Extreme Value Theorem on Compact Sets¶

Theorem¶

Let $E \subseteq \mathbb{R}^n\text{,}$ $f \colon E \to \mathbb{R}$ be a function, and $X \subseteq E$ be a nonempty compact set. If $f$ is continuous on $X\text{,}$ then $f$ attains both its maximum and minimum on $X\text{.}$

Interpretation¶

A continuous function on a compact set cannot sneak off to infinity or skip its highest and lowest values: it attains both a maximum and a minimum on that set.

A disc representing a compact set X crossed by parallel level lines of f; the maximum M and the minimum m are attained at two points on its edge, and f maps X onto the interval from m to M.

The level sets of $f$ are parallel lines, so $f$ increases steadily in one direction across the disc $X$ . The maximum $M$ is attained where the disc reaches furthest in that direction and the minimum $m$ on the opposite side. The image $f(X)$ is the closed interval $[m,M]$ : along the segment from the minimum point to the maximum point, $f$ takes every value in between by the intermediate value theorem.

Each hypothesis is needed. On the set $(0,1)$ , which is not closed, $f(x)=x$ has neither a maximum nor a minimum; on the unbounded set $[0,\infty)$ it has no maximum. Without continuity, $f$ can approach a value without reaching it: on $[0,1]$ , the function with $f(x)=x$ for $x<1$ and $f(1)=0$ has supremum $1$ but no maximum.

The proof combines two facts: the continuous image of a compact set is compact, and a nonempty compact subset of $\mathbb{R}$ contains its supremum and infimum. For $X=[a,b]$ this is the extreme value theorem of Math I.

The theorem guarantees existence only. The extreme points need not be unique or interior; as in the figure, they may lie on the boundary, where setting derivatives equal to zero does not find them. The theorem returns in Math IV as the existence argument for a global maximum and minimum on a compact feasible set.

Key Results¶

If the domain is not compact, the conclusion may fail. For example, $f(x)=1/x$ on $(0,1)$ is continuous but has no maximum (it blows up as $x\to0^+$ $0^{+}$ ) and no minimum (its values decrease towards $1$ as $x\to1^-\text{,}$ but $1$ is never attained, since $1\notin(0,1)$ $(0, 1)$ ).
This theorem is used to prove existence of solutions in optimization problems.
Together with the intermediate value theorem, it gives a complete picture of the range of a continuous real-valued function on a connected compact set: the range is the closed interval $[m,M]$ between the minimum $m$ and the maximum $M$ . (In $\mathbb{R}$ , a connected compact set is itself a closed bounded interval.)

Examples¶

The function $f(x,y)=x+y$ on the closed unit disc $\{(x,y)\in\mathbb{R}^2\colon x^2+y^2\leq 1\}$ attains maximum $\sqrt{2}$ at $(1/\sqrt{2},1/\sqrt{2})$ and minimum $-\sqrt{2}$ at $(-1/\sqrt{2},-1/\sqrt{2})$ .

Prerequisites¶

Sources¶

Math III: Analysis · Lecture 9

Symbols: element of, infinity, real numbers, square root, subset

Last updated: 20 October 2026 03:02