1 Introduction
Let \(\mathcal{H}\) be a separable Hilbert space, and \(\{x_{i}\}_{i \in I}\) be a countable sequence in \(\mathcal{H}\). It is called a frame for \(\mathcal{H}\) if there exist constants \(0< C_{1} \le C_{2}<\infty \) such that
where \(C_{1}\), \(C_{2}\) are called frame bounds. It is called a Bessel sequence in \(\mathcal{H}\) if the right-hand side inequality of (1.1) holds. And it is called a Riesz basis if it is a frame for \(\mathcal{H}\), while it ceases to be a frame for \(\mathcal{H}\) whenever an arbitrary element is removed. Given a frame \(\{x_{i}\}_{i\in I}\) for \(\mathcal{H}\), a sequence \(\{y_{i}\}_{i\in I}\) in \(\mathcal{H}\) is called a dual frame of \(\{x_{i}\}_{i\in I}\) if it is a frame for \(\mathcal{H}\) such that
It is easy to check that if \(\{y_{i}\}_{i\in I}\) is a dual frame of \(\{x_{i}\}_{i\in I}\), then \(\{x_{i}\}_{i\in I}\) is also a dual frame of \(\{y_{i}\}_{i\in I}\). So we say (\(\{x_{i}\}_{i\in I}\), \(\{y_{i}\} _{i\in I}\)) is a pair of dual frames in this case. It is well known that (\(\{x_{i}\}_{i\in I}\), \(\{y_{i}\}_{i\in I}\)) is a pair of dual frames for \(\mathcal{H}\) if and only if \(\{x_{i}\}_{i\in I}\) and \(\{y_{i}\} _{i\in I}\) are Bessel sequences in \(\mathcal{H}\) satisfying (1.2), and that a Bessel sequence (frame, Riesz sequence) is exactly the image of an orthonormal basis for \(\mathcal{H}\) under a linear bounded operator (bounded surjection, bounded bijection) on \(\mathcal{H}\). For basics on frames, see, e.g., [3, 18].
$$ C_{1} \Vert x \Vert ^{2}\le \sum _{i \in I} \bigl\vert \langle x, x_{i} \rangle \bigr\vert ^{2}\le C_{2} \Vert x \Vert ^{2} \quad{\text{for }}x\in {\mathcal{H}}, $$
(1.1)
$$ x=\sum_{i \in I}\langle x, x_{i}\rangle y_{i}\quad{\text{for }}x \in {\mathcal{H}}. $$
(1.2)
Throughout this paper, we denote by \(\mathcal{L}(\mathcal{H})\) the set of bounded linear operators on \(\mathcal{H}\). We define the operation “♠” as follows. For \(h_{1}, h_{2}\in {\mathcal{H}}\), we define the operator \(h_{1}\spadesuit h_{2}\) on \(\mathcal{H}\) by
It is well known that an arbitrary rank-one linear operator is always of this form. Herein, we use ♠ instead of ⊗ since ⊗ denotes rank-one antilinear operator throughout this paper.
$$ (h_{1}\spadesuit h_{2}) (x)=\langle x, h_{2}\rangle h_{1}\quad{\text{for }}x \in {\mathcal{H}}. $$
(1.3)
Anzeige
Now we turn to the tensor product of Hilbert spaces. There are several ways of defining the tensor product of Hilbert spaces. Folland in [6], Kadison and Ringrose in [9] represented the tensor product of Hilbert spaces \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\) as a certain linear space of operators. Let \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\) be complex separable Hilbert spaces. An operator T from \(\mathcal{H}_{2}\) to \(\mathcal{H}_{1}\) is said to be antilinear if
for x, \(y\in {\mathcal{H}}_{2}\) and α, \(\beta \in \Bbb {C}\). We consider the set of all bounded antilinear operators from \(\mathcal{H}_{2}\) to \(\mathcal{H}_{1}\). The norm of an antilinear operator \(T: {\mathcal{H}}_{2}\rightarrow \Bbb {\mathcal{H}}_{1}\) is defined as in the linear case:
By a standard argument as in [10], given a bounded antilinear operator \(T: {\mathcal{H}}_{2}\rightarrow \Bbb {\mathcal{H}}_{1}\), all series \(\sum_{j\in J}\|Tu_{j}\|^{2}\) associated with an orthonormal basis \(\{u_{j}\}_{j\in J}\) for \(\mathcal{H}_{2}\) take the same value (not necessarily finite).
$$ T(\alpha x+\beta y)=\overline{\alpha } Tx+\overline{\beta }Ty $$
(1.4)
$$\begin{aligned} \Vert T \Vert =\sup_{ \Vert g \Vert =1} \Vert Tg \Vert . \end{aligned}$$
(1.5)
Definition 1.1
Let \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\) be Hilbert spaces. Then the tensor product \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) of \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\) is the set of all bounded antilinear maps \(T: {\mathcal{H}}_{2}\rightarrow {\mathcal{H}}_{1}\) such that \(\sum_{j\in J}\|Tu_{j}\|^{2}<\infty \) for some, and hence every orthonormal basis \(\{u_{j}\}_{j\in J}\) of \(\mathcal{H}_{2}\). Moreover, for every \(T\in {\mathcal{H}}_{1}\otimes {\mathcal{H}}_{2}\), we set
$$ \|\!|T|\!\|^{2}=\sum_{j\in J} \Vert Tu_{j} \Vert ^{2}. $$
Recall from [6, Theorem 7.12], that \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) is a Hilbert space with the norm \(\|\!|\cdot |\!\|\) and the associated inner product
Let \(f\in {\mathcal{H}}_{1}\), \(g\in {\mathcal{H}}_{2}\), we define their tensor product
\(f\otimes g\) by
Obviously, \(f\otimes g\) belongs to \(\mathcal{H}_{1}\otimes {\mathcal{H}} _{2}\). By Lemma 2.2 below, a bounded antilinear operator T from \(\mathcal{H}_{2}\) to \(\mathcal{H}_{1}\) is rank-one if and only if \(T=f\otimes g\) for some \(0\ne f\in {\mathcal{H}}_{1}\) and \(0\ne g \in {\mathcal{H}}_{2}\). Also, by [6], for any \(f, f'\in {\mathcal{H}}_{1}\) and \(g, g'\in {\mathcal{H}}_{2}\),
Moreover, by Proposition 7.14 in [6], the tensor product of two orthonormal bases for \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\) is an orthonormal basis for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\). Throughout this paper, we always use \(\{e_{i}\}_{i\in I}\) and \(\{u_{j}\}_{j\in J}\) to denote orthonormal bases for \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively.
$$\begin{aligned} \langle Q, T\rangle =\sum_{j\in J} \langle Q u_{j}, T u_{j} \rangle \quad\text{for } Q, T\in { \mathcal{H}}_{1}\otimes {\mathcal{H}} _{2}. \end{aligned}$$
(1.6)
$$\begin{aligned} (f\otimes g) \bigl(g' \bigr)= \bigl\langle g, g' \bigr\rangle f\quad{\text{for }}g'\in { \mathcal{H}} _{2}. \end{aligned}$$
(1.7)
$$\begin{aligned} \bigl\langle f\otimes g, f'\otimes g' \bigr\rangle = \bigl\langle f, f' \bigr\rangle \bigl\langle g, g' \bigr\rangle . \end{aligned}$$
(1.8)
We now consider the tensor product of operators. Given \(T\in \mathcal{L}(\mathcal{H}_{1})\) and \(Q\in \mathcal{L}(\mathcal{H}_{2})\), define the tensor product \(T\otimes Q\) of T and Q by
By a standard argument, we have that \(T\otimes Q\in \mathcal{L}( \mathcal{H}_{1}\otimes {\mathcal{H}}_{2})\). The following proposition is repeated from [6, 7, 12].
$$ (T\otimes Q)B=TBQ^{\ast } \quad\text{for }B\in {\mathcal{H}}_{1} \otimes {\mathcal{H}}_{2}. $$
Anzeige
Proposition 1.1
Suppose
\(T, T'\in \mathcal{L}(\mathcal{H}_{1})\), \(Q, Q'\in \mathcal{L}(\mathcal{H}_{2})\), then
(a)
\(T\otimes Q\in \mathcal{L}(\mathcal{H}_{1}\otimes {\mathcal{H}} _{2})\)
and
\(\||T\otimes Q|\|=\|T\|\|Q\|\).
(b)
\((T\otimes Q)(f\otimes g)=Tf\otimes Qg\)
for all
\(f\in {\mathcal{H}} _{1}\), \(g\in {\mathcal{H}}_{2}\).
(c)
\((T\otimes Q)(T'\otimes Q')=(TT')\otimes (QQ')\).
(d)
If
T
and
Q
are invertible operators, then
\(T\otimes Q\)
is an invertible operator and
\((T\otimes Q)^{-1}=T^{-1}\otimes Q^{-1}\).
(e)
\((T\otimes Q)^{\ast }=T^{\ast }\otimes Q^{\ast }\).
(f)
Let
\(f, f'\in {\mathcal{H}}_{1}\backslash \{0\}\)
and
\(g, g' \in {\mathcal{H}}_{2}\backslash \{0\}\). If
\(f\otimes g=f'\otimes g'\), then there exist constants
a
and
b
with
\(ab=1\)
such that
\(f'=af\)
and
\(g'=bg\).
In [5, 8], it is proven that the tensor product of a sequence with itself is a frame if this sequence is a frame. In [10], it is proven that the tensor product of two frames must be a new frame for the corresponding tensor product space. In 2008, Bourouihiya in [1] and Upender in [17] obtained the following proposition which is generalized to the G-frame setting in [16]. For basics on G-frames, see, e.g., [13‐15] and the references therein.
Proposition 1.2
The sequence
\(\{f_{i}\otimes g_{j}: i\in I, j\in J\}\)
is a Bessel sequence (frame, Riesz basis) for
\(\mathcal{H}_{1}\otimes {\mathcal{H}} _{2}\)
if and only if
\(\{f_{i}\}_{i\in I}\)
and
\(\{g_{j}\}_{j\in J}\)
are Bessel sequences (frames, Riesz bases) for
\(\mathcal{H}_{1}\)
and
\(\mathcal{H}_{2}\), respectively.
A fundamental problem in the frame theory is to find dual frames with good properties. A pair of dual frames gives an expansion of the elements in the space which is like atomic decomposition in harmonic analysis. Unfortunately, for a given tensor product frame, little has been known to us except for its canonical dual in [10]. This study aims to investigate dual frames of a general tensor product frame \(\{f_{i}\otimes g_{j}\}_{(i, j)\in I\times J}\). Particularly, we are interested in dual frames with tensor product structure.
From the viewpoint of operators, [4, Theorem 3], demonstrates that, on an arbitrary ellipsoidal surface Δ in a separable Hilbert space \(\mathcal{H}\), there exists a sequence \(\{x_{i}\}_{i \in I}\) such that some multiple of the identity operator \(\lambda I _{\mathcal{H}}\) has the following rank-one decomposition:
where \(I_{\mathcal{H}}\) denotes the identity operator on \(\mathcal{H}\). For the case of \(\operatorname{dim}{\mathcal{H}}<\infty \), [2] characterizes sequences \(\{a_{i}\}_{i=1}^{M}\) such that
for some constant λ and \(\{x_{i}\}_{i=1}^{M}\), where \(\|x_{i}\|=a_{i}\) for \(1\leq i\leq M\). For the case of \(\operatorname{dim}{\mathcal{H}}= \infty \), [11, Proposition 7], presents another sufficient condition on \(\{a_{i}\}_{i=1}^{\infty }\) satisfying the following identity:
with \(\|x_{i}\|=\sqrt{a_{i}}\). Furthermore, the decomposition of more general operators was studied in [11]. For an arbitrary positive operator B with finite rank, [11, Theorem 2], characterizes a class of sequences \(\{a_{i}\}_{i=1}^{M}\) for which there exists \(\{x_{i}\}_{i=1}^{M}\) with \(\|x_{i}\|=\sqrt{a_{i}}\) such that
And for an arbitrary positive non-compact operator B, [11, Theorem 6], gives a sufficient condition on \(\{a_{i}\} _{i=1}^{\infty }\) for which there exists \(\{x_{i}\}_{i=1}^{\infty }\) with \(\|x_{i}\|=\sqrt{a_{i}}\) such that
In summary, these results focus on decomposing an bounded linear operator on \(\mathcal{H}\) into a sum of rank-one linear operators. Actually, a pair of tensor product dual frames also gives some rank-one decomposition of operators. Recall that every \(T\in {\mathcal{H}}_{1} \otimes {\mathcal{H}}_{2}\) is a bounded antilinear operator from \(\mathcal{H}_{2}\) to \(\mathcal{H}_{1}\) and that, if \(\{f_{i}\otimes g _{j}\}_{(i, j)\in I\times J}\) and \(\{\widetilde{f_{i}}\otimes \widetilde{g_{j}}\}_{(i, j)\in I\times J}\) form a pair of dual frames for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\), then
for \(T\in {\mathcal{H}}_{1}\otimes {\mathcal{H}}_{2}\), where \(a_{i, j}=\langle T, f_{i}\otimes g_{j}\rangle \). Therefore, the ♠-based decomposition presents a rank-one linear operator decomposition of bounded linear operators on \(\mathcal{H}\), while the ⊗-based (1.9) presents a rank-one antilinear operator decomposition of bounded antilinear operators from \(\mathcal{H}_{2}\) to \(\mathcal{H}_{1}\).
$$ \lambda I_{\mathcal{H}}=\sum_{i\in I}x_{i} \spadesuit x_{i}, $$
$$ \lambda I_{\mathcal{H}}=\sum_{i=1}^{M} x_{i}\spadesuit x_{i} $$
$$ I_{\mathcal{H}}=\sum_{i=1}^{\infty }x_{i} \spadesuit x_{i} $$
$$ B=\sum_{i=1}^{M} x_{i}\spadesuit x_{i}. $$
$$ B=\sum_{i=1}^{\infty }x_{i}\spadesuit x_{i}. $$
$$\begin{aligned} T=\sum_{(i, j)\in I\times J}a_{i, j} \widetilde{f_{i}}\otimes \widetilde{g_{j}} \end{aligned}$$
(1.9)
The rest of this paper is organized as follows. In Sect. 2, we characterize tensor product dual frames, give an explicit expression of all dual frames of tensor product frames, and prove the existence of non-tensor product dual frames. In Sect. 3, we present an explicit expression of all tensor product dual frames of tensor product frames. Some examples are also provided in Sect. 2 to illustrate the generality of the theory.
2 Dual characterization
This section is devoted to dual characterizations. We will derive a necessary and sufficient condition for two tensor product Bessel sequences to be a pair of dual frames; obtain an explicit expression of all duals of tensor product frames; and prove the existence of non-tensor product duals of tensor product frames. For this purpose, we first introduce some notations.
Let \(\mathbf{f}=\{f_{i}\}_{i\in I}\) and \(\mathbf{g}=\{g_{j}\}_{j \in J}\) be sequences in \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively. We write
Suppose that f and g are Bessel sequences in \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively. We denote by \(T_{\mathbf{f}}\), \(T_{\mathbf{g}}\), and \(T_{\mathbf{f}\otimes \mathbf{g}}\) the synthesis operators corresponding to f, g, and \(\mathbf{f}\otimes \mathbf{g}\), that is,
and
And we denote by \(S_{\mathbf{f}}\), \(S_{\mathbf{g}}\), \(S_{\mathbf{f} \otimes \mathbf{g}}\) the frame operators corresponding to f, g, and \(\mathbf{f}\otimes \mathbf{g}\), accordingly, i.e.,
$$ \mathbf{f}\otimes \mathbf{g}= \bigl\{ f_{i}\otimes g_{j}: (i, j)\in I\times J \bigr\} . $$
$$\begin{aligned} &T_{\mathbf{f}}c=\sum_{i\in I}c(i)f_{i} \quad\text{for }c\in l^{2}(I), \\ &T_{\mathbf{g}}c=\sum_{j\in J}c(j)g_{j} \quad\text{for }c\in l^{2}(J) \end{aligned}$$
$$ T_{\mathbf{f}\otimes \mathbf{g}}c=\sum_{(i, j)\in I\times J}c(i, j)f_{i} \otimes g_{j} \quad\text{for }c\in l^{2}(I\times J). $$
$$ S_{\mathbf{f}}=T_{\mathbf{f}}T^{\ast }_{\mathbf{f}},\qquad S_{\mathbf{g}}=T _{\mathbf{g}}T^{\ast }_{\mathbf{g}},\qquad S_{\mathbf{f}\otimes \mathbf{g}}=T_{\mathbf{f}\otimes \mathbf{g}}T^{\ast }_{\mathbf{f} \otimes \mathbf{g}}. $$
In particular, for the finite dimensional case, we also introduce the following notations. Let \(\{e_{i}\}_{i=1}^{m}\) and \(\{u_{j}\}_{j=1} ^{n}\) be orthonormal bases for \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively. Then each element of \(\mathcal{H}_{1}\) and \(\mathcal{H} _{2}\) can be uniquely represented in terms of \(\{e_{i}\}_{i=1}^{m}\) and \(\{u_{j}\}_{j=1}^{n}\), respectively. For \(f\in {\mathcal{H}}_{1}\) and \(g\in {\mathcal{H}}_{2}\), we define \(c(f)\in \Bbb {C}^{m}\) and \(c(g)\in \Bbb {C}^{n}\) by
and
with
Given sequences \(\mathbf{f}=\{f_{i}\}_{i=1}^{M}\) in \(\mathcal{H}_{1}\) and \(\mathbf{g}=\{g_{j}\}_{j=1}^{N}\) in \(\mathcal{H}_{2}\), we denote by \(M_{\mathbf{f}}\) and \(M_{\mathbf{g}}\) the following \(m\times M\) and \(n\times N\) matrices:
and
$$\begin{aligned} f= \begin{pmatrix} e_{1},&e_{2},&\cdots,&e_{m} \end{pmatrix} c(f) \end{aligned}$$
(2.1)
$$\begin{aligned} g= \begin{pmatrix} u_{1},&u_{2},&\cdots,&u_{n} \end{pmatrix} c(g) \end{aligned}$$
(2.2)
$$\begin{aligned} c(f)= \begin{pmatrix} c_{1}(f) \\ c_{2}(f) \\ \vdots \\ c_{m}(f) \end{pmatrix} \quad\text{and}\quad c(g)= \begin{pmatrix} c_{1}(g) \\ c_{2}(g) \\ \vdots \\ c_{n}(g) \end{pmatrix} . \end{aligned}$$
(2.3)
$$\begin{aligned} M_{\mathbf{f}}= \begin{pmatrix} c(f_{1}),&c(f_{2}),&\cdots,&c(f_{M}) \end{pmatrix} \end{aligned}$$
(2.4)
$$\begin{aligned} M_{\mathbf{g}}= \begin{pmatrix} c(g_{1}),&c(g_{2}),&\cdots,&c(g_{N}) \end{pmatrix} . \end{aligned}$$
(2.5)
The following lemma demonstrates that the synthesis (analysis) operator associated with the tensor product of two Bessel sequences is exactly the tensor product of their respective synthesis (analysis) operators.
Lemma 2.1
Let
\(\mathbf{f}=\{f_{i}\}_{i\in I}\)
and
\(\mathbf{g}=\{g_{j}\}_{j \in J}\)
be Bessel sequences in
\(\mathcal{H}_{1}\)
and
\(\mathcal{H}_{2}\), respectively. Then
$$ T_{\mathbf{f}\otimes \mathbf{g}}=T_{\mathbf{f}}\otimes T_{\mathbf{g}}\quad \textit{and}\quad T^{\ast }_{\mathbf{f}\otimes \mathbf{g}}=T^{\ast }_{ \mathbf{f}}\otimes T^{\ast }_{\mathbf{g}}. $$
Proof
Since f and g are Bessel sequences in \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively, \(T_{\mathbf{f}}\) and \(T_{\mathbf{g}}\) are all linear bounded operators. For arbitrary \(c\in l^{2}(I)\) and \(d\in l^{2}(J)\), by Proposition 1.1(b), we see that
Also observe that \(\{c\otimes d: c\in l^{2}(I), d\in l^{2}(J)\}\) is dense in \(l^{2}(I\times J)\). It follows that
whence
□
$$\begin{aligned} T_{\mathbf{f}\otimes \mathbf{g}} &=\sum_{(i, j)\in I\times J}c(i)d(j)f _{i}\otimes g_{j} \\ &=\sum_{(i, j)\in I\times J} \bigl(c(i)f_{i} \bigr) \otimes \bigl(d(j)g_{j} \bigr) \\ &=(T_{\mathbf{f}}c)\otimes (T_{\mathbf{g}}d) \\ &=(T_{\mathbf{f}}\otimes T_{\mathbf{g}}) (c\otimes d). \end{aligned}$$
$$ T_{\mathbf{f}\otimes \mathbf{g}}=T_{\mathbf{f}}\otimes T_{\mathbf{g}}, $$
$$ T^{\ast }_{\mathbf{f}\otimes \mathbf{g}}=T^{\ast }_{\mathbf{f}}\otimes T^{\ast }_{\mathbf{g}}. $$
Similarly to the case of “♠”, the following lemma shows that the tensor products take over all rank-one antilinear operators from \(\mathcal{H}_{2}\) to \(\mathcal{H}_{1}\).
Lemma 2.2
For
\(T\in {\mathcal{H}}_{1}\otimes {\mathcal{H}}_{2}\), \(\operatorname{dim}( \operatorname{range}(T))\leq 1\)
if and only if
for some
\(f\in {\mathcal{H}}_{1}\)
and
\(g\in {\mathcal{H}}_{2}\).
$$ T=f\otimes g $$
Proof
The sufficiency is obvious. Next we prove the necessity. Suppose \(\dim (\operatorname{range}(T))\leq 1\). If \(\dim (\operatorname{range}(T))=0\), then \(T=0\otimes g\) for an arbitrary \(g\in {\mathcal{H}}_{2}\). Next we suppose \(\dim (\operatorname{range}(T))=1\). Take \(0\neq f\in \operatorname{{range}}(T)\). Then, to every \(h\in {\mathcal{H}}_{2}\), there corresponds a unique number \(C_{h}\) such that
Define \(\varLambda: {\mathcal{H}}_{2}\rightarrow \Bbb {C}\) by
Then
equivalently,
for \(\alpha _{1}, \alpha _{2}\in \Bbb {C}\) and \(h_{1}, h_{2}\in {\mathcal{H}} _{2}\). Also observing that
we see that
This together with (2.6) leads to Λ is a linear bounded functional on \(\mathcal{H}_{2}\). So there exists unique \(g\in {\mathcal{H}} _{2}\) such that
It follows that
for \(h\in {\mathcal{H}}_{2}\), and thus \(T=f\otimes g\). □
$$ Th=C_{h}f. $$
$$ \varLambda h=\overline{C_{h}} \quad\text{for }h\in { \mathcal{H}}_{2}. $$
$$ \overline{(\alpha _{1}\varLambda h_{1}+\alpha _{2} \varLambda h_{2})}f=T(\alpha _{1}h_{1}+ \alpha _{2}h_{2})=\overline{\varLambda (\alpha _{1}h_{1}+ \alpha _{2}h_{2})}f, $$
$$\begin{aligned} \varLambda (\alpha _{1}h_{1}+ \alpha _{2}h_{2})=\alpha _{1}\varLambda h_{1}+ \alpha _{2}\varLambda h_{2} \end{aligned}$$
(2.6)
$$ \vert \varLambda h \vert \Vert f \Vert = \Vert Th \Vert \leq \Vert T \Vert \Vert h \Vert , $$
$$ \vert \varLambda h \vert \leq \frac{ \Vert T \Vert }{ \Vert f \Vert } \Vert h \Vert . $$
$$ \varLambda h=\langle h, g\rangle \quad\text{for }h\in {\mathcal{H}}_{2}. $$
$$ Th=\overline{\varLambda h}f=\langle g, h\rangle f=f\otimes g(h) $$
The following lemma gives an explicit expression of the tensor product of two finite-dimensional spaces. It will be used in Example 2.1 below.
Lemma 2.3
Let
\(\{e_{i}\}_{i=1}^{m}\)
and
\(\{u_{j}\}_{j=1}^{n}\)
be orthonormal bases for
\(\mathcal{H}_{1}\)
and
\(\mathcal{H}_{2}\), respectively. Then
$$\begin{aligned} {\mathcal{H}}_{1}\otimes { \mathcal{H}}_{2}={}& \Biggl\{ \mathcal{{A}}: {\mathcal{ {A}}} \begin{pmatrix} u_{1}, & u_{2}, & \cdots, & u_{n} \end{pmatrix} {z}= \begin{pmatrix} e_{1}, & e_{2}, & \cdots, & e_{m} \end{pmatrix} A \overline{z} \\ &\textit{for } z\in \Bbb {C}^{n}, A \textit{ is an }m\times n \textit{ matrix }\Biggr\} . \end{aligned}$$
(2.7)
Proof
By a simple computation, we have
for \(1\leq i\leq m\), \(1\leq j\leq n\), \(h\in {\mathcal{H}}_{2}\), where \(E_{i, j}\) denotes the \(m\times n\) matrix with \((i, j)\)-entry being 1 and others being zero, \(\overline{c(h)}\) denotes the conjugate of \(c(h)\). Since \(\{e_{i}\otimes u_{j}: 1\leq i\leq m, 1\leq j\leq n\}\) is an orthonormal basis for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\), observe that
This leads to (2.7) by (2.8). The proof is completed. □
$$\begin{aligned} (e_{i}\otimes u_{j}) (h) &= \overline{c_{j}(h)}e_{i} \\ &= \begin{pmatrix} e_{1},&e_{2},&\cdots,&e_{m} \end{pmatrix} E_{i, j} \overline{c(h)} \end{aligned}$$
(2.8)
$$ \mathcal{H}_{1}\otimes {\mathcal{H}}_{2}= \operatorname{{span}}\{e_{i}\otimes u_{j}: 1\leq i\leq m, 1 \leq j\leq n\}. $$
Next we turn to the main results of this section. The first theorem gives a necessary and sufficient condition for two tensor product Bessel sequences to form a pair of dual frames.
Theorem 2.1
Assume that
\(\mathbf{f}=\{f_{i}\}_{i\in I}\)
and
\(\mathbf{\widetilde{f}}=\{\widetilde{f}_{i}\}_{i\in I}\)
are Bessel sequences in
\(\mathcal{H}_{1}\), and
\(\mathbf{g}=\{g_{j}\}_{j\in J}\)
and
\(\mathbf{\widetilde{g}}=\{\widetilde{g}_{j}\}_{j\in J}\)
are Bessel sequences in
\(\mathcal{H}_{2}\). Then
\(\mathbf{f}\otimes \mathbf{g}\)
and
\(\mathbf{\widetilde{f}}\otimes \mathbf{\widetilde{g}}\)
form a pair of dual frames in
\(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\)
if and only if there exist constants
a
and
b
with
\(ab=1\)
such that
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\widetilde{\mathbf{f}}}=aI_{\mathcal{H}_{1}} \quad\textit{and}\quad T_{\mathbf{g}}T^{\ast }_{\widetilde{\mathbf{g}}}=bI_{\mathcal{H} _{2}}. \end{aligned}$$
(2.9)
Proof
Since f and \(\mathbf{\widetilde{f}}\) are Bessel sequences in \(\mathcal{H}_{1}\) and g and \(\mathbf{\widetilde{g}}\) are Bessel sequences in \(\mathcal{H}_{2}\), by the definition of dual frame, \(\mathbf{f}\otimes \mathbf{g}\) and \(\mathbf{\widetilde{f}}\otimes \mathbf{\widetilde{g}}\) are a pair of dual frames in \(\mathcal{H}_{1} \otimes {\mathcal{H}}_{2}\) if and only if
This is in turn equivalent to
due to the fact that \(\{e_{i}\otimes u_{j}: (i, j)\in I\times J\}\) is an orthonormal basis for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\). Therefore, to finish the proof, we only need to demonstrate that (2.10) is equivalent to (2.9). Next we do this. By Lemma 2.1 and Proposition 1.1(c), we have that
It follows that
by Proposition 1.1(b). So (2.10) can be rewritten as
equivalently, for each \((i, j)\in I\times J\),
by Proposition 1.1(f), where \(a_{i}, b_{j}\) are two constants. Fix \(i_{0}\in I\), then \(b_{j}=\frac{1}{a_{i_{0}}}\) for every \(j\in J\). It follows that all \(b_{j}\) take the same value b. Again from \(a_{i}b=1\) for every \(i\in I\), we see that all \(a_{i}\) take the same value a. This implies that (2.12) is equivalent to
due to the fact that \(\{e_{i}\}_{i\in I}\) and \(\{u_{j}\}_{j\in J}\) are orthonormal bases for \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively. The proof is completed. □
$$ T_{\mathbf{f}\otimes \mathbf{g}}T^{\ast }_{\mathbf{\widetilde{f}} \otimes \mathbf{\widetilde{g}}}=I_{\mathcal{H}_{1}\otimes {\mathcal{H}} _{2}}. $$
$$\begin{aligned} T_{\mathbf{f}\otimes \mathbf{g}}T^{\ast }_{\mathbf{\widetilde{f}} \otimes \mathbf{\widetilde{g}}}(e_{i} \otimes u_{j})=e_{i}\otimes u _{j} \quad \text{for }(i, j)\in I\times J \end{aligned}$$
(2.10)
$$\begin{aligned} T_{\mathbf{f}\otimes \mathbf{g}}T^{\ast }_{\mathbf{\widetilde{f}} \otimes \mathbf{\widetilde{g}}} &=(T_{\mathbf{f}}\otimes T_{\mathbf{g}}) \bigl(T ^{\ast }_{\mathbf{\widetilde{f}}}\otimes T^{\ast }_{ \mathbf{\widetilde{g}}} \bigr) \\ &= \bigl(T_{\mathbf{f}}T^{\ast }_{\mathbf{\widetilde{f}}} \bigr)\otimes \bigl(T_{ \mathbf{g}}T^{\ast }_{\mathbf{\widetilde{g}}} \bigr). \end{aligned}$$
$$ T_{\mathbf{f}\otimes \mathbf{g}}T^{\ast }_{\mathbf{\widetilde{f}} \otimes \mathbf{\widetilde{g}}}(e_{i}\otimes u_{j})= \bigl(T_{\mathbf{f}}T ^{\ast }_{\mathbf{\widetilde{f}}}e_{i} \bigr)\otimes \bigl(T_{\mathbf{g}}T^{ \ast }_{\mathbf{\widetilde{g}}}u_{j} \bigr) \quad\text{for }(i, j)\in I\times J $$
$$\begin{aligned} \bigl(T_{\mathbf{f}}T^{\ast }_{\mathbf{\widetilde{f}}}e_{i} \bigr)\otimes \bigl(T_{ \mathbf{g}}T^{\ast }_{\mathbf{\widetilde{g}}}u_{j} \bigr)=e_{i}\otimes u _{j} \quad\text{for }(i, j)\in I \times J, \end{aligned}$$
(2.11)
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\mathbf{\widetilde{f}}}e_{i}=a_{i}e_{i} \quad\text{and}\quad T_{\mathbf{g}}T^{\ast }_{\mathbf{\widetilde{g}}}u_{j}=b_{j}u _{j} \quad\text{with }a_{i}b_{j}=1 \end{aligned}$$
(2.12)
$$ T_{\mathbf{f}}T^{\ast }_{\mathbf{\widetilde{f}}}=aI_{\mathcal{H}_{1}}\quad\text{and} \quad T_{\mathbf{g}}T^{\ast }_{\mathbf{\widetilde{g}}}=bI_{\mathcal{H} _{2}}\quad \text{with }ab=1 $$
Observe that, given Bessel sequences \(\mathbf{f}=\{f_{i}\}_{i\in I}\), \(\mathbf{\widetilde{f}}=\{\widetilde{f}_{i}\}_{i\in I}\) in \(\mathcal{H}_{1}\) and \(\mathbf{g}=\{g_{j}\}_{j\in J}\), \(\mathbf{\widetilde{g}}=\{\widetilde{g}_{j}\}_{j\in J}\) in \(\mathcal{H}_{2}\), f and \(\widetilde{\mathbf{f}}\) (g and \(\widetilde{\mathbf{g}}\)) form a pair of dual frames in \(\mathcal{H}_{1}\) (\(\mathcal{H}_{2}\)) if and only if \(T_{\mathbf{f}}T ^{\ast }_{\widetilde{\mathbf{f}}}=I_{\mathcal{H}_{1}}\) (\(T_{ \mathbf{g}}T^{\ast }_{\widetilde{\mathbf{g}}}=I_{\mathcal{H}_{2}}\)). As a special case of Theorem 2.1, we have the following corollary. It shows that the dual property is preserved under the tensor product operation.
Corollary 2.1
Let
\(\mathbf{f}=\{f_{i}\}_{i\in I}\)
and
\(\mathbf{g}=\{g_{j}\}_{j \in J}\)
be frames for
\(\mathcal{H}_{1}\)
and
\(\mathcal{H}_{2}\), respectively. Assume that
\(\mathbf{\widetilde{f}}=\{\widetilde{f}_{i} \}_{i\in I}\)
and
\(\mathbf{\widetilde{g}}=\{\widetilde{g}_{j}\}_{j \in J}\)
are dual frames of
f
and
g, respectively. Then
\(\mathbf{\widetilde{f}}\otimes \mathbf{\widetilde{g}}\)
is a dual frame of
\(\mathbf{f}\otimes \mathbf{g}\).
Corollary 2.1 and Proposition 1.2 show that a tensor product frame always admits a tensor product dual. It is natural to ask whether all duals of a tensor product frame are of tensor product. The following theorem characterizes all duals of a general tensor product frame. Then using it we derive Theorem 2.3 which shows the existence of non-tensor product duals of tensor product frames.
Theorem 2.2
Let
\(\mathbf{f}\otimes \mathbf{g}\)
be a frame for
\(\mathcal{H}_{1} \otimes {\mathcal{H}}_{2}\)
with
\(\mathbf{f}=\{f_{i}\}_{i\in I}\)
and
\(\mathbf{g}=\{g_{j}\}_{j\in J}\). Then the dual frames of
\(\mathbf{f} \otimes \mathbf{g}\)
are precisely the families
where
\(\{w_{i, j}\}_{(i, j)\in I\times J}\)
is a Bessel sequence in
\(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\).
$$\begin{aligned} \biggl\{ S_{\mathbf{f}}^{-1}f_{i} \otimes S_{\mathbf{g}}^{-1}g_{j}+w_{i, j}- \sum _{(i', j')\in I\times J} \bigl\langle S_{\mathbf{f}}^{-1}f_{i}, f_{i'} \bigr\rangle \bigl\langle S_{\mathbf{g}}^{-1}g_{j}, g_{j'} \bigr\rangle w _{i', j'} \biggr\} _{(i, j)\in I\times J}, \end{aligned}$$
(2.13)
Proof
By Proposition 1.2, f and g are frames for \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively. So \(S_{ \mathbf{f}}\) and \(S_{\mathbf{g}}\) are well-defined, bounded, and invertible. By Theorem 6.3.7 in [3], the dual frames are precisely the families of the form
where \(\{w_{i, j}\}_{(i, j)\in I\times J}\) is a Bessel sequence in \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\). Also, by Proposition 3.2 of [10], \(S_{\mathbf{f}\otimes \mathbf{g}}^{-1}=S_{\mathbf{f}} ^{-1}\otimes S_{\mathbf{g}}^{-1}\). This implies that
for \((i, j), (i', j')\in I\times J\) by Proposition 1.1(b) and (1.8). Therefore, (2.14) can be written as (2.13). The proof is completed. □
$$\begin{aligned} \biggl\{ S_{\mathbf{f}\otimes \mathbf{g}}^{-1}(f_{i} \otimes g_{j})+w_{i, j}- \sum_{(i', j')\in I\times J} \bigl\langle S_{\mathbf{f}\otimes \mathbf{g}}^{-1}(f_{i}\otimes g_{j}), f_{i'}\otimes g_{j'} \bigr\rangle w _{i', j'} \biggr\} _{(i, j)\in I\times J}, \end{aligned}$$
(2.14)
$$\begin{aligned} &S_{\mathbf{f}\otimes \mathbf{g}}^{-1}(f_{i}\otimes g_{j})=S^{-1}_{ \mathbf{f}}f_{i} \otimes S^{-1}_{\mathbf{g}}g_{j}, \\ & \bigl\langle S_{\mathbf{f}\otimes \mathbf{g}}^{-1}(f_{i}\otimes g_{j}), f _{i'}\otimes g_{j'} \bigr\rangle = \bigl\langle S^{-1}_{\mathbf{f}}f_{i}, f_{i'} \bigr\rangle \bigl\langle S^{-1}_{\mathbf{g}}g_{j}, g_{j'} \bigr\rangle \end{aligned}$$
Lemma 2.2 shows that every element of \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) is of tensor product whenever one of \(\dim ( \mathcal{H}_{1})\) and \(\dim (\mathcal{H}_{2})\) equals 1. So a tensor product frame in \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) admits a non-tensor product dual only if both \(\dim (\mathcal{H}_{1})\) and \(\dim (\mathcal{H}_{2})\) are greater than 1. Next, with the help of Theorem 2.2, we will prove that this condition is also sufficient for a redundant tensor product frame to admit a non-tensor product dual.
Theorem 2.3
Let
\(\operatorname{dim}(\mathcal{H}_{1})\), \(\operatorname{dim}(\mathcal{H}_{2})>1\), and
\(\mathbf{f}\otimes \mathbf{g}\)
be a redundant frame for
\(\mathcal{H} _{1}\otimes {\mathcal{H}}_{2}\)
with
\(\mathbf{f}=\{f_{i}\}_{i\in I}\)
and
\(\mathbf{g}=\{g_{j}\}_{j\in J}\). Then
\(\mathbf{f}\otimes \mathbf{g}\)
admits at least one non-tensor product dual.
Proof
By Theorem 2.2 and Lemma 2.2, to finish the proof, we only need to find a Bessel sequence \(\{w_{i, j}\}_{(i, j)\in I\times J}\) in \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) such that
where
Since \(\mathbf{f}\otimes \mathbf{g}\) is not a Riesz basis for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\), there exist \((i_{1}, j _{1}), (i_{2}, j_{2})\in I\times J\) such that
by [3, Theorem 7.1]. By [10, Proposition 3.2], Proposition 1.1(b), and (1.8), we have
Hence, (2.16) is equivalent to
By a simple argument, we have that (2.17) holds if and only if one of the following three conditions happens:
$$\begin{aligned} \dim \bigl(\operatorname{{range}}(T_{i, j}) \bigr)\geq 2 \quad\text{for some }(i, j)\in I \times J, \end{aligned}$$
(2.15)
$$ T_{i, j}=S_{\mathbf{f}}^{-1}f_{i}\otimes S_{\mathbf{g}}^{-1}g_{j}+w _{i, j}-\sum _{(i', j')\in I\times J} \bigl\langle S_{\mathbf{f}} ^{-1}f_{i}, f_{i'} \bigr\rangle \bigl\langle S_{\mathbf{g}}^{-1}g_{j}, g_{j'} \bigr\rangle w_{i', j'}. $$
$$\begin{aligned} \bigl\langle S_{\mathbf{f}\otimes \mathbf{g}}^{-1}(f_{i_{1}} \otimes g_{j _{1}}), f_{i_{2}}\otimes g_{j_{2}} \bigr\rangle \neq \delta _{i_{1}, i_{2}} \delta _{j_{1}, j_{2}} \end{aligned}$$
(2.16)
$$ \bigl\langle S_{\mathbf{f}\otimes \mathbf{g}}^{-1}(f_{i_{1}}\otimes g_{j _{1}}), f_{i_{2}}\otimes g_{j_{2}} \bigr\rangle = \bigl\langle S_{\mathbf{f}} ^{-1}f_{i_{1}}, f_{i_{2}} \bigr\rangle \bigl\langle S_{\mathbf{g}}^{-1}g_{j_{1}}, g_{j_{2}} \bigr\rangle . $$
$$\begin{aligned} \bigl\langle S_{\mathbf{f}}^{-1}f_{i_{1}}, f_{i_{2}} \bigr\rangle \bigl\langle S_{ \mathbf{g}}^{-1}g_{j_{1}}, g_{j_{2}} \bigr\rangle \neq \delta _{i_{1}, i _{2}}\delta _{j_{1}, j_{2}}. \end{aligned}$$
(2.17)
Condition 1
\(\langle S_{\mathbf{f}}^{-1}f_{i_{1}}, f_{i _{1}}\rangle \langle S_{\mathbf{g}}^{-1}g_{j_{1}}, g_{j_{1}}\rangle \neq 1\), \(S_{\mathbf{f}}^{-1}f_{i_{1}}\otimes S_{\mathbf{g}}^{-1}g _{j_{1}}\neq 0\).
Condition 2
\(\langle S_{\mathbf{f}}^{-1}f_{i_{1}}, f_{i _{1}}\rangle \langle S_{\mathbf{g}}^{-1}g_{j_{1}}, g_{j_{1}}\rangle \neq 1\), \(S_{\mathbf{f}}^{-1}f_{i_{1}}\otimes S_{\mathbf{g}}^{-1}g _{j_{1}}=0\).
Condition 3
\((i_{1}, j_{1})\neq (i_{2}, j_{2})\), \(\langle S_{\mathbf{f}}^{-1}f_{i_{1}}, f_{i_{1}}\rangle \langle S _{\mathbf{g}}^{-1}g_{j_{1}}, g_{j_{1}}\rangle =1\), \(\langle S_{\mathbf{f}}^{-1}f _{i_{1}}, f_{i_{2}}\rangle \langle S_{\mathbf{g}}^{-1}g_{j_{1}}, g _{j_{2}}\rangle \neq 0\).
Next we prove that, whenever one of the above conditions is satisfied, we may construct a Bessel sequence \(\{w_{i, j}\}_{(i, j)\in I\times J}\) such that \(\operatorname{{range}}(T_{i_{1}, j_{1}})\) contains two orthogonal nonzero vectors. This leads to (2.15) with \((i_{1}, j _{1})\).
For Condition 1, choose unit vectors \(u\in {\mathcal{H}} _{1}\) and \(v\in {\mathcal{H}}_{2}\) such that
This can be done since \(\operatorname{dim}{\mathcal{H}}_{1}\), \(\operatorname{dim}{\mathcal{H}} _{2}>1\). Define \(\{w_{i, j}\}_{(i, j)\in I\times J}\) by
Then by a direct computation, we have
It follows from (2.18) that
which are two orthogonal nonzero vectors.
$$\begin{aligned} \bigl\langle u, S_{\mathbf{f}}^{-1}f_{i_{1}} \bigr\rangle = \bigl\langle v, S_{ \mathbf{g}}^{-1}g_{j_{1}} \bigr\rangle =0. \end{aligned}$$
(2.18)
$$ w_{i, j}= \textstyle\begin{cases} \frac{u\otimes v}{1-\langle S_{\mathbf{f}}^{-1}f_{i_{1}}, f_{i_{1}} \rangle \langle S_{\mathbf{g}}^{-1}g_{j_{1}}, g_{j_{1}}\rangle } &\text{if } (i, j)=(i_{1}, j_{1}); \\ 0& \text{otherwise}. \end{cases} $$
$$ T_{i_{1}, j_{1}}=S_{\mathbf{f}}^{-1}f_{i_{1}}\otimes S_{\mathbf{g}} ^{-1}g_{j_{1}}+u\otimes v. $$
$$ T_{i_{1}, j_{1}} \bigl(S_{\mathbf{g}}^{-1}g_{j_{1}} \bigr)= \bigl\Vert S_{\mathbf{g}}^{-1}g _{j_{1}} \bigr\Vert ^{2}S_{\mathbf{f}}^{-1}f_{i_{1}},\qquad T_{i_{1}, j_{1}}(v)=u, $$
For Condition 2, choose unit vectors \(u_{1}, u_{2}\in {\mathcal{H}}_{1}\) and \(v_{1}, v_{2}\in {\mathcal{H}}_{2}\) such that
Define \(\{w_{i, j}\}_{(i, j)\in I\times J}\) by
Then
It follows from (2.19) that
which are two orthogonal unit vectors.
$$\begin{aligned} \langle u_{1}, u_{2}\rangle = \langle v_{1}, v_{2}\rangle =0. \end{aligned}$$
(2.19)
$$ w_{i, j}=\textstyle\begin{cases} u_{1}\otimes v_{1}+u_{2}\otimes v_{2}& \text{if } (i, j)=(i_{1}, j _{1}); \\ 0& \text{otherwise}. \end{cases} $$
$$ T_{i_{1}, j_{1}}=w_{i_{1}, j_{1}}. $$
$$ T_{i_{1}, j_{1}}(v_{1})=u_{1},\qquad T_{i_{1}, j_{1}}(v_{2})=u_{2}, $$
For Condition 3, choose unit vectors \(u\in {\mathcal{H}} _{1}\) and \(v\in {\mathcal{H}}_{2}\) such that
Define \(\{w_{i, j}\}_{(i, j)\in I\times J}\) by
Then
It follows from (2.20) that
which are two orthogonal nonzero vectors. The proof is completed. □
$$\begin{aligned} \bigl\langle u, S_{\mathbf{f}}^{-1}f_{i_{1}} \bigr\rangle = \bigl\langle v, S_{ \mathbf{g}}^{-1}g_{j_{1}} \bigr\rangle =0. \end{aligned}$$
(2.20)
$$ w_{i, j}=\textstyle\begin{cases} -\frac{u\otimes v}{\langle S_{\mathbf{f}}^{-1}f_{i_{1}}, f_{i_{2}} \rangle \langle S_{\mathbf{g}}^{-1}g_{j_{1}}, g_{j_{2}}\rangle }& \text{if } (i, j)=(i_{2}, j_{2}); \\ 0& \text{otherwise}. \end{cases} $$
$$ T_{i_{1}, j_{1}}=S_{\mathbf{f}}^{-1}f_{i_{1}}\otimes S_{\mathbf{g}} ^{-1}g_{j_{1}}+u\otimes v. $$
$$ T_{i_{1}, j_{1}} \bigl(S_{\mathbf{g}}^{-1}g_{j_{1}} \bigr)= \bigl\Vert S_{\mathbf{g}}^{-1}g _{j_{1}} \bigr\Vert ^{2}S_{\mathbf{f}}^{-1}f_{i_{1}},\qquad T_{i_{1}, j_{1}}(v)=u, $$
Next we give an example of Theorem 2.3 in the finite dimensional case.
Example 2.1
Let \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\) be two finite dimensional Hilbert spaces, and \(\{e_{i}\}_{i=1}^{m}\) and \(\{u_{j}\}_{j=1}^{n}\) be orthonormal bases for \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively. Assume that \(\mathbf{f}\otimes \mathbf{g}\) is a redundant frame for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) with \(\mathbf{f}= \{f_{i}\}_{i=1}^{M}\) and \(\mathbf{g}=\{g_{j}\}_{j=1}^{N}\). Then, by a simple calculation, for any \(f\in {\mathcal{H}}_{1}\), \(g\in {\mathcal{H}} _{2}\), we have that
and
for \((i, j)\in \{1, 2, \ldots, M\}\times \{1, 2, \ldots, N\}\). By Lemma 2.3, an arbitrary sequence \(\{w_{i, j}\}_{i=1, j=1} ^{M, N}\) in \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) has the form
for \((i, j)\in \{1, 2, \ldots, M\}\times \{1, 2, \ldots, N\}\) and \(g\in {\mathcal{H}}_{2}\), where \(\varOmega _{i, j}\) is an \(m\times n\) matrix for every \((i, j)\in \{1, 2, \ldots, M\}\times \{1, 2, \ldots, N\}\). It follows by Theorem 2.2 that all dual frames of \(\mathbf{f}\otimes \mathbf{g}\) have the form
where \(\varOmega _{i, j}\) is an \(m\times n\) matrix for each \((i, j) \in \{1, 2, \ldots, M\}\times \{1, 2, \ldots, N\}\).
$$\begin{aligned} &c(S_{\mathbf{f}}f)=M_{\mathbf{f}}M^{\ast }_{\mathbf{f}}c(f),\qquad c(S _{\mathbf{g}}g)=M_{\mathbf{g}}M^{\ast }_{\mathbf{g}}c(g), \\ &c \bigl(S_{\mathbf{f}}^{-1}f \bigr)= \bigl(M_{\mathbf{f}}M^{\ast }_{\mathbf{f}} \bigr)^{-1}c(f),\qquad c \bigl(S_{\mathbf{g}}^{-1}g \bigr)= \bigl(M_{\mathbf{g}}M^{\ast }_{\mathbf{g}} \bigr)^{-1}c(g), \end{aligned}$$
$$\begin{aligned} & \bigl(S_{\mathbf{f}}^{-1}f_{i}\otimes S_{\mathbf{g}}^{-1}g_{j} \bigr)g \\ &\quad= \bigl(S_{\mathbf{f}}^{-1}f_{i}\otimes S_{\mathbf{g}}^{-1}g_{j} \bigr) \begin{pmatrix} u_{1},&\cdots,&u_{n} \end{pmatrix} c(g) \\ &\quad= \begin{pmatrix} e_{1},&\cdots,&e_{m} \end{pmatrix} \bigl(c \bigl(S_{\mathbf{f}}^{-1}{f_{i}} \bigr)\otimes c \bigl(S_{\mathbf{g}}^{-1} {g_{j}} \bigr) \bigr)c(g) \\ &\quad= \begin{pmatrix} e_{1},&\cdots,&e_{m} \end{pmatrix} \begin{pmatrix} c_{1}(S_{\mathbf{g}}^{-1}{g_{j}})c(S_{\mathbf{f}}^{-1}{f_{i}}), & \cdots, &c_{n}(S_{\mathbf{g}}^{-1}{g_{j}})c(S_{\mathbf{f}}^{-1}{f _{i}}) \end{pmatrix} \overline{c(g)} \\ &\quad= \begin{pmatrix} e_{1},&\cdots,&e_{m} \end{pmatrix} \varLambda _{i, j} \overline{c(g)}, \end{aligned}$$
(2.21)
$$\begin{aligned} w_{i, j}g=w_{i, j} \begin{pmatrix} u_{1}&\cdots &u_{n} \end{pmatrix} c(g)= \begin{pmatrix} e_{1}&\cdots &e_{m} \end{pmatrix} \varOmega _{i, j}\overline{c(g)} \end{aligned}$$
(2.22)
$$\begin{aligned} &\Biggl\{ T_{i, j}: T_{i, j}g= \begin{pmatrix} e_{1},&\cdots,&e_{m} \end{pmatrix} \Biggl[\varLambda _{i, j} +\varOmega _{i, j} \\ &\quad{}-\sum_{i'=1}^{M}\sum _{j'=1}^{N} \bigl\langle c \bigl(S_{ \mathbf{f}}^{-1}{f_{i}} \bigr), c(f_{i'}) \bigr\rangle \bigl\langle c \bigl(S_{\mathbf{g}} ^{-1}{g_{j}} \bigr), c(g_{j'}) \bigr\rangle \varOmega _{i', j'} \Biggr]\overline{c(g)} \mbox{ for }g\in { \mathcal{H}}_{2}\Biggr\} _{i=1, j=1}^{M, N}, \end{aligned}$$
(2.23)
Write
for \((i, j)\in \{1, 2, \ldots, M\}\times \{1, 2, \ldots, N\}\). Obviously, (2.23) is determined by \(\{\varOmega _{i, j}\}_{i=1, j=1} ^{M, N}\). Hence, by Lemma 2.2, constructing a non-tensor product dual frame of \(\mathbf{f}\otimes \mathbf{g}\) reduces to constructing \(\{\varOmega _{i, j}\}_{i=1, j=1}^{M, N}\) satisfying
Since \(\mathbf{f}\otimes \mathbf{g}\) is not a Riesz basis for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\), by [3, Theorem 7.1], we have that
for some \((i_{1}, j_{1}), (i_{2}, j_{2})\in \{1, 2, \ldots, M \}\times \{1, 2, \ldots, N\}\). According to the arguments in the proof of Theorem 2.3, we divide the following three cases to construct \(\{\varOmega _{i, j}\}_{i=1, j=1}^{M, N}\) satisfying (2.24).
$$ \varUpsilon _{i, j}=\varLambda _{i, j}+\varOmega _{i, j} -\sum_{i'=1} ^{M}\sum _{j'=1}^{N} \bigl\langle c \bigl(S_{\mathbf{f}}^{-1}{f_{i}} \bigr), c(f _{i'}) \bigr\rangle \bigl\langle c \bigl(S_{\mathbf{g}}^{-1}{g_{j}} \bigr), c(g_{j'}) \bigr\rangle \varOmega _{i', j'} $$
$$\begin{aligned} {\operatorname{{rank}}}(\varUpsilon _{i, j}) \geq 2 \quad\mbox{for some }(i, j) \in \{1, 2, \ldots, M\}\times \{1, 2, \ldots, N \}. \end{aligned}$$
(2.24)
$$\begin{aligned} \bigl\langle c \bigl(S_{\mathbf{f}}^{-1}{f_{i_{1}}} \bigr), c(f_{i_{2}}) \bigr\rangle \bigl\langle c \bigl(S_{\mathbf{g}}^{-1}{g_{j_{1}}} \bigr), c(g_{j_{2}}) \bigr\rangle &= \bigl\langle S_{\mathbf{f}}^{-1}f_{i_{1}}, f_{i_{2}} \bigr\rangle \bigl\langle S_{ \mathbf{g}}^{-1}g_{j_{1}}, g_{j_{2}} \bigr\rangle \\ &= \bigl\langle S_{\mathbf{f}\otimes \mathbf{g}}^{-1}(f_{i_{1}}\otimes g _{j_{1}}), f_{i_{2}}\otimes g_{j_{2}} \bigr\rangle \\ &\neq \delta _{i_{1}, i_{2}}\delta _{j_{1}, j_{2}} \end{aligned}$$
(2.25)
Case 1. \(\langle c(S_{\mathbf{f}}^{-1}{f_{i _{1}}}), c(f_{i_{1}}) \rangle \langle c(S_{\mathbf{g}} ^{-1}{g_{j_{1}}}), c(g_{j_{1}}) \rangle \neq 1\), \(c(S_{ \mathbf{f}}^{-1}{f_{i_{1}}})\otimes c(S_{\mathbf{g}}^{-1}{g_{j_{1}}}) \neq 0\).
Choose two linearly-independent nonzero vectors \(d_{1}, d_{2}\in \mathbb{C}^{m}\). Define \(\{\varOmega _{i, j}\}_{i=1, j=1}^{M, N}\) by \(\varOmega _{i, j}=\)
Then
by a simple computation. This implies that (2.24) holds for \((i_{1}, j_{1})\).
$$ \textstyle\begin{cases} \frac{ ( d_{1}-c_{1}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S_{\mathbf{f}}^{-1}{f _{i_{1}}}), d_{2}-c_{2}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S_{ \mathbf{f}}^{-1}{f_{i_{1}}}), 0,\ldots, 0 ) }{1- \langle c(S_{\mathbf{f}}^{-1}{f_{i_{1}}}), c(f_{i_{1}}) \rangle \langle c(S_{\mathbf{g}}^{-1}{g_{j_{1}}}), c(g_{j_{1}}) \rangle }&\text{if } (i,j)=(i_{1},j_{1}); \\ 0& \text{otherwise}. \end{cases} $$
$$ \varUpsilon _{i_{1}, j_{1}}= \begin{pmatrix} d_{1},&d_{2},&c_{3}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S_{\mathbf{f}} ^{-1}{f_{i_{1}}}),& \cdots,&c_{n}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S _{\mathbf{f}}^{-1}{f_{i_{1}}}) \end{pmatrix} $$
Case 2. \(\langle c(S_{\mathbf{f}}^{-1}{f_{i _{1}}}), c(f_{i_{1}}) \rangle \langle c(S_{\mathbf{g}} ^{-1}{g_{j_{1}}}), c(g_{j_{1}}) \rangle \neq 1\), \(c(S_{ \mathbf{f}}^{-1}{f_{i_{1}}})\otimes c(S_{\mathbf{g}}^{-1}{g_{j_{1}}})=0\).
Choose two linearly-independent nonzero vectors \(d_{1}, d_{2}\in \mathbb{C}^{m}\). Define \(\{\varOmega _{i, j}\}_{i=1, j=1}^{M, N}\) by
Then
This implies that (2.24) holds for \((i_{1}, j_{1})\).
$$ \varOmega _{i, j}=\textstyle\begin{cases} \begin{pmatrix} d_{1}, & d_{2}, & 0,&\cdots, & 0 \end{pmatrix} &\text{if } (i, j)=(i_{1}, j_{1}); \\ 0& \text{otherwise}. \end{cases} $$
$$ \varUpsilon _{i_{1}, j_{1}}=\varOmega _{i_{1}, j_{1}}. $$
Case 3. \((i_{1}, j_{1})\neq (i_{2}, j_{2})\), \(\langle c(S_{\mathbf{f}}^{-1}{f_{i_{1}}}), c(f_{i_{1}}) \rangle \langle c(S_{\mathbf{g}}^{-1}{g_{j_{1}}}), c(g_{j_{1}}) \rangle =1\), and \(\langle c(S_{\mathbf{f}}^{-1}{f_{i_{1}}}), c(f _{i_{2}}) \rangle\times \langle c(S_{\mathbf{g}}^{-1}{g_{j_{1}}}), c(g_{j_{2}}) \rangle \neq 0\).
Choose two linearly-independent nonzero vectors \(d_{1}, d_{2}\in \mathbb{C}^{m}\). Define \(\{\varOmega _{i, j}\}_{i=1, j=1}^{M, N}\) by \(\varOmega _{i, j}=\)
Then
by a simple computation. This implies that (2.24) holds for \((i_{1}, j_{1})\). The proof is completed.
$$ \textstyle\begin{cases} \frac{ (c_{1}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S_{\mathbf{f}}^{-1}{f_{i_{1}}})-d _{1}, c_{2}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S_{\mathbf{f}}^{-1}{f _{i_{1}}})-d_{2}, 0,\ldots, 0) }{ \langle c(S_{\mathbf{f}}^{-1}{f_{i_{1}}}), c(f_{i_{2}}) \rangle \langle c(S_{\mathbf{g}}^{-1}{g_{j_{1}}}), c(g_{j_{2}}) \rangle } &\text{if } (i,j)=(i_{2},j_{2}); \\ 0& \text{otherwise}. \end{cases} $$
$$ \varUpsilon _{i_{1}, j_{1}}= \begin{pmatrix} d_{1},&d_{2},&c_{3}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S_{\mathbf{f}} ^{-1}{f_{i_{1}}}),& \cdots,&c_{n}(S_{\mathbf{g}}^{-1}{g_{j_{1}}})c(S _{\mathbf{f}}^{-1}{f_{i_{1}}}) \end{pmatrix} $$
3 Tensor product dual expression
This section is devoted to expressing tensor product duals of tensor product frames. For this purpose, we first give the following lemma.
Lemma 3.1
Let
\(\{e_{i}\}_{i=1}^{m}\)
and
\(\{u_{j}\}_{j=1}^{n}\)
be orthonormal bases for
\(\mathcal{H}_{1}\)
and
\(\mathcal{H}_{2}\), respectively, \(\mathbf{f}=\{f_{i}\}_{i=1}^{M}\), \(\mathbf{\widetilde{f}}=\{ \widetilde{f}_{i}\}_{i=1}^{M}\)
be sequences in
\(\mathcal{H}_{1}\), and
\(\mathbf{g}=\{g_{j}\}_{j=1}^{N}\), \(\mathbf{\widetilde{g}}=\{ \widetilde{g}_{j}\}_{j=1}^{N}\)
be sequences in
\(\mathcal{H}_{2}\). Then, for any
\(f\in {\mathcal{H}}_{1}\)
and
\(g\in {\mathcal{H}}_{2}\),
and
where
\(T_{\mathbf{f}}\), \(T_{\widetilde{\mathbf{f}}}\), \(T_{\mathbf{g}}\), \(T_{\widetilde{\mathbf{g}}}\), \(M_{\mathbf{f}}\), \(M_{ \widetilde{\mathbf{f}}}\), \(M_{\mathbf{g}}\), and
\(M_{ \widetilde{\mathbf{g}}}\)
are defined as at the beginning of Sect. 2.
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\widetilde{\mathbf{f}}}f= \begin{pmatrix} e_{1},&e_{2},&\cdots,&e_{m} \end{pmatrix} M_{\mathbf{f}}M^{\ast }_{\widetilde{\mathbf{f}}}c(f) \end{aligned}$$
(3.1)
$$\begin{aligned} T_{\mathbf{g}}T^{\ast }_{\widetilde{\mathbf{g}}}g= \begin{pmatrix} u_{1},&u_{2},&\cdots,&u_{n} \end{pmatrix} M_{\mathbf{g}}M^{\ast }_{\widetilde{\mathbf{g}}}c(g), \end{aligned}$$
(3.2)
Proof
We only prove (3.1). (3.2) can be proved similarly. Arbitrarily fix \(1\leq i\leq m\). Since \(T^{\ast }_{ \widetilde{\mathbf{f}}}e_{i}=\{\overline{c_{i}(\widetilde{f_{l}})}\} _{l=1}^{M}\), we have
It follows that
for \(f\in {\mathcal{H}}_{1}\). The proof is completed. □
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\widetilde{\mathbf{f}}}e_{i} &= \begin{pmatrix} f_{1},&f_{2}&\cdots,&f_{m} \end{pmatrix} \begin{pmatrix} \overline{c_{i}(\widetilde{f_{1}})} \\ \overline{c_{i}(\widetilde{f_{2}})} \\ \vdots \\ \overline{c_{i}(\widetilde{f_{M}})} \end{pmatrix} \\ &= \begin{pmatrix} e_{1},&e_{2},&\cdots,&e_{m} \end{pmatrix} M_{\mathbf{f}} \begin{pmatrix} \overline{c_{i}(\widetilde{f_{1}})} \\ \overline{c_{i}(\widetilde{f_{2}})} \\ \vdots \\ \overline{c_{i}(\widetilde{f_{M}})} \end{pmatrix} . \end{aligned}$$
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\widetilde{\mathbf{f}}}f &=T_{\mathbf{f}}T ^{\ast }_{\widetilde{\mathbf{f}}} \begin{pmatrix} e_{1},&e_{2},&\cdots,&e_{m} \end{pmatrix} c(f) \\ &= \begin{pmatrix} e_{1},&e_{2},&\cdots,&e_{m} \end{pmatrix} M_{\mathbf{f}}M^{\ast }_{\widetilde{\mathbf{f}}}c(f) \end{aligned}$$
Theorem 3.1
Let
\(\{e_{i}\}_{i\in I}\)
and
\(\{u_{j}\}_{j\in J}\)
be orthonormal bases for
\(\mathcal{H}_{1}\)
and
\(\mathcal{H}_{2}\), respectively. Given sequences
\(\mathbf{f}=\{f_{i}\}_{i\in I}\), \(\mathbf{\widetilde{f}}=\{ \widetilde{f_{i}}\}_{i\in I}\)
in
\(\mathcal{H}_{1}\)
and
\(\mathbf{g}=\{g _{j}\}_{j\in J}\), \(\mathbf{\widetilde{g}}=\{\widetilde{g_{j}}\}_{j \in J}\)
in
\(\mathcal{H}_{2}\), \(\mathbf{f}\otimes \mathbf{g}\)
and
\(\mathbf{\widetilde{f}}\otimes \mathbf{\widetilde{g}}\)
form a pair of dual frames in
\(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\)
if and only if there exist constants
a, b
with
\(ab=1\), linear bounded operators
\(U, \widetilde{U}\)
on
\(\mathcal{H}_{1}\)
and
\(V, \widetilde{V}\)
on
\(\mathcal{H}_{2}\)
such that
and
$$\begin{aligned} &f_{i}=Ue_{i},\qquad \widetilde{f_{i}}=\widetilde{U}e_{i}\quad\textit{for }i \in I, \end{aligned}$$
(3.3)
$$\begin{aligned} &g_{j}=Vu_{j},\qquad \widetilde{g_{j}}=\widetilde{V}u_{j}\quad\textit{for }j \in J \end{aligned}$$
(3.4)
$$\begin{aligned} U{\widetilde{U}}^{\ast }=aI_{\mathcal{H}_{1}}, \qquad V{ \widetilde{V}}^{ \ast }=bI_{\mathcal{H}_{2}}. \end{aligned}$$
(3.5)
Proof
Observe that, if \(\operatorname{dim}{\mathcal{H}}_{1}<\infty (\operatorname{dim}{\mathcal{H}} _{2}<\infty )\), then \(\mathbf{f}\otimes \mathbf{g}\) and \(\mathbf{\widetilde{f}}\otimes \mathbf{\widetilde{g}}\) forming a pair of dual frames in \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) implies that f and \(\mathbf{\widetilde{f}}\) (g and \(\mathbf{\widetilde{g}}\)) are Riesz bases for \(\mathcal{H}_{1}\) (\(\mathcal{H}_{2}\)) due to \(\operatorname{card}(I)=\operatorname{dim}{\mathcal{H}}_{1}\) (\(\operatorname{card}(J)=\operatorname{dim}{\mathcal{H}}_{2}\)). This implies that there exist linear bounded and invertible operators U and Ũ (V and Ṽ) such that (3.3) ((3.4)) holds. For the case that \(\operatorname{dim}{\mathcal{H}}_{1}=\operatorname{dim}{\mathcal{H}}_{2}= \infty \), observe that a frame must be an image of an orthonormal basis under a linear bounded surjection, and that the bounded operators U, Ũ, V, and Ṽ satisfying (3.5) are surjections. We may as well assume that f, \(\mathbf{\widetilde{f}}\), g, and \(\mathbf{\widetilde{g}}\) have the following form:
where U and Ũ are linear bounded surjections on \(\mathcal{H}_{1}\), and V and Ṽ are linear bounded surjections on \(\mathcal{H}_{2}\). Observe that \(T_{\mathbf{f}}\), \(T _{\mathbf{\widetilde{f}}}\), \(T_{\mathbf{g}}\), and \(T_{ \mathbf{\widetilde{g}}}\) are all linear bounded operators. By Theorem 2.1, \(\mathbf{f}\otimes \mathbf{g}\) and \(\mathbf{\widetilde{f}}\otimes \mathbf{\widetilde{g}}\) are a pair of dual frames in \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) if and only if there exist constants a and b with \(ab=1\) such that
By (3.3) and (3.4), we have
for \(i\in I\). Similarly,
Therefore, (3.6) is equivalent to
for \(i\in I\) and \(j\in J\). This implies that (3.7) is equivalent to
due to the fact that \(\{e_{i}\}_{i\in I}\) and \(\{u_{j}\}_{j\in J}\) are orthonormal bases for \(\mathcal{H}_{1}\) and \(\mathcal{H}_{2}\), respectively. The proof is completed. □
$$\begin{aligned} &f_{i}=Ue_{i},\qquad \widetilde{f_{i}}= \widetilde{U}e_{i}\quad\text{for }i \in I, \\ &g_{j}=Vu_{j},\qquad \widetilde{g_{j}}= \widetilde{V}u_{j}\quad\text{for }j \in J, \end{aligned}$$
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\mathbf{\widetilde{f}}}=aI_{\mathcal{H}_{1}} \quad\text{and}\quad T_{\mathbf{g}}T^{\ast }_{\mathbf{\widetilde{g}}}=bI_{ \mathcal{H}_{2}}. \end{aligned}$$
(3.6)
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\mathbf{\widetilde{f}}}e_{i} &=\sum _{k\in I}\langle e_{i}, \widetilde{U}e_{k} \rangle Ue_{k} \\ &=U \biggl(\sum_{k\in I} \bigl\langle { \widetilde{U}}^{\ast }e_{i}, e_{k} \bigr\rangle e_{k} \biggr) \\ &=U{\widetilde{U}}^{\ast }e_{i} \end{aligned}$$
$$ T_{\mathbf{g}}T^{\ast }_{\mathbf{\widetilde{g}}}u_{j}=V{\widetilde{V}} ^{\ast }u_{j}\quad \text{for }j\in J. $$
$$\begin{aligned} U{\widetilde{U}}^{\ast }e_{i}=ae_{i} \quad\text{and}\quad V{\widetilde{V}}^{ \ast }u_{j}=bu_{j} \end{aligned}$$
(3.7)
$$ U{\widetilde{U}}^{\ast }=aI_{\mathcal{H}_{1}}\quad\text{and}\quad V{ \widetilde{V}} ^{\ast }=bI_{\mathcal{H}_{2}} $$
Remark 3.1
In Theorem 3.1, if \(\operatorname{dim}{\mathcal{H}}_{1}\), \(\operatorname{dim}{\mathcal{H}}_{2}<\infty \) in addition, then by the beginning argument in the proof of the theorem, f and g are Riesz bases for \(\mathcal{H}_{1}\) and \(\mathcal{H} _{2}\) respectively whenever \(\mathbf{f}\otimes \mathbf{g}\) is a frame for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\). Therefore, Theorem 3.1 cannot be applied to the case that \(\mathbf{f}\otimes \mathbf{g}\) is a redundant frame for \(\mathcal{H}_{1}\otimes {\mathcal{H}} _{2}\) with \(\operatorname{dim}{\mathcal{H}}_{1}\), \(\operatorname{dim}{\mathcal{H}}_{2}<\infty \). For the case that at least one of \(\operatorname{dim}{\mathcal{H}}_{1}\) and \(\operatorname{dim}{\mathcal{H}}_{2}\) is infinity, given a frame \(\mathbf{f} \otimes \mathbf{g}\) for \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) (at this time, U and V are determined by Proposition 1.2), Theorem 3.1 tells us that we may obtain all its tensor product dual frames by designing Ũ and Ṽ satisfying (3.3)–(3.5).
Remark 3.1 shows that Theorem 3.1 cannot be applied to general tensor product frames in a finite-dimensional setting. The following theorem presents an explicit expression of all tensor product dual frames in the finite-dimensional setting.
Theorem 3.2
Let
\(\{e_{i}\}_{i=1}^{m}\)
and
\(\{u_{j}\}_{j=1}^{n}\)
be orthonormal bases for
\(\mathcal{H}_{1}\)
and
\(\mathcal{H}_{2}\), respectively. Assume that
\(\mathbf{f}\otimes \mathbf{g}\)
is a frame for
\(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\)
with
\(\mathbf{f}=\{f_{i}\}_{i=1}^{M}\)
in
\(\mathcal{H}_{1}\)
and
\(\mathbf{g}=\{g_{j}\}_{j=1}^{N}\)
in
\(\mathcal{H}_{2}\). Then the dual frames of
\(\mathbf{f}\otimes \mathbf{g}\)
with the form of tensor product are precisely the families
$$\begin{aligned} &\bigl\{ \widetilde{\mathbf{f}}\otimes \widetilde{\mathbf{g}}: \mathbf{ \widetilde{f}}=\{\widetilde{f_{i}}\}_{i=1}^{M}, \mathbf{\widetilde{g}}=\{\widetilde{g_{j}}\}_{j=1}^{N} \textit{ and } M _{\mathbf{f}}M^{\ast }_{\widetilde{\mathbf{f}}}=aI, M_{\mathbf{g}}M ^{\ast }_{\widetilde{\mathbf{g}}}=bI \\ &\quad\textit{for some constants }a, b \textit{ satisfying }ab=1\bigr\} . \end{aligned}$$
Proof
Since \(\operatorname{dim}(\mathcal{H}_{1})=m\) and \(\operatorname{dim}(\mathcal{H}_{2})=n\), we have \(T_{\mathbf{f}}, T_{\mathbf{\widetilde{f}}}, T_{\mathbf{g}}\), and \(T_{\mathbf{\widetilde{g}}}\) are all linear bounded operators. By Theorem 2.1, \(\mathbf{f}\otimes \mathbf{g}\) and \(\mathbf{\widetilde{f}}\otimes \mathbf{\widetilde{g}}\) are a pair of dual frames in \(\mathcal{H}_{1}\otimes {\mathcal{H}}_{2}\) if and only if there exist constants a and b with \(ab=1\) such that
By Lemma 3.1, for any \(f\in {\mathcal{H}}_{1}\), \(g\in {\mathcal{H}} _{2}\), (3.8) is equivalent to
It follows that (3.9) is equivalent to
Then proof is completed. □
$$\begin{aligned} T_{\mathbf{f}}T^{\ast }_{\widetilde{\mathbf{f}}}=aI_{\mathcal{H}_{1}} \quad\mbox{and}\quad T_{\mathbf{g}}T^{\ast }_{\widetilde{\mathbf{g}}}=bI_{\mathcal{H} _{2}}. \end{aligned}$$
(3.8)
$$\begin{aligned} ( e_{1}, e_{2}, \ldots, e_{m} ) M_{\mathbf{f}}M^{\ast }_{\widetilde{\mathbf{f}}}c(f)=af \quad\mbox{and} \quad ( u_{1}, u_{2}, \ldots, u_{n} ) M_{\mathbf{g}}M^{\ast }_{\widetilde{\mathbf{g}}}c(g)=bg. \end{aligned}$$
(3.9)
$$ M_{\mathbf{f}}M^{\ast }_{\widetilde{\mathbf{f}}}=aI\quad\mbox{and}\quad M_{ \mathbf{g}}M^{\ast }_{\widetilde{\mathbf{g}}}=bI. $$
4 Conclusions
To construct dual frames with good structure is a fundamental problem in the theory of frames. The tensor product dual frames can provide a rank-one decomposition of bounded antilinear operators between two Hilbert spaces. This paper addresses tensor product dual frames. We characterize tensor product dual frames, give an explicit expression of all dual frames of tensor product frames, and prove the existence of non-tensor product dual frames. We present an explicit expression of all tensor product dual frames of tensor product frames.
Acknowledgements
The authors would like to thank the reviewers for their suggestions which greatly improve the readability of this article.
Competing interests
The authors declare that they have no competing interests.
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