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Characterizations on Mazur–Ulam theorem
Nonlinear Analysis 72 (2010) 1291–1297
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Characterizations on Mazur–Ulam theorem Jaeyoo Choy a,1 , Hahng-Yun Chu b,2 , Se-Hyun Ku c,∗ a
Department of Mathematics, Kyungpook National University, Sankyuk-Dong, Buk-Gu, Daegu 702-701, Republic of Korea
b
Department of Mathematical Sciences, Korea Advanced Institute of Science and Technology, 335, Gwahak-ro, Yuseong-gu, Daejeon 305-701, Republic of Korea
c
Department of Mathematics, Chungnam National University, Daehangno 79, Yuseong-Gu, Daejeon 305-764, Republic of Korea
article
info
Article history: Received 22 February 2009 Accepted 4 August 2009 MSC: primary 46B20 secondary 51M25 46S10 Keywords: 2-isometry Mazur–Ulam theorem Non-Archimedean 2-normed space
abstract We consider the area preserving 2-isometries in linear 2-normed spaces. In Chu (2007) [10], the Mazur–Ulam theorem was proved in linear 2-normed spaces with the assumption of the collinearity preserving. In this article, we prove the Mazur–Ulam theorem for the interior preserving mappings in linear 2-normed spaces. Recently Amyari, Sadeghi [16], in non-Archimedean 2-normed spaces, the Mazur–Ulam theorem was proved under the condition of strict convexity. In this paper, we also prove the theorem on non-Archimedean 2-normed spaces over a linear ordered non-Archimedean field without the strict convexity assumption. © 2009 Elsevier Ltd. All rights reserved.
1. Introduction Let X and Y be metric spaces with metrics dX and dY . A map f : X → Y is called a distance preserving mapping (isometry) if dY (f (x), f (y)) = dX (x, y) for any x, y ∈ X . Automatically, an isometry is injective. Two metric spaces X and Y are called isometric if there is an isometry from X to Y . For some fixed number r > 0, we call r a conservative (or preserved) distance for the mapping f if f preserves distance r; i.e., for all x, y in X with dX (x, y) = r, we have dY (f (x), f (y)) = r. In case of r = 1, especially, we call it a distance one preserving property. The fundamental problem of conservative distances, which is called Aleksandrov problem, is whether the existence of a single conservative distance for some mapping f from X into Y implies that f is an isometry. The Aleksandrov problem has been widely studied [1–6]. Mazur and Ulam [7], who stand at the beginning of the theory of isometric mappings, proved that every isometry f of a normed real vector space X onto another normed real vector space Y is a linear mapping up to translation (i.e., x 7→ f (x) − f (0) is linear, which amounts to the definition that f is affine). Also, Baker [8] proved that an isometry from one real normed linear space into a strictly convex real normed linear space is affine. Recently, Väisälä [9] proved that every bijective isometry f : X → Y between normed spaces is affine. Recently, Chu et al. [2] defined 2-isometry which is that a mapping preserves an area in a certain sense (see Section 2), and then proved the Aleksandrov problem in a linear 2-normed space. Chu [10], thereafter, also proved the Mazur–Ulam
∗
Corresponding author. E-mail addresses: [email protected] (J. Choy), [email protected] (H.-Y. Chu), [email protected] (S.-H. Ku).
1 The first author was supported by Korea Research Foundation Grant (KRF-2008-331-C00015). 2 The second author was supported by the second stage of the Brain Korea 21 Project, The Development Project of Human Resources in Mathematics, KAIST in 2008. 0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.08.017
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problem for a collinearity preserving mapping f in a linear 2-normed space, that is, if x, y, z are collinear then the f -images of x, y and z are also collinear. Much of this work in this area has been done; see Refs. [11–14]. On the other hand, Mazur–Ulam Theorem is no more valid in the context of non-Archimedean normed spaces, in general. Moslehian and Sadeghi [15] established the Mazur–Ulam theorem in the non-Archimedean strictly convex normed spaces. In addition, Amyari and Sadeghi [16] proved the Mazur–Ulam theorem under some conditions in non-Archimedean 2-normed spaces. In this paper, we consider an interior preserving mapping on linear 2-normed spaces, and then prove the image of the triangular barycenter is also the barycenter of the corresponding triangle. Due to this result, we are able to prove the Mazur–Ulam theorem for the interior preserving mappings on linear 2-normed spaces and on non-Archimedean 2-normed spaces. Note that our proof of the theorem on non-Archimedean 2-normed spaces does not require the condition for the strict convexity. 2. The Mazur–Ulam theorem on 2-normed spaces We begin this section with some definitions. Let X be a real linear space with dim X > 1 and k·, ·k : X 2 → [0, ∞) be a function. Then the pair (X , k·, ·k) is called a linear 2-normed space if
(2N1 ) kx, yk = 0 ⇔ x, y are linearly dependent (2N2 ) kx, yk = ky, xk (2N3 ) kα x, yk = |α| kx, yk (2N4 ) kx, y + z k ≤ kx, yk + kx, z k for all α ∈ R and x, y, z ∈ X . For x, y, z ∈ X , kx − z , y − z k is called an area of x, y, and z. We call f a 2-isometry if f is an area preserving mapping, that is,
kx − z , y − z k = kf (x) − f (z ), f (y) − f (z )k for all x, y, z ∈ X . For given points x, y and z in X , 4xyz denotes the triangle determined by x, y and z. A point 3 is called a barycenter of 4xyz. If p is a point of a set {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}, then p is called by an interior point of 4xyz. Define a mapping g between linear 2-normed spaces to be an interior preserving mapping of triangle if g carries an interior point in a triangle to an interior point in the corresponding triangle. In order to prove the Mazur–Ulam problem in linear 2-normed spaces, Chu [10] assumed the collinearity of the map. Here, distinct elements x, y, z of a real linear space X are collinear if y − z = t (x − z ) for some real number t. Moreover, he proved that the map carries the midpoint of two points to the midpoint of the corresponding two points. Using this property, he proved the Mazur–Ulam problem in linear 2-normed spaces. In this section, firstly we prove that the interior preserving mapping carries the barycenter of a triangle to the barycenter point of the corresponding triangle. Afterwards, using this result, we also prove a Mazur–Ulam problem on linear 2-normed spaces. x+y+z
Lemma 2.1 ([17]). For all γ ∈ R and all x, y ∈ X , kx, yk = kx, y + γ xk. Lemma 2.2. Let (X , k·, ·k) be a linear 2-normed space and x, y, z ∈ X . Then u :=
kx − y, x − uk = ky − z , y − uk = kz − x, z − uk =
1 3
x+y+z 3
is the unique member of X satisfying
kx − y , x − z k
with kx − y, x − z k 6= 0 and u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}. Proof. Let u :=
x+y+z . 3
It is clear that u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}. Applying the above lemma,
x+y+z
= 1 kx − y, 2x − y − z k = 1 kx − y, x − z k. k x − y , x − uk = x − y , x −
3
3
3
Similarly we can obtain that
k y − z , y − uk = k z − x , z − uk =
1 3 1 3
kx − y, x − z k, kx − y, x − z k.
For the uniqueness of u, assume that there exists another point v in X satisfying
kx − y, x − vk = ky − z , y − vk = kz − x, z − vk =
1 3
kx − y , x − z k
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with kx − y, x − z k = 6 0. Since v ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}, there exists s1 , s2 , s3 > 0 satisfying v = s1 x + s2 y + s3 z and s1 + s2 + s3 = 1. Then we obtain
kx − y, x − z k = = = = =
3kx − y, x − vk = 3kx − y, x − s1 x − s2 y − s3 z k 3kx − y, (s1 − 1)x + s2 y + (1 − s1 − s2 )z k 3kx − y, (s1 + s2 − 1)x + (1 − s1 − s2 )z k 3|s1 + s2 − 1|kx − y, x − z k 3|s3 |kx − y, x − z k.
Since kx − y, x − z k 6= 0, we have 3|s3 | = 1 and hence |s3 | = Consequently, we have s1 = s2 = s3 =
1 3
1 . 3
which completes the proof.
In the same way, we obtain that |s1 | = |s2 | =
1 . 3
Theorem 2.3. Suppose that X and Y are linear 2-normed spaces. If f : X → Y is an interior preserving 2-isometry, then the barycenter of triangle is f -invariant, i.e.,
f
x+y+z
=
3
f (x) + f (y) + f (z ) 3
for all x, y, z ∈ X with kx − y, x − z k 6= 0. Proof. Since f is a 2-isometry, we have
f (x) − f (y), f (x) − f x + y + z = x − y, x − x + y + z = 1 kx − y, 2x − y − z k
3 3 3 =
1 3
kx − y, x − z k =
1 3
kf (x) − f (y), f (x) − f (z )k.
Similarly we obtain that
f (y) − f (z ), f (y) − f x + y + z =
3
f (z ) − f (x), f (z ) − f x + y + z =
3
1 3 1 3
kf (x) − f (y), f (x) − f (z )k, kf (x) − f (y), f (x) − f (z )k.
Since f is an interior preserving mapping,
f
x+y+z
3
∈ {t1 f (x) + t2 f (y) + t3 f (z ) | t1 + t2 + t3 = 1, ti > 0 for all i}.
By Lemma 2.2, we have
f
x+y+z 3
=
f (x) + f (y) + f (z )
This completes the proof.
3
.
The next theorem is the Mazur–Ulam Theorem on linear 2-normed spaces. In the above lemma, we proved that the barycenter of the triangle maps to the barycenter of the corresponding triangle in linear 2-normed spaces. By this result, we prove the following theorem. Theorem 2.4. Let X and Y be linear 2-normed spaces. If f : X → Y is an interior preserving 2-isometry, then f is affine. Proof. Let g (x) := f (x) − f (0). It is clear that g (0) = 0 and g is a 2-isometry. For a, b, c ∈ X , let 4abc be a triangle determined by the points a, b, c, and x an interior point of 4abc. Since f is an interior preserving map, there exist positive numbers s1 , s2 , s3 with s1 + s2 + s3 = 1 such that f (x) = s1 f (a) + s2 f (b) + s3 f (c ). Then g (x) = f (x) − f (0)
= s1 f (a) + s2 f (b) + s3 f (c ) − f (0) = s1 (f (a) − f (0)) + s2 (f (b) − f (0)) + s3 (f (c ) − f (0)) = s1 g (a) + s2 g (b) + s3 g (c ), and hence g (x) is an interior point of 4g (a)g (b)g (c ). Therefore g is also an interior preserving mapping. g (x)+g (y)+g (z ) x+y+z Now let x, y, z ∈ X with kx − y, x − z k 6= 0. From Theorem 2.3, g = . 3 3
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Firstly, we are going to prove that g (rx) = rg (x) for any r ∈ R+ − {0}. For x ∈ X , since the dimension of X is greater than 1, there exists an element y of X satisfying kx, yk 6= 0. Since g is a 2-isometry, 0 = krx, xk = kg (rx), g (x)k. Thus g (x) and g (rx) are linearly dependent. So there exists a real number k depending on r such that g (rx) = kg (x). Then we obtain that r kx, yk = krx, yk = kg (rx), g (y)k = kkg (x), g (y)k = |k|kg (x), g (y)k = |k|kx, yk. Since kx, yk 6= 0, then we have k = r or k = −r. Suppose k = −r, then
|r − 1|kx, yk = = = = =
k(r − 1)x, yk = krx − x, y − 0k kg (rx) − g (x), g (y) − g (0)k k − rg (x) − g (x), g (y)k |r + 1|kg (x), g (y)k |r + 1|kx, yk.
By the fact kx, yk 6= 0, |r − 1| = |r + 1| and so r = 0 which is a contradiction. Hence k = r, i.e., g (rx) = rg (x) for every r ∈ R+ − {0}. Similarly, we can prove that g (rx) = rg (x) for every r ∈ R− − {0}. Let x, y ∈ X with kx, yk 6= 0. By the Theorem 2.3 and the above statements, g (x + y) = g
=
3x + 3y + 0 3
3g (x) + 3g (y) 3
=
g (3x) + g (3y) + g (0) 3
= g (x) + g (y).
Now we will show the additivity in the remaining case kx, yk = 0. Then x, y are linearly dependent. Thus there is a real number t such that y = tx. By the above statements, we have that g (x + y) = g (x + tx) = g ((1 + t )x) = (1 + t )g (x) = g (x) + tg (x) = g (x) + g (y). Therefore g (x) := f (x) − f (0) is R-linear, i.e., f is affine, which completes the proof.
3. The (additive) Mazur–Ulam theorem on non-Archimedean 2-normed spaces In general, the Mazur–Ulam Theorem is not valid on non-Archimedean normed spaces. Recently, Amyari and Sadeghi [16] offered the (additive) Mazur–Ulam theorem in non-Archimedean linear 2-normed spaces. They proved it under the condition of the strict convexity of codomain of a 2-isometric map but, here, we will prove the theorem without the condition for the strict convexity. A non-Archimedean (or ultrametric) valuation is a map | · | from a field K into [0, ∞) with the following properties: (i) |r | = 0 if and only if r = 0 (ii) |rs| = |r ||s| (iii) |r + s| ≤ max{|r |, |s|} for all r , s ∈ K . If every element in K carries a valuation then we call a field K a valued field, for a convenience, simply call it a field. Clearly |1| = |−1| = 1 and |n| ≤ 1 for all n ∈ N. An example of a non-Archimedean valuation is the map | · | taking everything but 0 into 1 and |0| = 0 (see [18]). Let X be a vector space over a valued field K . A function k · k : X → [0, ∞) is said to be a non-Archimedean norm if it satisfies the following conditions: (i) kxk = 0 if and only if x = 0 (ii) krxk = |r |kxk (r ∈ K , x ∈ X) (iii) the strong triangle inequality
kx + yk ≤ max{kxk, kyk} (x, y ∈ X). Then (X, k · k) is called a non-Archimedean space. Now, we expand above notions into a 2-normed spaces. Let X be a vector space of dimension greater than 1 over a field K with a non-Archimedean valuation | · |. A function k·, ·k : X × X → [0, ∞) is said to be a non-Archimedean 2-norm if it satisfies the following conditions: (i) (ii) (iii) (iv)
kx, yk = 0 if and only if x, y are linearly dependent kx, yk = ky, xk krx, yk = |r |kx, yk (r ∈ K , x, y ∈ X ) the strong triangle inequality
kx, y + z k ≤ max{kx, yk, kx, z k} (x, y, z ∈ X ). Then (X, k·, ·k) is called a non-Archimedean 2-normed space [16].
J. Choy et al. / Nonlinear Analysis 72 (2010) 1291–1297
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A non-Archimedean 2-normed space (X, k·, ·k) over a valued filed K is called strictly convex if for non-zero vectors x, y, z ∈ X with z 6∈ {α x + β y | α, β ∈ K }, kx + y, z k = max{kx, z k, ky, z k} and kx, z k = ky, z k then x = y. From now on, let x, y and z be given points in a non-Archimedean space X over a linearly ordered non-Archimedean field K . Then we can define a triangle, interior point, barycenter and interior preserving map as in the Section 2. Lemma 3.1 ([16]). Let (X, k·, ·k) is a non-Archimedean 2-normed space then kx, yk = kx, y + γ xk for all x, y ∈ X and all γ ∈ K. We denote the set of all elements of K whose norms are 1 by C , that is, C = {γ ∈ K | |γ | = 1}. Lemma 3.2. Let X be a non-Archimedean 2-normed space over a linear ordered non-Archimedean filed K with C = {3n |n ∈ Z} x+y+z and let x, y, z ∈ X. Then u := 3 is the unique member of X satisfying
kx − y, x − uk = ky − z , y − uk = kz − x, z − uk = kx − y, x − z k with kx − y, x − z k 6= 0 and u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti ∈ K , ti > 0, for all i}. Proof. Let u :=
x+y+z . 3
Then u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti ∈ K , ti > 0, for all i}. By Lemma 3.1 we have
2x − y − z x+y+z
kx − y, x − uk = x − y, x −
= x − y,
3 3 =
1
|3|
kx − y, 2x − y − z k = kx − y, 2x − y − z k = kx − y, x − z k.
Similarly, we have
ky − z , y − uk = ky − z , y − xk = kx − y, x − z k, kz − x, z − uk = kz − x, z − yk = kx − y, x − z k. For the uniqueness of u, assume that another v ∈ X satisfying
kx − y, x − vk = ky − z , y − vk = kz − x, z − vk = kx − y, x − z k. Since v is in an interior point of triangle given by the points x, y and z, there exist elements s1 , s2 and s3 of K such that s1 + s2 + s3 = 1, si > 0 and v = s1 x + s2 y + s3 z. In view of Lemma 3.1, we have
kx − y, x − z k = = = = = =
kx − y, x − vk kx − y, x − (s1 x + s2 y + s3 z )k kx − y, (s1 − 1)x + s2 y + (1 − s1 − s2 )z k kx − y, (s1 + s2 − 1)x + (1 − s1 − s2 )z k |s1 + s2 − 1|kx − y, x − z k |s3 |kx − y, x − z k.
Since kx − y, x − z k 6= 0, we have |s3 | = 1. Similarly we also have |s1 | = |s2 | = 1. Thus s1 + s2 + s3 = 1 and |s1 | = |s2 | = |s3 | = 1. From the definition, there exist integers k1 , k2 , k3 such that s1 = 3k1 , s2 = 3k2 , s3 = 3k3 . By the fact s1 + s2 + s3 = 1, ki < 0 for all i = 1, 2, 3. Without loss of generality, we can let s1 = 3−n1 , s2 = 3−n2 , s3 = 3−n3 and n1 ≥ n2 ≥ n3 . Assume that above one of inequalities holds. Then 1 = s1 + s2 + s3 = 3−n1 (1 + 3n1 −n2 + 3n1 −n3 ). So 3n1 = 1 + 3n1 −n2 + 3n1 −n3 . Then the number in left side is a multiple of 3, but the number in right side is not a multiple of 3. This is a contradiction. Hence s1 = s2 = s3 = 13 which means u = v . This completes the proof. The following theorem is the (additive) Mazur–Ulam Theorem on non-Archimedean 2-normed spaces. We prove the following theorem using the property that a barycenter point of the triangle maps to a barycenter point of the corresponding triangle. Theorem 3.3. Suppose that X and Y are non-Archimedean 2-normed spaces over a linear ordered non-Archimedean field K with C = {3n |n ∈ Z}. If f : X → Y is an interior preserving 2-isometry, then f (x) − f (0) is additive. Proof. Let g (x) := f (x) − f (0). Then g is a 2-isometry and g (0) = 0. By the proof of Theorem 2.4, g is also an interior preserving mapping. Let x, y, z ∈ X with kx − y, x − z k 6= 0. Since g is a 2-isometry, we have
g (x) − g (y), g (x) − g x + y + z = x − y, x − x + y + z
3 3
2x − y − z
=
x − y,
3
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= kx − y, 2x − y − z k = kx − y, x − z k = kg (x) − g (y), g (x) − g (z )k (x, y, z ∈ X) and similarly we can obtain
g (y) − g (z ), g (y) − g x + y + z = kg (x) − g (y), g (x) − g (z )k,
3
g (z ) − g (x), g (z ) − g x + y + z = kg (x) − g (y), g (x) − g (z )k.
3
x+y+z
Since 3 is an interior point of the triangle 4xyz and g is an interior preserving mapping, g t3 g (z ) | t1 + t2 + t3 = 1, ti ∈ K , ti > 0, for all i}. By Lemma 3.2
g
x+y+z 3
=
g (x) + g (y) + g (z ) 3
x+y+z 3
∈ {t1 g (x) + t2 g (y) +
,
for all x, y, z ∈ X with kx − y, x − z k 6= 0. This completes the proof.
We now consider a notion of collinearity. Let x, y, z ∈ X be distinct elements. Then x, y and z are called collinear if x − y = t (x − z ) for some t ∈ K . Lemma 3.4. Let X be a non-Archimedean 2-normed space over a linear ordered non-Archimedean field K with C = {2n |n ∈ Z} x +y and let x, y ∈ X. Then u := 2 is the unique member of X satisfying
kx − z , x − uk = ky − z , y − uk = kx − z , y − z k for some z ∈ X with kx − z , y − z k 6= 0 and u, x, y are collinear. x +y
Proof. We can easily show that u := 2 satisfies the above equations. For the uniqueness, assume that there is another v satisfying
kx − z , x − vk = ky − z , y − vk = kx − z , y − z k. Since x, y and v are collinear, we can find t ∈ K such that v = tx + (1 − t )y. Then
kx − z , y − z k = = = =
kx − z , x − vk kx − z , x − tx − (1 − t )yk |1 − t |kx − z , x − yk |1 − t |kx − z , y − z k
kx − z , y − z k = = = =
ky − z , y − vk ky − z , y − tx − (1 − t )yk |t |ky − z , y − xk |t |kx − z , y − z k.
Since kx − z , y − z k 6= 0, we have |1 − t | = 1 and |t | = 1. Then there exist elements k1 , k2 of K such that 1 − t = 2k1 , t = 2k2 . Since 2k1 + 2k2 = 1, ki < 0 for all i = 1, 2. With out loss of generality, we let 1 − t = 2−n1 , t = 2−n2 and n1 ≥ n2 . If n1 n2 then 1 = 2−n1 + 2−n2 = 2−n1 (1 + 2n1 −n2 ). So 2n1 = 1 + 2n1 −n2 . This is a contradiction because the left side is a multiple of 2 but the right side is not. Thus n1 = n2 , and hence t = 21 . This means that u = v , which completes the proof. In the proof of the next theorem, we use the midpoint of a two points maps to the midpoint of the corresponding two points. And one may consider that the following is another type of the (additive) Mazur–Ulam Theorem on non-Archimedean 2-normed spaces. Theorem 3.5. Suppose that X and Y are non-Archimedean 2-normed spaces over a linear ordered non-Archimedean field K with C = {2n |n ∈ Z}. Assume that f (x), f (y) and f (z ) are collinear when x, y and z are collinear. If f : X → Y is a 2-isometry, then f (x) − f (0) is additive. Proof. Let g (x) := f (x) − f (0) then g is a 2-isometry and g (0) = 0. It is clear that if x, y and z are collinear, then g (x), g (y) and g (z ) are also collinear. Since g is a 2-isometry, we obtain that
g (x) − g (z ), g (x) − g x + y = x − z , x − x + y
2
2
x−y
=
x − z , 2 = kx − z , x − yk = kx − z , y − z k = kg (x) − g (z ), g (y) − g (z )k,
J. Choy et al. / Nonlinear Analysis 72 (2010) 1291–1297
g (y) − g (z ), g (y) − g x + y = y − z , y − x + y
2 2 = ky − z , y − xk = kx − z , y − z k = kg (x) − g (z ), g (y) − g (z )k for all x, y ∈ X. x +y x +y Since 2 , x, y are collinear, g 2 , g (x), g (y) are also collinear. From Lemma 3.4, we get that g x, y ∈ X. This completes the proof.
1297
x +y 2
=
g (x)+g (y) 2
for all
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A.D. Aleksandrov, Mappings of families of sets, Soviet Math. Dokl. 11 (1970) 116–120. H.-Y. Chu, C.G. Park, W.-G. Park, The Aleksandrov problem in linear 2-normed spaces, J. Math. Anal. Appl. 289 (2004) 666–672. B. Mielnik, Th.M. Rassias, On the Aleksandrov problem of conservative distances, Proc. Amer. Math. Soc. 116 (1992) 1115–1118. Th.M. Rassias, Properties of isometric mappings, J. Math. Anal. Appl. 235 (1999) 108–121. Th.M. Rassias, On the A.D. Aleksandrov problem of conservative distances and the Mazur–Ulam theorem, Nonlinear Anal. 47 (2001) 2597–2608. Th.M. Rassias, P. Šemrl, On the Mazur-Ulam theorem and the Aleksandrov problem for unit distance preserving mappings, Proc. Amer. Math. Soc. 118 (1993) 919–925. S. Mazur, S. Ulam, Sur les transformationes isométriques d’espaces vectoriels normés, C. R. Acad. Sci. Paris 194 (1932) 946–948. J.A. Baker, Isometries in normed spaces, Amer. Math. Monthly 78 (1971) 655–658. J. Väisälä, A proof of the Mazur–Ulam theorem, Amer. Math. Monthly 110 (2003) 633–635. H.-Y. Chu, On the Mazur–Ulam problem in linear 2-normed spaces, J. Math. Anal. Appl. 327 (2007) 1041–1045. H.-Y. Chu, S.-H. Ku, D.S. Kang, Characterizations on 2-isometries, J. Math. Anal. Appl. 340 (2008) 621–628. C.R. Diminnie, S. Gähler, A. White, Remarks on strictly convex and strictly 2-convex 2-normed spaces, Math. Nachr. 88 (1979) 363–372. S. Gähler, 2-metrische Räume und ihre topologische Struktur, Math. Nachr. 26 (1963) 115–148. C.S. Lin, On strictly convex and strictly 2-convex 2-normed spaces, Math. Nachr. 149 (1990) 149–154. M.S. Moslehian, Gh. Sadeghi, A Mazur–Ulam theorem in non-Archimedean normed spaces, Nonlinear Anal. 69 (2008) 3405–3408. M. Amyari, G. Sadeghi, Isometries in non-Archimedean 2-normed spaces, preprint. Y.J. Cho, P.C.S. Lin, S.S. Kim, A. Misiak, Theory of 2-Inner Product Spaces, Nova Science Publ., New York, 2001. L. Narici, E. Beckenstein, Strange terrain—non-Archimedean spaces, Amer. Math. Monthly 88 (1981) 667–676.
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Characterizations on Mazur–Ulam theorem Jaeyoo Choy a,1 , Hahng-Yun Chu b,2 , Se-Hyun Ku c,∗ a
Department of Mathematics, Kyungpook National University, Sankyuk-Dong, Buk-Gu, Daegu 702-701, Republic of Korea
b
Department of Mathematical Sciences, Korea Advanced Institute of Science and Technology, 335, Gwahak-ro, Yuseong-gu, Daejeon 305-701, Republic of Korea
c
Department of Mathematics, Chungnam National University, Daehangno 79, Yuseong-Gu, Daejeon 305-764, Republic of Korea
article
info
Article history: Received 22 February 2009 Accepted 4 August 2009 MSC: primary 46B20 secondary 51M25 46S10 Keywords: 2-isometry Mazur–Ulam theorem Non-Archimedean 2-normed space
abstract We consider the area preserving 2-isometries in linear 2-normed spaces. In Chu (2007) [10], the Mazur–Ulam theorem was proved in linear 2-normed spaces with the assumption of the collinearity preserving. In this article, we prove the Mazur–Ulam theorem for the interior preserving mappings in linear 2-normed spaces. Recently Amyari, Sadeghi [16], in non-Archimedean 2-normed spaces, the Mazur–Ulam theorem was proved under the condition of strict convexity. In this paper, we also prove the theorem on non-Archimedean 2-normed spaces over a linear ordered non-Archimedean field without the strict convexity assumption. © 2009 Elsevier Ltd. All rights reserved.
1. Introduction Let X and Y be metric spaces with metrics dX and dY . A map f : X → Y is called a distance preserving mapping (isometry) if dY (f (x), f (y)) = dX (x, y) for any x, y ∈ X . Automatically, an isometry is injective. Two metric spaces X and Y are called isometric if there is an isometry from X to Y . For some fixed number r > 0, we call r a conservative (or preserved) distance for the mapping f if f preserves distance r; i.e., for all x, y in X with dX (x, y) = r, we have dY (f (x), f (y)) = r. In case of r = 1, especially, we call it a distance one preserving property. The fundamental problem of conservative distances, which is called Aleksandrov problem, is whether the existence of a single conservative distance for some mapping f from X into Y implies that f is an isometry. The Aleksandrov problem has been widely studied [1–6]. Mazur and Ulam [7], who stand at the beginning of the theory of isometric mappings, proved that every isometry f of a normed real vector space X onto another normed real vector space Y is a linear mapping up to translation (i.e., x 7→ f (x) − f (0) is linear, which amounts to the definition that f is affine). Also, Baker [8] proved that an isometry from one real normed linear space into a strictly convex real normed linear space is affine. Recently, Väisälä [9] proved that every bijective isometry f : X → Y between normed spaces is affine. Recently, Chu et al. [2] defined 2-isometry which is that a mapping preserves an area in a certain sense (see Section 2), and then proved the Aleksandrov problem in a linear 2-normed space. Chu [10], thereafter, also proved the Mazur–Ulam
∗
Corresponding author. E-mail addresses: [email protected] (J. Choy), [email protected] (H.-Y. Chu), [email protected] (S.-H. Ku).
1 The first author was supported by Korea Research Foundation Grant (KRF-2008-331-C00015). 2 The second author was supported by the second stage of the Brain Korea 21 Project, The Development Project of Human Resources in Mathematics, KAIST in 2008. 0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.08.017
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problem for a collinearity preserving mapping f in a linear 2-normed space, that is, if x, y, z are collinear then the f -images of x, y and z are also collinear. Much of this work in this area has been done; see Refs. [11–14]. On the other hand, Mazur–Ulam Theorem is no more valid in the context of non-Archimedean normed spaces, in general. Moslehian and Sadeghi [15] established the Mazur–Ulam theorem in the non-Archimedean strictly convex normed spaces. In addition, Amyari and Sadeghi [16] proved the Mazur–Ulam theorem under some conditions in non-Archimedean 2-normed spaces. In this paper, we consider an interior preserving mapping on linear 2-normed spaces, and then prove the image of the triangular barycenter is also the barycenter of the corresponding triangle. Due to this result, we are able to prove the Mazur–Ulam theorem for the interior preserving mappings on linear 2-normed spaces and on non-Archimedean 2-normed spaces. Note that our proof of the theorem on non-Archimedean 2-normed spaces does not require the condition for the strict convexity. 2. The Mazur–Ulam theorem on 2-normed spaces We begin this section with some definitions. Let X be a real linear space with dim X > 1 and k·, ·k : X 2 → [0, ∞) be a function. Then the pair (X , k·, ·k) is called a linear 2-normed space if
(2N1 ) kx, yk = 0 ⇔ x, y are linearly dependent (2N2 ) kx, yk = ky, xk (2N3 ) kα x, yk = |α| kx, yk (2N4 ) kx, y + z k ≤ kx, yk + kx, z k for all α ∈ R and x, y, z ∈ X . For x, y, z ∈ X , kx − z , y − z k is called an area of x, y, and z. We call f a 2-isometry if f is an area preserving mapping, that is,
kx − z , y − z k = kf (x) − f (z ), f (y) − f (z )k for all x, y, z ∈ X . For given points x, y and z in X , 4xyz denotes the triangle determined by x, y and z. A point 3 is called a barycenter of 4xyz. If p is a point of a set {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}, then p is called by an interior point of 4xyz. Define a mapping g between linear 2-normed spaces to be an interior preserving mapping of triangle if g carries an interior point in a triangle to an interior point in the corresponding triangle. In order to prove the Mazur–Ulam problem in linear 2-normed spaces, Chu [10] assumed the collinearity of the map. Here, distinct elements x, y, z of a real linear space X are collinear if y − z = t (x − z ) for some real number t. Moreover, he proved that the map carries the midpoint of two points to the midpoint of the corresponding two points. Using this property, he proved the Mazur–Ulam problem in linear 2-normed spaces. In this section, firstly we prove that the interior preserving mapping carries the barycenter of a triangle to the barycenter point of the corresponding triangle. Afterwards, using this result, we also prove a Mazur–Ulam problem on linear 2-normed spaces. x+y+z
Lemma 2.1 ([17]). For all γ ∈ R and all x, y ∈ X , kx, yk = kx, y + γ xk. Lemma 2.2. Let (X , k·, ·k) be a linear 2-normed space and x, y, z ∈ X . Then u :=
kx − y, x − uk = ky − z , y − uk = kz − x, z − uk =
1 3
x+y+z 3
is the unique member of X satisfying
kx − y , x − z k
with kx − y, x − z k 6= 0 and u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}. Proof. Let u :=
x+y+z . 3
It is clear that u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}. Applying the above lemma,
x+y+z
= 1 kx − y, 2x − y − z k = 1 kx − y, x − z k. k x − y , x − uk = x − y , x −
3
3
3
Similarly we can obtain that
k y − z , y − uk = k z − x , z − uk =
1 3 1 3
kx − y, x − z k, kx − y, x − z k.
For the uniqueness of u, assume that there exists another point v in X satisfying
kx − y, x − vk = ky − z , y − vk = kz − x, z − vk =
1 3
kx − y , x − z k
J. Choy et al. / Nonlinear Analysis 72 (2010) 1291–1297
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with kx − y, x − z k = 6 0. Since v ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti > 0 for all i}, there exists s1 , s2 , s3 > 0 satisfying v = s1 x + s2 y + s3 z and s1 + s2 + s3 = 1. Then we obtain
kx − y, x − z k = = = = =
3kx − y, x − vk = 3kx − y, x − s1 x − s2 y − s3 z k 3kx − y, (s1 − 1)x + s2 y + (1 − s1 − s2 )z k 3kx − y, (s1 + s2 − 1)x + (1 − s1 − s2 )z k 3|s1 + s2 − 1|kx − y, x − z k 3|s3 |kx − y, x − z k.
Since kx − y, x − z k 6= 0, we have 3|s3 | = 1 and hence |s3 | = Consequently, we have s1 = s2 = s3 =
1 3
1 . 3
which completes the proof.
In the same way, we obtain that |s1 | = |s2 | =
1 . 3
Theorem 2.3. Suppose that X and Y are linear 2-normed spaces. If f : X → Y is an interior preserving 2-isometry, then the barycenter of triangle is f -invariant, i.e.,
f
x+y+z
=
3
f (x) + f (y) + f (z ) 3
for all x, y, z ∈ X with kx − y, x − z k 6= 0. Proof. Since f is a 2-isometry, we have
f (x) − f (y), f (x) − f x + y + z = x − y, x − x + y + z = 1 kx − y, 2x − y − z k
3 3 3 =
1 3
kx − y, x − z k =
1 3
kf (x) − f (y), f (x) − f (z )k.
Similarly we obtain that
f (y) − f (z ), f (y) − f x + y + z =
3
f (z ) − f (x), f (z ) − f x + y + z =
3
1 3 1 3
kf (x) − f (y), f (x) − f (z )k, kf (x) − f (y), f (x) − f (z )k.
Since f is an interior preserving mapping,
f
x+y+z
3
∈ {t1 f (x) + t2 f (y) + t3 f (z ) | t1 + t2 + t3 = 1, ti > 0 for all i}.
By Lemma 2.2, we have
f
x+y+z 3
=
f (x) + f (y) + f (z )
This completes the proof.
3
.
The next theorem is the Mazur–Ulam Theorem on linear 2-normed spaces. In the above lemma, we proved that the barycenter of the triangle maps to the barycenter of the corresponding triangle in linear 2-normed spaces. By this result, we prove the following theorem. Theorem 2.4. Let X and Y be linear 2-normed spaces. If f : X → Y is an interior preserving 2-isometry, then f is affine. Proof. Let g (x) := f (x) − f (0). It is clear that g (0) = 0 and g is a 2-isometry. For a, b, c ∈ X , let 4abc be a triangle determined by the points a, b, c, and x an interior point of 4abc. Since f is an interior preserving map, there exist positive numbers s1 , s2 , s3 with s1 + s2 + s3 = 1 such that f (x) = s1 f (a) + s2 f (b) + s3 f (c ). Then g (x) = f (x) − f (0)
= s1 f (a) + s2 f (b) + s3 f (c ) − f (0) = s1 (f (a) − f (0)) + s2 (f (b) − f (0)) + s3 (f (c ) − f (0)) = s1 g (a) + s2 g (b) + s3 g (c ), and hence g (x) is an interior point of 4g (a)g (b)g (c ). Therefore g is also an interior preserving mapping. g (x)+g (y)+g (z ) x+y+z Now let x, y, z ∈ X with kx − y, x − z k 6= 0. From Theorem 2.3, g = . 3 3
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Firstly, we are going to prove that g (rx) = rg (x) for any r ∈ R+ − {0}. For x ∈ X , since the dimension of X is greater than 1, there exists an element y of X satisfying kx, yk 6= 0. Since g is a 2-isometry, 0 = krx, xk = kg (rx), g (x)k. Thus g (x) and g (rx) are linearly dependent. So there exists a real number k depending on r such that g (rx) = kg (x). Then we obtain that r kx, yk = krx, yk = kg (rx), g (y)k = kkg (x), g (y)k = |k|kg (x), g (y)k = |k|kx, yk. Since kx, yk 6= 0, then we have k = r or k = −r. Suppose k = −r, then
|r − 1|kx, yk = = = = =
k(r − 1)x, yk = krx − x, y − 0k kg (rx) − g (x), g (y) − g (0)k k − rg (x) − g (x), g (y)k |r + 1|kg (x), g (y)k |r + 1|kx, yk.
By the fact kx, yk 6= 0, |r − 1| = |r + 1| and so r = 0 which is a contradiction. Hence k = r, i.e., g (rx) = rg (x) for every r ∈ R+ − {0}. Similarly, we can prove that g (rx) = rg (x) for every r ∈ R− − {0}. Let x, y ∈ X with kx, yk 6= 0. By the Theorem 2.3 and the above statements, g (x + y) = g
=
3x + 3y + 0 3
3g (x) + 3g (y) 3
=
g (3x) + g (3y) + g (0) 3
= g (x) + g (y).
Now we will show the additivity in the remaining case kx, yk = 0. Then x, y are linearly dependent. Thus there is a real number t such that y = tx. By the above statements, we have that g (x + y) = g (x + tx) = g ((1 + t )x) = (1 + t )g (x) = g (x) + tg (x) = g (x) + g (y). Therefore g (x) := f (x) − f (0) is R-linear, i.e., f is affine, which completes the proof.
3. The (additive) Mazur–Ulam theorem on non-Archimedean 2-normed spaces In general, the Mazur–Ulam Theorem is not valid on non-Archimedean normed spaces. Recently, Amyari and Sadeghi [16] offered the (additive) Mazur–Ulam theorem in non-Archimedean linear 2-normed spaces. They proved it under the condition of the strict convexity of codomain of a 2-isometric map but, here, we will prove the theorem without the condition for the strict convexity. A non-Archimedean (or ultrametric) valuation is a map | · | from a field K into [0, ∞) with the following properties: (i) |r | = 0 if and only if r = 0 (ii) |rs| = |r ||s| (iii) |r + s| ≤ max{|r |, |s|} for all r , s ∈ K . If every element in K carries a valuation then we call a field K a valued field, for a convenience, simply call it a field. Clearly |1| = |−1| = 1 and |n| ≤ 1 for all n ∈ N. An example of a non-Archimedean valuation is the map | · | taking everything but 0 into 1 and |0| = 0 (see [18]). Let X be a vector space over a valued field K . A function k · k : X → [0, ∞) is said to be a non-Archimedean norm if it satisfies the following conditions: (i) kxk = 0 if and only if x = 0 (ii) krxk = |r |kxk (r ∈ K , x ∈ X) (iii) the strong triangle inequality
kx + yk ≤ max{kxk, kyk} (x, y ∈ X). Then (X, k · k) is called a non-Archimedean space. Now, we expand above notions into a 2-normed spaces. Let X be a vector space of dimension greater than 1 over a field K with a non-Archimedean valuation | · |. A function k·, ·k : X × X → [0, ∞) is said to be a non-Archimedean 2-norm if it satisfies the following conditions: (i) (ii) (iii) (iv)
kx, yk = 0 if and only if x, y are linearly dependent kx, yk = ky, xk krx, yk = |r |kx, yk (r ∈ K , x, y ∈ X ) the strong triangle inequality
kx, y + z k ≤ max{kx, yk, kx, z k} (x, y, z ∈ X ). Then (X, k·, ·k) is called a non-Archimedean 2-normed space [16].
J. Choy et al. / Nonlinear Analysis 72 (2010) 1291–1297
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A non-Archimedean 2-normed space (X, k·, ·k) over a valued filed K is called strictly convex if for non-zero vectors x, y, z ∈ X with z 6∈ {α x + β y | α, β ∈ K }, kx + y, z k = max{kx, z k, ky, z k} and kx, z k = ky, z k then x = y. From now on, let x, y and z be given points in a non-Archimedean space X over a linearly ordered non-Archimedean field K . Then we can define a triangle, interior point, barycenter and interior preserving map as in the Section 2. Lemma 3.1 ([16]). Let (X, k·, ·k) is a non-Archimedean 2-normed space then kx, yk = kx, y + γ xk for all x, y ∈ X and all γ ∈ K. We denote the set of all elements of K whose norms are 1 by C , that is, C = {γ ∈ K | |γ | = 1}. Lemma 3.2. Let X be a non-Archimedean 2-normed space over a linear ordered non-Archimedean filed K with C = {3n |n ∈ Z} x+y+z and let x, y, z ∈ X. Then u := 3 is the unique member of X satisfying
kx − y, x − uk = ky − z , y − uk = kz − x, z − uk = kx − y, x − z k with kx − y, x − z k 6= 0 and u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti ∈ K , ti > 0, for all i}. Proof. Let u :=
x+y+z . 3
Then u ∈ {t1 x + t2 y + t3 z | t1 + t2 + t3 = 1, ti ∈ K , ti > 0, for all i}. By Lemma 3.1 we have
2x − y − z x+y+z
kx − y, x − uk = x − y, x −
= x − y,
3 3 =
1
|3|
kx − y, 2x − y − z k = kx − y, 2x − y − z k = kx − y, x − z k.
Similarly, we have
ky − z , y − uk = ky − z , y − xk = kx − y, x − z k, kz − x, z − uk = kz − x, z − yk = kx − y, x − z k. For the uniqueness of u, assume that another v ∈ X satisfying
kx − y, x − vk = ky − z , y − vk = kz − x, z − vk = kx − y, x − z k. Since v is in an interior point of triangle given by the points x, y and z, there exist elements s1 , s2 and s3 of K such that s1 + s2 + s3 = 1, si > 0 and v = s1 x + s2 y + s3 z. In view of Lemma 3.1, we have
kx − y, x − z k = = = = = =
kx − y, x − vk kx − y, x − (s1 x + s2 y + s3 z )k kx − y, (s1 − 1)x + s2 y + (1 − s1 − s2 )z k kx − y, (s1 + s2 − 1)x + (1 − s1 − s2 )z k |s1 + s2 − 1|kx − y, x − z k |s3 |kx − y, x − z k.
Since kx − y, x − z k 6= 0, we have |s3 | = 1. Similarly we also have |s1 | = |s2 | = 1. Thus s1 + s2 + s3 = 1 and |s1 | = |s2 | = |s3 | = 1. From the definition, there exist integers k1 , k2 , k3 such that s1 = 3k1 , s2 = 3k2 , s3 = 3k3 . By the fact s1 + s2 + s3 = 1, ki < 0 for all i = 1, 2, 3. Without loss of generality, we can let s1 = 3−n1 , s2 = 3−n2 , s3 = 3−n3 and n1 ≥ n2 ≥ n3 . Assume that above one of inequalities holds. Then 1 = s1 + s2 + s3 = 3−n1 (1 + 3n1 −n2 + 3n1 −n3 ). So 3n1 = 1 + 3n1 −n2 + 3n1 −n3 . Then the number in left side is a multiple of 3, but the number in right side is not a multiple of 3. This is a contradiction. Hence s1 = s2 = s3 = 13 which means u = v . This completes the proof. The following theorem is the (additive) Mazur–Ulam Theorem on non-Archimedean 2-normed spaces. We prove the following theorem using the property that a barycenter point of the triangle maps to a barycenter point of the corresponding triangle. Theorem 3.3. Suppose that X and Y are non-Archimedean 2-normed spaces over a linear ordered non-Archimedean field K with C = {3n |n ∈ Z}. If f : X → Y is an interior preserving 2-isometry, then f (x) − f (0) is additive. Proof. Let g (x) := f (x) − f (0). Then g is a 2-isometry and g (0) = 0. By the proof of Theorem 2.4, g is also an interior preserving mapping. Let x, y, z ∈ X with kx − y, x − z k 6= 0. Since g is a 2-isometry, we have
g (x) − g (y), g (x) − g x + y + z = x − y, x − x + y + z
3 3
2x − y − z
=
x − y,
3
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= kx − y, 2x − y − z k = kx − y, x − z k = kg (x) − g (y), g (x) − g (z )k (x, y, z ∈ X) and similarly we can obtain
g (y) − g (z ), g (y) − g x + y + z = kg (x) − g (y), g (x) − g (z )k,
3
g (z ) − g (x), g (z ) − g x + y + z = kg (x) − g (y), g (x) − g (z )k.
3
x+y+z
Since 3 is an interior point of the triangle 4xyz and g is an interior preserving mapping, g t3 g (z ) | t1 + t2 + t3 = 1, ti ∈ K , ti > 0, for all i}. By Lemma 3.2
g
x+y+z 3
=
g (x) + g (y) + g (z ) 3
x+y+z 3
∈ {t1 g (x) + t2 g (y) +
,
for all x, y, z ∈ X with kx − y, x − z k 6= 0. This completes the proof.
We now consider a notion of collinearity. Let x, y, z ∈ X be distinct elements. Then x, y and z are called collinear if x − y = t (x − z ) for some t ∈ K . Lemma 3.4. Let X be a non-Archimedean 2-normed space over a linear ordered non-Archimedean field K with C = {2n |n ∈ Z} x +y and let x, y ∈ X. Then u := 2 is the unique member of X satisfying
kx − z , x − uk = ky − z , y − uk = kx − z , y − z k for some z ∈ X with kx − z , y − z k 6= 0 and u, x, y are collinear. x +y
Proof. We can easily show that u := 2 satisfies the above equations. For the uniqueness, assume that there is another v satisfying
kx − z , x − vk = ky − z , y − vk = kx − z , y − z k. Since x, y and v are collinear, we can find t ∈ K such that v = tx + (1 − t )y. Then
kx − z , y − z k = = = =
kx − z , x − vk kx − z , x − tx − (1 − t )yk |1 − t |kx − z , x − yk |1 − t |kx − z , y − z k
kx − z , y − z k = = = =
ky − z , y − vk ky − z , y − tx − (1 − t )yk |t |ky − z , y − xk |t |kx − z , y − z k.
Since kx − z , y − z k 6= 0, we have |1 − t | = 1 and |t | = 1. Then there exist elements k1 , k2 of K such that 1 − t = 2k1 , t = 2k2 . Since 2k1 + 2k2 = 1, ki < 0 for all i = 1, 2. With out loss of generality, we let 1 − t = 2−n1 , t = 2−n2 and n1 ≥ n2 . If n1 n2 then 1 = 2−n1 + 2−n2 = 2−n1 (1 + 2n1 −n2 ). So 2n1 = 1 + 2n1 −n2 . This is a contradiction because the left side is a multiple of 2 but the right side is not. Thus n1 = n2 , and hence t = 21 . This means that u = v , which completes the proof. In the proof of the next theorem, we use the midpoint of a two points maps to the midpoint of the corresponding two points. And one may consider that the following is another type of the (additive) Mazur–Ulam Theorem on non-Archimedean 2-normed spaces. Theorem 3.5. Suppose that X and Y are non-Archimedean 2-normed spaces over a linear ordered non-Archimedean field K with C = {2n |n ∈ Z}. Assume that f (x), f (y) and f (z ) are collinear when x, y and z are collinear. If f : X → Y is a 2-isometry, then f (x) − f (0) is additive. Proof. Let g (x) := f (x) − f (0) then g is a 2-isometry and g (0) = 0. It is clear that if x, y and z are collinear, then g (x), g (y) and g (z ) are also collinear. Since g is a 2-isometry, we obtain that
g (x) − g (z ), g (x) − g x + y = x − z , x − x + y
2
2
x−y
=
x − z , 2 = kx − z , x − yk = kx − z , y − z k = kg (x) − g (z ), g (y) − g (z )k,
J. Choy et al. / Nonlinear Analysis 72 (2010) 1291–1297
g (y) − g (z ), g (y) − g x + y = y − z , y − x + y
2 2 = ky − z , y − xk = kx − z , y − z k = kg (x) − g (z ), g (y) − g (z )k for all x, y ∈ X. x +y x +y Since 2 , x, y are collinear, g 2 , g (x), g (y) are also collinear. From Lemma 3.4, we get that g x, y ∈ X. This completes the proof.
1297
x +y 2
=
g (x)+g (y) 2
for all
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