Zeros of $ f''$ Let $ f : \mathbb{R} \to \mathbb{R} $ be a $C^2$ function such that 
$$ \lim_{x \to \pm \infty}{f(x)} = 0 $$
Prove that $f''$ has at least two zeros.
Assume $f$ is not a constant. Than $f$ must have a stationary point, $a$. Assume it's a max point. Than $f''$ must be negative in a neighborhood of that point. Now let's prove that $f''$ has at least one zero in $[-\infty, a ]$... From this my proof gets really messy...
 A: We can assume that $f$ is not constant, and without loss of generality we can assume that the maximum of $f$ is strictly positive (otherwise consider $-f$).
Let $a$ be a point where $f$ attains its maximum. The idea is to show that $f''$ has a zero in both open intervals $(-\infty, a)$ and $(a, \infty)$. 


*

*From $f(a) > 0$ and  $\lim_{x \to \infty}{f(x)} = 0$ it follows that there is a $x_0 > a$ with $f(x_0) < a$.

*Applying the mean-value theorem to $f$ on $[a, x_0]$ shows that there is a $x_1 \in (a, x_0)$  with $f'(x_1) < 0$.

*$f'(a) = 0$. Applying the mean-value theorem to $f'$ on $[a, x_1]$ shows that there is a $x_2 \in (a, x_1)$  with $f''(x_2) < 0$.

*Now assume that $f''$ has no zeros in $[x_2,  \infty)$. Then $f''$ is negative and $f'$ is decreasing on $[x_2,  \infty)$. In particular,
$$
 f(x) \le f(x_1) + f'(x_1) (x-x_1) 
$$
for $x >  x_1$, contradicting the assumption that $\lim_{x \to \infty}{f(x)} = 0$. 
This shows that $f''$ has necessarily a zero in $(a, \infty)$. The same argument can be used to demonstrate that $f''$ has a zero in  $(-\infty , a)$.
Remark: Since $f''$ has the mean-value property, it suffices to require that $f$ is twice differentiable. The second derivative need not be continuous for the above conclusions.
