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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?

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  • $\begingroup$ if $f'$ is (piecewise) continuous, no problem for writing $\int_{-A}^A f'(t) e^{-2 i \pi \xi t} dt = f(A) e^{-2 i \pi \xi A}-f(-A) e^{2 i \pi \xi A} + 2 i \pi \xi\int_{-A}^A f(t) e^{-2 i \pi \xi t} dt$. if $f',f \in L^1$ then $\int_{-A}^A f'(t) e^{-2 i \pi \xi t} dt $ and $\int_{-A}^A f'(t) e^{-2 i \pi \xi t} dt $ converge as $A \to \infty$, and it must exist a sequence $|A_n| \to \infty$ such that $f(\pm A_n) \to 0$ as $n \to \infty$ (because otherwise $f $ cannot be $\in L^1$) $\endgroup$
    – reuns
    Jun 11, 2016 at 0:27
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    $\begingroup$ @user1952009 what would be the problem if $f'$ was not piecewise continuous? $\endgroup$
    – Bananach
    Jun 11, 2016 at 0:30
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    $\begingroup$ @user1952009 I don't know much, but are you sure about the equivalence? I don't think it is necessary for riemann integrable functions to be piecewise continuous $\endgroup$
    – Bananach
    Jun 11, 2016 at 0:42
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    $\begingroup$ Bananach: you're spot-on. Consider the function $f$ that's nonzero only on numbers of the form $x = \frac{n}{2^k}$ in lowest terms ($n$ an integer, $k$ a nonnegative integer) with $f(x) = \frac{1}{2^k}$, is integrable, but discontinuous at all the places where it's nonzero. So integrable doesn't imply piecewise continuous. $\endgroup$
    – John Hughes
    Jun 11, 2016 at 0:58
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    $\begingroup$ On the interval $[0,1]$, the integral is zero. Clearly that's the lower sum for any partition. For upper sums, refine any partition by placing points to the left and right of $2^{-k}$, at a distance of, say, $2^{-k-s}$ (for some $s$). Then the upper sum for the interval containing $n/2^k$ is at most $2^{-k-s}*2^{-k}$. There are at most $2^k$ such intervals in the unit interval, so the sum is no more than $2^{-k-s}$. Summing this over all nonneg $k$ gives at most $2^{-s}$. Picking $s$ large makes this arbitrarily small. So...the integral's zero. $\endgroup$
    – John Hughes
    Jun 11, 2016 at 1:07

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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.

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