1 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 series-based 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 function-valued 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.
$$ A \Vert f \Vert ^{2}\leq\sum_{i\in I} \bigl\vert \langle f, f_{i}\rangle\bigr\vert ^{2}\leq B \Vert f \Vert ^{2} $$
Given \(a>1\), a measurable function f on \(\mathbb {R}_{+}\) is said to be a-dilation 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 a-dilation periodic if every \(f_{k}\) is a-dilation periodic. Let \(L^{2}(\mathbb {Z}\times[1, a))\) denote the Hilbert space equipped with the inner product
$$ L^{2}\bigl(\mathbb {Z}\times[1, a)\bigr)=\biggl\{ f=\{f_{k}\}_{k \in\mathbb {Z}}: \int^{a}_{1}\sum_{k \in{\mathbb {Z}}} \bigl\vert f_{k}(x) \bigr\vert ^{2}\,dx< \infty, \{f_{k}\}_{k\in\mathbb {Z}}{\mbox{ is $a$-dilation periodic}} \biggr\} $$
$$ \langle f, g\rangle_{L^{2}(\mathbb {Z}\times[1, a))} = \int_{1}^{a}\sum_{k \in{\mathbb {Z}}}f_{k}(x) \overline{g_{k}(x)}\,dx\quad{\mbox{for }}f,g\in L^{2}\bigl(\mathbb {Z}\times[1, a)\bigr). $$
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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 a-dilation 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.
$$ \langle f, g\rangle_{a}(\cdot)=\sum _{j\in\mathbb {Z}}a^{j}f\bigl(a^{j}\cdot\bigr) \overline{g\bigl(a^{j}\cdot\bigr)} \quad\biggl(\langle f, g \rangle_{a}(\cdot)=\sum_{k \in{\mathbb {Z}}}f_{k}( \cdot)\overline{g_{k}(\cdot)} \biggr) $$
(1.1)
Write and let \(\{\varLambda_{m}\}_{m\in\mathbb {Z}}\) denote the a-dilation 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 \(L^{2}(\mathbb {Z}\times[1, a))\).
$$B_{a}=\bigl\{ f\in L^{\infty}(\mathbb {R}_{+}):f {\mbox{ is }} a{ \mbox{-dilation periodic}}\bigr\} , $$
$$ \varLambda_{m}(\cdot)=\frac{1}{\sqrt{a-1}}e^{2\pi i\frac{m\cdot }{a-1}}\quad \mbox{on }[1, a). $$
(1.2)
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\(\varphi\in B_{a}\).
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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 right-hand side inequality of (1.7) holds.
$$ A \Vert f \Vert ^{2}_{a}(\cdot)\leq\sum _{k\in \mathbb{Z}} \bigl\vert \langle f, f_{k} \rangle_{a}(\cdot) \bigr\vert ^{2}_{a}\leq B \Vert f \Vert ^{2}_{a}(\cdot) \quad{\mbox{a.e. on }} [1, a), $$
(1.7)
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 \(L^{2}(\mathbb {R}_{+})\).
$$ F_{a}{\mbox{{{-span}}}}\{f_{n}\}= \biggl\{ \sum_{k, m\in\mathbb {Z}}c_{k, m}\varLambda_{m}f_{k}: c=\{c_{k, m}\}_{k, m\in\mathbb {Z}}\in l_{0}\bigl(\mathbb {Z}^{2}\bigr) \biggr\} , $$
(1.8)
$$\langle f, f_{k}\rangle_{a}(\cdot)=0 \quad{\mbox{a.e. on }}[1, a) {\mbox{ for }}k\in\mathbb {Z} $$
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 multi-window and vector-valued 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 a-factorable if
$$T(\varphi f ) = \varphi T(f) \quad{\mbox{for all }}f\in\mathcal{H} {\mbox{ and }} \varphi\in B_{a}. $$
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 a-factorable, bounded and invertible linear operator (a-factorable and unitary operator) T on \(L^{2}(\mathbb{R}_{+})\) such that
$$Tf_{k}=\tilde{f}_{k} \quad{\mbox{for }}k\in\mathbb {Z}. $$
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 a-factorable. The main result of this paper is as follows.
$$ D_{F}f=\bigl\{ \langle f, f_{k} \rangle_{a}\bigr\} _{k\in\mathbb {Z}} \quad{\mbox{for }}f\in L^{2}(\mathbb {R}_{+}) $$
(1.9)
$$ R_{F}g=\sum_{k\in\mathbb {Z}}g_{k}f_{k} \quad{\mbox{for }}g\in L^{2}\bigl(\mathbb {Z}\times[1, a)\bigr), $$
(1.10)
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)
2 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 a-dilation 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 pseudo-inverse of T, i.e.,
$$\begin{aligned}& \varphi V= \{ \varphi f: f\in V \}, \end{aligned}$$
(2.1)
$$\begin{aligned}& V(\varphi)= \bigl\{ \varphi f: f\in V, \varphi f\in L^{2}(\mathbb {R}_{+}) \bigr\} \quad{\mbox{if }}V\subset L^{2}(\mathbb {R}_{+}), \end{aligned}$$
(2.2)
$$\begin{aligned}& V(\varphi)= \bigl\{ \varphi f: f\in V, \varphi f\in L^{2}\bigl(\mathbb {Z}\times[1, a)\bigr) \bigr\} \quad{\mbox{if }}V\subset L^{2}\bigl(\mathbb {Z}\times[1, a)\bigr), \end{aligned}$$
(2.3)
$$\begin{aligned}& V^{\bot_{F_{a}}}= \{ f: f\bot_{F_{a}}g {\mbox{ for each }} g\in V \}, \end{aligned}$$
(2.4)
$$T^{\dagger}(y+z)= (T\mid_{(\ker(T))^{\bot}} )^{-1}y \quad{\mbox{for }} y\in \operatorname{range}(T) {\mbox{ and }}z\in\bigl(\operatorname{range}(T)\bigr)^{\bot}. $$
Example 2.1
Let \(a=2\). Define \(\{f_{k}\}_{k\in\mathbb {Z}}\) by
$$f_{k}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} 2^{\frac{k}{2}}\mathcal{X}_{[2^{-k}, 2^{-k+1})}(x) &\text{if $k\geq0$}; \\ \mathcal{X}_{[\frac{2}{3}, \frac{4}{3})}(x)+(2-x)^{\frac {1}{3}}\mathcal{X}_{[\frac{4}{3}, 2)}(x) &\text{if $k=-1$}; \\ [3pt] 2^{\frac{k+1}{2}}\mathcal{X}_{[2^{-k-1}, 2^{-k})}(x)& \text{if $k\leq-2$}. \end{array}\displaystyle \right . $$
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 \(f\in L^{2}(\mathbb{R}_{+})\),
$$\begin{aligned}& \sum_{0\neq k\in\mathbb{Z}} \bigl|\langle f, f_{k}\rangle_{a}(\cdot ) \bigr|^{2}=\left \{ \textstyle\begin{array}{l@{\quad}l} \Vert f \Vert ^{2}_{a}(\cdot) &\text{a.e. on $[1, \frac {4}{3})$}; \\ \sum_{0\neq j\in\mathbb{Z}}2^{j} \vert f(2^{j}\cdot ) \vert^{2}\\ \quad{}+ \vert2^{-1}f(2^{-1}\cdot)+f(\cdot)(2-\cdot)^{\frac {1}{3}} \vert^{2}& \text{a.e. on $[\frac{4}{3}, 2)$}, \end{array}\displaystyle \right . \end{aligned}$$
(2.5)
$$\begin{aligned}& \sum_{k\in\mathbb{Z}} \bigl\vert \langle f, f_{k}\rangle_{a}(\cdot) \bigr\vert ^{2}=\left \{ \textstyle\begin{array}{l@{\quad}l} \Vert f \Vert ^{2}_{a}(\cdot)+ \vert f(\cdot) \vert^{2} &\text{a.e. on $[1, \frac {4}{3})$}; \\ \Vert f \Vert ^{2}_{a}(\cdot)+ \vert2^{-1}f(2^{-1}\cdot)+f(\cdot )(2-\cdot)^{\frac {1}{3}} \vert^{2}& \text{a.e. on $[\frac{4}{3}, 2)$}. \end{array}\displaystyle \right . \end{aligned}$$
(2.6)
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 \(L^{2}(\mathbb{R}_{+})\). □
$$ \Vert f \Vert _{a}^{2}(\cdot)\leq\sum _{k\in\mathbb {Z}} \bigl\vert \langle f, f_{k} \rangle_{a}(\cdot) \bigr\vert ^{2}\leq3 \Vert f \Vert _{a}^{2}(\cdot)\quad{\mbox{a.e. on }}[1, 2). $$
$$\sum_{0\neq k\in\mathbb {Z}}\bigl|\langle f, f_{k}\rangle_{a}(\cdot)\bigr|^{2}=0\quad \text{a.e. on }[1, a)$$
$$ f(x)= \mathcal{X}_{[\frac{2}{3}, \frac {4}{3})}(x)-2^{-1}(2-x)^{-\frac{1}{3}} \mathcal{X}_{[\frac{4}{3}, 2)}(x). $$
$$ \Vert f \Vert _{a}^{2}(x)=2^{-1}+2^{-2}(2-x)^{-\frac {2}{3}} \quad{\mbox{for }}x\in\bigl[\frac{4}{3}, 2\bigr). $$
$$ A \Vert f \Vert ^{2}_{a}(\cdot)\leq\sum _{0\neq k\in\mathbb {Z}} \bigl\vert \langle f, f_{k} \rangle_{a}(\cdot) \bigr\vert ^{2}\quad{\mbox{a.e. on }}[1, 2). $$
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
$$ \mathcal{A}^{\dagger} = \mathcal{A}^{\ast}\bigl(\mathcal{A} \mathcal{A}^{\ast}\bigr)^{-1}. $$
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
$$ f=\sum_{k\in\mathbb{Z}}\bigl\langle f, S_{F}^{-1}f_{k} \bigr\rangle _{a}f_{k}\quad{\textit{for }}f\in L^{2}( \mathbb {R}_{+}). $$
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 arbitrarya-dilation 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 a-dilation 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. □
$$ V=\bigl(V^{\bot}\bigr)^{\bot}=\bigl(V^{\bot} \bigr)^{\bot _{F_{a}}}. $$
$$ V=\bigl(V^{\bot_{F_{a}}}\bigr)^{\bot_{F_{a}}} $$
(2.7)
$$ \langle f, \varphi g\rangle_{a}=\varphi\langle f, g \rangle_{a} $$
$$ \bigl[ \bigl(V(\varphi) \bigr)^{\bot_{F_{a}}} \bigr]^{\bot _{F_{a}}}\subset V $$
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\(m\in\mathbb{Z}\). Thenfor\(f\in L^{2}(\mathbb{R}_{+})\) (\(L^{2}(\mathbb{Z}\times[1, a)) \)) and\(\varphi\in B_{a}\).
$$ P_{V}(\varphi f)=\varphi P_{V}f $$
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. □
$$ P_{V}(\varphi P_{V^{\bot}} f)=0\quad{\mbox{for }}f \in L^{2}(\mathbb{R}_{+})\text{ }\bigl(L^{2}\bigl( \mathbb{Z}\times[1, a)\bigr) \bigr). $$
(2.8)
$$ P_{V}(\varphi P_{V}f)=\varphi P_{V}f $$
$$\begin{aligned} P_{V}(\varphi f)={}&P_{V}(\varphi P_{V}f + \varphi P_{V^{\bot}} f) \\ ={}&\varphi P_{V}f \end{aligned}$$
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 ana-factorable bounded linear operator fromVtoW. Then
(i)
\(T^{\ast}\), \(T^{\ast}T\)and\(TT^{\ast}\)area-factorable, and\(\langle Tf, g\rangle_{a}=\langle f, T^{\ast}g\rangle_{a}\)for\(f\in V\), \(g\in W\).
(ii)
\(T^{-1}\)isa-factorable 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
$$\langle Tf, g\rangle=\left \{ \textstyle\begin{array}{l@{\quad}l} \langle Tf, g\rangle_{L^{2}(\mathbb{R}_{+})} &\text{if $W\subset L^{2}(\mathbb{R}_{+})$}; \\ \langle Tf, g\rangle_{L^{2}(\mathbb{Z}\times[1, a))}& \text{if $W\subset L^{2}(\mathbb{Z}\times[1, a))$} \end{array}\displaystyle \right . $$
$$\bigl\langle f, T^{\ast}g\bigr\rangle =\left \{ \textstyle\begin{array}{l@{\quad}l} \langle f, T^{\ast}g\rangle_{L^{2}(\mathbb{R}_{+})} &\text{if $V\subset L^{2}(\mathbb{R}_{+})$}; \\ \langle f, T^{\ast}g\rangle_{L^{2}(\mathbb{Z}\times[1, a))}& \text{if $V\subset L^{2}(\mathbb{Z}\times[1, a))$}. \end{array}\displaystyle \right . $$
(i) If \(T^{\ast}\) is a-factorable, so are \(T^{\ast}T\) and \(TT^{\ast }\) since T is a-factorable. 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 a-factorable, 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 a-factorable. 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 a-factorable. It leads to (2.9) by the arbitrariness of E and [26, Theorem 1.40].
$$ \bigl\langle T(\varphi f), g\bigr\rangle =\langle\varphi Tf, g\rangle=\langle Tf, \overline{\varphi}g\rangle=\bigl\langle f, T^{\ast }(\overline{\varphi}g) \bigr\rangle $$
$$ \bigl\langle T(\varphi f), g\bigr\rangle =\bigl\langle \varphi f, T^{\ast}g \bigr\rangle =\bigl\langle f, \overline{\varphi}T^{\ast}g\bigr\rangle . $$
$$\bigl\langle f, T^{\ast}(\overline{\varphi}g)\bigr\rangle =\bigl\langle f, \overline{\varphi}T^{\ast}g \bigr\rangle , $$
$$ \langle Tf, g\rangle_{a}=\bigl\langle f, T^{\ast}g\bigr\rangle _{a}\quad{\mbox{for }}f\in V {\mbox{ and }}g\in W. $$
(2.9)
$$ \int_{1}^{a}\langle Tf, g \rangle _{a}(x)\,dx= \int_{1}^{a}\bigl\langle f, T^{\ast}g\bigr\rangle _{a}(x)\,dx\quad{\mbox{for }}f\in V {\mbox{ and }}g\in W. $$
(2.10)
$$ \int_{E}\langle Tf, g\rangle_{a}(x)\,dx= \int_{E}\bigl\langle f, T^{\ast}g\bigr\rangle _{a}(x)\,dx $$
(ii) Suppose T is invertible. For \(g\in W\) and \(\varphi\in B_{a}\), we have Since T is a-factorable. It follows that \(\varphi T^{-1}g=T^{-1}(\varphi g)\) for \(g\in W\). The proof is completed. □
$$ T\bigl(\varphi T^{-1}g\bigr)=\varphi TT^{-1}g=\varphi g $$
3 Proof of Theorem 1.1
Proof of Theorem 1.1
(i) Necessity. Suppose F and F̃ are \(F_{a}\)-equivalent. Then there exists an a-factorable, 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 \(\operatorname{range} (D_{F})=\operatorname{range}(D_{\widetilde{F}})\).
$$ Tf_{k}= \widetilde{f_{k}}\quad{\mbox{for }}k\in\mathbb{Z}. $$
$$ \langle f, \widetilde{f_{k}}\rangle_{a}=\bigl\langle T^{\ast}f, f_{k}\bigr\rangle _{a}\quad{ \mbox{for }}k\in\mathbb{Z}. $$
(3.1)
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 a-factorable. 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 a-factorable since \(R_{F}\) is a-factorable 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 a-factorable. 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 a-factorable 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 a-factorable 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).
$$\begin{aligned} \bigl\langle g, (R_{F}|_{V})^{\ast}f\bigr\rangle _{L^{2}(\mathbb{Z}\times [1, a))}={}&\bigl\langle (R_{F}|_{V})g, f \bigr\rangle _{L^{2}(\mathbb {R}_{+})} \\ ={}&\langle R_{F}g, f\rangle_{L^{2}(\mathbb{R}_{+})} \\ ={}&\bigl\langle g, R^{\ast}_{F}f\bigr\rangle _{L^{2}(\mathbb {Z}\times[1, a))}. \end{aligned}$$
$$ (R_{F}|_{V})^{\ast}f=R^{\ast }_{F}f \quad{\mbox{for }}f\in L^{2}(\mathbb{R}_{+}) $$
(3.2)
$$ V=\operatorname{range}\bigl(R^{\ast}_{F}\bigr)=\operatorname{range} \bigl(R^{\ast}_{F}R_{F}\bigr). $$
(3.3)
$$ \operatorname{range}(R_{F})=R_{F}\bigl[ \bigl(\operatorname{ker}(R_{F})\bigr)^{\bot }\bigr]=R_{F}(V)=\operatorname{range}(R_{F}|_{V}) $$
(3.4)
$$ V=\operatorname{range}\bigl[(R_{F}|_{V})^{\ast}(R_{F}|_{V}) \bigr]. $$
(3.5)
$$ T=R_{\widetilde{F}} \bigl[(R_{F}|_{V})^{\ast }(R_{F}|_{V}) \bigr]^{-1}D_{F}. $$
$$ L^{2}(\mathbb{R}_{+})=\operatorname{range} (R_{\widetilde {F}})=R_{\widetilde{F}}(V). $$
(3.6)
$$ V=\operatorname{range} \bigl( \bigl[(R_{F}|_{V} )^{\ast} (R_{F}|_{V})\bigr]^{-1}D_{F} \bigr). $$
$$ L^{2}(\mathbb{R}_{+})=\operatorname{range}(T) $$
(3.7)
$$\bigl[(R_{F}|_{V})^{\ast}(R_{F}|_{V}) \bigr]^{-1}D_{F} $$
$$\operatorname{range} \bigl(\bigl[(R_{F}|_{V})^{\ast}(R_{F}|_{V}) \bigr]^{-1}D_{F} \bigr)\subset V. $$
$$ \widetilde{f_{k}}=Tf_{k}\quad{\mbox{for }}k\in \mathbb{Z}. $$
(3.8)
$$e^{(k)}_{l}(\cdot)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0 &\text{if $l\neq k$}; \\ 1& \text{if $l=k$} \end{array}\displaystyle \right . $$
$$ f_{k}=R_{F} e^{(k)}= R_{F} \bigl(P_{V} e^{(k)}+P_{V^{\bot}} e^{(k)} \bigr)=R_{F}P_{V} e^{(k)} $$
$$ Tf_{k}=R_{\widetilde{F}} \bigl[(R_{F}|_{V})^{\ast }(R_{F}|_{V}) \bigr]^{-1}R^{\ast}_{F}R_{F}P_{V} e^{(k)}. $$
$$\begin{aligned} Tf_{k}={}&R_{\widetilde{F}} \bigl[(R_{F}|_{V})^{\ast }(R_{F}|_{V}) \bigr]^{-1}(R_{F}|_{V})^{\ast}(R_{F}|_{V})P_{V} e^{(k)} \\ ={}&R_{\widetilde{F}}P_{V} e^{(k)} \end{aligned}$$
$$ \widetilde{f_{k}}=R_{\widetilde{F}} e^{(k)}=R_{\widetilde{F}}P_{V} e^{(k)} $$
(ii) Necessity. Suppose F and F̃ are unitarily \(F_{a}\)-equivalent. Then there exists an a-factorable 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.
$$ Tf_{k}=\widetilde{f_{k}}\quad{\mbox{for }}k\in\mathbb{Z}. $$
$$ R_{\widetilde{F}}g=\sum_{k\in\mathbb {Z}}g_{k}Tf_{k}=TR_{F}g $$
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 a-factorable, 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 norm-preserving and thus is unitary. The proof is completed. □
$$ \operatorname{range}(D_{F})=\operatorname{range}(D_{\widetilde{F}}) $$
$$ T=R_{\widetilde{F}} \bigl[(R_{F}|_{V})^{\ast }(R_{F}|_{V}) \bigr]^{-1}D_{F} $$
$$ \widetilde{D}_{F}f=D_{F}f\quad{\mbox{for }}f\in L^{2}(\mathbb{R}_{+}). $$
$$\begin{aligned} \langle\widetilde{D}_{F}f, g\rangle_{L^{2}(\mathbb{Z}\times[1, a))}={}&\langle D_{F}f, g \rangle_{L^{2}(\mathbb{Z}\times[1, a))} \\ ={}&\bigl\langle f, D^{\ast}_{F}g\bigr\rangle _{L^{2}(\mathbb{R}_{+})} \\ ={}&\bigl\langle f, (R_{F}|_{V})g\bigr\rangle _{L^{2}(\mathbb {R}_{+})}. \end{aligned}$$
$$ (\widetilde{D}_{F})^{\ast}=R_{F}|_{V}. $$
(3.9)
$$ (\widetilde{D}_{F})^{\dagger}=(\widetilde{D}_{F})^{\ast } \bigl[\widetilde{D}_{F} (\widetilde{D}_{F})^{\ast} \bigr]^{-1} $$
$$\begin{aligned} (\widetilde{D}_{F})^{\dagger}={}&(R_{F}|_{V}) \bigl[(R_{F}|_{V})^{\ast }(R_{F}|_{V}) \bigr]^{-1} \\ ={}&R_{F} \bigl[(R_{F}|_{V})^{\ast}(R_{F}|_{V}) \bigr]^{-1} \end{aligned}$$
$$ (\widetilde{D}_{F})^{\dagger}{D}_{F}f=(\widetilde {D}_{F})^{\dagger} \widetilde{D}_{F}f=P_{(\operatorname{ker}(\widetilde{D}_{F}))^{\bot }}f=P_{(\operatorname{ker}({D}_{F}))^{\bot}}f $$
$$ R_{F} \bigl[(R_{F}|_{V})^{\ast}(R_{F}|_{V}) \bigr]^{-1}D_{F}f=P_{(\operatorname{ker}({D}_{F}))^{\bot}}f=f $$
$$\bigl[(R_{F}|_{V})^{\ast}(R_{F}|_{V}) \bigr]^{-1}D_{F}f $$
$$\begin{aligned} \Vert Tf \Vert _{L^{2}(\mathbb{R}_{+})}={}& \bigl\Vert R_{F} \bigl[(R_{F}|_{V})^{\ast} (R_{F}|_{V}) \bigr]^{-1}D_{F}f \bigr\Vert _{L^{2}(\mathbb {R}_{+})} \\ ={}& \Vert f \Vert _{L^{2}(\mathbb{R}_{+})}. \end{aligned}$$
4 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.
Acknowledgements
The authors would like to thank the referees for their comments, which improved the readability of this article.
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The authors declare that they have no competing interests.
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