A Note on the Górnicki-Proinov Type Contraction

In this paper, we propose a notion of the Górnicki-Proinov type contraction. Then, we prove the uniqueness and existence of the fixed point for such mappings in the framework of the complete metric spaces. Some illustrative examples are also expressed to strengthen the observed results.


Introduction and Preliminaries
The history of the fixed point theory goes back about a century. Banach's result initiated the metric fixed point theory in 1922 [1]. The first outstanding extension of this initial theorem was given by Kannan [2] in 1968. In this first generalization, Kannan [2] removed the necessity of the continuity of the contraction mapping. Recently, Górnicki [3] expressed an extension of Kannan type of contraction but the continuity condition was assumed. After then, Bisht [4] refined the result of Górnicki [3] by replacing the continuity condition for the considered mapping with orbitally continuity or p-continuity. Very recently, Górnicki [5] improved these two mentioned results by introducing new contractions, "Geraghty-Kannan type" and "ϕ-Kannan type." He proved the existence of a fixed point for such mappings. On the other hand, Proinov [6] discussed some existing results and noted that these results are particular cases of Skof [7]. He also proposed a very general fixed point theorem that also contains the result of Skof [7].
We first recall the pioneer theorem of Banach [1] and Kannan [2]. On a complete metric space ðX, dÞ, a mapping T : X → X admits a unique fixed point if there exists 0 ≤ K < 1 such that and d Tu, Tv for all u, v ∈ X. The inequality (1) belongs to Banach [1] and (2) belongs to Kannan [2]. By using the "asymptotic regularity" concept, Górnicki [3] proved an extension of Kannan Theorem 1.2. Before giving this interesting result, we recollect the interesting concepts: Let T be a self-mapping on a metric space ðX, dÞ and f T n ug be the Picard iterative sequence, for an initial point u ∈ X.
(o) The set OðT, uÞ = fT n u : n = 0, 1, 2, ⋯g is called the orbit of the mapping T at u.
The mapping T is said to be [3,5]: (o-c) orbitally continuous at a point w ∈ X if for any sequence fu n g in OðT, uÞ for some u ∈ X, lim n→∞ dðu n , wÞ = 0 implies lim n→∞ dðTu n , TwÞ = 0.
(a-r) asymptotically regular at a point u ∈ X if lim n→∞ d ðT n u, T n+1 uÞ = 0. If T is asymptotically regular at each point of X, we say that it is asymptotically regular. Remark 1. In [8], it is shown that p-continuity of T and the continuity of Tp are independent conditions for the case p > 1.
Theorem 2 (see [3,5]). On a complete metric space ðX, dÞ, a continuous asymptotically regular mapping T : X → X admits a unique fixed point if there exist 0 ≤ k < 1 and 0 ≤ K < +∞ such that for all u, v ∈ X.
Later, the assumption of continuity of the mapping T was replaced with weaker notions of continuity.
Theorem 3 (see [4]). On a complete metric space ðX, dÞ and a mapping T : X → X. Suppose that there exists 0 ≤ K < 1 such that for all u, v ∈ X. Then, T admits a unique fixed point if either T is (o-c) or (p-c) for p ≥ 1.
In [5], some generalizations of Theorems 2 and 3 are considered, by replacing the constant k with some real-valued functions.
Theorem 4 (see [5]). Let ðX, dÞ be a complete metric space and T : X → X be an (a-r) mapping such that there exist ψ : ½0,∞Þ → ½0,∞Þ and 0 ≤ K < ∞ such that for all u, v ∈ X. Suppose that: (i) ϕðθÞ < θ for all θ > 0 and ϕ is upper semicontinuous Then, T has a unique fixed point u * ∈ X and for each u ∈ X, T n u → u * as n → ∞.
Theorem 5 (see [5]). Let ðX, dÞ be a complete metric space and T : X → X be an (a-r) mapping such that there exist ς : ½0,∞Þ → ½0, 1Þ and 0 ≤ K < ∞ such that for all u, v ∈ X. Suppose that: (1) ςðθ n Þ → 1 ⇒ θ n → 0; (2) either T is (o-c) or T is (p-c) for some p ≥ 1 Then, T has a unique fixed point u * ∈ X and for each u ∈ X, T n u → u * as n → ∞.
On the other hand, very recently, Proinov announced some results which unify many known results [6].
Then, T admits a unique fixed point.
Lemma 8 (see [6]). Let ðu n Þ be a sequence in a metric space ðX, dÞ such that dðu n , u n+1 Þ → 0 as n → ∞. If the sequence ðu n Þ is not Cauchy, then there exist e > 0 and two subsequences fs k g, fr k g of positive integers such that Lemma 9 (see [6]). Let ðu n Þ be a sequence in a metric space ðX, dÞ such that dðu n , u n+1 Þ → 0 as n → ∞. If the sequence ðu n Þ is not Cauchy, then there exist e > 0 and two subsequences fs k g, fr k g of positive integers such that 2 Journal of Function Spaces In the end of this section, we recall the notions of α-orbital admissible and triangular α-orbital admissible mappings [9] with mention that these notions were extended in many directions, see, e.g., [10] and it could be potentially extended also to several approaches of recent developments in fixed point theory. See, for instance, [11][12][13][14][15][16][17][18][19][20][21].
On a metric space ðX, dÞ, a self-mapping T is called for any u, v ∈ X, where α : X × X → ½0,∞Þ (ii) triangular α-orbital admissible if it is α-orbital admissible and the following condition is satisfied for any u, v, w ∈ X Lemma 10. If for an triangular α-orbital admissible mapping T : X → X there exists u 0 ∈ X such that αðu 0 , Tu 0 Þ ≥ 1, then where the sequence fu n g is defined as u n+1 = Tu n .
Let ðX, dÞ be a metric space and the function α : X × X → ½0,∞Þ. The following conditions will be used further: R If for a sequence fu n g in X such that u n → u and αðu n , u n+1 Þ ≥ 1 for all n ∈ ℕ, then there exists a subsequence fu p k g of fu n g such that αðu p k , uÞ ≥ 1: ðUÞ For all u, v ∈ Fix X T = fz ∈ X : Tz = zg, we have αðu, vÞ ≥ 1.
Proof. Let u be any point (but fixed) in X and we build the sequence fu n g, where u 0 = u and u n = T n u for any n ∈ ℕ: If there exists m 0 ∈ ℕ such that T m 0 u = T m 0 +1 u = TðT m 0 uÞ, then T m 0 u is a fixed point of T. For this reason, we can suppose that T n u ≠ T n+1 u, for every n ∈ ℕ ∪ f0g and we claim that fu n g is Cauchy sequence. Assuming the contrary, that the sequence fu n g is not Cauchy, from Lemma 1, it follows that we can find e and two subsequences fs k g and fr k g of positive integers such that (9) holds. Letting u = u s k and v = u r k in (14), we have αðu s k , u r k Þ ≥ 1 (taking into account (1.8)), and then, or denoting ξ k = dðu s k +1 , u r k +1 Þ and ζ k = dðu s k , u r k Þ Taking into account the asymptotically regularity of T, from (9), it follows that ξ k → e + and ζ k → e: ð17Þ Thus, letting the limit in (16), we have This contradicts the assumption (a 2 ). Similarly, if we consider that the functions ψ, ϕ satisfy (a 3 ), the conclusion follows in the same way, but taking into account Lemma 2.

Journal of Function Spaces
Therefore, fu n g is a Cauchy sequence, and because the space ðX, dÞ is complete, there exists u * such that lim n→∞ u n = u * : We claim that u * is a fixed point of T: If T is orbitally continuous, then since fu n g ∈ OðT, uÞ and u n → u * , we have u n+1 = Tu n → Tu * as n → ∞. The uniqueness of the limit gives Tu * = u * : If T is p-continuous, for some p ≥ 1, by (19), we have lim n→∞ T p−1 u n = u * which implies lim n→∞ T p u n = Tu * (because T is p-continuous). Therefore, by uniqueness of the limit, we have Tu * = u * . Now, supposing that there exists v * ∈ X such that Tv * = v * ≠ u * = Tu * , from (14) and taking into account the property ðUÞ, we have which is a contradiction. Therefore, u * = v * .

Corollary 14.
Let ðT, dÞ be a complete metric space and T : X → X be an (a-r) mapping such that for each u, v ∈ X, where 0 ≤ K < ∞ and the function ς : ð0, ∞Þ → ð0, 1Þ is such that limsup θ→e+ ςðθÞ < 1 for any e > 0. If T is either (o-c) or (p-c) for some p ≥ 1, then T has a unique fixed point.
Taking ψðθÞ = θ and ϕðθÞ = k · θ, with k ∈ ½0, 1Þ Corollary 1 becomes: Corollary 15. Let ðT, dÞ be a complete metric space and T : X → X be an (a-r) mapping. If there exist k ∈ ½0, 1Þ and 0 ≤ K < ∞ such that for each u, v ∈ X, then T admits a unique fixed point provided that T is (o-c) or (p-c) for some p ≥ 1.

(iii) the mapping T is either (o-c) or (p-c)
Then, the mapping T possesses a fixed point. Moreover, the fixed point is unique, provided that property ðUÞ is satisfied.
Proof. Let fu n g be the sequence defined as in the previous theorem, as u n = T n u, where u ∈ X is arbitrary but fixed. Letting u = u s k and v = u r k in (2.7), we have and taking into account the assumptions (i), (ii), and Lemma 3, we get Setting ξ k = dðu s k +1 , u r k +1 Þ and ζ k = dðu s k , u r k Þ and since ϕðθÞ < ψðθÞ, we get On the other hand, from 1.5 that ξ k → e + , ζ k → e + and then, letting the limit as k → ∞ in the above inequality, since T is an (a-r) mapping and taking into account the second part of the assumption (i), we have 4 Journal of Function Spaces which is a contradiction. Thus, the sequence fu n g is Cauchy on a metric space, so there exists u * such that u n → u * as n → ∞ and following the lines of the previous proof, we get that u * is the unique fixed point of T: Again, letting αðu, vÞ = 1 for any u, v ∈ X we get the following: Theorem 17. Let ðX, dÞ be a complete metric space, and two functions ψ, φ ∈ Λ such that ( a 1 ) is satisfied. Let T : X → X be an (a-r) mapping. Suppose that there exists 0 ≤ K < ∞ such that for each u, v ∈ X with dðTu, TvÞ > 0, where ψ, ϕ ∈ Λ. Suppose also that (i) ψ is nondecreasing and limsup θ→e+ < ψðe + Þ for any e > 0

(ii) the mapping T is either (o-c) or (p-c)
Then, the mapping T possesses a unique fixed point.
Next, we consider mappings that satisfy a similar condition as (14), but for which the asymptotic regularity condition is not necessary.
Proof. Let fug be a sequence in X defined as where u 0 is an arbitrary but fixed point in X. Replacing in (31) and taking into account (11) or setting x n = dðu n−1 , u n Þ (we can suppose that x n > 0) and taking into account the condition (a 1 ) for any θ > 0, we get If the condition (C) holds, from the above inequality, we get x n < x n−1 , for every n ∈ ℕ. Consequently, being positive and strictly decreasing, the sequence fx n g is convergent and there is x ≥ 0 ∈ X such that x n → x. If we assume that x > 0, then from the above inequality, we have which is a contradiction. Thus, The aim for the next step is to prove that the sequence fu n g is Cauchy. Supposing by contradiction, the sequence