TheoremProved

Extreme value theorem

Statement

If $f:[a,b]\to\mathbb{R}$ is continuous on a closed bounded interval, then $f$ attains a maximum value and a minimum value on $[a,b]$ : there exist $x_1,x_2\in[a,b]$ with $f(x_1)\le f(x)\le f(x_2)$ for all $x\in[a,b]$ .

Why is it true?

A continuous graph drawn without lifting the pen, over a finite closed interval, must reach an actual highest point and an actual lowest point — it cannot merely approach a height without ever getting there, the way $f(x)=1/x$ approaches (but never reaches) a bound as $x$ nears an open endpoint.

Proof sketch

By the Bolzano–Weierstrass theorem and continuity, $f([a,b])$ is bounded: otherwise a sequence $x_n$ with $|f(x_n)|\to\infty$ would have a convergent subsequence $x_{n_k}\to x^*\in[a,b]$ , and continuity would force $f(x_{n_k})\to f(x^*)$ , a finite number, contradicting $|f(x_n)|\to\infty$ . Let $M=\sup f([a,b])$ (finite by boundedness); choosing $x_n$ with $f(x_n)\to M$ , a convergent subsequence $x_{n_k}\to x_2$ gives $f(x_2)=M$ by continuity, so the supremum is attained. The minimum follows by applying the same argument to $-f$ .

Topics that use this theorem

Continuous Functions

Related theorems

Intermediate value theorem Bolzano–Weierstrass theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Walter Rudin (1976). Principles of Mathematical Analysis
David M. Bressoud (2007). A Radical Approach to Real Analysis