Bounded convex function on $[0,\infty)$ interval It's a modification of a question regarding a proof for a bounded convex function on $\mathbb{R}$ to be constant. We have $f:[0,\infty)\longrightarrow\mathbb{R}$ that is twice differentiable, bounded and convex. One must show $f'(x)\leq 0 \ \ \forall_{x\in[0,\infty)}$
I'm trying to think out of the box, examine the definition, but nothing comes to my mind. It's rather simple, I'm sure of it. How do I start?

Hint: Note that for all $x,y$, there is $t\in [0,1]$ such that $$f'(x)-f'(y)=f''(tx+(1-t)y)(x-y). \tag{1}$$
Conclude from $(1)$ that $f'$ is an non decreasing function. Now assume that there is $z$ such that $f'(z)>0$. How do you get a contradiction?
