Decay of Fourier coefficients and smoothness

Remarks

For part 1, integration by parts over a period gives $\widehat{f'}(n) = in \hat f(n)$, since the boundary terms cancel by periodicity; induction gives the formula for $f^{(k)}$. The Riemann–Lebesgue lemma applied to $f^{(k)}$ then gives $\widehat{f^{(k)}}(n) \to 0$. For part 2, the series $\sum (in)^j \hat f(n) e^{inx}$ converge uniformly by the Weierstrass M-test, so their sums are continuous. This justifies term-by-term differentiation. The sum $g$ has the same Fourier coefficients as $f$, so $g = f$ almost everywhere, as in the uniform convergence of absolutely summable Fourier series. Part 3 follows from parts 1 and 2, since a bound $C_{m+2} (1 + \abs{n})^{-m-2}$ makes $\sum \abs{n}^m \abs{\hat f(n)}$ finite. Example: the sawtooth $f(x) = x$ on $(-\pi, \pi)$ has a jump, and $\abs{\hat f(n)} = 1/\abs{n}$ decays only like $1/\abs{n}$. The triangle wave $\abs{x}$ is continuous with a piecewise continuous derivative, and its coefficients decay like $1/n^2$. Part 1 gives only $o(\abs{n}^{-k})$ and part 2 needs slightly more than $O(\abs{n}^{-k-1})$, so there is a gap between the two directions for a fixed $k$.

Going deeper

For a student

This dictionary between regularity and decay is used constantly. It explains why Fourier series of smooth functions converge very fast, and why the Fourier series of a function with a jump converges slowly and shows the Gibbs phenomenon. It is the starting point of Sobolev spaces on the circle, where smoothness of order $s$ is measured by $\sum (1 + n^2)^s \abs{\hat f(n)}^2 < \infty$. For real analytic functions the coefficients decay exponentially, and conversely.