# About completeness of the Fourier series.

The Fourier series of a function is given by $$\frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos n \theta + \sum_{n=1}^\infty b_n \sin n \theta .$$ Here what does the statement " $\sum_{n=1}^\infty b_n \sin n \theta$ is complete" mean? And would you tell me how can I prove this statement?

• Where did you see this sentence? In which context? – Davide Giraudo Jul 26 '12 at 15:38

## 3 Answers

I don't think it makes sense to call a Fourier series complete.

But what you can call complete is the basis of your space.

For example, if the functions you're computing Fourier series of are in $L^2(G)$, where $G$ is a compact Abelian topological group, one can say that the characters $\chi$ of $G$ are complete and it means two things:

(i) that for every function $f$ in $L^2$ there exists a sequence of continuous characters $\chi_n \in \mathrm{Hom}(G, S^1)$ such that $\|f - \sum_{k=0}^n a_k \chi_k \|_{L^2} \xrightarrow{n \to \infty} 0$ and

(ii) that $\chi_n$ are orthonormal, that is, $\langle \chi_n , \chi_m \rangle = \delta_{nm}$ where $\langle \cdot, \cdot \rangle$ is the inner product of $L^2$, namely, $\langle f, g \rangle = \int_G f \;\overline{g}\; d \mu$ (where $\overline{g}$ denotes complex conjugation)

• To be more concrete: if your group is for example $G = \mathbb R / \mathbb Z$ then your characters are $\chi_n (x) = e^{i 2 \pi n x}$. To prove that these are orthonormal is not too difficult, it's a matter of integration by parts. On the other hand, proving claim (i) is rather involved as far as I know but I have not read a proof of it because it has been omitted from my lecture notes. – Rudy the Reindeer Jul 26 '12 at 15:43
• You want your topological group $G$ to be abelian and compact in order for this to make sense. – t.b. Jul 26 '12 at 16:48
• @t.b. I know it wants to be locally compact for the existence of a Haar measure. And for some reason, one needs a Haar measure to do Fourier series. Why can't I just define an inner product with a different measure? And why does it have to be compact and Abelian? – Rudy the Reindeer Jul 26 '12 at 17:04
• When proving the basic things about the Fourier transform, you're constantly exploiting that the measure is translation invariant (e.g. $\widehat{f \ast g} = \hat{f}\hat{g}$). If $G$ is not compact you have a Fourier transform $L^2(G) \to L^2(\hat{G})$ but no series. In your answer you're implicitly using that on a compact group the characters belong to $L^2$ (without compactness they don't) and that they form an o.n.B (enter commutativity). Sure, $L^2(\mu)$ and o.n. bases make sense for any measure, but they won't have much to do with $G$ as a group -- as opposed to a measure space. – t.b. Jul 26 '12 at 17:32

I doubt this statement makes sense.

What does make sense is the statement that the trigonometric polynomials ($\sin(kx), \cos(kx), k\in \mathbb{N}$ and the constant function 1) form a complete orthogonal system (orthonormal with correct normalization) of functions in $L^2([0,2\pi])$, say.

This then refers to the fact that each function in $L^2([0,2\pi])$ admits a unique representation like the one you wrote down with convergence w.r.t to the norm of $L^2$, i.e. $$||f(x) - \frac{a_0}{2} - \sum_1^n (a_n \cos (nx)+b_n \sin(nx))||_{L^2} \rightarrow 0 \,\, (n\rightarrow \infty)$$

A proof of this fact can be found in several standard texts on analysis, like e.g. Rudin's Principles of mathematical analysis.

Supposing you omit the $n=5$ term from the cosine series so $a_5\cos(5\theta)$ just isn't included. Why would the set of functions you're adding up be incomplete?

The point is that there would be some functions that cannot be approached by series that omit that term. One such function is $\cos(5\theta)$ itself.

"Approached" means convergence in the $L^2$ sense, i.e. $f_n\to f$ in $L^2$ means that $\displaystyle\int_0^{2\pi} |f_n(\theta) - f(\theta)|^2\,d\theta \to 0$ as $n\to\infty$.

And "some functions" means some $L^2$ functions, i.e. functions $f$ such that $\displaystyle\int_0^{2\pi} |f(\theta)|^2\,d\theta<\infty$.