A Spectral Multiplier Theorem Associated with a Schrödinger Operator

Abstract

We establish a Hörmander type spectral multiplier theorem for a Schrödinger operator $H=-\Delta +V(x)$ in $\mathbb {R}^3$, provided V is contained in a large class of short range potentials. This result does not require the Gaussian heat kernel estimate for the semigroup $e^{-tH}$, and indeed the operator H may have negative eigenvalues. As an application, we show local well-posedness of a 3d quintic nonlinear Schrödinger equation with a potential.

1 Introduction

1.1 Statement of the Main Theorem

In this paper, we establish a Hörmander type spectral multiplier theorem for a Schrödinger operator $H=-\Delta +V$ in $\mathbb {R}^3$, provided that V is contained in a large class of short range potentials. Precisely, we assume that V is contained in $\mathcal {K}_0\cap L^{3/2,\infty }$, where $\mathcal {K}_0$ is the norm closure of bounded, compactly supported functions with respect to the global Kato norm

$$\begin{aligned} \Vert V\Vert _{\mathcal {K}}:=\sup _{x\in \mathbb {R}^3} \int _{\mathbb {R}^3}\frac{|V(y)|}{|x-y|}dy, \end{aligned}$$

(1.1)

and $L^{3/2,\infty }$ is the weak $L^{3/2}$-space. We also assume that H has no eigenvalue or resonance on the positive real-line $[0,+\infty )$. By a resonance, we mean a complex number $\lambda$ such that the equation $\psi +(-\Delta -\lambda \pm i0)^{-1}V\psi =0$ has a slowly decaying solution $\psi \in L^{2,-s}\setminus L^2$ for any $s>\frac{1}{2}$, where $L^{2,s}=\{\langle x\rangle ^s f\in L^2\}$.

By the above assumptions, the operator H is self-adjoint on $L^2$. Moreover, its spectrum $\sigma (H)$ consists of purely absolutely continuous spectrum on the positive real-line $[0,+\infty )$ and at most finitely many negative eigenvalues [2]. Therefore, for a bounded Borel function $m:\sigma (H)\subset \mathbb {R}\rightarrow \mathbb {C}$, one can define a spectral multiplier m(H) as a bounded operator on $L^2$ via functional calculus.

A natural question is then to find a sufficient condition to extend boundedness of the multiplier m(H) to $L^p$ for $p\ne 2$. Such a condition is typically given in terms of regularity of symbols. To measure regularity of a symbol $m:\sigma (H)\rightarrow \mathbb {C}$, we define a Sobolev type norm by

$$\begin{aligned} \Vert m\Vert _{\mathcal {H}(s)}:=\sum _{\lambda _j: \mathrm{negative\;eigenvalues}} |m(\lambda _j)|+\sup _{t>0}\Vert \chi (\lambda )m((t\lambda )^2)\Vert _{W_\lambda ^{s,2}((0,+\infty ))}, \end{aligned}$$

(1.2)

where $\chi \in C_c^\infty (\mathbb {R})$ is a standard dyadic partition of unity function such that $\chi$ is supported in $[\frac{1}{2},2]$ and $\sum _{N\in 2^{\mathbb {Z}}}\chi (\tfrac{\cdot }{N})\equiv 1$ on $(0,+\infty )$, and $W^{s,2}$ is the $L^2$-Sobolev space of order s.

Our main result is the following.

Theorem 1.1

(Spectral multiplier theorem) Suppose that $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and $H=-\Delta +V$ has no eigenvalue or resonance on $[0,+\infty )$. We also assume that for $s>2$, the symbol $m:\sigma (H)\rightarrow \mathbb {C}$ satisfies $\Vert m\Vert _{\mathcal {H}(s)}<\infty$. Then, we have

$$\begin{aligned} \Vert m(H)\Vert _{L^p\rightarrow L^p}\lesssim \Vert m\Vert _{\mathcal {H}(s)},\quad \forall 1<p<\infty . \end{aligned}$$

(1.3)

When $V=0$, Theorem 1.1 is simply the classical Hörmander–Mikhlin multiplier theorem [4].

There are several ways to prove the spectral multiplier theorem for Schrödinger operators. For an operator A, we say that the semigroup $e^{-tA}$ satisfies the Gaussian heat kernel estimate if the kernel of $e^{-tA}$, denoted by $e^{-tA}(x,y)$, obeys

$$\begin{aligned} e^{-tA}(x,y)\lesssim t^{-3/2}e^{-\frac{|x-y|^2}{ct}},\quad \forall t>0 \end{aligned}$$

(1.4)

for some $c>0$. Gaussian upper bounds for the heat kernels have been used successfully to prove spectral multiplier theorems for rather general operators, not necessarily Schrödinger operators (see [4, 5, 16] and references therein). In the case of the Schrödinger operator $H=-\Delta +V$ in $\mathbb {R}^3$, if $V_+=\max (V,0)$ is in local Kato class, that is,

$$\begin{aligned} \lim _{r\rightarrow 0+}\sup _{x\in \mathbb {R}^3}\int _{|x-y|\le r}\frac{|V_+(y)|}{|x-y|}dy=0, \end{aligned}$$

(1.5)

and if $V_-=\min (V,0)\in \mathcal {K}_0$ and $\Vert V_-\Vert _{\mathcal {K}}<4\pi$, then it is known that the semigroup $e^{-tH}$ satisfies the Gaussian heat kernel estimate (1.4) [7, 20]. The spectral multiplier theorem for H then follows from [5, Theorem 3.1]. However, for Gaussian upper bounds (1.4), operators need to be positive definite, while the Schrödinger operator in Theorem 1.1 may have negative eigenvalues.

One can also use the wave operators to show the spectral multiplier theorem. The forward-in-time (backward-in-time, resp) wave operator of the Schrödinger operator $H=-\Delta +V$ is defined by

$$\begin{aligned} W_+=\underset{t\rightarrow +\infty }{s\text{- }\lim }e^{itH}e^{-it(-\Delta )}\quad \Big (W_-=\underset{t\rightarrow -\infty }{s\text{- }\lim }e^{itH}e^{-it(-\Delta )}{, \mathrm{resp}}\Big ). \end{aligned}$$

(1.6)

An important feature of wave operators is its intertwining property, that is, $P_cf(H)=W_\pm f(-\Delta )(W_\pm )^*$, where $P_c$ is the spectral projection to the continuous spectrum and $(W_\pm )^*$ is the dual of $W_\pm$. In [22], Yajima proved that the wave operators $W_\pm$ are bounded on $L^p$ for all $1\le p\le \infty$, provided that $|V(x)|\lesssim \langle x\rangle ^{-5-\epsilon }$ for $\epsilon >0$, and zero is not an eigenvalue or a resonance of H. Later, in [1], Beceanu extended this result to a larger space

$$\begin{aligned} B:=\left\{ V: \sum _{k=-\infty }^\infty 2^{k/2}\Vert V(x)\Vert _{L_x^2(2^k\le |x|<2^{k+1})}<\infty \right\} . \end{aligned}$$

(1.7)

The spectral multiplier theorem then follows immediately from the intertwining property and boundedness of wave operators and the classical Hörmander–Mikhlin multiplier theorem, since

$$\begin{aligned} \Vert P_cf(H)\Vert _{L^p\rightarrow L^p}&=\Vert W_\pm f(-\Delta )(W_\pm )^*\Vert _{L^p\rightarrow L^p} \nonumber \\&\lesssim \Vert f(-\Delta )(W_\pm )^*\Vert _{L^p\rightarrow L^p}\lesssim \Vert (W_\pm )^*\Vert _{L^p\rightarrow L^p}<\infty \end{aligned}$$

(1.8)

and $(I-P_c)f(H)$ is bounded on $L^p$ by Lemma 3.6. Theorem 1.1 improves the spectral multiplier theorem as a consequence boundedness of the wave operator, in that the potential class $\mathcal {K}_0\cap L^{3/2,\infty }$ is larger than the potential class B. Note that a potential having many singular points, such as $\sum _{k=1}^N 1_{|x-x_j|\le 1}\frac{1}{|x-x_j|^{2-\epsilon }}$ with $x_j\ne x_k$ and $\epsilon >0$, is contained in $\mathcal {K}_0\cap L^{3/2,\infty }$, but not in B.

Our proof of the spectral multiplier theorem is perturbative, and it relies heavily on the explicit integral representation of the kernel of the multiplier. We consider the spectral multiplier $m(H)P_c$ as a perturbation of the Fourier multiplier $m(-\Delta )$, and then we show that the difference $(m(H)P_c-m(-\Delta ))$ is bounded on $L^p$. In order to estimate the difference, we first decompose it into its dyadic pieces

$$\begin{aligned} \sum _{N\in 2^{\mathbb {Z}}}\chi \left( \tfrac{\sqrt{H}}{N}\right) \Big (m(H)-m(-\Delta )\Big ), \end{aligned}$$

(1.9)

where $\chi$ is the function given in (1.2). Then, we generate a formal series expansion for each dyadic piece to get explicit integral representations of kernels of terms in the series using the free resolvent formula

$$\begin{aligned} ((-\Delta -z)^{-1}f)(x)=\int _{\mathbb {R}^3}\frac{e^{i\sqrt{z}|x-y|}}{4\pi |x-y|} f(y)dy. \end{aligned}$$

(1.10)

We estimate these integral kernels. Summing them up, we prove the spectral multiplier theorem.

A key observation is that in spite of the singular integral nature of both $m(H)P_c$ and $m(-\Delta )$ as Calderon–Zygmund operators, the kernel of their difference is less singular than usual Calderon–Zygmund operators. This fact is essential in our analysis, since it allows us to avoid using the delicate classical Calderon–Zygmund theory for the complicated operator m(H) (see Remark 4.4). Instead, we just make use of the fractional integration inequality and Hölder inequality.

1.2 Application to NLS

The choice of the potential class in the main theorem is motivated by the following nonlinear application.

First, we recall the Strichartz estimates for the linear propagator $e^{-itH}$.

Proposition 1.2

(Strichartz estimates) If $V\in \mathcal {K}_0$ and H has no eigenvalue or resonance on $[0,+\infty )$, then

$$\begin{aligned} \Vert e^{-itH}P_cf\Vert _{L_t^q L_x^r}&\lesssim \Vert f\Vert _{L^2},\end{aligned}$$

(1.11)

$$\begin{aligned} \Big \Vert \int _0^t e^{-i(t-s)H}P_cF(s)ds\Big \Vert _{L_t^q L_x^r}&\lesssim \Vert F\Vert _{L_t^{2}L_x^{6/5}}, \end{aligned}$$

(1.12)

where $\frac{2}{q}+\frac{3}{r}=\frac{3}{2}$ and $2\le q,r\le \infty$.

Proof

Beceanu–Goldberg [2] proved the dispersive estimate

$$\begin{aligned} \Vert e^{-itH}P_c\Vert _{L^1\rightarrow L^\infty }\lesssim |t|^{-3/2}, \end{aligned}$$

(1.13)

where $P_c$ is the spectral projection to the continuous spectrum. Strichartz estimates then follow by the argument of Keel–Tao [15]. $\square$

Remark 1.3

The dispersive estimate of the form (1.13) was first proved by Journé–Soffer–Sogge under suitable assumptions on potentials [14]. The assumptions have been relaxed by Rodnianksi–Schlag [17], Goldberg–Schlag [10] and Goldberg [8, 9]. Recently, Beceanu–Goldberg established (1.13) for a scaling-critical potential class $\mathcal {K}_0$ [2].

An interesting question is then whether one can use the above Strichartz estimates to show the local well-posedness (LWP), for instance, for a 3d quintic nonlinear Schrödinger equation with a potential

$$\begin{aligned} iu_t+\Delta u-Vu\pm |u|^4u=0;\ u(0)=u_0 \qquad \qquad (({\mathrm{NLS}}_V)) \end{aligned}$$

assuming that V satisfies the conditions in Proposition 1.2. However, if one tries to show local well-posedness by the standard contraction mapping argument as in [4, 21], one will realize that there is a subtle problem, mainly because the linear propagator $e^{-itH}$ does not commute with the differential operators from the Sobolev norms.

We overcome this subtle problem by the two norm estimates lemma, whose proof relies on the spectral multiplier theorem.

Lemma 1.4

(Two norm estimates) If $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on the positive real-line $[0,+\infty )$, then

$$\begin{aligned} \Vert H^\frac{s}{2}P_c(-\Delta )^{-\frac{s}{2}}f\Vert _{L^r}&\lesssim \Vert f\Vert _{L^r},\end{aligned}$$

(1.14)

$$\begin{aligned} \Vert (-\Delta )^{\frac{s}{2}}H^{-\frac{s}{2}}P_cf\Vert _{L^r}&\lesssim \Vert f\Vert _{L^r}. \end{aligned}$$

(1.15)

for $0\le s\le 2$ and $1<r<\frac{3}{s}$.

Together with Strichartz estimates and the two norm estimates lemma, we prove local well-posedness.

Theorem 1.5

(LWP) Suppose that $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on the positive real-line $[0,+\infty )$. Then, $({\mathrm{NLS}}_V)$ is locally well-posed in $\dot{H}^1$.

Remark 1.6

• (i) The range of r in the two norm estimates lemma is sharp. See the counterexample in [19].

• (ii) The additional hypothesis $V\in L^{3/2,\infty }$, compared to Strichartz estimates, is from the two norm estimates lemma. In the proof of the two norm estimates lemma, we used this additional assumption.

1.3 Organization of the Paper

The outline of the proof of Theorem 1.1 is given in Sect. 2. We decompose the spectral representation of the difference $(m(H)P_c-m(-\Delta ))$ into the low, medium and high frequencies, and then analyze them separately in Sects. 4, 5 and 6. In Sect. 7, we establish LWP of a 3d quintic nonlinear Schrödinger equation with a potential.

1.4 Notations

For an integral operator T, its integral kernel is denoted by T(x, y). We denote by $A``=\text {''}B$ the formal identity which will be proved later.

2 Reduction to the Key Lemma

Suppose that $V\in \mathcal {K}_0$ and H has no eigenvalue or resonance on $[0,+\infty )$. Then, the spectrum of H, denoted by $\sigma (H)$, consists of purely continuous spectrum on the positive real-line $[0,+\infty )$ and at most finitely many negative eigenvalues. For $z\notin \sigma (H)$, we define the resolvent by $R_V(z):=(H-z)^{-1}$, and denote

$$\begin{aligned} R_V^\pm (\lambda ):=\underset{{\epsilon \rightarrow 0+}}{{\mathrm{s}-}\lim } R_V(\lambda \pm i\epsilon ). \end{aligned}$$

(2.1)

Let $P_c$ be the spectral projection on the continuous spectrum. Then, by the Stone’s formula, the spectral multiplier operator $m(H)P_c$ is represented by

$$\begin{aligned} m(H)P_c=\frac{1}{2\pi i}\int _0^\infty m(\lambda ) [R_V^+(\lambda )-R_V^-(\lambda )]d\lambda =\frac{1}{\pi }\int _0^\infty m(\lambda ){\text {Im}}R_V^+(\lambda )d\lambda . \end{aligned}$$

(2.2)

Applying the identity

$$\begin{aligned} R_V^+(\lambda )&=R_0^+(\lambda )(I+VR_0^+(\lambda ))^{-1} \nonumber \\&=R_0^+(\lambda )\Big (I-(I+VR_0^+(\lambda ))^{-1}VR_0^+(\lambda )\Big )\\&=R_0^+(\lambda )-R_0^+(\lambda )(I+VR_0^+(\lambda ))^{-1}VR_0^+(\lambda ),\nonumber \end{aligned}$$

(2.3)

we split $m(H)P_c$ into the pure and the perturbed parts,

$$\begin{aligned} m(H)P_c&=\frac{1}{\pi }\int _0^\infty m(\lambda ){\text {Im}}R_0^+(\lambda ) d\lambda \nonumber \\&\quad -\frac{1}{\pi }\int _0^\infty m(\lambda ){\text {Im}}[R_0^+(\lambda )(I+VR_0^+(\lambda ))^{-1}VR_0^+(\lambda )] d\lambda \\&=:m(-\Delta )+{\mathrm{Pb}},\nonumber \end{aligned}$$

(2.4)

where $m(-\Delta )$ is the Fourier multiplier such that $\widehat{m(-\Delta ) f}(\xi )=m(|\xi |^2)\hat{f}(\xi )$. For the pure part $m(-\Delta )$, it follows from the classical Hörmander–Mikhlin multiplier theorem [13] that for $s>\frac{3}{2}$,

$$\begin{aligned} \Vert m(-\Delta )\Vert _{L^p\rightarrow L^p}\lesssim \Vert m\Vert _{\mathcal {H}(s)},\quad \forall 1<p<\infty . \end{aligned}$$

(2.5)

Therefore, it suffices to show boundedness of the perturbed part. For the perturbed part ${\mathrm{Pb}}$, we further decompose it into dyadic pieces. Let $\chi$ be the smooth dyadic partition of unity function chosen in (1.2), and decompose

$$\begin{aligned} {\mathrm{Pb}}=\sum _{N\in 2^{\mathbb {Z}}}{\mathrm{Pb}}_N, \end{aligned}$$

(2.6)

where

$$\begin{aligned} {\mathrm{Pb}}_N:=-\frac{1}{\pi }\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}[R_0^+(\lambda )(I+VR_0^+(\lambda ))^{-1}VR_0^+(\lambda )]d\lambda . \end{aligned}$$

(2.7)

For a small dyadic number $N_0$ and a large dyadic number $N_1$ to be chosen later, we denote the low (high, resp) frequency part by

$$\begin{aligned} {\mathrm{Pb}}_{\le N_0}:=\sum _{N\le N_0}{\mathrm{Pb}}_N\ \left( {\mathrm{Pb}}_{\ge N_1}:=\sum _{N\ge N_1}{\mathrm{Pb}}_N{, \mathrm{resp}}\right) . \end{aligned}$$

(2.8)

In the next four sections, we will show the following lemma.

Lemma 2.1

(Key lemma) Suppose that $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on $[0,+\infty )$. Let $s>2$. Then, there exists $p>1$ but sufficiently close to 1 such that the following hold.

• (i) (High frequency) There exists $N_1=N_1(V)\gg 1$ such that

  $$\begin{aligned} \Vert {\mathrm{Pb}}_{\ge N_1}\Vert _{L^{p,1}\rightarrow L^{p,\infty }}\lesssim \Vert m\Vert _{\mathcal {H}(s)}, \end{aligned}$$

  (2.9)

  where $L^{p,1}$ and $L^{p,\infty }$ are the Lorentz spaces (see “Appendix”).

• (ii) (Low frequency) There exists $N_0=N_0(V)\ll 1$ such that

  $$\begin{aligned} \Vert {\mathrm{Pb}}_{\le N_0}\Vert _{L^{p,1}\rightarrow L^{p,\infty }}\lesssim \Vert m\Vert _{\mathcal {H}(s)}. \end{aligned}$$

  (2.10)

• (iii) (Medium frequency) For $N_0<N<N_1$,

  $$\begin{aligned} \Vert {\mathrm{Pb}}_N\Vert _{L^{p,1}\rightarrow L^{p,\infty }}\lesssim _{N_0,N_1} \Vert m\Vert _{\mathcal {H}(s)}. \end{aligned}$$

  (2.11)

Proof of Theorem 1.1, assuming Lemma 2.1

Let $p>1$ be sufficiently close to 1 as in Lemma 2.1. Summing the estimates in Lemma 2.1, we prove that ${\mathrm{Pb}}$ is bounded from $L^{p,1}$ to $L^{p,\infty }$. Then, it follows from the classical Hörmander–Mikhlin multiplier theorem that $m(H)P_c=m(-\Delta )+{\mathrm{Pb}}$ is bounded from $L^{p,1}$ to $L^{p,\infty }$. Moreover, by Lemma 3.6 (see below), $m(H): L^{p,1}\rightarrow L^{p,\infty }$ is bounded.

Recall that by functional calculus, m(H) is bounded on $L^2$. Thus, by the real interpolation lemma (Corollary 7.8), m(H) is bounded on $L^p$ for all $1<p\le 2$. Finally, applying the spectral multiplier theorem to the symbol $\bar{m}$ and the standard duality argument with $m(H)=\bar{m}(H)^*$, we conclude that m(H) is bounded on $L^p$ for $2<p<\infty$. $\square$

3 Preliminaries

3.1 Resolvent Estimates

Following Beceanu–Goldberg [2], we collect kernel estimates for $VR_0^+(\lambda )$, $V(R_0^+(\lambda )-R_0^+(\lambda _0))$, $(VR_0^+(\lambda ))^4$ and $(I+VR_0^+(\lambda ))^{-1}$, all of which will play as building blocks to analyze the kernel of ${\mathrm{Pb}}_N$.

Lemma 3.1

(Resolvent estimates) Suppose that $V\in \mathcal {K}_0$.

• (i) For $\lambda \ge 0$,

  $$\begin{aligned} \Vert VR_0^+(\lambda )f\Vert _{L^1}\le \frac{\Vert V\Vert _{\mathcal {K}}}{4\pi }\Vert f\Vert _{L^1}. \end{aligned}$$

  (3.1)

• (ii) Define the difference operator by

  $$\begin{aligned} B_{\lambda ,\lambda _0}:=V(R_0^+(\lambda )-R_0^+(\lambda _0)). \end{aligned}$$

  (3.2)

  For $\epsilon >0$, there exist $\delta >0$ and an integral operator $B: L^1\rightarrow L^1$ such that for $|\lambda -\lambda _0|\le \delta$ and $\lambda ,\lambda _0\ge 0$,

  $$\begin{aligned} |B_{\lambda ,\lambda _0}(x,y)|\le B(x,y){, \mathrm{and} }\Vert B(x,y)\Vert _{L_y^\infty L_x^1}\le \epsilon . \end{aligned}$$

  (3.3)

• (iii) For $\epsilon >0$, there exist $N_1\gg 1$ and an integral operator $D=D_\epsilon : L^1\rightarrow L^1$ such that for $\lambda \ge N_1$,

  $$\begin{aligned} |(VR_0^+(\lambda ))^4(x,y)|\le D(x,y){, \mathrm{and} }\Vert D(x,y)\Vert _{L_y^\infty L_x^1}\le \epsilon . \end{aligned}$$

  (3.4)

Proof

(i) By the free resolvent formula $R_0^+(\lambda )(x,y)=\frac{e^{i\sqrt{\lambda }|x-y|}}{4\pi |x-y|}$, the Minkowski inequality and the definition of the global Kato norm (1.1), we have

$$\begin{aligned} \Vert VR_0^+(\lambda )f\Vert _{L^1}\le \int _{\mathbb {R}^3} \Big \Vert \frac{|V(x)|}{4\pi |x-y|}\Big \Vert _{L_x^1} |f(y)|dy\le \frac{\Vert V\Vert _{\mathcal {K}}}{4\pi }\Vert f\Vert _{L^1}. \end{aligned}$$

(3.5)

(ii) For $\epsilon >0$, decompose $V=V_1+V_2$ such that $V_1$ is bounded and compactly supported and $\Vert V_2\Vert _\mathcal {K}\le \epsilon$. We choose $\delta >0$ such that $|\sqrt{\lambda }-\sqrt{\lambda _0}|\le \epsilon \Vert V_1\Vert _{L^1}^{-1}$ for all $\lambda ,\lambda _0\ge 0$ with $|\lambda -\lambda _0|\le \delta$. By the mean-value theorem,

$$\begin{aligned} |B_{\lambda ,\lambda _0}(x,y)|&\le \Big |\frac{V_1(x)(e^{i\sqrt{\lambda }|x-y|}-e^{i\sqrt{\lambda _0}|x-y|})}{4\pi |x-y|}\Big |+\Big |\frac{V_2(x)(e^{i\sqrt{\lambda }|x-y|}-e^{i\sqrt{\lambda _0}|x-y|})}{4\pi |x-y|}\Big | \nonumber \\&\le \frac{|V_1(x)||\sqrt{\lambda }-\sqrt{\lambda _0}|}{4\pi }+\frac{|V_2(x)|}{2\pi |x-y|}\\&\le \frac{\epsilon |V_1(x)|}{4\pi \Vert V_1\Vert _{L^1}}+\frac{|V_2(x)|}{2\pi |x-y|}=:B_\epsilon (x,y).\nonumber \end{aligned}$$

(3.6)

Then, we have

$$\begin{aligned} \Vert B_\epsilon (x,y)\Vert _{L_y^\infty L_x^1}\le \frac{\epsilon }{4\pi }+\frac{\Vert V_2\Vert _{\mathcal {K}}}{2\pi }\le \epsilon . \end{aligned}$$

(3.7)

(iii) Similarly, for $\epsilon >0$, decompose $V=V_1+V_2$ such that $V_1$ is bounded and compactly supported and $\Vert V_2\Vert _\mathcal {K}\le \epsilon \Vert V\Vert _{\mathcal {K}}^{-3}$. We then write

$$\begin{aligned} |(VR_0^+(\lambda ))^4(x,y)|\le |(V_1 R_0^+(\lambda ))^4(x,y)|+|(VR_0^+(\lambda ))^4(x,y)-(V_1 R_0^+(\lambda ))^4(x,y)|. \end{aligned}$$

(3.8)

For the first term, by the fractional integration inequalities, the Hölder inequalities in the Lorentz spaces (Lemma 7.5) and the free resolvent estimate $\Vert R_0^+(\lambda )\Vert _{L^{4/3}\rightarrow L^4}\lesssim \langle \lambda \rangle ^{-1/4}$ [11, Lemma 2.1], we get

$$\begin{aligned}&\Vert R_0^+(\lambda )(V_1 R_0^+(\lambda ))^3f\Vert _{L^\infty } \nonumber \\&\lesssim \Vert (V_1 R_0^+(\lambda ))^3f\Vert _{L^{3/2,1}}\le \Vert V_1\Vert _{L^{3,1}}\Vert R_0^+(\lambda )(V_1 R_0^+(\lambda ))^2f\Vert _{L^{3,\infty }} \nonumber \\&\lesssim \Vert (V_1 R_0^+(\lambda ))^2f\Vert _{L^1}\le \Vert V_1\Vert _{L^{4/3}} \Vert R_0^+(\lambda ) V_1 R_0^+(\lambda ) f\Vert _{L^4} \nonumber \\&\lesssim \langle \lambda \rangle ^{-1/4}\Vert V_1 R_0^+(\lambda )f\Vert _{L^{4/3}}\lesssim \langle \lambda \rangle ^{-\frac{1}{4}}\Vert V_1\Vert _{L^\infty }\Vert R_0^+(\lambda )f\Vert _{L_{x\in {\text {supp}}V_1}^{4/3}} \nonumber \\&\lesssim \langle \lambda \rangle ^{-\frac{1}{4}}\int _{\mathbb {R}^3}\Big \Vert \frac{1}{|x-y|}\Big \Vert _{L_{x\in {\text {supp}}V_1}^{4/3}}|f(y)| dy\lesssim \langle \lambda \rangle ^{-\frac{1}{4}}\Vert f\Vert _{L^1}. \end{aligned}$$

(3.9)

Taking $f\rightarrow \delta (\cdot -y)$, we obtain that $|R_0^+(\lambda )(V_1 R_0^+(\lambda ))^3(x,y)|\rightarrow 0$ as $\lambda \rightarrow +\infty$. Thus, there exists $N_1=N_1(\epsilon , V_1)\gg 1$ such that if $\lambda \ge N_1$, then

$$\begin{aligned} |(V_1 R_0^+(\lambda ))^4)(x,y)|\le \frac{\epsilon |V_1(x)|}{2\Vert V_1\Vert _{L^1}}=:D_1(x,y). \end{aligned}$$

(3.10)

Then, it is obvious that $\Vert D_1(x,y)\Vert _{L_{y}^\infty L_x^1}\le \frac{\epsilon }{2}$. For the second term, we split

$$\begin{aligned}&(VR_0^+(\lambda ))^4(x,y)-(V_1 R_0^+(\lambda ))^4(x,y) \nonumber \\&=(V_2R_0^+(\lambda )(VR_0^+(\lambda ))^3)(x,y)+(V_1 R_0^+(\lambda )V_2R_0^+(\lambda )(VR_0^+(\lambda ))^2)(x,y)\\&\quad +((V_1 R_0^+(\lambda ))^2V_2R_0^+(\lambda )VR_0^+(\lambda ))(x,y)+((V_1 R_0^+(\lambda ))^3V_2R_0^+(\lambda ))(x,y).\nonumber \end{aligned}$$

(3.11)

Since the kernel of $R_0^+(\lambda )$ is bounded by the kernel of $(-\Delta )^{-1}$, we have

$$\begin{aligned}&|(VR_0^+(\lambda ))^4(x,y)-(V_1 R_0^+(\lambda ))^4(x,y)| \nonumber \\&\le (|V_2|(-\Delta )^{-1}(|V|(-\Delta )^{-1})^3)(x,y) \nonumber \\&\quad +(|V_1| (-\Delta )^{-1}|V_2|(-\Delta )^{-1}(|V|(-\Delta )^{-1})^2)(x,y) \nonumber \\&\quad +((|V_1| (-\Delta )^{-1})^2|V_2|(-\Delta )^{-1}|V|(-\Delta )^{-1})(x,y) \nonumber \\&\quad +((|V_1| (-\Delta )^{-1})^3|V_2|(-\Delta )^{-1})(x,y) \nonumber \\&=:D_2(x,y). \end{aligned}$$

(3.12)

Then,

$$\begin{aligned} \Vert D_2(x,y)\Vert _{L_y^\infty L_x^1}&=\Vert D_2\Vert _{L^1 \rightarrow L^1} \nonumber \\&\le \Vert |V_2|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}\Vert |V|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}^3 \nonumber \\&\quad +\Vert |V_1|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}\Vert |V_2|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}\Vert |V|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}^2 \nonumber \\&\quad +\Vert |V_1|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}^2\Vert |V_2|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}\Vert |V|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}\nonumber \\&\quad +\Vert |V_1|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1}^3\Vert |V_2|(-\Delta )^{-1}\Vert _{L^1\rightarrow L^1} \nonumber \\&\le 4 \Big (\frac{\Vert V\Vert _{\mathcal {K}}+\Vert V_2\Vert _{\mathcal {K}}}{4\pi }\Big )^3 \frac{\Vert V_2\Vert _{\mathcal {K}}}{4\pi }\le \frac{\epsilon }{2}, \end{aligned}$$

(3.13)

where $D_2$ is an integral operator with kernel $D_2(x,y)$. Therefore, we conclude that

$$\begin{aligned} |(VR_0^+(\lambda ))^4(x,y)|\le D(x,y):=D_1(x,y)+D_2(x,y) \end{aligned}$$

(3.14)

and $\Vert D(x,y)\Vert _{L_y^\infty L_x^1}\le \epsilon$. $\square$

By algebra, the resolvent $R_V^+(\lambda )$ can be written as

$$\begin{aligned} R_V^+(\lambda )``=\text {''}R_0^+(\lambda )(I+VR_0^+(\lambda ))^{-1}. \end{aligned}$$

(3.15)

Let $\mathcal {L}(L^1)$ be the space of bounded operators on $L^1$. The following lemmas say that $(I+VR_0^+(\lambda ))$ is invertible in $\mathcal {L}(L^1)$ for $\lambda \ge 0$, its inverse $(I+VR_0^+(\lambda ))^{-1}$ is uniformly bounded in $\mathcal {L}(L^1)$, and is the sum of the identity map and an integral operator.

Lemma 3.2

(Invertibility of $(I+VR_0^+(\lambda ))$) If $V\in \mathcal {K}_0$ and H has no eigenvalue or resonance on $[0,+\infty )$, then $(I+VR_0^+(\lambda ))$ is invertible in $\mathcal {L}(L^1)$ for $\lambda \ge 0$.

Proof

If it is not invertible, there exists $\varphi \in L^1$, $\varphi \ne 0$, such that $(I+VR_0^+(\lambda ))\varphi =0$. Then, $\psi :=R_0^+(\lambda )\varphi$ solves the eigenvalue equation $(-\Delta +V)\psi =(\lambda +i0)\psi \Longleftrightarrow \psi +R_0^+(\lambda )V\psi =0$. Moreover, by the resolvent formula $R_0^+(\lambda )(x,y)=\frac{e^{i\sqrt{\lambda }|x-y|}}{4\pi |x-y|}$, if $s>\frac{1}{2}$, then

$$\begin{aligned} \Vert \langle x\rangle ^{-s}\psi \Vert _{L^2}&=\Vert \langle x\rangle ^{-s}R_0^+(\lambda )\varphi \Vert _{L^2}\le \int _{\mathbb {R}^3}\Big \Vert \frac{1}{\langle x\rangle ^s4\pi |x-y|}\Big \Vert _{L_x^2}|\varphi (y)|dy\lesssim \Vert \varphi \Vert _{L^1}. \end{aligned}$$

(3.16)

Hence, $\lambda$ is an eigenvalue or a resonance (contradiction!). $\square$

Lemma 3.3

(Uniform bound for $(I+VR_0^+(\lambda ))^{-1}$) If $V\in \mathcal {K}_0$ and H has no eigenvalue or resonance on $[0,+\infty )$, then $S_\lambda :=(I+VR_0^+(\lambda ))^{-1}: [0,+\infty )\rightarrow \mathcal {L}(L^1)$ is uniformly bounded.

Proof

Iterating the resolvent identity, we get the formal identity

$$\begin{aligned} (I+VR_0^+(\lambda ))^{-1}``=\text {''}(I-VR_0^+(\lambda )+(VR_0^+(\lambda ))^2-(VR_0^+(\lambda ))^3)\sum _{n=0}^\infty (VR_0^+(\lambda ))^{4n}. \end{aligned}$$

(3.17)

Indeed, by Lemma 3.1 (iii), $\Vert (VR_0^+(\lambda ))^4\Vert _{L^1\rightarrow L^1}<\frac{1}{2}$ for all sufficiently large $\lambda$. Hence, the formal identity (3.16) makes sense, and $(I+VR_0^+(\lambda ))^{-1}$ is uniformly bounded for all sufficiently large $\lambda$. Thus, it suffices to show that $(I+VR_0^+(\lambda ))^{-1}$ is continuous. To see this, we fix $\lambda _0\ge 0$ and write

$$\begin{aligned}&(I+VR_0^+(\lambda ))^{-1}-(I+VR_0^+(\lambda _0))^{-1}=(I+VR_0^+(\lambda _0)+B_{\lambda ,\lambda _0})^{-1}-S_{\lambda _0} \nonumber \\&=[(I+VR_0^+(\lambda _0)(I+S_{\lambda _0}B_{\lambda ,\lambda _0})]^{-1}-S_{\lambda _0}=(I+S_{\lambda _0}B_{\lambda ,\lambda _0})^{-1}S_{\lambda _0}-S_{\lambda _0}\nonumber \\ ``&=\text {''}\sum _{n=0}^\infty (-S_{\lambda _0}B_{\lambda ,\lambda _0})^nS_{\lambda _0}-S_{\lambda _0}``=\text {''}\sum _{n=1}^\infty (-S_{\lambda _0}B_{\lambda ,\lambda _0})^nS_{\lambda _0}. \end{aligned}$$

(3.18)

Then, by Lemma 3.1 (ii), we have

$$\begin{aligned}&\Vert (I+VR_0^+(\lambda ))^{-1}-(I+VR_0^+(\lambda _0))^{-1}\Vert _{L^1\rightarrow L^1}\le \sum _{n=1}^\infty \Vert S_{\lambda _0}\Vert _{L^1\rightarrow L^1}^{n+1}\Vert B_{\lambda ,\lambda _0}\Vert _{L^1\rightarrow L^1}^n \nonumber \\&=\frac{\Vert S_{\lambda _0}\Vert _{L^1\rightarrow L^1}^2\Vert B_{\lambda ,\lambda _0}\Vert _{L^1\rightarrow L^1}}{1-\Vert S_{\lambda _0}\Vert _{L^1\rightarrow L^1}\Vert B_{\lambda ,\lambda _0}\Vert _{L^1\rightarrow L^1}}\rightarrow 0\;{ \mathrm{as} }\;\lambda \rightarrow \lambda _0. \end{aligned}$$

(3.19)

Therefore, the formal identity (3.17) makes sense, and $(I+VR_0^+(\lambda ))^{-1}$ is continuous. $\square$

Lemma 3.4

If $V\in \mathcal {K}_0$ and H has no eigenvalue or resonance on $[0,+\infty )$, then $\tilde{S}_{\lambda }:=(S_\lambda -I)=(I+VR_0^+(\lambda ))^{-1}-I:[0,+\infty )\rightarrow \mathcal {L}(L^1)$ is not only uniformly bounded but also an integral operator with kernel $\tilde{S}_\lambda (x,y)$:

$$\begin{aligned} \tilde{S}:=\sup _{\lambda \ge 0}\Vert \tilde{S}_{\lambda }\Vert _{L^1\rightarrow L^1}=\sup _{\lambda \ge 0}\Vert \tilde{S}_{\lambda }(x,y)\Vert _{L_y^\infty L_x^1}<\infty . \end{aligned}$$

(3.20)

Proof

By algebra, we have

$$\begin{aligned} \tilde{S}_{\lambda }=(I+VR_0^+(\lambda ))^{-1}-I=-(I+VR_0^+(\lambda ))^{-1}VR_0^+(\lambda )=-S_\lambda VR_0^+(\lambda ). \end{aligned}$$

(3.21)

Since $\tilde{S}_{\lambda }: L^1\rightarrow L^1$ is bounded, sending $f_\epsilon \rightarrow \delta (\cdot -y_0)$ as $\epsilon \rightarrow 0$, we get

$$\begin{aligned} (\tilde{S}_{\lambda }f_\epsilon )(x)=(-S_\lambda VR_0^+(\lambda )f_\epsilon )(x)\rightarrow -S_\lambda \left( \frac{V(\cdot )e^{i\sqrt{\lambda }|\cdot -y_0|}}{4\pi |\cdot -y_0|}\right) (x)=:\tilde{S}_\lambda (x,y_0). \end{aligned}$$

(3.22)

Consider $F_V(x;y,\lambda ):=V(x)\frac{e^{i\sqrt{\lambda }|x-y|}}{4\pi |x-y|}$ as a function of x with parameters $y\in \mathbb {R}^3$ and $\lambda \in \mathbb {R}$. Then, $F_V(x;y,\lambda )$ is bounded in $L_x^1$ uniformly in y and $\lambda$. Therefore, by Lemma 3.3, we conclude that $\tilde{S}_\lambda (x,y)=-S_\lambda \Big (\frac{V(\cdot )e^{i\sqrt{\lambda }|\cdot -y|}}{4\pi |\cdot -y|}\Big )(x)$ is also bounded in $L_x^1$ uniformly in $\lambda$ and y. $\square$

3.2 Spectral Projections and Eigenfunctions

Let $\chi$ be the dyadic partition of unity function chosen in (1.2), and let $\tilde{\chi }_N(\lambda )\in C_c^\infty (\mathbb {R})$ such that $\tilde{\chi }_N(\lambda )=\chi (\tfrac{\sqrt{\lambda }}{N})$ if $\lambda \ge 0$; $\tilde{\chi }_N(\lambda )=0$ if $\lambda <0$. By functional calculus, we define the Littlewood-Paley projections by $P_N=\tilde{\chi }_N(H)$, $P_{\le N_0}=\sum _{N<N_0}P_N$, $P_{N_0<\cdot <N_1}=\sum _{N_0< N<N_1}P_N$ and $P_{\ge N_1}=\sum _{N\ge N_1}P_N$.

Lemma 3.5

Suppose that $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on $[0,+\infty )$. Let $\mathfrak {S}:=\{f\in L^1\cap L^\infty : P_cf= P_{N_0<\cdot <N_1} f for some N_0, N_1>0\}$. For $1<r<\infty$, $\mathfrak {S}$ is dense in $L^r$.

Proof

$L^1\cap L^\infty$ is dense in $L^r$. Fix $f\in L^1\cap L^\infty$. We claim that $\lim _{N_0\rightarrow 0}\Vert P_{<N_0}f\Vert _{L^r}=0$. By the spectral theory, $\lim _{N_0\rightarrow 0}\Vert P_{<N_0}f\Vert _{L^2}=0$. On the other hand, replacing $\tilde{\chi }_N$ by $\sum _{N<N_0}\tilde{\chi }_N$ in the proof of [12, Corollary 1.6], one can show that $\Vert P_{<N_0}f\Vert _{L^1}$ and $\Vert P_{<N_0}f\Vert _{L^\infty }$ are bounded uniformly in $N_0$. Hence the claim follows from interpolation. By the same argument, one can show that $\lim _{N_1\rightarrow \infty }\Vert P_{>N_1}f\Vert _{L^r}=0$. Thus, $\mathfrak {S}$ is dense in $L^r$. $\square$

Lemma 3.6

(Boundedness of eigenfunctions) Suppose that $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on $[0,+\infty )$. Let $\psi _j$ be an eigenfunction corresponding to the negative eigenvalue $\lambda _j$.

• (i) For all $1\le p<\infty$, $\psi _j\in L^p$ and $P_{\lambda _j}$ is bounded on $L^p$, where $P_{\lambda _j}$ is the spectral projection onto the point $\{\lambda _j\}$.

• (ii) $\nabla \psi _j\in L^r$ for $1<r<3$.

Proof

(i) We prove the lemma following the argument in [1]. We decompose $V=V_1+V_2$ such that $V_1$ is compactly supported and bounded, and $\Vert V_2\Vert _{\mathcal {K}}\le 1$. Then,

$$\begin{aligned}&\psi _j+R_0(\lambda _j)V\psi _j=\psi _j+R_0(\lambda _j)(V_1+V_2)\psi _j=0 \nonumber \\&\Rightarrow \psi _j=-(I+R_0(\lambda _j)V_2)^{-1}R_0(\lambda _j)V_1\psi _j=-\sum _{n=0}^\infty (-R_0(\lambda _j)V_2)^n R_0(\lambda _j)V_1\psi _j. \end{aligned}$$

(3.23)

Observe that, since $V_1$ is compactly supported, and $\lambda _j<0$, $R_0(\lambda _j)V_1\psi _j$ is exponentially decaying. To see this, we choose sufficiently small $\epsilon >0$ such that $\epsilon <\sqrt{-\lambda _j}$ for any negative eigenvalue $\lambda _j$. Indeed, there exists such $\epsilon$, since by the assumptions, there are at most finitely many negative eigenvalues (see [2]). Then, by the fractional integration inequality and the Hölder inequality in the Lorentz spaces (Lemma 7.5), we get

$$\begin{aligned} |e^{\epsilon |x|}(R_0(\lambda _j)V_1f)(x)|&\le e^{\epsilon |x|}\int _{\mathbb {R}^3}\frac{e^{i\sqrt{\lambda _j}|x-y|}}{4\pi |x-y|}|V_1(y)||\psi _j(y)|dy \nonumber \\&\le \int _{\mathbb {R}^3}\frac{e^{-(\sqrt{-\lambda _j}-\epsilon )|x-y|}}{4\pi |x-y|}e^{\epsilon |y|}|V_1(y)||\psi _j(y)|dy\\&\le \Vert e^{\epsilon |\cdot |}V_1\psi _j\Vert _{L^{3/2,1}}\lesssim \Vert e^{\epsilon |\cdot |}V_1\Vert _{L^{6,2}}\Vert \psi _j\Vert _{L^2}. \nonumber \end{aligned}$$

(3.24)

Similarly, one can check that $e^{\epsilon |\cdot |}R_0(\lambda _j)V_2e^{-\epsilon |\cdot |}$ is bounded on $L^\infty$ and its operator norm is strictly $<$1. Thus, we prove that

$$\begin{aligned} \Vert e^{\epsilon |\cdot |}\psi _j\Vert _{L^\infty }\le \left( \sum _{n=0}^\infty \Vert e^{\epsilon |\cdot |}R_0(\lambda _j)V_2e^{-\epsilon |\cdot |}\Vert _{L^\infty \rightarrow L^\infty }^n\right) \Vert e^{\epsilon |\cdot |}R_0(\lambda _j)V_1\psi _j\Vert _{L^\infty }<\infty . \end{aligned}$$

(3.25)

Therefore, $\psi _j\in L^p$ and $P_{\lambda _j}f=\langle \psi _j, f\rangle _{L^2}\psi _j$ is bounded on $L^p$ for all $1\le p\le \infty$.

(ii) Let $\delta _1,\delta _2>0$ be arbitrarily small numbers. Then, since $\lambda _j<0$, by the inhomogeneous Sobolev inequality, we get

$$\begin{aligned} \Vert \nabla \psi _j\Vert _{L^{\frac{1}{1-\delta _1}}}&=\Vert \nabla R_0^+(\lambda _j)V\psi _j\Vert _{L^{\frac{1}{1-\delta _1}}}\lesssim \Vert V\psi _j\Vert _{L^{\frac{1}{1-\delta _1}}} \nonumber \\&\le \Vert V\Vert _{L^{3/2,\infty }}\Vert \psi _j\Vert _{L^{\frac{3}{1-3\delta _1},1}}<\infty , \nonumber \\ \Vert \nabla \psi _j\Vert _{L^{\frac{3}{1+\delta _2}}}&=\Vert \nabla R_0^+(\lambda _j)V\psi _j\Vert _{L^{\frac{3}{1+\delta _2}}}\lesssim \Vert V\psi _j\Vert _{W^{-1,\frac{3}{1+\delta _2}}} \nonumber \\&\lesssim \Vert V\psi _j\Vert _{L^{\frac{3}{2+\delta _2}}}\le \Vert V\Vert _{L^{3/2,\infty }}\Vert \psi _j\Vert _{L^{\frac{3}{\delta _2},\frac{3}{2+\delta _2}}}<\infty . \end{aligned}$$

(3.26)

Thus, interpolation gives (ii). $\square$

4 High Frequency Estimate: Proof of Lemma 2.1 (i)

4.1 Construction of the Formal Series Expansion

For a large dyadic number $N_1$ to be chosen later, we construct a formal series for $Pb _{\ge N_1}$ as follows. First, iterating the resolvent identity

$$\begin{aligned} (I+VR_0^+(\lambda ))^{-1}=I-(I+VR_0^+(\lambda ))^{-1}VR_0^+(\lambda ), \end{aligned}$$

(4.1)

we generate a formal series expansion

$$\begin{aligned} (I+VR_0^+(\lambda ))^{-1}``=\text {''}\sum _{n=0}^\infty (-VR_0^+(\lambda ))^n. \end{aligned}$$

(4.2)

Plugging (4.2) into (2.7), we write

$$\begin{aligned} Pb _{\ge N_1}``=\text {''}-\sum _{N\ge N_1}\sum _{n=0}^\infty \frac{1}{\pi }\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}[R_0^+(\lambda )(-VR_0^+(\lambda ))^nVR_0^+(\lambda )]d\lambda . \end{aligned}$$

(4.3)

Then, writing the first and the last free resolvents explicitly by the free resolvent formula $R_0^+(\lambda )(x,y)=\frac{e^{i\sqrt{\lambda }|x-y|}}{4\pi |x-y|}$ and collecting terms having $\lambda$ by Fubini theorem, we write the kernel of $Pb _{\ge N_1}$ as

$$\begin{aligned}&Pb _{\ge N_1}(x,y) \nonumber \\ ``&=\text {''}\sum _{N\ge N_1}\sum _{n=0}^\infty \frac{(-1)^{n+1}}{\pi }\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }) \nonumber \\&\quad \quad \times {\text {Im}}\left[ \int _{\mathbb {R}^6}\frac{e^{i\sqrt{\lambda }|x-\tilde{x}|}}{4\pi |x-\tilde{x}|}(VR_0^+(\lambda ))^n(\tilde{x},\tilde{y})V(\tilde{y})\frac{e^{i\sqrt{\lambda }|\tilde{y}-y|}}{4\pi |\tilde{x}-y|}d\tilde{x}d\tilde{y}\right] d\lambda \nonumber \\&=\int _{\mathbb {R}^6}\frac{V(\tilde{y})}{16\pi ^3|x-\tilde{x}||\tilde{y}-y|}\left\{ \sum _{N\ge N_1}\sum _{n=0}^\infty (-1)^{n+1}Pb _N^n(x,\tilde{x},\tilde{y}, y)\right\} d\tilde{x}d\tilde{y}, \end{aligned}$$

(4.4)

where

$$\begin{aligned} Pb _N^n(x,\tilde{x},\tilde{y}, y)=\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}[e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}(VR_0^+(\lambda ))^n(\tilde{x},\tilde{y})]d\lambda . \end{aligned}$$

(4.5)

We note that the series (4.4) makes sense only formally at this moment, but it will be shown that the sum is absolutely convergent, and that it satisfies the bound we want to have.

4.2 Intermediate Kernel Estimates

We estimate the intermediate kernel $Pb _N^n(x,\tilde{x},\tilde{y}, y)$ in two ways. First, we show that the sum of $Pb _N^n(x,\tilde{x},\tilde{y}, y)$ in $N\ge N_1$ is absolutely convergent, and moreover each $Pb _N^n(x,\tilde{x},\tilde{y}, y)$ decays away from $x=\tilde{x}$ and $\tilde{y}=y$.

Lemma 4.1

(Summability in N) For $s_1,s_2\ge 0$, we have

$$\begin{aligned} |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|\lesssim \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}}{\langle N(x-\tilde{x})\rangle ^{s_1}\langle N(\tilde{y}-y)\rangle ^{s_2}}k_1^n(\tilde{x},\tilde{y}) \end{aligned}$$

(4.6)

and

$$\begin{aligned} \Vert k_1^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}\le \left( \frac{\Vert V\Vert _{\mathcal {K}}}{4\pi }\right) ^n. \end{aligned}$$

(4.7)

For the proof, we need the following lemma.

Lemma 4.2

(Oscillatory integral) For $s\ge 0$,

$$\begin{aligned} \left| \int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}(e^{i\sqrt{\lambda }\sigma }) d\lambda \right| \lesssim \frac{N^2}{\langle N\sigma \rangle ^s}\Vert m\Vert _{\mathcal {H}(s)}. \end{aligned}$$

(4.8)

Proof

By abuse of notation, we denote by $\chi$ the even extension of itself. Making change of variables $\lambda \mapsto N^2\lambda ^2$, we write

$$\begin{aligned} \int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}(e^{i\sqrt{\lambda }\sigma }) d\lambda&=N^2\int _0^\infty 2\lambda m(N^2\lambda ^2)\chi (\lambda )\sin (N\lambda \sigma ) d\lambda \nonumber \\&=N^2\int _\mathbb {R}\lambda m(N^2\lambda ^2)\chi (\lambda )e^{i\lambda N\sigma }d\lambda \nonumber \\&=N^2 \Big (m(N^2\lambda ^2)\lambda \chi (\lambda )\Big )^\vee (N\sigma ) \nonumber \\&=\frac{N^2}{\langle N\sigma \rangle ^s}\Big (\langle \nabla \rangle ^s(m(N^2\lambda ^2)\lambda \chi (\lambda ))\Big )^\vee (N\sigma ). \end{aligned}$$

(4.9)

Thus, it follows from Hausdorff–Young inequality and the fractional Leibniz rule that

$$\begin{aligned} \Big |\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}(e^{i\sqrt{\lambda }\sigma }) d\lambda \Big |&\le \frac{N^2}{\langle N\sigma \rangle ^s}\Vert m(N^2\lambda ^2)\lambda \chi (\lambda )\Vert _{W^{s,1}} \nonumber \\&\lesssim \frac{N^2}{\langle N\sigma \rangle ^s}\Vert m(N^2\lambda ^2)\chi (\lambda )\Vert _{W^{s,2}}\nonumber \\&\le \frac{N^2}{\langle N\sigma \rangle ^s}\Vert m\Vert _{\mathcal {H}(s)}. \end{aligned}$$

(4.10)

$\square$

Proof of Lemma 4.1

First, using the free resolvent formula, we write

$$\begin{aligned}&Pb _N^n (x,\tilde{x},\tilde{y},y) \nonumber \\&=\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}\Big \{\int _{\mathbb {R}^{3(n-1)}}\prod _{k=1}^n V(x_k)\frac{\prod _{k=0}^{n+1} e^{i\sqrt{\lambda }|x_k-x_{k+1}|}}{\prod _{k=1}^n 4\pi |x_k-x_{k+1}|}d\mathbf {x}_{(2,n)}\Big \} d\lambda \nonumber \\&=\int _{\mathbb {R}^{3(n-1)}}\frac{\prod _{k=1}^n V(x_k)}{\prod _{k=1}^n 4\pi |x_k-x_{k+1}|} \Big \{\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}(e^{i\sqrt{\lambda }\sigma _{n+1}}) d\lambda \Big \} d\mathbf {x}_{(2,n)}, \end{aligned}$$

(4.11)

where $x_0:=x$, $x_1:=\tilde{x}$, $x_{n+1}:=\tilde{y}$, $x_{n+2}:=y$, $d\mathbf {x}_{(2,n)}:=dx_2\ldots dx_n$ and $\sigma _n:=\sum _{j=0}^n|x_j-x_{j+1}|$. Then, by Lemma 4.2 with $s=s_1+s_2$ and the trivial inequality

$$\begin{aligned} |x_0-x_1|=|x-\tilde{x}|, |x_{n+1}-x_{n+2}|=|\tilde{y}-y|\le \sigma _{n+1}=\sum _{j=0}^{n+1}|x_j-x_{j+1}|, \end{aligned}$$

(4.12)

we obtain that

$$\begin{aligned} |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|{\lesssim } \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1{+}s_2)}}{\langle N(x-\tilde{x})\rangle ^{s_1}\langle N(\tilde{y}-y)\rangle ^{s_2}}\int _{\mathbb {R}^{3(n-1)}}\frac{\prod _{k=1}^n |V(x_k)|}{\prod _{k=1}^n 4\pi |x_k-x_{k+1}|} d\mathbf {x}_{(2,n)}. \end{aligned}$$

(4.13)

We define

$$\begin{aligned} k_1^n(\tilde{x},\tilde{y}):=\int _{\mathbb {R}^{3(n-1)}}\frac{\prod _{k=1}^n |V(x_k)|}{\prod _{k=1}^n 4\pi |x_k-x_{k+1}|} d\mathbf {x}_{(2,n)}. \end{aligned}$$

(4.14)

Then, by the definition of the global Kato norm, we have

$$\begin{aligned} \Vert k_1^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}&\le \sup _{x_{n+1}\in \mathbb {R}^3}\int _{\mathbb {R}^{3n}}\frac{\prod _{k=1}^n |V(x_k)|}{\prod _{k=1}^n 4\pi |x_k-x_{k+1}|} d\mathbf {x}_{(1,n)} \nonumber \\&\le \Big (\sup _{x_{n}\in \mathbb {R}^3}\int _{\mathbb {R}^{3(n-1)}}\frac{\prod _{k=1}^{n-1} |V(x_k)|}{\prod _{k=1}^{n-1} 4\pi |x_k-x_{k+1}|} d\mathbf {x}_{(1,n-1)}\Big ) \nonumber \\&\quad \quad \times \Big (\sup _{x_{n+1}\in \mathbb {R}^3}\int _{\mathbb {R}^3}\frac{|V(x_n)|}{4\pi |x_n-x_{n+1}|}dx_n\Big ) \nonumber \\&\le \Big (\sup _{x_{n}\in \mathbb {R}^3}\int _{\mathbb {R}^{3(n-1)}}\frac{\prod _{k=1}^{n-1} |V(x_k)|}{\prod _{k=1}^{n-1} 4\pi |x_k-x_{k+1}|} d\mathbf {x}_{(1,n-1)}\Big )\Big (\frac{\Vert V\Vert _{\mathcal {K}}}{4\pi }\Big ) \nonumber \\&\le \cdots (repeat )\cdots \le \Big (\frac{\Vert V\Vert _{\mathcal {K}}}{4\pi }\Big )^n. \end{aligned}$$

(4.15)

$\square$

Next, we show summability of the intermediate kernel in n.

Lemma 4.3

(Summability in n) For $\epsilon >0$, there exist $N_1=N_1(V,\epsilon )\gg 1$ and $k_2^n(\tilde{x},\tilde{y})\in L_{\tilde{y}}^\infty L_{\tilde{x}}^1$ such that for $N\ge N_1$,

$$\begin{aligned} |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|\lesssim \epsilon ^nN^2 \Vert m\Vert _{\mathcal {H}(0)}k_2^n(\tilde{x},\tilde{y}). \end{aligned}$$

(4.16)

and

$$\begin{aligned} \Vert k_2^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}\lesssim \epsilon ^n. \end{aligned}$$

(4.17)

Proof

By Lemma 3.1 (iii), given $\epsilon >0$, there exist $N_1\gg 1$ and an operator $D:L^1\rightarrow L^1$ such that $\Vert D(x,y)\Vert _{L_y^\infty L_x^1}\le \epsilon ^4$ and $|(VR_0^+(\lambda ))^4(x,y)|\le D(x,y)$. We also observe that

$$\begin{aligned} |(VR_0^+(\lambda ))(x,y)|=\left| V(x)\frac{e^{i\sqrt{\lambda }|x-y|}}{4\pi |x-y|}\right| =\frac{|V(x)|}{4\pi |x-y|}=\Big (|V|(-\Delta )^{-1}\Big )(x,y). \end{aligned}$$

(4.18)

We denote by $\lfloor a\rfloor$ the largest integer less than or equal to a. Then, we have

$$\begin{aligned} |Pb _N^n(x,\tilde{x},\tilde{y}, y)|&\le \int _0^\infty |m(\lambda )|\chi _N(\sqrt{\lambda })|(VR_0^+(\lambda ))^n(\tilde{x},\tilde{y})|d\lambda \nonumber \\&\le \int _0^\infty |m(\lambda )|\chi _N(\sqrt{\lambda })\Big |\Big (D^{\lfloor \frac{n}{4}\rfloor }(|V|(-\Delta )^{-1})^{n-4\lfloor \frac{n}{4}\rfloor }\Big ) (\tilde{x},\tilde{y})\Big |d\lambda \nonumber \\&\lesssim N^2\int _0^\infty |m(N^2\lambda )|\chi (\lambda )d\lambda \cdot \Big |\Big (D^{\lfloor \frac{n}{4}\rfloor }(|V|(-\Delta )^{-1})^{n-4\lfloor \frac{n}{4}\rfloor }\Big ) (\tilde{x},\tilde{y})\Big | \nonumber \\&\lesssim N^2 \Vert m\Vert _{\mathcal {H}(0)}\Big |\Big (D^{\lfloor \frac{n}{4}\rfloor }(|V|(-\Delta )^{-1})^{n-4\lfloor \frac{n}{4}\rfloor }\Big ) (\tilde{x},\tilde{y})\Big |. \end{aligned}$$

(4.19)

We define

$$\begin{aligned} k_2^n(\tilde{x},\tilde{y}):=\Big |\Big (D^{\lfloor \frac{n}{4}\rfloor }(|V|(-\Delta )^{-1})^{n-4\lfloor \frac{n}{4}\rfloor }\Big )(\tilde{x},\tilde{y})\Big |. \end{aligned}$$

(4.20)

Then, by Lemma 3.1, one can check (4.17). $\square$

4.3 Proof of Lemma 2.1 (i)

Let $\delta >0$ be a sufficiently small number to be chosen later. Let $\epsilon >0$ be a small number depending on $\Vert V\Vert _{\mathcal {K}}$ and $\delta >0$ [see (4.30)]. Then, we pick a large dyadic number $N_1$ from Lemma 4.3. We will show that $Pb _{\ge N_1}$ is bounded from $L^{\frac{3}{3-\delta },1}$ to $L^{\frac{3}{3-\delta },\infty }$.

Let $s=\frac{2}{1-\delta }>2$ and $\theta =\frac{2-\delta }{2}$ ($\Rightarrow 2\theta =2-\delta$, $(s-2)\theta >\delta$ and $s\theta >2$). Then, by Lemma 4.1 with $s_1=2$ and $s_2=s-2$ and Lemma 4.3, we get

$$\begin{aligned} |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|&=|Pb _N^n (x, \tilde{x}, \tilde{y}, y)|^{\theta }|Pb _N^n (x, \tilde{x}, \tilde{y}, y)|^{1-\theta } \nonumber \\&\lesssim \frac{N^2 \Vert m\Vert _{\mathcal {H}(s)}}{\langle N(x-\tilde{x})\rangle ^{2\theta }\langle N(\tilde{y}-y)\rangle ^{(s-2)\theta }}\Big (k_1^n(\tilde{x},\tilde{y})\Big )^{\theta }\Big (k_2^n(\tilde{x},\tilde{y})\Big )^{1-\theta }. \end{aligned}$$

(4.21)

We claim that

$$\begin{aligned} \sum _{N\in 2^{\mathbb {Z}}}\frac{N^2}{\langle Nx\rangle ^{2\theta }\langle Ny\rangle ^{(s-2)\theta }}\lesssim \frac{1}{|x|^{2-\delta }|y|^{\delta }}. \end{aligned}$$

(4.22)

Fix $x, y\in \mathbb {R}^3$, and consider the following four cases.

(Case 1 $N<\min (|x|^{-1}, |y|^{-1})$)

$$\begin{aligned} \sum _{Case 1 }\frac{N^2}{\langle Nx\rangle ^{2\theta }\langle Ny\rangle ^{(s-2)\theta }}\le \sum _{Case 1 }N^2\le \min \Big (\frac{1}{|x|}, \frac{1}{|y|}\Big )^2\le \frac{1}{|x|^{2-\delta }|y|^{\delta }}. \end{aligned}$$

(4.23)

(Case 2 $|x|^{-1}\le N<|y|^{-1}$)

$$\begin{aligned} \sum _{Case 2 }\frac{N^2}{\langle Nx\rangle ^{2\theta }\langle Ny\rangle ^{(s-2)\theta }}&\le \sum _{Case 2 }\frac{N^2}{|Nx|^{2\theta }}=\sum _{Case 2 }\frac{N^{2(1-\theta )}}{|x|^{2\theta }}=\sum _{Case 2 }\frac{N^\delta }{|x|^{2-\delta }} \nonumber \\&\le \frac{1}{|x|^{2-\delta }|y|^{\delta }}. \end{aligned}$$

(4.24)

(Case 3 $|y|^{-1}\le N<|x|^{-1}$)

$$\begin{aligned} \sum _{Case 3 }\frac{N^2}{\langle Nx\rangle ^{2\theta }\langle Ny\rangle ^{(s-2)\theta }}\le \sum _{Case 3 }\frac{N^2}{|Ny|^{\delta }}=\sum _{Case 3 }\frac{N^{2-\delta }}{|y|^{\delta }}\le \frac{1}{|x|^{2-\delta }|y|^{\delta }}. \end{aligned}$$

(4.25)

(Case 4 $N\ge \max (|x|^{-1}, |y|^{-1})$)

$$\begin{aligned} \sum _{Case 4 }\frac{N^2}{\langle Nx\rangle ^{2\theta }\langle Ny\rangle ^{(s-2)\theta }}&\lesssim \frac{1}{|x|^{2\theta }|y|^{(s-2)\theta }}\sum _{Case 4 }\frac{1}{N^{s\theta -2}} \nonumber \\&\le \frac{1}{|x|^{2-\delta }|y|^{(s-2)\theta }}|y|^{s\theta -2}\le \frac{1}{|x|^{2-\delta }|y|^{2-2\theta }}=\frac{1}{|x|^{2-\delta }|y|^{\delta }}. \end{aligned}$$

(4.26)

Collecting all, we prove the claim.

Applying (4.22) to (4.21) and summing in $N\ge N_1$, we obtain

$$\begin{aligned} \sum _{N\ge N_1}|Pb _N^n (x, \tilde{x}, \tilde{y}, y)|\lesssim \frac{\Vert m\Vert _{\mathcal {H}(s)}}{ |x-\tilde{x}|^{2-\delta }|\tilde{y}-y|^\delta }\Big (k_1^n(\tilde{x},\tilde{y})\Big )^{\theta }\Big (k_2^n(\tilde{x},\tilde{y})\Big )^{1-\theta }. \end{aligned}$$

(4.27)

Let

$$\begin{aligned} K(x,y)=\sum _{n=0}^\infty \Big (k_1^n(\tilde{x},\tilde{y})\Big )^{\theta }\Big (k_2^n(x,y)\Big )^{1-\theta }. \end{aligned}$$

(4.28)

Then, $K\in L_y^\infty L_x^1$, since if $N_1$ is large enough,

$$\begin{aligned} \Vert K(x,y)\Vert _{L_y^\infty L_x^1}&\le \sum _{n=0}^\infty \Big \Vert \Big (k_1^n(\tilde{x},\tilde{y})\Big )^{\theta }\Big (k_2^n(x,y)\Big )^{1-\theta }\Big \Vert _{L_y^\infty L_x^1} \nonumber \\&\le \sum _{n=0}^\infty \Vert k_1^n(\tilde{x},\tilde{y})\Vert _{L_y^\infty L_x^1}^\theta \Vert k_2^n(x,y)\Vert _{L_y^\infty L_x^1}^{1-\theta } \nonumber \\&\le \sum _{n=0}^\infty \Big (\frac{\Vert V\Vert _{\mathcal {K}}}{4\pi }\Big )^{n\theta }\epsilon ^n<\infty , \end{aligned}$$

(4.29)

where $\epsilon >0$ is chosen so that $(\frac{\Vert V\Vert _{\mathcal {K}}}{4\pi })^{\theta }\epsilon <1$ with $\theta =\frac{2-\delta }{2}$. Therefore, we obtain the kernel estimates for $Pb _{\ge N_1}(x,y)$,

$$\begin{aligned} |Pb _{N_1}(x,y)|&\le \int _{\mathbb {R}^6}\frac{V(\tilde{y})}{16\pi ^3|x-\tilde{x}||\tilde{y}-y|}\sum _{N\ge N_1}\sum _{n=0}^\infty |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|d\tilde{x} d\tilde{y} \nonumber \\&\lesssim \int _{\mathbb {R}^6}\frac{|V(\tilde{y})|}{|x-\tilde{x}|^{3-\delta }|\tilde{y}-y|^{1+\delta }}K(\tilde{x},\tilde{y})d\tilde{x}d\tilde{y}, \end{aligned}$$

(4.30)

with $K\in L_y^\infty L_x^1$.

Let $T_K$ be the integral operator with kernel K(x, y), which is bounded on $L^1$ by (4.29). By the fractional integration inequality and Hölder inequality in the Lorentz spaces (see “Appendix”), we conclude that

$$\begin{aligned}&\Vert Pb _{\ge N_1}f\Vert _{L^{\frac{3}{3-\delta },\infty }} \nonumber \\&\lesssim \Big \Vert \int _{\mathbb {R}^9}\frac{|V(\tilde{y})|}{|x-\tilde{x}|^{3-\delta }|\tilde{y}-y|^{1+\delta }}K(\tilde{x},\tilde{y})|f(y)|d\tilde{x}d\tilde{y}dy\Big \Vert _{L_x^{\frac{3}{3-\delta },\infty }} \nonumber \\&\lesssim \Vert |\nabla |^{-\delta }T_K(|V||\nabla |^{-(2-\delta )}(|f|))\Vert _{L^{\frac{3}{3-\delta },\infty }}\lesssim \Vert T_K(|V||\nabla |^{-(2-\delta )}(|f|))\Vert _{L_{x}^1}\\&\lesssim \Vert |V||\nabla |^{-(2-\delta )}(|f|)\Vert _{L_{x}^1}\le \Vert V\Vert _{L^{3/2,\infty }}\Vert |\nabla |^{-(2-\delta )}|f|\Vert _{L^{3,1}}\lesssim \Vert f\Vert _{L^{\frac{3}{3-\delta },1}}.\nonumber \end{aligned}$$

(4.31)

Remark 4.4

In (4.31), we only used the fractional integration inequality and the Hölder inequality. Note that after applying the fractional integration inequality, we always have the $L^{p,q}$-norm with smaller p on the right hand side, although we want to show the $L^{\frac{3}{3-\epsilon },1}-L^{\frac{3}{3-\epsilon },\infty }$ boundedness. Hence, one must have at least one chance to raise the number p to compensate the decrease of p caused by the fractional integration inequalities. In (4.31), the potential V plays such a role with the Hölder inequality. This is the main reason that we keep one extra potential term V in the spectral representation by considering the perturbation $m(H)P_c-m(-\Delta )$ instead of $m(H)P_c$, and introducing intermediated kernels $Pb _N^n(x,\tilde{x},\tilde{y},y)$, even though they look rather artificial.

5 Low Frequency Estimate: Proof of Lemma 2.1 (ii)

5.1 Construction of the Formal Series Expansion

We prove Lemma 2.1 (ii) by modifying the argument in Sect. 4. Note that for small N, the formal series expansion (4.2) may not be convergent, since $(VR_0^+(\lambda ))^4$ in (4.2) is not small anymore. Hence, we introduce a new series expansion for $(I+VR_0^+(\lambda ))^{-1}$,

$$\begin{aligned} (I+VR_0^+(\lambda ))^{-1}&=(I+VR_0^+(\lambda _0)+B_{\lambda ,\lambda _0})^{-1} \nonumber \\&=[(I+B_{\lambda ,\lambda _0}S_{\lambda _0})(I+VR_0^+(\lambda _0))]^{-1} \nonumber \\&=(I+VR_0^+(\lambda _0))^{-1}(I+B_{\lambda ,0}S_{\lambda _0})^{-1} \nonumber \\ ``&=\text {''}S_{\lambda _0}\sum _{n=0}^\infty (-B_{\lambda ,\lambda _0}S_{\lambda _0})^n, \end{aligned}$$

(5.1)

where $B_{\lambda ,\lambda _0}=V(R_0^+(\lambda )-R_0^+(\lambda _0))$ and $S_{\lambda _0}=(I+VR_0^+(\lambda _0))^{-1}$. Plugging the formal series (5.1) with $\lambda _0=0$ into (2.7), we write

$$\begin{aligned} Pb _N``=\text {''}\sum _{n=0}^\infty \frac{(-1)^{n+1}}{\pi }\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}[R_0^+(\lambda )S_{0}(B_{\lambda ,0}S_{0})^nVR_0^+(\lambda )]d\lambda . \end{aligned}$$

(5.2)

As in the previous section, writing the first and the last free resolvents explicitly by the free resolvent formula $R_0^+(\lambda )(x,y)=\frac{e^{i\sqrt{\lambda }|x-y|}}{4\pi |x-y|}$ and collecting terms having $\lambda$ by Fubini theorem, we write the kernel of $Pb _N$ as

$$\begin{aligned}&Pb _N(x,y) \nonumber \\ ``&=\text {''}\sum _{n=0}^\infty \frac{(-1)^{n+1}}{\pi }\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }) \nonumber \\&\quad \quad \quad \quad \times {\text {Im}}\Big [\iint _{\mathbb {R}^6}\frac{e^{i\sqrt{\lambda }|x-\tilde{x}|}}{4\pi |x-\tilde{x}|}[S_{0}(B_{\lambda ,0}S_{0})^n](\tilde{x},\tilde{y})V(\tilde{y})\frac{e^{i\sqrt{\lambda }|\tilde{y}-y|}}{4\pi |\tilde{x}-y|}d\tilde{x}d\tilde{y}\Big ]d\lambda \nonumber \\&=\iint _{\mathbb {R}^6}\frac{V(\tilde{y})}{16\pi ^3|x-\tilde{x}||\tilde{y}-y|}\Big [\sum _{n=0}^\infty (-1)^{n+1}Pb _N^n(x,\tilde{x},\tilde{y}, y)\Big ]d\tilde{x}d\tilde{y}, \end{aligned}$$

(5.3)

where

$$\begin{aligned} Pb _N^n(x,\tilde{x},\tilde{y}, y)=\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}[e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}[S_{0}(B_{\lambda ,0}S_{0})^n](\tilde{x},\tilde{y})]d\lambda . \end{aligned}$$

(5.4)

By Lemma 3.1 (ii), $B_{\lambda ,0}$ in (5.3) is small for sufficiently small N. This fact will guarantee the convergence of the formal series.

5.2 Intermediate Kernel Estimates

We will show the kernel estimates analogous to Lemmas 4.1 and 4.3. Then, Lemma 2.1 (ii) will follow from exactly the same argument in Sect. 4.3, thus we omit the proof.

Lemma 5.1

(Summability in N) There exists $k_1^n(\tilde{x},\tilde{y})$ such that for $s_1, s_2\ge 0$,

$$\begin{aligned} |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|\lesssim \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}}{\langle N(x-\tilde{x})\rangle ^{s_1}\langle N(\tilde{y}-y)\rangle ^{s_2}}k_1^n(\tilde{x},\tilde{y}) \end{aligned}$$

(5.5)

and

$$\begin{aligned} \Vert k_1^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}\le (\tilde{S}+1)^{n+1}\Big (\frac{\Vert V\Vert _{\mathcal {K}}}{2\pi }\Big )^n, \end{aligned}$$

(5.6)

where $\tilde{S}$ is the positive number given by (3.20).

Proof

First, splitting $B_{\lambda ,0}$ into $VR_0^+(\lambda )-VR_0^+(0)$ in

$$\begin{aligned} Pb _N^n(x,\tilde{x},\tilde{y}, y)=\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}[e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}[S_{0}(B_{\lambda ,0}S_{0})^n](\tilde{x},\tilde{y})]d\lambda , \end{aligned}$$

(5.7)

we write $Pb _N^n(x,\tilde{x},\tilde{y}, y)$ as the sum of $2^n$ copies of

$$\begin{aligned} \int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}[e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}[S_0VR_0^+(\alpha _1\lambda )S_0\cdot \cdot \cdot VR_0^+(\alpha _n\lambda )S_0](\tilde{x},\tilde{y})]d\lambda \end{aligned}$$

(5.8)

up to $\pm$, where $\alpha _k=0 or 1$ for each $k=1, \ldots , n$. Next, splitting all $S_0$ into I and $\tilde{S}_0$ in (5.8), we further decompose (5.8) into the sum of $2^{n+1}$ kernels.

Among them, let us consider the two representative terms,

$$\begin{aligned}&{\text {Im}}\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda })e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}[\tilde{S}_0VR_0^+(\alpha _1\lambda )\tilde{S}_0\ldots VR_0^+(\alpha _n\lambda )\tilde{S}_0](\tilde{x},\tilde{y})d\lambda ,\end{aligned}$$

(5.9)

$$\begin{aligned}&{\text {Im}}\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda })e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}[VR_0^+(\alpha _1\lambda )\ldots VR_0^+(\alpha _n\lambda )](\tilde{x},\tilde{y})d\lambda . \end{aligned}$$

(5.10)

For the first term, by the free resolvent formula (1.10), we write (5.9) in the integral form,

$$\begin{aligned}&{\text {Im}}\int _0^\infty \int _{\mathbb {R}^{6n}}m(\lambda )\chi _N(\sqrt{\lambda })\prod _{k=1}^{n+1}\tilde{S}_0(x_{2k-1}, x_{2k}) \nonumber \\&\quad \quad \times \prod _{k=1}^n V(x_{2k})\frac{\prod _{k=0}^{n+1} e^{i{\alpha _k\sqrt{\lambda }}|x_{2k}-x_{2k+1}|}}{\prod _{k=1}^n 4\pi |x_{2k}-x_{2k+1}|}d\mathbf {x}_{(2,2n+1)} d\lambda \nonumber \\&=\int _{\mathbb {R}^{6n}}\frac{\prod _{k=1}^{n+1}\tilde{S}_0(x_{2k-1}, x_{2k})\prod _{k=1}^n V(x_{2k})}{\prod _{k=1}^n 4\pi |x_{2k}-x_{2k+1}|} \nonumber \\&\quad \quad \times \left\{ \int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}(e^{i\sqrt{\lambda }\tilde{\sigma }_{n+1}})d\lambda \right\} d\mathbf {x}_{(2,2n+1)} \end{aligned}$$

(5.11)

where $x_0:=x$, $x_1:=\tilde{x}$, $x_{2n+2}:=\tilde{y}$, $x_{2n+3}:=y$, $d\mathbf {x}_{(2,n)}:=dx_2\cdot \cdot \cdot dx_n$, $\tilde{\sigma }_n:=\sum _{k=0}^{n+1} \alpha _k|x_{2k}-x_{2k+1}|$ and $\alpha _0=\alpha _{n+1}=1$. Then, by Lemma 4.2 with $s=s_1+s_2$ and $|x_0-x_1|, |x_{2n+2}-x_{2n+3}|\le \tilde{\sigma }_{n+1}$, we obtain that

$$\begin{aligned} \Big |\int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }){\text {Im}}(e^{i\sqrt{\lambda }\tilde{\sigma }_{n+1}})d\lambda \Big |\lesssim \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}}{\langle N(x_0-x_1)\rangle ^{s_1}\langle N(x_{2n+2}-x_{2n+3})\rangle ^{s_2}}. \end{aligned}$$

(5.12)

Applying (5.12) to (5.9), we get the arbitrary polynomial decay away from $x_0=x_1$,

$$\begin{aligned} |(5.9)|\lesssim \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}k_{(5.9)}^n(\tilde{x},\tilde{y})}{\langle N(x_0-x_1)\rangle ^{s_1}\langle N(x_{2n+2}-x_{2n+3})\rangle ^{s_2}}=\frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}k_{(5.9)}^n(\tilde{x},\tilde{y})}{\langle N(x-\tilde{x})\rangle ^{s_1}\langle N(\tilde{y}-y)\rangle ^{s_2}}, \end{aligned}$$

(5.13)

where

$$\begin{aligned} k_{(5.9)}^n(\tilde{x},\tilde{y}):&=\int _{\mathbb {R}^{6n}}\frac{\prod _{k=1}^{n+1}|\tilde{S}_0(x_{2k-1}, x_{2k})|\prod _{k=1}^n |V(x_{2k})|}{\prod _{k=1}^n 4\pi |x_{2k}-x_{2k+1}|}d\mathbf {x}_{(2,2n+1)} \nonumber \\&=[|\tilde{S}_0|(|V|(-\Delta )^{-1}|\tilde{S}_0|)^n](\tilde{x},\tilde{y}) \end{aligned}$$

(5.14)

and $|\tilde{S}_0|$ is the integral operator with kernel $|\tilde{S}_0(x,y)|$. We claim that

$$\begin{aligned} \Vert k_{(5.9)}^n(\tilde{x},\tilde{y})\Vert _{L_y^\infty L_{x_1}^1}\lesssim \tilde{S}^{n+1}(\Vert V\Vert _{\mathcal {K}}/4\pi )^n. \end{aligned}$$

(5.15)

Indeed, since $\Vert |\tilde{S}_0|(|V|(-\Delta )^{-1}|\tilde{S}_0|)^n f\Vert _{L^1}\le \tilde{S}^{n+1}(\frac{\Vert V\Vert _{\mathcal {K}}}{4\pi })^n\Vert f\Vert _{L^1}$ and $|\tilde{S}_0|(|V|(-\Delta )^{-1}|\tilde{S}_0|)^n$ is an integral operator, sending $f\rightarrow \delta (\cdot -y)$, we prove the claim.

Similarly, we write (5.10) as

$$\begin{aligned}&{\text {Im}}\int _0^\infty \int _{\mathbb {R}^{3n-3}} m(\lambda )\chi _N(\sqrt{\lambda })\prod _{k=1}^n V(x_{k})\frac{\prod _{k=0}^{n+1} e^{i{\alpha _k\sqrt{\lambda }}|x_k-x_{k+1}|}}{\prod _{k=1}^n 4\pi |x_{k}-x_{k+1}|}d\mathbf {x}_{(2,n)} d\lambda \nonumber \\&=\int _{\mathbb {R}^{3n-3}}\frac{\prod _{k=1}^n V(x_{k})}{\prod _{k=1}^n 4\pi |x_{k}-x_{k+1}|} \left\{ \int _0^\infty m(\lambda )\chi _N(\sqrt{\lambda }) {\text {Im}}(e^{i\sqrt{\lambda }\tilde{\tilde{\sigma }}_{n+1}})d\lambda \right\} d\mathbf {x}_{(2,n)} \end{aligned}$$

(5.16)

where $x_0:=x$, $x_1:=\tilde{x}$, $x_{n+1}:=\tilde{y}$, $x_{n+2}:=y$, $\alpha _0=\alpha _{n+2}=1$ and $\tilde{\tilde{\sigma }}_n:=\sum _{k=0}^n\alpha _k|x_k-x_{k+1}|$. Then, by Lemma 4.2 with $s=s_1+s_2$ and $|x_0-x_1|, |x_{n+1}-x_{n+2}|\le \tilde{\tilde{\sigma }}_{n+1}$, we obtain that

$$\begin{aligned} |(5.10)|\lesssim \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}k_{(5.10)}^n(\tilde{x},\tilde{y})}{\langle N(x_0-x_1)\rangle ^{s_1}\langle N(x_{n+1}-x_{n+2})\rangle ^{s_2}}= \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}k_{(5.10)}^n(\tilde{x},\tilde{y})}{\langle N(x-\tilde{x})\rangle ^{s_1}\langle N(\tilde{y}-y)\rangle ^{s_2}} \end{aligned}$$

(5.17)

where

$$\begin{aligned} k_{(5.10)}^n(\tilde{x},\tilde{y}):=\int _{\mathbb {R}^{3n-3}}\frac{\prod _{k=1}^n |V(x_{k})|}{\prod _{k=1}^n 4\pi |x_{k}-x_{k+1}|}d\mathbf {x}_{(2,n)}=(4\pi )^{-n}(|V|(-\Delta )^{-1})^n(\tilde{x},\tilde{y}). \end{aligned}$$

(5.18)

Then by the definition of the global Kato norm, we prove that

$$\begin{aligned} \Vert k_{(5.10)}^n(\tilde{x},\tilde{y})\Vert _{L_y^\infty L_{x_1}^1}\le (\Vert V\Vert _{\mathcal {K}}/4\pi )^n. \end{aligned}$$

(5.19)

Similarly, we estimate other kernels, and define $k_1^n(\tilde{x},\tilde{y})$ as the sum of all $2^{2n+1}$ many upper bounds including $K_{(5.9)}(\tilde{x},\tilde{y})$ and $K_{(5.10)}(\tilde{x},\tilde{y})$. Then, $k_1^n(\tilde{x},\tilde{y})$ satisfies (5.5) and (5.6). $\square$

Lemma 5.2

(Summability in n) For any $\epsilon >0$, there exist a small number $N_0=N_0(V,\epsilon )\ll 1$ and $k_2^n(\tilde{x},\tilde{y})\in L_{\tilde{y}}^\infty L_{\tilde{x}}^1$ such that for $N\le N_0$,

$$\begin{aligned} |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|\lesssim N^2 \Vert m\Vert _{\mathcal {H}(s)}k_2^n(\tilde{x},\tilde{y}) \end{aligned}$$

(5.20)

and

$$\begin{aligned} \Vert k_2^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}\le \epsilon ^n. \end{aligned}$$

(5.21)

Proof

Fix small $\epsilon >0$. Then, by Lemma 3.1 (ii), we choose small $N_0:=\delta =\delta (\epsilon )>0$ and an integral operator B such that $|B_{\lambda ,0}(x,y)|\le B(x,y)$ for $0\le \lambda \le N_0$, and

$$\begin{aligned} \Vert B\Vert _{L^1\rightarrow L^1}\le \epsilon (\tilde{S}+1)^{-1}, \end{aligned}$$

(5.22)

where $\tilde{S}$ is a positive number given from (3.20). We define

$$\begin{aligned} k_2^n(\tilde{x},\tilde{y}):=[(I+|\tilde{S}_0|)(B(I+|\tilde{S}_0|))^n](\tilde{x},\tilde{y}), \end{aligned}$$

where $|\tilde{S}_0|$ is the integral operator with $|\tilde{S}_0(x,y)|$ as kernel. Then, by definitions [see (5.4)], one can check that $k_2^n(\tilde{x},\tilde{y})$ satisfies (5.20). For (5.21), splitting $(I+|\tilde{S}_0|)$ into I and $|\tilde{S}_0|$ in $k_2^n(\tilde{x},\tilde{y})$, we get $2^{n+1}$ terms,

$$\begin{aligned} k_2^n(\tilde{x},\tilde{y})=[|\tilde{S}_0|(B|\tilde{S}_0|)^n](\tilde{x},\tilde{y})+\cdot \cdot \cdot +B^n(\tilde{x},\tilde{y}). \end{aligned}$$

(5.23)

For example, we consider $|\tilde{S}_0|(B|\tilde{S}_0|)^n$ and $B^n$. Since both $|\tilde{S}_0|$ and B are integral operators, by Lemma 3.4 and (5.12), we obtain

$$\begin{aligned} \Vert [|\tilde{S}_0|(B|\tilde{S}_0|)^n](\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}&=\Vert |\tilde{S}_0|(B|\tilde{S}_0|)^n\Vert _{L^1\rightarrow L^1}\le \tilde{S}^{n+1}\Big (\epsilon (\tilde{S}+1)^{-1}\Big )^{n}, \nonumber \\ \Vert B^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}&=\Vert B^n\Vert _{L^1\rightarrow L^1}\le \Big (\epsilon (\tilde{S}+1)^{-1}\Big )^{n}. \end{aligned}$$

(5.24)

Similarly, we estimate other $2^{n+1}-2$ terms. Summing them up, we prove (5.21). $\square$

6 Medium Frequency Estimate: Proof of Lemma 2.1 (iii)

The proof closely follows from that of Lemma 2.1 (ii), so we only sketch the proof. For $\epsilon >0$, we take $\delta =\delta (\epsilon )>0$ from Lemma 3.1 (ii). We choose a partition of unity function $\psi \in C_c^\infty$ such that ${\text {supp}}\psi \subset [-\delta , \delta ]$, $\psi (\lambda )=1$ if $|\lambda |\le \frac{\delta }{3}$ and $\sum _{j=1}^\infty \psi (\cdot -\lambda _j)\equiv 1$ on $(0,+\infty )$, where $\lambda _j=j\delta$.

Let $N_0$ and $N_1$ be dyadic numbers chosen in the previous sections. For $N_0\le N\le N_1$, we first decompose $\chi _N(\sqrt{\lambda })$ in $Pb _N$ [see (2.7)] into $\chi _N(\sqrt{\lambda })=\sum _{j=N/2\delta }^{2N/\delta }\chi _N^j(\lambda )$ where $\chi _N^j(\lambda )=\chi _N(\sqrt{\lambda })\psi (\lambda -\lambda _j)$. Plugging the formal series (5.1) with $\lambda _0=\lambda _j$ into each integral, we write the kernel of $Pb _N$ as

$$\begin{aligned} Pb _N(x,y)``=\text {''}\iint _{\mathbb {R}^6}\frac{V(\tilde{y})}{16\pi ^3|x-\tilde{x}||\tilde{y}-y|}\Big [\sum _{n=0}^\infty (-1)^{n+1}Pb _N^n(x,\tilde{x},\tilde{y}, y)\Big ]d\tilde{x}d\tilde{y}, \end{aligned}$$

(6.1)

where

$$\begin{aligned}&Pb _N^n(x,\tilde{x},\tilde{y}, y) \nonumber \\&=\sum _{j=N/2\delta }^{2N/\delta }\int _0^\infty m(\lambda )\chi _N^j(\sqrt{\lambda }){\text {Im}}[e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}[S_{\lambda _j}(B_{\lambda ,\lambda _j}S_{\lambda _j})^n](\tilde{x},\tilde{y})]d\lambda . \end{aligned}$$

(6.2)

By the arguments in the previous sections, for Lemma 2.1 (iii), it suffices to show the following two lemmas:

Lemma 6.1

(Summability in N) For $N_0<N<N_1$, there exists $k_{N,1}^n(\tilde{x},\tilde{y})$ such that for $s_1, s_2\ge 0$,

$$\begin{aligned} |Pb _N^n (x, \tilde{x}, \tilde{y}, y)|\lesssim \frac{N^2\Vert m\Vert _{\mathcal {H}(s_1+s_2)}k_{N,1}^n(\tilde{x},\tilde{y})}{\langle N(x-\tilde{x})\rangle ^{s_1}\langle N(\tilde{y}-y)\rangle ^{s_2}}, \end{aligned}$$

(6.3)

and

$$\begin{aligned} \Vert k_{N,1}^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}\le (\tilde{S}+1)^{n+1}\Big (\frac{\Vert V\Vert _{\mathcal {K}}}{2\pi }\Big )^n. \end{aligned}$$

(6.4)

Proof

For instance, consider

$$\begin{aligned} \int _0^\infty m(\lambda )\chi _N^j(\lambda ){\text {Im}}[e^{i\sqrt{\lambda }(|x-\tilde{x}|+|\tilde{y}-y|)}\{S_{\lambda _j}(B_{\lambda ,\lambda _j}S_{\lambda _j})^n\}(\tilde{x},\tilde{y})]d\lambda \end{aligned}$$

(6.5)

among O(N)-many similar integrals in (6.2). As we did in Lemma 4.2, we show that

$$\begin{aligned} \Big |\int _0^\infty m(\lambda )\chi _N^j(\lambda ){\text {Im}}(e^{i\sqrt{\lambda }\sigma })d\lambda \Big |\lesssim _{N_0, N_1}\frac{N\Vert m\Vert _{\mathcal {H}(s)}}{ \langle N\sigma \rangle ^s}. \end{aligned}$$

(6.6)

Repeating the proof of Lemma 5.1 [but replacing $S_0$ and $B_{\lambda ,0}$ by $S_{\lambda _j}$ and $B_{\lambda ,\lambda _j}$ and applying (6.6) instead of Lemma 4.2], one can find $k_{N,j,1}^n(\tilde{x},\tilde{y})$ such that for $s_1, s_2\ge 0$,

$$\begin{aligned} |(6.5)|&\lesssim \frac{N\Vert m\Vert _{\mathcal {H}(s_1+s_2)}k_{N,j,1}^n(\tilde{x},\tilde{y})}{\langle N(x-\tilde{x})\rangle ^{s_1}\langle N(\tilde{y}-y)\rangle ^{s_2}},\end{aligned}$$

(6.7)

$$\begin{aligned} \Vert k_{N,j,1}^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}&\le (\tilde{S}+1)^{n+1}\Big (\frac{\Vert V\Vert _{\mathcal {K}}}{2\pi }\Big )^n. \end{aligned}$$

(6.8)

Define

$$\begin{aligned} k_{N,1}^n(\tilde{x},\tilde{y}):=\frac{\delta }{N}\sum _{j=N/2\delta }^{2N/\delta }k_{N,j,1}^n(\tilde{x},\tilde{y}), \end{aligned}$$

then it satisfies (6.3) and (6.4). $\square$

Lemma 6.2

(Summability in n) Let $\epsilon >0$ be a small number chosen at the beginning of this section. For $N_0<N<N_1$, there exists $k_{N,2}^n(\tilde{x},\tilde{y})$ such that

(6.9)

and

$$\begin{aligned} \Vert k_{N,2}^n(\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}\le (1+\tilde{S})^{n+1}\epsilon ^n. \end{aligned}$$

(6.10)

Proof

Again, we consider (6.5). By the choice of $\epsilon$ and $\delta$ and Lemma 3.1 (ii), there exists an integral operator B such that $|B_{\lambda ,\lambda _j}(x,y)|\le B(x,y)$ for $|\lambda -\lambda _j|<\delta$, $\lambda ,\lambda _j\ge 0$, and $\Vert B\Vert _{L^1\rightarrow L^1}\le \epsilon$. Let $|\tilde{S}_{\lambda _j}|$ be the integral operator with integral kernel $|\tilde{S}_{\lambda _j}(x,y)|$. Then, we have

$$\begin{aligned} |(6.5)|\lesssim N\Vert m\Vert _{\mathcal {H}(s)}[(I+|\tilde{S}_{\lambda _j}|)(B(I+|\tilde{S}_{\lambda _j}|))^n](\tilde{x},\tilde{y}) \end{aligned}$$

(6.11)

and

$$\begin{aligned} \Vert [(I+|\tilde{S}_{\lambda _j}|)(B(I+|\tilde{S}_{\lambda _j}|))^n](\tilde{x},\tilde{y})\Vert _{L_{\tilde{y}}^\infty L_{\tilde{x}}^1}\le (1+\tilde{S})^{n+1}\epsilon ^n. \end{aligned}$$

(6.12)

Therefore, we define

$$\begin{aligned} k_2^n(\tilde{x},\tilde{y}):=\frac{\delta }{N}\sum _{j=N/2\delta }^{2N/\delta }[(I+|\tilde{S}_{\lambda _j}|)(B(I+|\tilde{S}_{\lambda _j}|))^n](\tilde{x},\tilde{y}), \end{aligned}$$

(6.13)

then it satisfies (6.9) and (6.10). $\square$

7 Application to the Nonlinear Schrödinger Equation

7.1 Two Norm Estimates

Following the argument in [6], we begin with proving the boundedness of the imaginary power operators. For $\alpha \in \mathbb {R}$, the imaginary power operator $H^{i\alpha }P_c$ is defined as a spectral multiplier of symbol $\lambda ^{i\alpha }1_{[0,+\infty )}$. We consider $H^{i\alpha }P_c$ instead of $H^{i\alpha }$ just for convenience’s sake. Indeed, by the assumptions, H has only finitely many negative eigenvalues, and the projection $P_{\lambda _j}$ is bounded on $L^r$ for any $1<r<\infty$ (see Lemma 3.6). Therefore, the boundedness of $H^{i\alpha }P_c$ implies that of $H^{i\alpha }=H^{i\alpha }P_c+\sum \lambda _j^{i\alpha }P_{\lambda _j}$, where $\lambda _j$’s are negative eigenvalues of H.

Lemma 7.1

(Imaginary power operator) If $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on $[0,+\infty )$, then for $\alpha \in \mathbb {R}$,

$$\begin{aligned} \Vert H^{i\alpha }P_c\Vert _{L^r\rightarrow L^r}\lesssim \langle \alpha \rangle ^3,\ 1<r<\infty . \end{aligned}$$

(7.1)

Proof

Since $\Vert \lambda ^{i\alpha }1_{[0,+\infty )}\Vert _{\mathcal {H}(3)}\lesssim \langle \alpha \rangle ^3$, the lemma follows from Theorem 1.1. $\square$

Proposition 7.2

(Two norm estimates) If $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on $[0,+\infty )$, then for $0\le s\le 2$ and $1<r<\frac{3}{s}$,

$$\begin{aligned} \Vert H^\frac{s}{2}P_c(-\Delta )^{-\frac{s}{2}}f\Vert _{L^r}&\lesssim \Vert f\Vert _{L^r}, \end{aligned}$$

(7.2)

$$\begin{aligned} \Vert (-\Delta )^{\frac{s}{2}}H^{-\frac{s}{2}}P_cf\Vert _{L^r}&\lesssim \Vert f\Vert _{L^r}. \end{aligned}$$

(7.3)

Proof

(7.2) Pick $f,g\in L^1\cap L^\infty$ such that ${\text {supp}}\hat{f}\subset B(0,R)\setminus B(0,r)$, $P_{n\le \cdot \le N}g=P_cg$ for some $R, r, N, n>0$. Note that by Lemma 3.5, the collection of such f (g, resp) is dense in $L^r$ ($L^{r'}$, resp). We define

$$\begin{aligned} F(z):=\langle H^zP_c(-\Delta )^{-z}f, g\rangle _{L^2}=\langle (-\Delta )^{-{\text {Re}}z-i{\text {Im}}z}f, H^{-i{\text {Im}}z}H^{{\text {Re}}z}g\rangle _{L^2}. \end{aligned}$$

(7.4)

Indeed, F(z) is well-defined, since $(-\Delta )^{-{\text {Re}}z-i{\text {Im}}z}f, H^{-i{\text {Im}}z}H^{{\text {Re}}z}g\in L^2$. Moreover, F(z) is continuous on $S=\{z: 0\le {\text {Re}}z\le 1\}\subset \mathbb {C}$, and it is analytic in the interior of S. We claim that $HP_c(-\Delta )^{-1}$ is bounded on $L^r$ for $1<r<\frac{3}{2}$. Indeed, by Lemma 3.6 (i),

$$\begin{aligned} \Vert HP_c(-\Delta )^{-1}f\Vert _{L^r}\lesssim \Vert (-\Delta +V)(-\Delta )^{-1}f\Vert _{L^r}\le \Vert f\Vert _{L^r}+\Vert V(-\Delta )^{-1}f\Vert _{L^r}. \end{aligned}$$

(7.5)

By the Hölder inequality (Lemma 7.5) and the Sobolev inequality in the Lorentz norms (Corollary 7.9), we have

$$\begin{aligned} \Vert V(-\Delta )^{-1}f\Vert _{L^r}\le \Vert V\Vert _{L^{3/2,\infty }}\Vert (-\Delta )^{-1}f\Vert _{L^{\frac{3r}{3-2r},r}}\lesssim \Vert f\Vert _{L^r}. \end{aligned}$$

(7.6)

Hence, by the claim and Lemma 7.1, we get

$$\begin{aligned} |F(1\!+\!i\alpha )|&\le \Vert H^{1+i\alpha }P_c (-\Delta )^{-1-i\alpha }f\Vert _{L^r}\Vert g\Vert _{L^{r'}}\!\lesssim \! \langle \alpha \rangle ^6\Vert f\Vert _{L^r}\Vert g\Vert _{L^{r'}},&(1\!<\!r\!<\!\tfrac{3}{2}),\end{aligned}$$

(7.7)

$$\begin{aligned} |F(i\alpha )|&\le \Vert H^{i\alpha }P_c(-\!\Delta )^{-i\alpha }f\Vert _{L^r}\Vert g\Vert _{L^{r'}}\!\lesssim \! \langle \alpha \rangle ^6\Vert f\Vert _{L^r}\Vert g\Vert _{L^{r'}},&(1\!<\!r\!<\!\infty ). \end{aligned}$$

(7.8)

Therefore (7.2) follows from the Stein’s complex interpolation theorem.

(7.3) Pick f and g as above, and consider

$$\begin{aligned} G(z):=\langle (-\Delta )^zH^{-z}P_c g, f\rangle _{L^2}. \end{aligned}$$

(7.9)

We claim that $(-\Delta )H^{-1}P_cg$ is bounded on $L^r$ for $1<r<\frac{3}{2}$. By the triangle inequality,

$$\begin{aligned} \Vert (-\Delta )H^{-1}P_cg\Vert _{L^r}=\Vert (H-V)H^{-1}P_cg\Vert _{L^r}\le \Vert P_cg\Vert _{L^r}+\Vert VH^{-1}P_cg\Vert _{L^r}. \end{aligned}$$

(7.10)

By Lemma 3.6 (i), $\Vert P_cg\Vert _{L^r}\lesssim \Vert g\Vert _{L^r}$. By the Hölder inequality in the Lorentz norms (Lemma 7.5) and the Sobolev inequality associated with H [12, Theorem 1.9], we get

$$\begin{aligned} \Vert VH^{-1}P_cg\Vert _{L^r}\le \Vert V\Vert _{L^{3/2,\infty }}\Vert H^{-1}P_cg\Vert _{L^{\frac{3r}{3-2r},r}}\lesssim \Vert V\Vert _{L^{3/2,\infty }}\Vert g\Vert _{L^r}. \end{aligned}$$

(7.11)

Repeating the above argument with the complex interpolation, we complete the proof. $\square$

7.2 Local Well-Posedness

Now we are ready to show the local well-posedness (LWP) of a 3d quintic nonlinear Schrödinger equation

$$\begin{aligned} iu_t+\Delta u-Vu\pm |u|^4u=0;\ u(0)=u_0.\qquad \qquad ({\mathrm{NLS}}_V) \end{aligned}$$

Theorem 7.3

(LWP) If $V\in \mathcal {K}_0\cap L^{3/2,\infty }$ and H has no eigenvalue or resonance on $[0,+\infty )$, then $(NLS _V)$ is locally well-posed in $\dot{H}^1$. Precisely, for $A>0$, there exists $\delta =\delta (A)>0$ such that for an initial data $u_0\in \dot{H}^1$ obeying

$$\begin{aligned} \Vert \nabla u_0\Vert _{L^2}\le A\quad and \quad \Vert e^{-itH} u_0\Vert _{L_{t\in [0,T_0]}^{10}L_x^{10}}<\delta , \end{aligned}$$

(7.12)

$(NLS _V)$ has a unique solution $u\in C_t(I; \dot{H}_x^1)$, with $I=[0,T)\subset [0,T_0]$, such that

$$\begin{aligned} \Vert \nabla u\Vert _{L_{t\in I}^{10}L_x^{30/13}}<\infty \quad and \quad \Vert u\Vert _{L_{t\in I}^{10}L_x^{10}}<2\delta . \end{aligned}$$

(7.13)

Proof

(Step 1 Contraction mapping argument) Let $\psi _j$ be the eigenfunction corresponding to the negative eigenvalue $\lambda _j$ normalized so that $\Vert \psi _j\Vert _{L^2}=1$. Choose small $T\in (0,T_0)$ such that $\Vert \psi _j\Vert _{L_{t\in I}^{10}L_x^{10}}, \Vert \psi _j\Vert _{L_{t\in I}^2L_x^2}\le 1$ for all j, where $I=[0,T]$ and $\psi _j(t,x)=\psi _j(x)$ for all $t\in I$. For notational convenience, we omit the time interval I in the norm $\Vert \cdot \Vert _{L_{t\in I}^p}$ if there is no confusion. Following a standard contraction mapping argument [3, 21], we aim to show that

$$\begin{aligned} \Phi _{u_0}(v)(t):=e^{-itH}u_0\pm i\int _0^t e^{-i(t-s)H}(|v|^4v(s))ds \end{aligned}$$

(7.14)

is a contraction map on the set

$$\begin{aligned} B_{a,b}:=\{v: \Vert v\Vert _{L_{t,x}^{10}}\le a,\ \Vert \nabla v\Vert _{L_t^{10}L_x^{30/13}}\le b\}, \end{aligned}$$

(7.15)

equipped with the metric $d(u,v)=\Vert u-v\Vert _{L_{t,x}^{10}}+\Vert \nabla (u-v)\Vert _{L_t^{10}L_x^{30/13}}$, where a, b and $\delta$ will be chosen later.

We claim that $\Phi _{u_0}$ maps from $B_{a,b}$ to itself. We write

$$\begin{aligned} \Vert \Phi _{u_0}(v)\Vert _{L_{t,x}^{10}}&\le \Vert e^{-itH}u_0\Vert _{L_{t,x}^{10}}+\Big \Vert \int _0^t e^{-i(t-s)H}P_c(|v|^4v(s))ds\Big \Vert _{L_{t,x}^{10}} \nonumber \\&\quad \quad +\sum _{j=1}^J\Big \Vert \int _0^t e^{-i(t-s)H}(\langle |v|^4v(s),\psi _j\rangle _{L^2}\psi _j)ds\Big \Vert _{L_{t,x}^{10}} \nonumber \\&=I+II+\sum _{j=1}^JIII_j. \end{aligned}$$

(7.16)

By assumption, $I\le \delta$. For II, by the Sobolev inequality associated with H [12, Theorem 1.6], Strichartz estimates (Proposition 1.2) and the two norm estimates, we get

$$\begin{aligned} II&\lesssim \Big \Vert \int _0^t e^{-i(t-s)H}P_cH^{1/2}(|v|^4v(s))ds\Big \Vert _{L_t^{10}L_x^{30/13}}\lesssim \Vert H^{1/2}P_c(|v|^4v)\Vert _{L_t^2L_x^{6/5}}\nonumber \\&\lesssim \Vert \nabla (|v|^4v)\Vert _{L_t^2L_x^{6/5}}\le 3\Vert (v^2\nabla v)(\bar{v})^2\Vert _{L_t^2L_x^{6/5}}+2\Vert (v^2\nabla v)(\bar{v})^2\Vert _{L_t^2L_x^{6/5}}\\&\lesssim \Vert v\Vert _{L_{t,x}^{10}}^4\Vert \nabla v\Vert _{L_t^{10}L_x^{30/13}}\le a^4b.\nonumber \end{aligned}$$

(7.17)

For the last term, by the Hölder inequality, the choice of T and (7.17), we obtain

$$\begin{aligned} III_j&=\Big \Vert \int _0^t e^{-i(t-s)\lambda _j}(\langle |v|^4v(s),\psi _j\rangle _{L^2}\psi _j)ds\Big \Vert _{L_{t,x}^{10}} \nonumber \\&\le \Big (\int _0^T |\langle |v|^4v(s),\psi _j\rangle _{L^2}|ds\Big )\Vert \psi _j\Vert _{L_{t,x}^{10}} \nonumber \\&\le \Vert \nabla (|v|^4v)\Vert _{L_t^2 L_x^{6/5}}\Vert |\nabla |^{-1}\psi _j\Vert _{L_t^2 L_x^6} \nonumber \\&\lesssim \Vert \nabla (|v|^4v)\Vert _{L_t^2 L_x^{6/5}}\Vert \psi _j\Vert _{L_t^2 L_x^2}\le a^4b. \end{aligned}$$

(7.18)

Therefore, we prove that

$$\begin{aligned} \Vert \Phi _{u_0}(v)\Vert _{L_{t,x}^{10}}\le \delta +Ca^4b. \end{aligned}$$

(7.19)

Next, we write

$$\begin{aligned} \Vert \nabla \Phi _{u_0}(v)\Vert _{L_t^{10}L_x^{30/13}}&\le \Vert \nabla P_c\Phi _{u_0}(v)\Vert _{L_t^{10}L_x^{30/13}}+\sum _{j=1}^J\Vert \nabla P_{\lambda _j}\Phi _{u_0}(v)\Vert _{L_t^{10}L_x^{30/13}} \nonumber \\&=\tilde{I}+\sum _{j=1}^J\tilde{II}_j. \end{aligned}$$

(7.20)

For $\tilde{I}$, by the two norm estimates, Strichartz estimates and (7.17), we obtain

$$\begin{aligned} \tilde{I}&\lesssim \Vert H^{1/2}P_c\Phi _{u_0}(v)\Vert _{L_t^{10}L_x^{30/13}} \nonumber \\&\lesssim \Vert H^{1/2}P_cu_0\Vert _{L^2}+\Vert H^{1/2}P_c(|v|^4v)\Vert _{L_t^2L_x^{6/5}}\\&\lesssim \Vert \nabla u_0\Vert _{L^2}+\Vert H^{1/2}P_c(|v|^4v)\Vert _{L_t^2L_x^{6/5}}\lesssim A+a^4b.\nonumber \end{aligned}$$

(7.21)

For $\tilde{II}$, by the Hölder inequality, (7.19) and Lemma 3.6, we get

$$\begin{aligned} \tilde{II}_j&\le \Vert \langle \Phi _{u_0}(v),\psi _j\rangle _{L^2}\Vert _{L_t^{10}}\Vert \psi _j\Vert _{L^{30/13}} \nonumber \\&\le \Vert \Phi _{u_0}(v)\Vert _{L_{t,x}^{10}}\Vert \psi _j\Vert _{L_x^{10/9}}\lesssim \delta +a^4b. \end{aligned}$$

(7.22)

Collecting all, we prove that

$$\begin{aligned} \Vert \nabla \Phi _{u_0}(v)\Vert _{L_t^{10}L_x^{30/13}}\le CA+Ca^4b. \end{aligned}$$

(7.23)

Let $b=2AC$, $a=\min ((2C)^{-\frac{1}{4}}, (2Cb)^{-\frac{1}{3}})$ and $\delta =\frac{a}{2}$ $(\Rightarrow Ca^4b\le AC$ and $Ca^3b\le \frac{1}{2})$. Then, by (7.19) and (7.23), $\Phi _{u_0}$ maps from $B_{a,b}$ to itself. Similarly, one can show that $\Phi _{u_0}$ is contractive in $B_{a,b}$. Thus, we conclude that there exists unique $u\in B_{a,b}$ such that

$$\begin{aligned} u(t)=\Phi _{u_0}(u)=e^{-itH}u_0+i\int _0^t e^{-i(t-s)H} (|u|^4u)(s)ds. \end{aligned}$$

(7.24)

(Step 2 Continuity) In order to show that $u(t)\in C_t(I; \dot{H}_x^1)$, we write

$$\begin{aligned} u(t)&=e^{-itH}\left( P_cu_0+\sum _{j=1}^JP_{\lambda _j}u_0\right) \nonumber \\&\quad \pm i\int _0^t e^{-i(t-s)H} \Big (P_c(|u|^4u)(s)+\sum _{j=1}^J P_{\lambda _j}(|u|^4u)(s)\Big ) ds \nonumber \\&=e^{-itH}P_cu_0+\sum _{j=1}^J e^{-it\lambda _j}P_{\lambda _j}u_0\pm i\int _0^t e^{-i(t-s)H} P_c(|u|^4u)(s)ds\\&\quad \pm i\sum _{j=1}^J \int _0^t e^{-i(t-s)\lambda _j}P_{\lambda _j}(|u|^4u)(s) ds \nonumber \\&=:I(t)+\sum _{j=1}^JII_j(t)+III(t)+\sum _{j=1}^JIV_j(t).\nonumber \end{aligned}$$

(7.25)

For I(t), by the two norm estimates and $L^2$-continuity of $e^{-itH}$, we have

$$\begin{aligned} \Vert I(t)-I(t_0)\Vert _{\dot{H}^1}\lesssim \Vert (e^{-itH}-e^{-it_0H})H^{1/2}P_cu_0\Vert _{L^2}\rightarrow 0\quad as t\rightarrow t_0, \end{aligned}$$

(7.26)

since $\Vert H^{1/2}P_cu_0\Vert _{L^2}\lesssim \Vert u\Vert _{\dot{H}^1}<\infty$. $II_j(t)$ is continuous in $\dot{H}^1$, since

$$\begin{aligned} \Vert P_{\lambda _j}u_0\Vert _{\dot{H}^1}=|\langle u_0,\psi _j\rangle _{L^2}|\Vert \psi _j\Vert _{\dot{H}^1}\lesssim \Vert u_0\Vert _{\dot{H}^1}\Vert \psi _j\Vert _{\dot{H}^{-1}}\lesssim \Vert u_0\Vert _{\dot{H}^1}\Vert \psi _j\Vert _{L^{6/5}}<\infty . \end{aligned}$$

(7.27)

For III(t), by the two norm estimates, Strichartz estimates and (7.17), we have

$$\begin{aligned} \Vert III(t)-III(t_0)\Vert _{\dot{H}^1}&\lesssim \Vert H^{1/2}(III(t)-III(t_0))\Vert _{L^2} \nonumber \\&\lesssim \Vert H^{1/2}P_c(|u|^4u)\Vert _{L_{s\in [t_0, t]}^2 L_x^{6/5}}\rightarrow 0\quad as t\rightarrow t_0. \end{aligned}$$

(7.28)

For $IV_j(t)$, by the Hölder inequality and (7.17), we write

$$\begin{aligned} \Vert IV_j(t)-IV_j(t_0)\Vert _{\dot{H}^1}&\le \Vert \psi _j\Vert _{\dot{H}^1}\Vert \nabla (|u|^4u)(s)\Vert _{L_{t\in [t_0,t]}^2L_x^{6/5}}\Vert |\nabla |^{-1}\psi _j\Vert _{L_{s\in [t_0,t]}^2L_x^6} \nonumber \\&\lesssim \Vert \nabla (|u|^4u)(s)\Vert _{L_{s\in [t_0,t]}^2L_x^{6/5}}\Vert \psi _j\Vert _{L_t^2L_x^2}\rightarrow 0\quad as t\rightarrow t_0. \end{aligned}$$

(7.29)

Collecting all, we conclude that u(t) is continuous in $\dot{H}^1$. $\square$

Appendix: Lorentz Spaces and Interpolation Theorem

Following [21], we summarize useful properties of the Lorentz spaces. Let $(X,\mu )$ be a measure space. The Lorentz (quasi) norm is defined by

(7.30)

Lemma 7.4

(Properties of the Lorentz spaces) Let $1\le p\le \infty$ and $1\le q, q_1, q_2\le \infty$.

• (i) $L^{p,p}=L^p$, and $L^{p,\infty }$ is the weak $L^p$-space.

• (ii) If $q_1\le q_2$, $L^{p,q_1}\subset L^{p,q_2}$.

Lemma 7.5

(Hölder inequality) If $1\le p, p_1, p_2, q,q_1,q_2\le \infty$, $\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}$ and $\frac{1}{q}=\frac{1}{q_1}+\frac{1}{q_2}$, then

$$\begin{aligned} \Vert fg\Vert _{L^{p,q}}\lesssim \Vert f\Vert _{L^{p_1,q_1}}\Vert fg\Vert _{L^{p_2,q_2}}. \end{aligned}$$

(7.31)

Lemma 7.6

(Dual characterization of $L^{p,q}$) If $1<p<\infty$ and $1\le q\le \infty$, then

$$\begin{aligned} \Vert f\Vert _{L^{p,q}}\sim \sup _{\Vert g\Vert _{L^{p',q'}}\le 1}\Big |\int _X f\bar{g} d\mu \Big |. \end{aligned}$$

(7.32)

A measurable function f is called a sub-step function of height H and width W if f is supported on a set E with measure $\mu (E)=W$ and $|f(x)|\le H$ almost everywhere. Let T be a linear operator that maps the functions on a measure space $(X,\mu _X)$ to functions on another measure space $(Y,\mu _Y)$. We say that T is restricted weak-type $(p,\tilde{p})$ if

$$\begin{aligned} \Vert Tf\Vert _{L^{\tilde{p},\infty }}\lesssim HW^{1/p} \end{aligned}$$

(7.33)

for all sub-step functions f of height H and width W.

Theorem 7.7

(Marcinkiewicz interpolation theorem) Let T be a linear operator such that

$$\begin{aligned} \langle Tf,g\rangle _{L^2}=\int _Y Tf \bar{g}d\mu _Y \end{aligned}$$

(7.34)

is well-defined for all simple functions f and g. Let $1\le p_0,p_1, \tilde{p}_0, \tilde{p}_1\le \infty$. Suppose that T is restricted weak-type $(p_i, \tilde{p}_i)$ with constant $A_i>0$ for $i=0,1$. Then,

$$\begin{aligned} \Vert Tf\Vert _{L^{\tilde{p}_\theta ,q}}\lesssim A_0^{1-\theta }A_1^\theta \Vert f\Vert _{L^{p_\theta ,q}}, \end{aligned}$$

(7.35)

where $0<\theta <1$, $\frac{1}{p_\theta }=\frac{1-\theta }{p_0}+\frac{\theta }{p_1}$, $\frac{1}{\tilde{p}_\theta }=\frac{1-\theta }{\tilde{p}_0}+\frac{\theta }{\tilde{p}_1}$, $\tilde{p}_\theta >1$ and $1\le q\le \infty$.

In this paper, we use the interpolation theorem of the following form.

Corollary 7.8

(Marcinkiewicz interpolation theorem) Let T be a linear operator. Let $1\le p_1<p_2\le \infty$. Suppose that for $i=0,1$, T is bounded from $L^{p_i,1}$ to $L^{p_i,\infty }$. Then T is bounded on $L^p$ for $p_1<p<p_2$.

Proof

The corollary follows from Theorem 7.7, since T is restricted weak-type $(p_i, p_i)$:

$$\begin{aligned} \Vert f\Vert _{L^{p_i,1}}=p_i\int _0^\infty \mu (|f|\ge \lambda )^{1/p_i} d\lambda \le p_i\int _0^H W^{1/p_i}d\lambda =p_i HW^{1/p_i}, \end{aligned}$$

(7.36)

for a sub-step function f of height H and width W. $\square$

Corollary 7.9

(Fractional integration inequality in the Lorentz spaces)

$$\begin{aligned} \Big \Vert \int _{\mathbb {R}^d}\frac{f(y)}{|x-y|^{d-s}}dy\Big \Vert _{L^{q,r}(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^{p,r}}, \end{aligned}$$

(7.37)

where $1<p<q<\infty$, $1\le r\le \infty$ and $\frac{1}{q}=\frac{1}{p}-\frac{s}{d}$. At the endpoints, we have

$$\begin{aligned} \Big \Vert \int _{\mathbb {R}^d}\frac{f(y)}{|x-y|^{d-s}}dy\Big \Vert _{L^{\frac{d}{d-s},\infty }(\mathbb {R}^d)}&\lesssim \Vert f\Vert _{L^1}, \Big \Vert \int _{\mathbb {R}^d}\frac{f(y)}{|x-y|^{d-s}}dy\Big \Vert _{L^\infty (\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^{d/s,1}}. \end{aligned}$$

(7.38)

Proof

(7.38) follows from [18, Theorem 1, p. 119] and duality. Then, (7.37) follows from Corollary 7.8. $\square$