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Carleson theorem (later extended by Hunt) states that given an $L^2$ function $f:{\mathbb R}/{\mathbb Z}\to{\mathbb C}$, the set of points $x$ where the Fourier series $$\lim_{n\to\infty}\sum_{k=-n}^n\hat f(k)e^{2\pi ik x}$$ does not converge to $f(x)$ has measure 0.

Kahane and Katznelson proved that given any measure zero set $E$ there is a continuous function $f:{\mathbb R}/{\mathbb Z}\to{\mathbb C}$ whose Fourier series diverges at all points of $E$.

These two results leave a little gap.

What is known about those sets $E$ for which there is an $L^2$ (or even continuous?) function $f$ whose Fourier series diverges at all points of $E$ and pointwise converges to $f$ at all points not in $E$?

There is some ambiguity with the question as currently stated, as it depends on the representative of the $L^2$-class of $f$ that one chooses. I would hope an answer would help clarify the effect of specific representatives. Let me point out that, once we pick representatives, not every measure zero set can be such an $E$. If $f$ is continuous, this is easy to see; in fact, $E$ must be Borel (of low complexity; and this of course seems related to this question). As pointed out below in a comment by Juris Steprans, just on cardinality grounds we know not every measure zero set can appear, even for $L^2$ functions. Hunt's extension of Carleson's result says that we may assume $f\in L^p$ for any $p\in(1,\infty)$; I do not even know whether the sets $E$ will vary with $p$.

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Doesn't a cardinality argument yield a negative answer to your last question for continuous functions? (Also for L2 functions actually.) – Juris Steprans Jun 6 '11 at 23:08
Hi Juris. Sure; in fact, for continuous functions, the set $E$ is Borel (of rather low complexity). But I do not even have a working conjecture on what the general answer should be (does it change as $p$ changes, where we require $f\in L^p$, for example?), or whether any kind of structure can be expected at all. – Andrés Caicedo Jun 7 '11 at 0:56
I imagine this question will forever remain impossibly difficult for finite mathematicians; and even if it is someday solved, I personally am sure I will never understand the proof [or the criteria on $E$]. Having said that, it would be nice to be proved wrong! I mean, I don't understand it, but I thought that the Carleson-Hunt theorem tells us nothing whatsoever about the detailed structure of $E$, apart from having zero measure. If even this horribly difficult theorem gives no information, I am not optimistic. The $L^p$ spaces cannot distinguish between different $E$ with $m(E) = 0$. – Zen Harper Jun 7 '11 at 1:02
@Zen: Yes, as far as I can see, the proofs I know of the C-H theorem do not give us an insight, but I imagine the question here can be answered by direct constructions that do not require the use or knowledge of the theorem. – Andrés Caicedo Jun 7 '11 at 1:09
@Andres: I've just noticed the other question "Behaviour of power series on their circle of convergence" you referred to above [together with your wonderful answer!] That question, of course, is just the special case of this question where the Fourier coefficients for negative $n$ are all zero. It seems like you've almost answered your own question! – Zen Harper Jun 7 '11 at 1:33
up vote 5 down vote accepted

I believe that the problem of characterizing the sets of divergence for classical Fourier series is more or less open for all interesting classes ($C$, $L^\infty$, $L^p$ with $p>1$).

The strongest result that I'm aware of is due to Buzdalin who showed that any null-set $E\in F_\sigma\cap G_\delta$ is a set of divergence for the Fourier series of some continuous complex-valued function ("Trigonometric Fourier series of continuous functions diverging on a given set", Math. USSR Sbornik, 24 (1974)).

The characterization problem is mostly solved however for several other orthogonal systems, including the Haar and Franklin systems. There is also a very recent paper by Karagulyan where it is proved, in particular, that

A necessary and sufficient condition for a set $E \subset [0, 1]$ to be a set of divergence for the sequence of $(C, \alpha)$-means ($\alpha>0$) of the Fourier series of some function $f \in L^\infty[0, 1]$ is that $E$ is a $G_{\delta\sigma}$-set of measure $0$.

(See G.A. Karagulyan, "Characterization of the sets of divergence for sequences of operators with the localization property", Sbornik: Mathematics, 202 (2011), pp. 9–33.)

To complicate things further, people tend to distinguish between the sets of divergence and unbounded divergence. A set $E \subset [0, 1]$ is said to be a set of divergence (resp. unbounded divergence) for a series of functions $$\sum_{n=1}^{\infty}f_n(x),\qquad x\in[0,1],$$ if the series diverges for $x ∈ E$ and converges for $x \in [0, 1] \backslash E$ (resp. diverges unboundedly for $x ∈ E$).

One may think of the two optimistic working conjectures.

  1. Every $G_{\delta\sigma}$-set $E $ of measure $0$ is a set of divergence for the Fourier series of some function $f \in C[0, 1]$.

  2. Every $G_{\delta}$-set $E$ of measure $0$ is a set of unbounded divergence for the Fourier series of some function $f \in C[0, 1]$.

Conjecture 2 was explicitly formulated by P.L. Ul'yanov in the late 1960s. Both conjectures seem to be open.

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Andrey, Many thanks! This is most likely the current state of affairs. I'll look at the references you indicate. – Andrés Caicedo Jun 7 '11 at 14:42
Andres, you're welcome. – Andrey Rekalo Jun 7 '11 at 15:08

The set where a sequence $S_n$ of continuous functions (and in particular a Fourier series) converges is always an $F_{\sigma\delta}$, i.e. a countable intersection of countable unions of closed sets.

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Yes, Robert. I added a link in the body of the question to a problem where this is used. Note here that we want convergence to $f(x)$, so we are taking a subset of the set of convergence. – Andrés Caicedo Jun 7 '11 at 1:15
Andres: I think it's the whole set of convergence: if $f$ is continuous at a point $x$ and $S_n(f)(x)$ has some limit, then the limit must be $f(x)$, because of Fejer's Theorem: the Cesaro means $\sigma_n(f)(x)$ always converge to $f(x)$ at points of continuity, and Cesaro means preserve limits. – Zen Harper Jun 7 '11 at 1:28
...Of course, for continuous $f$ at least. – Zen Harper Jun 7 '11 at 1:28
In the problem as stated, the series "diverges at all points of $E$ and pointwise converges to $f$ at all points not in $E$". So $E^c$ is the whole set of convergence. – Robert Israel Jun 7 '11 at 2:59

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