Resolvent bounds for repulsive potentials


Abstract

We prove limiting absorption resolvent bounds for the semiclassical Schrödinger operator with a repulsive potential in dimension \(n\ge 3\), which may have a singularity at the origin. As an application, we obtain time decay for the weighted energy of the solution to the associated wave equation with a short range repulsive potential and compactly supported initial data.

1 Introduction and statement of results↩︎

The goal of this paper is to establish limiting absportion resolvent bounds for the semiclassical Schrödinger operator with a repulsive potential in dimensions three and higher. The dimension one case was studied in [1]. As an application, we obtain time decay of a weighted energy for the solution to the associated wave equation with a short range repulsive potential and compactly supported initial data.

To fix some notation, let \(\Delta \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\sum_{j=1}^n \partial^2_j \le 0\) be the Laplacian on \(\mathbb{R}^n\), \(n \ge 3\). We use \((r, \theta) = (|x|, x/|x|) \in (0, \infty) \times \mathbb{S}^{n-1}\) for polar coordinates on \(\mathbb{R}^n \setminus \{0\}\). Put \(B(0,r_0) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\{x \in \mathbb{R}^n : |x| < r_0\}\). For a function \(u\) defined on a subset of \(\mathbb{R}^n\), we write \(u(r, \theta) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =u(r \theta)\), and denote the radial derivative by \(u' \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\partial_ru\). If \(E \subseteq \mathbb{R}^n\) is a Borel set, \(\mathbf{1}_E\) stands for its indicator function.

Our Schrödinger operator takes the form \[\label{P} P(h) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =-h^2 \Delta + V(x) : L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n),\qquad x \in \mathbb{R}^n,\tag{1}\] where \(h > 0\) is the semiclassical parameter. The conditions we place on the potential \(V : \mathbb{R}^n \to \mathbb{R}\) are as follows. We suppose \[\begin{gather} V \ge 0, \tag{2} \\ rV \mathbf{1}_{B(0,1)} \in L^\infty(\mathbb{R}^n), \tag{3} \\ V\mathbf{1}_{\mathbb{R}^n \setminus B(0,1)} \in L^\infty(\mathbb{R}^n) \tag{4}, \\ \text{for each \theta \in \mathbb{S}^{n-1}, (0, \infty) \ni r \mapsto V(r, \theta) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =V(r\theta) has bounded variation.} \tag{5} \end{gather}\]

Recall that function \(f\) of locally bounded variation on an interval \(I \subseteq \mathbb{R}\) has distributional derivative equal to a locally finite Borel measure, which we denote by \(df\); it satisfies \[\label{ftc} \int_{(a,b]} df = f^R(b) - f^R(a),\tag{6}\] for any interval \((a,b]\) contained in the interior of \(I\), where \(f^R(x) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\lim_{\delta \to 0^+} f(x + \delta)\).

The last condition we impose on \(V\) is that there exists \(C_V > 0\) so that for all \(\theta \in \mathbb{S}^{n-1}\), \[\label{V32prime32cond} dV( \cdot ,\theta) \le -C_V(r +1 )^{-1} V( \cdot ,\theta) dr.\tag{7}\] in the sense of Borel measures on \((0, \infty)\).

A prototype potential satisfying the above conditions is \[V(r, \theta) =g(\theta) \big( \mathbf{1}_{B(0,1)} r^{-1} + 2^{-1} \mathbf{1}_{\mathbb{R}^n \setminus B(0,1)} r^{-\delta} \big)\] for some \(\delta > 0\) and \(g\) a bounded function on \(\mathbb{S}^{n-1}\). Note that if \(V \in C^1(\mathbb{R}^n; [0, \infty))\), 7 implies each \(V(\cdot, \theta)\) is repulsive in sense of classical mechanics, i.e., that \(V(r, \theta) > 0\) implies \(V'(r, \theta) < 0\).

The condition 3 allows for potentials with an \(r^{-1}\)-singularity at the origin, most notably the repulsive Coulomb (\(V(x) = r^{-1}\)) and Yukawa (\(V(x) = e^{-r} r^{-1}\)) potentials [2]. Moreover, 3 and 4 imply 1 is self-adjoint with respect to the domain \(H^2(\mathbb{R}^n)\) [3]. We utilize \(H^2(\mathbb{R}^n)\) in a density argument in Appendix 6.

Our main results are the following weighted limiting absorption resolvent bounds for \(P\).

Theorem 1. Suppose \(n \ge 3\) and \(V\), satisfies 2 through 7 . Define \(P(h)\) by 1 , and equip it with the domain \(H^2(\mathbb{R}^n)\). For all \(s, \, s_1, \, s_2 > 1/2\) with \(s_1 + s_2 > 2\), there is \(C > 0\) such that for all \(z \in \mathbb{C}\setminus [0, \infty)\) and \(h > 0\), \[\begin{gather} \| (r + 1)^{-s} (P(h) - z)^{-1} (r + 1)^{-s} \|_{L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)} \le \frac{C}{h |z|^{1/2}}, \tag{8} \\ \| (r + 1)^{-s_1} (P(h) - z)^{-1} (r + 1)^{-s_2} \|_{L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)} \le \frac{C}{h^2}. \tag{9} \end{gather}\]

Christiansen and Datchev [1] obtained 8 and 9 for bounded, repulsive potentials on the half-line. Thus the novelty of our work is that it extends these bounds to higher dimensions for repulsive potentials which may have a singularity at \(r = 0\).

Remark 2. In Appendix 7, we recall how for the case \(V = 0\) and \(n = 3\), the conditions on \(s, \, s_1\) and \(s_2\) in Theorem 1, as well as the \(h\)- and \(z\)-dependencies of the right sides of 8 and 9 , are nearly optimal in a suitable sense.

We prove Theorem 1 by means of the so-called spherical energy method, which is a popular strategy for obtaining resolvent estimates (see, e.g., [4][6]). It relies on separation of variables and the well known identity \[\label{conjugation321} r^{\frac{n-1}{2}}(- \Delta) r^{-\frac{n-1}{2}} = -\partial^2_r + r^{-2} \Lambda,\tag{10}\] where \[\label{Lambda} \Lambda \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =-\Delta_{\mathbb{S}^{n-1}} + \frac{(n-1)(n-3)}{4},\tag{11}\] and \(\Delta_{\mathbb{S}^{n-1}}\) denotes the negative Laplace-Beltrami operator on \(\mathbb{S}^{n-1}\). We use in a crucial way that \(\Lambda \ge 0\) on \(L^2(\mathbb{S}^{n-1})\), see 26 . This is why our approach does not cover the case \(n = 2\) where the effective potential \(-1/(4r^2)\) has a strong negative singularity as \(r \to 0\). We expect that repulsive potentials in dimension two can be treated by adapting the Mellin transform methods used in [7], [8].

As an application of 9 , we prove weighted energy decay for the solution to the wave equation \[\label{wave32eqn32srp} \begin{cases} (\partial^2_t - \Delta + V(x))u(t,x) = 0 & (t, x) \in \mathbb{R}\times \mathbb{R}^n, \, n \ge 3, \\ u(0,x) = u_0(x), \, \partial_tu(0,x) = u_1(x) & x \in \mathbb{R}^n, \end{cases}\tag{12}\] where \(u_0 \in H^1(\mathbb{R}^n)\) and \(u_1 \in L^2(\mathbb{R}^n)\) have compact support. The potential \(V\) again obeys 2 through 7 , but also the extra short range condition \[\label{stronger32bd32V} |\mathbf{1}_{\mathbb{R}^n \setminus B(0,1)} V | \le C(r + 1)^{-\delta},\tag{13}\] for some \(C > 0\) and \(\delta > 0\) such that \[\label{delta32restrictions} \delta > \begin{cases} \frac{1}{2} + \frac{n + 3}{4} & n \neq 8, \\ \frac{1}{2} + 3 & n = 8. \end{cases}\tag{14}\]

Since \(P \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =P(1) = -\Delta + V\) is self-adjoint (and nonnegative) on \(L^2(\mathbb{R}^n)\), we may use the spectral theorem for self-adjoint operators to represent the solution to 12 by \[\label{soln32spect32thm} u(t, \cdot) = \cos(t \sqrt{P}) u_0 + \frac{\sin(t \sqrt{P})}{\sqrt{P}}u_1.\tag{15}\] For \(s > 0\) fixed, define the weighted energy of the solution \(u\) to 12 to be \[E_s[u](t) = E_s(t) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\int_{\mathbb{R}^n} \langle x \rangle^{-2s}( |\partial_t u(t,x)|^2 + | \nabla u(t,x)|^2 + |u(t,x)|^2 )dx.\] Set also \[E(0) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\|\nabla u_0 \|^2_{L^2} + \| u_1 \|^2_{L^2}.\]

Theorem 3. Suppose \(V\) satisfies 2 through 7 as well as 13 and 14 . Let \(u_0 \in H^1(\mathbb{R}^n)\) and \(u_1 \in L^2(\mathbb{R}^n)\) have compact support. For each \(s\) such that \[\label{s32restrictions} s > \begin{cases} \frac{n + 3}{4} & n \neq 8, \\ 3 & n = 8, \end{cases}\tag{16}\] there exists \(C_s > 0\) depending on \(s\) but independent of \(t\), \(u_0\), and \(u_1\) so that \[\label{LED} E_s(t) \le C_s\langle t \rangle^{-2} E(0),\tag{17}\] where \(\langle t \rangle \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =(1 + t^2)^{1/2}\).

Remark 4. Since \(u(-t, \cdot) = \cos(t \sqrt{P}) u_0 + (\sin(t \sqrt{P})/\sqrt{P})(-u_1)\), to prove 17 it suffices to suppose \(t \ge 0\).

For smooth, nonnegative potentials of compact support, the local energy \[E_{r_0}(t) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\int_{B(0,r_0)} |\partial_t u(t,x)|^2 + | \nabla u(t,x)|^2 + |u(t,x)|^2 dx, \qquad r_0 > 0,\] obeys \[\label{compact32support32case} E_{r_0}(t) = \begin{cases} O(e^{-ct}) \text{ for some } c> 0 & n \ge 3 \text{ odd}, \\ O(t^{-2n}) & n \ge 4 \text{ even}. \end{cases}\tag{18}\] Indeed, Vainberg showed 18 for compactly supported perturbations of the Laplacian satisfying the so called Generalized Huygens Principle (defined in [9]). From a propagation of singularities result of Melrose and Sjöstrand [10], [11], the Generalized Huygens Principle holds for a large class of the so called nontrapping pertubations of Laplacian, which includes smooth, nonnegative potentials compact support. The study of energy decay for nontrapping perturbations has a long history, going back to the works of Lax, Morawetz, and Phillips [12][14].

On the other hand, bounds similar to 17 were obtained in previous works for various classes of short range potentials. In [15], Zappacosta considered potentials \(V \in C^1(\mathbb{R}^3; (0,\infty))\) with \(\partial^\alpha_xV = O(\langle x \rangle^{- \delta - |\alpha|})\) for all \(0 \le |\alpha| \le 1\) and some \(\delta > 2\). For the each \(\chi \in C^\infty_0(\mathbb{R}^3)\), the bound
\(\| \chi \sqrt{V}(\sin(t \sqrt{P})/\sqrt{P}) \sqrt{V} \chi \|^2_{L^2 \to L^2} = O(t^{-2})\) was obtained. Vodev extended [15] in [16] by obtaining \(E_{r_0}(t) = O(t^{-2})\) in dimension \(n \ge 3\), where \(V \in C^1(\mathbb{R}^n ; [0,\infty))\) obeys \[\begin{gather} \partial^\alpha_x V = O(\langle x \rangle^{-\delta_0}) \qquad \text{for all 0 \le |\alpha| \le 1 and some \delta_0 > (n +1)/2, and} \tag{19} \\ 2V + r \partial_rV \le C\langle x \rangle^{-\delta} \qquad \text{for some C > 0 and some \delta > 1.} \tag{20} \end{gather}\] We note that if \(V \in C^1(\mathbb{R}^n ; [0,\infty))\) satisfies 7 with \(C_V \ge 2\), then 20 holds. Vodev also obtained weighted energy decay for a class of long range, nontrapping perturbations of the Laplacian that includes perturbation by a nonnnegative long range potential, provided the initial conditions are spectrally localized away from \([0, a]\) for \(a > 0\) sufficiently large [17].

The proof of Theorem 3 proceeds as in [16], with some modifications to address the possible singular behavior of \(V\) at the origin. It suffices to establish \[\label{wave32decay32outline} t^2 E_s(t) \le C t^2 \int_t^\infty E_s(\tau) d\tau \le CE(0), \qquad t \ge 1.\tag{21}\] For this, we use Duhamel’s formula and the Fourier transform, with \(t\) dual to \(\lambda\), to represent the Fourier transform of \(u\) in terms of \((P - \lambda^2)^{-1}\), see 58 . Since the initial data are compactly supported, finite speed of propagation holds, and we may freely introduce a cutoff function \(\eta\) into this expression. So Plancherel’s theorem gives a route to control \(E_s(t)\), thanks to the bound on \(\langle x \rangle^{-s} (P - \lambda^2)^{-1}\eta\) that comes from 9 , see also Lemma 1. However, due the factor of \(t^2\) in 21 , and since multiplication by \(t\) is dual to differentiation with respect to \(\lambda\), it is necessary to control \(\langle x \rangle^{-s} \tfrac{d}{d\lambda}(P - \lambda^2)^{-1}\eta\) too. We obtain bounds on this derivative under the additional conditions 13 , 14 , and 16 , see Lemma 3.

Another energy studied is the quantity \[E^{(1)}_K[u](t) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\int_{K} |\partial_t u(t,x)|^2 + | \nabla u(t,x)|^2 + V(x)|u(t,x)|^2 dx,\] where \(K\) is suitable subset of \(\mathbb{R}^n\). In [16], Vodev considered \(K = B(0, \gamma_0 t) \subseteq \mathbb{R}^n\) for \(n \ge 3\) and suitable \(0 < \gamma_0 < 1\). He showed \(E^{(1)}_K[u](t) = O(t^{-1})\) provided 19 and 20 hold. In [18], Ikehata considered exterior subdomains \(\Omega\) of \(\mathbb{R}^n\), \(n \ge 2\), that exclude the origin. For \(K\) a compact subset of \(\Omega\), he showed \(E^{(1)}_K[u](t) = O(t^{-1})\) provided \(V\) is nonnegative, \(C^1\), and obeys \(x\cdot \nabla V + 2V \le 0\). It seems \(O(t^{-1})\)-decay was first obtained first by Morawetz [19] for \(V = 0\) in the exterior of a three-dimensional star-shaped obstacle (and later improved to exponential decay in [12]).

As alluded to above, there is a large body literature on wave decay for higher order perturbations. See [20][26] for more historical background.

The rest of the paper is organized as follows. In Section 2, we prove Theorem 1. In Section 3, we apply Theorem 1 to prove norm bounds for \(\langle x \rangle^{-s}(P - \lambda^2)^{-1}\langle x \rangle^{-s}\) and its \(\lambda\)-derivative. In Section 4 we prove Theorem 3. Finally, we include several appendices of technical results that assist with the proofs of earlier sections.

We thank Kiril Datchev and Georgi Vodev for helpful discussions. J. S. and A. L-H. gratefully acknowledge support from NSF DMS-2204322. J. S. was also supported by a University of Dayton Research Council Seed Grant.

2 Proof of Theorem 1↩︎

In this section, we prove Theorem 1. Throughout this section, we take \(P(h)\) as in 1 , and assume the potential \(V\) satisfies 2 through 7 .

By 10 , \[\label{conjugation322} \begin{align} P^{\pm}(h) &\mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =r^{\frac{n-1}{2}}\left( P(h) - E \pm i\varepsilon \right) r^{-\frac{n-1}{2}}\\ &= -h^2\partial^2_r + h^2r^{-2} \Lambda + V - E \pm i\varepsilon. \end{align}\tag{22}\]

Let \(u \in r^{(n-1)/2} C^\infty_0(\mathbb{R}^n)\). Define a spherical energy functional \(F[u](r)\), \[\label{F} F(r) = F[u](r) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\|hu'(r, \cdot)\|^2 - \langle (h^2 r^{-2} \Lambda + V(r, \cdot) - E)u(r, \cdot), u(r, \cdot) \rangle,\tag{23}\] where \(\| \cdot \|\) and \(\langle \cdot, \cdot \rangle\) denote the norm and inner product on \(L^2(\mathbb{S}_\theta^{n-1})\). For a weight \(w(r)\) which is absolutely continuous, nonnegative, and increasing, one computes \((wF)'\) in the sense of distributions on \((0, \infty)\): \[\label{deriv32wF} \begin{align} (wF)' &= wF' + w'F\\ &= w(-2\mathop{\rm Re}\langle (-h^2\partial^2_r + h^2r^{-2}\Lambda + V - E)u , u' \rangle \\ &+2 h^2 r^{-3} \langle \Lambda u, u \rangle - \textstyle\int_{\mathbb{S}^{n-1}} |u(\theta, r)|^2 dV(r, \theta) d\theta ) \\ &+ w' (\|hu'\|^2 - \langle h^2 r^{-2} \Lambda u, u \rangle + \langle (E - V)u, u \rangle\\ &= -2 w \mathop{\rm Re}\langle P^{\pm}(h) u, u' \rangle \mp 2\varepsilon w \mathop{\rm Im}\langle u,u'\rangle + w'\|hu'\|^2\\ &+(2wr^{-1} - w') \langle h^2r^{-2}\Lambda u,u\rangle + Ew'\| u\|^2 \\ &-\textstyle\int_{\mathbb{S}^{n-1}} |u(\theta, r)|^2 (w(r) dV(r, \theta) + w'(r)V(r, \theta)) d\theta . \end{align}\tag{24}\]

First we show 8 . Since increasing \(s\) decreases the left side of 8 , without loss of generality we may take \(0 < \delta \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =2s - 1 < 1\). We will show the last line of 24 can be made to have a suitable lower bound, using \[\label{w32general32repulsive} w(r) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =1 - \frac{C_V}{C_V + \delta}(1 + r)^{-\delta}.\tag{25}\] For such \(w\), we clearly have \[w'(r) = \frac{\delta C_V}{C_V + \delta}(r + 1)^{-1 - \delta},\] Therefore, on the one hand \[2wr^{-1} - w' = 2r^{-1}\big(1 - \frac{C_V}{C_V + \delta}(r + 1)^{-\delta} \big[ 1 - \frac{\delta}{2} \frac{r}{r + 1} \big] \big) \ge 0,\] since \(\delta < 1\). On the other hand, in the sense of measures of Borel measures on \((0, \infty)\), using 7 \[wdV + w'V = \frac{\delta C_V V}{(C_V + \delta)(r + 1)^{1 + \delta}} + wdV \le \frac{C_V V}{1 + r} ( (r +1 )^{-\delta} - 1 ) \le 0.\]

Thus, the last two estimates and 24 imply, in the sense of distributions on \((0, \infty)\),

\[\label{deriv32wF32lwr32bd} \begin{align} (wF)' &\ge -2 w \mathop{\rm Re}\langle P^{\pm}(h) u, u' \rangle \mp 2\varepsilon w \mathop{\rm Im}\langle u,u'\rangle \\ &+ w' \| hu'\|^2 + E w'\| u\|^2. \end{align}\tag{26}\] Integrating 26 with respect to \(dr\) from \(r = r_0 > 0\) to \(r = \infty\) implies (because \(u\) is compactly supported) \[\label{pre32est32after32integrating} \int_{r_0}^\infty E w' \| u\|^2 + w' \| hu'\|^2 dr + w(r_0) F(r_0) \le \int_{r_0}^\infty 2 w \mathop{\rm Re}\langle P^{\pm}(h) u, u' \rangle \pm 2\varepsilon w \mathop{\rm Im}\langle u,u'\rangle dr.\tag{27}\]

Since \(u = r^{(n-1)/2} v\) for some \(v \in C^\infty_0(\mathbb{R}^n)\), we recognize that \[\label{Fr0} \begin{align} F(r_0) &= \|hu'(r_0, \cdot) \|^2 + r_0^{n-3} \langle h^2 \Delta_{\mathbb{S}^{n-1}}v(r_0, \cdot), v(r_0, \cdot) \rangle \\ &+(Er^{n-1}_0 - h^24^{-1}(n-1)(n-3)r_0^{n-3}) \|v(r_0, \cdot)\|^2 + r_0^{n-1} \langle V(r_0, \cdot)v(r_0, \cdot), v(r_0, \cdot) \rangle. \end{align}\tag{28}\] We rewrite the term in 28 involving \(\Delta_{\mathbb{S}^{n-1}}\) using the well known formula for the Laplacian in spherical coordinates: \[r^{-2}\Delta_{\mathbb{S}^{n-1}} = \Delta - \partial^2_r - (n-1)r^{-1} \partial_r.\] Therefore, \[\label{rewrite32laplace32beltrami} \begin{align} r_0^{n-3}& \langle h^2 \Delta_{\mathbb{S}^{n-1}}v(r_0, \cdot), v(r_0, \cdot) \rangle\\ &= h^2 r_0^{n-1}\langle (\Delta v)(r_0, \cdot), v(r_0, \cdot) \rangle \\ &- h^2r_0^{n-1}\langle (\partial^2_rv)(r_0, \cdot), v(r_0, \cdot) \rangle - h^2(n-1)r_0^{n-2}\langle (\partial_rv)(r_0, \cdot), v(r_0, \cdot) \rangle. \end{align}\tag{29}\] We can express the differential operators \(\partial_r\) and \(\partial^2_r\) with respect to the Euclidean coordinate system, \[\label{r32derivatives} \partial_r = r^{-1} \sum_{j=1}^n x_j \partial_{x_j}, \qquad \partial^2_r = r^{-2} \sum_{k=1}^n x_k \sum_{j=1}^n x_j \partial_{x_k}\partial_{x_j}.\tag{30}\] Thus by 29 and 30 , all terms in 28 tend to zero as \(r_0 \to 0\), except for possibly \(\|hu'(r, \cdot) \|^2\) in dimension three, which in that case tends to \(|v(0)|^2\int_{\mathbb{S}^{n-1}} d\theta\). We conclude \[\lim_{r_0 \to 0} w(r_0) F(r_0) = w(0) F(0) = \begin{cases} \omega_{n-1} w(0)|v(0)|^2 & n = 3, \\ 0 & n \ge 4,\end{cases}\] where \(\omega_{n-1}\) is the \((n-1)\)-dimensional volume of \(\mathbb{S}^{n-1}\).

Thus in view of 27 and \(0 < w \le 1\), \[\label{est32after32integrating} \begin{align} \int_{0}^\infty Ew' \| u\|^2 &+ w' \| hu'\|^2 dr \\ &\le 2 \big( \int_{0}^\infty \frac{1}{h^2w'} \| P^{\pm}(h) u \|^2 dr \big)^{1/2} \big(\int_{0}^\infty w'\| hu'\|^2 dr \big)^{1/2} \\ &+ \frac{2\varepsilon}{h} (\int_0^\infty \|u\|^2 dr)^{1/2} (\int_0^\infty \|hu'\|^2 dr)^{1/2}. \end{align}\tag{31}\] We now estimate, \[\begin{align} \int_0^\infty \|hu'\|^2dr &= \mathop{\rm Re}\int_0^\infty \langle u, -h^2u'' \rangle dr \\ &= \mathop{\rm Re}\big( \int_{0}^\infty \langle u, P^\pm(h)u \rangle dr + \int_0^\infty \langle u, (E - V - h^2 r^{-2}\Lambda)u \rangle dr \mp i \varepsilon\int^\infty_0 \|u\|^2 dr \big)\\ &= \mathop{\rm Re}\int_{0}^\infty \langle u, P^\pm(h)u \rangle dr + \int_0^\infty \langle u, (E - V - h^2 r^{-2}\Lambda)u \rangle dr \\ &\le \big( \int_0^\infty \frac{1}{w'} \|P^\pm(h)u\|^2dr \big)^{1/2} \big(\int_0^\infty w'\|u\|^2dr \big)^{1/2} + E\int_0^\infty \|u\|^2dr, \end{align}\] and \[\begin{align} \varepsilon\int_0^\infty \|u\|^2 dr &= \varepsilon\| v\|^2_{L^2} \\ &= |\mathop{\rm Im}\langle (P(h) - E \pm i\varepsilon)v, v \rangle_{L^2}| \\ &= \big| \mathop{\rm Im}\int_0^\infty \langle P^{\pm}(h)u, u \rangle dr \big| \\ &\le \big( \int_0^\infty \frac{1}{w'} \|P^\pm(h)u\|^2dr \big)^{1/2} \big(\int_0^\infty w'\|u\|^2dr \big)^{1/2}. \end{align}\] Combining these gives

\[\frac{\varepsilon^2}{h^2} \int_0^\infty \|u\|^2 dr \cdot \int_0^\infty \|hu'\|^2dr \le (E + \varepsilon) \int_0^\infty \frac{1}{h^2w'} \|P^\pm(h)u\|^2dr \cdot \int_0^\infty w'\|u\|^2dr.\] Plugging this into 31 yields

\[\label{est32after32integrating322} \begin{align} \int_{0}^\infty Ew' \| u\|^2 + w'\| hu'\|^2 dr &\le 2 \big( \int_0^\infty \frac{1}{h^2w'} \|P^\pm(h)u\|^2dr \big)^{1/2}\\ &\cdot \big( \big(\int_0^\infty w'\|hu'\|^2dr \big)^{1/2} + (E + \varepsilon)^{1/2} \big(\int_0^\infty w'\|u\|^2dr \big)^{1/2} \big). \end{align}\tag{32}\]

Completing the square in 32 , we find

\[\label{complete32the32square} \begin{align} \big( E^{1/2} \big( \int_0^\infty& w'\|u\|^2dr \big)^{1/2} - \frac{(E+\varepsilon)^{1/2}}{E^{1/2}} \big( \int_0^\infty \frac{1}{h^2w'} \|P^\pm(h)u\|^2dr \big)^{1/2} \big)^2 \\ &+ \big( \big( \int_0^\infty w'\|hu'\|^2dr \big)^{1/2} - \big( \int_0^\infty \frac{1}{h^2w'} \|P^\pm(h)u\|^2dr \big)^{1/2} \big)^2\\ &\le \frac{2E + \varepsilon}{E} \int_0^\infty \frac{1}{h^2w'} \|P^\pm(h)u\|^2dr. \end{align}\tag{33}\] Dropping the second term on the left side of 33 implies \[\label{E32and32ep32est} E^{1/2} \big( \int_0^\infty w'\|u\|^2dr \big)^{1/2} \le \big(\frac{(E+\varepsilon)^{1/2}}{E^{1/2}} + \frac{(2E + \varepsilon)^{1/2}}{E^{1/2}} \big) \big( \int_0^\infty \frac{1}{h^2w'} \|P^\pm(h)u\|^2dr \big)^{1/2}.\tag{34}\]

Next, consider the sector \(\{ z \in \mathbb{C}: |\mathop{\rm Im}z| < \alpha \mathop{\rm Re}z \}\) for \(0 < \alpha < 1\). Since
\(w' = C_V \delta (C_V + \delta)^{-1} (r + 1)^{-1-\delta}\) and \(\delta = 2s -1\), from 34 , we get for all \(h > 0\),
\(E \pm i \varepsilon\in \{ z \in \mathbb{C}: |\mathop{\rm Im}z| < \alpha \mathop{\rm Re}z \}\), and \(u \in r^{n-1/2}C^\infty_0(\mathbb{R}^n)\),

\[\label{alpha32est} \begin{align} |(E &\pm i \varepsilon)^{1/2}|\big( \int_0^\infty (r +1)^{-2s}\|u\|^2dr \big)^{1/2} \\ &\le h^{-1} (1 + \alpha^2)^{1/4} \big( \frac{1}{\delta} + \frac{1}{C_V} \big)((1 + \alpha)^{1/2} + (2 + \alpha)^{1/2}) \big( \int_0^\infty (r +1)^{2s} \|P^\pm(h)u\|^2dr \big)^{1/2}, \end{align}\tag{35}\] Here, our branch of the complex square root is chosen so that \(\mathop{\rm Im}(E \pm i\varepsilon)^{1/2} > 0\), and we used that \(|(E \pm i\varepsilon)|^{1/2} = (E^2 + \varepsilon^2)^{1/4} \le E^{1/2}(1 + \alpha^2)^{1/4}\) for \(E \pm i \varepsilon\in \{ z \in \mathbb{C}: |\mathop{\rm Im}z| < \alpha \mathop{\rm Re}z \}\).

Since \(u \in r^{(n-1)/2} C^\infty_0(\mathbb{R}^n)\), a standard density argument, which we review in Appendix 6, shows that 35 implies \[\label{resolv32est32in32sector} \begin{align} \|z^{1/2}(r + 1)^{-s}(P(h) &-z)^{-1}(r +1)^{-s}\|_{L^2 \to L^2} \\ &\le h^{-1}(1 + \alpha^2)^{1/4} \big( \frac{1}{\delta} + \frac{1}{C_V} \big)((1 + \alpha)^{1/2} + (2 + \alpha)^{1/2}), \end{align}\tag{36}\] on \(\{ z \in \mathbb{C}: |\mathop{\rm Im}z| < \alpha \mathop{\rm Re}z \}\) and for any \(0 < \alpha < 1\). To extend this bound to all \(z \in \mathbb{C}\setminus [0, \infty)\), we use the Phragmén-Lindelöf principle [27] in the following way. For \(u, v \in L^2(\mathbb{R}^n)\), put \[U(z) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =z^{1/2} \langle (r + 1)^{-s}(P(h) -z)^{-1}(r +1)^{-s}u, v \rangle_{L^2}.\] Then \(U(z)\) in analytic in \(\Omega_\alpha \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\{ z \in \mathbb{C}: \alpha \mathop{\rm Re}z < |\mathop{\rm Im}z| \}\). By 36 , on \(\partial \Omega_\alpha \setminus \{0 \}\) we have \[\label{U32on32bdry} |U(z)| \le h^{-1}(1 + \alpha^2)^{1/4} \big( \frac{1}{\delta} + \frac{1}{C_V} \big) ((1 + \alpha)^{1/2} + (2 + \alpha)^{1/2}) \|u\|_{L^2} \|v\|_{L^2}.\tag{37}\] On the other hand, in \(\Omega_\alpha\), we have the standard bound \[\label{std32bd} |U(z)| \le \frac{|z|^{1/2}\|u \|_{L^2} \|v\|_{L^2}}{\mathop{\rm dist}(z,[0, \infty))} = \begin{cases} \frac{\|u \|_{L^2} \|v\|_{L^2}}{|z|^{1/2}} & \mathop{\rm Re}z < 0, \\ \frac{|z|^{1/2}\|u \|_{L^2} \|v\|_{L^2}}{|\mathop{\rm Im}z|} & \mathop{\rm Re}z \ge 0, \, z \in \Omega_\alpha, \end{cases}\tag{38}\] where we used \[\frac{1}{\mathop{\rm dist}(z,[0, \infty))} = \frac{1}{ \inf_{r \ge 0} ((\mathop{\rm Re}z - r)^2 + (\mathop{\rm Im}z)^2)^{1/2}} = \begin{cases} \frac{1}{|z|} & \mathop{\rm Re}z <0 , \\ \frac{1}{|\mathop{\rm Im}z|} & \mathop{\rm Re}z \ge 0, \, z \in \Omega_\alpha. \end{cases}\]

Finally, define, \(g(z) = e^{i (z^{-1})^{1/2}}\), where our branch of the square root is as above. In \(\Omega_\alpha\), \(|g(z)| \le e^{-c|z|^{-1/2}}\) for some \(0 < c < 1\) depending on \(\alpha\). This with 38 says that for any \(\sigma > 0\), \[\label{sigma32bd} \limsup_{z \to 0, \, z \in \Omega_\alpha} |g(z)|^{\sigma} |U(z)| = 0.\tag{39}\] Therefore, from 37 and 39 , the Phragmén Lindelöf Theorem (Theorem 6 in Appendix 5) implies that 36 holds for all \(z \in \Omega_\alpha\) too. Sending \(\alpha \to 0^+\) completes the proof of 8 .

To prove 9 , start again at 33 and drop the first term on the left hand side. Still working on \(\{ z \in \mathbb{C}: |\mathop{\rm Im}z| < \alpha \mathop{\rm Re}z \}\), some manipulations give \[\label{uprimebound} \left(\int_0^\infty w'\|hu'\|^2\,dr\right)^{1/2}\leq (1+\sqrt{2+\alpha})\left(\int_0^\infty \frac{1}{h^2 w'} \|P^{\pm}(h)u\|^2 dr\right)^{1/2}.\tag{40}\] By integration by parts, \[\begin{align} \int_0^\infty (r +1)^{-3-\delta}\|u\|^2\,dr&=\frac{2}{2+\delta}\int_0^\infty (r +1)^{-2-\delta}\mathop{\rm Re}\langle u,u'\rangle\,dr\\ &\leq h^{-1}\left(\int_0^\infty(r +1)^{-1-\delta}\|hu'\|^2\,dr\right)^{1/2}\left(\int_0^\infty (r +1)^{-3-\delta}\|u\|^2\,dr\right)^{1/2}, \end{align}\] which implies \[\label{poincare} \left(\int_0^\infty (r + 1)^{-3-\delta}\|u\|^2\,dr\right)^{1/2}\leq h^{-1}\left(\int_0^\infty(r +1)^{-1-\delta}\|hu'\|^2\,dr\right)^{1/2}.\tag{41}\]

From 40 , 41 and \(w' = C_V \delta (C_V + \delta)^{-1} (r + 1)^{-1-\delta}\), \[\left(\int_0^\infty (r+1)^{-3-\delta}\|u\|^2\,dr\right)^{1/2}\leq h^{-2}\big( \frac{1}{\delta} + \frac{1}{C_V} \big) (1+\sqrt{2+\alpha}) \left(\int_0^\infty (r +1)^{1 + \delta}\|P^{\pm}(h)u\|^2\,dr\right)^{1/2}.\] Using again the density argument in Appendix 6, for \(0 < \delta < 1\), \[\label{resolv32est32in32sector32low32freq} \|(1+r)^{-\frac{3+\delta}{2}}(P(h)-z)^{-1}(1+r)^{-\frac{1+\delta}{2}} \|_{L^2 \to L^2 }\leq h^{-2}\left(\delta^{-1}+C_V^{-1}\right) (1+\sqrt{2+\alpha}),\tag{42}\] on \(\{ z \in \mathbb{C}: |\mathop{\rm Im}z| < \alpha \mathop{\rm Re}z \}\). Then, as above, 38 , the Phragmén Lindelöf Theorem, and sending \(\alpha \to 0^+\), imply \[\|(1+r)^{-\frac{3+\delta}{2}}(P(h)-z)^{-1}(1+r)^{-\frac{1+\delta}{2}} \|_{L^2 \to L^2 }\leq h^{-2}\left(\delta^{-1}+C_V^{-1}\right) (1+\sqrt{2}), \qquad z \in \mathbb{C}\setminus [0, \infty).\] Since the norm of an operator and its adjoint coincide, \[\|(1+r)^{-\frac{1+\delta}{2}} (P(h)-z)^{-1} (1+r)^{-\frac{3+\delta}{2}} \|_{L^2 \to L^2 }\leq h^{-2}\left(\delta^{-1}+C_V^{-1}\right) (1+\sqrt{2}), \qquad z \in \mathbb{C}\setminus [0, \infty).\] The three lines lemma then says that for fixed \(z \in \mathbb{C}\setminus [0, \infty)\), the analytic mapping \[\lambda \mapsto (1+r)^{-\frac{3+\delta}{2} +\lambda }(P(h)-z)^{-1}(1+r)^{-\frac{1+\delta}{2} - \lambda}, \qquad 0 < \mathop{\rm Re}\lambda < 1,\] (with values in the space of bounded operators \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\)) obeys \[\label{rest} \| (1+r)^{-\frac{3+\delta}{2} + \theta }(P(h)-z)^{-1}(1+r)^{-\frac{1+\delta}{2} - \theta} \|_{L^2 \to L^2} \le h^{-2}\left(\delta^{-1}+C_V^{-1}\right) (1+\sqrt{2}), \qquad \theta \in [0, 1].\tag{43}\]

Having established 43 , to finish, we need to see that we can choose \(\delta\) and \(\theta\) appropriately to arrive at 9 . That is, we need to attain the more general weights characterized by \(s_1, s_2 > 1/2\), \(s_1 + s_2 > 2\). However, because we have the restrictions \(\delta \in (0,1)\) and \(\theta \in [0,1]\), we first need to make reductions as follows. Since decreasing \(s_1\) or \(s_2\) in 9 increases the right side, it suffices to suppose \(s_1, \, s_2 > 1/2\), \(2< s_1 + s_2 < 3\). Furthermore, by taking the adjoint, it is no restriction to have \(s_1 \le s_2\). If we write \(s_1 = (1 + 2\delta_1)/2\) for some \(\delta_1 > 0\), then we may replace \(s_2\) by \(\min(s_2, (3 + \delta_1)/2 )\). Having made these reductions, 9 follows from 43 by setting \(\delta = s_1 + s_2 -2 < 1\), \(\theta = (s_2 - s_1 + 1)/2 \le (4 - \delta_1)/4 < 1\).

3 Resolvent bounds for wave decay↩︎

In this section, we consider the operator \(P = P(1) = -\Delta + V\), with \(P(h)\) as in 1 ; \(V\) obeys 2 through 7 . As a consequence of Theorem 1, we prove several more resolvent bounds for \(P\), which enable us in Section 4 to establish weighted energy decay for the solution to the wave equation 12 . Throughout this section, \(C\) denotes a positive constant whose precise value may change, but is always independent \(\lambda\), which plays the role of our spectral parameter.

Lemma 1. Fix \(s_1,\, s_2 > 1/2\) with \(s_1 + s_2 > 2\). There exist \(C > 0\) so that for all \(\lambda \in \mathbb{C}\) with \(0 < |\mathop{\rm Im}\lambda| \le 1\), and for all multiindices \(\alpha_1, \, \alpha_2\) with \(|\alpha_1| + |\alpha_2| \le 2\), \[\label{lap32A} \| \langle x \rangle^{-s_1} \partial^{\alpha_1}_x (P- \lambda^2)^{-1} \partial^{\alpha_2}_x \langle x \rangle^{-s_2} \|_{L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n) } \le C(1 + |\mathop{\rm Re}\lambda|)^{|\alpha_1| + |\alpha_2|-1 }.\tag{44}\]

Proof. Since \(((P - \lambda^2)^{-1})^* = (P- (\overline{\lambda})^2)^{-1}\), to prove 44 is suffices to assume \(\mathop{\rm Im}\lambda > 0\).

First, we treat the case \(|\alpha_2| = 0\). Using 8 if \(|\mathop{\rm Re}\lambda|>1\) or 9 if \(|\mathop{\rm Re}\lambda|\leq 1\), we get \[\label{L232to32L232bd32Helmholtz32resolvent} \|\langle x \rangle^{-s_1} (P- \lambda^2)^{-1} \langle x \rangle^{-s_2} \|_{L^2 \to L^2 } \le C(1 + |\mathop{\rm Re}\lambda|)^{-1}, \qquad 0 < \mathop{\rm Im}\lambda \le 1,\tag{45}\] Recall from standard elliptic theory that for all \(f \in H^2(\mathbb{R}^n)\) and all \(\gamma > 0\), \[\label{std32elliptic32thry} \begin{gather} \| f\|_{H^2} \le C( \| f\|_{L^2} + \| \Delta f\|_{L^2}), \\ \| f\|^2_{H^1} \le C \| f\|_{L^2} \|f \|_{H^2} \le C( \gamma^{-1} \| f\|^2_{L^2} + \gamma \| \Delta f\|^2_{L^2}). \end{gather}\tag{46}\] Therefore, for any \(f \in L^2(\mathbb{R}^n)\), \[\begin{align} \| \langle x \rangle^{-s_1}& (P- \lambda^2)^{-1} \langle x \rangle^{-s_2}f\|_{H^2(\mathbb{R}^n)} \\ &\le C ( \| \langle x \rangle^{-s_1} (P - \lambda^2)^{-1} \langle x \rangle^{-s_2}f\|_{L^2(\mathbb{R}^n)} + \| (-\Delta) \langle x \rangle^{-s_1} (P - \lambda^2)^{-1} \langle x \rangle^{-s_2}f \|_{L^2(\mathbb{R}^n)}) \\ &\le C ( \| \langle x \rangle^{-s_1} (P- \lambda^2)^{-1} \langle x \rangle^{-s_2}f\|_{H^1(\mathbb{R}^n)} + \| \langle x \rangle^{-s_1} (-\Delta) (P- \lambda^2)^{-1} \langle x \rangle^{-s_2}f \|_{L^2(\mathbb{R}^n)}) \\ &\le C(\gamma^{-1} + |\mathop{\rm Re}\lambda|^2) \| \langle x \rangle^{-s_1} (P- \lambda^2)^{-1} \langle x \rangle^{-s_2}f\|_{L^2(\mathbb{R}^n)} \\ &+ C\gamma \| \Delta \langle x \rangle^{-s_1} (P - \lambda^2)^{-1} \langle x \rangle^{-s_2}f\|_{L^2(\mathbb{R}^n)}\\ &+ C \| f \|_{L^2(\mathbb{R}^n)}. \end{align}\] Selecting \(\gamma\) sufficiently small depending on \(C\), and applying 45 yields \[\label{L232to32H232bd32Helmholtz32resolvent} \| \langle x \rangle^{-s_1}(P- \lambda^2)^{-1} \langle x \rangle^{-s_2}f\|_{H^2(\mathbb{R}^n)} \le C(1 + |\mathop{\rm Re}\lambda|) \| f \|_{L^2(\mathbb{R}^n)},\tag{47}\] as desired. This confirms 44 for \(|\alpha_1| = 2\). For \(|\alpha_1| = 1\) (still with \(|\alpha_2| = 0\)), combine 45 and 47 via the second line of 46 .

If \(|\alpha_2| > 0\), let \(f \in C^{\infty}_0(\mathbb{R}^n)\), and put \(u = \langle x \rangle^{-s_1} (P - \lambda^2)^{-1} \langle x \rangle^{-s_2} \partial^{\alpha_2}_x f\). We need to show \[\| u\|_{H^{|\alpha_2|}} \le C(1 + |\mathop{\rm Re}\lambda|)^{|\alpha_1| + |\alpha_2| -1} \| f\|_{L^2}, \qquad H^0 \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =L^2(\mathbb{R}^n).\] If \(|\alpha_1| = 0\), we use self-adjointness and \(\| \langle x \rangle^{-s_2} (P - \lambda^2)^{-1} \langle x \rangle^{-s_1} f \|_{H^1(\mathbb{R}^n)} \le C \| f \|_{L^2(\mathbb{R}^n)}\) to get \[\begin{align} \| u\|^2_{L^2} &= \langle u, \langle x \rangle^{-s_1} (P - \lambda^2)^{-1} \langle x \rangle^{-s_2} \partial^{\alpha_2}_x f \rangle_{L^2} \\ &\le \| \partial^{\alpha_2}_x \langle x \rangle^{-s_2} (P - (\overline{\lambda})^2)^{-1} \langle x \rangle^{-s_1} u \|_{L^2} \| f\|_{L^2} \\ &\le C(1 + |\mathop{\rm Re}\lambda|)^{|\alpha_2| -1} \| u \|_{L^2} \| f\|_{L^2}. \end{align}\] If \(|\alpha_1| = 1\), we recognize that \((P - \lambda^2)u = \langle x \rangle^{-s_1 - s_2} \partial^{\alpha_2}_x f + [-\Delta, \langle x \rangle^{-s_1}] \langle x \rangle^{s_1} u\). Then multiply by \(\overline{u}\), integrate over \(\mathbb{R}^n\), and integrate by parts as appropriate \[\begin{align} \| \nabla u \|_{L^2}^2 = \int (\lambda^2 - V) |u|^2 - \int \partial^{\alpha_2}_x (\langle x \rangle^{-s_1 - s_2} \overline{u}) f + \int \overline{u} [-\Delta, \langle x \rangle^{-s_1}] \langle x \rangle^{s_1} u. \end{align}\] Because both \[[-\Delta, \langle x \rangle^{-s_1}] \langle x \rangle^{s_1} = (-\Delta \langle x \rangle^{-s_1}) \langle x \rangle^{s_1} - 2 (\nabla \langle x \rangle^{-s_1}) \cdot \nabla \langle x \rangle^{s_1},\] and \(\partial^{\alpha_2}_x \langle x \rangle^{-s_1 - s_2}\) are first order differential operators with bounded coefficients, we conclude, for all \(\gamma > 0\), \[\begin{align} \| \nabla u \|^2_{L^2} &\le C_\gamma ( (1 + |\mathop{\rm Re}\lambda|)^2 \| u\|^2_{L^2} + \| f\|^2_{L^2} ) + \gamma \| \nabla u \|^2_{L^2} \\ &\le C_\gamma (1 + |\mathop{\rm Re}\lambda|)^{2} \| f\|^2_{L^2} + \gamma \| \nabla u \|^2_{L^2}, \end{align}\] for some \(C_\gamma > 0\) depending on \(\gamma\). Note also we used 3 and Lemma 7 to estimate \[\begin{align} \int V |u|^2 &\le \int \mathbf{1}_{B(0,1)} V |u|^2 + \int \mathbf{1}_{\mathbb{R}^n \setminus B(0,1)} V |u|^2 \\ & \le C\int \mathbf{1}_{B(0,1)} |r^{-1}u||u| + C\|u\|_{L^2}^2 \\ & \le C (\|u\|_{L^2} \|\nabla u \|_{L^2} + \|u\|_{L^2}^2). \end{align}\] Fixing \(\gamma\) small enough, we absorb the second term on the right side into the left side, confirming 44 when \(|\alpha_1| = |\alpha_2| = 1\).
 ◻

Next, we prove an estimate for the derivative of the weighted resolvent, which requires an extra short range condition on the potential. As input we need the following bound for the weighted square of the free resolvent, which we prove in Appendix [deriv free resolv appendix].

Lemma 2. Let \(n \ge 3\), \(j \in \{0,1\}\), and suppose \(s\) satisfies 16 . There exists \(C > 0\) such that for all \(\lambda \in \mathbb{C}\) with \(0 < |\mathop{\rm Im}\lambda| \le 1\), \[\label{lap32free32resolv32square} \| \lambda \langle x \rangle^{-s} \nabla^{j} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s}\|_{L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)} \le C(1 + |\lambda|)^{j -1}.\tag{48}\]

Remark 5. In [16], the estimate 48 is stated to hold in any dimension \(n \ge 3\) provided \(s > 3/2\). However, our proof of Lemma 2 in dimension \(n \ge 4\) needs \(s\) larger if 48 is to hold uniformly as \(|\lambda| \to 0\). In our approach, we use the integral kernel of \(\lambda \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s}\) to assess \(L^2\)-boundedness as \(|\lambda| \to 0\). The kernel is given in terms of the Macdonald function [28] of order \(n/2 - 2\), along with other factors. We are able to conclude boundedness on \(L^2(\mathbb{R}^n)\) for \(s\) as in 16 .

Lemma 3. Let \(n \ge 3\) and \(s\) as in 16 . Assume \(V\) obeys 2 through 7 as well as 13 and 14 . There exists \(C > 0\) so that for all \(\lambda \in \mathbb{C}\) with \(0 < | \mathop{\rm Im}\lambda | \le 1\), and all \(j_1, \, j_2 \in \{0, 1\}\) with \(j_1 + j_2 \le 1\), \[\label{lap32square} \| \tfrac{d}{d \lambda}\langle x \rangle^{-s} \lambda^{j_1} \nabla^{j_2} (P - \lambda^2)^{-1} \langle x \rangle^{-s} \|_{L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)} \le C.\tag{49}\]

Proof. Without loss of generality, we take \(s\) sufficiently close to \((n + 3)/4\) when \(n \neq 8\), or sufficiently close to \(3\) when \(n = 8\), so that by 14 we may fix \(s' > 1/2\) so that \(s + s' < \delta\).

We begin from the resolvent identity \[\label{resolv32id32for32deriv} (P - \lambda^2)^{-1} \langle x \rangle^{-s} (1 + K(\lambda)) = R_0(\lambda) \langle x \rangle^{-s},\tag{50}\] where \(K(\lambda) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =V(x) \langle x \rangle^{s + s'} \langle x \rangle^{-s'} R_0(\lambda) \langle x \rangle^{-s}\) and \(R_0(\lambda) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =(-\Delta - \lambda^2)^{-1}\).

It is well known that \(\langle x \rangle^{-s'} R_0(\lambda) \langle x \rangle^{-s} : L^2(\mathbb{R}^n) \to H^2(\mathbb{R}^n)\) has a continuous extension from either half-plane (\(\pm \mathop{\rm Im}\lambda > 0\)) to \(\mathbb{R}\) [29]. Let us denote this extension by \(R^{\pm}_{0, s',s}(\lambda)\) and put \(K^\pm(\lambda) = V(x) \langle x \rangle^{s + s'} R^{\pm}_{0, s',s}(\lambda)\)

We now show that \(K^\pm(\lambda)\) is a compact operator \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\). To see this, observe that we may write \(K^\pm(\lambda)\) as the sum \[K^\pm(\lambda) = (\chi \langle x \rangle^{s + s'} V) R^{\pm}_{0, s',s}(\lambda) + ((1 - \chi)V \langle x \rangle^{\delta}) \langle x \rangle^{s + s'- \delta} R^{\pm}_{0, s',s}(\lambda).\] where \(\chi \in C^\infty_0(\mathbb{R}^n ; [0,1])\) is identically one near the origin in \(\mathbb{R}^n\) and supported in \(B(0,1)\). The second operator on the right side is compact by [30]). The first operator on the right side is compact as follows: it may be viewed as the composition of bounded \(R^{\pm}_{0, s',s}(\lambda) : L^2(\mathbb{R}^n) \to H^2(\mathbb{R}^n)\) followed by multiplication by \(\chi \langle x \rangle^{s + s'} V\). Due to 3 and Lemma 7, we have \(\|\chi \langle x \rangle^{s + s'} V u\|_{L^2(\mathbb{R}^n)} \le C \| u \|_{H^1(B(0,1))}\) for some \(C > 0\) and all \(u \in H^2(\mathbb{R}^n)\). By the Kondrachov embedding theorem the inclusion \(H^2(B(0,1)) \to H^1(B(0,1))\) is compact. So compactness of \((\chi \langle x \rangle^{s + s'} V) R^{\pm}_{0, s',s}(\lambda)\) holds as desired.

We claim further that \(1 + K^\pm(\lambda)\) is invertible \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\) for all \(\lambda\) with \(\pm \mathop{\rm Im}\lambda \ge 0\). By compactness of \(K^\pm(\lambda)\) and the Fredholm alternative [31], we have that \(1 + K^\pm(\lambda)\) is invertible if we can show \((1 + K^\pm(\lambda))g = 0\) implies \(g = 0\). To this end, put \(u \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\langle x \rangle^{s'} R^{\pm}_{0, s',s}(\lambda)g\), which belongs to \(\langle x \rangle^{s'} H^2(\mathbb{R}^n)\). If we can show \(u = 0\), then in fact \(g = 0\). This is because \((-\Delta - \lambda^2)\langle x \rangle^{-s'} u = \langle x \rangle^{-s} g\) in the distributional sense.

Now let us show \(u = 0\). If \(\lambda^2 \in \mathbb{C}\setminus [0, \infty)\) (so that \(K^\pm(\lambda) = K(\lambda)\)), this follows immediately from \((P - \lambda^2) u = \langle x \rangle^{-s}g + V R_0(\lambda) \langle x \rangle^{-s} g = \langle x \rangle^{-s}(1 + K(\lambda))g = 0.\) If \(\lambda^2 \in [0, \infty)\), the idea is the same, but we incorporate a limiting step. Set \(u_{\pm, \varepsilon} = (-\Delta - \lambda^2 \pm i \varepsilon)^{-1} \langle x \rangle^{-s}g\). We have that [29] implies that \(\langle x \rangle^{-s'} u_{\pm, \varepsilon}\) converges to \(\langle x \rangle^{-s'}u\) in \(H^2(\mathbb{R}^n)\) as \(\varepsilon\to 0^+\). We also have that \[\begin{align} u_{\pm, \varepsilon} &= (-\Delta - \lambda^2 \pm i \varepsilon)^{-1} \langle x \rangle^{-s}g \\ &= (P - \lambda^2 \pm i\varepsilon)^{-1} (P - \lambda^2 \pm i\varepsilon) (-\Delta - \lambda^2 \pm i \varepsilon)^{-1} \langle x \rangle^{-s}g \\ &= (P - \lambda^2 \pm i\varepsilon)^{-1} \langle x \rangle^{-s} (I + V \langle x \rangle^{s}(-\Delta - \lambda^2 \pm i \varepsilon)^{-1} \langle x \rangle^{-s}) g. \end{align}\]

Therefore, by 9 , \[\begin{align} \| \langle x \rangle^{-s_1}u \|_{L^2} &= \lim_{\varepsilon\to 0^+} \| \langle x \rangle^{-s_1} u_{\pm, \varepsilon} \|_{L^2} \\ &\le C \lim_{\varepsilon\to 0^+} \| (I + V \langle x \rangle^{s}(-\Delta - \lambda^2 \pm i \varepsilon)^{-1} \langle x \rangle^{-s}) g \|_{L^2} \\ &= \| (1 + K^\pm(\lambda)) g \|_{L^2} = 0. \end{align}\] Thus we have demonstrated that \(I + K^\pm(\lambda)\) is invertible for \(\pm \mathop{\rm Im}\lambda \ge 0\). As \(\lambda \to \infty\), \(\|K(\lambda)\|_{L^2 \to L^2} \to 0\) thanks to 44 , hence we can compute \((I + K^\pm(\lambda))^{-1}\) by a Neumann series, thanks to 44 . Therefore \[\label{bd32I32plus32K32inv} \|(I + K^\pm(\lambda))^{-1} \|_{L^2 \to L^2} \le C.\tag{51}\]

Now for \(0 < |\mathop{\rm Im}\lambda| \le 1\) take the derivative of 50 with respect to \(\lambda\),

\[\label{deriv32resolv32id} \begin{align} \big( \tfrac{d}{d \lambda}\langle x \rangle^{-s} & \lambda^{j_1} \nabla^{j_2} (P - \lambda^2)^{-1} \langle x \rangle^{-s} \big) (I + K(\lambda)) \\ &=\tfrac{d}{d \lambda}\langle x \rangle^{-s} \lambda^{j_1} \nabla^{j_2} R_0(\lambda ) \langle x \rangle^{-s} \\ &-2 \langle x \rangle^{-s} \lambda^{j_1} \nabla^{j_2} (P - \lambda^2)^{-1} \langle x \rangle^{-s'} V \langle x \rangle^{s + s'} \lambda \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s}, \end{align}\tag{52}\] where we used \[\label{deriv32of32resolv} \tfrac{d}{d \lambda}\langle x \rangle^{-s'} \nabla^j R_0(\lambda ) \langle x \rangle^{-s} = 2 \lambda \langle x \rangle^{-s'} \nabla^j (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s}, \qquad j \in \{0,1\}.\tag{53}\] The operator norm \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\) of the term in the second line of 52 is bounded above by a constant due to 48 and 53 . As for the third line, \(\|\langle x \rangle^{-s} \lambda^{j_1} \nabla^{j_2} (P - \lambda^2)^{-1} \langle x \rangle^{-s'}\|_{L^2 \to L^2} \le C\) by 44 . Moreover \[\begin{align} \| V& \langle x \rangle^{s + s'} \lambda \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} \|_{L^2 \to L^2} \\ & \le C \| \lambda \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} \|_{H^1\to L^2} \end{align}\] since multiplication by \(V \langle x \rangle^{s + s'}\) is a bounded operator \(H^1(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\) (see 3 and Lemma 7). Finally, because \(\| \lambda \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} \|_{H^1\to L^2} \le C\) by 48 , the proof of 49 is complete.
 ◻

4 Proof of Theorem 3↩︎

In this section we prove Theorem 3 by combining the resolvent bounds of the previous section with an argument appearing in [16]. As before we use the notation \(P = -\Delta + V : L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\), \(n \ge 3\), where \(V\) obeys 2 through 7 along with 13 and 14 .

In several steps below, we use that for all \(0 \le \alpha \le 1\), there exists \(C > 0\) so that for any \(f \in H^1(\mathbb{R}^n)\), \[\label{Poincare32etc} \| V^\alpha f \|^2_{L^2} \le C (\| \nabla f \|^2_{L^2} + \|f\|^2_{L^2}) \le C\| \nabla f \|^2_{L^2}.\tag{54}\] The first inequality follows from 3 , 13 and Lemma 7, while the second follows from the Poincaré inequality (as we work in dimension \(n \ge 3\)).

Given \(s > 0\) and \(u\) as in 15 solving the wave equation 12 , with compactly supported initial conditions \(u(0, x) = u_0(x) \in H^1(\mathbb{R}^n)\), \(\partial_t u(0,x) = u_1(x) \in L^2(\mathbb{R}^n)\), define \[\begin{gather} E_s(t) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\int_{\mathbb{R}^n} \langle x \rangle^{-2s}( |\partial_t u(t,x)|^2 + |\nabla u(t,x)|^2 + |u(t,x)|^2)dx, \\ E(0) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\| \nabla u_0 \|^2_{L^2} + \| u\|^2_{L^2}. \end{gather}\]

Lemma 4. If \(s > 1/2\) and \(V\) satisfies 2 through 7 , there exists \(C > 0\) so that \[\label{E32integrable} \int_0^\infty E_s(\tau) d\tau \le CE(0).\tag{55}\] If in addition \(s\) satisfies 16 and \(V\) 13 and 14 , there exists \(C > 0\) so that for \(t \ge 1\), \[\label{t32minus32two32bd} \int_t^\infty E_s(\tau) d\tau \le Ct^{-2} E(0).\tag{56}\]

Proof. Choose \(\phi \in C^\infty(\mathbb{R})\), \(\phi \ge 0\), \(\phi(t) = 0\) near \((-\infty, 1/2]\), \(\phi(t) = 1\) near \([1, \infty)\). Since \((\partial^2_t + P)u = 0\), where \(P = -\Delta + V\), It holds that \[\label{v} (\partial^2_t + P) \phi u = (\phi'' + 2 \phi' \partial_t)u \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =v(t).\tag{57}\] Thus, by Duhamel’s formula for the solution to an inhomogeneous wave equation with zero initial conditions, \[\phi u(t) = \int^t_0 \frac{\sin(t - \tau)\sqrt{P}}{\sqrt{P}} v(\tau) d\tau.\] On the other hand, \[(P - (\lambda - i\varepsilon)^2)^{-1} = \int^\infty_0 e^{-it(\lambda - i\varepsilon)} \frac{\sin(t \sqrt{P})}{\sqrt{P}} dt, \qquad \varepsilon> 0.\] It follows from the last two identities that the Fourier transform \(\widehat{\phi u}\) of \(\phi u\) satisfies \[\label{FT32identity} \widehat{\phi u}(\lambda - i\varepsilon) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\int_{-\infty}^\infty e^{-it(\lambda - i\varepsilon)} \phi(t)u(\cdot, t) dt = (P - (\lambda - i\varepsilon)^2)^{-1} \hat{v}(\lambda - i\varepsilon).\tag{58}\]

By finite propagation speed for the wave equation, \(\mathop{\rm supp}_{x} v(t)\), and thus also \(\mathop{\rm supp}_{x} \hat{v}(\lambda)\), is contained in a compact set independent of \(t\). Choose \(\eta \in C^\infty_0(\mathbb{R}^n)\) such that \(\eta = 1\) near \(\mathop{\rm supp}_x {v}(t)\) for all \(t \in \mathbb{R}\). By 58 , \[\begin{gather} \langle x \rangle^{-s} \widehat{\phi u} (\lambda - i\varepsilon) = \langle x \rangle^{-s}(P - (\lambda - i\varepsilon)^2)^{-1} \eta \hat{v}(\lambda - i\varepsilon), \\ \langle x \rangle^{-s} \widehat{\partial_t (\phi u)}(\lambda - i\varepsilon) = \langle x \rangle^{-s} (\lambda - i\varepsilon) (P - (\lambda - i\varepsilon)^2)^{-1} \eta \hat{v}(\lambda - i\varepsilon), \\ \langle x \rangle^{-s} \nabla \widehat{\phi u}(\lambda - i\varepsilon) = \langle x \rangle^{-s} (\lambda - i\varepsilon) \nabla (P - (\lambda - i\varepsilon)^2)^{-1} \eta \hat{v}(\lambda - i\varepsilon). \end{gather}\] Therefore, by 44 , for \(s > 1/2\) and \(V\) obeying 2 through 7 , there is \(C > 0\) independent of \(\lambda\) and \(\varepsilon\), so that for all \(\lambda \in \mathbb{R}\), \(0 < \varepsilon\le 1\), we have \[\label{use32resolv32bds} \begin{align} \Big \| \frac{d^k}{d\lambda^k} & \langle x \rangle^{-s} \widehat{\partial_t (\phi u)}(\lambda - i\varepsilon) \Big \|_{L^2} + \Big \| \frac{d^k}{d\lambda^k} \langle x \rangle^{-s} \nabla \widehat{\phi u}(\lambda - i\varepsilon) \Big \|_{L^2}\\ &+ \Big \| \frac{d^k}{d\lambda^k} \langle x \rangle^{-s} \widehat{\phi u}(\lambda - i\varepsilon) \Big \|_{L^2} \le C\| \hat{v}(\lambda - i\varepsilon)\|_{L^2} + Ck\| \widehat{tv}(\lambda - i\varepsilon)\|_{L^2}. \end{align}\tag{59}\] for \(k = 0\). If in addition we suppose \(s\) satisfies 16 and \(V\) satisfies 13 and 14 , then by 49 , 59 holds for \(k \in \{0,1\}\). Note when \(k = 1\) we used the product rule and the identity \(\frac{d}{d\lambda}\hat{v}(\lambda-i\varepsilon)=-i\widehat{tv}(\lambda-i\varepsilon)\).

Next, by 59 and Plancherel’s theorem, there exist \(C_1, C_2, C_3, C > 0\) independent of \(\varepsilon\) so that \[\label{Plancherel} \begin{align} \int_{-\infty}^{\infty}& ( \| \langle x \rangle^{-s} \partial_t(\phi u)\|_{L^2}^2 + \| \langle x \rangle^{-s} \nabla (\phi u)\|_{L^2}^2 + \| \langle x \rangle^{-s} \phi u\|_{L^2}^2)e^{-2\varepsilon t} dt \\ &= C_1\int_{-\infty}^{\infty}( \| \langle x \rangle^{-s} \widehat{\partial_t(\phi u)}(\lambda - i\varepsilon)\|_{L^2}^2 + \| \langle x \rangle^{-s} \nabla \widehat{\phi u}(\lambda - i\varepsilon)\|_{L^2}^2 + \| \langle x \rangle^{-s} \widehat{\phi u}(\lambda - i\varepsilon)\|_{L^2}^2) d\lambda \\ &\le C_2 \int_{-\infty}^{\infty} \| \hat{v}(\lambda - i\varepsilon)\|^2_{L^2} d\lambda = C_3 \int_{-\infty}^{\infty} \| v(t)\|^2_{L^2} e^{-2\varepsilon t} dt \le C \sup_{t \in \mathbb{R}} \|v(t)\|^2. \end{align}\tag{60}\] The last constant \(C\) is independent of \(\varepsilon\) because \(v(t)\) has compact support in \(t\), see 57 . The proof of 55 is completed by sending \(\varepsilon\to 0\) in 60 and observing \[\label{apply32Poincare} \begin{align} \| v(t) \|_{L^2} &\le C(\| u_0 \|_{L^2} + \| \sqrt{P} u_0\|_{L^2} + \|u_1\|_{L^2}) \\ & \le C( \| \nabla u_0\|_{L^2} + \|u_1\|_{L^2}) = C\sqrt{E(0)}. \end{align}\tag{61}\] Between lines one and two of 61 , we used that for any \(f \in H^2(\mathbb{R}^n)\) (and thus any \(f \in H^1(\mathbb{R}^n)\), since \(H^2(\mathbb{R}^n)\) is dense in \(H^1(\mathbb{R}^n)\)), \[\begin{align} \| \sqrt{P} f \|^2_{L^2} &= \langle f, Pf \rangle_{L^2} = \| \nabla f \|^2_{L^2} + \| \sqrt{V} f\|^2_{L^2} \le C \|\nabla f\|^2_{L^2}, \end{align}\] with the second inequality due to 54 .

To prove 56 , we again use Plancherel’s theorem with 59 , so that for all \(0 < \varepsilon\le 1\) and \(T \ge 1\), \[\label{Plancherel32again} \begin{align} T^2\int_{T}^{\infty}& ( \| \langle x \rangle^{-s} \partial_t(\phi u)\|_{L^2}^2 + \| \langle x \rangle^{-s} \nabla (\phi u)\|_{L^2}^2 + \| \langle x \rangle^{-s} \phi u\|_{L^2}^2)e^{-2\varepsilon t} dt \\ & \le \int_{-\infty}^{\infty} ( \| \langle x \rangle^{-s} t \partial_t(\phi u)\|_{L^2}^2 + \| \langle x \rangle^{-s} t \nabla (\phi u)\|_{L^2}^2 + \| \langle x \rangle^{-s} t \phi u\|_{L^2}^2)e^{-2\varepsilon t} dt \\ &= C_1\int_{-\infty}^{\infty} \big( \big\| \frac{d}{d\lambda} \langle x \rangle^{-s} \widehat{\partial_t(\phi u)}(\lambda - i\varepsilon) \big\|_{L^2}^2 + \big\| \frac{d}{d \lambda} \langle x \rangle^{-s} \nabla \widehat{\phi u}(\lambda - i\varepsilon)\big\|_{L^2}^2 \\ &+ \big\| \frac{d}{d\lambda} \langle x \rangle^{-s} \widehat{\phi u}(\lambda - i\varepsilon)\big\|_{L^2}^2\big) d\lambda \\ &\le C_2 \int_{-\infty}^{\infty} \| \hat{v}(\lambda - i\varepsilon)\|^2_{L^2} + \|\widehat{tv}(\lambda - i \varepsilon) \|^2_{L^2}) d\lambda \\ &= C_3 \int_{-\infty}^{\infty} (\| v(t)\|^2_{L^2} + \|tv(t) \|^2_{L^2}) e^{-2\varepsilon t} dt \le C \sup_{t \in \mathbb{R}} \|v(t)\|^2 \le CE(0). \end{align}\tag{62}\] Once again sending \(\varepsilon\to 0^+\) concludes the proof of 56 .
 ◻

The local energy decay 17 follows from 56 and

Lemma 5. If \(s > 1/2\) and \(V\) satisfies 2 through 7 , there exists \(C> 0\) so that for all \(t \ge 1\), \[\label{bd32E32by32its32integral} E_s(t) \le C \int_t^\infty E_s(\tau) d\tau.\tag{63}\]

Proof. The strategy is the same as that of [16]. Computing \(\frac{d}{dt} E_s(t)\), one finds \[\label{deriv32of32wtd32E} \begin{align} \frac{d}{dt} E_s(t) &= -2 \mathop{\rm Re}\int_{\mathbb{R}^n} \partial_r u(t,x) \overline{\partial_t u(t,x)} \partial_r \langle x \rangle^{-2s}dx\\ &+ 2 \mathop{\rm Re}\int_{\mathbb{R}^n} (-Vu(t,x) \overline{\partial_t u(t,x)} + u(t,x) \overline{\partial_t u(t,x)} )\langle x \rangle^{-2s} dx. \end{align}\tag{64}\] By 54 , \[\begin{align} \| V \langle x \rangle^{-s} u(t,x) \|_{L^2} &\le C \|\nabla \langle x \rangle^{-s} u(t,x) \|_{L^2}\\ &\le C\|\langle x \rangle^{-s} \nabla u(t,x) \|_{L^2} + C\| \langle x \rangle^{-s} u(t,x) \|_{L^2}. \end{align}\] for \(C> 0\) independent of \(t\), and whose precise value may change between lines. Thus we can bound the right side of 64 from above by Cauchy-Schwarz, \[\begin{align} \frac{d}{dt} E_s(t) &\le C\| \langle x \rangle^{-s} \partial_r u(t,x) \|_{L^2} \| \langle x \rangle^{-s} \partial_t u(t,x)\|_{L^2} + C\| V \langle x \rangle^{-s} u(t,x) \|_{L^2} \| \langle x \rangle^{-s} \partial_t u(t,x)\|_{L^2} \\ &+ C\| \langle x \rangle^{-s} u(t,x) \|_{L^2} \| \langle x \rangle^{-s} \partial_t u(t,x)\|_{L^2} \le C E_s(t). \end{align}\] We then have, for all \(T > t \ge 1\), \[\label{penult32est} E_s(t) \le E_s(T) + C_s \int_t^T E_s(\tau) d \tau.\tag{65}\] From 55 , we also have a sequence \(T_j \to \infty\) so that \(\lim_{T_j \to \infty} E_s(T_j) =0\). So setting \(T = T_j\) in 65 and sending \(T_j \to \infty\) completes the proof of 63 .
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5 Phragmén-Lindelöf Theorem↩︎

In this appendix we recall the Phragmén Lindelöf Theorem. Let \(f(z)\) be a holomorphic function in a domain \(D\) of the complex plane with boundary \(\Gamma\). We say that \(f(z)\) does not exceed a number \(M \ge 0\) in modulus at a boundary point \(\zeta \in \Gamma\) if \(\limsup_{z \to \zeta, \, z \in D} |f(z)| \le M\).

Theorem 6 (Phragmén Lindelöf Theorem [27]). Suppose \(E \subseteq \Gamma\), and \(f\) analytic on \(D\) does not exceed \(M\) in modulus at any point of \(\Gamma \setminus E\). Suppose also there is a function \(g(z)\) with the following properties:

  1. \(g(z)\) is analytic in \(D\),

  2. \(|g(z)| < 1\) in \(D\),

  3. \(g(z) \neq 0\) in \(D\),

  4. For every \(\sigma > 0\), the function \(|g(z)|^\sigma |f(z)|\) does not exceed \(M\) is modulus at any \(\zeta \in E\).

Under these conditions, \(|f(z)| \le M\) everywhere in \(D\).

6 Density argument: proof of 36 and 42↩︎

In this appendix, we prove 36 and 42 as a consequence of

Lemma 6. Fix \(h, \, s_1 > 0\), \(0 < s_2 < 1\), and \(z \in \mathbb{C}\setminus [0, \infty)\). Let \(P(h)\) be as in 1 with \(V : \mathbb{R}^n \to \mathbb{R}\) obeying 3 and 4 (so that \(P(h)\) is self-adjoint with respect to the domain \(H^2(\mathbb{R}^n)\)). Suppose there exists \(C > 0\) so that for all \(v \in C^\infty_0(\mathbb{R}^n)\), \[\label{wtd32est32appendix} \| \langle x \rangle^{-s_1} v \|^2_{L^2} \le C \| \langle x \rangle^{s_2} (P(h) - z) v \|^2_{L^2}.\tag{66}\] Then \[\label{resolv32est32appendix} \| \langle x \rangle^{-s_1} (P(h) - z)^{-1} \langle x \rangle^{-s_2} \|_{L^2 \to L^2} \le C.\tag{67}\]

Proof. The operator \[[P(h), \langle x \rangle^{s_2}]\langle x \rangle^{-s_2} = \left(-h^2( \Delta \langle x \rangle^{s_2}) - 2h^2 (\nabla \langle x \rangle^{s_2}) \cdot \nabla \right) \langle x \rangle^{-s_2}\] is bounded \(H^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\). So, for \(v \in H^2(\mathbb{R}^n)\) such that \(\langle x \rangle^{s_2} v \in H^2(\mathbb{R}^n)\), \[\label{Czh} \begin{align} \|\langle x \rangle^{s_2}(P(h)-z)v\|_{L^2} &\le \|(P(h)-z)\langle x \rangle^{s_2} v \|_{L^2} + \|[P(h),\langle x \rangle^{s_2}]\langle x \rangle^{-s_2}\langle x \rangle^{s_2}v \|_{L^2} \\& \le C_{z,h} \| \langle x \rangle^{s_2}v \|_{H^2}, \end{align}\tag{68}\] for some constant \(C_{z, h} >0\) depending on \(z\) and \(h\).

Given \(f \in L^2(\mathbb{R}^n)\), the function \(u= \langle x \rangle^{s_2}(P(h)-z)^{-1}\langle x \rangle^{-s_2} f \in H^2(\mathbb{R}^n)\) because \[u = (P(h) - z)^{-1} (f + w), \qquad w = [P(h), \langle x \rangle^{s_2}] u,\] with \([P(h), \langle x \rangle^{s_2}]\) being bounded \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\) since \(s_2 < 1\).

Now, choose a sequence \(v_k \in C_{0}^\infty\) such that \(v_k \to \langle x \rangle^{s_2}(P(h)-z)^{-1}\langle x \rangle^{-s_2} f\) in \(H^2(\mathbb{R}^n)\). Define \(\tilde{v}_k \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\langle x \rangle^{-s_2}v_k\). Then, as \(k \to \infty\), \[\begin{align} \| \langle x \rangle^{-s_1} \tilde{v}_k - \langle x \rangle^{-s_1} (&P(h)-z)^{-1}\langle x \rangle^{-s_2}f \|_{L^2} \\ &\le \| v_k - \langle x \rangle^{s_2} (P(h)-z)^{-1}\langle x \rangle^{-s_2}f \|_{H^2} \to 0. \end{align}\] Also, applying 68 , \[\|\langle x \rangle^{s_2}(P(h)-z)\tilde{v}_k - f\|_{L^2} \le C_{z,h} \|v_k - \langle x \rangle^{s_2} (P(h)-E \pm i \varepsilon)^{-1} \langle x \rangle^{-s_2} f \|_{H^2} \to 0.\] Thus 67 follows by replacing \(v\) by \(\tilde{v}_k\) in 66 and sending \(k \to \infty\).
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7 Justification of Remark 2↩︎

In the setting of Theorem 1, consider the case of \(V = 0\) and \(n =3\). In that scenario the integral kernel of \((P(h) - z)^{-1} = (-h^2 \Delta -z)^{-1}\) with \(z \in \mathbb{C}\setminus [0, \infty)\) is given by \[R_0(x,y,z) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =h^{-2} \frac{e^{i \tfrac{\sqrt{z}}{h}|x -y|}}{4\pi |x -y|}, \qquad \mathop{\rm Im}\sqrt{z} > 0.\] We recall why having a bound like 8 on \(\langle \cdot \rangle^{-s_1}(-h^2 \Delta -z)^{-1}\langle \cdot \rangle^{-s_2}: L^2(\mathbb{R}^3) \to L^2(\mathbb{R}^3)\) requires \(s_1, \, s_2 > 1/2\).

Since the norm of an operator and its adjoint coincide, it suffices to show \(s_1 > 1/2\) is necessary. Use \(\sqrt{z}\) of the form \(\sqrt{z} = E + i\varepsilon\) for \(E > 0\) fixed and \(\varepsilon> 0\) tending to zero. Then, as calculated in the proof of [30], for \(f \in C^\infty_0(\mathbb{R}^3)\), \[\label{outgoing32asymptotic} \begin{align} \langle x \rangle^{-s_1} \int_{\mathbb{R}^3} &R_0(x,y,z) f(y) dy \\ &= h^{-2} \frac{\langle x \rangle^{-s_1}}{4\pi |x|} e^{\frac{i}{h}(E+ i\varepsilon)|x|} (\hat{f} \big( \tfrac{E}{h} \tfrac{x}{|x|} \big) + o(1)) + O(|x|^{-2}), \quad \text{as \varepsilon\to 0^+ and |x| \to \infty.} \end{align}\tag{69}\] If \(f\) is chosen so that \(|\hat{f}| > c\) for some \(c > 0\) on \(\{|x| = E/h\}\), 69 and \(s_1 \le 1/2\) imply
\(\| \langle x \rangle^{-s_1} \textstyle\int_{\mathbb{R}^3} R_0(x,y,z) f(y) dy \|_{L^2} \to \infty\) as \(\varepsilon\to 0^+\).

Next, supposing \(s_1, \, s_2 > 1/2\), we show why a bound like 9 on \(\langle \cdot \rangle^{-s_1}(-h^2 \Delta -z)^{-1}\langle \cdot \rangle^{-s_2}: L^2(\mathbb{R}^3) \to L^2(\mathbb{R}^3)\) requires additionally that \(s_1 + s_2 \ge 2\), which is nearly the condition we impose for 9 . Using \(\sqrt{z} = i \varepsilon\) for \(\varepsilon> 0\) tending to zero, and \(f_\eta(y) = \langle y \rangle^{-\eta- \frac{3}{2}}\), \(\eta > 0\), we see as in [32] that \[\begin{align} \langle x \rangle^{-s_1} \int_{\mathbb{R}^3} &R_0(x,y,z) \langle y \rangle^{-s_2} f(y) \\ &= h^{-2} \langle x \rangle^{-s_1} \int_{\mathbb{R}^3} \frac{e^{-\frac{\varepsilon}{h}|x-y|}}{4\pi |x- y|} \langle y \rangle^{-s_2} f(y) dy \\ &\gtrsim h^{-2} e^{-\frac{3\varepsilon}{2h} |x|} \langle x \rangle^{-s_1-1} \int_{|y| \le \frac{|x|}{2}} \langle y \rangle^{-s_2- \eta - \frac{3}{2}} dy \gtrsim h^{-2} e^{-\frac{3\varepsilon}{2h} |x|} \langle x \rangle^{-s_1 - s_2- \eta + \frac{1}{2}}, \end{align}\] where the implicit constants indicated by \(\gtrsim\) are independent of \(\varepsilon\) and \(\eta\). First sending \(\varepsilon\to 0^+\) gives \(s_1 + s_2 \ge 2 -\eta\), but since \(\eta > 0\) is arbitrary, we in turn get \(s_1 + s_2 \ge 2\).

To see that the \(O(|z|^{-\frac{1}{2}} h^{-1})\)-dependence of the right side of 8 is optimal, consider the function \(u = e^{i \frac{\sqrt{z}}{h} x_1} \chi\) for nontrivial \(\chi \in C^\infty_0(\mathbb{R}^3 ; [0, 1])\). We have \[\langle x \rangle^{s}(-h^2 \Delta -z)u = -i \sqrt{z} h \langle x \rangle^{s} \partial_{x_1} \chi - h^2 e^{i \frac{\sqrt{z}}{h}x_1} \langle x \rangle^s \Delta \chi =\mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}}f,\] whence \(\langle \cdot \rangle^{-s} (-h^2 \Delta - z)^{-1} \langle \cdot \rangle^{-s} f = \langle \cdot \rangle^{-s}u\) and thus, as \(h \to 0\), \[\frac{\|\langle \cdot \rangle^{-s} (-h^2 \Delta - z)^{-1} \langle \cdot \rangle^{-s} f\|_{L^2}}{\|f \|_{L^2} } = \frac{\| \langle \cdot \rangle^{-s} u\|_{L^2}}{\|f \|_{L^2} } \gtrsim |z|^{-\frac{1}{2}} h^{-1}.\]

Finally, we argue why the \(O(h^{-2})\)-dependence of the right side of 9 is sharp. As noted before [29] gives that \(R_{0, s_1, s_2}(\lambda) = \langle \cdot \rangle^{-s_1}(-h^2 \Delta -\lambda^2)^{-1}\langle \cdot \rangle^{-s_2}\) (\(s_1,\, s_2 > 1/2\), \(s_1 + s_2 > 2\)), has extends continuously from \(\mathop{\rm Im}\lambda > 0\) to \(\mathbb{R}\) in the space of bounded opeartors \(\mathbb{R}^3 \to \mathbb{R}^3\). In this case, we have, \[h^{-2}\big\| \langle x \rangle^{-s_1} \int_{\mathbb{R}^3} \frac{1}{4\pi |x -y|}\langle y \rangle^{-s_2}dy \big \|_{L^2 \to L^2} = \lim_{\varepsilon\to 0} \| R_{0, s_1, s_2}(i\varepsilon)\|_{L^2 \to L^2}.\]

8 Proof of Lemma 2↩︎

In this appendix we prove Lemma 2. The proof proceeds in two steps. First we treat the case \(|\lambda| \ge 1\), followed by \(|\lambda| \le 1\).

Proof of lemma 2. Initially we take \(j = 0\) in 48 , so may assume without loss of generality that \(\mathop{\rm Im}\lambda > 0\). We treat the \(j = 1\) case at the end of the proof. Observe that \(\tfrac{d}{d\lambda} \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-1} \langle x \rangle^{-s} = 2\lambda \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s}\), so we can bound the \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\) norm of either quantity.

If \(|\lambda| \ge 1\), begin from \[\label{identity32to32insert32Laplacian} \begin{align} -\lambda^2 \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} &= - \tfrac{1}{2} \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-1}(-2\Delta) (-\Delta - \lambda^2)^{-1} \langle x \rangle^{-s} \\ &+\langle x \rangle^{-s} (-\Delta - \lambda^2)^{-1} \langle x \rangle^{-s}. \end{align}\tag{70}\] By 44 , the \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\) norm of the second line of 70 is bounded by \(C(1 + | \mathop{\rm Re}\lambda|)^{-1}\). So it suffices to investigate the term on the right side of the first line of 70 . For notational brevity, put \(R_0(\lambda) \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =(-\Delta - \lambda^2)^{-1}\). We show that for all \(f \in C^\infty_0(\mathbb{R}^n)\), \[\label{A32inv32Delta32A32inv} \begin{align} \langle x \rangle^{-s} &R_0(\lambda)(-2 \Delta) R_0(\lambda) \langle x \rangle^{-s} f \\ &= -\langle x \rangle^{-s} R_0(\lambda) \partial_r (r \langle x \rangle^{-s} f) + \langle x \rangle^{-s} R_0(\lambda) \langle x \rangle^{-s}f \\ &+ \langle x \rangle^{-s} r \partial_r R_0(\lambda) \langle x \rangle^{-s}f \end{align}\tag{71}\] Since \(s > 3/2\) by 16 , the \(L^2(\mathbb{R}^n)\)-norm of the right side of 71 is bounded by \(C \| f\|_{L^2}\), thanks to 44 . So it remains to show 71 .

Recall the well known formula for the Laplacian in polar coordinates, \[\Delta = \partial^2_r + (n-1)r^{-1} \partial_r + r^{-2} \Delta_{\mathbb{S}^{n-1}},\] which implies the commutator identity \[\label{commutator32identity} [r \partial_r, \Delta] \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =r \partial_r (\Delta) - \Delta(r \partial_r) = -2 \Delta.\tag{72}\] Fix \(f \in C^\infty_0(\mathbb{R}^n)\), \(g \in L^2(\mathbb{R}^n)\), and put \(u \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =R_0(\lambda) \langle x \rangle^{-s} f \in H^2(\mathbb{R}^n)\). Let \(\{u_k\}_{k=1}^\infty \subseteq C^\infty_0(\mathbb{R}^n)\) be a sequence converging to \(u\) in \(H^2(\mathbb{R}^n)\). Starting from the left side of 71 and applying 72 , \[\label{long32calc321} \begin{align} \langle g, \langle x \rangle^{-s} R_0(\lambda)(-2 \Delta) &R_0(\lambda) \langle x \rangle^{-s}f \rangle_{L^2} \\ &= \lim_{k \to \infty} \langle g, \langle x \rangle^{-s} R_0(\lambda)[r\partial_r, \Delta] u_k \rangle_{L^2}\\ &= \langle g, \langle x \rangle^{-s} r \partial_r u \rangle_{L^2} \\ &- \lim_{k \to \infty} \langle g, \langle x \rangle^{-s} R_0(\lambda) r \partial_r (-\Delta - \lambda^2) u_k\rangle_{L^2}. \end{align}\tag{73}\]

The purpose of the following calculations is to show that the last line of 73 equals
\(-\langle g, \langle x \rangle^{-s} R_0(\lambda) r \partial_r \langle x \rangle^{-s} f \rangle_{L^2}\). First, for any \(v \in L^2(\mathbb{R}^n)\), \(r R_0(\lambda) \langle x \rangle^{-1} v \in H^1(\mathbb{R}^n)\). This holds because, if we put \(w \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =\langle x \rangle R_0(\lambda) \langle x \rangle^{-1} v\), then \(r R_0(\lambda) \langle x \rangle^{-1} v = r \langle x \rangle^{-1} w\) and \[\begin{gather} (-\Delta - \lambda^2) w = [- \Delta , \langle x \rangle] R_0(\lambda) \langle x \rangle^{-1} v + v \implies \\ w = R_0(\lambda)([- \Delta , \langle x \rangle] R_0(\lambda) \langle x \rangle^{-1} v + v) \in H^2(\mathbb{R}). \end{gather}\] Furthermore, for any \(w, v \in C^\infty_0(\mathbb{R}^n)\), \[\langle w, \partial_r v \rangle_{L^2} = \langle \partial_r^* w, v \rangle_{L^2} \mathrel{\vcenter{\baselineskip 0.5ex \lineskiplimit 0pt \scriptsize.\scriptsize.}} =(1-n) \langle r^{-1} w, v \rangle_{L^2} - \langle \partial_r w, v \rangle_{L^2}.\] Therefore, by the density of \(C^\infty_0(\mathbb{R}^n)\) in \(H^1(\mathbb{R}^n)\), and setting \(\tilde{u}_{k} = (-\Delta - \lambda) u_k\), we get \[\label{long32calc322} \begin{align} \lim_{k \to \infty} \langle g, \langle& x \rangle^{-s} R_0(\lambda) (r \partial_r) \tilde{u}_k \rangle_{L^2}\\ &= \lim_{k \to \infty} \langle (\partial_r)^* r R_0(\overline{\lambda}) \langle x \rangle^{-s} g, \tilde{u}_k \rangle_{L^2}\\ &= \langle (\partial_r)^* r R_0(\overline{\lambda}) \langle x \rangle^{-s} g, \langle x \rangle^{-s} f \rangle_{L^2}\\ &= \langle g, \langle x \rangle^{-s} R_0(\lambda) r \partial_r \langle x \rangle^{-s} f \rangle_{L^2}. \end{align}\tag{74}\] as desired. Taken together, 73 and 74 confirm 71 .

Now we turn to the case \(|\lambda | \le 1\), and utilize the integral kernel of the free resolvent, which is given by [33], \[(-\Delta-\lambda^2)^{-1}(|x-y|)=\frac{1}{2\pi}\left(\frac{-i\lambda}{2\pi|x-y|}\right)^{\frac{n}{2}-1}K_{\frac{n}{2}-1}(-i\lambda|x-y|), \qquad \mathop{\rm Im}\lambda > 0,\] where \(K_\nu(z)\) is the Macdonald function of order \(\nu\) [28]. Now, if \(n=3\), then the integral kernel of \(\langle x\rangle^{-s}\frac{d}{d\lambda}(-\Delta-\lambda^2)^{-1}\langle x\rangle^{-s}\) is given by \(i(4\pi)^{-1}\langle x\rangle^{-s}e^{i\lambda|x-y|}\langle y\rangle^{-s}\), which has Hilbert-Schmidt norm bounded uniformly in \(|\lambda| \le 1\) provided \(s>3/2.\) Moving on to \(n \ge 4\), by [28], \[\frac{d}{d\lambda}\left(\frac{-i\lambda}{2\pi|x-y|}\right)^{\frac{n}{2}-1}K_{\frac{n}{2}-1}(-i\lambda|x-y|)=\frac{-(-i)^{\frac{n}{2}}\lambda^{\frac{n}{2}-1}}{(2\pi)^{\frac{n}{2}-1}|x-y|^{\frac{n}{2}-2}}K_{\frac{n}{2}-2}(-i\lambda|x-y|).\] The Macdonald function satisfies [28] \[|K_{\nu}(z)| \le \begin{cases} C|z|^{-\nu} & 0 < |z| \le 1, \, \nu > 0, \\ C |\ln |z|| & 0 < |z| \le 1, \, \nu = 0, \\ C|z|^{-1/2} & |z| \ge 1, \, \mathop{\rm Re}z \ge 0, \end{cases}\] for \(C > 0\) a constant independent of \(z\). Therefore, for \(C > 0\) independent of \(\lambda\), \[\label{bd32deriv32kernel32low32freq} \begin{align} \Big| &\frac{\lambda^{\frac{n}{2}-1} \langle x \rangle^{-s} \langle y \rangle^{-s}}{|x-y|^{\frac{n}{2}-2}}K_{\frac{n}{2}-2}(-i\lambda|x-y|)\Big| \\ &\le \begin{cases} C |\lambda| \langle x \rangle^{-s} \langle y \rangle^{-s} |\ln (|\lambda||x-y|)| \mathbf{1}_{\{|\lambda||x-y|\leq 1\}}+C\frac{|\lambda|^{\frac{n}{2}-\frac{3}{2}} \langle x \rangle^{-s} \langle y \rangle^{-s}}{|x-y|^{\frac{n}{2}- \frac{3}{2}}} \mathbf{1}_{\{|\lambda||x-y|> 1\}} & n = 4, \\ C\frac{|\lambda|\langle x \rangle^{-s} \langle y \rangle^{-s}}{|x-y|^{n-4}} \mathbf{1}_{\{|\lambda||x-y|\leq 1\}}+C\frac{|\lambda|^{\frac{n}{2}-\frac{3}{2}}\langle x \rangle^{-s} \langle y \rangle^{-s}}{|x-y|^{\frac{n}{2}- \frac{3}{2}}} \mathbf{1}_{\{|\lambda||x-y|> 1\}} & n > 4. \end{cases} \end{align}\tag{75}\] As preparation for the conclusions we draw in the next paragraph, we observe that the first term in line three of 75 has the bound \[\label{HS32trick} \frac{|\lambda|\langle x \rangle^{-s} \langle y \rangle^{-s}}{|x-y|^{n-4}} \mathbf{1}_{\{|\lambda||x-y|\leq 1\}} = \frac{|\lambda| |x-y|^{\alpha}\langle x \rangle^{-s} \langle y \rangle^{-s}}{|x-y|^{n-4 + \alpha}} \mathbf{1}_{\{|\lambda||x-y|\leq 1\}} \le \frac{\langle x \rangle^{-s} \langle y \rangle^{-s}}{|x-y|^{n-4 + \alpha}} \mathbf{1}_{\{|\lambda||x-y|\leq 1\}}.\tag{76}\] for any \(0 < \alpha \le 1\).

In what follows we make repeated use of Lemma 9. In 75 , the second term in line two and the second term in line three are are uniformly bounded in Hilbert-Schmidt norm for \(|\lambda| \le 1\), provided \(s > (n + 3)/4\). This also holds for \[\begin{gather} \text{the first term in line two if s > 3/2,}\\ \text{the first term in line three if s > 3/2 and n = 5 (by \eqref{HS32trick} with \alpha = 1),}\\ \text{the first term in line three if s > 3/2 and n = 6 (by \eqref{HS32trick} with \alpha so that s > 2 - (\alpha/2), and}\\ \text{the first term in line three if s > 7/4 and n =7 (by \eqref{HS32trick} with \alpha so that s > 2 - (\alpha/2)).} \end{gather}\] Finally, if \(n > 8\), the first term in line three is uniformly bounded \(L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)\) for \(|\lambda| \le 1\) provided \(s > 3\). This is due to the Schur test, see Lemma 8.

We finish by resolving the \(j = 1\) case for 48 . By 48 in the \(j = 0\) case, and by 46 , we need to show \(\| \lambda \langle x \rangle^{-s}(-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} f \|_{H^2} \le O(1 + |\lambda|) \| f\|_{L^2}\). According to 77 below, \[\begin{align} \| \lambda& \langle x \rangle^{-s}(-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} f \|_{H^2} \\ &\le C \| \lambda \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} f \|_{L^2} + C\| \lambda \langle x \rangle^{-s}( -\Delta) (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} f \|_{L^2}\\ &= C \| f\|_{L^2} + C\| \lambda \langle x \rangle^{-s}( -\Delta) (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} f \|_{L^2}. \end{align}\] Then use \[\begin{align} \langle x \rangle^{-s}&( -\Delta) (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} f \\ &= \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-1} \langle x \rangle^{-s} f + \lambda^2 \langle x \rangle^{-s} (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s}f, \end{align}\] which in combination with 44 , as well as 48 in the \(j = 0\) case yields \[\| \lambda \langle x \rangle^{-s}( -\Delta) (-\Delta - \lambda^2)^{-2} \langle x \rangle^{-s} f \|_{L^2} \le C(1 + |\lambda|) \| f_{L^2},\] completing the proof.
 ◻

9 Useful lemmas↩︎

Lemma 7 ([34]). Let \(n \ge 3\). Then, \[\| r^{-1} u \|^2_{L^2} \le \Big( \frac{2}{n -2} \Big)^2 \| \nabla u \|^2_{L^2}, \qquad u \in H^1(\mathbb{R}^n).\]

Lemma 8 (Schur’s test [30]). Suppose that \(K(x,y)\) is measurable on \(\mathbb{R}^n \times \mathbb{R}^n\) and \[\sup_x \int |K(x,y)| dy, \, \sup_y \int |K(x,y)| dy \le C.\] Then the linear operator \[Tf(x) = \int K(x,y) f(y) dy,\] obeys the estimate \[\| T f \|_{L^2} \le C \|f \|_{L^2}.\]

Lemma 9 ([35]). The necessary and sufficient conditions for \[\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \langle x \rangle^{-s} \langle y \rangle^{-t} |x -y|^{-p} dx dy < \infty,\] are \[s + p > n, \quad t + p > n, \quad s + p + t > 2n, \quad p < n.\]

Lemma 10. Suppose \(T : L^2(\mathbb{R}^n) \to H^2(\mathbb{R}^n)\) is a bounded operator. For any \(s > 0\), there exists \(C > 0\) so that \[\label{recast32L232to32H232bd} \| \langle x \rangle^{-s} T \|_{L^2 \to H^2} \le C(\| \langle x \rangle^{-s} T \|_{L^2 \to L^2} + \| \langle x \rangle^{-s} \Delta T \|_{L^2 \to L^2}).\tag{77}\]

Proof. Let \(f \in L^2(\mathbb{R}^n)\) and put \(u = Tf\). By the first line of 46 , there exists \(C > 0\), whose precise value may change from line to line, so that \[\label{apply32std32elliptic32thry} \| \langle x \rangle^{-s} \tilde{u} \|_{H^2} \le C \| \langle x \rangle^{-s} \tilde{u} \|_{L^2} + C\| \Delta \langle x \rangle^{-s} \tilde{u} \|_{L^2}, \qquad \tilde{u} \in H^2(\mathbb{R}^n).\tag{78}\] Then use the second line of 46 , \[\begin{align} \| \Delta \langle x \rangle^{-s} u \|_{L^2} &\le \|[\Delta, \langle x \rangle^{-s}] u \|_{L^2} + \|\langle x \rangle^{-s} \Delta u \|_{L^2} \\ &\le C\|\langle x \rangle^{-s} u \|_{H^1} + \|\langle x \rangle^{-s} \Delta u \|_{L^2} \\ &\le C \gamma^{-1} \| \langle x \rangle^{-s} u \|_{L^2} + C\gamma \| \Delta \langle x \rangle^{-s} u \|_{L^2}) +\|\langle x \rangle^{-s} \Delta u \|_{L^2}, \qquad \gamma > 0. \end{align}\] Fixing \(\gamma\) small enough yields, \[\| \Delta \langle x \rangle^{-s} u \|_{L^2} \le C(\| \langle x \rangle^{-s} u \|_{L^2} +\|\langle x \rangle^{-s} \Delta u \|_{L^2}),\] which in combination with 78 implies 77 .
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