Proving that a function $f:[0,\infty)\to[0,\infty)$ satisfying these conditions is necessarily non-decreasing I have a function $f: [0, \infty) \to [0, \infty)$ which is smooth. I also have that

*

*$f(0) = 0$

*$f'(0) > 0$

*$f''(x) \leq 0$, for all $x \in [0, \infty)$
It intuitively makes sense then that $f$ on $[0, \infty)$ must be strictly non-decreasing (since if it was ever decreasing, we would have to eventually "pull up" and contradict clause 3). I want to prove this, and I was able to derive contradictions if at some point the slope was negative. If at point $c$ in $(0, infty)$, $f'(c) < 0$, take some point $c + h$ further on in $[0, \infty)$, $h > 0$, and three cases occur:
a. $f(c + h) = f(c)$ -> got a contradiction
b. $f(c + h) > f(c)$ -> got a contradiction
But for $f(c + h) < f(c)$, this leads nowhere since it can still technically happen. I am at a complete loss on how to go forward. I was able to show that $f$ can never intersect $0$.
Edit: My main goal is to claim that $f(a) \leq f(b)$ for any $a < b$ on $[0, \infty)$
 A: You have the key idea right, that if at any point $f'(c) < 0$ then the function must remain decreasing at this rate or more because of the third assumption. Thus eventually the function must become negative and leave the given codomain.
For your three cases: it's easy to derive a contradiction for your first two cases because $f'(c) < 0$ is inconsistent with the function later increasing or repeating a value given the third assumption. However, the third case is consistent with $f'(c) < 0 $, but both are inconsistent with the nonnegative codomain by the aforementioned argument.
Here's a full proof using the Mean Value Theorem:
Assume there is a $a>0$ with $f'(a) < 0 $. Then by the MVT for any $b \geq a$ and some $\xi_1$ between $a$ and $b$ $$\frac{f'(b)-f'(a)}{b-a}=f''(\xi_1) \leq 0$$ giving $f'(b) \leq f'(a)$. Applying the MVT again, for any $b \geq a$ with some $\xi_2$ between $a$ and $b$ $$\frac{f(b)-f(a)}{b-a}=f'(\xi_2) \leq f'(a)$$ giving $$f(b) \leq f'(a)(b-a)+f(a)$$
If  $b > a - \frac{f(a)}{f'(a)}$, then we must have $f(b) < 0 $, which is outside the given codomain of the function.
Thus, the original assumption of $f'(a) < 0$ must be wrong, and we must have $f'(a) \geq 0$ for all $a\geq 0$, that is $f$ is non-decreasing.
A: $f''(x) \le 0$ means that $f$ is concave. It is also non-negative, so for $a < b < c$ we have
$$
 f(b) \ge \frac{c-b}{c-a}f(a) + \frac{b-a}{c-a}f(c) \ge \frac{c-b}{c-a}f(a) \,.
$$
Now take the limit for $c \to \infty$ to conclude that $f(b) \ge f(a)$.
The conditions $f(0) = 0$ and $f'(0)> 0$ are not needed.
A: Take the function defined at $ [0,+\infty)$ by
$$(\forall x\ge 0)\;\; f(x)=x(1-x)$$
we have
$$f(0)=0$$
$$f'(0)=1>0$$
and
$$(\forall x\ge 0)\;\; f''(x)=-2\le 0$$
