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The Riemann-Lebesgue lemma states that the integral of a function like the above is small. The integral will approach zero as the number of oscillations increases.

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.

[edit] Statement

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

$\int^\infty_{-\infty} f(x) e^{izx}\,dx \rightarrow 0\text{ as } z\rightarrow \pm\infty.$

This says that the Fourier transform of ƒ tends to 0 as z tends to infinity. In fact, the same holds for the Laplace transform of ƒ if ƒ is supported on (0, ∞), i.e. the above holds as |z| → ∞, Im(z) ≥ 0 if ƒ(x) = 0 for x < 0.

If instead, ƒ is a periodic, integrable function, then we can conclude that the Fourier coefficients of ƒ tend to 0 as n → ±∞,

$\hat{f}_n \ \to \ 0 .$

(Indeed: extend ƒ on the entire real axis by defining it to be zero outside a single period [ 0, T ]).

[edit] 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.

[edit] 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

$\int_I e^{itx}\,dx \rightarrow 0$ as $\quad t\rightarrow \pm\infty$

for every interval I ⊂ [a, b]. The proposition is therefore true for all step functions with support in [a, b].

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 z to be real.

The case of non-real z. Assume first that f has a compact support on $(0,\infty)$ and that f is continuously differentiable. Denote the Fourier/Laplace transforms of f and $f'$ by F and G, respectively. Then $F (z) = G (z) / z$ , hence $F(z)\rightarrow 0$ as $|z|\rightarrow\infty$ . Because the functions of this form are dense in $L^1(0,\infty)$ , the same holds for every f.