Topological conjugacy between linear maps in dimension $1$

Question

Let $A_{\alpha}: \mathbb{R} \to \mathbb{R}$ denote the linear map

\begin{align} A_{\alpha}(x)=\alpha x. \end{align}

By definition two homeomorphisms $f:X \to X$ and $g:X \to X $ are topologically conjugate to each other if there is a homeomorphism $h:X \to X $ such that $hf=gh$.

I wish to prove that if $ 0 < \alpha < 1$ and $ 0 < \beta < 1 $ then $A_{\alpha}$ and $A_{\beta}$ are topologically conjugate. Could anyone help me find this homeomorphism?

Answer 1

Although it is "overkill" let us show how the more general result Any two orientation-preserving homeomorphisms of $[a,b]$ without fixed point in$(a,b)$ are topologically conjugate can be used.

Let us consider the maps $B_\alpha : [0,\infty) \to [0,\infty), B_\alpha(x) = \alpha x$. It suffices to show that all of them are conjugate:

In fact, each homeomorphism $h : [0,\infty) \to [0,\infty)$ must keep $0$ fixed because otherwise $h$ would map the connected set $(0,\infty)$ onto the non-connected set $[0,\infty) \setminus \{h(0)\}$. Thus $h$ has the odd extension $h' : \mathbb R \to \mathbb R$ given by $h'(x) = h(x)$ for $x \ge 0$ and $h'(x) = -h(-x)$ for $x \le 0$ which is a well-defined homeomorphism. Note that

1. $B'_\alpha = A_\alpha$.

2. $(h_1h_2)' = h'_1h'_2$.

Thus, given $h$ such that $h B_\alpha = B_\beta h$, then $h' A_\alpha = h' B'_\alpha = (h B_\alpha)' = (B_\beta h)' = B'_\beta h' = A_\beta h'$.

Now define $\phi : [0,\infty) \to [0,1), \phi(t) = \dfrac{t}{1+t}$. This is a homeomorphism with inverse $\phi^{-1}(t) = \dfrac{t}{1-t}$. This reduces our task to show that all $C_\alpha = \phi A_\alpha \phi^{-1} : [0,1) \to [0,1)$ are conjugate.

We have $C_\alpha(x) \to 1$ as $x \to 1$, thus $C_\alpha$ extends to a homeomorphism $D_\alpha : [0,1] \to [0,1]$ which keeps the boundary points $0,1$ fixed and therefore is orientation preserving. Moreover $D_\alpha$ has no fixed point in $(0,1)$: If $x \in (0,1)$ satisfies $D_\alpha(x) = C_\alpha(x) = x$, then $y = \phi^{-1}(x) \in (0,\infty)$ and $\alpha y = A_\alpha(\phi^{-1}(x)) = \phi^{-1}(x) =y$ which is impossible for $\alpha < 1$.

By the general result $D_\alpha$ and $D_\beta$ are conjugate and therefore also $C_\alpha$ and $C_\beta$ are conjugate.

Answer 2

You want to find $h$ such that $h(\alpha x)=\beta h(x)$. By symmetry, we may as well work on $[0,\infty)$.

Can you find $\gamma$ for which $h(x)=x^\gamma$ works? (Which means $h(x)=\mathrm{sgn}(x)|x|^\gamma$ on all of $\mathbb{R}$.)