Extreme value theorem

In calculus, the extreme value theorem states that if a function f(x) is continuous in the closed interval ^* then f(x) must attain its maximum and minimum value, each at least once.

That is, there exist numbers c, and d within the interval b such that for every value of x in b,

f(c) \le f(x) \le f(d).

A weaker version of this theorem is the boundedness theorem which states that a function f(x) continuous in the closed interval is bounded on the interval. That is, there exist numbers l and L such that for every value of x in [a, b,

l \le f(x) \le L.

The extreme value theorem enriches the boundedness theorem by saying that not only is the function bounded, but it also attains its least upper bound as its maximum and its greatest lower bound as its minimum.

The extreme value theorem is used to prove Rolle's theorem.

Proving the Theorems

We look at the proof for the maximum, the minimum is very similar. 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:

Prove the boundedness theorem.
Find a sequence so that its image converges to the supremum of f.
Show that there exists a subsequence that converges to a point in the domain (mathematics).
Use continuity to show that the image of the sequence converges to the supremum.

Proof of the boundedness theorem

Suppose f is not bounded. Then, by the Archimedean property of the real numbers, for every m, there exists an x in b such that f(x) > m. In particular, for every k in N, there exists an $x_k$ such that f( $x_k$ ) > k. This defines a sequence of $x_k$ s. Because b is bounded, by the Bolzano-Weierstrass theorem, there exists a convergent subsequence { $x_{n_k}$ } of { $x_k$ }. As b is closed, { $x_{n_k}$ } converges to some x in b. Because f(x) is continuous over b, we know that f( $x_{n_k}$ ) converges to f(x). But, f( $x_{n_k}$ ) > $n_k$ > k for every k, which implies f( $x_{n_k}$ ) diverges to infinity. Contradiction. Therefore, f(x) is bounded above.

Proof of the extreme value theorem

We will now show that f(x) has a maximum in b. As, by the boundedness theorem, f is bounded from above, there exists c the least upper bound (supremum) of f(x). It is necessary to find a $x_0$ in b such that $c=f({x_0})$ . Let n be a natural number. As c is the least upper bound, $c-1/n$ is not an upper bound for f(x). Therefore, there exists $x_n$ in b so that $c-1/n$ < f( $x_n$ ). This defines a sequence { $x_n$ }. Since c is an upper bound for f(x), $c-1/n$ < f( $x_n$ ) ≤ c for all n. Therefore, {f( $x_n$ )} converges to c.

The Bolzano-Weierstrass theorem tells us that { $x_{n_k}$ } exists in { $x_n$ } such that { $x_{n_k}$ } converges to some $x_0$ and, as b is closed, $x_0$ is in b. Since f(x) is continuous over b, {f( $x_{n_k}$ )} converges to f( $x_0$ ). But, {f( $x_{n_k}$ )} is a subsequence of {f( $x_n$ )} that converges to c, so $c = f(x_0)$ . Then $x_0$ is a maximum of f(x).

Examples

The following examples show why the function domain needs to be closed and bounded.

Bounded. f(x) = x defined over $[0,\infty)$ is not bounded from above.

Closed. f(x) = x defined over [0,1) never attains its least upper bound 1.

Topological formulation

In general topology, the extreme value theorem is follows from the general fact that compactness is preserved under continuity, and the fact that a subset of the real line is compact if and only if it is both closed and bounded.

External link

A Proof for extreme value theorem at cut-the-knot

Calculus | Mathematical theorems