What assumptions do we need for Fourier transform of derivative formula Suppose $f: \mathbb R \to\mathbb C$ is differentiable and $f$ and $f'$ are in $L^1(\mathbb R)$. Do we need further assumptions to have the formula:
$$\widehat{f'}(t) = (2\pi it)\hat f(t) $$
My textbook also assumes $f$ is continuously differentiable which I don't see why it is needed. Basically we want the limit of $f(x)$ to be zero when $x \to +$ or $- \infty$ which we can prove using the fundamental theorem of calculus, no?
 A: The minimum set of conditions needed can be expressed using the Lebesgue integral. Essentially you need $f$ to be absolutely continuous with derivative $f' \in L^1$. Absolutely continuous means $f$ is the integral of its derivative:
$$
              f(y)-f(x)=\int_{x}^{y}f'(t)dt.
$$
This all works out very nicely using the Lebesgue integral.
For the Riemann integral, you could assume that $f \in L^1$, and that there exists $g\in L^1$ such that
$$
               f(y)-f(x)=\int_{x}^{y}g(t)dt,\;\;\; x,y \in\mathbb{R}.
$$
Using the Riemann integral, you're more or less stuck with assuming that $g$ is Riemann integrable on every finite interval. You could allow a few isolated singularities by employing an improper Riemann integral, but that's about the least you can get away with, if you're going to use the Riemann integral. And it's enough to imply that
$$
            \widehat{f'}(s)=2\pi is\widehat{f}(s),\;\;\; s\in\mathbb{R},
$$
assuming the proper normalization for the Fourier transform.
