On Fixed Point Theory of Monotone Mappings with Respect to a Partial Order Introduced by a Vector Functional in Cone Metric Spaces

Remark 1. A total order minihedral cone P of a normed space E is certainly normal see [29].

Remark 2. Let (X, d) be a cone metric space over a normal cone P of a normed vector space E and {x_n} a sequence in (X, d). Then $$ if and only if lim _n→∞ d(x_n, x) = θ, and {x_n} is Cauchy if and only if lim _m,n→∞ d(x_n, x_m) = θ see [28].

Theorem 3. Let (X, d, ≺) be a complete partially ordered cone metric space over a total order minihedral and continuous cone P of a normed vector space E. Let φ : X → E be a sequentially continuous vector functional and let T : X → 2^X be a set-valued mapping such that Tx is compact for all x ∈ X. Assume that there exists x₀ ∈ X such that φ is bounded below on [x₀, +∞), T is increasing on [x₀, +∞), and Tx₀∩[x₀, +∞) ≠ ∅. Then T has a maximal fixed point x^* ∈ [x₀, +∞).

Proof. Since P is a total order minihedral cone and E is a normed space, then P is a normal cone by Remark 1. Set

Clearly, Q₁ is nonempty since x₀ ∈ Q₁. Let {x_α} _α∈Γ⊆Q₁ be an increasing chain, where Γ is a directed set. Then by (1) we have for all α, β ∈ Γ with α⪯β. This implies that {φ(x_α)} is a decreasing chain in E. Since P is total order minihedral and φ is bounded below on [x₀, +∞), then inf _α∈Γ φ(x_α) exists in E. Moreover, inf _α∈Γ ∥φ(x_α) − inf _α∈Γφ(x_α)∥ = 0 since P is continuous. Therefore there exists an increasing sequence $$ such that $$, that is, By (1) we have for all m ∈ ℕ such that m ≥ n, Let n → ∞, by (6) we have $$ and hence $$ by the normality of P. Moreover by Remark 2, $$ is a Cauchy sequence in X. Therefore by the completeness of X, there exists some $$ such that Note that $$ is an increasing sequence of Q₁, then by (1), we have for all n, And, for all n ≥ n₀, where n₀ is an arbitrary integer. Let n → ∞, then by (8) and the continuity of φ we have $$ and $$, that is, $$ and $$. Moreover the arbitrary property of n₀ forces that for all n. By $$, there exists $$ such that for all n. Since T is increasing on [x₀, +∞), then by (11) and $$, there exists $$ such that for all n. This together with (12) implies that for all n. Note that $$ is compact, and there exists a subsequence $$ and $$ such that From (14) we have $$ for all n_k and hence by (1), for all n_k. Let n_k → ∞, then by (8), (15), and the continuity of φ we have $$, that is, $$. This implies that $$ and hence $$ by $$.

For all α ∈ Γ, if there exists some n₀ such that $$, then $$ is an upper bound of {x_α} by (11). Otherwise, there exists some β ∈ Γ such that $$ for all n. Thus by (1) we have $$ for all n. Let n → ∞, by (6) we have inf _α∈Γ φ(x_α) − φ(x_β) ∈ P; that is, φ(x_β)⪯inf _α∈Γ φ(x_α). So we have φ(x_β) = inf _α∈Γ φ(x_α) and hence φ(x_β)⪯φ(x_α) for all α ∈ Γ. Note that {φ(x_α)} _α∈Γ is a decreasing chain, then β ≥ α for all α ∈ Γ. Moreover x_α≺x_β for all α ∈ Γ since {x_α} _α∈Γ is an increasing chain. Hence {x_α} _α∈Γ has an upper bound in Q₁. By Zorn′s lemma, (Q₁, ≺) has a maximal element x^*; that is, for all x ∈ Q₁, x^*≺x implies x = x^*. By x^* ∈ Q₁, there exists y^* ∈ Tx^* such that x^*≺y^*. Moreover by the increasing property of T on [x₀, +∞), there exists z^* ∈ Ty^* such that y^*≺z^*. Thus we have x^*≺z^* by x^*≺y^*. This indicates z^* ∈ Tx^*∩[x^*, +∞) and hence z^* ∈ Q₁. Finally the maximality of x^* in Q₁ forces that x^* = z^* ∈ Tx^*; that is, x^* is a maximal fixed point of T in [x₀, +∞). The proof is complete.

Theorem 4. Let (X, d, ≺) be a complete partially ordered cone metric space over a total order minihedral and continuous cone P of a normed vector space E. Let φ : X → E be a sequentially continuous vector functional and T : X → 2^X be a set-valued mapping such that Tx is compact for all x ∈ X. Assume that there exists y₀ ∈ X such that φ is bounded above on (−∞, y₀], T is quasi-increasing on (−∞, y₀], and Ty₀ ∩ (−∞, y₀] ≠ ∅. Then T has a minimal fixed point x_* ∈ (−∞, y₀].

Proof. Set

Clearly, Q₂ ≠ ∅. By the same method used in the proof of Theorem 3, we can prove that (Q₂, ≺) has a minimal element x_* which is also a minimal fixed point of T in (−∞, y₀]. The proof is complete.

Remark 5. If T : X → X is a single-valued mapping, then Tx is naturally compact for all x ∈ X. Hence both of Theorems 3 and 4 are still valid for a single-valued mapping.

Theorem 6. Let (X, d, ≺) be a complete partiallly ordered cone metric space over a total order minihedral and continuous cone P of a normed vector space E. Let φ : X → E be a sequentially continuous vector functional and let T : X → X be a single-valued mapping. Assume that there exists x₀ ∈ X such that φ is bounded below on [x₀, +∞), T is increasing on [x₀, +∞), and x₀≺Tx₀. Then T has a maximal fixed point x^* and a least fixed point x_* in [x₀, +∞) such that x_*≺x^*.

Proof. By Theorem 3 and Remark 5, T has a maximal fixed point x^* ∈ [x₀, +∞) and hence F = {x ∈ [x₀, +∞) : x = Tx} ≠ ∅. Set

Clearly, [x₀, +∞) ∈ S and hence S ≠ ∅. Define a relation ⊑ on S by for all I₁, I₂ ∈ S, then it is easy to check that ⊑ is a partial order on S.

Let {I_α} _α∈Γ be a decreasing chain of S, where I_α = [x_α, +∞). From (1), (18), and (19) we find that {x_α} _α∈Γ is an increasing chain of M, where

Set $$. Clearly, $$. Following the proof of Theorem 3, there exists $$ and an increasing sequence $$ satisfying (6) such that (8) and (11) are satisfied. From $$ we have that $$ for all x ∈ F and all n. Thus the increasing property of T on [x₀, +∞) implies that, for all x ∈ F and all n, and hence by (1), for all x ∈ F and all n. Let n → ∞, then by (8) and the continuity of φ we have $$; that is, for all x ∈ F. This together with $$ implies $$. Then in analogy to the proof of Theorem 3, by (6), (8), and $$ we can prove {x_α} _α∈Γ has an upper bound $$. By (18), we have $$. Note that $$ is an upper bound of {x_α} _α∈Γ in M, then $$ for all α ∈ Γ and hence by (19), for all α ∈ Γ. This means $$ is a lower bound of {I_α} _α∈Γ in S. By Zorn′s lemma, (S, ⊑) has a minimal element; denote it by I^* = [x_*, +∞). By (18) we have x₀≺x_*≺Tx_* and for all x ∈ F. By the increasing property of T, we have x₀≺x_*≺Tx_*≺T(Tx_*) and Tx_*≺Tx = x for all x ∈ F, which implies [Tx_*, +∞) ∈ S and [Tx_*, +∞)⊆I^*. Moreover by (19), [Tx_*, +∞)⊑I^*. The minimality of I^* in S forces that [Tx_*, +∞) = I^* and so we have x_* = Tx_*. Finally by (25), x_* is a least fixed point of T in [x₀, +∞) and x_*≺x^*. The proof is complete.

Theorem 7. Let (X, d, ≺) be a complete partially ordered cone metric space over a total order minihedral and continuous cone P of a normed vector space E. Let φ : X → E be a sequentially continuous vector functional and let T : X → X be a single-valued mapping. Assume that there exists x₀ ∈ X such that φ is bounded above on (−∞, y₀], T is increasing on (−∞, y₀], and Ty₀≺y₀. Then T has a minimal fixed point x_* and a largest fixed point in x^* in (−∞, x₀] such that x_*≺x^*.

Proof. By Theorem 4 and Remark 5, T has a minimal fixed point in x_* ∈ (−∞, y₀]. Set

Define a relation $$ on $$ as follows: for all $$, then $$ is a partial order on $$. In an analogy to the proof of Theorem 4, we can prove $$ has a minimal element (−∞, x^*] and x^* is a largest fixed point of T in (−∞, y₀]. The proof is complete.

Theorem 8. Let (X, d, ≺) be a complete partially ordered cone metric space over a total order minihedral and continuous cone P of a normed vector space E. Let φ : X → E be a sequentially continuous mapping and let T : X → X be a single-valued mapping. Assume that there exists x₀, y₀ ∈ X with x₀≺y₀ such that T is increasing on [x₀, y₀] and x₀≺Tx₀,  Ty₀≺y₀. Then T has a largest fixed point x^* and a least fixed point x_* in [x₀, y₀] such that x_*≺x^*.

Proof. For all x ∈ [x₀, y₀], by (1) we have φ(y₀)⪯φ(x)≺φ(x₀); that is, φ is bounded on [x₀, y₀]. In an analogy to the proof of Theorem 3, we can prove T has a maximal fixed point and a minimal fixed point in [x₀, y₀] by investigating the existence of maximal element and minimal element, respectively, in D₁ = {x ∈ [x₀, y₀] : x≺Tx} and D₂ = {x ∈ [x₀, y₀] : Tx≺x}. Let

where G = {x ∈ [x₀, y₀] : Tx = x} is nonempty. Define ⊑₁ on S₁ and ⊑₂ on S₂, respectively, by then it is easy to check that ⊑₁ and ⊑₂ are partial orders on S₁ and S₂, respectively. In an analogy to the proof of Theorem 4, we can prove (S₁, ⊑₁) has a minimal element I_* = [x_*, y₀] and (S₂, ⊑₂) has a minimal element J_* = [x₀, y^*]. By the definitions of S₁ and S₂, we have x_*, y^* ∈ [x₀, y₀], Moreover by (30) and the increasing property of T on [x₀, y₀], for all x ∈ G, we have and so by (31), From (32) and (33) we have that [Tx_*, y₀] ∈ S₁, [x₀, Ty^*] ∈ S₂, and which implies [Tx_*, y₀] = I_* and [x₀, Ty^*] = J_* by the minimality of I_* and J_*. This means that Tx_* = x_* and Ty^* = y^*. Hence x_* is the least fixed point and y^* is the largest fixed point of T in [x₀, y₀] by (31). The proof is complete.

Remark 9. Theorems 3–8 are extensions of [4, Theorems 3 and 4] and [2, Theorems 3, 4, and 5] to the case of cone metric spaces. It is worth mentioning that in Theorems 4, 7, and 8, not only the existence of maximal and minimal fixed points but also the existence of largest and least fixed points is obtained. Therefore Theorems 4, 7, and 8 are still new even in the case of metric space and hence they indeed improve [2, Theorems 3, 4, and 6].

Example 10. Let X = {1,2, 3,4}, E = ℝ² with the norm $$ for all u = (u₁, u₂) ∈ ℝ² and $$. Clearly, P is a strongly minihedral and continuous cone of E. Define a mapping d : ℝ × ℝ → P by

then (ℝ, d) is a complete cone metric space over P and hence (X, d) is a complete cone metric subspace of (ℝ, d). Define a vector functional φ : [1, +∞) → E by for all x ∈ [1, +∞). For arbitrary x ∈ [1, +∞), let {x_n}⊆[1, +∞) be a sequence such that $$, then $$ and hence ∥φ(x_n) − φ(x)∥→0, that is, lim _n→∞ φ(x_n) = φ(x). This means that φ : [1, +∞) → E is sequentially continuous; in particular, φ : X → E is sequentially continuous. From (35) and (36) it is easy to check that where ≺ is the partial order defined by (1). Let T : X → 2^X be a set-valued mapping such that Fix x₀ = 2, then [x₀, +∞) = {x ∈ X : 2≺x} = {2,3, 4} by (37), and so Tx₀∩[x₀, +∞) = {3} ≠ ∅. For x, y ∈ [x₀, +∞), if x≺y and x ≠ y, then we have only two cases: x = 2≺3 = y and x = 2≺4 = y by (37). Fix x = 2 and y = 3, for all u ∈ Tx, there exists v = 3,4 ∈ Ty such that u≺v. Fix x = 2 and y = 4, for all u ∈ Tx, there exists v = 3 ∈ Ty such that u≺v. This means that T : X → 2^X is increasing on [x₀, +∞). Therefore all the conditions of Theorem 3 are satisfied and hence T has a fixed point 3 ∈ [x₀, +∞).

Fix x = 4; for all y ∈ T4, we have x = 4⊀y by (37); that is, (2) is not satisfied. Therefore the existence of fixed points could not be obtained by generalized Caristi′s fixed point theorems in cone metric spaces of [1, 3].