Boundedness Theorem

The boundedness theorem is the most fundamental property of continuous functions on a closed interval; it guarantees that a continuous function on a closed interval never “runs off to infinity”.

Statement of the Theorem

Geometric Meaning

The geometric meaning of the boundedness theorem is that the graph of a continuous function on a closed interval is “sandwiched” between two horizontal lines.

More precisely:

• An upper bound exists: the function values never exceed some upper limit
• A lower bound exists: the function values never drop below some lower limit
• The graph is confined to a horizontal band

Proof Strategy

Proof by Contradiction

1. Assume the function is unbounded

   • Assume the function $f(x)$ is unbounded on $[a, b]$

2. Construct a sequence

   • For every positive integer $n$, there exists $x_n \in [a, b]$ with $|f(x_n)| > n$

3. Use the fact that a bounded sequence has a convergent subsequence

   • The sequence $\{x_n\}$ is bounded, so it has a convergent subsequence $\{x_{n_k}\}$
   • Let $\lim_{k \to \infty} x_{n_k} = x_0 \in [a, b]$

4. Use continuity

   • Since $f(x)$ is continuous at $x_0$
   • $\lim_{k \to \infty} f(x_{n_k}) = f(x_0)$

5. Derive a contradiction

   • But $|f(x_{n_k})| > n_k \to \infty$
   • This contradicts $f(x_{n_k}) \to f(x_0)$

6. Conclusion

   • The assumption fails, so the function must be bounded

Applied Examples

Example 1: Polynomial Function

Problem: prove that the function $f(x) = x^2$ is bounded on $[0, 2]$.

Solution:

• The function $f(x) = x^2$ is continuous on $[0, 2]$
• By the boundedness theorem, the function is bounded
• In fact, for $x \in [0, 2]$, we have $0 \leq f(x) = x^2 \leq 4$
• Hence we may take $M = 4$

Example 2: Trigonometric Function

Problem: prove that the function $f(x) = \sin x$ is bounded on $[0, \pi]$.

Solution:

• The function $f(x) = \sin x$ is continuous on $[0, \pi]$
• By the boundedness theorem, the function is bounded
• In fact, for $x \in [0, \pi]$, we have $0 \leq f(x) = \sin x \leq 1$
• Hence we may take $M = 1$

Example 3: Counterexample (Open Interval)

Problem: is the function $f(x) = \frac{1}{x}$ bounded on $(0, 1]$?

Solution:

• The function is continuous on $(0, 1]$
• But $(0, 1]$ is not a closed interval
• As $x \to 0^+$, $f(x) \to +\infty$
• Therefore the function is unbounded
• Note: the boundedness theorem applies only to closed intervals

Points to Note

1. The Interval Must Be Closed

The boundedness theorem applies only to closed intervals; for open or half-open intervals the conclusion need not hold.

Counterexample:

• $f(x) = \frac{1}{x}$ is unbounded on $(0, 1]$
• $f(x) = x$ is bounded on $(0, 1)$ but has no maximum value

2. The Function Must Be Continuous

If the function is not continuous on the closed interval, the boundedness theorem does not apply.

Counterexample:

• $f(x) = \begin{cases} \frac{1}{x}, & 0 < x \leq 1 \\ 0, & x = 0 \end{cases}$ is not continuous and is unbounded on $[0, 1]$

3. Bounded Does Not Mean Attaining Extrema

The boundedness theorem only guarantees that the function is bounded; it does not guarantee that a maximum and a minimum value are attained.

Example:

• $f(x) = x$ is bounded on $(0, 1)$ (that is, $0 < f(x) < 1$)
• But it cannot attain the supremum 1 or the infimum 0

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Exercises

Exercise 1

Prove that the function $f(x) = x^2 - 2x + 1$ is bounded on $[0, 3]$.

Exercise 2

Give an example of a function that is continuous but unbounded on $[0, 1)$, and explain why.

Exercise 3

Let $f(x)$ be continuous on $[a, b]$. Show whether $f(x)$ is necessarily bounded.

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Summary

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