Cross-Gram Matrix Associated to Two Sequences in Hilbert Spaces

Abstract

The conditions for sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ being Bessel sequences, frames or Riesz bases, can be expressed in terms of the so-called cross-Gram matrix. In this paper, we investigate the cross-Gram operator G, associated to the sequence $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$ and sufficient and necessary conditions for boundedness, invertibility, compactness and positivity of this operator are determined depending on the associated sequences. We show that invertibility of G is not possible when the associated sequences are frames but not Riesz Bases or at most one of them is Riesz basis. In the special case, we prove that G is a positive operator when $\{g_{k}\}_{k=1}^{\infty }$ is the canonical dual of $\{f_{k}\}_{k=1}^{\infty }$.

1 Introduction

The fundamental operators in frame theory are the synthesis, analysis and frame operators associated with a given frame. The ability of combining these operators to make a sensitive operator is indeed essential in frame theory and its applications. Time-invariant filter’s, i. e. convolution operators, are used frequently in applications. These operators can be called Fourier multipliers Benyi et al. (2005), Feichtinger and Narimani (2006). In the last decade Gabor filters which are beneficial tools to perform time-variant filters have very strong applications in psychoacoustics Balazs et al. (2010), computational auditory scene analysis Wang and Brown (2006), and seismic data analysis Margrave et al. (2005). For more information on these operators we refer to Balazs (2007), Faroughi et al. (2013), Stoeva and Balasz (2012).

In this paper for given two Bessel sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$, the synthesis operator of the sequence $\{f_{k}\}_{k=1}^{\infty }$ with the analysis operator of the sequence $\{g_{k}\}_{k=1}^{\infty }$ is composed and a fundamental operator is generated. This operator is called the cross-Gram operator associated with the sequence $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$ Balazs (2008), Pekalska and Duin (2005). This paper concerns this question that when can sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ in a Hilbert space H generate a cross-Gram operator with the properties of boundedness, invertibility and positivity. Vise versa if the cross-Gram operator has the above properties what can be expected of the sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$.

Let H be a complex Hilbert space. A frame for H is a sequence $\{f_{k}\}_{k=1}^{\infty }\subset H$ such that there are positive constants A and B satisfying

$$\begin{aligned} A\Vert f\Vert ^{2}\le \sum _{k=1}^{\infty }|\langle f, f_{k}\rangle |^{2}\le B\Vert f\Vert ^{2}, \quad f\in H. \end{aligned}$$

(1)

The constants A and B are called lower and upper frame bounds, respectively. We call $\{f_{k}\}_{k=1}^{\infty }$ a Bessel sequence with bound B, if we have only the second inequality in (1). Associated with each Bessel sequence $\{f_{k}\}_{k=1}^{\infty }$ we have three linear and bounded operators, the synthesis operator:

$$\begin{aligned} T:\ell ^{2}(N)\rightarrow H,\ \ T(\{c_{k}\}_{k=1}^{\infty })=\sum _{k=1}^{\infty }c_{k}f_{k}, \end{aligned}$$

the analysis operator which is defined by:

$$\begin{aligned} T^{*}:H\rightarrow \ell ^{2}(N);\ \ T^{*}f=\{\langle f, f_{k}\rangle \}_{k=1}^{\infty }, \end{aligned}$$

and the frame operator:

$$\begin{aligned} S:H\rightarrow H;\ \ Sf=TT^{*}f=\sum _{k=1}^{\infty }\langle f, f_{k}\rangle f_{k}. \end{aligned}$$

If $\{f_{k}\}_{k=1}^{\infty }$ is a Bessel sequence, we can compose the synthesis operator T and its adjoint $T^{*}$ to obtain the bounded operator

$$\begin{aligned} T^{*}T:\ell ^{2}(N)\rightarrow \ell ^{2}(N);\ \ T^{*}T\{c_{k}\}_{k=1}^{\infty }=\left\{ \left\langle \sum _{\ell =1}^{\infty }c_{\ell }f_{\ell }, f_{k}\right\rangle \right\} _{k=1}^{\infty }. \end{aligned}$$

Therefore the matrix representation of $T^{*}T$ is as follows:

$$\begin{aligned} T^{*}T=\{\langle f_{k}, f_{j}\rangle \}_{j, k=1}^{\infty }. \end{aligned}$$

The matrix $\{\langle f_{k}, f_{j}\rangle \}_{j, k=1}^{\infty }$ is called the matrix associated with $\{f_{k}\}_{k=1}^{\infty }$ or Gram matrix and it defines a bounded operator on $\ell ^{2}(N)$ when $\{f_{k}\}_{k=1}^{\infty }$ is a Bessel sequence.

In order to recognize that a sequence $\{f_{k}\}_{k=1}^{\infty }$ is a Bessel sequence or frame, we need to check (1) for all $f\in H$. But in practice this is not always so easy. The following two results give us a practical method to diagnose Bessel sequences or frames by the concept of Gram matrix or in other words just by calculating $\{\langle f_{k}, f_{j}\rangle \}_{j, k=1}^{\infty }$.

Lemma 1

Christensen (2016) Suppose that$\{f_{k}\}_{k=1}^{\infty }\subseteq H$. Then the following statements are equivalent:

1.:

$\{f_{k}\}_{k=1}^{\infty }$is a Bessel sequence with boundB.

2.:

The Gram matrix associated with$\{f_{k}\}_{k=1}^{\infty }$defines a bounded operator on$\ell ^{2}(N)$with norm at mostB.

Definition 1

A Riesz basis $\{f_{k}\}_{k=1}^{\infty }$ for H is a family of the form $\{Ue_{k}\}_{k=1}^{\infty }$, where $\{e_{k}\}_{k=1}^{\infty }$ is an orthonormal basis for H and $U:H\rightarrow H$ is a bounded bijective operator.

Proposition 1

Christensen (2016) A sequence$\{f_{k}\}_{k=1}^{\infty }$is a Riesz basis forHif and only if it is an unconditional basis forHand

$$\begin{aligned} 0<\inf \Vert f_{k}\Vert \le \sup \Vert f_{k}\Vert <\infty . \end{aligned}$$

Theorem 1

Christensen (2016) Suppose that$\{f_{k}\}_{k=1}^{\infty }\subseteq H$. Then the following conditions are equivalent:

1.:

$\{f_{k}\}_{k=1}^{\infty }$is a Riesz basis forH.

2.:

$\{f_{k}\}_{k=1}^{\infty }$is complete and its Gram matrix$\{\langle f_{k}, f_{j}\rangle \}_{j, k=1}^{\infty }$defines a bounded, invertible operator on$\ell ^{2}(N)$.

2 Cross-Gram Matrix

If $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are Bessel sequences, we compose the synthesis operator of the sequence $\{f_{k}\}_{k=1}^{\infty }$, $T_{f_{k}}$, and the analysis operator of the sequence $\{g_{k}\}_{k=1}^{\infty }$, $T^{*}_{g_{k}}$, to obtain a bounded operator on $\ell ^{2}(N)$

$$\begin{aligned} T^{*}_{g_{k}}T_{f_{k}}:\ell ^{2}(N)\rightarrow \ell ^{2}(N);\ \ T^{*}_{g_{k}}T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }=\left\{ \left\langle \sum _{\ell =1}^{\infty }c_{\ell }f_{\ell }, g_{k}\right\rangle \right\} _{k=1}^{\infty }. \end{aligned}$$

This operator, $G=T^{*}_{g_{k}}T_{f_{k}}$, is called the cross-Gram operator associated to $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$ Balazs (2008), Pekalska and Duin (2005).

If $\{e_{k}\}_{k=1}^{\infty }$ is the canonical orthonormal basis for $\ell ^{2}(N)$, the jk-th entry in the matrix representation for $T^{*}_{g_{k}}T_{f_{k}}$ is

$$\begin{aligned} \langle T^{*}_{g_{k}}T_{f_{k}}e_{k}, e_{j}\rangle =\langle T_{f_{k}}e_{k}, T_{g_{k}}e_{j}\rangle =\langle f_{k}, g_{j}\rangle . \end{aligned}$$

Therefore the matrix representation of $T^{*}_{g_{k}}T_{f_{k}}$ is as follows:

$$\begin{aligned} T^{*}_{g_{k}}T_{f_{k}}=\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }. \end{aligned}$$

The matrix $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$ is called the cross-Gram matrix associated to $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ Balazs (2008), Pekalska and Duin (2005). In the special case that the sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are biorthogonal, the cross-Gram matrix is the identity matrix.

The above discussion shows that the cross-Gram matrix is bounded above if $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are Bessel sequences. The following example shows that the inverse of the above assertion is not valid, in other words, the cross-Gram matrix associated to two sequences can be well-defined and bounded in the case that one of the sequences is not Bessel.

Example 1

Suppose that $\{e_{k}\}_{k=1}^{\infty }$ is the orthonormal basis for a Hilbert space H. Consider $\{f_{k}\}_{k=1}^{\infty }=\{\frac{1}{k}e_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }=\{ke_{k}\}_{k=1}^{\infty }$. A simple calculation shows that the cross-Gram matrix associated to these sequences is the identity matrix, but $\{g_{k}\}_{k=1}^{\infty }$ is not a Bessel sequence.

Definition 2

Let U be an operator on a Hilbert space H, and suppose that E is an orthonormal basis for H. We say that U is a Hilbert–Schmidt operator if

$$\begin{aligned} \Vert U\Vert _{2}=\left( \sum _{x\in E}\Vert Ux\Vert ^{2}\right) ^{\frac{1}{2}}<\infty . \end{aligned}$$

Theorem 2

Suppose that$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are sequences inHand$\{g_{k}\}_{k=1}^{\infty }$is a Bessel sequence with bound$B'$. Assume that there exists$M>0$such that$\sum _{k=1}^{\infty }\Vert f_{k}\Vert ^{2}\le M.$Then the cross-Gram operator associated to$\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$is a well-defined, bounded and compact operator.

Proof

Suppose that $G=\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$. For a given sequence $\{c_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ we have

$$\begin{aligned} \Vert G\{c_{k}\}_{k=1}^{\infty }\Vert ^{2}= & {} \sum _{j=1}^{\infty }\left| \sum _{k=1}^{\infty }c_{k}\langle f_{k}, g_{j}\rangle \right| ^{2}\\\le & {} \sum _{j=1}^{\infty }\sum _{k=1}^{\infty }|c_{k}|^{2}\sum _{k=1}^{\infty }|\langle f_{k}, g_{j}\rangle |^{2}\\= & {} \sum _{k=1}^{\infty }|c_{k}|^{2}\sum _{k=1}^{\infty }\sum _{j=1}^{\infty }|\langle f_{k}, g_{j}\rangle |^{2}\\\le & {} B'\sum _{k=1}^{\infty }|c_{k}|^{2}\sum _{k=1}^{\infty }\Vert f_{k}\Vert ^{2}\\\le & {} B'M\sum _{k=1}^{\infty }|c_{k}|^{2}. \end{aligned}$$

By above assertion, $G\{c_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ and therefore G is well-defined and bounded.

Now suppose that $\{e_{k}\}_{k=1}^{\infty }$ is the canonical orthonormal basis for $\ell ^{2}(N)$. Then

$$\begin{aligned} \left( \sum _{k=1}^{\infty }\Vert G(e_{k})\Vert ^{2}\right) ^{\frac{1}{2}}= & {} \left( \sum _{k=1}^{\infty }\sum _{j=1}^{\infty }|\langle f_{k}, g_{j}\rangle |^{2}\right) ^{\frac{1}{2}}\\\le & {} \sqrt{B'}\left( \sum _{k=1}^{\infty }\Vert f_{k}\Vert ^{2}\right) ^{\frac{1}{2}}\le \sqrt{B'M}. \end{aligned}$$

Therefore G is a Hilbert–Schmidt operator and so is compact Pedersen (1999).

Example 2

Let $\{e_{k}\}_{k=1}^{\infty }$ be an orthonormal basis for H. Consider $\{f_{k}\}_{k=1}^{\infty }=\{e_{1},\frac{1}{2}e_{2}, \frac{1}{3}e_{3}, \frac{1}{4}e_{4},\ldots \}$ and $\{g_{k}\}_{k=1}^{\infty }=\{\frac{1}{2}e_{1}, e_{2}, \frac{1}{2^{2}}e_{1}, e_{3},\ldots \}.$ Suppose that G is the cross-Gram operator associated to $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$. A simple calculation shows that

$$\begin{aligned} \sum _{k=1}^{\infty }\Vert G(e_{k})\Vert ^{2}=\sum _{k=1}^{\infty }\frac{1}{2^{2k}}+ \sum _{k=1}^{\infty }\frac{1}{k^{2}}. \end{aligned}$$

Therefore G is a Hilbert–Schmidt operator and so is compact.

If $\sup _{k}\Vert f_{k}\Vert <\infty$ (resp. $\inf _{k}\Vert f_{k}\Vert >0$), the sequence $\{f_{k}\}_{k=1}^{\infty }$ will be called norm-bounded above or NBA, (resp. norm-bounded below or NBB).

Theorem 3

Let$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$besequences forHandGbe the cross-Gram operator associated to$\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$. Then the following statements are satisfied:

1.:

Assume thatGis well-defined and bounded above and$\{g_{k}\}_{k=1}^{\infty }$is a frame with lower boundA. Then$\{f_{k}\}_{k=1}^{\infty }$is norm-bounded above.

2.:

Assume thatGis well-defined and bounded below and$\{g_{k}\}_{k=1}^{\infty }$is a Bessel sequence with upper boundB. Then$\{f_{k}\}_{k=1}^{\infty }$is norm-bounded below.

3.:

Assume thatGis well-defined and bounded above and$\{f_{k}\}_{k=1}^{\infty }$is an orthonormal basis forH. Then$\{g_{k}\}_{k=1}^{\infty }$is a Bessel sequence forH.

Proof

• 1. Since G is well-defined and bounded, there exists a constant $M>0$ such that

  $$\begin{aligned} \sum _{j=1}^{\infty }\left| \sum _{k=1}^{\infty }c_{k}\langle f_{k}, g_{j}\rangle |^{2}\le M\sum _{k=1}^{\infty }|c_{k}\right| ^{2}. \end{aligned}$$

  (2)

  Via (2) applied to the elements of the canonical orthonormal basis of $\ell ^{2}(N)$, we have

  $$\begin{aligned} \sum _{j=1}^{\infty }|\langle f_{k}, g_{j}\rangle |^{2}\le M,\ \ k\in N. \end{aligned}$$

  (3)

  Since $\{g_{k}\}_{k=1}^{\infty }$ is a frame for H, by (3) we have

  $$\begin{aligned} A\Vert f_{k}\Vert ^{2}\le \sum _{j=1}^{\infty }|\langle f_{k}, g_{j}\rangle |^{2}\le M,\ \ \ell \in N. \end{aligned}$$

  (4)

  Therefore

  $$\begin{aligned} \Vert f_{k}\Vert ^{2}\le \frac{M}{A},\ \ k\in N. \end{aligned}$$

• 2. By assumption there exist $M'>0$, such that

  $$\begin{aligned} M'\sum _{k=1}^{\infty }|c_{k}|^{2}\le \sum _{j=1}^{\infty }\left| \sum _{k=1}^{\infty }c_{k}\langle f_{k}, g_{j}\rangle \right| ^{2}. \end{aligned}$$

  (5)

  Similar to above discussion, since $\{g_{k}\}_{k=1}^{\infty }$ is a Bessel sequence, for each $k\in N$, via (5) applied to the elements of the canonical orthonormal basis of $\ell ^{2}(N)$, we have

  $$\begin{aligned} M' \le \sum _{j=1}^{\infty }|\langle f_{k}, g_{j}\rangle |^{2}\le B\Vert f_{k}\Vert ^{2}. \end{aligned}$$

  (6)

  Therefore

  $$\begin{aligned} \Vert f_{k}\Vert ^{2}\ge \frac{M'}{B},\ \ k\in N. \end{aligned}$$

• 3. Since G is well-defined and bounded above there exists $M''>0$, such that

  $$\begin{aligned} \sum _{j=1}^{\infty }\left| \sum _{k=1}^{\infty }c_{k}\langle f_{k}, g_{j}\rangle |^{2}\le M''\sum _{k=1}^{\infty }\right| c_{k}|^{2}. \end{aligned}$$

  (7)

  Since $\{f_{k}\}_{k=1}^{\infty }$ is an orthonormal basis for H, there exist a sequence $\{c_{k}\}\in \ell ^{2}(N)$, such that for each $f\in H$ we have

  $$\begin{aligned} \sum _{j=1}^{\infty }|\langle f, g_{j}\rangle |^{2}=\sum _{j=1}^{\infty }\left| \left\langle \sum _{k=1}^{\infty }c_{k}f_{k}, g_{j}\right\rangle \right| ^{2}. \end{aligned}$$

  Now by (7) we have

  $$\begin{aligned} \sum _{j=1}^{\infty }\left| \sum _{k=1}^{\infty }c_{k}\langle f_{k}, g_{j}\rangle \right| ^{2}\le M''\sum _{k=1}^{\infty }|c_{k}|^{2}, \end{aligned}$$

  Therefore

  $$\begin{aligned} \sum _{j=1}^{\infty }|\langle f, g_{j}\rangle |^{2}\le M''\sum _{k=1}^{\infty }|c_{k}|^{2}=M''\Vert f\Vert ^{2}. \end{aligned}$$

Proposition 2

Suppose that$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are Bessel sequences andGis the cross-Gram operator associated to$\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$. Then the following statements are satisfied:

1.:

If$\{f_{k}\}_{k=1}^{\infty }$is a Riesz basis and$\{g_{k}\}_{k=1}^{\infty }$is a frame forH, thenGis a bounded injective operator.

2.:

If$\{f_{k}\}_{k=1}^{\infty }$is a frame and$\{g_{k}\}_{k=1}^{\infty }$is a Riesz basis forH, thenGis a bounded surjective operator.

Proof

Since $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are Bessel sequences, we deduce that G is a well-defined and bounded operator.

• 1. Suppose that

  $$\begin{aligned} G\{c_{k}\}_{k=1}^{\infty }=G\{b_{k}\}_{k=1}^{\infty },\ \ \{c_{k}\}_{k=1}^{\infty }, \{b_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N). \end{aligned}$$

  Then $T^{*}_{g_{k}}T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }=T^{*}_{g_{k}}T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }.$ Since $\{g_{k}\}_{k=1}^{\infty }$ is a frame for H, $T^{*}_{g_{k}}$ is an injective operator and we have $T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }=T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }.$ Since $\{f_{k}\}_{k=1}^{\infty }$ is a Riesz basis, $T_{f_{k}}$ is invertible. So $\{c_{k}\}_{k=1}^{\infty }=\{b_{k}\}_{k=1}^{\infty }$ and we get the result.

• 2. Since $\{g_{k}\}_{k=1}^{\infty }$ is a Riesz basis, $T^{*}_ {g_{k}}$ is a bijective operator. Therefore for a given sequence $\{c_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$, there exist $h\in H$ such that $T^{*}_{g_{k}}h=\{c_{k}\}_{k=1}^{\infty }.$ Also by assumption $T_{f_{k}}$ is a surjective operator and there exists $\{b_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ such that $T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }=h.$ Therefore we have $T^{*}_{g_{k}}T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }=\{c_{k}\}_{k=1}^{\infty }$ and so $G\{b_{k}\}_{k=1}^{\infty }=\{c_{k}\}_{k=1}^{\infty }.$

Theorem 4

Suppose that$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are Riesz bases forH. Then$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are complete and the cross-Gram matrix associated to$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$defines a bounded invertible operator on$\ell ^{2}(N)$.

Proof

Since $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are Riesz bases, there exist bijective operators U and W such that $\{f_{k}\}_{k=1}^{\infty }=\{Ue_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }=\{We_{k}\}_{k=1}^{\infty }$, where $\{e_{k}\}_{k=1}^{\infty }$ is an orthonormal basis of H. For every $k, j\in N$ we have

$$\begin{aligned} \langle f_{k}, g_{j}\rangle =\langle Ue_{k}, We_{j}\rangle =\langle W^{*}Ue_{k}, e_{j}\rangle . \end{aligned}$$

i.e., the cross-Gram matrix associated to $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ representing the bounded invertible operator $W^{*}U$ in the basis $\{e_{k}\}_{k=1}^{\infty }$.

If the cross-Gram matrix associated to the sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ is invertible and the sequence $\{f_{k}\}_{k=1}^{\infty }$ is a Bessel sequence, then there is no need to sequence $\{g_{k}\}_{k=1}^{\infty }$ to be a Bessel sequence, see Example 1. Having in mind this result, now what can we say about the inverse of Theorem 4? By having the assumption of the invertibility of the cross-Gram matrix associated to $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ and completeness of the sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$, by Example 1, we deduce that there is no need to sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ to be Riesz bases. But what can we say in the case that $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are frames? In the following theorems we answer to this question by considering the assumption of being frame of both sequences.

Theorem 5

Suppose that$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are frames forHand the cross-Gram matrix associated to these sequences is bounded and invertible. Then$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are Riesz bases forH.

Proof

Suppose that G is the cross-Gram operator associated to $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$. So we have

$$\begin{aligned} G=T^{*}_{g_{k}}T_{f_{k}}. \end{aligned}$$

Since $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are frames for H, $T_{f_{k}}$ and $T_{g_{k}}$ are bounded and surjective operators. Now we want to show that $T_{f_{k}}$ is an injective operator. For the given sequences $\{c_{k}\}_{k=1}^{\infty }$, $\{b_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$, suppose that

$$\begin{aligned} T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }=T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }. \end{aligned}$$

Then we have

$$\begin{aligned} T^{*}_{g_{k}}T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }=T^{*}_{g_{k}} T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }. \end{aligned}$$

So

$$\begin{aligned} G\{c_{k}\}_{k=1}^{\infty }=G\{b_{k}\}_{k=1}^{\infty }. \end{aligned}$$

Since G is an invertible operator, we deduce that $\{c_{k}\}_{k=1}^{\infty }=\{b_{k}\}_{k=1}^{\infty }$ and therefore $T_{f_{k}}$ is an injective operator and so $\{f_{k}\}_{k=1}^{\infty }$ is a Riesz basis for H.

Now we want to show that $T_{g_{k}}$ is also a bijective operator. Since $N(T_{g_{k}})=R(T^{*}_{g_{k}})^{\perp }$, it is enough to show that $T^{*}_{g_{k}}: H\rightarrow \ell ^{2}(N)$ is a surjective operator. Since G is invertible, for a given sequence $\{c_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ there exists a sequence $\{b_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ such that

$$\begin{aligned} G\{b_{k}\}_{k=1}^{\infty }=\{c_{k}\}_{k=1}^{\infty }. \end{aligned}$$

So

$$\begin{aligned} T^{*}_{g_{k}}T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }=\{c_{k}\}_{k=1}^{\infty }, \end{aligned}$$

which shows that $T^{*}_{g_{k}}$ is a surjective operator. Therefore $\{g_{k}\}_{k=1}^{\infty }$ is a Riesz basis for H.

Corollary 1

Suppose that$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are frames but not Riesz bases, thenGcannot be invertible.

Example 3

Suppose that $\{e_{k}\}_{k=1}^{\infty }$ is an orthonormal basis for H. Consider the sequences $\{f_{k}\}_{k=1}^{\infty }=\{e_{1}, e_{1}, e_{2}, e_{3}, e_{4},\ldots \}$ and $\{g_{k}\}_{k=1}^{\infty }=\{e_{1}, e_{1}, e_{2}, e_{2}, e_{3}, e_{3},\ldots \}$. A simple calculation shows that these sequences are frames but not Riesz bases. We get the cross-Gram matrix associated to sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$:

$$\begin{aligned} G=\left[ \begin{array}{lllllll} 1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \cdots \\ 1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0 &{}\quad \cdots \\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad \cdots \\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad \cdots \\ 0&{}\quad 0\quad &{}\quad 0&{}\quad 1&{}\quad 0&{}\quad \cdots \\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \cdots \\ \end{array} \right] , \end{aligned}$$

We obtain that $\det ({G})=0$ and so G is not invertible.

Theorem 6

Suppose that $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are Bessel sequences for H and the cross-Gram matrix, associated to these sequences is bounded and invertible. Then the following statements are satisfied:

1.:

If$\{f_{k}\}_{k=1}^{\infty }$is a Riesz basis forH, then$\{g_{k}\}_{k=1}^{\infty }$is a Riesz basis forH.

2.:

If$\{f_{k}\}_{k=1}^{\infty }$is a frame forHand$\{g_{k}\}_{k=1}^{\infty }$is complete inH, then$\{g_{k}\}_{k=1}^{\infty }$is a frame forH.

Proof

• 1. Suppose that G is the cross-Gram operator associated to $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$. Then we have

  $$\begin{aligned} G=T^{*}_{g_{k}}T_{f_{k}}. \end{aligned}$$

  Since $\{f_{k}\}_{k=1}^{\infty }$ is a Riesz basis, $T_{f_{k}}$ has a bounded inverse and we can write

  $$\begin{aligned} T^{*}_{g_{k}}=G(T_{f_{k}}^{-1}). \end{aligned}$$

  Therefore $T^{*}_{g_{k}}$ is an invertible operator and we deduce that $\{g_{k}\}_{k=1}^{\infty }$ is a Riesz basis for H.

• 2. In order to show that $\{g_{k}\}_{k=1}^{\infty }$ is a frame for H it is enough to prove that $T_{g_{k}}$ is a surjective operator. Since $\{g_{k}\}_{k=1}^{\infty }$ is complete in H, we need to show that $T^{*}_{g_{k}}$ is injective. Suppose that

  $$\begin{aligned} T^{*}_{g_{k}}(f_{1})=T^{*}_{g_{k}}(f_{2}),\ \ f_{1}, f_{2}\in H. \end{aligned}$$

  Since $T_{f_{k}}$ is a surjective operator, there exist sequences $\{c_{k}\}_{k=1}^{\infty }$, $\{b_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ such that $f_{1}=T_{f_{k}}\{c_{k}\}_{k=1}^{\infty },\ \ f_{2}=T_{f_{k}}\{b_{k}\}_{k=1}^{\infty },$ and therefore we can write

  $$\begin{aligned} T^{*}_{g_{k}}T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }=T^{*}_{g_{k}}T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }. \end{aligned}$$

  Now by the invertibility of G we deduce that $\{c_{k}\}_{k=1}^{\infty }=\{b_{k}\}_{k=1}^{\infty },$ and so $T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }=T_{f_{k}}\{b_{k}\}_{k=1}^{\infty }.$ Hence we get the proof.

By changing the role of the sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ in above theorem we deduce the same results.

Corollary 2

Suppose that$\{f_{k}\}_{k=1}^{\infty }$is a Riesz basis and$\{g_{k}\}_{k=1}^{\infty }$is a Bessel sequence. Then the following statements are satisfied:

1.:

If$\{g_{k}\}_{k=1}^{\infty }$is not a frame, thenGcannot be invertible.

2.:

If$\{g_{k}\}_{k=1}^{\infty }$is not NBA or NBB, thenGcannot be invertible.

3 Dual Frames Associated to Cross-Gram Matrix

In this section, we investigate the cases when $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ are a pair of dual frames. Recall that for the Bessel sequences $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$, the pair $(\{f_{k}\}_{k=1}^{\infty }, \{g_{k}\}_{k=1}^{\infty })$ is a dual pair if for any $f\in H$ one of the following equivalent conditions holds:

• 1. $f=\sum _{k=1}^{\infty }\langle f, g_{k}\rangle f_{k}, f\in H.$

• 2. $f=\sum _{k=1}^{\infty }\langle f, f_{k}\rangle g_{k}, f\in H.$

• 3. $\langle f, g\rangle =\sum _{k=1}^{\infty }\langle f, f_{k}\rangle \langle g_{k}, g\rangle ,\ \ f, g\in H.$

Theorem 7

Suppose that$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$are Bessel sequences and$\langle f, g_{k}\rangle \ne 0$for each$k\in N$and$f\in H$. Assume thatG, the cross-Gram operator associated to$\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$, is a well-defined and bounded operator with boundMsuch that$0<M<1$. Then$\{f_{k}\}_{k=1}^{\infty }$and$\{g_{k}\}_{k=1}^{\infty }$cannot be a pair of dual frames.

Proof

Suppose that $\{f_{k}\}_{k=1}^{\infty }$ is a dual frame of $\{g_{k}\}_{k=1}^{\infty }$. Then for every $f\in H$,

$$\begin{aligned} f=\sum _{k=1}^{\infty }\langle f, g_{k}\rangle f_{k}. \end{aligned}$$

(8)

Since G is a bounded operator with bound M, for $\{c_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ we have

$$\begin{aligned} \sum _{j=1}^{\infty }\left| \sum _{k=1}^{\infty }c_{k}\langle f_{k}, g_{j}\rangle \right| ^{2}\le M\sum _{k=1}^{\infty }|c_{k}|^{2}, \end{aligned}$$

(9)

Now for each $f\in H$, by (8) and (9) we have

$$\begin{aligned} \sum _{j=1}^{\infty }|\langle f, g_{j}\rangle |^{2}= & {} \sum _{j=1}^{\infty }\left| \left\langle \sum _{k=1}^{\infty }\langle f, g_{k}\rangle f_{k}, g_{j}\right\rangle \right| ^{2}\\= & {} \sum _{j=1}^{\infty }\left| \sum _{k=1}^{\infty }\langle f, g_{k}\rangle \langle f_{k}, g_{j}\rangle \right| ^{2}\le M\sum _{k=1}^{\infty }|\langle f, g_{k}\rangle |^{2}, \end{aligned}$$

which is a contradiction. Therefore we get the proof.

Example 4

Let $\{e_{k}\}_{k=1}^{\infty }$ be an orthonormal basis for H. Consider $\{f_{k}\}_{k=1}^{\infty }=\{e_{1}, e_{2}, e_{3},e_{4},\ldots \}$ and $\{g_{k}\}_{k=1}^{\infty }=\{\frac{1}{2}e_{1}, \frac{1}{2}e_{1}, \frac{1}{3}e_{1}, \frac{1}{4}e_{1},\ldots \}$. Suppose that G is the cross-Gram operator associated to $\{\langle f_{k}, g_{j}\rangle \}_{j, k=1}^{\infty }$. Then G is a well-defined and bounded operator with bound $\sqrt{\frac{89}{100}}$. Suppose that $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ be a pair of dual frames. Then we have $e_{2}=\frac{1}{2}e_{1},$ which is a contradiction. Therefore $\{f_{k}\}_{k=1}^{\infty }$ and $\{g_{k}\}_{k=1}^{\infty }$ cannot be a dual pair.

Theorem 8

Suppose that$\{f_{k}\}_{k=1}^{\infty }$is a frame forHand$\{S^{-1}f_{k}\}_{k=1}^{\infty }$is its canonical dual frame. Assume thatGis the cross-Gram operator associated to$\{\langle f_{k}, S^{-1}f_{j}\rangle \}_{j, k=1}^{\infty }$. ThenGis self-adjoint and positive operator.

Proof

Since G is the cross-Gram operator associated to $\{\langle f_{k}, S^{-1}f_{j}\rangle \}_{j, k=1}^{\infty }$, we have

$$\begin{aligned} G=T^{*}_{s^{-1}f_{k}}T_{f_{k}}=T^{*}_{f_{k}}S^{-1}T_{f_{k}}. \end{aligned}$$

Therefore G is self-adjoint. Now we show that G is a positive operator. For $\{c_{k}\}_{k=1}^{\infty }\in \ell ^{2}(N)$ we have

$$\begin{aligned} \langle G\{c_{k}\}_{k=1}^{\infty }, \{c_{k}\}_{k=1}^{\infty }\rangle=\, & {} \langle T^{*}_{f_{k}}S^{-1}T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }, \{c_{k}\}_{k=1}^{\infty }\rangle \\ {}= & {} \langle S^{-1}T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }, T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }\rangle \\ {}= & {} \sum _{k=1}^{\infty }|\langle T_{f_{k}}\{c_{k}\}_{k=1}^{\infty }, S^{-1}f_{k}\rangle |^{2}. \end{aligned}$$

Corollary 3

If$\{f_{k}\}_{k=1}^{\infty }$is a Riesz basis in above theorem, thenGis the identity operator.

Example 5

Let $\{e_{k}\}_{k=1}^{\infty }$ be an orthonormal basis for H. Consider

$$\begin{aligned} \{f_{k}\}_{k=1}^{\infty }=\{e_{1}, e_{1}, e_{2}, e_{3},\ldots \}. \end{aligned}$$

The canonical dual frame is given by

$$\begin{aligned} \{S^{-1}f_{k}\}_{k=1}^{\infty }=\left\{ \frac{1}{2}e_{1}, \frac{1}{2}e_{1}, e_{2}, e_{3},\ldots \right\} . \end{aligned}$$

The cross-Gram matrix associated to these sequences is as follows:

$$\begin{aligned} G=\left[ \begin{array}{llllll} \frac{1}{2}&{}\quad \frac{1}{2}&{}\quad 0&{}\quad 0 &{}\quad 0 &{}\quad \cdots \\ \frac{1}{2}&{}\quad \frac{1}{2}&{}\quad 0&{}\quad 0 &{}\quad 0&{}\quad \cdots \\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad \cdots \\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \cdots \\ \end{array} \right] , \end{aligned}$$

Then

$$\begin{aligned} \langle G\{c_{k}\}_{k=1}^{\infty }, \{c_{k}\}_{k=1}^{\infty }\rangle= & {} \left\langle \left\{ \frac{1}{2}c_{1}+\frac{1}{2}c_{2}, \frac{1}{2}c_{1}+\frac{1}{2}c_{2}, c_{3}, c_{4},\ldots \right\} , \{c_{1}, c_{2}, c_{3}, c_{4},\ldots \}\right\rangle \\ {}& = {} \frac{1}{2}(c_{1}+c_{2})\overline{c_{1}}+\frac{1}{2}(c_{1}+c_{2})\overline{c_{2}}+ \sum _{k=3}^{\infty }|c_{k}|^{2}, \end{aligned}$$

which shows that G is a positive operator and $G^{2}=G$.