Redundancy in the Frequency Domain

Smooth well-localized Parseval wavelets based on wavelet sets in ℝ²

“…Note that these definitions satisfy all the requirements of the Lemma on the two preimages of C except (4). In particular, since one of the restrictions of star-simple is that ± 4 9 ω * l ∈ C, we have that ± 7 9 ω * l is in the component R with star-center ω * l , so that t(ω * l ) ∈ R for 7 9 < t < 11 9 , establishing condition (3) because of the rapidly vanishing at the center property ofh r for 2 3 of the distance from the center to the edge.…”

“…|β − α|, and the one-sided derivatives of all orders of f r,α,β (α) vanish at both α and β. By the Mean Value Theorem, we know that |f r,α,β (x)| < |x − α| r+2 for |x − α| < 2 3 |β − α|. We will first use the f r,α,β to define our preliminary filterh r on R = the connected component of ω * l in P , which we know (by the central axis requirement and by ω * l / ∈ C) to be given by R = ω * l + 1 2 C. Note that since 1 2 C ⊂ C, all of the boundary points of 1 2 C haveh r = 0, and thus all of the boundary points of this region R haveh r = 2.…”

“…Finally we show that m ≡the multiplicity function of the GMRA determined by φ is the same as m, the multiplicity function determined by the wavelet set (a.e.). By [2], m is the characteristic function of the support of Perφ. Since m and m both take on only the values 0 and 1, it will suffice to show their supports are the same up to a set of measure 0.…”

A non-MRA $$C^r$$ Frame Wavelet with Rapid Decay

“…Again, by applying the Mean Value Theorem to condition (5) in the definition of p, we see that this gives us a factor of jh i; j ð x 2 l Þj which is less than ð 1 8 n Þ rþ2 . Values of x 2 A n [ B n [ C n with a mixture of a j ¼ þ1 and a j ¼ À1 will be somewhere between these two extremes, but all will have one or more factors of pðxÞ among the jh i; j ð x 2 l Þj with x so close to 3 14 or 1 7 that the product of these factors is less than ð 1 8 n Þ rþ2 . Since the other factors are bounded by 1, we thus get that each term in b 0 i ðxÞ is bounded in absolute…”

“…We will show that whenever one of these discontinuities occurs as a factor in the infinite product, it is cancelled by a following factor that is 0 at the point of discontinuity. From the formula for the t i;j given in Lemma 1, we see that any term in the infinite product that contains a factor of h 2;1 ðn AE 3 7 Þ must also contain a factor of one of the forms h 1;1 ðn AE 2 7 Þ, h 1;1 ðn AE 3 14 Þ, h 1;2 ðn AE 3 14 Þ, or h 1;2 ðn AE 2 7 Þ. The last three possibilities are 0 in a neighborhood of the point in question, so if we have a discontinuous factor of h 2;1 , it is either cancelled out by a 0 factor, or we also have a factor of h 1;1 ðn AE 2 7 Þ with a smaller n. Similarly, any term in the infinite product that contains a factor of h 1;1 ðn AE 2 7 Þ must also contain a factor of one of the forms h 1;1 ðn AE 5 14 Þ, h 1;1 ðn AE 1 7 Þ, h 1;2 ðn AE 5 14 Þ, or h 1;2 ðn AE 1 7 Þ.…”