Abstract

The main result is a common fixed point theorem for a pair of multivalued maps on a complete metric space extending a recent result of Đorić and Lazović (2011) for a multivalued map on a metric space satisfying Ćirić-Suzuki-type-generalized contraction. Further, as a special case, we obtain a generalization of an important common fixed point theorem of Ćirić (1974). Existence of a common solution for a class of functional equations arising in dynamic programming is also discussed.

1. Introduction

Consistent with Nadler [1, page 620], (𝑋,𝑑) will denote a metric space and CL(𝑋), the collection of all nonempty closed subsets of 𝑋. For 𝐴,𝐵CL(𝑋) and 𝜀>0,𝐸𝑁(𝜀,𝐴)={𝑥𝑋𝑑(𝑥,𝑎)<𝜀forsome𝑎𝐴},𝐴,𝐵=𝐻{𝜀>0𝐴𝑁(𝜀,𝐵),𝐵𝑁(𝜀,𝐴)},(𝐴,𝐵)=inf𝐸𝐴,𝐵,if𝐸𝐴,𝐵+,if𝐸𝐴,𝐵=.(1.1) The hyperspace (CL(𝑋),𝐻) is called the generalized Hausdorff metric space induced by the metric 𝑑 on 𝑋.

For nonempty subsets 𝐴,𝐵 of 𝑋,𝑑(𝐴,𝐵) denotes the gap between the subsets 𝐴 and 𝐵, while 𝜌(𝐴,𝐵)=sup{𝑑(𝑎,𝑏)𝑎𝐴,𝑏𝐵},𝐵𝑁(𝑋)={𝐴𝐴𝑋andthediameterof𝐴isnite}.(1.2) As usual, we write 𝑑(𝑥,𝐵) (resp. 𝜌(𝑥,𝐵)) for 𝑑(𝐴,𝐵) (resp. 𝜌(𝐴,𝐵)) when 𝐴={𝑥}.

Let 𝑆,𝑇𝑋CL(𝑋). Then 𝑢𝑋 is a fixed point of 𝑆 if and only if 𝑢𝑆𝑢 and a common fixed point of 𝑆 and 𝑇 if and only if 𝑢𝑆𝑢𝑇𝑢.

Let 𝑆 and 𝑇 be maps to be defined specifically in a particular context, while 𝑥 and 𝑦 are the elements of a metric space (𝑋,𝑑): 𝑀(𝑆𝑥,𝑇𝑦)=max𝑑(𝑥,𝑦),𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦),𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑆𝑥)2.(1.3)

Recently Suzuki [2] and Kikkawa and Suzuki [3] obtained interesting generalizations of the Banach’s classical fixed point theorem and other fixed point results by Nadler [4], Jungck [5], and Meir and Keeler [6]. These results have important outcomes (see, e.g., [714]). The following result, due to Đorić and Lazović [9], extends and generalizes fixed point theorems from Ćirić [15], Kikkawa and Suzuki [3], Nadler [4], Reich [16], Rus [17], and others.

Theorem 1.1. Define a nonincreasing function 𝜑 from [0,1) onto (0,1] by 1𝜑(𝑟)=1𝑖𝑓0𝑟<211𝑟𝑖𝑓2𝑟<1.(1.4) Let 𝑋 be a complete metric space and 𝑇𝑋CL(𝑋). Assume there exists 𝑟[0,1) such that for every 𝑥,𝑦𝑋, 𝜑(𝑟)𝑑(𝑥,𝑇𝑥)𝑑(𝑥,𝑦)𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝐻(𝑇𝑥,𝑇𝑦)𝑟𝑀(𝑇𝑥,𝑇𝑦).(1.5) Then there exists 𝑧𝑋 such that 𝑧𝑇𝑧.

We remark that, for every 𝑥,𝑦𝑋, the generalized contraction 𝐻(𝑇𝑥,𝑇𝑦)𝑟𝑀(𝑇𝑥,𝑇𝑦), 0𝑟<1, was first studied by Ćirić [15]. The following important common fixed point theorem is due to Ćirić [18].

Theorem 1.2. Let 𝑋 be a complete metric space and 𝑆,𝑇𝑋𝑋. Assume there exists 𝑟[0,1) such that for every 𝑥,𝑦𝑋, 𝑑(𝑆𝑥,𝑇𝑦)𝑟𝑀(𝑆𝑥,𝑇𝑦).(1.6) Then 𝑆 and 𝑇 have a unique common fixed point.

For an excellent discussion on several special cases and variants of Theorem 1.2, one may refer to Rus [17]. However, the generality of Theorem 1.2 may be appreciated from the fact that (1.6) in Theorem 1.2 cannot be replaced by 𝑑(𝑆𝑥,𝑇𝑦)𝑟max{𝑑(𝑥,𝑦),𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦),𝑑(𝑥,𝑇𝑦),𝑑(𝑦,𝑆𝑥)}.(1.7) Indeed, Sastry and Naidu [19, Example  5] have shown that maps 𝑆 and 𝑇 satisfying (1.7) need not have a common fixed point on a complete metric space. Notice that the condition (1.7) with 𝑆=𝑇 is the quasicontraction due to Ćirić [20].

The main result of this paper (cf. Theorem 2.2) generalizes Theorems 1.1 and 1.2. Further, a corollary of Theorem 2.2 is used to obtain a unique common fixed point theorem for multivalued maps on a metric space with values in 𝐵𝑁(𝑋). As another application, we deduce the existence of a common solution for a general class of functional equations under much weaker conditions than those in [12, 14, 2124].

2. Main Results

We shall need the following result essentially due to Nadler [4] (see also [15, 25], [26, page 4], [27], [17, page 76]).

Lemma 2.1. If 𝐴,𝐵CL(𝑋) and 𝑎𝐴, then for each 𝜀>0, there exists 𝑏𝐵 such that 𝑑(𝑎,𝑏)𝐻(𝐴,𝐵)+𝜀.

Theorem 2.2. Let 𝑋 be a complete metric space and 𝑆,𝑇𝑋CL(𝑋). Assume there exists 𝑟[0,1) such that for every 𝑥,𝑦𝑋, 𝜑(𝑟)min{𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦)}𝑑(𝑥,𝑦)𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝐻(𝑆𝑥,𝑇𝑦)𝑟𝑀(𝑆𝑥,𝑇𝑦).(2.1) Then there exists an element 𝑢𝑋 such that 𝑢𝑆𝑢𝑇𝑢.

Proof. Obviously 𝑀(𝑆𝑥,𝑇𝑦)=0 iff 𝑥=𝑦 is a common fixed point of 𝑆 and 𝑇. So, we may take without any loss of generality that 𝑀(𝑆𝑥,𝑇𝑦)>0 for distinct 𝑥,𝑦𝑋. Let 𝜀>0 be such that 𝛽=𝑟+𝜀<1. Let 𝑢0𝑋 and 𝑢1𝑇𝑢0. Then by Lemma 2.1, their exists 𝑢2𝑆𝑢1 such that 𝑑𝑢2,𝑢1𝐻𝑆𝑢1,𝑇𝑢0+𝜀𝑀𝑆𝑢1,𝑇𝑢0.(2.2) Similarly, their exists 𝑢3𝑇𝑢2 such that 𝑑𝑢3,𝑢2𝐻𝑇𝑢2,𝑆𝑢1+𝜀𝑀𝑇𝑢2,𝑆𝑢1.(2.3) Continuing in this manner, we find a sequence {𝑢𝑛} in 𝑋 such that 𝑢2𝑛+1𝑇𝑢2𝑛,𝑢2𝑛+2𝑆𝑢2𝑛+1𝑑𝑢suchthat2𝑛+1,𝑢2𝑛𝐻𝑇𝑢2𝑛,𝑆𝑢2𝑛1+𝜀𝑀𝑇𝑢2𝑛,𝑆𝑢2𝑛1,𝑑𝑢2𝑛+2,𝑢2𝑛+1𝐻𝑆𝑢2𝑛+1,𝑇𝑢2𝑛+𝜀𝑀𝑆𝑢2𝑛+1,𝑇𝑢2𝑛.(2.4) Now, we consider two cases and show that for any 𝑛𝑁, 𝑑𝑢2𝑛+1,𝑢2𝑛𝑢𝛽𝑑2𝑛1,𝑢2𝑛.(2.5)Case 1. If 𝑑(𝑢2𝑛1,𝑆𝑢2𝑛1)𝑑(𝑢2𝑛,𝑇𝑢2𝑛), then 𝜑𝑑𝑢(𝑟)min2𝑛1,𝑆𝑢2𝑛1𝑢,𝑑2𝑛,𝑇𝑢2𝑛𝑢𝑑2𝑛1,𝑢2𝑛.(2.6) Therefore by the assumption, 𝐻𝑆𝑢2𝑛1,𝑇𝑢2𝑛𝑟𝑀𝑆𝑢2𝑛1,𝑇𝑢2𝑛.(2.7)Case 2. If 𝑑(𝑢2𝑛,𝑇𝑢2𝑛)𝑑(𝑢2𝑛1,𝑆𝑢2𝑛1), then 𝜑𝑑𝑢(𝑟)min2𝑛1,𝑆𝑢2𝑛1𝑢,𝑑2𝑛,𝑇𝑢2𝑛𝑢𝑑2𝑛1,𝑢2𝑛.(2.8) So by the assumption, 𝐻𝑆𝑢2𝑛1,𝑇𝑢2𝑛𝑟𝑀𝑆𝑢2𝑛1,𝑇𝑢2𝑛.(2.9) Hence in either case we obtain by (2.7) and (2.9), 𝑑𝑢2𝑛,𝑢2𝑛+1𝐻𝑆𝑢2𝑛1,𝑇𝑢2𝑛+𝜀𝑀𝑆𝑢2𝑛1,𝑇𝑢2𝑛𝑟𝑀𝑆𝑢2𝑛1,𝑇𝑢2𝑛+𝜀𝑀𝑆𝑢2𝑛1,𝑇𝑢2𝑛=𝛽𝑀𝑆𝑢2𝑛1,𝑇𝑢2𝑛𝑑𝑢=𝛽max2𝑛1,𝑢2𝑛𝑢,𝑑2𝑛1,𝑆𝑢2𝑛1𝑢,𝑑2𝑛,𝑇𝑢2𝑛,𝑑𝑢2𝑛1,𝑇𝑢2𝑛𝑢+𝑑2𝑛,𝑆𝑢2𝑛12𝑑𝑢𝛽max2𝑛1,𝑢2𝑛𝑢,𝑑2𝑛,𝑢2𝑛+1.(2.10) This yields (2.5). Analogously, we obtain 𝑑(𝑢2𝑛+2,𝑢2𝑛+1)𝛽𝑑(𝑢2𝑛+1,𝑢2𝑛), and conclude that for any 𝑛𝑁, 𝑑𝑢𝑛+1𝑢,𝑢𝑛𝛽𝑑𝑛,𝑢𝑛1.(2.11) Therefore{𝑢𝑛} is a Cauchy sequence and has a limit in 𝑋. Call it 𝑢.
Now we show that for any𝑦𝑋{𝑢}, 𝑑(𝑢,𝑇𝑦)𝑟max{𝑑(𝑢,𝑦),𝑑(𝑦,𝑇𝑦)},(2.12)𝑑(𝑢,𝑆𝑦)𝑟max{𝑑(𝑢,𝑦),𝑑(𝑦,𝑆𝑦)}.(2.13) Since 𝑢𝑛𝑢, there exists 𝑛0𝑁 (natural numbers) such that 𝑑𝑢,𝑢𝑛13𝑑(𝑢,𝑦)for𝑦𝑢andall𝑛𝑛0.(2.14) Then as in [2, page 1862], 𝜑𝑢(𝑟)𝑑2𝑛1,𝑆𝑢2𝑛1𝑢𝑑2𝑛1,𝑆𝑢2𝑛1𝑢𝑑2𝑛1,𝑢2𝑛𝑢𝑑2𝑛1,𝑢+𝑑𝑢,𝑢2𝑛231𝑑(𝑦,𝑢)=𝑑(𝑦,𝑢)3𝑢𝑑(𝑦,𝑢)𝑑(𝑦,𝑢)𝑑2𝑛1𝑢,𝑢𝑑2𝑛1.,𝑦(2.15) Therefore 𝜑𝑢(𝑟)𝑑2𝑛1,𝑆𝑢2𝑛1𝑢𝑑2𝑛1,𝑦.(2.16) Now either 𝑑(𝑢2𝑛1,𝑆𝑢2𝑛1)𝑑(𝑦,𝑇𝑦) or 𝑑(𝑦,𝑇𝑦)𝑑(𝑢2𝑛1,𝑆𝑢2𝑛1).
So in either case by (2.16), 𝜑𝑑𝑢(𝑟)min2𝑛1,𝑆𝑢2𝑛1𝑢,𝑑(𝑦,𝑇𝑦)𝑑2𝑛1,𝑦.(2.17) Hence by the assumption (2.1), 𝑑𝑢2𝑛,𝑇𝑦𝐻𝑆𝑢2𝑛1,𝑇𝑦𝑟𝑀𝑆𝑢2𝑛1𝑑𝑢,𝑇𝑦𝑟max2𝑛1𝑢,𝑦,𝑑2𝑛1,𝑆𝑢2𝑛1𝑑𝑢,𝑑(𝑦,𝑇𝑦),2𝑛1,𝑇𝑦+𝑑𝑦,𝑆𝑢2𝑛12.(2.18) Making 𝑛, 𝑑(𝑢,𝑇𝑦)𝑟max𝑑(𝑢,𝑦),𝑑(𝑢,𝑢),𝑑(𝑦,𝑇𝑦),𝑑(𝑢,𝑇𝑦)+𝑑(𝑦,𝑢)2𝑟max{𝑑(𝑢,𝑦),𝑑(𝑦,𝑇𝑦),𝑑(𝑢,𝑇𝑦)}.(2.19) This yields (2.12). Similarly, we can show (2.13).
Now, we show that 𝑢𝑆𝑢𝑇𝑢.
For 0𝑟<1/2, the following cases arise.
Case 1. Suppose 𝑢𝑆𝑢 and 𝑢𝑇𝑢. Then as in [8, page 6], let 𝑎𝑇𝑢 be such that 2𝑟𝑑(𝑎,𝑢)<𝑑(𝑢,𝑇𝑢),(2.20) and 𝑎𝑆𝑢 be such that 2𝑟𝑑(𝑎,𝑢)<𝑑(𝑢,𝑆𝑢).
Since 𝑎𝑇𝑢 implies 𝑎𝑢, we have from (2.12) and (2.13), 𝑑(𝑢,𝑇𝑎)𝑟max{𝑑(𝑢,𝑎),𝑑(𝑎,𝑇𝑎)},(2.21)𝑑(𝑢,𝑆𝑎)𝑟max{𝑑(𝑢,𝑎),𝑑(𝑎,𝑆𝑎)}.(2.22) On the other hand, since 𝜑(𝑟)𝑑(𝑢,𝑇𝑢)𝑑(𝑢,𝑇𝑢)𝑑(𝑎,𝑢), 𝜑(𝑟)min{𝑑(𝑎,𝑆𝑎),𝑑(𝑢,𝑇𝑢)}𝑑(𝑎,𝑢).(2.23) Therefore by the assumption (2.1), 𝑑(𝑆𝑎,𝑎)𝐻(𝑆𝑎,𝑇𝑢)𝑟max𝑑(𝑎,𝑢),𝑑(𝑢,𝑇𝑢),𝑑(𝑎,𝑆𝑎),𝑑(𝑢,𝑆𝑎)+𝑑(𝑎,𝑇𝑢)21=𝑟max𝑑(𝑎,𝑢),𝑑(𝑎,𝑆𝑎),2.𝑑(𝑢,𝑆𝑎)(2.24) This gives 𝑑(𝑎,𝑆𝑎)𝐻(𝑆𝑎,𝑇𝑢)𝑟𝑑(𝑎,𝑢)<𝑑(𝑎,𝑢).
So by (2.22), 𝑑(𝑆𝑎,𝑢)𝑟𝑑(𝑎,𝑢). Thus𝑑(𝑢,𝑇𝑢)𝑑(𝑢,𝑆𝑎)+𝐻(𝑆𝑎,𝑇𝑢)𝑟𝑑(𝑎,𝑢)+𝑟𝑑(𝑎,𝑢)=2𝑟𝑑(𝑎,𝑢)<𝑑(𝑢,𝑇𝑢)(bytheassumptionofCase1).(2.25) This contradicts 𝑢𝑇𝑢. Consequently 𝑢𝑇𝑢. Similarly 𝑢𝑆𝑢.
Case 2. Let 𝑢𝑆𝑢and𝑢𝑇𝑢. Then as in the previous case, let 𝑎𝑇𝑢 be such that 2𝑟𝑑(𝑎,𝑢)<𝑑(𝑢,𝑇𝑢).(2.26) Since 𝑎𝑢, we have from (2.13), 𝑑(𝑢,𝑆𝑎)𝑟max{𝑑(𝑢,𝑎),𝑑(𝑎,𝑆𝑎)}.(2.27) On the other hand, Since 𝜑(𝑟)𝑑(𝑢,𝑇𝑢)𝑑(𝑢,𝑇𝑢)𝑑(𝑎,𝑢), 𝜑(𝑟)min{𝑑(𝑎,𝑆𝑎),𝑑(𝑢,𝑇𝑢)}𝑑(𝑎,𝑢).(2.28) Therefore by the assumption (2.1), 𝑑(𝑆𝑎,𝑎)𝐻(𝑆𝑎,𝑇𝑢)𝑟max𝑑(𝑎,𝑢),𝑑(𝑢,𝑇𝑢),𝑑(𝑎,𝑆𝑎),𝑑(𝑢,𝑆𝑎)+𝑑(𝑎,𝑇𝑢)21=𝑟max𝑑(𝑎,𝑢),𝑑(𝑎,𝑆𝑎),2.𝑑(𝑢,𝑆𝑎)(2.29) This gives 𝑑(𝑎,𝑆𝑎)𝐻(𝑆𝑎,𝑇𝑢)𝑟𝑑(𝑎,𝑢)<𝑑(𝑎,𝑢).
So by (2.22), 𝑑(𝑆𝑎,𝑢)𝑟𝑑(𝑎,𝑢). Thus 𝑑(𝑢,𝑇𝑢)𝑑(𝑢,𝑆𝑎)+𝐻(𝑆𝑎,𝑇𝑢)𝑟𝑑(𝑎,𝑢)+𝑟𝑑(𝑎,𝑢)=2𝑟𝑑(𝑎,𝑢)<𝑑(𝑢,𝑇𝑢)(bytheassumptionofCase2).(2.30) This contradicts 𝑢𝑇𝑢. Consequently 𝑢𝑇𝑢.
Case 3. 𝑢𝑇𝑢 and 𝑢𝑆𝑢. As in the previous case, it follows that 𝑢𝑆𝑢.
Now we consider the case 1/2𝑟<1.
First we show that 𝐻(𝑆𝑥,𝑇𝑢)𝑟max𝑑(𝑥,𝑢),𝑑(𝑥,𝑆𝑥),𝑑(𝑢,𝑇𝑢),𝑑(𝑥,𝑇𝑢)+𝑑(𝑢,𝑆𝑥)2.(2.31) Assume that 𝑥𝑢. Then for every 𝑛𝑁, there exists 𝑧𝑛𝑆𝑥 such that 𝑑𝑢,𝑧𝑛1𝑑(𝑢,𝑆𝑥)+𝑛𝑑(𝑥,𝑢).(2.32) Therefore 𝑑(𝑥,𝑆𝑥)𝑑𝑥,𝑧𝑛𝑑(𝑥,𝑢)+𝑑𝑢,𝑧𝑛1𝑑(𝑥,𝑢)+𝑑(𝑢,𝑆𝑥)+𝑛𝑑(𝑥,𝑢).(2.33) Using (2.13) with 𝑦=𝑥, (2.33) implies 1𝑑(𝑥,𝑆𝑥)𝑑(𝑥,𝑢)+𝑟max{𝑑(𝑥,𝑢),𝑑(𝑥,𝑆𝑥)}+𝑛𝑑(𝑢,𝑥).(2.34) If 𝑑(𝑥,𝑢)𝑑(𝑥,𝑆𝑥),then (2.34) gives 1𝑑(𝑥,𝑆𝑥)𝑑(𝑥,𝑢)+𝑟𝑑(𝑥,𝑢)+𝑛=1𝑑(𝑢,𝑥)1+𝑟+𝑛𝑑(𝑥,𝑢).(2.35) Making 𝑛, 𝑑(𝑥,𝑆𝑥)(1+𝑟)𝑑(𝑥,𝑢).(2.36) Thus 𝜑(𝑟)𝑑(𝑥,𝑆𝑥)=(1𝑟)𝑑(𝑥,𝑆𝑥)(1/(1+𝑟))𝑑(𝑥,𝑆𝑥)𝑑(𝑥,𝑢).
Then 𝜑(𝑟)min{𝑑(𝑥,𝑆𝑥),𝑑(𝑢,𝑇𝑢)}𝑑(𝑥,𝑢),and by the assumption (2.1), 𝐻(𝑆𝑥,𝑇𝑢)𝑟max𝑑(𝑥,𝑢),𝑑(𝑥,𝑆𝑥),𝑑(𝑢,𝑇𝑢),𝑑(𝑥,𝑇𝑢)+𝑑(𝑢,𝑆𝑥)2.(2.37) If 𝑑(𝑥,𝑢)<𝑑(𝑥,𝑆𝑥),then (2.34) gives 1𝑑(𝑥,𝑆𝑥)𝑑(𝑥,𝑢)+𝑟𝑑(𝑥,𝑆𝑥)+𝑛𝑑(𝑢,𝑥),(2.38) that is, (1𝑟)𝑑(𝑥,𝑆𝑥)(1+(1/𝑛))𝑑(𝑥,𝑢).

Making 𝑛,𝜑(𝑟)𝑑(𝑥,𝑆𝑥)𝑑(𝑥,𝑢).(2.39) Then 𝜑(𝑟)min{𝑑(𝑥,𝑆𝑥),𝑑(𝑢,𝑇𝑢)}𝑑(𝑥,𝑢), and by the assumption, we get (2.37).

Taking 𝑥=𝑢2𝑛+1 in (2.37) and passing to the limit, we obtain 𝑑(𝑢,𝑇𝑢)𝑟𝑑(𝑢,𝑇𝑢).(2.40) This gives 𝑢𝑇𝑢. Analogously, 𝑢𝑆𝑢.

The following result generalizes Theorem 1.2.

Corollary 2.3. Let 𝑋 be a complete metric space and 𝑆,𝑇 maps from 𝑋 into 𝑋. Suppose there exists 𝑟[0,1) such that for every 𝑥,𝑦𝑋, 𝜑(𝑟)min{𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦)}𝑑(𝑥,𝑦)𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑑(𝑆𝑥,𝑇𝑦)𝑟𝑀(𝑆𝑥,𝑇𝑦).(2.41) Then 𝑆 and 𝑇 have a unique common fixed point.

Proof. For single-valued maps 𝑆 and 𝑇, it comes from Theorem 2.2 that they have a common fixed point. The uniqueness of the common fixed point follows easily.

Remark 2.4. Theorem 1.1 is obtained as a particular case of Theorem 2.2 when 𝑆=𝑇.

Now we derive the following result due to Đorić and Lazović [9, Corollary 2.3].

Corollary 2.5. Let 𝑋 be a complete metric space and 𝑇 a map from 𝑋 into 𝑋. Suppose there exists 𝑟[0,1) such that for every 𝑥,𝑦𝑋, 𝜑(𝑟)𝑑(𝑥,𝑇𝑥)𝑑(𝑥,𝑦)𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑑(𝑇𝑥,𝑇𝑦)𝑟𝑀(𝑇𝑥,𝑇𝑦).(2.42) Then 𝑇 has a unique fixed point.

Proof. It comes from Corollary 2.3 when 𝑆=𝑇.

The following example shows the generality of our results.

Example 2.6. Let 𝑋={(0,0),(0,4),(4,0),(0,5),(5,0),(4,5),(5,4)} be endowed with the metric 𝑑 defined by 𝑑𝑥1,𝑥2,𝑦1,𝑦2=||𝑥1𝑦1||+||𝑥2𝑦2||.(2.43) Let 𝑆 and 𝑇 be such that 𝑆𝑥1,𝑥2=𝑥1,0if𝑥1𝑥2(0,0)if𝑥1>𝑥2,𝑇𝑥1,𝑥2=𝑥2,0if𝑥1𝑥20,𝑥2if𝑥1>𝑥2.(2.44) Then 𝑆 and 𝑇 do not satisfy the condition (1.6) of Theorem 1.2 at 𝑥=(4,5),𝑦=(5,4). However, this is readily verified that all the hypotheses of Corollary 2.3 are satisfied for the maps 𝑆 and 𝑇.

Theorem 2.7. Let 𝑋 be a complete metric space and 𝑃,𝑄𝑋𝐵𝑁(𝑋). Assume there exists 𝑟[0,1) such that for every 𝑥,𝑦𝑋, 𝜑(𝑟)min{𝜌(𝑥,𝑃𝑥),𝜌(𝑦,𝑄𝑦)}𝑑(𝑥,𝑦)(2.45) implies 𝜌(𝑃𝑥,𝑄𝑦)𝑟max𝑑(𝑥,𝑦),𝜌(𝑥,𝑃𝑥),𝜌(𝑦,𝑄𝑦),𝑑(𝑥,𝑄𝑦)+𝑑(𝑦,𝑃𝑥)2.(2.46) Then there exsits a unique point 𝑧𝑋 such that 𝑧𝑃𝑧𝑄𝑧.

Proof. Choose 𝜆(0,1). Define single-valued maps 𝑆,𝑇𝑋𝑋 as follows. For each 𝑥𝑋, let 𝑆𝑥 be a point of 𝑃𝑥 which satisfies 𝑑(𝑥,𝑆𝑥)𝑟𝜆𝜌(𝑥,𝑃𝑥).(2.47) Similarly, for each 𝑦𝑋, let 𝑇𝑦 be a point of 𝑄𝑦 such that 𝑑(𝑦,𝑇𝑦)𝑟𝜆𝜌(𝑦,𝑄𝑦).(2.48) Since 𝑆𝑥𝑃𝑥 and 𝑇𝑦𝑄𝑦, 𝑑(𝑥,𝑆𝑥)𝜌(𝑥,𝑃𝑥),𝑑(𝑦,𝑇𝑦)𝜌(𝑦,𝑄𝑦).(2.49) So, (2.45) gives 𝜑(𝑟)min{𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦)}𝜑(𝑟)min{𝜌(𝑥,𝑃𝑥),𝜌(𝑦,𝑄𝑦)}𝑑(𝑥,𝑦),(2.50) and this implies (2.46). Therefore 𝑑(𝑆𝑥,𝑇𝑦)𝜌(𝑃𝑥,𝑄𝑦)𝑟𝑟𝜆𝑟max𝜆𝑑(𝑥,𝑦),𝑟𝜆𝜌(𝑥,𝑃𝑥),𝑟𝜆𝑟𝜌(𝑦,𝑄𝑦),𝜆𝑑(𝑥,𝑄𝑦)+𝑟𝜆𝑑(𝑦,𝑃𝑥)2𝑟1𝜆max𝑑(𝑥,𝑦),𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦),𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑆𝑥)2.(2.51) So (2.50), namely, 𝜑(𝑟)min{𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦)}𝑑(𝑥,𝑦) implies 𝑑(𝑆𝑥,𝑇𝑦)𝑟max𝑑(𝑥,𝑦),𝑑(𝑥,𝑆𝑥),𝑑(𝑦,𝑇𝑦),𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑆𝑥)2,(2.52) where 𝑟=𝑟1𝜆<1.
Hence by Theorem 2.2, 𝑆 and 𝑇have a unique point 𝑧𝑋 such that 𝑆𝑧=𝑇𝑧=𝑧. This implies 𝑧𝑃𝑧𝑄𝑧.

Corollary 2.8. Let 𝑋 be a complete metric space and 𝑃𝑋𝐵𝑁(𝑋). Assume there exists 𝑟[0,1) such that for every 𝑥,𝑦𝑋, 𝜌(𝑥,𝑃𝑥)(1+𝑟)𝑑(𝑥,𝑦)𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝜌(𝑃𝑥,𝑃𝑦)𝑟max𝑑(𝑥,𝑦),𝜌(𝑥,𝑃𝑥),𝜌(𝑦,𝑃𝑦),𝑑(𝑥,𝑃𝑦)+𝑑(𝑦,𝑃𝑥)2.(2.53) Then there exists a unique point 𝑧𝑋 such that 𝑧𝑃𝑧.

Proof. It comes from Theorem 2.7 when 𝑄=𝑃.

3. Applications

Throughout this section, we assume that 𝑌 and 𝑍 are Banach spaces, 𝑊𝑌 and 𝐷𝑍. Let 𝑅 denotes the field of reals, 𝑔1,𝑔2𝑊×𝐷𝑅and 𝐺1,𝐺2𝑊×𝐷×𝑅𝑅. Taking 𝑊 and 𝐷 as the state and decision spaces, respectively, the problem of dynamic programming reduces to the problem of solving functional equations:𝑝𝑖=sup𝑦𝐷𝑔𝑖(𝑥,𝑦)+𝐻𝑖𝑥,𝑦,𝑝𝑖(𝑥,𝑦),𝑥𝑊,𝑖=1,2.(3.1)

In the multistage process, some functional equations arise in a natural way (cf. [22, 23]; see also [21, 24, 28, 29]). In this section, we study the existence of common solution of the functional equations (3.1) arising in dynamic programming.

Let 𝐵(𝑊) denotes the set of all bounded real-valued functions on 𝑊. For an arbitrary 𝐵(𝑊), define =sup𝑥𝑊|(𝑥)|. Then (𝐵(𝑊),) is a Banach space. Suppose that the following conditions hold: (DP-1)𝐻1,𝐻2,𝑔1,and 𝑔2 are bounded.(DP-2) There exists 𝑟[0,1) such that for every (𝑥,𝑦)𝑊×𝐷,,𝑘𝐵(𝑊)and 𝑡𝑊,𝜑||(𝑟)min(𝑡)𝐴1||,||𝑘(𝑡)(𝑡)𝐴2𝑘||||||(𝑡)(𝑡)𝑘(𝑡)(3.2) implies||𝐻1(𝑥,𝑦,(𝑡))𝐻2||||||,||(𝑥,𝑦,𝑘(𝑡))𝑟max(𝑡)𝑘(𝑡)(𝑡)𝐴1||,||(𝑡)𝑘(𝑡)𝐴2||,||𝑘(𝑡)(𝑡)𝐴2||+||𝑘(𝑡)𝑘(𝑡)𝐴1||(𝑡)2,(3.3) where 𝐴1,𝐴2 are defined as follows:𝐴𝑖(𝑥)=sup𝑦𝐷𝐻𝑖(𝑥,𝑦,(𝑥,𝑦)),𝑥𝑊,𝐵(𝑊),𝑖=1,2.(3.4)

Theorem 3.1. Assume the conditions (DP-1) and (DP-2). Then the functional equations (3.1), 𝑖=1,2, have a unique common solution in 𝐵(𝑊).

Proof. For any ,𝑘𝐵(𝑊), let 𝑑(,𝑘)=sup{|(𝑥)𝑘(𝑥)|𝑥𝑊}. Then (𝐵(𝑊),𝑑) is a complete metric space.
Let 𝜆 be any arbitrary positive number and 1,2𝐵(𝑊). Pick 𝑥𝑊 and choose 𝑦1,𝑦2𝐷 such that 𝐴𝑖𝑖<𝐻𝑖𝑥,𝑦𝑖,𝑖𝑥𝑖+𝜆,(3.5) where 𝑥𝑖=(𝑥,𝑦𝑖),𝑖=1,2.
Further, 𝐴11𝐻1𝑥,𝑦2,1𝑥2𝐴,(3.6)22𝐻2𝑥,𝑦1,2𝑥1.(3.7) Therefore, the first inequality in (DP-2) becomes 𝜑||(𝑟)min1(𝑥)𝐴11||,||(𝑥)2(𝑥)𝐴22||||(𝑥)1(𝑥)2||(𝑥),(3.8) and this together with (3.5) and (3.7) implies 𝐴11𝐴22<𝐻1𝑥,𝑦1,1𝑥1𝐻2𝑥,𝑦,2𝑥1||𝐻+𝜆1𝑥,𝑦1,1𝑥1𝐻2𝑥,𝑦1,2𝑥1||𝐻+𝜆𝑟𝑀11,𝐻22+𝜆.(3.9) Similarly, (3.5), (3.6), and (3.8) imply 𝐴22(𝑥)𝐴11𝐴(𝑥)𝑟𝑀11,𝐴22+𝜆.(3.10) So, from (3.10) and (3.11), we obtain ||𝐴11(𝑥)𝐴22||𝐴(𝑥)𝑟𝑀11,𝐴22+𝜆.(3.11) Since this inequality is true for any 𝑥𝑊, and 𝜆>0 is arbitrary, on taking supremum, we find from (3.8) and (3.11) that 𝜑𝑑(𝑟)min1,𝐴11,𝑑2,𝐴22𝑑1,2(3.12) implies 𝑑𝐴11,𝐴22𝐴𝑟𝑀11,𝐴22.(3.13) Therefore, Corollary 2.3 applies, wherein 𝐴1 and 𝐴2 correspond, respectively, to the maps 𝑆 and 𝑇. So 𝐴1 and 𝐴2 have a unique common fixed point , that is, (𝑥) is the unique bounded common solution of the functional equations (3.1), 𝑖=1,2.

The following result generalizes a recent result of Singh and Mishra [12, Corollary  4.2] which in turn extends certain results from [21, 23, 24].

Corollary 3.2. Suppose that the following conditions hold.(i)𝐺 and 𝑔 are bounded.(ii)There exists 𝑟[0,1) such that for every 𝑥,𝑦𝑊×𝐷,,𝑘𝐵(𝑊) and 𝑡𝑊, ||||||||||||𝜑(r)(𝑡)𝐾(𝑡)(𝑡)𝑘(𝑡)𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝐺(𝑥,𝑦,(𝑡))𝐺(𝑥,𝑦,𝑘(𝑡))𝑟max𝑀(𝐾,(𝑡),𝑘(𝑡)),(3.14) where 𝐾 is defined as 𝐾(𝑡)=sup𝑦𝐷{𝑔(𝑡,𝑦)+𝐺(𝑡,𝑦,(𝑡,𝑦))},𝑡𝑊,𝐵(𝑊).(3.15) Then the functional equation (3.1) with 𝐻1=𝐻2=𝐺 and 𝑔1=𝑔2=𝑔 possesses a unique bounded solution in 𝑊.

Proof. It comes from Theorem 3.1 when 𝑔1=𝑔2=𝑔 and 𝐻1=𝐻2=𝐺.

Acknowledgments

The authors are grateful to all the three referees for their appreciation and valuable suggestions to improve upon the paper. They also thank Professor Yonghong Yao for his suggestions in this paper. The first author (S. L. Singh) acknowledges the support of the University Grants Commission, New Delhi under Emeritus Fellowship.