Example of a trigonometric series that is not fourier series?

My textbook doesn't give any example of this kind of series. Could you provide some?

Trigonometric series is defined in wikipedia as : $A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$

When

$A_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \cos{nx} dx\qquad (n=0,1,2,3 \dots)$

$B_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \sin{nx} dx\qquad (n=1,2,3, \dots)$

It is fourier series.

thanks.

I don't know the answer (I should have! -- see below), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight.

Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series. In other words, a trigonometric series is just any series of the form

$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$

This should be understood formally, i.e., this is a trigonometric series even if it doesn't converge anywhere. A fourier series is the trigonometric series associated to an $L^1$ function by taking $A_n$ and $B_n$ as above.

Just as a Taylor series need not need converge at any point except the central point, a Fourier series need not converge pointwise at any point. Thus the Fourier series need not be a function and in particular the Fourier inversion theorem need not apply.

A theorem of Borel asserts that given any sequence $(a_n)$ of real numbers, there exists a $C^{\infty}$-function on the real line whose Taylor series at $0$ is $\sum_{n=0}^{\infty} a_n x^n$. In particular, if the $a_n$'s grow too rapidly, the Taylor series will diverge away from zero and the function $f$ will not be analytic.

I interpret the question as asking whether the analogue of Borel's theorem is true for Fourier series: is every trigonometric series the Fourier series of some $L^1$ function (even if the trigonometric series does not converge to the function)?

P.S.: Boo to wikipedia for asserting the answer to this question without giving a reference.

Addendum: As Pierre-Yves Gaillard points out, the Fourier coefficients of any $L^1$ function $f$ are uniformly bounded (by $||f||_1$), so this answers the question as I have interpreted it.

A standard example is $$ f(t)= \sum_{n>1} \frac{\sin(nt)}{\log(n)}$$ The conjugate of $f$ is a Fourier series but $f\not\in L^1(\mathbb{T})$ and hence is no Fourier series. For further explanation see Katznelson's book page 85.

(Edit: If $f$ is not in $L^1(\mathbb{T})$ it is hard to define Fourier coefficients.

Added 29/8 - 2010 Here is a screen dump from Zygmund's book "Trigonometric Series Vol I" .

However, in order to be a Fourier series the coefficients are to be Fourier coefficients of some function $f$, these are well defined if $f\in L^1(\mathbb{T})$ (because these are calculated using integration against a bounded measurable function). Thus it is natural to define a trigonometric series if the coefficients are Fourier coefficients of some $L^1$ function. Hence, all answers given to this problem are in fact right!

Moreover, the Riemann-Lebesgue Lemma states that the Fourier coefficients of an $L^1$-function tends to $0$ as the index approach $\pm\infty$. Thus, it is easy to construct formal trigonometric series that are not Fourier series.

The example $f$ given above does in fact converge everywhere and hence this is a kind of non-trivial example of a trigonometric series that is not a Fourier series.

Theorem. If $a_n>0$ and $\sum_{n>0}\frac{a_n}{n}=\infty$. Then $\sum_1^\infty a_n\sin nt$ is not a Fourier series. (AN INTRODUCTION TO HARMONIC ANALYSIS, Yitzhak Katznelson)