Operator-Like Wavelet Bases of $L_{2}(\mathbb{R}^{d})$

Abstract

The connection between derivative operators and wavelets is well known. Here we generalize the concept by constructing multiresolution approximations and wavelet basis functions that act like Fourier multiplier operators. This construction follows from a stochastic model: signals are tempered distributions such that the application of a whitening (differential) operator results in a realization of a sparse white noise. Using wavelets constructed from these operators, the sparsity of the white noise can be inherited by the wavelet coefficients. In this paper, we specify such wavelets in full generality and determine their properties in terms of the underlying operator.

Appendix A: Discrete Sums

Let $X=\{\boldsymbol{x}_{k}\}_{k\in\mathbb{N}}$ be a countable collection of points in $\mathbb{R}^{d}$, and define

$$\begin{aligned} h_X &:= \sup_{\boldsymbol{x} \in\mathbb{R}^d} \inf_{k\in\mathbb{N}} \lvert\boldsymbol{x}-\boldsymbol{x}_k \rvert \\ q_X &:= \frac{1}{2} \inf_{k\neq k'} \lvert \boldsymbol{x}_k-\boldsymbol {x}_{k'} \rvert. \end{aligned}$$

Also, let B(x,r) denote the ball of radius r centered at x. Proving Riesz bounds for the wavelet spaces relies on the following propositions concerning sums of function values over discrete sets.

Proposition 5

If h _X <∞ and r>d/2, then there exists a constant C>0 (depending only on r and d, not h _X ) such that

$$ \sum_{ \lvert\boldsymbol{x}_k \rvert \geq2h_X} \lvert\boldsymbol {x}_k \rvert^{-2r} \geq C h_X^{-2r} $$

Proof

For |x _k |≥2h _X , we have

$$ \biggl\lvert \boldsymbol{x}_k -h_X \frac{\boldsymbol{x}_k}{ \lvert\boldsymbol {x}_k \rvert} \biggr\rvert \geq 2^{-1} \lvert\boldsymbol{x}_k \rvert, $$

which implies

$$ \lvert\boldsymbol{x}_k \rvert^{-2r} \geq2^{-2r} \biggl\lvert \boldsymbol{x}_k -h_X \frac{\boldsymbol{x}_k}{ \lvert\boldsymbol{x}_k \rvert} \biggr\rvert ^{-2r}. $$

Then

$$\begin{aligned} \sum_{ \lvert\boldsymbol{x}_k \rvert\geq2h_X} \lvert\boldsymbol {x}_k \rvert^{-2r} &\geq2^{-2r} \sum_{ \lvert\boldsymbol{x}_k \rvert\geq2h_X} \biggl\lvert \boldsymbol{x}_k -h_X \frac{\boldsymbol{x}_k}{ \lvert\boldsymbol {x}_k \rvert} \biggr\rvert ^{-2r} \frac{\text{Vol} (B(\boldsymbol{x}_k,h_X) )}{\text{Vol} (B(\boldsymbol{x}_k,h_X) )} \\ &\geq\frac{2^{-2r}}{\text{Vol} (B(\boldsymbol{0},h_X) )} \sum_{ \lvert\boldsymbol {x}_k \rvert \geq2h_X} \int _{B(\boldsymbol{x}_k,h_X)} \lvert\boldsymbol{x} \rvert^{-2r} {\rm d}\boldsymbol{x} \\ &\geq Ch_X^{-d} \int_{3h_X}^{\infty} t^{-2r+(d-1)}{\rm d}t \\ &\geq Ch_X^{-2r} \end{aligned}$$

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Proposition 6

If r>d/2 and |x _k |≥q _X /2 for all $k\in\mathbb {N}$, then there exists a constant C>0 (depending only on r and d, not q _X ) such that

$$ \sum_{k\in\mathbb{N}} \lvert\boldsymbol{x}_k \rvert^{-2r} \leq C q_X^{-2r} $$

Proof

Using the fact that |x _k |≥q _X /2, we can write

$$ \biggl\lvert \boldsymbol{x}_k +\frac{q_X}{4} \frac{\boldsymbol{x}_k}{ \lvert \boldsymbol{x}_k \rvert} \biggr\rvert \leq2 \lvert\boldsymbol{x}_k \rvert, $$

which implies

$$ \lvert\boldsymbol{x}_k \rvert^{-2r} \leq2^{2r} \biggl\lvert \boldsymbol {x}_k +\frac{q_X}{4} \frac{\boldsymbol{x}_k}{ \lvert\boldsymbol{x}_k \rvert} \biggr\rvert ^{-2r}. $$

We now have

$$\begin{aligned} \sum_{k\in\mathbb{N}} \lvert\boldsymbol{x}_k \rvert^{-2r} &\leq2^{2r} \sum_{k\in\mathbb{N}} \biggl\lvert \boldsymbol{x}_k +\frac {q_X}{4} \frac{\boldsymbol{x}_k}{ \lvert\boldsymbol{x}_k \rvert} \biggr\rvert ^{-2r} \frac{\text{Vol} (B(\boldsymbol{x}_k,q_X/4) )}{\text{Vol} (B(\boldsymbol {x}_k,q_X/4) )} \\ &\leq\frac{2^{2r}}{\text{Vol} (B(\boldsymbol{0},q_X/4) )} \sum_{k\in \mathbb{N}} \int _{B(\boldsymbol{x}_k,q_X/4)} \lvert\boldsymbol{x} \rvert^{-2r} {\rm d}\boldsymbol{x} \\ &\leq\frac{2^{2r}}{\text{Vol} (B(\boldsymbol{0},q_X/4) )} \int_{ \lvert\boldsymbol {x} \rvert>q_X/4} \lvert\boldsymbol{x} \rvert ^{-2r} {\rm d}\boldsymbol{x} \\ &\leq Cq_X^{-d} \int_{q_X/4}^{\infty} t^{-2r+(d-1)}{\rm d}t \\ &\leq C q_X^{-2r}. \end{aligned}$$

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