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The following statement makes sense intuitively, but is there a way to prove it mathematically? (This is something we make use of in applied optimization in calculus.)

If $f$ is continuous on an interval $I$ and $x_0$ is the only relative (local) extremum, then $x_0$ is actually an absolute (global) extremum on $I$.

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  • $\begingroup$ $(x+1)^2(x-1)^2$ on $\mathbb{R}$? This idea could even be adapted to a finite open interval. If you demand a closed, bounded interval, most people agree that endpoints can be defined as local extrema. $\endgroup$
    – user123641
    Aug 9, 2017 at 1:49

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Not sure if this is what you mean, but I'll give it a go.

If you consider the extrema to be the minima, you can say that an absolute minimum is always a relative minimum (because if it's not even a relative minimum, how can it be an absolute minimum?). That means that only relative minima are candidates for the absolute minimum. We could write: $$a = \min{(r_1,\ r_2,\ r_3,\ \ldots)},$$ where $a$ is the absolute minimum and $r_i$ is the $i$'th relative minimum. Hence, if we have only one minimum, we have: $$a = \min{(r_1)},$$ so that the absolute minimum automatically equals the relative minimum. The case with maxima instead of minima is analogous.

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  • $\begingroup$ It is possible for an absolute minimum to not be a relative minimum. For example, an endpoint of a closed interval can be an absolute extremum, but it is not considered as a local extremum. (x = -1 is the point of absolute minimum of f(x) = 2x - 3x^(2/3) on the interval [-1, 3], but it is NOT a point of local minimum!) $\endgroup$
    – Leponzo
    Nov 25, 2014 at 13:27
  • $\begingroup$ Why is it not a point of local minimum? It is the minimal value in the neighbourhood of $x=-1$... Just a matter of definition if you ask me. $\endgroup$
    – SPK.z
    Nov 27, 2014 at 14:17
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suppose I is closed. if $x_0$ is a local maximum, then if it is not an absolute maximum $\exists x_1. f(x_1) \gt f(x_0)$. if $x_0 \ne \sup\{x | x \in I\}$ we may assume w.l.o.g $x_1 \gt x_0$

since $I'=[x_0,x_1]$ is compact $f$ attains a minimum value on $I'$, say at $x'$, contradicting the assumption that the extremum at $x_0$ is unique.

other cases can be dealt with by slight modification of the same argument

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  • $\begingroup$ This is wrong: x' could be x_1, in which case it would not be an interior global extremum and hence would not necessarily be a local extremum. $\endgroup$ Aug 9, 2017 at 1:12
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Suppose $x_0$ is a local maximum. Assume $x_0$ is not an absolute maximum. Then $\exists x_1 \in I$ with $f(x_1) > f(x_1)$. Let $I'$ be the closed interval from $x_0$ to $x_1$. Let $x'$ be the global minimum of $f$ on $I'$.

Then we can choose $x' \ne x_0$, as $x_0$ is a local maximum for $f$, so there is a small open interval $V$ about $x_0$ in $I$ with $f(v) \le f(x_0)$ for $v \in V$.

Also, clearly $x' \ne x_1$.

So, $x'$ is an interior global minimum point for $f$ on $I'$, so $x'$ is a local minimum for $f$ on $I'$ and hence for $f$ on $I$, contradicting the status of $x_0$ as the unique local extremum for $f$ on $I$.

The case for $x_0$ a local minimum is similar.

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