Common Fixed Point Results Using Generalized Altering Distances on Orbitally Complete Ordered Metric Spaces

Definition 2.1. Let (𝒳, ≼) be a partially ordered set and 𝒮, 𝒯 : 𝒳 → 𝒳.

• (1)

  [23] The pair (𝒮, 𝒯) is called weakly increasing if 𝒮x≼𝒯𝒮x and 𝒯x≼𝒮𝒯x for all x ∈ 𝒳.

• (2)

  [24] The pair (𝒮, 𝒯) is called partially weakly increasing if 𝒮x≼𝒯𝒮x for all x ∈ 𝒳.

• (3)

  [24] The mapping 𝒮 is called a weak annihilator of 𝒯 if 𝒮𝒯x≼x for all x ∈ 𝒳.

• (4)

  [24] The mapping 𝒮 is called dominating if x≼𝒮x for each x ∈ 𝒳.

Example 2.2 (see [24].)Let 𝒳 = [0,1] be endowed with usual ordering.

Definition 2.3 (see [25], [26].)Let (𝒳, d) be a metric space and f, g : 𝒳 → 𝒳. The mappings f and g are said to be compatible if lim _n→∞d(fgx_n, gfx_n) = 0, whenever {x_n} is a sequence in 𝒳 such that lim _n→∞ fx_n = lim _n→∞ gx_n = t for some t ∈ 𝒳.

Definition 2.4. Let 𝒳 be a nonempty set. Then (𝒳, d, ≼) is called an ordered metric space if

• (i)

  (𝒳, d) is a metric space,

• (ii)

  (𝒳, ≼) is a partially ordered set.

Theorem 2.5. Let (𝒳, d, ≼) be an ordered metric space. Let 𝒮, 𝒯, ℛ : 𝒳 → 𝒳 be given mappings satisfying

for all $$ (for some x₀) such that ℛx and ℛy are comparable, where and  ψ₁ and ψ₂ are generalized altering distance functions (in ℱ₄) and Ψ₁(t) = ψ₁(t, t, t, t). We assume the following hypotheses:

• (i)

  (𝒮, 𝒯) is a.r. with respect to ℛ at x₀ ∈ 𝒳;

• (ii)

  𝒳 is (𝒮, 𝒯, ℛ)-orbitally complete at x₀;

• (iii)

  (ℛ, 𝒮) and (ℛ, 𝒯) are partially weakly increasing;

• (iv)

  𝒮 and 𝒯 are dominating maps;

• (v)

  𝒮 and 𝒯 are weak annihilators of ℛ;

• (vi)

  for a nondecreasing sequence {x_n}, x_n≼y_n for all n and y_n → u as n → ∞ imply that x_n≼u for all n ∈ ℕ.

Assume either

• (a)

  𝒮 and ℛ are compatible; 𝒮 or ℛ is orbitally continuous at x₀ or

• (b)

  𝒯 and ℛ are compatible; 𝒯 or ℛ is orbitally continuous at x₀.

Then 𝒮, 𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮, 𝒯, and ℛ in $$ is well ordered if and only if it is a singleton.

Proof. Since (𝒮, 𝒯) is a.r. with respect to ℛ at x₀ in 𝒳, there exists a sequence {x_n} in 𝒳 such that

By the given assumptions, x_2n−2≼𝒮x_2n−2 = ℛx_2n−1≼𝒮ℛx_2n−1≼x_2n−1, and x_2n−1≼𝒯x_2n−1 = ℛx_2n≼𝒯ℛx_2n≼x_2n. Thus, for all n ≥ 1, we have In view of (i), we have Now, we assert that {ℛx_n} is a Cauchy sequence in the metric space 𝒪(x₀; 𝒮, 𝒯, ℛ).

From (2.5), it will be sufficient to prove that {ℛx_2n} is a Cauchy sequence. We proceed by negation and suppose that {ℛx_2n} is not a Cauchy sequence. Then, there exists ε > 0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that for all positive integers i,

From (2.6) and using the triangular inequality, we get Letting i → ∞ in the above inequality and using (2.5), we obtain Again, the triangular inequality gives us Letting i → ∞ in the above inequality and using (2.5) and (2.8), we get Similarly, we have On the other hand, we have Then, from (2.5), (2.8), and the continuity of Ψ₁, we get by letting i → ∞ in (2.1) Now, using the considered contractive condition (2.1) for x = x_2m(i)−1 and y = x_2n(i), we have Then, from (2.5), (2.10), (2.11), and the continuity of ψ₁ and ψ₂, we get by letting i → ∞ in the above inequality Now, combining (2.13) with the above inequality, we get which implies that ψ₂(ε, 0,0, ϵ) = 0, which is a contradiction since ε > 0. Hence {ℛx_n} is a Cauchy sequence in 𝒪(x₀; 𝒮, 𝒯, ℛ). Since 𝒳 is (𝒮, 𝒯, ℛ)-orbitally complete at x₀, there exists some z ∈ 𝒳 such that ℛx_n → z as n → ∞.

Finally, we prove the existence of a common fixed point of the three mappings 𝒮, 𝒯, and ℛ.

We have

Suppose that (a) holds. Since {𝒮, ℛ} are compatible, we have Also, x_2n+1≼𝒯x_2n+1 = ℛx_2n+2. Now Assume that ℛ is orbitally continuous. Passing to the limit as n → ∞, we obtain so ψ₂(d(ℛz, z), 0,0, d(ℛz, z)) = 0, which implies that Now, x_2n+1≼𝒯x_2n+1 and 𝒯x_2n+1 → z as n → ∞, so by the assumption we have x_2n+1≼z and (2.1) becomes Passing to the limit as n → ∞ in the above inequality and using (2.21), it follows that which holds unless ψ₂(0, d(𝒮z, z), 0, (1/2)d(𝒮z, z)) = 0, so Now, since x_2n≼𝒮x_2n and 𝒮x_2n → z as n → ∞ implies that x_2n≼z, from (2.1) Passing to the limit as n → ∞, we have which gives that Therefore, 𝒮z = 𝒯z = ℛz = z, hence z is a common fixed point of ℛ, 𝒮, and 𝒯. The proof is similar when 𝒮 is orbitally continuous.

Similarly, the result follows when condition (b) holds.

Now, suppose that the set of common fixed points of 𝒮, 𝒯, and ℛ in $$ is well ordered. We claim that it cannot contain more than one point. Assume to the contrary that 𝒮u = 𝒯u = ℛu = u and 𝒮v = 𝒯v = ℛv = v but u ≠ v. By supposition, we can replace x by u and y by v in (2.1) to obtain

a contradiction. Hence, u = v. The converse is trivial.

Corollary 2.6. Let 𝒮, 𝒯, and ℛ satisfy the conditions of Theorem 2.5, except that condition (2.1) is replaced by the following: there exists a positive Lebesgue integrable function u on ℝ₊ such that $$ for each ε > 0 and that

Then, 𝒮, 𝒯, and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮, 𝒯, and ℛ in $$ is well ordered if and only if it is a singleton.

Remark 2.7. If we take

for k ∈ (0,1), then Ψ₁(t) = t for all t ≥ 0, and the contractive condition (2.1) becomes which corresponds to the contraction given by Theorem  2.1 in [24] by taking ψ(t) = t and φ(t) = (1 − k)t. Hence, the result of Abbas et al. [24] is covered by Theorem 2.5 for three maps.

Other results could be derived for other choices of ψ₁ and ψ₂.

Corollary 2.8. Let (𝒳, d, ≼) be an ordered metric space. Let 𝒯, ℛ : 𝒳 → 𝒳 be given mappings satisfying

for all $$ such that ℛx and ℛy are comparable, where ψ₁ and ψ₂ are generalized altering distance functions (in ℱ₄) and Ψ₁(t) = ψ₁(t, t, t, t). We assume the following hypotheses:

• (i)

  𝒯 is a.r. with respect to ℛ at x₀;

• (ii)

  𝒳 is (𝒯, ℛ)-orbitally complete at x₀;

• (iii)

  𝒯 or ℛ is orbitally continuous at x₀;

• (iv)

  (𝒯, ℛ) is partially weakly increasing;

• (v)

  𝒯 is a dominating map;

• (vi)

  𝒯 is a weak annihilator of ℛ;

• (vii)

  𝒯 and ℛ are compatible.

Let for a nondecreasing sequence {x_n} with x_n≼y_n  for all n, y_n → u as n → ∞ imply that x_n≼u for all n ∈ ℕ.

Then 𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒯 and ℛ in $$ is well ordered if and only if it is a singleton.

Corollary 2.9. Let (𝒳, d, ≼) be an ordered metric space. Let 𝒯 : 𝒳 → 𝒳 be a mapping satisfying

for all $$ such that x and y are comparable, where ψ₁ and ψ₂ are generalized altering distance functions (in ℱ₄) and Ψ₁(t) = ψ₁(t, t, t, t). We assume the following hypotheses:

• (i)

  𝒯 is a.r. at some point x₀ of 𝒳;

• (ii)

  𝒳 is 𝒯-orbitally complete at x₀;

• (iii)

  𝒯 is a dominating map.

Let for a nondecreasing sequence {x_n} with x_n≼y_n  for all n, y_n → u as n → ∞ imply that x_n≼u for all n ∈ ℕ.

Then 𝒯 has a fixed point. Moreover, the set of fixed points of 𝒯 in $$ is well ordered if and only if it is a singleton.

Example 2.10. Let the set 𝒳 = [0, +∞) be equipped with the usual metric d and the order defined by

Consider the following self-mappings on 𝒳: Take x₀ = 1/2. Then it is easy to show that and all the conditions (i)–(vi) and (a)-(b) of Theorem 2.5 are fulfilled. Take ψ₁(t₁, t₂, t₃, t₄) = max {t₁, t₂, t₃, t₄} and ψ₂(t₁, t₂, t₃, t₄) = (5/6)max {t₁, t₂, t₃, t₄}. Then contractive condition (2.1) takes the form for x, y ∈ 𝒪(x₀; 𝒮, 𝒯, ℛ). Using substitution y = tx, t > 0, the last inequality reduces to and can be checked by discussion on possible values for t > 0. Hence, all the conditions of Theorem 2.5 are satisfied and 𝒮, 𝒯, ℛ have a common fixed point (which is 0).

Remark 2.11. It was shown by examples in [22] that (in similar situations)

• (1)

  if the contractive condition is satisfied just on 𝒪(x₀; 𝒮, 𝒯, ℛ), there might not exist a (common) fixed point;

• (2)

  under the given hypotheses (common), fixed point might not be unique in the whole space 𝒳.