If $f:[0,\infty)\to\mathbb R$ is continuous and $\tau:=\inf\left\{t\ge0:|f|\ge\varepsilon\right\}\in(0,\infty)$, is $|f(\tau)|=\varepsilon$?

Question

Let $f:[0,\infty)\to\mathbb R$ be continuous and $$\tau:=\inf\left\{t\ge0:|f|\ge\varepsilon\right\}$$ for some $\varepsilon\ge0$. Assume $\tau\in(0,\infty)$. Is it possible to conclude $|f(\tau)|=\varepsilon$?

Clearly, since $\tau\in(0,\infty)$, there are $t_{<\varepsilon},t_{\ge\varepsilon}\ge0$ with $$|f(t_{<\varepsilon})|<\varepsilon$$ and $$|f(t_{\ge\varepsilon})|\ge\varepsilon$$ and hence $\tau\in(t_{<\varepsilon},t_{\ge\varepsilon}]$. If $|f(\tau)|>\varepsilon$, then there would be a $t_\varepsilon\in(t_{<\varepsilon},\tau)$ with $$|f(t_\varepsilon)|=\varepsilon;$$ in constrast to the definiton of $\tau$. So, $$|f(\tau)|\le\varepsilon.\tag0$$

Answer 1

Let $$I:=\left\{t\ge0:|f(t)|\ge C\right\}.$$ Assume $$|f(\tau)|<C.\tag1$$ Let $$\varepsilon:=C-|f(\tau)|>0.$$ $f$ is continuous at $\tau$ and $\tau>0$ $\Rightarrow$ $\exists\delta\in(0,\tau)$ with $$f(B_{\delta}(\tau))\subseteq B_{\varepsilon}(f(\tau))\tag2$$ and hence $$|f(t)|\le|f(\tau)|+|f(t)-f(\tau)|<C\;\;\;\text{for all }t\in B_\delta(\tau)\tag3.$$ Thus, $$I\subseteq[0,\infty)\setminus B_\delta(\tau)$$ and hence $$\forall t\in I:|t-\tau|\ge\delta\tag4.$$ $\tau$ is a lower bound for $I$ $\Rightarrow$ $$\forall t\in I:\tau\le t\tag5$$ and hence (since $t\le\tau-\delta<\tau$ is in contradiction to $(5)$ for all $t\in I$) $$\forall t\in I:t\ge\tau+\delta;\tag6$$ in contradiction to the definition of the infimum. By $(0)$, we obtain $$|f(\tau)|=C.$$

Answer 2

If $|f(\tau)|>\epsilon$ then there is a neighborhood $U\subset (0,\infty)$ of $\tau$ such that $|f(x)|>\epsilon$ for $x\in U$. Therefore there are points $<\tau$ in which $|f(x)|>\epsilon$.

Likewise, if $|f(\tau)|<\epsilon$ there is a neighborhood $U\subset (0,\infty)$ of $\tau$ in which $|f(\tau)|<\epsilon$ holds, contradicting the definition of $\tau$ because $\{t\geq0:\ |f(t)|\geq \epsilon\}\subset U^c$.