Assume the following is true for a continuous non-negative function $f(t)$ on $I=[0,T]$:
- $f(0) \le 1$
- For every $t \in I$, if $f(t) \le 1$ there is a neighborhood of $t$, $(t-\epsilon, t+\epsilon)$ such that $f \le 1$ on that whole neighborhood. And continuity implies that $f \le 1$ at $t \pm \epsilon$ also.
Does this imply $f(t) \le 1$ on the whole $I$?
Starting at $t=0$, then there is a neighborhood $[0, \epsilon_{1}]$ such that $f \le 1$. Then we can continue to start at $t = \epsilon_{1}$ where $f(\epsilon_{1}) \le 1$, again there is a closed interval $[\epsilon_{1}-\epsilon_{2}, \epsilon_{1}+\epsilon_{2}]$ such that $f \le 1$ there. We can do this continuously until reaching $t=T$?
My doubt is because every time we use the 2nd assertion, the neighborhood could get smaller and smaller such that we never reaches $t=T$.
If for some $t_0 \in[0,T]$ we have that $f(t_0)>1$, there is some $t_* \in [0, t_0)$ such that $f(t_*) = 1$ and $f(t)>1$ in $[t_*, t_*+\varepsilon)$. But this point $t_*$ cannot exist, due to the second assumption.