00 Fixed Point Theory and Applications
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Fixed point results for single and set-valued a-n-ty-contractive mappings
Nawab Hussain1, Peyman Salimi2 and Abdul Latif1*
Dedicated to Professor Wataru Takahashi on the occasion of his seventieth birthday
"Correspondence: alatif@kau.edu.sa 1 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia Full list of author information is available at the end of the article
Abstract
Samet etal. (Nonlinear Anal. 75:2154-2165,2012) introduced a-ty-contractive mappings and proved some fixed point results for these mappings. More recently Salimi etal. (Fixed Point Theory Appl. 2013:151, 2013) modified the notion of a-ty-contractive mappings and established certain fixed point theorems. Here, we continue to utilize these modified notions for single-valued Geraghty and Meir-Keeler-type contractions, as well as multi-valued contractive mappings. Presented theorems provide main results of Hussain etal. (J. Inequal. Appl. 2013:114, 2013), Karapinar etal. (Fixed Point Theory Appl. 2013:34,2013) and Asl etal. (Fixed Point Theory Appl. 2012:212, 2012) as corollaries. Moreover, some examples are given here to illustrate the usability of the obtained results. MSC: 46N40; 47H10; 54H25; 46T99
Keywords: modified Meir-Keeler-type contractions; triangular a-admissible map; a-n-ty-contractive map; metric space
ft Springer
1 Introduction and preliminaries
In metric fixed point theory, the contractive conditions on underlying functions play an important role for finding solution of fixed point problems. Banach contraction principle is a remarkable result in metric fixed point theory. Over the years, it has been generalized in different directions by several mathematicians (see [1-21]). In 2012, Samet et al. [15] introduced the concepts of a-ty-contractive and a-admissible mappings and established various fixed point theorems for such mappings in complete metric spaces. Afterwards, Karapinar and Samet [13] generalized these notions to obtain fixed point results. More recently, Salimi etal. [14] modified the notions of a-ty-contractive and a-admissible mappings and established fixed point theorems, which are proper generalizations of the recent results in [13, 15]. Here, we continue to utilize these modified notions for single-valued Geraghty and Meir-Keeler-type contractions, as well as multivalued contractive mappings. Presented theorems provide main results of Hussain etal. [9], Karapinar etal. [11] and Asl etal. [12] as corollaries. Moreover, some examples are given here to illustrate the usability of the obtained results.
Denote with ^ the family of nondecreasing functions ty : [0, +cc) ^ [0, +c) such that XXi tyn(t) < for all t >0, where tyn is the nth iterate of ty. The following lemma is obvious.
© 2013 Hussain 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.
Lemma 1.1 If f e ^, then f (t) < tfor all t > 0.
Samet etal. [15] defined the notion of a-admissible mappings as follows.
Definition 1.1 Let T be a self-mapping on X, and let a : X x X ^ [0, +œ) be a function. We say that T is an a-admissible mapping if
x,y e X, a(x,y) > 1 a(Tx, Ty) > 1.
Theorem 1.1 [15] Let (X, d) be a complete metric space, and let T be an a-admissible mapping. Assume that
a(x,y)d(Tx, Ty) < f (d(x,y)) (1.1)
for all x, y e X, where f e ^. Also, suppose that
(i) there exists x0 e X such that a(x0, Tx0) > 1;
(ii) either T is continuous or for any sequence {xn} in X with a(xn, xn+i) > 1 for all n e N U{0} and xn ^ x as n ^ we have a(xn,x) > 1 for all n e N U {0}.
Then T has a fixed point.
Very recently Salimi etal. [14] modified the notions of a-admissible and a-f -contractive mappings as follows.
Definition 1.2 [14] Let T be a self-mapping on X, and let a, n : X x X ^ [0, +œ) be two functions. We say that T is an a-admissible mapping with respect to n if
x,y e X, a(x,y) > n(x,y) a(Tx, Ty) > n(Tx, Ty).
Note that if we take n(x,y) = 1, then this definition reduces to Definition 1.1. Also, if we take a(x, y) = 1, then we say that T is n-subadmissible mapping.
The following result properly contains Theorem 1.1 and Theorems 2.3 and 2.4 of [13].
Theorem 1.2 [14] Let (X, d) be a complete metric space, and let T be an a-admissible mapping with respect to n. Assume that
x,y e X, a(x,y) > n(x,y) d(Tx, Ty) < ^(M(x,y)), (1.2)
where ^ e ^ and
M(x, y) = max|d(x, y), d(x, Tx) + d(y,Ty), d(x,Ty) + d(y,Tx)
Also, suppose that the following assertions hold:
(i) there exists xo e X such that a(xo, Txo) > n(xo, Txo);
(ii) either T is continuous or for any sequence {xn} in X with a(xn, xn+1) > n(xn, xn+1) for
all n e N U {o} and xn ^ x as n ^ +<x>, we have a ( xn , x ) > n(xn,x) for all n e N U{o}. Then T has a fixed point.
2 Modified a-ty-Geraghty type contractions
Our first main result of this section is concerning a-n-Geraghty-type [4] contractions.
Theorem 2.1 Let (X, d) be a complete metric space, and letf: X ^ X be an a-admissible mapping with respect to n. Assume that there exists a function j : [o, (x>) ^ [o, 1) such that for any bounded sequence {tn} of positive reals, j (tn) ^ 1 implies that tn ^ o and
x, y e X, a(x,fx)a (y,fy) > n (x,fx) n (y,fy)
d(fx,fy) < j(d(x,y)) ma^d(x,y),min{d(x,fx),d(y,fy)}}.
Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn ^ x, a(xn, xn+1) > n(xn, xn+1) for all n, then a(x,fx) > n(x,fx).
If there exists xo e X such that a(xo,fxo) > n(xo,fxo), thenf has a fixed point.
Proof Let xo e X such that a(xo,fxo) > n(xo,fxo). Define a sequence {xn} in X by xn = fnxo = fxn-1 for all n e N. If xn+1 = xn for some n e N, then x = xn is a fixed point for f, and the result is proved. Hence, we suppose that xn+1 = xn for all n e N. Since f is an a-admissible mapping with respect to n and a(xo,fxo) > n(xo,fxo), we deduce that a(x1,x2) = a(fxo,f2xo) > n(fxo,f2xo) = n(x1,x2). By continuing this process, we get a(xn,fxn) > n(xn,fxn) for all n e N U {o}. Then,
a(xn-1,fxn-1)a(xn,fxn) > n(xn-1,fxn-1)n(xn,fxn).
Now from (2.1), we have
d(xn,xn+1) < j(d(xn-1,xn)) ma^jd(xn-1,xn),min{d(xn-1,fxn-1),d(xn,fxn)}} = j (d(xn-1, xn)) ma^jd(xn-1, xn), min { d(xn-1, xn), d(xn, xn+i)}}.
Now, if d(xn-1, xn) < d(xn, xn+1) for some n e N, then
max{ d(xn-1, xn), min {d(xn-1, xn) , d (xn, xn+1)}} — d (xn— 1, xn).
Also, if d(xn,xn+1) < d(xn-1,xn) for some n e N, then
max{ d(xn-1, xn), min { d(xn-1, xn) , d (xn, xn+1)}} — d (xn— 1, xn).
That is, for all n e N,we have
max{ d(xn-1, xn), min { d(xn-1, xn) , d (xn, xn+1)}} — d (xn— 1, xn).
Hence,
d(xn,xn+1) < p(d(xn-1,xn))d(xn-1,xn) (2.2)
for all n e N, which implies that d(xn,xn+1) < d(xn-1,xn). It follows that the sequence {d(xn,xn+1)} is decreasing. Thus, there exists d e R+ such that limn—TO d(xn,xn+1) = d. We shall prove that d = 0. From (2.2), we have
^^ xn+l) < P (d(xn_i, xn)) < 1, d (x n_1, xn)
which implies that limn—TO p (d(xn-1, xn)) = 1. Regarding the property of the function p, we conclude that
lim d(xn,xn+1) = 0. (2.3)
Next, we shall prove that {xn} is a Cauchy sequence. Suppose, to the contrary, that {xn} is not a Cauchy sequence. Then there is e >0 and sequences {m(k)} and {n(k)} such that for all positive integers k, we have
n(k) > m(k) > k, d(xn(k),xm(k)) > e and d(xn(k),xm(k)-1) < e.
By the triangle inequality, we derive that
e < d(xn(k),xm(k)) < d(xn(k),xm(k)-1) + d(xm(k)-1,xffl(i)) < e + d(xm(k)-1, xm(k))
k e N. Taking the limit as k — in the inequality above, and regarding the limit in (2.3), we get
lim d(xn(k),xm(k)) = e. (.4)
Again, by the triangle inequality, we find that
d(xn(k), xm(k)) < d(xm(k), xm(k)+1) + d(xm(k)+1, xn(k)+1) + d(xn(k)+1, xn(k))
d(xn(k)+1,xm(k)+1) < d(xm(k),xm(k)+1) + d(xm(k),xn(k)) + d(xn(k)+1,xn(k)).
Taking the limit in inequality above as k — together with (2.3) and (2.4), we deduce that
lim d(xn(k)+1,xm(k)+1)=e. (.5)
Now, since
a(xn(k),fxn(k))a(xm(k),fxm(k)) > n(xn(k),fxn(k))n(xm(k),fxm(k)),
then from (2.1), (2.4) and (2.5), we have d(xn(k)+1, xm(k)+1)
< p(d(xn(k), xm(k))) ma^j d(xn(k), xm(k)), min{ d(xn(k),fxn(k)), d(xm(k),fxm(k))}} = j( d(xn(k), xm(k))) maxj d(xn(k), xm(k)), min{ d(xn(k), xn(k)+1), d(xm(k), xm(k)+1)}}.
Hence,
d(xn(k)+1, xm(k)+1)
< p(d(xn(k), xm(k)i) < 1.
max{d(xn(k),xm(k)), min{d(xn(k),xn(k)+1), d(xm(k),xm(k)+1)}} v n ' m Letting k — to in the inequality above, we get
lim j(d(xn(k),xm(k))) = 1.
That is, limk—TO d(xn(k),xm(k)) = o, which is a contradiction. Hence {xn} is a Cauchy sequence. Since X is complete, then there is z e X such that
xn ^ z. First, we suppose thatf
is continuous. Since f is continuous, then we have
fz = lim fxn = lim xn+1 = z.
n—n—
So z is a fixed point off. Next, we suppose that (b) holds. Then, a(z,fz) > n(z,fz), and so, a(z,fz)a(xn,fxn) > n(z,fz)n(xn,fxn). Now by (2.1), we have
dfz,xn+1) < j(d(z,xn)) maxjd(z,xn),min{d(z,fz),d(xn,x^)}},
and hence
dfz, z) < dfz, xn+1) +d(z, xn+1)
< j(d(z,xn)) max{d(z,xn),min{d(z,fz),d(xn,xn+1^} + d(z,xn+1).
Letting n —to in the inequality above, we get dfz, z) = o, that is, z = fz. □
If in Theorem 2.1 we take, n(x,y) = 1, then we have the following corollary.
Corollary 2.1 Let (X, d) be a complete metric space, and letf: X — Xbean a-admissible mapping. Assume that there exists a function j : [o, to) — [o, 1] such that for any bounded sequence {tn} of positive reals, j (tn) — 1 implies that tn — o and
x, y e X, a(x,fx)a(y,fy) > 1
dfx,fy) < j(d(x,y)) ma^d(x,y),min{d(x,fx),d(y,fy)\\.
Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, a(xn,xn+1) > 1 for all n, then a(x,fx) > 1. If there exists xo e X such that a(xo,fxo) > 1, thenf has a fixed point.
Corollary 2.2 Let (X, d) be a complete metric space, and letf : X ^ Xbean a-admissible mapping. Assume that there exists a function j : [0, c) ^ [0,1] such that for any bounded sequence {tn} of positive reals, j (tn) ^ 1 implies that tn ^ 0 and
{dfx,fy) + iY(xSx)a(ySy) < jj(d(x,y)) max{d(x,y), min{d(x,fx), d(y,fy)}} +1
for allx,y e X, where I >0. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, a(xn,xn+1) > 1 for all n, then a(x,fx) > 1. If there exists x0 e X such that a(x0,fx0) > 1, thenf has a fixed point.
Corollary 2.3 Let (X, d) be a complete metric space, and letf: X — Xbean a-admissible mapping. Assume that there exists a function p : [0, (x>) — [0,1] such that for any bounded sequence {tn} of positive reals, p (tn) — 1 implies that tn — 0 and
(a(x,fx)a(y,fy) + l)dfxfy) < 1&(d(x,y)) max(d(x,y),min(d(x/x),d(yfy)}}
for all x, y e X. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn ^ x, a(xn, xn+i) > 1 for all n, then a(x,fx) > 1. If there exists x0 e X such that a(x0,fx0) > 1, thenf has a fixed point.
Corollary 2.4 Let (X, d) be a metric space such that (X, d) is complete andf: X — X be an a-admissible mapping. Assume that there exists a function p : [0, <x>) — [0,1] such that for any bounded sequence {tn} of positive reals, p (tn) — 1 implies that tn — 0 and
a(x,fx)a(y,fy)d(fx,fy) < ß(d(x,y)) ma^jd(x,y),min{d(x,fx),d(y,fy)}}
for all x, y e X. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn ^ x, a(xn,fxn) > 1 for all n, then a(x,fx) > 1. If there exists xo e X such that a(xo,fxo) > 1, thenf has a fixed point.
Further, if in Theorem 2.1 we take a(x,y) = 1, then we have the following corollary.
Corollary 2.5 Let (X, d) be a complete metric space, and letf : X ^ Xbea n -subadmissible mapping. Assume that there exists a function j : [0, c) ^ [0,1] such that for any bounded sequence {tn} of positive reals, j (tn) ^ 1 implies that tn ^ 0 and
x, y e X, n (x,fx) n (y,fy) < 1
d(fx,fy) < ß(d(x,y)) max{d(x,y),min{d(x,fx),d(y,fy)\\.
Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn ^ x, n(xn, xn+1) < 1 for all n, then n(x,fx) < 1. If there exists xo e X such that n(xo,fxo) < 1, thenf has a fixed point.
Corollary 2.6 Let (X, d) be a complete metric space, and letf: X — X be a n-subadmis-sible mapping. Assume that there exists a function j : [o, to) — [o, 1] such that for any bounded sequence {tn} of positive reals, j(tn) — 1 implies thattn — o and
d(fx,fy) + l < [jj(d(x,y))max{d(x,y),min{d(x,fx),d(yfy)}} + ¿]n(xfx)n(yfy)
for allx,y e X, where I > o. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, n(xn,xn+i) < 1 for all n, then n(x,fx) < 1. If there exists xo e X such that n(xo,fxo) < 1, thenf has a fixed point.
Corollary 2.7 Let (X, d) be a complete metric space, and letf: X — Xbea n-subadmissible mapping. Assume that there exists a function j : [o, to) — [o, 1] such that for any bounded sequence {tn} of positive reals, j (tn) — 1 implies that tn — o and
2d(fxfy) < (n(x,fx)n(y,fy) + 1)j(d(x,y) max{d(x,y),min{d(xfx),d(yfy)}}
for all x, y e X. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, n(xn,xn+1) < 1 for all n, then n(x,fx) < 1. If there exists xo e X such that n(xo,fxo) < 1, thenf has a fixed point.
Corollary 2.8 Let (X, d) be a metric space such that (X, d) is complete, and letf: X — X be n-subadmissible mapping. Assume that there exists a function j : [o, to) — [o, 1] such that for any bounded sequence {tn} of positive reals, j (tn) — 1 implies that tn — o and
dfx,fy) < n(x,fx)n(y,fy)j(d(x, y)) max{ d(x, y), min{ d(x,fx), d(y,fy)}}
for all x, y e X. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, n(xn,fxn) < 1 for all n, then n(x,fx) < 1. If there exists xo e X such that n(xo,fxo) < 1, thenf has a fixed point.
From Corollary 2.1, we can deduce the following corollary.
Corollary 2.9 Let (X, d) be a complete metric space, and letf: X — Xbean a-admissible mapping. Assume that there exists a function j : [o, to) — [o, 1] such that for any bounded sequence {tn} of positive reals, j (tn) — 1 implies that tn — o and
x,y e X, a(x,fx)a(y,fy) > 1 d(fx,fy) < j(d(x,y))d(x,y).
Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, a(xn,xn+1) > 1 for all n, then a(x,fx) > 1. If there exists xo e X such that a(xo,fxo) > 1, thenf has a fixed point.
Also, from the corollary above, we can deduce the following corollaries.
Corollary 2.10 (Theorem 4 of [9]) Let (X, d) be a complete metric space, and letf: X — X be an a-admissible mapping. Assume that there exists a function p : [0, x) — [0,1] such that for any bounded sequence {tn} of positive reals, p (tn) — 1 implies that tn — 0 and
(d(fx,fy) + i)a{xfx)a{f < p(d(x,y))d(x,y) + £
for all x, y e X, where I > 1. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, a(xn,xn+1) > 1 for all n, then a(x,fx) > 1. If there exists x0 e X such that a(x0,fx0) > 1, thenf has a fixed point.
Corollary 2.11 (Theorem 6 of [9]) Let (X, d) be a complete metric space, and letf: X — X be an a-admissible mapping. Assume that there exists a function p : [0, x) — [0,1] such that for any bounded sequence {tn} of positive reals, p (tn) — 1 implies that tn — 0 and
(a(x,fx)a(y,fy) + 1)<Wy) < 2p(d(x*y)
for all x, y e X. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, a(xn,xn+1) > 1 for all n, then a(x,fx) > 1. If there exists x0 e X such that a(x0,fx0) > 1, thenf has a fixed point.
Corollary 2.12 (Theorem 8 of [9]) Let (X, d) be a metric space such that (X, d) is complete, and letf : X — X be an a-admissible mapping. Assume that there exists a function p : [0, x) — [0,1] such that for any bounded sequence {tn} of positive reals, p(tn) — 1 implies that tn — 0 and
a(x,fx)a(y,fy)d(fx,fy) < p(d(x,y))d(x,y)
for all x, y e X. Suppose that either
(a) f is continuous, or
(b) if {xn} is a sequence in X such that xn — x, a(xn,fxn) > 1 for all n, then a(x,fx) > 1. If there exists x0 e X such that a(x0,fx0) > 1, thenf has a fixed point.
Example 2.1 Let X = [0, x) be endowed with the usual metric d(x,y) = \x -y \ for all x,y e X, and letf: X — X be defined by
4x if x e [0,1],
ln(x2 + x + 3) ifx e (1, x). Define also a : X x X — [0, +x) and ty : [0, x) — [0, x) by
a(x, y) =
6 ifx,y e [0,1], 1
and p (t) = -. 0 otherwise 2
We prove that Corollary 2.9 can be applied to f, but Corollaries 2.10, 2.11 and 2.12 (Theorem 4, 6 and 8 of [9]) cannot be applied to f.
Clearly, (X, d) is a complete metric space. We show that f is an a-admissible mapping. Let x, y e X with a(x, y) > 1, then x, y e [o, 1]. On the other hand, for all x e [o, 1], we have fx < 1. It follows that a(fx,fy) > 1. Hence, the assertion holds. Also, a(o,fo) > 1. Now, if {xn} is a sequence in X such that a(xn,xn+1) > 1 for all n e N U {o} and xn — x as n — +to, then {xn} c [o, 1], and hence x e [o, 1]. This implies that a(xn,x) > 1 for all n e N.
Let a(x,y) > 1. Then x,y e [o, 1]. We get,
11 -x — y
= 4 |x -y| < 2 |x -y| = j(d(x,y))d(x,y).
d(fx,fy) = fy -fx| = That is,
a(x,y) > 1 d(fx,fy) < j(d(x,y))d(x,y),
then the conditions of Corollary 2.1 hold, and f has a fixed point. Let x = o, y = 1, and let I = 1, then
(d(f o,f 1) + 1)a(ofo)a(1f1) = (1/4 + 1)36 > 1/2 + 1 = jj(d(o, 1))d(o, 1) + 1.
That is, Corollary 2.1o (Theorem 4 of [9]) cannot be applied for this example. Let, x = o, and let y = 1, then
(a(o,f o)a(1,f 1) + 1)d(fof1) = \/37 > V2 = 2j(d(o,1))d(o,1).
That is, Corollary 2.11 (Theorem 6 of [9]) cannot be applied for this example. Let, x = o, and let y = 1, then
a(o,f o)a(1,f 1)d(fo,f 1) = 9 > 1/2 = j(d(o,1))d(o,1).
That is, Corollary 2.12 (Theorem 8 of [9]) cannot be applied for this example.
3 Modified -Meir-Keeler contractive mappings
Recently, Karapinar etal. [11] introduced the notion of a triangular a-admissible mapping as follows.
Definition 3.1 [11] Let f : X — X, and let a : X x X — (-to,+to). We say that f is a triangular a-admissible mapping if (Tl) a(x,y) > 1 implies that a(fx,fy) > 1, x,y e X;
I a (x, z) > 1,
_ ' implies that a(x,y) > 1. a(z, y) > 1
Lemma 3.1 [11] Letf be a triangular a-admissible mapping. Assume that there exists xo e X such that a(xo,fxo) > 1. Define sequence {xn} by xn = fnxo. Then
a(xm,xn) > 1 for all m,n e N withm < n.
Denote with ^ the family of nondecreasing functions ^ : [o, +to) — [o, +to) continuous at t = o such that
• ty (t) = 0 if and only if t =0,
• ty(t + s) < ty(t) + ty(s).
Definition 3.2 [11] Let (X, d) be a metric space, and let ty e ^. Suppose thatf: X — X is a triangular a-admissible mapping satisfying the following condition: for each e >0 there exists S >0 such that
e < ty(d(x,y)) < e + S implies that a(x,y)ty(d(fx,fy)) < e (3.1)
for all x,y e X. Thenf is called an a-ty-Meir-Keeler contractive mapping.
Now, we modify Definition 3.2 as follows.
Definition 3.3 Let (X, d) be a metric space, and let ty e ^. Suppose thatf: X — X is a triangular a-admissible mapping satisfying the following condition: for each e >0 there exists S >0 such that
e < ty(d(x,y)) < e + S implies that ty(d(fx,fy)) < e (3.2)
for all x,y e X with a(x,y) > 1. Thenf is called a modified a-ty-Meir-Keeler contractive mapping.
Remark 3.1 Let f be a modified a-ty-Meir-Keeler contractive mapping. Then
ty(d(fx,fy)) < ty (d(x,y))
for all x,y e X when x = y and a(x,y) > 1. Also, if x = y and a(x,y) > 1, then d(fx,fy) = 0, i.e.,
ty(d(fx,fy)) < ty(d(x,y)).
Theorem 3.1 Let (X, d) be a complete metric space. Suppose that f is a continuous modified a-ty -Meir-Keeler contractive mapping, and that there exists x0 e X such that a(xo,fxo) > 1, thenf has a fixed point.
Proof Let xo e X and define a sequence {xn} by xn = fnx0 for all n e N.If xn0 = xn0+1 for some n0 e N U {0}, then, obviously, f has a fixed point. Hence, we suppose that
xn = xn+1 (3.3)
for all n e NU{0}.Wehaved(xn, xn+1) > 0 for all n e NU{0}.Now, define sn = ty(d(xn, xn+1)). By Remark 3.1, we deduce that for all n e N U {0} ty(d(xn+1,xn+2)) = ty(d(fxn,fxn+1)) < ty(d(xn,xn+1)). By applying Lemma 3.1 for
a(xm, xn) > 1 for all m, n e N with m < n,
we have
t(d(xn+1,xn+2)) < t (d(xn,xn+1^.
Hence, the sequence {sn} is decreasing in R+, and so, it is convergent to s e R+. We will show that s = o. Suppose, to the contrary, that s > o. Note that
o < s < t (d(xn,xn+^) for all n e N U{o}. (.4)
Let e = s > o. Then by hypothesis, there exists a 5(e) > o suchthat (3.2) holds. On the other hand, by the definition of e, there exists no e N such that
e < sno = t (d(xno,xno+1)) < e + 5.
Now by (3.2), we have
sno+1 = t (d(xno+1,xno+2)) < t{dfx no,fxno+1)) < e = s,
which is a contradiction. Hence s = o, that is, limn—+TO sn = o. Now, by the continuity of t at t = o, we have limn—+TO d(xn,xn+1) = o. For given e > o, by the hypothesis, there exists a 5 = 5(e) > o such that (3.2) holds. Without loss of generality, we assume that 5 < e. Since s = o, then there exists N e N such that
sn-1 = t (d(xn-1,xn)) < 5 for all n > N. (.5)
We will prove that for any fixed k > No,
t(d(xk,xk+i)) < e for all l e N, (.6)
holds. Note that (3.6) holds for l =1 by (3.5). Suppose that condition (3.2) is satisfied for some m e N. For l = m + 1, by (3.5), we get
t (d(xk-1,xk+m)) < t(d(xk-1,xk) + d(xk,xk+m))
< t(d(xk-1, xk)) + t (d(xk, xk+m))
< e + 5. (3.7) If t (d(xk-1,xk+m)) > e, then by (3.2), we get
t(d(xk,xk+m+1)) = t (d(fxk-1,fxk+m)) < e,
and hence (3.6) holds. If t (d(xk-1,xk+m)) < e, by Remark 3.1, we get
t(d(xk, xk+m+1^ < t(d(xk-1, xk+m)) < e.
Consequently, (3.6) holds for l = m + 1. Hence, t (d(xk,xk+) < e for all k > No and l > 1, which means
d(xn,xm)<e for all m > n > No. (.8)
Hence {xn} is a Cauchy sequence. Since (X, d) is complete, there exists z e X such that xn — z as n — to. Now, sincef is continuous, then
fz = f ( lim xn) = lim xn+1 = z,
\n—TO / n—TO
that is, f has a fixed point. □
Corollary 3.1 (Theorem 1o of [11]) Let (X,d) be a complete metric space. Suppose thatf is a continuous a-t-Meir-Keeler contractive mapping, and that there exists xo e X such that a(xo,fxo) > 1, thenf has a fixed point.
Proof Let e < t(d(x,y)) < e + 5, where a(x,y) > 1. Then by e < t(d(x,y)) < e + 5 and Definition 3.2, we deduce that a(x,y)t(d(fx,fy)) < e. On the other hand, since a(x,y) > 1, then we have
t(d(fx,fy)) < a(x,y)t(d(fx,fy)) < e. That is, conditions of Theorem 3.1 hold, and f has a fixed point. □
Theorem 3.2 Let (X, d) be a complete metric space, and letf be a modified a-t -Meir-Keeler contractive mapping. If the following conditions hold:
(i) there exists xo e X such that a(xo,fxo) > 1,
(ii) if {xn} is a sequence in X such that a(xn,xn+1) > 1 for all n, andxn — x as n — +to, then a ( xn , x) > 1 for all n.
Thenf has a fixed point.
Proof Following the proof of Theorem 3.1, we say that a(xn,xn+1) > 1 for all n e N U {o}, and that there exist z e X such that xn — z as n — +to. Hence, from (ii) a(xn,z) > 1. By Remark 3.1, we have
t(d(fz, z)) < t (dfz,fxn) + dfxm z)) < t (dfz,fxn)) + t {dfxm z)) < t d(z, xn) + t d(xn+1, z) .
By taking limit as n — +to, in the inequality above, we get t(d(fz, z)) < o, that is, d(fz, z) = o. Hence fz = z. □
Corollary 3.2 (Theorem 11 of [11]) Let (X, d) be a complete metric space, and letf be a a-t -Meir-Keeler contractive mapping. If the following conditions hold:
(i) there exists xo e X such that a(xo,fxo) > 1,
(ii) if {xn} is a sequence in X such that a(xn,xn+1) > 1 for all n, andxn — x as n — +to, then a ( xn , x) > 1 for all n.
Thenf has a fixed point.
Example 3.1 Let X = [0, c), and let d(x,y)= \x -y\ be a metric on X.Definef : X ^ X by
and ty (t) = J- t. Clearly, (X, d) is a complete metric space. We show that f is a triangular a-admissible mapping. Let x,y e X, if a(x,y) > 1, then x,y e [0,1]. On the other hand, for all x, y e [0,1], we have fx < 1 and fy < 1. It follows that a(fx,fy) > 1. Also, if a(x, z) > 1 and a(z,y) > 1, then x,y, z e [0,1], and hence, a(x,y) > 1. Thus the assertion holds by the same arguments. Notice that a(0,f 0) > 1.
Now, if {xn} is a sequence in X such that a(xn,xn+1) > 1 for all n e N U {0}, and xn ^ x as n ^ +c, then {xn} c [0,1], and hence x e [0,1]. This implies that a(xn,x) > 1 for all n e N U {0}. Let a(x,y) > 1, then x,y e [0,1]. Without loss of generality, take x < y. Then
Clearly, by taking S = 4e, the condition (3.2) holds. Hence, conditions of Theorem 3.2 hold, and f has a fixed point. But if x, y e [0,1] and
e < d(x, y) <e + S,
where e >0 and S >0. Then
a(x,y)d(fx,fy) = 2\x -y\ = 2d(x,y) > 2e.
That is, Corollary 3.2 (Theorem 11 of [11]) cannot be applied for this example.
Denote with ^st the family of strictly nondecreasing functions tyst: [0, +x) — [0, +x) continuous at t =0 such that
• tyst(t) = 0 if and only if t = 0,
• tyst(t + s) < tyst (t) + tyst (s).
Definition 3.4 [11] Let (X, d) be a metric space, and let tyst e ^st. Suppose thatf: X — X is a triangular a-admissible mapping satisfying the following condition: for each e >0, there exists S >0 such that
e < tyst(M(x,y)) < e + S implies that a(x,y)tyst(d(fx,fy)) < e (3.9)
for all x, y e X, where
f if x e [0,1], xx2+1 x e (1, c),
and a(x, y)
10 if x, y e [0,1], -2 otherwise,
H Wy^irl),
( ) y x d(xy)) = j - J.
Thenf is called a generalized a-tyst-Meir-Keeler contractive mapping.
Definition 3.5 Let (X, d) be a metric space, and let e ^st. Suppose that/ : X — X is a triangular a-admissible mapping satisfying the following condition: for each e > 0 there exists S > 0 such that
Then/ is called a modified generalized a-tyst-Meir-Keeler contractive mapping.
Remark 3.2 Let / be a modified generalized a-tyst-Meir-Keeler contractive mapping. Then
^st(d/x,/yj) < fst{M(x,yj)
for all x,y e X, where a(x,y) > 1 whenM(x,y) > 0. Also, ifM(x,y) = 0 and a(x,y) > 1, then x = y, which implies that ty(d/x,/y)) = 0, i.e.,
fst{d/x,/y)) < fst{M(x,y)).
Proposition 3.1 Let (X, d) be a metric space, and let/ : X — X be a modified generalized -Meir-Keeler contractive mapping. I/there exists x0 e X such that a(x0,/x0) > 1, then limn—œ d(fn+1x0,/nx0) = 0.
Proof Define a sequence {xn} by xn = /nx0 for all n e N. Ifxn0 = xn0+i for some n0 e NU {0}, then, obviously, the conclusion holds. Hence, we suppose that
xn = xn+1 (3.11)
for all n e N U {0}. Then we have M(xn+i,xn) > 0 for every n > 0. Then by Lemma 3.1 and Remark 3.2, we have
tyst(d(xn+1,xn+2)) = fst(dtfxn,/xn+1)) < fst{M(xn , xn+1))
e < tyst(M(x,y)) < e + S implies that dtfx,/y)) < e
(3.10)
for all x, y e X, where a(x, y) > 1 and
d(xn, xn+1), d(/xn, xn), d(/xn+1, xn+1),
Now, since is strictly nondecreasing, then we get
d(xn+2,xn+1) < ma^jd(xn+1,xn), d(xn+2,xn+1)}.
Hence the case, where
max{ d(xn+l, xn), d(xn+2, xn+i)} = d(xn+2, xn+l),
is not possible. Therefore, we deduce that
d(x„+2, x„+i)< d(x„+i, x„) (.12)
for all n. That is, {d(xn+i,xn)j^=Q is a decreasing sequence in R+, and it converges to s 6 R+, that is,
lim fJd(xn+i,xn)) = lim fjM(xn+i,xn)) = ^St(s). (3.13)
n^œ ' n^œ '
Notice that s = inf{d(xn,xn+i) : n 6 N}. Let us prove that s = Q. Suppose, to the contrary, that s > Q. Then ^(s) > Q. Considering (3.i3) together with the assumption that f is a generalized a-^st-Meir-Keeler contractive mapping, for ^st(s), there exists S > Q and a natural number m such that
fst(s) < fst(M(xm, xm+i)) < fst(s) + S implies that
fst{d(xm+i, xm+2)) = fst(dfx m, fxm+i) < ^st(s).
Now, since ^st is strictly nondecreasing, then we get
d(xm+2,xm+i) < s,
which is a contradiction, since s = inf{d(xn,xn+i) : n 6 N}. Then s = Q, and so,
lim d(xn+i,xn) = Q. □
Theorem 3.3 Let (X, d) be a complete metric space, and letf : X ^ X be an orbitally continuous modified generalized a-^st -Meir-Keeler contractive mapping. If there exist xQ 6 X such that a(xQ,fxQ) > i, thenf has a fixed point.
Proof Define xn+i = fn+ixQ for all n > Q. We want to prove that limm,n^œ d(xn,xm) = Q. If this is not so, then there exist s > Q and a subsequence {xn(()} of {xn} such that
d(xn(i), xn(i+i))>2s. (3!4)
For this s > Q, there exists S > Q such that s < ^st(M(x,y)) < s + S implies that a(x,y)^st(d(fx,fy)) < s. Put r = min{s, S} and sn = d(xn,xn+i) for all n > i. From Proposition 3.i, there exists no such that
sn — d(xn, xn+i) < ~~ (3.i5)
for all n > n0. Let n(i) > n0. We get n(i) < n(i + 1) -1. If d(xn(),xn(i+1)_1) < e + z, then
d(Xn(i),Xn(i+1)) < d(Xn(i),Xn(i+l)_l) + d(Xn(i+l)_l,Xn(i+1))
< d(Xn(i), xn(i+1)_1) + d(xn(i+1)_1 xn(i+1))
< e + - + Sn(i+1)_1 < e + — < 2e,
which contradicts the assumption (3.14). Therefore, there are values of k such that n(i) < k < n(i +1) and d(xn(i),xk) > e + z. Now if d(xn(i),xn(i)+1) > e + z, then
r r r Sn(i) = d(xn(i), xn(i)+1) > e + -> r
which is a contradiction to (3.15). Hence, there are values of k with n(i) < k < n(i + 1) such that d(xn(i),xk) < e + z. Choose the smallest integer k with k > n(i) such that d(xn(i),xk) > e + z. Thus, d(xn(i), xk_1) < e + z, and so,
d(xn(i),xk) < d(xn(i),xk_1) + d(xk_1,xk)
< d(xn(i),xk_1) + d(xk_1,xk) < e + r + r = e + 3r.
Now, we can choose a natural number k satisfying n(i) < k < n(i + 1) such that
r ./ x 3r e ^ < d(xn(i),xk) < e + —. z 4
Therefore, we obtain
d(xn(i),xk) < e + < e + r, 4
d(xn(i), xn(i)+1) = dn(i) ^ < e + r,
0!7) (3.!8)
d(xk, xk+1) = dk < — < e + r.
Thus, we have
zz [d(xn(i), xk+1) + d(xn(i)+1, xk)] < zz [d(xn(i), xk) +d(xk, xk+1)
+ d(xn(i)+1, xn(i)) + d(xn(i), xk)]
< 1 [d(xn(i), xk) +d(xk, xk+1) + d(xn(i)+1, xn(i)) + d(xn(i), xk)]
= d(xn(i), xk) + z [Sk + Sn(i)]
< e + — + -4z
= e + r.
(3.20)
Now, inequalities (3.17)-(3.20) imply that M(xn(i), xk) < e + r < e + S, and so, ^st(M(xn(i), xk)) < ^st(e + S) < ^st(e) + ^st(S); the fact that f is a modified generalized a-^st-Meir-Keeler contractive mapping yields that
fst{d(Xn(i)+i,Xk+i)) < fst(e). Then d(xn(i)+1,xk+1) < e. We deduce
dfn(i)Xo,fkXo) < d{fn(i)xo,fn(i)+1xo) + d{fn(i)+1xo,fkxo)
< d{fn(i)xo,fn(i)+1xo) + d{fn(i)+1xo,fkxo)
< d{fn(i)xo,fn(i)+1xo) + d{fn(i)+1xo,fk+1xo) + d(fk+1Xo,fkXo).
From (3.16), (3.18) and (3.19), we obtain
d(xn(i)+i,Xk+1) > d(xn(i),Xk) - d(xn(i),Xn(i)+1) - d(xk,Xk+1)
> e +-----= e,
which is a contradiction. We obtained that limm,n^TO d(xn,xm) = o, and so, {xn = fnxo} is a Cauchy sequence. Since X is complete, then there exists z e X such that fnxo ^ z as n ^to. Asf is orbitally continuous, so z = fz. □
Corollary 3.3 (Theorem 17 of [11]) Let (X, d) be a complete metric space, and letf: X ^ X be an orbitally continuous generalized a-tyst-Meir-Keeler contractive mapping. If there exist xo e X such that a(xo,fxo) > 1, thenf has a fixed point.
Example 3.2 Let X = [o, to), and let d(x,y) = \x - y\ be a metric on X. Definef: X ^ X by
and ^st(i) = 71,
ifx e [0,1], x e (1, ()
a(x, y) =
28 if x, y e [0,1], -8 otherwise.
Clearly, f is a triangular a-admissible mapping, and it is orbitally continuous. Let a(x,y) > 1, then x,y e [0,1]. Without loss of generality, take x < y. Then
Mdff = £ -
^st(M(x,y)) = ^t^maxjy - x, 6y, x -1,y -
{y x 6 x y y x 1
---, — y,---,---}.
2 2 147 2 14 2 14 J
= maxi
Clearly, by taking S = 6e, the condition (3.10) holds. Hence, all conditions of Theorem 3.3 are satisfied, and f has a fixed point. But if x = 0 and y = 1
e < M(0,1) < S + e
for S >0 and e >0, then
e < 1 < S + e,
and so,
a(0,1)tyst(df 0,f 1)) =z> 1 > e. That is, Corollary 3.3 (Theorem 17 of [11]) cannot be applied for this example.
4 Modified «-^-contractive multifunction
Recently, Asl etal. [H] introduced the following notion.
Definition 4.1 Let T : X — zX, and let a : X x X — R+. We say that T is an a*-admissible mapping if
a(x,y) > 1 implies that a*(Tx, Ty) > 1, x,y e X, where
a* (A, B) = inf a(x, y).
xeA,yeB
We generalize this concept as follows.
Definition 4.2 Let T : X — zX be a multifunction, and let a, n : X x X — R+ be two functions, where n is bounded. We say that T is an a*-admissible mapping with respect to n if
a(x,y) > n(x,y) implies that a*(Tx, Ty) > n*(Tx, Ty), x,y e X, where
a*(A,B) = inf a(x,y) and n*(A,B) = sup n(x,y).
xeA,yeB xeAyeB
If we take n(x, y) = 1 for all x, y e X, then this definition reduces to Definition 4.1. In case a(x,y) = 1 for all x,y e X, then T is called an n*-subadmissible mapping.
Notice that ^ is the family of nondecreasing functions ty : [0, +c») — [0, such that
tyn(t) < for all t >0, where tyn is the nth iterate of ty. As an application of our new concept, we develop now a fixed point result for a multifunction, which generalizes Theorem 1.1.
Theorem 4.1 Let (X, d) be a complete metric space, and let T : X — zX be an a*-admissible, with respect to n, and closed-valued multifunction on X. Assume that for ty e ^,
x,y e X, a*(Tx, Ty) > n*(Tx, Ty) H(Tx, Ty) < ty(d(x,y)). (4.1)
Also, suppose that the following assertions hold:
(i) there exist x0 e X andx1 e Tx0 such that a(x0,x1) > n(x0,x1);
(ii) for a sequence {xn} c X converging to x e X and a(xn,xn+1) > n(xn,xn+1) for all n e N, we have a(xn,x) > n(xn,x) for all n e N.
Then T has a fixed point.
Proof Let x1 e Tx0 be such that a(x0,x1) > n(x0,x1). Since T is an a*-admissible mapping, then a*(Tx0, Tx1) > n*(Tx0, Tx1). Therefore, from (4.1), we have
H(Tx0, Tx\) < ty(d(x0,x1)). (.)
If x0 = x1, then x0 is a fixed point of T. Hence, we assume that x0 = x1. Also, if x1 e Tx 1, then x1 is a fixed point of T. Assume that x1 e Tx1 and q >1. Then we have
0 < d(x1, Ix1) < H(Tx0, Tx1) < qH(Tx0, Tx1),
and so, by (4.z), we get
0 < d(x1, Ix1) < qH(Tx0, Tx1) < qty(d(x0,x^).
This implies that there exists xz e Tx1 such that
0 < d(x1,xz) < qH(Tx0, Ix1) < qty(d(x0,x^). (.3)
Note that x1 = xz (since x1 e Tx1). Also, since a*(Tx0, Ix1) > n*(Tx0, Tx1), x1 e Tx0 and xz e Tx1, then a(x1,xz) > n(x1,xz). So a*(Tx1, Txz) > n*(Tx1, Txz). Therefore, from (4.1), we have
H(Tx1, Txz) < ty(d(x1,xz)). (4.4)
Put t0 = d(x0,x1). Then from (4.3), we have d(x1,xz) < qty(t0), where t0 > 0. Now, since ty is strictly increasing, then ty (d(x1, xz)) < ty (qty (t0)). Put
= ty (qty (tp)) ty (d(x1, xz))'
and so q1 > 1. If xz e Txz, then xz is a fixed point of T. Hence, we suppose that xz e Txz. Then
0 < d(xz, Txz) < H(Tx\, Txz) < q1H(Tx1, Txz).
So there exists x3 e Tx2 such that o < d(x2,x3) < q1H(Tx 1, Tx2), and then from (4.4), we get
o < d(x2,x3) < q1H(Tx1, Tx2) < q1f(d(x1,x2)) = f(qf(to)).
Again, since f is strictly increasing, then f (d(x2, x3)) < f (f (qf (to))). Put
= f (f (qf (to))) f (d(x2,X3)) .
So, q2 > 1. If x3 e Tx3, then x3 is a fixed point of T. Hence, we assume that x3 e Tx3. Then
o < d(x3, Tx3) < H(Tx2, Tx3) < q2H(Tx2, Tx3),
and so, there exists x4 e Tx3 such that
o < d(x3,x4) < H(Tx2, Tx3) < q2H(Tx2, Tx3). (4.5)
Clearly, x2 = x3. Also again, since a^(Tx1, Tx2) > n*(Tx1, Tx2), x2 e Tx1 and x3 e Tx2, then a(x2,x3) > n(x2,x3), and so, a^(Tx2, Tx3) > n*(Tx2, Tx3). Then from (4.1), we have
H(TX2, TX3) < f (d(X2, X3)),
and so, from (4.5), we deduce that
d(x3, X4) <q2H (TX2, TX3) < q2f (d(x2, X3)) = f(f( qf (to))).
By continuing this process, we obtain a sequence {xn} in X such that xn e Txn-1, xn = xn-1, a*(xn,xn+1) > n*(xn,xn+1) and d(xn,xn+1) < f n-1(qf (to)) for all n e N. Now, for all m > n, we can write
m-1 m-1
d(Xn, Xm) d(Xk, Xk+1) < ^ f k-1( qf (to)).
k=n k=n
Therefore, {xn} is a Cauchy sequence. Since (X, d) is a complete metric space, then there exists z e X such that xn ^ z as n ^to. Now, since a(xn,z) > n(xn,z) for all n e N, then a^(Txn, Tz) > n*(Txn, Tz), and so, from (4.1), we have
d(z, Tz) < H(Txn, Tz) + d(xn, z) < f (d(xn, z)) + d(xn, z)
for all n e N. Taking limit as n ^ to in the inequality above, we get d(z, Tz) = o, i.e., z e Tz. □
If in Theorem 4.1 we take n(x, y) = 1, we have the following corollary.
Corollary 4.1 Let (X, d) be a complete metric space, and let T : X — zX be an a*-admissible and closed-valued multifunction on X. Assume that
x,y e X, a*(Tx, Ty) > 1 H(Tx, Ty) < ty(d(x,y)).
Also, suppose that the following assertions hold:
(i) there exists x0 e X andx1 e Tx0 such that a(x0,x1) > 1;
(ii) for a sequence {xn}c X converging to xeX and a(xn, xn+1) > 1 for all n e N, we have a ( xn , x ) > 1 for all n N.
Then T has a fixed point.
If in Theorem 4.1 we take a(x, y) = 1, then we have the following result.
Corollary 4.2 Let (X, d) be a complete metric space, and let T : X — zX be an n*-subadmissible and closed-valued multifunction on X. Assume that
x,y e X, n*(Tx, Ty) < 1 H(Tx, Ty) < ty(d(x,y)).
Also, suppose that the following assertions hold:
(i) there exists x0 e X andx1 e Tx0 such that n(x0,x1) < 1;
(ii) for a sequence {xn} c X converging to xeX and n(xn, xn+1) < 1 for all n e N, we have n ( xn , x ) < 1 for all n N.
Then T has a fixed point.
Corollary 4.3 (Theorem H and z.3 of [H]) Let (X, d) be a complete metric space, and let T: X — zX bean a*-admissible and closed-valued multifunction on X. Assume that
a*(Tx, Ty)H(Tx, Ty) < ty(d(x,y)) (4.6)
for all x, y X. Also, suppose that the following assertions hold:
(i) there exists x0 e X andx1 e Tx0 such that a(x0,x1) > 1;
(ii) for a sequence {xn}c X converging to xeX and a(xn, xn+1) > 1 for all n e N, we have a ( xn , x ) > 1 for all n N.
Then T has a fixed point.
Proof Suppose that a*(Tx, Ty) > 1 for x,y e X. Then by (4.6), we have
H(Tx, Ty) < ty(d(x,y)).
That is, conditions of Corollary 4.1 hold, and T has a fixed point. □
Similarly, we can deduce the following corollaries.
Corollary 4.4 Let (X, d) be a complete metric space, and let T : X — zX be an a*-admissible and closed-valued multifunction on X. Assume that
(a*(Tx, Ty) + 1)H(Tx,Ty) < zty(d(x,y))
for all x, y e X. Also, suppose that the following assertions hold:
(i) there exists xo e X andx1 e Txo such that a(xo,x1) > 1;
(ii) for a sequence {xn}<zX converging to x e X and a(xn,xn+1) > 1 for all n e N, we have a ( xn , x ) > 1 for all n e N.
Then T has a fixed point.
Corollary 4.5 Let (X, d) be a complete metric space, and let T : X ^ 2X be an a*-admissible and closed-valued multifunction on X. Assume that
(H(Tx, Ty) + t)a*(Tx,Ty) < f (d(x,y)) + l
for all x, y e X, where l > o. Also, suppose that the following assertions hold:
(i) there exists xo e X andx1 e Txo such that a(xo,x1) > 1;
(ii) for a sequence {xn}<zX converging to x e X and a(xn,xn+1) > 1 for all n e N, we have a ( xn , x ) > 1 for all n e N.
Then T has a fixed point.
Corollary 4.6 Let (X, d) be a complete metric space, and let T : X ^ 2X be an n*-subadmissible and closed-valued multifunction on X. Assume that
H(Tx, Ty) < n*(Tx, Ty)f (d(x,y))
for all x, y e X. Also, suppose that the following assertions hold:
(i) there exists xo e X andx1 e Txo such that n(xo,x1) < 1;
(ii) for a sequence {xn} c X converging to x e X and n(xn,xn+1) < 1 for all n e N, we have n (Xn, x) < 1 for all n e N.
Then T has a fixed point.
Corollary 4.7 Let (X, d) be a complete metric space, and let T : X ^ 2X be an n*-subadmissible and closed-valued multifunction on X. Assume that
2H(Tx,Ty) < (n*(Tx, Ty) + 1)f(d(x,y)
for all x, y e X. Also, suppose that the following assertions hold:
(i) there exists xo e X andx1 e Txo such that n(xo,x1) < 1;
(ii) for a sequence {xn} c X converging to x e X and n(xn,xn+1) < 1 for all n e N, we have n ( xn , x ) < 1 for all n e N.
Then T has a fixed point.
Corollary 4.8 Let (X, d) be a complete metric space, and let T : X ^ 2X be an a*-admissible and closed-valued multifunction on X. Assume that
H(Tx, Ty) + l < (f (d(x,y)) + l)n*(Tx,Ty)
for all x, y e X, where l > o. Also, suppose that the following assertions hold: (i) there exists xo e X andx1 e Txo such that n(xo,x1) < 1;
(ii) for a sequence {xn} c X converging to xeX and n(xn, xn+1) < 1 for all n e N, we have n ( xn , x ) < 1 for all n N. Then T has a fixed point.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
Allauthors contributed equally and significantly in writing this article. Allauthors read and approved the finalmanuscript.
Author details
1 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia. 2Young Researchers
and Elite Club, Rasht Branch, Islamic Azad University, Rasht, Iran.
Acknowledgements
This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the first
and third authors acknowledge with thanks DSR, KAU for the financialsupport.
Received: 21 May 2013 Accepted: 23 July 2013 Published: 8 August 2013
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doi:10.1186/1687-1812-2013-212
Cite this article as: Hussain et al.: Fixed point results for single and set-valued a-q-ty-contractive mappings. Fixed Point Theory and Applications 2013 2013:212.
