2017
DOI: 10.1155/2017/2068163
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On the Power of Simulation and Admissible Functions in Metric Fixed Point Theory

Abstract: We investigate the existence and uniqueness of certain operators which form a new contractive condition via the combining of the notions of admissible function and simulation function contained in the context of complete -metric spaces. The given results not only unify but also generalize a number of existing results on the topic in the corresponding literature.

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Cited by 35 publications
(27 citation statements)
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“…, is an instantaneous example of a simulation function. For further and more interesting examples, we refer e.g., [5,[14][15][16][17][18] and relate references therein.…”
Section: Definition 5 ([5])mentioning
confidence: 99%
“…, is an instantaneous example of a simulation function. For further and more interesting examples, we refer e.g., [5,[14][15][16][17][18] and relate references therein.…”
Section: Definition 5 ([5])mentioning
confidence: 99%
“…It is known that b-metric space itself is a b-metric and, furthermore, b-metric turns out to be a standard metric for s = 1. As it is seen in [14,15,17,20], for different choices of σ ∈ Σ, we shall get more different consequences.…”
Section: Discussionmentioning
confidence: 95%
“…, is an instantaneous example of a simulation function. For further and more interesting examples, we refer to, e.g., [10][11][12][13][14][15][16][17] and relates references therein.…”
Section: Definition 3 ([10]mentioning
confidence: 99%
“…In the last decades, the renowned metric fixed point results of Banach [1] has been improved, extended, and generalized in several ways, see e.g., [2][3][4][5][6][7][8][9][10][11][12][13][14][15][16][17][18][19][20][21]. We first mention that the notion of E-contraction, defined by Fulga and Proca [10,11], is one of the interesting approach to improve the Banach mapping contraction.…”
Section: Introductionmentioning
confidence: 99%
“…Let M = {1,3,4, 5} and d : M → M, d(p, q) = |p − q|. We define the function S : M → M as, S1 = S3 = S4 = 3, S5 = 1, and set σ(t, s) = 1 2 s − t.One can easily get that d(3, 4) = d(4, 5) = 1, d(3, 5) = d(1, 3) = 2, d(1, 4) = 3, d(1, 5) = 4, d(S3, S4) = d(S1, S3) = d(S1, S4) = 0, d(S3, S5) = d(S4, S5) = d(S1, S5) = 2.Moreover, we have E(1, 4) = E(1, 3) = E(4, 5) = 4 and E(1, 5) = E(3, 5) = 6 and E(3, 4) = 2…”
mentioning
confidence: 99%