CUBO A Mathematical Journal Vol.10, N o ¯ 04, (67–72). December 2008 A Fixed Point Theorem for Certain Operators B.E. Rhoades Department of Mathematics, Indiana University, Bloomington, IN 47405-7106 email: rhoades@indiana.edu ABSTRACT We obtain a fixed point theorem for a class of operators. This result is an extension of a similar theorem of Constantin (1994). RESUMEN Obtenemos un teorema de punto fijo para una clase de operadores. Este resultado es una extensión de un teorema similar devido a Constantin (1994). Key words and phrases: Fixed point theorem. Math. Subj. Class.: 47H10. Most fixed point theorems are proved either by examining the successive iterates of the oper- ator, or by constructing an iteration scheme, such as that of Mann or Ishikawa. In this paper we consider the situation in which the operator is used on the successive iterates of a sequence. 68 B.E. Rhoades CUBO 10, 4 (2008) In a recent paper, Constantin [3] obtained a fixed point theorem for a class of operators which are selfmaps of a Banach space X, and which satisfy the condition ‖T x − T y‖ ≤ g(‖x − y‖, ‖x − T x‖, ‖y − T y‖) (1) for all x, y ∈ X, where g : R3+ → R+, g is continuous, nondecreasing in each variable, and is such that, if h(r) := g(r, r, r), then r − h(r) is nonnegative and strictly increasing on R+. A natural extension of (1) would be: Let Λ denote the set of all continuous functions g : R5 + → R+, nondecreasing in each variable and such that, if h(r) := g(r, r, r, r, r), then h(r) < r for each r > 0. Define T from X to X satisfying ‖T x − T y‖ ≤ g(‖x − y‖, ‖x − T x‖, ‖y − T y‖, (2) ‖x − T y‖, ‖y − T x‖) for all x, y ∈ X, some g ∈ Λ. However, a slightly more general extension of (1) is the following. Let X be a Banach space, g : R3+ → R+, g continuous, nondecreasng, and satisfying g(t) < t for each t > 0. Let T be a selfmap of X satisfying ‖T x − T y‖ ≤ g(M (x, y)), for all x, y ∈ X, (3) where M (x, y) := max{‖x − y‖, ‖x − T x‖, ‖y − T y‖, ‖x − T y‖, ‖y − T x‖). Theorem 1. Let A satisfy (3) and {xn} ⊂ X. Then the following are equivalent: (i) T xn − xn → 0 as n → ∞, (ii) {T xn − xn} is bounded, and {xn} converges to a point p which is the unique fixed point of T . Proof. (i) ⇒ (ii). Define yn = T xn − xn, αn = supn{‖xm − xn‖ : m ≥ n}, and βn = supn{‖ym‖ : m ≥ n}. Then {αn} and {βn} are nonincreasing nonnegative sequences. Hence lim αn = α ≥ 0 and, from the hypotheses, lim yn = 0 and {yn} is bounded. Assume that α > 0. From (3), with m ≥ n, ‖xm − xn‖ ≤ ‖T xm − ym − (T xn − yn)‖ ≤ ‖T xm − T xn‖ + ‖ym − yn‖ ≤ g(max{‖xm − xn‖, ‖ym‖, ‖yn‖, ‖xm − T xn‖, ‖xn − T xm‖}) + 2βn ≤ g(max{αn, βn, βn, αn + βn, αn + βn}) + 2βn. CUBO 10, 4 (2008) A Fixed Point Theorem for Certain Operators 69 Thus, αn ≤ g(αn +βn)+2βn. Taking the limit as n → ∞ yields α ≤ g(α) < α, a contradiction. Therefore α = 0 and {xn} is Cauchy, hence convergent to some point p in X. Since T xn − xn → 0 and xn → p, it follows that T xn → p. Again using (3), ‖T p − T xn‖ ≤ g(max{‖p − xn‖, ‖p − T p‖, ‖yn‖, ‖p − T xn‖, ‖xn − T p‖}). Taking the limit as n → ∞ yields ‖T p − p‖ ≤ g(max{0, ‖p − T p‖, 0, 0, ‖p − T p‖}) = g(‖p − T p‖), which implies that ‖p − T p‖ = 0, or T p = p. To prove uniqueness, suppose that q is also a fixed point of T . Then, from (3), ‖p − q‖ = ‖T p − T q‖ ≤ g(max{‖p − q‖, 0, 0, ‖p − q‖, ‖q − p‖}) = g(‖p − q‖), which implies that p = q. (ii) ⇒ (i) Using (3), ‖T xn − xn‖ = ‖T xn − T p + T p − xn‖ ≤ ‖T xn − T p‖ + ‖p − xn‖ ≤ g(max{‖xn − p‖, ‖xn − T xn‖, ‖p − T p‖, ‖xn − T p‖, ‖p − T xn‖}) + ‖p − xn‖. Taking the lim sup of both sides, since {yn} is bounded, one obtains, with the identification γ = lim sup ‖yn‖, γ ≤ g(max{0, γ, 0, 0, lim sup ‖p − T xn‖}) + 0. But lim sup ‖p − T xn‖ ≤ lim sup(‖p − xn‖ + ‖xn − T xn‖) = γ. Therefore we have γ ≤ g(γ), which implies that γ = 0. The special case of (3) with g(t) := kt for some 0 ≤ k < 1, and X a metric space is that of Ćirić [2], which was shown in [7] to be one of the most general contractive definitions for which a unique fixed point exists. In order to prove that a map satisfying (3) has a fixed point, it would be necessary to show that the orbit of some x ∈ X is bounded, which cannot be implied from (3). However, the following is true. Theorem 2. Let X be a complete metric space, T a selfmap of X satisfying d(T x, T y) ≤ g(M (x, y)), for each x, y ∈ X, (4) 70 B.E. Rhoades CUBO 10, 4 (2008) where M (x, y) = max{d(x, y), d(x, T x), d(y, T y), d(x, T y), d(y, T x)}. If there exists a point x0 ∈ X with bounded orbit, then T has a unique fixed point in X. Proof. For any n ∈ N, O(x, n) := {x, T x, T 2x, . . . , T nx}, and δ(A) denotes the diameter of a set A. Let m, n ∈ N, n < m. Then, from (4), with x = x0, d(T nx, T mx) = d(T (T n−1x), T (T m−1x)) ≤ g(max{d(T n−1x, T m−1x), d(T n−1x, T nx), d(T m−1x, T mx), d(T n−1 x, T m x), d(T m−1 x, T n x)} ≤ g(δ[O(T n−1x, n − m + 1)]) ≤ g(g(δ[O(T n−2x, n − m + 2)]) · · · ≤ gn(δ[O(x, m)]). (5) It is well known that the hypotheses on g imply that lim gn(t) = 0 for each t ≥ 0. Since the orbit of x = x0 is bounded, (5) implies that {T nx} is Cauchy, hence convergent to a point p ∈ X. Suppose that p 6= T p. Then, from (4), d(p, T p) ≤ d(p, T n+1x) + d(T n+1x, T p) ≤ d(p, T n+1x) + g(max{d(T nx, p), d(T nx, T n+1x), d(p, T p), d(T nx, p), d(p, T n+1x)}. Taking the limit of both sides of the above inequality as n → ∞ yields d(p, T p) ≤ g(d(p, T p)) < d(p, T p), a contradiction, and p = T p. Suppose that p and q are fixed points of T , with p 6= q. Then, using (4), d(p, q) = d(T p, T q) ≤ g(max({d(p, q), 0, 0, d(p, q), d(q, p)} = g(d(p, q)) < d(p, q), a contadiction. Therefore p = q. If T is continuous, then, even with X unbounded, Theorem 2 is a special case of Theorem 3.3 of [4] CUBO 10, 4 (2008) A Fixed Point Theorem for Certain Operators 71 If one replaces M (x, y) with m(x, y) := max{d(x, y), d(x, T x), d(y, T y), [d(x, T y) + d(y, T x)]/2}, in Theorem 2, then Theorem 2 is true without the boundedness assumption. See, e.g., Theorem 2.2 of [1]. Most of the recent papers on fixed ont theory, which do not involve fixed point iterations, deal with four maps. For a survey of these results the reader may wish to consult [6] and the references therein. Let F (T ) denote the fixed point set of a mapping T . In [5] it was conjectured that F (T n) = F (T ) for every map T which satisfies a contractive condition that does not include nonexpansive maps. That conjecture was verified in [5] for many such maps. We shall now show that the same is true for maps satisfying (4). Theorem 3. Let X be a metric space, T a selfmap of X satisfying (4) with F (T ) 6= ∅. Then F (T n) = F (T ) for every integer n ≥ 1. Proof. Since F (T ) 6= ∅. F (T n) 6= ∅. Clearly F (T ) ⊆ F (T n). Suppose that p ∈ F (T n), for some positive integer n. We shall assume that n > 1, since the case for n = 1 is trivial. Let i, j be integers, 0 ≤ i < j ≤ n. Then, using (4), d(T ip, T jp) ≤ g(M (T i−1p, T j−1p)) ≤ g(δ[(O(p, n)]). Suppose that δ[(O(p, n)] > 0. Then the above inequality implies that δ[(O(p, n)] ≤ g(δ[(O(p, n)]) < δ[(O(p, n)], a contradiction. Therefore δ[(O(p, n)] = 0, and p ∈ F (T ). Received: February 2008. Revised: March 2008. References [1] R.P. Agarwal, Donal O’Regan, and M. Sambandham, Random and deterministic fixed point theory for generalized contractive maps, Applicable Analysis 83 (2004), 711–725. [2] Lj. B. Ćirić, A generalization of Banach’s contraction principle, Proc. Amer. Math. Soc. 45 (1974), 27–273. [3] A. Constantin, On the approximation of fixed points of operators, Bull. Calcutta Math. Soc. 86 (1994), 323–326. 72 B.E. Rhoades CUBO 10, 4 (2008) [4] J. Jachymski, On common fixed point theorems for some families of maps, Int. J. Pure & Appl. Math. 25 (1994), 925–937. [5] G.S. Jeong and B.E. Rhoades, Maps for which F (T ) = F (T n), Fixed Point Theory and Appl. 6 (2003), 87–131. [6] Donal O’Regan, Naseer Shahzad, and Ravi P. Agarwal, Common fixed point theory for compatible maps, Nonlinear Analysis Forum 8 (2003), 179–22. [7] B.E. Rhoades, A comparison of various definitions of contractive mappings, Trans. Amer. Math. Soc. 226 (1977), 257–290. N5-FixedpointtheoremBCMS