In mathematics, the Riemann–Lebesgue lemma, named after Bernhard Riemann and Henri Lebesgue, states that the Fourier transform or Laplace transform of an L1 function vanishes at infinity. It is of importance in harmonic analysis and asymptotic analysis.
Let
f\inL1(\Rn)
f\colon\Rn → \C
\|f\| | |
L1 |
=
\int | |
\Rn |
|f(x)|dx<infty,
\hat{f}
f
\hat{f}\colon\Rn → \C, \xi\mapsto
\int | |
\Rn |
f(x)e-ix ⋅ \xidx.
\hat{f}
|\hat{f}(\xi)|\to0
|\xi|\toinfty
Because the Fourier transform of an integrable function is continuous, the Fourier transform
\hat{f}
n) | |
C | |
0(\R |
L1(\Rn)
n) | |
C | |
0(\R |
We will focus on the one-dimensional case
n=1
f
\xi ≠ 0
stylex\tox+
\pi | |
\xi |
\hat{f}(\xi)=\int\Rf(x)e-ix\xidx=\int\Rf\left(x+
\pi | |
\xi |
\right)e-ix\xie-i\pidx=-\int\Rf\left(x+
\pi | |
\xi |
\right)e-ix\xidx
\hat{f}(\xi)
|\hat{f}(\xi)|\le
1 | |
2 |
\int\R\left|f(x)-f\left(x+
\pi | |
\xi |
\right)\right|dx
f
\left|f(x)-f\left(x+\tfrac{\pi}{\xi}\right)\right|
0
|\xi|\toinfty
x\in\R
|\hat{f}(\xi)|
|\xi|\toinfty
If
f
L1
\varepsilon>0
g
\|f-g\| | |
L1 |
\leq\varepsilon
\limsup\xi → \pminfty|\hat{f}(\xi)|\leq\limsup\xi\to\pminfty\left|\int(f(x)-g(x))e-ix\xidx\right|+\limsup\xi → \pminfty\left|\intg(x)e-ix\xidx\right|\leq\varepsilon+0=\varepsilon.
\varepsilon>0
|\hat{f}(\xi)|\to0
|\xi|\toinfty
The Riemann–Lebesgue lemma holds in a variety of other situations.
f\inL1[0,infty)
f
infty | |
\int | |
0 |
f(t)e-tzdt\to0
as
|z|\toinfty
Re(z)\geq0
f
\hat{f}k
f
k\to\pminfty
f
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.