TITLE: What does Dini continuity mean?
QUESTION [9 upvotes]: What does Dini continuity (the integral condition) mean visually?
Description of Dini contuity:
https://en.wikipedia.org/wiki/Dini_continuity

REPLY [3 votes]: The Dini condition has no full characterization because there is no borderline case of $\omega$ where the integral converges. This condition arises out of the Dini convergence criterion for the Fourier series $S_{f}(x)$ of a function $f$ at $x$. You can write the difference between the truncated Fourier series $S_{f}^{N}(x)$ for $f$ on $[-\pi,\pi]$ and some potential limit $L$ as
$$
   S_{f}^{N}(x)-L = \frac{1}{\pi}\int_{0}^{\pi}\frac{f(\theta+\theta')+f(\theta-\theta')-2L}{\sin((\theta-\theta')/2)}\sin((N+\frac{1}{2})(\theta-\theta'))d\theta'.
$$
If $\theta$ is fixed and $\theta'=\theta+h$, then
$$
          \frac{f(\theta+\theta')+f(\theta-\theta')-2L}{\theta-\theta'}
$$
has the form $\omega(h)/h$. If this is absolutely integrable in a neighborhood of $0$, then then the Riemann-Lebesgue lemma gives $\lim_{N} (S_{f}^{N}(x)-L)=0$, meaning that the Fourier series for $f$ converges to $L$ at $\theta$.