Sharp Hölder continuity of the Lyapunov exponent of finitely differentiable quasi-periodic cocycles

Abstract

We show that if the base frequency is Diophantine, then the Lyapunov exponent of a $C^{k}$ quasi-periodic $SL(2,{\mathbb {R}})$ cocycle is 1 / 2-Hölder continuous in the almost reducible regime. As a consequence, we show that if the frequency is Diophantine, and the potential is small, then the integrated density of states of the corresponding quasi-periodic Schrödinger operator is 1 / 2-Hölder continuous.

5 Appendix: Proof of Lemma 3.1

In this appendix, we prove Lemma 3.1 by the following quantitative Implicit Function Theorem:

Theorem 5.1

[13, 22] Let X, Y, Z be Banach spaces, $U\subset X$ and $V\subset Y$ neighborhoods of $x_0$ and $y_0$ respectively. Fix $s,\delta >0$ and define $B_s(x_0)=\{x\in X\mid ||x-x_0||_ X\leqslant s \}, B_{\delta }(y_0)=\{y\in Y\mid ||y-y_0||_Y \leqslant \delta \}.$ Let $\Psi \in C^1(U\times V,Z)$ and $B_s(x_0)\times B_{\delta }(y_0)\subset U\times V$. Suppose also that $\Psi (x_0,y_0)=0$, and that $D_y\Psi (x_0,y_0)\in \mathcal {L}(Y,Z)$ is invertible. If

$$\begin{aligned}&\sup _{\begin{array}{c} \overline{B_s(x_0)} \end{array}}||\Psi (x,y_0)||_Z\leqslant \frac{\delta }{2||({D_y\Psi (x_0,y_0)}^{-1})||}, \end{aligned}$$

(5.1)

$$\begin{aligned}&\sup _{\begin{array}{c} \overline{B_s(x_0)}\times \overline{B_{\delta }(y_0)} \end{array}}||Id_Y-(D_y\Psi (x_0,y_0))^{-1}D_y\Psi (x,y) ||_{\mathcal {L}(Y,Y)}\leqslant \frac{1}{2}, \end{aligned}$$

(5.2)

then there exists $y\in C^1(B_s(x_0),\overline{B_{\delta }(y_0)})$ such that $\Psi (x,y(x))=0$.

With Theorem 5.1 in hand, now we can prove Lemma 3.1 easily. We construct the nonlinear functional

$$\begin{aligned} \Psi :\mathcal {B}_r^{nre}(\eta ) \times C^{\omega }_r({\mathbb {T}}^d,sl(2,{\mathbb {R}}))\rightarrow \mathcal {B}_r^{nre}(\eta ) \end{aligned}$$

by

$$\begin{aligned} \Psi (Y,g)=\mathbb {P}_{nre} \ln (e^{A^{-1}Y(\theta +\alpha )A}e^{g(\theta )}e^{-Y(\theta )}) \end{aligned}$$

Immediate check shows that

$$\begin{aligned} \Psi (0,0)=0, \ \ ||\Psi (0,g)||\leqslant |g|_r. \end{aligned}$$

and

$$\begin{aligned}&\Psi (Y+Y',g)-\Psi (Y,g)\\&\quad =\mathbb {P}_{nre} \ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-(Y(\theta )+Y'(\theta ))})\\&\qquad -\mathbb {P}_{nre} \ln (e^{A^{-1}Y(\theta +\alpha )A}e^{g(\theta )}e^{-Y(\theta )})\\&\quad =\mathbb {P}_{nre} \ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-(Y(\theta )+Y'(\theta ))})\\&\qquad -\mathbb {P}_{nre}\ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-Y(\theta )})\\&\qquad +\mathbb {P}_{nre}\ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-Y(\theta )})\\&\qquad -\mathbb {P}_{nre} \ln (e^{A^{-1}Y(\theta +\alpha )A}e^{g(\theta )}e^{-Y(\theta )}) \end{aligned}$$

To make further computations, we need the fact that if A, B, C are small $sl(2,{\mathbb {R}})$ matrices, then there exists $D,E\in sl(2,{\mathbb {R}})$ such that

$$\begin{aligned} e^{A}e^{B}e^{C}=e^{D+E} \end{aligned}$$

and

$$\begin{aligned} D=A+B+C \end{aligned}$$

where E is a sum of terms at least 2 orders in A, B, C. Also, the famous Baker-Campbell-Hausdorff Formula shows that

$$\begin{aligned} \ln (e^X e^Y)=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}([X,[X,Y]+[Y,[Y,X]])+\cdots , \end{aligned}$$

(5.3)

where $[X,Y]=XY-YX$ denotes the Lie Bracket and $\cdots$ denotes the sum of higher order terms.

Therefore, we can compute that

$$\begin{aligned}&\mathbb {P}_{nre} \ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-(Y(\theta )+Y'(\theta ))})\\&\qquad -\mathbb {P}_{nre}\ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-Y(\theta )})\\&\quad =\mathbb {P}_{nre} \ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-Y(\theta )}e^{-Y''(\theta )})\\&\qquad -\mathbb {P}_{nre}\ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-Y(\theta )})\\&\quad =\mathbb {P}_{nre} \ln (e^{D+E}e^{-Y''(\theta )})-\mathbb {P}_{nre}\ln (e^{D+E})\\&\quad =\mathbb {P}_{nre}(D+E-Y''+\frac{1}{2}[D+E,-Y'']+\cdots )-\mathbb {P}_{nre}(D+E)\\&\quad =\mathbb {P}_{nre}(-Y''+\frac{1}{2}[D+E,-Y'']+\cdots ) \end{aligned}$$

where

$$\begin{aligned} Y''(\theta )= & {} Y'(\theta )+\mathcal {O}(Y(\theta ))Y'(\theta ),\\ D(\theta )= & {} A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A+g(\theta )-Y(\theta ), \end{aligned}$$

and E is a sum of terms at least 2 orders in $A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A,g(\theta ),-Y(\theta )$.

Similarly, we have

$$\begin{aligned}&\mathbb {P}_{nre}\ln (e^{A^{-1}(Y(\theta +\alpha )+Y'(\theta +\alpha ))A}e^{g(\theta )}e^{-Y(\theta )})\\&\qquad -\mathbb {P}_{nre} \ln (e^{A^{-1}Y(\theta +\alpha )A}e^{g(\theta )}e^{-Y(\theta )})\\&\quad =\mathbb {P}_{nre}\ln (e^{Y'''}e^{A^{-1}Y(\theta +\alpha )A}e^{g(\theta )}e^{-Y(\theta )})\\&\qquad -\mathbb {P}_{nre} \ln (e^{A^{-1}Y(\theta +\alpha )A}e^{g(\theta )}e^{-Y(\theta )})\\&\quad =\mathbb {P}_{nre} \ln (e^{Y'''}e^{F+H})-\mathbb {P}_{nre}\ln (e^{F+H})\\&\quad =\mathbb {P}_{nre}(Y'''+F+H+\frac{1}{2}[Y''',F+H]+\cdots )-\mathbb {P}_{nre}(F+H)\\&\quad =\mathbb {P}_{nre}(Y'''+\frac{1}{2}[Y''',F+H]+\cdots ) \end{aligned}$$

where

$$\begin{aligned} Y'''(\theta +\alpha )= & {} A^{-1}Y'(\theta +\alpha )A+\mathcal {O}(A^{-1}Y(\theta +\alpha )A)\times A^{-1}Y'(\theta +\alpha )A,\\ F(\theta )= & {} A^{-1}Y(\theta +\alpha )A+g(\theta )-Y(\theta ) \end{aligned}$$

and H is a sum of terms at least 2 orders in $A^{-1}Y(\theta +\alpha )A,g(\theta ),-Y(\theta )$.

By the definition of Fréchet Differential, we only need to consider the linear terms of $\Psi (Y+Y',g)-\Psi (Y,g)$, thus we have

$$\begin{aligned} D_Y\Psi (Y,g)(Y')&=\mathbb {P}_{nre}(A^{-1}Y'(\theta +\alpha )A+\mathcal {O}(A^{-1}Y(\theta +\alpha )A)\times A^{-1}Y'(\theta +\alpha )A\\&\quad +\frac{1}{2}[Y''',F+H]+\cdots )\\&\quad +\mathbb {P}_{nre}(-Y'(\theta )-\mathcal {O}(Y(\theta ))Y'(\theta )+\frac{1}{2}[F+H',-Y'']+\cdots ), \end{aligned}$$

where $H'$ is a sum of terms at least 2 orders in $A^{-1}Y(\theta +\alpha )A,g(\theta ),-Y(\theta )$. Moreover, the first “$\cdots$” denotes the sum of terms which are at least 2 orders in $F+H$ but only 1 order in $Y'''$. The second “$\cdots$” denotes the sum of terms which are at least 2 orders in $F+H'$ but only 1 order in $Y''$.

Let $Y=0$ and $g=0$, then all the Lie Brackets vanish. So we immediately obtain

$$\begin{aligned} D_Y\Psi (0,0)(Y')&=\mathbb {P}_{nre}(A^{-1}Y'(\theta +\alpha )A)+\mathbb {P}_{nre}(-Y'(\theta ))\\&=A^{-1}Y'(\theta +\alpha )A-Y'(\theta ) \end{aligned}$$

Thus

$$\begin{aligned} ||D_Y\Psi (0,0)(Y')||&\geqslant |A^{-1}Y'(\theta +\alpha )A-Y'(\theta )|_r \\&\geqslant \eta |Y'|_r.\\ \end{aligned}$$

So we have

$$\begin{aligned} ||(D_Y\Psi (0,0))^{-1}||\leqslant \eta ^{-1}. \end{aligned}$$

For our purpose, we set $s=\epsilon ,\delta =\epsilon ^\frac{1}{2}$ and $\eta \geqslant 13||A||^2\epsilon ^{\frac{1}{2}}$. Then we have

$$\begin{aligned} 2\times \sup _{\begin{array}{c} \overline{B_{s}(0)} \end{array}}||\Psi (0,g)||\times ||(D_Y\Psi (0,0))^{-1}||\leqslant 2\times \epsilon \times \frac{1}{13||A||^2}\epsilon ^{-\frac{1}{2}}\leqslant \epsilon ^{\frac{1}{2}}=\delta , \end{aligned}$$

then (5.1) is fulfilled.

On the other hand, direct computation shows that

$$\begin{aligned}&D_Y\Psi (Y,g)(Y')-D_Y\Psi (0,0)(Y')\\&\quad =\mathbb {P}_{nre}(\mathcal {O}(A^{-1}Y(\theta +\alpha )A)\times A^{-1}Y'(\theta +\alpha )A+\frac{1}{2}[Y''',F+H]+\cdots )\\&\qquad +\mathbb {P}_{nre}(-\mathcal {O}(Y(\theta ))Y'(\theta )+\frac{1}{2}[F+H',-Y'']+\cdots ). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \sup _{\begin{array}{c} \overline{B_s(0)}\times \overline{B_{\delta }(0)} \end{array}}||D_Y\Psi (Y,g)(Y')-D_Y\Psi (0,0)(Y')||&\leqslant 6(||A||^2|Y|_r+||A||^2|g|_r)|Y'|\\&\leqslant 6(||A||^2\delta +||A||^2s)|Y'|\\&\leqslant 6(||A||^2 \epsilon ^{\frac{1}{2}}+||A||^2\epsilon )|Y'|, \end{aligned}$$

which implies

$$\begin{aligned} \sup _{\begin{array}{c} \overline{B_s(0)}\times \overline{B_{\delta }(0)} \end{array}}||D_Y\Psi (0,0)-D_Y\Psi (Y,g)||\leqslant 6(||A||^2 \epsilon ^{\frac{1}{2}}+||A||^2\epsilon ). \end{aligned}$$

Thus we have

$$\begin{aligned}&\sup _{\begin{array}{c} \overline{B_s(0)}\times \overline{B_{\delta }(0)} \end{array}}||Id_{\mathcal {B}_r^{nre}(\eta )}-(D_Y\Psi (0,0))^{-1}\times D_Y\Psi (Y,g) ||\\&\quad \leqslant \sup _{\begin{array}{c} \overline{B_s(0)}\times \overline{B_{\delta }(0)} \end{array}}||D_Y\Psi (0,0)-D_Y\Psi (Y,g)||\times ||(D_Y\Psi (0,0))^{-1}||\\&\quad \leqslant 6(||A||^2 \epsilon ^{\frac{1}{2}}+||A||^2\epsilon ) \times \frac{1}{13||A||^2}\epsilon ^{-\frac{1}{2}}\\&\quad \leqslant \frac{1}{2}, \end{aligned}$$

which satisfies (5.2). By Theorem 5.1, for $|g|_r\leqslant \epsilon$ and $\eta \geqslant 13||A||^2\epsilon ^{\frac{1}{2}}$, there exists $|Y|_r\leqslant \epsilon ^{\frac{1}{2}}$ such that $\Psi (Y,g)=0$, i.e.

$$\begin{aligned} e^{A^{-1}Y(\theta +\alpha )A}e^{g(\theta )}e^{Y(\theta )}=e^{g^{re}(\theta )}, \end{aligned}$$

which is equivalent to say

$$\begin{aligned} e^{Y(\theta +\alpha )}Ae^{g(\theta )}e^{Y(\theta )}=Ae^{g^{re}(\theta )} \end{aligned}$$

and it is easy to check $|g^{re} (\theta )|_r\leqslant 2\epsilon$. This finishes the proof of Lemma 3.1.