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?

  • $\begingroup$ This has been already done on this site, I am almost sure. If you find it I'll owe you a favour. $\endgroup$ – Giuseppe Negro Mar 2 '13 at 18:56
  • $\begingroup$ @GiuseppeNegro: I gave it a go already and was unsuccessful, which is why I pulled the trigger. $\endgroup$ – Ron Gordon Mar 2 '13 at 18:56
  • $\begingroup$ As stated, this is not true. Consider $f(x)=\frac{\sin x}{x}$, for example. All derivatives are continuous but since they are not in $L^2$, their transforms aren't either. // Maybe you meant functions that rapidly decay at infinity. $\endgroup$ – user53153 Mar 2 '13 at 19:02
  • $\begingroup$ @5pm: Hmmm. Forgive me, I am not quite following you - which doesn't mean I disagree with you! Your example produces a transform that is just $0$ at $\infty$. The other way, though, illustrates what I am saying perfectly: a rect function is discontinuous and it transform behaves as $1/x$ at $\infty$. Perhaps you could suggest a better way of expressing the problem statement, as I sort of suspected that mine was imperfect. $\endgroup$ – Ron Gordon Mar 2 '13 at 19:07
  • $\begingroup$ I guess my example isn't quite right. What I really meant was that: besides having a discontinuity, a derivative can contribute to the tail of the Fourier transform by having increasingly rapid oscillations toward the infinity... A jump discontinuity is a bunch of high frequencies localized at the same point in space. If a function has the same frequencies spread apart in space, it will not have a discontinuity and yet the transform will have a heavy tail. $\endgroup$ – user53153 Mar 2 '13 at 19:21

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

  1. $g'$ exists on $\mathbb R\setminus D$
  2. 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. $\Box$


  • 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.
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  • $\begingroup$ Many thanks for such a thorough statement that seems to capture what is going on. I will check through your solution in detail and see how well I understand. $\endgroup$ – Ron Gordon Mar 3 '13 at 1:14
  • $\begingroup$ Why does it suffice to work with $|\xi|^2$? $\endgroup$ – Ron Gordon Mar 4 '13 at 18:56
  • $\begingroup$ @rlgordonma We want to prove something about $|\xi|^{m+1}|\hat f(\xi)|$. Since $|\hat f(\xi)|=|\xi|^{m-1}|\hat g(\xi)|$, it follows that $|\xi|^{m+1}|\hat f(\xi)|=|\xi|^{2}|\hat g(\xi)|$, so we work with the latter. $\endgroup$ – user53153 Mar 4 '13 at 19:00
  • $\begingroup$ OK, not sure why that escaped me. Thanks. $\endgroup$ – Ron Gordon Mar 4 '13 at 19:05
  • $\begingroup$ Do you know a written source for this? Need it as a reference, but as op wrote I have not found anything. Any directions would be appreciated. $\endgroup$ – Raibyo Jun 2 '17 at 12:46

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