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).
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
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?