ISSN 2307-7743 http://scienceasia.asia
ON THE ALEKSANDROV-RASSIAS PROBLEM IN 2-NORMED SPACE
QIAN ZHANG, YUBO LIU, LIFANG CHANG, MEIMEI SONG
Abstract. LetXandY be 2-normed linear spaces. If a mapping
f :X×X →Y preserves two distances with a noninteger ratio,f
must be an isometry. In this paper, we provide some results of the Aleksandrov-Rassias problem for mappings which preserves three distances in 2-normed space.
.
1. INTRODUCTION
The theory of isometry had its beginning in the important paper by Mazur and Ulam in 1932. He proved that every isometry mapping of a normed real linear space onto a normed real linear space is a linear mapping up to translation. When the target space Y is a strictly convex real normed space, for into mapping, Baker[1] proved that every
isometry of a normed real linear space into a strictly convex normed real linear space is also a linear isometry up to translation. What happens if we require, instead of one conservative distance for a mapping between normed vector spaces, two conservative disrtances? Aleksandrov and Rassias give some results about this problem. Aleksandrov-Rassias problem has obtained some results in Hilbert spaces, X and Y are Hilbert spaces with dimX ≥ 2, if T :X → Y preserves two distances with a noninteger ratio, then T is linear isometry up to translation.
In 2001,Xiang Shuhuang[7] introduced that if f preserves two
dis-tances and X, Y are real normed vector spaces such that Y is strictly convex anddimY ≥2, it is an open problem whether or notf must be an isometry, however, if f preserves three distances, we have the result about isometry.
Aleksandrov-Rassias problem: If T preserves two distances with a noninteger ratio, and X and Y are real normed vector spaces such
2010Mathematics Subject Classification. 46B04.
Key words and phrases. Aleksandrov-Rassias problem; 2-normed space; AOPP.
c
2014 Science Asia
that Y is strictly convex and dimX ≥2,whether or not T must be an isometry?
Benz and Berens[2] proved the following theorem and pointed out
that the condition that Y is strictly convex can not be relaxed. Other authors proved Aleksandrov-Rassias problem in [3],[4]and[5].
Let X and Y be real normed vector spaces, assume thatdimX ≥2 andY is strictly convex, supposeT :X →Y satisfies that : T preserves the two distances ρ and λρ for some integer λ ≥ 2, that is, for all
x, y ∈ X with kx −yk = ρ, then kT(x)− T(y)k ≤ ρ, and for all
x, y ∈X with kx−yk =λρ, then kT(x)−T(y)k ≥ λρ, so T is linear isometry up to translation.
It is easy to verify rhat the condition in above result is equivalent to that T preserves the distances ρ and λρ. Next we need the following definitions we will use in our main result.
Definition 1.1. Let X be a real linear space with k·,·k : X2 → R,
then (X,k·,·k) is called a 2-normed space if
(N1) kx, yk= 0⇐⇒ x and y are linearly dependent.
(N2) kx, yk=ky, xk.
(N3) kαx, yk=|α|kx, yk.
(N4) kx, y+zk ≤ kx, yk+kx, zk.
For α ∈ R and x, y, z ∈ X.The function k·,·k is called the 2-norm onX.
Definition 1.2. [6] we calledf a 2-isometry ifkx−y, y−zk=kf(x)−
f(y), f(y)−f(z)k for all x, y, z ∈X.
Definition 1.3. [6] (AOPP) Let x, y, z ∈ X with kx−y, y−zk = 1 , then kf(x)−f(y), f(y)−f(z)k= 1.
Definition 1.4. [6] We callf a 2-Lipschitz mapping if there is a k≥0 such thatkf(z)−f(x), f(y)−f(x)k ≤kkz−x, y−xkfor allx, y, z ∈X, the smallest such k is called the 2-Lipschitz constant.
2. MAIN RESULTS
Lemma 2.1. [6] For b, c ∈ X, if b and c are linearly dependent with
the same direction, that is c = αb for some α > 0, then ka, b+ck = ka, bk+ka, ck for all a∈X.
Lemma 2.2. Let X, Y are 2-normed space and f : X×X → Y is a surjection and satisfied:
(1) kx−y, p−qk ≤1, thenkf(x)−f(y), f(p)−f(q)k ≤ kx−y, p−qk. (2) kx−y, p−qk ≥α, then kf(x)−f(y), f(p)−f(q)k ≥α.
Proof. (a) First, we proof kf(x)−f(y), f(p)−f(q)k ≤ kx−y, p−qk for all x, y, p, q ∈X.
Letkx−y, p−qk ≤ m
n, if m= 1, the result is obvious.
We suppose that m ≥2. Defineqi =q+ mi (p−q),(i= 0,1,2, . . . , m), and
qi+1−qi = 1
m(p−q), p−q =
m−1
X
i=0
(qi+1−qi)
then
kx−y, qi+1−qik=kx−y, 1
m(p−q)k=
1
mkx−y, p−qk ≤
1
n,
(i= 0,1,2, . . . , m−1),
kf(x)−f(y), f(p)−f(q) ≤
m−1
X
i=0
kf(x)−f(y), f(qi+1)−f(qi)k
≤
m−1
X
i=0
kx−y, qi+1−qik
≤ m
n.
By Lemma 2.1
kx−y, p−qk= m−1
X
i=0
kx−y, qi+1−qik
Thuskf(x)−f(y), f(p)−f(q)k ≤ kx−y, p−qk.
(b) We prooff perservesα. we supposekx−y, p−qk=α, Then there are m, n∈N, and α≤ m
n, by (a), we have
kf(x)−f(y), f(p)−f(q)k ≤ kx−y, p−qk
by the condition (2) kf(x)−f(y), f(p)−f(q)k ≥ kx−y, p−qk, so
kf(x)−f(y), f(p)−f(q)k=kx−y, p−qk=α.
(c) kf(x)−f(y), f(p)−f(q)k=kx−y, p−qk askx−y, p−qk< α. For α > 0, there must be m, n ∈ N, so α < mn, by (a), kf(x) −
f(y), f(p)−f(q)k ≤ kx−y, p−qk Assume that
Letz =x+ α
kx−y,p−qk(y−x), so
kz−x, p−qk=α,kz−y, p−qk=α− kx−y, p−qk.
Then by (b) and (a)
α = kf(z)−f(x), f(p)−f(q)k
≤ kf(z)−f(y), f(p)−f(q)k+kf(x)−f(y), f(p)−f(q)k
< α− kx−y, p−qk+kx−y, p−qk=α.
This is a contradiction, that implies thatkf(x)−f(y), f(p)−f(q)k= kx−y, p−qk.
(d) f preserve that the distance n2α.
Letkx−z, p−qk= n2α, by (a), then kf(x)−f(z), f(p)−f(q)k ≤ n
2α,
let
u=f(x) + α 2
f(z)−f(x)
kf(z)−f(x), f(p)−f(q)k
there exsits a v ∈X, such thatf(v) =u by f is a surjection. then
ku−f(x), f(p)−f(q)k= α 2 < α,
by (2), kv −x, p−qk < α. by (c), kv−x, p−qk=ku−f(x), f(p)−
f(q)k= α2. Then
ku−f(z), f(p)−f(q)k ≥ α
2(n−1). Otherwise if ku−f(z), f(p)−f(q)k< α
2(n−1), we can find a sequence
vi ∈X,(i= 1,2,· · · , n−1), such that v0 =v, vn−1 =z, which implies
kf(v)−f(z), f(p)−f(q)k< α
2(n−1), by f is surjection, there exsits
vi ∈X, then
f(vi) = f(v) +
i
n−1(f(z)−f(v)),(i= 0,1,2,· · · , n−1),
so f(vi)− f(vi+1) = n−11 (f(z)−f(v)) and f(vi)−f(vi+1) collinear,
hence
f(v)−f(z) = n−2
X
i=0
(f(vi)−f(vi+1))
by Lemma2.1,
kf(v)−f(z), f(p)−f(q)k= n−2
X
i=0
kf(vi)−f(vi+1), f(p)−f(q)k<
(n−1)α
So
kf(vi)−f(vi+1), f(p)−f(q)k<
α
2 < α,
by the condition (2), thus kvi−vi+1, p−qk< α(i= 0,1,2,· · · , n−1)
and by (c)
kvi−vi+1, p−qk=kf(vi)−f(vi+1), f(p)−f(q)k<
α
2(i= 0,1,2,· · · , n−1), that implies
kv−z, p−qk=k
n−1
X
i=0
(vi−vi+1), p−qk ≤
n−1
X
i=1
kvi−vi+1, p−qk
<
n−1
X
i=0
α
2 =
(n−1)α
2 .
by(c),
kv−x, p−qk=kf(v)−f(x), f(p)−f(q)k=ku−f(x), f(p)−f(q)k= α 2 Moreover
kx−z, p−qk ≤ kx−v, p−qk+kv−z, p−qk< α
2 +
n−1 2 α =
n
2α .
This is a contradiction with kx−z, p−qk= n2α. and
u−f(z) = f(x)−f(z) + α 2
f(z)−f(x)
kf(z)−f(x), f(p)−f(q)k
= (f(x)−f(z))(1− α
2kf(z)−f(x), f(p)−f(q)k)
then
ku−f(z), f(p)−f(q)k
= kf(x)−f(z), f(p)−f(q)k(1− α
2kf(z)−f(x), f(p)−f(q)k)
= kf(x)−f(z), f(p)−f(q)k − α 2 So
α
2(n−1)≤ ku−f(z), f(p)−f(q)k=kf(z)−f(x), f(p)−f(q)k −
α
2 that implies kf(z)−f(x), f(p)−f(q)k= α2n.
(e) f is isometry from X to Y. That is kf(x)−f(y), f(p)−f(q)k = kx−y, p−qk.
α
2n. For
α
2n, there exsit m, n0 ∈ N, and
α
2n <
m
n0, by (a), we have
kf(x)−f(y), f(p)−f(q)k ≤ kx−y, p−qk. Assume that
kf(x)−f(y), f(p)−f(q)k<kx−y, p−qk.
Let
z =x+
α
2n
kx−y, p−qk(y−x).
Sokz−x, p−qk= α2n,kz−y, p−qk= α2n− kx−y, p−qk. So by (d),
kf(z)−f(x), f(p)−f(q)k=kz−x, p−qk= α 2n. By (c),(d) and the assumption
α
2n = kf(z)−f(x), f(p)−f(q)k
≤ kf(z)−f(y), f(p)−f(q)k+kf(x)−f(y), f(p)−f(q)k
< α
2n− kx−y, p−qk+kx−y, p−qk=
α
2n
that is a contradiction, that implies kf(x)−f(y), f(p)−f(q)k=kx−
y, p−qk.
we can say that f preserves the two distance 1 andα in above Lem-ma.
Theorem 2.3. Let X and Y be real 2-normed space. Assume that Y is strictly convex, suppose f : X → Y satisfied AOPP and f is a 2-Lipschitz mapping with k = 1, that is kf(x)−f(y), f(p)−f(q)k ≤ kx−y, p−qk for all x, y, p, q ∈X. Then f is a 2-isometry.
Proof. Letx, y, p, q ∈X, and kx−y, p−qk= 12, set z =x+ 2(y−x) Then kx−z, p−qk= 1,kz−y, p−qk= 12
And by the condition 2-Lipschitz and AOPP,
kx−y, p−qk
≥ kf(x)−f(y), f(p)−f(q)k
≥ kf(z)−f(x), f(p)−f(q)k − kf(z)−f(y), f(p)−f(q)k
≥ 1− kz−y, p−qk ≥ 1 2
By the condition, kf(x)−f(y), f(p)−f(q)k ≤ kx−y, p−qk= 12 Hence
Similarly
kf(z)−f(y), f(p)−f(q)k= 1 2 And
kf(z)−f(x), f(p)−f(q)k
=kf(z)−f(y), f(p)−f(q)k+kf(y)−f(x), f(p)−f(q)k= 1
BecauseY is strictly convex, thenf(y) = f(x)+2f(z) andkf(y)−f(x), f(p)−
f(q)k= 12, so f preserves distances 1 and 12, sof is an isometry due to
lemma2.2.
Theorem 2.4. Let X and Y be real 2-normed spaces. Assume that dimX ≥ 2 and Y is strictly convex, suppose f :X×X → Y satisfies the property that f preserves the three distances 1,a and 1+a, where a is any positive constant. Then f is a 2-isometry.
Proof. (1) Letx, y ∈X,kx−y, p−qk= 2 +a, setx1 =x+2+1a(y−x),
x2 =x+1+2+aa(y−x)
Then
kx1 −x, p−qk= 1,kx1−x2, p−qk=a,
ky−x1, p−qk= 1 +a,kx2−x, p−qk= 1 +a,ky−x2, p−qk= 1
It follows that
kf(x1)−f(x), f(p)−f(q)k= 1,kf(x1)−f(x2), f(p)−f(q)k=a
kf(y)−f(x1), f(p)−f(q)k= 1 +a,kf(x2)−f(x), f(p)−f(q)k= 1 +a,
kf(y)−f(x2), f(p)−f(q)k= 1. Since Y is strictly convex, let
f(x1)−f(x) = α(f(x2)−f(x))(α >0)
Then
1 =kf(x1)−f(x), f(p)−f(q)k = αkf(x2)−f(x), f(p)−f(q)k
= α(a+ 1)
Soα = a+11 and f(x1)−f(x) = a+11 (f(x2)−f(x))
We have
f(x1) =f(x) +
1
1 +a(f(x2)−f(x))
And
f(x) = 1 +a
a f(x1)−
1
af(x2).
Since Y is strictly convex, let
f(y)−f(x2) =α
0
Then
1 =kf(y)−f(x2), f(p)−f(q)k = α
0
kf(x2)−f(x1), f(p)−f(q)k
= α0a
Soα0 = 1a and f(y)−f(x2) = a1(f(x2)−f(x1))
We have
f(x2) = f(x1) +
a
1 +a(f(y)−f(x1)
And
f(y) = 1 +a
a f(x2)−
1
af(x1).
Thuskf(x)−f(y), f(p)−f(q)k= 2 +a for all x, y ∈X with kx−y, p−qk= 2 +a, so f preserves the distance 2+a.
(2) Letkx−y, p−qk= 2a, setx1 =x+2+2a−1a(y−x), x2 =x+2+2aa(y−x)
Then
kx1−x, p−qk= 1 +a,kx1−x2, p−qk= 1,
ky−x1, p−qk= 1 +a,kx2−x, p−qk= 2 +a,ky−x2, p−qk=a,
sincef preserves distances 1,a,1+a and 2+a, in a similar way, we obtain that
kf(y)−f(x), f(p)−f(q)k= 2a.
By (1),(2) and Lemma 2.2, we have f preserves 1 and 2a, then f is a
2-isometry.
Corollary 2.5. Let X and Y be real 2-normed spaces. Assume that dimX ≥ 2 and Y is strictly convex, suppose f :X×X → Y satisfies the property that f preserves the three distances a,b and a+b, where a,b is any positive constant. Then f is a 2-isometry.
References
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