Global Journal of Mathematics and Mathematical Sciences (GJMMS). ISSN 0972-9836 Volume 6, Number 1 (2016), pp. 1–9 © Research India Publications http://www.ripublication.com/gjmms.htm
Decomposition of metrics and norms
Hernández J. M., Castañeda C. H., Álvarez L. C., Tochihuitl V. and Vázquez R. Instituto de Física y Matemáticas, Universidad Tecnológica de la Mixteca. E-mail: [email protected], [email protected], [email protected], [email protected], [email protected]
Abstract In this paper we describe as give a decomposition of a metric (norm) in two con- jugated quasi-metrics (asymmetric norm, respectively). After we introduce the notions of quasi-k-metrics and asymmetric k-norms and, under certain assump- tions also we give different ways to decompose a metric (norm) in two conjugated quasi-k-metrics (asymmetric k-norms, respectively). Finally we illustrate these with examples.
AMS Subject Classification: 46B20, 54E25, 54E55. Keywords: Asymmetric norm, quasi-metric, quasi-k-metric, asymmetric k-norm.
1. Introduction It is well known that, if one attempts to omit the requirement of symmetry in a semimetric or in a seminorm, then the structure of the space is drastically changed. Such unsymmet- rical spaces have been studied by Wilson [7] who used the term quasi-semimetric when the feature of symmetry is omitted in a semimetric, and Duffin and Karlovitz (1968) [4] who proposed the term asymmetric seminorm when the property of symmetry is not satisfied in a seminorm. Subsequently, many classical results of the analysis have been extended to such non-symmetric spaces, see for example [1], [5], [6], [8] and in [2] a compendium of works performed. Formally, a quasi-semimetric on a set X is a mapping ρ : X ×X −→ [ 0, ∞) satisfying the following conditions: (i) ρ(x,x) = 0, (ii) ρ(x,z) ≤ ρ(x,y) + ρ(y,z), for all x,y,z ∈ X. If, further, (iii) ρ(x,y) = ρ(y,x) = 0 implies x = y, then ρ is called a quasi-metric; while for a linear space X a func- tion p : X → R is an asymmetric norm if for all x,y ∈ X and yr ∈[0, ∞), (i) 2 Hernández J. M., Castañeda C. H., Álvarez L. C., Tochihuitl V. and Vázquez R. p(x) = p(−x) = 0 ⇒ x = 0, (ii) p(rx) = rp(x) and (iii) p(x + y) ≤ p(x) + p(y). If only (ii) and (iii) are satisfied, then p is called an asymmetric seminorm. A quasi- semimetric ρ determines another quasi-semimetric ρ defined by ρ(x,y) = ρ(y,x) and also a semimetric ρs given by ρs = max{ρ,ρ}; similarly, an asymmetric seminorm determines its conjugated asymmetric seminorm p(x) = p(−x) and its associated semi- norm ps = max{p, p}. Here we treat the problem: given a metric d (a norm ·), find a quasi-metric ρ (asymmetric norm p) such that d = ρs (· = ps), then we introduce the definitions of quasi-k-metric and asymmetric-k-norm with which, we try the similar problem: given a metric d (a norm ·), we find a quasi-k-metric ρ (asymmetric-k-norm p) with d = ρs (· = ps).It is good to mention that in [3] the author defines that an asymmetric normed linear space (E, q) is normable if there is a norm ||·||on the linear space E such that the topologies τ dq and τ d||·|| coincide on E. This is different from what we do in this work.1
2. Definitions In the following definitions, k is a real number greater than or equal to 1. Definition 2.1. Let X be a nonempty set, a k-metric on X is a function d : X × X → [0, ∞) which satisfies for all x,y,z ∈ X, the conditions:
(i) d(x,y) = 0 if and only if x = y, (ii) d(x,z) ≤ k (d(x,y) + d(y,z)) , (iii) d(x,y) = d(y,x).
The pair (X, d) is called a k-metric space. If only (ii) and (iii) are satisfied, then the pair is called a k-semimetric space. Definition 2.2. Let X be a real linear space. A function · k : X →[0, ∞) is a k-norm on X if for each x,y ∈ X and r ∈ R : (i) x k = 0 if and only if x = 0, | | (ii) rx k = r x k , + ≤ + (iii) x y k k x k y k .
(X, · k) is called a k-normed space. Also if only (ii) and (iii) are satisfied, then the pair is called a k-seminormed space.
Remark 2.3. A metric (semimetric) space is a 1-metric space (1-semimetric space) and a normed (seminormed) space is a 1-normed space (1-seminormed space).
1This work is partially supported by Programa de Mejoramiento del Profesorado (UTMIX-EXB-025). Decomposition of metrics and norms 3
Definition 2.4. A quasi-k-semimetric on an arbitrary nonempty set X is a function ρ : X × X →[0, ∞) that satisfies:
(i) ρ(x,x) = 0,
(ii) ρ(x,z) ≤ k (ρ(x,y) + ρ(y,z)) , for all x,y,z ∈ X. If, further,
(iii) ρ(x,y) = ρ(y,x) = 0 implies x = y, for all x,y ∈ X, then ρ is called a quasi-k-metric.
A quasi-metric (quasi-semimetric) is a quasi-1-metric (quasi-1-semimetric). x if x ≥ 0 R = Example 2.5. is a quasi-2-metric space, with ρ(x,y) x . − if x<0 2 Like a quasi-semimetric, given a quasi-k-semimetric ρ in a set X, the conjugate ρ of ρ is defined by ρ(x,y) = ρ(y,x), x,y ∈ X. It is easy to verify that ρ is also a quasi-k-semimetric, meanwhile the function ρs : X × X → R given by
ρs(x, y) = max {ρ(x,y),ρ(x,y)} , (2.1) provestobeak-semimetric. Even further, ρ is a quasi-k-metric if and only if ρs is a k-metric on X. The following inequality is trivially valid,
ρ(y,x) ≤ ρs(x, y) and ρ(x,y) ≤ ρs(x, y). (2.2)
Definition 2.6. Let X be a real linear space. A function p : X →[0, ∞) is an asymmetric k-norm on X, if for all x,y ∈ X and r ≥ 0,
(i) p(x) = p(−x) = 0 implies x = 0;
(ii) p(rx) = rp(x);
(iii) p(x + y) ≤ k (p(x) + p(y)) .
In this case (X, p) is called an asymmetrically k-normed space. If p only satisfied the conditions (ii) and (iii), then it is called an asymmetric k-seminorm, and the pair (X, p) is called an asymmetrically k-seminormed space. Also in some instances the value +∞ is allowed to p, in which case p is an extended asymmetric k-seminorm (k-seminorm) The conjugate p of the asymmetric k-seminorm p is defined by
p(x) = p(−x), x ∈ X. 4 Hernández J. M., Castañeda C. H., Álvarez L. C., Tochihuitl V. and Vázquez R.
In analogy to (2.1), the function ps : X →[0, ∞) given by ps(x) = max{p(x), p(x)},x∈ X, is a k-seminorm. An asymmetric k-seminorm p is an asymmetric k-norm if and only if ps is a k-norm on X. An asymmetric k-seminorm p induces a quasi-k-semimetric ρp on X, given by = − ∈ ρp(x, y) p(y x), x,y X. Here the inequalities in (2.2) become p(x) ≤ ps(x) and p(x) ≤ ps(x) for all x ∈ X. Definition 2.7. If d is a metric on X and for some quasi-k-metric ρ = d is satisfied that ρs = d, then the pair (ρ,ρ) is called a decomposition of the metric d. Similarly, if · is a norm on X and p = · is an asymmetric k-norm such that ps = · , then the pair (p, p) is called a decomposition of the norm · .
3. Results Now, we give some sufficient conditions to ensure that a metric space comes from a quasi-metric space. Theorem 3.1. Let (X, d) be a metric space. If f : X → R is a function satisfying d(x,z) ≥ d(y,z) and d(x,z) ≥ d(x,y) whenever f(x) ≤ f(y) ≤ f(z), then there exists a decomposition of d. d(x,y) if f(x)≤ f(y) Proof. Let ρ(x,y) = ρ (x, y) = . Let us prove that ρ is a f 0iff(x)>f(y) quasi-metric, we will start showing the triangular inequality by cases: given x,y,z ∈ X
Case 1 f (x) > f (y). ρ(x,y) = 0 ≤ ρ(x,z) + ρ(z,y).
Case 2 f(x)≤ f(y).There are 3 subcases. Subcase 1: f(x)≤ f (y) < f (z). ρ(x,z) + ρ(z,y) = d(x,z) ≥ d(x,y) = ρ(x,y). Subcase 2: f(x)≤ f(z)≤ f(y). ρ(x,y) = d(x,y) ≤ d(x,z) + d(z,y) = ρ(x,z) + ρ(z,y). Subcase 3: f(z) In addition, it is clear that ρ(x,y) ≥ 0, and if ρ(x,y) = ρ(y,x) = 0, then d(y,x) = 0 and so x = y. _ d(y,x) if f(y)≤ f(x) Furthermore, ρ(x,y) = ρ(y,x) = 0 if f(y)>f(x) thus, ρs(x, y) = max {ρ(x,y),ρ(y,x)} = d(x,y). Example 3.2. Let X = R, d(x, y) = |x − y| and f : R → R an increasing function, clearly, if f(x)≤ f(y)≤ f(z), d(x,z) ≥ d(z,y) and d(x,z) ≥ d(x,y), |x − y| if f(x)≤ f(y) then ρ(x,y) = is such that ρs = d. 0iff(x)>f(y) From this result we obtain Corollary 3.3. Let (X, ||·||) be a linear normed space. If f : X → R is a function satisfying ||z − x||≥||z − y|| and ||z − x||≥||y − x|| whenever f(x)≤ f(y)≤ f(z), then there exists a decomposition of ||·||. Proof. For the metric d p(x, y) =||y − x||, we obtain a quasi-metric ρ defined by ||y − x|| if f(x)≤ f(y) ρ(x,y) = ρ (x, y) = . Then we define the asymmetric f 0iff(x)>f(y) norm p given by p(x) = ρ(0,x),and this p is together with his conjugate p make up the desired decomposition of · . Theorem 3.4. Let (X, ·) be a normed space. For each f : X → R linear function and each k>1, there exist a decomposition of ·. x , f (x) ≥ 0, = Proof. Given such f, k, we define pf,k(x) x k ,f(x)<0, Let’s prove pf,k is an asymmetric k-norm. Clearly pf,k(x) = pf,k(−x) = 0 implies x = 0. Now, for λ ≥ 0, λx , f (λx) ≥ 0, λ x ,λf(x)≥ 0, = = = pf,k(λx) λx x λpf,k(x). k , f (λx) < 0, λ k ,λf(x)<0, Triangle inequality: Given x,y ∈ X, Case 1. If f(x)≥ 0, f (y) ≥ 0, then pf,k(x + y) = x + y ≤ x + y = pf,k(x) + pf,k(y). Case 2. If f(y)<0,f(x)<0, then x + y x + y x y p (x + y) = ≤ = + = p (x) + p (y). f,k k k k k f,k f,k 6 Hernández J. M., Castañeda C. H., Álvarez L. C., Tochihuitl V. and Vázquez R. Case 3. If f(x)≥ 0,f(y)<0, in this case we have two possibilities: 1. If f(x+ y) ≥ 0, then y p (x + y) = x + y ≤ x + y = x + k f,k k ≤ k(pf,k(x) + pf,k(y)). 2. If f(x+ y) < 0, then x + y x + y x y p (x + y) = ≤ = + f,k k k k k 1 = p (x) + p (y) ≤ p (x) + p (y). k f,k f,k f,k f,k Case 4. If f(x)<0, f (y) ≥ 0, is treated similarly to case 3. s = · : 1. So it remains to prove that pf,k . In order to do this, let’s identify pf,k −x ,f(−x) ≥ 0, x , f (x) ≤ 0, = − = = pf,k(x) pf,k( x) −x x ,f(−x) < 0, ,f(x)>0, k k s = = therefore, pf,k(x) max pf,k(x), pf,k(x) x . Example 3.5. Let X = C[a,b] with the supremum norm and let T : X → R given by T(f)= f(a),the quasi-k-metric f ∞ , f (a) ≥ 0, p (f ) = T,k f ∞ ,f(a)<0, k s = · is such that pT,k ∞ . Theorem 3.6. Let (X, d) be a metric space, if card(X) ≤ℵ1 then there exists a decomposition of d. Proof. If card(X) ≤ℵ1 then there exists f : X → R, an injective function. Given k>1, we define d(x,y), f(x) ≥ f(y), = = ρk(x, y) ρf,k(x, y) d(x,y) , f (x) < f (y), k Decomposition of metrics and norms 7 d(y,x) It’s clear that if ρ (x, y) = ρ (y, x) = 0, then d(x,y) = = 0ord(y,x) = k k k d(x,y) = 0 causing x = y. For x,y,z ∈ X, we have the following cases: k Case 1. If f(x)≥ f(y)≥ f(z),then ρ (x, y) = d(x,y) ≤ d(x,z) + d(z,y) = ρ (x, z) + kρ (z, y) k k k ≤ + k ρk(x, z) ρk(z, y) . Case 2. If f(x)>f(z)>f(y),then = ≤ + = + ρk(x, y) d(x,y) d(x,z) d(z,y) ρk(x, z) ρk(z, y). Case 3. If f(z)≥ f(x)≥ f(y),then = ≤ + ≤ + ρk(x, y) d(x,y) d(x,z) d(z,y) k(ρk(x, z) ρk(z, y)). Case 4. If f(x) Theorem 3.8. Let (X, ·) be a normed space and let f, g be functions as in Theorem 3.4 and k,l > 1. Then pf,k ≡ pg,l. Proof. Let x ∈ X; we may assume k x x Case 2. If f(x)<0 and g(x) < 0, then p (x) = > = p (x). f,k k l g,l x Case 3. If f(x)≥ 0 and g(x) < 0, then p (x) = x > = p (x). f,k l g,l x Case 4. If f(x)<0 and g(x) ≥ 0, then p (x) = < x = p (x). f,k k g,l Hence, kpf,k(x) satisfies for cases 1, 2 and 3 that kpf,k(x) ≥ pg,l(x), and case 4 that kpf,k(x) = pg,l(x). Likewise lpg,l(x) satisfies: for cases 1, 2 and 4 that lpg,l(x) ≥ pf,k(x), and case 3 that lpg,l(x) = pf,k(x), i.e. kpf,k(x) ≥ pg,l(x) and lpg,l(x) ≥ pf,k(x) for all x ∈ X. This is, pf,k ≡ pg,l. Theorem 3.9. Let (X, d) be a metric space with card(X) ≤ℵ1, let k,l > 1 and let : → R ≡ f, g X be injective functions. Then, ρf,k ρg,l (metrically). ∈ Proof. Let x,y X, we may assume k Hence, 1 kρ (x, y) ≥ ρ (x, y) ≥ ρ (x, y). f,k g,l l f,k ≡ Therefore, ρf,k ρg,l. References [1] Alegre C., I. Ferrando, L. M. García, and E. A. Sánchez-Pérez, Compactness in asymmetric normed spaces, Topology Appl., 155 (2008) 527–539. [2] Cobzas S., Functional Analysis in Asymmetric normed Spaces, Mathematics FA arXiv: 006.117v, 2010. [3] García-Raffi L. 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