Given $f: \mathbb{R} \to \mathbb{R}$ a $2\pi$-periodic function and continuously derivative such that \begin{align} \int_{0}^{2\pi}f(x)dx = 0 \end{align}
- Show that the derivative of $f$ is also $2\pi$-periodic.
- Given Parseval's Identity show that \begin{align} \int_{0}^{2\pi}(f'(x))^2 dx \geq \int_{0}^{2\pi} (f(x))^2 dx. \end{align}
Obviously the first one is trivial.
However I have no idea how to start for the second. Any thoughts?
The Parseval equevilance is:
$$\sum_{k=1}^\infty c_k \varphi_k \overset{L^2}{=} f \Longleftrightarrow \sum_{k=1}^\infty |c_k|^2 ||\varphi_k||_2^2 = ||f||_2^2$$ with $\Phi = (\varphi_k)_{k \in \mathbb{N}_0}$ an orthogonal system of functions in $L^2([a, b], \mathbb{K})$. I won't define every other concept that appear there, but don't hesitate to ask.
The Parseval identidy (see there) is quite different:
$$||f||_2^2 = \int_{0}^{2\pi} |f(x)|^2 \ dx = 2\pi \sum_{k = -\infty}^{+\infty}|c_k|^2 $$
Given we work in $\mathbb{R}$, $(f(x))^2 = |f(x)|^2$.
You basically want to prove $||f'||_2^2 \ge ||f||_2^2$, thus.
As $\int_0^{2\pi} f(x) \ dx = 0$, you also have $\int_0^{2\pi} f'(x) \ dx = 0$ (no need for Leibniz).
Let $c'_k$ be the Fourier coefficients of the derivative (NOT the derivative of $c_k$):
$$\sum_{k = -\infty}^{+\infty}|c'_k|^2 \ge \sum_{k = -\infty}^{+\infty}|c_k|^2$$
Let's use the orthogonal basis $\{1, \cos(nx), \sin(nx)\}$. We know that $a_k\sin'(kx) = k.a_k\cos(kx)$. As $k \ge 1$, $k.a_k \ge a_k$, and $||\cos(kx)||_2 = ||\sin(\pi/2 - kx)||_2= ||\sin(kx)||_2$.
We can conclude that $|c'_k|^2 > |c_k|^2$ given that $f$ et $f'$ have the same integral value between $0$ and $2\pi$ for all $k$. Note that the last step will maybe not be thorough enough for a mathematician. If not, you can find the last steps from here by your own. I hope it helped !