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I'm trying to proof the following, and I'm looking for a verification of my proof. If it is incorrect, I'm looking for some help towards a right proof.

If the omega limit set is $\omega(x_0)=\{x^*\}$, then $\lim_{t\to\infty}\phi(t;x_0)=x^*$

Suppose $\lim_{t\to\infty}\phi(t;x_0)$ exists and equals $x'$. Then $\omega(x_0)=\{x'\}$, so it suffices to proof that the limit exists. Now assume for the sake of contradiction that $\lim_{t\to\infty}\phi(t;x_0)$ does not exist. Because $x^*\in\omega(x_0)$, there exists a sequence $(t_n)_{n\in\mathbb{N}}$ such that $\lim_{n\to\infty}\phi(t_n;x_0)=x^*$. Because $\lim_{t\to\infty}\phi(t;x_0)$ does not exist, there exists a $\delta>0$ such that there is no $T$ such that $\phi(t;x_0)\in B_{\delta}(x^*)$ for all $t\geq T$ (otherwise the limit would exist). Fix this $\delta$. From continuity of the flow and existence of a sequence $(t_n)_{n\in\mathbb{N}}$ satisfying $\lim_{n\to\infty}t_n=\infty$ and $\lim_{n\to\infty}\phi(t_n,x_0)=x^*$, it follows that for an arbitrary $\epsilon>\delta$, there exists a sequence $(s_n)_{n\in\mathbb{N}}$ satisfying $\lim_{n\to\infty}s_n=\infty$ such that for all $n\in\mathbb{N}$, $\phi(s_n;x_0)\in\overline{B_{\epsilon}(x^*)}\setminus B_{\delta}(x^*)$, which is in particular a closed set. Now it follows from Bolzano-Weierstrass that $\left(\phi(s_n;x_0)\right)_{n\in\mathbb{N}}$ has a convergent subsequence $\left(\phi(s_{n_k};x_0)\right)_{k\in\mathbb{N}}$, belonging to $\overline{B_{\epsilon}(x^*)}\setminus B_{\delta}(x^*)$, because this set is closed, and thus the limit value is unequal to $x^*$; thus $\omega(x_0)\neq\{x^*\}$, which gives the desired contradiction.

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1 Answer 1

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Your proof is correct.​​​​​​​​​​​​​

Edit, in response to the comment: "Would you have any shorter proof in mind or do you think this is one of the simplest possible?".

This is the simplest proof that I know of. The presentation can be made very easy to read too.


Lemma. If $K$ is sequentially compact and $\phi(t_n;x_0)\in K$ for times $t_n\to\infty$, then $\omega(x_0)\cap K\neq\emptyset$.

Proof. Immediate just from the definitions.

If, by contradiction, $x^*$ is not the limit of $\phi(t;x_0)$, there is $\delta>0$ and $t_n\to\infty$ such that $\phi(t_n;x_0)\in B_\delta(x^*)^c$.

Since $x^*\in\omega(x_0)$, there are times $t'_n\to\infty$ such that $\phi(t'_n;x_0)\in B_{\delta/2}(x^*)$.

By continuity of $\phi(\,\cdot\,;x_0)$, there are times $t''_n\to\infty$ such that $\phi(t''_n;x_0)\in \overline{B_\delta(x^*)}\setminus B_{\delta/2}(x^*)=K$. Then the Lemma implies that $\omega(x_0)$ cannot be just $\{x^*\}$.

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  • $\begingroup$ Thanks for reading. Would you have any shorter proof in mind or do you think this is one of the simplest possible? $\endgroup$ Oct 29, 2018 at 18:06
  • $\begingroup$ Thank you for your edit. Indeed the idea is the same, but it is way shorther and better readable $\endgroup$ Oct 31, 2018 at 7:06
  • $\begingroup$ @LutzL you are totally right! thanks for spotting the typo $\endgroup$
    – Federico
    Nov 5, 2018 at 13:02

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