A Revisit to n-Normed Spaces through Its Quotient Spaces

TL;DR

This paper investigates properties of n-normed spaces via their quotient spaces, establishing equivalences of continuity types and proving a fixed point theorem for contractive mappings.

Contribution

It introduces a new approach to analyze n-normed spaces through quotient spaces, including fixed point results for contractive mappings.

Findings

All types of continuity are equivalent in this context

A fixed point theorem for contractive mappings is established

Properties of n-normed spaces are characterized via quotient space norms

Abstract

In this paper, we study some features of n-normed spaces with respect to norms of its quotient spaces. We define continuous functions with respect to the norms of its quotient spaces and show that all types of continuity are equivalent. We also study contractive mappings on n- normed spaces using the same approach. In particular, we prove a fixed point theorem for contractive mappings on a closed and bounded set in an n-normed space.

Equations

$$Y_{i_1,\dots ,i_m}^0:=\mathrm{span}Y\setminus \{y_{i_1},\dots ,y_{i_m}\}:=\left\{\sum _{i\notin \{i_1,\dots ,i_m\}}{\alpha }_i{y}_i;{\alpha }_i\in \mathbb{R}.\right\}.$$

$$\overline{u}:=\left\{u+\sum _{i\notin \{i_1,\dots ,i_n\}}{\alpha }_i{y}_i:{\alpha }_i\in \mathbb{R}\right\}.$$

$$X_{i_1,\dots ,i_m}^{\ast }:=X/Y_{i_1,\dots ,i_m}^0:=\{\overline{u}:u\in X\}.$$

$$\Vert \overline{u}{\Vert }_{i_1,\dots ,i_m}^{\ast }:=$$

$$+\Vert u,y_1,\dots ,y_{i_m-1},y_{i_m+1},\dots ,y_n\Vert .$$

$$\Vert \overline{u}{\Vert }_{i_1,\dots ,i_m}^{\ast }=\Vert \overline{u}{\Vert }_{i_1}^{\ast }+\cdots +\Vert \overline{u}{\Vert }_{i_m}^{\ast }.$$

$$\Vert \overline{x_k-x}{\Vert }_{i_1,\dots ,i_m}^{\ast }<ϵ,$$

$$\underset{n\to \infty }{\lim }\Vert \overline{x_k-x}{\Vert }_{i_1,\dots ,i_m}^{\ast }=0,$$

$$\Vert \overline{x_k-x_l}{\Vert }_{i_1,\dots ,i_m}^{\ast }<ϵ,$$

$$\underset{k,l\to \infty }{\lim }\Vert \overline{x_k-x_l}{\Vert }_{i_1,\dots ,i_m}^{\ast }=0,$$

$$\Vert \overline{x}{\Vert }_{i_1,\dots ,i_m}^{\ast }\le M,$$

$$\bigcup \{i_1,\dots ,i_m\}\supseteq \{1,\dots ,n\}.$$

$$\underset{x\to \infty }{\lim }T{x}_n=Tx.$$

$$\Vert \overline{Tx-Ty}{\Vert }_{i_1,\dots ,i_m}^{\ast }\le C\Vert \overline{x-y}{\Vert }_{i_1,\dots ,i_m}^{\ast },$$

$$\Vert \overline{Tx-Ty}{\Vert }_{i_1,\dots ,i_m}^{\ast }<ϵ,$$

$$\Vert \overline{Tx-Ty}{\Vert }_j^{\ast }\le C\Vert \overline{x-y}{\Vert }_j^{\ast },$$

$$\Vert \overline{Tx-Ty}{\Vert }_{i_1}^{\ast }$$

$$\Vert \overline{Tx-Ty}{\Vert }_{i_m}^{\ast }$$

$$\Vert \overline{Tx-Ty}{\Vert }_{i_1,\dots ,i_m}^{\ast }$$

$$\Vert \overline{Tx-Ty}{\Vert }_{i_1,\dots ,i_m}^{\ast }\le C\Vert \overline{x-y}{\Vert }_{i_1,\dots ,i_m}^{\ast },$$

$$(n-1)\Vert \overline{Tx-Ty}{\Vert }_{1,\dots ,n}^{\ast }\le C(n-1)\Vert \overline{x-y}{\Vert }_{1,\dots ,n}^{\ast },$$

$$\Vert \overline{Tx-Ty}{\Vert }_{1,\dots ,n}^{\ast }\le C\Vert \overline{x-y}{\Vert }_{1,\dots ,n}^{\ast },$$

$$x_k=T(x_{k-1})=T^k(x_0);k=1,2,\dots$$

$$\Vert \overline{T^2(x_0)-T^2(x_1)}{\Vert }_{i_1,\dots ,i_m}^{\ast }$$

$$\Vert \overline{T^k(x_0)-T^k(x_1)}{\Vert }_{i_1,\dots ,i_m}^{\ast }\le C^k\Vert \overline{x_0-x_1}{\Vert }_{i_1,\dots ,i_m}^{\ast },$$

$$\Vert \overline{x_k-x_l}{\Vert }_{i_1,\dots ,i_m}^{\ast }$$

$$\underset{k,l\to \infty }{\lim }\Vert \overline{x_k-x_l}{\Vert }_{i_1,\dots ,i_m}^{\ast }=0,$$

$$T(x)=\underset{k\to \infty }{\lim }T(x_k)=\underset{k\to \infty }{\lim }{x}_{k+1}=x.$$

$$\Vert \overline{x-x^{\prime }}{\Vert }_{i_1,\dots ,i_m}^{\ast }$$

Full text

1. Introduction

We will study some features of the $n$-normed spaces with its quotient spaces as a tool. In the 1960’s, the concept of $n$-normed spaces was investigated by S. Gähler [8, 9, 10, 11]. Let $n$ be a nonnegative integer and $X$ be a real vector space with $\dim(X)\geq n$. The pair $\left(X,\|\cdot,\dots,\cdot\|\right)$ is called an $\bm{n}$-normed space where $\|\cdot,\dots,\cdot\|:X^{n}\to\mathbb{R}$ is an $\bm{n}$-norm on $X$ which satisfies the following conditions:

• (i)

$\|x_{1},\dots,x_{n}\|\geq 0$; $\|x_{1},\dots,x_{n}\|=0$ if and only if $x_{1},\dots,x_{n}$ linearly dependent.

• (ii)

$\|x_{1},\dots,x_{n}\|$ is invariant under permutation.

• (iii)

$\|\alpha x_{1},\dots,x_{k}\|=|\alpha|\ \|x_{1},\dots,x_{k}\|$ for any $\alpha\in\mathbb{R}$.

• (iv)

$\|x_{1}+x_{1}^{\prime},x_{2},\dots,x_{k}\|\leq\|x_{1},x_{2},\dots,x_{k}\|+\|x_{1}^{\prime},x_{2},\dots,x_{k}\|$.

Some researchers studied further various aspects of these spaces [1, 2, 12, 13, 15, 7, 16, 17]. In particular, Brzdȩk and Ciepliński [5] proved a fixed point theorem for operators acting on some classes of functions, with values in $n$-Banach spaces. Ekariani et al. [6] also proved a fixed point theorem of a contractive mapping in $\ell^{p}$ as an $n$-normed space. They used a norm which is equivalent to the usual norm on $\ell^{p}$. The norm that they used is simpler than the normed used by Brzdȩk and Ciepliński. This norm is derived from the $n$-norm using a linearly independent set consisting of $n$ vectors.

In this paper, we provide a more general approach to prove a fixed point theorem in an $n$-normed space. First, we define some quotient spaces of an $n$-normed space. In each quotient space we define a norm which is derived from the $n$-norm in a certain way. This norm is simpler than the norm that Brzdȩk and Ciepliński or Ekariani et al. used. Using these norms, we investigate continuous mappings and contractive mappings in the $n$-normed space and prove a fixed point theorem for contractive mappings on a closed and bounded set in the $n$-normed space. Using this approach we will also prove a fixed point theorem in $\ell^{p}$ as an $n$-normed space. Compared to the approaches used by previous researchers [13, 15, 16, 5, 6], our approach provides a more general way which can be used to study properties of $n$-normed spaces.

2. Preliminaries

Let us begin with the construction of quotient spaces of an $n$-normed space. Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space, $m\in\{1,\dots,n\}$ and $Y=\{y_{1},\dots,y_{n}\}$ be a linearly independent set in $X$. For $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$, consider $Y\setminus\{y_{i_{1}},\dots,y_{i_{m}}\}$. We define the following subspace of $X$:

[TABLE]

For each $u\in X$, the corresponding coset in $X$ is defined by

[TABLE]

Then we have if $\overline{u}=\overline{v}$, then $u-v\in{\rm span}\,Y\setminus\{y_{i_{1}},\dots,y_{i_{m}}\}$. We define the quotient space

[TABLE]

The addition and scalar multiplication apply in this space. Moreover, we define the following norm on $X^{*}_{i_{1},\dots,i_{m}}$:

[TABLE]

Using the above construction, we get $\binom{n}{m}$ quotient spaces. Moreover, for an $m\in\{1,\dots,n\}$, the set that contains all quotient spaces constructed above is called class-$\bm{m}$ collection [4].

Observe that each term of (2.1) is a norm of quotient spaces in class-$1$ collection. Here (2.1) can be written as

[TABLE]

One can see that the norm we define in (2.1) is more simpler than the norm that Brzdȩk and Ciepliński used since we define it with respect to a linearly independent set consisting of $n$ vectors.

For an $m\in\{1,\dots,n\}$ we will use the phrase ’norms of class-$m$ collection’ which means ’all norms of each quotient space in the class-$m$ collection’. Furthermore, using the norms of class-$m$ collection, we have observed some topology characteristics of an $n$-normed space as presented in the following.

Definition 2.1**.**

[3] Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m\in\{1,\dots,n\}$. We say a sequence $\{x_{k}\}\subset X$ converges with respect to the norms of class-$\bm{m}$ collection to $x$ if for any $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that for $k\geq N$ we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. In this case we also say

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. If $\{x_{k}\}$ does not converge, we say it diverges.

By the above definition, we have the following theorem.

Theorem 2.2**.**

[3]* Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m\in\{1,\dots,n\}$. A sequence $\{x_{k}\}\subset X$ is convergent with respect to the norms of class-$1$ collection if and only if it is convergent with respect to the norms of class-$m$ collection.*

Corollary 2.3**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m_{1},m_{2}\in\{1,\dots,n\}$. Then, if a sequence in $X$ converges with respect to the norms class-$m_{1}$ collection, then it is also converges with respect to the norms of class-$m_{2}$ collection.

Definition 2.4**.**

[3] Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m\in\{1,\dots,n\}$. A sequence $\{x_{k}\}\subset X$ is called a Cauchy sequence with respect to the norms of class-$\bm{m}$ collection if for any $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that, for every $k,l\geq N$, we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. In other words

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$.

Theorem 2.5**.**

[3]* Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m\in\{1,\dots,n\}$. If $\{x_{k}\}$ is convergent with respect to the norms of class-$1$ collection, then $\{x_{k}\}$ is Cauchy with respect to the norms of class-$m$ collection.*

Corollary 2.6**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m_{1},m_{2}\in\{1,\dots,n\}$. Then, if a sequence in $X$ is a Cauchy sequence with respect to the norms class-$m_{1}$ collection, then it is also a Cauchy sequence with respect to the norms of class-$m_{2}$ collection.

Remark 2.7*.*

For $m_{1},m_{2}\in\{1,\dots,n\}$, we find that all types of convergent sequence with respect to the norms of class-$m_{1}$ collection and to the norms of class-$m_{2}$ colection are equivalent. The equivalence also applies to all types of Cauchy sequence with respect to the norms of class-$m_{1}$ collection and to the norms of class-$m_{2}$ collection. In this regard, we may simply use the word ’converges or Cauchy’ instead of ’converges or Cauchy with respect to the norms of class-$m$ collection’. Furthermore if every Cauchy sequence in $X$ converges, then $X$ is complete. By the word ’complete’, we mean ’complete with respect to the norms of class-$m$ collection’, for some $m\in\{1,\dots,n\}$.

Now let us move to the definition of closed sets with respect to the norms of class-$m$ collection, for any $m\in\{1,\dots,n\}$.

Definition 2.8**.**

[3] Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $K\subseteq X$. The set $K$ is called closed if for any sequence $\{x_{k}\}$ in $K$ that converges in $X$, its limit belongs to $K$.

Note that by saying ’closed’ we mean ’closed with respect to class-$m$ collection’, for some $m\in\{1,\dots,n\}$. Next is the definition of bounded sets.

Definition 2.9**.**

[3] Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space, $m\in\{1,\dots,n\}$ and $K\subseteq X$ be a nonempty set. The set $K$ is called bounded with respect to the norms of class-$\bm{m}$ collection if and only if for any $x\in K$ there exists an $M>0$ such that

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$.

We also have the following theorem which says that all types of bounded set are equivalent.

Theorem 2.10**.**

[3]* Let $(X,\|\cdot,\dots,\cdot\|)$ is an $n$-normed space, $m\in\{1,\dots,n\}$ and $K\subset X$ nonempty. The set $K$ is bounded with respect to the norms of class-$1$ collection if and only if it is bounded with respect to class-$m$ collection.*

Furthermore, for the completeness of the $n$-normed space with respect to the norms of class-$m$ collections we have the following theorem.

Theorem 2.11**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m\in\{1,\dots,n\}$. Then $X$ is complete with respect to the norms of class-$1$ collection if and only if $X$ is complete with respect to the norms of class-$m$ collection.

Proof.

Suppose that $X$ is complete with respect to the norms of class-$1$ collection and $m\in\{1,\dots,n\}$. Take any Cauchy sequence $\{x_{k}\}$ with respect to the norms of class-$m$ collection in $X$. Then by Theorems 2.2 and 2.5 we have $\{x_{k}\}$ is a convergent sequence with respect to the norms of class-$m$ collection in $X$. This tells us that $X$ is complete with respect to the norms of class-$m$ collection. The converse is similar. ∎

Remark 2.12*.*

As in Corollary 2.6, one can see that the equivalence also applies to boundedness and completeness of a set. Then, we will use the word ’bounded’ instead of the phrase ’bounded with respect to the norms of class-$m$ collection’, for some $m\in\{1,\dots,n\}$.

Remark 2.13*.*

For a fixed $m\in\{1,\dots,n\}$, the convergence of a sequence, the closedness and the boundedness of a set with respect to the norms of class-$m$ collection may be investigated with respect to some norms $\|\cdot\|^{*}_{i_{1},\dots,i_{m}}$ we choose such that

[TABLE]

Moreover, the least number of norms that can be used to investigate these notions is $\left\lceil\frac{n}{m}\right\rceil$. Next, by using a similar approach we will study continuous mappings, contractive mappings, and also prove a fixed point theorem of contractive mappings of a closed and bounded in an $n$-normed space.

3. Continuous Mappings with Respect to the Norms of Class-$m$ Collection

We shall now discuss the continuity of a mapping with respect to the norms of class-$m$ collection in an $n$-normed space.

Definition 3.1**.**

Let $(X,\|\cdot,\dots,\cdot\|^{*})$ be an $n_{1}$-normed space and $(Z,\|\cdot,\dots,\cdot\|^{**})$ be an $n_{2}$-normed space and $l\in\{1,\dots,n_{1}\}$, $m\in\{1,\dots,n_{2}\}$. Suppose that $f:X\to Z$.

1. (i)

We say that $f$ is continuous with respect to the norms of class-$\bm{(l,m)}$ collections at $\bm{a\in X}$ if and only if for any $\epsilon>0$ there exists a $\delta>0$ such that for $x\in X$ with $\|\overline{x-a}\|^{*}_{i_{1},\dots,i_{l}}<\delta$ for every $\{i_{1},\dots,i_{l}\}\subset\{1,\dots,n_{1}\}$, we have $\|\overline{f(x)-f(a)}\|^{**}_{j_{1},\dots,j_{m}}<\epsilon$ for every $\{j_{1},\dots,j_{m}\}\subset\{1,\dots,n_{2}\}$. 2. (ii)

We say that $f$ is continuous with respect to the norms of class-$\bm{(l,m)}$ collections on $\bm{X}$ if and only if $f$ continuous with respect to the norms of class-$(l,m)$ collection at each $x\in X$.

Moreover, if $Z=X$, then we say ’$f$ is continuous with respect to the norms of class-$m$ collections’ instead of ’continuous with respect to the norms of class-$(m,m)$ collections’. The norms $\|\cdot\|^{*}_{i_{1},\dots,i_{l}}$, $\|\cdot\|^{**}_{j_{1},\dots,j_{m}}$ are the norms of class-$(l,m)$ collections in $X$ and $Z$ respectively. Note that we define class-$(l,m)$ collections by using a linearly independent set consisting of $n_{1}$ and $n_{2}$ vectors in $X$ and $Z$ respectively. Based on the definition, we have the following theorem.

Theorem 3.2**.**

Let $(X,\|\cdot,\dots,\cdot\|^{*})$ be an $n_{1}$-normed space and $(Z,\|\cdot,\dots,\cdot\|^{**})$ be an $n_{2}$-normed space and $l\in\{1,\dots,n_{1}\}$, $m\in\{1,\dots,n_{2}\}$. A mapping $f:X\to Z$ is continuous with respect to class-$1$ collections if and only if $f$ is continuous with respect to class-$(l,m)$ collections.

Proof.

Let $f$ be a continuous mapping with respect to the norms of class-$1$ collections at $a\in X$, $l\in\{1,\dots,n_{1}\}$ and $m\in\{1,\dots,n_{2}\}$. For any $\epsilon>0$, there is $\delta>0$ such that for any $x\in X$ with $\|\overline{x-a}\|^{*}_{i}<\frac{\delta}{l}$ for every $i\in\{1,\dots,n_{1}\}$, we have $\|\overline{f(x)-f(a)}\|^{*}_{j}<\frac{\epsilon}{m}$ for every $j\in\{1,\dots,n_{2}\}$. Then for any $x\in X$ with $\|\overline{x-a}\|^{*}_{i_{1},\dots,i_{l}}<\delta$ for every $\{i_{1},\dots,i_{l}\}\subset\{1,\dots,n_{1}\}$, we have $\|\overline{f(x)-f(a)}\|^{*}_{j_{1},\dots,j_{m}}<\epsilon$ for every $\{j_{1},\dots,j_{m}\}\subset\{1,\dots,n_{2}\}$. This means $T$ is continuous with respect to the norms of class-$(l,m)$ collection.

The converse is obvious. ∎

Corollary 3.3**.**

Let $(X,\|\cdot,\dots,\cdot\|^{*})$ and $(Z,\|\cdot,\dots,\cdot\|^{**})$ be an $n_{1}$-normed space and an $n_{2}$-normed space respectively, $l_{1},l_{2}\in\{1,\dots,n_{1}\}$ and $m_{1},m_{2}\in\{1,\dots,n_{2}\}$. If a mapping $T:X\to Z$ is continuous with respect to class-$(l_{1},m_{1})$, then $T$ continuous with respect to the norms of class-$(l_{2},m_{2})$.

Remark 3.4*.*

By the above corollary, we can see that all types of continuity of a mapping with respect to the norms of class-$(l,m)$ collections are equivalent. Then from now on we will use the word ’continuous’ instead of ’continuous with respect to the norms of class-$(l,m)$ collection’.

Now we present a proposition that gives a relation between a convergent sequence and a continuous mapping with respect to the norms of class-$(l,m)$ collection.

Proposition 3.5**.**

Let $(X,\|\cdot,\dots,\cdot\|^{*})$ and $(Z,\|\cdot,\dots,\cdot\|^{**})$ be an $n_{1}$-normed space and an $n_{2}$-normed space respectively, $l\in\{1,\dots,n_{1}\}$, $m\in\{1,\dots,n_{2}\}$. Suppose that $T:X\to Z$ is continuous. If $\{x_{k}\}$ converges to $x$, then

[TABLE]

Proof. Let $l\in\{1,\dots,n_{1}\}$, $m\in\{1,\dots,n_{2}\}$, and $\epsilon>0$. Since $T$ is continuous, there is a $\delta>0$ such that if $\|\overline{y-x}\|_{i_{1},\dots,i_{m}}^{*}<\delta$ for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n_{1}\}$, then $\|Ty-T{x}\|_{j_{1},\dots,j_{l}}^{**}<\epsilon$. Since $\{x_{k}\}$ converges to $x$ with respect to norms of class-$m$, choose $N\in\mathbb{N}$ such that for $k\geq N$ we have $\|x_{k}-x\|_{i_{1},\dots,i_{m}}\leq\delta$. It then follows that $\|\overline{Tx_{k}-Tx}\|_{j_{1},\dots,j_{m}}^{**}<\epsilon$ whenever $k\geq N$. Therefore $Tx_{k}$ converges to $Tx$.

4. Fixed Point Theorem for Contractive Mappings with Respect to The Norms of Class-$m$ Collection

In this section, we shall discuss contractive mappings with respect to the norms of class-$m$ collection in an $n$-normed space and its fixed point theorem.

Definition 4.1**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m\in\{1,\dots,n\}$. A mapping $T:X\to X$ is called contractive with respect to the norms of class-$\bm{m}$ collection if there is a $C\in(0,1)$ such that for any $x,y\in X$ we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$ with $i_{1}<\dots<i_{m}$

Following the definitions of continuous mappings and contractive mappings, we have the following proposition and theorem.

Proposition 4.2**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space and $m\in\{1,\dots,n\}$. If $T$ is a contractive mapping with respect to the norms of class-$m$ collection, then $T$ is continuous.

Proof.

For an $m\in\{1,\dots,n\}$, let $T$ be a contractive mapping with respect to the norms of class-$m$ collection, then there is a $C\in(0,1)$ such that for any $x,y\in X$ we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$.

For any $\epsilon>0$, choose $\delta=\frac{\epsilon}{C}$. Then, for $x,y\in X$ where $\|\overline{x-y}\|^{*}_{i_{1},\dots,i_{m}}<\delta$ for every $\{i_{1},\dots,i_{m}\}\in\{1,\dots,n\}$, we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\in\{1,\dots,n\}$. Therefore, $T$ is continuous in $X$ ∎

Note that, for an $m_{1}\in\{1,\dots,n\}$ a contractive mapping with respect to the norms of class-$m_{1}$ collection are continuous with respect to the norms of class-$m$ collection, for any $m\in\{1,\dots,n\}$. Moreover, we have the following theorem.

Theorem 4.3**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space, $m\in\{1,\dots,n\}$ and $T:X\to X$. If $T$ is a contractive mapping with respect to the norms of class-$1$ collection, then $T$ is a contractive mapping with respect to the norms of class-$m$ collection.

Proof.

Suppose that $T$ is a contractive mapping with respect to the norms of class-$1$ collection, then there is a $C\in(0,1)$ such that for any $x,y\in X$ we have

[TABLE]

for every $j\in\{1,\dots,n\}$. Let $m\in\{1,\dots,n\}$, by (4.1) we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\in\{1,\dots,n\}$. It therefore follows from the above inequalities that

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. If we consider the norms of class-$m$ collection, then this means $T$ is a contractive mapping with respect to the norms of class-$m$ collection. ∎

We also have this following theorem which will be used later.

Theorem 4.4**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space, $m\in\{1,\dots,n\}$ and $T:X\to X$. If $T$ is a contractive mapping with respect to the norms of class-$m$ collection, then $T$ is also a contractive mapping with respect to the norm of class-$n$ collection.

Proof.

Let $T:X\to X$ be a contractive mapping with respect to the norms of class-$m$ collection for an $m\in\{1,\dots,n\}$. Then there is a $C\in(0,1)$ such that for every $x,y\in X$ we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. One can see that (4.2) contains $\binom{n}{m}$ inequalities. From these inequalities, we have

[TABLE]

or

[TABLE]

which means that $T$ is a contractive mapping with respect to the norm of class-$n$ collection. ∎

Finally, we provide a fixed point theorem for a contractive mapping in a closed and bounded set on an $n$-normed space, with respect to the norms of class-$m$ collection.

Theorem 4.5**.**

Let $(X,\|\cdot,\dots,\cdot\|)$ be an $n$-normed space, $m\in\{1,\dots,n\}$ and $K\subset X$ is nonempty, closed and bounded. Suppose that $X$ is complete (with respect to the norms of class-$m$ collection). If T: $K\to K$ is a contractive mapping with respect to the norms of class-$m$ collection, then $T$ has a unique fixed point.

Proof.

Fix an $m\in\{1,\dots,n\}$. Let $x_{0}\in K$ and $\{x_{k}\}$ be a sequence in $K$ such that

[TABLE]

Since $T$ is a contractive mapping, there is a $C\in(0,1)$ such that for $x_{0},x_{1}\in K$ we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. By using induction, we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. Now we show that $\{x_{n}\}$ is a Cauchy sequence in $K$. Let $k,l\in\mathbb{N}$. Without loss of generality, we take $l>k$ and $l=k+p$, with $p\in\mathbb{N}$. Then we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. Also, since $K$ is bounded, for any $x_{0},x_{1}\in K$ there exists an $M>0$ such that $\|\overline{x_{0}-x_{1}}\|^{*}_{i_{1},\dots,i_{m}}\leq M$, for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. Then we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. Since $C\in(0,1)$, we have

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$, which means that $\{x_{k}\}$ is a Cauchy sequence.

Moreover, since $X$ is complete with respect to the norms of class-$m$ collection and $K$ is closed then $x_{k}\to x$, with $x\in K$. Propositions 3.5 and 4.2 imply that

[TABLE]

Therefore, $T$ has a fixed point in $K$ with respect to the norms of class-$m$ collection. Next, we want to show the uniqueness of the fixed point with respect to the norms of class-$m$ collection. Assume that $x^{\prime}\in K$ is another fixed point of $T$. Because $T$ is a contractive mapping, there is a $C\in(0,1)$ such that

[TABLE]

for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. This is true only for $\|\overline{x-x^{\prime}}\|^{*}_{i_{1},\dots,i_{m}}=0$, for every $\{i_{1},\dots,i_{m}\}\subset\{1,\dots,n\}$. This means that $x=x^{\prime}$ or $T$ has a unique fixed point. Therefore, for any $m\in\{1,\dots,n\}$, a contractive mapping $T$ has a unique fixed point. ∎

5. Concluding Remarks

Let us consider the $p$-summable sequences $\ell^{p}$ (for $1\leq p\leq\infty$) containing all sequences of real numbers $x=(x_{j})$ for which $\sum|x_{j}|^{p}<\infty$. As in [14], one may equip this space with the following $n$-norm

[TABLE]

with $x_{i}=\left(\xi_{ij}\right)\in\ell^{p},i=1,\dots,n.$

Now, one can see that if $\{y_{1},\dots,y_{n}\}$ is a linearly independent set in $\ell^{p}$, the norm of class-$n$ collection $\|\cdot\|^{*}_{1,\dots,n}$ can be writen as

[TABLE]

This is precisely the norm that was used in [6] as a bridge to prove the fixed point theorem on $(\ell^{p},\|\cdot,\dots,\cdot\|_{p})$.

In this paper, we provide more alternatives to study $\ell^{p}$ as an $n$-normed space. Note that the usual norm on $\ell^{p}$ is defined by

[TABLE]

Furthermore, we have some equivalence relations between norm $\|\cdot\|_{p}$ dan $\|\cdot\|_{p}^{*}$.

Theorem 5.1**.**

[6]* Let $\{y_{1},\dots,y_{n}\}$ be a linearly independen set on $\ell^{p}$. Then the norm $\|\cdot\|_{p}^{*}$ defined in (5.2) is equivalent to the usual norm $\|\cdot\|_{p}$ on $\ell^{p}$. Precisely, we have*

[TABLE]

for every $x\in\ell^{p}$.

Since $(\ell^{p},\|\cdot\|_{p})$ is a Banach space, by Theorem 5.1 we have the following corollary.

Corollary 5.2**.**

[6]* The normed space $(\ell^{p},\|\cdot\|_{p}^{*})$ is a Banach space.*

With our approach, we have the proposition below.

Proposition 5.3**.**

Let $\{y_{1},\dots,y_{n}\}$ be a linearly independent set on $\ell^{p}$. Then the norm of class $n$-collection which is defined in (2.1) is equivalent with the norm $\|\cdot\|_{p}^{*}$ in (5.2). Precisely we have

[TABLE]

Proof.

Consider the $n$-normed space $(\ell^{p},\|\cdot,\dots,\cdot\|_{p})$ and $\{y_{1},\dots,y_{n}\}$ be a linearly independent set on $\ell^{p}$. For any $x\in X$, recall the norm of class-$n$ collection is

[TABLE]

and

[TABLE]

Then we have

[TABLE]

From Theorem (5.1) and by using Hölder inequality we have

[TABLE]

Then we have the inequality we want. ∎

Note that, we write $\|x\|_{1,\dots,n}^{*}$ instead of $\|\overline{x}\|_{1,\dots,n}^{*}$, because the element of class-$n$ collection of an $n$-normed space is the $n$-normed space itself. One might notice that the formula of the norm of class-$n$ collection is simpler than the norm $\|\cdot\|_{p}^{*}$ that Ekariani et al used and both norms are equivalent. Since the norm $\|\cdot\|_{p}^{*}$ equivalent to the usual norm $\|\cdot\|_{p}^{*}$ on $\ell^{p}$, we have the following corollaries.

Corollary 5.4**.**

Let $\{y_{1},\dots,y_{n}\}$ be a linearly independent set on $\ell^{p}$. Then the norm $\|\cdot\|_{1,\dots,n}^{*}$ is equivalent to the usual norm $\|\cdot\|_{p}$ on $\ell^{p}$. Precisely we have

[TABLE]

Corollary 5.5**.**

The normed space $(\ell^{p},\|\cdot\|_{1,\dots,n}^{*})$ is a Banach space.

Now, we present a more general fixed point theorem on $(\ell^{p},\|\cdot,\dots,\cdot\|_{p})$ as follows.

Theorem 5.6**.**

Consider the $n$-normed space $(\ell^{p},\|\cdot,\dots,\cdot\|_{p})$ and let $K\subseteq\ell^{p}$ be nonempty, closed and bounded. If $T:K\to K$ is a contractive mapping with respect to norms of class-$m$ collection for an $m\in\{1,\dots,n\}$, then $T$ has a unique fixed point.

Proof.

Fix an $m\in\{1,\dots,n\}$ and let $T:K\to K$ be a contractive mapping with respect to norms of class-$m$ collection on $\ell^{p}$. By Theorem 4.4, $T$ is contractive with respect to the norm of class-$n$ collection on $\ell^{p}$. By Corollary (5.5) $\ell^{p}$ is complete with respect to the norm of class-$n$ collection. Hence, $T$ has a unique fixed point. ∎

Remark 5.7*.*

We can also prove Theorem 5.6 using Theorem 4.5, noting the fact that $\ell^{p}$ is complete with respect to the norms of class-$m$ collection (see Theorem 2.11).

Acknowledgment

This work is supported by ITB Research and Innovation Program 2019. The first author is also supported by LPDP Indonesia.