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 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) = — + 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 I V y i x p(i) y i x p(i) k o < k < t H < M {Xi}k < 2~ ko+ 1 < £ fco 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 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). We take a maximal subset A c D such that M {Vk} = || Efce^yjfell- Here the maximality of A means that if M {yk} = ||E fc€ B |/fc|| and A C B C D, then A = B. By an application of e.g. Hahn-Banach theorem there exists an 77 in Sx * such that {Y^keAVk^) = II J2keAVk\\ (cf- e-S- Yosida [6 , page 108]). Hence we have M{yk} = J2keA{yk,v)- We maintain that A = {keD : yk,r])( > 0}. (8) Assume the existence of an i e A and yet (yi}rj) < 0. Then M{yk} = E (2/*’^) < E (yk,v) < E »* < M{yk}, keA &€A\{i} fce^4\{i} REVERSE TRIANGLE INEQUALITY 185 a contradiction. Conversely assume that ( yi,r7) > 0 and yet i ^ A. Then M{yk} = J2 (yk,v) < 2 - X ! yk < M{yk}, k£A fceAu{i} k&AU{i} which contradicts the maximality of A. We have thus established ( 8 ). In view of (8 ) we have the identity M{yk} = ^2 (yk,v) = ^2 (yk,v)+ = £ < ^ > + = J2 {xk,r])+m(Ek). keA keA keD keD For each keD, because of (a), we have \(xk,r])+ - (x,rj)+ 1 < \(xk,ri) - {x,rf)\ < Hr/H ||xfc - x|| < £ for every x e Ek and thus (xk,rj)+ < (x,r})+ +£ (xeEk). On integrating both sides of the above over Ek with respect to dm(x) we have (xk,r])+m(Ek) < / (x,r])+dm(x) + £m(Ek). JEk Thus we see by (b) and (d) that M{yk} < 5 3 ( / (x,rj)+dm(x) + £m(Ek) J = / (x,rj)+dm(x) + £m(Sx) keD \JEk J Jsx and therefore we have M{yk} < sup / (x,$>)+dm(x) + £m(Sx )- £eSx * JSx On the other hand, again by (d), we see that T{yk] =$ 3 W l= 1 3 = 1 3 m(E*0 = ™(sx)- keD keD keD By (c) we see that {yk}keD is a finite sequence in X \ {0} and a fortiori a(X) > T{yk}/M{yk} > ( sup 1 [ (x,£)+dm(x) + . \££SX. rn{bx) Jsx j Since £ > 0 is arbitrary, on letting £ j 0 in the above, (7) is deduced. □ 4. Computations of Best Constants We now apply Theorem 2 to compute a(lp(n)) (n e N ) for the space lp(n) (p = 1,2, oo). For the purpose we denote by B n and Sn~l the unit ball and the unit sphere in the Euclidean n-space E n. We denote by \Bn\ the Euclidean volume of B n and by |5n-1| the Euclidean surface area of S '71-1 so that \B2n\ = 7rn/n\, |B 2n~l \ = 22n7rn_1n!/(2n)!, IS"-1! = n\Bn\ (n e N ). (9) 1 86 MITSURU NAKAI AND TOSHIMASA TADA Here we follow the usual convention |5°| = 2 and we also set \B°\ = 1 for conve nience. We also need to consider the quantities Nn given by (n e N )y (10) where [ ] is the Gaussian symbol. For two sequences {am}meN and {&m} mejv of positive real numbers we write am >- bm or 6m -<: am if lim inf™ -^ am/6m > 0 and we also write am ~ bm if am y bm and am -< bm. We will write n-vectors x as x = (x1,... , x n). We have the following result. Theorem 3. The best constant cr(/p(n)) (n £ N) for three spaces lp(n) (p — 1,2, and oo) are given as follows: cr^^n)) = 1/Nn « y/n\ (11) a(l2 (n)) = n\Bn\/\Bn~l \ w y/n; (12) (r(/°°(n)) = 2n « n. (13) P roof. Recall that /1(n)* = l°°(n) and l2(n)* = l2 (n). Because of the finite dimensionality of /°°(n) we also have l°°(n)* = /1(n). We first prove (13). Let £k = (<5fci,.. • G Sji(n) = Si°°(n)* where <5^- is the Kronecker delta. Consider the Borel measure fi = X a n xGSto0(n) 2n ' k=1 so that we obtain cr(Z°°(n)) < 2 n. REVERSE TRIANGLE INEQUALITY 1 8 7 Conversely, we take ek = (Ski, • • • , <5/cn) £ Consider the Borel measure m = Yh <7(/°°(n)) 1 < sup €es|1(n) so that we obtain a(l°°(n)) > 2n. We have thus established (13). Next we prove (12). It is easy to see directly that (14) and similarly, by taking ds as dm in (7), we obtain - l (15) For any £ 6 Sn 1 there exists an orthogonal transformation P of En = l2 (n) such that Pen = £ where en = (0,... ,0,1) G En. By the change of variable Py = x (y = (y1,... ,yn)), on noting P~1(Sn~1) = 5 n_1, ds(Py) = ds(y), (Py,Pen) = (y,en) = yn and ds(y) = (1 /yn)dyl ... dyn~l on Sn_1 D {y n > 0} over 5 n_1, we deduce that f Sn-i (x, £)+ds(x) = |Bn-1|. This with (14) and (15) implies a (l1^)) 1 > inf f (x,£)+dn(Z) V {Sl°°(n)) Jsl0O{n) (1 6 ) J2 (x,ek)+- 188 MITSURU NAKAI AND TOSHIMASA TADA Take e = (1/n,... , 1/n) G Sti(n) and set = {fc 6 D : (e,Ek) > 0}. We maintain that i™n = = (17) Il(n) fceD fceo fceA Observe that k e A is equivalent to X^i = E (i = 1 ,... ,n). (18) keA Since {£k : k G D} and also Sji(n) are symmetric with respect to the coordinate hyperplane considered in ^(n), we only have to show that ^2(x,ek)+ > ^2(e,ek)+ (19) keD keD for every x G 5p(n) with xl > 0 (i = 1 ,... , n) to conclude the validity of (17). By (18) we infer as follows: E = J V e = E ^ e 1 = E = eJ 2x{ = ]^x*E i=l i=l 1=1 i=l = iy(E 4) = e (x>‘4) i=i VfceA / fceA \i=i / = ]T(£,£fc) < ^<£,£fc)+ < ]T(x,£fc) + keA keA keD so that (19) and hence (17) has been shown. By (16) and (17) we conclude that cr(^1(n))_1 > (20) keA Recall that k e A is equivalent to that the number of components e\ = 1 is greater than or equal to that of elk = —1. Hence the number j of components e\ = — 1 for k e A is at most t := [n/2]. We denote by Aj the set of k e A such that the number of components e\ = - 1 is j so that A = Uo E <«.«»> - sE(f>i) = ^e (e (e 4)) k£A keA \i=l / j=0 yfceAj \i=l / J 1 1 / / \ \ 1 [(^-t-l)/2]-l / v = ±z((:M e <»-«>(:)■ Hence by (10) and (20) we conclude that cr(/1(n)) < 1/A^n. REVERSE TRIANGLE INEQUALITY 189 We need to show that cr(/1(n)) > 1 /Nn to conclude the identity in (11). Using D = {1 ,2 ,3 ,... ,2n} above we consider the set {e k : fc G D} of 2n points ek in 5/i(n) such that e\ (i = 1,... , n) are either 1/n or —1/n , where ek = (e£,... , e£). Take the Borel measure m = YlkeD^ek on By (7) and ^(n)* = Z°°(n), we have a (Pin)) 1 < sup — -----r [ (x,£)+dm{x) £€Siccln) m (&,!(„)) J s ti(n) (21) = SUP 4 £ < e*>S>+ = £ ( e*’ ?>+ - Take e = (1,... , 1) G 5z«>(n) and set A = {fc € D : (ek,e) > 0}. We maintain that “a* £ (e*.?>+ = £ ( efc>«')+ = £ (et.e). (22) k€D keD * € A Since {e* : fc G D } and Sj°o(n) axe symmetric with respect to the coordinate hyper planes considered in /°°(n), we only have to show that £ ( e fc,e>+ < £(e*, £ 4 > 0 (i = l,...,n ). (24) keB Fix an arbitrary i = 1 ,... , n. If there is a fc G B such that ejj. = —1/n, then we consider an ek> for a fc' G D with the following properties: eJk, = 4 (j ¥> *) and ek' = 1/n. Since we have (ek>,£) = {ek,£) + 2£*/n > 0, we see that fc' G B. Therefore the number of fc G B with e\ = 1/n is not less than that of fc G B with e*. = —1/n for each fixed i = 1 ,... ,n. This assures the validity of (24). We next show that E(E4)>E(E4). (25) keA \ i = l / keB \ i = l / In fact, by the equivalence of fc G A and E i After these two preperations above we proceed to the proof of (23). By (25), 0 cr(^1(n) ) _1 < (26) keA Recall that k € A is equivalent to X)i E<^> = E(E4) =e (e (E4 keA keA \i—1 / j—0 ykeAj \i=l / - § (Q H 'H aH § <-wC Therefore by (10) and (26) we conclude that cr(/1(n)) > 1/-/Vn and the identity in (11) has been established. The proof of (11) is complete if we show that Nn « 1/y/n. We can easily see that Nn -< Nn+i. Hence we only have to show that N2n = 2~2n(2n)~1J 2 2(n - j ) f 2U) ~ l /V n . j =o V 3 ' On changing j to k by n — j = k in the above expression of N2n we have M = J - y /r(2n)Vn(2n)\y/n fr-tt A, A: 2n n .P n ^nn2o2n J.1 ( nJn fc=i ^ (n!)222nK 7 j=i 11 Vx n+? We use here the Wallis formula (2n)!i/n/(n!)222n | l/V^r (n | oo) to deduce n k n\fn N2n ~ IT (l - fc/(n + j))- fc=i j = l REVERSE TRIANGLE INEQUALITY 191 Hence we have to show that k ' k in order to assure N2n ~ 1/y/n. We divide the summation in the left side of (27) as the sum of an = and bn = Z)fc=[v^]+r T^en t^ie r a tio n (27) is derived if we can show that an ~ n and bn -< n. For the purpose we only have to show that cin + bn -< n and an y n. Clearly an < ^ = h/^Kh/ra] + l)/2 -< n. Observe that j= 1 x n + j j V 3 + n J J y j r i n + J n + 1 + k\ ( n + 1 xfc < exp —k log n + 1 J \n+l+k = 1 / 1 + —rr < 1 n + 1 / / n + 1 2(n + l)2 < 4(n + l)5 k3{k - 1) “ fc4 and therefore ^ pTl bn < 4(n + l)2 ^ 2 k~ 3 < 4(n + l ) 2 / x~3dx < 2(n + l)2 x [y/n] - Thus we have seen that an + bn -< n. We turn to the proof of an >- n. We use the inequality log(l — x) > (log4)x (0 < x < 1/2). If n > 4 and 1 < k < yfn, then kf(n + j) < y/n/n < 1/2 for j = 1,... , k. Therefore ft 0 - ^h) = exp (§ log 0 - ^7 > exp ( - ( lo g 4) ~^Tj j - exp ( ~ ( loS 4)fc loS 192 MITSURU NAKAI AND TOSHIMASA TADA Since 1 < k < y/n in the summation of an, we obtain for n > 4 that as required. The proof of (27) and thus that of Theorem 3 is herewith complete. □ 5. Order of cr(Zp(n)) For simplicity we write || • ||p = || •; lp(n)\\ (1 < p < oo) and we also denote by R n the n-dimensional linear space which is the base space for any lp(n) (1 < p < oo). Observe that p t-> n - 1 / p ||x||p for each fixed x in Rn is increasing on [1, oo]. Similarly P ^ Ik I Ip f°r each fixed x in Rn is decreasing on [1, oo]. Based on these two facts we can see that n1//pcr(Z1(n)) < a(lp(n)) < n1~1^pa(l1(n)) (1 < p < 2); (28) n1/2- 1/p(J(;2(n)) < a(lp{n)) < n1/p_1/,2cr(Z2(n)) (1 < p < 2); (29) a(lp(n)) < n1//2_1//pcr(/2(n)) (2 < p < oo); (30) n-1//p Using these inequalities we prove the following result. Theorem 4. The orders of the best constants cr(lp(n)) of lp(n) as n | oo are exactly given for 2 < p < oo as a (lp(n)) « n1-1/p (32) and are estimated for 1 < p < 2 as nm ax(l/p-l/2,l-l/p) ^ cr(lP(n)) ^ nmin(3/2-l/p,l/p) (33) Proof. Since a(l2 (n)) w y/n by (12), the above (30) implies that a{lp(n)) ^ n1_1/p (2 < p < oo). Similarly (13) and (31) yield n1_1/p -< a(lp(n)) (2 < p < oo). These two relations assure the validity of (32). By (11) and (28), we see that ni/P- i /2 ^ a(lp(n)) ^ n3/2_1/p (l Similarly, from (13) and (29) it follows that n i - i / p ^ a(lp(n)) n1//p (1 < p < 2). (35) Combining (34) and (35), we deduce (33). □ REVERSE TRIANGLE INEQUALITY 193 References 1. A. Dvoretzky nd C.A. Rogers, Absolute and conditional convergence in normed linear spaces, Proc. Nat. Acad. Sci. U.S.A. 36 (1950), 192-197. 2. J.T. Marti, Introduction to the Theory of Bases, Springer, 1969. 3. W. Rudin, Real and Complex Analysis, 2nd edition, McGraw-Hill, 1974. 4. W. Rudin, Real and Complex Analysis, 3rd edition, McGraw-Hill, 1987. 5. I. Singer, Bases in Banach Spaces I, Springer, 1970. 6. K. Yosida, Functional Analysis, Springer, 1965. Mitsuru Nakai Toshimasa Tada Nagoya Institute of Technology Daido Institute of Technology Gokiso Daido Showa Minami Nagoya 466 Nagoya 457 JAPAN JAPAN [email protected] t ada@ asuke. daido-it .ac.jp [email protected]