Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 229 https://internationalpubls.com Logic Density in Gabor Frame Structures Manal Yagoub Ahmed Juma M.juma@qu.edu.sa Department of mathematic, College of Science, Qassim University, Buraidah, Saudi Arabia Article History: Received: 28-07-2024 Revised: 17-09-2024 Accepted: 01-10-2024 Abstract: The Gabor system G(g,1+ε,ε-1)={e^2iπ(ε-1)nt g(t-(1+ε)m):m,n∈Z}is examined with respect to its frame property, uunder reasonable oversampling conditions, that is, when (ε+1,ε-1∈Q). An appropriate "rational" counterpart of the Ron-Shen Gramian is developed, establishing that for every peculiar pane function (g), The structure: G(g,ε+1,ε-1) fails to produce a frame if ε^2=(2n-1)/(n-1). A particular focus is placed on the initial Hermite function, h_1=te^(-πt^2 ). Key word: Gabor system, rational, window function, initial Hermite function. 1. Introduction One of the primary concerns with Gabor analysis is the following. Determine the set of lattice parameters (𝜀 > 1) for the Gabor system: 𝒢(𝑔, 𝜀 + 1, 𝜀 − 1) = {𝑒2𝑖𝜋(𝜀−1)𝑛𝑡𝑔(𝑡 − (𝜀 + 1)𝑚):𝑚, 𝑛 ∈ ℤ}; forms a frame in 𝐿2(ℝ) given a window function 𝑔 ∈ 𝐿2(ℝ): A frame for 𝐿2(ℝ)is given by: ℱ(𝑔) ≔ {(𝜀 + 1, 𝜀 − 1) ∈ ℝ+ 2 : 𝒢(𝑔, 𝜀 + 1, 𝜀 − 1) }. We evaluate some of the available information on the structure set ℱ(𝑔), which comes after [1] and [2] (in a more condensed format). In more relaxed circumstances ,if 𝑔 ∈ Feichtinger algebra 𝑀1, the set ℱ(𝑔) is open in ℝ+ 2 and includes the region around the origin. If 𝑔 ∈ Feichtinger algebra 𝑀1, the open set ℱ(𝑔) in ℝ+ 2 includes the region around the origin. In addition, ℱ(𝑔) ⊂ ∏+ ≔ {(𝜀 + 1, 𝜀 − 1) ∈ ℝ+ 2 : 𝜀 < √2} for 𝑔 ∈ 𝑀1, according to fundamental density theorems [3,4,5] and an adaptation of the uncertainly principle [6, 7]. Only a small number of functions for which ℱ = ∏+have been identified up until quite recently. The hyperbolic secant 𝑔(𝑡) = (𝑒𝑡 + 𝑒−𝑡)−1, while the Gaussian 𝑔(𝑡) = 𝑒−𝜋𝑡 2 [8,9,10], and the one- and two-sided exponential functions 𝑔(𝑡) = 𝑒−𝑖𝔦ℝ+ (here 𝜀 = √2 )also produces a frame) and 𝑔(𝑡) = 𝑒−𝜋𝑡 2 ) [11,12] are included in the list, along with their shifts and Fourier transformations. In [2], a significant discovery was made when the authors established that any fully positive function of finite type had this feature, this results in an infinite family of functions where ℱ(𝑔) = ∏+. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 230 https://internationalpubls.com However, [13] shows that even for a "simple" function like the characteristic function 𝑔 = 1𝑙 of an interval, the set ℱ(𝑔) may have a somewhat complex shape. When 𝑔 is concentrated in both time and frequency and ℱ(𝑔) ≠ ∏+, there are certain instances. First, we want to get the Hermite function ℎ1(𝑡) = 𝑡𝑒 −𝜋𝑡2. This conclusion is motivated by the uncertainty principle, which states that ℎ1 lowers the Heisenberg uncertainty of all functions orthogonal to the Gaussian, as well as present studies on vector-valued Gabor frames [14]. The paper is organized as following. Basic information from Gabor analysis and notation are presented in the next section. We examine the scenario of logical oversampling, where 𝜀2 = 𝑝 𝑞 + 1 ∈ ℚ .The union of hyperbolas is not contained in the set ℱ(𝑔) for any odd function 𝑔 ∈ 𝑀1 (especially ℎ1): 𝜀2 = 2𝑛 − 1 𝑛 − 1 ⇒ (𝜀 + 1, 𝜀 − 1) ∉ ℱ(𝑔), 𝑛 = 1,2, …. (1) The vector-valued Zak transform, which encodes the frame operator as matrix multiplication in a space of vector-valued functions, is examined in [15] and [4, ch. 8] to lay the groundwork for the proof. notice the following part, we factorize the vector-valued Zak transform matrix and extract a component that is a rational version of the well-known Ron-Shen Gramian (see [16] and [4]). We hypothesize that the only limitation on the set ℱ(ℎ1) is condition (1). ℱ(ℎ1) = {(𝜀 + 1, 𝜀 − 1) ∈ ∏+: 𝜀 2 ≠ 2𝑛 − 1 𝑛 − 1 , 𝑛 = 1,2, … . }. Regretfully, we are unable to fully verify this conjecture. We provide numerical verification for a larger collection of points and show it analytically only for a subset of ∏+. The rational equivalent of the Gramain provided in section 4 serves as the foundation for these creations. 2.Preliminaries The fundamental information from the Gabor analysis is reviewed in this part and will be applied later. For a more thorough explanation and the background information on the topic, we direct the reader to [4]. We investigate the lattice ∧= (𝜀 + 1)ℤ × (𝜀 − 1)ℤ, the Gabor system has an area function g and a value 𝜀 > 1. 𝒢(𝑔, 𝜀 + 1, 𝜀 − 1) = {𝜋𝜆: 𝜆 ∈∧} The standard time-frequency shift is indicated by the symbol: 𝜋𝜆: 𝑔 → 𝑒2𝜋𝑖𝑏𝑡𝑔(𝑡 − 𝑎) for 𝜆 = (𝑎, 𝑏). The definition of the Gabor frame operator 𝑆𝑔.∧: 𝐿 2(ℝ) → 𝐿2(ℝ) is: ∑𝑆𝑔,∧𝑓𝑖(𝑡) 𝑖 =∑(∑〈𝑓𝑖, 𝜋𝜆𝑔〉𝐿2(ℝ) 𝜆∈∧ 𝜋𝜆𝑔(𝑡)) 𝑖 , 𝑓𝑖 ∈ 𝐿 2(ℝ) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 231 https://internationalpubls.com The operation 𝒢(𝑔,∧) is bounded if and only if the Gabor frame operator is invertible and 𝑔 is in the modulation space 𝑀1(ℝ), as defined later in this section. In this case, the operator 𝑆𝑔,∧ can be written as a multiplication operator in a vector-valued function space. Assume that for reasonably prime 𝑝, 𝑞 ∈ ℕ, 𝜀2 = 𝑝 𝑞 + 1. Examine the following: the dimension of vector-valued function ℋ𝜀+1,𝑝 = 𝐿 2(𝑄𝜀+1,𝑝, ℂ 𝑝) ,and the rectangle 𝑄𝜀+1,𝑝 = [0, 𝜀 + 1 𝑝⁄ ) × [0, [0, 1 𝜀 + 1⁄ ). Recall that a function 𝑓𝑖 ∈ 𝐿 2(ℝ)has a Zak transform defined as: ∑𝒵𝜀+1𝑓𝑖(𝑥, 𝑤𝑖) 𝑖 =∑∑𝑓𝑖 𝑛∈ℤ (𝑥 − (𝜀 + 1)𝑛)𝑒2𝜋𝑖𝑛(𝜀+1)𝑤𝑖 . 𝑖 (2) We study the vector-valued Zak transform �⃗�𝜀+1: 𝐿 2(ℝ) → ℋ𝜀+1,𝑝 defind as follows, in accordance with [4,ch.8]: �⃗�𝜀+1∑𝑓𝑖(𝑥, 𝑤𝑖) 𝑖 = 𝒵𝜀+1∑((𝑥 + 𝜀 + 1 𝑝 𝑟, 𝑤𝑖)) 𝑖 𝑟=1 𝑝 , (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,𝑝. The unitary mapping between 𝐿2(ℝ) and ℋ𝜀+1,𝑝 is the vector-valued Zak transform, subject to normalization. Additionally, note: ∑𝐴𝑟 𝑠(𝑥, 𝑤𝑖) 𝑖 = (𝜀 + 1)(∑∑𝒵𝜀+1𝑔 (𝑥 + 𝜀 + 1 𝑝 𝑠,𝑤𝑖 − (𝜀 − 1)𝑗) ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅ 𝑞−1 𝑗=0 𝒵𝜀+1𝑔 (𝑥 + 𝜀 + 1 𝑝 𝑠,𝑤𝑖 − (𝜀 𝑖 − 1)𝑗) 𝑒2𝜋𝑖𝑗(𝑟−𝑠) 𝑞⁄ ), As an example, take into consideration the 𝑝 × 𝑝 matrix function 𝒜(𝑥,𝑤𝑖)=∑ ((𝐴𝑟 𝑠(𝑥, 𝑤𝑖))𝑟,𝑠=0 𝑝−1 )𝑖 , where (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,𝑝. Theorem (1.1): Zibulski and Zeevi. We obtain : �⃗�𝜀+1∑ ((𝑆𝑔,𝜀+1,(𝜀−1)𝑓𝑖)(𝑥, 𝑤𝑖))𝑖 = ∑ (𝒜(𝑥,𝑤𝑖)�⃗�(𝜀+1)𝑓𝑖(𝑥, 𝑤𝑖))𝑖 , for nearly all (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,𝑝, based on the aforementioned assumptions. We always assume that g is a member of the modulation space 𝑀1(ℝ) ,in the following. We merely restate the definition here, and readers are directed to [4] for a more thorough presentation. Examine the immediate Fourier transform, where 𝑓𝑖 acts as the aperture function, given a functions 𝑓𝑖 in the Schwartz class 𝑆(ℝ) ∑(𝑉𝑓𝑖𝑔(𝑥,𝑤𝑖)) 𝑖 = lim 𝑏→−∞ (∑(∫ 𝑔(𝑡) ∞ 𝑏 𝑓𝑖(𝑥 − 𝑡)𝑒 2𝜋𝑖𝑤𝑖𝑡𝑑𝑡) 𝑖 ). Definition (1.2) : The modulation space 𝑀1(ℝ) includes functions 𝑔 ∈ 𝐿2(ℝ) with ∑(∫∫|𝑉𝑓𝑖𝑔(𝑥,𝑤𝑖)| ℝ ℝ 𝑑𝑥𝑑𝑤𝑖) 𝑖 < ∞ For certain (or all) non-trivial functions, 𝑓𝑖is in 𝒮(ℝ). The following statement is the result of Theorem (1.1) [1]. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 232 https://internationalpubls.com Corollary (1.3) : The Gabor set 𝒢(𝑔,∧) is a structure in 𝐿2(ℝ) only if: 𝑑𝑒𝑡∑𝒜(𝑥,𝑤𝑖) 𝑖 ≠ 0 , (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,𝑝. (3) To make this condition more understandable, we factorize matrix 𝒜 A. Consider column vectors: ∑𝑋𝑗(𝑥, 𝑤𝑖) 𝑖 =∑(𝑋𝑟 𝑗(𝑥, 𝑤𝑖)) 𝑟=0 𝑝−1 𝑖 , 𝑗 = 0,1, …… . , 𝑞 − 1 In which: ∑𝑋𝑟 𝑗(𝑥, 𝑤𝑖) 𝑖 = 𝒵(𝜀+1)𝑔∑(𝑡 + (𝜀 + 1)𝑟 𝑝 ,𝑤𝑖 − (𝜀 − 1)𝑗) 𝑒 2𝜋𝑖𝑗𝑟 𝑞⁄ 𝑖 , (4) Including 𝑝 × 𝑞 matrix: ∑𝒬(𝑡,𝑤𝑖) 𝑖 = (𝑋𝑗) 𝑗=0 𝑞−1 =∑(𝒵(𝜀+1)𝑔 (𝑡 + (𝜀 + 1)𝑟 𝑝 ,𝑤𝑖 − (𝜀 − 1)𝑗) 𝑒 2𝜋𝑖𝑗𝑟 𝑞⁄ ) 𝑟=0,𝑗=0 𝑝−1,𝑞−1 𝑖 Clearly ∑𝒜(𝑥,𝑤𝑖) 𝑖 =∑𝒬(𝑥,𝑤𝑖)𝒬 𝑇(𝑥, 𝑤𝑖) 𝑖 . Here, as always,𝒬𝑇T represents the adjoint matrix of 𝒬. �⃗�(𝜀+1)∑𝒜(𝑥,𝑤𝑖)𝑓𝑖(𝑥, 𝑤𝑖) 𝑖 =∑(∑〈𝑋𝑗(𝑥, 𝑤𝑖), �⃗�(𝜀+1)𝑓𝑖(𝑥, 𝑤𝑖)〉 𝑞−1 𝑗=0 〈𝑋𝑗(𝑥, 𝑤𝑖), �⃗�(𝜀+1)𝑓𝑖(𝑥, 𝑤𝑖)〉) 𝑖 , For each (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,𝑝, the 𝑋𝑗(𝑥, 𝑤𝑖), 𝑗 = 0,1, …… . . 𝑞 − 1, must span the entire ℂ𝑝,to meet condition (3). Corollary (1.4): Let 𝜀2 = 𝑝 𝑞 + 1 ∈ ℚ, and 𝑔 ∈ 𝑀1(ℝ). If 𝒢(𝑔,∧) is a farm in 𝐿2(ℝ), rank ∑ 𝒬(𝑥,𝑤𝑖)𝑖 = must be equal to 𝑝 for any (𝑥, 𝑤𝑖)) in 𝑄𝜀+1,𝑝. In the following section, we apply this requirement to investigate Gabor systems formed by odd functions. 3. The Odd Functions that is Generate Gabor Frames The next part includes what comes next. Theorem (3.1): Assume 𝑔 is an odd function in 𝑀1(ℝ)) and 𝜀2 = 2𝑛−1 𝑛 , where 𝑛 = 1,2, …… ,Then, 𝒢(𝑔,∧) does not constitute a frame in 𝐿2(ℝ). Proof: It is sufficient to establish that 𝑟𝑎𝑛𝑘 𝒬(0,0) < 2𝑛 − 1 (5) The conclusion follows from Theorem (1.1) [2]. Equation (5) is calculated from the findings that for odd open spaces, the elements of the matrices 𝒬(0,0)have additional symmetries. We consider 𝜀 = Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 233 https://internationalpubls.com 0, this may always be accomplished by appropriately scaling 𝑔. We assume 𝜀 = 0, which can be done by appropriately scaling g. Lemma (3.2): Let 𝑔 be an odd function in 𝑀1(ℝ)with 𝜀 = 0, 𝜀 = 𝑝 𝑞 + 1 ∈ ℚ and 𝑀𝑠 𝑗 = 𝑀𝑠 𝑗(0,0), The functions such as 𝑀𝑠 𝑗(𝑥, 𝑤𝑖) are defined in [4]. Then: 𝑀𝑠 𝑗 = 𝑀𝑝−𝑠 𝑞−𝑗 , 𝑠 = 0,1, … . . 𝑝 − 1, 𝑗 = 0,1, … . . 𝑞 − 1 (6) The explanation of the Zak transformation, along with the fact that g is odd, makes the lemma simple to prove. We 𝑞 − 𝑝 = 1. Assume q is an even number; 𝑞 = 2𝑘 + 2, 𝑝 = 2𝑘 + 1. Q(0,0) denotes the (2𝑘 + 1) × (2𝑘 + 2) matrix. Additional relations must exist for the zero row, zero column, and (𝑘 + 1) − 𝑡ℎ(𝑘 + 1) -th column of 𝒬(0,0). In particular: 𝑋0 0 = 0, 𝑋0 𝑘+1 = 0, 𝑋0 𝑗 = −𝑋0 𝑞−𝑗 -zero row; (7) 𝑋𝑠 0 = −𝑋𝑝−𝑠 0 , 𝑠 = 1,…… , 𝑝 − 1 -zero column; (8) 𝑋𝑠 𝑘+1 = −𝑋𝑝−𝑠 𝑘+1, 𝑠 = 1, …… , 𝑝 − 1 -(𝑘 + 1)-th column. (9) As in theorem (1.1) [1], the connections are simply established by the Zak transform formula and the fact that 𝑔 is odd. R_s denotes the s-th row of 𝒬(0,0).. Consider the row vectors: 𝑒𝑙 = (𝑒𝑙 𝑗 ) 𝑗=0,1,……,2𝑘+2 𝑙 = 1,2, … . , 𝑘, where 𝑒𝑙 𝑗 = 0, for 𝑗 ≠ 𝑙 + 1,2𝑘 + 2 − 1, 𝑒𝑙 2𝑘+2−𝑙 = −1. According to the relations [6], [7], [8], and [9], all rows of 𝒬 belong to 𝑆 = 𝑠𝑝𝑎𝑛{{𝑅𝑠}𝑠=1 𝑘 ∪ {𝑒𝑙}𝑙=0 𝑘 }. Yes, the row 𝑅0 has the from : 𝑅 = (0, (𝜀 + 1)1, … . . , (𝜀 + 1)𝑘, 0, −(𝜀 + 1)𝑘, …… ,−(𝜀 + 1)1) (10) This vector can be spanned by {𝑒𝑙}𝑙=0 𝑘 for certain (𝜀 + 1)1, … . . , (𝜀 + 1)𝑘. The rows 𝑅𝑠, 𝑠 = 1,… , 𝑘, belong to the spanning set. To show that the rows 𝑅𝑝−𝑠, 𝑠 = 1,……𝑘, and the vectors: 𝑅𝑠 + 𝑅𝑝−𝑠 belong to 𝑠 . The latter is clear because [6] and [8] state that these vectors also contain [10]. This completes the example of the theorem for the case 𝑞 = 2𝑘 + 2, 𝑝 = 2𝑘 + 1. Let q be an odd number, 𝑝 = 2𝑘 and 𝑞 = 2𝑘 + 1. Again, in addition to the basic relation [6], we must have the relations for specific rows and columns: 𝑋0 𝑗 = −𝑋0 𝑞−𝑗 , -zero row; 𝑋𝑠 0 = −𝑋𝑝−𝑠 0 , -zero column; 𝑋𝑘 𝑗 = −𝑋𝑘 𝑞−𝑗 , -𝑘-th row. Take the rows 𝑒𝑙 = (𝑒𝑙 𝑗 ) 𝑗=0,1,….,2𝑘, 𝑙 = 1,2, … . . , 𝑘, with 𝑒𝑙 𝑗 = 0 ,if: 𝑗 ≠ 1, 𝑒𝑙 𝑗 = 1 and 𝑒𝑙 2𝑘−𝑙+1 = −1. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 234 https://internationalpubls.com Using the exact same values as before, we can observe that the set of 2𝑘 − 1 vectors {𝑅𝑠}𝑠=1,…..,𝑘−1 ∪ {𝑒𝑙}𝑙=1,…..,𝑘 spans all rows of the matrix 𝒬(0,0) This concludes the proving of Theorem (1.1) [1]. 4. Factorizing of Zibulskii-Zeevi Matrix This section focuses on the Zibulski-Zeevi matrix 𝒬(𝑥, 𝑤𝑖). Our objective is to reduce it to a simple 𝑝 × 𝑞 matrix with the same rank as 𝒬. This simpler matrix can be thought of as an analogue of the Ron-Shen Gramian [12] for rationally oversampled Gabor systems: similarly, to [12], the study of the frame property can be reduced to the study of a specific matrix generated by translates of the generating function; in our case, we consider the Zak transform of the generating function. We think this reduction is fascinating on its own, but it will also be utilized in the next part of our study Gabor frames generated by the first Harmonic function. Theorem (4.1): Assume the area of the function g from 𝑀1(ℝ) with 𝜀2 = 𝑝 𝑞 + 1 ∈ ℚ, where p and q are essentially prime. The system 𝒢(𝑔, 𝜀 + 1, 𝜀 − 1)) yields a frame in 𝐿2(ℝ) if and only if the matrix: ∑ 𝒫(𝑥, 𝑤𝑖)𝑖 = ∑ ((𝑍(𝜀+1)𝑞𝑔 (𝑥 + 𝜀+1 𝑝 (𝑡𝑝 + 𝑠𝑞), 𝑤𝑖)) 𝑠=0,𝑡=0 𝑝−1,𝑞−1 )𝑖 (11) Has 𝑟𝑎𝑛𝑘 = 𝑝 for all (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,𝑞. Proof: Fix (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,𝑞 and let 𝑋𝑠 𝑗 =∑𝒵(𝜀+1)𝑔 (𝑥 + 𝜀 + 1 𝑝 𝑠, 𝑤𝑖 − (𝜀 − 1)𝑗) 𝑒 2𝑖𝜋 𝑗𝑠 𝑞 𝑖 =∑𝑔(𝑥 + 𝜀 + 1 𝑝 (𝑠 − 𝑝𝑛)) 𝑒 2𝑖𝜋𝑗( 𝑠 𝑞 −(𝜀2−1)𝑛) 𝑛 𝑒 2𝑖𝜋𝑗( 𝑠 𝑞 −(𝜀2−1)𝑛) (12) Be the appropriate entry in the matrix 𝒬(𝑥,𝑤𝑖). Let 𝐿(𝑠) ≔ {𝑙: 𝑙 = 𝑠 − 𝑝 − 1, 𝑛 ∈ 𝕫}, 𝐿(𝑠, 𝑡) = {𝑙: 𝐿(𝑠): 𝑙 = 𝑡 + 𝑚𝑞,𝑚 ∈ 𝕫}and 1,… . , 𝑞 − 1. Setting 𝑙 = 𝑠 − 𝑝𝑛 in [12] yields: 𝑋𝑠 𝑗 =∑∑𝑔(𝑥 + 𝜀 + 1 𝑝 𝑙) 𝑒 2𝑖𝜋 𝑗𝑙 𝑞 +2𝑖𝜋(𝜀+1)𝑤𝑖 𝑠−1 𝑝 𝑛𝑖 =∑(∑𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖 𝑠 𝑝 𝑞−1 𝑡=0 ∑ 𝑔(𝑥 + 𝜀 + 1 𝑝 𝑙) 𝑙∈𝐿(𝑠,𝑡) 𝑒 −2𝑖𝜋(𝜀+1)𝑤𝑖 1 𝑝𝑒 2𝑖𝜋 𝑗𝑡 𝑞 ) 𝑖 (13) We picked 𝑘𝑡 ∈ {0,… , 𝑞 − 1} and 𝑚𝑠 ∈ {0,… , 𝑝 − 1} for each 𝑡 ∈ {0,… , 𝑞 − 1} and 𝑠 ∈ {0,… , 𝑝 − 1}, respectively, so that: 𝑘𝑡𝑝 = 𝑡 (𝑚𝑜𝑑𝑞), 𝑚𝑠𝑝 = 𝑠 (𝑚𝑜𝑑𝑝). Since 𝐿(𝑠, 𝑡) = {𝑘𝑡𝑝 +𝑚𝑠𝑞 − 𝑝𝑞𝑚: 𝑚 ∈ 𝕫}, (13) can be rewritten as: Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 235 https://internationalpubls.com 𝑋𝑠 𝑗 = ∑ 𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖 𝑠−𝑚𝑠𝑞 𝑝 ∑ 𝑒2𝑖𝜋(𝜀+1)(𝜀+1)𝑘𝑡 𝑞−1 𝑡=0 𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖𝑘𝑡𝑗 𝑝 𝑞 ∑ 𝑔 (𝑥 + 𝜀+1 𝑝 (𝑘𝑡𝑝 + 𝑚𝑠𝑞) − 𝑚(𝜀 + 1)𝑞) 𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖𝑚(𝜀+1)𝑞 𝑚∈𝕫`⏟ =∑ (𝑧(𝜀+1)𝑞𝑔(𝑥+ 𝜀+1 𝑝 (𝑘𝑡𝑝+𝑚𝑠𝑞),𝑤𝑖))𝑖 𝑖 (14) Because the integer 𝑘𝑡 passes through the set {0, … , 𝑞 − 1} as t runs through it, we may rewrite [14] as : 𝑋𝑠 𝑗 =∑𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖 𝑠−𝑚𝑠𝑞 𝑝 ∑𝑒2𝑖𝜋(𝜀+1)𝑤𝑖𝜏 𝑞−1 𝜏=0 𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖𝜏𝑗 𝑝 𝑞 ∑ 𝑔(𝑥 + 𝜀 + 1 𝑝 (𝜏𝑝 +𝑚𝑠𝑞) −𝑚(𝜀 + 1)𝑞) 𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖𝑚(𝜀+1)𝑞 𝑚∈𝕫`⏟ =𝑧(𝜀+1)𝑞𝑔(𝑥+ 𝜀+1 𝑝 (𝜏𝑝+𝑚𝑠𝑞),𝑤𝑖) 𝑖 Or ∑ 𝒬(𝑥,𝑤𝑖)𝑖 = 𝑑𝑖𝑎𝑔∑ {𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖 𝑠−𝑚𝑠𝑞 𝑝 } 𝑠=0 𝑝−1 �̃�(𝑥, 𝑤𝑖) 𝑑𝑖𝑎𝑔{𝑒 2𝑖𝜋(𝜀+1)𝑤𝑖𝜏} 𝜏=0 𝑞−1 𝑊𝑖 , Or ∑ �̃�(𝑥, 𝑤𝑖)𝑖 = ∑ (𝑍(𝜀+1)𝑞𝑔 (𝑥 + 𝜀+1 𝑝 (𝜏𝑝 + 𝑚𝑠𝑞), 𝑤𝑖)) 𝑠=0,𝜏=0 𝑝−1,𝑞−1 , 𝑊 = (𝑒 2𝑖𝜋𝜏𝑗 𝑝 𝑞) 𝜏,𝑗=0 𝑞−1 𝑖 . The matrix values �̃�(𝑥, 𝑤𝑖) and 𝒬(𝑥,𝑤𝑖) have the same rank. The matrices �̃�(𝑥, 𝑤𝑖) and 𝒬(𝑥,𝑤𝑖)difference only by row permutations, resulting in the same rank. The number 𝑚𝑠 spans the entire set of {0,1, … , 𝑝 − 1}. 5.Examples According to [5], the Gabor system 𝒢(ℎ1, 𝜀 + 1, 𝜀 − 1) is a frame if 𝜀2 < 3 2 , but not if 𝜀2 = 3 2 . Also, the authors present an example indicating that this result could be sharp. This section demonstrates that the system 𝒢(ℎ1, 𝜀 + 1, 𝜀 − 1)yields a frame in 𝐿2(ℝ),for 𝜀 + 1, 𝜀 − 1,with 𝜀2 > 3 2 . The proof is based on the matrix-function P from the previous paragraph and a result on diagonally dominant matrices. Theorem 5.1: Suppose that 𝜀2 = 8 5 and ℎ1(𝑡) = 𝑡𝑒 −𝜋𝑡2 , when the system: 𝒢(ℎ1, 𝜀 + 1, 𝜀 − 1) A structure in 𝐿2(ℝ). Proof: The structure of the matrix �̃�(𝑥, 𝑤𝑖) is defined as: ∑ 𝒫1(𝑥, 𝑤𝑖)𝑖 = ∑ ((𝑍8(𝜀+1)ℎ1(𝑥 + (𝜀 + 1)𝑡 + 𝑠 5(𝜀+1) 8 , 𝑤𝑖)))𝑠,𝑡=0 7,4 𝑖 . To prove rank ∑ �̃�(𝑥, 𝑤𝑖) = 5𝑖 , for all (𝑥, 𝑤𝑖) ∈ 𝑄𝜀+1,5, apply Theorem 2 and Corollary 2. We structured the proof into multiple steps. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 236 https://internationalpubls.com a) establish that 𝒢(ℎ1, 𝜀 + 1, 𝜀 − 1)) is a frame for 𝜀2 ≥ 2√2 5 + 1, simply show that 𝜀2 = 3 2 . Using the Fourier transform, the case 𝜀2 ≥ 2√2 5 + 1 is reduced to the prior one. b) Because the function ℎ1 decays quickly, we can approximate 𝑧5(𝜀+1)ℎ1(𝑥, 𝑤𝑖) using the series' maximal term. Specifically, the following holds true. Lemma 5.2: Let 0 ≤ |𝑥| < 5(𝜀+1) 2 . Then: |∑ (ℎ1(𝑥)) − 𝑍5(𝜀+1)ℎ1(𝑥, 𝑤𝑖)𝑖 | ≤ ℎ1(5𝜀 + 5 − |𝑥|), Where: 𝐶5(𝜀+1) = 2 + 1 ℎ(5𝜀 + 5) ∑ℎ1 𝑛≥2 (5𝑛𝜀 + 5𝑛) + ℎ1( (5𝜀 + 5)(2𝑛 − 1) 2 ) This lemma may be confirmed immediately. For practical purposes. We can suppose that 𝐶5(𝜀+1) = 0. c) We will demonstrate later in (𝑔), it is necessary for consideration: 0 ≤ 𝑥 ≤ 𝜀+1 6 . The Zak transform's quasi-periodicity and symmetry with ℎ1 will lead to this result. We split the interval 0 ≤ 𝑥 ≤ 𝜀+1 6 into two parts: : 0 ≤ 𝑥 < 𝜀+1 12 2 and 1 12 ≤ 𝑥 ≤ 𝜀+1 6 . d. Let 0 ≤ 𝑥 < 𝜀+1 12 . Assume the sub-matrix 𝒫1(𝑥, 𝑤𝑖)for : 𝑡 = 1,2,3, ……, When adjusting the second and third rows, this matrix assumes the following type: ( 𝑍5𝜀+5ℎ1(𝑥 + 𝜀 + 1,𝑤𝑖) 𝑍5𝜀+5ℎ1(𝑥 + 2𝜀 + 2,𝑤𝑖) 𝑍5𝜀+5ℎ1(𝑥 + 3𝜀 + 3,𝑤𝑖) 𝑍5𝜀+5ℎ1 (𝑥 + 13𝜀 + 13 3 , 𝑤𝑖) 𝑍5𝜀+5ℎ1 (𝑥 + 16𝜀 + 16 3 ,𝑤𝑖) 𝑍5𝜀+5ℎ1 (𝑥 + 19𝜀 + 19 3 ,𝑤𝑖) 𝑍5𝜀+5ℎ1 (𝑥 + 8𝜀 + 8 3 ,𝑤𝑖) 𝑍5𝜀+5ℎ1 (𝑥 + 11𝜀 + 11 3 ,𝑤𝑖) 𝑍5𝜀+5ℎ1 (𝑥 + 11𝜀 + 14 3 ,𝑤𝑖) ) We shall apply the following theorem to exponentially dominant matrices. Theorem (5.3): (refer to [15]). If (𝑎𝑖 𝑘), is an 𝑛 × 𝑛-matrix with complex members, then either: (i) |𝑎𝑖 𝑖| > ∑ |𝑎𝑘 𝑖 |𝑘≠𝑖 , 1 ≤ 𝑖 ≤ 𝑛. (ii) |𝑎𝑖 𝑖||𝑎𝑗 𝑗 | > (∑ |𝑎𝑘 𝑖 |𝑘≠𝑖 )(∑ |𝑎𝑘 𝑗 |𝑘≠𝑗 ) , 1 ≤ 𝑖, 𝑗 ≤ 𝑛, 𝑖 ≠ 𝑗 → |𝑎𝑖𝑘| ≠ 0. a. To establish the invertibility of the matrix in (d), we must use Theorem 5.3. We emphasized the essential steps of the proof. • Given that |𝑍5𝜀+5ℎ1(𝑥, 𝑤𝑖)| is (5𝜀 + 5 ),) periodic function, The absolute numbers of the matrix participants in (e) can be written as: ( |𝑍5𝜀+5ℎ1(𝑥 + 𝜀 + 1,𝑤𝑖) | |𝑍5𝜀+5ℎ1(𝑥 + 2𝜀 + 2,𝑤𝑖) | |𝑍5𝜀+5ℎ1(𝑥 + 2𝜀 − 2,𝑤𝑖) | |𝑍5𝜀+5ℎ1 (𝑥 − 2𝜀 + 2 3 ,𝑤𝑖)| |𝑍5𝜀+5ℎ1 (𝑥 + 𝜀 + 1 3 , 𝑤𝑖) | | 𝑍5𝜀+5ℎ1 (𝑥 + 4𝜀 + 4 3 , 𝑤𝑖)| |𝑍5𝜀+5ℎ1 (𝑥 − 7𝜀 + 7 3 ,𝑤𝑖)| |𝑍5𝜀+5ℎ1 (𝑥 − 4𝜀 + 4 3 ,𝑤𝑖) | |𝑍5𝜀+5ℎ1 (𝑥 − 𝜀 + 1 3 ,𝑤𝑖)| ) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 237 https://internationalpubls.com • For 0 ≤ 𝑥 ≤ 𝜀+1 12 ,and √ 3 5 − 1 ≤ 𝜀 ≤ 0 , the rule (ii) in Theorem (5.3) can now be verified effectively. • For 0 ≤ 𝑥 ≤ 𝜀+1 12 , and 𝜀 ≥ 0 , we consider the difference: 𝐻𝑖 𝑖 −∑𝐻𝑖 𝑘 𝑘≠𝑖 , 1 ≤ 𝑖 ≤ 3 If the above statement is positive, condition (i) in Theorem (5.3) is satisfied. Assume the first situation, when 𝑖 = 1. Then we have to verify: 𝐻1 1 − 𝐻1 2 − 𝐻1 3 > 0 (15) Let 𝑥 = 𝑦(𝜀 + 1) ,for 0 ≤ 𝑦 ≤ 1 12 . Then (15) is equivalent to: ℎ1((𝜀 + 1)(𝑦 + 1)) − 3ℎ1((𝜀 + 1)(𝑦 + 2)) − 3ℎ1((𝜀 + 1)(𝑦 − 2)) > ℎ1 ((𝜀 + 1) ( 1 2 + 1)) − 3ℎ1(2𝜀 + 2) − 3ℎ1 ((𝜀 + 1) (− 1 12 + 2)) > ℎ1 ( 13𝜀 + 13 12 ) − 6ℎ1 ( 23𝜀 + 23 12 ) − ℎ1 ( 13𝜀 + 13 12 ) (1 − 138 13 𝑒− 5𝜋(𝜀+1)2 2 ) This is clearly positive for any 𝜀 values that are greater than zero. The proofs for 𝑖 = 2 and 𝑖 = 3 follow the same method. f) The situations 1 12 ≤ 𝑥 ≤ 1 6 can be solved similarly to the previous one. The submatrix of 𝒫1(𝑥, 𝑤𝑖), the condition (i) of Theorem (3.5) has been confirmed by columns 𝑡 = 0,2,3,…. g) We show that the present circumstances 𝜀+1 6 ≤ 𝑥 ≤ 𝜀+1 3 ,can be simplified to the prior ones. By inserting 𝑥 = 𝜀+1 3 − 𝑦 ,we obtain: ∑|𝑍5𝜀+5ℎ1( 𝜀 + 1 3 − 𝑦 + (𝜀 + 1)𝑡 + 𝑟 5𝜖 + 5 3 ,𝑤𝑖)| 𝑖 =∑|𝑍5𝜀+5ℎ1( 𝜀 + 1 3 − 𝑦 + (𝜀 + 1)(𝑟 − 1) 5 3 + 5𝜖 + 5 3 ,𝑤𝑖)| 𝑖 =∑|𝑍5𝜀+5ℎ1(𝑦 − (𝜀 + 1)(𝑡 + 2) − (𝜀 + 1)(𝑟 − 1) 5 3 ,𝑤𝑖)| 𝑖 . The two manifestations are clearly exactly the same in terms of row/column permutations for 0 ≤ 𝑥 ≤ 𝜀+1 6 and 𝜀+1 6 ≤ 𝑥 ≤ 𝜀+1 3 . 6. Hypothesis Equations in the previous part become tedious for arbitrary 𝜀 + 1, 𝜀 − 1with 𝜀 ∈ 𝑄,𝜀 < √2. Numerous numerical calculations support the following statement. Conjecture (6.1): Suppose < √2 , and 𝜀2 ≠ 2𝑛−1 𝑛 , 𝑛 = 1 2 , …. Then 𝒢(𝐻1, 𝜀 + 1, 𝜀 − 1)in a frame in 𝐿2(ℝ).However, the researchers are at present unable to confirm or reject this conclusion, even if 𝜀 is Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 238 https://internationalpubls.com in 𝑄. Let 𝜀2 = 2𝑛−1 𝑛 , are the only not common issues. Figure 1 presents the most modest eigenvalue of �̃�(𝑥, 𝑤𝑖)�̃�(𝑥, 𝑤𝑖) 𝑇 for 0 ≤ 𝑥 ≤ 𝜀+1 2𝑝 , and 0 ≤ 𝑤𝑖 ≤ 1 𝜀+1 for 𝜀2 = 𝑛−𝑗 𝑛 + 1 , for 5 ≤ 𝑛 ≤ 201, and 1 ≤ 𝑗 ≤ 𝑛 − 1.Using these values for n and 𝑗, 𝜀 < √1.995.. In the experiments, we assumed 𝜀 = 0. The minimal eigenvalue of 𝒫𝒫𝑇for (a) 𝜀 < √1.98and (b)√1.98 < 𝜀 < √1.995. In the studies, we utilized 𝜀 = 0.It should be noted that the y-axis has been scaled differently. References [1] J. J. Benedetto, C. Heil, and D. F. Walnut. Differentiation and the Balian-Low theorem. J. Fourier Anal. Appl, 1:355–402, 1995. [2] W. Czaja and A. M. Powell. Recent developments in the Balian–Low theorem. In Harmonic Analysis and Applications. In Honor of John J. Benedetto, Applied and Numerical Harmonic Analysis. Birkh¨auser Boston, 2006. [3] I. Daubechies. Ten lectures on wavelets. Society for Industrial and Applied Mathematics, Philadelphia, PA, USA, 1992. [4] K. Gr¨ochenig. Foundations of Time-Frequency Analysis. Birkhuser, Boston, 2001. [5] K. Gr¨ochenig and Yu. Lyubarskii. Gabor (super)frames with Hermite functions. Math. Ann., 345(2):267–286, 2009.13 [6] K. Gr¨ochenig and J. St¨ockler. Gabor Frames and Totally Positive Functions. (arXiv:1104.4894), April 2011. [7] C. Heil. 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Density theorems for sampling and interpolation in the Bargmann-Fock space. II. J. Reine Angew. Math., 429:107–113, 1992. [15] O. Taussky. A recurring theorem on determinants. The American Mathematical Monthly, 56(10): pp. 672–676, 1949. [16] M. Zibulski and Y. Y. Zeevi. Analysis of Multiwindow Gabor-Type Schemes by Frame Methods. Applied and Computational Harmonic Analysis, 4(2):188 – 221, 1997.