Dislocated quasi cone b-metric space over Banach algebra and contraction principles with application to functional equations

Definition 3.1

Let 𝓐 be a Banach algebra, χ^q be a non empty set and d : χ^q × χ^q → 𝓐. For all p, q, r ∈ χ^q, consider the following conditions:

1. θ ⪯ d(p, q) and d(p, q) = d(q, p) = θ implies p = q

2. d(p, q) ⪯ s[d(p, r) + d(r, q)] for some s ∈ P, e ⪯ s.

3. d(p, q) = d(q, p)

If d satisfies conditions (dqCM1) and (dqCM2), then d is a dislocated quasi cone b-metric and (χ^q, d, 𝓐) will be called a dislocated quasi CbMS over Banach algebra 𝓐 (in short dqCbMS-𝓐). If d satisfies (dqCM1), (dqCM2) and (dqCM3) then (χ^q, d, 𝓐) is a dislocated CbMS over Banach algebra 𝓐 (in short dCbMS-𝓐). If (dqCM1) is replaced with θ ⪯ d(p, q) and d(p, q) = d(q, p) = θ if and only if p = q then the above two definitions reduces to quasi CbMS over Banach algebra 𝓐 (in short qCbMS-𝓐) and CbMS over Banach algebra 𝓐 (in short CbMS-𝓐) respectively.

Remark 3.2

In (Definition 2.1, [1]) the authors defined CbMS-𝓐 wherein they have taken s ≥ 1 a scalar. However in our definition of CbMS-𝓐 we take s ∈ P. Note that (Definition 2.1, [1]) implies definition 3.1(i) above in the sense that if d(p, q) ⪯ s[d(p, r) + d(r, q)] for some scalar s ≥ 1 then d(p, q) ⪯ s′[d(p, r) + d(r, q)] with s′ = se ∈ P where e is the identity element of the Banach algebra A.

Remark 3.3

(An open problem) It is not known whether CbMS-𝓐 in the sense of definition 3.1(i) above implies Definition (2.1) of [1].

Remark 3.4

In definition 3.1 above, if we replace the co-domain of the metric function d with a Banach space instead of a Banach algebra then (χ^q, d, 𝓐) will be a dislocated quasi cone b-metric space which is a generalisation of a cone metric space introduced by Huang and Zhang [14]. Many interesting fixed point theorems have been proved in a cone metric space. In recent years using scalarization method, many authors proved the equivalence of a metric space and a cone metric space and showed that the fixed point results in cone metric spaces were the consequences of their usual metric versions (see [9], [15, 16, 17]). Consequently Liu and Xu[11] introduced the concept of a cone metric space over Banach algebra, and proved some fixed point results in such spaces. They showed that thefixed point results in this new setting cannot be derived from their usual metric versions. Thus the cocept of a dislocated quasi cone b-metric space over a Banach algebra becomes more general than that of a dislocated quasi cone b-metric space.

Here after, throughout this paper, by (χ^q, d, 𝓐) and (χ^d, d, 𝓐) we mean a dqCbMS-𝓐 and dCbMS-𝓐 respectively, with coefficient s ∈ P (e ⪯ s).

Example 3.5

Let χ^q = ℝ. Consider the Banach algebra P_n+1 of all polynomials with complex coefficients and degree less than or equal to n, in which for any x(t) = α₀ + α₁ t + α₂t² + ⋯ + α_n tⁿ and y(t) = β₀ + β₁t + β₂t² + ⋯ + β_n tⁿ the norm of x(t) is given by ∥x(t)∥ = ∑i=0n | α_i | and the product by (xy)(t) = ∑i=0n a_kt^k where a_k = ∑_i+j=k α_iβ_j. The unit elemant e = 1 + 0t + 0t² ⋯ + 0tⁿ, zero elemant θ = 0 + 0t + 0t² + ⋯ + 0tⁿ and P = {x(t) ∈ P_n+1 : a_i ≥ 0, i = 0, 1, 2⋯ n} is a non normal cone in P_n+1. Define the function d : χ^q × χ^q → P_n+1 by

clearly (χ^q, d, P_n+1) is a dqCbMS − P_n+1.

Example 3.6

Let χ^d = ℝ⁺ ∪ {0} and consider the Banach algebra 𝓐 of all 3 × 3 matrices over the set ℝ with ∥a∥ = ∑_1≤i,j≤3|a_i,j| and solid cone P of all 3 × 3 matrices over the set ℝ⁺ ∪ {0}. Let d : χ^d × χ^d → 𝓐 be given by

then (χ^d, d, 𝓐) is a dCbMS − 𝓐 with s=200020002.

Proposition 3.7

Let (χ, d, 𝓐) be a qCbMS − 𝓐 (dqCbMS − 𝓐) and for all x, y ∈ χ, define d†(x,y)=d(x,y)+d(y,x)2. Then (χ, d^†, 𝓐) is a CbMS − 𝓐 (dCbMS − 𝓐).

Definition 3.8

Let (χ^q, d, 𝓐) be a dqCbMS − 𝓐, p ∈ χ^q and {p_n} be a sequence in χ^q.

1. {p_n} bi-converges to p if for each ε ∈ 𝓐, with θ ≪ ε, there exist n₀ ∈ ℕ such that d(p_n, p) ≪ ε and d(p, p_n) ≪ ε whenever n ≥ n₀. We write it as Lim_n→∞ p_n = p.

2. {p_n} is a L-Cauchy sequence (R-Cauchy sequence) provided for each ε ∈ 𝓐, with θ ≪ ε, there exist n₀ ∈ ℕ such that d(p_n, p_m) ≪ ε (d(p_m, p_n) ≪ ε) for all n > m ≥ n₀.

3. (χ^q, d, 𝓐) is L-complete (R-complete) dqCbMS provided every L-Cauchy sequence (R-Cauchy sequence) in (χ^q, d, 𝓐) is bi-convergent.

Proposition 3.9

Let (χ^q, d, 𝓐) be a dCbMS-BA over 𝓐, sequence {p_n} in χ^q. If {p_n} converges to p ∈ χ^q then

1. d(p_n, p) is a c -sequence.

2. d(p_n, p_n+r) is a c -sequence.

Proof

Follows from definitions 2.1, 3.1 and 3.8(i).

Proposition 3.10

If (χ^q, d, 𝓐) is L-complete or R-complete dqCbMS − 𝓐, then (χ^q, d^†) is a complete dCbMS − 𝓐.

Proof

We will show that a Cauchy sequence {p_n} in (χ^q, d^†, 𝓐) is always a L − Cauchy sequence and R − Cauchy sequence in (χ^q, d, 𝓐). Let ε ∈ 𝓐 with θ ≪ ε, then θ ≪ ε2 . Then by hypothesis, we can find n₀ ∈ ℕ such that d†(pn,pm)=d(pn,pm)+d(pm,pn)2≪ε2 whenever n, m ≥ n₀, that is d(p_n, p_m) ≪ ε and d(p_m, p_n) ≪ ε for all n, m ≥ n₀ and thus {p_n} is L - Cauchy sequence and R-Cauchy sequence in (χ^q, d, 𝓐). Now if (χ^q, d, 𝓐) is L - complete ( or R - complete), the L - Cauchy sequence (or R - Cauchy sequence) {p_n} bi-converges to some point p in χ^q and thus there exist n₀ ∈ ℕ such that d(p_n, p) ≪ ε and d(p, p_n) ≪ ε whenever n ≥ n₀. Thus d†(pn,p)=d(pn,p)+d(p,pn)2≪ε whenever n ≥ n₀ and hence {p_n} is convergent in (χ^q, d^†, 𝓐). Hence (χ^q, d^†, 𝓐) is a complete dCbMS − 𝓐.

For proving the uniqueness of the fixed point under α-admissible conditions, different hypothesis were used by different authors. In the sequel Popescu [18] considered the following condition:

(K) Whenever x ≠ y ∈ X, we can find w ∈ X satisfying α(x, w) ≥ 1, α(y, w) ≥ 1 and α(w, Tw) ≥ 1.

Definition 3.11

Let (X^d, d) be a dCbMS − 𝓐, T, S : X^d → X^d and α : X^d × X^d → [0, ∞) be mappings and x ∈ X^d. Then the pair (T, S) is

1. α-identical at x iff min{α(Tx, Sx), α(Sx, Tx)} ≥ 1. The pair (T, S) is α-identical on X^d iff (T, S) is α-identical at all x ∈ X^d.

2. Weak semi α-admissible iff x, y ∈ X^d and α(x, y) ≥ 1 ⇒ min{α(x, TSy), α(x, STy)} ≥ 1.

3. α-dominated at x iff min{α(x, Tx), α(x, Sx)} ≥ 1. The pair (T, S) is α-dominated on X^d iff (T, S) is α-dominated at all x ∈ X^d.

4. Satisfies condition (G_α) iff α(x, Tx) ≥ 1 and α(y, Sy) ≥ 1 implies α(x, y) ≥ 1 or α(Tx, Sy) ≥ 1 for any x, y ∈ X^d.

5. Satisfies condition (Gα′) iff whenever x ≠ y ∈ X with α(x, Tx) ≥ 1 and α(y, Sy) ≥ 1 there exists w ∈ X^d satisfying α(x, w) ≥ 1, α(y, w) ≥ 1, α(w, w) ≥ 1, α(w, Tw) ≥ 1 and α(w, Sw) ≥ 1.

6. (X^d, d) is α-regular iff for any sequence {x_p} in X^d with α(x_p, x_p+1) ≥ 1 and x_p → x_* as p → ∞, then α(x_p, x_*) ≥ 1

Example 3.12

Let χ = [0, ∞],

then the pair (T, S) is α-identical, satisfies condition (G_α) and (Gα′) but the pair (T, S) does not satisfy condition (K) and T is not α-dominated.

Example 3.13

Let χ = [−n, n] for some n ∈ ℕ,

then the pair (T, S) is α-identical, satisfies condition (G_α) and (Gα′) but the pair (T, S) does not satisfy condition (K) and T is not α-dominated.

Example 3.14

Let A = [−n, 0], B = [0, n] and χ = A ⋃ B for some n ∈ ℕ,

then α is not triangular and the pair (T, S) is not α-identical but (T, S) is weak semi α-admissible and α-dominted. T does not satisfy condition (G_α) but satisfies condition (Gα′) .

Example 3.15

Let A = [−n, 0), B = (0, n] and χ = A ⋃ {0} ⋃ B for some n ∈ ℕ,

then α is triangular and (T, S) is α-identical but (T, S) is not weak semi α-admissible and not α-dominted. T satisfy conditions (G_α) and (Gα′) but does not satisfy condition (K).

Lemma 3.16

In any non empty set X consider the functions T, S : X → X and α : X × X → [0, ∞). For some x₀ ∈ X consider the sequence {x_n} given by

with α(x₀, x₀) ≥ 1, α(x₀, Tx₀) ≥ 1. Then

1. If α is a triangular function and (T, S) is α-admissible then whenever n ≥ 1 and 0 ≤ p ≤ q ≤ n, we have α(x_p, x_q) ≥ 1.

2. If (T, S) is α-admissible and weak semi α-admissible, then for all n ≥ 1 and 0 ≤ p ≤ q ≤ n, α(x_p, x_q) ≥ 1.

Proof

For proof we will make use of principle of mathematical induction. Let P(n) denote the statement for all 0 ≤ p ≤ q ≤ n, we have α(x_p, x_q) ≥ 1.

1. Since α(x₀, x₀) ≥ 1, α(x₀, x₁) ≥ 1 and (T, S) is α-admissible we have α(x₁, x₁) = α(Tx₀, Tx₀) ≥ 1 and so P(1) holds. Again by α-admissibility of (T, S) we get α(x₁, x₂) = α(Tx₀, Sx₁) ≥ 1, α(x₂, x₂) = α(Sx₁, Sx₁) ≥ 1 and then since α is triangular we get α(x₀, x₂) ≥ 1. Thus P(2) holds. Suppose P(r) holds, i.e. α(x_p, x_q) ≥ 1 for all 0 ≤ p ≤ q ≤ r. We will show that P(r + 1) holds. Its enough to consider the case α(x_p, x_r+1), 0 ≤ p ≤ r + 1. By induction hypothesis since α(x_p, x_r) ≥ 1 for all 0 ≤ p ≤ r using α-admissibility of (T, S) we have α(x_p, x_r+1) ≥ 1 for all 1 ≤ p ≤ r + 1. Thus we have α(x₀, x₁) ≥ 1 and α(x₁, x_r+1) ≥ 1, hence by triangularity of function α we get α(x₀, x_r+1) ≥ 1 and thus P(r + 1) holds.

2. Since α(x₀, x₀) ≥ 1, α(x₀, x₁) ≥ 1 and (T, S) is α-admissible, α(x₁, x₁) = α(Tx₀, Tx₀) ≥ 1 and so P(1) holds. Again by α-admissibility of (T, S) we get α(x₁, x₂) = α(Tx₀, Sx₁) ≥ 1 and α(x₂, x₂) = α(Sx₁, Sx₁) ≥ 1. Since (T, S) is weak semi α-admissible and α(x₀, x₀) ≥ 1 we get α(x₀, x₂) = α(x₀, STx₀) ≥ 1. Thus P(2) holds. Suppose P(r) holds, i.e. α(x_p, x_q) ≥ 1 for all 0 ≤ p ≤ q ≤ r. We will show that P(r + 1) holds. Its enough to consider the case α(x_p, x_r+1), 0 ≤ p ≤ r + 1. By induction hypothesis and α-admissibility of (T, S) we have α(x_p, x_r+1) ≥ 1 for all 1 ≤ p ≤ r + 1. If r is even, then using α(x₀, x₁) ≥ 1 and repeated use of weak semi α-admissibility of T we get α(x₀, x_r+1) ≥ 1. If r is odd then using α(x₀, x₀) ≥ 1 and repeated use of weak semi α-admissibility of T we get α(x₀, x_r+1) ≥ 1. Thus P(r + 1) holds.

Lemma 3.17

[1] For any sequence {p_n} in a CbMS − 𝓐 (χ, d, 𝓐), if we can find α ∈ P with r(α) < 1r(s) , satisfying d(p_n, p_n+1) ⪯ α d(p_n−1, p_n) then {p_n} is a Cauchy sequence.

Lemma 3.18

For any sequence {p_n} in a CbMS − 𝓐 (χ, d, 𝓐), if we can find α ∈ P with r(α) < 1r(s) , satisfying d(p_n, p_n+1) ⪯ K α ⁿ then {p_n} is a Cauchy sequence.

Lemma 3.19

For any sequence {p_n} in (χ^d, d, 𝓐), if we can find β ∈ P with r(β) < 1, satisfying d(p_n, p_n+1) ⪯ β d(p_n−1, p_n) then {p_n} is a Cauchy sequence.

Proof

Let k ∈ N and k>log⁡r(s)−1log⁡r(β). Then we have

K = s^k β^k(e − β)⁻¹ d(p₀, p₁) ∈ 𝓐 and α = β^k. r(α) = r(β^k) ≤ r(β)^k and since k>log⁡r(s)−1log⁡r(β) we have r(β)^k < 1r(s) and hence by Lemma 3.18, sequence {p_kn} is a Cauchy sequence.

Now

Now since r(β) < 1, by lemma 2.5 and 2.6 skβk[nk] (e − β)⁻¹d(p₀, p₁) is a c-sequence and hence by lemma 2.4 skβk[nk] (e − β)⁻¹ d(p₀, p₁) → 0 as n → ∞. Thus we have

Now we have

Now using (3.2) and the fact that sequence {p_kn} is a Cauchy sequence, we conclude that sequence {p_n} is a Cauchy sequence.

Theorem 3.20

Let T, S : χ^d → χ^d. Let α : χ^d × χ^d → [0, ∞) be mappings such that

1. α is a triangular function or (T, S) is weak semi α-admissible.

2. (T, S) is α-admissible.

3. α(x₀, x₀) ≥ 1 and α(x₀, Tx₀) ≥ 1 for some x₀ ∈ χ^d

4. (χ^d, d) is α-regular

5. we can find λ, μ, ν ∈ P such that λ + μ + sν commutes with μ + 3sν, r(λ + μ + sν) + r(μ + 3sν) < 1 and for all x, y ∈ χ^d with α(x, y) ≥ 1 the following conditions are satisfied:

   d(Sx,Ty)⪯λd(x,y)+μ(d(x,Sx)+d(y,Ty))+ν(d(x,Ty)+d(y,Sx)) (3.3)

   d(Tx,Sy)⪯λd(x,y)+μ(d(x,Tx)+d(y,Sy))+ν(d(x,Sy)+d(y,Tx)) (3.4)

then Fix{T, S} ≠ ϕ. Further if (T, S) is α-identical or if (T, S) is α-dominated on χ^d, then d(u, u) = θ for any u ∈ Fix{T, S}.

Proof

Consider the iterative sequence {x_p} defined as in (3.1) and starting with x₀. Let d(x_p, x_p+1) = d_p and d(x_p, x_p) = d_p,p. Note that d_p,p ⪯ 2sd_p−1 and d_p,p ⪯ 2sd_p+1. By Lemma 3.16, α(x_2p, x_2p+1) ≥ 1 and α(x_2p+1, x_2p+2) ≥ 1. Hence using (3.4) we have

i.e

Theorem 3.21

Let T, S : χ^d → χ^q and α : χ^d × χ^d → [0, ∞) be mappings satisfying conditions (i), (ii), (iii), (iv), (v) of Theorem 3.20 and either of the following two conditions

1. T or S satisfy condition (G_α)

2. T or S satisfiy condition (Gα′)

then Fix{T, S} is a singleton set and the iterative sequence (3.1) converges to the unique u ∈ Fix{T, S}.

Proof

From Theorem 3.20, we see that Fix(T, S) ≠ ϕ and the iterative sequence (3.1) converges to some u ∈ Fix{T, S}. Suppose u, w ∈ Fix(T, S). Then again from Theorem 3.20 we have α(u, u) ≥ 1, α(w, w) ≥ 1, d(w, w) = θ and d(w, w) = θ. If T or S satisfy condition (G_α), we have α(u, w) ≥ 1 and then by (3.4)

Corollary 3.22

[Generalised α-admissible Kannan type contraction] Let T, S : χ^d → χ^d and α : χ^d × χ^q → [0, ∞) be mappings such that conditions (i), (ii), (iii), (iv) of Theorem 3.20, (iia) or (iib) of Theorem 3.21 and the following hold:

there exist μ, ∈ P such that μ commutes s, 2r(μ) < 1 and for all x, y ∈ χ^d with α(x, y) ≥ 1

Then the iterative sequence (3.1) converges to a unique common fixed point of S and T.

Corollary 2.32

[Generalised α-admissible Chatterjee type contraction] Let T, S : χ^d → χ^d and α : χ^d × χ^d → [0, ∞) be mappings such that conditions (i), (ii), (iii), (iv) of Theorem 3.20, (iia) or (iib) of Theorem 3.21 and the following hold:

there exist ν ∈ P such that ν commutes s, 4r(sν) < 1 and for all x, y ∈ χ^d with α(x, y) ≥ 1

Then the iterative sequence (3.1) converges to the unique u ∈ Fix{T, S}.

Corollary 3.24

[Generalised α-admissible Riech type contraction] Let T, S : χ^d → χ^d and α : χ^d × χ^d → [0, ∞) be mappings such that conditions (i), (ii), (iii), (iv) of Theorem 3.20, (iia) or (iib) of Theorem 3.21 and the following hold:

there exist λ, μ ∈ P such that ν commutes s, r(λ) + 2r(μ) < 1 and for all x, y ∈ χ^d with α(x, y) ≥ 1

Then the iterative sequence (3.1) converges to the unique u ∈ Fix{T, S}.

Remark 3.25

If α is a symmetric function, that is α(x, y) = α(y, x) for all x, y ∈ χ^q, then we require only either of the conditions (3.3) or (3.4), (3.6) or (3.7), (3.8) or (3.9) and (3.10) or (3.11) in Theorems 3.20, 3.21, 3.22 and 3.23 respectively.

Corollary 3.26

Let T, S : χ^d → χ^d. If there exist λ, μ, ν ∈ P such that s, λ, μ and ν commutes pairwise with each other, r(λ + μ + sν) + r(μ + 3sν) < 1 and for all x, y ∈ χ^d with

Then the iterative sequence (3.1) converges to the unique u ∈ Fix{T, S}.

Proof

The proof follows from Theorems 3.20 and 3.21 by taking α(x, y) = 1 for x, y ∈ χ^q.

Theorem 3.27

Let R, Q : χ^d → χ^d. Let k₁, k₂, k₃, k₄, k₅ ∈ P with r(k₁) + r(sk₂ + sk₃ + sk₄ + sk₅) + r(2sk₄ + 2sk₅) < 1. Suppose that k₁ and sk₄ + sk₅ commutes with sk₂ + sk₃ + sk₄ + sk₅ and f, g : χ^d → χ^d satisfy

for all x, y ∈ χ^d. If R(χ^d) ⊂ Q(χ^d), Q(χ^d) is a complete subspace of χ^d, and (R, Q) are weakly compatible pair of mappings, then Fix{R, Q} is a singleton set.

Proof

Let x₀ ∈ χ^d be an arbitrary point. Since R(χ^d) ⊂ Q(χ^q), there exists an x₁ ∈ χ^q such that Rx₀ = Qx₁. By induction, a sequence Rx_n can be chosen such that Rx_n = Qx_n+1(n = 0, 1, 2, …). Thus, by (3.13), for any natural number n, on one hand, we have

which implies that

On the other hand, we have

which means that

Adding (3.14) and (3.15) we see that

Put k = sk₂ + sk₃ + sk₄ + sk₅, k′ = sk₄ + sk₅. Then we get