We prove two general fixed theorems for maps in G-metric spaces and then show that these maps satisfy property P.. Introduction Metric fixed point theory is an important mathematical dis
Trang 1Volume 2010, Article ID 401684, 12 pages
doi:10.1155/2010/401684
Research Article
1 Department of Mathematics, Maharshi Dayanand University, Rohtak 124001, India
2 Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
Correspondence should be addressed to Renu Chugh,chughrenu@yahoo.com
Received 19 December 2009; Revised 1 May 2010; Accepted 13 May 2010
Academic Editor: Brailey Sims
Copyrightq 2010 Renu Chugh et al 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
We prove two general fixed theorems for maps in G-metric spaces and then show that these maps satisfy property P
1 Introduction
Metric fixed point theory is an important mathematical discipline because of its applications
in areas such as variational and linear inequalities, optimization, and approximation theory Generalizations of metric spaces were proposed by Gahler 1, 2 called 2-metric spaces and Dhage 3, 4 called D-metric spaces Hsiao 5 showed that, for every contractive
definition, with x n : Tn x0, every orbit is linearly dependent, thus rendering fixed point theorems in such spaces trivial Unfortunately, it was shown that certain theorems
involving Dhage’s D-metric spaces are flawed, and most of the results claimed by Dhage
and others are invalid These errors were pointed out by Mustafa and Sims in 6, among others They also introduced a valid generalized metric space structure, which
they call G-metric spaces Some other papers dealing with G-metric spaces are those in
7 11
Let T be a self-map of a complete metric space X, d with a nonempty fixed point set FT Then T is said to satisfy property P if FT FT n for each n ∈ N An interesting fact about maps satisfying property P is that they have no nontrivial periodic points Papers dealing with property P are those in 12–14
In this paper, we will prove two general fixed point theorems for maps in G-metric spaces and then show that these maps satisfy property P Throughout this paper, we mean
by N the set of all natural numbers.
Trang 2Definition 1.1see 8 Let X be a nonempty set, and let G : X × X × X → Rbe a function satisfying the following axioms:
G1 Gx, y, z 0 if x y z,
G2 0 < Gx, x, y for all x, y ∈ X with x / y,
G3 Gx, x, y ≤ Gx, y, z, for all x, y, z ∈ X, with z / y,
G4 Gx, y, z Gx, z, y Gy, z, x · · · symmetry in all three variables,
G5 Gx, y, z ≤ Gx, a, a Ga, y, z, for all x, y, z, a ∈ X rectangle inequality Then the function G is called a generalized metric, or, more specifically, a G-metric on X, and
the pairX, G is called a G-metric space.
Definition 1.2 see 8 Let X, G and X, G be G-metric spaces and let f : X, G →
X, G be a function, then f is said to be G-continuous at a point a ∈ X; if given ε > 0, there exists δ > 0 such that x, y ∈ X; Ga, x, y < δ implies that Gfa, fx, fy < ε A function f is G-continuous on X if and only if it is G-continuous at all a ∈ X.
Proposition 1.3 see 8 Let X, G, X, G be G-metric spaces, then a function f : X → Xis G-continuous at a point x ∈ X if and only if it is G-sequentially continuous at x; that is, whenever {x n } is G-convergent to x, {fx n } is G-convergent to fx.
Definition 1.4see 8 Let X, G be a G-metric space, and let {x n} be a sequence of points
of X; therefore, we say that {x n } is G-convergent to x if lim n,m → ∞ Gx, x n , x m 0; that is, for
any ε > 0, there exists N ∈ N such that Gx, x n , x m < ε, for all n, m ≥ N We call x the limit
of the sequence and write x n → x or lim x n x.
Proposition 1.5 see 8 Let X, G be a G-metric space Then the following are equivalent:
1 {x n } is G-convergent to x,
2 Gx n , x n , x → 0, as n → ∞,
3 Gx n , x, x → 0, as n → ∞,
4 Gx m , x n , x → 0, as m, n → ∞.
Definition 1.6see 8 Let X, G be a G-metric space A sequence {x n } is called G-Cauchy
if, for each ε > 0, there is N ∈ N such that Gx n , x m , x l < ε, for all n, m, l ≥ N; that is, Gx n , x m , x l → 0 as n, m, l → ∞.
Proposition 1.7 see 8 In a G-metric space X, G the following are equivalent
1 The sequence {x n } is G-Cauchy.
2 For every ε > 0, there exists N ∈ N such that Gx n , x m , x m < ε, for all
n, m ≥ N.
Proposition 1.8 see 8 Let X, G be a G-metric space Then the function Gx, y, z is jointly continuous in all three of its variables.
Definition 1.9see 8 A G-metric space X, G is called a symmetric G-metric space if
G
x, y, y
Gy, x, x
∀x, y ∈ X. 1.1
Trang 3Proposition 1.10 see 8 Every G-metric space X, G defines a metric space X, d G by
d G
x, y
Gx, y, y
Gy, x, x
∀x, y ∈ X. 1.2
Note that, if X, G is a symmetric G-metric space, then
d G
x, y
2Gx, y, y
, ∀x, y ∈ X. 1.3
However, if X, G is not symmetric, then it holds by the G-metric properties that
3
2G
x, y, y
≤ d G
x, y
≤ 3Gx, y, y
, ∀x, y ∈ X. 1.4
In general, these inequalities cannot be improved.
Proposition 1.11 see 8 A G-metric space X, G is G-complete if and only if X, d G is a complete metric space.
Proposition 1.12 see 8 Let X, G be a G-metric space Then, for any x, y, z, a ∈ X, it follows that
1 if Gx, y, z 0, then x y z,
2 Gx, y, z ≤ Gx, x, y Gx, x, z,
3 Gx, y, y ≤ 2Gy, x, x,
4 Gx, y, z ≤ Gx, a, z Ga, y, z,
5 Gx, y, z ≤ 2/3{Gx, a, a Gy, a, a Gz, a, a}.
Theorem 1.13 see 15 Let T be a self-map of a metric space X such that X is T-orbitally complete Suppose that T satisfies
d
Tx, Ty
≤ k maxd
x, y
, dx, Tx, dy, Ty
, d
x, Ty
, d
y, Tx
, 1.5
where k is a real number satisfying 0 ≤ k < 1 Then T has a unique fixed point u ∈ X Moreover, for each x ∈ X, lim T n x u and
dT n x, u ≤ q n
1− q dx, Tx. 1.6
Trang 42 Fixed Point Theorems
Theorem 2.1 Let X, G be a complete G-metric space, and let T be a self-mapof X satisfying, for all
x, y, z ∈ X,
G
Tx, Ty, Tz
≤ k max
G
x, y, z
, Gx, Tx, Tx, Gy, Ty, Ty
, Gz, Tz, Tz,
G
x, Ty, Ty
Gz, Tx, Tx
G
x, Ty, Ty
Gy, Tx, Tx
G
y, Tz, Tz
Gz, Ty, Ty
2 , Gx, Tz, Tz Gz, Tx, Tx
2
,
2.1
where k is a constant satisfying 0 ≤ k < 1 Then T has a unique fixed point (say p) and T is G-continuous at p.
Proof Let x0 ∈ X and define the sequence {x n } by x n T n x0 We may assume that x n / x n1
for each n ∈ N ∪ {0} For, if there exists an N such that x N x N1 , then x Nis a fixed point of
T.
From2.1, with x x n−1 , y z x n,
Gx n , x n1 , x n1 ≤ k max Gx n−1 , x n , x n , Gx n−1 , x n , x n , Gx n , x n1 , x n1 ,
Gx n , x n1 , x n1 , Gx n−1 , x n1 , x n1 0
2 , Gx n−1 , x n1 , x n1 0
Gx n , x n1 , x n1,Gx n−1 , x n1 , x n1 0
2
,
2.2
Gx n , x n1 , x n1 ≤ kM n, say
Suppose that, for some n ∈ N, M n Gx n , x n1 , x n1 Then we have
Gx n , x n1 , x n1 ≤ kGx n , x n1 , x n1 , 2.3
which is a contradiction, since x n’s are distinct
Suppose that there is an n ∈ N for which M n Gx n−1 , x n1 , x n1 /2 Using property
G5,
Gx n−1 , x n1 , x n1 ≤ Gx n−1 , x n , x n Gx n , x n1 , x n1 , 2.4 and one obtains
Gx n , x n1 , x n1 ≤ k
2{Gx n−1 , x n , x n Gx n , x n1 , x n1 }, 2.5
Trang 5which leads to
Gx n , x n1 , x n1 ≤ k
2− k Gx n−1 , x n , x n < kGx n−1 , x n , x n , since k < 1. 2.6 Thus, we get
Gx n , x n1 , x n1 ≤ kGx n−1 , x n , x n ≤ · · · ≤ k n Gx0, x1, x1. 2.7
For every m, n ∈ N, m > n, using G5,
Gx n , x m , x m ≤ Gx n , x n1 , x n1 · · · Gx m−1 , x m , x m
≤ k n · · · k m−1 Gx0, x1, x1 ≤ k n
1− k Gx0, x1, x1. 2.8
Therefore{x n } is G-Cauchy, hence G-convergent, since X is G-complete Call the limit p.
From2.1 with x x n , y z p,
G
x n1 , Tp, Tp
≤ k max G
x n , p, p
, Gx n , x n1 , x n1 , Gp, Tp, Tp
, G
p, Tp, Tp
,
G
x n , Tp, Tp
Gp, x n1 , x n1
G
x n , Tp, Tp
Gp, x n1 , x n1
G
p, Tp, Tp
,
G
x n , Tp, Tp
Gp, x n1 , x n1
2
.
2.9
Taking the limit of both sides of2.9 as n → ∞ yields
G
p, Tp, Tp
≤ kGp, Tp, Tp
which implies that Gp, Tp, Tp 0 and hence p Tp.
Suppose that q is also a fixed point of T Then, from 2.1 with x p, y z q,
G
p, q, q
≤ k max
G
p, q, q
, 0, 0, 0,
G
p, q, q
Gq, p, p
G
p, q, q
Gq, p, p
2 , 0,
G
p, q, q
Gq, p, p
2
,
2.11
which implies that
G
p, q, q
≤ k
2− k G
q, p, p
Trang 6Using2.1 again, this time with x q, y z p, one obtains
G
q, p, p
≤ k max
G
q, p, p
, 0, 0, 0,
G
q, p, p
Gp, q, q
G
q, p, p
Gp, q, q
2 , 0,
G
q, p, p
Gp, q, q
2
,
2.13
which implies that
G
q, p, p
≤ k
2− k G
p, q, q
Combining2.12 and 2.14 gives
G
p, q, q
≤
k
2− k
2
G
p, q, q
Therefore, p q, since k/2 − k < 1.
Let{y n } ⊂ X be any sequence with limit p Using 2.1 with x z y n , y p,
G
Ty n , Tp, Ty n
≤ k max
G
y n , p, y n
, G
y n , Ty n , Ty n
, 0, G
y n , Ty n , Ty n
,
G
y n , p, p
Gy n , Ty n , Ty n
G
y n , p, p
Gp, Ty n , Ty n
2
,
2.16 That is,
G
Ty n , p, Ty n
≤ k max
G
y n , p, y n
, G
y n , Ty n , Ty n
,
G
y n , p, p
Gy n , Ty n , Ty n
G
y n , p, p
Gp, Ty n , Ty n
2
.
2.17 Using the fact that, fromG5,
G
y n , Ty n , Ty n
≤ Gy n , p, p
Gp, Ty n , Ty n
,
G
Ty n , p, Ty n
≤ kL, say. 2.18
Trang 7If, for some n, L is equal to Gy n , p, y n, then we have
G
Ty n , p, Ty n
≤ kGy n , p, y n
If, for some n, L is equal to Gy n , Ty n , Ty n, then, using G5,
G
Ty n , p, Ty n
≤ kGy n , Ty n , Ty n
≤ Gy n , p, p
Gp, Ty n , Ty n
, 2.20 which implies that
G
Ty n , p, Ty n
≤ k
1− k G
y n , p, p
If, for some n, L is equal to Gy n , p, p Gy n , Ty n , Ty n /2, then, using G5,
G
Ty n , p, Ty n
≤ k 2
G
y n , p, p
Gy n , p, p
Gp, Ty n , Ty n
, 2.22 which implies that
G
Ty n , p, Ty n
≤ 2k
2− k G
y n , p, p
If, for some n, L is equal to Gy n , p, p Gp, Ty n , Ty n /2, then, using G5,
G
Ty n , p, Ty n
≤ k 2
G
y n , p, p
Gp, Ty n , Ty n
, 2.24 which implies that
G
Ty n , p, Ty n
≤ k
2− k G
y n , p, p
Therefore, for all n, lim Gp, Ty n , Ty n 0 and T is G-continuous at p.
Special cases of Theorem2.1are Theorem 2.1 of 9 and Theorems 2.1, 2.4, 2.6, and 2.8
of10
Theorem 2.2 Let X, G be a complete G-metric space, and let T be a self-map of X satisfying, for all
x, y, z ∈ X,
G
Tx, Ty, Tz
≤ k maxG
x, y, z
, Gx, Tx, Tx, Gy, Ty, Ty
,
G
x, Ty, Ty
, G
y, Tx, Tx
, Gz, Tz, Tz,
2.26
Trang 8G
Tx, Ty, Tz
≤ k maxG
x, y, z
, Gx, x, Tx, Gy, y, Ty
,
G
x, x, Ty
, G
y, y, Tx
, Gz, z, Tx,
2.27
where k is a constant satisfying 0 ≤ k < 1 Then T has a unique fixed point (call it p) and T is G-continuous at p.
Proof Suppose that T satisfies 2.26 Using 2.26 with z y, we have
G
Tx, Ty, Ty
≤ k maxG
x, y, y
, Gx, Tx, Tx, Gy, Ty, Ty
, G
x, Ty, Ty
, G
y, Tx, Tx
.
2.28
Suppose thatX, G is symmetric.
From Proposition 1.10, d G , defined by d G x, y 2Gx, y, y makes, X, d G into a metric space Substituting into2.28 and then multiplying by 2 yield
d G
Tx, Ty
≤ k maxd G
x, y
, d G x, Tx, d G
y, Ty
, d G
x, Ty
, d G
y, Tx
. 2.29
From Theorem1.13, T has a unique fixed point.
Suppose thatX, G is not symmetric Define
A nG
T i x, T j x, T j x : 0≤ i, j ≤ n,
δ n max
Then δ n GT i x, T m x, T m x for some i, m satisfying 0 ≤ i, m ≤ n.
Suppose that i > 0 Then, from 2.26,
δ n Gx i , x m , x m
≤ k max{Gx i−1 , x m−1 , x m−1 , Gx i−1 , x i , x i , Gx m−1 , x m , x m , Gx i−1 , x m , x m , Gx m−1 , x i , x i}
≤ kδ n ,
2.31
a contradiction Therefore, i 0.
Thus, for some m satisfying 0 ≤ m ≤ n, using property G5 and 2.26,
δ n Gx0, x m , x m ≤ Gx0, x1, x1 Gx1, x m , x m
≤ Gx0, x1, x1 k max{Gx0, x m−1 , x m−1 , Gx0, x1, x1,
Gx m−1 , x m , x m , Gx0, x m , x m , Gx m−1 , x1, x1}
≤ Gx0, x1, x1 kδ n ,
2.32
Trang 9which implies that
δ n≤ 1
1− k Gx0, x1, x1, 2.33
and δ n is bounded in n Call this bound δ.
Define x n Tx n−1 Without loss of generality, we may assume that x n / x n1 for each n For, if there exists an N for which x N x N1 , then x N1 Tx N and x N is a fixed point of T.
Again from2.26,
Gx n , x n1 , x n1
≤ k max{Gx n−1 , x n , x n , Gx n−1 , x n , x n , Gx n , x n1 , x n1 , Gx n−1 , x n1 , x n1 , 0}
k max{Gx n−1 , x n , x n , Gx n−1 , x n1 , x n1}
≤ k max{Gx n−1 , x n , x n , δ}
≤ · · · ≤ k nmax{Gx0, x1, x1, δ} ≤ k n δ.
2.34
For any m, n ∈ N; m > n,
Gx n , x m , x m ≤ Gx n , x n1 , x n1 Gx n1 , x n2 , x n2
· · · Gx m−1 , x m , x m
≤ k n k n1 · · · k m−1 δ ≤ k
n δ
1− k .
2.35
Therefore, lim Gx n , x m , x m 0 as m, n → ∞ and {x n } is G-Cauchy, hence G-convergent, since X is G-complete Call the limit p.
From2.26,
G
x n , Tp, Tp
≤ k maxG
x n−1 , p, p
, Gx n , x n1 , x n1 , Gp, Tp, Tp
,
G
x n−1 , Tp, Tp
, G
p, x n , x n
.
2.36
Taking the limit of both sides of2.36 as n → ∞ yields
G
p, Tp, Tp
≤ kGp, Tp, Tp
which implies that p Tp.
Suppose that q is another fixed point of T with p / q Then, from 2.26,
G
p, q, q
≤ k maxG
p, q, q
, 0, 0, G
p, q, q
, G
q, p, p
kGq, p, p
. 2.38
Trang 10Again using2.26,
G
q, p, p
≤ k maxG
q, p, p
, 0, 0, G
q, p, p
, G
p, q, q
kGp, q, q
. 2.39
Combining2.36 and 2.38 gives Gp, q, q ≤ k2Gp, q, q, a contradiction Therefore p q
and the fixed point is unique
Now let{y n } ⊂ X with lim y n p Using 2.26,
G
Ty n , p, Ty n
≤ k maxG
y n , p, y n
, G
y n , Ty n , Ty n
, 0, G
y n , p, p
,
G
p, Ty n , Ty n
, G
y n , Ty n , Ty n
But fromG5, we have
G
y n , Ty n , Ty n
≤ Gy n , p, p
Gp, Ty n , Ty n
Therefore,2.40 reduces to
G
Ty n , p, Ty n
≤ max kG
y n , p, y n
, k
1− k G
y n , p, p
. 2.42
Taking the limit of both sides of the above equation as n → ∞ gives lim GTy n , p, Ty n 0,
which implies that lim Ty n p, and T is G-continuous at p.
The proof using2.27 is similar Special cases of Theorem2.2are Theorems 2.5, 2.8, and 2.9 of9
3 Property P
In this section we shall show that maps satisfying2.1 or 2.26 possess property P.
Theorem 3.1 Under the conditions of Theorem 2.1 , T has property P
Proof From Theorem2.1, T has a fixed point Therefore FT n / ∅ for each n ∈ N Fix n > 1 and assume that p ∈ FT n We wish to show that p ∈ FT.
Trang 11Suppose that p / Tp Using 2.1,
G
p, Tp, Tp
G T n p, T n1 p, T n1 p
≤ k max G
T n−1 p, T n p, T n p , G
T n−1 p, T n p, T n p , G
T n p, T n1 p, T n1 p ,
G
T n p, T n1 p, T n1 p ,
G
T n−1 p, T n1 p, T n1 p
0
G
T n−1 p, T n1 p, T n1 p
0
G
T n p, T n1 p, T n1 p
GT n p, T n1 p, T n1 p
G
T n−1 p, T n1 p, T n1 p
0 2
kG T n−1 p, T n p, T n p ≤ k2G
T n−2 p, T n−1 p, T n−1 p
≤ · · · ≤ k n G
p, Tp, Tp
,
3.1
a contradiction
Therefore p ∈ FT and T has property P
Theorem 3.2 Under the conditions of Theorem 2.2 , T has property P
Proof From Theorem2.2, T has a fixed point Therefore FT n / ∅for each n ∈ N Fix n > 1 and assume that p ∈ FT n Using 2.26 and assuming that p / Tp, we have
G
p, Tp, Tp
G T n p, T n1 p, T n1 p
≤ k maxG
T n−1 p, T n p, T n p , G
T n−1 p, T n p, T n p ,
G
T n p, T n1 p, T n1 p , G
T n−1 p, T n1 p, T n1 p , 0, 0
.
3.2
Define B n {GT i p, T j p, T j p : 0 ≤ i, j ≤ n} Then
δ n max
Then, δ n GT i p, T m p, T m p for some 0 ≤ i, m ≤ n.
... G < /p>T n−1 p, T n p, T n p , G < /p>
T n p, T n1 p, T n1 p , < /p>
G < /p>
T n p, ... < /p>
G < /p>
q, p, p< /i> < /p>
Gp, q, q < /p>
2 < /p>
< /p>
, < /p>
2.13 < /p>
which implies that < /p>
G < /p>
q, p, p< /i> < /p>
≤... n1 p, T n1 p< /i> < /p>
≤ k maxG < /p>
T n−1 p, T n p, T n p , G < /p>
T n−1 p, T n p,