Condition for Fourier series I read that


Any "well-behaved" function of period $2\pi$ can be expressed as a Fourier series. 


What qualifies as "well-behaved"? Any examples of functions that cannot be expressed as a Fourier series?
Thanks.
 A: I'm assuming that you are interested about the point-wise convergence of the Fourier series of a function. For this we have the Dini test:
Suppose that $f \in L^1[-\pi,\pi]$ is a $2\pi$-periodic function and consider the point $x_0 \in \mathbb{R}$. If
$$\int_{0}^\pi \left|\frac{f(x_0+t) + f(x_0-t)}{2} - f(x_0)\right| \frac{dt}{t} < \infty,$$
then the limit
$$\lim_{N \to \infty} S_N f (x_0) = \lim_{N \to \infty} \sum_{n=-N}^N \widehat{f}(n) e^{inx_0} = \sum_{n=-\infty}^\infty \widehat{f}(n)e^{inx_0}$$
exists and is equal to $f(x_0)$.
Using this one can prove for example that if $f$ is differentiable at the point $x_0$, then the Fourier series converges at that point.
A: You should definitively take a look at Accessible Proof of Carlesons $L^2$-theorem.
Note that: 


*

*The classical definition of Fourier coefficient of a function $f$ works for $f\in L^1(-\pi,\pi)$. 

*$L^p(-\pi,\pi)\subset L^1(-\pi,\pi),$ $\ $ provided $1\le p \le \infty$.

*Carleson's (Hunt's) theorem say that the Fourier series of an $L^2(-\pi,\pi)$-function ($L^p$ where $p>1$ in Hunt's case) converges pointwise almost everywhere.

*For $L^1(-\pi,\pi)$ Kolmogorov constructed a function that diverges everywhere.
EDIT
Note that if a function $f$ is piecewise continuous and if $\int |f|^2<\infty$ then the function belongs to $L^2$.
A: I'll give you one of the easy parts of the answer.  There's the question of what is "well-behaved" and there's the question of what is meant by "expressed as".
Suppose the $n$th partial sum, i.e. the sum of the first $n$ terms of the Fourier series of the function $f$, is $S_n$.  Do you want $\lim\limits_{n\to\infty} S_n = f(x)$?  For every value of $x$?  That's a pretty strong demand, and it can be satisfied for sufficiently well behaved functions $f$.
But suppose you only want $S_n$ to approach $f$ in a weaker sense: you want $\lim\limits_{n\to\infty}\displaystyle\int_0^{2\pi} \left|S_n(x) - f(x)\right|^2\;dx=0.$  That will happen if $f$ is quadratically integrable, i.e. if $\displaystyle\int_0^{2\pi} \left|f(x)\right|^2\;dx<\infty$.
There are lots of more difficult results to prove and some of them had to wait well over a century to get proved.
