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 Journal of Applied Mathematics and Physics, 2013, 1, 65-70 Published Online November 2013 (http://www.scirp.org/journal/jamp) http://dx.doi.org/10.4236/jamp.2013.15010 Open Access JAMP The Modified Heinz’s Inequality Takashi Yoshino Mathematical Institute, Tohoku University, Sendai, Japan Email: yoshi no@math.tohoku.ac.jp Received August 23, 2013; revised September 24, 2013; accepted September 28, 2013 Copyright © 2013 Takashi Yoshino. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. ABSTRACT In my paper [1], we aimed to determine the best possible range of such that the modified Heinz’s inequality ()(AAA ABA) holds for any bounded linear operators A and on a Hilbert space such as and for any given B some 0AB I and such as 0 and 0 . But the counter-examples pre- pared in [1] and also in [2] were not sufficient and, in this paper, we shall constitute the sufficient counter-examples which will satisfy all the lacking parts. Keywords: Heinz’s Inequality By the same way as in the proof of Theorem ([1]), we have the following. Lemma For any ,,,abx and such as 1 0,0,0ab xa and 0Oa b where )( O is a function of such that 0 lim O is bounded, let 0 and . 0 aab x AB b ab b If AAA ABA for any real numbers , and such as 0 and 0 , then we have the following inequality (1). Theorem ([1]) The region of such that the opera- tor inequality AAA ABA holds for any bounded linear operators A and on a Hilbert space such as B som eAB I 0 and for any given and such as 0 and 0 is as follows: 1) 010, 1 , 2) 011, 2 11 max,min 0, 222 11 3) 1 012 , , 0 4) 012, 21 max 2121 21 min , , 2121 1,0 and 5) 1 , 1 x0, 21 ma 22 22 2220 21 222 2 2 22 lim () . axbabxb ab b xb axb aabbax bax (1)  T. YOSHINO 66 Proof Since the sufficiency of our range for the modi- fied Heinz’s inequality is already proved in [1], we have only to constitute counter-examples of A and in the outside of our ranges. For any B ,,,abx and such as in Lemma, let 0 and 0 x b ab b . Let aa AB b 1aa aba xb 1 ,0 1by b b a , and 1 b a . Then we have A BbI because 0,abxbyb , 11 0 11 aaab babaab a ax ab ab 1 ecause and b 1 1 11 1 11 1 ax ab bx ax abxaxb a 0. aa xab abaayab a Also if b , then, by multiply- ing 1 2 bax lity (1) in Lemma, we h to the both sides of the ine- quaave AAA ABA 22 21 222 1 1 1 aby ab b a ay b y 2122 22 2 22 aabbay abybay (2) 212222 2 22 =1 aab aby aby ab y (3) 22 22 2 22 1 1 aa bbay abyb ay 2 (4) 2122 22 . bbay (5) 2 aa ay Let 22 1ab y b 1, 01xaba and 1a . Then we have A BbI 0, aba x because 1b and because 11 0. axabbx ax bb Also if AAA ABA 1 ax , then, by m ing ultiply- to the both sides of the inequality (1) in Lemma, we have the following inequality 22 22 222 21222 2 2 22 1 1 1 . 1 ba bab b ba ba b aabba aba y multiplyi (6) And, bng b 22 1 to the both sides of ality, we the above inequhave 122 2121 222 2 2 2 11 1 1 1 11 bbab ab b a ba b ab b a 21 22 22. 11 a a b a (7) Let 2and 0xb . Then we have 2 A BbI Open Access JAMP  T. YOSHINO 67 because 2 0 x bba ab b AAA and because Also if 0ax x . ABA e both sides 22 1 b to th a, we have the , then, by m plying of the inequality (1) in Lemmfollowing inequality ulti- 21 222 2 11abbba 2 22 22 2 22 1 11 . b ba ba ba 21 2 2 1 1 b aa b (8) For 2 , let 1 1 0b2 . Then 1 11 0bb 2 because 1 01 1 . And let 1 1 1 1 2 a b 11, 1 bb 11 1 0 1 xb b ybby ab and 1 1aa 0 . Then we have A BxI because 211 11 11 11 11 11 11 11 1 11 1 11 1 12 10, 1 babbaba bx ab babbb ba ab babbba ab bbabb ab 11 1 11 11 1 11 0 1 aaab baba ax ab abaab b ab and because 0axab bx . Also if AAA ABA , then, by Lemma, we have the following inequality 11 22 22 11 21 122 1 1 1 22 2 21 2 1 22 1 abbybbbybab abby b bbyb aabbabby aby 21 bb . Since 11 11,b bybbbyb 2 1 112 2 22 2 111 ,babbybb aby 1 221 2 12 12 1 babby bb aby 12 22 11 , and since by multiplying 2 1 b to the both sides of the ab ve, for ove inequality, we ha1 1 2, 02 b and 1 1 1 bb 1 1 1 12 a b , Open Access JAMP  T. YOSHINO 68 2 12 1 1 2 2 1 2 22 1 22 21 21 2 222 1 22 22 1 1 ab byb by b byba b aby b a baby ab b 2.aby and, if (9) Case 1 Let 0,1,0 . Then 22 21 222 21 2 1 lim 1 1 21 2 1 aby ab b a ay b b because 2 ay 1aa lim lim2 1 aa yab and 212 222 2 22 lim . 1 a aababy abyab y This contradicts (3). Case 2 Let 01,0, . Then 22 21 222 21 2 1 lim 1 1 21 21 a aby ab b a ay b y b because If lim 2 ay . 20 , then we have 22 22 2 22 1 lim 0 1 a aa bbay abybay 20 , then we have also 2122 22 2 22 lim 1 0,2 10 ,210 0 ,21 0 a a abbay abybay b and hence, by (4) and (5), we have and this contradicts the fact that all 21 0 21 0 for 1 . 1 0, , Case 3 Let 0 . Then 22 21 222 2 1 lim 1 1 21 a aby ab b a ay b y and 212 222 2 22 2 lim 1 12 12 a aababy ab y ab y because lim 2 ay . , we have By (3)021 0 and this contradicts the fact that 21 for all 1 . Case 4 1 1,0 1,0 max0, 21 Let 1 021 Then and and hence 21 1 221 0 . fore we have There 22 221 222 01 b 1 lim 0 1 bb abab b a ba b and 2 21222 2 2 022 lim 1 0. 1 b aabba aba a This contradicts (6). 2 a 2 Open Access JAMP  T. YOSHINO 69 Case 5 Let 01,1 , 21 max , 2121 In this case Then, in the case where 21max1, 0 d21 max1,0 and he n cbec ause1. an e20 1 02 , we have 21 and hence 21 0 . Therefore we have 22 22 222 22 2 1 lim 1 1 1 1 a ba bab b ba ba b bb b b b and 212 22 aab a 2 2 22 lim 1 a b aba ontradicts (6). And, in the case where . This c11 2 , we have 21 21 and hence Therefore we have 212 10 . 2121 2 0 122 22 lim 1 11 0 1 1 b b b bbab ab a ba and 212222 2 11 11 1 b ba a This contradicts (7). Let 2 02 bab a 21 0. a lim aa Case 61 01,12,2 . Then 0212 2 2120 21 . e And, by (8), we hav 21 212 2 2 0 22 22 21 22 2 022 2 2 0lim 1 11 1 lim 1 1 1 0 b b abb b bab ab aab ab ab a and this is a contradiction. Case 7 Let 1 01,1, 21 0 . Then 1 1 212111 0 and 20 because 1 . Therefore we have 22 221 1 m 0 bb aab b 22 2 0 li 1 1 b b a ba b and 212222 2 022 22 lim 1 0. 1 b aabba aba a a This contradicts (6). se 8 Let Ca 2 01,2, 21 0 . 2 22 1 0 . Then Since 11aa 1 lim lim1 1 aa bybab and since 10blim a in the case where 1 , 2 11 1 1 11 2, 01 12 bb ba b because 12 11 1 1 11 11 1 10 12 12 bb bb a bb , and Open Access JAMP  T. YOSHINO Open Access JAMP 70 we have 2 12 1 1 2 12 1 2 22 1 22 21 lim 1 0 a ab byb by b byba b aby b and 21 22 2 12 lim a22 1 1. aab ba by baby 22 12 completed the proof of the best pos- sianges in our theorem. REFERENCES [1] T. Yoshino, “A Modified Heinz’s Inequality,” Linear bra and its Applications, Vol. 420, p. http://dx.doi.org/10.1016/j.laa.2006.08 This contradicts (9). Therefore we bility of the r Alge No. 2-3, 2007, p 686-699. .031 [2] T. Yssible Range of a Modified Heinz’s Ine quality ,” International Journal of Funct. Anal. Oper. Theory Appl., Vol. 3, No. 1, 2011, pp. 1-7. oshino, “The Best Po |