Extreme Value Theorem

A plane takes off, climbs, cruises, descends, and lands. The altitude is a continuous function of time on a closed interval. It must reach both a highest point and a lowest point somewhere during the flight. The Extreme Value Theorem says this is always true — not just for aircraft, but for any continuous function on a closed bounded interval.

Statement

Theorem (Extreme Value Theorem, EVT). If $f : [a, b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and minimum on $[a, b]$: there exist $x_{\max}, x_{\min} \in [a, b]$ such that

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

Proof

The proof has two stages: first show $f$ is bounded above, then show the supremum is actually attained.

Stage 1: $f$ is bounded above

Suppose for contradiction that $f$ is not bounded above on $[a, b]$. Then for each $n \in \mathbb{N}$ there exists $x_n \in [a, b]$ with $f(x_n) > n$. The sequence $(x_n)$ lies in the bounded interval $[a, b]$, so by the Bolzano–Weierstrass theorem it has a convergent subsequence $(x_{n_k})$ with $x_{n_k} \to c \in [a, b]$.

Since $f$ is continuous at $c$:

$$\lim_{k \to \infty} f(x_{n_k}) = f(c).$$

But $f(x_{n_k}) > n_k \to \infty$, contradicting convergence to the finite value $f(c)$. Therefore $f$ is bounded above.

By the same argument applied to $-f$, the function $f$ is also bounded below.

Stage 2: $f$ attains its supremum

Let $M \coloneqq \sup_{x \in [a,b]} f(x)$, which is finite by Stage 1. By the definition of supremum, for each $n \in \mathbb{N}$ there exists $y_n \in [a, b]$ with

$$M - \frac{1}{n} < f(y_n) \leq M.$$

By Bolzano–Weierstrass, $(y_n)$ has a convergent subsequence $y_{n_k} \to x_{\max} \in [a, b]$. Continuity gives

$$f(x_{\max}) = \lim_{k \to \infty} f(y_{n_k}) = M.$$

So $f$ attains its maximum at $x_{\max}$. The minimum follows by applying the argument to $-f$. $\square$

Why each hypothesis is necessary

All three hypotheses — continuity, closedness, and boundedness of the interval — are essential. Remove any one and the conclusion can fail.

Without continuity (discontinuous function on $[0,1]$)

Define

$$f(x) \coloneqq \begin{cases} x & x \in (0, 1] \\ 0.5 & x = 0. \end{cases}$$

Then $\sup_{x \in [0,1]} f(x) = 1$, but $f(x) = 1$ has no solution in $[0, 1]$ — the supremum is not attained.

Without closedness (open interval)

The function $f(x) = x$ on the open interval $(0, 1)$ is continuous, but $\sup f = 1$ and $\inf f = 0$ are never attained — neither endpoint belongs to the domain.

Without boundedness (unbounded interval)

The function $f(x) = x$ on $[0, \infty)$ is continuous on a closed (but unbounded) set, and it is not bounded above, so no maximum exists.

The image of $[a, b]$ is a closed interval

The EVT says $f$ attains a minimum $m$ and maximum $M$. Combined with the Intermediate Value Theorem — which guarantees $f$ hits every value strictly between any two values it takes — the image of $f$ on $[a, b]$ is exactly the closed interval $[m, M]$.

Summary

• Extreme Value Theorem: a continuous function on a closed bounded interval $[a, b]$ attains both a maximum value and a minimum value.
• Proof idea: near-supremum points form a sequence in $[a, b]$; Bolzano–Weierstrass extracts a convergent subsequence; continuity forces its limit to equal the supremum.
• All three hypotheses are sharp: dropping continuity, closedness, or boundedness each gives a counterexample where no extremum is attained.
• Corollary: the image $f([a, b])$ is the closed interval $[\min f,\, \max f]$, by the EVT and the Intermediate Value Theorem together.