TITLE: Does there exist some strictly positive function such that both the derivative and second derivative are strictly negative?
QUESTION [5 upvotes]: If $f(x) > 0$ over the reals is it possible to have $f'(x) < 0$ and $f''(x)< 0 $ over the reals? Assuming $f$ can be differentiated twice.

REPLY [5 votes]: It can be easily seen that such a function does not exist. Since $f''$ is negative the derivative $f'$ is strictly decreasing. And further $f'$ is negative so it follows that it either tends to a negative limit $L$ or to $-\infty$ as $x\to\infty$. Next the original function $f$ is strictly decreasing and bounded below by $0$ therefore it tends to a finite limit $M$. Now there is an obvious contradiction if we take limit as $x\to\infty$ in the following equation (which is a consequence of  Mean Value Theorem) $$f(x+1)-f(x)=f'(\xi)\text{ for some } \xi\in(x, x+1)$$ because LHS tends to $M-M=0$ and RHS tends to a negative number $L$ or to $-\infty$.