Analysis of F-contractions in function weighted metric spaces with an application

Definition 1.1

[8] Let ℱ be the collection of functions f :(0,  ∞ ) → R such that the following conditions hold:

• ( F 1) f is non-decreasing, i.e., 0 < a < u implies f ( a ) ≤ f ( u ) ;

• ( F 2) For any sequence ( ω n ) ⊂ (0,  ∞ ) , we have

Definition 1.2

[8] Suppose U is a non-empty set and d : U × U → [0,  ∞ ) is a given function. Suppose that there exists ( f , σ ) ∈ ℱ × [ 0 , ∞ ) such that

• ( D 1) ( a , u ) ∈ U × U , d ( a , u ) = 0 if and only if a = u ;

• ( D 2) for all ( a , u ) ∈ U × U , d ( a , u ) = d ( u , a ) ;

• ( D 3) for every ( a , u ) ∈ U × for each n ′ ∈ N , n ′ ≥ 2 and for every ( ω m ) m = 1 n ′ ⊂ X with ( ω 1 , ω n ′ ) = ( a , u ) , we have

The function d is known as an F-metric on U, and the pair ( U , d ) is called an F-metric space.

Example 1.1

[8] Let U = N and d : U × U → (0,  ∞ ) be defined as

for all ( a , u ) ∈ U × U with f ( t ) = ln t , t > 0 and α = ln 3 , then d is a function weighted metric on U.

Example 1.2

[8] Let U = N and d : U × U → (0,  ∞ ) be defined as

for all ( a , u ) ∈ U × U . Then, d is a function weighted metric on U.

Definition 1.3

[8] Let d be an ℱ -metric space. Suppose ( a n ) is a sequence in U. Then,

1. ( a n ) is ℱ -convergent to a point a ∈ U if lim n → ∞ d ( a n , a ) = 0 ;

2. ( a n ) is ℱ -Cauchy if lim n → ∞ d ( a n , a m ) = 0 ;

3. the space ( U , d ) is ℱ -complete if every Cauchy sequence ( a n ) ⊂ U is convergent to a point a ∈ U .

Definition 1.4

[8] Let ( U , d ) be a function weighted metric space. A subset O of U is said to be F-open if for every a ∈ O , there is some r ′ > 0 such that B ( a , r ′ ) ⊂ O , where

We say that a subset C of U is F-closed if U \ C is F-open.

Proposition 1.1

[8] Let ( U , d ) be a function weighted metric space and V be a non-empty subset of U. Then, the following statements are equivalent:

1. V is ℱ -closed.

2. For any sequence ( a n ) ⊂ V , we have

implies a ∈ V .

Theorem 1.1

[8] Suppose ( U , d ) is a complete function weighted metric space, and let g : U → U be a given mapping. Suppose that there exists k ∈ (0,1) such that

Then, g has a unique fixed point a ∗ ∈ U . Moreover, for any a 0 ∈ U , the sequence ( a n ) ⊂ U defined by a n + 1 = g ( a n ) , n ∈ N is convergent to a ∗ .

Theorem 1.2

[26] Suppose U is a complete metric space (d is the metric) and let g : U → U be a function such that

for all a , u ∈ U , where α , β , γ are non-negative and satisfy α + β + γ < 1 . Then, g has a unique fixed point. Next, let W : ( − ∞ , 0] → R be given continuous bounded initial function. Let C be the space of all continuous functions from R to R and define the set B ( W ) by B ( W ) = { ϕ : R → R , ϕ ( t ) = W ( t ) i f t ≤ 0 , ϕ ( t ) → 0 a s t → ∞ , ϕ ∈ C } . Then, equipped with the supremum norm ∥ ⋅ ∥ , S ξ is a Banach space.

Lemma 1.1

[27] The Banach space B ( W ) , ∥ ⋅ ∥ endowed with the metric d defined by

for h ∈ B ( W ) , is a function weighted metric space.

Definition 2.1

Let ( U , d ) be a function weighted metric space and ( F , τ ) , ( f , σ ) ∈ ℱ × [0,  ∞ ) with σ < τ . Let S , T : U → U be self-mappings. Then, the pair ( T , S ) is called a rational type F-contraction if there exist α , β , γ , δ , λ , μ , η ∈ [0,  ∞ ) such that α + 2 β + 2 γ + δ + μ < 1 for all ( a , u ) ∈ U × U , we have

where

Theorem 2.1

Let U , d be a complete function weighted metric space and S , T : U → U be self-mappings such that ( T , S ) is a rational type F-contraction. Then, S and T have at most one common fixed point in U.

Proof

Suppose s 0 is an arbitrary point. Define a sequence ( s n ) by

Using (1) and (2) and letting a = s 2 v and u = s 2 v + 1 , we can write

By ( D 3 ) ,

By hypothesis and ( F 1) , we can write

That is,

where α + β + γ + μ 1 − β − γ − δ = ω .

Note that ω < 1 . Similarly,

Hence,

It yields that

i.e.,

Using the above inequality, we can write

Since

for any δ > 0 , there exists some n ′ ∈ N such that

Furthermore, suppose ( f , α ) ∈ ℱ × [ 0 , ∞ ) such that ( D 3 ) is satisfied. Suppose ε > 0 is fixed. By ( F 2) there is some δ > 0 such that

Using (3) and (4), we write

Using ( D 3) and the above equation, we obtain

which shows

Therefore, it is proved that the sequence ( s n ) is Cauchy in U. Since ( U , d ) is complete, there exists s ∗ ∈ U such that ( s n ) is convergent to s ∗ , i.e.,

To prove that s ∗ is a fixed point of S, assume that d ( S s ∗ , s ∗ ) > 0 . Then,

Using (2) and ( F 1) and letting j → ∞ , we get

It is a contradiction. Hence, d ( s ∗ , S s ∗ ) = 0 , i.e., S s ∗ = s ∗ . Following the same steps, we get T s ∗ = s ∗ . Hence, S s ∗ = s ∗ = T s ∗ .

Uniqueness: say s ∗ ∗ is another common fixed point of S and T, i.e., d ( s ∗ , s ∗ ∗ ) > 0 . Then,

Using ( F 1) , we write

which is a contradiction. Hence, d ( s ∗ ∗ , s ∗ ) = 0 , i.e., s ∗ ∗ = s ∗ .□

Example 2.1

Suppose U = N , F ( a ) = ln a = f ( a ) and S , T : U → U and d : U × U → R are defined as

and

for ( a , u ) ∈ U × U . Observe that f and F satisfy ( F 1) − ( F 2) and d is the function weighted metric on U. Fix α = β = γ = λ = δ = μ = 0 and η = e 2 . Now, if d ( S a , T u ) > 0 , then

and

and σ can be chosen according to τ . Therefore (1) holds. Moreover, it is clear that 1 is the unique common fixed point of S and T. Taking S = T in Theorem 2.1, the following result for single-valued mappings is obtained.

Corollary 2.1

Suppose ( U , d ) is a complete function weighted metric space and ( F , τ ) , ( f , σ ) ∈ ℱ × [0,  ∞ ) with σ < τ . Let T : U → U be a self-mapping and there exist α , β , γ , δ , λ , μ , η ∈ [0,  ∞ ) with α + 2 β + 2 γ + δ + μ < 1 such that for all ( a , U ) ∈ U → U ,

where

provided that d ( T a , T u ) > 0 . Then, T has at most one fixed point in U.

Put δ = λ = μ = η = 0 in Theorem 2.1, the following result for Hardy-Rogers-type F-contraction is obtained.

Corollary 2.2

Suppose ( U , d ) is a complete function weighted metric space and ( F , τ ) , ( f , σ ) ∈ ℱ × [0,  ∞ ) with σ < τ . Let S , T : U → U be a self-mapping and there exist α , β , γ ∈ [0,  ∞ ) with α + 2 β + 2 γ < 1 such that for all ( a , U ) ∈ U → U ,

where

provided that d ( T a , T u ) > 0 . Then, S and T have at most one common fixed point in U. Similarly, by putting α = β = 0 , α = γ = 0 and γ = 0 , we can obtain fixed point theorems of Chatterjea-type F-contractions, Kannan-type F-contractions and Reich-type F-contractions, respectively.

Definition 3.1

Suppose ( U , d ) is a complete function weighted metric space and ( F , τ ) , ( f , σ ) ∈ ℱ × [0,  ∞ ) with σ < τ . Let S , T : B ( a 0 , r ′ ) → U be a self-mapping. Then, the pair ( T , S ) is called a rational type F-contraction on B ( a 0 , r ′ ) if there exist α , β , γ , δ , λ , μ , η ∈ [0,  ∞ ) with α + 2 β + 2 γ + δ + μ < 1 such that for all ( a , u ) ∈ B ( a 0 , r ′ ) × B ( a 0 , r ′ ) ,

where F ( M 1 ( a , u ) ) is given in Definition 2.1.

Theorem 3.1

Let ( U , d ) be a complete function weighted metric space and S , T : U → U be self-mappings such that the pair ( T , S ) is a rational type F-contraction on B ( s 0 , r ′ ) . Let r ′ > 0 so that the following conditions hold:

1. B ( s 0 , r ′ ) is ℱ -closed.

2. d ( s 0 , s 1 ) ≤ (1 − ω ) r ′ , for s 1 ∈ U and ω = α + β + γ + μ 1 − β − γ − δ .

3. There exists 0 < ε < r ′ such that B ( s 0 , r ′ ) f ( ( 1 − ω k + 1 ) r ′ ) ≤ f ( ε ) − α , where k ∈ N . Then, S and T have at most one common fixed point in B ( s 0 , r ′ ) .

Proof

Suppose s 0 is an arbitrary point. Define a sequence ( s n ) by

We need to show that ( s n ) is in B ( s 0 , r ′ ) for all n ∈ N . It is clear from definition that ω < 1 , therefore, by using mathematical induction and by (b), we write

Therefore s 1 ∈ B ( s 0 , r ′ ) . Suppose s 2 , … , s k ∈ B ( s 0 , r ′ ) for some k ∈ N . Now, if s 2 v + 1 ≤ s k , then by 5, we write

Using (D3), we write

Using ( F 1) we write

Similarly, if s 2 v ≤ s k , then

Therefore, from inequalities (6) and (7), we write

and

From (8) and (9), we write

Now, using the above equation, we write