Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 2 '13 at 18:56
    
@GiuseppeNegro: I gave it a go already and was unsuccessful, which is why I pulled the trigger. –  Ron Gordon Mar 2 '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. –  user53153 Mar 2 '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 2 '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. –  user53153 Mar 2 '13 at 19:21
show 2 more comments

1 Answer

up vote 7 down vote accepted

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.
share|improve this answer
    
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 3 '13 at 1:14
    
Why does it suffice to work with $|\xi|^2$? –  Ron Gordon Mar 4 '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. –  user53153 Mar 4 '13 at 19:00
    
OK, not sure why that escaped me. Thanks. –  Ron Gordon Mar 4 '13 at 19:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.