In what sense does the Laplacian on compact intervals converge to one on all of $\mathbb{R}$?

Question

I guess this topic may have been addressed somewhere but I cannot really find a reference myself, so I ask here.

For each $N \in \mathbb{N}$, consider the Laplacian $\Delta$ on the interval $[-N,N]$ subject to the periodic boundary conditions. Then, it is well-known that it is a self-adjoint operator on $L^2[-N,N]$ with discrete spectrum. Yet, the "gaps" between neighboring eigenvalues become "smaller" with larger $N$.

Now, the Laplacian $\Delta$ on whole $\mathbb{R}$ has the continuous spectrum $(-\infty,0]$.

My question is that:

In what sense does the Laplacian $\Delta$ on $[-N,N]$ converge to one on the whole real line as $N \to \infty$?

Moreover, does the boundary condition on $[-N,N]$ matter as long as $\Delta$ remains self-adjoint?

Could anyone please clarify for me?

Answer 1

A general concept that fits this well is strong resolvent convergence. As the name suggests, $T_n\to T$ in this sense means that $(T_n-i)^{-1}\to (T-i)^{-1}$ strongly.

In your case, we first of all have to make sure that all operators act on the same Hilbert space, and the natural way of doing this is to take $H=L^2(\mathbb R)$, $T=\Delta$, $T_n=\Delta\oplus 0$, where this last sum refers to the decomposition $H=L^2(-n,n)\oplus L^2((-n,n)^c)$. Then $T_n\to T$ in strong resolvent sense, as can be verified by working out the resolvents (by variation of constants, they are integral operators with kernels built from the solutions of $y''=iy$), or, more conveniently, by observing that for any $u\in C_0^{\infty}(\mathbb R)$, we also have $u\in D(T_n)$ eventually and (trivially) $T_nu\to Tu$. This condition implies strong resolvent convergence; see for example Weidmann, Linear operators in Hilbert spaces, Theorem 9.16.

This then also shows that the boundary conditions don't matter. Finally, strong resolvent convergence implies strong convergence of the spectral projections.