STATIONARY POINTS FOR SET-VALUED MAPPINGS ON TWO METRIC SPACES

We give stationary point theorems of set-valued mappings in complete and compact metric spaces. The results in this note generalize a few results due to Fisher. 2000 Mathematics Subject Classification. 54H25.


Introduction and preliminaries.
In [2,4], Fisher and Popa proved fixed point theorems for single-valued mappings on two metric spaces.The purpose of this note is to generalize these results from single-valued mappings into set-valued mappings.In this note, we show stationary point results of set-valued mappings in complete and compact metric spaces.
Let (X, d) and (Y , ρ) be complete metric spaces and B(X) and B(Y ) be two families of all nonempty bounded subsets of X and Y , respectively.The function δ(A, B) with A and B in B(X) is defined as follows: A sequence of sets in B(X), {A n : n = 1, 2,...} converges to the set A in B(X) if (i) each point a in A is the limit of some convergent sequence {a n ∈ A n : n = 1, 2,...}; (ii) for arbitrary > 0, there exists an integer N such that A n ⊂ A , for n > N, where A is the union of all open spheres with centers in A and radius .Let T be a set-valued mapping of The following Lemmas 1.1 and 1.2 were proved in [1,3], respectively.
Lemma 1.1.If {A n } and {B n } are sequences of bounded subsets of a complete metric space (X, d) which converge to the bounded subsets A and B, respectively, then the sequence {δ(A n ,B n )} converges to δ(A, B).Lemma 1.2.Let {A n } be a sequence of nonempty subsets of X and let x be a point of X such that lim n→∞ δ(A n ,x) = 0. Then the sequence {A n } converges to the set {x}.
From the above inequalities, we obtain immediately (2.9) which implies that δ(z, X n ) → 0 as n → ∞.Similarly, there exists w in Y such that By the continuity of T and Lemma 1.1, we have δ (w, T z) → 0 as n → ∞.From Lemma 1.2 it follows that T z = {w}.Note that (2.11) Letting n tend to infinity, we have which implies that ST z = {z} = Sw.Similarly, we can show that w is a stationary point of T S. This completes the proof of the theorem.
Theorem 2.2.Let (X, d) be a complete metric space, and let S and T be continuous mappings of X into B(X) and map bounded set into bounded set.If S and T satisfy the inequalities Proof.Let x be an arbitrary point in X. Define a sequence of sets {X n } by T (ST ) n−1 x = X 2n−1 , (ST ) n x = X 2n for n ≥ 1 and X 0 = {x}.Now suppose that {δ(X n )} is unbounded.Then the real-valued sequence {a n } is unbounded, where a 2n−1 = δ(X 2n−1 ,X 3 ), a 2n = δ(X 2n ,X 2 ) for n ≥ 1 and so there exists an integer k such that (2.16) Suppose that k is even.Put k = 2n.From (2.15) and (2.16) we have (2.17)
Let M = sup{δ(X r ,X s ) : r , s = 0, 1, 2,...} < ∞.For arbitrary > 0, choose a positive integer N such that c N M < .Thus for m, n greater than 2N with m and n both even or both odd, from (2.13) and (2.14) we have (2.24) That is, δ(z, X 2n ) → 0 as n → ∞.Similarly {x 2n+1 } converges to some point w in X and δ(w, X 2n+1 ) → 0 as n → ∞.Since δ(w, T X 2n ) = δ(w, X 2n+1 ), by the continuity of T and Lemma 1.1, we have δ(w, T z) → 0 as n → ∞.From Lemma 1.2 it follows that T z = {w}.In view of (2.13), we obtain that which implies that as n → ∞.Since c < 1, δ(ST z, z) = 0. Therefore ST z = {z} = Sw and T Sw = T z = {w}.Now suppose that z = w and that z is the second common stationary point of S and T .Using (2.1) Proof.Suppose that the right-hand sides of inequalities (2.28) and (2.29) are positive for all distinct x, y in X and distinct x ,y in Y .Define the real-valued function f (x,y) in X × X as follows: f (x,y) = δ(ST x,ST y) max δ(x, y), δ(x, ST x), δ(y, ST y), δ (T x, T y) . (2.30) Since S and T are continuous, f is continuous and achieves the maximum value s on the compact metric space X × X. Inequality (2.28) implies s <  Remark 2.7.Theorem 4 of Fisher [2] is a particular case of our Theorem 2.6 if the set-valued mappings in Theorem 2.6 are replaced by single-valued mappings.

( 2 .Remark 2 . 3 .Remark 2 . 4 .
27)So z = z and this completes the proof of the theorem.If we use single-valued mappings in place of set-valued mappings in Theorems 2.1 and 2.2, Theorems 2 and 3 of Fisher[2] can be attained.The following example demonstrates that the continuity of S and T in Theorems 2.1 and 2.2 is necessary.Example 2.5.Let X = {0} ∪ {1/n : n ≥ 1} = Y with the usual metric.Define mappings S, T by T 0 = {1}, T (1/n) = {1/2n} for n ≥ 1 and S = T .It is easy to prove that all the conditions of Theorems 2.1 and 2.2 are satisfied except that the mappings S and T are continuous.But ST and T S have no stationary points.Now we give the following theorem for the compact metric spaces.Theorem 2.6.Let (X, d) and (Y , ρ) be compact metric spaces.If T is a continuous mapping of X into B(Y ) and S is a continuous mapping of Y into B(X) satisfying the following inequalities:δ(ST x,ST y) < max δ(x, y), δ(x, ST x), δ(y, ST y), δ (T x, T y) ,(2.28) δ T Sx ,T Sy < max δ x ,y ,δ x ,T Sx ,δ y ,T Sy ,δ Sx ,Sy , (2.29) for all distinct x, y in X and distinct x ,y in Y , then ST has a stationary point z and T S has a stationary point w.Further T z = {w} and Sw = {z}.
33) which implies {z} = {z } = ST z and T z = T z , a singleton, {w}, say.Therefore we have ST z = sw = {z},T Sw = T z = {w}.If there exist w, w in Y such that max δ w, w ,δ (w, T Sw), δ w ,T Sw ,δ Sw, Sw = 0. (2.34)Similarly, we also have ST z = Sw = {z} and T Sw = T z = {w}.This completes the proof of the theorem.

2. Stationary point results. Now
we prove the following theorem for set-valued mappings.
Let x be an arbitrary point in X. Define sequences {x n } and {y n } in B(X) and B(Y ), respectively, by choosing a point x n x, y in X.It is obvious that (2.31) is also true for x = y.Similarly, there exists t < 1 such that δ T Sx ,T Sy ≤ t max δ x ,y ,δ x ,T Sx ,δ y ,T Sy ,δ Sx ,Sy (2.32) for all x ,y in Y .So Theorem 2.6 follows immediately from Theorem 2.1.Now suppose that there exist z, z in X such that max δ z, z , δ(z, ST z), δ z ,ST z ,δ T z,T z = 0, (2.