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In this article, we establish some common fixed point results for two pairs of compatible mappings satisfying Meir-Keeler type contractive conditions in metric space and dislocated metric space which extend and improve some similar fixed point results in the literature.

The word metrics plays a dominant role in most of the fields like, geometry, economics, statistics, graph theory, probability theory, coding theory, game theory, pattern recognition, computer graphics, theory of information and computer semantics, molecular biology etc. M Frechet in 1906, first time introduced the notion of metric space which is now a very useful topic in mathematical analysis. In 1912, L. E. J. Brouwer [

In 1969, A. Meir and E. Keeler [

There exists a vast literature which generalizes the result of Meir and Keeler. Maiti and Pal [

ε ≤ max { d ( x , y ) , d ( x , T y ) , d ( y , T y ) } < ε + δ ⇒ d ( T x , T y ) < ε

Park-Rhoades [

ε ≤ max { d ( S x , S y ) , d ( S x , T x ) , d ( S y , T y ) , 1 2 [ d ( S x , T y ) + d ( S y , T x ) ] } < ε + δ

⇒ d ( T x , T y ) < ε

In 1986, Jungck [

The study of common fixed points satisfying contractive type conditions has been a very active field of research activity. The most general common fixed point theorems for four mappings, say A, B, S and T of a metric space ( X , d ) use either a.

Banach type contractive condition of the form

1) d ( A x , B y ) ≤ h M ( x , y ) , 0 ≤ h < 1

where

M ( x , y ) = max { d ( S x , T y ) , d ( A x , S x ) , d ( B y , T y ) , 1 2 [ d ( A x , T y ) + d ( B y , S x ) ] }

a Meir-Keeler type contractive definition

2) given ϵ > 0 there exists a δ > 0 such that

ϵ ≤ M ( x , y ) < ϵ + δ d ( A x , B y ) < ϵ

or ϕ -contractive condition of the form

3) d ( A x , B y ) ≤ ϕ ( M ( x , y ) )

where ϕ : ℝ + → ℝ + is such that ϕ ( t ) < t for each t > 0

The contractive condition (2) ensures the existence of a fixed point, only when f satisfies some additional conditions.

The following conditions on the function ϕ have introduced and employed by various authors for the establishment of fixed point

1) ϕ ( t ) is non decreasing and t t − ϕ ( t ) is non increasing [

2) ϕ ( t ) is non decreasing and l i m n → ∞ ϕ n ( t ) = 0 for each t > 0 [

3) ϕ is upper semi-continuous [

4) ϕ is non decreasing and continuous from the right [

In 1985, S. G. Matthwes [

In this paper, we establish some common fixed point theorems for two pairs of compatible and weakly compatible mappings satisfying Meir-Keeler type contractive condition in dislocated metric space.

We start with the following definitions and theorems.

Definition 1. [

1) d ( x , x ) = 0

2) d ( x , y ) = 0 implies x = y .

3) d ( x , y ) = d ( y , x )

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

If d satisfies the conditions 1 - 4 Then d is called the metric on X and the pair (X, d) is called the metric space. If d satisfies the conditions 2 - 4, then d is called the d-metric on X and the pair (X, d) is called the dislocated metric space.

Definition 2. [

ε ≤ d ( x , y ) < ε + δ ⇒ d ( T x , T y ) < ε (1)

Theorem 1. [

Definition 3. [

l i m n → ∞ d ( S T x n , T S x n ) = 0

whenever { x n } is a sequence in X such that lim n → ∞ S x n = lim n → ∞ T x n = x for some x ∈ X

Definition 4. [

Definition 5. [

If A and S both are continuous they are obviously reciprocally continuous but the converse is not true.

Now we establish the following lemma in dislocated metric space.

Lemma 1. Let (X, d) be a dislocated metric space. Let A , B , S , T : X → X be mappings satisfying the conditions

A ( X ) ⊆ T ( X ) and B ( X ) ⊆ S ( X ) (2)

Assume further that given for each ε > 0 there exists δ > 0 such that for all x , y ∈ X

ε ≤ M ( x , y ) < ε + δ ⇒ d ( A x , B y ) ≤ ε (3)

and

d ( A x , B y ) < M ( x , y ) whenever M ( x , y ) > 0 (4)

where

M ( x , y ) = max { d ( S x , T y ) , d ( A x , S x ) , d ( B y , T y ) , 1 2 d ( S x , B y ) , 1 2 d ( A x , T y ) }

then for each x 0 ∈ X , the sequence { y n } in X defined by the rule

y 2 n = A x 2 n = T x 2 n + 1 , y 2 n + 1 = B x 2 n + 1 = S x 2 n + 2

is a Cauchy sequence.

Proof. let x 0 ∈ X , then by condition (2) we can define a sequence

{ A x 0 , B x 1 , A x 2 , B x 3 , A x 2 n , B x 2 n + 1 } such that

y 2 n = A x 2 n = T x 2 n + 1 and y 2 n + 1 = B x 2 n + 1 = S x 2 n + 2 for n ∈ ℕ ∪ { 0 } (5)

Now we have

d ( y 2 n + 1 , y 2 n + 2 ) = d ( A x 2 n + 1 , B x 2 n + 2 ) < max { d ( S x 2 n + 1 , T x 2 n + 2 ) , d ( A x 2 n + 1 , S x 2 n + 1 ) , d ( B x 2 n + 2 , T x 2 n + 2 ) , 1 2 d ( S x 2 n + 1 , B x 2 n + 2 ) , 1 2 d ( A x 2 n + 1 , T x 2 n + 2 ) } = max { d ( y 2 n , y 2 n + 1 ) , d ( y 2 n + 1 , y 2 n ) , d ( y 2 n + 2 , y 2 n + 1 ) , 1 2 d ( y 2 n , y 2 n + 2 ) , 1 2 d ( y 2 n + 1 , y 2 n + 1 ) }

But,

1 2 d ( y 2 n , y 2 n + 2 ) ≤ 1 2 [ d ( y 2 n , y 2 n + 1 ) + d ( y 2 n + 1 , y 2 n + 2 ) ] ≤ max { d ( y 2 n , y 2 n + 1 ) , d ( y 2 n + 1 , y 2 n + 2 ) }

and,

1 2 d ( y 2 n + 1 , y 2 n + 1 ) ≤ 1 2 [ d ( y 2 n + 1 , y 2 n ) + d ( y 2 n , y 2 n + 1 ) ] ≤ max { d ( y 2 n + 1 , y 2 n ) , d ( y 2 n + 1 , y 2 n ) } = d ( y 2 n + 1 , y 2 n )

Hence, d ( y 2 n + 1 , y 2 n + 2 ) < max { d ( y 2 n , y 2 n + 1 ) , d ( y 2 n + 2 , y 2 n + 1 ) }

which implies that, d ( y 2 n + 1 , y 2 n + 2 ) < d ( y 2 n , y 2 n + 1 )

[ d ( y 2 n + 1 , y 2 n + 2 ) < d ( y 2 n + 1 , y 2 n + 2 ) is impossible.]

Similarly, we can show that

d ( y 2 n + 1 , y 2 n ) < d ( y 2 n , y 2 n − 1 )

Consequently, we conclude that d ( y n + 1 , y n ) < d ( y n , y n − 1 ) . For n even or odd, the last inequality implies that { y n } is a Cauchy sequence.

Now, we establish a common fixed point theorem for two pairs of compatible mappings in metric space.

Theorem 2. Let (X, d) be a complete metric space. Let A , B , S , T : X → X such that the pairs (A, S) and (B, T) be compatible mappings which satisfy the following conditions

A ( X ) ⊂ T ( X ) , B ( X ) ⊂ S ( X ) (6)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , B y ) ≤ ε

where

M ( x , y ) = max { d ( S x , T y ) , d ( A x , S x ) , d ( B y , T y ) , 1 2 [ d ( A x , T y ) + d ( B y , S x ) ] } (7)

d ( A x , B y ) < k { d ( S x , T y ) d ( S x , A x ) d ( S x , B y ) + d ( S x , T y ) + d ( A x , S x ) + d ( B y , T y ) + d ( A x , T y ) + d ( B y , S x ) + d ( B y , T y ) d ( A x , T y ) d ( S x , B y ) } (8)

0 < k ≤ 1 3 . Suppose that the mappings in one of the pairs (A, S) or (B, T) are

reciprocally continuous, then A, B, S and T have a unique common fixed point.

Proof:

let x 0 ∈ X be any point in X. Define sequences { x n } and { y n } in X given by the rule

y 2 n = A x 2 n = T x 2 n + 1 and y 2 n + 1 = B x 2 n + 1 = S x 2 n + 2 for n ∈ ℕ ∪ { 0 } (9)

then by Jachymski’s lemma [

y 2 n = A x 2 n = T x 2 n + 1 → z and y 2 n = A x 2 n = T x 2 n + 1 → z (10)

Suppose that the pair ( A , S ) is reciprocally continuous, then A S x n → A z and S A x n → S z . Since the pair ( A , S ) is compatible so

lim n → ∞ S x 2 n = lim n → ∞ A x 2 n = z and lim n → ∞ S x 2 n = lim n → ∞ A x 2 n = z (11)

This implies that d ( A z , S z ) = 0 . Hence A z = S z .

Since, A ( X ) ⊂ T ( X ) there exists a point w ∈ X such that A z = T w .We claim that B w = T w .If B w ≠ T w ,then by using condition (8) we get,

d ( A z , B w ) < k { d ( S z , T w ) d ( S z , A z ) d ( S z , B w ) + d ( S z , T w ) + d ( A z , S z ) + d ( B w , T w ) + d ( A z , T w ) + d ( B w , S z ) + d ( B w , T w ) d ( A z , T w ) d ( S z , B w ) } = k { 0 + 0 + 0 + d ( B w , A z ) + 0 + d ( B w , A z ) + 0 } = 2 k d ( A z , B w )

which is a contradiction. So d ( A z , B w ) = 0 A z = B w . Hence A z = S z = T w = B w .

Since the compatible maps commute at their coincidence point, we get A S z = S A z . This further implies that A A z = A S z = S A z = S S z . If A z ≠ A A z , by condition (8) we get

d ( A z , A A z ) = d ( B w , A A z ) = d ( A A z , B w ) < k ( d ( S A z , T w ) d ( S A z , A A z ) d ( S A z , B w ) + d ( S A z , T w ) + d ( A A z , S A z ) + d ( B w , T w ) + d ( A A z , T w ) + d ( B w , S A z ) + d ( B w , T w ) d ( A A z , T w ) d ( S A z , B w ) } = k { 0 + d ( A A z , B w ) + 0 + 0 + d ( A A z , B w ) + d ( B w , A A z ) + 0 } = 3 k d ( A A z , B w ) = 3 k d ( A A z , A z )

which is a contradictions, so d ( A z , A A z ) = 0 ⇒ A z = A A z . Hence, A z = A A z = S A z . Thus Az is the common fixed point of the mappings A and S. Similarly we obtain B w ( = A z ) is the common fixed point of the mappings B and T.

Uniqueness:

If possible, let u and v ( u ≠ v ) are two common fixed points of the maps A, B, S and T. Now by virtu of (8)

d ( u , v ) = d ( A u , B v ) < k { d ( S u , T v ) d ( S u , A u ) d ( S u , B v ) + d ( S u , T v ) + d ( A u , S u ) + d ( B v , T v ) + d ( A u , T v ) + d ( B v , S u ) + d ( B v , T v ) d ( A u , T v ) d ( S u , B v ) } = k { d ( u , v ) d ( u , u ) d ( u , v ) + d ( u , v ) + d ( u , u ) + d ( v , v ) + d ( u , v ) + d ( v , u ) + d ( v , v ) d ( u , v ) d ( u , v ) } = 3 k d ( u , v )

which is a contradiction. This shows that d ( u , v ) = 0 ⇒ u = v

The proof is similar when the mappings B and T are assumed compatible and reciprocally continuous. This completes the proof of the theorem.

Now, on the light of above theorem, one can establish the following corollaries easily.

Corrollary 1. Let (X, d) be a complete metric space. Let A , B , S : X → X such that the pairs (A, S) and (B, S) be compatible mappings which satisfy the following conditions

A ( X ) , B ( X ) ⊂ S (X)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , B y ) ≤ ε

where

M ( x , y ) = max { d ( S x , S y ) , d ( A x , S x ) , d ( B y , S y ) , 1 2 [ d ( A x , S y ) + d ( B y , S x ) ] }

d ( A x , B y ) < k { d ( S x , S y ) d ( S x , A x ) d ( S x , B y ) + d ( S x , S y ) + d ( A x , S x ) + d ( B y , S y ) + d ( A x , S y ) + d ( B y , S x ) + d ( B y , S y ) d ( A x , S y ) d ( S x , B y ) }

0 < k ≤ 1 3 suppose that the mappings in one of the pairs (A, S) or (B, S) are

reciprocally continuous, then A, B and S have a unique common fixed point.

Corrollary 2. Let (X, d) be a complete metric space. Let A , S , T : X → X such that the pairs (A, S) and (A, T) be compatible mappings which satisfy the following conditions

A ( X ) ⊂ T ( X ) , A ( X ) ⊂ S (X)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , A y ) ≤ ε

where

M ( x , y ) = max { d ( S x , T y ) , d ( A x , S x ) , d ( A y , T y ) , 1 2 [ d ( A x , T y ) + d ( A y , S x ) ] }

d ( A x , A y ) < k { d ( S x , T y ) d ( S x , A x ) d ( S x , A y ) + d ( S x , T y ) + d ( A x , S x ) + d ( A y , T y ) + d ( A x , T y ) + d ( A y , S x ) + d ( A y , T y ) d ( A x , T y ) d ( S x , A y ) }

0 < k ≤ 1 3 . Suppose that the mappings in one of the pairs (A, S) or (A, T) are

reciprocally continuous, then A, S and T have a unique common fixed point.

Corrollary 3. Let (X, d) be a complete metric space. Let A , S : X → X such that the pair (A, S) be compatible mappings which satisfy the following conditions

A ( X ) ⊂ S (X)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , A y ) ≤ ε

where

M ( x , y ) = max { d ( S x , S y ) , d ( A x , S x ) , d ( A y , S y ) , 1 2 [ d ( A x , S y ) + d ( A y , S x ) ] }

d ( A x , A y ) < k { d ( S x , S y ) d ( S x , A x ) d ( S x , A y ) + d ( S x , S y ) + d ( A x , S x ) + d ( A y , S y ) + d ( A x , S y ) + d ( A y , S x ) + d ( A y , S y ) d ( A x , S y ) d ( S x , A y ) }

0 < k ≤ 1 3 . Suppose that the pair (A, S) is reciprocally continuous, then A, and

S have a unique common fixed point.

Corrollary 4. Let (X, d) be a complete metric space. Let A , B , I X : X → X such that the pairs ( A , I X ) and ( B , I X ) be compatible mappings which satisfy the following conditions

A ( X ) , B ( X ) ⊂ I X

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , B y ) ≤ ε

where

M ( x , y ) = max { d ( x , y ) , d ( A x , x ) , d ( B y , y ) , 1 2 [ d ( A x , y ) + d ( B y , x ) ] }

d ( A x , B y ) < k { d ( x , y ) d ( x , A x ) d ( x , B y ) + d ( x , y ) + d ( A x , x ) + d ( B y , y ) + d ( A x , y ) + d ( B y , x ) + d ( B y , y ) d ( A x , y ) d ( x , B y ) }

0 < k ≤ 1 3 . Suppose that the mappings in one of the pairs ( A , I X ) or ( B , I X )

are reciprocally continuous, then A, B, I_{X} have a unique common fixed point.

Now, we establish a common fixed point theorem for two pais of compatible mappings in dislocated metric space.

Theorem 3. Let (X, d) be a complete dislocated metric space. Let A , B , S , T : X → X such that the pairs (A, S) and (B, T) be compatible mappings which satisfy the following conditions

A ( X ) ⊂ T ( X ) , B ( X ) ⊂ S ( X ) (12)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , B y ) ≤ ε

where

M ( x , y ) = max { d ( S x , T y ) , d ( A x , S x ) , d ( B y , T y ) , 1 2 d ( A x , T y ) , 1 2 d ( B y , S x ) } (13)

d ( A x , B y ) < ϕ ( k { d ( S x , T y ) + d ( A x , S x ) + d ( B y , T y ) + d ( A x , T y ) + d ( B y , S x ) } ) (14)

where, 0 < k ≤ 1 8 and ϕ : ℝ + → ℝ + is such that ϕ ( t ) < t for each t > 0 . If

one of the mappings A, B, S or T be continuous then A, B, S and T have a unique common fixed point.

Proof. Let x 0 ∈ X be any point in X. Define sequences { x n } and { y n } in X given by the rule

y 2 n = A x 2 n = T x 2 n + 1 and y 2 n + 1 = B x 2 n + 1 = S x 2 n + 2 for n ∈ ℝ ∪ { 0 } (15)

then by above lemma (1), { y n } is a Cauchy sequence. Sine X is complete there exists a point z ∈ X such that y n → z . Also the sequences

y 2 n = A x 2 n = T x 2 n + 1 → z and y 2 n + 1 = B x 2 n + 1 = S x 2 n + 2 → z (16)

Suppose that S is continuous, then S S x 2 n → S z and S A x 2 n → S z . Since the pair ( A , S ) is compatible so

A S x 2 n → S z (17)

Since A ( X ) ⊂ T ( X ) , corresponding to each value of n, there exists a sequence z 2 n ∈ X such that A S x 2 n = T z 2 n .

Thus A S x 2 n = T z 2 n → S z and S S x 2 n → S z .

We assert that

lim n B z 2 n = S z

If not, there exists a subsequence B z 2 m of B z 2 n , a number r > 0 and a positive number N such that for each m ≥ N , we have

d ( A S x 2 m , B z 2 m ) ≥ r and d ( B z 2 m , S z ) ≥ r

Moreover, by virtu of (14) for large m we obtain,

d ( A S x 2 m , B z 2 m ) < ϕ ( k { d ( S S x 2 m , T z 2 m ) + d ( A S x 2 m , S S x 2 m ) + d ( B z 2 m , T z 2 m ) + d ( A S x 2 m , T z 2 m ) + d ( B z 2 m , S S x 2 m ) } )

now taking limit as m → ∞ we have

lim m d ( S z , B z 2 m ) < lim m ϕ ( k { d ( B z 2 m , S z ) + d ( B z 2 m , S z ) } )

lim m d ( S z , B z 2 m ) < lim m ϕ ( 2 k d ( B z 2 m , S z ) )

lim m d ( S z , B z 2 m ) < 2 k lim m d ( B z 2 m , S z )

which is a contradiction, Hence

lim m d ( S z , B z 2 m ) = 0

This represents that,

lim m B z 2 n = S z

We claim that A z = S z

If A z ≠ S z , then by virtu of (14) for large n, We obtain

d ( A z , B z 2 n ) < ϕ ( k { d ( S z , T z 2 n ) + d ( A z , S z ) + d ( B z 2 n , T z 2 n ) + d ( A z , T z 2 n ) + d ( B z 2 n , S z ) } )

Now letting n → ∞ we obtain,

d ( A z , S z ) < ϕ ( k { d ( S z , S z ) + d ( A z , S z ) + d ( S z , S z ) + d ( A z , S z ) + d ( S z , S z ) } ) ≤ ϕ ( k 8 d ( A z , S z ) ) < 8 k d ( A z , S z )

which is a contradiction. Therefore, d ( A z , S z ) = 0 ⇒ A z = S z .

Since, A ( X ) ⊂ T ( X ) there exists a point w ∈ X such that A z = T w . We claim that B w = T w . If B w ≠ T w by using condition (14) we get,

d ( A z , B w ) < ϕ ( k { d ( S z , T w ) + d ( A z , S z ) + d ( B w , T w ) + d ( A z , T w ) + d ( B w , S z ) } ) = ϕ ( k { d ( A z , A z ) + d ( A z , A z ) + d ( B w , A z ) + d ( A z , A z ) + d ( B w , A z ) } ) = ϕ ( k { 3 d ( A z , A z ) + 2 d ( A z , B w ) } ) < 8 k d ( A z , B w )

which is a contradiction. So d ( A z , B w ) = 0 ⇒ A z = B w . Hence A z = S z = T w = B w .

Sice the compatible maps commute at their coincidence point, we get A S z = S A z . This further implies that A A z = A S z = S A z = S S z .

If A z ≠ A A z , by condition (14) we get

d ( A z , A A z ) = d ( B w , A A z ) = d ( A A z , B w ) < ϕ ( k { d ( S A z , T w ) + d ( A A z , S A z ) + d ( B w , T w ) + d ( A A z , T w ) + d ( B w , S A z ) } ) = ϕ ( k { d ( A A z , A z ) + d ( A A z , A A z ) + d ( A z , A z ) + d ( A A z , A z ) + d ( A z , A A z ) } ) = ϕ ( k { 3 d ( A A z , A z ) + d ( A A z , A A z ) + d ( A z , A z ) } ) ≤ ϕ ( 7 k d ( A A z , A z ) ) < 7 k d ( A A z , A z )

which is a contradictions, so d ( A z , A A z ) = 0 ⇒ A z = A A z . Hence, A z = A A z = S A z . Thus Az is the common fixed point of the mappings A and S.

Similarly we obtain B w ( = A z ) is the common fixed point of the mappings B and T when T is supposed to be continuous.

Uniqueness:

Let u and v ( u ≠ v ) are two common fixed points of the mappings A, B, S and T. Now by virtu of (14),

d ( u , v ) = d ( A u , B v ) < ϕ ( k { d ( S u , T v ) + d ( A u , S u ) + d ( B v , T v ) + d ( A u , T v ) + d ( B v , S u ) } ) = ϕ ( k { d ( u , v ) + d ( u , u ) + d ( v , v ) + d ( u , v ) + d ( v , u ) } ) = ϕ ( k { 3 d ( u , v ) + d ( u , u ) + d ( v , v ) } ) ≤ ϕ ( 7 k d ( u , v ) ) < 7 k d ( u , v )

which is a contradiction. Hence d ( u , v ) = 0 ⇒ u = v .

Moreover, the proof follows on similar lines when A or B is assumed to be continuous since A ( X ) ⊂ T ( X ) and B ( X ) ⊂ S ( X ) .

This completes the proof of the theorem.

Now, in the light of the above theorem one can establish the following corollaries easily.

Corrollary 5. Let (X, d) be a complete dislocated metric space. Let A , B , S : X → X such that the pairs (A, S) and (B, S) be compatible mappings which satisfy the following conditions

A ( X ) and B ( X ) ⊂ S ( X ) (18)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , B y ) ≤ ε

where

M ( x , y ) = max { d ( S x , S y ) , d ( A x , S x ) , d ( B y , S y ) , 1 2 d ( A x , S y ) , 1 2 d ( B y , S x ) } (19)

d ( A x , B y ) < ϕ ( k { d ( S x , S y ) + d ( A x , S x ) + d ( B y , S y ) + d ( A x , S y ) + d ( B y , S x ) } )

where, 0 < k ≤ 1 8 and ϕ : ℝ + → ℝ + is such that ϕ ( t ) < t for each t > 0 . If

one of the mappings A, B or S be continuous then A, B and S have a unique common fixed point.

Corrollary 6. Let (X, d) be a complete dislocated metric space. Let A , S , T : X → X such that the pairs (A, S) and (A, T) be compatible mappings which satisfy the following conditions

A ( X ) ⊂ T ( X ) and S ( X ) (20)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , A y ) ≤ ε

where

M ( x , y ) = max { d ( S x , T y ) , d ( A x , S x ) , d ( A y , T y ) , 1 2 d ( A x , T y ) , 1 2 d ( A y , S x ) } (21)

d ( A x , A y ) < ϕ ( k { d ( S x , T y ) + d ( A x , S x ) + d ( A y , T y ) + d ( A x , T y ) + d ( A y , S x ) } )

where, 0 < k ≤ 1 8 and ϕ : ℝ + → ℝ + is such that ϕ ( t ) < t for each t > 0 . If

one of the mappings A, S or T be continuous then A, S and T have a unique common fixed point.

Corrollary 7. Let (X, d) be a complete dislocated metric space. Let A , S : X → X such that the pair (A, S) be compatible mappings which satisfy the following conditions

A ( X ) ⊂ S ( X ) (22)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , B y ) ≤ ε

where

M ( x , y ) = max { d ( S x , S y ) , d ( A x , S x ) , d ( A y , T y ) , 1 2 d ( A x , S y ) , 1 2 d ( A y , S x ) } (23)

d ( A x , A y ) < ϕ ( k { d ( S x , S y ) + d ( A x , S x ) + d ( A y , S y ) + d ( A x , S y ) + d ( A y , S x ) } )

where, 0 < k ≤ 1 8 and ϕ : ℝ + → ℝ + is such that ϕ ( t ) < t for each t > 0 . If

one of the mappings A or S be continuous then A and S have a unique common fixed point.

Corrollary 8. Let (X, d) be a complete dislocated metric space. Let A , B , I X : X → X such that the pairs (A, I_{X}) and (B, I_{X}) be compatible mappings which satisfy the following conditions

A ( X ) and B ( X ) ⊂ I X (24)

Given, ε > 0 , ∃ δ > 0 , such that

ε < M ( x , y ) < ϵ + δ ⇒ d ( A x , B y ) ≤ ε

where

M ( x , y ) = max { d ( x , y ) , d ( A x , x ) , d ( B y , y ) , 1 2 d ( A x , y ) , 1 2 d ( B y , x ) } (25)

d ( A x , B y ) < ϕ ( k { d ( x , y ) + d ( A x , x ) + d ( B y , y ) + d ( A x , y ) + d ( B y , x ) } )

where, 0 < k ≤ 1 8 and ϕ : ℝ + → ℝ + is such that ϕ ( t ) < t for each t > 0 . If

one of the mappings A, B, or I_{X} be continuous then A, B and I_{X} have a unique common fixed point.

Remarks: Our results extend and improve the results of Meir-Keeler [

This work is carried out under Small Research Development and Innovation Grant SRDIG-73/74-S&T-08 supported by University Grants Commission, Nepal.

Panthi, D. (2018) Some Common Fixed Point Theorems Satisfying Meir-Keeler Type Contractive Conditions. Open Journal of Discrete Mathematics, 8, 35-47. https://doi.org/10.4236/ojdm.2018.82004