Local extremum

Extreme Value Theorem

Let (𝑋,𝑑) be a metric space and let 𝑓 :𝑋 →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 ∈𝐾 such that 𝑓(𝑎) ≤𝑓(𝑥) ≤𝑓(𝑏) for all 𝑥 ∈𝐾. anal

Proof

proof

Euclidean spaces

The Extreme Value Theorem is stated as follows¹

Let 𝐷 be a non-empty closed and bounded subset of ℝ𝑛 and let 𝑓 :𝐷 →ℝ be a continuous function. Then 𝑓 is bounded and there exist ⃗𝐚 ∈𝐷 and ⃗𝐛 ∈𝐷 such that (∀⃗𝐱 ∈𝐷)[𝑓(⃗𝐚) ≤𝑓(⃗𝐱) ≤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

1. At critical points, i.e. ∇𝑓(⃗𝐯) =⃗𝟎∗ and det(𝐻𝑓(⃗𝐯)) ≥0.
2. At extrema along the boundary 𝜕𝐷.
3. At extrema along the boundary of the boundary 𝜕2𝐷
   &c.

Example

In the case of a rectangular domain in ℝ2

𝐷={\vtwo𝑥𝑦∈ℝ2∣𝑎≤𝑥≤𝑏∧𝑐≤𝑥≤𝑑}

this involves checking for local extrema in 𝐷

∇𝑓(⃗𝐯)=⃗𝟎∗det(𝐻𝑓(⃗𝐯))≥0

and then the extrema on the boundary 𝜕𝐷

𝑓𝑥(𝑥,𝑐)=0∧𝑓𝑥𝑥(𝑥,𝑐)≠0𝑓𝑥(𝑥,𝑑)=0∧𝑓𝑥𝑥(𝑥,𝑑)≠0𝑓𝑦(𝑎,𝑦)=0∧𝑓𝑦𝑦(𝑎,𝑦)≠0𝑓𝑦(𝑏,𝑦)=0∧𝑓𝑦𝑦(𝑏,𝑦)≠0

and then the extrema on the boundary of the boundary 𝜕2𝐷, 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 min𝐷𝑓 and max𝐷𝑓.

---

tidy | SemBr | en

Footnotes

1. 2022. MATH1011: Multivariable Calculus, p. 59 ↩