This post will be something about the distribution of eigenvalues of one dimensional discrete Schrodinger operators in absolutely continuous spectrum region. Namely, given a triple 
For eigenfunction equations 
Let 
![[0, n-1]](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5B0%2C+n-1%5D&bg=ffffff&fg=000000&s=0&c=20201002)
![N(E)=\int^{E}_{-\infty}dN:\mathbb R\rightarrow [0,1]](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=N%28E%29%3D%5Cint%5E%7BE%7D_%7B-%5Cinfty%7DdN%3A%5Cmathbb+R%5Crightarrow+%5B0%2C1%5D&bg=ffffff&fg=000000&s=0&c=20201002)
Lemma 1:
Thus by notes 3 we know, 
Proof: To see why 
Thus 
Then we note the following facts:
(1) For 
(2) 
(3)
Thus as 
Now I would like to Sketch a rough proof the following main theorem of this post, which is contained in the paper ‘Bulk Universality and Clock Spacing of Zeros for Ergodic Jacobi Matrices with A.C. Spectrum‘ by Artur Avila, Yoram Last and Barry Simon.
Theorem 2: Assume 
And 
Remark: This theorem tells that if we look at a window around 
Proof: Note that by the proof of Lemma 1, we know that 
Thus
where 
Assume that 
In fact, replacing 
(1) 
(2) Complexifying 
(3) Then 
(4) 
(5) By passing to a suitable subsequence, we get that
for all 
(6) Thus we can write it as 
Finally, we want to show that 
The above formula lies in
Since 
Let 
Since for
where 
Here we used the fact that 
that 
is 
-valued cocycles. Namely, there is a measurable map 
such that 
and 
for almost every 
Let 
which is finite almost everywhere by the Maximal Ergodic Theorem. WLOG, we can assume that 
can be holomorphically extended to 
which is also Lipschitz in 
Then we have the following lemma.
such that for almost every 
we have for every 
and 
is bounded in 
In fact, 
for almost every 
Let 
be such a point. Note 
If we let 
then by Lipschitz condition 
Obviously, 
Hence by induction we have
and 
Here for example, 
if it is a Lebesgue density point of 
for every 
It’s standard result that this is a full measure set since 
By definition, it’s easy to see the portion of 
is close to 
is getting larger and larger in smaller and smaller neighborhood of 
Let 
which is a full measure condition. Then Lemma 1 implies the following estimate
there exist a 
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'|}](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5C%7CA_%7B%28-1%29%5Enq_n%7D%28x%29%5C%7C%5Cleq+%5Cinf%5Climits_%7Bx%27-x_0%5Cin+%5B-%5Cfrac%7Bd%7D%7Bq_n%7D%2C+%5Cfrac%7Bd%7D%7Bq_n%7D%5D%7DC%28x_0%29e%5E%7BC%28x_0%29q_n%7Cx-x%27%7C%7D&bg=ffffff&fg=000000&s=0&c=20201002)
![x'\in [x_0-\frac{d}{q_n}, x_0+\frac{d}{q_n}],](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=x%27%5Cin+%5Bx_0-%5Cfrac%7Bd%7D%7Bq_n%7D%2C+x_0%2B%5Cfrac%7Bd%7D%7Bq_n%7D%5D%2C&bg=ffffff&fg=000000&s=0&c=20201002)
we can find some 
with 
and such that 
are close to 
and 
are close to 
WLOG, we can assume 
Since 
and 
Combined these together we get the estimate we want. For 
and
Then we have the following obvious corollary from Lemma 2
Then![\|A^{(n,i)}(x)\|\leq\inf\limits_{x'\in [-d,d]}Ce^{C|x-x'|},](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5C%7CA%5E%7B%28n%2Ci%29%7D%28x%29%5C%7C%5Cleq%5Cinf%5Climits_%7Bx%27%5Cin+%5B-d%2Cd%5D%7DCe%5E%7BC%7Cx-x%27%7C%7D%2C&bg=ffffff&fg=000000&s=0&c=20201002)
where 
and 
are precompact in 
So we can take limit along some subsequence. Denote the limit by 
Then by estimate in Corollary 3 we get 
for 
This is given by the following lemma
be as above, then for every 
and every 
there exists a 
such that if 
and 
then
is 
close to ![x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}].](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=x%5Cin%5Bx_0-%5Cfrac%7Bd%7D%7Bn%7D%2C+x_0%2B%5Cfrac%7Bd%7D%7Bn%7D%5D.&bg=ffffff&fg=000000&s=0&c=20201002)
![x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}],](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=x%5Cin%5Bx_0-%5Cfrac%7Bd%7D%7Bn%7D%2C+x_0%2B%5Cfrac%7Bd%7D%7Bn%7D%5D%2C&bg=ffffff&fg=000000&s=0&c=20201002)
as in proof of Lemma 2, we can find some 
which is close to 
are 
. Then the same argument of proof of Lemma 2 implies that 
and 
are 
is 
But this is clear since we can furthermore choose 
and we’ve already had 
are 
where 
satisfying 
is an entire function. Thus 
must be linear. We’ve basically proved the following theorem
is 
there exists 
and a sequence of affine functions with bounded coefficients 
such that 
and 
Then there exists a sequence of renomalization representatives 
and 
such that
in 
as 
and 
be as in Theorem 5. Let 
Then 
is 
is 
where 
By Lemma 1 of Notes 10, we know there exists 
which is 
is a normalizing map for 
and 
is 
where 
is linear. Since the way we get renormalization representatives preserving homotopic relation and the degree of n-th renormalization representative of 
is 
we get that degree of 
is 
Thus the linear coefficient of 
and 
must be close to 
for some 
satisfying some arithematic properties, we can then apply standard KAM theorem to get reducibility.
Then given a SL(2,R) cocycle dynamics 
is equivalent to a 
action 
such that 
around 
such that 
where 
is rescaling operator and 
is the basing change operator. By definition we have
and 
to be 
for 
Because then the correponding 
is well-defined on 
by commuting relation. Secondly, we will explain the relation between the dynamics of original system and these of renormalized systems.
with 
we can 
conjugate it to be that 
Here 
Here 
is the projection to the first coordinate.
Then it suffices to find some 
such that 
Then automatically, 
is well-defined on 
Obviously, 
is equivalent to 
Thus it’s sufficient to define 
on an arbitrary interval with length biger than 1. One way to do it is to let 
around a small interval around 0, hence 
around 1, then ![[0,1].](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5B0%2C1%5D.&bg=ffffff&fg=000000&s=0&c=20201002)
It’s easy to see that if 
is 
then we can choose 
such that so is 
for some 
needs a little bit more computation. We will first deal with case 
is close to identity near a neighborhood of 
in 
It’s enough to find some 
where 
and 
(this obviously imlies that D’ is holomorphic).
Indeed, we have 
By analyticity, we get 
Combined with 
we get
And we can of course 
for some 
It’s easy to see that 
where 
is the Cauchy transform.
inverts 
A direct computation show that 
and Thus 
is invertible near zero function. Thus if 
Then 
can be our choice and 
smallness of 
and then approximate 
in 
Then 
is 
and 
Let 
be the normalizing map such that 
Let 
Then
.
conjugate to rotations. Here again 
and
for every 
then it’s not difficult to see that the above choice of 
for every 
which in turn implies that 
for every 
and 
A simpler version of proof of Lemma 1 implies that we can choose some 
for 
Thus if we set 
we have 
be the continued fraction expansion. Thus 
be the n’th step approximant. We let 
hence 
grows sufficiently fast (see my third post for introduction of Liouville number). Let the potential 
be a very small real analytic function. Combining averaging procedure with some renormalization and KAM procedure, 
cocycles with Liouvillean frequencies’ the following theorem
be analytic and close to a constant. For every 
there exists a positive measure set of 
such that 
is conjugate to a cocycle of rotations.
Namely, we consider one-parameter family of cocycles over fixed point. We have two different ways to consider the spectrum, which is the following
for at least one 
for all 
For simplicity, set 
and we omit the dependence on 
Let 
be the invariant direction of 
Let 
be that 
Thus 
for some 
analytic. Thus there exists a constant 
such that
inside the interior of 
we can choose large 
such that 
is again elliptic. Hence similarly, there is some 
which is 
close to identity and 
such that
Hence
for some constant 
is conjugate to 
Iterating 
where 
is the n-th Birkhoff sume 
It’s not difficult to see that 
is close to its averaging 
Thus except those bad 
for some 
stay ellipitc. By monotonicity of the Schrodinger cocycle with respect to energy and suitable choice of 
these bad 
This will be a operator defined on the space of all cocycle dynamics. More concretely, it’s defined on the space of 
is to see it as 
which commutes with 
Because it’s easy to check that
Then 
is orientation preserving diffeomorphism if and only if it can be lifted to 
and commutes with 
where 
More generally, consider a commuting pair 
where G has no fixed point. Then 
define a dynamical systems on 
Because 
preserving the equivalent relation. Here 
is diffeomorphic to 
but not in a canonical way.
as a 
with 
and 
be subgroup of 
Then for 
satisfying 
we can define 
which is given by
For 
it’s given by
be the Gauss map 
Denote 
Thus ![a_n=[\alpha_{n-1}^{-1}],](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=a_n%3D%5B%5Calpha_%7Bn-1%7D%5E%7B-1%7D%5D%2C&bg=ffffff&fg=000000&s=0&c=20201002)
where 
comes from the continued fraction expansion. Define a map![U(x)=\begin{pmatrix}[x^{-1}]&1\\1&0\end{pmatrix}](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=U%28x%29%3D%5Cbegin%7Bpmatrix%7D%5Bx%5E%7B-1%7D%5D%261%5C%5C1%260%5Cend%7Bpmatrix%7D&bg=ffffff&fg=000000&s=0&c=20201002)
for 
Then it’s easy to see that we also have 
is given by
where 
and
it looks like we glue 
and 
via 
Then by commuting relation 
define a dynamical systems on 
Then we rescale the first coordinate to be 
again. Finally we get the new dynamics 
Then it’s easy to see that
so that 
and do some computation concerning the limit of the renormalzation.
where 
is a compact Cantor group, 
is a minimal translation and 
is Haar measure. Let 
be continuous. Then this potential 
It’s limit periodic if it can be approximated in 
by periodic sequence. In fact if so , the hull of 
in 
Where the topology is induced from product topology. Taking minimal translation 
Note in this topology 
is dense and 
as 
of period 
Let’s start with some 
Define 
by induction as follows
and
is sufficiently small. Thus
then 
remains 
for suitable 
Here 
can be any compact set. On the other hand, in the interior of each band we have no control for two types of points: boundary points, where the matrices is parabolic; these 
Because then 
and small perturbation may lead to the appearance of new gaps. But we can always ignore a small interval around these 
so that 
as 
Eventually, we can get spectrum 
such that 
which may also be a Cantor set.
let 
be the invariant direction of 
Then it’s easy to see 
are invariant section of cocylce 
Let 
be the projection. By above induction procedure, it’s not difficult to see that for each 
and 
there are some 
such that
Obviously, 
is an invariant section of 
which takes values in 
Thus Lyapunov exponents stay zero through 
Thus by Kotani Theory we get absolutely continous spetrum.
As in the Corollary of Notes 6, 
is the frequency. Let 
which is the quasiperiodic potential. Let’s recall the operator is for 
is
where 
is defined as 
and 
is the coupling constant. Here are some of the Theorems concerning this model
for all 
and 
and 
for al 
as 
Recall in Notes 3 we define functon
such that 
is the unstable invariant direction of 
then 
As the above cases, 
But 
(eigenvalue) is invariant under conjugacy. Thus we can always move the computation to unit disk to avoid 
Then for above matrices, we get
(note for each fixed 
is well-defined but not uniquely determined).
and the function 
Let’s use 
to move to unit disk 
Then 
Let’s consider the function 
(see notes 3) instead of 
It’s easy to see that for nonelliptic matrices, 
is not well-defined. But it’s well-defined as 
By the proof of Lemma 2 of Notes 3, we know when 
so is some 
Thus the change of correponding angle of the point in 
is exactly 
This implies that 
WLOG, we can set 
Thus 
Combining the equivalent description of spectrum in Notes 2 give the following description of the corresponding operator:![[v-2,v+2];](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Bv-2%2Cv%2B2%5D%3B&bg=ffffff&fg=000000&s=0&c=20201002)
![\rho|_{(-\infty,v-2]}=\frac{1}{2},](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Crho%7C_%7B%28-%5Cinfty%2Cv-2%5D%7D%3D%5Cfrac%7B1%7D%7B2%7D%2C&bg=ffffff&fg=000000&s=0&c=20201002)
where 
and 
hence 
we have for any 
This implies that
of 
is independent of 
Denote it by 
is the spectral radius of 
and 
Elliptic,
stands for trace. For the second case we also have 
(which is also 
for elliptic case). It’s obviously a polynomial in 
It has exactly 
critical points with critical values 
satisfying 
We have in fact the following description of the spectrum 
, we have the following easy facts for 
as a function on real line:
-valued outside the interior of spectrum thus constant in each connected components of the resolvent set;
and 
and nonincreasing.![\psi^{-1}[-2,2]](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Cpsi%5E%7B-1%7D%5B-2%2C2%5D&bg=ffffff&fg=000000&s=0&c=20201002)
consists of 
on 
These are labelings of spectral gaps. 
are simple. Let’s give another description of these polynomials. Let’s restrict the operator 
with three different type of boundary conditions. Let 
![[j,j+n-1]\subset[0,q-1]](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Bj%2Cj%2Bn-1%5D%5Csubset%5B0%2Cq-1%5D&bg=ffffff&fg=000000&s=0&c=20201002)
and consider Dirichlet boundary condition, i.e. 
Denote it by 
. In this case it can be represented as a 
symmetric matrix
Then by induction it’s easy to see that
In case 2 and 3 we will only restrict to ![[0,q-1].](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5B0%2Cq-1%5D.&bg=ffffff&fg=000000&s=0&c=20201002)
Denote it by 
then it is
Denote it by 
then it is
and 
and 
are simple. An easy case is that assume 
are sufficiently large and distinct. Then after scaling, all nondigonal coefficients of 
and 
Then spectrum bands are![[\mu_j,\lambda_j], j](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5B%5Cmu_j%2C%5Clambda_j%5D%2C+j&bg=ffffff&fg=000000&s=0&c=20201002)
odd; ![[\lambda_j,\mu_j], j](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5B%5Clambda_j%2C%5Cmu_j%5D%2C+j&bg=ffffff&fg=000000&s=0&c=20201002)
even.
even; 
odd.
Then it’s easy to see that it can be analytically extends through interior of each bands and through gaps. But they cannot be globally defined, since there are nontrivial winding of the invariant direction 
around each 
This winding comes from the parabolicity. Indeed, if we instead consider the eigenvalue 
of 
then
and 
continous.
there are more and more bands which are also thiner and thiner. Thus in quasiperiodic potential case, it’s natural to expect the spectrum is Cantor set under some assumption. We will give an example in next post.
class. Obviously, it’s more difficult to obtain density results in higher regularity class.
assume 
and 
to denote the Lyapunov exponent of the corresponding cocyle map 
Let’s first introduce a concept to state the main theorem.
is ample if there exists some dense vector space 
, endowed with some finer (than uniform) topological vector space structure, such that for every 
for every 
and the map 
from 
to 
is continous.
then 
is ample. Namely we can take 
be ample. Then the Lyapunov exponent is positive for a dense subset of 
denote the sup norm in 
and 
and for 
let 
and 
be the correponding 
-balls. Then
such that if 
is 
-close to 
then for 
and every 
the map 
and for 
is 
provided 
is 
such that 
where 
Notice that 
Let’s denote 
Once we have these facts, by pluriharmonic theorem in the post`Proof of HAB formula‘ and mean value formula for harmonic functions, we have
for 
by fact 2; thus
but
is pluriharmonic for 
we define function 
and check that 
at 
points inside 
for 
in facts 1 or 2. Thus 
for any 
and 
This is similar to the cases in Kotani theory or HAB formula, where when we complexify 
we get 
there exists a 
sufficiently close to 
in 
Then by subharmonicity of 
, more concretely, by upper semicontinuity and sub-mean value property, we can choose suitable closed path to see that if 
then 
and 
is sufficient close to ![t\in [-1,1].](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=t%5Cin+%5B-1%2C1%5D.&bg=ffffff&fg=000000&s=0&c=20201002)
Then by above observation and Corollary of last post we can find some 
such that 
is ample and the analyticity of the map 
we can assume 
such that 
is analyic in ![s\in [-1,1].](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=s%5Cin+%5B-1%2C1%5D.&bg=ffffff&fg=000000&s=0&c=20201002)
Since 
, we have for every sufficiently small s>0, 
Thus we can choose sufficiently small 
and some ![t_0\in[-1,1]](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=t_0%5Cin%5B-1%2C1%5D&bg=ffffff&fg=000000&s=0&c=20201002)
such that 
. 
where 
and 
endowed with some inductive limit topology via subspace 
Here 
and 
is the space of real analytic cocycle maps which can be extended to 
is jointly continuous.
Lyapunov exponent is positive for an open and dense subset of 
Almost Periodicity of the base dynamics.
and 
such that 
operators’. It is another application of Kotani Theory, which also has its own interests. I will break the proof of main theorem down to several lemmas and point out the key ideas. I will also try to carry out all the details. I am always grateful to Artur who is always willing to explain to me ideas and details whenever I want.
to denote the Lyapunov exponents of the system 
The main theorem is the following
for any 
). Then for generic 
(generic means residual), 
for a.e. 
for a.e. 
where 
are nonpositive integers and 
stands for bounded real-valued sequence on 
is the ball around zero function in 
with radius 
Now let’s state and prove our lemmas.
is bounded. Here boundedness in 
and 
for some 
Let
and 
of which the latter is a bounded set in 
to denote the map 
We can of course take limit along even integers. Thus the inclusion 
is obvious. For boundedness it’s easy to see that for any 
and 
![v^{(l)},v\in [-c,c]^{\mathbb Z^-}](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=v%5E%7B%28l%29%7D%2Cv%5Cin+%5B-c%2Cc%5D%5E%7B%5Cmathbb+Z%5E-%7D&bg=ffffff&fg=000000&s=0&c=20201002)
for some 
. Let 
as 
pointwisely. Then 
in compact open topology as functions on 
converge to 
in compact open topology and the convergence is independent of ![v\in [-c,c]^{\mathbb Z^-}](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=v%5Cin+%5B-c%2Cc%5D%5E%7B%5Cmathbb+Z%5E-%7D&bg=ffffff&fg=000000&s=0&c=20201002)
. This is due to the same reason of the proof of Theorem 2 of last post. Namely, 
are holomorphic functions takes value in 
on any compacta in 
for any 
large enough and 
the function 
is continuous.
such that 
and pointwisely as 
we have the 
converges to 
pointwisely as 
By Lemma 1 this implies that 
converges to 
for a.e. 
Hence by Bounded Convergence Theorem, we have
in Lemma 3 are just uniform boundedness and pointwise convergence.![[-N, N]\subset\mathbb R.](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5B-N%2C+N%5D%5Csubset%5Cmathbb+R.&bg=ffffff&fg=000000&s=0&c=20201002)
Then the function 
is continuous. 
behave badly when 
. The only thing that is always true is that it’s upper semicontinuous and plurisubharmonic. This Lemma shows that after taking an appropriate integral, it behaves nicely in 
sense.![\gamma=[-N,N]\cup S^+(N),](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Cgamma%3D%5B-N%2CN%5D%5Ccup+S%5E%2B%28N%29%2C&bg=ffffff&fg=000000&s=0&c=20201002)
where 
is upperhalf part of the circle centered at origin with radius 
Pick a point 
Then there is a harmonic measure 
on 
and 
are continuous. Thus ![\int_{[-N,N]}L(E,v)d\nu_{E_0}](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Cint_%7B%5B-N%2CN%5D%7DL%28E%2Cv%29d%5Cnu_%7BE_0%7D&bg=ffffff&fg=000000&s=0&c=20201002)
is also continuous. Via the conformal transformation which transform the unit disk to the region inside 
it’s not difficult to see that ![\int_{[-N,N]}L(E,v)dE](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Cint_%7B%5B-N%2CN%5D%7DL%28E%2Cv%29dE&bg=ffffff&fg=000000&s=0&c=20201002)
is also continuous. 
We define a function 
such that 
. Then![E\in[-2-r,2+r]=I_r.](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=E%5Cin%5B-2-r%2C2%2Br%5D%3DI_r.&bg=ffffff&fg=000000&s=0&c=20201002)
Because outside this interval 
such that 
and any 
there exists a 
and 
we have 
By definition of 
we can choose a 
such that 
where 
Then by Lemma 4 we can choose 
such that 
with 
Indeed we then have
on 
thus 
This obviously implies what we want to show. 
such that if 
then
takes finitely many values;
is not periodic for a.e. 
let’s define 
to be the period of the sequence 
Then by ergodicity for a.e. 
Again by ergodicity it’s not difficult to see that if 
then 
Thus by our assumption on the triple 
sufficiently close to 
in 
such that 
is dense in 
in 
By Lemma 6, we can choose 
such that 
By Theorem 4 of last post, we have 
By standard real analysis theorem and by Lemma 5, we can choose 
such that 
is sufficiently close to 
we have by Lemma 5 and Lemma 7 
is both open and dense. Thus 
is residual, which completes the proof. 
such that for each 
is the rotation matrix with rotation angle 
hence Lemma 5 in terms of the map 
On the other hand, Kotani theory can be carried over to this one parameter family 
thus we can have similar Theorem 4 of last post, hence similar result of Lemma 7 for 
Then the conclusion follows easily. 
, 
the unstable (stable) direction. We further assume that 
Then the first theorem is that
for a.e. 
and 
which are measurable and integrable in some sense.
Indeed, we only need to replace 
by 
Then it’s easy to see that 
and everything follows from the same argument of the existence of 
.
is real, we see obviously that 
and 
are also invariant sections. If 
for a positive measure of 
send 
to 
for some positive 
This implies that 
fix both 
and 
under mobius transformation. This happens only when 
in 
is monotonic in 
for a positive measure of 
be an open interval (then it’s also bounded by boundedness of the 
then there exists a map 
such that it’s contiuous on both variable, holomophic in 
and 
The generalized eigenfunctions are all wave like (not just in some 
such that
and
for a.e. 
.
extends through 
On the other hand, if we conformal transform 
we see that 
is a normal family. Thus for any compacta 
depends only on the compacts such that
for any 
we can for any 
converging to 
On the other hand, for 
we know that 
is continous in 
such that 
Thus 
is a probability measure on 
is the hull in the product topology. For example,
is periodic then the hull is finite;
dimensional rational independent frequency, then hull the 
.
where 
is the left shift. Namely 
Then, we have the following theorem
Then for all 
in hull, if 
for all 
then 
for a.e. 
where 
is as in Theorem 2. This follows from Theorem 2 and Fatou’s theorem via a contradiction argument.
determines 
This is not difficult to see by a contradiction argument.
are uniquely determined by their limits on a positive measure set of 
obviously determines 
But 
determines 
by fact (a) and (c). Finally, 
determines 
by fact (b). 
is said to be deterministic if for all 
is determined by 
Otherwise it is nondeterministic. Obviously, I.I.d. process are far away from deterministic. In particular, if the process 
is i.i.d, then for a.e. 
then 
as last time, then for a.e. 
implies that 
for small 
is holomophic. We are going to show this 
and is continuous up to 
.
is monotonic in 
and Cauchy-Riemann equation will imply the conclusion of our theorem.
. Thus
thus
is holomorphic.
(Here 
is well-defined. Because by last post, more concretely proof of Lemma 2, it’s easy to see that 
. Thus there is no nontrivial loop around origin). For obvious reason, it’s convenient to instead consider 
(so 
is holomorphic functon). By Birkhoff Ergodic Theorem, we have for a.e. x
upperhalf plane 
via the following matrix
And 
, where 
is the subgroup of 
preserving the unit disk in 
under Mobius transformation. For
let![\hat A=QAQ^*=\begin{pmatrix}\frac{1}{2}[(a+d)+(b-c)i],&\frac{1}{2}[(a-d)-(b+c)i]\\\frac{1}{2}[(a-d)+(b+c)i],&\frac{1}{2}[(a+d)-(b-c)i]\end{pmatrix}=\begin{pmatrix}\hat u&\hat v\\v&u\end{pmatrix}.](https://cdn.statically.io/img/s0.wp.com/latex.php?latex=%5Chat+A%3DQAQ%5E%2A%3D%5Cbegin%7Bpmatrix%7D%5Cfrac%7B1%7D%7B2%7D%5B%28a%2Bd%29%2B%28b-c%29i%5D%2C%26%5Cfrac%7B1%7D%7B2%7D%5B%28a-d%29-%28b%2Bc%29i%5D%5C%5C%5Cfrac%7B1%7D%7B2%7D%5B%28a-d%29%2B%28b%2Bc%29i%5D%2C%26%5Cfrac%7B1%7D%7B2%7D%5B%28a%2Bd%29-%28b-c%29i%5D%5Cend%7Bpmatrix%7D%3D%5Cbegin%7Bpmatrix%7D%5Chat+u%26%5Chat+v%5C%5Cv%26u%5Cend%7Bpmatrix%7D.&bg=ffffff&fg=000000&s=0&c=20201002)
i.e. 
then 
and 
then 
Let’s denote this class by 
then 
are multiplicative.
The Lemma is
and is continuous on 
.
or equivalently, 
, we can define 
(not necessary invariant).
be that
is the unstable invariant section of 
, thus 
can be replaced by any continuous section 
for all 
then
with distance at least 
Thus the above estimate follows easily (Note this is not true for Lyapunov exponents, i.e. 
cannot necessary be bounded). Now we may fix constant section 
to do the remaing computation.
, so
Then the above formula implies that
and obviously 
as 
is unform continuous on 
So we can choose 
such that 
are equi-uniform continuous. Now for any 
for all 
we have 
Thus
for 
large. Since 
for 
.
Again we denote it by 
sufficient close to 
:
is nonincreasing.
consider the function in 
, consider
Indeed,
. Thus
. So
and 
Then the relation between 
and 
Thus it’s not difficult to see that
we obviously have 
and
for all 
and 
which completes the proof. 
is Poisson integral of 
is again harmonic. Since 
Then Fatou’s Theorem tells us for Leb a.e. 
Thus the above discussion shows that for a.e. E with 
we have
in the future. As I said in the last post, 
is relatively easy. It lies in the fact that the generalized eigenfunctions of absolutely continuous spectrum grow at most polynomially fast, which obviously contradicts with positive Lyapunov exponents. And the other part due to Kotani theory has already been contained in these two posts. Let me go back to this in the future.
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