C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat. Volum e 67, N umb er 2, Pages 306–316 (2018) D O I: 10.1501/C om mua1_ 0000000884 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
FIXED POINTS FOR GENERALIZED
(F; h; ; ) CONTRACTIONS IN b METRIC SPACES
VILDAN OZTURK, DURAN TURKOGLU, AND ARSLAN HOJAT ANSARI
Abstract. In this paper, we de…ned (F; h; ; ) contractions using pair of (F; h) upper class functions for admissible and subadmissible map-pings.We proved some …xed point theorems for this type contractive mappings in b metric spaces. Our results generalize admissible results in the litera-ture.
1. Introduction
De…nition 1. ([9]) Let X be a nonempty set and s 1 be a given real number. A function d : X X ! [0; 1) is a b-metric if, for all x; y; z 2 X, the following conditions are satis…ed:
(i) d (x; y) = 0 if and only if x = y; (ii) d (x; y) = d (y; x) ;
(iii) d (x; z) s [d (x; y) + d (y; z)] .
In this case, the pair (X; d) is called a b-metric space.
It should be noted that, the class of b-metric spaces is e¤ectively larger than that of metric spaces, every metric is a b-metric with s = 1:
Example 1. ([1]) Let (X; d) be a metric space and (x; y) = (d (x; y))p, where p > 1 is a real number. Then is a b metric with s = 2p 1:
However, if (X; d) is a metric space, then (X; ) is not necessarily a metric space. For example, if X = R is the set of real numbers and d (x; y) = jx yj is usual Euclidean metric, then (x; y) = (x y)2is a b metric on R with s = 2:But is not a metric on R:
De…nition 2. ([7]) Let fxng be a sequence in a b-metric space (X; d).
Received by the editors: May 08, 2017; Accepted: September 09, 2017. 2010 Mathematics Subject Classi…cation. 54H25, 47H10.
Key words and phrases. Fixed point, generalized contraction, b metric space. c 2 0 1 8 A n ka ra U n ive rsity. C o m m u n ic a tio n s Fa c u lty o f S c ie n c e s U n ive rs ity o f A n ka ra -S e rie s A 1 M a t h e m a tic s a n d S t a tis tic s . C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra -S é rie s A 1 M a t h e m a tic s a n d S t a tis t ic s .
(a) fxng is called b convergent if and only if there is x 2 X such that d (xn; x) !
0 as n ! 1:
(b) fxng is a b Cauchy sequence if and only if d (xn; xm) ! 0 as n; m ! 1:
A b-metric space is said to be complete if and only if each b Cauchy sequence in this space is b convergent.
Proposition 1. ([7]) In a b metric space (X; d) ; the following assertions hold: (p1) A b convergent sequence has a unique limit.
(p2) Each b convergent sequence is b Cauchy. (p3) In general, a b metric is not continuous.
On the other hand the notion of -contractive type mapping was introduced by Samet et al. [11],[17]. Also, see ([10],[12],[13-15])
Now we give some de…nitions that will be used throughout this paper.
A mapping : [0; 1) ! [0; 1) is called a comparison function if it is increasing and limn!1 n(t) = 0 for all t > 0:
Lemma 1. ([5]) Let : [0; 1) ! [0; 1) is a comparison function then (a) each iterate n of , n 1, is also a comparison function, (b) is continuous at t = 0,
(c) (t) < t for all t > 0.
De…nition 3. ([5]) A function : [0; 1) ! [0; 1) is said to be a (c)-comparison function if
(c1) is increasing,
(c2) there exists k02 N, a 2 (0; 1) and a convergent series of nonnegative terms 1
P
k=1
vk, such that k+1(t) a k(t) + vk, for k k0 and any t 2 [0; 1) :
De…nition 4. ([6]) Let s 1 be a real number. A function : [0; 1) ! [0; 1) is said to be a (b)-comparison function if
(b1) is monotonically increasing,
(b1) there exists k0 2 N, a 2 (0; 1) and a convergent series of nonnegative
terms P1
k=1
vk, such that sk+1 k+1(t) ask k(t) + vk, for k k0 and any
t 2 [0; 1) :
When s = 1, (b)-comparison function reduces to (c)-comparison function. We denote by b for the class of (b)-comparison function.
Lemma 2. ([4]) If : [0; 1) ! [0; 1) is a (b)-comparison function then one has the following:
(i) P1
k=0
sk k
(ii) the function bs : [0; 1) ! [0; 1) de…ned by bs(t) = 1
P
k=0
sk k(t) ; t 2 R+, increasing and continuous at 0.
Any (b)-comparison function is a comparison function.
De…nition 5. ([17])For any nonempty set X; let T : X ! X and : X X ! [0; 1) be mappings. T is called -admissible if for all x; y 2 X;
(x; y) 1 ) (T x; T y) 1:
De…nition 6. ([16])Let T : X ! X, : X X ! R+. We say T is an
-subadmissible mapping if
x; y 2 X; (x; y) 1 =) (T x; T y) 1:
Bota et. al. in ([8]) gave the de…nition of contractive mapping of type (b) in b metric space which is a generalization of De…nition 9.
De…nition 7. Let (X; d) be a b metric space and T : X ! X be a given mapping. T is called an contractive mapping of type (b), if there exists two functions
: X X ! [0; 1) and 2 b such that
(x; y) d (T x; T y) (d (x; y)) ; 8x; y 2 X:
De…nition 8. ([2],[3])We say that the function h : R+ R+ ! R is a function of
subclass of type I, if x 1 =) h(1; y) h(x; y) for all y 2 R+.
Example 2. ([2],[3])De…ne h : R+ R+! R by:
(a) h(x; y) = (y + l)x; l > 1; (b) h(x; y) = (x + l)y; l > 1; (c) h(x; y) = xny, n 2 N; (d) h(x; y) = y; (e) h(x; y) = 1 n+1 Pn i=0xi y, n 2 N; (f) h(x; y) =hn+11 Pni=0xi + liy; l > 1; n 2 N
for all x; y 2 R+. Then h is a function of subclass of type I.
De…nition 9. ([2],[3]) Let h; F : R+
R+ ! R, then we say that the pair (F; h) is an upper class of type I, if h is a function of subclass of type I and: (i) 0 s 1 =) F(s; t) F(1; t), (ii) h(1; y) F(1; t) =) y t for all t; y 2 R+.
Example 3. ([2],[3]) De…ne h; F : R+ R+! R by:
(a) h(x; y) = (y + l)x; l > 1 and F(s; t) = st + l; (b) h(x; y) = (x + l)y; l > 1 and F(s; t) = (1 + l)st; (c) h(x; y) = xmy, m 2 N and F(s; t) = st; (d) h(x; y) = y and F(s; t) = t; (d) h(x; y) = 1 n+1 Pn i=0xi y; n 2 N and F(s; t) = st;
for all x; y; s; t 2 R+. Then the pair (F; h) is an upper class of type I.
De…nition 10. ([2],[3])We say that the function h : R+ R+ R+ ! R is a
function of subclass of type II, if x; y 1 =) h(1; 1; z) h(x; y; z) for all z 2 R+:
Example 4. ([2],[3])De…ne h : R+ R+ R+! R by:
(a) h(x; y; z) = (z + l)xy; l > 1;
(b) h(x; y; z) = (xy + l)z; l > 1;
(c) h(x; y; z) = z;
(d) h(x; y; z) = xmynzp; m; n; p 2 N;
(e) h(x; y; z) = xm+xn3yp+yqzk; m; n; p; q; k 2 N
for all x; y; z 2 R+: Then h is a function of subclass of type II.
De…nition 11. ([2],[3])Let h : R+ R+ R+ ! R and F : R+ R+ ! R; then
we say that the pair (F; h) is an upper class of type II, if h is a subclass of type II and: (i) 0 s 1 =) F(s; t) F(1; t), (ii) h(1; 1; z) F(s; t) =) z st for all s; t; z 2 R+.
Example 5. ([2],[3]) De…ne h : R+ R+ R+! R and F : R+ R+! R by:
(a) h(x; y; z) = (z + l)xy; l > 1; F(s; t) = st + l;
(b) h(x; y; z) = (xy + l)z; l > 1; F(s; t) = (1 + l)st; (c) h(x; y; z) = z; F (s; t) = st;
(d) h(x; y; z) = xmynzp; m; n; p 2 N; F(s; t) = sptp
(e) h(x; y; z) = xm+xn3yp+yqzk; m; n; p; q; k 2 N; F(s; t) = sktk
for all x; y; z; s; t 2 R+. Then the pair (F; h) is an upper class of type II.
2. Main results
De…nition 12. ([13])Let (X; d) be a b metric space and T : X ! X be a given mapping. T is called generalized contractive mapping of type (I), if there exists two functions : X X ! [0; 1) and 2 b such that for all x; y 2 X
(x; y) d (T x; T y)) (Ms(x; y))
where,
Ms(x; y) = max d (x; y) ; d (T x; x) ; d (T y; y) ;
d (T x; y) + d (x; T y)
2s :
Theorem 1. ([13])Let (X; d) be a complete b metric space. Suppose that T : X ! X be a generalized contractive mapping of type (I) and satis…es:
(i) T is admissible
(ii) there exists x02 X such that (x0; T x0) 1
(iii) T is continuous. Then T has a …xed point.
De…nition 13. ([13])Let (X; d) be a b metric space and T : X ! X be a given mapping. T is called generalized contractive mapping of type (II), if there exists two functions : X X ! [0; 1) and 2 b such that for all x; y 2 X
(x; y) d (T x; T y)) (Ns(x; y)) where, Ns(x; y) = max d (x; y) ; d (T x; x) + d (T y; y) 2s ; d (T x; y) + d (T y; x) 2s :
De…nition 14. Let (X; d) be a b metric space and T : X ! X be a given mapping. T is called generalized (F; h; ; ) contractive mapping of type (I), if there exists two functions ; : X X ! [0; 1) and 2 b such that for all x; y 2 X
h( (x; y) ; d (T x; T y)) F( (x; y) ; (Ms(x; y))) (2.1)
where,pair (F; h) is an upper class of type I and Ms(x; y) = max d (x; y) ; d (T x; x) ; d (T y; y) ;
d (T x; y) + d (x; T y)
2s :
Theorem 2. Let (X; d) be a complete b metric space. Suppose that T : X ! X be a generalized (F; h; ; ) contractive mapping of type (I) and satis…es:
(i) T is admissible and -subadmissible
(ii) there exists x02 X such that (x0; T x0) 1; (x0; T x0) 1
(iii) T is continuous. Then T has a …xed point.
Proof. By assumption (ii), there exists x02 X such that (x0; T x0) 1; (x0; T x0)
1. De…ne the sequence fxng in X by xn+1= T xn for all n 2 N. If xn= xn+1for
some n 2 N, then xn is a …xed point of T:
Assume that xn 6= xn+1for all n 2 N.
Since T is admissible, then
(x0; x1) = (x0; T x0) 1 =) (T x0; T x1) = (x1; x2) 1:
(x0; x1) = (x0; T x0) 1 =) (T x0; T x1) = (x1; x2) 1:
By induction, we get for all n 2 N,
(xn; xn+ 1) 1 ; (xn; xn+ 1) 1: (2.2) Using (2.1) and (2.2) h(1; d (xn; xn+1)) = h(1; d (T xn 1; T xn)) h( (xn 1; xn) ; d (T xn 1; T xn)) F( (xn 1; xn) ; (Ms(xn 1; xn))) F(1; (Ms(xn 1; xn))) =)
d (xn; xn+1) (Ms(xn 1; xn)): (2.3) where Ms(xn 1; xn) = max d (xn 1; xn) ; d (T xn 1; xn 1) ; d (T xn; xn) ; d(T xn 1;xn)+d(T xn;xn 1) 2s ; = max d (xn 1; xnd(x) ; d (xn;xn)+d(xn; xn 1n+1;x) ; d (xn 1)n+1; xn) ; 2s ; = max d (xs[d(xn 1n+1; x;xnn) ; d (x)+d(xnn+1;xn; x1)]n) ; 2s max fd (xn 1; xn) ; d (xn+1; xn)g :
If Ms(xn 1; xn) = d (xn; xn+1) , then from (2.3) and de…nition of ,
d (xn; xn+1) (d (xn; xn+1)) < d (xn; xn+1)
a contradiction. Thus Ms(xn 1; xn) = d (xn 1; xn). Hence,
d (xn; xn+1) (d (xn 1; xn)) < d (xn 1; xn)
for all n 1. If operations are continued in this way,
d (xn; xn+1) n(d (x0; x1)) : (2.4)
Thus, for all p 1;
d (xn; xn+p) sd (xn; xn+1) + s2d (xn+1; xn+2) + ::: +sp 1d (xn+p 2; xn+p 1) + spd (xn+p 1; xn+p) s n(d (x0; x1)) + s2 n+1(d (x0; x1)) + ::: +sp 1 n+p 2(d (x0; x1)) + sp n+p 1(d (x0; x1)) = 1 sn 1[s n n(d (x 0; x1)) + sn+1 n+1(d (x0; x1)) + ::: +sp n 2 p n 2(d (x0; x1)) + sp+n 1 p+n 1(d (x0; x1))]: Denoting Sn= 1 P k=n sk k (d (x0; x1)) ; n 1; we obtain d (xn; xn+p) 1 sn 1[Sn+p 1 Sn 1] (2.5)
for n 1, p 1. From Lemma 2, we conclude that the series P1
k=0
sk k(d (x 0; x1))
is convergent. Thus, there exists
S = limn!1Sn2 [0; 1) :
Regarding s 1 and by (2.5) fxng is a Cauchy sequence in b metric space (X; d) :
Since (X; d) is complete, there exists x 2 X such that xn ! x as n ! 1: Using
continuity of T ,
as n ! 1: By the uniqueness of the limit, we get x = T x . Hence x is a …xed point of T:
De…nition 15. Let (X; d) be a b metric space and T : X ! X be a given mapping. T is called generalized (F; h; ; ) contractive mapping of type (II), if there exists two functions : X X ! [0; 1) and 2 b such that for all x; y 2 X
h( (x; y) ; d (T x; T y)) F( (x; y) ; (Ns(x; y))) (2.6)
where,pair (F; h) is an upper class of type I and Ns(x; y) = max d (x; y) ;
d (T x; x) + d (T y; y)
2s ;
d (T x; y) + d (T y; x)
2s :
Theorem 3. Let (X; d) be a complete b metric space. Suppose that T : X ! X be a generalized (F; h; ; ) contractive mapping of type (II) and satis…es:
(i) T is admissible, -subadmissible
(ii) there exists x02 X such that (x0; T x0) 1; (x0; T x0) 1
(iii) T is continuous. Then T has a …xed point.
Proof. By assumption (ii), there exists x02 X such that (x0; T x0) 1; (x0; T x0)
1. De…ne the sequence fxng in X by xn+1= T xn for all n 2 N. If xn= xn+1for
some n 2 N, then xn is a …xed point of T:
Assume that xn 6= xn+1for all n 2 N.
Since T is admissible, then
(x0; x1) = (x0; T x0) 1 =) (T x0; T x1) = (x1; x2) 1;
(x0; x1) = (x0; T x0) 1 =) (T x0; T x1) = (x1; x2) 1:
By induction, we get for all n 2 N,
(xn; xn+ 1) 1 ; (xn; xn+ 1) 1: Using (2.6) h(1; d (xn; xn+1)) = h(1; d (T xn 1; T xn)) h( (xn 1; xn) ; d (T xn 1; T xn)) F( (xn 1; xn) ; (Ns(xn 1; xn))) F(1; (Ns(xn 1; xn))) =) d (xn; xn+1) (Ns(xn 1; xn)) (Ms(x; y)):
The rest of proof is evident due to Theorem 2.
In the following two theorems we are able to remove the continuity condition for the contractive mappings of type (I) and type (II).
Theorem 4. Let (X; d) be a complete b metric space. Suppose that T : X ! X be a generalized contractive mapping of type (I) and satis…es:
(i) T is admissible,
(ii) there exists x02 X such that (x0; T x0) 1
(iii) if fxng is a sequence in X such that (xn; xn+1) 1 for all n and xn !
x 2 X;as n ! 1, then there exists a subsequence xn(k) of fxng such
that xn(k); x 1; for all k:
Then T has a …xed point.
Proof. Following the proof of Theorem 2 , we know that the sequence fxng de…ned
by xn+1= T xn for all n 0, is Cauchy and converges to some u 2 X.
We shall show that T u = u. Suppose on the contrary that d (T u; u) > 0. From (2.2) and (iii), there exists a subsequence xn(k) of fxng such that xn(k); u 1
for all k. By (2.1) h(1; d xn(k)+1; T u ) = h(1; d T xn(k); T u ) h( xn(k); u ; d T xn(k); T u ) F( xn(k); u ; (Ms xn(k); u )) F(1; (Ms xn(k); u )) =) d xn(k)+1; T u Ms xn(k); u ; (2.7) where Ms xn(k); u = max ( d xn(k); u ; d T xn(k); xn(k) ; d (T u; u) ; d(T xn(k);u);d(T u;xn(k)) 2s ) : As k ! 1; limk!1 Ms xn(k); u = d (T u; u) : In (2.7), as k ! 1
d (u; T u) (d (u; T u)) < d (u; T u)
which is a contradiction. Hence, u = T u and u is a …xed point of T:
Theorem 5. Let (X; d) be a complete b metric space. Suppose that T : X ! X be a generalized contractive mapping of type (II) and satis…es:
(i) T is admissible,
(ii) there exists x02 X such that (x0; T x0) 1;
(iii) if fxng is a sequence in X such that (xn; xn+1) 1 for all n and xn !
x 2 X;as n ! 1, then there exists a subsequence xn(k) of fxng such
that xn(k); x 1; for all k:
Proof. Following the proof of Theorem 2.5, we know that the sequence fxng de…ned
by xn+1= T xn , for all n 0, is Cauchy and converges to some u 2 X.
We shall show that T u = u. Suppose on the contrary that d (T u; u) > 0. From (2.2) and (iii), there exists a subsequence xn(k) of fxng such that xn(k); u 1
for all k. Applying (2.6),
h(1; d xn(k)+1; T u ) = h(1; d T xn(k); T u ) h( xn(k); u ; d T xn(k); T u ) F( xn(k); u ; (Ns xn(k); u )) F(1; (Ns xn(k); u )) =) d xn(k)+1; T u Ns xn(k); u (2.8) where Ns xn(k); u = max 8 < : d xn(k); u ; d(T xn(k);xn(k))+d(T u;u) 2s ; d(T xn(k);u);d(T u;xn(k)) 2s 9 = ;: As k ! 1; limk!1 Ns xn(k); u = d(T u;u)2s , for s 1:
In (2.8), as k ! 1
d (u; T u) d (T u; u)
2s <
d (T u; u) 2s
which is a contradiction. Hence, u = T u and u is a …xed point of T: Example 6. Let X = (0; 1) endowed with b metric
d : X X ! R+; d (x; y) = (x y)2
with constant s = 2: (X; d) is a complete b metric space. Let the functions T : X ! X , : X X ! [0; 1) and : X X ! [0; 1) be de…ned by T (x) = x+1 4 ; x 2 (0; 1] 2x; x > 1 , (x; y) = 1; x 2 (0; 1] 0; otherwise; (x; y) = 1 2; x 2 (0; 1] 1; otherwise:
Clearly, T is admissible, continuous and subadmissible. Let h,F : R+ R+! R be de…ned by;
h(x; y) = (y + l)x; l > 1 and F(s; t) = st + l. (F,h; ; ) contraction of type
Let x; y 2 X if (x; y) 1 and (x; y) 1, then x; y 2 (0; 1] :Thus h ( (x; y) d (T x; T y)) = h 1; x + 1 4 y + 1 4 2! = 1 16(x y) 2 + l 1 2: 1 2(x y) 2 + l = F ( (x; y) ; (d (x; y))) F( (x; y) ; (Ms(x; y)) :
Then all conditions of Theorem 5 are satis…ed. 13 is …xed point of T .
Corollary 1. Let (X; d) be a complete b metric space and T : X ! X be contin-uous mapping. Suppose that there exists a function 2 b such that
d (T x; T y) (Ms(x; y))
for all x; y 2 X, then T has a …xed point.
Similarly, be taken (x; y) = 1 in Theorem 4, the following result is obtained. Corollary 2. Let (X; d) be a complete b metric space and T : X ! X be contin-uous mapping. Suppose that there exists a function 2 b such that
d (T x; T y) (Ns(x; y))
for all x; y 2 X, then T has a …xed point. References
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Current address : Vildan Ozturk: Department of Mathematics and Sciences Education, Faculty of Education, Artvin Coruh University, Artvin, Turkey
E-mail address : vildanozturk84@gmail.com
ORCID Address: http://orcid.org/0000-0001-5825-2030
Current address : Duran Turkoglu: Department of Mathematics, Faculty of Sciences, Gazi University, Ankara, Turkey.
E-mail address : dturkoglu@gazi.edu.tr
ORCID Address: http://orcid.org/0000-0002-8667-1432
Current address : Arslan H. Ansari: Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran.
E-mail address : analsisamirmath2@gmail.com
