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I'm interested in estimates on dimension of spectral projection subspaces of some limit operator. I recently asked a related question in the thread Dimension of spectral projection subspaces under strong convergence of operators. I was shown that the answer is no given just strong convergence, but I was wondering whether additional structure might imply what I want.

Let $H_n$ be a sequence of bounded self-adjoint operators on $\ell^2(\mathbb{Z}^2)$. For each $q\in \mathbb{N}$, I can consider $D_q:=\{ -q,...,q \}^2$ and the projection of $\ell^2(\mathbb{Z}^2)$ to functions supported on $D_q$, which I denote by $P_q$. I assume that there exists a sequence of natural $q_n\to \infty$, and there exist matrices $M_n $ such that $ P_{q_{N}} H_n P_{q_{N}}=M_N$ for $n\geq N$(This condition is a later edit). If I assume that there exists some $m$ such that,

$$ \dim \Big(\text{Im}\big(\chi_{(a-\epsilon,a+\epsilon)}(M_n) \big) \Big) \geq m \quad \text{for all} \quad n, $$

does it follow that $\dim \Big(\text{Im}\big(\chi_{[a-\epsilon,a+\epsilon]}(H_\infty) \big) \Big)\geq m $, where $H_\infty$ is the strong operator limit of $H_n$?

I hope that if this is not the case, someone will once again have a clever simple argument why this is not true. My question is motivated by a sequence of operators $H_n= \Delta +V_n$, for diagonal operators $V_n$ that agree around bigger and bigger boxes around the origin and $\Delta$ being the discrete Laplacian. These $V_n$ converge strongly to a limit operator $V_\infty$, which gives us $H_\infty=\Delta+V_\infty$. These sort of operators do not allow counter examples as in my previous question in the other thread.

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This runs into similar problems as before. The Laplacian has spectrum $\sigma(\Delta)=[-4,4]$ in dimension $2$, and any finitely supported potential $V\ge 0$, $V\not\equiv 0$ will give $\Delta+V$ an eigenvalue $E>4$ that will disappear in the limit if we shift the potentials out to infinity.

In more concrete style, let $H_n=\Delta+V_n$, with $V_n(n,n)=100$, $V_n(x,y)=0$ otherwise. Then $H_n\to\Delta$ strongly, the $V_n$ all agree on any box about $(0,0)$ for large $n$ (they are zero), $\dim \chi_{\{ E\}}(H_n)\ge 1$, but $\dim \chi_{\{ E\} }(\Delta)=0$.

We can also project on $|n|\le q_n=2^n$, say, then still $\dim \chi_{[5,200]}(P_{q_n}H_nP_{q_n})\ge 1$ (the eigenvalue will have moved a bit, but of course will stay in that generously sized interval).

I thought about the edited version some more, and I now think this is also false. At the very least we will need more than a (potentially sparse) sequence $q_N\to\infty$, though even then I'm skeptical now.

Things become especially clear if we modify and simplify your problem as follows: Consider $H=-d^2/dx^2+V(x)$ on $[-L_n,L_n]$ with Dirichlet boundary conditions, with an even periodic potential $V$, and focus on a spectral gap $(a,b)\cap\sigma(H)=\emptyset$ above the first band. Then the solutions of $-y''+Vy=Ey$, with $a<E<b$, will have infinitely many zeros, by oscillation theory, since we have essential spectrum below $E$. Thus $E=(a+b)/2$ will be an eigenvalue on $[-L,L]$ for suitable $L=L_n\to\infty$, and again your statement fails.

One can do things along these lines, but I think the quantity to consider is a (suitable) spectral measure $\rho$. Then $\rho_{H_n}(I)>0$ (even for a single $n$) will also give suitable lower bounds on $\rho_H(I)$. See Theorem 4.1 of my paper [18] here and the references quoted there for such statements.

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  • $\begingroup$ Thanks for your answer. I was asking that $H_n$ projected onto the finite box, what I denoted by $M_n$, have $E$ in its spectrum with multiplicity bounded from below. It seems like in the example you wrote, $\dim \chi_{\{ E\}}(M_n)= 0$ eventually. $\endgroup$ Commented Feb 13, 2024 at 14:58
  • $\begingroup$ @Keen-ameteur: This doesn't change anything essentially. I only need to take $q_n$ sufficiently large and will then still have an eigenvalue $\simeq E$. Admittedly, it will have moved a bit, but we then have $\dim\chi_{(E-\epsilon,E+\epsilon)}(M_n)\ge 1$, $(E-\epsilon,E+\epsilon)\cap\sigma (\Delta)=\emptyset$. $\endgroup$ Commented Feb 13, 2024 at 15:06
  • $\begingroup$ It is very likely that I am confused, but $H_n$ have to stabilize near the origin. Namely, if $H_n\to 0$ then $M_n\equiv 0$. I think that for all $n\geq n_0$, we have $P_{q_{n_0}}H_nP_{q_{n_0}} = M_{n_0}$. So if $H_n\to 0$ that must mean that $M_{n_0}\equiv 0$, doesn't it? $\endgroup$ Commented Feb 13, 2024 at 15:19
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    $\begingroup$ @Keen-ameteur: Sorry, it's not completely clear to me what you are saying in your last comment, but I edited my answer and I hope this will address everything. Also, $PH_nP$ is not the zero operator in my example for a fixed projection, it's the (restricted) Laplacian. $\endgroup$ Commented Feb 13, 2024 at 15:35
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    $\begingroup$ I think I may have misstated the condition I wanted. I've edited the question. Under my original intention and my corrected edit, $q_n\leq n-1$ for your proposed $V_n$. I'm sorry for the confusion. $\endgroup$ Commented Feb 13, 2024 at 16:35

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