Re: (un)stable fixed points

On 3 Dec 2003 09:06:36 GMT, [EMAIL PROTECTED] (Robert Israel) wrote:

>In article <[EMAIL PROTECTED]>,
>billy d. <[EMAIL PROTECTED]> wrote:
>>f:R-->R differentiable, continuous derivative, and f(a)=a for some a
>>in R.
>
>>if |f'(a)|<1, then the sequence x_n=f(x_n-1) converges to a when x_0
>>is sufficiently close to a.
>
>>if |f'(a)|>1, then there exists c_0>0 s.t. for all x_0=/=a,
>>|x_N-a|>c_0 for some positive integer N.
>
>>the first part was coming along nicely. it seems that i am close to
>>showing that f is a contraction from [a-d,a+d] for some d>0.
>
>Not true.  

Counterexample? It seems true to me - I hesitate to post what
seems like the easy proof since we're just giving hints.
Possibly I'm being stupid again (or possibly you missed
the fact that f is _continuously_ differentiable?)

>But it will have the property that, for some c with
>0 < c < 1, |f(x) - a| < c |x - a| if x is sufficiently close to a.
>
>> the other
>>part seems more difficult. i think i just need a hint. thanks
>
>Hint: There is C > 1 such that |f(x) - a| > C |x - a| if x is sufficiently
>close to a.
>
>Robert Israel                                [EMAIL PROTECTED]
>Department of Mathematics        http://www.math.ubc.ca/~israel 
>University of British Columbia            
>Vancouver, BC, Canada V6T 1Z2
> 

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

David C. Ullrich