For simplicity, let us set $V\_n:=\{-1+1/2^n, \dotsc, 0 ,\dotsc, 1-1/2^n, 1\}$ with the periodic boundary conditions for each $n \in \mathbb{N}$ and think of the following vector space over this lattice:


\begin{equation}
M(V\_n):=\{f:V\_n \to \mathbb{C} \mid f \text{ is the restriction of some Schwartz function on }
\mathbb{R} \text{ to } V\_n \}.\end{equation}


Then, clearly, $M(V\_n)$ is a well-defined finite dimensional vector space over $\mathbb{C}$. Moreover, we can think of the direct limit of the collection $\{ M(V\_n)\}\_n$, which may be identified as the restriction of Schwartz space to the domain $(-1,1]$.


So, here are my questions. They are divided into two parts.


1. $M(V\_n)$ is set-theoretically identical to the "set of all mappings from $V\_n$ to $\mathbb{C}$". However, the "set of all mappings from $V\_n$ to $\mathbb{C}$" cannot form a direct limit of the above kind since they can approximate singular functions such as the Heaviside function. What makes the crucial difference?
2. A finite symmetric difference operator $\partial\_n$ on $M(V\_n)$ defined by
\begin{equation}
\partial\_n f(x):=2^{n+1}[f(x+1/2^n)-f(x-1/2^n)]
\end{equation}
is a linear operator. As $n \to \infty$, in what sense does this $\partial\_n$ converge to the ordinary continuum differential operator?


These two questions seem quite subtle to me…. Could anyone please clarify?



In what sense does the finite difference operator converge to the continuum differential operator in the hydrodynamic limit?
"In what sense does this $\partial\_n$ converge to the ordinary continuum differential operator?"


It converges in the sense that there exist injective linear maps $\iota\_n:M(V\_n)\to C^\infty\_\text{periodic}([-1,1])$ that take the $2^{n+1}$ eigenspaces of $\partial\_n$ to the first $2^{n+1}$ eigenspaces of $d/dx$, such that the $j$-th eigenvector of $\partial\_n$ converges pointwise on $[-1,1]\cap \mathbb Z[\tfrac12]$ to the $j$-th eigenvector of $d/dx$, as $n\to \infty$.  
  
  



Here, the sentence "first $2^{n+1}$ eigenspaces of $d/dx$" requires some clarification:


The eigenspaces of $d/dx$ are ordered by increasing absolute value of the eigenvalue, with the (arbitrary) convention that $+a$ comes before $-a$ whenever $a>0$.


