E extracta mathematicae Vol. 10, N´um.2, 146 – 151 (1995) Isometries of Finite-Dimensional Normed Spacesy Felix´ Cabello Sanchez´ and Jesus´ M.F. Castillo Dpto. de Matem´aticas, Univ. de Extremadura, 06071-Badajoz, Spain AMS Subject Class. (2000): 46B04, 46B25 Received July 10, 1995 A fundamental result in Functional Analysis establishes that no matter which norm is defined on a finite-dimensional space, the underlying topological space is the same. A different question is the equality of the underlying metric spaces. The basic examples of norms in Kn, K = R or C, are the p-norms, 1 · p · 1, which generalize the euclidean modulus (p=2): à ! 1 Xn p p kxkp = jxij ; p < 1; i=1 kxk1 = max jxij; p = 1: n n We denote by `p the space K endowed with the k kp norm. Here we present n n several proofs that the spaces `p and `q are not isometric when p is differ- ent from q. This result is certainly well-known to specialists and it appears mentioned in several books on real analysis and/or functional analysis. Nev- ertheless, it is not easy to find an explicit proof. For instance, in [3], it appears as an exercise to prove the cases p = 1; 2; 1; and in [13, p. 280, Prop. 37.6] n n it is established that the Banach-Mazur distance between `p and `p¤ (the only case that matters, as we shall see) is proportional to n1=p¡1=2; the proof there presented, using Khintchine’s and Kahane’s inequalities, has little over- lap with ours. Besides this, Pelczynski [2] attributes to Gurarii, Kadec and n Macaev [5, 6] the exact calculus of the Banach-Mazur distance between `p spaces: If either 1 · p < q · 2 or 2 · p < q · 1, then d(`n; `n) = n1=p¡1=q. p p q If 1 · p < 2 < q · 1, then ( 2¡1) d(`n; `n) · max(n1=p¡1=2; n1=2¡1=q) · p p q n n 2 d(`p ; `q ). ySupported in part by DGYCIT project PB94-1052-C02-02. 146 isometries 147 It is enough to consider the case of linear isometries since, by an old theo- rem of Ulam and Mazur [10], an isometry of a real normed space that carries 0 to 0 must be linear (cf. [1, p. 166]). In fact, if f is an isometry between normed spaces then for some linear isometry T one has that f(x) = T (x)+f(0) (see [4, p. 107, Ex. 3(b)]). The set fx 2 E : kxk = 1g will be termed the unit sphere of k k. From now on, the unit sphere of the scalar field shall be denoted D. An isometry between the normed spaces (E; k k1) and (F; k k2) is a linear application T : E ! F such that, for all x 2 E, kT xk2 = kxk1. It is clear that an isometry transforms the unit sphere of one space into exactly the unit sphere of the other. Let Sp be the unit sphere of k kp. Our first proof is based on the idea: how many “peaks” has Sp? n n Theorem. If p is different from q, the spaces `p and `q are not linearly isometric except in the case: K = R, p; q 2 f1; 1g, and n = 2. Let us start with: An obvious case: K = R, p; q 2 f1; 1g, and n = 2. The isometry is an easy consequence of the equality 2 maxfjaj; jbjg = ja + bj + ja ¡ bj. An impossible case: p 2 f1; 1g and q 62 f1; 1g (or viceversa). In this case, Sp contains segments, which are preserved by linear applications, while Sq does not. We now calculate the points where Sp intersects the smallest sphere ¹S2 that containis it. An intermission: comparison with the k k2 norm. It is a direct consequence 1=p¡1=2 of H¨older’sinequality that k k2 · k kp · n k k2, if 1 · p < 2, and that 1=2¡1=q k kq · k k2 · n , if 2 < q < 1. Besides, one easily verifies: (1 < p < 2) The norms k kp and k k2 coincide exactly on the points x = σei, 1=p¡1=2 P σ 2 D. Moreover, kxkp = n kxk2 if and only if x = σiei, σi 2 D. (2 < q < 1) The norms k kq and k k2 coincide exactly on the points x = σei, 1=2¡1=q P σ 2 D. Moreover kxk2 = n kxkq if and only if x = σiei, σi 2 D. For the proof of the second parts of these assertions just verify that if ® < ¯ then the minimum of kxk¯ over the unit sphere S® is attained if and only if all coordinates are equal in modulus. 148 f. cabello, j.m.f. castillo The Proof The first maybe not-entirely-trivial step is to show that n n ¤ Claim 1. An isometry between `p and `q necessarily implies q = p . Proof. To see this, let 1 < p 6= q < 1. Without loss of generality we can n n assume that p < q. Let T : `p ! `q be an isometry represented by a matrix (aij) with respect to the natural basis (ei) and (ej), 1 · i; j · n. It is clear that ¤ n n ¤ the transposed application T : `q¤ ! `p¤ with 1=r + 1=r = 1, r = p; q, must also be an isometry. Since 1 = keikp = kT eikq, and 1 = keikq¤ = kT eikp¤ , one obtains the equalities Xn Xn q p¤ 1 = jaijj ; (1 · i · n) and 1 = jajij ; (1 · j · n): j=1 i=1 Summing all equations one obtains X X q p¤ n = jaijj = jaijj : i;j i;j It is clear that jaijj · 1. The case jaijj 2 f0; 1g directly leads to an application T having the form T x = (rix¼(i)), where jrij = 1 and ¼ is a permutation of f1; : : : ; ng, and this yields p = q. Otherwise, the last equation is only consistent when q = p¤ (and, therefore, p < 2). To complete the proof, the idea is quite simple: why, in the real case, S1 and S1 cannot be (except in the case n = 2) linearly isometric?: Because S1 has 2n “peaks”, and a linear application must transform “peaks” into “peaks”. n Put it otherwise, let [nD be the disjoint union of n copies of D and let D be the product of n copies of D. The vertices of S1 form the set [nD and the n vertices of S1 form the set D . In the real case, [nD has 2n elements and n n D has 2 . In the complex case, [nD is a one-dimensional (real) manifold with n connected components (n circumferences) and Dn is a connected n- dimensional manifold (an n-torus). Exception made of the obvious case n = 1 (and, perhaps, n = 2 real) they cannot be continuously transformed one into the other. Thus, what we want to make is to mimic this proof and make it work with other p. To carry that program through we consider as “peaks” of the norm k kp the points where its unit sphere intersects the smallest ellipsoid ¹S2 that contains it. These points have been calculated in the preceding section. Now, the core of our argumentation appears: isometries 149 n n 1=p¡1=2 2 Claim 2. Let 1 < p < 2. If T : `p ! `p¤ is an isometry, then n T : `n 2 ! `n is also an isometry. Proof. Let T be such an isometry. If x1; : : : ; xn are points in Sp¤ , q > 2, then X 1=p min k §xikp¤ · n ; § (consequence of the parallelogram law plus Holder’s inequality) with strict ¡1=p¤ P inequality except if xi = n ij σijei for all i (with a similar argument to that of the second parts of the assertions in the intermission). Since p X X p n = k σieikp = k σiT eikp¤ it follows that X ¡1=p¤ T ei = n σijej; ¡1 ¡1 which, taking into account the form of T and that (ei; ej) = (T T ei; ej) = ¡1 ¤ 1=p¤¡1=2 (T ei; (T ) ej) = ±ij,P yields (T ei; Tp ejP) = n ±ij. 2 2 1=p¡1=2 Hence kT xk2 = k xiT eik2 = jxij kT eik2 = n kxk2. The immediate effect all this has is that: 1=p¡1=2 T (S2 [ Sp) = TS2 [ Sp = n S2 [ Sp¤ : 1=p¡1=2 n And the lasting surprise: S2 [ Sp = \nD, while n S2 [ Sp¤ = D . Epilogue. There is one case overlooked: K = R and n = 2; here, the only possibility for an operator to be isometry is to be T (x; y) = 2¡1=p¤ (x+y; x¡y). That it is not can be seen as follows: let p < 2 < p¤ and put d = p¤=p; this makes p¤ = d + 1 and p = (1 + d)=d. Consider points (1; r) with r > 1. The equality k(1; r)kp = kT (1; r)kp¤ implies the equality ³ ´d 2 1 + rd+1=d = (r + 1)d+1 + (r ¡ 1)d+1: If f(r) denotes the function on the left and g(r) denotes the function on 0 0 the right it is an elementary matter of calculus that limr!1 g (r)=f (r) = 0. A second proof after claim 1. Our second proof starts once it has been n n ¤ shown that an isometry between `p and `q implies q = p . If T is an isometry n n 1=p between `p and `q then since kT eikp = 1 and kT (ei +ej)kp = 2 it should be 1=p possible to find three points a, b and c such that kc ¡ akq = 2 , kb ¡ akq = 150 f.