Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuously differentiable with $h(0) = 0$ and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ Here $1_\Omega$ means the corresponding multiplication operator. Are there any known results?
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$\begingroup$ Is it even clear that the operator on the RHS (for $h(x)=x$) is compact? $\endgroup$– Christian RemlingCommented Nov 7 at 22:41
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1$\begingroup$ @ChristianRemling If $h(x) = x$, then the RHS operator is trace class, see e.g. Barry Simon: Trace ideals, Theorem 4.5. For general $h$, I think we might need continuously differentiable (and obviously $h(0) = 0$); I updated the question. $\endgroup$– lasik43Commented Nov 8 at 10:20
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