Relation between function discontinuities and Fourier transform at infinity I have made the following assertion a few times in this space without ever having provided a proof:
Let $m$ be the smallest number such that a function $f \in L^2(\mathbb{R})$ has a discontinuity in its $m$th derivative.  (That is, the $(m-1)$th and lower derivatives of $f$ are continuous.)  Then $\hat{f}(k) \sim A k^{-(m+1)}$ as $k \rightarrow \infty$, where
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) e^{i k x}$$
is the Fourier transform of $f$, and $A$ is a constant.
I have looked for a proof of this statement without success.  Does anyone know of such a proof, or if it is not true, a counterexample?
 A: Proposition. Suppose that $f$ is $(m-1)$ times continuously differentiable, and let $g=f^{(m-1)}$. Suppose that $g$ tends to $0$ at infinity, and there exists a finite nonempty set $D$ such that 


*

*$g'$ exists on $\mathbb R\setminus D$

*There exists $h\in L^1(\mathbb R)$ such that $g'(x)-\int_{-\infty}^xh(t)\,dt$ is constant on each connected component of $\mathbb R\setminus D$.


Then $|\xi|^{m+1}|\hat f(\xi)|$ is bounded as $\xi\to\infty$, but does not tend to zero.

Proof. Since $|\hat f(\xi)|$ is equal to $|\xi|^{m-1}|\hat g(\xi)|$ up to some constant factor, it suffices to work with $ |\xi|^{2}|\hat g(\xi)|$. Split the integral defining $\hat g$ into integrals over connected components of $\mathbb R\setminus D$, denoted $(a_k,a_{k+1})$ below, $-\infty=a_0<\dots<a_n=+\infty$. And integrate by parts:
$$
\hat g(\xi)=\sum_k \int_{a_k}^{a_{k+1}} e^{-i\xi x} g(x)\,dx = 
\frac{1}{i\xi}\sum_k \int_{a_k}^{a_{k+1}} e^{-i\xi x} g'(x)\,dx
\tag1 $$
No boundary terms appear because $g$ is continuous and vanishes at infinity. But they do appear when we integrate again, turning (1) into 
$$
\frac{-1}{\xi^2}\sum_k \int_{a_k}^{a_{k+1}} e^{-i\xi x} h(x)\,dx -\frac{1}{\xi^2} \sum_{k=1}^{n-1} e^{-i\xi a_k} (g'(a_k-)-g'(a_k+))
\tag2 $$
By the Riemann-Lebesgue lemma, the first integral in (2) tends to zero as $\xi\to\infty$. Therefore, 
$|\xi|^2|\hat g(\xi)|=o(1)+|P(\xi)|$, where $P$ is a nonzero trigonometric polynomial. $\Box$

Remarks 


*

*In the special case when $D$ consists of one point, $|P|$  is constant and therefore $|\xi|^{m+1}|\hat f(\xi)|$ has a finite nonzero limit at infinity.

*Condition 2 can be expressed by saying that $g'$ is absolutely continuous on $\mathbb R\setminus D$  with integrable derivative. 

