Definition 1. (see [5], [8].)Suppose X is a nonempty set and s ≥ 1 is a constant, the mapping $$ is said to be a cone b-metric if

(d1) θ ≤ d(ϖ, η) for all ϖ, η ∈ X and d(ϖ, η) = θ if and only if ϖ = η

(d2) d(ϖ, η) = d(η, ϖ) for all ϖ, η ∈ X

(d3) d(ϖ, η) ≤ s[d(ϖ, ζ) + d(ζ, η)] for all ϖ, η, ζ ∈ X

The pair (X, d) is called a cone b-metric space over Banach algebra $$.

Definition 2. (see [8].)Suppose (X, d) is a cone b-metric space over Banach algebra $$, ϖ ∈ X, and {ϖ_n} is a sequence in X, we say

• (i)

  {ϖ_n} converges to x if for each $$ with c ≫ θ, there is an integer N ≥ 1 such that d(ϖ_n, ϖ) ≪ c for all n > N

• (ii)

  {ϖ_n} is a Cauchy sequence if for each $$ with c ≫ θ, there is an integer N ≥ 1 such that d(ϖ_n, ϖ_m) ≪ c for all n, m > N

• (iii)

  (X, d) is complete if each Cauchy sequence in X is convergent

Example 3. Take $$ with a norm ‖ϖ₁, ϖ₂‖ = |ϖ₁| + |ϖ₂|. For any ϖ = (ϖ₁, ϖ₂) and η = (η₁, η₂) in $$, set the multiplication as

Let P = {(ϖ₁, ϖ₂) ∈ ℝ² : ϖ₁, ϖ₂ ≥ 0}. Then, $$ is a real Banach algebra owing the unit element e = (1, 0), and the cone $$ is normal. Set X = ℕ ∪ {∞} × ℕ ∪ {∞} and $$ be defined as

In the above definition, (ϖ₁, ϖ₂) is odd if both ϖ₁ and ϖ₂ are odd; (ϖ₁, ϖ₂) is ∞ if both ϖ₁ and ϖ₂ are ∞. We can check that (X, d) is a cone b-metric space over Banach algebra $$ where s = 5/2.

Let ζ_m = (4m − 1, 4m + 1), m ∈ ℕ⁺. We have

which indicates ζ_m⟶∞ but

So, we have showed that the cone b-metric is discontinuous in the case of normal cone.

Definition 4. (see [25].)Suppose P is a solid cone, $$. A sequence {σ_n} ⊂ P is a c-sequence if for any c ≫ θ, there is n₀ ∈ ℕ such that σ_n ≪ c for all n ≥ n₀.

Lemma 5. (see [8].)Suppose P is a solid cone, $$. If α, β ∈ P, {ϖ_n}, and {η_n} are c-sequences in $$, then {αϖ_n + βη_n} is a c-sequence in $$.

Lemma 6. (see [31].)Suppose $$ is a Banach algebra with a unit e and $$, if the spectral radius ρ(ϖ) of ϖ satisfies

then e − ϖ is invertible. Actually, $$

Definition 7. Suppose (X, d) is a cone b-metric space over Banach algebra $$ and Φ : X⟶X, take ϖ ∈ X and O_Φ(ϖ) = {ϖ, Φϖ, Φ²ϖ, Φ³ϖ, ⋯}, namely, the orbit of ϖ under Φ.

Example 8. Suppose $$ with a norm,

For any ϖ = (ϖ₁, ϖ₂) and η = (η₁, η₂) in $$, set the multiplication as

Let $$. It follows that there is a unit element e = (1, 0) in the real Banach algebra $$. Let X = [0, 4] × [0, 4] and define $$ by

for any ϖ = (ϖ₁, ϖ₂), η = (η₁, η₂) ∈ X. It is obvious that (X, d) is complete and s = 2. Suppose Φ : X⟶X is defined as

For any ζ₀ = (ϖ₁, ϖ₂) ∈ X, if ζ_n = Φζ_n−1, n ∈ ℕ, then ζ_n⟶θ implies Φζ_n⟶Φθ = θ while θ = (0, 0). Clearly, Φ is an orbitally continuous mapping rather than continuous. Moreover, notice that Φ is 2-continuous but not continuous, that is, 2-continuity of Φ does not imply continuity of Φ². Furthermore, for each integer k ≥ 2, Φ^k is discontinuous while Φ is k-continuous. This indicates that k-continuity of Φ does not imply continuity of Φ^k in usual situation.

Definition 9. The space (X, d) is named Φ-orbitally complete, if each Cauchy sequence included in O_Φ(ϖ) for some ϖ ∈ X converges in X. Each complete space (X, d) is Φ-orbitally complete for any Φ but not the converse.

Lemma 10. Assume the mapping Φ : X⟶X is a generalized b-quasicontraction in the space (X, d), for each ϖ₀ ∈ X, let ϖ_n = Φϖ_n−1. Then, for any integers i, j ≥ 1, it holds that

Proof. At first, we show that for any n ≥ 1 and 1 ≤ i ≤ n,

If n = 1, the result is trivial. Assume n = 2, i = 1, then

where

Obviously, u(ϖ₀, ϖ₁) ≠ d(ϖ₁, ϖ₂) and u(ϖ₀, ϖ₁) ≠ d(ϖ₁, ϖ₁); otherwise, there is a contradiction.

If u(ϖ₀, ϖ₁) = d(ϖ₀, ϖ₁), then

If u(ϖ₀, ϖ₁) = d(ϖ₀, ϖ₂), then

which implies that d(ϖ₁, ϖ₂) ≤ sr(e − sr)⁻¹d(ϖ₀, ϖ₁). Thus, (13) is true. Now, we suppose for all integers n ≥ 2 and 1 ≤ i ≤ n,

We have to prove

for n + 1 ≥ 2, 1 ≤ i ≤ n + 1. Now, we prove

By (10), we obtain

where

If u(ϖ₀, ϖ_n) = d(ϖ₀, ϖ_n), then

If u(ϖ₀, ϖ_n) = d(ϖ₀, ϖ₁), then d(ϖ₁, ϖ_n+1) ≤ rd(ϖ₀, ϖ₁) ≤ sr(e − sr)⁻¹d(ϖ₀, ϖ₁).

If u(ϖ₀, ϖ_n) = d(ϖ₀, ϖ_n+1), then

which yields (e − sr)d(ϖ₁, ϖ_n+1) ≤ srd(ϖ₀, ϖ₁). That is, d(ϖ₁, ϖ_n+1) ≤ sr(e − sr)⁻¹d(ϖ₀, ϖ₁).

If u(ϖ₀, ϖ_n) = d(ϖ₁, ϖ_n), then

At last, we only check that (20) is true when u(ϖ₀, ϖ_n) = d(ϖ_n, ϖ_n+1). That is,

where

Clearly, u(ϖ_n−1, ϖ_n) ≠ d(ϖ_n, ϖ_n+1) and u(ϖ_n−1, ϖ_n) ≠ d(ϖ_n, ϖ_n). If u(ϖ_n−1, ϖ_n) = d(ϖ_n−1, ϖ_n), then

If u(ϖ_n−1, ϖ_n) = d(ϖ_n−1, ϖ_n+1), then

where

Similarly, we also have u(ϖ_n−2, ϖ_n) ≠ d(ϖ_n, ϖ_n+1). If u(ϖ_n−2, ϖ_n) equals to one of d(ϖ_n−2, ϖ_n), d(ϖ_n−2, ϖ_n−1) and d(ϖ_n−1, ϖ_n), then by the assumption (18), we have

It remains to check (20) when u(ϖ_n−1, ϖ_n) = d(ϖ_n−2, ϖ_n+1); that is,

By a similar analysis, we can deduce that

where

Since u(ϖ₁, ϖ_n) ≠ d(ϖ_n, ϖ_n+1) and u(ϖ₁, ϖ_n) ≠ d(ϖ₁, ϖ_n+1), we know u(ϖ₁, ϖ_n) equals to one of d(ϖ₁, ϖ_n), d(ϖ₁, ϖ₂) and d(ϖ_n, ϖ₂). Therefore, we finally obtain

or or

Hence, (20) is always true. Moreover, by (18), (20), and (33), we know

For 2 ≤ i ≤ n + 1, we have

while

If u(ϖ_i−1, ϖ_n) equals to one of d(ϖ_i−1, ϖ_n), d(ϖ_i−1, ϖ_i), d(ϖ_n, ϖ_i) and d(ϖ_n, ϖ_n+1), then by (18) and (38),

If u(ϖ_i−1, ϖ_n) = d(ϖ_i−1, ϖ_n+1), by (18) and (20), (33), and (38), we conclude that

Therefore, (19) is true. The proof is finished.

Theorem 11. Suppose Φ : X⟶X is a generalized b-quasicontraction mapping in the Φ-orbitally complete space (X, d), if ρ(r) < 1/s, then the mapping Φ possesses one and only one fixed point ζ ∈ X, and the sequence {Φⁿϖ₀} converges to ζ for each ϖ₀ ∈ X.

Proof. For each ϖ₀ ∈ X, take ϖ_n = Φϖ_n−1. If there is some n ∈ ℕ such that ϖ_n = ϖ_n+1 = Φϖ_n, then ϖ_n is the fixed point. Hence, we assume ϖ_n ≠ ϖ_n+1 for all n ∈ ℕ. Let us show that {ϖ_n} is a Cauchy sequence. For any n > m > 1, write E(m, n) = {d(ϖ_i, ϖ_j): m ≤ i < j ≤ n}. From the concept of generalized b-quasicontraction, for any u ∈ E(m, n), there exits v ∈ E(m − 1, n) satisfying u ≤ rv. Thus, it follows that

where u_m−1 ∈ E(m − 1, n), u_m−2 ∈ E(m − 2, n), ⋯, u₁ ∈ E(1, n). The last inequality is obtained by Lemma 10. Since ‖sr^m(e − sr)⁻¹d(ϖ₀, ϖ₁)‖⟶0 as m⟶∞(‖r^m‖⟶0 as m⟶∞). Therefore, for any $$ with c ≫ θ, there exists an integer N ≥ 1 satisfying

That is, {ϖ_n} is a Cauchy sequence. By the Φ-orbital completeness of (X, d), we have ζ ∈ X such that ϖ_n⟶ζ as n⟶∞. Next, we check Φζ = ζ. By (10), we see

where

There are the following three cases:

Case 1. If u(ϖ_n, ζ) equals to one of d(ϖ_n, ζ), d(ϖ_n, ϖ_n+1) and d(ζ, Φϖ_n), then d(ϖ_n+1, Φζ) is a c-sequence.

Case 2. If u(ϖ_n, ζ) = d(ζ, Φζ), we get

that is, d(ϖ_n+1, Φζ) ≤ sr(e − sr)⁻¹d(ϖ_n+1, ζ). Thus, d(ϖ_n+1, Φζ) is a c-sequence.

Case 3. If u(ϖ_n, ζ) = d(ϖ_n, Φζ), we gain

that is, d(ϖ_n+1, Φζ) ≤ sr(e − sr)⁻¹d(ϖ_n, ϖ_n+1). Then, d(ϖ_n+1, Φζ) is a c-sequence.

In summary, d(ϖ_n+1, Φζ) is always a c-sequence. This gives ϖ_n⟶Φζ as n⟶∞. According to the uniqueness of the limit, we know ζ = Φζ.

It remains to prove the uniqueness of ζ. We assume there exists another fixed point $$ such that $$, and then

where

It is a contradiction. In conclusion, the fixed point ζ is unique, and the sequence {Φⁿϖ₀} converges to ζ. The proof is finished.

Corollary 12. Suppose T : X⟶X is a generalized quasicontraction mapping in Φ-orbitally complete cone metric space over Banach algebra $$ with a unit e, if ρ(r) < 1, then the mapping Φ possesses one and only one fixed point ζ ∈ X, and the sequence {Φⁿϖ₀} converges to ζ for every ϖ₀ ∈ X.

Remark 13. Theorem 11 greatly improves Theorem 3.1 in [20], while Theorem 3.1 in [20] depends strongly on the continuity of b-metric. It also generalizes the condition sh < 1 (i.e. h < 1/s) of Theorem 2.13 in [23] to ρ(r) < 1/s. The assumption of completeness in Theorem 3.1 of [11] and Theorem 2.13 in [23] is relaxed by Φ-orbital completeness. Corollary 12 mainly improves and generalizes Theorem 9 in [28], while the results rely on the conditions that the cone is normal and the d is continuous.

Lemma 14. Suppose {ϖ_n} is a sequence in (X, d) satisfying

for some r ∈ P with ρ(r) < 1 and n ∈ ℕ. Then, {ϖ_n} is a Cauchy sequence in (X, d).

Theorem 15. Suppose the space (X, d) is Φ-orbitally complete, Assume the mapping Φ : X⟶X satisfies

for all ϖ, η ∈ X and ρ(r) ∈ [(1/s), 1), where

If Φ is k-continuous for some k ≥ 1 or orbitally continuous, then the mapping Φ possesses one and only one fixed point ζ ∈ X, and the sequence {Φⁿϖ₀} converges to ζ for every ϖ₀ ∈ X.

Proof. From Theorem 11, we obtain a sequence {ϖ_n} by ϖ_n+1 = Φϖ_n and suppose ϖ_n ≠ ϖ_n+1 for all n ∈ ℕ and n ≥ 0. In view of (52), we see that

where

We immediately get u(ϖ_n−1, ϖ_n) ≠ d(ϖ_n, ϖ_n+1) and u(ϖ_n−1, ϖ_n) ≠ θ. If u(ϖ_n−1, ϖ_n) = d(ϖ_n−1, ϖ_n), then

If u(ϖ_n−1, ϖ_n) = d(ϖ_n−1, ϖ_n+1)/2s, then

which gives

In fact,

So, the conditions of Lemma 14 are satisfied for each case. By Lemma 14, {ϖ_n} is a Cauchy sequence. In view of Φ-orbital completeness of (X, d), we have ζ ∈ X such that ϖ_n⟶ζ.

We are now in a position to show that Φζ = ζ. If Φ is k-continuous, then Φ^kϖ_n⟶Φζ by k-continuity of Φ and Φ^k−1ϖ_n⟶ζ as n⟶∞. According to the uniqueness of the limit, we have Φζ = ζ.

If Φ is orbitally continuous, then Φϖ_n⟶Φζ due to the fact that ϖ_n⟶ζ. This yields ζ = Φζ.

Uniqueness of the fixed point follows immediately from (52).

Theorem 16. Suppose (X, d) is the same as in Theorem 15, assume the mapping Φ : X⟶X that satisfies

for all ϖ, η ∈ X and ρ(r) ∈ [(1/s), 1), where

Then, the mapping Φ possesses one and only one fixed point ζ ∈ X, and the sequence {Φⁿϖ₀} converges to ζ for every ϖ₀ ∈ X.

Proof. The proof is analogous to Theorem 15. We at first gain a sequence ϖ_n = Φϖ_n−1 = Φⁿϖ₀, n ≥ 1 and suppose that ϖ_n ≠ ϖ_n+1, ∀n ∈ ℕ. By an analogous analysis with Theorem 15, ϖ_n⟶ζ for some ζ ∈ X. We proceed to show that ζ = Φζ. By (60), we see that

where

There are the following three cases.

Case 1. If u(ϖ_n−1, ζ) equals to one of d(ϖ_n−1, ζ), d(ϖ_n−1, ϖ_n) and d(ζ, ϖ_n), then {d(ϖ_n, Φζ)} is a c-sequence.

Case 2. If u(ϖ_n−1, ζ) = d(ζ, Φζ)/s, we have

that is, d(ϖ_n, Φζ) ≤ r(e − r)⁻¹d(ϖ_n, ζ). Thus, {d(ϖ_n, Φζ)} is a c-sequence.

Case 3. If u(ϖ_n−1, ζ) = d(ϖ_n−1, Φζ)/2s, we have

that is, d(ϖ_n, Φζ) ≤ r(2e − r)⁻¹d(ϖ_n−1, ϖ_n). Thus, {d(ϖ_n, Φζ)} is a c-sequence.

In summary, we always deduce that {d(ϖ_n, Φζ)} is a c-sequence. This gives ϖ_n⟶Φζ as n⟶∞. Since the limit is unique, we know ζ = Φζ. The remaining proof is analogous to Theorem 11.

Remark 17. In Theorem 15, we complement and perfect Theorem 11 in [8], which obtained the conclusions under the condition ρ(r) < 1/s in complete spaces (X, d). Moreover, the conditions in Theorem 15 are much weaker than Theorem 2.1 in [20], since we obtain the results by k-continuity for some k ≥ 1 or orbital continuity of the mapping, without appealing the continuity of cone b-metric or the mapping Φ.

According to the proof of Theorem 16 and the symmetry of the cone b-metric d, we see at once that (61) can be replaced by

for all ϖ, η ∈ X.

Example 18. Set $$ with a norm $$, for any $$. The multiplication in $$ is taken as the pointwise multiplication. We conclude that $$ is a Banach algebra owing the unit element e = 1. Let X = [0, 2). For all ϖ, η ∈ X, define

where $$. Note that the cone P is nonnormal, and the cone b-metric d is discontinuous. Indeed, let ϖ_n = 1 + (1/n), n ∈ ℕ, and then

So, ϖ_n⟶1. However,

that is,

Thus, the cone b-metric d is discontinuous.

The mapping Φ : X⟶X is defined as

It suffices to show that Φ is orbitally continuous rather than continuous (one also can check that Φ is k-continuous for each integer k ≥ 2 but Φ^k is discontinuous for each k ≥ 1). In addition, (X, d) is Φ-orbitally complete but not complete cone b-metric space over Banach algebra $$ with the coefficient s = 4. In fact, for ϖ_n = 2 − 1/n, n ∈ ℕ, we get

but there exists no ϖ ∈ X satisfying d(ϖ_n, ϖ)⟶0. Hence, (X, d) is not complete. Take r(z) = (z/4) + 1/2. We can calculate that ρ(r) = 3/4 ∈ [(1/s), 1). Now, there are the following three cases to verify the inequality (52).

For all ϖ, η ∈ [0, 1], we get d(Φϖ, Φη)(z) = |ϖ/3 − η/3|²ψ and

Then, (52) holds by taking u(ϖ, η)(z) = |ϖ − η|²ψ.

For all ϖ ∈ [0, 1], η ∈ (1, 2), we gain d(Φϖ, Φη)(z) = |ϖ/3 − 0|²ψ = (ϖ²/9)ψ and

Then, (52) holds by taking u(ϖ, η)(z) = (4ϖ²/9)ψ.

For all ϖ, η ∈ (1, 2), then Φϖ = Φη = 0. We observe that d(Φϖ, Φη)(z) = 0 and

So, (52) holds trivially. In the same manner, we can prove that (52) is true for all ϖ ∈ (1, 2), η ∈ [0, 1]. Therefore, Φ possesses a unique fixed point 0 ∈ X, and the sequence {Φⁿϖ} converges to 0 for each ϖ ∈ X by Theorem 15.

Furthermore, there is no r ∈ P with ρ(r) ∈ [0, 1) satisfying (ϖ²/9)ψ ≤ 2r|ϖ − η|²ψ for

in Case 2, which means that Φ is not a Banach-type b-contraction. The work from b-metric space, cone b-metric space, or cone b-metric space over Banach algebra which requires completeness, continuity, or Banach-type b-contraction is not applicable here (see [6, 8, 11, 18–20, 23, 26]). Due to the continuity of metric and cone metric, the corresponding theorems from such spaces (see [1–4, 7, 22, 24, 28]) cannot be used in this example, either.

Definition 19. The mapping Φ : X⟶X is named a Ćirić-type b-quasicontraction in (X, d), if for all ϖ, η ∈ X with ϖ ≠ η, we have

where

Definition 20. Suppose the mapping Φ : X⟶X, the set X is Φ-orbitally compact, if each sequence in O_Φ(ϖ) has a convergent subsequence for all ϖ ∈ X. Clearly, every Φ-orbitally compact cone b-metric space over Banach algebra does not need to be complete.

Example 21. Suppose the Banach algebra $$ and cone P are the same as in Example 8, take X = [0, 2) × [0, 2) and Φ : X⟶X be given by Φ(ϖ₁, ϖ₂) = ((ϖ₁/3), (ϖ₂/5)). Then, X is Φ-orbitally compact rather than complete.

Theorem 22. Suppose (X, d) is Φ-orbitally compact, if Φ : X⟶X is a Ćirić-type b-quasicontraction and orbitally continuous, then the mapping Φ possesses one and only one fixed point ζ ∈ X.

Proof. For any ϖ₀ ∈ X, set ϖ_n = Φϖ_n−1 = Tⁿϖ₀, n ≥ 1. Note that ϖ_n ≠ ϖ_n+1 for all n ∈ ℕ. In fact, if ϖ_n = ϖ_n+1 = Φϖ_n for some n ∈ ℕ, then ϖ_n is a fixed point of Φ. Let l_n = d(ϖ_n, ϖ_n+1) for every n ∈ ℕ. From (77), we know

where

Indeed, u(ϖ_n−1, ϖ_n) ≠ d(ϖ_n, ϖ_n+1); so, it remains the following two cases.

Case 1. When u(ϖ_n−1, ϖ_n) = d(ϖ_n−1, ϖ_n), then d(ϖ_n, ϖ_n+1) < d(ϖ_n−1, ϖ_n).

Case 2. When u(ϖ_n−1, ϖ_n) = d(ϖ_n−1, ϖ_n+1) + d(ϖ_n, ϖ_n)/2s, then

which means d(ϖ_n, ϖ_n+1) < d(ϖ_n−1, ϖ_n). Thus, by repeating the above process, we deduce that

By the regularity of the cone, there exists $$ such that l_n⟶c₀ as n⟶∞. Because X is Φ-orbitally compact, we have a convergent subsequence $$ of {ϖ_n} and a point ζ ∈ X satisfying $$, according to orbital continuity of Φ, $$.

When c₀ > θ, we gain

Moreover, since the cone is regular, we see that

where

It is evidence to get

a contradiction. So, c₀ = θ and ζ = Φζ, namely, ζ is a fixed point.

Finally, let us prove that the fixed point is unique by (77). Otherwise, if there is another fixed point η, then

where

If u(η, ζ) = d(η, ζ) or u(η, ζ) = θ, then this is a contradiction. If u(η, ζ) = [d(η, Φζ) + d(ζ, Φη)]/2s, then

a contradiction too. The result follows.