# Problem of the Week #66 - September 2nd, 2013

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Establish the Riemann-Lebesgue Theorem: If $f$ is integrable, then
$\lim_{n\to\infty}\int_{\mathbb{R}}f(x)\cos nx\,dm = 0.$

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#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. You can find my solution below.

We will need to use the following fact that I will state without proof (I may add a proof for this statement in the next couple days):

Lemma: Let $f$ be integrable over $\mathbb{R}$, and let $\epsilon>0$. Then there is a step function $\psi$ on $\mathbb{R}$ which vanishes outside a closed, bounded interval and $\displaystyle\int_{\mathbb{R}} |f-\psi|<\epsilon$.

Proof
: Suppose $f$ is integrable on $\mathbb{R}=(-\infty,\infty)$. By the Lemma, given $\epsilon>0$, there is a step function $\psi$ such that $\int_{\mathbb{R}}|f-\psi|< \epsilon/2$. Now,
$\left|\int_{\mathbb{R}} f(x)\cos nx\,dx\right|\leq \int_{\mathbb{R}} |(f(x)-\psi(x))\cos nx|\,dx + \int_{\mathbb{R}} |\psi(x)\cos nx|\,dx<\epsilon/2 + \int_{\mathbb{R}} |\psi(x)\cos nx|\,dx.$
Integrating $|\psi(x)\cos nx|$ over each interval on which $\psi$ is constant, we see that $\int_{\mathbb{R}} |\psi(x)\cos nx|\,dx \rightarrow 0$ as $n\to\infty$. Thus, there exists $N$ such that $\int_{\mathbb{R}}|\psi(x)\cos nx|\,dx<\epsilon/2$ for $n\geq N$ so $\left|\int_{\mathbb{R}} f(x)\cos nx\,dx\right|<\epsilon$ for $n\geq N$, i.e.
$\lim_{n\to\infty}\int_{\mathbb{R}} f(x)\cos nx \,dx = 0.$

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