Some Common Fixed Point Theorems Satisfying (ψ φ) Maps in Partial Metric Spac...

Reza Arab

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Some Common Fixed Point Theorems Satisfying (ψ φ) Maps in Partial Metric Spaces

Reza Arab

Department of Mathematics, Sari Branch, Islamic Azad University, Sari, Iran

Abstract

In this paper, we present some coincidence and common fixed point results for infinite families of contractive maps satisfying a new class of pairs of generalized contractive type mappings defined in partial metric spaces. Our results extend and generalize many known results in the literature. Also, we introduce an example to support the validity of our results.

Cite this article:

  • Arab, Reza. "Some Common Fixed Point Theorems Satisfying (ψ φ) Maps in Partial Metric Spaces." Turkish Journal of Analysis and Number Theory 3.2 (2015): 53-60.
  • Arab, R. (2015). Some Common Fixed Point Theorems Satisfying (ψ φ) Maps in Partial Metric Spaces. Turkish Journal of Analysis and Number Theory, 3(2), 53-60.
  • Arab, Reza. "Some Common Fixed Point Theorems Satisfying (ψ φ) Maps in Partial Metric Spaces." Turkish Journal of Analysis and Number Theory 3, no. 2 (2015): 53-60.

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1. Introduction and Preliminaries

In 1994, in [8] Matthews introduced the notion of a partial metric space and proved the contraction principle of Banach in this new framework. The Existence of a fixed point for the contraction type mappings in partially metric spaces and its applications have been considered recently by many authors [1,2,3,6,7,9,12,13,15]. Consistent with [4, 8], the following de_nitions and results will be needed in the sequel.

Definition 1.1. [8] A partial metric on a nonempty set is a function such that for all

(P1) x = y p(x, x) = p(x, y) = p(y, y),

(P2) p(x, x) ≤ p(x, y),

(P3) p(x, y) = p(y, x),

(P4) p(x, y) ≤ p(x, z) + p(z, y) − p(z, z).

A partial metric space is a pair (X, p) such that X is a nonempty set and p is a partial metric on X.

It is clear that, if p(x, y) = 0, then, we get x = y from (P1) and (P2). But if x = y, p(x, y) may be a positive number other than 0.

Example 1.2. [8] Let a function p : R+×R+ R+ be defined by p(x, y) = max{x, y} for any x, y R+. Then, (R+, p) is a partial metric space.

Example 1.3. [8] If X = {[a, b] : a, b R, a b}, then p : X × X R+ de_ned by p([a, b], [c, d]) =max{b, d} − min{a, c} is a partial metric on X.

If p is a partial metric on X, then the function ps : X × X R+ given by

(1.1)

is a metric on X.

Definition 1.4. [8, 10, 11] Let (X, p) be a partial metric space. Then

(i) A sequence {xn} in a partial metric space (X, p) converges to a point xX if and only if

(ii) A sequence {xn} in a partial metric space (X, p) is called a Cauchy sequence if there exists (and is finite

(iii) A partial metric space (X, p) is said to be complete if every Cauchy sequence {xn} in X converges to a point x X, that is

It is easy to see that, every closed subset of a complete partial metric space is complete.

Lemma 1.5. [8, 9, 10] Let (X, p) be a partial metric space. Then

(a) {xn} is a Cauchy sequence in (X, p) if and only if it is a Cauchy sequence in the metric space (X, ps).

(b) A partial metric space (X, p) is complete if and only if the metric space (X, ps) is complete. Further-more,

if and only if

Lemma 1.6. [2] A mapping f : X −→ X is said to be continuous at a X, if for every ϵ > 0, there existsδ > 0 such that f(B(a, δ)) B(f(a), ϵ).

The following result is easy to check.

Lemma 1.7. Let (X, p) be a partial metric space. T : X X is continuous if and only if given a sequence {xn} X and x X such that then

Definition 1.8. Let X be a set, T and g are selfmaps of X. A point x in X is called a coincidence point of T and g if Tx = gx. We shall call w = Tx = gx a point of coincidence of T and g.

Definition 1.9. [14] Let (X, p) be a partial metric space and T, g : X → X are mappings of X into itself. We say that the pair {T, g} is partial compatible if the following conditions is held:

(b1) if p(x, x) = 0 then p(gx, gx) = 0,

(b2) whenever {xn} is a sequence in X such that Txn t and gxn t for some t X.

Example 1.10. Let X = [0,) endowed with the usual partial metric p de_ned by p : X × X X with p(x, y) = max{x, y}. Suppose T, g : X X such that Tx = 5x3 and gx = 2x3 for all x X. It is easy to check that the pair {T, g} is partial compatible.

Matthews [8] obtained the following Banach fixed-point theorem on complete partial metric spaces.

Theorem 1.11. (Matthews[8]) Let f be a mapping of a complete dualistic partial metric space (X, p) into itself such that there is a real number c with 0 c < 1, satisfying

(1.2)

for all x, y X, then f has a unique fixed point.

Recently, I. Altun and A. Erduran [5] obtained the following nice result, which generalizes Theorem 1.11 of Matthews.

Theorem 1.12. [5] Let (X, p) be a complete partial metric space and let f : X X be a map such that

(1.3)

for all x, y X, where ϕ : [0,) [0,) is a continuous, non-decreasing function such that ϕ(t) < t for each t > 0. Then f has a unique fixed point.

Theorem 1.13. [3] Let (X, p) be a complete partial metric space and let F : X X be a map such that

(1.4)

for all x, y X, where ϕ : [0,) −→ [0,) is continuous, nondecreasing function such that ϕ(t) < t for each t > 0, then F has a unique fixed point.

2. Main Results

In this section, we fixed the set of functions by ψ, φ : [0,+) [0,+) such that

(a) ψ is increasing;

(b) φ(t) < ψ(t) for each t > 0, φ(0) = ψ(0) = 0;

(c) φ(t) and ψ(t) are continuous functions.

Define = {(ψ, φ) : ψ and φ satisfy (a),(b) and (c)}.

Now, we establish an existence of common fixed point of a family mappings satisfying contractive condition involving ψ φ functions in the setup of partial metric spaces. Our results generalize Theorems 1.11,1.12 and 1.13.

Theorem 2.1. Let (X, p) be a complete partial metric space. Suppose {Tα: X −→ X : α I} be a family of mappings and g : X X be a self map. Also there exists i0 I such that , gX is closed in (X, p) and assume that there exists (ψ, φ) such that

(2.1)

for all x, y X, where

(2.2)

Then, there exists a unique x ∈ X such that gx = Tix for all i ∈ I, that is, g and {Ti: i ∈ I} have a unique coincidence point in X. Moreover, any coincidence point of g and is a coincidence point of g and {Ti: i ∈ I}. Also, we have p(gx, gx) = 0.

Proof. Let x0 be an arbitrary point in X. Construct the sequence {xn} such that (xn) = g(xn+1) for each n = 0, 1, 2, .... which is possible since X gX. Now by (2.1), we have

(2.3)

where

By (p4), we have

Therefore,

If we suppose that

and (xn) = (xn+1) for some n, then by (xn) = g(xn+1), (xn+1) = g(xn+1), that is, and g have a coincidence at x = xn+1, and so the proof is completed, otherwise if p((xn), (xn+1)) > 0, then

It is impossible.

If we get

(2.4)

If for some n, then by that is, and g have a coincidence at x = xn, and so the proof is completed. Since we suppose that and by the monotone property of ψ function, we have . Therefore, the sequence of real numbers is monotone decreasing. Hence there exists a real number r 0 such that,

We claim that r = 0. On the contrary, assume that

(2.5)

Since ψ and φ are continuous then from (2.4) and (2.5), we have

and so r = 0, a contradiction. Thus

(2.6)

From p((xn), Ti0 (xn)), p((xn+1), (xn+1)) p((xn), (xn+1)) and (2.6), we have

(2.7)

From (2.6), (2.7) and (1.1), we have

(2.8)

Next, we claim that {} is a Cauchy sequence in the metric space (X, ps). Assume the contrary. Then there exists an ϵ > 0 and subsequences {xn(k)} and {xm(k) } of {xn} with n(k) > m(k) > k such that p((xn(k)), (xm(k))) ϵ and p( (xn(k)-1), (xm(k))) < ϵ.

Then we have

(2.9)

Taking k → ∞ in (2.9) and using (2.6) and (2.7) we get

(2.10)

Thus from the definition of p we have

(2.11)
(2.12)

Taking k → ∞ in (2.11) and (2.12) and using (2.6), (2.7) and (2.10) we get

(2.13)

Also,

(2.14)

Taking k → ∞ in (2.14) and using (2.6), (2.7) and (2.13) we get

(2.15)

Now using inequality (2.1), we have

(2.16)

where

Letting k → ∞ in the above inequality and using (2.6),(2.10), (2.13) and (2.15), we obtain

As k → ∞, inequality (2.16) becomes,

which is a contradiction by virtue of a property of φ.

Then, we deduce that {(xn)} is a cauchy sequence in (X, ps). Hence, we have

Now, from the definition of ps and from (2.7), we have

(2.17)

Since X is a complete partial metric space, then, from lemma 1.5, the sequence {} converges in the metric space (X, ps), so there exist z in X such that

Again, from Lemma 1.5, we get

(2.18)

But, from (2.17) and (2.18) we have

(2.19)

Since and gX is closed, there exists xX such that z = gx. Now, we claim that x is a coincidence point of Ti0 and g. We have

On taking limit as n → ∞ in the above inequality, we have

(2.20)

By property of ψ and using (2.20), we have

(2.21)

Indeed,

we deduce, taking limit as n → ∞, that

Now suppose that x X is a coincidence point of and g. Then for any i I, from (2.1), we have

where

So,

therefore p(gx, Ti(x)) = 0, which implies that Ti(x) = gx, that is, x is a coincidence point of {Ti : i∈ I} and g.

Assume that z, y are coincidence points of {Ti : i I} and g in X such that y z. Then there exist t1, t2 in X such that, for all i I, Ti(t1) = g(t1) = z and Ti(t2) = g(t2) = y. Using (2.1), we have

where

So

which is a contradiction which proves the uniqueness of point of coincidence.

Example 2.2. Let X = [0, 1] and p(x, y) = max{x, y}, then it is clear that (X, p) is a complete partial metric space. Let I = {1, 2, 3, ...}, i0 = 1 and for every i I, Ti, g : X X, and ψ, φ: [0,) [0,) be given by Clearly (ψ, φ) . We show that condition (2.1) is satisfied.

If x, y X, then we have

Note that, {Ti} and g satisfy all the conditions given in Theorem 2.1. Moreover, 0 is a unique common fixed point of {Ti} and g.

As immediate consequences of Theorem 2.1 are the following fixed point results.

Corollary 2.3. Let (X, p) be a complete partial metric space. Suppose T1, T2, g : X X be three self mappings such that , gX is closed in (X, p) and for all x, y X

where

and (ψ, φ) ∈. Then, there exists a unique x ∈ X such that gx = T1x = T2x. Moreover, we have p(gx, gx) = 0.

Proof. Considering {TI : I I} = {T1, T2} in Theorem 2.1, we have the required proof.

Remark 2.4. Corollary 2.3 extends and generalizes many existing fixed point theorems.

If we take ψ(t) = t and Ti = T in Theorem 2.1, we have the following corollary.

Corollary 2.5. Let (X, p) be a complete partial metric space. Suppose T: X X be a self mapping such that T(X)g(X), gX is closed in (X, p) and for all x, y X

where φ: [0,) [0,) is continuous, nondecreasing function such that ϕ(t) < t for each t > 0 and

Then, there exists a unique x X such that gx = Tx. Moreover, we have p(gx, gx) = 0.

If we take g = I and φ(t) = λ t for λ [0, 1) in Corollary 2.5, we have the following corollary.

Corollary 2.6. Let (X, p) be a complete partial metric space. Suppose T : X X be a self mapping and g: X X. Also T(X) g(X), gX is closed in (X, p) and for all x, y X

Then, there exists a unique x X such that x = Tx. Moreover, we have p(x, x) = 0.

Now, we will prove the following result.

Theorem 2.7. Let (X, p) be a complete partial metric space. Suppose {Ti: X→ X : i ∈ I} be a family of mappings and g : X −→ X be a self map. Also there exists i0 ∈ I such that (X) g(X) and

(2.22)

where

(2.23)

and (ψ, φ) . Also suppose , g are partial compatible and continuous in (X, p). Then, there exists a unique x X such that gx = Tix for all i I. Moreover, we have p(gx, gx) = 0.

Proof. Using the same arguments in the proof of Theorem 2.1, we deduce that {(xn)} is a Cauchy sequence in the complete metric space (X, ps), and therefore, there exists zX such that

(2.24)

Now we show that z is a coincidence point of and g.

Since and g are continuous, from (2.24) and using Lemma 1.7, we get

(2.25)

Since and g are partial compatible mappings, this implies that

(2.26)

The condition (p4), we obtain

Letting n → ∞ in the above inequality and using (2.24), (2.25) and (2.26), we have

(2.27)

Now, we will prove that = 0. Suppose that this is not the case. Then, from (2.22) with i = i0, y = x = z, we get

where

Therefore, from (2.27) and the above inequality, we have

a contradiction. Hence p(gz, ) = 0 which implies that = gz, that is, z is a coincidence point of and g. If w is a coincidence point of and g other then z, then putting i = i0, x = z, y = w in (2.22) we have

where

Therefore,

which is possible only when z = w. Hence z is the unique coincidence point of and g. Similar Theorem 2.1 can be shown that z is a coincidence point of {Ti} and g for any i I.

We give in the following a sufficient condition for the uniqueness of the common fixed point of the mappings {Ti} and g.

Theorem 2.8. Adding to the hypotheses of Theorem 2.1 (resp. Theorem 2.7) the condition:

and g commute at their coincidence points, we obtain the uniqueness of common fixed point of {Ti} and g.

Proof. Suppose that and g commute at x. Set y = x = gx. Then

(2.28)

from (2.1) we get

(2.29)

where

Suppose that p(y, y) > 0, from (2.29), we get

which is a contradiction. Hence p(y, y) = 0. Therefore,

(2.30)

Thus we proved that and g have a common fixed point.

Uniqueness: Let v and w be two common fixed points of and g. (i.e) v = v = gv and w = w = gw. Using inequality (2.1), we have

where

Therefore,

which is possible only when w = v. Hence and g have an unique common fixed point. Now suppose that x X is a common fixed point of and g. Then x = gx = x.

For any i ∈ I, from inequality (2.1), we have

(2.31)

where

Suppose that p(x, Tix) > 0, from (2.31), we get

which is a contradiction. Hence p(x, Tix) = 0. So, we have

that is, x is a common fixed point of g and {Ti : i I}.

Example 2.9. Let X = [0, 1] and p(x, y) = max{x, y}, then it is clear that (X, p) is a complete partial metric space. Let I = {1, 2, 3, ...}, i0 =1 and for every i I, Ti, g : XX, and ψ, φ : [0,) [0,) be given by Clearly (ψ, φ) . We show that condition (2.1) is satisfied.

If x, y X, then we have

Note that, {Ti} and g satisfy all the conditions given in Theorem 2.7 and 2.8. Moreover, 0 is a unique common fixed point of {Ti} and g.

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