Some Properties of Threshold Eigenstates and Resonant States of Discrete Schrödinger Operators

Abstract

We consider for discrete Schrödinger operators with dimension three or larger on the square lattice. We prove that the asymptotic behavior at infinity of resonances at elliptic thresholds is similar to those of continuous Schrödinger operators and that resonances are absent at hyperbolic thresholds. Moreover, we show the limiting absorption principle near the hyperbolic thresholds with the optimal weight.

Appendices

Appendix A: Lorentz Space

For a measure space $(X,\mu )$, $L^{p,r}(X,\mu )$ denotes the Lorentz space for $1\le p\le \infty$ and $1\le r\le \infty$:

$$\begin{aligned}&\Vert f\Vert _{L^{p,r}(X)}={\left\{ \begin{array}{ll} p^{\frac{1}{r}}\left( \int _0^{\infty }\mu (\{x\in X\,|\, |f(x)|>\alpha \})^{\frac{r}{p}} \alpha ^{r-1}\mathrm{d}\alpha \right) ^{\frac{1}{r}},\quad &{}r<\infty ,\\ \sup _{\alpha>0}\alpha \mu (\{x\in X\mid |f(x)|>\alpha \})^{\frac{1}{p}},\quad &{}r=\infty , \end{array}\right. }\\&L^{p,r}(X,\mu )=\{f:X\rightarrow {\mathbb {C}}\mid f:\text {measurable},\, \Vert f\Vert _{L^{p,r}(X)}<\infty \}. \end{aligned}$$

Moreover, we denote $L^{p,r}({\mathbb {R}}^d)=L^{p,r}({\mathbb {R}}^d, \mu _L)$ and $l^{p}_r({\mathbb {Z}}^d)=L^{p,r}({\mathbb {Z}}^d, \mu _c)$, where $\mu _L$ is the Lebesgue measure on ${\mathbb {R}}^d$ and $\mu _c$ is the counting measure on ${\mathbb {Z}}^d$. For a detail, see [1]. In this section, we state some fundamental properties of the Lorentz spaces. Note that $L^{p,p}(X,\mu )=L^{p}(X)$.

Lemma A.1

(The Young inequalities in the Lorentz spaces). Let $1<p_i<\infty$, $1\le q_i\le \infty$ with $\frac{1}{r}=\frac{1}{p_1}+\frac{1}{p_2}-1>0$ and $s\ge 1$ with $\frac{1}{q_1}+\frac{1}{q_2}\ge \frac{1}{s}$. Then, we have

$$\begin{aligned} \Vert f*g\Vert _{l^{r}_s({\mathbb {Z}}^d)}\le C\Vert f\Vert _{l^{p_1}_{q_1}({\mathbb {Z}}^d)}\Vert g\Vert _{l^{p_2}_{q_2}({\mathbb {Z}}^d)}. \end{aligned}$$

Lemma A.2

(The Hölder inequalities in the Lorentz spaces). If $1\le p_1,p_2,q_1,q_2\le \infty$ and $1\le r\le \infty$ satisfy

$$\begin{aligned} \frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{r}<1, \end{aligned}$$

then

$$\begin{aligned} \Vert fg\Vert _{l^{r}_{\min (q_1,q_2)}({\mathbb {Z}}^d)}\le \Vert f\Vert _{l^{p_1}_{q_1}({\mathbb {Z}}^d)}\Vert g\Vert _{l^{p_2}_{q_2}({\mathbb {Z}}^d)}. \end{aligned}$$

For these proofs, see [6].

Appendix B: Harmonic Analysis

Proposition B.1

Let $m\in C^{\infty }({\mathbb {R}}^d{\setminus }\{0\})$ be a function on ${\mathbb {R}}^d$ which is compactly supported, $C^{\infty }$ for $\xi \ne 0$ and satisfies for a $0\le k<d$ that

$$\begin{aligned} |\partial _{\xi }^{\alpha }m(\xi )|\le C_{\alpha }|\xi |^{-k-|\alpha |},\quad x\in {\mathbb {R}}^d \end{aligned}$$

(B.1)

for $|\alpha |\le d-k+1$. Then, if we set

$$\begin{aligned} I=\int _{{\mathbb {R}}^d}e^{-2\pi ix\cdot \xi }m(\xi )\mathrm{d}\xi , \end{aligned}$$

then $|I|\le C\langle x \rangle ^{-d+k}$.

Proof

Since m is compactly supported, we may assume $|x|\ge 1$. Take $\chi \in C_c^{\infty }({\mathbb {R}}^d)$ such that $\chi =1$ on $|\xi |\le 1$ and $\chi =0$ on $|\xi |\ge 2$. Set ${\bar{\chi }}=1-\chi$. For $\delta >0$, we have

$$\begin{aligned} I=\int _{{\mathbb {R}}^d}(\chi (\xi /\delta )+{\bar{\chi }}(\xi /\delta ))e^{-2\pi ix\cdot \xi }m(\xi )\mathrm{d}\xi =:I_1+I_2. \end{aligned}$$

Since m is integrable on ${\mathbb {R}}^d$, we have

$$\begin{aligned} |I_1|\le \int _{|\xi |\le 2\delta }|\chi (\xi /\delta )||\xi |^{-k}\mathrm{d}\xi \le C\delta ^{d-k}. \end{aligned}$$

By integrating by parts, for $N> d-k$, we have

$$\begin{aligned} |I_2|&\le C|x|^{-N}\sum _{|\alpha |= N}\left| \int _{{\mathbb {R}}^d}e^{-2\pi ix\cdot \xi } D_{\xi }^{\alpha }({\bar{\chi }}(\xi /\delta )m(\xi )) \mathrm{d}\xi \right| \\&\le C|x|^{-N}\sum _{|\alpha |= N}\left| \sum _{\beta \le \alpha }\int _{{\mathbb {R}}^d}e^{-2\pi ix\cdot \xi } D_{\xi }^{\beta }({\bar{\chi }}(\xi /\delta ))\partial _{\xi }^{\alpha -\beta }m(\xi ) \mathrm{d}\xi \right| \\&\le C|x|^{-N}\sum _{|\alpha |\le N}\sum _{\beta \le \alpha }\int _{{\mathbb {R}}^d} \delta ^{-|\beta |}{\bar{\chi }}^{(\beta )}(\xi /\delta )|\xi |^{-k-(N-|\beta |)} \mathrm{d}\xi . \end{aligned}$$

For $\beta =0$,

$$\begin{aligned} \int _{{\mathbb {R}}^d}{\bar{\chi }}(\xi /\delta )|\xi |^{-k-N}\mathrm{d}\xi \le C\delta ^{d-k-N} \end{aligned}$$

follows and for $\beta \ne 0$,

$$\begin{aligned} \int _{{\mathbb {R}}^d} \delta ^{-|\beta |}{\bar{\chi }}^{(\beta )}(\xi /\delta )|\xi |^{-k-(N-|\beta |)} \mathrm{d}\xi&\le C\int _{\delta \le |\xi |\le 2\delta }\delta ^{-|\beta |}|\xi |^{-k-N+|\beta |}\mathrm{d}\xi \\&\le C\delta ^{d-k-N}. \end{aligned}$$

These imply $|I_2|\le C|x|^{-N}\delta ^{d-k-N}$. We set $\delta =|x|^{-1}$ and obtain $|I|\le C|x|^{-d+k}$ for $|x|\ge 1$. $\square$

Corollary B.2

Let $d\ge 1$, $0<l<d$ and $K_{l}$ be defined by

$$\begin{aligned} K_l(x)=\int _{{\mathbb {T}}^d}e^{2\pi ix{{\dot{\xi }}}}h_0(\xi )^{-l/2}\mathrm{d}\xi . \end{aligned}$$

Then, we have a pointwise bound $|K_l(x)|\le C\langle x \rangle ^{-d+l}$.

Proof

By the Morse lemma, we have $|\partial _{\xi }^{\alpha }h_0(\xi )^{-l/2}|\le C_{\alpha }|\xi |^{-l-|\alpha |}$ near $\xi =0$ for any multi-index $\alpha$. Moreover, it follows that $h_0(\xi )^{-l/2}$ is smooth away from $\xi =0$. Applying Proposition B.1, we obtain $|K_l(x)|\le C\langle x \rangle ^{-d+l}$. $\square$

Now we define operators $H_0^{-l/2}$ for $0<l<d$ by

$$\begin{aligned} H_0^{-l/2}u(x)=\sum _{y\in {\mathbb {Z}}^d}K_l(x-y)u(y),\,\, u\in \bigcap _{s>0}l^{2,s}({\mathbb {Z}}^d). \end{aligned}$$

It is easily seen that $H_0^{-l/2}$ is a continuous linear operator:

$$\begin{aligned} H_0^{-l/2}:\bigcap _{s>0}l^{2,s}({\mathbb {Z}}^d)\rightarrow \bigcup _{s\in {\mathbb {R}}}l^{2,s}({\mathbb {Z}}^d). \end{aligned}$$

The next corollary implies that $H_0^{-1}$ can be uniquely extended to the continuous linear operator from $l^{2,\alpha }({\mathbb {Z}}^d)$ to $l^{2,-\beta }({\mathbb {Z}}^d)$ for $\alpha ,\beta >1/2$ with $\alpha +\beta \ge 2$.

Corollary B.3

(Discrete version of the HLS inequality). Let $d\ge 1$ and $0<l<d$. Then, $H_0^{-l/2}$ is bounded from $l^{p}_r({\mathbb {Z}}^d)$ to $l^{q}_r({\mathbb {Z}}^d)$ if $1< p< q < \infty$ satisfies

$$\begin{aligned} \frac{1}{p}-\frac{1}{q}=\frac{l}{d} \end{aligned}$$

(B.2)

and $1\le r\le \infty$.

Moreover, if $W_1\in l^{r_1}_{\infty }({\mathbb {Z}}^d)$ and $W_2\in l^{r_2}_{\infty }({\mathbb {Z}}^d)$ with $1/r_1+1/r_2=l/d$ with $r_1,r_2>2$. Then, we have

$$\begin{aligned} W_1H_0^{-l/2}W_2\in B(l^2({\mathbb {Z}}^d)). \end{aligned}$$

In particular, $\langle x \rangle ^{-\alpha }H_0^{-1}\langle x \rangle ^{-\beta }\in B(l^2({\mathbb {Z}}^d))$ if $\alpha +\beta \ge 2$ and $\alpha ,\beta >0$ if $d\ge 4$ and $\alpha +\beta \ge 2$ and $\alpha ,\beta >1/2$ if $d=3$.

Remark B.4

This corollary gives $H_0^{-l/2}\langle x \rangle ^{-l}\in B(l^2({\mathbb {Z}}^d))$ for $0<l<d$. In fact,

$$\begin{aligned} \left\| H_0^{-l/2}\langle x \rangle ^{-l}f\right\| _{l^2({\mathbb {Z}}^d)}\le C\left\| H_0^{-l/2}\right\| _{B\left( l^{\frac{2l}{l+2d}}_{\infty }({\mathbb {Z}}^d), l^2({\mathbb {Z}}^d)\right) } \Vert \langle x \rangle ^{-l}\Vert _{l^{\frac{d}{l}}_{\infty }({\mathbb {Z}}^d)} \Vert f\Vert _{l^2({\mathbb {Z}}^d)}. \end{aligned}$$

These are exactly the discrete Hardy inequalities.

Appendix C: Restriction Theorem for a Lipschitz Manifold

In this appendix, we prove the $L^2$-restriction theorem for a Lipschitz manifold. Its proof is standard; however, we give its proof for readers’ convenience.

Lemma C.1

Let $f\in H^{1}({\mathbb {R}}^d)$ and g be a real-valued Lipschitz function on ${\mathbb {R}}^{d-1}$. Then, it follows that $k(\xi )=f(\xi ', \xi _d+g(\xi '))$ belongs to $H^1({\mathbb {R}}^d)$ and there exists $C>0$ which depends only on the dimension d and $\Vert \partial _{\xi }g\Vert _{L^{\infty }({\mathbb {R}}^d)}$ such that

$$\begin{aligned} \Vert k\Vert _{H^1({\mathbb {R}}^d)}\le C\Vert f\Vert _{H^1({\mathbb {R}}^d)}. \end{aligned}$$

(C.1)

Proof

It is evident that $\Vert k\Vert _{L^2({\mathbb {R}}^d)}=\Vert f\Vert _{L^2({\mathbb {R}}^d)}$. For $j=1,\dots d-1$, we have

$$\begin{aligned} \partial _{\xi _j}(k(\xi ',\xi _d+g(\xi ')))&=(\partial _{\xi _j}k)(\xi ',\xi _d+g(\xi '))+(\partial _{\xi _j}g)(\xi ')(\partial _{\xi _d}k)(\xi ',\xi _d+g(\xi ')),\\ \partial _{\xi _d}(k(\xi ',\xi _d+g(\xi ')))&=(\partial _{\xi _d}k)(\xi ',\xi _d+g(\xi ')). \end{aligned}$$

Using this computation, we obtain (C.1). $\square$

Proposition C.2

Under the assumption of Lemma C.1, we have

$$\begin{aligned} \Vert \langle D_{\xi '} \rangle ^{1/2}(f(\xi ', g(\xi ')))\Vert _{L^2({\mathbb {R}}^{d-1})}\le C\Vert f\Vert _{H^1({\mathbb {R}}^d)}. \end{aligned}$$

Proof

In the following, we denote the Fourier transform of f by ${\hat{f}}$. By using Fourier inversion formula and by using Schwarz’s inequality, we have

$$\begin{aligned} \left| \int _{{\mathbb {R}}^{d-1}}f(\xi ',g(\xi '))e^{-2\pi ix'\cdot \xi '}\mathrm{d}\xi '\right|&=\left| \int _{{\mathbb {R}}^d} {\hat{k}}(x)\mathrm{d}x_d\right| \\&\le \left( \int _{{\mathbb {R}}}\langle x \rangle ^{-2}\mathrm{d}x_d\right) ^{1/2} \left( \int _{{\mathbb {R}}}|\langle x \rangle {\hat{k}}(x)|^2\mathrm{d}x_d\right) ^{1/2}\\&\le C\langle x' \rangle ^{-1/2}\left( \int _{{\mathbb {R}}}|\langle x \rangle {\hat{k}}(x)|^2\mathrm{d}x_d\right) ^{1/2}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \Vert \langle D_{\xi '} \rangle ^{1/2}(f(\xi ', g(\xi ')))\Vert _{L^2({\mathbb {R}}^{d-1})}^2&=\Vert \langle x' \rangle ^{1/2}\widehat{f(\xi ',g(\xi '))}(x)\Vert _{L^2({\mathbb {R}}^{d-1})}^2\\&\le C^2\Vert \langle x \rangle {\hat{k}}\Vert _{L^2({\mathbb {R}}^d)}^2\\&=C^2\Vert k\Vert _{H^1({\mathbb {R}}^d)}^2. \end{aligned}$$

This computation with Lemma C.1 completes the proof. $\square$