# Fourier coefficients of exponential function: why don't they decay exponentially?

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?

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

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.