Research Article On Sharp Triangle Inequalities in Banach Spaces II

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1 Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2010, Article ID , 17 pages doi: /2010/ Research Article On Sharp Triangle Inequalities in Banach Spaces II Ken-Ichi Mitani 1 and Kichi-Suke Saito 2 1 Department of Applied Chemistr and Biotechnolog, Facult of Engineering, Niigata Institute of Technolog, Kashiwaaki , Japan 2 Department of Mathematics, Facult of Science, Niigata Universit, Niigata , Japan Correspondence should be addressed to Kichi-Suke Saito, saito@math.sc.niigata-u.ac.p Received 4 November 2009; Accepted 2 March 2010 Academic Editor: V Khoi Le Copright q 2010 K.-I. Mitani and K.-S. Saito. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in an medium, provided the original work is properl cited. Sharp triangle inequalit and its reverse in Banach spaces were recentl showed b Mitani et al In this paper, we present equalit attainedness for these inequalities in strictl conve Banach spaces. 1. Introduction In recent ears, the triangle inequalit and its reverse inequalit have been treated in 1 5 see also 6, 7.Kato et al. 8 presented the following sharp triangle inequalit and its reverse inequalit with n elements in a Banach space X. Theorem 1.1 see 8. For all nonero elements 1, 2,..., n in a Banach space X, n min 1 n n ma. 1 n 1.1 These inequalities are useful to treat geometrical structure of Banach spaces, such as uniform non-l n 1 -ness see 8. Moreover,Hsu et al. 9 presented these inequalities for strongl integrable functions with values in a Banach space.

2 2 Journal of Inequalities and Applications Mitani et al. 10 showed the following inequalities which are sharper than Inequalit 1.1 in Theorem 1.1. Theorem 1.2 see 10. For all nonero elements 1, 2,..., n in a Banach space X with 1 2 n, n 2, k k k1 k2 k k2 n k 1 ( n k n k 1, where 0 n1 0. In this paper we first present a simpler proof of Theorem 1.2. Todothisweconsider the case 1 > 2 > > n, as follows. Theorem 1.3. For all nonero elements 1, 2,..., n in a Banach space X with 1 > 2 > > n,n 2, k k k1 k2 k k2 n k 1 ( n k n k 1, where 0 n1 0. From this result we can easil obtain Theorem 1.2. Moreover we consider equalit attainedness for sharp triangle inequalit and its reverse inequalit in strictl conve Banach spaces. Namel, we characterie equalit attainedness of Inequalities 1.4 and 1.5 in Theorem Simple Proofs of Theorems 1.2 and 1.3 Proof of Theorem 1.3. According to Theorem 1.1 Inequalities 1.4 and 1.5 hold for the case n 2 cf. 3. Therefore let n 3. We first prove 1.4 b the induction. Assume that 1.4

3 Journal of Inequalities and Applications 3 holds true for all n 1 elements in X. Let 1, 2,..., n be an n elements in X with 1 > 2 > > n > 0. Let u ( n, 2.1 for all positive numbers with 1 n. Then n u 2.2 and u 1 > u 2 > > u n 1 > 0. B assumption, u u u k u uk u k1 k2 2.3 holds, where u n 0. Since u k u k1 k k1,from2.2 and 2.3, n u n u n n 1 u u k u uk u k1 k2 n n 1 ( n k k k1 k2 k k k1 k2 2.4 and hence 1.4.Thus1.4 holds true for all finite elements in X. Net we show Inequalit 1.5.Let v ( 1 n n n, 1 n

4 4 Journal of Inequalities and Applications Then 1 v 2.6 and v 1 > > v n 1 > 0. Appling Inequalit 1.4 to v 1,...,v n 1, 1 v 1 n 1 v v k v vk v k1 k2 1 n 1 ( 1 n k n n n k1 n k k2 k k2 n k 1 n k1 n k, 2.7 where v n 0. Thus we obtain 1.5. This completes the proof. Proof of Theorem 1.2. Let 1, 2,..., n be an nonero elements in X with 1 n. For all positive numbers m with m>nlet k,m ( 1 k k, k 1, 2,...,n. 2.8 m Then 1,m > 2,m > > n,m > 0. Appling Theorem 1.3 to 1,m,..., n,m,,m k,m,m k,m k1,m k2,m,m k k2 n k 1,m (,m n k,m n k 1,m, 2.9 where 0,m n1,m 0 for all positive numbers m with m>n.asm, we have Inequalities 1.4 and 1.5.

5 Journal of Inequalities and Applications 5 3. Equalit Attainedness in a Strictl Conve Banach Space In this section we consider equalit attainedness for sharp triangle inequalit and its reverse inequalit in a strictl conve Banach space. Kato et al. in 8 showed the following. Theorem 3.1 see 8. Let X be a strictl conve Banach space and 1, 2,..., n nonero elements in X. Let 0 min{ :1 n} and 1 ma{ :1 n}. LetJ 0 { : 0, 1 n}.then n min k k n if and onl if either a 1 2 n or b / 1 / 1 for all J c 0 and n / n / 1 / 1. Theorem 3.2 see 8. Let X be a strictl conve Banach space and 1, 2,..., n nonero elements in X. Let 0 min{ :1 n} and 1 ma{ :1 n}. LetJ 1 { : 1, 1 n}.then n ma n if and onl if either a 1 2 n or b / 0 / 0 for all J c 1 and n n 0 / 0. We present equalit attainedness for 1.4 and 1.5 in Theorem 1.2. The following lemma given in 8 is quite powerful. Lemma 3.3 see 8. Let X be a strictl conve Banach space. Let 1, 2,..., n be nonero elements in X. Then the following are equivalent. i n α n α holds for an positive numbers α 1,α 2,...,α n. ii n α n α holds for some positive numbers α 1,α 2,...,α n. iii 1 / 1 2 / 2 n / n. Theorem 3.4. Let X be a strictl conve Banach space and, nonero elements in X with >. Then ( if and onl if there eists a real number α with 0 <α<1 satisfing ±α.

6 6 Journal of Inequalities and Applications Proof. Assume that 3.3 is true. B Theorem 3.1, the equalit 3.3 is equivalent to Equalit. 3.4 Put ( α Then α. Since >, weobtain0< α < 1. Conversel, if α where 0 < α < 1, then ( 1 α α. 3.6 B 1 α/ α 0, we have 3.4. Thus we get 3.3. Net we consider the case n 3. Theorem 3.5. Let X be a strictl conve Banach space and,, nonero elements in X with > >. Then (3 ( 2 ( 3.7 if and onl if there eist α, β with 0 <β<α<1satisfing one of the following conditions: a α, ±β, b α, β. Proof. Assume that 3.7 is true. Put u, v (. 3.8 Then u > v > 0and ( u v. 3.9 Note that u v / 0. As in the proof of Theorem 1.2 given in 10, we have 3.7 if and onl if we have the equalities ( u v u v, 3.10 ( u v u v 2 u u v v v. 3.11

7 Journal of Inequalities and Applications 7 B Theorem 3.4, Equalit 3.11 implies that v ±γu 3.12 for some γ with 0 <γ<1. B 3.8 we have ±α 3.13 for some 0 <α<1. On the other hand, b Lemma 3.3, Equalit 3.10 implies u v u v Hence, b using 3.8, 3.12, and3.13 we have β for some real number β. Since > > > 0, we have 0 < β <α<1. We consider the following two cases Case 1. α. Equalit 3.14 implies β β β β α 2 β. α 2β 3.15 Hence we have 2 β β 2 β β 1 α 2 β 1 α 2 β B 2 β/ β 0and1 α 2 β 1 β α β 0, Equalit 3.16 is valid for all real numbers β with β / 0. Case 2. α. Equalit 3.14 implies β β β β α α So we have β β β β 1 α 1 α β β Hence β>0. Thus holds.

8 8 Journal of Inequalities and Applications Conversel, assume that there eist α, β with 0 < β < α < 1 satisfing one of the conditions a and b. Then it is clear that 3.7 holds. Thus we obtain. Moreover we consider general cases. For each m with 1 m n, weputi m {1, 2,...,m}. For α α 1,α 2,...,α n R n and 1 m n, we define I mα {k I m : α k > 0}, I mα {k I m : α k < 0} For a finite set A, the cardinal number of A is denoted b A. Lemma 3.6. Let α 1 > α 2 > > α n > 0. If I mα I mα for all m with 1 m n, then α > Proof. Let I n α {m 1,m 2,...,m l } ( I m l α, I n α {n 1,n 2,...,n k } ( I n k α, 3.21 where m 1 <m 2 < <m l and n 1 <n 2 < <n k. B the assumption, we have l>k.wefirst show m <n for all with 1 k. It is clear that m 1 <n 1. Assume that m i <n i for all i with 1 i. We will show m <n. Suppose that m >n.bm <n <n <m,we have I n α I m α { m 1,m 2,...,m }, I n α { n 1,n 2,...,n } Hence we have I n α < I n α, which is a contradiction. Therefore we have m <n. Namel, m <n for all with 1 k. From this result, we obtain α m α n α m α n > 0. Hence α l α m α n (α m α n l α m > 0. k Theorem 3.7. Let X be a strictl conve Banach space and 1, 2,..., n nonero elements in X with 1 > 2 > > n.then k k k k2

9 Journal of Inequalities and Applications 9 if and onl if there eists α α 1,α 2,...,α n R n with 1 α 1 > α 2 > α 3 > > α n such that m α m 1, 3.25 I mα I m α 3.26 for ever m with 1 m n. Proof. : According to Theorems 3.4 and 3.5, Theorem 3.7 is valid for the cases n 2, 3. Therefore let n 4. We will prove Theorem 3.7 b the induction. Assume that this theorem holds true for all nonero elements in X less than n. Let 1 > 2 > > n and assume that Equalit 3.24 holds. Let u ( n 3.27 for positive integer with 1 n 1. As in the proof of Theorem 1.3, Equalit 3.24 holds if and onl if n u n u, 3.28 u u i u k u uk u k k2 hold, where u n 0. Hence, b assumption, there eists β β 1,...,β n 1 R n 1 with 1 β 1 > β 2 > > β n 1 > 0 such that ( u β u 1, 1 n 1, ( ( 3.30 I m β I m β, 1 m n 1. Since n 1 β > 0 b Lemma 3.6, we have u β u 1 / Since ( 1 ( n β 1 n b the definition of u, we have α 1, ( 1 n 1, 3.33

10 10 Journal of Inequalities and Applications where 1 n α β n Since I mα I m( β, I m α I m( β, 3.35 we have I mα I m α 3.36 for all m with 1 m n 1. B Lemma 3.3, Equalit 3.28 implies u u Hence there eists α n R such that n α n 1.Also, α 1 α 1 ( I n 1 α I n 1 α α n 1 α n 1, u n 1 β u Since n 1 β > 0, we have from 3.37 and 3.38, I n 1 α I n 1 α α n α n I n 1 α I n 1 α α n α n, 3.40 which implies I n 1 α I n 1 α α n α n If I n 1 α > I n 1 α, then it is clear that I n α I n α. If I n 1 α I n 1 α, then, b 3.41, we have α n/ α n 0. Hence α n > 0. Thus we obtain I n α > I n α.

11 Journal of Inequalities and Applications 11 : Letα α 1,α 2,...,α n R n with 1 α 1 > α 2 > > α n > 0 satisfing 3.25 and 3.26, andletα n1 0. From 3.25 we have k k k1 k2 α α 1 k 1 α 1 αk 1 α k1 1 k2 α α k α αk α k1 1. k B Lemma 3.6, α α I n α α I n α α Let In α {k 1,k 2,...,k m }, where 1 <k 1 <k 2 < <k m <n.from3.26 and I α k α k we have I k α k α αk α k1 k2 ( k I k α I k α α k α k1 k2 2 I k α αk α k1 2 2 k2 ( k1 1 I k α αk α k1 k2 ( k kk 1 I k α αk α k1 3 1 α k α k1 2 α k α k1 m α k α k1 kk 1 kk 2 kk m α k1 α k2 2 α k2 α k3 I k α αk α k1 kk m m 1 α km 1 α km m α km α kn1 2 α. I n α 3.44 Thus we obtain This completes the proof.

12 12 Journal of Inequalities and Applications In what follows, we characterie the equalit condition of Inequalit 1.3 in Theorem 1.2. For α α 1,α 2,...,α n R n and positive integer m with 2 m n 1we define J m {n m 1,...,n 1,n},J mα { J m : α > 0}, andj mα { J m : α < 0}. Lemma 3.8. Let α α 1,α 2,...,α n R n with α 1 > α 2 > > α n 1 >α n 1. If J mα J m α 3.45 for all positive integers m with 2 m n 1, then one has α 1 ( J n 1 α J n 1 α α Proof. Let J n 1 α {m 1,m 2,...,m l,m l1,...,m k } and J n 1 α {n 1,n 2,...,n l }, where n m 1 > >m k 2andn>n 1 > >n l 2. As in the proof of Lemma 3.6, we have m >n for all.so α m < α n for all. Hence α 1 ( J n 1 α J n 1 α α 2 α α J n 1 α J n 1 α α 1 ( J n 1 α J n 1 α α α J n 1 α J n 1 α ( α1 ( α α1 α J n 1 α J n 1 α α 1 ( J n 1 α J n 1 α ( α 1 α m ( α 1 α m l1 l ( α 1 α n l ( αn αm This completes the proof. Theorem 3.9. Let X be a strictl conve Banach space and 1, 2,..., n nonero elements in X with 1 > 2 > > n.then k k2 n k 1 ( n k n k

13 Journal of Inequalities and Applications 13 holds if and onl if there eists α α 1,α 2,...,α n R n with α 1 > α 2 > > α n 1 >α n 1, α n for all positive integers with 1 n satisfing J mα J m α 3.49 for all positive integers m with 2 m n 1 and n α 0. Proof. :Let 1 > 2 > > n and k k2 n k 1 ( n k n k For positive integers with 2 n we put v ( Note that 0 < v 2 < < v n and 1 v Then Equalit 3.48 holds if and onl if v v, 2 v v n 1 v v2 2 k k2 2 n k 1 2 v ( v v n k 1 v n k v Thus, b the equalit condition of sharp triangle inequalit with n 1 elements, there eists β β 1,...,β n R n such that v β v n 3.55 for all positive integers with 2 n, 0 < β2 < β3 < < βn 1 <βn 1, J m ( β J m ( β 3.56

14 14 Journal of Inequalities and Applications for all positive integers m with 2 m n 1. Since ( ( β n 1 n 3.57 for each with 2 n, we have α n, where 1 n α β 1 n Note that J mα J mβ and J mα J mβ.b3.53, v v Hence there eists α 1 R such that 1 α 1 n. We also have α 1 > α 2 > > α n 1 >α n 1 and v 2 ( α α1 α α n 2 α1 ( J n 1 α J n 1 α α n Hence we have α α1 ( J n 1 α J n 1 α α 2 α 1 ( J n 1 α J n 1 α α α From 3.3, weobtain n α 0. Thus we have. : Assume that there eists α α 1,α 2,...,α n R n with α 1 > α 2 > > α n 1 > α n 1, α n for all positive integers with 1 n satisfing 3.49 for all positive

15 Journal of Inequalities and Applications 15 integers m with 2 m n 1and n α 0. B Theorem 3.7, we have Fromthe assumption, α n, v α1 ( J n 1 α J n 1 α α n B n α proof. 0 and Lemma 3.8, weobtain3.53. Thus we have. This completes the If n 2, then we have the following corollar. Corollar Let X be a strictl conve Banach space and, nonero elements in X with >. Then ( if and onl if there eists a real number α with α>1 such that α. If n 3, then we have the following corollar. Corollar Let X be a strictl conve Banach space and,, nonero elements in X with > >. Then (3 ( 2 ( 3.64 if and onl if there eist α, β with α > β > 1 such that α, β and α β Remark In this section we consider equalit attainedness for sharp triangle inequalit in a more general case, that is, the case without the assumption that 1 > 2 > > n.let us consider the case n 3. Proposition 4.1. Let X be a strictl conve Banach space, and,, nonero elements in X. i If, then the equalit ( 3 ( 2 ( 4.1 alwas holds.

16 16 Journal of Inequalities and Applications ii If >, then the equalit 4.1 holds if and onl if there eists a real number α satisfing α / and α. iii If >, then the equalit 4.1 holds if and onl if there eists a real number α satisfing α / and α. Proof. i Is clear. ii Assume that 4.1 holds. B, 4.1 implies ( From Theorem 3.1, we have. 4.3 Hence α for some α R. The following ( 1 α, 4.4 implies 1 α 1 α. 4.5 Hence 1 α / 0andsoα /. The converse is clear. iii Assume that 4.1 holds. Put u / and v /. As in the proof of Theorem 3.5, we have ( u v u v 4.6 Since, we have α for some α R. The following ( 1 α 4.7 implies 1 α 1 α. 4.8 Hence α /. The converse is clear.

17 Journal of Inequalities and Applications 17 Conecture 1. What is the necessar and sufficient condition when Equalit 3.24 (resp. Equalit 3.48 holds for n elements 1, 2,..., n with 1 2 n in Theorem 3.7 (resp. Theorem 3.9? Acknowledgment The second author was supported in part b Grants-in-Aid for Scientific Research no , Japan Societ for the Promotion of Science. References 1 J. B. Dia and F. T. Metcalf, A complementar triangle inequalit in Hilbert and Banach spaces, Proceedings of the American Mathematical Societ, vol. 17, pp , S. S. Dragomir, Reverses of the triangle inequalit in Banach spaces, Journal of Inequalities in Pure and Applied Mathematics, vol. 6, no. 5, article 129, pp. 1 46, H. Hudik and T. R. Landes, Characteristic of conveit of Köthe function spaces, Mathematische Annalen, vol. 294, no. 1, pp , D. S. Mitrinović, J. E. Pečarić,andA.M.Fink,Classical and New Inequalities in Analsis, vol. 61 of Mathematics and Its Applications (East European Series, Kluwer Academic Publishers, Dordrecht, The Netherlands, J. Pečarić and R. Raić, The Dunkl-Williams inequalit with n elements in normed linear spaces, Mathematical Inequalities & Applications, vol. 10, no. 2, pp , C. F. Dunkl and K. S. Williams, A simple norm inequalit, The American Mathematical Monthl, vol. 71, no. 1, pp , J. L. Massera and J. J. Schäffer, Linear differential equations and functional analsis. I, Annals of Mathematics, vol. 67, pp , M. Kato, K.-S. Saito, and T. Tamura, Sharp triangle inequalit and its reverse in Banach spaces, Mathematical Inequalities & Applications, vol. 10, no. 2, pp , C.-Y. Hsu, S.-Y. Shaw, and H.-J. Wong, Refinements of generalied triangle inequalities, Journal of Mathematical Analsis and Applications, vol. 344, no. 1, pp , K.-I. Mitani, K.-S. Saito, M. Kato, and T. Tamura, On sharp triangle inequalities in Banach spaces, Journal of Mathematical Analsis and Applications, vol. 336, no. 2, pp , 2007.