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Here's my question and suggestion to solve it. Please let me know if I'm wrong.

Let $f$ be a continuous function in $\Bbb R$, where $f(a), f(b)$ are two Maxima and Minima points. $f(a), f(b)$ are the only Maxima and Minima Points of $f$ in $\Bbb R$.

Assuming $$f(a)<f(b), a<b$$ Prove that $f(a)$ is a local minimum point.

Solution

Since $f(a),f(b)$ are Maxima and Minima points, there is some $\varepsilon$ such that in the neighborhood of $a,b$, such that $$a,b\in[a-\varepsilon,b+\varepsilon]$$.

Since the function is continuous, according the Extreme Value Theroem, the function receives minimum and maximum in $[a-\varepsilon,b+\varepsilon]$. Since $f(a)<f(b)$, we conclude that $f(a)$ is a local minimum.

Am I wrong?

Thanks,

Alan

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  • $\begingroup$ I don't see how you have proved that f(a) is a local minimum. Why can't f(a) and f(b) both be local maxima? $\endgroup$ – David Hill Sep 15 '15 at 1:07
  • $\begingroup$ @DavidHill You are right, I don't really know. I assume because there are only two maxima and Minima In the whole interval. $\endgroup$ – Alan Sep 15 '15 at 3:29
  • $\begingroup$ Try to draw a continuous function with two local maxima. Can you do it without also having a local minimum between them? Also, a rigorous proof would be easier if $f$ is differentiable. Is this assumed? $\endgroup$ – David Hill Sep 15 '15 at 15:37
  • $\begingroup$ @DavidHill thank you. It is not assumed that $f$ is differentiable, but isn't maxima and minima points imply differentiality? $\endgroup$ – Alan Sep 15 '15 at 15:44
  • $\begingroup$ @DavidHill also, the question says only two maxima and minima in whole $\Bbb R$... I don't understand how a function have two maxima and no minima in whole $\Bbb R$ $\endgroup$ – Alan Sep 15 '15 at 15:46
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Assume the $f(a)$ and $f(b)$ are both local maxima. By the extreme value theorem, $f$ attains both a maximum and minimum value on $[a,b]$. Since $f(a)<f(b)$, $f(b)$ is not the absolute minimum on $[a,b]$. But, we have assumed that $f(a)$ is a local maximum, so there exists $\epsilon>0$ such that $f(x)<f(a)$ for all $a<x<a+\epsilon$. Hence, $f(a)$ is not the absolute minimum either. It follows that the absolute minimum occurs in $(a,b)$, which must be a local minimum of $f$ on $\mathbb{R}$. This contradicts the assumption that $f(a)$ and $f(b)$ are the only local extrema.

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  • $\begingroup$ Thanks. I will try with your answer. $\endgroup$ – Alan Sep 15 '15 at 16:33

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