# Fourier Transform of self-similar functions

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.

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 $$\sin(2 \xi)=2\sin(\xi)\cos(\xi)$$.

• Wow, that's really helpful, Thanks. Commented May 14, 2020 at 23:58