Riemann–Lebesgue lemma

Statement

If ƒ is L¹ integrable on R^d, that is to say, if the Lebesgue integral of |ƒ| is finite, then the Fourier transform of ƒ satisfies

f ^ ( z ) := ∫ R d f ( x ) exp ⁡ ( − i z ⋅ x ) d x → 0  as  | z | → ∞ .

Other versions

The Riemann–Lebesgue lemma holds in a variety of other situations.

If ƒ is L¹ integrable and supported on (0, ∞), then the Riemann–Lebesgue lemma also holds for the Laplace transform of ƒ. That is,

as |z| → ∞ within the half-plane Re(z) ≥ 0.

A version holds for Fourier series as well: if ƒ is an integrable function on an interval, then the Fourier coefficients of ƒ tend to 0 as n → ±∞,

This follows by extending ƒ by zero outside the interval, and then applying the version of the lemma on the entire real line.

A version holds for abstract measure spaces and replaces the use of exponential function by abstract functions. Yet the proof is not complicated. Please refer to the references listed at the end of the article.

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

We'll focus on the one-dimensional case, the proof in higher dimensions is similar. Suppose first that ƒ is a compactly supported smooth function. Then integration yields

| ∫ f ( x ) e − i z x d x | = | ∫ 1 i z f ′ ( x ) e − i z x d x | ≤ 1 | z | ∫ | f ′ ( x ) | d x → 0  as  z → ± ∞ .

If ƒ is an arbitrary integrable function, it may be approximated in the L¹ norm by a compactly supported smooth function g. Pick such a g so that ||ƒ − g||_L¹ < ε. Then

lim sup z → ± ∞ | f ^ ( z ) | ≤ lim sup z → ± ∞ | ∫ ( f ( x ) − g ( x ) ) e − i x z d x | + lim sup z → ± ∞ | ∫ g ( x ) e − i x z d x | ≤ ε + 0 = ε ,

and since this holds for any ε > 0, the theorem follows.