Many other posts have discussed the standard result that the smoothness of a function is related to the rate at which its Fourier coefficients decay. For example, there are proofs that show that if $f$ is of class $C^k$ then $|c_n|$ tends to zero faster than $|n|^{-k}$ as $n\to\pm\infty$.

Is there a intuitive geometric picture to explain this? If $f$ is not differentiable imagine slowly changing the function so that it becomes twice differentiable, then the Fourier coefficients $c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta) e^{-in\theta}d\theta$ should change their rate of decay, but how is that obvious from the definition of the Fourier coefficients.

What about considering $c_n(\theta) = \frac{1}{2\pi}\int_{-\pi}^{\theta}f(\phi) e^{-in\phi}d\phi$. Is there some kind of bad behavior in $c_n(\theta)$ when it encounters a discontinuity in $f^{(k)}(\theta)$ that would prevent $c_n(\theta)$ from being small.

  • $\begingroup$ One way to think about this intuitively is to consider that $\sin(nx)$ and $\cos(nx)$ get "spikier" for larger $n$ values. To represent abrupt changes in a discontinuous function (or a function which is less smooth), we might need high frequency components. $\endgroup$
    – Anu
    Commented Aug 31, 2023 at 18:55

1 Answer 1


It really boils down to integration by parts.. the $n$th Fourier coefficient of ${d^k f \over dt^k}$ is given by $(in)^k \hat{f}(n)$, which you prove by integrating by parts $k$ times. If $f$ is $C^k$, then the Fourier coefficients of the continuous function ${d^k f \over dt^k}$ go to zero by the Riemann-Lebesgue lemma, so you get that $|n|^k|\hat{f}(n)|$ goes to zero or that $|\hat{f}(n)|$ is $o({1 \over |n|^k})$.

If you have a small discontinuity in the $k$th derivative, you might get a small error term like $\epsilon {1 \over |n|^{k-{1 \over 2}}}$ just to give an example, and as the discontinuity goes to zero the $\epsilon$ will go to zero... but for any $\epsilon > 0$ it will still be the dominant term if $|n|$ is large enough. So slowly changing the function will gradually make this term go away in one sense, but it will still dominate in another sense regardless of what $\epsilon$ is.


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