This is maybe not really research level, but I have not found anything in the literature, and asking on math.stackexchange wasn't successful either.

Fourier series define an isometry $L^2(\mathbb{Z}) \rightarrow L^2(S^1), (a_k)_{k \in \mathbb{Z}} \mapsto \sum a_k z^k \colon S^1 \rightarrow \mathbb{C}$ (all Hilbert spaces are complex). It is known (and easy) that, when we start with a sequence $a_k$ which is actually in $L^1(\mathbb{Z})$, we end up with a continuous function on $S^1$. My question is: What more can we say about sequences $a_k$ whose associated Fourier series is a continuous function on $S^1$? In particular, is there an easily checkable if and only if-criterion? I would guess that not, but I have never seen such a question discussed in the literature.

  • $\begingroup$ Reading through Katznelson's Intro to Harmonic Analysis will give lots of circumstantial evidence that the answer to your question is "not really, and probably not ever". Certainly it is not enough to look at the sequence $(|a_n|)$ $\endgroup$ – Yemon Choi Jun 14 '13 at 20:19
  • $\begingroup$ (Probably even more is in Zygmund, but I find those books hard to carry, let alone read) $\endgroup$ – Yemon Choi Jun 14 '13 at 20:22
  • $\begingroup$ I would have been somewhat disappointed in mathematics if there were such a criterion, so everything is fine. $\endgroup$ – Fabian Lenhardt Jun 15 '13 at 20:04

Maybe you are already aware of the following if-and-only-if condition, though I wouldn't say it counts as "easily checkable": let $(a_k)$ be the coefficient sequence, let $s_N$ denote the partial sums $$ s_N(z) = \sum_{k=-N}^N a_k z^k $$ and let $\sigma_N(z)$ denote the Cesaro means $$ \sigma_N(z)=\frac{1}{N+1}\sum_{n=0}^N s_N(z). $$ Then $(a_k)_{k\in\mathbb Z}$ is the Fourier series of a continuous function on the circle if and only if the sequence $\sigma_N$ is uniformly Cauchy on the circle, in which case it converges to a continuous function $f$ with $\hat{f}(k)=a_k$. (The sufficiency of this condition is clear from term-by-term integration, the necessity is Fejer's theorem.)

  • 1
    $\begingroup$ Mike, with all due respect, this is a condition on functions, not on the Fourier coefficients themselves. $\endgroup$ – Yemon Choi Jun 14 '13 at 20:17
  • $\begingroup$ No, I haven't been aware of this, thanks! $\endgroup$ – Fabian Lenhardt Jun 15 '13 at 20:04

This is really a glorified comment, but the formatting in answer boxes is easier (and I think the information deserves greater visibility). Apologies to the OP if he already knew this.

From Katznelson's introduction to Harmonic Analysis (the Dover version, 2nd corrected edition). Everything takes place on the circle ${\bf T}$.

Chapter IV, Theorem 2.4.

(a) There exists a continuous function $f$ such that for all $p<2$, $\sum \vert\widehat{f}(n)\vert^p = \infty$.

(b) [paraphrased by YC] There exists a sequence $(a_n)_{n\in{\bf Z}}$ which belongs to $\ell^p$ for every $p>2$, but which is not the Fourier series of any finite Borel measure on ${\bf T}$.

Quite a lot has been done, in the classical days of the subject, on how smoothness/Hölder conditions on a function are reflected in the rate at which its Fourier series decays: see Chapter I, Section 4 of the same book.

On a positive note, if the Fourier series is lacunary, then it is the Fourier series of a continuous function iff it belongs to $\ell^1$. (Chapter V, paragraph 1.4; this is sometimes stated as "lacunary sets are Sidon sets")

The circumstantial evidence, as I understand it, is that there is no way to characterize the Fourier series of continuous functions by means of a naive "sequence space" condition on the sequence. However, as Mike Jury has pointed out, one can use Fejer's theorem to get some kind of algorithm-flavoured criterion.


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