Behavior of a periodic function

Question

Can a periodic function satisfy $f''(x)f(x)>0, x\in \mathbb{R}$

My intuition says no. Any thoughts on how to approach this?

Answer 1

I will try to prove that $\forall x\in\Bbb R(f''(x)f(x)>0) \implies f$ is not periodic.

From $f''(x)f(x)>0$ we can see that neither $f$ nor $f''$ can be $0$ at any point. Because $f$ is differentiable, it is continuous, and so it can never pass $0$ and must therefore be either positive everywhere or negative everywhere. $f''$ must have the same sign as $f$ everywhere.

Without loss of generality, assume that $f''$ and $f$ are always positive. So, because $f''$ is always positive, $f'$ is strictly increasing. If $f'$ is always negative, $f$ is strictly decreasing and $f(x)=f(x+p)$ can't hold. If $f'$ is always positive, $f$ is strictly increasing and $f(x)=f(x+p)$ can't hold. If $f'$ passes $0$, $f$ has exactly one minimum at some $x_0$, and $f(x_0)=f(x_0+p)$ can't hold. So $f$ can not be periodic.