Artificial intelligent assistant

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

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:

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

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