Show Question
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 $$X \subseteq \mathbb{R}$$ and let $$f : X \to \mathbb{R}$$ be a function.

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

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

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 $$f : [a,b] \to \mathbb{R}$$ be a function continuous on $$[a, b]$$. Then $$f$$ is a bounded function.

Obtains maximum/minimum

Let $$f: X \to \mathbb{R}$$ be a function, and $$x_0 \in X$$.

We say that $$f$$ obtains its maximum at $$x_0$$ iff $$f(x_0) \ge f(x) \text{ for all } x \in X$$.

We say that $$f$$ obtains its minimum at $$x_0$$ iff $$f(x_0) \le f(x) \text{ for all } x \in X$$.

Maximum principle

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

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

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

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

1. $$f$$ 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).

2. $$a < b$$ ⇒ $$[a, b]$$ is non-empty ⇒ $$\{f(x) : x \in [a, b] \}$$ is non-empty ⇒
3. $$m := \sup(\{f(x) : x \in [a, b] \})$$ exists.
4. Define a sequence: $$S := (a_n)_{n=1}^{\infty}$$ where $$a_n := \text{ some x such that } m - \frac{1}{n} < f(x)$$ (requires axiom of choice)
5. 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).
6. 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:
$$f(a_{n_j}) > m - \frac{1}{n_j} \ge m - \frac{1}{j}$$
(I find it easier to conceptualize the above by replacing the inequality with $$f(a_{n_j}) = m - \frac{1}{j} + \delta$$, for some $$\delta > 0$$)
7. $$\lim_{j \to \infty} f(a_{n_j}) = \lim_{j \to \infty} m - \frac{1}{j} + \delta = m + \delta = m \text{ (the only valid case is for } \delta = 0 \text{)}$$
8. 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 $$[a, b]$$.

p236-237