Decay rate of Fourier coefficients of a continuous function with discontinuous derivative Can you prove or disprove the following:
The Fourier coefficients of a continuous function with discontinuous derivative decay like $ 1/n^2$.
 A: In stated generality this is false. Take  some periodic function with behaves like $f(x)=x^{1/3}$ near $0$. Since $f'(x)\notin L^2$, the Fourier coefficients of $f'$ are not in $\ell^2$.  In particular, they cannot decay like $1/n$. 
The following  (adapted from here) is true. 
Claim. Suppose $f $ is piecewise differentiable, $f'$ has bounded variation on each "piece" $(a_k,a_{k+1})$ and   has nonzero jump discontinuity at some $a_k$.  Then the sequence $n^2\widehat f(n)$ is bounded but does not tend to $0$. 
Proof: $n^2 \widehat f(n)$ has the same size as $\widehat \mu(n)$, where $\mu$ is the signed measure such that $f'=\int \,d\mu$. (That is, $\mu$ is the weak/distributional derivative of $f'$). By  the assumptions on $f'$, the measure $\mu$ consists of absolutely continuous component and some  point masses. The Fourier series for the absolutely continuous part has coefficients that tend to $0$, by the Riemann-Lebesgue lemma. The Fourier series for the sum of point masses is a trigonometric polynomial; therefore it is bounded but does not tend to $0$. $\quad \Box$
With some extra machinery, Claim can be generalized: 
Claim 2. Suppose $f $ is absolutely continuous, $f'$ has bounded variation and is not continuous. Then the sequence $n^2\widehat f(n)$ is bounded but does not tend to $0$. 
Proof goes as above, but the sticky point is to show that $\widehat \mu(n)\not\to 0$. A measure such that $\widehat \mu(n) \to 0$ is called a Rajchman measure. From the survey Seventy years of Rajchman measures by Lyons we find that a Rajchman measure has no atoms (proved by Neder in 1920). But $f'$ is not continuous, which means the measure $\mu$ has atoms. $\quad \Box$
