Continuous Wavelet Transform on the Hyperboloid - UniversitÃ© de ...

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Integrating over ϕ and using the rotation invariance of ψ, we obtain the following approximation for I 1 : I 1 ∼ ∫ σ 0 α(a)a 2 da ∣ ∫ ˜χ 0 [ ] 1 sinh 2pχ sinh χ ′ dχ ′ ′ 2 ψ(χ ′ )∣ ∣ ∣2 . (76) sinh χ ′ The second subintegral (I 2 ) is straightforward, since the operator D a is strongly continuous and thus the integrand is bounded on [σ, 1 σ ]. Consider now the inequality : I 3 ≤ × ∫ +∞ 1 σ (∫ ˜χ ∫ 2π 0 α(a)da× 0 dµ(χ ′ ,ϕ) λ 1 2 (a −1 ,χ ′ ) |ψ(χ ′ )|| cosh χ ′ a − sinh χ′ a cos ϕ|−1/2 ) 2 . The term | cosh χ ′ a − sinh χ′ a cos ϕ|−1/2 is bounded from above and from below by : Now, we have e − χ′ a 2 ≤|cosh χ ′ a − sinh χ′ a cos ϕ|−1/2 ≤ e χ′ a 2 . (77) e χ′ a 2 = ( ) 1 ] 1 e pχ′ a 2p = [√1+a 2 sinh 2 pχ ′ + a sinh pχ ′ 2p , and so we get the asymptotic behavior of this upper bound at large scale: e χ′ a 2 ∼ a 1 2p (sinh pχ ′ ) 1 2p . Again using the explicit form of χ ′ a large scale a →∞: and the following asymptotic behaviors at cosh pχ a ∼ a sinh pχ, sinh χ a ∼ a 1 1 p (sinh pχ) p , we reach the following majoration of I 3 : I 3 ≤ ∫ +∞ 1 σ (∫ ˜χ 2 α(a)a 2 p da dχ ′ (sinh χ ′ ) 1 2 (sinh pχ ′ ) 1 p − 1 2 (cosh pχ ′ ) 1 2 |ψ(χ )|) ′ . 0 Since the hyperbolic functions involved in the integration on the χ ′ variable are increasing, we finally end with the estimate : ∫ +∞ I 3 ≤∼ (sinh ˜χ) 1 2 (sinh p˜χ) ( 1 p − 1 2 ) (cosh p˜χ) 1 2 ‖ψ‖ 2 1 α(a)a 2 p da. 1 σ 22
and so α(a) should behave at least like a −β with β> 2 +1fora →∞. p The convergence of I 1 and I 3 clearly depends on the choice of measure in the integral over scales. Restricting ourselves to homogeneous measures α(a) = a −β and to the range p>0, one can distinguish the following cases : • β 2 +1: in thiscaseI p 3 does not converge and there are no admissible wavelets. • β> 2 +1:InthiscaseI p 1 diverges except when ∫ H ψ [ ] 1 sinh 2pχ 2 =0. + 2 sinh χ 6.3 An example of Hyperbolic <strong>Wavelet</strong> Let us present here a class of admissible vectors which satisfy the admissibility condition. We restrict ourself to the simplest case p = 1 . Let us first state a 2 preliminary result. Proposition 6 Let ψ ∈ L 2 (H+ 2 , dµ). Then ∫ ∫ D a ψ(χ, ϕ)dµ(χ, ϕ) =a ψ(χ, ϕ)dµ(χ, ϕ). (78) H+ 2 H+ 2 Proof: We have to compute the following integral ∫ ∫ I = D a ψ(χ, ϕ)dµ(χ, ϕ) = λ 1 2 (a, χ)ψ(χ 1 ,ϕ)dµ(χ, ϕ). H+ 2 a H 2 + By change of variable χ 1 a ∫ I = ∫ = H 2 + H 2 + = χ ′ ,weget λ 1 2 (a, χ ′ a ) ψ(χ ′ ,ϕ) λ(a −1 ,χ ′ )dµ(χ ′ ,ϕ) λ 1 2 (a −1 ,χ ′ )ψ(χ ′ ,ϕ)dµ(χ ′ ,ϕ), and having λ 1 2 (a −1 ,χ ′ )=a, which follows directly from (45), we get ∫ I = a ψ(χ ′ ,ϕ)dµ(χ ′ ,ϕ), H+ 2 which proves the proposition. Using this result, we can build the hyperbolic “difference” wavelet (differenceof-Gaussian, or DOG wavelet). Given a square-integrable function ψ, we define f ϑ ψ(χ, ϕ) =ψ(χ, ϕ) − 1 ϑ D ϑψ(χ, ϕ), ϑ > 1. 23
Page 1 and 2: Continuous <strong
Page 3 and 4: and on the sphere, it is natural to
Page 5 and 6: 0 x 0 C 2 + H 2 + r 0 x 2 x 1 Fig.
Page 7 and 8: The action of a motion on a point x
Page 9 and 10: 4 p=0.5 4 p=1 3.5 3.5 3 3 2.5 2.5 2
Page 11 and 12: x 0 H 2 + a N x 1 a S x 2 H 2 - Fig
Page 13 and 14: E ν,ξ (x)=(ξ · x) − 1 2 −i
Page 15 and 16: get the more elaborate expression
Page 17 and 18: We now have all the basic ingredien
Page 19 and 20: that 0
Page 21: By performing the change of variabl
Page 25 and 26: Fig. 9. The hyperbolic DOG wavelet
Page 27 and 28: Fig. 11. The hyperbolic DOG wavelet
Page 29 and 30: lim ˆψ ρ (ν, ξ)= 1 ∫ ψ(⃗x
Page 31: April 2004. I. Tosic, I. Bogdanova,

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