Frames in $L^2$. Fix $0<\beta<1$ and for all $f \in L^2(\mathbb{T})$, where $\mathbb{T}= [0,1)$ show that we have:

*

*$$ \sum_{n \in \mathbb{Z}} |\langle f, e^{2 \pi i n \beta x}\rangle|^2 = \frac{1}{\beta} \|f\|_2^2, $$

*$$ f = \beta \sum_{n \in \mathbb{Z}} \langle f, e^{2 \pi i n \beta x}\rangle e^{2 \pi i n \beta x}.$$
This question was a bit inspired by this one here. Now it seems to me that the spirit of this problem is try to reduce the set of $\{e^{2 \pi i n \beta x}\}$ to the set $\{e^{2 \pi i n x}\}$ via some change a varibles. The reason we would like to do this is because we are familiar with everything in $L^2(\mathbb{T})$ in terms of  $\{e^{2 \pi i n x}\}$. However, it seems I cannot find the correct choice of variables to reduce this nicely, it gets quite messy.
I am unsure in what direction I should be heading, any help is appreciated.
Thanks! :)
 A: For any $f \in L^2(\mathbb{T})$, we have $f = \sum_{n \in \mathbb{Z}} \langle  f, e_n\rangle e_n $ where $e_n(x) = e^{2\pi i n x}$ and $\langle f, g\rangle = \langle f, g \rangle_{L^2[0,1]}  $.  Define $D_{\beta}f(x) := f(\beta x)$.
We will show
$$ f = \beta \sum_{n \in \mathbb{Z}} \langle f, e_{\beta n} \rangle e_{\beta n} $$
This is the second claim in your post. The first claim follows from the second easily by taking the inner product with $f$. Let  $\bar{f} := \chi_{[0,\beta]}(D_{1/\beta }f)$. Then
\begin{align}
\int_0^1 |\bar{f}(x)|^2 & = \int_0^{\beta} |f(\frac{x}{\beta} )|^2  \ dx \\
& = \beta \int_0^1 |f(u)|^2 du \\
& < \infty
\end{align}
so $\bar{f} \in L^2[0,1]$. Therefore we may express $\bar{f}$ as a Fourier series $\bar{f} = \sum \langle \bar{f}, e_n \rangle e_n  $. Observe, that
\begin{align} 
\langle  \bar{f} , e_n \rangle   = &  \langle D_{1/\beta}  f, e_n \rangle_{L^2[0,\beta]}  \\
& = \beta \langle f , e_{\beta n}\rangle _{L^2[0,1]} 
\end{align}
Lastly, we have $(D_{\beta} \bar{f}) = f$. So, we have
\begin{align}
\bar{f}  & =   \beta \sum  \langle e_{\beta n}, f\rangle _{} e_{ n} \\
\text{applying} \  D_{\beta}\Longrightarrow f & = \beta \sum  \langle e_{\beta n}, f\rangle _{} e_{\beta n}
\end{align}
There is a subtly here in that you need $D_{\beta}: L^2[0,1] \to L^2[0,1] $ to be a continuous operator, but verifying this should be straightfoward.
