0 Fixed Point Theory and Applications

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Some fixed point results for multi-valued mappings in partial metric spaces

Jamshaid Ahmad1, Cristina Di Bari2, Yeol Je Cho3* and Muhammad Arshad4

"Correspondence: yjcho@gnu.ac.kr 3Department of Mathematics Education and RINS, Gyeongsang NationalUniversity, Jinju, 660-701, Korea

Full list of author information is available at the end of the article

Abstract

In this paper, we obtain some fixed point results for multi-valued mappings in partial metric spaces. Our results unify, generalize and complement various known comparable results from the current literature. An example is also included to illustrate the main result in the paper. MSC: 46S40; 47H10; 54H25

Keywords: partial Hausdorff metric; multi-valued mapping; fixed point; completeness

ft Springer

1 Introduction and preliminaries

The fixed point theory is one of the most powerful and fruitful tools in nonlinear analysis. Its core subject is concerned with the conditions for the existence of one or more fixed points of a mapping T from a topological space X into itself, that is, we can find x e X such that Tx = x. The Banach contraction principle [1] is the simplest and one of the most versatile elementary results in fixed point theory. Moreover, being based on an iteration process, it can be implemented on a computer to find the fixed point of a contractive mapping. It produces approximations of any required accuracy, and, moreover, even the number of iterations needed to get a specified accuracy can be determined. Recently, Samet et al. [2] introduced a new concept of a-contractive type mappings and established various fixed point theorems for such mappings in complete metric spaces. The presented theorems extend, generalize and improve several results on the existence of fixed points in the literature.

In 1994, Matthews [3] introduced the concept of a partial metric space and obtained a Banach-type fixed point theorem on complete partial metric spaces. Later on, several authors (see, for example, [4-31]) proved fixed point theorems in partial metric spaces. After the definition of the partial Hausdorff metric, Aydi etal. [28] proved the Banach-type fixed point result for set-valued mappings in complete partial metric spaces.

The aim of this paper is to generalize various known results proved by Nadler [32], Kikkawa and Suzuki [33], Mot and Petrusel [34], Dhompongsa and Yingtaweesittikul [35] to the case of partial metric spaces and give one example to illustrate our main results.

We start with recalling some basic definitions and lemmas on partial metric spaces. The definition of a partial metric space is given by Matthews [3] (see also [7, 29, 30]) as follows.

© 2013 Ahmad et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the originalwork is properly cited.

Definition 1 A partial metric on a nonempty set X is a functionp: X x X — [0, +to) such that the following conditions hold: for all x, y, z e X,

(Pi) p(x, x) = p(y,y) = p(x,y) if and only if x = y,

(P2) p(x,x) <p(x,y),

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

(P4) p(x,z) <p(x,y)+p(y,z)-p(y,y).

The pair (X,p) is then called a partial metric space.

If (X,p) is a partial metric space, then the function ps : X x X — [0, +to) given by ps(x, y) = 2p(x, y) - p(x, x) - p(y, y) for all x,y e X is a metric on X.

A basic example of a partial metric space is the pair ([0, +to),p), wherep(x, y) = max{x, y} for all x,y e [0, +to).

Lemma 1 Let (X, p) be a partial metric space, then we have the following:

(1) A sequence {xn} in a partial metric space (X, p) converges to a point x e X if and only if limn^+TO p(x, xn) =p(x, x).

(2) A sequence {xn} in a partial metric space (X, p) is called a Cauchy sequence if the limn,m—p(xn, xm) exists and is finite.

(3) A partial metric space (X, p) is said to be complete if every Cauchy sequence {xn} in X converges to a point x e X, that is, p(x, x) = limn,m—p(xn, xm).

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

(5) A partial metric space (X, p) is complete if and only if the metric space (X, ps) is complete. Furthermore, limn—ps(xn, z) = 0 if and only if

p(z, z) = lim p(xn,z)= lim p(xn,xm).

n—n,m—+TO

Remark 1 ([7], Lemma 1) Let (X,p) be a partial metric space and let A be a nonempty set in (X,p), then a e A if and only if

p(a, A) = p(a, a),

where A denotes the closure of A with respect to the partial metric p. Note A is closed in (X, p) if and only if A = A.

Now, we state the following definitions and propositions of a very recent paper of Aydi etal. [28].

Let CBp(X) be a collection of all nonempty closed and bounded subsets of X with respect to the partial metric p. For any A e CBp(X), we define

p(a,A) = inf{p(a,x) :x e A}.

On the other hand, for any A, B e CBp(X), we define

Sp(A,B) = supjp(a,B): a e A}, SP(B, A) = sup{p(b, A): b e B}

Hp(A, B) = max {5p (A, B), Sp(B, A)}.

Proposition 1 [28] Let (X,p) be a partial metric space. For any A, B, C e CBp(X), we have

(1) Sp(A, A) = sup{p(a, a): a e A}.

(2) Sp(A,A) < Sp(A,B).

(3) Sp(A, B) = 0 implies that A c B.

(4) Sp(A,B) < Sp(A, C) + SP(C,B) - infceCp(c, c).

Proposition2 [28] Let (X,p) be a partial metric space. ForanyA, B, C e CBp(X), wehave

(1) Hp (A, A) < Hp (A, B).

(2) Hp (A, B)=Hp(B, A).

(3) Hp (A, B) < Hp (A, C) + Hp(C, B) - infceC p(c, c).

Lemma 2 [28] Let A and B be nonempty closed and bounded subsets of a partial metric space (X,p) and h >1. Then, for all a e A, there exists b e B such thatp(a, b) < hHp(A,B).

The following result was proved by Aydi etal. in [28].

Theorem 1 Let (X,p) be a partial metric space. IfT: X — CBp(X) is a multi-valued mapping such that, for all x, y e X,

Hp(Tx, Ty) < kp(x,y),

where k e (0,1). Then T has a fixed point.

2 Main results

Now, we characterize the celebrated theorem of Kikkawa and Suzuki [33] in the framework of partial metric spaces.

Theorem 2 Define a strictly decreasingfunction © from [0,1) onto (2,1] by ®(r) = i+r. Let (X,p) be a complete partial metric space and F: X — CBp(X) be a multi-valued mapping. Assume that there exists r e [0,1) such that

©(r)p(x,Fx) <p(x,y) =}■ Hp(Fx,Fy) < rp(x,y) (2.1)

for all x, y e X. Then there exists u e X such that u e Fu.

Proof Let x0 e X be arbitrarily chosen. For all x1 e Fx0, we have

©(r)p(x0,Fx0) < ©(r)p(x0,xO < p(x0,xO

and, by the condition (2.1), we get

Hp(Fx0,Fx1) < rp(x0,x1).

Let h e (1,1), by Lemma 2, there exists x2 e Fx1 such thatp(x\,x2) < hHp(Fx0,Fx1). Using the previous inequality, we obtain

p(x1,x2) < hHp(Fx0,Fx1) < hrp(x0,x1).

Now, we have

©(r)p(x1,Fx1) < ©(r)p(x1,x2) <p(x1,x2)

and, by the condition (2.1), we get

Hp(Fxi,FX2) < rp(x1,X2).

By Lemma 2, there exists x3 e Fx2 such that

p(x2,x3) < hHp(Fx\,Fx2) < hrp(x1,x2) < (hr)2p(x0,x1).

Continuing in this way, we can generate a sequence {xn} in X such that xn+1 e Fxn and

p(xn,Xn+1) < knp(x0,X1) (2.2)

for all n e N, where k = hr <1.

Now, we show that {xn} is a Cauchy sequence. Using (2.2) and the triangle inequality for partial metrics (p4), for all n, m e N, we have

p(xn, xn+m) < p(xn, xn+1) + p(xn+1, xn+m) —p(xn+1, xn+1)

< p(xn, xn+1) + p(xn+1, xn+m)

< p(Xn, Xn+1) + p(Xn+1, Xn+2) + p(Xn+2,Xn+m) -p(Xn+2, Xn+2)

< p(Xn, Xn+1) + P(Xn+1, Xn+2) + P(Xn+2, xn+m).

Inductively, we have

p(xn, xn+m) < p(xn, Xn+1) + P(Xn+1, Xn+2) + ••• + p(Xn+m-1, Xn+m)

< knp(X0,X1) + kn+1p(X0,X1) + ••• + kn+m-1p(X0,X1)

< k + kn+1 + ••• + kn+m-1)p(X0,X1)

< -—7p(X0,X1) ^ 0 1 - k

as n ^ since 0 < k < 1. By the definition ofps, we get

p (Xn, Xn+m) < 2p(Xn, Xn+m) ^ 0

as n ^ which implies that {Xn} is a Cauchy sequence in (X,ps). Since (X,p) is complete, by Lemma 1, the corresponding metric space (X, ps) is also complete. Therefore, the sequence {Xn} converges to some u e X with respect to the metric ps, that is,

limn—ps(xn, u) = 0. Again, by Lemma 1, we have

p(u, u) = lim p(xn, u) = lim p(xn,xm) = 0. (2.3)

n—n,m—+^

Next, we show that p(u, Fx) < rp(u, x)

for all x e X\{u}. Sincep(xn, u) — 0 as n — there exists n0 e N such that

p(xn, u) < 3p(u,x) for all n e N with n > n0. Then we have

®(r)p(xn,Fxn) < p(xn,Fxn) <p(xn,xn+1) <p(xn, u) + p(u,xn+1) -p(u, u)

= p(xn, u) +p(u, xn+1) 2

< -p(u,x) < p(u,x) -p(xn, u)

< p(xn, x)

and hence Hp(Fxn,Fx) < rp(xn,x). Since

p(u, Fx) < p(u,xn+1) + p(xn+1, Fx)

< p(u, xn+1) +Hp(Fxn, Fx)

< p(u, xn+1) + rp(xn, x),

letting n — we obtain

p(u, Fx) < rp(u, x) (.4)

for all x e X\{u}. Next, we prove that

Hp(Fx, Fu) < rp(x, u)

for all x e X with x = u. For all n e N,we choose vn e Fx such that

p(u, vn) <p(u,Fx) + -p(x, u).

Then, using (2.4) and the previous inequality, we get

p(x, Fx) < p(x, vn) < p(x, u) + p(u, vn) -p(u, u) = p(x, u) + p(u, vn)

< p(x, u) + p(u, Fx) + 1 p(x, u)

< p(x, u) + rp(u, x) + 1p(x, u)

= (1 + r + — Ip(x, u) n

for all n e N.As n ^ we obtain ^p^,Fx) <p(x, u). From the assumption, we have Hp(Fx,Fu) < rp(x, u).

Finally, if, for some n e N,we have xn = xn+1, then xn is a fixed point of F. Assume that xn = xn+1 for all n e N. This implies that there exists an infinite subset J of N such that xn = u for all n e J. From

p(u, Fu) < p(u, xn+1) + p(xn+1, Fu)

< p(u, xn+1) +Hp(Fxn, Fu)

< p(u, xn+1) + rp(xn, u),

letting n ^ with n e J,we get p(u, Fu) = 0 = p(u, u).

By Remark 1, we deduce that u e Fu and hence u is a fixed point of F. This completes the proof. □

It is obvious that Theorem 1 of Aydi et al. follows directly from Theorem 2. The following theorem is a result of Reich type [36] as well as a generalization ofKikkawa and Suzuki type in the framework of partial metric spaces.

Theorem 3 Let (X,p) be a complete partial metric space and let F : X ^ CBp(X) be a multi-valued mapping satisfying the following:

0p(x,Fx) < p(x,y) Hp(Fx,Fy) < ap(x,y)+bp(x,Fx) + cp(y,Fy) (2.5)

for allx,y e X, nonnegative numbers a, b, c with a + b + c e [0,1) and 0 = —-+-c. Then F has a fixed point.

Proof Let h e (1, a+b+c) and x0 e X be arbitrary. Let x1 e Tx0. By Lemma 2, there exists x2 e Fx1 such that

p(x1,x2) < hHp(Fx0,Fx1).

Since 0p(x0, Fx0) < 0p(x0,x1) < p(x0,x1), we have

p(x1,x2) < hHp(Fx0,Fx1) < h(ap(x0,x1) + bp(x0,Fx0) + cp(x1,Fx1))

< h(a + b)p(x0,x1) + hcp(x1,x2) h(a + b)

-p(X0, X1).

Continuing in a similar way, we can obtain a sequence {xn} of successive approximations for F, starting from x0, satisfying the following:

(a) xn+1 e Fxn for all n e N;

(b) p(xn,xn+1) < knp(x0,x1) for all n e N,

where k = < 1. Now, proceeding as in the proof of Theorem 2, we deduce that the sequence {xn} converges to some u e X with respect to the metric ps, that is, limn—ps(xn, u) = 0. Moreover, (2.3) holds by Lemma 2. First, we show that

p(u, Fx) < la + — Jp(u, x) + cp(x, Fx)

for all x e X\{u}. Since p(xn, u) — 0 as n — under the metric p, there exists n0 e N such that

p(xn, u) < 3p(u,x)

for each n > n0. Then we have

0p(xn, Fxn) < p(xn, Fxn) < p(xn,xn+1)

< p(xn, u) + p(u,xn+1) -p(u, u)

= p(xn, u) + p(u, xn+1) 2

< -p(u,x) < p(u,x) -p(xn, u)

< p(xn,x),

which implies that

Hp(Fxn,Fx) < ap(xn,x) + bp(xn,Fxn) + cp(x,Fx) b

< ap(xn,x) + -p(xn,x) + cp(x,Fx)

( b \ = la + o jp(xn, x) + cp(x, Fx)

for all n > n0. Thus we have

p(u, Fx) < p(u,xn+1) + p(xn+1, Fx)

< p(u,xn+1) +Hp(Fxn, Fx)

< p(u,xn+1) + I a + o jp(xn,x) + cp(x, Fx)

for all n > n0. Letting n — we get ( b \

p(u, Fx) < la + o j p(u, x) + cp(x, Fx)

for all x e X\{u}.

Next, we show that

Hp(Fx,Fu) < + — Jp(x, u) + cp(u,Fu)

for all x e X with x = u. Now, for all n e N, there exists yn e Fx such that

p(u,yn) <p(u,Fx) + —p(x, u).

p(x, Fx) < p(x,yn) < p(x, u) + p(u,yn) -p(u, u) = p(x, u) + p(u, yn)

< p(x, u) + p(u, Fx) + —p(x, u)

< p(x, u)+ (a + — \p(u, x) + cp(x, Fx) + —p(x, u)

V 0\ n

= | 1 + a + b + — | p(x, u) + cp(x, Fx) \ 0 n/

for all n e N, it follows that, as n ^

( b \ (1 - c)p(x, Fx) < I 1 + a + — Jp(x, u)

and so 0p(x, Fx) < p(x, u). Thus we have

Hp(Fx,Fu) < ap(x, u) + bp(x,Fx) + cp(u,Fu)

( b \ < la + $ j p(x, u) + cp(u, Fu)

for all x e X \ {u}.

Finally, if, for some n e N,we have xn = xn+1, then xn is a fixed point of F. Assume that xn = xn+— for all n e N. This implies that there exists an infinite subset J of N such that xn = u for all n e J. Now, for all n e J,we have

p(u, Fu) < p(u,xn+—) + p(xn+—,Fu)

< p(u, xn+—) +Hp(Fxn, Fu)

< p(u,xn+—) W a + $ jp(xn, u) + cp(u, Fu).

Letting n ^ with n e J,we get p(u, Fu) = 0 = p(u, u).

By Remark 1, we deduce that u e Fu and hence u is a fixed point of F. This completes the proof. □

The following theorem is a generalization of a result of Dhompongsa and Yingtaweesit-tikul [35] to the setting of partial metric space.

Theorem 4 Let (X,p) be a complete partial metric space and let F : X — CBp(X) be a multi-valued mapping such that

0p(x,Fx) <p(x,y) Hp(Fx,Fy) < X(p(x,Fx)+p(y,Fy)) + ¡p(x,y) (2.6)

for all x,y e X, where 0 = 2X+1 ^ with X, ¡i nonnegative real numbers and 0 < 2k + ¡i < 1. Then F has a fixed point.

Proof Let h e (1, 2x+x) and x0 e X be arbitrary. Following the same proof of Theorem 3, by replacing 0 = in the proof by 0 = 2X+l +v we can obtain a sequence {xn} such that

(a) xn+1 e Fxn for all n e N;

(b) p(xn,xn+1) < knp(x0,x1) for all n e N, where k = h-hX < 1.

Now, proceeding as in the proof of Theorem 2, we deduce that the sequence {xn} converges to some u e X with respect to the metric ps, that is, limn—ps(xn, u) = 0. Again, from Lemma 2, we have

p(u, u) = lim p(xn, u) = lim p(xn,xm) = 0. (2.7)

n— + TO n— +TO

Next, we show that

p(u, Fx) < ¡xp(u,x) + Xp(x,Fx)

for all x e X\{u}. Sincep(xn, u) — 0 as n — there exists n0 e N such that p(xn, u) < 3p(u,x) for all n > n0. We have

0p(xn, Fxn) < p(xn, Fxn) < p(xn,xn+1)

< p(xn, u) + p(u,xn+1) -p(u, u)

= p(xn, u) + p(u, xn+1) 2

< 3p(u,x) < p(u,x) -p(xn, u) < p(xn,x). Now, using the conditions (2.6) and (2.7), we obtain

p(u, Fx) < p(u,xn+1) + p(xn+1, Fx)

< p(u,xn+1) +Hp(Fxn, Fx)

< p(u,xn+1) + Xp(xn, Fxn) + Xp(x, Fx) + ¡xp(xn,x)

< p(u,xn+1) + Xp(xn,xn+1) + Xp(x, Fx) + ¡xp(xn,x)

for all n > n0. Letting n — we get p(u, Fx) < Xp(x, Fx) + ¡¡p(u, x),

as desired.

Next, we show that

Hp(Fx, Fu) < Xp(x, Fx) + Xp(u, Fu) + fip(x, u) for all x e X \ {u}. By Lemma 2, for all n e N, there exists yn e Fx such that

p(u,yn) <p(u,Fx) + —p(u,x).

Clearly, we have

p(x, Fx) < p(x,yn) < p(x, u) + p(u,yn) —p(u, u) = p(x, u) + p(u, yn)

< p(x, u) + p(u, Fx) + — p(x, u)

< p(x, u) + Xp(x, Fx) + /p(u, x) + —p(x, u)

< | 1 + / + — |p(x, u) + Xp(x, Fx)

for all n e N. Hence, as n ^ we get

(1 - X)p(x, Fx) < (1 + /)p(x, u) and so &p(x, Fx) < p(x, u) since © < —. Now, using the condition (2.6), we obtain

Hp(Fx,Fu) < kp(x,Fx) + Xp(u,Fu) + /p(x, u) for all x e X \ {u}.

Finally, if, for some n e N,we have xn = xn+1, then xn is a fixed point of F. Assume that xn = xn+— for all n e N. This implies that there exists an infinite subset J of N such that xn = u for all n e J. From

p(u, Fu) < p(u,xn+—) + p(xn+—,Fu)

< p(u, xn+—) +Hp(Fxn, Fu)

< p(u, xn+—) + Xp(xn, Fxn) + Xp(u, Fu) + /p(xn, x)

< p(u, xn+—) + X11 + X p(xn, x) + Xp(u, Fz) + /p(xn, x),

1 — X

letting n ^ with n e J,we get p(u, Fu) = 0 = p(u, u).

By Remark 1, we deduce that u e Fu and hence u is a fixed point of F. This completes the proof. □

Now, we give one example to illustrate Theorem 3.

Example 1 Let X = {2,3,4} andp: X x X ^ [0, be a partial metric on X defined by

p(2,2) = p(3,3) = 0, p(4,4) = ^, p(2,3) = p(3,2) = 2,

p(2,4) = p(4,2) = 2-, p(3,4) = p(4,3) =

Let F: X ^ CBp(X) be defined by

Fx = j {2} if x e {2,3}, I {2,3} otherwise.

It is easy to see that {2} and {2,3} are closed in X with respect to the partial metric p. Now, we have

Hp(F2,F2) = Hp(F3,F3) = Hp(F2,F3) = Hp({2}, {2}) = 0; Hp(F4, F4)= Hp({2,3}, {2,3}) =0;

Hp(F 2, F4) = Hp(F 3, F4) = Hp({2}, {2,3}) = 5; p(2, F 2) = p(2, {2}) = 0; p(3, F3) = p(3, {2}) = 2; p(4, F 4)= p(4, {2,3}) =

If we choose a = b = | and c = 10, the multi-valued mapping F satisfies the hypotheses of Theorem 3 and so has a fixed point. To such end, it is enough to show that (2.5) is satisfied in the following cases. Case 1. x = 2 and y = 4. Now, 0p(2, F2) < p(2,4), where 0 = § and

Hp(F2, F4) = 2 < 50 < 4P(2,4) + ^ F2) + ¿P(4, F4). Case 2. x = 3 andy = 4. Now, 0p(3, F3) < p(3,4) and

Hp(F3, F4) = 2 < 20 < 4P(3,4) + 3P(3, F3) + ^(4, F4). Case 3. x = 4 and y =3. Now, 0p(4, F4) < p(4,3) and

Hp(F4, F3) = 2 < 20 < |p(4,3) + 3P(4,F4) + i0p(3,F3).

Case 4. x = 4 and y = 2. Now, 0p(4, F4) < p(4,2) and

2 43 3 1 1

Hp(F4, F2) = - <-< —p(4,2) + -p(4, F4) + — p(2, F2).

py 5 " 100 " 4F ' ' r 10^

Thus all the conditions of Theorem 3 are satisfied. Here x = 2 is a fixed point of F. On the other hand, the metric ps induced by the partial metric p is given by

ps(1,1) = ps(2,2) = ps(3,3) = 0, ps(2,3) = ps(3,2) = 4,

Ps (4,3) = ps (3,4) = 330, Ps (4,2) = ps (2,4) = 20.

Note that, in the case of an ordinary Hausdorff metric, the given mapping does not satisfy the condition (2.5). Indeed, for x = 2 and y = 4, the condition Qps (2, F2) < ps(2,4) is satisfied. But the condition H(F2, F4) < aps (2,4) + bps (2, F2) + cps (4, F4) is not satisfied. In fact, we have

Competing interests

The authors declare that they have no competing Interests. Authors' contributions

Allauthors read and approved the finalmanuscript. Author details

1 Department of Mathematics COMSATS, Institute of Information Technology, Chack Shahzad, Islamabad 44000, Pakistan. 2Dipartimento di Matematica e Informatica, Université degli Studi di Palermo, Via Archirafi 34, Palermo, 90123, Italy. 3Department of Mathematics Education and RINS, Gyeongsang NationalUniversity, Jinju, 660-701, Korea. 4Department of Mathematics, InternationalIslamic University, Islamabad, Pakistan.

Acknowledgements

The second author was supported by Université degli Studi di Palermo, LocalUniversity Project R. S. ex 60% and the third author was supported by the Basic Science Research Program through the NationalResearch Foundation of Korea funded by the Ministry of Education, Science and Technology (NRF-2012-0008170).

Received: 6 February 2013 Accepted: 9 June 2013 Published: 3 July 2013

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doi:10.1186/1687-1812-2013-175

Cite this article as: Ahmad et al.: Some fixed point results for multi-valued mappings in partial metric spaces. Fixed PointTheory and Applications 2013 2013:175.

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