Wavelet Frame Bijectivity on Lebesgue and Hardy Spaces

Abstract

We prove a sufficient condition for frame-type wavelet series in L ^p, the Hardy space H ¹, and BMO. For example, functions in these spaces are shown to have expansions in terms of the Mexican hat wavelet, thus giving a strong answer to an old question of Meyer.

Bijectivity of the wavelet frame operator acting on Hardy space is established with the help of new frequency-domain estimates on the Calderón–Zygmund constants of the frame kernel.

Appendices

Appendix A: Geometric Sums

In order to extend our decay estimates from K ₀ to the full kernel K, earlier in the paper, we summed over dilation scales $j \in \mathbb{Z}$. Here we collect elementary estimates used in that task.

Lemma A.1

Assume $g : \mathbb{R}\to\mathbb{R}$ and let σ,τ>0.

• (i) If |g(z)|≤min{σ,τ/|z|²} for all z≠0, then

  $$\sum_{j \in\mathbb{Z}} \bigl|A^j g\bigl(A^j z\bigr)\bigr| \leq\frac{2|A|}{|A|-1} \frac{\sqrt{\sigma\tau}}{|z|} , \quad z \neq0 . $$

• (ii) If |g(z)|≤min{σ,τ/|z|³} for all z≠0, then

  $$\sum_{j \in\mathbb{Z}} \bigl|A^{2j} g\bigl(A^j z\bigr)\bigr| \leq\frac{|A|(2|A|+1)}{|A|^2-1} \frac{\sqrt[3]{\sigma \tau^2}}{|z|^2} , \quad z \neq0 . $$

Proof of Lemma A.1

(i) Fix z≠0 and let J be any integer. We split the sum into two pieces and estimate each piece using the assumption on g:

by the geometric series. Choose J to be the smallest integer satisfying $|A|^{J} |z| > \sqrt{\tau/\sigma}$, so that $|A|^{J-1} |z| \leq\sqrt{\tau/\sigma}$. (To motivate this choice, notice that σ≤τ/|z|² if and only if $|z| \leq\sqrt{\tau/\sigma}$.) Then

$$\sum_{j \in\mathbb{Z}} \bigl|A^j g\bigl(A^j z\bigr)\bigr| \leq\frac{|A|}{|A|-1} \bigl( \sqrt{\tau/\sigma} |z|^{-1} \sigma+ \sqrt{\sigma/\tau} |z|^{-1} \tau\bigr) , $$

from which part (i) of the lemma follows.

(ii) Adapt part (i), except choose J to be the smallest integer satisfying $|A|^{J} |z| > \sqrt[3]{\tau/\sigma}$. □

Appendix B: Weak–Strong Interpolation Between L ¹ and L ²

The Marcinkiewicz interpolation theorem implies in particular that if a sublinear operator satisfies weak-L ^r and weak-L ² bounds, then strong-L ^p bounds hold for p between r and 2. By strengthening the second hypothesis to strong-L ² and calling also on Riesz–Thorin interpolation (which requires the operator to be linear), we will obtain an explicit estimate on the L ^p norm such that equality holds when p=2.

The bound will involve the following constants, for 1≤r<p≤2:

Notice p=2 gives c(2)=1, and indeed c(2,r)=1 for each r. At the other extreme of p-values, we see c(p,r) blows up like (p−r)^−1/r as p↓r, and c(p) blows up like (p−1)⁻¹ as p↓1. See Fig. 4.

Fig. 4

The logarithm of the L ^p constant in Proposition B.1. Note c(2)=1 (Color figure online)

Proposition B.1

Let (X,μ) and (Y,ν) be measure spaces, and 1≤r<2. Assume Z is a linear operator from L ^r(X)+L ²(X) to the space of measurable complex-valued functions on Y.

If Z is weak type (r,r) and strong type (2,2), then Z is strong type (p,p) for r<p≤2, with

$$\Vert Z \Vert_{L^p(X) \to L^p(Y)} \leq c(p,r) \Vert Z \Vert _{L^r(X) \to\mathrm{weak}\mbox{-}L^r(Y)}^{ [(2/p) - 1 ]/ [(2/r)-1 ]} \Vert Z \Vert_{L^2(X) \to L^2(Y)}^{ [ (2/r) - (2/p) ]/ [ (2/r)-1 ]} . $$

(Equality holds for p=2.) In particular, when r=1 the conclusion says for 1<p≤2 that

$$\Vert Z \Vert_{L^p(X) \to L^p(Y)} \leq c(p) \Vert Z \Vert_{L^1(X) \to\mathrm{weak}\text{-}L^1(Y)}^{(2/p) - 1} \Vert Z \Vert_{L^2(X) \to L^2(Y)}^{2 - (2/p)} . $$

A similar result holds if we assume strong type (q,q), but the proposition focuses on strong type (2,2) because so many operators in harmonic analysis are bounded on L ².

The case r=1 of the proposition improves by a factor of about 8 (when p is close to 2) on the best bound we found in the literature, which is an exercise in the monograph by Grafakos [19, Exercise 1.3.2]. We follow the same approach as Grafakos, except in the proof below we employ the harmonic mean of 1 and p instead of the arithmetic mean; the harmonic mean yields simpler formulas.

Proof of Proposition B.1

Suppose r<q<p≤2. Then Marcinkiewicz interpolation applied with r<q<2 (see [19, Theorem 1.3.2]) implies boundness of Z on L ^q, with

$$ \Vert Z \Vert_{L^q \to L^q} \leq2 \biggl( \frac{q}{q-r} + \frac{q}{2-q} \biggr)^{\! \! 1/q} \Vert Z \Vert_{L^r \to\text{weak-}L^r}^{ [(2/q) - 1 ]/ [ (2/r)-1 ]} \Vert Z \Vert_{L^2 \to L^2}^{ [ (2/r) - (2/q) ]/ [ (2/r)-1 ]} . $$

(35)

Next, Riesz–Thorin interpolation applied with q<p≤2 (see [19, Theorem 1.3.4]) says that

$$\Vert Z \Vert_{L^p \to L^p} \leq\Vert Z \Vert_{L^q \to L^q}^{ [ (2/p)-1 ]/ [ (2/q)-1 ]} \Vert Z \Vert_{L^2 \to L^2}^{ [ (2/q)-(2/p) ]/ [ (2/q)-1 ]} . $$

We substitute the Marcinkiewicz bound (35) into this last Riesz–Thorin bound, giving that

(36)

We would like to choose q∈(r,p] to minimize the right side of bound (36). That coefficient seems too complicated for analytical minimization to be feasible, but numerical work suggests that a good (i.e., within a factor of about 2 of being minimal) choice for q is the harmonic mean of r and p:

$$\frac{1}{q} = \frac{1}{2} \biggl( \frac{1}{r} + \frac{1}{p} \biggr) . $$

Substituting this choice of q into (36) yields the constant c(p,r) claimed in the proposition.

Notice equality holds when p=2, because equality holds in the Riesz–Thorin bound. □