of the
MALAYSIAN MATHEMATICAL SOCIETY
Common Fixed Points for Compatible Mappings of Type (A)
1ZEQING LIU,1FENGRONG ZHANG AND 2JIANFENG MAO
1Department of Mathematics, Liaoning Normal University, Dalian, Liaoning 116029 People’s Republic of China
2Department of Mathematics, Xianning Teachers’College, Xianning, Hubei 437005 People’s Republic of China
Abstract. In this paper we give criteria for the existence of a common fixed point of a pair of mappings in 2-metric spaces and establish common fixed and coincidence point theorems for certain classes of contractive type mappings. The results presented here generalize, improve and unify a number of fixed point theorems given by Cho [1], Imdad et al. [5], Khan and Fisher [11], Kubiak [14], Murthy et al. [17], Rhoades [25], Singh et al. [34]and others.
1. Introduction
Gähler [4] introduced the concept of 2-metric spaces. A 2-metric space is a set X with a function d :X ×X × X →
[
0,∞)
satisfying the following conditions:
(G1) for two distinct points x,y∈X, there exists a point z∈ X such that ,
0 ) , ,
(x y z ≠
d
(G2) d(x,y,z) =0 if at least two of x,y,z are equal, (G3) d(x,y,z) = d(x,z,y) = d(y,z,x),
(G4) d(x,y,z) ≤ d(x,y,u)+d(x,u,z)+d(u,y,z) for all x,y,z,u∈X.
It has been shown by Gähler [4] that a 2-metricd is a continuous function of any one of its three arguments but it need not be continuous in two arguments. If it is continuous in two arguments, then it is continuous in all three arguments. A 2-metricdwhich is continuous in all of its arguments will be called continuous.
Iséki [7], for the first time, established a fixed point theorem in 2-metric spaces.
Since then a quite number of authors ([1]-[3], [5]-[36]) have extended and generalized the result of Iséki and various other results involving contractive and expansive type mappings. Especially, Murthy et al. [17] introduced the concepts of compatible mappings and compatible mappings of type (A) in 2-metric spaces, derived some relations between these mappings and proved common fixed point theorems for compatible mappings of type (A) in 2-metric spaces.
On the other hand, Cho [1], Constantin [2], Khan and Fisher[11] and Kubiak [14]
established some necessary and sufficient conditions which guarantee the existence of a common fixed point for a pair of continuous mappings in 2-metric spaces.
In this paper we establish criteria for the existence of a common fixed point of a pair of mappings in 2-metric spaces and establish common fixed and coincidence point theorems for certain classes of contractive type mappings. The results presented here generalize, improve and unify the corresponding results of Cho [1], Imdad et al. [5], Khan and Fisher [11], Kubiak [14], Murthy et al. [17], Rhoades [25], Singh et al. [34]
and others.
2. Preliminaries
Throughout this paper, Nandωdenote the sets of positive and nonnegative integers, respectively. Let R+ =[0,∞) and
{
: :( )
5 satisfiesconditions(a1)and(a2)}
,1
+
+ →
= ϕ ϕ R R
Φ
{
: :( )
11 satisfiesconditions(a1)and(a3)}
,2
+
+ →
= ϕ ϕ R R
Φ
where conditions (a1), (a2) and (a3) are as follows:
(a1) ϕ is upper semicontinuous, nondecreasing in each coordinate variable, (a2) b(t)= max
{
ϕ(t,0,0,t,t),ϕ(t,t,t,2t,0),ϕ(t,t,t,0,2t)}
< t for all t > 0, (a3) c(t) = max{
ϕ(t,t,t,0,2t,t,0,2t,0,2t,0),ϕ(t,0,0,t,t,0,0,0,0,0,t),ϕ(t,t,t,2t,0,t,2t,0,2t,0,0)
}
< t for allt > 0.Lemma 2.1. [37] For every t > 0 ,c(t)<t if and only if ( )= 0,
∞
→ c t lim n
n where cn
denotes the n-times composition of c .
Definition 2.1. A sequence {xn}n∈N in a 2-metric space (X,d) is said to be convergent to a point x∈X if ( , , ) =0
∞
→ d x x a
lim n
n for all a∈X. The point x is called the limit of the sequence {xn}n∈Nin X .
Definition 2.2. A sequence {xn}n∈N in a 2-metric space (X,d) is said to be a Cauchy
sequence if ( , , ) 0
, =
∞
→ d x x a
lim m n
n
m for all a∈X.
Definition 2.3. A 2-metric space(X,d) is said to be complete if every Cauchy sequence inX is convergent.
Note that, in a 2-metric space (X,d), a convergent sequence need not be a Cauchy sequence, but every convergent sequence is a Cauchy sequence when the 2-metric d is continuous on X ([19]).
Definition 2.4. Let f and g be mappings from a 2-metric spaces (X,d) into itself.
f and g are said to be compatible if
(
, ,)
= 0∞
→ d fgx gfx a
lim n n
n
for all a∈X, whenever {xn}n∈N ⊂ X such that lim fx lim gxn t
n n
n = =
∞
→
∞
→ for some
t∈X; f and g are said to be compatible of type (A) if
(
, ,)
= →∞(
, ,)
=0∞
→ d fgx ggx a lim d gfx ffx a
lim n n
n n n n
for all a∈X, whenever {xn}n∈N ⊂ X such that lim fx lim gxn t
n n
n = =
∞
→
∞
→ for some
. X t∈
Definition 2.5. A mapping f from a 2-metric space (X,d) into itself is said to be continuous at x∈X if for every sequence {xn}n∈N ⊂ X such that
0 ) , ,
( =
∞
→ d x x a
lim n
n for all a∈X, ( , , )= 0.
∞
→ d fx fx a
lim n
n f is called continuous on
X if it is so at all points of X.
Lemma 2.2. [1] Let f and g be compatible mappings of type (A) from a 2-metric spaces(X,d) into itself. If ft = gtfor some t∈X, then fgt = ggt = gft = fft.
Lemma 2.3. [1] Let f and g be compatible mappings of type (A) from a 2-metric spaces (X,d) into itself. If f is continuous at some t∈X and if
, t gx lim fx im
l n
n n
n = =
∞
→
∞
→ then lim gfxn ft.
n =
∞
→
Lemma 2.4. [1] Let f and g be compatible mappings from a 2-metric spaces )
,
(X d into itself. If f and g are continuous, then they are compatible of type (A).
3. Characterizations of common fixed points
Our main results are as follows:
Theorem 3.1. Let (X,d) be a complete 2-metric space with d continuous on X and let h and t be two mappings from X into itself. Then the following conditions are equivalent:
(1) h and t have a common fixed point;
(2) there exist r∈(0,1) ,f :X →t(X) and g :X →h(X) such that
(b1) the pairs f,h and g,t are compatible of type (A), (b2) one of f,g,h and t is continuous,
(b3) d
(
fx,gy,a)
≤ rmax{
d(
hx,ty,a) (
,d hx, fx,a) (
,d ty,gy,a)
, 21[
d(
hx,gy,a)
+d(
ty, fx,a) ] }
for all x,y,a∈X;(3) there exist ϕ∈Φ1, f :X →t(X) and g :X →h(X) satisfying conditions (b1), (b2) and (b4):
(b4) d
(
fx,gy,a)
≤ϕ(
d(
hx,ty,a) (
,d hx,fx,a) (
,d ty,gy,a) (
,d hx,gy,a) (
,d ty,fx,a) )
for all x y a, , ∈X;
(4) there exist ϕ∈Φ2, f :X →t(X) and g :X →h(X) satisfying conditions (b1), (b2) and (b5):
d2
(
fx,gy,a)
≤ ϕ(
d2(
hx,ty,a) (
,d hx,ty,a) (
d hx, fx,a)
, d(
hx,ty,a) (
d ty,gy,a)
,(b5)
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
(
ty gy a) (
d ty fx a) (
d hx gy a) (
d ty fx a))
d
a gy hx d a gy ty d
a fx ty d a fx hx d a gy hx d a fx hx d
a gy ty d a fx hx d
a fx ty d a ty hx d a gy hx d a ty hx d
, , , , , , , , ,
, , , ,
,
, , , , , , , , ,
,
, , , , ,
, , , , , , , , , ,
for all x y a, , ∈X;
Proof. (1)⇒(2) and (4). Let z be a common fixed point of h and t. Define )
(
:X t X
f → and g :X →h(X) by fx=gx=z for allx∈X. Then (b1) and (b2) hold. For each r∈(0,1) and ϕ ∈Φ2,(b3) and (b5) also hold.
) 3 ( ) 2
( ⇒ Takeϕ
(
u,v,w,x,y)
= rmax{
u,v,w,21(
x+ y) }
for all u v w x y, , , , ∈R+.Then ϕ ∈Φ1 and (b3) implies (b4).
) 1 ( ) 3
( ⇒ By using the method of Cho [1], we can similarly show that (3) implies (1).
) 1 ( ) 4
( ⇒ Let x0 be an arbitrary point in X. Since f(X)⊂t(X) and ,
) ( )
(X h X
g ⊂ there exist sequences {xn}n∈ω and {yn}n∈ω in X satisfying
1 2 2 2 1 2 2 1 2
2n = tx n+ = fx n, y n+ = hx n+ = gx n+
y for all n∈ω. Define
(
y y a)
d a
dn( ) = n, n+1, for all a∈X and n∈ω. We claim that for any i, j,k∈ω
(
yi,yj,yk)
=0.d (3.1) Suppose that d2n(y2n+2)> 0. Using (b5), we have
(
, ,) (
2(
2 2, 2 1, 2)
,2 1 2 2 2 2
n n n n
n
n gx y d hx tx y
fx
d + + ≤ϕ + +
d
(
hx2n+2,tx2n+1,y2n) (
d hx2n+2,fx2n+2,y2n)
,
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
(
,, ,,) (
,, ,,))
, ,, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
2 2 2 1 2 2 1 2 2 2
2 2 2 1 2 2 1 2 1 2
2 1 2 2 2 2 1 2 1 2
2 2 2 1 2 2 2 2 2 2
2 1 2 2 2 2 2 2 2 2
2 1 2 1 2 2 2 2 2 2
2 2 2 1 2 2 1 2 2 2
2 1 2 2 2 2 1 2 2 2
2 1 2 1 2 2 1 2 2 2
n n n n n n
n n n n n n
n n n n n n
n n n n n n
n n n n n n
n n n n n n
n n n n n n
n n n n n n
n n n n n n
y fx tx d y gx hx d
y fx tx d y gx tx d
y gx hx d y gx tx d
y fx tx d y fx hx d
y gx hx d y fx hx d
y gx tx d y fx hx d
y fx tx d y tx hx d
y gx hx d y tx hx d
y gx tx d y tx hx d
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
which implies that
( ) (
0,0,0,0,0,0,0,0,0,0,0) ( ( ) ) (
2 2)
,2 2 2 2 2 2 2
2 2
2n y n+ ≤ ≤ c d n y n+ <d n y n+
d ϕ
which is a contradiction. Henced2n(y2n+2) =0. Similarly, we have d2n+1(y2n+3)= 0. Consequently, dn(yn+2)= 0for all n∈ω. Note that
(
y ,y 2,a)
d(
y 2)
d( )
a d 1( )
a d( )
a d 1( )
a.d n n+ ≤ n n+ + n + n+ = n + n+ (3.2)
By (b5) and (3.2) we have
( ) ( )
( )
( ( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
(
, ,) (
, ,))
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , , ,
,
, , ,
, ,
2 2 1 2 1
2 2 2
2 2 1 2 1
2 1 2
1 2 2 2 1
2 1 2
2 2 1 2 2
2 2 2
1 2 2 2 2
2 2 2
1 2 1 2 2
2 2 2
2 2 1 2 1
2 2 2
1 2 2 2 1
2 2 2
1 2 1 2 1
2 2 2
2 2 2 2 1
2 2 2
1 2 2 2 2
1 2 2 2 2 2
1 2
a fx tx d a gx hx d
a fx tx d a gx tx d
a gx hx d a gx tx d
a fx tx d a fx hx d
a gx hx d a fx hx d
a gx tx d a fx hx d
a fx tx d a tx hx d
a gx hx d a tx hx d
a gx tx d a tx hx d
a fx hx d a tx hx d
a tx hx d
a gx fx d a d
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n n n
n n n
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+
+ + +
+ + +
+ + +
≤
= ϕ
( [ ]
[ ]
( ) [
( ) ( )]
,0)
.
, 0 , ) ( )
( ) (
, 0 , ) ( , ) ( )
( ) ( , 0
, ) ( , ) ( ) ( , ) (
1 2 2
2
1 2 2
1 2
2 1 2 1 2 2
2
2 2 1 2 2 2 2
a d a d a d
a d a d a d
a d a d a d a d
a d a d a d a d
n n
n
n n
n
n n
n n
n n
n n
+ + +
+ +
+
+ + +
=ϕ
Suppose that d2n+1(a)> d2n(a). Then
(
( ), ( ), ( ),0,2 ( ),)
( 2 12 2 12 2 12 2 12
2 1
2 a d a d a d a d a
d n+ ≤ϕ n+ n+ n+ n+
)
0 , ) ( 2 , 0 , ) ( 2 , 0 , )
( 2 12 2 12
2 1
2 a d a d a
d n+ n+ n+
(
( ))
2 12( ), 21
2 a d a
d
c n+ < n+
≤
which is a contradiction. Hence d2n+1(a) ≤d2n(a) and so d2n+12(a)≤ c(d2n2(a)). Similarly, we have d2n(a) ≤d2n−1(a) and d2n2(a) ≤c(d2n−12(a)). That is, for all
N n∈
(
( ))
.) ( ), ( )
( 12 2
1 a d a d a c d a
dn+ ≤ n n+ ≤ n
Hence for all n∈N
(
( )) (
( )) (
( ))
.)
( 2 2 12 1 02
2
1 a c d a c d a c d a
dn+ ≤ n ≤ n− ≤L≤ n+ (3.3)
It follows from (3.3) and Lemma 2.1 that 0 )
1( →
+ a
dn as n → ∞. (3.4)
Let n,m be in .ω If n ≥ m, then 0 =dm(ym)≥ dn(ym); if n < m, then
. 0 ) ( )
( ) (
) ( ) ( ) (
) ( )
( )
( ) (
1 2
1
1 1
1 1 1
1
=
≤
≤
≤
≤
+ +
≤
+ +
≤
+
−
−
+
−
+
−
−
−
n n m
n m
n
n n n n m
n
n m n m m
n m n
y d y
d y
d
y d y d y
d
y d y d y
d y d
L Thus, for any n,m∈ω
. 0 )
( m =
n y
d (3.5) For all i,j,k∈ω, we may, without loss of generality, assume that i < j. It follows from (3.5) that
. 0 ) ( )
, , (
) , , ( ) , , (
) , , ( ) ( ) ( ) , , (
1 1
2 1
1
=
=
≤
≤
≤
=
+ +
≤
−
−
+ +
+
k j k j j
k j i k
j i
k j i k
i j i k j i
y d y y y d
y y y d y y y d
y y y d y d y d y y y d
L Therefore (3.1) holds.
In order to show that {yn}n∈ω is a Cauchy sequence, by (3.4), it is sufficient to show that {y2n}n∈ω is a Cauchy sequence. Suppose that {y2n}n∈ω is not a Cauchy sequence. Then there exist ε > 0 and a∈ X such that for each even integer 2k, there are even integers2m(k) and 2n(k) with 2m(k) >2n(k) >2k and
. ) , ,
(y2 ( ) y2 ( ) a ≥ε d mk n k
For each even integer2k,let 2m(k) be the least even integer exceeding 2n(k) satisfying the above inequality, so that
(
y2 ( )−2,y2 ( ),a)
≤ε,d mk nk d
(
y2m(k),y2n(k),a)
> ε. (3.6) For each even integer 2k, by (3.1) and (3.6) we haveε < d
(
y2m(k),y2n(k),a)
( ) ( )
( )
( ) ( )
(
y y a)
d
a y
y d y
y y
d
y y y d
a y
y d a y y
d
k m k m
k m k
m k
m k m k m
k m k n k m
k m k m k
n k m
, ,
, ,
, ,
, ,
, ,
, ,
) ( 2 1 ) ( 2
1 ) ( 2 2 ) ( 2 1
) ( 2 ) ( 2 2 ) ( 2
2 ) ( 2 ) ( 2 ) ( 2
2 ) ( 2 ) ( 2 )
( 2 2 ) ( 2
−
−
−
−
−
−
−
−
+
+ +
≤ +
+
≤
ε
=ε +d2m(k)−2(a)+d2m(k)−1(a)
which implies that
(
, ,)
.lim 2 ( ) 2 ( ) =ε
∞
→ d y mk y n k a
k (3.7) It follows from (3.7) that
(
y y a) (
d y y a)
d nk , mk , nk , mk ,
0 < 2 ( ) 2 ( ) − 2 ( ) 2 ( )−2
≤d
(
y2m(k)−2,y2m(k),a)
≤d2m(k)−2(a)+d2m(k)−1(a). In view of (3.5) and (3.7) we immediately obtain that
(
, ,)
.lim 2 ( ) 2 ( )−2 =ε
∞
→ d y nk y mk a
k (3.8) Note that
(
y2n(k),y2m(k) 1,a) (
d y2n(k),y2m(k),a)
d2m(k) 1(a) d2m(k) 1(
y2n(k))
,d − − ≤ − + −
d
(
y2n(k)+1,y2m(k),a) (
−d y2n(k),y2m(k),a)
≤d2n(k)(a)+d2n(k)(
y2m(k))
,d
(
y2n(k)+1,y2m(k)−1,a) (
−d y2n(k),y2m(k)−1,a)
≤ d2n(k)(a)+d2n(k)(
y2m(k)−1)
.It is easy to see that
(
y y a)
d(
y y a)
d nk mk
k k m k
klim→∞ 2n( ), 2 ( )−1, = lim→∞ 2 ( )+1, 2 ( ),
(
, ,)
.lim 2 ( ) 1 2 ( ) 1 =ε
= + −
∞
→ d y nk y mk a
k (3.9)
It follows from (b5) that
( ) ( )
( )
( ( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
(
, ,) (
, ,)
,
, , ,
, ,
, , ,
, ,
, , ,
, ,
, , ,
, ,
, , ,
, ,
, , ,
, ,
, , ,
, ,
, ,
) ( 2 1 ) ( 2 )
( 2 ) ( 2
1 ) ( 2 ) ( 2 )
( 2 ) ( 2
1 ) ( 2 1 ) ( 2 )
( 2 ) ( 2
) ( 2 1 ) ( 2 1
) ( 2 ) ( 2
1 ) ( 2 ) ( 2 1
) ( 2 ) ( 2
1 ) ( 2 1 ) ( 2 1
) ( 2 ) ( 2
) ( 2 ) ( 2 1
) ( 2 ) ( 2
1 ) ( 2 ) ( 2 2
1 ) ( 2 ) ( 2 1
) ( 2 ) ( 2
a fx tx
d a fx hx
d
a gx
hx d a fx hx
d
a gx
tx d a fx hx
d
a fx tx
d a tx
hx d
a gx
hx d a tx
hx d
a gx
tx d a tx
hx d
a fx hx
d a tx
hx d
a tx
hx d
a gx
fx d a y
y d
k m k
n k
m k m
k n k m k
m k m
k n k
n k
m k m
k m k
n k
n k m
k n k m k
n k m
k n k
n k
n k m
k m k m k
n k m
k n k m
k n k m k
n k m
+
+ + +
+ +
+ +
+ +
+ + + + +
≤
= ϕ
( ) ( )
( ) ( )
(
hx gx a) (
dtx fx a))
d
a fx tx
d a gx
tx d
a gx
hx d a gx
tx d
k m k
n k
n k m
k m k
n k
n k
n
k n k m k
n k
n
, ,
, ,
, , ,
, ,
, , ,
, ,
) ( 2 1 ) ( 2 1
) ( 2 ) ( 2
) ( 2 1 ) ( 2 1
) ( 2 1 ) ( 2
1 ) ( 2 ) ( 2 1
) ( 2 1 ) ( 2
+ +
+ +
+
+ +
+
( )
( ( )
( )
(
, ,) (
, ,)
,
, ) ( ,
,
, ) ( ,
, ,
, ,
1 ) ( 2 1 ) ( 2 ) ( 2 1 ) ( 2
) ( 2 ) ( 2 1 ) ( 2
1 ) ( 2 ) ( 2 1 ) ( 2 )
( 2 1 ) ( 2 2
a y
y d a y y
d
a d a y y
d
a d
a y y
d a y y
d
k n k m k
n k m
k n k n k m
k m k
n k m k
n k m
+
−
−
−
−
−
=ϕ −
d
(
y2m(k)−1,y2n(k),a) (
d y2(k),y2m(k),a)
,
( )
( )
( ) (
, ,)
,, , , )
(
, , ,
) (
, ) ( ) (
1 ) ( 2 1 ) ( 2 )
( 2
) ( 2 ) ( 2 1
) ( 2
1 ) ( 2 1 ) ( 2 1
) ( 2
) ( 2 1 ) ( 2
a y
y d a d
a y y d a d
a y
y d a d
a d a d
k n k m k
n
k m k n k
m
k n k m k
m
k n k
m
+
−
−
+
−
−
−
( ) (
( )) ( )
(
2 ( ), 2 ( ), ,))
, , , , ,1 ) ( 2 1 ) ( 2 2
) ( 2 )
( 2
a y y d
a y
y d a y y d a d
k m k n
k n k m k
m k n k
n − +
Letting k → ∞, by (3.9), (3.7) and (3.5) we have
(
2,0,0, 2, 2,0,0,0,0,0, 2) ( )
2 2,2 ϕ ε ε ε ε ε ε
ε ≤ ≤c <
which is a contradiction. Therefore {y2n}n∈ω is a Cauchy sequence inX. It follows from completeness of (X,d)that {yn}n∈ω converges to a point u∈X .
Now, suppose that t is continuous. Since g and t are compatible of type (A) and
ω
∈ + n
gx n }
{ 2 1 and {tx2n+1}n∈ω converge to the point ,u by Lemma 2.3 we get that tu
ttx
gtx2n+1, 2n+1 → as n→ ∞. In virtue of (b5), we have
( ) ( ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ))
.
, , , ,
, , , , ,
,
2 1 2 1
2 2
2 1 2 1
2 1 2
1 2 2 1
2 1 2
2 1 2 2
2
1 2 2 2
2
1 2 1 2 2
2
2 1 2 1
2 2
1 2 2 1
2 2
1 2 1 2 1
2 2
2 2 1
2 2 1
2 2 2
1 2 2
,a ,fx ttx d ,a ,gtx hx d
, ,a ,fx ttx d ,a ,gtx ttx d
, ,a ,gtx hx d ,a ,gtx ttx d
, ,a ,fx ttx d ,a ,fx hx d
, ,a ,gtx hx d ,a ,fx hx d
, ,a ,gtx ttx d ,a ,fx hx d
, ,a ,fx ttx d ,a ,ttx hx d
, ,a ,gtx hx d ,a ,ttx hx d
, ,a ,gtx ttx d ,a ,ttx hx d
a fx hx d a ttx hx d a ttx hx d a
gtx fx d
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n
n n n
n n
n n
n
+ +
+ +
+
+ +
+
+ +
+ +
+ +
+ +
+ + +
+ +
+ ≤ϕ