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
- \( f \) 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 \( a_n := \text{ some x such that } m - \frac{1}{n} < f(x) \) (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:
\( f(a_{n_j}) > m - \frac{1}{n_j} \ge m - \frac{1}{j} \)
(I find it easier to conceptualize the above by replacing the inequality with \( f(a_{n_j}) = m - \frac{1}{j} + \delta \), for some \( \delta > 0 \)) - \( \lim_{j \to \infty} f(a_{n_j}) = \lim_{j \to \infty} m - \frac{1}{j} + \delta = m + \delta = m \text{ (the only valid case is for } \delta = 0 \text{)} \)
- 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 \( [a, b] \).