Math and science::Analysis::Tao::09. Continuous functions on R

The maximum principle

Continuous functions whose domain is a closed enjoy two useful properties:

the maximum principle
the intermediate value theorem

This card covers the first. If a function is continuous and it's domain is a closed interval, then it has a maximum value (one or more values of the domain map to a maximum).

Bounded functions

Let $X \subseteq R$ and let $f : X \to R$ be a function.

We say that $f$ is bounded from above iff there exists an $M \in R$ such that $f (x) \leq M for all x \in X$ .

We say that $f$ is bounded from below iff there exists an $M \in R$ such that $f (x) > M for all x \in X$ .

We say that $f$ is bounded iff it is both bounded from above and bounded from below.

Continuous functions on a closed interval are bounded

Let $a < b$ be real numbers. Let $f : [a, b] \to R$ be a function continuous on $[a, b]$ . Then $f$ is a bounded function.

Obtains maximum/minimum

Let $f : X \to R$ be a function, and $x_{0} \in X$ .

We say that $f$ obtains its maximum at $x_{0}$ iff $f (x_{0}) \geq f (x) for all x \in X$ .

We say that $f$ obtains its minimum at $x_{0}$ iff $f (x_{0}) \leq f (x) for all x \in X$ .

Maximum principle

Let $a < b$ be real numbers. Let $f : [a, b] \to R$ be a function continuous on $[a, b]$ .

Then there is an $x_{m a x} \in [a, b]$ such that $f (x_{m a x}) \geq f (x) for all x \in [a, b]$ ( $f$ obtains its maximum at $x_{m a x}$ ).

Similarly, there is an $x_{m i n} \in [a, b]$ such that $f (x_{m i n}) \leq f (x) for all x \in [a, b]$ ( $f$ obtains its minimum at $x_{m i n}$ ).

Proof

The proof of the maximum principle is good to study, as it brings together many ideas, including:

Supremum
Axiom of choice
Heine-Borel theorem
Function convergence and continuity
Sequences and limits of sequences
Sequences and their limits
Correspondence between function limits and sequences

Outline

$f$ is continuous with a closed interval domain ⇒ $f$ is bounded. (by: continuous functions on a closed interval are bounded)
(proof for this lemma: proof by contradiction—identify an unbounded sub-sequence, then show that it contradicts with Heine-Borel).
$a < b$ ⇒ $[a, b]$ is non-empty ⇒ ${f (x) : x \in [a, b]}$ is non-empty ⇒
$m := sup ({f (x) : x \in [a, b]})$ exists.
Define a sequence: $S := (a_{n})_{n = 1}^{\infty}$ where $a_{n} := some x such that m - \frac{1}{n} < f (x)$ (requires axiom of choice)
There must exist a subsequence of $S$ , $(a_{n_{j}})_{j = 1}^{\infty}$ , that converges to an element of $[a, b]$ (by Heine-Borel theorem).
If the subsequence converges to $c \in [a, b]$ . What is the value of $f (c)$ ? We should take a limit of the sequence $(f (a_{n_{j}}))_{j = 1}^{\infty}$ . To do so, view the elements:
$f (a_{n_{j}}) > m - \frac{1}{n_{j}} \geq m - \frac{1}{j}$
(I find it easier to conceptualize the above by replacing the inequality with $f (a_{n_{j}}) = m - \frac{1}{j} + δ$ , for some $δ > 0$ )
$lim_{j \to \infty} f (a_{n_{j}}) = lim_{j \to \infty} m - \frac{1}{j} + δ = m + δ = m (the only valid case is for δ = 0)$
Thus, we have a sequence of elements from $[a, b]$ that converges to an element $c \in [a, b]$ and that $f (c) = m$ , the maximum of all values of $f$ in $[a, b]$ .

Source

p236-237

17.03.2020