Riemann-Lebesgue lemma

Question

How can I prove the following result? Let $([-1,1],M,m)$ a measure space, where $m$ is the Lebesgue measure in $[-1,1]$. If $f$ is Lebesgue integrable, then $$\lim_{n\to\infty}\int_{-1}^1f(x)\cos(nx)dm=0$$

Thanks !

Answer 1

There are several ways to do this. Most involve approximation (e.g. using smooth bump functions of the "standard machine").

Here is another proof of the lemma that is more direct, but also ultimately relies on approximation.

First note that this is just a special case of the fact that $|\hat{f}(t)|\to0\;\text{as}\;t\to\pm\infty$, and so we will prove this instead (recall $e^{-ixt}=\cos(xt)-i\sin(xt)$).

We will use the trick of writing $\hat{f}(t)$ in two different ways to get an unexpected expression that yields the result.

To that end, noting $e^{i\pi}=-1$, we have (with several steps included for your understanding) $$\begin{align*} \hat{f}(t)&=\int_{-\infty}^{\infty}f(x)e^{-ixt}\;dx\\ &=-e^{i\pi}\int_{-\infty}^{\infty}f(x)e^{-ixt}\;dx\\ &=-\int_{-\infty}^{\infty}f(x)e^{-i(xt+\pi)}\;dx\\ &=-\int_{-\infty}^{\infty}f(x)e^{-it(x+\pi/t)}\;dx\\ &=-\int_{-\infty}^{\infty}f(x-\pi/t)e^{-ixt}\;dx&\text{(change of variable $x\mapsto x-\pi/t$)}. \end{align*}$$ If we add the first and second line, we get $$2\hat{f}(t)=\int_{-\infty}^{\infty}\left(f(x)-f(x-\pi/t)\right)e^{-ixt}\;dx=\int_{-\infty}^{\infty}(f(x)-f_{\pi/t}(x))e^{-ixt}\;dx.$$

The notation $f_{h}(x)$ is shorthand for $f(x-h)$.

Taking absolute values and dividing by $2$, we get $$|\hat{f}(t)|\leq\frac{1}{2}\int_{-\infty}^{\infty}|f(x)-f_{\pi/t}(x)||e^{-ixt}|\;dx=\frac{1}{2}||f-f_{\pi/t}||_{L^{1}(\mathbb{R})}.$$

Sending $t\to\infty$ yields the claim since translation is continuous with respect to $||\cdot||_{1}$. This is easily verified with a continuous function, and the claim follows by density of continuous functions in $L^{1}$.

Answer 2

We had the following lemma in class:

[Riemann-Lebesgue Lemma] Let $h \in \mathcal{C}(\mathbb{R})$ be a $T$-periodic function, with mean value $m = \frac{1}{T} \int_{0}^{T} h(x) \, dx.$ Then for all $g \in L^1(\mathbb{R})$ it holds $\displaystyle\lim_{\lambda \to \infty} \int_{\mathbb{R}} g(x) h(\lambda x) \, dx = m \int_{\mathbb{R}} g(x) \, dx.$

Then taking $h(x):= cos(x)$ which has mean value $0$ on $[-1, 1]$ yields the result directly.