By Gitta Kutyniok

In wavelet research, abnormal wavelet frames have lately come to the vanguard of present examine because of questions about the robustness and balance of wavelet algorithms. a massive trouble within the learn of those structures is the hugely delicate interaction among geometric houses of a chain of time-scale indices and body houses of the linked wavelet systems.

This quantity offers the 1st thorough and finished therapy of abnormal wavelet frames through introducing and making use of a brand new inspiration of affine density as a powerful software for studying the geometry of sequences of time-scale indices. a number of the effects are new and released for the 1st time. issues contain: qualitative and quantitative density stipulations for life of abnormal wavelet frames, non-existence of abnormal co-affine frames, the Nyquist phenomenon for wavelet structures, and approximation houses of abnormal wavelet frames.

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**Extra resources for Affine density in wavelet analysis**

**Sample text**

A similar argument proves the remaining claim. , they are equal in a qualitative sense concerning the upper density. 4. 5 is used), we omit them. 12. If Λ ⊆ A and w : Λ → R+ , then the following conditions are equivalent. (i) D+ (Λ, w) < ∞. (ii) There exists h > 0 such that sup(x,y)∈A #w (Λ ∩ Qh (x, y)) < ∞. (iii) For every h > 0 we have sup(x,y)∈A #w (Λ ∩ Qh (x, y)) < ∞. 13. Let Λ1 , . . , ΛL ⊆ A with associated weight functions w : Λ → R+ for = 1, . . , L be given. Then we have D+ ({(Λ , w )}L=1 ) < ∞ ⇐⇒ D+ (Λ , w ) < ∞ for all = 1, .

Let {yn }n∈N ⊆ R+ be such that the sets yn [e− 2 , e 2 ), n ∈ N, are mutually disjoint. 1 1 (i) Suppose that yn → 0 as n → ∞. Then the function ψ ∈ L2 (R) deﬁned by ψˆ = n∈N 1 1 χ −1 n yn [e 2 ,e 2 ) satisﬁes ψˆ ∈ WR∗ (L∞ , L2 ). (ii) Suppose that yn → ∞ as n → ∞. Then the function ψ ∈ L2 (R) deﬁned by 1 1 ψˆ = √ χ −1 n yn yn [e 2 ,e 2 ) n∈N satisﬁes ψˆ ∈ WR∗ (L∞ , L2 ). Proof. (i) Suppose that yn → 0 as n → ∞. It is easy to check that ψˆ ∈ L2 (R), hence ψ ∈ L2 (R). We next observe that for each k ∈ Z and x ∈ R+ , we have 1 1 1 1 ek [e− 2 , e 2 ) ∩ x[e− 2 , e 2 ) = ∅ if and only if ln x − 1 ≤ k ≤ ln x + 1.

Assume that Λ1 , . . , ΛL ⊆ A with associated weight functions w : Λ → R+ for = 1, . . , L are given such that D+ ({(Λ , w )}L=1 ) = ∞. We L will show that =1 W(ψ , Λ , w ) does not possess an upper frame bound for any collection of analyzing wavelets ψ1 , . . , ψL ∈ L2 (R) \ {0}. Fix some f ∈ L2 (R) with f 2 = 1. 4(i) there exists 0 ∈ {1, . . , L} with D+ (Λ 0 , w 0 ) = ∞. Now since Wψ 0 f is nonzero and continuous, there must exist some (c, d) ∈ A and some h > 0 such that Wψ 0 f does not vanish on the closure of Qh (c, d), and consequently, |Wψ 0 f (x, y)| = δ > 0.