Dini test

Remarks

Since the Dirichlet kernel is even with mean $1$, one has $S_N f(x) - s = \frac{1}{2\pi} \int_0^{\pi} \big(f(x + t) + f(x - t) - 2s\big) D_N(t) \, dt$. Write $D_N(t) = \frac{\sin((N + \frac{1}{2}) t)}{\sin(t/2)}$; the hypothesis says that $t \mapsto \frac{f(x + t) + f(x - t) - 2s}{\sin(t/2)}$ is integrable on $[0, \pi]$, because $\sin(t/2) \ge t/\pi$ there. The Riemann–Lebesgue lemma then sends the integral to $0$. Example: if $f$ is differentiable at $x$, or only satisfies $\abs{f(x + t) - f(x)} \le C \abs{t}^\alpha$ for some $\alpha > 0$ and small $t$, then the hypothesis holds with $s = f(x)$, so the Fourier series converges to $f(x)$. Continuity of $f$ at $x$ alone does not suffice, since there are continuous functions whose Fourier series diverge at a point.

Going deeper

For a student

The test is local: it depends only on $f$ near $x$, in line with the Riemann localization principle. At a jump with one-sided Hölder conditions, it applies with $s = \frac{1}{2}(f(x^+) + f(x^-))$. It is independent of the Dirichlet–Jordan test: some functions of bounded variation fail the Dini condition, and some functions satisfying it at every point are not of bounded variation.