On the Frame Set for the 3-Spline

Abdou Ganiou Débayo Atindehou

Département de Mathématiques, Faculté des Sciences et Techniques (FAST), Université d’Abomey-Calavi (UAC), Cotonou, Benin.

DOI: 10.4236/am.2022.135026                

Abstract

This paper investigates the fine structure of the Gabor frame generated by the B-spline B₃. In other words, one extends the known part of the Gabor frame set for the 3-spline with the construction of the compactly supported dual windows. The frame set of the function B₃ is the subset of all parameters (a,b) ∈ R²₊ for which the time-frequency shifts of B₃ along aZ × bZ form a Gabor frame for L²(R).

Keywords

Gabor Frames, Frame Set, 3-Splines, Compactly Dual Frame

1. A New Set of Points in the Frame Set of the 3-Spline

The Gabor frame generated by $g\in L^2\left(ℝ\right)$ and $a,b>0$ is the set of functions

$\mathcal{G}\left(g\mathrm{,}a\mathrm{,}b\right)=\left\{M_{\mathcal{l}b}{T}_{ka}g={\text{e}}^{2\pi i\mathcal{l}b\cdot }g\left(\cdot -ka\right)\mathrm{<mo>\; :\; </mo>}\left(\mathcal{l}\mathrm{,}k\right)\in {\mathbb{Z}}^2\right\}$

for which there exist $A,B>0$ such that

$A{\Vert f\Vert }_2^2\le {\displaystyle \underset{\mathcal{l}\mathrm{,}k\in \mathbb{Z}}{\sum }}{\left|\langle f\mathrm{,}{M}_{\mathcal{l}b}{T}_{ka}g\rangle \right|}^2\le B{\Vert f\Vert }_2^2\mathrm{,}$

for all $f\in L^2\left(ℝ\right)$, see, [1] [2] for details. The mystery of the fine structure of Gabor frames [3] and application requirements have obliged many researchers to investigate the characterization of Gabor frame set for some functions having good time-frequency concentration. For example, in the hyperbolic secant functions or different splines windows [4] [5] [6], the functions satisfy the partitions of unity [7] [8] and the sign-changing functions with compact support [9] [10] [11] are already been studied. More generally, the frame set $\mathcal{F}\left(g\right)$ of $g\in L^2\left(ℝ\right)$ is known only for a few classes of functions, see [3] [12] - [22]. For instance, the determination of the frame set of B-splines for $N\ge 2$ is listed as one of the six problems in frame theory [12]. Recently, we have investigated the frame set of a class of compactly supported functions that include the B-splines. In particular, we have extended and put in a more general framework some of the known results on the frame set for this class of functions [23].

In this paper, we investigate the set of parameters $\left(a\mathrm{,}b\right)\in {ℝ}_{+}^2$ such that $\mathcal{G}\left(B_3\mathrm{,}a\mathrm{,}b\right)$ is a Gabor frame where

$B_3\left(x\right)={\chi }_{\left[-1/2,1/2\right]}\ast {\chi }_{\left[-1/2,1/2\right]}\ast {\chi }_{\left[-1/2,1/2\right]}\left(x\right)=\{\begin{array}{ll}\frac{1}{2}{x}^2+\frac{3}{2}x+\frac{9}{8}\hfill & x\in \left[-\frac{3}{2}\mathrm{,}-\frac{1}{2}\right]\hfill \\ -x^2+\frac{3}{4}\hfill & x\in \left[-\frac{1}{2}\mathrm{,}\frac{1}{2}\right]\hfill \\ \frac{1}{2}{x}^2-\frac{3}{2}x+\frac{9}{8}\hfill & x\in \left[\frac{1}{2}\mathrm{,}\frac{3}{2}\right]\hfill \end{array}$

is the 3-spline. This set is called the frame set of $B_3$ and is given by

$\mathcal{F}\left(B_3\right)=\left\{\left(a\mathrm{,}b\right)\in {ℝ}_{+}^2\mathrm{<mo>\; :\; </mo>}\mathcal{G}\left(B_3\mathrm{,}a\mathrm{,}b\right)\text{is}\text{a}\text{frame}\right\}\mathrm{.}$

It is not difficult to see that $\mathcal{F}\left(B_3\right)\ne \varnothing$. Indeed, for any $a\in \left(\mathrm{0,3}\right)$ there exists $b_0>0$ such that for all $0<b\le b_0$, $\left(a\mathrm{,}b\right)\in \mathcal{F}\left(B_3\right)$ [ [24], Theorem 4.18]. Furthermore, $\mathcal{F}\left(B_3\right)$ is an open set in ${ℝ}_{+}^2$ [25] [26]. This is a consequence of the fact that $B_3$ belongs to the modulation space $M^1\left(ℝ\right)$ [2]. However, the complete characterization of $\mathcal{F}\left(B_3\right)$ is still unknown.

To the best of our knowledge [1] [5] [8] [11] [23] [27] [28], the biggest subset known in $\mathcal{F}\left(B_3\right)$ is the connected set

$\left\{\left(a,b\right)\in {ℝ}_{+}^2:ab<1,0<a<3,0<b\le \underset{a}{\max }\left(\frac{2}{3},\frac{4}{3+3a}\right)\right\}.$

In this note, we complete this last subset by showing that $\Gamma$ belongs to $\mathcal{F}\left(B_3\right)$ where $\Gamma ={\displaystyle \underset{m=3}{\overset{\infty }{\cup }}{\Gamma }_m}$ with for $m\ge 4$,

$\begin{array}{l}{\Gamma }_m:=\{\left(a,b\right)\in {ℝ}_{+}^2:a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-1\right)}{2m-1}\right],\\ b\in \left(\frac{2\left(m-1\right)}{3+\left(2m-3\right)a},\underset{a}{{\displaystyle \min }}\left(\frac{2m}{3+\left(2m-1\right)a},\frac{4}{3+2a}\right)\right],b>\frac{2}{3}\}\end{array}$ (1)

and

${\Gamma }_3:=\left\{\left(a,b\right)\in {ℝ}_{+}^2:a\in \left[\frac{3}{4},\frac{6}{5}\right],b\in \left(\frac{4}{3+3a},\frac{6}{3+5a}\right],b>\frac{2}{3}\right\}.$ (2)

More specifically, our main result gives the frame properties on $\Gamma$.

Theorem 1. For $m\ge 3$, let $\left(a\mathrm{,}b\right)\in {\Gamma }_m$. Then, the Gabor system $\mathcal{G}\left(B_3\mathrm{,}a\mathrm{,}b\right)$ is a frame for $L^2\left(ℝ\right)$, and there is a unique dual window $H_3\in L^2\left(ℝ\right)$ such

that $supp{H}_3\subseteq \left[-\frac{2m-1}{2}a\mathrm{,}\frac{2m-1}{2}a\right]$. Furthermore, for each $\left(a\mathrm{,}b\right)\in \Gamma$, the Gabor system $\mathcal{G}\left(B_3\mathrm{,}a\mathrm{,}b\right)$ is a frame for $L^2\left(ℝ\right)$.

Theorem 1 is proved by using the following well-known necessary and sufficient condition, which is well developed in [ [23], Proposition 2], for two Bessel Gabor systems to be dual of each other.

For $m\ge 1$, let $0<a<3$ and $\frac{2\left(m-1\right)}{3+\left(2m-3\right)a}<b\le \frac{2m}{3+\left(2m-1\right)a}$. Assume that $H_3$ is a bounded real-function with support on $\left[-\frac{2m-1}{2}a\mathrm{,}\frac{2m-1}{2}a\right]$. Then, the Gabor systems $\mathcal{G}\left(B_3\mathrm{,}a\mathrm{,}b\right)$ and $\mathcal{G}\left(H_3\mathrm{,}a\mathrm{,}b\right)$ are dual frames for $L^2\left(ℝ\right)$ if and only if

${\displaystyle \underset{k\mathrm{<mo>\; =\; </mo>1}-m}{\overset{m-1}{\sum }}}{B}_3\left(x-\mathcal{l}/b+ka\right)H_3\left(x+ka\right)=b{\delta }_{\mathcal{l}\mathrm{,0}}\mathrm{,}\left|\mathcal{l}\right|\le m-1\mathrm{,}\text{for}\text{a}\text{.e}\mathrm{.}x\in \left[-\frac{a}{2}\mathrm{,}\frac{a}{2}\right]\mathrm{.}$ (3)

We can rewrite (3) as a matrix-vector equation:

${\hat{E}}_m\left(x\right){\hat{F}}_m^t\left(x\right)=B^t\text{for}\text{a}\text{.e}\text{.}x\in \left[-\frac{a}{2}\mathrm{,}\frac{a}{2}\right]\mathrm{,}$ (4)

where ${\hat{F}}_m\left(x\right)$ and $B$ are the row vectors given respectively by ${\hat{F}}_m\left(x\right)={\left[H_3\left(x+ka\right)\right]}_{\left|k\right|\le m-1}$ and $B={\left[b{\delta }_{\mathcal{l}\mathrm{,0}}\right]}_{\left|\mathcal{l}\right|\le m-1}$ ; ${\hat{E}}_m\left(x\right)$ is the $\left(2m-1\right)\times \left(2m-1\right)$ matrix-valued function defined by ${\hat{E}}_m\left(x\right)={\left[B_3\left(x-\frac{\mathcal{l}}{b}+ka\right)\right]}_{1-m\le \mathcal{l},k\le m-1}$. Using the fact that the 3-spline $B_3$ is supported by $\left[-\frac{3}{2}\mathrm{,}\frac{3}{2}\right]$ and $\left(a\mathrm{,}b\right)\in {\Gamma }_m$ ; one observes easily that for all $x\in \left[-\frac{a}{2}\mathrm{,0}\right]$, not only the all ${\hat{E}}_m\left(x\right)$ ’s on-diagonal entries are positive but also the matrix-valued function ${\hat{E}}_m\left(x\right)$ can be rewritten as the following block matrix:

${\hat{E}}_m\left(x\right)=\left(\begin{array}{cc}{\hat{G}}_{m-1}\left(x\right)& {\hat{I}}_{m-1}\left(x\right)\\ O& {\hat{J}}_{m-1}\left(x\right)\end{array}\right),$ (5)

where $O$ is a $\left(m-2\right)\times \left(m+1\right)$ matrix of 0’s, ${\hat{I}}_{m-1}\left(x\right)$ is a $\left(m+1\right)\times \left(m-2\right)$ matrix, ${\hat{J}}_{m-1}\left(x\right)$ is a $\left(m-2\right)\times \left(m-2\right)$ upper triangular matrix and ${\hat{G}}_{m-1}\left(x\right)$ is the $\left(m+1\right)\times \left(m+1\right)$ tridiagonal matrix given by

${\hat{G}}_{m-1}\left(x\right)={\left[B_3\left(x-\frac{\mathcal{l}}{b}+ka\right)\right]}_{1-m\le \mathcal{l}\mathrm{,}k\le 1}\mathrm{.}$ (6)

Figure 1 shows the connected set known in $\mathcal{F}\left(B_3\right)$.

The rest of the paper is organized as follows. In Section 2 we prove our main results by studying the invertibility of the matrix ${\hat{G}}_{m-1}\left(x\right)$.

2. Invertibility of ${\hat{G}}_{m-1}\left(x\right)$ for $\left(a,b\right)\in {\Gamma }_m$

To prove Theorem 1, we only need to show that (3) has a unique solution ${\hat{F}}_m\left(x\right)$. This is equivalent to proving respectively that the $\left(2m-1\right)\times \left(2m-1\right)$ matrix-valued function ${\hat{E}}_m\left(x\right)$ is invertible for a.e. $x\in \left[-\frac{a}{2}\mathrm{,}\frac{a}{2}\right]$. In other

Figure 1. A sketch of $\mathcal{F}\left(B_3\right)$. No frame property in the red region, the green region is from [29] and the yellow region is the result from [11]. The blue and the magenta regions are respectively from [28] and [27]. The cyan region is the result from [23] completed by theorem 1.

words, from the block form for the matrix ${\hat{E}}_m\left(x\right)$, we need to prove that the matrix ${\hat{G}}_{m-1}\left(x\right)$ is invertible.

Throughout the paper, we use the fact that

$\underset{a}{\min }\left(\frac{2m}{3+\left(2m-1\right)a},\frac{4}{3+2a}\right)=\{\begin{array}{ll}\frac{4}{3+2a}\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ \frac{2m}{3+\left(2m-1\right)a}\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}$

The following lemma gives the behavior of the entries of ${\hat{G}}_{m-1}\left(x\right)$.

Lemma 1. For $m\ge 3$, let $\left(a\mathrm{,}b\right)\in {\Gamma }_m$, $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$ and consider $B_3$. Then the following hold:

1) $B_3\left(x+\frac{k}{b}-ka\right)>0$, for all $k\in \left\{1-m\mathrm{,}\cdots \mathrm{,}m-1\right\}$.

2) $B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)>0$ for $k\in \left\{\mathrm{0\; <mo>\; ;\; </mo>1}\right\}$ and

$\begin{array}{l}{B}_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)\\ =\{\begin{array}{ll}{B}_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)>0\hfill & \text{if}x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)\hfill \\ 0\hfill & \text{if}x\in \left[\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\mathrm{<mo>\; ;\; </mo>0}\right]\hfill \end{array}\end{array}$

for all $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$.

3) $B_3\left(x+\frac{m-2}{b}-\left(m-1\right)a\right)>0$ and

$\begin{array}{l}{B}_3\left(x+\frac{k}{b}-\left(k+1\right)a\right)\\ =\{\begin{array}{ll}0\hfill & \text{if}x\in \left[-\frac{a}{2};-\frac{3}{2}-\frac{k}{b}+\left(k+1\right)a\right]\hfill \\ B_3\left(x+\frac{k}{b}-\left(k+1\right)a\right)>0\hfill & \text{if}x\in \left(-\frac{3}{2}-\frac{k}{b}+\left(k+1\right)a;0\right]\hfill \end{array}\end{array}$

for all $\left\{k=-\mathrm{1,}\cdots \mathrm{,}m-3\right\}$.

Proof. 1) Firstly, let $k\in \left\{1-m\mathrm{,}\cdots \mathrm{,}-1\right\}$. We have

$\begin{array}{l}x+\frac{k}{b}-ka\le \frac{k}{b}-ka\\ \le \{\begin{array}{ll}k\left(\frac{3+2a}{4}\right)-ka=k\left(\frac{3-2a}{4}\right)\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ k\left(\frac{3+\left(2m-1\right)a}{2m}\right)-ka=\frac{3k}{2m}-\frac{k}{2m}a\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}\\ \le \{\begin{array}{ll}\frac{3k}{4}-\frac{k}{2}\times \frac{3\left(m-2\right)}{2\left(m-1\right)}=\frac{3k}{4\left(m-1\right)}\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ \frac{3k}{2m}-\frac{k}{2m}\times \frac{3\left(m-1\right)}{2m-1}=\frac{3k}{2\left(2m-1\right)}\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}\\ <0\end{array}$

and

$\begin{array}{c}x+\frac{k}{b}-ka\ge \frac{k}{b}-\frac{2k+1}{2}a\\ >k\left(\frac{3+\left(2m-3\right)a}{2\left(m-1\right)}\right)-\frac{2k+1}{2}a=\frac{3k}{2\left(m-1\right)}-\frac{m-1+k}{2\left(m-1\right)}a\\ \ge \frac{3k}{2\left(m-1\right)}-\frac{m-1+k}{2\left(m-1\right)}\times \frac{3\left(m-1\right)}{2m-1}=\frac{3}{2}\left(\frac{k}{m-1}-\frac{m-1+k}{2m-1}\right)\\ \ge -\frac{3}{2}\left(\text{because}\frac{k}{m-1}-\frac{m-1+k}{2m-1}\ge -1\right).\end{array}$

Therefore for all $k\in \left\{1-m\mathrm{,}\cdots \mathrm{,}-1\right\}$, $-\frac{3}{2}<x+\frac{k}{b}-ka<0$. Thus $B_3\left(x+\frac{k}{b}-ka\right)>0$. Secondly, let $k\in \left\{\mathrm{0,}\cdots \mathrm{,}m-1\right\}$. We have

$\begin{array}{c}x+\frac{k}{b}-ka\le \frac{k}{b}-ka<k\left(\frac{3+\left(2m-3\right)a}{2\left(m-1\right)}\right)-ka=\frac{3k}{2\left(m-1\right)}-\frac{ka}{2\left(m-1\right)}\\ \le \frac{3k}{2\left(m-1\right)}-\frac{k}{2\left(m-1\right)}\times \frac{3\left(m-3\right)}{2\left(m-2\right)}=\frac{k}{4\left(m-2\right)}\le \frac{3}{2}\end{array}$

and

$\begin{array}{l}x+\frac{k}{b}-ka\ge -\frac{a}{2}+\frac{k}{b}-ka=\frac{k}{b}-\frac{2k+1}{2}a\\ \ge \{\begin{array}{ll}k\left(\frac{3+2a}{4}\right)-\frac{2k+1}{2}a=\frac{3k}{4}-\frac{k+1}{2}a\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)}\mathrm{,}\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ k\left(\frac{3+\left(2m-1\right)a}{2m}\right)-\frac{2k+1}{2}a=\frac{3k}{2m}-\frac{m+k}{2m}a\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)}\mathrm{,}\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}\\ \ge \{\begin{array}{l}\frac{3k}{4}-\frac{k+1}{2}\times \frac{3\left(m-2\right)}{2\left(m-1\right)}=-\frac{3}{2}\left(\frac{\left(m-2\right)-k}{2\left(m-1\right)}\right)\\ \frac{3k}{2m}-\frac{m+k}{2m}\times \frac{3\left(m-1\right)}{2m-1}=-\frac{3}{2}\left(\frac{\left(m-1\right)-k}{2m-1}\right)\end{array}\\ >-\frac{3}{2}\end{array}$

Then for all $k\in \left\{\mathrm{0,}\cdots \mathrm{,}m-1\right\}$, $-\frac{3}{2}<x+\frac{k}{b}-ka<\frac{3}{2}$. Thus $B_3\left(x+\frac{k}{b}-ka\right)>0$.

2) Let $k\in \left\{\mathrm{0,}\cdots \mathrm{,}m-1\right\}$. We have

$\begin{array}{c}x+\frac{k}{b}-\left(k-1\right)a\le \frac{k}{b}-\left(k-1\right)a<k\left(\frac{3+\left(2m-3\right)a}{2\left(m-1\right)}\right)-\left(k-1\right)a\\ =\frac{3k}{2\left(m-1\right)}+\frac{2\left(m-1\right)-k}{2\left(m-1\right)}a\\ \le \frac{3k}{2\left(m-1\right)}+\frac{2\left(m-1\right)-k}{2\left(m-1\right)}\times \frac{3\left(m-1\right)}{2m-1}\\ =\frac{3}{2}\left(\frac{k}{m-1}+\frac{2\left(m-1\right)-k}{2m-1}\right)\end{array}$

and

$\begin{array}{l}x+\frac{k}{b}-\left(k-1\right)a\ge -\frac{a}{2}+\frac{k}{b}-\left(k-1\right)a=\frac{k}{b}-\frac{2k-1}{2}a\\ \ge \{\begin{array}{ll}k\left(\frac{3+2a}{4}\right)-\frac{2k-1}{2}a=\frac{3k}{4}-\frac{k-1}{2}a\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ k\left(\frac{3+\left(2m-1\right)a}{2m}\right)-\frac{2k-1}{2}a=\frac{3k}{2m}+\frac{m-k}{2m}a\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}\\ \ge \{\begin{array}{ll}\frac{3k}{4}-\frac{k-1}{2}\times \frac{3\left(m-2\right)}{2\left(m-1\right)}=\frac{3}{2}\left(\frac{m+k-2}{2\left(m-1\right)}\right)\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ \frac{3k}{2m}+\frac{m-k}{2m}\times \frac{3\left(m-2\right)}{2\left(m-1\right)}=\frac{3}{2}\left(\frac{m+k-2}{2\left(m-1\right)}\right)\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}\end{array}$

Then for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$ ;

$0<\frac{3}{2}\left(\frac{m+k-2}{2\left(m-1\right)}\right)\le x+\frac{k}{b}-\left(k-1\right)a<\frac{3}{2}\left(\frac{k}{m-1}+\frac{2\left(m-1\right)-k}{2m-1}\right).$

Therefore for all $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$,

$\begin{array}{l}{B}_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)\\ =\{\begin{array}{ll}{B}_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)>0\hfill & \text{if}x\in \left[-\frac{a}{2};\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)\hfill \\ 0\hfill & \text{if}x\in \left[\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a;0\right]\hfill \end{array}\end{array}$

and using the fact that $b>\frac{2}{3}$, we have easily $B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)>0$ for $k\in \left\{\mathrm{0\; <mo>\; ;\; </mo>1}\right\}$.

3) Let $k\in \left\{\mathrm{0,}\cdots \mathrm{,}m-2\right\}$. We have

$\begin{array}{c}x+\frac{k}{b}-\left(k+1\right)a\le \frac{k}{b}-\left(k+1\right)a<k\left(\frac{3+\left(2m-3\right)a}{2\left(m-1\right)}\right)-\left(k+1\right)a\\ =\frac{3k}{2\left(m-1\right)}-\frac{2\left(m-1\right)+k}{2\left(m-1\right)}a\\ \le \frac{3k}{2\left(m-1\right)}-\frac{2\left(m-1\right)+k}{2\left(m-1\right)}\times \frac{3\left(m-3\right)}{2\left(m-2\right)}\\ =\frac{3}{2}\left(\frac{k-2m+6}{2\left(m-2\right)}\right)=\frac{3}{2}\left(\frac{\left(k-m+2\right)+\left(4-m\right)}{2\left(m-2\right)}\right)\le 0\end{array}$

as far as $m\ge 4$ and

$\begin{array}{l}x+\frac{k}{b}-\left(k+1\right)a\ge -\frac{a}{2}+\frac{k}{b}-\left(k+1\right)a\mathrm{<mo>\; =\; </mo>}\frac{k}{b}-\frac{2k+3}{2}a\\ \ge \{\begin{array}{ll}k\left(\frac{3+2a}{4}\right)-\frac{2k+3}{2}a=\frac{3k}{4}-\frac{k+3}{2}a\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ k\left(\frac{3+\left(2m-1\right)a}{2m}\right)-\frac{2k+3}{2}a=\frac{3k}{2m}-\frac{3\left(m-1\right)+k+3}{2m}a\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}\\ \ge \{\begin{array}{ll}\frac{3k}{4}-\frac{k+3}{2}\times \frac{3\left(m-2\right)}{2\left(m-1\right)}=\frac{3}{2}\left(\frac{k-3\left(m-2\right)}{2\left(m-1\right)}\right)\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ \frac{3k}{2m}-\frac{3\left(m-1\right)+k+3}{2m}\times \frac{3\left(m-1\right)}{2m-1}=\frac{3}{2}\left(\frac{k-3\left(m-1\right)}{2m-1}\right)\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}\end{array}$

Then for all $k\in \left\{\mathrm{0,}\cdots \mathrm{,}m-2\right\}$ ;

$\{\begin{array}{l}\frac{3}{2}\left(\frac{k-3\left(m-2\right)}{2\left(m-1\right)}\right)\\ \frac{3}{2}\left(\frac{k-3\left(m-1\right)}{2m-1}\right)\end{array}\le x+\frac{k}{b}-\left(k+1\right)a<\frac{3}{2}\left(\frac{\left(k-m+2\right)+\left(4-m\right)}{2\left(m-2\right)}\right)\le 0$

as far as $m\ge 4$ and

$-\frac{3}{2}-\frac{3\left({\left(m-1\right)}^2+1\right)}{2\left(m-1\right)\left(2m-1\right)}<x-\frac{1}{b}\le \{\begin{array}{ll}-\frac{3\left(2m-5\right)}{4\left(m-2\right)}\hfill & \text{if}a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)},\frac{3\left(m-2\right)}{2\left(m-1\right)}\right]\hfill \\ -\frac{3\left(2m-3\right)}{4\left(m-1\right)}\hfill & \text{if}a\in \left[\frac{3\left(m-2\right)}{2\left(m-1\right)},\frac{3\left(m-1\right)}{2m-1}\right]\hfill \end{array}<0.$

Thus $B_3\left(x+\frac{m-2}{b}-\left(m-1\right)a\right)>0$ and

$\begin{array}{l}{B}_3\left(x+\frac{k}{b}-\left(k+1\right)a\right)\\ =\{\begin{array}{ll}0\hfill & \text{if}x\in \left[-\frac{a}{2};-\frac{3}{2}-\frac{k}{b}+\left(k+1\right)a\right]\hfill \\ B_3\left(x+\frac{k}{b}-\left(k+1\right)a\right)>0\hfill & \text{if}x\in \left[-\frac{3}{2}-\frac{k}{b}+\left(k+1\right)a;0\right]\hfill \end{array}\end{array}$

for $k=-1,\cdots ,m-3$.

The following remark gives the trivial two different partitions of $\left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$ which can be derived from the definition of ${\Gamma }_m\left(m\ge 3\right)$ and will be used to compute the determinant of the matrix ${\hat{G}}_{m-1}\left(x\right)$.

Remark 1. a) If $3-\frac{3}{b}+a\le 0$ and

· $a\in \left[\frac{3\left(m-3\right)}{2\left(m-2\right)}\mathrm{,}\frac{6m-9}{4m-3}\right]$, then $-\frac{3}{2}-\frac{m-3}{b}+\left(m-2\right)a\le -\frac{a}{2}$. Hence

$\left[-\frac{a}{2},0\right]={\displaystyle \underset{k=1}{\overset{m-2}{\cup }}}\left(Q_k\cup T_k\right)\cup T_{m-1}$

where $Q_k=\left[\frac{3}{2}-\frac{k+1}{b}+ka,-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a\right]$ and $T_k=\left(-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a,\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$ with the convention that $T_{m-1}=\left[-\frac{a}{2},\frac{3}{2}-\frac{m-1}{b}+\left(m-2\right)a\right)$ and $T_1=\left(-\frac{3}{2}+\frac{1}{b},0\right]$.

· $a\in \left[\frac{3\left(m-2\right)}{2m-3}\mathrm{,}\frac{3\left(m-1\right)}{2m-1}\right]$, then $-\frac{a}{2}\le -\frac{3}{2}-\frac{m-3}{b}+\left(m-2\right)a$. Thus

$\left[-\frac{a}{2},0\right]={\displaystyle \underset{k=1}{\overset{m-1}{\cup }}}\left(Q_k\cup T_k\right)$

where $Q_k=\left[\frac{3}{2}-\frac{k+1}{b}+ka,-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a\right]$ and $T_k=\left(-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a,\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$ with the convention that $Q_{m-1}=\left[-\frac{a}{2},-\frac{3}{2}-\frac{m-3}{b}+\left(m-2\right)a\right]$ and $T_1\mathrm{<mo>\; =\; </mo>}\left(-\frac{3}{2}+\frac{1}{b}\mathrm{,0}\right]$.

· $a\in \left(\frac{6m-9}{4m-3}\mathrm{<mo>\; ;\; </mo>}\frac{3\left(m-2\right)}{2m-3}\right)$ for $m\ne 3$, then $-\frac{3}{2}-\frac{m-3}{b}+\left(m-2\right)a+\frac{a}{2}$ is sign-changing. So, we use both different previous partitions.

b) If $3-\frac{3}{b}+a>0$, then a is only in $\left[\frac{3\left(m-3\right)}{2\left(m-2\right)}\mathrm{,}\frac{6m-9}{4m-3}\right]$. Hence

$\left[-\frac{a}{2},0\right]=\left({\displaystyle \underset{k=1}{\overset{m-1}{\cup }}}{\tilde{Q}}_k\right)\cup \left({\displaystyle \underset{k=2}{\overset{m-1}{\cup }}}{\tilde{T}}_k\right)$

where ${\tilde{Q}}_k=\left[\frac{3}{2}-\frac{k+1}{b}+ka,-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a\right]$ and ${\tilde{T}}_k=\left(-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a,\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$ with the convention that ${\tilde{Q}}_{m-1}=\left[-\frac{a}{2},-\frac{3}{2}-\frac{m-4}{b}+\left(m-3\right)a\right]$ and ${\tilde{Q}}_1=\left[\frac{3}{2}-\frac{2}{b}+a,0\right]$.

NB: Specially for $m=3$, one substitutes $\frac{3\left(m-3\right)}{2\left(m-2\right)}$ by $\frac{3}{4}$. The different cases considered in proving this result are illustrated for the cases $m=3$ and $m=4$ in Figure 2.

Let $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, $\mathcal{l}=2,3,4$ and denote $A_{\mathcal{l}\mathcal{l}}^k\left(x\right)$ the $\mathcal{l}\times \mathcal{l}$ sub-matrix of ${\hat{G}}_{m-1}\left(x\right)$, defined respectively by

$A_{22}^k\left(x\right)=\left(\begin{array}{cc}{B}_3\left(x+\frac{k}{b}-ka\right)& B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)\\ B_3\left(x+\frac{k-1}{b}-ka\right)& B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)\end{array}\right)\mathrm{,}$

$A_{33}^k\left(x\right)=\left(\begin{array}{ccc}{B}_3\left(x+\frac{k}{b}-ka\right)& B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)& 0\\ B_3\left(x+\frac{k-1}{b}-ka\right)& B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-1}{b}-\left(k-2\right)a\right)\\ 0& B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)\end{array}\right)$

Figure 2. The off-diagonal of ${\hat{G}}_{m-1}\left(x\right)$ for $m=3$ and $m=4$ when $x\in \left[-\frac{a}{2}\mathrm{,0}\right]$.

and

$A_{44}^k\left(x\right)=\left(\begin{array}{cccc}{B}_3\left(x+\frac{k}{b}-ka\right)& B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)& 0& 0\\ B_3\left(x+\frac{k-1}{b}-ka\right)& B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-1}{b}-\left(k-2\right)a\right)& 0\\ 0& B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)& B_3\left(x+\frac{k-2}{b}-\left(k-3\right)a\right)\\ 0& 0& B_3\left(x+\frac{k-3}{b}-\left(k-2\right)a\right)& B_3\left(x+\frac{k-3}{b}-\left(k-3\right)a\right)\end{array}\right).$

The following Lemma indicates that the matrices $A_{\mathcal{l}\mathcal{l}}^k\left(x\right)$ are invertible.

Lemma 2. Given $m\ge 3$, let $\left(a\mathrm{,}b\right)\in {\Gamma }_m$ and $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$. The following hold:

a) if $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, then $\left|A_{\mathcal{l}\mathcal{l}}^k\left(x\right)\right|>0$ for all $\mathcal{l}=2,3$.

b) $\left|A_{44}^k\left(x\right)\right|>0$, for all $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$.

This result is proved in Appendix 4.

The following Proposition gives an explicit expression for the determinant of the sub-matrix ${\hat{G}}_{m-1}\left(x\right)$ when $m\ge 3$, $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$ and under the hypotheses of Theorem 1.

Proposition 1. For $m\ge 3$, let $\left(a\mathrm{,}b\right)\in {\Gamma }_m$ and $x\in \left[-\frac{a}{2}\mathrm{,0}\right]$. The following statements hold:

$\forall x\in D,\left|{\hat{G}}_{m-1}\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|$

where $D\mathrm{,}I\mathrm{,}i$ and j are given in the following cases:

1) If $3-\frac{3}{b}+a\le 0$ and

a) $\frac{3\left(m-3\right)}{2\left(m-2\right)}\le a\le \frac{6m-9}{4m-3}$, then

i) for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-2\right\}$, $D=Q_k$, $I=\left\{-1,\cdots ,m-1\right\}\backslash \left\{k,k-1\right\}$, $i=k$, and $j=2$.

ii) for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, $D=T_k$, $I=\left\{-\mathrm{1,}\cdots \mathrm{,}m-1\right\}\backslash \left\{k\mathrm{,}k-\mathrm{1,}k-2\right\}$, $i=k$ and $j=3$.

b) $\frac{3\left(m-2\right)}{2m-3}\le a\le \frac{3\left(m-1\right)}{2m-1}$, then for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, one has:

i) for $D=Q_k$, $I=\left\{-1,\cdots ,m-1\right\}\backslash \left\{k,k-1\right\}$, $i=k$, and $j=2$.

ii) for $D=T_k$, $I=\left\{-\mathrm{1,}\cdots \mathrm{,}m-1\right\}\backslash \left\{k\mathrm{,}k-\mathrm{1,}k-2\right\}$, $i=k$ and $j=3$.

c) $\frac{6m-9}{4m-3}<a<\frac{3\left(m-2\right)}{2m-3}$ when $m\ne 3$, then one uses i) and ii) obtained in a) or b).

2) If $3-\frac{3}{b}+a>0$, then

a) For all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, $D={\tilde{Q}}_k$, $I=\left\{-\mathrm{1,}\cdots \mathrm{,}m-1\right\}\backslash \left\{k\mathrm{,}k-\mathrm{1,}k-2\right\}$, $i=k$ and $j=3$.

b) For all $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$, $D={\tilde{T}}_k$, $I=\left\{-\mathrm{1,}\cdots \mathrm{,}m-1\right\}\backslash \left\{k,k-1,k-2,k-3\right\}$, $i=k$ and $j=4$.

Moreover the matrix ${\hat{E}}_m\left(x\right)$ is invertible.

Proof. We prove the result by induction on m by using the different partitions of $\left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$ obtained in Remark 1. For $m=3$, the matrix ${\hat{G}}_2\left(x\right)$ given by

${\hat{G}}_2\left(x\right)=\left(\begin{array}{cccc}{B}_3\left(x+\frac{2}{b}-2a\right)& B_3\left(x+\frac{2}{b}-a\right)& 0& 0\\ B_3\left(x+\frac{1}{b}-2a\right)& B_3\left(x+\frac{1}{b}-a\right)& B_3\left(x+\frac{1}{b}\right)& 0\\ 0& B_3\left(x-a\right)& B_3\left(x\right)& B_3\left(x+a\right)\\ 0& 0& B_3\left(x-\frac{1}{b}\right)& B_3\left(x-\frac{1}{b}+a\right)\end{array}\right)\mathrm{.}$

Suppose that $3-\frac{3}{b}+a\le 0$ and $\frac{3}{4}\le a\le 1$, then $\left[-\frac{a}{2};0\right]=Q_1\cup T_1\cup T_2$ with $T_2=\left[-\frac{a}{2};\frac{3}{2}-\frac{2}{b}+a\right)$, $Q_1=\left[\frac{3}{2}-\frac{2}{b}+a;-\frac{3}{2}+\frac{1}{b}\right]$ and $T_1=\left(-\frac{3}{2}+\frac{1}{b};0\right]$.

So, it is easy to see that for all $x\in Q_1$, $\left|{\hat{G}}_2\left(x\right)\right|=B_3\left(x+\frac{2}{b}-2a\right)B_3\left(x-\frac{1}{b}+a\right)\cdot \left|A_{22}^1\left(x\right)\right|$ ; $x\in T_1$, $\left|{\hat{G}}_2\left(x\right)\right|=B_3\left(x+\frac{2}{b}-2a\right)\cdot \left|A_{33}^1\left(x\right)\right|$ and $x\in T_2$, $\left|{\hat{G}}_2\left(x\right)\right|=B_3\left(x-\frac{1}{b}+a\right)\cdot \left|A_{33}^2\left(x\right)\right|$. This establishes the part (a) for the base case $m=3$.

Suppose that $3-\frac{3}{b}+a\le 0$ and $1<a\le \frac{6}{5}$, then $\left[-\frac{a}{2};0\right]=Q_2\cup T_2\cup Q_1\cup T_1$ with $Q_2=\left[-\frac{a}{2};-\frac{3}{2}+a\right]$, $T_2=\left(-\frac{3}{2}+a;\frac{3}{2}-\frac{2}{b}+a\right)$, $Q_1=\left[\frac{3}{2}-\frac{2}{b}+a;-\frac{3}{2}+\frac{1}{b}\right]$ and $T_1=\left(-\frac{3}{2}+\frac{1}{b};0\right]$. So, it is easy to see that

$\forall x\in D=Q_k\left(k=1,2\right),\left|{\hat{G}}_2\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|$

where $I=\left\{-\mathrm{1,0,1,2}\right\}\backslash \left\{k\mathrm{,}k-1\right\}$, $i=k$, and $j=2$ ; and

$\forall x\in D=T_k\left(k=1,2\right),\left|{\hat{G}}_2\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|$

where $I=\left\{-\mathrm{1,0,1,2}\right\}\backslash \left\{k\mathrm{,}k-\mathrm{1,}k-2\right\}$, $i=k$ and $j=3$. Hence (b) holds. Similarly, (2) holds.

Suppose that (1) and (2) hold for $m-1\ge 3$ and let us prove that they hold for m. So, one compute the determinant of the matrix ${\hat{G}}_m\left(x\right)$ given by

${\hat{G}}_m\left(x\right)={\left[B_3\left(x-\frac{\mathcal{l}}{b}+ka\right)\right]}_{-m\le \mathcal{l}\mathrm{,}k\le 1}\mathrm{.}$

Suppose $3-\frac{3}{b}+a>0$. From Remark 1, we have $\left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]=\left({\displaystyle \underset{k=1}{\overset{m}{\cup }}}{\tilde{Q}}_k\right)\cup \left({\displaystyle \underset{k=2}{\overset{m}{\cup }}}{\tilde{T}}_k\right)$ with ${\tilde{Q}}_m=\left[-\frac{a}{2};-\frac{3}{2}-\frac{m-3}{b}+\left(m-2\right)a\right]$ ; ${\tilde{Q}}_k=\left[\frac{3}{2}-\frac{k+1}{b}+ka;-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a\right]$, for all $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$ ; ${\tilde{Q}}_1=\left[\frac{3}{2}-\frac{2}{b}+a;0\right]$ and ${\tilde{T}}_k=\left(-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a;\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$, for all $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m\right\}$.

Let $x\in D={\tilde{Q}}_m$. Thus $B_3\left(x+\frac{k}{b}-\left(k+1\right)a\right)=0$ for all $k\in \left\{-\mathrm{1,}\cdots \mathrm{,}m-3\right\}$. Hence,

$\left|{\hat{G}}_m\left(x\right)\right|=\left({\displaystyle \underset{\mathcal{l}=-1}{\overset{m-3}{\prod }}}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\right)\cdot \left|A_{33}^m\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|$

where $I=\left\{-\mathrm{1,}\cdots \mathrm{,}m-3\right\}$, $i=m$ and $j=3$.

Let $x\in D={\tilde{Q}}_{m-1}$. Therefore $B_3\left(x+\frac{m}{b}-\left(m-1\right)a\right)=0$ and $B_3\left(x+\frac{k}{b}-\left(k+1\right)a\right)=0$ for all $k\in \left\{-\mathrm{1,}\cdots \mathrm{,}m-4\right\}$. Consequently,

$\left|{\hat{G}}_m\left(x\right)\right|=\left({\displaystyle \underset{\begin{array}{c}\mathcal{l}=-1\\ \mathcal{l}\ne m-3,m-2,m-1\end{array}}{\overset{m}{\prod }}}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\right)\cdot \left|A_{33}^{m-1}\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|$

where $I=\left\{-1,\cdots ,m\right\}\backslash \left\{m-3,m-2,m-1\right\}$, $i=m-1$ and $j=3$.

Let $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-2\right\}$ and $x\in D={\tilde{Q}}_k$. Thus $B_2\left(x+\frac{m}{b}-\left(m-1\right)a\right)=0$. Hence, $\left|{\hat{G}}_m\left(x\right)\right|=B_3\left(x+\frac{m}{b}-ma\right)\cdot \left|{\hat{G}}_{m-1}\left(x\right)\right|$ and by the induction assumption, we have

$\left|{\hat{G}}_m\left(x\right)\right|=\left({\displaystyle \underset{\begin{array}{c}\mathcal{l}=-1\\ \mathcal{l}\ne k,k-1,k-2\end{array}}{\overset{m}{\prod }}}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\right)\cdot \left|A_{33}^k\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|$

where $I=\left\{-1,\cdots ,m\right\}\backslash \left\{k,k-1,k-2\right\}$, $i=k$ and $j=3$.

Let $x\in D={\tilde{T}}_m$. Thus $B_3\left(x+\frac{k}{b}-\left(k+1\right)a\right)=0$ for all $k\in \left\{-\mathrm{1,}\cdots \mathrm{,}m-4\right\}$. Hence,

$\left|{\hat{G}}_m\left(x\right)\right|=\left({\displaystyle \underset{\begin{array}{c}\mathcal{l}=-1\\ \mathcal{l}\ne m-3,m-2,m-1,m\end{array}}{\overset{m}{\prod }}}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\right)\cdot \left|A_{44}^m\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|$

where $I=\left\{-1,\cdots ,m\right\}\backslash \left\{m-3;m-2,m-1,m\right\}$, $i=m$ and $j=4$.

Let $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$ and $x\in D={\tilde{T}}_k$. Then $B_3\left(x+\frac{m}{b}-\left(m-1\right)a\right)=0$. Thus $\left|{\hat{G}}_m\left(x\right)\right|=B_3\left(x+\frac{m}{b}-ma\right)\cdot \left|{\hat{G}}_{m-1}\left(x\right)\right|$ and by the induction assumption, we have

$\left|{\hat{G}}_m\left(x\right)\right|=\left({\displaystyle \underset{\begin{array}{c}\mathcal{l}=-1\\ \mathcal{l}\ne k,k-1,k-2,k-3\end{array}}{\overset{m}{\prod }}}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\right)\cdot \left|A_{44}^k\left(x\right)\right|={\displaystyle \underset{l\in I}{\prod }}{B}_3\left(x+\frac{\mathcal{l}}{b}-\mathcal{l}a\right)\cdot \left|A_{jj}^i\left(x\right)\right|.$

where $I=\left\{-1,\cdots ,m\right\}\backslash \left\{k,k-1,k-2,k-3\right\}$, $i=k$ and $j=4$.

Together, (2) holds for the case m. Similarly, one proves that (1) holds for the case m.

To end the proof, we observe that by using the block decomposition of ${\hat{E}}_m\left(x\right)$, we have for all $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$,

$\left|{\hat{E}}_m\left(x\right)\right|=\left({\displaystyle \underset{1-m}{\overset{-2}{\prod }}}{B}_3\left(x+\frac{k}{b}-ka\right)\right)\cdot \left|{\hat{G}}_{m-1}\left(x\right)\right|.$

We know that $B_3\left(x+\frac{k}{b}-ka\right)>0$, $\forall k\in \left\{1-m,\cdots ,m-1\right\}$ and $\left|{\hat{G}}_{m-1}\left(x\right)\right|>0$ because $\left|A_{\mathcal{l}\mathcal{l}}^k\left(x\right)\right|>0$ from Lemma 2. Thus we conclude that $\left|{\hat{E}}_m\left(x\right)\right|>0$ for all $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$, and by symmetry $\left(\left|{\hat{E}}_m\left(-x\right)\right|=\left|{\hat{E}}_m\left(x\right)\right|\right)$ this holds for all $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>}\frac{a}{2}\right]$.

We are now ready to prove Theorem 1.

Proof of Theorem 1. By Proposition 1 we know that ${\hat{E}}_m\left(x\right)$ is invertible. Let ${\hat{F}}_m\left(x\right)$ be defined on $ℝ$ as follows. For $x\in ℝ\backslash \left[-\frac{2m-1}{2}a\mathrm{,}\frac{2m-1}{2}a\right]$, let ${\hat{F}}_m\left(x\right)=0$ and for $x\in \left[-\frac{2m-1}{2}a\mathrm{,}\frac{2m-1}{2}a\right]$ let ${\hat{F}}_m\left(x\right)$ be defined by ${\hat{F}}_m^t\left(x\right)=b{\left({\hat{E}}_m^{-1}\left(x\right)\right)}_m$, where ${\hat{F}}_m\left(x\right)={\left[H_3\left(x+ka\right)\right]}_{\left|k\right|\le m-1}$ and ${\left({\hat{E}}_m^{-1}\left(x\right)\right)}_m$ is the m^th column vector of the matrix ${\hat{E}}_m^{-1}\left(x\right)$. Consequently, $H_3$ is a compactly supported and bounded function for which $\mathcal{G}\left(H_3\mathrm{,}a\mathrm{,}b\right)$ is a Bessel sequence. By construction, it also follows that $B_3$ and $H_3$ are dual windows.

3. Conclusion

We studied the fine structure of the Gabor frame generated by the B-spline $B_3$. It should be remembered that this structure determines all pairs $\left(a\mathrm{,}b\right)\in {ℝ}_{+}^2$ for which the Gabor system $\mathcal{G}\left(B_3\mathrm{,}a\mathrm{,}b\right)$ forms a frame. We presented the known results of the Gabor frame set $\mathcal{F}\left(B_3\right)$ and we added a new set of points in the frame set of the 3-spline while building the compactly supported dual windows of $B_3$. All our results are obtained thanks to the partitioning of the domain in which the frame set is sought. This partitioning allowed us to establish a global approach, based on the study of the invertibility of a square matrix specific to each sub-domain, and which will be used to find other points belonging to the frame set of the B-spline $B_3$.

Appendix

Consider the first-order difference ${\Delta }_a{B}_3$ and the second-order difference ${\Delta }_a^2{B}_3$ given respectively by

${\Delta }_a{B}_3\left(x\right)=B_3\left(x\right)-B_3\left(x-a\right)\text{and}{\Delta }_a^2{B}_3\left(x\right)=B_3\left(x\right)-2B_3\left(x-a\right)+B_3\left(x-2a\right)$

The following Lemma completes the Lemma 2.2 of [27] in the case of $B_3$.

Lemma 3. Let $0<a<3$. Then ${\Delta }_a^2{B}_3\left(x\right)>0$, for all $x\in \left(-\frac{3}{2}+a\mathrm{,}\frac{3}{2}-\frac{2}{b}+a\right)$.

Proof. By definition of the functional space $V_{a}g$ and the Lemma 2.2 in [27], it is known that ${\Delta }_a^2{B}_3\left(x\right)>0$ for all $x\in \left[-\frac{3}{2}\mathrm{,}-\frac{3}{4}+\frac{3a}{4}\right]$.

In particular, for $a\ge 1$ or $b\le \frac{8}{9+a}$, $\left(-\frac{3}{2}+a\mathrm{,}\frac{3}{2}-\frac{2}{b}+a\right)\subset \left[-\frac{3}{2}\mathrm{,}-\frac{3}{4}+\frac{3a}{4}\right]$ while for $a<1$ and $b>\frac{8}{9+a}$, $-\frac{3}{4}+\frac{3a}{4}<\frac{3}{2}-\frac{2}{b}+a$. In other words, to have that ${\Delta }_a^2{B}_3\left(x\right)>0$ on $\left(-\frac{3}{2}+\frac{3a}{4}\mathrm{,}\frac{3}{2}-\frac{2}{b}+a\right)$, it suffices to show that, for $a<1$ and $b>\frac{8}{9+a}$,

${\Delta }_a^2{B}_3\left(x\right)>0,\forall x\in \left(-\frac{3}{4}+\frac{3a}{4},\frac{3}{2}-\frac{2}{b}+a\right).$

Otherwise, we only show that for $\frac{3}{4}\le a<1$ and $\frac{8}{9+a}<b\le \frac{4}{3+2a}$ ;

${\Delta }_a^2{B}_3\left(x\right)>0,\forall x\in \left(-\frac{3}{4}+\frac{3a}{4},\frac{3}{2}-\frac{2}{b}+a\right)\subset \left(-\frac{3}{16},0\right].$

Let $f\left(x\right):={\Delta }_a^2{B}_3\left(x\right)$. It is easy to see that for all $x\in \left(-\frac{3}{4}+\frac{3a}{4}\mathrm{,}\frac{3}{2}-\frac{2}{b}+a\right)$,

$\begin{array}{c}f\left(x\right)=B_3\left(x\right)-2B_3\left(x-a\right)\\ =-x^2+\frac{3}{4}-{\left(x-a\right)}^2-3\left(x-a\right)-\frac{9}{4}\\ =-2x^2+\left(2a-3\right)x-a^2+3a-\frac{3}{2}.\end{array}$

One has $f^{\prime }\left(-\frac{3}{16}\right)=2a-\frac{9}{4}<0$ et $f\left(0\right)=-a^2+3a-\frac{3}{2}>0$. Thus $f\left(x\right)>0$, $\forall x\in \left(-\frac{3}{16},0\right]$. Consequently,

$\forall x\in \left(-\frac{3}{4}+\frac{3a}{4},\frac{3}{2}-\frac{2}{b}+a\right),{\Delta }_a^2{B}_3\left(x\right)>0$

as wanted.

The following Lemma shows the invertibility of the matrix $A_{44}^2\left(x\right)$.

Lemma 4. Let $a\in \left[\frac{3}{4}\mathrm{,}\frac{3}{2}\right]$, $b\in \left[\frac{3}{3+a}\mathrm{,}\frac{4}{3+2a}\right]$ and $x\in \left(-\frac{3}{2}+\frac{1}{b}\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{2}{b}+a\right)$. Then $\left|A_{44}^2\left(x\right)\right|>0$.

Proof. Let $L_2\left(x\right):=\left|A_{44}^2\left(x\right)\right|$ and $L_2^{i\mathrm{,}j}\left(x\right)$ denote the ij^th minor of $L_2\left(x\right)$, the determinant of the matrix obtained by removing the i^th row and the j^th column from $L_2\left(x\right)$. We have respectively

(1) $B_3\left(x\right)>B_3\left(x-a\right)$ and $B_3\left(x\right)>B_3\left(x+a\right)$ ;

(2) $L_2^{3,2}\left(x\right)>0$ and $L_2^{3,4}\left(x\right)>0$ ;

(3) $L_2^{33}\left(x\right)>L_2^{3,2}\left(x\right)+L_2^{3,4}\left(x\right)$.

Combining (1), (2) and (3), we have for all $x\in \left(-\frac{3}{2}+\frac{1}{b}\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{2}{b}+a\right)$,

$\begin{array}{c}{L}_2\left(x\right)=-B_3\left(x-a\right)\cdot L_2^{3,2}\left(x\right)+B_3\left(x\right)\cdot L_2^{33}\left(x\right)-B_3\left(x+a\right)\cdot L_2^{3,4}\left(x\right)\\ >\left[B_3\left(x\right)-B_3\left(x-a\right)\right]\cdot L_2^{3,2}\left(x\right)+\left[B_3\left(x\right)-B_3\left(x+a\right)\right]\cdot L_2^{3,4}\left(x\right)>0.\end{array}$

This implies that for all $x\in \left(-\frac{3}{2}+\frac{1}{b};\frac{3}{2}-\frac{2}{b}+a\right)$, $\left|A_{44}^2\left(x\right)\right|>0$.

We have for all $x\in \left(-\frac{3}{2}+\frac{1}{b}\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{2}{b}+a\right)$, $x-a<x<0<x+a$ and $-x-a<x$. Thus $B_3\left(x-a\right)<B_3\left(x\right)$ and $B_3\left(x\right)>B_3\left(x+a\right)$. On the other hand,

$\begin{array}{l}{L}_2^{3,2}\left(x\right)=B_3\left(x+\frac{2}{b}-2a\right)B_3\left(x+\frac{1}{b}\right)B_3\left(x-\frac{1}{b}+a\right)>0\\ \text{and}{L}_2^{3,4}\left(x\right)=B_3\left(x-\frac{1}{b}\right)\left|A_{22}^2\left(x\right)\right|>0.\end{array}$

Then (1) and (2) hold.

One has $\begin{array}{c}{L}_2^{33}\left(x\right)-L_2^{3,2}\left(x\right)-L_2^{3,4}\left(x\right)=\left[B_3\left(x-\frac{1}{b}+a\right)-B_3\left(x-\frac{1}{b}\right)\right]\left|A_{22}^2\left(x\right)\right|\\ -B_3\left(x+\frac{2}{b}-2a\right)B_3\left(x+\frac{1}{b}\right)B_3\left(x-\frac{1}{b}+a\right)\end{array}$ where $B_3\left(x-\frac{1}{b}\right)=\frac{1}{2}{\left(x-\frac{1}{b}\right)}^2+\frac{3}{2}\left(x-\frac{1}{b}\right)+\frac{9}{8}$, $B_3\left(x+\frac{1}{b}\right)=\frac{1}{2}{\left(x-\frac{1}{b}\right)}^2-\frac{3}{2}\left(x+\frac{1}{b}\right)+\frac{9}{8}$ and

$B_3\left(x-\frac{1}{b}+a\right)=\{\begin{array}{ll}\frac{1}{2}{\left(x-\frac{1}{b}+a\right)}^2+\frac{3}{2}\left(x-\frac{1}{b}+a\right)+\frac{9}{8}\hfill & \text{if}a\in \left[\frac{3}{4}\mathrm{,1}\right]\hfill \\ -{\left(x-\frac{1}{b}+a\right)}^2+\frac{3}{4}\hfill & \text{if}a\in \left[\mathrm{1\; <mo>\; ;\; </mo>}\frac{3}{2}\right]\mathrm{.}\hfill \end{array}$

Let $g\left(x\right):=\left[B_3\left(x-\frac{1}{b}+a\right)-B_3\left(x-\frac{1}{b}\right)\right]-B_3\left(x+\frac{1}{b}\right)$. It is easy to remark that the function g is strictly increasing on $\left(-\frac{3}{2}+\frac{1}{b}\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{2}{b}+a\right)$ and $g\left(-\frac{3}{2}+\frac{1}{b}\right)>0$ where

$g\left(-\frac{3}{2}+\frac{1}{b}\right)=\{\begin{array}{ll}\frac{1}{2}\left(-3+a+\frac{2}{b}\right)\left(3+a-\frac{2}{b}\right)\hfill & \text{if}a\in \left[\frac{3}{4}\mathrm{,1}\right]\hfill \\ -a^2+3a-\frac{2}{b^2}+\frac{6}{b}-6\hfill & \text{if}a\in \left[\mathrm{1\; <mo>\; ;\; </mo>}\frac{3}{2}\right]\mathrm{.}\hfill \end{array}$

Consequently $B_3\left(x-\frac{1}{b}+a\right)-B_3\left(x-\frac{1}{b}\right)>B_3\left(x+\frac{1}{b}\right)$.

Let $h\left(x\right)\mathrm{<mo>\; :\; </mo>}=\left|A_{22}^2\left(x\right)\right|-B_3\left(x+\frac{2}{b}-2a\right)B_3\left(x-\frac{1}{b}+a\right)$ where

$A_{22}^2\left(x\right)=\left(\begin{array}{cc}{B}_3\left(x+\frac{2}{b}-2a\right)& B_3\left(x+\frac{2}{b}-a\right)\\ B_3\left(x+\frac{1}{b}-2a\right)& B_3\left(x+\frac{1}{b}-a\right)\end{array}\right).$

We obtain

$\underset{x}{\min }h\left(x\right)=h\left(\frac{3}{2}-\frac{2}{b}+a\right)=B_3\left(\frac{3}{2}-a\right)\left[B_3\left(\frac{3}{2}-\frac{1}{b}\right)-B_3\left(\frac{3}{2}-\frac{3}{b}+2a\right)\right]>0$

because $\frac{3}{2}-\frac{3}{b}+2a<-\frac{3}{2}+\frac{1}{b}\Rightarrow B_3\left(\frac{3}{2}-\frac{3}{b}+2a\right)<B_3\left(\frac{3}{2}-\frac{1}{b}\right)$. Consequently (3) holds.

Therefore $\left|A_{22}^2\left(x\right)\right|>B_3\left(x+\frac{2}{b}-2a\right)B_3\left(x-\frac{1}{b}+a\right)$. Hence $L_2^{33}\left(x\right)>L_2^{3,2}\left(x\right)+L_2^{3,4}\left(x\right)$.

Proof of Lemma 2. Throughout this proof, we use the Lemma 1 and Remark 1 without notify it.

(a) Let $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$. In the first time, we consider the matrix $A_{22}^k\left(x\right)$ given by

$A_{22}^k\left(x\right)=\left(\begin{array}{cc}{B}_3\left(x+\frac{k}{b}-ka\right)& B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)\\ B_3\left(x+\frac{k-1}{b}-ka\right)& B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)\end{array}\right)\mathrm{.}$

We prove that for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, $B_3\left(x+\frac{k}{b}-ka\right)>B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)$. For this, we know that $-\frac{3}{2}<x+\frac{k}{b}-ka<\frac{3}{2}$ and $x+\frac{k}{b}-\left(k-1\right)a>0$, therefore we have two different cases.

* If $x+\frac{k}{b}-ka>0$, we have $x+\frac{k}{b}-ka<x+\frac{k}{b}-\left(k-1\right)a$ and using the strict decreasing of $B_3$ on $\left[\mathrm{0,}\frac{3}{2}\right]$, one has $B_3\left(x+\frac{k}{b}-ka\right)>B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)$.

* If $x+\frac{k}{b}-ka\le 0$, then $-x-\frac{k}{b}+ka\ge 0$ and

$-x-\frac{k}{b}+ka-\left(x+\frac{k}{b}-\left(k-1\right)a\right)=-2x-\frac{2k}{b}+\left(2k-1\right)a\le 2k\left(a-\frac{1}{b}\right)<0.$

Thus $-x-\frac{k}{b}+ka<x+\frac{k}{b}-\left(k-1\right)a$ and using the fact that $B_3$ is symmetric around the origin and strict decreasing on $\left[\mathrm{0,}\frac{3}{2}\right]$, one has $B_3\left(x+\frac{k}{b}-ka\right)>B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)$.

Next, we prove that for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-2\right\}$, $B_3\left(x+\frac{k-1}{b}-ka\right)<B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)$. We also know that $-\frac{3}{2}<x+\frac{k-1}{b}-\left(k-1\right)a<\frac{3}{2}$ and $x+\frac{k-1}{b}-ka<0$, therefore we can consider the following two cases.

* If $x+\frac{k-1}{b}-\left(k-1\right)a\le 0$, on has $x+\frac{k-1}{b}-ka<x+\frac{k-1}{b}-\left(k-1\right)a$. Therefore by the strict increasing of $B_3$ on $\left[-\frac{3}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$, we obtain $B_3\left(x+\frac{k-1}{b}-ka\right)<B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)$.

* If $x+\frac{k-1}{b}-\left(k-1\right)a>0$, then $-x-\frac{k-1}{b}+\left(k-1\right)a\le 0$ and $-x-\frac{k-1}{b}+\left(k-1\right)a-\left(x+\frac{k-1}{b}-ka\right)=-2x-\frac{2\left(k-1\right)}{b}+\left(2k-1\right)a\ge 0$. Indeed,

$\begin{array}{l}-2x-\frac{2\left(k-1\right)}{b}+\left(2k-1\right)a\ge -\frac{2\left(k-1\right)}{b}+\left(2k-1\right)a\\ \ge -\left(k-1\right)\frac{3+\left(2m-3\right)a}{m-1}+\left(2k-1\right)a=-\frac{3\left(k-1\right)}{m-1}+\frac{k+m-2}{m-1}a\\ \ge -\frac{3\left(k-1\right)}{m-1}+\frac{k+m-2}{m-1}\times \frac{3\left(m-3\right)}{2\left(m-2\right)}=\frac{3\left(m-2-k\right)}{2\left(m-2\right)}\ge 0.\end{array}$

Thus $x+\frac{k-1}{b}-ka\le -x-\frac{k-1}{b}+\left(k-1\right)a$ and using the fact that $B_3$ is symmetric around the origin and strict increasing on $\left[-\frac{3}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$, one has $B_3\left(x+\frac{k-1}{b}-ka\right)\le B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)$.

All in all, we conclude that $\left|A_{22}^k\left(x\right)\right|>0$ for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-2\right\}$.

For $k=m-1$, we consider the matrix $A_{22}^{m-1}\left(x\right)$ given by

$A_{22}^{m-1}\left(x\right)=\left(\begin{array}{cc}{B}_3\left(x+\frac{m-1}{b}-\left(m-1\right)a\right)& B_3\left(x+\frac{m-1}{b}-\left(m-2\right)a\right)\\ B_3\left(x+\frac{m-2}{b}-\left(m-1\right)a\right)& B_3\left(x+\frac{m-2}{b}-\left(m-2\right)a\right)\end{array}\right).$

We know respectively that $B_3\left(x+\frac{m-1}{b}-\left(m-1\right)a\right)>0$, $B_3\left(x+\frac{m-2}{b}-\left(m-1\right)a\right)>0$, $B_3\left(x+\frac{m-2}{b}-\left(m-2\right)a\right)>0$, $B_3\left(x+\frac{m-1}{b}-\left(m-1\right)a\right)>B_3\left(x+\frac{m-1}{b}-\left(m-2\right)a\right)$ and

$\begin{array}{l}{B}_3\left(x+\frac{m-1}{b}-\left(m-2\right)a\right)\\ =\{\begin{array}{ll}{B}_3\left(x+\frac{m-1}{b}-\left(m-2\right)a\right)>0\hfill & \text{if}x\in \left[-\frac{a}{2};\frac{3}{2}-\frac{m-1}{b}+\left(m-2\right)a\right)\hfill \\ 0\hfill & \text{if}x\in \left[\frac{3}{2}-\frac{m-1}{b}+\left(m-2\right)a;0\right]\hfill \end{array}\end{array}$

It is easy to see that if $x\in \left[\frac{3}{2}-\frac{m-1}{b}+\left(m-2\right)a\mathrm{<mo>\; ;\; </mo>0}\right]$, then $\left|A_{22}^{m-1}\left(x\right)\right|>0$. To end, we prove that for all $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{m-1}{b}+\left(m-2\right)a\right)$, $\left|A_{22}^{m-1}\left(x\right)\right|>0$.

We also know that $-\frac{3}{2}<x+\frac{m-2}{b}-\left(m-2\right)a<\frac{3}{2}$ and $x+\frac{m-2}{b}-\left(m-1\right)a<0$, therefore we can consider the following two cases.

*If $x+\frac{m-2}{b}-\left(m-2\right)a\le 0$, one has $x+\frac{m-2}{b}-\left(m-1\right)a<x+\frac{m-2}{b}-\left(m-2\right)a$. Therefore by the strict increasing of $B_3$, we obtain $B_3\left(x+\frac{m-2}{b}-\left(m-1\right)a\right)<B_3\left(x+\frac{m-2}{b}-\left(m-2\right)a\right)$.

*If $x+\frac{m-2}{b}-\left(m-2\right)a>0$, then $-x-\frac{m-2}{b}+\left(m-2\right)a\le 0$ and $-x-\frac{m-2}{b}+\left(m-2\right)a-\left(x+\frac{m-2}{b}-\left(m-1\right)a\right)$

$=-2x-\frac{2\left(m-2\right)}{b}+\left(2m-3\right)a>-3+\frac{2}{b}+a\ge 0$. Thus $x+\frac{m-2}{b}-\left(m-1\right)a<-x-\frac{m-2}{b}+\left(m-2\right)a$ and using the fact that $B_3$ is symmetric around the origin and strict increasing, one has

$B_3\left(x+\frac{m-2}{b}-\left(m-1\right)a\right)\le B_3\left(x+\frac{m-2}{b}-\left(m-2\right)a\right)\mathrm{.}$

In the second time, we consider, for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, the matrix $A_{33}^k\left(x\right)$ given by

$A_{33}^k\left(x\right)=\left(\begin{array}{ccc}{B}_3\left(x+\frac{k}{b}-ka\right)& B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)& 0\\ B_3\left(x+\frac{k-1}{b}-ka\right)& B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-1}{b}-\left(k-2\right)a\right)\\ 0& B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)\end{array}\right)$

For $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>}-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a\right]\cup \left[\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\mathrm{<mo>\; ;\; </mo>0}\right]$, we have $B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)=0$ or $B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)=0$ and hence

$\begin{array}{l}\left|A_{33}^k\left(x\right)\right|=B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)\left|A_{22}^k\left(x\right)\right|>0\\ \text{or}\left|A_{33}^k\left(x\right)\right|=B_3\left(x+\frac{k}{b}-ka\right)\left|A_{22}^{k-1}\left(x\right)\right|>0.\end{array}$

Let $x\in \left(-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$.

We proved previously that $B_3\left(x+\frac{k}{b}-ka\right)>B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)$ for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$.

Nest, we prove that $B_3\left(x+\frac{k-1}{b}-ka\right)<B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)$ and $B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)>B_3\left(x+\frac{k-1}{b}-\left(k-2\right)a\right)$.

We know for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$, $-\frac{3}{2}<x+\frac{k-1}{b}-\left(k-1\right)a<\frac{3}{2}$, $x+\frac{k-1}{b}-ka<0$ and $x+\frac{k-1}{b}-\left(k-2\right)a>0$, therefore we have different following cases.

*If $x+\frac{k-1}{b}-\left(k-1\right)a\le 0$, on has $x+\frac{k-1}{b}-ka<x+\frac{k-1}{b}-\left(k-1\right)a$. Therefore by the strict increasing of $B_3$ on $\left[-\frac{3}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$, we obtain $B_3\left(x+\frac{k-1}{b}-ka\right)<B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)$.

*If $x+\frac{k-1}{b}-\left(k-1\right)a>0$, then $-x-\frac{k-1}{b}+\left(k-1\right)a<0$ and $-x-\frac{k-1}{b}+\left(k-1\right)a-\left(x+\frac{k-1}{b}-ka\right)=-2x-\frac{2\left(k-1\right)}{b}+\left(2k-1\right)a>0$. Indeed,

$\begin{array}{l}x\in \left(-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a;\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)\\ \Rightarrow -2x-\frac{2\left(k-1\right)}{b}+\left(2k-1\right)a>-3+\frac{2}{b}+a>0.\end{array}$

Therefore $x+\frac{k-1}{b}-ka<-x-\frac{k-1}{b}+\left(k-1\right)a$ and consequently we have

$B_3\left(x+\frac{k-1}{b}-ka\right)<B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right).$

*If $x+\frac{k-1}{b}-\left(k-1\right)a>0$, we have $x+\frac{k-1}{b}-\left(k-1\right)a<x+\frac{k-1}{b}-\left(k-2\right)a$ and then $B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)>B_3\left(x+\frac{k-1}{b}-\left(k-2\right)a\right)$.

*If $x+\frac{k-1}{b}-\left(k-1\right)a\le 0$, then $-x-\frac{k-1}{b}+\left(k-1\right)a\ge 0$ and $-x-\frac{k-1}{b}+\left(k-1\right)a-\left(x+\frac{k-1}{b}-\left(k-2\right)a\right)=-2x-\frac{2\left(k-1\right)}{b}+\left(2k-3\right)a<0$ because

$x\in \left(-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a;\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)\Rightarrow -2x-\frac{2\left(k-1\right)}{b}+\left(2k-3\right)a<3-\frac{2}{b}-a<0.$

Thus $-x-\frac{k-1}{b}+\left(k-1\right)a<x+\frac{k-1}{b}-\left(k-2\right)a$ and using the fact that $B_3$ is symmetric around the origin and strict decreasing on $\left[\mathrm{0,}\frac{3}{2}\right]$, one has

$B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)>B_3\left(x+\frac{k-1}{b}-\left(k-2\right)a\right).$

Let us prove that $B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)<B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)$ for all $k\in \left\{\mathrm{1,}\cdots \mathrm{,}m-1\right\}$.

One has $B_3\left(x-\frac{1}{b}\right)<B_3\left(x-\frac{1}{b}+a\right)$ because $-\frac{3}{2}<x-\frac{1}{b}<x-\frac{1}{b}+a<0$.

Let $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$. We known that $-\frac{3}{2}<x+\frac{k-2}{b}-\left(k-2\right)a<\frac{3}{2}$ and $x+\frac{k-2}{b}-\left(k-1\right)a<0$, therefore we have two cases.

*If $x+\frac{k-2}{b}-\left(k-2\right)a\le 0$, then $x+\frac{k-2}{b}-\left(k-1\right)a<x+\frac{k-2}{b}-\left(k-2\right)a$ and therefore $B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)<B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)$.

*If $x+\frac{k-2}{b}-\left(k-2\right)a>0$, then $-x-\frac{k-2}{b}+\left(k-2\right)a<0$ and $-x-\frac{k-2}{b}+\left(k-2\right)a-\left(x+\frac{k-2}{b}-ka\right)$

$=-2x-\frac{2\left(k-2\right)}{b}-\left(2k-3\right)a>-3+\frac{4}{b}-a>0$. Therefore $x+\frac{k-2}{b}-ka<-x-\frac{k-2}{b}+\left(k-2\right)a$ and then

$B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)<B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right).$

Let

$\begin{array}{c}p\left(x\right)=\left|\begin{array}{cc}{B}_3\left(x+\frac{k}{b}-ka\right)& 0\\ 0& B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)\end{array}\right|\\ -\left|\begin{array}{cc}{B}_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)& 0\\ B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)\end{array}\right|\end{array}$

$-\left|\begin{array}{cc}{B}_3\left(x+\frac{k}{b}-ka\right)& B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)\\ 0& B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)\end{array}\right|.$

A direct computation shows that

$\begin{array}{l}p\left(x\right)=B_3\left(x+\frac{k}{b}-ka\right)B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)-B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)\\ \times B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)-B_3\left(x+\frac{k}{b}-ka\right)B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)\\ =-{\Delta }_a{B}_3\left(x+\frac{k}{b}-\left(k-1\right)a\right){\Delta }_a{B}_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)\\ +{\Delta }_a{B}_3\left(x+\frac{k}{b}-\left(k-2\right)a\right){\Delta }_a{B}_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)\\ ={\Delta }_a{B}_3\left(-x-\frac{k}{b}+ka\right){\Delta }_a{B}_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)-{\Delta }_a{B}_3\left(-x-\frac{k}{b}+\left(k-1\right)a\right)\\ \times {\Delta }_a{B}_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)\left(\text{one}\text{uses}{\Delta }_a{B}_3\left(x\right)=-{\Delta }_a{B}_3\left(-x+a\right)\right)\end{array}$

Hence to have $p\left(x\right)>0$, it suffices to show that for all $x\in \left(-\frac{3}{2}-\frac{k-2}{b}+\left(k-1\right)a\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$,

$\begin{array}{l}\{\begin{array}{l}{\Delta }_a{B}_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)>{\Delta }_a{B}_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)\\ {\Delta }_a{B}_3\left(-x-\frac{k}{b}+ka\right)>{\Delta }_a{B}_3\left(-x-\frac{k}{b}+\left(k-1\right)a\right)\end{array}\\ \iff \{\begin{array}{l}{\Delta }_a^2{B}_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)>0\\ {\Delta }_a^2{B}_3\left(-x-\frac{k}{b}+ka\right)>0\end{array}\end{array}$

This means precisely that ${\Delta }_a^2{B}_3\left(x\right)>0$, $x\in \left(-\frac{3}{2}+a\mathrm{,}\frac{3}{2}-\frac{2}{b}+a\right)$ which is obtained in Lemma 3.

b) Let $k\in \left\{\mathrm{2,}\cdots \mathrm{,}m-1\right\}$ and consider the matrix

$A_{44}^k\left(x\right)=\left(\begin{array}{cccc}{B}_3\left(x+\frac{k}{b}-ka\right)& B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)& 0& 0\\ B_3\left(x+\frac{k-1}{b}-ka\right)& B_3\left(x+\frac{k-1}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-1}{b}-\left(k-2\right)a\right)& 0\\ 0& B_3\left(x+\frac{k-2}{b}-\left(k-1\right)a\right)& B_3\left(x+\frac{k-2}{b}-\left(k-2\right)a\right)& B_3\left(x+\frac{k-2}{b}-\left(k-3\right)a\right)\\ 0& 0& B_3\left(x+\frac{k-3}{b}-\left(k-2\right)a\right)& B_3\left(x+\frac{k-3}{b}-\left(k-3\right)a\right)\end{array}\right).$

*If $3-\frac{3}{b}+a\le 0$, then $\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\le -\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a$ and therefore $\forall x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>0}\right]$, $B_3\left(x+\frac{k-3}{b}-\left(k-2\right)a\right)=0$ or $B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)=0$. Thus

$\begin{array}{l}\left|A_{44}^k\left(x\right)\right|=B_3\left(x+\frac{k-3}{b}-\left(k-3\right)a\right)\cdot \left|A_{33}^k\left(x\right)\right|>0\\ \text{or}\left|A_{44}^k\left(x\right)\right|=B_3\left(x+\frac{k}{b}-ka\right)\cdot \left|A_{33}^{k-1}\left(x\right)\right|>0.\end{array}$

*If $3-\frac{3}{b}+a>0$, then $-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a<\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a$.

When $x\in \left[-\frac{a}{2}\mathrm{<mo>\; ;\; </mo>}-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a\right]\cup \left[\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\mathrm{<mo>\; ;\; </mo>0}\right]$, then $B_3\left(x+\frac{k-3}{b}-\left(k-2\right)a\right)=0$ or $B_3\left(x+\frac{k}{b}-\left(k-1\right)a\right)=0$ and therefore

$\begin{array}{l}\left|A_{44}^k\left(x\right)\right|=B_3\left(x+\frac{k-3}{b}-\left(k-3\right)a\right)\cdot \left|A_{33}^k\left(x\right)\right|>0\\ \text{or}\left|A_{44}^k\left(x\right)\right|=B_3\left(x+\frac{k}{b}-ka\right)\cdot \left|A_{33}^{k-1}\left(x\right)\right|>0.\end{array}$

We finish by proving that for all $x\in \left(-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$, $\left|A_{44}^k\left(x\right)\right|>\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; 0}$. We observe that for $k=3$ and $x\in \left(-1+a\mathrm{<mo>\; ;\; </mo>1}-\frac{3}{b}+2a\right)$, then $\left|A_{44}^3\left(x\right)\right|=\left|A_{44}^2\left(x+\frac{1}{b}-a\right)\right|$ and for $k\in \left\{\mathrm{4,}\cdots \mathrm{,}m-1\right\}$ and $x\in \left(-\frac{3}{2}-\frac{k-3}{b}+\left(k-2\right)a\mathrm{<mo>\; ;\; </mo>}\frac{3}{2}-\frac{k}{b}+\left(k-1\right)a\right)$, then $\left|A_{44}^k\left(x\right)\right|=\left|A_{44}^3\left(x+\frac{k-3}{b}-\left(k-3\right)a\right)\right|>0$. So we only prove that for all $x\in \left(-\frac{3}{2}+\frac{1}{b};\frac{3}{2}-\frac{2}{b}+a\right)$, $\left|A_{44}^2\left(x\right)\right|>0$. Using the condition $3-\frac{3}{b}+a>0$, we consider $a\in \left[\frac{3}{4}\mathrm{,}\frac{3}{2}\right]$ and $b\in \left[\frac{3}{3+a}\mathrm{,}\frac{4}{3+2a}\right]$. In other words, by the Lemma 4, we have this result.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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