If $f:[0,\infty)\to\mathbb R$ is continuous and $\tau=\inf\left\{t>0:f(t)>\varepsilon\right\}$, then $f(\tau\wedge t)\le\varepsilon$ for all $t\ge0$

Question

Let $f:[0,\infty)\to\mathbb R$ be continuous, $\varepsilon>0$ and $$\tau:=\inf\left\{t>0:f(t)>\varepsilon\right\}.$$ Are we able to show that $f(\tau\wedge t)\le\varepsilon$ for all $t\ge0$ (where $x\wedge y:=\min(x,y)$ for $x,y\in\mathbb R$)?

I'm struggling, since $\left\{t>0:f(t)>\varepsilon\right\}=(0,\infty)\cap f^{-1}\left(\left(\varepsilon,\infty\right)\right)$ is open.

Answer 1

This is possible to show if $\tau > 0$. If $\tau = 0$, it's possible to have $f(0) > \varepsilon$, and so $f(0 \land t) = f(0) > \varepsilon$.

If $\tau > 0$, then for $0 \le x < \tau$, we have $f(x) \le \varepsilon$, otherwise $\tau$ would not be a lower bound for $\{ t > 0 : f(t) > \varepsilon\}$. In particular, if we take $x_n \to \tau$, with $x_n < \tau$ (e.g. $x_n = \tau(1 - 1/n)$), then by continuity, we must have $f(x_n) \to f(\tau)$, but given $f(x_n)$ lies in the closed set $[0, \varepsilon]$, so does $f(\tau)$. That is, $f(\tau) \le \varepsilon$.

(We could go further and actually prove $f(\tau) = \varepsilon$, but this is not necessary!)

So, $\tau \land t \le \tau$ for all $t$, so either $\tau \land t = \tau$ or $0 \le \tau \land t < \tau$. In either case, we have proven $f(\tau \land t) \le \varepsilon$.