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Journal of Inequalities and Applications

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Open Access 01.12.2013 | Research

Common fixed point results for weak contractive mappings in ordered b-dislocated metric spaces with applications

verfasst von: Nawab Hussain, Jamal Rezaei Roshan, Vahid Parvaneh, Mujahid Abbas

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2013

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Abstract

We first introduce a new concept of b-dislocated metric space as a generalization of dislocated metric space and analyze different properties of such spaces. A fundamental result for the convergence of sequences in b-dislocated metric spaces is established and is employed to prove some common fixed point results for four mappings satisfying the generalized weak contractive condition in partially ordered b-dislocated metric spaces. Moreover, some examples and applications to integral equations are given here to illustrate the usability of the obtained results.

MSC:47H10, 54H25.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

1 Introduction and preliminaries

The Banach contraction principle is one of the simplest and most applicable results of metric fixed point theory. It is a popular tool for proving the existence of solution of problems in different fields of mathematics. There are several generalizations of the Banach contraction principle in literature on metric fixed point theory [1‐10]. Hitzler and Seda [11] introduced the concept of dislocated topologies and named their corresponding generalized metric a dislocated metric. They have also established a fixed point theorem in complete dislocated metric spaces to generalize the celebrated Banach contraction principle. The notion of dislocated topologies has useful applications in the context of logic programming semantics (see [12]). Further useful results can be seen in [13‐23].

Definition 1.1 [11]

Let X be a nonempty set. A mapping

d_{l} : X \times X \to [0, \infty)

is called a dislocated metric (or simply

d_{l}

-metric) if the following conditions hold for any

x, y, z \in X

(i)

d_{l} (x, y) = 0

, then

x = y

;

(ii)

d_{l} (x, y) = d_{l} (y, x)

;

(iii)

d_{l} (x, y) \leq d_{l} (x, z) + d_{l} (z, y)

The pair

(X, d_{l})

is called a dislocated metric space or a

d_{l}

-metric space. Note that when

x = y

d_{l} (x, y)

may not be 0.

Example 1.2 If

X = R^{+} \cup {0}

, then

d_{l} (x, y) = x + y

defines a dislocated metric on X.

Definition 1.3 [11]

A sequence

{x_{n}}

in a

d_{l}

-metric space is called: (1) a Cauchy sequence if, given

ε > 0

, there exists

n_{0} \in N

such that for all

n, m \geq n_{0}

, we have

d_{l} (x_{m}, x_{n}) < ε

{lim}_{n, m \to \infty} d_{l} (x_{n}, x_{m}) = 0

, (2) convergent with respect to

d_{l}

if there exists

x \in X

such that

d_{l} (x_{n}, x) \to 0

n \to \infty

. In this case, x is called the limit of

{x_{n}}

and we write

x_{n} \to x

d_{l}

-metric space X is called complete if every Cauchy sequence in X converges to a point in X.

Definition 1.4 A nonempty set X is called an ordered dislocated metric space if it is equipped with a partial ordering ⪯ and there exists a dislocated metric

d_{l}

on X.

Definition 1.5 Let

(X, ⪯)

be a partially ordered set. Then

x, y \in X

are called comparable if

x ⪯ y

y ⪯ x

holds.

Definition 1.6 [1]

Let

(X, ⪯)

be a partially ordered set. A self-mapping f on X is called dominating if

x ⪯ f x

for each x in X.

Example 1.7 [1]

Let

X = [0, 1]

be endowed with the usual ordering, and let

f : X \to X

be defined by

f x = \sqrt[n]{x}

. Since

x \leq x^{\frac{1}{n}} = f x

for all

x \in X

, therefore f is a dominating map.

Definition 1.8 [1]

Let

(X, ⪯)

be a partially ordered set. A self-mapping f on X is called dominated if

f x ⪯ x

for each x in X.

Example 1.9 [1]

Let

X = [0, 1]

be endowed with the usual ordering, and let

f : X \to X

be defined by

f x = x^{n}

for some

n \in N

. Since

f x = x^{n} \leq x

for all

x \in X

, therefore f is a dominated map.

In the following, we give the definition of a b-dislocated metric space.

Definition 1.10 Let X be a nonempty set. A mapping

b_{d} : X \times X \to [0, \infty)

is called a b-dislocated metric (or simply

b_{d}

-metric) if the following conditions hold for any

x, y, z \in X

and

s \geq 1

(

b_{d 1}

) If

b_{d} (x, y) = 0

, then

x = y

;

(

b_{d 2}

)

b_{d} (x, y) = b_{d} (y, x)

;

(

b_{d 3}

)

b_{d} (x, y) \leq s (b_{d} (x, z) + b_{d} (z, y))

The pair

(X, b_{d})

is called a b-dislocated metric space or a

b_{d}

-metric space. It should be noted that the class of

b_{d}

-metric spaces is effectively larger than that of

d_{l}

-metric spaces, since a

b_{d}

-metric is a

b_{l}

-metric when

s = 1

Here, we present an example to show that in general a b-dislocated metric need not be a

b_{l}

-metric.

Example 1.11 Let

(X, d_{l})

be a dislocated metric space, and

b_{d} (x, y) = {(b_{l} (x, y))}^{p}

, where

p > 1

is a real number. We show that

b_{d}

is a b-dislocated metric with

s = 2^{p - 1}

Obviously, conditions (

b_{d 1}

) and (

b_{d 2}

) of Definition 1.10 are satisfied.

1 < p < \infty

, then the convexity of the function

f (x) = x^{p}

(

x > 0

) implies that

{(\frac{a + b}{2})}^{p} \leq \frac{1}{2} (a^{p} + b^{p})

. Hence,

{(a + b)}^{p} \leq 2^{p - 1} (a^{p} + b^{p})

holds. Thus, for each

x, y, z \in X

, we obtain that

\begin{array}{rcl} b_{d} (x, y) & = & {(d_{l} (x, y))}^{p} \leq {[d_{l} (x, z) + d_{l} (z, y)]}^{p} \\ \leq & 2^{p - 1} [{(d_{l} (x, z))}^{p} + {(d_{l} (z, y))}^{p}] \\ = & 2^{p - 1} [b_{d} (x, z) + b_{d} (z, y)] . \end{array}

So, condition (

b_{d 3}

) of Definition 1.10 is also satisfied and

b_{d}

is a

b_{d}

-metric.

However, if

(X, d_{l})

is a dislocated metric space, then

(X, b_{d})

is not necessarily a dislocated metric space. For example, if

X = R

is the set of real numbers, then

d_{l} (x, y) = | x | + | y |

is a dislocated metric, and

b_{d} (x, y) = {(| x | + | y |)}^{2}

is a b-dislocated metric on ℝ with

s = 2

, but not a dislocated metric on ℝ.

Recently, Sarma and Kumari [15] established the existence of a topology induced by a dislocated metric which is metrizable with a family of sets

{B (x, ε) \cup {x} : x \in X, ε > 0}

as a base, where

B (x, ε) = {y \in X : d_{l} (x, y) < ε}

for all

x \in X

and

ε > 0

. Also,

\bar{B (x, ε)} = {y \in X : d_{l} (x, y) \leq ε}

is a closed ball.

On the similar lines, we show that each b-dislocated metric space on X generates a topology

τ_{b_{d}}

whose base is the family of open

b_{d}

-balls

B_{b_{d}} (x, ε) = {y \in X : b_{d} (x, y) < ε} .

Definition 1.12 We say that a net

(x_{α} : α \in Δ)

in X converges to x in

(X, b_{d})

and write

{lim}_{α \in Δ} x_{α} = x

{lim}_{α \in Δ} b_{d} (x_{α}, x) = 0

Note that the limit of a net in

(X, b_{d})

is unique. For

A \subseteq X

, we write

D (A) = {x \in X : x is a limit of a net in (A, b_{d})}

Proposition 1.13 If

A, B \subseteq X

, then

(i)

D (A) = \emptyset

A = \emptyset

(ii)

D (A) \subseteq D (B)

A \subseteq B

(iii)

D (A \cup B) = D (A) \cup D (B)

(iv)

D (D (A)) \subseteq D (A)

Proof To prove (i), (ii) and (iii), we refer to [15]. To prove (iv), let

x \in D (D (A))

. Suppose that for each α in Δ,

(x_{α_{β}} : β \in Δ (α))

is a net in A such that

x_{α} = {lim}_{β \in Δ (α)} x_{α_{β}}

. Thus, for each positive integer i, there is

α_{i} \in Δ

such that

b_{d} (x_{α_{i}}, x) < \frac{1}{2 i s}

, and

β_{i} \in Δ (α_{i})

such that

b_{d} (x_{α_{i_{β_{i}}}}, x_{α_{i}}) < \frac{1}{2 i s}

. Take

α_{i_{β_{i}}} = γ_{i}

for each i, then

{γ_{1}, γ_{2}, γ_{3}, \dots}

is a directed set

γ_{i} < γ_{j}

i < j

, and

b_{d} (x_{γ_{i}}, x) \leq s (b_{d} (x_{γ_{i}}, x_{α_{i}}) + b_{d} (x_{α_{i}}, x)) < \frac{1}{i}

. This implies that

x \in D (A)

. □

As a corollary, we have the following.

Corollary 1.14 Let, for all

A \subset X

\bar{A} = A \cup D (A)

. Then the operation

A \to \bar{A}

P (X)

satisfies Kuratowski’s closure axioms [2]:

(i)

\bar{\emptyset} = \emptyset

(ii)

A \subset \bar{A}

(iii)

\bar{A} = \bar{\bar{A}}

(iv)

\bar{A \cup B} = \bar{A} \cup \bar{B}

Consequently, we have the following.

Theorem 1.15 Let ϒ be the family of all subsets A of X for which

\bar{A} = A

and

τ_{b_{d}}

are the complements of members of ϒ. Then the

τ_{b_{d}}

is a topology for X and the

τ_{b_{d}}

-closure of a subset A of X is

\bar{A}

Definition 1.16 The topology

τ_{b_{d}}

obtained in Theorem 1.15 is called the topology induced by

b_{d}

and simply referred to as the

b_{d}

-topology of X; and it is denoted by

(X, b_{d}, τ_{b_{d}})

Now we state some propositions and corollaries in

(X, b_{d}, τ_{b_{d}})

which can be proved following similar arguments to those given in [15].

Proposition 1.17 Let

A \subseteq X

. Then

x \in D (A)

iff for every

δ > 0

B_{δ} (x) \cap A \neq \emptyset

Corollary 1.18

x \in \bar{A} ⟺ x \in A

B_{δ} (x) \cap A \neq \emptyset

\forall δ > 0

Corollary 1.19 A set

A \subseteq X

is open in

(X, b_{d}, τ_{b_{d}})

if and only if for every

x \in A

, there is

δ > 0

such that

{x} \cup B_{δ} (x) \subseteq A

Proposition 1.20 If

x \in X

and

δ > 0

, then

{x} \cup B_{δ} (x)

is an open set in

(X, b_{d}, τ_{b_{d}})

Corollary 1.21 If

x \in X

and

V_{r} (x) = B_{r} (x) \cup {x}

for

r > 0

, then the collection

{V_{r} (x) | x \in X}

is an open base at x in

(X, b_{d}, τ_{b_{d}})

. If

b_{d}

is a b-metric and

V = B (x)

, then

τ_{b_{d}}

coincides with the metric topology.

Proposition 1.22

(X, b_{d}, τ_{b_{d}})

is a Hausdorff space.

Proof If

x, y \in X

and

\frac{b_{d} (x, y)}{2 s} = r > 0

, then

V_{r} (x) \cap V_{r} (y) = \emptyset

. □

Corollary 1.23 If

x \in X

, then the collection

{V (x) | x \in X}

is an open base at x for

(X, b_{d}, τ_{b_{d}})

. Hence,

(X, b_{d}, τ_{b_{d}})

is first countable.

Remark 1.24 The above corollary enables us to deal with sequences instead of nets.

Motivated by Proposition 3.2 in [11], we have the following proposition for the b-dislocated metric space.

Proposition 1.25 Let

(X, b_{d})

be a b-dislocated metric space. The following three conditions are equivalent:

(i)

For all

x \in X

, we have

b_{d} (x, x) = 0

(ii)

b_{d}

is a b-metric.

(iii)

For all

x \in X

and all

r > 0

, we have

B_{r} (x) \neq \emptyset

Proof We show that (iii) implies (i). Since

B \frac{r}{2 s} (x) \neq \emptyset

for all

r > 0

, there exists some

y \in X

with

b_{d} (x, y) < \frac{r}{2 s}

. But for all

y \in X

, we have

b_{d} (x, x) \leq 2 s b_{d} (x, y)

. Therefore,

b_{d} (x, x) < r

for all

r > 0

. Hence,

b_{d} (x, x) = 0

. □

(X, b_{d})

is a b-dislocated metric space, then

(X^{'}, b_{d})

, where

X^{'} = {x \in X | b_{d} (x, x) = 0}

is a b-metric space. Indeed,

(X^{'}, b_{d})

is a b-dislocated metric space, so assertion now follows immediately from the above proposition.

Definition 1.26 A sequence

{x_{n}}

in a b-dislocated metric space

(X, b_{d})

converges with respect to

b_{d}

(

b_{d}

-convergent) if there exists

x \in X

such that

b_{d} (x_{n}, x)

converges to 0 as

n \to \infty

. In this case, x is called the limit of

{x_{n}}

, and we write

x_{n} \to x

Proposition 1.27 Limit of a convergent sequence in a b-dislocated metric space is unique.

Proof Let x and y be limits of the sequence

{x_{n}}

. By properties (

b_{d 2}

) and (

b_{d 3}

) of Definition 1.10, it follows that

b_{d} (x, y) \leq s (b_{d} (x_{n}, x) + b_{d} (x_{n}, y)) \to 0

. Hence,

b_{d} (x, y) = 0

, and by property (

b_{d 1}

) of Definition 1.10 it follows that

x = y

. □

Definition 1.28 A sequence

{x_{n}}

in a b-dislocated metric space

(X, b_{d})

is called a

b_{d}

-Cauchy sequence if, given

ε > 0

, there exits

n_{0} \in N

such that for all

n, m \geq n_{0}

, we have

b_{d} (x_{m}, x_{n}) < ε

{lim}_{n, m \to \infty} b_{d} (x_{n}, x_{m}) = 0

Proposition 1.29 Every convergent sequence in a b-dislocated space is

b_{d}

-Cauchy.

Proof Let

{x_{n}}

be a sequence which converges to some x, and

ε > 0

. Then there exists

n_{0} \in N

with

b_{d} (x_{n}, x) < \frac{ε}{2 s}

for all

n \geq n_{0}

. For

m, n \geq n_{0}

, we obtain

b_{d} (x_{n}, x_{m}) \leq s (b_{d} (x_{n}, x) + b_{d} (x_{m}, x)) < 2 s \frac{ε}{2 s} = ε

. Hence,

{x_{n}}

b_{d}

-Cauchy. □

Definition 1.30 A b-dislocated metric space

(X, b_{d})

is called complete if every

b_{d}

-Cauchy sequence in X is

b_{d}

-convergent.

The following example shows that in general a b-dislocated metric is not continuous.

Example 1.31 Let

X = N \cup {\infty}

and

b_{d} : X \times X \to R

be defined by

b_{d} (m, n) = {\begin{matrix} \frac{1}{m} + \frac{1}{n} & if m, n are even or m n = \infty, \\ 5 & if m and n are odd and m \neq n, \\ 2 & otherwise . \end{matrix}

Then it is easy to see that for all

m, n, p \in X

, we have

b_{d} (m, p) \leq 5 (b_{d} (m, n) + b_{d} (n, p)) .

Thus,

(X, b_{d})

is a b-dislocated metric space. Let

x_{2 n} = 2 n

for each

n \in N

. Then

b_{d} (2 n, \infty) = \frac{1}{2 n} \to 0 as n \to \infty,

that is,

x_{n} \to \infty

, but

b_{d} (x_{n}, 1) = 2 ↛ b_{d} (\infty, 1)

n \to \infty

We need the following simple lemma about the

b_{d}

-convergent sequences in the proof of our main results.

Lemma 1.32 Let

(X, b_{d})

be a b-dislocated metric with parameter

s \geq 1

. Suppose that

{x_{n}}

and

{y_{n}}

are

b_{d}

-convergent to x, y, respectively. Then we have

\frac{1}{s^{2}} b_{d} (x, y) \leq \underset{n \to \infty}{lim inf} b_{d} (x_{n}, y_{n}) \leq \underset{n \to \infty}{lim sup} b_{d} (x_{n}, y_{n}) \leq s^{2} b_{d} (x, y) .

In particular, if

b_{d} (x, y) = 0

, then we have

{lim}_{n \to \infty} b_{d} (x_{n}, y_{n}) = 0 = b_{d} (x, y)

. Moreover, for each

z \in X

, we have

\frac{1}{s} b_{d} (x, z) \leq \underset{n \to \infty}{lim inf} b_{d} (x_{n}, z) \leq \underset{n \to \infty}{lim sup} b_{d} (x_{n}, z) \leq s b_{d} (x, z) .

In particular, if

b_{d} (x, z) = 0

, then we have

{lim}_{n \to \infty} b_{d} (x_{n}, z) = 0 = b_{d} (x, z)

Proof Using the triangle inequality in a b-dislocated metric space, it is easy to see that

b_{d} (x, y) \leq s b_{d} (x, x_{n}) + s^{2} b_{d} (x_{n}, y_{n}) + s^{2} b_{d} (y_{n}, y)

and

b_{d} (x_{n}, y_{n}) \leq s b_{d} (x_{n}, x) + s^{2} b_{d} (x, y) + s^{2} b_{d} (y, y_{n}) .

Taking the lower limit as

n \to \infty

in the first inequality and the upper limit as

n \to \infty

in the second inequality, the result follows. Similarly, using again the triangle inequality, the last assertion follows. □

Definition 1.33 [3]

Let f and g be two self-maps on a nonempty set X. If

w = f x = g x

, for some x in X, then x is called a coincidence point of f and g, where w is called a point of coincidence of f and g.

Definition 1.34 [3]

Let f and g be two self-maps defined on a set X. Then f and g are said to be weakly compatible if they commute at every coincidence point.

Definition 1.35 Let

(X, b_{d})

be a b-dislocated metric space. Then the pair

(f, g)

is said to be compatible if and only if

{lim}_{n \to \infty} b_{d} (f g x_{n}, g f x_{n}) = 0

, whenever

{x_{n}}

is a sequence in X so that

{lim}_{n \to \infty} f x_{n} = {lim}_{n \to \infty} g x_{n} = t

for some

t \in X

2 Common fixed point results

Suppose that

\begin{aligned} Ψ = & {ψ : [0, \infty) \to [0, \infty) | ψ is a continuous non-decreasing function with \\ ψ (t) = 0 \Leftrightarrow t = 0} \end{aligned}

and

\begin{aligned} Φ = & {φ : [0, \infty) \to [0, \infty) | φ is a lower semi-continuous function with \\ φ (t) = 0 \Leftrightarrow t = 0} . \end{aligned}

Theorem 2.1 Let

(X, b_{d}, ⪯)

be an ordered complete b-dislocated metric space, and let f, g, S and T be four self-maps on X such that

(f, g)

and

(S, T)

are dominated and dominating maps, respectively, with

f X \subseteq T X

and

g X \subseteq S X

. Suppose that for all two comparable elements

x, y \in X

ψ (2 s^{4} b_{d} (f x, g y)) \leq ψ (M_{s} (x, y)) - φ (M_{s} (x, y))

(2.1)

is satisfied, where

M_{s} (x, y) = max {b_{d} (S x, T y), b_{d} (f x, S x), b_{d} (g y, T y), \frac{b_{d} (S x, g y) + b_{d} (f x, T y)}{4 s}},

(2.2)

ψ \in Ψ

and

φ \in Φ

. If for every non-increasing sequence

{x_{n}}

and a sequence

{y_{n}}

with

y_{n} ⪯ x_{n}

, for all n such that

y_{n} \to u

, we have

u ⪯ x_{n}

and either

(a₁)

(f, S)

are compatible, f or S is continuous and

(g, T)

is weakly compatible, or

(a₂)

(g, T)

are compatible, g or T is continuous and

(f, S)

is weakly compatible,

then f, g, S and T have a common fixed point. Moreover, the set of common fixed points of f, g, S and T is well ordered if and only if f, g, S and T have one and only one common fixed point.

Proof Let

x_{0}

be an arbitrary point in X. We define inductively the sequences

{x_{n}}

and

{y_{n}}

in X by

y_{2 n + 1} = f x_{2 n} = T x_{2 n + 1}, y_{2 n + 2} = g x_{2 n + 1} = S x_{2 n + 2}, n = 0, 1, 2, \dots .

This can be done as

f X \subseteq T X

and

g X \subseteq S X

. By given assumptions,

x_{2 n + 1} ⪯ T x_{2 n + 1} = f x_{2 n} ⪯ x_{2 n}

and

x_{2 n} ⪯ S x_{2 n} = g x_{2 n - 1} ⪯ x_{2 n - 1}

. Thus, we have

x_{n + 1} ⪯ x_{n}

for all

n \geq 0

. We will show that

{y_{n}}

b_{d}

-Cauchy. Suppose that

b_{d} (y_{2 n}, y_{2 n + 1}) > 0

for every n. If not, then for some k,

b_{d} (y_{2 k}, y_{2 k + 1}) = 0

, and from (2.1), we obtain

\begin{array}{rcl} ψ (b_{d} (y_{2 k + 1}, y_{2 k + 2})) & \leq & ψ (2 s^{4} b_{d} (y_{2 k + 1}, y_{2 k + 2})) \\ = & ψ (2 s^{4} b_{d} (f x_{2 k}, g x_{2 k + 1})) \\ \leq & ψ (M_{s} (x_{2 k}, x_{2 k + 1})) - φ (M_{s} (x_{2 k}, x_{2 k + 1})), \end{array}

(2.3)

where

\begin{aligned} M_{s} (x_{2 k}, x_{2 k + 1}) = & max {b_{d} (S x_{2 k}, T x_{2 k + 1}), b_{d} (f x_{2 k}, S x_{2 k}), b_{d} (g x_{2 k + 1}, T x_{2 k + 1}), \\ \frac{b_{d} (S x_{2 k}, g x_{2 k + 1}) + b_{d} (f x_{2 k}, T x_{2 k + 1})}{4 s}} \\ = & max {b_{d} (y_{2 k}, y_{2 k + 1}), b_{d} (y_{2 k + 1}, y_{2 k}), b_{d} (y_{2 k + 2}, y_{2 k + 1}), \\ \frac{b_{d} (y_{2 k}, y_{2 k + 2}) + b_{d} (y_{2 k + 1}, y_{2 k + 1})}{4 s}} \\ = & max {0, 0, b_{d} (y_{2 k + 1}, y_{2 k + 2}), \frac{b_{d} (y_{2 k}, y_{2 k + 2}) + b_{d} (y_{2 k + 1}, y_{2 k + 1})}{4 s}} \\ = & b_{d} (y_{2 k + 1}, y_{2 k + 2}), \end{aligned}

(2.4)

since

\begin{aligned} \frac{b_{d} (y_{2 k}, y_{2 k + 2}) + b_{d} (y_{2 k + 1}, y_{2 k + 1})}{4 s} & \leq \frac{s b_{d} (y_{2 k}, y_{2 k + 1}) + s b_{d} (y_{2 k + 1}, y_{2 k + 2}) + 2 s b_{d} (y_{2 k}, y_{2 k + 1})}{4 s} \\ = \frac{s b_{d} (y_{2 k + 1}, y_{2 k + 2})}{4 s} < b_{d} (y_{2 k + 1}, y_{2 k + 2}) . \end{aligned}

So, from (2.3) and (2.4), we obtain that

ψ (b_{d} (y_{2 k + 1}, y_{2 k + 2})) \leq ψ (b_{d} (y_{2 k + 1}, y_{2 k + 2})) - φ (b_{d} (y_{2 k + 1}, y_{2 k + 2})),

which gives

φ (b_{d} (y_{2 k + 1}, y_{2 k + 2})) \leq 0

and so

y_{2 k + 1} = y_{2 k + 2}

, which further implies that

y_{2 k + 2} = y_{2 k + 3}

. Thus,

{y_{n}}

becomes a constant sequence, hence,

y_{n}

is a Cauchy sequence.

Now, take

b_{d} (y_{2 n}, y_{2 n + 1}) > 0

for each n. As

x_{2 n}

and

x_{2 n + 1}

are comparable, so from (2.1) we have

\begin{aligned} ψ (b_{d} (y_{2 n + 1}, y_{2 n + 2})) & \leq ψ (2 s^{4} b_{d} (y_{2 n + 1}, y_{2 n + 2})) \\ = ψ (2 s^{4} b_{d} (f x_{2 n}, g x_{2 n + 1})) \\ \leq ψ (M_{s} (x_{2 n}, x_{2 n + 1})) - φ (M_{s} (x_{2 n}, x_{2 n + 1})) \\ \leq ψ (M_{s} (x_{2 n}, x_{2 n + 1})) . \end{aligned}

(2.5)

Hence

b_{d} (y_{2 n + 1}, y_{2 n + 2}) \leq M_{s} (x_{2 n}, x_{2 n + 1}),

(2.6)

where

\begin{aligned} M_{s} (x_{2 n}, x_{2 n + 1}) = & max {b_{d} (S x_{2 n}, T x_{2 n + 1}), b_{d} (f x_{2 n}, S x_{2 n}), b_{d} (g x_{2 n + 1}, T x_{2 n + 1}), \\ \frac{b_{d} (S x_{2 n}, g x_{2 n + 1}) + b_{d} (f x_{2 n}, T x_{2 n + 1})}{4 s}} \\ = & max {b_{d} (y_{2 n}, y_{2 n + 1}), b_{d} (y_{2 n + 1}, y_{2 n}), b_{d} (y_{2 n + 2}, y_{2 n + 1}), \\ \frac{b_{d} (y_{2 n}, y_{2 n + 2}) + b_{d} (y_{2 n + 1}, y_{2 n + 1})}{4 s}} \\ \leq & max {b_{d} (y_{2 n}, y_{2 n + 1}), b_{d} (y_{2 n + 1}, y_{2 n + 2}), \\ \frac{s b_{d} (y_{2 n}, y_{2 n + 1}) + s b_{d} (y_{2 n + 1}, y_{2 n + 2}) + 2 s b_{d} (y_{2 n}, y_{2 n + 1})}{4 s}} \\ = & max {b_{d} (y_{2 n}, y_{2 n + 1}), b_{d} (y_{2 n + 1}, y_{2 n + 2}), \\ \frac{3 s b_{d} (y_{2 n}, y_{2 n + 1}) + s b_{d} (y_{2 n + 1}, y_{2 n + 2})}{4 s}} \\ = & max {b_{d} (y_{2 n}, y_{2 n + 1}), b_{d} (y_{2 n + 1}, y_{2 n + 2})} . \end{aligned}

If for some n,

b_{d} (y_{2 n + 1}, y_{2 n + 2}) \geq b_{d} (y_{2 n}, y_{2 n + 1}) > 0

, then (2.6) gives that

M_{s} (x_{2 n}, x_{2 n + 1}) = b_{d} (y_{2 n + 1}, y_{2 n + 2})

and from (2.1) we have

\begin{aligned} ψ (b_{d} (y_{2 n + 1}, y_{2 n + 2})) & \leq ψ (2 s^{4} b_{d} (y_{2 n + 1}, y_{2 n + 2})) \\ \leq ψ (M_{s} (x_{2 n}, x_{2 n + 1})) - φ (M_{s} (x_{2 n}, x_{2 n + 1})) \\ = ψ (b_{d} (y_{2 n + 1}, y_{2 n + 2})) - φ (b_{d} (y_{2 n + 1}, y_{2 n + 2})), \end{aligned}

which yields that

φ (b_{d} (y_{2 n + 1}, y_{2 n + 2})) \leq 0

, or, equivalently,

b_{d} (y_{2 n + 1}, y_{2 n + 2}) = 0

, a contradiction.

Hence,

M_{s} (x_{2 n}, x_{2 n + 1}) \leq b_{d} (y_{2 n}, y_{2 n + 1})

. Since

M_{s} (x_{2 n}, x_{2 n + 1}) \geq b_{d} (y_{2 n}, y_{2 n + 1})

, therefore,

b_{d} (y_{2 n + 1}, y_{2 n + 2}) \leq M_{s} (x_{2 n}, x_{2 n + 1}) = b_{d} (y_{2 n}, y_{2 n + 1})

. Following similar arguments to those given above, we have

b_{d} (y_{2 n + 2}, y_{2 n + 3}) \leq M_{s} (x_{2 n + 1}, x_{2 n + 2}) = b_{d} (y_{2 n + 1}, y_{2 n + 2}) .

(2.7)

Therefore,

{b_{d} (y_{n}, y_{n + 1})}

is a non-increasing sequence and so there exists

r \geq 0

such that

lim_{n \to \infty} b_{d} (y_{n - 1}, y_{n}) = lim_{n \to \infty} M_{s} (x_{n}, x_{n + 1}) = r .

Suppose that

r > 0

. As

\begin{array}{rcl} ψ (b_{d} (y_{2 n + 1}, y_{2 n + 2})) & \leq & ψ (2 s^{4} b_{d} (y_{2 n + 1}, y_{2 n + 2})) \\ \leq & ψ (M_{s} (x_{2 n}, x_{2 n + 1})) - φ (M_{s} (x_{2 n}, x_{2 n + 1})), \end{array}

by taking the upper limit as

n \to \infty

, we obtain

\begin{array}{rcl} ψ (r) & \leq & ψ (r) - \underset{n \to \infty}{lim inf} φ (M_{s} (x_{2 n}, x_{2 n + 1})) \\ = & ψ (r) - φ (\underset{n \to \infty}{lim inf} M_{s} (x_{2 n}, x_{2 n + 1})) \\ = & ψ (r) - φ (r), \end{array}

a contradiction. Hence

lim_{n \to \infty} b_{d} (y_{n - 1}, y_{n}) = 0 .

(2.8)

Now, we prove that

{y_{n}}

is a

b_{d}

-Cauchy sequence. To do this, it is sufficient to show that the subsequence

{y_{2 n}}

b_{d}

-Cauchy in X. Assume on the contrary that

{y_{2 n}}

is not a

b_{d}

-Cauchy sequence. Then there exists

ε > 0

for which we can find subsequences

{y_{2 m_{k}}}

and

{y_{2 n_{k}}}

{y_{2 n}}

so that

n_{k}

is the smallest index for which

2 n_{k} > 2 m_{k} > k

b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) \geq ε

(2.9)

and

b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 2}) < ε .

(2.10)

Using the triangle inequality and (2.10), we obtain that

ε \leq b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) \leq s b_{d} (y_{2 m_{k}}, y_{2 m_{k} + 1}) + s b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k}}) .

Taking the upper limit as

k \to \infty

and using (2.8), we obtain

\frac{ε}{s} \leq \underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k}}) .

(2.11)

Using the triangle inequality and (2.10), we have

\begin{aligned} ε & \leq b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) \\ \leq s b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 2}) + s^{2} b_{d} (y_{2 n_{k} - 2}, y_{2 n_{k} - 1}) + s^{2} b_{d} (y_{2 n_{k} - 1}, y_{2 n_{k}}) \\ < ε s + s^{2} b_{d} (y_{2 n_{k} - 2}, y_{2 n_{k} - 1}) + s^{2} b_{d} (y_{2 n_{k} - 1}, y_{2 n_{k}}) . \end{aligned}

Taking the upper limit as

k \to \infty

and using (2.8), we obtain

ε \leq \underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) \leq ε s .

(2.12)

Also,

ε \leq b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) \leq s b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 1}) + s b_{d} (y_{2 n_{k} - 1}, y_{2 n_{k}}) .

Hence

\frac{ε}{s} \leq \underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 1}) .

On the other hand, we have

b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 1}) \leq s b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) + s b_{d} (y_{2 n_{k}}, y_{2 n_{k} - 1}) .

So, from (2.8) and (2.12), we have

\underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 1}) \leq s \underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) \leq ε s^{2} .

Consequently,

\frac{ε}{s} \leq \underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 1}) \leq ε s^{2} .

(2.13)

Similarly,

\frac{ε}{s^{2}} \leq \underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k} - 1}) \leq ε s^{2} .

(2.14)

x_{2 m_{k}}

and

x_{2 n_{k} - 1}

are comparable, from (2.1) we have

\begin{aligned} ψ (2 s^{4} b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k}})) & = ψ (2 s^{4} b_{d} (f x_{2 m_{k}}, g x_{2 n_{k} - 1})) \\ \leq ψ (M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})) - φ (M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})), \end{aligned}

where

\begin{aligned} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1}) = & max {b_{d} (S x_{2 m_{k}}, T x_{2 n_{k} - 1}), b_{d} (f x_{2 m_{k}}, S x_{2 m_{k}}), b_{d} (g x_{2 n_{k} - 1}, T x_{2 n_{k} - 1}), \\ \frac{b_{d} (S x_{2 m_{k}}, g x_{2 n_{k} - 1}) + b_{d} (f x_{2 m_{k}}, T x_{2 n_{k} - 1})}{4 s}} \\ = & max {b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 1}), b_{d} (y_{2 m_{k} + 1}, y_{2 m_{k}}), b_{d} (y_{2 n_{k}}, y_{2 n_{k} - 1}), \\ \frac{b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) + b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k} - 1})}{4 s}} . \end{aligned}

Taking the upper limit and using (2.8) and (2.12)-(2.13), we get

\begin{aligned} \frac{ε + \frac{ε}{s^{2}}}{4 s} = & min {\frac{ε}{s}, \frac{ε + \frac{ε}{s^{2}}}{4 s}} \\ \leq & \underset{k \to \infty}{lim sup} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1}) \\ = & max {\underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k}}, y_{2 n_{k} - 1}), 0, 0, \\ \frac{{lim sup}_{k \to \infty} b_{d} (y_{2 m_{k}}, y_{2 n_{k}}) + {lim sup}_{k \to \infty} b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k} - 1})}{4 s}} \\ \leq & max {ε s^{2}, \frac{ε s + ε s^{2}}{4 s}} = ε s^{2} . \end{aligned}

Hence, we have

\frac{ε + \frac{ε}{s^{2}}}{4 s} \leq \underset{k \to \infty}{lim sup} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1}) \leq ε s^{2} .

(2.15)

Similarly, we can obtain

\frac{ε + \frac{ε}{s^{2}}}{4 s} \leq \underset{k \to \infty}{lim inf} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1}) \leq ε s^{2} .

(2.16)

\begin{aligned} ψ (2 s^{4} b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k}})) & = ψ (2 s^{4} b_{d} (f x_{2 m_{k}}, g x_{2 n_{k} - 1})) \\ \leq ψ (M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})) - φ (M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})), \end{aligned}

so, by taking the upper limit as

k \to \infty

, and from (2.11) and (2.15), we obtain

\begin{aligned} ψ (2 ε s^{3}) & = ψ (2 s^{4} \frac{ε}{s}) \\ \leq ψ (2 s^{4} \underset{k \to \infty}{lim sup} b_{d} (y_{2 m_{k} + 1}, y_{2 n_{k}})) \\ \leq ψ (\underset{k \to \infty}{lim sup} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})) - \underset{k \to \infty}{lim inf} φ (M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})) \\ \leq ψ (ε s^{2}) - φ (\underset{k \to \infty}{lim inf} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})) \\ \leq ψ (2 ε s^{3}) - φ (\underset{k \to \infty}{lim inf} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})), \end{aligned}

which implies that

φ (\underset{k \to \infty}{lim inf} M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1})) = 0,

lim inf M_{s} (x_{2 m_{k}}, x_{2 n_{k} - 1}) = 0

, a contradiction to (2.16). Hence

{y_{2 n}}

is a

b_{d}

-Cauchy sequence in X. Since X is complete, there exists

y \in X

such that

lim_{n \to \infty} f x_{2 n} = lim_{n \to \infty} T x_{2 n + 1} = lim_{n \to \infty} g x_{2 n + 1} = lim_{n \to \infty} S x_{2 n} = y .

Now, we show that y is a common fixed point of f, g, S and T.

Assume that (a₁) holds and S is continuous. Then

lim_{n \to \infty} S^{2} x_{2 n + 2} = S y and lim_{n \to \infty} S f x_{2 n} = S y .

Using the triangle inequality, we have

b_{d} (f S x_{2 n}, S y) \leq s (b_{d} (f S x_{2 n}, S f x_{2 n}) + b_{d} (S f x_{2 n}, S y)) .

Since the pair

(f, S)

is compatible,

{lim}_{n \to \infty} b_{d} (f S x_{2 n}, S f x_{2 n}) = 0

. So, by taking the limit when

n \to \infty

in the above inequality, we have

lim_{n \to \infty} b_{d} (f S x_{2 n}, S y) \leq s (lim_{n \to \infty} b_{d} (f S x_{2 n}, S f x_{2 n}) + lim_{n \to \infty} b_{d} (S f x_{2 n}, S y)) = 0 .

Hence,

{lim}_{n \to \infty} f S x_{2 n} = S y

. As

S x_{2 n + 2} = g x_{2 n + 1} ⪯ x_{2 n + 1}

, from (2.1) we obtain

ψ (2 s^{4} b_{d} (f S x_{2 n + 2}, g x_{2 n + 1})) \leq ψ (M_{s} (S x_{2 n + 2}, x_{2 n + 1})) - φ (M_{s} (S x_{2 n + 2}, x_{2 n + 1})),

(2.17)

where

\begin{aligned} M_{s} (S x_{2 n + 2}, x_{2 n + 1}) \\ = max {b_{d} (S^{2} x_{2 n + 2}, T x_{2 n + 1}), b_{d} (f S x_{2 n + 2}, S^{2} x_{2 n + 2}), b_{d} (g x_{2 n + 1}, T x_{2 n + 1}), \\ \frac{b_{d} (S^{2} x_{2 n + 2}, g x_{2 n + 1}) + b_{d} (f S x_{2 n + 2}, T x_{2 n + 1})}{4 s}} . \end{aligned}

Now, by using Lemma 1.32, we get

\begin{aligned} \underset{n \to \infty}{lim sup} M_{s} (S x_{2 n + 2}, x_{2 n + 1}) & \leq max {s^{2} b_{d} (S y, y), 0, 0, \frac{s^{2} b_{d} (S y, y) + s^{2} b_{d} (S y, y)}{4 s}} \\ = s^{2} b_{d} (S y, y) . \end{aligned}

Hence, by taking the upper limit in (2.17) and using Lemma 1.32, we obtain

\begin{array}{rcl} ψ (2 s^{2} b_{d} (S y, y)) & = & ψ (2 s^{4} \frac{1}{s^{2}} b_{d} (S y, y)) \leq ψ (s^{2} b_{d} (S y, y)) - φ (s^{2} b_{d} (S y, y)) \\ \leq & ψ (2 s^{2} b_{d} (S y, y)) - φ (s^{2} b_{d} (S y, y)) \end{array}

which gives

φ (s^{2} b_{d} (S y, y)) \leq 0

, or, equivalently,

S y = y

Now, since

g x_{2 n + 1} ⪯ x_{2 n + 1}

and

g x_{2 n + 1} \to y

n \to \infty

, then

y ⪯ x_{2 n + 1}

and from (2.1) we have

ψ (2 s^{4} b_{d} (f y, g x_{2 n + 1})) \leq ψ (M_{s} (y, x_{2 n + 1})) - φ (M_{s} (y, x_{2 n + 1})),

(2.18)

where

\begin{aligned} M_{s} (y, x_{2 n + 1}) = & max {b_{d} (S y, T x_{2 n + 1}), b_{d} (f y, S y), b_{d} (g x_{2 n + 1}, T x_{2 n + 1}), \\ \frac{b_{d} (S y, g x_{2 n + 1}) + b_{d} (f y, T x_{2 n + 1})}{4 s}} . \end{aligned}

Taking the upper limit as

n \to \infty

in (2.18) and using Lemma 1.32, we have

\begin{aligned} ψ (2 s^{3} b_{d} (f y, y)) & = ψ (2 s^{4} \frac{1}{s} b_{d} (f y, y)) \leq ψ (b_{d} (f y, y)) - φ (b_{d} (f y, y)) \\ \leq ψ (2 s^{3} b_{d} (f y, y)) - φ (b_{d} (f y, y)), \end{aligned}

which implies that

φ (b_{d} (f y, y)) \leq 0

, so

f y = y

Since

f (X) \subseteq T (X)

, there exists a point

v \in X

such that

f y = T v

. Suppose that

g v \neq T v

. Since

v ⪯ T v = f y ⪯ y

, from (2.1) we have

ψ (b_{d} (T v, g v)) = ψ (b_{d} (f y, g v)) \leq ψ (M_{s} (y, v)) - φ (M_{s} (y, v)),

(2.19)

where

\begin{aligned} M_{s} (y, v) & = max {b_{d} (S y, T v), b_{d} (f y, S y), b_{d} (g v, T v), \frac{b_{d} (S y, g v) + b_{d} (f y, T v)}{4 s}} \\ = b_{d} (g v, T v) . \end{aligned}

So, from (2.19) we have

ψ (b_{d} (T v, g v)) \leq ψ (b_{d} (g v, T v)) - φ (b_{d} (g v, T v)),

a contradiction. Therefore

g v = T v

. Since the pair

(g, T)

is weakly compatible,

g y = g f y = g T v = T g v = T f y = T y

and y is the coincidence point of g and T. Since

S x_{2 n} ⪯ x_{2 n}

and

S x_{2 n} \to y

n \to \infty

, it implies that

y ⪯ x_{2 n}

and from (2.1) we obtain

ψ (2 s^{4} b_{d} (f x_{2 n}, g y)) \leq ψ (M_{s} (x_{2 n}, y)) - φ (M_{s} (x_{2 n}, y)),

(2.20)

where

\begin{aligned} M_{s} (x_{2 n}, y) = & max {b_{d} (S x_{2 n}, T y), b_{d} (f x_{2 n}, S x_{2 n}), b_{d} (g y, T y), \\ \frac{b_{d} (S x_{2 n}, g y) + b_{d} (f x_{2 n}, T y)}{4 s}} . \end{aligned}

(2.21)

Taking the upper limit as

n \to \infty

in (2.21) and using Lemma 1.32, we have

\begin{aligned} max {\frac{1}{s} b_{d} (y, g y), b_{d} (g y, T y), \frac{2 s}{4 s} b_{d} (y, g y)} \\ \leq \underset{n \to \infty}{lim inf} M_{s} (x_{2 n}, y) \\ \leq \underset{n \to \infty}{lim sup} M_{s} (x_{2 n}, y) \\ \leq max {s b_{d} (y, g y), b_{d} (g y, T y), \frac{2 s}{4 s} b_{d} (y, g y)} \\ = max {s b_{d} (y, g y), b_{d} (g y, g y)} \\ \leq max {s b_{d} (y, g y), 2 s b_{d} (y, g y)} \\ = 2 s b_{d} (y, g y) . \end{aligned}

(2.22)

Taking the upper limit as

n \to \infty

in (2.20) and using Lemma 1.32 and (2.22), we have

\begin{aligned} ψ (2 s^{3} b_{d} (y, g y)) & = ψ (2 s^{4} \frac{1}{s} b_{d} (y, g y)) \\ \leq ψ (\underset{n \to \infty}{lim sup} M_{s} (x_{2 n}, y)) - \underset{n \to \infty}{lim inf} φ (M_{s} (x_{2 n}, y)) \\ \leq ψ (2 s b_{d} (y, g y)) - φ (\underset{n \to \infty}{lim inf} M_{s} (x_{2 n}, y)) \\ \leq ψ (2 s^{3} b_{d} (y, g y)) - φ (\underset{n \to \infty}{lim inf} M_{s} (x_{2 n}, y)), \end{aligned}

which implies that

{lim inf}_{n \to \infty} M_{s} (x_{2 n}, y) = 0

, so we have

y = g y

. Therefore,

f y = g y = S y = T y = y

The proof is similar when f is continuous.

Similarly, if (a₂) holds, then the result follows.

Now, suppose that the set of common fixed points of f, g, S and T is well ordered. We show that they have a unique common fixed point. Assume on the contrary that

f u = g u = S u = T u = u

and

f v = g v = S v = T v = v

, but

u \neq v

. By assumption, we can apply (2.1) to obtain

\begin{aligned} ψ (2 s b_{d} (u, v)) & = ψ (2 s b_{d} (f u, g v)) \\ \leq ψ (2 s^{4} b_{d} (f u, g v)) \leq ψ (M_{s} (u, v)) - φ (M_{s} (u, v)), \end{aligned}

where

\begin{aligned} M_{s} (u, v) & = max {b_{d} (S u, T v), b_{d} (f u, S u), b_{d} (g v, T v), \frac{b_{d} (S u, g v) + b_{d} (f u, T v)}{4 s}} \\ = max {b_{d} (u, v), b_{d} (u, u), b_{d} (v, v), \frac{b_{d} (u, v) + b_{d} (u, v)}{4 s}} \\ = max {b_{d} (u, v), b_{d} (u, u), b_{d} (v, v)} \\ \leq max {b_{d} (u, v), 2 s b_{d} (u, v), 2 s b_{d} (u, v)} \\ \leq 2 s b_{d} (u, v) . \end{aligned}

Hence

ψ (2 s b_{d} (u, v)) \leq ψ (2 s b_{d} (u, v)) - φ (M_{s} (u, v)) .

So, we have

M_{s} (u, v) = 0

, a contradiction. Therefore

u = v

. The converse is obvious. □

In the following theorem, we omit the continuity assumption of f, g, T and S and replace the compatibility of the pairs

(f, S)

and

(g, T)

by weak compatibility of the pairs, and we show that f, g, S and T have a common fixed point on X.

Theorem 2.2 Let

(X, b_{d}, ⪯)

be an ordered complete b-dislocated metric space, and f, g, S and T be four self-maps on X such that

(f, g)

and

(S, T)

are dominated and dominating maps, respectively, with

f X \subseteq T X

and

g X \subseteq S X

, and TX and SX are

b_{d}

-closed subsets of X. Suppose that for all two comparable elements

x, y \in X

ψ (2 s^{4} d (f x, g y)) \leq ψ (M_{s} (x, y)) - φ (M_{s} (x, y))

(2.23)

is satisfied, where

M_{s} (x, y) = max {b_{d} (S x, T y), b_{d} (f x, S x), b_{d} (g y, T y), \frac{b_{d} (S x, g y) + b_{d} (f x, T y)}{4 s}},

ψ \in Ψ

and

φ \in Φ

. If for every non-increasing sequence

{x_{n}}

and a sequence

{y_{n}}

with

y_{n} ⪯ x_{n}

, for all n such that

y_{n} \to u

, we have

u ⪯ x_{n}

, and the pairs

(f, S)

and

(g, T)

are weakly compatible, then f, g, S and T have a common fixed point. Moreover, the set of common fixed points of f, g, S and T is well ordered if and only if f, g, S and T have one and only one common fixed point.

Proof Following the proof of Theorem 2.1, there exists

y \in X

such that

lim_{k \to \infty} b_{d} (y_{k}, y) = 0 .

(2.24)

Since

T (X)

b_{d}

-closed and

{y_{2 n + 1}} \subseteq T (X)

, therefore

y \in T (X)

. Hence, there exists

u \in X

such that

y = T u

and

lim_{n \to \infty} b_{d} (y_{2 n + 1}, T u) = lim_{n \to \infty} b_{d} (T x_{2 n + 1}, T u) = 0 .

(2.25)

Similarly, there exists

v \in X

such that

y = T u = S v

and

lim_{n \to \infty} b_{d} (y_{2 n}, S v) = lim_{n \to \infty} b_{d} (S x_{2 n}, S v) = 0 .

(2.26)

Now we prove that v is a coincidence point of f and S.

Since

T x_{2 n + 1} \to y = S v

n \to \infty

, so, by assumption,

T x_{2 n + 1} ⪯ S v

. Therefore, from (2.23) we have

ψ (2 s^{4} b_{d} (f v, g x_{2 n + 1})) \leq ψ (M_{s} (v, x_{2 n + 1})) - φ (M_{s} (v, x_{2 n + 1})),

(2.27)

where

\begin{aligned} M_{s} (v, x_{2 n + 1}) = & max {b_{d} (S v, T x_{2 n + 1}), b_{d} (f v, S v), b_{d} (g x_{2 n + 1}, T x_{2 n + 1}), \\ \frac{b_{d} (S v, g x_{2 n + 1}) + b_{d} (f v, T x_{2 n + 1})}{4 s}} \\ = & max {b_{d} (T u, T x_{2 n + 1}), b_{d} (f v, y), b_{d} (g x_{2 n + 1}, T x_{2 n + 1}), \\ \frac{b_{d} (S v, y_{2 n + 2}) + b_{d} (f v, T x_{2 n + 1})}{4 s}} . \end{aligned}

Taking the upper limit as

n \to \infty

and using (2.25)-(2.26) and Lemma 1.32, we obtain that

\begin{matrix} max {b_{d} (f v, y), \frac{1}{s^{2}} b_{d} (y, y), \frac{1}{4 s^{2}} b_{d} (y, y)} \\ \leq \underset{n \to \infty}{lim inf} M_{s} (v, x_{2 n + 1}) \\ \leq \underset{n \to \infty}{lim sup} M_{s} (v, x_{2 n + 1}) \\ \leq max {0, b_{d} (f v, y), s^{2} b_{d} (y, y), \frac{0 + s^{2} b_{d} (f v, y)}{4 s}} \\ \leq max {b_{d} (f v, y), 2 s^{2} b_{d} (f v, y)} \\ = 2 s^{2} b_{d} (f v, y) . \end{matrix}

(2.28)

Taking the upper limit as

n \to \infty

in (2.27) and using (2.28) and Lemma 1.32, we obtain that

\begin{array}{rcl} ψ (2 s^{3} d (f v, y)) & = & ψ (2 s^{4} \frac{1}{s} d (f v, y)) \\ \leq & ψ (2 s^{2} b_{d} (f v, y)) - φ (\underset{n \to \infty}{lim inf} M_{s} (v, x_{2 n + 1})) \\ \leq & ψ (2 s^{3} b_{d} (f v, y)) - φ (\underset{n \to \infty}{lim inf} M_{s} (v, x_{2 n + 1})), \end{array}

which implies that

{lim inf}_{n \to \infty} M_{s} (v, x_{2 n + 1}) = 0

, so from (2.28) we obtain

f v = y = S v

As f and S are weakly compatible, we have

f y = f S v = S f v = S y

. Thus, y is a coincidence point of f and S.

Similarly, it can be shown that y is a coincidence point of the pair

(g, T)

. Now, we show that

f y = g y

. From (2.23) we have

ψ (2 s^{4} d (f y, g y)) \leq ψ (M_{s} (y, y)) - φ (M_{s} (y, y)),

where

\begin{array}{rcl} M_{s} (y, y) & = & max {b_{d} (S y, T y), b_{d} (f y, S y), b_{d} (g y, T y), \frac{b_{d} (S y, g y) + b_{d} (f y, T y)}{4 s}} \\ = & max {b_{d} (f y, g y), b_{d} (f y, f y), b_{d} (g y, g y), \frac{b_{d} (f y, g y) + b_{d} (f y, g y)}{4 s}} \\ = & max {b_{d} (f y, g y), b_{d} (f y, f y), b_{d} (g y, g y)} \\ \leq & max {b_{d} (f y, g y), 2 s b_{d} (f y, g y), 2 s b_{d} (f y, g y)} \\ = & 2 s b_{d} (f y, g y) . \end{array}

So, we have

\begin{array}{rcl} ψ (2 s^{4} d (f y, g y)) & \leq & ψ (M_{s} (y, y)) - φ (M_{s} (y, y)) \\ \leq & ψ (2 s b_{d} (f y, g y)) - φ (M_{s} (y, y)) \\ \leq & ψ (2 s^{4} b_{d} (f y, g y)) - φ (M_{s} (y, y)), \end{array}

which implies that

M_{s} (y, y) = 0

, so we have

f y = g y

. Therefore,

f y = g y = S y = T y

Now, similar to the proof of Theorem 2.1, indeed from (2.20)-(2.22), we have

g y = y

. Therefore,

f y = g y = S y = T y = y

, as required. The last conclusion follows similarly as in the proof of Theorem 2.1. □

Now, we give an example to support our result.

Example 2.3 Let

X = [0, \infty)

be equipped with the b-dislocated metric

b_{d} (x, y) = {(x + y)}^{2}

where

s = 2

and suppose that ‘⪯’ is the usual ordering ≤ on X. Obviously,

(X, b_{d}, \leq)

is an ordered complete b-dislocated metric space. Let

f, g, S, T : X \to X

be defined as

\begin{matrix} f (x) = ln (1 + \frac{x}{4}), g (x) = ln (1 + \frac{x}{5}), \\ S (x) = e^{5 x} - 1, T (x) = e^{4 x} - 1 . \end{matrix}

For each

x \in X

, we have

1 + \frac{x}{4} \leq e^{x}

and

1 + \frac{x}{5} \leq e^{x}

, so

f (x) = ln (1 + \frac{x}{4}) \leq x

g (x) = ln (1 + \frac{x}{5}) \leq x

x \leq e^{5 x} - 1 = S (x)

and

x \leq e^{4 x} - 1 = T (x)

. Thus, f and g are dominated and T and S are dominating with

f (X) = g (X) = S (X) = T (X) = [0, \infty)

. Also, the pair

(g, T)

is compatible, g is continuous and

(f, S)

is weakly compatible. Let the control functions

ψ, φ : [0, \infty) \to [0, \infty)

be defined as

ψ (t) = b t

and

φ (t) = (b - 1) t

, for all

t \in [0, \infty)

, where

1 < b \leq \frac{400}{32}

. Note that

\begin{array}{rcl} ψ (2 s^{4} b_{d} (f (x), g (y))) & = & 32 b {(f (x) + g (y))}^{2} \\ = & 32 b {(ln (1 + \frac{x}{4}) + ln (1 + \frac{y}{5}))}^{2} \\ \leq & 32 b {(\frac{x}{4} + \frac{y}{5})}^{2} = \frac{32}{400} b {(5 x + 4 y)}^{2} \\ \leq & {(e^{5 x} - 1 + e^{4 y} - 1)}^{2} \\ = & b_{d} (S (x), T (y)) \\ \leq & M_{2} (x, y) = ψ (M_{2} (x, y)) - φ (M_{2} (x, y)), x, y \in X . \end{array}

Thus, f, g, S and T satisfy all the conditions of Theorem 2.1. Moreover, 0 is a unique common fixed point of f, g, S and T.

Corollary 2.4 Let

(X, b_{d}, ⪯)

be an ordered complete b-dislocated metric space, and let f and g be two dominated self-maps on X. Suppose that for every two comparable elements

x, y \in X

ψ (2 s^{4} b_{d} (f x, g y)) \leq ψ (M_{s} (x, y)) - φ (M_{s} (x, y))

is satisfied, where

M_{s} (x, y) = max {b_{d} (x, y), b_{d} (f x, x), b_{d} (g y, y), \frac{b_{d} (x, g y) + b_{d} (f x, y)}{4 s}},

ψ \in Ψ

and

φ \in Φ

. If for every non-increasing sequence

{x_{n}}

and a sequence

{y_{n}}

with

y_{n} ⪯ x_{n}

, for all n such that

y_{n} \to u

, we have

u ⪯ x_{n}

, then f and g have a common fixed point. Moreover, the set of common fixed points of f and g is well ordered if and only if f and g have one and only one common fixed point.

Proof Taking S and T as identity maps on X, the result follows from Theorem 2.2. □

Corollary 2.5 Let

(X, b_{d}, ⪯)

be an ordered complete b-dislocated metric space. Let f and g be dominated self-maps on X. Suppose that for every two comparable elements

x, y \in X

2 s^{4} b_{d} (f x, g y) \leq M_{s} (x, y) - φ (M_{s} (x, y))

is satisfied, where

M_{s} (x, y) = max {b_{d} (x, y), b_{d} (f x, x), b_{d} (g y, y), \frac{b_{d} (x, g y) + b_{d} (f x, y)}{4 s}},

and

φ \in Φ

. If for every non-increasing sequence

{x_{n}}

and a sequence

{y_{n}}

with

y_{n} ⪯ x_{n}

, for all n such that

y_{n} \to u

, it implies that

u ⪯ x_{n}

, then f and g have a common fixed point. Moreover, the set of common fixed points of f and g is well ordered if and only if f and g have one and only one common fixed point.

Proof If we take S and T as the identity maps on X and

ψ (t) = t

for all

t \in [0, \infty)

, then from Theorem 2.2 it follows that f and g have a common fixed point. □

Remark 2.6 As corollaries we can state partial metric space as well as b-metric space versions of our proved results in a similar way, which extends recent results in these settings.

3 Existence of a common solution for a system of integral equations

Consider the following system of integral equations:

\begin{aligned} x (t) = \int_{a}^{b} K_{1} (t, r, x (r)) d r, \\ x (t) = \int_{a}^{b} K_{2} (t, r, x (r)) d r, \end{aligned}

(3.1)

where

b > a \geq 0

. The purpose of this section is to present an existence theorem for a solution to (3.1) that belongs to

X = C [a, b]

(the set of continuous real functions defined on

[a, b]

) by using the obtained result in Corollary 2.4.

Here,

K_{1}, K_{2} : [a, b] \times [a, b] \times R \to R

. The considered problem can be reformulated in the following manner.

Let

f, g : X \to X

be the mappings defined by

\begin{matrix} f x (t) = \int_{a}^{b} K_{1} (t, r, x (r)) d r, \\ g x (t) = \int_{a}^{b} K_{2} (t, r, x (r)) d r \end{matrix}

for all

x \in X

and for all

t \in [a, b]

Then the existence of a solution to (3.1) is equivalent to the existence of a common fixed point of f and g. According to Example 1.11, X equipped with

b_{d} (u, v) = max_{t \in [a, b]} {(| u (t) | + | v (t) |)}^{p}

for all

u, v \in X

, is a complete b-dislocated metric space with

s = 2^{p - 1}

We endow X with the partial ordering ⪯ given by

x ⪯ y ⟺ x (t) \leq y (t)

for all

t \in [a, b]

. Moreover, in [4], it is proved that

(X, ⪯)

is regular.

Now, we will prove the following result.

Theorem 3.1 Suppose that the following hypotheses hold:

(i)

K_{1}, K_{2} : [a, b] \times [a, b] \times R \to R

are continuous;

(ii)

for all

t, r \in [a, b]

and

x \in X

, we have

x (t) \leq min {\int_{a}^{b} K_{1} (t, r, x (r)) d r, \int_{a}^{b} K_{2} (t, r, x (r)) d r};

(iii)

for all

r, t \in [a, b]

and

x, y \in X

with

x ⪯ y

, we have

(| K_{1} (t, r, x (r)) | + | K_{2} (t, r, y (r)) |) \leq ξ (t, r) ln (1 + {(| x (r) | + | y (r) |)}^{p}),

where ξ is a continuous function satisfying

sup_{t \in [a, b]} (\int_{a}^{b} ξ {(t, r)}^{p} d r) < \frac{1}{2^{4 p^{2} - 3 p} {(b - a)}^{p - 1}} .

Then the integral equations (3.1) have a common solution

x \in X

Proof From condition (ii), f and g are dominated self-maps on X.

Let

1 \leq p, q < \infty

with

\frac{1}{p} + \frac{1}{q} = 1

Now, let

x, y \in X

be such that

x ⪰ y

. From condition (iii), for all

t \in [a, b]

, we have

\begin{aligned} {(2^{4 p - 3} (| f x (t) | + | g y (t) |))}^{p} \\ \leq 2^{4 p^{2} - 3 p} {(\int_{a}^{b} (| K_{1} (t, r, x (r)) | + | K_{2} (t, r, x (r)) |) d r)}^{p} \\ \leq 2^{4 p^{2} - 3 p} [{(\int_{a}^{b} 1^{q} d r)}^{\frac{1}{q}} {{(\int_{a}^{b} {(| K_{1} (t, r, x (r)) | + | K_{2} (t, r, x (r)) |)}^{p} d r)}^{\frac{1}{p}}]}^{p} \\ \leq 2^{4 p^{2} - 3 p} {(b - a)}^{\frac{p}{q}} (\int_{a}^{b} ξ {(t, r)}^{p} {(ln (1 + {(| x (r) | + | y (r) |)}^{p}))}^{p} d r) \\ \leq 2^{4 p^{2} - 3 p} {(b - a)}^{\frac{p}{q}} (\int_{a}^{b} ξ {(t, r)}^{p} {(ln (1 + b_{d} (x, y)))}^{p} d r) \\ \leq 2^{4 p^{2} - 3 p} {(b - a)}^{\frac{p}{q}} (\int_{a}^{b} ξ {(t, r)}^{p} {(ln (1 + M_{s} (x, y)))}^{p} d r) \\ = 2^{4 p^{2} - 3 p} {(b - a)}^{p - 1} (\int_{a}^{b} ξ {(t, r)}^{p} d r) {(ln (1 + M_{s} (x, y)))}^{p} \\ < {(ln (1 + M_{s} (x, y)))}^{p} \\ = M_{s} {(x, y)}^{p} - (M_{s} {(x, y)}^{p} - {(ln (1 + M_{s} (x, y)))}^{p}) . \end{aligned}

Hence,

\begin{aligned} {(2 s^{4} b_{d} (f x, g y))}^{p} & = 2 s^{4} sup_{t \in [a, b]} {(| f x (t) | + | g y (t) |)}^{p} \\ \leq M_{s} {(x, y)}^{p} - (M_{s} {(x, y)}^{p} - {(ln (1 + M_{s} (x, y)))}^{p}) . \end{aligned}

Taking

ψ (t) = t^{p}

and

φ (t) = t^{p} - {(ln (1 + t))}^{p}

in Corollary 2.4, there exists

x \in X

, a common fixed point of f and g, that is, x is a solution for (3.1). □

Acknowledgements

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the first author acknowledges with thanks DSR, KAU for financial support.

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License ( https://creativecommons.org/licenses/by/2.0 ), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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Titel: Common fixed point results for weak contractive mappings in ordered b-dislocated metric spaces with applications
verfasst von: Nawab Hussain
Jamal Rezaei Roshan
Vahid Parvaneh
Mujahid Abbas
Publikationsdatum: 01.12.2013
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2013
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2013-486

Springer Professional

Abstract

Competing interests

Authors’ contributions

1 Introduction and preliminaries

2 Common fixed point results

3 Existence of a common solution for a system of integral equations

Acknowledgements

Competing interests

Authors’ contributions

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