$H_p^s$ is a standard $L^2$ Sobolev space of periodic functions on $[0, 2\pi]$, with $\|f\|^2_{H_p^s} = \sum_{j = 0}^s \|f^{(j)}\|_{L^2}^2$. Let
$$f_N(x) = \sum_{|n| < N} \hat{f}_n \frac{1}{\sqrt{2\pi}} e^{inx}$$
Prove that
$$\|f - f_N\|_{H_p^j} \leq N^{j - s} \|f\|_{H_p^s}$$
I have seen the proof for the $L^2 = H_p^0$ case and want to imitate it. My argument has gotten this far without getting stuck:
\begin{align} \|f - f_N\|_{H_p^j}^2 &= \sum_{k = 0}^j\|D_x^k(f - f_N) \|^2_{L_2} \\ &=\sum_{k = 0}^j\left\|\left(f^{(k)} - \left(f^{(k)}\right)_N\right) \right\|^2_{L_2} \\ &=\sum_{k = 0}^j \sum_{|n| \geq N} \left|\hat{f}_n^{(k)}\right|^2 \end{align}
Then I run into problems. I want to say $\sum_{k = 0}^j \sum_{|n| \geq N} \left|\hat{f}_n^{(k)}\right|^2 = \sum_{k = 0}^j \sum_{|n| \geq N} n^{k - s}\left|\hat{f}_n^{(s)}\right|^2 \leq \sum_{k = 0}^j \sum_{|n| \geq N} N^{k - s}|\hat{f}_n^{(s)}|^2$ but this ends up with a bound not as tight as the one requested. What should I do next?
UPDATE: My instructor said that the correct statement may be
$$\|f - f_N\|_{H_p^j} \leq (j+1) N^{j - s} \|f\|_{H_p^s}$$
In this case the proof is basically done. We don't know if the original statement without this factor is correct (it might not be), so I'll leave it as an open problem to prove or disprove whether that original bound is correct.