Some fixed point theorems using expansive type mappings in 2-Banach space

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Adv. Fixed Point Theory, 5 (2015), No. 2, 158-167 ISSN: 1927-6303

SOME FIXED POINT THEOREMS USING EXPANSIVE TYPE MAPPINGS IN 2-BANACH SPACE

SARLA CHOUHAN1,∗_{, PRABHA CHOUHAN}2_{, MRIDULA DUBEY}1 1_{Barkatullah University, Bhopal (M.P.) 462026, India}

2_{Scope College of Engineering Bhopal, Bhopal (M.P.) 462026, India}

Copyright c2015 Chouhan, Chouhan and Dubey. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In present paper, we define expansive mappings in 2-Banach space and prove some common unique fixed point theorems.

Keywords:2-normed space; 2-Banach space; Expansive mapping; Fixed point. 2010 AMS Subject Classification:47H10, 54H25.

1. Introduction

The research about fixed points of expansive mapping was initiated by Machuca [6]. Later Jungck discussed fixed points for other forms of expansive mapping; see [5]. In 1982, Wanget al. [12] presented some interesting work on expansive mappings in metric spaces which cor-respond to some contractive mapping in [10]. In order to generalize the results on fixed points of nonlinear operators, Zhang [14] studied fixed point problems for expansive mapping. As applications, he also investigated the existence of solutions of equations for locally condensing mapping and locally accretive mappings. On the other hand, Gahler ([2],[3]) investigated the

∗_{Corresponding author}

E-mail addresses: [email protected] (S. Chouhan), [email protected] (P. Chouhan ), [email protected] (M. Dubey)

Received December 11, 2014

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idea of 2-metric and 2-Banach spaces and proved the same results. Subsequently, several au-thors including Iseki [4], Rhoades [8], White [13], Panja and Baisnab [7] and Sahaet al. [11] studied various aspects of fixed points and proved fixed point theorems in 2-metric spaces and 2-Banach spaces. Recently, the study about fixed points for expansive mapping is deeply ex-plored and has extended too many others directions. Motivated and inspired by the above work, we investigate fixed points for expansive mapping in 2-Banach spaces.

2. Preliminaries

Definition 2.1. LetX be a real linear space andk., .kbe a non-negative real valued function defined onX×X satisfying the following conditions:

i)kx,yk=0 if and only ifxandyare linearly dependent inX, ii)kx,yk=ky,xk, ∀x,y∈X

iii)kx,ayk=|a| kx,yk,∀x,y∈X

iv)kx,y+zk=kx,yk+kx,zk,∀x,y,z∈X.

Thenk., .kis called a 2-norm and the pair(X,k., .k)is called a linear 2-normed space. Some of the basic properties of 2-norms are that they are non negative satisfyingkx,y+axk=

kx,yk, for allx,y∈X and all real numbersa.

Definition 2.2. A sequence {xn} in a linear 2-normed space (X,k., .k) is called Cauchy se-quence if limm,n→∞kxm−xn,yk=0∀y∈X.

Definition 2.3. A sequence{xn}in a linear 2-normed space (X,k., .k) is said to be convergent if there is a pointx inX such that limn→∞kxn−x,yk=0 ∀y∈X.If {xn}converges tox, we write{xn} →xasn→∞.

Definition 2.4. A linear 2-normed spaceX is said to be complete if every Cauchy sequence is convergent to an element ofX. We then callX to be a 2-Banach space.

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Example 2.6. Let X is R3and consider the following 2-norm on X as

kx,yk=

det     

i j k x₁ x₂ x₃ y₁ y₂ y₃

     ,

where x= (x₁,x₂,x₃)and y= (y₁,y₂,y₃). Then (X,k., .k) is a 2-Banach space.

Example 2.7.Let P_ndenotes the set of all real polynomials of degree≤n, on the interval [0,1]. By considering usual addition and scalar multiplication, Pn is a linear vector space over the reals. Let{xo,x1, ...,x2n} be distinct fixed points in [0,1] and define the following 2-norm on P_n:k f,gk=∑2_kn=0| f(xk)g(xk)|, whenever f and g are linearly independent andk f,gk=0, if

f , g are linearly dependent. Then (P_n,k., .k) is a 2-Banach space.

Example 2.8.Let X is Q3, the field of rational number and consider the following 2-norm on X as:

kx,yk=

det     

i j k x₁ x₂ x₃ y1 y2 y3

     ,

where x= (x₁,x₂,x₃) and y= (y₁,y₂,y₃). Then (X,k., .k) is not a 2-Banach space but a 2-normed space.

Definition 2.9.Let (X,k., .k) be a 2-Banach space with 2-normk., .k. A mappingT ofXinto itself is said to be expansive if there exists a constanth>1 such thatkT x−Ty,ak≥hkx−y,ak,

∀x,y∈X.

Example 2.10.Let X =R2and consider the following2-norm on X askx,yk=|x₁y₂−x₂y₁|,

where x= (x1,x2)and y= (y1,y2). Then(X,k., .k)is a2-Banach space. Define a self map T

on X as follows T x=βx, whereβ >1for all x∈X , clearly T is an expansive mapping.

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Theorem 3.1. Let (X,k., .k) be a 2-Banach space and T be a surjective mapping of X into itself such that for every x,y∈X,

kT x−Ty,ak ≥ α1

kx−T x,ak ky−Ty,ak kx−y,ak + α2

[1+kx−T x,ak] kx−Ty,ak kx−y,ak

+ α3ky−Ty,ak kx−Ty,ak

kx−y,ak + β1 kx−T x,ak +β2 ky−Ty,ak

+ β3 kx−Ty,ak +β4 kx−y,ak,

(3.1.1)

whereα1,α2,α3,β1,β2,β3,β4≥0,α1+β1+β2+β4>1andβ2+β4>0. Then T has a fixed

point.

Proof. Let x_o be an arbitrary point in X. There is x₁ inX such that T(x₁) =x_o. In this way we define a sequence{xn}as follows.

x_n=T x_n₊₁ f or n =0,1,2, ...

(3.1.2)

Ifx_n=x_n₊₁ for somenthen we see thatx_n is a fixed point ofT, therefore we suppose that no two consecutive terms of sequence{xn}are equal. Now, we consider

kx_n−x_n+1,ak=kT xn+1−T xn+2,ak

≥ α1

kx_n+1−T xn+1,ak kxn+2−T xn+2,ak

kx_n₊₁−x_n₊₂,ak

+α2[1+ kxn+1−T xn+1,ak] kxn+1−T xn+2,ak

kx_n+1−xn+2,ak

+ α3

kx_n₊₂−T x_n₊₂,ak kx_n₊₁−T x_n₊₂,ak kxn+1−xn+2,ak

+β1 kxn+1−T xn+1,ak +β2 kxn+2−T xn+2,ak

+β3 kxn+1−T xn+2,ak +β4 kxn+1−xn+2,ak

= α1

kx_n₊₁−x_n,ak kx_n₊₂−x_n₊₁,ak kx_n+1−xn+2,ak

+ α2

[1+kx_n₊₁−x_n,ak] kx_n₊₁−x_n₊₁,ak kx_n+1−xn+2,ak

+ α3

kx_n+2−xn+1,ak kxn+1−xn+1,ak

kx_n₊₁−x_n₊₂,ak + β1 kxn+1−xn,ak +β2 kxn+2−xn+1,ak

+ β3 kxn+1−xn+1,ak +β4 kxn+1−xn+2,ak

= (α1+β1) kxn−xn+1,ak + (β2+β4) kxn+1−xn+2,ak

⇒kx_n₊₁−x_n₊₂,ak ≤

1−(α1+β1)

β2+β4 k

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So, in general, we have

kxn−xn+1,ak≤k kxn−1−xn,ak f or n=1,2,3, ..,

wherek=1−(α1+β1)

β2+β4 <1 [Asα1+β1+β2+β4>1]

⇒kxn−xn+1,ak ≤kn kx0−x1,ak

(3.1.3)

Now we shall prove that{xn}is a Cauchy sequence. For this,for every positive integer p, we have

kx_n−x_n+p,ak ≤ kxn−xn+1,ak + kxn+1−xn+2,ak +...+ kxn+p−1−xn+p,ak

≤ (kn+kn+1+kn+2+...+kn+p−1) kx₀−x₁,ak [By(3.1.3)] = kn(1+k+k2+...+kp−1) kx₀−x₁,ak

< k

n

(1−k) kx0−x1,ak.

As n→ ∞,k xn−xn+p,ak→0, it follows that {xn} is a Cauchy sequence in X. As X is a complete Banach space, so there exist a pointx ∈ X such that{xn} →x.

Existence of fixed points Since T is a surjective self map and hence there exist pointy in X such that

(3.1.4) x=Ty.

Consider

kxn−x,ak=kT xn+1−Ty,ak

≥ α1kxn+1−T xn+1,ak ky−Ty,ak

kx_n₊₁−yk +α2

[1+ kxn+1−T xn+1,ak] kxn+1−Ty,ak

kx_n₊₁−y,ak

+ α3

ky−Ty,ak kx_n₊₁−Ty,ak kxn+1−y,ak

+β1 kxn+1−T xn+1,ak +β2 ky−Ty,ak

+β3 kxn+1−Ty,ak +β4 kxn+1−y,ak

= α1

kx_n₊₁−x_n,ak ky−Ty,ak kxn+1−y,ak

+ α2

[1+kx_n₊₁−x_n,ak] kx_n₊₁−Ty,ak kxn+1−y,ak

+ α3

ky−Ty,ak kxn+1−Ty,ak

kx_n₊₁−y,ak + β1 kxn+1−xn,ak +β2 ky−Ty,ak

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Since{x_n+1}is a subsequence of{xn}, so{xn} →x,{xn+1} →xwhenn→∞

0≥ α1

kx−x,ak ky−x,ak kx−y,ak + α2

[1+kx−x,ak] kx−x,ak kx−y,ak

+ α3

ky−x,ak kx−x,ak

kx−y,ak + β1 kx−x,ak +β2 ky−x,ak

+ β3 kx−x,ak +β4 kx−y,ak.

0≥(β2+β4)kx−y,ak ⇒kx−y,ak=0.Hence

(3.1.5) x=y.

The fact (3.1.5) along with (3.1.4) shows thatxis a common fixed point of T. This completes the proof of the theorem 3.1.

Theorem 3.2. Let (X,k., .k) be a 2-Banach space and T be a mapping of X into itself such that for every x,y,a∈X,

kT x−Ty,ak≥qmax

n

kx−y,ak,kx−T x,ak ky−Ty,ak

kx−y,ak ,

1+kx−T x,ak

kx−Ty,ak kx−y,ak ,

ky−Ty,ak kx−Ty,ak kx−y,ak

o

(3.1.6)

and q>1. Then T has a fixed point.

Proof. Construct a sequence {xn} as in proof of Theorem 3.1. We claim that the inequality (3.1.6) forx=x_n₊₁andy=x_n₊₂implies that

kT x_n+1−T xn+2,ak ≥ qmax n

kx_n+1−xn+2,ak,

kxn+1−T xn+1,ak kxn+2−T xn+2,ak

kx_n+1−xn+2,ak

,

[1+ kx_n₊₁−T x_n₊₁,ak] kx_n₊₁−T x_n₊₂,ak kx_n+1−xn+2,ak

,

kx_n+2−T xn+2,ak kxn+1−T xn+2,ak

kx_n₊₁−x_n₊₂,ak

o

⇒kxn−xn+1,ak ≥ qmax n

kxn+1−xn+2,ak,

kx_n₊₁−x_n,akkx_n₊₂−x_n₊₁,ak kx_n+1−xn+2,ak

,

[1+ kx_n₊₁−x_n,ak]kx_n₊₁−x_n₊₁,ak kxn+1−xn+2,ak

,kxn+2−xn+1,ak kxn+1−xn+1,ak

kxn+1−xn+2,ak

o

=qmaxkx_n+1−xn+2,ak,kxn+1−xn,ak . Case I:

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kx_n−x_n+1,ak≥q kxn+1−xn+2,ak

⇒kx_n₊₁−x_n₊₂,ak≤ 1_q kx_n−x_n₊₁,ak ⇒kxn+1−xn+2,ak≤k kxn−xn+1,ak h

wherek=1_q <1 (As q>1)i. So, in general, we have

kx_n−x_n+1,ak≤k kxn−1−xn,ak f or n=1,2,3, ...

⇒kx_n−x_n₊₁,ak ≤ kn kx₀−x₁,ak.

(3.1.7)

We find that {xn} is a Cauchy sequence using (3.1.7) as proved in theorem 3.1. As X is a complete Banach space, so there exist a pointx ∈ X such that{xn} →x.

Existence of fixed pointSinceT is a surjective self map and hence there exist pointyinXsuch that

(3.1.8) x=Ty

Considerkx_n−x,ak=kT x_n₊₁−Ty,ak

≥ qmaxnkx_n₊₁−y,ak, kxn+1−T xn+1,ak ky−Ty,ak

kx_n₊₁−y,ak ,

[1+ kx_n+1−T xn+1,ak] kxn+1−Ty,ak

kx_n₊₁−y,ak , ky−Ty,ak kx_n₊₁−Ty,ak

kx_n+1−y,ak

o

⇒kx_n−x,ak ≥ qmax

n

kx_n+1−y,ak,

kx_n+1−xn,akky−x,ak kx_n₊₁−y,ak ,

[1+ kx_n₊₁−x_n,ak]kx_n₊₁−x,ak kx_n+1−y,ak

, ky−x,ak kxn+1−x,ak

kx_n+1−y,ak

o

.

Since{x_n+1}is a subsequence of{xn}, so{xn} →x,{xn+1} →xwhenn→∞.

⇒kx−x,ak ≥ qmax

n

kx−y,ak,kx−x,akky−x,ak

kx−y,ak ,

[1+ kx−x,ak]kx−x,ak kx−y,ak ,

ky−x,ak kx−x,ak kx−y,ak

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0≥qkx−y,ak ⇒kx−y,ak=0x=y(3.1.9)The fact (3.1.9) along with (3.1.8) shows thatx is a common fixed point ofT. This completes the proof of Theorem 3.2.

Theorem 3.3. Let (X,k., .k) be a 2-Banach space and T be a mapping of X into itself such that for every x,y,a∈X,

kT x−Ty,ak2≥ a kx−T x,ak kx−y,ak +b ky−Ty,ak kx−y,ak

+ c kx−T x,ak ky−Ty,ak (3.1.10)

for all distinct x,y∈X where a,c≥0, b>0and a+b+c>1. Then T has a fixed point. Proof. Construct a sequence {xn} as in proof of Theorem 3.1. We claim that the inequality (3.1.10) forx=xn+1andy=xn+2implies that

kT x_n₊₁−T x_n₊₂,ak2≥ a kx_n₊₁−T x_n₊₁,ak kx_n₊₁−x_n₊₂,ak

+ b kx_n₊₂−T x_n₊₂,ak kx_n₊₁−x_n₊₂,ak

+ c kxn+1−T xn+1,ak kxn+2−T xn+2,ak

=a kx_n+1−xn,ak kxn+1−xn+2,ak

+ b kx_n₊₂−x_n₊₁,ak kx_n₊₁−x_n₊₂,ak

+ c kx_n₊₁−x_n,ak kx_n₊₂−x_n₊₁,ak kx_n−x_n₊₁,ak2_{≥ k}_x

n+1,xn+2,ak (a+b+c) min{kxn,xn+1,ak,kxn+1,xn+2,ak}

Case I:

kx_n−x_n₊₁,ak2≥(a+b+c) kx_n₊₁−x_n₊₂,ak kx_n−x_n₊₁,ak

⇒kx_n+1−xn+2,ak≤

1

a+b+c kxn−xn+1,ak ⇒kx_n+1−xn+2,ak≤k1 kxn−xn+1,ak,

h

where k₁= 1

a+b+c <1(As a+b+c>1)

i

.

Case II:

kx_n−x_n+1,ak2≥(a+b+c) kxn+1−xn+2,ak kxn+1−xn+2,ak

⇒kxn+1−xn+2,ak2≤

1

a+b+c kxn−xn+1,ak

2

⇒kx_n+1−xn+2,ak≤

1 a+b+c

1₂

kx_n−x_n+1,ak

⇒kx_n₊₁−x_n₊₂,ak≤k₂ kx_n−x_n₊₁,ak,

h

where k₂= 1

a+b+c

1₂

<1(As a+b+c>1)

i

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Letk=max{k₁,k₂}. Hencek<1. So, in general

kx_n−x_n₊₁,ak≤k kx_n₋₁−x_n,ak f or n=1,2,3...

kxn−xn+1,ak ≤kn kx0−x1,ak.

(3.1.11)

We can prove that{xn}is a Cauchy sequence using (3.1.3) as proved in theorem 3.1. AsX is a complete Banach space, so there exist a pointx ∈ X such that{xn} →x.

Existence of fixed pointSinceT is a surjective self map and hence there exist pointyinXsuch that

(3.1.12) x=Ty

Considerkxn−x,ak=kT xn+1−Ty,ak

kT x_n+1−Ty,ak2≥ a kxn+1−T xn+1,ak kxn+1−y,ak

+ b ky−Ty,ak kx_n+1−y,ak

+ c kx_n₊₁−T x_n₊₁,ak ky−Ty,ak

kxn−x,ak2≥ a kxn+1−xn,ak kxn+1−y,ak

+ b ky−x,ak kx_n+1−y,ak

+ c kx_n+1−xn,ak ky−x,ak.

Since{xn+1}is a subsequence of{xn}, so{xn} →x,{xn+1} →xwhenn→∞

kx−x,ak2≥ a kx−x,ak kx−y,ak

+ b ky−x,ak kx−y,ak

+ c kx−x,ak ky−x,ak

0≥bkx−y,ak2 ⇒kx−y,ak=0. This shows that x is a common fixed point of T. This completes the proof of Theorem 3.3.

Conflict of Interests

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