Fixed Point Theory and Applications
Volume 2010 (2010), Article ID 390183, 7 pages
doi:10.1155/2010/390183
Research Article

Fixed Point Theorems for Set-Valued Contraction Type Maps in Metric Spaces

A. Amini-Harandi1 and D. O'Regan2

1Department of Mathematics, University of Shahrekord, Shahrekord, 88186-34141, Iran
2Department of Mathematics, National University of Ireland, Galway, Ireland

Received 13 August 2009; Revised 15 October 2009; Accepted 13 January 2010

Academic Editor: Mohamed A. Khamsi

Copyright © 2010 A. Amini-Harandi and D. O'Regan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We first give some fixed point results for set-valued self-map contractions in complete metric spaces. Then we derive a fixed point theorem for nonself set-valued contractions which are metrically inward. Our results generalize many well-known results in the literature.

1. Introduction and Preliminaries

Let ( 𝑋 , 𝑑 ) be a metric space and let CB ( 𝑋 ) denote the class of all nonempty bounded closed subsets of 𝑋 . Let 𝐻 be the Hausdorff metric with respect to 𝑑 , that is,

𝐻 ( 𝐴 , 𝐵 ) = m a x s u p 𝑢 ∈ 𝐴 𝑑 ( 𝑢 , 𝐵 ) , s u p 𝑣 ∈ 𝐵 𝑑 ( 𝑣 , 𝐴 ) ( 1 . 1 ) for every 𝐴 , 𝐵 ∈   CB ( 𝑋 ) , where 𝑑 ( 𝑢 , 𝐵 ) = i n f { 𝑑 ( 𝑢 , 𝑦 ) ∶ 𝑦 ∈ 𝐵 } . In 1969, Nadler [1] extended the Banach contraction principle [2] to set-valued mappings.

Theorem 1.1 (Nadler [1]). Let ( 𝑋 , 𝑑 ) be a complete metric space and let 𝑇 ∶ 𝑋 → CB ( 𝑋 ) be a set-valued map. Assume that there exists 𝑟 ∈ [ 0 , 1 ) such that 𝐻 ( 𝑇 𝑥 , 𝑇 𝑦 ) ≤ 𝑟 𝑑 ( 𝑥 , 𝑦 ) ( 1 . 2 ) for all 𝑥 , 𝑦 ∈ 𝑋 . Then 𝑇 has a fixed point.

Mizoguchi and Takahashi [3] proved the following generalization of Theorem 1.1.

Corollary (Mizoguchi and Takahashi [3]). Let ( 𝑋 , 𝑑 ) be a complete metric space and let 𝑇 ∶ 𝑋 → CB ( 𝑋 ) be a set-valued map satisfying 𝐻 ( 𝑇 𝑥 , 𝑇 𝑦 ) ≤ 𝛼 ( 𝑑 ( 𝑥 , 𝑦 ) ) 𝑑 ( 𝑥 , 𝑦 ) , f o r e a c h 𝑥 , 𝑦 ∈ 𝑋 , ( 1 . 3 ) where 𝛼 ∶ [ 0 , ∞ ) → [ 0 , 1 ) satisfies l i m s u p 𝑠 → 𝑡 + 𝛼 ( 𝑠 ) < 1 for each 𝑡 ∈ [ 0 , ∞ ) . Then 𝑇 has a fixed point.

Also, Reich [4] has proved that if for each 𝑥 ∈ 𝑋 , 𝑇 𝑥 is nonempty and compact, then the above result holds under the weaker condition l i m s u p 𝑠 → 𝑡 + 𝛼 ( 𝑠 ) < 1 for each 𝑡 > 0 . To set up our results in the next section, we introduce some definitions and facts.

Definition. Throughout the paper, let Ψ be the family of all functions 𝜓 ∶ [ 0 , ∞ ) → [ 0 , ∞ ) satisfying the following conditions: (a) 𝜓 ( 𝑠 ) = 0 ⇔ 𝑠 = 0 ; (b) 𝜓 is lower semicontinuous and nondecreasing; (c) l i m s u p 𝑠 → 0 + ( 𝑠 / 𝜓 ( 𝑠 ) ) < ∞ .

Theorem 1.4 (Bae [5]). Let ( 𝑀 , 𝜌 ) be a complete metric space, 𝜙 ∶ 𝑀 → [ 0 , ∞ ) a lower semicontinuous function, and 𝜑 ∶ [ 0 , ∞ ) → [ 0 , ∞ ) a lower semicontinuous function such that 𝜑 ( 𝑡 ) > 0 for 𝑡 > 0 and l i m s u p 𝑠 → 0 + 𝑠 𝜑 ( 𝑠 ) < ∞ . ( 1 . 4 ) Let 𝑔 ∶ 𝑀 → 𝑀 be a map such that for any 𝑥 ∈ 𝑀 , 𝜌 ( 𝑥 , 𝑔 𝑥 ) ≤ 𝜙 ( 𝑥 ) and 𝜑 ( 𝜌 ( 𝑥 , 𝑔 𝑥 ) ) ≤ 𝜙 ( 𝑥 ) − 𝜙 ( 𝑔 ( 𝑥 ) ) ( 1 . 5 ) hold. Then 𝑔 has a fixed point in 𝑀 .

Definition. Let ( 𝑋 , 𝑑 ) be a complete metric space and 𝐷 be a nonempty closed subset of 𝑋 . (i)Set M I 𝐷 𝑑 ( 𝑥 ) = { 𝑧 ∈ 𝑋 ∶ 𝑧 = 𝑥 o r t h e r e e x i t s 𝑦 ∈ 𝐷 s a t i s f y i n g 𝑦 ≠ 𝑥 , ( 𝑥 , 𝑧 ) = 𝑑 ( 𝑥 , 𝑦 ) + 𝑑 ( 𝑦 , 𝑧 ) } . ( 1 . 6 ) Then M I 𝐷 ( 𝑥 ) is called the metrically inward set of 𝐷 at 𝑥 (see [5]); (ii)Let 𝑇 ∶ 𝐷 → C B ( 𝑋 ) be a set-valued map. 𝑇 is said to be metricaly inward, if for each 𝑥 ∈ 𝐷 , 𝑇 𝑥 ⊆ M I 𝐷 ( 𝑥 ) . ( 1 . 7 )

In Section 2 we generalize Corollary 1.2 and Theorem 1.4.

2. Extension of Mizoguchi-Takahashi’s Theorem

In the first result of this section, we use the technique in [6] to extend Corollary 1.2.

Theorem 2.1. Let ( 𝑋 , 𝑑 ) be a complete metric space and let 𝑇 ∶ 𝑋 → CB ( 𝑋 ) be a set-valued map satisfying 𝜓 ( 𝐻 ( 𝑇 𝑥 , 𝑇 𝑦 ) ) ≤ 𝛼 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) ) 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) , f o r e a c h 𝑥 , 𝑦 ∈ 𝑋 , ( 2 . 1 ) where 𝛼 ∶ [ 0 , ∞ ) → [ 0 , 1 ) satisfies l i m s u p 𝑠 → 𝑡 + 𝛼 ( 𝑠 ) < 1 for each 𝑡 ∈ [ 0 , ∞ ) and 𝜓 ∈ Ψ . Then 𝑇 has a fixed point.

Proof. Define a function 𝛽 ∶ [ 0 , ∞ ) → [ 0 , 1 ) by 𝛽 ( 𝑡 ) = ( 𝛼 ( 𝑡 ) + 1 ) / 2 . Then 𝛼 ( 𝑡 ) < 𝛽 ( 𝑡 ) and l i m s u p 𝑠 → 𝑡 + 𝛽 ( 𝑠 ) < 1 for all 𝑡 ∈ [ 0 , ∞ ) . Since 𝜓 is nondecreasing, then from (1.3), for each 𝑥 ≠ 𝑦 , we have m a x s u p 𝑢 ∈ 𝑇 𝑥 𝜓 ( 𝑑 ( 𝑢 , 𝑇 𝑦 ) ) , s u p 𝑣 ∈ 𝑇 𝑦 𝜓 𝜓 ( 𝑑 ( 𝑣 , 𝑇 𝑥 ) ) = m a x s u p 𝑢 ∈ 𝑇 𝑥 𝑑 ( 𝑢 , 𝑇 𝑦 ) , 𝜓 s u p 𝑣 ∈ 𝑇 𝑦 𝑑 ( 𝑣 , 𝑇 𝑥 ) = 𝜓 ( 𝐻 ( 𝑇 𝑥 , 𝑇 𝑦 ) ) < 𝛽 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) ) 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) . ( 2 . 2 ) Hence for each 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑇 𝑥 , there exists an element 𝑧 ∈ 𝑇 𝑦 such that 𝜓 ( 𝑑 ( 𝑦 , 𝑧 ) ) ≤ 𝛽 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) ) 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) . Thus we can define a sequence { 𝑥 𝑛 } in 𝑋 satisfying 𝑥 𝑛 + 1 ∈ 𝑇 𝑥 𝑛 𝑑 𝑥 , 𝜓 𝑛 + 1 , 𝑥 𝑛 + 2 𝜓 𝑑 𝑥 ≤ 𝛽 𝑛 , 𝑥 𝑛 + 1 𝜓 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 , ( 2 . 3 ) for each 𝑛 ∈ ℕ . Let us show that { 𝑥 𝑛 } is convergent. Since 𝛽 ( 𝑡 ) < 1 for each 𝑡 ∈ [ 0 , ∞ ) , then { 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ) } is a nonincreasing sequence of non-negative numbers and so is convergent to a real number, say 𝑟 0 . Since l i m s u p 𝑠 → 𝑟 0 + 𝛽 ( 𝑠 ) < 1 and 𝛽 ( 𝑟 0 ) < 1 , there exist 𝑟 ∈ [ 0 , 1 ) and 𝜖 > 0 such that 𝛽 ( 𝑠 ) ≤ 𝑟 for all 𝑠 ∈ [ 𝑟 0 , 𝑟 0 + 𝜖 ] . We can take 𝑛 0 ∈ ℕ such that 𝑟 0 ≤ 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ) ≤ 𝑟 0 + 𝜖 for all 𝑛 ∈ ℕ with 𝑛 ≥ 𝑛 0 . Since 𝜓 𝑑 𝑥 𝑛 + 1 , 𝑥 𝑛 + 2 𝜓 𝑑 𝑥 ≤ 𝛽 𝑛 , 𝑥 𝑛 + 1 𝜓 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 𝑑 𝑥 ≤ 𝑟 𝜓 𝑛 , 𝑥 𝑛 + 1 ( 2 . 4 ) for all 𝑛 ≥ 𝑛 0 , then we have 𝑟 0 ≤ 𝑟 𝑟 0 and so 𝑟 0 = 0 (note that 𝑟 < 1 ). If 𝑑 ( 𝑥 𝑚 , 𝑥 𝑚 + 1 ) = 0 for some 𝑚 ∈ ℕ , then 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) = 0 for each 𝑛 ≥ 𝑚 (note that { 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ) } is nonincreasing). Thus { 𝑥 𝑛 } is eventually constant, so we have a fixed point of 𝑇 (note that 𝑥 𝑛 + 1 ∈ 𝑇 𝑥 𝑛 ). Now, we assume that 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ≠ 0 for each 𝑛 ∈ ℕ . Since { 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ) } is decreasing and 𝜓 is nondecreasing, then the nonnegative sequence 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) converges to some nonnegative real number 𝜏 . Since 𝜓 is nondecreasing and 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) is nonincreasing, then 𝜓 ( 𝜏 ) ≤ 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ) for each 𝑛 ∈ ℕ . Thus 𝜓 ( 𝜏 ) ≤ l i m 𝑛 → ∞ 𝜓 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 = 𝑟 0 = 0 . ( 2 . 5 ) Thus 𝜏 = 0 (note that 𝜓 ( 𝜏 ) = 0 implies 𝜏 = 0 ). Also we have (note 𝜓 ( 𝑑 ( 𝑥 𝑛 + 1 , 𝑥 𝑛 + 2 ) ) ≤ 𝑟 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ) for 𝑛 ≥ 𝑛 0 ) ∞ 1 𝜓 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 ≤ 𝑛 0 1 𝜓 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 + ∞ 1 𝑟 𝑛 𝜓 𝑑 𝑥 𝑛 0 , 𝑥 𝑛 0 + 1 < ∞ . ( 2 . 6 ) Since l i m s u p 𝑛 → ∞ 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 𝜓 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 ≤ l i m s u p 𝑠 → 0 + 𝑠 𝜓 ( 𝑠 ) < ∞ , ( 2 . 7 ) then ∑ ∞ 1 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) < ∞ . Hence { 𝑥 𝑛 } is a Cauchy sequence. Since 𝑋 is complete, { 𝑥 𝑛 } converges to some point 𝑥 0 ∈ 𝑋 . Since 𝜓 is lower semicontinuous and nondecreasing (recall also from above that l i m 𝑛 → ∞ 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 𝑛 + 1 ) ) = 0 ), then 𝜓 𝑑 𝑥 0 , 𝑇 𝑥 0 ≤ l i m i n f 𝑛 → ∞ 𝜓 𝑑 𝑥 𝑛 + 1 , 𝑇 𝑥 0 ≤ l i m i n f 𝑛 → ∞ 𝜓 𝐻 𝑇 𝑥 𝑛 , 𝑇 𝑥 0 ≤ l i m i n f 𝑛 → ∞ 𝛽 𝜓 𝑑 𝑥 𝑛 , 𝑥 0 𝜓 𝑑 𝑥 𝑛 , 𝑥 0 ≤ l i m i n f 𝑛 → ∞ 𝜓 𝑑 𝑥 𝑛 , 𝑥 0 = l i m 𝑠 → 0 + 𝜓 ( 𝑠 ) = l i m 𝑛 → ∞ 𝜓 𝑑 𝑥 𝑛 , 𝑥 𝑛 + 1 = 0 , ( 2 . 8 ) and this with 𝑇 𝑥 0 closed and (a) of Definition 1.3 implies 𝑥 0 ∈ 𝑇 𝑥 0 .

Corollary. Let ( 𝑋 , 𝑑 ) be a complete metric space and let 𝑇 ∶ 𝑋 → CB ( 𝑋 ) be a set-valued map satisfying 𝜓 ( 𝐻 ( 𝑇 𝑥 , 𝑇 𝑦 ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) − 𝜑 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) ) , f o r e a c h 𝑥 , 𝑦 ∈ 𝑋 , ( 2 . 9 ) where 𝜓 ∈ Ψ and 𝜑 ∶ [ 0 , ∞ ) → [ 0 , ∞ ) satisfying l i m i n f 𝑠 → 𝑡 + ( 𝜑 ( 𝑠 ) / 𝜓 ( 𝑠 ) ) > 0 for each 𝑡 ∈ [ 0 , ∞ ) . Then 𝑇 has a fixed point.

Proof. Let 𝛼 ( 𝑠 ) = 1 − 𝜑 ( 𝑠 ) / 𝜓 ( 𝑠 ) and apply Theorem 2.1.

In the following, we present a fixed point theorem for nonself set-valued contraction type maps which are metrically inward.

Theorem. Let 𝐷 be a nonempty closed subset of a complete metric space ( 𝑋 , 𝑑 ) and 𝑇 ∶ 𝐷 → CB ( 𝑋 ) be a set-valued map satisfying 𝜓 ( 𝐻 ( 𝑇 𝑥 , 𝑇 𝑦 ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) − 𝜑 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑦 ) ) ) , f o r e a c h 𝑥 , 𝑦 ∈ 𝑋 , ( 2 . 1 0 ) for which 𝜓 ∈ Ψ is continuous and 𝜓 ( 𝑟 − 𝑠 ) + 𝜓 ( 𝑠 + 𝑡 ) ≤ 𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) , f o r e a c h 0 ≤ 𝑠 ≤ 𝑟 ≤ 𝑠 + 𝑡 . ( 2 . 1 1 ) Assume that 𝜑 ∶ [ 0 , ∞ ) → [ 0 , ∞ ) is a lower semicontinuous function satisfying l i m i n f 𝑠 → 0 + ( 𝜑 ( 𝑠 ) / 𝜓 ( 𝑠 ) ) > 0 and 𝜑 ( 𝑠 ) > 0 for 𝑠 > 0 . Suppose that 𝑇 is metrically inward on 𝐷 . Then 𝑇 has a fixed point in 𝐷 .

Proof. We first show that l i m s u p 𝑠 → 0 + ( 𝑠 / 𝜑 ( 𝑠 ) ) < ∞ . On the contrary, we assume that there exists a sequence 𝑠 𝑛 → 0 + for which l i m s u p 𝑛 → ∞ 𝑠 𝑛 𝑠 𝜑 𝑛 = l i m s u p 𝑛 → ∞ 𝑠 𝑛 𝑠 / 𝜓 𝑛 𝑠 𝜑 𝑛 𝑠 / 𝜓 𝑛 = ∞ . ( 2 . 1 2 ) Since l i m i n f 𝑛 → ∞ ( 𝜑 ( 𝑠 𝑛 ) / 𝜓 ( 𝑠 𝑛 ) ) > 0 , then we get l i m s u p 𝑛 → ∞ ( 𝑠 𝑛 / 𝜓 ( 𝑠 𝑛 ) ) = ∞ , which contradicts our assumption on 𝜓 . Let 𝑀 = { ( 𝑥 , 𝑦 ) ∶ 𝑥 ∈ 𝑋 , 𝑦 ∈ 𝑇 𝑥 } be the graph of 𝑇 . Let 𝜌 ∶ 𝑀 × 𝑀 → [ 0 , ∞ ) be given by 𝜌 ( ( 𝑥 , 𝑧 ) , ( 𝑢 , 𝑣 ) ) = m a x { 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) , 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) ) } . ( 2 . 1 3 ) We show that ( 𝑀 , 𝜌 ) is a complete metric space. First note that since 𝜓 ( 𝑠 ) = 0 ⇔ 𝑠 = 0 then 𝜌 ( ( 𝑥 , 𝑧 ) , ( 𝑢 , 𝑣 ) ) = 0 ⇔ ( 𝑥 , 𝑧 ) = ( 𝑢 , 𝑣 ) . Clearly, 𝜌 ( ( 𝑥 , 𝑧 ) , ( 𝑢 , 𝑣 ) ) = 𝜌 ( ( 𝑢 , 𝑣 ) , ( 𝑥 , 𝑧 ) ) . Now we show the triangle inequality. From (2.11), we have 𝜓 ( 𝑟 + 𝑡 ) ≤ 𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) , ∀ 𝑟 , 𝑡 ≥ 0 . Hence, 𝜌 ( ( 𝑥 , 𝑧 ) , ( 𝑟 , 𝑠 ) ) + 𝜌 ( ( 𝑟 , 𝑠 ) , ( 𝑢 , 𝑣 ) ) = m a x { 𝜓 ( 𝑑 ( 𝑥 , 𝑟 ) ) , 𝜓 ( 𝑑 ( 𝑧 , 𝑠 ) ) } + m a x { 𝜓 ( 𝑑 ( 𝑟 , 𝑢 ) ) , 𝜓 ( 𝑑 ( 𝑠 , 𝑣 ) ) } ≥ m a x { 𝜓 ( 𝑑 ( 𝑥 , 𝑟 ) ) + 𝜓 ( 𝑑 ( 𝑟 , 𝑢 ) ) , 𝜓 ( 𝑑 ( 𝑧 , 𝑠 ) ) + 𝜓 ( 𝑑 ( 𝑠 , 𝑣 ) ) } ≥ m a x { 𝜓 ( 𝑑 ( 𝑥 , 𝑟 ) + 𝑑 ( 𝑟 , 𝑢 ) ) , 𝜓 ( 𝑑 ( 𝑧 , 𝑠 ) + 𝑑 ( 𝑠 , 𝑣 ) ) } ≥ m a x { 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) , 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) ) } = 𝜌 ( ( 𝑥 , 𝑧 ) , ( 𝑢 , 𝑣 ) ) . ( 2 . 1 4 ) To prove the completeness of 𝜌 , we first need to show that 𝑇 is Hausdorff continuous. To prove this, let ( 𝑥 𝑛 ) be a sequence in 𝐷 such that 𝑥 𝑛 → 𝑥 ∈ 𝐷 . Since 𝜓 is continuous at 0 , then l i m 𝑛 → ∞ 𝜓 ( 𝑑 ( 𝑥 𝑛 , 𝑥 ) ) = 𝜓 ( 0 ) = 0 . Hence from (2.10), we get l i m 𝑛 → ∞ 𝜓 ( 𝐻 ( 𝑇 𝑥 𝑛 , 𝑇 𝑥 ) ) = 0 . We claim that l i m 𝑛 → ∞ 𝐻 ( 𝑇 𝑥 𝑛 , 𝑇 𝑥 ) = 0 (and then we are finished). On the contrary, assume that there exist 𝜖 > 0 and a subsequence 𝑥 𝑛 𝑘 such that 𝐻 ( 𝑇 𝑥 𝑛 𝑘 , 𝑇 𝑥 ) ≥ 𝜖 , 𝑘 =1,2,3, …   . Since 𝜓 is nondecreasing, then 𝜓 ( 𝐻 ( 𝑇 𝑥 𝑛 𝑘 , 𝑇 𝑥 ) ) ≥ 𝜓 ( 𝜖 ) > 0 , a contradiction. Now, let ( 𝑥 𝑛 , 𝑧 𝑛 ) be a Cauchy sequence in 𝑀 with respect to 𝜌 . Then { 𝑥 𝑛 } and { 𝑧 𝑛 } are Cauchy sequences in the complete metric space ( 𝑋 , 𝑑 ) . Then there exist 𝑥 , 𝑧 ∈ 𝑋 such that 𝑑 ( 𝑥 𝑛 , 𝑥 ) → 0 and 𝑑 ( 𝑧 𝑛 , 𝑧 ) → 0 . Since 𝑧 𝑛 ∈ 𝑇 𝑥 𝑛 and 𝑇 is Hausdorff continuous, then 𝑧 ∈ 𝑇 𝑥 . Thus ( 𝑥 , 𝑧 ) ∈ 𝑀 and 𝜌 ( ( 𝑥 𝑛 , 𝑧 𝑛 ) , ( 𝑥 , 𝑧 ) ) → 0 . Therefore, ( 𝑀 , 𝜌 ) is a complete metric space. Suppose that 𝑇 has no fixed point. Then for each ( 𝑥 , 𝑧 ) ∈ 𝑀 , we have 𝑥 ≠ 𝑧 . Since 𝑧 ∈ 𝑇 𝑥 ⊆ M I 𝐷 ( 𝑥 ) , we can choose 𝑢 ∈ 𝐷 such that 𝑢 ≠ 𝑥 and 𝑑 ( 𝑥 , 𝑧 ) = 𝑑 ( 𝑥 , 𝑢 ) + 𝑑 ( 𝑢 , 𝑧 ) . ( 2 . 1 5 ) Since 𝑇 satisfies (2.10) and 𝜓 is continuous, then we can choose 𝑣 ∈ 𝑇 𝑢 such that 1 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) − 2 𝜑 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) ) . ( 2 . 1 6 ) Let 𝜑 ( 𝑡 ) = 𝜑 ( 𝑡 ) / 2 . Then by combining (2.15) and (2.16), we get 𝜑 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) − 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) ) = 𝜓 ( 𝑑 ( 𝑥 , 𝑧 ) − 𝑑 ( 𝑢 , 𝑧 ) ) − 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) ) . ( 2 . 1 7 ) From (2.11), we have (note that 𝜓 is nondecreasing) 𝜓 ( 𝑑 ( 𝑥 , 𝑧 ) − 𝑑 ( 𝑢 , 𝑧 ) ) − 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑧 ) ) − 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) + 𝑑 ( 𝑢 , 𝑧 ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑧 ) ) − 𝜓 ( 𝑑 ( 𝑢 , 𝑣 ) ) . ( 2 . 1 8 ) Thus (2.17) and (2.18) yield 𝜑 ( 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑧 ) ) − 𝜓 ( 𝑑 ( 𝑢 , 𝑣 ) ) . ( 2 . 1 9 ) Since 𝜌 ( ( 𝑥 , 𝑧 ) , ( 𝑢 , 𝑣 ) ) = m a x { 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) , 𝜓 ( 𝑑 ( 𝑧 , 𝑣 ) ) } = 𝜓 ( 𝑑 ( 𝑥 , 𝑢 ) ) ≤ 𝜓 ( 𝑑 ( 𝑥 , 𝑧 ) ) ≡ 𝜙 ( 𝑥 , 𝑧 ) , by defining 𝑔 ∶ 𝑀 → 𝑀 by 𝑔 ( 𝑥 , 𝑧 ) = ( 𝑢 , 𝑣 ) , from Theorem 1.4, 𝑔 must have a fixed point, say ( 𝑥 0 , 𝑧 0 ) . Then ( 𝑥 0 , 𝑧 0 ) = 𝑔 ( 𝑥 0 , 𝑧 0 ) = ( 𝑢 0 , 𝑣 0 ) . Hence 𝑥 0 = 𝑢 0 . This is a contradiction. Therefore, 𝑇 has a fixed point.

Remark. Note that Theorem 2.3 does not follow from Theorem 3.3 of Bae [5] by replacing the metric 𝑑 by 𝜓 ( 𝑑 ) . In Theorem 2.3, we assume 𝑇 is metrically inward with respect to 𝑑 but to apply Theorem 3.3 of [5] with 𝜓 ( 𝑑 ) rather than 𝑑 , we need 𝑇 to be metrically inward with respect to 𝜓 ( 𝑑 ) .

Letting 𝜓 ( 𝑠 ) = 𝑠 for each 𝑠 ∈ [ 0 , ∞ ) , we get the following corollary due to Bae [5].

Corollary. Let 𝐷 be a nonempty closed subset of a complete metric space ( 𝑋 , 𝑑 ) and 𝑇 ∶ 𝐷 → CB ( 𝑋 ) be a set-valued map satisfying 𝐻 ( 𝑇 𝑥 , 𝑇 𝑦 ) ≤ 𝑑 ( 𝑥 , 𝑦 ) − 𝜑 ( 𝑑 ( 𝑥 , 𝑦 ) ) , f o r e a c h 𝑥 , 𝑦 ∈ 𝑋 , ( 2 . 2 0 ) for which 𝜑 ∶ [ 0 , ∞ ) → [ 0 , ∞ ) is a lower semicontinuous function satisfying l i m i n f 𝑠 → 0 + ( 𝜑 ( 𝑠 ) / 𝑠 ) > 0 . Suppose that 𝑇 is metrically inward on 𝐷 . Then 𝑇 has a fixed point in 𝐷 .

Example 2.6. Let 𝜓 ∶ [ 0 , ∞ ) → [ 0 , ∞ )  be a differentiable function with 𝜓 ( 0 ) = 0   such that   𝜓   is  positive and decreasing in   ( 0 , ∞ )   and   l i m 𝑠 → 0 + 𝜓 ( 𝑠 ) = ∞ .  Now we show that   ( 𝜓 )   satisfies all the conditions of Theorem 2.3. Obviously, 𝜓   is continuous and increasing. Since   l i m 𝑠 → 0 + ( 1 / 𝜓 ( 𝑠 ) ) = 0 ,  then by L’Hopital’s rule   l i m 𝑠 → 0 + ( 𝑠 / 𝜓 ( 𝑠 ) ) = 0 . Thus l i m s u p 𝑠 → 0 + ( 𝑠 / 𝜓 ( 𝑠 ) ) < ∞ .  Now we prove that for each   0 ≤ 𝑡 ≤ 𝑟 ,   𝜓 ( 𝑟 + 𝑡 ) ≤ 𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) . To show this let   ℎ ( 𝑡 ) = 𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) − 𝜓 ( 𝑟 + 𝑡 ) for 0 ≤ 𝑡 ≤ 𝑟 . Then   ℎ ( 𝑡 ) = 𝜓 ( 𝑡 ) − 𝜓 ( 𝑟 + 𝑡 ) > 0 . Since   ℎ ( 0 ) = 0   and ℎ is increasing, we get   ℎ ( 𝑡 ) ≥ 0   for each   0 ≤ 𝑡 ≤ 𝑟   and we are done. Finally, we show that for each   0 ≤ 𝑠 ≤ 𝑟 ≤ 𝑠 + 𝑡 ,  we have   𝜓 ( 𝑟 − 𝑠 ) + 𝜓 ( 𝑠 + 𝑡 ) ≤ 𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) .  Let   𝑘 ( 𝑠 ) = 𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) − 𝜓 ( 𝑟 − 𝑠 ) + 𝜓 ( 𝑠 + 𝑡 )   for   0 ≤ 𝑠 ≤ 𝑟 .  Then   𝑘 ( 𝑠 ) = 𝜓 ( 𝑟 − 𝑠 ) − 𝜓 ( 𝑠 + 𝑡 ) .  If   𝑟 ≤ 𝑡 ,  then   𝑘 ( 𝑠 ) > 0 .  Since   𝑘 ( 0 ) = 0 ,  we obtain   𝑘 ( 𝑠 ) ≥ 0   for each   0 ≤ 𝑠 ≤ 𝑟   and  we  are finished. In the case,   𝑟 > 𝑡 , 𝑘 ( 𝑠 ) = 0   if and only  if   𝑠 = ( 𝑟 − 𝑡 ) / 2 .  Since   𝑘 ( 𝑠 ) > 0   for   0 < 𝑠 < ( 𝑟 − 𝑡 ) / 2   and   𝑘 ( 𝑠 ) < 0   for   ( 𝑟 − 𝑡 ) / 2 < 𝑠 ≤ 𝑡 ,  then   i n f 0 ≤ 𝑠 ≤ 𝑟 𝑘 ( 𝑠 ) = m i n ( 𝑘 ( 0 ) , 𝑘 ( 𝑟 ) ) = m i n ( 0 , 𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) − 𝜓 ( 𝑟 + 𝑡 ) ) = 0 ,  and we  are finished (note that we proved above that   𝜓 ( 𝑟 ) + 𝜓 ( 𝑡 ) − 𝜓 ( 𝑟 + 𝑡 ) ≥ 0 ).

Acknowledgments

The authors would like to thank the referees for careful reading and giving valuable comments. This work was supported in part by the Shahrekord University. The first author would like to thank this support.

References

  1. S. B. Nadler Jr., “Multi-valued contraction mappings,” Pacific Journal of Mathematics, vol. 30, pp. 475–488, 1969.
  2. S. Banach, “Sur les opération dans les ensembles et leur application aux équations intégrales,” Fundamenta Mathematicae, vol. 3, pp. 133–181, 1922.
  3. N. Mizoguchi and W. Takahashi, “Fixed point theorems for multivalued mappings on complete metric spaces,” Journal of Mathematical Analysis and Applications, vol. 141, no. 1, pp. 177–188, 1989.
  4. S. Reich, “Fixed points of contractive functions,” Bollettino dell'Unione Matematica Italiana, vol. 5, no. 4, pp. 26–42, 1972.
  5. J. S. Bae, “Fixed point theorems for weakly contractive multivalued maps,” Journal of Mathematical Analysis and Applications, vol. 284, no. 2, pp. 690–697, 2003.
  6. T. Suzuki, “Mizoguchi-Takahashi's fixed point theorem is a real generalization of Nadler's,” Journal of Mathematical Analysis and Applications, vol. 340, no. 1, pp. 752–755, 2008.