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

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    @Rajesh D: Use `\to` instead of `->`; also, `\colon` instead of `:` tends to produce better spacing for functions.2010-11-27
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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 http://en.wikipedia.org/wiki/Convergence_of_Fourier_series2010-11-27
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    @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 ?2010-11-27
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    @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: http://mathoverflow.net/questions/165/does-the-continuous-locus-of-a-function-have-any-nice-properties2010-11-27
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    @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*.2010-11-27
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    If $f\in L^1_\text{loc}(\mathbb{R})$ then you can define the Fourier coefficients on any interval $I=(a,b)$, also the corresponding Fourier series will be periodic with period $b-a$.2010-11-27
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    @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).2010-11-27
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    @Arturo Magidin: Thank you for the reference to the result on $G_\delta$(in $\mathbb{R}$) and continuity of a function. After knowing this,I am interested in functions that have discontinuities at a countable subset of $\mathbb{R}$.2010-11-27
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    @Arturo Magidin: I am not interested in any convergence.2010-11-27
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    @Rajesh D. The simple answer is that the series exists if and only if $f(x)\cos(nx)$ and $f(x)\sin(nx)$ are integrable for every $n$ on the interval in question. A necessary and sufficient condition for the Riemann integral to exist is that the functions be bounded in the interval and that the set of discontinuities in the interval have Lebesgue measure zero.2010-11-27
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    @Arturo Magidin: Does a 'Lebesgue measure zero' implies the set is countable and the converse ?2010-11-27
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    @Rajesh D.: Adding to Arturo's comment, a bounded function discontinuous only on a countable set will be fine.2010-11-27
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    @Rajesh D.: No, the converse is true (countable implies measure zero) but Lebesgue measure zero sets can have cardinality as big as $\mathbb{R}$. E.g., look up the Cantor set. So for example, there are bounded functions discontinuous only at each point in the Cantor set, and these will be Riemann integrable.2010-11-27
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    @Rajesh D: A countable set always has Lebesgue measure zero, but not every set of Lebesgue measure zero is countable: the Cantor set is uncountable and of measure zero. Since you have apparently never encountered the notion, you might have tried doing some searching on your own instead of reflexively asking for an answer; e.g., http://en.wikipedia.org/wiki/Lebesgue_measure. Note that a function *can* be discontinuous at uncountably many points; the key is that the set of of points where it is continuous has to be $G_{\delta}$ (see http://en.wikipedia.org/wiki/G-delta for the definition)2010-11-27
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    @Arturo Magidin,@Jonas Meyer: It may not be relevant here, but does a function which is continuous everywhere but not differentiable only at rational points exist ?2010-11-27
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    @Rajesh D.: I think you will get an example by integrating the function Arturo gives at the following link of a monotone function discontinuous at the rationals: http://math.stackexchange.com/questions/7821/constructing-continuous-functions-at-given-points/7825#7825. In any case, how about working on focusing your present question?2010-11-27

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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.2010-11-27
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    @Jonas Meyer: What is that "Day's votes"? :)2010-11-27
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    We can each vote 30 times per day. Today I was on a voting spree and used them all up.2010-11-27
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    @Jonas Meyer: Ok, I did not know that2010-11-27
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    In fact, Kolmogorov improved his example to $S[f]$ diverges everywhere.2010-11-27
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    @TCL: Yes, I know - just a year later or so :)2010-11-27
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    @Jay Thanks. + some letters ;)2013-07-18
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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 $\mathbb{R}$.