Show that if $X:[0,T]\to\mathbb R$ is continuous, then $\inf\left\{t\in\overline I:|X(t)|\geε\right\}\le t$ iff $|X(s)|\geε$ for some $s\in[0,t]$

Question

Let

• $T>0$
• $I:=(0,T]$
• $X:\overline I\to\mathbb R$ be continuous with $X(0)=0$
• $\varepsilon>0$
• $\tau:=\inf\left\{t\in\overline I:|X(t)|\ge\varepsilon\right\}$
• $t\in\overline I$

How can we show that $\tau\le t$ if and only if $|X(s)|\ge\varepsilon$ for some $s\in[0,t]$?

Clearly, the "if" part is trivial. However, I don't know how we need to argue in the "only if" part. For example, why isn't it possible that $\tau=t$, $|X(s)|<\varepsilon$ for all $s\in[0,t]$ and $|X(s)|\ge\varepsilon$ for all $s\in(t,T]$?

Answer 1

It suffices to show that $|X(\tau)| = \epsilon$.

Observe that $\tau \in I$. Suppose $|X(\tau)|<\epsilon$. Take $\eta = \frac12(\epsilon - X(\tau))$. The continuity of $X$ gives a $\delta$-neighbourhood of $\tau$ so that $X(t') \in (X(\tau)-\eta, X(\tau)+\eta) \subseteq (X(\tau)-\eta, \epsilon)$ for all $t' \in (\tau-\delta, \tau+\delta)$. This contradicts the definition of $\tau$ as a greatest lower bound of $\{|X| \ge \epsilon\}$ (because any $t' \in (\tau, \tau + \delta)$ is a greater lower bound for $\{|X| \ge \epsilon\}$).

For this question, we're done with $|X(\tau)| \ge \epsilon$. (It's given that $\tau \le t$, so $s = \tau$ satisfies the question.) The rest is left for exercise.

Remarks: In probability-theory, $\tau$ is called the hitting time of the event $\{|X| \ge \epsilon\}$, where $\Omega = [0,T]$ equipped with the usual Borel $\sigma$-algebra. Then $X$ is a random variable starting from $0$. It's quite intuitive that $|X(\tau)| = \epsilon$ due to the continuity of $X$.

Answer 2

Let $A=\{t\in [0,T]: |X(t)|\geq \varepsilon\}.$

If $t\in [0,T]$ and $|X(s)|< \varepsilon$ for all $s\in [0,t]$ then $A\subset [t,T].$ So if $A$ is not empty then $\tau =\inf A\geq \inf [t,T]=t .$

If $t\in [0,T]$ and $|X(s)| \geq \varepsilon$ for some $s\in [0,t]$ then $s\in A$ so $\tau =\inf A\leq s\leq t.$