Coupled fixed point theorems in d-complete topological spaces

Research Article

Coupled fixed point theorems in d-complete topological spaces

K. P. R. Rao^a,∗, S. Hima Bindu^b, Md. Mustaq Ali^a

aDepartment of Applied Mathematics, Acharya Nagarjuna University - Dr. M.R. Appa Row Campus, Nuzvid-521 201, Krishna Dt., A.P., India.

bDepartment of Mathematics, CH. S. D. St. Theresa’s Junior College for women, Eluru- 534 001, W. G. Dt., A. P., India.

This paper is dedicated to Professor Ljubomir ´Ciri´c Communicated by Professor V. Berinde

Abstract

In this paper, we obtain prove two common coupled fixed point theorems in Hausdorffd- complete topological spaces. c2012 NGA. All rights reserved.

Keywords: Coupled fixed points, d-complete space ,weakly compatible maps.

2010 MSC: Primary 47H10; Secondary 54H25.

1. Introduction and Preliminaries

In 1975, Kasahara [11, 12] introduced the notion of d-complete topological spaces as a generalization of complete metric spaces.

Definition 1.1. [11, 12]. Let (X,T) be a topological space. Supposed:X×X →[0,∞) satisfies (i) d(x, y) = 0 if and only ifx=y ,

(ii) for any sequence{x_n}in X,Σ^∞_n=1d(xn, xn+1)<∞ implies {x_n}is convergent in (X,T).

Then the triplet (X,T, d) is called ad- complete topological space.

∗Corresponding author

Email addresses: [email protected](K. P. R. Rao),s.hima [email protected](S. Hima Bindu), [email protected](Md. Mustaq Ali)

Received 2011-6-13

For details on d-complete topological spaces, we refer to Iseki [10] and Kasahara [11, 12, 13]. Hicks [6]

and Hicks and Rhoades [7, 8] proved several fixed point theorems in d- complete topological spaces. Hicks and Saliga [9] and Saliga [19] obtained fixed point theorems for non - self maps ind- complete topological spaces.

In 2006, Bhaskar and Lakshmikantham [3] introduced the notion of a coupled fixed point in partially ordered metric spaces, also discussed some problems of 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(Refer to [1, 4, 5, 14, 15, 16, 17, 18, 20, 21, 22, 23] ).

In this paper ,we prove a common coupled fixed point theorem for four mappings in d - complete topological spaces.

Definition 1.2. ([3]).Let X be a nonempty set. An element (x, y)∈X×X is called a coupled fixed point of the mappingF :X×X →X ifx=F(x, y) and y=F(y, x).

Definition 1.3. ([1]).Let X be a nonempty set. An element (x, y)∈X×X is called

(i) a coupled coincidence point ofF :X×X →X and g:X→X ifgx=F(x, y) and gy=F(y, x).

(ii)a common coupled fixed point of F :X×X → X and g : X → X ifx =gx =F(x, y) and y = gy = F(y, x).

Definition 1.4. ([1]).Let X be a nonempty set. The mappingsF :X×X→X and g:X →X are called W-compatible ifg(F(x, y)) =F(gx, gy) andg(F(y, x)) =F(gy, gx) whenevergx=F(x, y) andgy=F(y, x) for some (x, y)∈X×X.

In this paper,we obtain two common coupled and common fixed point theorems for two and four mappings satisfying a Berinde [2] type weak contraction conditions in Hausdorff d- complete topological spaces.

2. Main Results

Theorem 2.1. Let (X, τ, d) be a Hausdorff topological space.Let F, G :X×X → X and f, g :X → X be mappings satisfying

(2.1.1) d(F(x, y), G(u, v))≤h max

d(f x, gu), d(f y, gv), d(F(x, y), f x), d(G(u, v), gu) +L min

d(f x, gu), d(f y, gv), d(F(x, y), f x), d(G(u, v), gu), d(F(x, y), gu), d(G(u, v), f x)

,

for allx, y, u, v ∈X, where 0≤h <1 and L≥0, (2.1.2)F(X×X)⊆g(X), G(X×X)⊆f(X), (2.1.3)one of f(X) and g(X) is d- complete,

(2.1.4)the pairs (F, f) and (G, g) are W-compatible, (2.1.5)d(x, y) =d(y, x) for allx, y∈X and

(2.1.6)for each y∈X, d(x_n, y)→d(x, y), whenever {x_n} ⊆X, x∈X such that xn→x.

ThenF, G, f andghave a unique common coupled fixed point in X×X and also they have a unique common fixed point in X.

Proof. Letx₀ and y₀ be inX.

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) andf y2 =G(y1, x1). Continuing this process, we can construct the sequences{x_n},{y_n} ,{z_n} and {p_n}in X such that

gx_2n+1=F(x_2n, y_2n) =z_2n, say ; gy2n+1 =F(y2n, x2n) =p2n, say ;

f x2n+2=G(x2n+1, y2n+1) =z2n+1, say ; and

f y_2n+2=G(y_2n+1, x_2n+1) =p_2n+1, say ; forn= 0,1,2, .... Now d(z2n, z2n+1) =d(F(x2n, y2n), G(x2n+1, y2n+1))

≤h max{d(z_2n−1, z2n), d(p2n−1, p2n), d(z2n, z2n−1), d(z2n+1, z2n)}

+L min

d(z2n−1, z_2n), d(p2n−1, p_2n), d(z_2n, z2n−1), d(z_2n+1, z_2n), d(z_2n, z_2n), d(z_2n+1, z2n−1)

=h max{d(z2n−1, z2n), d(p2n−1, p2n)}. Also

d(p2n, p2n+1) =d(F(y2n, x2n), G(y2n+1, x2n+1))

≤h max{d(p_2n−1, p2n), d(z2n−1, z2n), d(p2n, p2n−1), d(p2n+1, p2n)}

+L min

d(p2n−1, p_2n), d(z2n−1, z_2n), d(p_2n, p2n−1), d(p_2n+1, p_2n), d(p_2n, p_2n), d(p_2n+1, p2n−1)

=h max{d(p_2n−1, p2n), d(z2n−1, z2n)}.

Thus max{d(z_2n, z_2n+1), d(p_2n, p_2n+1)} ≤h max{d(p_2n−1, p_2n), d(z2n−1, z_2n)}.

d(z2n−1, z2n) =d(G(x2n−1, y2n−1), F(x2n, y2n))

=d(F(x2n, y2n), G(x2n−1, y2n−1))

≤h max{d(z2n−1, z2n−2), d(p2n−1, p2n−2), d(z2n, z2n−1), d(z2n−1, z2n−2)}

+L min

d(z2n−1, z2n−2), d(p2n−1, p2n−2), d(z2n, z2n−1), d(z2n−1, z2n−2), d(z_2n, z2n−2), d(z2n−1, z2n−1)

=h max{d(z2n−2, z2n−1), d(p2n−2, p2n−1)}.

d(p2n−1, p_2n) =d(G(y2n−1, x2n−1), F(y_2n, x_2n)) =d(F(y_2n, x_2n), G(y2n−1, x2n−1))

≤h max{d(p_2n−1, p2n−2), d(z2n−1, z2n−2), d(p_2n, p2n−1), d(p2n−1, p2n−2)}

+L min

d(p2n−1, p2n−2), d(z2n−1, z2n−2), d(p2n, p2n−1), d(p2n−1, p2n−2), d(p2n, p2n−2), d(p2n−1, p2n−1)

=h max{d(p_2n−2, p2n−1), d(z2n−2, z2n−1)}.

Thus max{d(z_2n−1, z2n), d(p2n−1, p2n)} ≤h max{d(z_2n−2, z2n−1), d(p2n−2, p2n−1)}. Hence max{d(z_n, zn+1), d(pn, pn+1)} ≤h max{d(zn−1, zn), d(pn−1, pn)}

≤h² max{d(z_n−2, zn−1), d(pn−2, pn−1)}

: :

≤hⁿ max{d(z₀, z₁), d(p₀, p₁)}. Since

∞

P

n=1

hⁿis convergent, it follows that

∞

P

n=1

d(zn, zn+1) and

∞

P

n=1

d(pn, pn+1) are convergent. Henced(zn, zn+1)→ 0, d(pn, pn+1)→0 as n→ ∞.

Supposef(X) is d-complete.Then {z_2n+1}={f x_2n+2} ⊆ f(X) and{p_2n+1}={f y_2n+2} ⊆ f(X) converge to someα and β inf(X) respectively.

Hence there exist x and y in X such that α = f x and β = f y. Also the subsequences {z_2n}and {p_2n} converge to α andβ respectively.

d(F(x, y), z2n+1) =d(F(x, y), G(x2n+1, y2n+1))

≤h max{d(f x, z_2n), d(f y, p_2n), d(F(x, y), f x), d(z_2n+1, z_2n)}

+L min

d(f x, z2n), d(f y, p2n), d(F(x, y), f x), d(z_2n+1, z_2n), d(F(x, y), z_2n), d(z_2n+1, f x)

. Lettingn→ ∞and using (2.1.5) and (2.1.6), we get

d(F(x, y), f x)≤h d(F(x, y), f x) +L(0).

HenceF(x, y) =f x=α.

d(F(y, x), p_2n+1) =d(F(y, x), G(y_2n+1, x_2n+1))

≤h max{d(f y, p_2n), d(f x, z_2n), d(F(y, x), f y), d(p_2n+1, p_2n)}

+L min

d(f y, p_2n), d(f x, z_2n), d(F(y, x), f y), d(p2n+1, p2n), d(F(y, x), p2n), d(p2n+1, f y)

. Lettingn→ ∞, we getd(F(y, x), f y)≤h d(F(y, x), f y) +L(0).

HenceF(y, x) =f y=β.

Since the pair (f, S) is W-compatible, we have F(α, β) = F(f x, f y) = f(F(x, y)) = f α and F(β, α) = F(f y, f x) =f(F(y, x)) =f β.

Consider

d(F(α, β), z2n+1) =d(F(α, β), G(x2n+1, y2n+1))

≤h max{d(f α, z_2n), d(f β, p2n), d(F(α, β), f α), d(z2n+1, z2n)}

+L min

d(f α, z2n), d(f β, p2n), d(F(α, β), f α), d(z_2n+1, z_2n), d(F(α, β), z_2n), d(z_2n+1, f α)

. Lettingn→ ∞, we getd(f α, α)≤h max{d(f α, α), d(f β, β)}. Also consider

d(F(β, α), p2n+1) =d(F(β, α), G(y2n+1, x2n+1))

≤h max{d(f β, p_2n), d(f α, z2n), d(F(β, α), f β), d(p2n+1, p2n)}

+L min

d(f β, p2n), d(f α, z2n), d(F(β, α), f β), d(p_2n+1, p_2n), d(F(β, α), p_2n), d(p_2n+1, f β)

. Lettingn→ ∞, we getd(f β, β)≤h max{d(f β, β), d(f α, α)}.

Thusmax{d(f α, α), d(f β, β)} ≤h max{d(f α, α), d(f β, β)}. Hencef α=α and f β =β.

Thusα=f α=F(α, β)...(I) and β =f β=F(β, α)...(II)

Since F(X×X)⊆gX, there existγ, δ∈X such that gγ=F(α, β) =f α=α andgδ=F(β, α) =f β=β. Now

d(gγ, G(γ, δ)) =d(F(α, β), G(γ, δ))

≤h max{0,0,0, d(G(γ, δ), gγ)}

+L min{0,0,0, d(G(γ, δ), gγ),0, d(G(γ, δ), gγ)}

=h d(G(γ, δ), gγ).

HenceG(γ, δ) =gγ. Also

d(gδ, G(δ, γ)) =d(F(β, α), G(δ, γ))

≤h max{0,0,0, d(G(δ, γ), gδ)}

+L min{0,0,0, d(G(δ, γ), gδ),0, d(G(δ, γ), gδ)}

=h d(G(δ, γ), gδ).

Hence G(δ, γ) =gδ. Since the pair (G, g) is W-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). Now consider

d(z_2n, G(α, β)) =d(F(x_2n, y_2n), G(α, β))

≤h max{d(z_2n−1, gα), d(p2n−1, gβ), d(z_2n, z2n−1),0}

+L min

d(z2n−1, gα), d(p2n−1, gβ), d(z_2n, z2n−1), 0, d(z2n, gα), d(G(α, β), z2n−1)

.

Lettingn→ ∞, we getd(α, gα)≤h max{d(α, gα), d(β, gβ)}. Also consider d(p2n, G(β, α)) =d(F(y2n, x2n), G(β, α))

≤h max{d(p2n−1, gβ), d(z2n−1, gα), d(p2n, p2n−1),0}

+L min

d(p2n−1, gβ), d(z2n−1, gα), d(p2n, p2n−1), 0, d(p_2n, gβ), d(G(β, α), p2n−1)

.

Lettingn→ ∞, we getd(β, gβ)≤h max{d(β, gβ), d(α, gα)}. Hence gα=α and gβ=β.

Thusα=gα=G(α, β)...(III) and β=gβ=G(β, α)....(IV).

From (I),(II) (III) and (IV), we have

f α=gα=α=F(α, β) =G(α, β)...(V) and f β=gβ=β =F(β, α) =G(β, α)...(VI) Thus (α, β) is a common coupled fixed point ofF, G, f and g.

Suppose (α1, β1)∈X×X is another common coupled fixed point of F, G, f and g.

d(α₁, α) =d(F(α₁, β₁), G(α, β))≤h max{d(α₁, α), d(β₁, β),0,0}

+L min{d(α₁, α), d(β₁, β),0,0, d(α₁, α), d(α, α₁)}

=h max{d(α₁, α),d(β1, β)}. Also

d(β₁, β) =d(F(β₁, α₁), G(β, α))≤h max{d(β₁, β), d(α₁, α),0,0}

+L min{d(β₁, β), d(α₁, α),0,0, d(β₁, β), d(β, β₁)}

=h max{d(α₁, α), d(β₁, β)}. Thus max{d(α₁, α), d(β1, β)} ≤hmax{d(α₁, α), d(β1, β)}.

Henceα₁ =α and β₁ =β.

Thus (α, β) is the unique common coupled fixed point ofF, G, f and g.

Now, we will show that α=β.

d(α, β) =d(F(α, β), G(β, α))

≤h max{d(α, β), d(α, β),0,0}+L min{d(α, β), d(α, β),0,0, d(α, β), d(β, α)}

=h d(α, β).

Thus α =β. Hence α is a common fixed point of F, G, f and g. Using (2.1.1), we can show that α is the unique common fixed point ofF, G, f and g.

The following example illustrates Theorem 2.1.

Example 2.2. Let X= [0,1] andd(x, y) =|x²−y²|,∀x, y∈X.

DefineF(x, y) =Sin(^x2+y₄ ²) =G(x, y) and f x=x=gx,∀x∈X . Then d(F(x, y), G(u, v)) =d(Sin(x²+y²

4 ), Sin(u²+v² 4 ))

=

Sin²(x²+y²

4 )−Sin²(u²+v²

4 )

=

Sin(x²+y²

4 +u²+v²

4 )Sin(x²+y²

4 −u²+v²

4 )

≤ 1 4(

x²−u² +

y²−v² )

≤ 1

2max{d(f x, gu), d(f y, gv)}

≤ 1

2 max

d(f x, gu), d(f y, gv), d(F(x, y), f x), d(G(u, v), gu) +L min

d(f x, gu), d(f y, gv), d(F(x, y), f x), d(G(u, v), gu), d(F(x, y), gu), d(G(u, v), f x)

,

where L= 0.

One can verify all the other conditions easily. (0,0) is the unique common fixed point ofF, G, f and g.

Note: In Example 2.2, it is clear that (X, d) is a d-complete topological space and (X, d) is not a complete metric space .

Now we give another theorem for a pair of Jungck type maps without using symmetry of d.

Theorem 2.3. Let (X, τ, d) be a Hausdorff topological space.Let F : X ×X → X and f : X → X be mappings satisfying

(2.3.1) d(F(x, y), F(u, v))≤h max

d(f x, f u), d(f y, f v), d(f x, F(x, y), d(f u, , F(u, v) +L min

d(f x, f u), d(f y, f v), d(f x, F(x, y)), d(f u, F(u, v)), d(f x, F(u, v)), d(F(x, y), f u)

,

for allx, y, u, v ∈X, where 0≤h <1 and L≥0, (2.3.2)F(X×X)⊆f(X),

(2.3.3)f(X) is d- complete,

(2.3.4)the pair (F, f) is W-compatible,

(2.3.5)for each y∈X, d(x_n, y)→d(x, y),whenever {x_n} ⊆X, x∈X such that x_n→x.

Then the mappings F and f have a unique common coupled fixed point in X ×X and also they have a unique common fixed point in X.

Proof. Letx0 andy0be inX. SinceF(X×X)⊆f(X), we can choosex1, y1 ∈X such thatf x1=F(x0, y0) and f y₁ =F(y₀, x₀).

Continuing this process, we can construct the sequences{x_n},{y_n},{z_n} and{p_n}inX such thatf x_n+1= F(xn, yn) =zn, say andf yn+1 =F(yn, xn) =pn, say forn= 0,1,2, ...

Consider

d(zn, zn+1) =d(F(xn, yn), F(xn+1, yn+1))

≤h max{d(zn−1, zn), d(pn−1, pn), d(zn−1, zn), d(zn, zn+1)}

+L min

d(zn−1, zn), d(pn−1, pn), d(zn−1, zn), d(z_n, z_n+1), d(zn−1, z_n+1), d(z_n, z_n)

=h max{d(zn−1, z_n), d(pn−1, p_n)}. d(p_n, p_n+1) =d(F(y_n, x_n), F(y_n+1, x_n+1))

≤h max{d(p_n−1, pn), d(zn−1, zn), d(pn−1, pn), d(pn, pn+1)}

+L min

d(pn−1, p_n), d(zn−1, z_n), d(pn−1, p_n), d(pn, pn+1), d(pn−1, pn+1), d(pn, pn)

=h max{d(p_n−1, pn), d(zn−1, zn)}. Thus

max{d(z_n, zn+1), d(pn, pn+1)} ≤h max{d(p_n−1, pn), d(zn−1, zn)}

≤h² max{d(p_n−2, pn−1), d(zn−2, zn−1)}

: :

≤hⁿ max{d(p₀, p1), d(z0, z1)}. Since

∞

P

n=1

hⁿis convergent, it follows that

∞

P

n=1

d(zn, zn+1) and

∞

P

n=1

d(pn, pn+1) are convergent. Henced(zn, zn+1)→ 0, andd(pn, pn+1)→0 asn→ ∞. Supposef(X) isd-complete. Then there existαandβ inf(X) such that {z_n} and{p_n} coverge to α and β respectively. Hence there exist x, y∈X such thatα =f xand β =f y.

d(zn, F(x, y)) =d(F(xn, yn), F(x, y))

≤h max

d(zn−1, f x), d(pn−1, f y), d(zn−1, z_n), d(f x, F(x, y))

+L min

d(zn−1, f x), d(pn−1, f y), d(zn−1, z_n), d(f x, F(x, y)), d(zn−1, F(x, y)), d(zn, f x)

. Lettingn→ ∞, we get

d(f x, F(x, y))≤h max{0,0,0, d(f x, F(x, y))}

+L min{0,0,0, d(f x, F(x, y)), d(f x, F(x, y)),0}

=hd(f x, F(x, y)).

HenceF(x, y) =f x=α.

d(pn, F(y, x)) =d(F(yn, xn), F(y, x)))

≤h max

d(pn−1, f y), d(zn−1, f x), d(pn−1, p_n), d(f y, F(y, x))

+L min

d(pn−1, f y), d(zn−1, f x), d(pn−1, pn), d(f y, F(y, x)), d(pn−1, F(y, x)), d(pn, f y)

. Lettingn→ ∞, we getd(f y, F(y, x))≤h d(f y, F(y, x)),

so thatF(y, x) =f y=β.

Since (F, f) is W-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).

d(zn, f α) =d(F(xn, yn), F(α, β))

≤h max{d(zn−1, f α), d(pn−1, f β), d(zn−1, z_n),0}

+L min

d(zn−1, f α), d(pn−1, f β), d(zn−1, zn) 0, d(zn−1, f α), d(z_n, f α)

.

Lettingn→ ∞, we getd(α, f α)≤h max{d(α, f α), d(β, f β)}.

d(p_n, f β) =d(F(y_n, x_n), F(β, α))

≤h max{d(p_n−1, f β), d(zn−1, f α), d(pn−1, pn),0}

+L min

d(pn−1, f β), d(zn−1, f α), d(pn−1, p_n) 0, d(pn−1, f β), d(pn, f β)

.

Lettingn→ ∞, we getd(β, f β)≤h max{d(β, f β), d(α, f α)}.

Thusmax{d(α, f α), d(β, f β} ≤h max{d(α, f α), d(β, f β)}

so thatf α=α and f β =β.

Thusα=f α=F(α, β)——(I) andβ=f β=F(β, α)—–(II).

Using (2.3.1) we can show that (α, β) is the unique pair inX×X satisfying (I) and (II) . Now we will show thatα=β.

d(α, β) =d(F(α, β), F(β, α))

≤h max{d(α, β), d(β, α),0,0}+L(0)

=h max{d(α, β), d(β, α)}. d(β, α) =d(F(β, α), F(α, β))

≤h max{d(β, α), d(α, β)}.

Hence, max{d(α, β), d(β, α)} ≤h max{d(α, β), d(β, α)}. Thusα=β. Henceα is a common fixed point of F and f. Using (2.3.1), we can show thatα is the unique common fixed point ofF and f.

The following example illustrates Therorem 2.3.

Example 2.4. Let X = {0,1} and d(x, y) = |x² −y|,∀x, y ∈ X. Define F(x, y) = 1,∀x, y ∈ X and f1 = 1, f0 = 0.

Acknowledgements:

The authors are thankful to the referee for his valuable suggestions.

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