Ann. Funct. Anal. 3 (2012), no. 2, 107–114
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL: www.emis.de/journals/AFA/
COMMON COUPLED FIXED POINT THEOREMS IN
d-COMPLETE TOPOLOGICAL SPACES
K. P. R. RAO1∗ , K. R. K. RAO1 AND ERDAL KARAPINAR2
Communicated by H.-K. Xu
Abstract. In this paper, we give two unique common coupled fixed point theorems for mappings satisfying a generalized condition in d-complete topological
spaces.
1. Introduction and preliminaries
In 1992, Hicks [5] introduced the notion of d-complete topological spaces as a
generalization of complete metric spaces.
Definition 1.1. (See [5]) Let (X, τ ) be a topological space and d : X × X →
[0, ∞) satisfy
(i) d(x, y) = 0 if and only if x = y ,
(ii) for any sequence {xn } in X,
∞
X
d(xn , xn+1 ) < ∞ ⇒ {xn } is convergent in (X, τ ).
n=1
Then the triplet (X, τ, d) is called a d-complete topological space.
For details on d-complete topological spaces, we refer to Iseki [9] and Kasahara
[10, 11, 12]. Hicks [5] and Hicks and Rhoades [6, 7] proved several fixed point
theorems in d- complete topological spaces. Hicks and Saliga [8] and Saliga [18]
obtained fixed point theorems for non-self maps in d-complete topological spaces.
In 2006, Bhaskar and Lakshmikantham [2] introduced the notion of a coupled
fixed point in partially ordered metric spaces, also discussed some problems of
Date: Received: 3 July 2011; Revised: 30 October 2011; Accepted: 7 April 2012.
∗
Corresponding author: K.P.R.Rao.
2010 Mathematics Subject Classification. Primary 47H10; Secondary 54H25, 54E50.
Key words and phrases. Coupled fixed point, d-complete topological spaces, weakly
compatible.
107
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K.P.R. RAO, K.R.K. RAO, E. KARAPINAR
the uniqueness of a coupled fixed point and applied their results to the problems
of the existence and uniqueness of a solution for the periodic boundary value
problems.
Later several authors proved coupled fixed and common coupled fixed point
theorems in partial ordered metric spaces, partially ordered cone metric spaces
and cone metric spaces for two maps (see e.g. [1, 3, 4],[13]-[17],[19]-[22].)
In this paper, we prove a common coupled fixed point theorem for four mappings in d-complete topological spaces.
Definition 1.2. (See [2]) Let X be a nonempty set. An element (x, y) ∈ X × X
is called a coupled fixed point of the mapping F : X × X → X if x = F (x, y)
and y = F (y, x).
Definition 1.3. (See [1]) Let X be a nonempty set. An element (x, y) ∈ X × X
is called
(i) a coupled coincidence point of F : X × X → X and g : X → X if
gx = F (x, y) and gy = F (y, x).
(ii) a common coupled fixed point of F : X × X → X and g : X → X if
x = gx = F (x, y) and y = gy = F (y, x).
Definition 1.4. (See [1]) Let X be a nonempty set. The mappings F : X ×
X → X and g : X → X are called weakly-compatible if g(F (x, y)) = F (gx, gy)
whenever gx = F (x, y) and gy = F (y, x) for some (x, y) ∈ X × X.
2. An implicit relation
Let Φ7 be the family of all continuous mappings φ : R7+ → R satisfying the
following conditions :
(φ1 ) there exists 0 ≤ h1 < 1 such that u, v, w, p ≥ 0 with
(φa ) φ(u, v, w, u, v, p, 0) ≤ 0 or,
(φb ) φ(u, v, w, v, u, 0, p) ≤ 0
implies u ≤ h1 max{v, w},
(φ2 ) there exists 0 ≤ h2 < 1 such that u, v ≥ 0 with
φ(u, u, v, 0, 0, u, u) ≤ 0 ⇒ u ≤ h2 v.
Example 2.1.
φ(t1 , t2 , t3 , t4 , t5 , t6 , t7 ) = t1 −
α
(t2 + t3 ) − β(t4 + t5 ),
2
where α, β ≥ 0 with α + 2β < 1.
If u = 0, then (φ1 ) and(φ2 ) are satisfied. Assume that u > 0.
φ(u, v, w, u, v, p, 0) ≤ 0 ⇒ u − α2 (v + w) − β(v + u) ≤ 0
⇒ u ≤ α2 (v + w) + β(v + u)
⇒ (1 − β)u ≤ α max{v, w} + β max{v, w}
⇒ u ≤ h1 max{v, w},
α+β
< 1.
where h1 = α−β
Similarly φ(u, v, w, v, u, 0, p) ≤ 0 ⇒ u ≤ h1 max{v, w}. φ(u, u, v, 0, 0, u, u) ≤
α/2
0 ⇒ u ≤ h2 v, where h2 = 1−α/2
< 1.
COMMON COUPLED FIXED POINT THEOREMS
109
Example 2.2.
φ(t1 , t2 , t3 , t4 , t5 , t6 , t7 ) = t1 − α max{t2 , t3 , t4 , t5 } − β min{t6 , t7 },
where α, β ≥ 0 with α + β < 1.
Example 2.3.
φ(t1 , t2 , t3 , t4 , t5 , t6 , t7 ) = t1 −
α
(t2 + t3 ) − L min{t4 , t5 , t6 , t7 },
2
where 0 ≤ α < 1 and L ≥ 0.
Example 2.4.
φ(t1 , t2 , t3 , t4 , t5 , t6 , t7 ) = t1 − αt2 − βt3 − L min{t4 , t5 , t6 , t7 },
where α, β ≥ 0 such that α + β < 1 and L ≥ 0.
Example 2.5.
α 2
(t + t23 ) − βt4 t5 − γ t6 t7 ,
2 2
where α, β, γ ≥ 0 with α + β < 1 and α + γ < 1.
φ(t1 , t2 , t3 , t4 , t5 , t6 , t7 ) = t21 −
Example 2.6.
φ(t1 , t2 , t3 , t4 , t5 , t6 , t7 ) = t21 −
t2 + t25
α 2
(t2 + t23 ) − β 4
,
2
t6 + t7 + 1
where α, β ≥ 0 with α + 2β < 1.
3. Main results
Theorem 3.1. . Let (X, τ ) be a Hausdorff topological space. Let F, G : X ×X →
X and f, g : X → X be mappings satisfying
d(F (x, y), G(u, v)), d(f x, gu), d(gv, f y), d(F (x, y), f x),
≤0
(3.1)
φ
d(G(u, v), gu), d(F (x, y), gu), d(G(u, v), f x)
for all x, y, u, v ∈ X, where φ ∈ Φ7 ,
(a) F (X × X) ⊆ g(X), G(X × X) ⊆ f (X),
(b) one of (f (X), τ, d) and (g(X), τ, d) is d-complete,
(c) the pairs (F, f ) and (G, g) are weakly compatible,
(d) d(x, y) = d(y, x) for all x, y ∈ X and
(e) for each y ∈ X, d(xn , y) → d(x, y), whenever {xn } ⊆ X, x ∈ X such that
xn → x.
Then there exists a unique (α, β) ∈ X × X such that
f α = gα = α = F (α, β) = G(α, β) and f β = gβ = β = F (β, α) = G(β, α).
Proof. Let x0 and y0 be in X. Since F (X × X) ⊆ g(X), we can choose x1 , y1 ∈ X
such that
gx1 = F (x0 , y0 ) and gy1 = F (y0 , x0 ).
Since G(X × X) ⊆ f (X), we can choose x2 , y2 ∈ X such that
f x2 = G(x1 , y1 ) and f y2 = G(y1 , x1 ).
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K.P.R. RAO, K.R.K. RAO, E. KARAPINAR
Continuing this process, we can construct the sequences {xn } and {yn } in X such
that
gx2n+1 = F (x2n , y2n ) = z2n+1 ,
gy2n+1 = F (y2n , x2n ) = p2n+1 ,
f x2n+2 = G(x2n+1 , y2n+1 ) = z2n+2 ,
f y2n+2 = G(y2n+1 , x2n+1 ) = p2n+2 .
Putting x = x2n , y = y2n , u = x2n+1 , v = y2n+1 in (3.1), we get
d(z2n+1 , z2n+2 ), d(z2n , z2n+1 ), d(p2n+1 , p2n ), d(z2n+1 , z2n ),
φ
≤0
d(z2n+2 , z2n+1 ), 0, d(z2n+2 , z2n )
From (φb ), we have
d(z2n+1 , z2n+2 ) ≤ h1 max{d(z2n , z2n+1 ), d(p2n+1 , p2n )}
where 0 ≤ h1 < 1 . Also putting x = y2n , y = x2n , u = y2n+1 , v = x2n+1 in (3.1)
and using (φb ), we get
d(p2n+1 , p2n+2 ) ≤ h1 max{d(z2n , z2n+1 ), d(p2n+1 , p2n )}.
Thus
max{d(z2n+1 , z2n+2 ), d(p2n+1 , p2n+2 )} ≤ h1 max{d(z2n , z2n+1 ), d(p2n+1 , p2n )}.
Similarly, using (3.1) and (φa ), we can show that
max{d(z2n , z2n+1 ), d(p2n+1 , p2n )} ≤ h1 max{d(z2n , z2n−1 ), d(p2n−1 , p2n )}.
Hence
max{d(zn , zn+1 ), d(pn , pn+1 )} ≤ h1 max{d(zn−1 , zn ), d(pn−1 , pn )}.
Inductively we have
max{d(zn , zn+1 ), d(pn , pn+1 )} ≤ hn1 max{d(z0 , z1 ), d(p0 , p1 )}.
∞
P
P
P
n
d(zn , zn+1 ) and ∞
Since ∞
n=1 d(pn , pn+1 )
n=1 h1 is convergent, it follows that
n=1
are convergent. Hence d(zn , zn+1 ) → 0 and d(pn , pn+1 ) → 0 as n → ∞.
Suppose (f (X), τ, d) is d-complete. Then {z2n+2 } = {f x2n+2 } ⊆ f (X) and
{p2n+2 } = {f y2n+2 } ⊆ f (X) converge to α and β respectively for some α, β ∈
f (X). Hence there exist x and y in X such that α = f x and β = f y. Also the
subsequences {z2n+1 } and {p2n+1 } converge to α and β, respectively.
Putting x = x, y = y, u = x2n+1 , v = y2n+1 in (3.1), we get
d(F (x, y), z2n+2 ), d(f x, z2n+1 ), d(p2n+1 , f y), d(F (x, y), f x),
φ
≤0
d(z2n+2 , z2n+1 ), d(F (x, y), z2n+1 ), d(z2n+2 , f x)
Letting n → ∞, we get
φ(d(F (x, y), f x), 0, 0, d(F (x, y), f x), 0, d(F (x, y), f x), 0) ≤ 0.
From (φa ), we have F (x, y) = f x = α. Replacing x, y, u, v with y, x, y2n+1 , x2n+1
respectively in (3.1) and letting n → ∞ and by using (φa ), we get
F (y, x) = f y = β.
COMMON COUPLED FIXED POINT THEOREMS
111
Since the pair (F, f ) is weakly compatible, we have
f α = f (F (x, y)) = F (f x, f y) = F (α, β) and
f β = f (F (y, x)) = F (f y, f x) = F (β, α).
Putting x = α, y = β, u = x2n+1 , v = y2n+1 in (3.1), we get
d(f α, z2n+2 ), d(f α, z2n+1 ), d(p2n+1 , f β), 0,
φ
≤0
d(z2n+2 , z2n+1 ), d(f α, z2n+1 ), d(z2n+2 , f α)
Letting n → ∞, we get
φ(d(f α, α), d(f α, α), d(f β, β), 0, 0, d(f α, α), d(f α, α)) ≤ 0.
From (φ2 ), we have d(f α, α) ≤ h2 d(f β, β), where 0 ≤ h2 < 1.
Putting x = β, y = α, u = y2n+1 , v = x2n+1 in (3.1) and letting n → ∞ and then
using (φ2 ), we get
d(f β, β) ≤ h2 d(f α, α).
Hence f α = α and f β = β. Thus
α = f α = F (α, β) and β = f β = F (β, α).
(3.2)
Since F (X × X) ⊆ g(X), there exist γ, δ ∈ X such that
gγ = F (α, β) = f α = α and gδ = F (β, α) = f β = β.
Putting x = α, y = β, u = γ, v = δ in (3.1),we get
φ(d(gγ, G(γ, δ)), 0, 0, 0, d(G(γ, δ), gγ), 0, d(G(γ, δ), gγ)) ≤ 0.
From (φb ), we have gγ = G(γ, δ).
Putting x = β, y = α, u = δ, v = γ in (3.1) and using (3.2) and (φb ), we have
gδ = G(δ, γ). Since the pair (G, g) is weakly compatible, we have
gα = g(gγ) = g(G(γ, δ)) = G(gγ, gδ) = G(α, β) and
gβ = g(gδ) = g(G(δ, γ)) = G(gδ, gγ) = G(β, α).
Putting x = x2n , y = y2n , u = α, v = β in (3.1) and letting n → ∞ and then
using (φ2 ), we get
d(α, gα) ≤ h2 d(gβ, β).
Similarly, by putting x = y2n , y = x2n , u = β, v = α in (3.1) and letting n → ∞
and then using (φ2 ), we get
d(gβ, β) ≤ h2 d(α, gα).
Hence gα = α and gβ = β.
α = gα = G(α, β) and β = gβ = G(β, α)
(3.3)
From (3.2) and (3.3), we have
f α = gα = α = F (α, β) = G(α, β) and
f β = gβ = β = F (β, α) = G(β, α).
Suppose there exists (α1 , β1 ) ∈ X × X such that
f α1 = gα1 = α1 = F (α1 , β1 ) = G(α1 , β1 ) and
f β1 = gβ1 = β1 = F (β1 , α1 ) = G(β1 , α1 ).
(3.4)
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K.P.R. RAO, K.R.K. RAO, E. KARAPINAR
By putting x = α1 , y = β1 , u = α, v = β and x = β1 , y = α1 , u = β, v = α in (3.1)
and using (φ2 ), one can show that α1 = α and β1 = β. Thus (α, β) is the unique
pair satisfying (3.4).
The following example illustrates Theorem 3.1.
Example 3.2. Let X = [0, 1] and d(x, y) = |x2 − y 2 | for all x, y ∈ X.
Define F (x, y) = 21 = G(x, y) and
 1
 1
 2 , if x ∈ [0, 12 ],
 2 , if x ∈ [0, 12 ],
fx =
, gx =


1
x, if x ∈ ( 2 , 1],
1, if x ∈ ( 12 , 1].
Now, we prove a unique common coupled fixed point theorem for a pair of
Jungck type maps without using symmetry of d.
Theorem 3.3. . Let (X, τ ) be a Hausdorff topological space. Let F : X ×X → X
and f : X → X be mappings satisfying
d(F (x, y), F (u, v)), d(f x, f u), d(f y, f v), d(f u, F (u, v)),
φ
≤0
(3.5)
d(f x, F (x, y)), d(f x, F (u, v)), d(F (x, y), f u)
for all x, y, u, v ∈ X, where φ ∈ Φ7 without (φb ),
(a) F (X × X) ⊆ f (X),
(b) (f (X), τ, d) is d-complete,
(c) the pair (F, f ) is weakly compatible,
(d) for each y ∈ X, d(xn , y) → d(x, y), whenever {xn } ⊆ X, x ∈ X such that
xn → x.
Then the mappings F and f have a unique common coupled fixed point.
Proof. Let x0 and y0 be in X.
Since F (X × X) ⊆ f (X), we can choose x1 , y1 ∈ X such that
f x1 = F (x0 , y0 ) and f y1 = F (y0 , x0 ).
Continuing this process, we can construct two sequences {xn } and {yn } in X such
that (for n = 0, 1, 2, · · · )
f xn+1 = F (xn , yn ) = zn and f yn+1 = F (yn , xn ) = pn .
Putting x = xn , y = yn , u = xn+1 , v = yn+1 in (3.5), we get
d(zn , zn+1 ), d(zn−1 , zn ), d(pn−1 , pn ), d(zn , zn+1 ),
φ
≤0
d(zn−1 , zn ), d(zn−1 , zn+1 ), 0
From (φa ), there exists h1 ∈ [0, 1) such that
d(zn , zn+1 ) ≤ h1 max{d(zn−1 , zn ), d(pn−1 , pn )}.
Similarly, by putting y = xn , x = yn , v = xn+1 , u = yn+1 in (3.5) and using (φa ),
we get
d(pn , pn+1 ) ≤ h1 max{d(zn−1 , zn ), d(pn−1 , pn )}.
COMMON COUPLED FIXED POINT THEOREMS
113
Thus
max{d(zn , zn+1 ), d(pn , pn+1 )} ≤ h1 max{d(zn−1 , zn ), d(pn−1 , pn )}
≤ h21 max{d(zn−2 , zn−1 ), d(pn−2 , pn−1 )}
·
·
·
≤ hn1 max{d(z0 , z1 ), d(p0 , p1 )}.
Since
∞
P
hn1 is convergent, it follows that
n=1
P∞
n=1 d(zn , zn+1 ) and
∞
P
d(pn , pn+1 ) are
n=1
convergent. Hence d(zn , zn+1 ) → 0 and d(pn , pn+1 ) → 0 as n → ∞.
Suppose (f (X), τ, d) is d-complete. Then there exist α and β in f (X) such
that {zn } and {pn } converge to α and β, respectively. Hence there exist x, y ∈ X
such that α = f x and β = f y.
Replacing x, y, u, v with xn , yn , x, y respectively in (3.5), we get
d(zn , F (x, y)), d(zn−1 , f x), d(pn−1 , f y), d(f x, F (x, y)),
φ
≤0
d(zn−1 , zn ), d(zn−1 , F (x, y)), d(zn , f x)
Letting n → ∞ and using (d), we get
φ(d(f x, F (x, y)), 0, 0, d(f x, F (x, y)), 0, d(f x, F (x, y)), 0) ≤ 0.
Now, from (φa ), we have F (x, y) = f x.
Similarly, replacing x, y, u, v with yn , xn , y, x in (3.5) and letting n → ∞
and using (d), (φa ), we can show that F (y, x) = f y.
Since (F, f ) is a weakly compatible pair, we have
f α = f f x = f (F (x, y)) = F (f x, f y) = F (α, β) and
f β = f f y = f (F (y, x)) = F (f y, f x) = F (β, α).
Putting x = xn , y = yn , u = α, v = β and x = yn , y = xn , u = β, v = α in (3.5)
and letting n → ∞ and using (φ2 ), we can show that f α = α and f β = β.
Thus
α = f α = F (α, β) and β = f β = F (β, α).
(3.6)
Using (3.5) and (φ2 ), one can show that (α, β) is the unique pair in X × X
satisfying (3.6).
The following example illustrates Theorem 3.3.
Example 3.4. Let X = {0, 1} and d(x, y) = |x2 − y| for all x, y ∈ X. Define
F (x, y) = 1 and f 0 = 0 and f 1 = 1.
Acknowledgement. The authors are grateful to the referee for his valuable
suggestions.
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K.P.R. RAO, K.R.K. RAO, E. KARAPINAR
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1
Department of Applied Mathematics, Acharya Nagarjuna University-Dr.M.R.Appa
Row Campus, Nuzvid-521 201,Andhra Pradesh,India
E-mail address: kprrao2004@yahoo.com,krkr08@gmail.com
2
Department of Mathematics, Atilim University 06836, İncek, Ankara, Turkey.
E-mail address: erdalkarapinar@yahoo.com