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On equivalent conditions of two sequences to be R-dual

Journal of Inequalities and Applications20152015:10

https://doi.org/10.1186/s13660-014-0529-8

  • Received: 28 September 2014
  • Accepted: 13 December 2014
  • Published:

Abstract

The concept of R-duals was introduced by Casazza, Kutyniok, and Lammers in 2004. In this paper, we give a condition when a Parseval frame can be dilated to an orthonormal basis of a given separable Hilbert space H. This is advantageous for deriving a condition for a sequence \(\{\omega_{j}\}_{j\in J} \) to be an R-dual of a given frame \(\{f_{j}\}_{j\in J} \).

Keywords

  • Orthonormal Basis
  • Separable Hilbert Space
  • Riesz Base
  • Gabor Frame
  • Infinite Dimension

1 Introduction

The concept of R-duals was first introduced by Casazza et al. in [1]. Although it is defined in general frame theory, there is a natural connection with Gabor frame theory. And it is still an open problem whether the duality principle in Gabor analysis actually can be derived from the theory of the R-dual. Lots of scholars have done much research in this area. Reference [2] introduces various alternative R-duals and shows their relations with Gabor frames. References [3] and [4] consider R-dual in Banach space. In [5], the authors give an equivalent condition for a sequence \(\{\omega_{j}\}_{j\in J} \) to be an R-dual of a given frame \(\{f_{j}\}_{j\in J} \). However, we think there is a mistake in their proof. The correction of it will be discussed in Section 3.

The dilation viewpoint on frames is introduced by Larson and Han in [6], which has a natural relation with the R-dual. They point out that any Parseval frame can be dilated to an orthonormal basis. But given a Hilbert space H and a Parseval frame of a subspace of H, can the Parseval frame be dilated to an orthonormal basis for H? This will be discussed in Section 2.

In the entire paper, we let H denote a separable Hilbert space, with the inner product \(\langle\cdot, \cdot\rangle\), and J be a countable index set.

Definition 1

A sequence \(\{f_{j}\}_{j\in J} \) of elements in H is a frame for H if there exist constants \(A , B>0 \) such that
$$A\Vert {f}\Vert ^{2} \leq\sum_{j \in J} \bigl| \langle f, f_{j} \rangle \bigr|^{2}\leq B\Vert {f} \Vert ^{2},\quad f\in H. $$
The constants A, B are called a lower and upper frame bounds for the frame. A frame is A-tight, if \(A=B \). If \(A=B=1 \), it is called a Parseval frame (a normalized tight frame in [6]).

Definition 2

A sequence \(\{\omega_{j}\}_{j\in J} \) in H is a Riesz sequence if there exist constants \(C , D>0 \) such that
$$C\sum_{j \in J} |c_{j}|^{2} \leq \biggl\Vert {\sum_{j \in J} c_{j} \omega_{j}}\biggr\Vert ^{2} \leq D \sum _{j \in J} |c_{j}|^{2} $$
for all finite sequence \(\{c_{j}\}_{j\in J} \). The numbers C, D are called Riesz bounds. A Riesz sequence is a Riesz basis for H if it is complete in H.

For more information as regards frames and Riesz bases we refer to the monograph [7]. We now state the definition of the R-dual sequence.

Definition 3

[1]

Let \(\{e_{i}\}_{i\in J} \) and \(\{h_{i}\}_{i\in J} \) denote two orthonormal bases for H, and let \(\{f_{i}\}_{i\in J} \) be any sequence in H for which
$$\sum_{i \in J} \bigl| \langle f_{i}, e_{j} \rangle \bigr|^{2}< \infty,\quad\forall j\in J. $$
The R-dual of \(\{f_{i}\}_{i\in J} \) with respect to the orthonormal bases \(\{e_{i}\}_{i\in j} \) and \(\{h_{i}\}_{i\in J} \) is the sequence \(\{\omega_{j}\}_{j\in J} \) given by
$$ \omega_{j}=\sum_{i \in J} \langle f_{i}, e_{j} \rangle h_{i},\quad j\in J. $$
(1.1)
It is well known from [1] that \(\{f_{i}\}_{i\in J} \) is a frame for H with bounds A, B if and only if \(\{\omega_{j}\}_{j\in J} \) is a Riesz sequence in H with bounds A, B. But given two sequence \(\{f_{i}\}_{i\in J} \) and \(\{\omega_{j}\}_{j\in J} \), under what conditions can we find orthonormal bases \(\{e_{i}\}_{i\in J} \) and \(\{h_{i}\}_{i\in J} \) for H such that (1.1) holds? This is the main question we want to answer in this paper. It will be discussed in Section 3 explicitly. Assume that \(\{f_{i}\}_{i\in J} \) is a frame for H. Define a sequence \(\{n_{i}\}_{i\in J} \) by
$$ n_{i}=\sum_{k \in J} \langle e_{k}, f_{i} \rangle\tilde{\omega}_{k},\quad i \in J, $$
(1.2)
where \(\{\tilde{\omega}_{j} \}_{j\in J} \) is the canonical dual Riesz sequence of \(\{\omega_{j} \}_{j\in J} \). The construction of \(\{n_{i}\}_{i\in J} \) comes from [5]. It plays an important role in this paper.

Proposition 1

[5]

Let \(\{\omega_{j}\}_{j\in J} \) be a Riesz basis for the subspace W of H, with dual Riesz basis \(\{\tilde{\omega}_{k}\}_{k\in J} \). Let \(\{e_{i}\}_{i\in J} \) be an orthonormal basis for H. Given any sequence \(\{f_{i}\}_{i\in J} \) in H, the following hold:
  1. (i)
    There exists a sequence \(\{h_{i}\}_{i\in J} \) in H such that
    $$ f_{i}=\sum_{j \in J} \langle \omega_{j}, h_{i} \rangle e_{j},\quad\forall i\in J. $$
    (1.3)
     
  2. (ii)
    The sequence \(\{h_{i}\}_{i\in J} \) satisfying (1.3) is characterized as
    $$ h_{i}=m_{i}+n_{i}, $$
    (1.4)
    where \(n_{i} \) is given by (1.2) and \(m_{i}\in W^{\perp}\).
     
  3. (iii)
    If \(\{\omega_{j}\}_{j\in J} \) is a Riesz basis for H, then (1.3) has the unique solution
    $$h_{i}=n_{i},\quad i\in J. $$
     

In [5], Christensen et al. give a solution to the main question.

Theorem 1

[5]

Let \(\{\omega_{j}\}_{j\in J} \) be a Riesz sequence spanning a proper subspace W of H and \(\{e_{i}\}_{i\in J} \) an orthonormal basis for H. Given any frame \(\{f_{i}\}_{i\in J} \) for H, the following are equivalent:
  1. (i)

    \(\{\omega_{j}\}_{j\in J} \) is an R-dual of \(\{f_{i}\}_{i\in J} \) w.r.t. \(\{e_{i}\}_{i\in J}\) and some orthonormal basis \(\{h_{i}\}_{i\in J}\).

     
  2. (ii)

    There exists an orthonormal basis \(\{h_{i}\}_{i\in J}\) for H satisfying (1.3).

     
  3. (iii)

    The sequence \(\{n_{i}\}_{i\in J}\) in (1.2) is a Parseval frame.

     

We point out that, in fact, (iii) is not equivalent to the other items in Theorem 1. In order to clarify this, we need the following proposition from [6].

Proposition 2

[6]

Let J be a countable (or finite) index set. Suppose that \(\{x_{n} : n\in J\}\) is a Parseval frame for W. Then there exist a Hilbert space \(K\supseteq W \) and an orthonormal basis \(\{e_{n} :n\in J \} \) for K such that \(Pe_{n}=x_{n} \), where P is the orthogonal projection from K onto W.

2 A dilation theorem

In this section, a dilation theorem is given, which will be used in Section 3. Firstly, we give an example to show that Theorem 1 is not strictly right.

Example 1

In this example, we choose the index set \(J=\mathbb{N} \), the natural number set. Suppose \(\{z_{i}\}_{i\in J} \) is an orthonormal basis for H. Define \(f_{i}=2z_{i} \) and \(\omega_{i}=2z_{2i} \) for all \(i\in J \). Then the sequence \(\{f_{i}\}_{i\in J} \) is a Parseval frame with frame bounds 2 and \(\{\omega_{j}\}_{j\in J} \) is a Riesz sequence with bounds 2 as well. The canonical dual \(\{\tilde{\omega}_{j}\}_{j\in J} \) of \(\{\omega _{j}\}_{j\in J} \) equals \(\{\frac{1}{2}z_{2j}\}_{j\in J} \). Let
$$n_{i}=\sum_{k \in J} \langle z_{k}, f_{i} \rangle \tilde{\omega}_{k}=\sum _{k \in J} \langle z_{k}, 2z_{i} \rangle\frac{1}{2}z_{2k}=z_{2i}. $$
Obviously, \(\{n_{i}\}_{i\in J}\) is a Parseval frame, but \(\{\omega_{j}\}_{j\in J} \) cannot be an R-dual of \(\{f_{i}\}_{i\in J} \). If not, by (ii) of Proposition 1, an orthonormal basis \(\{h_{i}\}_{i\in J} \) for H can be characterized by
$$h_{i}=m_{i}+n_{i}, $$
where \(m_{i}\in W^{\perp}\) for all \(i\in J \). Since \(n_{i}\in W \), we have
$$1=\Vert {h_{i}}\Vert ^{2}=\Vert {m_{i}+n_{i}} \Vert ^{2}=\Vert {m_{i}}\Vert ^{2}+\Vert {n_{i}}\Vert ^{2}=\Vert {m_{i}}\Vert ^{2}+\Vert {z_{2i}}\Vert ^{2}. $$
Since \(\Vert {z_{2i}}\Vert =1 \), one has \(m_{i}=0 \) for all \(i\in J \). Therefore \(h_{i}=n_{i}=z_{2i} \). This contradicts \(\{h_{i}\}_{i\in J} \) being an orthonormal basis for H. Thus (iii) of Proposition 1 is not right.

In fact, given any orthonormal sequence (of course a Parseval frame), it cannot be dilated to any orthonormal basis but itself. Generally, we have the following theorem.

Theorem 2

Given two separable Hilbert spaces \(H \supseteq M \), suppose that \(\{x_{n} \}_{n\in J} \) is a Parseval frame for W. Then there exists an orthonormal basis \(\{e_{n}\}_{n\in J} \) for H s.t. \(Pe_{n}=x_{n} \) if and only if
$$ \dim(\ker T)=\dim \bigl(W^{\perp}\bigr), $$
(2.1)
where P is an orthogonal projection from H onto W, T is the synthesis operator of \(\{x_{i}\}_{i\in J} \).

Proof

First we treat sufficiency. Since
$$\sum_{i\in J}c_{i}x_{i}=\sum _{i\in J} c_{i}Pe_{i}=P\sum _{i\in J} c_{i}e_{i}, $$
for any \(\{c_{i}\}_{i\in J}\in\ell^{2}(J)\), a sequence \(\{c_{i}\}_{i\in J} \in\ker T\) if and only if \(\sum_{i \in J}c_{i}e_{i}\in W^{\perp}\). So (2.1) holds.
Now we treat necessity. Suppose (2.1) holds, from the proof of the Proposition 2, there exist a Hilbert space \(K=\ell^{2}(J) \), an orthogonal projection P, and an orthonormal basis \(\{e_{i}\}_{i\in J} \) for K, such that
$$ Pe_{i}=\theta(x_{i}), $$
(2.2)
where θ is the analysis operator of \(\{x_{i}\} \). Since θ is injective, it has inverse restricted to \(\theta (W) \). For simplicity, we just denote it by \(\theta^{-1} \).
For any \(\{c_{i}\}_{i\in J} \in\ell^{2}(J)\), since
$$\sum_{i\in J}c_{i} \langle x, x_{i} \rangle= \biggl\langle x, \sum_{i \in J} \overline{c}_{i}x_{i} \biggr\rangle , $$
we have
$$ \dim\ker T=\dim \bigl(\theta(W) \bigr)^{\perp}. $$
(2.3)
Together with (2.1), we have
$$\dim W^{\perp}=\dim(\ker T)= \dim \bigl(\theta(W) \bigr)^{\perp}. $$
Therefore, there is an unitary operator η from \(W^{\perp}\) onto \((\theta(W))^{\perp}\). Combining with θ, we can define a unitary operator U from H onto K:
$$Ut=U(t_{1}+t_{2})=\theta t_{1}+\eta t_{2},\quad t_{1}\in W,t_{2}\in W^{\perp}. $$
One can easily get
$$U^{-1}y=U^{-1}(y_{1}+y_{2})= \theta^{-1}y_{1}+\eta^{-1} y_{2},\quad y_{1}\in\theta(W),y_{2}\in\theta(W)^{\perp} . $$
Therefore, \(U^{*}=U^{-1} \). In fact, for \(t\in H \) and \(y\in K \),
$$\begin{aligned} \langle Ut, y \rangle =& \bigl\langle U(t_{1}+t_{2}), y_{1}+y_{2} \bigr\rangle \\ =& \langle\theta t_{1}, y_{1} \rangle+ \langle\eta t_{2}, y_{2} \rangle \\ =& \bigl\langle t_{1}, \theta^{-1}y_{1} \bigr\rangle + \bigl\langle t_{2}, \eta^{-1}y_{2} \bigr\rangle \\ =& \bigl\langle t, \theta^{-1}y_{1}+\eta^{-1}y_{2} \bigr\rangle \\ =& \bigl\langle t, U^{-1}y \bigr\rangle \\ =& \bigl\langle t, U^{*}y \bigr\rangle , \end{aligned}$$
where the third equation is due to the Parseval frame property of \(\{x_{n}\}_{n\in J} \) and unitarity of η. Because of the unitarity of U, also \(\epsilon_{i}= U^{-1}e_{i} \) is an orthonormal basis for H.
Now, taking \(U^{-1} \) on the two sides of (2.2), we have
$$U^{-1}Pe_{i}=U^{-1}PUU^{-1}e_{i}=U^{-1}PU \epsilon_{i}=x_{i}. $$
We claim that \(U^{-1}PU \) is also an orthogonal projection. In fact, by the properties of U and P, we have
$$\bigl(U^{-1}PU \bigr)^{2}=U^{-1}P^{2}U=U^{-1}PU $$
and
$$\bigl(U^{-1}PU \bigr)^{*}= \bigl(U^{*}PU \bigr)^{*}=U^{*}PU=U^{-1}PU. $$
Thus we get as desired the complete proof. □

3 Conditions of R-dual

In this section, we discuss under what conditions \(\{\omega_{i}\}_{i\in J} \) can be an R-dual of \(\{f_{i}\}_{i\in J} \). At first, we give two lemmata which will be used later.

Lemma 1

Let \(\{n_{i}\}_{i\in J}\) be defined as (1.2), W the close span of \(\{\omega_{j}\}_{j\in J} \), then \(\overline {\operatorname {span}}\{n_{i}\}_{i\in J}=W \).

Proof

Since \(n_{i}=\sum_{k \in J} \langle e_{k}, f_{i} \rangle\tilde{\omega}_{k}\), we have
$$\overline {\operatorname {span}}\{n_{i}\}_{i\in J}\subseteq \overline {\operatorname {span}}\{\tilde{ \omega}_{i}\}_{i\in J} =W. $$
In the opposite direction, since \(\{f_{i}\}_{i\in J} \) is a frame for H, there exists a sequence \(\{c_{\ell}\}_{\ell\in J}\in\ell^{2}(J) \) such that \(e_{m}=\sum_{\ell \in J}c_{\ell}f_{\ell}\) for \(m\in J \). Then one has
$$\sum_{\ell \in J}\overline{c}_{\ell}n_{\ell}= \sum_{\ell \in J} \overline{c}_{\ell}\sum _{k \in J} \langle e_{k}, f_{\ell}\rangle \tilde{\omega}_{k} =\sum_{k \in J} \biggl\langle e_{k}, \sum_{\ell \in J}c_{\ell}f_{\ell}\biggr\rangle \tilde{\omega}_{k}= \sum _{k \in J} \langle e_{k}, e_{m} \rangle \tilde{ \omega}_{k}=\tilde{\omega}_{m}. $$
Thus \(W\subseteq \overline {\operatorname {span}}\{n_{i}\}_{i\in J}\). We have the desired result. □
Define \(S_{\omega}f=\sum_{k \in J} \langle f, \omega_{k} \rangle \omega_{k} \) and \(S_{\tilde{\omega}}f= \sum_{k \in J} \langle f, \tilde{\omega}_{k} \rangle\tilde{\omega }_{k}\), for \(f\in W \). Then \(S_{\tilde{\omega}}^{-\frac{1}{2}} \tilde{\omega}_{k} \) is an orthonormal basis for W. Since \(\langle\omega_{k}, S^{-1}_{\omega}\omega_{\ell}\rangle=\delta_{k,\ell } \) by [7], one has \(\tilde{\omega}_{k}=S^{-1}_{\omega}\omega_{k} \). Furthermore, we have
$$S_{\tilde{\omega}}f=\sum_{k \in J} \bigl\langle f, S^{-1}_{\omega } \omega_{k} \bigr\rangle S^{-1}_{\omega} \omega_{k} =S^{-1}_{\omega}S_{\omega}S^{-1}_{\omega}f=S^{-1}_{\omega}f, \quad\forall f\in W . $$
This means the operator equation \(S_{\tilde{\omega}}=S^{-1}_{\omega } \) holds.
Let \(\epsilon_{k}=S^{-\frac{1}{2}}_{\tilde{\omega}}\tilde{\omega} \), then \(\tilde{\omega}_{k}=S^{\frac{1}{2}}_{\tilde{\omega}}\epsilon _{k} \). Let \(\{e_{k}\}_{k\in J} \) be an orthonormal basis for H, define an antiunitary operator \(\Lambda:H\longrightarrow W \) by
$$\Lambda f=\Lambda\biggl(\sum_{k \in J}c_{k}e_{k} \biggr)= \sum_{k \in J} \overline{c}_{k} \epsilon_{k}, \quad\mbox{for } f=\sum_{k \in J} c_{k}e_{k} \in H. $$
Obviously, the inverse of Λ is also an antiunitary operator and
$$\Lambda^{-1}g=\Lambda^{-1} \biggl(\sum _{k \in J}c_{k} \epsilon_{k} \biggr)=\sum _{k \in J}\overline{c}_{k} e_{k},\quad \forall g\in W. $$
Furthermore, the antiunitary operator Λ has the following property.

Lemma 2

Let Λ be defined as above, then \(\langle\Lambda f, g \rangle= \langle\Lambda ^{-1}g, f \rangle\) for any \(f\in H \) and \(g\in W \).

Proof

By the definition of Λ, one has
$$ \langle\Lambda f, g \rangle= \biggl\langle \sum_{k \in J} \langle e_{k}, f \rangle\epsilon_{k}, g \biggr\rangle \ = \sum_{k \in J} \langle e_{k}, f \rangle \langle \epsilon_{k}, g \rangle= \biggl\langle \sum _{k \in J} \langle\epsilon_{k}, g \rangle e_{k}, f \biggr\rangle = \bigl\langle \Lambda^{-1}g, f \bigr\rangle . $$
 □

Theorem 3

There exists an orthonormal basis \(\{e_{i}\}_{i\in J} \) such that \(\{n_{i}\}_{i\in J}\) is a Parseval frame if and only if there exists an antiunitary operator Λ such that \(S_{\omega} =\Lambda S\Lambda^{-1}\), where S is the frame operator of \(\{f_{i}\}_{i\in J} \).

Proof

By the definition of \(\{n_{i}\}_{i\in J}\) and Lemma 2, we have
$$\begin{aligned} \sum_{i \in J} \bigl\vert \langle f, n_{i} \rangle \bigr\vert ^{2} =& \sum _{i \in J} \biggl\vert \biggl\langle f, \sum _{k \in J} \langle e_{k}, f_{i} \rangle \tilde{\omega}_{k} \biggr\rangle \biggr\vert ^{2} \\ =&\sum_{i \in J} \biggl\vert \sum _{k \in J} \langle f_{i}, e_{k} \rangle \langle f, \tilde{\omega}_{k} \rangle \biggr\vert ^{2} \\ =&\sum_{k \in J}\sum _{\ell \in J} \biggl(\sum_{i \in J} \langle f_{i}, e_{k} \rangle \langle e_{\ell}, f_{i} \rangle \biggr) \langle f, \tilde{\omega}_{k} \rangle \langle\tilde{ \omega}_{\ell}, f \rangle \\ =&\sum_{k \in J}\sum _{\ell \in J} \langle e_{\ell}, Se_{k} \rangle \bigl\langle f, S^{\frac{1}{2}}_{\tilde{\omega}}\Lambda e_{k} \bigr\rangle \bigl\langle S^{\frac{1}{2}}_{\tilde{\omega}}\Lambda e_{\ell}, f \bigr\rangle \\ =& \sum_{k \in J}\sum _{\ell \in J} \langle e_{\ell}, Se_{k} \rangle \bigl\langle e_{k}, \Lambda^{-1}S^{\frac{1}{2}}_{\tilde {\omega}}f \bigr\rangle \bigl\langle \Lambda^{-1}S^{\frac{1}{2}}_{\tilde {\omega}}f, e_{\ell}\bigr\rangle \\ =& \sum_{k \in J} \bigl\langle e_{k}, \Lambda^{-1}S^{\frac {1}{2}}_{\tilde{\omega}}f \bigr\rangle \bigl\langle \Lambda^{-1}S^{\frac{1}{2}}_{\tilde {\omega}}f, Se_{k} \bigr\rangle \\ =& \biggl\langle \sum_{k \in J} \bigl\langle \Lambda^{-1}S^{\frac {1}{2}}_{\tilde{\omega}}f, Se_{k} \bigr\rangle e_{k} , \Lambda^{-1}S^{\frac{1}{2}}_{\tilde{\omega}}f \biggr\rangle \\ =& \bigl\langle S\Lambda^{-1}S^{\frac{1}{2}}_{\tilde{\omega}}f, \Lambda^{-1}S^{\frac{1}{2}}_{\tilde{\omega}}f \bigr\rangle \\ =& \bigl\langle f, S^{\frac{1}{2}}_{\tilde{\omega}} \Lambda S\Lambda ^{-1}S^{\frac{1}{2}}_{\tilde{\omega}}f \bigr\rangle . \end{aligned}$$
(3.1)
Suppose \(\{n_{i}\}_{i\in J}\) is a Parseval frame; then we have
$$ \sum_{i \in J}\bigl| \langle f, n_{i} \rangle\bigr|^{2}=\Vert {f}\Vert ^{2},\quad \forall f\in W . $$
(3.2)
By (3.1), it becomes
$$ \sum_{i \in J} \bigl\vert \langle f, n_{i} \rangle \bigr\vert ^{2}= \bigl\langle f, S^{\frac{1}{2}}_{\tilde{\omega}} \Lambda S \Lambda^{-1}S^{\frac{1}{2}}_{\tilde{\omega}}f \bigr\rangle = \langle f, f \rangle. $$
(3.3)
For arbitrary complex numbers a and b, we have
$$\Lambda S\Lambda^{-1}(af+bg)= \Lambda S \bigl(\overline{a} \Lambda^{-1} f + \overline{b}\Lambda^{-1} g \bigr)=a\Lambda S \Lambda^{-1} f+b\Lambda S\Lambda^{-1} g. $$
Thus \(\Lambda S\Lambda^{-1} \) is a linear operator, so is the operator \(S^{\frac{1}{2}}_{\tilde{\omega}} \Lambda S\Lambda^{-1}S^{\frac{1}{2}}_{\tilde{\omega}} \). This means \(S^{\frac{1}{2}}_{\tilde{\omega}} \Lambda S\Lambda^{-1}S^{\frac{1}{2}}_{\tilde{\omega}}=I \) by (3.3), i.e.
$$S_{\omega}=\Lambda S \Lambda^{-1}. $$
On the other hand, assume there exists an antiunitary operator Λ such that \(S_{\omega}=\Lambda S\Lambda^{-1} \). Define \(e_{j}=\Lambda^{-1}\epsilon_{j}=\Lambda^{-1} S^{-\frac{1}{2}}_{\omega}\tilde{\omega}_{k} \), then (3.1) means
$$n_{i}=\sum_{k \in J} \langle e_{k}, f_{i} \rangle \tilde{\omega_{k}} $$
is a Parseval frame. □

Theorem 4

Suppose \(\{f_{i}\}_{i\in J} \) is a frame for a separable Hilbert space H and \(\{\omega_{j}\}_{j\in J} \) is a Riesz sequence in H. \(\{f_{i}\}_{i\in J} \) is an R-dual of \(\{\omega_{j}\}_{j\in J} \) if and only if the following two conditions hold:
  1. (i)

    there exists an antiunitary operator Λ s.t. \(S_{w}=\Lambda S\Lambda^{-1} \);

     
  2. (ii)

    \(\dim(\ker T)=\dim(W^{\perp}) \).

     

Proof

By Proposition 1, \(\{f_{i}\}_{i \in J} \) is an R-dual of \(\{\omega_{j}\}_{j\in J} \) if and only if \(\{n_{i}\}_{i\in J}\) can be dilated to an orthonormal basis for H. By Theorem 2, this is equivalent to \(\{n_{i}\}_{i\in J}\) being a Parseval frame and (ii) holding. Using Theorem 3, we see that \(\{f_{i}\}_{i\in J} \) is an R-dual of \(\{\omega_{j}\}_{j\in J} \) if and only (i) and (ii) hold. □

We appreciate one reviewer having pointed out that Theorem 4 is of exactly the same type as the characterizations of type II/III in [2]. In the special case, if \(\{f_{i}\}_{i\in\mathbb{N}} \) is an A-tight frame for a separable Hilbert space H with infinite dimension and \(\{\omega_{j}\}_{j\in\mathbb{N}} \) is an A-tight Riesz sequence where denotes the natural number set, then there must be an antiunitary operator Λ form H onto W. So we have \(S=AI_{H} \), \(S_{W}=AI_{W} \), and
$$S_{W}=AI_{W}=\Lambda A I_{H} \Lambda^{-1}= \Lambda S \Lambda^{-1}. $$
Thus the condition (i) of Theorem 4 holds automatically. And we get the following corollary, first given in [2].

Corollary 1

[2] Let \(\{f_{i}\}_{i\in J} \) be a tight frame for H and let \(\{\omega_{j}\}_{j\in J} \) be a tight Riesz sequence in H with the same bound. Denote the synthesis operator for \(\{f_{i}\}_{i\in J} \) by T. Then \(\{\omega_{j}\}_{j\in J} \) is an R-dual of \(\{f_{i}\} _{i\in J}\) if and only if \(\dim(\ker T)=\dim(W^{\perp}) \) holds.

Remark 1

Since \(S_{W}f=\sum_{j\in J} \langle f, \omega_{j} \rangle \omega_{j}\) and
$$\Lambda S \Lambda^{-1}f= \sum_{j \in J} \bigl\langle f_{j}, \Lambda^{-1}f \bigr\rangle \Lambda f_{j}=\sum_{j \in J} \langle f, \Lambda f_{j} \rangle\Lambda f_{j}, $$
(i) of Theorem 4 is equivalent to there existing an antiunitary operator such that
$$\sum_{j \in J} \langle f, \omega_{j} \rangle \omega_{j} =\sum_{j \in J} \langle f, \Lambda f_{j} \rangle\Lambda f_{j}. $$

Remark 2

For parameters \(a, b \in \mathbb{R}\), define the operators \(T_{a} \) and \(E_{b} \) on \(L_{2}(\mathbb{R})\)by \(T_{a}f(x)=f(x-a) \) and \(E_{b}f(x)=e^{2\pi ibx}f(x) \), respectively. From [8], we know that if \(ab<1 \) and \(\{E_{mb}T_{na}g\}_{m,n\in\mathbb{Z}} \) is a frame, then \(\{E_{mb}T_{na}g\}_{m,n\in\mathbb{Z}} \) has an infinite excess. If \(ab>1 \), then \(\{E_{mb}T_{na}\}_{m,n\in \mathbb{Z}} \) has an infinite deficit. This demonstrates that, if we want to solve the open problem, we only need (i) of Theorem 4 to hold. By Remark 1, this is equivalent to finding an antiunitary operator Λ such that
$$\sum_{m,n} \langle f, \Lambda E_{mb}T_{na}g \rangle\Lambda E_{mb}T_{na}g =\sum _{m,n} \biggl\langle f, \frac{1}{\sqrt{ab}} E_{m/a}T_{n/b}g \biggr\rangle \frac{1}{\sqrt{ab}} E_{m/a}T_{n/b}g. $$

Declarations

Acknowledgements

This study was partially supported by the National Science Foundation for Young Scientists of China (Grant No. 11401433). The authors would like to express their thanks to the reviewers for their helpful comments and suggestions.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
College of Mathematics and Information Science, North China University of Water Resources and Electric Power, Zhengzhou, 450011, P.R. China
(2)
Department of Mathematics, Tianjin Polytechnic University, Tianjin, 300387, P.R. China

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