Fixed Points of Proinov E -Contractions

: In this paper, we consider a new type of Proinov contraction on the setting of a symmetrical abstract structure, more precisely, the metric space. Our goal is to expand on some results from the literature using admissible mappings and the concept of E -contraction. The considered examples indicate the validity of the obtained results.


Introduction and Preliminaries
Fixed point theory is one of the most dynamic research topics of the last two decades. New and interesting results are obtained, following especially two directions: changing the frame (the structure of the abstract space-e.g., b-metric, delta symmetric quasi-metric or non-symmetric metric space, etc.) or changing the property of the operators.
The notion of E-contraction was introduced by Fulga and Proca [1]. Later, this concept has been improved by several authors, e.g., [2][3][4]. Undoubtedly, one of the most interesting, most original, most impressive fixed point theorem published in the last two decades is the result of Proinov [5]. By using certain auxiliary functions, Proinov [5] obtained interesting fixed point theorems that generalize, extend, and unify several recent fixed point results in the literature.
In this paper, we shall propose a new type of contraction, namely, Proinov type E-contraction, which combines the Proinov approach and the E-contraction setting.
First, we recall the basic results and definitions. Definition 1. Let (X, d) be a metric space and the functions ϑ, θ : (0, ∞) → R. A mapping T : X → X is said to be a Proinov type contraction if for all x, y ∈ X with d(T x, T y) > 0. Theorem 1 ([5]). Let (X, d) be a complete metric space and T : X → X be a Proinov type contraction, where the functions ϑ, θ : (0, ∞) → R are such that the following conditions are satisfied: (1) ϑ is non-decreasing; (2) θ(s) < ϑ(s) for any s > 0; (3) lim sup s→s 0 + θ(s) < ϑ(s 0 +) for any s 0 > 0, then T admits a unique fixed point.

Remark 1.
Notice that in [5], the author did not put the completeness assumption of the metric space; but, he used in the proof.
If the sequence {x m } is not Cauchy, then there exist e > 0 and the subsequences {m i } and {p i } of positive integers such that In 2012, Karapinar-Samet proposed [6] generalized α − ψ-contraction. In this consideration, the role of α admissible mapping is to combine the problem of fixed point of "cyclic contractions" with the same problem in the framework of metric spaces endowed with a"partially ordered set", see, e.g., [7][8][9][10][11][12]. Later, this notion is refined by Popescu [13]. Let (X, d) be a metric space, T : X → X be a mapping and a given mapping α : X × X → [0, ∞). We say that T is triangular α-orbital admissible (on short α-t.o.a.) [13] if the following two conditions are satisfied α(x, y) ≥ 1 and α(y, T y) ≥ 1 ⇒ α(x, T y) ≥ 1 for any x, y ∈ X. Lemma 2 ([13]). Let {x m } be a sequence on a non-empty set X defined by x m = T x m−1 for any m ∈ N, where T : X → X is an α-t.o.a.mapping. If there exists x 0 ∈ X such that α(x 0 , T x 0 ) ≥ 1, then α(x n , x m ) ≥ 1, for all n, m ∈ N.

Main Results
Throughout this section, we will consider that ϑ, θ : (0, ∞) → R are two functions such that holds for every distinct x, y ∈ X such that d(T x, T y) > 0, where Theorem 3. Let (X, d) be a complete metric space and T : X → X be a (α, ϑ, θ)-E contraction such that: (m 1 ) ϑ is lower semi-continuous and non-decreasing; (m 2 ) lim sup Then, T possesses a fixed point.
Proof. Let {x m } be the sequence in X defined as where x 0 is an arbitrary fixed point in X such that α(x 0 , T x 0 ) ≥ 1. We can assume that d(x m , x m+1 ) > 0 for every m ∈ N, because, on the contrary, if we can find m l ∈ N such that x m l = x m l +1 , due to the definition of the sequence x m l we have d(T x m l , x m l ) = d(x m l +1 , x m l ) = 0, which means T x m l = x m l . Therefore, letting x = x m and y = x m+1 in (4), using (m 0 ) and taking Lemma 2 into account, we have where and the inequality (6) yields and the inequality (6) becomes Therefore, by (m 1 ), we get so that, the sequence {d(x m , x m+1 )} is a decreasing and bounded below by 0. Thus, we can find δ ≥ 0, such that lim in fact, δ = 0. Supposing on the contrary, that δ > 0, letting m → ∞ in the first part of (6), we have but this contradicts the assumption (m 2 ). Consequently, δ = 0, that is Supposing that {x n } is not a Cauchy sequence, from Lemma 1, we can find two Therefore, applying (4) for x = x m i and y = x p i , and using Lemma 2, we get Thereupon, considering the limit superior in the above inequality, we have which contradicts (m 2 ). Consequently, {x m } is a Cauchy sequence on a complete metric space, which guarantees that {x m } is a convergent sequence. Denoting by x * the limit of this sequence, we will show that under the assumption (m 4 ) this point is a fixed point for the mapping T . Indeed, if not, then we have (here, we used (m 2 )), where (13) and keeping in mind the lower-semicontinuity of ϑ, we obtain lim inf which is a contradiction. Then, x * = T x * .

Theorem 4.
Adding the condition to the hypotheses of Theorem 3, one obtains uniqueness of the fixed point.
Proof. Supposing that y * ∈ X is such that T y * = y * = x * , by (4) we have which is a contradiction. Therefore, x * = y * . )) = (x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 for any U 1 , U 2 ∈ X, U 1 = (x 1 , y 1 ), U 2 = (x 2 , y 2 ). Let, also, the mapping T : X → X be defined as First of all, we can remark that neither Theorem 1, nor Theorem 2 cannot be applied, because letting, for example, x = A and y = C, we have Thus, replacing in (1) and (2) we are leading to some contradictions, since 13 4 ). Now, we consider the following functions: , (A 7 , A 0 )|}. We can easily see that the assumptions (m 1 )-(m 5 ) hold. Thus, we have to prove that the mapping T is an (α, ϑ, θ)-E contraction. So, consider the following cases: • Therefore, the mapping T has a unique fixed point, that is A = (4, 0).

Corollary 1.
Let (X, d) be a complete metric space and T : X → X be a mapping such that holds for every distinct x, y ∈ X, such that d(T x, T y) > 0, where and ϑ, θ : (0, ∞) → R. Suppose that: (m 0 ) θ(s) < ϑ(s), for all s > 0; (m 1 ) ϑ lower semi-continuous and non-decreasing; Then, T possesses a unique fixed point.

Corollary 2.
On a complete metric space (X, d) let T : X → X be a mapping such that holds for every distinct x, y ∈ X with d(T x, T y) > 0, where E (x, y) is defined by (15), κ ∈ [0, 1) and ϑ : (0, ∞) → (0, ∞) is left-continuous and non-decreasing. Then T possesses a unique fixed point.

Definition 4. A mapping
holds for every distinct x, y ∈ X, such that d(T 2 x, T 2 y) > 0, where Theorem 5. Let (X, d) be a complete metric space and T : X → X be a (α, ϑ, θ)-E 2 contraction such that (m 1 ) ϑ is lower semi-continuous and non-decreasing; (m 2 ) lim sup Then, T possesses a fixed point.
o.a., by Lemma 2, we get α(x m , x m+1 ) ≥ 1, for any m ∈ N and then Furthermore, the monotonicity of ϑ implies that where Thus, the sequence {d(x m , x m+1 )} is strictly decreasing and bounded below by 0. Hence, there exist δ ≥ 0 such that lim m→∞ d(x m+1 , x m+2 ) = δ. Suppose that δ > 0. Since lim m→∞ E 2 (x m , x m+1 ) = lim m→∞ d(x m+1 , x m+2 ) = δ > 0, letting the limit superior in (19), as n → ∞ and taking (m 2 ) into account, we obtain which is a contradiction. Therefore, and we claim that, in these conditions, the sequence {x m } is Cauchy. If not, from Lemma 1 we can find e > 0 and two subsequences {x m i } and x p i of {x m }, such that the equalities in (3), hold. Moreover, we remark that and then, using (21) we get lim i→∞ E 2 (x m i , x p i ) = e. On the other hand, letting i → ∞ in the below inequality we find that lim i→∞ d(x m i +2 , x p i +2 ) = e. Replacing in (17), x by x m i and y by x p i , and since, by Thus, which contradicts (m 2 ). Consequently, {x m } is a Cauchy sequence on a complete metric space, so that, there exists x * , such that lim m→∞ d(x m , x * ) = 0. We claim that x * is a fixed point of T , under the assumptions (m 6 ). Now, since T 2 is continuous, we have and then, we derive that T 2 x * = x * . By reductio ad absurdum, we assume that T x * = x * . Thus, by (17), and keeping in mind (m 1 ) and (m 6 ), we have which is a contradiction. Therefore, d(x * , T x * ) = 0, so that, x * is a fixed point of T .

Theorem 6.
Adding the condition to the hypotheses of Theorem 5, one obtains uniqueness of the fixed point.
Proof. In order to prove the uniqueness of the fixed point, we suppose, by reductio ad absurdum, that there exists x * , y * ∈ {x ∈ X : T x = x} such that x * = y * . Then, by (17) and (m 5 ) we obtain which is a contradiction. Therefore, the mapping T possesses a unique fixed point.

Remark 2.
In case of the mapping T is continuous, we get the same result without the assumption (m 6 ).
Then, T possesses a unique fixed point.

Corollary 4.
On a complete metric space (X, d) let T : X → X be a mapping, such that ϑ(d(T 2 x, T 2 y)) ≤ κϑ(E 2 (x, y)), holds for every distinct x, y ∈ X with d(T 2 x, T 2 y) > 0, where