On semi-Lipschitz functions with values in a quasi-normed linear space

@Applied General Topology c Universidad Polit´ecnica de Valencia Volume 6, No. 2, 2005 pp. 217-228 On semi-Lipschitz functions with values in a quasi-normed linear space Jos´ e Manuel S´ anchez- ´ Alvarez∗ Abstract. In a recent paper, S. Romaguera and M. Sanchis discussed 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 motivated, 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, quasidistance, right K-complete. 1. Introduction and preliminaries Motivated, in part, by some problems from computer science and their applications (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 complexity analysis of programs. Recently, it was introduced in [14] the dual complexity 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 quasimetric 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. S´anchez-´ Alvarez 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 Nwill denote the set of nonnegative 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 Xis 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 dcan take the value ∞then it is called a quasi-distance on X. Given a quasi-metric don X, the function d−1defined on X×Xby d−1(x, y) = d(y, x), is also a quasi-metric on X, called the conjugate of d, and the function dsdefined on X×Xby ds(x, y) = d(x, y)∨d−1(x, y), is a metric on X. If d is a quasi-distance, then d−1and dsare a quasi-distance and a distance on X, respectively. A quasi-metric space is a pair (X, d) such that Xis a (nonempty) set and dis a quasi-metric on X. Each quasi-distance don Xinduces a topology T(d) on Xwhich 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 T1topology. A quasi-metric dis said to be bicomplete if dsis 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+×Xinto 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 k · k :X→R+such that for all x, y ∈Xand r∈R+: (i)x=0if and only if there is −x∈Xand kxk= 0 = k − xk, (ii)kr·xk=rkxk, and (iii)kx+yk ≤ kxk+kyk. If the quasi-norm qsatisfies: (i′)kxk= 0 if and only if x= 0, then qis called a norm on the cone (X, +,·). A (quasi-)normed cone is a pair (X, k · k) such that Xis a cone and k · k is a (quasi-)norm on X. If (X, +,·) is a linear space and k · k is a quasi-norm on X, then the pair (X, k · k) is called a quasi-normed linear space. Note that in this case, the function k · k−1:X→R+given by kxk−1=k − xkis also a quasi-norm on X and the function k · ks:X→R+given by kxks=kxk ∨ kxk−1is 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 respectively. A function f:X−→ Yis 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 kis called a semi-Lipschitz constant for f. On semi-Lipschitz functions with values in a quasi-normed linear space 219 A function fon 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 YX (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 T1quasi-metric space, then every function from Xto Yis ≤(d,q)-increasing. If for each f, g ∈YX (d,q)and a∈R+we define f+gand af in the usual way, then it is a routine to show that (YX (d,q),+,·) is a cone. Example 2.1. Let X=Z3. Let dbe the quasi-metric on Xgiven by d(x, y) = (1 if x > y, 0 if x≤y. Let Y=R,q(x) = x∨0 and take fsuch that f(0) = 0, f(1) = 1 and f(−1) = −2. It is easy to see that f∈YX (d,q)but −f /∈YX (d,q). Thus YX (d,q)is not a linear space. A simple but interesting example of a semi-Lipschitz function is the following: 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/ ∞ P n=0 2−n(f(n)∨0) <∞} and q(f) = ∞ P n=0 2−n(f(n)∨0). Let now F: (N, d)→(B∗, q) be the function defined by: F(0) = 0, F (n) = fnsuch that n < m implies fn> fm, where the order is given by fn> fmif and only if fn(x)> fm(x) for all x∈ω. Clearly Fis 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∈YX (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 (YX (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. S´anchez-´ Alvarez 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 0we 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 f6=0. Consequently, the nonnegative function k · k(d,q)defined on SL0(d, q) by kfk(d,q)=ρ(d,q)(f, 0) is a norm on SL0(d, q). Therefore (SL0(d, q),k · k(d,q)) is a normed cone. Example 1.1 of [11] provides an instance of a T1quasi-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 x0is a fixed point in the quasi-metric space (X, d), then the set SL0(d−1, q) = {f∈YX (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),k · k(d−1,q)) is a normed cone, where kfk(d−1,q)=ρ(d−1,q)(f, 0), i.e., kfk(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 quasinormed 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 xby yhence 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), kfk(d,q)=k − fk(d−1,q). Thus the normed cones (SL0(d, q),k · k(d,q)) and (SL0(d−1, q),k · k(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∈YX (d,q)∩YX (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),k · kB) is a normed linear space, where k · kB is the norm defined by kfkB= 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 k · kB=k · k(d,q)∨ k · k(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≥0such 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 YX (d,q)∩YX (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∈Xsatisfying 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 dbe the quasi-metric on Xgiven by d(x, y) = y−xif x≤yand 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−→ Ysatisfies 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∈Xsuch that SL0(d, q) = SL0(d−1, q), then SL1(d, q) = SL1(d−1, q)for each x1∈X. 222 J. M. S´anchez-´ Alvarez Proof. Let f∈ SL1(d, q). Define a function gon Xby 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 ∈Xwe 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),k · kB) is a Banach space. Theorem 3.1. Let (X,d), (Y,q) be a quasi-metric and a quasi-normed bicomplete space respectively. Then (SL0(d, q)∩ SL0(d−1, q),k · kB)is a Banach space. Proof. Let {fn}be a Cauchy sequence in (SL0(d, q)∩SL0(d−1, q),k·kB). Then, given ε≥0 there is n0∈Nsuch 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=x0then fn(x) = 0 for all n∈N. Let x6=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∈Nsuch that if n, m ≥n′ 0then qs(fn(x)−fm(x)) < ε. Therefore, fn(x) is a Cauchy sequence in (Y, qs). Case 2. d(x, x0) = 0. Then from remark 2.7 fn(x) = fn(x0) and fm(x) = fm(x0). Therefore qs(fn(x)−fm(x)) = 0 < ε. Consequently, fn(x) is a Cauchy sequence in (Y, qs) and {fn(x)}converges to an element f(x) in (Y, qs) for all x∈X. Moreover, {fn}converges to fin 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 n0such 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 n0is 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), k · kB). Then there exists a convergent sequence {kn}in (R, Tu)such that kn is a semi-Lipschitz constant for fn, where Tuis 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),k · k(d,q)=k · k(d−1,q). (3) (SL0(d, q),k · k(d,q))is a Banach space. Then: (1) ⇒(2) ⇔(3) Proof. (1) ⇒(2) Trivial. (2) ⇒(3) Trivial. (3) ⇒(2) Suppose that (SL0(d, q),k · k(d,q)) is a Banach space. Then SL0(d, q) is a group, so SL0(d, q) = SL0(d−1, q). Moreover k · k(d,q)is a norm on SL0(d, q), so that kfk(d,q)=k − fk(d,q)for all f∈ SL0(d, q). Since −f∈ SL0(d, q) it follows that k − fk(d,q)=kfk(d−1,q). We conclude that k · k(d,q)=k · k(d−1,q) on (SL0(d, q).  To see that in general (3) ⇒(1) is not true we take Y=0for a quasi-metric space that is not a metric space. Corollary 3.4. If qis a bicomplete quasi-norm on Ythen 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 bicomplete 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∈Nsuch 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=x0then fn(x) = 0 for all n∈N. Let x6=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∈Nsuch that if n, m ≥n′ 0then q(fn(x)−fm(x)) < ε and if we 224 J. M. S´anchez-´ Alvarez change nand 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, qs), thus {fn(x)}converges in (Y, qs) and we define fsuch that {fn(x)} −→ f(x) in (Y, q). Moreover, {fn} converges to fin (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 n0such 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 n0is 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),k · k(d,q))there exists a convergent sequence {kn}in (R, Tu)such that knis 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 context. In fact, they have been successfully applied to different fields from hyperspaces 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 δ0such that d(xδ2, xδ1)< ε for all δ1≥δ2≥δ0, and {xδ} ⊂ Xis called right K-Cauchy provided that for each ε > 0 there exists δ0such that d(xδ1, xδ2)< ε for all δ1≥δ2≥δ0. A quasi-metric ρis 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 bicomplete 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 δ0such that (⋆) sup d(x,y)6=0 q((fδ1−fδ2)(x)−(fδ1−fδ2)(y)) d(x, y)< ε for all δ1≥δ2≥δ0. Let x6=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′ 0such that if δ1≥δ2≥δ′ 0then 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≥δ0and q(−fδ(x)) = 0, since qs(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≥δ0then qs(fδ1(x)−f(x)) = qs(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 2because {fδ′ 1(x)}is a convergent net. Given δ′ 1≥δ1, such that fδ′ 1is 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 2because {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).