In these notes (https://www.math.ucla.edu/~tao/preprints/fourier.pdf), a simple prototype for the fourier transform over the multiplicative group $\{1,-1\}$ is given as, \begin{equation} f_{even}(x) = \frac{f(x)+f(-x)}{2} \hspace{1cm} f_{odd}(x) = \frac{f(x)-f(-x)}{2} \end{equation} Building on this, when the group is roots of unity and the function is $f:\mathbb{C}\rightarrow\mathbb{C}$, and it is assumed that it can be expanded in harmonics $f_j$ of order $j$, where $f_j(e^{2\pi i/n} z)=e^{2\pi i j/n} f_j(z)$, we have, \begin{equation} f=\sum_j f_j \hspace{1cm} f_j(z)=\frac{1}{n} \sum_k f(e^{2\pi i k/n} z) e^{-2\pi i j k/n} \ \ \ (\star) \end{equation} This theme continues for fourier transforms with expansion like $f(z)=\sum_n c_n z^n$. Finally when we get to the regular fourier transform and the group is the translation group it is not written with the group action so explicit like in the other prototypical fourier transforms. Can the translation operation on $f(x)$ be somehow made more explicit and then we arrive at the usual formulas of the Fourier transform?
I tried this but it looks too trivial and silly. Assuming that $f(x)$ is a continuous sum of some functions $f_{\xi} (x)=\hat{f} (\xi) e^{2\pi i x \xi}$. Then similar to $(\star)$ we can write (with $T_a(x)=x+a$),
\begin{align} f_\xi(x)&=\int_a f\left(T_a(x)\right) e^{-2\pi i\xi a} da\\ &=\hat{f}(\xi) e^{2\pi ix \xi} = \int_a f\left(T_a(x)\right) e^{-2\pi i\xi a} da \end{align} Then I can set $x=0$, and arrive at the usual formula for the fourier transform, \begin{equation} \hat{f} (\xi) = \int_a f(T_a(0)) e^{-2\pi i\xi a} da = \int_a f(a) e^{-2\pi i\xi a} da \end{equation} Its silly and trivial because its just relabelling of some variables, but my aim was to make more explicit that the fourier transform is exploiting how the various functions $f_\xi(x)$ transform under translations to isolate their form. Are there any mistakes here or is there a better way to do this?