The canonical partial metric and the uniform convexity on normed spaces

@Applied General Topology c Universidad Polit´ecnica de Valencia Volume 6, No. 2, 2005 pp. 185-194 The canonical partial metric and the uniform convexity on normed spaces S. Oltra, S. Romaguera and E. A. S´ anchez-P´ erez∗ Abstract. In this paper we introduce the notion of canonical partial metric associated to a norm to study geometric properties of normed spaces. In particular, we characterize strict convexity and uniform convexity of normed spaces in terms of the canonical partial metric defined by its norm. We prove that these geometric properties can be considered, in this sense, as topological properties that appear when we compare the natural metric topology of the space with the non translation invariant topology induced by the canonical partial metric in the normed space. 2000 AMS Classification: 54E35, 46B04. Keywords: Partial metric, convexity, normed spaces. 1. Introduction S. G. Matthews introduced in [6] the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks, and obtained, among other results, a nice relationship between partial metric spaces and the so-called weightable quasi-metric spaces. Partial metrics were also used in the context of the complexity analysis of algorithms and programs (see [6, 8, 7]). The domain of words which appears in a natural way by modelling the streams of information in G. Kahn’s model of parallel computation provides a wellknown example of partial metric space (see [4] and [6]). Other motivations for exploring partial metrics can be found in [6]. In this paper we present an application of the theory of partial metrics to the framework of the normed space theory. Other applications in this direction can be found in [9]. Let (X, k.k) be a normed space. Our aim is to show that it ∗The authors acknowledge the support of the Generalidad Valenciana, grant GV04B-371, the Spanish Ministry of Science and Technology, Plan Nacional I+D+I, grant BFM2003- 02302, and FEDER, and the Polytechnical University of Valencia. 186 S. Oltra, S. Romaguera and E. A. S´anchez-P´erez is possible to define a partial metric pk.kin Xin a canonical way that implicitly contains the information about the convexity properties of the space. However, the topology that defines this partial metric is not standard at all, since it fails to satisfy some of the basic properties that topologies on topological linear spaces use to have. In particular, we obtain a topology τpwith the following properties. (1) τpis not T1. (2) τpis not translation invariant. (3) τpcoincides with the norm topology when restricted to the unit sphere of X. We present these results in three sections. Section 2 is devoted to define and prove the basic properties of the canonical partial metric. In Section 3 we characterize when (X, k.k) is strictly convex in terms of the natural base of neighborhoods of (X, τp), providing also several examples. Finally, in Section 4 we present the results concerning uniform convexity of (X, k.k) and the characterization of this property in terms of a particular class of neighborhoods of the elements of the unit sphere of (X, k.k). In what follows we introduce the basic definitions and results on partial metrics. In this direction, our main references are [6, 7, 8]. Each partial metric defines a quasi-metric that generates the same topology. Therefore, we start by recalling several definitions on quasi-metrics. Our basic references for quasimetric spaces are [3] and [4], and for general topological questions [2]. Let us recall that a quasi-pseudo-metric on a set Xis a nonnegative real valued function don X×Xsuch that for all x, y, z ∈X, (i)d(x, x) = 0; (ii)d(x, y)≤d(x, z) + d(z, y). By a quasi-metric on a set Xwe mean a quasi-pseudo-metric don Xthat satisfies also the following condition. (iii)d(x, y) = d(y, x) = 0 ⇔x=y. Aquasi-metric space is a pair (X, d) such that Xis a (nonempty) set and d is a quasi-metric on X. Each quasi-metric don Xgenerates a T0-topology T(d) on Xwhich has as a base the family of open d-balls {Bd(x, ε) : x∈X, ε > 0}, where Bd(x, ε) = {y∈X:d(x, y)< ε}for all x∈Xand ε > 0. We will write R+for the set of nonnegative real numbers. A partial pseudometric on a (nonempty) set Xis a function p:X×X→R+such that for all x, y, z ∈X, (i)x=y⇒p(x, x) = p(x, y) = p(y, y); (ii)p(x, x)≤p(x, y); (iii)p(x, y) = p(y, x); (iv)p(x, z)≤p(x, y) + p(y, z)−p(y, y). By a partial metric on a set Xwe mean a partial pseudo-metric pon Xthat satisfies the following condition, (i)′x=y⇔p(x, x) = p(x, y) = p(y, y). The canonical partial metric and the uniform convexity on normed spaces 187 Apartial (pseudo-)metric space is a pair (X, p) such that Xis a (nonempty) set and pis a partial (pseudo-)metric on X. Each partial metric pon Xdefines a quasi-metric dpon Xby means of the formula, dp(x, y) := p(x, y)−p(x, x), x, y ∈X, and the topology given by pis the one generated by dp. Consequently, each partial metric generates a T0-topology T(p) on Xand is the one given by the base B={Vε,p(x) : x∈X, ε > 0}, where Vε,p(x) = {y∈X:p(x, y)< ε +p(x, x)}for all x∈Xand ε > 0. The following lemma is a direct consequence of Lemma 2.2 in [6] and gives information about the sets Vε,p(x) that will be useful in this paper. Lemma 1.1. Let (X, p)be a partial metric space and x∈X. Then B= {Vε,p(x) : ε > 0}is a base of open neighborhoods of xfor the topology T(p). Let us finish this section by introducing several definitions of Functional Analysis. The basic references for these definitions are [10, 11, 5]. We use standard Banach space notation. If (X, k.k) is a normed space, we denote by SXthe corresponding unit sphere {x∈X:kxk= 1}. If x∈Xand ε > 0, we denote by Bε,k.k(x) the set Bε,k.k(x) := {y∈X:kx−yk< ε}. If x, y ∈X, we denote by [x, y] the segment {z=θx + (1 −θ)y: 0 ≤θ≤1}. If x∈X,hxidenotes the linear span of xinto the linear space X. Although we present the results of the paper for normed spaces, completeness is not necessary. Therefore, our main results are valid for normed spaces. The same holds for the following definitions. Definition 1.2. Let (X, k.k)be a normed space. We say that the norm k.k (equivalently, the normed space (X, k.k)) is strictly convex if for every pair of norm one elements x, y ∈X,kx+y 2k= 1 implies x=y. It can be proved that this definition is equivalent to the following one. The norm k.k(equivalently, the normed space (X, k.k)) is strictly convex if for every x, y ∈X, if x6= 0 and kx+yk=kxk+kyk, then y∈ hxi. The proof of this equivalence can be found in Proposition 1, Part 3, Ch.I of [1]. Throughout the paper we will use both definitions. Definition 1.3. Let (X, k.k)be a normed space. The norm k.k(equivalently, the normed space (X, k.k)) is uniformly convex if for every ε > 0there is a δ > 0-only depending on εsuch that if kxk= 1 = kykand kx−yk ≥ ε, then kx+y 2k ≤ 1−δ, for every x, y ∈X. We will use the following notation. Let n∈Nand consider the corresponding n-dimensional linear space Rn. Let x= (x1, ..., xn)∈Rn. We will write kxk2for kxk2:= ( n X i=1 |xi|2)1 2, 188 S. Oltra, S. Romaguera and E. A. S´anchez-P´erez and kxk1for kxk1:= n X i=1 |xi|. 2. Partial metrics in normed spaces In this section we construct the canonical partial metric pk.kassociated to the norm of a normed space (X, k.k), and we prove the main results that will be used in the following two sections. The relations between the elements of the base of neighborhoods given by Lemma 1.1 of the topological space space (X, τpk.k) and the translation invariant topology associated to the norm k.kgives the characterization of the convexity properties of the normed space (X, k.k). Definition 2.1. Let (X, k.k)be a normed space. We define the nonnegative function pk.k:X→Rby the formula pk.k(x, y) := kx−yk+kxk+kyk, x, y ∈X. A related construction has been done at [6] Definition 2.2. Let τbe a topology on a linear space X. We say that a norm k.kis 0-compatible with the topology τif the balls Bε,k.k(0) = {x∈X:kxk< ε}, ε > 0, define a base of neighborhoods of 0for the topology τ. Proposition 2.3. If (X, k.k)is a normed space, the function pk.kis a partial metric that satisfies 1) pk.k(x+y, 0) ≤pk.k(x, 0) + pk.k(y, 0), 2) pk.k(λx, λy) = |λ|pk.k(x, y)for every x, y ∈Xand λ∈R, 3) pk.k(x, y) = 0 if and only if x=y= 0, 4) The norm k.kis 0-compatible with the topology τpk.k. Proof. The following calculations show that pk.kis a partial metric. To prove the condition (i)′in the definition of partial metric (Section 1), let x, y ∈X such that pk.k(x, x) = pk.k(x, y) = pk.k(y, y), hence 2kxk=kx−yk+kxk+kyk= 2kyk so that kx−yk+kxk − kyk= 0 and kx−yk − kxk+kyk= 0. This clearly implies kx−yk= 0, and then x=y. Now suppose that x=y; the equalities above gives directly pk.k(x, x) = pk.k(x, y) = pk.k(y, y), since kx−yk= 0. Now let us show that pk.k(x, x)≤pk.k(x, y) for every x, y ∈X. But this is a direct consequence of the triangular inequality for the norm k.k, since pk.k(x, x) = 2kxk ≤ kx−yk+kxk+kyk=pk.k(x, y). The definition of pk.kclearly gives pk.k(x, y) = pk.k(y, x) for every x, y ∈X. To see the last condition of partial metric, consider x, y, z ∈X. Then pk.k(x, y) + pk.k(z, z) = kx−yk+kxk+kyk+ 2kzk The canonical partial metric and the uniform convexity on normed spaces 189 ≤ kx−zk+kxk+kzk+kz−yk+kyk+kzk=pk.k(x, z) + pk.k(y, z). Now let us show 1). If x, y ∈X, then pk.k(x+y, 0) = kx+yk+kx+yk ≤ 2kxk+ 2kyk=pk.k(x, 0) + pk.k(y, 0). Condition 2) is a consequence of the homogeneity of the norm. pk.k(λx, λy) = kλx−λyk+kλxk+kλyk=|λ|(kx−yk+kxk+kyk) = |λ|pk.k(x, y) for every x, y ∈Xand λ∈R. 3) is also given directly by the definition. If x∈X, obviously pk.k(x, y) = kx−yk+kxk+kyk= 0 if and only if x=y= 0. Finally, let us show that k.kis 0-compatible with τpk.k. It is enough to write explicitly the basic neighborhoods Vε,pk.k(x) for the case x= 0. Vε,pk.k(0) = {x∈X:pk.k(x, 0) < ε+p(0,0)}={x∈X: 2kxk< ε}=Bε 2,k.k(0)  We will call the function pk.kthe canonical partial metric associated to k.k. Consider ε > 0 and x∈X. The basic neighborhood of x,Vε,k.k(x) is given in this case by the particular expression Vε,pk.k(x) := {y∈X:pk.k(x, y)< pk.k(x, x)+ε}={y∈X:kx−yk+kyk−kxk< ε}. This description of the neighborhood Vε,pk.k(x) will be useful in the following sections. 3. Strict convexity and the canonical partial metric Let (X, k.k) be a normed space. In this section we characterize when the norm k.kis strictly convex in terms of the base of neighborhoods for the topology τpk.kgiven by Lemma 1.1 and described at the end of Section 2 for the particular case of the canonical partial metric. Lemma 3.1. Let (X, k.k)be a normed linear space. For every x∈X,[0, x]⊂ ∩ε>0Vε,pk.k(x). Proof. Let y∈[0, x] and let ε > 0. Then there exists an αsuch that 0 ≤α≤1 and y=αx. Thus, p(x, y) = kx−yk+kyk+kxk= (1−α)kxk+αkxk+kxk= 2kxk=p(x, x)< p(x, x)+ε. This proves the lemma, since implies that y∈Vε,pk.k(x) for every ε > 0.  Note that the only point that satisfies that the intersection of all its neighborhoods is the same point is 0. As a direct consequence, we obtain that the topology generated by the canonical partial metric only satisfy the separation axiom T0. Remark 3.2. The topology defined by the canonical partial metric in a non trivial normed space is not T1. To prove this, consider x∈X− {0}and define y=1 2x. Suppose that there is a neighborhood of x,Vsuch that ydo not belong to V. Since Vis a neighborhood of x, by Lemma 1.1 there exists ε > 0 such that Vε,pk.k(x)⊂V. But y∈Vε,pk.kas a consequence of Lemma 3.1, and then y∈V, a contradiction. 190 S. Oltra, S. Romaguera and E. A. S´anchez-P´erez Let us discuss in what follows the situation for the converse inclusion that the one given in Lemma 3.1. The first one shows that the 2-dimensional Euclidean space satisfies also this inclusion. Example 3.3. Consider the 2-dimensional Euclidean space R2 2:= (R2,k.k2), and let x0∈R2 2. Then \ ε>0 Vε,pk.k(x0) = \ ε>0 {y∈R2:kx0−yk2+kyk2< ε +kx0k2}= ={y:kx0−yk2+kyk2=kx0k2}, since the inequality kxk ≤ kx−yk+kykalways holds for every x, y ∈Xand every norm. In terms of the Euclidean distance d2in R2 2, the above condition can be written as d2(y, x0) + d2(0, y) = d2(0, x0), which only holds when y=αx0for some α∈[0,1]. Therefore, in this case Tε>0Vε,pk.k(x0) = [0, x0]. Example 3.4. Consider the 2-dimensional space R2 1:= (R2,k.k1), and the element x0:= (1/2,1/2). Then Vε,pk.k(x0) = {(y1, y2) : k(1/2−y1,1/2−y2)k1+k(y1, y2)k1< ε+k(1/2,1/2)k1}, and then Tε>0Vε,pk.k(x0) = {(y1, y2) : k(1/2−y1,1/2−y2)k1+k(y1, y2)k1= k(1/2,1/2)k1}. Consider now any element (a, b)∈[0,1/2] ×[0,1/2]. Then the condition that appears in Tε>0Vε,pk.k(x0) can be written as |1 2−a|+|1 2−b|+|a|+|b|=1 2−a+1 2−b+a+b= 1 that obviously holds for every (a, b)∈[0,1/2]×[0,1/2]. Then [0,1/2]×[0,1/2] ⊂ Tε>0Vε,pk.k(x0), which implies that Tε>0Vε,pk.k(x0) is not contained in [0, x0]. Lemma 3.5. A normed space (X, k.k)is strictly convex if and only if for every x, y ∈SX, if k2x−yk= 1 then x=y. Proof. Suppose that (X, k.k) is strictly convex, and consider two elements x, y ∈ SXsuch that k2x−yk= 1. Then k2x−yk+kx+yk ≥ k3xk= 3, and so kx+yk ≥ 2. Moreover, 2kx+yk=k2x+ 2yk ≤ k2x−yk+k3yk= 4, and then kx+yk ≤ 2. Since (X, k.k) is strictly convex, we obtain that x=y. Conversely, suppose that the second property in the statement of the lemma holds, and consider two elements x, y ∈SXsatisfying kx+yk= 2. Then we define z1=x+y 2and z2=y, that obviously satisfy kz1k=kz2k= 1 and k2z1−z2k= 1. Thus the property gives z1=z2, and then x+y 2=y, which clearly implies x=y. The canonical partial metric and the uniform convexity on normed spaces 191 Theorem 3.6. A normed linear space (X, k.k)is strictly convex if and only if ∩ε>0Vε,pk.k(x) = [0, x]. Proof. First let us prove that ∩ε>0Vε,pk.k(x) = [0, x] implies that the normed linear space is strictly convex. Suppose that kxk= 1. Then ∩ε>0Vε,pk.k(2x)∩SX= [0,2x]∩SX={x}. Since ∩ε>0Vε,pk.k(2x)∩SX={y:kyk= 1,k2x−yk= 1}, we obtain the result as a consequence of Lemma 3.5. Conversely, we use the characterization given by Proposition 1 (p. 175 of [1]) of the strict convexity of (X, k.k) that we have referred in Section 1. Consider x∈SXand suppose that y∈ ∩ε>0Vε,pk.k(x) but yis not in [0, x]. Then kx−yk+kyk=kxk= 1 and in particular kyk ≤ 1 and kx−yk ≤ 1. These two inequalities and the fact that yis not an element of [0, x] clearly imply that yis not a linear combination of x. If we write z1=x−yand z2=y, we have kz1k+kz2k= 1 = kz1+z2k, and then the proposition quoted above gives that there is a λ6= 0 such that z1=λz2, and then x−y=λy. Thus x= (1 + λ)y, a contradiction.  4. Uniform convexity and the canonical partial metric Strict convexity of the norm can be understood, in a certain sense, as a limit case of the uniform convexity. After the result obtained in the theorem above, we will prove in this section that it is also possible to give a characterization of the uniform convexity of a normed space (X, k.k) in terms of a particular class of neighborhoods of the points of Xfor the topology τpk.k. Let us fix a normed space (X, k.k). Let x∈SXand δ > 0. We define the set Wδ,k.k(x) = {y∈X:kxk+kyk ≤ kx+yk+δ}. Lemma 4.1. Let (X, k.k)be a linear normed space. For every x∈Xand ε > 0,Vε,pk.k(x)⊂Wε,k.k(x). In particular, Wε,k.k(x)is a neighborhood of x for the topology τpk.k. Proof. Fix x∈Xand consider an element y∈Vε,pk.k(x). Then 2kxk+kyk ≤ kx+yk+kx−yk+kyk ≤ kx+yk+kxk+ε, Thus kxk+kyk<kx+yk+ε, and so y∈Wε,k.k(x).  Lemma 4.2. For every x∈SXand every ε > 0,Vε,pk.k(x)∩SX=Bε,k.k(x)∩ SX. Proof. If kxk= 1 and y∈Vε,pk.k(x)∩SX, kx−yk+kyk=kx−yk+ 1 ≤ kxk+ε= 1 + ε. Then kx−yk< ε, and thus y∈Bε,k.k(x). The same argument shows the opposite inclusion.  192 S. Oltra, S. Romaguera and E. A. S´anchez-P´erez Definition 4.3. Let (X, τ)be a topological space. For every element x∈X, consider two sets of neighborhoods of xboth of them indexed by ε∈R+; Vx={Vε(x) : ε > 0}and Wx={Wε(x) : ε > 0}. Consider now the families of neighborhoods V={Vx:x∈X}and W={Wx:x∈X}. We say that Vand Ware uniformly equivalent if they verify the following relations: (i) for every ε > 0there is a δ > 0(only depending on ε) such that Wδ(x)⊂Vε(x)for all x∈X, and (ii) for every ε′>0there is a δ′>0(only depending on ε′) such that Vε′(x)⊂Wδ′(x)for all x∈X. For the following theorem, we consider the particular families of neighborhoods V={Vx:x∈SX},W={Wx:x∈SX}and B={Bx:x∈SX}, given by Vx={Vε,pk.k(x)∩SX:ε > 0}, Wx={Wε,k.k(x)∩SX:ε > 0} and Bx={Bε,k.k(x)∩SX}. Theorem 4.4. Let (X, k.k)be a normed space. The following are equivalent. (1) (X, k.k)is uniformly convex. (2) For every ε > 0there is a δ > 0(only depending on ε) such that Wδ,k.k(x)∩SX⊂Vε,pk.k(x), for all x∈SX. (3) Wand Vare uniformly equivalent families of neighborhoods. (4) Wand Bare uniformly equivalent families of neighborhoods. Moreover, if any of the statements (1) to (4) holds, then the family Wxdefines a local base in the topological space (SX, τpk.k|SX)(equivalently, in the topological space (SX, τk.k|SX)) for every x∈SX. Proof. Let us prove first that (1) implies (2). Suppose now that (X, k.k) is uniformly convex. It follows that for all x, y ∈SXand εthere is a δ > 0 such that if kx+yk ≥ 2−2δit is true that kx−yk< ε. Let us show that Wδ,k.k(x)∩SX⊂Vε,pk.k(x). To check this, consider y∈ Wδ,k.k(x)∩SX. Note that Wδ,k.k(x)∩SX={y∈SX:kxk+kyk ≤ kx+yk+δ}={y∈SX: 2−δ≤ kx+yk}. Hence 2−2δ≤2−δ≤ kx+yk, and since Xis uniformly convex this implies that kx−yk< ε. Consequently, it follows that kx−yk+kyk ≤ kxk+ε. This completes the proof of (1) implies (2). The canonical partial metric and the uniform convexity on normed spaces 193 To deduce (1) from (2), suppose that for every ε > 0 there is a δ > 0 such that Wδ,k.k(x)∩SX⊂Vε,p(x) for every x∈SX. Consider ε > 0 and δ′=δ 2. If y∈Wδ,k.k(x)∩SX={y∈SX: 2 −δ≤ kx+yk}, we have that y∈Vεpk.k(x), and then kx−yk< ε. This condition is equivalent to the following one. For every x, y ∈SX, if kx−yk ≥ ε, then 2−2δ′≥ kx+yk. Hence, for such x, y ∈SX, we have kx+yk<2−δ= 2 −2δ′, which can be written as kx+yk 2<1−δ′. This shows that Xis uniformly convex. Let us prove now the equivalence between (2) and (3). As a consequence of Lemma 4.1, we only need to prove that for each ε > 0 there is a δ > 0 (only depending on ε) such that Wδ,k.k(x)∩SX⊂Vε,pk.k(x) for all x∈SX. But this is what the statement (2) of the theorem assures, so Vand Ware uniformly equivalent. The converse is obvious. It follows easily that (3) if and only if (4), as a consequence of Lemma 4.2. Finally, we have to check that {Wδ,k.k(x)∩SX:δ > 0}defines for every x∈SXa local base for the topological space (SX, τpk.k|SX). It is clear, using Lemma 4.2, that this is equivalent to the fact that {Wδ,k.k(x)∩SX:δ > 0}a local base for the topological space (SX,k.k|SX) for each x∈SX. Obviously {Wδ,k.k(x)∩SX:δ > 0} 6=φsince x∈ {Wδ,k.k(x)∩SX:δ > 0}. It is also clear that given Wδ1,k.k(x)∩SXand Wδ2,k.k(x)∩SXwe obtain x∈Wδ3,k.k(x)∩SX⊂Wδ1,k.k(x)∩SX∩Wδ2,k.k(x)∩SX, where δ3:= min{δ1, δ2}. Consider Wδ,k.k(x)∩SX. Since as Vand Ware uniformly equivalent families of neighborhoods, there is ε > 0 such that Vε,pk.k(x)∩SX⊂Wδ,k.k(x)∩SX. Since Vε,pk.k(x)∩SXis a local base, then there is a Vε1,pk.k(x)∩SX⊂Vε,pk.k(x)∩SX such that for all y∈Vε1,pk.k(x)∩SXthere is an ε2>0 such that Vε2,pk.k(y)∩ SX⊂Vε,pk.k(x)∩SX. Given Vε1(x) we can find a δ1>0 such that it verifies that Wδ1,k.k(x)∩ SX⊂Vε1,pk.k(x)∩SX⊂Wδ,k.k(x)∩SX. Now we have to show that for all y∈Wδ1,k.k(x)∩SXthere is a δ2>0 such that Wδ2,k.k(y)∩SX⊂Wδ,k.k(x)∩SX. We know that for all y∈Wδ1,k.k(x)∩SX⊂Vε1,pk.k(x)∩SXthere is Vε2,pk.k(y)⊂Vε,pk.k(x). Then for each y∈Wδ1,k.k(x) there exists Vε2,pk.k(y)⊂ Vε(x), and given Vε2,pk.k(y) there is δ2such that Wδ2,k.k(y)⊂Vε2,pk.k(y)⊂ Wδ,k.k(x). Therefore, we have proved that the family {Wδ,k.k(x)∩SX:δ > 0} defines for every x∈SXa local base for the topological space (SX, τpk.k|SX). This finishes the proof.  The ideas exposed in this paper allow to consider the convexity properties of normed spaces as sequential properties, since they can be characterized using the topology defined by the canonical partial metric. This provides a new framework for the study of certain geometric properties of normed spaces and in the particular case of the Banach spaces.