Q:

Fourier transform of an aperiodic function

Given a square-integrable function $f(t)\in \mathbb{R}$ defined on $t \in \mathbb R$ which is both 

Aperiodic in the sense that $f(t)=f(t+T)$ if and only if $T=0$
Discretely valued in the sense that $f(t) \in \{f_1, f_2 ... \}$ 

where the $f_i$ are either (a) a finite number or (b) a countably infinite number of distinct real values.
My expectation is that, in both cases (a) and (b), the Fourier transform $\hat{f}(\omega)$ will be decaying as $\hat{f}(\omega)\sim O(\omega^{-1})$ for large $|\omega|$, arguing simply that $f(t)$ must have at least one discontinuity, and the Fourier transform of the step function decays this way.
Is this intuitive picture sufficient to argue this? Or are there cases where it is incorrect.

A:

(b) If countably many values are allowed, the tail of the Fourier transform can be heavier than $|\omega|^{-1}$. Indeed, if $f(x) = |x|^{-\alpha}$ with $0<\alpha<1/2$ near zero (and is truncated at infinity), the Fourier transform will behave like $|\omega|^{\alpha-1}$ at infinity. (Why? The decay of Fourier transform is controlled by $\|\tau_{1/\omega} f - f\|_1$ where $\tau$ is the translation operator. Translating $|x|^{-\alpha}$ by $1/\omega$ and subtracting cancels off some of it, but it leaves a neighborhood of size $1/|\omega|$ around $0$ without much cancellation, and this neighborhood contributes $|1/\omega|^{-\alpha+1}$ to the $L^1$ norm.)
A countably-valued function can have this type of singularity at $0$ too, just discretize  $|x|^{-\alpha}$  at the scales $2^{-k} < |x| < 2^{1-k}$. 
(a) If a function takes on finitely many values on interval, that is, $f(x) = \sum a_k \chi_{I_k}$ where each $I_k$ is an interval, then the explicit form of the Fourier transform can be found, and it indeed decays like $|\omega|^{-1}$.  
But if you allow general measurable sets, i.e, $f(x) = \sum a_k \chi_{E_k}$ for some disjoint measurable sets $E_k$, then again there's a problem because the decay of $\|\tau_{1/\omega}\chi_{E_k} - \chi_{E_k}\|_1$ can be slower than $1/|\omega|$; I think it can actually be arbitrarily slow.

