When is a function "equal to" its Fourier series?

kloptok

First of all - a bit unsure where this post fits in, there seems to be no immediately appropriate subforum.

So I'm a physics student and currently looking at what it takes for a Fourier series to converge. I've looked at wiki (http://en.wikipedia.org/wiki/Converg...Fourier_series ) and this probably should tell me everything I need to know, if i only were fluent in the language of convergence. I don't really know the significance of the different types of convergence (uniform, pointwise etc.) and since I'm a physicist I suspect that this might not be of very much importance since we usually assume all functions are "nice" in physics ("nice" being the appropriate simplification in the problem at hand). I vaguely remember something about L^2 functions being important for this stuff - does this have any significance? Something to compare with is perhaps analytic functions and Taylor series - what would be the analog of analytic functions in the case of Fourier series?

So what I'm asking is: when is a function "equal to" its Fourier series? Is "equal to" the same as some form of convergence? Are there "analytic functions" for Fourier series?

What I'm interested in is if there is some simple criteria which will almost always be satisfied for problems in physics. Please be gentle with me, I have forgotten a lot of this stuff and I know I'm far from an expert.

jbunniii

There are several ways in which a function can "equal" its Fourier series. The simplest is pointwise convergence. This means that the following equality holds for all [itex]x[/itex]:
[tex]f(x) = \sum_n a_n e^{i2\pi n x/T}[/tex]
where [itex]a_n[/itex] is the [itex]n[/itex]'th Fourier coefficient and [itex]T[/itex] is the period of [itex]f[/itex]. In other words, for any [itex]x[/itex], the series converges to the value [itex]f(x)[/itex].

When working in spaces such as [itex]L^1[/itex] or [itex]L^2[/itex], it is no longer possible to talk about pointwise convergence. Indeed, the elements of these spaces not functions at all, but equivalence classes of functions. Two functions are considered one and the same if they differ only on a set of measure zero. A set of measure zero can be infinite, and even uncountably infinite (e.g. Cantor set), so the functions can be quite different pointwise and yet still considered "the same" in these spaces. So, pointwise convergence is replaced by "almost everywhere" convergence, meaning that the above equality holds except possibly on a set of measure zero.

There are other useful notions of convergence in these spaces, notably convergence in the norm. For example, in [itex]L^1[/itex], the norm of a function [itex]f \in L^1[/itex] is defined as
[tex]||f||_1 = \int |f(x)| dx[/tex]
where the integration is taken over [itex]\mathbb{R}[/itex] or whatever the underlying domain is. Then we say that a sequence of functions, say [itex]g_n[/itex], converges in norm to [itex]f[/itex] if
[tex]||f - g_n||_1 \rightarrow 0[/tex]
as [itex]n \rightarrow \infty[/itex]. There is a similar notion for [itex]L^2[/itex] and indeed any [itex]L^p[/itex] space.

If we take [itex]g_n[/itex] to be the n'th partial sum of the Fourier series of [itex]f[/itex], we thereby obtain another sense in which [itex]f[/itex] may "equal" its Fourier series. However, it's important to note that convergence in the norm can occur even in the absence of "almost everywhere" convergence, so in that sense it's a weaker form of convergence.

homeomorphic

As far as physics goes, it is often said that Dirichlet solved the problem for most situations of physical interest:

http://en.wikipedia.org/wiki/Dirichlet_conditions

Developing stronger methods to deal with nastier functions that came up, for example, in number theory, played a big role in the development of modern mathematics (set theory, understanding the real numbers more deeply, Riemann sums, measure and integration, analytic number theory).

kloptok

Thanks a lot! Particularly the Dirichlet conditions seem appropriate for me, but thanks for a great post anyway jbuniii.

So am I right in my understanding that the Dirichlet conditions give pointwise convergence , which in turn means that for any x the series converges to f(x)? And for discontinuities the value is the average of left and right limits.

I must confess I did not entirely follow the stuff on [itex]L^p[/itex] spaces. If equivalence classes are given by functions with equal measure, is the measure in [itex]\left( \int |f(x)-g(x)|^p dx\right)^{1/p}[/itex] the equivalence relation? And is convergence given by the fact that the difference in measure between a function and the partial sum goes to zero as N goes to infinity, i.e.,
[tex]\lim_{N\rightarrow \infty} ||f(x)-s_N||=\left( \int |f(x)-s_N|^p dx \right)^{1/p} =0 [/tex]
means convergence? So that would imply that it is the equivalence class of functions which converges and not the function itself, since the equivalent functions can't be distinguished?

I'm sorry if I'm rambling incoeherently and/or use incorrect terminology. Haven't really taken formal math courses in this subject.

jbunniii

Yes, I could see that you were mainly looking for a simple criterion for pointwise convergence, but you also asked about the other types of convergence and L^2 spaces, so I tried to give a sketch of how these work. You can see that there are a lot of technicalities involved.

Slight correction: the equivalence classes are given by functions which are equal except on a set of measure zero. For example, let's consider what functions of a real variable are equivalent to the zero function, i.e. the function defined by f(x) = 0 for all real x.

In order to understand this, you have to know what is meant by a set of measure zero. This definition is actually quite simple: it is a set that can be covered by a collection of intervals of arbitrarily small total length. Any finite set has measure zero. Any countably infinite set has measure zero. This includes the set of integers and the set of rationals. The Cantor set is an uncountably infinite set with measure zero.

So, the following functions are equivalent to the zero function:
* g(x) = 1 if x is rational, 0 if x is irrational
* h(x) = 1 if x is in the Cantor set, 0 if it is not

The equivalence relation is: f ~ g if and only if [itex]m(\{x : f(x) \neq g(x)\}) = 0[/itex], where [itex]m[/itex] denotes the measure (generalized notion of length) of a set. This is equivalent to [itex]||f - g||_p = 0[/itex].

By the way, the reason for defining the [itex]L^p[/itex] spaces as spaces of equivalence classes instead of spaces of functions is so that the norm [itex]||f||_p[/itex] will satisfy [itex]||f||_p = 0[/itex] if and only if [itex]f = 0[/itex] (i.e. if and only if f is equivalent to zero). Otherwise we would have many elements of [itex]L^p[/itex] with zero "norm", which makes the space harder to work with.

Yes this is convergence in the norm, which is a weaker notion that pointwise convergence, but still very useful in many contexts.

Yes, that's right. The integral in your expression above will have the exact same value if you replace [itex]f[/itex] by another member of its equivalence class, so there's no way to distinguish between them using this notion of convergence.

For a function in [itex]L^p[/itex] with [itex]p > 1[/itex], it is actually known that the Fourier series converges almost everywhere, which is a stronger notion than convergence in norm. It means that there is pointwise convergence at all points except possibly in a set of measure zero. This result is called the Carleson-Hunt theorem and was only proven in the 1960s; the proof is highly technical and understanding it requires a lot more knowledge than I have. This result is not true for [itex]L^1[/itex]; indeed, there is a counterexample of an [itex]f \in L^1[/itex] whose Fourier series diverges at all points.

I think you have understood the basic idea pretty well. Perhaps you'll find some of this intriguing enough to want to learn more. One or two courses in real analysis will probably get you there.

jbunniii

P.S. The answer to all of these question is yes. Note that the Dirichlet conditions are sufficient but not necessary.

kloptok

Just wanted to throw in another thank you for your effort! Sure this is interesting stuff, perhaps some time I'll have time to take a real analysis course or at least look into it on my own a bit.