If a continuous function has one relative extremum, then it is actually an absolute extremum

Question

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 $I$.

Answer 1

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.

Answer 2

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

Answer 3

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.