Some sharp discrete inequalities in normed linear spaces are obtained.

New reverses of the generalised triangle inequality are also given

Dragomir, Sever S

English

Victoria University Eprints Repository

ON SOME DISCRETE INEQUALITIES IN NORMED LINEARSPACESSEVER S. DRAGOMIRAbstract. Some sharp discrete inequalities in normed linear spaces are ob-tained. New reverses of the generalised triangle inequality are also given.1. IntroductionLet (X, ‖·‖) be a normed linear space over the real or complex number field K.The mapping f : X → R, f (x) = 12 ‖x‖2 is obviously convex on R and then thereexists the following limits:〈x, y〉i := limt→0−‖y + tx‖2 − ‖y‖22t,〈x, y〉s := limτ→0−‖y + τx‖2 − ‖y‖22τfor any two vectors in X. The mapping 〈·, ·〉s (〈·, ·〉i) will be called the superiorsemi-inner product (inferior semi-inner product) associated to the norm ‖·‖ .The following fundamental calculus rules are valid for these semi-inner products(see for instance [4, p. 27–32]):〈x, x〉p = ‖x‖2 for x ∈ X;(1.1)〈λx, y〉p = λ 〈x, y〉p for λ ≥ 0 and x, y ∈ X;(1.2)〈x, λy〉p = λ 〈x, y〉p for λ ≥ 0 and x, y ∈ X;(1.3)〈λx, y〉p = λ 〈x, y〉q for λ ≤ 0 and x, y ∈ X;(1.4)〈αx, βy〉p = αβ 〈x, y〉p for α, β ∈ R with αβ ≥ 0 and x, y ∈ X;(1.5)〈−x, y〉p = 〈x,−y〉p = −〈x, y〉q for x, y ∈ X;(1.6)where p, q ∈ {s, i} and p 6= q.The following inequality is valid:‖y + tx‖2 − ‖y‖22t≥ 〈x, y〉s ≥ 〈x, y〉i(1.7)≥ ‖y + sx‖2 − ‖y‖22s,for any x, y ∈ X and s < 0 < t.Date: 4 August, 2005.2000 Mathematics Subject Classification. Primary 46B05, 26D15.Key words and phrases. Inequalities in normed spaces, Semi-inner products, Analyticinequalities.12 SEVER S. DRAGOMIRAn important result is the following Schwarz inequality :(1.8)∣∣∣〈x, y〉p∣∣∣ ≤ ‖x‖ ‖y‖ for each x, y ∈ X.Also, the following properties of sub(super)-additivity should be noted:(1.9) 〈x1 + x2, y〉s(i) ≤ (≥) 〈x1, y〉s(i) + 〈x2, y〉s(i)for each x1, x2, y ∈ X.Another important property of “quasi-linearity” holds as well:(1.10) 〈αx+ y, x〉p = α ‖x‖2 + 〈y, x〉pfor any x, y ∈ X and α a real number, where p = s or p = i.Finally, we mention the continuity property:(1.11)∣∣∣〈y + z, x〉p − 〈z, x〉p∣∣∣ ≤ ‖y‖ ‖x‖for each x, y, z ∈ X and p = s or p = i.One of the most used inequalities in normed spaces is the triangle inequality forseveral vectors, i.e.,(1.12)∥∥∥∥∥∥n∑j=1xj∥∥∥∥∥∥ ≤n∑j=1‖xj‖for any xj ∈ X, j ∈ {1, . . . , n} .The main aim of this paper is to point out some inequalities for norms of thevectors xj and∑nj=1 xj , including some reverses of the triangle inequality in themultiplicative form, i.e., lower bounds for the quantity∥∥∥∑nj=1 xj∥∥∥∑nj=1 ‖xj‖,provided that not all xj are zero and satisfy some appropriate conditions.For classical results related to the reverse of the triangle inequality in normedspaces see [3], [7], [9] and [8]. For more recent results, see [5], [6], [1] and [2].2. The ResultsThe following lemma is of interest itself as well.Lemma 1. Let (X, ‖·‖) be a normed linear space. If x, a ∈ X, then(2.1) 〈x, a〉i ≥12(‖a‖2 − ‖x− a‖2).If ‖a‖ > ‖x− a‖ , then the constant 12 cannot be replaced by a larger quantity.Proof. Utilising the semi-inner product properties, we have by (1.7) that〈x, a〉i = lims→0−‖a+ sx‖2 − ‖a‖22s≥ ‖a+ (−1)x‖2 − ‖a‖22 (−1) =‖a‖2 − ‖x− a‖22and the inequality (2.1) is proved.Now, assume that ‖a‖ > ‖x− a‖ and there exists a C > 0 with the propertythat(2.2) 〈x, a〉i ≥ C(‖a‖2 − ‖x− a‖2).INEQUALITIES IN NORMED LINEAR SPACES 3Obviously a 6= 0, and if we choose x = εa, ε ∈ (0, 1) , then ‖a‖ > ‖x− a‖ since‖x− a‖ = (1− ε) ‖a‖ . Replacing x in (2.2) we getε ‖a‖2 ≥ C(‖a‖2 − (1− ε)2 ‖a‖2)givingε ≥ C (2ε− ε2) ,for any ε ∈ (0, 1) . This is in fact 1 ≥ C (2− ε) and if we let ε → 0+, we getC ≥ 12 .Remark 1. As a coarser, but maybe more useful inequality, we can state that(2.3) 〈x, a〉i ≥12‖x‖ (‖a‖ − ‖x− a‖) ,provided ‖a‖ ≥ ‖x− a‖ .We observe that (2.3) follows from (2.1) since, for ‖a‖ ≥ ‖x− a‖ , the triangleinequality gives:12(‖a‖2 − ‖x− a‖2)=12(‖a‖ − ‖x− a‖) (‖a‖+ ‖x− a‖)≥ 12(‖a‖ − ‖x− a‖) ‖x‖ .It is an open question whether the constant 12 in (2.3) is sharp.The following result may be stated.Theorem 1. Let (X, ‖·‖) be a normed space and xj ∈ X, j ∈ {1, . . . , n}, a ∈X\ {0} . Then for any pj ≥ 0, j ∈ {1, . . . , n} with∑nj=1 pj = 1 we have(2.4)∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ ‖a‖+ 12n∑j=1pj ‖xj − a‖2 ≥ 12 ‖a‖2.The constant 12 in the right hand side of (2.4) is best possible in the sense that itcannot be replaced by a larger quantity.Proof. We apply Lemma 1 on stating that〈xj , a〉i +12‖xj − a‖2 ≥ 12 ‖a‖2for each j ∈ {1, . . . , n} .Multiplying with pj ≥ 0 and summing over j from 1 to n, we get(2.5)n∑j=1pj 〈xj , a〉i +12n∑j=1pj ‖xj − a‖2 ≥ 12 ‖a‖2n∑j=1pj .Utilising the superadditivity property of the semi-inner product 〈·, ·〉i in the firstvariable (see [4, p. 29]) we have(2.6)〈n∑j=1pjxj , a〉i≥n∑j=1pj 〈xj , a〉i .4 SEVER S. DRAGOMIRBy the Schwarz inequality applied for∑nj=1 pjxj and a, we also have(2.7)∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ ‖a‖ ≥〈n∑j=1pjxj , a〉i.Therefore, by (2.5) – (2.7) we deduce the desired inequality (2.4).Now assume that there exists a D > 0 with the property that(2.8)∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ ‖a‖+ 12n∑j=1pj ‖xj − a‖2 ≥ D ‖a‖2 ,for any n ≥ 1, xj ∈ X, pj ≥ 0, j ∈ {1, . . . , n} with∑nj=1 pj = 1 and a ∈ X\ {0} .If in (2.8) we choose n = 1, p1 = 1, x1 = εa, ε ∈ (0, 1) , then we getε ‖a‖2 + 12(1− ε)2 ‖a‖2 ≥ D ‖a‖2 ,givingε+12(1− ε)2 ≥ D,for any ε ∈ (0, 1) . Letting ε→ 0+, we deduce D ≤ 12 and the proof is complete.The following result may be stated as well:Proposition 1. Let xj , a ∈ X with a 6= 0 and ‖xj − a‖ ≤ ‖a‖ for each j ∈{1, . . . , n} . Then for any pj ≥ 0, j ∈ {1, . . . , n} with∑nj=1 pj = 1 we have(2.9)∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ ‖a‖+ 12n∑j=1pj ‖xj‖ ‖xj − a‖ ≥ 12 ‖a‖n∑j=1pj ‖xj‖ .Proof. From (2.3) we have〈xj , a〉i +12‖xj‖ ‖xj − a‖ ≥ 12 ‖a‖ ‖xj‖for any j ∈ {1, . . . , n} .The proof follows in the same manner as in Theorem 1 and we omit the details.The following reverse of the generalised triangle inequality may be stated:Theorem 2. Let xj ∈ X\ {0} and a ∈ X\ {0} such that ‖a‖ ≥ ‖xj − a‖ for eachj ∈ {1, . . . , n} . Then for any pj ≥ 0, j ∈ {1, . . . , n} with∑nj=1 pj = 1 we have(2.10)∥∥∥∑nj=1 pjxj∥∥∥∑nj=1 pj ‖xj‖≥ 12min1≤j≤n{‖a‖2 − ‖a− xj‖2‖xj‖ ‖a‖}(≥ 0) .The constant 12 is best possible in (2.10).Proof. Let us denoteρ := min1≤j≤n{‖a‖2 − ‖a− xj‖2‖xj‖}.From Lemma 1 we have〈xj , a〉i‖xj‖ ≥12· ‖a‖2 − ‖xj − a‖2‖xj‖ ≥12ρINEQUALITIES IN NORMED LINEAR SPACES 5for each j ∈ {1, . . . , n} . Therefore〈xj , a〉i ≥12ρ ‖xj‖ , j ∈ {1, . . . , n} .Multiplying with pj and summing over j from 1 to n we obtain(2.11)n∑j=1pj 〈xj , a〉i ≥12ρn∑j=1pj ‖xj‖ ,and since:(2.12)∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ ‖a‖ ≥〈n∑j=1pjxj , a〉i≥n∑j=1pj 〈xj , a〉i ,hence by (2.11) and (2.12) we deduce the desired result (2.10).Now, assume that there exists a constant E > 0 such that(2.13)∥∥∥∑nj=1 pjxj∥∥∥∑nj=1 pj ‖xj‖≥ E · min1≤j≤n{‖a‖2 − ‖a− xj‖2‖xj‖ ‖a‖},provided ‖a‖ ≥ ‖xj − a‖ , j ∈ {1, . . . , n} .If we choose x1 = · · · = xn = εa, ε ∈ (0, 1) , and p1 = ... = pn = 1n , then we get1 ≥ E · ‖a‖2 − (1− ε)2 ‖a‖2ε ‖a‖2 ,giving1 ≥ E (2− ε)for any ε ∈ (0, 1) . Letting ε→ 0+, we deduce E ≤ 12 and the proof is complete.The following result may be stated as well:Proposition 2. Let xj , a ∈ X\ {0} , j ∈ {1, . . . , n} such that ‖xj − a‖ ≤ ‖a‖ .Then for any pj ≥ 0, j ∈ {1, . . . , n} with∑nj=1 pj = 1 we have(2.14)∥∥∥∑nj=1 pjxj∥∥∥∑nj=1 pj ‖xj‖≥ (‖a‖ −max1≤j≤n ‖xj − a‖)2 ‖a‖ (≥ 0) .Proof. From (2.3) we have〈xj , a〉i‖xj‖ ≥12(‖a‖ − ‖xj − a‖)≥ 12min1≤j≤n(‖a‖ − ‖xj − a‖)=12(‖a‖ − max1≤j≤n‖xj − a‖).Now the proof follows the same steps as in that of Theorem 1 and the details areomitted.Remark 2. If ‖a‖ = 1 and ‖xj − a‖ ≤ 1, then (2.10) has a simpler form:(2.15)∥∥∥∑nj=1 pjxj∥∥∥∑nj=1 pj ‖xj‖≥ 12min1≤j≤n{1− ‖xj − a‖2‖xj‖}(≥ 0) ,6 SEVER S. DRAGOMIRwhile (2.14) becomes(2.16)∥∥∥∑nj=1 pjxj∥∥∥∑nj=1 pj ‖xj‖≥ 12(1− max1≤j≤n‖xj − a‖)(≥ 0) .A different approach for bounding the semi-inner product is incorporated in thefollowing:Lemma 2. Let (X, ‖·‖) be a normed space. If x, a ∈ X, then(2.17) 〈x, a〉i ≥ ‖a‖ (‖a‖ − ‖x− a‖) .The inequality (2.17) is sharp.Proof. If a = 0, then obviously (2.17) holds with equality. For a 6= 0, considerτ− (x, a) := lims→0−‖a+ sx‖ − ‖a‖s.Observe that〈x, a〉i = lims→0−‖a+ sx‖2 − ‖a‖22s(2.18)= τ− (x, a) lims→0−[‖a+ sx‖+ ‖a‖2]= τ− (x, a) ‖a‖ .On the other hand, since the function R 3 s 7−→ ‖a+ sx‖ ∈ R+ is convex on R,hence(2.19) τ− (x, a) ≥ ‖a+ (−1)x‖ − ‖a‖(−1) = ‖a‖ − ‖x− a‖ .Consequently, by (2.18) and (2.19) we get (2.17).Now, let x = εa, ε ∈ (0, 1) , a 6= 0. Then〈x, a〉i = ε ‖a‖2 , ‖a‖ − ‖x− a‖ = ‖a‖ − (1− ε) ‖a‖ = ε ‖a‖ ,which shows that the equality case in (2.17) holds true for the nonzero quantitiesε ‖a‖2. The proof is complete.The following reverse of the generalised triangle inequality may be stated.Theorem 3. Let a, xj ∈ X\ {0} for j ∈ {1, . . . , n} with the property that ‖a‖ ≥‖xj − a‖ for j ∈ {1, . . . , n} . Then for any pj ≥ 0, j ∈ {1, . . . , n} with∑nj=1 pj = 1we have(2.20)∥∥∥∑nj=1 pjxj∥∥∥∑nj=1 pj ‖xj‖≥ min1≤j≤n{‖a‖ − ‖xj − a‖‖xj‖}(≥ 0) .The inequality (2.20) is sharp.Proof. On making use of Lemma 2, we have:〈xj , a〉i‖xj‖ ≥ ‖a‖(‖a‖ − ‖xj − a‖‖xj‖)≥ ‖a‖ η,for each j ∈ {1, . . . , n} , whereη := min1≤j≤n{‖a‖ − ‖xj − a‖‖xj‖}.INEQUALITIES IN NORMED LINEAR SPACES 7Now utilising the same argument explained in the proof of Theorem 2, we get thedesired inequality (2.20).If we choose in (2.20) x1 = · · · = xn = εa, ε ∈ (0, 1) , a 6= 0, and p1 = ... = pn = 1then we have equality, and the proof is complete.Remark 3. The above result may be stated in a simpler way, i.e., if ρ ∈ (0, 1) , aand xj ∈ X\ {0} , j ∈ {1, . . . , n} are such that(2.21) (‖xj‖ ≥) ‖a‖ − ‖xj − a‖ ≥ ρ ‖xj‖ (≥ 0)for each j ∈ {1, . . . , n} , then(2.22)∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ ≥ ρn∑j=1pj ‖xj‖ .3. Other Related Results to the Triangle InequalityThe following result may be stated:Theorem 4. Let (X, ‖·‖) be a normed linear space and x1, . . . , xn nonzero vectorsin X and pj ≥ 0 with∑nj=1 pj = 1. If x¯p :=∑nj=1 pjxj 6= 0 and there exists a r > 0with(3.1)〈xj , x¯p〉i‖xj‖ ‖x¯p‖ ≥ r for each j ∈ {1, . . . , n}then(3.2)∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ ≥ rn∑j=1pj ‖xj‖ .If pj > 0 for each j ∈ {1, . . . , n} , then the equality holds in (3.2) if and only if theequality case hold in (3.1) for each j ∈ {1, . . . , n} .Proof. From (3.1) on multiplying with pi ≥ 0 we have〈pjxj , x¯p〉i ≥ rpj ‖x¯p‖ ‖xj‖for any j ∈ {1, . . . , n} .Summing over j from 1 to n and taking into account the superadditivity propertyof the interior semi-inner product, we have(3.3)〈n∑j=1pjxj , x¯p〉i≥n∑j=1〈pjxj , x¯p〉i ≥ r ‖x¯p‖n∑j=1pj ‖xj‖and since 〈n∑j=1pjxj , x¯p〉i=∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥26= 0hence by (3.3) we get (3.2).The equality case is obvious and the proof is complete.8 SEVER S. DRAGOMIRFor the system of vectors x1, . . . , xk ∈ X, we denote by x¯ its gravity center, i.e.,x¯ :=1nn∑i=1xi.The following corollary is obvious.Corollary 1. Let x1, . . . , xn ∈ X\ {0} be such that x¯ 6= 0. If there exists a r > 0such that(3.4)〈xj , x¯〉i‖xj‖ ‖x¯‖ ≥ r for each j ∈ {1, . . . , n} ,then the following reverse of the generalised triangle inequality holds:(3.5)∥∥∥∥∥∥n∑j=1xj∥∥∥∥∥∥ ≥ rn∑j=1‖xj‖ .The equality holds in (3.5) if and only if the case of equality holds in (3.4) for eachj ∈ {1, . . . , n} .The following refinements of the generalised triangle inequality may be stated aswell:Theorem 5. Let xi, x¯p, pi, i ∈ {1, . . . , n} be as in Theorem 4. If there exists aconstant R with 1 > R > 0 and such that(3.6) R ≥ 〈xj , x¯p〉s‖xj‖ ‖x¯p‖ for each j ∈ {1, . . . , n} ,then(3.7) Rn∑j=1pj ‖xj‖ ≥∥∥∥∥∥∥n∑j=1pjxj∥∥∥∥∥∥ .If pj > 0 for each j ∈ {1, . . . , n} , then the equality holds in (3.7) if and only if theequality case holds in (3.6) for each j ∈ {1, . . . , n} .The proof is similar to the one in Theorem 4 on taking into account that thesuperior semi-inner product is a subadditative functional in the first variable.Corollary 2. Let xj , j ∈ {1, . . . , n} be as in Corollary 1. If there exists an R with1 > R > 0 and(3.8) R ≥ 〈xj , x¯〉s‖xj‖ ‖x¯‖ for each j ∈ {1, . . . , n} ,then the following refinement of the generalised triangle inequality holds:(3.9) Rn∑j=1‖xj‖ ≥∥∥∥∥∥∥n∑j=1xj∥∥∥∥∥∥ .The equality hold in (3.9) if and only if the case of equality holds in (3.8) for eachj ∈ {1, . . . , n} .INEQUALITIES IN NORMED LINEAR SPACES 9References[1] A.H. ANSARI and M.S. MOSLEHIAN, Refinements of the triangle inequal-ity in Hilbert and Banach spaces, Preprint: ArXiv Math. FA/0502010,http://front.math.ucdavis.edu/math.FA/0502010.[2] A.H. ANSARI and M.S. MOSLEHIAN, More on reverse triangle in-equality in inner product spaces, Preprint: ArXiv Math. FA/0506198,http://front.math.ucdavis.edu/math.FA/0506198.[3] J.B. DIAZ and F.T. METCALF, A complementary triangle inequality in Hilbert and Banachspaces, Proc. Amer. Math. Soc., 17(1) (1966), 88-97.[4] S.S. DRAGOMIR, Semi-Inner Products and Applications, Nova Science Publishers, Inc., NewYork, 2004.[5] S.S. DRAGOMIR, Advances in Inequalities of the Schwarz, Gru¨ss and Bessel Type in InnerProduct Spaces, Nova Science Publishers, Inc., New York, 2004.[6] S.S. DRAGOMIR, Advances in Inequalities of the Schwarz, Triangle and HeisenbergType in Inner Product Spaces, RGMIA Monographs, Victoria University, 2005. (ONLINE:http://rgmia.vu.edu.au/monographs/).[7] S.M. KHALEELULA, On Diaz-Metcalf’s complementary triangle inequality, Kyungpook Math.J., 15 (1975), 9-11.[8] P.M. MILICˇIC´, On a complementary inequality of the triangle inequality (French), Mat. Ves-nik, 41(2) (1984), 83-88.[9] D.S. MITRINOVIC´, J.E. PECˇARIC´ and A.M. FINK, Classical and New Inequalities in Anal-ysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.School of Computer Science and Mathematics, Victoria University, PO Box 14428,Melbourne City, Victoria 8001, Australia.E-mail address: sever.dragomir@vu.edu.auURL: http://rgmia.vu.edu.au/dragomir

A complementary triangle inequality in Hilbert and Banach spaces,

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On Some Discrete Inequalities in Normed Linear Spaces

On Some Discrete Inequalities in Normed Linear Spaces

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