\begin{lemma}
Let $f:\mathbb{R}\to \mathbb{C}$, $f\in L^1$. Assume $f$ is absolutely continuous, and that the total variation $|f'|_{TV}$ of $f'$ is finite. Then
$|\widehat{f}(t)|\leq \frac{|f'|_{TV}}{(2\pi |t|)^2}$ for all $t\ne 0$.
\end{lemma}
\begin{proof}
By integration by parts, applied twice. You may need to be careful the second
time, using Lebesgue-Stieltjes (if you have it) or Riemann-Stieltjes.

You will need that $f'(x)\to 0$ as $x\to \pm\infty$ (because $f'$ has bounded
total variation, hence finite limits as $x\to \pm\infty$, and a non-zero limit would contradict $f\in L^1$) and
$f(x)\to 0$ as $x\to \pm\infty$ (since $f'(x)\to 0$ as $x\to \pm\infty$,
$f'$ is bounded; hence, you could not have $f\in L^1$ if you did not have
$f'(x)\to 0$).
\end{proof}