I want to prove the following:
Let $f:\mathbb{R}\rightarrow \mathbb{C}$ so that $f \big |_{[0,2\pi]}$ is integrable. Let $V$ be the vectorspace of all $2\pi$-periodic functions and $U \subset V$ be the real-subspace of real functions of $V$. Let further $x \in \mathbb{R}$ and $k \in \mathbb{Z}$.
Prove that: If $f(x) = f(-x)$ and if $F_n(f)$ in the quadratic mean approaches $f$ one has
$$\frac{a_0^2}{2} + \sum_{k=1}^n a_k^2 = \frac{2}{\pi} \int_0^{2\pi} |f(x)|^2 dx$$
With $$a_k = c_k + c_{-k} = \frac{1}{\pi} \int_0^{2\pi} f(x) cos(kx) dx$$ $$F_n(f) = \frac{a_0}{2} + \sum_{k=1}^n a_k cos(kx)$$ $$c_{-k} = \overline{c_k}$$
I really don't have much yet because this one is pretty tough to me... Here is what I've got so far:
$$\frac{a_0^2}{2} + \sum_{k=1}^n a_k^2 = \frac{1}{2\pi^2} \int_0^{2\pi} f(x)^2dx + \frac{1}{\pi^2} \int_0^{2\pi} f(x)^2cos^2(kx)dx$$
The product of two uneven function results in an even function. And the produnct of an even function with another even function results in an even function aswell. Thus we get that the right integral is even and we can write:
$$\frac{1}{\pi^2} \int_0^{2\pi} f(x)^2cos^2(kx)dx = \frac{1}{2\pi^2} \int_{-2\pi}^{2\pi} f(x)^2cos^2(kx)dx$$
But I don't really know what this should help me... Should I just go for it and calculate the right side? I don't think that is necessary here....
Thank you very much for your help and sorry for not providing any helpful own thoughts.
FunkyPeanut