The Continuous Wavelet Transform and Window Functions

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 143, Number 11, November 2015, Pages 4759–4773 http://dx.doi.org/10.1090/proc/12590 Article electronically published on July 24, 2015

THE CONTINUOUS WAVELET TRANSFORM AND WINDOW FUNCTIONS

J. N. PANDEY AND S. K. UPADHYAY

(Communicated by Ken Ono)

Abstract. We define a window function ψ as an element of L2(Rn)satisfying certain boundedness properties with respect to the L2(Rn)normand prove that it satisfies the admissibility condition if and only if the integral of ψ(x1,x2, ··· ,xn) with respect to each of the variables x1,x2, ··· ,xn along the real line is zero. We also prove that each of the window functions is an element of L1(Rn). A function ψ ∈ L2(Rn) satisfying the admissibility condition is a wavelet. We define the wavelet transform of f ∈ L2(Rn)(which is a window function) with respect to the wavelet ψ ∈ L2(Rn) and prove an inversion formula interpreting convergence in L2(Rn). It is also proved that at a point of continuity of f the convergence of our wavelet inversion formula is in a pointwise sense.

1. Introduction

The necessity for the inversion formula for the wavelet transform in dimensions higher than 1 has long been felt. Before the inversion formula for a two-dimensional wavelet transform was known, workers in image processing used to separate two- dimensional wavelets into the product of two one-dimensional wavelets, thereby making the job very simple [12]. But even two-dimensional wavelets are sometimes hard to separate and are sometimes impossible. Consider the wavelet ψ(x1,x2)in R2: −(x2+1)(x2+1) 2 ψ(x1,x2)=x1x2e 1 2 ∈ S(R ). This cannot be separated into the product of two one-dimensional wavelets so the use of the inversion formula for the wavelets in dimensions higher than 1 was needed by workers in image processing. Consider another wavelet in R2 given by − 1 1−(x2+x2) x1x2e 1 2 , |x| < 1, ψ(x1,x2)= 0, |x|≥1, | | 2 2 x = x1 + x2.

Received by the editors April 17, 2014 and, in revised form, July 2, 2014. 2010 Mathematics Subject Classiﬁcation. Primary 46F12; Secondary 46F05, 46F10. Key words and phrases. Continuous wavelet transform, Fourier inversion theory, inverse wavelet transform.

c 2015 American Mathematical Society 4759

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This wavelet is a wavelet of compact support in R2 and cannot be separated. Now consider ⎧ ⎨ ≤ ≤ 1 1, 0 x 2 , (1.1) ψ(x)= −1, 1

(1.2) ψ(x1,x2)=ψ(x1)ψ(x2). So one can work with the one-dimensional wavelet inversion formula in this case but not in the aforesaid previous two cases, as those are not the wavelets of separable type [12, p. 331]. Many authors worked on the inversion formula for the n-dimensional wavelet transform such as Daubechies [5], Meyer [7], Pathak [11], Keinert [6] and others. But the most notable amongst them are the works of Dauchechies [5, pp. 33–34] and Meyers [7, pp. 125–126]. We have proved the following wavelet inversion formula for f ∈ L2(Rn). Our wavelet is a window function whose integral along each of the axes is zero, and our inversion formula is as follows: − 1 | |−1/2 x b da db a ψ( )Wf (a, b) 2 = f, Cψ Rn Rn a |a| where 1 t − b Wf (a, b)= f(t),ψ( ) |a| a

and |a| = |a1 · a2 ···an|. If we take x =(x1,x2,...,xn), Λ=(λ1,λ2,...,λn)and |Λ| = |λ1 · λ2 ···λn|,then | ˆ |2 n ψ(Λ) ∞ Cψ =(2π) | | dΛ < (admissibility condition), Λn Λ where ˆ ˆ 2 n ψ(λ1,λ2, ··· ,λn)=ψ(Λ), the Fourier transform of ψ(x) ∈ L (R ). The convergence is interpreted in the L2(Rn) sense and at a point of continuity of f the convergence is in the pointwise sense. Note that in our derivation as shown later, we take a =(a1,a2,...,an) whereas Daubechies and Meyer take a =(a,a,...,a) and we do not choose ψ as spherically symmetric. In our case a1,a2,...,an are all non-zero real numbers, whereas in their cases a>0. We have used the Fourier inversion theorem to prove the wavelet inversion theorem. The advantage of this method is that it is simple and to prove the pointwise convergence only the continuity of f is required and the continuity of the wavelet ψ at a point x = x0 concerned is not required. To prove the pointwise convergence using the Hilbert space technique, we require the continuity of the function f and the wavelet ψ both at a point x = x0 concerned [4, p. 63]. It sounds quite strange but that is the way it is! Our inversion formula is valid over Rn × Rn whereas the formulas derived by Daubechies and Meyer are valid only at R × Rn. Thus our formula is more general than their inversion formula.

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We will give a characterisation of a subclass of functions belonging to L2(Rn) ∩ L1(Rn) in the proof of the Corollary to Theorem 3.3, which are wavelets. This makes the construction of wavelets useful to applied scientists much easier.

2. Definitions and preliminaries Definition 2.1. A function f ∈ L2(Rn) is called a window function if it satisfies the following conditions: 2 n (1) x1f,x2f,...,xnf all belong to L (R ). 2 n (2) xixj f ∈ L (R ) for all i, j =1, 2,...,n, i= j. 2 n (3) xixj xkf ∈ L (R ) for all i, j, k =1, 2,...,n, i = j = k = i.Notethat i = j = k = i implies that i, j, k are all different, i = j = k may imply that i and k could be equal. Finally we have 2 n (n) x1x2 ...xn f ∈ L (R ). Here, the lower suffixes in a term are all different. Let us illustrate this definition with reference to n =2.Soiff ∈ L2(R2)wemusthave 2 2 x1f,x2f ∈ L (R ), 2 2 x1x2f ∈ L (R ). Example 2.2. Define ⎧ ⎨ ≤ ≤ 1 1, 0 x 2 , ψ(x)= −1, − 1

−(x2+x2+···+x2 ) 2 n Example 2.3. ψ(x)=(x1,x2...xn)e 1 2 n is a window function in L (R ), and is a wavelet.

The wavelet ψ that we have chosen satisfies the following conditions: (1) ψ ∈ L2(Rn). (2) ψ is a window function as defined above. ∞ ∀ (3) −∞ ψ(x)dxi =0, i =1, 2,...,n. We will show that under the set of conditions (1), (2) and (3) the wavelet kernel ψ satisfies the admissibility condition | ˆ |2 ψ(Λ) ∞ | | dΛ < , Λn Λ which enables us to prove the aforesaid wavelet inversion formula.

3. Orthants and pseudo-orthants Let Rn stand for the n-dimensional Euclidean space. Then Rn = R×R×···×R, n times (Cartesian product): n Rn R ∪ R (3.1) = |xi|≥a |xi|≤a ,a>0, i=1 R ∈ Rn | |≥ |xi|≥a = x : xi a, x =(x1,x2,...,xn) ,

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where i =1, 2,...,n. | |≥ | |≥ | |≥ | |≤ | |≤ | | Let k = x : xj1 a, xj2 a,..., xjk a; xjk+1 a, xjk+2 a,..., xjn ≤ a ,wherej1,j2,...,jn are permutations of 1, 2, 3,...,n; k is known as a pseudo- Rn orthant and when a =0, k is an orthant in . Rn n n ∈ Now = k=0 k . Then the inner union has k terms for a ﬁxed k n n ··· (0 ,1, 2,...,n). Varying k from 0, 1, 2,...,n, we see that there are 0 + 1 + + n n Rn n =2 pseudo-orthants. The total number of pseudo-orthants in can be easily ﬁgured out from the representation (3.1). There are n factors in the representation (3.1) of Rn and there are two elements in each of the factors so there will be 2n pseudo-orthants in Rn. From now on we will choose a = 1 as there is no loss of generality in doing so. Theorem 3.1. Let f ∈ L2(Rn) be a window function on Rn.Thenf ∈ L1(Rn).

Proof. By using Holder’s inequality one can see that |f(x)|dx k | ··· | 1 = xj1 xj2 xjk f(x) dx |x x ···x | k j1 j2 jk 1 1 1 ≤ | ··· |2 2 2 xj1 xj2 xjk f(x) dx 2 dx |x x ···x | k k j1 j2 jk 1 1 1 ≤ | ··· |2 2 |2 2 xj1 xj2 xjk f(x) dx dx Rn |x x ···x k j1 j2 jk 1 1 ≤ ··· 2 xj1 xj2 xjk f(x) 2 2 2 2 dx x x ···x k j1 j2 jk −1 +∞ 1 1 ≤ ··· 2 xj1 xj2 xjk f 2 + 2 dxj1 −∞ 1 x j1 −1 +∞ −1 +∞ 1 1 1 1 × + dx 2 ··· + dx 2 2 j2 2 jk −∞ 1 x −∞ 1 x j2 jk 1 1 1 1 × ··· 2 dxjk+1 dxjk+2 dxjk+n−k −1 −1 −1 k n−k ≤ ··· 2 · 2 xj1 xj2 xjk f 22 2 n ≤ 2 ··· 2 xj1 xj2 xjk f 2, and n | | ≤ 2 ··· ··· f(x) dx 2 xj1 xj2 xji xjk f 2, k n where the number of terms in the R.H.S. summation is k . Therefore, n | | | | f(x) dx = f(x) dx Rn k=0 k n n n (3.2) ≤ 2 2 f 2 + xif 2 + xixj f 2 + ···+ x1x2 ···xnf 2 . i=1 i,j=1,i= j

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Since f is a window function, each of the terms in (3.2) is bounded. Therefore | | ∈ 1 Rn Rn f(x) dx is bounded; i.e. f L ( ). For the details of the proof one can see [10].

2 n ˆ Corollary. Let f ∈ L (R ) be a window function. Then f(λ1,λ2,...,λn) is a ˆ continuous function of λ1,λ2, ··· ,λn,wheref is the Fourier transform of f.

Note that Theorem 3.1 cannot be proved by breaking Rn into the unit sphere with the centre at the origin and a complement of it with respect to Rn, hence there lies the importance of our window function technique.

n 2 n ˆ Theorem 3.2. Let f : R → C be a L (R ) window function. Let f(λ1,λ2,...,λn) be the Fourier transform of f deﬁned by fˆ(λ ,λ ,...,λ ) 1 2 n 1 − ··· = 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 two statements are equivalent. ˆ (a) f(λ1,λ2,...,λn) =0. λ =0 ∞ j

(b) f(x1,x2,...,xj ,...,xn)dxj =0, j =1, 2, 3,.... −∞

Proof. The proof is very simple and so is omitted.

Plancherel formula ([3, p. 107] and [1, p. 75]). Suppose that f and g ∈ L2. −1 −1 Then Ff, g = f,Fg, Ff, FgL2 = f,gL2 and F f, F gL2 = f,gL2 . In particular

Ff L2 = f L2 . ∞ −1 Here, u, v L2 = −∞ u(t)v(t)dt. F, F are deﬁned below.

Fourier inversion theorem. Let x =(x1,x2,...,xn) and ω =(ω1,ω2,...,ωn) be elements of Rn and f ∈ L2(Rn).Then 1 N F −1(fˆ)(x)=l.i.m. fˆ(ω)eiω·xdω = f(x). →∞ n/2 N (2π) −N

This deﬁnes convergence in L2(Rn) and is called the limit in the mean (l.i.m.) [1, p. 75]. Here fˆ(ω) is the Fourier transform of f deﬁned by 1 N fˆ(ω)=l.i.m. f(x)e−iω·xdx, →∞ n/2 N (2π) −N

where ω · x = ω1x1 + ω2x2 + ···+ ωnxn, N =(N1,N2,...,Nn)andN →∞implies that each of the components of N tend to ∞ independently of each other. This is a well-known result, which was proved in several books such as [1,9]. At the point of continuity of f the convergence in the above inversion formula is proved in the pointwise sense. The proof is given using the Plancherel formula Ff,Fg = f,g, ∀ f,g ∈ L2(Rn),

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{ }∞ →∞ and replacing g by the elements of the sequence gm(x) m=1 and letting m ; elements g are deﬁned as follows: m n || − || ≤ 1 m if xm x0 1 2m , gm(x)= || − || 1 0ifxm x0 1 > 2m ,

1 2 n x0 =(x0,x0,...,x0 ), | | | | ··· | | x 1 =max( x1 , x2 , , xn ), 1 1 1 x = x1 + ,x2 + , ··· ,xn + . m 0 2m 0 2m 0 2m || || | | 2 2 ··· 2 The norm 1 and x = x1 + x2 + + xn = x generate the same topology over Rn as x ≤ x ≤n x . 1 1 1 iωx0 The Fourier inversion formula (2π)n/2 Rn F (ω)e dw = f(x0)withpointwisecon- vergence at x = x0 follows letting m →∞in Ff,Fgm = f,gm. We assume n that f is continuous at x0 ∈ R . Details of the proof are omitted and the proof is left as a simple and interesting exercise for the readers. Admissibility condition. Let f ∈ L2(Rn). We say that f satisﬁes the admissi- ˆ bility condition if its Fourier transform f(λ1,λ2,...,λn) satisﬁes the condition ∞ ∞ ∞ ˆ 2 |f(λ1,λ2,...,λn)| (3.3) ··· dλ1dλ2 ...dλn < ∞, −∞ −∞ −∞ |λ1λ2 ...λn| and then the function f is said to be a basic wavelet.

We know that the Fourier transform fˆ(λ1,λ2,...,λn) of a window function f is a continuous function of n variables λ1,λ2,...,λn. A window function belonging to L2(Rn) is a regular (basic) wavelet function if it satisﬁes the admissibility condition | ˆ |2 f(λ1,λ2,...,λn) ∞ (3.4) | | dλ1dλ2 ...dλn < . Λn λ1λ2 ...λn ˆ So we see that f should be zero at each of the points (0,λ1,λ2,...,λn), (λ1, 0,λ2,...,λn), ...,(λ1,λ2,...,λn−1, 0), i.e. ˆ ˆ (3.5) f(0,λ2,λ3,...,λn), f(λ1, 0,λ3,λ4,...,λn) ... and ˆ f(λ1,λ2,...,λn−1, 0) are all zero. By Theorem 3.2 ∞ ˆ f(0,λ2,λ3,...,λn)=0⇔ f(x1,x2,...,xn)dx1 =0, −∞ ∞ ˆ (3.6) f(λ2, 0,λ3,...,λn)=0⇔ f(x1,x2,...,xn)dx2 =0, −∞ ∞ ˆ f(λ2,λ3,...,λn, 0) = 0 ⇔ f(x1,x2,...,xn)dxn =0. −∞ If the window function f is a regular (basic) wavelet, then it satisﬁes (3.4) ⇒ (3.5) ⇒ (3.6). So we propose the following theorem.

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Theorem 3.3. Let f ∈ L2(Rn) be a window function. Assume also that ∞

f(x1,x2,...,xi,xn)dxi =0∀ i =1, 2,...,n. −∞ Then f satisﬁes the admissibility condition | ˆ |2 f(λ1,λ2,...,λn) ··· ∞ (3.7) | | dλ1dλ2 dλn < . Λn λ1λ2 ...λn More precisely we have ˆ 2 |f(λ1,λ2,...,λn)| dλ1dλ2 ...dλn n |λ λ ...λ | Λ 1 2 n n n ≤ 2 1 2 2 2 (3.8) f 2 +2 xif 2 +2 xixj f 2 i=1 i,j=1,i= j ∞ 3 || ||2 ··· n 2 +2 xixj xkf 2 + +2 x1x2 ...xn)f 2 . i,j,k=1 i= j= k= i

Note that all the terms in the R.H.S. of (3.8) are bounded by virtue of the fact that f is a window function. Let us recall some notation as follows:

Λ=(λ1,λ2,λ3, ··· ,λn)and|Λ| = |λ1λ2 ···λn|. ˆ ˆ 2 n f(λ1,λ2, ··· ,λn)=f(Λ), the Fourier transform of f(x) ∈ L (R ). n Here x =(x1,x2,...,xn)andΛ will stand for the n-dimensional Euclidean space where (λ1,λ2, ··· ,λn)representsthecoordinatesofapointofitingeneral. We also introduce the symbols Λ0, Λ1, Λ2, ··· , Λk, ··· , Λn as follows: | |≤ ··· | |≥ ··· Λk =[(λ1,λ2,...,λn): λji 1,i=1, 2, ,k; λji 1,i= k +1,k+2, ,n] , | |≥ ··· ··· Λ0 =[(λ1,λ2,...,λn): λji 1,i=1, 2, 3, ,k, ,n] , | |≤ ··· Λn =[(λ1,λ2,...,λn): λji 1,i=1, 2, 3, ,n] . n Taking all variations of suﬃxes of λji we can see that Λk can represent k pseudo-orthants. The regions (pseudo-orthants) Λ0, Λ1, Λ2, ··· , Λk, ··· , Λn are all contained in Λn. It is easy to show that n Λ =Λ0 ∪ ( Λ1) ∪ ( Λ2) ∪···∪( Λk) ∪···∪Λn.

The symbol dΛ stands for dλ1dλ2 ···dλk ···dλn.Now |fˆ(Λ)|2dΛ |fˆ(Λ)|2 |fˆ(Λ)|2dΛ (3.9) | | = | | dΛ+ | | Λn Λ Λ Λ Λ Λ 0 1 | ˆ |2 | ˆ |2 f(Λ) ··· f(Λ) + | | dΛ+ + | | dΛ Λ Λ Λ Λ 2 n |fˆ(Λ)|2 |fˆ(Λ)|2 = dΛ+ dΛ Λ0 | | Λ1 | | n Λ n Λ (0)=1 term (1) terms

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4766 J. N. PANDEY AND S. K. UPADHYAY |fˆ(Λ)|2 |fˆ(Λ)|2 + dΛ+···+ dΛ Λ2 |Λ| Λk |Λ| (n) terms (n) terms 2 k |fˆ(Λ)|2 + ···+ dΛ, Λn | | n Λ (n)=1 term |fˆ(Λ)|2 |fˆ(Λ)|2 dΛ= dΛ ≤ |fˆ(Λ)|2dΛ | | | |≥ | | Λ Λ λi 1 Λ Λn 0 i=1,2,...,n = |f(x)|2dx Rn || ||2 = f 2 [Plancherel’s Theorem]. So | ˆ |2 f(Λ) ≤|| ||2 (3.10) | | dΛ f 2. Λ0 Λ

| ˆ |2 Next we can find a bound for f(Λ) dΛ as follows. The function Λn |Λ| fˆ(λ1,λ2, ··· ,λi−1,λi,λi+1, ··· ,λn)vanishesatpointsλi =0,i =1, 2, 3, ··· ,n. Therefore all the first partials of this function with respect to λ1,λ2, ··· ,λi−1, λi+1, ··· ,λn vanish. Therefore, λn λ2 λ1 n ˆ ··· ∂ f(Λ) f(Λ) = ··· dΛ. 0 0 0 ∂λ1∂λ2 ∂λn Now by Holder’s inequality we have 2 λn λ2 λ1 ∂nfˆ(Λ) λn λ2 λ1 | ˆ |2 ≤ ··· ··· 2 f(Λ) ··· dΛ 1 dΛ , 0 0 0 ∂λ1∂λ2 ∂λn 0 0 0 2 |fˆ(Λ)|2 λn λ2 λ1 ∂nfˆ(Λ) ≤ ··· | | ··· dΛ Λ 0 0 0 ∂λ1∂λ2 ∂λn 2 ∂nfˆ(Λ) ≤ dΛ Rn ∂λ1∂λ2 ···∂λn 2 ≤||x1x2 ···xnf(x)|| [Plancherel’s Theorem], (3.11) |fˆ(Λ)|2 dΛ ≤ x x ···x f(x) 2 × dΛ=||xf(x)||22n. | | 1 2 n 2 2 Λn Λ Λn We now proceed to find a bound for |fˆ(Λ)|2 | | dΛ, 1

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Using Holder’s inequality we get 2 λj λj λj k k 2 1 ∂ fˆ(Λ) |fˆ(Λ)|2 ≤ ··· dλ dλ ···dλ ··· j1 j2 jk 0 0 0 ∂λj1 ∂λj2 ∂λjk λ λ λ jk j2 j1 × ··· 12dλ dλ ···dλ . j1 j2 jk 0 0 0

So |fˆ(Λ)|2 |λ λ ···λ | j1 j2 jk 2 λj λj λj λj k k k−1 2 1 ∂ fˆ(Λ) ≤ ··· dλ dλ ···dλ , ··· j1 j2 jk 0 0 0 0 ∂λj1 ∂λj2 ∂λjk |fˆ(Λ)|2 |λ λ ···λ λ ···λ | j1 j2 k jk+1 jn 2 λj λj λj k k 2 1 ∂ fˆ(Λ) ≤ ··· dλ dλ ···dλ ··· j1 j2 jk 0 0 0 ∂λj1 ∂λj2 ∂λjk | |≥ ··· as λji 1fori = k +1,k+2, ,n,

|fˆ(Λ)|2 ··· dλ dλ ···dλ |Λ| jk+1 jk+2 jn | |≥ λji 1 i=k+1,k+2,··· ,n 2 λj λj λj λj k k k−1 2 1 ∂ fˆ(Λ) ≤ ··· ··· ··· dΛ 0 0 0 0 ∂λj1 dλj2 ∂λjk | |≥ λji 1 i=k+1,k+2,··· ,n 2 ∂kfˆ(Λ) ≤ ··· ∂λj1 ∂j2 ∂λjk 2 2 ≤ ··· (3.12) xj1 xj2 xjk f(x) , 2 |fˆ(Λ)|2 ··· ··· dλ dλ ···dλ dλ dλ ···dλ |Λ| j1 j2 jk jk+1 jk+2 jn | |≤ | |≥ λji 1 λji 1 i=1,2,3,··· ,k i=k+1,k+2,··· ,n 2 ≤ ··· ··· ··· xj1 xj2 xjk f(x) dλj1 dλj2 dλjk , 2 |λi|≤1 i=1,2,··· ,k |fˆ(Λ)|2 2 ≤ k ··· dΛ 2 xj1 xj2 xjk f(x) . | | 2 Λk Λ

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Now using (3.9) and the other foregoing results, we get | ˆ |2 n n f(Λ) ≤|| ||2 || ||2 2 || ||2 dΛ f(x) 2 +2 xif(x) 2 +2 xixj f(x) 2 n |Λ| Λ i=1 i,j=1 i= j 2 n (3.13) + ···+2 x1x2 ···xnf(x) . 2 Corollary. Let f ∈ L2(Rn) be a window function. Then f satisﬁes the admissibility ∞ condition if and only if −∞ f(x1,x2,...,xi,...,xn)dxi =0,i=1, 2,...,n.

Proof. Suﬃciency follows from Theorem 3.3. The necessity follows by virtue of the fact that if f is a window function, then its Fourier transform is a continuous function of λ1,λ2,...,λn. Therefore fˆ must vanish at any point in λ-coordinates whose one component is zero to maintain the admissibility condition (3.7). ∞

f(x1,x2,...,xi ...,xn)dxi =0 −∞ ˆ ⇔ f(λ1,λ2,...,λi−1, 0,λi+1,...,λn)=0,λi =0∀i =1, 2,...,n.

−|t| t t Examples of window functions which are wavelets. te , 1+|t|3 , 1+t4 in n n −t2 ti 2 n L2(R), t e i , , in L (R ). i 1+t4 i=1 i=1 i

4. Wavelet inversion formula for functions in L2(Rn) Let ψ ∈ L2(Rn) be a window function satisfying the additional conditions ∞ −∞ ψ(x1,x2,x3,...,xi,...,xn)dxi =0,i=1, 2, 3,...,n.Thenψ is called a regular or basic wavelet function since it then also satisﬁes the admissibility condition ∞ ∞ ∞ 2 |ψˆ(λ1,λ2,λ3,...,λn)| (4.1) ··· dλ1dλ2 ...dλn < ∞. −∞ −∞ −∞ |λ1λ2 ...λn|

2 n Let ψˆ be the Fourier transform of ψ ∈ L (R ) as described above and let Cψ = ˆ 2 n |ψ(λ1,λ2,λ3,...,λn)| (2π) n dλ1dλ2 ...dλn. The integral in one dimension of a func- R |λ1λ2...λn| tion f on R can be written as ∞ M ∞ −N f(x)dx = f(x)dx + f(x)dx + f(x)dx −∞ −N M −∞ provided the integral of f(x) on the real line is convergent. So ∞ M f(x)dx = f(x)dx + R(M,−N), −∞ −N where R(M,−N) is the remainder term. If f(x, y) is integrable over the XY -plane, then ∞ ∞ M2 M1 f(x, y)dxdy = f(x, y)+R(M1,M2, −N1, −N2). −∞ −∞ −N2 −N1

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The remainder term R(M1,M2, −N1, −N2) can be represented by R(M, −N)for simplicity and it can be worked out that R(M, −N) = the sum of eight integrals and if f(x, y) ∈ L(R2), then R(M, −N) → 0asM, N →∞independently of each other. If f ∈ L(Rn) the integral Mn M2 M1 f(x)dx = ··· f(x)dx + R(M, −N). n R −Nn −N2 −N1 The remainder term R(M, −N) consists of finitely many, i.e. (3n −1), integrals and the remainder R(M, −N) → 0asM1,M2,...,Mn,N1,N2,...,Nn all go to infinity independently of each other. We will make use of this fact in proving the following. Theorem 4.1. Let f ∈ L2(Rn) and define the wavelet transform of f by (4.2) Wf (b1,b2,...,bn,a1,a2,...,an) 1 x1 − b1 x2 − b2 xn − bn = f(x1,x2,...,xn), ψ , ,..., , |a1a2 ...an| a1 a2 an where ψ is a basic wavelet as defined above in Section 2, i.e. ψ is a window function satisfying ∞

ψ(x)dxi =0 ∀i =1, 2,...n. −∞

Let Fb1,b2,...,bn be the Fourier transform operator with respect to variables b1,b2, ...,bn.Then F W (b ,b ,...,b ,a ,a ,...,a ) b1,b2,...,bn f 1 2 n 1 2 n − − − ψ x1 b1 , x2 b2 ,..., xn bn ¯ a1 a2 an = f(x1,x2,...,xn), Fb1,b2,...,bn . |a1a2 ...an|

Proof. x1 − b1 x2 − b2 xn − bn Fx1,x2,...,xn ψ , ,..., (λ1,λ2,...,λn) a1 a2 an − ··· ˆ i(λ1b1+λ2b2+ +λnbn) = |a1a2 ...an|ψ(a1λ1,a2λ2,...,anλn)e

and x1 − b1 x2 − b2 xn − bn (4.3) f(x1,x2,...,xn),ψ , ,..., a1 a2 an x1 − b1 x2 − b2 xn − bn = Ff(x1,x2,...,xn),Fψ , ,..., a1 a2 an [by Plancherel’s lemma] ˆ ˆ = f(λ1,λ2,...,λn), ψ(a1λ2,a2λ2,...,anλn)|a1a2 ...an| − ··· × e i(λ1b1+λ2b2+ +λnbn) ˆ ˆ = |a1a2 ...an| f(λ1,λ2,...,λn)ψ(a1λ1,...,anλn) Rn

i(λ1b1+λ2b2+···+λnbn) × e dλ1dλ2 ...dλn,

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ˆ n ˆ 2 n ˆˆ where ψ(λ1,λ2,...,λn) is bounded uniformly on R and f ∈ L (R ), so fψ ∈ L2 . Therefore the above integral in (4.2) belongs to L2 by the λ1,λ2,...,λn b1,b2,...,bn property of the inverse Fourier transform. Now let − − − ¯ x1 b1 x2 b2 xn bn I = f(x1,x2,...,xn)ψ , ,..., dx1dx2 ...dxn Rn a a a 1 2 n Mn Mn−1 M2 M1 = ··· f(x1,x2,...,xn) − − − Nn Nn−1 N2 N1 − − − ¯ x1 b1 x2 b2 xn bn × ψ , ,..., dx1dx2 ...dxn a1 a2 an + R(M,−N,b) M x − b = f(x)ψ¯ dx + R(M, −N, b) −N a = I(M, −N, b)+R(M, −N, b),

F I = F I(M, −N, b)+F R(M, −N, b) b b b M x − b = f(x)Fbψ¯ dx −N a (4.4) +FbR(M, −N, b) [by Fubini’s theorem]. R(M, −N, b)asafunctionofb goes to zero in L2(Rn)asM, N →∞independently 2 n of each other; therefore FbR(M, −N, b) also goes to zero in L (R ) as a function of Λ. Hence, letting M, N →∞in (4.4) we obtain the desired result. The desired result could have been obtained by switching the order of integration in FbI, but ¯ x−b we did not do so as neither of the iterated integrals Rn Rn f(x)Fbψ a dxdb ¯ x−b and Rn Rn f(x)Fbψ a dbdx is absolutely convergent, nor is there any other way of proving that the double integral exists. So we applied Fubini’s theorem as shown above in (4.3).

Theorem 4.2. Let ψ be a basic or regular wavelet function in L2(Rn) satisfying the admissibility condition (3.1) and assume that f ∈ L2(Rn). Then the following inversion formula holds: − − − 1 −1/2 x1 b1 x2 b2 xn bn f(x)= |a1.a2.....an| ψ , ,..., Cψ Rn a1 a2 an × db1db2 ...dbn ...da1da2 ...dan (4.5) Wf (a1,a2,...,an,b1,b2,...,bn) 2 , |a1a2 ...an|

where Wf (a1,a2,...,an,b1,b2,...,bn) is deﬁned to be the integral wavelet transform of f ∈ L2(Rn) by −1/2 |a1.a2.....an| f(x1,x2,...,xn) Rn − − − ¯ x1 b1 x2 b2 xn bn × ψ , ,..., dx1dx2 ...dxn. a1 a2 an

The ai’s and bi’s are real numbers, and the ai’s are non-zero. When ai is zero the above wavelet transform of f is deﬁned to be zero. In duality notation we write it

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in the form (4.6) W (a ,a ,...,a ,b,b ,...,b ) f 1 2 n 1 2 n − − − ψ x1 b1 , x2 b2 ,..., xn bn a1 a2 an = f(x1,x2,...,xn), . |a1a2 ...an|

The inversion formula (4.5) is proved interpreting convergence in L2(Rn); at the point of continuity of f the convergence in (4.5) takes place in the pointwise sense, as indicated in section 3 immediately preceding 3.3.

Proof. Our proof will be sketchy, leaving the veriﬁcation of the lengthy computa- tions to the readers. Step A. In the one-dimensional case the proof of this step is given in the book of Bogess and Narcovich [3]. This result is x − b F ψ¯ = |a|e−iλx ψˆ(aλ) . b a Therefore, x − b (4.7) F ψ¯ (λ)=|a|eiλxψˆ(aλ). b a

xi−bi In the n-dimensional case using substitutions = vi, i =1, 2, 3,...,n,andthe ai same trick of calculation as used by Bogess and Narcovich to prove (4.5), we get − − − ¯ x1 b1 x2 b2 xn bn Fb1,b2,...,bn ψ , ,..., a1 a2 an − ··· i(λ1x1+λ2x2+ +λnxn) ˆ (4.8) = |a1a2 ...an|e ψ(a1λ1,a2λ2,...,anλn). Step B. So, using a similar technique we get n | | ˆ ˆ Fb1,b2,...,bn Wf (a1,a2,...,an,b1b2,...,bn) = 2π ai ψf(λ1,λ2,...,λn). i=1 n n Here the wavelet transform of f is deﬁned as Wf : R × R → C: W (a ,a ,...,a ,b ,b ,...,b ) f 1 2 n 1 2 n f(x1,x2,...,xn) x1 − b1 x2 − b2 xn − bn = ψ¯ , ,..., dx1dx2 ...dxn Rn |a1a2 ...an| a1 a2 an and we also use the fact proved in the previous theorem that F W (a ,a ,...,a ,b ,b ,...,b ) b1,b2,...,bn f 1 2 n 1 2 n − − − f(x1,x2,...,xn) x1 b1 x2 b2 xn bn = Fb1,b2,...,bn ψ , ,..., dx1dx2 ...dxn Rn |a a ...a | a1 a2 an 1 2 n f(x ,x ,...,x ) 1 2 n −i(λ1x1+λ2x2+···+λnxn) = |a1a2 ...an|e Rn |a1a2 ...an| ˆ ×ψ(a1λ1,a2λ2,...,anλn)dx1dx2 ...dxn n 1/2 ˆ ˆ = |2πai| f(λ1,λ2,...,λn)ψ(a1λ1,a2λ2,...,anλn). i=1

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Fb1,b2,...,bn Wf (a1,a2,...,an,b1,b2,...,bn)(λ1,λ2,...,λn) n ˆ ˆ (4.9) = 2π|ai| ψ(a1λ1,a2λ2,...,anλn)f(λ1,λ2,...,λn). i=1 Step C. Final derivation of the inversion formula. Now take − − − 1 −1/2 x1 b1 x2 b2 xn bn F (x)= |a1a2 ...an| ψ , ,..., Cψ Rn Rn a1 a2 an × da1da2 ...dan (4.10) Wf (a1,a2,...,an,b1,b2,...,bn)db1db2 ...dbn 2 . |a1a2 ...an| Using Plancherel’s formula we get 1 da da ...da F (x)= 1 2 n 2 Cψ Rn |a1a2 ...an| (a1a2 ...an) − − − × x1 b1 x2 b2 xn bn Fb1,b2,...,bn ψ , ,..., (λ1,λ2,...,λn) Rn a1 a2 an × F W (a ,a ,...,a ,b,b ,...,b )(λ ,λ ,...,λ )dλ dλ ...dλ b1,b2,...,bn f 1 2 n 1 2 n 1 2 n 1 2 n 1 da1da2 ...dan = |a1a2 ...an| Cψ Rn |a a ...a |(a a ...a )2 1 2 n 1 2 n

Cψ = f(x1,x2,...,xn) Cψ

(4.11) = f(x1,x2,...,xn).

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Acknowledgement

The authors express their gratefulness to Professor Mike Moore of the School of Mathematics and Statistics, Carleton University, Ottawa, who carefully went through the manuscript and gave us some valuable suggestions. The authors also express their gratefulness to a referee for his constructive criticisms.

References

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School of Mathematics and Statistics, Carleton University, Ottawa, Canada E-mail address: [email protected]

Department of Mathematical Sciences, Indian Institute of Technology, DST-CIMS Banaras Hindu University, India E-mail address: sk [email protected]

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