Resolvent estimates for the magnetic Schrödinger operator in dimensions $\ge 2$

Abstract

It is well known that the resolvent of the free Schrödinger operator on weighted $L^2$ spaces has norm decaying like $\lambda ^{-\frac{1}{2}}$ at energy $\lambda$. There are several works proving analogous high frequency estimates for magnetic Schrödinger operators, with large long or short range potentials, in dimensions $n \ge 3$. We prove that the same estimates remain valid in all dimensions $n \ge 2$.

A Appendix

We now show that H is self-adjoint with form domain $H^1$. We define the sesquilinear form $q_H(u,v) := (u,Hv)$ for $u \in H^1$ and $v \in C^\infty _c$. Under these assumptions, by integration by parts one can show that

$$\begin{aligned} q_H(u,v) = (Du, D v) + (Du,Wv) + (Wu,Dv) + (u,(V + |W|^2) v). \end{aligned}$$

(A.1)

Then, since $W \in L^\infty ({\mathbb {R}}^n, {\mathbb {R}}^n)$, $q_H(u,v)$ makes sense for all $u,v\in H^1$.

Proposition A.1

Let $W \in L^\infty ({\mathbb {R}}^n, {\mathbb {R}}^n)$ and $V \in L^\infty ({\mathbb {R}}^n, {\mathbb {R}})$. Then there is a unique self-adjoint operator H with form domain $H^1$, such that (A.1) holds for all $u,v \in H^1$.

Proof

The proof follows from [12, Theorem X.17]. Thanks to this theorem, it is enough to show that the form $q_H$ is relatively bounded with respect to the form associated to the negative Laplacian, that is $q_{-\varDelta }(u,v) = (Du, D v)$. This is immediate by Young’s inequality. If $u\in H^1$, for every $\varepsilon >0$ one has

$$\begin{aligned}&|(Du,Wu) + (Wu,Du) + (u,(V + |W|^2) u)| \\&\qquad \le \varepsilon ||\nabla u ||^2 + ((\varepsilon ^{-1}+1)||W ||_{L^\infty }^2 +||V ||_{L^\infty }) ||u ||^2 , \end{aligned}$$

so actually the relative bound is zero. $\square$

We now give the proof of a couple of auxiliary results used in the paper.

Proof of Lemma 2.3

We have that

$$\begin{aligned} {{\tilde{\varphi }}}_0(r)&= \langle r \rangle , \qquad \qquad \qquad \quad \ {{\tilde{\varphi }}}_0'(r) = r \langle r \rangle ^{-1},\\ \psi _0'(r)&= (2\delta -1) r \langle r \rangle ^{-2\delta -1}, \quad \psi _0''(r) = (2\delta -1) \langle r \rangle ^{-2\delta -3}\left( 1-2\delta r^2 \right) . \end{aligned}$$

First, we combine the first and third terms on the right hand side of (2.4) and show that

$$\begin{aligned} \left( e^{\tau \psi } ({{\tilde{\varphi }}}_0 \psi _0''(r) + 2 \tilde{\varphi }_0'(r) \psi _0'(r)) \partial _r u,\partial _r u \right) > \alpha ( e^{\tau \psi } \langle r \rangle ^{-2\delta } \partial _r u, \partial _r u), \end{aligned}$$

(A.2)

for $\alpha = (2-2\delta )(2\delta -1)$. This follows from the fact that

$$\begin{aligned} {{\tilde{\varphi }}}_0(r) \psi _0''(r) + 2{{\tilde{\varphi }}}_0'(r) \psi _0'(r)&= (2\delta -1) \langle r \rangle ^{-2\delta -2}(1+(2-2\delta )r^2) \\&> (2-2\delta )(2\delta -1) \langle r \rangle ^{-2\delta }, \end{aligned}$$

since $0<2-2\delta <1$. Then, using (A.2) in (2.4) we obtain that

$$\begin{aligned} (\varphi ''(x) \nabla u,\nabla u)\ge & {} \alpha \tau ( e^{\tau \psi } \langle r \rangle ^{-2\delta } \partial _r u, \partial _r u) \nonumber \\&+\, \tau ^2 (e^{\tau \psi } {{\tilde{\varphi }}}_0 (\psi _0'(r))^2 \partial _r u, \partial _r u) + \tau ( e^{\tau \psi } {{\tilde{\varphi }}}_0 \frac{\psi _0'}{r} \nabla ^{\perp } u,\nabla ^{\perp } u) .\nonumber \\ \end{aligned}$$

(A.3)

Therefore, using that ${{\tilde{\varphi }}}_0 \frac{\psi _0'}{r} = (2\delta -1) \langle r \rangle ^{-2\delta }$, and that $\alpha < (2\delta -1)$ we get

$$\begin{aligned} (\varphi ''(x) \nabla u,\nabla u)\ge & {} \alpha \tau ( e^{\tau \psi } \langle r \rangle ^{-2\delta } \nabla u, \nabla u) \nonumber \\&+\, \tau ^2 (e^{\tau \psi } {{\tilde{\varphi }}}_0 (\psi _0'(r))^2 \partial _r u, \partial _r u). \end{aligned}$$

(A.4)

This yields (2.5). (2.6) follows by direct computation. $\square$

Proof of Lemma 4.3

The proof is similar to [15, Lemma 3.2]. First we define ${{\widetilde{\psi }}} := \tau ^{-1}k \log w +\frac{1}{2}\psi$, so that we have $w^k e^{\frac{1}{2}\tau \psi }=e^{\tau {{\widetilde{\psi }}}}$. By the conditions assumed on $\varphi$ and since $\tau \ge 1$, we have that $|\nabla {{\widetilde{\psi }}}|, |\varDelta {{\widetilde{\psi }}}| \le C(k)$, where we remark that $C(k) \ge 1$ can be chosen to be independent of h and $\tau$. By direct computation we get

$$\begin{aligned} w^k(G_{h,\tau }-i)w^{-k}v&= (e^{\tau \widetilde{\psi }}h^2\,(P+2W^L\cdot D -i\mathrm {div}( W^L) +V^L)\,e^{-\tau \widetilde{\psi }}- i)\, v \\&= (h^2P-i)v + Q_{h,\tau } v, \end{aligned}$$

where $Q_{h,\tau }$ is a first order operator defined by

$$\begin{aligned} \begin{aligned} Q_{h,\tau } v&= h^2 \left( - \tau ^2 | \nabla {\widetilde{\psi }} |^2 + \tau \,\varDelta {{\widetilde{\psi }}} +2i \tau W^L\cdot \nabla {{\widetilde{\psi }}} - i\mathrm {div}(W^L) + V^L \right) v \\&\quad + 2\, h(\tau \nabla {{\widetilde{\psi }}} -i W^L) \cdot h\nabla v. \end{aligned} \end{aligned}$$

(A.5)

Using the Fourier transform, one can easily check that

$$\begin{aligned} ||(h^2P\, -i)^{-1}v ||_{H^{1+a}_{scl}}\le 2|| v ||_{H^{-1+a}_{scl}}. \end{aligned}$$

We also consider the resolvent identity

$$\begin{aligned} (h^2 P -i + Q_{h,\tau } )^{-1}=(h^2 P -i)^{-1} +(h^2 P -i)^{-1} Q_{h,\tau } (h^2 P -i + Q_{h,\tau } )^{-1}, \end{aligned}$$

which allows us to show that

$$\begin{aligned} \begin{aligned} || (h^2 P -i + Q_{h,\tau } )^{-1}v ||_{H_{scl}^{1+a}}&\le 2|| v ||_{H^{-1+a}_{scl}} + 2|| Q_{h,\tau } ||_{ {\mathcal {L}}(H_{scl}^{1+a},\, H_{scl}^{-1+a})} \\&\qquad \times || (h^2 P -i + Q_{h,\tau } )^{-1}v ||_{H_{scl}^{1+a}}. \end{aligned} \end{aligned}$$

(A.6)

Then, since $a=0,1$, taking the $L^2$ norm of (A.5) we obtain

$$\begin{aligned} || Q_{h,\tau } ||_{ {\mathcal {L}}(H_{scl}^{1+a},\, H_{scl}^{-1+a})} \le || Q_{h,\tau } ||_{ {\mathcal {L}}(H_{scl}^{1+a},\, L^2)}\le \frac{1}{4}, \end{aligned}$$

whenever

$$\begin{aligned} h< \tau ^{-1} 18C(k)^2(1+|| W^L ||_{L^\infty } + || V^L ||_{L^\infty } + || \nabla \cdot W^L ||_{L^\infty })^{-1}. \end{aligned}$$

This implies the desired result by absorbing the second term on the right hand side of (A.6) in the left. $\square$