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


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?

  • $\begingroup$ The standard setting that covers most interesting examples is when $X$ is a compact metric space. In that category, the shadowing property is certainly invariant under topological conjugacy. I suspect the same will be true if $X$ is only metric, as long as $f$ is uniformly continuous and/or you consider only uniformly continuous conjugacies. You should check. $\endgroup$ – Blackbird Aug 29 '18 at 17:38
  • $\begingroup$ The significance of knowing that a property is invariant under conjugacies is that we can pass to conjugate systems when we want to verify OR use the property. This sometimes makes the life easier. Plus, mathematics is all about abstractions. If we know property (X) is invariant under topological conjugacies and we manage to show that system (S) satisfies (X), then we have automatically shown that EVERY system conjugate to (S) satisfies (X). $\endgroup$ – Blackbird Aug 29 '18 at 17:46
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    $\begingroup$ @Blackbird I understand the significance of a property being preserved by conjugacy, but was unable to show that shadowing is conjugacy invariant. Could you show some details on why shadowing is invariant? $\endgroup$ – Analysis Aug 31 '18 at 0:07
  • $\begingroup$ See the answer below. $\endgroup$ – Blackbird Sep 1 '18 at 3:23

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

  1. 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 $x,x'\in X$,

  1. $d(x,x')<\delta$ 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 $(X,f)$.


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 $f$.

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    $\begingroup$ Is there supposed to be a comma right before "that is" $\endgroup$ – mathworker21 Sep 1 '18 at 13:37
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    $\begingroup$ Do you know a counterexample if we drop the uniform continuity? $\endgroup$ – Severin Schraven Sep 2 '18 at 0:37
  • $\begingroup$ @SeverinSchraven: See the update. $\endgroup$ – Blackbird Sep 2 '18 at 1:45
  • $\begingroup$ Unfortunately, one can only upvote once! Great counterexample $\endgroup$ – Severin Schraven Sep 2 '18 at 11:43

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