1. Introduction
Mazur and Ulam in [1] proved that every surjective isometry U between X and Y is a affine, also states that the mapping with
, then U is linear. Let X and Y be normed spaces, if the mapping
satisfying that
It was called isometry. About it’s main properties in sequences spaces, Tingley, D, Ding Guanggui, Fu Xiaohong in [2] [3] [4] [5] [6] proved. So, we give a new definition that if there is a function
such that
is a linear isometry. we can say the mapping
is phase equivalent to J.
If the two spaces are Hilbert spaces, Rätz proved that the phase isometries
are precisely the solutions of functional equation in [7] . If the two spaces are not inner product spaces, Huang and Tan [8] gave a partial answer about the real atomic
spaces with
. Jia and Tan [9] get the conclusion about the
-type spaces. In [6] , xiaohong Fu proved the problem of isometry extension in the s space detailedly.
In this artical, we mainly discuss that all mappings
or
also have the properties, that are solutions of the functional equation
(1)
All metric spaces mentioned in this artical are assumed to be real.
2. Results about s
First, let us introduction some concepts. The s space in [10] , which consists of all scalar sequences and for each elements
, the F-norm of x is defined by
. Let
denote the set of all elements of the form
with
. where
. We denote the support of x by
, i.e.,
For all
, if
, we say that x is orthogonal to y and write
.
Lemma 2.1. Let
be a sphere with radius
and center 0 in s. Suppose that
is a mapping satisfying Equation (1). Then for any
, we have
Proof: Necessity. Choosing
,
that satisfying
. We can suppose
,
. And we also have
.
So
or
Thus
That means
(2)
It is easy to know
is strictly increasing. And
. We can get the result
.
For
, similarty to the above
. It is
. Sufficiency. For
, that is,
, so (2) holds, and we have
so, it must have
.
or
as the same
. It follows that
(3)
Similarly to the proof of necessity, we get
.
Lemma 2.2. Let
be a sphere with radius
in the finite dimensional space
, where
. Suppose that
is an phase isometry. Let
, then there is a unique real
with
, such that
.
Proof: We proof first that for any
, there is a unique
and a unique real
with
such that
(because the assumption of
implies
). To this end, suppose on the contrary that
and
. In view of Lemma 1, we have
.
Hence, by the “pigeon nest principle” (or Pigeonhole principle) there must exist
such that
, which leads to a contradiction.
Next, if
,
, where
, then
and
. Indeed, if
, we have
or
and
(4)
a contradiction which implies
. From this
follows. Finally, there is a unique
with
such that
. Indeed, if
, by the result in the last step, we have
, thus
and
(5)
So, we get
and we also have
through the two equalities of above
In the end,
(6)
The proof is complete.
Lemma 2.3. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1) and
as in Lemma 2.2. Then for any lement
, we have
, where
for any
.
Proof: Note that the defination of
, we can easily get
. For any
, write
, where
. we can write
, where
. we have
On the other hand, we have
Combiniing the two equations, we obtain that
As
and
, we have
Therefore,
Analysis of the equation, according to the monotony of the function, that is
(7)
The proof is complete.,
The next result shows that a mapping satisfying functional Equation (1) has a property close to linearity.
Lemma 2.4. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1). there exist two real numbers
and
with absolute 1 such that
for all nonzero vectors x and y in X, x and y are orthogonal.
Proof: Let x and y be nonzero orthogonal vectors in X, we write
,
.
,
,
where
and
. We infer from Equation (1) that
Through the above equation we can get
or
. This implies that
, and similarly
. The proof is complete.,
Lemma 2.5. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1). Then V is injective and
for all
.
Proof: Suppose that V is surjective and
for some
. Putting
in the Equation (1), this yields
if and only if
. Assume that
choose
such that
, using the Equation (1) for
, we obtain
This yields
. If
, then
, which is a contradiction. So we obtain
, and we must have
. For otherwise we get
and
This lead to the contradiction that
.
Theorem 2.6. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.
Proof: Fix
, and let
. By Lemma 2.4 we can write
for any
. Then, we can define a mapping
as follows:
,
for
. The J is phase equivalent to V. So it is easily to know that J satisfies functional Equation (1). For any
, and
,
That means
,
for any
, and
.
That yields
That means
. On the other hand,
for
, It follows that
for all
, by assumed conditions, so J is a surjective isometry.,
Theorem 2.7. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.
Proof: According to [10] Theorem 1, Theorem 2 the author presents some results of extension from some spheres in the finite dimensional spaces
. And also we have the above Theorem 2.6, so we can get the result easily.
3. Results about
In this part, we mainly introduce the space
, where H is a Hilbert space. In [11] mainly discussed the isometric extension in the space
. For each element
, the F-norm of x is defined by
. Let
denote the set of all elements of the form
with
. where
.
Some notations used:
, where
.
Specially, when
, we have
.
Next, we study the phase isometry between the space
to
, that if V is a surjective phase isometry, then V is phase equivalent to a linear isometry J.
Lemma 3.1. If
, then
if and only if
where
.
Proof: It has a detailed proof process in [11] .
Lemma 3.2. Let
be a sphere with radius
in the finite dimensional space
, where
. Defined
is an phase isometry, then we can get
.
Proof: “Þ” Take any two elements
,
, let
,
. Then we have
or
(8)
at the same time, we have
(9)
That means
, it is
. “Ü” The proof of sufficiency is similar to the Lemma 2.1.
Lemma 3.3. Let
be as in Lemma 3.2,
, and
.
. Then there exists
, such that
.
Proof: We prove first that, for any
, there exist
and
such that
. And then prove
. It is the same an Lemma 2.2.
Finally, we assert that, there exists
such that
. Indeed, if
, by the result in the last step, we have
,
Therefore,
or
(10)
So, we can get
. And
, that means
.
Lemma 3.4. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1). there exist two real numbers
and
with absolute 1 such that
for all nonzero vectors x and y in X, x and y are orthogonal. Proof: Let
and
be nonzero orthogonal vectors in X.
,
,
where
and
. We infer from Equation (1) that
Through the above equation we can get
or
. The proof is complete.,
Lemma 3.5. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1). Then V is injective and
for all
.
Proof: Suppose that V is surjective and
for some
. Putting
in the Equation (1), this yields
if and only if
. Assume that
choose
such that
, using the Equation (1) for
, we obtain
This yields
. If
, then
, which is a contradiction. So we obtain
, and we must have
. For otherwise we get
and
This lead to the contradiction that
.
Theorem 3.6. Let
and
. Suppose that
is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.
Proof: Fix
, and let
. By Lemma 3.4 we can write
for any
. Then, we can define a mapping
as follows:
,
for
. The J is phase equivalent to V. So it is easily to know that J satisfies functional Equation (1). For any
, and
,
That means
,
for any
, and
.
That yields
That means
. On the other hand,
for
, It follows that
for all
, by assumed conditions, so J is a surjective isometry.,
4. Conclusion
Through the analysis of this article, we can get the conclusion that if a surjective mapping satisfying phase-isometry, then it can phase equivalent to a linear isometry in the space s and the space
.
Acknowledgements
The author wish to express his appreciation to Professor Meimei Song for several valuable comments.