I've seen many questions and answers about the relation between the smoothness of some function and the rate at which its Fourier transform decays. However, I haven't found the answer to the following question:
If $f \in C^1(\mathbb{R}) \cap L^1(\mathbb{R})$, i. e. $f$ is continuously differentiable and integrable on $\mathbb{R}$, is it true that its Fourier transform is $\mathcal{O}(t^{-1})$ at infinity?
Note that we don't require $f'$ to be integrable.
If $f'$ were integrable this could be shown by the relation $\hat{f}(t) = it \widehat{f'}(t)$ which in turn follows from integration by parts. In my situation however the fourier transform of $f'$ does not even exist (or does is it in some sense?).
I've thought about approximating the original function $f$ by functions with compact support or Schwartz-functions but the best I can get by this method is that ${f}(t) \leq C \cdot t^{-1} + \epsilon$ for every $\epsilon > 0$ and for some constant $C > 0$ which depends on $\epsilon$.