This question is with respect to Theorem 7.1 of Mallat's Wavelet Tour text. It is a follow-up of sorts to a previous question.


Suppose I have a set $\{\theta(t-n)\}_{n \in \mathbb{Z}}$ which is a Riesz basis for a space $V \subset L^2(\mathbb{R})$. For some $\varphi \in V$, I can thus expand:

$$ \varphi(t) = \sum_{n\in \mathbb{Z}} a_n \theta(t-n) $$


$$ \hat{\varphi}(w) = \hat{a}(w)\hat{\theta}(w) $$ where $\hat{f}$ denotes the Fourier transform of $f$, and $\hat{a}(w) = \sum_n a_n e^{-iwn}$.

By letting $$ \hat{a}(w) = \left( \sum_{k \in \mathbb{Z}} |\hat{\theta}(w+2\pi k)|^2 \right)^{-1/2} $$

we actually determine a $\varphi$ such that

$$\langle \varphi(t-m), \varphi(t-n) \rangle = \delta[m-n] \tag{1}$$

That is, integer translates of $\varphi$ form an orthonormal set on $V$. This follows since condition $(1)$ is equivalent to

$$\sum_k |\hat{\varphi}(w+2\pi k)|^2 = 1$$


It is said that this orthonormal set also forms an orthonormal basis for the space $V$, but I can't seem to see why this should be the case. Here is my reasoning:

Suppose we have some arbitrary $f \in V$. Then: $$ f(t) = \sum_{n\in \mathbb{Z}} b_n \theta(t-n) $$ and $\hat{f}(w) = \hat{b}(w) \hat{\theta}(w)$ where $\hat{b}(w) = \sum_n b_n e^{-iwn}$.

Now set $$ g(t) = \sum_{n \in \mathbb{Z}} \langle f(t), \varphi(t-n)\rangle \varphi(t-n) = \sum_{n \in \mathbb{Z}} c_n \varphi(t-n) $$ where we have $\hat{g}(w) = \hat{c}(w)\hat{\varphi}(w)$ with $\hat{c}(w) = \sum_n c_n e^{-iwn}$.

If $\{\varphi(t-n)\}_{n \in \mathbb{Z}}$ is indeed a basis, it should follow that $\hat{f}(w) = \hat{g}(w)$ (and hence $f(t)=g(t)$) or

$$\hat{b}(w) = \hat{c}(w) \hat{a}(w)$$

since $\hat{\varphi}(w) = \hat{a}(w)\hat{\theta}(w)$. My problem is that I can't seem to verify this equality! Is there a more obvious way to show this?


I will expand on Chris's answer below.

Suppose we found coefficients, $b_n$, such that $\theta = \sum_n b_n \varphi_n$. Then

$$ \theta_n = \sum_m b_m \varphi_{m+n} = \sum_m b_{m-n} \varphi_m $$

so that $$ f\in V \Rightarrow f = \sum_n c_n \theta_n = \sum_n \sum_m c_n b_{m-n} \varphi_m $$

We'd like to show these coefficients, $b_n$ exist and that $||\theta - \sum_nb_n\varphi_n||^2 = 0$

Now, $$ ||\theta - \sum_nb_n\varphi_n|| = \frac{1}{2\pi} ||\hat{\theta} - \hat{b}\hat{\varphi}|| = \frac{1}{2\pi} ||\hat{\theta} (1 - \hat{b}\hat{a})|| $$

Consider letting $b_n = \langle \theta, \varphi_n \rangle$. Or rather,

$$ b[n] = (\theta * \tilde{\varphi})[n] $$

where $\tilde{\varphi} = \varphi(-t)^*$ (time reversed conjugate). So then

\begin{align} \hat{b}(w) &= \sum_k \hat{\theta}(w + 2\pi k) \varphi(w + 2\pi k)^* \\ &= \sum_k \hat{\theta}(w + 2\pi k) \theta(w + 2\pi k)^* \hat{a}(w+2 \pi k)^* \\ &= \hat{a}(w) \sum_k |\hat{\theta}(w + 2\pi k)|^2 \\ & = \frac{1}{\hat{a}(w)} \end{align}

So that $||\hat{\theta} (1 - \hat{b}\hat{a})|| = 0$ ( the second to last line owing to the fact that $\hat{a}(w)$ is real and $2\pi$ periodic by definition ).


1 Answer 1


We know that the family $\{T_k\theta\}_{k\in \mathbb Z}$ spans $V$. Therefore, if we can write any of the functions in in this family as a superposition of the functions $\{T_n\varphi\}_{k\in \mathbb Z}$ we can conclude that this family also spans $V$. Therefore we need to find $b_{n,k}$, such that $$ \theta(t-k) = \sum_{n \in \mathbb Z} b_{n,k} \varphi(t-n). \tag{1} $$ Looking at Equation (1) we see through an index shift that it suffices to find $b_{n, 0}$. Taking Fourier transforms we need to construct the sequence, such that $$ \hat \theta(\omega) = \hat b(\omega) \hat \varphi(\omega). \tag{2} $$ Using your formulas from above we see that $\hat b(\omega) = 1/\hat a(\omega)$. So we can inverse Fourier transform Equation (2) to actually get the sequence $b_n$. I have not thought about convergence of all the involved transforms but that should be the correct rationale...

  • $\begingroup$ Thanks, I've added a follow up using your idea. Something still feels a little...funny, but I think you are right that this is the basic gist of it. $\endgroup$
    – Chester
    Feb 27, 2015 at 20:04
  • $\begingroup$ I am glad that you could follow the reasoning behind my answer... it was quite confusing, I was in a bit of a rush when I wrote it. I did some editing to improve it a bit. I hope your funny feeling will subside at a certain point because I think the resoning is true. ;) $\endgroup$
    – Chris
    Mar 2, 2015 at 8:56

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