The extreme value theorem is the reason optimization problems on closed intervals always have answers. Its proof uses Bolzano–Weierstrass to catch a maximizing point, and every one of its hypotheses matters. It pairs with the intermediate value theorem: together they say a continuous function maps $[a, b]$ onto a closed interval $[m, M]$.
Idea clave
Extreme value theorem. A continuous function on a closed, bounded interval $[a, b]$ is bounded and attains a maximum and a minimum: there are $c, d \in [a, b]$ with $f(d) \le f(x) \le f(c)$ for all $x \in [a, b]$.
Por qué funciona
Bounded. If not, choose $x_n$ with $|f(x_n)| > n$. By Bolzano–Weierstrass some subsequence $x_{n_k} \to c$, and $c \in [a, b]$ because the interval is closed. By continuity $f(x_{n_k}) \to f(c)$, a finite number, contradicting $|f(x_{n_k})| > n_k$. Attained. Let $s = \sup f([a, b])$, finite now. Choose $x_n$ with $f(x_n) > s - \frac{1}{n}$. A subsequence converges to some $c \in [a, b]$, and $f(c) = \lim f(x_{n_k}) = s$. $\blacksquare$ Closed is used to keep $c$ in the interval, bounded to apply Bolzano–Weierstrass, and continuity to pass to the limit.
Worked examples
Ejemplo resuelto: A closed interval
Find the maximum of $f(x) = x^{3} - 12x$ on $[-1, 4]$.
- $f'(x) = 3x^{2} - 12 = 0$ at $x = \pm 2$; only $x = 2$ is in the interval.
- Values: $f(-1) = 11$, $f(2) = -16$, $f(4) = 16$.
Respuesta: The maximum is $16$, at $x = 4$.
Ejemplo resuelto: An open end
Find $\sup f$ on $(1, 3)$ for $f(x) = (x - 1)^{2}$, and decide whether it is attained.
- $f$ increases on $(1, 3)$ toward $f(3) = 4$.
- Every value is less than $4$, but values come as close to $4$ as you like.
Respuesta: $\sup f = 4$, not attained: no maximum. The interval is not closed.
Ejemplo resuelto: Which hypothesis fails?
$f(x) = x - \lfloor x \rfloor$ on $[0, 2]$ takes values in $[0, 1)$ and has no maximum. Why does the EVT not apply?
- The interval is closed and bounded.
- But $f$ jumps at $x = 1$ and $x = 2$: it is not continuous.
Respuesta: Continuity fails.
Error común
Using the theorem on the wrong set. On $(0, 1)$, $f(x) = x$ has supremum $1$ and no maximum; on $[0, \infty)$, it has no maximum at all; and $\frac{1}{x}$ (with $f(0) = 0$) on $[-1, 1]$ is not even bounded. When a problem asks whether a maximum exists, check all three: continuous, closed, bounded. When it does exist, find it among the critical points inside and the two ends.