I. Altun and D. Turkoglu Department of Mathematics

FIXED POINT AND HOMOTOPY RESULT FOR MAPPINGS SATISFYING AN IMPLICIT RELATION

I. Altun and D. Turkoglu Department of Mathematics

Faculty of Science and Arts, Gazi University 06500 Teknikokullar, Ankara, Turkey

e-mail: ialtun@gazi.edu.tr e-mail: dturkoglu@gazi.edu.tr

Abstract

In this paper, we prove some fixed point theorems for single valued mappings satisfying an implicit relation on space with two metrics. In addition we give a homotopy result using our theorems.

Keywords and phrases: fixed point, implicit relation, homotopy result.

2000 Mathematics Subject Classification: Primary 54H25;

Secondary 47H10.

1. Introduction and preliminaries

This article presents new fixed point and homotopy results for single valued mappings satisfying an implicit relation on space with two metrics. Our fixed point theorem improves that in [1, 2, 3]. In Section 2, we give an implicit relation and some examples. In Section 3, we prove two fixed point theorems and in Section 4, we give a homotopy result.

2. Implicit relation

An implicit relation on metric spaces was used in many articles (see [4, 5, 6, 7]).

Let R ₊ denote the nonnegative real numbers and let T be the set of all

continuous functions T : R ₊ ⁶ → R satisfying the following conditions:

T ₁ : T (t ₁ , . . . , t ₆ ) is decreasing in variables t ₂ , . . . , t ₆ .

T ₂ : there exists a continuous and nondecreasing function f : R ₊ → R ₊ , f (0) = 0, f (t) < t for t > 0, such that for u ≥ 0,

T (u, v, v, u, u + v, 0) ≤ 0 or

T (u, v, 0, 0, v, v) ≤ 0 implies u ≤ f (v).

T ₃ : T (u, 0, 0, u, u, 0) > 0, ∀u > 0.

Example 1. T(t ₁ , . . . , t ₆ ) = t ₁ − α max{t ₂ , t ₃ , t ₄ } − (1 − α)[at ₅ + bt ₆ ], where 0 ≤ α < 1, 0 ≤ a < ¹ ₂ , 0 ≤ b < ¹ ₂ , α + a + b < 1.

T ₁ : Clearly satisfied.

T ₂ : Let u > 0 and T (u, v, v, u, u + v, 0) = u − α max{u, v} − (1 − α) a(u + v) ≤ 0. If u ≥ v, then (1 − a)u ≤ av which implies a ≥ ¹ ₂ , a con- tradiction. Thus u < v and u ≤ ^α _{1−(1−α)a} ^+(1−α)a v = βv. Similarly, let u > 0 and T (u, v, 0, 0, v, v) = u − αv − (1 − α)(a + b)v ≤ 0, then u ≤ (α(1 − a − b) + a + b)v = γv. If u = 0, then u ≤ γv. Thus T ₂ is satisfied with f (t) = max{β, γ}t.

T ₃ : T (u, 0, 0, u, u, 0) = u − αu − (1 − α)au = (1 − α)(1 − a)u > 0, ∀u > 0.

Example 2. T (t ₁ , . . . , t ₆ ) = t ₁ −k max{t ₂ , t ₃ , t ₄ , ¹ ₂ (t ₅ +t ₆ )}, where k ∈ (0, 1).

T ₁ : Clearly satisfied.

T ₂ : Let u > 0 and T (u, v, v, u, u + v, 0) = u − k max{u, v} ≤ 0. If u ≥ v, then u ≤ ku, which is a contradiction. Thus u < v and u ≤ kv. Similarly, let u > 0 and T (u, v, 0, 0, v, v) = u − kv ≤ 0, then we have u ≤ kv. If u = 0, then u ≤ kv. Thus T ₂ is satisfied with f (t) = kt.

T ₃ : T (u, 0, 0, u, u, 0) = u − ku > 0, ∀u > 0.

Example 3. T(t ₁ , . . . , t ₆ ) = t ₁ − φ(max{t ₂ , t ₃ , t ₄ , ¹ ₂ (t ₅ + t ₆ )}), where φ : R ₊ → R ₊ continuous nondecreasing and φ(0) = 0, φ(t) < t for t > 0.

T ₁ : Clear.

T ₂ : Let u > 0 and T (u, v, v, u, u + v, 0) = u − φ(max{u, v}) ≤ 0. If u ≥ v, then u − φ(u) ≤ 0, which is a contradiction. Thus u < v and u ≤ φ(v). Similarly, let u > 0 and T (u, v, 0, 0, v, v) = u − φ(v) ≤ 0, then we have u ≤ φ(v). If u = 0, then u ≤ φ(v). Thus T 2 is satisfied with f = φ.

T ₃ : T (u, 0, 0, u, u, 0) = u − φ(u) > 0, ∀u > 0.

Example 4. T (t ₁ , . . . , t ₆ ) = t ² ₁ − t ₁ (at ₂ + bt ₃ + ct ₄ ) − dt ₅ t ₆ , where a > 0, b, c, d ≥ 0, a + b + c < 1 and a + d < 1.

T ₁ : Clear.

T ₂ : Let u > 0 and T (u, v, v, u, u + v, 0) = u ² − u(av + bv + cu) ≤ 0.

Then u ≤ ( ^a _1−c ^+b )v = h ₁ v. Similarly, let u > 0 and T (u, v, 0, 0, v, v) = u ² − auv − dv ² ≤ 0, then we have u ≤ ^{a + √ 4d+a} ₂

v = h ₂ v. If u = 0, then u ≤ h ₂ v.

Thus T 2 is satisfied with f (t) = max{h 1 , h ₂ }t.

T ₃ : T (u, 0, 0, u, u, 0) = (1 − c)u ² > 0, ∀u > 0 and T (u, u, u, 0, u, u) = u ² (1 − a − b − d) > 0, ∀u > 0.

Example 5. T (t ₁ , . . . , t ₆ ) = t ³ ₁ − α ^{3 t}

2 3

t

4

+t

t

6

t

+t

+t

+1 , where α ∈ (0, 1).

T ₁ : Clear.

T ₂ : Let u > 0 and T (u, v, v, u, u + v, 0) = u ³ − u ^α +2v+1

^v

^u

≤ 0, which implies u ≤ _{u +2v+1} ^α

^v

. But _{u +2v+1} ^α

^v

≤ α ³ v, thus u ≤ α ³ v. Similarly, let u > 0 and T (u, v, 0, 0, v, v) = u ³ − α ^{3 v} _{v +1}

≤ 0, then we have u ≤ αv. If u = 0, then u ≤ αv. Thus T ₂ is satisfied with f (t) = αt.

T ₃ : T (u, 0, 0, u, u, 0) = u ³ > 0, ∀u > 0.

Example 6. T (t ₁ , . . . , t ₆ ) = t ₁ −ψ(t ₂ , t ₃ , t ₄ , t ₅ , t ₆ ), where ψ : R ⁵ ₊ → R ₊ con- tinuous and nondecreasing in each coordinate variables such that ψ(t, t, t, at, bt) < t, for each t > 0 and a, b ≥ 0 with a + b ≤ 2.

T ₁ : Clear.

T ₂ : Let u > 0 and T (u, v, v, u, u + v, 0) = u − ψ(v, v, u, u + v, 0) ≤ 0.

If u ≥ v then u ≤ ψ(u, u, u, 2u, 0) < u, which is a contradiction. Thus u < v and u ≤ ψ(v, v, v, 2v, 0). Similarly, let u > 0 and T (u, v, 0, 0, v, v) = u − ψ(v, 0, 0, v, v) ≤ 0 then again we have u ≤ ψ(v, 0, 0, v, v). Now let f (t) = max{ψ(t, t, t, 2t, 0), ψ(t, 0, 0, t, t)}, for each t > 0 and f (0) = 0. Thus, if u > 0, then u ≤ f (v) and if u = 0, then u ≤ f (v).

T ₃ : T (u, 0, 0, u, u, 0) = u − ψ(0, 0, u, u, 0) ≥ u − ψ(u, u, u, u, u) > 0,

∀u > 0.

3. Fixed point theorem

This section presents fixed point results for single valued mappings satisfying an implicit relation on space with two metrics. The theorems extend fixed point results of [1, 2, 3]. Throughout this section (X, d ⁰ ) will be a complete metric space and d will be another metric on X. If x 0 ∈ X and r > 0 let

B(x 0 , r) = {x ∈ X : d(x, x 0 ) < r},

and we let B(x ₀ , r) ^d

denote the d ⁰ -closure of B(x ₀ , r).

Now, we give our first theorem.

Theorem 1. Let (X, d ⁰ ) be a complete metric space, d another metric on X, x ₀ ∈ X, r > 0 and F : B(x ₀ , r) ^d

→ X. Suppose F is satisfying the following conditions ;

(a) There exists T ∈ T such that

(3.1) T(d(F x, F y), d(x, y), d(x, F x), d(y, F y), d(x, F y), d(y, F x)) ≤ 0 for all x, y ∈ B(x ₀ , r) ^d

,

(b)

(3.2) r − d(x ₀ , F x ₀ ) = γ > 0 and

∞

X

n =1

f ⁿ (d(x ₀ , F x ₀ )) < γ,

where f is defined by T ₂ and f ⁿ denotes the composition of f n−times with itself.

(c)

(3.3) If d  d ⁰ assume F is uniformly continuous from (B(x ₀ , r), d) into (X, d ⁰ ).

(d)

(3.4) If d 6= d ⁰ assume F is continuous from (B(x ₀ , r) ^d

, d ⁰ ) into (X, d ⁰ ).

Then F has a fixed point. That is, there exists x ∈ B(x ₀ , r) ^d

with x = F x.

P roof. Let x ₁ = F x ₀ . Now d(x ₀ , x ₁ ) < r, so x ₁ ∈ B(x ₀ , r). Next we let x n = F x _n−1 for n ∈ {2, 3, . . .}. Now, we claim

x n ∈ B(x ₀ , r) for n ∈ {1, 2, . . .} and {x ⁿ } is a Cauchy sequence with respect to d .

At first, we show x n ∈ B(x ₀ , r) for n ∈ {1, 2, . . .}. The proof is by induction.

We know x 0 , x ₁ ∈ B(x 0 , r). Thus we can use the condition (3.1). Therefore we have

T (d(F x ₀ , F x ₁ ), d(x ₀ , x ₁ ), d(x ₀ , F x ₀ ), d(x ₁ , F x ₁ ), d(x ₀ , F x ₁ ), d(x ₁ , F x ₀ )) ≤ 0 and so

T (d(x 1 , x ₂ ), d(x 0 , x ₁ ), d(x 0 , x ₁ ), d(x 1 , x ₂ ), d(x 0 , x ₂ ), 0) ≤ 0.

From T ₂ , there exists a continuous and nondecreasing function f : R ₊ → R ₊ , f (0) = 0, f (t) < t for t > 0, such that

d(x ₁ , x ₂ ) ≤ f (d(x ₀ , x ₁ )).

Thus from (3.2) we have

d(x ₀ , x ₂ ) ≤ d(x ₀ , x ₁ ) + d(x ₁ , x ₂ ) ≤ d(x ₀ , x ₁ ) + f (d(x ₀ , x ₁ ))

< d(x ₀ , x ₁ ) + γ = r

and so x ₂ ∈ B(x ₀ , r). Now suppose there exists k ∈ {2, 3, . . .} with x m ∈ B(x 0 , r) for m ∈ {1, 2, . . . , k}. We must show x ^k ₊₁ ∈ B(x 0 , r). Since x _m−1 , x m ∈ B(x ₀ , r) for m ∈ {1, 2, . . . , k}, we can use the condition (3.1).

Therefore we have

T (d(F x _m−1 , F x m ), d(x _m−1 , x m ), d(x _m−1 , F x _m−1 ), d(x ^m , F x m ), d(x _m−1 , F x m ), d(x ^m , F x _m−1 )) ≤ 0 and so

T (d(x ^m , x m +1 ), d(x _m−1 , x m ), d(x _m−1 , x m ), d(x ^m , x m +1 ), d(x _m−1 , x m +1 ), 0) ≤ 0.

From T ₂ , we have

d(x ^m , x m +1 ) ≤ f (d(x _m−1 , x m )) for m ∈ {1, 2, . . . , k}. Therefore we have

d(x ^m , x m +1 ) ≤ f ^m (d(x ₀ , x ₁ )) for m ∈ {1, 2, . . . , k}. On the other hand, since

d(x ₀ , x k +1 ) ≤ d(x ₀ , x ₁ ) +

k

X

m =1

d(x ^m , x m +1 )

for k ∈ {1, 2, . . .}, then from (3.2) we have

d(x ₀ , x k +1 ) ≤ d(x ₀ , x ₁ ) +

k

X

m =1

f ^m d(x ₀ , x ₁ )

≤ d(x ₀ , x ₁ ) +

∞

X

m =1

f ^m d(x ₀ , x ₁ )

< d(x ₀ , x ₁ ) + γ = r .

Thus we have x ^k ₊₁ ∈ B(x ₀ , r). Therefore x ⁿ ∈ B(x ₀ , r) for n ∈ {1, 2, . . .}.

Also note that

(3.5) d(x n , x n +1 ) ≤ f ⁿ (d(x ₀ , x ₁ )) for n ∈ {1, 2, . . .}.

Next we show that {x ⁿ } is a Cauchy sequence with respect to d. Suppose it is not true. Then we can find a δ > 0 and two sequences of integers {m(k)}, {n(k)}, m(k) > n(k) ≥ k with

(3.6) r k = d(x n (k) , x m (k) ) ≥ δ for k ∈ {1, 2, . . .}.

We may also assume

(3.7) d(x m (k)−1 , x n (k) ) < δ

by choosing m(k) to be the smallest number exceeding n(k) for which (3.6) holds. Now (3.5), (3.6) and (3.7) imply

δ ≤ r k ≤ d(x m (k) , x m (k)−1 ) + d(x m (k)−1 , x n (k) )

≤ f ^{m (k)−1} (d(x ₀ , x ₁ )) + δ and so

(3.8) lim

k→∞ r k = δ.

Also, since

δ ≤ r k ≤ d(x n (k) , x n (k)+1 ) + d(x m (k) , x m (k)+1 ) + d(x n (k)+1 , x m (k)+1 )

we have from (3.5) that

(3.9) δ ≤ r ^k ≤ f ^{n (k)} (d(x ₀ , x ₁ )) + f ^{m (k)} (d(x ₀ , x ₁ )) + d(F x n (k) , F x m (k) ).

On the other hand, since x n (k) , x m (k) ∈ B(x ₀ , r) we can use the condition (3.1). Therefore we have

T (d(F x n (k) , F x m (k) ), d(x n (k) , x m (k) ), d(x n (k) , F x n (k) ), d(x m (k) , F x m (k) ), d(x n (k) , F x m (k) ), d(x m (k) , F x n (k) )) ≤ 0 and so

T (d(F x n (k) , F x m (k) ), r ^k , f ^{n (k)} (d(x ₀ , x ₁ )), f ^{m (k)} (d(x ₀ , x ₁ )), r k + f ^{m (k)} (d(x ₀ , x ₁ )), r ^k + f ^{n (k)} (d(x ₀ , x ₁ ))) ≤ 0.

Now letting k → ∞ and using (3.8) we have, by continuity of T, that T



k→∞ lim d(F x n (k) , F x m (k) ), δ, 0, 0, δ, δ



≤ 0.

From T 2 , we have lim _k→∞ d(F x n (k) , F x m (k) ) ≤ f (δ). Therefore letting k → ∞ in (3.9) we have δ ≤ f (δ). This is a contradiction since f (t) < t for t > 0. Thus {x n } is a Cauchy sequence with respect to d.

We now claim that {x n } is a Cauchy sequence with respect to d ⁰ . If d ≥ d ⁰ , it is trivial. Next suppose d  d ⁰ . Let ε > 0 be given. Now (3.3) guarantees that there exists δ > 0 such that

(3.10) d ⁰ (F x, F y) < ε whenever x, y ∈ B(x ₀ , r) and d(x, y) < δ.

From above we know that there exists N ∈ {1, 2, . . .} with (3.11) d(x ⁿ , x m ) < δ whenever n, m ≥ N.

Now (3.10) and (3.11) imply

d ⁰ (x n +1 , x m +1 ) = d ⁰ (F x n , F x m ) < ε whenever n, m ≥ N,

and as a result, {x ⁿ } is a Cauchy sequence with respect to d ⁰ . Now since (X, d ⁰ ) is complete there exists x ∈ B(x ₀ , r) ^d

with d ⁰ (x ⁿ , x) → 0 as n → ∞.

We claim that x = F x.

If x = F x we are finished. First, consider the case when d 6= d ⁰ . Notice d ⁰ (x, F x) ≤ d ⁰ (x, x n ) + d ⁰ (F x _n−1 , F x).

Let n → ∞ and use (3.4) to obtain d ⁰ (x, F x) = 0, so x = F x. Next suppose d = d ⁰ . Since x, x _n−1 ∈ B(x 0 , r) we can use the condition (3.1) and we have

T (d(F x _n−1 , F x), d(x _n−1 , x), d(x _n−1 , F x _n−1 ), d(x, F x), d(x _n−1 , F x), d(x, F x _n−1 )) ≤ 0 and letting n → ∞ we have

T (d(x, F x), 0, 0, d(x, F x), d(x, F x), 0) ≤ 0.

From T ₃ , we have d(x, F x) = 0, so x = F x. This completes the proof.

If we take

(3.12) d(x ₀ , F x ₀ ) < r − f (r)

instead of the condition (3.2), then we have the following result.

Theorem 2. Let (X, d ⁰ ) be a complete metric space, d another metric on X, x ₀ ∈ X, r > 0 and F : B(x ₀ , r) ^d

→ X. Suppose F satisfies the conditions (3.1), (3.3), (3.4) and (3.12). Then F has a fixed point. That is, there exists x ∈ B(x ₀ , r) ^d

with x = F x.

P roof. Let x ₁ = F x ₀ . Now d(x ₀ , x ₁ ) < r, so x ₁ ∈ B(x ₀ , r). Next we let x n = F x _n−1 for n ∈ {2, 3, . . .}. Now, we claim

x n ∈ B(x ₀ , r) for n ∈ {1, 2, . . .} and {x n } is a Cauchy sequence with respect to d.

At first, we show x n ∈ B(x ₀ , r) for n ∈ {1, 2, . . .}. The proof is by induction.

We know x ₀ , x ₁ ∈ B(x ₀ , r). Thus we can use the condition (3.1). Therefore we have

T (d(F x 0 , F x ₁ ), d(x 0 , x ₁ ), d(x 0 , F x ₀ ), d(x 1 , F x ₁ ), d(x 0 , F x ₁ ), d(x 1 , F x ₀ )) ≤ 0

and so

T (d(x ₁ , x ₂ ), d(x ₀ , x ₁ ), d(x ₀ , x ₁ ), d(x ₁ , x ₂ ), d(x ₀ , x ₂ ), 0) ≤ 0.

From T ₂ , there exists a continuous and nondecreasing function f : R ₊ → R ₊ , f (0) = 0, f (t) < t for t > 0, such that

d(x ₁ , x ₂ ) ≤ f (d(x ₀ , x ₁ )).

Thus from (3.12) we have

d(x ₀ , x ₂ ) ≤ d(x ₀ , x ₁ ) + d(x ₁ , x ₂ ) ≤ d(x ₀ , x ₁ ) + f (d(x ₀ , x ₁ ))

< [r − f (r)] + f (r) = r

and so x ₂ ∈ B(x ₀ , r). Now suppose there exists k ∈ {2, 3, . . .} with x m ∈ B(x 0 , r) for m ∈ {1, 2, . . . , k}. We must show x ^k ₊₁ ∈ B(x 0 , r). Since x _m−1 , x ^m ∈ B(x ₀ , r) for m ∈ {1, 2, . . . , k}, we can use the condition (3.1).

Thus we have

T (d(F x _m−1 , F x m ), d(x _m−1 , x m ), d(x _m−1 , F x _m−1 ), d(x ^m , F x m ), d(x _m−1 , F x m ), d(x ^m , F x _m−1 )) ≤ 0 and so

T (d(x m , x m +1 ), d(x _m−1 , x m ), d(x _m−1 , x m ), d(x m , x m +1 ), d(x _m−1 , x m +1 ), 0) ≤ 0.

From T ₂ , we have

(3.13) d(x ^m , x m +1 ) ≤ f (d(x _m−1 , x m )) for m ∈ {1, 2, . . . , k}. Therefore we have

d(x m , x m +1 ) ≤ f ^m (d(x ₀ , x ₁ )) for m ∈ {1, 2, . . . , k}. Now, we show that

(3.14) d(x ₁ , x k ) ≤ f (r).

If k = 2, this is immediate since (3.13) implies d(x ₁ , x ₂ ) ≤ f (d(x ₀ , x ₁ )) ≤ f (r). If k = 3, we have d(x ₁ , x ₃ ) = d(F x ₀ , F x ₂ ). On the other hand, x ₀ , x ₂ ∈ B(x 0 , r), we can use (3.1). Thus we have

T (d(F x ₀ , F x ₂ ), d(x ₀ , x ₂ ), d(x ₀ , F x ₀ ), d(x 2 , F x ₂ ), d(x 0 , F x ₂ ), d(x 2 , F x ₀ )) ≤ 0 and

T (d(x ₁ , x ₃ ), d(x ₀ , x ₂ ), d(x ₀ , x ₂ ), d(x ₁ , x ₃ ), d(x ₀ , x ₃ ), 0) ≤ 0 and so

T (d(x ₁ , x ₃ ), r, r, d(x ₁ , x ₃ ), r + d(x ₁ , x ₃ ), 0) ≤ 0.

From T ₂ , we have

d(x ₁ , x ₃ ) ≤ f (r).

If k = 4, we have d(x ₁ , x ₄ ) = d(F x ₀ , F x ₃ ). On the other hand, since x ₀ , x ₃ ∈ B(x 0 , r), we can use (3.1). Thus we have

T (d(F x ₀ , F x ₃ ), d(x ₀ , x ₃ ), d(x ₀ , F x ₀ ), d(x ₃ , F x ₃ ), d(x ₀ , F x ₃ ), d(x ₃ , F x ₀ )) ≤ 0 and so

T (d(x ₁ , x ₄ ), d(x ₀ , x ₃ ), d(x ₀ , x ₃ ), d(x ₁ , x ₄ ), d(x ₀ , x ₄ ), 0) ≤ 0 and

T (d(x ₁ , x ₄ ), r, r, d(x ₁ , x ₄ ), r + d(x ₁ , x ₄ ), 0) ≤ 0.

From T ₂ , we have

d(x ₁ , x ₄ ) ≤ f (r).

Thus (3.14) is true if k = 4. Continue this process to obtain (3.14) if k ∈ {5, 6 . . .}.

Now, we show that x ^k ₊₁ ∈ B(x 0 , r). Since x 0 , x k ∈ B(x 0 , r) we can use (3.1). Thus we have

T (d(F x 0 , F x k ), d(x 0 , x k ), d(x 0 , F x ₀ ),

d(x ^k , F x k ), d(x ₀ , F x k ), d(x ^k , F x ₀ )) ≤ 0

and so

T (d(x ₁ , x k +1 ), d(x ₀ , x k ), d(x ₀ , x k ), d(x ₁ , x k +1 ), d(x ₀ , x k +1 ), 0) ≤ 0 and from (3.14) we have

T (d(x ₁ , x k +1 ), r, r, d(x ₁ , x k +1 ), r + d(x ₁ , x k +1 ), 0) ≤ 0.

From T ₂ , we have

d(x ₁ , x k +1 ) ≤ f (r).

Thus we have

d(x 0 , x k +1 ) ≤ d(x 0 , x ₁ ) + d(x 1 , x k +1 ) < [r − f (r)] + f (r) = r

and so we have x ^k ₊₁ ∈ B(x ₀ , r). Consequently, x ⁿ ∈ B(x ₀ , r) for n ∈ {1, 2, . . .}.

We can complete the proof as in Theorem 1.

Essentially, the reasoning as in Theorem 2 guarantees the following global result.

Theorem 3. Let (X, d ⁰ ) be a complete metric space, d another metric on X, and F : X → X. Suppose F satisfies the following conditions;

(a) There exists T ∈ T such that

T (d(F x, F y), d(x, y), d(x, F x), d(y, F y), d(x, F y), d(y, F x)) ≤ 0 for all x, y ∈ X,

(b) If d  d ⁰ assume F is uniformly continuous from (X, d) into (X, d ⁰ ).

(c) If d 6= d ⁰ assume F is continuous from (X, d ⁰ ) into (X, d ⁰ ).

Then F has a fixed point.

Remark 1.

(1) Theorem 2.1 of [1] follows from Example 2 and Theorem 1.

(2) If we take d = d ⁰ in Theorem 3, then the main theorem of [3] follows

from Example 2 and Theorem 3.

(3) If we take d = d ⁰ in Theorem 2, then Theorem 2.1 of [2] follows from Example 3 and Theorem 2.

(4) If we take d = d ⁰ in Theorem 3, then Theorem 2.2 of [2] follows from Example 3 and Theorem 3.

4. Homotopy result

In this section we present a homotopy result for the maps discussed in Sec- tion 3. Our theorem improves the results in [1, 2].

Theorem 4. Let (X, d ⁰ ) be a complete metric space and let d be another metric on X. Let Q ⊆ X be d ⁰ -closed and let U ⊆ X be d-open and U ⊆ Q.

Suppose H : Q × [0, 1] → X satisfied the following properties : (a) x 6= H(x, λ) for x ∈ Q\U and λ ∈ [0, 1],

(b) There exists T ∈ T such that

T(d(H(x, λ), H(y, λ)), d(x, y), d(x, H(x, λ)) d(y, H(y, λ)), d(x, H(y, λ)), d(y, H(x, λ))) ≤ 0 for all x, y ∈ Q and λ ∈ [0, 1] ,

(c) H(x, λ) is continuous in λ with respect to d, uniformly for x ∈ Q, (d) If d  d ⁰ , assume H is uniformly continuous from U × [0, 1] endowed

with the metric d on U into (X, d ⁰ ),

(e) If d 6= d ⁰ , assume H is continuous from Q × [0, 1] endowed with the metric d ⁰ on Q into (X, d ⁰ ).

In addition, assume H ₀ has a fixed point. Then for each λ ∈ [0, 1] we have that H λ has a fixed point x λ ∈ U (here H λ (·) = H(·, λ)).

P roof. Let

A = {λ ∈ [0, 1] : H(x, λ) = x for some x ∈ U }.

Now since H ₀ has a fixed point (and (a) holds) we have that 0 ∈ A, so A is

nonempty. We will show A is both closed and open in [0, 1], and so by the

connectedness of [0, 1] we are finished since A = [0, 1].

First, we show A is closed in [0, 1]. Let (λ k ) be a sequence in A with λ k → λ ∈ [0, 1] as k → ∞. By definition for each k, there exists x k ∈ U with x k = H(x ^k , λ k ). We claim

(4.1) {x k } is a Cauchy sequence with respect to d.

Suppose (4.1) is not true. Then we can find a δ > 0 and two sequences of integers {m(k)}, {n(k)}, m(k) > n(k) ≥ k with

(4.2) r k = d(x n (k) , x m (k) ) ≥ δ for k ∈ {1, 2, . . .}.

Notice also that

δ ≤ r k ≤ d(H(x m (k) , λ m (k) ), H(x m (k) , λ)) + d(H(x m (k) , λ), H(x n (k) , λ)) + d(H(x n (k) , λ), H(x n (k) , λ n (k) )).

and letting k → ∞ and using (c) we have

(4.3) δ ≤ lim

k→∞ r k ≤ lim

k→∞ d(H(x m (k) , λ), H(x n (k) , λ)).

On the other hand, using (b), we have

T (d(H(x m (k) , λ), H(x n (k) , λ)), d(x m (k) , x n (k) ), d(x m (k) , H(x m (k) , λ)), d(x n (k) , H(x n (k) , λ)), d(x m (k) , H(x n (k) , λ)), d(x n (k) , H(x m (k) , λ))) ≤ 0 and so

T(d(H(x m (k) , λ), H(x n (k) , λ)), r ^k , d(H(x m (k) , λ m (k) ), H(x m (k) , λ)), d(H(x n (k) , λ n (k) ), H(x n (k) , λ)), r ^k + d(H(x n (k) , λ n (k) ), H(x n (k) , λ)), r k + d(H(x m (k) , λ m (k) ), H(x m (k) , λ))) ≤ 0.

Now letting k → ∞ and using (c) we have T



k→∞ lim d H (x m (k) , λ), H(x n (k) , λ)
, lim

k→∞ r k , 0, 0, lim

k→∞ r k , lim

k→∞ r k



≤ 0.

From T ₂ we have lim

k→∞ d(H(x m (k) , λ), H(x n (k) , λ)) ≤ f lim

k→∞ r k

and using (4.3) we have

δ ≤ lim

k→∞ r k ≤ lim

k→∞ d H(x m (k) , λ), H(x n (k) , λ)
≤ f lim

k→∞ r k
.

which is a contradiction. Thus (4.1) holds. We now claim that (4.4) {x k } is a Cauchy sequence with respect to d ⁰ . If d ≥ d ⁰ , this is trivial. If d  d ⁰ , then

d ⁰ (x ⁿ , x m ) = d ⁰ (H(x ⁿ , λ n ), H(x ^m , λ m ))

and (d) guarantee that (4.4) is true. Now since (X, d ⁰ ) is complete there exists an x ∈ Q with d ⁰ (x ⁿ , x) → 0 as n → ∞. We next claim that

(4.5) x = H(x, λ).

We consider first the case d 6= d ⁰ . Now

d ⁰ (x, H(x, λ)) ≤ d ⁰ (x, x n ) + d ⁰ (x n , H(x, λ))

≤ d ⁰ (x, x ⁿ ) + d ⁰ (H(x ⁿ , λ n ), H(x, λ))

together with (e) guarantee that d ⁰ (x, H(x, λ)) = 0, so (4.5) holds. Next suppose that d = d ⁰ . Now using (b) we have

T (d(H(x ⁿ , λ n ), H(x, λ ⁿ )), d(x ⁿ , x), d(x ⁿ , H(x ⁿ , λ n )), d(x, H(x, λ ⁿ )), d(x ⁿ , H(x, λ ⁿ )), d(x, H(x ⁿ , λ ⁿ ))) ≤ 0 and so

T(d(H(x ⁿ , λ n ), H(x, λ ⁿ )), d(x ⁿ , x), 0, d(x, H(x, λ)) + d(H(x, λ), H(x, λ ⁿ )), d(x ⁿ , x) + d(x, H(x, λ)) + d(H(x, λ), H(x, λ ⁿ )), d(x, x ⁿ )) ≤ 0

and letting n → ∞ and use (c) we have

T (d(x, H(x, λ)), 0, 0, d(x, H(x, λ)), d(x, H(x, λ)), 0) ≤ 0.

From T ₃ , we have d(x, H(x, λ)) = 0, so (4.5) holds when d = d ⁰ . Now from (4.5) and (a) we have x ∈ U. Consequently, λ ∈ A so A is closed in [0, 1].

Next we show A is open in [0, 1]. Let λ 0 ∈ A and x 0 ∈ U with x 0 = H(x ₀ , λ ₀ ). Since U is d-open there exists a d-ball B(x ₀ , δ) = {x ∈ X : d(x, x ₀ ) < δ}, δ > 0, with B(x ₀ , δ) ⊆ U. Now (c) guarantees that there exists η = η(δ) > 0 with

d(x ₀ , H(x ₀ , λ)) = d(H(x ₀ , λ ₀ ), H(x ₀ , λ)) < δ − f (δ)

for λ ∈ [0, 1] and |λ − λ ₀ | ≤ η, where f is defined by T ₂ . Now (b), (d) and (e) together with Theorem 2 (in this case r = δ and F = H ^λ ) guarantee that there exists x λ ∈ B(x ₀ , δ) ^d

⊆ Q with x λ = H λ (x λ ) for λ ∈ [0, 1] and

|λ − λ ₀ | ≤ η. Consequently A is open in [0, 1].

Remark 2.

(1) Theorem 3.1 of [1] follows from Example 2 and Theorem 4.

(2) If we take d = d ⁰ in Theorem 4, then Theorem 2.4 of [2] follows from Example 3 and Theorem 4.

References

[1] R.P. Agarwal and D. O’Regan, Fixed point theory for generalized contraction on space with two metrics, J. Math. Anal. Appl. 248 (2000), 402–414.

[2] R.P. Agarwal, D. O’Regan and M. Sambandham, Random and deterministic fixed point theory for generalized contractive maps, Appl. Anal. 83 (7) (2004), 711–725.

[3] G.E. Hardy and T.G. Rogers, A generalization of a fixed point theorem of Riech, Canad. Math. Bull. 16 (1973), 201–206.

[4] M. Imdad, S. Kumar and M.S. Khan, Remarks on some fixed point theorems satisfying implicit relations, Rad. Math. 11 (1) (2002), 135–143.

[5] V. Popa, A general coincidence theorem for compatible multivalued mappings satisfying an implicit relation, Demonstratio Math. 33 (1) (2000), 159–164.

[6] V. Popa, Some fixed point theorems for compatible mappings satisfying an im- plicit relation, Demonstratio Math. 32 (1) (1999), 157–163.

[7] S. Sharma and B. Desphande, On compatible mappings satisfying an implicit relation in common fixed point consideration, Tamkang J. Math. 33 (3) (2002), 245–252.

Received 15 September 2006