A question about lebesgue integration

Question

Let $f$ be a Lebesgue-integrable map over $\mathbb{R}$. Please show that: $$\lim_{n\rightarrow \infty}\int_{\mathbb{R}}f(x)\cos(nx)\,\mathrm{d}x=0.$$ Thanks in advance.

Answer 1

Hint. Check the desired property for characteristic functions, then deduce it by linearity for simple functions. Conclude by density.

Answer 2

For $f \in C_c^\infty(\mathbb R)$, we see by integration by parts $$\left \lvert \int_{\mathbb R} f(x) \cos(nx) dx \right \rvert = \frac{1}{n} \left \lvert \int_\mathbb R f'(x) \sin(nx) dx \right \rvert \le \frac 1 n \int_\mathbb R \lvert f'(x) \rvert dx \to 0$$ as $n\to \infty$ since $\int_{\mathbb R} \lvert f'(x) \rvert dx$ is some finite number. Thus the property holds for $f \in C_c^\infty(\mathbb R)$.

Let $\epsilon> 0$. Since $C_c^\infty(\mathbb R)$ is dense in $L^1(\mathbb R)$, for any $f \in L^1(\mathbb R)$, we can find $g \in C_c^\infty(\mathbb R)$ such that $$\int_\mathbb R \lvert f(x) -g(x) \rvert dx < \epsilon. $$ By the above, for $n$ sufficiently large, we have $$\left \lvert \int_\mathbb R g(x) \cos(nx) dx \right \rvert < \epsilon.$$ Thus for such $n$, we have \begin{align*} \left \lvert \int_\mathbb R f(x) \cos(nx) dx \right\rvert &\le \left \lvert \int_\mathbb R g(x) \cos(nx) dx \right \rvert + \left \lvert \int_\mathbb R [f(x) - g(x)]\cos(nx) dx \right \rvert \\ &< \epsilon + \int_{\mathbb R} \lvert f(x) - g(x) \rvert dx \\ &< 2\epsilon. \end{align*} This shows that $$\lim_{n\to \infty} \int_\mathbb R f(x) \cos(nx) dx = 0$$ for all $f \in L^1(\mathbb R)$.

As a side note, this is a version of the Riemann-Lebesgue lemma which states that for any $f\in L^1(\mathbb R)$, we have $\lim_{\lvert\xi \rvert \to \infty} \hat f(\xi) = 0$ where $\hat f$ is the Fourier transform of $f$.