If $f^{\prime \prime} (x) \geq c > 0$ on $[0, \infty )$, then $f$ is unbounded Assume that there is a constant c > 0 such that $f''(x) \geq c$ for all $x \geq 0$. Show that $f(x)$ is not
bounded above on $[0,\infty)$.
My idea is very simple. Since the second derivative function is always positive, that means the first derivative function is keeping increasing. This means our $f(x)$ must be increasing and increasing (steeper and steeper). Thus, $f(x)$ cannot be bounded.
But how does one give a rigorous proof?
 A: You have some correct ideas, but you need to be careful. Consider $f(x) = e^{-x}$. The second derivative $f''(x) = e^{-x}$ is always positive, and the first derivative $f'(x) = -e^{-x}$ is increasing. However, $f(x) = e^{-x}$ is not "increasing and increasing (steeper and steeper)", and in fact, $f(x) = e^{-x}$ is bounded above by $1$ on $[0,\infty)$. So you will need to use more of the given information in the problem to prove that $f(x)$ cannot be bounded above on $[0,\infty)$.
To prove the theorem rigorously, you can use the fundamental theorem of calculus to obtain a lower bound on the derivative of the function: $$f'(y) = f'(0)+\displaystyle\int_{0}^{y}f''(z)\,dz \ge f'(0)+\displaystyle\int_{0}^{y}c\,dz = cy+f'(0).$$ Then, use the fundamental theorem of calculus again to obtain a lower bound on the function itself: $$f(x) = f(0)+\int_{0}^{x}f'(y)\,dy \ge f(0)+\int_{0}^{x}(cy+f'(0))\,dy \ge \ \ldots.$$ Finally, use this to show that for any $M > 0$, there exists an $x \in [0,\infty)$ such that $f(x) > M$.
