Common Fixed Point Theorems for Nonlinear Contractive Mappings in Dislocated Metric Spaces

Definition 1 (see [11].)Let R₊ = [0, ∞). Let $$, $$, satisfying the following:

• (1)

  if w ≤ G₁(u, v), then there exists c ∈ (0,1), such that w ≤ cmax {u, v};

• (2)

  if w ≤ G₂(u, v, r), then there exists c ∈ (0,1), such that w ≤ cmax {u, v, r}.

Theorem 2 (see [11].)Let (X, d) be complete metric spaces, let f, g : X → X be continuous mappings, and for all x, y ∈ X, such that

or then f,  g have a unique common fixed point.

Definition 3 (see [6].)Let X be a nonempty set and let d : X × X → [0, ∞) be a function called a distance function. If for all x, y, z ∈ X,

• (1)

  nonnegativity: d(x, y) ≥ 0;

• (2)

  faithful: d(x, y) = 0⇔x = y;

• (3)

  the triangle inequality: d(x, y) ≤ d(x, z) + d(z, x),

So, here d is a quasimetric on X, and (X, d) is called a quasimetric space.

Definition 4 (see [6].)Let X be a nonempty set and let d : X × X → [0, ∞) be a function called a distance function. If for all x, y, z ∈ X,

• (1)

  nonnegativity: d(x, y) ≥ 0;

• (2)

  indistancy implies equality: d(x, y) = 0 = d(y, x) implies x = y;

• (3)

  the triangle inequality: d(x, y) ≤ d(x, z) + d(z, x),

so, here d is called a dislocated quasimetric or d_q-metric on X, and (X, d) is called a dislocated quasimetric space.

Definition 5 (see [6].)Let X be a nonempty set and let d : X × X → [0, ∞) be a function called a distance function. If for x, y, z ∈ X,

• (1)

  nonnegativity: d(x, y) ≥ 0;

• (2)

  indistancy implies equality: d(x, y) = 0 = d(y, x) implies x = y;

• (3)

  symmetry: d(x, y) = d(y, x);

• (4)

  the triangle inequality: d(x, y) ≤ d(x, z) + d(z, x),

so, here d is called a dislocated metric or d-metric on X and the pair (X, d) is called a dislocated metric space.

Definition 6 (see [6].)A sequence {x_n} in d_q-metric space (dislocated quasimetric space) (X, d) is called a Cauchy sequence, if for a given ϵ > 0, there exists n₀ ∈ N such that d(x_m, x_n) < ϵ or d(x_n, x_m) < ϵ; that is, min {d(x_m, x_n), d(x_n, x_m)} < ϵ for all m, n ≥ n₀.

Definition 7 (see [6].)A sequence {x_n} in d-metric spaces (X, d) is said to be d-converged to x ∈ X provided that

In this case, x is called a d-limit of {x_n} and we write x_n → x.

Definition 8 (see [6].)A d-metric space (X, d) is called d-complete if every d-Cauchy sequence in X converges with respect to x in X.

Lemma 9. Every converging sequence in a d-metric space is a Cauchy sequence.

Proof. Let {x_n} be a sequence which converges to some x, and let ϵ > 0 be arbitrarily given. Then there exists n₀ ∈ N with d(x_n, x) < ϵ/2 for all n ≥ n₀. For m, n ≥ n₀, then we obtain that d(x_m, x) + d(x, x_n) < ϵ/2 + ϵ/2 = ϵ. Hence {x_n} is a Cauchy sequence.

Lemma 10. Limits in dislocated metric spaces are unique.

Proof. Let x and y be limits of the sequence {x_n}. Then d(x_n, x) → 0 and d(x_n, y) → 0 as n → ∞. By the triangle inequality of Definition 5, we conclude that d(x, y) ≤ d(x, x_n) + d(x_n, y) → 0 as n → ∞. Hence d(x, y) = 0 and using the properties (2) of Definition 5, we conclude that x = y.

Lemma 11. Limits in dislocated quasimetric spaces are unique.

Proof. Let x and y be limits of the sequence {x_n}. Then x_n → x and x_n → y as n → ∞. By the triangle inequality, it follows that d(x, y) ≤ d(x, x_n) + d(x_n, y). As n → ∞, we have d(x, y) ≤ d(x, x) + d(y, y). Similarly, d(y, x) ≤ d(x, x) + d(y, y). Hence |d(x, y) − d(y, x)| ≤ 0; that is, d(x, y) = d(y, x). Also d(x, y) ≤ d(x, x_n) + d(x_n, y) → 0 as n → ∞. That is, d(x, y) = 0, and d(x, y) = d(y, x) = 0. By the property (2) of Definition 4, we conclude that x = y.

Example 12. Let X = R⁺. Define d : X × X → R⁺ by d(x, y) = |x − y|. Then the pair (X, d) is a dislocated metric space. We define an arbitrary sequence {x_n} in X; if k > N, there exists an positive integer N that satisfies |x_k − a | < ϵ/2. Then, for any m, n > N, we have d(x_n, x_m) = |x_n − x_m | ≤|x_n − a | +|x_m − a | < ϵ/2 + ϵ/2 = ϵ. Thus, {x_n} is a Cauchy sequence in X. Also as n → ∞, then {x_n} → ∞ ∈ X. Hence, every Cauchy sequence in X is convergent with respect to d. Thus, (X, d) is a complete dislocated metric space.

Definition 13 (see [9].)There exist ϕ(t) that satisfy the condition ϕ^′, if one lets ϕ : [0, +∞]→[0, ∞) be nondecreasing and non-negative, then lim ϕ_n(t) = 0, for a given t > 0.

Lemma 14 (see [9].)If ϕ satisfy the condition ϕ^′, then ϕ(t) < t, for a given t > 0.

Lemma 15 (see [11].)Let $$, and satisfy the condition ϕ^′; for all u, v ≥ 0, if u ≤ F(v, v, u) or u ≤ F(v, u, v) or u ≤ F(u, v, v), then u ≤ ϕ(v).

Theorem 16. Let (X, d) be a complete dislocated metric space, and let f, g : X → X be self-mapping, if

• (1)

  either f or g is continuous;

• (2)

  there exists F satisfying the condition ϕ^′, for all x, y ∈ X, such that

  $$ ()

then f,  g have unique common fixed points.

Proof. Let g be continuous, x₀ arbitrary in X, x_n, y_n the sequence of X, and

Obviously, By the given condition, we have By Lemma 15, we have Also Therefore By Lemma 14, we have Hence, by induction, for all n ∈ N, we obtain Similarly If n ≥ 2, we have Note that by the condition ϕ^′ we know, for n, m ∈ N such that m > n, we have

Hence, d(x_n, x_m) → 0 as m, n → ∞. This forces that {x_n} is a Cauchy sequence in X. But X is a completely dislocated metric space; hence, {x_n} is d-converges. Call the d-limit x_* ∈ X. Then, x_n → x_* as n → ∞. By the continuity of g, as n → ∞, y_n = y_* = g(x_*). So as n → ∞, d(x_n+1, y_n+1) ≤ d(x_*, y_*) ≤ 0, x_* is the fixed point of g.

By the given condition, we have

Hence, d(f(x_*), x_*) ≤ ϕ(0) = 0⇒f(x_*) = x_*, so x_* is a common fixed point of f, g.

Uniqueness. Let y_* be another common fixed point of f, g. Then by the given condition, we have

Since d(x_*, y_*) ≤ ϕ(0) = 0⇒x_* = y_*, x_* is the unique fixed point of f; similarly, we prove that   x_* is also the unique fixed point of g. Thus the fixed point of f, g is unique, and we prove the theorem.

Theorem 17. Let (X, d) be a complete dislocated metric space; and let f, g : X → X be continuous mapping, if

• (1)

  there exists F satisfying the condition ϕ^′, for all x, y ∈ X, if x ≠ y, such that

  $$ ()

• (2)

  there exists x₀ ∈ X such that {(fg) ⁿ(x₀)} have a condensation point, then f, g have a unique common fixed point.

Proof. Let x_n, y_n be the sequence of X, and for all n, x_n ≠ y_n,

Obviously,

Suppose that x_* is the condensation point of {x_n}; there exists the subsequence $$ of {x_n} such that $$. Since g is continuous, $$.

Consider

Also consider

Thus

Hence, we know that {d(x_n+1, y_n+1)} is decreasing. Let lim d(x_n+1, y_n+1) = ϵ and $$. Since $$ is the subsequence of {x_n}, we have

Hence, we conclude that d(x_*, y_*) ≤ d(f(y_*), y_*), a contradiction. So x_* = y_*, y_* is the fixed point of g. Similarly, y_* is the fixed point of f.

Uniqueness. Let y^′ be another common fixed point of f, g. Then by the given condition, we have

Since d(y_*, y^′) ≤ ϕ(0) = 0⇒y_* = y^′, y_* is the unique fixed point of f. Similarly, we prove that y_* is also the unique fixed point of g. Thus the fixed point of f, g is unique.