If we assume that $f$ is a real-valued and continously differentiable. Let's also assume that $f(x) = f(x+2\pi) \forall x \in \mathbb{R}$. Can we then deduce something about the following statements:
- $f \in \mathcal{L^2}(-\pi, \pi)$
- $f' \in \mathcal{L^2}(-\pi, \pi)$
- $\langle f , f' \rangle$
My attempt:
If $f$ is continously differentiable, then both $f$ and $f'$ are continous on the reals. If $f$ is continous on a bounded interval $[-\pi, \pi]$, then it attains a maximum $M_f$ and likewise $M_{f'}$ for $f'$.
From the definition of $\mathcal{L^2}(-\pi, \pi)$, we can bound both $f$ and $f'$ such that:
$$\int_{-pi}^\pi |f|^2 \leq 2 M_f^2 \pi $$
and correspondingly with $f'$. So both of them must be in $\mathcal{L^2}(-\pi, \pi)$.
For the last statement, we use the definition:
$\langle f , f' \rangle := \int_{-\pi}^\pi f f' dx$
Since $f'$ and $f$ are continous, $f'f$ is also continous on $[-\pi, \pi]$, which means that we can deduce that this integral is finite.
Is this a correct way of thinking about these problems? Thank you in advance.
Minor: you have to pick $M_f$ as maximum of $|f|$.
As to (3): we have $(f^2)' = 2ff'$, so $$ \int_{-\pi}^\pi ff' \ dx = \frac12( f(\pi)^2 - f(-\pi)^2) = 0 $$ because of periodicity.