Re: circle homeomorphism

On Sun, 16 Oct 2005 11:46:21 +0100, "Thomas Novascott"
<Thomas.novascott@xxxxxxxxxxxxxxxxxx> wrote:

>> It seems we have an increasing homeomorphism F of R,
>> with period 2\pi, such that
>>
>> f(eit(t)) = eit(F(t))
>>
>> (writing eit(t) = exp(2 Pi i t)); F is a "lift"
>> of f (ie a lifting of f to R wrt the covering
>> map eit:R -> S^1.)
>>
>> and the rotation number is the limit of
>>
>> (*) (F^n(t) - t)/n,
>>
>> (where F^n denotes iterated composition).
>>
>> It follows that
>>
>> eit(F(F(t)) = f(eit(F(t)) = f(f(eit(t));
>>
>> that is, F^2 is a lift of f^2. If we're willing
>> to believe that the limit (*) exists the result
>> you ask about follows easily...
>
>
>I was reading this thread with great interest, but i do not see how the
>limit exists?

Had a thought about that at lunch but no pencil and paper -
let's see if this works:

We will use without comment the facts that F is increasing
and F(t + 1) = F(t) + 1.

Say N is a large integer. Choose an integer k so that

k <= F^N(0) <= k+1.

Then there is an interval I of length 1, containing 0, so
that

k <= F^N(t) <= k+1

for all t in I. Hence

F(0) + k <= F^(N+1)(t) <= F(0) + k+1,

and in fact

F^n(0) + k <= F^(N+n)(t) <= F^n(0) + k+1

for all t in I. In particular

2k <= F^N(0) + k <= F^(2N)(t) <= F^N(0) + k + 1 <= 2(k+1)

and in general

jk <= F^(jN)(t) <= j(k+1).

So (insert here either "as above" or "since
F is certainly unformly continuous", as you wish)
if 0 <= i < N then

jk - c <= F^(jN + i)(t) <= j(k+1) + c,

where c depends only on N. Divide by jN + i,
and you see that

(limsup - liminf) F^(n)(t)/t <= 1/N,

uniformly for t in I. So the limit exists,
uniformly on I (and in fact the limit is independent
of t.)

Of course F^n(t)/n and (F^n(t)-t)/n have the same
limit. The difference is that the second function
is _periodic_ (in t), so that uniform convergence
on I implies uniform convergence on all of R
(which I mention because I used it the other day.)

>Many thanks
>Thomas
>


************************

David C. Ullrich
.