Aleksandrov Problem, Positively Homogeneous and Affine Maps

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 following (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 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 proposition.

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. (i) If distance one is contractive by f then for each x, y ∈ X we have f(x) − f(y)≤x − y, and so f is a Lipschitz map with k =1. 180 M. Ettefagh and M. Rouhbakhsh (ii) If distance one is extensive by f then for each x, y ∈ X we have f(x) − f(y)≥x − y. (iii) If f has DOPP, then it is an isometry. (iv) If f has DnPP for some n>0(n =1) , then it is an isometry.

Proof. Take x, y ∈ X.Ifx = y, the inequalities in (i) and (ii) are abvious. If x = y we know that x y − . x − y x − y =1 Since distance one is contractive by f in part (i) we have x y f − f ≤ , (x − y) (x − y) 1 and by positively homogeneous property for f f(x) − f(y)≤x − y, and this proves part (i). The proof of part (ii) is similar. Finally, we know that f has DOPP if and only if distance one is contractive and extensive by f. Hence part (iii) is a consequence of parts (i) and (ii). Part (iv)isa consequence of part (iii) and corollary 2.4.

Corollary 2.6. Let X and Y be two metric spaces with metrics induced by norms. Suppose that f : X −→ Y be a positively homogeneous and Lipcshitz function with k =1. If distance one is extensive by f, then f is an isometry.

Proof. By part (ii) of proposition 2.5, for every x, y ∈ X we have f(x) − f(y)≥x − y, and by Lipcshitz property of f with k = 1 we have f(x) − f(y)≤x − y. So for each x, y ∈ X f(x) − f(y) = x − y. This proves that f is an isometry.

Corollary 2.7. Let X and Y be two metric spaces with metrics induced by norms, and let f : X −→ Y be a positively homogeneous function with DOPP. Then (i) f is norm preserving, (ii) if also f is surjective then it is linear. Aleksandrov problem 181 Proof. Since f is positively homogeneous then f(0) = 0. Part (i) is a consequence of proposition 2.5 and the first part of (i) in lemma 2.3. Part (ii)isa consequence of proposition 2.5 and Mazur-Ulam theorem.

3. Aleksandrov problem in discrete metric According to proposition 2.5 it is important that we consider metrics that are not induced by norms in Aleksandrov problem. For example in this section we investigate this problem with discrete metric. First we show that a discrete d X metric on a set defined by 1(x = y) d(x, y)= 0(x = y) can not be induced by a norm. Because if there is a norm . on X such that d(x, y)=x − y, then for each λ ∈ R and x = y we have 1=d(λx, λy)=λx − λy = |λ|x − y = |λ|d(x, y)=|λ|, that is a contradiction.

Proposition 3.1. Let (X, dX ) and (Y,dY ) be two discrete metric spaces. Then following conditions are equivalent for f : X −→ Y . (i) f is injective, (ii) f has DOPP, (iii) f is an isometry.

Proof. Suppose that f is injective and x, y ∈ X.Ifx = y then f(x)=f(y) and so dX (x, y)=dY (f(x),f(y)) = 0. If x = y then f(x) = f(y) and so dX (x, y)=dY (f(x),f(y)) = 1. Hence f has DOPP and it is an isometry. Also if f is an isometry then it is injective by lemma 2.2 and obviously it has DOPP.

Proposition 3.2. Let (X, dX ) be a discrete metric space and (Y,dY ) be an arbitrary metric space. Suppose that the map f : X −→ Y has DOPP then it is an injective isometry. If in addition f is surjective, then dY will be a discrete metric.

Proof. Take x, y ∈ X.Ifx = y then f(x)=f(y) and

dX (x, y)=dY (f(x),f(y)) = 0.

If x = y then dX (x, y) = 1, so dY (f(x),f(y)) = 1 because f has DOPP. This proves that f is an isometry. Now f is injective by lemma 2.2, and if f is also surjective we conclude that dY should be a discrete metric. 182 M. Ettefagh and M. Rouhbakhsh The following example shows that the converse of above proposition is not true.

Example 3.3. Let (X, d1) be a discrete metric space and choose x, y ∈ X(x = y), we can make a metric d2 on X such that d2(x, y) =1.Now it is easy to check that f(x)=x(f :(X, d1) −→ (X, d2)) is an injective map but it has not DOPP, because d1(x, y) = 1 and d2(f(x),f(y)) = d2(x, y) =1, and so it is not an isometry.

Proposition 3.4. Let (X, dX ) be an arbitrary metric space and (Y,dY ) be a discrete metric space and consider the map f : X −→ Y . (i) If f is injective then it has DOPP. (ii) If f is an isometry then dX is a discrete metric.

Proof. (i) Take x, y ∈ X such that dX (x, y) = 1, so x = y. Since f is injective, we have f(x) = f(y). Hence dY (f(x),f(y)) = 1 and f has DOPP. (ii)Forx, y ∈ X such that x = y we have f(x) = f(y) and so

dY (f(x),f(y)) = 1 = dX (x, y), that shows dX is discrete metric.

In the part (i) of following example we show that the inverse of (i) of above proposition is not true.

Example 3.5. (i) There is a function f with hypotesis of above proposition such that it has DOPP but it is not injective. (ii) There is a function f with hypotesis of above proposition such that it is injective but it is not an isometry. 1 1 For (i) take X =[− , ] with Euclidean metric and Y = R with discrete 4 4 metric. The function f(x)=|x|(f : X −→ Y ) is not injective but it has DOPP. For (ii) take f(x)=x(f :(X, d1) −→ (X, d2)) in which d2 is discrete metric. Choose x, y ∈ X(x = y) we can make d1 such that d1(x, y) =1.Now it is easy to check that f is injective but it is not an isometry because

d1(x, y) =1= d2(x, y)=d2(f(x),f(y)).

4. Isometry and affine maps Definition 4.1. The function f : X −→ Y on two real vector spaces X and Y is called an affine map if for all x, y ∈ X and λ ∈ [0, 1], f(λx +(1− λ)y)=λf(x)+(1− λ)f(y). (1) Aleksandrov problem 183 When Y = R, we say that f is convex if for all x, y ∈ X and λ ∈ [0, 1], f(λx +(1− λ)y) ≤ λf(x)+(1− λ)f(y). (2) Equivalantly, f is affine if the map T : X −→ Y , defined by T (x)=f(x)−f(0), is linear. Also f is affine if and only if there is a linear map T : X −→ Y such that f(x)=T (x)+(x + a). So if f is affine with f(0) = 0, then f will be a linear map. Sometimes we can define an affine map by following equality for all x, y ∈ X and all positive scalers α, β with α + β =1, f(αx + βy)=αf(x)+βf(y). It is obvious that if the map f : X −→ R is positively homogeneous and subadditive [additive], then it is convex [affine] function. In the next lemmas we consider the convers problems. Lemma 4.2. Let X be a vector space and the f : X −→ R be a convex (or affine) map such that f(0) ≤ 0. Then for λ ∈ [0, 1], f(λx) ≤ λ.f(x). Proof. Let y = 0 in inequality (2), then we have f(λx)=f(λx +(1− λ)0) ≤ λf(x)+(1− λ)f(0) ≤ λf(x).

Lemma 4.3. Let X be a vector space. The positively homogeneous map f : X −→ R is convex if and only if it is subadditive, and it is affine if and only if it is additive. Proof. First, suppose that f is convex, then for x, y ∈ X, 1 1 1 1 f(x + y)=f( x + y) ≤ (f(x)+f(y)), 2 2 2 2 that shows subadditivity of f. Now assume that f is subadditive, then for λ ∈ [0, 1], f(λx +(1− λ)y) ≤ f(λx)+f((1 − λ)y)=λf(x)+(1− λ)f(y), this proves that f is convex. The rest of proof for affine maps is similar. In spacial case that X is an interval in R we obtain the following corollary. Corollary 4.4. Let X be an interval in R, and suppose that f : X −→ R be a convex (or affine) function such that f(0) ≤ 0. Then f is superadditive, that is for x, y ∈ X with x, y ≥ 0, f(x)+f(y) ≤ f(x + y). 184 M. Ettefagh and M. Rouhbakhsh Proof. We know by lemma 4.2 that for λ ∈ [0, 1] and x ∈ X, f(λx) ≤ λf(x). Now for x, y ∈ X with x, y > 0 x y f(x)+f(y)=f((x + y) )+f((x + y) ) x + y x + y x y ≤ f(x + y)+ f(x + y) x + y x + y = f(x + y).

An isometry need not be affine (or linear), see an example in [8]. By Mazur- Ulam theorem, evrey surjective isometry on real normed spaces is affine. In the next proposition we show that when an affine map can be an isometry.

Proposition 4.5. Let X and Y be two vector spaces with metrics dX and dY , respectively. Suppose that f : X −→ Y be an aﬀine map such that it preserves all distances m<1. Then f is an isometry.

Proof. Let x, y ∈ X, since the vector space X is convex then for all λ ∈ [0, 1], λx+(1−λ)y ∈ X. We can choose λi ∈ [0, 1](i =0, 1, ···,n) such that 0=λ0 <λ1 < ···<λn−1 <λn =1, and

d(zi−1,zi) < 1(i =1, ··· ,n), where zi = λix +(1− λi)y, (i =0, 1, ··· ,n). Thus n d(x, y)= d(zi−1,zi). i=1 Since f preserves all distances m<1, we have for all i =1, ··· ,n

d(f(zi−1),f(zi)) = d(zi−1,zi). On the other hand f is aﬀine so for i =0, 1, ···,n

f(zi)=λif(x)+(1− λi)f(y), that shows f(zi) is a point on the segment between f(x) and f(y)inY,orf preserves line segments. Because λ0 <λ1 < ···<λn−1 <λn, we can write n n d(f(x),f(y)) = d(f(zi−1),f(zi)) = d(zi−1,zi)=d(x, y). i=1 i=1 This proves that f is an isometry. Aleksandrov problem 185 References

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Received: May, 2012