# Riemann–Lebesgue lemma

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.

## Statement

If f is L1 integrable on Rd, that is to say, if the Lebesgue integral of |f| is finite, then the Fourier transform of f satisfies

$\hat{f}(z):=\int_{\mathbb{R}^{d}} f(x) e^{-iz \cdot x}\,dx \rightarrow 0\text{ as } |z|\rightarrow \infty.$

## Other versions

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

• If f is L1 integrable and supported on (0, ∞), then the Riemann–Lebesgue lemma also holds for the Laplace transform of f. That is,
$\int_0^\infty f(t) e^{-tz}\,dt \to 0$
as |z| → ∞ within the half-plane Re(z) ≥ 0.
$\hat{f}_n \ \to \ 0 .$
This follows by extending f by zero outside the interval, and then applying the version of the lemma on the entire real line.

## 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 f is a compactly supported smooth function. Then integration by parts in each variable yields

$\left| \int f(x) e^{-izx}dx\right|=\left|\int \frac{1}{iz} f'(x)e^{-izx}dx\right| \leq \frac{1}{|z|}\int|f'(x)|dx \rightarrow 0 \mbox{ as } z\rightarrow\pm\infty.$

If f is an arbitrary integrable function, it may be approximated in the L1 by a compactly supported smooth function g. Pick such a g so that ||f-g||L1. Then

$\limsup_{z\rightarrow\pm\infty} |\hat{f}(z)| \leq \limsup_{z\to\pm\infty} \left|\int (f(x)-g(x))e^{-ixz}dx\right| + \limsup_{z\rightarrow\pm\infty} \left|\int g(x)e^{-ixz}dx\right| \leq \varepsilon+0=\varepsilon,$

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

The case of non-real t. 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 $F(t)=G(t)/t$, hence $F(z)\rightarrow 0$ as $|t|\rightarrow\infty$. Because the functions of this form are dense in $L^1(0,\infty)$, the same holds for every f.