Editing Fourier transform (section)

=== Uniform continuity and the Riemann–Lebesgue lemma ===
[[File:Rectangular function.svg|class=skin-invert-image|thumb|The [[rectangular function]] is [[Lebesgue integrable]].]]
[[File:Sinc function (normalized).svg|class=skin-invert-image|thumb|The [[sinc function]], which is the Fourier transform of the rectangular function, is bounded and continuous, but not Lebesgue integrable.]]
The Fourier transform may be defined in some cases for non-integrable functions, but the Fourier transforms of integrable functions have several strong properties.

The Fourier transform <math>\hat{f}</math> of any integrable function <math>f</math> is [[uniformly continuous]] and{{sfn|Katznelson|2004|p=134}}{{sfn|Stein|Weiss|1971|p=2}}
<math display="block">\left\|\hat{f}\right\|_\infty \leq \left\|f\right\|_1</math>

By the ''[[Riemann–Lebesgue lemma]]'',<ref name="Stein-Weiss-1971">{{harvnb|Stein|Weiss|1971}}</ref>
<math display="block">\hat{f}(\xi) \to 0\text{ as }|\xi| \to \infty.</math>

However, <math>\hat{f}</math> need not be integrable. For example, the Fourier transform of the [[rectangular function]], which is integrable, is the [[sinc function]], which is not [[Lebesgue integrable]], because its [[improper integral]]s behave analogously to the [[alternating harmonic series]], in converging to a sum without being [[absolutely convergent]].

It is not generally possible to write the ''inverse transform'' as a [[Lebesgue integral]]. However, when both <math>f</math> and <math>\hat{f}</math> are integrable, the inverse equality
<math display="block">f(x) = \int_{-\infty}^\infty \hat f(\xi) e^{i 2\pi x \xi} \, d\xi</math> holds for almost every {{mvar|x}}.  As a result, the Fourier transform is [[injective]] on {{math|[[Lp space|''L''<sup>1</sup>('''R''')]]}}.