AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi 10.1155/2018/1327691 1327691 Research Article Fixed Point Theorems for L -Contractions in Generalized Metric Spaces http://orcid.org/0000-0001-6208-3276 Cho Seong-Hoon 1 Jeribi Aref Department of Mathematics Hanseo University Chungnam 356-706 Republic of Korea hanseo.ac.kr 2018 2122018 2018 21 08 2018 08 10 2018 04 11 2018 2122018 2018 Copyright © 2018 Seong-Hoon Cho. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this paper, the notion of L -contractions is introduced and a new fixed point theorem for such contractions is established.

Hanseo University
1. Introduction and Preliminaries

Branciari  introduced the notion of generalized metric spaces and obtained a generalization of the Banach contraction principle, whereafter many authors proved various fixed point results in such spaces, for example,  and references therein. Also, Suzuki et al.  and Abtahi et al.  studied ν -generalized metric spaces and proved the Banach and Kannan contraction principles in such spaces, and Mitrović et al.  introduced the notion of b ν ( s ) -generalized metric spaces and Banach and Reich contraction principles in such spaces.

In particular, Jleli and Samet  introduced the notion of θ -contractions and gave a generalization of the Banach contraction principle in generalized metric spaces, where θ : ( 0 , ) ( 1 , ) is a function satisfying the following conditions:

θ is nondecreasing;

{ t n } ( 0 , ) , (1) lim n θ t n = 1 lim n t n = 0 + ;

r ( 0,1 ) l ( 0 , ) : (2) lim t 0 + θ t - 1 t r = l .

Also, Ahmad et al.  extended the result of Jleli and Samet  to metric spaces by applying the following simple condition ( θ 4) instead of ( θ 3).

θ is continuous on ( 0 , ) .

Recently, Khojasteh et al.  introduced the notion of Z -contractions by defining the concept of simulation functions. They unified some existing metric fixed point results. Afterward, many authors ( and references therein) obtained generalizations of the result of .

In the paper, we introduce the concept of a new type of contraction maps, and we establish a new fixed point theorem for such contraction maps in the setting of generalized metric spaces.

Let L be the family of all mappings ξ : [ 1 , ) × [ 1 , ) R such that

ξ ( 1,1 ) = 1 ;

ξ ( t , s ) < s / t s , t > 1 ;

for any sequence { t n } , { s n } ( 1 , ) with t n s n n = 1,2 , 3 , (3) lim n t n = lim n s n > 1 lim n sup ξ t n , s n < 1 .

We say that ξ L is a L -simulation function.

Note that ξ ( t , t ) < 1 t > 1 .

Example 1.

Let ξ b , ξ w , ξ : [ 1 , ) × [ 1 , ) R be functions defined as follows, respectively:

ξ b ( t , s ) = s k / t t , s 1 , where k ( 0,1 ) ;

ξ w ( t , s ) = s / t ϕ ( s ) t , s 1 , where ϕ : [ 1 , ) [ 1 , ) is nondecreasing and lower semicontinuous such that ϕ - 1 ( { 1 } ) = 1 ; (4) ξ t , s = 1 i f s , t = 1,1 , s 2 t i f s < t , s λ t o t h e r w i s e ,

s , t 1 , where λ ( 0,1 ) .

Then ξ b , ξ w , ξ L .

We recall the following definitions which are in .

Let X be a nonempty set, and let d : X × X [ 0 , ) be a map such that for all x , y X and for all distinct points u , v X , each of them is different from x and y :

d ( x , y ) = 0 if and only if x = y ;

d ( x , y ) = d ( y , x ) ;

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

Then d is called a generalized metric on X and ( X , d ) is called a generalized metric space.

Let ( X , d ) be a generalized metric space, let { x n } X be a sequence, and x X .

Then we say that

{ x n } is convergent to x (denoted by l i m n x n = x ) if and only if l i m n d ( x n , x ) = 0 ;

{ x n } is Cauchy if and only if l i m n , m d ( x n , x m ) = 0 ;

( X , d ) is complete if and only if every Cauchy sequence in X is convergent to some point in X .

Let ( X , d ) be a generalized metric space.

A map T : X X is called continuous at x X if, for any V τ containing T x , there exists U τ containing x such that T U V , where τ is the topology on X induced by the generalized metric d . That is, (5) τ = U X : x U , B β , x B U , β = B x , r : x X , r > 0 , B x , r = y X : d x , y < r .

If T is continuous at each point x X , then it is called continuous.

Note that T is continuous if and only if it is sequentially continuous, i.e., lim n d ( T x n , T x ) = 0 for any sequence { x n } X with lim n d ( x n , x ) = 0 .

Remark 2 (see [<xref ref-type="bibr" rid="B23">6</xref>]).

If d is a generalized metric on X , then it is not continuous in each coordinate.

Lemma 3 (see [<xref ref-type="bibr" rid="B13">20</xref>]).

Let ( X , d ) be a generalized metric space, let { x n } X be a Cauchy sequence, and x , y X . If there exists a positive integer N such that

x n x m n , m > N ;

x n x n > N ;

x n y n > N ;

lim n d ( x n , x ) = lim n d ( x n , y ) ,

then x = y .

2. Fixed Point Theorems

We denote by Θ the class of all functions θ : ( 0 , ) ( 1 , ) such that conditions ( θ 1) and ( θ 2) hold.

A mapping T : X X is called L -contraction with respect to ξ if there exist θ Θ and ξ L such that, for all x , y X with d ( T x , T y ) > 0 , (6) ξ θ d T x , T y , θ d x , y 1 .

Note that if T is L -contraction with respect to ξ , then it is continuous. In fact, let x X be a point and let { x n } X be any sequence such that (7) lim n d x n , x = 0 + , d T x n , T x > 0 n = 1,2 , 3 , .

Then from ( θ 2 ) lim n θ ( d ( x n , x ) ) = 1 .

It follows from (6) and ( ξ 2 ) that (8) 1 ξ θ d T x n , T x , θ d x n , x < θ d x n , x θ d T x n , T x , which implies (9) θ d T x n , T x < θ d x n , x .

Since θ is nondecreasing, we have (10) d T x n , T x < d x n , x ,

and so (11) lim n d T x n , T x = 0 . Hence T is continuous.

Now, we prove our main result.

Theorem 4.

Let ( X , d ) be a complete generalized metric space, and let T : X X be a L -contraction with respect to ξ .

Then T has a unique fixed point, and for every initial point x 0 X , the Picard sequence { T n x 0 } converges to the fixed point.

Proof.

Firstly, we show uniqueness of fixed point whenever it exists.

Assume that w and u are fixed points of T .

If u z , then d ( w , u ) > 0 , and so it follows from (6) that (12) 1 ξ θ d T w , T u , θ d w , u = ξ θ d w , u , θ d w , u < θ d w , u θ d w , u . Hence (13) θ d w , u < θ d w , u which is a contradiction.

Hence w = u , and fixed point of T is unique.

Secondly, we prove existence of fixed point.

Let x 0 X be a point. Define a sequence { x n } X by x n = T x n - 1 = T n x 0 n = 1,2 , 3 .

If x n 0 = x n 0 + 1 for some n 0 N , then x n 0 is a fixed point of T , and the proof is finished.

Assume that (14) x n - 1 x n n = 1,2 , 3 .

It follows from (6) and (14) that n = 1,2 , 3 , (15) 1 ξ θ d T x n - 1 , T x n , θ d x n - 1 , x n = ξ θ d x n , x n + 1 , θ d x n - 1 , x n < θ d x n - 1 , x n θ d x n , x n + 1 .

Consequently, we obtain that (16) θ d x n , x n + 1 < θ d x n - 1 , x n n = 1,2 , 3 , which implies (17) d x n , x n + 1 < d x n - 1 , x n n = 1,2 , 3 , .

Hence { d ( x n - 1 , x n ) } is a decreasing sequence, and so there exists r 0 such that (18) lim n d x n - 1 , x n = r .

We now show that r = 0 .

Assume that r 0 .

Then it follows from ( θ 2 ) that (19) lim n θ d x n - 1 , x n 1 , and so (20) lim n θ d x n - 1 , x n > 1 .

Let s n = θ ( d ( x n - 1 , x n ) ) and t n = θ ( d ( x n , x n + 1 ) ) n = 1,2 , 3 , .

From ( ξ 3 ) we obtain (21) 1 lim n sup ξ t n , s n < 1 which is a contradiction.

Thus we have (22) lim n d x n - 1 , x n = 0 and so (23) lim n θ d x n - 1 , x n = 1 .

We show that (24) lim n d x n - 1 , x n + 1 = 0 .

We consider three cases.

Case 1. x n x n + 2 n = 1,2 , 3 , .

From (6) and (14) we obtain that n = 1,2 , 3 , (25) 1 ξ θ d T x n - 1 , T x n + 1 , θ d x n - 1 , x n + 1 = ξ θ d x n , x n + 2 , θ d x n - 1 , x n + 1 < θ d x n - 1 , x n + 1 θ d x n , x n + 2 , and so (26) θ d x n , x n + 2 < θ d x n - 1 , x n + 1 n = 1,2 , 3 , which implies (27) d x n , x n + 2 < d x n - 1 , x n + 1 n = 1,2 , 3 , . Hence { d ( x n - 1 , x n + 1 ) } is decreasing.

In a manner similar to that which proved (22), we have (28) lim n d x n - 1 , x n + 1 = 0 .

Case 2. There exists n 0 1 such that x n 0 = x n 0 + 2 .

From the first term to the n 0 th term shall be removed, and let x n = x n 0 + n n = 1,2 , 3 , .

Then x n x n + 2 n = 1,2 , 3 , . By Case 1, we have (29) lim n d x n - 1 , x n + 1 = 0 .

Case 3. x n = x n + 2 n = 0,1 , 2 , .

We have (30) d x n - 1 , x n + 1 = 0 n = 1,2 , 3 , .

Hence (31) lim n d x n - 1 , x n + 1 = 0 .

In all cases, (24) is satisfied.

Now, we show that { x n } is bounded.

If { x n } is not bounded, then there exists a subsequence { x n ( k ) } of { x n } such that n 1 = 1 and k = 1,2 , 3 , , n ( k + 1 ) is the minimum integer greater than n ( k ) with (32) d x n k + 1 , x n k > 1 , d x l , x n k 1 for n ( k ) l n ( k + 1 ) - 1 .

Then we have (33) 1 < d x n k + 1 , x n k d x n k + 1 , x n k + 1 - 2 + d x n k + 1 - 2 , x n k + 1 - 1 + d x n k + 1 - 1 , x n k d x n k + 1 , x n k + 1 - 2 + d x n k + 1 - 2 , x n k + 1 - 1 + 1 .

By letting k in the above, we obtain (34) lim k d x n k + 1 , x n k = 1 .

By using (22), (34), and condition (d3), we deduce that (35) lim k d x n k + 1 - 1 , x n k - 1 = 1 .

It follows from ( θ 2 ) , (34), and (35) that (36) lim k θ d x n k + 1 , x n k > 1 , (37) lim n θ d x n k + 1 - 1 , x n k - 1 > 1 .

From (6) and (32) we infer that (38) 1 ξ θ d T x n k + 1 - 1 , T x n k - 1 , θ d x n k + 1 - 1 , x n k - 1 = ξ θ d x n k + 1 , x n k , θ d x n k + 1 - 1 , x n k - 1 < θ d x n k + 1 - 1 , x n k - 1 θ d x n k + 1 , x n k which implies (39) θ d x n k + 1 , x n k < θ d x n k + 1 - 1 , x n k - 1 .

Let (40) s n = θ ( d x n k + 1 - 1 , x n k - 1 , t n = θ d x n k + 1 , x n k n = 1,2 , 3 , .

Then t k < s k n = 1,2 , 3 , and lim k t k = lim k s k > 1 .

It follows from ( ξ 3 ) that (41) 1 lim k sup ξ t k , s k < 1 which is a contradiction.

Thus { x n } is bounded.

Now, we show that { x n } is a Cauchy sequence.

Let (42) M n = sup d x i , x j : i , j n .

Clearly, (43) 0 M n + 1 M n < n = 1,2 , 3 , and so there exists M 0 such that (44) lim n M n = M .

Assume that M > 0 .

It follows from (42) that k = 1,2 , 3 , there exist n ( k ) , m ( k ) k with (45) M k - 1 k < d x m k , x n k M k .

So (46) lim k d x m k , x n k = lim k M k = M .

It follows from (6) and (14) that (47) ξ θ d T x m k - 1 , T x n k - 1 , θ d x m k - 1 , x n k - 1 = ξ θ d x m k , x n k , θ d x m k - 1 , x n k - 1 < θ d x m k - 1 , x n k - 1 θ d x m k , x n k which implies (48) θ d x m k , x n k < θ d x m k - 1 , x n k - 1 . Hence we have (49) d x m k , x n k < d x m k - 1 , x n k - 1 d x m k - 1 , x m k + d x m k , x n k + d x n k , x n k - 1 .

Letting k in the above inequality, we obtain (50) lim n d x m k - 1 x n k - 1 = M .

Let (51) s k = θ d x m k - 1 , x n k - 1 , t k = θ d x m k , x n k k = 1,2 , 3 , .

Then t k < s k k = 1,2 , 3 , . Since M > 0 , (52) lim k t k = lim k s k > 1 .

Thus we have (53) 1 lim k sup ξ t k , s k < 1 which is a contradiction.

Hence M = 0 , and hence { x n } is a Cauchy sequence.

Since X is complete, there exists z X such that (54) lim n d x n , z = 0 .

Because T is continuous, (55) lim n d x n , T z = lim n d T x n - 1 , T z = 0 .

By Lemma 3, z = T z .

We give an example to illustrate Theorem 4.

Example 5.

Let X = { 1,2 , 3,4 } and define d : X × X [ 0 , ) as follows: (56) d 1,2 = d 2,1 = 3 , d 2,3 = d 3,2 = d 1,3 = d 3,1 = 1 , d 1,4 = d 4,1 = d 2,4 = d 4,2 = d 3,4 = d 4,3 = 4 , d x , x = 0 x X .

Then ( X , d ) is a complete generalized metric space, but not a metric space (see ).

Define a map T : X X by (57) T x = 3 x 4 , 1 x = 4 . And define a function θ : ( 0 , ) ( 1 , ) by (58) θ t = e t .

We now show that T is a L -contraction with respect to ξ b , where ξ b ( t , s ) = s k / t t , s 1 , k = 1 / 2 .

We have (59) d T x , T y = d 1,3 = 1 x = 4 , y 4 , d 1,1 = 0 x = 4 , y = 4 , d 3,3 = 0 x 4 , y 4 so (60) d T x , T y > 0 x = 4 , y 4 .

We have, for x = 4 and y 4 , (61) d x , y = 4 , d T x , T y = 1 .

We deduce that, for all x , y X with d ( T x , T y ) > 0 , (62) ξ b θ d T x , T y , θ d x , y = θ d x , y k θ d T x , T y = e 4 1 / 2 e 1 = e > 1 .

Thus all hypotheses of Theorem 4 are satisfied, and T has a fixed point x = 3 .

Note that Banach’s contraction principle is not satisfied with the usual metric ρ x , y = x - y x , y X . In fact, if x = 2 , y = 4 , then (63) ρ T 2 , T 4 k ρ 2,4 , k 0,1 which implies (64) k 1 .

Also, note that the θ -contraction condition  does not hold.

Let θ ( t ) = e t , t > 0 .

Then θ ( t ) satisfies conditions ( θ 1 ) , ( θ 2 ) , and ( θ 4 ) .

If (65) θ ρ T 2 , T 4 θ ρ 2,4 k , w h e r e k 0,1 then (66) e 2 e 2 k and so k 1 . Hence T is not θ -contraction map.

By taking ξ = ξ b in Theorem 4, we obtain Corollary 6.

Corollary 6.

Let ( X , d ) be a complete generalized metric space, and let T : X X be a mapping such that for all x , y X with d ( T x , T y ) > 0 (67) θ d T x , T y θ d x , y k where θ Θ and k ( 0,1 ) .

Then T has a unique fixed point.

Remark 7.

Corollary 6 is a generalization of Theorem 2.1 of  without condition ( θ 3 ) and Theorem 2.2 of  without condition ( θ 4 ).

By taking ξ = ξ w in Theorem 4, we obtain Corollary 8.

Corollary 8.

Let ( X , d ) be a complete generalized metric space, and let T : X X be a mapping such that for all x , y X with d ( T x , T y ) > 0 (68) θ d T x , T y θ d x , y ϕ θ d x , y where θ Θ and ϕ : [ 1 , ) [ 1 , ) is nondecreasing and lower semicontinuous such that ϕ - 1 ( { 1 } ) = 1 .

Then T has a unique fixed point.

Corollary 9.

Let ( X , d ) be a complete generalized metric space, and let T : X X be a mapping such that for all x , y X with d ( T x , T y ) > 0 (69) d T x , T y d x , y - φ d x , y where φ : [ 0 , ) [ 0 , ) is nondecreasing and lower semicontinuous such that φ - 1 ( { 0 } ) = 0 .

Then T has a unique fixed point.

Proof.

Condition (69) implies T is continuous.

Let θ ( t ) = e t , t > 0 .

From (69) we have that, for all x , y X with d ( T x , T y ) > 0 , (70) θ d T x , T y = e d T x , T y e d x , y - φ d x , y = e d x , y e φ d x , y .

Let φ ( t ) = ln ( ϕ ( θ ( t ) ) ) , t 0 , where ϕ : [ 1 , ) [ 1 , ) is nondecreasing and lower semicontinuous such that ϕ - 1 ( { 1 } ) = 1 .

Then φ is nondecreasing and lower semicontinuous, and (71) φ t = 0 ϕ θ t = 1 θ t = e t = 1 t = 0 .

It follows from (70) that, for all x , y X with d ( T x , T y ) > 0 , (72) θ d T x , T y θ d x , y e ln ϕ θ d x , y = θ d x , y ϕ θ d x , y .

By Corollary 8, T has a unique fixed point.

By taking θ ( t ) = 2 - 2 / π arctan ( 1 / t α ) , where α ( 0,1 ) , t > 0 in Corollary 8, we obtain the following result.

Corollary 10.

Let ( X , d ) be a complete generalized metric space, and let T : X X be a mapping such that for all x , y X with d ( T x , T y ) > 0 (73) 2 - 2 π arctan 1 d T x , T y α 2 - 2 / π arctan 1 / d x , y α ϕ 2 - 2 / π arctan 1 / d x , y α where α ( 0,1 ) and ϕ : [ 1 , ) [ 1 , ) is nondecreasing and lower semicontinuous such that ϕ - 1 ( { 1 } ) = 1 .

Then T has a unique fixed point.

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The author express his gratitude to the referees for careful reading and giving variable comments. This research was supported by Hanseo University.

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