Suppose that $f : [0, 1] \to \mathbb{R}$ is an element of the order-$k$ Sobolev space,
\begin{multline} f^{(k - 1)}~\text{is absolutely continuous},\quad \|f\|_{W^k}^2 := \int_0^1 f^{(k)}(x)^2 dx < \infty, \\\ \quad \mbox{and} \quad f^{(j)}(0) = f^{(j)}(1) = 0,~\mbox{for}~j = 0,\dots k -1. \end{multline}

I am wondering what can be said about the approximation of the norm $\|f\|_{L^2}^2 = \int_0^1 f^2(x) dx$ on a finite grid of points. Namely fix points, $x_1, \dots, x_n \in [0, 1]$. I am wondering what can be said about the unit Sobolev ball on this grid, $$ \Psi_k(x_1, \dots, x_n) := \sup_{ \|f\|_{W^k}^2 \leq 1} \big\{ \|f\|_{L^2}^2 - \frac{1}{n} \sum_{i=1}^n f^2(x_i)\big\}. $$ I know that there is an interpolant $\tilde f$ such that $\|f - \tilde f\|_{L^2}^2 \leq C h^{-2k}$ where $h$ is the largest spacing between nearest points in the sequence $\{x_i\}_{i=1}^n$, and $C$ is a constant.

Can anything concrete be said of $\Psi_k(x_1, \dots, x_n)$?

Suppose that $f : [0, 1] \to \mathbb{R}$ is an element of the order-$k$ Sobolev space,
\begin{multline} f^{(k - 1)}~\text{is absolutely continuous},\quad \|f\|_{W^k}^2 := \int_0^1 f^{(k)}(x)^2 \, dx < \infty, \\\ \quad \text{and} \quad f^{(j)}(0) = f^{(j)}(1) = 0, \text{ for } j = 0,\dots, k -1. \end{multline}

I am wondering what can be said about the approximation of the norm $\|f\|_{L^2}^2 = \int_0^1 f^2(x) \, dx$ on a finite grid of points. Namely fix points, $x_1, \dots, x_n \in [0, 1]$. I am wondering what can be said about the unit Sobolev ball on this grid, $$ \Psi_k(x_1, \dots, x_n) := \sup_{ \|f\|_{W^k}^2 \leq 1} \left\{ \|f\|_{L^2}^2 - \frac{1}{n} \sum_{i=1}^n f^2(x_i) \right\}. $$ I know that there is an interpolant $\tilde f$ such that $\|f - \tilde f\|_{L^2}^2 \leq C h^{-2k}$ where $h$ is the largest spacing between nearest points in the sequence $\{x_i\}_{i=1}^n$, and $C$ is a constant.

Can anything concrete be said of $\Psi_k(x_1, \dots, x_n)$?

Suppose that $f : [0, 1] \to \mathbb{R}$ is an element of the order-$k$ Sobolev space,
\begin{multline} f^{(k - 1)}~\text{is absolutely continuous},\quad \|f\|_{W^k}^2 := \int_0^1 f^{(k)}(x)^2 dx < \infty, \\\ \quad \mbox{and} \quad f^{(j)}(0) = f^{(j)}(1) = 0,~\mbox{for}~j = 0,\dots k -1. \end{multline}

I am wondering what can be said about the approximation of the norm $\|f\|_{L^2}^2 = \int_0^1 f^2(x) dx$ on a finite grid of points. Namely fix points, $x_1, \dots, x_n \in [0, 1]$. I am wondering what can be said about the unit Sobolev ball on this grid, $$ \Psi_k(x_1, \dots, x_n) := \sup_{ \|f\|_{W^k}^2 \leq 1} \big\{ \|f\|_{L^2}^2 - \frac{1}{n} \sum_{i=1}^n f^2(x_i)\big\}. $$ I know that there is an interpolant $\tilde f$ such that $\|f - \tilde f\|_{L^2}^2 \leq h^{-2k}$ where $h$ is the largest spacing between nearest points in the sequence $\{x_i\}_{i=1}^n$.

Can anything concrete be said of $\Psi_k(x_1, \dots, x_n)$?

Suppose that $f : [0, 1] \to \mathbb{R}$ is an element of the order-$k$ Sobolev space,
\begin{multline} f^{(k - 1)}~\text{is absolutely continuous},\quad \|f\|_{W^k}^2 := \int_0^1 f^{(k)}(x)^2 dx < \infty, \\\ \quad \mbox{and} \quad f^{(j)}(0) = f^{(j)}(1) = 0,~\mbox{for}~j = 0,\dots k -1. \end{multline}

I am wondering what can be said about the approximation of the norm $\|f\|_{L^2}^2 = \int_0^1 f^2(x) dx$ on a finite grid of points. Namely fix points, $x_1, \dots, x_n \in [0, 1]$. I am wondering what can be said about the unit Sobolev ball on this grid, $$ \Psi_k(x_1, \dots, x_n) := \sup_{ \|f\|_{W^k}^2 \leq 1} \big\{ \|f\|_{L^2}^2 - \frac{1}{n} \sum_{i=1}^n f^2(x_i)\big\}. $$ I know that there is an interpolant $\tilde f$ such that $\|f - \tilde f\|_{L^2}^2 \leq C h^{-2k}$ where $h$ is the largest spacing between nearest points in the sequence $\{x_i\}_{i=1}^n$, and $C$ is a constant.

Can anything concrete be said of $\Psi_k(x_1, \dots, x_n)$?