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Let $(X,d)$ is a metric space and $X$ has no isolated points, $T:X\rightarrow X$ is a continuous self-map.

Def1. $T$ is strongly expansive if there exist $\varepsilon>0$, for any $x,y\in X$, $x\neq y$, we can find a number $n\in\mathbb{N}$, such that $d(T^nx,T^ny)>\varepsilon$.

Def2. $T$ is an expanding map if there exist a constant $c>1$ and a positive number $\varepsilon>0$, for any $x,y\in X$, if $d(x,y)<\varepsilon$, we have $d(Tx,Ty)>cd(x,y)$.

Of course, an expanding map must be weakly expansive. (The definition can be found from Are these two kinds of definitions about expansivity equivalent?)

My quesiton is: must an expanding map be strongly expansive? If not, does there exist a counterexample?

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  • $\begingroup$ Consider $x + $ a perturbation that decays at infinity to see why this is false. $\endgroup$ – user296602 Dec 31 '15 at 8:36
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They are equivalent if $T$ is injective. Suppose that $\epsilon$ and $c$ witness that $T$ is an expanding map. Let $\delta=\frac{1}2\epsilon$, and let$x$ and $y$ be distinct points of $X$. If $d(Tx,Ty)>\delta$, we're done. Otherwise, $d(Tx,Ty)\le\delta<\epsilon$, and therefore $d(T^2x,T^2y)>cd(Tx,Ty)$. As long as $d(T^nx,T^ny)<\epsilon$, we must have $d(T^{n+1}x,T^{n+1}y)>cd(T^nx,T^ny)$, so by an easy induction we have $d(T^{n+1}x,T^{n+1}y)>c^nd(Tx,Ty)$. Since $c^n$ increases without bound as $n$ increases, at some point we must have $d(T^nx,T^ny)>\delta$, and it follows that $\delta$ witnesses the strong expansiveness of $T$.

Otherwise, however, they are not. Let $X=\{x\in\Bbb R:|x|\ge1\}$ with the usual metric. Let $Tx=2|x|$ for $x\in X$. Let $x,y\in X$ with $x\ne y$. If $0<d(x,y)<1$, then $d(Tx,Ty)>\frac{3}2d(x,y)$, so $T$ is an expanding map. In this answer, however, I showed that $T$ is not strongly expansive.

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  • $\begingroup$ Why $Tx\neq Ty$? $\endgroup$ – James Chan Dec 31 '15 at 11:45
  • $\begingroup$ @James: Oversight on my part: I'm used to dealing with injective maps in the context of dynamical systems and unconsciously assumed that $T$ was injective. I'm headed for bed now; when I'm up again I'll either fix it or delete it. $\endgroup$ – Brian M. Scott Dec 31 '15 at 12:15
  • $\begingroup$ @James: As it turned out, I realized that I already had the fix. $\endgroup$ – Brian M. Scott Dec 31 '15 at 12:25

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