High-Dimensional Periodic Framelet with Only One Symmetric Generator

: 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 .

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 (R d ) 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 ([0, 1] d ).Compared with framelets in L 2 (R d ), 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 ([0, 1] d ).Moreover, we can determine the optimal bounds of the derived periodic frame.

Preliminary
Let {ϕ n } n∈Z + be a sequence of elements in the Hilbert space H.If there exists a positive constant B, such that for all f ∈ H, then {ϕ n } n∈Z + is called a frame with lower bound C 1 and upper bound C 2 in 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 {ϕ n } n∈Z + and { ϕ n } n∈Z + be two frames in H.If for all f ∈ H, Then {ϕ n } n∈Z + and { ϕ n } n∈Z + are called a pair of dual frames in H.
A framelet can generate a frame in L 2 (R d ) through its dyadic dilations and integer translates.In detail, for {ψ µ } µ=1,2,...τ ⊂ L 2 (R d ), if the associated affine system is a frame in L 2 (R d ), then {ψ µ } µ=1,2,...τ are called a framelet, and each ψ µ 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.

Construction of High-Dimensional Periodic Frames
Let ψ(t) ∈ L 2 (R d ) be a real-valued, even, and symmetric function, such that its Fourier transform satisfies the following three conditions: and In this section, we will give the following theorem: Theorem 2. The system {Ψ per , Ψ per } denotes a pair of dual periodic frames in L 2 ([0, 1] d ).
where C 1 and C 2 are constants.
Proof.Noticing that (2), it is enough to prove that the following inequalities hold: For any ( , we see that D(ω) has only finite nonzero terms in any closed interval which does not contain the point ω = 0. Since D(ω) has positive lower bound A, by where t = (t 1 , . . ., t d ).
Denote the left-hand side of the first inequality in (3) by K( f ), i.e., where From this and (5), we obtain K( f ) ≤ 2(J 1 + J 2 ), where From here on, "O" only depends on the dimension d.Let p = n − 2 m k, we further have 5), we have Parseval identity of the Fourier series and supp ψ( For J 2 , let f * (t) be a 1-periodic function and Applying the Cauchy's inequality gives .
Again by (4), we have where k = (k 1 , k 2 , . . ., k d ), t = (t 1 , t 2 , . . ., t d ), n = (n 1 , n 2 , . . ., n d ).Furthermore Combining these and (9) gives We begin to estimate the core term in J 2 We consider the following d cases: Case 1.We consider the index k satisfying Case 2. We consider the index k satisfying that k 1 = 0 and k i > 0 (i = 1).It follows that Similarly, we can obtain

Case p. We consider the index k satisfying that
Similarly, we can obtain

Case d.
We consider the index k satisfying that k i = 0 and k j > 0 (j = i).Similar to the argument of Case p, we can deduce that By the combination of all d cases and using the inequality (|a Again by the symmetry of the index k, by (11), we can deduce that Combining this and ( 10)-( 11) gives From this and ( 8) and ( 5), noticing that K( f ) ≤ 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 ∈ L 2 ([0, 1] d ), we have Proof.First, we compute By (4), it is clear that Noticing that using the inequality ab ≤ 1 2 (a 2 + b 2 ) and ( 3), we obtain From this and ( 13), we know that and this series is convergent absolutely.By the rearrangement of terms, we obtain Let both f * and g * be 1-periodic functions and f Define two sequences of functions as Noticing that we obtain Since supp ψ = supp ψ ⊂ [−π, π] d , applying the Parseval identity of the Fourier transforms gives Noticing that { e i n•ω ) and g l (ω) ψ( ω 2 m ) into Fourier series, where the associated Fourier coefficients are, respectively, Comparing ( 16) with ( 17), we have Applying the Parseval identity of the Fourier series gives From this and ( 18), and supp Again, by (15), we obtain Noticing that it is clear that From this and ( 19), applying the Lebegue dominant convergence theorem gives Secondly, we compute Take an auxiliary function Since ψ(ω) = 0 (ω ∈ (− , ) d ), for m ≥ 0, we have ψ ω 2 m = 0 (ω ∈ (− , ) d ).From this and (20), we have Again, by η(0) = √ A, we have ) has only finitely many non-zero terms in any closed interval which does not contain the point ω = 0.In the beginning of the proof of Lemma 1, we have shown Again by ( 23) and ( 24), we have Take the other auxiliary function η satisfying three conditions: Then, by ( 25) and ( 23), we have From this and ( 22) and (25), we obtain Applying the Parseval identity of the Fourier series gives where the Parseval identity of the Fourier transform is used in the last equality.Similarly, we have By (26), we obtain By the definitions of f 0 and g k , we have Since η(ω), η(ω) ∈ C ∞ (R d ), we know that η(t), η(t) decay very fast, so the series ( 27) is absolutely convergent and then The Poisson summation formula indicates that the Fourier coefficients of . Again, by (27), we obtain , by (22), we obtain Combining this and ( 21) and ( 20), we obtain By the definitions of f 0 and g k , we obtain Again, by the definitions of M and N, we obtain 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 {Ψ per , Ψ per } are a pair of dual periodic frames, i.e., Theorem 2 is proved.

Optimal Frame Bounds
In Section 3, we constructed a periodic frame with only one symmetric generator.Now we will provide its optimal frame bounds.
Combining ( 34) and ( 35) and (33) gives This implies by (33) that A and B are the lower and upper frame bounds for the periodic frame Ψ per .We begin to show that these frame bounds are optimal.By (34), we know that for any > 0, there exists an l = 0 such that D(2πl) ≥ B − .Take f (t) = e 2πil•t .By (33), it follows that This means that B is the optimal upper bound of Ψ per .Now we take f (t) = 1.By (33), we have This implies that A is the optimal lower bound of Ψ per .Similarly to the above arguments, we can deduce that B −1 and A −1 are the optimal bounds of Ψ per .Finally, Theorem 3 is proved.

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 ≤ B < +∞, we construct a symmetric periodic frame Ψ per generated by integertranslates, dyadic-dilations, and periodization of a single symmetric generator ψ and its optimal frame bounds are just A and B. Since the generator ψ is symmetric, smooth, and bandlimited, the derived periodic frame Ψ per is also symmetric and smooth.Moreover, we present the construction of its dual frame, which has the same desirable properties as Ψ per .In the future, we will extend our results to periodic framelets with matrix dilation.
Z) and the point set { 2kπ 2 l (k ∈ R d \{0}, l ∈ Z) is dense in R d and D(ω) is continuous on R d \ {0}, we obtain sup For any f ∈ L 2 ([0, 1] d ), we have