Using Bessel's inequality to prove the Riemann-Lebesgue lemna

Question

Let $f$ and $f'$ be piecewise continuous function on $[-L,L]$. Use Bessel's inequality to show that $$\lim_{ n\to \infty} \int_{-L}^L f(x)\cos \bigg(\frac{n \pi x}{L}\bigg) dx=\lim_{n\to \infty} \int_{-L}^L f(x) \sin\bigg(\frac{n \pi x}{L}\bigg) dx=0$$

Progress

I wrote the fourier expansion of $f(x)$ then multiplied both sides by $f(x)$ and tried to integrate by parts, but was not able to get the desired result.

Answer 1

For example, for the first term, using integration by part

$$\int_{-L}^L f(x) \cos \bigg(\frac{n\pi x}{L}\bigg) dx = - \frac{L}{n\pi} \int_{-L}^L f'(x) \cos \bigg(\frac{n\pi x}{L}\bigg) dx $$

Note that now you have an $n$ at the bottom.

The integral on the right can be bounded:

$$\bigg| \int_{-L}^L f'(x) \cos \bigg(\frac{n\pi x}{L}\bigg) dx \bigg| \leq \int_{-L}^L \bigg| f'(x) \cos \bigg(\frac{n\pi x}{L}\bigg)\bigg| dx \leq \int_{-L}^L | f'(x)| dx$$

Answer 2

These expressions are $a_n$ and $b_n$ in the Fourier Series (times L). If these limits did not equal 0 then the sum in Bessel's Inequality would diverge by the divergence test and the equality would be false.