$L^p$-Boundedness of Wave Operators for 2D Schrödinger Operators with Point Interactions

Abstract

For two dimensional Schrödinger operator H with point interactions, we prove that wave operators of scattering for the pair $(H,H_0)$, $H_0$ being the free Schrödinger operator, are bounded in the Lebesgue space $L^p({{\mathbb {R}}}^2)$ for $1<p<\infty$ if and only if there are no generalized eigenfunctions of $Hu(x)=0$ which satisfy $u(x)= C|x|^{-1}+ o(|x|^{-1})$ as $|x|\rightarrow \infty$, $C\not =0$. Otherwise they are bounded for $1<p\le 2$ and unbounded for $2<p<\infty$.

Appendix

In this appendix we show the following lemma:

Lemma A

For any $\varepsilon >0$ there exits a constant $C_\varepsilon >0$ such that

$$\begin{aligned} \int _{{{\mathbb {R}}}^4}e^{ix\xi -ip y} \frac{\chi _{\le {\varepsilon }}(|\xi |)\chi _{\le \varepsilon }(|p|)}{|\xi |+ |p|} d\xi {dp} {\le _{|\, \cdot \, |}\, }\frac{C_\varepsilon }{({\langle x \rangle }+ {\langle y \rangle })^3}\log \left( \frac{({\langle x \rangle }+ {\langle y \rangle })^2}{{\langle x \rangle }{\langle y \rangle }}\right) .\nonumber \\ \end{aligned}$$

(4.55)

Proof

If we use the identity

$$\begin{aligned} \frac{1}{|\xi |+|p|} = \int _0^\infty e^{-t(|\xi |+ |p|)} \hbox {d}t \end{aligned}$$

and Fubini’s theorem, then the left side of (4.55) becomes

$$\begin{aligned} \int _0^\infty \left( \frac{1}{2\pi } \int _{{{\mathbb {R}}}^2}e^{ix\xi -t|\xi |} \chi _{\le {\varepsilon }}(|\xi |)\hbox {d}\xi \right) \left( \frac{1}{2\pi } \int _{{{\mathbb {R}}}^2}e^{-ip y-t|p|} \chi _{\le \varepsilon }(|p|){\hbox {d}p} \right) \hbox {d}t.\nonumber \\ \end{aligned}$$

(4.56)

The functions inside parentheses are convolutions of the Poisson kernel with bump functions ${{\mathcal {F}}}\chi _{\le {\varepsilon }}(x)$ and ${{\mathcal {F}}}\chi _{\le {\varepsilon }}(y)$, respectively, (see [25], p. 61). They are bounded by $C_1 t({\langle x \rangle }^2+ t^2)^{-3/2}$ and $C_2 t({\langle y \rangle }^2+ t^2)^{-3/2}$, respectively. It follows by changing variable t to ${\langle x \rangle }^{1/2}{\langle y \rangle }^{1/2} t$ that

$$\begin{aligned} (4.56)&{\le _{|\, \cdot \, |}\, }C \int _0^\infty \frac{t^2 \hbox {d}t}{({\langle x \rangle }^2+ t^2)^{3/2}({\langle y \rangle }^2+ t^2)^{3/2}} \nonumber \\&= \frac{C}{{\langle x \rangle }^{3/2}{\langle y \rangle }^{3/2}} \int _0^\infty \frac{t^2\hbox {d}t}{(t^4 + s^2 t^2 +1)^{3/2}}, \quad s^2= {\frac{{\langle x \rangle }^2+{\langle y \rangle }^2}{{\langle x \rangle }{\langle y \rangle }}}. \end{aligned}$$

(4.57)

We estimate the integral in the right hand side of (4.57) by slitting $(0,\infty )$ into intervals into (0, 1/s), (1/s, s) and $(s,\infty )$ where the denominator is bounded from below by 1, $s^3 t^3$ and $t^6$, respectively. Then

$$\begin{aligned} \int _0^\infty \frac{t^2\hbox {d}t}{(t^4 + s^2 t^2 +1)^{3/2} } \le \int _0^{1/s}{t^2\hbox {d}t}+ \int _{1/s}^s \frac{dt}{s^3 t} + \int _{s}^\infty \frac{\hbox {d}t}{t^4} = \frac{2}{3s^3}(1+3 \log {s}) . \end{aligned}$$

Since $s\ge \sqrt{2}$, the right side may be further estimated by $C s^{-3}\log {s^2}$ and $s^2\ge ({\langle x \rangle }+{\langle y \rangle })^2/2{\langle x \rangle }{\langle y \rangle }$. Combining this with (4.57), we obtain the lemma. $\square$

For applications in the text we need only the following weaker version which trivially follows from Lemma A.

Lemma B

For any $\varepsilon >0$ there exits a constant $C_\varepsilon >0$ such that

$$\begin{aligned} \frac{1}{(2\pi )^2} \int _{{{\mathbb {R}}}^4}e^{ix\xi -ip y} \frac{\chi _{\le {2\varepsilon }}(\xi )\chi _{\le 2 \varepsilon }(|p|)}{|\xi |+ |p|} \hbox {d}\xi {\hbox {d}p} {\le _{|\, \cdot \, |}\, }\frac{C_\varepsilon }{{\langle x \rangle }{\langle y \rangle }({\langle x \rangle }+ {\langle y \rangle })}. \end{aligned}$$

(4.58)