How to prove that $\lim\limits_{n\to \infty} \int\limits_{-\pi}^{\pi} |f(x) - S_n(f;x)|\,\mathrm dx = 0$?

Question

Let $f$ be bounded and integrable on $[-\pi, \pi]$. Denote the partial sum of Fourier series of $f$ by $S_n(f;x)$. Prove that $$\lim_{n\to \infty}\int\limits_{-\pi}^{\pi} |f(x) - S_n(f;x)|\,\mathrm dx = 0.$$

I want to apply Parseval's theorem. Since $f$ is integrable, we have:

$$\lim_{n\to \infty} \int\limits_{-\pi}^{\pi} |f(x) - S_n(f;x)|^2\,\mathrm dx = 0$$

But I don't know how to derive the required argument through algebra in the $L_2$ norm.

Answer 1

Thanks for the comment.

Applying the Cauchy-Schwartz Inequality:

let $g(x) = 1$ then we have

$\int\limits_{-\pi}^{\pi} |f(x) - S_n(f;x)|^2\, \mathrm dx*\int\limits_{-\pi}^{\pi}1\,\mathrm dx \ge (\int\limits_{-\pi}^{\pi} |f(x) - S_n(f;x)|\, \mathrm dx)^2$

taking the limit then we will get the result.