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Consider a certain function $f : [-\pi,\pi] \to \mathbb{R}$, such that it is differentiable $p$ times, $f \in C^p[-\pi,\pi]$. It is usually said, when writing down its Fourier series, that the Fourier coefficients will scale, for large $n$, as $c_n \sim 1/n^p$. As a consequence, infinitely differentiable functions will have Fourier coefficients that decay exponentially with $n$.

However, if I consider the simple function $f(x) = e^{-x}$, its Fourier coefficients are given by

$ c_n = \sqrt{\frac{2}{\pi}} (-1)^n \frac{\sinh(\pi)}{1+i n} \,, $

which is definitely not exponential.

My intuition is that this is related to the fact that the requirement on the $p$-th derivative is understood for the function extended to the whole real axis as a $2\pi$-periodic function.

Is this correct? If yes, how does this fact enter in the proof regarding the scaling of the Fourier coefficients?

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    $\begingroup$ Because you should look at the periodic function on $\Bbb R$, and this one is discontinuous $\endgroup$
    – LL 3.14
    Commented Apr 5, 2023 at 7:13

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It is known that if $f \in C^p(\mathbb{R}/2\pi\mathbb{Z})=C^p(\mathbb{S}^1)$ (not $C^p[-\pi, \pi]$), then $|c_n| \le \frac{a}{|n|^p}$ ($a$ is some constant). Regarding your function, It's not even continuous in $ C^0(\mathbb{R}/2\pi\mathbb{Z})$! Discontinuity occurs at $\pi \sim -\pi$. Also, infinitely differentiable functions do not necessarily have Fourier coefficients that decay exponentially with $n$, although the decaying speed is faster than any polynomial.

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