Double Controlled Metric Type Spaces and Some Fixed Point Results

Thabet Abdeljawad 1 ID , Nabil Mlaiki 1, Hassen Aydi 2,* ID and Nizar Souayah 3 1 Department of Mathematics and General Sciences, Prince Sultan University, P.O. Box 66833, Riyadh 11586, Saudi Arabia; tabdeljawad@psu.edu.sa (T.A.); nmlaiki@psu.edu.sa (N.M.) 2 Department of Mathematics, College of Education in Jubail, Imam Abdulrahman Bin Faisal University, P.O. Box 12020, Jubail 31961, Saudi Arabia 3 Department of Natural Sciences, Community College Al-Riyadh, King Saud University, Riyadh 4545, Saudi Arabia; nizar.souayah@yahoo.fr * Correspondence: hmaydi@iau.edu.sa or hassen.aydi@isima.rnu.tn; Tel.: +966-530894964


Introduction
One of the generalizations of metric spaces was studied by Bakhtin [1] and Czerwik [2] who introduced the notion of b-metric spaces.Since then, many authors obtained several fixed point results for single valued and multivalued operators in the setting of b-metric spaces, for instance, see [3][4][5][6][7][8][9][10][11][12][13][14][15][16].Among the generalizations of b-metric spaces, we cite the work of Kamran et al. [17] (see also [18][19][20][21]) who introduced extended b-metric spaces by controlling the triangle inequality rather than using control functions in the contractive condition.Proving extensions of Banach contraction principle from metric spaces to b-metric spaces and hence to controlled metric type spaces is useful to prove existence and uniqueness theorem for different types of integral and differential equations.Some nice applications can be found for example in the recent article [22].In fact, the authors in [17] gave a slightly modified application of a proven fixed point result.However, finding serious applications to integral equations and dynamical systems is still of interest.In this article, we have been only motivated theoretically to relax the triangle inequality of b-metric spaces by using two controlled functions rather than using one.Definition 1. [17] Given a function θ : X × X → [1, ∞), where X is a nonempty set.The function d : for all x, y, z ∈ X.
for all u, v, w ∈ X.Then, ρ is called a controlled metric type and (X, ρ) is called a controlled metric type space.
Now, we introduce a more general b-metric space.
for all u, v, w ∈ X.Then, q is called a double controlled metric type by α and µ.

Remark 1.
A controlled metric type is also a double controlled metric type when taking the same function(s).The converse is not true in general (see Examples 1 and 2).
(ii): Otherwise, first (q3) is verified in the case that x = y.Consider the case that x = y, hence we get that x = y = z.In the subcases (x ≥ 1 and y ∈ [0, 1)) and (y ≥ 1 and x ∈ [0, 1)), it is easy to see that (q3) holds.
On the other hand, we have q(0, ).
This leads us to say that q is not an extended b-metric when considering the same function µ = α.
Thus, q is not a controlled metric type for the function α.
The topological concepts as continuity, convergent and Cauchy on double controlled metric type spaces are given in the following.Definition 4. Let (X, q) be a double controlled metric type space by one or two functions.
(1) The sequence {u n } is convergent to some u in X, if for each positive ε, there is some integer N ε such that q(u n , u) < ε for each n ≥ N ε .It is written as lim n→∞ u n = u.
(2) The sequence {u n } is said Cauchy, if for every ε > 0, q(u n , u m ) < ε for all m, n ≥ N ε , where N ε is some integer.
(3) (X, q) is said complete if every Cauchy sequence is convergent.Definition 5. Let (X, q) be a double controlled metric type space by either one function or two functions-for u ∈ X and k > 0.
(ii) The self-map T on X is said to be continuous at u in X if for all δ > 0, there exists k > 0 such that T(B(u, k)) ⊆ B(Tu, δ).
Note that if T is continuous at u in (X, q), then u n → u implies that Tu n → Tu when n tends to ∞.
In this paper, we present some fixed point theorems in double controlled metric type spaces.The first one is the related Banach contraction principle.The second one concerns with a nonlinear case involving a function φ satisfying suitable conditions.The last one is the related Kannan type result.The given concepts and theorems are illustrated by some examples.

Main Results
Our first fixed point result is the following: Theorem 1.Let (X, d) be a complete double controlled metric type space by the functions α, µ : X × X → [1, ∞).Suppose that T : X → X satisfies q(Tx, Ty) ≤ kq(x, y), for all x, y ∈ X, where k ∈ (0, 1).For u 0 ∈ X, choose u n = T n u 0 .Assume that In addition, for each u ∈ X, suppose that lim n→∞ α(u, u n ) and lim n→∞ µ(u n , u) exist and are finite.
Then, T has a unique fixed point.
Proof.Consider the sequence {u n = T n u 0 } in X that satisfies the hypothesis of the theorem.By using label (1), we get q(u n , u n+1 ) ≤ k n q(u 0 , u 1 ) for all n ≥ 0.
Let n, m be integers such that n < m.We have We used α(x, y) ≥ 1.Let Hence, we have The ratio test together with (2) imply that the limit of the real number sequence {S n } exits, and so {S n } is Cauchy.Indeed, the ration test is applied to the term so the sequence {u n } is Cauchy.Since (X, q) is a complete double controlled metric type space, there exists some ξ ∈ X such that lim n→∞ q(u n , ξ) = 0.
It is a contradiction, so ξ = η.Hence, ξ is the unique fixed point of T.

Remark 2.
The assumption (3) in Theorem 1 above can be replaced by the assumptions that the mapping T and the double controlled metric d are continuous.Indeed, when u n → ξ, then Tu n → Tξ and hence we have lim n→∞ q(Tu n , Tξ) = 0 = lim n→∞ q(Tu n+1 , Tξ) = q(ξ, Tξ), and hence Tξ = ξ.
Theorem 1 is illustrated by the following examples.
Example 4. Let X = [0, 4].Consider the double controlled metric q and functions α and µ given in Example 1. Choose Tx = 1 for all x ∈ X.Let u 0 = 1 and k = 1 2 .We have that is, (2) holds.In addition, for each u ∈ [0, 4], we have That is, (3) holds.All hypotheses of Theorem 1 are satisfied and ξ = 1 is the unique fixed point.Definition 6.Given u 0 ∈ X, the orbit O(u 0 ) of u 0 is defined as O(u 0 ) = {u 0 , Tu 0 , T 2 u 0 , ...}, where T is a self-map on the set X.The operator G : X −→ R is called T-orbitally lower semi-continuous at η ∈ X if when {u n } in O(u 0 ) such that lim n→∞ q(u n , η) = 0, we get that G(η) ≤ lim n→∞ inf G(u n ).
Proceeding similarly as [17] and using Definition 6, we have the following corollary generalizing Theorem 1 in [24].
Corollary 1.Let T be a self-map on (X, q) a complete double controlled metric type space by two mappings α, µ.Given u 0 ∈ X.Let k ∈ (0, 1) be such that q(Tz, T 2 z) ≤ kq(z, Tz), f or each z ∈ O(u 0 ). ( Take u n = T n u 0 and suppose that Then, lim n→∞ q(u n , ξ) = 0. We also we have that Tξ = ξ if and only if the operator x → q(x, Tx) is T−orbitally lower semi-continuous at u.
Our next fixed point result concerns with the nonlinear case using a control function of Matkowski [25].
Proof.Let {u n } and u 0 be as in the statement of the theorem.If, for some m, we have u m = u m+1 = Tu m , then clearly u m is the fixed point.Now, suppose that u n+1 = u n for each n.From condition (10), where clearly If, for some n, we accept that Λ(u n−1 , u n ) = q(u n , u n+1 ), then from (12) and that φ(t) < t, ∀t > 0, we have which leads to a contradiction.Hence, for all n, we must have If we proceed inductively, we deduce that for each n ≥ 0, we have 0 < q(u n , u n+1 ) ≤ φ n (q(u 0 , u 1 )).
To show that {u n } is Cauchy, we proceed as in the proof of Theorem 1.For all m > n, we may get The assumption (11) by means of the ratio test applied to the series derived from the right-hand side of ( 14), as in the proof of Theorem 1, will lead to the sequence {u n } being Cauchy.Since (X, d) is complete, there exists η ∈ X such that lim n→∞ q(u n , η) = 0.That η is a fixed point is shown as in Remark 2. To prove the uniqueness of the fixed point, assume z is such that Tz = z and z = η.By (10), we have 0 < q(η, z) = q(Tη, Tz) ≤ φ(Λ(η, z)) = φ(q(η, z)) < q(η, z), which is a contradiction.
In the following theorem, we propose the related fixed point result of Kannan [26].
Theorem 3. Let (X, d) be a complete double controlled metric type space by the functions α, µ : X × X → [1, ∞).Let T : X → X be a Kannan mapping defined as follows: q(Tx, Ty) ≤ a[q(x, Tx) + q(y, Ty)], for all x, y ∈ X, where a ∈ (0, ).For u 0 ∈ X, take u n = T n u 0 .Suppose that For each u ∈ X, assume that Then, there exists a unique fixed point of T.
Proof.Let {u n = Tu n−1 } in X be such that the hypotheses ( 17) and ( 18) hold.From ( 16), we obtain Then, q(u n , u n+1 ) ≤ a 1 − a q(u n−1 , u n ).By induction, we get Now, let us prove that {u n } is a Cauchy sequence.Using the triangle inequality, for all n, m ∈ N, we obtain Similar to the proof of Theorem 1, we get ) which allows us to proceed as in the proof of Theorem 1 and we deduce that {u n } is a Cauchy sequence in the complete double controlled metric space (X, d).Thus, there exists u ∈ X as a limit of {u n } in (X, d).Assume that Tu = u.We have 0 < q(u, Tu) ≤ α(u, u n+1 )q(u, u n+1 ) + µ(u n+1 , Tu)q(u n+1 , Tu) ≤ α(u, u n+1 )q(u, u n+1 ) + µ(u n+1 , Tu)[aq(u n , u n+1 ) + aq(u, Tu)].
Passing to the limit on both sides of (20) and making use of the condition (18), we deduce that 0 < q(u, Tu) < q(u, Tu), which is a contradiction.Hence, Tu = u.To prove the uniqueness of the fixed point u, suppose that T has another fixed point v.Then, q(u, v) = q(Tu, Tv) ≤ a[q(u, Tu) + q(v, Tv)] = a[q(u, u) + q(v, v)] = 0.
Therefore, u = v and T has a unique fixed point.

1.
Condition (18) in Theorem 3 can be replaced by the continuity of the double controlled metric d and the mapping T as it was done in Theorem 2.

2.
Continuity of the double controlled metric d and the mapping T in Theorem 2 can be replaced by the following condition: For each u ∈ X, we have lim n→∞ α(u, u n ) < ∞ and lim n→∞ µ(u n , Tu)φ(q(u, Tu)) < q(u, Tu).

Perspectives
It is an open question to treat the cases of the related Chatterjea, Hardy-Rogers, Ćirić and Suzuki contraction types.Moreover, it is always of great interest to find real applications for the proven fixed point theorems in metric type spaces.A future work in this direction will be highly recommended.