Some Theorems on Fixed Point

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Some Theorems on Fixed Point

Sampada Navshinde¹ and J. Achari²

1Asst. Prof. SGGSI&T, Nanded-431605, (Maharashtra) India E-mail: [email protected]

2Retired H.O.D.Mathematics, N.E.S.’Science College, Nanded-431605,(Maharashtra) India

(Received: 24-12-10/ Accepted: 23-3-11) Abstract

Fixed point theorems for a class of mappings using rational symmetric ex- pression involving four points of the space under consideration have been stud- ied.

Keywords: Clouser, Common fixed point, commuting mappings.

1 Introduction

The chief aim of this paper is to introduce a class of mappings by using rational symmetric expression and which involve four points of the space under con- sideration. A fixed point theorem with this mapping has been proved. Finally some related results with this type of mappings have been proved.

Let (M,d) be a complete metric space. Letψ_i : ˜P →[0,∞) (P is the range of d and ˜P is the closure of P) be an upper semicontinuous function from the right on P and satisfies the condition

ψ_i(t)< t

3 f or t >0 and ψ_i(0) = 0 i= 1,2,3. (1.1)

Also, let f be a mapping of M into itself such that

d(f u₁, f u₂)≤ ψ₁(d(u₂, f u₄))[1 +ψ₁(d(u₁, f u₃))]

1 +ψ₁(d(u₁, u₂))

+ ψ₂(d(u₁, f u₄))[1 +ψ₂(d(u₂, f u₃))]

1 +ψ₂(d(u₁, u₂))

+ ψ₃(d(u₁, f u₃))[1 +ψ₃(d(u₂, f u₄))]

1 +ψ₃(d(u₁, u₂)) (1.2) f or u₁, u₂, u₃, u₄ ∈ M.

2 The Main Results

Theorem 2.1. If f be mapping of M into itself satisfying (1.2), then f has a unique fixed point.

Proof. Let x, y ∈M and we define u₁ =f y, u₂ =f x, u₃ =x, u₄ =y Then (1.2) takes the form

d(f(f y), f(f x))≤ ψ1(d(f x, f y))[1 +ψ1(d(f y, f x))]

1 +ψ₁(d(f y, f x))

+ ψ₂(d(f y, f y))[1 +ψ₂(d(f x, f x))]

1 +ψ₂(d(f y, f x))

+ ψ₃(d(f y, f x))[1 +ψ₃(d(f x, f y))]

1 +ψ3(d(f y, f x))

≤ ψ₁(d(f x, f y)) + ψ₃(d(f x, f y)) (2.1) Letx₀ ∈M be arbitrary and construct a sequence {x_n} defined by

f xn−1 =x_n, f x_n =x_n+1, f x_n+1, n= 1,2,· · · Let us put x=xn−1, y =x_n in (2.1), then we have,

d(f(f x_n), f(f xn−1))≤ ψ₁(d(f xn−1, f x_n)) + ψ₃(d(f xn−1, f x_n))

i.e. d(x_n+1, x_n+2)≤ ψ₁(d(x_n, x_n+1)) + ψ₃(d(x_n, x_n+1)) (2.2) Now setC_n =d(xn−1, x_n). Then

C_n+2 =d(x_n+1, x_n+2)

≤ψ₁(d(x_n, x_n+1)) + ψ₃(d(x_n, x_n+1))

≤ψ₁(C_n+1) + ψ₃(C_n+1) (2.3) From (2.3) it follows that C_n decreases with n and hence C_n → C say as n → ∞ . Then since ψ_i is upper semicontinous we obtain in the limit as n→ ∞

C≤ψ₁(C) +ψ₃(C)< 2 3C

which is impossible unless C = 0.

Next, we shall show that the sequence {x_n} is Cauchy. Suppose that it is not so. Then there exist an > 0 and sequence of integers {m(k)} ,{n(k)}

with m(k) > n(k) ≥ k such that

d_k =d(x_m(k), x_n(k))≥, k = 1,2,3,· · · (2.4) If m(k) is the smallest integer exceding n(k) for which (2.4) holds, then from the well ordering principle we have,

d(xm(k)−1, x_n(k))≤ (2.5) Thendk =d(x_m(k), xm(k)−1) +d(xm(k)−1, x_n(k))≤C_m(k)+ < Ck+

which implies thatd_k → as n → ∞.

Also we have,

dk =d(xm, xn)

≤d(x_m, x_m+1) +d(x_m+1, x_n+1) +d(x_n+1, x_n)

≤C_m+1+C_n+1+d(f x_n, f x_m)

≤C_m+1+C_n+1+ψ₁(d(x_m, f xm−1))[1 +ψ₁(d(x_n, f xn−1))]

1 +ψ₁(d(x_n, x_m)) + ψ₂(d(x_n, f xm−1))[1 +ψ₂(d(x_m, f xn−1))]

1 +ψ2(d(xn, xm))

+ ψ₃(d(x_n, f x_n−1))[1 +ψ₃(d(x_m, f x_m−1))]

1 +ψ₃(d(x_n, x_m))

(By putting u₁ =x_n, u₂ =x_m, u₃ =xn−1, u₄ =xm−1)

d_k =d(x_m, x_n)≤C_m+1+C_n+1+ψ₂(d(x_nx_m)) +ψ₃(d(x_n, x_m))

≤C_m+1+C_n+1+ψ₂(d_k) +ψ₃(d_k) letting k→ ∞ we have

≤ ψ₂() +ψ₃() < 2 3 which is a contradiction if > 0.

This leads us to conclude that {x_n} is a Cauchy sequence and since M is complete, there exists a point z ∈ M such that x_n → z as n → ∞ . We shall show thatz is a fixed point of f.

Now puttingu₁ =xn−1, u₂ =z, u₃ =x_n+1, u₄ =x_n in (1.2) we have, d(f xn−1, f z)

≤ ψ₁(d(z, f x_n))[1 +ψ₁(d(xn−1, f x_n+1))]

1 +ψ₁(d(xn−1, z))

+ ψ₂(d(xn−1, f x_n))[1 +ψ₂(d(z, f x_n+1))]

1 +ψ2(d(xn−1, z))

+ ψ₃(d(x_n−1, f x_n+1))[1 +ψ₃(d(z, f x_n))]

1 +ψ₃(d(xn−1, z))

≤ ψ₁(d(z, x_n+1))[1 +ψ₁(d(xn−1, x_n+2))]

1 +ψ₁(d(xn−1, z))

+ ψ₂(d(xn−1, x_n+1))[1 +ψ₂(d(z, x_n+2))]

1 +ψ2(d(xn−1, z))

+ ψ3(d(xn−1, xn+2))[1 +ψ3(d(z, xn+1))]

1 +ψ₃(d(xn−1, z)) (2.6) Letting n → ∞ we get d(z, f z) ≤ 0 which implies z = f z . Thus z is a fixed point of f.

If possible, let there be another fixed point w(6= z) , then putting u₁ =u₄ = z, u₂ =u₃ =w in (1.2) we get,

d(z, w) = d(f z, f w)

≤ ψ₁(d(w, f z))[1 +ψ₁(d(z, f w))]

1 +ψ₁(d(z, w)) + ψ₂(d(z, f z))[1 +ψ₂(d(w, f w))]

1 +ψ₂(d(z, w)) + ψ₃(d(z, f w))[1 +ψ₃(d(w, f z))]

1 +ψ₃(d(z, w))

≤ψ₁(d(z, w)) +ψ₂(d(z, w)) < 2

3d(z, w) which is impossible. Hencez =w.

Theorem 2.2. Let (M,d) be a complete metric space and f_k

(k = 1,2,· · · , n) be a family of mappings of M into itself. If f_k (k =1,2,· · · ,n) satisfies

(i) f_kf_m=f_mf_k (m, k = 1,2,· · ·n)

(ii) there is a system of positive integers m₁, m₂,·m_n such that d(f₁^m1f₂^m2· · ·f_n^mnu1, f₁^m1f₂^m2· · ·f_n^mnu2)

≤ ψ₁(d(u₂, f₁^m1f₂^m2· · ·f_n^mnu₄))[1 +ψ₁(d(u₁, f₁^m1f₂^m2· · ·f_n^mnu₃))]

1 +ψ1(d(u1, u2))

+ψ₂(d(u₁, f₁^m1f₂^m2· · ·f_n^mnu₄))[1 +ψ₂(d(u₂, f₁^m1f₂^m2· · ·f_n^mnu₃))]

1 +ψ₂(d(u₁, u₂))

+ψ₃(d(u₁, f₁^m1f₂^m2· · ·f_n^mnu₃))[1 +ψ₃(d(u₂, f₁^m1f₂^m2· · ·f_n^mnu₄))]

1 +ψ₃(d(u₁, u₂))

for u₁, u₂, u₃, u₄ ∈ M and ψ_i(t) satisfies (1.1) , then f_k (k = 1,2,· · ·n) have a unique common fixed point.

Proof. Let f =f₁^m1f₂^m2· · ·f_n^mn . Then (ii) takes the form (iii)

d(f u₁, f u₂)≤ ψ₁(d(u₂, f u₄))[1 +ψ₁(d(u₁, f u₃))]

1 +ψ₁(d(u₁, u₂))

+ψ₂(d(u₁, f u₄))[1 +ψ₂(d(u₂, f u₃))]

1 +ψ₂(d(u₁, u₂))

+ψ₃(d(u₁, f u₃))[1 +ψ₃(d(u₂, f u₄))]

1 +ψ₃(d(u₁, u₂))

Then by Theorem [2.1], f has a unique fixed point z in M. Therefore f z =z, then we have,

f_k(f z) =f_kz, k = 1,2,· · ·n By commutativity off_k we have,

f(f_kz) =f_kz, k = 1,2,· · ·n

Since f has a unique common fixed point z, we obtain f_kz, k = 1,2,· · ·n.

Hence z is a common fixed point of the family f_k. Let z,w be common fixed point off_k, then by (ii) we have by puttingu₁ =u₄ =z, u₂ =u₃ =w

d(z, w) = d(f z, f w)

≤ ψ₁(d(w, f z))[1 +ψ₁(d(z, f w))]

1 +ψ₁(d(z, w)) + ψ₂(d(z, f z))[1 +ψ₂(d(w, f w))]

1 +ψ₂(d(z, w)) + ψ₃(d(z, f w))[1 +ψ₃(d(w, f z))]

1 +ψ₃(d(z, w))

≤ψ₁(d(z, w)) +ψ₂(d(z, w)) < 2

3d(z, w) which impliesz =w. Hence the proof.

Theorem 2.3.Let M be a metric space with d and p andf_k (k= 1,2,· · · , n) be a family of mappings of M into itself.

Suppose that

(i) d(x, y) ≤ p(x, y) to all x,y ∈ M (ii) M is f-orbitally complete w.r.t. d.

(iii) f_kf_m =f_mf_k (m, k = 1,2,· · ·n)

(iv) there is a system of positive integers m₁, m₂,·m_n such that

p(f₁^m1f₂^m2· · ·f_n^mnu₁, f₁^m1f₂^m2· · ·f_n^mnu₂)

≤ ψ₁(p(u₂, f₁^m1f₂^m2· · ·f_n^mnu₄))[1 +ψ₁(p(u₁, f₁^m1f₂^m2· · ·f_n^mnu₃))]

1 +ψ1(p(u1, u2))

+ ψ₂(p(u₁, f₁^m1f₂^m2· · ·f_n^mnu₄))[1 +ψ₂(p(u₂, f₁^m1f₂^m2· · ·f_n^mnu₃))]

1 +ψ₂(p(u₁, u₂))

+ ψ₃(p(u₁, f₁^m1f₂^m2· · ·f_n^mnu₃))[1 +ψ₃(p(u₂, f₁^m1f₂^m2· · ·f_n^mnu₄))]

1 +ψ₃(p(u₁, u₂))

for u₁, u₂, u₃, u₄ ∈ M and ψ_i(t)< ₃^t f or t >0 and ψ_i(0) = 0

, i= 1,2,3 Then f_k (k = 1,2,· · · , n) have a unique common fixed point.

Proof. As in Theorem [2.1] put f = f₁^m1f₂^m2· · ·f_n^mn then(iv) takes the form

p(f u1, f u2)≤ ψ₁(p(u₂, f u₄))[1 +ψ₁(p(u₁, f u₃))]

1 +ψ₁(p(u₁, u₂))

+ψ₂(p(u₁, f u₄))[1 +ψ₂(p(u₂, f u₃))]

1 +ψ₂(p(u₁, u₂))

+ψ₃(p(u₁, f u₃))[1 +ψ₃(p(u₂, f u₄))]

1 +ψ3(p(u1, u2)) f or u₁, u₂, u₃, u₄ ∈ M

Following the lines of arguments of the proof of Theorem [2.1], it can be shown that the sequence of iterates {x_n} is Cauchy with respect to p. Since d(x, y) ≤ p(x, y) for all x,y ∈ M , so {x_n} is Cauchy with respect to d also.

Again M being f-orbitally complete with respect to d, so we have {x_n} has a limit u in M. From the proof of Theorem [2.1] it can be easily shown thatu is the unique common fixed point of the familyf_k .

Acknowledgements

The authors thank the referee for his/her suggestions and comments.

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