Topologically conjugate function

[Mentallic]

Homework Statement

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))

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?

 

[pasmith]

Homework Statement

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))

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?

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)?

 

[Mentallic]

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}.

Thanks! This analysis helped a lot!

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

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

 

[Mentallic]

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