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Suzuki type fixed point theorems for generalized multi-valued mappings in b-metric spaces

Abstract

In this paper, we obtained a new condition for a multi-valued mapping in a b-metric space, which guarantees the existence of its fixed point.

MSC:47H10, 54H25, 54E50.

1 Introduction

Let (X,d) be a metric space, and let CB(X) be a collection of all non-empty closed and bounded subsets of X. For every A,BCB(X), a Hausdorff metric H induced by the metric d of X is given by

H(A,B)=max { sup a A d ( a , B ) , sup b B d ( b , A ) } ,

where d(a,B)=inf{d(a,x),xB}.

For a multi-valued mapping T:X 2 X , a point xX is called a fixed point of T if xTx. We denote the set of fixed points of T by Fix(T).

Banach’s fixed point theorem is extended to the following result of Nadler [1] from the single-valued mappings to the multi-valued contractive mappings.

Theorem 1.1 [1]

Let (X,d) be a complete metric space, and let T:XCB(X) be a set-valued α-contraction, that is, a mapping, for which there exists a constant α(0,1) such that H(Tx,Ty)αd(x,y), x,yX. Then T has at least one fixed point.

The following remarkable generalization of the classical Banach contraction theorem due to Suzuki [2], states the following.

Theorem 1.2 [2]

For a metric space (X,d), define a nonincreasing function θ from [0,1) onto (1/2,1] by

θ(r)={ 1 if  0 r ( 5 1 ) / 2 , ( 1 r ) r 2 if  ( 5 1 ) / 2 r 2 1 / 2 , ( 1 + r ) 1 if  2 1 / 2 r < 1 .

The following are equivalent:

  1. (i)

    X is complete.

  2. (ii)

    Every mapping T on X such that there exists r[0,1), θ(r)d(x,Tx)d(x,y) implies that d(Tx,Ty)rd(x,y) for all x,yX has a fixed point.

Theorem 1.2 has been generalized to multi-valued mappings by Kikkawa and Suzuki [3], Mot and Petrusel [4], Dhompongsa and Yingtaweesittikul [5], Singh and Mishra [6], Shahzad and Bassindowa [7], and Aleomraninejad et al. [8].

The concept of a b-metric space was introduced by Czerwik (see [9] and [10]). We recall from [9] the following definition.

Definition 1.3 [9]

Let X be a set, and let s1 be a given real number. A function d:X×X R + is said to be a b-metric if and only if for all x,y,zX, the following conditions are satisfied:

  1. 1.

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

  2. 2.

    d(x,y)=d(y,x).

  3. 3.

    d(x,z)s[d(x,y)+d(y,z)].

A pair (X,d) is called a b-metric space.

We remark that a metric space is evidently a b-metric space. However, Czerwik (see [9, 10]) has shown that a b-metric on X need not be a metric on X.

We cite the following lemmas from Czerwik [911] and Singh et al. [6]

Lemma 1.4 Let (X,d) be a b-metric space. For any A,B,CCB(X) and any x,yX,

  1. 1.

    d(x,B)d(x,b) for any bB,

  2. 2.

    d(x,B)H(A,B) for any xA,

  3. 3.

    d(x,A)s[d(x,y)+d(y,A)].

Lemma 1.5 Let (X,d) be a b-metric space, and let A,BC(X). Then for each α>0 and for all bB, there exists aA such that

d(a,b)H(A,B)+α.

Some examples of b-metric spaces and some fixed point theorems for single-valued and multi-valued mappings in b-metric spaces can also be found in Czerwik [9], Boriceanu et al. [12], Boriceanu et al. [13], Aydi and Bota [14], Bota et al. [15], and Bota [16].

Theorem 1.6 [9]

Let (X,d) be a b-complete metric space, and let T:XCB(X) be a multi-valued mapping such that T satisfies the inequality

H(Tx,Ty)rd(x,y)for all x,yX,

where 0<r< 1 s . Then T has a fixed point.

Theorem 1.7 [16]

Let (X,d) be a b-complete metric space, and let T:XCB(X) be a multi-valued mapping. Suppose that there exist a,b,c>0 with c<1 and a + b 1 c < 1 s such that T satisfies the inequality

H(Tx,Ty)ad(x,y)+bd(x,Tx)+cd(y,Ty)

for all x,yX. Then T has a fixed point.

Theorem 1.8 [14]

Let (X,d) be a b-complete metric space, and let T:XCB(X) be a multi-valued mapping such that for all x,yX,

H(Tx,Ty)rmax { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) } ,

where r< 1 s 2 + s . Then T has a fixed point.

In 2011, Aleomraninejad et al. [17] gave a new condition for multi-valued mappings in a metric space, which guarantees the existence of its fixed point.

Consider a continuous function g: [ 0 , ) 5 [0,) satisfying the following conditions:

  1. (i)

    g(1,1,1,2,0)=g(1,1,1,0,2)=h(0,1).

  2. (ii)

    g is subhomogeneous, that is, for all ( x 1 , x 2 , x 3 , x 4 , x 5 ) [ 0 , ) 5 , α>0

    g(α x 1 ,α x 2 ,α x 3 ,α x 4 ,α x 5 )αg( x 1 , x 2 , x 3 , x 4 , x 5 ).
  3. (iii)

    If x i , y i [0,), x i < y i for i=1,,4, then

    g( x 1 , x 2 , x 3 , x 4 ,0)<g( y 1 , y 2 , y 3 , y 4 ,0)andg( x 1 , x 2 , x 3 ,0, x 4 )<g( y 1 , y 2 , y 3 ,0, y 4 ).

Theorem 1.9 [17]

Let (X,d) be a complete metric space, and let F,G:XCB(X) be two multi-valued mappings. Suppose that there exist α(0,1) and gR such that α(h+1)1 and αd(x,Fx)d(x,y) or αd(y,Gy)d(x,y) implies that

H(Fx,Gy)g ( d ( x , y ) , d ( x , F x ) , d ( y , G y ) , d ( x , G y ) , d ( y , F x ) )

for all x,yX. Then Fix(F)=Fix(G) and Fix(F) is non-empty.

The aim of this paper is to apply the concept of this function g to b-metric spaces.

Let s1 be fixed, and let R s be the set of all continuous functions g: [ 0 , ) 5 [0,) satisfying the conditions (ii), (iii) and

  1. (iv)

    g(1,1,1,2s,0)=g(1,1,1,0,2s)= h s (0,1/s).

Following the proofs in [18] and [17] with minor modification, we get the following results, respectively.

Lemma 1.10 If g R s and u,v[0,) are such that

u max { g ( v , v , u , s ( v + u ) , 0 ) , g ( v , v , u , 0 , s ( v + u ) ) , g ( v , u , v , s ( v + u ) , 0 ) , g ( v , u , v , 0 , s ( v + u ) ) } ,

then u h s v.

Proof Without loss of generality, we can suppose that ug(v,v,u,s(v+u),0).

If v<u, then

ug ( v , v , u , s ( v + u ) , 0 ) g(u,u,u,2us,0)ug(1,1,1,2s,0)= h s u<u

which is a contradiction. Thus uv. So,

ug ( v , v , u , s ( u + v ) , 0 ) g(v,v,v,2vs,0)vg(1,1,1,2s,0)= h s v.

 □

Lemma 1.11 Let (X,d) be a b-complete metric space, and let F,G:XCB(X) be two multi-valued mappings. Suppose that there exist α(0,) and g R s such that αd(x,Fx)d(x,y) or αd(y,Gy)d(x,y) implies that

H(Fx,Gy)g ( d ( x , y ) , d ( x , F x ) , d ( y , G y ) , d ( x , G y ) , d ( y , F x ) )

for all x,yX. Then Fix(F)=Fix(G).

Proof Let xFix(F), then αd(x,Fx)=0=d(x,x). Thus,

d ( x , G x ) H ( F x , G x ) g ( d ( x , x ) , d ( x , F x ) , d ( x , G x ) , d ( x , G x ) , d ( x , F x ) ) g ( 0 , 0 , d ( x , G x ) , d ( x , G x ) , 0 ) g ( 0 , 0 , d ( x , G x ) , s d ( x , G x ) , 0 ) .

Using Lemma 1.10, we have d(x,Gx) h s 0=0. So, xFix(G).

Hence Fix(F)Fix(G). Similarly, we can obtain Fix(G)Fix(F). □

2 Main results

Theorem 2.1 Let (X,d) be a b-complete metric space, and let F,G:XCB(X) be two multi-valued mappings. Suppose that there exist α(0,1) and g R s such that sα( h s +1)1 and αd(x,Fx)d(x,y) or αd(y,Gy)d(x,y) implies that

H(Fx,Gy)g ( d ( x , y ) , d ( x , F x ) , d ( y , G y ) , d ( x , G y ) , d ( y , F x ) )

for all x,yX. Then Fix(F)=Fix(G) and Fix(F) is non-empty.

Proof The main idea of the proof follows from Theorem 1.9.

By Lemma 1.11, Fix(F)=Fix(G). Let r( h s , 1 s ) and x 0 X. If x 0 is not a fixed point, choose x 1 F x 0 such that αd( x 0 ,F x 0 )<d( x 0 , x 1 ). Thus,

d ( x 1 , G x 1 ) H ( F x 0 , G x 1 ) g ( d ( x 0 , x 1 ) , d ( x 0 , F x 0 ) , d ( x 1 , G x 1 ) , d ( x 0 , G x 1 ) , d ( x 1 , F x 0 ) ) g ( d ( x 0 , x 1 ) , d ( x 0 , x 1 ) , d ( x 1 , G x 1 ) , s [ d ( x 0 , x 1 ) + d ( x 1 , G x 1 ) ] , 0 ) .

By Lemma 1.10, we have d( x 1 ,G x 1 ) h s d( x 0 , x 1 )<rd( x 0 , x 1 ). If x 1 is not a fixed point, there exists x 2 G x 1 such that d( x 1 , x 2 )<rd( x 0 , x 1 ). Since αd( x 1 ,G x 1 )<d( x 1 , x 2 ),

d ( x 2 , F x 2 ) H ( F x 2 , G x 1 ) g ( d ( x 1 , x 2 ) , d ( x 2 , F x 2 ) , d ( x 1 , G x 1 ) , d ( x 2 , G x 1 ) , d ( x 1 , F x 2 ) ) g ( d ( x 1 , x 2 ) , d ( x 2 , F x 2 ) , d ( x 1 , x 2 ) , 0 , s [ d ( x 1 , x 2 ) + d ( x 2 , F x 2 ) ] ) .

By Lemma 1.10, we have d( x 2 ,F x 2 ) h s d( x 1 , x 2 )<rd( x 1 , x 2 ).

Similarly, there exists x 3 F x 2 such that d( x 2 , x 3 )<rd( x 1 , x 2 )< r 2 d( x 0 , x 1 ).

By continuing this process, we obtain a sequence { x n } in X such that

x 2 n 1 F x 2 n 2 , x 2 n G x 2 n 1 , d ( x n , x n + 1 ) r n d ( x 0 , x 1 ) d ( x 2 n , F x 2 n ) h d ( x 2 n 1 , x 2 n ) and d ( x 2 n 1 , G x 2 n 1 ) h d ( x 2 n 2 , x 2 n 1 ) .

We prove next that the sequence { x n } is Cauchy,

d ( x n , x n + p ) s [ d ( x n , x n + 1 ) + d ( x n + 1 , x n + p ) ] = s d ( x n , x n + 1 ) + s d ( x n + 1 , x n + p ) s d ( x n , x n + 1 ) + s 2 [ d ( x n + 1 , x n + 2 ) + d ( x n + 2 , x n + p ) ] = s d ( x n , x n + 1 ) + s 2 d ( x n + 1 , x n + 2 ) + s 2 d ( x n + 2 , x n + p ) s d ( x n , x n + 1 ) + s 2 d ( x n + 1 , x n + 2 ) + s 3 d ( x n + 2 , x n + 3 ) + + s p 1 d ( x n + p 2 , x n + p 1 ) + s p 1 d ( x n + p 1 , x n + p ) s r n d ( x 0 , x 1 ) + s 2 r n + 1 d ( x 0 , x 1 ) + s 3 r n + 2 d ( x 0 , x 1 ) + + s p 1 r n + p 2 d ( x 0 , x 1 ) + s p 1 r n + p 1 d ( x 0 , x 1 ) , d ( x n , x n + p ) s r n ( 1 + s r + s 2 r 2 + + s p 2 r p 2 + s p 2 r p 1 ) d ( x 0 , x 1 ) s r n ( 1 + s r + s 2 r 2 + + s p 2 r p 2 + s p 1 r p 1 ) d ( x 0 , x 1 ) = s r n [ 1 ( s r ) p 1 s r ] d ( x 0 , x 1 ) .

Notice that

s r n [ 1 ( s r ) p 1 s r ] d( x 0 , x 1 )0as n.

So { x n } is Cauchy, and x n x for some xX.

Now, we claim that for each n1,

αd( x 2 n ,F x 2 n )d( x 2 n ,x)orαd( x 2 n + 1 ,G x 2 n + 1 )d( x 2 n + 1 ,x).

If αd( x 2 n ,F x 2 n )>d( x 2 n ,x) and αd( x 2 n + 1 ,G x 2 n + 1 )>d( x 2 n + 1 ,x) for some n1, then

d ( x 2 n , x 2 n + 1 ) s [ d ( x 2 n , x ) + d ( x 2 n + 1 , x ) ] < s [ α d ( x 2 n , F x 2 n ) + α d ( x 2 n + 1 , G x 2 n + 1 ) ] s [ α d ( x 2 n , x 2 n + 1 ) + α h d ( x 2 n , x 2 n + 1 ) ] = s ( α + α h ) d ( x 2 n , x 2 n + 1 ) = s α ( h s + 1 ) d ( x 2 n , x 2 n + 1 ) .

Thus, we get sα( h s +1)>1, which is a contradiction. By using the assumption, for each n1, either

H(F x 2 n ,Gx)g ( d ( x 2 n , x ) , d ( x 2 n , F x 2 n ) , d ( x , G x ) , d ( x 2 n , G x ) , d ( x , F x 2 n ) )

or

H(Fx,G x 2 n + 1 )g ( d ( x , x 2 n + 1 ) , d ( x , F x ) , d ( x 2 n + 1 , G x 2 n + 1 ) , d ( x , G x 2 n + 1 ) , d ( x 2 n + 1 , F x ) ) .

Therefore, one of the following cases holds.

  1. (a)

    There exists an infinite subset IN such that

    d ( x 2 n + 1 , G x ) H ( F x 2 n , G x ) g ( d ( x 2 n , x ) , d ( x 2 n , F x 2 n ) , d ( x , G x ) , d ( x 2 n , G x ) , d ( x , F x 2 n ) )

    for all nI.

  2. (b)

    There exists an infinite subset JN such that

    d ( F x , x 2 n + 2 ) H ( F x , G x 2 n + 1 ) g ( d ( x , x 2 n + 1 ) , d ( x , F x ) , d ( x 2 n + 1 , G x 2 n + 1 ) , d ( x , G x 2 n + 1 ) , d ( x 2 n + 1 , F x ) )

    for all nJ.

In case (a), we obtain

d ( x , G x ) s [ d ( x , x 2 n + 1 ) + d ( x 2 n + 1 , G x ) ] s [ d ( x , x 2 n + 1 ) + g ( d ( x 2 n , x ) , d ( x 2 n , F x 2 n ) , d ( x , G x ) , d ( x 2 n , G x ) , d ( x , F x 2 n ) ) ] s [ d ( x , x 2 n + 1 ) + g ( d ( x 2 n , x ) , d ( x 2 n , x 2 n + 1 ) , d ( x , G x ) , s [ d ( x 2 n , x ) + d ( x , G x ) ] , d ( x , x 2 n + 1 ) ) ]

for all nI. Since g is continuous, d(x,Gx)s(g(0,0,d(x,Gx),sd(x,Gx),0)). Using Lemma 1.10, d(x,Gx)=0. We have xGx.

In case (b), we obtain

d ( x , F x ) s [ d ( x , x 2 n + 2 ) + d ( x 2 n + 2 , F x ) ] s [ d ( x , x 2 n + 2 ) + g ( d ( x , x 2 n + 1 ) , d ( x , F x ) , d ( x 2 n + 1 , G x 2 n + 1 ) , d ( x , G x 2 n + 1 ) , d ( x 2 n + 1 , F x ) ) ] s [ d ( x , x 2 n + 2 ) + g ( d ( x , x 2 n + 1 ) , d ( x , F x ) , d ( x 2 n + 1 , x 2 n + 2 ) , d ( x , x 2 n + 2 ) , s [ d ( x 2 n + 1 , x ) + d ( x , F x ) ] ) ]

for all nJ. Since g is continuous, d(x,Fx)s(g(0,d(x,Fx),0,0,sd(x,Fx))). Using Lemma 1.10, d(x,Fx)=0. We have xFx. This completes the proof. □

Remark 2.2 Taking s=1 in Theorem 2.1 (case of metric spaces), we recover Theorem 1.9.

The following result is a consequence of Theorem 2.1.

Corollary 2.3 Let (X,d) be a b-complete metric space, and let T:XCB(X) be a multi-valued mapping. Suppose that there exist α(0,1) and g R s such that sα( h s +1)1 and αd(x,Tx)d(x,y) implies that

H(Tx,Ty)g ( d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) )

for all x,yX. Then T has a fixed point.

Corollary 2.4 Let (X,d) be a b-complete metric space, and let T:XCB(X) be a multi-valued mapping. Suppose that there exists r(0, 1 s ) such that 1 s ( r + 1 ) d(x,Tx)d(x,y) implies

H(Tx,Ty)rmax { d ( x , y ) , d ( x , T x ) , d ( y , T y ) }

for all x,yX. Then T has a fixed point.

Proof Let g R s by g( x 1 , x 2 , x 3 , x 4 , x 5 )=rmax{ x 1 , x 2 , x 3 }, where r< 1 s . Put α= 1 s ( r + 1 ) . Since h s =r< 1 s and sα( h s +1)1, by using Corollary 2.3, T has a fixed point. □

Remark 2.5 Corollary 2.4 is an extension of Theorem 1.6.

Corollary 2.6 Let (X,d) be a b-complete metric space, and let T:XCB(X) be a multi-valued mapping. Suppose that there exists a,b[0,1) and a+2b< 1 s such that 1 s ( 1 + a + 2 b ) d(x,Tx)d(x,y) implies that

H(Tx,Ty)ad(x,y)+bd(x,Tx)+bd(y,Ty)

for all x,yX. Then T has a fixed point.

Proof Let g R s be g( x 1 , x 2 , x 3 , x 4 , x 5 )=a x 1 +b( x 2 + x 3 ), where a+2b< 1 s . Put α= 1 s ( 1 + a + 2 b ) . Since h s =a+2b< 1 s and sα( h s +1)1, by using Corollary 2.3, T has a fixed point. □

The following examples show that we can apply Corollary 2.3 but cannot apply Theorem 1.8.

Example 2.7 Let X=[0,1] and d(x,y)= | x y | 2 for all x,yX. It is obvious that d is a b-metric on X with s=2 and (X,d) is complete. Also, d is not a metric on X. Define T:XCB(X) by

Tx={ { 1 3 , 2 3 } if  0 x < 1 , { 1 3 } if  x = 1 .

Let x,yX. Without loss of generality, take xy.

If x=y or x,y<1, then Tx=Ty. Hence H(Tx,Ty)=0.

If x<1 and y=1, then

H(Tx,Ty)= 1 9 4 27 = 1 3 4 9 = 1 3 d(y,Ty)rmax { d ( x , y ) , d ( x , T x ) , d ( y , T y ) } ,

where r= 1 3 < 1 2 = 1 s . So all the conditions of Corollary 2.4 are satisfied. Moreover, 1 3 and 2 3 are the two fixed points of T.

On the other hand, if we choose x= 1 3 and y=1, then

H ( T x , T y ) = 1 9 > 1 6 4 9 = 1 s 2 + s max { 4 9 , 0 , 4 9 , 0 , 1 9 } = 1 s 2 + s max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) } .

So we could not apply Theorem 1.8.

Example 2.8 Let X=[1,) and d(x,y)= | x y | 2 for all x,yX. Then (X,d) is a complete b-metric space with s=2. Define T:XCB(X) by

Tx= [ 1 , 1 + x 2 ] for all xX.

Consider H(Tx,Ty)= 1 4 ( x y ) 2 = 1 4 d(x,y)rmax{d(x,y),d(x,Tx),d(y,Ty)}, where r= 1 4 < 1 2 = 1 s for all x,yX. So all the conditions of Corollary 2.4 are satisfied. Moreover, 1 and 2 are the two fixed points of T.

On the other hand, if we choose x=1 and y=2, then

H ( T x , T y ) = 1 4 > 1 6 = 1 s 2 + s max { 1 , 0 , 0 , 0 , 1 4 } = 1 s 2 + s max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) } .

So we could not apply Theorem 1.8.

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Acknowledgements

The author would like to thank referees for their helpful comments and suggestions and Professor Sompong Dhompongsa for his appreciation and suggestion regarding this work. This work was supported by the Faculty of Science, Chiang Mai University, Chiang Mai, Thailand.

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Correspondence to Hatairat Yingtaweesittikul.

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Keywords

  • fixed point
  • b-metric space
  • multi-valued mappings