Extreme value theorem

From Exampleproblems

In calculus, the extreme value theorem states that if a function f(x) is continuous in the closed interval [a,b] 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 [a, b] such that for every value of x in [a, b],

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

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

1 Proving the Theorem
- 1.1 Sketch of Proof
- 1.2 Proof
2 External link

Proving the Theorem

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

Sketch of Proof

Before proving the theorem, we outline the basic steps involved:

1. Show f(x) is bounded if it is continuous over a closed interval.

2. Find a sequence that converges to the supremum of f.

3. Use sequential compactness to show that there exists a subsequence that converges to a point in the domain.

4. Use continuity to show that the image of the sequence converges to the supremumW

Proof

Let D denote the domain of f(x). Also note that everything in the proof is done within the context of the real numbers. First, we will show that f(x) is bounded. Suppose it is not bounded. Then for every m, there exists an x in D such that f(x) > m. This follows from the Archimedian property of the real numbers. In particular, for every k we choose, there exists an $x k$ such that f( $x k$ ) > k. This defines a sequence of $x k$ s. Because D is closed and bounded, it is sequentially compact. Therefore, there exists a convergent subsequence { $x_{n_k}$ } of { $x k$ } with { $x_{n_k}$ } converging to some x in D. Because f(x) is continuous over D, 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.

We will now show that f(x) has a maximum in D. Select c to be the supremum of f(x). It is necessary to find a $x 0$ in D such that $c = f (x 0)$ . Let n be a natural number. Then $c - 1 / n$ is not an upper bound for f(x). Choose this point and label it $x n$ . 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.

Sequential compactness tells us that { $x_{n_k}$ } exists in { $x n$ } such that { $x_{n_k}$ } converges to $x 0$ in D. Since f(x) is continuous over D, {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 maximizer of f(x).