If $f(t)$ is continuous for $t$ $\in [0,1]$, $f' > 0$ and $f''(t) > 0$ for $t \in (0,1)$, do we have that $f'(t)$ is strictly increasing on $[0,1]$.

Question

If $f(t)$ is continuous for $t$ $\in [0,1]$, $f' > 0$ and $f''(t) > 0$ for $t \in (0,1)$, do we have that $f'(t)$ is strictly increasing on $[0,1]$?

Here is what i think:

Since $f(t)$ is continuous for $t$ $\in [0,1]$, and $f' > 0$ for $t \in (0,1)$, it follows that $f$ is strictly increasing on $[0,1]$.

I want to use this method again to prove $f'(t)$ is strictly increasing on $[0,1]$. But I don't know if $f'(t)$ it is continuous on $[0,1]$.

Is there any other way to prove the statement? or "$f'(t)$ is strictly increasing on $[0,1]$" is just a wrong statement?

thanks for your help~

Answer 1

Let $t,s \in [0,1]$ and $s<t$. By the mean value theorem there is $c \in (s,t) \subseteq (0,1)$ such that

$f'(t)-f'(s)=(t-s)f''(c) >0.$ Hence $f'(s)<f'(t).$

Answer 2

First, if a function is differentiable, then it is continuous!

if you talk about $f'$ and $f''$, we can assume $f$ is at least differentiable twice. So you don’t have to precise that it is continuous, and you know $f'$ is continuous too.

Moreover, the way you enunciate your theorem is wrong, it should be :

If a function $g$ is differentiable on any interval $I$, and on that interval, $g'>0$, then $g$ is strictly increasing on that interval.

(notice differentiable and not continuous)

By applying that, you get that since $f''>0$ on $[0,1]$, then $f'$ increases on $[0,1]$.

By the way, this proves that $f$ is strictly convex on that interval.