I was reading an answer (the only one so far) to this question.
I am wondering about the following statement:
$f$ is absolute continous and of bounded variation (hence has an antidrivative $g$), $f$ has total variation of $\int |g| dx<\infty$ (and hence $g$ is in $L^1$). That's all clear so far.
Then one can apply the Riemann-Lebesgue Lemma and show $$ \widehat g(u)=o(1) $$
but how do we can conclude this statement:
$$|\widehat f(u) |=\frac{1}{|u|}|\widehat g(u)| = o(|u|^{-1})$$
The second equality is clear, but the first one stays a mystery to me. Any insights?