Let g be a function in L², bandlimited, i.e., whose Fouriere transform has compact support, Supp(g^) < [A,B] with A<1/a and B>a. Suppose that there exists a function h such that

Notice that using the notation of the scaled function, F_a(x) = 1/sqrt(a) F(x/a), the wavelet theorem can be stated

This is the main wavelet formula. Due to the compact support of g^ only a finite number of k appear in this sum. g and h form a wavelet pair. This formula says that they provide a resolution of the identity (the delta function).

N.B. In this page we use a symmetric definition of the Fourier transform, F^(v) = 1/sqrt(2 π) Int e^{-i v x} F(x). This differs from the definition in the Section Fourier Transform by the distribution of the constant factors.

Examples

1. g^(v)=cos( π/2 log_a|v| ) and h^(v)=cos( π/2 log_a|v| )
2. g^(v)= cos²( π/2 log_a|v| ) and h^(v)=1. Thus h(x)=sqrt(2/π) [sin(ax) - sin(x/a)]/x.

Discrete Wavelet Theorem

If g^ and h^ are supported in (-π a^k/D, +π a^k/D) then

Sketch of the proof. The proof consists of writing g() and h() in the sum in terms of their Fourier transforms, and using the identity ∑_j e^{i(v-w)D a-k j} = 2 π ∑_n d((v-w)D a^-k + 2 π n). Next, due to the support property of g^ only one n remains in the sum. The double integral is so reduced to a single integral which equals the convolution of the two functions.

Wavelet Transform

Let g and h a pair of functions satisfying the conditions of the Wavelet theorem. Since g^ is supported in (-π a^k/D, +π a^k/D), (F*g)^ is also supported in the same interval, because the support of the Fourier transform of the convolution of two functions is contained in the intersection of the supports of the Fourier transforms of the two functions. (h_a^-k)^=(h^)_a^k is supported in (-a^kB, +a^kB), and, provided π/D>B,

Proof. The proof of this result relies on the Wavelet theorem:

Wavelet Numerics

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