Bounded function whose second derivative is non negative Is it true that a twice continuously differentiable bounded function from R to R with non negative second derivative for all x in R is necessarily a constant? If not give a counter example. The given function is convex throughout R since it’s second derivative is non negative. And its boundedness geometrically implies it is a constant. How should I rigorously prove the result? Or is my geometric intuition wrong? Help me please. 

If $f''\ge 0$ then $f'$ is increasing. Because if $x<y$ and $f'(x)>f'(y)$ then by the MVT there exists $z\in (x,y)$ such that $0>\frac {f'(y)-f'(x)}{y-x}=(f')'(z)=f''(z)\ge 0,$ which is absurd.
If $f$ is differentiable and not constant then $f'$ is not everywhere $0.$ For if $f(x)\ne f(y)$ then by the MVT there exists $z$ between $x$ and $y$ with $0\ne \frac {f(y)-f(x)}{y-x}=f'(z).$
Therefore:
If $f''\ge 0$ and $f'(z)>0$ then for $x>z$ we have $$f(x)=f(z)+\int_z^xf'(t)dt\ge f(z)+\int_z^xf'(z)dt=f(z)+(x-z)f'(z)$$ which ( for a fixed $z$) is unbounded above as $x\to \infty.$
If $f''\ge 0$ and $f'(z)<0$ then for $x<z$ we have $$f(x)=f(z)+\int_z^x f'(t)dt=f(z)+\int_x^z(-f'(t))dt\ge f(z)+\int_x^z(-f'(z))dt=$$ $$= f(z)+(z-x)(-f'(z))$$ which (for a fixed $z$) is unbounded above as $x\to -\infty.$
