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Let $f:\mathbb{R} \rightarrow\mathbb{R}$ be twice differentiable and bounded, and suppose $f$ gets minimum at $x_0$. Prove there exists $c\in\mathbb{R}$ such that $f^\ {''}(c)=0$.

$f$ is bounded $\Rightarrow \exists M>0 \ s.t \ |f(x)|\leq M, \forall x\in\mathbb{R}$

So by MVT I got that the derivative is positive for all $x>x_0$ and negative for all $x<x_0$, and from that I concluded that $f(x)\rightarrow M$ when $x\rightarrow\pm\infty$.

Not sure how to continue from here.

Any help appreciated.

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  • $\begingroup$ "So by MVT I got that the derivative is positive for all $x>x_0$ and negative for all $x<x_0$" - that is wrong. $f(x) = cos(x), x_0 = \pi$ satisfies given conditions. $\endgroup$ – Abstraction Mar 2 '17 at 13:04
  • $\begingroup$ @Abstraction But, of course, if the derivative conditions are not satisfied, then there is a point where $f''(c)=0$. So, perhaps the OP is eliminating that case. $\endgroup$ – Michael Burr Mar 2 '17 at 13:13
  • $\begingroup$ Hint: Are the derivatives increasing or decreasing for $x>x_0$ (or $x<x_0$)? $\endgroup$ – Michael Burr Mar 2 '17 at 13:15
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We know that $f'(x_0) = 0$ and $f''(x_0) \geq 0$, since there's a minimum. Suppose for contradiction that $f''$ is non-zero everywhere. Then, $f''(x_0) > 0$. Now, we may state that $f'(x) > 0$ when $x > x_0$.

Pick an $x_1 > x_0$; note that $f'(x_1) > 0$. Now, select an $R > 0$ large enough so that $$ M - f(x_1) < f'(x_1)(R - x_1) $$ We note that $f$ is bounded above by $M$ and increasing on $[x_1,R]$. However, we have $$ |f(R) - f(x_1)| = f(R) - f(x_1) \leq M - f(x_1) < f'(x_1)(R - x_1) $$ by the mean value theorem, there exists an $x_2 \in [x_1,R]$ such that $f'(x_2) = \frac{f(R) - f(x_1)}{R - x_1} < f'(x_1)$. By the MVT again, there exists an $x_3 \in [x_1,x_2]$ such that $f''(x_3) < 0$ (recall: $x_0 < x_1 < x_3 < x_2 < R$). This contradicts our supposition.

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  • $\begingroup$ How do you get $f'(x) >0$ for $x >x_0$ without knowing if $f''$ is continuous at $x_0$? $\endgroup$ – Stefano Mar 2 '17 at 13:20
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    $\begingroup$ $f''$ might not be continuous, but it necessarily satisfies the intermediate value property (see "Darboux's theorem"). If $f''(x_0) > 0$ and $f'' \neq 0$ everywhere, then $f'' > 0$ everywhere. $\endgroup$ – Omnomnomnom Mar 2 '17 at 13:22
  • $\begingroup$ Could you please explain how did you $M−f(x_1)<f^{'}(x_1)(R−x_1)$? $\endgroup$ – Itay4 Mar 2 '17 at 13:24
  • $\begingroup$ Do you understand the sentence "Now, select an $R > 0$ large enough so that..."? $\endgroup$ – Omnomnomnom Mar 2 '17 at 13:25
  • $\begingroup$ I'm afraid not.. $\endgroup$ – Itay4 Mar 2 '17 at 13:27

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