A Fixed Point Theorem in Orbitally Complete Partially Ordered Metric Spaces

Theorem 1 (see [8].)Let (X, d) be a complete metric space, φ : X → [0, ∞) a lower semicontinuous function, and T : X → X a given mapping. Suppose that, for any 0 < a < b < ∞, there exists 0 < γ(a, b) < 1 such that a ≤ d(x, y) + φ(x) + φ(y) ≤ b implies

for all x, y ∈ X. Then, T has a unique fixed point x^* ∈ X. Moreover, one has φ(x^*) = 0.

Theorem 2 (see [8].)Let (X, d) be a complete metric space, φ : X → [0, ∞) a lower semi-continuous function and T : X → X a given mapping. Suppose that there exists a constant γ ∈ (0,1/2) such that

for all x, y ∈ X. Then, T has a unique fixed point x^* ∈ X. Moreover, one has φ(x^*) = 0.

Theorem 3 (see [8].)Let (X, d) be a complete metric space, φ : X → [0, ∞) a lower semi-continuous function and T : X → X a given mapping. Suppose that, there exist α, β, γ ∈ (0,1) with α + β + γ < 1 such that for all x, y ∈ X,

Then, T has a unique fixed point x^* ∈ X. Moreover, one has φ(x^*) = 0.

Definition 4. Let (X, ⪯) be a partially ordered set. A map T : X → X is said to be non-decreasing if, for any x, y ∈ X with x⪯y, one has Tx⪯Ty.

Theorem 5 (see [9].)Let (X, ⪯) be a partially ordered set and d be a metric on X such that (X, d) is a complete metric space. Suppose that T : X → X is a continuous map and there exists k ∈ (0,1) such that

for each x, y ∈ X with x⪯y. If there exists x₀ ∈ X such that x₀⪯Tx₀, then T has a fixed point in X.

Definition 6. Let X be a nonempty set and T : X → X. Let x₀ ∈ X. The orbit of x₀ is defined by 𝒪(x₀) = {x₀, Tx₀, T²x₀, …}.

Definition 7. Let (X, d) be a metric space and T : X → X. X is said to be T-orbitally complete if every Cauchy sequence in 𝒪(x), x ∈ X, converges to a point in X.

Definition 8. Let (X, d) be a metric space and T : X → X. T is said to be orbitally continuous at z ∈ X if Tx_n → Tz as n → ∞ whenever x_n → z as n → ∞.

Definition 9. Let X be a nonempty set. Let {T_i/T_i : X → X, i = 1,2, 3, …} be a sequence of self-maps. An element x ∈ X is said to be a common fixed point of {T_i} if T_i(x) = x for each i = 1,2, 3, ….

Lemma 10 (see [13].)Suppose that (X, d) is a metric space. Let {x_n} be a sequence in X such that d(x_n, x_n+1) → 0 as n → ∞. If {x_n} is not a Cauchy sequence, then there exist ϵ > 0 and sequences of the positive integers m(k) and n(k) with m(k) > n(k) > k such that d(x_m(k), x_n(k)) ≥ ϵ, d(x_m(k)−1, x_n(k)) < ϵ and lim _k→∞d(x_m(k), x_n(k)) = ϵ and lim _k→∞d(x_m(k)+1, x_n(k)+1) = ϵ.

Theorem 11. Let (X, ⪯) be a partially ordered set and d a metric on X. Suppose that T : X → X is a non-decreasing map and x₀ ∈ X such that x₀⪯Tx₀. Suppose that there exist a lower semicontinuous function φ : X → [0, ∞) and ψ ∈ Ψ such that the following condition holds.

“For each 0 ≤ a < b < ∞, there exists γ(a, b)∈[0,1) such that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b implies

where for each $$ with x⪯y.”

Assume that X is T-orbitally complete. Then, the sequence {x_n} defined by x_n+1 = Tx_n, n = 0,1, 2, …, is Cauchy in X. Let lim _n→∞x_n = z, z ∈ X. Suppose that either

•  

  (a) T is orbitally continuous at z

  •  

    or

•  

  (b) if {x_n} is a nondecreasing sequence converging to x ∈ X, then x_n⪯x, for all n.

Then, z is a fixed point of T and φ(z) = 0.

Proof. Let x₀ ∈ X such that x₀⪯Tx₀. Choose x₁ ∈ X such that x₁ = Tx₀. Then, x₀⪯x₁. Since T is non-decreasing, we have x₁ = Tx₀⪯Tx₁. Now, we choose x₂ ∈ X such that x₂ = Tx₁. Hence, x₁ = Tx₀⪯Tx₁ = x₂.

Continuing the same procedure, we obtain a sequence $$ such that

If there exists an integer n₀ such that $$, then, $$ is a fixed point of T and hence we are through. Without loss of generality, we assume that x_n ≠ x_n+1 for each n.

We denote

We show that u_n+1 ≤ u_n for each n = 0,1, 2, …. Suppose, if possible, there exists n₀ such that $$. That is, Then, from (6), there exists $$ such that where Hence, from (11), we get that is, a contradiction since $$. Therefore, u_n+1 ≤ u_n for each n = 0,1, 2, …. That is, the sequence {u_n} is a decreasing sequence of real numbers. Hence, there exists u ≥ 0 such that lim _n→∞u_n = u and u ≤ u_n for each n = 0,1, 2, ….

We now show that u = 0. Suppose that u > 0. Since 0 < u ≤ u_n ≤ u₀ for all n, from (6), there exists γ(0, u₀)∈[0,1) such that

where Thus, from (15), we get Letting n → ∞, we get a contradiction. Hence, u = 0.

Therefore

Thus, we have lim _n→∞ψ(d(x_n, x_n+1)) = 0 and lim _n→∞φ(x_n) = 0.

By the property of ψ, we have

We now show that {x_n}⊆𝒪(x₀) is a Cauchy sequence in X. Suppose if possible that {x_n} is not a Cauchy sequence in X. Then, there exist ϵ > 0 and sequences of the positive integers {m(k)} and {n(k)} with m(k) > n(k) > k such that

We choose m(k), the least positive integer satisfying (21). Then, we have m(k) > n(k) > k with Then, by Lemma 10, we have Now, since lim _k→∞d(x_m(k), x_m(k)−1) = 0, there exists a number K such that Hence, Since ψ is non-decreasing, we have Now, using (22), we get That is, 0 < a ≤ ψ(d(x_m(k), x_n(k))) + φ(x_m(k)) + φ(x_n(k)) < b for each k ≥ K.

Therefore, from (6), there exists γ(a, b)∈[0,1) such that

where Letting k → ∞, we get Now, letting k → ∞ in (28), we get a contradiction. Hence, {x_n}⊆𝒪(x₀) is a Cauchy sequence in X. Since X is T-orbitally complete, there exists z ∈ X such that lim _n→∞x_n = z.

Since φ is lower semi-continuous, we have

Hence, φ(z) = 0.

Now, suppose that (a) holds. That is, T is orbitally continuous at z. Then, x_n+1 = Tx_n → Tz. Hence,  Tz = z.

Now, suppose that (b) holds. Then, we have x_n⪯z for all n.

If x_n = z for infinitely many n, then x_n+1 = Tx_n = Tz. But x_n+1 → z. Hence, Tz = z. So, without loss of generality, we assume that d(x_n, z) > 0 for infinitely many n. That is, there exists a subsequence {x_n(k)} of {x_n} such that d(x_n(k), z) > 0 for each k = 0,1, 2, ….

As the sequence {d(x_n(k), z)} is convergent, it is bounded. Hence, there exists l ≥ 0 such that d(x_n(k), z) ≤ l for each k = 0,1, 2, ….

Since ψ is non-decreasing, we have

Now, for each k = 0,1, 2, …, we have Hence, from (6), there exists γ(0, b)∈[0,1) such that where and lim _k→∞M(x_n(k), z) = ψ(d(z, Tz)) + φ(Tz), since φ(z) = 0.

Letting k → ∞ in (35), we get

Thus, ψ(d(z, Tz)) + φ(Tz) = 0. That is, Tz = z.

Corollary 12. Let (X, ⪯) be a partially ordered set. Suppose that T : X → X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x₀ ∈ X such that x₀⪯Tx₀. Suppose that there exist a lower semi-continuous function φ : X → [0, ∞), ψ ∈ Ψ, and k ∈ [0,1) such that

where for each $$ with x⪯y.

Then, the sequence {x_n} defined by x_n+1 = Tx_n, n = 0,1, 2, …, is Cauchy in X. Let lim _n→∞x_n = z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z) = 0.

Corollary 13. Let (X, ⪯) be a partially ordered set. Suppose that T : X → X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete metric space. Assume that there exists x₀ ∈ X such that x₀⪯Tx₀. Suppose that there exist a lower semi-continuous function φ : X → [0, ∞), ψ ∈ Ψ such that the following condition holds.

“For each 0 ≤ a < b < ∞, there exists γ(a, b)∈[0,1) such that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b implies

for each $$ with x⪯y.”

Then, the sequence {x_n} defined by x_n+1 = Tx_n, n = 0,1, 2, …, is Cauchy in X. Let lim _n→∞x_n = z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z) = 0.

Proof. Since inequality (40) implies inequality (6); the conclusion of the corollary follows from Theorem 11.

Corollary 14. Let (X, ⪯) be a partially ordered set. Suppose that T : X → X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x₀ ∈ X such that x₀⪯Tx₀. Suppose that there exist a lower semi-continuous function φ : X → [0, ∞), ψ ∈ Ψ such that the following condition holds.

“For each 0 ≤ a < b < ∞, there exist α(a, b), β(a, b)∈[0,1) with α(a, b) + β(a, b) < 1 such that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b implies

for each $$ with x⪯y.”

Then, the sequence {x_n} defined by x_n+1 = Tx_n, n = 0,1, 2, …, is Cauchy in X. Let lim _n→∞x_n = z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z) = 0.

Proof. Inequality (41) implies inequality (6). Hence, the conclusion follows from Theorem 11.

Corollary 15. Let (X, ⪯) be a partially ordered set. Suppose that T : X → X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x₀ ∈ X such that x₀⪯Tx₀. Suppose that there exist a lower semi-continuous function φ : X → [0, ∞), ψ ∈ Ψ, and α, β ∈ [0,1) with α + β < 1 such that

for each $$ with x⪯y.

Then, the sequence {x_n} defined by x_n+1 = Tx_n, n = 0,1, 2, …, is Cauchy in X. Let lim _n→∞x_n = z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z) = 0.

Corollary 16. Let (X, ⪯) be a partially ordered set. Suppose that T : X → X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x₀ ∈ X such that x₀⪯Tx₀. Suppose that there exist a lower semi-continuous function φ : X → [0, ∞), ψ ∈ Ψ such that the following condition holds.

“For each 0 ≤ a < b < ∞ there exist α(a, b), β(a, b), and γ(a, b)∈[0,1) with α(a, b) + β(a, b) + γ(a, b) < 1 such that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b implies

for each $$ with x⪯y.”

Then, the sequence {x_n} defined by x_n+1 = Tx_n, n = 0,1, 2, …, is Cauchy in X. Let lim _n→∞x_n = z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z) = 0.

Proof. The conclusion follows from Theorem 11, since inequality (43) implies inequality (6).

Corollary 17. Let (X, ⪯) be a partially ordered set and d a metric on X. Suppose that T : X → X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x₀ ∈ X such that x₀⪯Tx₀. Suppose that there exist a lower semi-continuous function φ : X → [0, ∞), ψ ∈ Ψ and there exist α, β, and γ ∈ [0,1) with α + β + γ < 1 such that

for each $$ with x⪯y.

Then, the sequence {x_n} defined by x_n+1 = Tx_n, n = 0,1, 2, …, is Cauchy in X. Let lim _n→∞x_n = z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z) = 0.

Remark 18. By choosing ψ(t) = t in Corollaries 13, 15, and 17, we obtain partially ordered versions of Theorems 1, 2, and 3, respectively.

Remark 19. By choosing φ ≡ 0 and ψ(t) = t, t ≥ 0, in Corollary 12, we obtain Theorem 5 as a corollary to Corollary 12, which in turn Theorem 5 follows as a corollary to Theorem 11.

Theorem 20. Let (X, ⪯) be a partially ordered set. Suppose that T₁ : X → X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T₁-orbitally complete. Assume that there exists x₀ ∈ X such that x₀⪯T₁x₀. Let {T_i/T_i : X → X, i = 1,2, 3, …} be a sequence of maps and suppose that there exist a lower semi-continuous function φ : X → [0, ∞), ψ ∈ Ψ, and k ∈ [0,1) such that

where for each $$ with x⪯y and j = 1,2, 3, ….

If either (a) or (b) of Theorem 11 holds, then the family of maps {T_j/j = 1,2, 3, …} has a common fixed point.

Proof. If j = 1, then T₁ has a fixed point by Corollary 12. Let that fixed point be u (say). Then, φ(u) = 0. We show that T_ju = u for each j = 2,3, 4, …. Let j > 1 be fixed such that T_ju ≠ u.

Now, using (45), we get

where Then, from (47), we obtain a contradiction. Hence, T_ju = u for each j = 1,2, 3, ….

Thus, the conclusion of the theorem follows.

Example 21. Let X = [0,2) with the usual metric. We define a partial order ⪯ on X by

We define T : X → X, ψ : [0, ∞)→[0, ∞), and φ : X → [0, ∞) by Let x₀ = 1/4; then, 𝒪(x₀) = {(1/2ⁿ)/n ≥ 2, n ∈ Z⁺} and $$.

And Tx₀ = 1/8 so that x₀⪯Tx₀.

We now verify inequality (6).

In this example, we choose γ(a, b) = 3/4 for each a, b ∈ [0, ∞), 0 ≤ a < b satisfying a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b for all $$.

Let $$.

Case (i) (x = 1/2ⁿ, y = 1/2^m, and n ≥ m ≥ 2). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b. That is, a ≤ (1/2)[(1/2^m) − (1/2ⁿ)] + (1/2ⁿ⁺¹) + (1/2^m+1) ≤ b. That is, a ≤ 1/2^m ≤ b. Now,

Case (ii) (x = 1/2ⁿ, n ≥ 2, and y = 0). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b; that is, a ≤ 1/2ⁿ ≤ b. Now,

Case (iii) (x = y = 1/2ⁿ, n ≥ 2). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b; that is, a ≤ 1/2ⁿ ≤ b. Now,

Hence the verification of inequality (6) is the same as in Case (ii).

Case (iv) (x = y = 0). In this case, ψ(d(Tx, Ty)) + φ(Tx) + φ(Ty) = 0 so that inequality (6) holds trivially.

Example 22. Let X = [0,1) with the usual metric. We define a partial order ⪯ on X by

“x⪯y if and only if x ≥ y for each x, y ∈ X.”

We define T : X → X, ψ : [0, ∞)→[0, ∞), and φ : X → [0, ∞) by

Let x₀ = 2/3. Then, Tx₀ = 1/4 so that x₀⪯Tx₀. Consider We now verify inequality (6).

We choose γ(a, b) = 12/69 for each 0 ≤ a < b < ∞ satisfying a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b for all $$.

Let $$.

Case (i) (x = 2/3, y = 1/4). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b; that is, a ≤ 11/8 ≤ b. Now,

Case (ii) (x = 2/3, y = 3/16). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b; that is, a ≤ 69/48 ≤ b. Now,

so that

Case (iii) (x = 1/4, y = 3/16). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b; that is, a ≤ 3/16 ≤ b. Now,

Hence, (6) holds trivially.

Case (iv) (x = y = 2/3). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b; that is, a ≤ 23/24 ≤ b. Now,

Then,

Case (v). (let x = y = 1/4). Suppose that a ≤ ψ(d(x, y)) + φ(x) + φ(y) ≤ b; that is, a ≤ 1/8 ≤ b. Now,

Hence, (6) holds trivially.