Uniform convergence of absolutely summable Fourier series

Remarks

Since $\abs{\hat f(n) e^{inx}} = \abs{\hat f(n)}$, the Weierstrass M-test shows that $S_N f$ converges uniformly to a continuous $2\pi$-periodic function $g$. Uniform convergence allows integration term by term, so $\hat g(n) = \hat f(n)$ for all $n$. Then $f - g$ is continuous with all Fourier coefficients zero, and its Fejér means, which converge uniformly to $f - g$ by Fejér's theorem, are all zero; hence $f = g$. Example: for $f(x) = \abs{x}$ on $[-\pi, \pi]$, extended periodically, $\hat f(0) = \frac{\pi}{2}$ and $\hat f(n) = \frac{(-1)^n - 1}{\pi n^2}$ for $n \ne 0$, which is summable. Without continuity the conclusion fails: if $f$ is only integrable with summable coefficients, the series converges uniformly to a continuous function that equals $f$ only almost everywhere. Every $f \in C^1$ periodic satisfies the hypothesis, by the Cauchy–Schwarz inequality and Parseval's identity applied to $f'$.

Going deeper

For a student

This is the simplest sufficient condition for uniform convergence, and it is easy to check from decay of the coefficients, for instance $\abs{\hat f(n)} \le C \abs{n}^{-1-\delta}$ with $\delta > 0$. The functions with summable Fourier coefficients form the Wiener algebra $A(\mathbb{T})$. Bernstein's theorem shows that Hölder continuity of order $\alpha > \frac{1}{2}$ is enough to be in $A(\mathbb{T})$. The condition is far from necessary: there are continuous functions whose Fourier series converge uniformly but not absolutely.