It is well-known that on $L^2[0,1]$, the differential operator $i\frac{d}{dx}$ is self-adjoint with domain of differentiable, periodic functions on $[0,1]$.
Now, let us discretize $[0,1]$, say $B_N:=\{ 0, \frac{1}{N} , \frac{2}{N} , \cdots , 1\}$ for some very large $N$. Next define \begin{equation} M(B_N):= \{ f: B_N \to \mathbb{C} \mid f(0)=f(1) \} \end{equation} Then, obviously, $M(B_N)$ is a finite dimensional Hilbert space with the inner product given by \begin{equation} \langle f , g \rangle:= \frac{1}{N}\sum_{x \in B_N} f(x) \overline{g(x)} \end{equation}
Using the summation by parts formula and the periodicity conditions, it is not hard to show that the linear operator $i\partial^N : M(B_N) \to M(B_N)$ defined by \begin{equation} (i\partial^N f)(x):=iN[f(x+1/N)-f(x)] \end{equation} has the adjoint $i\partial^{-N}$ defined by \begin{equation} (i\partial^{-N} f)(x):=iN[f(x)-f(x-1/N)] \end{equation} with respect to the inner product above.
Thus, in the finite lattice, the difference operator is not exactly self-adjoint, but I see that it is 'almost' self-adjoint for all large $N$'s.
Does this mean that it is possible to have some "almost" orthonormal basis of $M(B_N)$ that are the eigenvectors of $i\partial^N$?
I wonder if this is up to the level of a research question, but have not found any relevant reference myself. Could anyone please help me?
