Riemann-Lebesgue Lemma

Lemma

Let $f \in L^1$.

Then:

$\displaystyle \lim_{n \to \infty} \int f(x)e^{inx} dx = 0$

That is, the Fourier transform of an $L^1$ function vanishes at infinity.

Proof

First suppose that $f(x) = \chi_{(a,b)}(x)$, the characteristic function of an interval.

Then:

$\displaystyle \int f(x)e^{inx} dx = \int_a^b e^{inx} dx = \frac{e^{inb} - e^{ina}}{in} \to 0$ as $n \to \infty$

So the theorem is true for characteristic functions of intervals.

Now let $\displaystyle f = \sum_1^N c_n \chi_{(a_i, b_i)} $, a simple function.

Then the same argument shows that the theorem holds for all such simple functions which are dense in $L^1$.

So let $\epsilon > 0$ and choose a simple function of the form above, call it $g_N(x)$, such that $\displaystyle \int \left|{f - g_N}\right| < \epsilon$.

Now:

$\displaystyle \left|{\int f(x)e^{inx} dx}\right| \leq \int \left|{f(x) - g_N(x)}\right| dx + \left|{\int g_N(x)dx}\right| \leq \epsilon + \epsilon$

for sufficiently large $N$.

Since $\epsilon > 0$ is arbitrary, we are done.

$\blacksquare$

Source of Name

This entry was named for Georg Friedrich Bernhard Riemann and Henri Léon Lebesgue.