Extreme Value Theorem

Let $(𝑋, 𝑑)$ be a metric space and let $𝑓 : 𝑋 \to ℝ$ be a continuous function. Then for every compact subspace $𝐾$ of $𝑋$ the function $𝑓$ has an absolute maximum and absolute minimum on $𝐾$ , i.e. there exists $𝑎, 𝑏 \in 𝐾$ such that $𝑓 (𝑎) \leq 𝑓 (𝑥) \leq 𝑓 (𝑏)$ for all $𝑥 \in 𝐾$ . #m/thm/anal

Proof

#missing/proof

Euclidean spaces

The Extreme Value Theorem is stated as follows¹

Let $𝐷$ be a non-empty closed and bounded subset of $ℝ 𝑛$ and let $𝑓 : 𝐷 \to ℝ$ be a continuous function. Then $𝑓$ is bounded and there exist $⃗ 𝐚 \in 𝐷$ and $⃗ 𝐛 \in 𝐷$ such that $(\forall ⃗ 𝐱 \in 𝐷) [𝑓 (⃗ 𝐚) \leq 𝑓 (⃗ 𝐱) \leq 𝑓 (⃗ 𝐛)]$ . #m/thm/calculus

As a consequence of this, it is possible to determine the absolute extrema of a function on such a domain $𝐷$ by narrowing our view to the specific set of circumstances in which an absolute extrema can occur. These are

At critical points, i.e. $\nabla 𝑓 (⃗ 𝐯) = ⃗ 𝟎 *$ and $d e t (𝐻 𝑓 (⃗ 𝐯)) \geq 0$ .
At extrema along the boundary $𝜕 𝐷$ .
At extrema along the boundary of the boundary $𝜕 2 𝐷$
&c.

Example

In the case of a rectangular domain in $ℝ 2$

𝐷={\vtwo𝑥𝑦∈ℝ2∣𝑎≤𝑥≤𝑏∧𝑐≤𝑥≤𝑑}

this involves checking for local extrema in $𝐷$

\nabla 𝑓 (⃗ 𝐯) = ⃗ 𝟎 * d e t (𝐻 𝑓 (⃗ 𝐯)) \geq 0

and then the extrema on the boundary $𝜕 𝐷$

𝑓 𝑥 (𝑥, 𝑐) = 0 \land 𝑓 𝑥 𝑥 (𝑥, 𝑐) \neq 0 𝑓 𝑥 (𝑥, 𝑑) = 0 \land 𝑓 𝑥 𝑥 (𝑥, 𝑑) \neq 0 𝑓 𝑦 (𝑎, 𝑦) = 0 \land 𝑓 𝑦 𝑦 (𝑎, 𝑦) \neq 0 𝑓 𝑦 (𝑏, 𝑦) = 0 \land 𝑓 𝑦 𝑦 (𝑏, 𝑦) \neq 0

and then the extrema on the boundary of the boundary $𝜕 2 𝐷$ , i.e. the corners

𝑓 (𝑎, 𝑐) 𝑓 (𝑏, 𝑐) 𝑓 (𝑎, 𝑑) 𝑓 (𝑏, 𝑑)

and then determining which of these values are indeed $m i n 𝐷 𝑓$ and $m a x 𝐷 𝑓$ .

#state/tidy | #SemBr | #lang/en

2022. MATH1011: Multivariable Calculus, p. 59 ↩

Extreme Value Theorem

Euclidean spaces

Footnotes