On surrounding quasi-contractions on non-triangular metric spaces

Definition 1.1

[11] We say that ϱ is a JS-metric on X if it satisfies the following conditions:

(a₁) for each pair ( x , y ) ∈ X × X , we have

(a₂) for each pair ( x , y ) ∈ X × X , we have

(a₃) there exists κ > 0 such that for all x , y ∈ X , if { x n } ∈ ℳ ( ϱ , X , x )

In this case, we say the pair ( X , ϱ ) is a JS-metric space by modulus κ .

Definition 1.2

[11] Let ( X , ϱ ) be a JS-metric space.

(b₁) We say that x n ϱ -converges to x if { x n } ∈ ℳ ( ϱ , X , x ) ,

(b₂) if { x n } ϱ -converges to x and ϱ - converges to y, then x = y ,

(b₃) { x n } is a ϱ -Cauchy sequence if lim m , n → ∞ ϱ ( x n , x m ) = 0 ,

(b₄) ( X , ϱ ) is said to be ϱ -complete if every ϱ -Cauchy sequence in X is convergent to some element in X.

Definition 1.3

Let X be a non-empty set and let ρ : X × X → ℝ + be a mapping. We say that ρ is a non-triangular metric on X if it satisfies the following conditions:

(c₁) ρ ( x , x ) = 0 for all x ∈ X ;

(c₂) If { x n } ∈ ℳ ( ρ , X , x ) ∩ ℳ ( ρ , X , y ) , then x = y for all x , y ∈ X .

Definition 1.4

Let ( X , ρ ) be a non-triangular metric space.

(d₁) We say that { x n } ρ -converges to x if lim n → ∞ ρ ( x n , x ) = 0 ,

(d₂) { x n } is a ρ -Cauchy sequence if lim sup n → ∞ ρ ( x n , x m ) : m ≥ n = 0 ,

(d ₃) ( X , ρ ) is said to ρ -complete if every ρ -Cauchy sequence in X is ρ - convergent to some element in X.

Definition 1.5

Let ( X , ρ ) be a non-triangular metric space and T : X → X be a mapping. A mapping T is ℳ -continuous in x ∈ X if

Remark 1.6

Note that, if T is a contraction, i.e., there exists k ∈ [ 0 , 1 ) such that

for all x , y ∈ X , then T is ℳ -continuous at each point x in X.

Example 1.7

Let X = [ 0 , + ∞ ) , define

Condition ( c 1 ) is trivially satisfied. We need to verify condition ( c 2 ). For this, let x , y ∈ X and { x n } be a sequence in X such that ρ ( x n , x ) → 0 and ρ ( x n , y ) → 0 as n → ∞ . It implies that

and these hold if and only if lim n → ∞ x n = − x = − y in ℝ and so x = y . Hence, condition ( c 2 ) is true. Therefore, ( X , ρ ) is a non-triangular metric space. On the other hand, condition ( a 3 ) does not hold. For all n ∈ ℕ and for each y ∈ X ,

Since { x n } is a convergent sequence to zero. If there exists C ≥ 1 such that

we have C ≥ y 2 + 1 2 y ≥ y 2 . Therefore, there is no bound for C, by which,

Definition 2.1

Let ( X , ϱ ) be a non-triangular metric space. The mapping T : X →   X is said to be a surrounding quasi-contraction with respect to Θ if there exists α ∈ [ 0 , 1 ) such that for all x , y ∈ X ,

where Θ : ℝ 4 → ℝ is a mapping such that Θ ( t , s , z , w ) ≤ max { z , w } and

Theorem 2.2

Let ( X , ϱ ) be a ϱ -complete non-triangular metric space and T : X → X be a surrounding quasi-contraction with respect to Θ such that δ 1 ( T , x 0 ) < ∞ for some x 0 ∈ X . Then { T n x 0 } ϱ converges to ω ∈ X . Moreover, if T is ℳ -continuous in ω , then ω is a fixed point of T.

Proof

Suppose that { x n } is a sequence defined by x n + 1 = T x n , n = 0 , 1 , … . Note that 0 ≤ δ n + 1 ( T , x 0 ) ≤ δ n ( T , x 0 ) . Therefore, { δ n ( T , x 0 ) } is a monotone bounded sequence from below and so is convergent. Thus, there exists δ ≥ 0 such that lim n → ∞ δ n ( T , x 0 ) = δ . We shall show that δ = 0 . If δ > 0 , then by the definition of δ n ( T , x 0 ) , for every k ∈ ℕ there exists n k , m k such that m k > n k ≥ k and

Hence,

Also, we have

If we let k → ∞ get it δ ≤ α δ . Thus, α ≥ 1 and this is a contradiction. Therefore, we deduce that δ = 0 and so { x n } is a ϱ -Cauchy sequence. Since ( X , ϱ ) is ϱ - complete, there exists some ω ∈ X such that { x n } is ϱ -convergent to ω . Since { x n } ∈ ℳ ( ϱ , X , ω ) and T is a ℳ -continuous we have that { T x n } ∈ ℳ ( ϱ , X , T ω ) , so we conclude that

From condition ( c 2 ) , we obtain that ω = T ω , so ω is a fixed point of T. Condition (1) implies that ω is a unique fixed point.□

Corollary 2.3

Let ( X , ϱ ) be a ϱ - complete non-triangular metric space and T : X →   X be a mapping. If there exists k ∈ [ 0 , 1 ) such that T satisfies

for all x , y ∈ X and δ 1 ( T , x 0 ) < ∞ for some x 0 ∈ X , then T has a fixed point w in X.

Corollary 2.4

Let ( X , ϱ ) be a ϱ - complete non-triangular metric space and T : X →   X be a ℳ -continuous mapping. If there exists k ∈ [ 0 , 1 ) such that T satisfies

for all x , y ∈ X and δ 1 ( T , x 0 ) < ∞ for some x 0 ∈ X , then T has a fixed point w in X.

Proof

It suffices to consider Θ ( t , s , z , w ) = max { z , w } t + s + 1 and apply Theorem 2.2.□

Corollary 2.5

Let ( X , ϱ ) be a ϱ - complete non-triangular metric space and T : X →   X be a ℳ -continuous mapping. If there exists k ∈ [ 0 , 1 ) such that T satisfies

for all x , y ∈ X and δ 1 ( T , x 0 ) < ∞ for some x 0 ∈ X , then T has a fixed point w in X.

Proof

It suffices to consider Θ ( t , s , z , w ) = z + w 2 and apply Theorem 2.2.□

Corollary 2.6

Let ( X , ϱ ) be a ϱ - complete non-triangular metric space and T : X →   X be a ℳ -continuous mapping. If there exists k ∈ [ 0 , 1 ) such that T satisfies

for all x , y ∈ X and δ 1 ( T , x 0 ) < ∞ for some x 0 ∈ X , then T has a fixed point w in X.

Proof

It suffices to consider Θ ( t , s , z , w ) = max { z , w } and then apply Theorem 2.2.□

Example 2.7

Let X = [ 0 , 1 2 ] ∪ [ 2 3 , 1 ] be endowed with the Euclidean metric. Obviously, ( X , ρ ) is a complete non-triangular metric space, where ρ ( x , y ) = | x − y | for each x , y ∈ X . Let T : X → X be defined by

Note that for each x ∈ [ 0 , 1 2 ] and for each y ∈ [ 2 3 , 1 ] , we have | T x − T y | = 1 4 . Since

thus, if we take 1 > α ≥ 27 32 (for example, α = 7 8 ), we have

Therefore,

Similar argument holds for the other cases with the same α . It is easy to see that T is ℳ -continuous and δ 1 ( T , 0 ) < 1 . Therefore, T is satisfied in the conditions of Corollary 2.4 and so it has a fixed point in X.

Theorem 2.8

Let ( X , ϱ ) be a ϱ - complete non-triangular metric space and T : X →   X be a mapping. Let there exists α ∈ 0 , 1 2 such that

Proof

We will use the same technique as the proof of Theorem 2.2. Let x n be the Picard sequence based at x 0 . We show that { x n } is a Cauchy sequence. Note that 0 ≤ δ n + 1 ( T , x 0 ) ≤ δ n ( T , x 0 ) . Therefore, δ n ( T , x 0 ) is a monotone bounded sequence and so is convergent. Thus, there exists δ ≥ 0 such that lim n → ∞ δ n ( T , x 0 ) = δ . We shall show that δ =0 . If δ >0 , then by the definition of δ n ( T , x 0 ) for every k ∈ ℕ there exist n k , m k such that m k > n k ≥ k and

Hence,

Also, we have

Taking limit as k → ∞ we get δ ≤ 2 α δ . Thus, α ≥ 1 2 and this is a contradiction. Therefore, we deduce that δ = 0 and so { x n } is a ϱ -Cauchy sequence. Since ( X , ϱ ) is ϱ -complete, there exists some ω ∈ X such that { x n } is ϱ -convergent to ω . Proof that ω is a unique fixed point of T is similar to that in the proof of Theorem 2.2.□

Corollary 2.9

Let ( X , ϱ ) be a ϱ -complete non-triangular metric space and T : X →   X be a mapping. Let there exists α ∈ 0 , 1 2 such that

for all x , y ∈ X . If δ 1 ( T , x 0 ) < ∞ for some x 0 ∈ X , then { T n x 0 } converges to some ω ∈ X . Also, if T ℳ -continuous in ω , then ω is a fixed point of T.

Theorem 3.1

Let ( X , d ) be a complete b-metric space with coefficient s and T : X → X be a mapping satisfying:

for all x , y ∈ X , where 0 ≤ λ < 1 . Then T has a unique fixed point x ⁎ , and for every x 0 ∈ X , the sequence { T n x 0 } converges to x ⁎ .

Proof

In view of Corollary 2.3 it suffices to prove that

Since for j > i and for each x 0 we have

so we need to show that there is a constant C > 0 such that d ( x 0 , T n x 0 ) ≤ C for all n ∈ ℕ . We know that there exists n 0 ∈ ℕ such that λ n 0 < 1 s 2 . Let x 0 ∈ X be arbitrary. Define the sequence { x n } by x n + 1 = T n x 0 for all n ≥ 0 . Then (8) implies that

and

Applying the triangle-type inequality (3) for b-metric space to triples, we have

Using (9) and (10) and the fact that λ n < 1 , for each n ∈ ℕ , we obtain

Hence, we have

Now, using Corollary 2.3 T has a unique fixed point in X.□

Theorem 3.2

Let ( X , d ) be a complete b-metric space with coefficient s ≥ 1 and T : X → X be a mapping satisfying:

for all x , y ∈ X , where 0 ≤ λ < 1 2 and s λ < 1 . Then T has a unique fixed point x ⁎ , and for every x 0 ∈ X , the sequence { T n x 0 } converges to x ⁎ .

Proof

Take Θ ( x , y , z , w ) = z + w 2 . Obviously, T is a surrounding quasi-contraction with respect to Θ . In view of Theorem 2.2, it suffices to prove that δ 1 ( T , x 0 ) < 1 for some x 0 ∈ X . Let x 0 ∈ X be arbitrary. Define the sequence { x n } by x n + 1 = T n x 0 for all n ≥ 0 . Then (11) implies that

and

so,

Applying the triangle-type inequality for b-metric space, and from (11) and (12), we have

So, we can put

Now we obtain that x ⁎ is the unique fixed point of T. Let n ∈ ℕ be arbitrary, we have

Since lim n → ∞ d ( x n + 1 , x n ) = 0 and lim n → ∞ d ( x ⁎ , x n + 1 ) = 0 , we have

Since λ s < 1 , then d ( x ⁎ , T x ⁎ ) = 0 , i.e., T x ⁎ = x ⁎ .

For uniqueness, let y ⁎ be another fixed point of T. Then it follows from (11) that

Therefore, we must have d ( x ⁎ , y ⁎ ) = 0 , i.e., x ⁎ = y ⁎ . □

Remark 3.3

Using the technique from the proofs of Theorem 3.1 and Theorem 3.2 we can also get the main result in [22], as well as the main result in [23], also in this way we can also get a whole series of known results in b-metric, rectangular metric and b-rectangular metric spaces.