Let $f\in C^1[-\pi ,\pi]$ be such that $f(-\pi)=f(\pi)$. Define
$$a_n=\int _{-\pi}^\pi f(t)\cos nt \, dt, n\in \mathbb N$$
Prove that:
- $\{a_n\}$ is bounded,
- $\{na_n\} $ is convergent to $0$,
- $\sum n^2|a_n|^2$ is convergent.
Solution:
Since every continuous function on a compact set is bounded so $|f(x)|\leq M$ which implies that $$|f(x)\cos nx|\leq |f(x)||\cos nx|\leq M \cdot 1 = M$$ is bounded on $[-\pi,\pi]$. Hence $|a_n|\leq |\int _{-\pi}^\pi f(t)\cos nt \, dt|\leq 2M\pi$.
How should I go on $2,3$. Practically I am not getting any idea.
Hint: For $2$ use integration by parts and the famous Riemann-Lebesgue Lemma and for $3$ recall Bessel's inequality.
Edit1: As $f'$ is continuous on $[-\pi,\pi]$ implies $f'\in L^2([-\pi,\pi])$, Apply Bessel's Inequality to $f'$ hence $$ \sum_{n\geq 1} n^2 a_n^2 <+\infty $$
Edit2: Recall that we have the (complete) orthonormal trigonometric system (basis) $\;\{u_n\}_{n\in\Bbb N}\;,\;\;u_n=\{1,\sin x,\cos x, \sin 2x,\cos2x,\ldots\}\;$ with which we work in Fourier series, so then denoting by $\;\langle,\rangle\;$ the usual inner product we get(ofcourse using Bessel's inequality)
$$\sum_{n=1}^\infty \langle f'(x),u_n\rangle^2\le||f'||^2$$
But using the usual notation from Fourier series the above is just
$$||f'||^2\ge\sum_{n=1}^\infty n^2 a_n^2 $$
Note that here we are using that fourier coefficients of $f'$ are $na_n$ which you must have noticed in part $2$.