ABSTRACT

In recent times the fixed point resulting in partially ordered metric spaces has greatly developed. In this paper we prove common fixed point results for multivalued and singlevalued mappings in partially ordered metric space. Our theorems generalized the theorem in [1] and extended much more recent results in such spaces.

1. Introduction and Preliminaries

Throughout this paper, let be a metric space unless mentioned otherwise and is the set of all non-empty bounded subsets of. Let and be the functions defined by

for all A, B in. If A is a singleton i.e., we write

and

If B is also a singleton i.e., we write

and

It is obvious that. For all . The definition of yields the following:

and

.

Several authors used these concepts of weakly contraction, compatibility, weak compatibility to prove some common fixed point theorems for set valued mappings (see [2-8]).

Definition 1.1. [9] A sequence of subsets of X is said to be convergent to a subset A of X if 1) Given, there is a sequence in X such that for and converges to a.

2) Given, there exists a positive integer N such that for where is the union of all open spheres with centers in A and radius.

Lemma 1.1. [9,10] If and are sequences in converging to A and B in, respectively, then the sequence converges to.

Lemma 1.2. [9] Let be a sequence in and y a point in X such that. Then the sequence converges to the set in.

In [11], Jungck and Rhoades extended definition of compatibility to set valued mappings setting as follows:

Definition 1.2. The mapping and are δ-compatible if , whenever is a sequence in X such the for some.

Recently, the following definition is given by Jungck and Rhoades [12].

Definition 1.3. The mapping and are weakly compatible if for each point u in X such that, we have.

It can be seen that any δ-compatible mappings are weakly compatible but the converse is not true as shown by an example in [13]. We will use the following relation between two nonempty subsets of a partially ordered set.

Definition 1.4. [3] Let A and B be two nonempty subsets of a partially ordered set. The relation between A and B is denoted and defined as follows:, if for every there exists such that.

We will utilize the following control function which is also referred to as altering distance function.

Definition 1.5. [14] A function is called an Altering distance function if the following properties are satisfied:

1) is monotone increasing and continuous2) if and only if

For the use of control function in metric fixed point theory see some recent references ([15,16]).

2. Main Result

Recently fixed point theory in partially ordered metric spaces has greatly developed. Choudhury and Metiya [17] proved certain fixed point theorems for multi valued and single valued mappings in partially ordered metric spaces. They proved the following:

Theorem 2.1. Let be a partially ordered set and suppose that there exists a metric d on X such that is a complete metric space. Let be a multi valued mappings such that the following conditions are satisfied:

There exists such that1) For implies

2) If is a non decreasing sequence in X, then, for all n3)for all comparable, where and is an Altering distance function. Then T has a fixed point.

We prove the following theorem for four single-valued and multivalued mappings:

Theorem 2.2. Let be a partially ordered set and suppose that there exists a metric d on X such that is a complete metric space. Let be single valued and be multivalued mappings such that the following conditions are satisfied:

1)

2) and are weakly compatible3) If is a strictly decreasing sequence in X, then, for all n4)for all comparable, , where and is an Altering distance function and suppose that one of or is complete. Then there exists a unique point such that

Proof: Let be an arbitrary point of X. By 1) we choose a point such that. For this point, there exists a point such that

, and so on. Continuing in this manner we can define a sequence as follows

(2.1)

We claim that is a Cauchy sequence. For which two cases arise, either for some n, or, for each n.

Case I. If for some n then, for each. For instance suppose. Then. Otherwise using 3), we get

Since

It follows that

(2.2)

Suppose that if, for some positive integer n, then from (2.2), we have

which implies that

Hence Similarly implie Proceeding in this manner, it follows that for each, so that for each, for some n, and is a Cauchy sequence.

Case II. When for each n. In this case, using 3), we obtain

Since

It follows that

(2.3)

Now if for each positive integer n, then from (2.3), we have

which implies that contradicting our assumption that, for each n. Therefore for all and is strictly decreasing sequence of positive numbers and therefore tends to a limit. If possible suppose r > 0. Then for given, there exists a positive integer N such that for each, we have

(2.4)

Taking the limit in (2.3) and using the continuity of, we have or

which is a contradiction unless. Hence

(2.5)

Next we show that is a Cauchy sequence. Suppose it is not, then there exists an and since

there exists two sequences of positive numbers and such that for all positive integers k, and

. Assuming that is the smallest positive integer, we get

Now,

i.e.

(2.6)

Taking the limit as in (2.6) and using (2.5), we have

(2.7)

Again

and

Taking the limit as and using (2.6) and (2.7), we have

(2.8)

Again we have

and

Letting and using (2.6) and (2.7), we have

(2.9)

Similarly, we have.

For each positive integer k, and are comparable. Now using the monotone property of in 4), we have

Letting and using (2.6)-(2.9), and the continuity of, we have, which is a contradiction by virtue of property of. Therefore and hence any subsequence thereof, is a Cauchy sequence.

Suppose is complete. Since

is a subsequence of, by the above is Cauchy and, for some.

We now show. For suppose

Since and therefore,. But

is a subsequence of the strictly decreasing sequence which tends to the lim r = 0. Therefore

tends to limit r = 0 and hence

implying. Thus. Now using, we have

or

which is a contradiction. Consequently

as.

In the same manner, it follows that as We now show. For this, in view of, we have

implies

or

which is a contradiction. Consequently, as. Hence. Since there exists some such that. Hence. We now show. For this, first we prove. Suppose then . Then in accordance with such that

implies while . Therefore a contradiction arises. Hence. But then, which, by, implies

Therefore Fu is a singleton. Since and Fu is a singleton,. Hence

Since the pair and are weakly compatible,

and 

From the above, it is clear that Fp and Gp are singletons and

We now show that. For instance, suppose then from, we have

Implies as above as. Hence and therefore

We now show. For, suppose. For this let in, we have

or

which is a contradiction. Consequently as Therefore and hence

Let be any point satisfying

Suppose then from, we have

in view of  Hence.

Corollary 2.1. Let I be a self mapping of a metric space and a set valued mapping satisfying 1)'

2)' are weakly compatible3)'for all comparable, where and is an altering distance function. If is complete subspace of X, there exists a unique point such that

Proof: Taking I = J and in Theorem 2.2.

Taking I = identity mapping in Corollary 2.1, we get the new corollary as follows:

Corollary 2.2. Let be a complete metric space and a set valued mapping satisfying

Then f has a unique fixed point in X.

Proof. Obvious.

Corollary 2.3. Let be a partially ordered set and suppose that there exists a metric d on X such that is a complete metric space. Let be single valued and be multivalued mappings such that the following conditions are satisfied:

1)''

2)'' and are weakly compatible3)'' if is a strictly decreasing sequence in X, then, for all n4)''for all comparable, , where and is an Altering distance function and suppose that one of or is complete. Then there exists a unique point such that

Example 2.1. Let be a sub set of with the order defined as for

if and only if. Let be given as

for.

The is a complete metric space with the required properties of Theorem 2.2.

Let, be defined as follows:

Let defined as, and. Then all the conditions in the Theorem 2.2 satisfied. Without loss of generality, we assume that, we discuss the following cases.

1) If, , then and

2) If then, and

3) If then, and

4) If then, and

5) If then and

In all above cases, it is clearly shown that

Hence the conditions of Theorem 2.2 are satisfied and shown that is a fixed point of I, J, F, and G.

3. Acknowledgements

Dedicated to Professor H. M. Srivastava on his 71^st Birth Anniversary.

REFERENCES