1
$\begingroup$

I am struggling with a problem for a Fourier Analysis course I am doing.

I am given a definition that a function $\phi\in L^2(\mathbb{R})$ is self similar if there is a sequence $\{h_k\}_{k=-\infty}^{\infty}$ such that $\frac{1}{2}\phi\left(\frac{x}{2}\right)=\sum_{k=-\infty}^{\infty}h_k\phi(x-k)$

And I need to show that $\phi$ satisfies this definition if and only if its Fourier transform, $\hat{\phi}$ satisfies

$\hat{\phi}(2\xi)=m(\xi)\hat{\phi}(\xi)$

for some 1-periodic function $m$ with $\int_{0}^{1}|m(\xi)|^2d\xi<\infty$

I have never encountered self-similarity before now just to be clear and when I looked it up I didn't see how it linked to the definition I have. So far I have tried manipulating just to get the forward direction of the iff statement but I just don't feel like I am getting anywhere. Any hints or other help would be appreciated.

$\endgroup$

1 Answer 1

3
$\begingroup$

The initial equation

$$\frac{1}{2}\phi\left(\frac{x}{2}\right)=\sum_{k=-\infty}^{\infty}h_k\phi(x-k) \tag{1}$$

is equivalent (using classical properties of this isometric transform) :

$$\hat{\phi}(2\xi)=\sum_k h_k \left(\hat{\phi}(\xi) e^{2i \pi k \xi}\right)\tag{2}$$

$$\hat{\phi}(2\xi)=\hat{\phi}(\xi)\underbrace{\sum_k h_ke^{2i \pi k \xi}}_{m(\xi)}\tag{3}$$

You recognize in the summation the complex form of the Fourier series (not transform) of a certain function $m$ and the work is done under the condition (which isn't given in the question...) that

$$S:=\sum_k |h_k|^2 < \infty$$

in which case it is well known (Parseval formula) that $S=\int_{0}^{1}|m(\xi)|^2d\xi$

Remark : we have to justify that the Fourier Transform of the sum is the sum of its Fourier Transforms.


Edit: An example illustrating formulas (1) and (2)

Le us take for $f$ the tent function defined by

$$f(x)=\begin{cases}1-|x|& \text{if } \ x \in [-1,1] \\ 0&\text{otherwise}\end{cases}$$.

It is easy to show geometrically (see figure) that $\frac12 f(x/2)$ which is a flattened enlarged tent can be written as the combination of three (still smaller) tents/shifted tents :

$$\frac12 f(x/2)=\color{red}{\frac14 f(x+1)}+\color{blue}{\frac12 f(x)} + \color{green}{\frac14 f(x-1)}$$

with $$m(\xi)=\color{red}{\frac14} e^{-2i\pi \xi}+\color{blue}{\frac12}+\color{green}{\frac14 e^{2i\pi \xi}=}\frac12+\frac14\left(e^{2i\pi \xi}+e^{-2i\pi \xi}\right)=\frac12\left(1+\cos(2\pi \xi)\right)=\cos(\pi \xi)^2$$

How can we check that (2) is true ?

If you happen to know that the Fourier transform of the tent function $f$ is sinc$^2(\xi):=\dfrac{\sin(\pi \xi)}{\pi \xi}$ (the square of the cardinal sine), we have just to verify that :

$$\text{sinc}^2(2 \xi)=\text{sinc}^2(\xi)\cos^2(\pi \xi) \ \iff \ \frac{\sin^2(2 \pi \xi)}{(2 \pi \xi)^2}=\frac{\sin^2(\pi \xi)}{(\pi \xi)^2}\cos^2(\pi \xi) $$

which is true, due to relationship $\sin(2 \xi)=2\sin(\xi)\cos(\xi)$.

enter image description here

$\endgroup$
1
  • $\begingroup$ Wow, that's really helpful, Thanks. $\endgroup$ May 14, 2020 at 23:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.