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!