If $x:[0,T]\to E$ is continuous and $Ω⊆E$ is open with $x(0)\in Ω$, how are $\inf\{t∈[0,T]:x(t)\not\in Ω\}$ and $\sup\{t∈(0,T]:x([0,t])⊆Ω\}$ related?

Question

Let $T>0$, $(E,d)$ be a metric space, $x:[0,T]\to E$ be continuous, $\Omega\subseteq E$ be open, $$\tau:=\inf\underbrace{\{t\in[0,T]:x(t)\not\in\Omega\}}_{=:\:\tilde I}$$ and $$\sigma:=\sup\underbrace{\{t\in(0,T]:x([0,t])\subseteq\Omega\}}_{=:\:I}.$$

Question: How are $\sigma$ and $\tau$ related?

We can show the following:

Lemma 1: If $I\ne\emptyset$, then

1. $\tau\in\tilde I$ and hence $x(\tau)\not\in\Omega$;
2. if $x(0)\in\Omega$, then $\tau>0$ and $x(\tau)\in\partial\Omega$.

Let's try to figure out the relationship to $I$ and $\sigma$. Let $x_0:=x(0)$. Since $x_0\in\Omega^\circ$, there is a $\varepsilon>0$ with $$B_\varepsilon(x_0)\subseteq\Omega\tag1.$$ Since $x$ is continuous at $0$, there is a $\delta\in(0,T]$ with $$x([0,\delta])\subseteq B_\varepsilon(x_0)\tag2.$$ By $(1)$ and $(2)$, $I\ne\emptyset$ and $\sigma\in(0,T]$. By Lemma 1, $$\tau\in\tilde I\cap(0,T]\tag3$$ and $$x(\tau)\in\partial\Omega.\tag4$$

Now, I'm unsure whether we can conclude $\sigma\in\tilde I$ and $\sigma<\tau$ or $\sigma=\tau$.

Answer 1

Answer: $\tau = \sigma$ unless $I=\emptyset$, in which case $\tau > \sigma$.

Actually, one can do without Lemma 1. If $x(0) \notin \Omega$, $0 \in \tilde I$ and $I = \emptyset$, resulting in $\tau = 0 > -\infty = \sigma$. Now, suppose $x(0) \in \Omega$.

• "$\sigma > 0$". By continuity, $x((-\delta, \delta)\cap[0,T]) \subseteq \Omega \cap B_1(x(0))$ for some $\delta > 0$. Then, $t := \min(\delta,T)/2 > 0$ is contained in $I$, as $x([0,t]) \subseteq x((-\delta, \delta)\cap[0,T]) \subseteq \Omega \cap B_1(x(0)) \subseteq \Omega$. Thus, $\sigma = \sup I \geq t > 0$.
• "$\tau \geq \sigma$". Suppose $\tau < \sigma$. Since $\sigma = \sup I$, there exists $t \in I$ such that $0 < \sigma - t < \sigma - \tau$, i.e., $\tau < t < \sigma$. Since $x([\tau,t)) \subseteq x([0,t]) \subseteq \Omega$, we have the contradiction $\tau \neq \inf\tilde I$ where $x(\tilde I) = \Omega^c$.

If $\sigma = T$, then $T = \sigma \leq \tau \leq T$, meaning that $\tau = \sigma$. Now, suppose $\sigma < T$.

• "$x(\sigma) \notin \Omega$". Suppose $x(\sigma) \in \Omega$. By continuity, $x((\sigma-\delta, \sigma+\delta)\cap[0,T]) \subseteq \Omega \cap B_1(x(\sigma))$ for some $\delta > 0$. Then, $t := \sigma + \min(\delta, T - \sigma)/2 > \sigma$ is contained in $I$, as $x([\sigma,t]) \subseteq x((\sigma-\delta, \sigma+\delta)\cap[0,T]) \subseteq \Omega \cap B_1(x(\sigma)) \subseteq \Omega$. This leads to the contradiction $\sigma < t \leq \sup I = \sigma$.

Since $x(\sigma) \notin \Omega$, $\sigma \in \tilde I$. Thus, $\tau = \inf \tilde I \leq \sigma \leq \tau$, meaning that $\tau = \sigma$.