I am currently dealing with a problem in functional analysis where I want to show the following.
Let $\phi_{k+1}(x):=\sqrt{2} \sum_{n \in \mathbb{Z}} h(n) \phi_k(2x-n)$ and $\phi_0= \chi_{[0,1]},$ if we define $a_k(n):=\langle \phi_k(.),\phi_k(.-n) \rangle$ and take the operator
$$P:L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$$ such that $$P\hat{f}(\xi) = \frac{1}{2} \left( |\hat{h}(\frac{\xi}{2})|^2 \hat{f}(\frac{\xi}{2})+|\hat{h}(\frac{\xi}{2}+\pi)|^2 \hat{f}(\frac{\xi}{2}+\pi) \right) $$
then this one also satisfies $\hat{a}_{k+1}= P \hat{a}_k$.
Here I give a few thoughts about this, but maybe they are just misleading and the exercise is much simpler: Now I tried to show this, but somehow I could not get anywhere. $h$ is taken in such a way that $\hat{h} \in L^{\infty}$ where $\hat{h}(\xi):= \sum_{n \in \mathbb{Z}} h(n)e^{-in \xi}$ and $\phi_{k+1},a_{k+1} \in L^2$ again. Otherwise, $h$ is arbitrary. Furthermore $\hat{a_{k}}:=\sum_{n \in \mathbb{Z}} \langle \phi(.),\phi(.-n)\rangle e^{-in \xi}.$
It is useful to see ( I guess) that $$\hat{\phi}_{k+1}(\xi) = \frac{1}{\sqrt{2}}\hat{h}(\frac{\xi}{2}) \phi_{k}(\frac{\xi}{2}) = \Pi_{i=1}^{k+1} \left( \frac{\hat{h}(2^{-p}\xi)}{\sqrt{2}} \right) \cdot \hat{\phi_0}(\frac{\xi}{2}).$$
Furthermore by Plancherel and the definition of the Fourier transform we get $\langle \phi(.), \phi(.-n) \rangle= \langle \hat{\phi} , e^{-in(.)} \hat{\phi} \rangle =\mathcal{F}(|\hat{\phi}|^2)(n).$
$\hat{a_{k+1}}(\xi) = \sum_{n \in \mathbb{Z}} \langle \phi_{k+1}(.), \phi_{k+1}(.-n) \rangle e^{-i n \xi}$ which can be rewritten using Plancherel in the inner-product as
$$\hat{a_{k+1}}(\xi) = \frac{1}{2} \sum_{n \in \mathbb{Z}} \langle |\hat{h}(\frac{.}{2})|^2 |\phi_{k}(\frac{.}{2})|^2 ,e^{-in(.)} \rangle e^{-i n \xi}$$
I guess these are all the simple calculations one would do, but somehow this got me nowhere.
Probably you need to use that $\hat{\phi}_0(x) = \frac{i(1-e^{ix})}{x}.$
If anything is unclear, please let me know.