Introduction and preliminaries

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THEOREMS HARDY-ROGERS-TYPE IN K-METRIC SPACES.
Paunović Lj.
University of Pristina, Faculty of Teacher Education in Prizren-Leposavic,Serbia
The purpose of the present paper is to establish coincidence point theorem for two mappings and fixed point
theorem for one mapping in K-metric spaces which satisfy contractive conditions of Hardy-Rogers type. Results
generalize fixed point theorems from K-metric spaces to symmetric spaces and metric spaces.Two examples are
given to illustrate the usability of results.
Introduction and preliminaries
Let E be a real Banach space. A subset P of E is called a cone if and only if
(a) P is closed, nonempty and P   ;
(b) a, b  R, a, b  0 , and x, y  P imply ax  by  P;
(c) P  ( P)  {}.
Given a cone P  E , we define a partial ordering  with respect to P by x  y if and only
if y  x  P , while x  y whill stand for y  x  int P (interior of P). A cone P  E , is
called normal if there is a number K  0 such that for all x, y  E ,
  x  y implies x  K y .
(1)
The least positive number satisfying the above inequality is called the normal constant of P. It
is clear that K  1 . Most of ordered Banach spaces used in applications posses a cone with
the normal constant K  1 , and if this is the case, proofs of the corresponding results are
much alike as in the metric setting. If K  1 , this is not the case. We know that there exists
ordered Banach space E with cone P which is not normal but with int P   .
The cone P is called regular if every increasing sequence in E which is bounded from above is
convergent. Every regular cone is normal, but the converse is not true.
Definition 1.1. Let X be a nonempty set. Suppose that the mapping d : X  X  E satisfies
(d1)   d ( x, y ) for all x, y  X and d ( x, y )   if and only if x  y ;
(d2) d ( x, y )  d ( y, x) for all x, y  X ;
(d3) d ( x, y )  d ( x, z )  d ( z, y ) for all x, y, z  X .
Then d is called a cone metric on X and (X,d) is called a cone metric space.
Definition 1.2. Let (X,d) be a cone metric space. We say that {x n } is
(1) a Cauchy sequence if for every c in E with   c , there is an N such that for all
n, m  N , d ( xn , xm )  c ;
(2) a convergent sequence if for every c in E with   c , there is an N such that for all
n>N, d ( xn , x)  c for some fixed x in X.
A cone metric space X is said to be complete if every Cauchy sequence in X is convergent in
X.
For the given cone metric space (X,d), one can construct a symmetric space (X,D) where
“symmetric” D : X  X  R is given by D( x, y)  d ( x, y) . In the case when (X,d) is
cone metric space with a normal cone P, then (1) implies
D( x, y)  K D( x, z)  D( z, y) ,
K is a normal constant of P.
x, y , z  X
Definition 1.3. Let (X,d) be a cone metric space and P a cone with nonempty interior.
Suppose that the mappings f , g : X  X are such that the range of g contains the range of
f, and f(X) or g(X) is a complete subspace of X. In this case we will say that the pair (f,g) is
Abbas and Jungck’s pair, or shortly AJ’s pair.
Definition 1.4. Let f and g be self-maps of a set X. If   fx  gx for some x in X, then x
is called a coincidence point of f and g, and  is called a point of coincidence of f and g.
Self-maps f and g are said to be weakly compatible if they commute at their coincidence
point, that is, if fx  gx for x  X , then fgx  gfx.
Proposition 1.1. Let f and g be weakly compatible self-maps of a set X. If f and g have a
unique point of coincidence   fx  gx , then  is the unique common fixed point of f
and g.
Main results
Theorem 2.1. Let (X,d) be a complete cone metric space, P a normal cone with normal
constant K. Suppose that (f,g) is AJ’s pair, and that there exist nonnegative constants a i ,
i  1,5 satisfying

3
i 1
ai  K (a4  a5 )  1 such that, for every x, y  X ,
(2)
D( fx, fy)  a1 D( gx, gy)  a2 D( gx, fx)  a3 D( gy, fy)  a4 D( gx, fy)  a5 D( gy, fx)
Then, f and g have a unique coincidence point in X. Moreover, if f and g are weakly
compatible, f and g have a unique common fixed point.
Proof. Let x0  X be arbitrary and let x1  X be chosen such that y0  f ( x0 )  g ( x1 ) .
f ( X )  g ( X ) . Let
This can be done, since
be such that
x2  X
y1  f ( x1 )  g ( x2 ). Continuing this process, having chosen x n  X , we choose x n 1 in X
such that y n  f ( xn )  g ( xn1 ). We have to show that
D( y n , y n1 )  D( y n1 , y n ) for some   [0,1), n  1.
(3)
From
D( y n , y n1 )  D( fxn , fxn1 )  a1 D( gxn , gxn1 )  a2 D( gxn , fxn )  a3 D( gxn1 , fxn1 )
 a4 D( gxn , fxn1 )  a5 D( gxn1 , fxn )
 a1 D( y n1 , y n )  a2 D( y n1 , y n )  a3 D( y n , y n1 )  a4 D( y n1 , y n1 )
 a 5 D( y n , y n )
 (a1  a2  Ka4 ) D( y n1 , y n )  (a3  Ka4 ) D( y n , y n1 ),
and from
D( y n1 , y n )  D( fxn1 , fxn )  a1 D( gxn1 , gxn )  a2 D( gxn1 , fxn1 )  a3 D( gxn , fxn )
 a4 D( gxn1 , fxn )  a5 D( gxn , fxn1 )
 a1 D( y n , y n1 )  a2 D( y n , y n1 )  a3 D( y n1 , y n )  a4 D( y n , y n )
 a5 D( y n1 , y n1 )
 (a1  a3  Ka5 ) D( y n1 , y n )  (a2  Ka5 ) D( y n , y n1 ),
we obtain
2D( y n1 , y n )  2a1  a2  a3  K (a4  a5 )D( y n , y n1 )  a2  a3  K (a4  a5 )D( y n1 , y n ),
that is,
2a  a  a3  K (a4  a5 )
n  1,2,...
D( y n1 , y n )  D( y n , y n1 ),
 1 2
 1,
2  (a2  a3  K (a4  a5 ))
Further, (3) implies that
D( y n , y n 1 )  D( y n 1 , y n 2 )  ...  n 1 D( y1 , y 0 ).
Now we shall show that
have
(4)
{ y n } is a Cauchy sequence. By the triangle inequality, for n>m we
d ( y n , y m )  d ( y n , y n1 )  d ( y n1 , y n2 )  ...  d ( y m1 , y m )
Hence, as P is a normal cone, we have
D( yn , ym )  d ( yn , ym )  K  d ( yn , yn1 )  d ( yn1 , yn2 )  ...  d ( ym1 , ym ) 
 K  d ( yn , yn1  d ( yn1 , yn2 )  ...  d ( ym1 , ym ) 
= KD( y n , y n1 )  KD( y n1 , y n2 )  ...  KD( y m1 , y m ).
Now by (4), it follow that
D( y n , y m )  K (n 1  n 2  ...  m ) D( y1 , y 0 )
Km
D( y1 , y 0 )  0, as m   .
1 
From [1] follows that { y n }  { fxn }  {gxn1} is a Cauchy sequence. Since g(X) is complete,

there exists a q in g(X) such that y n  q as n  . Consequently, we can find p in X
such that g(p) = q. We shall show that f(p) = q. Substituting x = p, y = x n
in (2), we get
D( fp, fxn )  a1 D(q, q)  a2 D( gp, fp)  a3 D( gxn , fxn )  a4 D( gp, fxn )  a5 D( gxn , fp).
According to [1], it follows
D( fp, q)  a1 D(q, q)  a2 D(q, fp)  a3 D(q, q)  a4 D(q, q)  a5 D(q, fp)
 (a2  a5 ) D( fp, q)  D( fp, q),
because a2  a5  i 1 ai  K (a4  a5 )  1. Now, if we suppose that fp  q , then we have
3
a contradiction. Hence, gp = fp = q. We shall show that f and g have a unique point of
coincidence. For this, assume that there exists another point of coincidence q1  q in X such
that fp1  gp1  q1 . Now,
D(q, q1 )  D( fp, fp1 )  a1 D( gp, gp1 )  a2 D( gp, fp)  a3 D( gp1 , fp1 )  a4 D( gp, fp1 )
 a5 D( gp1 , fp)
 a1 D(q, q1 )  a2 D(q, q)  a3 D(q1 , q1 )  a4 D(q, q1 )  a5 D(q1 , q)
 (a1  a4  a5 ) D(q1, q)  D(q1, q).
As
a1  a4  a5  i 1 ai  K (a4  a5 )  1 , we get D(q, q1 )  0 , that is, q  q1 . From the
3
Proposition 1.1, it follows that f and g have a unique common fixed point.
Theorem 2.2. Let (X,d) be a sequentially compact cone metric space, P a regular cone and
f : X  X a continuous mapping such that
(5)
d ( fx, fy)  a1d ( x, y)  a2 d ( x, fx)  a3 d ( y, fy)  a4 d ( x, fy)  a5 d ( y, fx)
for all x, y  X ,
where ai  [0,1) , i  1,5 and
x y

5
i 1
ai  1. Then, f has a unique
fixed point.
In order to prove Theorem 2.2., we shall need the following lemma.
Lemma 2.1. Let (X,d) be a cone metric space, f : X  X a mapping satisfying (5) for all
x, y  X , x  y where ai  [0,1) , i  1,5 and

5
i 1
ai  1. Then
d ( f 2 x, fx)  d ( x, fx) , for each x  X with x  fx .
Proof. Putting y  fx in (5), we have
d ( fx, f 2 x)  a1 d ( x, fx)  a 2 d ( x, fx)  a3 d ( fx, f 2 x)  a 4 d ( x, f 2 x)  a5 d ( fx, fx)
 (a1  a 2  a 4 )d ( x, fx)  (a3  a 4 )d ( fx, f 2 x),
that is,
a1  a2  a4
(6)
d ( x, fx) .
1  a3  a 4
By symmetry in (5), we have
(7)
d ( fy, fx)  a1d ( y, x)  a2 d ( y, fy)  a3 d ( x, fx)  a4 d ( y, fx)  a5 d ( x, fy)
i.e., putting y  fx in (7), we obtain
d ( f 2 x, f x)  a1 d ( fx, x)  a 2 d ( fx, f 2 x)  a3 d ( x, f x)  a 4 d ( fx, f x)  a5 d ( x, f 2 x)
d ( fx, f 2 x) 
 (a1  a3  a5 )d ( x, fx)  (a 2  a5 )d ( fx, f 2 x),
that is,
d ( f 2 x, fx) 
a1  a3  a5
d ( x, fx).
1  a 2  a5
(8)
 a  a 2  a4 a1  a3  a5 
k  min  1
,
  [0,1), then
 1  a3  a 4 1  a 2  a5 
d ( fx, f 2 x)  kd( x, fx)  d ( x, fx).
If
(9)
Proof of the Theorem 2.2. First, if u and v are two different fixed points of f,
according to (5), we have
d (u, v)  d ( fu, fv)  a1d (u, v)  a2 d (u, fu)  a3 d (v, fv)  a4 d (u, fv)  a5 d (v, fu)
 a1d (u, v)  a4 d (u, v)  a5 d (u, v)  (a1  a4  a5 )d (u, v)  d (u, v).
This is a contradiction.
Let x0  X . We define the sequence x n  f n x0 , n  0,1,2,... If xn1  xn for some n,
then x n is a fixed point of f. Suppose that xn1  xn for each n. Then, we have
d ( x n , x n 1 )  d ( f n x0 , f
 a1 d ( f
n 1
 a4 d ( f
n 1
n 1
x0 )  d ( ff n1 x0 , ff n x0 )
x0 , f n x0 )  a 2 d ( f
x0 , f
n 1
n 1
x0 , f n x0 )  a3 d ( f n x0 , f
n 1
x0 )
x0 )  a5 d ( f x0 , f x0 )
n
n
 a1d ( xn1 , xn )  a2 d ( xn1 , xn )  a3 d ( xn , xn1 )  a4 d ( xn1 , xn )
 a 4 d ( xn , xn1 ),
that is,
d ( x n 1 , x n )  d ( f
n 1
x0 , f n x0 )  d ( ff n x0 , ff n 1 x0 )
 a1 d ( f n x0 , f
n 1
x0 )  a 2 d ( f n x0 , f
 a 4 d ( f n x0 , f n x0 )  a5 d ( f
n 1
x0 , f
n 1
x0 )  a3 d ( f
n 1
n 1
x0 , f n x0 )
x0 )
 a1d ( xn1 , xn )  a2 d ( xn , xn1 )  a3 d ( xn1 , xn )  a5 d ( xn1 , xn )
 a5 d ( xn , xn1 ),
Now, we obtain
2d ( xn , xn1 )  (2a1  a2  a3  a4  a5 )d ( xn1 , xn )  (a2  a3  a4  a5 )d ( xn , xn1 ),
i.e.
2a  a2  a3  a4  a5
a 1
d ( xn , xn1 )  1
d ( xn1 , xn )  1
d ( xn1 , xn )  d ( xn1 , xn ) .
2  a 2  a3  a 4  a5
a1  1
Hence, the sequence d n  d ( xn , xn1 ) is strictly decreasing bounded below by  . Since P is
regular, there is d *  E such that
d n  d * (n  ). From the sequence compactness of
(X,d), there are subsequence { x ni } of {x n } and x *  X such that
xni  x * (i  ). Since mappings f and f
2
are continuous, we have
fxni  fx* and f 2 xni  f 2 x* .
By using [1] , we have
d ( fxni , xni )  d ( fx* , x * ) (i  ) and d ( f 2 xni , fxni )  d ( f 2 x* , fx* )
It is obvious that
(i  ) and
d ( fxni , xni )  d ni  d *  d ( fx* , x* )
d ( f 2 xni , fxni )  d ni 1  d *  d ( f 2 x * , fx* )
(i   ).
(i   ).
(10)
Now we shall prove that fx  x . If fx  x , then d  0. From (10) and according to
Lema 2.1., it follows
d *  lim d ni 1  lim d ( f 2 xni , fxni )  d ( f 2 x * , fx* )  d ( fx* , x * )  d * .
*
i 
*
*
*
*
i 
We have a contradiction, so fx*  x * . That is, x * is a fixed point of f. This completes
the proof of Theorem 2.2.
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