A NOTE ON LINEAR FUNCTIONAL NORMS

A NOTE ON LINEAR FUNCTIONAL NORMS YIFEI PAN AND MEI WANG Abstract. For a vector u in a normed linear space, Hahn-Banach Theorem provides the existence of a linear functional f, f(u) = u such that f = 1. In this paper we prove the existence of functionals with f( ) = for linearly independent vectors and characterize the norm-one property f = 1 in terms of the triangle inequality. Key words: Hahn-Banach Theorem; triangle inequality Let (X, ) be a normed vector space over a field K (K = R or C) and X be the space of K-valued, continuous linear functionals on X. For any u X \ {0}, Hahn-Banach Theorem [1] gives the existence of f X, f(u) = u with the property f = 1. In this paper we prove the existence of f X, f( ) = for linearly independent vectors, j = 1,, k and investigate the implications of the norm-one property of such functionals. This paper is motivated by the study of Schwarz lemma for higher order derivatives of holomorphic mappings from unit ball to itself in C n. The following are the results. Theorem 1. Let (X, ) be a normed vector space over K. Let u 1,, u k X \ {0} be linearly independent. Then f X such that f( ) =, j = 1,, k. Remarks. Here the condition of linear independence is necessary (see Example 1 below). The functional in Theorem 1 has f 1 by definition. The following shows that in most situations, f > 1. 2000 Mathematics Subject Classification. 46B20. 1

2 YIFEI PAN AND MEI WANG Proposition 1. Let (X, ) be a normed vector space over K, u 1,, u k X \ {0}. If f X such that f = 1 and (1) f( ) =, j = 1,, k, then the triangle equality holds, i.e., (2) =. Furthermore, 1 1 + + m m = m, {j 1,, j m } {1,, k}. Remarks. The following figure is an example of the triangle equality case under maximum norm. u1 u2 0 Figure 1. Triangle equality case for k = 2 under maximum norm. Our proofs are based on applications of the Hahn-Banach Theorem [1]. Here we quote a corollary of the Hahn-Banach Theorem as a lemma.

Lemma 1. Let L be a linear subspace of a normed vector space X over K. Then for each u o X with dist(u o, L) = inf u L u o u > 0, f X such that 3 f(u) = 0, u L, f(u o ) = dist(u o, L), f = 1. Proof of Theorem 1. Let L 1 = span{u 2,, u k }, v 1 = 1 u 1 u 1. L 1 is a finite dimensional subspace thus L 1 is closed. Since v 1 is independent of u 2,, u k, dist(u 1, L 1 ) > 0 so By Lemma 1, h X such that d 1 = dist(v 1, L 1 ) = 1 u 1 dist(u 1, L 1 ) > 0. h = 1, h(v 1 ) = d 1, h( ) = 0, j = 2,, k (since L 1 ). Let f 1 = 1 d 1 h = Then f 1 X and f 1 = Define u 1 dist(u 1, L 1 ) h = u 1 dist(u 1, span{u 2,, u k }) h. u 1 dist(u 1, L 1 ), f 1( ) = 0, j = 2,, k, f 1 (u 1 ) = 1 d 1 h( u 1 v 1 ) = u 1. L m = span{, j {1,, k} \ {m}}, m = 2,, k. We can construct f m X in a similar way such that m = 2,, k, f m = Now let u m dist(u m, L m ), f m( ) = 0, j {1,, k} \ {m}, f m (u m ) = u m. f = f 1 + + f k.

4 YIFEI PAN AND MEI WANG Then f X and f( ) = for j = 1,, k. This completes the proof of Theorem 1. Corollary 1. Let (X, ) be a normed vector space over K. Let u 1,, u k X \ {0} be linearly independent. Then f X and θ 1,, θ k [0, 2π) such that f(u j) = u j for u j = e iθ j, j = 1,, k. Proof. By Theorem 1, there exists a f X such that f( ) =, j = 1,, k. Thus there are θ 1,, θ k [0, 2π) such that for j = 1,, k, f( ) = e iθ j. Consequently, f(u j) = f(e iθ j ) = e iθ j f( ) = e iθ j e iθ j = = e iθ j = u j. The following example shows that Theorem 1 and Corollary 1 may not hold if the vectors are linearly dependent. Example 1. Let X = R 2 with the Euclidean norm. Let u 1 = (1, 0), u 2 = (0, 1) and u 3 = (1, 1). If there were a functional f X such that f( ) = for j = 1, 2, 3, then f(u i ) = ±1 for i = 1, 2. By linearity f(u 3 ) = f(u 1 + u 2 ) = f(u 1 ) + f(u 2 ) = 0 or 2, which contradicts f(u 3 ) = u 3 = 2. Therefore f( ) = may not exist for linearly dependent s.

Proof of Proposition 1. From (1), f( ) = for j = 1,, k. If f = 1, by the linearity of f and the definition of functional norm, u 1 + + u k u 1 + + u k = u 1 + + u k u 1 + + u k. = f(u 1 + + u k ) u 1 + + u k sup f(u) u X \{0} u = 1 5 The triangle inequality implies that the equality holds. Consequently, by applying the above result to subsets of the s we obtain 1 + + m = 1 + + m, {j 1,, j m } {1,, k}. Consider w j = for j = 1,, k. Then f(w j) = w j = 1. Applying the above result to the w j s we have = w j = w j = k, and 1 1 + + m m = w j 1 + + w jm = w j1 + + w jm = m for any {j 1,, j m } {1,, k}. This completes the proof of Proposition 1. Remark. Notice that our result is consistent with [2], where the sharpness of the triangle inequality was discussed thoroughly and extensively. Corollary 2. Let (X, ) be a normed vector space over K and u 1,, u k X \ {0}. If f X such that (3) f( ) =, j = 1,, k, <,

6 YIFEI PAN AND MEI WANG then f > 1. Proof. By the definition of functional norm and (3), f = f(u) sup u X \{0} u f(u 1 + + u k ) u 1 + + u k = u 1 + + u k u 1 + + u k > 1. The example below illustrates that, when a functional f with f( ) = for k linearly independent vectors exists (Theorem 1) and the triangle equality = holds (ref. (2) in Proposition 1), f is not necessarily of norm one. As indicated in Proposition 1, f = 1 is a very restricted condition. In fact the corresponding f in X can be arbitrarily large. Example 2. Let X = C 0 = { u = (a 1, a 2, ) = (a n ) : a n K, } lim a n = 0 n, (a n ) = max a n. n (X, ) is a complete normed vector space (Banach space) with the dual space where X = C 0 = l 1 = { f = {α 1, α 2, } = {α n } : α n K, ( ) f(u) = f {αn} u(an) = α n a n, f = } α n < α n. For k > 0 and a 1,, a k such that 0 < a 1 < a 2 < < a k, let u 1 = (a 1, 0, 0, ), u 2 = (a 2, a 1, 0, ),, u k = (a k, a k 1,, a 1, 0, ).

7 Then u 1,, u k are linearly independent vectors in C 0. For r (0, 1), consider f has the property (as in Theorem 1) f = { 1, 0,, 0, r, r 2, r 3, } C }{{} 0 = l 1. k 1 times f( ) = a j =, j = 1,, k. Furthermore, = a j, j = 1,, k, u 1 + +u k = a 1 + +a k = =. However f = α n = 1 1 r > 1. Furthermore, for any u = (a 1, a 2, ) C o with u = 1, we have a k 1 for all k and a j < 1 for some j. Let v = (1, 1, ). Then Thus f(u) = α n a n < α n = f(v) = 1 1 r = f. f(u) < f, u with u = 1, which implies that the functional f is not norm-attaining, that is, f(u) u < f, u X \ {0}. Remark. An immediate corollary of the above results is that, if X is strictly convex and u 1,, u k X \ {0} are linearly independent, then a functional f with f( ) =, j = 1,, k, f = 1 implies f(u 1 ) u 1 = f(u 2) u 2 = = f(u k) u k.

8 YIFEI PAN AND MEI WANG References [1] Zeidler, E. (1995) Applied Functional Analysis: Main Principles and Their Applications. Applied Mathematical Sciences Vol. 109. Springer-Verlag, New York. [2] Mitani, K., Saito, K., Kato, M. and Tamura, T., On sharp triangle inequalities in Banach spaces. J. Math. Anal. Appl. 336 (2007) 1178-1186. Yifei Pan, Department of Mathematical Sciences, Indiana University - Purdue University Fort Wayne, Fort Wayne, IN 46805-1499 School of Math and Informatics, Jiangxi Normal University, Nanchang, China E-mail address: pan@ipfw.edu Mei Wang, Department of Statistics, University of Chicago, Chicago, IL 60637 E-mail address: meiwang@galton.uchicago.edu