ON THE TRIANGLE INEQUALITY FOR THE STANDARD 2-NORM Hendra Gunawan and Wono Setya-Budhi Abstract. We shall show here that the triangle inequality for the standard 2-norm is equivalent to a generalized Cauchy-Schwarz inequality, as is that for norms to the Cauchy-Schwarz inequality. Besides its direct proof, we also present two alternative proofs through its equivalent inequality. 1. Introduction. Let X be a real-vector space, equipped with an inner-product h¢; ¢i 1 together with its induced norm kxk = hx; xi 2 . De¯ne the function k¢; ¢k : X £ X ! R by n o 1 2 kx; yk := kxk2kyk2 ¡ hx; yi2 ; which is equal to twice the area of the triangle having vertices 0, x, and y (or the area of the parallelogram spanned by the vectors x and y) in X. It can be shown that the above function de¯nes a 2-norm on X, which satis¯es the following four properties: (i) kx; yk = 0 i® x and y are linearly dependent; (ii) kx; yk = ky; xk; (iii) kx; ayk = jaj kx; yk; a 2 R; (iv) kx; y + zk · kx; yk + kx; zk: The space X, being equipped with the 2-norm, is thus a 2-normed space. The concepts of 2-normed spaces (and 2-metric spaces) were intially introduced by Gahler [G1], [G2], [G3] in 1960's. A standard example of a 2-normed space is R2 equipped Keywords: 2-normed spaces, inner-product spaces, generalized Cauchy-Schwarz inequality 1 with the following 2-norm kx; yk := the area of the triangle having vertices 0; x; and y: The 2-norm above on X is a just generalization of this standard example. For recent results on 2-normed spaces, see for example [GM]. As it usually happens with norms, given a candidate for a 2-norm, the hardest part is to check the property (iv) | better known as the triangle inequality. In this note, we shall show that the triangle inequality for the 2-norm above is equivalent to a generalized Cauchy-Schwarz inequality, just as is that for norms to the Cauchy-Schwarz inequality. Besides its direct proof, we also present two alternative proofs through its equivalent inequality. 2. The Inequality. For our 2-norm above on X, we have the following fact: Fact. The triangle inequality is equivalent to kxk2hy; zi2 + kyk2hx; zi2 + kzk2hx; yi2 · kxk2kyk2kzk2 + 2hx; yihx; zihy; zi: Proof. Observe that kx; y + zk2 = kxk2ky + zk2 ¡ hx; y + zi2 ¡ ¢ ¡ ¢ = kxk2 kyk2 + 2hy; zi + kzk2 ¡ hx; yi2 + 2hx; yihx; zi + hx; zi2 ¡ ¢ = kx; yk2 + 2 kxk2hy; zi ¡ hx; yihx; zi + kx; zk2: The triangle inequality is thus equivalent to kxk2hy; zi ¡ hx; yihx; zi · kx; yk kx; zk: Replacing y + z by y ¡ z, we ¯nd that the triangle inequality is equivalent to ¯ ¯ ¯kxk2hy; zi ¡ hx; yihx; zi¯ · kx; yk kx; zk: 2 Squaring both sides, we get kxk4hy; zi2 ¡ 2kxk2hx; yihx; zihy; zi + hx; yi2hx; zi2 ¡ ¢ · kxk4kyk2kzk2 ¡ kxk2 kyk2hx; zi2 + kzk2hx; yi2 + hx; yi2hx; zi2: Canceling hx; yi2hx; zi2 and then dividing both sides by kxk2 (assuming that x 6= 0), we obtain the desired inequality. Note. It is clear from the proof that the equality kxk2hy; zi2 + kyk2hx; zi2 + kzk2hx; yi2 = kxk2kyk2kzk2 + 2hx; yihx; zihy; zi holds i® kx; y + zk = kx; yk + kx; zk or kx; y ¡ zk = kx; yk + kx; zk holds. Let us take a closer look at the inequality. First note that the equality holds when x or y or z equals 0. One may also observe that the equality holds when x = §y or x = §z or y = §z. Further, if z ? spanfx; yg and z 6= 0, then the inequality becomes hx; yi2 · kxk2kyk2; which is the Cauchy-Schwarz inequality. Hence the inequality may be viewed as a gener- alized Cauchy-Schwarz inequality. For X = R, the equality obviously holds. For X = R2, the equality also holds. To see this, assume x; y; z 6= 0. Dividing both sides by kxk2kyk2kzk2, we obtain hx; yi2 hx; zi2 hy; zi2 hx; yi hx; zi hy; zi + + · 1 + 2 : kxk2kyk2 kxk2kzk2 kyk2kzk2 kxkkyk kxkkzk kykkzk Next, assuming that kxk = kyk = kzk = 1, the inequality becomes hx; yi2 + hx; zi2 + hy; zi2 · 1 + 2hx; yihx; zihy; zi: Now put ® = 6 (x; y); ¯ = 6 (x; z), and γ = 6 (y; z). Then ® + ¯ + γ = 2¼, and we have the equality cos2 ® + cos2 ¯ + cos2 γ = 1 + 2 cos ® cos ¯ cos γ: 3 Alternatively, assume x 6= §y, so that spanfx; yg = R2. Writing z = ax + by for some a; b 2 R, one may check that both sides are equal to 1 + 2abhx; yi + (1 + a2 + b2)hx; yi2: In general, it can be shown that the equality holds i® spanfx; y; zg is at most two dimen- sional. 3. The Proof. We shall now prove the inequality. There are at least three ways to do it. First, if X is separable, then we can verify the triangle inequality for the 2-norm directly. Let (ei) be an orthonormal basis for X (indexed by a countable set). Then, by Parseval's formula and polarization identity, we have n³X ´³X ´ ³X ´2o 1 2 2 2 kx; yk = hx; eii hy; eji ¡ hx; eiihy; eii i j i n o 1 1 X X¡ ¢2 2 = hx; e ihy; e i ¡ hx; e ihy; e i : 2 i j j i i j The triangle inequality then follows easily. Second, whether or not X is separable, we can always prove its equivalent inequality as follows. As argued earlier, under the assumption kxk = kyk = kzk = 1, we only need to show hx; yi2 + hx; zi2 + hy; zi2 · 1 + 2hx; yihx; zihy; zi: Assuming x 6= §y and z is not perpendicular to spanfx; yg, we may write z = z1 + z2 where z1 2 spanfx; yg, that is z1 = ax + by for some a; b 2 R, and z2 ? spanfx; yg. As in R2, we then have the equality 2 2 2 hx; yi + hx; n1i + hy; n1i = 1 + 2hx; yihx; n1ihy; n1i 2 where n1 = z1=kz1k. Multiplying both sides by kz1k , we get 2 2 2 2 2 kz1k hx; yi + hx; zi + hy; zi = kz1k + 2hx; yihx; zihy; zi; 4 since hx; z1i = hx; zi and hy; z1i = hy; zi. Hence 2 2 2¡ 2¢ 2 hx; zi + hy; zi ¡ 2hx; yihx; zihy; zi = kz1k 1 ¡ hx; yi · 1 ¡ hx; yi ; 2 since kz1k · kzk = 1 and 1 ¡ hx; yi ¸ 0; and the equality holds i® kz1k = 1 (and consequently z2 = 0), that is i® z 2 spanfx; yg. Third, one may observe that our inequality is actually equivalent to det(M) ¸ 0 where M is the Gram matrix 0 1 hx; xi hx; yi hx; zi M = @ hy; xi hy; yi hy; zi A : hz; xi hz; yi hz; zi Since M is positive semide¯nite, the inequality follows immediately. It is also easy to see here that the equality holds i® span fx; y; zg is at most two dimensional (see [HJ, pp. 407-408). References [G1] S. GÄahler,\2-metrische RÄaumeund ihre topologische Struktur", Math. Nachr. 26 (1963/64), 115-148. [G2] S. GÄahler,\Lineare 2-normietre RÄaume", Math. Nachr. 28 (1965), 1-43. [G3] S. GÄahler,\Uber der Uniformisierbarkeit 2-metrische RÄaume", Math. Nachr. 28 (1965), 235-244. [GM] H. Gunawan and Mashadi, \On ¯nite dimensional 2-normed spaces", Preprint (No- vember 1999). [HJ] R. Horn and C. Johnson, Matrix Analysis, Cambridge Univ. Press, New York/Cam- bridge, 1985. Department of Mathematics Bandung Institute of Technology Jl. Ganesa 10 Bandung 40132 Indonesia Email [email protected] [email protected] 5.