A Proposal to the Study of Contractions in Quasi-Metric Spaces

Definition 1. Let X be a nonempty set and let q : X × X → [0, +∞) be a given function which satisfies

• (1)

  q(x, y) = 0 if and only if x = y;

• (2)

  q(x, y) ≤ q(x, z) + q(z, y) for any points x, y, z ∈ X.

Then q is called a quasi-metric and the pair (X, q) is called a quasi-metric space.

Definition 2. Let (X, q) be a quasi-metric space and let {x_n} be a sequence in X and x ∈ X. The sequence {x_n} converges to x if

Remark 3. A convergent sequence in a quasi-metric space has a unique limit.

Remark 4. If {x_n} converges to x in a quasi-metric space (X, q), then

In other words, q is a continuous mapping on its first argument. This property follows from q(x_n, y) ≤ q(x_n, x) + q(x, y) and q(x, y) ≤ q(x, x_n) + q(x_n, y). Therefore,

Definition 5 (see, e.g., [1, 2]). Let (X, q) be a quasi-metric space and let {x_n} be a sequence in X. We say that {x_n} is left-Cauchy if, for every ε > 0, there exists a positive integer N = N(ε) such that q(x_n, x_m) < ε for all n ≥ m > N.

Definition 6 (see, e.g., [1, 2]). Let (X, q) be a quasi-metric space and let {x_n} be a sequence in X. We say that {x_n} is right-Cauchy if, for every ε > 0, there exists a positive integer N = N(ε) such that q(x_n, x_m) < ε for all m ≥ n > N.

Definition 7 (see, e.g., [1, 2]). Let (X, q) be a quasi-metric space and let {x_n} be a sequence in X. We say that {x_n} is Cauchy if, for every ε > 0, there exists a positive integer N = N(ε) such that q(x_n, x_m) < ε for all m, n > N.

Remark 8. A sequence {x_n} in a quasi-metric space is Cauchy if and only if it is left-Cauchy and right-Cauchy.

Definition 9 (see, e.g., [1, 2]). Let (X, q) be a quasi-metric space. We say that

• (1)

  (X, q) is left-complete if each left-Cauchy sequence in X is convergent;

• (2)

  (X, q) is right-complete if each right-Cauchy sequence in X is convergent;

• (3)

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

Definition 10 (see [4].)A simulation function is a mapping $$ satisfying the following conditions:

•  

  ζ₁ ζ(0,0) = 0;

•  

  ζ₂ ζ(t, s) < s − t for all t, s > 0;

•  

  ζ₃ if {t_n}, {s_n} are sequences in (0, ∞) such that lim⁡_n→∞t_n = lim⁡_n→∞s_n > 0 then

  $$ ()

Let $$ be the family of all simulation functions $$.

Definition 11 (Khan et al. [5]). An altering distance function is a continuous, nondecreasing mapping ϕ : [0, ∞) → [0, ∞) such that ϕ⁻¹({0}) = {0}.

Example 12. Let ϕ and ψ be two altering distance functions such that ψ(t) < t ≤ ϕ(t) for all t > 0. Then the mapping

is a simulation function.

Example 13. If φ : [0, ∞) → [0, ∞) is a lower semicontinuous function such that φ⁻¹(0) = {0} and we define $$ by

then ζ_R is a simulation function.

Example 14. If φ : [0, ∞) → [0, ∞) is a continuous function such that φ(t) = 0⇔t = 0 and we define

then ζ_R is a simulation function.

Example 15. Let f, g : [0, ∞)→(0, ∞) be two continuous functions with respect to each variable such that f(t, s) > g(t, s) for all t, s > 0 and define

Then ζ is a simulation function.

Example 16. If φ : [0, ∞) → [0,1) is a function such that $$ for all r > 0 and we define

then ζ_T is a simulation function.

Example 17. If η : [0, ∞) → [0, ∞) is an upper semicontinuous mapping such that η(t) < t for all t > 0 and η(0) = 0 and we define

then ζ_BW is a simulation function.

Example 18. If ϕ : [0, ∞) → [0, ∞) is a function such that $$ exists and $$, for each ε > 0, and we define

then ζ_K is a simulation function.

Example 19. Let h : [0, ∞) × [0, ∞) → [0, ∞) be a function such that h(t, s) < 1 for all t, s > 0 and limsup⁡_n→∞h(t_n, s_n) < 1 provided that {t_n} and {s_n}⊂(0, +∞) are two sequences such that lim⁡_n→∞t_n = lim⁡_n→∞s_n > 0, and we define

and then ζ_U is a simulation function.

Proposition 20. Let $$ be a function such that η(0,0) = 0 and there exists $$ verifying that η(t, s) ≤ ζ(t, s) for all s, t ≥ 0. Then $$.

Proof. For all t, s > 0, η(t, s) ≤ ζ(t, s) < s − t. If {t_n} and {s_n} are sequences in (0, ∞) such that lim⁡_n→∞t_n = lim⁡_n→∞s_n = δ > 0, then limsup⁡_n→∞η(t_n, s_n) ≤ limsup⁡_n→∞ζ(t_n, s_n) < 0.

Proposition 21. Let $$. Then the following statements hold.

• (a)

  For each $$, the function $$ defined by

  $$ ()

•  

  is a simulation function (i.e., $$ for any $$).

• (b)

  For each $$, the function $$ defined by

  $$ ()

•  

  is a simulation function (i.e., $$ for any $$).

Proof. Since $$ for all t, s > 0, the conclusion (a) is a direct consequence of Proposition 20. Next, we prove the conclusion (b). Let $$ be given. It is obvious that $$ for all s, t > 0 because

Let {t_n}, {s_n}⊂(0, +∞) be two sequences such that lim⁡_n→∞t_n = lim⁡_n→∞s_n = δ > 0. For any $$, we have

Definition 22. Let (X, q) be a quasi-metric space. We will say that a self-mapping T : X → X is a $$-contraction if there exists $$ such that

Remark 23. By axiom (ζ₃), it is clear that a simulation function must verify ζ(r, r) < 0 for all r > 0. Consequently, if T is a $$-contraction with respect to $$, then

In other words, if T is a $$-contraction, then it cannot be an isometry.

Lemma 24. If a $$-contraction in a quasi-metric space has a fixed point, then it is unique.

Proof. Let (X, q) be a quasi-metric space and let T : X → X be a $$-contraction with respect to $$. We are reasoning by contradiction. Suppose that there are two distinct fixed points u, v ∈ X of the mapping T. Then q(u, v) > 0. By (18), we have

which is a contradiction due to Remark 23.

Definition 25. We will say that a self-mapping T : X → X on a quasi-metric space (X, q) is

• (i)

  asymptotically right-regular at a point x ∈ X if lim⁡_n→∞q(Tⁿx, Tⁿ⁺¹x) = 0;

• (ii)

  asymptotically left-regular at a point x ∈ X if lim⁡_n→∞q(Tⁿ⁺¹x, Tⁿx) = 0;

• (iii)

  asymptotically regular if it is both asymptotically right-regular and asymptotically left-regular.

Lemma 26. Every $$-contraction on a quasi-metric space is asymptotically regular.

Proof. Let x be an arbitrary point of a quasi-metric space (X, q) and let T : X → X be a $$-contraction with respect to $$. If there exists some $$ such that T^px = T^p−1x, then y = T^p−1x is a fixed point of T; that is, Ty = y. Consequently, we have that Tⁿy = y for all $$, so

for sufficient large $$. Thus, we conclude that Similarly, lim⁡_n→∞q(Tⁿ⁺¹x, Tⁿx) = 0, so T is asymptotically regular at x. On the contrary, suppose that Tⁿx ≠ Tⁿ⁻¹x for all $$; that is, On what follows, from (18) and (ζ₂), we have that, for all $$,

In particular,

The above inequality yields that {q(Tⁿx, Tⁿ⁻¹x)} is a monotonically decreasing sequence of nonnegative real numbers. Thus, there exists r ∈ [0, ∞) such that lim⁡_n→∞q(Tⁿx, Tⁿ⁺¹x) = r ≥ 0. We will prove that r = 0. Suppose, on the contrary, that r > 0. Since T is $$-contraction with respect to $$, by (ζ₃), we have which is a contradiction. Thus, r = 0 and this proves that lim⁡_n→∞q(Tⁿx, Tⁿ⁺¹x) = 0. Hence, T is an asymptotically right-regular mapping at x. Similarly, it can be demonstrated that T is asymptotically left-regular at x.

Remark 27. In the proof of the previous result we have proved that if T : X → X is a $$-contraction on a quasi-metric space (X, q) and {x_n = Tⁿ⁻¹x₁} is a Picard sequence of T, then either there exists $$ such that $$ is a fixed point of T (i.e., $$) or

Lemma 28. Let (X, q) be a quasi-metric space and let T : X → X be a $$-contraction with respect to ζ. If {x_n} is a Picard sequence generated by T, then $$ is bounded.

Proof. Let x₀ ∈ X be arbitrary and let {x_n} be defined iteratively by x_n+1 = Tx_n for all n ≥ 0. If there exists some n ≥ 0 and p ≥ 1 such that x_n+p = x_n, then the set $$ is finite, so it is bounded. Hence, assume that x_n+p ≠ x_n for all n ≥ 0 and p ≥ 1. In this case, by Remark 27, we have that

Notice that by Lemma 26, In particular, there exists $$ such that We will prove that $$ is bounded reasoning by contradiction. We distinguish between right and left boundedness. Suppose that the set is not bounded. Then we can find n₁ > n₀ such that $$. If n₁ is the smallest natural number, greater than n₀, verifying this property, then we can suppose that Again, as D is not bounded, there exists n₂ > n₁ such that Repeating this process, there exists a partial subsequence $$ of {x_n} such that, for all k ≥ 1, Therefore, by the triangular inequality, we have that, for all k, Letting k → ∞ in (35) and using (29) we obtain By (28), we have $$. Therefore using the triangular inequality we obtain Letting k → ∞ and using (29) we obtain Owing to the fact that T is a $$-contraction with respect to $$, we deduce from (ζ₃) that, for all k, which is a contradiction. This proves that D = {q(x_m, x_n) : m > n} is bounded. Similarly, it can be proved that D^′ = {q(x_m, x_n) : m < n} is also bounded. Therefore, the set $$ is bounded.

Theorem 29. Every $$-contraction on a complete quasi-metric space has a unique fixed point. In fact, every Picard sequence converges to its unique fixed point.

Proof. Let (X, q) be a complete quasi-metric space and let T : X → X be a $$-contraction with respect to ζ. Take x₀ ∈ X and consider the Picard sequence {x_n = Tⁿx₀} _n≥0. If {x_n} contains a fixed point of T, the proof is finished. In other case, Lemma 26 and Remark 27 guarantee that

We are going to show that {x_n} is a left Cauchy sequence. For this purpose, taking into account that Lemma 28 guarantees that $$ is bounded, we can consider the sequence {C_n} ⊂ [0, ∞) given by It is clear that the sequence {C_n} is a monotonically nonincreasing sequence of nonnegative real numbers. Therefore, it is convergent; that is, there exists C ≥ 0 such that lim⁡_n→∞C_n = C. Let us show that C = 0 reasoning by contradiction. If C > 0 then, by definition of C_n, for every $$ there exists $$ such that m_k > n_k ≥ k and Hence, By using (40) and the triangular inequality, we have, for all k, Letting k → ∞ in the above inequality and using (41) and (44), we derive that Due to fact that T is a $$-contraction with respect to $$ and by using (ζ₃), (18), (44), and (46), we have which is a contradiction. This contradiction concludes that C = 0 and, hence, {x_n} is a left Cauchy sequence. Similarly, it can be proved that {x_n} is a right Cauchy sequence. Therefore, {x_n} is a Cauchy sequence. Since (X, q) is a complete quasi-metric space, there exists u ∈ X such that lim⁡_n→∞x_n = u.

We will show that the point u is a fixed point of T reasoning by contradiction. Suppose that Tu ≠ u; that is, q(u, Tu) > 0. By Remark 4,

Therefore, there is $$ such that In particular, Tx_n ≠ Tu. This also means that x_n ≠ u for all n ≥ n₀. As q(Tx_n, Tu) > 0 and q(x_n, u) > 0, axiom (ζ₂) and property (18) imply that, for all n ≥ n₀, In particular, 0 ≤ q(Tx_n, Tu) ≤ q(x_n, u) for all n ≥ n₀, which means that Similarly, it can be proved that lim⁡_n→∞q(Tu, x_n+1) = 0. Therefore, {x_n} converges, at the same time, to u and to Tu. By the unicity of the limit, u = Tu, which contradicts Tu ≠ u. As a consequence, u is a fixed point of T. Notice that the uniqueness of the fixed point follows from Lemma 24.

Corollary 30 (see, e.g., [1]). Let (X, q) be a complete quasi-metric space and let T : X → X be a mapping such that

where λ ∈ [0,1). Then T has a unique fixed point in X.

Proof. The result follows from Theorem 29 taking into account that T is a $$-contraction with respect to $$, where ζ_B is defined by ζ_B(t, s) = λs − t for all s, t ∈ [0, ∞) (see (6)).

Example 31. Let α, β, k ∈ (0,1) be such that α ≤ k. Let X = [0,1] and q : X × X → [0, ∞) be a function defined by

Then (X, q) is a complete quasi-metric space (but it is not a metric space). Consider the mapping T : X → X defined as Tx = αx for all x ∈ X. It is clear that it is a $$-contraction with respect to $$, where Indeed, if x ≥ y, then Tx ≥ Ty. Hence, we get that If x < y then Tx < Ty. Hence, we get that Notice that all conditions in Theorem 29 are satisfied and T has a unique fixed point, which is x = 0.

Corollary 32 (Rhoades type). Let (X, q) be a complete quasi-metric space and let T : X → X be a mapping satisfying the following condition:

where φ : [0, ∞) → [0, ∞) is a lower semicontinuous function and φ⁻¹(0) = {0}. Then T has a unique fixed point in X.

Proof. The result follows from Theorem 29 taking into account that T is a $$-contraction with respect to $$, where ζ_R is defined by ζ_R(t, s) = s − φ(s) − t for all s, t ∈ [0, ∞) (see Example 13).

Remark 33. Note that Rhoades assumed in [8] that the function φ was continuous and nondecreasing and it verified lim⁡_t→∞φ(t) = ∞. In Corollary 32, we replace these conditions by the lower semicontinuity of φ, which is a weaker condition. Therefore, our result is stronger than Rhoades’ original version.

Corollary 34. Let (X, q) be a complete quasi-metric space and let T : X → X be a mapping. Suppose that for every x, y ∈ X,

for all x, y ∈ X, where φ : [0, +∞) → [0,1) is a function such that $$ for all r > 0. Then T has a unique fixed point.

Proof. The result follows from Theorem 29 taking into account that T is a $$-contraction with respect to $$, where ζ_T is defined by ζ_T(t, s) = s  φ(s) − t for all s, t ∈ [0, ∞) (see Example 16).

Corollary 35. Let (X, q) be a complete quasi-metric space and let T : X → X be a mapping. Suppose that, for every x, y ∈ X,

for all x, y ∈ X, where η : [0, +∞) → [0, +∞) is an upper semicontinuous mapping such that η(t) < t for all t > 0 and η(0) = 0. Then T has a unique fixed point.

Proof. The result follows from Theorem 29 taking into account that T is a $$-contraction with respect to $$, where ζ_BW is defined by ζ_BW(t, s) = η(s) − t for all s, t ∈ [0, ∞) (see Example 17).

Corollary 36. Let (X, q) be a complete quasi-metric space and let T : X → X be a mapping satisfying the following condition:

where ϕ : [0, ∞) → [0, ∞) is a function such that $$ exists and $$, for each ϵ > 0. Then T has a unique fixed point in X.

Proof. The result follows from Theorem 29 taking into account that T is a $$-contraction with respect to $$, where ζ_K is defined by

(see Example 18).

Corollary 37. Let (X, q) be a complete quasi-metric space and let T : X → X be a mapping satisfying the following condition:

where h : [0, ∞) × [0, ∞) → [0, ∞) is a function such that h(t, s) < 1 and limsup⁡_n→∞h(t_n, s_n) < 1 provided that {t_n} and {s_n}⊂(0, +∞) are two sequences such that lim⁡_n→∞t_n = lim⁡_n→∞s_n. Then T has a unique fixed point in X.

Proof. The result follows from Theorem 29 taking into account that T is a $$-contraction with respect to $$, where ζ_U is defined by ζ_U(t, s) = sh(t, s) − t for all s, t ∈ [0, ∞) (see Example 19).

Example 38. The following example is inspired by Remark 3 in Boyd and Wong [9]. Let X = [0,1]∪{2,3, 4, …} and let us define

It is apparent that (X, q) is a complete quasi-metric space, but it is not a metric space (for instance, q(1,2) ≠ q(2,1)). Let us consider the mappings T : X → X, $$, and $$ defined by Although η is not an upper semicontinuous mapping, it is easy to show that ζ is a simulation function (if {t_n} → δ > 0 and {s_n} → δ, then lim⁡sup⁡_n→∞ζ(t_n, s_n) ≤ max⁡(−δ²/4, −1/2) < 0). Furthermore, it can be proved that Therefore, T is a $$-contraction with respect to ζ. Using Theorem 29, T has a unique fixed point, which is x = 0.

As Boyd and Wong pointed out in [9], as

there can be no decreasing function φ with φ(t) < 1 for t > 0 and for which (58) holds. Furthermore, since there is no increasing function φ with φ(t) < 1 for t > 0 and for which (58) holds.

Example 39. Let X = [0, ∞) (it is also possible to consider X = [0, A], where A > 0) and let us define

It is clear that (X, q) is a complete quasi-metric space, but it is not a metric space since q(1,2) ≠ q(2,1). Let us define T : X → X and $$ by Then $$ and T is a $$-contraction with respect to ζ. Therefore, T has a unique fixed point, which is x = 0.

Definition 40 (Mustafa and Sims [3]). A generalized metric (or a G-metric) on a nonempty set X is a mapping G : X × X × X → [0, ∞) satisfying the following properties for all x, y, z, a ∈ X:

•  

  G₁ G(x, y, z) = 0  if  x = y = z;

•  

  G₂ 0 < G(x, x, y)  for all  x, y ∈ X  with  x ≠ y;

•  

  G₃ G(x, x, y) ≤ G(x, y, z)  for all  x, y, z ∈ X  with  y ≠ z;

•  

  G₄ G(x, y, z) = G(x, z, y) = G(y, z, x) = ⋯ (symmetry in all three variables);

•  

  G₅ G(x, y, z) ≤ G(x, a, a) + G(a, y, z)  (rectangle inequality).

In such a case, the pair (X, G) is called a G-metric space.

Lemma 41. If (X, d) is a metric space and we define G_max⁡, G_sum : X × X × X → [0, +∞), for all x, y, z ∈ X, by

then G_max⁡ and G_sum are G-metrics on X.

Example 42. Let X = [0, ∞). The function G : X × X × X → [0, +∞), defined by

for all x, y, z ∈ X, is a G-metric on X.

Lemma 43. Let (X, G) be a G-metric space and let us define $$, for all x, y ∈ X, by

Then q_G and $$ are quasi-metrics on X and $$ and $$ are metrics on X.

Definition 44. Let (X, G) be a G-metric space, and let {x_n} be a sequence of points of X. We say that {x_n} is G-convergent to x ∈ X if

that is, for any ε > 0, there exists $$ such that G(x, x_n, x_m) < ε, for all n, m ≥ N. We call x the limit of the sequence and write {x_n} → x or lim⁡_n→∞x_n = x.

Proposition 45. If (X, G) is a G-metric space, then the following statements are equivalent:

• (1)

  {x_n} is G-convergent to x;

• (2)

  {G(x_n, x_n, x)} → 0 as n → ∞;

• (3)

  {G(x_n, x, x)} → 0 as n → ∞.

Definition 46. Let (X, G) be a G-metric space. A sequence {x_n} is called a G-Cauchy sequence if, for any ε > 0, there exists $$ such that G(x_n, x_m, x_l) < ε for all m, n, l ≥ N; that is, {G(x_n, x_m, x_l)} → 0 as n, m, l → +∞.

Proposition 47. Let (X, G) be a G-metric space. Then the following are equivalent:

• (1)

  the sequence {x_n} is G-Cauchy,

• (2)

  for any ε > 0, there exists $$ such that G(x_n, x_m, x_m) < ε, for all m, n ≥ N.

Definition 48. A G-metric space (X, G) is called G-complete if every G-Cauchy sequence is G-convergent in (X, G).

Lemma 49 (Agarwal et al. [12]). Let (X, G) be a G-metric space and let us consider the quasi-metrics q_G and $$ as in Lemma 43. Then the following statements hold.

• (1)

  $$ for all x, y ∈ X.

• (2)

  In (X, q_G) and in $$, a sequence is right-convergent (resp., left-convergent) if and only if it is convergent. In such a case, its right-limit, its left-limit, and its limit coincide.

• (3)

  In (X, q_G) and in $$, a sequence is right-Cauchy (resp., left-Cauchy) if and only if it is Cauchy.

• (4)

  In (X, q_G) and in $$, every right-convergent (resp., left-convergent) sequence has a unique right-limit (resp., left-limit).

• (5)

  If {x_n}⊆X and x ∈ X, then $$.

• (6)

  If {x_n}⊆X, then {x_n} is G-Cauchy ⇔  {x_n} is q_G-Cauchy ⇔  {x_n} is $$-Cauchy.

• (7)

  (X, G) is complete ⇔(X, q_G) is complete $$ is complete.

Corollary 50. Let (X, G) be a complete G-metric space and let T : X → X be a mapping such that there exists $$ verifying

Then T has a unique fixed point in X. Furthermore, every Picard sequence generated by T converges to the unique fixed point of T.

Proof. Since (X, G) is complete, then item 7 of Lemma 49 guarantees that (X, q_G) is a complete quasi-metric space, and T is a $$-contraction in (X, q_G) with respect to ζ.

Corollary 51. Let (X, G) be a complete G-metric space and let T : X → X be a mapping satisfying the following condition:

where λ ∈ [0,1). Then T has a unique fixed point in X.

Corollary 52 (see, e.g., [13]). Let (X, G) be a complete G-metric space and let T : X → X be a mapping satisfying the following condition:

where φ : [0, ∞) → [0, ∞) is lower semicontinuous function and φ⁻¹(0) = {0}. Then T has a unique fixed point in X.

Corollary 53 (see, e.g., [14]). Let (X, G) be a complete G-metric space and let T : X → X be a mapping. Suppose that, for every x, y ∈ X,

for all x, y ∈ X, where φ : [0, +∞) → [0,1) is a mapping such that $$, for all r > 0. Then T has a unique fixed point.

Corollary 54 (cf. [15]). Let (X, G) be a complete G-metric space and let T : X → X be a mapping. Suppose that, for every x, y ∈ X,

for all x, y ∈ X, where η : [0, +∞) → [0, +∞) is an upper semicontinuous mapping such that η(t) < t for all t > 0 and η(0) = 0. Then T has a unique fixed point.

Corollary 55. Let (X, G) be a complete G-metric space and let T : X → X be a mapping satisfying the following condition:

where φ : [0, ∞) → [0, ∞) is a function such that $$ exists and $$, for each ϵ > 0. Then T has a unique fixed point in X.