I am taking a mathematical methods course where I encountered a contradiction about the Fourier transform of a function's derivative.

(The $\tilde{}$ denotes the Fourier Transform)

For Fourier Transforms, we have the following properties:

  1. Linearity: $h(x) = \lambda f(x) \iff \tilde{h}(k) = \lambda \tilde{f}(k)$.
  2. Derivative Rule: $h(x) = f'(x) \iff \tilde{h}(k) = ik \tilde{f}(k)$.

However, we also have the following functions and their Fourier Transforms:

Note that $f,g$ are defined for $|x|\le c$, and zero otherwise.


  1. $f(x) = \sin ax \iff \tilde{f}(k) = -i(\frac{\sin((a-k)c)}{a-k}-\frac{\sin((a+k)c)}{a+k})$
  2. $g(x) = \cos ax \iff \tilde{g}(k) = \frac{\sin((a-k)c)}{a-k}+\frac{\sin((a+k)c)}{a+k}$


According to properties 1 and 2, we should have $\tilde{g}(k) = (ik/a)\tilde{f}(k)$, but this is not the case when we try to verify it using 3 and 4.

My thoughts

Problematic causes might be:

  • (Possible) jump discontinuities of f and g

However, these suspicions seem unfounded, because they don't violate the derivation of Rule 2. Any help is appreciated. Thank you.

Appendix Derivation of 2 (integrate by parts): $\tilde{h}(k) = \int_{-\infty}^{\infty}f'(x)e^{-ikx}dx = ik\int_{-\infty}^{\infty}f(x)e^{-ikx}dx + [f(x)e^{-ikx}]_{-\infty}^{\infty}$, and the second term is zero for well-behaved functions (those tending to zero as $|x| \rightarrow \infty$).

In particular, this derivation is not violated for 3 and 4, because they are defined to be zero outside a closed interval.


I think the issue is that the discontinuity of $ f $ and $ g $ invalidates the derivation of property 2 in the appendix, because the derivation assumes that the function in question is differentiable. Even if f and g are continuous at $ \pm c $ (which they might be for certain values of $ c $), they aren't differentiable. That might seem like a technicality, but I think it really matters when we're dealing with Fourier transforms, where even a very localized function (i.e. a delta function) is sent to an infinite sine wave, and intuitions about what a function's Fourier transform should look like don't always hold.

On a side note, you wouldn't expect property 2 to hold for functions like this, because you can apply it repeatedly to get:

$ h(x) = f^{n}(x) \implies \tilde{h}(x) = (ik)^{n} \tilde{f}(x) $

and if you applied this (with n=2) to the original function $ f(x) = \sin{ax} $ (restricted to $ -c < x < c $), you'd find that:

$ -a^{2}\tilde{f}(k) = -k^{2}\tilde{f}(k) $

which clearly makes no sense! This example shows that when you apply property 2, you need the function to be very well behaved: not just to tend to $ 0 $ at $ \pm \infty $, but also to be differentiable everywhere.

To add in another helpful analogy (although at this point I might just be making it more confusing!), we learnt earlier in the course that we can't always differentiate Fourier series, because we could get an infinite sum that doesn't converge. The function has to be sufficiently well-behaved for us to be allowed to differentiate it - I think that's analogous to this case.

I hope something that I've said here has been helpful!

  • $\begingroup$ Thank you. This is really helpful. The complicating thing here for me is that although differentiability is a sufficient condition, it is not a necessary one. For example, if you substitute $a = 1$ and $c = \pi / 2$ then you get $\tilde{f}(k) = (2\cos (k \pi / 2))/(1-k^2)$, and $\tilde{g}(k)$ is actually $ik$ times this. $\endgroup$ – Benjamin Wang Nov 27 '19 at 21:31
  • 1
    $\begingroup$ Oh yeah, I hadn't considered that as a special case - maybe the problem is discontinuity rather than undifferentiability then. That would still make sense with the example I gave using n=2, because for a fixed value of a, you couldn't have both $ f(x) $ and $ g(x) $ continuous, meaning even if you could apply property 2 once you couldn't apply it a second time. $\endgroup$ – cm976 Nov 27 '19 at 21:49

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