If $f \in C^{(2)}([-L,L])$, then $\exists$ a constant M > 0 that satisfy: $$|c_{n}(f)| \leq \frac{M}{n^2}.$$
where $c_{n}(f)= \frac{1}{2L}\int_{-L}^{L} f(x) \overline {g_{n}(x)}dx,$ and $g_{n}(x) = e^{in\pi x/L},$ for$ L>0$ and $n \in \mathbb{Z}.$
Could anyone give me a hint?
Suppose that $f \in C^2[-L,L]$ and that either $(i)$ $f$ is periodic $[$so that $f(-L) = f(L)$ and $f'(-L) = f'(L)]$ or $(ii)$ $f$ has compact support inside $[-L,L]$. Then by integrating by parts, we see \begin{align*} c(f) &= \frac{1}{2L}\int_{-L}^L f(x) e^{-in\pi x /L} dx \\ &= -\frac{1}{2L}\int^L_{-L} f'(x) \frac{e^{-in\pi x/L}}{-in\pi/L} dx + \frac{1}{2L}\left[ f(x) \frac{e^{-in\pi x/L}}{-in\pi/L} \right]_{x=-L}^{x=L}\\ &= \frac{1}{2L} \int_{-L}^L f''(x) \frac{e^{-in\pi x/L}}{(-in\pi/L)^2}dx -\frac{1}{2L}\left[ f'(x) \frac{e^{-in\pi x/L}}{(-in\pi/L)^2} \right]_{x=-L}^{x=L} + \frac{1}{2L}\left[ f(x) \frac{e^{-in\pi x/L}}{-in\pi/L} \right]_{x=-L}^{x=L}. \end{align*} Now $$\frac 1{2L}\left[ f(x) \frac{e^{-in\pi x/L}}{-in\pi/L} \right]_{x=-L}^{x=L} = \frac{1}{-2i\pi n}[f(L) e^{-in\pi} - f(-L)e^{in\pi}] = \frac{(-1)^n}{-2i\pi n}[f(L) - f(-L)] = 0$$ by our assumption and similarly for the other boundary term. Thus \begin{align*}\lvert c(f) \rvert &= \left \lvert\frac 1 {2L(-in\pi/L)^2} \int^L_{-L} f''(x) e^{-in\pi x/L} dx \right\rvert \\ &\le \frac{L}{2\pi^2 n^2} \int^L_{-L} \lvert f''(x) \rvert dx. \end{align*} Thus we can take $$M = \frac{L}{2\pi^2}\int^L_{-L} \lvert f''(x) \rvert dx $$ $($which is finite since $f \in C^2[-L,L])$ and we have $$\lvert c(f) \rvert \le \frac M{n^2}.$$