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Consider $f:\, \mathbb{R} \rightarrow \mathbb{R}$ be a univariate, real-valued function with continuous derivative. Show that if $f$ has a local minimizer that is not global minimizer, then f must have another critical point?

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  • $\begingroup$ Just draw a picture and keep in mind that positive derivative means increasing function (negative derivative means decreasing function respectively). Use proof by contradiction (or prove the reciprocal if you want) as suggested in my answer below keeping this in mind. No need for formulas! $\endgroup$ – Pedro Sep 26 '18 at 1:50
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Let $x$ be be the local minimum which is not global, there exists $u<x<v$ such that $f(u)>f(x), f(v)>f(x)$ and $f(x)$ is minimum on $f([u,v])$ since $x$ is a local minimum. Let $y$ such that $f(y)<f(x)$, suppose that $y>v$, since $f$ is continuous, $f([v,z])$ is an interval, we deduce that there exists $z\in [v,y]$ such that $f(z)=f(x)$, $f(z)-f(x)=0=f'(c)(z-x)$, where $c\in (x,z)$ implies that $f'(c)=0$. If $z<u$, $f(u)-f(z)=f'(c)(u-z)=0$ implies that $f'(c)=0$.

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  • $\begingroup$ Can you explain the last sentence? why the last implication implies $f'(c)=0$? $\endgroup$ – Saeed Sep 26 '18 at 1:16
  • $\begingroup$ because $u\neq z$. $\endgroup$ – Tsemo Aristide Sep 26 '18 at 1:18
  • $\begingroup$ Shouldn't be $f(u)-f(z)=f'(c)(u-z)$ instead of what you have written? $\endgroup$ – Saeed Sep 26 '18 at 1:21
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If $f$ has no other critical point, the derivative only vanishes at the local minimum. Since it is a local minimum, the derivative is negative to its left and positive to the right. Since the derivative is continuous and has no other zero, it always remains positive to the right and negative to the left (here we apply Bolzano's theorem). So the function always decreases to the left and increases to the right of the point, hence it is a global minimum.

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