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 be an integrable function, i.e. is a measurable function such that
and let be the Fourier transform of , i.e.
Then vanishes at infinity: as .
Because the Fourier transform of an integrable function is continuous, the Fourier transform is a continuous function vanishing at infinity. If denotes the vector space of continuous functions vanishing at infinity, the Riemann–Lebesgue lemma may be formulated as follows: The Fourier transformation maps to .
We will focus on the one-dimensional case , the proof in higher dimensions is similar. First, suppose that is continuous and compactly supported. For , the substitution leads to
This gives a second formula for . Taking the mean of both formulas, we arrive at the following estimate:
Because is continuous, converges to as for all . Thus, converges to 0 as due to the dominated convergence theorem.
If is an arbitrary integrable function, it may be approximated in the norm by a compactly supported continuous function. For , pick a compactly supported continuous function such that . Then
Because this holds for any , it follows that as .
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.
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 be an integrable function, i.e. is a measurable function such that
and let be the Fourier transform of , i.e.
Then vanishes at infinity: as .
Because the Fourier transform of an integrable function is continuous, the Fourier transform is a continuous function vanishing at infinity. If denotes the vector space of continuous functions vanishing at infinity, the Riemann–Lebesgue lemma may be formulated as follows: The Fourier transformation maps to .
We will focus on the one-dimensional case , the proof in higher dimensions is similar. First, suppose that is continuous and compactly supported. For , the substitution leads to
This gives a second formula for . Taking the mean of both formulas, we arrive at the following estimate:
Because is continuous, converges to as for all . Thus, converges to 0 as due to the dominated convergence theorem.
If is an arbitrary integrable function, it may be approximated in the norm by a compactly supported continuous function. For , pick a compactly supported continuous function such that . Then
Because this holds for any , it follows that as .
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.