On a fixed point theorem with application to functional equations

Theorem 1

[18] Let (ℑ, ⪯, d) be a complete ordered metric space and T : ℑ → ℑ be a continuous and nondecreasing mapping such that there exist α, β ≥ 0 with α + β < 1 satisfying

for all σ, ς ∈ ℑ with σ ⪯ ς. If there exists σ₀ ∈ ℑ such that σ₀ ⪯ T(σ₀), then T has a unique fixed point.

Definition 1

[2] A partial metric on a non-empty set ℑ is a function P : ℑ × ℑ → [0, ∞) satisfying the following axioms, for all σ, ς, ν ∈ ℑ,

1. σ = ς ⇔ P(σ, σ) = P(ς, ς) = P(σ, ς);

2. P(σ, σ) ≤ P(σ, ς);

3. P(σ, ς) = P(ς, σ);

4. P(σ, ν) ≤ P(σ, ς) + P(ς, ν) – P(ς, ς).

The pair (ℑ, P) is called a partial metric space.

Definition 2

[3] A dualistic partial metric on a non-empty set ℑ is a function η : ℑ × ℑ → ℝ satisfying the following properties, for all σ, ς, ν ∈ ℑ,

1. σ = ς ⇔ η(σ, σ) = η(ς, ς) = η(σ, ς);

2. η(σ, σ) ≤ η(σ, ς);

3. η(σ, ς) = η(ς, σ);

4. η(σ, ν) ≤ η(σ, ς) + η(ς, ν) – η(ς, ς).

The pair (ℑ, η) is called a dualistic partial metric space.

Remark 1

It is obvious that every partial metric is a dualistic partial metric but the converse is not true. To support this comment, define η_m : ℝ × ℝ → ℝ by

Clearly, η_m is a dualistic partial metric (see [4]) but η_m is not a partial metric. Indeed, for all σ < 0, ς < 0 implies η_m(σ, ς) < 0 ∉ R0+ .

Definition 3

Let (ℑ, η) be a dualistic partial metric space and T : ℑ → ℑ be a mapping. We say that T has a convergence comparison property (CCP) if for every sequence {σ_n} in ℑ such that σ_n → σ, T satisfies

Example 1

Let ℑ = ℝ. Define η_m as in Remark 1. Consider any sequence {σ_n} converging to σ in (ℑ, η_m). Define T : ℑ → ℑ by T(σ) = e^σ. We have σ ≤ e^σ for any σ ∈ ℑ, that is, η_m(σ, σ) ≤ η_m(T(σ), T(σ), i.e., T satisfies (CCP).

Example 2

If (ℑ, d) is a metric space and c ∈ ℝ is arbitrary constant, then

defines a dualistic partial metric on ℑ.

Example 3

[3] Let (ℑ, P) be a partial metric space. The mapping η : ℑ × ℑ → ℝ defined by

satisfies the conditions (η₁) – (η₄) and hence defines a dualistic partial metric on ℑ. We note that η(σ, ς) may have negative values.

Example 4

Let ℑ = ℝ. Define the mapping η : ℑ × ℑ → ℝ by

The axioms (η₁), (η₂) and (η₃) can be proved immediately. We prove axiom (η₄) in details.

1. If σ ≠ ς = ν, then

   η(σ, ν) ≤ η(σ, ς) + η(ς, ν) – η(ς, ς) implies |σ – ν| = |σ – ς|.

2. If σ = ς ≠ ν, then

   η(σ, ν) ≤ η(σ, ς) + η(ς, ν) – η(ς, ς) implies |σ – ν| = |ς – ν|.

3. If σ = ς = ν, then

   η(σ, ν) ≤ η(σ, ς) + η(ς, ν) – η(ς, ς) is obious.

4. If σ ≠ ς ≠ ν, then

   η(σ, ν) ≤ η(σ, ς) + η(ς, ν) – η(ς, ς) implies |σ – ν| ≤ |σ – ς| + |ς – ν| + b.

Thus, the axiom (η₄) holds in all cases. Hence (ℑ, η) is a dualistic partial metric space.

Definition 4

[11] Let (ℑ, η) be a dualistic partial metric space.

1. A sequence {σ_n}_n∈ℕ converges to a point σ in (ℑ, η) if

   limn→∞η(σn,σ)=η(σ,σ).

2. A sequence {σ_n}_n∈ℕ in (ℑ, η) is called a Cauchy sequence if

   limn,m→∞η(σn,σm) exists and is finite.

3. A dualistic partial metric space (ℑ, η) is said to be complete if every Cauchy sequence {σ_n}_n∈ℕ in ℑ converges, with respect to 𝓣[η], to a point σ ∈ ℑ such that

   η(σ,σ)=limn,m→∞η(σn,σm).

Remark 2

For a sequence {σ_n} ⊂ ℑ, convergence wrt(with respect to) metric space (ℑ, d) may not implies convergence wrt dualistic partial metric space (ℑ, η).

Lemma 1

[11] Let (ℑ, η) be a dualistic partial metric space.

1. Every Cauchy sequence in (ℑ, dηs ) is also a Cauchy sequence in (ℑ, η).

2. A dualistic partial metric (ℑ, η) is complete if and only if the induced metric space (ℑ, dηs ) is complete.

3. A sequence {σ_n}_n∈ℕ in ℑ converges to a point υ ∈ ℑ with respect to 𝓣[( dηs )] if and only if

   limn→∞η(υ,σn)=η(υ,υ)=limn,m→∞η(σn,σm).

Definition 5

Let (ℑ, ⪯) be a partially ordered set and (ℑ, η) be a dualistic partial metric space. A mapping T : ℑ → ℑ is said to be a dualistic contraction of rational type if there exist α, β ≥ 0 with α + β < 1 such that:

for all σ, ς ∈ 𝓓.

Remark 3

The contractive condition (3.1) has some differences with (1.1). Since, for a metric d, d(σ, σ) = 0 for any σ ∈ ℑ which ensures that (1.1) holds for all σ such that σ = T(σ) and conversely. However, from definition of dualistic partial metric we know that, in general, η(σ, σ) ≠ 0 for any σ ∈ ℑ. For if σ is a fixed point of T then from (3.1) one can follow that η(σ, σ) = 0. So if a self-mapping T has a fixed point σ such that η(σ, σ) ≠ 0, then σ ⪯ σ but σ does not satisfy (3.1).

Theorem 2

Let (ℑ, ⪯) be a partially ordered set and (ℑ, η) be a complete dualistic partial metric space. If T is a nondecreasing, dualistic contraction of rational type satisfying the following conditions:

1. there exists σ₀ ∈ ℑ such that σ₀ ⪯ T(σ₀);

2. if {σ_n} is a nondecreasing sequence in ℑ such that {σ_n} → υ, then σ_n ⪯ υ for all n ∈ ℕ.

Then T has a fixed point.

Proof

Let σ₀ be an initial point of ℑ and let us define Picard iterative sequence {σ_n} by

If there exists a positive integer i such that σ_i = σ_i+1, then σ_i = σ_i+1 = T(σ_i). So σ_i is a fixed point of T. In this case, the proof is complete.

On the other hand, if σ_n ≠ σ_n+1 for all n ∈ ℕ, then σ_n ⪯ σ_n+1. Indeed by σ₀ ⪯ T(σ₀), we obtain σ₀ ⪯ σ₁. Since T is nondecreasing, σ₀ ⪯ σ₁ implies T(σ₀) ⪯ T(σ₁) and thus σ₁ ⪯ σ₂. Continuing in this way, we get

Since σ_n ⪯ σ_n+1, using (3.1), we have

If we set λ=β1−α, then 0 < λ < 1 and so

Further, generalizing the above inequality, we have

To find self distance, since σ_n ⪯ σ_n for all n ∈ ℕ, again from (3.1), we have

where |η(σ₀, σ₁)| = h and |η(σ₀, σ₀)| = η₀. Repeating the process, we get

This implies that

Now since

where

Further, generalization implies

Next step is to show that {σ_n} is a Cauchy sequence in (ℑ, η). For this purpose, first we show that {σ_n} is a Cauchy sequence in (ℑ, dηs ). Since d_η is a quasi metric, by using triangular property and (3.3), we have

Note that each term in μⁿ contains either βⁿ or λⁿ and since β < 1, λ < 1 so each term in μⁿ vanishes as n → ∞. Thus, lim_n→∞ λⁿ = 0 and lim_n→∞ μⁿ = 0 and hence

Replacing the positions of σ_n and σ_n+1, we get

Further, generalization implies

Since d_η is a quasi metric, by using triangular property and (3.5), we have

Since lim_n→∞ λⁿ = 0 and lim_n→∞ μⁿ = 0, thus, lim_n→∞ d_η(σ_n+k, σ_n) = 0. By definition of dηs , we deduce that lim_n→∞ dηs (σ_n, σ_n+k) = 0. Hence {σ_n} is a Cauchy sequence in (ℑ, dηs ). Since (ℑ, η) is a complete dualistic partial metric space, by Lemma 1, (ℑ, dηs ) is also a complete metric space. Thus {σ_n} converges to a point υ ∈ ℑ, i.e., lim_n→∞ dηs (σ_n, υ) = 0. By Lemma 1, we get

Now from (3.4) and for m = n + k

Consequently, lim_n,m→∞ η(σ_n, σ_m) = 0 and so {σ_n} is a Cauchy sequence in (ℑ, η). From (3.6), we obtain

This shows that {σ_n} converges to υ in (ℑ, η) and η(υ, υ) = 0. By hypothesis (2) and (η₄), we have

Letting n → ∞, we get η(υ, T(υ)) ≤ 0 and by (2.1) we have η(υ, T(υ)) ≥ 0. Thus, η(υ, T(υ)) = 0. By (CCP), we have

By (η₂), we get

These arguments imply that η(T(υ), T(υ)) = 0 and hence

By (η₁), we have υ = T(υ).□

Remark 4

Usually the range of a dualistic partial metric η is (–∞, ∞) but if we replace (–∞, ∞) by [0, ∞), then η is identical to a partial metric P and hence Theorem 2 is applicable in the setting of partial metric space. The Theorem 2 also generalizes the results in [18, Theorem 1] [22].

Example 5

Let ℑ = {0, –1} and consider the partially ordered dualistic partial metric space (ℑ, η_m, ⪯) with 0 ⪯ 0, –1 ⪯ –1 and –1 ⪯ 0. It is clear that (ℑ, η_m) is complete. Define the increasing mapping T : ℑ → ℑ by T(0) = 0 and T(–1) = –1. Clearly, –1 ⪯ T(–1). A straightforward computation shows that

for all x, y ∈ ℑ with x ⪯ y and η_m(σ, ς) ≠ –1. Besides if a sequence {σ_n} is increasing in (ℑ, ⪯) and σ_n → υ then σ_n ⪯ υ for all n. It is obvious that given σ, ς ∈ ℑ we always have that σ, ς ⪯ 0. Nevertheless, T has two fixed points.

Theorem 3

Let T : ℑ → ℑ be defined on (ℑ, ⪯, η) and satisfies conditions assumed in Theorem 2. If there exists an element ω ∈ ℑ such that it is comparable with every fixed point of T, then T has a unique fixed point in ℑ.

Proof

From Theorem 2, it follows that the set of fixed points of T is non-empty. We suppose that υ₁ is also a fixed point of T. Here two cases arise, first, υ and υ₁ are comparable and second, υ and υ₁ are not comparable.

In Case 1, υ ⪯ υ₁ and using (3.1),

Therefore, (1 – β)|η(υ₁, υ₁)| ≤ 0, which is only true if η(υ₁, υ₁) = 0 and η(υ, υ₁) = 0 = η(υ, υ) = η(υ₁, υ₁). Hence υ = υ₁.

In Case 2, there exists an element ω ∈ ℑ such that it is comparable with υ, υ₁. Without any loss of generality, we assume that ω⪯ υ and ω ⪯ υ₁. Since T is nondecreasing, T(ω) ⪯ T(υ) and T(ω) ⪯ T(υ₁). Moreover, T^n–1(ω) ⪯ T^n–1(υ) and T^n–1(ω) ≤ T^n–1(υ₁). Thus

This implies |η(Tⁿ(ω), υ)| ≤ β |η(υ, T^n–1(ω))|.

Now using (3.7), we get lim_n→∞ η(υ, Tⁿ(ω)) = 0.

Similarly, we can show that lim_n→∞ η(υ₁, Tⁿ(ω)) = 0. By η₄, we have

Letting n → ∞, we obtain that η(υ₁, υ) ≤ 0. Now η_η(υ, υ₁) = η(υ, υ₁) – η(υ, υ) implies that η(υ, υ₁) ≥ 0. Hence η(υ, υ₁) = 0, which gives that υ = υ₁.□

Example 6

Let ℑ = (–∞, 0]². Define η_m : ℑ × ℑ → ℝ by η_m(x, y) = σ₁ mς₁ = max{σ₁, ς₁}, where x = (σ₁, σ₂) and y = (ς₁, ς₂). Note that (ℑ, η_m) is a complete dualistic partial metric space. Let T : ℑ → ℑ be given by

In ℑ, we define the relation ⪯ in the following way:

Clearly, ⪯ is a partial order on ℑ and T is a nondecreasing mapping and satisfies (CCP) with respect to ⪯. Moreover, (–1, 0) ⪯ T(–1, 0).

We shall show that for all x, y ∈ ℑ with x ⪯ y, (3.1) is satisfied. For this, consider

Set α=13,β=12 and for all x, y ∈ ℑ, we find that

holds for x ⪯ y if and only if 6|ς₁| |1 + ς₁| ≤ |ς₁|(2 + |σ₁|) + 6|ς₁| |1 + ς₁|.

Thus, all the conditions of Theorem 2 are satisfied. Moreover, (0, 0) is a fixed point of T.

Example 7

Let ℑ = (–∞, –2]² ∪ {O}. Define η : ℑ × ℑ → ℝ by

where σ = (σ₁, σ₂) and ς = (ς₁, ς₂). Then (ℑ, η) is a complete dualistic partial metric space. Let T : ℑ → ℑ be given by

In ℑ, we define the relation ⪯ in the following way

Clearly, ⪯ is a partial order on ℑ and T is a nondecreasing mapping and satisfies (CCP) with respect to ⪯. Moreover, (–1, 0) ⪯ T(–1, 0). Also we have the following information for all σ ≠ ς (either σ ⪯ ς or ς ⪯ σ)

Thus, the contractive condition (3.1) takes the form

For β = 23 , there exists α such that α + β < 1 which satisfies inequality (3.8).

For all σ = ς we have

Here, the contractive condition (3.1) takes the following form

It can be noticed that there exist α and β with α + β < 1 satisfying (3.9) for all σ ∈ ℑ. Thus T is the required self-mapping having fixed point O and which fulfills requirements of Theorem 2.

We note that under the dualistic partial metric defined in this example, fixed point theorem established by Oltra and Valero in [11] fails to have fixed point. Indeed, for all σ = ς

which is a contradiction to definition of c used in Said theorem [11].

Example 8

Define the mapping T₀ : ℝ⁺ → ℝ by

Clearly, for all comparable σ, ς ∈ ℝ, the following contractive condition without absolute value function is satisfied

where η_m is a complete dualistic partial metric on ℝ. Here, T has no fixed point. Thus a fixed point free mapping satisfies this contractive condition.

On the other hand, for all 0 < α + β < 1, we have

Thus the contractive condition (3.1) does not hold. This situation happens always in dualistic partial metric space and that is why we place absolute value function in contractive condition taken under dualistic partial metric.

Corollary 1

[11] Let (ℑ, η) be a complete dualistic partial metric space and T : ℑ → ℑ be a mapping such that there exists c ∈ [0, 1[ satisfying

for all σ, ς ∈ ℑ with σ ⪯ ς. Then T has a unique fixed point σ^* ∈ ℑ. Moreover, η(σ^*, σ^*) = 0 and the Picard iterative sequence {Tⁿ(σ₀)}_n∈ℕ converges to σ^* with respect to τ( dηs ) for every x ∈ ℑ.

Remark 5

Corollary 1 generalizes the result presented by Oltra and Valero in [11] to ordered dualistic partial metric spaces.

Corollary 2

Let the self-mapping T defined on an ordered complete dualistic partial metric space (ℑ, ⪯, η) be nondecreasing and there exists α ∈ [0, 1) such that,

for all σ, ς ∈ 𝓓. Moreover, if

1. there exists σ₀ ∈ ℑ such that σ₀ ⪯ T(σ₀),

2. if {σ_n} is a nondecreasing sequence in ℑ such that {σ_n} → υ, then σ_n ⪯ υ for all n ∈ ℕ,

then T has a fixed point υ such that η(υ, υ) = 0.

Proof

Set β = 0 in the statement of Theorem 2. □

Lemma 2

[28] Let G, H : 𝓢 → ℝ be two bounded functions. Then

Lemma 3

[28] Assume that

1. g, F are bounded functions;

2. there exists k > 0 such that for all t, r ∈ ℝ, σ ∈ 𝓢 and ς ∈ 𝓦

   |F(σ,ς,t)−F(σ,ς,r)|≤k|t−r|.

Then the operator R : 𝓑(S) → 𝓑(S) defined by

is well-defined.

Theorem 4

Let all the conditions of Lemma 3 be satisfied and

for all u, v ∈ 𝓑(S) with u ⪯ v and η(u, v) ≠ –1. Then the functional equation (4.1) has a unique solution.

Proof

Let R : 𝓑(S) → 𝓑(S) be an operator as defined in Lemma 3. Then R is continuous and nondecreasing with respect to ⪯. We shall show that R satisfies the contractive condition (3.1). Indeed, by Lemma 2, for all u, v ∈ 𝓑(S) with u ⪯ v,

Therefore,