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The following higher order estimate for periodic Sobolev function $ f \in H_\text{per}^r(0,2\pi) $ is a really practical result in numerical analysis.

Let $ P_n : L^2 ( 0, 2\pi ) \longrightarrow \mathcal{T}_n \subset L^2(0, 2\pi ) $ be an orthogonal projection operator, where \begin{equation*} \mathcal{T}_n := \left\{ \xi_0 \cdot \frac{1}{\sqrt{2\pi}} + \sum^n_{k=1}\xi_k \frac{\cos kt}{\sqrt{\pi}} + \sum^n_{k=1} \eta_k \frac{\sin kt}{\sqrt{\pi}} : \ \xi_0, \xi_k , \eta_k \in \mathbb{R} \right\}. \end{equation*} Then $ P_n $ is given as follows: \begin{equation} (P_n x)(t) = \xi_0 \frac{1}{\sqrt{2\pi}} + \sum^n_{ k = 1 }\xi_k \frac{\cos k t}{\sqrt{\pi}} + \sum^n_{ k=1 } \eta_k \frac{\sin k t}{\sqrt{\pi}} , \end{equation} where \begin{equation*} \xi_0 = \frac{1}{\sqrt{2\pi}} \int^{2\pi}_0 x(t) \, dt ,\ \xi_k = \int^{2\pi}_0 x(t) \frac{ \cos k t}{\sqrt{\pi}} \, dt, \end{equation*} \begin{equation*} \eta_k = \int^{2\pi}_0 x(t) \frac{ \sin k t}{\sqrt{\pi}} \, dt, \ \ k \in \overline{1,2,\ldots,n} \end{equation*} are the Fourier coefficients of $ x $. Furthermore, the following estimate holds: \begin{equation*} \Vert x- P_n x \Vert_{L^2} \leq \frac{1}{ n^r } \Vert x \Vert_{H^r} \text{ for all } x \in \ H_\text{per}^r(0,2\pi), \end{equation*} where $ r \geq 0 $.

Now I was wondering if there exist a similar estimate for a non-periodic Sobolev function $ f \in H^r(0,2\pi) $?

Here notice that the precipitation of a term $ \frac{1}{n^r} $ is the key point I mostly need.

Thanks in advance!

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  • $\begingroup$ I'd expect the answer is no though I can't give a precise reference for that. Fourier series for a non periodic function should be the same as Fourier series for its periodic extension which will be discontinuous and thus will have no good rate of convergence. $\endgroup$
    – VorKir
    Apr 6, 2019 at 23:43
  • $\begingroup$ @ VorKir I also trend to the negative answer. I tried to examine the concrete case when $ r = 1 $, that is, $ y \in H^1 (0,2\pi) $. In comparsion between $ \Vert (I-P_n) y \Vert_{L^2}$ and $ \Vert y' \Vert_{L^2} $, I found the latter represented in Fourier coefficients would induce a term as $ y(2\pi) - y(0) $ which makes estimate hard to move on. This is the discontinnuity you specify. But i still feel suspicious if there exists some trick to move on or with a new condition, like evaluation at one endpoint can make this estimate move on? $\endgroup$
    – Yidong Luo
    Apr 7, 2019 at 0:53

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