Riemann-Lebesgue lemma
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2008-06-07T22:48:16Z
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[[Image:Highly oscillatory function.png|right|frame|The Riemann-Lebesgue lemma states that the integral of a function like the above is small.]]
In [[mathematics]], the '''Riemann-Lebesgue lemma''' (one of its special cases is also called '''Mercer's theorem'''), is of importance in [[harmonic analysis]] and [[asymptotic analysis]]. It is named after [[Bernhard Riemann]] and [[Henri Lebesgue]].
The lemma says that the Fourier or [[Laplace transform]] of an [[Lp_space|''L''<sup>1</sup> function]] vanishes at infinity.
Intuitively, the lemma says that if a function oscillates rapidly around zero, then the integral of this function will be small. The integral will approach zero as the number of oscillations increases.
==Definition==
Let ''f'':'''R''' → '''C''' be a [[measurable function]]. If ''f'' is '''L'''<sup>1</sup> integrable, that is to say if the Lebesgue integral of |''f''| is finite, then
:<math>\int^{\infty}_{-\infty} f(x) e^{izx}\,dx \rightarrow 0 </math> as <math>\quad z\rightarrow \pm\infty</math>.
This says that the Fourier transform of ''f'' tends to 0 as ''z'' tends to infinity.
In fact, the same holds for the Laplace transform of ''f'' if ''f'' is supported on <math>(0,\infty)</math>,
i.e., the above holds as <math>|z|\rightarrow +\infty,\ \text{im}\,z\ge0\ </math> if <math>\ f(x)=0\ (x\le0)</math>.
If, instead, ''f'' is a periodic, integrable function, then we can conclude that the [[Fourier coefficient]]s of ''f'' tend to 0 as ''n'' → ± ∞ ,
:<math>\hat{f}_n \ \to \ 0 .</math>
(Indeed: extend ''f'' on the entire real axis by defining it to be zero outside a single period <math>[0,T]</math>).
==Applications==
The Riemann-Lebesgue lemma can be used to prove the validity of asymptotic approximations for integrals. Rigorous treatments of the [[method of steepest descent]] and the [[method of stationary phase]], amongst others, are based on the Riemann-Lebesgue lemma.
==Proof==
The proof of the last special case can be organized into 3 steps; the 4th step extends the result to the first special case.
''Step 1''. An elementary calculation shows that
:<math>\int_I e^{itx}\,dx \rightarrow 0</math> as <math>\quad t\rightarrow \pm\infty</math>
for every interval ''I'' ⊂ [''a'', ''b'']. The proposition is therefore true for all step functions with support in [''a'', ''b''].
''Step 2''. By the [[monotone convergence theorem]], the proposition is true for all positive functions, integrable on [''a'', ''b''].
''Step 3''. Let ''f'' be an arbitrary measurable function, integrable on [''a'', ''b'']. The proposition is true for such a general ''f'', because one can always write ''f'' = ''g'' − ''h'' where ''g'' and ''h'' are positive functions, integrable on [''a'', ''b''].
''Step 4''. Because functions with finite support are dense in
'''L'''<sup>1</sup>('''R'''),
this special case extends to the general result if we require ''z'' to be real.
''The case of non-real z''.
Assume first that ''f'' has a compact support on <math>(0,\infty)</math> and that ''f'' is continuously differentiable.
Denote the Fourier/Laplace transforms of ''f'' and <math>f'</math> by ''F'' and ''G'', respectively.
Then <math>F(z)=G(z)/z</math>, hence <math>F(z)\rightarrow 0</math> as <math>|z|\rightarrow\infty</math>.
Because the functions of this form are dense in <math>L^1(0,\infty)</math>, the same holds for every ''f''.
==References ==
*{{cite book | author =[[Salomon Bochner|Bochner S.]],Chandrasekharan K. | title=Fourier Transforms | publisher= Princeton University Press | year=1949}}
* {{mathworld|urlname=Riemann-LebesgueLemma|title=Riemann-Lebesgue Lemma}}
[[Category:Asymptotic analysis]]
[[Category:Harmonic analysis]]
[[Category:Lemmas]]
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