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Edit: A more precise formulation of my question follows the separation line.

The Schwartz space of test functions $\mathcal{S}(\mathbb{R})$ is isomorphic to $\mathfrak{s}$ the space of sequences of real numbers with faster than power-like decay. Likewise, the space $\mathcal{D}(\mathbb{R})$ of compactly supported test functions is isomorphic to $\oplus_{\mathbb{N}} \mathfrak{s}$.

My question: is it possible to find a Schauder basis which realizes such isomorphisms and has a wavelet-like structure, namely, the elements of this basis are scaled copies of one mother function?

Of course, this is related to this other MO question.

The only explicit Schauder basis for such Valdivia-Vogt isomorphisms that I have seen is in this article by Bargetz. But it does not look wavelet-like to me.


A more detailed formulation: (Aug 2017 edit)

Let $f(x)$ be a function on $\mathbb{R}$. For $(m,k)\in\mathbb{Z}^2$ let me define its dyadic translates $f_{m,k}$ by $$ f_{m,k}(x)=2^{\frac{m}{2}}f(2^m x-k)\ . $$ For a Schwartz distribution $T$ in $\mathcal{D}'(\mathbb{R})$ let me also define its translates by the same formula " $T_{m,k}(x)=2^{\frac{m}{2}}T(2^m x-k)$ ", except that it has to be interpreted, as usual, in the sense of distributions. Namely, for every test function $f\in\mathcal{D}(\mathbb{R})$, the space of smooth compactly supported functions, I define $$ \langle T_{m,k},f\rangle=\langle T, g\rangle $$ with $$ g(x)=2^{-\frac{m}{2}}f(2^{-m}(x+k))\ . $$ Let me call a pair $(f,T)\in\mathcal{D}(\mathbb{R})\times\mathcal{D}'(\mathbb{R})$ an unobtainium biorthogonal pair if for all $k,l,n,m$ in $\mathbb{Z}$, one has (using Kronecker deltas) $$ \langle T_{m,k},f_{n,l}\rangle=\delta_{m,n}\delta_{l,k}\ . $$

My question is: does there exist unobtainium biorthogonal pairs?

This is an "extreme sport" version of biorthogonal wavelets.

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  • $\begingroup$ In the case of the Schwartz space, the eigenfunctions for the Schrodinger/Hamiltonian operator $-{d^2\over dx^2}+x^2$ constitute a nice basis, and they are obtained from the "bottom" eigenfunction $e^{-x^2/2}$ by repeatedly applying the "raising operator", and so on. Rate of decay of coefficients with respect to this basis distinguishes Schwartz functions, etc. Is this (fairly well-known business) at all relevant to your interests? $\endgroup$ – paul garrett Sep 10 '15 at 16:41
  • $\begingroup$ Thanks Paul. I am familiar with this Schauder basis and I do use it a lot in my research interests. Unfortunately, it is not wavelet-like. It is not good for the kind of microlocal harmonic analysis I would like to do. $\endgroup$ – Abdelmalek Abdesselam Sep 10 '15 at 19:58
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    $\begingroup$ Interesting question! For $\mathscr{D}$ the answer, if positive, should not be trivial - for instance, it cannot be an orthonormal basis, for there is no orthonormal set of infinitely differentiable wavelets with exponential decay, let alone of compact support (Corollary 5.5.3 of Ingrid Daubechies's "Ten Lectures on Wavelets", apparently independently due to Battle and Meyer). As for $\mathscr{S}$, the Littlewood-Paley wavelets as defined in Meyer's book "Wavelets and Operators" should do the job, thanks to the $L^\infty$ Bernstein inequalities. $\endgroup$ – Pedro Lauridsen Ribeiro Sep 12 '15 at 18:35
  • $\begingroup$ @Pedro: indeed othogonality would have to be sacrificed. Thanks for pointing out the LP wavelets from Meyer's book. I will have a look. $\endgroup$ – Abdelmalek Abdesselam Sep 14 '15 at 13:20
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This is not at all a complete answer, but rather an expanded update on my above comment. I shall start with a few general considerations:

  • If $\{f_n\ |\ n\in\mathbb{N}\}$ is any Schauder basis for either $\mathscr{S}(\mathbb{R})$ or $\mathscr{D}(\mathbb{R})$, at least one $f_n$ must have a non-vanishing moment of order zero, otherwise we run into a contradiction with the fact that $u(f)=\int^{+\infty}_{-\infty}f(x)\mathrm{d}x$ is a continuous linear functional on both spaces. This means that one must employ inhomogeneous wavelet bases (i.e. starting from both a "mother wavelet" and a "father wavelet" in Meyer's terminology), unless one works in "homogeneous" subspaces defined by a certain number of vanishing moments (e.g. $\mathscr{S}_0(\mathbb{R})\subset\mathscr{S}(\mathbb{R})$, discussed in more detail below).

  • Secondly, as I remarked in my comment, Battle and Meyer have shown that there is no countably infinite orthonormal set of smooth wavelets with exponential decay, let alone of compact support, so any wavelet Schauder basis of $\mathscr{D}(\mathbb{R})$ must be non-orthogonal.

  • Thirdly, since the Fourier transform of any $f\in\mathscr{D}(\mathbb{R})$ is real analytic, no such $f$ can have all its moments vanishing unless it is identicaly zero, therefore any would-be wavelet basis of $\mathscr{D}(\mathbb{R})$ can have vanishing moments of positive order only up to a finite order $m\in\mathbb{N}$. This is not a problem for $\mathscr{S}(\mathbb{R})$, though (e.g. any $f\in\mathscr{S}(\mathbb{R})$ whose Fourier transform vanishes in a neighborhood of the origin has vanishing moments at all orders).

  • Finally, as for the form of the wavelet coefficients one should expect from members of $\mathscr{D}(\mathbb{R})$, the compact support restriction together with the group action of $\mathbb{Z}$ on any wavelet Schauder basis by translations should entail that "many" of these coefficients would be zero altogether. This scenario is encouragingly consistent with the Valdivia-Vogt characterization of $\mathscr{D}(\mathbb{R})$. Unfortunately, non-orthogonal wavelets are far more unwieldy than their orthonormal counterparts, so I do not have any more ideas to go on at the moment regarding this case.

That being said, it was shown by K. Saneva and J. Vindas (Wavelet Expansions and Asymptotic Behavior of Distributions, J. Math. Anal. Appl. 370 (2010) 543-554) that the (one-dimensional) homogeneous Littlewood-Paley wavelet basis is an orthonormal wavelet Schauder basis for the "homogeneous" version of $\mathscr{S}(\mathbb{R})$ - namely, the closed subspace $\mathscr{S}_0(\mathbb{R})\subset\mathscr{S}(\mathbb{R})$ consisting of all tempered test functions on $\mathbb{R}$ all of whose moments are vanishing: $$\mathscr{S}_0(\mathbb{R})=\bigg\{\phi\in\mathscr{S}(\mathbb{R})\ \bigg|\ \int^{+\infty}_{-\infty}x^n\phi(x)\mathrm{d}x=0\ ,\,\forall n\in\mathbb{N}\cup\{0\}\bigg\}\ .$$

Now, to go a bit on the details: recall as in e.g. Chapter 3 of the book of Y. Meyer, Wavelets and Operators, Cambridge University Press, 1993, that the homogeneous Littlewood-Paley wavelet basis is the countable orthonormal subset of $\mathscr{S}_0(\mathbb{R})$ given by $$\mathrm{LP}_0(\mathbb{R})=\big\{\psi_{m,n}(x)=2^{\frac{m}{2}}\psi(2^m x-n)\ \big|\ m,n\in\mathbb{Z}\big\}\ ,$$ where the "mother wavelet" $\psi\in\mathscr{S}_0(\mathbb{R})$ is given by its Fourier transform $\hat{\psi}$ as $$\hat{\psi}(\xi)=e^{-i\xi/2}\sqrt{\hat{\phi}^2(\xi/2)-\hat{\phi}^2(\xi)}=e^{-i\xi/2}\theta_1(\xi)$$ and the "father wavelet" $\phi\in\mathscr{S}(\mathbb{R})$ has a Fourier transform $\hat{\phi}\in\mathscr{D}(\mathbb{R})$ with the following properties:

  • $\hat{\phi}$ is real-valued and even;

  • $\hat{\phi}(\xi)=1$ if $0\leq\xi\leq 2\pi/3$ and $\hat{\phi}(\xi)=0$ if $\xi\geq 4\pi/3$;

  • $\hat{\phi}(\xi)$ is strictly decreasing for $2\pi/3\leq\xi\leq4\pi/3$;

  • $\hat{\phi}^2(\xi)+\hat{\phi}^2(2\pi-\xi)=1$ for all $\xi\in[0,2\pi]$.

Conversely, it is possible to recover $\phi$ from $\psi$ by noticing that $$\hat{\phi}(\xi)=\begin{cases} 1 & (|\xi|\leq 2\pi/3) \\ \sqrt{1-\theta_1^2(\xi)} & (2\pi/3\leq|\xi|\leq 4\pi/3) \\ 0 & (|\xi|\geq 4\pi/3) \end{cases}\ .$$ It is not difficult to find $\hat{\phi}$ with the above properties: for instance, set $$\hat{\phi}(\xi)=\sqrt{\frac{\chi(4\pi/3-\xi)}{\chi(\xi-2\pi/3)+\chi(4\pi/3-\xi)}}\ ,$$ where $$\chi(\xi)=\begin{cases} e^{-\frac{1}{\xi}} & (\xi>0) \\ 0 & (\xi\leq 0) \end{cases}\ .$$

It is shown in the aforementioned paper that the expansion of $f\in\mathscr{S}_0(\mathbb{R})$ in this wavelet basis $$f=\sum_{m,n\in\mathbb{Z}}c^\psi_{m,n}(f)\psi_{m,n}\ ,\,c^\psi_{m,n}(f)=\int^{+\infty}_{-\infty}\overline{\psi_{m,n}(x)}f(x)\mathrm{d}x$$ converges in the induced topology from $\mathscr{S}(\mathbb{R})$ and yields a topological isomorphism $f\mapsto(c^\psi_{m,n}(f))_{m,n\in\mathbb{Z}}$ of $\mathscr{S}_0(\mathbb{R})$ with the so-called space of dyadic rapidly decreasing (double) sequences $$\mathscr{W}=\big\{(c_{m,n})_{m,n\in\mathbb{Z}}\ \big|\ \|(c_{m,n})\|^{\mathscr{W}}_l<+\infty\ ,\,\forall l\in\mathbb{N}\cup\{0\}\big\}\ , \\ \|(c_{m,n})\|^{\mathscr{W}}_l\doteq\sup_{m,n\in\mathbb{Z}}|c_{m,n}|(1+|n|)^l(2^m+2^{-m})^l\ .$$ The above rapid decay of the wavelet coefficients easily follows from Parseval's formula and the support properties of $\hat{\psi}_{m,n}$. Particularly, $\mathrm{LP}_0(\mathbb{R})$ is an absolute (hence Schauder) basis of $\mathscr{S}_0(\mathbb{R})$.

Presumably, the argument of Saneva and Vindas can be adapted to the "inhomogeneous" space $\mathscr{S}(\mathbb{R})$ if we use instead the inhomogeneous Littlewood-Paley wavelet basis $$\mathrm{LP}(\mathbb{R})=\big\{\phi_n(x)=\phi(x-n)\ \big|\ n\in\mathbb{Z}\big\}\cup\big\{\psi_{m,n}(x)=2^{\frac{m}{2}}\psi(2^m x-n)\ \big|\ m\in\mathbb{N}\cup\{0\}\ ,\,n\in\mathbb{Z}\big\}\ ,$$ which is also orthonormal. It can be seen from the form of the seminorms $\|\cdot\|^{\mathscr{W}}_l$ that the outcome should be a Valdivia-Vogt-like isomorphism for $\mathscr{S}(\mathbb{R})$ after a suitable rearrangement of the wavelet coefficients into a single sequence.

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  • $\begingroup$ Thanks, even if that's a long time ago. I was indeed aware of this work by Vindas and coauthors (and also the more detailed book by Holschneider) because Abelian/Tauberian theory for distributions is related to my recent work arxiv.org/abs/1604.05259 But what I really need here is the compact support together with smoothness. $\endgroup$ – Abdelmalek Abdesselam Aug 20 '17 at 19:57

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