In this note we establish a general inequality valid in metric spaces

that is related to the polygonal inequality and admits also a natural geometrical

interpretation. Particular instances of interest holding in normed linear spaces

and inner product spaces are pointed out as well

Dragomir, Sever S

Goşa, Anca C

English

In this note we establish a general inequality valid in metric spaces that is related to the polygonal inequality and admits also a natural geometrical interpretation. Particular instances of interest holding in normed linear spaces and inner product spaces are pointed out as well

Victoria University Eprints Repository

An inequality in metric spaces

Victoria University Institutional Repository (VUIR)

An Inequality in Metric Spaces

AN INEQUALITY IN METRIC SPACESSEVER S. DRAGOMIR AND ANCA C. GOS¸AAbstract. In this note we establish a general inequality valid in metric spacesthat is related to the polygonal inequality and admits also a natural geometricalinterpretation. Particular instances of interest holding in normed linear spacesand inner product spaces are pointed out as well.1. IntroductionLet X be a nonempty set. A function d : X × X → [0,∞) is called a distanceon X if the following properties are satisfied:(d) d (x, y) = 0 if and only if x = y;(dd) d (x, y) = d (y, x) for any x, y ∈ X (the symmetry of the distance);(ddd) d (x, y) ≤ d (x, z)+ d (z, y) for any x, y, z ∈ X (the triangle inequality).The pair (X, d) is called in the literature a metric space.Important examples of metric spaces are normed linear spaces. We recall that,a linear space E over the real or complex number field K endowed with a function‖·‖ : E → [0,∞), is called a normed space if ‖·‖ , the norm, satisfies the properties(n) ‖x‖ = 0 if and only if x = 0;(nn) ‖αx‖ = |α| ‖x‖ for any scalar α ∈ K and any vector x ∈ E;(nnn) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for each x, y ∈ E (the triangle inequality).Further, we recall that, the linear space H over the real or complex number fieldK endowed with an application 〈·, ·〉 : H ×H → K is called an inner product space,if the function 〈·, ·〉 , called the inner product, satisfies the following properties:(i) 〈x, x〉 ≥ 0 for any x ∈ H and 〈x, x〉 = 0 if and only if x = 0;(ii) 〈αx+ βy, z〉 = α 〈x, z〉+β 〈y, z〉 for any scalars α, β and any vectors x, y, z;(iii) 〈y, x〉 = 〈x, y〉 for any x, y ∈ H.It is well know that the function ‖x‖ :=√〈x, x〉 defines a norm on H and thusan important example of normed spaces are the inner product spaces.A fundamental inequality in metric spaces, which obviously follows by the trian-gle inequality and mathematical induction, is the generalised triangle inequality, orthe polygonal inequality which states that: for any points x1, x2, ..., xn−1, xn (n ≥ 3)in a metric space (X, d) , we have the inequality(1.1) d (x1, xn) ≤ d (x1, x2) + ...+ d (xn−1, xn) .Date: March 22, 2004.2000 Mathematics Subject Classification. Primary 51Fxx, 46B20; Secondary 26D15, 26D10.Key words and phrases. Metric Spaces, Polygonal Inequality, Triangle Inequality, Inequalitiesfor Norms.12 SEVER S. DRAGOMIR AND ANCA C. GOS¸AThe main aim of this note is to point out a general inequality valid in metricspaces that is related to the polygonal inequality and admits also a natural geo-metrical interpretation. Particular instances of interest holding in normed linearspaces and inner product spaces are pointed out as well.2. The ResultsThe following result in the general setting of metric spaces holds.Theorem 1. Let (X, d) be a metric space and xi ∈ X, pi ≥ 0 (i ∈ {1, ..., n}) with∑ni=1 pi = 1. Then we have the inequality(2.1)∑1≤i<j≤npipjd (xi, xj) ≤ infx∈X[n∑i=1pid (xi, x)].The inequality is sharp in the sense that the multiplicative constant c = 1 in frontof ” inf ” cannot be replaced by a smaller quantity.Proof. Using the triangle inequality, we have for any x ∈ X and i, j ∈ {1, ..., n},that(2.2) d (xi, xj) ≤ d (xi, x) + d (x, xj) .If we multiply (2.2) with pipj ≥ 0 and sum over i and j from 1 to n, then we deduce(2.3)n∑i,j=1pipjd (xi, xj) ≤n∑i,j=1pipj [d (xi, x) + d (x, xj)] .However, by the symmetry of distance,n∑i,j=1pipjd (xi, xj) = 2∑1≤i<j≤npipjd (xi, xj)andn∑i,j=1pipj [d (xi, x) + d (x, xj)] = 2[n∑i=1pid (xi, x)]therefore, by (2.3), we deduce(2.4)∑1≤i<j≤npipjd (xi, xj) ≤n∑i=1pid (xi, x) ,for any x ∈ X.Taking the infimum over x in (2.4), we deduce the desired inequality (2.1).Now, suppose that (2.1) holds with a constant c > 0, i.e.,(2.5)∑1≤i<j≤npipjd (xi, xj) ≤ c infx∈X[n∑i=1pid (xi, x)].Then, on choosing n = 2, p1 = p, p2 = 1− p, p ∈ (0, 1) , we deduce(2.6) p(1− p)d (x1, x2) ≤ c [pd (x1, x) + (1− p)d (x, x2)]for any x ∈ X and p ∈ (0, 1) . If in this inequality we let x = x1, then we getpd (x1, x2) ≤ cd (x1, x2)AN INEQUALITY IN METRIC SPACES 3for any x1, x2 ∈ X and p ∈ (0, 1) which implies that c ≥ 1, and the proof iscomplete.The following particular case holds.Corollary 1. Let (X, d) be a metric space and xi ∈ X (i ∈ {1, ..., n}) . Then wehave the inequality(2.7)∑1≤i<j≤nd (xi, xj) ≤ n infx∈X[n∑i=1d (xi, x)].The proof is obvious from the above theorem on choosing pi = 1n , i ∈ {1, ..., n}.The above corollary has an interesting geometrical interpretation:Proposition 1. The sum of all edges and diagonals of a polygon with n verticesin a metric space is less than n-times the sum of the distances from any arbitrarypoint in the space to its vertices.The following corollary holds as well.Corollary 2. Let (X, d) be a metric space and xi ∈ X, (i ∈ {1, ..., n}) . If thereexists a closed ball of radius r > 0 centered in a point x containing all the pointsxi, i.e., xi ∈ B (x, r) := {y ∈ X : d (x, y) ≤ r}, then for any pi ≥ 0 (i ∈ {1, ..., n})with∑ni=1 pi = 1 we have the inequality(2.8)∑1≤i<j≤npipjd (xi, xj) ≤ r.The proof is obvious from the above Theorem 1 and we omit the details.3. ApplicationsIf (E, ‖·‖) is a normed linear space and xi ∈ E, (i ∈ {1, ..., n}) , pi ≥ 0 (i ∈ {1, ..., n})with∑ni=1 pi = 1, then by (2.1) we have the inequality(3.1)∑1≤i<j≤npipj ‖xi − xj‖ ≤ infx∈X[n∑i=1pi ‖xi − x‖].In particular, for the uniform distribution pi = 1n , we have(3.2)∑1≤i<j≤n‖xi − xj‖ ≤ n infx∈X[n∑i=1‖xi − x‖].We can state the following results as well.Proposition 2. Let (E, ‖·‖) be a normed linear space and xi ∈ E, (i ∈ {1, ..., n}) ,pi ≥ 0 (i ∈ {1, ..., n}) with∑ni=1 pi = 1. Denote xp :=∑ni=1 pixi. Then we have theinequalities(3.3)12n∑i=1pi ‖xi − xp‖ ≤∑1≤i<j≤npipj ‖xi − xj‖ ≤n∑i=1pi ‖xi − xp‖ .The constant 12 is best possible in the sense that it cannot be replaced by a largerquantity.4 SEVER S. DRAGOMIR AND ANCA C. GOS¸AProof. The second inequality is obvious by (3.1).By the generalised triangle inequality we have∑1≤i<j≤npipj ‖xi − xj‖ = 12n∑i,j=1pipj ‖xi − xj‖≥ 12n∑i=1pi∥∥∥∥∥∥xi −n∑j=1pjxj∥∥∥∥∥∥ = 12n∑i=1pi ‖xi − xp‖ ,proving the first part of (3.3).Now, assume that the first inequality holds with a constant k > 0, i.e.,(3.4) kn∑i=1pi ‖xi − xp‖ ≤∑1≤i<j≤npipj ‖xi − xj‖under the hypothesis of the proposition stated above. Then, by (3.4) for n = 2 andp1 = p2 = 12 we deduce12k ‖x1 − x2‖ ≤ 14 ‖x1 − x2‖ ,for any x1, x2 ∈ E, implying k ≤ 12 , and the proposition is proved.Remark 1. It is an open question whether the multiplicative constant c = 1 in thesecond part of (3.3) is sharp or not in the general setting of normed linear spaces.The following particular case with a simple geometric interpretation holds.Corollary 3. Let (E, ‖·‖) be a normed linear space and xi ∈ E, (i ∈ {1, ..., n}) . Ifx :=x1 + ...+ xnndenotes the gravity center of the vectors xi, i ∈ {1, ..., n}, then we have the inequality(3.5)12nn∑i=1‖xi − x‖ ≤∑1≤i<j≤n‖xi − xj‖ ≤n∑i=1‖xi − x‖ .The constant 12 in the first inequality is sharp.Remark 2. Geometrically, the inequality (3.5) means that: the sum of all edgesand diagonals of a polygon with n vertices in a normed linear space is less thann-times the sum of the distances from the gravity center to its vertices and greaterthan n2 -times the same quantity.Finally, in the case of inner product spaces, we may point out an upper boundas follows.Proposition 3. Let (H, 〈·, ·〉) be an inner product space, xi ∈ H, (i ∈ {1, ..., n})and assume that there exists the vectors a,A ∈ H so that eitherRe 〈A− xi, xi − a〉 ≥ 0, for i ∈ {1, ..., n},or, equivalently, ∥∥∥∥xi − a+A2∥∥∥∥ ≤ 12 ‖A− a‖ , for i ∈ {1, ..., n}.AN INEQUALITY IN METRIC SPACES 5Then for any pi ≥ 0 (i ∈ {1, ..., n}) with∑ni=1 pi = 1 one has the inequality(3.6)∑1≤i<j≤npipj ‖xi − xj‖ ≤ 12 ‖A− a‖ .The proof is obvious by Corollary 2 and we omit the details.Remark 3. It is an open problem if 12 in (3.6) is the best possible constant in thegeneral case of inner product spaces.For other classical and recent results related to the triangle and polygonalinequality, see the papers [1]- [3], [5], Chapter XVII of the book [4] and the referencestherein.References[1] J. B. DIAZ and F. T. METCALF, A complementary triangle inequality in Hilbert and Banachspaces, Proc. Amer. Math. Soc., 17 (1966), 88–97.[2] S.M. KHALEELULLA, On Diaz-Metcalf’s complementary triangle inequality, KyungpookMath. J., 15 (1975), 9–11.[3] P.M. MILICˇIC´, On a complementary inequality of the triangle inequality (French), Mat. Ves-nik, 41 (1989), no. 2, 83–88.[4] D.S. MITRINOVIC´, J.E. PECˇARIC´ and A.M. FINK, Classical and New Inequalities in Anal-ysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.[5] D. K. RAO, A triangle inequality for angles in a Hilbert space, Rev. Colombiana Mat., 10(1976), no. 3, 95–97.School of Computer Science & Mathematics, Victoria University, PO Box 14428, MC8001 Melbourne City, Victoria, AustraliaE-mail address: sever@matilda.vu.edu.auURL: http://rgmia.vu.edu.au/SSDragomirWeb.htmlCollege No. 12 Res¸it¸a, Jud. Caras¸-Severin, R0-1700, Res¸it¸a, RomaniaE-mail address: ancagosa@hotmail.com

A complementary triangle inequality in Hilbert and Banach spaces,

A triangle inequality for angles in a Hilbert space,

Classical and New Inequalities in Analysis,

On a complementary inequality of the triangle inequality

On Diaz-Metcalf’s complementary triangle inequality,

In this note we establish a general inequality valid in metric spaces\ud
that is related to the polygonal inequality and admits also a natural geometrical\ud
interpretation. Particular instances of interest holding in normed linear spaces\ud
and inner product spaces are pointed out as well

An Inequality in Metric Spaces

Abstract

Similar works

Full text

Available Versions

Victoria University Eprints Repository

Victoria University Institutional Repository (VUIR)

Victoria University Eprints Repository

Victoria University Institutional Repository (VUIR)

Name not available

Name not available