SanchezAlvarezAGT.dvi @ Applied General Topology c© Universidad Politécnica de Valencia Volume 6, No. 2, 2005 pp. 217-228 On semi-Lipschitz functions with values in a quasi-normed linear space José Manuel Sánchez-Álvarez ∗ Abstract. In a recent paper, S. Romaguera and M. Sanchis dis- cussed several properties of semi-Lipschitz real valued functions. In this paper we analyze the structure of the space of semi-Lipschitz functions that are valued in a quasi-normed linear space. Our approach is moti- vated, in part, by the fact that this structure can be applied to study some processes in the theory of complexity spaces. 2000 AMS Classification: 54E25, 54E50, 54C35. Keywords: Semi-Lipschitz function, normed cone, bicomplete space, quasi- distance, right K-complete. 1. Introduction and preliminaries Motivated, in part, by some problems from computer science and their appli- cations (see for instance [5, 6, 13, 15, 16, 17]), the theories of completeness have received a certain attention in the recent years (see, among other contributions, [1, 2, 3, 9, 10, 18, 19]). These advances have also permitted the development of generalizations, to the nonsymmetric case, of classical mathematical theories: hyperspaces, function spaces, etc. The complexity quasi-metric space was introduced in [16] to study comple- xity analysis of programs. Recently, it was introduced in [14] the dual com- plexity space. Several quasi-metric properties of the complexity space were obtained via the analysis of the dual complexity space. In [15] Romaguera and Schellekens show that the structure of a quasi-normed semilinear space provides a suitable setting to carry out an analysis of the dual complexity space. This paper is a contribution to the study of semi-Lipschitz functions from a nonsymmetric point of view. We show that this set defined on a quasi- metric space, that are valued in a quasi-normed linear space and that vanish ∗The author is supported by a grant FPI from the Spanish Ministry of Education and Science. 218 J. M. Sánchez-Álvarez at a fixed point can be endowed with the structure of a quasi-normed linear space. We show that this space is bicomplete and we also study other types of completeness. Throughout this paper the letters R+ and N will denote the set of nonneg- ative real numbers and the set of positive integers numbers, respectively. Our basic reference for quasi-metric spaces is [4]. A quasi-metric on a (nonempty) set X is a function d : X × X → R+ such that for all x, y, z ∈ X: (i) d(x, y) = d(y, x) = 0 ⇔ x = y, and (ii) d(x, y) ≤ d(x, z) + d(z, y). If d can take the value ∞ then it is called a quasi-distance on X. Given a quasi-metric d on X, the function d−1 defined on X×X by d−1(x, y) = d(y, x), is also a quasi-metric on X, called the conjugate of d, and the function ds defined on X × X by ds(x, y) = d(x, y) ∨ d−1(x, y), is a metric on X. If d is a quasi-distance, then d−1 and ds are a quasi-distance and a distance on X, respectively. A quasi-metric space is a pair (X, d) such that X is a (nonempty) set and d is a quasi-metric on X. Each quasi-distance d on X induces a topology T (d) on X which has as a base the family of balls {Bd(x, r) : x ∈ X, r > 0} where Bd(x, r) = {y ∈ X : d(x, y) < r}. We remark that the topology T (d) is T0. Moreover, if condition (i) above is replaced by (i′) d(x, y) = 0 ⇔ x = y, then T (d) is a T1 topology. A quasi-metric d is said to be bicomplete if d s is a complete metric. For more information about quasi-metric spaces see [4] and [8]. Following [7], a cone is a triple (X, +, ·) such that (X, +) is an abelian semigroup with neutral element 0 and · is a function from R+ × X into X which satisfies for all a, b ∈ R+ and x, y ∈ X: (i) a·(b·x) = (ab)·x, (ii) (a+b)·x = (a·x)+(b·x), (iii) a·(x+y) = (a·x)+(a·y) and (iv) 1 · x = x. A quasi-norm on a cone (X, +, ·) is a function ‖ · ‖ : X → R+ such that for all x, y ∈ X and r ∈ R+: (i) x = 0 if and only if there is −x ∈ X and ‖x‖ = 0 = ‖ − x‖, (ii) ‖r · x‖ = r‖x‖, and (iii) ‖x + y‖ ≤ ‖x‖ + ‖y‖. If the quasi-norm q satisfies: (i′) ‖x‖ = 0 if and only if x = 0, then q is called a norm on the cone (X, +, ·). A (quasi-)normed cone is a pair (X, ‖ · ‖) such that X is a cone and ‖ · ‖ is a (quasi-)norm on X. If (X, +, ·) is a linear space and ‖ · ‖ is a quasi-norm on X, then the pair (X, ‖ · ‖) is called a quasi-normed linear space. Note that in this case, the function ‖ · ‖−1 : X → R+ given by ‖x‖−1 = ‖ − x‖ is also a quasi-norm on X and the function ‖ · ‖s : X → R+ given by ‖x‖s = ‖x‖ ∨ ‖x‖−1 is a norm on X. 2. On the structure of the set of semi-Lipschitz functions Let (X, d), (Y, q) be a quasi-metric space and a quasi-normed space respec- tively. A function f : X −→ Y is said to be a semi-Lipschitz function if there exists k ≥ 0 such that q(f(x)−f(y)) ≤ kd(x, y) for all x, y ∈ X. The number k is called a semi-Lipschitz constant for f. On semi-Lipschitz functions with values in a quasi-normed linear space 219 A function f on a quasi-metric space (X, d) with values in a quasi-normed linear space (Y, q) is called ≤(d,q)-increasing if q(f(x) − f(y)) = 0 whenever d(x, y) = 0. By Y X (d,q) we shall denote the set of all ≤(d,q)-increasing functions from (X, d) to (Y, q). It is clear that if (X, d) is a T1 quasi-metric space, then every function from X to Y is ≤(d,q) -increasing. If for each f, g ∈ Y X (d,q) and a ∈ R + we define f + g and af in the usual way, then it is a routine to show that (Y X(d,q), +, ·) is a cone. Example 2.1. Let X = Z3. Let d be the quasi-metric on X given by d(x, y) = { 1 if x > y, 0 if x ≤ y. Let Y = R, q(x) = x ∨ 0 and take f such that f(0) = 0, f(1) = 1 and f(−1) = −2. It is easy to see that f ∈ Y X (d,q) but −f /∈ Y X (d,q) . Thus Y X (d,q) is not a linear space. A simple but interesting example of a semi-Lipschitz function is the follo- wing: Example 2.2. Let (N, d) be a quasi-metric space where: d(x, y) = { 1 if y > x, 0 if y ≤ x. Then, the dual complexity space, is the quasi-normed space (B∗, q), with B ∗ = {f : ω → R/ ∞ ∑ n=0 2−n(f(n) ∨ 0) < ∞} and q(f) = ∞ ∑ n=0 2−n(f(n) ∨ 0). Let now F : (N, d) → (B∗, q) be the function defined by: F(0) = 0, F(n) = fn such that n < m implies fn > fm, where the order is given by fn > fm if and only if fn(x) > fm(x) for all x ∈ ω. Clearly F is a semi-Lipschitz function. Given a quasi-metric space (X, d) and quasi-normed space (Y, q), fix x0 ∈ X and put SL0(d, q) = {f ∈ Y X (d,q) : sup d(x,y) 6=0 q(f(x) − f(y)) d(x, y) < ∞ , f(x0) = 0}. Then SL0(d, q) is exactly the set of all semi-Lipschitz functions that vanishes at x0, and it is clear that (SL0(d, q), +, ·) is a subcone of (Y X (d,q) , +, ·). Now let ρ(d,q) : SL0(d, q) × SL0(d, q) −→ [0, ∞] defined by ρ(d,q)(f, g) = sup d(x,y) 6=0 q((f − g)(x) − (f − g)(y)) d(x, y) for all f, g ∈ SL0(d, q). Then ρ(d,q) is a quasi-distance on SL0(d, q). However ρ(d,q) is not a quasi-metric in general, as Example 1.1 of [11] shows. 220 J. M. Sánchez-Álvarez Furthermore, it is clear that for each f, g, h ∈ SL0(d, q) and each r > 0, ρ(d,q)(f + h, g + h) = ρ(d,q)(f, g) and ρ(d,q)(rf, rg) = rρ(d,q)(f, g) i.e., ρ(d,q) is an invariant quasi-distance. Moreover, it is easy to check that ρ(d,q)(f, 0) = 0 if and only if f = 0, where by 0 we denote the function that vanishes at every x ∈ X. Moreover, we can see that, by example 2.1, there exists f ∈ SL0(d, q) such that ρ(d,q)(0, f) = 0 but f 6= 0. Consequently, the nonnegative function ‖ · ‖(d,q) defined on SL0(d, q) by ‖f‖(d,q) = ρ(d,q)(f, 0) is a norm on SL0(d, q). Therefore (SL0(d, q), ‖ · ‖(d,q)) is a normed cone. Example 1.1 of [11] provides an instance of a T1 quasi-metric space (X, d) such that (SL0(d, q), +) is not a group for some x0 ∈ X. This example suggests the question of characterizing when (SL0(d, q), +) is a group. In order to give an answer to this question note that if x0 is a fixed point in the quasi-metric space (X, d), then the set SL0(d −1, q) = {f ∈ Y X(d−1,q) : sup d(y,x) 6=0 q(f(x) − f(y)) d(y, x) < ∞ , f(x0) = 0} has also a structure of a cone and (SL0(d −1, q), ‖ · ‖(d−1,q)) is a normed cone, where ‖f‖(d−1,q) = ρ(d−1,q)(f, 0), i.e., ‖f‖(d−1,q) = sup d(y,x) 6=0 q(f(x) − f(y)) d(y, x) for all f ∈ SL0(d −1, q). Proposition 2.3. Let (X,d), (Y ,q) be a quasi-metric space and a quasi- normed space respectively. Then f ∈ SL0(d, q) if and only if −f ∈ SL0(d −1, q). Proof. Let f ∈ SL0(d, q) then there exists k ∈ R + such that q(f(x) − f(y)) ≤ kd(x, y) for all x, y ∈ X. We change x by y hence q(f(y) − f(x)) ≤ kd(y, x) and q(−f(x)−(−f(y))) ≤ kd−1(x, y) then −f ∈ SL0(d −1, q). The converse is analogous. � Corollary 2.4. Let (X,d), (Y ,q) be a quasi-metric space and a quasi-normed space respectively.Then (SL0(d, q) ∩ SL0(d −1, q), +, ·) is a linear space. Proof. It follows from Proposition 2.3 that f ∈ SL0(d, q) ∩ SL0(d −1, q) if and only if −f ∈ SL0(d, q) ∩ SL0(d −1, q). � Remark 2.5. Note that for each f ∈ SL0(d, q), ‖f‖(d,q) = ‖ − f‖(d−1,q). Thus the normed cones (SL0(d, q), ‖ · ‖(d,q)) and (SL0(d −1, q), ‖ · ‖(d−1,q)) are isometrically isomorphic by the bijective map φ : SL0(d, q) −→ SL0(d −1, q) defined by φ(f) = −f. Furthermore, we have SL0(d, q) ∩ SL0(d −1, q) = {f ∈ Y X(d,q) ∩ Y X (d−1,q) : On semi-Lipschitz functions with values in a quasi-normed linear space 221 sup d(x,y) 6=0 q(f(x) − f(y)) ∨ q(f(y) − f(x)) d(x, y) < ∞, f(x0) = 0}. Hence (SL0(d, q)∩SL0(d −1, q), ‖ ·‖B) is a normed linear space, where ‖ ·‖B is the norm defined by ‖f‖B = sup d(x,y) 6=0 q(f(x) − f(y)) ∨ q(f(y) − f(x)) d(x, y) , for all f ∈ SL0(d, q) ∩ SL0(d −1, q). Observe that ‖ · ‖B = ‖ · ‖(d,q) ∨ ‖ · ‖(d−1,q) on SL0(d, q) ∩ SL0(d −1, q). The next result, whose proof is very easy, is a characterization that will be useful. Proposition 2.6. f ∈ SL0(d, q) ∩ SL0(d −1, q) if and only if f(x0) = 0 and there exists k ≥ 0 such that qs(f(x) − f(y)) ≤ kd(x, y). Remark 2.7. It is straightforward to see that f : (X, d) −→ (Y, q) belongs to Y X (d,q) ∩ Y X (d−1,q) if and only if f(x) = f(y) whenever d(x, y) = 0. Example 2.8. Let (X, d), (Y, q) be a quasi-metric and a quasi-normed space such that there is x0 ∈ X satisfying d(x, x0) ∧ d(x0, x) = 0 for all x ∈ X. Then SL0(d, q) ∩ SL0(d −1, q) = {0}. Example 2.9. Let X = [0, 1] and let d be the quasi-metric on X given by d(x, y) = y−x if x ≤ y and d(x, y) = 1 otherwise. Clearly T (d) is the restriction of the Sorgenfrey topology to [0, 1]. Let (Y, q) be a quasi-normed space and put x0 = 0. Then, a function f : X −→ Y satisfies f ∈ SL0(d, q) ∩ SL0(d −1, q) if and only if there is k ≥ 0 such that q(f(x) − f(y)) ∨ q(f(y) − f(x)) ≤ k(d(x, y) ∧ d(y, x)) for all x, y ∈ X. Theorem 2.10. Let (X,d), (Y ,q) be a quasi-metric and a quasi-normed space respectively. Then the following assertions are equivalent: (1) SL0(d, q) = SL0(d −1, q). (2) SL0(d, q) is a group. (3) SL0(d −1, q) is a group. (4) SL0(d, q) ⊂ SL0(d −1, q). (5) SL0(d −1, q) ⊂ SL0(d, q). Proof. (1) ⇒ (2) By corollary 2.4 (SL0(d, q) ∩ SL0(d −1, q), +, ·) is a linear space. If SL0(d, q) = SL0(d −1, q) then (SL0(d, q), +) is a group. (2) ⇒ (3) Let f ∈ SL0(d −1, q). By proposition 2.3 −f ∈ SL0(d, q), since SL0(d, q) is a group, f ∈ SL0(d, q), by proposition 2.3 −f ∈ SL0(d −1, q). (3) ⇒ (4) The proof is similar to the proof of (2) ⇒ (3). (4) ⇒ (5) Let f ∈ SL0(d −1, q). Then −f ∈ SL0(d, q) ⊂ SL0(d −1, q) hence −f ∈ SL0(d −1, q).Thus f ∈ SL0(d, q). (5) ⇒ (1) is the same that (4) ⇒ (5). � Proposition 2.11. Let (X,d), (Y ,q) be a quasi-metric and a quasi-normed space respectively. If there exists x0 ∈ X such that SL0(d, q) = SL0(d −1, q), then SL1(d, q) = SL1(d −1, q) for each x1 ∈ X. 222 J. M. Sánchez-Álvarez Proof. Let f ∈ SL1(d, q). Define a function g on X by g(x) = f(x) − f(x0) for all x ∈ X. It easy to check that g ∈ SL0(d, q). Thus, g ∈ SL0(d −1, q). Since g(x) − g(y) = f(x) − f(y) for all x, y ∈ X we obtain that f ∈ SL1(d −1, q). � 3. Completeness properties In this section, we discuss the completeness properties of the semi-Lipschitz function space. The following result allows us to prove that if (Y , q) is a biBanach space then (SL0(d, q) ∩ SL0(d −1, q), ‖ · ‖B) is a Banach space. Theorem 3.1. Let (X,d), (Y ,q) be a quasi-metric and a quasi-normed bi- complete space respectively. Then (SL0(d, q) ∩ SL0(d −1, q), ‖ · ‖B) is a Banach space. Proof. Let {fn} be a Cauchy sequence in (SL0(d, q)∩SL0(d −1, q), ‖·‖B). Then, given ε ≥ 0 there is n0 ∈ N such that (∗) sup d(x,y) 6=0 qs((fn − fm)(x) − (fn − fm)(y)) d(x, y) < ε for all n, m ≥ n0. If x = x0 then fn(x) = 0 for all n ∈ N. Let x 6= x0. We consider the following cases: Case 1. d(x, x0) 6= 0. Then, we deduce from (∗) that given ε d(x,x0) there exists n′0 ∈ N such that if n, m ≥ n ′ 0 then q s(fn(x) − fm(x)) < ε. Therefore, fn(x) is a Cauchy sequence in (Y, q s). Case 2. d(x, x0) = 0. Then from remark 2.7 fn(x) = fn(x0) and fm(x) = fm(x0). Therefore q s(fn(x) − fm(x)) = 0 < ε. Consequently, fn(x) is a Cauchy sequence in (Y, q s) and {fn(x)} converges to an element f(x) in (Y, qs) for all x ∈ X. Moreover, {fn} converges to f in SL0(d, q) ∩ SL0(d −1, q). Indeed, given ε, since {fn(x)} converges to f(x) for all x ∈ X, for each x, y there exists n′ such that if m′ ≥ n′ then qs(f(x) − f′m(x) − (f(y) − f ′ m(y))) d(x, y) < ε 2 . Since {fn} is a Cauchy sequence, we can also find n0 such that if n, m ≥ n0 then qs(fn(x) − fm(x) − (fn(y) − fm(y))) d(x, y) < ε 2 for all x, y ∈ X. Thus we have ε > qs(f(x) − f′m(x) − (f(y) − f ′ m(y))) d(x, y) ≥ qs(f(x) − fn(x) − (f(y) − fn(y))) d(x, y) − qs(f′m(x) − fn(x) − (f ′ m(y) − fn(y))) d(x, y) and hence qs(f(x) − fn(x) − (f(y) − fn(y))) d(x, y) ≤ ε 2 + ε 2 = ε. On semi-Lipschitz functions with values in a quasi-normed linear space 223 Since n0 is independent of x, y, we obtain sup d(x,y) 6=0 qs(f(x) − fn(x) − (f(y) − fn(y))) d(x, y) < ε, for all n ≥ n0. � Corollary 3.2. Let {fn} be a Cauchy sequence in (SL0(d, q) ∩ (SL0(d −1, q), ‖ · ‖B). Then there exists a convergent sequence {kn} in (R, Tu) such that kn is a semi-Lipschitz constant for fn, where Tu is the usual topology. Theorem 3.3. Let (Y, q) be a bi-Banach space. (1) (X, d) is a metric space. (2) SL0(d, q) = SL0(d −1 , q), ‖ · ‖(d,q) = ‖ · ‖(d−1,q). (3) (SL0(d, q), ‖ · ‖(d,q)) is a Banach space. Then: (1) ⇒ (2) ⇔ (3) Proof. (1) ⇒ (2) Trivial. (2) ⇒ (3) Trivial. (3) ⇒ (2) Suppose that (SL0(d, q), ‖ · ‖(d,q)) is a Banach space. Then SL0(d, q) is a group, so SL0(d, q) = SL0(d −1, q). Moreover ‖ · ‖(d,q) is a norm on SL0(d, q), so that ‖f‖(d,q) = ‖ − f‖(d,q) for all f ∈ SL0(d, q). Since −f ∈ SL0(d, q) it follows that ‖ − f‖(d,q) = ‖f‖(d−1,q). We conclude that ‖ · ‖(d,q) = ‖ · ‖(d−1,q) on (SL0(d, q). � To see that in general (3) ⇒ (1) is not true we take Y = 0 for a quasi-metric space that is not a metric space. Corollary 3.4. If q is a bicomplete quasi-norm on Y then ρ(d,q) is a bicomplete quasi-metric in SL0(d, q) ∩ SL0(d −1, q). The following result allow us to prove that if (Y, q) is a bicomplete space then (SL0(d, q), ρ(d,q)) is a bicomplete space: Theorem 3.5. Let (X, d), (Y, q) be a quasi-metric and a quasi-normed bicom- plete spaces respectively. Then (SL0(d, q), ρ(d,q)) is a bicomplete space. Proof. Let {fn} be a Cauchy sequence in (SL0(d, q), ρ(d,q)). Then, given ε ≥ 0 there is n0 ∈ N such that (∗) sup d(x,y) 6=0 q((fn − fm)(x) − (fn − fm)(y)) d(x, y) < ε for all n, m ≥ n0. If x = x0 then fn(x) = 0 for all n ∈ N. Let x 6= x0.We consider the following cases. Case 1. d(x, x0) 6= 0. Then, we deduce from (∗) that given ε d(x,x0) there exists n′0 ∈ N such that if n, m ≥ n ′ 0 then q(fn(x) − fm(x)) < ε and if we 224 J. M. Sánchez-Álvarez change n and m q(fm(x) − fn(x)) < ε. Therefore, fn(x) is a Cauchy sequence in (Y, qs). Case 2. d(x, x0) = 0. Then d(x0, x) 6= 0 so q(fm(x) − fn(x)) < ε and q(fn(x) − fm(x)) < ε. Consequently, fn(x) is a Cauchy sequence in (Y, q s), thus {fn(x)} converges in (Y, qs) and we define f such that {fn(x)} −→ f(x) in (Y, q). Moreover, {fn} converges to f in (SL0(d, q), ρ(d,q)). Indeed, given ε, since fn(x) converges to f(x) for all x ∈ X, for each x, y there exists n′ such that if m′ ≥ n′ then qs(f(x) − fm′(x) − (f(y) − fm′(y))) d(x, y) < ε 2 and since {fn} is a Cauchy sequence, we can also find n0 such that if m ′, n ≥ n0 then qs(fm′(x) − fn(x) − (fm′(y) − fn(y))) d(x, y) < ε 2 for all x, y ∈ X. Thus we have ε > qs(f(x) − fm′(x) − (f(y) − fm′(y))) d(x, y) ≥ qs(f(x) − fn(x) − (f(y) − fn(y))) d(x, y) − qs(f′m(x) − fn(x) − (f ′ m(y) − fn(y))) d(x, y) and hence qs(f(x) − fn(x) − (f(y) − fn(y))) d(x, y) ≤ ε 2 + ε 2 = ε. Since n0 is independent of x, y sup d(x,y) 6=0 qs(f(x) − fn(x) − (f(y) − fn(y))) d(x, y) < ε, for all n ≥ n0. Consequently (SL0(d, q), ρ(d,q)) is a bicomplete space. � Corollary 3.6. Let {fn} be a Cauchy sequence in (SL0(d, q), ‖ · ‖(d,q)) there exists a convergent sequence {kn} in (R, Tu) such that kn is a semi-Lipschitz constant for fn. 4. Another completeness properties In this section, we discuss another completeness properties of the semi- Lipschitz function space. Let us recall that right K-completeness and left K-completeness constitute very useful extensions of the notion of completeness to the nonsymmetric con- text. In fact, they have been successfully applied to different fields from hyper- spaces and function spaces to topological algebra and theoretical computer science. On semi-Lipschitz functions with values in a quasi-normed linear space 225 Let (X, d) be a quasi-metric space. A net {xδ} ⊂ X, δ ∈ Λ, is called left K-Cauchy provided that for each ε > 0 there is δ0 such that d(xδ2, xδ1) < ε for all δ1 ≥ δ2 ≥ δ0, and {xδ} ⊂ X is called right K-Cauchy provided that for each ε > 0 there exists δ0 such that d(xδ1, xδ2 ) < ε for all δ1 ≥ δ2 ≥ δ0. A quasi-metric ρ is called left K-complete (resp. right K-complete) if each left K-Cauchy net (resp. right K-Cauchy net) converges. The following result allow us to prove that if (Y, q) is a biBanach and finite dimensional space then (SL0(d, q) ρ(d,q)) is a right K-complete space: Theorem 4.1. Let (X, d), (Y, q) be a quasi-metric and a quasi-normed bicom- plete finite dimensional space respectively. Then ρ(d,q) is right K-complete. Proof. Let {fδ} be a right K-Cauchy net in (SL0(d, q), ρ(d,q)). Then, given ε ≥ 0 there is δ0 such that (⋆) sup d(x,y) 6=0 q((fδ1 − fδ2)(x) − (fδ1 − fδ2)(y)) d(x, y) < ε for all δ1 ≥ δ2 ≥ δ0. Let x 6= x0.We consider the following cases. Case 1. d(x, x0) 6= 0 and d(x0, x) 6= 0. Then, we deduce from (⋆) that given ε d(x,x0) and ε d(x0,x) respectively there exists δ′0 such that if δ1 ≥ δ2 ≥ δ ′ 0 then qs(fδ1(x) − fδ2(x)) < ε. Therefore{fδ(x)} is a Cauchy net in (Y, q). Case 2. d(x0, x) = 0 and d(x, x0) 6= 0. Then q(fδ1(x)) − q(fδ2(x)) ≤ q(fδ1(x) − fδ2(x)) ≤ εd(x, x0) for all δ1 ≥ δ2 ≥ δ0 and q(−fδ(x)) = 0, since q s(fδ1(x)) ≤ εd(x, x0) + q(fn0(x)) thus {fn(x)} is a bounded net in the finite dimensional space (Y, q), there exists a convergent subnet {fδ′(x)} in (Y, q). Given ε > 0 there exists δ0 ∈ Λ such that if δ1 ≥ δ ′ 2 ≥ δ0 then qs(fδ1(x) − f(x)) = q s(fδ1(x) − fδ′ 2 (x) + fδ′ 2 (x)) ≤ qs(fδ1(x) − fδ′ 2 (x)) + qs(fδ′ 2 (x) − f(x)). Now qs(fδ′ 2 (x)−f(x)) < ε 2 because {fδ′ 1 (x)} is a convergent net. Given δ′1 ≥ δ1, such that fδ′ 1 is in the subnet then q(fδ′ 2 (x) − fδ1(x)) = q(fδ′ 2 (x) − fδ′ 1 (x) + fδ′ 1 (x) − fδ1(x)) ≤ q(fδ′ 2 (x) − fδ′ 1 (x)) + q(fδ′ 1 (x) − fδ1(x)) < ε 2 since {fδ′ 1 (x)} converges and {fδ} is right K-Cauchy. On the other hand q(fδ1(x) − fδ′2(x)) < ε 2 because {fδ2} is a right K-Cauchy net. Thus {fδ(x)} is a convergent net. Case 3. d(x0, x) 6= 0 and d(x, x0) = 0. Then q(−fδ1(x)) ≤ εd(x0, x) + q(−fδ0(x)) and q(fδ(x)) = 0 respectively, since {fδ} is a bounded sequence on the finite dimensional space (Y, q). 226 J. M. Sánchez-Álvarez Consequently {fδ(x)} is a convergent net in (Y, q s), and we define f such that {fn(x)} converges to f(x) for each x ∈ X. Let {fδ} a right K-Cauchy net in (SL0(d, q), ρ(d,q)). Let us see that { q(fδ(x)−fδ(y)) d(x,y) } converges to q(f(x)−f(y)) d(x,y) for all x, y ∈ X such that d(x, y) 6= 0. Since {fδ} is a right K-Cauchy net, given ε > 0 there exists δ0 such that if δ1 ≥ δ2 ≥ δ0 then sup d(x,y) 6=0 q((fδ1 − fδ2)(x) − (fδ1 − fδ2)(y)) d(x, y) < ε 2 . Since fn(x) converges to f(x)∀x ∈ X, for each x, y there exists δ ′ 0 such that if δ′1 ≥ δ ′ 0 then qs(f(x) − fδ′ 1 (x) − (f(y) − fδ′ 1 (y))) d(x, y) < ε 2 . Thus given ε > 0, for all δ′ ≥ δ0 and for each x, y ∈ X : d(x, y) 6= 0 and we take δ1 ≥ (δ ′ ∨ δ′0) q(f(x) − fδ′(x) − (f(y) − fδ′(y))) d(x, y) = q(f(x) − fδ′(x) − fδ1(x) + fδ1(x) − (f(y) − fδ′(y) − fδ1(y) + fδ1(y))) d(x, y) ≤ q(f(x) − fδ1(x) − (f(y) + fδ1(y))) d(x, y) + q(fδ1(x) − fδ′(x) − (fδ′(y) + fδ1(y))) d(x, y) < ε. for all x, y such that d(x, y) 6= 0 sup d(x,y) 6=0 q(f(x) − fδ′(x) − (f(y) − fδ′(y))) d(x, y) < ε, for all δ′n ≥ δ0. � Corollary 4.2. Let (X, d), (Y, q) be a quasi-metric and a quasi-normed space respectively. Let {fδ} be a right K-Cauchy in (SL0(d, q), ρ(d,q)). If for each x ∈ X {fδ(x)} converges to f(x) in (Y, qs) then {fδ} converges to f in (SL0(d, q), ρ(d,q)). Theorem 4.3. Let (X, d), (Y, q) be a quasi-metric T1 and a quasi-normed bicomplete space respectively. Then ρ(d,q) is right K-complete. Proof. Let {fδ} be a right K-Cauchy net in (SL0(d, q), ρ(d,q)). Then, given ε ≥ 0 there is δ0 such that (♦) sup d(x,y) 6=0 q((fδ1 − fδ2)(x) − (fδ1 − fδ2)(y))) d(x, y) < ε for all δ1 ≥ δ2 ≥ δ0. Let x 6= x0. On semi-Lipschitz functions with values in a quasi-normed linear space 227 Since (X, d) is T1, d(x, x0) 6= 0 and d(x0, x) 6= 0 then, we deduce from (♦) that given ε d(x,x0) and ε d(x0,x) there exists δ′0 such that if δ1 ≥ δ2 ≥ δ ′ 0 then qs(fδ1(x) − fδ2(x)) < ε. Thereforefn(x) is a Cauchy net in (Y, q) for all x ∈ X. Thus {fδ(x)} is a convergent net in (Y, q s) and we define f such that {fn(x)} converges to f(x) for each x ∈ X. Let {fδ} a right K-Cauchy net in (SL0(d, q), ρ(d,q)). Let us see that { q(fδ(x)−fδ(y)) d(x,y) } converges to q(f(x)−f(y)) d(x,y) for all x, y ∈ X such that d(x, y) 6= 0. Since {fδ} is a right K-Cauchy net, given ε > 0 there exists δ0 such that if δ1 ≥ δ2 ≥ δ0 then sup d(x,y) 6=0 q((fδ1 − fδ2)(x) − (fδ1 − fδ2)(y)) d(x, y) < ε 2 . Since fn(x) converges to f(x)∀x ∈ X, for each x, y there exists δ ′ 0 such that if δ′1 ≥ δ ′ 0 then qs(f(x) − fδ′ 1 (x) − (f(y) − fδ′ 1 (y))) d(x, y) < ε 2 . Thus given ε > 0, for all δ′ ≥ δ0 and for each x, y ∈ X : d(x, y) 6= 0 and we take δ1 ≥ (δ ′ ∨ δ′0) q(f(x) − fδ′(x) − (f(y) − fδ′(y))) d(x, y) = q(f(x) − fδ′(x) − fδ1(x) + fδ1(x) − (f(y) − fδ′(y) − fδ1(y) + fδ1(y))) d(x, y) ≤ q(f(x) − fδ1(x) − (f(y) + fδ1(y))) d(x, y) + q(fδ1(x) − fδ′(x) − (fδ′(y) + fδ1(y))) d(x, y) < ε. for all x, y such that d(x, y) 6= 0 sup d(x,y) 6=0 q(f(x) − fδ′(x) − (f(y) − fδ′(y))) d(x, y) < ε, for all δ′n ≥ δ0. � References [1] J. Deák, A bitopological view of quasi-uniform completeness, I, Studia Sci. Math. Hun- gar. 30 (1995), 389–409; II 30 (1995), 411-431; 31 (1996), 385–404. [2] D. Doitchinov, On completeness in quasi-metric spaces, Topology Appl. 30 (1988), 127– 148. [3] P. Flecher and W. Hunsaker, Completeness using pairs of filters, Topology Appl. 44 (1992), 149–155. [4] P. Fletcher and W. F. 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