The Haar system of wavelets forms a basis of $L^2[0,1]$. What other wavelets are there that also form bases of $L^2[0,1]$ (or $L^2[0,a]$ in general)?. Thanks
What other wavelets (besides the Haar system) form a basis of $L^2(0,1)$?
1 Answers
You can take a wavelet basis $\{\psi_{j,n}\}_{j,n \in \mathbb Z}$ for $L^2(\mathbb R)$ and transform it into a wavelet basis for $L^2[0,1]$ by periodizing each $\psi_{j,n}$, as follows:
$\psi^{\mathrm per}_{j,n} (t) = \frac{1}{\sqrt{2^j}} \sum\limits_{k\in \mathbb Z} \psi\left( \frac{t - 2^j n +k}{2^j}\right)$.
That the resulting collection forms a basis for $L^2[0,1]$ is Theorem 7.16 in Mallat, A Wavelet Tour of Signal Processing (1998).
However, you end up with "boundary wavelets" which have some of their support near the 0 boundary and some near the 1 boundary, and these have no vanishing moments, so you get large coefficients on these terms in your wavelet decomposition. There are ways to deal with this. See section 7.5 (op. cit.) for more details. Specifically, on page 289 Mallat explains how to construct a Daubechies basis for $L^2[0,1]$.