---
tag: F5PHA35M50
kind: statement
role: theorem
name: Dini test
msc: [42A20]
uses: [G12KXDHPPS, 3JY4BVE4DW, WVKDC011KP, 0ATF6Q0633]
about: [G12KXDHPPS]
sources:
  - url: https://en.wikipedia.org/wiki/Dini_test
same_as:
  wikidata: Q898252
  wikipedia: Dini test
---

## Statement

Let $f \colon \R \to \C$ be a [[periodic function|0ATF6Q0633]] of period $2\pi$ that is integrable on $[-\pi, \pi]$, let $x \in \R$ and let $s \in \C$.
Suppose that
$$\int_0^{\pi} \frac{\abs{f(x + t) + f(x - t) - 2s}}{t} \, dt < \infty.$$
Then the partial sums $S_N f(x)$ of the [[Fourier series|G12KXDHPPS]] of $f$ converge to $s$ as $N \to \infty$.

## Remarks

Since the [[Dirichlet kernel|3JY4BVE4DW]] 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|WVKDC011KP]] 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.

## Explanations

### L1 human

The Dini test checks convergence of a Fourier series at one point by looking only at how fast the function approaches a value near that point.
If the symmetric difference $f(x + t) + f(x - t) - 2s$ goes to zero fast enough that dividing by $t$ still leaves something integrable, the series converges to $s$.
A small power of $t$ is enough, so all differentiable and all Hölder continuous functions pass.

### L2 human

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.