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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?


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Depends how you mean "expressed as a Fourier series". Convergence in which mode? In $L^2$? Pointwise (almost everywhere)? –  Jonas Teuwen Oct 20 '11 at 12:24
@JonasTeuwen: So there are different degrees...? I would be grateful if you could elaborate on the differences, that would be a good ans to this (vaguish) question. Thanks. –  Huh Oct 20 '11 at 12:30
If the function has an infinite number of jump discontinuities, then that's a problem. –  Unreasonable Sin Oct 20 '11 at 12:44
@Huh: much ink has been used in discussing the different modes of convergence. For an introductory (but by no means exhaustive) account, you can look at chapters 2 and 3 or Stein & Shakarchi, Fourier Analysis. –  Willie Wong Oct 20 '11 at 13:39

3 Answers 3

You should definitively take a look at Accessible Proof of Carlesons $L^2$-theorem.

Note that:

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

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

  3. 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.

  4. For $L^1(-\pi,\pi)$ Kolmogorov constructed a function that diverges everywhere.


Note that if a function $f$ is piecewise continuous and if $\int |f|^2<\infty$ then the function belongs to $L^2$.

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Well,the Carleson-Hunt theorem is highly non-trivial and I would think that someone that doesn't know that there are different modes of convergence would have a lot of trouble understanding the proof. Nevertheless, the post is quite informative hence +1 ;-). –  Jonas Teuwen Oct 20 '11 at 17:33
@JonasTeuwen Thanks and yes, you are right. However, even though the results are deep their conclusions are not. –  AD. Oct 20 '11 at 18:19

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.

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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.

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Should you use $x_0$ instead of $x$ in the line with the limit? –  Thomas Andrews Oct 20 '11 at 13:58
@Thomas Andrews: Yes, absolutely. Thanks! –  J. J. Oct 20 '11 at 14:01

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