# Fixed points of conjugate functions

1. Apr 18, 2012

### razmtaz

1. The problem statement, all variables and given/known data

suppose f and g are conjugate

show that if p is an attractive fixed point of f(x), then h(p) is an attractive fixed point of g(x).

2. Relevant equations

f and g being conjugate means there exist continuous bijections h and h^-1 so that h(f(x)) = g(h(x))

a point p is an attractive fixed point of there exists an interval I = (p-a,p+a) such that for all x in I the iterates of f(x) tend to p as the number of iterations tends to infinity

3. The attempt at a solution

so far I can show that if p is a fixed point of f then h(p) is a fixed point of g:
h(f(p)) = g(h(p)) and we know f(p) = p so simplify to get
h(p) = g(h(p)) and this part is now done.

Also, I know that if x is in I, then h(x) is in h(I)

what I want to show is that for all x in h(I), gn(x) -> h(p)
(that is, the iterates of x under g converge to h(p))
and thats as far as Ive gotten. How can I proceed?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 18, 2012

### sunjin09

h being continuous means for any convergent sequence {x_n}→x, h(x_n)→h(x). Now you have a sequence of f(x_n) that converges to f(x), what about their image sequence under h?

3. Apr 18, 2012

### razmtaz

Sorry Sunjin I might have missed what you were getting at but heres an attempt:

we have a sequence {x, f(x), f(f(x)), ...} which has x as some point in the basin of attraction of a fixed point p, but not equal to p, that converges to f(p)=p. Similarly, if we apply h to every element in this sequence, we get {h(x), h(f(x)), ...} which converges to h(f(p)) = h(p).

I think that on a test I could make a somewhat convincing argument that goes like this, but is this what you were thinking of? it seems right to me, because we know it works for ANY x in the basin of attraction of the fixed point, and since h is bijective and we have that h(f(x)) = g(h(x)) then every point in the sequence has an image in the space that J is in and eventually converges to h(p), the image under h of the limit of the sequence

Is this strong enough? have I missed the point?

Thanks alot : )

4. Apr 18, 2012

### razmtaz

you mentioned that h is continuous. Is this the reason that we can find a neighbourhood N around our fixed point such that the sequence of iterates of any point in the neighbourhood converged to the fixed point? And hence there is an analogous neighbourhood h(N) where the same is true for iterates of g(h(x)) except that they converge to the fixed point g(h(p)) = h(p)

5. Apr 19, 2012

### sunjin09

A continuous funcation preserves convergent sequences is the key. (This applies to functions defined on a first countable space. R is certainly first countable, but this theorem in real analysis is proved using ordinary definition of limit and convergence.)