ON A FIXED POINT PROBLEM OF REICH 1. Introduction Let X be a

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 124, Number 10, October 1996

ON A FIXED POINT PROBLEM OF REICH CHEN YU-QING (Communicated by Palle E. T. Jorgensen) Abstract. In this paper, we give an affirmative answer to a fixed point problem of S. Reich.

1. Introduction Let X be a metric space. CB(X) stands for the set of all non-empty closed bounded subsets of X. CB(X) is a metric space with the Hausdorff metric H. In [6], S. Reich presented the following Problem. Let (X, d) be a complete metric space. Suppose that F : X → CB(X) satisfies H(F (x), F (y)) ≤ K(d(x, y))d(x, y) for all x, y in X, x 6= y, where K : (0, +∞) → [0, 1) and limr→t+ sup K(r) < 1 for all 0 < t < +∞. Does F have a fixed point? In fact, this problem was raised by S. Reich in [4]. S. Reich [5] also gives an affirmative answer to this problem when F x is non-empty compact for x ∈ X. In this paper, we give an affirmative answer to this problem. We have the following results. Theorem 1. Let all the conditions of the above problem be satisfied. Then F has a fixed point if and only if there exists a closed subset Y ⊆ X, F x ∩ Y 6= ∅ for all x ∈ Y , such that for each closed subset Z ⊆ Y , if F x ∩ Z 6= ∅ for all x ∈ Z, then d(x, F (x) ∩ Z) = d(x, F (x)), ∀x ∈ Z. Remark. When F is single valued, let Y = X; then for each subset Z ⊆ X, such that F x ∈ Z for all x ∈ Z, we must have d(x, F (x)) = d(x, F (x) ∩ Z). Theorem 2. Let (X, d) be a complete metric space. F : X → CB(X) satisfies the following conditions. (1) If Y ⊆ X is a non-empty closed bounded subset, and F x ∩ Y 6= ∅ for all x ∈ Y , then d(x, F x) = d(x, F x ∩ Y ) for all x ∈ Y ; (2) H(F (x), F (y)) ≤ K(d(x, y))d(x, y), ∀x, y ∈ X, x 6= y, where K : (0, +∞) → [0, 1) and limr→t+ sup K(r) < 1 for all 0 < t < +∞. Then F has a fixed point. Received by the editors September 14, 1994 and, in revised form, January 10, 1995 and March 27, 1995. 1991 Mathematics Subject Classification. Primary 47H06, 47H10; Secondary 54H25. Key words and phrases. Fixed point, multivalued contraction, Hausdorff metric. c

1996 American Mathematical Society

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2. Proofs We first prove Theorem 2, then we can prove Theorem 1 similarly to the proof of Theorem 2. Proof of Theorem 2. Take tn ∈ (0, +∞), n = 1, 2, . . . , and t1 > t2 > · · · > tn → 0. Since limr→t+ sup K(r) < 1, there exist 0 ≤ kn < 1 and δn > 0, such that K(r) ≤ n kn , ∀r ∈ (tn , tn + δn ), n = 1, 2, . . . . Let ηn = min{ δ4n , n1 }, εn = tn + ηn ; then K(r) ≤ kn , ∀r ∈ [εn − η2n , εn + ηn ], n = 1, 2, . . . . It is easy to see εn → 0+ as n → ∞. Step 1. For ε1 > 0, we prove there exists x1 ∈ X, such that (2.1)

F x ∩ B1 6= ∅,

∀x ∈ B1 = {x|d(x, x1 ) ≤ ε1 }.

Suppose (2.1) is not true. Then for each x ∈ X, there exists x0 ∈ X, d(x, x0 ) ≤ ε1 , but d(x, y) > ε1 , Case (a). If d(x, x0 ) < ε1 −

η1 2 ,

∀y ∈ F x0 .

then

d(x, F x) ≥ d(x, F x0 ) − H(F x0 , F x) ≥ ε1 − K(d(x0 , x))d(x0 , x) η1 > ε1 − d(x, x0 ) > . 2 Case (b). If d(x, x0 ) ≥ ε1 −

η1 2 ,

then

d(x, F x) ≥ d(x, F x0 ) − H(F x0 , F x) ≥ ε1 − K(d(x0 , x))d(x0 , x) ≥ ε1 − k1 ε1 = (1 − k1 )ε1 . From Cases (a) and (b), we have nη o 1 d(x, F x) ≥ min , (1 − k1 )ε1 > 0, (2.2) 2 Now, fix x0 ∈ X and x1 ∈ F x0 . Since

∀x ∈ X.

d(x1 , F x1 ) ≤ H(F x0 , F x1 ) ≤ K(d(x0 , x1 ))d(x0 , x1 ) < d(x0 , x1 ), there exists x2 ∈ F x1 , such that d(x1 , F x1 ) −

1 ≤ d(x1 , x2 ) ≤ d(x0 , x1 ) 22

and d(x1 , x2 ) ≤ K(d(x0 , x1 ))d(x0 , x1 ) +

1 . 22

By induction, we get xn ∈ F xn−1 , n ≥ 3, such that d(xn−1 , F xn−1 ) −

1 ≤ d(xn−1 , xn ) ≤ d(xn−2 , xn−1 ) 2n

and d(xn−1 , xn ) ≤ K(d(xn−2 , xn−1 ))d(xn−2 , xn−1 ) +

1 . 2n

By the construction of {xn }, we know limn→∞ d(xn−1 , xn ) = S0 exists.

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ON A FIXED POINT PROBLEM OF REICH

Suppose S0 > 0. Then



lim d(xn−1 , xn ) ≤ lim

n→∞

n→∞

1 K(d(xn−2 , xn−1 ))d(xn−2 , xn−1 ) + n 2

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≤ lim+ K(r) lim d(xn−2 , xn−1 ). r→S0

n→∞

So S0 ≤ lim r→S + K(r)S0 < S0 , a contradiction. 0 Hence we have limn→∞ d(xn−1 , xn ) = 0. This implies that   1 lim d(xn−1 , F xn−1 ) ≤ lim d(xn−1 , xn ) + n = 0, n→∞ n→∞ 2 a contradiction to (2.2). So (2.1) is true. Step 2. For ε2 > 0, we prove there exists x2 ∈ B1 , such that (2.3)

F x ∩ B2 6= ∅,

∀x ∈ B2 = {x ∈ B1 |d(x, x2 ) ≤ ε2 }.

Suppose (2.3) is not true. For each x ∈ B1 , there exists y0 ∈ B1 , such that d(x, y0 ) ≤ ε2 , but d(x, y) > ε2 ,

∀y ∈ F y0 .

With the same argument of Cases (a) and (b) in Step 1, we get o nη 2 (2.4) d(x, F x) ≥ min , (1 − k2 )ε2 , ∀x ∈ B1 . 2 Now, fix x0 ∈ B1 , x1 ∈ F x0 ∩ B1 . By assumption (1), d(x1 , F x1 ∩ B1 ) = d(x1 , F x1 ) ≤ H(F x0 , F x1 ) ≤ K(d(x0 , x1 ))d(x0 , x1 ). So there exists x2 ∈ F x1 ∩ B1 , such that 1 d(x1 , F x1 ) − 2 ≤ d(x1 , x2 ) ≤ d(x0 , x1 ) 2 and d(x1 , x2 ) ≤ K(d(x0 , x1 ))d(x0 , x1 ) + 1/22 . Generally, we get xn ∈ F xn−1 ∩ B1 , n ≥ 3, such that 1 d(xn−1 , F xn−1 ) − n ≤ d(xn−1 , xn ) ≤ d(xn−2 , xn−1 ) 2 and 1 d(xn−1 , xn ) ≤ K(d(xn−2 , xn−1 ))d(xn−2 , xn−1 ) + n . 2 So limn→∞ d(xn−1 , xn ) exists and equals zero. We get limn→∞ d(xn−1 , F xn−1 ) = 0, a contradiction to (2.4). So (2.3) is true. Step 3. By induction, we get xn+1 ∈ Bn , such that (2.5)

F x ∩ Bn+1 6= ∅,

∀x ∈ Bn+1 = {x ∈ Bn |d(x, xn+1 ) ≤ εn+1 }, n ≥ 2.

It is obvious that B1 ⊇ B2 ⊇ B3 ⊇ · · · ⊇ Bn ⊇ . . . , and lim diam(Bn ) = 0. T So there exists only one point x ∈ n≥1 Bn , and x ∈ F x. This completes the proof. n→∞

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CHEN YU-QING

Proof of Theorem 1. Necessity: If F has a fixed point, let Y = {x ∈ X|x ∈ F x}. Then Y 6= ∅ is closed and it is the desired subset. Sufficiency: Suppose Y ⊆ X is non-empty closed, F x ∩ Y 6= ∅, ∀x ∈ Y , and for each closed subset Z ⊆ Y , if F x ∩ Z 6= ∅, ∀x ∈ Z, then d(x, F (x)) = d(x, F x ∩ Z), ∀x ∈ Z. Let {εn } be as in the proof of Theorem 2. Step 1. Take B1 = Y ; then d(x, F (x)) = d(x, F (x) ∩ B1 ), ∀x ∈ B1 . Step 2. With the same argument of Step 2 in the proof of Theorem 2, we get x2 ∈ B1 , such that F x ∩ B2 6= ∅, ∀x ∈ B2 = {x ∈ B1 |d(x, x2 ) ≤ ε2 }. By induction, we get xn+1 ∈ Bn , such that So

T

F x ∩ Bn+1 6= ∅,

n≥1

∀x ∈ Bn+1 = {x ∈ Bn |d(x, xn+1 ) ≤ εn+1 }, n ≥ 2.

Bn has only one point; it is the fixed point of F . Acknowledgement

I am grateful to the referee for his (or her) valuable suggestions. References 1. T. Hu, Fixed point theorems for multi-valued mappings, Canad. Math. Bull. 23 (1980), 193– 197. MR 82d:54052 2. H. M. Ko and Y. H. Tsai, Fixed point theorems with a localized property, Tamkang. Ja. Math. 8 (1977), 81–85. MR 57:7566 3. S. B. Nadler, Some results on multi-valued contraction mappings, Lect. Notes. Math., vol. 171, Springer-Verlag, 1970, pp. 64–69. MR 43:1148 4. S. Reich, Some fixed point problems, Atti Acad. Naz. Lincei. 57 (1974), 194–198. MR 53:1346 5. S. Reich, Fixed points of contractive functions, Boll. Un. Mat. Ital. 5 (1972), 26–42. MR 46:8206 6. S. Reich, Some problems and results in fixed point theory, Contemporary Math. 21 (1983), 179–187. MR 85e:47082 7. S. Reich, A fixed point theorem for locally contractive multi-valued functions, Rev. Roumaine. Math. Pures. Appl. 17 (1972), 569–572. MR 47:7721 Department of Mathematics, Sichuan University, Chengdu, People’s Republic of China

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