0 Fixed Point Theory and Applications

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Fixed and coupled fixed points of a new type set-valued contractive mappings in complete metric spaces

A Amini-Harandi*

Correspondence: aminih_a@yahoo.com Department of Pure Mathematics, University of Shahrekord, Shahrekord, 88186-34141, Iran SchoolofMathematics, Institute for Research in FundamentalSciences (IPM), P.O. Box 19395-5746,Tehran, Iran

Abstract

In this paper, motivated by the recent work ofWardowski (Fixed Point Theory Appl. 2012:94, 2012), we introduce a new concept of set-valued contraction and prove a fixed point theorem which generalizes some well-known results in the literature. As an application, we derive a new coupled fixed point theorem. Some examples are also given to support our main results. MSC: 47H10

Keywords: fixed point; coupled fixed point; set-valued contractions

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1 Introduction

In the literature, there are plenty of extensions of the famous Banach contraction principle [1], which states that every self-mapping T defined on a complete metric space (X, d) satisfying

d(Tx, Ty) < kd(x,y) for each x,y e X, (1)

where k e [0; 1), has a unique fixed point, and for every x0 e X, the sequence {T"x0}„eN is convergent to the fixed point. Some of the extensions weaken the right side of the inequality in the condition (1) by replacing k with a mapping; see, e.g., [2-4]. In other results, the underlying space is more general; see, e.g., [5-8]. In 1969, Nadler [9] extended the Banach contraction principle to set-valued mappings. For other extensions of the Banach contraction principle, see [10-21] and the references therein.

Recently, Wardowski [22] introduced a new concept of contraction and proved a fixed point theorem which generalizes the Banach contraction principle in a different way than in the known results from the literature. In this paper, we present an improvement and generalization of the main result ofWardowski [22]. To set up our results, in the next section, we introduce some definitions and facts.

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) = max] sup d(u, B), sup d(v, A)

^ueA veB

ringer

for every A, B e CB(X), where d(u, B) = inf{d(u, y):y e B}.

© 2012 Amini-Harandi; 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 original work is properly cited.

Theorem 1.1 (Nadler [9]) Let (X, d) be a complete metric space and letT: X — CB(X) be a set-valued map. Assume that there exists k e [0,1) such that

H(Tx, Ty) < kd(x,y) for each x,y e X. (2)

Then T has a fixed point.

In 1989 Mizoguchi and Takahashi [13] proved the following generalization of Theorem 1.1.

Theorem 1.2 (Mizoguchi and Takahashi [13]) Let (X, d) be a complete metric space and letT: X — CB(X) be a set-valued map satisfying

H(Tx, Ty) < a(d(x,y))d(x,y) for each x,y e X,

where a : [0, to) — [0,1) satisfies limsupt—r+ a(t) < 1 for each r e [0, to). Then T has a fixed point.

2 Main results

Let F :(0, to) — R and 0 : (0, to) — (0, to) be two mappings. Throughout the paper, let A be the set of all pairs (F, 0) satisfying the following:

(51) 0(tn) — 0 for each strictly decreasing sequence {tn};

(52) F is strictly increasing;

(53) For each sequence {an}neN of positive numbers, limn—TO an = 0 if and only if limn-to F (an) = -to;

(54) If tn | 0 and 0 (tn) < F(tn) - F(tn+1) for each n e N, then Y,TO=1 tn < to.

Example 2.1 Let 01(t) = t for each t e (0, to), where t > 0 is a constant, and let F1 : (0, to) — R be a mapping satisfying limx—0+ xkF(x) = 0 for some k e (0,1) where F : (0, to) — R is strictly increasing. Then the proof of the main result in [22] shows that (F1,01) e A. We give the details for completeness. Using (54), the following holds for every n e N:

F(tn) < F(tn-1) - t < F(tn-2) - 2t < ••• < F(tc) - nT. (3)

By (3), the following holds for every n e N:

tlF(tn) - tlF(t0) < tl(F(t0) - nT) - (*F(t0) = -tknT < 0. (4)

Since limn—TO t^F(tn) = 0, letting n — to in (4), we obtain limn—TO ntn = 0. Then there exists n1 e N such that ntn < 1 for n > n1. Consequently, we have tn < for all n > n1.

Thus^to=1 tn < to (note that £^ ^r < to).

Example 2.2 Let F2(t) = lnt and let 02(t) = -ln(a(t)) for each t e (0,to), where a : (0, to) — [0,1) satisfying

limsupa(t)<1 foreachre [0,to).

t—r+

Now, we show that (F2,e2) e A. It is easy to see that F2 and d2 satisfy (5i)-(53). To show (54), assume that tn | 0 and

- ln (a (tn)) < ln tn - ln tn+1 Vn e N.

Then tn+1 < a(tn)tn for each n e N. Since limsupt^0+ a(t) < 1, then there exist n0 e N and 0 < r <1 such that a(tn) < r for n > n0. Thus, tn+1 < rtn for each n > n0, and so ^tn < to.

Example 2.3 Let F3(t) = lnt + t and let e3(t) = t for each t e (0,to), where t > 0 is a constant. Now, we show that (F3, &3) e A. We only show (54). Suppose that tn ^ 0 and

t < (ln tn + tn) - (ln tn+1 + tn+1) Vn e N.

sn+1 < e-T sn Vn e N,

where sn = tnetn. Since e-T < 1, then from the above we get ^TO1 sn < to, and so ^TO1 tn < to (note that tn < sn for each n e N).

Now, we state the main result of the paper.

Theorem 2.4 Let (X, d) be a complete metric spaces, let T : X ^ CB(X) be a set-valued mapping and let (F, |) e A. Assume that either T is compact valued or F is continuous from the right. Furthermore, assume that

e (d(x,y)) + F(H(Tx, Ty)) < F(d(x,y)) Vx,y e X with Tx = Ty. (5)

Then T has a fixed point.

Proof Let x0 e X and x1 e Tx0. If Tx0 = Tx1,thenx1 e Tx0 = Tx1 and x1 is a fixed point of T. So, we may assume that Tx0 = Tx1. Since either T is compact valued or F is continuous from the right, x1 e Tx0 and F(d(xu Txi)) < F(H(Tx0, TxO) + e(d(x20,x1)) then there exists x2 e Tx1 such that (note that F is increasing)

F(d(x1,x2)) < F(H{Tx0, Tx1)) + e{d{x°x1)). (6)

From (5) and (6), we have

e (d(x0, x1)) + F(d(x1, x2))

-niM W riucr TW e(d(x0,x1)) ^ riM W e(d(x0,x1)) < e(d(x0,x1)) + F(H(Tx0, Tx1)) +-2-< F(d(x0,x1)) +-2-,

and so

(d(xo,xi)) + f^d(x1,x2)) < F(d(xo,xi)).

We may also assume that Tx1 = Tx2 (otherwise, x2 e Tx2 = Tx1). Proceeding this manner, we can define a sequence {xn} in X satisfying

xn+1 e Txn, (^ < F(tn) - F(tn+i), for each n e N, (8)

where tn = d(xn,xn+1). Since 0(tn) > 0 then from (8), we have F(tn) > F(i«+i) for each n e N. Since F is strictly increasing, then we deduce that {tn} is a nonnegative strictly decreasing sequence and so is convergent to some r > 0, limn—TO tn = r. Now we show that r = 0. On the contrary, assume that r >0. From (8), we get

- J2 0 t) ^ F(ti) - F(tn+i) for each n e N. (9)

Since {tn} is strictly decreasing, then from (51) we get 0 (tn) — 0. Thus, ^°=10 (ti) = to, and then from (9), we have limn—TO F(tn) = -to. Then by (53), tn — 0, a contradiction. Hence,

lim tn = 0. (10)

n—>to

From (8), (10) and (54), we have

^ ti = ^2, d(xi,Xi+1) < to. 1=1 i=1

Then, by the triangle inequality, {xn} is a Cauchy sequence. From the completeness of X, there exists x e X such that limn^TO xn = x. Now, we prove that x is a fixed point of T. To prove the claim, we may assume that Txn = Tx for sufficiently large n e N. On the contrary, assume that Txni = Tx for each i e N. Since Tx is closed, xni+1 e Txni = Tx and xni+1 — x, then x e Tx, and we are finished.

From (5), we have (note that xn+1 e Txn and Txn = Tx for n > N)

F(d(xn+1, Tx)) < 0(d(xn,x)) + F(d(xn+1, Tx))

< 0 (d(xn,x)) + F(H(Txn, Tx)) < F(d(xn,x)). (11)

Since d(xn,x) — 0, then (11) together with (53) imply that

d(x, Tx) = lim d(xn+1, Tx) = 0, and so d(x, Tx) = 0. Hence, x e Tx (note that Tx is closed). □

Remark 2.5 By Example 2.1, Theorem 2.4 is an extension and improvement of Theorem 2.1 of Wardowski [22]. From Example 2.2, we infer that Theorem 2.4 is a generalization of the above mentioned Theorem 1.2 of Mizoguchi and Takahashi.

Now, we illustrate our main result by the following example.

Example 2.6 Consider the complete metric space (X = {0,1,2,3,...},d), where d is defined as

d(x, y) =

0, x = y, x + y, x = y.

Let T : X ^ CB(X) be defined as

{0,1,2,3,...}, x = 0, {x-1,x,x + 1,...}, x >0.

Let f : X ^ X be given by

0, x = 0, x -1, x > 0.

Now, we show that T satisfies (5), where e (t) = 1 for each t e (0, to) and F(x) = ln x + x for each x e (0, to). To show the claim, notice first that H(Tx1, Tx2) = d(fx1,fx2) for each x1,x2 e X. Now let x1,x2 e X withfx1 =fx2. Since d(fx1,fx2) - d(x1,x2) < -1, then we have

dfc1,Jx2} edfx1fx2)-d(x1x2) < e-1, for each x1,x2 e X with/»1 =fx2. d(x1, x2)

Thus, from the above, we have

1 < [ln d(x1,x2) + d(x1,x2)] - [ln d(fx1,fx2) + d(fx1,fx2)] = F(d(x1,x2)) - F(d(fx1,fx2^.

Therefore, (note that H(Tx1, Tx2) = d(fx1,fx2))

1 < F(d(xltx2)) - F(H(Tx1, Tx2)).

Then, by Theorem 2.4, T has a fixed point.

Now, we show that T does not satisfy the condition of Nadler's theorem. On the contrary, assume that there exists a function k e [0,1) such that

H(Tx1, Tx2) < kd(x1,x2)

for all x1, x2 e X. Then

d(fx1,fx2) < kd(x1,x2).

Then, for each x1 > 1 and x2 = x1 + 1, we have

2x1 -1 < k(2x1 +1), for each x1 > 1.

Hence,

! T -X1-1 „ , 1 = lim -< k,

X1—TO 2x1 + 1

a contradiction.

Example 2.7 For each t e (0, to), let F4(t) = -1 and let

04(i) =

- ^, 0< t <1, 1, 1 < t.

Then it is easy to see that (F4,04) e A, but F4 does not satisfy the condition (F3) of the definition of F-contraction in [22].

Now, by using the technique in [23], we present a new coupled fixed point result. For more details on coupled fixed point theory, see [23-25] and the references therein.

Corollary 2.8 Let (M, p) be a complete metric space and let (F, |) e A. Letf: M x M — M be a mapping satisfying

e(p (x, u) + p(y, v)) + F (pf (x, y),f (u, v)) + pf (y, x),f (v, u)))

< F( p (x, u) + p (y, v)) (12)

for each x, y, u, v e M. Thenf has a coupled fixed point (x0, y0), that is, f (x0, y0) = x0 and f (yo, xo) =yo.

Proof Let X = M x M and let d be the metric on M which is defined by

d( (x, y), (u, v)) = p(x, u) + p(y, v).

Then it is straightforward to show that (X, d) is a complete metric space. Let T : X — X be defined by T(x,y) = (f (x,y),f (y,x)). From (12), we get

0(d((x,y), (u, v))) + F(d(T(x,y), T(u, v))) < F(d((x,y),(u, v)))

for each (x,y), (u, v) e X. Then from Theorem 2.4 we deduce that T has a fixed point u0 = (x0, y0). Then (x0, y0) is a coupled fixed point off. □

Competing interests

The author declares that they have no competing interests. Acknowledgements

The author is gratefulto the referees for their helpfulcomments leading to improvement of the presentation of the work. This work was supported by the University of Shahrekord and by the Center of Excellence for Mathematics, University of Shahrekord, Iran. This research was also in part supported by a grant from IPM (No. 91470412). This research is partially carried out in the IPM-Isfahan Branch.

Received: 20 June 2012 Accepted: 7 November 2012 Published: 26 November 2012

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doi:10.1186/1687-1812-2012-215

Cite this article as: Amini-Harandi: Fixed and coupled fixed points ofa new type set-valued contractive mappings in complete metric spaces. Fixed PointTheory and Applications 2012 2012:215.

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