tailieunhanh - A new aspect to Picard operators with simulation functions
In the present paper, considering the simulation function, we give a new class of Picard operators on complete metric spaces. We also provide a nontrivial example that shows the aforementioned class properly contains some earlier such classes. | Turk J Math (2016) 40: 832 – 837 ¨ ITAK ˙ c TUB ⃝ Turkish Journal of Mathematics doi: Research Article A new aspect to Picard operators with simulation functions ¨ ˙ ¸ ER, Tu˘ Murat OLGUN∗, Ozge BIC gc ¸e ALYILDIZ Department of Mathematics, Faculty of Science, Ankara University, Ankara, Turkey Received: • Accepted/Published Online: • Final Version: Abstract: In the present paper, considering the simulation function, we give a new class of Picard operators on complete metric spaces. We also provide a nontrivial example that shows the aforementioned class properly contains some earlier such classes. Key words: Fixed point, Picard operators, simulation functions 1. Introduction Let (X, d) be a metric space and T : X → X be a mapping; then T is called a Picard operator on X , if T has a unique fixed point and the sequence of successive approximation for any initial point converges to the fixed point. The concept of Picard operators is closely related to that of contractive-type mappings on metric spaces. It is well known that almost all contractive-type mappings are Picard operators on complete metric spaces. (See for more details [2–6]). In the present paper, considering the simulation function, we give a new class of Picard operators on complete metric spaces. The concept of simulation functions is given by [8] in fixed point theory. Let ζ : [0, ∞)×[0, ∞) → R be a mapping; then ζ is called a simulation function if it satisfies the following conditions: (ζ1 ) ζ(0, 0) = 0 (ζ2 ) ζ(t, s) 0 (ζ3 ) If {tn }, {sn } are sequences in (0, ∞) such that lim tn = lim sn > 0, then n→∞ n→∞ lim sup ζ(tn , sn ) 0. Therefore, if T is a Z -contraction with respect to ζ ∈ Z then d(T x, T y) 0 for all n ∈ N and define dn = d(xn , xn+1 ). Then, since { M (xn , xn−1 ) = max d(xn , xn−1 ), d(xn , xn+1 ), d(xn−1 , xn ), 1 2 [d(xn , xn ) } + d(xn−1 , xn+1 )] = max {dn−1 , dn } 833 OLGUN
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