Need help with a twice differentiable and bounded function. Let $f : \mathbb{R} \to \mathbb{R} $ a bounded and twice diffentiable function so that $\begin{equation} \forall x \in \mathbb{R}, f"(x) \geq 0 \end{equation}. $
My point is to prove that $f$ is constant.
So if I can prove that $\forall x \in \mathbb{R}, f'(x) = 0$ then my job is done.
So far, I only know that $f'$ is increasing but I don't know how I can use that to end the proof. 
Can anyone help me please? Thanks
 A: If there is an $a\in\mathbb R$ with $f'(a)>0$, then for all $x>a$ we have $f(x)\geq f(a)+(x-a)f'(a)$. [Proof: By the mean value theorem, we have $f(x)= f(a)+(x-a)f'(y)$ for some $y$ between $a$ and $x$. And $f'(y)\geq f'(a)$ because, as you noticed, $f'$ is increasing.] This contradicts the assumption that $f$ is bounded; specifically, $f(x)$ won't be bounded above as $x\to+\infty$.
If there is an $a$ with $f'(a)<0$, then a similar argument shows that $f(x)$ cannot be bounded above as $x\to-\infty$. 
So $f'(a)=0$ for all $a$, as required.
A: This is what I did, if anything's wrong just hit me up.
Let $a < b $ so that $f(a) < f(b)$.
By the mean-value theoreme, $c \in ]a,b[ $ exists  so that $f'(c)= \frac{f(b)-f(a)}{b-a}<0$.
Let $x \geq b$.
Since $f$ is convex, we have $f(x) \geq f'(b)(x-b)+ f(b).$
But $f'$ is increasing and $b \geq c$ thus $f'(b) \geq f'(c)$.
Finally, $f(x) \geq \frac{f(b)-f(a)}{b-a}(x-b)+f(b).$
So $\forall x \geq b, f(x) \geq \frac{f(b)-f(a)}{b-a}(x-b)+f(b).$
As $x \to +\infty$, we can see that $f$ is not bounded hence the contradiction.
As a conclusion, $f$ is constant.
