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I make the following observation:

Let $\Delta^{(n)}$ be the discrete Laplacian on $\mathbb{C}^n$ (ie the $n\times n $ matrix with diagonal $-2$ and upper/lower diagonal $1$.)

This one has eigenvalues and eigenfunctions

$$ \lambda_j = -4 \sin^2\left(\frac{\pi j}{2(n + 1)}\right)$$ and

$$v_{j,i} = \sqrt{\frac{2}{n+1}} \sin\left(\frac{i j \pi}{n+1}\right).$$

I now made the following observation:

Consider the Dirichlet Laplacian on $[0,1]$ this one has eigenvalues and eigenfunctions

$$\lambda_j = -j^2 \pi^2$$ and $$\phi_j(x) = \sqrt{2} \sin\left(j \pi x\right)$$

We consider the scaling $ v_j^{(n)}(t) = \sqrt{n} v_j(\lfloor tn\rfloor)$ on $L^2([0,1])$. I think we have the following heuristic $$ n^2\Delta^{(n)} \rightarrow \Delta$$ $$ v_j^{(n)} \rightarrow \phi_j$$

Question: How does one show this convergence quantitatively and rigorously and in what sense does it hold?-From the expressions, we already expect convergence in $L^{\infty}$ for fixed $j$ as $n$ to infinity. I am looking somewhat for an abstract argument that implies this that does not rely on the concrete expressions.

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    $\begingroup$ Well, your heuristic, as you call it, only works when $j \ll n$, i.e., for the bottom of the spectrum. The discrete laplacian is an approximation of the Laplacian when the oscillations are bounded. That's because, as you know, the discrete Laplacian corresponds to the Laplacian (in $1d$: a second order derivative really) with Dirichlet b. c. ($u(0)=u(n+1)=0$) up to an error which you can compute by a Taylor expansion at $x= \frac{j}{n}$ (of order $n^{-3} \| u^{(3)} \|_{L^\infty}$ possibly?) . Thus provided you impose to limit yourself to lower eigenvalues, everything falls into place nicely. $\endgroup$
    – username
    Commented Apr 20, 2021 at 20:38
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    $\begingroup$ If one has an operator $T$ on a space $V$, to find a discrete approximation one considers a finite-dimensional subspace $V_h\subset H$ and an operator $T_h$ defined to approximate $T$ on $V_h$. Here $h$ is a "fineness" parameter describing how discrete the discretization is, usually related to mesh size. Then you study how quickly solutions to equations involving $T_h$ tend to solutions to equations involving $T$ as $h\to 0$. I wrote more about this here once upon a time. $\endgroup$
    – Neal
    Commented Apr 20, 2021 at 21:01
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    $\begingroup$ Your approximate operator converges uniformly to the actual operator on the ball where the appropriate derivative is bounded. Then it is not so hard to show that eigenpairs converge to eigenpairs, (it works particularly well when all eigenvalues are distinct). I am saying the same thing as @Neal, he said it better in a more enlightening way. $\endgroup$
    – username
    Commented Apr 20, 2021 at 21:28

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I would consider this a "homogenization problem," or at least it can be thought of as very similar to one. In fact, I think the analysis will be essentially the same as (or maybe easier than) the case of periodic homogenization for uniformly elliptic equations.

In the case of periodic elliptic homogenization, the quantitative convergence of the eigenvalues has been considered in works by Carlos Kenig, Fanghua Lin & Zhongwei Shen (see https://arxiv.org/abs/1209.5458). Note: your parameter $1/n$ is analogous to their parameter $\epsilon$.

What they are interested in is in quantitative estimates on just how small the small scale $\epsilon$ need to be before the eigenvalues high up in the spectrum start to converge. Their main result translated in your setup, if I am not mistaken, would be $$ \left| \lambda_j^{(n)} - \lambda_j \right| \leq \frac{C}{n} \lambda_j^{3/2} . $$ Here $\lambda^{(n)}_j$ is the $j$th eigenvalue of the $\Delta^{(n)}$, the discrete Laplacian with grid size $n^{-1}$. Since $\lambda_j$ is of order $j^2$, it looks like you would need $n\gg j^3$ before you would see the $j$th eigenvalue start to converge.

This estimate is interesting (if I am remembering correctly) because if we instead had $\frac{C}{n} \lambda_j^{2}$ on the right side, then the estimate would following easily from scaling and the convergence of the discrete Laplacian to the Laplacian. (Maybe this fact is what you really wanted to know, but if so, they also explain this in the paper as well.) The point of their paper is that they had to fight to reduce the exponent by $1/2$.

Once you have the eigenvalues converging, you can get the eigenfunctions converging easily by a quantification of the closeness of the discrete to the continuous operator.

You should be able to do at least as well as they did, using similar ideas, in the discrete setup. It is possible you can do even better, because the boundary layers might be more tame, especially in one dimension, and some of the analogues of the "correctors" might vanish. Anyway, from their paper I think you can find some inspiring arguments that can get you going. (Note: you probably don't have to know about homogenization to get most of the main idea out of their paper: see Section 3 in particular.)

(I am sorry if this is not really a complete answer, but it was too long for a comment. I'm also sorry if this is not what you meant by "quantitative," because I wasn't sure--"quantitative" and "abstract argument" are pulling in opposite directions.)

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