More on Reverse Triangle Inequality in Inner Product. Spaces

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1 More on Reverse Triangle Inequality in Inner Product arxiv:math/ v1 [math.fa] 10 Jun 005 Spaces A. H. Ansari M. S. Moslehian Abstract Refining some results of S. S. Dragomir, several new reverses of the generalized triangle inequality in inner product spaces are given. Among several results, we establish some reverses for the Schwarz inequality. In particular, it is proved that if a is a unit vector in a real or complex inner product space (H;.,. ), r,s > 0,p (0,s],D = {x H, rx sa p},x 1,x D {0} α r,s = min{ r x k p +s rs x k 1 Introduction. x 1 x Re x 1,x ( x 1 + x ) α r,s. : 1 k }, then It is interesting to know under which conditions the triangle inequality went the other way in a normed space X; in other words, we would like to know if there is a constant c with the property that c n x k n x k for any finite set x 1,,x n X. M. Nakai T. Tada [7] proved that the normed spaces with this property are precisely those of finite dimensional. The first authors investigating reverse of the triangle inequality in inner product spaces were J. B. Diaz F. T. Metcalf [] by establishing the following result as an extension of an inequality given by M. Petrovich [8] for complex numbers: 000 Mathematics Subject Classification. Primary 46C05; Secondary 6D15. Key words phrases. Triangle inequality, reverse inequality, Schwarz inequality, inner product space. 1

2 Diaz-Metcalf Theorem. Let a be a unit vector in an inner product space (H;.,. ). Suppose the vectors x k H,k {1,,n} satisfy where equality holds if only if 0 r Re x k,a,k {1,,n} x k r x k x k. x k = r x k a. Inequalities related to the triangle inequality are of special interest; cf. Chapter XVII of [6]. They may be applied to get interesting inequalities in complex numbers or to study vector-valued integral inequalities [3],[4]. Using several ideas following the terminology of [3] [4] we modify or refine some results of Dragomir ours [1] get some new reverses of triangle inequality. Among several results, we show that if a is a unit vector in a real or complex inner product space (H;.,. ),x k H {0},1 k n, α = min{ x k : 1 k n},p (0, α +1),max{ x k a : 1 k n} p β = min{ x k p +1 x k x k : 1 k n}, then x k 1 β β Re n x k,a. We also examin some reverses for the celebrated Schwarz inequality. In particular, it is proved that if a is a unit vector in a real or complex inner product space (H;.,. ), r,s > 0,p (0,s],D = {x H, rx sa p},x 1,x D {0} α r,s = min{ r x k p +s rs x k : 1 k }, then x 1 x Re x 1,x ( x 1 + x ) α r,s Throughout the paper (H;.,. ) denotes a real or complex inner product space. We use repeatedly the Cauchy-Schwarz inequality without mentioning it. The reader is referred to [9],[5] for the terminology on inner product spaces.

3 Reverse of Triangle inequality. We start this section by pointing out the following theorem of [1] which is a modification of theorem 3 of [4]. Theorem.1 Let a 1,...,a m be orthonormal vectors in the complex inner product space (H;.,. ). Suppose that for 1 t m,r t,ρ t R that the vectors x k H,k {1,,n} satisfy 0 rt x k Re x k,r t a t,0 ρ t x k Im x k,ρ t a t,1 t m (1) m ( (rt +ρ t)) 1 x k x k () t=1 the equality holds in () if only if m x k = x k (r t +iρ t )a t. (3) t=1 The following theorem is a strengthen of Corollary 1 of[4] a generalization of Theorem of [1]. Theorem. Let a be a unit vector in the complex inner product space (H;.,. ). Suppose that the vectors x k H {0},k {1,,n} satisfy max{ rx k sa : 1 k n} p,max{ r x k is a : 1 k n} q where r,r,s,s > 0 p ((rα) +s ) 1,q ((r α) +s ) 1 α = min{ x k : 1 k n}. Let α r,s = min{ r x k p +s rs x k : 1 k n}, 3

4 β r,s = min{r x k q +s r s x k : 1 k n} (αr,s +β r,s )1 x k x k the equality holds if only if x k = (α r,s +iβ r,s ) n x k a. Proof. From the first inequality above we infer that rx k sa,rx k sa p r x k +s p Re rx k,sa. Similarly consequently r x k +s p x k Re x k,a. rs x k r x k q +s x r s k Im x k,a x k α r,s x k Re x k,a β r,s x k Im x k,a. Applying Theorem.1 for m = 1,r 1 = α r,s ρ 1 = β r,s we deduce desired inequality. The next result is an extension of Corollary 3 of [1]. Corollary.3 Let a be a unit vector in the complex inner product space (H;.,. ). Suppose that x k H,k {1,,n},max{ rx k sa : 1 k n} r,max{ rx k isa : 1 k n} s where r > 0,s > 0 α = min{ x k : 1 k n}. The equality holds if only if rα s x k x k. x k = rα (1+i) s 4 x k a.

5 Proof. ApplyTheorem.withr = r,s = s,p = r,q = s. Notethatα r,s = rα s = β r,s. Theorem.4 Let a be a unit vector in H;.,. ). Suppose that the vectors x k H,k {1,,n} satisfy max{ rx k sa : 1 k n} p < ((rα) +s ) 1 where r > 0,s > 0 α = min 1 k n x k. Let α r,s = min{ r x k p +s rs x k : 1 k n}. n α r,s x k x k. Moreover, the equality holds if only if n x k = α r,s x k a. Proof. Proof is similar to that of Theorem. in which we use Theorem.1 with m = 1,ρ 1 = 0. Theorem.5 Let a be a unit vector in (H;.,. ). Suppose that r,s > 0, vectors x k H {0},k {1,,n} satisfy x k = 0. r α +s max{ rx k sa : 1 n} where α = min 1 k n x k. 5

6 Proof. Let p = max{ rx k sa : 1 k n}. If p < r α +s, then using Theorem.4 we get n α r,s x k x k = 0. Hence α r,s = 0. On the other h p s r < α, so α r,s = min{ r x k p +s rs x k : 1 k n} > 0 a contradiction. Theorem.6 Let a 1,...,a m be orthonormal vectors in the complex inner product space (H;.,. ),M t m t > 0,L t l t > 0,1 t m x k H {0},k {1,,n} such that or equivalently Re M t a t x k,x k m t a t 0,Re L t ia t x k,x k l t ia t 0 x k m t +M t for all 1 k n 1 t m. Let then a t M t m t, x k L t +l t ia t L t l t α mt,m t = min{ x k +m t M t (m t +M t ) x k : 1 k n},1 t m α lt,l t = min{ x k +l t L t (m t +M t ) x k : 1 k n},1 t m m ( αm t,m t +αl t,l t ) 1 t=1 The equality holds if only if that x k x k. m x k = ( x k ) (α mt,m t +iα lt,l t )a t. t=1 Proof. Given 1 t m all 1 k n, it follows from x k mt+mt a t Mt mt x k +m t M t (m t +M t )Re x k,a t. 6

7 so x k +m t M t (m t +M t ) x k x k Re x k,a t α mt,m t x k Re x k,a t. Similarly from the second inequality we deduce that α lt,l t x k Im x k,a t. Applying Theorem.5 for r t = α mt,m t ρ t = α lt,l t, we obtain the required inequality. We will need theorem 7 of [3]. We mention it for the sake of completeness. Theorem.7 Let a be a unit vector in (H;.,. ), x k H {0},k {1,,n}. If r k 0,k {1,,n} such that x k Re x k,a r k then x k x k r k. The equality holds if only if x k r k x k = ( x k r k )a. Theorem.8 Let a be a unit vector in (H;.,. ) x k H {0},k {1,,n}. Let α = min{ x k : 1 k n},p (0, α +1),max{ x k a : 1 k n} p β = min{ x k p +1 x k x k : 1 k n}. x k 1 β β Re n x k,a. 7

8 The equality holds if only if x k 1 β β Re n x k,a x k = ( x k 1 β n β Re x k,a )a. Proof. Since max{ x k a : 1 k n} p, we have x k a,x k a p for all k {1,,n}. x k +1 p Re x k,a x k p +1 x k Re x k,a x k β x k Re x k,a x k 1 β Re x k,a x k Re x k,a 1 β β Re x k,a,k {1,,n}. Applying Theorem.7 for r k = 1 β β Re x k,a,k {1,,n}, we deduce the desired inequality. As a corollary, we obtain a result similar to Theorem 9 of [3]: Corollary.9 Let a be a unit vector in (H;.,. ) x k H {0},k {1,,n}. Let max{ x k a : 1 k n} 1 α = min{ x k : 1 k n}. x k x k α α Re n x k,a. 8

9 The equality holds if only if x k α α Re n x k,a x k = ( x k α n α Re x k,a )a. Proof. Apply Theorem.8 with β = α. Theorem.10 Let a be a unit vector in (H;.,. ), M m > 0 x k H {0},k {1,,n} such that Re Ma x k,x k ma 0 or equivalently Let x k m+m a M m. α m,m = min{ x k +mm (m+m) x k : 1 k n}. x k The equality holds if only if x k 1 α m,m α m,m x k 1 α m,m α m,m Re x k,a. Re x k,a x k = ( x k 1 α m,m α m,m Re x k,a )a. Proof. For each 1 k n, it follows from the inequality x k m+m a M m that x k m+m a,x k m+m a ( M m ). 9

10 Hence So that x k +mm (m+m)re x k,a. α m,m x k Re x k,a consequently x k Re x k,a 1 α m,m α m,m Re x k,a. Now apply Theorem.7 for r k = 1 α m,m α m,m Re x k,a,k {1,,n}. 3 Reverses of Schwarz inequality. In this section we provide some reverses of the Schwarz inequality. The first theorem is an extension of Proposition 5.1 of [3]. Theorem 3.1 Let a be a unit vector in (H;.,. ). Suppose that r,s > 0,p (0,s] D = {x H, rx sa p}. If 0 x 1 D,0 x D, then x 1 x Re x 1,x 1 x 1 p +s ( x 1 + x ) (1 (r ) ) rs x 1 or x 1 x Re x 1,x 1 x p +s ( x 1 + x ) (1 (r ) ) rs x Proof. Put α r,s = min{ r x k p +s rs x k : 1 k }. By Theorem.4, we obtain α r,s ( x 1 + x ) x 1 +x. α r,s( x 1 + x 1 x + x ) x 1 +Re x 1,x + x. 10

11 Set α r,s = 1 t. namely x 1 x Re x 1,x ( x 1 + x ) 1 t, x 1 x Re x 1,x ( x 1 + x ) 1 (1 α r,s ). Corollary 3. Let a be a unit vector in (H;.,. ). Suppose that r,s > 0 D = {x H, rx sa s}. If x,y D 0 < x < y, then x y Re x,y ( x + y ) 1 (1 (r x s ) ). Proof. In the notation of the proof of Theorem 3.1 we get from p = s,x 1 = x,x = y that α r,s = r x. Now apply Theorem 3.1. s Corollary 3.3 Let a be a unit vector in (H;.,. ), M m > 0 x k H {0},k = 1, such that Re Ma x k,x k ma 0 or equivalently, or x k m+m x 1 x Re x 1,x ( x 1 + x ) x 1 x Re x 1,x ( x 1 + x ) a M m. 1 (1 ( x 1 +mm (m+m) x 1 ) ) 1 (1 ( x +mm (m+m) x ) ). Proof. Put r = 1,s = m+m,p = M m,x = x 1 y = x in Theorem 3.1. Acknowledgement. The authors would like to thank the referees for their valuable suggestions. 11

12 References [1] A. H. Ansari M. S. Moslehian, Refinements of reverse triangle inequalities in inner product spaces, to appear in J. Inequal. Pure Appl. Math. [] J.B. Diaz F.T. Metcalf, A complementary triangle inequality in Hilbert Banach spaces, Proceedings Amer. Math. Soc., 17(1) (1966), [3] S.S. Dragomir, Reverses of the triangle inequality in inner product spaces. Aust. J. Math. Anal. Appl. 1 (004), no., Art. 7, 14 pp. [4] S.S. Dragomir, Some reverses of the generalized triangle inequality in complex inner product spaces, arxiv:math.fa/ [5] S.S. Dragomir, Discrete inequalities of the Cauchy-Bunyakovsky-Schwarz type. Nova Science Publishers, Inc., Hauppauge, NY, 004. [6] D.S. Mitrinović, J.E. Pečarić A.M. Fink, Classical New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, [7] M. Nakai T. Tada, The reverse triangle inequality in normed spaces, New Zeal J. Math. 5 (1996), no., [8] M. Petrovich, Module dune somme, L Ensignement Mathématique, 19 (1917), [9] Th.M. Rassias, Inner Product Spaces Applications, Chapman-Hall,1997. Arsalan Hojjat Ansari Mohammad Sal Moslehian Dept. of Math., Ferdowsi Univ., P. O. Box 1159, Mashhad 91775, Iran moslehian@ferdowsi.um.ac.ir Home: moslehian/ 1