Extreme value theorem

A continuous function $f(x)$ on the closed interval $[a,b]$ showing the absolute max (red) and the absolute min (blue).

In calculus, the extreme value theorem states that if a real-valued function $f$ is continuous on the closed and bounded interval $[a,b]$, then $f$ must attain a maximum and a minimum, each at least once. That is, there exist numbers $c$ and $d$ in $[a,b]$ such that: $$f(c)\le f(x)\le f(d)\mathrm{\forall }x\in [a,b].$$

The extreme value theorem is more specific than the related boundedness theorem, which states merely that a continuous function $f$ on the closed interval $[a,b]$ is bounded on that interval; that is, there exist real numbers $m$ and $M$ such that: $$m\le f(x)\le M\mathrm{\forall }x\in [a,b].$$

This does not say that $M$ and $m$ are necessarily the maximum and minimum values of $f$ on the interval $[a,b],$ which is what the extreme value theorem stipulates must also be the case.

The extreme value theorem is used to prove Rolle's theorem. In a formulation due to Karl Weierstrass, this theorem states that a continuous function from a non-empty compact space to a subset of the real numbers attains a maximum and a minimum.

History

The extreme value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. Both proofs involved what is known today as the Bolzano–Weierstrass theorem.

Functions to which the theorem does not apply

The following examples show why the function domain must be closed and bounded in order for the theorem to apply. Each fails to attain a maximum on the given interval.

1. $f(x)=x$ defined over $[0,\mathrm{\infty })$ is not bounded from above.
2. $f(x)=\frac{x}{1+x}$ defined over $[0,\mathrm{\infty })$ is bounded from below but does not attain its least upper bound $1$.
3. $f(x)=\frac{1}{x}$ defined over $(0,1]$ is not bounded from above.
4. $f(x)=1-x$ defined over $(0,1]$ is bounded but never attains its least upper bound $1$.

Defining $f(0)=0$ in the last two examples shows that both theorems require continuity on $[a,b]$.

Generalization to metric and topological spaces

When moving from the real line $ℝ$ to metric spaces and general topological spaces, the appropriate generalization of a closed bounded interval is a compact set. A set $K$ is said to be compact if it has the following property: from every collection of open sets $U_{\alpha }$ such that $\bigcup U_{\alpha }\supset K$, a finite subcollection $U_{{\alpha }_1},\dots ,U_{{\alpha }_n}$can be chosen such that ${\bigcup }_{i=1}^n{U}_{{\alpha }_i}\supset K$. This is usually stated in short as "every open cover of $K$ has a finite subcover". The Heine–Borel theorem asserts that a subset of the real line is compact if and only if it is both closed and bounded. Correspondingly, a metric space has the Heine–Borel property if every closed and bounded set is also compact.

The concept of a continuous function can likewise be generalized. Given topological spaces $V,W$, a function $f:V\to W$ is said to be continuous if for every open set $U\subset W$, $f^{-1}(U)\subset V$ is also open. Given these definitions, continuous functions can be shown to preserve compactness:

In particular, if $W=ℝ$, then this theorem implies that $f(K)$ is closed and bounded for any compact set $K$, which in turn implies that $f$ attains its supremum and infimum on any (nonempty) compact set $K$. Thus, we have the following generalization of the extreme value theorem:

Slightly more generally, this is also true for an upper semicontinuous function. (see compact space#Functions and compact spaces).

Proving the theorems

We look at the proof for the upper bound and the maximum of $f$. By applying these results to the function $-f$, the existence of the lower bound and the result for the minimum of $f$ follows. Also note that everything in the proof is done within the context of the real numbers.

We first prove the boundedness theorem, which is a step in the proof of the extreme value theorem. The basic steps involved in the proof of the extreme value theorem are:

1. Prove the boundedness theorem.
2. Find a sequence so that its image converges to the supremum of $f$.
3. Show that there exists a subsequence that converges to a point in the domain.
4. Use continuity to show that the image of the subsequence converges to the supremum.

Proof from first principles

Statement      If $f(x)$ is continuous on $[a,b]$ then it attains its supremum on $[a,b]$

Extension to semi-continuous functions

If the continuity of the function f is weakened to semi-continuity, then the corresponding half of the boundedness theorem and the extreme value theorem hold and the values –∞ or +∞, respectively, from the extended real number line can be allowed as possible values.Template:Clarify

A function $f:[a,b]\to [-\mathrm{\infty },\mathrm{\infty })$ is said to be upper semi-continuous if $$\underset{y\to x}{\mathrm{lim\; sup}}f(y)\le f(x)\mathrm{\forall }x\in [a,b].$$

Applying this result to −f proves a similar result for the infimums of lower semicontinuous functions. A function $f:[a,b]\to [-\mathrm{\infty },\mathrm{\infty })$ is said to be lower semi-continuous if $$\underset{y\to x}{\mathrm{lim\; inf}}f(y)\ge f(x)\mathrm{\forall }x\in [a,b].$$

A real-valued function is upper as well as lower semi-continuous, if and only if it is continuous in the usual sense. Hence these two theorems imply the boundedness theorem and the extreme value theorem.

Further reading

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External links

• A Proof for extreme value theorem at cut-the-knot
• Extreme Value Theorem by Jacqueline Wandzura with additional contributions by Stephen Wandzura, the Wolfram Demonstrations Project.
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• Mizar system proof: http://mizar.org/version/current/html/weierstr.html#T15