Spectral convergence of coefficients of a Fourier series

Question

I have seen claims that if a smooth function $f(x)$ is represented by its Fourier series, $f(x)=\sum_{n=-\infty}^\infty a_ne^{i(nt)}$, then as $|n|\rightarrow\infty$, then $|a_n|\rightarrow 0$ "exponentially fast", also known as spectral convergence. I have heard some heuristic and empirical arguments, particularly in the context of numerical analysis, but rigorous treatments in the language of functional analysis and group theory were surprisingly hard to find. I was wondering if someone could give a higher view of proving the spectral convergence or point to the right references.

Answer 1

Smoothness of a function and the decay of $a_n$ at infinity are connected.

For smooth functions ( $C^{\infty}$), $a_n$ coverges to $0$ faster than any power of $1/n$, that is

$$\lim_{n \to \infty} n^k \cdot a_n = 0$$

or $|a_n| = O(\frac{1}{n^k})\ \forall k>0$.

For real analytic functions ( equivalent to : restriction to the circle of a holomorphic function defined on an open set containing $S^1$) there exist $q < 1$ and $C>0$ so that

$$|a_n| \le C\cdot q^n$$ that is $|a_n| = O(q^n)$ ( exponential decay).

${\bf Added:}$

The estimates for smooth functions are based on the following facts.

For continuous functions $f$ we have $a_n \to 0$ ( in fact it is true for $L^1$ functions - this is called Riemann-Lebesgue lemma).

If $f$ is a function of class $C^1$ with continuous derivative $f'$ then we have using integration by parts ( note that the function has period $2\pi$) $$a_n= \frac{1}{2\pi}\int_0^{2\pi}f(t)e^{-int}\, dt = -\frac{1}{2\pi}\int_0^{2\pi} f'(t) \left(\frac{1}{-i n } e^{-int}\right) dt =\\ = \frac{-i}{n}\frac{1}{2\pi} \int_0^{2\pi}f'(t) e^{-int} dt = \frac{-i}{n} a'_n$$ where $a'_n$ is the Fourier coefficient of the function $f'$. One proves by induction now for function of class $C^k$ that $$a_n= \left(\frac{-i}{n}\right)^k a^{(k)}_n$$ where $a^{(k)}_n$ is the $n$ -th Fourier coefficient of $f^{(k)}$. Since $a^{(k)}_n\to 0$ as $n\to \infty$, we get the estimate $$a_n = o(\frac{1}{n^k})$$

Conversely: if we have a sequence $(a_n)_{n \in \mathbb{Z}}$ so that $\sum_n |a_n| < \infty$ then the function $f(t) = \sum a_n e^{int}$ is continuous and has Fourier series $(a_n)$. If we have a sequence $(a_n)$ with $a_n = o(\frac{1}{n^k})$ for all $k$ then $\sum |a_n| < \infty$ and the function $f(t) = \sum_n a_n e^{int}$ is smooth ( differentiate term by term as many times as wanted).