Localization for Almost-Periodic Operators with Power-law Long-range Hopping: A Nash-Moser Iteration Type Reducibility Approach

Abstract

In this paper we develop a Nash-Moser iteration type reducibility approach to prove the (inverse) localization for some d-dimensional discrete almost-periodic operators with power-law long-range hopping. We also provide a quantitative lower bound on the regularity of the hopping. As an application, some results of Sarnak (Comm Math Phys 84(3):377–401, 1982), Pöschel (Comm Math Phys 88(4):447–463, 1983), Craig (Comm Math Phys 88(1):113–131, 1983) and Bellissard et al. (Comm Math Phys 88(4):465–477, 1983) are generalized to the power-law hopping case.

Appendix A

In this appendix we prove Lemma 3.1.

Proof of Lemma 3.1

The proof is standard and is based on the Hölder inequality. For any $X\in \mathcal {M}^s$, recall that

$$\begin{aligned} X_\textbf{k}=(X_\textbf{k}(\textbf{i}))_{\textbf{i}\in \mathbb Z^d},\ X_\textbf{k}(\textbf{i})=X_{\textbf{i}, \textbf{i}-\textbf{k}}. \end{aligned}$$

We first show if $Z=XY,$ then

$$\begin{aligned} Z_\textbf{k}=\sum _{\textbf{j}\in \mathbb Z^d}X_\textbf{j}(\sigma _\textbf{j}Y_{\textbf{k}-\textbf{j}}). \end{aligned}$$

(6.6)

In fact, we have

$$\begin{aligned} Z_\textbf{k}(\textbf{i})=Z_{\textbf{i},\textbf{i}-\textbf{k}} =\sum _{\textbf{j}\in \mathbb Z^d}X_{\textbf{i},\textbf{i}-\textbf{j}}Y_{\textbf{i}-\textbf{j}, \textbf{i}-\textbf{k}}=\sum _{\textbf{j}\in \mathbb Z^d}X_\textbf{j}(\textbf{i})(\sigma _{\textbf{j}}Y_{\textbf{k}-\textbf{j}})(\textbf{i}), \end{aligned}$$

which implies (A.1).

Next, from (2.2), we obtain since (A.1)

$$\begin{aligned} \Vert Z\Vert _s^2=\sum _{\textbf{k}\in \mathbb Z^d}\Vert Z_\textbf{k}\Vert _\mathfrak {B}^2\langle \textbf{k}\rangle ^{2s}&=\sum _{\textbf{k}\in \mathbb Z^d}\Vert \sum _{\textbf{j}\in \mathbb Z^d}X_\textbf{j}(\sigma _\textbf{j}Y_{\textbf{k}-\textbf{j}})\Vert _\mathfrak {B}^2\langle \textbf{k}\rangle ^{2s}\\&\le \sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathbb Z^d}\Vert X_\textbf{j}\Vert _\mathfrak {B}\Vert \sigma _\textbf{j}Y_{\textbf{k}-\textbf{j}}\Vert _\mathfrak {B}\right) ^2\langle \textbf{k}\rangle ^{2s}\\&= \sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathbb Z^d}\Vert X_\textbf{j}\Vert _\mathfrak {B}\Vert Y_{\textbf{k}-\textbf{j}}\Vert _\mathfrak {B}\right) ^2\langle \textbf{k}\rangle ^{2s}, \end{aligned}$$

where in the last equality we use the translation invariance of $\mathfrak {B}$. Let $a_\textbf{k}=\Vert X_\textbf{k}\Vert _\mathfrak {B},\ b_{\textbf{k}}=\Vert Y_{\textbf{k}}\Vert _\mathfrak {B}$. It suffices to study the sum $\left( \sum _{\textbf{j}\in \mathbb Z^d}a_\textbf{j}b_{\textbf{k}-\textbf{j}}\right) ^2\langle \textbf{k}\rangle ^{2\,s}.$ We have the following two cases.

Case 1.:

$\textbf{j}\in \mathcal {I}_\textbf{k}:=\{\textbf{j}\in \mathbb Z^d:\ {\langle \textbf{k}\rangle ^{2\,s}}{\langle \textbf{k}-\textbf{j}\rangle ^{-2\,s}}\le 10\}.$ In this case we have

$$\begin{aligned} {\langle \textbf{k}\rangle ^{2s}}{\langle \textbf{k}-\textbf{j}\rangle ^{-2s}}\langle \textbf{j}\rangle ^{-2\alpha _0}\le 10\langle \textbf{j}\rangle ^{-2\alpha _0}. \end{aligned}$$

(6.7)

Hence we have by using the Hölder inequality

$$\begin{aligned} \sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathcal {I}_\textbf{k}}a_\textbf{j}b_{\textbf{k}-\textbf{j}}\right) ^2\langle \textbf{k}\rangle ^{2s}&\le \sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathcal {I}_\textbf{k}}a_\textbf{j}^2\langle \textbf{j}\rangle ^{2\alpha _0}\right) \left( \sum _{\textbf{j}\in \mathcal {I}_\textbf{k}}b_{\textbf{k}-\textbf{j}}^2\langle \textbf{j}\rangle ^{-2\alpha _0}\right) \langle \textbf{k}\rangle ^{2s}\\&\le \Vert X\Vert _{\alpha _0}^2\sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathcal {I}_\textbf{k}}b_{\textbf{k}-\textbf{j}}^2\langle \textbf{k}-\textbf{j}\rangle ^{2s}\langle \textbf{j}\rangle ^{-2\alpha _0}\langle \textbf{k}\rangle ^{2s}\langle \textbf{k}-\textbf{j}\rangle ^{-2s}\right) \\&\le 10\Vert X\Vert _{\alpha _0}^2\sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathcal {I}_\textbf{k}}b_{\textbf{k}-\textbf{j}}^2\langle \textbf{k}-\textbf{j}\rangle ^{2s}\langle \textbf{j}\rangle ^{-2\alpha _0}\right) \ (\mathrm{since (A.2)})\\&\le 10\Vert X\Vert _{\alpha _0}^2\sum _{\textbf{j}\in \mathbb Z^d}\langle \textbf{j}\rangle ^{-2\alpha _0}\left( \sum _{\textbf{k}\in \mathbb Z^d}b_{\textbf{k}-\textbf{j}}^2\langle \textbf{k}-\textbf{j}\rangle ^{2s}\right) \\&\le M_0^2\Vert X\Vert _{\alpha _0}^2\Vert Y\Vert _{s}^2, \end{aligned}$$

where $M_0=\sqrt{10\sum \limits _{\textbf{k}\in \mathbb Z^d}\langle \textbf{k}\rangle ^{-2\alpha _0}}<\infty$ since $\alpha _0>d/2.$

Case 2.:

$\textbf{j}\notin \mathcal {I}_\textbf{k}.$ In this case we must have $\textbf{k}\ne \textbf{0}.$ Then

$$\begin{aligned} \langle \textbf{k}\rangle&>10^{\frac{1}{2s}}\langle \textbf{k}-\textbf{j}\rangle >10^{\frac{1}{2s}}|\textbf{k}-\textbf{j}|\ge 10^{\frac{1}{2s}}(|\textbf{k}|-|\textbf{j}|)\ge 10^{\frac{1}{2s}}(\langle \textbf{k}\rangle -\langle \textbf{j}\rangle ), \end{aligned}$$

which yields

$$\begin{aligned} \langle \textbf{j}\rangle ^{-2s}\le (1-10^{-\frac{1}{2s}})^{-2s}\langle \textbf{k}\rangle ^{-2s}. \end{aligned}$$

(A.1)

Thus using again the Hölder inequality implies

$$\begin{aligned} \sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\notin \mathcal {I}_{\textbf{k}}}a_\textbf{j}b_{\textbf{k}-\textbf{j}}\right) ^2\langle \textbf{k}\rangle ^{2s}&\le \sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathbb Z^d}b_{\textbf{k}-\textbf{j}}^2\langle \textbf{k}-\textbf{j}\rangle ^{2\alpha _0}\right) \left( \sum _{\textbf{j}\notin \mathcal {I}_\textbf{k}}a_{\textbf{j}}^2\langle \textbf{k}-\textbf{j}\rangle ^{-2\alpha _0}\right) \langle \textbf{k}\rangle ^{2s}\\&\le \Vert Y\Vert _{\alpha _0}^2\sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\notin \mathcal {I}_\textbf{k}}a_{\textbf{j}}^2\langle \textbf{j}\rangle ^{2s}\langle \textbf{k}-\textbf{j}\rangle ^{-2\alpha _0}\langle \textbf{j}\rangle ^{-2s}\langle \textbf{k}\rangle ^{2s}\right) \\&\le (1-10^{-\frac{1}{2s}})^{-2s}\Vert Y\Vert _{\alpha _0}^2\sum _{\textbf{k}\in \mathbb Z^d}\left( \sum _{\textbf{j}\in \mathbb Z^d}a_{\textbf{j}}^2\langle \textbf{j}\rangle ^{2s}\langle \textbf{k}-\textbf{j}\rangle ^{-2\alpha _0}\right) \ (\mathrm{since (A.3)})\\&\le (1-10^{-\frac{1}{2s}})^{-2s}\Vert Y\Vert _{\alpha _0}^2\sum _{\textbf{j}\in \mathbb Z^d}a_{\textbf{j}}^2\langle \textbf{j}\rangle ^{2s}\left( \sum _{\textbf{k}\in \mathbb Z^d}\langle \textbf{k}-\textbf{j}\rangle ^{-2\alpha _0}\right) \\&\le M_1^2(s)\Vert Y\Vert _{\alpha _0}^2\Vert X\Vert _{s}^2, \end{aligned}$$

where $M_1(s)=(1-10^{-\frac{1}{2\,s}})^{-s}\sqrt{\sum \limits _{\textbf{k}\in \mathbb Z^d}\langle \textbf{k}\rangle ^{-2\alpha _0}}<\infty$ since $\alpha _0>d/2.$

Combining Case 1 and Case 2 implies

$$\begin{aligned} \Vert XY\Vert _s&\le \sqrt{2M^2_0(\alpha _0)\Vert X\Vert _{\alpha _0}^2\Vert Y\Vert _{s}^2+2M^2_1(s)\Vert Y\Vert _{\alpha _0}^2\Vert X\Vert _{s}^2}\\&\le K_0(\alpha _0)\Vert X\Vert _{\alpha _0}\Vert Y\Vert _{s}+K_1(s)\Vert X\Vert _{s}\Vert Y\Vert _{\alpha _0}, \end{aligned}$$

which proves Lemma 3.1, where $K_0=\sqrt{2}M_0,\ K_1(s)=\sqrt{2}M_1(s).$