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The equivalence of \(F_{a}\)frames
Journal of Inequalities and Applications volume 2020, Article number: 59 (2020)
Abstract
Structured frames such as wavelet and Gabor frames in \(L^{2}(\mathbb {R})\) have been extensively studied. But \(L^{2}(\mathbb{ R}_{+})\) cannot admit wavelet and Gabor systems due to \(\mathbb{R}_{+}\) being not a group under addition. In practice, \(L^{2}(\mathbb{R}_{+})\) models the causal signal space. The functionvalued inner productbased \(F_{a}\)frame for \(L^{2}(\mathbb{R}_{+})\) was first introduced by Hasankhani Fard and Dehghan, where an \(F_{a}\)frame was called a functionvalued frame. In this paper, we introduce the notions of \(F_{a}\)equivalence and unitary \(F_{a}\)equivalence between \(F_{a}\)frames, and present a characterization of the \(F_{a}\)equivalence and unitary \(F_{a}\)equivalence. This characterization looks like that of equivalence and unitary equivalence between frames, but the proof is nontrivial due to the particularity of \(F_{a}\)frames.
Introduction
An at most countable sequence \(\{e_{i}\}_{i\in I}\) in a separable Hilbert Space \(\mathcal{H}\) is called a frame for \(\mathcal{H}\) if there exist constants \(0< A\leq B<\infty\) such that
for \(f\in\mathcal{H}\). It was first introduced by Duffin and Schaeffer in [5] to study nonharmonic Fourier series, but had not attracted much attention until Daubechies, Grossman and Meyer published their joint work [4] in 1986. Now the theory of frames has seen great achievements in abstract spaces as well as in function spaces ([3, 10, 11, 13, 14, 18, 25]). In particular, structured frames in \(L^{2}(\mathbb {R})\) such as wavelet and Gabor frames have been extensively studied. However, structured frames in \(L^{2}(\mathbb {R}_{+})\) with \(\mathbb{R}_{+}=(0, \infty)\) have not. It is because \(\mathbb{R}\) is a group under addition but \(\mathbb{R}_{+}\) is not. This results in nonexistence of wavelet and Gabor systems in \(L^{2}(\mathbb {R}_{+})\). In practice, the time variable is nonnegative, and \(L^{2}(\mathbb {R}_{+})\) models the causal signal space. Motivated by this observation, some mathematicians studied Walsh seriesbased wavelet analysis in \(L^{2}(\mathbb {R}_{+})\) using Cantor group operation on \(\mathbb {R}_{+}\) ([1, 6–9, 16, 17]). Recently, Hasankhani Fard and Dehghan in [12] introduced the notion of functionvalued frame in \(L^{2}(\mathbb {R}_{+})\) which is referred to as “\(F_{a}\)frame” in our papers. Let us first recall and extend some related notions.
Given \(a>1\), a measurable function f on \(\mathbb {R}_{+}\) is said to be adilation periodic if \(f(a\cdot)=f(\cdot)\) a.e. on \(\mathbb {R}_{+}\), and a sequence \(\{f_{k}\} _{k\in {\mathbb {Z}} }\) of measurable functions on \(\mathbb {R}_{+}\) is said to be adilation periodic if every \(f_{k}\) is adilation periodic. Let \(L^{2}(\mathbb {Z}\times[1, a))\) denote the Hilbert space
equipped with the inner product
The following definition is an extension of [12, Definition 2.1], and that in [23] which only dealt with functions in \(L^{2}(\mathbb {R}_{+})\). It is slightly different from [12, Definition 2.1], even for functions in \(L^{2}(\mathbb {R}_{+})\), but it is more convenient for our purpose. By [12, Theorem 2.2], the \(F_{a}\)inner product herein has many properties similar to those of inner products.
Definition 1.1
Given \(a>1\), for f, \(g\in L^{2}(\mathbb {R}_{+})\) (\(L^{2}(\mathbb {Z}\times[1, a)) \)), the \(F_{a}\)inner product \(\langle f, g\rangle_{a}\) of f and g is defined as the adilation periodic function on \(\mathbb {R}_{+}\) given by
a.e. on \([1, a)\). The \(F_{a}\)norm \(\f\_{a}\) of f is defined as \(\f\_{a}(\cdot)=\sqrt{\langle f, f\rangle_{a}(\cdot)}\). And f and g are said to be \(F_{a}\)orthogonal if \(\langle f, g\rangle_{a}(\cdot)=0\) a.e. on \([1, a)\). In symbols, \(f\bot _{F_{a}}g\). It is to distinguish from the orthogonality “⊥” with respect to inner products.
Write
and let \(\{\varLambda_{m}\}_{m\in\mathbb {Z}}\) denote the adilation periodic function sequence on \(\mathbb {R}_{+}\) satisfying
The following proposition is taken from [23, Lemma 2.3] which dealt with \(L^{2}(\mathbb {R}_{+})\). A similar argument shows that it is true for \(L^{2}(\mathbb {Z}\times[1, a))\).
Proposition 1.1

(i)
\(\int_{[1, a)}f(x)^{2}\,dx=\sum_{m\in\mathbb {Z}} \vert\langle f, \varLambda_{m}\rangle_{L^{2}[1, a)} \vert^{2} \)for\(f\in L^{1}[1, a)\).

(ii)
For\(f, g\in L^{2}(\mathbb {R}_{+})\) (\(L^{2}(\mathbb {Z}\times[1, a)) \)) and\(\varphi\in B_{a}\), we have
$$\begin{aligned}& \langle f, g\rangle_{a}\in L^{1}[1, a),\qquad \langle f, \varphi g\rangle_{a}=\overline{\varphi} \langle f, g \rangle_{a}, \end{aligned}$$(1.3)$$\begin{aligned}& \langle f, g\rangle_{L^{2}(\mathbb {R}_{+})} = \int_{1}^{a}\langle f, g\rangle_{a}(x)\,dx \quad{\textit{if }}f, g\in L^{2}(\mathbb {R}_{+}), \end{aligned}$$(1.4)$$\begin{aligned}& \langle f, g\rangle_{L^{2}(\mathbb {Z}\times[1, a))}= \int_{1}^{a}\langle f, g\rangle_{a}(x)\,dx \quad{\textit{if }}f, g\in L^{2}\bigl(\mathbb {Z}\times[1, a)\bigr), \end{aligned}$$(1.5)$$\begin{aligned}& \Vert f+g \Vert _{a}^{2}(\cdot)= \Vert f \Vert _{a}^{2}(\cdot)+ \Vert g \Vert _{a}^{2}(\cdot)\quad{\textit{a.e. on }}[1, a) \textit{ if } f \perp_{F_{a}}g. \end{aligned}$$(1.6) 
(iii)
\(\sum_{m\in\mathbb {Z}} \vert\langle f, \varLambda_{m} g\rangle _{L^{2}(\mathbb {R}_{+})} \vert^{2}=\int_{1}^{a} \vert\langle f, g\rangle _{a}(x) \vert^{2}\,dx\)for\(f, g\in L^{2}(\mathbb {R}_{+})\), and\(\sum_{m\in\mathbb {Z}} \vert\langle f, \varLambda_{m} g\rangle _{L^{2}(\mathbb {Z}\times[1, a))} \vert^{2}=\int_{1}^{a} \vert\langle f, g\rangle _{a}(x) \vert^{2}\,dx \)for\(f, g\in L^{2}(\mathbb {Z}\times[1, a))\).

(iv)
For\(f, g\in L^{2}(\mathbb {R}_{+})\) (\(L^{2}(\mathbb {Z}\times[1, a)) \)), \(f\perp_{F_{a}}g\)if and only if\(f\perp\varLambda_{m}g\)for\(m\in \mathbb {Z}\).

(v)
For\(f, g\in L^{2}(\mathbb {R}_{+})\) (\(L^{2}(\mathbb {Z}\times[1, a)) \)), if\(f\perp_{F_{a}}g\), then\(f\perp\varphi\varLambda_{m}g\)for\(m\in\mathbb {Z}\)and\(\varphi\in B_{a}\).
The following definition is taken from [12, Definition 4.5] or [23, Definition 1.5].
Definition 1.2
A sequence \(\{f_{k}\}_{k\in\mathbb {Z}}\) in \(L^{2}(\mathbb {R}_{+})\) is called an \({F_{a}{\mbox{{{frame}}}}}\) for \(L^{2}(\mathbb {R}_{+})\) if there exist constants \(0< A\leq B<\infty\) such that, for each \(f\in L^{2}(\mathbb {R}_{+})\),
where A and B are called frame bounds. It is called a Parseval (tight) \(F_{a}\)frame for \(L^{2}(\mathbb {R}_{+})\) if \(A=B=1\) (\(A=B\)) in (1.7). And it is called an \({F_{a}{\mbox{{Bessel sequence}}}}\) in \(L^{2}(\mathbb {R}_{+})\) with Bessel bound B if the righthand side inequality of (1.7) holds.
For a sequence \(\{f_{k}\}_{k\in\mathbb {Z}}\) in \(L^{2}(\mathbb {R}_{+})\), its \(F_{a}\)span is defined by
and \(\overline{F_{a}\mbox{{{span}}}}\{f_{k}\}\) denotes the closure of \(F_{a}\)span\(\{f_{k}\}\) in \(L^{2}(\mathbb {R}_{+})\), where \(l_{0}(\mathbb {Z}^{2})\) is the set of finitely supported sequences on \(\mathbb {Z}^{2}\). We say \(\{ f_{k}\}_{k\in\mathbb {Z}}\) is \(F_{a}\)complete in \(L^{2}(\mathbb {R}_{+})\) if \(\overline{F_{a}\mbox{{{span}}}}\{f_{n}\}=L^{2}(\mathbb {R}_{+})\). By [23, Lemma 2.6], \(\{f_{k}\}_{k\in\mathbb {Z}}\) is \(F_{a}\)complete in \(L^{2}(\mathbb {R}_{+})\) if and only if \(f=0\) is a unique solution to
in \(L^{2}(\mathbb {R}_{+})\). And \(\{f_{k}\}_{k\in\mathbb {Z}}\) is called an \(F_{a}\)orthonormal system in \(L^{2}(\mathbb {R}_{+})\) if \(\langle f_{k}, f_{k'}\rangle_{a}(\cdot)=\delta_{k, k'}\) a.e. on \([1, a)\) for \(k, k'\in\mathbb {Z}\), and called an \(F_{a}\)orthonormal basis if it is an \(F_{a}\)orthonormal system and \(F_{a}\)complete in \(L^{2}(\mathbb {R}_{+})\).
Recall from [23, Theorem 2.2] and [12, Theorem 4.8] that a sequence \(\{f_{k}\}_{k \in\mathbb {Z}}\) in \(L^{2}(\mathbb {R}_{+})\) is an \(F_{a}\)Bessel sequence (\(F_{a}\)frame sequence, \(F_{a}\)frame) in \(L^{2}(\mathbb {R}_{+})\) if and only if \(\{\varLambda_{m}f_{k}\}_{m, k\in \mathbb {Z}}\) is a Bessel sequence (frame sequence, frame) in \(L^{2}(\mathbb {R}_{+})\) with the same bounds. Also by a standard argument, a sequence \(\{f_{k}\}_{k \in\mathbb {Z}}\) in \(L^{2}(\mathbb {R}_{+})\) is an \(F_{a}\)orthonormal system (\(F_{a}\)orthonormal basis) in \(L^{2}(\mathbb {R}_{+})\) if and only if \(\{\varLambda_{m}f_{k}\}_{m, k\in\mathbb {Z}}\) is an orthonormal system (orthonormal basis) in \(L^{2}(\mathbb {R}_{+})\). According to this, using “\(F_{a}\)”language we can say that \(F_{a}\)frames \(\{f_{k}\} _{k \in\mathbb {Z}}\) of the form \(f_{k}(\cdot)=a^{\frac{k}{2}}\psi (a^{k}\cdot)\) with \(\psi\in L^{2}(\mathbb {R}_{+})\) have been studied more. Li and Zhang in [22] characterized \(F_{a}\)frames, \(F_{a}\)dual frames and Parseval \(F_{a}\)frames for \(L^{2}(\mathbb {R}_{+})\) of the form \(\{ a^{\frac{k}{2}}\psi(a^{k}\cdot)\}_{k\in\mathbb{Z}}\), and as a special case, Li and Wang studied \(F_{a}\)frame sets in [21]. Its multiwindow and vectorvalued cases and another variation were studied in [20, 23, 24, 27]. By [22, Corollary 3.1], for \(0\ne\psi\in L^{2}(\mathbb {R}_{+})\), the following are equivalent:
 (i)
\(\{a^{\frac{k}{2}}\psi(a^{k}\cdot)\}_{k\in\mathbb{Z}}\) is a Parseval \(F_{a}\)frame for \(L^{2}(\mathbb {R}_{+})\).
 (ii)
\(\{a^{\frac{k}{2}}\psi(a^{k}\cdot)\}_{k\in\mathbb{Z}}\) is an \(F_{a}\)orthonormal basis.
 (iii)
\(\{a^{\frac{k}{2}}\psi(a^{k}\cdot)\}_{k\in\mathbb{Z}}\) is an \(F_{a}\)orthonormal system.
Obviously, we do not have a similar result for frames. On the other hand, recall from [3, Theorem 5.4.7] that removing one vector from a frame leaves either a frame or an incomplete set. Example 2.1 below in Sect. 2 tells us that a similar conclusion does not hold for \(F_{a}\)frames. It shows that removing one vector from an \(F_{a}\)frame possibly leaves an \(F_{a}\)complete set which is not an \(F_{a}\)frame.
From the above discussion, there exist essential differences between frames and \(F_{a}\)frames. This paper focuses on general \(F_{a}\)frames. Two frames \(\{f_{i}\}_{i\in I}\) and \(\{\widetilde {f_{i}}\}_{i\in I}\) for a separable Hilbert space \(\mathcal{H}\) are said be equivalent (unitarily equivalent) if there exists a bounded and invertible linear operator (unitary operator) T on \(\mathcal{H}\) such that \(\widetilde{f_{i}}=Tf_{i}\) for \(i\in I\). The following proposition is taken from [2, 11, 15].
Proposition 1.2
Let\(\{f_{i}\}_{i\in I}\)and\(\{ \widetilde{f_{i}}\}_{i\in I}\)be frames for a separable Hilbert space\(\mathcal{H}\). Then
 (i)
\(\{f_{i}\}_{i\in I}\)and\(\{\widetilde{f_{i}}\}_{i\in I}\)are equivalent if and only if their analysis operators have the same range, i.e.,
$$ \bigl\{ \bigl\{ \langle f, f_{i}\rangle\bigr\} _{i\in I}: f\in \mathcal{H} \bigr\} = \bigl\{ \bigl\{ \langle f, \widetilde {f_{i}}\rangle \bigr\} _{i\in I}: f\in\mathcal{H} \bigr\} . $$  (ii)
\(\{f_{i}\}_{i\in I}\)and\(\{\widetilde{f_{i}}\}_{i\in I}\)are unitarily equivalent if and only if
$$ \biggl\Vert \sum_{i\in I}c_{i}f_{i} \biggr\Vert = \biggl\Vert \sum_{i\in I}c_{i} \widetilde{f_{i}} \biggr\Vert \quad{\textit{for }}c\in l^{2}(I). $$
A natural question is whether Proposition 1.2 can be extended to “\(F_{a}\)frame” setting. This paper gives an affirmative answer. For this purpose, we first need to introduce “(unitary) equivalence” between \(F_{a}\)frames. It is different from that of frames due to the particularity of \(F_{a}\)frames.
Definition 1.3
Let \(\mathcal {H}\), \(\mathcal{K}=L^{2}(\mathbb {R}_{+})\) or \(L^{2}(\mathbb {Z}\times [1, a))\), a bounded linear operator \(T: \mathcal {H}\to\mathcal {K}\) is said to be afactorable if
Definition 1.4
Two \({F_{a}{\mbox{{{frames}}}}}\)\(F=\{ f_{k}\}_{k\in\mathbb{Z}}\) and \(\widetilde{F}=\{\tilde{f}_{k}\}_{k\in \mathbb{Z}}\) for \(L^{2}(\mathbb {R}_{+})\) are said to be \(F_{a}\)equivalent (unitarily \(F_{a}\)equivalent) if there exists an afactorable, bounded and invertible linear operator (afactorable and unitary operator) T on \(L^{2}(\mathbb{R}_{+})\) such that
Let \(F=\{f_{k}\}_{k \in\mathbb {Z}}\) be an \(F_{a}\)Bessel sequence in \(L^{2}(\mathbb {R}_{+})\). Define the \(F_{a}\)analysis operator\(D_{F}:L^{2}(\mathbb {R}_{+})\to L^{2}(\mathbb {Z}\times[1, a))\) and the \(F_{a}\)synthesis operator\(R_{F}:L^{2}(\mathbb {Z}\times[1, a))\to L^{2}(\mathbb {R}_{+})\) by
and
respectively. By [23, Theorem 2.1] they are well defined and bounded, and \(D_{F}^{*}=R_{F}\). The \(F_{a}\)frame operator\(S_{F}\) of F is defined by \(S_{F}=R_{F}D_{F}\). Obviously, these three operators are all afactorable. The main result of this paper is as follows.
Theorem 1.1
Let\(F=\{f_{k}\}_{k\in\mathbb{Z}}\)and\(\widetilde{F}=\{\widetilde{f_{k}}\}_{k\in\mathbb{Z}}\)be\(F_{a}\)frames for\(L^{2}(\mathbb{R}_{+})\). Then
 (i)
FandF̃are\(F_{a}\)equivalent if and only if
$$ \operatorname{range}(D_{F})=\operatorname{range} (D_{\widetilde{F}}). $$(1.11)  (ii)
FandF̃are unitarily\(F_{a}\)equivalent if and only if
$$ \Vert R_{F}g \Vert _{L^{2}(\mathbb{R}_{+})}= \Vert R_{\widetilde{F}}g \Vert _{L^{2}(\mathbb{R}_{+})}\quad{\textit{for }}g\in L^{2} \bigl(\mathbb{Z}\times[1, a)\bigr). $$(1.12)
The rest of this paper is organized as follows. Section 2 makes preparation for Theorem 1.1. Section 3 is devoted to proving Theorem 1.1.
Some preliminaries
This section is an auxiliary one. On one hand, we give an example that is an \(F_{a}\)frame, but when removing some element, it leaves an \(F_{a}\)complete set which is not an \(F_{a}\)frame for \(L_{2}(\mathbb {R}_{+})\). It is well known that removing an element from a frame leaves either a frame or an incomplete set. This demonstrates that \(F_{a}\)frames are very different from frames. On the other hand, we give some lemmas for later use. For this purpose, we first introduce some notations which are frequently used through the paper.
For a set E, we denote by \(\mathcal{X}_{E}\) the characteristic function of E. Given \(f_{0}\in L^{2}(\mathbb {R}_{+})\) (\(L^{2}(\mathbb {Z}\times[1, a))\)), a nonempty subset V of \(L^{2}(\mathbb {R}_{+})\) (\(L^{2}(\mathbb {Z}\times[1, a))\)) and an adilation periodic measurable function φ on \(\mathbb {R}_{+}\), \(f_{0}\bot_{F_{a}}V\) means that \(f_{0}\bot_{F_{a}}g\) for each \(g\in V\), φV, \(V(\varphi)\) and \(V^{\bot_{F_{a}}}\) denote the sets
respectively. Observe that \(\varphi V=V(\varphi)\) if \(\varphi\in B_{a}\). Let \(\mathcal{H}\) and \(\mathcal{K}\) be Hilbert spaces, and V be a closed linear subspace of \(\mathcal{H}\). We denote by \(V^{\bot}\) and \(P_{V}\) the orthogonal complement of V in \(\mathcal{H}\) and the orthogonal projection from \(\mathcal{H}\) onto V, respectively. For a bounded linear operator T from \(\mathcal{H}\) to \(\mathcal{K}\), we denote by \(T\mid_{V}\), \(T^{*}\), \(\operatorname{range} (T)\) and \(\ker (T)\) its restriction onto V, its adjoint operator, its range and its kernel, respectively. If T is also of closed range, we denote by \(T^{\dagger}\) the pseudoinverse of T, i.e.,
Example 2.1
Let \(a=2\). Define \(\{f_{k}\}_{k\in\mathbb {Z}}\) by
Then
 (i)
\(\{f_{k}\}_{k\in\mathbb{Z}}\) is an \(F_{a}\)frame for \(L^{2}(\mathbb{R}_{+})\).
 (ii)
\(\{f_{k}\}_{0\neq k\in\mathbb{Z}}\) is not an \(F_{a}\)frame, but it is \(F_{a}\)complete in \(L^{2}(\mathbb{R}_{+})\).
Proof
Obviously, \(\{f_{k}\}_{k\in\mathbb{Z}}\) is a sequence in \(L^{2}(\mathbb{R}_{+})\). By a standard computation, we have, for each \(f\in L^{2}(\mathbb{R}_{+})\),
From (2.6), it follows that, for each \(f\in L^{2}(\mathbb{R}_{+})\),
Thus (i) holds. Next we prove (ii). By (2.5) it follows that, for \(f\in L^{2}(\mathbb{R}_{+})\),
implies that \(f=0\). This shows that \(\{f_{k}\}_{0\neq k\in\mathbb {Z}}\) is \(F_{a}\)complete in \(L^{2}(\mathbb{R}_{+})\). Take \(f\in L^{2}(\mathbb{R}_{+})\) by
Then
But \(\sum_{0\neq k\in\mathbb{Z}}\langle f, f_{k}\rangle _{a}(x)^{2}=2^{1}\) for \(x\in[\frac{4}{3}, 2)\) by (2.5). Observe that \(\lim_{x\rightarrow2}\f\^{2}_{a}(x)=\infty\). It follows that there exists no positive constant A such that
Therefore, \(\{f_{k}\}_{0\neq k\in\mathbb{Z}}\) is not an \(F_{a}\)frame for \(L^{2}(\mathbb{R}_{+})\). □
By a standard argument, we have the following.
Lemma 2.1
Let\(\mathcal{A}\)be a bounded linear surjection from a Hilbert space\(\mathcal{H}\)onto another Hilbert space\(\mathcal{K}\). Then
By a standard argument similar to the case of frame, we have the following lemma, which is also a special case of [19, Lemma 2.5].
Lemma 2.2
Let\(\{f_{k}\}_{k\in\mathbb{Z}}\)be an\({F_{a}{\textit{{{frame}}}}}\)for\(L^{2}(\mathbb {R}_{+})\)with frame boundsAandB, and\(S_{F}\)be its frame operator. Then\(S_{F}\)is a bounded and invertible linear operator on\(L^{2}(\mathbb {R}_{+})\), \(\{ S_{F}^{1}f_{k}\} _{k\in\mathbb{Z}}\)is an\({F_{a}{\textit{{{frame}}}}}\)for\(L^{2}(\mathbb {R}_{+})\)with frame bounds\(B^{1}\)and\(A^{1}\), and
The following lemma demonstrates that the orthogonal complement operation preserves unimodular factor product invariant property of initial sets.
Lemma 2.3
Given\(\varphi\in B_{a}\)with\(\varphi=1\)and a nonempty subsetVof\(L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb {Z}\times[1, a)) \)), let\(\varphi V=V\). Then\(\varphi V^{\bot }=V^{\bot}\).
Proof
Observe that, for \(f\in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)), \(f \bot\varphi V\) if and only if \(\overline{\varphi}f\bot V\). It follows that \((\varphi V)^{\bot }=\varphi V^{\bot}\). On the other hand, \((\varphi V)^{\bot}=V^{\bot }\) if \(\varphi V=V\). Therefore, \(\varphi V^{\bot}=V^{\bot}\). □
The following lemma is an extension of [19, Lemma 2.3] which dealt with the subspaces of \(L^{2}(\mathbb{R}_{+})\). The proof herein is simpler than that of [19, Lemma 2.3].
Lemma 2.4
LetVbe a closed linear subspace of\(L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)). Then the following are equivalent:
 (i)
\(\Lambda_{m}V=V\)for\(m\in\mathbb{Z}\).
 (ii)
\(V^{\bot_{F_{a}}}=V^{\bot}\).
 (iii)
\(V(\varphi)\subset V\)for an arbitraryadilation periodic measurable function on\(\mathbb{R}_{+}\).
Proof
By Proposition 1.1(iv), for \(f\in L^{2}(\mathbb {R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)), \(f \bot _{F_{a}}V\) if and only if \(f \bot\Lambda_{m}V\) for each \(m\in\mathbb {Z}\). On the other hand, (i) is equivalent to \(V^{\bot}=(\Lambda _{m}V)^{\bot}\) for each \(m\in\mathbb{Z}\). It follows that (i) is equivalent to (ii). Since (i) is equivalent to \(\Lambda_{m}V\subset V\) for each \({m\in\mathbb{Z}}\), (iii) implies (i). Next we prove (ii) implies (iii) to finish the proof. Suppose (ii) holds. Observe that \(V^{\bot}\) is a closed subspace, and \(V^{\bot}=\Lambda_{m}V^{\bot}\) for each \(m\in\mathbb{Z}\) by Lemma 2.3. Applying the equivalence between (i) and (ii) to \(V^{\bot}\), we obtain
It follows that
by (ii). On the other hand,
for \(g\in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)) and adilation periodic measurable functions φ on \(\mathbb{R}_{+}\) satisfying \(\varphi g \in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)). It follows that \(V^{\bot _{F_{a}}}\subset(V(\varphi) )^{\bot_{F_{a}}}\), and thus
by (2.7). This leads to (iii) by the fact that \(V(\varphi) \subset[ (V(\varphi) )^{\bot_{F_{a}}} ]^{\bot_{F_{a}}}\). The proof is completed. □
Lemma 2.5
LetVbe a closed linear subspace of\(L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)) satisfying\(\Lambda_{m}V=V\)for each\(m\in\mathbb{Z}\). Then
for\(f\in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)) and\(\varphi\in B_{a}\).
Proof
Fix \(\varphi\in B_{a}\). Then \(V(\varphi)= \varphi V\). By Lemma 2.3, \(V^{\bot}\) is also a closed linear subspace of \(L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)) satisfying \(\Lambda_{m} V^{\bot}=V^{\bot}\) for each \(m\in\mathbb{Z}\). Applying Lemma 2.4 to \(V^{\bot}\) leads to \(\varphi V^{\bot }\subset V^{\bot}\). It follows that \(\varphi P_{V^{\bot}} f \in V^{\bot }\) for \(f\in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)), and thus
By Lemma 2.4, we have \(\varphi P_{V}f \in V\) which implies that
for \(f\in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)). This together with (2.8) leads to
for \(f\in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)). The proof is completed. □
Lemma 2.6
LetVandWbe closed subspaces of\(L^{2}(\mathbb{R}_{+})\)or\(L^{2}(\mathbb{Z}\times[1, a))\)satisfying\(\Lambda_{m}V=V\)and\(\Lambda_{m}W=W\)for each\(m\in\mathbb{Z}\), and\(T:V\rightarrow W\)be anafactorable bounded linear operator fromVtoW. Then
 (i)
\(T^{\ast}\), \(T^{\ast}T\)and\(TT^{\ast}\)areafactorable, and\(\langle Tf, g\rangle_{a}=\langle f, T^{\ast}g\rangle_{a}\)for\(f\in V\), \(g\in W\).
 (ii)
\(T^{1}\)isafactorable ifTis invertible.
Proof
For simplicity, for \(f\in V\) and \(g\in W\), we use \(\langle Tf, g\rangle\) and \(\langle f, T^{\ast}g\rangle\) to denote the inner products of Tf and g, and f and \(T^{\ast}g\) in the corresponding spaces, i.e.,
and
(i) If \(T^{\ast}\) is afactorable, so are \(T^{\ast}T\) and \(TT^{\ast }\) since T is afactorable. Arbitrarily fix \(f\in V\), \(g\in W\) and \(\varphi\in B_{a}\). Then \(\varphi V\subset V\) and \(\varphi W \subset W\) by Lemma 2.4. Since T is afactorable,
and
It follows that
and thus \(T^{\ast}(\overline{\varphi}g)=\overline{\varphi}T^{\ast }g\) by the arbitrariness of f. And again by the arbitrariness of φ and g, \(T^{\ast}\) is afactorable. Next we prove that
Observe that, for \(f\in V\) and \(g\in W\), \(\langle Tf, g\rangle=\langle f, T^{\ast}g\rangle\). By Proposition 1.1(ii), it may be rewritten as
Given an arbitrary \(E\subset[1, a)\) with \(E>0\), replace f by \(\mathcal{X}_{\bigcup_{j\in\mathbb{Z}}a^{j}E}f\) in (2.10) (this can be done by Lemma 2.4). Then we have
due to the fact that T is afactorable. It leads to (2.9) by the arbitrariness of E and [26, Theorem 1.40].
(ii) Suppose T is invertible. For \(g\in W\) and \(\varphi\in B_{a}\), we have
Since T is afactorable. It follows that \(\varphi T^{1}g=T^{1}(\varphi g)\) for \(g\in W\). The proof is completed. □
Proof of Theorem 1.1
Proof of Theorem 1.1
(i) Necessity. Suppose F and F̃ are \(F_{a}\)equivalent. Then there exists an afactorable, bounded and invertible linear operator T on \(L^{2}(\mathbb{R}_{+})\) such that
By Lemma 2.6, it follows that, for each \(f\in L^{2}(\mathbb{R}_{+})\),
Since T is bounded and invertible, so is \(T^{\ast}\). This implies that \(\operatorname{range} (T^{\ast})=L^{2}(\mathbb{R}_{+})\). Therefore, (3.1) implies that \(\operatorname{range} (D_{F})=\operatorname{range}(D_{\widetilde{F}})\).
Sufficiency. Suppose \(\operatorname{range} (D_{F})=\operatorname{range}(D_{\widetilde{F}})=V\). Obviously, \(\Lambda_{m}V=V\) for \(m\in\mathbb{Z}\). By Lemma 2.2 and [23, Theorem 2.1], \(\operatorname{range}(R_{F})=L^{2}(\mathbb{R}_{+})\). This implies that V is closed due to the fact that \(D_{F}=R^{\ast}_{F}\). Let \(R_{F} _{V}\) be the restriction of \(R_{F}\) on V. We first claim that \((R_{F} _{V} )^{\ast} (R_{F} _{V} )\) is bounded and invertible, and its inverse \([(R_{F} _{V})^{\ast}(R_{F} _{V}) ]^{1}\) is afactorable. Let us check it. For \(g\in V\) and \(f\in L^{2}(\mathbb{R}_{+})\), we have
This implies that
by the arbitrariness of g. Since \(\operatorname{range}(R_{F})=L^{2}(\mathbb{R}_{+})\) and \(R^{\ast}_{F}=D_{F}\),
Also observe that
due to \((\operatorname{ker}(R_{F}))^{\bot}=V\). Collecting (3.2)–(3.4) gives
Since \(R_{F} _{V}\) is injective, so is \((R_{F} _{V} )^{\ast} (R_{F} _{V} )\). This together with (3.5) leads to \((R_{F} _{V} )^{\ast} (R_{F} _{V} )\) being a bounded bijection on V. On the other hand, \(R_{F} _{V}\) is afactorable since \(R_{F}\) is afactorable and \(\Lambda_{m}V=V\) for \(m\in\mathbb{Z}\). By Lemma 2.6, \((R_{F} _{V} )^{\ast} (R_{F} _{V} )\) and \([(R_{F} _{V})^{\ast}(R_{F} _{V}) ]^{1}\) are both afactorable. We have proved the claim. Now we define \(T:L^{2}(\mathbb{R}_{+})\rightarrow L^{2}(\mathbb{R}_{+})\) by
Then it is well defined and bounded. Next we prove that T is an afactorable bijection satisfying \(\widetilde{f_{k}}=Tf_{k}\) for \(k\in\mathbb{Z}\) to finish the proof of sufficiency. By Lemma 2.2 and [23, Theorem 2.1] and the fact that \((\operatorname{ker}(R_{\widetilde{F}}) )^{\bot}=V\), we have
Also observing \((R_{F} _{V} )^{\ast} (R_{F} _{V} )\) being a bijection on V leads to
It follows that
by (3.6). Since F is an \(F_{a}\)frame for \(L^{2}(\mathbb {R}_{+})\), \(D_{F}\) is injective, and thus
is injective. Also \(R_{\widetilde{F}}\) is injective when restricted on V, and
It follows that T is injective. Therefore, T is bijective. Since \(R_{\widetilde{F}}\), \(D_{F}\) and \([(R_{F}_{V})^{\ast }(R_{F}_{V}) ]^{1}\) are all afactorable by Lemma 2.6, so is T. Finally, we prove that
For \(k\in\mathbb{ Z}\), define \(e^{(k)}=\{e^{(k)}_{l}(\cdot)\}_{l\in \mathbb{Z}}\in L^{2}(\mathbb{Z}\times[1, a))\) by
on \([1, a)\). Then, for \(k\in\mathbb{Z}\),
due to \(V^{\bot}=\operatorname{ker}(R_{F})\), and thus
It follows that
by (3.2). Also observing that
leads to (3.8).
(ii) Necessity. Suppose F and F̃ are unitarily \(F_{a}\)equivalent. Then there exists an afactorable and unitary operator T on \(L^{2}(\mathbb{R}_{+})\) such that
It follows that
for \(g\in L^{2}(\mathbb{Z}\times[1, a))\), and thus (1.12) holds by the unitarity of T.
Sufficiency. Suppose (1.12) holds. Then \(\operatorname{ker}(R_{F})=\operatorname{ker}(R_{\widetilde{F}})\). On the other hand, \(\operatorname{range}(D_{F})\) and \(\operatorname{range}(D_{\widetilde{F}})\) are closed by the arguments in (i). It follows that
due to the fact that \(R^{\ast}_{F}=D_{F}\) and \(R^{\ast }_{F}=D_{\widetilde{F}}\). Therefore, F and F̃ are \(F_{a}\)equivalent, and
is an afactorable, bounded bijection on \(L^{2}(\mathbb{R}_{+})\) satisfying \(\widetilde{f_{k}}=Tf_{k}\) for \(k\in\mathbb{Z}\) by (i) and its proof. Next we prove that T is unitary to finish the proof. Write \(V=\operatorname{range}(D_{F})=\operatorname{range}(D_{\widetilde{F}})\), and define \(\widetilde{D}_{F}:L^{2}(\mathbb{R}_{+})\rightarrow V\) by
Observe that \(\widetilde{D}_{F}\) is different from \({D}_{F}\) since V need not be equal to \(L^{2}(\mathbb{Z}\times[1, a))\) and \({D}_{F}\) is from \(L^{2}(\mathbb {R}_{+})\) to \(L^{2}(\mathbb{Z}\times[1, a))\). Obviously, it is well defined. For \(f\in L^{2}(\mathbb{R}_{+})\) and \(g\in V\), we have
It follows that
Since \(\widetilde{D}_{F}\) is surjective, we have
by Lemma 2.1. Thus
by (3.9). Also observe that
for \(f\in L^{2}(\mathbb{R}_{+})\). It follows that
for \(f\in L^{2}(\mathbb{R}_{+})\) due to \(D_{F}\) being injective. Again substituting g for
in (1.12), we obtain
This shows that T is normpreserving and thus is unitary. The proof is completed. □
Conclusions
The space \(L^{2}(\mathbb {R}_{+})\) does not admit wavelet and Gabor systems due to \(\mathbb {R}_{+}\) being not a group under addition. This paper addresses the \(F_{a}\)frame for \(L^{2}(\mathbb {R}_{+})\). We introduce the notions of \(F_{a}\)equivalence and unitary \(F_{a}\)equivalence between \(F_{a}\)frames, and characterize the \(F_{a}\)equivalence and unitary \(F_{a}\)equivalence. This characterization looks like that of equivalence and unitary equivalence between frames, but the proof is nontrivial due to the particularity of \(F_{a}\)frames.
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Hussain, T., Li, YZ. The equivalence of \(F_{a}\)frames. J Inequal Appl 2020, 59 (2020). https://doi.org/10.1186/s1366002002331x
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DOI: https://doi.org/10.1186/s1366002002331x
MSC
 42C15
 42C40
 47A80
Keywords
 Frame
 \(F_{a}\)Frame
 \(F_{a}\)Equivalence