Bernstein's theorem on absolutely convergent Fourier series

Remarks

Proof idea: for $h > 0$ the function $x \mapsto f(x + h) - f(x - h)$ has Fourier coefficients $2i \sin(nh) \hat f(n)$. By Parseval's identity and the Hölder bound, $\sum_n 4 \sin^2(nh) \abs{\hat f(n)}^2 \le K^2 (2h)^{2\alpha}$. Take $h = \pi / (3 \cdot 2^j)$; for $2^j \le \abs{n} < 2^{j+1}$, $\sin^2(nh) \ge 3/4$, so the sum of $\abs{\hat f(n)}^2$ over this block is $O(2^{-2j\alpha})$. The block has fewer than $2^{j+2}$ terms, so by Cauchy–Schwarz the sum of $\abs{\hat f(n)}$ over it is $O(2^{j(1/2 - \alpha)})$. These bounds form a convergent geometric series when $\alpha > 1/2$.

Example: every continuously differentiable periodic function is Hölder continuous with exponent $1$, so its Fourier series converges absolutely.

The bound $\alpha > 1/2$ is sharp: there are Hölder continuous functions with exponent $1/2$ whose Fourier series does not converge absolutely. Mere continuity is far from enough, since there are continuous functions whose Fourier series even diverges at a point.

Going deeper

For a student

The proof is a model of a standard technique: estimate the coefficients in dyadic blocks with Parseval's identity, then pass from $\ell^2$ to $\ell^1$ with Cauchy–Schwarz. The loss of a factor $2^{j/2}$ in the last step explains the threshold $1/2$. The theorem gives a large class of functions in the Wiener algebra. A related criterion of Zygmund says that a function of bounded variation that is Hölder continuous with some positive exponent also has an absolutely convergent Fourier series.