On equivalent conditions of two sequences to be R-dual

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 $\{f_{j}\}_{j\in J}$.

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:

• (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)

• (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 $m_{i}\in W^{\perp}$.

• (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:

• (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 $\{h_{i}\}_{i\in J}$.

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

• (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:

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

• (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 ℕ 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

$$\sum_{m,n} \langle f, \Lambda E_{mb}T_{na}g \rangle\Lambda E_{mb}T_{na}g =\sum _{m,n} \biggl\langle f, \frac{1}{\sqrt{ab}} E_{m/a}T_{n/b}g \biggr\rangle \frac{1}{\sqrt{ab}} E_{m/a}T_{n/b}g. $$