# Riemann-Lebesque lemma

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The result is named for mathematicians Riemann and Henri Lebesgue, and is important in our understanding of Fourier Series and the Fourier Transform.

## The Riemann-Lebesgue Lemma

If $f(t)$ is $L_{1}$ integrable, which is to say that

$\int _{a}^{b}|f(t)|\;dt<\infty$ then $|{\hat {f}}(\omega )|\rightarrow 0$ as $|\omega |\rightarrow \infty$ .

In English, this means that the Fourier transform of an absolutely integrable function vanishes at infinity. This result is important for resolving the issues of convergence of Fourier series and the asymptotic behavior of Fourier transforms and Fourier-like integrals, such as those encountered in seismic imaging.

## Proof , Case 1 - infinite limits of integration  =

Consider the forward Fourier transform

${\hat {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{i\omega t}\;dt$ and make use of the fact that $-1=e^{i\pi }$ ${\hat {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{i\omega t}\;dt=-\int _{-\infty }^{\infty }f(t)e^{i\omega t+i\pi }\;dt.$ Rewriting this expression slightly,

${\hat {f}}(\omega )=-\int _{-\infty }^{\infty }f(t)e^{i\omega (t+\pi /\omega )}\;dt,$ and changing variables by replacing $t+\pi /\omega$ by $t$ , we obtain

${\hat {f}}(\omega )=-\int _{-\infty }^{\infty }f(t-\pi /\omega )e^{i\omega t}\;dt,$ We may write, combining this result with the initial representation of the Fourier transform, as

${\hat {f}}(\omega )={\frac {1}{2}}\left[\int _{-\infty }^{\infty }f(t)e^{i\omega t}\;dt-\int _{-\infty }^{\infty }f(t-\pi /\omega )e^{i\omega t}\;dt\right].$ Applying the Cauchy-Schwartz inequality, we may write

$|{\hat {f}}(\omega )|=\left|{\frac {1}{2}}\int _{-\infty }^{\infty }\left[f(t)-f(t-\pi /\omega )\right]e^{i\omega t}\;dt\right|\leq {\frac {1}{2}}\int _{-\infty }^{\infty }\left|f(t)-f(t-\pi /\omega )\right|\;dt\rightarrow 0$ as $|\omega |\rightarrow \infty .$ Thus we may write, in terms of the Bachman-Landau order symbol that $f(\omega )=o(1)$ as $|\omega |\rightarrow \infty .$ ## Rate of Convergence

Of course, this does not tell us anything about the rate at which $|{\hat {f}}(\omega )|$ tends to zero. We only know that the Fourier transform of an $L_{1}$ -integrable function vanishes as $|\omega |\rightarrow \infty .$ If $f(t)$ vanishes $C_{0}^{N}$ smoothly, which is to say that $f(t)$ is a continuous function that is $N$ -times differentiable and that $f(t)$ and all of its derivatives vanish as $|t|\rightarrow \infty .$ Then we may perform repetitive integration by parts $N$ -times to yield

${\hat {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{i\omega t}\;dt=\left({\frac {-1}{i\omega }}\right)^{N}\int _{-\infty }^{\infty }f^{(N)}(t)e^{i\omega t}\;dt$ The function $f^{N}(t),$ which is the $N$ -th derivative of $f(t)$ is also $L_{1}$ -integrable, meaning that we can estimate the integral as being $o(1)$ by the Riemann-Lebesgue Lemma, and write that

${\hat {f}}(\omega )=O(|\omega |^{-N})o(1)=o(|\omega |^{-N}).$ If $f(t)$ is $C_{0}^{\infty }$ then ${\hat {f}}(\omega )$ decays faster than any power of $|\omega |.$ 