Continuous Wavelet Transform of Schwartz Distributions in DL2'(Rn), N1

S S symmetry

Article Continuous Wavelet Transform of Schwartz Distributions in D0 ( n) ≥ L2 R , n 1

Jagdish N. Pandey

School of Mathematics and Statistics, Carleton University, Ottawa, ON K1S 5B6, Canada; [email protected]

n ∞ n Abstract: We define a testing function space DL2 (R ) consisting of a class of C functions defined on R , n ≥ 1 whose every derivtive is L2(Rn) integrable and equip it with a topology generated by a separating ∞ n (k) n { } D 2 ( ) | | = ( ) = k k ∈ D 2 ( ) collection of seminorms γk |k|=0 on L R , where k 0, 1, 2, ... and γk φ φ 2, φ L R . D0 ( n) ≥ We then extend the continuous wavelet transform to distributions in L2 R , n 1 and derive the corresponding wavelet inversion formula interpreting convergence in the weak distributional sense. n n The kernel of our wavelet transform is defined by an element ψ(x) of DL2 (R ) ∩ DL1 (R ), n ≥ 1 which, when integrated along each of the real axes X1, X2, ... Xn vanishes, but none of its moments R xmψ(x)dx is zero; here xm = xm1 xm2 ··· xmn , dx = dx dx ··· dx and m = (m , m , ... m ) and Rn 1 2 n 1 2 n 1 2 n n each of m1, m2,... mn is ≥ 1. The set of such wavelets will be denoted by DM(R ).

Keywords: wavelet transform; continuous wavelet transform; window functions; integral transform of generalized functions; Schwartz distributions

MSC: Primary: 42C40; 46F12; Secondary: 46F05; 46F10

Citation: Pandey, J.N. Continuous 1. Introduction Wavelet Transform of Schwartz (a) Wavelet analysis has now entered into almost every walk of human life [1–5]. Distributions in D0 (Rn), n ≥ 1. L2 There are applications of wavelets in areas such as audio compression, communication, Symmetry 2021, 13, 1106. https:// de-noising, differential equations, ECG compression, FBI ﬁngerprinting, image compres- doi.org/10.3390/sym13071106 sion, radar, speech and video compression, approximation, and so on. Discrete wavelet

Academic Editor: Roman Ger transform has many applications in Engineering and Mathematical sciences. Most notably, it is used for signal coding. Continuous wavelet transform is used in image processing. It Received: 10 May 2021 is an excellent tool for mapping the changing properties of non-stationary signals. Accepted: 9 June 2021 Another recent example of an interesting application of wavelets was the LIGO Published: 22 June 2021 experiment that detected gravitational waves using wavelets for signal analysis. See the paper submitted on 11 February 2016 to “General Relativity and Quantum Cosmology” [6]. Publisher’s Note: MDPI stays neutral These types of works on wavelets sparked the research on continuous wavelet trans- with regard to jurisdictional claims in form of functions and generalized functions. The generalized function space that we chose 0 n published maps and institutional afﬁl- D ( ) for this work is the space L2 R [7,8]. 0 iations. My earlier work in this connection was on the generalized function space D (Rn), n ≥ 1. The disadvantage in this space was that two functions having the same wavelet transform may differ by a constant even though all the moments of the wavelets are non- D0 ( n) ≥ zero. The space L2 R , n 1 does not contain a non-zero constant so that kind of D0 ( n) Copyright: © 2021 by the author. difﬁculty is not encountered with this space. Besides, the space L2 R is different from 0 n Licensee MDPI, Basel, Switzerland. the generalized function space D (R ), n ≥ 1. This article is an open access article (b) The wavelet that we will be dealing with is a variation of one dimensional wavelet 2 distributed under the terms and xe−x . All the even order moments of this wavelet are zero and so two functions having the conditions of the Creative Commons same wavelet transform may differ by a polynomial. The kernel of the wavelet transform Attribution (CC BY) license (https:// √1 x−b is generated by this wavelet with the formula ψ a , a, b real and a 6= 0 where creativecommons.org/licenses/by/ |a| −x2 4.0/). ψ(x) = xe . In order to remove the above mentioned problem we construct a wavelet

Symmetry 2021, 13, 1106. https://doi.org/10.3390/sym13071106 https://www.mdpi.com/journal/symmetry Symmetry 2021, 13, 1106 2 of 11

( + − 2) −x2 R ∞ m( + − 2) −x2 6= ≥ 1 x 2x e such that all the moments −∞ x 1 x 2x e dx 0, m 1 R ∞ 2 −x2 and −∞(1 + x − 2x )e dx = 0. Many other wavelets satisfying this condition can be generated and some examples will be given in the coming section. An interesting point is that this wavelet is the union of a symmetric and an anti-symmetric wavelet as follows 2 2 2 2 (1 + x − 2x2)e−x = xe−x + (1 − 2x2)e−x . The wavelet xe−x is antisymmetric and the 2 wavelet (1 − 2x2)e−x is symmetric and therefore this paper is very well suited to the journal “Symmetry”. (c) In the foregoing deﬁnition of the wavelet transform the kernel of the wavelet transform is 1 x − b p ψ , |a| a

2 where ψ(x) = xe−x . We generalize the kernel ψ to dimensions n > 1 as ψ(x) = −|x|2 1 x−b x1x2 ... xne and the corresponding kernel of the wavelet transform √ ψ . Clearly |a| a n n ψ(x) ∈ DL2 (R ) ∩ DL1 (R ). Therefore, two functions having the same wvelet transform may differ by a polynomial. We now illustrate this fact as follows. Let

f1 = f h t − b mi f = f + . 2 a These two distributions have the same wavelet transform but they may differ by a polynomial involving a constant term. See the calculation below:

t − b m t − b Z t − b m t − b , ψ = ψ dt. a a Rn a a Put Z t − b m x = = |a|x ψ(x)dx, |a| = |a1a2 ... an| a Rn = 0,

when at least one of components m1, m2 ... mn is even and is 6= 0 when each of m1, m2, ... mn is odd. t−b m So f1 and f2 will differ by the polynomial a . Therefore, in order that the uniqueness theorems may hold for the inversion formula for this wavelet transform, we have to select the kernel of our wavelet transform such that none of its moments of order m is zero, m = (m1, m2,... mn), mi ≥ 1 ∀ i = 1, 2, . . . n. The dual of D(Rn) contains a non-zero constant. The wavelet kernel that we are n n choosing belongs to DL1 (R ) ∩ DL2 (R ); as for example x1x2 ... xnφ(x) belongs to this space. ( ) R ∞ ( ) = = The kernel ψ x of our wavelet transform should be such that −∞ ψ x dxi 0, i 1, 2, ... n but none of its moments of order m = (m1, m2, ... mn) where each of m1, m2, ... mn ≥ ( ) = ( ) ∈ D( n) R ∞ ( ) = is 1, is zero. If we take ψ x x1x2 ... xnφ x R then −∞ ψ x dxi 0 but all its R m moments of order m, ie. Rn x ψ(x)dx will not be non-zero. We therefore seek our kernel n ψ(x) ∈ DL2 (R ) such that Z ∞ ψ(x)dxi = 0, i = 1, 2, 3, . . . n −∞

and Z m x ψ(x)dx 6= 0, m = (m1, m2,... mn). Rn In Section3 we will show how such functions are selected or constructed. Symmetry 2021, 13, 1106 3 of 11

2. Background Results It is assumed that the readers are familiar with the elementary theory of distributions. Details of the theory may be found in [9–17]. 2 n A function f ∈ L (R ) is called a window function [18–20] if xi f (x), xixj f (x), xixjxk f (x) i6=j i6=j6=k6=i 2 n ... x1x2 ... xn f (x) belong to L (R ), i, j, k ... all assume values 1, 2, 3, ... n. It is known that such a window function also belongs to L1(Rn). A function f ∈ L2(Rn) is said to be a basic wavelet if it satisﬁes the admissibility condition

Z | fˆ(Λ)|2 dΛ is bounded, (1) Λn |Λ|

where fˆ(Λ) is the Fourier transform of f (x) deﬁned by

1 Z Nn Z Nn−1 Z N2 Z N1 fˆ( ) = ··· f (t)e−iΛ·tdt Λ lim n/2 N1,N2...Nn→∞ (2π) −Nn −Nn−1 −N2 −N1 1 Z ∞ Z ∞ Z ∞ = ··· f (t)e−iΛ·tdt (2π)n/2 −∞ −∞ −∞

Λ = (λ1, λ2, ... λn), t = (t1, t2 ... tn), dt = dt1dt2 ... dtn and the limit is interpreted in the L2(Rn) sense ([21], p. 75). 2 n −(x2+x2+...+x2 ) In L (R ), let us take f (x) = x1x2 ··· xne 1 2 n , then

n λ λ ··· λni fˆ(Λ) = 1 2 e−(λ1+λ2+···+λn)/4. 23n/2 Therefore Z | fˆ(Λ)|2 1 = < dΛ n ∞. Rn |Λ| 4 So f (x) is a basic wavelet in Rn. We now describe some results proved in [20] which will be used in the sequel. These results are being stated for the convenience of our readers.

Theorem 1. Let f ∈ L2(Rn) be a window function on Rn. Then f ∈ L1(Rn) ([19], Theorem 3.1).

n 2 n Theorem 2. Let f : R → C be a L (R ) window function. Let f (λ1, λ2, ... , λn) be the Fourier transform of f deﬁned by

fˆ(λ1, λ2,..., λn) 1 Z = ( f (x x x )e−i(x1λ1+x2λ2+···+xnλn)dx dx dx n/2 1, 2 ... n 1 2 ... n (2π) Rn

Then the following statements are equivalent

(a) fˆ(λ1, λ2,..., λn) = 0 λj=0 Z ∞ (b) f (x1, x2,..., xj,..., xn)dxj = 0, j = 1, 2, 3 . . . n ([19], Theorem 3.2). −∞

Theorem 3. Let f ∈ L2(Rn) be a window function. Assume also that

Z ∞ f (x1, x2,... xi,... xn)dxi = 0 ∀i = 1, 2, . . . , n. −∞ Symmetry 2021, 13, 1106 4 of 11

Then, f satisﬁes the admissibility condition

Z 2 | f (λ1, λ2 ··· λn)| dλ1dλ2, ··· dλn < ∞. Λn |λ1λ2 ··· λn| ([19], Theorem 3.3).

More precisely we have

Z 2 | fˆ(λ1, λ2,..., λn)| dλ1dλ2 ··· dλn Λn |λ1λ2,..., λn| " n n 2 1 2 2 2 ≤ k f k2 + 2 ∑ ||xi f ||2 + 2 ∑ ||xixj f ||2 i=1 i,j=1,i6=j n # 3 2 n 2 + 2 ∑ ||xixjxk f || + ··· + 2 ||x1x2 ··· xn f ||2 . i,j,k=1

Theorem 4. Let f ∈ L2(Rn) be a window function. Then f satisﬁes the admissibility condition if R ∞ and only if −∞ f (x1, x2,..., xn)dxi = 0, i = 1, 2, 3, . . . , n.

This is a corollary to the previous results.

n n Theorem 5. Let φ ∈ DL2 (R ) ∩ DL1 (R ); then φ satisﬁes the admissibility condition if and only if Z ∞ φ(x1, x2,..., xi,..., xn)dxi = 0 ∀i = 1, 2, . . . , n. −∞ Now let us deﬁne a function ψ(x) as follows q −|x|2 2 2 2 ψ(x) = x1x2 ... xne , |x| = x1 + x2 + ··· + xn.

Clearly ψ ∈ L1(Rn). Then ψ(x) is a window function belonging to L2(Rn) and satisfying

Z ∞ ψ(x)dxi = 0 ∀i = 1, 2, . . . n. −∞

Therefore in view of the foregoing results ψ(x) is a wavelet.

∈ D0 ( n) Therefore, we deﬁne the wavelet transform of f L2 R by * + 1 t − b Wf (a, b) = f (t), p ψ , ai 6= 0, i = 1, 2, . . . |a| a

= 0, when ai = 0, for any i = 1, 2, . . . n.

Here a = (a1, a2,... an), b = (b1, b2,... bn). q q |a| = |a1 a2 ... an| t − b t − b t − b t − b ψ = ψ 1 1 , 2 2 ,... n n . a a1 a2 an (c) In the foregoing deﬁnition of the wavelet transform the kernel of the wavelet transform is

1 x − b p ψ , |a| a Symmetry 2021, 13, 1106 5 of 11

2 where ψ(x) = xe−x . We generalize the kernel ψ to dimensions n > 1 as ψ(x) = −|x|2 1 x−b x1x2 ... xne and the corresponding kernel of the wavelet transform as √ ψ . |a| a Clearly ψ(x) ∈ L2(Rn) ∩ L1(Rn). Therefore, two functions having the same wavelet transform may differ by a polynomial. We now illustrate this fact as follows. Let

f1 = f

h t − b mi f = f + . 2 a These two distributions have the same wavelet transform but they differ by a polynomial involving a constant term. See the calculation below:

t − b m t − b Z t − b m t − b , ψ = ψ dt. a a Rn a a Put Z t − b m x = = |a|x ψ(x)dx, |a| = |a1a2 ... an| a Rn = 0,

when at least one of components m1, m2, ... mn is even and is 6= 0 when each of m1, m2, ... mn is odd. m t−b So f1 and f2 will differ by the polynomial a . Therefore, in order that the uniqueness theorems may hold for the inversion formula for this wavelet transform, we have to select the kernel of our wavelet transform such that none of its moments of order m is zero, m = (m1, m2,... mn), mi ≥ 1 ∀ i = 1, 2, . . . n. n The dual of DL2 (R ) does not contain a non-zero constant. The wavelet kernel that n n −|x|2 we are choosing belongs to DL2 (R ) ∩ DL1 (R ); as for example x1x2 ··· xn e belongs n 2 xi n n to this space, but ∏ q does not belong to the space DL2 (R ) ∩ DL1 (R ). i=1 6 1 + xi ( ) R ∞ ( ) = = The kernel ψ x of our wavelet transform should be such that −∞ ψ x dxi 0, i 1, 2, ... n but none of its moments of order m = (m1, m2, ... mn) where each of m1, m2, ... mn ≥ ( ) = −|x|2 R ∞ ( ) = is 1, is zero. If we take ψ x x1 x2 ... xne then −∞ x dxi 0 but all its moments R m of order m, i.e., Rn x ψ(x)dx will not be non-zero. We therefore seek our kernel ψ(x) ∈ n n DL2 (R ) ∩ DL1 (R ) such that Z ∞ ψ(x)dxi = 0, i = 1, 2, 3, . . . n. −∞

and Z m x ψ(x)dx 6= 0, m = (m1, m2,... mn) Rn and mi ≥ 1, i = 1, 2, . . . n. In Section3 we will show how such a wavelet kernel is constructed.

n 3. Construction of Functions in the Space D 2 (R ) Which Is a Wavelet Such That Z L m φ(x)x dx 6= 0, m = (m1, m2, ··· mn); each mi ≥ 1, i = 1, 2, . . . n Rn n I.e., Construction of functions ψ(x) ∈ DM(R ), n ≥ 1. In dimension n = 1 one such function is given as:

2 φ(x) = (1 + x − kx2)e−x .

The constant k is so selected that Symmetry 2021, 13, 1106 6 of 11

Z ∞ φ(x)dx = 0. −∞

Therefore, Z ∞ 2 e−x dx k = −∞ = 2 = k . Z ∞ 2 0 x2e−x dx −∞ A somewhat trivial construction of such a function φ(x) in n dimension can be a function ψ(x) given by n 2 2 −xi ψ(x) = ∏(1 + xi − 2xi )e . i=1 One can see that Z ∞ ψ(x)dxi = 0 ∀i = 1, 2, . . . n −∞

and Z ψ(x)xmdx 6= 0. Rn

for m1, m2,... mn ≥ 1, m = (m1, m2, ··· mn). We now give a non-trivial construction of such a wavelet as follows: n = 2 2 2 h i −(x1+x2) 2 2 ψ(x) = e 1 + x1 + x2 + x1x2 − k0(x1 + x2) .

We select the same constant k0 as before, i.e., k0 = 2. Clearly, integration along X1 and X2 respectively gives Z ∞ ψ(x)dxi = 0, i = 1, 2. −∞ Veriﬁcation of the fact that Z ∞ Z ∞ m1 m2 ψ(x)x1 x2 dx1dx2; m1 ≥ 1, m2 ≥ 1 −∞ −∞

is non-zero is easy and is done as follows:

(i) m1, m2 both even, Z ∞ Z ∞ m1 m2 ψ(x)x1 x2 dx1dx2 −∞ −∞ Z ∞ Z ∞ 2 h i −|x| m1 m2 2 2 m1 m2 = e x1 x2 − k0(x1 + x2)x1 x2 dx 6= 0. −∞ −∞

(ii) m1, m2 both odd: Z ∞ Z ∞ m1 m2 ψ(x)x1 x2 dx1dx2 −∞ −∞ Z ∞ Z ∞ 2 2 m +1 m +1 −(x1+x2) 1 2 = e x1 x2 dx1dx2 > 0. −∞ −∞

(iii) m1 even and m2 is odd Z Z ∞ Z ∞ m m 2 2 m m +1 1 2 −(x1+x2) 1 2 ψ(x) x1 x2 dx1dx2 = e x1 x2 dx1dx2 > 0. R2 −∞ −∞

when m1 odd and m2 even Z Z m m 2 2 m m +1 1 2 −(x1+x2) 2 1 ψ(x) x1 x2 dx1dx2 = e x1 x2 dx1dx2 > 0. R2 R2

n = 3. We then deﬁne ψ(x) as Symmetry 2021, 13, 1106 7 of 11

2 2 2 h i −(x1+x2+x3) 2 2 2 ψ(x) = e 1 + x1 + x2 + x3 + x1x2 + x2x3 + x3x1 + x1x2x3 − k0(x1 + x2 + x3) .

The integral of ψ(x) with respect to x1, x2, x3 along the axes X1, X2, X3 respectively is zero. Z m ψ(x)x dx 6= 0 for each m1, m2, m3 ≥ 1. R3 This fact can be veriﬁed similarly as in the case n = 2. Proceeding this way, a non- trivial construction of the function ψ(x) can be done in any dimension n > 1.

4. Main Theorem We hereby quote a theorem proved in ([11], p. 51, [17], p. 137) which plays a crucial role in the proof of our main theorem.

0 n ∞ n Theorem 6. Let f ∈ D (R ). We can find a sequence { fk(x)}k=1 of functions in D(R ) such that

n lim h fk(x), φ(x)i = h f , φi, ∀φ ∈ D(R ). k→∞

0 This fact is expressed by saying that D(Rn) is dense in D (Rn). If is well known that by identifying 0 φ(x) ∈ D as a regular distribution, D(Rn) ⊂ D (Rn), ([11], p. 51, [17], p. 137).

∈ D0 ( n) { ( )}∞ Corollary 1 (Corollary to Theorem 6). Let f L2 R . We can ﬁnd a sequence fk x k=1 of functions in D(Rn) such that

n lim h fk(x), φ(x)i = h f (x), φ(x)i ∀ φ ∈ D(R ). k→∞ D( n) D0( n) D0 ( n) D0 ( n) ⊂ This fact is expressed by saying that R is dense in R and so in L2 R as L2 R 0 D (Rn) with identiﬁcation similar to given in Theorem6.

n n Lemma 1. Let ψ be a wavelet belonging to the space DL2 (R ) ∩ DL1 (R ), n ≥ 1 and f ∈ D0 ( n) ( ) L2 R then the wavelet transform F a, b of the distribution f with respect to wavelet function x−b ψ a is deﬁned by * + 1 x − b F(a, b) = f (x), p ψ , |a| a n x, a, b ∈ R , a 6= 0. i.e., each |ai| > 0, i = 1, 2, ··· n.

We wish to prove that F(a, b) is a continuous function of a, b ; a, b ∈ Rn. D E Proof. We can write F(a, b) = √1 f (x), ψ( x−b ) so it is enough to prove the continuity |a| a x−b of h f (x), ψ( a )i = G(a, b) (say)

x − (b + ∆b) x − b G(a + ∆a, b + ∆b) − G(a, b) = f (x), ψ − ψ . a + ∆a a

x−(b+∆b) x−b n So we need only show that ψ a+∆a − ψ( a ) → 0 in the topology of DL2 (R ) as ∆a, ∆b → 0 independently of each other. Now

x − (b + ∆b) x − b Dk ψ − ψ = I (say). x a + ∆a a

Therefore 1 x − (b + ∆b) 1 x − b I = ψ(k) − ψ(k) . (a + ∆a)k (a + ∆a ak a Symmetry 2021, 13, 1106 8 of 11

−( + ) Note that x−b is x1−b1 , x2−b2 , ··· xn−bn and a similar explanation for x b ∆b . a a1 a2 an a+∆a

x = (x1, x2,... xn)

a = (a1, a2,... bn)

b = (b1, b2,... bn)

k = (k1, k2,... kn).

In view of the mean value theorem of differential calculus of n-variables there exists a number 0 < θ < 1 such that ([22], p. 483) " n ψ(k+1) x − (b + θ∆b) x − (b + θ∆b ) I = − i i i ∆a ∑ k + ( + )2 i i=1 (a + θ∆a) a θ∆a ai θ∆ai x − (b + θ∆b) k 1 − k i a ψ k ∆ i a + θ∆a (a + θ∆a) (ai + θ∆ai) # ψ(k+1) x − (b + θ∆b) ∆b − i k . (a + θ∆a) a + θ∆a (ai + θ∆ai)

n → 0 in DL2 (R ) as ∆a and ∆b → 0 independently of each other. Convergence is with n respect to x in the topology of DL2 (R ). D0 ( n) D ( n) The dual space L2 R of L2 R does not contain a non-zero constant, therefore 0 n 0 n we will not use the notation D 2 (R ); this notation will also mean the space D 2 (R ). LF L Our main theorem is stated and proved as follows.

∈ D0 ( n) D ( n) ∩ D ( n) Theorem 7. Let f L2 R and ψ be a wavelet belonging to the space L2 R L1 R ∈ D0 ( n) ∈ D ( n) ∩ then the wavelet transform of f L2 R with respect to the wavelet kernel ψ L2 R n DL1 (R ) is deﬁned as * + 1 t − b Wf (a, b) = f (t), p ψ |a| a or

Wf (a1, a2,..., an, b1, b2,..., bn) * + 1 t1 − b1 t2 − bn tn − bn = f (t1, t2,..., tn), p ψ , ,..., |a1a2 ··· an| a1 a2 an

R ∞ ai and bi are real and ai 6= 0 ∀ i = 1, 2, ... , n. It is asumed that −∞ ψ(x)dxi = 0 ∀ i = 1, 2, . . . n and R xmφ(x)dx 6= 0, for m = (m , m ,..., m ) and m ≥ 1, i = 1, 2, . . . n. Rn 1 2 n i Then − − ψ( t b ) R A + R η R B 1 f (t) √ a x−b dbda√ (x) η −A −B C , ψ a 2 , φ ψ |a| a |a| (2) −→ h f (x), φ(x)i as A, B → ∞ and η → 0+

Here, when we say B → ∞, it means that all the components B1, B2, ... Bn of B → ∞ independently of each other and similar notation for A → ∞ and η → 0+ means that all the components η1, η2,... ηn of η → 0 independently of each other. In (2) db = db1db2 ... dbn and the integration is being performed with respect to R Bn R B2 R B1 variables b , b2, ... bn with the corresponding limit terms being ··· and da = 1 −Bn −B2 −B1 da1da2 ... dan and the integration is being performed with respect to variables a1, a2, ... an with the corresponding limit terms being Symmetry 2021, 13, 1106 9 of 11

Z An Z −ηn Z A2 Z −η2 Z A1 Z −η1 + ··· + + . ηn −An η2 −A2 η1 −A1

D( n) D0 ( n) { ( )}∞ D( n) Proof. Since R is dense in L2 R we can find a sequence fn t n=1 in R such that n lim h fk(t), φ(t)i = h f (t), φ(t)i ∀φ ∈ DL2 (R ) k→∞

Now Z A Z −η 1 Z B t − b x − b dbda + ( ) ( ) fk t , ψ ψ 2 , φ x (3) η −A Cψ −B a a a |a|

Z 2 n |ψˆ(Λ)| Cψ = (2π) dΛ < ∞ (admissibility condition) Λn |Λ| 2 n ψ(λ1, λ2,... λn) = ψˆ(Λ), the Fourier transform of a window function ψ(x) ∈ L (R ).

1 Z A Z −ηZ B Z C x−b t−b dxdbda = ( ) + ( ) ¯ fk t , φ x ψ ψ 2 (4) Cψ η −A −B −C a a a |a|

[using Fubini’s Theorem]

Here also dx = dx1dx2 ... dxn and integration is being performed with respect to vari- Z Cn Z C2 Z C1 ables x1, x2, ... xn with the corresponding limit terms being ··· −Cn −C2 −C1 C = (C1, C2, ... Cn) and C → ∞ means all the components C1, C2 ... Cn of C → ∞ independently of each other. Now letting η → 0+, A, B, C → ∞ we see that D E ( ) ( ) 1 R ∞ R ∞ R ∞ ( ) ¯ x−b t−b dxdbda fk t , P C −∞ −∞ −∞ φ x ψ a ψ a 2| | ψ a a (5) = h fk(t), φ(t)i

Z ∞ Z ∞ Z ∞ Z ∞ ([18], Theorem 4.2). Each of the integral sign means ··· and similar −∞ −∞ −∞ −∞ n times) Z ∞ meaning to the integral sign from now on. Therefore, from (3) and (5) we get −∞

1 Z A Z −ηZ B t−b x−b dbda + ( ) ( ) lim fk t , ψ ψ 2 , φ x (6) A, B → ∞ Cψ η −A −B a a a |a| η → 0+

= h fk(x), φ(x)i. 2 n The integral in (6) converges to fk(x) in L (R ), ([19], Theorem 4.2). So D D E E ( ) 1 R ∞ R ∞ ( ) t−b x−b dbda ( ) P C −∞ −∞ fk t , ψ a ψ a 2| | , φ x ψ a a (7) = h fk(x), φ(x)i. 2 n The integral in (7) converges to fk(x) in L (R ). Now letting k → ∞ we get (P) 1 R ∞ R ∞ f (t) √1 t−b x−b dbda√ (x) C −∞ −∞ , ψ a ψ a 2 , φ ψ |a| a |a| (8) = h f (x), φ(x)i.

([19], Theorem 4.2). The L.H.S. expressions in (2) and (8) are meaningful in view of Lemma1. Symmetry 2021, 13, 1106 10 of 11

5. Conclusions In order to deal with the wavelet transform of elements of D0 we have to ﬁnd wavelet R ∞ ( ) = function ψ(x) in D satisfying the condition −∞ ψ x dx 0. It turned out that with φ(x) = − 1 e 1−x2 |x| < 1 R ∞ we construct a function ψ(x) = xφ(x), [18]. Then − ψ(x)dx = 0 0 |x| ≥ 1 ∞ and so it is a wavelet in view of results given in Section2. The corresponding wavelet transform kernel will be √1 ψ x−b , a, b real and a 6= 0. All even order moments of ψ(x) |a| a is zero so two functions having the same wavelet transform will differ by a polynomial plus a constant; therefore we construct wavelet φ(x) in D such that none of its moments of order ≥ 1 is zero. It turned out that one such wavelet is (1 + x − 2x2)φ(x) and the result is generalized in n dimension n > 1, many other functions were constructed. The disadvantage with this wavelet was that two functions having the same wavelet transform could differ by a constant. Bearing with the fault in our technique we generalized this result to dimension n > 1 and corrected this fault by deleting all non-zero constants from 0 the space D (Rn), n ≥ 1 [18]. D0 ( n) If we look into the generalized function space L2 R we ﬁnd that this space does not contain a non-zero constant. For this reason, this space is quite interesting and using technique similar to that used for the space D0(Rn), n ≥ 1 we construct wavelet function n φ ∈ DL2 (R ) whose all moments of order m = (m1, m2, ... mn) each of m1, m2, ... mn is ≥ 1 D0 ( n) ≥ are non-zero. We then proved the wavelet inversion formula for the space L2 R , n 1 using these results derived. Uniqueness theorem for the inversion formula then follows. There are many applications of wavelets and continuous wavelet transforms which are mentioned in the beginning part of the introduction. The author expresses his gratitude to two referees for their constructive criticisms.

Funding: This research received no external funding. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Conﬂicts of Interest: The author declares no conﬂict of interest.

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