One of the properties I have seen for isotropic tight wavelet frames is

$$\sum_{i\in\mathbb{Z}} \left|h(2^i\omega)\right|^2 = 1$$

where $h(\omega)$ is the frequency spectrum of the original wavelet. See page 8 in A unifying parametric framework for steerable wavelet transforms.

I have always assumed that this enables the downsampling often seen in wavelet decompositions, which start off with a wavelet satisfying $h(\omega) = 1\, \forall \omega > \pi/2$, meaning that the top half of the spectrum is covered and thus subsampling of the low pass remainder is possible because there are no longer any frequencies above $\pi/2$.

  • Is this correct?
  • Or can you perform downsampling still having frequencies above $\pi/2$?
  • If one does not need downsampling or wants to have different partitions, is the following condition sufficient, $$\sum_{i=1}^{I} \left|h_i(\omega)\right|^2 = 1$$ where $\{h_i(\omega)\}$ are the frequency responses of the wavelets in the frame?
  • $\begingroup$ @LaurentDuval If you have set of wavelets, and the addition of their squared frequency response = 1 for all frequencies (ie unit transform) is that sufficient to be called a wavelet frame or are there some other conditions. And what are the conditions of a wavelet that enable subsampling when scaling by 2. $\endgroup$ May 13, 2016 at 1:46
  • $\begingroup$ @LaurentDuval I am coming to wavelets from normal filter banks, making use of the tight frame property to reconstruct images from archetypical image features. Most wavelet decompositions have the wavelets spaced octaves apart in the spectrum and then do subsampling, eg. simoncelli pyramid. However, some times it is handy to have the wavelets spaced tighter, e.g. four per octave. In this case the first condition doesnt hold, so I'm wondering if the second is sufficient. For a discrete bounded signal it obviously is but I'm not sure about the general continuous case. $\endgroup$ May 14, 2016 at 11:19
  • $\begingroup$ Does the admissibility condition only apply to the CWT? $\endgroup$ May 14, 2016 at 11:19
  • $\begingroup$ Second question I guess is if you can do a pyramidal wavelet decomposition (subsampling by 2) does that imply that all frequencies above $\pi/2$ are zero for the low pass part. $\endgroup$ May 14, 2016 at 11:23
  • $\begingroup$ I am sorry, I am a discrete filter bank guy, very applied, so I am off from my confidence base, just trying to share what I still can handle $\endgroup$ May 14, 2016 at 11:29

1 Answer 1


I am not completely confident with the full theory of wavelet frames, however I will share some pointers and my beliefs, do not take any of the following for a truth. The property you show is (partly) a characterization of an orthogonal system. In Hernandez & Weiss, A first course on wavelets, 1996, you find Theorem 2.12:

If $\psi\in L^2(\mathbb{R})$ is a band-limited function such that $\hat{\psi}$ is zero in a neighborhood of the origin and $\{ \psi_{j,k}: j,k\in \mathbb{Z}\}$ is an orthonormal system, then this system is coomplete if and only if $$\sum_{j\in \mathbb{Z}} |\hat{\psi}(2^j \xi)|^2=1 \quad \text{a.e. on }\mathbb{R}-\{0\}$$ and $$\sum_{j=0}^{\infty} |\hat{\psi}(2^j \xi)\hat{\psi}(2^j (\xi+2k\pi))| =0\quad \text{a.e. on }\mathbb{R},\quad k\in 2\mathbb{Z}+1$$ with the system defined as: $$ \psi_{j,k}(x)= 2^{\frac{j}{2}}\psi(2^j x-k)\,.$$ So the conditions are a little bit stricter for an orthonormal system. So it seems rather to me a characterization in 1D (f band-limited wavelets), than can help define radial 2D wavelets, than a characterization of radial wavelet per se. So if I understand your first question well, this imply the possibility of discrete wavelets and downsampling.

Since it is a characterization, I (very unsure) would say that you can downsample, but you have no guarantee about the completeness. I do not understand your condition $$h(\omega)=1, \forall \omega>\pi/2, h(\omega)=1, \forall\omega>\pi/2,$$ but I can add that in Chapter 3.4 of the above book, you have characterization of wavelets with support in $[-\frac{8}{3}\pi,-\frac{2}{3}\pi] \cup [\frac{2}{3}\pi,\frac{8}{3}\pi] $, so I would say yes, as $\frac{8}{3}> \frac{1}{2}$. I would say that the limit on $\pi/2$ is not an issue because of the dilations of the mother wavelet.

For the last question, without further information on the $h_i$, it is unlikely that you generate a wavelet system (possibly because of refinement equations), but as long as you cover the whole spectrum, who really cares? You get a decomposition that is energy preserving, but I am not sure yet it suffices to define a tight frames.

Additional lectures:

  • $\begingroup$ Thanks for the answer. I just had a paper accepted where I had a short section introducing steerable wavelets and I added in the second relaxed condition as a small comment, just assuming it was true as finer spacing is done in other papers. I might remove that line for the final version... unfortunately I have no one to ask about these things except on here. I can't think of any partition of the spectrum that satisfies the relaxed condition that wouldn't be a frame, but I'm not confident in that. $\endgroup$ May 14, 2016 at 11:55
  • $\begingroup$ By the way, what is the canonical definition of a wavelet that separates it from a regular filter kernel? $\endgroup$ May 14, 2016 at 11:58
  • $\begingroup$ @geometrikal I would say refinement conditions. Just dropping a phd thesis I looked at when try to answer your question alpha.math.uga.edu/~mjlai/papers/Nam_Kyunglim_200508_phd.pdf $\endgroup$ May 14, 2016 at 12:04
  • $\begingroup$ Thanks, I will give it a read, it looks like a good reference. I might just remove the couple of sentences from the paper as they make no difference to the results. $\endgroup$ May 14, 2016 at 12:23
  • $\begingroup$ I'll send it to you, that would be great. $\endgroup$ May 14, 2016 at 12:41

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