3

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...

José Carlos Santos
  • 427,504
astrobarrel
  • 217
  • 1
    What have you tried? – Dayton Apr 19 '19 at 08:30
  • 1
    Where are you stuck ? –  Apr 19 '19 at 08:30
  • 2
    Hint: either $f$ is constant, or it is not. In the latter case, it must have a stationary point. –  Apr 19 '19 at 08:32
  • I actually think I did it, but in a very inelegant way. I was wondering if there was a more sofisticated idea that could help to solve this problem. – astrobarrel Apr 19 '19 at 08:32
  • @astrobarrel: so we are deemed to guess what you did ? –  Apr 19 '19 at 08:33
  • I just added some of what I did – astrobarrel Apr 19 '19 at 08:45
  • @astrobarrel If the nonconstant $f$ has three stationary points, you're done. So you can reduce to at most two stationary points $a$ and $b$, with $a\le b$ (note that one stationary point surely exists). Study what happens over $(-\infty,a]$ and $[b,\infty)$. – egreg Apr 19 '19 at 10:48

1 Answers1

2

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.

Martin R
  • 113,040