Asymptotic fixed points for nonlinear contractions
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Recently, W. A. Kirk proved an asymptotic fixed point theorem for nonlinear contractions by using ultrafilter methods. In this paper, we prove his theorem under weaker assumptions. Furthermore, our proof does not use ultrafilter methods. 1. Introduction There are many papers in the literature that discuss the asymptotic fixed point theory, in which the existence of the fixed points is deduced from the assumption on the iterates of an operator (e.g., [1, 6] and the references therein). Recently, Kirk [5] studied an asymptotic fixed point theorem concerning nonlinear contractions. He proved the following theorem [5, Theorem 2.1] by appealing to ultrafilter methods. Theorem 1.1. Let (M,d) be a complete metric space. Let T : M → M be a continuous mapping such that
d T n x,T n y ≤ φn d(x, y)
(1.1)
for all x, y ∈ M, where φn : [0, ∞] → [0, ∞] and limn→∞ φn = φ uniformly on the range of d. Suppose that φ and all φn are continuous, and φ(t) < t for t > 0. If there exists x0 ∈ M which has a bounded orbit O(x0 ) = {x0 ,Tx0 ,T 2 x0 ,...}, then T has a unique fixed point x∗ ∈ M such that limn→∞ T n x = x∗ for all x ∈ M. In this paper, we prove Theorem 1.1 under weaker assumptions without the use of ultrafilter methods. 2. Main results We need the following recursive inequality (cf. [2, Lemmas 2.1 and 3.1], [3, Lemmas 2.1 and 2.4], and [4, Lemma 1]). Lemma 2.1. Let φ : R+ → R+ be upper semicontinuous, that is, limsupt→t0 φ(t) ≤ φ(t0 ) for all t0 ∈ R+ , and φ(t) < t for t > 0. Suppose that there exist two sequences of nonnegative real Copyright © 2005 Hindawi Publishing Corporation Fixed Point Theory and Applications 2005:2 (2005) 213–217 DOI: 10.1155/FPTA.2005.213
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Asymptotic fixed points for nonlinear contractions
numbers {un } and {n } such that
u2n ≤ φ un + n ,
(2.1)
where n → 0 as n → ∞. Then either supun = ∞ or liminf un = 0. Proof. Suppose that b = sup{un } < ∞. Assume that liminf un = 0. Then there exist m > 0 and N1 > 0 such that un > m for all n > N1 . Since φ is upper semicontinuous, φ(t)/t is upper semicontinuous on [m,b] and so that Lm = max{φ(t)/t, t ∈ [m,b]} < 1 due to the facts that φ(t) < t for t > 0 and that φ(t)/t achieves its maximum on [m,b]. Let > 0. By (2.1), there exists N2 > N1 such that
u2n ≤ φ un + ≤ Lm un +
(2.2)
for all n > N2 . Note that the contraction mapping f (x) = Lm x + has a unique fixed point /(1 − Lm ) and limn→∞ f n (x) = /(1 − Lm ) for any real number x. Now for any n > N2 ,
u22 n ≤ φ u2n + ≤ Lm u2n + ≤ Lm f un + = f 2 un .
(2.3)
By induction, u2k n ≤ f k (un ) for all k, so that m ≤ f k (un ). Letting k → ∞, m ≤ /(1 − Lm ). This is impossible since > 0 can be arbitrarily chosen. Theorem 2.2. Let (M,d) be a complete metric space. Let T : M → M be such that
d T n x,T n y ≤ φn d(x, y)
(2.4)
for all x, y ∈ M, where φn : [0, ∞] → [0, ∞] and limn→∞ φn = φ uniformly on any bounded interval [0,b]. Suppose that φ is upper semicontinuous and φ(t) < t for t > 0. Furthermore, suppose there exists a positi
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