I am self-studying a functional analysis course and am stuck at one of the problems that introduces Sobolev spaces in relation to the Fourier series and Hilbert spaces. I am having trouble with
Let $s \geq 0$. Then $f \in L^2([-\pi, \pi])$ is an element of the Sobolev space $H^s(\mathbb{T})$ of order $s$ if \begin{equation} \lim_{N\rightarrow \infty} \sum_{|n| \leq N} |\hat{f}(n)|^2 (1+|n| ^2)^s < \infty \end{equation} Suppose that $f \in C^k([-\pi, \pi])$ and $f(\pi) = f(-\pi)$, $f'(\pi) = f'(-\pi)$, ..., $f^{(k-1)}(\pi) = f^{(k-1)}(-\pi)$. Show $f \in H^s(\mathbb{T})$ for $0\leq s \leq k$.
I have tried using Parseval's identity on $L^2$ to show that $ \| f \|_{L^2}^2 = 2\pi \sum_{n\in\mathbb{Z}} |\hat{f}(n)|^2 < \infty$ and Abel's convergence test to show that the series of the products converges but run into the issue that $\{(1+|n|^2)^s\}_{n\in \mathbb{Z}}$ is not bounded. I have also tried to use Fejér's theorem about the uniform convergence of the Cesario-Fourier mean to continuous periodic functions, but am not sure how to proceed from there.
Let $0\leq s\leq k$.
Because $f^{(s)}$ is continuous on $[-\pi, \pi]$, it is in $L^2(-\pi, \pi])$. Thus, by Parseval, $$\sum_{n\in\mathbb Z} |\widehat {f^{(s)}}(n)|^2 = \frac 1 {2\pi}\int_{-\pi}^{\pi}|f^{(s)}|^2dt < +\infty\tag{1}$$
If you compute the Fourier coefficients of $f^{(s)}$ by repeatedly integrating by parts, noting that the brackets vanish because of the boundary conditions: $$\begin{split} \widehat {f^{(s)}}(n) &= \frac 1 {2\pi} \int_{-\pi}^\pi f^{(s)}(t)e^{-2i\pi nt}dt\\ &=\underbrace{\left[ -2i\pi ne^{-2i\pi nt}f^{(s)}(t)\right]_{-\pi}^\pi}_{=0}+\frac {2i\pi n} {2\pi} \int_{-\pi}^\pi f^{(s-1)}(t)e^{-2i\pi nt}dt\\ &\vdots\\ &=(2i\pi n)^s \cdot\frac 1 {2\pi} \int_{-\pi}^\pi f(t)e^{-2i\pi nt}dt\\ &= (2i\pi n)^s \cdot \widehat f(n) \end{split}$$ Connecting with $(1)$, you get that $$\sum_{n\in\mathbb Z} |n|^{2s}|\widehat {f(n)}|^2 <+\infty$$ which is equivalent to $$\sum_{n\in\mathbb Z} (1+|n|^2)^{s}|\widehat {f(n)}|^2 <+\infty$$