Fourier inversion theorem

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:

:$(mathcal\{F\}f)(t)=int\_\{-infty\}^infty\; f(x),\; e^\{-itx\},dx.$

Then it is asserted that

:$f(x)=frac\{1\}\{2pi\}int\_\{-infty\}^infty\; (mathcal\{F\}f)(t),\; e^\{itx\},dt.$

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:

:

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

:$2sin(at)/t.$

In such a case, the integral in the Fourier inversion theorem above must be taken to be an improper integral (Cauchy principal value)

:$lim\_\{b\; ightarrowinfty\}frac\{1\}\{2pi\}int\_\{-b\}^b\; (mathcal\{F\}f)(t)\; e^\{itx\},dt$

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.

Fourier transforms of quadratically integrable functions

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

:

Then the Fourier transform is another quadratically integrable function.

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

:$widehat\{f\}(n)=frac\{1\}\{2pi\}int\_\{-pi\}^pi\; f(x),e^\{-inx\},dx.$

The Fourier inversion theorem might then say that

:$sum\_\{n=-infty\}^\{infty\}\; widehat\{f\}(n),e^\{inx\}=f(x).$

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

:$lim\_\{N\; ightarrowinfty\}int\_\{-pi\}^pileft|f(x)-sum\_\{n=-N\}^\{N\}\; widehat\{f\}(n),e^\{inx\}\; ight|^2,dx=0.$

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

:$f(x)=lim\_\{N\; ightarrowinfty\}sum\_\{n=-N\}^\{N\}\; widehat\{f\}(n),e^\{inx\}.$

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

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

References

*Lennart Carleson (1966). On the convergence and growth of partial sums of Fourier series. "Acta Math." 116, 135&ndash;157.