# Finiteness of Schatten $p$-norm of truncated free resolvent

Consider the resolvent operator $$R(z) := (-\Delta - z)^{-1}$$ of the Laplace operator on $$L^2(\mathbb R^d)$$, where $$z\in \rho(-\Delta) = \mathbb C \setminus \mathopen [0, \infty)$$.

For $$p \geq 1$$, let $$\lVert \cdot \rVert_p$$ denote the Schatten $$p$$-norm on the space of compact operators and let $$1_{\Gamma_n}(x)$$ denote multiplication by the indicator function of some cube $$\Gamma_n := n + [0,1]^d, n \in \mathbb Z^d$$. For a given $$E>0$$, I am interested whether $$\sup_{\substack{z = E+i\epsilon \\ -1 \leq \epsilon \leq 1}}\lVert 1_{\Gamma_n}(x) R(z) 1_{\Gamma_n}(x) \rVert_p < \infty \tag{1}$$ for some suitable $$p=p(d)$$, presumably all $$p>d/2$$. So far I could not find any reference proving this, the problem being the supremum in front: In the book Trace Ideals and their Applications by Barry Simon, one can find the bound $$\lVert f(x) g(-i\nabla) \rVert_p \leq C \lVert f \rVert_p \lVert g \rVert_p.$$ Noting that $$\lVert 1_{\Gamma_n}(x) R(z) 1_{\Gamma_n}(x) \rVert_p \leq \lVert 1_{\Gamma_n}(x) R(z) \rVert_p$$ and applying the above inequality to $$f(x) := 1_{\Gamma_n}(x)$$ and $$g_z(x):= \frac{1}{\lvert x \rvert^2 - z}$$ yields $$\lVert 1_{\Gamma_n}(x) R(z) 1_{\Gamma_n}(x) \rVert_p \leq C \lVert g_z \rVert_p,$$ which is finite if $$p>d/2$$. Unfortunately, this is not enough for $$(1)$$ since the expression blows up as $$\operatorname{Im }z = \epsilon \to 0$$.

Still, I think $$(1)$$ should be true. For example, if $$d=3$$, we know that the kernel of $$R(z)$$ is given by $$R(x,y;z) = \frac{1}{4\pi \lvert x - y \rvert} e^{-\sqrt{-z} \lvert x -y \rvert}$$ so that we can explicitly compute the Hilbert-Schmidt norm $$\lVert 1_{\Gamma_n}(x) R(z) 1_{\Gamma_n}(x) \rVert_2^2 = \int_{\Gamma_n} \int_{\Gamma_n} \lvert R(x,y;z) \rvert^2 \, dx dy \leq C \int_{\Gamma_n} \int_{\Gamma_n - y} \frac{1}{\lvert x \rvert^2} \, dx dy < \infty$$ uniformly in $$z$$. However, this does not work in $$d>3$$, where a higher $$p$$-norm would be needed (I assume $$p>d/2$$ as suggested by the above). Any help is appreciated!

• @ChristianRemling exactly, I have worked quite a bit with limiting absorption principles valid for a class of Schrödinger operators. The problem however is that these are always concerned with the operator norm of the corresponding operators, not any Schatten norms. I could not work out any connection so far. In addition, in the above question we have the very special operator $-\Delta$ and I am not sure whether the inequality in question is true for general Schrödinger operators. Jul 3, 2021 at 12:51
• Yes, it's certainly not the same question. (I naively thought it might be a useful hint, if you hadn't been familiar with LAP already.) Jul 3, 2021 at 13:07