Keywords:Affine Frame; Oversampling
1. Introduction
Let 
denote, as usual, the space of all complex-valued square integrable functions on the real line with inner product 
and norm
. For any
, we will use the notation
(1)
where 
and
. A function 
is said to generate an affine frame
(2)
of
, with frame bounds 
and
, where
, if it satisfies
(3)
The frame (2) of 
is called a tight frame, if (3) holds with
, see [1] and [2]. In 1993, C. K.Chui and X. L. Shi [3] proved the following oversampling theorem:
Theorem A. Let 
be any positive integer and
. Also, let 
generate a frame 
with frame bounds 
and 
as given by (3). Then for any positive integer 
which is relatively prime to
, the family
(4)
remains a frame of 
with the same bounds. If
, this result does not hold. But they only gave a countexample for the case where 
as in [4]. For other positive integer 
and 
which satisfy
, they did not prove. The aim of this paper is to establish the inverse proposition of Theorem A, and then we following:
Theorem 1.1. Let 
be any positive integer and
. Also, let 
be any affine frame of 
with frame bounds 
and
. The family (4) remains a frame of 
with the same bounds: that is,
(5)
if and only if 
and 
are relatively prime.
2. Proofs
The sufficiency has been included in the theorem 4 of [3]. In the following we will prove the necessary part of the theorem.
Suppose for any affine frame (2) of 
with frame bounds 
and
, the family (4) is also a frame of 
with the same bounds. Then when (1) forms an orthonormal basis, the family (4) forms a tight frame with frame bound
. So we just need to prove that there exists a function 
such that the family (1) forms the orthonormal basis, but for any two positive integers 
and 
which satisfy
, there exist two functions 
and 
such that
Doesn’t equal
.
Let
, then 
forms an orthonormal basis, which is called Haar basis. Set
and 
We prove that if
, then
(6)
and
Denote
. We have
where
and
In order to prove the theorem, we have three cases.
Case 1. When
.
We have 
if
. Thus, if 
is an even integer, we can get
So, we have
If 
is an odd integer, we have
So, we have
Case 2. When
.
If 
is an even integer, we have
Thus
If 
is an odd integer, we can get 
because of 
As in the case
, we also have
So, we get
Case 3. When
.
If 
is an even integer. Let
and
When
, there exists an integer 
satisfying
. Therefore we have
where
. When
, we have 
and
. Thus we have
Therefore
When
, similar to the case
, we also have
So we have
If 
is an odd integer. We have
where
A familiar calculation shows
Since 
and
, we have
. Also when 
and
, we have
When 
and
, obviously we have
When
,
. So we have 
in this case. This completes the proof of the theorem.
Acknowledgements
The authors would like to thank anonymous reviewers for their comments and suggestions. The authors are partially supported by project 11226108, 11071065, 11171306 funded by NSF of China, and project Y201225301. Project 20094306110004 funded by RFDP of high education of China.