In this paper we prove the reverse of triangle inequality via Selberg's  inequalities in the framework of Hilbert  C*-modules

Abdellatif Abdellatif Chahbi

Nordine Bounader

Samir Kabbaj

International Journal of Analysis and Applications

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International Journal of Analysis and ApplicationsISSN 2291-8639Volume 9, Number 1 (2015), 29-38http://www.etamaths.comREVERSE OF THE TRIANGLE INEQUALITY IN HILBERTC∗-MODULESNORDINE BOUNADER, ABDELLATIF CHAHBI∗ AND SAMIR KABBAJAbstract. In this paper we prove the reverse of triangle inequality via Sel-berg’s inequalities in the framework of Hilbert C∗-modules.1. IntroductionIn 1966, Diaz and Matcalf [4] proved the following reverse triangle inequality insetting of Hilbert spaces as follows .Theorem 1.1. Let x1, ..., xn be vectors in a Hilbert space H. If e is a unit vectorof H such that 0 ≤ r ≤ Re〈xi,e〉||xi|| for some r ∈ R and each 1 ≤ i ≤ n, thenrn∑j=1‖xi‖ ≤∥∥∥∥∥n∑i=1xi∥∥∥∥∥A number of mathematicians have represented several refinements of the reversetriangle inequality in Hilbert spaces and normed spaces, see[1, 2, 5, 8, 9, 12, 13]Recently, M. Khosravi, H. Mahyar and M.S. Moslehian [12] obtained the follow-ing reverse of the triangle inequality in the framework of Hilbert C∗-modules.Theorem 1.2. Let X be a Hilbert A-module and e1, ..., em ∈ X be a family ofvectors with 〈ei, ej〉 = 0 (1 ≤ i 6= j ≤ m) and ||ei|| = 1 (1 ≤ i ≤ m). If the vectorsx1, . . . , xn in X satisfy the conditionsRe 〈ek, xj〉 ≥ ρk ‖xj‖ , Im 〈ek, xj〉 ≥ µk ‖xj‖for j ∈ {1, . . . , n} , k ∈ {1, . . . ,m} , where ρk, µk ∈ [0,∞) k ∈ {1, . . . ,m} , then(m∑k=1(ρ2k + µ2k))12n∑j=1‖xj‖ ≤∥∥∥∥∥∥n∑j=1xj∥∥∥∥∥∥ .In [3] we obtained an extension of Selberg’s inequality in the framework of HilbertC∗-modules.The goal of this paper is to show the reverse of triangle inequality via a extensionof Selberg’s inequality in the framework of Hilbert C∗-modules. Our results are ex-tensions of theorem 2.1 and Corollary 2.3 obtained by Dragomir in [5] and theorem9 obtained by Fujii and Nakamoto see [9] in the setting of Hilbert C∗-modules.2010 Mathematics Subject Classification. Primary 46L08; Secondary 26D15, 46L05.Key words and phrases. triangle inequality; reverse inequality; Hilbert C∗−module;C∗−algebra.c©2015 Authors retain the copyrights of their papers, and allopen access articles are distributed under the terms of the Creative Commons Attribution License.2930 BOUNADER, CHAHBI AND KABBAJ2. PreliminariesIn this section we briefly recall the definitions and examples of Hilbert C∗-modules. For information about Hilbert C∗-module, we refer to [10, 13]. Ourreference for C∗-algebras as [15].Let A be a C∗-algebra (not necessarily unitary) and X be a complex linear space.Definition 2.1. A pre-Hilbert A-module is a right A-module X equipped with asesquilinear map 〈., .〉 : X × X → A satisfying(1) 〈x, x〉 ≥ 0; 〈x, x〉 = 0 if and only if x = 0 for all x in X ,(2) 〈x, αy + βz〉 = α 〈x, y〉+ β(〈x, z)〉 for all x, y, z in X , α, β in C,(3) 〈x, y〉 = 〈y, x〉∗ for all x, y in X ,(4) 〈x, y.a〉 = 〈x, y〉 a for all x, y in X , a in A.The map 〈., .〉 is called an A-valued inner product of X and for x ∈ X , we define||x|| = ||〈x, x〉||12 as a norm on X , where the latter is a norm in the C∗−algebra A.This norm makes X into a right normed module over A. A pre-Hilbert module Xis called a Hilbert A-module if it is complete with respect to its norm. Two typicalexamples of Hilbert C∗-modules are as follows:(I) Every Hilbert space is a Hilbert C∗-module.(II) Every C∗algebra A is a Hilbert A -module via 〈a, b〉 = a∗b (a, b ∈ A).Notice that the inner product structure of a C∗-algebra is essentially more com-plicated than complex numbers. One may define an A -valued norm |.| by |x| =〈x, x〉12 . Clearly, ‖x‖ = ‖|x|‖ for each x ∈ X .It is known that |.| does not satisfy the triangle inequality in general (see [[13],p.4]). We also use the elementary C∗-algebra theory, we use the following property:if a ≤ b then a 12 ≤ b 12 ,where a, b are positive elements of C∗-algebra A, and therelation 12 (aa∗ + a∗a) = Re(a)2 + Im(a)2 where a is an arbitrary element of A3. Main resultLet X be a right Hilbert A-module, which is an algebraic left A-module satisfy-ing:〈x, ay〉 = a 〈x, y〉 for all x, y ∈ X and a ∈ A.For example if A is a unital C∗-algebra and I is a commutative right ideal of A,then I is a right Hilbert module over A and〈x, ay〉 = x∗(ay) = ax∗y (x, y ∈ I, a ∈ A).For a reverse of triangle inequality, we use the following lemma.Lemma 3.1. Let X be a right Hilbert A- module which is an algebraic left A-moduleand y1, . . . , ym be a non zero vectors in X . If x ∈ X then(3.1)m∑j=1|〈yj , x〉|2∑mk=1 ‖〈yj , yk〉‖≤ |x|2and(3.2)m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖≤ |x|2REVERSE OF THE TRIANGLE INEQUALITY 31Proof. Inequality (3.1) is proved in [[3], theorem 3.1 ].Next we prove the inequality (3.2). Let αj ∈ A, j = 1, . . . , n. We know that0 ≤∣∣∣∣∣∣x−m∑j=1αjyj∣∣∣∣∣∣2=〈x−m∑j=1αjyj , x−m∑j=1αjyj〉= 〈x, x〉 −〈x,m∑j=1αjyj〉−〈m∑j=1αjyj , x〉+〈m∑j=1αjyj ,m∑j=1αjyj〉= |x|2 −m∑j=1αj 〈x, yj〉 −m∑j=1〈yj , x〉α∗j +m∑j,k=1αj 〈yk, yj〉α∗k= |x|2 −m∑j=1αj 〈x, yj〉 −m∑j=1〈yj , x〉α∗j+12m∑j,k=1(αj 〈yk, yj〉α∗k + αk 〈yj , yk〉α∗j ).It follows from [[6], lemma 3.2] thatαj 〈yj , yk〉α∗k + αk 〈yk, yj〉α∗j ≤∣∣α∗j ∣∣2 ‖〈yj , yk〉‖+ |α∗k|2 ‖〈yj , yk〉‖ ,so∣∣∣∣∣∣x−m∑j=1αjyj∣∣∣∣∣∣2≤ |x|2 −m∑j=1αj 〈x, yj〉 −m∑j=1〈yj , x〉α∗j+12m∑j,k=1(|α∗j |2| 〈yj , yk〉 |+ |α∗k|2 ‖〈yk, yj〉‖),Takeαj =〈yj , x〉∑mk=1 ‖〈yj , yk〉‖,A simple calculation shows that∣∣∣∣∣∣x−m∑j=1αjyj∣∣∣∣∣∣2≤ |x|2 −m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖−m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖+12m∑j,k=1|〈x, yj〉|2 ‖〈yj , yk〉‖(∑mk=1 ‖〈yj , yk〉‖)2+12n∑j,k=1|〈x, yj〉|2 ‖〈yj , yk〉‖(∑mk=1 ‖〈yj , yk〉‖)2.32 BOUNADER, CHAHBI AND KABBAJSince|x|2 −m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖−m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖+12m∑j,k=1|〈x, yj〉|2 ‖〈yj , yk〉‖(∑mk=1 ‖〈yj , yk〉‖)2+12m∑j,k=1|〈x, yj〉|2 ‖〈yj , yk〉‖(∑mk=1 ‖〈yj , yk〉‖)2= |x|2 − 2m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖+n∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖= |x|2 −m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖.It follows that ∣∣∣∣∣∣x−m∑j=1αjyj∣∣∣∣∣∣2≤ |x|2 −m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖,hence|x|2 −m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖≥ 0.The proof is then completed. Corollary 3.2. Let X be a right Hilbert A− module which is an algebraic leftA-module, then :| 〈y, x〉 |2 ≤ ‖y‖2 |x|2and| 〈x, y〉 |2 ≤ ‖y‖2 |x|2 .Corollary 3.3. Let X be a right Hilbert A− module which is an algebraic left A-module. If y1, . . . , yn is a sequence of unit vectors in X such that 〈yj , yk〉 = 0 for1 ≤ j 6= k ≤ n. Thenm∑j=1|〈yj , x〉|2 ≤ |x|2 .andm∑j=1|〈x, yj〉|2 ≤ |x|2 .Theorem 3.4. Let X be a right Hilbert A- module which is an algebraic left A-module, x1, . . . , xn and y1, . . . , ym be a non zero vectors in X such that there existthe non-negative real numbers ρj , µj , j ∈ {1, . . . ,m} with(3.3) Re 〈yj , xi〉 ≥ ρj ‖xi‖ ‖yj‖ , Im 〈yj , xi〉 ≥ µj ‖xi‖ ‖yj‖for each i ∈ {1, . . . , n} , j ∈ {1, . . . ,m} . Then(m∑j=1(ρ2j + µ2j ) ‖yj‖2∑mk=1 ‖〈yj , yk〉‖)12n∑i=1|xi| ≤∣∣∣∣∣n∑i=1xi∣∣∣∣∣ .REVERSE OF THE TRIANGLE INEQUALITY 33Proof. By (3.3), we have(n∑i=1Re 〈yj , xi〉)2 + (n∑i=1Im 〈yj , xi〉)2 ≥ ρ2j |yj |2(n∑i=1|xi|)2 + µ2kj ‖yj‖2(n∑i=1‖xi‖)2= (ρ2j + µ2j )|yj |2(n∑i=1|xi|)2.By combining the above inequality and this equality12 (|〈∑ni=1 xi, yj〉|2+ |〈yj ,∑ni=1 xi〉|2) = (∑ni=1Re 〈yj , xi〉)2 + (∑ni=1 Im 〈yj , xi〉)2,we deduce12∑mj=1|〈yj ,∑ni=1 xi〉|2∑mk=1‖〈yj ,yk〉‖+ 12∑mj=1|〈∑ni=1 xi,yj〉|2∑mk=1‖〈yj ,yk〉‖≥ (∑mj=1(ρ2j+µ2j )‖yj‖2∑mk=1‖〈yj ,yk〉‖)(∑ni=1 ‖xi‖)2.Apply lemma 3.1, we get12m∑j=1|〈yj ,∑ni=1 xi〉|2∑mk=1 ‖〈yj , yk〉‖+12m∑j=1|〈∑ni=1 xi, yj〉|2∑mk=1 ‖〈yj , yk〉‖≤ |n∑i=1xi|2.Then(m∑j=1(ρ2j + µ2j ) ‖yj‖2∑mk=1 ‖〈yj , yk〉‖)(n∑i=1||xi| |)2 ≤∥∥∥∥∥n∑i=1xi∥∥∥∥∥2.And since |x| ≤ ‖x‖ and |x|2 ≤ ‖x‖2 for all x ∈ X , then(m∑j=1(ρ2j + µ2j ) ‖yj‖2∑mk=1 ‖〈yj , yk〉‖)(n∑i=1|xi|)2 ≤∣∣∣∣∣n∑i=1xi∣∣∣∣∣2.The desired result follows by taking the square roots. Remark 3.5. If the first condition of (3.3) is the only one available, then(m∑j=1ρ2j ‖yj‖2∑mk=1 ‖〈yj , yk〉‖)12n∑i=1|xi| ≤∣∣∣∣∣n∑i=1xi∣∣∣∣∣ .Corollary 3.6. Let X be a right Hilbert A−module which is an algebraic left A-module, x1, . . . , xn and y1, . . . , ym be a non zero vectors in X such that there existthe non-negative real numbers ρj , µj , j ∈ {1, . . . ,m} withRe 〈yj , xi〉 ≥ ρj ‖xi‖ ‖yj‖ , Im 〈yj , xi〉 ≥ µj ‖xi‖ ‖yj‖ .Then(m∑j=1(ρ2j + µ2j )|yj |2max1≤j≤m |yj |2 + (m− 1) maxk 6=j | 〈yj , yk〉 |)12n∑i=1|xi| ≤∣∣∣∣∣n∑i=1xi∣∣∣∣∣ .Proof. It is easy to show thatm∑k=1‖〈yj , yk〉‖ ≤ max1≤j≤m‖yj‖2 + (m− 1) maxj 6=k‖〈yj , yk〉‖ .We thus have that1max1≤j≤m ‖yj‖2 + (m− 1) maxj 6=k ‖〈yj , yk〉‖≤ 1∑mk=1 ‖〈yj , yk〉‖,34 BOUNADER, CHAHBI AND KABBAJand(ρ2j + µ2j ) ‖yj‖2max1≤j≤m ‖yj‖2 + (m− 1) maxj 6=k ‖〈yj , yk〉‖≤(ρ2j + µ2j ) ‖yj‖2∑mk=1 ‖〈yj , yk〉‖.Consequently(∑nj=1(ρ2j+µ2j )‖yj‖2max1≤j≤n‖yj‖2+(m−1)maxj 6=k‖〈yj ,yk〉‖)12∑ni=1 |xi| ≤ (∑nj=1(ρ2j+µ2j )‖yj‖2∑nk=1‖〈yj ,yk〉‖)12∑ni=1 |xi| .We apply the theorem 3.4 to get the result The next corollary is the theorem 2.5 in [12].Corollary 3.7. Let X be a Hilbert A-module, x1, . . . , xn be a family of vectors inX and ej be a unitary orthogonal vectors for j ∈ {1, . . . ,m} in X such that thereexist the real numbers ρj , µj , j ∈ {1, . . . ,m} withRe 〈ρjxi, ej〉 ≥ ρ2j ‖xi‖ , Im 〈µjxi, ej〉 ≥ µ2j ‖xi‖for each i ∈ {1, . . . , n} , j ∈ {1, . . . ,m} . Then(m∑j=1(ρ2j + µ2j ))12n∑j=1‖xi‖ ≤∥∥∥∥∥n∑i=1xi∥∥∥∥∥ .Corollary 3.8. Let X be a right Hilbert A−module which is an algebraic left A-module, x1, . . . , xn and y1 . . . , ym be a non zero vectors in X , such that there existthe non-negative real number in [0; 1] pj , qj , j ∈ {1, . . . ,m} with(3.4) ||xi||2 − 2Re 〈yj , xi〉+ ||yj ||2 ≤ p2j ≤ ||yj ||2and(3.5) ||xi||2 − 2 Im 〈yj , xi〉+ ||yj ||2 ≤ p2j ≤ ||yj ||2for each i ∈ {1, . . . , n} , j ∈ {1, . . . ,m} . Then(m∑j=1(2 ‖yj‖2 − p2j − qj2)∑mk=1 ‖〈yj , yk〉‖)12n∑i=1|xi| ≤∣∣∣∣∣n∑i=1xi∣∣∣∣∣ .Proof. By the inequality (3.4), we get||xi||2 + ||yj ||2 − p2j ≤ 2Re 〈yj , xi〉 .Since ||yj ||2 − p2j ≥ 0, then2||xi||√||yj ||2 − p2j ≤ ||xi||2 + ||yj ||2 − p2j ≤ 2Re 〈yj , xi〉 .This implies thatRe 〈yj , xi〉 ≥√||yj ||2 − p2j||yj ||||yj ||||xi||.Even from (3.5), we getIm 〈yj , xi〉 ≥√||yj ||2 − q2j||yj ||||yj ||||xi||,and if we let ρj =√||yj ||2−p2j||yj and µj =√||yj ||2−p2j||yj in theorem 3.4,then by simple computation, we get the desired result. REVERSE OF THE TRIANGLE INEQUALITY 35The following lemma gives a refinement of Selberg’s inequality in a right HilbertA− module which is an algebraic left A-module.Lemma 3.9. Let X be a right Hilbert A− module which is an algebraic left A-module and y1, . . . , ym be a non zero vectors in X . If x ∈ X then(3.6) | 〈y, x〉 |2 +m∑j=1|〈yj , x〉|2∑mk=1 ‖〈yj , yk〉‖||y||2 ≤ |x|2 ||y||2,(3.7) | 〈x, y〉 |2 +m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖||y||2 ≤ |x|2 ||y||2,(3.8) | 〈x, y〉 |2 +m∑j=1|〈yj , x〉|2∑mk=1 ‖〈yj , yk〉‖||y||2 ≤ |x|2 ||y||2,and(3.9) | 〈y, x〉 |2 +m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖||y||2 ≤ |x|2 ||y||2.Proof. Inequality ( 3.6) is proved in [ [3],theorem 3.3].Now we prove the inequality (3.7), letu = x−m∑j=1αjyjwhereαj =〈yj , x〉∑nk=1 ‖〈yj , yk〉‖.According to the proof of lemma 3.1, we have|u|2 = |x−m∑j=1αjyj |2≤ |x|2 −m∑j=1| 〈yj , x〉 |2∑mk=1 | 〈yj , yk〉 |.Hence it follows that‖y‖2|x|2 − m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖ ≥ ‖y‖2 |u|2 .Applying Cauchy Schwartz inequality, we get‖y‖2 |u|2 ≥ |〈u, y〉|2 .and since 〈y, yj〉 = 0, so|〈u, y〉|2 =∣∣∣∣∣∣〈x−m∑j=1αjyj , y〉∣∣∣∣∣∣2= |〈x, y〉|2 .36 BOUNADER, CHAHBI AND KABBAJIt follows that‖y‖2|x|2 − m∑j=1|〈x, yj〉|2∑mk=1 ‖〈yj , yk〉‖ ≥ |〈x, y〉|2 ,which completes the proof of the inequality (3.7).Similarly, we can get inequalities (3.8) and (3.9) . Lemma 3.10. Let X be a Hilbert A - module, y1, · · · , ym, y be non zero vectors inX and x1, · · · , xn ∈ X such that such that there exist the real numbers ρj , µj , j ∈{1, · · · ,m} with(3.10) 0 ≤ ρj ‖xi‖ ||yj || ≤ Re 〈yj , xi〉 , 0 ≤ µj ‖xi‖ ||yj || ≤ Im 〈yj , xi〉and 〈y, yj〉 = 0, for each i ∈ {1, . . . , n} , j ∈ {1, . . . ,m} . Then(3.11) |(y,n∑i=1xi)|2 + (m∑k=1ρ2j + µ2jcj||y2j )(n∑i=1|xi|)2||y||2 ≤ |n∑i=1xi|2||y||2,and(3.12) |(n∑i=1xi, y)|2 + (m∑j=1ρ2j + µ2jcj||yj ||2)(n∑i=1|xi|)2||y||2 ≤ |n∑i=1xi|2||y||2,where cj =∑mk=1 ‖〈yj , yk〉‖ .Proof. Let x =∑ni=1 xi, from (3.10), we get||y||2{|x|2 −∑mj=1ρ2j+µ2jcj(∑ni=1 ||xi||)2}≥ ||y||2{|x|2 −∑mj=1Re〈x,yj〉2+Im〈x,yj〉2cj},Since||y||2{|x|2−∑mj=1Re〈x,yj〉2+Im〈x,yj〉2cj} = ‖y‖2 {|x|2− 12∑mj=1|〈x,yj〉|2cj− 12∑mj=1〈yj ,x〉2cj}.Then, from (3.6) and (3.9), we get‖y‖2 {|x|2 − 12m∑j=1| 〈x, yj〉 |2cj− 12m∑j=1〈yj , x〉2cj} ≥ | 〈y, x〉 |2,it follows that||y||2{|x|2 −m∑j=1ρ2j + µ2jcj(‖x1‖+ · · ·+ ‖xn‖)2} ≥ |〈y, x〉|2 .By using (3.7) and (3.8) and by similar argument, we get (3.12). Theorem 3.11. Let X be a Hilbert A - module, y1, · · · , ym be non zero vec-tors in X and x1, · · · , xn ∈ X such that such that there exist the real numbersa, b, ρj , µj , j ∈ {1, · · · ,m} with(3.13) 0 ≤ ρj ‖xi‖ ||yj || ≤ Re 〈yj , xi〉 , 0 ≤ µj ‖xi‖ ||yj || ≤ Im 〈yj , xi〉 ,(3.14) 0 ≤ a ‖xi‖ ||y|| ≤ Re 〈y, xi〉 , 0 ≤ b ‖xi‖ ||yj || ≤ Im 〈y, xi〉REVERSE OF THE TRIANGLE INEQUALITY 37and 〈y, yj〉 = 0, for each i ∈ {1, . . . , n} , j ∈ {1, . . . ,m} . Then(a2 + b2 +m∑j=1ρ2j + µ2jcj)||y2j ||12 (|x1|+ · · ·+ |xn|) ≤ |n∑i=1xi|.where cj =∑mk=1 ‖〈yj , yk〉‖ .Proof. From (3.11) and (3.12), we get12 (| 〈∑ni=1 xi, y〉 |2 + | 〈y,∑ni=1 xi〉 |2)+(∑mj=1ρ2j+µ2jcj||y||2)(∑ni=1 |xi|)2||y||2≤ |∑ni=1 xi|2||y||2,since12 (| 〈∑ni=1 xi, y〉 |2 + | 〈y,∑ni=1 xi〉 |2) ≥ (Re 〈y, x〉)2 + (Im 〈y, x〉)2applying (3.13) and (3.14) and taking the square root, the desired result follows. Remark 3.12. If in Theorem 3.11 y1, . . . , ym is a sequence orthonormal vectors,then(a2 + b2 +m∑j=1(ρ2j + µ2j ))12 (n∑i=1|xi|) ≤∣∣∣∣∣n∑i=1xi∣∣∣∣∣ .This inequality is an extension of Diaz-Metcalf [4] inequality in C∗-module.References[1] A.H. Ansari and M.S.Moslehian, Refinement of reverse triangle inequality in inner productspaces.J. Inequalities in Pure and Applied Math., 6(2005). Article ID 64.[2] A.H. Ansari and M.S.Moslehian, More on reverse triangle inequality in inner product spaces.Inter J. Math. 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Reverse of the Triangle Inequality in Hilbert C*-Modules

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