Common Fixed Point Theorems In G-Metric Spaces for Weakly Mapping by Using Contraction Condition

Abstract

In this paper, we studied common fixed point theorems in G-metric spaces. We establish several fixed point theorems for single and multi-valued mappings in G-metric spaces, including Banach, Kannan, Chatterjea type contractions. The main result is fixed point theorem in compact G- metric spaces for weakly mapping by using contraction condition. Our results provide a unified framework for studying the existence and uniqueness of fixed points in G-metric spaces, and have applications in various fields such as mathematics, physics, and engineering.

Keywords

G-metric spaces, fixed point theorems, Banach contraction, Kannan contraction, Chatterjea contraction

Introduction

Mustafa defined new approach of Generalized metric spaces [5][6]. Renu defines property P in G- metric space [8]. Aydi & Binayak proved some fixed-point theorems in G- metric spaces [1][2]. Shantanawi explained Fixed point theorems for nonlinear contraction in G- metric space [9]. Dhage proved fixed point theorems in G- metric space [3][4]. Preeti Bhardwaj defines fixed point theorem in G- metric space for auxiliary functions [7]. The study of fixed-point theory has been a vibrant area of research in mathematics, with applications in various fields such as physics, engineering, and economics. In recent years, the concept of G-metric spaces has gained significant attention, as it provides a more general framework for studying metric spaces.

PRELIMINARIES:

Definition 1: A G
-metric space is a triple (X, G, ?)
, where X
 is a non-empty set, G: X³ → R?
 is a function satisfying the following conditions:

1. G(x, y, z) = 0 if and only if x = y = z

2. G(x, x, y) ≤ G(x, y, z) for all x, y, z ∈ X

3.G(x, y, z) = G(x, z, y) = G(y, z, x) = ...
(symmetry)

4.G(x, y, z) ≤ G(x, a, a) +G(a, y, z) for all x, y, z, a ∈ X

Definition 2:  A mapping T: X → X
 is said to be G
-continuous if for every x ∈ X
 and every ε > 0,
 there exists δ > 0
 such that:

G(Tx, Ty, Tz) < ε
 , whenever G(x, y, z) < δ

Definition 3: A mapping T: X → X
 is said to be a G-contraction if there exists k ∈ (0, 1)
 such that: G(Tx, Ty, Tz) ≤ kG(x, y, z)
 for all x, y, z ∈ X

 Definition 4: A point x ∈ X
 is said to be a common fixed point of two mappings T, S: X → X
 if  Tx = Sx = x

Definition 5: Two mappings T, S: X → X
 are said to commute if

TSx = STx
 , for all x ∈ X.

Some Properties of G
-metric space: Some properties of G- metric spaces those who have used in showing proof of theorems.

1)  Continuity: - G(x, y, z)
 is continuous in each of its variables.

2) Homogeneity: G(cx, cy, cz) = |c|G(x, y, z)
 for all x, y, z ∈ X
 and c ∈ R.

3) Translation Invariance:  G(x + a, y + a, z + a) = G(x, y, z)
 for all x, y, z, a ∈ X.

4)  Scaling Property: G(αx, αy, αz) = αG(x, y, z)
 for all x, y, z ∈ X
 and α ≥ 0.

5)  Metric Property: G(x, y, z) = 0
 if and only if x = y = z.

6) G-
Completeness: A G
-metric space (X, G)
 is said to be G-
complete if every Cauchy sequence in X
 converges to a point in X.

These properties are used to establish various fixed point theorems in G-metric spaces, such as the Banach fixed point theorem, Kannan fixed point theorem, and others.

Theorem 1: Banach Contraction Principle in G-Metric Spaces

Let (X, G)
 be a complete G-metric space, and let T: X → X
 be a contraction mapping, i.e., there exists k ∈ (0, 1)
 such that G(Tx, Ty, Tz) ≤ kG(x, y, z)
 for all x, y, z ∈ X
. Then T
has a unique fixed point.

Proof: Let x? ∈ X be arbitrary. Define a sequence {x?} in X by x? = Tx??? for all n ∈ N.

Then, for all n ∈ N, we have:

G(x?,x???,x???) = G(Tx???,Tx?,Tx?) ≤ kG(x???,x?,x?)

Using the property of G-metric, we get:

G(x?,x???,x???) ≤ k?G(x?,x?,x?)

Since k ∈ (0,1), we have k? → 0 as n → ∞. Therefore, {x?} is a Cauchy sequence in X. Since X is complete, there exists x ∈ X such that x? → x as n → ∞. Now, we have:

G(x,Tx,Tx) ≤ G(x,x?,x?) + G(x?,Tx,Tx)

Using the continuity of G, we get:

G(x,Tx,Tx) ≤ G(x,x,x) + kG(x???,x?,x?)

Letting n → ∞, we get G(x,Tx,Tx) = 0, which implies that Tx = x. Therefore, x is a fixed point of T.

Uniqueness: suppose that y ∈ X is another fixed point of T. Then, we have:

G(x,y,y) = G(Tx,Ty,Ty) ≤ kG(x,y,y)

Since k ∈ (0,1), we must have G(x,y,y) = 0, which implies that x = y. Therefore, the fixed point of T is unique.

Theorem 2: Kannan Fixed Point Theorem in G-Metric Spaces

Let (X, G) be a complete G-metric space, and let T: X → X be a mapping such that G(Tx,Ty,Tz) ≤ k(G(x,Tx,Tx) + G(y,Ty,Ty)) for all x,y,z ∈ X, where k ∈ (0,1/2). Then T has a fixed point.

Proof: Let x? ∈ X be arbitrary. Define a sequence {x?} in X by x? = Tx???, for all n ∈ N.

Then G(x?,x???,x???) = G(Tx???,Tx?,Tx?)

 ≤ k (G (x???, Tx???, Tx???) + G(x?,Tx?,Tx?))

 = k (G(x???, x?,x?) + G(x?,x???,x???))

    ≤ k (G(x???, x???,x?) + G(x???,x?,x?) + G(x?,x???,x???))

    ≤ 2kG (x???, x?,x?) + kG(x?,x???,x???)

    ≤ 2kG (x???, x???, x?) + 2kG(x?,x?,x???)

    ≤ 4k²G (x???, x???, x?) + 2kG(x?,x?,x???)

    ≤ 4k²G(x???, x???,x?) + 4k²G(x?,x?,x???)

    ≤ 8k³G(x???, x???,x?) ≤ ...  ≤ (2k) ?G(x?,x?,x?)

Since k ∈ (0,1/2), we have (2k) ? → 0 as n → ∞.

Therefore G(x?,x???,x???) → 0 as n → ∞.

Hence G(x?,x???,x???) < ε   for all n ≥ N, where ε > 0 is arbitrary.

This shows that {x?} is a Cauchy sequence in X.

Since X is complete, there exists x ∈ X such that x? → x   as n → ∞.

Now G(Tx,x,x) ≤ G(Tx,Tx?,Tx?) + G(Tx?,x,x)

      ≤ k(G(x,Tx,Tx) + G(x?,Tx?,Tx?)) +       G(Tx?,x,x)

      ≤ k(G(x,Tx,Tx) + G(x?,x?,x)) + G(Tx?,x,x)

      ≤ 2kG(x,Tx,Tx) + G(Tx?,x,x)

Since Tx? → Tx as n → ∞, we have:

G(Tx, x,x) → G(Tx,x,x)   as n → ∞.

Therefore    G(Tx,x,x) ≤ 2kG(x,Tx,Tx)

Since k ∈ (0,1/2), we have:

G(Tx,x,x) = 0   which implies that Tx = x.

Hence?x is a fixed point of T.

Example 1: Let X = R and G(x,y,z) = |x - y| + |y - z| + |z - x|. Let T: X → X be defined by Tx = x/2. Then T is a contraction mapping with k = 1/2. Therefore, T has a unique fixed point, which is x = 0.

Theorem 3: Chatterjea Fixed Point Theorem in G-Metric Spaces

Let (X,G) be a complete G-metric space, and let T: X → X be a mapping such that G(Tx,Ty,Tz) ≤ k max{G(x,Tx,Tx),G(y,Ty,Ty)} for all x,y,z ∈ X, where

k ∈ (0,1). Then T has a fixed point.

Proof: Let x? ∈ X be arbitrary. Define a sequence {x?} in X by x? = Tx??? for all n ∈ N.

Then G(x?,x???,x???) = G(Tx???,Tx?,Tx?)

         ≤ k max{G(x???,Tx???,Tx???),G(x?,Tx?,Tx?)}

         = k max{G(x???,x?,x?),G(x?,x???,x???)}

         ≤ kG(x???,x?,x?) ≤ k²G(x???,x???,x???)

         ≤ ...   ≤ k?G(x?,x?,x?)

Since k ∈ (0,1), we have k? → 0 as n → ∞. Therefore:

G(x?,x???,x???) → 0  as n → ∞.

Hence G(x?,x???,x???) < ε   for all n ≥ N, where ε > 0 is arbitrary.

This shows that {x?} is a Cauchy sequence in X.

Since X is complete, there exists x ∈ X such that: 

x? → x as n → ∞.

Now G(Tx,x,x) ≤ G(Tx,Tx?,Tx?) + G(Tx?,x,x)

        ≤ k max{G(x,Tx,Tx),G(x?,Tx?,Tx?)}          G(Tx?,x,x) ≤ kG(x,Tx,Tx) + G(Tx?,x,x)

Since Tx? → Tx as n → ∞, we have:

       G(Tx, x,x) → G(Tx,x,x) as n → ∞.

Therefore G(Tx,x,x) ≤ kG(x,Tx,Tx)

Since k ∈ (0,1), we have:   G(Tx,x,x) = 0 which implies that Tx = x.

Hence:  X is a fixed point of T.

Lemma: Every compact G-metric space is complete.

Proof: Let (X, G) be a compact G-metric space. We need to show that every Cauchy sequence in X converges to a point in X.

Let {x?} be a Cauchy sequence in X. Since X is compact, there exists a subsequence {x??} of {x?} that converges to a point x ∈ X.

We claim that x? → x as n → ∞. Suppose not. Then there exists ε > 0 such that for all N ∈ N, there exists n ≥ N such that G(x, x,x) ≥ ε.

Since {x?} is Cauchy, there exists N ∈ N

such that   G(x?,x?,x?) < ε/2 for all n,m ≥ N.

Since x?? → x asks → ∞, there exists K ∈ N

such that   G(x, x,x) < ε/2 for all k ≥ K.

Choose k ≥ K such that n? ≥ N. Then:

     G(x, x,x) ≤ G(x?,x??,x??) + G(x??,x,x)< ε/2 + ε/2= ε

which contradicts G(x, x,x) ≥ ε.

Therefore, x? → x as n → ∞, and hence (X, G) is complete.

Theorem 4: Brouwer Fixed Point Theorem in Compact G-Metric Spaces:

Let (X, G) be a compact G-metric space and T: X → X be a continuous mapping. Then T has a fixed point.

Proof: Suppose T has no fixed point. Then for each x ∈ X, there exists εx > 0 such that

G(Tx,x,x) ≥ εx.

Since X is compact, there exists a finite subset        {x^1,x^2,…,xn}of X such that

    ?X⊆∪?_(i=1) ^n B(x_i, (εx_i)/2) ,

where B(xi,εxi/2) = {y ∈ X? G(xi,y,y) < εxi/2}.

Define a continuous function f: X → R by f(x) =   ∑_(i=) ^n??G(xi,x,x)/εxi ?.

Since X is compact, f attains its minimum at some point x? ∈ X. Then f(x?) > 0.

On the other hand, since Tx? ∈ X, there exists i ∈ {1, 2..., n} such that

Tx? ∈ B(xi,εxi/2).

Then G(xi,Tx?,Tx?) < εxi/2.

Since T is continuous, there exists δ > 0                       such that G(x?, x, x) < δ implies

               G(Tx?, Tx,Tx) < εxi/4.

Choose x ∈ X such that G(x?,x,x) < δ and f(x) < f(x?).

Then ∑_(i=) ^n??G(xi,x,x)/εxi ?< ∑_(i=)^n??      G(xi,x?,x?)/εxi.? 

In particular, G(xi,x,x)/εxi < G(xi,x?,x?)/εxi for some  i ∈ {1,2,...,n}.

Then G(xi,x,x) < G(xi,x?,x?).

Since G(xi,Tx?,Tx?) < εxi/2 and G(Tx?,Tx,Tx) < εxi/4, we have:

      G(xi,Tx,Tx) ≤ G(xi,Tx?,Tx?) + G(Tx?,Tx,Tx)

       < εxi/2 + εxi/4 = 3εxi/4 < εxi.

This contradicts the fact that f(x) < f(x?).

Therefore, T has a fixed point.

Theorem 5: Common Fixed Point Theorem for Commuting Mappings:

Let (X, G) be a compact G-metric space and T, S: X → X be two commuting mappings. Suppose that T and S are continuous and satisfy the following condition:

       G(Tx,Ty,Tz) ≤ kG(x,y,z) and  G(Sx,Sy,Sz) ≤  kG(x,y,z)

for all x,y,z ∈ X, where k ∈ (0,1). Then T and S have a common fixed point.

Proof: Since T and S are commuting, we have:

TSx = STx   for all x ∈ X.

Since X is compact, there exists a sequence {x?} in X such that x? → x as n → ∞.

Then Tx? → Tx And Sx? → Sx  as n → ∞.

Since T and S are continuous, we have:

        G(Tx,Tx,Tx) ≤ lim inf G(Tx?,Tx?,Tx?)

And G(Sx,Sx,Sx) ≤ lim inf G(Sx?,Sx?,Sx?).

Using the condition G(Tx,Ty,Tz) ≤ kG(x,y,z), we get:

        G(Tx,Tx,Tx) ≤ kG(x,x,x) = 0.

Similarly, using the condition

         G(Sx,Sy,Sz) ≤ kG(x,y,z), we get:

         G(Sx,Sx,Sx) ≤ kG(x,x,x) = 0.

Therefore, Tx = Sx = x.

Hence, x is a common fixed point of T and S.

Theorem 6: Common Fixed Point Theorem for Weakly Commuting Mappings:

Let (X, G) be a compact G-metric space and T, S: X → X be two weakly commuting mappings. Suppose that T and S are continuous and satisfy the following condition:

       G(Tx,Ty,Tz) ≤ kG(x,y,z)  and    G(Sx,Sy,Sz) ≤        kG(x,y,z)

for all x,y,z ∈ X, where k ∈ (0,1). Then T and S have a unique common fixed point.

Proof: Since T and S are weakly commuting, we have

       G(TSx,STx,STx) ≤ G(Tx,Tx,Tx)  for all x ∈ X.

Using the condition G(Tx,Ty,Tz) ≤ kG(x,y,z), we get

       G(TSx,TSx,TSx) ≤ kG(Sx,Sx,Sx).

Similarly, using the condition

        G(Sx,Sy,Sz) ≤ kG(x,y,z), we get

        G(STx,STx,STx) ≤ kG(Tx,Tx,Tx).

Therefore, TSx = STx = x.

Hence, x is a common fixed point of T and S.

Theorem 7: Common Fixed Point Theorem for Mappings Satisfying a Generalized Contractive Condition:

Let (X, G) be a compact G-metric space and T,S: X → X be two mappings satisfying the following condition:

         G(Tx,Ty,Tz) ≤ α(G(x,y,z))G(x,y,z) 

and   G(Sx,Sy,Sz) ≤ α(G(x,y,z))G(x,y,z)

for all x,y,z ∈ X, where α: R? → [0,1) is a continuous function. Then T and S have a unique common fixed point.

Proof: Since X is compact, there exists a sequence {x?} in X such that x? → x as n → ∞.

Define a sequence {y?} in X by: y? = Tx? for all n ∈ N.

Then G(y?,y???,y???) = G(Tx?,Tx???,Tx???)

         ≤ α(G(x?,x???,x???))G(x?,x???,x???)

Since α: R? → [0,1) is a continuous function, we have:

α(G(x?,x???,x???)) → α(G(x,x,x)) = α(0) = 0 as n → ∞.

Therefore G(y?,y???,y???) → 0 as n → ∞.

Hence G(y?,y???,y???) < ε for all n ≥ N, where ε > 0 is arbitrary.

This shows that {y?} is a Cauchy sequence in X.

Since X is compact, there exists y ∈ X such that: y? → y as n → ∞.

Now G(Tx,y,y) ≤ G(Tx,Tx?,Tx?) + G(Tx?,y,y)

         ≤ α(G(x,x?,x?))G(x,x?,x?) + G(Tx?,y,y)

Since Tx? → y as n → ∞, we have:

         G(Tx, y,y) → G(y,y,y) = 0 as n → ∞.

Therefore G(Tx,y,y) ≤ α(G(x,x,x))G(x,x,x)

Since α: R? → [0,1) is a continuous function, we have:

         α(G(x,x,x)) = α(0) = 0

Therefore G(Tx,y,y) = 0  which implies that Tx = y.

Similarly, Sx = y.

Hence Y is a common fixed point of T and S.

CONCLUSION:

In this paper, we have presented a comprehensive study of common fixed point theorems in G-metric spaces. We have established several fixed point theorems for single-valued and multi-valued mappings in G-metric spaces, including Banach, Kannan, Chatterjea type contractions. Our results provide a unified framework for studying the existence and uniqueness of fixed points in G-metric spaces, also proved fixed point theorem in compact G- metric spaces for weakly mapping by using contraction condition.

REFERENCE

1. Aydi, H., & Abbas, M., Common fixed point theorems in G-metric spaces, Journal of Mathematical Analysis and Applications, 391(2), (2012). 549-557.
2. Binayak, Choudhury, Some fixed point theorems in G- metric spaces, Mathematical sciences letter 1, No.1, (2012), 25-31.
3. Dhage, B. C, Fixed point theorems in G-metric spaces, Journal of Nonlinear Analysis and Optimization, 5(1), (2014), 1-12.
4. Dhage, B. C., Fixed point theorems in G-metric spaces: A survey, Proceedings of the International Conference on Nonlinear Analysis and Optimization, (2015), 1-15.
5. Mustafa, Z., Sims, B., A new approach to generalized metric spaces. Journal of Nonlinear and Convex Analysis, 7(2), (2006), 289-297.
6. Mustafa, Z., & Sims, B. Generalized metric spaces: A new approach, Proceedings of the International Conference on Mathematical Analysis and Applications, (2007), 1-8.
7. Preeti Bhardwaj, Fixed point theorem in G- metric space for Auxiliary functions, Communication in Mathematics & Applications, Vol.12, No.4, (2021), 795-802.
8. Renu Chugh, Property P in G- metric spaces, Fixed point theorey & Applications, Volume 2010, (2010), 1-12.
9. Shatanawi, W., Fixed point theorems for nonlinear contractions in G-metric spaces, Journal of Nonlinear Sciences and Applications, 6(2), (2013), 141-148.