nLab extreme value theorem

Idea

The classical extreme value theorem states that a continuous function on the bounded closed interval $[0,1]$ with values in the real numbers does attain its maximum and its minimum (and hence in particular is a bounded function). This is a special case in analysis of the more general statement in topology that continuous images of compact spaces are compact.

Although the Extreme Value Theorem (EVT) is often stated as a theorem about continuous maps, it's really about semicontinuous maps.

Recall that the lower real numbers are more general than the real numbers in just the way needed to guarantee that any inhabited set of lower reals has a supremum (by including $\infty$, and in constructive mathematics by additionally relaxing the requirements of locatedness). Consequently, the range of any lower-real-valued function with an inhabited domain must have a supremum.

The Extreme Value Theorem states that such a range must also have an infimum when certain conditions are met. Specifically, we move to the realm of topology, where the natural lower-real-valued functions are the lower semicontinuous ones. So long as the domain of a lower semicontinuous map is compact (and inhabited), says the EVT, the range has an infimum, and furthermore this infimum is attained, becoming a minimum.

Statements

We first discuss the statement

• for continuous functions

and then the general case

• for semicontinuous function.

For continuous functions

Proposition

(extreme value theorem)

Let $C$ be a compact topological space, and let

$$f \;\colon\; C \longrightarrow \mathbb{R}$$

be a continuous function to the real numbers equipped with their Euclidean metric topology.

Then $f$ attains its maximum and its minimum, i.e. there exist $x_{min}, x_{max} \in C$ such that for all $x \in C$ it is true that

$$f(x_{min}) \leq f(x) \leq f(x_{max}) \,.$$

Proof

Since continuous images of compact spaces are compact the image $f([a,b]) \subset \mathbb{R}$ is a compact subspace.

Suppose this image did not contain its maximum. Then $\{(-\infty,x)\}_{x \in f([a,b])}$ were an open cover of the image, and hence, by its compactness, there would be a finite subcover, hence a finite set $(x_1 \lt x_2 \lt \cdots \lt x_n)$ of points $x_i \in f([a,b])$, such that the union of the $(-\infty,x_i)$ and hence the single set $(-\infty, x_n)$ alone would cover the image. This were in contradiction to the assumption that $x_n \in f([a,b])$ and hence we have a proof by contradiction.

Similarly for the minimum.

Example

(classical extreme value theorem)

Let

$$f \;\colon\; [a,b] \longrightarrow \mathbb{R}$$

be a continuous function from a bounded closed interval ($a \lt b \in \mathbb{R}$) regarded as a topological subspace of real numbers to the real numbers, with the latter regarded with their Euclidean metric topology.

Then $f$ attains its minimum $f(x_{min})$ and maximum $f(x_{max})$ and the image of $f$ is the closed interval

$$f([a,b]) = [f(x_{min}), f(x_{max})] \,.$$

Proof

Since continuous images of compact spaces are compact the image $f([a,b]) \subset \mathbb{R}$ is a compact subspace.

By the Heine-Borel theorem the image, being compact, is a bounded closed subset, hence a finite union of bounded closed intervals and singleton subsets. By continuity of $f$, this union cannot be disjoint, for if $f([a,b])$ were a disjoint union $C_1 \sqcup C_2$ of closed inhabited subsets, then also the pre-image were a disjoint union of closed inhabited subsets $f^{-1}(C_1) \sqcup f^{-1}(C_2)$, contradicting the fact that the pre-image of the image is the connected interval $[a,b]$.

For semicontinuous functions

If $I$ is a compact space, and if $f$ is an upper semicontinuous function from $I$ to the upper real numbers $[{-\infty,\infty}[$, then the range of $f$ is not only bounded above and not only has a finite supremum, but it actually has a maximum value (unless $I$ is empty).

Similarly, and consequently (by replacing $f$ with $-f$), if $f$ is lower semicontinuous to the lower real numbers $]{-\infty,\infty}]$, then $\ran f$ is bounded below by a finite infimum which is its minimum value. Consequently, if $f$ is continuous to the real numbers $]{-\infty,\infty}[$, then $\ran f$ is bounded and has both a maximum and a minimum.)

In constructive mathematics, this statement is correct in locale theory (in a sense that we should explain here!), but if we are speaking of pointwise functions defined on a real interval, then it fails in general. However, we have approximate versions; in particular, it is constructive that $\ran f$ is bounded and has (if real-valued and pointwise-continuous) a real infimum and supremum. However, the EVT interacts subtly with the fan theorem; the fan theorem is equivalent to the statement that whenever a pointwise-continuous real-valued function $f$ on $[0,1]$ satisfies $m \lt f \lt M$, then $m \lt \inf f$ and $\sup f \lt M$, which (once the existence of $\inf f$ and $\sup f$ is granted) may be viewed as a contrapositive form of the EVT. (In fact, the fan theorem is separately equivalent to the statement that every continuous function on $[0,1]$ is uniformly continuous and to the statement that every uniformly continuous function on $[0,1]$ has this contrapositive property.) But without the fan theorem, even this contrapositive form cannot be proved (in marked contrast to similar theorems such as the intermediate value theorem and the mean value theorem), and in fact it is false in Russian constructivism (give a counterexample?). (Does the full semicontinuous version follow from the fan theorem?)

Related concepts

• continuous images of compact spaces are compact

References

• Wikipedia, Extreme value theorem

• Karin U. Katz, Mikhail G. Katz, Taras Kudryk, Toward a clarity of the extreme value theorem (arXiv:1404.5658)