Some Common Fixed-Point Theorems for Generalized-Contractive-Type Mappings on Complex-Valued Metric Spaces

Theorem 1 (see [6], Theorem 1.)If S and T are self-mappings defined on a complete complex-valued metric space (X, d) satisfying the condition

for all x, y ∈ X where λ, μ, and γ are nonnegative with λ + μ + γ < 1, then S and T have a unique common fixed point.

Definition 2. Let X be a nonempty set. Suppose that the mapping d : X × X → ℂ satisfies the following conditions:

• (d₁)

  0≾d(x, y) for all x, y ∈ X;

• (d₂)

  d(x, y) = 0 if and only if x = y for all x, y ∈ X;

• (d₃)

  d(x, y) = d(y, x) for all x, y ∈ X;

• (d₄)

  d(x, y)≾d(x, z) + d(z, y) for all x, y, z ∈ X.

Then, d is called a complex-valued metric on X, and (X, d) is called a complex valued metric space.

Definition 3. Let (X, d) be a complex-valued metric space.

• (i)

  A point x ∈ X is called interior point of a set B⊆X whenever there exists 0≺r ∈ ℂ such that N(x, r): = {y ∈ X : d(x, y)≺r}⊆B.

• (ii)

  A point x ∈ X is called limit point of a set B⊆X whenever for every 0≺r ∈ ℂ such that N(x, y)∩(X − B) ≠ ∅.

• (iii)

  A subset B⊆X is called open whenever each element of B is an interior point of B.

• (iv)

  A subset B⊆X is called closed whenever each limit point of B belongs to B.

• (v)

  The family F = {N(x, r) : x ∈ X, 0≺r} is a subbasis for a topology on X. We denote this complex topology by τ_c. Indeed, the topology τ_c is Hausdorff.

Definition 4 (see [2].)Let (X, d) be a complex-valued metric space, and let {x_n} be a sequence in X and x ∈ X.

• (i)

  If for every c ∈ ℂ, with 0≺c there is N ∈ ℕ such that for all n > N, d(x_n, x)≺c, then {x_n} is said to be convergent, {x_n} converges to x and x is limit point of {x_n}. We denote this by x_n → x as n → ∞ or lim _n→∞x_n = x.

• (ii)

  If for every c ∈ ℂ, with 0≺c there is N ∈ ℕ such that for all n > N, d(x_n, x_n+m)≺c, where m ∈ ℕ, then {x_n} is said to be Cauchy sequence.

• (iii)

  If every Cauchy sequence in X is convergent, then (X, d) is said to be a complete complex-valued metric space.

Lemma 5 (see [2].)Let (X, d) be a complex-valued metric space, and let {x_n} be a sequence in X. Then, {x_n} converges to x if and only if |d(x_n, x)| → 0 as n → ∞.

Lemma 6 (see [2].)Let (X, d) be a complex-valued metric space, and let {x_n} be a sequence in X. Then, {x_n} is a Cauchy sequence if and only if |d(x_n, x_n+m)| → 0 as n → ∞, where m ∈ ℕ.

Definition 7. Two families of self-mappings $$ and $$ are said to be pairwise commuting if:

• (i)

  T_iT_j = T_jT_i, i, j ∈ {1,2, …, m}.

• (ii)

  S_iS_j = S_jS_i, i, j ∈ {1,2, …, n}.

• (iii)

  T_iS_j = S_jT_i, i ∈ {1,2, …, m}, j ∈ {1,2, …, n}.

Definition 8. Let S and T be self-mappings of a nonempty set X.

• (i)

  A point x ∈ X is said to be a fixed point of T if Tx = x.

• (ii)

  A point x ∈ X is said to be a common fixed point of T and S if Tx = Sx = x.

Remark 9. We obtain that the following statements hold:

• (i)

  If z₁≾z₂ and z₂≾z₃, then z₁≾z₃.

• (ii)

  If z ∈ ℂ,  a, b ∈ ℝ, and a ≤ b, then az≾bz.

• (iii)

  If 0≾z₁≾z₂, then |z₁ | ≤|z₂|.

Theorem 10. If S and T are self-mappings defined on a complete complex valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where A, B, C, D, and E are nonnegative with A + B + C + 2D + 2E < 1, then S and T have a unique common fixed point.

Proof. Let x₀ be an arbitrary in X. Since S(X)⊆X and T(X)⊆X, we construct the sequence {x_k} in X such that x_2k+1 = Sx_2k and x_2k+2 = Tx_2k+1 for all k ≥ 0. From the definition of {x_k} and (3), we obtain that

Since x_2k+1 = Sx_2k implies that d(x_2k+1, Sx_2k) = 0; therefore, by Remark 9 and |1 + d(x_2k, x_2k+1)|>|d(x_2k, x_2k+1)|, we have From (6) and Definition 2, we have it follows that |d(x_2k+1, x_2k+2)|<((A + D)/(1 − B − D)) | d(x_2k, x_2k+1)|.

Similarly, we get

Since x_2k+2 = Tx_2k+1 implies that d(x_2k+2Tx_2k+1) = 0; therefore, by Remark 9 and |1 + d(x_2k+2, x_2k+1)|>|d(x_2k+2, x_2k+1)|, we have From (10) and Definition 2, we have it follows that |d(x_2k+2, x_2k+3)|<((A + E)/(1 − B − E)) | d(x_2k+1, x_2k+2)|.

Putting k = max {((A + D)/(1 − B − D)), ((A + E)/(1 − B − E))}, we obtain that

Thus, for any n ∈ ℕ, we have it follows that |d(x_n, x_m)|≤(kⁿ/(1 − k)) | d(x₀, x₁)| → 0 as n → ∞.

By Lemma 6, the sequence {x_n} is a Cauchy. Since X is compete, there exists a point z ∈ X such that x_n → z as n → ∞.

Next, we will show that Sz = z. By the notion of a complete complex-valued metric d, we have

which implies that Taking k → ∞, we have |d(z, Sz)| = 0; it is obtained that d(z, Sz) = 0. Thus, Sz = z. It follows that similarly Tz = z. Therefore, z is common fixed point of S and T.

Finally, to prove the uniqueness of common fixed point, let z^* ∈ X be another common fixed point of S and T such that Sz^* = Tz^* = z^*. Consider

so that Since |1 + d(z, z^*)|>|d(z, z^*)|, therefore |d(z, z^*)| < A | d(z, z^*)| + C | d(z^*, z)| = (A + C) | d(z, z^*)|.

This is contradiction to A + C < 1. Hence, z = z^*. Therefore, z is a unique common fixed point of S and T.

Corollary 11. If T is a self-mapping defined on a complete complex-valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where A, B, C, D, and E are nonnegative with A + B + C + 2D + 2E < 1, then T has a unique fixed point.

Proof. We can prove this result by applying Theorem 10 by setting T = S.

Corollary 12. If S and T are self-mappings defined on a complete complex valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where A, B, C, and D are nonnegative with A + B + C + 2D < 1, then S and T have a unique common fixed point.

Proof. We can prove this result by applying Theorem 10 by setting E = 0.

Corollary 13. If T is a self-mapping defined on a complete complex valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where A, B, C, and D are nonnegative with A + B + C + 2D < 1, then T has a unique fixed point.

Proof. We can prove this result by applying Corollary 12 by setting T = S and E = 0.

Corollary 14. If S and T are self-mappings defined on a complete complex valued metric space (X, d) satisfying the condition

for all x, y ∈ X where A, B, C, and E are nonnegative with A + B + C + 2E < 1, then S and T have a unique common fixed point.

Proof. We can prove this result by applying Theorem 10 by setting D = 0.

Corollary 15. If T is a self-mapping defined on a complete complex valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where A, B, C, and E are nonnegative with A + B + C + 2E < 1, then T has a unique fixed point.

Proof. We can prove this result by applying Corollary 14 by setting T = S.

Remark 16. (i) By choosing D = E = 0 in Theorem 10, we get Theorem 1 of [6].

(ii) By choosing D = E = 0 and S = T in Theorem 10, we get Corollary 3 of [6].

(iii) By choosing C = D = E = 0 in Theorem 10, we get Theorem 4 of Azam et al. [2].

(iv) By choosing C = D = E = 0 and S = T in Theorem 10, we get Corollary 5 of Azam et al. [2].

Theorem 17. If $$ and $$ are two finite pairwise commuting finite families of self-mapping defined on complete complex-valued metric space (X, d) such that the mappings T and S with T = T₁T₂ ⋯ T_m and S = S₁S₂ ⋯ S_n satisfy condition (3), then the component maps of the two families $$ and $$ have a unique common fixed point.

Proof. By Theorem 10, one can infer that T and S have a unique common fixed point Z  (i.e.,  Tz = Sz = z). Now, we will show that z is a common fixed point of all the component maps of both families. In view of pairwise commutativity of the families $$ and $$, for every 1 ≤ k ≤ m, we can write

It implies that T_kz (∀k) is also a common fixed point of T and S. By using the uniqueness of common fixed point, we have T_kz = z (∀k). Hence, z is a common fixed point of the family $$. Similarly, we can show that z is a common fixed point of the family $$. This completes the proof of the theorem.

Corollary 18. If F and G are self-mappings defined on a complete complex-valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where A, B, C, D, and E are nonnegative with A + B + C + 2D + 2E < 1, then F and G have a unique common fixed point.

Proof. We can prove this result by applying Theorem 17 by setting T₁ = T₂ = ⋯ = T_m = F and S₁ = S₂ = ⋯ = S_n = G.

Corollary 19. If T is a self-mapping defined on a complete complex valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where A, B, C, D, and E are nonnegative with A + B + C + 2D + 2E < 1, then T has a unique fixed point.

Proof. We can prove this result by applying Corollary 18 by setting F = G = T.

Remark 20. (i) By choosing D = E = 0 in Theorem 17, we get Theorem 1 of [6].

(ii) By choosing D = E = 0 in Corollary 18, we get Corollary 6 of [6].

(iii) By choosing D = E = 0 in Corollary 19, we get Corollary 7 of [6].

(iv) By choosing C = D = E = 0 in Corollary 19, we get Corollary 6 of Azam et al. [2].

Corollary 21 (see [5].)If  T : X → X is a mapping defined on a complete complex-valued metric space (X, d) satisfying the condition

for all x, y ∈ X, where λ is nonnegative reals λ < 1, then T has a unique fixed point.

Example 22. Let X = ℂ be the set of complex numbers. Define d : ℂ × ℂ → ℂ as

where z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂. Then, (ℂ, d) is a complete complex-valued metric space. Define T : ℂ → ℂ as Now, for $$ and y = 0, we get Thus, $$, which is a contradiction as 0 ≤ λ < 1. However, notice that T²z = 0, so that 0 = d(T²z₁, T²z₂)≾λd(z₁, z₂), which shows that T² satisfies the requirement of Bryant theorem and z = 0 is the unique fixed point of T.