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!

  • $\begingroup$ @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. $\endgroup$
    – user271621
    Jul 3, 2021 at 12:51
  • 1
    $\begingroup$ 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.) $\endgroup$ Jul 3, 2021 at 13:07

1 Answer 1


I found a positive answer to this question in this paper of Frank and Sabin: Restriction theorems for orthonormal functions, Strichartz inequalities, and uniform Sobolev estimates, Theorem 12.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.