Is "shadowing" invariant under topological conjugacy?

Question

A homeomorphism $f: X \rightarrow X$ of a metric space is said to have the shadowing property if for all $\varepsilon > 0$ there is a $\delta>0$ such that for every sequence $(x_n)_{-\infty}^{\infty}$ with

$$d(x_{n+1},f(x_n))<\delta$$

there is a $y \in X$ with

$$d(f^n(y), x_n)<\varepsilon$$

Is this property invariant under topological conjugacy? I think I can prove it assuming the conjugating homeomorphsim is Lipshitz.

If it's not invariant under conjugacy, why is it relavant in dynamics?

Answer 1

Let $f:X\to X$ and $g:Y\to Y$ be homeomorphisms of metric spaces. Let $\Phi:X\to Y$ be a uniform conjugacy, that is a bijection such that

• both $\Phi$ and $\Phi^{-1}$ are uniformly continuous,
• $f\circ\Phi=\Phi\circ g$.

Claim. If $g$ has the shadowing property, so does $f$.

Proof. Let $\varepsilon>0$. By the uniform continuity of $\Phi^{-1}$, there exists an $\varepsilon'>0$ such that for $y,y'\in Y$,

1. $d(y,y')<\varepsilon'$ implies $d(\Phi^{-1}y,\Phi^{-1}y')<\varepsilon$.

By the shadowing property of $(Y,g)$, there is a $\delta'>0$ such that

2. for every sequence $(y_n)_{-\infty}^\infty$ in $Y$ such that $d(y_{n+1},g(y_n))<\delta'$, there is a point $\tilde{y}\in Y$ such that $d(g^n(\tilde{y}),y_n)<\varepsilon'$. [In words: every $\delta'$-chain in $(Y,g)$ is $\varepsilon'$-shadowed by an orbit of $g$*.]

By the uniform continuity of $\Phi$, there exists a $\delta>0$ such that for $x,x'\in X$,

3. $d(x,x')<\delta$ implies $d(\Phi x,\Phi x')<\delta'$.

Now, let $(x_n)_{-\infty}^\infty$ be an $\delta$-chain in $(X,f)$, that is, $d(x_{n+1},f(x_n))<\delta$. By (1), $d(\Phi x_{n+1},g(\Phi x_n))<\delta'$, that is, $(\Phi x_n)_{-\infty}^\infty$ is a $\delta$-chain in $(Y,g)$. By (2), this chain is $\delta'$-shadowed by an orbit of $g$, that is, there exists a $\tilde{y}\in Y$ such that $d(g^n(\tilde{y}),\Phi x_n)<\varepsilon'$. By (3), the latter implies that $d(f^n(\Phi^{-1}\tilde{y}),x_n)<\varepsilon$, that is, $(x_n)_{-\infty}^\infty$ is $\varepsilon$-shadowed by the orbit of $\Phi^{-1}\tilde{y}$ in $(X,f)$.

Q.E.D.

For homeomorphisms on metric spaces, the shadowing property is not invariant under non-uniform conjugacies. Here is an example:

Counter-example when the conjugacy is not uniform
Consider $X:=\{2^n: n\in\mathbb{Z}\}\subseteq\mathbb{R}$ and $Y:=\mathbb{Z}\subseteq\mathbb{R}$ as metric spaces with the Euclidean metric. Define $f(x):=2x$ on $X$ and $g(y):=y+1$ on $Y$. Then $f$ and $g$ are conjugate homeomorphisms with non-uniform conjugacy map $\Phi x:= \log_2 x$.

Observe that $(Y,g)$ has the shadowing property for the simple reason that every $\delta$-chain in $(Y,g)$ with $\delta\leq 1$ is in fact an orbit. On the other hand, $(X,f)$ does not have the shadowing property. In fact, for every $\delta>0$ (no matter how small), there is a $\delta$-chain $(x_n)_{-\infty}^\infty$ in $(X,f)$ that is not $\varepsilon$-shadowed by any orbit for any $\varepsilon>0$ (no matter how large). Namely, choose $m\in\mathbb{Z}$ such that $d(2^m,2^{m+1})<\delta$ (i.e., $m<\log_2\delta$) and set $x_n:= 2^m$ for every $n\in\mathbb{Z}$. Then, $(x_n)_{-\infty}^\infty$ is a $\delta$-chain but cannot be shadowed by any orbit of $f$.