I believe the function must be continuous and periodic, and the Fourier series must converge uniformly and absolutely. Is this correct?

Would $f(x) = |x|$ equal its Fourier series?

  • $\begingroup$ Do you want necessary conditions or only sufficient ones? $\endgroup$ Mar 31 at 14:26
  • $\begingroup$ There are tons of examples of Fourier series converging uniformly without converging absolutely; also there are examples where the Fourier series converges pointwise everywhere but not uniformly $\endgroup$
    – Conrad
    Mar 31 at 14:49

2 Answers 2


This is not an easy question - it turns out pointwise convergence of Fourier series is a very fiddly topic.

There are many theorems giving sufficient conditions, which cover a wide class of functions. One useful theorem is:

Theorem (Dirichlet, 1829) Suppose $f$ is bounded, piecewise continuous (has finitely many points of discontinuity), and piecewise monotonous (increases or decreases between a finite number of local minima and maxima). Then for every $x$, the Fourier series converges at $x$, to the value $$\frac{f(x^{+})+f(x^-)}{2}$$ where $f(x^\pm)$ represent the limits from the right and left, respectively.

If we let $f(x)$ be the periodic extension of $\text{abs}(x)$ on, say, $[-\pi, \pi]$ then all the conditions are satisfied. Moreover $f$ is continuous, so the left and right limits are always equal to $f(x)$, thus the Fourier series converges to $f$ everywhere.

From this theorem we can also see it is not necessary for $f$ to be continuous, nor for the Fourier series to converge uniformly. Take, for example

$$ f(x) = \begin{cases}1 & 0 < x \le \pi \\ 1/2 & x= 0 \\ 0 & -\pi \le x < 0 &\end{cases}$$

By Dirichlet's Theorem this function is equal to its Fourier series. However $f$ is discontinuous. And this discontinuity gives rise to the Gibbs phenomenon, preventing uniform convergence.

For a undergraduate-level guide to the many results about the pointwise convergence of Fourier series, I recommend Fourier Analysis by T. W. Körner - specifically the first 19 chapters.


Here is a supplement to @FlipTack's answer. A bit advanced, but the general ideas can be sketched.

  1. As he says, pointwise convergence of a Fourier series for $f$ to all values of the function does not demand continuity, at least not continuity everywhere. [Such a function would, however, have to be continuous at "most" points, i.e., at the points of a dense $\cal G_\delta$.]

  2. Conversely continuity is not enough. The Fourier series of a continuous function can diverge on a large set. You evidently need continuity plus some strong other conditions.

  3. Dirichelet [as the other answer indicates] brought attention to functions with finitely many jump discontinuities and that behaved very nicely and appropriately for the 1800's: monotonic in a finite number of pieces. At a jump discontinuity $x_0$ you could replace the value of the function $f(x_0)$ with the average value $\frac12(f(x_0)+f(x_0-))$, and then you would have convergence to the function everywhere.

  4. A good setting for this question then is not "continuous functions" but regulated functions. These are defined to be those functions with right and left hand limits everywhere. Such functions have only a countable number of jump discontinuities. Dirichelet worked with only a finite number of jump discontinuities and had to assume those strong requirments about monotonicity.

So maybe this is the question which enlarges the scope from continuous functions to the more natural regulated functions:

What extra properties must a regulated function $f$ have so that the Fourier series of $f$ converges pointwise at any $x_0$ to $f(x_0)$ at every point of continuity and to the average $\frac12(f(x_0)+f(x_0-))$ at any of the jump discontinuities.

There is a very interesting key to working with this.

If the Fourier series of a regulated function $f$ diverges at some point, then there is a sequence of pairwise disjoint intervals $\{(a_n,b_n)\}$ such that $$\sum_{n=1}^\infty \frac1n |f(b_n)−f(a_n)|=\infty.$$

This observation is due to Casper Goffman and Dan Waterman from the 1960s. When you look at it you are "reminded" of Dirichelet's condition and can appreciate why he needed some kind of assumption to avoid this possibility.

Here is the paper to read for technical details (allowing me to pay tribute to a late friend who was an inspiration and influence on a generation of real analysts):

Goffman, Casper. Everywhere convergence of Fourier series. Indiana Univ. Math. J. 20 (1970/71), 107–112.

From the Math Review: The author obtains results on the everywhere convergence of the Fourier series of regulated functions that satisfy certain generalized conditions of bounded variation. (Regulated functions are those for which right and left limits exist at every point.) By basing his proofs on a single fundamental fact, the author unifies known results involving different types of generalized bounded variation conditions. Also, known results are extended from continuous functions to regulated functions, although everywhere convergence takes the place of uniform convergence.

The proofs are based on the fact that, if the Fourier series of a regulated function diverges at some point, then there is a sequence of pairwise disjoint intervals $(a_n,b_n)$ such that

$$∑_{n=1}^\infty n^{-1}|f(b_n)−f(a_n)|=∞.$$

The proof of this fact given in the paper depends on a generalization to regulated functions of a theorem of the author and D. Waterman [Proc. Amer. Math. Soc. 19 (1968), 80–86; MR0221193], although the author has since discovered a more direct proof.

$f$ is said to be of bounded $Φ$-variation if it satisfies the usual definition of bounded variation, except that the numbers $|f(x_i)−f(x_{i−1})|$ are replaced by
$Φ\left(|f(x_i)−f(x_{i−1})|\right)$. R. Salem [Essais sur les séries trigonométriques, Hermann, Paris, 1940; MR0002656; correction, MR 2, p. 419] has proved that if $f$ is continuous and of bounded $Φ$-variation and $∑Ψ(1/n)<∞$, where $Φ$ and $Ψ$ are associated in the sense of W. H. Young, then the Fourier series of $f$ converges (uniformly) everywhere. This result is extended to regulated functions by an easy application of a theorem involving Köthe spaces. The Köthe space theorem is an immediate consequence of the basic fact of the paper.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.