In mathematics, the Riemann–Lebesgue lemma, named after Bernhard Riemann and Henri Lebesgue, is of importance in harmonic analysis and asymptotic analysis.
The lemma says that the Fourier transform or Laplace transform of an L1 function vanishes at infinity.
Contents |
Let ƒ:R → C be a measurable function. If ƒ is L1 integrable, that is to say if the Lebesgue integral of |ƒ| is finite, then
This says that the Fourier transform of ƒ tends to 0 as z tends to infinity.
The Riemann–Lebesgue lemma holds in a variety of other situations.
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.
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
for every interval I ⊂ [a, b]. The proposition is therefore true for all step functions with support in [a, b] (similarly, the proposition is also true for a slightly larger class of functions known as simple function).
Step 2. By the dominated 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 compact support are dense in L1(R), this special case extends to the general result if we require t to be real.
The case of non-real t. Assume first that f has a compact support on and that f is continuously differentiable. Denote the Fourier/Laplace transforms of f and by F and G, respectively. Then , hence as . Because the functions of this form are dense in , the same holds for every f.