Convergence of sequence involving integral Let $f$ be continuously differentiable real valued function in $[-\pi,\pi]$ such that $f(-\pi)=f(\pi)$.
Define
$\{a_n\} = \int_{-\pi}^{\pi}f(t)\,\cos(nt) dt, n\in\, N$
Which of the following statements are true?


*

*The sequence $\{a_n\}$is bounded. 

*The sequence $\{na_n\}$ converges to zero as $n\to \infty$.

*The sequence $\sum_{n=1}^\infty n^2 {|a_n|}^2 $ is convergent.


Try 1 is correct as $f(t) \cos(nt)$ is continuous on a compact set hence it is closed and bounded. Therefore the integral is bounded but how to prove that 2 and 3 is correct along with 1?
 A: Suppose $g \in L^2[-\pi,\pi] $ and $b_k$ are an orthonormal set, then
it is straightforward to check that
$\langle g - \sum_k \langle b_k, g \rangle b_k, b_i\rangle = 0$, from
which we get
$\|g\|^2 = \|g - \sum_k \langle b_k, g \rangle b_k \|^2 + \| \sum_k \langle b_k, g \rangle b_k\|^2 \ge \| \sum_k \langle b_k, g \rangle b_k\|^2 = \sum_k | \langle b_k, g \rangle|^2$
(Bessel's inequality).
Note that $b_k = {1 \over \sqrt{\pi}} \sin (kt)$ is an orthonomal set and
$f'$ is continuous hence $f' \in L^2[-\pi,\pi]$, so we have
$\|f'\|^2 \ge {1 \over \pi} \sum_k |g_k|^2$, where 
$g_k = \int_{-\pi}^{\pi} f'(t) \sin (kt) dt$.
Since $g_k = f(t)\sin(kt) |_{-\pi}^\pi -k\int_{-\pi}^{\pi} f(t) \cos (kt) dt = -k a_k$, we see that
$\sum_k |k a_k|^2 $ is bounded from which answers to 3, 2 & 1 follow.
A: We will show that 2) is true. Note that by integration by parts,
$$
a_n=\int_{-\pi}^{\pi}f(t)\cos(nt)dt=\left[\frac{f(t)\sin(nt)}{n}\right]_{-\pi}^{\pi}-\frac{1}{n}\int_{-\pi}^{\pi}f'(t)\sin(nt)dt=-\frac{1}{n}\int_{-\pi}^{\pi}f'(t)\sin(nt)dt.
$$
Since $f'$ is continuous then $na_n\to 0$ by Riemann-Lebesgue Lemma.
Note that for 2) the condition $f(-\pi)=f(\pi)$ is irrelevant.
