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An (orthonormal dyadic) wavelet set on ${\mathbb R}$ is a subset $E \subset {\mathbb R}$ such that
- $\chi_E \in L^2({\mathbb R})$ (since $\|\chi_E\| = \sqrt{m(E)}$ , this implies $m(E) < \infty$ ).
- $\frac{\chi_E}{\sqrt{m(E)}}$ is the Fourier transform of an orthonormal dyadic wavelet,
where $\chi_E$ is the characteristic function of $E$ , and $m(E)$ is the Lebesgue measure of $E$ .
$E \subset {\mathbb R}$ is a wavelet set iff
- $\{E + 2\pi n\}_{n\in {\mathbb Z}}$ is a measurable partition of $\mathbb R$ ; i.e. ${\mathbb R}\backslash \bigcup_{n\in \mathbb Z} \{ E + 2\pi n\}$ has measure zero, and $\bigcap_{n=i,j} \{E+2\pi n\}$ has measure zero if $i\neq j$ . In short, $E$ is a $2\pi$ -translation ``tiler'' of $\mathbb R$
- $\{2^n E\}_{n\in \mathbb Z}$ is a $2$ -dilation ``tiler'' of $\mathbb R$ (once again modulo sets of measure zero).
There are higher dimensional analogues to wavelet sets in $\mathbb R$ , corresponding to wavelets in higher dimensions. Wavelet sets can be used to derive wavelets-- by creating a set $E$ satisfying the conditions given above, and using the inverse Fourier transform on $\chi_E$ , you are guaranteed to recover a wavelet. A particularly interesting open question is: do all wavelets contain wavelet sets in their frequency support?
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"wavelet set" is owned by swiftset.
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Cross-references: support, open question, wavelets, measure zero, measurable partition, iff, Lebesgue measure, characteristic function, orthonormal dyadic wavelet, Fourier transform, implies, subset, dyadic, orthonormal
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This is version 4 of wavelet set, born on 2004-06-27, modified 2007-09-27.
Object id is 5971, canonical name is WaveletSet2.
Accessed 2773 times total.
Classification:
| AMS MSC: | 46C99 (Functional analysis :: Inner product spaces and their generalizations, Hilbert spaces :: Miscellaneous) | | | 65T60 (Numerical analysis :: Numerical methods in Fourier analysis :: Wavelets) |
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Pending Errata and Addenda
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