Derivative of monotone maximum function Say $f:[0,\infty)\to \mathbb{R}$ is smooth. Define
$$F(t)=\max_{s\le t}f(s).$$
Clearly since $F$ is monotonic, $F'$ exists for almost all $t\ge 0$. Say $t>0$ is one of those times for which $F'(t)$ exists and say we're given that $F'(t)>0$.
Then I'd like to claim that $F'(t)=f'(t)$. What I have so far is the following. First, $F'(t)>0$ should imply that $F(t)>F(s)$ for all $s<t$. This is clearly true because $F$ is monotonic and so if $F(t)=F(s)$ is is true for some $s<t$, then the left sided derivative of $F$ at $t$ would be zero (but we know $F'(t)>0$).
An immediate consequence is that $F(t)=f(t)$ for otherwise if $F(t)>f(t)$, then we must have $F(t)=f(s)$ for some $s<t$, but then we'd have $F(t)=F(s)$, which we ruled out.
Then, it follows that for $s<t$,
$$F(t)-F(s)=f(t)-F(s)\le f(t)-f(s)$$
so that after dividing by $t-s$ and taking the limit, we find
$$F'(t)\le f'(t)$$
That's one direction. How does one see the opposite direction? Is it even true?
 A: For the other direction you can argue similarly, now with $s > t$:
$$
 f(s) - f(t) \le F(s) - f(t) = F(s) - F(t) \, ,
$$
then divide by $s - t > 0$ and take the limit $s \to t+$. It follows that
$f'(t) \le F'(t)$.
A: Here is a tedious answer:
Let $S(t) = \{ s \in [0,t] | F(t)=f(s) \}$ and suppose $f$ is differentiable.
Then $F$ is differentiable at $t$ iff one of the following conditions holds:

*

*$t \notin S(t)$,

*$S(t)= \{t\}$ and $f'(t) \ge 0$,

*$ t\in S(t)$, $S(t)$ is not a singleton and $f'(t) = 0$.

Furthermore, if $F$ is differentiable at $t$ then $F'(t) = f'(t)$ if $ t \in S(t)$ and $F'(t) = 0$ otherwise.
Note that we always have $F(t) \ge f(t)$.
Suppose $F$ is differentiable at $t$ (since $F$ is non decreasing, we must have $F'(t) \ge 0$).
If $t \notin S(t)$ then $f(t) < F(t)$ and $F$ is constant in a neighbourhood of $t$ and hence $F'(t) = 0$.
Suppose $t \in S(t)$, then for $s>t$ we have
${F(s)-F(t) \over s-t} \ge {f(s)-f(t) \over s-t}$ and so $F'(t) \ge f'(t)$.
If $s<t$ we have ${F(s)-F(t) \over s-t} \le {f(s)-f(t) \over s-t}$ and so $F'(t) \le f(t)$. Combining gives $F'(t) = f'(t)$.
If $t \in S(t)$ and $S(t)$ is not a singleton, then $F$ must be constant on $[\xi,t]$ (where $\xi \in S(t) \setminus \{t\}$) and so $F'(t) = 0$ and from the previous remark we have $f'(t) = 0$.
For the reverse direction, suppose 1. is true. Then $f(t) < F(t)$ and $F$ is constant in a neighbourhood of $t$ and hence $F'(t) = 0$.
If 2. is true, note that $F(t) = f(t)$ and that $f(s) <f(t)$ for $s<t$ and so $f'(t) \ge 0$. We also have $F(s) < F(t)$ for $s < t$.
First suppose $f'(t) = 0$. Then for any $\epsilon>0$ we can find a neighbourhood $U$ of $t$ such that
$f(t)-\epsilon |s-t| \le f(s) \le f(t)+\epsilon |s-t|$ for $s \in U$. It
is not hard to see that we have $f(s) \le F(s) \le f(t)+\epsilon |s-t|$ for $s >t$ (and in $U$) and so
${f(s)-f(t) \over s-t} \le {F(s)-F(t) \over s-t} \le \epsilon$.
For $s<t$ we have $f(s) \le F(s) \le F(t) =f(t)$ and so
${f(s)-f(t) \over s-t} \ge {F(s)-F(t) \over s-t} \ge 0$. Since $\epsilon>0$ was arbitrary, it follows that
$F$ is differentiable at $t$ and $F'(t) = f'(t) = 0$.
Otherwise we have $f'(t) >0$. As above, choose $\epsilon>0$ and find a neighbourhood $U$ such that $f(s) \le f(t)+f'(t)(s-t)+\epsilon |s-t|$. We need to satisfy a little technicality
in order to find an upper bound for $F(s)$. Let $U(s)=f(t)+f'(t)(s-t)+\epsilon |s-t|$ and choose some $s^* < t$ in $U$ (we must have $F(s^*) < F(t)$). Now choose some $\tilde{s} \in [s^*,t)$ such that $F(\tilde{s}) \le U(\tilde{s})$ and let $\tilde{U} = U \cap (\tilde{s},\infty)$. Then for
$s \in \tilde{U}$ we have
$f(s) \le F(s) \le f(t)+f'(t)(s-t)+\epsilon |s-t|$. Similar to above, for $s>t$ we have
${f(s)-f(t) \over s-t} \le {F(s)-F(t) \over s-t} \le f'(t) + \epsilon$ and for
$s<t$ we have
${f(s)-f(t) \over s-t} \ge {F(s)-F(t) \over s-t} \ge f'(t) - \epsilon$.
Since $\epsilon>0$ was arbitrary, it follows that
$F$ is differentiable at $t$ and $F'(t) = f'(t)$.
If 3. is true, then $F$ must be constant on some non empty interval of the form $[s^*,t]$. Since $f'(t) = 0$, as above we have
$f(s) \le F(s) \le f(t)+\epsilon |s-t|$ for $s>t$ (and in the relevant $U$), and
similarly, we have $0 \le {F(s)-F(t) \over s-t} \le \epsilon$ and hence
$F'(t) = 0$.
Note The only way you can have $F'(t) >0$ is if $S(t) = \{t\}$ and $f'(t) >0$ in which case we have $F'(t)=f(t)$.
