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

Applied Mathematics

2152-7385

Scientific Research Publishing

10.4236/am.2022.135026

AM-117220

Articles

Physics&Mathematics

On the Frame Set for the 3-Spline

Abdou

Ganiou Débayo Atindehou

₁^*

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

19052022

13053774002, March 202217, May 2022 20, May 2022

2014

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

This paper investigates the fine structure of the Gabor frame generated by the B-spline B ₃. 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 B ₃ is the subset of all parameters ( a, b) ∈ R ² ₊for which the time-frequency shifts of B ₃ along aZ × bZ form a Gabor frame for L ²(R).

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 ∈ L 2 ( ℝ ) and a , b > 0 is the set of functions

G ( g , a , b ) = { M l b T k a g = e 2 π i l b ⋅ g ( ⋅ − k a ) : ( l , k ) ∈ ℤ 2 }

for which there exist A , B > 0 such that

A ‖ f ‖ 2 2 ≤ ∑ l , k ∈ ℤ | 〈 f , M l b T k a g 〉 | 2 ≤ B ‖ f ‖ 2 2 ,

for all f ∈ L 2 ( ℝ ) , 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 F ( g ) of g ∈ L 2 ( ℝ ) 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 ≥ 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 ( a , b ) ∈ ℝ + 2 such that G ( B 3 , a , b ) is a Gabor frame where

B 3 ( x ) = χ [ − 1 / 2 , 1 / 2 ] ∗ χ [ − 1 / 2 , 1 / 2 ] ∗ χ [ − 1 / 2 , 1 / 2 ] ( x ) = { 1 2 x 2 + 3 2 x + 9 8 x ∈ [ − 3 2 , − 1 2 ] − x 2 + 3 4 x ∈ [ − 1 2 , 1 2 ] 1 2 x 2 − 3 2 x + 9 8 x ∈ [ 1 2 , 3 2 ]

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

F ( B 3 ) = { ( a , b ) ∈ ℝ + 2 : G ( B 3 , a , b ) is a frame } .

It is not difficult to see that F ( B 3 ) ≠ ∅ . Indeed, for any a ∈ ( 0,3 ) there exists b 0 > 0 such that for all 0 < b ≤ b 0 , ( a , b ) ∈ F ( B 3 ) [ [24], Theorem 4.18]. Furthermore, F ( B 3 ) 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 ( ℝ ) [2]. However, the complete characterization of F ( B 3 ) is still unknown.

To the best of our knowledge [1] [5] [8] [11] [23] [27] [28], the biggest subset known in F ( B 3 ) is the connected set

{ ( a , b ) ∈ ℝ + 2 : a b < 1 , 0 < a < 3 , 0 < b ≤ max a ( 2 3 , 4 3 + 3 a ) } .

In this note, we complete this last subset by showing that Γ belongs to F ( B 3 ) where Γ = ∪ m = 3 ∞ Γ m with for m ≥ 4 ,

Γ m : = { ( a , b ) ∈ ℝ + 2 : a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 1 ) 2 m − 1 ] , b ∈ ( 2 ( m − 1 ) 3 + ( 2 m − 3 ) a , min a ( 2 m 3 + ( 2 m − 1 ) a , 4 3 + 2 a ) ] , b > 2 3 } (1)

and

Γ 3 : = { ( a , b ) ∈ ℝ + 2 : a ∈ [ 3 4 , 6 5 ] , b ∈ ( 4 3 + 3 a , 6 3 + 5 a ] , b > 2 3 } . (2)

More specifically, our main result gives the frame properties on Γ .

Theorem 1. For m ≥ 3 , let ( a , b ) ∈ Γ m . Then, the Gabor system G ( B 3 , a , b ) is a frame for L 2 ( ℝ ) , and there is a unique dual window H 3 ∈ L 2 ( ℝ ) such

that s u p p H 3 ⊆ [ − 2 m − 1 2 a , 2 m − 1 2 a ] . Furthermore, for each ( a , b ) ∈ Γ , the Gabor system G ( B 3 , a , b ) is a frame for L 2 ( ℝ ) .

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 ≥ 1 , let 0 < a < 3 and 2 ( m − 1 ) 3 + ( 2 m − 3 ) a < b ≤ 2 m 3 + ( 2 m − 1 ) a . Assume that H 3 is a bounded real-function with support on [ − 2 m − 1 2 a , 2 m − 1 2 a ] . Then, the Gabor systems G ( B 3 , a , b ) and G ( H 3 , a , b ) are dual frames for L 2 ( ℝ ) if and only if

∑ k = 1 − m m − 1 B 3 ( x − l / b + k a ) H 3 ( x + k a ) = b δ l ,0 , | l | ≤ m − 1 , for a .e . x ∈ [ − a 2 , a 2 ] . (3)

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

E ^ m ( x ) F ^ m t ( x ) = B t for a .e . x ∈ [ − a 2 , a 2 ] , (4)

where F ^ m ( x ) and B are the row vectors given respectively by F ^ m ( x ) = [ H 3 ( x + k a ) ] | k | ≤ m − 1 and B = [ b δ l ,0 ] | l | ≤ m − 1 ; E ^ m ( x ) is the ( 2 m − 1 ) × ( 2 m − 1 ) matrix-valued function defined by E ^ m ( x ) = [ B 3 ( x − l b + k a ) ] 1 − m ≤ l , k ≤ m − 1 . Using the fact that the 3-spline B 3 is supported by [ − 3 2 , 3 2 ] and ( a , b ) ∈ Γ m ; one observes easily that for all x ∈ [ − a 2 ,0 ] , not only the all E ^ m ( x ) ’s on-diagonal entries are positive but also the matrix-valued function E ^ m ( x ) can be rewritten as the following block matrix:

E ^ m ( x ) = ( G ^ m − 1 ( x ) I ^ m − 1 ( x ) O J ^ m − 1 ( x ) ) , (5)

where O is a ( m − 2 ) × ( m + 1 ) matrix of 0’s, I ^ m − 1 ( x ) is a ( m + 1 ) × ( m − 2 ) matrix, J ^ m − 1 ( x ) is a ( m − 2 ) × ( m − 2 ) upper triangular matrix and G ^ m − 1 ( x ) is the ( m + 1 ) × ( m + 1 ) tridiagonal matrix given by

G ^ m − 1 ( x ) = [ B 3 ( x − l b + k a ) ] 1 − m ≤ l , k ≤ 1 . (6)

Figure 1 shows the connected set known in F ( B 3 ) .

The rest of the paper is organized as follows. In Section 2 we prove our main results by studying the invertibility of the matrix G ^ m − 1 ( x ) .

2. Invertibility of G ^ m − 1 ( x ) for ( a , b ) ∈ Γ m

To prove Theorem 1, we only need to show that (3) has a unique solution F ^ m ( x ) . This is equivalent to proving respectively that the ( 2 m − 1 ) × ( 2 m − 1 ) matrix-valued function E ^ m ( x ) is invertible for a.e. x ∈ [ − a 2 , a 2 ] . In other

words, from the block form for the matrix E ^ m ( x ) , we need to prove that the matrix G ^ m − 1 ( x ) is invertible.

Throughout the paper, we use the fact that

min a ( 2 m 3 + ( 2 m − 1 ) a , 4 3 + 2 a ) = { 4 3 + 2 a if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] 2 m 3 + ( 2 m − 1 ) a if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ]

The following lemma gives the behavior of the entries of G ^ m − 1 ( x ) .

Lemma 1. For m ≥ 3 , let ( a , b ) ∈ Γ m , x ∈ [ − a 2 ; 0 ] and consider B 3 . Then the following hold:

1) B 3 ( x + k b − k a ) > 0 , for all k ∈ { 1 − m , ⋯ , m − 1 } .

2) B 3 ( x + k b − ( k − 1 ) a ) > 0 for k ∈ { 0 ; 1 } and

B 3 ( x + k b − ( k − 1 ) a ) = { B 3 ( x + k b − ( k − 1 ) a ) > 0 if x ∈ [ − a 2 ; 3 2 − k b + ( k − 1 ) a ) 0 if x ∈ [ 3 2 − k b + ( k − 1 ) a ; 0 ]

for all k ∈ { 2, ⋯ , m − 1 } .

3) B 3 ( x + m − 2 b − ( m − 1 ) a ) > 0 and

B 3 ( x + k b − ( k + 1 ) a ) = { 0 if x ∈ [ − a 2 ; − 3 2 − k b + ( k + 1 ) a ] B 3 ( x + k b − ( k + 1 ) a ) > 0 if x ∈ ( − 3 2 − k b + ( k + 1 ) a ; 0 ]

for all { k = − 1, ⋯ , m − 3 } .

Proof. 1) Firstly, let k ∈ { 1 − m , ⋯ , − 1 } . We have

x + k b − k a ≤ k b − k a ≤ { k ( 3 + 2 a 4 ) − k a = k ( 3 − 2 a 4 ) if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] k ( 3 + ( 2 m − 1 ) a 2 m ) − k a = 3 k 2 m − k 2 m a if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ] ≤ { 3 k 4 − k 2 × 3 ( m − 2 ) 2 ( m − 1 ) = 3 k 4 ( m − 1 ) if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] 3 k 2 m − k 2 m × 3 ( m − 1 ) 2 m − 1 = 3 k 2 ( 2 m − 1 ) if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ] < 0

and

x + k b − k a ≥ k b − 2 k + 1 2 a > k ( 3 + ( 2 m − 3 ) a 2 ( m − 1 ) ) − 2 k + 1 2 a = 3 k 2 ( m − 1 ) − m − 1 + k 2 ( m − 1 ) a ≥ 3 k 2 ( m − 1 ) − m − 1 + k 2 ( m − 1 ) × 3 ( m − 1 ) 2 m − 1 = 3 2 ( k m − 1 − m − 1 + k 2 m − 1 ) ≥ − 3 2 ( because k m − 1 − m − 1 + k 2 m − 1 ≥ − 1 ) .

Therefore for all k ∈ { 1 − m , ⋯ , − 1 } , − 3 2 < x + k b − k a < 0 . Thus B 3 ( x + k b − k a ) > 0 . Secondly, let k ∈ { 0, ⋯ , m − 1 } . We have

x + k b − k a ≤ k b − k a < k ( 3 + ( 2 m − 3 ) a 2 ( m − 1 ) ) − k a = 3 k 2 ( m − 1 ) − k a 2 ( m − 1 ) ≤ 3 k 2 ( m − 1 ) − k 2 ( m − 1 ) × 3 ( m − 3 ) 2 ( m − 2 ) = k 4 ( m − 2 ) ≤ 3 2

and

x + k b − k a ≥ − a 2 + k b − k a = k b − 2 k + 1 2 a ≥ { k ( 3 + 2 a 4 ) − 2 k + 1 2 a = 3 k 4 − k + 1 2 a if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] k ( 3 + ( 2 m − 1 ) a 2 m ) − 2 k + 1 2 a = 3 k 2 m − m + k 2 m a if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ] ≥ { 3 k 4 − k + 1 2 × 3 ( m − 2 ) 2 ( m − 1 ) = − 3 2 ( ( m − 2 ) − k 2 ( m − 1 ) ) 3 k 2 m − m + k 2 m × 3 ( m − 1 ) 2 m − 1 = − 3 2 ( ( m − 1 ) − k 2 m − 1 ) > − 3 2

Then for all k ∈ { 0, ⋯ , m − 1 } , − 3 2 < x + k b − k a < 3 2 . Thus B 3 ( x + k b − k a ) > 0 .

2) Let k ∈ { 0, ⋯ , m − 1 } . We have

x + k b − ( k − 1 ) a ≤ k b − ( k − 1 ) a < k ( 3 + ( 2 m − 3 ) a 2 ( m − 1 ) ) − ( k − 1 ) a = 3 k 2 ( m − 1 ) + 2 ( m − 1 ) − k 2 ( m − 1 ) a ≤ 3 k 2 ( m − 1 ) + 2 ( m − 1 ) − k 2 ( m − 1 ) × 3 ( m − 1 ) 2 m − 1 = 3 2 ( k m − 1 + 2 ( m − 1 ) − k 2 m − 1 )

and

x + k b − ( k − 1 ) a ≥ − a 2 + k b − ( k − 1 ) a = k b − 2 k − 1 2 a ≥ { k ( 3 + 2 a 4 ) − 2 k − 1 2 a = 3 k 4 − k − 1 2 a if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] k ( 3 + ( 2 m − 1 ) a 2 m ) − 2 k − 1 2 a = 3 k 2 m + m − k 2 m a if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ] ≥ { 3 k 4 − k − 1 2 × 3 ( m − 2 ) 2 ( m − 1 ) = 3 2 ( m + k − 2 2 ( m − 1 ) ) if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] 3 k 2 m + m − k 2 m × 3 ( m − 2 ) 2 ( m − 1 ) = 3 2 ( m + k − 2 2 ( m − 1 ) ) if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ]

Then for all k ∈ { 1, ⋯ , m − 1 } ;

0 < 3 2 ( m + k − 2 2 ( m − 1 ) ) ≤ x + k b − ( k − 1 ) a < 3 2 ( k m − 1 + 2 ( m − 1 ) − k 2 m − 1 ) .

Therefore for all k ∈ { 2, ⋯ , m − 1 } ,

B 3 ( x + k b − ( k − 1 ) a ) = { B 3 ( x + k b − ( k − 1 ) a ) > 0 if x ∈ [ − a 2 ; 3 2 − k b + ( k − 1 ) a ) 0 if x ∈ [ 3 2 − k b + ( k − 1 ) a ; 0 ]

and using the fact that b > 2 3 , we have easily B 3 ( x + k b − ( k − 1 ) a ) > 0 for k ∈ { 0 ; 1 } .

3) Let k ∈ { 0, ⋯ , m − 2 } . We have

x + k b − ( k + 1 ) a ≤ k b − ( k + 1 ) a < k ( 3 + ( 2 m − 3 ) a 2 ( m − 1 ) ) − ( k + 1 ) a = 3 k 2 ( m − 1 ) − 2 ( m − 1 ) + k 2 ( m − 1 ) a ≤ 3 k 2 ( m − 1 ) − 2 ( m − 1 ) + k 2 ( m − 1 ) × 3 ( m − 3 ) 2 ( m − 2 ) = 3 2 ( k − 2 m + 6 2 ( m − 2 ) ) = 3 2 ( ( k − m + 2 ) + ( 4 − m ) 2 ( m − 2 ) ) ≤ 0

as far as m ≥ 4 and

x + k b − ( k + 1 ) a ≥ − a 2 + k b − ( k + 1 ) a = k b − 2 k + 3 2 a ≥ { k ( 3 + 2 a 4 ) − 2 k + 3 2 a = 3 k 4 − k + 3 2 a if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] k ( 3 + ( 2 m − 1 ) a 2 m ) − 2 k + 3 2 a = 3 k 2 m − 3 ( m − 1 ) + k + 3 2 m a if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ] ≥ { 3 k 4 − k + 3 2 × 3 ( m − 2 ) 2 ( m − 1 ) = 3 2 ( k − 3 ( m − 2 ) 2 ( m − 1 ) ) if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] 3 k 2 m − 3 ( m − 1 ) + k + 3 2 m × 3 ( m − 1 ) 2 m − 1 = 3 2 ( k − 3 ( m − 1 ) 2 m − 1 ) if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ]

Then for all k ∈ { 0, ⋯ , m − 2 } ;

{ 3 2 ( k − 3 ( m − 2 ) 2 ( m − 1 ) ) 3 2 ( k − 3 ( m − 1 ) 2 m − 1 ) ≤ x + k b − ( k + 1 ) a < 3 2 ( ( k − m + 2 ) + ( 4 − m ) 2 ( m − 2 ) ) ≤ 0

as far as m ≥ 4 and

− 3 2 − 3 ( ( m − 1 ) 2 + 1 ) 2 ( m − 1 ) ( 2 m − 1 ) < x − 1 b ≤ { − 3 ( 2 m − 5 ) 4 ( m − 2 ) if a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 3 ( m − 2 ) 2 ( m − 1 ) ] − 3 ( 2 m − 3 ) 4 ( m − 1 ) if a ∈ [ 3 ( m − 2 ) 2 ( m − 1 ) , 3 ( m − 1 ) 2 m − 1 ] < 0.

Thus B 3 ( x + m − 2 b − ( m − 1 ) a ) > 0 and

B 3 ( x + k b − ( k + 1 ) a ) = { 0 if x ∈ [ − a 2 ; − 3 2 − k b + ( k + 1 ) a ] B 3 ( x + k b − ( k + 1 ) a ) > 0 if x ∈ [ − 3 2 − k b + ( k + 1 ) a ; 0 ]

for k = − 1 , ⋯ , m − 3 .

The following remark gives the trivial two different partitions of [ − a 2 ; 0 ] which can be derived from the definition of Γ m ( m ≥ 3 ) and will be used to compute the determinant of the matrix G ^ m − 1 ( x ) .

Remark 1. a) If 3 − 3 b + a ≤ 0 and

· a ∈ [ 3 ( m − 3 ) 2 ( m − 2 ) , 6 m − 9 4 m − 3 ] , then − 3 2 − m − 3 b + ( m − 2 ) a ≤ − a 2 . Hence

[ − a 2 , 0 ] = ∪ k = 1 m − 2 ( Q k ∪ T k ) ∪ T m − 1

where Q k = [ 3 2 − k + 1 b + k a , − 3 2 − k − 2 b + ( k − 1 ) a ] and T k = ( − 3 2 − k − 2 b + ( k − 1 ) a , 3 2 − k b + ( k − 1 ) a ) with the convention that T m − 1 = [ − a 2 , 3 2 − m − 1 b + ( m − 2 ) a ) and T 1 = ( − 3 2 + 1 b , 0 ] .

· a ∈ [ 3 ( m − 2 ) 2 m − 3 , 3 ( m − 1 ) 2 m − 1 ] , then − a 2 ≤ − 3 2 − m − 3 b + ( m − 2 ) a . Thus

[ − a 2 , 0 ] = ∪ k = 1 m − 1 ( Q k ∪ T k )

where Q k = [ 3 2 − k + 1 b + k a , − 3 2 − k − 2 b + ( k − 1 ) a ] and T k = ( − 3 2 − k − 2 b + ( k − 1 ) a , 3 2 − k b + ( k − 1 ) a ) with the convention that Q m − 1 = [ − a 2 , − 3 2 − m − 3 b + ( m − 2 ) a ] and T 1 = ( − 3 2 + 1 b ,0 ] .

· a ∈ ( 6 m − 9 4 m − 3 ; 3 ( m − 2 ) 2 m − 3 ) for m ≠ 3 , then − 3 2 − m − 3 b + ( m − 2 ) a + a 2 is sign-changing. So, we use both different previous partitions.

b) If 3 − 3 b + a > 0 , then a is only in [ 3 ( m − 3 ) 2 ( m − 2 ) , 6 m − 9 4 m − 3 ] . Hence

[ − a 2 , 0 ] = ( ∪ k = 1 m − 1 Q ˜ k ) ∪ ( ∪ k = 2 m − 1 T ˜ k )

where Q ˜ k = [ 3 2 − k + 1 b + k a , − 3 2 − k − 3 b + ( k − 2 ) a ] and T ˜ k = ( − 3 2 − k − 3 b + ( k − 2 ) a , 3 2 − k b + ( k − 1 ) a ) with the convention that Q ˜ m − 1 = [ − a 2 , − 3 2 − m − 4 b + ( m − 3 ) a ] and Q ˜ 1 = [ 3 2 − 2 b + a , 0 ] .

NB: Specially for m = 3 , one substitutes 3 ( m − 3 ) 2 ( m − 2 ) by 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 ∈ { 1, ⋯ , m − 1 } , l = 2 , 3 , 4 and denote A l l k ( x ) the l × l sub-matrix of G ^ m − 1 ( x ) , defined respectively by

A 22 k ( x ) = ( B 3 ( x + k b − k a ) B 3 ( x + k b − ( k − 1 ) a ) B 3 ( x + k − 1 b − k a ) B 3 ( x + k − 1 b − ( k − 1 ) a ) ) ,

A 33 k ( x ) = ( B 3 ( x + k b − k a ) B 3 ( x + k b − ( k − 1 ) a ) 0 B 3 ( x + k − 1 b − k a ) B 3 ( x + k − 1 b − ( k − 1 ) a ) B 3 ( x + k − 1 b − ( k − 2 ) a ) 0 B 3 ( x + k − 2 b − ( k − 1 ) a ) B 3 ( x + k − 2 b − ( k − 2 ) a ) )

and

A 44 k ( x ) = ( B 3 ( x + k b − k a ) B 3 ( x + k b − ( k − 1 ) a ) 0 0 B 3 ( x + k − 1 b − k a ) B 3 ( x + k − 1 b − ( k − 1 ) a ) B 3 ( x + k − 1 b − ( k − 2 ) a ) 0 0 B 3 ( x + k − 2 b − ( k − 1 ) a ) B 3 ( x + k − 2 b − ( k − 2 ) a ) B 3 ( x + k − 2 b − ( k − 3 ) a ) 0 0 B 3 ( x + k − 3 b − ( k − 2 ) a ) B 3 ( x + k − 3 b − ( k − 3 ) a ) ) .

The following Lemma indicates that the matrices A l l k ( x ) are invertible.

Lemma 2. Given m ≥ 3 , let ( a , b ) ∈ Γ m and x ∈ [ − a 2 ; 0 ] . The following hold:

a) if k ∈ { 1, ⋯ , m − 1 } , then | A l l k ( x ) | > 0 for all l = 2 , 3 .

b) | A 44 k ( x ) | > 0 , for all k ∈ { 2, ⋯ , m − 1 } .

This result is proved in Appendix 4.

The following Proposition gives an explicit expression for the determinant of the sub-matrix G ^ m − 1 ( x ) when m ≥ 3 , x ∈ [ − a 2 ; 0 ] and under the hypotheses of Theorem 1.

Proposition 1. For m ≥ 3 , let ( a , b ) ∈ Γ m and x ∈ [ − a 2 ,0 ] . The following statements hold:

∀ x ∈ D , | G ^ m − 1 ( x ) | = ∏ l ∈ I B 3 ( x + l b − l a ) ⋅ | A j j i ( x ) |

where D , I , i and j are given in the following cases:

1) If 3 − 3 b + a ≤ 0 and

a) 3 ( m − 3 ) 2 ( m − 2 ) ≤ a ≤ 6 m − 9 4 m − 3 , then

i) for all k ∈ { 1, ⋯ , m − 2 } , D = Q k , I = { − 1 , ⋯ , m − 1 } \ { k , k − 1 } , i = k , and j = 2 .

ii) for all k ∈ { 1, ⋯ , m − 1 } , D = T k , I = { − 1, ⋯ , m − 1 } \ { k , k − 1, k − 2 } , i = k and j = 3 .

b) 3 ( m − 2 ) 2 m − 3 ≤ a ≤ 3 ( m − 1 ) 2 m − 1 , then for all k ∈ { 1, ⋯ , m − 1 } , one has:

i) for D = Q k , I = { − 1 , ⋯ , m − 1 } \ { k , k − 1 } , i = k , and j = 2 .

ii) for D = T k , I = { − 1, ⋯ , m − 1 } \ { k , k − 1, k − 2 } , i = k and j = 3 .

c) 6 m − 9 4 m − 3 < a < 3 ( m − 2 ) 2 m − 3 when m ≠ 3 , then one uses i) and ii) obtained in a) or b).

2) If 3 − 3 b + a > 0 , then

a) For all k ∈ { 1, ⋯ , m − 1 } , D = Q ˜ k , I = { − 1, ⋯ , m − 1 } \ { k , k − 1, k − 2 } , i = k and j = 3 .

b) For all k ∈ { 2, ⋯ , m − 1 } , D = T ˜ k , I = { − 1, ⋯ , m − 1 } \ { k , k − 1 , k − 2 , k − 3 } , i = k and j = 4 .

Moreover the matrix E ^ m ( x ) is invertible.

Proof. We prove the result by induction on m by using the different partitions of [ − a 2 ; 0 ] obtained in Remark 1. For m = 3 , the matrix G ^ 2 ( x ) given by

G ^ 2 ( x ) = ( B 3 ( x + 2 b − 2 a ) B 3 ( x + 2 b − a ) 0 0 B 3 ( x + 1 b − 2 a ) B 3 ( x + 1 b − a ) B 3 ( x + 1 b ) 0 0 B 3 ( x − a ) B 3 ( x ) B 3 ( x + a ) 0 0 B 3 ( x − 1 b ) B 3 ( x − 1 b + a ) ) .

Suppose that 3 − 3 b + a ≤ 0 and 3 4 ≤ a ≤ 1 , then [ − a 2 ; 0 ] = Q 1 ∪ T 1 ∪ T 2 with T 2 = [ − a 2 ; 3 2 − 2 b + a ) , Q 1 = [ 3 2 − 2 b + a ; − 3 2 + 1 b ] and T 1 = ( − 3 2 + 1 b ; 0 ] .

So, it is easy to see that for all x ∈ Q 1 , | G ^ 2 ( x ) | = B 3 ( x + 2 b − 2 a ) B 3 ( x − 1 b + a ) ⋅ | A 22 1 ( x ) | ; x ∈ T 1 , | G ^ 2 ( x ) | = B 3 ( x + 2 b − 2 a ) ⋅ | A 33 1 ( x ) | and x ∈ T 2 , | G ^ 2 ( x ) | = B 3 ( x − 1 b + a ) ⋅ | A 33 2 ( x ) | . This establishes the part (a) for the base case m = 3 .

Suppose that 3 − 3 b + a ≤ 0 and 1 < a ≤ 6 5 , then [ − a 2 ; 0 ] = Q 2 ∪ T 2 ∪ Q 1 ∪ T 1 with Q 2 = [ − a 2 ; − 3 2 + a ] , T 2 = ( − 3 2 + a ; 3 2 − 2 b + a ) , Q 1 = [ 3 2 − 2 b + a ; − 3 2 + 1 b ] and T 1 = ( − 3 2 + 1 b ; 0 ] . So, it is easy to see that

∀ x ∈ D = Q k ( k = 1 , 2 ) , | G ^ 2 ( x ) | = ∏ l ∈ I B 3 ( x + l b − l a ) ⋅ | A j j i ( x ) |

where I = { − 1,0,1,2 } \ { k , k − 1 } , i = k , and j = 2 ; and

∀ x ∈ D = T k ( k = 1 , 2 ) , | G ^ 2 ( x ) | = ∏ l ∈ I B 3 ( x + l b − l a ) ⋅ | A j j i ( x ) |

where I = { − 1,0,1,2 } \ { k , k − 1, k − 2 } , i = k and j = 3 . Hence (b) holds. Similarly, (2) holds.

Suppose that (1) and (2) hold for m − 1 ≥ 3 and let us prove that they hold for m. So, one compute the determinant of the matrix G ^ m ( x ) given by

G ^ m ( x ) = [ B 3 ( x − l b + k a ) ] − m ≤ l , k ≤ 1 .

Suppose 3 − 3 b + a > 0 . From Remark 1, we have [ − a 2 ; 0 ] = ( ∪ k = 1 m Q ˜ k ) ∪ ( ∪ k = 2 m T ˜ k ) with Q ˜ m = [ − a 2 ; − 3 2 − m − 3 b + ( m − 2 ) a ] ; Q ˜ k = [ 3 2 − k + 1 b + k a ; − 3 2 − k − 3 b + ( k − 2 ) a ] , for all k ∈ { 2, ⋯ , m − 1 } ; Q ˜ 1 = [ 3 2 − 2 b + a ; 0 ] and T ˜ k = ( − 3 2 − k − 3 b + ( k − 2 ) a ; 3 2 − k b + ( k − 1 ) a ) , for all k ∈ { 2, ⋯ , m } .

Let x ∈ D = Q ˜ m . Thus B 3 ( x + k b − ( k + 1 ) a ) = 0 for all k ∈ { − 1, ⋯ , m − 3 } . Hence,

where I = { − 1, ⋯ , m − 3 } , i = m and j = 3 .

Let x ∈ D = Q ˜ m − 1 . Therefore B 3 ( x + m b − ( m − 1 ) a ) = 0 and B 3 ( x + k b − ( k + 1 ) a ) = 0 for all k ∈ { − 1, ⋯ , m − 4 } . Consequently,

where I = { − 1 , ⋯ , m } \ { m − 3 , m − 2 , m − 1 } , i = m − 1 and j = 3 .

Let k ∈ { 1, ⋯ , m − 2 } and x ∈ D = Q ˜ k . Thus B 2 ( x + m b − ( m − 1 ) a ) = 0 . Hence, | G ^ m ( x ) | = B 3 ( x + m b − m a ) ⋅ | G ^ m − 1 ( x ) | and by the induction assumption, we have

where I = { − 1 , ⋯ , m } \ { k , k − 1 , k − 2 } , i = k and j = 3 .

Let x ∈ D = T ˜ m . Thus B 3 ( x + k b − ( k + 1 ) a ) = 0 for all k ∈ { − 1, ⋯ , m − 4 } . Hence,

where I = { − 1 , ⋯ , m } \ { m − 3 ; m − 2 , m − 1 , m } , i = m and j = 4 .

Let k ∈ { 2, ⋯ , m − 1 } and x ∈ D = T ˜ k . Then B 3 ( x + m b − ( m − 1 ) a ) = 0 . Thus | G ^ m ( x ) | = B 3 ( x + m b − m a ) ⋅ | G ^ m − 1 ( x ) | and by the induction assumption, we have

| G ^ m ( x ) | = ( ∏ l = − 1 l ≠ k , k − 1 , k − 2 , k − 3 m B 3 ( x + l b − l a ) ) ⋅ | A 44 k ( x ) | = ∏ l ∈ I B 3 ( x + l b − l a ) ⋅ | A j j i ( x ) | .

where I = { − 1 , ⋯ , m } \ { k , k − 1 , k − 2 , k − 3 } , 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 E ^ m ( x ) , we have for all x ∈ [ − a 2 ; 0 ] ,

| E ^ m ( x ) | = ( ∏ 1 − m − 2 B 3 ( x + k b − k a ) ) ⋅ | G ^ m − 1 ( x ) | .

We know that B 3 ( x + k b − k a ) > 0 , ∀ k ∈ { 1 − m , ⋯ , m − 1 } and | G ^ m − 1 ( x ) | > 0 because | A l l k ( x ) | > 0 from Lemma 2. Thus we conclude that | E ^ m ( x ) | > 0 for all x ∈ [ − a 2 ; 0 ] , and by symmetry ( | E ^ m ( − x ) | = | E ^ m ( x ) | ) this holds for all x ∈ [ − a 2 ; a 2 ] .

We are now ready to prove Theorem 1.

Proof of Theorem 1. By Proposition 1 we know that E ^ m ( x ) is invertible. Let F ^ m ( x ) be defined on ℝ as follows. For x ∈ ℝ \ [ − 2 m − 1 2 a , 2 m − 1 2 a ] , let F ^ m ( x ) = 0 and for x ∈ [ − 2 m − 1 2 a , 2 m − 1 2 a ] let F ^ m ( x ) be defined by F ^ m t ( x ) = b ( E ^ m − 1 ( x ) ) m , where F ^ m ( x ) = [ H 3 ( x + k a ) ] | k | ≤ m − 1 and ( E ^ m − 1 ( x ) ) m is the m^th column vector of the matrix E ^ m − 1 ( x ) . Consequently, H 3 is a compactly supported and bounded function for which G ( H 3 , a , b ) 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 ( a , b ) ∈ ℝ + 2 for which the Gabor system G ( B 3 , a , b ) forms a frame. We presented the known results of the Gabor frame set F ( B 3 ) 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 .

Conflicts of Interest

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

Cite this paper

Atindehou, A.G.D. (2022) On the Frame Set for the 3-Spline. Applied Mathematics, 13, 377-400. https://doi.org/10.4236/am.2022.135026

Appendix

Consider the first-order difference Δ a B 3 and the second-order difference Δ a 2 B 3 given respectively by

Δ a B 3 ( x ) = B 3 ( x ) − B 3 ( x − a ) and Δ a 2 B 3 ( x ) = B 3 ( x ) − 2 B 3 ( x − a ) + B 3 ( x − 2 a )

The following Lemma completes the Lemma 2.2 of [27] in the case of B 3 .

Lemma 3. Let 0 < a < 3 . Then Δ a 2 B 3 ( x ) > 0 , for all x ∈ ( − 3 2 + a , 3 2 − 2 b + a ) .

Proof. By definition of the functional space V a g and the Lemma 2.2 in [27], it is known that Δ a 2 B 3 ( x ) > 0 for all x ∈ [ − 3 2 , − 3 4 + 3 a 4 ] .

In particular, for a ≥ 1 or b ≤ 8 9 + a , ( − 3 2 + a , 3 2 − 2 b + a ) ⊂ [ − 3 2 , − 3 4 + 3 a 4 ] while for a < 1 and b > 8 9 + a , − 3 4 + 3 a 4 < 3 2 − 2 b + a . In other words, to have that Δ a 2 B 3 ( x ) > 0 on ( − 3 2 + 3 a 4 , 3 2 − 2 b + a ) , it suffices to show that, for a < 1 and b > 8 9 + a ,

Δ a 2 B 3 ( x ) > 0 , ∀ x ∈ ( − 3 4 + 3 a 4 , 3 2 − 2 b + a ) .

Otherwise, we only show that for 3 4 ≤ a < 1 and 8 9 + a < b ≤ 4 3 + 2 a ;

Δ a 2 B 3 ( x ) > 0 , ∀ x ∈ ( − 3 4 + 3 a 4 , 3 2 − 2 b + a ) ⊂ ( − 3 16 , 0 ] .

Let f ( x ) : = Δ a 2 B 3 ( x ) . It is easy to see that for all x ∈ ( − 3 4 + 3 a 4 , 3 2 − 2 b + a ) ,

f ( x ) = B 3 ( x ) − 2 B 3 ( x − a ) = − x 2 + 3 4 − ( x − a ) 2 − 3 ( x − a ) − 9 4 = − 2 x 2 + ( 2 a − 3 ) x − a 2 + 3 a − 3 2 .

One has f ′ ( − 3 16 ) = 2 a − 9 4 < 0 et f ( 0 ) = − a 2 + 3 a − 3 2 > 0 . Thus f ( x ) > 0 , ∀ x ∈ ( − 3 16 , 0 ] . Consequently,

∀ x ∈ ( − 3 4 + 3 a 4 , 3 2 − 2 b + a ) , Δ a 2 B 3 ( x ) > 0

as wanted.

The following Lemma shows the invertibility of the matrix A 44 2 ( x ) .

Lemma 4. Let a ∈ [ 3 4 , 3 2 ] , b ∈ [ 3 3 + a , 4 3 + 2 a ] and x ∈ ( − 3 2 + 1 b ; 3 2 − 2 b + a ) . Then | A 44 2 ( x ) | > 0 .

Proof. Let L 2 ( x ) : = | A 44 2 ( x ) | and L 2 i , j ( x ) denote the ij^th minor of L 2 ( x ) , the determinant of the matrix obtained by removing the i^th row and the j^th column from L 2 ( x ) . We have respectively

(1) B 3 ( x ) > B 3 ( x − a ) and B 3 ( x ) > B 3 ( x + a ) ;

(2) L 2 3 , 2 ( x ) > 0 and L 2 3 , 4 ( x ) > 0 ;

(3) L 2 33 ( x ) > L 2 3 , 2 ( x ) + L 2 3 , 4 ( x ) .

Combining (1), (2) and (3), we have for all x ∈ ( − 3 2 + 1 b ; 3 2 − 2 b + a ) ,

L 2 ( x ) = − B 3 ( x − a ) ⋅ L 2 3 , 2 ( x ) + B 3 ( x ) ⋅ L 2 33 ( x ) − B 3 ( x + a ) ⋅ L 2 3 , 4 ( x ) > [ B 3 ( x ) − B 3 ( x − a ) ] ⋅ L 2 3 , 2 ( x ) + [ B 3 ( x ) − B 3 ( x + a ) ] ⋅ L 2 3 , 4 ( x ) > 0.

This implies that for all x ∈ ( − 3 2 + 1 b ; 3 2 − 2 b + a ) , | A 44 2 ( x ) | > 0 .

We have for all x ∈ ( − 3 2 + 1 b ; 3 2 − 2 b + a ) , x − a < x < 0 < x + a and − x − a < x . Thus B 3 ( x − a ) < B 3 ( x ) and B 3 ( x ) > B 3 ( x + a ) . On the other hand,

L 2 3 , 2 ( x ) = B 3 ( x + 2 b − 2 a ) B 3 ( x + 1 b ) B 3 ( x − 1 b + a ) > 0 and L 2 3 , 4 ( x ) = B 3 ( x − 1 b ) | A 22 2 ( x ) | > 0.

Then (1) and (2) hold.

One has L 2 33 ( x ) − L 2 3 , 2 ( x ) − L 2 3 , 4 ( x ) = [ B 3 ( x − 1 b + a ) − B 3 ( x − 1 b ) ] | A 22 2 ( x ) | − B 3 ( x + 2 b − 2 a ) B 3 ( x + 1 b ) B 3 ( x − 1 b + a ) where B 3 ( x − 1 b ) = 1 2 ( x − 1 b ) 2 + 3 2 ( x − 1 b ) + 9 8 , B 3 ( x + 1 b ) = 1 2 ( x − 1 b ) 2 − 3 2 ( x + 1 b ) + 9 8 and

B 3 ( x − 1 b + a ) = { 1 2 ( x − 1 b + a ) 2 + 3 2 ( x − 1 b + a ) + 9 8 if a ∈ [ 3 4 ,1 ] − ( x − 1 b + a ) 2 + 3 4 if a ∈ [ 1 ; 3 2 ] .

Let g ( x ) : = [ B 3 ( x − 1 b + a ) − B 3 ( x − 1 b ) ] − B 3 ( x + 1 b ) . It is easy to remark that the function g is strictly increasing on ( − 3 2 + 1 b ; 3 2 − 2 b + a ) and g ( − 3 2 + 1 b ) > 0 where

g ( − 3 2 + 1 b ) = { 1 2 ( − 3 + a + 2 b ) ( 3 + a − 2 b ) if a ∈ [ 3 4 ,1 ] − a 2 + 3 a − 2 b 2 + 6 b − 6 if a ∈ [ 1 ; 3 2 ] .

Consequently B 3 ( x − 1 b + a ) − B 3 ( x − 1 b ) > B 3 ( x + 1 b ) .

Let h ( x ) : = | A 22 2 ( x ) | − B 3 ( x + 2 b − 2 a ) B 3 ( x − 1 b + a ) where

A 22 2 ( x ) = ( B 3 ( x + 2 b − 2 a ) B 3 ( x + 2 b − a ) B 3 ( x + 1 b − 2 a ) B 3 ( x + 1 b − a ) ) .

We obtain

min x h ( x ) = h ( 3 2 − 2 b + a ) = B 3 ( 3 2 − a ) [ B 3 ( 3 2 − 1 b ) − B 3 ( 3 2 − 3 b + 2 a ) ] > 0

because 3 2 − 3 b + 2 a < − 3 2 + 1 b ⇒ B 3 ( 3 2 − 3 b + 2 a ) < B 3 ( 3 2 − 1 b ) . Consequently (3) holds.

Therefore | A 22 2 ( x ) | > B 3 ( x + 2 b − 2 a ) B 3 ( x − 1 b + a ) . Hence L 2 33 ( x ) > L 2 3 , 2 ( x ) + L 2 3 , 4 ( x ) .

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

(a) Let k ∈ { 1, ⋯ , m − 1 } . In the first time, we consider the matrix A 22 k ( x ) given by

A 22 k ( x ) = ( B 3 ( x + k b − k a ) B 3 ( x + k b − ( k − 1 ) a ) B 3 ( x + k − 1 b − k a ) B 3 ( x + k − 1 b − ( k − 1 ) a ) ) .

We prove that for all k ∈ { 1, ⋯ , m − 1 } , B 3 ( x + k b − k a ) > B 3 ( x + k b − ( k − 1 ) a ) . For this, we know that − 3 2 < x + k b − k a < 3 2 and x + k b − ( k − 1 ) a > 0 , therefore we have two different cases.

* If x + k b − k a > 0 , we have x + k b − k a < x + k b − ( k − 1 ) a and using the strict decreasing of B 3 on [ 0, 3 2 ] , one has B 3 ( x + k b − k a ) > B 3 ( x + k b − ( k − 1 ) a ) .

* If x + k b − k a ≤ 0 , then − x − k b + k a ≥ 0 and

− x − k b + k a − ( x + k b − ( k − 1 ) a ) = − 2 x − 2 k b + ( 2 k − 1 ) a ≤ 2 k ( a − 1 b ) < 0.

Thus − x − k b + k a < x + k b − ( k − 1 ) a and using the fact that B 3 is symmetric around the origin and strict decreasing on [ 0, 3 2 ] , one has B 3 ( x + k b − k a ) > B 3 ( x + k b − ( k − 1 ) a ) .

Next, we prove that for all k ∈ { 1, ⋯ , m − 2 } , B 3 ( x + k − 1 b − k a ) < B 3 ( x + k − 1 b − ( k − 1 ) a ) . We also know that − 3 2 < x + k − 1 b − ( k − 1 ) a < 3 2 and x + k − 1 b − k a < 0 , therefore we can consider the following two cases.

* If x + k − 1 b − ( k − 1 ) a ≤ 0 , on has x + k − 1 b − k a < x + k − 1 b − ( k − 1 ) a . Therefore by the strict increasing of B 3 on [ − 3 2 ; 0 ] , we obtain B 3 ( x + k − 1 b − k a ) < B 3 ( x + k − 1 b − ( k − 1 ) a ) .

* If x + k − 1 b − ( k − 1 ) a > 0 , then − x − k − 1 b + ( k − 1 ) a ≤ 0 and − x − k − 1 b + ( k − 1 ) a − ( x + k − 1 b − k a ) = − 2 x − 2 ( k − 1 ) b + ( 2 k − 1 ) a ≥ 0 . Indeed,

− 2 x − 2 ( k − 1 ) b + ( 2 k − 1 ) a ≥ − 2 ( k − 1 ) b + ( 2 k − 1 ) a ≥ − ( k − 1 ) 3 + ( 2 m − 3 ) a m − 1 + ( 2 k − 1 ) a = − 3 ( k − 1 ) m − 1 + k + m − 2 m − 1 a ≥ − 3 ( k − 1 ) m − 1 + k + m − 2 m − 1 × 3 ( m − 3 ) 2 ( m − 2 ) = 3 ( m − 2 − k ) 2 ( m − 2 ) ≥ 0.

Thus x + k − 1 b − k a ≤ − x − k − 1 b + ( k − 1 ) a and using the fact that B 3 is symmetric around the origin and strict increasing on [ − 3 2 ; 0 ] , one has B 3 ( x + k − 1 b − k a ) ≤ B 3 ( x + k − 1 b − ( k − 1 ) a ) .

All in all, we conclude that | A 22 k ( x ) | > 0 for all k ∈ { 1, ⋯ , m − 2 } .

For k = m − 1 , we consider the matrix A 22 m − 1 ( x ) given by

A 22 m − 1 ( x ) = ( B 3 ( x + m − 1 b − ( m − 1 ) a ) B 3 ( x + m − 1 b − ( m − 2 ) a ) B 3 ( x + m − 2 b − ( m − 1 ) a ) B 3 ( x + m − 2 b − ( m − 2 ) a ) ) .

We know respectively that B 3 ( x + m − 1 b − ( m − 1 ) a ) > 0 , B 3 ( x + m − 2 b − ( m − 1 ) a ) > 0 , B 3 ( x + m − 2 b − ( m − 2 ) a ) > 0 , B 3 ( x + m − 1 b − ( m − 1 ) a ) > B 3 ( x + m − 1 b − ( m − 2 ) a ) and

B 3 ( x + m − 1 b − ( m − 2 ) a ) = { B 3 ( x + m − 1 b − ( m − 2 ) a ) > 0 if x ∈ [ − a 2 ; 3 2 − m − 1 b + ( m − 2 ) a ) 0 if x ∈ [ 3 2 − m − 1 b + ( m − 2 ) a ; 0 ]

It is easy to see that if x ∈ [ 3 2 − m − 1 b + ( m − 2 ) a ; 0 ] , then | A 22 m − 1 ( x ) | > 0 . To end, we prove that for all x ∈ [ − a 2 ; 3 2 − m − 1 b + ( m − 2 ) a ) , | A 22 m − 1 ( x ) | > 0 .

We also know that − 3 2 < x + m − 2 b − ( m − 2 ) a < 3 2 and x + m − 2 b − ( m − 1 ) a < 0 , therefore we can consider the following two cases.

*If x + m − 2 b − ( m − 2 ) a ≤ 0 , one has x + m − 2 b − ( m − 1 ) a < x + m − 2 b − ( m − 2 ) a . Therefore by the strict increasing of B 3 , we obtain B 3 ( x + m − 2 b − ( m − 1 ) a ) < B 3 ( x + m − 2 b − ( m − 2 ) a ) .

*If x + m − 2 b − ( m − 2 ) a > 0 , then − x − m − 2 b + ( m − 2 ) a ≤ 0 and − x − m − 2 b + ( m − 2 ) a − ( x + m − 2 b − ( m − 1 ) a )

= − 2 x − 2 ( m − 2 ) b + ( 2 m − 3 ) a > − 3 + 2 b + a ≥ 0 . Thus x + m − 2 b − ( m − 1 ) a < − x − m − 2 b + ( m − 2 ) a and using the fact that B 3 is symmetric around the origin and strict increasing, one has

B 3 ( x + m − 2 b − ( m − 1 ) a ) ≤ B 3 ( x + m − 2 b − ( m − 2 ) a ) .

In the second time, we consider, for all k ∈ { 1, ⋯ , m − 1 } , the matrix A 33 k ( x ) given by

For x ∈ [ − a 2 ; − 3 2 − k − 2 b + ( k − 1 ) a ] ∪ [ 3 2 − k b + ( k − 1 ) a ; 0 ] , we have B 3 ( x + k − 2 b − ( k − 1 ) a ) = 0 or B 3 ( x + k b − ( k − 1 ) a ) = 0 and hence

| A 33 k ( x ) | = B 3 ( x + k − 2 b − ( k − 2 ) a ) | A 22 k ( x ) | > 0 or | A 33 k ( x ) | = B 3 ( x + k b − k a ) | A 22 k − 1 ( x ) | > 0.

Let x ∈ ( − 3 2 − k − 2 b + ( k − 1 ) a ; 3 2 − k b + ( k − 1 ) a ) .

We proved previously that B 3 ( x + k b − k a ) > B 3 ( x + k b − ( k − 1 ) a ) for all k ∈ { 1, ⋯ , m − 1 } .

Nest, we prove that B 3 ( x + k − 1 b − k a ) < B 3 ( x + k − 1 b − ( k − 1 ) a ) and B 3 ( x + k − 1 b − ( k − 1 ) a ) > B 3 ( x + k − 1 b − ( k − 2 ) a ) .

We know for all k ∈ { 1, ⋯ , m − 1 } , − 3 2 < x + k − 1 b − ( k − 1 ) a < 3 2 , x + k − 1 b − k a < 0 and x + k − 1 b − ( k − 2 ) a > 0 , therefore we have different following cases.

*If x + k − 1 b − ( k − 1 ) a ≤ 0 , on has x + k − 1 b − k a < x + k − 1 b − ( k − 1 ) a . Therefore by the strict increasing of B 3 on [ − 3 2 ; 0 ] , we obtain B 3 ( x + k − 1 b − k a ) < B 3 ( x + k − 1 b − ( k − 1 ) a ) .

*If x + k − 1 b − ( k − 1 ) a > 0 , then − x − k − 1 b + ( k − 1 ) a < 0 and − x − k − 1 b + ( k − 1 ) a − ( x + k − 1 b − k a ) = − 2 x − 2 ( k − 1 ) b + ( 2 k − 1 ) a > 0 . Indeed,

x ∈ ( − 3 2 − k − 2 b + ( k − 1 ) a ; 3 2 − k b + ( k − 1 ) a ) ⇒ − 2 x − 2 ( k − 1 ) b + ( 2 k − 1 ) a > − 3 + 2 b + a > 0.

Therefore x + k − 1 b − k a < − x − k − 1 b + ( k − 1 ) a and consequently we have

B 3 ( x + k − 1 b − k a ) < B 3 ( x + k − 1 b − ( k − 1 ) a ) .

*If x + k − 1 b − ( k − 1 ) a > 0 , we have x + k − 1 b − ( k − 1 ) a < x + k − 1 b − ( k − 2 ) a and then B 3 ( x + k − 1 b − ( k − 1 ) a ) > B 3 ( x + k − 1 b − ( k − 2 ) a ) .

*If x + k − 1 b − ( k − 1 ) a ≤ 0 , then − x − k − 1 b + ( k − 1 ) a ≥ 0 and − x − k − 1 b + ( k − 1 ) a − ( x + k − 1 b − ( k − 2 ) a ) = − 2 x − 2 ( k − 1 ) b + ( 2 k − 3 ) a < 0 because

x ∈ ( − 3 2 − k − 2 b + ( k − 1 ) a ; 3 2 − k b + ( k − 1 ) a ) ⇒ − 2 x − 2 ( k − 1 ) b + ( 2 k − 3 ) a < 3 − 2 b − a < 0.

Thus − x − k − 1 b + ( k − 1 ) a < x + k − 1 b − ( k − 2 ) a and using the fact that B 3 is symmetric around the origin and strict decreasing on [ 0, 3 2 ] , one has

B 3 ( x + k − 1 b − ( k − 1 ) a ) > B 3 ( x + k − 1 b − ( k − 2 ) a ) .

Let us prove that B 3 ( x + k − 2 b − ( k − 1 ) a ) < B 3 ( x + k − 2 b − ( k − 2 ) a ) for all k ∈ { 1, ⋯ , m − 1 } .

One has B 3 ( x − 1 b ) < B 3 ( x − 1 b + a ) because − 3 2 < x − 1 b < x − 1 b + a < 0 .

Let k ∈ { 2, ⋯ , m − 1 } . We known that − 3 2 < x + k − 2 b − ( k − 2 ) a < 3 2 and x + k − 2 b − ( k − 1 ) a < 0 , therefore we have two cases.

*If x + k − 2 b − ( k − 2 ) a ≤ 0 , then x + k − 2 b − ( k − 1 ) a < x + k − 2 b − ( k − 2 ) a and therefore B 3 ( x + k − 2 b − ( k − 1 ) a ) < B 3 ( x + k − 2 b − ( k − 2 ) a ) .

*If x + k − 2 b − ( k − 2 ) a > 0 , then − x − k − 2 b + ( k − 2 ) a < 0 and − x − k − 2 b + ( k − 2 ) a − ( x + k − 2 b − k a )

= − 2 x − 2 ( k − 2 ) b − ( 2 k − 3 ) a > − 3 + 4 b − a > 0 . Therefore x + k − 2 b − k a < − x − k − 2 b + ( k − 2 ) a and then

B 3 ( x + k − 2 b − ( k − 1 ) a ) < B 3 ( x + k − 2 b − ( k − 2 ) a ) .

Let

p ( x ) = | B 3 ( x + k b − k a ) 0 0 B 3 ( x + k − 2 b − ( k − 2 ) a ) | − | B 3 ( x + k b − ( k − 1 ) a ) 0 B 3 ( x + k − 2 b − ( k − 1 ) a ) B 3 ( x + k − 2 b − ( k − 2 ) a ) |

− | B 3 ( x + k b − k a ) B 3 ( x + k b − ( k − 1 ) a ) 0 B 3 ( x + k − 2 b − ( k − 1 ) a ) | .

A direct computation shows that

p ( x ) = B 3 ( x + k b − k a ) B 3 ( x + k − 2 b − ( k − 2 ) a ) − B 3 ( x + k b − ( k − 1 ) a ) × B 3 ( x + k − 2 b − ( k − 2 ) a ) − B 3 ( x + k b − k a ) B 3 ( x + k − 2 b − ( k − 1 ) a ) = − Δ a B 3 ( x + k b − ( k − 1 ) a ) Δ a B 3 ( x + k − 2 b − ( k − 2 ) a ) + Δ a B 3 ( x + k b − ( k − 2 ) a ) Δ a B 3 ( x + k − 2 b − ( k − 1 ) a ) = Δ a B 3 ( − x − k b + k a ) Δ a B 3 ( x + k − 2 b − ( k − 2 ) a ) − Δ a B 3 ( − x − k b + ( k − 1 ) a ) × Δ a B 3 ( x + k − 2 b − ( k − 1 ) a ) ( one uses Δ a B 3 ( x ) = − Δ a B 3 ( − x + a ) )

Hence to have p ( x ) > 0 , it suffices to show that for all x ∈ ( − 3 2 − k − 2 b + ( k − 1 ) a ; 3 2 − k b + ( k − 1 ) a ) ,

{ Δ a B 3 ( x + k − 2 b − ( k − 2 ) a ) > Δ a B 3 ( x + k − 2 b − ( k − 1 ) a ) Δ a B 3 ( − x − k b + k a ) > Δ a B 3 ( − x − k b + ( k − 1 ) a ) ⇔ { Δ a 2 B 3 ( x + k − 2 b − ( k − 2 ) a ) > 0 Δ a 2 B 3 ( − x − k b + k a ) > 0

This means precisely that Δ a 2 B 3 ( x ) > 0 , x ∈ ( − 3 2 + a , 3 2 − 2 b + a ) which is obtained in Lemma 3.

b) Let k ∈ { 2, ⋯ , m − 1 } and consider the matrix

*If 3 − 3 b + a ≤ 0 , then 3 2 − k b + ( k − 1 ) a ≤ − 3 2 − k − 3 b + ( k − 2 ) a and therefore ∀ x ∈ [ − a 2 ; 0 ] , B 3 ( x + k − 3 b − ( k − 2 ) a ) = 0 or B 3 ( x + k b − ( k − 1 ) a ) = 0 . Thus

| A 44 k ( x ) | = B 3 ( x + k − 3 b − ( k − 3 ) a ) ⋅ | A 33 k ( x ) | > 0 or | A 44 k ( x ) | = B 3 ( x + k b − k a ) ⋅ | A 33 k − 1 ( x ) | > 0.

*If 3 − 3 b + a > 0 , then − 3 2 − k − 3 b + ( k − 2 ) a < 3 2 − k b + ( k − 1 ) a .

When x ∈ [ − a 2 ; − 3 2 − k − 3 b + ( k − 2 ) a ] ∪ [ 3 2 − k b + ( k − 1 ) a ; 0 ] , then B 3 ( x + k − 3 b − ( k − 2 ) a ) = 0 or B 3 ( x + k b − ( k − 1 ) a ) = 0 and therefore

| A 44 k ( x ) | = B 3 ( x + k − 3 b − ( k − 3 ) a ) ⋅ | A 33 k ( x ) | > 0 or | A 44 k ( x ) | = B 3 ( x + k b − k a ) ⋅ | A 33 k − 1 ( x ) | > 0.

We finish by proving that for all x ∈ ( − 3 2 − k − 3 b + ( k − 2 ) a ; 3 2 − k b + ( k − 1 ) a ) , | A 44 k ( x ) | > 0 . We observe that for k = 3 and x ∈ ( − 1 + a ; 1 − 3 b + 2 a ) , then | A 44 3 ( x ) | = | A 44 2 ( x + 1 b − a ) | and for k ∈ { 4, ⋯ , m − 1 } and x ∈ ( − 3 2 − k − 3 b + ( k − 2 ) a ; 3 2 − k b + ( k − 1 ) a ) , then | A 44 k ( x ) | = | A 44 3 ( x + k − 3 b − ( k − 3 ) a ) | > 0 . So we only prove that for all x ∈ ( − 3 2 + 1 b ; 3 2 − 2 b + a ) , | A 44 2 ( x ) | > 0 . Using the condition 3 − 3 b + a > 0 , we consider a ∈ [ 3 4 , 3 2 ] and b ∈ [ 3 3 + a , 4 3 + 2 a ] . In other words, by the Lemma 4, we have this result.

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