# Zhenghe's Blog

## April 4, 2011

### Notes 11: Renormalization (III)-Convergence of Renormalization

I’ve been back to Evanston from Toronto. But I guess I still have 10 more notes to post. It will take me a very long time to finish.

In this post we will discuss the convergence of renormalization. As pointed out in last post, it’s necessary that we should assume zero Lyapunov to get convergence. In fact, we will assume $L^2$-conjugacy to rotations. This is somehow natural because Kotani theory tells us that in the Schrodinger cocycle case, for almost every energy, zero Lyapunov exponent implies $L^2$-conjugacy. More concretely, we assume for $A\in C^{\omega}(\mathbb R/\mathbb Z, SL(2,\mathbb R)), \alpha\in\mathbb R\setminus\mathbb Q$ that $(\alpha,A)$ is $L^2$-conjugate to $SO(2,\mathbb R)$-valued cocycles. Namely, there is a measurable map $B:\mathbb R/\mathbb Z\rightarrow SO(2,\mathbb R)$ such that $\int_{\mathbb R/\mathbb Z}\|B(x)\|^2dx<\infty$ and $B(x+\alpha)A(x)B(x)^{-1}\in SO(2,\mathbb R)$ for almost every $x\in\mathbb R/\mathbb Z.$  Let $S(x)=\sup_{n\geq 1}\frac{1}{n}\sum^{n-1}_{k=0}\|B(x+k\alpha)\|^2,$ which is finite almost everywhere by the Maximal Ergodic Theorem. WLOG, we can assume that $A$ can be holomorphically extended to $\Omega_{\delta}=\{z\in\mathbb C/\mathbb Z: |\Im z|<\delta\}$ which is also Lipschitz in $\Omega_{\delta}.$ Then we have the following lemma.

Lemma 1: There exists $C>0$ such that for almost every $x_0,$ we have for every $n\geq 1$ and $x\in\Omega_{\delta}$

$\|A_n(x_0)^{-1}(A_n(x)-A_n(x_0))\|\leq Ce^{C\|B(x_0)\|^2S(x_0)n|x-x_0|}.$

Proof: As explained in Notes 2, $A_k(x)$ is bounded in $L^1.$ In fact, $\|A_k(x)\|\leq\|B(x+k\alpha)\|\|B(x)\|$ for almost every $x.$ Let $x_0$ be such a point. Note $A_n(x_0)^{-1}A_n(x)=A(x_0)^{-1}\cdots A(x_0+(n-1)\alpha)^{-1}A(x+(n-1)\alpha)\cdots A(x).$ If we let $A(x_0+k\alpha)^{-1}A(x+k\alpha)=H_k(x)+id,$ then by Lipschitz condition $H_k(x)\leq C|x-x_0|.$ Obviously, $A_n(x_0)^{-1}A_n(x)=A_{n-1}(x_0)^{-1}H_{n-1}(x)A_{n-1}(x)+A_{n-1}(x_0)^{-1}A_{n-1}(x).$ Hence by induction we have

$A_n(x_0)^{-1}A_n(x)=id+\sum^{n-1}_{k=0}A_k(x_0)^{-1}H_k(x)A_k(x).$

This obviously implies that

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq e^{\sum^{n-1}_{k=0}\|A_k(x_0)\|^2\|H_k(x)\|}-1.$

Thus we obtain

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq e^{\sum^{n-1}_{k=0}\|B(x_0+k\alpha)\|^2\|B(x_0)\|^2\|H_k(x)\|}-1.$

Hence,

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq Ce^{S(x_0)n\|B(x_0)\|^2C|x-x_0|},$

which completes the proof. $\square$

Assume further that $x_0$ is a measurable continuity point of $S$ and $B.$ Here for example, $x_0$ is a measurable continuity point of $S,$ if it is a Lebesgue density point of $S^{-1}(S(x_0)-\epsilon, S(x_0)+\epsilon)$ for every $\epsilon.$ It’s standard result that this is a full measure set since $S$ is measurable and almost everywhere finite. Same definition can be applied to $\|B(x)\|.$ By definition, it’s easy to see the portion of $x$ that $S(x)$ is close to $S(x_0)$ is getting larger and larger in smaller and smaller neighborhood of $x_0.$ Let $x_0$ be that $S(x_0)<\infty,$ which is a full measure condition. Then Lemma 1 implies the following estimate

Lemma 2:  Let $x_0$ be as above. Then for for every $d>0,$ there exist a $n_0(d)>0$ such that

$\|A_{(-1)^nq_n}(x)\|\leq \inf\limits_{x'-x_0\in [-\frac{d}{q_n}, \frac{d}{q_n}]}C(x_0)e^{C(x_0)q_n|x-x'|}$

as long as $n\geq n_0(d).$

Proof: If $n$ is sufficiently large, measurable continuity hyperthesis implies that for every $x'\in [x_0-\frac{d}{q_n}, x_0+\frac{d}{q_n}],$ we can find some $x_0'$ with $|x_0'-x'|\leq\frac{1}{q_n}$ and such that $B(x_0'), B(x_0'+\beta_n)$ are close to $B(x_0)$ and $S(x_0'), S(x_0'+\beta_n)$ are close to $S(x_0).$ WLOG, we can assume $n$ is even. Then Lemma 1 together with our choice of $x_0$ implies that

$\|A_{q_n}(x_0')^{-1}(A_{q_n}(x)-A_{q_n}(x_0'))\|\leq Ce^{C\|B(x_0)\|^2S(x_0)n|x-x_0'|}.$

And we can of course assume $x_0'$ such that $\|A_{q_n}(x_0')\|\leq\|B(x_0')\|\|B(x_0'+\beta_{n})\|.$  Since $\|A^{-1}B\|\geq\|B\|\|A\|^{-1}$ and $|x_0'-x'|<\frac{1}{q_n}.$ Combined these together we get the estimate we want. For $n$ odd, we apply the same discussion to $\|A_{-q_n}(x_0')^{-1}(A_{-q_n}(x)^{-1}-A_{-q_n}(x_0')^{-1})\|.$ $\square$

If we renormalize around $x_0,$ we know from last post that

$\mathcal R^n(\Phi)(1,0)=(1,A_{(-1)^{n-1}q_{n-1}}(x_0+\beta_{n-1}(\cdot)))=(1,A^{(n,0)})$ and
$\mathcal R^n(\Phi)(0,1)=(\alpha_n, A_{(-1)^{n}q_{n}}(x_0+\beta_{n-1}(\cdot)))=(\alpha_n, A^{(n,1)}).$

Note $q_{n-1} Then we have the following obvious corollary from Lemma 2

Corollary 3: Let $\Omega_{\delta/\beta_{n-1}}(\mathbb R)=\{z\in\mathbb C: |\Im z|<\delta/\beta_{n-1}\}.$ Then

$\|A^{(n,i)}(x)\|\leq\inf\limits_{x'\in [-d,d]}Ce^{C|x-x'|},$ where $i=0,1$ and $x\in\Omega_{\delta/\beta_{n-1}}(\mathbb R).$

Thus by homorphicity, $A^{(n,i)}$ are precompact in $C^{\omega}.$ So we can take limit along some subsequence. Denote the limit by $\tilde{A}.$ Then by estimate in Corollary 3 we  get $\|\tilde A(z)\|\leq Ce^{|\Im z|}.$

We in fact also have that $B(x_0)\tilde A(x) B(x_0)^{-1}\in SO(2,\mathbb R)$ for $x\in \mathbb R.$ This is given by the following lemma

Lemma 4: Let $A, x_0$ be as above, then for every $d>0$ and every $\epsilon>0,$ there exists a $n_0(d,\epsilon)$ such that if $n>n_0(d, \epsilon)$ and $\|\alpha n\|_{\mathbb R/\mathbb Z}\leq\frac{d}{n},$ then

$B(x_0)A_n(x)B(x_0)^{-1}$ is $\epsilon$ close to $SO(2,\mathbb R)$ for every $x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}].$

Proof: For $n$ sufficiently large, for every $x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}],$ as in proof of Lemma 2, we can find some $x'$ $\frac{\epsilon}{n}$ which is close to $x$ and $S(x'), B(x'), B(x'+n\alpha))$ are $\epsilon$ close to $S(x_0), B(x_0)$. Then the same argument of proof of Lemma 2 implies that $A_n(x)$ and $A_n(x')$ are $\epsilon$ close.

Thus we can reduce the proof to the case $B(x_0)A_n(x')B(x_0)^{-1}$ is $\epsilon$ close to $SO(2,\mathbb R).$ But this is clear since we can furthermore choose $x'$ such that $B(x'+\alpha n)A_n(x')B(x')^{-1}\in SO(2,\mathbb R),$ and we’ve already had $B(x'), B(x'+n\alpha)$ are $\epsilon$ close to $B(x_0).$ $\square$

Thus we can write $\tilde A(z)=B(x_0)^{-1}R_{\phi (z)}B(x_0),$ where $\phi:\mathbb C\rightarrow\mathbb C$ satisfying $|\Im{\phi(z)}|\leq C+C|\Im z|$ is an entire function.  Thus $\phi$ must be linear. We’ve basically proved the following theorem

Theorem 5: If the real analytic cocycle dynamics $(\alpha, A)$ is $L^2$ conjugate to rotations, then for almost every $x_0\in\mathbb R/\mathbb Z$ there exists $B(x_0)\in SL(2,\mathbb R),$ and a sequence of affine functions with bounded coefficients $\phi^{(n,0)}, \phi^{(n,1)}:\mathbb R\rightarrow\mathbb R$ such that

$R_{-\phi^{(n,0)}(x)}BA^{(n,0)}(x)B^{-1}\rightarrow id$ and $R_{-\phi^{(n,1)}(x)}BA^{(n,1)}(x)B^{-1}\rightarrow id.$

as $n\rightarrow\infty.$

To conclude, Theorem 5 implies the following final version of renormalization convergence theorem

Theorem 6: Let $(\alpha, A)$ be as in Theorem 5; let deg be the topological degree of map $A.$ Then there exists a sequence of renomalization representatives $(\alpha_n, A^{(n)})$ and $\theta_n\in\mathbb R,$ such that

$R_{-\theta_n-(-1)^ndeg x}A^{(n)}(x)\rightarrow id$ in $C^{\omega}$ as $n\rightarrow\infty$

Proof:  Let $B, \phi^{(n,0)}(x)=a_{n,0}x+b_{n,0}$ and $\phi^{(n,1)}(x)=a_{n,1}x+b_{n,1}$ be as in Theorem 5. Let $n$ be large and let $\tilde B(x)=R_{a_{n,0}\frac{x^2-x}{2}+b_{n,0}x}B.$ Then $\tilde A(x)=\tilde B(x+1)A^{(n,0)}(x)\tilde B(x)^{-1}$ is $C^{\omega}$ close to identity and $\tilde B(x+\alpha_n)A^{(n,1)}(x)\tilde B(x)^{-1}$ is $C^{\omega}$ close to $R_{\psi^n(x)},$ where $\psi^n(x)=(a_{n,0}\alpha_n+a_{n,1})x+\frac{\alpha_{n}^2-\alpha_n}{2}+b_{n,0}+b_{n,1}.$ By Lemma 1 of Notes 10, we know there exists $C\in C^{\omega}(\mathbb R, SL(2,\mathbb R))$ which is $C^{\omega}$ close to identity such that $C(x+1)\tilde A(x)C(x)^{-1}=id.$

Thus $B^{(n)}=C\tilde B$ is a normalizing map for $A^{(n,0)}$ and $A^{(n)}(x)=B^{(n)}(x+\alpha_n)A^{(n,1)}(x)B^{(n)}(x)^{-1}$ is $C^{\omega}$ close to some $R_{\psi_n(x)},$ where $\psi_n$ is linear. Since the way we get renormalization representatives preserving homotopic relation and the degree of n-th renormalization representative of $(\alpha, R_{deg x})$ is $(-1)^n deg,$ we get that degree of $(\alpha_n, A_n)$ is $(-1)^ndeg.$ Thus the linear coefficient of $\psi_n$ must be close to $(-1)^ndeg$ and $A^{(n)}$ must be close to $R_{\theta_n-(-1)^ndeg x}$ for some $\theta_n\in\mathbb R.$ $\square$

I’ve finished the serial posts about renormalization. It’s a powerful technique in the way that we can use it to reduce global problem to local problem and apply perturbation theory like KAM thoerem. More precisely, we can start with $L^2$ conjugacy to rotations and end up as $C^{\omega}$ close to rotations. If degree is zero and $\alpha$ satisfying some arithematic properties, we can then apply standard KAM theorem to get reducibility.

For these posts, I am following Artur‘s course and he and Krikorian‘s papers. Here I only do the $C^{\omega}$ case while they’ve considered smooth cases in there papers.

Next post will be something about distribution of eigenvalues of the our old friend: One dimensional discrete Schrodinger operator:)