# Fixed Point Theorems for a Class of Weakly C-Contractive Mappings in a Setting of 2-Banach Space

Authors: M. Saha; Anamika Ganguly
Journal: Journal of Mathematics
 (2013)

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## Abstract

In this paper, the authors have proved some existence theorems of fixed points for a class of weakly C-contractive mappings in a setting of 2-Banach space. The authors have supported the results with the help of suitable examples.


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## Body

## 1. Introduction

The theory of 2-Banach spaces was investigated by Gähler in [1] and Iseki in [2] who proved some fixed point theorems in such spaces and these new spaces have subsequently been studied by many authors. For example, we refer to [3, 4] where authors have dealt with mappings that are of contractive nature. Our findings as presented here have added further findings in fixed point theory in a 2-Banach space by working with a class of mappings so general that take care of those as found in [5, 6].We now state some definitions before presenting our main results.


## 2. Preliminaries

Definition 1.
LetX be a real linear space and  ∥·,·∥  a non-negative real valued function defined on X satisfying the following conditions:(i)
∥x,y∥=0 if and only if  x and y are linearly dependent,(ii)
∥x,y∥=∥y,x∥ for all x,y∈X,(iii)
∥x,αy∥=|α|∥x,y∥, α being real for all x,y∈X,(iv)
∥x,y+z∥≤∥x,y∥+∥x,z∥  for all x,y,z∈X.Then ∥·,·∥ is called a 2-norm and the pair (X,∥·,·∥) is called a linear 2-normed space. So a 2-norm ∥x,y∥ always satisfies ∥x,y+αx∥=∥x,y∥ for all x,y∈X and all scalars α.Definition 2.
A sequence{xn} in a linear 2-normed space (X,∥·,·∥) is said to be a Cauchy sequence if limn,m→∞∥xn-xm,a∥=0 for all a in X.Definition 3.
A sequence{xn} in a linear 2-normed space (X,∥·,·∥) is said to be convergent in X if there exists a point x in X such that limn→∞∥xn-x,a∥=0 for all a in X. If {xn} converges to x, we write xn→x as n→∞.Definition 4.
A linear 2-normed spaceX is said to be complete if every Cauchy sequence is convergent to an element of X. A complete linear 2-normed space X is called a 2-Banach space.
The main purpose of our paper is to find out the existence of fixed point for a class of weakly C-contractive results due to Choudhury [7] in a setting of 2-Banach space. We have also established a common fixed point theorem for such type of mappings with a supporting example. Some allied theorems have also been presented here.


## 3. Main Results

Definition 5.
A mappingT:(X,∥·,·∥)→(X,∥·,·∥) is said to be weakly C-contractive or a weak C-contraction if for all x,y,a in X,
(1)∥T(x)-T(y),a∥≤12[∥x-T(y),a∥+∥y-T(x),a∥]-ϕ[∥x-T(y),a∥,∥y-T(x),a∥],
where ϕ:[0,∞)2→[0,∞) is a continuous mapping such that ϕ(x,y)=0 if and only if x=y=0.Theorem 6.
Let(X,∥·,·∥) be a 2-Banach space and let T:X→X be a weak C-contraction. Then T has a unique fixed point in X.Proof.
Letx0∈X. Let  xn+1=T(xn), n=1,2,….
Ifxn=xn+1=T(xn), then T has a fixed point xn.
Assume thatxn≠xn+1. Now for any a∈X,
(2)∥xn-xn+1,a∥=∥T(xn-1)-T(xn),a∥≤12[∥xn-1-T(xn),a∥+∥xn-T(xn-1),a∥]-ϕ[∥xn-1-T(xn),a∥,∥xn-T(xn-1),a∥]≤12[∥xn-1-xn,a∥+∥xn-xn+1,a∥]-ϕ[∥xn-1-xn+1,a∥,0].
Therefore,
(3)∥xn-xn+1,a∥≤∥xn-1-xn,a∥.
So {∥xn-xn+1,a∥} is a monotone decreasing sequence of real numbers and hence {∥xn-xn+1,a∥} is convergent and let ∥xn-xn+1,a∥→r (say) as n→∞
We now show thatr=0. If not, then
(4)∥xn-xn+1,a∥=∥T(xn-1)-T(xn),a∥≤12∥xn-1-xn+1,a∥≤12[∥xn-1-xn,a∥+∥xn-xn+1,a∥].
Taking n→∞, we get
(5)r≤12limn→∞∥xn-1-xn+1,a∥+12limn→∞∥xn-xn+1,a∥≤12r+12r,
implying that
(6)limn→∞∥xn-1-xn+1,a∥=2r.
Also from (2) and using continuity of ϕ, we get r≤r-ϕ[2r,0] showing that ϕ[2r,0]≤0, a contradiction, unless r=0. So, ∥xn-xn+1,a∥→0 as n→∞, that is
(7)limn→∞∥xn-xn+1,a∥=0.
We shall now show that {xn} is Cauchy sequence in X.
If not, then there exists aε>0 and for each positive integer k, there exist integers n(k) and m(k) with k<m(k)<n(k) such that ∥xm(k)-xn(k),a∥≥ε and ∥xm(k)-xn(k)-1,a∥<ε. Then
(8)ε≤∥xm(k)-xn(k),a∥=∥T(xm(k)-1)-T(xn(k)-1),a∥≤12[∥xm(k)-1-xn(k),a∥+∥xn(k)-1-xm(k),a∥]-ϕ[∥xm(k)-1-xn(k),a∥,∥xn(k)-1-xm(k),a∥].
Also,
(9)ε≤∥xm(k)-xn(k),a∥≤∥xm(k)-xn(k)-1,a∥+∥xn(k)-1-xn(k),a∥<ε+∥xn(k)-1-xn(k),a∥.
So by using (7), we obtain
(10)limn→∞∥xm(k)-xn(k),a∥=ε.
Now
(11)∥xm(k)-xn(k)-1,a∥≤∥xm(k)-xm(k)-1,a∥+∥xm(k)-1-xn(k),a∥+∥xn(k)-xn(k)-1,a∥.
Also
(12)∥xm(k)-1-xn(k),a∥≤∥xm(k)-1-xm(k),a∥+∥xm(k)-xn(k),a∥.
Therefore,
(13)limk→∞∥xm(k)-1-xn(k),a∥=ε.
Similarly, it can be shown that(14)limk→∞∥xn(k)-1-xm(k),a∥=ε.
So from (8), we get
(15)ε≤12(ε+ε)-ϕ[ε,ε]
implying that ϕ[ε,ε]≤0, a contradiction.
Hence{xn} is a Cauchy sequence and therefore {xn} is a convergent in X and let xn→z∈X (say) as n→∞. Now for any a∈X,
(16)∥z-T(z),a∥≤∥z-xn+1,a∥+∥xn+1-T(z),a∥≤∥z-xn+1,a∥+∥T(xn)-T(z),a∥≤∥z-xn+1,a∥+12[∥z-T(xn),a∥+∥xn-T(z),a∥]-ϕ[∥z-T(xn),a∥,∥xn-T(z),a∥].
So by the continuity of ϕ and taking n→∞, we get
(17)∥z-T(z),a∥≤12[∥z-T(z),a∥-ϕ[0,∥z-T(z),a∥]]≤12∥z-T(z),a∥
implying that z=T(z).
We shall now show that the fixed pointz is unique. If possible let z1∈X be another fixed point of T, then for any a∈X(18)∥z-z1,a∥=∥T(z)-T(z1),a∥≤12[∥z-T(z1),a∥+∥z1-T(z),a∥]-ϕ[∥z-T(z1),a∥,∥z1-T(z),a∥]≤∥z-z1,a∥-ϕ[∥z-z1,a∥,∥z-z1,a∥],
which by property of ϕ is a contradiction unless ∥z-z1,a∥=0 for all a∈X that is z=z1.
This completes the proof.Example 7.
LetX be ℝ3 and consider the following 2-norm on X as
(19)∥x,y∥=|det(ijkx1x2x3y1y2y3)|,
where x=(x1,x2,x3), y=(y1,y2,y3)∈X. Then (X,∥·,·∥) is a 2-Banach space (see [8, 9]).
T
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→
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Now setting(20)ϕ((t,t′))=12[t+t′];(t,t′)∈[0,∞)×[0,∞).ϕ:[0,∞)2→[0,∞) is a continuous mapping such that ϕ((t,t′))=0 if and only if t=t′=0. Now T(x)=T(x1,x2,x3)=(0,0,1) and T(y)=T(y1,y2,y3)=(0,0,1). Take a=(a1,a2,a3)∈X.
Now(21)∥T(x)-T(y),a∥=|det(ijk000a1a2a3)|=0.
Also,
(22)∥T(x)-T(y),a∥≤12[∥x-T(y),a∥+∥y-T(x),a∥]-ϕ[∥x-T(y),a∥,∥y-T(x),a∥].
All the conditions of Theorem 6 are satisfied.
HenceT has a unique fixed point (0,0,1) in ℝ3.Theorem 8.
LetF and G be two self-mappings of a 2-Banach space X. Suppose that for any a∈X,
(23)∥F(x)-G(y),a∥≤12[∥x-G(y),a∥+∥y-F(x),a∥]-ϕ[∥x-G(y),a∥,∥y-F(x),a∥],
where ϕ:[0,∞)2→[0,∞) is a continuous mapping such that ϕ(x,y)=0 if and only if x=y=0. Then F and G have a unique common fixed point in X.Proof.
Forx0∈X, define {xn}n≥1 by x2n+1=F(x2n),x2n+2=G(x2n+1). So for any a∈X,
(24)∥x2n+1-x2n+2,a∥=∥F(x2n)-G(x2n+1),a∥≤12[∥x2n-G(x2n+1),a∥+∥x2n+1-F(x2n),a∥]-ϕ[∥x2n-G(x2n+1),a∥,∥x2n+1-F(x2n),a∥]≤12[∥x2n-x2n+2,a∥]≤12[∥x2n-x2n+1,a∥+∥x2n+1-x2n+2,a∥].Hence
(25)∥x2n+1-x2n+2,a∥≤∥x2n-x2n+1,a∥.Similarly
(26)∥x2n-x2n+1,a∥≤∥x2n-1-x2n,a∥.
Thus {∥xn-xn+1,a∥} is a monotone decreasing sequence of real numbers and hence is convergent and
Let(27)limn→∞∥xn-xn+1,a∥=r≥0(say)
We now claim thatr=0.
If not,(28)∥x2n+1-x2n+2,a∥≤12∥x2n-x2n+2,a∥≤12[∥x2n-x2n+1,a∥+∥x2n+1-x2n+2,a∥].
So,
(29)limn→∞∥x2n-x2n+2,a∥=2r
from (24) taking n→∞(30)r≤r-ϕ[2r,0]
implying that ϕ[2r,0]≤0, a contradiction.
Therefore,r=0.
Now for any integerp>0(31)∥xn-xn+p,a∥≤∥xn-xn+1,a∥+∥xn+1-xn+2,a∥+⋯+∥xn+p-1-xn+p,a∥→0asn→∞.
So{xn} is Cauchy in X and hence by completeness of X, let
(32)limn→∞xn=z∈X.
Now for all a∈X,
(33)∥z-F(z),a∥≤∥z-xn+2,a∥+∥x2n+2-F(z),a∥≤∥z-x2n+2,a∥+∥F(z)-G(x2n+1),a∥.
By routine check up, we can see that for sufficiently large values of n, Hence, z=Fz.
Similarly we can show thatz=G(z).
We shall now show thatz is unique.
If not letz1∈X be another fixed point of F and G. Then for any a∈X(34)∥z-z1,a∥=∥F(z)-G(z1),a∥≤12[∥z-z1,a∥+∥z1-z,a∥]-ϕ[∥z-z1,a∥,∥z1-z,a∥].
So∥z-z1,a∥≤∥z-z1,a∥-ϕ[∥z-z1,a∥,∥z1-z,a∥], showing that ϕ[∥z-z1,a∥,∥z1-z,a∥]≤0. So z-z1=0 implies z=z1.Example 9.
LetX be ℝ3 and consider the following 2-norm on X as
(35)∥x,y∥=|det(ijkx1x2x3y1y2y3)|,
where x=(x1,x2,x3), y=(y1,y2,y3)∈X. Then (X,∥·,·∥) is a 2-Banach space (see [8, 9]).
DefineF,G:X→X by F(x,y,z)=((1/4)x,(1/4)y,(1/4)z) for all (x,y,z)∈X and G(x,y,z)=((1/3)x,(1/3)y,(1/3)z) for all (x,y,z)∈X.
Now setting(36)ϕ((t,t′))=12max{t,t′};(t,t′)∈[0,∞)×[0,∞).ϕ:[0,∞)2→[0,∞) is a continuous mapping such that ϕ((t,t′))=0 if and only if t=t′=0. Take a=(a1,a2,a3)∈X.
Now(37)∥F(x)-G(y),a∥=|det(ijk(14x1-13y1)(14x2-13y2)(14x3-13y3)a1a2a3)|∥x-G(y),a∥=|det(ijk(x1-13y1)(x2-13y2)(x3-13y3)a1a2a3)|,∥y-F(x),a∥=|det(ijk(y1-13x1)(y2-13x2)(y3-13x3)a1a2a3)|.
By routine calculation, it can be seen that the condition in Theorem 8(38)∥F(x)-G(y),a∥≤12[∥x-G(y),a∥+∥y-F(x),a∥]-ϕ[∥x-G(y),a∥,∥y-F(x),a∥]
is satisfied. Also F and G have a unique common fixed point (0,0,0) in X.Theorem 10.
Let(X,∥·,·∥) be a 2-Banach space with T:X→X and let {Tn:X→X} be a sequence of mappings such that(i)
for allu∈X and n=1,2,…(39)∥Tn(x)-Tn(y),u∥≤12[∥x-Tn(y),u∥+∥y-Tn(x),u∥]-ϕ[∥x-Tn(y),u∥,∥y-Tn(x),u∥],whereϕ:[0,∞)2→[0,∞) is a continuous mapping such that ϕ(x,y)=0 if and only if x=y=0,(ii)
limn→∞Tn(x)=T(x) for each x∈X. Then T has a unique fixed point z in X such that z=limn→∞zn where zn is the fixed point of Tn(n=1,2,…).Proof.
(40)
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Now
(41)∥x-Tn(y),u∥≤∥x-T(y),u∥+∥Tn(y)-T(y),u∥→∥x-T(y),u∥asn→∞.
Similarly
(42)∥y-Tn(x),u∥→∥y-T(x),u∥asn→∞.
As ϕ is continuous,
(43)ϕ[∥x-Tn(y),u∥,∥y-Tn(x),u∥]→ϕ[∥x-T(y),u∥,∥y-T(x),u∥].
So from (40) taking limit as n→∞, we obtain
(44)∥T(x)-T(y),u∥≤12[∥x-T(y),u∥+∥y-T(x),u∥]-ϕ[∥x-T(y),u∥,∥y-T(x),u∥].
Hence by Theorem 6, T has a unique fixed point z in X.
If possible, let(45)limn→∞zn≠z∈X.
Now for u∈X let
(46)limn→∞∥zn-z,u∥=α>0(47)∥z-zn,u∥=∥T(z)-Tn(zn),u∥≤∥T(z)-Tn(z),u∥+∥Tn(z)-Tn(zn),u∥.
Now
(48)∥Tn(z)-Tn(zn),u∥≤12[∥z-Tn(zn),u∥+∥zn-Tn(z),u∥]-ϕ[∥z-Tn(zn),u∥,∥zn-Tn(z),u∥].
Now
(49)∥zn-Tn(z),u∥≤∥zn-z,u∥+∥z-T(z),u∥+∥T(z)-Tn(z),u∥=∥zn-z,u∥+∥T(z)-Tn(z),u∥.
So from (46), (47), (48), and (49), we get
(50)limn→∞∥z-zn,u∥≤limn→∞∥Tn(z)-Tn(zn),u∥≤limn→∞12[∥z-zn,u∥+∥zn-z,u∥]-limn→∞ϕ[∥z-zn,u∥,∥zn-z,u∥]
Hence,α≤(1/2)(2α)-ϕ[α,α] showing that ϕ[α,α]≤0 which is possible only when α=0
Hencelimn→∞∥z-zn,u∥=0, implying that limn→∞zn=z.Theorem 11.
LetX be a 2-Banach space, {Tn} a sequence of mappings of X onto itself with fixed points zn. Let T be a mapping of X into itself satisfying the condition of Theorem 6 with fixed point z, such that Tn→T uniformly on {zn:n=1,2,…}. Then limn→∞zn=z∈X.Proof.
Fixε>0. From the uniform convergence of {Tn} on {zn:n=1,2,…}, there exists an integer N such that for all n≥N, and for all u∈X,  ∥T(zn)-Tn(zn),u∥<ε for all zn.
Now(51)∥z-zn,u∥=∥T(z)-Tn(zn),u∥≤∥T(z)-T(zn),u∥+∥T(zn)-Tn(zn),u∥.
If possible let limn→∞zn≠z∈X that is there exists u∈X,
(52)limn→∞∥zn-z,u∥=α>0.
Now for all u∈X,
(53)∥T(z)-T(zn),u∥≤12[∥z-T(zn),u∥∥zn-T(z),u∥]-ϕ[∥z-T(zn),u∥,∥zn-T(z),u∥].
So for u∈X satisfying (52)
(54)∥z-T(zn),u∥≤∥z-zn,u∥+∥Tn(zn)-T(zn),u∥.
Therefore
(55)limn→∞∥z-T(zn),u∥≤ε+limn→∞∥z-zn,u∥=ε+α.
As ε>0 is arbitrary,
(56)limn→∞∥z-T(zn),u∥≤α.
Again
(57)∥z-zn,u∥≤∥z-T(zn),u∥+∥T(zn)-Tn(zn),u∥.
Hence
(58)limn→∞∥z-zn,u∥≤limn→∞∥z-T(zn),u∥+ε.
As ε>0 is arbitrary,
(59)α≤limn→∞∥z-T(zn),u∥.
Therefore
(60)limn→∞∥z-T(zn),u∥=α.
Hence
(61)limn→∞ϕ[∥z-T(zn),u∥,∥zn-T(z),u∥]→ϕ[α,α].
Taking n→∞ in (53), we get α≤(1/2)(2α)-ϕ[α,α], showing that ϕ[α,α]≤0 which  possible only when α=0, a contradiction to (52).
Hencelimn→∞∥zn-z,u∥=0, implying that limn→∞zn=z.Theorem 12.
Let(X,∥·,·∥) be a 2-Banach space and for any x∈X, {Tn(x)}n=1,2,… a sequence of mappings of X onto itself. Suppose there exists a sequence of nonnegative integers {mn} such that for all x,y,u∈X, and every pair i,j with i≠j satisfying
(62)∥Timi(x)-Tjmj(y),u∥≤12[∥x-Tjmj(y),u∥+∥y-Timi(x),u∥∥x-Tjmj(y),u∥]-ϕ[∥x-Tjmj(y),u∥,∥y-Timi(x),u∥∥x-Tjmj(y),u∥],
where ϕ:[0,∞)2→[0,∞) is a continuous mappings such that ϕ(x,y)=0 if and only if x=y=0. Then {Tn} have a unique common fixed point in X.Proof.
DefineSi=Timi,i=1,2,….
Then(63)∥Si(x)-Sj(y),u∥≤12[∥x-Sj(y),u∥+∥y-Si(x),u∥∥x-Sj(y),u∥]-ϕ[∥x-Sj(y),u∥,∥y-Si(x),u∥∥x-Sj(y),u∥].
Pickx0∈X and define xn=Sn(xn-1),n=1,2,….
Now(64)∥xn-xn+1,u∥=∥Sn(xn-1)-Sn+1(xn),u∥≤12[∥xn-1-Sn+1(xn),u∥+∥xn-Sn(xn-1),u∥]-ϕ[∥xn-1-Sn+1(xn),u∥,∥xn-Sn(xn-1),u∥].
By routine calculation, we can see that
(65)∥xn-xn+1,u∥≤∥xn-1-xn,u∥.
Implying that {∥xn-xn+1,u∥} is a monotone decreasing sequence of real numbers and hence is convergent and let
(66)limn→∞∥xn-xn+1,u∥=r>0.
We shall now show that r=0.
If not,(67)∥xn-xn+1,u∥≤12[∥xn-1-xn+1,u∥]≤12[∥xn-1-xn,u∥+∥xn-xn-1,u∥].
Taking n→∞,
(68)r≤12limn→∞[∥xn-1-xn+1,u∥]≤12[limn→∞∥xn-1-xn,u∥+limn→∞∥xn-xn+1,u∥]=r.
Hence
(69)limn→∞[∥xn-1-xn+1,u∥]=2r.
Taking n→∞ in (64), we get r≤r-ϕ[2r,0], showing that ϕ[2r,0]≤0 which is possible only when r=0.
Now for any integerp>0(70)∥xn-xn+p,u∥≤∥xn-xn+1,u∥+∥xn+1-xn+2,u∥+⋯+∥xn+p-1-xn+p,u∥→0asn→∞.
Therefore {xn} is Cauchy in X and hence by completeness of X, let
(71)limn→∞xn=z∈X.
Now for any integerm,
(72)∥z-Sm(z),u∥≤∥z-xn,u∥+∥xn-Sm(z),u∥=∥z-xn,u∥+∥Sn(xn-1)-Sm(z),u∥≤∥z-xn,u∥+12[∥xn-1-Sm(z),u∥+∥z-Sn(xn-1),u∥]≤∥z-xn,u∥+12[∥xn-1-z,u∥]+12[∥z-Sm(z),u∥]+∥z-xn,u∥.
Hence
(73)12∥z-Sm(z),u∥≤2[∥z-xn,u∥]+12[∥xn-1-z,u∥]→0asn→∞
Therefore ∥z-Sm(z),u∥=0 implying that Sm(z)=z for all m.
Hence{Sm} has a common fixed point z.
Ifz1∈X is another fixed point of {Sn}, then
(74)∥z-z1,u∥=∥Sn(z)-Sn(z1),u∥≤12[∥z-z1,u∥+∥z-z1,u∥]-ϕ[∥z-z1,u∥,∥z-z1,u∥].
Implying that ϕ[∥z-z1,u∥,∥z-z1,u∥]≤0. Hence z-z1=0 implying that z=z1.
Now for eachm,
(75)Tn(z)=Tn(Sn(z))=Tn(Tnmn(z))=Tnmn(Tn(z)),
showing that Tn(z) is the fixed point of Sn.
AlsoSn(z)=z and Sn(Tn(z))=Tn(z). Hence z=Tn(z) for all n.



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*Source: 434205-2013-01-17.xml*