# 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.

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What is the definition of the term "trigonometric series"? – Rasmus Aug 27 '10 at 11:26
Sometimes the adjective "lacunary" has to be used. – J. M. Aug 27 '10 at 11:28
tan, cot, sec, csc? – kennytm Aug 27 '10 at 11:41
Kenny: an infinite series of those would surely have a nasty fence of poles... – J. M. Aug 27 '10 at 11:51
@JMangaldan: you need to put an @ in front of the username to have him notified – Tobias Kienzler Aug 27 '10 at 12:05

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.

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@Pete L. Clark: Please check the post. – AD. Aug 27 '10 at 19:51
@Pete L. Clark: Sure, I just want you to fix the LaTeX, the renderings is not ok. (When I first wrote this very comment I accidentally wrote $Pete L. Clark... It is not easy :) – AD. Aug 28 '10 at 4:39 The Fourier coefficients of an L^1 function are bounded. – Pierre-Yves Gaillard Aug 28 '10 at 11:25 In fact the sequence of Fourier coefficients of an L^1 function vanishes at infinity, giving a Banach algebra morphism from L^1(S^1) to C_0(Z), which is a very particular case of Gelfand transform: en.wikipedia.org/wiki/… – Pierre-Yves Gaillard Aug 28 '10 at 19:58 @PYG: Right, the Riemann-Lebesgue Lemma. Sigh -- it's been too many years since my last real variables course. Perhaps I need to teach one to get my bearings straight. – Pete L. Clark Aug 29 '10 at 20:25 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. - You can define the Fourier series (FS) of a distribution, or even of a hyperfunction. The FS sum of the c_n e^{inx} of a hyperfunction is such that the power series with coefficients c_n and c_{-n} converge on |z|<1. And any such FS is the FS of a unique hyperfunction. – Pierre-Yves Gaillard Aug 29 '10 at 7:03 @Pierre-Yves Gaillard: Of course I am not speaking about high generalizations here. In classical analysis we are usually talking about some function that we want to investigate using Fourier analysis and then we need some integrability assumption on the function - I do not say we can extend this idea to 'other shelfs' as well. :) – AD. Aug 29 '10 at 8:23 When all the coefficients are 1, the series does not converge to a function. (It does converge in the sense of a distribution, to the Dirac Delta.). - Using Riemann-Lebesgue lemma we can give the example of a trigonometric series which is not a Fourier series. e.g.$\sum_{n=1}^{\infty}\cos nx$is not a Fourier series as its coefficient does not tends to zero. For better understand on this topics i would suggest you to prefer Bary( Vol.1) book. - 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) - Take a look at this wikipedia stub. A trigonometric series, which is not a Fourier series, must therefore be one that is defined by $$f(x) = A_{0}+\displaystyle\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$$ for which the relations $$A_{0}=\frac{1}{2\pi}\displaystyle\int^{2 \pi}_0 f(x)\,dx$$ $$A_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \cos{nx} \,dx\qquad (n=1,2,3 \dots)$$ $$B_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \sin{nx}\, dx\qquad (n=0,1,2,3, \dots)$$ do not hold. But since {$\sin(nx),\cos(nx)$} forms a complete orthonormal basis, I cannot imagine one. Except for a non-convergent one perhaps... edit As J. Mangaldan pointed out: Lacunary trigonometric series might be the answer you search, i.e. series that cannot be analytically be continued. - Probably an interesting question: if we treat the expressions for the Fourier series coefficients as integral transforms, are there functions$f(x)$that cannot be expressed as inverse transforms (whatever the form of the inverse may be)? – J. M. Aug 27 '10 at 11:55$f(x) = 0, 0 < x < \pi; f(x) = 1, \pi < x \le 2\pi, f(\pi) = 1/3$is not an inverse transform (because the inverse would have to have a value of 1/2 at$\pi$). – whuber Aug 27 '10 at 14:23 the whole problem here is that f need not exist in any meaningful sense. – Qiaochu Yuan Aug 27 '10 at 15:40 As you see from the definition, you need that$f\$ is integrable in some sense. – AD. Aug 27 '10 at 20:00