AroElAGT.dvi @ Applied General Topology c© Universidad Politécnica de Valencia Volume 9, No. 2, 2008 pp. 185-188 A note on a fixed point theorem for ray oriented maps A. Arockiasamy and A. Anthony Eldred Abstract. In this paper, we will prove a fixed point theorem for a ray-oriented map defined on a nonempty closed bounded convex subset of a Banach space. 2000 AMS Classification: 47H10 , 54H25. Keywords: Fixed point, ray - oriented map. Notations Let X be a Banach space and K be a non-empty subset of X. Let T : K → K be a mapping. Let R x be the ray passing through the segment < x, T x > and so R x := {(1 − λ)x + λT x : λ ∈ R}. Let < x, y > be defined to be as {(1 − λ)x + λy : λ ∈ [0, 1]} and (x, y) := {(1 − λ)x + λy : λ ∈ (0, 1)}. For any x1, x2 ∈ Rx, we say that x1 ≤ x2 whenever λ1 ≤ λ2 where x1 = (1 − λ1)x + λ1T x and x2 = (1 − λ2)x + λ2T x for some λ1 , λ2 ∈ R. 1. Introduction Let X be a normed linear space and let K be a nonempty closed bounded convex subset of X. Suppose T : K → K is a mapping satisfying the following conditions: (i). For some element x0 of K, Rx0 ⋂ K is invariant under T and (ii). For each element x ∈ R x0 ⋂ K, T | < x, T x > ⋂ K is continuous. Then, we will prove that there exists y0 ∈ Rx0 ⋂ K such that < y0, T y0 >⊆ Rx0 ⋂ K is invariant under T . Moreover,the above theorem will be followed by a corollary as in the following: Suppose T : [a, b] → [a, b] is a mapping where a, b ∈ ℜ. If for each x ∈ [a, b], the map T restricted to the segment joining x and T x is continuous. Then we will prove that there exists an invariant interval under T and so it will have a fixed point in [a, b]. This result extends one dimensional Brouwer’s result for a larger class of mappings which need not be continuous. Also one can find some similar treatment for the convergence of fixed point in the real line by Beardon [1]. For further important fixed point results one can refer to [2]. 186 A. Arockiasamy and A. Anthony Eldred 2. Main Results Theorem 2.1. Let X be a normed linear space and let K be a nonempty closed bounded convex subset of X. Suppose T : K → K is a mapping satisfying the following conditions: (1) For some element x0 of K, Rx0 ⋂ K is invariant under T and (2) For each element x ∈ R x0 ⋂ K, T | < x, T x > ⋂ K is continuous Then, T has a fixed point in R x0 ⋂ K. Note: When we say T | < x, T x > is continuous, we mean that T is right continuous at x and left continuous at T x if x < T x. Proof. Assume that the conclusion of the theorem is false. That is, T does not have a fixed point in R x0 ⋂ K. Therefore, for every b ∈ R x0 ⋂ K, < b, T b > is not invariant under T. Fix y0 ∈ Rx0 ⋂ K and let x0 ∈< y0, T y0 > such that T x0 /∈< y0, T y0 > . Let G x0 = R x0 ⋂ K. Now we can easily prove that A = {λ ∈ R : (1 − λ)x0 + λT x0 ∈ K} is bounded. Let α = inf A and β = sup A. Let a = (1 − α)x o + αT x o and b = (1 − β)x o + βT x o . Therefore, there exists a sequence {α n } ∈ A such that {α n } converges to α. Hence (1 − α n )x o + α n T x o converges to (1 − α)x o + αT x o . Therefore it is easy to see that a ∈ G xo and b ∈ G xo . Hence G xo = {(1−λ)a + λb : 0 ≤ λ ≤ 1}. Now, define a map g : G x0 −→ G x0 by g(z) =    x0 if z ≤ x0, z if z ∈ (x0, T x0), T x0 if z ≥ T x0, . Since g and T are continuous, goT :< x0, T x0 >−→< x0, T x0 > is also continuous. Hence the map goT has a fixed point z0 ∈< x0, T x0 > . Case 1: z0 = x0 Then x0 = z0 = goT (z0) = goT (x0) = g(T x0) = T x0. Hence x0 = T x0, contradicting our assumption. Case 2: z0 ∈ (x0, T x0). If T z0 ≤ x0, then z0 = (goT )(z0) = g(T z0) = x0, contradicting z0 ∈ (x0, T x0]. If T z0 ∈< x0, T x0 >, then z0 = (goT )(z0) = g(T z0) = T z0, again contradicting our assumption, < z0, T z0 > is not invariant under T . Therefore, (2.1) T z0 ≥ T x0 A note on a fixed point theorem for ray oriented maps 187 That is , (2.2) z0 = goT (z0) = g(T z0) = T x0 Substituting (2.2) in (2.1) we get (2.3) T 2x0 ≥ T x0 Now let us construct B = {x ∈ R x0 ⋂ K : x < T x < T 2x}. Moreover it is bounded above and so it must have a least upper bound. There- fore let u be the least upper bound of B. Then there exists x n ∈ B such that x n → u. Suppose T u < u, then there exists a positive integer N such that for all n ≥ N , x n ∈< u, T u > . Then since T | < u, T u > is continuous, T x n → T u. Since x n < T x n , u ≤ T u, a contradiction. Therefore, u ≤ T u. Since T | < u, T u > is not invariant, by 2.3 we have T 2u ≥ T u. Therefore, u < T u < T 2u . Hence u ∈ B. But again, T 3u ≥ T 2u. Therefore, u < T u < T 2u < T 3u. Hence T u ∈ B, which is a contradiction. Therefore there exists a y0 ∈ Rx0 ⋂ K such that T | < y0, T y0 > is invariant. Hence T has a fixed point in < y0, T y0 > . � Corollary 2.2. Suppose T : [a, b] → [a, b] is a mapping where a, b ∈ ℜ. For each element x ∈ [a, b], T | < x, T x > is continuous. Then T has a fixed point in [a, b]. Remark 2.3. There exists a discontinuous mapping T satisfying the conditions of corollary 2.2. ( T : [0, 1] → [0, 1] by T (0) = 0 and T (x) = 1 for 0 < x ≤ 1). Acknowledgement The authors are indebted to Dr. S. Romaguera for his valuable comments on this paper. Moreover, the authors are grateful to Dr. P. Veeramani, Depart- ment of Mathematics, Indian Institute of Technology Madras, Chennai - 600 036 (India) and Dr. S. Somasundaram, Department of Mathematics, Manon- maniam Sundaranar University, Tirunelveli - 627 012 (India) for the knowledge they have acquired from the meaningful discussions with them. 188 A. Arockiasamy and A. Anthony Eldred References [1] A. F. Beardon, Contractions of the Real line, The Mathematical Association of America, Monthly113(June-July 2006), 557-558. [2] M. A. Khamsi and W. A. Kirk, An Introduction to Metric spaces and Fixed Point Theory, A Wiley-interscience Publication, 2001. Received February 2007 Accepted October 2007 A. Arockiasamy (arock − sj@rediffmail.com) Department of Mathematics, St.Xavier’s College, Palayamkottai, Tirunelveli- 627 002, India. Anthony Eldred Department of Mathematics,Indian Institute of Technology Madras, Chennai- 600 036, India.