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The Fourier transform of periodic function $f$ yields a $l^2$-series of the functions coefficients when represented as countable linear combination of $\sin$ and $\cos$ functions.

  • In how far can this be generalized to other countable sets of functions? For example, if we keep our inner product, can we obtain another Schauder basis by an appropiate transform? What can we say about the bases in general?

  • Does this generalize to other function spaces, say, periodic functions with one singularity?

  • What do these thoughts lead to when considering the continouos FT?

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5 Answers 5

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It is not what you want, but may be worth mentioning. There is a huge branch of abstract harmonic analysis on (abelian) locally compact groups, which generalizes Fourier transformation on reals and circle. The main point about sin and cos (or rather complex exponent $e^{i n x}$) is that it is a character (continuous homomorphism from a group to a circle) and it is not hard to see that those are the only characters of the circle. That what makes Fourier transform so powerful. If you generalize it along the direction which drops characters, you'll probably get a much weaker theory.

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You need the orthogonality condition to get such an integral representation for the coefficients; otherwise it would probably be more complicated.

The Fourier series of any $L^2$ function converges not only in the norm (which follows from the fact that $\{e^{inx}\}$ is an orthonormal basis) but also almost everywhere (the Carleson-Hunt theorem). Both these assertions are also true in any $L^p,p>1$ but at least the first one requires different methods than Hilbert space ones. In $L^1$, by contrast, a function's Fourier series may diverge everywhere.

There are many conditions that describe when a function's Fourier series converges to the appropriate value at a given point (e.g. having a derivative at that point should be sufficient). Simple continuity is insufficient; one can construct continuous functions whose Fourier series diverge at a dense $G_{\delta}$. The problem arises because the Dirichlet kernels that one convolves with the given function to get the Fourier partial sums at each point are not bounded in $L^1$ (while by contrast, the Fejer kernels or Abel kernels related respectively to Cesaro and Abel summation are, and consequently it is much easier to show that the Fourier series of an $L^1$ function can be summed to the appropriate value using either of those methods). Zygmund's book Trigonometric Series contains plenty of such results.

There is a version of the Carleson-Hunt theorem for the Fourier transform as well.

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Any compact normal operator on a Hilbert space has an orthonormal basis of eigenvectors. If I remember correctly then the standard Fourier series comes from the second derivative operator on L^2(0,2pi) with boundary conditions f(0)=f(2pi) and f'(0)=f'(2pi). This operator is not compact, but its inverse is (and has the same eigenvectors). Using other compact normal operators (usually inverses of differential operators with certain boundary conditions) you obtain other orthonormal bases.

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There are certainly many other basis for spaces of functions on an interval, if we eliminate the periodicity condition. The more widely used are orthogonal polynomials. Given an interval $I\subset\mathbb{R}$ and a weight $w\colon I\to (0,\infty)$, there is a sequence of polynomials $\{P_n\}$ orthogonal with respect the weight $w$: $$\int_I P_m(x)P_n(x)w(x)\,dx=0,\quad m\ne n.$$ They are a basis of $L^2(I)$. A classical reference is Gabor Szego (1939). Orthogonal Polynomials. Colloquium Publications - American Mathematical Society.

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If you are interested in more general Fourier transforms, then the two things which spring immediately to my mind are:

  1. Titchmarsh's book Fourier Integrals contains a detailed treatment of what he calls "generalized kernels", which vaguely are pairs of functions $h(x),k(x)\in L^2(\mathbb{R})$ such that

    $\int_{0}^{\infty}k(xy)\int_{0}^{\infty}h(yw)f(w)dwdy=f(x)$.

  2. There is a lovely theory of "wavelets" due to Daubechies et al, which are described in many places.

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agh, can someone make the integral signs work right? –  David Hansen Nov 27 '09 at 21:54
    
They work for me! –  Ilya Nikokoshev Nov 27 '09 at 23:47

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