Prove that $\inf_{t>0}\frac{ω(t)}t=\lim_{t\to\infty}\frac{ω(t)}t$ if $ω:[0,\infty)\toℝ$ be bounded above on every finite interval and subadditive

Question

Let $\omega:[0,\infty)\to\mathbb R$ be bounded above on every finite interval and subadditive. I want to show that $$\omega_0:=\inf_{t>0}\frac{\omega(t)}t=\lim_{t\to\infty}\frac{\omega(t)}t\;.\tag 1$$ Notice that $\omega_0\in[-\infty,\infty)$. I've found a proof which starts by choosing some $\gamma>\omega_0$ and stating that there exists a $t_0>0$ with $$\frac{\omega(t_0)}{t_0}<\gamma\tag 2\;.$$ Why does such a $t_0$ exist?

Answer 1

I'm sorry, but the answer to my question is trivial: If $\omega_0>-\infty$, then $\gamma=\omega_0+\varepsilon$ for some $\varepsilon>0$ and we all know that, by definition of the infimum, there is some $t_0>0$ with $$\frac{\omega(t_0)}{t_0}<\omega_0+\varepsilon\;.$$ And if $\omega_0=-\infty$, then $$\frac{\omega(t_0)}{t_0}<-\left|\theta\right|\le\theta\;.$$ for some $t_0>0$, again, by definition of the infimum.

Answer 2

Let $t>0$ and $k \in \mathbb{N}$. Then let $s = k\cdot t\geq t$. We have

$$ \frac{\omega(s)}{s} = \frac{\omega(k\cdot t)}{k \cdot t} \leq \frac{k \cdot \omega(t)}{k \cdot t} \leq \frac{\omega(t)}{t}.$$

That is the quotient $\frac{\omega(t)}{t}$ decreases as $t$ increases. Since we can write all numbers $t>1$ as $n \cdot s$ for some $n \in \mathbb{N}, s \in [0,1]$, we have the desired conclusion.