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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?

Ron Gordon
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  • This has been already done on this site, I am almost sure. If you find it I'll owe you a favour. – Giuseppe Negro Mar 02 '13 at 18:56
  • @GiuseppeNegro: I gave it a go already and was unsuccessful, which is why I pulled the trigger. – Ron Gordon Mar 02 '13 at 18:56
  • 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. –  Mar 02 '13 at 19:02
  • @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. – Ron Gordon Mar 02 '13 at 19:07
  • 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. –  Mar 02 '13 at 19:21
  • @5pm: perhaps there's something there with the oscillations. I am not interested in those, just the modulating behavior. But if the oscillations have something to say about behavior at infinity, then please expand some. Put it in the answers and I'll at least upvote it if it adds to the question. – Ron Gordon Mar 02 '13 at 19:27
  • @AntonioVargas: I have very good references on MSP. But how would you apply it to this problem? – Ron Gordon Mar 04 '13 at 18:29

1 Answers1

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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. $\Box$

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.
  • 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. – Ron Gordon Mar 03 '13 at 01:14
  • Why does it suffice to work with $|\xi|^2$? – Ron Gordon Mar 04 '13 at 18:56
  • @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. –  Mar 04 '13 at 19:00
  • OK, not sure why that escaped me. Thanks. – Ron Gordon Mar 04 '13 at 19:05
  • 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. – Raibyo Jun 02 '17 at 12:46