Let $f$ be a twice-differentiable function on $\mathbb{R}$ such that $f''$ is continuous. Prove that $f(x) f''(x) < 0$ cannot hold for all $x.$ Let $f$ be a twice-differentiable function on $\mathbb{R}$ such that $f''$ is continuous. Prove that
$f(x) f''(x) < 0$ cannot hold for all $x.$
I have been able to think of specific examples of $f(x)$ in which $f(x)f''(x) <0$ does not hold, but I have not been able to come up with specific values of $x$ for which $f(x)f''(x)<0$ does not hold.
Any help is greatly appreciated!

Sketch of a proof. Details remain to be filled. The devil is in the details of course, but the idea should work.

*

*Since f is continuous, it must be $f>0$ or $f<0$ everywhere (why?). wlog consider $f>0$. The hypothesis then implies $f''<0$ everywhere.


*Now $f$ cannot be constant, therefore there must be $x_0$ where $f'(x_0) \ne 0$


*Consider the case $f'(x_0)<0$. Prove that the function must remain below the straght line $y=f(x_0)+f'(x_0)(x-x_0)$ for $x>x_0$. This implies that at a certain point $f(x)$ must change sign and become negative. Contradiction.


*If $f'(x_0)>0$ make a similar argument for $x<x_0$
