What are the conditions for existence of fourier series expansion of a function $f\colon\mathbb{R}\to\mathbb{R}$

What are the conditions for existence of fourier series expansion of a function $f\colon\mathbb{R}\to\mathbb{R}$

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@Rajesh D: In a sense, the Fourier series always exists. Are you asking for conditions under which it converges to the original function? See en.wikipedia.org/wiki/Convergence_of_Fourier_series – Arturo Magidin Nov 27 '10 at 6:23
@Arturo Magidin: For example a function which could have a discontinuity at every irrational point and continuous at every rational point. Does it have a Fourier series ? – Rajesh Dachiraju Nov 27 '10 at 6:30
@Rajesh: There is no such function. The set of points where a function is continuous is a $G_\delta$. (And conversely, each $G_\delta$ is the set of points where some function is continuous.) For example, see the answer of Eric Wofsey (and that of Alon Amit) here: mathoverflow.net/questions/165/… – Jonas Meyer Nov 27 '10 at 6:34
@Rajesh D: Any function for which the appropriate integrals are defined has a Fourier series; if you change the integrals from Riemann to more general kinds, you extend the kind of functions for which you can define the Fourier series (though convergence to the original function is something else entirely. In any case, you cannot have a function that is continuous at the rationals and discontinuous at the irrationals. – Arturo Magidin Nov 27 '10 at 6:39
@Rajesh D: But the question is: are you interested only in whether the series can be defined (that is, whether the Fourier series expansion exists), or are you interested in whether the series can be defined and converges (pointwise, uniformly, almost everywhere, or in some other sense) to the original function? They are different questions. Think about Taylor series: any infinitely differentiable function has a Taylor series expansion, but the Taylor series doesn't always converge to the original function ($e^{-1/x^2}$ for $x\neq 0$, $0$ for $x=0$ is the standard example). – Arturo Magidin Nov 27 '10 at 6:51

If $f\in L^1_\text{loc}(\mathbb{R})$, then on an interval $I=(a,b)$ we can define $$\hat{f}(n)=\frac{1}{b-a}\int_a^b f(x)e^{-2\pi inx/(b-a)}dx.$$ However, in order for the formal Fourier series $$S[f](x)=\sum_{-\infty}^{\infty} \hat{f}(n)e^{2\pi inx/(b-a)}$$ to converge we need more conditions on $f$. Kolmogorov proved in 1925 that there is $f\in L^1(0,2\pi)$ such that $S[f]$ diverges almost everywhere. In 1966 Carleson proved that $S[f]$ converges almost everywhere provided $f\in L^2(0,2\pi)$.

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This is a nice answer, and I'd upvote if I had any of my day's votes left. – Jonas Meyer Nov 27 '10 at 7:21
@Jonas Meyer: What is that "Day's votes"? :) – AD. Nov 27 '10 at 7:42
We can each vote 30 times per day. Today I was on a voting spree and used them all up. – Jonas Meyer Nov 27 '10 at 7:53
@Jonas Meyer: Ok, I did not know that – AD. Nov 27 '10 at 8:08
In fact, Kolmogorov improved his example to $S[f]$ diverges everywhere. – TCL Nov 27 '10 at 12:05

In addition to Carleson's theorem (stated by AD above), which gives a sufficient condition for pointwise convergence almost everywhere, one might also consider the following theorem about uniform convergence:

Suppose $f$ is periodic. Then, if $f$ is $\mathcal{C}^0$ and piecewise $\mathcal{C}^1$, $S_N(f)$ converges uniformly to $f$ on $\mathbb{R}$.

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