As the question suggests I am looking for a (not very restrictive) condition on a function of bounded variation so that its Fourier transform is $o(|u|^{-1})$ as $|u| \to \infty$. Let me elaborate on my actual question:
Let $f: \mathbb{R} \to \mathbb{R}$ be bounded, integrable (i.e. from $L^1(\mathbb{R})$) and of bounded variation. Then, by basic Fourier analysis, $|\mathcal{F}[f](u)|=O( |u|^{-1})$ as $|u|\to \infty$, where \begin{equation} \mathcal{F}[f](u):= \int_{\mathbb{R}}e^{iux}f(x)dx. \end{equation} My question is, are there any not very restrictive conditions we can put on $f$ so that not only the above is true but furthermore $|\mathcal{F}[f](u)|=o(|u|^{-1})$ as $|u|\to \infty$?
The function I am dealing with is, essentially, of the form $f(x)=e^{-|x|}-g(x)$, where $g$ is non-negative, it is in $L^p(\mathbb{R})$ for any $p\in [1,\infty]$, is such that $\lim_{|x|\to 0} g(x)=1$ and it is unimodal at $0$ (i.e. it is non-increasing on $\mathbb{R}^+$ and non-decreasing on $\mathbb{R}^-$). Hence $f$ satisfies the assumptions above and also $\lim_{|x|\to 0} f(x)=0$.
This last property of $f$ seems to be crucial to obtain the $o(|u|^{-1})$ decay in the Fourier transform because \begin{align} \lim_{|u|\to \infty} |u\mathcal{F}[f](u)|= | \, lim_{|u|\to \infty} \int_{\mathbb{R}}e^{iy}f(y/u)dy \, |= | \, lim_{|x|\to 0} \int_{\mathbb{R}}e^{iy}f(xy)dy \, |. \end{align}
The dominated and the monotone convergence theorems can be used to impose fairly general conditions on $f$ to obtain the aforementioned conclusion, but these conditions are too restrictive if we look at the form of my function $f$. Any thoughts on other conditions on $g$ such that the above display is zero? I suppose asking $g$ to have a $(1+\epsilon)$-weak derivative ($\epsilon>0$ arbitrarily small) would be enough, but doesn't my function $g$ have enough structure to imply the desired decay on the Fourier transform of $f$?
A jump discontinuity contributes a term that is not $o(|u|^{-1})$. Indeed, suppose that $f$ is absolutely continuous except for a jump discontinuity at $x=a$. Then the distributional derivative $f'$ consists of an $L^1$ function plus a multiple of $\delta_a$. The transform of an $L^1$ function is $o(1)$ at infinity, but the transform of $\delta_a$ is $e^{-ia u}$ (up to normalizations). Hence, $|\hat{f'}(u)|$ has a nonzero limit at infinity, which implies that $|u|\,|\hat f(u)|$ has a nonzero limit at infinity.
On the other hand, if your function is absolutely continuous (in addition to being BV), then its Fourier transform is $o(|u|^{-1})$. Indeed, absolute continuity implies that $f$ is an antiderivative of a locally integrable function $g$, and $\int|g|$ is the total variation of $f$ (hence, $g\in L^1$). By the Riemann-Lebesgue lemma, $\widehat g(u)=o(1)$ as $u\to\infty$. Hence, $|\widehat f(u) |=\frac{1}{|u|}|\widehat g(u)| = o(|u|^{-1})$.