(c.f. January 2007 #3a, June 2010 #2a) Prove Theorem 3.2 for a function \(f:K\subset\mathbb R\to\mathbb R\). (The proof is nearly unchanged when \(n\) or \(m\) is greater than 1.)
Solution
1. Reduce to the one-dimensional extreme value theorem.Proof: let \(f: K \to \mathbb{R}\) be continuous with \(K \subseteq \mathbb{R}\) compact.We show \(f\) attains a maximum; the minimum follows by applying the result to \(-f\).
2. \(f(K)\) is bounded above.Proof: suppose not; then for each \(n\) there is \(x_n \in K\) with \(f(x_n) \ge n\).Since \(K\) is compact (sequentially compact in \(\mathbb{R}\)), a subsequence \(x_{n_j} \to x \in K\).By continuity \(f(x_{n_j}) \to f(x)\), but \(f(x_{n_j}) \ge n_j \to \infty\), contradiction.
3. The supremum is attained.Proof: let \(M = \sup_{x \in K} f(x) < \infty\); choose \(x_n \in K\) with \(f(x_n) \to M\) (e.g. \(f(x_n) \ge M - 1/n\)). A subsequence \(x_{n_j} \to x^* \in K\) (compactness), and by continuity \(f(x^*) = \lim_j f(x_{n_j}) = M\).Hence \(f\) attains its maximum at \(x^* \in K\).
4. Q.E.D. Proof: \(f\) attains both its maximum and (applied to \(-f\)) its minimum on \(K\).
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Dependencies
T-RA-WORKSHOP-D4-3-2Continuous images of compact sets and the extreme value theoremuses