Question. Let $~f\in C^1[-\pi,\pi]$ be such that $f(-\pi)=f(\pi)$. Define $$a_n=\int_{-\pi}^{\pi}f(t)\cos nt~dt~, ~n\in \mathbb{N}$$ Which of the following statements are true?
a. The sequence $\{a_n\}$ is bounded.
b. The sequence $\{na_n\}$ converges to zero as $n \to \infty$
c. The series $\sum_{n=1}^{\infty}n^2|a_n|^2$ is convergent.
My Solution.
(a) True. $|a_n|\le \int_{-\pi}^{\pi}|f(t)||\cos nt|dt \le \int_{-\pi}^{\pi}|f(t)|dt, ~ \text{for all} ~n \in \mathbb{N}$.
But I can't prove/disprove the other two options. Please help..thank you.
For b) and c) you just have to integrate by parts. $a_n=-\frac 1 n \int_{-\pi} ^{\pi} f'(t) \sin (nt) \, dt$ so $na_n$ are the coefficients of the sine series of $g=-f'$ which is continuous and periodic. By standard results in the theory of Fourier series b) and c) are both true. In fact $\pi \sum n^{2}|a_n|^{2}\leq\int |f'(t)|^{2} \, dt $