The proof of Bessel's inequality

Question

Let $f \in L^2[-\pi, \pi]$, we are to prove that $$\sum_{-N}^{N}|c_n|^2 \le \frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f^2(x)\mbox{dx} \iff 2\pi\sum_{-N}^{N}|c_n|^2 \le \int\limits_{-\pi}^{\pi}f^2(x)\mbox{dx}$$ where $c_n$ is the Fourier's coefficient.
Let $S_n(x) = \sum_{n=-N}^{N}c_ne^{inx}$. I started with an estimation $$\int\limits_{-\pi}^{\pi}|f(x) - S_n(x)|^2\mbox{dx} = \int\limits_{-\pi}^{\pi}f^2(x)\mbox{dx} - \int\limits_{-\pi}^{\pi}f(x)\overline{S_n(x)} + f(x){S_n(x)}\mbox{dx} + \int\limits_{-\pi}^{\pi}S_n(x)\overline{S_n(x)}\mbox{dx}$$ The third integral is equal to $2\pi \sum_{n=-N}^{N}|c_n|^2$ (it's quite easy to show). That means that the second integral should be equal to $4\pi \sum_{n=-N}^{N}|c_n|^2$.
I tried to show that in such a way $$ \int\limits_{-\pi}^{\pi}f(x)\overline{S_n(x)} + f(x){S_n(x)}\mbox{dx} = \int\limits_{-\pi}^{\pi}f(x)\overline{S_n(x)}\mbox{dx} + \int\limits_{-\pi}^{\pi}f(x){S_n(x)}\mbox{dx} = \int\limits_{-\pi}^{\pi}f(x)\sum_{-N}^{N}\overline{c_n}e^{-inx}\mbox{dx} + \int\limits_{-\pi}^{\pi}f(x)\sum_{-N}^{N}{c_n}e^{inx}\mbox{dx} = \sum_{-N}^{N}\overline{c_n}\int\limits_{-\pi}^{\pi}f(x)e^{-inx}\mbox{dx} + \sum_{-N}^{N}{c_n}\int\limits_{-\pi}^{\pi}f(x)e^{inx}\mbox{dx}$$ Now I think that using some properties of Dirichlet kernel or maybe to use Riemann's - Lebesgue Lemma can help but I don't know how.

Answer 1

Let $g(x)=f(x)-S_N(x)$ where $S_N(x)=\sum_{-N}^N c_n\exp(inx)$. The point is that $g$ is orthogonal to $S_N$, that is $\int_{-\pi}^\pi g(x)\overline{S_N(x)}\,dx=0$. This follows from $\int_{-\pi}^\pi g(x)\overline{e^{inx}}\,dx=0$ (for $|n|\le N$) which is effectively the definition of the Fourier coefficient $c_n$.

Then $$\int_{-\pi}^\pi|f(x)|^2\,dx=\int_{-\pi}^\pi|g(x)|^2\,dx +\int_{-\pi}^\pi|S_N(x)|^2\,dx$$ by this orthogonality, giving $$\int_{-\pi}^\pi|f(x)|^2\,dx\ge\int_{-\pi}^\pi|S_N(x)|^2\,dx =\sum_{n=-N}^N2\pi|c_n|^2.$$