I am self-taught numerical method for PDE using MIT's public graduate course(18.336) material. While I was doing the accompanying homework set, there is a problem that doesn't quite make sense to me. I would very much appreciate if I could have some hint here.
The question is about the approximation of properties of finite differences for functions with limited differentiability. Define any piecewise polynomials for some $n \geq 0$ as $$ u(x) = x_+^n = \begin{cases}x^n &, x \geq 0 \\ 0 &, x<0 \end{cases} $$ Two contentious difference are defined as $$ D_cu(-\epsilon) = \frac{u(-\epsilon +h) - u(-\epsilon -h)}{2h},\quad D_cu(-\epsilon+h) = \frac{u(-\epsilon+2h) - u(-\epsilon)}{2h} $$ for some $0 < \epsilon < h$. $u(x)$ is defined on $[-1,1]$. We could also consider $D_cu(-\epsilon+mh)$ for all related values fo the integer $m$.
The question asks
(a)For which $n \geq 0$ is $D_c$ consistent for $u(x)$ in the maximum(uniform, $l_\infty$) norm? what is the order of accuracy of $D_c$ as a function of $n \geq 0$? How about for the $l_1$ norm $\left(||E||_1 = h\sum_{j}|E_j| \right)$?
(b)For $n=3$, is the Taylor expansion argument for the error of $D_c$(in the max norm) for cubic splines too optimistic/too pessimistic/just right?
I plug in the Taylor expansion of $u(-\epsilon+h)$ and $u(-\epsilon-h)$ into $D_cu(-\epsilon)$ and get
$D_cu(-\epsilon) = u'(-\epsilon) + O(h^2)$
which shows it is second-order accuracy, but I am not sure how to justify the consistency of $D_c$.
Following the textbook(Finite Difference Methods for Ordinary and Partial Differential Equations, LeVeque) and the lecture notes, the consistency is discussed under the set up of boundary value problems, in which we set up a system of linear equations and build matrices. But here we don't have matrices but only a scaler $D_cu(-\epsilon) = u'(-\epsilon) + O(h^2) \in \mathbb{R}$ in the analysis. How do we even discuss it under different norms($l_1$ and $l_{infty}$ are the same)?
I am very new to this topic. Please feel free to point out if I made any mistakes. Thanks all in advance.