4
$\begingroup$

Let $T : L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ be a possibly unbounded self adjoint operator. Let $R_\varepsilon$ denote the resolvent $(T - i \varepsilon)^{-1}$, $\varepsilon > 0$. Suppose that, for all $\varepsilon \in (0, 1]$, there exists some $M > 0$ so that the weighted operator norm

$$\| (1 + |x|)^{-1} R_{\varepsilon}(1 + |x|)^{-1} \|_{L^2 \to L^2} \le M .$$

Then, is it true that $(1 + |x|)^{-1} R_{\varepsilon}(1 + |x|)^{-1}$ has a limit in the uniform topology as $ \varepsilon \to 0?$ More generally, if we replace the weight $(1 + |x|)^{-1}$ by an arbitrary bounded operator $B$, do we have that $\lim_{\varepsilon \to 0} BR_\varepsilon B $ exists in the uniform topology?

This statement is some type of "weighted" limiting absorption principle. I have been trying to prove this using the resolvent formula:

$$R_{\varepsilon_2} - R_{\varepsilon_1} = R_{\varepsilon_1}(i\varepsilon_1 - i\varepsilon_2)R_{\varepsilon_2},$$

To show that sequences $(1 + |x|)^{-1} R_{\varepsilon_k}(1 + |x|)^{-1}$ are Cauchy when $\varepsilon_k \to 0$. The idea is to use the resolvent formula to get to a place where we can take advantage of the bound $M$:

$$\| (1 + |x|)^{-1} R_{\varepsilon_2}(1 + |x|)^{-1} - (1 + |x|)^{-1} R_{\varepsilon_1}(1 + |x|)^{-1} \| = \| (1 + |x|)^{-1} R_{\varepsilon_1}(i\varepsilon_1 - i\varepsilon_2)R_{\varepsilon_2}(1 + |x|)^{-1}\| = \\ \| (1 + |x|)^{-1} R_{\varepsilon_1}(1 + |x|)^{-1}(1 + |x|)^{1}(i\varepsilon_1 - i\varepsilon_2)(1 + |x|)^{1}(1 + |x|)^{-1}R_{\varepsilon_2}(1 + |x|)^{-1}\|. $$

But I am stuck at this point because I have had to introduce the unbounded operator $(1 + |x|)^{1}$, and even though $(i\varepsilon_1 - i\varepsilon_2)$ can be small, I have little control over what $(1 + |x|)^{1}(i\varepsilon_1 - i\varepsilon_2)(1 + |x|)^{1}$ does to $L^2$ functions.

Solutions are greatly appreciated!

$\endgroup$

1 Answer 1

1
$\begingroup$

Take a look at Lemma 3.3 in this paper, which we wrote exactly for the purpose of proving a limiting absorption principle in the operator norm between weighted spaces; it is sufficient to have some weak convergence to get the stronger result. Maybe this can be applied also in your concrete situation.

$\endgroup$

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.