Pure Mathematical Sciences, Vol. 1, 2012, no. 4, 175 - 185 Aleksandrov Problem, Positively Homogeneous and Affine Maps Mina Ettefagh and Maryam Rouhbakhsh Department of Mathematics, Tabriz Branch Islamic Azad University, Tabriz, Iran [email protected], [email protected] [email protected] Abstract. In this paper, Aleksandrov problem of conservative distances is solved for positively homogeneous map. Also it is shown that when an affine map can be an isometry. Mathematics Subject Classification: 51K05, 39B82 Keywords: affine maps, Aleksandrov problem, distance preserving map, isometry, positively homogeneous map 1. Introduction Let (X, dX ) and (Y,dY ) be two metric spaces. The map f : X −→ Y is said to be a metric-isometry if for all x, y ∈ X dY (f(x),f(y)) = dX (x, y). Consider the condition (distance one preserving property) for f as following (DOPP) if x, y ∈ X with dX (x, y) = 1, then dY (f(x),f(y)) = 1, and the condition (distance n preserving property) for f as following (DnPP) if x, y ∈ X with dX (x, y)=n, then dY (f(x),f(y)) = n. Now let X, Y be two real normed spaces and consider the map f : X −→ Y . The mapping f is called an normed-isometry if for all x, y ∈ X, f(x) − f(y) = x − y, 176 M. Ettefagh and M. Rouhbakhsh and it is called a norm preserving map if for all x ∈ X, f(x) = x. The mapping f satisfies the (strong distance one preserving property) as fol- lowing (SDOPP) for all x, y ∈ X with x − y = 1, then f(x) − f(y) = 1 and conversely, and the (strong distance n preserving property) as following (SDnPP) for all x, y ∈ X with x − y = n, then f(x) − f(y) = n and conversely. A distance n>0 is said to be contractive [extensive]byf : X −→ Y if x − y = n implies f(x) − f(y)≤n [f(x) − f(y)≥n]. We say n is conservative (or preserved) by f if n is contractive and extensive by f, simultaneously. It is obvious that if f is metric [normed] isometry then it has DnPP [SDnPP] for each n>0. The theory of isometry had its begining in important paper by Mazur and Ulam in 1932. They proved the following resulet [5]. Theorem 1.1 (Mazur-Ulam). Let X and Y be real normed spaces. Every surjective isometry f : X −→ Y is a linear mapping up to translation (or is an affine map.) A. D. Aleksandrov [1] posed the following problem: Aleksandrov problem: Under what conditions is a mapping of metric space X into itself preserving a distance n>0 an isometry? This problem has been solved for f : En −→ En(2 ≤ n<∞), where En is a finit-dimentional real Euclidean space by F. S. Bekman and D. A. Quarles [2]. For non-Euclidean spaces the similar result has been obtained by A. Guc [3] and A. V. Kuz’minyh [4]. Also B. Mielnik and T. M.Rassias [6] have proved that every homeomorphism f : En −→ En(2 <n≤∞) with nontrivial conservative distance n>0 is an isometry. Also the following result was proved in [7] by T. M. Rassias and P. Sˇemrl. Theorem 1.2. Let X and Y be real normed spaces such that one of them has dimention greater than one. Suppose that f : X −→ Y is a surjective and Lipschitz map with k =1, and it satisfies (SDOPP). Then f is an isometry. Aleksandrov problem 177 We recall that if X is a normed space with norm ., then it becomes a metric space with metric induced by the norm, defined by d(x, y)=:x − y for x, y ∈ X. We mention some easy assertions in following lemma. Lemma 1.3. Let X and Y be two metric spaces whose metrics are induced by norms, and consider the map f : X −→ Y . Then (i) f is (normed) isometry if and only if f is (metric) isometry. (ii) if n ≥ 1 and f has SDnPP then f has DnPP. In section 2, we solve the Aleksandrov problem for positively homogeneous functions f : X −→ Y with metrics induced by norms on X and Y .Asa consequence we show that if also f is Lipschitz with k = 1 and distance one is extensive by f, then f is an isometry (corollary 2.6). In section 3, we consider the relation between DOPP and isometry of the map f : X −→ Y , in which X or Y is a discrete metric space. Finally, in section 4 we consider the invers of Mazur-Ulam theorem, and we show when an affine map can be an isometry. 2. Aleksandrov problem in positively homogeneous functions First we need the following definitions and lemmas that we will use in our main result. Definition 2.1. Let X and Y be two real normed spaces (or metric and real vector spaces). The map f : X −→ Y is called positively homogeneous if for each x ∈ X and each λ ≥ 0, f(λx)=λf(x), and it is called a Lipschitz map if there is a constant k>0 such that for all x, y ∈ X, d (f(x),f(y)) ≤ kd (x, y). Lemma 2.2. Let X and Y be real normed [metric] spaces and f : X −→ Y be a normed [metric] isometry then f is continuous injection. Proof. By the definition of an isometry, the continuity of f is obvious. Suppose that f is not injective, so there are x, y ∈ X such that x = y and f(x)=f(y). Thus x − y =0[ d(x, y) = 0] and f(x) − f(y) =0[d(f(x),f(y)) = 0] that is a contradiction. Lemma 2.3. Let f : X −→ Y be a function on real normed spaces X and Y . (i) If f is an isometry and f(0) = 0, then f is norm preserving. If f is linear and norm preserving map then it is an isometry. 178 M. Ettefagh and M. Rouhbakhsh (ii) If f is an isometry then it has SDOPP. If f has SDOPP and it is positively homogeneous then it is an isometry. (iii) If f has SDOPP and f(0) = 0 then f preserves the unit balls. (iv) Suppose that f is positively homogeneous. If it has SDnPP for some n>0(n =1) then f has SDOPP. If f has SDOPP then f has SDnPP for every n>0. (v) f has SDOPP then for all x, y ∈ X with x = y x y f − f . (x − y) (x − y) =1 (vi) Suppose that f is positively homogeneous. If there is a distance n>0(n =1) that is contractive [extensive] by f, then f contracts [ex- tends] distance one. If distance one is contractive [extensive] by f, then every distance n>0 is contractive [extensive] by f. Proof. (i)Iff is an isometry with f(0) = 0 for x ∈ X we have f(x) = f(x) − f(0) = x − 0 = x, that shows f is norm preserving. If f is a linear norm preserving then for x, y ∈ X we have f(x) − f(y) = f(x − y) = x − y, that shows f is an isometry. (ii) The first part is obvious. For the second part let x, y ∈ X,ifx = y the assertion is held, and if x = y we have x y − , x − y x − y =1 so by SDOPP of f x y f − f , (x − y) (x − y) =1 and since f is positively homogeneous we have 1 f x − f y , x − y ( ) ( ) =1 hence f(x) − f(y) = x − y. This proves that f is an isometry. (iii)Forx ∈ X with x = 1 we have x − 0 =1,sof(x) − f(0) = f(x) =1. Aleksandrov problem 179 (iv) For the first part suppose that x, y ∈ X with x − y = 1. Then for n>0(n = 1) in the hypothesis we have nx − ny = n. By SDnPP of f we have f(nx) − f(ny) = n, since f is positively homogeneous we have nf(x) − f(y) = n −→ f(x) − f(y) =1, similarly the equation f(x) − f(y) = 1 implies x − y =1. For the second part, suppose that n>0 and x, y ∈ X such that x − y = n, x y − f so n n = 1, and by SDOPP of we have x y f − f , (n) (n) =1 since f is positively homogeneous we have f(x) − f(y) = n, similarly the equation f(x) − f(y) = n implies x − y = n. (v) This is a consequence of the following equality that holds for each x, y ∈ X such that x = y x y − . x − y x − y =1 (vi) The proof is similar to proof of (iv). The following corollary is similar to part (iv) of lemma 2.3, in which we have metrics induced by norms. Corollary 2.4. Let X and Y be two metric spaces whose metrics are induced by norms, and let f : X −→ Y be a positively homogeneous function. If it has DnPP for some n>0(n =1) then it has DOPP. If f has DOPP then it has DnPP for every n>0. Proof. These are obvious from (iv) of lemma 2.3 and the fact that in metrics induced by norms, SDnPP is equivalant to DnPP, and (norm) isometry is equivalant to (metric) isometry. Now we can solve the Aleksandrov problem in part (iii) of following propo- sition. Proposition 2.5. Let X and Y be two metric spaces with metrics induced by norms, and let f : X −→ Y be a positively homogeneous function.