![THE REVERSE TRIANGLE INEQUALITY in NORMED Spacest](https://data.docslib.org/img/3a60ab92a6e30910dab9bd827208bcff-1.webp)
NEW ZEALAND JOURNAL OF MATHEMATICS Volume 25 (1996), 181-193 THE REVERSE TRIANGLE INEQUALITY IN NORMED SPACESt MlTSURU NAKAI* AND TOSHIMASA TADA* (Received January 1995) Abstract. The validity of the reverse triangle inequality in a normed spaceX is characterized by the finiteness of what we call the best constantcr(X) associ­ ated with X. The validity of the reverse triangle inequality inX, i.e. cr(X) < oo, is completely determined algebraically by d im X < oo. Certain practical es­ timates of cr(X ) from above and from below for general normed space X are given. These are applied to our main purpose of this paper of describing the order of <j(lp(n)) as n —* oo for n-dimensional lp space (1 < p < oo). A typical result is that cr(lp(n)) « n 1 -1 / p as n —► oo for 2 < p < n. 1. Introduction Consider a real normed space X equipped with the the norm || • || = || • ; X ||. The triangle inequality, which is one of the defining properties of norms, may be formulated as follows: for any finite sequence {x/c}fceL> (D = {1,... , h}) in X (1) We say that the reverse triangle inequality is valid in X if there exists a space constant C = C x with the following property: for,any finite sequence {xk}keD in X there exists a subset A c D depending on {xk}k&D such that The smallest possible constant of such C ’s is said to be the best constant for the reverse triangle inequality in X and will be denoted by cr(X). For convenience we set cr(X) = oo if the reverse triangle inequality is invalid in X. In this paper we first remark that the reverse triangle inequality is valid in X , i.e. cr(X) < oo, if and only if X is finite dimensional, i.e. dim X < oo (Theorem 1). The main purpose of this paper is to determine the order of a(lp(n)) as n f oo, where lp(n) is the finite n dimensional real linear space with lp norm (1 < p < oo). For the purpose we first give certain general practical estimates of cr(X) from above and also from below for general normed spaces X (Theorem 2). These are used to determine the constants a(lp(n)) exactly for p = 1,2, and oo (Theorem 3), which 1991 AMS Mathematics Subject Classification: Primary 46B20; Secondary 46B45. t Dedicated to our old friend Professor Heppe O ’Malla on his sixtieth birthday. * The authors were supported in part by Grant-in-Aid for Scientific Research, No. 06640227, Japanese Ministry of Education, Science and Culture. 182 MITSURU NAKAI AND TOSHIMASA TADA implies a complete result a{lp(n)) « n1 1//p for 2 < p < oo and also an estimate of the order of a(lp(n)) for 1 < p < 2 (Theorem 4). 2. Finite Dimensional Normed Spaces It is convenient to associate two quantities with a finite sequence {x*;} = {xkjkeD (D = {1 ,... ,/i}) as follows. The first is the total length T{xk} = J2ktD ll^fcll {xfc} and the second is the maximal length M{xk} of {x*;} given by M{xk\ = max (3) L 3 a c d E i4 keA In terms of these two quantities the triangle inequality (1) is formulated as ini(T{xk}/M{xk}) = 1 where the infimum is taken with respect to {x*;} which runs over all finite sequences in X \{0}. In contrast the best constant cr(X) for the reverse triangle inequality (2) in X is given by sup {T{xk}/M{xk}) = cr(X) (4) where again the supremum is taken with respect to {xfc} which runs over all finite sequences in X \{0}. Thus the reverse triangle inequality is valid in X if and only if a(X) < oo. The property cr(X) < oo for X is completely characterized by the linear space structure of X as follows: Theorem 1. The reverse triangle inequality is valid in a normed space X if and only if X is finite dimensional. Proof. Since cr(X) = <r(X) for the completion X of X , we may assume that X is a Banach space. We first show that <x(X) = oo if dim X = oo. By the theorem of Dvoretzky and Rogers [1] (see also e.g. Marti [2, page 27] and Singer [5]), there exists a series J2ieN x i = {1 ,2 ,... }) in X which is not absolutely convergent, i.e. Y^ieN 11^*11 = °°> but unconditionally convergent, i.e. J2 ieNxp(i) convergent in X for every permutation p of N. We can assume here that {x^} C X\{0}. The unconditional convergence of J2 ieN Xi is equivalent to its unordered convergence (cf. e.g. [2 , page 20]), i.e. setting x := YhieN x*> we ^ave x = lim7er where T is the family of finite subsets of N, so that there exists a 70 € T such that k - E i67 ' < 1 for every 7 G T with 7 D 70- Let K — 1 + ||x|| + 7o which lies in (1,00). Then we see that — + x — Xi < K E Xf E x i~ x ie 7 yiS7U7o y ie 70 \7 for every 7 € T. From this it follows that M {x ;}i< i< m < K for any m € N so that E ||xj|| —>■ 00 (m —> 00), l< i< m which implies that cr(X) = 00 as desired. REVERSE TRIANGLE INEQUALITY 183 We next show that cr(X) < oo if dim X < oo (cf. Rudin [3, page 126], [4, page 118]). Contrary to the assertion assume that a(X) = oo. Then there are a sequence {xijiejv m and a strictly increasing sequence {^(A;)}*;^ in N with i/(l) = 1 such that T{Xi}k > 2kM{Xi}k {keN ) (5) where {x j}fc := {xi : v(k) < i < u(k + 1)} (fc G JV). On replacing Xj by ('T{xi}k)~1xj (^(fc) < j < v(k + 1)), if necessary, we can assume that T{xi}k = 1 so that M{xi]k < 2~k (k E N ). By virtue of the fact that E in i = E = E 1 = °°- ieN keN TMk keN the series J2ieN x i is not absolutely convergent in X. On the other hand, the series J2ieN Xi *s unconditionally convergent in X. To see this take an arbitrary permutation p on JV. Let e be any positive number. Fix a ko E N such that 2-fco+i < £m There exists an m E N such that p(i) > u(ko) for any i > m. Then, for any y > m and s E N there exists at E N such that p(i) < u(t) (y < i < /x + s) so that w VI y i x p(i) y i x p(i) k o < k < t H<i<fj,+s,i/(k)<p(i)<i>(k+l) < M {Xi}k < 2~ ko+ 1 < £ fco<fc<oo for any y > m and s E N. This means that J2ieNxp{i) is convergent in X and thus J^ieN x i *s unconditionally convergent. This contradicts the Riemann theorem (c.f. e.g. [2, page 22]) which asserts the equivalence of absolute convergence of a series and unconditional convergence of a series in a finite dimensional Banach space X . □ 3. Upper and Lower Estimates We denote by X* the dual space of a normed space X. The value of £ E X* at x E X will be denoted by (x,£). We denote by Sx (Sx*, resp.) the unit sphere in X (X *, resp.). For any A E R we write A+ = max(A, 0). We maintain the following Theorem 2. For any finite nonzero Borel measure y on Sx* the following upper estimate holds: \ (6) tsx . for any finite nonzero Borel measure m on Sx the following lower estimate holds: - l a{X) > ( sup 77; r [ {x,£)+ dm{x) (7) \tesx* m[bX) Jsx P roof. First we prove (6). Choose an arbitrary sequence {xk}keD (.D = {1 ,... ,h}) in X \ {0}. Since the function £ i—» YlkeD(xki£)+ is weakly* continuous on Sx• which is weakly* compact by the Banach-Alaoglu theorem, 184 MITSURU NAKAI AND TOSHIMASA TADA there exists an 7? in S x * such that J2keD(xkiV)+ — m&x-$esx * ^2keD(xkiO+ so that Y^keD(xk^v)+ > J2keD(xki£)+ f°r every £ in Sx*- Integrating both sides of the above over Sx* with respect to dfi(£) we obtain ^ i*k<o+ M «)• keD X ’ keDJSx* On setting A = {k e D : (xk, rj) > 0} we see that M {xk} > J 2Xk > (J2xk,v) = ^2(xk,v) = ^2(xk,v)+= ^2ixk,vY keA \keA / keA keA keD Hence we deduce M[Xk] - ) (\\xk\\ 1xk,^)+d ^ ) IM > so that we conclude that -1 T{xk)/U {xk} < Sx* On taking the supremum in the above with respect to {xfc} C X \ {0} we obtain (6). We proceed to the proof of (7). By Theorem 1, a(X) = 00 if dimX = 00, in which case (7) is trivially valid. We can thus assume that dim X < 00. Then Sx is strongly compact so that we can find for an arbitrary positive number e a finite family {Ek}keD {D = {1 ,... ,/i}) of Borel subsets Ek C Sx with the following properties: (a) the diameter of Ek is less than e for every keD; (b) Ei D Ej = 0 (i ± j ); (c) m(Ek) > 0 for every k e D; (d) m(Sx\ UkeD Ek) = 0. Choose and then fix an arbitrary point x k in Ek for each k e D and set yk = m(Ek)xk (k € D).
Details
-
File Typepdf
-
Upload Time-
-
Content LanguagesEnglish
-
Upload UserAnonymous/Not logged-in
-
File Pages14 Page
-
File Size-