# Topologically conjugate function

1. Aug 29, 2014

### Mentallic

1. The problem statement, all variables and given/known data
Are $f(x)=2x, x\in R$ and $g(x)=x^2, x>0$ topologically conjugate?

i.e. does there exist an h(x) such that

$$h(g(x))=f(h(x))$$

3. The attempt at a solution

My professor gave one example in class about finding such a function h which was by guessing it to be equal to xn and subsequently solving and finding the value of n. However, when I tried to apply the same idea to this problem, I come off short.

$$h(g(x)) = h(x^2)$$

$$f(h(x)) = 2h(x)$$

If we let $h(x)=x^n$ then we want to solve for n in

$$x^{2n}=2x^n$$

$$x^n=2 = h(x)$$

Hence I find h(x)=2 but this doesn't work. Are there other guesses I could make? Or better yet, is there a more systematic approach to these sorts of problems? Does there even exist an h?

2. Aug 29, 2014

### pasmith

You need $h(x^2) = 2h(x)$ for all $x \in \mathbb{R}$. The left hand side is an even function, so $h$ must itself be even. We can ensure that by taking $h(x) = p(|x|)$ for some $p : \{x \in \mathbb{R} : x > 0\} \to \mathbb{R}$.

Do you know of any functions defined on $x > 0$ with the property that $p(x^\alpha) = \alpha p(x)$?

3. Aug 30, 2014

### Mentallic

Thanks! This analysis helped a lot!

Ahh well when you put it that way, evidently $p(x)=log_k(x)$ for k>0 would work.

So does it suffice to choose h to be

$$h(x) = |\log_k(x)| : x\in \mathbb{R} \backslash {0} , h(x) = 0 : x=0$$

4. Aug 30, 2014

### Mentallic

Actually, I only need to find one such h so ln(x) would suffice.