Let $f \in C^{n}(\mathbb R)$, $n \in \mathbb N$, be a $2\pi$-periodical function with fourier coefficients $c_{k}$.
Show that: There is a $C>0$ that $\forall k \in \mathbb Z\backslash\{0\}$ the following inequality holds:
$|c_{k}|\leq C|k|^{-n}$
My proof:
$c_{k}=\frac{1}{2\pi}\int^{2\pi}_{0}f(x)e^{-ikx}dx=\frac{1}{2\pi}[f(x)(-\frac{1}{ik})e^{-ikx}\bigg\rvert^{2\pi}_{0}-\int^{2\pi}_{0}f^{'}(x)\frac{e^{-ikx}}{-ik}dx=...=(\frac{1}{-ik})^{n}(\frac{1}{2\pi})\int^{2\pi}_{0}f^{(n)}(x)e^{-ikx}dx$
This means that $|c_{k}|=\frac{1}{2\pi}(\frac{1}{k})^{n}|\int^{2\pi}_{0}f^{n}(x)e^{-ikx}dx|\leq\frac{1}{2\pi}\frac{1}{k^{n}}\int^{2\pi}_{0}|f^{(n)}(x)|dx$
So if we let $C:= \frac{1}{2\pi}\int^{2\pi}_{0}|f^{n}(x)|dx$ then we have the inequality:
$|c_{k}|\leq\frac{1}{k^{n}}C$. My question pertains to my selection of $C$, since $f$ is $2\pi$-periodic does that not mean that $\int^{2\pi}_{0}|f^{n}(x)|dx=0$, and therefore $C=0$. But we need to find a $C>0$.
$\int^{2\pi}_{0}|f^{n}(x)|dx=0$ would imply that $f^n(x)=0$ by continuity, but then integrating would yield a polynomial, which can only be $2\pi$-periodic if $f$ is constant.
Even if we would have that $C=0$ (for example if $f\equiv0)$, you may still take $C=1$, or any $C>0$ you like to make the inequality hold (note that $k=0$ needs to be calculated separately).
Anyway, $C:= \frac{1}{2\pi}\int^{2\pi}_{0}|f^{n}(x)|dx$ is correct and doesn't have to be equal to $0$. Take for example $f(x)=\sin(x)$.