Riemann-Lebesgue Lemma

Lemma
Let $f \in L^1$.

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} \int \map f x e^{i n x} \rd x = 0$

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

Proof
First suppose that $\map f x = \map {\chi_{\tuple {a, b} } } x$, the characteristic function of an open interval.

Then:
 * $\displaystyle \int \map f x e^{i n x} \rd x = \int_a^b e^{i n x} \rd x = \frac {e^{i n b} - e^{i n a} } {i n} \to 0$ as $n \to \infty$

By additivity of limits, the same holds for an arbitrary simple function.

That is, for any function $f$ of the form:
 * $\displaystyle f = \sum_{n \mathop = 1}^N c_n \chi_{\tuple {a_i, b_i} }, \qquad c_n \in \R, \qquad a_i \le b_i \in \R$

we have that:
 * $\displaystyle \lim_{n \mathop \to \infty} \displaystyle \int \map f x e^{i n x} \rd x = 0$

Finally, let $f \in L^1$ be arbitrary.

Let $\epsilon \in \R_{>0}$ be fixed.

Since the simple functions are dense in $L^1$, there exists a simple function $g$ such that:


 * $\displaystyle \int \size {f - g} < \epsilon$

By our previous argument and the definition of a limit of a complex function, there exists $N \in \N$ such that for all $n > N$:
 * $\displaystyle \size {\int \map g x e^{i n x} \rd x} < \epsilon$

By Integral of Integrable Function is Additive:
 * $\displaystyle \int \map f x e^{i n x} \rd x = \int \paren {\map f x - \map g x} e^{i n x} \rd x + \int \map g x e^{i n x} \rd x$

By the triangle inequality for complex numbers, the triangle inequality for integrals, multiplicativity of the absolute value, and Euler's Formula:


 * $\displaystyle \size {\int \map f x e^{i n x} \rd x} \le \int \size {\map f x - \map g x} \rd x + \size {\int \map g x e^{i n x} \rd x}$

For all $n > N$, the is bounded by $2 \epsilon$ by our previous arguments.

Since $\epsilon$ was arbitrary, this establishes:
 * $\displaystyle \lim_{n \mathop \to \infty} \int \map f x e^{i n x} \rd x = 0$

for all $f \in L^1$.