I am considering two Schrodinger operators on $\mathbb{Z}^2$ and compare their essential spectrum. The operators are both of the form $H=A+V$ where $A$ is the adjacency operator on the $\mathbb{Z}^2$-lattice, and $V$ is a diagonal operator of the form $[ V\psi ](n,m)=u(n,m)\cdot \psi(n,m)$ for any function $\psi\in \ell^2(\mathbb{Z}^2)$ and $(n,m)\in \mathbb{Z}^2$, where $u:\mathbb{Z}^2\to \mathbb{R}$ is some function.
Concretely, I am trying to compare $\sigma_{ess}(H_1)$ and $\sigma_{ess}(H_2)$, where $u_1\equiv 0$ and
$$ u_2(m_1,m_2)= \begin{cases} 16 &;m_1=0, m_2\geq 0 \\ 0 &;\text{else} \end{cases} \; . $$
It is easy to show that $\Vert H_1 \Vert= 4$ and $ \sqrt{20} \leq \Vert H_2\Vert\leq 20$. Since the operators are self-adjoint, we know that $\lambda'\in \sigma(H_2)\setminus \sigma(H_1)$ for $\lambda':= \Vert H_2\Vert $. I am now trying to show that there exists $\lambda \in \sigma_{ess}(H_2)\setminus \sigma_{ess}(H_1)$ with $\lambda\geq 12$, hopefully $\lambda'=\lambda$ itself works.
I want to find a sequence of normalized functions $\psi_n\in \ell^2(\mathbb{Z}^2)$ such that $\Vert (H_2-\lambda I)\psi_n\Vert \to 0$ and $\psi_n\to 0$ weakly. It seems like the spectral projection onto $\chi_{[12,20]}(H_2)$ should be infinite dimensional, but I can't manage to show this directly. I thought to look at increasing finite sections around the origin which would each have a strictly growing dimension of projection onto $[12,20]$ using a Gershgorin circle argument, but this hasn't worked so far.
It feels like there should be a simple argument I am missing that someone well versed in spectral theory would notice. I would appreciate any input on this question.
Best Answer
What is clear here is that $H_2$ has spectrum above $4$, and that $\lambda\in\sigma_{\textrm{ess}}(H_2)$ also, with $\lambda:=\max\sigma(H_2)=\|H_2\|>4$. What $\lambda$ is actually equal to seems a rather delicate question. Certainly $\lambda<20$, and I'm also not at all sure that $\lambda\ge 12$; that may already be too ambitious.
To discuss these claims, let me switch to the continuous counterpart of your problem, the operator $H=-\Delta-V(x,y)$ on $L^2(\mathbb R^2)$. Everything I'm going to say has a precise analog in your (discrete) setting, but the details become considerably less tedious in the continuous case. Even so, I will only sketch most of them.
We now assume that $V(x,y)=16$ on $|x|\le 1$, $y\ge 0$, and $V=0$ otherwise. We are interested in the part of the spectrum below zero. (We clearly have $[0,\infty)\subseteq\sigma (H)$.)
First of all, there actually is negative spectrum. This would not be clear in dimension $3$ or higher, but here it still works. Compare my answer here. We can find $\lambda=\min\sigma (H)$ as the infimum of the quadratic form $\lambda=\inf Q(f)$, $$ Q(f)=\int_{\mathbb R^2} \left(|\nabla f|^2-V|f|^2\right) , \tag{1} $$ taken over $f\in C_0^{\infty}$ (say) with $\|f\|_{L^2}=1$. Clearly, we can only hope to make $Q(f)$ small if we concentrate $f$ where $V=16$, but then $Q(f)>-16$ because we can not avoid paying a (potentially steep) price in the first term. A more chunky potential that is equal to $16$ on a large disk rather than a thin strip would be more effective at pulling down the bottom of the spectrum. In any event, we can try concrete test functions in (1) and get upper bounds on $\lambda$, but finding $\lambda$ exactly seems a formidable task.
Finally, there are $f\in L^2$ that make $\|(H-\lambda)f\|$ arbitrarily small. A carefully cut off version (compare the discussion in the comments to the linked answer) of $f$ will still keep $(H-\lambda)f$ small, and then we obtain a Weyl sequence by also shifting and considering $g(x,y)=f(x,y-L)$, $L\gg 1$.