Fourier inversion theorem

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In mathematics, Fourier inversion recovers a function from its Fourier transform. Several different Fourier inversion theorems exist.

Sometimes the following identity is used as the definition of the Fourier transform:

<math>(\mathcal{F}f)(t)=\int_{-\infty}^\infty f(x)\, e^{-itx}\,dx.</math>

Then it is asserted that

<math>f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty (\mathcal{F}f)(t)\, e^{itx}\,dt</math>

In this way, one recovers a function from its Fourier transform.

However, this way of stating a Fourier inversion theorem sweeps some more subtle issues under the carpet. One Fourier inversion theorem assumes that f is Lebesgue-integrable, i.e., the integral of its absolute value is finite:

<math>\int_{-\infty}^\infty\left|f(x)\right|\,dx<\infty.</math>

In that case, the Fourier transform is not necessarily Lebesgue-integrable; it may be only "conditionally integrable". For example, the function f(x) = 1 if −a < x < a and f(x) = 0 otherwise has Fourier transform

<math>-2i\sin(t)/t.</math>

In such a case, the integral in the Fourier inversion theorem above must be taken to be an improper integral

<math>\lim_{b\rightarrow\infty}\frac{1}{2\pi}\int_{-b}^b (\mathcal{F}f)(t) e^{itx}\,dt</math>

rather than a Lebesgue integral.

By contrast, if we take f to be a tempered distribution -- a sort of generalized function -- then its Fourier transform is a function of the same sort: another tempered distribution; and the Fourier inversion formula is more simply proved.

One can also define the Fourier transform of a quadratically integrable function, i.e., one satisfying

<math>\int_{-\infty}^\infty\left|f(x)\right|^2\,dx<\infty.</math>

[How that is done might be explained here.]

Then the Fourier transform is another quadratically integrable function.

In case f is a quadratically integrable periodic function on the interval then it has a Fourier series whose coefficients are

<math>\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\,e^{-inx}\,dx.</math>

The Fourier inversion theorem might then say that

<math>\sum_{n=-\infty}^{\infty} \hat{f}(n)\,e^{inx}=f(x).</math>

What kind of convergence is right? "Convergence in mean square" can be proved fairly easily:

<math>\lim_{N\rightarrow\infty}\int_{-\pi}^\pi\left|f(x)-\sum_{n=-N}^{N} \hat{f}(n)\,e^{inx}\right|^2\,dx=0.</math>

What about convergence almost everywhere? That would say that if f is quadratically integrable, then for "almost every" value of x between 0 and 2π we have

<math>f(x)=\lim_{N\rightarrow\infty}\sum_{n=-N}^{N} \hat{f}(n)\,e^{inx}.</math>

This was not proved until 1966 in (Carleson, 1966).

For strictly finitary discrete Fourier transforms, these delicate questions of convergence are avoided.

Reference

• Lennart Carleson (1966). On the convergence and growth of partial sums of Fourier series. Acta Math. 116, 135–157.