Extreme Value Theorem

The Extreme Value Theorem (EVT) states that a continuous function on a closed bounded interval attains its maximum and minimum values. This fundamental result guarantees that optimization problems have solutions and is essential in calculus, economics, and physics. The proof relies on compactness via the Heine-Borel theorem.

Statement

Theorem4.1Extreme Value Theorem

Let $f : [a, b] \to \mathbb{R}$ be continuous. Then $f$ attains its maximum and minimum on $[a, b]$ . That is, there exist $c, d \in [a, b]$ such that

$f(c) \leq f(x) \leq f(d) \quad \text{for all } x \in [a, b].$

Proof

We prove $f$ attains its maximum; the minimum case is similar (or apply the result to $-f$ ).

Step 1: $f([a, b])$ is bounded above. Since $[a, b]$ is compact and $f$ is continuous, $f([a, b])$ is compact (continuous images of compact sets are compact). By Heine-Borel, $f([a, b])$ is closed and bounded. Thus $M = \sup f([a, b])$ exists and is finite.

Step 2: $M \in f([a, b])$ . Since $M = \sup f([a, b])$ , for each $n \in \mathbb{N}$ , there exists $x_n \in [a, b]$ with $f(x_n) > M - 1/n$ (otherwise $M - 1/n$ would be a smaller upper bound). Thus $f(x_n) \to M$ .

The sequence $(x_n)$ lies in $[a, b]$ , which is compact. By Bolzano-Weierstrass, $(x_n)$ has a convergent subsequence $x_{n_k} \to d$ for some $d \in [a, b]$ (since $[a, b]$ is closed).

By continuity, $f(x_{n_k}) \to f(d)$ . But $f(x_{n_k}) \to M$ , so $f(d) = M$ . Thus the maximum is attained at $d$ .

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Applications

ExampleMaximum of a polynomial

Let $f(x) = -x^2 + 4x$ on $[0, 3]$ . By EVT, $f$ attains its maximum. By calculus, $f'(x) = -2x + 4 = 0$ at $x = 2$ , and $f(2) = 4$ . Checking endpoints: $f(0) = 0$ , $f(3) = 3$ . Thus $\max_{[0,3]} f = 4$ (attained at $x = 2$ ).

ExampleOptimization problems

In economics, to maximize profit $P(x)$ over a feasible region $[a, b]$ , the EVT guarantees an optimal solution exists. Without continuity or compactness, optimal solutions may not exist.

ExampleDistance to a compact set

If $K \subseteq \mathbb{R}$ is compact and $x \in \mathbb{R}$ , the function $d(y) = |x - y|$ is continuous on $K$ . By EVT, there exists $y_0 \in K$ minimizing $d(y)$ . Thus $\text{dist}(x, K) = \inf_{y \in K} |x - y|$ is attained.

EVT fails without compactness

ExampleFailure on open intervals

Let $f(x) = x$ on $(0, 1)$ . Then $\sup_{(0,1)} f = 1$ and $\inf_{(0,1)} f = 0$ , but neither is attained (since $0, 1 \notin (0, 1)$ ). The interval must be closed.

ExampleFailure on unbounded intervals

Let $f(x) = e^{-x}$ on $[0, \infty)$ . Then $\sup_{[0,\infty)} f = 1$ (attained at $x = 0$ ), but $\inf f = 0$ is not attained. The interval must be bounded.

ExampleFailure for discontinuous functions

Let $f(x) = 1/x$ on $(0, 1]$ . Then $f$ is continuous on $(0, 1]$ but unbounded as $x \to 0^+$ . No maximum exists. The function must be continuous on the entire closed bounded interval.

Summary

The Extreme Value Theorem guarantees extrema for continuous functions on compact sets:

$f$ attains max and min on $[a, b]$ if $f$ is continuous.
Proof uses compactness and Bolzano-Weierstrass.
Essential for optimization, distance problems, and variational calculus.
Fails without continuity or compactness.

See Heine-Borel and Compactness for the underlying theory.