Absolutely convergent Fourier series

Remarks

If the series converges absolutely, then $\sum_n \hat f(n) e^{inx}$ converges uniformly on $\R$ to a continuous function $g$. The function $g$ has the same Fourier coefficients as $f$, so $g = f$ almost everywhere. One often says simply that $f$ has an absolutely convergent Fourier series.

Example: let $f(x) = \abs{x}$ for $-\pi \le x \le \pi$, extended with period $2\pi$. Integration by parts gives $\hat f(0) = \pi/2$ and $\hat f(n) = \frac{(-1)^n - 1}{\pi n^2}$ for $n \ne 0$. So $\abs{\hat f(n)} \le 2/(\pi n^2)$, and the sum $\sum_n \abs{\hat f(n)}$ is finite.

Non-example: the square wave equal to $1$ on $(0, \pi)$ and $-1$ on $(-\pi, 0)$ has $\abs{\hat f(n)} = \frac{2}{\pi \abs{n}}$ for odd $n$. The sum of these over odd $n$ diverges like the harmonic series. This also follows from the remark above: the square wave is not equal almost everywhere to a continuous function.