High-Dimensional Periodic Framelet with Only One Symmetric Generator

Abstract

Various symmetric framelets and periodic framelets are widely utilized in data analysis due to their resilience to background noise, avoidance of linear phase distortion, and the stability of redundant representation. At present, the number of generators in known periodic framelets in high-dimensional space is infinite. It is natural to ask whether a periodic framelet exists with only one generator in high-dimensional space. In this study, for any given positive numbers, A and B, we will construct one symmetric framelet generator. This generator’s integer translate, dyadic dilation, and periodization can produce a periodic frame with optimal bounds A and B.

1. Introduction

Many data processing systems concern redundant characters since redundancy can make the systems optimally robust to erasures [1,2]. The redundant representation requirement leads to the introduction of frames [3,4,5]. Among various types of frames, well-known framelets can generate frames in $L^2\left({\mathbb{R}}^d\right)$ through their dyadic dilations and integer translates [6,7,8,9,10].

Framelets can be directly constructed from the structure of frame multiresolution analysis (FMRA) [11,12,13,14,15,16]. Benedetto and Li [11,12] and Kim and Lim [13] introduced the theory of one-dimensional FMRA, a foundational concept in framelet theory. Later on, Mu et al. [14] established the theory of high-dimensional FMRA and determined the number of generators needed in derived framelets. Zhang and Saito [16] further extended these results to the case of generalized MRAs.

Symmetry is pivotal in the development of framelet theory since symmetrical framelets have linear phases, which can prevent data distortion during data decomposition and reconstruction. A lot of known symmetric framelets for $L^2\left({\mathbb{R}}^d\right)$ have been constructed, mostly in one-dimensional spaces, with a few in higher-dimensional spaces [17,18,19,20,21]. In one-dimensional space, the number of generators in a framelet can easily be one, but in high-dimensional spaces, the number of generators in a framelet is generally more than one [17,18,19,20,21].

Periodic framelets can generate frames in $L^2\left({[0,1]}^d\right)$. Compared with framelets in $L^2\left({\mathbb{R}}^d\right)$, periodic framelets are more difficult to be constructed. By imitating the idea of generalized MRA, Goh et al. [22] and Lebedeva and Prestin [23] introduced the theory of periodic MRA and constructed some tight periodic framelets. Unfortunately, these periodic framelets have an infinite number of generators. It is natural to ask whether there exists a periodic framelet with only one generator in high-dimensional space. In this study, we will construct one symmetric framelet generator, such that its integer translate, dyadic dilation, and periodization can generate a frame in $L^2\left({[0,1]}^d\right)$. Moreover, we can determine the optimal bounds of the derived periodic frame.

2. Preliminary

Let ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ be a sequence of elements in the Hilbert space $\mathcal{H}$. If there exists a positive constant B, such that

$$\sum _{n\in {\mathbb{Z}}_{+}}|{(f,{\phi }_n)}_{\mathcal{H}}{|}^2\le B{\Vert f\Vert }_{\mathcal{H}}^2\mathrm{for} r\mathrm{all}f\in \mathcal{H},$$

then ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ is called a Bessel sequence in $\mathcal{H}$; if there exist two positive constants $C_1,C_2$, such that

$$C_1{\Vert f\Vert }_{\mathcal{H}}^2\le \sum _{n\in {\mathbb{Z}}_{+}}|{(f,{\phi }_n)}_{\mathcal{H}}{|}^2\le C_2{\Vert f\Vert }_{\mathcal{H}}^2\mathrm{for} \mathrm{all}f\in \mathcal{H},$$

then ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ is called a frame with lower bound $C_1$ and upper bound $C_2$ in $\mathcal{H}$ [5,18]. The largest lower bounds and the smallest upper bounds are called the optimal lower and upper bounds, especially when the optimal lower bound is the same as the optimal upper bound. Such a frame is called a tight frame [5,18].

Let ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ and ${\left\{{\tilde{\phi }}_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ be two frames in $\mathcal{H}$. If for all $f\in \mathcal{H}$,

$$f=\sum _{n\in {\mathbb{Z}}_{+}}{(f,{\phi }_n)}_{\mathcal{H}}{\tilde{\phi }}_n=\sum _{n\in {\mathbb{Z}}_{+}}{(f,{\tilde{\phi }}_n)}_{\mathcal{H}}{\phi }_n.$$

Then ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ and ${\left\{{\tilde{\phi }}_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ are called a pair of dual frames in $\mathcal{H}$. The following theorem [1,2,18] gives a characterization of dual frames:

Theorem 1. 

Two sequences ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ and ${\left\{{\tilde{\phi }}_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ are a pair of dual frames in $\mathcal{H}$ if and only if

(i) 

Both ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ and ${\left\{{\tilde{\phi }}_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ are Bessel sequences for $\mathcal{H}$

(ii) 

For any $f,g\in \mathcal{H}$, ${(f,g)}_{\mathcal{H}}={\sum }_{n\in {\mathbb{Z}}_{+}}{(f,{\phi }_n)}_{\mathcal{H}}\overline{{(g,{\tilde{\phi }}_n)}_{\mathcal{H}}}$.

A framelet can generate a frame in $L^2\left({\mathbb{R}}^d\right)$ through its dyadic dilations and integer translates. In detail, for ${\left\{{\psi }_{\mu }\right\}}_{\mu =1,2,\dots \tau }\subset L^2\left({\mathbb{R}}^d\right)$, if the associated affine system

$${\psi }_{\mu ,m,\mathbf{n}}\left(\mathbf{t}\right):=2^{\frac{md}{2}}{\psi }_{\mu }(2^m\mathbf{t}-\mathbf{n}),m\in \mathbb{Z};\mathbf{n}\in {\mathbb{Z}}^d;\mu =1,2,\dots ,\tau .$$

is a frame in $L^2\left({\mathbb{R}}^d\right)$, then ${\left\{{\psi }_{\mu }\right\}}_{\mu =1,2,\dots \tau }$ are called a framelet, and each ${\psi }_{\mu }$ is called a framelet generator. The number τ is just the number of generators in this framelet. In the one-dimensional space, τ can easily be one, while in high-dimensional space, τ is generally larger than one.

Similar to the construction of these framelets, some periodic framelets have also been constructed [22,23]. Unfortunately, these periodic framelets have an infinite number of generators.

3. Construction of High-Dimensional Periodic Frames

Let $\psi \left(\mathbf{t}\right)\in L^2\left({\mathbb{R}}^d\right)$ be a real-valued, even, and symmetric function, such that its Fourier transform satisfies the following three conditions:

(a)

$\widehat{\psi }\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$;

(b)

$\mathrm{supp}\widehat{\psi }\left(\mathit{\omega }\right)\subset {[-\pi ,\pi ]}^d\setminus {(-ϵ,ϵ)}^d(ϵ>0)$;

(c)

$\underset{\mathit{\omega }\in {\mathbb{R}}^d\setminus \left\{0\right\}}{min}D\left(\mathit{\omega }\right)=A$ and $\underset{\mathit{\omega }\in {\mathbb{R}}^d\setminus \left\{0\right\}}{max}D\left(\mathit{\omega }\right)=B$, where $D\left(\mathit{\omega }\right)={\sum }_{m\in \mathbb{Z}}{|\widehat{\psi }\left(2^m\mathit{\omega }\right)|}^2$,

We define $\tilde{\psi }\left(\mathbf{t}\right)\in L^2\left({\mathbb{R}}^d\right)$ as

$$\widehat{\tilde{\psi }}\left(\mathit{\omega }\right)=\frac{\widehat{\psi }\left(\mathit{\omega }\right)}{D\left(\mathit{\omega }\right)}(\mathit{\omega }\ne \mathbf{0});\widehat{\tilde{\psi }}\left(\mathbf{0}\right)=0.$$

(1)

For $m\in \mathbb{Z}$ and $\mathbf{n}\in {\mathbb{Z}}^d$, let ${\psi }_{m,\mathbf{n}}\left(\mathbf{t}\right)=2^{\frac{md}{2}}\psi (2^m\mathbf{t}-\mathbf{n}),{\tilde{\psi }}_{m,\mathbf{n}}\left(\mathbf{t}\right)=2^{\frac{md}{2}}\tilde{\psi }(2^m\mathbf{t}-\mathbf{n})$ and

$${\displaystyle \begin{array}{ccc}{\psi }_{m,\mathbf{n}}^{per}\left(\mathbf{t}\right)\hfill & =\hfill & {\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}{\psi }_{m,\mathbf{n}}\left(\mathbf{t}+\mathbf{k}\right),\hfill \\ {\tilde{\psi }}_{m,\mathbf{n}}^{per}\left(\mathbf{t}\right)\hfill & =\hfill & {\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}{\tilde{\psi }}_{m,\mathbf{n}}\left(\mathbf{t}+\mathbf{k}\right),\hfill \end{array}}$$

and

$${\displaystyle \begin{array}{c}{\mathbf{\Psi }}^{per}=\{\sqrt{A},{\psi }_{m,\mathbf{n}}^{per},m\ge 0,\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d\},\hfill \\ {\tilde{\mathbf{\Psi }}}^{per}=\{\frac{1}{\sqrt{A}},{\tilde{\psi }}_{m,\mathbf{n}}^{per},m\ge 0,\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d\}.\hfill \end{array}}$$

In this section, we will give the following theorem:

Theorem 2. 

The system $\{{\mathbf{\Psi }}^{per},{\tilde{\mathbf{\Psi }}}^{per}\}$ denotes a pair of dual periodic frames in $L^2\left({[0,1]}^d\right)$.

The proof of Theorem 2 is very long. With the help of Theorem 2, we first prove that both ${\mathbf{\Psi }}^{per}$ and ${\tilde{\mathbf{\Psi }}}^{per}$ are Bessel sequences in $L^2\left({[0,1]}^d\right)$, i.e.,

Lemma 1. 

For any $f\in L^2\left({[0,1]}^d\right)$, we have

$${\displaystyle \begin{array}{c}|{(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{|}^2+{\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}|{(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}{|}^2\le C_1{\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2,\hfill \\ |{(f,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}{|}^2+{\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}|{(f,{\tilde{\psi }}_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}{|}^2\le C_2{\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2,\hfill \end{array}}$$

(2)

where $C_1$ and $C_2$ are constants.

Proof. 

Noticing that

$$|{(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}|\le \sum _{\mathbf{k}\in {\mathbb{Z}}^d}|{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}|,$$

$$|{(f,{\tilde{\psi }}_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}|\le \sum _{\mathbf{k}\in {\mathbb{Z}}^d}|{(f,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}|,$$

and $|{(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{|}^2\le A{\Vert f\Vert }_{L^2{([0,1]}^d}^2$ and $|{(f,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}{|}^2\le \frac{1}{A}{\Vert f\Vert }_{L^2{([0,1]}^d}^2$, in order to prove (2), it is enough to prove that the following inequalities hold: For any $f\in L^2\left({[0,1]}^d\right)$,

$${\displaystyle \begin{array}{c}{\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}|{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}|\right)}^2\le C_1{\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2,\hfill \\ {\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}|{(f,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}|\right)}^2\le C_2{\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2.\hfill \end{array}}$$

(3)

By $\widehat{\psi }\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$ and $\mathrm{supp}\widehat{\psi }\subset {[-\pi ,\pi ]}^d\backslash {(-ϵ,ϵ)}^d(ϵ>0)$, we see that $D\left(\mathit{\omega }\right)$ has only finite nonzero terms in any closed interval which does not contain the point $\mathit{\omega }=\mathbf{0}$. Since $D\left(\mathit{\omega }\right)$ has positive lower bound A, by $\widehat{\psi }\left(\mathit{\omega }\right)=0(\mathit{\omega }\in {[-ϵ,ϵ]}^d)$, we see that $\widehat{\tilde{\psi }}\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$. Combining this and $\widehat{\psi }\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$ gives

$$\psi \left(\mathbf{t}\right)=O\left(\prod _{i=1}^d\frac{1}{(1+|t_i{|)}^{3d}}\right),\tilde{\psi }\left(\mathbf{t}\right)=O\left(\prod _{i=1}^d\frac{1}{(1+|t_i{|)}^{3d}}\right),$$

(4)

where $\mathbf{t}=(t_1,\dots ,t_d)$.

Denote the left-hand side of the first inequality in (3) by $K\left(f\right)$, i.e.,

$$K\left(f\right)=\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2,$$

(5)

where $P_{m,\mathbf{n}}^{\mathbf{k}}=|{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}|$. Applying the inequality formula ${(a+b)}^2\le 2(a^2+b^2)$ gives

$${\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2\le 2\left({\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2+{\left(\sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2\right).$$

From this and (5), we obtain $K\left(f\right)\le 2(J_1+J_2)$, where

$${\displaystyle \begin{array}{ccc}{J}_1\hfill & =\hfill & {\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d\bigcap {[-2,2]}^d}}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2,\hfill \\ J_2\hfill & =\hfill & {\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2.\hfill \end{array}}$$

For $J_1$, noticing that ${\psi }_{m,\mathbf{n}}\left(\mathbf{t}+\mathbf{k}\right)={\psi }_{m,\mathbf{n}-2^m\mathbf{k}}\left(\mathbf{t}\right)$, we obtain

$${\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2=O\left(1\right)\sum _{\mathbf{k}\in {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{\left(P_{m,\mathbf{n}}^{\mathbf{k}}\right)}^2=O\left(1\right)\sum _{\mathbf{k}\in {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{\left|{\int }_{{[0,1]}^d}f\left(\mathbf{t}\right){\overline{\psi }}_{m,\mathbf{n}-2^m\mathbf{k}}\left(\mathbf{t}\right)d\mathbf{t}\right|}^2.$$

From here on, “O” only depends on the dimension d. Let $\mathbf{p}=\mathbf{n}-2^m\mathbf{k}$, we further have

$${\displaystyle \begin{array}{ccc}{J}_1\hfill & =\hfill & O\left(1\right){\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d\bigcap {[-2,2]}^d}}{\left|{\int }_{{[0,1]}^d}f\left(\mathbf{t}\right){\overline{\psi }}_{m,\mathbf{n}-2^m\mathbf{k}}\left(\mathbf{t}\right)d\mathbf{t}\right|}^2\hfill \\ & =\hfill & O\left(1\right){\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{p}\in {\mathbb{Z}}^d\bigcap {[-2^{m+1},2^{m+1}+2^m-1]}^d}}{\left|{\int }_{{[0,1]}^d}f\left(\mathbf{t}\right){\overline{\psi }}_{m,\mathbf{p}}\left(\mathbf{t}\right)d\mathbf{t}\right|}^2\hfill \\ & =\hfill & O\left(1\right){\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{p}\in {\mathbb{Z}}^d}}{\left|{\int }_{{[0,1]}^d}f\left(\mathbf{t}\right){\overline{\psi }}_{m,\mathbf{p}}\left(\mathbf{t}\right)d\mathbf{t}\right|}^2.\hfill \end{array}}$$

(6)

Denote $Q\left(\mathbf{t}\right)=f\left(\mathbf{t}\right){\chi }_{{[0,1]}^d}\left(\mathbf{t}\right)$, where ${\chi }_{{[0,1]}^d}\left(\mathbf{t}\right)$ is the characteristic function of ${[0,1]}^d$. By (5), we have

$$J_1=O\left(1\right)\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d}{\left|{\int }_{{\mathbb{R}}^d}Q\left(\mathbf{t}\right){\overline{\psi }}_{m,\mathbf{n}}\left(\mathbf{t}\right)d\mathbf{t}\right|}^2.$$

(7)

By $\mathrm{supp}\widehat{\psi }\left(2^m\mathit{\omega }\right)\subset {[-\frac{\pi }{2^m},\frac{\pi }{2^m}]}^d$, we have

$$J_1=O\left(1\right)\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d}{\left|\frac{2^{-\frac{dm}{2}}}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}\widehat{Q}\left(\mathit{\omega }\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)e^{i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}d\mathit{\omega }\right|}^2=O\left(1\right)\sum _{m\in \mathbb{Z}}\sum _{\mathbf{n}\in {\mathbb{Z}}^d}{\left|\frac{2^{-\frac{dm}{2}}}{2^d{\pi }^d}{\int }_{{[-\frac{\pi }{2^m},\frac{\pi }{2^m}]}^d}\widehat{Q}\left(\mathit{\omega }\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)e^{i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}d\mathit{\omega }\right|}^2.$$

Noticing that ${\{\frac{e^{i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}}{\sqrt{2^d{\pi }^d}{2}^{\frac{md}{2}}}\}}_{\mathbf{n}\in {\mathbb{Z}}^d}$ is an orthonormal basis for $L^2\left({[-2^m\pi ,2^m\pi ]}^d\right)$, applying Parseval identity of the Fourier series and $\mathrm{supp}\widehat{\psi }\left(2^m\mathit{\omega }\right)\subset {[-\frac{\pi }{2^m},\frac{\pi }{2^m}]}^d$ gives

$${\displaystyle \begin{array}{ccc}{J}_1\hfill & =\hfill & O\left(1\right){\displaystyle \sum _{m\ge 0}}\frac{1}{2^d{\pi }^d}{\int }_{{[-\frac{\pi }{2^m},\frac{\pi }{2^m}]}^d}{\left|\widehat{Q}\left(\mathit{\omega }\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\right|}^{2}d\mathit{\omega }\hfill \\ & =\hfill & O\left(1\right){\displaystyle \sum _{m\ge 0}}\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\left|\widehat{Q}\left(\mathit{\omega }\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\right|}^{2}d\mathit{\omega }=O\left(1\right)\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\left|\widehat{Q}\left(\mathit{\omega }\right)\right|}^2{\displaystyle \sum _{m\ge 0}}{\left|\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\right|}^{2}d\mathit{\omega }\hfill \end{array}}$$

By ${\displaystyle \sum _{m\in \mathbb{Z}}}{|\widehat{\psi }\left(2^m\mathit{\omega }\right)|}^2=D\left(\mathit{\omega }\right)<B$, it follows that

$$J_1=O\left(1\right)\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\left|\widehat{Q}\left(\mathit{\omega }\right)\right|}^{2}d\mathit{\omega }={O\left(1\right)\Vert Q\Vert }_{L^2\left({\mathbb{R}}^d\right)}^2=O\left(1\right){\Vert f\Vert }_{L^2{([0,1]}^d}^2.$$

(8)

For $J_2$, let $f^{*}\left(\mathbf{t}\right)$ be a 1-periodic function and $f^{*}\left(\mathbf{t}\right)=f\left(\mathbf{t}\right)(\mathbf{t}\in {[0,1]}^d)$. Then

$$P_{m,\mathbf{n}}^{\mathbf{k}}=\left|{\int }_{{[0,1]}^d+\mathbf{k}}{f}^{*}\left(\mathbf{t}\right){\overline{\psi }}_{m,\mathbf{n}}\left(\mathbf{t}\right)d\mathbf{t}\right|.$$

Applying the Cauchy’s inequality gives

$$\left|\sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{P}_{m,\mathbf{n}}^{\mathbf{k}}\right|=O\left(1\right)\sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{\Vert f\Vert }_{L^2{\left([0,1]\right)}^d}{\left({\int }_{{[0,1]}^d+\mathbf{k}}{|{\psi }_{m,\mathbf{n}}\left(\mathbf{t}\right)|}^{2}d\mathbf{t}\right)}^{\frac{1}{2}}.$$

Again by (4), we have

$${\int }_{{[0,1]}^d+\mathbf{k}}{|{\psi }_{m,\mathbf{n}}\left(\mathbf{t}\right)|}^{2}d\mathbf{t}=O\left(1\right)\prod _{i=1}^d\underset{k_i}{\overset{k_i+1}{\int }}\frac{2^m}{(1+|2^m{t}_i-n_i{|)}^{6d}}d{t}_i$$

where $\mathbf{k}=(k_1,k_2,\dots ,k_d)$, $\mathbf{t}=(t_1,t_2,\dots ,t_d)$, $\mathbf{n}=(n_1,n_2,\dots ,n_d)$. Furthermore

$$J_2=O\left(1\right){\Vert f\Vert }_{L^2{\left([0,1]\right)}^d}^2\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{\left(\prod _{i=1}^d\underset{k_i}{\overset{k_i+1}{\int }}\frac{2^m}{(1+|2^m{t}_i-n_i{|)}^{6d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2$$

(9)

When $\mathbf{k}=(k_1,k_2,\dots ,k_d)\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d$, there exists a $k_p$ such that $|k_p|>2$. In the case of $k_p>2$, when $0\le n_p\le 2^m-1$, it follows that $2^m\le 1+|2^m{t}_p-n_p|(t_p\in [k_p,k_p+1])$ and then

$$\frac{1}{(1+|2^m{k}_p-n_p{|)}^{6d}}\le \frac{1}{(1+|2^m{k}_p-n_p{|)}^{4d}}·\frac{1}{(1+|2^m{k}_p-n_p{|)}^{2d}}\le \frac{2^{-2md}}{(1+|2^m{k}_p-n_p{|)}^{4d}}$$

In the case of $k_p<-2$, when $0\le n_p\le 2^m-1$, it follows that $2^m\le 1+|2^m{t}_p-n_p|$ and then

$$\frac{1}{(1+|2^m{k}_p-n_p{|)}^{6d}}\le \frac{2^{-2md}}{(1+|2^{m}a\left(k_p\right)-n_p{|)}^{4d}}$$

For $i\ne p$, it is clear that

$$\frac{1}{(1+|2^m{k}_i-n_i{|)}^{6d}}\le \frac{1}{(1+|2^m{k}_i-n_i{|)}^{4d}}·\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}\le \frac{1}{(1+|2^m{k}_i-n_i{|)}^{4d}}$$

Combining these and (9) gives

$$J_2=O\left(1\right){\Vert f\Vert }_{L^2{\left([0,1]\right)}^d}^2\sum _{m\ge 0}{2}^{-md}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}\prod _{i=1}^d{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2$$

(10)

We begin to estimate the core term in $J_2$

$$I:=\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}\prod _{i=1}^d{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2$$

(11)

We consider the following d cases:

Case 1. 

We consider the index $\mathbf{k}$ satisfying $k_i>0$ for $i=1,2,\dots ,d$. When $0\le n_i\le 2^m-1$, we have $2^m{t}_i-n_i>0$, and then

$${\displaystyle \begin{array}{c}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i>0,i=1,2,\dots ,d}}}{\displaystyle \prod _{i=1}^d}{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2\hfill \\ =O\left(1\right){\left({\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i>0,i=1,2,\dots ,d}}}{\displaystyle \prod _{i=1}^d}{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2\hfill \\ =O\left(1\right){\left({\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i>0,i=1,2,\dots ,d}}}{\displaystyle \prod _{i=1}^d}{\left(\frac{1}{(1+|2^m{k}_i-n_i{|)}^{4d}}\right)}^{\frac{1}{2}}\right)}^2=O\left(1\right){\left({\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i>0,i=1,2,\dots ,d}}}{\displaystyle \prod _{i=1}^d}\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}\right)}^2\hfill \\ =O\left(1\right){\left(\underset{{\mathbb{R}}^d}{\int }{\displaystyle \prod _{i=1}^d}\frac{1}{(1+|t_i{|)}^{2d}}d\mathbf{t}\right)}^2=O\left(1\right).\hfill \end{array}}$$

Case 2. 

We consider the index $\mathbf{k}$ satisfying that $k_1=0$ and $k_i>0(i\ne 1)$. It follows that

$${\displaystyle \begin{array}{c}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_1=0,k_i>0,i=2,\dots ,d}}}{\displaystyle \prod _{i=1}^d}{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2\hfill \\ ={\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_1=0,k_i>0,i=2,\dots d}}}{\displaystyle \prod _{i=2}^d}{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\times {\left(\underset{0}{\overset{1}{\int }}\frac{1}{(1+|2^m{t}_1-n_1{|)}^{4d}}d{t}_1\right)}^{\frac{1}{2}}\right)}^2\hfill \\ =O\left(1\right){\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_1=0,k_i>0,i=2,\dots d}}}{\displaystyle \prod _{i=2}^d}{\left(\frac{1}{(1+|2^m{k}_i-n_i{|)}^{4d}}\right)}^{\frac{1}{2}}\times {\left(\underset{-n_1}{\overset{2^m-n_1}{\int }}\frac{2^{-m}}{{(1+|u|)}^{4d}}du\right)}^{\frac{1}{2}}\right)}^2\hfill \\ =O\left(1\right){\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{2}^{-m}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_1=0,k_i>0,i=2,\dots ,d}}}{\displaystyle \prod _{i=2}^d}\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}{\left(\underset{\mathbb{R}}{\int }\frac{1}{{(1+|u|)}^{4d}}du\right)}^{\frac{1}{2}}\right)}^2\hfill \\ =O\left(1\right)\left({\displaystyle \sum _{n_1\in \mathbb{Z}\bigcap [0,2^m-1]}}{2}^{-m}\underset{\mathbb{R}}{\int }\frac{1}{{(1+|u|)}^{4d}}du\right)\times \left({\displaystyle \sum _{(n_2,\dots ,n_d)\in {\mathbb{Z}}^{d-1}\bigcap {[0,2^m-1]}^{d-1}}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_1=0,k_i>0,i=2,\dots ,d}}}{\displaystyle \prod _{i=2}^d}\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}\right)}^2\right)\hfill \\ =O\left(1\right){\left({\displaystyle \sum _{(n_2,\dots ,n_d)\in {\mathbb{Z}}^{d-1}\bigcap {[0,2^m-1]}^{d-1}}}{\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_1=0,k_i>0,i=2,\dots ,d}}}{\displaystyle \prod _{i=2}^d}\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}\right)}^2=O\left(1\right){\left(\underset{{\mathbb{R}}^{d-1}}{\int }{\displaystyle \prod _{i=2}^d}\frac{1}{(1+|t_i{|)}^{2d}}d{t}_2\cdots d{t}_d\right)}^2=O\left(1\right).\hfill \end{array}}$$

Similarly, we can obtain

$$\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_j=0,k_i>0,i\ne j}}\prod _{i=1}^d{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2=O\left(1\right).$$

$$⋮$$

$$⋮$$

Case p. 

We consider the index $\mathbf{k}$ satisfying that $k_i=0(i\le p-1)$ and $k_i>0(i>p-1)$. It follows that

$${\displaystyle \begin{array}{c}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i=0(i\le p-1),k_i>0(i>p-1)}}}{\displaystyle \prod _{i=1}^d}{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2\hfill \\ ={\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i=0(i\le p-1),k_i>0(i>p-1)}}}{\displaystyle \prod _{i=p}^d}{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\times {\left(\prod _{i=1}^{p-1}\underset{0}{\overset{1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2\hfill \\ =O\left(1\right){\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i=0(i\le p-1),k_i>0(i>p-1)}}}{\displaystyle \prod _{i=p}^d}{\left(\frac{1}{(1+|2^m{k}_i-n_i{|)}^{4d}}\right)}^{\frac{1}{2}}\times {\left(\prod _{i=1}^{p-1}\underset{-n_i}{\overset{2^m-n_i}{\int }}\frac{2^{-m}}{{(1+|u|)}^{4d}}du\right)}^{\frac{1}{2}}\right)}^2\hfill \\ =O\left(1\right){\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{2}^{-(p-1)m}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i=0(i\le p-1),k_i>0(i>p-1)}}}{\displaystyle \prod _{i=p}^d}\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}{\left(\underset{\mathbb{R}}{\int }\frac{1}{{(1+|u|)}^{4d}}du\right)}^{\frac{p-1}{2}}\right)}^2\hfill \\ =O\left(1\right)\left({\displaystyle \sum _{(n_1,\dots ,n_{p-1})\in {\mathbb{Z}}^{p-1}\bigcap {[0,2^m-1]}^{p-1}}}{2}^{-(p-1)m}{\left(\underset{\mathbb{R}}{\int }\frac{1}{{(1+|u|)}^{4d}}du\right)}^{p-1}\right)\times \hfill \\ \left({\displaystyle \sum _{(n_p,\dots ,n_d)\in {\mathbb{Z}}^{d-p+1}\bigcap {[0,2^m-1]}^{d-p+1}}}{\left({\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i=0(i\le p-1),k_i>0(i>p-1)}}}{\displaystyle \prod _{i=p}^d}\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}\right)}^2\right)\hfill \\ =O\left(1\right){\left({\displaystyle \sum _{(n_p,\dots ,n_d)\in {\mathbb{Z}}^{d-p+1}\bigcap {[0,2^m-1]}^{d-p+1}}}{\displaystyle \sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i=0(i\le p-1),k_i>0(i>p-1)}}}{\displaystyle \prod _{i=p}^d}\frac{1}{(1+|2^m{k}_i-n_i{|)}^{2d}}\right)}^2\hfill \\ =O\left(1\right)\underset{{\mathbb{R}}^{d-p+1}}{\int }{\displaystyle \prod _{i=p}^d}\frac{1}{(1+|t_i{|)}^{2d}}d{t}_p\cdots d{t}_d=O\left(1\right).\hfill \end{array}}$$

Similarly, we can obtain

$$\begin{gathered} \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_{j_1}=0,\dots ,k_{j_{p-1}}=0,k_i>0,i\ne j_1,\dots ,j_{p-1}}}\prod _{i=1}^d{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2=O\left(1\right). \\ ⋮ \\ ⋮ \end{gathered}$$

Case d. 

We consider the index $\mathbf{k}$ satisfying that $k_i=0$ and $k_j>0(j\ne i)$. Similar to the argument of Case p, we can deduce that

$$\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i>0,k_j=0(j=1,2,\dots d,j\ne i)}}\prod _{i=1}^d{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2=O\left(1\right).$$

By the combination of all d cases and using the inequality $(|a_1|+\cdots +|a_{2^d}{|)}^2\le 2^d(a_1^2+\cdots +a_{2^d}^2)$, it follows that

$$\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\stackrel{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}{k_i\ge 0,i=1,2,\dots ,d}}\prod _{i=1}^d{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2=O\left(1\right)$$

Again by the symmetry of the index $\mathbf{k}$, by (11), we can deduce that

$$I:=\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{\left(\sum _{\mathbf{k}\notin {\mathbb{Z}}^d\bigcap {[-2,2]}^d}\prod _{i=1}^d{\left(\underset{k_i}{\overset{k_i+1}{\int }}\frac{1}{(1+|2^m{t}_i-n_i{|)}^{4d}}d{t}_i\right)}^{\frac{1}{2}}\right)}^2=O\left(1\right).$$

Combining this and (10)–(11) gives

$$J_2={O\left(1\right)\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2\sum _{m\ge 0}{2}^{-md}=O\left(1\right){\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2$$

From this and (8) and (5), noticing that $K\left(f\right)\le 2(J_1+J_2)$, we obtain the first inequality of (3). Since the arguments of two inequalities of (3) are similar, we can deduce the second inequality of (3). Finally, Lemma 1 is proved. □

Next, we prove that the Parseval identity holds, i.e.,

Lemma 2. 

For any $f,g\in L^2\left({[0,1]}^d\right)$, we have

$${(f,g)}_{L^2\left({[0,1]}^d\right)}=\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}{(g,{\tilde{\psi }}_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}^{-}+{(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{(g,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}^{-}.$$

(12)

Proof. 

First, we compute $M:={\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}{(g,{\tilde{\psi }}_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}^{-}$.

By (4), it is clear that

$$M=\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}\right){\left(\sum _{\mathbf{l}\in {\mathbb{Z}}^d}{(g,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}\right)}^{-}.$$

(13)

Noticing that

$$\begin{gathered} \sum _{\mathbf{k}\in {\mathbb{Z}}^d}\sum _{\mathbf{l}\in {\mathbb{Z}}^d}|{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}{(g,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}| \\ =\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d}|{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}|\right)\left(\sum _{\mathbf{l}\in {\mathbb{Z}}^d}|{(g,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}|\right), \end{gathered}$$

using the inequality $ab\le \frac{1}{2}(a^2+b^2)$ and (3), we obtain

$${\displaystyle \begin{array}{c}{\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}{\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}|{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}{(g,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}|\hfill \\ \le \frac{1}{2}{\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}\left({\left({\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}{|(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))|}_{L^2\left({[0,1]}^d\right)}\right)}^2+{\left({\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}{|(g,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{l}))|}_{L^2\left({[0,1]}^d\right)}\right)}^2\right)<\infty .\hfill \end{array}}$$

From this and (13), we know that

$$M=\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}\sum _{\mathbf{k}\in {\mathbb{Z}}^d}\sum _{\mathbf{l}\in {\mathbb{Z}}^d}{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}{(g,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}^{-}$$

and this series is convergent absolutely. By the rearrangement of terms, we obtain

$$M=\sum _{\mathbf{l}\in {\mathbb{Z}}^d}\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}{(g,{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{k}+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}^{-}.$$

(14)

Let both $f^{*}$ and $g^{*}$ be 1-periodic functions and $f^{*}\left(\mathbf{t}\right)=f\left(\mathbf{t}\right),g^{*}\left(\mathbf{t}\right)=g\left(\mathbf{t}\right)(\mathbf{t}\in {[0,1]}^d)$. Define two sequences of functions as

$$f_{\mathbf{k}}\left(\mathbf{t}\right)=f^{*}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d+\mathbf{k}}\left(\mathbf{t}\right),g_{\mathbf{k}}\left(\mathbf{t}\right)=g^{*}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d+\mathbf{k}}\left(\mathbf{t}\right).$$

So

$${\displaystyle \begin{array}{c}{(f,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}={(f_{\mathbf{0}},{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)},\hfill \\ {(g,{\psi }_{m,\mathbf{n}}(·+\mathbf{k}+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}={(g_{\mathbf{l}},{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}.\hfill \end{array}}$$

Noticing that

$${\psi }_{m,\mathbf{n}}\left(\mathbf{t}+\mathbf{k}\right)={\psi }_{m,\mathbf{n}-2^m\mathbf{k}}\left(\mathbf{t}\right),{\tilde{\psi }}_{m,\mathbf{n}}\left(\mathbf{t}+\mathbf{k}\right)={\tilde{\psi }}_{m,\mathbf{n}-2^m\mathbf{k}}\left(\mathbf{t}\right),$$

we obtain

$${\displaystyle \begin{array}{ccc}M\hfill & =\hfill & {\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}{\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}{(f_{\mathbf{0}},{\psi }_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}{(g_{\mathbf{l}},{\tilde{\psi }}_{m,\mathbf{n}}(·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}^{-}\hfill \\ & =\hfill & {\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}{\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{\displaystyle \sum _{\mathbf{k}\in {\mathbb{Z}}^d}}{(f_{\mathbf{0}},{\psi }_{m,\mathbf{n}-2^m\mathbf{k}})}_{L^2\left({\mathbb{R}}^d\right)}{(g_{\mathbf{l}},{\tilde{\psi }}_{m,\mathbf{n}-2^m\mathbf{k}})}_{L^2\left({\mathbb{R}}^d\right)}^{-}\hfill \\ & =\hfill & {\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}\left({\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d}}{(f_{\mathbf{0}},{\psi }_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)}{(g_{\mathbf{l}},{\tilde{\psi }}_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)}^{-}\right).\hfill \end{array}}$$

(15)

Since $\mathrm{supp}\widehat{\psi }=supp\widehat{\tilde{\psi }}\subset {[-\pi ,\pi ]}^d$, applying the Parseval identity of the Fourier transforms gives

$${\displaystyle \begin{array}{c}{(f_{\mathbf{0}},{\psi }_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)}=\frac{2^{-\frac{dm}{2}}}{2^d{\pi }^d}\underset{{[-2^m\pi ,2^m\pi ]}^d}{\int }{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)e^{i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}d\mathit{\omega },\hfill \\ {(g_{\mathbf{l}},{\tilde{\psi }}_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)}=\frac{2^{-\frac{dm}{2}}}{2^d{\pi }^d}\underset{{[-2^m\pi ,2^m\pi ]}^d}{\int }{\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)\overline{\widehat{\tilde{\psi }}}\left(\frac{\mathit{\omega }}{2^m}\right)e^{i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}d\mathit{\omega },\hfill \end{array}}$$

(16)

Noticing that ${\left\{\frac{e^{i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}}{\sqrt{2^d{\pi }^d}{2}^{\frac{md}{2}}}\right\}}_{\mathbf{n}\in {\mathbb{Z}}^d}$ is an orthonormal basis for $L^2\left({[-2^m\pi ,2^m\pi ]}^d\right)$, we expand ${\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)$ and ${\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)\overline{\widehat{\tilde{\psi }}}\left(\frac{\mathit{\omega }}{2^m}\right)$ into Fourier series, where the associated Fourier coefficients are, respectively,

$${\displaystyle \begin{array}{c}{c}_{\mathbf{n}}=\frac{2^{-\frac{dm}{2}}}{\sqrt{2^d{\pi }^d}}\underset{{[-2^m\pi ,2^m\pi ]}^d}{\int }{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)e^{-i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}d\mathit{\omega },\hfill \\ d_{\mathbf{n}}=\frac{2^{-\frac{dm}{2}}}{\sqrt{2^d{\pi }^d}}\underset{{[-2^m\pi ,2^m\pi ]}^d}{\int }{\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)\overline{\widehat{\tilde{\psi }}}\left(\frac{\mathit{\omega }}{2^m}\right)e^{-i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}d\mathit{\omega },\hfill \\ \end{array}}$$

(17)

Comparing (16) with (17), we have

$$c_{\mathbf{n}}=\sqrt{2^d{\pi }^d}{(f_{\mathbf{0}},{\psi }_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)},d_{\mathbf{n}}=\sqrt{2^d{\pi }^d}{(g_{\mathbf{l}},{\tilde{\psi }}_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)}.$$

(18)

Applying the Parseval identity of the Fourier series gives

$$\sum _{\mathbf{n}\in {\mathbb{Z}}^d}{c}_{\mathbf{n}}{\overline{d}}_{\mathbf{n}}=\underset{{[-2^m\pi ,2^m\pi ]}^d}{\int }{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)d\mathit{\omega }.$$

From this and (18), and $\mathrm{supp}\widehat{\psi }=\mathrm{supp}\widehat{\tilde{\psi }}\subset {[-\pi ,\pi ]}^d$, it follows that

$$\sum _{\mathbf{n}\in {\mathbb{Z}}^d}{(f_{\mathbf{0}},{\psi }_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)}{(g_{\mathbf{l}},{\tilde{\psi }}_{m,\mathbf{n}})}_{L^2\left({\mathbb{R}}^d\right)}^{-}=\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)d\mathit{\omega }.$$

Again, by (15), we obtain

$$M=\sum _{\mathbf{l}}\left(\frac{1}{2^d{\pi }^d}\sum _{m\ge 0}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)d\mathit{\omega }\right).$$

(19)

Noticing that

$$\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\ge 0,\sum _{m\in \mathbb{Z}}\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)=1(\omega \in {\mathbb{R}}^d\backslash \left\{\mathbf{0}\right\}),$$

(20)

it is clear that

$$\sum _{m\ge 0}{\int }_{{\mathbb{R}}^d}\left|{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)\right|d\mathit{\omega }={\int }_{{\mathbb{R}}^d}|{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)|\sum _{m\ge 0}\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)d\mathit{\omega }\le {\int }_{\mathbb{R}}|{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)|d\mathit{\omega }<\infty .$$

From this and (19), applying the Lebegue dominant convergence theorem gives

$$M=\sum _{\mathbf{l}\in {\mathbb{Z}}^d}\left(\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\sum _{m\ge 0}\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)d\mathit{\omega }\right).$$

(21)

Secondly, we compute $N:=(f,\sqrt{A}){(g,\frac{1}{\sqrt{A}})}^{-}$. Let

$$Q=\sum _{\mathbf{l}\in {\mathbb{Z}}^d}\left(\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\sum _{m<0}\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)d\mathit{\omega }\right).$$

(22)

Take an auxiliary function $\eta \left(\mathbf{t}\right)\in L^2\left({\mathbb{R}}^d\right)$ such that

$${\displaystyle \left\{\begin{array}{ccc}\widehat{\eta }\left(\mathit{\omega }\right)\hfill & =\hfill & \sqrt{A}{\displaystyle \sum _{m<0}}\overline{\widehat{\tilde{\psi }}}\left(\frac{\mathit{\omega }}{2^m}\right)\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)(\mathit{\omega }\in {\mathbb{R}}^d\backslash \left\{\mathbf{0}\right\}),\hfill \\ \widehat{\eta }\left(\mathbf{0}\right)\hfill & =\hfill & \sqrt{A}.\hfill \end{array}\right.}$$

(23)

Since $\widehat{\psi }\left(\mathit{\omega }\right)=0(\mathit{\omega }\in {(-ϵ,ϵ)}^d)$, for $m\ge 0$, we have $\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)=0(\mathit{\omega }\in {(-ϵ,ϵ)}^d)$. From this and (20), we have

$$\widehat{\eta }\left(\mathit{\omega }\right)=\sum _{m<0}\overline{\widehat{\tilde{\psi }}}\left(\frac{\mathit{\omega }}{2^m}\right)\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)=\sum _{m\in \mathbb{Z}}\overline{\widehat{\tilde{\psi }}}\left(\frac{\mathit{\omega }}{2^m}\right)\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)=\sqrt{A}for\mathit{\omega }\in {(-ϵ,ϵ)}^d\backslash \left\{\mathbf{0}\right\}.$$

Again, by $\widehat{\eta }\left(\mathbf{0}\right)=\sqrt{A}$, we have

$$\widehat{\eta }\left(\mathit{\omega }\right)=\sqrt{A}(\mathit{\omega }\in {(-ϵ,ϵ)}^d).$$

(24)

By $\mathrm{supp}\widehat{\psi }\subset {[-\pi ,\pi ]}^d{\setminus (-ϵ,ϵ)}^d$, we deduce that ${\sum }_{m<0}\overline{\widehat{\tilde{\psi }}}\left(\frac{\mathit{\omega }}{2^m}\right)\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)$ has only finitely many non-zero terms in any closed interval which does not contain the point $\mathit{\omega }=\mathbf{0}$. In the beginning of the proof of Lemma 1, we have shown $\widehat{\psi }\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$ and $\widehat{\tilde{\psi }}\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$. Again by (23) and (24), we have $\widehat{\eta }\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$. By $\mathrm{supp}\widehat{\psi }\left(\mathit{\omega }\right)=\mathrm{supp}\widehat{\tilde{\psi }}\left(\mathit{\omega }\right)\subset {[-\pi ,\pi ]}^d$, we have

$$\mathrm{supp}\widehat{\eta }\left(\mathit{\omega }\right)\subset {\left[-\frac{\pi }{2},\frac{\pi }{2}\right]}^d.$$

(25)

Take the other auxiliary function $\tilde{\eta }$ satisfying three conditions: $\widehat{\tilde{\eta }}\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right),\mathrm{supp}\widehat{\tilde{\eta }}\left(\mathit{\omega }\right)\subset {[-\pi ,\pi ]}^d$, and $\widehat{\tilde{\eta }}\left(\mathit{\omega }\right)=\frac{1}{\sqrt{A}}(\mathit{\omega }\in {[-\frac{\pi }{2},\frac{\pi }{2}]}^d)$. Then, by (25) and (23), we have

$$\widehat{\tilde{\eta }}\left(\mathit{\omega }\right)\overline{\widehat{\eta }}\left(\mathit{\omega }\right)=\sum _{m<0}\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)(\mathit{\omega }\in {\mathbb{R}}^d\setminus \left\{\mathbf{0}\right\}).$$

From this and (22) and (25), we obtain

$$Q=\sum _{\mathbf{l}}\left(\frac{1}{2^d{\pi }^d}{\int }_{{[-\pi ,\pi ]}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\widehat{\tilde{\eta }}\left(\mathit{\omega }\right)\overline{\widehat{\eta }}\left(\mathit{\omega }\right)d\mathit{\omega }\right).$$

Applying the Parseval identity of the Fourier series gives

$$Q=\sum _{\mathbf{l}\in {\mathbb{Z}}^d}\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d}\left(\frac{1}{2^d{\pi }^d}{\int }_{{[-\pi ,\pi ]}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right)\overline{\widehat{\eta }}\left(\mathit{\omega }\right)e^{-i\mathbf{k}·\mathit{\omega }}d\mathit{\omega }\right)\overline{\frac{1}{2^d{\pi }^d}{\int }_{{[-\pi ,\pi ]}^d}{\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)\overline{\widehat{\tilde{\eta }}}\left(\mathit{\omega }\right)e^{-i\mathbf{k}·\mathit{\omega }}d\mathit{\omega }}\right).$$

(26)

By $\mathrm{supp}\widehat{\eta }\left(\mathit{\omega }\right)\subset {[-\frac{\pi }{2},\frac{\pi }{2}]}^d$, we obtain

$$\frac{1}{2^d{\pi }^d}{\int }_{{[-\pi ,\pi ]}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right)\overline{\widehat{\eta }}\left(\mathit{\omega }\right)e^{-i\mathbf{k}·\mathit{\omega }}d\mathit{\omega }=\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right)\overline{\widehat{\eta }}\left(\mathit{\omega }\right)e^{-i\mathbf{k}·\mathit{\omega }}d\mathit{\omega }={(f_{\mathbf{0}},\eta (·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)},$$

where the Parseval identity of the Fourier transform is used in the last equality. Similarly, we have

$$\frac{1}{2^d{\pi }^d}{\int }_{{[-\pi ,\pi ]}^d}{\widehat{g}}_{\mathbf{l}}\left(\mathit{\omega }\right)\overline{\widehat{\tilde{\eta }}}\left(\mathit{\omega }\right)e^{-i\mathbf{k}·\mathit{\omega }}d\mathit{\omega }={(g_{\mathbf{l}},\tilde{\eta }(·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}^{-}.$$

By (26), we obtain

$$Q=\sum _{\mathbf{l}\in {\mathbb{Z}}^d}\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{(f_{\mathbf{0}},\eta (·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}{(g_{\mathbf{l}},\tilde{\eta }(·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}^{-}.$$

By the definitions of $f_{\mathbf{0}}$ and $g_{\mathbf{k}}$, we have

$${\displaystyle \begin{array}{ccc}{(f_{\mathbf{0}},\eta (·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}\hfill & =\hfill & {\int }_{{\mathbb{R}}^d}{f}^{*}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d}\left(\mathbf{t}\right)\overline{\eta }\left(\mathbf{t}+\mathbf{k}\right)d\mathbf{t}={\int }_{{[0,1]}^d}f\left(\mathbf{t}\right)\overline{\eta }\left(\mathbf{t}+\mathbf{k}\right)d\mathbf{t}={(f,\eta (·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)},\hfill \\ \\ {(g_{\mathbf{l}},\tilde{\eta }(·+\mathbf{k}))}_{L^2\left({\mathbb{R}}^d\right)}\hfill & =\hfill & {\int }_{{\mathbb{R}}^d}{g}_{\mathbf{l}}\left(\mathbf{t}\right)\overline{\tilde{\eta }}\left(\mathbf{t}+\mathbf{k}\right)d\mathbf{t}={\int }_{{\mathbb{R}}^d}{g}^{*}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d+\mathbf{l}}\left(\mathbf{t}\right)\overline{\tilde{\eta }}\left(\mathbf{t}+\mathbf{k}\right)d\mathbf{t}\hfill \\ \\ & =\hfill & {\int }_{{\mathbb{R}}^d}{g}^{*}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d}\left(t\right)\overline{\tilde{\eta }}\left(\mathbf{t}+\mathbf{k}+\mathbf{l}\right)d\mathbf{t}={\int }_{{[0,1]}^d}g\left(\mathbf{t}\right)\overline{\tilde{\eta }}\left(\mathbf{t}+\mathbf{k}+\mathbf{l}\right)d\mathbf{t}={(g_{\mathbf{l}},\tilde{\eta }(·+\mathbf{k}+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}.\hfill \end{array}}$$

So

$$Q=\sum _l\sum _k{(f,\eta (·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}{(g,\tilde{\eta }(·+\mathbf{k}+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}^{-}.$$

(27)

Since $\widehat{\eta }\left(\mathit{\omega }\right),\widehat{\tilde{\eta }}\left(\mathit{\omega }\right)\in C^{\infty }\left({\mathbb{R}}^d\right)$, we know that $\eta \left(\mathbf{t}\right),\tilde{\eta }\left(\mathbf{t}\right)$ decay very fast, so the series (27) is absolutely convergent and then

$$Q=\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{(f,\eta (·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}\right){\left(\sum _{\mathbf{l}\in {\mathbb{Z}}^d}{(g,\tilde{\eta }(·+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}\right)}^{-}$$

The Poisson summation formula indicates that the Fourier coefficients of ${\eta }^{per}\left(\mathbf{t}\right)={\sum }_{\mathbf{k}}\eta \left(\mathbf{t}+\mathbf{k}\right)$ are

$${\displaystyle {\int }_{{[0,1]}^d}{\eta }^{per}\left(\mathbf{t}\right)e^{-2\pi i\mathbf{k}·\mathbf{t}}d\mathbf{t}=\widehat{\eta }\left(2\mathbf{k}\pi \right)=\left\{\begin{array}{ccc}\sqrt{A},\hfill & \mathbf{k}=\mathbf{0},\hfill & \\ 0,\hfill & \mathbf{k}\ne \mathbf{0}.\hfill \end{array}\right.}$$

So ${\sum }_{\mathbf{k}}\eta \left(\mathbf{t}+\mathbf{k}\right)=\sqrt{A}$. Similarly, ${\sum }_{\mathbf{k}}\tilde{\eta }\left(\mathbf{t}+\mathbf{k}\right)=\frac{1}{\sqrt{A}}$. Again, by (27), we obtain

$$Q=\left(\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{(f,\eta (·+\mathbf{k}))}_{L^2\left({[0,1]}^d\right)}\right){\left(\sum _{\mathbf{l}\in {\mathbb{Z}}^d}{(g,\tilde{\eta }(·+\mathbf{l}))}_{L^2\left({[0,1]}^d\right)}\right)}^{-}={(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{(g,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}^{-}.$$

Noticing that $N={(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{(g,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}^{-}$, by (22), we obtain

$$N=\sum _{\mathbf{l}}\left(\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\sum _{m<0}\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)d\mathit{\omega }\right).$$

Combining this and (21) and (20), we obtain

$${\displaystyle \begin{array}{ccc}M+N\hfill & =\hfill & {\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}\left(\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)\left({\displaystyle \sum _{m\in \mathbb{Z}}}\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\overline{\widehat{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)\right)d\mathit{\omega }\right)\hfill \\ & =\hfill & {\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}\left(\frac{1}{2^d{\pi }^d}{\int }_{{\mathbb{R}}^d}{\widehat{f}}_{\mathbf{0}}\left(\mathit{\omega }\right){\overline{\widehat{g}}}_{\mathbf{l}}\left(\mathit{\omega }\right)d\mathit{\omega }\right)={\displaystyle \sum _{\mathbf{l}\in {\mathbb{Z}}^d}}{\int }_{{\mathbb{R}}^d}{f}_{\mathbf{0}}\left(\mathbf{t}\right){\overline{g}}_{\mathbf{l}}\left(\mathbf{t}\right)d\mathbf{t}.\hfill \end{array}}$$

By the definitions of $f_{\mathbf{0}}$ and $g_{\mathbf{k}}$, we obtain

$$M+N=\sum _{\mathbf{l}\in {\mathbb{Z}}^d}{\int }_{{\mathbb{R}}^d}{f}^{*}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d}\left(\mathbf{t}\right)\overline{g^{*}}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d+\mathbf{l}}\left(\mathbf{t}\right)d\mathbf{t}={\int }_{{\mathbb{R}}^d}{f}^{*}\left(\mathbf{t}\right)\overline{g^{*}}\left(\mathbf{t}\right){\chi }_{{[0,1]}^d}\left(\mathbf{t}\right)d\mathbf{t}={\int }_{{[0,1]}^d}f\left(\mathbf{t}\right)\overline{g}\left(\mathbf{t}\right)d\mathbf{t}={(f,g)}_{L^2\left({[0,1]}^d\right)},$$

Again, by the definitions of M and N, we obtain

$${(f,g)}_{L^2\left({[0,1]}^d\right)}=\sum _{m\ge 0}\sum _{n=0}^{2^m-1}{(f,{\psi }_{m,n}^{per})}_{L^2\left({[0,1]}^d\right)}{(g,{\tilde{\psi }}_{m,n}^{per})}_{L^2\left({[0,1]}^d\right)}^{-}+{(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{(g,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}^{-},$$

i.e., Lemma 2 is proved. □

Finally, we present the proof of Theorem 2 as follows:

Proof of Theorem 1. 

Applying Theorem 2, the combination of Lemma 1 and Lemma 2 implies that $\{{\mathbf{\Psi }}^{per},{\tilde{\mathbf{\Psi }}}^{per}\}$ are a pair of dual periodic frames, i.e., Theorem 2 is proved. □

4. Optimal Frame Bounds

In Section 3, we constructed a periodic frame with only one symmetric generator. Now we will provide its optimal frame bounds.

Theorem 3. 

Under the conditions of Theorem 2, the optimal lower and upper frame bounds of ${\mathbf{\Psi }}^{per}$ are A and B, respectively, and those of ${\tilde{\mathbf{\Psi }}}^{per}$ are $\frac{1}{B}$ and $\frac{1}{A}$, respectively.

Proof. 

Noticing that both ${\psi }_{m,\mathbf{n}}^{per}={\sum }_{\mathbf{k}\in {\mathbb{Z}}^{\mathbf{d}}}{\psi }_{m,\mathbf{n}}\left(\mathbf{t}+\mathbf{k}\right)$ and ${\tilde{\psi }}_{m,\mathbf{n}}^{per}={\sum }_{\mathbf{k}\in {\mathbb{Z}}^{\mathbf{d}}}{\tilde{\psi }}_{m,\mathbf{n}}\left(\mathbf{t}+\mathbf{k}\right)$ are one-periodic functions, then we expand them into Fourier series ${\psi }_{m,\mathbf{n}}^{per}\left(\mathbf{t}\right)={\sum }_{\mathbf{k}\in {\mathbb{Z}}^d}{c}_{\mathbf{k}}^{(m,\mathbf{n})}{e}^{2\pi i\mathbf{k}·\mathbf{t}}$ and ${\tilde{\psi }}_{m,\mathbf{n}}^{per}\left(\mathbf{t}\right)={\sum }_{\mathbf{k}\in {\mathbb{Z}}^d}{\tilde{c}}_{\mathbf{k}}^{(m,\mathbf{n})}{e}^{2\pi i\mathbf{k}·\mathbf{t}}$, by the Poisson summation formula, the Fourier coefficients of ${\psi }_{m,\mathbf{n}}^{per}$ and ${\tilde{\psi }}_{m,\mathbf{n}}^{per}$ are, respectively, □

$$c_{\mathbf{k}}^{(m,\mathbf{n})}={\widehat{\psi }}_{m,\mathbf{n}}\left(2\mathbf{k}\pi \right),{\tilde{c}}_{\mathbf{k}}^{(m,\mathbf{n})}={\widehat{\tilde{\psi }}}_{m,\mathbf{n}}\left(2\mathbf{k}\pi \right)(\mathbf{k}\in {\mathbb{Z}}^d).$$

(28)

By $D\left(\mathit{\omega }\right)={\sum }_{l\in \mathbb{Z}}{|\widehat{\psi }\left(2^l\mathit{\omega }\right)|}^2$, we have $D\left(\frac{\mathit{\omega }}{2^m}\right)=D\left(\mathit{\omega }\right)$. Again, by (1), we obtain

$$\widehat{\tilde{\psi }}\left(\frac{\mathit{\omega }}{2^m}\right)=\frac{\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)}{D\left(\frac{\mathit{\omega }}{2^m}\right)}(\mathit{\omega }\ne \mathbf{0}).$$

By ${\widehat{\psi }}_{m,\mathbf{n}}\left(\mathit{\omega }\right)=2^{-\frac{md}{2}}\widehat{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)e^{-i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}$ and ${\widehat{\tilde{\psi }}}_{m,\mathbf{n}}\left(\mathit{\omega }\right)=2^{-\frac{md}{2}}\tilde{\psi }\left(\frac{\mathit{\omega }}{2^m}\right)e^{-i\frac{\mathbf{n}·\mathit{\omega }}{2^m}}$, we obtain

$${\widehat{\tilde{\psi }}}_{m,\mathbf{n}}\left(\mathit{\omega }\right)=\frac{{\widehat{\psi }}_{m,\mathbf{n}}\left(\mathit{\omega }\right)}{D\left(\mathit{\omega }\right)}(\mathit{\omega }\ne \mathbf{0}).$$

From this and (28), we obtain

$${\tilde{c}}_{\mathbf{k}}^{(m,\mathbf{n})}=\frac{c_{\mathbf{k}}^{(m,\mathbf{n})}}{D\left(2\mathbf{k}\pi \right)}(\mathbf{k}\ne \mathbf{0}).$$

(29)

By $\widehat{\psi }\left(\mathbf{0}\right)=\widehat{\tilde{\psi }}\left(\mathbf{0}\right)=0$, we have ${\tilde{c}}_{\mathbf{0}}^{(m,\mathbf{n})}=c_{\mathbf{0}}^{(m,\mathbf{n})}=0$.

Denote Fourier coefficients of $f\left(\mathbf{t}\right)\in L^2\left({[0,1]}^d\right)$ by $c_{\mathbf{k}}\left(f\right)$. Take a function $h\left(\mathbf{t}\right)\in L^2\left({[0,1]}^d\right)$, such that its Fourier coefficients $c_{\mathbf{k}}\left(h\right)$ satisfy

$$c_{\mathbf{k}}\left(h\right)=c_{\mathbf{k}}\left(f\right)D\left(2\mathbf{k}\pi \right)(\mathbf{k}\ne \mathbf{0}),c_{\mathbf{0}}\left(h\right)=0.$$

(30)

From this and $D\left(2\mathbf{k}\pi \right)\le B(\mathbf{k}\ne \mathbf{0})$, we know that ${\left\{c_{\mathbf{k}}\left(h\right)\right\}}_{\mathbf{k}\in {\mathbb{Z}}^d}\in l^2\left({\mathbb{Z}}^d\right)$, and then $h\left(\mathbf{t}\right)$ is well-defined. Using the Parseval identity of the Fourier series, by (28)–(30), we obtain

$${(h,{\tilde{\psi }}_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}=\sum _{\mathbf{k}}{c}_{\mathbf{k}}\left(h\right)\overline{{\tilde{c}}_{\mathbf{k}}^{(m,\mathbf{n})}}=\sum _{\mathbf{k}}{c}_{\mathbf{k}}\left(f\right)\overline{c_{\mathbf{k}}^{(m,\mathbf{n})}}={(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}(m\ge 0;\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d).$$

Since $\{{\mathbf{\Psi }}^{per},{\tilde{\mathbf{\Psi }}}^{per}\}$ are a pair of dual frames, we have

$${\displaystyle \begin{array}{ccc}{(f,h)}_{L^2\left({[0,1]}^d\right)}\hfill & =\hfill & {\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}{(h,{\tilde{\psi }}_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}^{-}+{(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{(h,\frac{1}{\sqrt{A}})}_{L^2({[0,1]}^d)}^{-}\hfill \\ & =\hfill & {\displaystyle \sum _{m\ge 0}}{\displaystyle \sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}}{|{(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}|}^2+{(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{(h,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}.\hfill \end{array}}$$

From ${(h,\frac{1}{\sqrt{A}})}_{L^2\left({[0,1]}^d\right)}=\frac{1}{\sqrt{A}}{\int }_{{[0,1]}^d}h\left(\mathbf{t}\right)d\mathbf{t}=\frac{1}{\sqrt{A}}{c}_{\mathbf{0}}\left(h\right)=0$, it follows that

$${(f,h)}_{L^2\left({[0,1]}^d\right)}=\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}{|(f,{\psi }_{m,\mathbf{n}}^{per})|}_{L^2\left({[0,1]}^d\right)}^2.$$

(31)

Using the Parseval identity, by (30), we obtain

$${(f,h)}_{L^2\left({[0,1]}^d\right)}=\sum _{\mathbf{k}\in {\mathbb{Z}}^d}{c}_{\mathbf{k}}\left(f\right){\overline{c}}_{\mathbf{k}}\left(h\right)=\sum _{\mathbf{k}\ne \mathbf{0}}{|c_{\mathbf{k}}\left(f\right)|}^{2}D\left(2\mathbf{k}\pi \right).$$

(32)

Noticing that ${(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}={\int }_{{[0,1]}^d}\sqrt{A}f\left(\mathbf{t}\right)d\mathbf{t}=\sqrt{A}{c}_{\mathbf{0}}\left(f\right)$, by (31) and (32), we obtain

$$L\left(f\right):=\sum _{m\ge 0}\sum _{\mathbf{n}\in {\mathbb{Z}}^d\bigcap {[0,2^m-1]}^d}|{(f,{\psi }_{m,\mathbf{n}}^{per})}_{L^2\left({[0,1]}^d\right)}{|}^2+|{(f,\sqrt{A})}_{L^2\left({[0,1]}^d\right)}{|}^2=\sum _{\mathbf{k}\ne \mathbf{0}}|c_{\mathbf{k}}\left(f\right){|}^{2}D\left(2\mathbf{k}\pi \right)+A{|c_{\mathbf{0}}\left(f\right)|}^2.$$

(33)

Since $D\left(2\mathbf{k}\pi \right)=D\left(\frac{2\mathbf{k}\pi }{2^l}\right)(\mathbf{k}\in {\mathbb{R}}^d\backslash \left\{\mathbf{0}\right\},l\in \mathbb{Z})$ and the point set $\{\frac{2\mathbf{k}\pi }{2^l}(\mathbf{k}\in {\mathbb{R}}^d\backslash \left\{\mathbf{0}\right\},l\in \mathbb{Z})$ is dense in ${\mathbb{R}}^d$ and $D\left(\mathit{\omega }\right)$ is continuous on ${\mathbb{R}}^d\setminus \left\{\mathbf{0}\right\}$, we obtain

$$\underset{\mathbf{k}\ne \mathbf{0}}{sup}D\left(2\mathbf{k}\pi \right)=\underset{\stackrel{\mathbf{k}\in {\mathbb{Z}}^d\setminus \left\{\mathbf{0}\right\}}{l\in \mathbb{Z}}}{sup}D\left(\frac{2\mathbf{k}\pi }{2^l}\right)=\underset{\mathit{\omega }\in {\mathbb{R}}^d\setminus \left\{0\right\}}{sup}D\left(\mathit{\omega }\right)=B.$$

(34)

Similarly, we have

$$\underset{\mathbf{k}\ne \mathbf{0}}{inf}D\left(2\mathbf{k}\pi \right)=\underset{\stackrel{\mathbf{k}\in {\mathbb{Z}}^d\setminus \left\{\mathbf{0}\right\}}{l\in \mathbb{Z}}}{inf}D\left(\frac{2\mathbf{k}\pi }{2^l}\right)=\underset{\mathit{\omega }\in {\mathbb{R}}^d\setminus \left\{\mathbf{0}\right\}}{inf}D\left(\mathit{\omega }\right)=A.$$

(35)

Combining (34) and (35) and (33) gives

$$A\sum _{\mathbf{k}\ne \mathbf{0}}|c_{\mathbf{k}}{\left(f\right)|}^2+A|c_{\mathbf{0}}{\left(f\right)|}^2\le L\left(f\right)\le B\sum _{\mathbf{k}\ne \mathbf{0}}|c_{\mathbf{k}}{\left(f\right)|}^2+A{|c_{\mathbf{0}}\left(f\right)|}^2.$$

Since ${\sum }_{\mathbf{k}\ne \mathbf{0}}|c_{\mathbf{k}}{\left(f\right)|}^2+|c_{\mathbf{0}}{\left(f\right)|}^2={\sum }_{\mathbf{k}\in {\mathbb{Z}}^d}|c_{\mathbf{k}}{\left(f\right)|}^2={\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2$, we have

$${A\Vert f\Vert }_{L^2{\left({[0,1]}^d\right)}^2}\le L\left(f\right)\le B{\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2.$$

This implies by (33) that A and B are the lower and upper frame bounds for the periodic frame ${\mathbf{\Psi }}^{per}$.

We begin to show that these frame bounds are optimal. By (34), we know that for any $ϵ>0$, there exists an $\mathbf{l}\ne \mathbf{0}$ such that $D\left(2\pi \mathbf{l}\right)\ge B-ϵ$. Take $f\left(\mathbf{t}\right)=e^{2\pi i\mathbf{l}·\mathbf{t}}$. By (33), it follows that

$$L\left(f\right)\ge (B-ϵ)|c_{\mathbf{l}}{\left(f\right)|}^2=B-ϵ=(B-ϵ){\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2.$$

This means that B is the optimal upper bound of ${\mathbf{\Psi }}^{per}$. Now we take $f\left(\mathbf{t}\right)=1$. By (33), we have

$$L\left(f\right)=A|c_{\mathbf{0}}{\left(f\right)|}^2=A=A{\Vert f\Vert }_{L^2\left({[0,1]}^d\right)}^2.$$

This implies that A is the optimal lower bound of ${\mathbf{\Psi }}^{per}$.

Similarly to the above arguments, we can deduce that $B^{-1}$ and $A^{-1}$ are the optimal bounds of ${\tilde{\mathbf{\Psi }}}^{per}$. Finally, Theorem 3 is proved.

5. Conclusions

Due to their resilience to background noise, stability in sparse reconstruction, and ability to capture local time-frequency information, periodic framelets are widely utilized in data analysis. Existing periodic framelets in high-dimensional space always require an infinite number of generators. Moreover, it is very difficult to construct periodic framelets with given optimal bounds. In this study, for any given constants, A and B, satisfying $0<A\le B<+\infty$, we construct a symmetric periodic frame ${\mathrm{\Psi }}^{per}$ generated by integer-translates, dyadic-dilations, and periodization of a single symmetric generator $\psi$ and its optimal frame bounds are just A and B. Since the generator $\psi$ is symmetric, smooth, and bandlimited, the derived periodic frame ${\mathrm{\Psi }}^{per}$ is also symmetric and smooth. Moreover, we present the construction of its dual frame, which has the same desirable properties as ${\mathrm{\Psi }}^{per}$. In the future, we will extend our results to periodic framelets with matrix dilation.