Prove there is no homeomorphism that makes two functions conjugate

[razmtaz]

Homework Statement

let f(x) = x³ and g(x) = x - 2x³. Show there is no homeomorphism h such that h(g(x)) = f(h(x))

Homework Equations

Def let J and K be intervals. the function f:J->K is a homeomorphism of J onto K if it is one to one, onto, and both f and its inverse are continuous

Def let f:J->J and g:K->K, then f is conjugate to g if there is a homeomorphism h:J->K such that h(f(x)) = g(h(x)). (Then use the inverse of h as the homeomorphism that makes g conjugate to f, as in the given question)

Thm. let f conjugate to g by h. then:
1. h(fⁿ) = gⁿ(h) for n >=1
2. if x* is a periodic point of f, then h(x*) is a periodic point of g
3. if f has a dense set of periodic points, so does g


Another question in the book goes: let f(x) = x³ and g_$\mu$(x) = x - $\mu$x³. Show there is no nontrivial polynomial h such that f is conjugate to g by h.

The Attempt at a Solution

I know nothing of topology so this question is difficult for me to start.

I want to show there is no h(x) such that h(x)^3 = h(x-2x³). this is from the definition/question statement, but doesn't really seem useful in helping me solve the problem.

My next intuition is that certain properties are preserved by conjugacy, so if I could show that one of them is violated then I would have a solution.
for example, periodic points are preserved by conjugacy, so if I can find a periodic point x* of g(x) for which h(x*) is not a periodic point of f(x), then i'd be in business. I used maple to determine that g(x) has period 2 points (1 and -1) and no period 3 points. I also know that 1 and -1 are periodic points of f(x). Can I use this in any way? The trouble I have here is that I don't know what h is going to "do", so I can't really use the theorem I quoted to claim that h(x*) won't be a periodic point for f(x).

Any ideas would be excellent. I found this other thread on proving no homeomorphism exists (https://www.physicsforums.com/showthread.php?t=423411&highlight=no+homeomorphism) but it seems a little different in that theyre showing two spaces arent homeomorphic to one another whereas I am showing that two functions arent conjugate... also much of the language of topology (compactness, connectedness..) is supposedly outside the scope of the course I am taking and not needed to solve this problem at all. so any hints/help?

thanks alot

 

[sunjin09]

Homework Statement

let f(x) = x³ and g(x) = x - 2x³. Show there is no homeomorphism h such that h(g(x)) = f(h(x))

Homework Equations

Def let J and K be intervals. the function f:J->K is a homeomorphism of J onto K if it is one to one, onto, and both f and its inverse are continuous

Def let f:J->J and g:K->K, then f is conjugate to g if there is a homeomorphism h:J->K such that h(f(x)) = g(h(x)). (Then use the inverse of h as the homeomorphism that makes g conjugate to f, as in the given question)

Thm. let f conjugate to g by h. then:
1. h(fⁿ) = gⁿ(h) for n >=1
2. if x* is a periodic point of f, then h(x*) is a periodic point of g
3. if f has a dense set of periodic points, so does g


Another question in the book goes: let f(x) = x³ and g_$\mu$(x) = x - $\mu$x³. Show there is no nontrivial polynomial h such that f is conjugate to g by h.

The Attempt at a Solution

I know nothing of topology so this question is difficult for me to start.

I want to show there is no h(x) such that h(x)^3 = h(x-2x³). this is from the definition/question statement, but doesn't really seem useful in helping me solve the problem.

My next intuition is that certain properties are preserved by conjugacy, so if I could show that one of them is violated then I would have a solution.
for example, periodic points are preserved by conjugacy, so if I can find a periodic point x* of g(x) for which h(x*) is not a periodic point of f(x), then i'd be in business. I used maple to determine that g(x) has period 2 points (1 and -1) and no period 3 points. I also know that 1 and -1 are periodic points of f(x). Can I use this in any way? The trouble I have here is that I don't know what h is going to "do", so I can't really use the theorem I quoted to claim that h(x*) won't be a periodic point for f(x).

Any ideas would be excellent. I found this other thread on proving no homeomorphism exists (https://www.physicsforums.com/showthread.php?t=423411&highlight=no+homeomorphism) but it seems a little different in that theyre showing two spaces arent homeomorphic to one another whereas I am showing that two functions arent conjugate... also much of the language of topology (compactness, connectedness..) is supposedly outside the scope of the course I am taking and not needed to solve this problem at all. so any hints/help?

thanks alot

All you have to prove is that h such that h(x)^3=h(x-2x^3) cannot be one-to-one, I think you can easily find a pair of x's that lead to the same h(x) by making use of the fact that x-2x^3 is not one-to-one

 

[razmtaz]

Thanks! a lot simpler than what I was attempting to do.

 

[razmtaz]

Okay, so I ended up doing this question differently: since its a homeomorphism then it preserves the fixed points, and so h(g(x)) has a certain number of fixed points which implies that f(h(x)) has the same number, but f only has 1 fixed point and h(g(x)) has 3 i think, so that's the contradiction and thus h cannot exist.

however I still have a question about your method sunjin09: I can find two points that lead to the same h(x), but the thing is they arent different in the domain of h(x). what i mean is, I can find different points that make g(x) = 0 which are x = 0 and +/-(1/sqrt(2)), but putting these into h(x) would give h(0) each time, so yes, I am starting with different values (0, 1/sqrt(2), -1/sqrt(2)) but they are all coming down to h(0), so h still maps 0 to one value. how would I show that h isn't one-to-one?

just as another example, i can find two points that g(x) maps to the same value (cause its not one-to-one), so i can put each of these points into h(g(x)). but say g(x) maps the value to a number a, then we have h(a) each time, so this doesn't really tell us anything about whether h is injective or not. I don't know if youre still subscribed to this thread but if you are I would appreciate some clarification, cause I am still confused.

Thanks

 

[sunjin09]

Okay, so I ended up doing this question differently: since its a homeomorphism then it preserves the fixed points, and so h(g(x)) has a certain number of fixed points which implies that f(h(x)) has the same number, but f only has 1 fixed point and h(g(x)) has 3 i think, so that's the contradiction and thus h cannot exist.

however I still have a question about your method sunjin09: I can find two points that lead to the same h(x), but the thing is they arent different in the domain of h(x). what i mean is, I can find different points that make g(x) = 0 which are x = 0 and +/-(1/sqrt(2)), but putting these into h(x) would give h(0) each time, so yes, I am starting with different values (0, 1/sqrt(2), -1/sqrt(2)) but they are all coming down to h(0), so h still maps 0 to one value. how would I show that h isn't one-to-one?

just as another example, i can find two points that g(x) maps to the same value (cause its not one-to-one), so i can put each of these points into h(g(x)). but say g(x) maps the value to a number a, then we have h(a) each time, so this doesn't really tell us anything about whether h is injective or not. I don't know if youre still subscribed to this thread but if you are I would appreciate some clarification, cause I am still confused.

Thanks

Sorry about the late reply. If g(x1)=g(x2), then h(x)^3=h(g(x)) → h(x)=h(g(x))^(1/3) is the same for x1 and x2, because x^(1/3) is one to one, not h(g(x)). This is what I originally meant.