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


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.


1 Answer 1


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

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

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