ is a fixed point of $\eta$.

The following theorem gives an important characterization of finite normed cyclic group that possess expansive mapping.

Theorem 3.15. Let $\eta$ be an expansive mapping of a finite normed cyclic group $G$ into itself, then

(a) $\eta$ is bi-Lipschitz equivalent.

(b) $\eta$ has a unique fixed point $g_0\in G$ such that $\eta (g_0)=g_0$.

Proof. (a). Because $G$ is a normed cyclic group, therefore we have $\parallel g_1{g}_2\parallel =\parallel g_2{g}_1\parallel$ for any $g_1,g_2\in G$. Assume that $|G|=m$, then define a mapping

$\eta (g)=g^n,n\in \mathbb{N},g\in G,$

where $gcd(n,m)=1$ for $n\in \mathbb{N}$.

Further, it is obvious that $\parallel g^n\parallel \le n\parallel g\parallel$ for $n\in \mathbb{N}$.

Considering an expansive mapping $\eta$ and $\gamma >1$, then for any $g_1,g_2\in G$, proof of (a) consists of the following simple computation:

$\begin{array}{cc}\gamma \parallel g_1{g}_2^{-1}\parallel & \le \parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \\ \hspace{0.17em}& \le \parallel g_1^n{[g_2^n]}^{-1}\parallel \\ \hspace{0.17em}& \le \parallel {[g_1{g}_2^{-1}]}^n\parallel \\ \gamma \parallel g_1{g}_2^{-1}\parallel & \le \parallel \eta (g_1)[\eta (g_2{)]}^{-1}\parallel \le n\parallel g_1{g}_2^{-1}\parallel .\end{array}$

(b). Since $\eta$ be an expansive mapping and $\gamma >1$, then for every $g_1,g_2\in G$, we can obtain

$\gamma \parallel g_1{g}_2^{-1}\parallel \le \parallel \eta (g_1)[\eta (g_2){]}^{-1}\parallel .$

If $g_1^n=g_2^n$ then $g_1^n{g}_2^{-n}=e$, therefore, we have

$\begin{array}{cc}\gamma \parallel g_1{g}_2^{-1}\parallel & \le \parallel g_1^n{g}_2^{-n}\parallel \\ \gamma \parallel g_1{g}_2^{-1}\parallel & =\parallel e\parallel \\ \gamma \parallel g_1{g}_2^{-1}\parallel & =0\\ g_1& =g_2,\end{array}$

which infers that $\eta$ is injective. So $\eta$ is bijective, then ${\eta }^{-1}$ exists. Also, by expansive mapping

$\begin{array}{cc}\gamma \parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2{)]}^{-1}\parallel & \le \parallel \eta ({\eta }^{-1}(g_1))[\eta ({\eta }^{-1}(g_2{))]}^{-1}\parallel \\ \gamma \parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2{)]}^{-1}\parallel & \le \parallel g_1{g}_2^{-1}\parallel \\ \parallel {\eta }^{-1}(g_1)[{\eta }^{-1}(g_2{)]}^{-1}\parallel & \le \frac{1}{\gamma }\parallel g_1{g}_2^{-1}\parallel ,\end{array}$

thus ${\eta }^{-1}$ is also contraction mapping for $\gamma >1$. Since $G$ is finite, therefore it is a complete normed group. According to the fixed point theorem, there exists $g_0\in G$ such that ${\eta }^{-1}(g_0)=g_0$. Hence $g_0$ is a fixed point of $\eta$.

4. Conclusion

Thus, in this paper, we have proposed and investigated the general properties of Lipschitz mapping of a normed group $G$ to itself and concluded that every isomorphic mapping is 1-Lipschitz. Moreover, we have also demonstrated that the conjugate Lipschitz mapping ${\eta }_g$ has an equivalent norm. It is observed that the left-shift ($L_g$) under left-invariant metric ($d_L$) and right-shift ($R_g$) under right-invariant metric ($d_R$) are 1-Lipschitz, and also $L_g$ and $R_g$ mappings have equivalent norms. Particularly, for a normed finite group $G$, a continuous self-mapping $\eta$ has an isomorphic norm.

We have also shown that for homomorphic Lipschitz mapping of the abelian Banach group, there is contraction mapping and isomorphic mapping such that the inverse Lipschitz mapping also exists. We have formulated some fixed point theorems for contraction mappings of a Banach group $G$ and determined unique fixed points following certain conditions. It is also proved that the contraction mapping of a normed finite group $G$ to itself has a unique fixed point in $G$. We have concluded that an expansive mapping of a finite normed cyclic group $G$ has a unique fixed point in $G$ and is also bi-Lipschitz equivalent.

Conflicts of Interest

The author(s) declare(s) that there is no conflict of interest regarding the publication of this paper.

Acknowledgements

The authors are grateful to the referees and the editors for valuable comments and suggestions, which have improved the original manuscript greatly.

Additional information

Funding

This work was supported by the National Natural Science Foundation of China [11971493].

Notes on contributors

Yongjin Li

Dr. Yongjin Li is working as a professor at the School of Mathematics, Sun Yat-sen University. His area of specialization is Functional Analysis, Differential Equations. Under his guidance, 04 students are working for a Ph.D., and 03 post Doctors are working in his group. He has published 150 Research papers, 06 books and completed 03 projects for the National Natural Science Foundation of China.

Muhammad Sarfraz is currently a Ph.D. Research Scholar under the supervision of Professor Dr. Yongjin Li in School of Mathematics, Sun Yat-sen University. His area of specialization is Functional Analysis and Cryptography. He has published 7 Research papers.

Fawad Ali is currently a Ph.D. Research Scholar with the School of Mathematics and Statistics Xi’an Jiaotong University, Xi’an 710049, P. R. China. He has been working in algebraic graph theory since 2017.