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>
Perhaps surprisingly, although this result is true, it was not proved until 1966 in (Carleson, 1966).
For strictly finitary discrete Fourier transforms, these delicate questions of convergence are avoided.
Reference
Carleson (1966). On the convergence and growth of partial sums of Fourier series. Acta Math. 116, 135–157.